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WITH ANALYTICGEOMETRY edition g third

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LoulsLelthold UNIVERSITY

OF SOUTHERN

CALIFORNIA

HARPER & ROW, PUBLISHERS New York, Hagerstoutn,SanFrancisco,London

Sponsoring Editor: GeorgeJ. Telecki Project Editor: Karen A. fudd Designer: Rita Naughton Production Supervisor:FrancisX. Giordano Compositor: ProgressiveTypographers Printer and Binder: Kingsport Press Art Studio: J & R TechnicalServicesInc. Chapter opening art: "Study Light,, by patrick Caulfield THE CALCULUS WITH ANALYTIC GEOMETRY, Copyright @ 1.968,1,972,t976 by Louis Leithold

Edition

All rights reserved. Printed in the United States of America. No part of this book may be used or reproduced in any manner whatsoever without written permission except in the case of brief quotations embodied in critical articles and reviews. For information address Harper & Row, Publishers, Inc., 10 East 53rd street, New york, N.y. 10022. Library of Congress Cataloging in Publication Data Leithold, Louis. The calculus, with analytic geometry. Includes index. L. Calculus. 2. QA303.L428 7975b ISBN 0-06-043951-3

Geometry, Analytic, L Title. 515'.1s 75-26639

To Gordon Marc

Gontents

Preface

Chapter 1 REAL NUMBERS, INTRODUCTION TO ANALYTIC GEOMETRY, AND FUNCTIONS page 1

Chapter 2 LIMITS AND CONTINUIT} paSe oc

Chapter 3 THE DERIVATIVE page110

)cu,

1.1 1.2 1.3 1.4 1.5 1.6 L.7 1.8

2 Sets,Real Numbers, and Inequalities 14 Absolute Value 2L The Number Plane and Graphs of Equations 28 Formula Distance Formula and Midpoint 33 Equations of a Line 43 The Circle 48 Functions and Their Graphs Function Notation, Operations on Functions, and Types of Func56 tions

2.1, 2.2 2.3 2.4 2.5 2.6

66 The Limit of a Function 74 Theorems on Limits of Functions 85 One-Sided Limits 88 Infinite Limits Continuity of a Function at a Number 1-0L Theorems on Continuity

97

1-1-i3.1 The Tangent Line 11,5 3.2 InstantaneousVelocity in Rectilinear Motion L213.3 The Derivative of a Function L26 3.4 Differentiability and Continuity 3.5 Some Theorems on Differentiation of Algebraic Functions L38 3.6 The Derivative of a Composite Function 3.7 The Derivative of the Power Function for Rational Exponents 1-45 3.8 Implicit Differentiation 1.50 3.9 The Derivative as a Rate of Change 154 3.10 RelatedRates 157 3.1L Derivatives of Higher Order

1.30 L42

Viii

CONTENTS

-T

t I

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Chapter 4 TOPICS ON LIMITS, CONTINUITY, AND THE DERIVATIVE page164

Chapter 5 ADDITIONAT APPLICATIONS OF THE DERIVATIVE page204

Chapter 5 THE DIFFERENTIAL AND ANTIDIFFERENTIATION page243

Chapter 7 THE DEFINITE INTEGRAL page275

Chapter 8 APPLICATIONS OF THE DEFINITE INTEGRAL page323

4.L 4.2 4.3 4.4 4.5 4.6

Limits at Infinity 765 Horizontal and Vertical Asymptotes L71 Additional Theorems on Limits of Functions 1.74 Continuity on an Interval 177 Maximum and Minimum Values of a Function 19L Applications Involving an Absolute Extremum on a Closed Interval L89 4.7 Rolle's Theorem and the Mean-Value Theorem lgs 5.1 Increasing and Decreasing Functions and the First-Derivative Test 205 5.2 The second-Derivative Test for Relative Extrema 2i.1 5.3 Additional Problems Involving Absolute Extrema 213 5.4 Concavity and Points of Inflection 220 5.5 Applications to Drawing a Sketchof the Graph of a Function 227 5.6 An Application of the Derivative in Economics 280 5.1 The Differential 244 6.2 Differential Formulas 249 5.3 The Inverse of Differentiation 253 6.4 Differential Equations with Variables Separable 261 6.5 Antidifferentiation and Rectilinear Motion 265 6.6 Applications of Antidifferentiation in Economics 269

7.L 7.2 7.3 7.4 7.5 7.6

The Sigma Notation 276 Area 28L The Definite Integral 288 Properties of the Definite Integral 2gG The Mean-Value Theorem for Integrals 306 The Fundamental Theorem of the Calculus SLi.

8.1 Area of a Region in a Plane 324 8.2 Volume of a Solid of Revolution: Circular-Disk and Circular-Ring Methods 330 8.3 Volume of a Solid of Revolution: Cylindrical-Shell Method 3J6 8.4 Volume of a Solid Having Known Parallel Plane Sections 341 8.5 Work 344 8.6 Liquid Pressure 348 8.7 Center of Mass of a Rod 35L 8.8 Center of Mass of a Plane Region 356 8.9 Center of Mass of a Solid of Revolution 366 8.10 Length of Arc of a Plane Curve 372

CONTENTS Chapter 9 LOGARITHMIC AND EXPONENTIAL FUNCTIONS

page38L

ChaPterL0 TRIGONOMETRIC FUNCTIONS page430

382 9.1 The Natural Logarithmic Function 9.2 The Graph of the Natural Logarithmic Function 395 9.3 The Inverse of a Function 405 9.4 The Exponential Function 9.5 Other Exponential and Logarithmic Functions 420 9 . 6 Laws of Growth and DecaY

391'

4L4

431 10.1 The Sine and Cosine Functions 438 t0.2 Derivatives of the Sine and Cosine Functions 447 10.3 Integrals Involving Powers of Sine and Cosine 452 Functions and Cosecant Secant, Cotangent, Tangent, 10.4 The of a to the Slope Function Tangent the of 10.s An Application

10.5 L0.7 10.8 10.9

461, Line Integrals Involving the Tangent, Cotangent, Secant, and Co466 secant 47L InverseTrigonometricFunctions 477 Functions Trigonometric Derivatives of the Inverse 484 Functions Integrals Yielding Inverse Trigonometric

492 Introduction 493 Integration by Parts 498 Integration by Trigonometric Substitution Integration of Rational Functions by Partial Fractions. CasesL and 504 2: The Denominator Has Otly Linear Factors 11.5 Integration of Rational Functions by Partial Fractions. Cases3 and 51.2 4: The Denominator Contains Quadratic Factors 516 LL.6' Integration ofRational Functions of Sine and Cosine 51-9 Ll..7 MiscellaneousSubstitutions 521, 1.L.8 The Trapezoidal Rule 'n.9 526 Simpson'sRule

Chapter LL 1 1 . 1 TECHNIQUES OF LL.2 INTEGRATION 1 1 . 3 page 497 Lt.4

536 Chapter L2 Lz.L The Hyperbolic Functions 543 Functions Hyperbolic Inverse The I2,2 HYPERBOTIC FUNCTIONS page535 12.3 Integrals Yielding Inverse Hyperbolic Functions 555 Chapter13 13.1 The Polar Coordinate System 560 Polar in Coordinates Equations Graphs of T3.2 POLAR COORDINATES page 554 13.3 Intersection of Graphs in Polar Coordinates 567 57113.4 Tangent Lines of Polar Curves 573 13.5 Area of a Region in Polar Coordinates 579 Chapter 1.4 LA.l The Parabola 583 Axes of L4.2 Translation THE CONIC SECTIONS page 578 588 14.3 Some Properties of Conics 592 L4.4 Polar Equations of the Conics

548

X

CONTENTS

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L4.5 14.5 L4.7 14.8

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1

chapter L5 INDETERMINATE FORMS, IMPROPER INTEGRALS, AND TAYLOR'S FORMULA 628 Page

Cartesian Equations of the Conics The Ellipse 606 The Hyperbola 613 Rotation of Axes 620

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15.1 The IndeterminateForm 0/0 629 1,5.2 Other IndeterminateForms 636 15.3 Improper Integralswith Infinite Limits of Integration lS.4 Other Improper Integrals 647 15.5 Taylor's Formula 6s7

Chapter 15 1,5.1 Sequences 5G0 INFINITE SERIES 1'6.2 Monotonic and Bounded Sequence s 667 page65e 16.g Infinite Series of Constant Terms 673 1,6.4 Infinite Seriesof Positive Terms 6g4 1,6.5 The Integral Test 694 15.6 Infinite series of Positive and Negative Terms L6 7 Power Series 707 L6.8 Differentiation of Power Series 713 16.9 Integration of Power Series 722 15.10Taylor Series 729 L6.ll The Binomial Series 738 Chapter 17 VECTORS IN THE PLANE AND PARAMETRIC EQUATIONS page745

Chapter 18 VECTORS IN THREEDIMENSIONAL SPACE AND SOLID ANALYTIC GEOMETRY page 810

77.1, 77.2 L7.3 17.4 77.5 77.6 t7.7 I7.8

641

697

Vectors in the Plane 746 Properties of Vector Addition and Scalar Multiplication 751 Dot Product 756 vector-Valued Functions and Parametric Equations 763 Calculus of Vector-Valued Function s 772 Length of Arc 779 Plane Motion 785 The Unit Tangent and Unit Normal Vectors and Arc Length as a Parameter 792 L7.9 Curvature 796 17.10 Tangential and Normal Components of Acceleration g04

18.1 R3,The Three-Dimensional Number Space 8L1 L8.2 Vectors in Three-Dimensional Space 818 18.3 The Dot Product in V, 825 18.4 Planes 829 18.5 Lines in R3 836 L8.6 Cross Product 842 1,8.7 Cylinders and Surfaces of Revolution BS2 18.8 Quadric Surfaces 858 18.9 Curves in R3 864 18.10 Cylindrical and Spherical Coordinates 872

CONTENTS

Chapter 19 CALCULUS DIFFERENTIAL OF FUNCTIONS OF SEVERAL VARIABLES page 880

Chapter 20 DIRECTIONAL DERIVATIVES, GRADIENTS, APPTICATIONS OF PARTIAL DERIVATIVES, AND LINE INTEGRALS page 944

ChaPter2| MULTIPTE INTEGRATION page 1001

APPENDIX pageA-1

88L L9.']-. Functions of More Than One Variable 889 lg.2 Limits of Functions of More Than one Variable 900 tg.3 Continuity of Functions of More Than One Variable 905 19.4 Partial Derivatives 91'3 19.5 Differentiability and the Total Differential 926 19.6 The Chain Rule 934 19.7 Higher-Order Partial Derivatives 20.L 20.2 20.3 20.4 20.5 20.6 20.7

945 Directional Derivatives and Gradients 953 Surfaces to Normals Tangent Planes and 956 Variables Two Extrema of Functions of Some Applications of Partial Derivatives to Economics 975 Obtaining a Function from Its Gradient 981 Line Integrals 989 Line Integrals Independent of the Path

21.1, 2'1..2 21.3 21.4 21.5 21.6 21.7

L002 The Double Integral 1008 Evaluation of Double Integrals and Iterated Integrals 1'0L6 Center of Mass and Moments of Inertia 1'022 The Double Integral in Polar Coordinates 1028 Area of a Surface 1034 The Triple Integral L039 The Triple Integral in Cylindrical and Spherical Coordinates

A-2 Table 1. Powers and Roots A-3 Table 2 Natural Logarithms A-5 Table 3 Exponential Functions A-L2 Table 4 Hyperbolic Functions A-13 Table 5 Trigonometric Functions A-1-4 Table 6 Common Logarithms A-15 Table 7 The Greek Alphabet ANSWERS TO ODD-NUMBERED EXERCISES INDEX

4.45

A.1.7

967

ACKNOWLEDGMENTS

Reviewers of rheCatcutu

Professor William D. Professor Archie D. l Professor Phillip Cla Professor Reuben W ProfessorJacob Golil ProfessorRobert K. Goodrich, University of Colorado Professor Albert Herr, Drexel University ProfessorJamesF. Hurley, University of Connecticut Professor Gordon L. Miller, Wisconsin State Universitv Professor William W. Mitchell, Jr., phoenix College Professor Roger B. Nelsen, Lewis and Clark Colle-ge Professor Robert A. Nowlan, southern connectictit state college Sister Madeleine Rose, Holy Names College Professor George W. Schultz, St. petersbuig Junior College Professor Donald R. Sherbert, Universitv of Illinois Professor]ohn V_adney,Fulton-Montgomery community college Professor David Whitman, San Diego State College

Production Staff at Harper & Row GeorgeTelecki, MathematicsEditor Karen fudd, Project Editor Rita Naughton, Designer Assistantsfor Answers to Exercises JacquelineDewar, Loyola Marymount University Ken Kast, Logicon, Inc. |ean Kilmer, West Covina Unified SchoolDistrict Cover and ChapterOpenins Artist Patrick Cadlfield,Londin, England To these _peopleand to all the users of the first and second editions who have suggested changes,I expressmy deep appreciation. L. L.

rl]

FTETAGE

This third edition of THE CALCULUS WITH ANALYTIC GEOMETRY, like the other two, is designed for prospective mathematics majors as well as for students whose primary interest is in engineering, the physical sciences, or nontechnical fields. A knowledge of high-school algebra and geometry is assumed. The text is available either in one volume or in two parts: Part I consists of the first sixteen chapters, and Part II comprises Chapters 16 through 21 (Chapter L6 on Infinite Series is included in both parts to make the use of the two-volume set more flexible). The material in Part I consists of the differential and integral calculus of functions of a single variable and plane analytic geometrlz, and it may be covered in a one-year course of nine or ten semester hours or twelve quarter hours. The second part is suitable for a course consisting of five or six semester hours or eight quarter hours. It includes the calculus of several variables and a treatment of vectors in the plane, as well as in three dimensions, with a vector approach to solid analytic geometry. The objectives of the previous editions have been maintained. I have endeavored to achieve a healthy balance between the presentation of elementary calculus from a rigorous approach and that from the older, intuitive, and computational point of view. Bearing in mind that a textbook should be written for the student, I have attempted to keep the presentation geared to a beginner's experience and maturity and to leave no step unexplained or omitted. I desire that the reader be aware that proofs of theorems are necessary and that these proofs be well motivated and carefully explained so that they are understandable to the student who has achieved an average mastery of the preceding sections of the book. If a theorem is stated without proof ,I have generally augmented the discussion by both figures and examples, and in such cases I have always stressed that what is presented is an illustration of the content of the theorem and is not a proof. Changes in the third edition occur in the first five chapters. The first

xlv

PREFACE

section of Chapter I has been rewritten to give a more detailed exposition of the real-number system. The introduction to analytic geometry in this chapter includes the traditional material on straight lines as well as that of the circle, but a discussion of the parabola is postponed to Chapter 14, The Conic Sections. Functions are now introduced in Chapter 1. I have defined a function as a set of ordered pairs and have used this idea to point up the concept of a function as a correspondence between sets of real numbers. The treatment of limits and continuity which formerly consisted of ten sections in Chapter 2 is now in fwo chapters (2 and 4), with the chapter on the derivative placed between them. The concepts of limit and continuity are at the heart of any first course in the calculus. The notion of a limit of a function is first given a step-by-step motivation, which brings the discussion from computing the value of a function near a number, through an intuitive treatment of the limiting process, up to a rigorous epsilon-delta definition. A sequence of examples progressively graded in difficulty is included. All the limit theorems are stated, and some proofs are presented in the text, while other proofs have been outlined in the exercises. In the discussion of continuity, I have used as examples and counterexamples "common, everyday" functions and have avoided those that would have little intuitive meaning. In Chapter 3, before giving the formal definition of a derivative, I have defined the tangent line to a curve and instantaneous velocity in rectilinear motion in order to demonstrate in advance that the concept of a derivative is of wide application, both geometrical and physical. Theorems on differentiation are proved and illustrated by examples. Application of the derivative to related rates is included. Additional topics on limits and continuity are given in Chapter 4. Continuity on a closed interwal is defined and discussed, followed by the introduction of the Extreme-Value Theorem, which involves such functions. Then the Extreme-Value Theorem is used to find the absolute extrema of functions continuous on a closed interval. Chapter 4 concludes with Rolle's Theorem and the Mean-Value Theorem. Chapter 5 gives additional applications of the derivative, including problems on curve sketching as well as some related to business and economics. The antiderivative is treated in Chapter 6. I use the term "antidifferentiation" instead of indefinite integration, but the standard notation ! f (x) dx is retained so that you are not given a bizarre new notation that would make the reading of standard references difficult. This notation will suggest that some relation must exist between definite integrals, introduced in Chapter 7, and antiderivatives, but I see no harm in this as long as the presentation gives the theoretically proper view of the definite integral as the limit of sums. Exercises involving the evaluation of definite integrals by finding limits of sums are given in Chapter 7 to stress that this is how they are calculated. The introduction of the definite inte-

PREFACE

gral follows the definition of the measure of the area under a curye as a limit of sums. Elementary properties of the definite integral are derived and the fundamental theorem of the calculus is proved. It is emphasized that this is a theorem, and an important one, because it provides us with an alternative to computing limits of sums. It is also emphasized that the definite integral is in no sense some special type of antiderivative. In Chapter 8 I have given numerous applications of definite integrals. The presentation highlights not only the manipulative techniques but also the fundamental principles involved. In each application, the definitions of the new terms are intuitively motivated and explained. The treatment of logarithmic and exponential functions in Chapter 9 is the modern approach. The natural logarithm is defined as an integral, and after the discussion of the inverse of a function, the exponential function is defined as the inverse of the natural logarithmic function. An irrational power of a real number is then defined. The trigonometric functions are defined in Chapter 10 as functions assigning numbers to numbers. The important trigonometric identities are derived and used to obtain the formulas for the derivatives and integrals of these functions. Following are sections on the differentiation and integration of the trigonometric functions as well as of the inverse trigonometric functions. Chapter LL, on techniques of integration, involves one of the most important computational aspects of the calculus. I have explained the theoretical backgrounds of each different method after an introductory motivation. The mastery of integration techniques depends upon the examples, and I have used as illustrations problems that the student will certainly meet in practice, those which require patience and persistence to solve. The material on the approximation of definite integrals includes the statement of theorems for computing the bounds of the error involved in these approximations. The theorems and the problems that go with them, being self-contained, can be omitted from a course if the instructor so wishes. A self-contained treatment of hyperbolic functions is in Chapter 12. This chapter may be studied immediately following the discussion of the circular trigonometric functions in Chapter L0, if so desired. The geometric interpretation of the hyperbolic functions is postponed until Chapter L7 because it involves the use of parametric equations. Polar coordinates and some of their applications are given in Chapter 13. In Chapter 1.4,conics are treated as a unified subject to stress their natural and close relationship to each other. The parabola is discussed in the first two sections. Then equations of the conics in polar coordinates are treated, and the cartesian equations of the ellipse and the hyperbola are derived from the polar equations. The topics of indeterminate forms, improper integrals, and Taylor's formula, and the computational techniques involved, are presented in Chapter L5. I have attempted in Chapter 16 to give as complete a treatment of

xvr

PREFACE

infinite series as is feasible in an elementary calculus text. In addition to the customary computational material, I have included the proof of the equivalence of convergence and boundedness of monotonic sequences based on the completeness property of the real numbers and the proofs of the computational processes involving differentiation and integration of power series. The first five sections of Chapter L7 on vectors in the plane can be taken up after Chapter 5 if it is desired to introduce vectors earlier in the course. The approach to vectors is modern, and it serves both as an introduction to the viewpoint of linear algebra and to that of classical vector analysis. The applications are to physics and geometry. Chapter 18 treats vectors in three-dimensional space, and, if desired, the topics in the first three sections of this chapter may be studied concurrently with the corresponding topics in Chapter 17. Limits, continuity, and differentiation of functions of several variables are considered in Chapter L9. The discussion and examples are applied mainly to functions of two and three variables; however, statements of most of the definitions and theorems are extended to functions of n variables. In Chaptet 20, a section on directional derivatives and gradients is followed by a section that shows the application of the gradient to finding an equation of the tangent plane to a surface. Applications of partial derivatives to the solution of extrema problems and an introduction to Lagrange multipliers are presented, as well as a section on applications of partial derivatives in economics. Three sections, new in the third edition, are devoted to line integrals and related topics. The double integral of a function of two variables and the triple integral of a function of three variables, along with some applications to physics, engineering, and geometry/ are given in Chapter 2/... New to this edition is a short table of integrals appearing on the front and back endpapers. However, as stated in Chapter L1, you are advised to use a table of integrals only after you have mastered integration. Louis Leithold

Realnumberq,introduction to analytic geometry and functions

REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY. AND FUNCTIONS

1.1 SETS, REAL NUMBERS, AND INEQUALITIES

The idea of "set" is used extensively in mathematics and is such a basic concept that it is not given a formal definition. We can say that a set is a collection of objects, and the objects in a set are called the elements of a set. We may speak of the set of books in the New York Public Library, the set of citizens of the United States, the set of trees in Golden Gate Park, and so on. In calculus, we are concerned with the set of real numbers. Before discussing this set, we introduce some notation and definitions. We want every set to be weII defined; that is, there should be some rule or property that enables one to decide whether a given object is or is not an element of a specific set. A pair of braces { } used with words or symbols can describe a set. If S is the set of natural numbers less than 6,we can write the set S as {L,2,3, 4,5} We can also write the set S as {r, such that r is a natural number less than 6} where the symbol. "x" is called a"variable." A aariable is a symbol used to represent any element of a given set. Another way of writing the above set S is to use what is called set-builder notation, where a vertical bar is used in place of the words "such that." Using set-builder notation to describe the set S, we have {rlr is a natural number less than 6} which is read "the set of all r such that r is a natural number less than G." The set of natural numbers will be denoted by N. Therefore, we may write the set N as

{7,2,3,

I .l

where the three dots are used to indicate that the list goeson and on with no last number. With set-builder notation the set N may be written as {rlr is a natural number}. The symbol " e " is used to indicate that a specificelementbelongs to a set. Hence, we may write 8 € N, which is read "8 is an elementof N.,, The notation a,b e S indicates that both a andb are elementsof S. The symbol fi is read "is not an elementof." Thus, we read * € w as "+ is not an elementof N." We denote the set of all integers by /. Becauseevery element of N is also an element of / (that is, every natural number is an integer),we say that N is a "subset" of /, written N E /.

:',.t

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The set S is a subsetof the set T, written S e T,if and only if every element of S is also an element of T.If , in addition, there is at least one element of T

1.1 SETS,REAL NUMBERS,AND INEQUALITIES

which is not an elementof S, then S is a propersubsetof T, and it is writtenS C T. Observe from the definition that every set is a subsetof itself, but a set is not a proper subset of itself. In Definition 1.1.1,the "if and only if" qualification is used to combine two statements:(i) "the set S is a subset of the set T if every element of S is also an element of T"; and (ii) "the set S is a subset of set T only if every element of S is also an element of 7," which is logically equivalent to the statement"if. S is a subset of.T, then every elementof S is also an element of.T." o rLLUsrRArroN1: Let N be the set of natural numbers and let M be the set denoted by {rlr is a natural number less than 1,0}.Becauseevery element of M is also an element of N, M is a subset of N and we write M e N. Also, there is at least one element of N which is not an element of M, and so M is a proper subsetof N and we may write M C N. Furthermore, because{5} is the set consisting of the number 5, {6} C M, which statesthat the set consisting of the single element 5 is a Proper subset of the set M. We may also write 5 € M, which statesthat the number 6 is o an element of the set M. Consider the set {xlzx * L:0, and x e I}. This set containsno elements because there is no integer solution of the equation 2x * 1.: 0. Such a set is called the "empty set" or the "null set." 1.t.2 Definition

The empty set (or null sef) is the set that contains no elements.The emPty set is denoted by the symbol A. The concept of "subset" may be used to define what is meant by two sets being "equal."

1.1.3 Definition

Two setsA and B are said to be equal,written A: andB e A.

B, if and only if A e B

Essentially, this definition states that the two sets A and B are equal if and only if every element of A is an element of B and every element of B is an element of A, that is, if the sets A and B have identical elements. There are two operations on sets that we shall find useful as we proceed. These operations are given in Definitions 1.1.4 and 1.1.5. 1.1.4 Definition

Let A and B be two sets. The union of A and B, denoted by A U B and read "A union 8," is the set of all elements that are in A or in B or in both A and B.

REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS

E x A M P LE1:

Le t A :

{ 2 , 4 , 6 ,8 ,

10,LzI, B : {1, 4, 9, '1.5} , and C - { 2 , 1 0 } .F i n d (a)AuB (c)BUC

(b)Auc (d)AuA 1..1..5 Definition

nxelvrpr.E 2: If A, B, and C are the sets defined in Example1, find

@)AnB (c)BnC

SOLUTION:

(a)Au (b)Au (c)BU (d)Au

BC: C: A:

1 ,, 21, , 6 } {1,2,4,6,9,9, "1.0 { 2 , 4 , 6 , 8 , 1 , 0 1, 2 } { 1 , 2 ,4 , 9 , 1 , 0 , 1 , 6 } {2, 4, 6, 8, I0,12} : 4

Let A and B be two sets.The intersectionof A and B, denoted by A n B and tead " A intersection8," is the set of all elementsthat are in both A and B. SOLUTION:

( a )A n g : { 4 } ( c )B f i C : A

(b)An C: {2, 1.0} ( d ) A n A : { 2 , 4 , 6 , 8 , t 0 , 7 2 1: 4

(b)Anc @)AnA

The real number system consists of a set of elements called real numbers and two operations called addition and multiplication The set of real numbers is denoted by Rt. The operation of addition is denoted by the symbol "*", and the operation of multiplication is denoted by the symbol',.',. rf a, b c Rt, a * b denotes the sum of a and b, and a - b (or ab) denotes their product. We now present seven axioms that give laws governing the operations addition of and multiplication on the set Rl. The word axiom is used to indicate a formal statement that is assumed to be true without proof.

1.1.6Axiom (Closure and UniquenessLaws)

1.1.7 Axiom

rf a ,b e R t, then si b number.

i s a uni que real number, andab i s a unique r eal

If a, b e Rr, then

(CommutatiaeLaws)

a*b:b*a

L.L.8 Axiom (AssociatiaeLaws)

1."1,.9 Axiom

and ab:ba

If. a, b, c € Rl, then a + (b* c) :

(a + b) + c

and

a(bc):

(ab)c

l f a , b , c C R 1 ,t h e n

(DistributiaeLaw)

a(b+c):ab*ac 1..L.10 Axiom

There exist two distinct real numbers 0 and 1 such that for anv real num-

(Existence of IdentityElements) ber A,

ail:a

J

and A.t:a

AND INEOUALTTIES 1.1 SETS,REAL NI.JMBERS,

L.1.11 Axiom For every real number a, there exists a real number called the negatiae lnaerse) of a (or additiae inaerse of a), denoted by -a (read "the negative of a"), or Additiae (Existence of Negatiae such that a*(-s):Q

'1,.1.12 Axiom

For every real number a, except 0, there exists a real number called the

(Existenceof Reciprocalor reciprocat of a (or multiplicatiae inaerse of a), denoted by a-t, such that Multiplicatiae lnuerse)

a ' a-r,: t

Axioms 1,.1,.6through 1,.I.L2 are called field axioms because if these axioms are satisfied by a set of elements, then the set is called afield under the two operations involved. Hence, the set R1 is a field under addition and multiplication. For the set / of integers, each of the axioms t.1.6 is not satisfied (for instance, through 1.1.11 is satisfied, but Axiom L.1,.1,2 it /). Therefore, the set of inverse no multiplicative has the integer 2 multiplication. and field under addition a integers is not

Definition 1..1..13

lf a, b € Rl, the operation of subtraction assigns to a andb areal number, denoted by a - b (read "a minus b"), called the differenceof a and b, where (1)

a-b:A*(-b) Equality (1) is read "A minus b equals a plus the negatle 'i,.1.14

Definition

of-b."

I f . a , b G R l , a n d b * 0 , t h e o p e r a t i o no f d i a i s i o na s s i g n s t o a a n d b a r c a l number, denoted by o + b (read "a divided by b"), called the quotient of a and b, where a:b:A'b-r Other notations for the quotient of a and b ate !

b

and

alb

By using the field axioms and Definitions 1.1.13 and 1.1.14, we can derive properties of the real numbers from which follow the familiar algebraic operations as well as the techniques of solving equations, factoring, and so forth. In this book we are not concerned with showing how such properties are derived from the axioms. Properties that can be shown to be logical consequences of axioms are theorems. In the statement of most theorems there are two parts: the "if" part, called the hypothesis, and the "then" part, called the conclusion. The argument verifying a theorem is a proof. A proof consists of showing that the conclusion follows from the assumed truth of the hypothesis.

REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

The concept of a real number being "positive" lowing axiom.

is given in the fol-

L.1..L5Axiom In the set of real numbers there exists a subset called the positiae numbers (OrderAxiom) such that (i) if a € Rr, exactly one of the following three statements holds: -a i s posi ti ve. a: 0 a i s posi ti ve (ii) the sum of two positive numbers is positive. (iii) the product of two positive numbers is positive. Axiom 1.1.15 is called the order axiom because it enables us to order the elements of the set Rr. In Definitions 'J,.7.77 and 1.1.18 we use this axiom to define the relations of "greater than" and "less than" on Rr. The negatives of the elements of the set of positive numbers form the set of "negative" numbers, as given in the following definition.

1*1.16Definition

The real number a is negatiae iI and only if -a is positive. From Axiom 1.1.15and Definition 7.1..76it follows that a real number is either a positive number, a negative number, or zero. Ary real number can be classified as a rqtional number or an irrational number. A rational number is any number that can be expressed as the ratio of two integers. That is, a rational number is a number of the form plq, where p a:nd,q are integers and q + 0.The rational numbers consist of the following: The integers (positive, negative, and zero) . , -5, -4, -3, -2, -L, 0, 1, 2, 3, 4, 5,

.

The positive and negative fractions such as

+-#+ The positive and negative terminating decimar.ssuch as

-0.003251 :

3,251 L,000,000

2s6:ffi

The positive and negative nonterminatingrepeatingdecimalssuch as 0.333.

:+

-0549s49s49...:-ft

The real numbers which are not rational numbers are called irrational numbers. These are positive and negative nonterminating, nonrepeating decimals, for example,

rt:

t.732.

n : 3.L4'1,59.

tan 140o: -0.839L.

The field axioms do not imply any ordering of the real numbers. That is, by means of the field axioms alone we cannot state that 1 is less than


2,2isless than 3, and so on. However, we have introduced the order axiom (Axiom 1.1.15),and because the set Rr of real numbers satisfies the order axiom and the field axioms, we say that Rl is an ordered field. We use the concept of a positive number given in the order axiom to define what we mean by one real number being "less than" another. 11*\ 7 D e fi n i ti o n

If a ,b € R t, th e n a i s l essthan b (w ri tten a< b) positive.

i f and onl y i f b-ars

and 5 is positive. (b) o rt,r,usrRArIbrs 2: (a) 3 b) than a; with symbols we write

if andonlyif b isless

albifandonlyifb
Definition

The symbols = (" is less than or equal to") and = (" is greater than or equal to") are defined as follows: (1 ) a - b i f and onl y i f ei ther a b if. and only if either a > b ot a: bT h e s ta te m entsa 1b, a) b, a < b, anda = b are cal l ed i nequal i ti es. -4, -3> -6, Some examplesare 2 <7, -5 < 6, -S 1-4,"].'4> 8,2> -10 -7.In < > b are called strict inequalparticular, a 1b and a 5 = 3, -b anda= b are called nonstrict inequalities.. ities, whereas a The following theorems can be proved by using L.L.6 through 1.1'.19.

"1..1.20Theorem

1 . 7 . 2 1T h e o r e m

(i) (ii) (1ii) (iv)

a a a a

> < > <

0 0 0 0

if if if if

If.a
and and and and

only only only only

if if if if

a is positivq. a is negative. -a < 0. -a > 0.

thena1c.

o rLLUsrRArroN 3: 3 < 5 and 6 <'1.4; so 3 < 14. 1 . 1 . 2 2T h e o r e m

lf a< b,then a* c<'b+ o rLLUsrRATroN4: 41

1.1.23 Theorem

c,and a-c
cif c isanyrealnumber.

s o4 * 5 < 7 * 5 ; a n d 4 - 5 < 7 - 5 .

If. a < b and c I d, then a * c < b + d.

5: 2 <6 and -3 < 1; so 2+ (-3) < 6 + L. o rLLUsrRArIoN

REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS 1.1.24 Theorem

lf a < b, and c is any positive number, then ac I o TLLUSTRATToN6: 2 I

L.l.25 Theorem

bc.

5; so 2 . 4 < s

rf a < b, and c is any negative number, then ac ) bc. o rLLUSrRArroN7: 2 < 5; so 2(-4) > 5(-4).

1 , . 1 . 2 T6 h e o r e m I f 0 < a < b

and0(c(

d,thenac
o r L L U S r R A r r o N 80: < 4 < 7

and0 < 8 < 9;so4(8) <7(9).

Theorem 1'.1.24states that if both members of an inequality are multiplied by a positive number, the direction of the inequality remains unchanged, whereas Theorem 1.1.25 states that if both members of an inequality are multiplied by a negative number, the direction of the inequality is reversed. Theorems 7.7.24 and 1.1..25also hold for division, because dividing both members of an inequality by a number d is equivalent to multiplying both members by lld. To illustrate the type of proof that is usually given, we present a proof of Theorem 1.L.23: By hypothesis a I b. Then b - a is positive (by Definition 1..1,.17). By hypothesis c I d. Then d - c is positive (by Definition 1.1.17). H e n ce, (b - a) + (d- c) i s posi ti ve (by A xi om 1.1.15(i i )). T h erefore, (b t d) - (a * c) i s posi ti ve (by A xi oms 1.1.gand l. l. T and Definition 1.1.13). Therefore, A * c b and b ) c, then a ) c. . r L L U s r R A r r o N 9 : 8> 4 a n d 4 > - 2 ;

ljl..28 Theorem

so8 )-2.

rf a > b , then a* c > b + c, and a- c > b-

.

c i f c i s any real num ber .

o rLLUSrRArroN 10: 3 = -5; so 3 - 4 > -5 - 4. 1.1.29 Theorem

If a > b and c ) d, then a I c > b + d. o r L L U S r R A r r otN ' 1 , : 7> 2 a n d 3

L.1.30Theorem

>-5;

so7*3>2+(-5).

If a > b and if c is any positive number, then ac ) bc. o rLLUsrRArroN 12: -3 > -7; so (-3)

> eDA.

o


1.1.31Theorem lf. a > b and if c is any negative number, then ac I bc. o rLLUsrRArroN 13: -3 > -7; so (-3)(-4)

< (-7)(-4).

o

1.1.32Theorem If a > b > }and c > d > 0, then ac > bd. o r L L U S r R A r r o1N4 : 4 > 3 > 0 a n d 7

n

79

tz24

lllttrttllrll -4t -2

0 |

zt

4

F i g u r e1 . 1 . 1

>5 ) 0; so 4(7)>3(6).

o

So far we have required the set Rl of real numbers to satisfy the field axioms and the order axiom, and we have stated that because of this requirement Rr is an ordered field. There is one more condition that is imposed upon the set R1. This condition is called the axiom of completeness (Axiom 16.2.5). We defer the statement of this axiom until Section 16.2 because it requires some terminology that is best introduced and discussed later. However, we now give a geometric interpretation to the set of real numbers by associating them with the points on a horizontal line, called an axis. The axiom of completeness guarantees that there is a oneto-one correspondence between the set Rl and the set of points on an axis. Refer to Figure 1.1.1. A point on the axis is chosen to represent the number 0. This point is called the origin A unit of distance is selected. Then each positive number x is represented by the point at a distance of r units to the right of the origin, and each negative number x is represented by the point at a distance of -x units to the left of the origin (it should be noted that if r is negative, then -r is positive). To each real number there corresponds a unique point on the axis, and with each point on the axis there is associated only one real number; hence, we have a one-to-one correspondence between Rl and the points on the axis. So the points on the axis are identified with the numbers they represent, and we shall use the same symbol for both the number and the point representing that number on the axis. We identify Rl with the axis, and we call Rl the real number line. We see that a b if and only if the point representing a is to the right of the point representing b. For instance, the number 2 is less than the number 5 and the point 2 is to the left of the point 5. We could also write 5 ) 2 and say that the point 5 is to the right of the point 2. A numberx is between a andbif a ( rand x
(2)

The set of all numbers x satisfying the continued inequality (2) is called an "open interyal." 1.L.33 Definition

The open interaal from a to b, denoted by (o, b), is defined by (a,b):{rlalx
10

REALNUMBERS, INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS The "closed interval" from 4 to b is the open interval (a, ,) together with the two endpoints c and b.

1.1.34Definition -

The closedinteraalfrom a to b, denoted by lo, bl, is defined by b,bl:{xla-x-b}

a

F i g u r e1 . 1 . 2

b

a

Figure 1,.1,.2 illustrates the open interval (a,b) and Fig. 1.1.3illustrates the closed interval la, bl. The "interval half-open on the lef.t" is the open interval (a, b) together with the right endpoint b.

F i g u r e1 . 1 . 3 "

1.1.35 Definition

The interaal half-open on the left, denoted by (o, bl, is defined by (a,bl:{xla
'1,.1.36 Definition

The interaal half-open on the right, denoted by lo, b), is defined by la,b):{rl

ab

Figure 1'.1.4illustrates the interval (a,bl and Fig. 1.1.5 illustrates the interval la, b). we shall use the symbol +@ ("positive infinity") and the symbol -m ("negative infinity"); however, care must be taken not to confuse these symbols with real numbers, for they do not obey the properties of the real numbers

F i g u r e1 . 1 . 4 f--L ab F i g u r e1 . 1 . 5

7.1.37 Definition

(i) (a, +-) : (ii) (-.o, b): (iii) la, +*7 : (iv) (-m, bl: (v) (-*,

a

**)

{xlx > {xlx < {xlx > {xlx :

a} b} a} b}

Rl

Figure 1,.1,.6illustrates the interval (a, *a), and Fig. r.7.T illustrates the interval (-*, b). Note that (-m,1oo) denotes the set of all real numbers. For each of the intervals (a,b), [a, b], lq, b), and (a, b] the numbers a

F i g u r e1 . 1 . 6

b F i g u r e1 . 1 . 7

a=x
1.1 SETS,REAL NUMBERS,AND INEQUALITIES 11

of its endpoints, and a closed interval can be regarded as one which conis considered tains all of its endpoints. Consequently, the intewalla,ao). to be a closed interval because it contains its only endpoint a. Similarly, (--, b] is a closed interval, whereas (a, **) and (-oo, b) are oPen. The intervals la, b) and (a, bl are neither open nor closed. The interval (--, *-) has no endpoints, and it is considered both oPen and closed. Intervals are used to represent "solution sets" of inequalities in one variable. The solution set of such an inequality is the set of all numbers that satisfy the inequality.

EXAMPLE 3: Find the solution set of the inequality 2*3x<5r*8 Illustrate it on the real number line.

solurroN: lf x is a number such that 2*3x<5r*8 then 2 t 3x - 2 < 5x * 8 - 2 (by Theorem 1'.1'.22) or, equivalently, 3x<5xI6 Then, adding -5r to both membersof this inequality, we have -2x < 6 (by Theorem7.7.22) Dividing on both sides of this inequality by -2 and reversing the direction of the inequality, we obtain r ) -3

(by Theorem1.1.25)

What we have proved is that if 2*3x<5r*8 then x>-3 Each of the steps is reversible; that is, if we start with x>-3 we multiply on each side by -2, reversethe direction of the inequality, and obtain -2x<6 Then we add 5x and 2 to both membersof the inequality, in which case we get 2i3x<5x*8

12

REAL NUMBERS;INTRODUCTION TO ANALYTICGEOMETRY,AND FUNCTIONS

-30 F i g u r e1 . 1 . 8

Therefore, we can conclude that

2*3x<5r*8

i f a n d o n l y i fx > - 3

So the solutionsetof the given inequalityis the interval(-3, +-;, which is illustrated in Fig. 1.1.8.

ExAMPLE4: Find the solution set of the inequality 4<3x-2<'1,0 Illustrate it on the real number line.

0 F i g u r e1 . 1 . 9

rxauprn 5: Find the solution set of the inequality 722 x Illustrate it on the real number line.

solurroN:

Adding 2 to each member of the inequalit|, we obtain

6<3x<12 Dividing each member by 3, we get 21x<4 Each step is reversible; so the solution set is the interval (2,41, as is illusstrated in Fig. 7.1,.9.

solurroN: we wish to multiply both members of the inequality by x. However, the direction of the inequality that results will depend upon whether x is positive or negative.So we must consider two cases. Case7: r is positive;that is, x ) 0. Multiplying on both sides by *, we obtain 7>2x Dividing on both sidesby 2,we get E > x or, equivalently, x < t Therefore,since the abovestepsare reversible,the solution set of Case1 is {xlx > 0} n {xlx < E} ot, equivalently, {rl0 < 7s E Again, becausethe above stepsare reversible,the solution set of Case2 is {rl*< 0} n {x>g}:A. From Cases1 and 2 we concludethat the solution set of the given inequality is the open interval (0,2), which is illustratedin Fig. 1.1.10.

1.1 SETS,REAL NUMBERS,AND INEQUALITIES 13

EXAMPLE6: Find the solution set of the inequality

x=14 x-5 Illustrate it on the real number line.

0 F i g u r e1 , 1 .11

solurroN: To multiply both members of the inequality by x-3, must consider two cases,as in Example5. Case'l: x- 3 > 0; that is, x ) 3. Multiplying on both sides of the inequality by x- 3, we get

we

x14x-12 Adding -4x to both members,we obtain -3x < -12 Dividing on both sidesby-3 and reversingthe direction of the inequality, we have x) 4 Thus the solution set of Case1 is {rlx > 3} n {xlr > 4} or, equivalently, {xlx > 4}, which is the interval (4, +oo;. Case2: x-3 < 0; thatis,x <-3. Multiplying on both sides by * - 3 and reversing the direction of the inequ ality,we have x) 4x-L2 or, equivalently, -3x > -12 or, equivalently, x14 Therefore,.x must be less than 4 and also less than 3. Thus, the solution set of Case2 is the interval (-*, 3). If the solution sets for Cases 1 and 2 are combined, we obtain (--, 3) U (4, **), which is illustrated in Fig. 1.1.11.

ExavrplE 7: Find the solution set of the inequality

(r+3)(x+4) > 0

solurroN: The inequality will be satisfied when both factors have the s a m e s i g n , t h a t i isf ,x t 3 > 0 a n d x + 4 > 0 , o r i f x t 3 < 0 a n d r + 4 < 0 . Let us consider the two cases. C a s e l - : x t 3 > 0 a n dx t 4 x>-3

) 0.Thatis,

and x)-4

Thus, both inequalities hold if x > -3, which is the interval (-3, +*;. C a s e 2 : x - 1 3 < 0 a n dx * 4 < 0 . T h a t i s , and x 1-4 x 1-3 Both inequalities hold if x < -4, which is the interval (-*,-4). If we combine the solution setsfor Cases1 and 2,we have (-*,-4)

(-3, +-;.

U

14


Exercises7.L In Exercises1 through 10,list the elementsof the given set if A: { 0 , 3 ,6 , 9 } .

{0,2,4,6,8}, B : { 1 , z , 4 , B } , C : { 1 , i , s , 7 , 9 } , a n d D :

I,.AUB

2.CUD

4.CnD

5.BUD

6. AUC

8.AnC

g.

7.BnD

3. ANB

@n D)u (Bnc)

10.(AuB)n(cuD) -ilP';3,t"i l1"t*T"-:11

,T:tyo.n

1.'1..5x+2>x-6 ? 1 4 . 3 x - 5- 2x-3>5

1,6.2 < 5 - 3 x

i14

,f8.I .9

-7 { r : i1 - 3 > Z x "x

r) 20r i-L - x = 1. \-r

2 1 ,*. > 4

22. f < g

23.(x-3)(r+s)>0

2 4 . x 2- 3 r *

26.**3r*1>0

27.4fI9x<9

28. 2 f - 6 x * 3 < 0

29.

x*'j,x ? " -n' 2 - r - 3 * x

31,.

32.

33. Prove Theorem 1,.1.27.

94. Prove Theorem 1,.1,.22.

17.2>-3-3x>-7

35. Prove 36. lf

2> 0

25. l - x - 2 x 2 > 0 74 3x-7 3-2x

Theorems 1.1,.24and 1.1.25.

n > b = O,prove that a2 > W.

37. lf. a

and b are nonnegative

numbers, and az ) br, prove that a > b.

38. Prove

that if a > 0 and b > 0, then A2: b2 if and only if a:

39.Pncve

that if b > a > 0 and c ) 0, then

a* c b + t'

b.

40. Prove that if x 1 !, then r < t(x + y) < y.

a b

1.2 ABSOLUTE VALUE

1.2|t Definition

The conceptof the "absolute value" ofa number is used in some important definitions in the study of calculus. Furtherrnore, you will need.tb work with inequalities involving "absolute value." The absoluteaalue of x, denoted by lrl, is defined by ifx>0 l*l-** ltl :-t

l 0 l: 0

ifx
1.2ABSOLUTE VALUE 1 5

l-u - a: la br|

Thus, the absolute value of a positive number or zero is equal to the number itself. The absolute value of a nggative numloer is the corresponding positive number because the negative of a negative number

is positive. . I L L U s T R A T I1o:Nl 3 l : 3 ; l - 5 1 : - ( - 5 ) : 5 ;

l,-b=la-bl1 lb F igure1.2.1

-L4l: l-61:-(-6):6. 18

We see from the definition that the absolute value of a number is either a positive number or zero;that is, it is nonnegative. In terms of geometry, the absolute value of a number x is its distance from 0, without regard to direction. In general, lo - bl is the distance between a and b without regard to direction, that is, without regard to which is the larger number. Refer to Fig. 1..2.1'. We have the following properties of absolute values.

1.2.2 Theorem l"l < a lf. and only if -a 1x I a,where a > "1..2.3 Corollary Irl = a if. and only if -a < )c < a, where a > \.2.4 Theorem ltl > a if andonly if x > a or x I -a,where 7.2.5 Corollary ltl = a if andonly if x > a or x s -a,where

0. 0. a > 0. a ) 0.

The proof of a theorem that has ant"if and only if " qualification requires two parts, as illustrated in the following proof of Theorem 1.2.2. plnr l.: Prove that lrl < a if.-a < x I a,where a ) O.Here,we haveto consider two cases:.x > 0 and r < 0. Casel: x=0. Then ltl : x. Becauser < a,we concludethat lrl < a. C a s e 2 :x < 0 . and obtain Then lrl : -r. Because-a < x, we aPPlyTheorem L.1,.25 a)-x or, equivalently,-x < a.But because-x: lrl, we have l*l < o. In both cases,then, wherea>0 0. Here we panr 2: Prove that lrl < a only if must show that whenever the inequality ltl < a holds, the inequality -a < x 1a also holds. Assume that lrl < a and consider the two cases x>0andr(0. CaseT: x>0. Then ltl : r. Because lxl < a,we conclude that r < a. Also, because -a 10 a ) 0, it follows from Theorem 1.1.31 that-a < 0. Thus/ we have
C a s e 2 :x < 0 . Then ltl :

Because ltl < a, we conclude that -x

( a. Also, be-

16


cause x I 0, it follows from Theorem l.r.zs that 0 < -x. Therefore,we have -a < 0 ( -r < a, or -a <-x < a, which by applying Theorem 1..7.25 yields-a < x 1a. In both cases, onlyif -a
l*l
wherea>0

I

The proof of Theorcm7.2.4is left as an exercise(seeExercise27). The following examples illustrate the solution of equations and inequalitiesinvolving absolutevalues.

EXAMPLE l-:

Solve for x:

lar+ 2l:5.

solurroN:

This equation will be satisfiedif either 3x*2:5 or -(3r*2):5

Considering each equation separatellr w€ have x: L and x: -& which are the two solutions to the given equation. uxluplr

2:

Solve for x:

lz*- rl: l4x+ 31.

solurroN:

This equation will be satisfiedif either

2x-L-4x*3

or 2x-1--Gx+j\

Solving the first equation, w€ have x: -2; solving the second,we get x: -*, thus giving us two solutions to the original equation. Exarnrpt,n3:

Solve f.or x:

l 5 r + 4 1: - 3 . rxr.vrpu 4: Find the solution set of the inequality

l x - 5 1< + Illustrate it on the real number line.

soLUTroN: Because the absolute value of a number may never be negative, this equation has no solution.

solurroN:

If r is a number such that

lr-51
< 4

(by Theorem' 1,.2.2)

Adding 5 to each member of the preceding inequality, we obtain 7
ifandonlyif

L
So, the solution set of the given inequality is (1, 9), which is illustrated in Fig. 1,.2.2.

1.2ABSOLUTE VALUE 17

EXAMPLE 5: Find the solution set of the inequatity

=n l3=-,,*l l2+xl-'

By Corollary

solurroN

'1..2.3, the given inequality is equivalent to

- ! ,- -< . 3 =. 2 ' - 4 2* x If we multiply by 2 * x, we must consider two cases,depending uPon whether 2 t x is positive or negative. C a s e ' l - :2 * r > 0 o r x > . - 2 . Then we have -4(2*r) =3-2x=4(2+x) or, equivalently, -8- 4x <3-2x

<8*4x

So, if x ) -2, then also-8 - 4x < 3- 2x andS-2x < 8 + 4x.We solve these two inequalities. The first inequality is -8-

4x <3-2x

Adding 2x * 8 to both members gives -2x < LL Dividing both members by .2 and reversing the inequality sign, we obtain x>-# The second inequality is 3-2x<8t4x Adding -4x - 3 to both members gives -6x <5 Dividing both members by -5 and reversing the inequality sign, we obtain x>-E Therefore, if x ) -2, x>-#andr>-8.

then the

inequality

holds if and only if

Becauseall three inequalities r ) -2, )c > -+, and r > -B must be satisfied by the same value of x, we have x > -8, or the interval [-8, +-).

Case2: 2*x(0orx<-2. Thus, we have -4(2*r)

=3-2x>4(2*x)

or, equivalently, -8-4x>3-2x>814x

18

REALNUMBERS,INTROEUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

Considering the left inequality, we have -8-4x>3-2x or, equivalently, -2x >'l-.'1, or, equivalently,

rs-+ From the right inequality we have 3-2x>8*4x or, equivalently, -5x>5 or, equivalently, r s-8 Therefore,if x 1-2, and r = -8.

the original inequality holds if and only if x = -t

Because all three inequalities must be satisfied by the same value of x, we have x =-+, or the interval (-*, -u, l. Combining the solution sets of Cases 1 and 2, we have as the solution

set(-o, -+l u [-8, **).

rxaprprr 6: Find the solution set of the inequality

l 3 r + 2 1> 5

soLUrIoN' By Theorem L.2.4,the given inequality is equivarent to 3x*2>5

or 3x*2<-s

(1)

That is, the givgn inequality will be satisfied if either of the inequalities in (1) is satisfied. Considering the first inequality, we have 3x*2>5 or, €guivalently, x)l Therefore, every number in the interval (1, +oo; is a solution. From the second inequality, we have 3xt2<-5 or, equivalently, x<-rs Hence, every number in the interval (-*, -+) is a solution.

1.2 ABSOLUTEVALUE

The solution set of the given inequality

is therefore

(1,+*;.

19

_+)u

You may recall from algebra that the symbol {a, where A > 0, is defined as the unique nonnegfltloe number x such that x2: A. We read tfi as "the principal square root of fl." For example,

fr:2

\6:0

G:E

NorE: tfr + -2; -2 is a square root of.4, bfi {4 denotesonly t]nepositiae square root of 4. Becausewe are concerned only with real numbers in this book, \/i is not defined lf. a < 0. From the definition of fr, it follows that

G:lxl -{5zl':3. 5 and .... For example, {*: The following theorems about absolute value will be useful later. 1.2.6 Theorem

lf. a and b are any numbers, then

l a b l :l o l' l u l Expressedin words, this equation states that the absolute value of the product of two numbers is the product of the absolute values of the two numbers. PROOF:

labl:'@ :{ffi

:G'\E : l o l' l b l 7.2.7 Theorem

lf a is any number and b is any number except 0,

That is, the absolute value of the quotient of two numbers is the quotient of the absolute values of the two numbers. The proof of Theorem 1.2.7 is left as an exercise (see Exercise 28). 1..2.8Theorem The Triangle lnequality

If. a and b are any numbers, then

la+bl=lal+ lbl

REALNUMBERS,]NTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS pRooF: By Definition '1..2.'1., either a :

or a: _lol; thus

-lol=a-lal Furthermore,

-lbl =b-lbl

(3)

From inequalities (2) and (3) and Theorem 1.1.23, we have

- ( l a l + l b l )= a + b < + lal lul Hence, from Corollary '/...2.3, it follows that

l a + b l= l a l + l b l

I

Theorem r.2.8 has two important and prove. 1.2.9 Corollary

corollaries which we now

If a and b are any numbers, then

lo-bl =lal+ lbl pRooF:lo- bl: la * (-b)l = lol+ l(-b) l: lal+ 1.2.10 Corollary

If a and b arc any numbers, then

l o l l b ls l a - b l pRooF: lol: l(r- b) + bl = lo - bl + lbl; thus, subtractinglbl both membersof the inequality,we have l o l - l b l= l o - b l '.Exercises1,.2**

st , T -tarL,

In Exercises1 through L0, solve fo, J

l. l4x* 3l:7 4 . 1 4 +3 r l: 1 7.lTxl:4-x

,olffil:n

Q ) l s t - 3 1: l 3 r + 5 1 g.2x*3:lar+51

3. 15- 2xl:11 6 . l x - 2 l: 1 3- 2 x l

g.l'+31 :s zl lx-

In ExercisesL1.through 14, find all the values of r for which the number is real. At.i s+

,:: ,j

state

.t v

'

Uttt.-l

.

1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS 21

>),23 fi-')fitu5

i tn

In Exercises15 through 26, find the solution set of the given inequality, and illustrate the solution on the real number line.

1 5 .l r + 4 1 < 7 1 8 . 1 6- 2 x l > 7

!. lr+41= lzx-61 -) -a' l.1 3o + x- s l -r2=l 1 27. Prcve Theorem 1.2.4.

16. l2x-sl < 3 1 9 .l 2 x - s l > 3 22. l3rl > 16- }xl v*?l .n 2s.llffil

1 7 .l 3 x - 4 1- z 2 0 .1 3 + 2 x l < l a - r l 23.le- zxl =- l4xl rc I l1_l

=1,-rr 26. !ffiI

28. Prove Theorem 1'.2.7.

In Exercises29 through 32, solve for r and use absolute value bars to write the answer.

29. 31..

12'8') thenlc-bl = l4l +lbl.(HINI:Writea-basa'+(-b) anduseTheorem 33. prcvethatif , andb areany nurnberg, 34. prcve that if a and b are any numben, then lal lbl < la bl. (Hrr.rr:Let lal: l(a b) + bl, and useTheorem1.2.8.) 35. What singleinequalityis equivatentto the followingtwo inequalities:s>b+ ca d a> b- c?

1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONS writing them in parentheses with a comma separating them as (I/ y). Note that the ordered pair (3, 7) is different from the ordered pair (7,3).

Definition 1..3.1.

The set of all ordered pairs of real numbers is called the number Plane' and each ordered Pafi (x, D is called a point in the number plane. The number plane is denoted by R2. fust as we can identify Rl with points on an axis (a one-dimensional space), we can identify R2 with points in a geometric plane (a two-dimendional space). The method we use with R2 is the one attributed to the French mathematician Rene Descartes (1595-1650), who is credited with the invention of analytic geometry in L637. A horizontal line is chosen in the geometric plane and is called the r axis. A vertical line is chosen and is called lhe y axis. The point of intersection of the r axis and tlne y axis is called the origin and is denoted by the letter O. A unit of length is chosen (usually the unit length on each axis is the same). We establish the positive direction on the r axis to the right of the origin, and the positive direction on the y axis above the origin.


F i g u r e1 . 3 . 1

We now associate an ordered pair of real numbers (x, y) with a point P in the geometric plane. The distance of P from the y axis (considered as positive if P is to the right of the y axis and negative if P is to the left of the y axis) is called the abscissa(or x coordinate) of P and is denoted by *.The distance of P from the r axis (considered as positive if P is above r the x axis and negative if P is below the x axis) is called the ordinate (or y coordinate)of P and is denoted by y. The abscissa and the ordinate of a point are called the rectqngular cartesian coordinatesof the point. There is a one-to-one correspondence between the points in a geometric plane 7) and R2; that is, with each point there corresponds a unique ordered pair (x,y), and with each ordered pafu (x, y) there is associated only one point. This one-to-one correspondence is called a rectangular cartesian coordinate system, Figure 1.3.1 illustrates a rectangular cartesian coordinate system with some points plotted. The x and y axes are called the coordinateaxes,They divide the plane into four parts, called quadrants, The first quadrant is the one in which the abscissa and the ordinate are both positive, that is, the upper right quadrant. The other quadrants are numbered in the counterclockwise direction, with the fourth, for example, being the lower right quadrant. Because of the one-to-one correspondence, we identify R2 with the geometric plane. For this reason we call an ordered pair (x, y) a point. Similarly, we refer to a "line" in R2 as the set of all points corresponding to a line in the geometric plane, and we use other geometric terms for sets of points in R2. Consider the equation A:x2-2

(1)

where (x,y) is a point inR2. we call this an equation inR2. By u solution of this equation, we mean an ordered pair of numbers, one for r and one for y, which satisfies the equation. Foi example, if r is re p l a c e d by 3 i n E q. (1), w e see that A :7; thus, x:3 and,y:7 const itutes a solution of this equation. If any number is substituted for x in the right side of Eq. (1), we obtain a corresponding value for y.It is seen, then, that Eq. (1) has an unlimited number of solutions. Table 1.3.1 gives a few such solutions. T a b l e1 .3 .7

x

l0

a:x2-zl-z

F i g u r e1 . 3 . 2

L2

-1

3

-3-4

4-1.-Z

2 7 74 -1

2

7

14

If we plot the points having as coordinates the number pairs (x,y) satisfying Eq. (1), we have a sketch of the graph of the equation. In Fig. 1'.3.2we have plotted points whose coordinates are the number pairs ob-

1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS

tained frorn Table 1.3.1. These points are connected by u smooth curve. Ary point (x, y) on this curye has coordinates satisfying Eq. (L). Also, the coordinates of any point not on this curve do not satisfy the equation. We have the following general definition. 1.g.2 Definition

The graph of an equation in I€ is the set of all points (x, y) in R2 whose coordinates are numbers satisfying the equation. We sometimes call the graph of an equation the locus of the equation. The graph of an equation in R2 is also called a curae. Unless otherrryise stated, an equation with two unknowns, x and y, is considered an equation in R2.

sxelrpr-n L: Draw a sketch of the graph of the equation' Y'-x-2:0

Q)

soLUrIoN:

Solving Eq. (2) for !, we have

(3)

y:!\tX+2 Equations (3) are equivalent to the two equations

(4 )

Y:lx+2

(s)

Y-_\'x-2

The coordinates of all points that satisfy Eq. (3) will satisfy either Eq. (4) or (5), and the coordinates of any point that satisfieseither Eq. (4) or (5) will satisfy Eq. (3). Table 1.3.2 gives some of these values of x and y. Table1-.3.2

x

00

v \/2 -\/2

1

\/i

"1. 2

-\/5

2

2 -2

3

3-L-1-2

\/5 -\tr

1, -L

0

Note that for any value of r <-2 there is no real value for y. Also, for each value of x > -2 there are two values for y. A sketch of the graph of Eq. (2) is shown in Fig. 1.3.3.The graph is a parabola.


2: Draw sketchesof ExAMPLE the graphs of the equations y:\/x+2 and

soLUrIoN: Equation (5) is the same as Eq. (a). The value of y is nonnegative; hence, the graph of Eq. (6) is the upper half of the graph of Eq. (3). A sketch of this graph is shown in Fig. 1..3.4. (5) Similarly, the graph of the equation

Y:-\E+2

Y:-1/Yt'2

a sketch of which is shown in Fig. 1.3.5, is the lower half of the parabola o f F i g . 1 .3.3.

F i g u r e1 . 3 . 4

F i g u r e1 . 3 . 5

rxltrpr.n 3: Draw a sketch of the graph of the equation

soLUTroN: From the definition have

y:lx*31

0)

y:x*3

if

of the absolute value of a number, we

r*3>0

and

y--(x+3)

if

r*3<0

or, equivalently, y:x*3

if

x>-3

and y--(x+3)

if

x1-3

Table 1.3.3 gives some values of x and y satisfying Eq. (Z). T a b l e1 .3.3 Figure1.3.6

x

0

L

2

3

v

3

4

5

6

-L 2

-2 1

-3 0

-4 1,

-5 2

-6

-7

3

-8

4

5

A sketch of the graph of Eq. (7) is shown in Fig. 1,.9.6. rxetvrplr 4: Draw a sketch of the graph of the equation (x-2y+3)(V-x'):0

(8)

soLUTroN: By the property of real numbers that ab:O a : 0 o r b :0, w e have from E q. (8) x-2y*3:0

if and

-9 6

1.3 THE NUMBERPLANE AND GRAPHSOF EQUATIONS

y-xz:o

(10)

The coordinates of all points that satisfy Eq. (8) will satisfy either Eq. (9) or Eq. (L0), and the coordinates of any point that satisfies either Eq. (9) or (10) will satisfy Eq. (8). Therefore, the graph of Eq. (8) will consist of the graphs of Eqs. (9) and (10).Table 1..3.4gives some values of r and y satisfying Eq. (9), and Table 1.3.5 gives some values of x and y satisfying Eq. (10).A sketchof the graph of Eq. (8) is shown in Fig. 7.3.7. L.3.4 Table

F i g u r e1 . 3 . 7

x

0L23-1-2-3-4-5

v

821r31,+o-+-1

T a b l e1 .3 .5

1.3.3 Definition

x

01,23-7-2-3

v

0L49149

An equation of a graph is an equation which is satisfied by the coordinates of those, and only those, points on the graph. is an equation whose graph consists of o ILLUSTRATIoN1: In R2, y:8 those points having an ordinate of 8. This is a line which is parallel to the o x axis, and 8 units above the r axis. In drawing a sketch of the graph of an equation, it is often helpful to consider properties of symmetry of a graph.

1.3.4 Definition

(3'2\o

F i g u r e1 . 3 . 8

Two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ. Two points P and Q are said to be symmetric with respectto a thirdpoint if and only if the third point is the midpoint of the line segment PQ. o rLLUsrRArroN 2: The points (3,2) and (3,-2) are symmetric with respect to the x axis, the points (3, 2) and (-3, 2) are symmetric with respect to the y axis, and the points (3,2) and (-3, -2) are symmetricwith o respect to the origin (see Fig. 1.3.8). In general, the points (x, y) and (x, -y) are symmetric with respect to the x axis, the points (r, y) and (-x, y) are symmetric with respect to the y axLS,and the points (x,y) and (-r, -y) arc symmetric with respect to the origin.


1.3.5 Definition

are symmetric with respect to R. From Definition 1.3.5 it follows that if a point (x, y) is on a graph which is symmetric with respect to the x axis, then the point (x,-y) also must be on the graph. And, if both the points (x,y) and (x,-y) are on the graph, then the graph is symmetric with respect to the x axis. Therefore, the coordinates of the point (x,-y) as well as (x, y) must satisfy an equation of the graph. Hence, we may conclude that the graph of an equation in r and y is symmetric with respect to the r axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation. We have thus proved part (i) in the following theorem. The proofs of parts (ii) and (iii) are similar. L.3.6 Theorem (Testst'or Symmetry)

The graph of an equation in r and y is (i) symmetric with respect to the x axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation; (ii) symmetric with respect to the y axis if and only if an equivalent equation is obtained when x is replaced by -x in the equation; (iii) symmetric with respect to the origin if and only if an equivalent equation is obtained when r is replaced by -x and y is replaced by -y in the equation. The graph in Fig. 1.3.2 is symmetric with respect to the y axis, and for Eq. (1) an equivalent equation is obtained when x is replaced by -r. In Example 1 we have Eq. (2) for which an equivalent equation is obtained when y is replaced by-y, and its graph sketched inFig. L.3.3 is symmetric with respect to the r axis. The following example gives a graph which is symmetric with respect to the origin.

rxlrvrpr.r 5: Draw a sketch of the graph of the equation xy:1

solurroN: We see that if in Eq. (1,1)x is replaced by -x and y is replaced by -y,an equivalent equation is obtained; hence,by Theorem 1.3.6(iii) the graph is symmetric with respect to the origin. Table L.3.6 gives some (11) values of x and y satisfying Eq. (11). T a b l eL .3.6

1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS

'l.lx. We seethat as r increasesthrough From Eq. (11) we obtain y : positive values and gets closer and positive values, y decreasesthrough positive values,y increasesthrough closerto zero.As x decreasesthrough positive values and gets larger and larger. As I increasesthrough negative values (i.e.,r takeson the values-4, -3, -2, -t, -+, etc.),y takes on negative values having larger and larger absolute values. A sketch of the graph is shown in Fig. 1..3.9.

F i g u r e1 . 3 . 9

1.3 Exercises In Exercises 1 through 6, plot the dven Point P and such of the following PoinB as may aPPly: (a) The point O suitr ttrit ttre tnJ through Q and P is perpendicular to the r axis and is bisected by it. Give the coordinates of Q. (b) The poiniR such that the line through P and R is perpmdicular to and i6 bisected by the y axis. Give the coordinates of R. (c) The point s such that the line through P and 5 iE bisected by the origrn. Give the_coordin"F-"-9f s.by the als' line though the origin 1ai fn" poi"t T such that the line *Eough P and 7 is perpendicular to and is bisected of T. coordinates bisecting the first and thtd quadrants. Give the 1 . P ( 1, - 2 ) 4. P(-2, -2)

2. P(-2,2) 5 . P ( - 1, - 3 )

3. P(2,2) 5 . P ( 0 ,- 3 )

In Exercises7 through2S, draw a sketch of the graph of the equation.

7'Y:2x*5 l0.y:-nffi 1 3 .x : - J 15.y:-lx+21 1 9 .4 f * 9 Y 2 : 3 6 22. Y' : 41cs 25. xz*y',:O 2 8 . ( y 2- x * z ) ( V + t E -

8'Y:4x-3 11.y':x-3 'I.. y2 * 1 7 .y : l r l - 5 2 0 . 4 x 2- 9 Y ' : 3 6 L4. x:

23. 4x2- Az:0 2 6 .( 2 x + V - l ) ( 4 V + r 2 ) : 0

Y: \81 y:5 y:lx-51 y--lxl+z A:4x3 3x2- L3xy- L0y2: g xa- 5x2yI 4y2: g

+):O 29. Draw a sketch of the graph of each of the following equations:

(a) V: lzx

(b\ v : -{2x

(c) y' :2x

30. Draw a sketch of the graph of each of the following equations:

(a) Y: !-2x

(b\ y : -!-2x

(c\ y' : -2N


3L. Draw a sketch of the graph of each of the following equations:

( a )r * 3 y : 0

(c) r' - 9y' :0

(b)r-3y:0

32. (a) Write an equation whose gtaPh is the r axis. (b) Write an equation whose graph is the y axis. (c) Write an equafion whose graph is the set of all points on either the .x axis or the y axis. 33. (a) Write a,n__equ{iol whose graph consists of all points having an abscissa of 4. O) Write an equation whose graph consistE of all points having an ordinate of -3. 34. Prove that a graPh that i5 symmetric with respect to both coordinate axes is also symmetric with rcspect to the origin. 35' Pro-eethat a graPh that is symmeFic with rcspect to any two perpendicular lines is also sym.rretric with respect to their point of intersection-

1.4 DISTANCE FORMULA AND MIDPOINT FORMULA

If A is the point (x,.,y) and B is the point (xr,y) (i.e., A and B havethe same ordinate but different abscissas), then the directed distance from A to B, denoted by AB, is defined ds x2 - x1. o ILLUSTRATToN 1: Refer to Fig. 1.4.1(a),(b), and (c).

u B(r,z)

A(4,2)

AB:'/.,4 (b)

Figur1 e.4.1 If .4 is the point (3, 4) and B is the point (9, 4), then AB :9 - 3: 6. rf A is the point (-8, 0) and B is the point (6,0), then B :6 - (-s) : 14. If A is the point (4, 2) and B is the point (L,2), then AB : 1.- 4: -3. we see that AB is positive if B is to the right of A, and AB is negative if B is to the left of A. o

v

D(-2,4)

C(-2, -g) CD: -6 (a)

Figure 1.4.2

CD:7 (b)

If C is the point (x,yr) and D is the point (x, yr), then the directed distance from C to D, denoted by CD, is defined as!z!r. o rLLUsrRArroN 2: Refer to Fig. La.2@) and (b). If C is the point (L, -2) and D is the point (1, -8), then CD : -g -J-2)--6. If C is the point (-2, -3) and D is the point (-2,4), then (-3) :7. Tl r.enumber C p i s posi ti vei f D i s above C , and CD C D :4 is negative if D is below C. o We consider a directed distance AB as the signed distance traveled by u particle that starts at A(rr, y) and travels to B(x2, !). ln such a case, the abscissa of the particle changes from 11 to x2, dfld we use the notation Ar ("delta x") to denote this change; that is, A,x: xz- xt Therefore, AB:

Lx.

1.4 DISTANCEFORMULAAND MIDPOINTFORMULA

Pr(x", vz)

It is important to note that the symbol Ar denotes the difference between the abscissa of B and the abscissa of A, and it doesnot mean "delta multiplied by x." Similarly, if we consider a particle moving along a line parallel to the y axis from a point C(x, yr) to a point D(x, yr), then the ordinate of the particle changes from Ar to Az.We denote this change by Ly ot

Ly: az- ar Thus, CD: Ly. Now let Pr(r1, yr) and Pr(xr, a) be any two points in the plane. we wish to obtain a formula for finding the nonrtegative distance between these two points. We shatl denote this distanceby lPtPrl. W" use absolute-value bars becausewe are concernedonly with the length, which is a nonnegative number, of the line segmentbetween the two points P1and Pr. To derive the formula, we note that lffil is the length of the hypotThis is illustrated in Fig. 1.4.3for P1and of a right triangle PLMP2. "rrrrr" quadrant. first are in the Pr, both of which we have theorem, Using the Pythagorean

F i g u r e1 . 4 . 3

+llvl' EF"f: lArl2 So

lF-nl:\MV+WP That is,

(1)

lF:P,l:

Formula (1) holds for all possible positions of Pt and P, in all four qqadrants. The length of the hypotenuse will always be lFfrl, and the lengths of the two legs will always be [Ar[ and lAyl (seeExercisesL and 2). We state this result as a theorem. '1,.4.1 Theorem The undirected distance between the two points Pr(xr, Ar) and Pr(xr, yr) is given by

EXAMPLEl.: If a point P(x, U) is such that its distance from A(3,2) is always twice its distance from B(-4, L), find an equation which the coordinates of P must satisfy.

solurroN:

From the statement of the problem

lPTl:2lPBl Using formula (1), we have

@:z@ Squaring on both sides, we have x 2- 6 x + 9 + y ' - 4 y I 4 :

4(xz*8r

or, equivalently, 3 x 2* 3 y ' * 3 8 r - 4 y + 5 5 : 0

* t6 + y2- 2y * 1)


ExAMPLE2: Show that the triangle with verticesat A(-2, 4), B(-5, L), and C(-6,5) is isosceles.

SOLUTION:

The triangle is shown in Fig. L.4.4.

lBel: @ : \ 4 + 1 6 : f r : @:\tr6+1:{lT la-e; IEAI:W : \ / N : j f i Therefore,

A(-2,4)

Hence, triangle ABC is isosceles.

B(-5, i.)

Figure1.4.4

nxervrrrn3: Prove analytically that the lengths of the diagonals of a rectangleare equal.

C(0, b)

(0,0) F i g u r e1 . 4 . 5

B(a,b)

SOLUTION:Draw a generalrectangle.Becausewe can choosethe coordinate axes anywhere in the plane, and becausethe choice of the position of the axes does not affect the truth of the theorem, we take the origin at one vertex, the x axis along one side, and the y axis along another side. This procedure simplifies the coordinates of the vertices on the two axes. Referto Fig. 1..4.5. Now the hypothesis and the conclusion of the theorem can be stated. Hypothesis:OABC is a rectanglewith diagonalsOB and AC. Conclusion: IO-B| : lrel .

l o B:lf f i : r t + 6 2 l A C: lW : \ f r ] A t Therefore,

lml : ld-cl Let P, (xr, yr) and P, (xr, yr) be the endpoints of a line segment. We shall denote this line segment by PtPr. This is not to be confused with the notation PrPr, which denotes the directed distance from P, to Pr. That is, PrP, denotes a number, whereas PrP2is a line segment. Let P(x, y) be the midpoint of the line segment P,Pr. Refer to Fig. L.4.6. In Fig. I.4.6 we see that triangles P'RP andPTPrare congruent. Therefore, l-PtRl: lPTl, and so r - xr: xz- x, giving us

1.4 DISTANCEFORMULAAND MIDPOINTFORMULA

Similarly, lRPl :

lTPzl.Then A Ar: Az- A, andtherefore (3)

Hence, the coordinates of the midpoint of a line segment are, respectively, the average of the abscissasand the average of the ordinates of the endpoints of the line segment. P z( x z ,v z )

P(x,y)

u__ R(t, yt)

T(xr, Y)

__JS(rr,yr)

F i g u re1 .4 .6 In the derivation of formulas (2) and (3) it was assumed that x, ) xt and y, ) yr.The same formulas are obtained by using any orderings of these numbers (see Exercises 3 and 4).

nxaurr,E 4: Prove analytically that the line segments ioining the midpoints of the opposite sides of any quadrilateral bisect each other.

B(b,c)

F i g u r e1 . 4 . 7

Draw a general quadrilateral. Take the origin at one vertex solurroN: and the x axis along one side. This method simplifies the coordinates of the two vertices on the r axis. See Fig. L.4.7. Hypothesis: OABC is a quadrilateral. M is the midpoint of OA, N is the midpoint of CB, R is the midpoint of OC, and S is the midpoint of AB. Conclusion:MN and RS bisect each other. pRooF:

To prove that two line segments bisect each other, we show that they have the same midpoint. Using formulas (2) and (3), we obtain the coordinates of M, N, R, and S. M is the point Ga, 0), N is the point ( + ( b + d ) , t r ( c * e ) ) ,R i s t h e p o i n t 1 ! d , t e ) , a n d S i s t h e p o i n t G @ + b ) , i c ) . + d). T h e a b s c i s s ao f t h e m i d p o i n t o f M N i s i l L a + + ( b + d ) l : * ( a + b : e). The ordinate of the midpoint of MN is +[0 + Lk + e)f ik + Therefore, the midpoint of MN is the point (ifu + b + d), ik + e)). + d). T h e a b s c i s s ao f t h e m i d p o i n t o f R S i s i l i a + * @ + b ) l : * ( a + b The ordinate of the midpoint of RS is tlie + *cf : Ik + e). Therefore, the midpoint of RS is the point (ifu + b + d), i(c + e)). Thus, the midpoint of MN is the same point as the midpoint of RS. I Therefore, MN and RS bisect each other.

92


Exercises 7.4 1. Derive distance formula (1) if Pr is in the third quadrant and P, is in the second quadrant. Draw a figure. 2. Derive distance formula (1) if Pr is in the second quadrant and P, is in the fourth quadrant. Draw a figure. 3. Derive midpoint formulas (2) and (3) if Pr is in the first quadrant and p, is in the third quadrant. 4. Derive midpoint formulas (2) and (3) iI4 (rr, yr) and &(rr, yr) are both in the second quadrant and x, > xrand,y, > yr. 5. Find the length of the medians of the rriangle having vertices A(2,3),8(3,-g), and C(-1,-1). 5. Find the midpoints of the diagonals of the quadrilateral whose vertices are (0,0), (0,4), (3,5), and (3, 1). (.\Prove that the trianSle with vertices A(3,-6), B(8,-2), and C(-1, -1) is a right triangle. Find the areaof the triangle. \'-l(HrNr: Use the converseof the Pythagoreantheorem.) 8. Prove that the Points A(6, -lS) , B(-2, 2), C(13, 10), and D(21, -5) are the vertices of a square. Find the tength of a diagonal. 9. By using distance formula (1), prove that the points (-3, 2) , (1, -2), and (9, -10) lie on a line. 10. If one end of a line segment is the point (-4, 2) and the midpoint is (3, -L), find the coordinates of the other end of the line segment.

11. The abscissa of a Point iB -6, and its distance from the point (1, 3) is V-74. Find the ordinate of the point. 12. Determine whether or not the poifis (14,7) , (2, 2) , and,(-4, -1) lie on a line by using distance formula (1). /;:\ '\3.llf two vertices of an equilateral hiangle are (-4,3) and (0,0), find the third vertex. d Findan equation that must be satisfied by the coordinatesof any point that is equidistant from the two points (-3, 2) and (4,6). 15. Find an equation that must be satisfied by the coordinates of any point whose distance from the point (5, 3) is atways two units Feater than it8 distance frcm the point (-4, -2). 16. Giver the two Points d(-3,4) andB(2,5), find the coordinates of a point P on the line through A and B such thatp is (a) twice as far from 4 as frcm B, and (b) twice as far frcm B as from _A. 17. Find the coordinates of the three points that divide the tine segment ftom l(-5,

3) to 8(6, g) into four equal parts.

18. If rr. and r, are positiv€_intgg:ers,-provethat the coordinates of the point p(r, y), which divides the line segment plp, : r,/rr-are given by in the ratio rr/r2-that i3, IP-,,PUlPrEI ,:

(t" - rt)x, + rrxu fz"t2

ur4 , :

(r, - ,r)y, + ,rv"

In Exercises 19 through 23, use the formulas of Exercise 18 to find the coordinates oI point p. 19' The Point P is on the line segment between points Pr (1 ,3) ?flrd,P26,2) and is three times as far from P, as it b ftom pr. 20. The Point P is on the line segmentbetween points Pl(1 ,3) and Pr(6,2) and is three times as far from Pr as it is frcmpr. 21. The Point P i8 on the line through P' and P, and is three times as far ftom Pr(6, 2) as it is from P, (1,3) but is not between Pr and Pr. 22. The Point P is on the Une though P, and P, and is three times as far from P1(1, 3) as it is ftom Pr(6,2) but is not between P1and P2.

23. The point P is on the line through Pr(-3,5) and Pr(-L,2) so that ppr : + . p.,pr. 24. Find an equation whose graph is the circle that is the set of all points that are at a distance of 4 units from the point (1, 3).

1.5 EQUATIONSOF A LINE

25. (a) Find an equation whose graph consist8 of all points equidistant frcm the Points (-1,2) and (3,4). (b) Draw a skekh of the gnph of the equation found in (a)' 26. prove analytically that the sum of the squares of the distanc$ of any Point from two opposite vertic€s of any rcctangle is equal to the sum of the squares of its distances from the other two verticeE' 22. prcve analytically that the line segment ioining the midpoints of two opposite sides of any quadrilateral and the line segment joining the midpoints of the diagonala of the quadrilateral bieect each other' 28. prove analytically that the midpoint of the hypotenuse of any right triangle is equidistant frcm each of the three vettices. 29. Prove analytically that if the lengths of two of the medians of a triangle arc equal, the triangle is isoeceles.

1..sEQUATIONS OF A LINE

Letl be a nonverticalline and Pr(x' U) andP"(xr, A)be any two distinct points on l. Figure L.5.1shows such a line. In the figure, R is the point Qc,Ur),and the points Pr, Pu and R are vertices of a right triangle; furthermore, P,.R: xz-xr and R-Pr: Az-Ar The number Az-Ar gives the measure of the change in the ordinate from Pr to Pr, and it may be positive, negative, or zero.The number x2- xl gives the measureof the change in the abscissafrom P, to Pr, and it may be positive or negative.The number x, - rr may not be zero becaus?x2 * r, since the line I is not vertical. For all choices of the points P, and P, on l, the quotient

R(rz,yJ

Az- Ur Xz- xr

,

is constanq this quotient is called the "slope" of the line. Following is the formal definition.

F i g u r e1 . 5 . 1

1.S.1 Definition

If Pr(xr, Ar) and Pr(xr, yr) arc any two distinct points on line l, which is not parallel to the y axis, then the slopeof l, denoted by m, is given by

; : y EXz *

-at P, (xr, y, R(xr,Yr)

Pr(7r,V,)

(1) ?Cr

In Eq. (L), x, * x1 since / is not parallel to the y axis. The value of m computed from Eq.(t) is independent of the choice of the two pointr_P, and P2 on L To show this, suppose we choose two different points, tr(n, yr) and Pr(ir,yr), and compute a number nr from Ee. (1). ar *m- :_- A - zxz- xt

7z,Ai x

We shall show tlnilt n - m. Refer to Fig. 1.5.2. Triangles PrRPz and P1RP2 are similar, so the lengths of corresponding sides are proportional. Therefore,

F i g u r e1 . 5 . 2

V:-

Y_t- Az

Xz-

Xr

or m:m

-

Xz-

Ar Xt

34

REAL NUMBERS,INTRODUCTION T O ANALYTICGEOMETRY, AND FUNCTIONS

Hence, we conclude that the value of m computedfrom Eq. (1) is the same number no matter what two points on I are selected. In Fig. 1.5.2, x2 ) x1, Uz) Ar, x, > n, and y2 > fr. The discussion above is valid for any ordering of these pairs of numbers since Definition 1.5.1holds for any ordering. In sec. 1.4 we defined Ly:uz-ur and Ar:xz rr. substituting thesevalues into Eq. (L), we have F i g u r e1 . 5 . 3

Multiplying on both sidesof this equationby Lx,we obtain Ly: mLx

e)

It is seen from Eq. (2) that if we consider a particle moving along line l, the change in the ordinate of the particle is proportional to the change in the abscissa, and the constant of proportionality is the slope of the line. If the slope of a line is positive, then as the abscissa of a point on the line increases, the ordinate increases. Such a line is shown in Fig. 1.5.3. In Fig. "1..5.4, we have a line whose slope is negative. For this line, as the abscissa of a point on the line increases, the ordinate decreases. Note that if the line is parallel to the x axis, then Uz: Ur and so m:0. If the line is parallel to the y axis, xz: xr, thus, Eq. (1) is meaningless because we cannot divide by zerc. This is the reason that lines paiallel to the y axis, or vertical lines, are excluded in Definition 1.5.L. We Jay that a vertical line does not have a slope.

F i g u r e1 . 5 . 4

. rLLUSrRArroNl.: Let I be the line through the points P,,(2,3) and p2(4,7). The slope of I, by Definition "!..5.'!., is given by

*:7n!:z L

Pr $ , 9 )

Referto Fig. 1.5.5.rf P(x,y) and Q@+ L*,y + Lil are any two points on I, then L A_ . , Ax-' Ay :2

P:-(2,3)

P+(-1

F i g u r e1 . 5 . 5

A,x

Thus, if a particle is moving along the line l, the change in the ordinate is twice the change in the abscissa. That is, if the particle is at pr(4,7) and the abscissa is increased by one unit, then the ordinate is increased by two units, and the particle is at the point Pr(5,9). Similarly, if the particle is at Pt(2,3) and the abscissa is decreased by three units, then the ordinate is decreased by six units, and the particle is at Pr(-L, -3). o Since two points Pr(xr, yr) and Pr(xr, y)

determine a unique line, we

1.5 EQUATIONS OF A LINE

should be able to obtain an equation of the line through these two points. Consider P(x,y) any point on the line. We want an equation that is satisfied by r and y if and only rf P(x, D is on the line through P1(rr, yr) and Pr(xr, Ar).We distinguish two cases. C a s e7 : x z : x r . In this case the line through P, and P, is parallel to the y axis, and all points on this line have the same abscissa. So P(x,,U) ts any point on the line if and only if (3)

X: Xr

Equation (3) is an equation of a line parallel to the y axis. Note that this equation is independent of y; that is, the ordinate may have any value whatsoever, and the point P(x,y) is on the line whenever the abscissais rr. Case2: x2 * x1. The slope of the line through Pt and P, is given by Ar ry:Uzxz- xr lf P(x, g) is any point on the line except (\, given by

(4) !r), the slope is also

At 1--A x- xr

(s)

The point P will be on the line through P, and Pz if and only if the value of m fuomEq. (4) is the same as the value of m ftom Eq. (5), that is, if and only if U A t- A z - A r X-

Xr

Xz- Xr

Multiplying on both sides of this equation by (x - xr),we obtain (6) Equation (5) is satisfied by the coordinates of P, as well as by the coordinates of any other point on the line through P1 and Pr. Equation (6) is called tkte two-point fonn of an equation of the line. It gives an equation of the line if two points on the line are known. -3) o rLLUsrRArroN 2: An equation of the line through the two points (5, and (-2,3) is

y-(-3):*(r-5)

y+3--2@-6) 3x*4Y:6

REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

If in Eq. (6) we replace(y, - !r)l(x, - x) by *, we get I i.:r )i' I.'

+ l- o :

t I

';

...

} i Yfl:gqt{,&Es'*"atu

e)

Equation (7) is called the point-slope form of an equation of the line. It gives an equation of the line if a point Pr(xr, a) on the line and the slope m of the line are known. . rLLUSrRArroN 3: An equation of the line through the point (-4,-5) and having a slope of 2 is

y - (-s) :Zlx - (-4)l 2x-y*3:0

o

If we choosethe particular point (0, b) (i.e., the point where the line intersectsthe y axis) for the point (h, A) in Eq. (7), we have y-b:m(x-0) or, equivalently,

+b

(8)

The number b, which is the ordinate of the point where the line intersects the y axis, is called the y intercept of the line. Consequently, Eq. (8) is called the slope-intercept form of an equation of the line. This is especially important because it enables us to find the slope of a _form line from its equation. It is also important becauseit expressesttre y coordinate explicitly in terms of the r coordinate. rxevrpln L: Given the line having the equation 3r I 4y :7, find the slope of the line.

sol.urroN:

Solving the equation f.or !, we have

,:-fix*I Comparing this equation with Therefore, the slope is -*.

Eq. (8), we see that

and b:*.

Another form of an equation of a line is the one involving the intercepts of a line. We define the r intercept of a line as the abscissa of the point at which the line intersects the r axis. The r intercept is denoted by o. If the r intercept a and the y intercept b arc given, we have two points (a, 0) and (0, b) on the line. Applying Eq. (6), the two-point form, we have h- 0 .

Y-o:6ik-

-aY:bx-ab b x* a A : a b

a)

OF A LINE 1.5 EOUATIONS

by ab, if. a * 0 and b + 0, we obtain

Dividing

(e) Equation (9) is called the intercept fonn of an equation of the line. Obviously it does not apply to a line through the origin, becausefor such a line both a and b arc zero. nxlupln 2: The point (2,3) bisects that portion of a line which is cut off by the coordinate axes.Find an equation of the line.

Refer to Fig. 1.5.6. lf. a is the r intercept of the line and b is the y intercept of the line, then the point (2, 3) is the midPoint of the line segment joining (a, 0) and (0, b). By the midpoint formulas/ we have soLUrIoN:

A

andt:ry

atl

A:4

and b:6

The intercept form, Eq. (9), gives us x +!-t 4'6 or, equivalently, F i g u r e1 . 5 . 6

3x*2y-12 t.5.2 Theorem The graph of the equation (10)

Ax*By*C:0

where A, B, and C are constants and where not both A and B are zero, is a straight line. PRooF: Consider the two cases B + 0 and B : 0. CaseL: B+0. Because B + 0, we divide on both sides of Eq. (10) by B and obtain AC

(11)

y--i*-,

Equation (11) is an equation of a straight line becauseit is in the slope-- -CIB. intercept form, where m: -AlB and b C a s e2 : B : 0 . BecauseB:0, Ax*C:0

we may concludethat A * 0 and thus have


or/ equivalently,

*:- aC

(12)

Equation (12) is in the form of Eq. (3), and so the graph is a straight line parallel to the y axis. This completes the proof. r Because the graph of Eq. (10) is a straight line, it is called a linear equation.Equation (10) is the general equation of the first degree in x andy. Because two points determine a line, to draw a sketch of the graph of a straight line from its equation we need only determine the cooidit utet of two points on the line, plot the two points, and then draw the line. Aty two points will suffice, but it is usually convenient to plot the two points where the line intersects the two axes (which are given by the intercepts). rxeupm 3: Given line lr, having the equation 2x - 3y : 12, and line Ir, having the equation 4x * 3y:6, draw a sketch of each of the lines. Then find the coordinates of the point of intersection of l, and Ir.

solurroN: To draw a sketch of the graph of lr, we find the intercepts a and b. In the equation of Ir, we substitute 0 for r and get b - -4.In the equation of Ir, we substitute 0 for y and get a: 6. Similarly, we obtain the intercepts a and b for lr, and for l, we have a:8 and,b: 2. The two lines are plotted in Fig. 1..5.7. To find the coordinates of the point of intersection of /, and Ir, we solve the two equations simultaneously. Because the point must lie on both lines, its coordinates must satisfy both equations. If both equations are put in the slope-intercept form, we have

a:?x-4

and

-tx*2

Eliminating y gives 3x-4:-gx*2 2x-L2:-4x#6 x:3 So

y:BG)_4

F i g u r e1 . 5 . 7

Therefore, the point of intersection is (3, -2). 1'5'3 Theorem

If /r and l, are two distinct nonvertical lines having slopes m, and mr, respectively, then /, and l, are parallel if and only if mr: //t2. PRooF: Let an equation of /, be : tftrx * br, and let an equation of l, A be y : rnzx * br. Because there is an "if and only if " qualiiication, the proof consists of two parts. panr 1: Prove that l, and I, are parallel if ftir:

tnz.

1.5E QU A TIONOF S A LIN E Assume that I, and I, are not parallel. Let us show that this assumPtion leads to a contradiction. If I, and I, are not parallel, then they intersect. Catl this point of intersection P(*o, /o).The coordinates of P must satisfy the equations of 11and Ir, and so we have ls: But mr: ls:

!o:

fti2xo* b2

!o:

nttfio* b2

rnlxs* b1 and rrt2rwhich gives mfis * b1 and

b2,both from which it follows thatbl: bz.Thus, becauserrtr: mrandbl: lines 11and /, have the same equation , A : mrx * br, and so the lines are the same. But this contradicts the hypothesis that l, and I, are distinct lines. Therefore, our assumption is false. So we conclude that /r and 12 are parallel. pARr 2: Prove that l, and I, are parallel only if mt: lftz. Here we must show that if Ir and l, are parallel, then rflt: rtrz. Assume that m, * mr. Solving the equations for \ and I, simultaneously, we get, upon eliminating V, m t x * b r : m 2 x* b 2 from which it follows that ( m , - m r ) x : b z- b t Because we have assumed that m, * *r, this gives - - _ _b r - b ,

L _

t7\-

ffiz

Hence, 11and l, have a point of intersection, which contradicts the hypothesis that l, and 12are parallel. So our assumption is false, and thereI fore mr: nflz. In Fig. 1.5.8, the two lines Ir and l, are perpendicular. We state and prove the following theorem on the slopes of two perpendicular lines. 1.5.4 Theorem

If neither line /, nor line l, is vertical, then Ir and 12are perpendicular if and only if the product of their slopes is -1.. That is, if m, is the slope of 11and m, is the slope of. Ir, then Ir and I, are perpendicular if and only if n tfl tz - -1 . PROOF:

- -1' Panr 1: Prove that l, and 12are perPendicular only if mtm2 Let L, be the line through the origin parallel to l, and let L, be the line through the origin parallel to Ir. See Fig. 1.5.8. Therefore,by Theorem 1..5.3,the slope of line L, ts m, and the slope of line L, is m2.Because neither


L, nor L, rs vertical, these two lines intersect the line x: ! at points p, and Pr, respectively.The abscissaof both P, and P, is L. Let be the ordi/ nate of Pt. Since Lr contains the points (0,0) and (1.,y) and its slopeis mr, we have from Definition 1.5.1 ttL:

a-0 i=

and so y = mr.Similarly, the ordinate of P, is shown to be mr. Applying the Pythagorean theorem to right triangle prop2, we get

l o t r ,+; zl @ f : l P , k l , F i g u r e1 . 5 . 8

(1 3 )

By applyingthe distanceformula,we obtain l o - 4 l r : ( 1- 0 ) ,f - ( m r - 0 ) ,: 7 * m ] l O t r l r : ( 1- o ) , * ( m r - o ) , : r * m z 2 l F J r P : ( 1 - 1 ) , * ( m r - m r ) ,- n t r z- 2 m r m r t m r , Substituting into Eq. (13)gives (l + mrz)+ (1 + m22)- ftizz- 2mrm,I.mrt 2:

-2mrffiz

tltinz - -1 penr 2: Prove that I, and l, are perpendicular if.mrm" - -1. Starting with lt|fitz - -'1, we can reverse the steps of the proof of Part 1 in order to obtain

lO-&l'+ l1trl': lPEl'z from which it follows, from the converse of the Pythagorean theorem, that Lt andL, are pe{pendicular; hence, /, and l, are perpendicular. I Theorem I.5.4 states that if two lines are perpendicular and neither one is vertical, the slope of one of the lines is the negative reciprocal of the slope of the other line. o rLLUsrRArroN 4: If line /t is perpendicular to line I, and the slope of /, is 3, then the slope of /, must be -9. o

sxarvrpr,n 4: Prove by meahs of slopes that the four points A(5,2), B(9,6), C(4,g), and DQ,4) are the verticesof a rectangle.

solurroN:

See Fig. 1.5.9. Let

6-2 ntr: the slope of AB - 8 - 6 8-6 m, J the slope of BC -

4- I

1.5 EQUATIONSOF A LINE

/fi.:the sloPe of DC:ft:

C(4,8)

B(8,6)

41

''

rn.:the slopeof.AD:=:-+. Because

D

Tnr: mr, AB ll DC. /n2: ma,BC ll AD. ntrttrz--'1, AB L BC.

F i g u r e1 . 5 . 9

Therefore, quadrilateral ABCD has opposite sides parallel and two adjacent sides perpendicular, and we conclude that ABCD is a rectangle.

EXAMPLE5: Given the line / having the equation

Because the required line is perpendicular to line l, its slope solurroN: must be the negative reciprocal of the slope of l. We find the slope of lby putting its equation into the slope-intercept form. Solving the given' equation for y, we obtain

2x*3Y-5:0

find an equation of the line perpendicular to line I and passing through the point

Aet,3).

y--tx** Therefore, the slope of I is -3, and the slope of the required line is 9. gecausewe also know that the required line contains the point (-L, 3), we use the point-slope form, which gives y-3:$(x*t) 2y-6:3x*3 3x-2Y+9:0

1-.5 Exercises In Exercises L through 4, find the slope of the line through the given points.

1 . ( 2 ,- 3 ) , ( - 4 , 3 )

2 . ( 5 , 2 ), ( - 2 , - 3 )

3.(+,+),eE,&)

4 . ( - 2 . T , 0 . 3 ), ( 2 . 3 , 1 . 4 )

In Exercises5 through14, find an equation of the line satisfying the given conditions. 5. The slope is 4 and through the point (2,-3). 6. Through the two points (3, L) and (-5,4). 7. Through the point (-3, -4) and parallel to the y axis. 8. Through the point (I,-7) and parallelto the x axis. 9. The x intercept is -3, and the y intercept is 4.

TO ANALYTICGEOMETRY. REAL NUMBERS,INTRODUCTION AND FUNCTIONS

10. Through (1, 4) and parallel to the line whose equation is 2r - 5y + 7:0. 11. Through (-2, -5) and having a slope of l/5. 12. Through the origin and bisecting the angle between the axes in the first and third quadranb, 13. Through the origin and bisecting the angle between the axes in the second and fou h quadrants. 14. The slope is -2, and the r intercept is 4. the points (1, 3) and (2, -2), and put the equation in the intercept form. equation of the line through the points (3, -5) and (1, -2), and put the equation in the slope-interceptform. dg. Fitrd "tr 17. Show by means of slopes that the points (-4, -1) , (3, +) , (8, -4) , and (2, -9) are the vertices of a trapezoid. 15. Find an equation of the line thouth

18. Thrce consecutivevertices of a parallelogramare (-4, 1), (2,3),and (8,9). Find the coordinatesof the fou h vertex. 19. For each of the following sets of three point6, detemine by means of slopes if the points are on a line: (a) (2, 3), (-4,-7), (s,8);(b) (-3,6), (3,2), (e,-2); (cl (2,-7), (7,7), (3,t); and (d) (1,6), (1,2), (-s,-4). 20. Prove by means of slopes that the three points A(3, 1), B(6,0), and C(4, ) are the vertices of a right triangle, and find the area of the triangle. /\2T. Given the line I having the equation 2y - 3r : 4 and the point P(1, -3), find (a) an equation of the line through P " and perpendicul to l/(b) the shortest distance from P to line l. \ 22. lf A,B, C, and D arc cdhstarts, show that (a) the lines Ax + By+ C:0 a dAx+By+D:0 are parallel and (b) the ljmesAr+ By + C:0 and Bx- Ay * D:0 areperpendicular. 23. Given the line l, having the equation Ar + By + C: O, B + 0, find (a) the slope, (b) the y intercept, (c) the x intercept, (d) an equation of the line through the origin perpendicular to L 24. Find an equation of the line which has equal intercepts and which passesthrouth the point (8, -6). 25. Find equations o{ the three medians of the triangle having vertices A(3, -2) , BG, a), and C(-l,1), they m€et in a point.

and prove that

26. Find equations of the perpendicular bisectors of the sides of the triande having vertices A(-l , -3) , B (5, -3), and C(5,5), and prove that they meet in a point. 27. Find an equation of each of the lines through the point (3,2), which forms with the coordinateaxesa triangle of area12. 28. Let il be the line having the equation.r4'r* Bly+ C1:0,andlet I'be the line having the equationArx+ 82! + Ca:0. If ll is not parallel to L and iJ & is any constant,the equation A'x 'l Bry I C' * k(A"x * Bry * C") :0 represents an unlimited number of lines. Prove that eadr of these lines contains the point of intersection df lr and t. 29. Given an equation ot It is 2x 'f 3y - 5: 0 and an equation of l, is 3x + 5y - 8 : 0, by using Exercise 28 and without finding the coordinates of the point of intersection of ,r and lr, find an equation of the line through this point of intersection and (a) passing through the point (1, 3); (b) parallel to the r axis; (c) parallel to the y axis; (d) having slope-2; (e) perpendicular to the line having the eqtatio\ 2x + y :7; (f) forming an isoscelestriangle with the coordinate axes. 30. Find an equation of each straight line that is perpendicular to the line having the equation 5r - y: I and that forms with the coordinate axes a triantle having an area of measure 5. 31. Prove analytically that the diatonals of a rhombus are perpendicular. 32. Prove analytically that the line segments joining conEecutivemidpoints of the 6ide8 of any quadrilateral form a parallelogram.

1.6THECIBCLE 43 33. Prove analytically that the diagonals of a parallelogram bisect each other. 34. Prove analytically that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

1.5 THE CIRCLE The simplest curve that is the graph of a quadratic equation in two variables is the "citcle," which we now define.

'1,.6.1 Definition

1.6.2 Theorem

A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is called the center of the circle, and the measure of the constant equal distance is called the radius of the circle. The circle with center at the point C(h, k) and radius r has as an equation ';

l -,: i

:

;,(x *'h}P, * (g1= k)2,* 7zi'1

(1)

pRooF: The point P(x, y) lies on the circle if and only if

l n c l :, that is, if and only if

@:r This is true if and only if (x-h)'+

(y-k)z--tz

which is Eq. (1). Equation (1) is satisfied by the coordinates of those and only those points which lie on the given circle. Hence, (1) is an equation I of the circle. From Definition L.3.2, it follows that the graph of Eq. (1) is the circle with center at (h, k) and radius r. If the center of the circle is at the origin, then h: k: 0; therefore, its equation is x' + y' : 12.

' :-LJ-'

C(2, - 3)

o rLLUsrRArroN 1: Figure 1.6.1 shows the circle with center at (2, -3) and radius equal to 4. For this circle h:2,k:-3, and r:4. We obtain an equation of the circle by substituting these values into Eq. (1) and we obtain (x-2)'+ly

- (-3)12:42

(x-2)'+ (y+3)':t6 Squaring and then combining terms, we have x2-4x+4+y'*5y*9:t6

F i g u r e1 . 6 . 1

)c'+y'-4x+6y-3:0

REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

EXAMPLE1:

Given the equation

x'+ y'* 6x-2y - 15: o prove that the graph of this equation is a circle, and find its center and radius.

solurroN:

The given equation may be written as

(xr+6x)t(yr-2y):ts Completing the squares of the terms in parentheses by adding 9 and 1 on both sides of the equation, we have

(x' * 6x*9) + (y' - 2y I1) : 15+ 9 + 1, (r*3)2+(V-1)':25 Comparing this equation with Eq. (1), we see that this is an equation of a circle with its center at (-3, I) and a radius of 5.

In Eq. (L), removing parenthesesand combining terms gives x ' + y ' - Z h x- Z k y+ ( h , + k 2- r r ) : 0

Q)

Equation (2) is of the form )e+y'*Dx+Ey*F:0

(3) -2h, -2k, : : : where D E and F h2* k2 f. Equation (3) is called the general form of an equation of a circle, whereas Eq. (1) is called the center-rsdiusform. Becauseevery circle has a center and a radius, its equation can be put in the center-radius form, and hence into the general form. We now consider the question of whether or not the graph of every equation of the form x'+y,*Dx*Ey*F:0 is a circle. To determine this, we shall attempt to write this equation in the center-iadius form. We rewrite the equation as x'+y,*Dx*Ey--F and completethe squaresof the terms in parenthesesby adding]nDzand iE' on both sides, thus giving us

--F + +Dz* iE, (x' * Dxt +D')+ (y' * Ey++E'z) or, €guivalently, (x + LD)2+ (y + +E)2: i(D' + Ez- 4F)

( 4)

Equation (a) is in the form of Eq. (1) if and only if

+(D,+E2-4F):rz We now consider three cases,namely, (D'+ E2- 4F) as positive, zero, and negative. Case7: (D, + E2- 4F) > 0. Then r': *(D' + E2- 4F), and so Eq. (a) is an equation of a circle having a radius equal to it/FE= 4F and its center at (-*D,-+E).

1.6 THE CIRCLE

C a s e2 : D z + E z- 4 F : 0 . Equation ( ) is then of the form (x + tD)z + (Y * iE)z :0

(5)

- -L2D and Becausethe only real values of r and y satisfying Eq. (5) are v -+E). Comparing Eq. (5) with y:-+E, the graph is the point (-+D, and r - 0. Thus, this point can be Eq. (1),we see that h:-+D, k:-iE, called a point-circle. Case3: (Dt + E2- 4F) < 0. Then Eq. (4) has a negative number on the right side and the sum of the squaresof two real numbers on the left side. There are no real values of r and y that satisfy such an equation; consequently,we say the graph is the empty set. Before stating the results of these three casesas a theorem, we observe that an equation of the form Axz*Ay'+Dx*Ey*

F:0

w h e r eA + 0

(6)

can be written in the form of Eq. (3) by dividing by A, thereby obtaining F r c ' + y ' +Dt r *, *E= A,U +i:0 Equation (6) is a special case of the general equation of the second degree: Axz * Bxy * Cy' t Dx * Ey * F : 0 in which the coefficients of x2 and yz are equal and which has no xy term. We have, then, the following theorem. 1.6.3 Theorem

The graph of any second-degree equation in R2 in r and U, in which the coefficients of x2 and y2 are equal and in which there is no xy term, is either a circle, a point-circle, or the empty set. o rLLUSrRArroN 2: The equation 2x2*2y'*LZx-By+31:0 is of the form (6), and therefore its graph is either a circle, a point-circle, or the empty set. If the equation is put in the form of Eq. (1), we have

x'+y't6x-4y-F#:o ( x ' + 6 x )+ ( y ' - 4 y ) : - T ( x ' * 6 x* 9 ) + ( y ' - 4 y + 4 ) : - # + (r*3)2 + (Y-2)':-E Therefore,the graphis the empty set.

9+ 4


ExAMPLE2: Find an equation of the circle through the three points A(4,5), B(3, -2), and

c(1,,-4).

soLUrIoN:

The general form of an equation of the circle is

x'+y,tDx-fEy*F:0 Because the threepoints A, B,and C must lie on the circle,the coordinates of these points must satisfy the equation. So we have

L6+25+4D+5E+F:0 9+ 4+3D-2E*F:0 1+16+ D-4E*F:0 or, €guivalently, 4D+ 5E+F:-47 3D-2E+F:-L3 D - 4 E* F : - 1 7 Solving these three equations simultaneously, we get D:7

E:-5

F:-44

Thus, an equation of the circle is

x'+y'*7x-5y-44:0

In the following example we have a line which is tangent to a circle. The definition of the tangent line to a general curve at a specific point is given in Sec. 3.L. However, for a circle we use the definition from plane geometry which states that a tangent line at a point P on the circle is the line intersecting the circle at only the point P.

nxeuprr 3: Find an equation of the circle with its center at the point C(1,6) and tangent to the line / having the equation x-v-1:0.

soLUrIoN: See Fig. L.6.2. Given that h: L and k: 6, if we find r, we can obtain an equation of the circle by using the center-radius form. Let I, be the line through C and the point P, which is the point of tangency of line I with the circle.

r: lPCl Hence, we must find the coordinates of P. We do this by finding an equation of /, and then finding the point of intersection of /1 with /. Since 11 is along a diameter of the circle, and / is tangent to the circle , I, is perpendicular to l. Because the slope of / is L, the slope of 11is -L. Therefore, using the point-slope form of an equation of a line, we obtain as an equation of I,

y-5--1(r-1)

1,6THE CIRCLE

or, equivalently,

x+Y-7:0 Solving this equation simultaneouslywith the given equation of l, namely, X- A- L:0 we get x: 4 and y: 3. Thus, P is the point (4, 3). Therefore,

,:l-PCl:ffi or, equivalently,

r: !18 So, an equation of the circle is

(x-l)'+

(Y-6)':

(\ft)'

or, equivalently,

F i g u r e1 . 6 . 2

x ' + y ' - 2 x - 1 2 y* L 9 : 0

1.6 /- )Exercises In Exercises 1 through 4, find an equation of the circle with center at C and radius r. Write the equation in both the centerradius forrr and the gmeral form. ' , . c ( Q ,0 ) , r : 8 r . c 1 4 , - 3 ) ,r : S 3. C(-5, -12), f :3

4. C(-l,l),r:2

In Exercises5 through 10, find an equation of the circle satisfying the given condition$. 5. Center is at (1, 2\ and through the point (3, -1). 5. Center is at (-2,5) and tangent to the line x:7. is at (-3, -5) and tangent to the line l}x t 5y - 4:0. C"n /Z) w "r 8. Through the three points (2, 8) , (7, 3), and (-2,0). 9. Tangent to the line 3x * y * 2:0 at (-L, 1) and through the point (3, 5). 10. Tangent to the line 3x * 4y - 16:0 at (4,1.) and with a radius of 5. (Two possiblecircles.) In ExercisesL1 through 14, find the center and radius of each circle, and draw a sketch of the graph. ll. f*y'-6x-8y*9:0 13. 3f * 3y' * 4y - 7 :0

12.2f*2y'-2x*2y*7:0 1 4 .* * y ' -

10r-lOy+25:0

In ExercisesL5 through 20, determine whether the graph is a circle, a point-circle, or the empty set. 16. 4* l- 4y' * 24x - 4y t L : 0 1 5. 12* y'- 2x * LO y* L9:0 1 7 .f * y ' -

10r* 6y*36:0

1 8 .x 2 * y ' * 2 x - 4 y

*5:0

48

REALNUMBERS, INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

19. *+ f -{f.x+36v-719:0 20. W 1 9 f I 6 x - 6 y + 5 - 0 21. Find an equation of the comrnon chord_of the two cirdes f*

- 6y - 12:0 and f* 2yrg:o. 3F+rlt l+gr(rrrrvr: If the coordinates of a Point gatisfy two different equations, then thjcoordinates also satisfy the difierence of the two equatione.)

22. Find the points of intersection of the two cirdes in Exercise 21. i\

Find an equation of the line which is tangent to the circle .d + y, - 4t + 6y - 12: O at the point (5, 1). 24. Findanequationofeachofthetwolineshavingslope-jgwhicharetangenttothecirclef*l*2x-W-g:0. I

25. From the origin, dbrda of the cirde t' * I * 4r: 0 are drawn. Pr€ve that the set of midpoinb of these chords is a circle. 26. Prove analytically that a line from th€ center of any circle biEecting any drord is perpendicular to the chord. 27. Prcve analytically that an angle inscribed in a semicircle is a right angle. 28. Given the line y: flr + b tangent to the circle rP * y, = r2, find an equation involving m, b, and t

1.7 FUNCTIONS AND We intuitively consider y to be a function of r if there is some rule by THEIR GRAPHS which a unique value is assignedto y by a correspondingvalue of x. Familiar examples of such relationships are given by equations such as y:2f+5

(1)

Y:tP-g

(2)

and It is not necessary that r and y be related by an equation in order for a functional relationship to exist between them, as shown in the following illustration. o ILLUsTRATIoN1,: lf y is the number of cents in the postage of a domestic first class letter, and if r is the number of ounces in the weight of the letter, then y is a function of x. For this functional relationship, there is no equation involving r and /; however, the relationship between r and y may be given by means of a table, such as Table 1.7.L. . T a b l e1 . 7 . 1

r: number of ounces in the weight of the letter

x <3

y: number of cents in first classpostage The formal definition 1.7.1 Definition

makes the concept of a function precise.

A function is a set of ordered pairs of numbers (r, fi in which no two

A N D T H E I RG R A P H S 1.7 FUNCTIONS

distinct ordered pairs have the same first number. The set of all possible values of r is called the domain of the function, and the set of all possible values of y is called the range of the function. In Definition 1.7.1, the restriction that no two distinct ordered pairs can have the same first number assures us that y is unique for a specific value of r.

:2x2+ 5} f : { ( x ,y ) l y The numbers x and,y are called aariables.Because for the function / values are assigned to x and because the value of y is dependent upon the choice of x, we call x the independent aariable andy the dependentaariable' The domain of the function is the set of all possible values of the independent variable, and the range of the function is the set of all possible values of the dependent variable. For the function / under consideration, the domain ii tne set of all real numbers which can be denoted with interval notation as (-m, +€). The smallest value that y can assume is 5 (when The range of f is then the set of all positive numbers greater than r:0). or equal to 5, which is [5, +m). o rLLUsrRArroN 2: Let g be the function which is the set of all ordered pairs (x,y) defined by Eq. (2); that is,

g : { @ ,Y ) l y : t / T

vY

Because the numbers are confined to real numbers, ! is a function of r only for x > 3 or x <-3 (orsimply lrl = 3) because forany rsatisfying either of these inequalities a unique value of y ts determined. However, if x is in the intenial (-9, g), a square root of a negative number is obtained, and hence no real number y exists. Therefore, we must restrict r, and so

g : { G , y ) l y : t / 7 - g a n dl r l > 3 } The domain of g is (-o, -31 U [3, *-),

and the range of g is [0, +-;'

o

It should be stressed that in order to have a function there must be exactly one aalue of the dependent variable for a value of the independent variable in the domain of the function. 1.7.2 Definition

If f is a function, then the graph of f is the set of all points (x,y) in R2 for which (x,y) is an ordered Pafu in f .

REALNUMBERS,INTRODUCTION TO ANALYTIC GEOMETRY, AND FUNCTIONS Hence, the graph of a function is a curve which is the set of all points in R2 whose cartesian coordinates are given by the ordered pairs of numbers (x, y). Because for each value of r in the domain of the function there corresPonds a unique value of y, no vertical line can intersect the graph of the function at more than one point. o rLLUSrRArroN 3: Let f : {(x, y)ly : \6=}. A sketch of the graph of f is shown in Fig. 1.7.1,.The domain of is the set of all real numbers less / than or equal to 5, which is (-*,5], and the range of is the set of all non_ / negative real numbers, which is [0, +oo). o ' rLLUSrRArroN 4: Let I be the function which is the set of all ordered pairs (x,y) such that

F i g u r e1 . 7 . 1

-3

y:[

1 4

if x <-1. if -1 < x <2 if2
The domain of g is (-m, *m), while the range of g consists of the three numbers -3, 1, and 4. A sketch of the graph is shown in Fig. 1,.7.2. o observe in Fig. r.z.z that there is a break at x:-L and another at x: 2. we say that g is discontinuous at -7 and 2. Continuous and discontinuous functions are considered in Sec. 2.5. o TLLUSTRATToN 5: Consider the set

{ ( x ,y ) l * ,+ y , : 2 5 } This set of ordered pairs is not a function because for any x in the interval (-5, 5) there are two ordered pairs having that number as a first element.

EXAMPLE

Let

h : { ( x ,y ) l y: l * l }

solurroN: The domain of h is (--, +*), and the range of h is [0, +*;. A sketch of the graph of h is shown in Fig. 1,.7.9. v

Find the domain and range of h, and draw a sketch of the graph of h.

F i g u r e1 . 7 . 3

1.7 FUNCTIONSAND THEIRGRAPHS 51

EXAMPLE2: Let F be the function which is the set of all ordered pairs (x, Y) such that

(3x-2 U:l*

A sketchof the graph of F is shown in Fig. l.7.A.The domain of F is (-m , **), and the range of F is (--, +m).

SOLUTION:

if.x<"1' ifL
Find the domain and range of-F, and draw a sketch of the graPh of F.

F i g u r e1 . 7 . 4

nxnvrrr,n 3: Let G be the function which is the set of all ordered pairs (x, Y) such that

u :"

x2-9 x-

soLUrIoN: A sketch of the graph is shown in Fig. 1'.7.5-Because a value for y is determined for each valrue of r except x: 3, the domain of G conboth the numerator and sists of all real numbers except 3. When i:3, undefined. denominator are zero, and 0/0 is - 3) (r + 3), we obtain Factoring the numerator into (r

J

Find the domain and range of' G, and draw a sketch of the graPh of.G.

Y-

(x-3)(r+3)

e1l

or y: x* 3, provided that r # 3.In other words, the function G consists of all ordered pairs (x,y) such that y-x+3

and x*3

The range of G consists of all real numbers excePt6. The graph consists of all points on the line y : x * 3 except the point (3, 6).

F i g u r e1 . 7 . 5

REAL NUMBERS,INTRODUCTION TO ANALYTICGEOMETBY, AND FUNCTIONS

4: Let H be the funcEXAMPLE tion which is the set of all ordered pairs (x, y) such that

lx*3 A:lz

ifx*3 ifx_3

solurroN: A sketch of the graph of this function is shown in Fig. 1.2.6. The graph consists of the point (9, 2) and all points on the hne y - x * 3 exceptthe point (3,5). Function H is defined for all valuesof x, and.therefore the domain of H is (-co,+*). The range of H consistsof all real numbers except6. u

Find the domain and range of H, and draw a sketch of the graph of H.

F i g u r e1 . 7 . 6

nxenapr,n5: Let d be the function which is the set of all ordered pairs (x,y) such that ,r , _ ( x 2 * 3 x - 4 ) ( f - 9 ) (x'+x-12)(r+3) Find the domain and range of O, and draw a sketch of the graph of 0. v

soLUrIoN: A sketch of the graph of this function is shown in Fig. 7.7.7. Factoring the numerator and denominator, we obtain ,' , _

(x * 4) (r- 1)(x - 3)(x + 3) (x*4)(x-3)(r+3)

We see that the denominator is zero f.or x: -4,-3, and 3; therefore, j, @ is undefined for these three values of x. For values of.x * -4,-g,or we may divide numerator and denominator by the common factors and obtain

!:x-I

if x

-3, ot 3

Therefore, the domain of f is the set of all real numbers except, -3, and 3, and the range of Q is the set of all real numbers except those values of (x - 1,) obtained by replacing r by -4,-3, or 3, that is, all real numbers except -5,-4, and 2. The graph of this function is the straight line y - x - 1, with the points (-4, -5), (-3 -4), and (3, 2) deleted.

F i g u r e1 . 7 . 7

1.7 FUNCTIONSAND THEIRGRAPHS

EXAMPLE6: Let f be the function which is the set of all ordered pairs (x,y) such that (x'

A:lz

tfx*2

soLUrIoN: A sketch of the graph of / is shown in Fig. 1'7'8' The graph : consistsof the point (2, 7) and all points on the parabola A x2, except the point (2,A)."Function/ is defined for all values of.x, and so the domain of / is (-*, +*). The range of f consistsof all nonnegirtivereal numbers.

ifx-2

Find the domain and runge of t', and draw a sketch of the graPh

of f.

F i g u r e1 . 7 . 8

7' Let h be the funcnx.s.lrpr-n tion which is the set of all ordered pairs (x,y) such that I x-L a:lzx+t

If x<3 if3
The domain solurroN: A sketchof the graph of.h ts shown in Fig. 1,.7.9. of h is (-*, +oo).The values of y are either less than 2 ot greaterthan or equal to 7. So the range of h is (-*,2) U 17,+*7 or, equivalently,all real numbers not in 12,7). u

Find the domain and range of h, and draw a sketch of the graPh of.h.

F i g u r e1 . 7 . 9


rxaupr.n 8: Let

8: { @,y)ly: t/i64} Find the domain and range of g, and draw a sketch of the graph of 8.

solurroN: Because\RFT is not a rcal number when x(x - 2) < 0, the d-omainof g consistsof the values of x for which x(x - 2) > 0. This in_

equality will be satisfied when one of the following . x> 0 a n d r - 2 > 0 ; o r x < } a n d x - 2 < 0 .

two cases holds:

CaseT: r>0andr-Z>0. That is, x>0

and x>2

Both inequalities hold if x > 2, which is the interval [2, +*;. Case2: x<0andx-2<0. That is, x<0

and x<2

F i g u r e1 . 7 . 1 0

In the next illustration, the symbol [r]] is used. [r]l is defined as the greatest integer which is less than or equal to r; that is,

[*n-n

ff n
where n is an integer

This function is called thegreatestinteger function It follows that [1]l : 1,

[1.3n:1, [+]l:0, [-4.2n:-5, [-8]l :-8, [9.8n:9, and.so on.

o rLLUSrRArroN 6: Ler F be the greatest integer function; that is, F: { (x , y ) l y : fi rn]. A sketch of the graph of F i s show n i n Fi g. 1.7. 11.The following values are used to draw the sketch.

5 4

2

-4
-5-4 -3-Z-7 -1 -2 -3 -4

-J F i g u r e1 . 7 . 1

lrn flxn: firn : [x]l -

-5
3

-D

2345

-3
-s -4 -3 -z

0

[x]J : -t

0
1.

fxn: o

7
2

l[r]l :

1

2
3

[t]l :

2

3
4

["n: 3

4
5

[r]l -

4

A N D T H E I RG R A P H S 1.7 FUNCTIONS

The domain of F is the set of all real numbers. The range of F consists of all the integers. SOLUTION:

EXAMPLE9: Let

G:{(x,y)ly:firn-x] Draw a sketch of the graPh of G. State the domain and range of G.

v

0; ["]l : : L fitn : 2 ; [tJl fixn:-1;

If 0 < x 1!, If L < x <-2, If 2 < x 13, lf-l
so so so so

G(r) :-x' G(x):"1- x' G(r) :2- x' G(x):-l-x'

and so on. A sketchof the graph is shown in Fig. L.7.L2.The domain of G is the set of all real numbers,and the rangeof G is (-1,0]

F i g u r e1 . 7 . 1 2

1 s. 7 t . ' l Exercise the gfaPh of the function' In Exercises 1 through 10, find. the domain and range of the given function, and draw a sketch of 3. F: {(x,y)ly:3x' - 6} 2.g:{(x,y)ly:x'+2} y ) l y: 3 x - 1 } 1. :

f {(x, 4. G: {(x, y)ly : t/iT17 7 . g : { @ ,y ) l y : t / P - + l (

S . h : { ( x ,y ) l y: f t x - + } 8. H: {(x,y\ly:l*-31}

6 .f : { ( x ,y ) l V : { a ; F } 9 . 6 : { G ,y ) 1:, l 3 x+ 2 l }

4t'-11

r o .F : I e , y ) t v : 2 * r J Find the domain In Exercises11 through 34, the function is the set of all ordered pairs (r, y) satisfying the given equation the functlon' and range of the function, ard draw a sketch of the graPh of It

G:Y:l2

(-2

t>

ifr<3 if 3
[x*5 1 , 4g. : V : \ \ t r 5 = 7 Lr-5 1 . -:.@ - ( "r ;+ t ) ( f I /n. t :, a

ifr<-5 if -s
2 0 .h : u : { F 6 4 23.h: y:

'

f*5x2-6x-30 x*5

26g . :y:lxl 'lr-11

l-4 ifx<-2 1f-2
15.H:v:lir_i

if3
( v 2 r -e r - 4 ) ( f

zL. 8iu:

ifx*2 if x :2

fzx-t lo

[6x*7 ro'8:Y:I4-x

ifr<3

18.G:V:ffi

. 1 5 .I i ' :

-5x*6)

xr - 2x2 *_2

if'x=-2 if-2
1 9 .f : y :

22.f: y:

x*3

24.F:y:T

25.f: y: lxl+ lr - 1l

2 7 .G : y : x - f l t n

28F . :!:[x+2n

xa*x3-9x2-3rf18


29.h:y: [x'n

30.H: y: lxl + flrn

32.f: U: (x- [rn)'

3 3 .f : U : 2 +

( - 1 ) " , w h e r en :

[,xn

1.8 FUNCTIONNOTATION, If f is the function having as its domain values of x and as its range values OPERATIONS ON FUNCTIONS, of y ,the symb ol t'@) (read "f oI r") denotes the particular value of y which AND TYPESOF FUNCTIONS corresponds to the value of x. o ILLUSTRATToN 1: In Illustration 3 of Sec. "1..7,

f : { ( x ,y ) l y: l i - , } Thus, we may write f (x) - \6=. Becausewhen )c: l, ti=:2, we write f(1):2. Similarly,f(-6): \/TT, t'e): t6, andso on. . When defining a function, the domain of the independent variable must be given either explicitty or implicitly. For instance,if we are given that f is defined by f(x):3x2-5x*2 it is implied that r can be any real number. However, if we are given that / is defined by f (x):3x2-5x*Z

I < x < 10

then the domain of / consists of all real numbers between and including 1 and 10. Similarly, if g is defined by the equation / \ 5x-2 8(x/: x+4 it is implied that x # -4, becausethe quotient is undefined for x: hence, the domain of g is the set of all real numbers except--4. If we are given

-4;

h(x) - \/e -E it is implied that r is in the closed interval -3 < x = J, because t is undefined (i.e., not a real number) for x ) 3 or tr < -3. So the domain of h is [-3, 3], and the range of h is [0, 3].

Exevrpr.n1: Given that / is the function defined by f (x) - tsz I3x - 4, find: (a) f (0); (b) f (2); (c) f (h); (d) f Qh); (e) f (2x);

(t) f(x + h);(d f @)+ f(h).

SOLUTION:

(a) /(0) :02 * 3 . 0 - 4 - - 4 (b) f(2):221-3' 2-4:6 (c)f(h):h2+3h -4

AND TYPESOF FUNCTIONS ON FUNCTIONS, 1.8 FUNCTIONNOTATION,OPERATIONS

( d ) f ( z h ) : ( 2 h ) 2+ 3 ( 2 h ) - t - 4 h 2+ 6 h - 4 (e) f(zx) : (zx)z * 3(2x) - 4 - 4x2t 6x - 4 ( f ) f ( x + h ) : ( r + h ) ' * 3 ( . r+ h ) - 4 :x2*2hx*h2*3x+3h-4 : ) c zI ( 2 h + 3 ) x + ( h ' + 3 h - 4 )

4) (g)/(r)+f(h)::9'lrtJ* ,nll:I':3r-

SOLUTION:

s G + h ) - g , k ) \MT6=1-\ffi h

h

\tri1)6ffi=+ 6ffi58=uNWT.+tF)

h+0

(3x+3h-L)-

(3x-L)

Vgr-1)

h(\M.+

:

t6_1)

3h

h(\triT5T=+ ..ffi) 3

16;agy-r+ \E-

In the second step of this solution, the numerator and denominator were multiplied by the conjugate of the numerator in order to rationalize the numerator, and this gave a common factor of h tn the numerator and the denominator.

We now consider operations (i.e., addition, subtraction, multiplication, division) on functions. The functions obtained from these operathe sum, the difference, the product, and the quotient of tions-called defined as follows. the original functions-are L.8.1 Definition

Given the two functions f and g: (i) their sltm, denoted by f + g, is the function defined bV U +SXr)

:f(x)+g(x);

(ii) their difference,denoted by

the function

defined bv

(f-g(x):f(x)-B@);

(iii) their product,denotedby f '9, is the function definedbV (/' gXr)

:f(x)'B@);

(iv) their quotient,denoted by flg, is the function defined by (flil@)

: f (x)lg(x).

TO ANALYTICGEOMETRY, REAL NUMBERS,INTRODUCTION AND FUNCTIONS

In each case the domain of the resulting function consists of those values of x common to the domains of.f and g, with the exception that in case(iv) the values of x for which g(r) - 0 are excluded. rxeupr.n 3: Given that / is the function defined by f(x):\/x+Landgistne function defined by

4, find: (a)(f + gXr); B@): {r (b)(/ - s)(r);(c)(/ ' s)(x); G) (flS)(x).In each case/determine the domain of the resulting function.

SOLUTION:

( a )( f + i l ( r ) : \ E + 1 + \ t x - 4 ( b )( f - i l k ) : \ / x + I - . / x - 4 ( c )( f . i l ( r ) : I x + 1 . { x - 4 \/# r (d) (fls)(r)-

\tr-4

The domain of f is 11,, **), and the domain of g is [4, +*;. So in parts (a), (b), and (c) the domain of the resulting function is [4, *m). In part (d) the denominator is zero when x: 4; thus, 4 is excluded from the domain, and the domain is thereforc (4, +oo).

To indicate the product of a function / multiplied by itself, or f . f , we write f 2. For example,if f is defined by f (x) : 3x, then f it the function defined by f '(x): (3x)(3r):9x2. In addition to combining two functions by the operations given in Definition L.8.L,we shall consider the "composite function" of two given functions. 1.8.2 Definition

Given the two functions f and g, the composite function, denoted by f " g, is defined by

(/"s)(r): fk?)) and the domain of f . g is the set of all numbers r in the domain of g such that g(x) is in the domain of f . rxervrrr.n 4: Given that / is defined by f (x) : \G and g is defined by g(x) :2x - 3, find F(x) if F : f o g, and find the domain of F.

rxarvrprE5: Given that f is defined by f (r): lx and g is defined by S@)- f -'/.., find: (a)f"f;(b)g"g; (c)/.9;(d) g"f. Also find the domain of the composite function in each part.

soLUrIoN:F(r): ff " S)(r): f@(x)) :f(2x_3) : f2x-3 The domain of g is (-m, *oo), and the domain of / is [0, +m). So the domain of F is the set of real numbers for which 2x - 3 > 0 or, equivalently, [9, **). solurroN:

The domain of / is [0, *-),

and the domain of g is (-*,

( a )( f " f ) ( x ) : f ( f ( x ) ) : f ( { - * ) : { T x : 9 i The domain of f " f i" [0, +*;. (b) (g" g)(r):Sk(r)): g(*' - 1): (x'-l)'-"1,:xa-2x2 The domain of I ' I is (-co, +*).

+oo).

AND TYPESOF FUNCTIONS ON FUNCTIONS, 1.8 FUNCTIONNOTATION,OPERATIONS

( c )( f " i l ( r ) : f k G D : f e - 1 ) : { P - r The domain of f " git (--,-1] not in (-1, 1).

( d ) ( s " i l ( x ) : S U ( x ) ): 8 ( { i ) :

U [1, +*; or, equivalently,all r

( f r ) ' - 1- x - L

'l' is deThe domain of I " f is [0, +*). Note that even though x '1..8.2, fined for all values of x, the domain of I " f , by Definition is the set of all numbers I in the domain of / such that f (x) is in the domain of g.

1.8.3 Definition

(i) A function / is said to be an eaen function if for every r in the domain of f , f(-x): /(r). (ii) A function / is said to be an odd function if for every r in the -f(x). domain of f , f(-x): In both parts (i) and (ii) it is understood that -r is in the domatn of f whenever x is. .

ILLUSTnATTON

2:

3(-x)n - 2(-x)' * 7 :3xa (a) If f(x): 3xa- 2x2+ 7, then f(-x): 2x2* Z : f(x).Therefore, f is an even function. (b) If g(r) : 2x5I Sxs- 8x, then S(-x) : 2(-x)5 * 5(-r)B - 8(-x) : -2x5 - 5x3* 8r : -(2xs * 5x3- 8x): -g(r). Thus, g is an odd function. *5(-r)3 - (-x)' * 8: ( c ) I f h ( x ) : 2 x a * 5 r 3 - x 2* 8 , h ( - x ) : 2 ( - x ) n 2xa 5x3 x2 * 8. We see that the function h is neither even o nor odd. From the definition of an even function and Theorem 1.3.6(ii) it follows that the graph of an even function is symmetric with respect to tlne y axis. From Theorem 1.3.5(iii) and the definition of an odd function, we see that the graph of an odd function is symmetric with respect to the origin. If the range of a function / consists of only one number, then / is called a constant function So if f(x): c, and if c is any real number, then / is a constant function and its graph is a straight line parallel to the r axis at a directed distance of c units from the x axis. If a function f is defined by f ( x ) : a s x n* a 1 x 6 nI - ra 2 x n -*z'

+ a n - 1 xI a n

where n is a nonnegative integer, and ao, ar, . . . , an ate real numbers (ao * 0), then / is called a polynomial function of degree n. Thus, the function / defined by f(x):3x5-x2+7x-r is a polynomial function of degree 5.

REALNUMBERS, INTRODUCTION TO ANALYTIC GEOMETRY. AND FUNCTIONS If the degree of a polynomial function is 7, then the function is called a linear function; if the degree is 2, the function is called a quad-

'r!!,"':::_:

if the degreeis 3' thefunctionis careda cubicfunc-

The function / defined by f (x) : 3x * 4 is a linear function. The function g defined by S@) :5x2 - 8x * 1 is a quadratic function. The function h defined by h(x):8x3 - x * 4 is a cubic function. o If the degree of a polynomial function is zero, the function is a constant function. The general linear function is defined by

f(x):mx*b where m and b are constants and m # 0. The graph of this function is a straight line having m as its slope and b aSits y intercept. The particular linear function defined by

f(x):x 1S

called the identity function, The general qLadratic function is defined

by

f(x):af*bxrc where a, b, and c are constantsand a * 0. If a function can be expressedas the quotient of two polynomial functions, the function is called a rational function For example, the function / defined by -s-x'za5 : \--/ Jf(r) _9 f

is a rational function, for which the domain is the set of all real numbers except 3 and -3. An algebraicfunction ts a function formed by a finite number of algebraic operations on the identity function and the constant function. These algebraic operations include addition, subtraction, multiplication, division, raising to powers, and extracting roots. An example of an algebraic function is the function f defined by

f(x) : In addition to algebraicfunctions, transcendental functionsare considered in elementary calculus.Examplesof transcendentalfunctions are trigonometric functions, logarithmic functions, and exponential functions, which are discussed in later chapters.

ANDTYPESOF FUNCTIONS ON FUNCTIONS, OPERATIONS NOTATION, 1.8 FUNCTION

Exercises7.8 / 1.)Given f (x) :Zxz + 5x - 3, fnd:

(a) f(-z)

(c) /(o)

(d) /(3)

(e)/(h + 1)

G)f @'- s)

(h\ f(x+h)

(i) /(r) + f(h) 2. Givengk) :3x2 - 4, find: (a) g(-4)

(b)

(e)g(x - h)

(f)

(c)

8(+) s@)- s(h)

(d) g(3xz- a)

8e)

(s)s ( x + hh) - s ( x ), h + 0

F(x) : \/zxT 3, i3-..,iGiven (a) F(-1 )

(b) F(4)

(d) F(30)

(e) F(2r + 3)

4. Given G(x) : lET7,

frnd:

(a) G(-2)

(b ) G(0 )

(c)

c(+)

(d) G(+)

(e \ G(Z x ' z -l )

(f)

G(r+h)-G(x) h+0 h

5. Given

ilrl o"*o

t\x):1x

[1 if t:o find:(a)/(1); (b)/(-t); (4 f G); @) f|a); @)f (-r), (t) f(x + 7); (g)l(f); (n)/(-f). l a r r l - i r l- a

6. Given l(r)

expressf(t) without absolute-value bars if (a) f >0;(b)-3

=t <0; (c) f <-3.

g arc defined. In each problem define the following functions and determine In Exercises 7 through fz, tn" for,"tlon, 1 "nd ( a ) f f u n c t i o n : + g ; (b)f -g;G)f .S;@)flg;Grglh(Df "g;Grg"f. o f t h e r e s u l t i n g thedomain 7.f(x)-x-5;S@):x2-7

8. f(x) : tfi; g(r) : xz+ 1

s.f(*):fit sQ):*

10./(x): \/i -'2, SG):1

rr. f (x) : \/iz - r; 8@) : {i]

12.f (x) : lxl;g(r) : lr - 3l

L3. For each of the following functions, determine whether f is even, odd, or neither. (a) /(x) :2xa - 3xz I 1 (d) /(r) : .16- L tt3

G)fQ):ffi

(b) /(x) :Sxs - 7x

( c )/ ( s ) : s 2 * 2 s t 2

(e ) f (t) :5 t7 * ' 1 .,

(f) f(x): lxl

v-1

(h)/(r):ffi

1.4.Prove that if f and g arc both odd functions, then (/ * g) and (f - il are also odd functions.

d2

REALNUMBERS, INTBODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS

15. Prove that if I and g are both odd functions, then / . g and /g are both even functions. 16. Prove that any function can be expressed as the sum of an even function and an odd function by writing

f (r) :trlf (x)+ l(-x)l + *If k) - l(-x)l and showint that the function having Iunction valuestr[l(r) + l(-r)] is an even function and the function having function values{ff(r) - f(-r)l is an odd function. 17. Usethe result of Exercise16to exprcssthe functions defined by the followint equationsasthe sumof an evenfunction a n da no d df u n c t i o n( :a ) l ( x ) : * + 2 ; b ) f k ) : t C - L ( c r f ( x ) : f + x a- x + 3 ;( d )l ( r ) : 1 l x ,( e )f ( r ) : ( x - r \ l ( x + 7 ) ;

(f)l(r) : lrl + lr - 11.

18. There is one function that is both even and odd. What is it? 19. Determine whether the comPositefunction I " g is odd or even in each of the following cases:(a)land g are both even; (b) I and I are both odd; (c) / is even ard g i8 odd; (d) I is odd and g is even. 20. The function I is defned by g(r) =f.

Define a function f such that (/"g)(r):r

if (a) r > 0; (b) r < 0.

In Exercises 21 through 34, draw a sketch of the graph of the given function. In Exercises 23 through 34, the function U is the unit step function defined in Exertise 21 and the function sgn is the signum function defined in Exercise 22.

21.tI isthetunction defined by,(t :

{l

if l: 3

This function U is called the u t step f nctiofl. 22. The signum functiotr (w sign functioz), denoted by sgn, is defined by sgnr:l

l-1 0 [ 1

ifx<0 itx:0 ifr>0

sgn r is read "signum of x." 2 3 .f ( x ) : U ( x - 1 )

2a. gQ): U(r) - U(x - 1)

2 5 . g ( x ) : s g n ( x+ 1 ) - s g n ( x- 1 )

26. f(x):

2 7 .h ( x ) : x ' U ( x ) 2 9 . G ( x ) : ( r f 1 , ). U ( x + 1 ) - x . l l ( x )

2 8 .F ( x ) : ( r * 1 ) . U ( x + 1 ) 30.F(r) -x-2sgnx

31,.h(x):sgnr-U(x)

32.f(x):sgnx'U(x+1,)

3 3 .g ( r ) : s g n x * x . U ( x )

3 4 .G ( r ) : s g n ( U ( r ) )

bi.

fi"a formulas tor (f " g)(r) if ifr<0 [0 if07

lr

and g(x):]Lx tt

ifr<0 if01.

36. Find formulas for (g " fXr) for the functions of Exercise 35. 37. Find formulas for (f . g)(r) if [0 f(*):lf [0

ifr1

[1, and g(x):lZx Ll

ifr<0 if0
sgn x2- sgnx

REVIEWEXERCISES 63 v2,find two functionsg for which (/ " S)(r) : 4x2-12x*9. f i n d t w o f u n c t i o n sg f o r w h i c h ( f " i l ( r ) : x 2 - 4 x 1 5 . 39. I t f ( x ) : x 2 * 2 x 1 2 ,

38. It f(x):

(Chapter1,) ReaieutExercises In Exercises 1 thrcugh 6, find the solution set of the given inequality, and illustrate the solution on the real number line. 1.8<5r*4<10 ?7

4 .x++<4 +

r-5

,#.,L

3. Zxz* r < 3

s. 13+Sxl-9

6- ' lz-gxl 1 l3+rl=a

7. Define the foUowing sets of points by either ar equation or an inequality: (a) the point cide (3, -5); (b) the set of all points whose distancefrom the point (3, -5) is lessthan 4; (c) the set of all points whose distancefrom the point (3, -5) is at least 5. 8. Ptove that the points (1, -l) , (3, 2), and (7, 8) ar€ collinear in two ways: (a) by usint the distance formula; (b) by ueing slopes. 9. Find equations of the lines passing through the origin that are tangent to the circle having its center at (2, 1) and a radius of 2. L0. Prov€ that the quaddlateral having vertices at (1, 2); (5, -1\ , (71,7) , and (7, 10) is a rectangle. and 5t - y + ,l:0 are equations of the same line. 12. Shovvthat the tdangle with vertices at (-8, 1), (-f, -6) and (2, 4) is isoscelesand find its area. 1 3. Two vertices of a parallelogram are at (-3, 4) and (2, 3) and its center is at (0, -1). Find the other two vertices. 14. Two opposite verticrs of a square are at (3, -4) and (9, -4). Find the other two vertices. 1 1 . Determine the values of & and h ifSr + ky + 2:0

15. DetemrineallvaluesoftforwNchthegraphsofthetwoequations**f:kandx+y:fintersect. t6. Prcve that iI r i8 any real number, lrl < f + 1. 1 7 . Find an equation of the circle circumscribedabout the triangle having sideson the lines r - 3y * 2 :0,3x and2r+y-3:0.

- 2y * 6:0,

18. Find an equation of the circle having as its diameter the common chord of the two circles rl + y'z+ 2x - 2y - 74: O andt'+!P-4r+4y-2:O. 19. Find an equation of the line through the point of intersection of the lines 5r + 6y - 4: Oa d'x- 3y*2:0andperpendicular to the line x - 4y - 20:0 without finding the point of intersection of the two lines. (nrtr: See Exercise 28 in Exercisee1.5.) +20:0, x+2y-15=0, and 3r- y * 10: 0. 20. The sides of a parallelogramare on the lines r+ 2y- 10:0,3r-y Find the equations of the diagonals urithout finding the vertic$ of the parallelogram, (rrrm: See Exercise 28 in Exerciees1.5.) In Exercises 21 through 24, the function is the set of all ordered pairs (r, y) satisfying the given equation. Find the domain and range o{ the function, and draw a sketch of the graph of the function.

2 1 .f .: y : \ E x + 5

2 2g. : y : #

64

REALNUMBERS,INTRODUCTION TO ANALYTICGEOMETRY, AND FUNCTIONS | *+3 24. F: "al : 1\/9= 7 I x-3

ifx(-3 i f-3 < x < 3 if3
In Exetcises 25 through 28, for the given functions f and g, define the following functions and determine the domain of the resultingtunction: (a\ f+ g; (b)f - gt 1c)f'g; (d) flg; (e)Slf; (D f. S; G) S" f. E. f ( x ) : x z- 4a n d g (r) : 4 x - 3 . 26. f(x) : \/x+ 2and g(r) : f * 4. 27. f ( x ) : t l( x - 3 ) a n d t@ ) : x l k + 1 ,). 28. f (x) : t/i and S@) : Uxr. 29. Prove that r sgn r:

l.xl. 30. Provethat the two lines.Afi+Bg +C1:O

andAzx + B2V+C,:0

are parallelif and only if ArBr-Are:0.

31. Prove analytically that the three medians of any t angle meet in a point. 32. Plt)ve analytically that the line segment joining the midpoints of any two sides of a triangle is panllel to the third side and that its length is one-half the length of the third side. 33. Prove analytically that if the diagonals of a rectangle 6re perpendicular, then the rectangle is a square. 34. Prove analytically that the set of points equidistant from two given points is the perpendicular bisector of the line segment joining the two points.

35. In a triangle the point of intersection of the medians, the point of intersection of the altitudes, and the center of the circumscribed cirde are collinear. Find these three points and prove they are collinear for the triangle having vertices at (2,8), (5,-L), and (6,5\.

lcontinuity

LIMITSAND CONTINUITY

2.1 THE LIMIT OF A FUNCTION

Consider the function / defined by the equation (2x+3)(x-1) lt \l -x. )\ _: n -1;-

(1)

/ is defined for all values of x except x: l. Furthermore, if x * r, the numerator and denominator can be divided by (r - 1) to obtain f(x):2x*3

x*t

e)

T a b l e2 . 1 . 1

x

0 0.25 0.5 0.75 0.9 0.99 0.999 0.gggg o.ggggg

f(x) -2x*3 (x + 1,)

3 3.5

4

4.5

4.8 4.98 4.998 4.9998 4.gggg8

.r approach L, through values that are greater than 1; that is, let r take on the values2, 1.75,'1,.5, 1.25,7.L,7.01, 1.001,,1.0001, L.0000L,and so on. Referto Table 2.1.2. Table 2.7.2 x

2 L.75 1.5 1,.25 1.1 1.01 1.001 1.0001 1.00001

f(r):2xi3 (x+I)

7 6.5

6.0 5.5

5.2 5.02 5.002 5.0002 5.00002

We seefrom both tables that as r gets closerand closerto l,/(r) gets closerand closerto 5; and the closerr is to 1, the closer (x) is to 5. For inf stance,from Table2.1.7,when x:0.9, f (x): 4.8; that is, when r is 0.j. less than 1, f (x) is 0.2lessthan 5. When x : 0.999,f(x) : 4.998;that is, when r is 0.001less than 1,f(r) is 0.002less than 5. Furthermore,when x:0.9999, f(x) : 0.49998;that is, when x is 0.0001lessthan r, f(x) is 0.0002lessthan 5. Table2.1..2shows that when x: 1,.r,f(x):5.2, that is, when r is 0.1 greaterthan 7, f(*) is 0.2 greaterthan 5. when r: r.001, f(x): 5.002;that is, when r is 0.001greater than L, f (x) is 0.002greater than 5. when r: L.0001, f(*),:5.0002; that is, when r is 0.0001greaterthan l,f (x) is 0.0002 greater than 5. Therefore,from the two tables, we see that when r differs from l by -f 0.001(i.e., r :0.999 or x: 1.001),/(r) differs from 5 by -10.002[i.e., f !!).:4.998 or f (x):5.0021. And when r differs from l by -+0.0001, /(r) differs from 5 bv +0.0002.

2.1THELIMITOFA FUNCTION67 Now, looking at the situation another wdlt we consider the values of f (x\ first. We see that we can make the value of f (x) as close to 5 as we please by taking x close enough to 1. Another way of saying this is that *" .ut make the absolute value of the difference between f (x\ and 5 as small as we please by making the absolute value of the difference between r and L small enough. That is, lf (x) 5l can be made as small as we pleaseby making l, - Ll small enough. A more precise way of noting this is by using two symbols for these small differences.The symbols usually used are € (epsilon) and 6 (delta)So we state that If(x) - 5 | will be less than any given positive number e whenever lr - Ll is less than some appropriatelychosenpositive number

that lf(*)-51
w h e n e v e r0 < l r - 1 1

(3)

<6

We seefrom the two tablesthat lf (x)- 5l :0.2 when lx So,given e:0.2, we take 6:0.L and statethat

1l:0.1.

- 5l < 0.2 whenever 0 < lt - 1l < 0.1 This is statement(3),with e: 0.2 and 6 : 0.1. Also, lf (*) - 5l : 0.002when lx - Ll : 0.00t.Hence,if e : 0.002,we take 6: 0.001,and then lf (x)

l f ( x ) - 5 1 < 0 . 0 0 2 w h e n e v e r0 < l r - l l

<0.001

This is statement(3),with e: 0.002and 6: 0.001. Similarly, if e: 0.0002,we take D : 0.0001and statethat lf (x)

- 5l < 0.0002 whenever 0 < l*-11 < 0.0001

This is statement(3),with e: 0.0002and 6: 0.000L. We could go on and give € any small positive value, and find a suitable value for 6 such thai lf(x) - 5l will be less than e whenever lx- 1,1 i s l e s st h a n 6 a n d x # t ( o r l r - 1 1 > 0 ) .N o w , b e c a u s e f o r a n y e) 0 w e can tind a E > 0 such that If(x) -51 < e whenever 0 < lx- 1l < 6, we state that the limit of f (x) as r approaches1 is equal to 5 or, exPressedin symbols, (4) trm/(r):s

1-6 F i g u r e2 . 1 . 1

1+E

You will note that we state 0 < lr - 11.This condition is imposed becausewe are concernedonly with values of f (x) for x closeto L, but not f.orx: L. As a matter of fact,this function is not defined for x:L. Let us see what this means geometrically for the particular function defined by Eq. (1). Figure 2.1..1illustrates the geometric significance of

LIMITSAND CONTINUIW

€ and 6. We see that /(x) on the vertical axis will lie between 5 - e and 5 * e whenever x on the horizontal axis lies between 1 - 6 and 1 * 6; or lf(x)-51 <.

whenever 0(l*-11

Another way of stating this is that f (x) on the vertical axis can be restricted to lie between 5 - e and 5 -l eby restricting r on the horizontal axis to lie between L - 6 and 1 + 6. Note that the values of e are chosen arbitrarily and can be as small as desired, and that the value of a 6 is dependent on the e chosen. We should also point out that the smaller the value of e, the smaller will be the corresponding value of 6. Summing up for this example, we state that lim f (*):5 because for any e ) 0, however small, there exists a 6 ) 0 such that

lf(*)-51
w h e n e v e r0 < l r - 1 1 < D

We are now in a position to define the limit of a function in general. 2.1.1 Definition

Let f be a function which is defined at every number in some open interval 1 containing a, except possibly at the number a itself. The timit of f (x) ns x approachesa is L, written as

ti2f(c):L

(s)

if for any € ) 0, however small, there exists a E ) 0 such that lf!)-Ll

whenever 0
<6

(6)

In words, Definition 2.1.L states that the function values f (x) approach a limit L as x approaches a number a if the absolute value of the difference between /(x) and L can be made as small as we please by taking x sufficiently near a,but not equal to a. It is important to realize that in the above definition nothing is mentioned about the value of the function when x: a. That is, it is not necessary that the function be defined for x: A in order for the lim /(x) to exist.

In particular, we saw in our example that ,.

(2x*3)(r-1)

TTGI)__c but that

(2x+ 3) (x - 1,) (r- 1) is not defined f.or x: L. However, the first sentence in Definition 2.1.1 requires that the function of our example be defined at all numbers except f. in some open interval containing 1.

LIMIT OF A FUNCTION

ExAMPLEl-: Let the function / be defined by the equation f(x):4x-1' Given lim f(r) --ll, find a D for

e: 0.0Lsuchthat lf(*) - 111< 0.01

SOLUTION:

'

l f ( * )- 1 1 1 : l @ x -1 ) 1 1 1 - l4x- 121 -Alx-31 Therefore/ we want

4lx- 3l < 0.01 whenever 0 ( lr - 3l < 6 or, equivalently,

whenever

0 < lr-31 < 6

lr

- 3l < 0.0025 whenever 0 <

If we take 6: 0.0025,we have

l(ax- 1) - 111< 0.01 whenever0 (

< 0.0025

(7)

It is important to realize that in this example any positive number less than 0.0025 can be used in place of 0.0025 as the required 6. That is, if 0 < y < 0.0025 and statement (7) holds, then

- 1) - 111< 0.01 whenever 0 ( l*- 3l < y

(8)

because every number r satisfying the inequality 0 < satisfiesthe inequality 0 < lx - 3l < 0.0025.

< 7 also

l(+*

The solution of Example L consisted of finding a 6 for a specific e. If for any € we can find a 6 that will work, we shall have established that the value of the limit is LL. We do this in the next example.

nxervrrr.n2: Using Definition z.L.L,prove that

L'T tnt

- 1): 1.1

solurroN: We must showthat for any€ ) 0 thereexistsa E > 0 suchthat r
l@x-1)-11l:lax-Dl Therefore, we want

alx-31
w h e n e v e r0 <

<6

w h e n e v e r0 < l x - 3 1

<6

or, equivalently, l"-31 <|e

So if we choose6: *e, we have alx- 3l < 46 whenever 0 < lx- 3l < 6

LIMITSAND CONTINUIry

or/ equivalently,

al*- 3l< 4

*e

whenever

0 (

<6

or, equivalently,

4lx- 3l < . giving us

l ( a r- 1 )

whenever

<€

0 (

whenever 0(

<6

<6

if 6: *e. This provesthat lim (4x-- 1) : 11. In particular, if e : 0.0L, then we take E : 0.01/4, or 0.0025,which correspondsto our result in Example 1. A.y positive number 6' ( *e can be used.in place of ie as the required 6 in this example. rxauprr 3: Using Definition 2.1.7, prove that limx2:4

solurroN: wemustshowthatforany€) 0thereexistsa6 ) 0suchthat l * ' - 4 1 < , w h e n e v e r0 < l x - Z l < a Factoring, we get

lx,-41:lx-2l .lx+zl we want to show that ll - 4l is small when r is closeto 2. To do this, we first find an upper bound for the factor + zl.lf x is closeto 2, we lx know that the factor I* - 2l is small, and that the factor * 2l is closeto 4. Belx causewe are considering values of r close to 2, we can concernourselves with only those values of x for which l* - zl < 1; that is, we are requiring the 6, for which we are looking, to be lessthan or equal to 1. The inequalit!

l x - 2 1< r is equivalent to -L < x-2<'!, which is equivalent to 1,
2.1THELIMITOF A FUNCTION

< 6, it follows that lr - 2l < *e and lx + 2l < 5 (becausethis is true when ' that lx 2l < 1) and so lr2 4l < (*e1 (5). Therefore, we conclude l r ' - 4 1 < e w h e n e v e r0 < l t - 2 1 < 6 if E is the smaller of the two numbers L and *e, which we write as 6 - min(1, *e).

4: Using Definition ExAMPLE prove that 2.1.'/.,, 8 lim*:2 t-7 t-

solurroN:

We must show that for any€ > 0 there existsaD ) 0 suchthat

ls'=-21 I I -l
v-3

c

Now

w h e n e v e r0 ( l t - z l

:l#l t*-21

:# . ).m =lt-71 We wish to show that l8/(f - 3) 2l is small when f is close to 7. We - 31.By requiring proceedto find some upper bound for the fraction zlv ln" A for which we are looking to be less than or equal to 1",we can say that whenever lt - Zl ( 6, then certainly lt 7l < L. The inequality

It-zl
3 < lt-31< s we haveshown wheneverlt - 7l < l,lt - 3l > 3, and because Therefore, that 18/(t 3) - 2l: lt - 7l ' zllt - 3',,we have

l*-21

: V-71 h<

lr-71
We want then V-71 'A < € or, equivalently,V-71 <*e. Consequently, we take 6 as the smaller of L and Be, which assuresus that whenever


I t - z l < 6 , t h e nl t - z l 3 (becausethis is true when

whenever 0 ( lt_ zl < D, and where 6: min(!, te). we have therbfore proved that lim l8l(t - 3) f : 2.

The following theorem states that a function cannot approach two different limits at the same time. It is called, a uniquenesstheirem because it guarantees that if the limit of a function exists, it is unique. 2.'1,.2Theorem

U

fT

fk):

L, and li^ f(r):

Lr, then Lr.: Lz.

pRooF: we shall assume that L, * L, and show that this assumption leads to a contradiction. Because lt : Lr, it follows from Definition f@ 2.1'.1that for any € > 0 there exists a 6r ) 0 such that

r (lr-al lf(x)-Ltl
< l L ,- f ( x ) l+ V @ )- L , l

(11)

so from (9), (10), and (11) we may conclude that for any € ) 0 there exists a 6r ) 0 and a 6z ) 0 such that lLr-Lrl< e*e

w h e n e v e r0 < l * - - o l < 8 , a n d 0 < l x - a l < 6 , ( t z ) If 6 is the smaller of E, and 6r, then E = Er and I S 6r, and (12)statesthat for any e ) 0 there exists a D > 0 such that w h e n e v e r0 (

(13) l*-ol <6 : However, if we take e tlL, hl, then (13) states that there exists a 6>0suchthat - Lrl < lt, lLr- Lrl whenever 0 ( lral < D (14)

lLr-Lrl<2e

2.1 THE LIMIT OF A FUNCTION

73

Because(1a) is a contradiction, our assumption is false. So Lt : Lr, and I the theorem is proved.

2.1 Exercises In Exercises1 though 8, we are given l(r)' a' and L' as well as lim l(x) : L' Deternine a number 6 for the given e such that ll(r) - Ll ( e whenever0 ( h- al < 6. ' 3' lin (3-ttr):7,e :o'o2 2. lim (4r-5):3;€:0.m1 t : ' t t ' ^ ( Z r + l ) : t 0 ;e : 0 . 0 1

n . o i ' , r * u t , : - 8 ; e : 0 . 0 @ ' t ' ; ; : * -t-8s ; , : o . o o s

e'-t-'"r6:z;':o'oos

t--2

q*-1

ar-A. -4; € = o,o1 8. lim ff=t' : 2; 6: 9.91 '7. tm I-:: nB5x- L ;:-, x+2 In Exercises9 through 29, establishthe lirrit by using Definition 2.1.1;that i8, for any € > 0, find a 6 > 0, such that l f ( r \ - L l ( e w h e n e v e 0r < l r - c l < 6 ' ri.#:6 J0. lim (7-2r):11 9 , { m ( 5 x - 3 ) :' Z 'r r .

,-s x-3

.-1

,7;i'

p. \^{:=i2:+ ,-t

I-1

+

.

J9"$,*:e

!?,urn*-r

, / r ' . -------= t 2 - - ; -lim ;r-r 15.

-76. - ; -nm=:2 ;x-3

1 18. lim ----:-==-l

19. lim (f -3r):10

21. lim (5-r-f):-1

22. 6- 11-4:9 t

r--1x

+ 3

17.lim-!--;:2 ".rr-4 20' Itm (f +2x-l):7

5+E

zg' llm r/r+5-:g '-1

t-t]25-I

r--a

11

24. lim =#:)

r _ 1v 5 - r

z

25. Prove that lim f:

at i4 cis any Positive number'

26. Prcve that lim f :a2if 27. Prove that litrr t/i:

ais any negative number.

la if a is a y Positive number.

28. Prove that u'm Vi :'Va. 29. Prove that iI lim l(r)

(mxr: qs- bs: (a - b\ (d + ab + 8).)

exists and is L, e*

liT

ll(r) | exists and is lll.

30. Prove that, if ^t) : 8(t) for all values of r except r : a, then lim ir) : t(d if the limits exist. ItT 31. Prcve that, if l(x) : g(r) for all values of x except r:

n, then if lim 8(x) does not exist, then lim /(r) does not exist.

(Hrlfn Show that the assumPtion that lim l(x) does exist leads to a contadiction.)

74

LIMITS ANDCONTINUITY 2.2 THEOREMS ON LIMITS In order to find limits of functions in a straightforward manner, we shall OF FUNCTIONS need some theorems. The proofs of the theorems are based on Definition 2.1'.1.These theorems, as well as other theorems on limits of functions appearing in later sections of this chapter, will be labeled "limit theorems" and will be so designated as they are presented.

2.2.1.Limit theorem 1

If m and b are any constants,

tu(*r+b)-ma*b PRooF: To prove this theorem, we use Definition2.1,.1. For any e ) 0, we must prove that there exists a D ) 0 such that

w h e n e v e r0 ( l * - o l < 6 (1) l(**+b)-(ma*b)l 0 suchthat w h e n e v e r0 < l r - a l < 6 l*l .l*-al
l*- ol.*

lml

w h e n e v e rO ( l r - a l

<6

e)

Statement (2) will hold if we take E : ellml, and thereforewe conclude that l(** + b) - (ma* b)l < e whenever 0 <

<6

if

6:+l m l

This proves the theorem for Case L. C a s e2 : m : 0 . I f m : 0 , t h e n l ( * * + b ) - ( m a* b ) l : O f o r a l l values of x. So we take 6 to be any positive number, and statement (L) holds. This proves the theorem for Case2. I . rLLUsrRArroN1: From Limit theorem 1, it follows that lim(3r+5)-3.2+5 t-2

2.2.2 Limit theorem 2

If c is a constant, then for any number a, lim c:

c

t-A

pRooF: This follows immediately from Limit theorem 1 by taking m : andb-c.

0 I

E;,.

2,2 THEOREMSON LIMITSOF FUNCTIONS

2.2.3 Limit theorem 3

lim x:

75

a

I-A

PRooF: This also follows immediately from Limit theorem 1 by taking I tn:Landb:0. O ILLUSTRATION

2:

From Limit theorem 2,

limT -7 t-5

and from Limit theorem 3,

]1 2.2.4 Limit theorem 4

'--6

If lim f (x) : L and lim g(r) : M, then t-A E-A

}l

tf@) I g(r)l : L -+M

pRooF: We shall Prove this theorem using the plus sign. Given

tuf('):r

(3)

*l 8G):M

(4 )

we wish to prove that (5)

l i m [ / ( x )+ S ( r ) l : L + M

To prove Eq. (5), we must use Definition 2.1.1;.thatis, for any e ) 0 we must prove that there exists a 6 ) 0 such that

llf(x)+g(r)l-(L+M)l

w h e n e v e r0 ( l * - o l

<6

(6 )

Becausewe are given Eq. (3), we know from the definition of a limit that for ie > 0 there exists a 6r ) 0 such that lf(*)-Ll<*,

w h e n e v e r0 ( l r - a l

<6'

Similarly, from Eq. (4), for ie ) 0 there exists a Dz) 0 such that -Ml
lg(r)-Ml
w h e n e v e r0 ( l r - o l

<6

LIMITSAND CONTINUITY Hence, we have

l t / ( r )+ s ( r ) l - ( L+ M ) l: l ( / ( r )- L )+ ( s ( r )- M ) l

: j:'?;:'..53;I'o.,x-a, <6 ln this warlr we have obtained statement (6), thereby proving that

lr

lf(x)+S(r)l:L+M

The proof of Limit theorem4 using the minus sign is left as an exercise (see Exercise 24).

I

Limit theorem 4 can be extended to any finite number of functions. 2.2.5Limit theorem5

If lim fr(x) : Lr, lim fr(x) : Lr, .,

"-irf

and fu

f r ( r ) : L , , ,t h e n

-,f,(x)J : L,* L,*

t iia t" lf,(x)

. -{-Ln

This theorem may be prove dby applying Limit theorem 4 and mathematical induction (see Exercise 25). 2.2.5 Limit theorem 6

If lim f(*):

L and lim g(x) : M, then

limf(x)'Sk):L.M The proof of this theorem is more sophisticated than those of the preceding theorems, and it is often omitted from a beginning calculus text. We have outlined the proof in Exercises 27 and 28. o rLLUSrRArroN3: From Limit theorem r, tiT x:3,

and from Limit theo-

rem L, lim (2x + 1) - 7. Thus, from Limit theorem 6, we have

lim x(2r + 1) : lim r ' lim (2x + I) t-3

t-3

J'-3

Limit theorem 6 also can be extended to any finite number of functions by applying mathematical induction. 2.2.7 Limit theorem 7

If lim fr(x):Lr,limfr(x):Lr,

lim [/r G)fr(x)

.

.

, andlimfn(x):

. f"(x)f : L,L,

Ln

The proof is left as an exercise (see Exercise 29).

L,,,then

2.2 THEOREMSON LIMITSOF FUNCTIONS

2.2.8 Limit theorem 8 U

Q) : L and n is any positive integer, then lT f

tiq t/(r) l" : L The proof follows immediately from Limit theorem 7 by taking ' ' _ L, f"(x) and L:Lr-Lr: f(x):f'(x) -f2(x):' -3. Therefore, 4: From Limit theorem L, lim (5r * 7): . TLLUSTRATIoN from Limit theorem 8, it follows that lim (5r + 7)n: I lim (5x + 7)la &--2

:

(-3)n

:81.

Z.Z.gLimit theorem 9 If lim f(x):

L and tr1 S(x) : M, and M # 0, then

n#:h is also frequently omitted from The proof, based on Definition 2.L."1., a beginning calculus text. However, a proof is given in Sec.2.6. Although we are postponing the proof, we apply this theorem when necessary. o rLLUsrRArroN5: From Limit theoremr, tj1 x:4, and from Limit theorem 1.,hm (-7x + 1) - -27. Therefore,from Limit theorem 9, lim r s-4

lim (-7x + t) t-4

_4 -27 __

4 27

2.2.70Limit theorem L0 tt tji f @) : L, then

n{f@:{L if L > 0 and rais any positive integer, or if L < 0 and n ts a positive odd integer. A proof of this theorem is given in Sec.2.6, and as is the casewith the preceding theorem, we apply it when necessary.

F

LIMITSAND CONTINUIW

o ILLUSTRATToN 6: From the result of Illustration 5 and Limit theorem 1.0, it follows that

lim e-4

J?#

L-

Following are some examplesillustrating the application of the above _ theorems. To indicate the limit theorem being ,s"d, we use the abbreviation "L.T." followed by the theorem number; for example, ,'L.T. 2,, refers to Limit theorem 2. EXAMPLE L:

lT

Find

(* +7x-5)

SOLUTION:

lim (f *7x-

5) : lim f * lim 7x-lim 5 .1'-3

and, when applicable, indicate the limit theorems which are being used.

(1.r.s)

I-B

: lim x . Iim r* t-3

."-3

lim 7 . limr-

lim 5

:r-S

r_g

-3'3+7.3-5

t-B

(L.T.6)

G.T.3andL.T.2)

:9*2L-5

It is important, at this point, to realize that the limit in Example 1 was evaluated by direct application of the theorems on limits. For the f u n c t i o n/ d e f i n e db y f ( x ) : x 2 i r x - S , w e s e et h a t f ( g ) : g z + 7 . g - 5 - 25, which is the same as lim (x' + 7x - 5). It is not always true that we havet_i1/(r) : f (a) (seeExample4). In ExampleL, rim (x) :/(3) f cause the function / is continuous at x:3. continuous functions in Sec. 2.5.

be-

we discuss the meaning of

rxatvrpr,n 2: Find

tFzxc+3

VT+s

and when applicable indicate the limit theorems being used.

(1.T.10) lim (r3 * 2x t 3)

(L.r.e)

lim 12 * lim 5 t-2

u-2

(L.r.s)

2.2 THEOREMSON LIMITSOF FUNCTIONS

imx)3+lim2'limr*lim3

(L.T.7 and L.T. 8) (L.T. 3 and L.T. 2)

8+4+3

EXAMPLE 3:

Find

-27 r. x3 llm-^ x-B x- J

and, when applicable, indicate the limit theoremsbeing used'

solurroN: Here we have a more difficult problem since Limit theorem 9 - 3) because lim (r - 3) - g. cannot be applied to the quotient (rt - 27)l(x However, factoring the numerator, we obtain -xs : : 27 -

(x - 3) (x'z* 3x -f 9)

x-3

x-3

This quotient is (r2*3x+9) if x*3 (sinceif x* 3 we can divide the numerator and denominator by (r - 3)). When evaluatingItT [(xt - 27)l(x - 3)], we are consideringvaluesof r close to 3, but not equal to 3. Therefore, it is possible to divide the numerator and denominator by (x- 3). The solution to this problem takes the following form: xs_27

..

ttfr-3:itTT :

lT

(r-3)(f+3x*9) (x' i 3x + 9)

: lim

: (lim r)2 + 1.8

dividing numerator and denominator by (, - 3) since x * 3

(1.r.4) (L.T. 8 and L.T. 1)

I-3

:32 * L8

(L.r.3)

:27

Note that in Example 3 (rt - 27)l(r - 3) is not defined when x:3, but lim [(rt - 27)l(x - 3)l exists and is equal to 27.

NY

80


ExAMPLE4: Given that / is the function defined by

'f[(cx ):

(* -3

li

ifx*4 ifx:4

soLUrIoN:

when

evaluating lim f (*), we are considering values o f x

closeto 4 but not equalto 4. Tf;il, we have lim /(r) : lim (x - 3)

find lim f (x).

:t

(1.r.1)

I n E x a m p l e4 l ; r y f @ ) : t

b u t f ( 4 ) : 5 ' t h e r e f o r e ,t T f @ ) + f ( ) .

This is an example of a function which is discontinuousat x:4. In terms of geometry, this means that there is a break in the graph of the function at the point where x: 4 (seeFig. 2.2.1).The graprr or ttre function consistsof the isolatedpoint (4,5) and the straight line whose equa_ tion is y : x - 3, with the point (4, L) deleted. There are two limits that we need in order to prove Limit theorems 9 and 10 in Sec.2.6. They are given in the following two theorems.

Figure2.2.1

2.2.11Theorem If. a is any number except zero, ,.

1

llm-:t-aX

1 a

pRooF: We need to consider two cases: a > 0 and a < 0. We prove the theorem if a > 0 and leave the proof for a 1 O as an exercise (see Exercise 30). so, if a is any positive number, from Definition 2.L.j, we must show that for any € ) 0 there exists a 6 > 0 such that

Ir

1l

li-il
w h e n e v e r0 < I r - o l < a

1 1 t l ___ l ta_ - x l

lx- al

li-d- I * I:loll|* and because a ) 0, we have

: tx- ot-L 11-ll al alxl

lx

(6)

We proceed to find some upper bound for the fraction Llalxl. If we require the D for which we are looking to be less than or equal to La, then certainly whenever lx- al < 6, we know l*- ol < ia. The inequality l*-ol<*a

2.2 THEOREMSON LIMITSOF FUNCTIONS 81

is equivalent to ba
t a < l x l< 8 a Hence,whenever lx - al < ta,l"l > ta. So from Eq' (6) we have

-v-'t'h,
l*- rl ta (becausethis is true when l*- ol < 6, then lrol <*a'e and Jrf - ol < ia).This gives us l,

:tx-'t'h
Il -t- - l 1 l < lx

al

€

whenever 0 < lx - ,l ( 6, and where 6 - min(# a, iaze). Hence, we have I proved that lim (U x) : Ll a if a is any positive number' The proof of the next theorem rnakes use of the following formula, where n is any positive integer. A n - b n : ( a - b ) ( s " - r a o n - 2 f ua+n - 3 b 2 + . . + a 2 b n - 3a* b " - z + b " - r ) Q ) The proof of Eq. (7) follows. a ( a n - rl a " - z b + ' ' '

* a b " - z+ b " - ' ) : a nl a " - r b* " '

b ( s n - r* a n - z b + ' '

* a b " - z+ b " - r ) - a ' - r b + ' '

+ 0 ; 2 b n*- 2a b " - r ( 8 )

+ f l i z b n*- za b - r * b

subtracting terms of Eq. (9) from terms of Eq. (8) gives us Eq. (7). 2.2.12 Theorem

lf n is a positive integer, then lim ,trx: \fr

(9)

ilv

LIMITSANDCONTINUITY

if either (i) a is any positive number or (ii) a is a negative number or zero and n is odd. pRooF: We prove part (i) and leave the proof of paft (ii) as an exercise (seeExercise31). Letting a be any positive number and using Definition 2.1.1, we must show that for any € ) 0 there exists a 6 ) 0 such that w h e n e v e r0 < l r - a l

liT-{il<,

<6 To expressl{G {al in terms of lx- al, we use Eq. (z). we have l{i-

q' t." ).1(x-' t" )"(7rtn)n-zsrtn .-' * + ' {i l :l (xtt" Wl l"-rl

Applying

l{i-

Eq,. (7) to the numerator, we have

ffil: lxrl

l'{"-rlt" + /(n-z)tnartn +

a artng@-2)tn + A@-r)tnl

(10)

We wish to find an upper bound for the fraction on the right side of Eq. (10).If we require the 6 for which we are looking to be less than ot "q.rilis to a, then whenever lx- al < E, we know that lrrl < a, which equivalent to -a < x- a < a or 0
Flence, whenever lx - al 1 a, we have

l\Ti- til .l*- ol # - ol < a@--r)tne. we want, then, l* - ol ' 1,fa@-r1rnI e or, equivalently, l, Thus, we take 6 as the smaller of a and 6@-r)tnr,which assures us that whenever lx- al < 6, then l*- al q s(n-r)rne and r ) 0. Therefore,

l{i - {Gl -lx-al ly(r-r)tn |

1(n-z)lnortnJ

a yrtng(n-z)tn+ A@-L)tnl

"l[t_

2,2 THEOREMSON LIMITSOF FUNCTIONS

1

< a@ -r)tne. +

A\n-r)ln

:e whenever 0 < l* - ol < 6, where 6 : min( a, 6(n-r)tne).We have therefore I proved that lim ffi - \fr, if a is any positive number.

7: From part (i) of Theorem2.2.'l'2 it follows thatlim li: o TLLUSTRATIoN 8-5

rG and \im '{i:

2; frompart (ii) it follows that lim {i:

-3. We do not

n'-27

tr-16

consider a limit such as lim Vi because fi

is not defined on any oPen

interval containin g - 4.

2.2 Exercises In Exercises 1 through 17, find the value of the limit. and when applicable indicate rl l used. I 1 . l i m (x't 2x - 1. \ 4. -

"t.

lrfl

;;

,

2xtL t-3x*4

d' Irm

tt'

10.

r-o

,1f,--r c - tE

(y' - 2y' * 3y - 4)

3.

u3+8

6.

5. trm-u--zVtz

---------;-

,. x2+5x+6 rlm--------.;-; f -x-12

13. lim

:t:,

3x2-l7x*20 4x'z- zsx+'J6

r--E

]'-T,!9.

1 " t'i

ig. L2.

r, . s

(nrwr: Rationalize the numerator.)

X

r s _ x z- r -* 1 0 L 6 .l i m - - - : ^ x--z x'+ cx I 2

2 x 3 - S x 2- 2 x - 3 i'"7 ' i i m # 13x2I 4x - 3 4x3 ;:; i:r

s h o w t h a t l i mf ( x ) : f ( 2 ) .

1 8 .I f f ( x ) : x 2 + 5 x - 3 , 1 9. If F(r) : 2x 3*7x - 1,

s h o w th a t l i m F (r) : F (-1 ).

{,\t

"!

20. If g@) : (x' - 0lQ - 2) , show that lim 8(r) : 4 but that g(2) is not defined. 21,.lf h(x) : 6trT9

- 3)lx, show that lim h(x) : * but that h(0) is not defined.

the limit

theorems being

..uF

84

LIMITS AND CONTINUIry

22. Given that f is the function defined by

f ( x ) : {[?r ' (a) Find lim/(r), . -t^:\

t-2

1 i rx r 2 i f .x : 2 and show that lim f(x) + f(Z). x-2

(b)- praw a sketch of the graph of f .

23. Given that / is the function defined by (-z-9 ifx*-j f ( x ) : lLi 4 ifr:-3 (a) Find,li1 /(r), and show that,lim f (x) + /(-gl. (b) Draw a sketch of the graph of /. 24. Using Definition 2.'1,.1, prove that if lim /(r) : 1 and lim g(x) : 1'4 then

tq t/(r)- s?)l: L- M 25- Prove Limit theorem 5 by applying 26. Prove that if lim /(r)

Limit theorem 4 and mathematical

induction.

: L, then lim (/(x) - L) : 0.

27. Using Definition 2.'/-..L, prove that if lim /(x) : 1 and lim g(r) : g then tim /(r) ' g(r) : 0 (rflNT:Inordertoprovethatliml(.r)'g(r):0,wemustshowthatforanye>0thereexistsa6>0suchthat < 6 . F i " " t s h o w t h a t t h e r e i s a d 1> 0 s u c h t h a t l / ( x ) l < 1 + ll(r)'sk)l o-suchthat kr"lil.ltr * lrl) whenever 0 < lx- al < 8", by applying Definition 2.1.1 to lim g(r) : 0. By taking 6 as the smalhr of the two nunbers A and 6r, the theorem i. p-""ay' 28. Prove Limit theot€m 6: If lim /(r) = L and lim g(r) = M, then

rimU(x)'s(xll:L.M ' (ym, _writg-{1) 8e)=[l(x) cises26 and,27.)

-t]s(r)+tk(r)

- Ml+L.M.

Apply Limit theorem5 and the resultsof Exer-

29. ProveLinit theorcm 7 by applying Limit theor€m 6 and mathematicalinduction. 3). ProveTheorem2.2.1.1if a < 0. 31. ProveTheorem2.2.12(ii).

"*ry,".

2.3 ONE-SIDEDLIMITS

2.3 ONE-SIDED LIMITS

Definition 2.9."1.

85

f @) we are concernedwith values of r in an oPen lrT interval containing abut not at a itself, that is, at values of r closeto a and either greater than a or less than a. However,_supPose,for example, Because/(r) doesnot that we have the function / for which f (x) : {R' containing 4. Hence, interval on any open < not defined 4, is x exist if f is restricted to values x if. 4. However, lim {x consider cannot we greater than 4, the"utlrJ of fr-x - +can be made as closeto 0 as we please by taking r sufficiently closeto 4 but greater than 4.ln such a caseas this, we let r approach 4 from the right and consider the one-sidedlimit from the right or the right-handlimit, which we now define. When considering

Let f be a function which is defined at every number in some open ina from the right, is L, terval (a, c).Then the limit of f (x), as x npproaches written

: L |'T,/t') if for any € ) 0, however small, there exists a 6 > 0 such that lf(x)

x-

-Ll <.

w h e n e v e r0 < x - a 1 6

(1)

Note that in statement (1) there are no absolute-valuebars around a s i n c e x - a ) 0, becausex ) a. It follows from Definition 2.3.1 that

hm Vr - 4:0 I-4+

If when considering the limit of a function the independent variable r is restricted to values less than a number a,we say that x a|Ptoachesa from the left; the limit is called the one-sidedlimit from the left ot the lefthandlimit. 2.9.2 Definition

Let f be a function which is defined at every number in some open ina from the left, is L, written terval (d, a).Then thelimit of f (x), asx approaches

'"^tk):L if for any € > 0, however small, there exists a D ) 0 such that -Ll <. w h e n e v e r- D < x - a 1 0 lf (*) limit, or the undirectedlimit, We can refer ,o I,,T f @) as the tuso-sided to distinguish it from the one-sided limits. Limit theorems t-10 given in Sec. 2.2 remain unchanged when 4 a" is replacedby "x + a*" ot "x + a-." "x


EXAMPLEL:

Let f be defined by

f-t ifr
solurroN: A sketch of the graph is shown in Fig. 2 . 3 . 7 l.i m - -1 /(r) since, if r is any number less than 0, f(x) has the value -1. Similarly, tt1/(r) : *1. v

(a) Draw a sketch of the graph of f . (b) Determine Iim f (x) it it exists. (c) Determine lim f (x) if it exists. F i g u r e2 . 3 . 1

In the aboveexamplelim /(r) * Lim (x). Becausethe left-hand limit f the right-hand limit J# r",o, say that the rwo-sided limit, 1nd "qdi,';e ttl, does not exist.The concept of the two-sided limit failing to exist f ftq becausethe two one-sided limits are unequal is stated in the following theorem. 2'3'3 Theorem

@) exists and is equal to L if and only if nf lin_r/(r) and lim /(r) both exist and both are equal to L. The proof of this theorem is left as an exercise(seeExercise16).

Let g be defined by (t-t

g ( r ): { I ' l1

ifx#0

solurroN:

lim g(r) : lim (-r)

ifx:0

(a) Draw a sketch of the graph of g. (b) Find lim g(r) if it exists.

A sketch of the graph is shown in Fig. Z.g.Z. : 0

and

lim g(r) lim g(r) : lim lim r: Jr-o+

0

J-O+

Therefore,by Theorem 2.9.9lim g(r) exists and is equal to 0. Note that 8(0): 2, which has no effectJ"ii- S@).

Figure2.3.2

LIMITS 2.3 ONE-SIDED

Let h be defined by

EXAMPLE3:

(4- xz

. h ( x ): 1 '

i f .x = t

, 1zr xz if.L < x

sol,urroN: A sketch of the graph is shown in Fig. 2-3.3. lim h(r) - lim (4 - f) :3 t-l-

(a) Draw a sketch of the graph of h. (b) Find each of the following limits if they exist: lim h (r),

r-l-

tim h(r) : lim (Z + f7 :3 t-l+

Therefore, by Theorem 2.3.3 lim h(r) exists and is equal to 3. Note that

lim h(r), lim h(r). c-l

Figure2.3.3

2.3 Exercises In Exercises 1 through 10, draw a skekh oI the graph and find the indicated limit if it exists; if the limit does not odst, give the reason.

[ 2 if x<1'l r. f (x): 1-1 if x :1[; (a)tim /(r); (b) lim f (x); (c)lim /(r) r-rr'r '-1+ L-3 1f.| < xJ

2.g(s): {;l:

(c) st'l (u) ii'-;;1}; (a),riq.s(s); "!1_s(s);111

s.h(x): {?t!i lf ; : l}, ,n I*nr,t' or }:Th(x\;(c)rimft(x) a. Fk):

tt'1,tu) F(x);(c)rimF(r) l{ - r, il;:2,}; @llT }11'_

(2r+3 s. f (r) : li 17

= 2r

ifr<1l if r: r [; (a) trmf (r); (b) lim f (r); (c)lim /(r)

if 7 < r)

r''-

r-t-

r'r

y-1

11 \1't5 ,1

LIMITSAND CONTINUITY [3 +t2 if r<-2] 6 . 8(f):19.

- f., :f"r=-21, it-Z < t)

Ul

(a).lims(r);(b)lim g(f);(c)lims(f) t--2+ t--z-"

t--2

F(x): lx- sl;f") F(x);trl F(x);(c)limF(r) l'-T l'_? 8. f (x):3 + l2x- ah(a)tryf @; (b) (x);(c)rr\ f (x) l'* f 7.

9. f(x) :f;

l-l

(a)rim f (x); (b) lim /(r); (c)lim f(x)

I

r-O+

ft-O-

t-O

10. The absolute value of the signum function (see Exercise 22 in Exercises 1.g).

(") rl; to) lsgnrl; rc) lsgnrl. 1'_T"lsgn llT. I'i

1 1 . F is the function of Exercise30 in Exercises1.g. Find, if they exist:

(") F(x);(b) F(x);G) F(r). 11[ ]1T1T 12. lt is the functionof Exercise31 in Exer€ises 1.8.Find, if they exis* (a) lim h(r), (b) lim_ft(r); (c)lim l(r). 13. Find, il they exisr (a) liq [r]l; (b) tim [r]l; (c) tim l[rj. ,-2a_2

14' Let G(t) : ["n + [a - x]1.Draw a sketchof the graphof G.Find,if theyexish(a)rim G(r);(b) lim G(r);(c)lim G(r). 15. Given if x=l xl:[f +g if rsl . . , a. tr**\r_- 1l f, [x*1 if r>1-'if r>1 (a) show that lim /(r) and lim /(r) both exist but ar€ not equal, and hencelim /(r) doesnot exist. r_l-

-

I-l+

t_l

(b) Show that lim g(x) and lim g(r) both exist but are not equal, and hence u lim g(r) mg I- 1t-l+

.ifil. -E\rl. (g) Find formulastor ; } k J _ r . r r r q a v l r t . u r q r l ( r l I \ L t f(x) --7(G)"Frove (

\--.-

-.1'r"r .X 7'{ * r l "I * f - j . - '1: , 'fr, ),:. '. i",,,t

, _ . * , ' , . ' , Ii .l "i 'f. i .: : that lim f (x) .g(r) existsby' sfiowing that tim /(x) . S(r) : Iim (x) . S(r). fr-l ,_r_

does not exist.

t-l

'

l

Ir- -l +1 +

L5. Prove Theorem2.3.9. 17. Let rf \( 'x ) : l

( - l' i f x < 0 L 1 ifx>0

show that lim /(x) does not exist but that lim l/(r)l does exist. .z'--0 r-O

2.4 INFINITE LIMITS

Let f be the tunction defined by

i@'):& A sketch of the graph of this function is We investigate the function values of / when .r is close to 2. Letting

2.4 INFINITELIMITS

x approach 2 from the right, w€ have the values of f (x) given in Table 2.4.L. From this table we intuitively see that as r gets closer and closer T a b l e2 .4 .7

x

3

t T

3

t2

I

5

4

27

48

2l TO

300

20r T0o-

30,000

2001 T0T-6

3,000,000

to 2 through values greater than 2, f (*) increaseswithout bound; in other words, we can make f (x) as large as we please by taking r close enough to 2. To indicate that /(x) increases without bound as x approaches 2 through values greater than 2, we write

Figure2.4.1

?

l i m f t : -r - *

.r-z* (I

Z)

If we let r approach 2 from the left, we have the values of f (x) given in Table 2.4.2.'We intuitively see from this table that as r gets closer and T a b l e2 .4 .2

x

1

3

.)

B

4

3

t2

27

48

D

l9 10

300

199

T00.

30,000

1999

TIITT'

3,000,000

closer to 2, through values less than 2, f (x) increaseswithout bound; so we write f i m $ : *-c c

,-z- \X

z)-

Therefore/as )c approaches2 fuomeither the right or the left, f (x) increases without bound, and we write h m j -T: *-z \x

*oo

We have the following definition. 2.4.1 Definition

Let f be a function which is defined at every number in some open interval I containing a, exceptpossibly at the number a itself. As x approaches a, f (x) increaseswithout bound,which is written lim /(r) I-A

: *co

(1)


if foranynumberN > 0 thereexistsa 6 > 0 suchthat/(r) > Nwhenever

0 < l r - a l< D . In words, Definition 2.4.1states that we can make f (x) as large as we please (i.e., greater than any positive number N) by taking r sufficiently close to a. NorE: It should be stressed again (as in Sec. 1.1) that *m is not a real number; hence, when we write it does not have the same |2fO:*a, meaning ur l,ti f@) : L, where L is a real number. Equation (1) can be read as "the limit of f(x) as r approaches a is positive infinity." In such a case the limit does not exist, but the symbolism r'+@" indicates the behavior of the function values f(x) as x gets closer and closer to a. In an analogous manner, we can indicate the behavior of a function whose function values decrease without bound. To lead up to this, we consider the function g defined by the equation

-3

/\

g\x):lJ--2]l A sketchof the graph of this function is in Fig.2.a.2. The function valuesgiven by g(r):-31(x-Z)'are the negativesof the function valuesgiven by f (x)-31(x-2)'. So for the functiong as r approaches 2, either from the right or the left, g(r) decreaseswithout bound, and we write

Figure 2.4.2

-3

#W:-q) In general,we have the following definition. 2.4.2 Definition

Let / be a function which is defined at every number in some open interval / containing a, exceptpossibly at the number a itself.As x approaches a, f(x) decreases without bound,which is written

rim f(x)

( 2)

t-A

if for any number N < 0 t h e r e e x i s t s a 6 ) 0 s u c h t h a t / ( x ) < N w h e n e v e r <6. 0
and if for any numberN > 0 thereexistsa D ) 0 suchthat/(x) > N when-

2.4 INFINITELIMITS

91

a < 6. Similar definitions can be given if lim f(x):an lim /(x) _ -@, and lim f (x) : -a. "-o* *o* suppose ,t"i n is the function defined by the equation

ever 0 < x-

,2x h(x):-

(3)

A sketch of the graph of this function is in Fig.2.4.3. By referring to Figs. 2.4.I, 2.4.2, and 2.4.3, note the difference in the behavior of the function whose graph is sketched in Fig. 2.4.3 from the functions of the other two figures. We see that

2x

lrfll r-t X-

Figure2.4.3

:

--r,

L

and 2x lim -;: L 2-vX-

aco

That is, for the function defined by Eq. (3), as r approaches1 through values less than L, the function values decreasewithout bound, and as x approachesL through values greater than L, the function values increase without bound. Before giving some examples,we need two limit theorems involving "infinite" limits. 2.4.3 Limit theorem 11

If r is any positive integer, then (i) lim *: ,t-O+

**

4

...\ ,. L [-o if r is odd (tt, lt_TF: l+.o if r ts even pRooF: We prove part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 19). We must show that for any N > 0 there existsa6)0suchthat 1 ;tN

whenever 0(r<6

or, equivalently, because r > 0 and N ) 0, 1

t"aft

w h e n e v e r0 ( r < 6

or, equivalently, since r ) 0,

x<

whenever 0(x(6

92

LIMITSAND CONTINUITY The above statement holds if E: 1 ;t

N

w henever

(l/Nr;tt'. We conclude that

0 ( x ( 6,i f 6:

/1\r/r (N /

I

o ILLusrRArroN L: From Limit theorem 11(i), it follows that h m 4:

t-O* )P

* oo

and

1i m 1: ,.o* f

* cc

From Limit theorem 11(ii), we have rim 4:

r-o-f,

-@

and hm 1: t-o- X*

*oo

2.4.4 Limit theorem 1.2 lf. a is any real number, and if lim /(x) : 0 and lim g(r) : c, where c is a t-A

constant not equal to 0, then (0 if c ) 0 andif f(x)-

0throughpositivevaluesof f(x),

'jT#:*m (ii) if c ) 0 andif f (x)-

0 throughnegativevaluesof f (x),

Ir#:-co (iii) if c < 0 andif f(x) -

0throughpositivevalues of.f(x),

n#:-m (iv) if c < 0 and if f(x) -

0 through negative values of f(x),

n8:*oo The theorem is also valid if "x + a" is replaced by "x + a*" ot "x +

a-."

pRooF: We prove part (i) and leave the proofs of the other parts as exercises (see Exercises 20-22). To prove that ,. , q(r) lim i*: *co " -" l \x) we must show that for anv N ) 0 there exists a D ) 0 such that

#>N

w h e n e v eor < l r - o l < o

(4)

Sincelim g(r) - c > 0,by taking e: tc in Definition 2.1.1,it follows that t-A

2.4 INFINITELIMITS

93

there exists a 0r ) 0 such that lg(r)

- cl < trc whenever 0

By applying Theorcm 1.2.2 to the above inequality, it follows that there existsaDr)0suchthat -tc
w h e n e v e r0 ( l * - o l

So there exists a Dr ) 0 such that (5) whenever 0 < lx - al < 6' Now lt^ f (*): 0. Thus, for any € ) 0, there exists a 6z ) 0 such that

g(x) > ic

l/(r)l
w h e n e v e r0 < l r - a l

<6'

Since f (x) is approaching zero through positive values of f (x), we can remove the absolute-value bars around /(x); hence, for any € ) 0 there existsaEz)0suchthat f(x)<,

w h e n e v e r0 < l * - o l

<0,

(5)

From statements(5) and (6) we can conclude that for any € > 0 there exist a Dr ) 0 and a 6z ) 0 such that g_2rit

f(x)

'

Hence, if e :

€ w h e n e v e r0 < l r - a l cl2N and E:

&) t--r p *t N:N M

<6,

and 0
min(Dt, Er), then

w h e n e v e r0 < l r - - o l < S

which is statement (4). Hence, part (i) is proved.

t

SOLUTION:

x2+x+2 ,. (u)llTm:l,TIm

x2*x*2

The limit of the numerator rs L4, which can be easily verified. lim (r-3)(r+

1): lim (r-3)'

lim (r+ 1)

The limit of the denominator is 0, and the denominator is approaching 0 through positive values. Then applying Limit theorem Lz(i), we obtain

94


)c2+x+2

(r-3)(r+t) As in part (a), the limit of the numerator is L4.

li?

('- 3)(r + 1):

llT

('- 3) ' lim (r + 1)

=0'4:0 In this case, the limit of the denominator is zero, but the denominator is approaching zero through negative values. Applying Limit theorem lz(ii), we have

s o r , u r r o N : ( a )B e c a u sxe+ 2 + ,x - 2

,rlilh. ^ \--------E - 4--- , :: , -rlli:i. - @x-2* x-2

*-*

: lim I-2+

) 0,andso.r- Z:

{@-2Y. Thus,

!(x-2\2

\/x-2\/x+2 !x-2

lx-2

Yx*2 ,r ::'-Tffi

The limit of the numerator is 2. The limit of the denominator is 0, and the denominator is approaching 0 through positive values. Therefore, by Limit theorem 12(i) it follows that

,. {7-4

IIIIIt-2+

X-

^ Z

--T€

(b) Because x+2-, Therefore,

\/4=7 : r l l f.. ,. r: rill:r------t-z- x-'2

x-2<0,

and so x-2--(2-x):-\/Q=T

\/r-

r-2--V2Y2* ,r :j'_T:ffi

\/r+ x x Y2-

x

x

The limit of the numerator is2. The limit of the denominator is 0, and the denominator is approaching 0 through negative values. Hence, by Limit theorem lz(ii), we have

\/4=T

l r m . 7 - : -- € Z .x-2- X

2.4 INFINITELIMITS

ExAMPLE 3:

sor,urroN: lim [r]l :3., Therefore, lim ([rl - 4): -t. Furthermore, (t - 4) : 0, and x - 4 is approaching 0 through negative values. l:? Hence, from Limit theorem lZ(iv), we have

Find:

u* l[4];4 x-4 r-4-

u * f lXr- n+ l 4 : * o r-4-

Remember that because *co and -oo are not numbers, the Limit theorems 1-10 of Sec.2.2 do not hold for "infinite" limits. However/ we do have the following properties regarding such limits. The proofs are left as exercises(seeExercises23-25). (i) If lim /(r) : *@, and lim g(r) : c, where c is any constant, then

2.4.5 Theorem

t^lf(x) +g(r)I:+* (ii) If lim /(r) : -Q, and lim 8@) : c, where c is any constant, then t-A

r^ lf (x)+ g(r)l: -a" is replaced by "x+

The theorem is valid if. "x+

fl*" ot "x+

A-."

:1, it follows o rLLUSrRArroN 2: Because lim +:- ''t' *m and lim + 2-2+x * 2 + s-2+x from Theorem 2.4.5(i) that ,'

1 I

r

r m tI _ _ _ _ _ + _ + l : * m i-i;Lx-2' x*21

2.4.6Theorem If l;af{x):*.o

and

.

fiB@):

c, where c is any constantexcept0,

then

(i) if c ) 0,I\f O>. s@): *oo. (ii) if c I 0,l\f(x).

g(r) : -o.

The theorem is valid if. "x -> a" is replaced by . rr,r,usrRArrorq 3: q

l i m . . * f -u : * o 5) r_s tx

r

r.

x*4

and lim *-s x-

. --7 4

Therefore, from Theorem 2.4.6(ii), we have S i. I ' x*Af 'iT LG-3F x-41:-€'

96

LIMITSAND CONTINUITY 2.4.7 Theorem

If t;yf@)_-oo

and limg(x):c,

where c is any constant except 0,

then (i ) i f c ) O,Li ^ f (x)' S (r) : -oo. (i i ) i f c I 0,l t^ f (x)' g(r): T h e th e orem i s val i d i f " x+

* o.

a" i s repl aced by " )c+ a* " ot" x+

a- . "

o ILLUsrRArroN 4: In Example 2(b) we showed

{44 ,llm. -----:

-oo

;:;- x-2 Furthermore, ,. x-3 llm -- _ ;:;--x + 2

t 4

Thus, from Theorem 2.4.7(ii), it follows that

.lfm . l$=7

l--:;:;-L x-2

Exercises 2.4 ,,Q

In Exercises1 through"l.4, evaluatethe limit. 1. lim -x2-a+X

'

4

t+)

4.rimffi ,. t/s+*

/. rlm .r*o-

X'

'\R 10. lim r-4-

X'- 4

Dm(i-+) 16. Prove that lim 17. Prove that lim

?

ffi: -) ffi:

*o by using Definition 2.4.1. -@ by using Definition 2.4.2.

18. Use Definition 2.4.1,to prove that

jlils+'l :*,.

15-rl

x - 3 1| :

x*21

r€

o

AT A NUMBER OF A FUNCTION 2.5 CONTINUIry 19. Prove Theorem 2.4.3(ii).

20. Prove Theorem 2.4.4(1i).

21. Prove Theorem 2.4.4(iii).

22. Prove Theorem Z.a.a(iv\.

23. Prove Theorem 2.4.5.



2.5 CONTINUITY OF A FUNCTION AT A NUMBER v

In Sec.2.1 we considered the function / defined by the equation ,.,-.\ (2x * 3) (x - 1) I\x): x_.1 We noted that / is defined for all values of r except 1. A sketch of the graph consisting of all points on the line y :2x * 3 except (1, 5) is shown in Fig. 2.5.1.There is a break in the graph at the point (1,5), and we state at the number 1.. that the function f is discontinuous for instance, the function is defined for all If we define f(l):2, values of x, but there is still a break in the graph, and the function is still however, there is no break in discontinuous at L. If we define f(l):5, the graph, and the function / is said to be continuousat all values of x. We have the following definition.

2.5.1 Definition

The function / is said to be continuousat the number a if and only if the following three conditions are satisfied: (i) f (a) exists. (ii) /(x) exists. L'i (iii) lim f (x): f (o). If one or more of these three conditions fails to hold at a, the function f at a. is said to be discontinuous We now consider some illustrations of discontinuous functions. In each illustration we draw a sketch of the graph, determine the points where there is a break in the graph, and show which of the three conditions in Definition 2.5.1fails to hold at each discontinuity. L: Let / be defined as follows: . rLLUSrRArroN ?r. [(2x+3)(l-1) ,-1 f(x):j lz

ifx*t if x:'1.

A sketch of the graph of this function is given in Fig. 2.5.2.We see that there is a break in the graph at the point where x : 1, and so we investi-

LIMITSAND CONTINU]TY

gate there the conditions of Definition 2.5.1.. f (1) :2; thercfore,condition (i) is satisfied. LtT fttl Itm f (*):

:5; therefore,condition (ii) is satisfied. 5, but f (L) :2; therefore,condition (iii) is not satisfied.

We conclude that f is discontinuous at L.

o

Note that if in Illustration I f (l) is defined to be 5, then lim /(r) : f (1) and / would be continuous at 1.. o rLLUsrRArroN2: Let f be defined bv 1

,f,\( xt \ : +x _ 2

Figure2.5.3

A sketch of the graph of f is given in Fig. 2.s.2. There is a break in the graph at the point where x : 2, and so we investigate there the conditions of Definition 2.5.1.. f (2) is not defined; therefore, condition (i) is not satisfied. We conclude that f is discontinuous at 2. o o rLLUsrRArroN 3: Let g be defined by

rJ

8(r): l[3

Figure2.5.4

if x # 2 ifx-2

A sketch of the graph of g is shown in Fig. 2.5.4.Investigatingthe three conditions of Definition 2.5.1,at 2 we have the following. S(2):3; therefore,condition (i) is satisfied. : *mi therefore, condition (ii) is not liT g(x) : -m, and ]i1 S(r) satisfied. Thus, g is discontinuousat 2. . . rLLUsrRArroN4: Let h be defined by 7., [3*x n\x):tg_,

if.x="]. ifj,
A sketchof the graph of h is shown in Fig. 2.5.5.Becausethere is a break in the graph at the point where x:1, we investigate the conditions of Definition 2.5.1at L. We have the following. h(t1 :4; therefore,condition (i) is satisfied. lim h(r) : lim (3 + r1 :4 t-7lim h(r) - lim (3- x):2

a-1"

Figure2.5.5

,t-l+

AT A NUMBER 99 OF A FUNCTION 2.5 CONTINUIW Because limh(x)

+limh(r),

we conclude that lim h(x) does not exis|

lr-l+

t-l-

t'l

therefore, condition (ii) fails to hold at 1. Hence, h is discontinuous at 1. 5: LCt F bC dCfiNCdbY o ILLUSTRATION

F(r) Figure2.5.6

flr-sl L2

ifx*3 i f .x : 3

A sketch of the graph of F is shown in Fig. 2.5.5. We investigate the three conditions of Definition 2.5.1at the point where x: 3. We have F(3) :2; therefore, condition (i) is satisfied.

nttl : 0. So lim F(r) exists and is 0; therefore, tttl : 0 and l,T_ l.T condition (ii) is satisfied. tim F(r) :0 but F(3) :2; therefore,condition (iii) is not satisfied. c'3

Therefore, F is discontinuous at 3.

o

It should be apparent that the geometric notion of a break in the graph at a certain point is synonymous with the analytic concept of a function being discontinuous at a certain value of the independent variable. After Illustration L we mentioned that it f(1) had been defined to be 5, then / would be continuous at 1. This illustrates the concept of.aremovable discontinuity. In general, suppose that f is a function which is discontinuous at the number a,but for which f(tl exists. Then either f(a) + Tq fOl or else /(a) does not exist. Such a discontinuity is called a refi movable discontinuity because if f wete redefined at a so that f(a) l;a f@), / becomes continuous at a.If. the discontinuity is not removable it is called an essentialdiscontinuita.

nxervrpr.El-: In each of the Illustrations 1 through 5, determine if the discontinuity is removable or essential.

solurroN:

In Illustration 1 the function is discontinuous at 1 but lilf

(x):

5. By redefining f0 :5, we have lim /(x) : f (t), and so the discontir-r nuity is removable. In Illustration 2, the function / is discontinuous at 2. lim /(r) does not r'2 exist; hence, the discontinuity is essential. In Illustration 3, lim g(r) does not exist, and so the discontinuity is I-2 essential.

In Illustration 4, the function h is discontinuousbecauselim h(r) does t-r not exist, and so again the discontinuity is essential. F(x):0, but F(3): 2, and so F is discontinuousat In IllustrationU, TT 3. However, if F(3) is redefined to be 0, then the function is continuous at 3, and so the discontinuity is removable.

lOO


Exercises 2.5

y ,1

In Exercises 1 through 22, draw a sketch of the graph of the function; then by observing wher€ there are breals in the graph, detemrine the values of the independent variable at which the fun*ion is discontin;ous and show why Definition 2.5.1 is not satisfied at each discontinuitv.

4 l . f .( ,x )\ : t , fx+ -4 Lt

irx#4 i f .x : 4

ra_16 a.h@):fr

. r+3 7.h(x):Vfrj_6

I t+x 1 0 .H ( r ) : l 2 - x l2x-1

x2+x-6 3. F(r) : x*3

6. f(x):

f-2x2-llx*12 x2-5x*4

e.f(x):l

t1, if x < 0 0 if x :0 I r if 0 < x

v2-4

8.g(r):;d

if x<-2 it-Z1x<2 if21x

13. f(x\:ffi 't [t-+*-t !6.f(x\:l ,c+2

irx#-Z

2 . g ( x.) : llx' + * 2 if x: -2 1.0 (x2+x-6 ifx*-3 5.s(r):{ r+3 ifx:-3 Lt

+ sl i f x + - t

1 , 1f ,( .x ) : l Z x + 5 1

1 2 .G k \ :

f_r-? r a .F ( r ) : l l ' - 3 T i r x # 3 ifx:3 L0

1 s s. ( r ) : . 18 - 3 r

{l2r LJ

f2x*j I x*3

if.x:-* iIx
1f'x*-2 ifx:-2

l-3

17. The greatest inteter function. (See lllustration 5 in Sec. 1.7.) 18. The unit step function, (SeeExercise21 in Exercises1.8.) 19. The signum function. (See Exercise 22 in Exercises 1.8.) 20. The function of Exercise 24 in Exercises 1.8. 21.. The function of Exercise 33 in Exercises 1.8. 22. The function of Exercise 32 in Exercises 1.8.

In Exercises 23 through 30, prove that the function is discontinuous at the number c. Then detemrine if the discontinuity is rcrrovable or eaeential. If the discontinuity is removable, define l(a) so that the discontinuity is removed.

23. f(x):'ft, o:l

2a. f (s): {# [0

- '' 2s. f (x): {f

ifx#3) ifx:3J;a:c

ir s *-u},

,: -g

ifs:-5)

r,\ [*-ax-+3 26.f(x):l x-3 [5

ze.f0):#,a:o

i f x * g )- l ; a : 3 if x :3)

ON CONTINUITY101 2,6 THEOREMS ( t z - L- i f t < 2 I . " : , a:

zs.f(t): ll

{rt riJ,

z

3 0 .f ( x )

a:0

31. Let / be the function defined bY

rt'r- [rXl 1 'f ( x ): lt1 tr-[x+lnl

if fixn is even if flrnisodd

Draw a sketch of the graph of l. At what numbers is f discontinuous?

2.6 THEOREMS ON CONTINUITY 2.6.t Theorem

By applying Definition 2.5.1. and limit theorems, we have the following theorem about functions which are continuous at a number. lf f and,g are two functions which are continuous at the number a, then (i) (ii) (iiD (iv)

/ + g is continuous at a. / 8 is continuous at a. /' I is continuous at a. 0' flg is continuous at a, provided that g(a) *

To illustrate the kind of proof required for each part of this theorem, we prove part (i): Because f and,g are continuous at a, from Definition 2.5.1 we have (1)

lim/(r):f(a) 8-&

and

ljT st'l : s(a)

Q)

Therefore, from Eqs. (1) and (2) and Limit theoretrr 4, we have (3)

l i m [ / ( r ) + s ( r ) l : f f u )+ g ( a ) fi'A

Equation (3) is the condition that / *g is continuous at a, which furnishes the proof of (i). The proofs of parts (ii), (iii), and (iv) are left as exercises (see Exercises 1-3). 2.6.2 Theorem

A polynomial function is continuous at every number. To prove this theorem, consider the polynomial function / defined by f ( x ) : b s x "* b 1 Y n -*r b 2 x " - *2' '

+ b n - r x* b n

b 0* 0

where n is a nonnegative integer, and bo, br, . , bn are real numbers' By successive applications of limit theorems, w€ can show that if a is

102


any number II

tt*l : boan* brqn-r t bran-z+

. . + bn_ra* b,

from which it follows that Iim f(x) : f(a) thus establishing the theorem. The details of the proof are left as an exercise (see Exercise 4). 2'6'3 Theorem

A rational function is continuous at every number in its domain. PRooF: If is a rational function, it can be expressed as the quotient of / two polynomial functions. So f can be defined by o( v\

/f (\ x \: 9h ()x*)a where g and h are two polynomial functions, and the domain of consists / of all numbers except those for which h(x): g. If a is any number in the domain of I then h(a) + 0; and so by Limit theorem 9, l i m g(r) limf(x):-im h(x) .r-o

( 4)

I-A

Becauseg and h are polynomial functions, by Theorem 2.6.2 they are continuous at A, and so lim g(r) : g(a) and lim h(x): h(a). Consequently, t-A

from Eq. (4) we have

o! l i m f ('"" r ') : { \h(a) i-i;' Thus, we can conclude that / is continuous domain.

at every number

in its I

Definition 2.5.'l'statesthat the function / is continuous at the number a iI f(a) exists,if l:\f(x) exists,and if ri^ f(r):

(s)

f(q)

Definition2.7.1 statesthat lim f (x):Lif x'a 6 > 0 such that lf(x)-Ll
w h e n e v e r0 < 1 " - a l

for any € ) 0 there existsa <6

Applying this definition to Eq. (5), we have lim /(r) : f (a) if for any

2.6THEOREMS ONCONTINUITY 103 € > 0 there exists a 6 > 0 such that lf(x)-f(a)l
w h e n e v e r0 ( l x - o l

(5)

If f is continuous at at we know thatf (a) exists;thus, in statement(6) it is not necessarythat lr - al > 0 becausewhen x: a, statement(6)obviously holds. We have, then, the following definition of continuity of a function by using e and 6 notation.

2.6.4Dehnition

The function f is said to be continuous at the number a if / is defined on some open interval containing a and if for any € > 0 there exists a 6 > 0 such that lf(x)-f(a)l
wheneverlx-41 <6

This alternate definition is used in proving the followirtg important theorem regarding the limit of a composite function. 2.6.5 Theorem If lim g(r) - b and if the function / is continuous at b,

(/'s)@):f(b) or, equivalently, rim f (g(x) ) : /(lim S(r) ) PRooF: Because / is continuous at b, we have the following statement from Definition2.6.4. For any e1 ) 0 there exists a 61 ) 0 such that (7) <6' whenever

lf@-f(Ull<.,

lY-bl

Because lim g(x) : b, for any 6, ) 0 there exists a 62 > 0 such that

lg(r)

- bl < 6, whenever 0 < lx- al < 6,

(8)

Whenever 0 < l* - ol < Dr,we replacey in statement(7) by g(x) and obtain the following: For any €r > 0 there exists a 6r ) 0 such that - (b) < ., whenever lS(t) - bl < D' (e) lf QQ)) | f From statements (9) and (8), we conclude that for any er ) 0 there exists a 6z ) 0 such that ( l* - ol < D' IfGGD f(b) | < ., whenever 0 from which it follows that lim/(s(r)):/(b)

104


or, equivalently,

) : /(tims(r)) fif G<*)

T

An immediate application of Theorem 2.6.5 is in proving Limit theorems 9 and 10, the proofs of which were deferred until now. We restate these theorems and prove them, but before the statement of each theorem a special case is given as an example. EXAMpLE1: By using Theorem 2.6.5 and not Limit theorem 9 (limit of a quotient), find

solurroN:

Let the functions / and g be defined by a n d g ( r ) : x 2- t 2 x t S f(x) :4x-3

We consider /(r) lSk) as the product of /(r) and U S@) and use Limit theorem 6 (limit of a product). First of all, though, we must find lim Llg(x)

4x-3 trTfr]x+s

r.

and this we do by considering l,lg(x) as a composite function value. If h is the function defined by h(x) - 1lx, then the composite function h " B is the function defined by h(g(x)11: llg(x). Now

: t"tg(x'+ 2x-r 5) : 13 }T sftl To use Theorem 2.6.5,h must be continuous at L3, which follows from Theorem 2.2.L7.We have, then, r.:-

1'

,.

|\Nzx+s:f'3m

1,

: t"T h(sk)) - n\\S(r))

(by Theorem2.6.5)

: h(13)

:+ Therefore, by using Limit theorem G,we have

- 3)' t"'3\4x Ig v+fi

4x-Z ,. r;+ affi:

-5.* - 1 75

Limit theorem 9 nf@: 7.fk\L lfm ---:

L and tf

]1g-g(;r):

M and,M + 0, then

_

M "-i 8@) PRooF: Let h be the function defined by h(x):.l.lx. Then the composite

2.6 THEOREMSON CONTINUITY 105

function h " I is defined by h(S@D: Ug@).The function h is continuous everywhere except at 0, which follows from Theorem 2.2.7t. Hence, 1 r. lim ,,:limh(S@)) t-a *-" 8\x)

- h(lim S(r) )

(by Theorem2.6.5)

,,;;

-h Applying Limit theorem 6 and the above result, we have r.

L r. t / \ 'lim r. lim f(x) ^ ' t-e' ,-" 8\x)

fk)

lim *: *-" 8\x)

_L M

rxavrrr.n 2: BY using Theorem 2.6.5and not Limit theorem L0, find

11^flffi

solurroN;

Let the functions / and h be defined by

and h(fi:ffi

f(x):3x-5

5 The compositefunction h. f is defined by h(f(x)) :9gxf (x) *g '1.6; lrtg tgr - 5): so, to use Theorem 2.5.5 for the composite function h " f ,h must be continuousat 76, which follows from Theorem2.2.12(1). The solution, then, is as follows. fim,vex - 5: lim h(f (x))

- nQ)) (x)) f

(by rheorem2.6.5)

: h(L6)

:% -/

Limit theoremLo U

Q) : L, then fT f lim1@:\/L

if. L > 0 and n is any positive integer, or if L < 0 and n is a positive odd integer.

106


pRooF: Let h be the function defined by h(t) : Vr. Then the composite function h . f ts defined by h(f (x)) : \yf6. From Theorem 2.2.t2 it follows that h is continuous at L If L > 0 and n is any positive integer or if L < 0 and n is a positive odd integer. Therefore, we have lim tf@:

lim h(f (x))

: ,-ri;i f @)) (byrheorem2.6.5) : h(L)

:"W Another application of Theorem 2.6.5is in proving thal a continuous function of a continuousfunction is continuozs.This is now stated as a theorem. 2.6.(, Theorem. If the function g is continuous at a and the function is continuous at / g@),then the compositefunction / . g is continuous at a. pROoF: Becauseg is continuous at a we have

gf'l : sk) LT

(10)

Now / is continuous at g(a); thus, we can apply Theorem 2.G.5 to the composite function f o g, thereby giving us lim (/. g) (r) : lim f (g(x))

:",r;*s(r)) : f (s(a)) (by Eq. (10)) : (f '8) (a) which proves that f . g is continuous at a. Theorem 2.6.6 enables us to determine the numbers for which a particular function is continuous. The following example illustrates this. EXAMPLE 3:

Given

h(x): \FT' determinethe valuesof x tor which h is continuous.

s o r T - r r r o NI:f . g ( x ) : 4 xz andf (x): \6, then h(x): (f " il(r). Because g is a polynomial function, g is continuous everywhere. Furthermore/ / is continuous at every positive number. Therefore, by Theorem 2.6.6,ft is continuous at every number x for which g(x) > 0, that is, when 4 - xz > 0. Hence, h is continuous at every number in the open interval (-2,2).

2.6 THEOREMSON CONTINUITY

1O7

In Sec.4.4 (Continuity on an interval) we extend the concept of continuity to include continuity at an endpoint of a closed interva.l. We define "right-hand continuity" and "left-hand continuity" and with these definitions we show that the function h of Example3 is continuous on the ciosed interval [-2, 2].

Exercises 2.6 1. Prove Theorem 2.6.1(ii). 2. Prove Theorem 2.6.1(iii). 3. Prove Theorem 2.6.1(iv). 4. Prove Theorem 2.6.2, showing step by steP which limit theorems are used. In Exercises 5 through 10, show the application of Theorem 2.6.5 to find the limit. 7. lim l{/3y + + U'-4

10. rim €=F J-3

Y

t

4

In Exercises11.through 26, determine all values of.x f.or which the given function is continuous. Indicate which theorems you apply. w3-1

1 1 ' f. ( x ) :

1 , 2f.( x ) : ( x - 5 ) ' ( r ' + 4 ) '

x 2 ( x+ 3 ) ' -J-2

x 2+ 2 x - 8 x3* 6* t 5x -'1,2

1 3 .g ( x ) : ; d

16. SQ\ : 7/P a 4

14.h(x): +x3-7x-6

1 5 . F (x ) :

1 7 .f ( x ) : f f -

ls. F(x): {L6 -E

t9.h(x\: lfi

. tT4 20.f(x): {ffi

2 1 ,f.( x ) : l r - 5 l

22.g(x\ : l9 - xzl

+ x-'\t2)s zs.f (x)- yz(:s-z

2a.G(x): ffi,-

15

It +4

1)"'

2 5 .g ( x ) : [ \ R \

26.F(x)- 1 -r + [r]l - [1 - r] Prove that if the function / is continuous at t, then lim f (t h) : f (t). prove that if f is continuous at a and g is discontinuous at a, then f + g is discontinuous at a. If(

fr'r:E'lf;:3

.

and g o (\ -r -) l: { ^

tt ifr
iff>0

prove that f and I ar€ both discontinuous at 0 but that the Product f ' I is continuous at 0' 30. Give an example of two functionE that are both discontinuous at a number a' but the sum of the two functions fu continuous at 4.

LIMITSANO CONTINUITY

108

31. Give an example to show that the product of tr,yo functions f and g may be continuous at a number c where / is continuous at 4 but I is discontinuous at 4. 32. If the function g is continuous at a number c and the function f is discontinuous at 4, is it possible for the quotient of the two functions, /9, to be continuous at 4? Prove )'our answer.

(Chapter2) ReaiewExercises In Exercises 1 through 4, evaluate the limit, and when applicable indicate the limit theoremsbeing used.

\/9-t-3

1. lim (3* - 4r * 5)

J.

t-2

rrlrr -

,-o

t-

n

---14

Jlltih

In Exercises5 through 8, draw a sketch of the graph and discuss the continuity of the function. x-3

5.

6' 8G):

7.

8.F(r):#

4-x

ln Exercises9 through 14, establish the limit by using Definition 2.1.1; that is, for any € > 0 find a E ) 0 such that l f ( x ) - L l < . w h e n e v e r 0< l x - a l < 6 . 9. limr(8 - 3x) :14

10. lim (2x, - x - 6) :9

11.lim l./T-3:7 t-4

3 12. , : _ _

n 2 v- t : -

1

15. Prove that lim {Fl:0

.^ ,.

t''

5*3r

I T 2 ** , : 2

1,4. lim 13: -$ I--2

by showing that for any € } 0 there exists a E > 0 such that \/F4

( e whenever

0 < t - 2 < 6. 1,6.If f (x) : ( lxl - x) lx, evaluate: (a) lim f (x); (b) lim f(x); (c) lim f (x). .T-0+ "u-0

In Exercises 17 through 24, evaluate the limit if it exists.

,+7t+ t j-

,

(nrm: Write --t

24. ri*[',4_-$4E .r-l+

X'

L

25. D r a w a s k e t c h o f t h e g r a p ho f f i t f ( x ) : [ 7 - x " n

a n d - 2 < r c < 2 . ( a ) D o e sl i m / ( x ) e x i s t ? ( b ) I s / c o n t i n u o u s a t 0 ?

REVIEWEXERCISES 109 26. Dlaw a sketchof the graph of g if g@) : (x- 1)[r]l,and 0 < x < 2. (a) Does

ItT

Sttl exist? (b) Is g continuous at L?

27. Give an example of a function for which lim l/(r) | exists but lim /(r) does not exist. r-0

J-0

28. Give an example of a function f that is discontinuous at I for whidr (a) lim l(r) exists but /(1) does not exiEq (b) /(1) exists but tim l(r) does not exist; (c) lin /(r) and l(1) both exist but are not equal. r-l

29. Letlbe the tunctiondefinedby f1 if r is an inteser [0 if r is not an integer (a) Draw a sketchof the graph of f. (b) For what valuesof c doestim f(r) extut?(c)At what numbersis f cutinuous? rNhy? 30. If the function g is continuous at a and f is continuous at g(a), is the composite function f 'g continuous at a? ( b ) S h o w t h a t t h e c o n v e r s e ot h f etheoremin thenlti/(r+h):|tT/(r-h). 31. ( a )P r o v e t h a t i f ItT/(r+h):f(r) (a)isnottruebygivinganexampleofafunctionforwhichtjryf(:+h):lim/(r-h)but{iq/(x+h)+f(x).

' : 32. If the domain of f is the set of all real nu.urbers,and I is continuousat 0, prove that if l(c * b) f(a) f(b\ for all a and b, then f is continuousat every numb€r. 33. If the domain of I is the set of all real numbers and / is continuous at 0, Prove that tl f(a + b) = l(a) + f(b) tor dl a and b, thin / is continuous at every number.

The derivative

3.1THETANGENT LINE 111 3.1 THE TANGENT LINE

Many of the important problems in calculus depend on the problem of finding the tangent line to a given curye at a specific point on the curue. If the curve is a circle, we know from plane geometry that the tangent line at a point P on the circle is defined as the line intersecting the circle at only one point P. This definition does not suffice for a curve in general. For example, in Fig. 3.1.1 the line which we wish to be the tangent line to the curve at point P intersectsthe curye at another point Q. In this section we arrive at a suitable definition of the tangent line to the graph of a function at a point on the graph. We proceed by considering how we should define the slope of the tangent line at a point, becauseif we know the slope of a line and a point on the line, the line is determined.

Q(xr, f(xr))

1)

left side of P. Let us denote the difference of the abscissasof Q and P by Ar so that At: xz- xr

P ( x 1 ,f @ 1 ) )

Ar may be either positive or negative. The slope of the secantline PQ then is given by f(xr) f(x') Ittpe:

LX

provided that line PQ is not vertical. Becaus€ 12 : Xt * At, we can write the above equation as T|flpe:

f(x'+Ar)-f(x') LX

Now, think of point P as being tixed, and move point Q along the curve toward P; that is, Q approaches P. This is equivalent to stating

Ar approaches zero the line PQ approaches the line through P which is p*uU"t to the y axis.In this case we would want the tangent line to the graph at P to be the line r : xr.The preceding discussion leads us to the following definition.

3.1.1 Definition

If the function

f is continuous

at )cu then the tangent line to the graph of

112

THE DERIVATIVE

/ at the point P(*r, /(r,)) is (i) the line through P having slope m(xr), given by

-"_f(xr*Ar) -f(*r) rrc*ixrl*fiat% ar*o

(1)

AX

if this limit exists; (ii) the line x : x, if ,t^f(xt+

L{) - f(xr):*oo

or-oo

If neither (i) nor (ii) olDefinr,r"^"r:.1 holds, then there is no tangent line to the graph of f at the point p(xr, (xr)). f EXAMPLE1: Find the slope of the tangent line to the curve a: x2 4x * 3 at the point (xr, yt).

s o l -u rro N , f(x) - x2- 4x* 3; therefore, f(xr) - xr2- 4xr* J, : (r, Ax) * Lx), 4(xr+ A r) * 3. From E q. (l ).w e have f(x r+ lim f(xr+Ax)-f(xr)

m(x):

Ar-O

:

A,x

-f Lx)z- 4(x, -t A,x)* 3f lxrz - 4x, * jl lim l(x, Ar

ar-0

: lim

Ax

At:-O

:

Iim Ar-0

2xrA,x*(A,x)z_-4Ax A^x

BecauseAx * 0, we can divide the numerator and the denominator bv Ar and obtain m(xr) : lim ( 2 x 1 - tA x - 4 )

m(x):2h-4

(2)

3.1 THE TANGENTLINE

113

T a b l e3 . 7 . 1

2 1 0 -1 3 4 5

-1 0-2 3-4 8-5 02 34 85

0

F i g u r e3 . 1 . 3

2: Find an equation of EXAMPLE the tangent line to the curve of Example L at the Point (4, 3).

solurroN:

Because the slope of the tangent line at any point (xt, Ut) is

given by m(xr):2h-

4

- 4:4' the slope of the tangent l i n e a t t h e p o i n t ( 4 , 3 ) L s m ( 4 ) : 2 ( 4 ) Therefore, an equation of the desired line, if we use the point-slope form, is y-3:4(x-4) 4x-y-L3-0 g.l.z Definition

nxevrpr-n 3: Find the normal line to 3 which y : {F theline 5x*3Y-

an equation of the curye is Parallel to 4:0.

The normal line to a curve at a given point is the line perpendicular to the tangent line at that Point. write its soLUrIoN: Let I be the given line. To find the slope of l, we is which form, equation in the slope-intercept

y--Zx++ -2, and the slope of the desired normal line Therefore, the slope of I is -2 becausethe two lines are parallel' is also To find the slope of the tangent line to the€iven curve at any point -3, and we have (xr, Ar),we apply plfinition 3.l.iwith f (x) \F

m(x):

lrP.

To evaluate this limit, we rationaltze the numerator.

m(xr): lr"T.

114

THE DERIVATIVE

Dividing

numerator and denominator by Ax (since A,x * 0), we obtain

m(x,):

[1

1_ 3

:--L

2Y x, - 3

Because the normal line at a point is perpendicular to the tangent line at that point, the product of their slopes is -1.. Hence, the slope of the normal line at (x1, U) is given by

-z\/xr - 3 As shown above, the slope of the desired line is -2. so we solve the equation -z\/xt-3--2 giving us xt: 4 Therefore, the desired line is the line through point (4, l) on the curve and has a slope of -2. using the point-slope form oi ur,'equation of a line, we obtain y-L--2@-a)

F i g u r e3 . 1 , 4

2x*y-9:0 to Fig. 3.L.4, which shows a sketch of the curve together with - _Refer the line /, the normal line pN at (4,1), and the tangent line pi at (4, r).

Exercises 3.1 In Exercises 1 through 8, find the slope of ,the-tangent line to the $aph at-the point (xr, y1). Make a table of values of r, y, and m at vadous Points on the graphl and includeln the taule all ioiits wheteih" grifl, ha, h-irorriJ turrg"r,t. o."* skekh of the graph. " " l'Y:9-* 2.y:x2-6x*9 3.y:7-5x-x2 4. y: lx2 5.y:x3-3x 6.y:x3-xcz-r*L0 7.y:4x3-13x2 I4x-3

8'Y:\/x+t

In Exercises 9 through 18, find an equation of the tangent line and an equation of the normal line to the given curve at the indicated point Draw a skekh of the curve together with the resulting tangent line and normal line. "

9.y:x.'-1x-5;(-2,7) 12.y :2x - x3;(-2, 4) 15. y :

6

|;

13.Y: \/q4i; R

(3, 2)

18.y:Vfi;

1 0 . y : x z * 2x * 1,;(1, 4)

e4,5)

15.y:-+; (4,-4) e3,2\

1 1 .y : t x 3 ;( 4 , 8 ) 14. y: \/+i=; (3, 3) 17.y : {i;

@,2)

MOTION .II5 VELOCIryIN BECTILINEAR 3.2 INSTANTANEOUS + 3 that is Parallelto the line 8r - y * 3: 0' that is parallel to the linezx+78y-9=o' 20. Find an equation of the nomral line to the curve y:t3-3r is PerPendicularto the line x+2y-77:0' equation of the tangent line to the curve y: \/b=-1that f|' Fird "r, -2) that is tantent to the curve y: r'- 7' 22. Find an equation of each line through the point (3, -6) that is tangent to the curve y: 3xu- 8' 23. Find an equation of each lin€ through th€ Point (2, 24. Prcve analytically that therc is no line through the Point (1, 2) that is talSent to the curve y: 4 x,. 19. Find an equation of the tantent line to the curve y:21

3.2 INSTANTANEOUS VELOCITY IN RECTILINEAR MOTION

Consider a particle moving along a straight line. Such a motion is called rectilinear motion. One diiection is chosen arbitrarily as positive, and the opposite direction is negative. For simplicity, w€ assll'me that the motion of the particle is along a horizontal line, with distance to the right as positirru u.d distance to the left as negative. Select some point on the line and denote it by the letter O. Let / be the function determining the directed distance of the particle from O at any particular time. To be more specific, let s feet be the directed distance of the particle from O at t r".o1d, of time. Then / is the function defined by the equation

(1)

s: f (t)

which gives the directed distance from the point O to the particle at a particular instant of time. Equation (1) is called an equationof motion of' the Particle. O ILLUSTRATION

1: LCt

s--P+2t-3 Then, when t:0, s: -3; therefore,the particle is 3 ft to the left of point O when f : 0. When t: I, s : 0; so the particle is at point O at l- sec. When t: 2, s : 5; so the particle is 5 ft to the right of point O at 2 sec. When t:3, s: !2; so the particleis 12 ft to the right of point O at 3 sec. Figure g.z.L illustrates the various positions of the particle for specific values of f.

the particle moves from the Between the time fot t: L and t-3, point where s : 0 to the point where s:12; thus, in the 2-sec interval ihe change in the directed distance from O is L2 ft. The average velocity of the pirti.t" is the ratio of the change in the directed distance from a

116

THEDERIVATIVE fixed point to the change in the time. So the number of feet per second in the average velocity of the particle from t: ! to f : 3 is la: 6. From f : 0 to f : 2, the change in the directed distance from O of the particle is 8 ft, and so the number of feet per second in the average velocity of the particle ' in this 2-sec interval is $: 4. . In Illustration 1, the average velocity of the particle is obviously not constanU and the average velocity supplies no specific information about the motion of the particle at any particular instant. For example, if a person is driving a car a distance of 70 miles in the same direction, and it takes him 2 hr, we say his average velocity in traveling that distance is 35 mi/hr. However, from this information we cannot determine the speedometer reading of the car at any particular time in the 2-hr penod. The speedometer reading at a specific time is referred to as the iistantaneous aelocity, The following discussion enables us to arrive at a definition of what is meant by "instantaneous velocity.,, I9t Eq. (1) define s (the number of feet in the directed distance of the particle from point O) as a function of f (the number of seconds in the time)' When t: tr, s: sr. The change in tire directed distance from O is (s - st) ft over the interv al of time (t - tr) sec, and the number of feet per second in the average velocity of the particle over this interval of time is given by s-sr t-t, or, because s:

f(t) - f(t,) t-

f (t) and st: f (t), the average verocity is found from

tt

Now the shorter the interval is from t, to t, the closer the average velocity will be to what we would intuitively think of as the instantaneous veloclty at tr. For example, if the speedometer reading of a car as it passes a point Pt is 40 mi/hr and if a point p is, for instance, 100 ft from pr, then the average velocity of the car as it travels this 100 ft will very likely be close to 40 mi/hr because the variation of the velocity of the car along this short stretch is probably slight. Now, if the distance irorn p, to p were shortened to 50 ft, the average velocity of the car in this interyal would be even closer to the speedometer reading of the car as it passes pr. We can continue this process, and the speedometer reading ui ê, ."n b" represented as the limit of the average velocity between P, ind pas p approaches pr. We have, then, the following definition. 3.2.7 Definition

rf f is a function given by the equation s: f(t)

(1)

and a particle is moving along a straight line so that s is the number of

MOTION 117 VELOCIWIN RECTILINEAR 3.2 INSTANTANEOUS units in the directed distance of the particle from a fixed point on the line at f units of time, then the instantaneousuelocity of the particle at f1 units of time is u(f1) units of velocity where

if this limit exists. Because t + tr, we can write (3)

t:h+At and conclude that

"t -+ tl'

is equivalent to

"Lt + 0"

(4)

From Eqs. (2) and (3) and statement (4), we obtain the following expression for a(tt): + Ll

+af)-

(f')

(5)

if this limit exists. Formula (5) can be substituted for formula (2) in the definition of instantaneous velocity. The instantaneous velocity may be either positive or negative, depending on whether the particle is moving along the line in the positive -or the negative direction. When the instantaneous velocity is zero, the particle is at rest. The speed of a particle at any time is defined as the absolute value of the instantaneous velocity. Hence, the speed is a nonnegative number. The terms "speed" and "instantaneous velOcity" are often cOnfused. It should be noted that the speed indicates only how fast the particle is moving, whereas the instantaneous velocity also tells us the direction of motion. rxavrrr.r L: A Particle is moving along a straight line according to the equation of motion s:2t3-4t2+2t-L Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine the instant when the particle reverses its direction.

s o l u r r o N f U ) : Z t B - 4 t 2+ 2 t - t ' Applying formula (2) in the definition of the instantaneousvelocity of the particle al tr, we have a(t):

lim f t-tt

@ -- f - ( t ) L

Ll

- (zttg- Attz+ 2h - 1) : Iim (2t3 4t2J 2t 1) t-tt

2(t3 - tf) - 4(t2 - ttz) + 2(t - t)

: lim t-tr

2 ( t- t ) l ( t z + t h + t f ) - 2 ( t+ t ) + L f

118

THEDERIVATIVE

Dividing the numerator and the denominator by t - t, (becauset + t), we obtain a(tr) : lim z(tz + ttt + tr, - 2t - 2tr + 't) I _tt

: 2 ( t r 2 * t r ,I t r ,- 2 t ,- 2 t 1 + l ) :2(3tt2-4h+1,) : 2 ( 3 t , - 1 )( f r - 1 )

I T a b l e3 . 2 . 1

tr- 1

f'<* L -l Ll-g

zr(fr) is negative, and the particle is moving to the left "l'

11t,

Table3.2.2

-1 0

+

1 2

-g -L2

1,6

-#o -1 310

0

-*

a(tr) is positive, and the particle is moving to the right u(tr) is zero, and the particle is changing direction from right to left

*
Conclusion

-10-5o510

Figure3.2.2

o(fr) is zero, and the particle is changing direction from left to right zr(fr) is positive, and the particle is moving to the right

IN RECTILINEAR MOTION 119 VELOCITY 3.2 INSTANTANEOUS In the above example, u(ft) can also be found by applying formula (5). (See Exercise 1.)

EXAMPLE2: A ball is thrown vertically upward from the ground with an initial velocitY of 64 ftlsec.If the positive direction of the distance from the starting point is up, the equation of motion is

solurroN: a(t) is the number of feet per second in the instantaneous velocity of the ball at f, sec. where f (t) : -16t2 + 64t s: f (t) Applying formula (2), we find that

a(tr\:tr-Tfry

s: -L 6t z + 64t If f is the number of seconds in the time which has elaPsed since the ball was thrown, and s is the number of feet in the distance of the ball from the starting Point at f sec, find: (a) the instantaneous velocity of the ball at the end of L sec; (b) the instantaneous velocitY of the ball at the end of 3 sec; (c) how many seconds it takes the ball to reach its highest point; (d) how high the ball will go; (e) the sPeed of the ball at the end of 1 sec and at the end of 3 sec; (f ) how many seconds it takes the ball to reach the ground; (g) the instantaneous velocity of the ball when it reaches the ground. At the end of L sec is the ball rising or falling? At the end of 3 sec is the ball rising or falling?

: lim

-16t2 + 54t- (-L6tf + 64t) t -

t-tt

: lim

tt

-1'6(t2- tf) + 64(t - t) t -

t-tt

tt

Dividing the numerator and the denominator by t h (becauset + t), we obtain o(t'): tt-T [-16(r + 11)+ 64]: -32h + 54 ( a ) o ( 1 ) : - 3 2 ( 1 ) + 5 4 : 3 2 ; s o a t t h e e n d o f L s e c t h e b a l li s r i s i n g with an instantaneousvelocity of 32 ft/sec. (b) u(3) : -g2(3) + 64: -32; so at the end of 3 secthe baII is falling with an instantaneousvelocity of -32 ftlsec. (c) The ball reaches its highest point when the direction of motion changes,that is, when o(tt): 0. Settinga(tt) - 0, we obtain -32' * 54:0 Thus, h:2 (d) when t:2, s: 64; therefore,the ball reachesa highest point of 64 ft above the starting Point. (e) lr(tr)lis the number of feet per secondin the speedof theball at f1 sec. l a ( 1 )l : 3 2

a n d l u ( 3 )l : 3 2

(f) The ball will reach the ground when s: 0. Setting 5: 0, we have -15t2 + 64t: 0, from which we obtain f : 0 and t: 4. Therefore,the ball will reach the ground in 4 sec. -64; when the ball reachesthe ground, its instantaneous G) u(4) : velocitv is -64 ftlsec.

1 20

THE DERIVATIVE

Table 3.2.3gives values of s and a for some specificvalues of f. The motion of the ball is indicated in Fig. 3.2.3.The motion is assumedto be in a straight vertical line, and the behavior of the motion is indicated to the left of the line. (r:S)

T a b l e3 . 2 . 3

0 ,I 1 2 3 , 4

G:zt

0 28 48 64 48 28 0

64 48 32 0 -32 -48 -64

(t:4)

Figure3.2.3

Exercises3.2

t

1. Apply formula (5) to find 2,(t1) for the rectilinear motion in Example 1. 2. Apply formula (5) to find r(tr) {or the rectilinear motion in Example 2. In Exercises 3-through 8, a particle is rnoving along a horizontal lirle according to the given equation of motion, rgher€ s ft b the directed distance of the pa*ide fron a pointb at I sec. the instantineo"" .iG,iiis-eclt find r, sec; and ttren find z(f,) for t}le pa icular value of t, given. "lfoatf 3 . s : 3 P+ l ; t r : 3 .11, o' s:

4 t ;h : ,

4. s:8-t2;

fr:5

5. s: GT;

,52

,J,':m)h:2

tr: j

8. s: Vl-+Z\:6

In Exerogel 9 thmugh 12, the motion of a particle is along a horizontal line according !o the given equation of motion, where s ft is the directed distance of.the particle from a point o at t sec. The positive direction is to the right. Determine the intervals of time when the Particle is moving to the ight and when it is moving to the left. Also deterErine when the Partide rcveres its direction. Show the behavior of the iotion by a figure similar to Fig. g,2.2, choosing values of, at random but induding the values of f when the particle reverses it; direction.

. . 9 .s : t 3 + 3 t 2 - g t + 4 "r I..'

L 0 .s : 2 t 3 - g t z - l z t + g

l+t

s: 1-,+t r

1)

c:

t-

L+t2 $. rf an obiect fa s ftom rest, its equation of motion i6 s : -161e, where f is the number of seconds in the tirne that has ' d"Pl"l since the.obiect left the starting point, s is the number of feet in tre digtance oith" oi;o rro gr" Ii I Point at , 8ec, and the Positive direction is upward. If a stone b dropped from a bui in! zleiirigh, ri"rd: (a)"r"*ir,g the instantaneou6 velocity of the stone I sec after it is dropped; (b) the initantaneous velociti of the stine 2 sec after it is

3.3 THE DERIVATIVE OF A FUNCTION 121 dropped; (c) how long it takes the stone to reach the ground; (d) the in8tantaneous velocity of the stone when it reaches the ground. is thrown vedically upward from the ground with an initial velocity of 32 ft/sec, the equation of motion i8 hi. It u ',., "t*" -l32t, where t sec is the time that has elapsed since the Etonewas thrown, s ft is the distance of the stone s: -l6P from the starting point at t sec, and the positive direction is upward. Find: (a) the average velocity of the 6tone during thetimeinterval:*-t-*(b)theinstantareousvelocityofthestoneat*secandat*sec;(c)thesPeedofthestone at I sec and at I sec; (d) the average velocity of the stone during the time interval: t = t = *, (e) how many seconds it rvill take the stone to reach the highest point; (f ) how high the stone will go; (g) how many seconds it will take the stone to reach the gFound; (h) the instantaneous velocity of the stone when it readles the ground. Show the behavior of the motion by a figur€ similar to Fig. 3.2.3. 15. If a ball is given a push so that it has an initial velocity ot 24 ftlsec down a certain inclined plane, then s : 241* 1012, where s ft is the distance of the ball from the starting point at t Bec and the Positive direction i3 down the inclined plane. (a) What is the instantaneous velocity of the batl at I, sec? (b) How long does it take for the velocity to increase to tl8 ft/sec? : 16, A rocket i6 fired vertically upward, and it is s ft above the ground t sec after being fired, where s 560t 16tsand the long it takes for positive dircction is upward. Find (a) the velocity of the rocket 2 sec after being fired, and (b) how the rocket to reach its maximum height.

3.3 THE DERIVATIVE OF A FUNCTION

In Sec. 3.1 the slope of the tangent line to the graph of y: point (xr, f (x')) is defined by 1 \ ftt\xrl :

,.

lrm ar-o

fkt+Ar)-f(xt)

f (x) at the (1)

Lx

if this limit exists. In Sec. 3.2 we learned that if a particle is moving along a straight line so that its directed distance s units from a fixed Point at f units of time is given by r : /(f), then if r:(tr) units of velocity is the instantaneous velocity of the particle at f1 units of time,

a(t):lli'ry

(2)

The limits in formulas (1) and (2) are of the same form. This type of limit occurs in other problems too, and it is given a sPecific name. 3.3.1 Definition

',such that The deriuatiae of the function / is that function, denoted by f its value at any number x in the domain of / is given by

yi+5xYtgry,

(3)

'(x) is read "f prime of x.") if this limit exists. (f is read "f prime," and f '(x) is D"f(x), which is read Another symbol that is used instead of f "the derivative of f of x with resPectto x." If y - f (x), then f '(x) is the derivative of y with resPectto x, and we use the notation D*A. The notaliony'is also used for the derivative of y

122

THE DERIVATIVE

with respect to an independent variable, if the independent variable is understood. If r, is a particular number in the domain of f, then ( 4)

if this limit exists. Comparing formulas (1) and (4), note that the slope of the tangent line to the graph of y: f (x) at the point (xr, (xr)) is precisely f the derivative of / evaluated at xr. If a particle is moving along a straight line according to the equation of motion s_ /(/), then upon comparing formulas (2) utra 1+;, it is seen that a(tt) in the definition of instantaneous velocity of the particle at f, is the derivative of f evaluated at t, or, equivalently, the deiivative of s with respect to f evaluated at fr. EXAMPLE1: Given f (x): hcz+ L2, find the derivative of. f.

SOLUTION:

If x is any number in the domain of f , ftom Eq. (3) we obtain

,,* f(x+Ar)-f(x1 f ' ( x ) - i}l [3(r + Lx)' + 121- Gr, + t2) ,:_ IIITI AX

: lim 3x2l6x A,x* 3(Ar)' + 12- Jxz L2 A,x

Ar_o

:

lim

6x A'x* 3(A'x)2 :

Lx

Ar-o

111

(6xi 3 Ar)

:5X '(x): Therefore, the derivative of f is the function 6r. The /' defined by f domain of f is the set of all real numbers, which is the same as the domain of f. In Sec. 3.2 we showed that the limit

t:,nfW is equivalent to

,, f(t) - f(t')

rrr-n--t-tr

t -

tt

Therefore, an alternative formula to (4) for f,(rr) is given by

(r) - f (x,)

3.3 THE DERIVATIVE OF A FUNCTION

EXAMPLE 2: For the function / of Example 1, find the derivative of f at 2 in three ways: (a) apply formula ( ); (b) apply formula (5); (c) substitute 2 for r in the expressionfor f '(x) in Example1.

(a) Applying

SOLUTION:

f

' ( 2 ):

129

formula (4), we have

lim Ar-0

lim A.z-0

lim Ar-O

1 !.__ :um - - 12 A ,x+ 3(A x)' z

^fi : (n+'^', :LZ

(b) From formula (5) we obtain

f,(2):LTry :

(3x' + tZ) - 24 ,,,llr-I1 ,'...'

:llmx-2

,*:,;, x-

/.

(x-2)(x+2) -Jlllrl-

:r;;

(.+2;

:12 ' (c) Because,from ExampleL, f ' (x) : 6x, we obtain f (2) : 12.

If the function / is given by the equation y:

f(x), we can let

Ly:f(*+Lx)-f(x) '(x), so that from formula (3) we have and writs DrA in place of f

(6) A derivative is sometimes indicated by the notation dyldx. But we avoid this symbolism until we have defined what is meant by dY and dx.

124

THE DERIVATIVE

EXAMPLE

3:

Given

2* x Y: g= find D;y.

SOLUTION:

DrA: li- &

or-u Ar

:

lim An-0

:

lim A.r-0

Ar-0

5 (3-x-Ar)(3-r)

lim Ar-0

5

:

rxavrpln 4: find /'(r).

Given /(x)

( 2 + x + A x ) l G - x - A x )- ( 2 + x ) l G - x ) A,x

5Ax Ar(3-x-Ar)(3-r)

: lim :

f(x+Lx)-f(x) Ax

- r;' 13,

SOLUTION:

f@ + a'x) f(x), f' (x): lim

.

( x * A r ) z t -s x 2 t 3

o"-oT:it"Tof

We rationalize the numerator in order to obtain a common factor of Ar in the numerator and the denominator; this yields

f'(x): ltl :

lim Ar-0

(x*Lx)z-f Ar[ (r + Ax) nt7+ (x + Ar) zts*zre -; *er1

_ r:* :ili :

lim As-o

:

lim Ar-0

x2 * Zx(A,x) * (Lx)z - x2

2x(A,x)* (Ar),

Axl(x + Ar) nt7+ (x * Ar) zt%czts + x4t3l

2x* Ax (x * Ar) 4t3*

(x + Ar) 2tlx2t3+ x4t3

2x X4l3+)C2l3X2l3+X413

3.3 THE DERIVATIVEOF A FUNCTION

125

_2x

w

_2 Sxus

-B -6 - 4-zO

Note that /'(0) does not exist even though f is continuous at 0. A sketch of the graph of / is shown in Fig. 3.3.1.

F i g u r e3 , 3 . 1

If for the function of Example 4 we evaluate ,l i. m ^ -ft(' x r t L x ) f ( x r ) Ia t r l : O Lx

Ar-o+

we have

..

ufii

;;:i

f(0 + Ar) - f(0)

,.

(Ar;"' - O

A'x

;;:;.

Ax

-:lilll

If we evaluate

,,_ -f(x, + Ax) f(r')

lilfl

Aa-o-

Ax

at .r1:0

we obtain

+ Ar) - f(0) _ _@ ,._ r l m - -1(0 *" .#lude that the tangent line to the graph of f attheorigin ,n"*ir", is the y axrs. Example4 shows thatf ' (x) can exist for somevalues of r in the domain of f but fail to exist for other values of r in the domain of f .W" have the following definition. 3.3.2 Definition

The function / is said to be differentiable at xt if.f '(r1) exists. From Definition 3.3.2it follows that the function of Example4 is differentiable at every number except 0.

3.3.3 Definition

A function is said to be differentiableif it is differentiable at every number in its domain.

Exercises 3.3 In ExercisesL through L0, tind f '(r) for the given function by applying formula (3) of this section. l.f@):4**5r*3

2.f(x):vz

3.f(x):{i

126

THEDERIVATIVE

a.f@): v3;E 1

7.f(x\:ax'

5 .f ( , ) : ; -

,

8./(r):#

x

r0.f(x):ffii In Exercises L1through16,find f '(a) for the given valueoI a by applyingformula(a) of this section.

i!. ltrl -1- x2;a:t

1 , 2f (. x \ : f i , a : 2

u.f(x):+-L;a:4

15.f (x) : 1/fi;

)

r J .f ( x \ : f i ; a : 6 n: s

1 . 6f.( x ) : ] *

x*x2;a:-j

In Exercises 17 through 22, find f '(a) for the given value of a by applying formula (5) of this section.

1 7 .f ( x ) : 3 r * 2 ; a : - 3

L 8 .f ( x ) - x 2 - x * 4 ; a : 4

1 9 .f ( x ) - 2 - - x 3 ; a : - 2

20. f (x) : \/TT ei; a:7

2r.f (x): !+TE, a: l

2 2 .'fY ( xx) : + - x ; a - - 8

'l

zs. Givenl(r):gi=T, findl'(r). Is / di{ferentiableat 1?Draw a sketchof the graphof I : +/(E=T),, fil:.dl'(r). Is I differenriableat f ? Dnw a sketchot the gaph of u. ctven l(x) l. 25. If g is continuousat a andl(r) : (r - a)g(r), find l,(c). (nrr.rr:UseEq. (5).) . Let f be a function whose domain is the set of all real numbers andf(c * b) : f(s) . f(b) for all a and D.Furthermore, supposethat/(0) : 1 andl'(0) odsts.Provethat/, (r) existsfor all r and that l,(r) : l'(O) . /(r). 27. If I is differentiable at a, prove that ,. f(a + Ax)- f(a - Ax) / {4) : rlm --2l;(rruvr:l(c * Ax) - f (a - Lr) : f (a + Ax)- f (a) + f (a) - f (d - Ax).)

3.4 DIFFERENTIABITITY The function of Example 4 of Sec. 3.3 is continuous at the number zero AND CONTINUITY but is not differentiable there. Hence, it may be concluded that continuity of a function at a number does not imply differentiability of the function at that number. However, differentiability does imply continuiry, which is given by the next theorem.

3.4.'1,Theorem If a function f is differentiable dt xy then / is continuous at xr PRooF: To prove that f is continuous aI xr, we must show that the three conditions of Definition 2.5.1, hold there. That is, w€ must show that (i) /(r') exists; (ii) lim /(r) exists; and (iii) lim f (x): f (xr'1.

By hypothesis,/ is differentiableat )+ Therefore,f ' (xr)exists.Because

by formula (5) of Sec. 3.3

f

'(*r): lim f @;::f;!

(1)

ANDCONTINUITY 127 3.4 DIFFERENTIABILIW we conclude fhat f(xr) must exist; otherwise, the above limit has no meaning. Therefore, condition (i) holds at xt. Now, let us consider

t.ty,lf(x) - f(x')l We can write

r

f'r)-f(r')l

''fr] t/(r)-f(x,)l:IT, L('-x') ]1m

(2)

Because

(r- rr): o and,,+,fff: l1T,

f'(xr)

we apply the theorem on the limit of a product (Theorem 2.2.5\ to the right side of Eq. (2), and we have r(r) - f (xr) - n

- f(x,)l: lim (x- x,) ]lt'H

!\,lf(*)

0' f' (x')

so that

l.r\,Ufr)-f(x')l:o Then we have

t"\'f(*) : ;*' r',' lli'r,,,, o',,'r'-i r* r*o'u',' :0*f(xr)

which givesus lim/(r):f(x')

(3)

&-Ir

From Eq. (3) we may conclude that conditions (ii) and (iii) for continuity t of f at 11 hold. Therefore, the theorem is proved. As previously stated, Example 4 of Sec. 3.3 shows that the converse of the above theorem is not true. Before giving an additional example of a function which is continuous at a number but which is not differentiable there, the concept of a one-sidedderiaatiae is introduced.

g.4.2 Defirrition

If the function / is defined at rr, then the deriaatiae from the right of f at 11, denoted by f '*(xr), is defined by (4)

128

THE DERIVATIVE or, equivalently,

., fk) -f(xr) ' - , ' ' , . : tim fitrr} ,', rt=.rr+ X-Xt

(5)

if the limit exists. 3.4.3 Definition

If the function / is defined at rr, then the deriaatiuefrom the left of f at xr, denoted by f -(xr), is defined by

lim f k' f -(x,): ai,-o_

+ L9-- f (x) t)r

(6)

or, equivalently,

f

'-(x,): lim f k), f=!x) -, X

7r-nr-

Xt

(7)

if the limit exists.

EXAMTLE 1: Let f be the function defined by (,t*-1 ifx<3

f(x): t-i _,

if3< r

(a) Draw a sketch of the graph of f . (b) Prove that / is continuous at 3. (c) Find f '-(3) and f 'ap). (d) Is / differentiable at 3? v

solurroN: (a) A sketchof the graph is shown in Fig. 9.4.1,. (b) To Prove that f is continuous at 3, we verify the three conditions for continuity at a number: (i) /(3) :5 (ii) Iim f(x) : lim ( 2 x - 1 ) : 5 &-3-

!E-3-

lim /(r) : lim ( 8 - r ) : 5

t-3+

t-3+

Therefore, lim /(x) : 5 fi-3 (iii) lim f(x) : f(3) E-3

Becauseconditions (i), (ii), and (iii) all hold at 3, f is continuous at 3.

(c)/t(3):^"1 # ,.

urrr-

lzQ+Ar)-1.l -5 AX

A..r-o-

:

lim Ax-o-

Figure3.4.1

6t2-A'x-6 LX

2 A,x -;llm d,r-o- AX 1.

:lim2 Ae:-O-

-2

AND CONTINUITY 129 3.4 DIFFERENTIABILIW

f

f(3+Lx)-f(3)

'+(3)= lim

L,x

Ar-0+

:

[8 - (3 + Ax)] - s Ar

lim Ar-0+

: lim 8 - 3 - A r - 5 Ar-.0+

:lim+

-Av

Ar-0,* L)C

:

lim (-1) AJr ' 0+

--l (d) Because

,.

utl-

Af -0-

f ( 3 +a r ) - / ( 3 ) Lx

we conclude that

-/(3) ,tlm . -f G + A r )

A?..n

LX

does not exist. Hence, / is not differentiable at 3. However, the ilerivative from the left and the derivative from the right both exist at 3.

The function in Example 1 giveb us another illustration of a function which is continuous at a number but not differentiable there.

3.4 Exercises conIn Exercises1 through 14, do each of the following: (a) Draw a sketchof the graph of the function; (b) determine if f is tinuous at :r; (c) find l1(r,) and l1(rr) if they exi6u (d) determine if I is differcntiable at Ir. ( 3-2x if x<2 ( n r *' 1 if.x<-4 7 L . tf\k*\' : 4

l_x_6 xt: -4

:

ifx>_4

2.f(x):ls*-i'

irx>2

xt:2

3 . f ( x ) : l r - 3 l ;x , : 3

a.fk):L*lx+2lr;lh:-2

?t s.f(x):trl1

6.f(x):lbll; == 3

ifro

rr:0

xr:0 (r2

i.f

<

7 . f ( x ) : F r , i rxr r o0 rr:0

^ ., \, tf-+ 6. l\x): ltffi xt:2

Lf.x<2 ifx>2

130

THE DERIVATIVE

e. f (x):

{tr-a

.

ifx
1,1'. f (x) : Y-x+'/-.;xt: -7 ( ^ - 6 x' l i l-4-x, xt:3

ifx<-l

ifx>-1,

xl ,: -l

xt: I

1 3 .rf \( .x ) : l

l'2

1 0 ./f \( x ): { ^ , (-7-2x

.t {1i'_ ,1, ir x >

12.f(x) : (x - 2)-2; xr:2

ifr<3 if r>3

M . f (/ x\ ) : {

(-*2t3 ifx<0 -: ifr>0 lxr," Ir:0

15. Given l(r) : lrl, draw a sketch of the graph of I Prove that is continuous at 0. Prove that is not differentiabte / at O, f but that f'(r) : lrllx lor all x + 0. (HrNr: Let lrl : r/r1) t0 Given l(r) : \g' Prove that neither the derivative from the right at -3 nor the derivative from the left at 3 exist. Draw a sketch of the graph. 17. Given l(r) : d/r, ptove that fi(o) exist8 and find its value. Draw a sketch of the graph. (l- i')3t2, Prove that /'(r) exists for all values of r in the open interval (-1, 18 Givm/(r): 1), and that both /-i(-1) and fl(1) exist. Draw a sketdr of the graph ofl on [-1, l]. ^..a f 19-rFind the values of a and b so thatl'(1) exists if

'

itx 7 z0' Given l(r) : [tn, find /'(xr) if xr is not an integer. Prove by applying Theorem 3.4.1 that l,(r,) does not exist if ,1 is an inteter. lf x, is arl integer, what can you say about f1(x,) li(r,)Z 21. G i v e nf ( x ) : ( r - 1 ) [ x ] . D r a w a s k e t c h o ft h e g r a p ho f f o r x i""a n [ 0 , 2 ] . F i n d :( a ) f_ ( r ) ; ( b )f l ( l ) ; ( c ) f , ( r ) . f 22. Given

f@):{0., lf;;3 where n ]s a Positive integer. (a) For what values of zr is differentiable for all values of x? (b) For what values of n is / /' continuous for all values of x?

2 3. G iv enf ( x ) :

s g n r. (a ) P ro v eth a t/i (0 ):+ o o

and,f' -(0):+ * . (b) provethat l i m /' (r) : 0 and l i m f ' (r) : 0. ' 'O+ "r

24. Let the function / be defined by ( g ( x )- g ( a ) . . tr x f a f (x) : l=--T:;if x: a Lg'(a)

J.-O-

Prove that if g'@) exists,/ is continuous at a.

3'5 SOME THEOREMS ON DIFFERENTIATION oF ALGEBRAIC FUNCTIONS

The operation of finding the derivative of a function is called differentiation, which can be peiformed by applying Definition 3.3.1. However, because this proc"r, i, usually rather lengthy, we state and prove some theorems which enable us to find the derivative of certain functions more easily. These theorems are proved by applying Definition 3.3.1. Following the proof of each theorem, we state the corresponding formula of differentiation.

OF ALGEBRAICFUNCTIONS 131 3.5 SOME THEOREMSON DIFFERENTIATION

3.5.1 Theorem

If c is a constant and if f (x): c for all x, then

f'(r) - o PROOF:

f'(x):ltg'ry :lt'T,T limO

, I

:Q

ffiffiffi

(1)

The deriaatiaeof a constantis zero. . rLLUsrRArIoN1: If f(x):5,

then

f(x):o g.5.2 Theorem

lt n is a positive integer and if f (x): ro, then f'(x) : nYn-r PROOF:

(tv+Lx)-f(x)

f'(x):llq'tr

:lt"T,ry Applying the binomial theorem to (r +Ar)n, wQ have

xn + nxn-tO, *W

,n-z([y)2 *

* nx(Lx)"-l

* (Ar)n

lim

f'(r):

'

Ar-0

:

lim

nxn-r Lx*n(n;l) "4 2l

,o-z1yx\z+ . . . *nx(Ax)"-t+

(Ar)'

Aa-0

Dividing the numerator and the denominator by Ax, we have

f,(x):l,*|n**_,*w*n-z1Lx)+...*nx(Lx)n+(Ar),-']

132

THE DERIVATIVE

Every term, except the first, has a factor of A,x;therefore, every term, except the first, approacheszero as Ar approacheszero. So we obtain

I

f

'(x) nx6n-r

(2) o ILLUSTRATToN2: (a) lf f (x) :

x8, then

f'(x) : 8x7 (b) If f(x): x, then ' f'(x): L ro :L'1. -1 3'5'3 Theorem If f is a function, c is a constant, and g is the function defined by 8(r)-c.f(x) then if f '(x) exists, g'(r)-c'f'(*) PROOF:

8'k):lt*ry

:l:Try : lin

lf@+ Ar) - f(x)1

o"lorL:j

:c.t.-/(r+AI)-/(r) Arr-.o

A)C

- cf'(x)

i

ffi##iF;rn**l

(3)

The deriaatiaeof a constanttimes a function is the constanttimesthe deriaatiaeof the function if this deriaatiae exists. By combining Theorems 3.5.2 and 3.5.3 we obtain the following result: If f (x) : ocn, where n is a positive integer and c is a constant, ' f (x): sn26n-r (4)

FUNCTIONS 133 OF ALGEBRAIC ON DIFFERENTIATION 3.5 SOMETHEOREMS o rLLUSrRArroN3: If f(x) :5x7, then :5 ' 7xG f'(x) : 3516

3.5.4 Theorem If / and g are functions and if h is the function defined by

h(x):f(x)+s@) theniI f ' (x) andg'(r) exist, h'(x):1'(x)*g'(r) PROOF:

h'(x): :

:

h ( x* L x )- h ( x ) ll* t/(r + ar) + s(r +Ar)l [/(r) + s(r)] lr,T.

Ax)- s(r)l

l'"T.

:1'"T'ry*hlry - 7 '@ )* 8 ' ( r )

I

(s) The deriuative of the sum of two functions is the sum of their deriratiaes if these deriaatiaesexist. The result of the preceding theorem can be extended to any finite number of functions, by mathematical induction, and this is stated as another theorem. 3.5.5 Theorem

The derivative of the sum of a finite number of functions is equal to the sum of their derivatives if these derivatives exist. From the preceding theorems the derivative of any polynomial func' tion can be found easilY.

Exevrpr,r1: Given f(x):7xa-2x3*8x*5 hnd f ' (x).

SOLUTION:

f' (x) : D*(7xa 2x3* 8x + 5) : D*(7xa)+ D*(-Zx") * D*(8r) + D'(5) :28xs-6x2+8

DERIVATIVE

3.5.5 Theorem If f andg are functions and if ft is the function defined by

h ( x ): f ( x ) s ! ) then if.f '(x) and g'(x) exist, h ' ( x ) : f ( x ) g '( r ) + g ( x ) f ' ( x ) PROOF:

h'(x):l'nW . _- lim fG + A'x)' S@+ A,x) f(x) g(x) A,x We subtract and add f (x * Ar) . S(r) in the numerator, therebv

grvlng us

h'(x\ :

Ln . g(r+Ar)-g(x)+g(r).f(x+ t4- r
: lim f (x+ Ar). lim A.r- 0

Ar- 0

Ax .trrn /(x + AI)-/(x) * lim s@) " A,r .0

Ar. .0

ax

Because / is differentiable at x, by Theorem 3.4.1, i" continuous at r; f therefore, lim f(x+ Ar) : f(*). Also,

+ AI).- g(r) _ 6R,(x) ,o, 8(r \ Ar Ar-o

HW:f'(x) and

: llq st'l 8(r) thus giving us h ' ( x ) : f ( x ) g '( x ) + g ( x ) f , ( x )

r

OF ALGEBRAICFUNCTIONS 135 3.5 SOME THEOREMSON DIFFERENTIATION

the deriaatiaeof the secondfunctionplus the secondfunction timesthe deriaatiae of tlte first functionif thesederiaatiaesexist. EXAMPLE2: h(x) :

Given

SOLUTION:

(2*t - 4x2)(3x5 * x')

find h' (r).

h' (x) : (2x3- 4x',)('l.Sxa+ 2x) * (3r5 t x',)(6x2 8x) - (30r?- 50rG* 4xa- 8rt) * (18r?-24xG * 5xa- 8rt) :48x7-84x6*1014-L6xB In Example 2, note that if we multiply first and then perform the differentiation, the same result is obtained. Doing this, we have h (x ) :6 x B - L2x7* Zxs- 4xa "Thus, h' (x) :

g.5.7 Theorem

'l'5x3 48x7- 84x6* L0ra -

If f andg are functions and if h is the function defined by f( v\

h(x\ :

wherexU) * o 8\x) thenrf f ' (x) andg'(r) exist, o ( * ) f , ( * )- f @ ) SG ) h,(x) : { [s (r) ], PROOF:

h'(x):lt1W f (x + A,x)_ I@-

:l'"T'ry :l'n

Subtracting and adding f (x) 'g(r)

h,(x):Inf

in the numerator, we obtain

136

THEDERIVATIVE

e+4i3. : l:l'r'r llt

lli rr'r llt e.r*Fs!{

]t"qst'l H s(r * Ar)

_g(x)' f'(x)- f(x)- g'@) s ( r )' s ( r ) _s@)f'(x)-f(x)s'(x) [s(x)l

(7)

The deriaatiaeof the quotientof two functionsis the fraction haaingas its denominatorthe squareof the original denominator,and as its numerator the denominatortimes the deriuatiaeof the numeratorminus the numerator timesthe deriaatiaeof the denominatorif thesederiaatiaesexist. nxaupln 3: Given

SOLUTION:

h,(x) - k2 ax+ r)-(6!') (2r1! D Qx a)

2x3+4

h(x) :

(x'-4x*1,)z

xz- 4x *'1,

find h'(x).

3.5.8 Theorem

rf f (x) - x-n, where -n is a negative integer and.x * 0, then ' (x) - -nx-n-r f PRooF: If -n is a negative integer, then n is apositive integer. We write

f(x) -1

1

Applying Theorem 3.5.7,we have -n.0-1.nxn-r _nxn-r '

f (x) :

nxauprn 4: Given f(x\ :4 hnd f ' (x).

ffi

- -nx'|i-r-2it

: -T

solurroN, f (x) :3x-5. Hence,

?

f

'(x):

3(-5r-u): -15r-6: - g

)c6

OF ALGEBRAICFUNCTIONS 137 3.5 SOME THEOREMSON DIFFERENTIATION If r is any

positive

or negative

integer

it follows

from

Theorems

3.5.2 and 3.5.8 that D*(xn) : rxr-r and from Theorems 3.5.2,3.5.3,and 3.5.8 we obtain D*(cxr') + s77;r-r

Exercises3.5 In ExercisesL throu gh 26, differentiate the given function by applying the theorems of this section. 3. f(x) - $74- xa 2. f(x) :3xa - 5x2* 1 l.f(x):x3-3x2,I5x-2 a. g(x) : )c7- 2x5+Sxs - 7x

5. F(f) : t.to- it'

6. H(x):*r3-x*2

7. a(r) : t.nr'

8.C(y):y'ot7y"-y"+l

9. F(x):x2*Zr+

10. f(x) :

- 5 * x-2 I 4x-a .x4

13.f(s) : t6(s'- s') L6.g(x) : (4xz+ 3)'

r s f. ( x ) : h 2 2 .g ( x ) :

xa*2x2*5r*L

zs.f (x):'#

(3r- 1)

1 1 .g ( r ) :

35

12.H(x) -

,r+ j

)

*

1,a.S@) : (2x2+ 5) (ar - 1) #+).x+l

15.f(x) : (2xa- 1) (5x3* 6x)

17.H\x):'ffi

18.F(y)-'#

20.f (x) : (x' - 3x * 2) (2x3+ 1')

2r.h(x):{*

r*,

2a-fG):m

-3-R.

23.f(x):;"fr v3+1 * r) 26.g(x): ?fr (x'- 2x-1

\,22. V f, g, andh are functionsand if O(x):f(x.)'8e) 'ft(r), prove that it f'(x)' 8'(r)' and l'(r) exist,d'(r): l(i):S@).h'(x)+f(x).g'(x).h(r)+f'(x).gG).h(r).(rru.rr:ApplyTheor€m3.5.6twice.) Use the result of Exercide27 to differentiate the functions in Exercises28 through 31' 2 8 .f ( x ) : ( * + s ) ( 2 x - s ) ( s x+ 2 ) 3 0 .g ( r ) : ( 3 r s+ r - 3 )( r + 3 ) ( f - s )

2 9 .h ( x ) : ( 3 x + 2 ) ' ( f - 1 ' ) 3 1 .d ( x ) : ( 2 f + x + r ) z

32. Find arl equation of the tangentline to the curve y : 8/ (f + 4) at the Point (2, 1)' 33. Find an equationof eachof the lines throughthe point (-1,2) which is a tangentline to the curvey: (r-1)/(r*3). +6r+4, which is parallelto the line 34. Find an equation of each of the tangmt lines to the curve 3y:x"-3f 2x-y+3:0. - : 35. Find an equation of eachof the normal lines to the cur-sey: it 4x which is Parallelto the line x + 8y 8 0. 36. An obiectis moving alonga straightline accordingto the equationof motion s:3r/(f + 9), with t > 0'where6 ft i8 the directed distan& of the obje* from the starting point atJ 8€c.(a) What is the instantaneousvelocity of the object at ,1 sec?(b) What is the instantaneousvelocity at 1 iec? (c) At what time i5 the instantaneousveloclty zeto?

138

THEDERIVATIVE 3.6 THE DERMTIVE OF A COMPOSITE FUNCTION

Suppose that y is a function of u and u, it'r.turn, is a function of t. For example,let

Y : f ( u ): u 5

(1)

u:8@):2x3-5x2+4

e)

and

Equations (1) and (2) together define y as a function of x because if we replace u in (1) by the right side of (2) we have

A : h ( x ) : f \ e ) 1 : ( 2 x B _ 5 x+24 ) ' where h ts a composite function which was defined previously (refer to Definition 1.8.2). We now state and prove a theorem for finding the derivative of a composite function. This theorem is known as the chain rule. 3.6.1 Theorem If y is a function of u, defined by y: f (u), and Duy exists, and if u is a ChainRute function of x, defined by r: g@i, ind'Dru exists, tiren y is afunction of r and Dry exists and is given by D rA:

D uy ' D ru

pRooF: From Eq. (6) of Sec. 3.3, we have

Duu:Iin#, Hence, when l\"l difference

is small (close to zero but not equal to zero) the

Au fi'- o"v is numerically small. Denoting the difference by n, we have

A,u n:ft,-DuA

ifLu#0

The above equation defines T as a function of Lu, provided that Lu * 0. Letting 11: F(Lu), we obtain Att

F(Az):ffi-

Dua

if Lu * o

(3)

Equation (3) defines F(Aa) provided that Az * 0. In a later part of this proof , we want the function F to be continuous at 0. Thus, *" d"fi.re F(0) to be lim F(Ar.r).From Eq. (3) we have

: H-'î\n,, Hllnro') llT

OF A COMPOSITEFUNCTION 3.6 THE DERIVATIVE

139

Duu- D"a : Hence, we define F(0) : 0, and we have

, lN-D",u F ( A u )- ' l A u " u r L0

irLu*o if\u-0

Solving Eq. (3) for LY, we obtain Ly: DuU' Lu * Lu ' F(A'u)

(4 )

1f Lu * 0

Note that Eq. (4) also holds rf Lu:0 becausewe have Ly:0 (remember t h a tL y : f ( u + A u ) - f ( u ) ) . Dividing on both sides of Eq. (4) by Ar, where Ax * 0, we obtain La = *:Dua'

Att

htt

6+Ô'r(Au)

Taking the limit on both sides of the above equation as Ar approaches zero and applying limit theorems, we have

F(Au) Hence, DrA:

DuA ' Dru * Dru ' Lim F(Lu)

(5)

Ar-0

We now show that lim F(Az):0 by making use of Theorem 2.6-5.We Atr-O

first expressAtt as a function of Ax. Because

rim *

ar-o AI

: D,u

it follows that when lAxl is small and Ax * 0, Lul Lr differs from D "u by a small number which dePends on Ar, which we call d(Ar). Hence, we write Lu

ffi-- D,u* 0(Ar)

if a'x* o

and multiplying by Lx we obtain Lu:D*il' Lx* d(Ar) 'Ar

if Lx # 0

The above equation expressesAz as a function of Ar. Calling this function G, we have G(Ar) : D,il' Ar * d(Ar) ' Ar

ifA,x*0

THEDERIVATIVE

140

H e n c e , F(A z) : F(G(A 1) ), and

lim F(Az) :

l''1

F(G(Ax))

:0 and F is continuous at 0 (rememberthat we made Because ln "(Ax) it so), we can apply Theorem2.6.5,to the right side of the aboveequation and we have

lim F(Au): F(l',Tc(Ax)) : F(0) -0 So in Eq. (5) if we replace F(Az) by 0, we obtain olim D,A:DuA.Dru*Dru.0 Thus, D*A: Duy . Dru

I

In the following example, we apply the chain rule to the function given at the beginning of this section. EXAMPLEL: ,:

Given

( 2x s- Sx z+ 4 )t

soLUTroN: Considering y as a function of z, where u is a function of x, we have U : tts

find D,y.

w here u:2xr

- S xzI 4

Therefore, from the chain rule, DrA : DuA . Dru:

5ua(6x2- 10x)

: 5(2x3 - 5x2* 4)a(6x2- 10r) rxavrprn 2:

f(x) :

Given

4x3*5x2-7xt8

solurroN: to obtain

write f (x) :

(4x3 + 5x2- zx * 8)-t, and apply the chain

- -1,(4xB * 5x2- Zx * 8)-2(1,2x2* llx - T\ f' (x)

-1.2x2-1,0x*7 (4x"*5x2-7x+g)2 nx,q.upr.n 3:

Given

f(x):e#)^ find f '(x).

solurroN:

Applying the chain rule, we have

(3r - 1)(2J- (2{ + 1)(3) ,(x\ /f \ / : 4- PJI l)3 (3r-1), \3r-1l

OF A COMPOSITEFUNCTION 3.6 THE DERIVATIVE

ExAMPLE 4:

Given

: (3x2t 2)'(x' - 5r)' f(x) find f '(x).

141

Consider f as the product of the two functionsg and h, where : (x2- Sx)B 8(r) : (3x2* 2)z and h(x) Using Theorem 3.5.5 for the derivative of the product of two functions, we have ' ( x ) : g @ ) h ' ( x )+ h ( x ) g ' ( x ) f We find h' (x) and g'(r) by the chain rule, thus giving us i.J 'i i i! t ' ( x ) : ( 3 x z* Z ) ' I g k ' 5x)'(2x- 5)l + (x'- 5r)3[2(3r'+ 2)(6r)] f : 3 ( 3 f * 2 ) ( x z- 5 r ) ' [ ( 3 x 2* 2 ) ( 2 x - 5 ) + 4 x ( f - 5 r ) ] : 3(3x2* 2) (x2- 5r)'16x" - L1xzt 4x- 10 * 4xs- 20x21

solurroN:

: 3(3f * 2) (x2- 5x)'(10r3- 35xz* 4x- 10)

3.6 Exercises In Exercises1 through20, find the derivative of the given function. 2. f(x): (10- Sx)n 1. F(r) : (x'* 4r - 5)3 a. S(r) : (zta* 8rz + 1)' 7. h(u) :

( 3 u 2* 5 ) 3 ( 3 u _1 ) '

5 . f ( x ) : ( x* 4 ) - ' 8. f(x): (4xzr7)2(2x3+ 1)'

/)t2 +

l\2 ' 1_2. g(t): (fr,,117

1;-r 10.f (x): (f - n*-z1z1xz*

i'-r-l. f(il:

13.f(x):*fu

M h(x):Gj!#*6)'

(F)''

- 4) : t6.f(il : (y*s)'(sy+ L)2(3y2 -r7.f(z) ffi,

Tg.G(x):ffi

g. f(t) : (2t4- 7tB+ 2t - l)2 5. H(z) : (2"- 322* t1-e 9. S (r) : (2x - 5)-' (4x * 3\-2

(\t

+ s)' 15.f(r) : (r?+ 7)3(1r 18. g(r) : (2x - 9)t(x" t 4x - 5)3

-

20.F(x):ffi-

g)-2

21. A particle i5 moving along a straight line according to the equation of motion s: [(t2 l)/(rc + 1)]" with t = 0, velocity of the instantaneous What i3 the (a) t sec. the origin at particle frorn oif the where s ft is the dire-cted distance particle at t1 sec? (b) Whai is the instantaneous velocity of the partide at 1 sec? (c) What is the instantaneous velocity at t sec? at each of the followint Points: (0'l.)' (1'&)' (2'+)' 22. Find an equation of the tangent line to the curve y:2/(4-r)r (g,2), (5,2), (6, *). Draw J sketch of the graph and setments of the tangent lines at the given points. - 2x - 4)2 at the point (3.:2). 23. Find an equation of the normal line to the curve y : ?l(x' Find an equation of the tangent line to the curve y: (x'- 4)'l(3x 5)2 at the point (1, *). ' (x ' ); (b ) g ' (r). 25. Gi ve nf ( x ) : 13and S @ ) : f (f).F i n d : (a )f Fi nd the deri vati veof f " g i n tw o w ays: (a) by fi rst fi ndi n g a n d g (r): (x + 7 )l (r-1 ). 26. Gi ven f( u) : u2*S u*5 U. il(r) and then finding U " 8)'@); (b) by using the chain rule.

a

i

?

142

THE DEFIIVATIVE

27. Suppose that I and g are two functio,nssuch that (i) g'(rl) and l'(g(r,)) terval containing xr, g(r) - g(rr) + 0. Then

exist and (ii) for all x * 11in some open rn-

( / " 9 ) ( r )- ( / . 9 ) ( r , )_ ( / . s ) ( r ) - ( / . g ) ( r , ) g ( r )- s ( x , ) . x- xr x- xr s(r) - g(x')

(6)

(a) Provethat as , - *r, gk) - gk) and hencethat (f . s)'(rr): f' (g(xr))g'(xr) thus simplifying the proof of the chain rule under the additional hypothesis (ii); (b) Show that the proof of the chain rule given in part (a) applies it f(u): LP arrdg(x) : xs,but that it ioes not apply'ii (u) : u2 and,gtr) : sgn r. f 28. Use the chain rule to Prove that (a) the derivative of an even function is an odd function, and (b) the derivative of an odd function is an even function, provided that these derivatives exist. 29.use the result of Exercise28(a) to Prove that if g is an even function and g'G) exists, then if h(x) : (f " s) @) and,f is differentiableeverywhere,h'(0) : 0.

30. Supposethat/ and g are functionssuch thatf' (x): llx and,f(S@)):r. prove that if g,@) existstheng,(r):S(r). 31.luppose that y is a function of a, ts is a function oI u, and u is a function of.x, and,thatthe derivative

sDrA,Duv,and,

Dru all exist. Prove the chain rule for three functions: D,A : (D,y) (D"a) (D"u)

3.7 THE DERIVATIVE oF THE The tunction defined by / POWER FUNCTION FOR x' I\x) RATTONAL EXpONENTS

(1)

is called the poaw function In Sec. 3.5 we obtained the following formula for the derivative of this function when r is a positive or negatirre irrt"gen f

' (x ) : p6r-r

( 2)

We now show that this formula holds when r is a rational number, with

certain stipulations when x: 0. We first consider x * 0, and r : Llq, where q is a positive integer. Equation (1) then can be written f (x) : 26tro

From Definition 3.3.1, we have

f,(x):111 w

(3)

To evaluate the limit in Eq. (3) we must rationalize the numerator. To do this, we use formula (7) inSec.2.2, which is an-bn:

(a-b)(a"-t

so we rationalize

* a " - z b+ a n - s b + z .

the numerator

+ a z b n - i* a b " - z+ b " - r )

of the fraction

(4)

in Eq. (3) by applying

'.qw;

FORRATIONAL EXPONENTS1 4 3 FUNCTION OF THEPOWER 3.7 THEDERIVATIVE Eq. (4), where a: (r + Ar)tto, b: x7rt8,and n: q. So we multiPly the numerator and denominator by + ' + [ (r + Ar) rtqf(q-2)xrtq [ (r + Lx)rta1
'(x) equals

[- '( r + A x ) 'u c - x v q ] [ ( x + A r ) ( c - r l r a *( r * L,xl(x*Ax)(a-trto+ (xl

1;

L4v"'orrtq+ ''

Lx){a-zttqxrtq+' '+

'+ x@-r\tql

xG-\tql

(s)

and now applying Eq. (4) to the numerator we get (r * Ar)nts )csts, which is Ar. So from (5) we obtain t, f ' ( x ) : Al ri-m 0 Ar[ (r * Ar)tc-t)ta a ,* * Ar)ta-z)ta2srto

+ xg-r)tqf

1

: lim

(r * Ar)(q-l)/s + (r * Ar)ta-z)tqxrts+

Ar-0

f

a(a-r)/o

1

-

y(a-t)tu j

xg{s-tyu t,

+ x9-r)tq

Because there are exactlY q terms in the denominator of the above fraction, we have 1

' f'\x):fuo qx' or, equivalently, ,(x) 'fL l

-

1 ! 26ua-t

(5)

which is formula (2) with r - Llq. Now, in Eq. (1) with x # o,let r: plq, whete p is any nonzero integer, and.q is any positive integer; that is, r is any rational number except zero. Then Eq. (1) is written as f (x):

frq

or, equivalently, f(x)

:

(xrraln

Because p is either a positive or negative integer, we have, from the chain rule and Theorems 3.5.2 and 3.5.8, ' (x ) : ' P (x tto)e-r D r(xrtq) f Applying

formula (5) for D,(x''o), we get

' '+ xuq-l 'q f (x) : P(x'to)'-'

"'sjr'

144 THEDEBIVATIVE

f

f

,(x\ -P

q

,nh-uq+uq-l

'(x) -L vnrc-t 'qs

(7)

1*

Formula (7) is the same as forrnula (2) with r: plq. l f r:0, and x * 0, E q. (1) becomesf (x): r0: 1..In thi s case / , ( x) : 0, which can be written as f ' (r): 0 . r0-1. Therefore, formula (Z) holds if r: 0 with x * 0. We have therefore shown that formula (2) holds when r is any rational number and x * 0. Now 0 is in the domain of the power functionf if and only if r is a positive number because when r = 0, /(0) is not defined. Hence, we wish to determine for what positive values of r, f ' (0) will be given by formula (2). We must exclude the numbers in the interval (0, 1.1because for those values of r, x'-r is not a real number when r: 0. Suppose, then, that r ) 1. By the definition of a derivative

f '(o):tjTH: When r ) l,lim :f -0

ljT''-'

r'-r exists and equals 0 provided that r is a number such

that x"-1 is defined on some open interval containing 0. For example, if r: i, x''-1: x:ttz-r:. vtt2,which is not defined on any open interval containing 0 (since r1l2 does not exist when r < 0). However, if. r x,-l - x,ts-l : x2t3, which "a, is defined on every open interval containing 0. Hence, we conclude that formula (2) gives the derivative of the power function when x: 0 provided that r is a number for which r'-r is dlfined on some oPen interval containing 0. Thus, we have proved the following theorem. 3.7.1,Theorem

rf f is the power function, where r is any rational number (i.e., (x) : x,), f then f

'(x) : 716r-l

For this formula to give f '(0), r must be a number such that x'-1 is defined on some open interval containing 0. An immediate consequence of Theorem 3.7.1 and the chain rule is the following theorem. 3.7.2 Theorem

rf f and g are functions such that /(x) : number, and if g'@) exists, then

[g(x)]',

where r is any rational

' f (x): rlB@)l'-'g'G) For this formula to give f '(0), r must be a number such that [g(r)]"-t defined on some open interval containing 0.

is

-. OF THE POWERFUNCTIONFOR RATIONALEXPONENTS 3.7 THE DERIVATIVE

EXAMPLE 1:

Given

f (x) :4VP findf '(x).

solurroN, f (x):4xzt3. Applying Theorem 3.7.1,we find

f

'(x) - l,' :9

I

ilxzt}-rl

t-r,, 38 3*u3

8 :5ffi

Exauprn 2:

Given

h(x)-\ffiB find h'(r).

solurroN h(x): (ZxB- 4x * 5;trz. Applying Theorem 3.7.2,we have - 4) h' (x) - $(Zxs- 4x* 5)-t'215x2 3x2-2

EXAMPLE

Given

soLUrIoN:

Writing the given fraction as a product we have

g@):xl(3xz_L)-us Using Theorems 3.5.6 and 3.7.2, we have g ' G) :3 x 2 ( 3f - L)-tts - i (3,x2 1' )-e' a16x)(xB ) : x2(3xz- L)-ais;3(3xz- t) - 2x2f - 3)-- *(7xz(3 f L)ri s

3.7 Exercises In Exercises1 through 18, find the derivative of the given function.

2'f(s):ffi

3. s(r) :

4.h(t):{F1

5.f(x)-{16112+5x-rt2

6. sfu) : (y, * 3)tralrs- L)ttz

7. F(x):9Fffi

s.g(x):{@ffi

1. f(x):

( 3 x* 5 ; z B

10. f(x) - Ja2t3- 6xtt3 a xs-rtz

\F 1 1F . (r):?

'1,9. 12.

' .'-f;F-

146

THE DERIVATIVE

M. f (s):

( s nl S s z* l ) - z r s

17.f(x): \/s+m 19. Find an equation of the tangent line to the cur.ve y : {f + 9 at the point (4, 5). 20. Find an equation of the tansent rine to the {urve y: (6- zx)Ltxat each of the folrowing points: (-1, 2), (L,.{y4), (3,01, 15,-94r. (7, -2). Driw a skekh of the gt"irr se-jirents of the tangent rines at the given ponts. ""a 21. Find an equation of the normal line to the cuwe y : a1/l$ ap at the origin. 2, Find, aa equation of the tantent line to the curve y:1/$?r=Twhich is perpendicular to th e line 12x_7y I2_O. 23' Alr obiect i8 moYing along a straitht line according_to the equation of motion s : l4FT3, with t > 0. Find the values of I for which the measure of rhe instantaneous vilocity is 0; (b) U (c) Z. ?a) 24'Givenf(u):llu2andg(r):{xl{2filal,findthederivativeof/"gintwoways:(a)byfirstfinding(l.g)(r) and then finding U " d,e); (b) by using the chain rule. In Exercises 25 through 28, find the derivative of the given function. (nrwr: jal : fii.y

2 5 .f ( x ) : l x z- 4 l 2 7 .g ( x ) : l r l '

2 6 .g ( x ) : r l r l 28. h(x): VltlE

29. Supposeg(x) : lf (x) l. prove that if ,(x) and g,(x) exist,then ,(x)1. f lg,(r) | : lf : 30' Supposethat 8(r) \/FE and ft(x) : f kQ)) where / is differentiableat 3. prove that h,(0; : g. 31. If g and h are functions and if / is the function defined by f(x): [s(x)]'[h(x)]' where r and s are rationar numbers, prove that if g,@) and,h, (r) exist '(x): [ g ( r ) ] , - , l h ( x ) 1 s - r [ r. h ( x ) g ,( r ) + s . g ( r ) h , ( x ) ] f In Exercises32 through 35, use the result of Exercise3L to find the derivative 32. g(x) :

(4x + 3)rt2(4 -

x2)rt3

u. f(r):(#)'{r*n;,,,

3.8 IMPLICIT DIFFERENTIATION

of the given function.

33. f (x):

( 3 x - t 2 ) a ( x 2- l ) z r s

35. F(f) :

( t ' - 2 t + 1 ) z t z ( 1 2 +f + 5 ) 1 / 3

If f - {(x, y)ly : g*" * 5x + 1}, then the equation a:3x2*5x*L

( 1) defines the function / explicitly. However, not all functions are defined explicitly. For example, if we have the equation

f-2x-3yu*y'-y'

e)

we cannot solve for y in terms of r; however, there may exist one or more functions / such that if y: f(x), Eq. (2) is satisfied, that is, such that the equation

x B- 2 x : i l f ( x ) l u+ [ / ( x ) ] ' - l f ( x ) 1 ,

.aie9s .

147 DIFFERENTIATION 3.8 IMPLICIT is true for all values of r in the domain of f . In this case we state that the function / is defined implicitly by the given equation. Using the assumption that Eq. (2) defines y as one or more differentiable functions of x, we can find the derivative of y with respect to xby the process called implicit differentiation, which we now do. The left side of Eq. (2) is a function of x, and the right side is a function of y. Let F be the function defined by the left side of (2), and let G be the function defined by the right side of (2). Thus, F(r) - x6 - 2x

(3)

G(y)-3yu*y"-Y'

(4)

and

where y is a function of.x, saY,

Y: f(x) So we write Eq. (2) as

(s)

F(r) - G(f(x))

Equation (5) is satisfied by all values of x in the domain of / for which

exists. Glf(x)l '

Then for all values of x fot which / is differentiable, we have

D,lxu - 2xl - D,l3yu* y' - y'l

(6)

The derivative on the left side,of Eq. (5) is easily found, and we have V)

D*lxu-2xl:6x5-2

we find the derivative on the right side of Eq. (5) by the chain rule, giving us D'lSyu l y'- a'l:

t}y' ' D*a * Syn' D,V - 2a ' D,a

(8)

substituting the values from (7) and (8) into (6), we obtain 6x5- 2:

(LB!' t Syn- 2y)D"y

Solving for D rY, we get

D,U

6x5-2 L8y5*Syn-2y

Equation (2) is a special type of equation involvrng x and y because it can be written so that all the terms involving x are on one side of the equation and all the terms involving y are on the other side. In the following illustration we use the method of implicit differentiation to findD*y from a more general tyPe of equation.

148

THE DERIVATIVE

o rLLUSrRArroN 1: Consider the equation 3*nY'-7xy":4-8U

(9)

and assumethat thereexist one or more differentiablefunctions such that / if y: f (x), Eq. (9) is satisfied. Differentiating on both sides of Eq. (9) (bearing in mind that y is one or more differentiable functions of r), and applying the theorems for the derivative of a product, the derivative of a power, and the chain rule, we obtain 72x3y2 * 3xa(2yD*a)- 7yt - 7x(3y2D,y) - 0 - gD,a Solving for Dry, we have D,y(6xay - 27xy2+ 8) : 7yt - l2xsy2

DrA:ffi

o

Remember that we assumed that both Eqs. (2) and (9) define y as one or more differentiable functions of x. It may be that an equation in x and y does not imply the existence of any real-valued function, as is the case for the equation

x'+y'*4:0 which is not satisfied by any real values of r and y. Furtherrnore,it is possible that an equation in I and V may be satisfied by many different functions, some of which are differentiable and some of which are not. A general discussion is beyond the scope of this book, but can be found in an advancedcalculustext. In subsequentdiscussions,when we statethat an equation in x and y definesy implicitly as a function of x, we assume that one or more of these functions is differentiable. Example 3, which follows, illustrates the fact that implicit differentiation gives the derivative of every differentiable function defined by the given equation. EXAMPLE

(x * Y)'-

Given ( x - y ) 2 : x a* A a

find D,y.

solurroN:

Differentiating implicitly with respect to x, we have

2 ( x * V ) 0 * D , y ) - 2 ( +- V ) ( t - D * y ) : A x s - r A y B D , y from whicir we obtain 2x * 2y + (2x + 2y)D"y - 2x + 2y * (2x - 2y)D,y : 4xB*

yBD,y

D"y(Ax-4y'):4x3-4y D

rxevrrrE 2: Find an equation of the tangent line to the curye f t y":9 at the point (1,,2).

solurroN:

xg-tl

,!:;=T

Differentiating implicitly with respect to x, we obtain

3x2*3y2D,y:g

I

3 . 8 I M P L I C I TD I F F E R E N T I A T I O N1 4 9

Hence, v2

D,u:-+ @r

y"

Therefore,at the point (L, 2), is then

An equation of the tangent line

y-2:-I(x-1) (a) Differentiating implicitly, we find ExAMPLE3: Given the equation solurroN: x' * y' :9, find: (a) D"Y bY -X 2x t 2YD*Y: 0 and so D-Y: implicit differentiation; @) rwo functions defined bY the equa(b) Solving the given equation for y, we obtain tion; (c) the derivative of each of and Y : -!9 - xz the functions obtained in Part (b) y -- ttr=p by explicit differentiation. Let fl be the function for which (d) Verify that the result obtained in part (a) agreeswith the results fr(x):6zT obtained in Part (c). and f, be the function for which

frv '

'i,'.,4

f,(x): -\E=i'

;+Y i

*

,J,- r'

'r/ '' *J

"'l

(c) Because/, (r) : (9 - xr)tt', bY using the chain rule we obtain - x')-trz1-2x) : f ,.'(x) i(9

I

C,,t

ii.1f f

t/9 - x' r0'

Similarly, we get -x

f''\x):ffi; fr(x), where f ,(x)

(d) For a:

-- \tr=T,

we have from part (c)

--x

f " ( \"/ -t) .,I

y

{r4

which agrees with the answer in part (a). Fot y: -fr=E, we have from Part (c)

fr'(x) : -r678 v/

&

__L

v

which also agrees with the answer in part (a)'

: fr(x), where fr(x)

150

THEDERIVATIVE

Exercises 3.8 In Exercises L through L6, find D,y by implicit

1 , .f + A 2 : 1 6

, i T , ' * : '- + ? Y \.J

XC Zy

7. \/i * t/y:4

differentiation.

(z., zr'y * 3xys: 5

3 . x s+ y t : 8 x y

s.l+1:t xy

6.x-4u:x

8.y+{xy:g*"

9. ,'y':

v

x2i A2

t O .y t f f i t x f f y : y

11..(x * y)'- (x - A)": x" + y"

1 2 . ( 2 xt 3 ) n : 3 y n

13. Y : 2 1 x 2 x-y

1.4.t/y + VV + {y:

15.l*V + 2x: fr

1 6. fyt:

*

xn - yn

In Exercises 77 through 20, consider y as the independent variable and, find, Dox. 1 7. xn * yn: L2x2y 18.y:2x3 - 5x t9. x " y * 2 y n - x a : 0 2 0 .y f r - x t / y : 9 2L. Find an equation of the tangent line to the curve "L6xaI at Aa:32

the point (1,2).

22. There are two lines through the point (-r, g) which are tangent to the curve x2+4y2-4x-8y*3:0 Find an equation of each of these lines. Prove that the sum of the r and y intercepts of any tangent line to the curve {tz a yuz: If. x"y*: (r * y)n+*, prove that r . DrA: A.

2 5 -y 2 : X t - g ; l r : g

2 7 .x 2 - y ' : 9 ; x r : - 5 29.f * y' - 2x - 4y - 4- 0; xr:'!.

3.9 THE DERIVATIVE AS A RATE OF CHANGE

ku2 isconstant and equal to k.

2 6 .f + \ e : Z l ; x t : A 28. y' - x2:' ).6; x1- -3 30. x2* 4y, * 6x - 40y* 9Z : 0; xr: -2

The concept of velocity in rectilinear motion coffesponds to the more general concept of instantaneous rate of change. For eiample, ifa particle is moving along a straight line according to the-equation of motion s: f(t), we have seen that the velocity of the particle at f units of time is determined by the derivative of s .with respect to f. Because velocity can be interpreted as a rate of change of distance per unit change in time, we see that the derivative of s with respect to f is the rate of c"hange of s per unit change in f. In a simirar way, if a quantity y is a function of a quantity x, wemay exPress the rate of change of y per unit change in r. The discussion is analogous to the discussions of the slope of a tangent line to a graph and

AS A RATEOF CHANGE 151 3.9 THE DERIVATIVE the instantaneous velocity of a particle moving along a straight line. If the functional relationship between y and r is given by

Y: f(x)

f (x,+ Lx) f (x) :Ly Lx A,x

(1)

If the limit of this quotient exists as Ax e 0, this limit is what we intuitively think of as the instantaneous rate of change of.y per unit change in r at lr. Accordingly, we have the following definition. 3.9|t Definition

rate of changeof y per unit changein x at xt lf y - f (x), tirreinstantaneous derivative of y with respect to r at xr, if it the it7'drj or, equivalently, exists there. The instantaneousrate of changeof y per unit change in r may be interpreted as the change in y causedby a change of one unit in x if the rate of changeremains constant.To illustrate this geometrically,letl'(rr) be the instantaneousrate of change of.y per unit changein x at xr' Then if we multi ply f' (rr) by Ar (the change in x), we have the change that would occur t" y it'the point (x,y) were to move along the tangent line at (x1,y) of the graph of y:/(r). 3ee Fig. 3.9.1,.The averagerate of change o{ y per unil.h"r,g" i" r is given by ihe fraction in Eq. (L), and if this is multiplied by Ax, we have

:Lu n. ^*- Ly Figure3.9.1

which is the actual change in y caused by a change of Ar in r when the point (x,y) moves along the graPh.

sxevlprE L: LetV cubic inches be the volume of a cube having an edge of length e inches. Find the average rate of change of the volume per inch change in the length of the edge as e changes from (a) 3.00 to 3.20; (b) 3.00 to 3.10; (c) 3.00 to 3.01. (d) What is the instantaneous rate of change of the volume Per inch change in the length of the edge when e:3?

Because the formula for finding the volume of a cube is solurroN: : e3. Then the average rate v: e3, let f be the function defined by f (e) from e, to et * Ae is changes e e as in change unit of chang e of V per

f (e'+ A'e) f (e') A,e

- 0.2,and'=s@: 9Jf i, A,e (a)or-fllfi/Q) (b)e,: 3,A,e:0.1; ".a

: g}}9

:#

: 28.8

:'# : 27'e

152

THE DERIVATIVE

( 3 ' 0 1 ) 3- 3 3 - 0 ' 2 7 1 ' : 27 (c) €r:3, A,e:0.01, undl(3'01.)- /(3) 't 0.0L 0.01 0J01 In part (a) we see that as the length of the edge of the cube changes from 3.00 inches to 3.20 inches, the change in the volume is S.7T cubic inches and the average rate of change of the volume is 28.8 cubic inches per inch change in the length of the edge. There are similar interpretations of parts (b) and (c). (d) The instantaneous rate of change of V per unit change in e at 3 is /'(3). '(e) :3e2 f Hence,

f

' ( 3 ): 2 7

Therefore, when the length of the edge of the cube is 3 inches, the instantaneous rate of change of the volume is 27 cubic inches per inch change in the length of the edge.

ExAMPLE 2:

The annual gross earnings of a particular corporation f years after January I,'1974, is p millions of dollars and

P:€F+2t+10 Find: (a) the rate at which the gross earnings were growing January 7, 1976; (b) the rate at which the gross earnings should be growing January 1,1980.

solurroN: (a) On January '1.,'j.976,

DtP:tt+z

hence, we find Dp when

,,r],:, -3+ 2:3.G

s9.9" I^!!?rv 1,1976,the gross earnings were growing at the rate of 3.6 million dollars per year. (b) On January 1,,lg10, t: 6 and. I Dtpl :+*Z:6.8 Jt:e

Therefore, on January l, 7980,the gross earnings should be growing at the rate of 6.8 million dollars per year.

The results of Example 2 are meaningful only if they are compared to the actual earnings of the corporation. For eximple, ir o. fanuary 1, 1975, it was found that the earnings of the coqporation for the year lg14 had been 3 million dollars, then the rate of growth on fanuary 7, !9T6, of 3.4 million dollars annually would have b-een excellent. Howev er, if the earnings in 7974 had been 300 million dollars, then the growth rate on fanuary 1,"1,976,would have been very poor. The measire used to comPare the rate of change with the amount of the quantity which is being changed is called the relatiue rate. 3.9.2 Definition

y -- f (r), the relatiae rate of changeof y per unit changein x at xris given lf by f'(xr)lf(xr) or, equivalently, O.VIV' evaluated at xJ xr.

3.9 THE DERIVATIVE AS A RATE OF CHANGE

153

If the relative rate is multiplied by 100, we have the percent rate of change. nxaupln 3: Find the relative rate of growth of the gross earnings on Januaryt,'1.976,and January l, L980,for the corPoration of Example2.

solurroN: (a) When t:2, P : +(4) + 2(2) + L0 : L5.6.Hence,on Januaty 1, 1976,the relative rate of growth of the corporation's annual gross earningswas

: : 23.'m ry1,:,:ffi 0.231 (b) When t:6, p: ?(36)+ 2(6) + 10 :36.4. Therefore,on January 1980, the relative rate of growth of the corporation's annual gross earnings should be '!,,

%P.l : 6'R P lt:e

967:

0'1'87:'l'8'7%

Note that the growth rate of 6.8 million dollars for ]anuary 1, 1980, is greater than the 3.6 million dollars for January 1,1976; however, the rela'J.8.7Vo for January 1, 1980, is less than the relative tive growth rate of growth rate of 23.LVofor January L, L976.

3.9 Exercises l. If A in., is the area of a square and s in, i5 the length of a side of the square, find the average rate of change of A with respectto s as s changesfrom (a) 4.00to 4.50;(b) 4.00to 4.30;(c) 4.00to 4.10.(d) What is the instantaneousrate of change of d vrith respect to s when s is 4.00? 2. Supposea right-circular cylinder has a constant height of 10.00in. If V in.3 is the volume of the right-circular cylinder, ani r in. is the radius of its base, find the average rate of change of V with rcsPect to / as t changes from (a) 5.00 to 5.40; (b) 5.00 to 5.10;(c) 5.00 to 5.01.(d) Find the instantaneousrate of change of y with resPectto r when r is 5.00. 3. Let r be the reciprocal of a number r. Find the instantaneous rate of drante of / with resPect to t and the relative rate of change of r per unit change in n when z is (a) 4 and (b) 10' 4. Let s be the principal square root of a number l. Find the instantaneous rate of change of s with resPect to r and the relative rate of drange of s Per unit change in x when r is (a) 9 and (b) a. 5. If water is being drained from a swimming pool and y gal is the volume of water in the Pool I min after the draining starts, where y-= 250(4O- l)1, find (a) the averagerate at which the water leaves the pool during the first 5 min, and (b) how faet the water ie flowing out of the pool 5 min after the draining starts. 6, The supply equation for a certain kind of pencil is r: 3f I 2p whete p cents fu the price P,erPencil when 10mt PenAt are suppfea. (a) Find the average rate of drange of the supply per I cent drange in the Pdce when the Price is increased &om 10 cents to 11 cents. (b) Find the instantareous (or marginal) rate of drange of the suPPly Per 1 cent change in the price whm the price is 10 cents. - O.Zf. 7. The profit of a rctail store is 100y dollars when r dolLars are spent daily on advertiBing a\d V : 25,0O+ Xx if the currmt increased Use the derivative to detemrine ii it would be profitable for the daily advertising budget to be (b) $100 daily advertising budget is (a) $60 ard

154

THE DERIVATIVE

8' A balloon maintains the shape of a spherc as it is being inflated. Find the rate of change of the surface area with respect to the radius at the instant when the radius is 2 in. 9. In an electdc circuit, if E volt8 is the electromotive force, R ohms is the resistance, and I amperes is the current, Ohm,8 l,aw states that IR: E. Assuming that E is constant, show that R decreasesat a rate that is-proportional to the inverse square of L 10. Boyle's lgw fo1 tle expansion of a gas is PV: C, where P is the number of pounds per square unit of pressure, y is -change the numbet of cubic units in the_volume of the gas, and C is a constant. Fina the insiantaneous rate of of the volune per change of one pound per square unit in the pressure when p : 4 and V: g. 11' A bomber is flying parallel to the ground at an altitude of 2 mi and at a speed of 4l mi/rrin. If the bomber flies directly over a target, at what rate is the line-of-Eight distance between the bomber and the target changing 20 sec later? 12. At 8 a.M. a ship sailing due north at 24 knots (nauticat miles per hour) is at a point P. At 10 A.M. a second ship sailing due east at 32 knots i8 at P. At what rate is the distance between the two ships changing at (a) 9 e.r.,r.and @!1 e.r,a.?

3.10 RELATED

RATES

'

ExAMrLE1: A ladder ZSft long is leaning against a veftical wall. If the bottom of the ladder is pulled horizont ally away from the wall at 3 ftlsec, how fast is the top of the ladder sliding down the wall, when the bottom is 15 ft from the wall?

There are many problems concerned with the rate of change of two or more related variables with respect to time, in which it is not necessary to express each of these variables directly as functions of time. For examPle, supPose that we are given an equation involving the variables r and y, and that both x and y arc functions of a third variable f, where t sec denotes time. Then, because the rate of change of r with respect to f and the rate of change of y with respect to f are given by D,x aid. D,y, resPectively, we differentiate on both sides of the given equation with respect to t by applying the chain rule and proceed as below. SoLUTIoN: Let f : the number of seconds in the time that has elapsed since the ladder started to slide down the wall; a: the number of feet in the distance from the ground to the top of the ladder at f sec; r: the number of feet in the distance from the bottom of the ladder to the wall at f sec. See Fig. 3.10.1.Because the bottom of the ladder is pulled horizontally away from the wall at 3 ftlsec,D.x:3. We wish to find Dry whenx:.1_5. From the Pythagorean theorem, we have U2:625

- xz

(1)

Because r and y are functions of t, we differentiate on both sides of Eq. (t) with respect to f and obtain 2y Dty : -2x Dtx giving us

Dta: -1o,* u F i g u r e3 . 1 0 . 1

When x:

lS,it follows from Eq. (1) thaty :

(2) 2f.BecauseDg:3, we

3.10 RELATEDRATES

155

get from (2)

o,rf":,o Therefore, the top of the ladder is sliding down the wall at the rate of 2* ftlsec when the bottom is 15 ft from the wall. (The significance of the minus sign is that y is decreasing as f is increasing.)

EXAMPLE2: A tank is in the form of an inverted cone, having an altitude of L6 ft and a base radius of 4 ft. Water is flowing into the.tank at the rate of 2 ftglri;rir. How fast is the water level rising when the water is 5 ft deep?

Let f : the number of minutes in the time that has elapsed since water started to flow into the tank; the number of feet in the height of the water level at h f min; r: the number of feet in the radius of the surface of the water at f min; V: the number of cubic feet in the volume of water in the tank at f min. At anv time, the volume of water in the tank may be expressed in terms of the volume of a cone (see Fig. 3.1'0.2).

solurroN:

y:

(3)

lnrzh

V, r, and h are all functions of f. Becausewater is flowing into the tank at the rate of 2 ft}lmtn, D1V: 2.We wish to find Dfu when h: 5. To express r in terms of.h, we have from similar triangles

r:ih

;:+

Substituting this value of r into formula (3), we obtain V : in(*h)'(h) 3.10. 2 F ig ure

or

Y : |gnhg

( 4)

Differentiating on both sides of Eq. (a) with respect to f, we get D1V: L*rhz D1h Substituting 2 for D1V and solving for D1h, we obtain

D -, h : 1

?)

Tn-

Therefore,

*o,']o:, :# We conclude that the water level is rising at the rate of 32125znftlmin when the water is 5 ft deeP.

156

THE DERIVATIVE

EXAMPLE3: Two cars, one going due east at the rate of 37.5 mi/hr and the other going due south at the rate of 30.0 mi/hr, are traveling toward an intersection of the two roads. At what rate are the two cars approaching each other at the instant when the first car is 400 ft and the second car is 300 ft from the intersection?

solurroN: Refer to Fig. 3.10.3, where the point P is the intersection of the two roads. Let x: the number of feet in the distance of the first car from P at f sec; y: the number of feet in the distance of the second car from P at f sec; z: the number of feet in the distance between the two cars at f sec. Because the first car is approaching P at the rate of jT.S mi/nr : 37.5' rcuftlsec:55 ft/sec and because r is decreasing as f is increasing, .# f t lsec: i t fo l l o w s that D fi :-55. S i mi l arl y, because 30 mi /hr:30 4 4 ttl s e c, D tA :-44. W e w i sh to fi nd D rz w hen tr:400 and U: 200. From the Pythagorean theorem we have z 2 : x 2* y 2

(5)

Differentiating on both sides of Eq. (5) with respect to f, we obtain 2z DP:2x

Dg I 2U D,Y

and so (5)

F i g u r e3 . 1 0 . 3

when x: 400 and y: 300,it follows from Eq. (5) that z: 500.In Eq. (6), substitutingDrx: -55, DtA: -44, x: 400,y : 300, and,z: 500,we get I : (4oo)(-5s) + (3oo)(-44) :_- t 0n, .4 D,zl 500 Jz:soo Therefore,at the instant in question the cars are approachingeachother at the rate of 70.4 ftlsec.

Exercises 3.10 1. A kite is llying at a height of 40 ft. A boy flying it so that it is moving hodzontally at a rate of 3 ft/sec. If the string is taut, at what rate is the stdnt being paid! out when the length of the i-t ing releasei is 50 ft? 2 A spherical balloon is being inflated so that its volume is increasing at the rate of 5 ftslmin. At what rate is the diameter incteasing when the diameter is 12 ft? 3' A spherical snowball i5 being made so that its volume is incr€asing at the rate of 8 ft3lmin. Find the rate at which the radius is increasing when the snowball is 4 ft in diameter. 4' suPPose that when the diameter is 6 ft the snowball in Exercise 3 stopped growing and started to melt at the rate of I ft3lmin. Find the rate at which the radius is changing when the radiu; is tft. 5' Sand is being dropped at the rate of 10 ft3lmin onto a conical pile. ft the height of the pile is always twice the base radius, at what rate is the height increasing when the pile is a fi nign? 6.

l_llght is hunt 15 ft above a 6tr-aight horizontal path. ff a man 6 ft tall is walking away ftom the litht at therateof 5 ft/eec, how {ast is his shadow lengthening?

3.11 DERIVATIVES OF HIGHERORDER 157 7. .In Exercise 6 at what rate is the tip of the man's shadow moving? / b. A man 6 ft tall is walking toward a building at the rate of 5 ft/sec. If there is a light on the ground 50 ft from the buitding, how fast is the man's shadow on the building growing shotter when he is 30 ft from the building? 9. A water tank in the form of an inverled cone ie being emptied at the rate of 6 ft8/min, The dtitude of the cone i6 24 ft, and the base radius is 12 ft. Find how fa6t the water level is lowerint when the water is 10 ft deep. 10. A trough is 12 ft long and its ends are in the forrr of inverted isoscelea triangles having an altitude of 3 ft and a bas€ of 3 ft. Water is flowing into the trough at the rate of 2 ft3/min. How fa6t is the water level rising when the water is 1,ft deep? 11, Boyle's law for the expansion of gas ig PV: C, where P is the number of pounds per square unit of pressure, y is the number of cubic units of volume of the gas, and C is a constant. At a certain instant the pr€sEure is 3000 lb/fP, the vol: ume is 5 fC, and the volume is increasing at the rate of 3 fP/min. Find the rate of change of the pressure at this instant. 12. The adiabatic law (no gain or loss of heat) for the expansion of air is PIl ' : C, where P is the number of pounds per square unit of pressure, y is the number of cubic unit8 of volume, and C is a constant. At a specific instant, the pressure is 40 lb/in.2 ard is increasing at the rate of 8 lb/in.'2 each second. What i6 the rate of change of volume at this instant? '

13, An autourobile traveung at a rate of 30 ft/sec is approaching an intersection. When the automobile is 120 ft from the i' intersection, a truck traveling at'the rate of atoft/sec crosseqthe inter8ection, The automobile and the truck are on roads that are at right angles to each other. How fast are the automobile and the truck separating 2 sec after the truck leaveg the intersection? 14. A man on a dock is pnlling in a boat at the rate of 50 ft/min by means of a rope attached to the boat at water level. If the man's hands are 16 ft above the water level, how fast is the boat approadring the dock when the amount of rope out is 20 ft? 15. A ladder 20 ft long is leaning against an embankment inclined 50' to the horizontat. If the bottom of the ladder i8 being uroved horizontally toward the embar*ment at L fvsec, how fast is the top of the ladder moving when the bottom is 4 ft from the embankment? 16. A horizontal trough is 16 ft lon& and its ends are isosceles trapezoids with an altitude of 4 ft, a lower base of 4 ft, and an upper base oI 5 ft. Water is being poured into the trough at the rate of l0 felmin. How fast is the water level rising when the water i6 2 ft deep? 17. In Exercise 16 if the water level is decreasing at the rate of I ft/min when the water i8 3 ft deeP, at what rate is water being drawn from the trough? 18. Water is being pour€d at the rate of 8 ft3/min into a tank in the forrr of a cone. The cone is 20 ft deep and 10 ft in diameter at the top. If there iB a leak in the bottom, and the water level is rising at the rate of 1 in./rnin, when the water is 16 ft deep, how fast is the watet leaking?

OF 3.11 DERMTIVES HIGHER ORDER

lf f ' rs the derivative of the function I then /' is also a function, and it is the first deriaatiae of.f . It is sometimes referred to as the first deriaed function. If the derivative of f ' exists, it is called the secondderiaatiae of f , or the second derived function, and can be denotedby f" (read as "f double prime"). Similarly, we define the third deriaatiae of f , or the third derived function, as the first derivative of f " if it exists. We denote the third derivative of f by f "' (read as "f triple prime"). The nth deriaatiae of the function /, where n is a positive integer greater than L, is the first derivative of the (n - l)st derivative of /. We denote the nth derivative of f by f{"t. Thus, if. f{"t denotes the nth derived

THEDERIVATIVE

158

function, we can denote the function / itself by fo.Another symbol for the nth derivative of f is D *"f . If the function / is defined by the equation y: f(x), we can denote the nth derivative of f by D,"y. EXAMPLE 1: Find all the derivatives of the function / defined by f(x):8ra*5r3-

SOLUTION:

f'(x) :32x8 * LSxz 2x f" (x) :96x2 * 3ox 2

x2+7

f "'(x) :192x + 30 f+t(x) : tg2 / ( 5 ) ( r ): 0 n>S

f{n)(y)-0

Becausef '(x) gives the rate of change of f (x) per unit change in x, ' '(r) per f " (x), being the derivative of f (x), gives the rate of change of f unit change in r. The second derivative f " (x) is expressed in units of f'(x) per unit of x, which is units of f(x) per unit of x, per unit of r. For example, in straight-line motion, it f (t) feet is the distance of a particle from the origin at f seconds, then f '(t) feet per second is the velocity of the particle at f seconds, and f " (t) feet per second per second is the instantaneous rate of change of the velocity at f seconds. In physics, the instantaneous rate of change of the velocity is called the instantaneousacceleration Therefore, if a particle is moving along a straight line according to the equation of motion s : f (t), where the instantaneous velocity at f sec is given by a ftisec and the instantaneous acceleration is given by a ftlsecz, then a is the first derivative of z with respect to t or, equivalently, the second derivative of s with respect to f; that is, z t:.DtS rxavrprn 2: A particle is moving along a straight line according to the equation of motion '1 4t ,, :;

-5

A : D rts: D tzs

SOLUTION:

r,t:D6:f

i-'T7--

2-'t+t

where s ft is the directed distance of the particle from the origin at f sec. If u ftlsec is the instantaneous velocity at f sec and a ft/secz is the instantaneous acceleration at f sec, find f, s, and a when a: 0.

and

A: Dp: Se tti n g a :0,

*' , t ! = , . (t + 1.)2 r -.,- 1 D' -t 2 w e have

( r + 1 ) 3 _8 _ (f + 1)3

o

or (f + 1)t: 8

8 (t+1)t

3.11DERIVATIVES OF HIGHER ORDER 159 from which the only real value of t is obtained from the principal cube root of 8, so that

t*l:2 When t:

or I

,:|(1)'+ #:2+ ll:

7

lf (x, y) is any point on the graph of y : /(x), then D"y gives the slope of the tangent line to the graph at the point (x, y).Thus, D"'A is the rate of changeof the slopeof the tangentline with respectto r at the point (x,y). rxavrprr 3: Let m(x) be the slope of the tangent line to the c u r v eA : x 3 - 2 x 2 * r a t t h e point (x,y).Find the instantaneous rate of change of.m per unit change in r at the point Q,2).

'1,. soLUTroN m: Dr! : 3xz - 4x * The instantaneous rate of change of. m per unit change in x is given by Drm or, equivalently, Dr'A. D rffi - D r' A :6x

- 4

At the point (2,2), D,'A:8. is 8 times the change in r.

Hence, at the point (2,2), the change in m

Further applications of the second derivative are its uses in the second-derivative test for relative extrema (Sec. 5.2) and the sketching of the graph of a function (Secs. 5.4 and 5.5). An important application of other higher-ordered derivatives is to determine infinite series as shown in Chapter 76. The following example illustrates how the second derivative is found for functions defined implicitly. rxlvrprs

4:

Given

4xz * 9Y2:36 find D,2y by implicit differentiation.

sot.urroN: Differentiating implicitly with respect to x, we have 8x *'l,8yD"y - 0 so that

Dr!:-fr

(1)

To find Dr'A, we find the derivative of a quotient and keep in mind that y Lsa function of r. So we'have (2)

THE DERIVATIVE

Substituting the value of D,y from Eq. (1) into Eq. (2), we get -36y'-

l6x2

8Ly"

Thus, -L(gyz * 4x2)

Dr'y:ft

(3)

Becauseany values of r and y satisfying Eq. (3) must also satisfy the original equation, we can replace (9y'* 4xz) by 36 and obtain

Drra:-#:-# Exercises 3.71 In Exercises1 through 10, find the first and second derivative of the function defined by the given equation. t.f(x):x5-2x3*x 2. F(x) :7xs - 8x2 3. S(s) :2sa - 4s3* 7s- 7 4. G(t):f3- P+t 5. f(x): \Eq 6.h(V):gffi 7. F(x): xztG-5x

8.g(t):lr+W

1

1

9 .G \ x ) : f f i r z

2- \n ru./(I) : rT yE 1 1 .F i n d D , 3 y i fA : f

-2f*r-5.

12. Find Dfs if s: @T.. {3.

Find D,"f (x)' it f(x) : i: (1-x)" "'

' ..1

,,

o-/ffIn. Find /ra)(r) if /(r) : i=T. 15. Find DraY if. A : x7t2- 2x5t2+ xrt2

15. Find D,su if u: aG2.. -')

17. Given x" + y": 1, show that Drzy:;118. Given xtg * yLt2:2, show that D"'y :

v

$.

19. Given y' + l:4' (a is a constant),find D,? in simplest form. 20. Given F* a2f = FV (a allrdb are constants), find D"!r in simplest fomr. 2t. Find the slope of the tangent line at each point of the graph of y : t' + f - 3f wnet€ the rate of change of the slope iB zerc, In B
BEVIEWEXERCISES 161 the direct€d distance of the particle from the origin at t 6ec. If ftlsec is the velocity and c ftlseC is the acceleration of the " partide at t sec, find ? and 4 in terms of l. Also find when the accelemtion is zero and the intervals of tine when the particle is moving toward the origin and $'hen it is moving away ftom the origin. 23. s: *tn - ztg+ 4t2 22. s: t3- 9t2+ l5t In Exercises 24 through 27, a patnde is moving along a Btraight Iine according to the Biven equation of motion, wherc s ft is the directed distance of the particle from the origin at t sec. Find the tiure when the instantan€ous acceleration is zeto, and then find the directed distance of the particle from the origin and the instantaneous velocity at this instant.

>o

2 ss. : f f i - z u f , f

2 4 .s = 2 t 3 - 6 t 2 + 3 t - 4 , f > 0 -

{2?. s: $Ssrz + 2trt2,f > 0

2 6 .s : g t t + 2 \ / 2 t + l , t > 0

I n E xe rci ses 28t hr ough31 .,fi n d fo rm u l al os rf' (r) a n d f" (x ) andstatethedomai nsof f' and.f" (x'

t.rx# o zB. f(x):IFTi f x : 0 L0

ze. f(x): ff

3 0 .f ( x ) : l r l '

31.

lf; : 3

32. For the function of Exercise30 find f "' (x) when it exists. 33. For the function of Exercise31 find f "'(x) when it exists. 34. Show that if xA:1, then D,zy ' Dozx: 4. expressh"(r) in terms of the derivatives offand g. 35. fi f', g', f", and8" exist and if tt:/"9, 36. If / and g are two functions such that their first and secondderivatives exist and if h is the function definedby h(x): ' f(x) sG) ' Prove that

n" G\ : f(x) . g" (x) + 2f'(x) ' g'(x) + f" (x) ' g(x)

37. lI V = t, 38. If

whete z is any positive integer, Prove by mathematical induction that D'?:

n!

1, -

L-

2X

prcve by mathematical induction that D

nt :

t,rl -------:-i::-

(l - 2x't"*l

Reaiew Exercises(Chapter3) In Exercises1.through 14, find D"y. +2

l.y:fra

2.

3 . 4 x z* 4 y ' - U 3 : 0

4.y:\GV+\Fx

5.

6.

v:

x2 (x*2)2(4r-5)

162

THE DERIVATIVE

7.y:{tr t4

8.*y'*2y": x-2U

10. XztB* yzts:

g2t3

lL. y:

xffi t4-y: -7*.,

lrt* (xn*x)'ft

13.Y:f*

l ) s t z ( x z- 4 ) r r z

(x'-

15. A parttcle is movint in a straight line according to the equation of motion s: r3- llP + 24t + 100 where s ft i8 the direct€d distance of the particle fiom the starting point at , sec. (a) The partide is at the starting poht when t: 0. For what other values of t is the particle at the starting point? (b) Determine the velocity of the partide at each instant that it is at the startint point, and interpret the sign of the velocity in each case. + 4.C - r that have the slope l. 17. Find equations of the tangent line and normal line to the curve 2/ * 2y3 - 9xy : g at the point (2, 1). 16. Find equations of the tangent lines to the curve y:2f

18. An obiect is sliding down an hclined plane according to the equation of motion s : 121,* 5t wherc s ft is the dirccted distance of the obiect ftom the top I sec after stafting. (a) Find the velocity 3 sec after the start. (b) Find the initial velocity. 19. Using only the definition of a derivative find l'(r) it f (x') : \/-4x - 3. 20. Using only the definition of a derivative find /'(5) it f(x) : 19iiT1. 2 7 - F i n d f, ' ( x ) i f l ( r ) : ( l r + 1 1 - l r l ) , . 2 2 . F i n df , ( x \ i I f ( x ) : x - [ x \ . : 23. Given /(r) lrl3. 1a)Draw a sketch of the graph of f (b) Find lirn l(r) if ir existe. (c) Find l,(0) if it exists. 24. Given l(r) : t' sgn x. (a) Discuss the differentiability

of. f. (b) Is f' continuous on its domain?

25.Findl'(-3) if /(r): [r + ]lV9r.

26.Findt,(-3) iff(r):(lrl

27- Ptove tha! the line tantent to the curve y:-y' point, and find this point

- x)fl-sx.

+ Zf + r at the point (1, 2) is also tangent to the curve at another

28. Find an equation of the normal line to the eurve r - y:

at the point 13, l). fETl 29. A ball is thrown vertically uPward from the top of a houee 112 ft high. Its equation of motion is s : -15f + l)5t wherc s ft is the dirccted distance of the ball from the starting point at t sec. Find: (a) the in8tantaneous velocity of the batl at 2_sec; O) how high the ball will go; (c) how long it takes for the ball to reach the ground; (d) the instantaneous velocity of the ball when it reaches the ground. 30. Prove that the tangent lines to the curves 4Vs- t1y - x + sy:g

and,/ - 4y3+'Sxt y:0

at the origin are perpmdicular.

31. Giv€n (qf+b ((t\

:1

1

Ei

if r =l

ifx>1

Find the values of c and, 8o that/'(1)

exists.

32. Suppose IVJ:

(f lot"+bt +,

if x<1 if , =t

Find the values of a, ,, and c so that f"(1) exi8ts. 33. A sltip leavea a Port at 12 noon and havels due west at 20 knots. At 12 noon the next day, a second ship leaves the same

REVIEWEXERCISES 163

port and travels northwest at 15 knots. How fast are the two ships separating when the second ship has traveled 90 nautical miles? 34. A particle is moving in a straight line according to the equation of motion s: liT bF, whete a and D are positive constants. Prove that the measure of the acceleration of the particle is inveBely proportional to f for any ,.

35. A funnel in the form of a cone is 10 in. across the top and 8 in. deep. Water is flowing into the funnel at the rate of 12 in.3/sec,and out at the rate of4 in.8/sec.How fast is the surfaceof the waterrising when it i8 5 in. deep?

36. As the last car of a train passes under a bridge, an automobile cro8sesthe bridge on a rcadway pe4rendicular to the track and 30 ft above it. The train is traveling at the rate of 80 ft/sec and the automobile is traveling at the rate of 40 ftl6ec. How fast are the train and the automobile separating after 2 sec? 37. A partide is moving along the cuwe rl - / : 9, and the ordinate of the point of its position is inqeasing at the rate of 3 units per second.How fast is the abscissachanging at the point (5, -4)?

38. Supposey is the number of workers in the labor force needed to produce r units of a certain commodity, and x: 4yz. If the production of the commodity this year is 250,0Q0units and the prcduction is increa6ing at the rate of 18,000units per year, what is the current rate at which the labor force should be increased? 39. Using Boyle's law for the expansion of gas (Exercise 10 in Exercises 3.9), find the rate of change of the pressur€ of a certain gas at the instant when the pressure is 8 lb/in.? and the volume in 7(X)ff if the volume of the gas is increasint at the rate of 3 ff/min. 40. If the two functionE f and 8 arc differ€ntiable at the number .rr, is the composite function f. g necessarily differentiabl€ at 11?If your answer G yes, prove it. If your answer is no, tive a connters(ample. + lrl and g(r) : tr - *lrl. Prove that neith€r f'(0) norg'(0) exists but that (l " g)'(0) does exist. 41. Supposel(r):3r 42. Give an example of two functions f and g for which I is differentiable at g(0), 8 is not differentiable at 0, and l . g is diff€r€ntiable at 0. 43. Give an example of two functions f and g for whidr I is not diffet€ntiable at g(0), I is differentiable at 0, and I . g is differentiable at 0. ,14. In Exercise 27 of Exercises 3.3, you are to prcve that if / is differentiable at 4, then + Lt) - f(a - Ax) .. f(s ----------i;t (4, : lrm Show by using the absolute value function that it is possible for the limit in the above equation to exist even though f'(c) does not exist. 45. ll f' (x) exists, prove that

,^4k,):

tl?) : fe) _ xl.,(x)

46. Let I and g be two functions whose domains are the set of all real numbers. Furthermore, suppose that (i) 8(r) : x/(r) + 1; (ii) g(a + D) : gb) ' 8@) lor all a and.b; (iii) lim /(r) : 1. Prove that 8'(r) : 8(x). 47. The remainder theorem of elenientary algebra states that iI P(r) is a polynomial in .r and r is any real number, then thert is a polynomial Q(r) such that P(r) = Q(i) (r - r) + P(r). What is lim O(r)? tl8. Suppose g(r) : ll(r) l. If lt't(a)

exists and /(r)

{^t(*):ffif@G)

+ 0, prove that

Topicsonlimits, continuity and the derivative

4.1 LIMITSAT INFINITY 1 6 5

4.1 LIMITS AT INFINITY

Consider the function / defined by the equation

v

)'v2

rf \(-x- /\ : +X z+ . 1 , A sketch of the graph of this function is shown in Fig. 4.L.7.Let x take on the values 0, L, 2, 3, 4, 5, 10, L00, 1000, and so on, allowing x to increase without bound. The corresponding functionvalues are given inTable 4.L.1..

F i g u r e4 . 1 . 1

T a b l e4 . 1 . 1

10

100

1000

x

0

1

2

3

f(x)

û

.t

g 5

18 32 50 200 2o,ooo 2,000,000 10 rr % 101 1o"oo1 lpoo"ool

4

5

We see from Table 4.1..1thatas r increasesthrough positive values, the function values f (x) get closer and closer to 2. In particular, when x:4, A2x2n322

z-

r:

x2+l:z-

17:O

Therefore, the difference between 2 and /(r) is # when x: 100, 2f

4. When

20,000

z - x z + r - zA- L o , o o L10,00L : Hence, the difference between 2 and /(r) is 21t0,001when r: L00. Continuing on, we intuitively see that we can make the value of f (x'1 as close to 2 as we please by taking r large enough. In other words, we can make the difference between 2 andf (x) as small as we please by taking r large enough; or going a step further, for any e ) 0, however small, we can find a number N ) 0 such that If(x) - 2l < e whenever r ) N. When an independent variable r is increasing without bound through positive values, we write "x + *q." From the illustrative example above, then, we can say that

hm4:2 r-*oftL

In general, we have the following definition. 4.'1,.1,Definition

Let f be a function which is defined at every number in some intenral (a, +c,c'1. The limit of f (x), as x increaseswithout bound, is L, written

(1)

Jltf?):t

if for any € > 0, however small, there exists a number N ) 0 such that lf(*)-Ll

w h e n e v e rr ) N

166

TOPICSON LIMITS,CONTINU|ry,AND THE DERIVATIVE

NorE: When we write x + *m, it does not have the same meaning as, for instance, x --+ 1000. The symbolism "x -->*a" indicates the behavior of the variable x. However, w€ can read Eq. (1) as "the limit of f(x) as x approaches positive infinity is L," bearing in mind this note. Now let us consider the same function and let r take on the values 0, -L, -2, -3, -4, -5, -10, -L00, -1000, and so on, allowing x to decrease through negative values without bound. Table 4.7.2 gives the corresponding function values of f (x). Table 4.7.2

0 -1 ') v2

f(*):ft+I

-2

-3

-4

-5

8183250

5

10 tr%

-10

-100

-1000

200 20,000- 2,000,000 101 L0,001. 1,000,001

Observe that the function values are the same for the negative numbers as for the corresponding positive numbers. So we intuitively see that as x decreases without bound, f (x) approaches 2, and formally we say that for any e ) 0, however small, we can find a number N < 0 such that lf (x) 2l < , whenever x < N. Using the symbolism "x -->-@" to denote that the variable r is decreasing without bound, we write 2x2

1. jl*i+-2

In general, we have the following definition.

4.1.2 Definition

Let f be a function which is defined at every number in some interval (-*, a). The limit of f(x), as x decreasesraithoutbound,is L, written

f?):t ,t1T-

Q)

if for any € ) 0, however small, there exists a number N < 0 such that -Ll <. whenever r(N lf(*) NorE: As in the note following Definition4.r.L, the symbolism "x + only indicates the behavior of the variable x, but we can read Eq. "the limit of f (x) as r approaches negative infinity is L." Limit theorems 2, 4, 5, 6,7, 8, 9, and 10 given in Sec. 2.2 and theorems 11 and 12 given in Sec. 2.4 remain unchanged when "x + replaced by r-++co or x--->-oo. We have the following additional theorem. 4.1.3 Limit theorem 1,3 If r is any positive integer, then

(t) j: o "t1T_

(ii),11a4:o

-6" (2) as Limit a" is limit

4.1 LIMITSAT INFINITY

pRooF oE (i): To prove part (i), we must show that Definition 4.1,.1, holds tor f (x): tlx' and L: 0; that is, we must show that for any e ) 0 there exists a number N ) 0 such that lr

I

l+-01
w h e n e v e rr ) N

or, equivalently, 1

lrl"=Z

wheneverrlN

or, equivalently, since r ) 0, /1\rb

whenever r)N

ltl t(;/

In order for the above to hold, take N : (lle)rtr. Thus, we can concludethat rr | /l\ur l+ - 0l < e whenever x ) N, if N: l: I ter I lx' r This proves part (i). The proof of part (ii) is analogous and is left as an exercise (see Exercise 22).

rxlvrplr r. rm 'j-+@

L:

Find

4x-3 ;J= L* | J


sol,urroN: To use Limit theorem 1.3,we divide the numerator and the denominator by r, thus giving us 4x-3

"11

,.

4-3lx

zrc+s:"t1t7-+i|jlim (4 - 3lx)

-

(L.r. e)

t-lo

lim (2 + slx) t-l@

lim 4_

t-+6

Iim (3/r) t-*o

j1t2+"lit $tx') lim 4-

lim 3' lim (Ux\

lim 2 * lim 5 ' lim (Ux) J| -*@

_4-3.0 2+5.0 -2

(L.r.4)

Jf-j-o

(L.r. 6)

t-*@

(L.T.2 and L.T. 13)

168

TOPICS ON LIMITS, CONTINUITY, ANDTHEDERIVATIVE

EXAMPLE2: Find r.

2x2-r*5

,'14 4,,*l and when applicable indicate the limit theorems being used.

soLUrIoN: To use Limit theorem 13, we divide the numerator and the denominator by the highest power of x occurring in either the numerator or denominator, which in this case is 13. So we have r:* 2x2-x*5_

rj_ 2lx-'l,lxz*Slx}

"'14 Axs-L:"'1a@ lim (2lx - tlx2 + Slx")

-

t'-@

(L.r.e)

(4- ttxs)

Ji*

lim (zlx) - lim (Ux') * lim (5/r') 04-@

I+-@

lim 4 -

(1.r.5)

t+-@

lim (Ux')

Q--6

2 ' l i m (ux) - lim (Llf) * lim 5 . lim (Llx") I--@

I--@

4 - lim (Ux')

!t--6

(1.r.6)

fi--@

2. 0- 0 + 5 . 0

(L.T.2 and L. T. 13)

4-0

EXAMPLE

,.

riii:t

Find --

3x*4

r-l.a V2*

- 5


soLUrIoN:

We divide the numerator and the denominator of the fraction by r. In the denominator we let r: t/7 since we are considering only positive values of r.

r:_

3x*4

3+4lx

,:-

"'11ffi,:,t1tffilw

lim G + alx) lim 1D=W

(L.r.e)

I-lq

I--lq

lim G + alx) t-*a

lim

,(L.r. 10)

(2 - 5lx2)

f-*m

]g3*_li1 @tx) .T-*€

(1.r.4)

t-*a

lim 3* lim 4. lim (Tlx) t-*a

Vlim 2-

I-l@

,f,-*co

lim 5. lim (tlfl

(L.r. 6)

t-'ta

3+4.0

(L.T.2 andL.T.13)

4.1 LIMITSAT INFINITY 1 6 9

The function is the same as the one in Example 3; however, solurroN. -x or, since we are considering negative values of x, in this case #: equivalen tly, -t/P : r. Hence, in the first step when we divide numerin the denominator, and we ator and denominator bv x, we let x :-!* have ,.

3xt4

,:--

Ja6-s:*'11

3*4lx

ffi

lim (3 + alx) t--@

lim 1-t/ffi) lim 3* lim (alx) Jf--@

c--@

-@

The remaining steps are similar to those in the solirtion of Example 3, and in the denominator we obtain -tfZ instead of t/-2. Hence, the limit is

4trt.

"Infinite" limits at infinity can be considbred. There are formal definitions for each of the following. l i m /(r) : * m

l i m /(r) : *m

t--@

t-l@

l i m /(r) : - m

lim

/(x)

: -@

iX--@

fi-l@

For exampte, f@): *m if the function/is defined on some interval "lit (a, +*1 and if for any number N > 0 there exists an M > 0 such that f (x) > N whenever r > M. The other definitions are left as an exercise (seeExercise17). Find

EXAMpLE 5: ,2

lim-*

rr-*o X t

L

Dividing

solurroN; obtain

the numerator

Y2

= *rlim t.1 t w x ' lim _*: r - * c ox f L

and the denominatot

by x2, we

1

Evaluating the limit of the denominator, we have

l i m f t * +f ) : H m1X * r i m4 :o*o:o T

r-+-

\I

/

r-+*

r-+*

Therefore, the limit of the denominator is 0, and the denominator is approaching 0 through positive values. The limit of the numerator is 1, and so by Limit theorem lz(i) (2.4.4) it follows that ryz

lim _*: r-*a

X+

L

*cc

170

TOPICSON LIMITS,CONTINUIW,AND THE DERIVATIVE

EXAMPLE 6:

soLU'oN, ,rit#-*litf,ffi

Find

,l r.m . -2 x - *

We consider the limits of the numerator and the denominator separately.

x-*@ 5X + 5

/)\t

lim I'r-+o

1l:

\I

/

lim /q* 4): f/

r-+*

\I

lim:,-+*

lim 1:0-1:-1

X

r-+cc

tim !+ ti* i": o * o: o f

. . r - + c oI

*-+*

Therefore, we have the limit of a quotient in which the limit of the numerator is - L and the limit of the denominator,is 0, where the denominator is approaching 0 through positive values. By Limit theorem 1.2(iii) it follows that r.

llm r-+€

2x- xz

--:-fr, JX +

5

Exercises 4.1 In ExercisesL through'l'4, find the limits, and when applicable indicate the limit theorems being used. 4

a. r. u mr-+a

2x*l JX -

,^ . _r .l l f l4's32J* 3- I

L

{FT4

*-Zx*5

.,' X,i.T x s + x + 1

5. trm u**o y+4

,. 4f+2f-5 '',11 w

to.

,{1 fffi

x- x)

^ ,. 3xa-7f +2 o' 2x4+ l "1T. 11. lim (\7F; - fF + 1)

9. lim ,3-*@

12. lim [-

\FT

*o

r s .n ^ ?5 "y -* 34 y-a-

For the functions defined in Ex€rrises 15 and 16, prove that lim l(x) :1 by applying Definition 4.1.1; that is, for any e ) 0, show that there exists a number N > 0 such that lf (*l - 1l < . whenever x ) N.

r s .f ( x ) : *

16.f(x):H

17. Give a definition for each of the following:

(a) lim /(x) - -@; (b) lim f (x):

*o;

,f--6

18. Prove tlut lim

(f -4):+€

(c) lim /(r) : -oo f--@

by showing that for anyN > 0 there exists anM > 0 such t}at (f -4) > N whenever

x> M. 19. Prove that lim (6 - I - t') : -6 by applying the definition in Exercise 17(a).

l 4.2 HORIZONTALAND VERTICALASYMPTOTES

171

20. Prove part (i) of Limit theorem 12 (2.4.4) if "x + a" is replaced by "x + *a." 21. Prove that

,._ 8r*3 _n j1i 2'- 1 = by showing that for any e ) 0 there exists a number N < 0 such that

- l <. lry_+l lzx-1 whenever r ( N. 22. Prove part (ii) of Limit theorem 13 (4.1.3). fr,

c

fne function / is defined bY

f(x):]n#* Draw a sketch of the graph of f . At what values of r is / discontinuous?

4.2 HORIZONTAL AND An aid in drawing the sketch of the graph of a function is to find, if there VERTICAL ASYMPTOTES are any, the "horizontal and vertical asymptotes" of the graph. Consider the function / defined by

v

(1)

r

asymptote.

Figure4.2.1

y

A sketch of the graph of.f is in Fig. 4.2.1,.Any line parallel to and above the x axis will intersect this graph in two points: one point to the left of the line x: a and one point to the right of this line. Thus, for any k > 0, no matter how large, the line y: k will intersect the graph of f in two points; the distance of these two points from the line x: a gets smaller and smaller as k gets larger and larger. The line r : A is called a "vertical asymptote" of the graph of /. Following is the definition of a vertical

4.2.1 Definition

The line x: A is said to be a aerticalasymptoteof the graph of the function f if at least one of the following statementsis true: (i) lim f ( x ) : 4 a t-d+

(ii) lim 8'd+

(iii) lim tr-A-

f(x):-* f@)-a*^.

(iv) lim f ( x ) : - a . fi'A-

For the function defined by Eq. (1), both parts (i) and (iii) of the above definition are true. If g is the function defined by Figure 4.2.2

. q\ (. ,x/ ) : o

---1

(X _

A),

172

TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE

then both parts (ii) and (iv) are true, and the line r : A isa vertical asymptote of the graph of g. This is shown in Fig. 4.2.2. A "hotizontal asymptote" of a graph is a line parallel to the x axis. 4.2.2 Definition

The line A: b is said to be a horizontalasymptoteof the graph of the function f tf at least one of the following statementsis true:

(i) ,lit f @):u.

(ii) lim f (x):to. ExAMPLE L: Find the horizontal asymptotes of the graph of the function / defined by r/

r

I\x):ffi

I

and draw a sketch of the graph.

solurroN:

First we consider,lim

f (x), and we have

lirfe)-J1*# To evaluate this limit we write x: t/x' (x > 0, because r -> +co) and then divide the numerator and the denominator, under the radical sign, by xr. x t:--llm ....--=: r-*co Vxz * 1

1. --- \/P ltm .-J; Vf + | :

lim ,T-l@

(Ltf) -1

l

Therefore,by Definition a.2.2(1), the line y:

L is a horizontal asymptote. Now we consider (x); in this case we write x- - \/P because ,lim f if x --+-@, x < 0. So we have -l./&

lim/(r):,tiaffi :limt--a

t I * Llxz

(uf) --1 Figure4.2.3

Accordingly,by Definition 4.2.2(ii), the line y - -1 is a horizontal asymptote. A sketch of the graph is in Fig.4.2.2.

Exavprn 2: Find the vertical and horizontal asymptotes of the graph of the equation

solurroN:

y:t2

Solving the given equation for y, we obtain

4.2 HORIZONTALAND VERTICALASYMPTOTES

xy' - 2y' - 4x -- 0, and draw a

Equation (2) defines two functions:

sketch of the graph. U

v

: f ,(x)

where fi is defined by

f ,(x)

: *2

{-

and

- -ztlh A: fr(x) wheref2 is definedby fr(x) The graph of the given equation is composed of the graphs of the two functions f, and fr. The domains of the two functions consist of those > 0. By using the result of Example 8, values of x f.or which xl(x-z) we see that the domain of /r and fz is )c:2, excluding 1.7, and Sec.

(-*,0] u (2,+m). Figure 4.2.4

Now, consider /r necause

Ir2f,(x): ri^.21#:

*oo

by Definition 4.2.1,(i)the line x - 2 is a vertical asymptote of the graph of ft'

2rl#:,t11 zyftO:z ,tit f,@):*t1t Figure4.2.5

so by Definition 4.2.2(1)the line y : 2 is a horizontal asymptoteof the graph of ft. similarly,li1; fr@):2. A sketch of the graph of f is shown in Fig. 4.2.4.

Hence, by Definition 4.2.1,(ii)the line x: graph of fz.

2 is a vertical asymptote of the

so by Definition a.2.2(1)the line y : -2 is a horizontal asymptoteof the graph of fr. -2. A sketchof the graph of fris shown in Fig.4.2.5. Also, 11* fr(x): The graph of the given equation is the union of the graphs of.ft and fz, and a sketchis shown in Fig. 4.2.6.

4.2 Exercises In Exercises I through 14, find the horizontal and vertical asymptotes of the graph of the function defined by the givm equation, and draw a sketch of the graPh.

l.f(x):#

2. f (x):;,

_a

3. g(x):

_?

G;zy

174

TOPICS ON LIMITS, CONTINU|ry, ANDTHEDERIVATIVE

4. F(x):

x2*8x*-16 4x2

s .f ( x ) : # x _ 6 8./(r):;a

/ . h \ x ): 7 ; J

10.s(r):ffi6 /Ly2

6. G(r):

6x2*11.r-10

y2

11.F(x):A;+

A.f(x):w#fr

13.tf(x)-+r\, \/x"_2

In ExercisesL5 through 21, find the horizontal and vertical asymptotes of the graph of the given equation, and draw a sketch of the graph. 1 5 . 3 x y- 2 x - 4 y - 3 : 0 '/..8. Zxyz* 4y, - Jx:0 21. x2y * 6xy - f + 2x * 9y * 3 : 0

4'3 ADDITIONAL THEOREMS ON LIMITS OF FUNCTIONS

4'3'1'Theorem

1 6 .Z x y* 4 x - 3 y * 6 : 0 1 9 .( y ' - 1 ) ( x - 3 ) : 6

17. x2y2- x2+ 4yr:0 20. xyzI 3y, - 9r: 0

We now discuss five theorems which are needed to prove some rmportant theorems in later sections.After the statementof each theorem, a graphicalillustration is given. If liry /(x) existsand is positive, then there is an open interval containing a such that f (x) > 0 for every x # a in the interval. . rLLUsrRArroN1: Consider the function defined bv f

f(x) : 2 x - l A sketchof the graph of / is in Fig. 4.9.r. Becausetiq /(4 : L, and 1 > 0, according to Theorem 4.3.1 there is an open interval containing 3 such that /(r) ) 0 for every x * 3 in the interval. such an interval is (2, 4). Actually, any open interval (a,b)for which i = a< 3 and b > gwill do. o pRooFoF rHEoRa''4.3.7: Let L: tjT f@).By hypothesis,L > 0. Apply_ ing Definition z.r.r and taking e:!L, we know thereis a 6 > 0 such that

lf (*) - Ll < +f whenever 0 < l* - ol < a (1) Also, lf (x)- LI < lL is equivalentto -*L < f (x)- L < tf (refer to Theorcm'/-..2.2),which in turn is equivalent to

Figure4,3.1

tt
4.3 ADDITIONALTHEOREMSON LIMITSOF FUNCTIONS 175

in turn is equivalent to but

a-61x1a*D

x*a

which is equivalent to stating that r is in the open interval (a - 6, a * 6)

but

x * a

(3)

From statements (2) and (3), we can replace (1) by the statement +L < f(x)
when r is in the open intenral (a- 6, a * E) butx * a

Since L > 0,we have the conclusion that/(r) the open interval (a - 6,4 * 6). 4.3.2 Theorem

> 0 for every x * a in

tt tti /(r) exists and is negative, there is an open interval containing a such that /(x) < 0 for every x * a in the interval. The proof of this theorem is similar to the proof of Theorem4.3.Land is left as an exercise(SeeExerciseL). . rLLUsrRATroN 2: Let 6-x , \ g e c ): g _ , h - -4 ( 0; hence, Figure 4.3.2 shows a sketch of the graph of 8. ljT Sttl by Theorem 4.3.2there is an oPen interval containing 2 such that g(r) < 0 for every x # 2 in the interval. Such an interval is (8,3). Any oPen interval o (a,b) for which t = a <2 and} < b = 6 will suffice. theotem. The following theoremis sometimesreferredto as the squeeze

Figure4.3,2

4.3.3 Theorem

Suppose that the functions f , g, and h are defined on some oPen interval / containinga exceptpossiblyat a itself, and that /(x) < g(r) 'h(x) for all x in I for which x # a. Also suPPosethat lim f (x) andr11\r(r) both exist and are equal to L. Then li* S(t) also exists and is equal to L. The proof is left as an exercise(seeExercise2). o rLLUsrRArIoN3: Consider the functions f , 8, and h defined by -4(x - 2)''+ 3 f (x) : ^/-\_(x-2)(x2-4x1-7) 8\X):x_2 h(x):4(x-2)2+3 The graphs of the functions / and h are parabolas having their vertex at

176


fore satisfied, and it follows that lim g(r) - 3. c-2

o

Figure 4.3.3 4.3.4 Theorem

Supposethat the function / is defined on some open interval / containing a, except possibly at a. Also suppose that there is some number M fir which thereisaE ) 0suchthat/(x) <. llrlwhenever0( lrol < 6.Then, tt f Q) existsand is equal to L, L < M. IrT o TLLUSTRATIoN 4: Figure 4.3.4shows a sketch of the graph of a function / satisfying the hypothesis of Theorem 4.9.4. From the figure we see that /(1) is not defined, but / is defined on the open interval G, il exceptat 1.. Furthermorc,f(x) s I whenever 0 < lr - 1l < *. Thus, we can conclude from Theorem 4.3.4that if lim /(r) exists and is L, then L = *. From the figure we observe that th"rJii an L and it is 2. . PRooFoF rHEoRnrv4.3.4: We assumethat M < L and show that this as-

ON AN INTERVAL 177 4.4 CONTINUITY sumption leads to a contradiction.If M < L, there is some e ) 0 such that

M + e: L. Because fif lf(x)-Ll

w h e n e v e r0 < l r - a l

which is equivalent to L-e
w h e n e v e r0 < l x - a l

ReplaceL by M + e; it follows that there exists a 6r ) 0 such that (M *e)-e
w h e n e v e r0 ( l * - o l

<6'

or, equivalently, M
w h e n e v e r0 ( l * - o l

(4)

But, by hypothesis, there is a 6 such that (s) f (x) = tvt whenever 0 < l* ol < 6 Statements(4) and (5) contradict each other. Hence, our assumption that I . herefore,L'M. M < L is falseT

Figure4.3.4

4.3.5 Theorem

Supposethat the function / is defined on some open interval I containing a, exceptpossibly at a. Also suppose that there is some number M fot which there is a a ) 0 such that /(x) > M whenever 0 < l*- ol < E. > Then tt /(r) exists and is equal to L, L M' lT The proof is left as an exercise(SeeExercise3)' 5: Figure 4.3.4 also illustrates Theorem 4.3.5.From the o rLLUSrRArroN - 1l < i; and, becauseas figure we see that /(r) = I whenever 0 < lt (*, B) exceptat L, Theointerval pieviously stated, f is dehnedon the open o L > +' L, then rem 4.3.5statesthat if lim /(r) exists and is

4.3 Exercises 1. Prove Theorem 4.3.2. - cl ts sufficiently small, and € > 0, the following inequalities 2. prove Theorem 4.3.3. (rrrNr: Ftust show that when lr m u s t h o l d : L - e < f ( x ) < I * e , a n d L - e < h ( x )< L * ' e ' \ 3. Prove Theorem 4.3.5. and that l'(0) : 0. (IIrNr: UEeTheo4. Ler f be a function such that ll(r) | = I for all r. Prove that f is differentiable at 0 rem 4.3.3.)

ON 4.4 CONTINUITy INTERVAL AN

In Example 3 of Sec. 2.6 we showed that the function h, for which is continuous at every number in the oPen interval h(x): {4-P,

178

TOPICSON LIMITS, CONTINUIry, AND THEDERIVATIVE

' 4'4'l Definition

ExAMPLE1: If f (x) : U (x - g) , on what open intervals is f continuous?

(-2, 2). Because of this fact, we say that h is continuous on the open interval (-2, 2).Following is the general definition of continuity on an open interval. A function is said to be continuoLtson an open interaalif and only if it is continuous at every number in the open interval.

solurroN: The function / is continuous at every number except3. Hence, by Definition 4.4.1,/ is continuous on every open interval which does not contain the number 3.

We refer again to the function hfor which h(x) - {E4.We know that h is continuous on the open interval (-2,2). However, becauseh is not defined on any oPen interval containing either -2 or 2, we cannot consider lim h(r) or lim h(x).Hence, to discussthe question of the r--2

r-2

continuity of h on the closedinterval [-2,21, we must extend.the concept continuity to include continuity at an endpoint of a closed interval. _of We do this by first defining right-handcontinuity and,left-handcontinuity. 4'4'2 Definition

The function / is said to be continuous from the right at the numbera if. and only if the following three conditions are satisfied: (t) f (a) exists.

(ii)

exists. liT /(r)

(iii) lim f(x):f(o). 4.4.3 Definition

Th9 fungtion / is said to be continuous from the teft at the numbera if and. only if the following three conditions are satisfied: (i) f(a) exists.

Gt)I'T /(x) exists. (iii) lim f(x):f(a).

4.4.4 Definition

rxaupr.n 2: Prove that the function h, f.or which h(x): l/4- xr, is continuouson the closedinterval l-2,21.

A function whose domain includes the closed interval la, bl is said to be continuous on fa,b] if and only if it is continuous on the open interval (a, b), as well as continuous from the right at a and,continuous from the left at b.

soLUTroN: The function h is continuouson the open interval (-2,2) and

\/4-7:0: "ti1.

h(-2)

4.4 CONTINUIW ON AN INTERVAL

tim GE

179

- o: h(2)

I-2-

Thus, by Definition 4.4.4,h is continuous on the closedinterval l-2,2f . 4.4.5 Definition

nxavrprr 3: Given / is the function defined bY

f(x) : determine whether f is continuous or discontinuous on each of the following intervals:

(-3,2),?3,21,[-3,2),(-3,21.

(i) A function whose domain includes the interval half-open on the

right la, b) is said to be continuous on la, b) if and only if it is continuous on the open interval (a, b) and continuous from the rig}:rt at a. (ii) A function whose domain includes the interval half-open on the left (a, bl is said to be continuouson (a,bl if and only if it is continuous on the oPen interval (a,b) and continuous from the left at b. solurroN: We first determine the domain of /. Th" domain of / is the is nonnegative. Thus, any set of all numbers for which (2-x)l(3*x) of this fraction the denominator and numerator the which r for values of changes numerator The domain. the from excluded are signs have opposite -3. We : I : when sign changes denominator artd the 2, I sign when *ik" use of Table 4.4.L to determine when the fraction is positive, negative, zero,or undefined, from which we are able to determine the values of.x for which /(x) exists. The domain of / is then the interual half-open on the left (-3, 21. The function / is continuous on the oPen interval (-3,z).It is continuous from the left at 2 because

t2-'

l'_t{ffi:s:f(2) -3 because However, f i" not continuous from the right at b--

,ri1.rl'#:*rc We concludethat f is continuous on (-3, 2] and discontinuouson [-3,2] and [-3, 2). Table4.4.7

2- x

3*x

2- x 3+x

f(x)

x<-3 x: -3

+ 5

0

undefined

does not exist

-3 1x

+

+

+

+

x:2

0

5

0

0

2
does not exist

+

does not exist

180


ExAMrLE4: Given g(r) : firn and L < x < 3. Discussthe continuity of g.

solurroN: Figure 4.4.1,shows a sketch of the graph of g. Supposethat a is any number in (1, 2). Then f f u n : 1 , a n d l i m [t]l : L. Hence, if

t 1 u . 2,

fig(r)

limg(r):lim

: g(u), and so g is continuouson [r]l :f

(r,2).

:g(1)

Therefore,g is continuous from the right at 1..But lim g(r) : lim firn : t I-2-

r-2'

and g(2) :2. So 3 is not continuous from the left at 2. The function g is thereforecontinuous on the interval half-open on the right [1,2). similarly, g is continuous at every number u for *ni.r, 2 < u < j. At 2, g is continuous from the right since lim g(r) : g(z), but g is not

Figure4.4.1

continuous from the left at 3 becauselim Siij.:2 g is also continuous on the interv ul tZ., ii.

and g(3) : 3. Hence,

Exercises4.4 In Exercises 1 through 16, determine whether the function is continuous or discontinuous on each of the indicated intervals. ,

t. f (i :

;i-s; $, 7)' l-6, 4J.(-@,o), (-s, +-), I-s, +-), t-10,-s). r)-1

z. f(r):Flt

(0' 41,(-2,2), (--,-zl, (2,+@),l-4,4f, (-2,21. 3. s(r) : t@-- s;1--,-s1, (--, -31, (3,+oo),[s,+6), (_3,3). a . f @ ) : [ x n ; e r , b , G ,D , 0 , , 2 ) , 1 1 , 2 )( .7 , 2 ] . tr-ll

s. f(t)-fr

(-@,r),(--,1), t-r,tl, (-1,+e),(1,+@).

(2x-3 rrx<-2 l 6. h(x): ] | - 5 lr,-2- x - 1|l, (--, r), (-2, +*), (-2.'t), t-2, 1), l-2. Ll l3-x ifl
'*. (-2,2),t-2,21, t-2,2),(-2,21,(-@,-21,(2,+-).

6=-

e. t('l : ,,l'r:i e2,2), l-2.21,l-2,2), (-2,21,(--,-2), 12,+-).

e'r{il:6fu;?1,s), 10./(r) : ITIE=E; to_J

n. c(x):li_|;

(-r,31. t-1,31, t-1,3),

(-1,3), t-1,31,t-1,3), (-1,31.

(--.-3). (-3,3),t_3.31, (3,41, l_s,s),ls.41, t4,+*),(4,+_).

4 . 5 M A X I M U MA N D M I N I M U MV A L U E SO F A F U N C T I O N 1 8 1

t z + T ( - * , - 5 ) , (, - * , - 5 1, 1 - 5 , - 2 ) , 1 - 2 , 5 1 , 1 - 2 ,(5-)2, , 5 1(,- 2 , 5 ) ,[ s , + @ )( ,5 , + - ; ' t z .S @ ) :l # + ; In Exercises 13 through 20, determine the intervals on which the given function is continuous.

1 6 .F ( x ) :

1 3f .( x ) : h

tFE

Vr+5

17. The function of Exercise25 in Exercises1.8. 1.8. 18. The functionof Exercise26 in Exercises 1.7. 31 in Exercises 19. The functionof Exerciee 20. The function of Exercise34 in Exercises1.7. In Exercises21 through 28, functions f and g are defined. In eachexetcisedefine I " g, and determine all valuesof r for which f . I is continuous. 21.f (x) : x3;g(x) : li

22. f(x) : x2;8(x) : x2- 3

2s.f(x): t7;s(x): #

2 a .f @ ) : : . f x ; 8 @ ) : x + " l '

25. f (x): fi, s@): {i

26.f (x) : fl-xi8(x) : \tr + 1

v*1

sG): t/i

2 7f.( x ) : f i ,

2s.f (x\ : \ETI; g@): {x

(-oo, +6) and In Exercise8 29 through 32, find the values of the constants c and lr that male the function continuous on function' resulting the graph of draw a sketch of the lfx<2 if.2 < x

zsf(x):{*:T il;2i fx gr.f(x):lcx*k l-2x

ifr
3o/(r):

32.f(x):

ifx<-2 if-z
33. Given that f is defined by

f(x):{f,[;]il|-=::i If g is continuous on [c, b), and ft is continuous on [b, c], can we conclude thatf is continuous on [4, c] ? If your answ_er is-yes, prove it. If your answer is no, what additionat condition or conditions would assute continuity of f on la' c)? U. 1I f is the function defined by /(x) : 9!;, terval [0, +@).

where n and n are positive integers, prove that f is continuous on the in-

4.5 MAXIMUM AND We have seen that the geometrical interpretation of the derivative of a MINIMUM VALUES function is the slope of the tangent line to the graph of a function at a OF A FUNCTION point. This fact enables us to apply derivatives as an aid in sketching graphs. For example, the derivative may be used to determine at what points the tangent line is horizontal; these are the points where the derivative is zero.Also, the derivative may be used to find the intervals for


which the graph of a function lies above the tangent line and the intervals for which the graph lies below the tangent line. Beforeapplying the derivative to draw sketchesof graphs, we need some definitionr u.,d theorems. 4.5.1 Definition

The function f is said to have a relatiaemaximumaalue at c if there exists an open interval containing c, on which / is defined, such that /(c) > f (x) for all I in this interval. Figures 4.5.1,and 4.s.2 each show a sketch of a portion of the graph of a function having a relative maximum value at c.

x

4.5.2 Definition

Figure4.5.2

The function / is said to have a relatioe minimum aalue at c if there exists an open interval containing c, on which is defined, such that (c) = (x) / f f for all r in this interval. Figures 4.5.3 and 4.5.4 each show a sketch of a portion of the graph of a function having a relative minimum value at c.

ac

Figure4.5.3

Figure 4.5.4

If the function / has either a relative maximum or a relative minimum value at c, then / is said to have a relatiue extremum at c. (The plurals of maximum and minimum are maxima and minima; the plural of extremum is extrema.) The following theorem enables us to locate the possible values of c for which there is a relative extremum.

4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION

183

4.5.3 Theorem lf f(x) exists for all values of x in the open interval (a,b) and if / has a relative extremum at c, where a < c 1 b, then It f' (c) exists,f' (c) : 0. pRooF: The proof will be given for the casewhen / has a relative minimum value at c. tf f '(c) exists,from formula (5) of Sec.3.3 we have

(1)

f,(,):tjTfff Because / has a relative minimum existsaD)0suchthat

value at c, by Definition 4.5.2, there

w h e n e v e r(o l x - c l < 6

f(x)-f(c) =o

If r is approachingc from the right, x - c ) 0, and therefore f(c) = o f(x) x-c

whenevero ( r - c ( D

By Theorem 4.3.5,if the limit exists, ,.

fk)-f(c)

lrrll->U n-c+

)C

-

- n

(2)

C

Similarly, if x is approaching c from the left, x - c ( 0, and therefore f(x) f(c) -o x-c

whenever-6 < x- c < o

so that by Theorem 4.3.4,if the limit exists, Ilm-

;-c-

f(x) - f(c) x-

c

Sv

(3)

Becausef '(c) exists, the limits in inequalities (2) and (3) must be '(c). So from (2) we have equal, and both must be equal to f

f'(c) =o

(4)

and from (3),

f'(c)'o

(5)

Becauseboth (4) and (5) are taken to be true, we conclude that

f ' ( c ): o which was to be proved. The proof for the case when / has a relative maximum value at c is I similar and is left as an exercise (see Exercise 37). The geometrical interpretation of Theorem 4.5.3 is that if / has a rela-

184


tive extremum at c and if f ' (c) exists,then the graph of y : /(x) must have a horizontal tangent line at the point where x: c. If / is a differentiable function, then the only possible values of x for which f canhave a relative extremum are thosefor which f ' (x): 0. However, f ' (x) can be equal to zero for a specific value of x, and yet f may not have a relative extremum there, as shown in the following illustration. 1: Consider the function f defined bv r rLLUSrRArroN f(x):(r-1;' A sketch of the graph of this function is shown in Fig. 4.5.5. ,(x): f 3 ( x - L ) 2 ,a n d s o f ' ( 1 ) : 0 . H o w e v e r f, ( x ) < 0 , i f x . - - ' 1 ,a, n d / ( r ) j 0 , if. x > 1,.So / does not have a relative extremum at L. o A function / may have a relative extremum at a number and f, may fail to exist there. This situation is shown in Illustration 2.

Figure4.5.5

o rLLUSrRArroN2: Let the function f be defined as follows:

f(x): {';_: lf; .=i

x

A sketch of the graph of this function is shown in Fig. 4.5.6.The function / has a relative maximum value at 3. The derivativ" irorr, the left at 3 is given by f -(3) : 2, and the derivative from the right at 3 is given by f *(3) - -1. Therefore,we concludethat /'(3) does notlxist o Illustration 2 demonstrateswhy the condition ,'f ,(c) exists,,must be included in the hypothesis of Theorem 4.5.3. In summary, then, if. a function / is defined at a number c, anecessary condition fot f to have a relative extremum there is that either ,(c) :'O f or f '(c) does not exist. But we have noted that this condition is not sufficient.

Figure4.5.6

4.5.4 Definition

If c is a number in the domain of the function and if either , (c): f f f'(c) does not exist, then c is called a critical number of f.

0 or

Because of this definition and the previous discussion, we can conclude that a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number.

ExAMPLE1: Find the critical numbers of the function / defined by f (x) - x4ts+ 4xrt3.

SoLUTIoN:f,(x)-!*,,,+t*_,,":t*_,,'1x*|):w '(x): 0 when r_ -L, and f '(x) doesnot existwhen t: 0. Both-1 f and 0 are in the domain of f therefore, the critical numbers of f are -L and 0.

AND MINIMUM VALUES 4.5 MAXIMUM OF A FUNCTION 185 We are frequently concerned with a function which is defined on a given interval, and we wish to find the largest or smallest function value on the interual. These intervals can be either closed, open, or closed at one end and open at the other. The greatest function value on an interval is called the "absolute maximum vallJe," and the smallest function value on an interval is called the "absolute minimum value." Following are the precise definitions. 4.5.5 Definition

The function / is said to have an absolute maximum aalue on an interaal if there is some number c in the interval such that /( c) > f (x) tor all r in the interval. In such a case, f(c) is the absolute maximum value of / on the interval.

4.5.6 Definition

The function / is said to have an absoluteminimum aalue on an interaal if there is some number c in the interval such that /(c) = f (x) for all r in the interual. In such a case, f(c) is the absolute minimum value of / on the interval. An absolute extremum of. a function on an interval is either an absolute maximum value or an absolute minimum value of the function on the interval. A function may or may not have an absolute extremum on a given interval. In each of the following illustrations, a function and an interval are given, and we determine the absolute extrema of the function on the interval if there are any. o rLLUsrRArroN 3: Suppose / is the function defined by

Figure4.5.7

f(x) -2x A sketchof the graphof / on ll, 4) is shownin Fig.4.5.7.The function/ has an absoluteminimum valueof 2 on 11,4).Thereis no absolutemaxilim /(r) : 8, but /(r) is alwayslessthan mum valueof / on lL, 4),because 8 on the given interval. o rLLUsrRArIoN 4: Consider the function / defined by

f(x):-* A sketch of the graph of f on (-3,2f is shown in Fig. 4.5.8.The function / has an absolute maximum value of 0 on (-3' 21. There is no absolute minimum value of / on (-3,21, because lim /(r) -9,but, f(x) is always t'-Bo greater than -9 on the given interval. o TLLUSTRATIoN 5: The function / defined by Figure4.5.8

f(x) :

L-

x2

186

TOPICSON LIMITS,CONTINUITY,AND THE DERIVATIVE

has neither an absolute maximum value nor an absolute minimum value on (-1, 1). A sketch of the graph of / on (-1, 1) is shown in Fig. 4.s.9. Observe that ,t1T. f

@) : -a

and l_tT_fOl:

*oo

o

o rLLUSrRArroN 6: Let f be the function defined by

f(x): {X,*_rr* *,

ifx<1 'l-, if
A sketchof the graph of f on [-5, 4] is shown in Fig. 4.5.10.The absolute maximum value of / on [-5, 4] occurs at ']-.,and /(1) :2; the absolute minimum value of / on l-5, 4l occursat -5, and /(-5) : -4. Note that / has a relative maximum value at 1 and a relative minimum value at 3. Also, note that 1 is a critical number of f because/' doesnot exist at 1, and 3 is a critical number of f because/'(3) : 0. o

Figure4.5.9

F i g u re 4 .5.10 . rLLUSrRArroN 7: The function f defined bv 1

tf(x) \,/ : x_3 has neither an absolute maximum value nor an absolute minimum value o n [ L , 5 ] . s e e F i g . 4 . 5 . 1 1f o r a s k e t c h o f t h e g r a p h o f /.,11p f@):-a; so /(x) can be made less than any negative number by taking (3 - r) > 0 -- *a) ro and less than a suitable positive 6. Also, 1ry can be Ir1/tr) made greater than any positive number by taking (x - 3) > 0 and less than a suitable positive D. We may speak of an absolute extremum of a function when no interval is specified. In such a case we are referring to an absolute extremum of the function on the entire domain of the function.

F i g u r e4 . 5 . 1 1

4.5.7 Definition

/(c) is said to be the absolute maximum aalue of the function f if c is in the domain of f and it f (c) > f (*) for all varues of r in the domain of f.

4.5 MAXIMUMAND MINIMUMVALUESOF A FUNCTION 187

4.5.8 Definition

/(c) is said to be the absoluteminimumaalueof the function f if c is in the domain of / and if f (c) = f (x) for all values of r in the domain of /. . rLLUsrRArIoN 8: The graph of the function / defined by f(x):x2-4x*8 is a parabola, and a sketch is shown in Fig. 4.5.12. The lowest point of the parabola is at (2, 4), and the parabola oPens upward. The function has an absolute minimum value of 4 at 2. There is no absolute maxio mum value of /. Referring back to Illustrations 3-8, we see that the only case in which there is both an absolute maximum function value and an absolute minimum function value is in Illustration 5, where the function is continuous on the closed interval l-5,4]. In the other illustrations, either we do not have a closed interval or we do not have a continuous function. If a function is continuous on a closed interval, there is a theorem, called the extreme-oalue theorem, whrch assures us that the function has both an absolute maximum value and an absolute minimum value on the interval. The proof of this theorem is beyond the scope of this book, but we can state it without proof. You are referred to an advanced calculus text for the

F i g u r e4 . 5 . 1 2

proof.

4.5.9 Theorem Extreme-Vglue Theorem

If the function / is continuous on the closed interval la, bf , then / has an absolute maximum value and an absolute minimum value on la, blAn absolute extremum of a function continuous on a closed interval must be either a relative extremum or a function value at an endpoint of the interval. Because a necessary condition for a function to have a relative extremum at a number c is for c to be a critical number, we can determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval la,bl by the following procedure: (1) (2) (3) .

rxeuplr 2:

Given

f(x):x3+x2-x*l find the absolute extrema of f on

[_2,+1.

find the function values at the critical numbers of.f on la, bl; find the values of f(a) and f(b); the largest of the values from steps (1) and (2) is the absolute maximum value, and the smallest of the values from steps (L) and (2) is the absolute minimum value.

sor,urroN: Because/ is continuous on l-2, +1,the extreme-valuetheorem applies. To find the critical numbers of f *e first find /': '(x) :3x2 * 2x -'l' f ' will f (x) exists for all real numbers, and so the only critical numbers of /

188


be the valuesof x for which f '(x):0. ( 3 x - 1 ) ( x* 1 ) : s

Settingf ' (x) - 0, we have

from which we obtain and

The critical numbers of f are-1 and *, and each of these numbers is in the given closed interval l-2, +l.We find the function values at the critical numbers and at the endpoints of the interval, which are given in Table 4.5.7. The absolute maximum value of f on l-2, +l is therefore 2, which occurs at-L, and the absoluteminimum value of /on l-2,+l is-1, which occurs at the left endpoint -2. A sketch of the graph of this function on [-2, +l is shown in Fig. 4.5.73.

I

Table4.5.1 F i g u r e4 . 5 . 1 3

EXAMPLE 3:

Given

f (x) : (x - 21rr" find the absolute extrema of f on [1,5].

solurroN: applies.

Because / is continuous on Il, sl, the extreme-value theorem

rf ' (\ "x/ ) : *3J( x - 2-1-u a There is no value of x for which f'(x):0. However,becausef,(x) does not exist at 2, we conclude that 2 is a critical number of , so that the f absolute extrema occur either at 2 ot at one of the endpoints of the interval. The function values at these numbers are given in tabte 4.s.2. From the table we conclude that the absolute minimum value of / ot [L5] is 0, occurring at2, and the absolutemaximum value of /on [1,5J is179, occurringat 5-A sketchof the graph of this function on [1,5] is shown in Fig. 4.5.14. Table4.5.2

Exercises4.5 In ExercisesI through 10, find the critical numbers of the given function. 1. f (x) : x3I7x2 - 5x 2. f (x) :2x3 - 2x2- 1.6x-f 1.

3. f(x) : xa * 4x3- 2x2- 12x

INVOLVINGAN ABSOLUTEEXTREMUM ON A CLOSEDINTERVAL 189 4.6 APPLICATIONS

a. f@) : v7t3+ x4tl- 3xrt3 7. f (x): 1 0 .f t x ) :

(* - A)ztz

6. f(x) : xa* 1L13* 34* + 75x- 2 x

5.f(x)-x6t5-12rrts 8. f (x\:

(r' - 3f * 4;tte

e.flx):74

r*1 f _Sx+4

In Exercises 11 through 24, find the absolute extrema of the given function on the given interval' if there are any' and find the values of r at which the absolute extrema occur, Draw a eketch of the graph of the function on the interval. ,l

1,2.f (x) : x2- 2x -l 4; (-*, **)

73,f (x):|; ?L3J

1'a. f k) :;; [2,3)

15.f (x) : \/t4;

[-3, +-1

16. f(x):fi,

4 17. f (x) : ---^\,,; 12,51 J)'

18./(r) : \84';

e2,2)

1 9 f. ( x ) : l x - 4 l + r ; ( 0 , 6 )

r]1,'f (x)

- 4 - 3x;(-1, 2f 1

\x-

[2

]

irx* s[; t3,s] ifr:5J

20. f (x) : 14- 12l;(-co, 1m;

z r .f ( x ) : l r - 5 L2

23.f (x) : )c- [rn; (1,3)

2 a .f @ ) : U ( x )- U ( x - 1 ) ; ( - 1 , 1 )

?3,2)

22f(x):{r *'' ;i:=_l};e, t

In Exercises25 through 36, find the absolute maximum value and the absolute minimum value of the given function on the indicated intervul by the method used in Examples2 and 3 of this section. Draw a sketch of the graph of the function on the interval. 26. f (x): rs * 3x2- 9x; l-4, 4l 2 5.f (x): 13* 5x - 4; [ - 3, -1 ] 'l'6; 28. f (x) : x4- 8x2+ 16; l-1, 4l 27. f (x) : x4- 8x2+ l-4, 0l 30. f (x) : 74- 8xzI 16; l-3 ' 2l 2 9 .f ( x ) : x 4- 8 x 2* 1 6 ; 1 0 , 3 f

31.f (x): f* ,, l-1, 2)

t-l-6 32.f (x): fi; l-s,2l

3 3 .f ( x ) : ( r * 1 . ) 2 t 3 ; l - 2 , " 1 . 1

3 a .f @ ) - L -

35f (x): {'J-t lf;'=; :: t},r-3,3r

(x-31't'' l-5,41

36f (x): {rl- [; I ?]; ;l-:-=:=oa];r-0,o1

37. Prove Theorem 4.5.3 for the case when f has a relative maximum value at c.

4.S APPLICATIONS INVOLVING AN ABSOLUTE EXTREMUM ON A CLOSED INTERVAL

We consider some problerns in whidr the solution is air absolute extremum oI a function on a dosed interval. Use is made of the extremevalue theorem, which assures us that both an absolute maximum value and an absolute minimum value of a function exist on a closed interval if the function is continuous on that closed interval. The procedure is illustrated by some examples.

TOPICSON LIMITS, CONTINUITY, AND THEDERIVATIVE

L: A cardboard box EXAMPLE manufacturer wishes to make open boxes from pieces of cardboard 12 in. square by cutting equal squares from the four corners and turning up the sides. Find the length of the side of the square to be cut out in order to obtain a box of the largest possible volume.

i ft N

I N ri

rn.

r in. (L2 -zr) in.

solurroN: Let x:

the number of inches in the length of the side of the square to be cut out; V: the number of cubic inches in the volume of the box. The number of inches in the dimensions of the box are then r, (12 - 2x), and (12- 2r). Figurc 4.6.t representsa given piece of cardboard, and Fig. 4.6.2representsthe box. The volume of the box is the product of the three dimensions, and so V is a function of x, and we write V(x) : xc(Lz- 2x)(12 - 2x) (1) lf. x : 0, l/ :0, and if.x: 6, V :0. The value of r that we wish to find is in the closedinterval l0,6l.BecauseV is continuous on the closedinterval 10, 61,it follows from the extreme-valuetheorem that V has an absolute maximum value on this interval. We also know that this absolute maximum value of V must occur either at a critical number or at an endpoint of the interval. To find the critical numbers of V, we findV, (r), and then find the values of x for which either V' (x) : 0 or V, (x) does not exist. From Eq. (1), we obtain V (x) : LMx - 48x2* 4xg Thus, V' (x) : lM - 96x * l,xz

Figure4.6.1

V' (x) exists for all values of r. Setting V, (r) - 0, we have l2(*-8r*12):g from which we obtain x: 6 and x:2

k-trz

-2r) in.

Figure 4.6.2

The critical numbers of V are2 and 5, both of which are in the closed interval [0, 6]. The absolute maximum value of v on [0, 6] must occur at either a critical number or at an endpoint of the interval. BecauseV(0) - g and v(6) : 0, while v(2) : 128,we concludethat the absolutemaximum value of V on [0, 6] is L28,occurring at 2. Therefore, the largest possible volume is l2g in.B, and this is obtained when the length of the side of the square cut out is 2 in. We should emphasize that in Example 1 the existence of an absolute maximum value of I/ is guaranteed by the extreme-value theorem. In the following example, the existence of an absolute minimum value is guaranteed by the same theorem.

nxaprprr 2: An island is at point 4,6 miles offshore from

S O L U T I O N : Refer to Fig. 4.6.3.Let P be the point on the beach where the man lands. Therefore, the man rows from A to P and walks from p to c.

4.6 APPLICATIONSINVOLVINGAN ABSOLUTEEXTREMUMON A CLOSED INTERVAL

the nearest point B on a straight beach. A store is at point C,7 miles down the beach from B. If a man can row at the rate of 4 mi/hr and walk at the rate of 5 mi/hr, where should he land in order to go from the island to the store in the least possible time?

191

the number of miles in the distance from B to P. the number of hours in the time it takes the man to make the trip from A to C. Then T: the number of hours in the time to go from A to P plus the number of hours in the time to go from P to C. Because time is obtained by dividing distance by rate, we have Let r: T:

: _5 lAPl , -, 4

lPCl

(2)

From Fig. 4.6.3,we seethat triangle ABP. Therefore,

is the length of the hypotenuse of right

lAPl: \/F+ 35 We also see from the figure that lPe | : 7 - x. So from Eq. (2) T can be expressedas a function of x, and we have 6 miles

\456,7-x

T(x):--+?

Becausethe distance from B to C is 7 miles and becauseP can be any point on the line segment BC, we know that r is in the closed interual Figure4.6.3

n. 10,

We wish to find the value of x for which T has an absolute minimum value on [0, Zl. Because T is a continuous function of x on [0, 71,we know that such a value exists. The critical numbers of T are found by first computing T (x):

?- "\*'' (r):-L-L 4\/i, +-56 5 T' (x) existsfor all valuesof x. Setting T'(r) - 0 and solving for x, we have (3)

'zi::2':,:*' 5x:4\m

36

x2:64 r:

+8

The number -8 is an extraneous root of Eq. (3), and 8 is not in the interval l0,7l.Therefore, there are no critical numbers of T in [0, 7]. The absolute minimum value of T on 10,7) must therefore occur at an endpoint of the interval. Computing T(0) and T(7), we get

T(0): i3 and T(7): +\65

192

TOPICSON LIMITS, AND THEDERIVATIVE CONTINUIW,

Since +\65 < H, the absolute minimum value of T on [0, 7] is +\/85, occurring when x:7. Therefore, in order for the man to go from the island to the store in the least possible time, he should row directly there and do no walking. EXAMPLE3: A rectangular field is to be fenced off along the bank of a river and no fence is required along the river. If the material for the fence costs $4 per running foot for the two ends and $6 per running foot for the side parallel to the river, find the dimensions of the field of largest possible area that can be enclosed with $1800 worth of fence.

soLUrIoN: Let r: the number of feet in the length of an end of the field; a: the number of feet in the length of the side parallel to the river; A: the number of square feet in the area of the field. Hence, A: )cA

(4)

Since the cost of the material for each end is g4 per running foot and the length of an end is x ft, the total cost for the fence for each end is 4r dollars. Similarly, the total cost of the fence for the third side is 5y dollars. We have, then, 4x * 4x * 6y :1800

(s)

To expressA in terms of a single variable, we solve Eq. (5) for y in terms of x and substitute this value into Eq. (4), yielding A as a function of x, and A ( x ) : r ( 3 0 0- t x ) (6) If y - 0, x: 225,and if x: 0, A :300. Becauseboth r and y must be nonnegative, the value of x that will make A an absolute maximum is in the closed interval 10, 2251.BecauseA is continuous on the closed interval 10,2251,we conclude from the extreme-valuetheorem that A has an absolute maximum value on this interval. From Eq,.(6),we havei A ( x ) : 3 0 0 r- t x ' Hence, A ' ( x ) : 3 0 0- t x BecauseA'(x) exists for all x, the critical numbers of A will be found bv settingA'(x):0, which gives x:'l,t2t The only critical number of A is ILz+, which is in the closed interval 10,2251.Thus, the absolute maximum value of A must occur at either 0, tl2+, or 225.Because,4(0) :0 and A(225):0, while A(1,1,2t):16,875, we conclude that the absolute maximum value of A on [0, 22Slis L6,BT| occurring when x: LL2i and y: L50 (obtained from Eq. (5) by substi-

INVOLVINGAN ABSOLUTEEXTREMUMON A CLOSEDINTERVAL 193 4.6 APPLICATIONS

ExAMPLE4: In the planning of a coffee shop it is estimated that if there are places for 40 to 80 people, the weekly profit will be $8 per place. However, if the seating capacity is above 80 places, the weekly profit on each place will be decreasedbY 4 cents times the number of Placesabove 80. What should be the seating capacity in order to Yield the greatestweekly profit?

SoLUrroN: Let r: the number of placesin the seating capacity; P: the number of dollars in the total weekly profit. The value of P depends uPon r. and it is obtained by multiplying r by the number of dollars in the profit per place. When 40 < r < 80, $8 is the profit per place, and so P :8x. However, when r ) 80, the numthus giving ber of dollars in the profit per place is [8-0.04(r-80)], P - r[8 - 0.04(r - 80)] : 11.20x- 0-04x2.so we have if40 280,1L.20r- 0.04x2is negative. Even though x, by definition, is an integer, to have a continuous function we let x take on all real values in the intewall40,280l. Note that there is continuity at 80 because

llt

P(r): l11_t': 640

: jt1. P(x): l1T. Gt.20x 0.04x2\ 640 :540: from which it follows that the two-sided limit lim P(r)

P(80).

So P is continuous on the closed interval 140,2801andthe extreme-value theorem guaranteesan absolute'maximum value of P on this interval. :lL'20 When 40 < r ( 80,P'(r):8, and when 80 < x 1280, P'(x) - 0.08r.P'(80) doesnot exist sincePl(80) - 8 and Pi(80) : 4.80.Setting P'(x) : 0, we have 11.20-0.08r-0 x:'1.40 The critical numbers of P are then 80 and 1.40.Evaluating P(x) at the endpoints of the intenral [40,280]and at the critical numbers, we have P(40) : 920,P(ao;: 540,P(140):784, and P(280):0. The absolutemaximum value of.P, then, is 784 occurring when x:140The seating capacity should be L40places,which gives a total weekly profit of$784.

rxavpr,n 5: Find the dimensions of the right-circular cYlinder of greatest volume which can be inscribed in a right-circular cone with a radius of 5 in. and a height of L2 in.

solurroN:

Let r: h: v:

the number of inches in the radius of the cylinder; the number of inches in the height of the cylinder; the number of cubic inches in the volume of the

cylinder. the rylinder inscribed in the cone, and illustrates 4.6.4 Figure through the axis of the cone. plane a section 4.6.5illustrates

194

TOPICSON LIMITS,CONTINUITY, AND THE DERIVATIVE

lf r:0 and h: t2, we have a degeneratecylinder, which is the axis of the cone. rf r:5 and h: O,we alsohave a degeneratecylinder, which is a diameter of the base of the cone. We conclude that r is in the closed interval [0, 5] and h is in the closed interval 10,tzl. The following formula expressesv in terms of r and h: V:

nrzh

(7)

To expressV in terms of a single variable we need another equation involving r and h. From Fig. 1.6.s, and using similar triangles,w€ have 12-h:p r5 Figure4.6,4

Thus, , 50-L2r ":---

Substituting from Eg. (8) into formula (7), we obtain V as a function of r and write

I

V(r): #n(Sf - d)

N r N F{

rln

5 in.

Figure4.6.5

( 8)

with r in [0, 5]

(e)

Because V is continuous on the closed interval [0, 5], it follows from the extreme-value theorem that V has an absolute maximum value on this interval. The values of r and h that give this absolute maximum value for V are the numbers we wish to find. V'(r):En(l}r-3.rz) To find the critical numbers of V,we set V,(r):0

and solve for r:

r(1,0-3r):g from which we obtain

and BecauseV'(r) exists for all values of r,rhe only critical numbers of Vare 0 and *, both of which are in the closedintervil [0,5J.The absolutemaximum value of v on [0, 5] must occur at either 0, +, or 5. From Eq. (9) we obtain v(0) -0,v(+):#r, and y(5):0. we thereforeconcludethat the absolute maximum value of.v is a8szr.,and this occurs when r: *. When r: #, we find from Eq. (8) that h: 4. . ^Thtt, lh" greatestvolume of an inscribed cylinder in the given cone is "8n"t in.3, which occurs when the radius is # in. and the heiiht is 4 in.

Exercises 4.6 1. Find the area of the largest rectangle having a perimeter of 200 ft.

THEOREM 1 9 5 4.7 ROLLE'STHEOREMAND THE MEAN-VALUE 2. Find the area of the largest isosceles triangle having a perimeter of 18 in. 3 . A manufacturer of tin boxe8 wishes to make use of pieces of tin rvith dimeNions 8 in. by 15 in. by cutting equal Equar€8 from the four comers and tuming up the sides. Find the length of the side of the squarc to be cut out if an oPen box having the largest possible volume is to be obtained from each Piece of tin. 4 . A r€ctangular plot of ground is to be endosed by a fmce and then divided down the middle by another fence' If the fmce down thi middle costs $1.pet running foot and the other fence costs $2.50 Per running foot, find the dimen8ions of the plot of latgest possible arca that can be mcloEed with $al8oworth of fence. 5 . Points /{ and B are opposite each other on shoresof a Etraightriver that is 3 mi wide. Point C i9 on the sameshoreas B but 6 mi down the river from 8. A telephone courpany wishes to lay a cable from A to C. If the cost per nile of the cable is 25% nore under the water than it is on land, what line of eable would be least exp€nsive for the comPany? 6. golve Exercisa 5 if point C i8 only 3 mi down the river from B. 7. Solve Example 2 of this section if the store i8 9 mi down the beach from Point B. : u1/o7*_7 8 . Example 2 and Exercises 5, 6, and.7 are special cases of the following more general problem. Let f(r) milimum value of absolute for the > > that in order r, 0. Show f to occur at a + o(i- t), where r is in [0, b] and tl < na b\/F=e' must be satisfied: (0, inequality following b) the nuurber in the open interval 9 . Find the dimensions of the right-circular cylinder of greatest lateral sur{ace area that can be inscribed in a sPherc with a radius of 6 in. 10. Find the dimeneions of the right-circular cylinder of greatest volume that can be inscribed in a sphere with a radius of 6 in. point on the 1 1 .Given the cirrJe haviirg the equation f + 4 : 9, 6tt4 ,", the shortest distance from the Point (4' 5) to a the circle' on (4, 5) to a from the longist distance Point circle, and (b) the Point more than 8OOitems are Produced each week. The Profit item if not each profit of on a $20 12. A manufacturer can make to have declreases2 C€nts per item over 800. How many items shpuld the manufacturer produce eadl week in order the grcatest Profit? 13. A school-sponsored trip will cost each student $15 if not morc than 150 student8 make the lriP; however, the cost Per student wiil be reduced Qcents for each student in excessof 150. How many students should make the trip in order fur the school to receive the Large8tSross income? L4. Solve Exerciee 13 iI the reduction Per student in excess of 150 i5 7 cent8. 15. A pdvate club charges annual memhrship dues of $lfi).per qrember less 50 cents for each member over 600 and plus S0'cents for eactr mJmber less than 600. How many members would give the club the most revenue from annual dues? 15. Suppoae a weight is to be held 10 ft below a horizontal line AB by a wire in the shape of a Y If the Points A and B art 8 ft lpa*, wttai is the shortest total length of wire that can be used? 12. -' A piece of wire 10 ft long ie cut into two pieces. One piece i8 bent into the sh4pe of a circle and th€ other into the shaPe i; shouli the wire be cui eo that (a)-the combined area of the two fiSures i8 as small as possible and ;;;;;;;(b) the combined area of the two figues is as large as Possible? lg, Solve Exercise 17 if one piece of wire is bent into the shape of an equilateral tdangle and the other Piece is bent into the ehape of a square.

4.7 ROLLE,S THEOREM AND TIIE MEAN-VALUE THEOREM .

Letl be a function which is continuous on the closed interval [4, b], differ€ntiable on the open interval (a, b), and such thatf(a):0 andl(b):0. The French mathematician Michel Rolle (1652-1779) proved that if a

196


x

function / satisfies these conditions, there is at least one number c between a and b for which f '(c) : 0. Let us see what this means geometrically. Figure 4.7.1,shows a sketch of the graph of a function / that satisfies the conditions in the preceding ParagtaPh. We intuitively see that there is at least one point on the curve between the points (a,0) and (b,0) where the tangent line is parallelto the r axis; that is, the slope of the tangent line is zero. This situation is illustrated in Fig. 4.7.7 at the point P. So the abscissa of P would be the c such thaLf ' (c): 0. The function, whose graph is sketched in Fig. 4.7,1, not only is differentiable on the open interval (a, b) but also is differentiable at the end-

It is necessary, however, that the function be continuous at the endpoints of the interval to guarantee a horizontal tangent line at an interior point' Figure 4.7.3 shows a sketch of the graph of a function that is continuous on the interval [a, b) but discontinuous at b; the function is differentiable on the oPen interval (a, b), and the function values are zero at both a and b. However, there is no point at which the graph has a horizontal tangent line. We now state and prove Rolle,s theorem.

Figure 4.7.3

4,7.'1,Theorem Rolle's Theorem

Let f be a function such that (i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the open interval (a, b); (iii) f ( a ): f ( b ): 0 .

Then there is a number c in the open interval (a, b) such that

f ' ( c ): o PRooF: We consider two cases. Case1-: f (x) :0 for all r in la, bl. Then f '(x): 0 for all r in (a, b); therefore, any number between a and b can be taken f.or c. Case2: /(r) is not zero for some value of r in the open interval (a,b). Because / is continuous on the closed interval fa, bl, we know by _ Theorem 4.5.9 that f has an absolute maximum value on la, bl and an a b s o l u te mi ni mum val ue on l a, bl . B y hypothesi s, (r):-f(b):0. Fur _ f thermore , f (x) is not zero f.or some x in (a, b). Hence, we can conclude that / will have either a positive absolute maximum value at some c, in (a, b) or a negative absolute minimum value at some c, in (a, b), or both. Thus,

THEOREM 197 ANDTHEMEAN-VALUE THEOREM 4.7 ROLLE'S tot c: Cr, or c: cz as the case may be, we have an absolute extremum at an interior point of the interval la, bl.Therefore, the absolute extrernum ' f (c) is also a relative extremum , and because f (c) exists by hypothesis, it follows from Theorem 4.5.3 that f' (c) :0. This proves the theorem. I It should be noted that there may be more than one number in the open interval (a, b) for which the derivative of / is zero. This is illustrated geometricalty in Fig. 4.7.4, where there is a horizontal tangent line at the ' point where X: Crand also at the point where *: ,r, so that both f (ct) :0 '(tr) :0. a n df The converse of Rolle's theorem is not true. That is, we cannot conc l u d e th a t i f a functi on/i s such thatf' (c) -0, w i th a 1c ( b, then th e conditions (i), (ii), and (iii) must hold. (See Exercise 40.)

Figure 4.7.4

EXAMPLE

Given

f(x):4x3-9x verify that conditions (i), (ii), and (iii) of the hyoothesis of Rolle's theorem are satisfied for each of the following intervals: [-8, 01, [0,3r],and [-8, B].Then find a suitable value for c in each of these intervals for which

f

' ( c ): o '

sor,urroN: f'(x):12*-9; f'(x) existsfor all valuesof x, and so / is differentiable on (-oo, 1m) and thereforecontinuous on (-*, +*). Conditions (i) and (ii) of Rolle's theorem thus hold on any interval. To determine on which intervals condition (iii) holds, we find the values of x fot which f (x) - 0. Settingf (x) :0, we have Ax(f - *) :0 which gives us - _

rL-

_3

2

-_ 'r-

3 2

and b:0, we see that Rolle's theorem holds on [-8, 0]. Taking a:-E Rolle's theorem holds on [0, srland I-9, gl. Similarly, To find the suitable values for c, we set /'(r) - 0 and get - 9:0 12x2 which gives us

x - -+\E

and *: *15

-tfi. In the Therefore, in the interv al l-ar,0l a suitable choice for c is gl, possiare two there the interval interval lO, we take c: +\/3.In l-9, *i bilities for c: either -+\/g or t{5. We apply Rolle's theorem to prove one of the most important theorems in calculus-that known as the mean-aalue theorem (or law of the mean). The mean-value theorem is used to prove many theorems of both differential and integral calculus. You should become thoroughly familiar with the content of this theorem.

4.7.2 Theorem Let f be a function such that

Mean-Value Theorem

(i) it is continuous on the closed interval la, bl; (ii) it is differentiable on the oPen interval (a,b).

198

TOPICS ON LIMITS, CONTINUIry, ANDTHEDERIVATIVE Then there is a number c in the open interval (a,b) such that

-f(u) - f(')

' f'(c)

0-a

r r n l_ f @ )

f'(c):tff

Refer to Fig. 4.7.5. By taking the r axis along the line segment AB,we observethat the mean-value theorem is a generalization of Rolle's theorem, which is used in its proof. p,(b, f(b)) pRooF: An equation of the line through A and B in Fig. 4.7.s is

y-f(a):fffk_a) or, equivalently,

f ( b )- f ( a ) ( x a )+ f ( a ) Y:'-ff Figure4.7.5

Now if F(r) measuresthe verticaldistancebetweena point (x,/(r)) on lhe graph of the function / and the corresponding point on the secant line through A and B, then

F ( r ) :f ( x ) - f % P k - a ) - f ( o )

(1)

We show that this function F satisfies the three conditions of the hypothesis of Rolle's theorem.

But

(u|--lfu) F'(x): f '(x)-f

THEOREM199 ANDTHEMEAN-VALUE THEOREM 4.7ROLLE'S Thus

F'(c)- 1'G)- f(u)r=fJ'l Therefore, there is a number c rn (a, b) such that

o:f,(c)_fW or, equivalently, r,,.\

f(b)-f(a)

!f l\ c- ,t : - - - -

b_a

which was to be proved.

2: Given nxarurPr,n f(x):x3-5x2-3x verify that the hlryothesis of the mean-value theorem is satisfied for a: 1 and b : 3. Then find all numbers c in the oPen interval (1,3) such that

f'(c\:fW

solurroN: Because/ is a polynomial function, / is continuous and differentiable for all values of r. Therefore,the hypothesis of the mean-value theorem is satisfied for any a and b. - Lor - 3 f' (x) :3xz -z and (3) : -22 f f (t) : Hence,

_-27 _(-7):_l.o 2

3-1

' Settingf (c): -1.0, we obtain 3c2-10c-3:-L0 3c2-10c*7:0 (3c-7)(c-1):0 which gives us c:&

and c:t

Because1.is not in the open interval (1,3), the only possible value for c is 6.

nxavrpr.n3: Given

solurroN:

A sketch of the graph of f is shown in Fig. 4.7.6.

f (x) : xzrs draw a sketch of the graPh of f . Show that there is no number c in the open interval (-2,2) such that

Figure4.7.6

TOPICS ON LIMITS, CONTINUlry, ANDTHEDERIVATIVE

Ic ,\/c^)\:--f (22=) -l f ( - 2 ) Which condition of the hypothesis of the mean-value theorem fails to hold for f when a: -2 a n db : 2 ?

f (2)- f(-2)

4u3_ 4u3

2- (-2)

4 -0

There is no number c for which 2l3crts: e. The function / is continuous on the closedinterval l-2,21; however, is not differentiable on the open interval (-2,2) because / f,bl does not exist. Therefore, condition (ii) of the hypothesis of the mean-value theorem fails to hold for f when a: -2 and,b : 2.

Exercises 4.7 In Exercises1 through 4, verify that conditions (i), (ii), and (iii) of the hypothesis of Rolle's theorem are satisfied by the given function on the indicated interval' Then find a suitable value for i that satisfies the conclusion of Rolle,s theorem. r.f(x)-*-4x*3; : [1,3] 2.

f(x)

x3 2xz x * 2; lt, Zl

3. f(x) : x3- 2xz- x * 2; l-1, 2f a. f Q): .r3- 1.6x;l-4,01 5' For the function f defined bv f(r) : Ltt l 12* - x- 3, determine three sets of values for 4 and D so that conditions (i), (ii)' and (iii) of the hypotheiis of Rolle's *t r*L rina a suttabt".,,ar." to" ir, of tlr" th*" op.r. intervale (a, b) for which /, (c) : 0. "",itii"llir-",' " """t In 6 thtough 13, verify that the hypothesis of the mean-value theorem ,Exercises is satisfied for the given function on the indicated interval. Then find a ;uitable valu-e for c that sati"ri"" ti" *".ro.ior, of tle mean-value theorcm.

6.f(x):xs**-x;l-2,tl

7 . f (x ) : xz l 2x - l ; [0, 1]

9. f (x) = x2t3; [0, 1]

1,0.f (x) : ffifi' -];

12. f(x):ff,

1,J. f(x):ffi,?r,4f

12,61

l-6, 8l

8.f(x)-tc-r+!_y[*,s] 1,r.f (x): \ETz; 14,61

In Ex€rcises 14 through 23, (a) draw a sketch olthe_graph of the given function on the indicated interval; (b) test the three 6:T {itt) .oj 1ty hypothesis of _nol-e,itheorem ind determine which conditions are satisfied and rdhich, fi:l'l: !i)1 iia, if any, are not satisfied; and (c) if the three conditions in part (b) are satisfied, detenni i ne a point at which there is a hori-

zontal tangent line.

M. f (x) : y3t4- 2xtt4. 10, 4l

15. f(x) : y4t! - 3xrt3. [0, 3]

16. f(x):'#;F+,sl

if x=21. 1 7 .f ( x ) : [ x + l-3'71 -3

1,8. r(x): {t::X il,;:l.}; ?2, +l

lz * irz<;J;

THEOREM 4.7 ROLLE'STHEOREMAND THE MEAN.VALUE

fx ' - 5x * 4

2 0f.( x ) : l

,-L L0

i rx * , ] , ifx:U

t r ,n l

2 2 .f ( x ) : | - l r l ; [ - 1 , 1 ]

201

21,', f (x): {:.-t ;i::'*}, ?r, +) 23.f (x) : 19- 4x'l; l-*, *l

2a. lt f(x): (2r - 7)[Qx- 4) and if c:1 and b:2, show that there is no number c in the oPen interval (c, b) that satis- fiis the conclusion of the mean-value theorem. Which part of the hypothesis of the mean-value theorerr fails to hold? Draw a sketch of the graph of I on [1, 2]. The geometric interprctation of the mean-value theorem is that for a suitable c in the oPen interval- (4, b), the tangent (a,l(a)), and (b' f(b)).ln tine io the curve y =l(r) -30, at the point (c,l(c)) is parallel to the secantline through the points find a vllue of c satisfying the conclusion of the mean-value theorem, draw a sketch of the graph Exercises 25 through on the dosed interv al la , bl , and show the tangent line and secant line. 2 : 2 7 .f ( x \ : =;a:3.'!.,b:3.2 2 6 .f ( x ) : x 2 ) a : 2 , b : 4 79''f(x) x2;a: 3, b 5 x- 3

28.f(x) : x - t; a : "l'0,b : 26

Z !. f(x ) : x s- 9 x * l ; a: -3, b : 4

30. f (x) :

) fi;

a: 3.01,b : 3.02

For each of the functions in Exercises 31 through 34, there is no number c in the open interval (a, ,) that satisfies the conclusion of the mean-value theorem. In each exercise, det€mrine which pafi of the hypothesis of the mean-value theorem fails to hold. Draw a skekh of the 6taph ot y : f(x) and the line through the points (a, f(t)) and (b, f(b)).

ffJ f

( x ):

a: \, b:6 {, - sy,,

n i f @ ) : 3 ( r - 4 ) u e 'a : - 4 , b : 5

3 2f.( x ) : ' #

a:1,b:2

3afQ): {?;:tr.lf; : '.},,: -!, b:5

-6*+4t-7. Prove by Rolle's theorcm that the equation 4ti-6f : t - 2xr * ?l - x, then l'(t):4f 35, '' If f(r) (0, 1). open interval + +t* t:O has at least one r€al root in the 36, prove by Rolle,g theorem that the equation f + 2r + c = O,where c is any constant, cannot have more than one rcal root 37. Use Rolle,s theorem to prove that the equation4rl + 3d * 3r 2 : 0 has exactly one root that lies in the interval (0, 1). (0, 1) that is a root of the equation. Tlten assumethat there is number in least one is at that ihere First show (HrNr: more than one root of the equation in (0, 1) and show that this leads to a contradiction ) 38. Suppose s : /(t) is an equation of motion of a particle moving in a sttaight line where f_satisfies the-hypothesis of the meiir-value iheorem. Show that the condusion of the mean-value theorcm assures us that there will be some instant durint any time intel\'al when the instantan€ou8 velocity will equal the average velocity during that time interval. 39. Supposethat the tunction f is continuous on [a, D] and /'(r) : 1 for all r in (4, ]). Prove thatf(x): x- a + f(a) lot all x in [a, bl. ,l(). The converse of Rolle's theorem is not true, Make up an example of a function for which the condusion of Rolle's theorem is true and for which (a) condition (i) i8 not satisfied but conditions (ii) and (iii) are Eatisfied; (b) condition (D is not satisfied but conditions (i) and (iii) are satisfied; (c) condition (iii) is not satisfied but conditions (i) and (ii) are satisfied. Draw a skekh of the graph showing the horizontal tangent Une for each case. 41., Use Rolle,s theorem to prove that if every polynomial of the fourth degree has at most Iour real_roots, then every Poly(HINr; Assume a polynomial of the fifth degree has six real roots nomial of the fifth degree has at most tiiJ""it "oote. contradiction.) leads to a that this and show 42. Use the method of Exercise 41 and mathematical induction to prove that a polynomial of the nth degree has at most z real rcots.

202


ReaiewExercises (Chapter4) In Exercises1 through4, evaluate the limit. r'

,.

8r3-Sf+3

:1

2r+?x-4

3. tr- ffi .r--€

2x - 4

'A' , tr1. 1 4.

3x2-4x-l

6f +x'z+4

:im O/F+t-\/F+4)

In Exercises 5 and 6, find the horizontal and vertical asymptotes o{ the graph of the function defined by the grven equatron, and draw a sketch of the graph.

s.f(x):# In Exercises 7 thmugh 10, functions I and g are defined. In each exercise, define . g, and determine all values of x for which I is continuous.

f"g

7. f(x) : t/-x - 3; g(r) : x * 2 9. f(x):

sgn x; gQ) - x2- |

: li 8.f (x):;ft, s@) 10. f (x):

sgn x; gQ) : xz- ,c

In Exetcises 11 thrcugh 16, find the absolute extrema of the given function on the given interval, if there are any, and find the values of t at which the absolute extt€ma ocour. Draw a sketch of the graph oi the function on the interval.

rr. f (x): \6 + r; [-5,+*1 1 3f.( x ) : l e- * l ; [ - 2 , 3 ] 1 5. f ( x ) :

x 4- 12;z+ 3 5 ; [0 ,5 ]

12.f(x): t$- rz; (-3, 3) 1,a.f Q) : x4 -'l.2xz * 36; l-2,61

1 6rf\.( x' ) : f : : 1

lxz1_4

=':1I;?2,21 :i f' .L :< 'x < 2 )

17. Find two nonnegative numbers whose sum is 12 such that the sum of their squares is an absolute minimum. 18. A pieceof wire 80 in. long is bent to forrr a rectangle. Find the dimensions of the rectangle so that its area is as large as possible. 19. In order for a packageto be mailed by parcel post, the sum of the length and girth (the perimeter of a cro6ssection)must not be greater than 100 in. If a package is to be in the shape of a rectangulir box with a square cross section, find the dimensions of the package having the greatest possible volume that can be sent by parcel post. 20. If lis_ the function defined byJ(r) : l2x - 4l - 6, ttenl(-l) Rolle's theorem does not hold,

: l(5) : 0. However, /, never has the value 0. Show why

21,.(a) lf. f is a polynomial tunction and f(a): f(b) : f'(a):/,(D) : 0, prove that there are at least two numbers in the open interval (s, b) that are loot8 of the equation f,'(x) : O- (b) Show that the function defined by the equation - a)' satisfiespart (a,. f(t) : $ 22. If I is a polynomial function, ehow that between any two consecutive roots of the equation l,(r) one root of the equation l(r) : 0.

: 0 there is at most

23. SuPPose that / and I ar€ two func'tions that satisfy the hypothe8is of the mean-value theorem on [a, D]. Furthermore, - g
REVIEWEXERCISES 203

: gla) and f(b) : g,(b). Prove that there exists a number c in the open interval (4, b) such that l' (c) : 8' (c) ' f(a) (HrNr: Consider the function f - g') zs. lI f(x) : zttlP=

6i4

, find the horizontal and vertical asymptotes of the graph of l, and draw a sketch of the graph.

26. Let

[3x + 6a l f x- 3< - <3 x < 3 f(x):l3ax-7b if lx - 12b i f 3 < x a sketch of the Find the values of the constants a and D that make the function f continuous on (-€, +'o), and draw gnph of l. of a straight 27. Two towns A and B are to tet their water suPPly from the EamePumPing station to be located on_the bank mi aPartand A B are 20 to and river that is 15 mi from town.A and 10 mi fromiown B. If the points on the river nearest of least amount that the located so be A and B are on the same side of the river, where should the pumping station piping is required? table on the entire 28. A manufacturer offers to deliver to a dealer 300 tables at $90 Per table and to reduce the Price Per possible transaction largest in the order by Z5 cents for each additional table over 3fl). Find the dolar total involved these circumstances' between the manufactuter and the dealer under at every number 29. Give an example of a function whose domain is the set of all real numbers and that (a) i6 continuous from the discontinuous left at I but the from (c) continuous 0 and 1; number except at every except 0; (b) continuous right at 1.

Additional applications of the derivative

TEST AND THE FIRST-DERIVATIVE FUNCTIONS AND DECREASING 5.1 INCREASING

205

function f 5.1 INCREASING AND Suppose that Fig. 5.1.1.representsa sketch of the graph of a we have sketch this drawing In )crl. interval lxr, DECREASING FUNCTIONS foi itt r in the closed rr]. on [rr, AND THE FIRST. assumed that f is continuous DERIVATIVE TEST : f(x) G(xr, yr) D ( x + , Yn ) F 1 x 6y, 6 )

B ( x z, y z )

A(xt, y)

E(ru,Yt)

C(xt, Yt)

5 .' 1 .1 F i g u re Figure 5.1.1 shows that as a point moves along the curve from A to B, the function values increase as the abscissa increases, and that as a point moves along the curve from B to C, the function values decrease as the abscissa increases. we say, then, that / is "increasing" on the closed interval lxr, xr] and that f is"d,ecreasing" on the closed interval lxr, x"l' Following are the precise definitions of a function increasing or decreasing on an interval. 5.1.L Definition

A function / defined on an irrterval is said to be increasing on that interval if and only if f (*r) < f (xr)

whenevet x, I )c,

where r, and )cz ate any numbers in the interval' The function of Fig. 5.1.1 is increasing on the following closed intervals: lxr, xrf; lxr, xnl; lxr, xuf; lxu, xrl; lxr, xrls.1,2 Definition

A function / defined on an interval is said to be decressingon that interval if and only if f (*r) > f (rr)

whenevet xt I x,

where rr and x2 are any numbers in the interval. The function of Fig. 5.1.1 is decreasing on the following closed intervals: lxr, xt); fxn, x"f.

ADDITIONALAPPLICATIONS OF THE DERIVATIVE

If a function / is either increasing on an interval or decreasing on an interval, then I is said to be monotonicon the interval. Before stating a theorem that gives a test for determining if a given function is monotonic on an interval, let us see what is happining geometrically. Referring to Fig. 5.1.1,we observe that when tfre itope oittre tangent line is positive the function is increasing, and when the slope of the tangent-line is negative the function is decreasing. Because /,i:) is the slope of the tangent line to the cuwe y: f(x), iJ increasing when / f'(x) - 0, and f is decreasing when f'(x) < 0. Also, becausel'(rj is the rate of changeof the function valuesl(r) with respectto r, whenl,(x) > 0, the function values are increasing as r increases; and when /,(r) < 0, the function values are decreasing as .r increases. We have the iollowing theorem. .

5.1.3 Theorem

Let the function f be continuous on the closed interval la, bl and differentiable on the open interval (a, b): (i) { f'@) > 0 for all x n (a, b), then f is increasing on [a, b]; (ii) if l'(r) < 0 for all x in (a, b), then / is decreasingon [a, b]. pRooF oF (i): Let r, and r, be any two numbers in [a, b] such that r, < :r. Then / is continuous on [rr, rr] and differentiable on (xr, rr). From the mean-value theorem it follows that there is some number c in (xr, x2) such that

1' 1r'':f(x,)

-

f(x')

Because:, a rr, tn"^ rz- rr )0. Atso, f,(c) > 0 by hypothesis. Theretore f(x) - f(rr) > 0, and so f(xr) > f(xr). We have shown, then, that f(xr) < f(h) whenever x, < xr, where x., and r, are any numbers in the interval- [4, b]. Therefore, by Definition 5.1.1, it follows ti.rat is increasing / on fa, bl. The proof of part (ii) is similar and is left as an exercise(see Exer_ cise 34). I An immediate application of Theorem 5.1.1,is in the Droof of what is known as the first-deriaatioetestfor relafiaeextremaof a funcuon. 5.1'4 Theorem Let the function f be continuous at all points of the opm interval (a, b) Fitst-Derioatioe Test Extrena containing the number c, and suppose that lor Relatioe l' exists at'aii points of (a, b) except possibly at c: (1) if f' (x) > 0 for all values of r in some open intewal having c as its right endpoint, and if f'(x) < O for all values of r in iome open interval having c as its left endpoint, then f has a relative maximum value at c;

TEST ANDTHEFIRST-DERIVATIVE FUNCTIONS AND DECREASING 5 . 1 INCREASING

207

(ii) if f '(x) < 0 for all values of r in some oPen interval having c as its right endpoint, and if f ' (x) ) 0 for all values of x in some open interval having c as its left endpoint, then /"has a relative minimum value at c.

F i g u r e5 . 1 . 2

F i g u r e5 . 1 . 3

pRooF or (i): Let (d, c) be the interval having c as its right endpoint for which /'(r) > 0 for all r in the interval. It follows from Theorem 5.1.3(i) that / is increasing on ld, c). Let (c, e) be the interval having c as '(x) < 0 for all x in the interval. By Theorem its left endpoint for which f 5.1.3(ii) / is decreasing on lc, ef. Because/is increasing onld, c], it follows from DefinitionS.l..L that if x, is tnld, c] and xr* c, then f(xt) < f (c) .Also, because / is decreasing on fc, ef , it follows from Definition1.'1..2 that if x, is in [c, e] and xr* c, then f(c) > f(rr). Therefore, from Definition 4.5.L, / has a relative maximum value at c. The proof of part (ii) is similar to the proof of part (i), and it is left I as an exercise (see Exercise 35). The first-derivative test for relative extrema essentially states that if '(x) changes algebraic sign from positive to is continuous at c and f / negative as r increases through the number c, then / has a relative maxi'(x) changes algebraic sign from negative to posimum value at c; and tf f tive as r increases through c, then / has a relative minimum value at c. Figures 5.1.2 and 5.1.3 illustrate parts (i) and (ii), respectively, of Theorem 5.1,.4when /'(c) exists. Figure 5.1,.4shows a sketch of the graph '(c) of a function / that has a relative maximum value at a numbet c,but f d o e s n o te x i s t; how ever,f' (x) ) 0w hen x l c,andf' (r) < 0w hen x ) c . In Fig. 5.1.5, we show a sketch of the graph of a function / for which c is (r) <0whenxlc; acriticalnumber,andf'(x) < 0when xlc,andf' c. at a relative extremum not have does /

F i g u r e5 . 1 . 4

F i g u r e5 . 1 . 5

Further illustrations of Theorem 5.1,.4occur in Fig. 5.1.1. At x, and xo the function has a relative maximum value, and at 13 and rr the function

ADDITIONALAPPLICATIONS OF THEDERIVATIVE

has a relative minimum value; even though x6 is a critical number for the function, there is no relative extremum at 16. In summary, to determine the relative extrema of a function f: (1) Find f '(x). (2) Find the critical numbers of f, that is, the values of x for which ' (x): 0 or for w hi ch f ,(x) does not exi st. f (3) Apply the first-derivative test (Theorem 5.1.4). The following examples illustrate this procedure.

EXAMPLE

Given

f(x):x3-5x2*9x*L find the relative extrema of f by applying the first-derivative test. Determine the values of x at which the relative extrema occur, as well as the intervals on which / is increasing and the intervals on which / is decreasing.Draw a sketch of the graph.

solurroN, f' (r) - Bx2-I2x*9. '(x) exists for all values of r. Settingf , (r) : 0, we have f 3x2-'1,2x*9:0 3(r-3)(r-1):0 which gives us and Thus, the critical numbers of f are 1 and 3. To determine whether has a / relative extremum at either of these numbers, we apply the first-derivative test. The results are summarized in Table 5.1.1. We see from the table that 5 is a relative maximum value of occur/ ring at L, and 1 is a relative minimum value of occurring at 3. A sketch f of the graph is shown in Fig. 5.1.6. T a b l e5 . 1 . 1

f'(x)

Exeupln 2:

Given (-2-4 ifx<3 ,f(x) t \ - - /: l*^ if3
find the relative extrema of f by applying the first-derivative test.

Conclusion

x <'/.. x: "L

/ has a relative maximum value

t
/ is decreasing

x--3

/ has a relative minimum

3
/ is increasing

/ is increasing

value

s o r u r r o x : I f x 1 3 , f ' ( x ) : 2 x . I f x ) 3 , , ( x )- - 1 . B e c a u s e f f_(j):6 and f *(3):-L, we conclude that f' (3) does not exist. Therefore,3 is a critical number of /. Becausef'(x):0 when x:0, it follows that 0 is a critical number of f . Applying the first-derivative test, we summ arize the results in Table 5.L.2. A sketch of the graph is shown in Fig. S.I.Z.

AND DECREASING FUNCTIONSAND THE FIRST-DERIVATIVE TEST 5 . 1INCREASING Determine the values of x at which the relative extrema occur, as well as the intervals on which / is increasing and the intervals on which / is decreasing. Draw a sketch of the graph.

T a b l e5 . 1 . 2

(r) I

x:0

I -+

0
: x4ts+ 4'cu3 f (x) find the relative extrema of /, determine the values of x at which the relative extrema occur/ and determine the intervals on which / is increasing and the intervals on which / is decreasing. Draw a sketch of the graph.

I

3
|

'(x) -

r(0

x:3

f

5

Conclusion /isdecreasing

0

f has a relative minimum value

+

/isincreasing

does not exist -

/ has a relative maximum value /isdecreasing

+ 1). SOLUTIoN f '(x)- Lxrts* $v-us: $v'2t3(x '(r) - 0 when x: -t, Becausef ' (x) doesnot exist when x: 0, andf -1 and 0. We apply the first-derivative test the critical numbers of f are and summarizethe results in Table 5.1.3.A sketchof the graph is shown in Fig. 5.1.8. T a b l e5 . L . 3

(r) x1-l x: -"L

'(x)

Conclusion / is decreasing

-3

0

/ has a relative minimum value

-1
+

/ is increasing

r:0

does not exist

0
+

/ does not have a relative extremum at x: 0 / is increasing

210


Exercises 5.'L In Fxetcises 1 through 30, do each oI the following: (a) find the relative extrema of by applying the first-derivauve test / -on O) deteflnine the values of t at whidt the relative extrema occur; (c) determine tit" i"te*ar which I is increasing; (d) determine the intervals on which / is decreasing; (e) draw a sketch of the graph.

r.f(x)-f -4x-r

2 . f ( x ) : 1 3- 9 x zl 1 5 r - 5

3. f (x) :2x3 - x2+ 3x -"1.

a.fQ):x4+4x

5.f(x):x5-

6.f(x):2x+fi

1

5r3-20x-2

7.f(x): \/x-ffi1

y -

' )

8 . 'f ( x ) : " x-r z

9 . f ( x ) : 2x t / 3 - x 1 1 .f ( x ) : ( 1 , - x ) z ( l * r ) '

10.f(x): x\/fr L 2 .f ( x ) : ( r * 2 ) ' ( x- 1 ) '

13. f ( x ) : 2 - 3 ( x - A ) z t s

1 , af .@ ) : l -

15.

f(x):{T:: itr;=:

if x<-2 1 7 ./f \(.x ) : [ 2 1 + s Lx"+l if_2
1 , 6 .tf \( x /) :

(r- 1;'r'

( q-2x .l^" [3x-10

if x<3 if 3
(r * 5)'z if x < -4 1 , 8 .f ( x ) : { , : - (x -f 1)2 if.-4 < x rLz

ir5 < r

--. 20../ f(x\ \--': [!tr=T*W U2_ xz

if r < -3 if _3 < r

[3x+5 ifx<-t 2 1 ,f .( x ) : ] x 2 * ' t i f - 1 < x < 2

(r-6 2 2 .f ( x ) : J - \ [ 4 = l t = ) r 120-2x

ifx<6 if 6
112-(x*S)2 2a.fQ):J5-r

ifx<-3 if-3(x<-L

f

tiL *rr | 7- x

if 2 < x

((r+9)r-s

23.f(x): t-t/tr= 1@ t(x-2)r-r

i f x < -7 if -z < . x = 0 if0< x

tvmo--T=ry if -1 < r

2 5 .f ( x ) : y s t 4 + 1 0 x t t 4

26. f(x):y;tl-l1x2ts

2 7 .f ( x ) : ( r * l ) 2 t 3 ( x - 2 ) r r c ' 29. f(x) - rrrs(x* 41-ue'

28. f(x):

(4x- 6)trt(2x- a)ztt

30.f (x):

( r - a ) z r s* ,

31. Find 4 and b so that the function defined byl(r):13+

af*

b will have a relative extrernum at (2,3).

32. Find 4, ,, and c so that the funcdon defined by l(r) : af + bx+ c will have a relative maximum value of Zatlandthe graph oI y : f(x) will go through the point (2, -2). 33. Find c, b, c, and d so that th€ function defined byf(t) : af ibf f u I dwill have relative exhema at (1,2) and (2,3). 34. Prove Theorem 5.1.3(ii). prcve 35. Theorem 5.1.4(ii). 36. Given that the function f is continu""s 1o. values o{r,/(3) :2,f'(x) < 0 ifx < 3, and,f'(x) >0 if r > 3, draw a sketch of a Possible graph of I in each of_the. "l following casis, where the additional condition is satisfied: (a) is con/, t i n u o u sa t 3 ; ( b ) / ' ( x ) : - 1 . i l t < 3 , a n d f , ( x ) : 1 i f r > 3 ; ( c ) h 1 f , e ) : - 1 , I i - m /,(r) :1,andf,(a) + f''6j;lo * U. 32

/(1.- r)1 whe.re P gd 4 are positive integers greater than 1, prove each of the folowing: (a) if p is even, 9I* {1,1 : / has a relative minimum value at 0; (b) if 4 is evm, / has a relative rrinimum value at 1; (c) I has a Ghtive rraxirnum value at pl(p + q) whether p and 4 are odd or even.

TESTFORRELATIVE 5,2THESECOND-DERIVATIVE EXTREMA 211 38. Prcvethat if I is incrcasingon [a, b] andif 8 is increasingon lf(a'1,l(b\1, thenrt g. f existson [a, b], g . I is increasing on fa, bl. 39. The functionI is inqeasingon the intervalI. Prcve:(a) if g(x): -f(x\, then 3 is decreasingon 4 (b) il h(x):1'lf(t) andf(x) > 0 on I, then ft is decreasingon I. 40. The functionf i8 differcntiableat eachnumberin the closedinterval [a, b]. Provethat if f'(a) ' f'(b) < 0, therei8 a number c in the opn interval (c, b) such that l'(c) : 0. 5.2 THE SECOND- In Sec. 5.1. we learned how to determine whether a function f has a relaDERIVATIVE TEST FOR tive maximum value or a relative minimum value at a critical number c ' RELATIVE EXTREMA by checking the algebraic sign of f ut numbers in intervals to the left and right of c. Another test for relative extrema is one that involves only the critical numb er c, and often it is the easier test to apply. It is called the second-deriuatiae test for relatiae extrema, and it is stated in the following theorem. and let f ' exist for all values of r in some open interval containing c. Then if f " (c) exists and

'(c):0, 5.2.1Theorem Let c be a critical number of a function / at which f Second-DeriuatiueTest for Relatiae Extremct

(i) if f "(c) ( 0, then / has a relative maximum value at c; (ii) if f "(c) ) 0, then / has a relative minimum value at c. pRooF or (i):

By hypothesis, f " (c) exists and is negative; so we have

f,,(c):}Tfff.o Therefore, by Theorem 4.3.2, there is an oPen interval I containing c such that

.

f'(x)-f'(c)
(1)

for every x * c in the interval. Let I' be the open interval containing all values of r in I for which X 1 c; therefore, c is the right endpoint of the open interval /'. Let l" be the open interval containing all values of x in I for which x ) c; so c is the left endpoint of the open interval 1". Then if r is in I' , (x - c) ( 0, and it follows from inequality (1) that lf'(x) ',( x ) > f ' ( c ) . I f - x i s i n 1 " , ( x c ) > f ' ( c ) l > 0 o r , e q u i v a l e n t l yf '(x) '(c)l < 0 or, equivalently, 0, and it follows from (1) that lf f < f'(x) f'(r). '(r) > 0, and if But becausef '(c) : 0, we concludethat if x is in l' , f '(r) '(r) < 0. Therefore,f changesalgebraicsign from positive r is in 1" , f to negativeas r increasesthrough c, and so, by Theorem5.L.4,/hasa relative maximum value at c. The proof of part (ii) is similar and is left as an exercise (see ExerI cise L8).

212

ADDITIONAL APPLICATIONS OF THEDERIVATIVE

EXAMPLE

Given

SOLUTION:

f(x):xa*$x3-4x2 find the relative maxima and the relative minima of f by applying the second-derivativetest.

f'(x) :4xs * 4xz 8x f"(x):12x2*8r-8 Settingf '(x):0, we have

4 x ( x+ 2 )( x - 1 ) which gives

x:L

ff:0

Thus, the critical numbers of f are-2,0, and L. We determine whether or not there is a relative extremum at any of these critical numbers by finding the sign of the second derivative there. The results are summ arized. in Table 5.2.1. T a b l e5 .2 .1

f'(x)

-+

0

0

0


0

/ has a relative minimum value

_5

x

f" (*)

f(x)

3

T

Conclusion t' has a relative minimum

value

rf f " (c) : 0, as well as f '(r) :0, nothing can be concluded regarding a relative extremum of f at c. The following three illustrations lustify thii statement. .IL L U ST R A TToN7: If f(x)- x4, then f ' (x):4x3 and ,,(x):' !,2x2. Thus, f '(0), and ''(0) all have the value zero. By applying the first-derivative f (0), f f test we see that / has a relative minimum value at 0. A sketch of the graph of / is shown in Fig. 5.2.1. o

Figure5.2.1

Figure5.2.2

o IL L U S TR A TION 2: If 8(x):-xn, then g' (r) :-4x3 and g,,(r): - 12x2. Hence, g(0) : g'(0) : g"(0) : 0. In this case,g has a rerative maximum value at 0, as can be seen by applying the first-derivative test. A sketch of the graph of g is shown in Fig. 5.2.2. o . rLLUsrRArroN 3: If h(x) : r3, then h'(*) : 3x2 and,h" (*) : 6x; so h(0) : h'(0) : h" (0): 0. The function h does not have a relative extremum at 0 because if r < 0, h(x) < h(0); and if x ) o, h(x) > h(0). A sketch of the graph of h rs shown in Fig. 5.2.9. o

Figure5.2,3

In Illustrations 7,2, and 3 we have examples of three functions, each of which has zero for its second derivative at a number for which its first derivative is zero; yet one function has a relative minimum value at that

5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA

213

number, another function has a relative maximum value at that number, and the third function has neither a relative maximum value nor a relative minimum value at that number.

5.2 Exercises In Exercises 1 through 14, find the relative o(trema of the given function by using the second-derivative test, if it can be applied. If the second-derivative test cannot be applied, us€ the first-derivative test. 2.s@):13-sx*6 5. f(x) : (x - 4)2 8 . f(* ) : x (x - 1)'

1.f(x):3x2-2x*l a. h@) :2xs - %c2+ 27 7. G(x) : ( r - 3) a 1 0.f(x):

x\/R

13.F ( r ) : 6 x r t 3 - x z t l

11'.f(x) - lx6rtz+ x-rtz

3.f(x):-4x3+3f *'8r 6. G(x): (x * 2)3 9. h(x): xfr$ Ov2

1 2 .S U ) : ; * ;

- 4)' M. G(x) - xszts(a

15.G i v e n f ( x ) : x 3 + s r x * 5 , p r o v e t h a t ( a ) i f r ) 0 , / h a s n o r e l a t i v e e x t r e m a ; ( b ) i f r / - 0 , / h a s b o t h a r e l a t i v e m a x i m u m value and a relative minimum value. 16. Givenf(r):x'-1.r+ k, where r > 0 and r + 1, prove that (a) if 0 < r< 1,/hasa relativemaximumvalue at 1; (b) if r > 1.,f ha8 a relative minimwn valu€ at l. 77. Given f(x) : f * rr-1, prove that regardless of the value of r, f has a relative qrinimum value and no rclative maximurn value. 18. Prove Theorem 5.2.1(ii).

5.3 ADDITIONAL

The extreme-value thmrem (Iheorem 4.5.9) guarantees an absolute maximum value and an absolute minimum value for a function which is continuous on a dosed interval. We now consider some functions defned on intervals for which the extreme-value theorem does not apply and which may or may not have absolute extrema.

Given

solurroN: f is continuous on the interval [0, 6) becausethe only discontinuity of f is at 6, which is not in the interval. t, t ^"\_zx(ffi _ xz L2x* 27 _ (r 3) (r 9)

PROBLEMS INVOLVING ABSOLUTE EXTREMA

EXAMpLE l,:

f(x\:

t\re,

yz_27 x_5

find the absolute extrema of f on the interval [0, 5) if there are any.

I\x)

@:

1r_5y::-ffiy

' '(r) : 0 when x: 3 or 9; so f (*) exists for all values of r in 10,6), and f the only critical number of / in the interval [0, 5) is 3. The first-derivative test is applied to determine if f has a relative extremum at 3. The results are summarizedin Table 5.3.1. Because/ has a relative maximum value at 3, and f is increasing on the interval [0, 3) and decreasingon the interval (3,6), we conclude that

214


on [0, 6) f has an absolute maximum value at 3, and it is f(3), which is 6. Noting that lim f (x) : -m, w€ conclude that there is no absolute mini-

mum value of / on [0, 6). T a b l e5 .3 . 1

Conclusion

EXAMPLE

Given

f ( x \ : r r(-X., *

t vrz

*

6)2

find the absolute extrema of f on (0, +m; if there are any.

0
/ is increasing

x:3


3
/ is decreasing

solurroN:

f'(x)

l l

f is continuous for all values of x.

-"*'*

-r Q)? 4r'Q'* A --k2

e2*6)a

* 6) * 4x2- _3x'- 6

1xriffi:ft7y

f'g) existsfor all valuesof x. Settingf'(x):0, we obtain x:-+{2; so V2 is the only critical number of f in (0, +*;. The first-derivative test is applied at f2, and the results are summarizedin Table s.g.2. Because/ has a relative minimum value at f2 and because is de/ creasingon (0, {21 and increasinAon ({2,**), we conclude that f has an absolute minimum value at f1 on (0, +oo).The absolute minimum value is-#{2. There is no absolutemaximum value of f on (0, +*;. T a b l e5 . 3 . 2

f(x)

f'(r)

0
nxl.trapr-n3: Find the shortest distance from the point A(2, l) to a point on the parabola A : x', and find the point on the parabola that is closest to A.

Conclusion / is decreasing

_#\/i

o +

/ has a relative minimum

value

/ is increasing

soLUTroN: Let z: the number of units in the distance from the point AQ, +) to a point P(x, U) on the parabola. Refer to Fig. 5.3.L. (1) Because P(x, y) is on the parabola, its coordinates satisfy the equation A: x2. Substituting this value of y into Eq. (1), we obtain z as a function of x and write

z(x):@

5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA

z(x)- \E=ET+

215

(2)

BecauseP can be any point on the parabola, x is any real number. We wish to find the absoluteminimum value of z on (-*, +oo). -tt2:Zf' z' (x) : t(4x3- 4) (xn- 4x+i +) 4t

t, y)

A(r'+) x Figure5.3.1

- U t'" + - + tl \F@

z' (x) exists for all values of r because the denominator is z(x), which is never zero. Consequently, the critical numbers are obtained by setting z'(x): 0. The only real solution is 1; thus, L is the only critical number of z. We apply the first-derivative test and summarize the results in Table 5.3.3. T a b l e5 .3 .3

z(x)

z'(x)

-@
x:L Ilxafo

Conclusion z is decreasing

+f5

0

z has a relative minimum value

+

z is increasing

Because z has a relative minimum value at l, and becausez is decreasing on (-m, L) and increasing_on (L, +*;, w€ concludethat z has an absolute minimum value of tt/S at L. So the point on the parabola closest to A is (1, t).

We can show that the point AQ, +) lies on the normal line at (L, t) of the graph of the parabola. Becausethe slope of the tangent line at any point (x, y) of the parabola is 2x, the slope of the tangent line at (1, 1) is 2. Therefore, the slope of the normal line at (1, 1) is -*, which is the same as the slope of the line through A(2, *) and (1, 1).

In the next two examples, where the second-derivative test is used to determine the relative extrema, we use the following theorem to determine the absolute extrema.

S.3.1 Theorem

Let the function f b" continuous on the interval I containing the number c. It f(c) is a relative extremum of f on / and c is the only number in I for which / has a relative extremum, then /(c) is an absolute extremurnof / on I. Furthermore, (i) if /(c) is a relative maximum value of / on I, then /(c) is an absolute maximum value of f on I;

216


(ii) if /(c) is a relative minimum value of / on I, then /(c) is an absolute minimum value of f on I. PRooF: We prove part (i). The proof of part (ii) is similar. Because/(c) is a relative maximum value of on I, then by Definition / 4.5.1. there is an open interval I where I c I, and where / contains c, such that

f(c) = f(x)

for all x e I

Because c is the only number in I for which f has a relative maximum value, it follows that

f(c)>f(k)

ifkelandk*c

(3)

In order to show that /(c) is an absolute maximum value of on l, f we show that if d is any number other than c in l, then (c) > f f(d). we assumethat

f(c) -f(d)

e)

and show that this assumption leads to a contradiction. Because d + c, then either c 1 d or d < c. we consider the case that c < d (the proof is similarifd (d) f if d e I and d * c, and consequently /(c) is an absolute maximum value offonl. I

nxavpr.n 4: A closed box with a square base is to have a volume of 2000 cubic inches. The material for the top and bottom of the box is to cost $3 per square inch, and the material for the sides is to cost $L.50 per square inch. If the cost of the material is to be the least, find the dimensions of the box.

solurroN:

Let r:

the number of inches in the length of a side of the square base; A : the number of inches in the depth of the box; C: the number of dollars in the cost of the material. The total number of square inches in the combined area of the top and bottom is 2x2, and for the sides it is 4xy; so we have C:

3 (2xr) + E(4xV)

(5)

Because the volume of the box is the product of the area of the base and the depth, we have

x'Y:2ooo

5.3ADDITIONAL PROBLEMS INVOLVING ABSOLUTE EXTREMA 217 solving Eq. (6) for y in terms of r and substituting into Eq. (5) we get c:

6xz+

L2'ooo xQ)

x is in the interval (0, **), and C is a function of r, which is continuous on (0, +oo).From Eq. (Z) we obtain DrC: l2x -

12,000

(8)

xc2

DrC does not exist when x:0, but 0 is not in (0, +*). Hence, the only critical numbers will be those obtained by setting D,C: 0. The only real solution is L0; thus, 10 is the only critical number. To determine if x: 10 makes C a relative minimum we apply the second-derivative test. From Eq. (8) it follows that

Dr'C: !2 +24'q00 x3

The results of the second-derivative test are summarized in Table 5.9.4. Table5.3.4

r:

L0

D rC

D *,C

Conclusion

0

+

C has a relative minimum value

From Eq. (7) we see that C is a continuous function of x on (0, *m). Because the one and only relative extremum of C on (0, +m; is at x: 10, it follows from Theorem 5.3.1(ii) that this relative minimum value of C is the absolute minimum value of C. Hence, we conclude that the total cost of the material will be the least when the side of the square base is 10 in. and the depth is 20 in. In the preceding examples and in the exercisesof Sec.4.6, the variable for which we wished to find an absolute extremum was expressed as a function of only one variable. Sometimes this procedure is either too difficult or too laborious, or occasionally even impossible. Often the given information enables us to obtain two equations involving three variables. Instead of eliminating one of the variables, it may be more advantageous to differentiate implicitly. The following example illustrates this method.

nxarvrlr 5: If a closed tin can of specific volume is to be in the form of a right-circular cylinder, find the ratio of the height to the base radius if the least amount of material is to be used in its manufacture.

8:-, ' .

We wish to find a relationship between the height and the solurroN: base radius of the right-circular cylinder in order for the total surface area to be an absolute minimum for a fixed volume. Therefore, we consider the volume of the cylinder a constant. Let v: the number of cubic units in the volume of a cylinder (a constant). We now define the variables:

218

OF THE DERIVATIVE ADDITIONALAPPLICATIONS

r - the number of units in the base radius of the cylinder; 0 ( r < +e; h : the number of units in the height of the cylinder; 0 < h < +oc; S - the number of square units in the total surface area of the cylinder. We have the following equations: S:2nr2

(e)

* 2nrh

V : rrzh

(10)

BecauseV is a constant,we could solveEq. (10)for either r or h in terms of the other and substitute into Eq. (9), which will give us S as a function of one variable. The alternative method is to consider S as a function of two variables r and h; however, r and h arenot independent of eachother. That is, if we choose r as the independent variable, then S depends on r; also,h depends on r. Differentiating S and V with respect to r and bearing in mind that h is a function of. r, we have D,S:

(1 1 )

4nr * 2nh * 2nr D,h

and DrV :2nrh

(12)

* nrz Drh

Because V is a constant, DrV :0;

therefore, from Eq. (12) we have

Z n r h* n r 2 D r h : O with r # 0, and we can divide by r and solve f.or Drh, thus obtaining D,h: -;

)h

(13)

Substituting from Eq. (13)into Eq. (11),we obtain tD,s:zolzr+h+r(+)) DrS: 2n(2r - h) To find when S has a relative minimum obtain 2r * h: 0, which gives us

(14) value, we set D r S : 0

and

r: th To determine if this relationship between r and h makes S a relative minimum, w€ apply the second-derivative test. Then from Eq. (14) we find that D,'S:2n(2-D,h)

(1s)

5.3 ADDITIONALPROBLEMSINVOLVINGABSOLUTEEXTREMA 2 1 9

Substituting from Eq. (13)into (15),we get D,2 5 : 2 r

olrL

e)l:2n(r.+)

The results ofrf the second-derivative test are summarized in Table 5.3.5. Table5.3.5

r:*h

I

I

DrS

Dr'S

0

+

Conclusion

Shasarelativeminimumvalue

From Eqs. (9) and (10) we see that S is a continuous function of r on (0, +"o;. Because the one and only relative extremum of s on (0, *co) is at r: +h, we conclude from Theorem 5.3.1(ii) that S has an absolute minimum value at r: ih. Therefore, the total surface area of the tin can will be least for a specific volume when the ratio of the height to the base radius is 2.

Exercises5.3 In Exercises 1 through8, find the

absolute extrema of the given function on the given interval if there are any.

L. f (x) : x2)(-3,2l

2 . g (x ): 1 3* 2 x2- 4x * L; ( - 3 , 2 )

3.F(r):#,

a.f@):fi3,?4,-rl

5. S(r) : 4x2- 2x * 1; (-* , + @ )

6. G(x) : (x - S )zrs'(--, * * )

7. f (x) :

B.f(x):ffi,F*,4)

k";

+-; 4yrr,[0,

V4,41

9. A rectangular field, having an area of 27C0ydz, is to be enclosed by a fence, and an additional fence is to be used to divide the field down the middle. If the cost of the fence down the middle is 92 per rururing yard, and the fence alont the sides costs $3 per running yard, find the dimensions of the field so that the cost of the fencing will be the least. 10. A rectangular open tank i8 to have a squarc base, and its volume is to be 125 yd8. The cost per square yard for the bottom is $8 and for the sides is $4. Find the dimensione of the tank in order for the cost oflhe material to be the leaet. 1 1 . A box manufacturer is to produce a closed box of specific volume whose base is a rectangle having a length that is three times its width. Find the most economical dimensions.

t2. Solve

Exercise 11 if the box is to have an open top.

13.A

funnel of specific volume is to be in the shape of a right-circular cone. Find fhe ratio of the height to the base radius if the least amount of material is to be used in its manufacture.

't4. A

Norman window consists of a rectangle surmounted by a semicircle. Find the shape of such a window admit the most light for a given perimeter.

that will

22O

ADDITIONAL APPLICATIONS OF THE DERIVATIVE

15, Solve Exercise 14 if the window is such that the Bemicircle transmits only half as much light per square unit of area as the rectantle. 16, The strength of a rectantulat beam is proportional to the breadth and the square of its depth. Find the dimensions of the strong*t beam that can be cut from a log in the shape of a right-circular cylinder of radius c in. 17. The stifbnees of a Ectangular beam fu proportional to the breadth and the cube of its depth. I4rhat is the shape ofthe stiffest beam that can be cut from a log in the shape of a dght-circular cylinder? L8. A page of print is to contain 24 in.z of printed area, a margin of 1+in. at the top and bottom, and a rrargin of 1 in. at the eides. What are the dimeruions oJ the snallest pate that wou.ld fi.ll tlese requirements? 19. A one-story building having a rectangular floor space of 13,200 fe is to be constructed where a 22-ft easerrent is re. quircd in the front and back and a 15-ft easement iE required on eadr side, Find the dimensions of the lot havint the least area on which this building can be located. 20. Find the point on the curve yr - t':1

that is do8est to the point (2,0).

21. A direct current generator has an electromotive force of E volts and an intemal rcsistance of / ohms, E and r are mnstants. If R ohms is the external resistance, the total resistance is (r * R) ohms and iI P wafts is the power, then

-^

ER

(r+x1r

What odemal resi8tance will consume the most powe!? 22. A ritht-cfuculal cone is to be inscribed in a sphere of givm radius. Find the ratio of the altitude to the bage radius of the cone of largest possible volume. 23. A right-circular cone is to be circumacribed about a sphere of given radius. Find the ratio of the altitude to the base radius of the cone of least possible volume. 24. Prove by the method of this section that the shortest distance from the point P1(r1, yr) to the line l, having the equation Ar+_By+ C: O,is lA4*Byl*' CV\/TTF. (nrnr: If sis the number oi uniis trom p. to a point p[.r, y) on t, then s will be an absolute ininimum when d is an absolute minimum.)

5'4 CONCAVITY AND POINTS OF INFLECTION

Figure 5.4.1,shows a sketch of the graph of a function / whose first and second derivatives exist on the closJd interval lxr, xrl. Becauseboth and f ' ' f are differentiable there, / and f are continuous on lxr, xzl. If we consider a point P moving along the graph of Fig. s.4.l from A to G, then the position of P varies as we increasex from r, to xr. As P B(x Uz)

G(xy vr) C(x", y t)

A ( xr , y r )

F ( x u ,V 6 ) E(xr,y t)

Figure5.4.1

D(xn,vn)

5.4 CONCAVITY AND POINTSOF INFLECTION 22'I moves along the graph from A to B, the slope of the tangent line to the graph is positive and is decreasing; that is, the tangent line is turning clockwise, and the graph lies below the tangent line. When the point F is at B, the slope of the tangent line is zero and is still decreasing. As P moves along the graph from B to C, the slope of the tangent line is negative and is still decreasing; the tangent line is still tuming clockwise, and the graph is below its tangent line. We say that the graph is "concave downward" from A to c. As P moves along the graph from c to D, the slope of the tangent line is negative and is increasing; thatis, the tangent line is turning counterclockwise, and the graph is above its tangent line. At D, the slope of the tangent line is zero and is still increasing. From D to E, the slope of the tangent line is positive and increasing; the tangent line is still turning counterclockwise, and the graph is above its tangent line. we say that the graph is "concave upward" from c to E. At the point c, the graph changes from concave downward to concave upward. Point C is called a "point of inflection." we have the following definitions. 5.4J1'Definition

The graph of a function / is said to be concaveupward at the point (c,f (c)) if f '(c) exists and if there is an open interval 1 containing c such that for all values of x * c in I the point (x,f (x)) on the graph is above the tangent line to the graph at (c, f (c)).

Definition

The graph of a function / is said to be concaae downward at the point (c,f (c)) if f'(c) exists and if there is an open interval lcontaining c such that for all values of x * c in l the point (x,f (x)) on the graph is below the tangent line to the graph at (c, f (t)).

l)

o ILLUSTRATToN 1: Figure 5.4.2 shows a sketch of a portion of the graph of a function / that is concave upward at the point (c, f(c)), and Fig. s.4.9 shows a sketch of a portion of the graph of a function / that is concave downward at the point (c,f(c)). o

Figure5.4.2

The graph of the function f of Fig. 5.4.1,is concave downward at all points (x, f(x)) for which x is in either of the following open intervals: (xt, x") or (xr, x6). Similarly, the graph of the function / in Fig. 5.4.i. is concave upward at all points (x, f(x)) for which r is in either (xs, xr) or (xu, xr). The following theorem gives a test for concavity.

Figure5.4.3

5.4.3 Theorem

Let f be a function which is differentiable taining c. Then

on some open interval con-

(i ) i f f" (c ) ) 0, the graph of /i s concaveupw ard at (c,f(t)); (ii) if f " (c) ( 0, the graph of / is concave downward at (c, f (c)).

OF THE DERIVATIVE ADDITIONALAPPLICATIONS PRooF or (i):

f,,(c):*Tfry Becausef"(c)>0, ,. f'(x)- f'(cl rmJ)0 tr-c

x-

c

Then, by Theorem 4.3.1, there is an open interval I containing c such that

f'(*)*f'(c) >o

(r)

x- c

foreveryx*cinl. Now consider the tangent line to the graph of f at the point (c, f (c)). An equation of this tangent line is

y : f ( c ) + f ' ( c )( x- c ) (c' f(c))

iQ(',f('))

Figure5.4.4

(2)

Let x be a number in the interval I such that r * c, and let Q be the point on the graph of / whose abscissa is r. Through Q draw a line parallel to the y axis, and let T be the point of intersection of this line with the tangent line (see Fig. 5.4.4). To prove that the graph of / is concave upward at (c, f (c)), we must show that the point Q is above the point T or, equivalently, that the directed distance TQ > 0 for all values of x * c in 1. TQ equals the ordinate of Q minus the ordinate of T. The ordinate of Q is f (x), and the ordinate of T is obtained from Eq. (2); so we have

r Q : f k ) - l f ( ' ) + f ' ( r ) ( x* c ) l r Q : l f ( x )- f ( c ) J- f ' ( c )( r - c )

(3)

From the mean-value theorem there exists some number d between x and c such that

f

, ( d )- f ( x ) f ( c ) ' x- c

That is, f(x)-f(c):f'(d)(x-c)

for some d between x and c

Substituting from Eq. (4) into (3), we have TQ- f'(d)(x- c)- f'(c)(x- c) TQ: k-c)lf'(d)-f'(c)l Because d is between x and c, d is in the interval /, and so by taking x:

(s) d

5.4 CONCAVIryAND POINTSOF INFLECTION in inequality (1), we obtain

f ' ( d ) _ -f ' ( r ) > d-

c

o

(6)

To prove that TQ ) 0, we show that both of the factors on the right side of Eq. (5) have the same sign. rf (x - c) > 0, then x ) c. And because d is between r and c, then d > c; therefore, from inequality (6), lf'(d)-f_'(c)l >0. If (r-c) <0, then x1c, andso d
Prove. The proof of part (ii) is similar and is left as an exercise (see Exerctse 27). t

{

T -t{ i

Figure5.4.6

5.4.4 Definition

The converse of Theorem 5.4.i is not true. For example, if / is the function defined by f(x) :1s4, the graph of / is concave upward at the point (0, 0) but f" (0) :0 (see Fig. s.2.1). Accordingly, a sufficient condition for the graph of a function / to be concave upward at the point (c, f (c)) is that f " (c) ) 0, but this is not a necessary condition. similarly, a sufficient-but not a necessary-condition that the graph of a function ,,(c) < 0. / be concave downward at the point (c, f(c)) is that f If there is a point on the graph of a function at which the sense of concavity changes, then the graph crosses its tangent line at this point, as shown in Figs. 5.4.5, 5.4.6, and 5.4.7. Such a point is called a "point of inflection." The point (c, f (c)) is apoint of inflection of the graph of the function/if the graph has a tangent line there, and if there exists an open interval / containing c such that if r is in /, then either ( i ) f " ( r ) < 0 i f x ( c a n d .f " ( x ) > 0 i f x ) c ; o r (ii) f"(r) > 0 if x ( c and f" (x) < 0 if x ) c o rLLUsrRArroN 2: Figure 5.4.5 illustrates a point of inflection where condition (i) of Definition 5.4.4 holds; in this case,the graph is concave downward at points immediately to the left of the point of inflection, and the graph is concave upward at points immediately to the right of the point of inflection. Condition (ii) is illustrated in Fig. 5.4.6, where the sense of concavity changes from upward to downward at the point of inflection. Figure 5.4.7 is another illustration of condition (i), where the sense of concavity changes from downward to upward at the point of inflection. Note that in Fig. 5.4.7 there is a horizontal tangent line at the point of inflection. .

OF THEDERIVATIVE APPLICATIONS ADDITIONAL For the graph in Fig. 5.4.1, there are points of inflection at C, E, and F. Definition 5.4.4 indicates nothing about the value of the second derivative of f at a point of inflection. The following theorem states that if the second derivative exists al a point of inflection, it must be zero there. 5.4.5 Theorem

If the function / is differentiable on some open interval containing c, and fi (c, f(c)) is a point of inflection of the graph of f, then if f" (c) exists,

/" (c)- o'

pRooF: Let g be the function such that g(r) - 1' k); then g'(x) : f " (x). Because(c, f (c)) is a point of inflection of the graph of /, then f " (x) changessign at c and so 8'(r) changessign at c. Therefore,by the firstderivative,test(Theorem5.1.4),g has a relative extremum at c, and c is a critical number of g. Becauseg' G) : f " (c) and sinceby hypothesisf " (c) exists, it follows that g'k) exists.Therefore,by Theorem4.5.3 g'G) :0 and f" (c) : 0, which is what we wanted to prove. I The converse of Theorem 5.4.5 is not true. That is, if the second derivative of a function is zero at a number c, it is not necessarily true that the graph of the function has a point of inflection at the point where x: c. This fact is shown in the following illustration. o rLLUsrRArroN 3: Consider the function / defined by f (x) : x4.f '(x) : 4x3 a n d f" (x):L2x2. Further, f " (0):0; but becausef" (x) > 0 i f x ( 0 and f"(x) > 0 if x) 0, the graph is concave upward at points on the graph immediately to the left of (0, 0) and at points immediately to the right of (0, 0). Consequently, (0, 0) is not a point of inflection. In Illustration 1 of Section 5.2 we showed that this function / has a relative minimum value at zero. Furthermore, the graph is concave upward at the point (0, 0) (see F i g . 5 .2 .1). o The graph of a function may have a point of inflection at a point, and the second derivative may fail to exist there, as shown in the next illustration. If / is the function defined by f (x) : tr1l3,then * $y-l tl f' (x ) : !x6-zt? and f" (x) :

. rLLUSrRerrox 4:

r ,rr6,r* rc /f\

-/l

Figure5.4.8

I

4--

(r) < 0. f " ( 0 ) d o e s n o t e x i s t ;b u t i f x < 0 , f " ( x ) > 0 , a n d t f x ) 0 , f " Hence, / has a point of inflection at (0, 0). A sketch of the graph of this function is shown in Fig. 5.4.8. Note that for this function /'(0) also fails to exist. The tangentline to the graph at(0,0) is the y axis. . In drawing a sketch of a graph having points of inflection, it is helpful to draw a segment of the tangent line at a point of inflection. Such a tangent line is called an inflectional tangent.

5.4 CONCAVITYAND POINTSOF INFLECTION

EXAMPLE 1: For the function in Example1 of Sec.5.1, find the points of inflection of the graph of the function, and determine where the graph is concaveupward and where it is concave downward.

SOLUTION:

f(x):x3-6x2*9x*1' f ' ( x ) : 3 x 2 r 2 x* 9 f " ( x ): 5 x - L 2 f " (x) existsfor all valuesof x; so the only possiblepoint of inflection is where f " (x): 0, which occurs at x:2. To determine whether there is a point of inflection at x:2, we must check to see if f " (x) changes sign; at the same time, we determine the concavity of the graph for the respective intervals. The results are summanzed in Table 5.4.1. Table5.4,1

f (x) -oo<

f

'(x)

Conclusion graph is concave downward

x12

x:2

-3

3

2{x(foo

Figure5.4.9

f " (x)

0

graph has a point of inflection

+

graph is concave upward

In Example L of Sec.5.1 we showed that / has a relative maximum value at L and a relative minimum value at 3. A sketch of the graph showing a segmentof the inflectional tangent is shown in Fig. 5.4.9.

Exruprn 2: If f (x) : (1 - 2x)t, find the points of inflection of the graph of f and determine where the graph is concave upward and where it is concave downward. Draw a sketch of the graph of f. v

SOLUTION:

f

'(x) - -6(1. * 2x)2

- 2x) f " (x) :24(l Becausef " (x) exists for all values of x, the only possible point of inflection is where f " (x) : 0, that is, at x: *.By using the results summarized in Table 5.4.2,we see that /"(x) changessign from rt+" to "-" at x: L, and so the graph has a point of inflection there. Note also that becausef '(+): 0, the graph has a horizontal tangent line at the point of inflection. A sketchof the graph is shown in Fig. 5.4.10. Table 5.4.2

f (x)

'(x)

f " (x)

Conclusion

-m< x
T

graph is concave upward

x:*

0


i
f

graph is concave downward


5.4 Exercises ln Exercises I through 10, determine whele the Faph of the given function is concave upward, where it is concave downward, and find the point8 of inflection if there are any. L.f(x\:x3+9x 3. F(r) : x4- 8xB+ 24xz

2.sG):.r3+3x2-3x-3

c.g(rr:7q

6 .G k ) : (x' , ^+? *41tn

a. f Q) :"l.6xa * 32xs* 24x2- 5x - 2o

F/rI

8. F(r) : (2x - 6)srz* t (*z if.x l

7.f(x):(x-))rrs (*2 :ff<0 9 . f/ (\ x ) : 1 ^ :' l-* " ifx>0

17. Il f(x) : sp a 65, detennine 4 and D so that the graph o{ / will have a point of inflection at (1, 2). 12. Itf(x):sf abf * cr, determine a, b, and c ao that the gaph of I will have a point of inflection at (1,2) and so that the slope of the inflectional tangent there w.ill be -2. 13. Ilf(r): af tb*l cr* d, detemdne a, b, c, a\d. d 80 that I will have a rel,ativeextremum at (0, g) and so that the graph ofl will have a point of inflection at (1, -1). 7a. fif(x): al +bf +cf + dr+ e, deternine the values of a, b, c, d, and.e so the graph of will have a point of inflec/ tion at (1, -1), have the origin on it, and be symmetric with respect to the y axis. In Exercises 15 through 24, draw a skekh of a portion of the graph ot a function through the point where r: c if the I given conditions are satisfied. If the conditions are incomplete or inconsistent, explain. It is assumed that is continuous f on some open interval containing c. 1 5 .f ' ( x ) > 0 r t x < c ; f ' ( x )

< 0 ifr > c;f"(x) < 0itx < c;f"(r)<

0if x> c.

1 6 .f ' ( x ) > 0 i t x < c ; f ' ( x ) > 0 i f r > c ; f " ( x ) > 0 i t x < c ; f , , ( t ) < 0 i f r > c . 7 7 .f " ( c ) : o rf t ( c ) : o ; 1 " ( x ) > 0 i f r < c ; 1 " ( x ) > o i I x > c . 1 8 .f ' ( c ) : 0 ; f ' ( x ) > 0 i f r < c;f"(x) > 0ifr>c. 79. f' (c) 0; f' (x) < 0 if r < c; f" (x) > 0 iI x > c.

m . f " ( c ): 0 ;f ' ( c ): * ; 1 " ( r ) > 0 i f r < c ; f , , ( x ) < o i t r > c . 21.l'(c) doesnot exist;f" (r\ >0itr < c;f,(x\ > 0 ifr > c. 22. l'(c) doesnot exisf /" (c) doesnot eist f" (x) < 0 ifr < c; f,'(r) > 0 ifr > c. 23. lim f'(x) - +co,lim l'(t) : o; f" (x\ > 0 ifr < c;f"(x) < 0 ifr > c. 24. lim f'(x):*@j

limf' (r):

-*;f"

(x) > 0 if r < c;f" (x) > 0 if x > c.

E. Draw a Bketchof the graph of a functionf for which l(x), f'(r), andf,,(r) exist and are positive for all r. 26. II f(t\ :3f + xlxl, prove that l" (0) does not exist but the graph of f is concaveupward everyrvhere. 27. Prove Theorem 5.4.3(ii). 28. SuPPos€ that I is a function for which l" (x) exists for all values of r in some open interval I and that at a number c in I, f" (r) : O and'l"' (c) €xists and is not zero. Prove that the point (c, l(c) ) is a point of inflection oI the graph of /. (HrNr; The proof is similar to the prcof of the second-derivative test (Theorem 5.2.1.).)

TO DRAWING A SKETCHOF THE GRAPHOF A FUNCTION 5.5 APPLICATIONS

5.5 APPLICATIONS TO DRAWING A SKETCH OF THE GRAPH OF A FUNCTION

nxanaptnL: Given f(x):x3-3**3 find: the relative extrema of f the points of inflection of the graph of f the intervals on which f is increasing;the intervals on which / is decreasing;where the graph is concaveupward; where the graph is concavedownward; and the slope of any inflectional tangent. Draw a sketch of the graph.

We now apply the discussionsin Secs.5.'1.,5.2,and 5.4 to drawing a sketch of the graph of a function. If we are given f (x) and wish to draw a sketchof the graph of.f ,we proceedas follows. First, findf' (r) and f"(x). Then the critical numbers of f arc the values of r in the domain of f for which either f '(x) does not exist or f ' (r):0. Next, apply the first-derivative test (Theorem 5.1.4)or the second-derivative test (Theorem 5.2.1)to determine whether we have at a critical number a-relativemaximum value, a relative minimum v3lue, or neither. To determine the intervals on which '(x) is positive; to de/ is increasing, we find the values.of.x for which f termine the intervals on which / is decreasing,we find the values of r for which f '(x) is negative. In determining the intervals on which / is monotonic, we also check the critical numbers at which / does not have a relative extremum. The values of r forwhich f"(*):0 or f"(x) does not exist give us the possible points of inflection, and we check to seeif f " (x) changessign at eachof these values of r to determine whether we actually have a point of inflectioh. The values of r for which f " (x) is positive and those for which f" (r) is negative will give us points at which the graph is concaveupward and points at which the graph is concavedownward. It is also helpful to find the slope of each inflectional tangent. It is suggestedthat all the information so obtained be incorporated into a table, as illustrated in the following examples.

S e t t i n gf ' ( x ) : 0 , w e o b sol,urroN f' (r)-3x2-5x; f" (r):5x-5. tain r : 0 and x: 2. Setting f " (x) : 0, we obtain x: t. In making the table,we considerthe points at which x:0, x:L, and r:2, and the intervals excluding these values of.x: -oo
0
t
2
Using the information in Table 5.5.1 and plotting a few obtain the sketch of the graph shown in Fig. 5.5.1.

Figure5.5.1

ADDITIONALAPPLICATIONS OF THE DERIVATIVE T a b l e5 . 5 . 1

f (x)

f

'(x)

-@
/ has a relative maximum value; graph is concave downward

0
/ is decreasing; graph is concave downward

-3

1
21x(*m

f (x) : Sxzt7 )c5t3 find: the relative extrem a of f; the points of inflection of the graph of f, the intervals on which / is increasing; the intervals on which / is decreasing; where the graph is concave upward; where the graph is concave downward; and the slope of any inflectional tangent. Draw a sketch of the graph.

Conclusion / is increasing; graph is concave downward

x:1

nxarvrplr 2: Given

f " (x)

/ is decreasing; graph has a point of inflection / is decreasing; graph is concave upward

-1

/ has a relative minimum value; graph is concave upward

+

/ is increasing; graph is concave upward

SOLUTION:

f'(x)'

- +x._rts and f" (x) : _-ff1s=ats ' f (x) doesnot exist when x : 0. Settingf ' (x): 0, we obtain x: 2. Therefore, the critical numbers of f are0 and z. f " (r) doesnot existwhen r : 0.

F i g u r e5 . 5 . 2

5.5 APPLICATIONS TO DRAWINGA SKETCHOF THE GRAPHOF A FUNCTION

setting f " (x): 0, we obtain x: -'1.. In making the table, we consider the points at which r: -L, x:0, and x:2, and the following intervals: -m< x<-1 -1 <.t<0 0
f(x)

f'(x)

f" (x)

-oo
/ is decreasing; graph is concave upward

x: -l

-)

/ is decreasing; graph has a point of inflection is decreasing; graph is / concave downward

-1
l:0

does not exist

+

0
x:2

Conclusion

21x({m

does not exist

/ has a relative minimum value / is increasing; graph is concave downward / has a relative maximum value; graph is concave downward / is decreasing; graph is concave downward

5.5 Exercises For each of the following functions find: the relative extrema of t the point8 of inflection of the graph ol f; the intervals on which f is inseasin& the intervals on which f is decreasing; where the graph is concave upward; where the traph is concave downward; the slope of any hflectional tangent. Draw a sketch of the graph. 2.f(x):x3 lx2-5x 1. f (x) :2)c3 - 6x -f L 3.f(x):)ca-2x3 -: a. f k) 1 3+ 5 x 2+ 3x- 4 3xa* 2x3 5 . /(r) 6. f(x) : 2x3- tx' - l2x * 1 7.f(x):2s4- 3x3*3*+l 8. /(r) : v4- 4x3+ L6x 9. f (x) : ixn - *r$ - x2+'l.. (v2 ifr<0 (-# ifx<0 x): {1 1 __ . 1.tf. (.\--, 1 2 .!f \( x') : l 10. f (x) :3xa * 4x3* 6x2- 4 : ,n l /x ," i f x > 0 Lxr if x>0 (-va

1v3. .r f\k- - \/ : l L

.* [rn

(tt*-11t

i.fx<0

:

if x >

0

ra.f Q): t?)^_,l-

ifx
if x > I

ls. /(r) : (r * 1)*(r -2)'

ADDITIONAL APPLICATIONS OF THE DERIVATIVE

1 6 .f ( x ) : * ( x + 4 ) ' 1 9 ,f ( x ) - 3 x 2 t 3 - 2 x

1 7 .f ( x ) : 3 x E * S x a 20- f(x) : t7rt3- , 2 3 .f ( x ) : 2 * ( x - 3 ) t r a

18. f(x):3xs * 5r3

X. f (x):2 * (r - 31'r' 28.f(x):3 * (x + r7'r'

26. f (x) :2 * (x - 31z' 2 9 .f ( x ) : 7 2 \ P y

2 1 .f ( x ) : 7 r t 3 a 2 r t t s 2a. f @) :2 t (x - 31nrc (x+1;uru 2 7 .f ( x ) : 3 * 30. f(x) v11{- rz

3t.f(x):H

32. f(x):h

33.f(x):

22.f(x):Jall-4*

(r * 2) \R

U.f(x):ffi 5'6 AN APPLICATION OF THE DERIVATIVE IN ECoNoMIcS

In economicsthe variation of one quantity with respectto another can be described by either an aaerageconceptor a margi;a/ concept. The average concePtexPressesthe variation of one quantity over a specified range of values of a second quantity, whereas the marginal concept is the instantaneous change in the first quantity that results from a very small unit change in the second quantity. For example, in our discussion of rectilinear motion (Sec.3.2) we considered both an averageconcept and a marginal concept. The averagevelocity in traveling a distanceof J miles in f hours is s/f mi/hr, whereas the instantaneous velocity is a marginal concePt as it gives the rate of change of s with respect to a small unit change in f at a particular instant of time. We begin our examplesin economics with the definitions of averagecost and marginal cosi. It should be clear that to define a marginal concept precisely we must use the notion of a limit, and this will lead to the derivative. Supposethat C(r) dollars is the total cost of producing r units of a commodity. The function C is called a total cost function. In normal circumstances,r and C(r) are nonnegative. If zero is in the domain of C, C(0) is called the oaerheadcost of production. Note that since r representsthe number of units of a commodity, r must be a nonnegative inieger. However, t9 apply the calculuswe assumethat r is a nonnegative reil number (and then round off any noninteger values of x to th" ,rearestinteger), thus giving us the continuity requirements for the function C. The aaeragecost of producing each unit of a commodity is obtained by dividing the total cost by the number of units produced. Letting e(r) be the number of dollars in the averagecost, *e hane C(x) Q (r), xand Q is called an aaeragecostfunction. Nory,,let us suPPosethat the number of units in a particular output is rt and this is changed by A,x.Then the changein the total cost is given by C(xt + Ar) - C(xt), and the averagechangl in the total cost wiih re-

5.6 AN APPLICATIONOF THE DERIVATIVEIN ECONOMICS

231

spect to the change in the number of units produced is given by C(xr*Ax)-C(r,) Ax

(1)

Economists use the term marginal cost for the limit of the quotient in (1) as At approacheszero, provided the limit exists. Being the derivative of C at xr, this limit gives us the following definition. 5.5.1 Definition

If C(r) is the number of dollars in the total cost of producing r units of a commodity, then the marginalcost,when *:rr., is given by C'(rt) if it exists. The function C' is called the marginqlcostfunction. In the above definition , C'(xr) can be interpreted as the rate of change of the total cost per unit change in the quantity produced, when 11 units are produced. o rLLUsrRArroNL: Supposethat C(r) is the number of dollars in the total cost of producing x picture frames (x > 10) and

C ( x ) : 1 5 * 8 r +x.r y (a) The marginal cost function is C' and

:8 -# C'(x) (b) The marginal cost when I : 20 is C'(20) and

C ' ( 2 0-) 8 - ' # : 8

- 0 . s 0: 7 . 5 0

(c) The number of dollars in the cost of producing the twenty-first frame is C(21) - C(20) and C(21) - C(20) :192-52 - 1'85:7.52 Observe discrepancy of change of approximate frame.

that the answers in parts (b) and (c) differ by 0.02. This occurs because the marginal cost is the instantaneous rate C(r) with respect to a unit change in r. Hence, C'(20) is the number of dollars in the cost of producing the twenty-first .

Reasoning similar to that preceding Definition following definition. 5.5.2 Definition

5.5.1 leads to the

lf Q@) is the number of dollars in the average cost of producing one unit of r units of a commodity, then the marginal average cost, when x: x1,is given by Q'(r1) if it exists, and Q' is called the marginal aaerage cost function.

ADDITIONALAPPLICATIONS OFTHEDERIVATIVE

The graphs of the total cost function, the marginal cost function, and the averagecost function are called the total costcurae (labeled TC), the marginalcost curae (labeled MC), and the nrreragecost curae (labeled AC), respectively. Q'(r1) gives the slope of the tangent line to the averagecost curve at the point where x: xr. Figure5.6,1

o rlr,usrRArroN 2: Consider a linear total cost function. C(x):mx*b

Figure5.6.2

Note that b rePresentsthe overheadcost (the total cost when.r: 0). The marginal cost is given by C'(r) : m. If Q is the averagecost function, Qk) : m * blx, and the marginal averagecostis given by e,k) : -blxr. Refer to Fig. 5.6.1'for sketchesof the total cost curve and the average cost curve. The total cost curve is a segment of a straight line in the first quadrant having slope m and y-intercept b.The averagecost curve is a branch of an equilateral hyperbola in the first quadrant having as horizontal asymptote the line y : m. BecauseQ'(r) is always negatirre, the averagecost function is always decreasinB,and as r increases,the value of Q@) gets closer and closer to m. . o rtlustRATIoN

suppose that we have a quadratic total cost function

C(r) -- ax2* bx * c where a and c are positive. Here c is the number of dollars in the overhead cost. The total cost curve is a parabola opening upward. Because c'(x):2Ax*b, a critical number of c is-blza. We distinguish two cases:b>0andb<0.

Figure5.6.3

rxeivrr.r 1: Supposethat C(x) dollars is the total cost of producing 100r units of a commodity and C(r) : Lxz- 2x * 5. Find the function giving (a) the average cost; (b) the marginal cost; (c) the marginal averagecost. (d) Find the absolute minimum average unit cost. (e) Draw sketchesof the total cost curve, the average

Case7: b > 0. -bl2a is then either negative or zero, and the vertex of the parabola is either to the left of the y axis or on the y axis. Hence, the domain of C is the set of all nonnegative numbers. A sketch of TC for which b > 0 is shown in Fig. 5.6.2. Case2: b < O.-bl2a is positive; so the vertexof the parabolais to the right of the y axis, and the domain of C is restricted to numbers in the interval l-bl2a, +oo).A sketchof TC for which b < 0 is shown in Fig. s.G.g. . SOLUTION:

(a) If Q is the average cost function, (_(*) 1 ^ , 5

QQ):Y:i.r-2*;

(b) The marginal cost function is C,, and

c'(x)


cost curve, and the marginal cost curve on the same set of axes.

(c) Q' is the marginal

average cost function,

and

Q',k):+-* (d) Setting Q' @): 0, we obtain \[0 as a critical number of e, and Q(V10) : VLO- / - L.L6.BecauseQ" k) :101x3, Q" (\m) ) 0, and so Q has a relative minimum value of approximately '1,.L6at x: {fr. From the equation definirg Q(r) we seethat Q is continuous on (0, +oo). Becausethe only relative extremum of Q on (0, aoo) is at x: \m, it follows from Theorem5.3.1(ii)that Q has an absoluteminimumvalue there. when x: t/fr : 3.L6,100r : 3L6,and so we concludethat the absolute minimum averageunit cost is $1.15when 3L6 units are produced. The sketchesof the curves TC, AC, and MC are shown in Fig. s.6.4.

Figure5.6.4

In Fig. 5.6.4, note that the lowest point on curve AC occurs at the point of intersection of the curves AC and MC, and this is the point where the marginal average cost is zero. This is true in general and it follows from the fact that the value of r which causes the marginal average cost to be zero is a critical number of the function Q. That is, since Q(r) : C(x)lx, Q',k)

-

*r_,(x)_ 4v

and so Q'k):

C(x)

)c2

0 when xC'(x)- C(x) : 0 or, equivalently, when

"r r) c'(x):aYou should note the economic significance that when the marginal cost and the average cost are equal, the commodity is being produced at the very lowest average unit cost. In normal circumstances, when the number of units of the commodity produced is large, the marginal cost will be eventually increasing or zero; hence, C" (x) > 0 for r greater than some positive number N. So unless C" (x) : 0, the graph of the total cost function is concave upward tor x > N. However, the marginal cost may decrease for some values of x; hence, for these values of x, C" (x) < 0, and therefore the graph of the total cost function will be concave downward for these values of x. The following example involving a cubic cost function illustrates the case in which the concavity of the graph of the total cost function changes.

234

ADDITIONALAPPLICATIONSOF THE DERIVATIVE

2: Draw a sketch of ExAMPLE the graph of the total cost function C for which C(r):x3-6x2*

13r*1'

Determine where the graph is concaveupward and where it is concavedownward. Find any points of inflection and an equation of any inflectional tangent. Draw a segment of the inflectional tangent.

SOLUTION:

C'(x) :3xz - 12x+ 13 C"(x):6x-t2 C'(x) can be written as 3(r - 2)2 + 1. Hence, C'(x) is never zero. C" (x): 0 when r: 2. To determine the concavity of the graph for the intervals (0, 2) and (2, am) and if the graph has a point of inflection at x:2, we use the resultssummarizedin Table 5.6.1. T a b l e5 . 6 . 1

C(x)

C'(r)

C" (x)

Conclusion

0

2
graph is concaveupward

An equation of the inflectional tangent is x - y + 9: 0. A sketch of the graph of the total cost function together with a segment of the inflectional tangent is shown in Fig. 5.6.5.

Figure5.6.5

Consider now an economic sifuation in which the variables are the price and quantity of the commodity demanded. Let p dollars be the price of one unit of the commodity and x be the number of units of the commodity. Upon reflection, it should seem reasonablethat the amount of the commodity demanded in the marketplaceby consumersdepends on the price of the commodity. As the price falls, consumers generally demand more of the commodity. Should the price rise, the opposite occurs. Consumers demand less. Thus, if p, and p2 are the number of dollars in the prices of x, and r, units, respectively, of a commodity, then pt ) pz if and only if

x, 1 x,

(2)

An equation giving the relationship between the amount, given by x, of a commodity demanded and the price, given by p, is called a demand equation.If the equation is solved for p, we have P: P(x) The function P is called the pricefunction and P(r) is the number of dollars in the unit price for which r units of the commodity are demanded at that price. Becauseof statement (2),P is a decreasingfunction. In a normal eco-


nomic situation, r and P(r) are nonnegative numbers. Even though r should be an integer (becauser is the number of units of a commodity) we assumeonly that r is a real number in order that P may be a continuous function so that we can apply the calculus. Another function important in economicsis the total reaenuefunction, which we denote by R, and R(x) - xP(x) where P(x) dollars is the price of each unit and r is the number of units sold. When x * 0, from the above equation we obtain R(r) _ p(r) x which shows that the revenue per unit (the average revenue) and the price per unit are equal. 5.6.3 Definition

If R(r) is the number of dollars in the total revenue obtained when x units of a commodity are demanded, then the marginal reaenue,when *: *r, is given by R'(rt) , if.it exists.The function R' is called the marginalreaenue function. R'(rr) can be positive, negative, or zero, and it can be interpreted as the rate of changeof the total revenueper unit changein the demand when rt units are demanded. The graphs of the functions R and R' are called the total rer)enuecurae(IabeledTR) and the marginalreaenuecurve(labeled MR), respectively.The graph of the demand equation is called the demand curTJe.

ExAMPLE3: The demand equation for a particular commodity is p' * x - 12: 0. Find the price function, the total revenue function, and the marginal revenue function. Draw sketches of the demand curve, the total revenue curve, and the marginal revenue curye on the same set of axes.

solurroN: If the demand equation is solved for p,we find p : !\EV. Since P(r) > 0, we have P(r):Ex R(x) : x{FV R'(r) :

A-3x 2\/12 - x

Setting R'(r) :0, we obtain r: 8. Using the information in Table 5.6.2, we seethat the required sketchesare drawn as shown in Fig. 5.6.6. T a b l e5 . 6 . 2

x

P(x)

R(r)

R'(r)

0\n0\m. 33e8 82t60 11. 1 12 0

L1 0

-g does not exist


demand curve

Figure5.6.6

we have seen that for a given demand equation the amount demanded by the consumer depends only on the price of the commodity. Under a mlnopoly (which means that there is only one producer of a certain commodity), price and, hence, demand can be controlled by regulating the quantity of the commodity produced. The producer undei a monopoly is called a monopolist,and he wishes to control the quantity produced and, hence,the price per unit so that the profit will be as largl as possible. The profil earned by u business is the difference between the total revenue and the total cost. That is, if 5(x) dollars is the profit obtained by producing r units of a commodity, then 5(r) - R(x) - C(r) (3) where R(x) dollars is the total revenue and C(x) dollars is the total cost. S is calledthe profit function.From Eq. (3),if we differentiatewith respect to x, we obtain S ' ( r ) : R ' ( r ) - C '( x )

(4)

S" (x) : R" (r) - C" (x)

(5)

and It follows from Eq. (4) that s'(r) > 0 if and only if R'(r) > c, (x); therefore, the profit is increasing if and only if the marginal revenue is greater than the marginal cost. Let us determine what level of production is necessary to obtain the greatestprofit. S has a relative maximum function value at a number x for which S'(r) : 0 and S" (x) ( 0. From Eqs. ( ) and (5)

5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS

(r, R(r))

Figure5.6.7

4: Supposethat the ExAMPLE demand equation for a certain commodityis P:4 - 0.0002r, where x is the number of units produced each week and p dollars is the price of each unit. The number of dollars in the total cost of producing x units is 600 + 3x. If the weekly profit is to be as large as possible, find: (a) the number of units that should be produced each week; (b) the price of each unit; (c) the weekly profit.

we observe that this will occur at a value of x for which the marginal revenue equals the marginal cost and R" (r) < C" (x). The values of r will be restricted to a closed interval, in which the left endpoint is 0 (since r = 0) and the right endpoint (the largest permissible value of x) is determined from the demand equation. At each of the endpoints there will be no profit, and so the absolute maximum value of 5 occurs at a value of r where S has a relative maximum value. To illustrate this geometrically, refer to Fig. 5.6.7, where both the x total revenue curve and the total cost curye are drawn on the same set of axes. The vertical distance between the two curves for a particular value of r is 5(r), which gives the profit corresponding to that value of x. When this vertical distance is largest, S(r) is an absolute maximum. This is the distance an in the figure where A and B are the points on the two curves where the tangent lines are parallel, and hence C'(x) - R'(r).

sor,urroN: The price function is given by P (r) - 4 - 0.0002x,and r is in the closed interval [0, 20,000]becauser and P(r) must be nonnegative. R(r) : xP(x), and so R(x) :4x - 0.000212 r in [0, 20,000]

(6)

The cost function is given by C ( x ) : 6 0 0* 3 r

(7)

If S(r) dollarsis the profit, S(r): S(r):r-0.0002f

-600

R(r) - C(x), and so r in [0,20,000]

From Eqs. (6) and (7) we obtain R'(r) - 4- 0.0004x c'(r) : 3 Also, R"(r) : -0.0004 and C"(x):0. 4 - 0.0004x: 3

EquatingR'(r) and C'(r) we get

x:2500 BecauseR" (r) is always less than C" (x) , S" (x): R" (x) - C" (x) < 0, and so we conclude that 5(2500) is an absolute maximum value. (Note that S(x) is negative for r:0 and r:20,000.) P(2500):3.50 and s(2500):650. Therefore, to have the greatest weekly profit, 2500 units should be produced each week to be sold at $3.50each for a total profit of $550.

238


ExAMPLE 5: Solve Example 4 if a tax of 20 cents is levied by the govemment on the monopolist for each unit produced.

soLUrIoN: With the added tax, the total cost function is now given by

C(x):(500*3r) *0.20x and so the marginal cost function is given by C'(x):3.20 EquatingR'(r) and C'(r), we get 4 - 0.0004x:3.20

As in Example4 it follows that when x: 2000,S has an absolute maximum value on [0, 20,000]. From P(x) - 4- 0.0002r and S(x) : O.gx - 0.0002x2 - 600,we have P(2000): 3.60and S(2000): 200. Therefore, if the tax of 20 cents per unit is levied, only 2000units should be produced each week to be sold at $3.60 in order to attain a maximum total weekly profit of $200. It is interesting to note that in comparing the results of Examples 4 and 5, the entire 20 cents rise should not be passedon to the consumer to achieve the greatestweekly profit. That is, it is most profitable to raise the unit price by only L0 cents. The economic significance of this result is that consumers are sensitive to price changes, which prohibits the monopolist from passing on the tax completely to the consumer. A further note of interest in the two examplesis how the fixed costs of a company do not affect the determination of the number of units to be produced or the unit price such that maximum profit is obtained. The fixed costof a company is the cost that does not change as the company's output changes.Regardlessof whether anything is produced, the fixed cost must be met. In Examples 4 and 5, becauseC(r) : d00+ 3r and 600 + 3.2x, respectively, the fixed cost is $600. If the 600 in the expressions for C(r) is replacedby any constantk, C'(x) is not affected;hence, the value of x for which the marginal cost equals the marginal revenue is not affected by any such change. Of course, a change in fixed cost affects the unit cost and hence the actual profit; however, if a company is to have the greatestprofit possible, a change in its fixed cost will not affect the number of units to be produced nor the price per unit.

Exercises5.6 1. The number of dollars in the total cost of manufacturing .x watches in a certain plant is given by C(r) :1500 + 30r + 20/r' Find (a) the matginal cost function, (b) the marginal cost when x: 40, and (c) the cost of manufacturing the forty-first watch. 2. lf C(r) dollars is the total cost of manufacturing r toys and C(r) : 110 + 4r * 0.02f, find (a) the marginal cost function, (b) the marginal cost when r: 10, and (c) the cost of manufaduring the eleventh toy.

5.6 AN APPLICATION OF THE DERIVATIVE IN ECONOMICS 239

3. Suppose a liquid is produced by a certain dremical process and the total cost function C is given by C(x) :514rrr, where C(r) dollars is the total cost of producing t gallons of the liquid. Find (a) the marginal cost when 16 gal are prcduced and (b) the number of gallons produced when the marginal cost is 40 cents per gal. 4. The number of dollaB in the total cost of prcducing x units of a certain commodity is C(r) : & + 3x + 9{2t. Fi^d, (a) the marginal cost when 50 units are produced and (b) the nurnber of units produced when the marginal cost is $4.50. 5. T'he number of dollars in the total cost of prcducing r units of a commodity is C(r) : ae14r * * Find the function grving (a) the average cost, (b) the marginal cost, and (c) the marginal average cost (d) Find the absolute ninimum average unit cost. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same Bet of axes. Verify that the average cost and marginal cost are equal when the average cost has it8 least value. - 6t + 4, find (a) the average cost 6. II C(x) dollars is the total cost of producing t units of a commodity and C(x):3f function, (b) the marginal cost function, arld (c) the marginal average cost function. (d) What is the range of C? (e) Find the absolute minimum average unit cost. (f) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axes. Verify that the average cost and marginal cost are equal when the average cost has its least value. 7. The total cost function C is given by C(x) = lat - 2a2+ 5x + 2. (a) Determine the range of C. (b) Find the marginal co9t function, (c) Find the interval on whidr the martinal cost function is decreaaing and the interval on which it is incr€asing. (d) Draw a sketch of the graph of the total cost function; detemdne where the graph is concave upward and where it is concave downward, and find the points of inflection and an equation of any inflectional tangent. 8. If C(r) dollars b the total cost of producing t units of a commodity and C(r) = 2f - 8r + 18, find (a) the domain and range of C, (b) the average cost function, (c) the absolute minimum average unit cost, and (d) the marginal cost tunction. (e) Draw sketches of the total cost, average cost, and marginal cost curves on the same set of axe8. 9. The fixed overhead expense of a manufactuer of childrcnls toys is $4fi) per week, and other costs amount to $3 fot each toy ploduced. Find (a) the total cost function, (b) the average cost function, and (c) the marginal cost function. (d) Show thaf there is no absolute minimum average unit cost. (e) What is the smallest number of toys that mu8t be Produced so that the average cost per toy is le8s than $3.42?(f) Draw sketches of the graphs of the functions in (a), (b), and (c) on the 6ame Bet of axes. 10. The number of hundreds of dollars in the total cost of producing 100xradios per day in a certain factory is C (x) : 4x + 5' Find (a) the average cost function, (b) the marginal cost function, and (c) the martinal average cost function. (d) Show that there i8 no absolute minimum average urlit cost. (e) What is the smallest number of radios that the factory must produce in a day so that the averagecost per radio is lessthan $7?(f) Draw sketchesof the total cost, averageco6t, and martinal cost curves on the same set of axes. 11. If the demand equation for a particular commodlty is 3r I 4p : 12, find (a) the price function, O) the total r€venue function, and (c) the marginal revenue function, Draw sketches of the demand, total revenue, and marginal revenue curves on the same set of axes. Verify that the marginal revenue curve intersects the t axis at the Point whose abscissa is the value of r for whidr the total levenue iB greate8t and that the demand curve interEects the t axis at the Point whose absci66a is twice that. 12. The derrand equation for a particular commodity is pf + ? - 18:0 wherc p dollars is the price per unit when 100r units are demanded. Find (a) the price function, (b) the total revenue function, and (c) the marginal revenue function. (d) Find the absolute maximum total revenue. 13. Follow the instructiona of Exercise 12 if the demand equation is t' + Pt - 36: 0' 14. Let R(x) dollars be the total revenue obtained when r rrnits of a comrnodity are demanded and R(r):2+31i1, where I is in the closed interval [2, 17]. Find (a) the demand equation, O) the Price function, and (c) the marginal revenue function. (d) Find the absolute maxtnum total levenue. (e) Draw skekhes of the demand, total revenue, and marginal revenue curves on the Sameset of axe8. 1.5.The demand equation for a certain commodity is r + p: 14, where r is the numbet of units Produced daily andp is the number of hundreds of dollars in the price of each unii. The number of hundre& of dollars in the total cost of Pro-

240


ducing x units is given by C(x) : * - 2x I 2, and r is in the closed interval [1, 1a]. (a) Find the profit function and drav, a sketch of its graph. (b) On a set of axes different from that in (a), draw sketches of the total revenue and total cost curves and show the Seometrical interpretation of the profit function. (c) Find the maximum daily profit. (d) Find the marginal tevenue and rrarginal cost functions. (e) Draw sketches of the graphs of the marginal revinue and margi nal cost functions on the same set of axes and show that they intersect at the point for which the value of r makes the prcfit a maximum, 16. Follow the instructions of Exercise15 if the demand equation is f + p :32 and C(x) :5r. 17' The demand equation for a ce*ain commodity is p : (x - 8),, and the total cost function is given by C(r) : l8t - f , where C(r) dollars is the total cost when x units are purchased. (a) Detemine the permissible vaiues of r. (b) Find margin-al revenue and marginal co6t functions. (c) Find the value oI r which yields the maximum piofit. (d) Draw $e sketches of the marginal revenue and marginal cost functions on the same set of axes. 18. A monoPotst d€termines that if C(r) cents is the total cost of producing x units of a certain commodity, then C(r) : 25r + 20/000. The demand equation is x + sOp: 5000, where x units are demanded each week when the unit Drice tj.P If the weekly profit is to be maximized, find (a) the numbe! of units that should be produced each week, -"".b: O) the price of each unit, and (c) the weekly profit. 19. solve Exercise 18 if the govemment levies a tax on the monopolist of 10 cents per unit produced. 20, Solve Exercise18 if the govemment imposes a 10% tax based upon the consume/s price, 21. For the monopoligt of Exerciee 18, determine the amount of tax that shoutd be levied by the government on each unit prcduced in order fo. the tax revenue received by the government to be maxirnized. 22' Find the maximum tax revenue that can be received by the government if an additive tax for each unit produced is leyied on a monopolist for which th_edeaTd equation is xllp:75, where.' units are demanded when p dollars is the price of one unit, and c(r) :3x r 1e0,where c(r) dollars is the total cost of producint r units. 23. The de-mand equation for a certain commodity pmduced by a monopolist is p : a - 6r, ,ne total cost, C(r) dollars, of Prcducing :r units is detennined by C1r; : r a dx, where a, b, c, aid, d arc positive constants. "rrd If the government levies a tax on_the monopolist of f dollars per unit produced, show that in order ior the monopolist to nJximize his profits he should pass on to the consumer only one-half of the tax; that is, he should increase his unit price bv it dollars.

ReaiewExercises (Chnpter5) 1. Find the shortest distance from the point (*, 0) to the curve y:

t[.

2. Prove that among all the rectangleshaving a given perimeter, the square has the greatestarea. 3. Prove that among all the rectangles of a given area, the square has the least perimeter. 4. The demand in a cerlain market for a particular kind of breakfastcereal is given by the demand equation p, + 25p 2000: 0, where P cents is the price of one box and x thousandso{ boxes is the quantity demandedper week. If the current Price of the cerealis 40 cents per box and the pdce per box is increasing at th€ rate of 0.2 cent each week, find the rate of change in the demand. 5. Two Particles start theit motion at the same time. one particle is moving along a horizontal line and rts equation of motion i6 t = t: - 2t, where r ft is the directed distance of the particle from the o;igin at t sec.The otherparticle is moving along a vedical line that intersectsthe horizontal line at the odgin, and its equation of motion is y : p - 2, where y ft is the directed distance of the partide from the odgin at t sec. Find when the directed distance Letween the two particles is least, and their velocities at that time.

REVIEWEXERCISES 241 the intervals on which f In Exercises 5 through 9, find: the relative extremi of f the points of inflection of the traph of t glaph i6 concave downwhete the upward; graph is concave the tf," i.tlrvals on which I is decreasing witere i'i""r""ri"g graph' of the Draw a sketch tangmt. ward; the sl--opeof any inflectional 6.f(x):(x*2)at3

8. /(r) : x\/FT

7.f(x): (r- 4)'(x+2)' 9 . f ( x ) : ( r - 3 ) s r*r t

10. If I\x):

x*L xr+1

gxaPh. prove that the graph of I ha8 three points of inflection that are collinear. DrarY a Eketch of the 11. If l(r) : 1111,show that the graph of I has a Point oI inflection at the origin' and only xo where r is a positive integer. Prove that the gnph of I has a point oJ inflection at th€ origin if at 0. value relative minimum has a is wen, , > 1. Furthirmore show that if I / Jd int"g". "nd (f * ar)!, where p is a rational numb er and'p * 0, prove that if P < ]the graph ofl has two Point8 of inflectg. fif(x\: ti6n and if p > * the graph of f has no Pointg of inflection' what can you condude, if an''ll1ing, about (a) the L4. Suppose the glaph of a function has a point of inflection at l: c. of continuity (c) c; the of at f" at c? continuity of / ai c; (b) the continuity /' P is the numbef of millions 15. Suppose c is the number of millions of dollars in the caPitalization of a certain colPoration, - 0.004c'. r its capitalization is incleasing at the rate of : P 0-.05c profit, and annual of dollars in the corpofation's profit if its current caPitalization is (a) $3 million $400,000per year, find the rate of chanie of the corporation's annual and (b) $6 million. when r aPartmentsare rented and 16. A property development company rents each apartment at p dollars per month is zero? revenue marginal How many apartments must be rented before the p :2(;\/ffi=E -!2. Let fe\: ;; f;

about a right-circular cylinder L7 . Find the dimensions of the right-circular cone of leastvolume that can be circumscribed of radius r in. and altitude h in. radius to the measure of the altitude for a 1g. A tent is to be in the shape of a cone. Find the mtio of the measure of the tent of given volume to require the least material' :l18' whete P dollars is the Price per unit when 19. The demand equation for a particular commodity is (P + 4)(r + 3) the marginal revenue function' lg0r units are demanded. Find (a) the price functi;n, 16) the total rwenue function, (c) and margiral revenue revenue, total the demand, of (d) Find the absolute maxrsrum iotal revenue. (e) Draw sketches cuw$ on the same set of axes. $ and the total cost function is given by c(r) : 2r 1' 20. The demand equation for a certain commodity i8 p + 2V?J: -(c) Find the functions. cost marginal permissible values of r. iul ritta the marginal revenue and t"l-oJ""rr""'ae value of r which yields the maximum profit' 21. The demand equation for a certain cpmmodity is l16px:70s - 2. 1.03r* 18 ' 10'3f 6f unit, and I > 100. The number of where r is the number oI units produced weekly and p dollars is the price of each ' : 10k-1' Find the number of units * 11 24 dolLars in the avenge cost of producing each unit is given by QQ) "\7for the weekly unit in oldet price of each Profit to be marirnized. that should be produced each week and the

242

ADDITIONALAPPLICATIONS OF THE DEBIVATIVE

22. When 1.000rboxes of a c€rtain*ind of material are produced, the-nurnber of dollars in the total cost of production is givenbv c(r) - 135r1i3 + ft50.Find (a)the marginicost when8000br""r ifi'r,*uu. ot uor"" produced whm the marginal cost "i"-p-au""a?ffi Grer thousand boxes) is g20. 23' A ladder is to rcach over a fence ' ft high to a watl ur ft behind the fence. Find the length of the shortest ladder that may be used. 24. Dtaw a sketch of the graph of.a function on the interval I in eachcase: (a) I is the open interval (0, 2) ard/is f con_ tinuous on r' At 1, I has a relative maximum value but /'(1) does not exirt. 1iy i i, th" to, zl. rrru ror,""io"eJiiiu*"i

-i',iio- ua,,"Jt7;;f 0.6ilil; Hll,Ti,""tiffi?ffifrTff""j,1'.but theabsorute

olenintervar (0.z),

25. (a) It t'@):3lrl + alx - tl, prove that/has an absolute minimum value of 3. (b) If g(r) =alll + 3lr _ 11,prove that 8 ha8 an absolute minimum value o_f3. (c) rl h(tc): alxl + tp - il, where d > 0 and > b o, plove that , has an abso_ lute minimum value that is the smalter of the t*,o ,rr.-t"""'o f. "r,i 26' rl f(x): lxl" ' lr - tlb, where c and b are positive rational numbers, prove that has a relative maximum / value of L'bbl@.+ bl+b.

The differential and antidifferentiation

THE DIFFERENTIAL AND ANTIDIFFERENTIATION

6.1 THE DIFFERENTIAL

Supposethat the function / is defined by Y:f(x) Then, when f ,(x) exists, A/ '(x) lim /f \ - ar;-o A'r

( 1)

where Ly:f(x+ Lx) - f(x). From (1),itfollows thatfor any€ ) 0there existsa6>0suchthat

l + - f ' '( x ) 1 I ..

lAr

w h e n e v e( r ol A r l< 6

which is equivalent to

l\l,-f'

(x) Arl lA;:<.

whenever0( lArl
This means that lAy -f'(*) Arl is small comparedto lAxl. That is, for a sufficiently small lArl ,f '(x) Ax is a good approximation to the value of Ay. using the symbol :, meaning "is approximatelyequal to,,, we say then that Ly:f,(x)Ax if lArl is sufficiently small. The right side of the above expression is defined to be the ,,differential,, of y. 6'L'1 Definition

If the function / is defined by y : (x), then the differentialof y, d,enotedby f dy, is given by (1) where r is in the domain of ' and,Ar is an arbitrary increment f of x. This concept of the differential involves a special type of function of two variables (a detailed study of such functions will be girr"t in Chapter 19)' The notation df may be used to represent this funltion, where the symbol df is tegallud as a single entity. The variable r can be any number in the domain of f ' , andLx can be any number whatsoever.To state that df is a function of the two independent variablesr and Ar means that to each ordered paft (x, Ar) in the domain of df therc correspondsone and only one number in the range of df, and this number .u., b" represented by df (*, Ax), so that

df (r, Ar) : f '(x) Ax (2) ComparingEqs.(1) and (2) we seethat when y : (x), dy and df (x, A,x) f

6.1 THE DIFFERENTIAL 245

are two different notations f.orf '(x) Lx. We shall use the "dy" symbolism in subsequentdiscussions. o T L L U S T R A Tt I: oINf y : 4 x 2 - r , t h e nf ( * ) : A x z - x , a n d s o / ' ( r ) : 8 r - 1 " . Therefore,from Definition 6.1.7, dy: (8r-1) Ax In particular, tf.x:2,

then dy : 1'5Ax.

o

When y : f (r), Definition 6.1,.Lindicates what is meant by dy, the differential of the dependent variable. We also wish to define the differential of the independent variable, or dx. To arrive at a suitable definition for dx that is consistent with the definition of dy, we consider the identity '(x): L and function, which is the function / defined by f (x) - r. Then f ' : A : x; so dy L Ar : Ar. Becausey x, we want dx to be equal to dy for this function we want dx: Ax. It is this is, for that this particular function; definition. following to the leads us that reasoning 6.1.2 Definition

If the function /is defined by y : f (x), then the differentialof x, denotedby dx, ts given by (3) where Ar is an arbitrary increment domain of f'.

of. x, and r is any number in the

From Eqs. (1) and (3) we obtain

(4)

$ Dividing on both sides of the above by dx, we have

#:f'k)

tfdx*0

(5)

Equation (5) expresses the derivative as the quotient of two differentials. The notation dyldr often denotes the derivative of y wlth respect to x-a symbolism we have avoided until now.

nxlvpln L: Given y--4x'-3x*I find Ly, dy, and Ly - dy for (a) any )cand Ax; (b) r :2, Lx: 0.1; ( c )x : 2 , A x : 0 . 0 1 ;( d ) x : 2 , Ar: 0.001.

solurroN: (a) Becausey :4x2 - 3x * L, we have y + Ly :4(x * Ar)2 - 3(x * Ar) * 1' (4x' - 3x * 1) + Ly : 4x2* 8x A,x* 4(A'x)2- 3x- 3 Ar * 1 Ly : (8r - 3) Ar * A(A,x)z Also, dy: f'(x) dx

THE DIFFERENTIAL AND ANTIDTFFERENTIATION

Thus, (8r - 3) dx :- (8r - 3) Ar Ay-dy-A(A,x)z The resultsfor parts (b), (c),yd (d) aregiven in Table5.1.-l,,where : Ly (8r - 3) Ar + 4(A,x)zand,dy: (8r - 3) Ar. T a b l e6 . 1 . 1

1..34 0.1304 0.013004

0.04 0.0004 0.000004

Note from Table 5-1.1,that the closer Ax is to zero, the smaller is the difference between Ly and dy. Furthermore, observe that for each value of Ax, the corresponding value of A,y - dy is smaller than the value of Ar. More generally, dy is an approximation of ay when Ar is small, and the approximation is of better accuracy than the size of Ar. For a fixed value of. x, say xo, dy - f '(xs) dx (6) That is, dy is a linear function of dx; consequently dy is usually easier to compute than Ay (this was seen in Example i;. Because (xo + Ar) - /(ro) : f Ly, we have

f(xo+Ar): f(x) + Ly Thus,

(ro * Ar, f(xo+ ar))Q R(re * Ax, flr)

+ dy)

AT,

M (xo+ Ax,f (xJ)

f(xo+ Ax) - f(x) + dy (7) We illustrate our results in Fig. 5.1.1.The equation of the curve in the figure is y : /(r). The line pT is tangent to the curye at p(xs, (xo));Ax and f dx ate equal and are represented by the directed distarr..-pfr, where M is the point (ro + a'x,f (xs))._w" ret Q be the point (ro + ax, (xs* Ar)), and the directed distance MA is Ly or, equivalently, (x, f f f @o) L!4_ The slop=f PT is f '(x): dyldx. Aiso, the slope of'pi'i; ffiJm, becausePM: dx, we have dy: MR and Re: Ay - dy. Note that ""a the smaller the value of dx (i.e., the closer the point e i to tire point p), then the smaller will be the value of Ay - dy (i.e., the smaller wilibe the length of the line segmentRQ). An equation of the tangent line pT is y-f(ro)+f'(x)(r-ro)

6.1 THE DIFFERENTIAL 247

Thus, if y is the ordinate of R, then (8)

y: f(xi + dy

comparing Eqs. (7) and (8), we see that when using /(xo) + dy to ap' proximate the ,ruir" of f:1xo* Ar) we are approximating the ordinate ff tn" point Q(ro * A,x,i@o + Ar)) on the curve by the ordinate of the point filxo + Lx, f (xs)+ dV'l on the line that is tangent to the curve at P(ro,/(ro)). Observethat in Fig. 6.1.1,the graph is concaveupward' In Exercise30 you arc asked to draw a similar figure if the graph is concave downward'

(*): 2: Find an aPProximate soLUrIoN: Consider the function / defined by f Hence, value for \l/28 without using 3r \/x tables. EXAMPLE

{x and let y : f (x)'

and dy: f'(x) dx :i*ut7 , ax The nearest perfect cube to 28 is 27. Thus, we comPute dY with and dx: Ax : L.

dy:#wrrt By applyingformula(7)with xs: 27, f(27+ 1) : f(27) +*

1.,and dy : #, we have

iln+7: W+# i/2s: g+ ;,

Therefor",{28:3.037. : From Table 1 in the Appendix, we get t/28 3.037.Thus, the apdecimal three Places. proximation is accurateto rxarrpr.n 3: Find the approximate volume of a sPherical shell whose inner radius is 4 in. and whose thickness is t in.

We consider the volume of the spherical shell as an increment solurroN: of the volume of a sPhere. Let r: the number of inches in the radius of a sphere; V: the number of cubic inches in the volume of a sphere; LV: the number of cubic inches in the volume of a spherical shell. Y:

$nrs

Substitutingr

so that

= 4 and dr:

dV - 4rr(4)z # :4n

dV : 4nr2 dr # into the above, we obtain

248

THE DIFFERENTIAL AND ANTIDIFFERENTIATION Therefore, LV : 4rt, and we conclude that the volume of the spherical shell is approximately 4n in.3.

Exercises 5.1 In ExercisesL through6, find (a) Ly; (b) dy; (c) Ay - dV. l' Y: x" 2. y:4x2 - 3x* 1

4.y:h

5. Y:2f

+ 3*

In Exercises7 through 12, find for the given values: qa) ay; (b) dv; (c) Ay - dy. 7.y:*-3x;x:2; Ax:0.03 8. y : x2- 3x; x: -l; Ar : 0.02

10.y : xs* 7; x: -7; Ax: 0.1

1 1 .y : i ,

1

,:2;

3.

v : \E

6.

v:

9.y:

72.y :

Ax: 0.01

I Yx

1 3 * L ;x : 1 ; A r : - 0 . 5 1

ir,

x: -3; Ar: -0.1

In Exercises 13 through 20, use differentials to find an approximate value for the given quantity. Express each answer to three significant digits.

13.\/m

14.v.s

$. vn

12.fd00b%

18. \/M.

19.

1 YL2O

16. \/{2

20.

1

ffiF

21.

n:asureTent of an edge of a cube is found to be 15 in. with a possible error of 0.01 in. Using differ€ntials find lle aPproxrmate tne er()r in comPutins ftom this measuremmt (a) the volume; (b) the area of one of the faces. 22' The altitude of a right-circular con€ is tvdce the radius of the base. The altitude is measured as 12 in., with a possible error of 0.005 in, Find the approdmate error in the calcul,atedvolume of the cone. 23' An-open rylindrical tank is to have an out8ide coating oI thickness I in. If the inner radius is 6 ft and the altih.rde is 10 ft, find by differentials the apprcximate amount of coating material to be used. 24 A metal box in the form of a cube is to have an interior volune of 64 in.s. The six sides are to be madeof metaltin. thick. If the cost of the metal io be used js I cents per cubic inch, use differentiats to find the app;imate cost of the metal to be u6ed in the manufacturc of the box. 25 If the possible error in the measurement of the volurne of a gas is 0.1 fte and the allowable error in the pressure is 0.001Clb/ff' find the size of the smallest container for which !o/e's law (Exerase rO in p*erc-ises i.ry no6". ' A conh'actor agrees to paint on both sides-of 1000_ circular signs each of radius 3 ft. Upon receiving the signs, it is discover€d that the radius i8 I in' too large. use differentials to- find the approximat" p'.i*"i it of pilni ttrat wltt be needed. "*"'"" 27' The measure of the electrical reefoblT o{." wire is proportional to the measure of its length and inversely proportional to the square of the measure of its diameter. Suppose the resistance of a wite of givei length is computed from a measurement of the diaureter l4iith a possible 2% i:rror. Find the possible percent Jror in th-e computed value of the resistance. *

the for.?ne c-oTplete swing of a simple pendulum of length I ft, then 4?l t: gp, whereg: 32.2.A clock y ls]e ij: "a Pendulum naung of lmgth 1 fi gains 5 min eadr day, Find the approximate amount by whictr the pendulum should b€ Lengthen€d in order to correct the inaccuracv.

6.2 DIFFERENTIAL FORMULAS

29. For the adiabatic Law for the expansion of air of Exercise12 in Sec.3.10,prove that dPJP:-1.4

dVlV.

30. Draw a figure sirnilar to Fig. 6.1.1, if the graph is concave downward. Lrdicate the line setments whose lengths represmt the following quantities: A:, Ay, dt, and dy.


Supposethat y is a function of r and that r, in turn, is a function of a third variable f; that is, y:f(x)

and

x-gG)

(1 )

The two equations in (1) together define y as a function of f. For example, suPPosethat y : r! and x:2tz - 1..Combining these two equations,we get y - (2t2- 1)t. In general,if the two equationsin (1) are combined,we obtain

v: f k(t))

(2)

The derivative of y with respect to f can be found by the chain rule, which yields DtA : D;y Dp

(3)

Equation (3) expressesDg as a function of r and f becauseD"y is a function of r and Dg is a function of f. o rLLUsrRATroN 1,:If.y: rl and x:2t2 - 1.,then DtA : Dry DP :3x2(4t) -- LLxzt

o

BecauseEq. (2) defines y as a function of the independent variable t, the differential of y is obtained from Definition 6.1.1: dy - D,y dt

(4)

Equation (4) expressesdy as a function of t and dt. Substituting from (3) into (4), we get dy: D,y D6 dt

(s)

Now because r is a function of the independent variable t, Definition 5.1,.Lcan be applied to obtain the differential of.x, and we have dx - Dtx dt

(6 )

Equation (5) expressesdr as a function of t and df. From (5) and (5), we get dY: DrY dx

(7 )

You should bear in mind that in Eq. (4 dy is a function of f and dt, and that '(r), we have dx is a function of t and dt. Tf.D;y in Eq. (7) is replacedby f

dy: f'(x) dx

(8)


Equation (8) resembles Eq. (a) of Sec. 6.1. However, in that equation r is the independent variable, and dy is expressed in terms of r and dx, whereas in Eq. (8) f is the independent variable, and both dy and dx are expressed in terms of f and dt. Thus, we have the following theorem.

6.2.L Theorem

If y : /(r), then when f '(x) exists, dy: f'(x) dx whether or not r is an independent variable. If on both sides of Eq. (8) we divide by dx (provided dx + 0), w€ obtain

f ' ( x ): -dY dx

dx*O

(e)

Equation (9) statesthat if y: f (r), then f '(x) is the quotient of the two differentials dy and dx, even though r may not be an independent variable. We now proceed to write the chain rule for differentiation by expressingthe derivativesas quotients of differentials.If y: f (u) andDuy exists,and if u : g(x) and D, rzexists,then the chain rule gives D r A : D u yD * u

(10)

But D*y:dyldx rf dx*0; Duy:dyldu rf du*0; dx * 0. From (10),therefore,we have

andD*u:duldx if

x-{#){#)

if du#0and dx*0

Observe the convenient form the chain rule takes with this notation. The German mathematician Gottfried Wilhelm Leibniz (7646-1716) was the first to use the notation dyldx for the derivative of y with respect to r. The concept of a derivative was introduced, in the seventeenth century, almost simultaneously by Leibniz and Sir Isaac Newton (1,642-7727), who u/ere working independently. Leibniz probably thought of dx and dy as small changes in the variables r and y and of the derivative of y with respect to I as the ratio of dy to dx as dy and dx become small. The concept of a limit as we know it today was not known to Leibniz. Corresponding to the Leibniz notation dyldx for the first derivative of y with respect to x, we have the symbol dzyldxz for the second derivative of y withrespect to x. However, d2yldxz must not be thought of as a quotient because in this text the differential of a differential is not considered. Similarly, d"yldx" is a notation for the nth derivative of. y with respect to r. The symbolism f ' , f ", f " ', and so on, for the successive


251

derivatives of a function / was introduced by the French mathematician Joseph Louis Lagrange (1736-1,813) in the eighteenth century. Earlier in Chapter 3, we derived formulas for finding derivatives. These formulas are now stated using the Leibniz notation. Along with the formula for the derivative, we shall give a corresponding formula for the differential. In these formulas, u and T are functions of. x, and it is understood that the formulas hold if Dru and Dra exist. When c aPPears, it is a constant.

r dax! r ) : o ,, !#_

I ' d ( c ): g Il' d(x") - nrn-r dx

nlcn-1

( " u :) , 4 u l d dx ax

Ill'd(cu):cdu

( u: *Ea-) E :d, ,, rU , -dT

*d,

IV'd(u*v):du*da

(Ya):u4+o4 v d dx flx ilc __ Yr

. /u\ dt-t

\al _ dx

du an-u a

u2

run-,d!-

-, !#-

da d,

V'd(ua):uda*adu

vr o(+):a d u - u d a VII' d(u") - nun-r du

The operation of differentiation is extended to include the process of : finding ttre differential as well as finding the derivative.If y f (x) , dy can ' be found either by applying formulas I'-VII' or by finding f (x) and multiplying itbY dx.

rxluplr

Y: find dy.

1:

Given

Applying formula VI', we obtain

solurroN:

\m

a,., y -:

z;+l

(zx+ D a(F+ t) - {V + t a2x+ t) n

From formula VII'

d6/V+

L ) : t ( x ' * 1 ) - r r z2 x d x : x ( x 2* L ) - r r z6 *

(L2)

and d(2x*L)-24, Substituting values from (12) and (13) into (L1'),we get

(1 3)

AND ANTIDIFFERENTIATION THE DIFFERENTIAL

x ( Z x* t ) ( r ' z* t ) - t r z d x - 2 ( x z* t ) r r zO , a) -y. _: _ (2x2+ x) dx 2oc2+ 1,) dx ( 2 x* L ) ' ( x ' * 1 ; t r z -

EXAMPLE 2:

Given

Zx'y'- 3x3* 5y" I 6xy2:5 where r and y arefunctions of a third variable t, find dyldx by finding the differential term by term.

x-2 ,(2x]-7)z{x;4 1*^

solurroN: This is a problem in implict differentiation.Taking the differential term by term, we get 4xy2 dx * 4xzy dy - 9x2dx * 15y2dy * 6y2 dx * 1,2xydy : O Dividing by dx, if dx * 0, we have (4x'y t 15yz+ 12xy)

#:

-4*y' * 9x2- 6y'

dy_: 9xz- 6y2- 4xy2 dx 4x2y11,5y2* I2xy

Exercises 6.2 In Exercises1 through 8, find dy. l.y: +, 3. yx :

(3x,-2x*l)3

2'Y:\/4-F

3.y:*g2x+3

xr+2

5- . y :

6,y:

/.y:ffi

2x

lx_l Vr+1

(x+2)ttlbc-2)2t3

8. y: $x+ 4f/71

In Exercises9 through 76, x and y are functions of a third variable f. Find dyldx by finding the differential term by term (seeExample2). 9. 3xs* 4yz-- 48 L 0 . 8 x 2- ! 2 : 3 2 11.t/i+ {y:+ 1 2 . L x z y- 3 x y " * 6 y 2 : 1 1 3 .x a - 3 x " y * 4 x y 3 * y n : 2

.1,4.

L5. 3r3 - x'y * 2xy2- y3 - 3x2* A2:'!. In Exercises17 through 24, find dyldt.

'1.6. xzI y': +,C+ V

17. y : 3r 3- 5x 2i l ; x : P - 1 .

1 8 .y :

19. y : x2- 3x t- 1; 2s: \/F=j

+ 4

x2t3 I

yzt:t _ fl2t3

4v-)

x : ( 2 t- 1 ) ' ff; 20.y: S6;- 1;x: t/2tTJ

21. y - x2- 5x * t;r : s3- 2s * 1; s : t/T+-t

22.y:ft,r:ffi;s:tz-4t+5

23. x3- 3x'y * y" : 5; xs: 4tz + 1.

24. 3xzy- 4xy' l7y":

0; 2x3- 3xtzI t3:'J.

6.3 THEINVERSE OF DIFFERENTIATION 253

5.3 THE INVERSE OF You are already familiar with inuerse operations. Addition and subDIFFERENTIATION traction are inverse operations; multiplication and division are also inverse operations, as well as raising to powers and extracting roots. The inverse operation of differentiation is called antidifferentiation.

5.3.L Definition

A function F is called an antideriaatiae of a function f on an interval I if F' (x) : f (x) for every value of r in /. o rL L U s rR A rIo N 1: If F i s defi ned by F(x):4xs]-x2+ 5, then F' (x)l2x2 * 2x. Thus,if f is the function defined by f (x) : L2f * 2x, we state that f is the derivative of F and that F is an antiderivative of f .lf G is the function defined by G(r) - 4xB* x2 - t7, then G is also an antiderivative of f because G'(r) : l2x2 * 2x. Actually any function whose function value is given by 4x3* x2 * C, where C is any constant, is an antiderivao tive of /. In general, if a function F is an antiderivative of a function / on an interval I and if G is defined by G(r):F(r)+C where C is an arbitrary constant, then G' (x) : F' (x) : f (x) and G is also an antiderivative of / on the interval I. We now proceed to prove that if F is any particular antiderivative of / on an interval /, then all possible antiderivatives of / on I are defined by F(x) + C, where C is an arbitrary constant. First, two preliminary theorems are needed.

5.3.2 Theorem If f is a function such that f ,(x) : 0 for all values of r in the interval I, then / is constant on I. pRooF: Let us assume that is not constant on the interval I. Then there / exist two distinct numbers rr and x, in I whetl x1( r, such that / (xt) * : 0 for all r in I, then f' (x) - 0 for all r f (xr). Because,by hypothesis, f' (x) in the closed interval lxr, xr]. Hence, / is differentiable at all x in lxr, x2l and/ is continuous on [xr, rz]. Therefore, the hypothesis of the meanvalue theorem is satisfied, and we conclude that there is a number c, with x, 1c 1*r, such that (1)

'(c): 0, and But because/'(r) : 0 for all x in the interval lxr, rr], then f : from Eq. (1) it follows that f (rt) f (xr). Yet our assumption was that (*t) (xr). and so / is constanton f, a contradiction, Hence, there is + f f which is what we wished to prove.


6.3.3 Theorem

If f andg are two functions such that f , (x) : g' (x) for all valuesof r in the interval I, then there is a constant K such that f(x) :g(r)

+ K

for all r in I

pRooF: Let h be the function defined on I b v h(x): f (x)- SG) so that for all values of r in l we have h'(x):f' (r) -g'(x) But, by hypothesis,f '(r) : g'k) for all values of r in /. Therefore, h'(x) : g for all valuesof r in I Thus, Theorem 6.3.2 applies to the function h, and there is a constant K such that h(x) : Y

for all values of x in I

Replacingh(x) by f (x) - S Q), w e have for all values of r in / f (x) : g(x) + K and the theorem is proved. The next theorem follows immediately from Theorem 6.9.2. 6'3'4 Theorem

If F is any particular antiderivative of / on an interval l, then the most general antiderivative of / on I is given by

F ( r )+ C

e)

where C is an arbitrary constant,and all antiderivatives of on/can be obf tained from (2) by assigning particular values to C. PRooF: Let G be any antiderivative of on f G'(x) : f (x) on /

then

Because F is a particular antiderivative of / on .1,we have F'(x):f(x)

onl

@

From Eqs. (3) and (4), it follows that G'(x) -- F'(x) on I Therefore, from Theorem 6.9.9, there is a constant K such that G(x) : F(r) + K for all x in I Because G is any antiderivative of / on 1, it follows that all antiderivatives of f can be obtained from F(x) + c, where c is an arbitrary constant. Hence, the theorem is proved. I

6.3 THEINVERSE OF DIFFERENTIATION 255 If F is an antiderivative of f, then F'(x) : f (x), and so d(F(x)):f

(x) dx

Antidifferentiation is the process of finding the most general antiderivative of a given function. The symbol

t

J

denotes the operation of antidifferentiation,

and we write

f

(s)

I f (r) dx: F(r) + C J where

F'(r) : f (x) or, equivalently, d(F(x)) : f (x) dx From Eqs. (5) and (6) / we can write f

(6)

v)

ld(F(x)):F(x)+c J

Equation (7) statesthat when we antidifferentiate the differential of a function we obtain that function plus an arbitrary constant. So we can think of the f symbol for antidifferentiation as meaning that operation which is the inverse of the operation denoted by d for finding the differential. Becauseantidifferentiation is the inverse operation of differentiation, we can obtain antidifferentiation formulas from differentiation formulas. We use the following formulas that can be proved from the corresponding formulas for differentiation. 6 . 3 . 5F o r m u l a l I d * - x f

C

6.9.6Formula 2 I af(x) dx: alf@) dx,whete a is a constant Formula 2 states that to find an antiderivative of a constant times a function, first find an antiderivative of the function, and then multiply it by the constant. 6.s.7Formula3

I lf'(x) + f"(x)f dx: If'k)

dx* [fr(x) dx

Formula 3 states that to find an antiderivative of the sum of two functions, find an antiderivative of each of the functions separatelY, and then add the results. It is understood that both functions must be defined on the same interval. Formula 3 can be extended to any finite number of functions. Combining Formula 3 with Formula 2, we have Formula 4.


5 . 3 . 8F o r m u l a4

6.3.9 Formula 5

+ c " f " ( x ) fd x I [ c t f t ( x )+ c r f r ( x )* ' ' ' : h[ f t(x) dx * cr[fr(x) dx + " :;

' + c"[ f"(x) dx

'tr'*? d**dfo'+ c irn* *r

As stated above, these formulas follow from the corresponding formulas for finding the differential. Following is the proof of Formula 5: / ryn+1

"!.)xn

(n I

_\

7

d l - : - + C l : \ '!)#: d x n*1-a \/rJ-, :xndx Applications examples.

EXAMPLE

1:

Evaluate

f

| (st +5) dx J

of the above formulas are illustrated in the following

SOLUTION:

rfr | (3r + 5) dx: 3 | x dxt 5 | dx (by Formula4)

JJJ

/vz

:3 (i

\z

\

+ C , I + 5 ( r + C r ) ( b y F o r m u l a s5 a n d 1 ) /

: Ex'* 5r + 3Cr+ 5c2 Because3Cl + 5C2 is an arbitrary constant, it may be denoted by C, and so our answer is *x2+5xtC This answer may be checked by finding the derivative. Doing this, we have D * ( z r x*25 r + C ) : 3 r * 5

rxauplr

2:

Evaluate

solurroNt I vP dx: | ,',, 4* JJ

Igpa*

:

J

)c213+r

3.. ,

+ C (from Formula 5)

: P,xst3 + C

nxawrrr,n 3:

Evaluate

I(++S\a.

solurroN '

I $.

+)

dx-- a-^ * *-rr+)dx [ y-{-r

I

y-l r I t a

14+ |

-4+1.'-*+r

'

6.3 THE INVERSEOF DIFFERENTIATION r-3,x314,^ - J 4^ l 3 l \ .

:-

1 =q ! * t , n a g 3x3' 3^

Many antiderivatives cannot be found directly by applying formulas. However, sometimes it is possible to find an antiderivative by the formulas after changing the variable. 2: Suppose that we wish to find . TLLUSTRATTON f_

J

If we make the substitution u: f

J

(8)

zxt/t + x'zdx "L* x2, then du:2x

dx, and (8) becomes

uttzdu

which by Formula 5 gives tustz* C Then, replacingu by (1 + r2), we have as our result o

3 ( 1+ r ? ) B I Z + C

fustification of the procedure used in Illustration 2 is provided by the following theorem, which is analogousto the chain rule for differentiation, and hence may be called the chain rule for antidifferentiation. 6.g.'l,0Theorem Letg be a differentiable function of r, and let the range of g be an interval I. ChninRule Supposethat / is a function defined on / and that F is an antiderivativeof / forAntidifterentiation on /. Then if u: gG) , ff

I f G k D y . l - )d x : JI' f f u ) d u : F ( u )+ c - F ( s ( r ) )+ c J' u: g(r), then z is in PRooF: Because tive of f on l, it follows that

and because F is an antideriva-

d F ( u \- . / 'f'(\ u ) du

Irut

F ( u )+ C

Also, dF(sk)) _ dF(u) dx dx

(10)

( 1 1)

THE DIFFERENTIAL AND ANTIDIFFERENTIATION Applying the chain rule for differentiation to the right side of Eq. (11), obtain

d F ( g ( x ) )_ d F ( u ) . d u dx du dx Substituting from Eq. (9) into (12) gives

dF(sk))_ .. ' Edu tlu) dx Because u:

gQ), from Eq. (13) we have

d F ( s . @:) ) f e(r)) . 8,(r) dx from which we conclude that

r ' 8 ' G )d x : F ( 8 ( r )+) c J f k@) Becauseu:

g(x), we have

F(g(x))+C:F(u)+C Therefore,from (10), (I4), and (15), we have fr

dx: J f G @) g ' ( r ) J f O t d u : F ( u )+ C : F ( g ( x ) )+ C

which is what we wished to prove.

I

As a particular case of Theorem 6.3.10, from Formula 5 we have the generalized power formula for antiderivatives, which we state as Formula 5. 6.3.1,1Formula 6

If g is a differentiable function, then if u:

ffi;ffi

gG) ,

[*{*1"*'+c Tff'

where n # --L. Examples4, 5, and 5 illustrate the application of Formula 6. rxavrprr

r_

4:

Evaluate

I t/3x+4 dx

J

soLUrIoN: To apply Formula6,we makethe substitution u:Bx * 4; then du : 3 dx, or tdu: dx. Hence,

Ia,+t dx-[u,,,+ :

+ / u' Itzdr' t

6.3 THE INVERSEOF DIFFERENTIATION 259 1 u 3- +1a 2 , ^

-a3l\/

J2

: $ystz* C : g(Jx * 41trz* ,

The details of the solution of Example 4 can be shortened by not explicitly stating the substitution of z. The solution then takes the following form. JJ

dx - + [ fg* -r 41r,g o* (3x*4)3t2 *" _ 1 . ---=-

-

- fi(3x * 41zrz * , EXAMPLE

5:

Evaluate

f

I t(s + 3tz)8dt

J

sor-urroN: Since d(5 + 3t2)- 6t dt, we write rf I r ( s + g t 2 ) 8 d t : i | ( 5 + 3 t ' z ) 8 5dtt JJ ( 5 + 3 P ) ' g+ c _ - 6! . 9 : #(5 + 3t2)e+ c

rxavrpr.r 6:

Evaluate

f -

l l\Y7-4x3dx J

r-f xrr/7=B J

dx: -#

: -*.

nxenaprn 7: f_

Evaluate

I x2t/'j.,+ x dx J

J

(7 - 4x3)u5(-\hc2)dx

e -

xs)o/s-r-g

then t:2:'L * x. Hence, r : sor,urroN: Let a : GT; dx:2u fut. Making these substitutions,we have .a . (2a da) | *r ,6,, * dx: | @r- 1), JJ :2 _ J[ a6da - 4 | t:adu + 2 | a2da J J

:?a7-fu'+fut+C : ?(t * x)ztz- +(1 * x)srz+ ?(1 * x)zrza ,

and


260

The answers to each of the above examples can be checked by finding the derivative (or the differential) of the answer. o TLLUSTRATToN 3: In Example 5, we have

r

I t(S + 3tz)Bdt : # (5 + 3tz)e+ C

J

Checking by differentiation gives D r [ # ( 5 + 3 t 2 ) e ]: ; b '

9 ( 5 + 3 t 2 ) 8' 6 t

:f(5+3t2)t

.

o TLLUSTRATToN 4: In Example 7, we have f_

I x 2 ! 1 , * x d x : + ( 1 + x ) z n - * ( 1 * x ) s r z + 3 ( 1* x ) s r z * , J Checking by differentiation

gives

D"l+(L* x)zrz-e(1* x)srz+3(1* x)erz1 : ( 1 * x ) s r -z2 ( l * x ) s r z+ ( 1 * x ) t r z : ( 1 * x ) r r z[ ( 1 + x ) r - 2 ( t + r ) + 1 ] : ( 1+ _ A : . l L + 2 x * x z - 2 - 2 x + t l -*zfay

Exercises6.3 In Exercises L through26, of your answer.

find the most general antiderivative.

In Exercises 1 through 10, check by finding the derivative

f

1,. 3xadx J

4.

f

J@*+bx*c)

z.

to.

f

J

f

J

(x'- 4x+ 41ut4,

F-

t.JV3-xxzdx f_

8I%+d,

dx

3.

22. V3 + s(s* 1,)2ds J

"Iwh ,n.

t.

xa!3x3-5 dx

,,I# ,,

t .

f_

J

I

(r, * 3)rn*so*

,'\ffi,,

r dt JG-zt+t2)

6 J[t:#1 vy

^-_

xV(4- x2)2dx

B[J;4# ,u.

' IG***u) *

dx

,l(*-#\*

IUr"-3xz*6x-r)

o,

r_ ,'\/x3 -'l".dx J f-

J

!5r*r dr

f

15. x'zG- xz)rdx J

" Itlr.,1 | Q't 2x) dx -'' J !x3 1- 3x2+ 1

24.

f

J

{zt't llrrzssO,

,u[ff* 2T.Evaluate!(2xtl)sdxbytwomethods:(a)Expand(2r*1)gbythebinomialtheorem,andapplyForrrulasl,4,and in (a) and (b)' 5; (b) maie the substituiion u = 2x + L ixplain the difference in appearanceof the answersobtained o = V7- 1. 2g. EvaluateJ \/r - | * d.rby two methods:(a) Make the substitution u: x 7; (b) rrake the substitution 29. Letf(x):1 for all x in (-r, 1) andlet Fl. il-1 < r s0 8(rr:11 if0
r and F (x ):

l rl .Sh o w th a t F ' (r):

w h enever3' existsin (-1, L). However,f(x) + g(r) + K forr in of /on (-m , +-)? Explain. f(x ) i tx * O.Is F an anti deri vati ve

31. Let f (x): lrl and let F be definedby (-lf ifr
F('):l+o

ifr>o

Show that F is an antiderivative

of / on (-m, +oo)'

1.8) does not have an antiderivative on (-6' +@)' (HIvr: 32. show that the unit step function u (Exercise 21 in Exercises is obtained by showing that it fouows from Assume that U has an antiderivative F on (--, +-), and a contradiction thegrean-valuetheoregrthatthereexistsan.umber'&suchthatF('):'+kifr>0,andF(r):kifr<0.)

5.4 DIFFERENTIAL If F is a function defined by the equation EQUATIONS WITH a:F(x) vARIABLES sEPARABLE and f is the derivative of F, then

fu--r(x\ ''\

(1)

(2)

dx

and F is an antiderivative of /' Writing Eq' (2) using differentials, we have

(3) dy: f(x) dx Equations (2) and (3) are very simple types of differential,equations' They aie differential equations of the firstorder (becauseonly derivatives To of the first order are involved) for which the variables are separable' : the that so G(r) which for G functions A solve Eq. (3) we must find all G are equatioi is satisfied. So if F is an antiderivative of /, all functions is' if That constant' arbitrary : an C is where C, F(x) + aefined by G(r) solution complete the called is what (x) then dx, d(G(x)) d(F(x) + C)'- f of Eq. (3) is given bY y-F(x)+C

(4)


Equation (4) represents a family of functions depending on an arbitrary constant C. This is called a one-parameter amily. The graphs of these f functions form a one-parameter famity of curves in the plane, ind through any particular point (xr,yr) there passes just one curye of the famiiy. o ILLUSTRATTON 1: Suppose we wish to find the complete solution of the differential equation dy -2x

dx

(5)

The most general antiderivative for the left side of Eq. (5) is (y * C,) and the most general antiderivative of 2x is (r2 * Cr).Thus, we have A * Ct:

x 2+ C ,

Because(C" - Cr) is an arbitrary constant if C, and C1 are arbitrary, we can replace (Cz - Cr) by C, thereby obtaining 1 Q: CC-

F i g u r e6 . 4 . 1

o -1 -4

Y:x2+C

(6)

which is the complete solution of the given differential equation. Equation (6) represents a one-parameter family of functions. Figure 6.4.1' shows sketches of the graphs of the functions corresponding to C:-4, C:-L,C:0, C:1, and C:Z. Often in problems involving differential equations it is desired -. to find particular solutions which titirfy certain .onditior.,s called boundary conditions ot initiql conditians. For example, if differential equation (3) is given as well as the boundary condition thaty: y, whe rr x: xllthen after the complete solution (4) is found, if r and y in $) are replaced by r, and Au a particular value of C is determined. When this value of C is substituted back into Eq. (4), a particular solution is obtained. ' rLLusrRArIoN 2,: To find the particular solution of differential equation (5) satisfying the boundary condition that y: 6 when : x Z,we substitute th e s e v a l u es i n E q. (6) and sol ve for c, gi vi ng 6:4+ c, or c:2. substituting this value of C in (6), we obtain U:x2+2 which is the particular solution desired. Another type of differential equation is dra

ffi:ftx)

(7)

Equation (71isa differential equation of the secondorder,Two succes. sive antidifferentiations are necessaryto solve Eq. (T), and two arbitrary constantsoccur in the complete solution. The complete solution of Eq. 1i1 thereforerePresentsa two-parameter family of functions,and the graphsof these functions form a two-parameter family of curves in the plane.

EQUATIONSWITH VARIABLESSEPARABLE 6.4 DIFFERENTIAL

L: Find the complete EXAMPLE solution of the differential equation d2u *:4x*3 ax'

solurroN:

Becausedzyldx'* dy'ldr, the given equation can be written

AS

du' ax Writing this in differential form, we have dy': (4x+3) dx Antidifferentiatin 8, w e have rf ldy':l(4x+3)dx J Jfrom which we get

i

,

,

A' :2x2 * 3x* Ct Becausey' -- dyldx, we make this substitution in the above equation and get

+ax: 2 x 2 + 3 x * c , Using differentials, we have dU:

(2x2+ 3x * Cr) dx

Antidifferentiatin

g, w e obtain

ff

I dv- l (zx'* 3r * c,) dx

J-

J

from which we get A:3x3+hcz*Crxi_C, which is the comPlete solution.

BxauprE 2: Find the Particular solution of the differential equation in ExamPle L for which A :2 and y' - -3 when tc: L.

soLUrIoN: Because y' : 2x2+ 3x * C1,we substitute-3 for y' andl for x' - -8. Substituting this value of Ct into the giving -3 - 2 + g * Cr, or Cr complete solution gives

U:&x3+tx'z-8x*C, Becausey :2 when x: L, we substitute these values in the above equag- 8 + C2,fromwhich we obtaif|Cz: atz.The partiction and get2:3+ ular solution desired, then, is

A:3x3+*x2-8r++


form

A more general type of differential equation of the first order is of the

-dv : dx

f(x) sQ)

(8)

For such equations,the variablescan be separatedby multiplyirrg on both sidesof the equationby g(y) dx; thus, Eq.(g) can be writienias

g(y) dy : f (x) dx The following example gives us an equation of this type. ExAMPLE3: The slope of the tangent line to a curve at any point (x, y) on the curye is equal to 3*'y'. Find an equation of the curve if it contains the point (2, l) .

G";i.h" ,""

soLUTroN: Because the (x, y) is the value of the derivative at that point, we have

l

du r' : 3x'u' a)c

This is a first-order differential equation, in which the varia plescan be separated,and we have du

u' errtiAifferentiating, we have

rr

lY-'dy-l3x'dx

JJ

from which we obtain

-;:1

x3+ c

T-his rrt'r'DrD is 4rr an equarlon equation or of a one-parameter one-parameter tamily family of curyes. curves. To To fin{ the the particpartic-

ular curve of this family which containsthe point (2,1),*; ;;br;itute 2 for -LI _: Og T* Lc,, Ifrom r-and 1 for Jy,' which r v L . L - L . gives us LrD rOm 6rvsD W which nl we- obthin - - "f' C: -9. Therefore, Therefore, the the required required curve curve has has the the equation equation

-+:r-e

Exercises 6,4 ln Exercises 1 through 8, find the complete solution of the given differential equation. -dv du -

L.

d x : 3 x z* 2 x 7

Z.

#:

( 3 x+ 1 ) t

3.

du E:3xy2

AND RECTILINEAB MOTION 6.5 ANTIDIFFERENTIATION

ds dt:5Vs

, - dy

d2a 7.-?+:5x'l7

8. #:

4.

J .

d2u

dr

du

3xt/TT!

t/i+

6' dx:fr-

1

axv

x

v

YZx-3

In Exercises 9 through 14, for each of the differential equations find the particular solution determined by the given boundarv conditions. f.

n

du i:x2-2x-4;y:-6 dx

4da

;:t;y:-z

tO #:

w h e nx : 3

( x* 1 )( r + 2 ) ;y : - + w h e nx : - 3

t.H:ffi;Y:owhen

w h e nx : 4

,!2,, y : -1andy': -2 whenx: -1 s. ffi: ut )- 3x)2;

dzU

u |rt:Azx-t;v:2ar.dv':5

x:L whenx:3

2r - 3. Find an 15. The point (3, 2) is on a cuwe, and at any point (n y) on the curve the tangent line has a slope equal to equation of the curve. ) an equation 16! The slope of the tangent line at any point (r, y) on a curve is 3Vi. If the Point (9, 4) is on the curve, find -.cf the curve. :2 - 4x Find an equation of 17j'fn" poir,t" (-1, 3) and (0, 2) are on a curve, and at ary Point (r, y) on the curve D ,'y -' the curve. -1: y) on the curve, D"'zy : 5a, 18. An equation of the tangent line to a curve at the point (1, 3) is y a 2. If at any Point (t, find an equation of the curve. and an equation of the tangent line to the curve at the Point (1, 1) is 19. At any point (r, y) on a (j(rwe,Dx2y:1-i, ! : 2 t- Find an equation of the curve' the slope of the inflectional tangent 20. At any point (r, y) on a cuwe, D"? : 2, and (1, 3) is a Point of inflection at which is -2. Find an equation of the curve. of another one-parameter 21. The equation f :4oy represents a one-parametef family of parabolas. Find an equation and the tangent lines to the two it through family cuwe oi eadr y) there is a point (rl at any family of curves such-thai any Point (r, y), not on the line at of the tangent the sloPe that (nrr.rr: First show p"rp".rdlcrrlu". u"e point at this curves y axis, of the paraboLa of the given family through that point is 2ylx ) y": a"' 22. Solve Exercise 21 if the given one-Parameter family of cuwes has the equation '" +

6.5 ANTIDIFFERENTIATION AND RECTILINEAR MOTION

We learned in Secs.3.2 and 3.11 that in considering the motion of a particle along a straight line, when an equation of motion, s: f (t), is given, then the instantaneous velocity and the instantaneous acceleration can be determined from the equations u:Dls-f'(t) and a:Dtzs:Dtu-f"(t) Hence, if we are given o or fl aS a function of f , as well aS Some boundary conditions, it is possible to determine the equation of motion by antidifferentiation. This is illustrated in the following example.


ExAMPLE 1.: A particle is moving on a straight line; s is the number of feet in the directed distance of the particle from the origin at f sec of time, u is the number of ftlsec in the velocity of the particle at f sec, and a is the number of ft/sec2 in the acceleration of the particle . I andv :3, a t f s e c .l f n : 2 t and s : 4 when t: t, express u, and s as functions of f.

soLUTroN: Becausea:

D{s, we have

d'-Jt-1 dtr which,

expressed

in differential

form,

gives

dzs: (2t - t) dt Antidifferentiatin g, we have

rr IJ J d u : I e t - r ) d t from which we get a:P-t+Cl Substitutinga: 3 and t:

(1)

L, we have

3:1,-L+Cl

:il*rT*€

cr :3. Therefore, substitutingthis valueof c, intf Eq. (1),we

u:t2-f+3

(2)

which expresses? as a function of f. Now, letting u : D 7sin Eq. e), *e have

+d t :e LP - t + Z and writing this in differential form gives ds:(tr_f+3)dt Antidifferentiating, we have rt dt: | (tr-t+s) dt I JJ s:*ft-ttr+gt+Cz

(3)

Letting s:4 and f : f. in Eq. (3) and solving for Cr, we o$tain Cz: t. Therefore, by substituting 6 fot C, in Eq. (3f we have s exfressed as a function of t, which is s:*fa-+tr+gt++ If an object is moving fueelyin a vertical line and is leing pulled toward the earth by a force of gravity, the accelerationof grorlty, denoted by g ftfsecz,varies with the distanceof the object from ti" dur,t"r of the earth. However, for small changesof distancesthe acceleratiqnof gravity

AND RECTILINEAR MOTION 6.5 ANTIDIFFERENTIATION

is almost constant, and an approximate value of g, if the object is near sea level, is 32. EXAMrLE 2: A stone is thrown vertically upward from the ground with an initial velocity of L28 ftlsec. If the only force considered is that attributed to the acceleration of gravity, find how high the stone will rise and the sPeed with which it will strike the ground. Also, find how long it will take for the stone to strike the ground.

so1,urroN: The motion of the stone is illustrated in Fig. 6.5.1.The positive direction is taken as upward. Let t: the number of seconds in the time that has elapsed since the stone was thrown; s: the number of feet in the distance of the stone from the ground at / sec of time; u : the number of feet per second in the velocity of the stone at f sec of time; lrl: the number of feet per second in the speed of the stone at f sec of time. The stone will be at its highest point when the velocity is zero. Let s be the particular value of s when o : 0. When the stone strikes the ground, s : 0 . L e t f a n d a b e t h e p a r t i c u l a r v a l u e s o ff a n d p w h e n 5 : 0 , a n d t + 0 . Table 6.5.1 is a table of boundary conditions for this problem. The positive direction of the stone from the starting point is taken as upward. Because the only acceleration is due to gravity, which is in the -32 ft/sec2. downward direction, the acceleration has a constant value of Because the acceleration is given by the derivative of o with respect to f,

r'T{:=: we have da dt

-32

Writing this in differential form, we have da : -32 dt

f,:7

1r:o

lo: a

F i g u r e6 . 5 . 1

Antidifferentiatingr w€ have

I a,: [

-2,a'

from which we obtain a:-32t*Ct Becauseu : L28 when t : 0, we substitute these values in the above and get C1:'l'28. Therefore, we have a:

-32t + L28

(4)

Because u is the derivative of s with respect to f, then a : dsldt, and so T a b l e6 . 5 . 1

t s 7)

OT 0s0 I280a

dc -32t + 128 :dr:

which in differential form is ds:

(-32t + L28) dt


Antidifferentiating gives

I*:l

(-gzt + I28) dt


s: -L6t2 + 128t* C, B e c a u sse: 0 w h e n t : A , we get Cz:0, and substituting 0 for C2in the above equation gives s: -L6t2 + L28t

Substituting,tfor t and 0 for s, we get 0: -L6l(i- 8) from which we obtain l: 0 and f : g. However, the value 0 occurswhen the stone is thrown; so we concludethat it takes8 secfor the stone to strike the ground. To obtain a, substitute 8 for f and a for a in Eq. (4), which gives 6: (-32X8) + 128,from which we obtain a: -L29. So l7l : L2g.There_ fore, the stone strikes the ground with a speed of 1.2gftlsec. To find i we first find the value of t for which a is 0. From Eq. (4), we obtain t:4 when u:0.In Eq. (5) we substitute4 for f and s fbr s and obtain s:256. Thus, the stone will rise 256 ft.

Exercises6,5 In Exercises 1 through 4, a particle is moving on a slraight line, s ft is the_directed distance of the paticle from the origin at I 8ec of time, a ft/eec is rhe velocity of ttre palticle ati sec, and4 ft/s€c ;;;;;";1;;;;ir,1*p'"1"t "*,.". 1. a:5-Zt; a:2 and s:0 when t:0. Express? and s in terms of ,. 2. a-3tf,; p:f 4rd s:1when t:1. Expressz and s in terns of t. 3. c=8@; rr:20whens:1.

Find an equationinvolving - o

"rrd

r. /"rr.rr, \ dt dsdt "=+:4*:"@\

-

ds) Find an equation involving u and s. (nrr.rr:SeeHint for Exercise 3.) In Exercises 5 throush 10' the orilv force considered is that due to the acceleration of gravity, which we tak e as 32 ftlsec in the downward dir=ectiorr4. a:2s + L u:2

when s:1.

5' A stone i: vertically upward from the gound with an initial velocity of 20 ft/eec. lf*Y" How long will it take the t9--etrike the ground, and with what speid witt it strike? How tong r,"ilt th. :tot-rc g.lrrj and how high wil it go? "t"""i" "p-"ra, 6' A ball.is dropped from the top of the was_hington monument, which is 555 ft high. How long will it take the ball to readt the groutld, and with what speed will it strike the glound? 7' A man in a balloon droDs his binoculars when it is l5o ft above the gound and rising at the rate of l0 ttlsec. How long will it take the bin&dare to strike the gouna, -fr"t i, if,Ar Jpeea on i-paci ".a

IN ECONOMICS 269 OF ANTIDIFFERENTIATION 6.6 APPLICATIONS

8 . A stone is thrown vertically upward from the top of a house 60 ft above the ground with an initial velocity of 40 ftlsec. At what time will the stone reach its greatest height, and what is its greatest height? How long will it take the stone to pass the top of the house on its way down , and what is its velocity at that instant? How long will it take the stone to strike the ground, and with what velocity does it strike the ground?

9. A ball is thrown vertically upward with an initial velocity of 40 ftlsec from a point 20 ft above the ground. lf a ftlsec is the velocity of the ball when it is s ft from the starting point, express o in terms of s. What is the velocity of the ball when it is 36 ft from the ground and rising? (Hrnr: See Hint for Exeicise 3.)

1 0 . If a ball is rolled across level ground with an initial velocity of 20 ftlsec and if the speed of the ball is decreasing at the rate of 6 ftlsecz due to friction, how far will the ball roll?

1 1 . If the driver of an automobile wishes to increase his speed from 20 milhr to 50 mi/hr while traveling a distance of 528 ft, what constant acceleration should he maintain?

L2, What constant acceleration (negative) will enable a driver to decrease his speed from 60 miihr to 20 mi/hr while traveling a distance of 300 ft?

13. If the brakes are applied on a car traveling 50 mi/hr and the brakes can give the car a constant negative acceleration of 20 ftlsecz,how long^will it take the car to come to a stop? How far will the car travel before stopping? a downward L4. A ball started upward from the bottom of an inclined plane with an initial velocity of 5 ftlsec. If there is acceleration of 4 ftlsecz, how far up the plane will the ball go before rolling down?

greatest speed it may 15. If the brakes on a car can give the car a constant negative acceleration of 20 ft|secz, what is the be going if it is necessary [o be able to stop the car within

80 ft after the brake is applied?

and the marginal 6.6 APPLICATIONS OF In Sec. 5.6 we learned that the marginal cost function function and the ANTIDIFFERENTIATION revenue function are the first derivatives of the total cost function IN ECONOMICS total revenue function, respectively. Hence, if the marginal cost

and the marginal revenue function are known, the total cost function and the total ,errerrnu function are obtained from them by antidifferentiation. In finding the total cost function from the marginal cost function, the arbitrary constant can be evaluated if we know either the fixed cost (i.e., the cost when no units are produced) or the cost of production of a specific number of units of the commodity. Because it is generally true that the total revenue is zero when the number of units produced is zero, this fact may be used to evaluate the arbitrary constant when finding the total revenue function from the marginal revenue function.

nxarvrprE 1: If the marginal revenue is given by 27 - 12x * x2, hnd the total revenue function and the demand equation. Also determine the permissible values of x. Draw sketches of the demand curve, the total revenue curve, and the marginal revenue curve on the same set of axes.

solurroN: If R is the total revenue function, the marginal revenue func' tion is R', and R'(r) :27 - 12xI x2 Thus,

ntt) :

/

(27 - LZx* x2) dx

:27x-6x2**r3*C


BecauseR(0)

R(r) :27x - 6x2+ 1#t

J5

If P is the price function, R(x) : xP(r), and so P(r) :27 - 6x + *x2

30 25 20

we get C : 0. Hence,

demand curve

15 10 5

o -5 -10 F i g u r e6 . 6 . 1

rxavrprE 2: The marginal cost function C' is given by C'(r) : 4x - 8. If the costof producing 5 units is $20,find the total cost function. Determine the permissible values of r and draw sketchesof the total cost curve and the marginal cost curve on the same set of axes.

If we let p dollars be the price of one unit of the commodity when r units are demanded,and becausep : P(x), the demand equation is 3p :81 - 1'8x* f To determine the permissible values of x, we use the facts that r > 0, p > 0, and P is a decreasingfunction. Because P'(x)

3x

P is_decreasing when x <9 (i.e., when p'(x) < 0). Also, when x:9, : 0, and so the permissible values of r are the numbers in the closedinP terval [0,9]. The required sketchesare shown in Fig. 6.6.t. solurroN: The marginal cost must be nonnegative. Hence, 4x- 8 = 0, and therefore the permissible v a l u e so f . x a r e x > 2 . C'(x):4x-8 Therefore,

r C(x):1 (4x-8) dx J

:2tc2-8x*k BecauseC(5) : 20, we get k: I0. Hence, C(r) :2x2 - 8x+ 10 (x = 2) The required sketchesare shown in Fig. b.6.2. v

Figure6.6.2

IN ECONOMICS OF ANTIDIFFERENTATION 6.6 APPLICATIONS

3: After experiEXAMPLE mentation, a certain manufacturer determined that if r units of a Particular commodity are produced per week, the marginal cost is given by 0.3r - 17,where the Production cost is in dollars.If the selling price of the commodity is fixed at $19per unit, and the fixed costis $100per week, find the maximum weekly profit that can be obtained.

Let C(r) : the number of dollars in the total production cost of r units; : R(r) the number of dollars in the total revenue ob-

solurroN:

tained by selling r units; the number of dollars in the profit obtained by selling x units. Becauser units are sold at $19 per unit, we have

s(x)-

R(r) : l9x The marginal revenue then is given by R'(r) : t9 C'. is the marginal cost function, and we are given that C'(r):0.3r-11 The profit will be a maximum when marginal revenue equals marginal costlsee Sec.5.5).Equating R'(r) and C'(x),ws get x:100. So 1'00units should be produced each week for maximum profit. (0.3r - 11,)dx

C(r)

- LLx * k : 0.1.512 Becausethe fixed cost is $100, C(O;: 100, and hence fore, C(r) :0.15yz- \Lx + L00

L00. There-

SettingS(r) - R(r) - C(x), we have - ILx + 100) S(r) : 19x- (0.15x2 - -0.1512+ 30x- 100 Hence, 5(100) :1400. Therefore, the maximum weekly profit is $1400, which is obtained if L00 units are produced weekly'

nxlrvrpln 4: A business firm has made an analysis of its Production facilities and its Personnel. With the present equiPment and number of workers, the firm can produce 3000 units Per daY. It was estimated that without anY change in investment, the rate of change of the number of units produced per day with resPect to a

Let U:

solurroN:

+: dx

the number of units produced per day' Then

Bo- 6xttz

from which we get dy:

(80 - 5*rrz) dx

Antidifferentiating U :8 0 x

gives us

- 4 xgt2+ C

272


change in the number of additional workers is 80 - 6x1t2,where r is the number of additional workers. Find the daily production if 25 workers are added to the labor force.

BecauseU :300A when x: A, we get C : 3000.Hence, y -80x- 4x3t2 + 3000 Letting Azsbe the value of y when r : 25, we have Azs:2000 - 500+ 3000 Therefore,4500units are produced per day if the labor force is increased by 25 workers.

Exercises 6.6 1' Find an equation of the total revenue cule of company iI its slope at any point is given by 12 - 3r and p(4) = 6, I where P is the price function. Draw sketches of tlte toLl revenue curve and the derrand curve on the game set of axes. 2' The rate oJ change of the slope the totakost curve of a particular company is the constant 2, and the total cost curve _of contains the points (2,72) and (3, 18). Find the total cost function. 3' Find the demand equation for a commodity for which the marginal revenue function is given by 10/(r + 5), - 4. 4. The rrarginal cost function i8 given by 3llET-q,.If the fixed cost is zero, find the total cost function. 5. The urarginal revenue function is given by 16 - 3rp. Find the permissible values of t, the total revenue function, and the demand equation. Draw sketches of the demand curve, tire total revenue curye,-u"a ,f," _"[in"f ."venue curye on the Bameset of axes. 6. Find the total revenue function, the demard equation, and the permissible values of x if the marginal revenuefunction is given by frf - 10r + 12. Also draw sketches of the dem*a trt"1t-,a-""ri"r" revenue curve on the same set of axes. "ir*", "ir*i,"""i"ri.ri--ginal 7. The marginal cost function iB given by 3rl + 8r + 4, and the fixed cost is g6. If C(r) dollars is the total cost of , units, find the total cost function, and draw iketches of the total cost curve and the marginal cost curve on the same set of axes. 8' The marginal cost function c' is defined by C'(r) :6r, where c(r) is the number of hundreds of do ars in the total cost of r hundred units of a certain-comm'odity. u ttt"'*"i J irb units is g2000, find the total cost function and the fixed cost. Draw sketches of the totar cost curve ana trr" -*6"J cost culve on the same set of axes. 9' The cost of a certain piece of machinery is $700 and its value depreciates with time according to the formula D,y: -500(' + l)-'z where v dollars is its value t years after its purchase.what is its value three years afte! its purchase? 10' suPpose that a particular company estimates it' growth in income ftom sales by the formula D,s : 2(, - 1)r/3, where s millions of dollars i8 the incone from saLs r y""* i"ii". r trr" g-"" ii..o^" iro,,'-it ye-,, s"t"" is "'ossexpected gross irr.o*Jlrorn 98 million, what should be ihe "'lrri"r,t ,"i". t*o years from now?

ReoiewExercises (Chapter6) In Exercises 1 through 5, find the most general antiderivative.

,L. f Vxt , J iF-ax +.

,f$-g)a,

f_

Jk3*x)!x2*3dx

5.

f_

J

\/4x* 3(* + 1,)dx

3. 6.

REVIEWEXERCISES 273 In Exercises 7 through 10, find the complete solution of the given differential equation.

7. x'tt ' 9: tlx

fu' - 1)'

9. y' dx* Y zdY : dY

du "' dx

R -:-

10.

d2u

#:

xIL \/x lZxz- 30x

11.If.y:80r-'J.6x2,findthedifferenceAy-dVif(a)r:2,Lx:0.1;(b)r:4,Lx:-0.2. 12. Use differentials to find an apProximatevalue of fl126. -1) is on the curve. Find an 1 3. The slope of the tangent line at any point (x, y) on a curve is L0 4x and the point (1, equation of the curve. f and an equation of the tangent line to the curve at the Point (1, -1) is L4. At any point (r, y) on a curve D,"y:4: 2r 3y 3. Find an equation of the curve. 1 5. ll xs * yB- 3xy?-f l: 0, find dy at the point (1, 1) in fu: 16. Find d.yld.tiI y: rs - 2x + 1 and xl + ts - 3l: 0. 17. Fnd dyl dt it 8f + 27f - 4xy : S and F * fx - f : 4.

O.1,-

18. Evaluate J (.ra)04d dr as Juo du and as l4x2r dx and compale the results. 19. Evaluate J(ri * l)2* dx by two methods: (a) Make the substitution u: f'l l; (b) first expand (f * 1)'. ComPare the results and exptain the difference in the appearance of the answers obtained in (a) and (b). 20. Find the particular solution of the differential equation f dy : y3 dx fot whiclj. y:

1 when r = 4.

21,. FindtheparticularsolutionofthedifferentialequationD'2y:\rQltrforwhichy:3andy':2whent:4' 22. A ball is threwn vertically upward from the top of a house 64 ft above the ground, and the initial velocity is r18ft/sec. How long will it take the ball to reach its greatest height, and what iB it6 greatest height? How long will it take the ball to strike the ground and with what velocity does it strike the ground? 23. Suppose the ball in Exercise 22 is thrown downward with an initial velocity of 48 ftlsec. How long will it take the ball to strike the ground and with what velocity does it strike the 8lound? 24. Suppose the ball in Exercise 22 is dropped from the top of the house. Determine how long it takes for the ball to stiike the ground and the velocity when it strikes the ground. 25. Neglecthg air resistance, if an obiect is dropped from an airplane at a height of 30,000 ft above the ocean, how long would it take the obiect to strike the water? 26. Suppose a bullet is lired direcfly downward from the airplane in Exetcise 25 with a muzzle velocity of 2500 ft/sec. If air resistance is neglected, how long would it take the bullet to reach the ocean? 27. A container in the form of a cube having a volume of 1000in.3 is to be made by using 6 equal squares of material costing 2 cents per square inch. How accurately must the side of each square be nade so that the total cost of the material shau be corect to within 50 cents? 28. The measure of the radius of a right-circular cone is * times the measure of the altitude. How accurately must the altitude be measured if the error in the computed volume is not to exceed 3%? 's 29. If t sec is the time for one complet€ swing of a simple pendulurn of length I 11,gt"t1 4721: gP, where 8: 32.2. What the effect upon the tim€ if an eror of 0.01 ft is made in measuring the lentth of the pendulun? 30. An automobile traveung at a constant speed of 60 mi/hr along a straight'highway fails to stoP at a stoP sign. If 3 sec l,ater a highway patrol car starts from rest from the stop sign and maintains a constant acceleration of 8 ft/geC, how long will it take the patrol car to overtake the automobile, and how far from the stoP sign will this occur? Also deteF mine th€ speed of the Pabol car when it overtakes the automobile.

271


37. If l(x) = r * lr - tl, and Ftr\:[x lt'-x+t

if r<1 if r>1

show that F i8 an antiderivative of I on (-c., +ce). 32. Let / and I be two tunctions sucl_rth31_f91-alt il and g,(r) : *l(r). Furthermore, suppose I !_-, +-), f,(!:g(r) thatl(0) : 0 andS(0) : 1. Provethar_[(r)]'+.[g(r)]': t. lnrwr: Consi'derth" r.i iur". r'"ii'C wiere r(r) : t/(r)1, artd c(:) : -[g(r)]2 and show rhat,F,(x) : c,G) for all x.i 33' The marginal revenue function is given by abl(x * b)2- c. Find the total revenue function and the demand equation. 34' A company has determined that the marginal cost function for the production of a particular commodity is given by C'(r) : 125'tr-1gt - " where C(x) dotlars i, *"'"or, ing 15 units?

ot p-ao"ing

r units of the commodity. If the fixed cost is $250,what is the cost of produc-

The definite integral

THE DEFINITE INTEGRAL

7'1 THE SIGMA NOTATION

In this chapter we are concemed with the sums of many terms, and so a notation called the sigma notation is introduced to faciliiate writing these sums. This notation involves the use of the symbol ), the capital silma of the Greek alphabet, which correspondsto our letter S. Some examplesof the sigma notation are given in the following illustration. o rLLUsrRATroN 1: 5

\F:72+22f32+42+52

(3t+2): t3(-2)+21+ t3(-1)+ 2l + 13'0 + zl 1>= - ' + [ 3 . 1 + 2 J+ 1 3 . 2 + 2 ] : ( - 4 )+ ( - 1 )+ 2 + s + 8 n

*ns

)it:Ls+23+33+ j:1

};:f++.+*t*+.t We have, in general, n

F ( i ): F ( m )+ F ( m+ 1 ) + F ( m* 2 ) +

+ F(n)

i:m

(1)

where m and n are integers, and m < n. The right side of formula (1) consists of the sum of (n - m * 1) terms, the first of which is obtained by replacing i by m in F(i), the second by rePlacing iby (m * 1) in F(i), and so on, until the last term is obtained bv replacing i by n in F(i). The number m is called the lower timit of the sum, and.n is called the upper limit. The symbol i is called the index of summation. rt is a ,,dummy,, symbol because any other letter can be used. For example, 5

k2:32+42+52 k:3

is equivalent to 5

i2:32+42+52 i:3

rLLUSrRArroN 2: By using the formula of Eq. (1), we have $ t r _ 3 r _ 4 r _ 5 r , 6''

ei+

1

3+1

4+']..'5+1 ' 6+']-.

Sometimes the terms of a sum involve subscripts, as shown in the next illustration.

7.1 THE SIGMA NOTATION 3:

O ILLUSTRATION n

'*A'

)a':A'+A2+'

j:'

gbs pûon: 4b++ 5b5+ 6b6+ 7b7+ 8bB+ a

2fQ)

A x : f ( x , ) A r * f ( x r )A x * f ( x r )A r * f ( * ) A r * f ( x ' ) L x .

,n" n"tt"wing propertiesof the sigmanotation are useful,and they are easily proved. cn where c is any constant

,:

7.'t.1,Property 1, f

The proof is left as an exercise(seeExercise9).

property2 7..t..2

ii : l r. F(i): r i

ftO wherec is any constant

PROOF: n

F(1)+c'F(2)*c'F(3)+

T c'F(l) LI

ic'F(n)

i:l

: c [ F ( l ) + F ( 2 )+ F ( 3 )+ ' ' '

+F(n)]

=r ; r 1 ; 1 i:r

7.1..3Property 3

i trrr)+ c(t)l:2 rtil +; cr;l

.L/

-

i:1

L/

i:l

.?

i:7

The proof is left as an exercise (see Exercise 10). Property 3 can be extended to the sum of any number of functions.

7.1.4 Property 4

n t F ( t )- F ( ,- 1 ) l - - F ( n )- F ( 0 ) > i:1

PROOF:

n - F ( 1 ) ] + [ r ( 3 )- F ( 2 ) ] + ' tZ-/ t- r t i l - F ( i - 1 ) l : [ F ( 1 )- F ( 0 ) ] + [ F ( 2 )

i:1

+ l F ( n - 1 ) - F ( n - 2 ) l + l F ( n ) - F ( n - 1 )l

THE DEFINITEINTEGRAL

: - F ( 0 ) + [ F ( 1 )_ r ( r ) ] + [ F ( 2 )_ r 1 z ; 1 + . . . + l F ( n - 1 ) - F ( n- 1 ) l + F ( n ) :-F(0) +0+0+ . . . + 0+ F(n) : F(n) - F(0)

I

The following formulas, which are numbered for future reference, are also useful. n

7.'l.,.5Formula L

. n(n*71 ,:T

i:r n

7.1.6 Formula 2

. r-_b n ( n * 1 ) ( 2 n + 1 )

i:r n

7.'1,.7Formula 3

,, r - -f l- '-(4n-+ 7 ) '

i:r 'n

7.'1,.8Formula 4 i:1

n(n r t) (6nBt 9n2+ n - 1) ,+_ 30

Formulas 1 through 4 can be proved with or without using mathematical induction. The next illustration shows how Formula L can be proved without using mathematical induction. The proof of Form ula 1,by mathematical induction is left as an exercise (see Exercise 11). . ILLUSTRATToN 4: We prove Formula 1. n

) l:1,+2+3+

. . . -t (n-t) *n

and n

)i:n*(n-1)+

(n-2)+..

+2+t

,it*" add thesetwo equationsterm by term, the left side is n

2>i ""0

.;;he

right side are n terms, each having the value (n + 1). Hence,

n

2 > i : ( n + 1 ) + ( n * L ) + ( n+ 1 ) + ' i:r

:n(n*t) Therefore,

S\ : , _ n ( n,)+ t )

z.,l i=l

'

'+(n+I)

nterms

7.1 THE SIGMA NOTATION

EXAMrLE1: Prove Formula 2bY mathematical induction.

solurroN:

We wish to Prove that

*,, _n(n+L)(Zn+I)

>r:

6

'L.The left sideis then2 ,': l'. When ,rrr,ln" formulais verifiedfor n: '2'3-li6- l. Theren : L , t h er i g h ts i d ei s [ 1 ( 1+ 1 ) ( 2+ D ) t e - ( 1 fore, the formula is true when n: 1. Now we assume that the formula is true for n: k, where k is any positive integer; and with this assumption we wish to prove that the formula is also true for n:k + 1. If the formula is true for tt: k, we have k

S ./-/ i , : i:1

Wh e n n : k+l

f

k(k + 1,)(2k+ 1,) 6

(2)

k * L , w e have

i r : 1 2 * 2 2 + 9 2 + . . . + k 2+ ( k + 1 ) '

i:l k

:)i2+(k+1)' + L) (k 1), (by applyingEq. (2)) _ k(k + L)_(zk + + 6

_ k ( k+ L )( 2 k+ L ) + 6 ( k+ L ) ' z 6 -_- ( k + 1 )[ k ( 2 k+ 1 ) + 6 ( k+ 1 ) ] _ (k+L)(zk'z+7k+6) 5 _ (k+1)(k+2)(2k+3) 6 - ( / c +1 ) [ ( k + 1 ) + 1 ] [ 2 ( k +1 ) + 1 ] 6 Therefore, the formula is true f.ot n: k * L. We have proved that the formula holds f.or n: 1, and we have also proved that when the formula holds f.or n: k, the formula also holds for n: k * 1. Therefore, it follows that the formula holds when n is any positive integer.

A proof of Formula 2, without using mathematical induction, is left as an exercise. The proofs of Formulas 3 and 4 are also left as exercises (see Exercises L2 to 76).


ExAMpLE 2: n

Evaluate

soLUTroN: From Property 4, where F(i) : 4i,1t follows that n

(4'- 4i-')- 4 - 4 0 > i:r

(4' - 4i-')

t

.{-J

i:r

:4n-I

nxaivrPrn3:

Evaluate

SOLUTION:

n

i(3i- 2) > i:r

n

r(3t- 2) > i:r

(3i2 - 2i)

by using Properties 1-4 and Formulas 1-4.

(3i')

:3

+> (-2i)

i'- Z2 i > i:7

(by Property 3)

(by Property2) n(n+ 7) 2 (by Formulas2 and 1 )

2nBl3nzIn-2n2-2n

2 2n3*n2-n

Exercises7.1 In Exercises 1 through 8, find the given sum. 6"

1 .> ( s i - 2 ) i=l

t $

2.>(i+1.)' i:r

2

KFz

-i . + , -( - t ; u * r ) , k F

3

U ,D-,''

3.

it

?

i-1,

s1

6. z ' - l + i 2

3 b 8 " '. y t I ,ro-rk+3

9. Prove Property 1 (7.1,.1,). 10. Prove Property 3 (7.1.3). 11. Prove Formula 7 (7.1.5)by mathematicalinduction. L2. Prove Formula 2 (7.1.6\without using mathematicalinduction. (HrNr: iB- (i - 7)":3i ' - 3i + L, so that nn

)

t ; '- ( i - 1 ) ' l : )

tgt,- 3t+ 1)

On the left side of the above equation, use Property 4; on the right side, use Properties '1,,Z, and 3 and Formula 1.)

7.2 AREA 281 13. Prove Formula 3 (7.1.7)without using mathematicalinduction. (HrNr: ia- (i - qa:4p - rO * 4i- 1, and use a method similar to the one for Exercise12.) 14. Prove Formula 4 (7.1.8)without using mathematical induction (seehints for Exercises12 and 13 above). 15. Prove Formula 3 (7,1.4 by mathematical induction. 16, Prove Formula 4 (7.1.8)by mathematicalinduction. In Exercises17 through 25, evaluatethe indicated sum by using Properties 1 thlough 4 and Formulas 1 through 4.

t

20

1 8 .> 3 i ( i 2 + 2 )

2zi(i-1,)

i:r n

n

20. > (2k-t-zk)

19.>(10i*'-10,)

k:r

40 22.>3/T+r -\Dl-l

--'f:,Lk k+L) ,r.''flff-=+-l ,t

24iz(i

i:r n

- z\

24.>\i(L+i2) i--l lr

n

25. > [(3-o - 3o)'- (3t'-t+ 3-k-r)2]

f

/i\21r12

2 6 . P r o v e :) l t - { - } | \nl J ,?nL

:2}

n

? = ,[ ' -

(*)']''' *,

Prope rty4 toshowthat

,.i;Use

) F ( n )- F ( 1 )- r ( 0 ) > ttt, + 1)- F(i- 1)l : F(nr-1'+

Lli:1

(b) Prove that n

t

,1J

t(t+1)'-(i-'

1 ) ' l: ( n * 1 , ) 3 * n 3 - L

i:r

(c) Prove that nnn

)

t t ; + 1 ) ' - ( i - 1 ) ' l : 2 ( 0 i '+ 2 ) : 2 n* 6 )

i2

i:r

(d) Using the results of parts (b) and (c), prove Formula 2. 28. Use the method of Exercise 27 to prove Formula 3. n )t'

n

29. If X:El - , pr ov e n

t ha t )

(x , - X)' :2

i:r

7.2 AREA

n

i=l

r.

r,' - X 2 * , i--L

We use the word mensure extensively throughout the book. A measure refers to a number (no units are included). For example, if the area of a triangle is L0 irt.z, we say that the measure of the area of the triangle is 10. You probably have an intuitive idea of what is meant by the measure of the area of certain geometrical figures; it is a number that in some way


Figure7.2.1

Figure7.2.2

gives the size of the region enclosed by the figure. The area of a rectangle is the product of its length and width, and the area of a triangle is half the product of the lengths of the base and the altitude. The area of a polygon can be defined as the sum of the areas of triangles into which it is decomposed, and it can be proved that the area thus obtained is independent of how the polygon is decomposed into triangles (see Fig. 7.2.1). However, how do we define the measure of the area of a region in a plane if the region is bounded by a curve? Are we even certain that such a region has an area? Let us consider a region R in the plane as shown in Fig. 7.2.2. The region R is bounded by the r axis, the lines x: a and x: b, and the curve having the equation y -- f(x), where f is a function continuous on the closed interval [a,b]. For simplicity, we take f (x) > 0 for all x in [a,bl.We wish to assign a number A to be the measure of the area of R. We use a limiting process similar to the one used in defining the area of a circle: The area of a circle is defined as the limit of the areas of inscribed regular polygons as the number of sides increases without bound. We realize intuitively that, whatever number is chosen to represent A, that number must be at least as great as the measure of the area of any polygonal region contained in R, and must be no greater than the measure of the area of any polygonal region containing R. We first define a polygonal region contained in R. Divide the closed interval [a, b] into resubintervals. For simplicity, we shall now take each of these subintervals as being of equal length, for instance, Ar. Therefore, Ar: (b - a)ln. Denote the endpoints of these subintervals by xo, xt, x 2 , . . , x n - r ,r r , w h e r € f f o : f l , x t : a i L x , . . . , x i : A * i L , x , . . , xn-r: a -l (n - t) A,x,xn: b. Let the ith subinterval be denoted by lxr-r, xrf. Because / is.continuous on the closed interval la, bf, it is continuous on each closed subinterval. By the extreme-value theorem (4.5.9), there is a number in each subinterval for which / has an absolute minimum value. In the lth subinterval, let this number be ci, so that /(c) is the absolute minimum value of f on the subinterval fx,-r, r1]. Consider n rectangles, each having a width Ax units and an altitude f (r,) units (see Fig. 7.2.3). Let the sum of the areas of these n rcctangles be given by S, square units; then

S":f(cr) Ax*f(rr) Ax*'

'+f(c,) Ax.r-

* f(c") Lx

or, with the sigma notation, n

sn:\

f (c) Ax

(1)

The summation on the right side of Eq. (L) gives the sum of the measuresof the areasof n inscribed rectangles.Thus, however we define A, rt must be such that A= S,

7.2 AREA

Figure7.2.3

Figure7.2.4

In Fig. 7.2.3 the shaded region has an area of 5n square units. Now, let n increase. Specifically, multiply nby 2; then the number of rectangles is doubled, and the width of each rectangle is halved. This is illustrated in Fir.7.2.4, showing twice as many rectangles as Fig. 7.2.9. By comparing the two figures, notice that the shaded region in Fig. 7.2.4 appears to


approximate the region R more nearly than that of Fig.7.2.3. So the sum of the measures of the areas of the rectangles in Fig. 7.2.4 is closer to the number we wish to represent the measure of the area of R. As n increases, the values of s, found from Eq. (1) increase, and successive values of 5, differ from each other by amounts that become arbitrarily small. This is proved in advanced calculus by a theorem which states that if / is continuous on fa, b], then as n increases without bound, the value of S, given by (1) approaches a limit. It is this limit that we take as the definition of the measure of the area of region R. 7.2.'l' Definition

Suppose that the function/ is continuous on the closed interval la,bl,with f (x) > 0 for all x in fa, b), and that R is the region bounded by the curve y - f(x), the x axis, and the lines x: a and r: b. Divide the interval la,bI into n subintervals, each of length Ax: (b - a) ln, and denote the ith subinterval by lxt-t, xil. Then if f(c) is the absolute minimum function value on the ith subinterval, the measure of the area of region R is given by

rim ) /(r,) ar

A*+€b {*1

(2)

.

Equation (2) means that, for any e ) 0, there is a number N ) 0 such that

-ol l>f,',,a,x . .

whenever rz ) N

and n is a positive integer. We could take circumscribed rectangles instead of inscribed rectangles. In this case, we take as the measures of the altitudes of the rectangles the absolute maximum value of / on each subinterval. The existence of an absolute maximum value of / on each subinterval is guaranteed by the extreme-value theorem (4.5.9). The corresponding sums of the measures of the areas of the circumscribed rectangles are at least as great as the measure of the area of the region R, and it can be shown that the limit of these sums as n increases without bound is exactly the same as the limit of the sum of the measures of the areas of the inscribed rectangles. This is also proved in advanced calculus. Thus, we could define the measure of the area of the region R by

A:

n

lim 2 f @) Lx

(3)

wheref (";.;"'.]1" urrotutemaximumvatueof / on lx,-,, x,l. The measure of the altitude of the rectangle in the lth subinterval actually can be taken as the function value of any number in that subinterval, and the limit of the sum of the measures of the areas of the rectangles is the same no matter what numbers are selected. This is also proved in advanced calculus, and later in this chapter we extend the definition of the measure of the area of a region to be the limit of such a sum.

7.2 AREA EXAMPLEL: Find the area of the region bounded by the curve A : x', the r axis, and the line x: 3 by taking inscribed rectangles.

solurroN: Figure 7.2.5 shows the region and the ith inscribed rectangle. We apply Definition 7.2.L. Divide the closed interval [0, 3] into n subintervals, each of length A,x: xs: 0, xr: Lx, xz: 2 Ax, . . ., xi: i Lx, . . r x n -r: (n - t) A,x,xn: 3. ^

-

rJ,,t,

3-0 nn

3

The function value /(x) -- xz and because / is increasing on [0, 3], the absolute minimum value of f on the ith subinterval lxr-r, r;] is f (xr-r). Therefore, from Eq. (2) n

A:

lim D fk,-r) Lx

Because ::-,-::;-

( 4)

1) Ax and /(x) : ,sz,we have

f (x,-,)- [(t 1) Ar]'z Therefore, nn

) f (x,-') Ar: > (t 1)'z(Ar)g 8". ;; : ztn,ro tnut: n n s ( t - L ) ' -Ln:,2 7 n A r : / ( r ' ' ) ) > sL i:r i:r

:4[$;,-2$ no

t+$

r), ']

,?:, ,?:, f4 and using Formulas 2 and L and Property L from Sec.7.'1., we get

._27 ln(n+t)(2n+L) ,: 2fQ'-')Ax:lirt 27

_2.ry!*"1

2n3*3n2*n-6n2-6n*6n

n3

9 2

2n2-3n*L ___v-

Then, from Eq. (4), we have

(,-1.#) :*(2_0+0) -9 Therefore, the area of the region is 9 square units.


2: Find the areaof the ExAMPLE region in ExampleL by taking circumscribed rectangles.

solurroN: we take as the measureof the altitude of the lth rectangle the absolute maximum value of / on the lth subintervaI lxi4, 11] which is f (x,). From Eq. (3), we have n

A: lim 2 f k,) tx Because [!

r'ir,thenf (x,): (i a,x)z,and so

n

n

Ar: 2f(',) i:r

]

L/ i:1

)7n

i2(A,x)S:

+n"s?- i , l = l

27 l n ( n+ 7 )( 2 n+ 1 ) ' l n3

L6l

2n2*3n*L _____7Therefore, from Eq. (5), we obtain q

A:

Itm n-+6

,

.(r*q+4) \

/r

n"/

( a s in Example1)

uxaupru 3: Find the area of the trapezoid which is the region bounded by the line 2x * y : 8, the x axis, and the lines x : 1 and x: 3. Take inscribed rectangles.

soLUTroN: The region and the ith inscribed rectangle are shown in Fig. 7.2.6.The closed interval [1, 3] is divided into n subintervals, each of length A,x; xs: L , xr: "1.* Lx, xz: I * 2 A,x, xi:t*i Lx,

1.-t (n - L) L,x,xn 3-1

Solving the equation of the line for y, we obtainy:-2x * B. Therefore, f (x): -2x* 8, and because / is decreasing on [1,3], the absolute minimum value of f on the ith subinterval lxi_r, ri] is f (rr). Because ri: 1 + i Ax a nd f(* ) : -2x* 8, then f(x,) : -2(L+ i A x) * 8 : 6 - zi Ax. Fr om Eq,. (2), we have

f(x) A,x

lim fl-*@

: lim i tu 2i a,x)ax n-+@ i:l F i g u r e7 . 2 , 6

:

n

lim )

2-+@

i:l n

i:r

I0 Ax - Zi(A,x)'z1

,,G)',1 I'G)-

7.2 AREA

:

lim ll-lo

287

l+y-i-E'l

we have Using Property 1,(7.1.1)and Formula l, (7.1,.5),

: lim ft-A) nl - --\

h-*@

-8 Therefore,the area is 8 square units. Using the formula from plane geometry for the area of a trapezoid,A:ih(b1*b2), where h,br, andb, are, respectively,the number of units in the lengths of the altitude and the two bases,we get A: +(2)(6+ 2): 8, which agreeswith our result.

7,2 Exercises In Exercises 1 through 14, use the method oI this section to find the area of the given region; use inscribed or circumscribed rectantles as indicated. For each exercise, draw a figure showing the redon and the ith rectangle. The region bounded by y: r', the x axis, and the line r = 2; inscribed rectangles, J. 2. The region of Exercise 1; cirormscribed rectangles. ..\ The re6on boun dedby y :2x,

1 and r : 4; circumscribed rectangles.

tte r axis, and the lines r:

4. The region of Exercise 3; insclibed rectangles. 5l The region above the r axis and to the right of the line r:1 ''' y=4-f, inscribed rectangles.

bounded by the r axis, the line r= 1, and the curve

6, The region of Exercise 5; circumsoibed rectangles. 7i The region lying to the left of the line : = 1 bounded by the curve and lines of Exercise 5; circumscribed rectangles. 8. The region of Exercise 7; inscribed rectangles. 9. The region bounded by y : 3/, dre r axis, and the line r:

1; inscribed rectangles.

10. The region of Exercise 9i circumscribed rectantles. 11. The region bounded by y:

ri, the r axis, and the lines r:

-L and r:

2; inscribed rectangles.

12. The region of Exercise lli circumscribed rectantles. li.lnre region bou nded by y: ' rectangles.

mx, with tt > 0, the r axis, and the lines l:

a and r:

b, with b ) a > 0; circumscribed

14. The region of Exercbe 13; inscribed rectangles. 15. Use the rrethod of this s€ction to find the ar€a of an isosceles trapezoid whoee bases have measr:res b. and D"and whose altitude has measure l.

284

INTEGRAL THE DEFINITE

-4 tox:4lorm 76. The Snph of y:4l.zland the r axis hom x: area of this triangle.

a triarrgle. Use the method of this section to lind the

In Exercises 17 throtgb 22, tind the area of the region by taking as the measure of the altitude of the ith rectangle l(nr), where z1 is the midpoint of the ith subinterval. (rrrwr: n1: !(4 a 4 1).) 17. The region of Example 1..

18. The region of Exercise L.

19. The region of Exercise 3.




7.3 THE DEFINITE INTEGRAL

In the preceding section, the measure of the area of a region was defined as the following limit: n

lim ) f(c1) A,x

n++6

(1)

i:l

To lead up to this definition, we divided the closed interval fa, bl into subintervals of equal length and then took ci as the point in the ith subinterval for which / has an absolute minimum value. We also restricted the function values f (x) to be nonnegative on fa,bl and further required f tobe continuous on la, bl. To define the definite integral, we need to consider a new kind of limiting process/ of which the limit given in (1) is a special case. Let / be a function defined on the closed interval [a, b].Divide this interval into n subintervals by choosing any (n - 1.)intermediate points between a andb. Let t xn-r be the intermediate points so that ro(xr1xr1

1 xn_r I

x,

The points x6,xr, xz, . r xn-rrx?rarenot necessarilyequidistant.Let Arr be the length of the first subinterval so that Arr : xr - xil let Arr be the length of the secondsubinterval so that Azr - x2- xr; and so forth, so that the length of the lth subinterval is A1r, and a.g: xi A set of all such subintervals of the interval [a, b] is called apartition of the interval [a, b].Let A be such a partition. Figure 7.3.1 illustrates one such partition A, of la, bl. I a:ro

I

|

|

I

x7

)t2

13

x4

I xn_L

l.>r In:o

F i g u re7 .3 .1 The partition A contains n subintervals. One of these subintervals is longest; however, there may be more than one such subinterval. The length of the longest subinterval of the partition A, called the norm of the partition, is denoted by llAll. Choose a point in each subinterval of the partition A: Let fr be the point chosen in [xo, xr] so that ro < €, < xr. Let f, be the point chosen in

7.3 THE DEFINITEINTEGRAL

v

lxr, xrf so that xr s tz = x2, afld so forth, so that fi is the point chosenin lxu-r,x-], and xr-r 3 tt - x.. Form the sum

0

' 'tf({") A'"x

f ( ( , )A , r * f ( ( , )L , x t ' ' ' + f ( ( , ) L p t ' \

or n

> f (€,) Lix \

Su.nl sum is called a Riemannsum,named for the mathematicianGeorg Friedrich Bernhard Riemann (1826-1.866). o ILLUSTRATIoN 1: Suppose/(x) :1.0 - x2, with i = x < 3. We will find the Riemann sum for the function f on [i, e1 for the partition A: xo:i, x.t:1, xz:li, xs:Lt, xq:2*, xs:3, andf1: t, tz:Lt, ts:11, €q:2,

t": 21.

Figure 7.3.2 shows a sketch of the graph of / on l*, Z1 and the five rectangles,the measuresof whose areasare the terms of the Riemannsum.

5

5

2f tt,l L&: f(€,)A,r * f(il A,rx*f(€') A'r * f(t^) Anr*/((') A'r

- 1)+ f@(L*- r+) : fE)(l- +)+ fG)0+ + f (2)(2+-1*)+ f (+)(3- 2+) : (e*)(+)+ (8+)(+)+ (6€)(+)+ (5)(+)+ (2+)(+) : 18#

.

Thenorm of A is the lengthof the longestsubinterval.Hence,llAll:*.

o

xot,

Figure7.3.2

xrtrl

xr\xn

l, t, !.

€rx

Because the function values f (x) are not restricted to nonnegative values, some of the /(fi) could be negative. In such a case/ the geometric interpretation of the Riemann sum would be the sum of the measures of the areas of the rectangles lying above the r axis plus the negatives of the measures of the areas of the rectangles lying below the r axis. This situation is illustrated in Fig.7.3.3. Here

v

a\

y : f(x)

J

li

IC

Arl

|.i,

la'

dt, \

tn

t, *rl i4 trv

xott \

€z O

x2

,{N

A^ 'l

: i:f: :tt

As i/

\

i

{o xo

.xs

f,l f' I ,/,,

t1

b .{rc

;-\ Aa

t F i g u r e7 . 3 . 3

fto r,s

A,I

T I

t

1

rto


j

ftg,l Lix: Ar* A2_ As_A4- Au* Au* A7- As- An- Aro

because f(€"), fG), fG), f$r), f$n), and /(f,o) are negative numbers. We are now in a position to definewhat is meantby a function / being "integrable" on the closedinterval la, bl.

7.3.1 Definition

Let f be a function whose domain includesthe closedinterval [a, b]. Then / is said to be integrableon fa, bl if there is a number L satisfyingthe condition that, for every e ) 0, there exists a 6 ) 0 such that I Lix- tl ' ' fte,l l) ln

for every partition A for which llAll < 6, and for any fi in the closed interval f x r _ r x, 1 l i, : 1 , 2 , . . . , n .

In words, Definition7.3.L states that, for a given function / defined on the closed interval la, bl, we can make the values of the Riemann sums as close to L as we please by taking the norms llAllof all partitions A of [a, b] sufficiently small for all possible choices of the numbers f1 for which xi-r s €i = xi.If Definition 7.3.1 holds, we write n

(2)

lim ) /(f,) A,ix: L llAll-0 t:r

The above limiting process is different from that discussed in Chapter the number L in (2) exists if for every e ) 0 there 2. From Definition 7.3.'I.., existsaD)0suchthat

lf rtr,l Lix- rlI . . lr-:r' for every partition A for which llAll< 6, and for any f, in the closedinterval , n. f x r - r ,x i l ,i : L , 2 , In Definition 2.1.1we had the following: (3)

lim/(r):1 if for every e > 0 there exists a 6 > 0 such that lf(x)- Ll < e

w h e n e v e r 0< l r - a l

< 6

In limiting process(2), for a particular 6 > 0 there are infinitely many partitions A having norm l[ll < 6. This is analogousto the fact that in limiting process (3), for a given 6 > 0 there are infinitely many values of x for which 0 < lr - al < 6. However, in limiting process(2) , for eachpartition

7.3 THE DEFINITE INTEGRAL 291 A there are infinitely many choices of ti.It is in this respect that the two limiting processes differ. In Chapter 2 (Theorem 2.1,.2) we showed that if the number L in limiting process (3) exists, it is unique. In a similar manner we can show that if there is a number L satisfying Definition 7.3.1, then it is unique. Now we can define the "definite integral,."

7.9.2 Definition

If /is a function defined on the closed interval [a,bf, then the definite integral of / from a to b, denoted bV Il f @) dx, is given by

u,' fl'f;w'n*t' ,tiil,:/tut

(4)

if the limit exists. Note that the statement"the function / is integrableon the closedinterval la, bf" is synonymouswith the statement"the definite integral of / from atob exists." In the notation for the definite integral I! f @) dx, f (x) is called the integrand,a is called the lower limit, and b is called the upper limit. T}l.e symbol

I

J

is called an integral sign. The integral sign resembles a capital 5, which is appropriate because the definite integral is the limit of a sum. It is the same symbol we used in Chapter 6 to indicate the operation of antidifferentiation. The reason for the common symbol is that a theorem (7.6.2), called the fundamental theorem of the calculus, enables us to evaluate a definite integral by finding an antiderivative (also called an indefinite integral). A question that now arises is as follows: Under what conditions does a number L satisfying Definition 7.3.Lexis| that is, under what conditions is a function f integrable? An answer to this question is given by the following theorem. 7.3.3 Theorem

If a function / is continuous on the closed interval lq, bf, then/ is integ ra b l e o n l a , b l . The proof of this theorem is beyond the scope of this book and is given in advanced calculus texts. The condition that / is continuous on la,bl,while being sufficient to guarantee that/is integrable on la,bl,is not a necessary condition for the existence of [l f @) dx. That is, if / is continuous on la,bT,then Theorem 7.3.3 assures us that Il f G) dx exists; however,


it is possible for the integral to exist even if the function is discontinuous at somenumbers in la, b]. The following examplegives a function which is discontinuous and yet integrable on a closed interval.

EXAMPLE1: defined by

Let f be the function

,,, [o ifx*o l \ x ) : I "Ltr r i f r _ o Letfa, bl be any interval such that a 1 0 < b. Show that/is discontinuous on la, bl and yet integrable on [a, bl.

solurroN:

Becauselim f (x) :0

+ f (0),/is discontinuous at 0 and hence

discontinuous on [a, b]. To prove that / is integrable on la, bl we show that Definition 7.3.7 is satisfied. Consider the Riemann sum n

) /(f,) l,' If none of the numbers (r, €r, zero.Supposethat $: 0. Then

is zero, then the Riemann sum is

n

. > /(f,) Aix: I L,x

,r, "u:,iur.ur"

{rl s llall lf, rr*,, Hence,

If, f rc,l

Lic_ol . .

whenever llAll< e

Comparingthe abovewith DefinitionT.3.'1, where D: e and L:0, that f is integrable on la, bl.

we see

In Definition7.3.2, the closed interval [a, b] is given, and so we assume Ihat a < b. To consider the definite integral of a function from a ta b / when o s b, or when fl: b, we have the following definitions. 7.3.4 Definition

I f a > . , b, then rb

J"f(x) dx:it tf f( x) 7.3.5 Definition

dx [: f(x)

dx exists.

It f Q) dx 0 if f (a) exists. At the beginning of this section, we stated that the limit used in Definition 7.2.1to define the measure of the area of a region is a special caseof


the limit used in Definition7.3.2 to define the definite integral. In the discussion of area, the intervalfa, b] was divided into n subintervals of equal length. Such a partition of the interval [a, bj is called a regular partition.If Ar is the length of each subinterval in a regular partition, then each Lix: Ar, and the norm of the partition is Ar. Making these substitutions in Eq. (4) , we have fbn

I f- tr>dx: Alim ) f ((,) Lx Ja r ,o i:l Furthermore, b-a n and

Lx So from Eq. (6) lim lL-

(8)

-t@

and from (7),because b > a and Ar approaches zero through positive valu e s (b e c a u s eA x > 0), lim n:

*oo

(e)

Ar-0

From limits (8) and (9), we concludethat Ar --+0 is equivalent to n -->+@

(10)

Thus, we have from Eq. (5) and statement(10), fbn

)i : l f (€,)Lx I f t*l dx: nlim Ja _*o

(11)

It should be remembered that fi can be any point in the ith subinterval lxi-t, xrl. In applications of the definite integral, regular partitions are often used; therefore, formulas (5) and (L1) are especially important. Comparing the limit used in Definition 7.2.L, which gives the measure of the area of a region, with the limit on the right side of Eq. (11), we have in the first case,

lim ) f(c) Lx

(t2)

n-+@ i:l

where/(ci) is the absolute minimum function value on lxi-t, x1).In the sec-


ond case, we have n

rim ) f(€,) Lx n-*q i--t

(13)

where (1 is any number in [r1-r,r,]. Becausethe function / is continuous on fa,bl, by Theorem 7.9.9, Il f @ dx exists;therefore,this definite integral is the limit of all Riemann sums of / on la, bl including those in (12) and (13).Becauseof this, we redefine the area of a region in a more general way. 7.3.6 Definition f(x)

Letthe function/becontinuousonfa, bl and f(x) > 0forall rin la,bl.Let R be the region bounded by the curve y - f (x), the r axis, and the lines x: fr and r : b. Then the measure of the area of region R is given by

o'* P,rtr'lo":I)f(x) dx ,,1il,

F i g u r e7 . 3 . 4

ExAMpLE2: Find the exact value of the definite integral f3

I x2dx

Jt

Interpret the result geometrically.

The above definition states that if f (x) > 0 for all r in Ia, bl, the definite integral [! f (x) dx can be interpreted geometrically as the measure of the area of the region R shown in Fig. 7.i.4. Equation (11) can be used to find the exact value of a definite integral as illustrated in the following example.

soLUrIoN: Consider a regular partition of the closedinterval [1, 3] into n subintervals.Then A^x: 2ln. If we choose ti as the right endpoint of each subinterval, we have

€ r : L + 1 , € z : 1 * r G ) , t s : L+3G),

€i:1*'(1), tn:1

*"(t

Becausef (x): x',

f c,): (r* +)' : (! j-4)' Therefore,by using Eq. (11) and applying properties and formulas from Sec.7.'1,, we get

I' *' dx:lri) :

(+)'t"

)n

lim h-*a

+s n"?

1:l

(n' t 4ni -l 4i2)


:

lim ll-

*6

#l*z

'l., t 4n

295

n

n

i+4

i'] i:r

i:r

:Ii hl"'"

-,r,+ n ' n ( n * L ) ,- T4l n ( n+ t ) ( 2 n + 1 ) l Z

:,li #l**2nst2n2.Wl

:.llils*f*8n2 -;+ L 2n + I Jn'

:,Illo*f* 8g -, 4i -,

4f g"')

:6*O+8+0+0 _83 (1, L)

Figure7.3.5

We interpret the result geometrically. Because xz = 0 for all x in the x axis, and the lines [1,3], the region bounded by the curve A:x', : 3 has an area of 83 square units. The region is shown x: 1 and x in Fig. 7.3.5.

Exercises7.3 In Exercises1 through 6, iind the Riemannsum for the function on the interval, using the given partition A and the given valuesof $. Draw a sketchof the graph of the function on the given interval, and show the rectanglesthe measureof whose areasare the terms of lhe Riemannsum. (SeeIllustration 1 and Fig. 7,3,2.) 1. f (x) : f , O s r = 3; for A: xo: 0, xr: tr,xz: 7*,rs : 2i, xe: 3; : L ta : 1' &: lt, €n= 2i 2.f(x):f,g
J;J,oo*

8

I:xzdx

s'

L'

x3dx


,o

,r.

dx I:,xa

I" e,*4x*s)

dx

t

dx t e,*x-6)

tn. (4x'- Bxz) dx I:,

".1s: ,[_,(r,* 1,)dx

In Exercises 15 through 20, find the exact area of the region in the following way: (a) Express the measure of the area as the limit oI a Riemann sum with regular Partitions; (b) express this limit as a definite integral; (c) evaluate the definite integal by the method of this section and a suitable choice of fr. Draw a figure showing the region. 15. Boundedby the line y:2a-

1,,he r axis,and the lines r:1 and r:5. - 5, *e r axis, and the lines x:4 arrdr:2. 17. Bounded by the culve y:4 the r axis, and the lines r:1 and r:2. "e, 18. Bounded by the curve y : (r + 3)r, the r axis, and the lines r : -3 and .r: 0. 19. Bounded by the c:.rrvey :12 - , - f, the r axis, and the lines .r: -3 and r:2. 20. Bounded by the c'lrrrcy:6r+ f - d, the r axis, and the lines r: -1 and r:3. 16. Bounde{ by the line y:2r

21. ExpresEaEa definite integral: iim j 22. Expressa8 a definite inteFal

(8ir/n3). (HrNr: Consider the function I for which l(r) : d.) j

Um f71n*i). -1',"

(HrNr: Consider the tunction f for which f(x):1lx

on [1, 2].)

23. Expressas a definitet.:"O",

(n/i:). (HrNr:Considerthe tuncrionI for which l(r) : Ul.) ,g : 24,Letla,blbeanyintervalsuchthat4<0
25. Prove that the signum function is discontinuous !r sgnx dx:0.

on [-1,

1] and yet integrable on

1]. Furthermore show that

26. Prcve that the greatest integer function is discontinuous on [0, *] and yet integrable on [0, sl. Furthermore, show that

ftp [xn dx:i.

7.4 PROPERTIES OF THE DEFINITE INTEGRAL

7.4j1, Theorem

Evaluating a definite integral from the definition, by actuatly finding the limit of a sum as we did in S,ec.7.3,is usually quite tedious and frequently almost impossible. To establish a much simpler method we first need to develop some properties of the definite integral. First the following two theorems about Riemann sums are needed. If A is any partition of the closed interval [a, bl, then lim llall-o

b-a

PROOF: n

L&-(b-a):(b-a)-(b-a) i:r

OF THE DEFINITEINTEGRAL 7.4 PROPERTIES

297

Hence, for any e ) 0 any choice of E > 0 guaranteesthe inequality ln

lU

I or - (b- o,l . .

and so by Definition 7.3.1, n

I

l i m -) A r r : b - a

llAll-o i:1

7.4.2 Theorem

If f is defined on the closed interval la, bl and if

li* f, f G,)o,,

ilA|-0 r:1

exists, where A is any partition of [a, b], then if k is any constant, nll

lim ) k/(f,) L#: k lim 2 fG) L,x l l a l l - ot : r l l a l l - o3 r The proof of this theorem is left as an exercise (see Exercise 1). We now state and prove some theorems that give important properties of the definite integral.

7.4.g Theorem

If the function / is integrable on the closed interval la, bl and if k is any constant, then

pRooF: Because / is integrableon la,r],

if ,,t^1il.

ClAix exists,and so by

Theorem 7.4.2 nn

lim >,kf(€,) Lix: k lim 2 fG) t,, Ft -J

-

l l a l l - oF r

-'

-

llall*o

Therefore, fb

rb

Ja

Ja'

I kf(x)dx:kl

7.4.4 Theorem

f@)dx

I

If the functions f and 8 are integrable on La,bf, then f + S is integrable on

298


PRooF: The functions f and g are integrable on la, bl; thus, let fb rb

J" f

@)dx: M and

dx: N J" SQ)

To prove that / * g is integrable on la, bl and that IP, lf (x) + g(r) f dx : M+ N, we must show that for any € > 0 there exists a 6 > 0 such that ln r + s(f,)la#- (M+ N)l. . t/(f,) l> li--r I t-

forall partitions A for which llAll< 6 and for any 4i in lxr_r,xil. Because

M:,Jim t fG,) o,* and IY: fim XS(f,) a,t l l a l l- o t : t llall-oi:r it follows that for any e ) 0 there exist a 6, ) 0 and a 6z ) 0 such that

.*z and r(ri)Aix-rl.f - Air- utl r,*,, l> l7:r' | lp-, for all partitions A for which llAll< 61 and llAll< 6r, and for any fi in fxr-r, ri]. Therefore,if 5 : min(6r, 6r), then for any e ) 0

Air-,l*l> r(ri)Air-Nl.f ' 2+i.:, l>rte,l le' I l?r"" |

(1)

for all partitions A for which llAll< 6 and for any fi in lxi_r, xif. By the triangle inequalit!, we have \ /n \l Aix- *) * (=>,t(fi) Air- t)l l(p fre,l l/n

= fG,)Lic-rl * lI

,^, s(fi) Air- Nl

From inequalities (L) and (2), we have lln

m

\

I

lx* ) s(f,)t,r) - (M+N)l . . l() ftr'l

(3)

From Property 3 (7.1,.3) of the sigma notation, we have nnn

> /(4') A,r* ) s((,) Lix: > t/(f,) + s(f,)l Lix ," ;; ,"bstitutin ,: n"^(4) e)0 ln

(4)

into'O, *" are able to concludethat for any I

+s(f,)lL&- (M+N)l.. l}_,ffrc; for all partitions A for which llAll< 6, where 6: min(6r, 5r) and for any€i

OF THE DEFINITEINTEGRAL 7.4 PROPERTIES r2l. This proves

in lxia,

that f * g is integrable

299

on [a, bl and that

dx:| ,o, dx* [' su)a* f rru>+ s(r)I Theorem 7.4.4 can be extended to any number of functions. That is, if the functions fr, fr, . . , fn are all integrable on la, b), then (ft+ fr* ' ' ' + f,) ,s integrable on [a,bf and rb IJ a lf,(x) + f"(x) + . . . + f"(x)l dx

: lo f,(x)dx* fuf,(*) dx* ' ' ' + [' f,(x) dx Jct Ja Jo'"

The plus sign in the statementof Theorem7.4.4canbe replacedby a -'l'. minus sign as a result of applying Theorem 7.4.3,where k: 7.4.5 Theorem If the function / is integrableon the closedintervals la, bf,la, c], andlc, bf,

f rr.tdx:["r,,,tdx* ['

fr*>o*

wherea
= llall llA'll If in the partition A' the intervalla, cl is divided into r subintervals and the interv il lc, bl is divided into (n - r) subintervals, then the part of the partition A' from a to c gives a Riemann sum of the form r

2fG) ^,x other part of the partition A', from c to b,gives a Riemann sum of ""0;" the form n


Using the definition of the definite integral and properties of the sigma notation, we have fbn

I .f0) dx: lim ). 'f(f,) A'r

Ja

l l a l l - . oG i

:

[> rtc,lA;x* i, /tr,to,'] ,,t^1T^

llall-o ui:r

:

i:r*l

f, rrclAir* ,*il. i,f {t) o'' ,,Til,

Since0 < llA'lls llAll,we canreplacelAll- 0 by llA'll- 0, giving us fbrn

'f @ dx: JI a

lim > /(6,) - A;x* lim

)

l l a , l-lo i ? * r

l l a , l l -A o'

f (€,)Lt

Applying the definition of the definite integral to the right side of the above equation, we have y: f(x)

fb

fc

fb

I f@ dx: JIa ' fG) dx* J|e t'Q)dx Ja' o rLLUSrRArroN 1: We interpret Theorem 7.4.5 geometrically. If (x) > 0 f for all x rn fa, b], then Theorem 7.4.5 states that the measure of the area of the region bounded by the curve y : f (x) and the x axis from a to b is equal to the sum of the measures of the areas of the regions from a to c and from c to b. See Fig. 7.4.1. o

F i g u r e7 . 4 . 1

The result of Theorem7.4.5 is true for any ordering of the numbers a, b, and c. This is stated as another theorem. 7.4.6 Theorem

lf f is integrable on a closed interval containing the three numbers a,b, and c, then

Irutdx:f" rul dx* f' f ul n,

(5/

regardless of the order of.a, b, and c. PRooF: If a,b, and c are distinct, there are six possible orderings of these t h r e e n u m b e r sa: < b I c , a 1 c < b , b 1 a 1 c , b 1 c 1 a , c 1 a 1 b , a n d c < b < a. The second orderi ng, a I c 1 b,i sTheorem7.4.5.W e makeuse of Theorem 7.4.5 in proving that Eq. (5) holds for the other orderings. Suppose that a < b < c; then from Theorem 7.4.5 we have fb

fe

fe

Ju-

Jtr'

Ja'

I fQ) dx* | f(x) dx: I fG) dx

(6)

INTEGRAL OF THEDEFINITE 7.4 PROPERTIES From Definition 7.3.4 lc

fb

dx | f(x) '

I f(x) dx:*

Jc

Ja'

Substituting from (7) into (5), we obtain fb

fb

fc

fb

fc

fb

Ja'

Ja'

I f(*) dx- JIc ' fk) dx: JIa f(x) dx Ja' Thus,

I fU) dx: I f(r) dx* J|e f(x) dx

which is the desired result. The proofs for the other four orderings are similar and left as exercises(seeExercises2 to 5). There is also the possibility that two of the three numbers are equal; for example,a: c 1 b. Then rc

fa

I' f trl dx: J|a /trl dx:0 (by Definition7.3.5) Ja' Al s o , b e c a u s ea : rb

c, fb

I f @ ) d x : Jl a f ( x ) d x Je' Therefore, fc

fb

fb

I f U ) d x * lJ " ' f @ l d x : o + fJ a 'f ( x ) d x

Jo'

I

which is the desired result. 7.4.T Theorem

If k is any constant and tf.f isa function such that /(r) : k for all r in la, bf, then fb rb I f t l . )d x : | r ca x : k ( b a )

Ja'

Ja

PRooF: From Definition 7.3.2, fbn

I f (*) dx: lim -)i : r/({,) A,r

Ja'

llall-o

:

n

l i m Ll!

kAox

l l a l l - oi : r

: k ,.lim )i : r O,t

(by Theorcm7.4.2)

llall-0

: k(b - a)

(by Theorem7.4.1)


2 : A geometric interpretation of Theorem 7.4.7 when 0 is given en by Fig. 7.4.2. The definite integral fil t dx gives the measure of the area of theel sha ded region, which is a rectangle whose dimensions are k units an ar md (lb7 - Q ) units. . o r ILLUSTRATIO IION 7-: n/ >

7.4.8 Theorem

IIff tthe functioons / and 8 are integrable on the closed interval [a, bl and if fk' ) > g(r) foor all xr i n [a, b7,then f( fb

f 8 (x) dx I rutdx >- l" pRooF: Because and g are integrableonfa,bl, f [! f (x) dx and ff g(x) dx both exist. Therefore(by Theorem 7.4.3), fb,fbrbfb

I f f r l d x - JIa -s ( " )d x : lJ o ' f t * l d x * JIn t - s e ) l d x

Jo'

Figure7.4.2

rb

:

J"

[f {*) - S G)] dx (by Theorem 7-4.4)

LeI h be the function defined by

h(x):f(x)-sG) Then h(x) - 0 for all r in la,bl becausef (x) > g(x) for all x in la, bl. We wish to prove that Il lf G) - s?)l dx = 0. fb

I lf (x)- s(r)l d x : Ja

fbn

I h(x) dx:

Ja

lim ). ft(f,) Air

l l a l l' o F r

(8)

Assume that

(e)

l i m L,l) h ( { , ) A i x : L ( 0 llAll-0 i=1

Then by Definition 7.3.1, with e:-L,

,(fi) Air- .l . -t l> li:l I t ? ' l

there exists a 6 > 0 such that

whenever llAll< 6

But because n

ln h(t) A,*- L = l t h(t) Aix- Ll > i:r l-t li:l

from inequality (10)we ha VC n

tt(il A,ix- L < - L 2 i:r

wheneverllAll< D

(10;


n

h(il A;r ( 0

wheneverllAll< S

(1 1 )

i:r

But statement(11) is impossible becauseevery h((1) is nonnegative and every L,g 7 0; thus, we have a contradiction to our assumption (9). Therefore, (9) is false, and lim

). ft(f,) A1x > 0

(J2)

LJ

llall*o t:t

From (8) and (L2),we have fb IJ a t/tt) - s(r)I dx> 0

Hence,

["r
dx-

dx>o [' sal

and so

f(x) g(r)

x

dx> f sroa' f tal

. rLLUSrRArroN 3: Figure 7.4.3 gives a geometric interpretation of Theorem 7.4.8 when f(x) > g(x) > 0 for all r in la,bl. Il f @ dx gives the measure of the area of the region bounded by the curye y: f (x)'the x axis, and the lines x: A and r : b. Ig g@) dxgives the measure of the area of the region bounded by the curve y : 8(x), the r axis, and the lines x: a and r : b.In the figure we see that the first area is greater than the second area.

Figure7.4.3

7.4.9 Theorem

Supposethat the function / is continuous on the closedinterval la, bl.If m and M are, respectively, the absolute minimum and absolute maximum function values of f on la, bl so that m-f(x)=M

m ( b- a )

fora-x=b

=r

f(x) dx < M(b a)

pRooF: Because / is continuous on la, bl, the extreme-value theorem (4.5.9) guarantees the existence of m and M.


By Theorem 7.4.7 fb

J,

* d x : m ( b- a )

(13)

, dx: M(b- s)

(14)

and fb

J"

Because/ is continuous on fa, bl, itfollows from Theorem 7.3.j that/ is integrable on fa, bl. Then because f (*) > m for all r in [a, b], we have from Theorem 7.4.8 fb

fb

I f ( * ) d x = lJ n m d x Ja' which from (13) gives fb

I f' @ d x > m ( b - a ) Ja

(15)

Similarly, becauseM = f (x) for all .r in la, bf, it follows from Theorem7.4.8 that

f,,,dx>[)roo, which from (14) gives

M(b_-a)=-f) fato-

(16)

Combining inequalities (15) and (16) we have

v: f@)

Figure7.4.4

nxevrprn1: Apply Theorem7.4.9 to find a smallestand a largest possiblevalue of r4 |r l 2 ( r ' - 6 x 2 * 9 x + 1 . ) d x

J

m(b-a)=['fildx=

M(b - a)

o rLLUSrRArroN4: A geometric interpretation of Theorem 7.4.9 is given in Fig.7.4.4, where f (x) > 0 for all x in la, b). The integral t!, (x) dx gives the f measure of the area of the region bounded by the curve y : (x),the r axis, f and the lines x: A and x : b. This area is greater than that of the rectangle whose dimensions are m and (b - a) and less than that of the rectangle whose dimensions are M and (b - a). o

solurroN: Referring to Example 1, Sec.5.1, we see that has a relative / minimum value of 1 at x:3 and a relative maximum value of 5 at x:"1.. and f $):5. Hence, the absolute minimum value of / on 1i, 41isI, f e):# and the absolute maximum value is 5. Taking m: '!. and M : 5 in Theorem 7.4,.9,we have

L(4-+) = f- (rr- 6xz*9x*t) dx=5(a-i) I rtz


Use the results of ExamPleL, Sec.5.1.

and so

t=l

r4

(x'-6x2* 9x*t)

dx <3rE

J rtz

In Examplel , 9ec.7.6,the exact value of the definite integral is shown to be 10*?. From Example 3, sec. 5.1, we see that/has a relative minimum solurroN: -3 Furthermore, f (l ):5. H ence, on [-L, 1] the abso a t x :-L. value of lute minimum value of / is -3 and the absolute maximum value is 5. So, taking m: -3 and M : 5 in Theorem 7.4.9, we have

Exevrpr-E2: Apply Theorem 7.4.9 to find a smallest and a largest possible value of

rr |

(*n,t * nrtrs) dx

(-3)[1 - (-11] = f' (r4l3+ Ax't")dx
J-t

Use the results of ExamPle3, Sec.5.L.

Therefore, _6 < |rr (*n," + Axttt) dx < t0 Jt g The exact value of the definite integral is as shown in Example 2,Sec'7.6'

Exercises7.4 L. Prove Theorem7.4'2result of Theorem7.4.5. In Exercises2 througlnT, Prove that Theotem7.4.6is valid; in each caseuse the 4. c
7. a:b < c

6. a < c:b

5. c
largest possible value of the given integral. In Exercises8 thro ugh 17, apPly Theorem 7.4.9 to find a smallest and a

, tt.

t

kxdx r6

,r.

t/3 + x dx J_,

f,(xn

- vxz* 16)dx

'ur_,fia,

'nr,ffio. 17 .

tx2

itx

tO.

lr

J_,

(r + 1)ztsdx

r0 tt.

J_r(xn

- 8xz i L6) dx

f2

dx tU. J_rt/xz+5

I w-zldx

1 8. Show that if / is continuous on [-L, 2], then

dx+ rel ax+ [-' f (x)dx:, dx+ I' f ,r<.1 [' f k) 19. show that

I

x dx >

I'

*, dx but

f'

* a* =

Ii

o dx. Do not evaluatethe definiteintegrals.

306


20. If / is continuous on la, b], prove that

l['-lfro*l=I:lf(x)ldx

(Hrur:

(x)l = f (x) s l/(x)l.)

7.5 THE MEAN-VALUE Before stating and proving the mean-value theorem for integrals, w€ THEOREM FOR INTEGRALS discuss an important theorem about a function that is continuous on a closedinterval. It is called the intermediate-aalue theorem,and we need to use it to prove the mean-value theorem for integrals. 7.5.1 Theorem I nt erm edint e-V ahte Tlrcor ent

v f(b) k f(a)

F i g u r e7 . 5 . 1

y : f(x)

If the function / is continuous on the closedinterval [a, bl andif f (a) + f (b), for any number k between (a) and (b) there exists a number c f fhen f between a and b such that f (c) : 16. The proof of this theorem is beyond the scope of this book; it can be found in an advancedcalculustext. However, *" dir..rss the geometricinterpretation of the theorem. In Fig.7.s.1, (0,k) is any point-on the y axis between the points (0,f (a)) and (0,f (b)).TheoremT.s.j.statesthat the line y : k must intersect the curve whose equation is y : (x) at the point f (c'k), where c lies between a and b. Figure 7.5.1.shows ihir it,t"rsection. Note that for some values of k there may be more than one possible value f.ot c. The theorem statesthat there is always at least or" lrulue of.c, but that it is not necessarilyunique. Figure 7.5.2 showsthree possible values of c (cr, cr, and c.r)for a particular k. Theorem 7.s.j,statesthat if the function / is continuous on a closed interval [a, bf, then / assumesevery value between f (a) andf (b) as.r assumesall values between a and.b.The importanceof the continuity of /on la,bl is demonstratedin the following illustration. . ILLUSTRATTON 1: Consider the function I defined by tf \('x- \/ : [ x , - 1 lx"

ifo
A sketch of the graph of this function is shown in Fig. 7.s.g. The function / is discontinuous at 2, which is in the closed interval I0' 3l;/(0) : -L and f (3):9. rf k is any number between 1 and 4, there is no value of c such that f(c)-k because there are no function values between L and 4. . Figure7.5.2

Another function for which Theorem 7.5.7 does not hold is given in the next illustration. . rLLUSrRArroN 2: Let the function g be defined by

8(r):

x- 4

FORINTEGRALS 307 THEOREM 7.5 THEMEAN-VALUE A sketch of the graph of this function is shown in Fig. 7.5.4. The function g is discontinuous at 4, which is in the closed interval -1 and 2, there is -1 and 5l; 8(5) : z.If k is any number between 12, g(2) : : no value of c between 2 and 5 such that g(c) k. In particulat,lf k: 1, then ' 8(6) : L, but 6 is not in the interval 12,51.

Figure7.5.4

Figure7,5.3

nxevmn 1: Given the function / defined by f(x):4*3x-x2

2=x<5

verify the intermediate-value theorem if k: L.

- -6' s o L U rIo N : f (2 ) : 6; f (5) T o fi n d c , s et f (c): k:1; that i s, 4*3c-c2:t c2-3c-3:0 This gives v-

g!

\E 2

because this number is outside the interval We reject + (3 - \E) and 12,51.The number* G + {n ) is in the interval [ 2 , 5 ] ,

rF+!n\:t /\ z / We are now ready to state and prove the mean-value theorem for integrals.

7.5.2 Theorem MeturVslttc Theoretn for lntegrnls

If the function / is continuous on the closed interval la, bl, then there exists a number X such that a = X < b, and fb IJ a ' f (*) dx: /(x) (b a)

308


pRooF: Because is continuous on / fa, bl, fuom the extreme-value theorem (4.5.9)/ has an absolute maximum value and an absolute minimum value on [a, b]. Let m be the absolute minimum value occurring at x: x^. Thus, A=x*
f(x*):m

(1)

Let M be the absolute maximum value, occurring at x: a
f(x")-M

rr. Thus, (2)

We have, then,

m =f(x) -M

forall xtnla,bl

From Theorem 7.4.9it follows that m ( b- a ) = [ ' f f * l d x - M ( b - a ) Ja(b - a) andnoting that (b - a) is positive because b ) a,we

Dividingby get lb

m

I fQ) a' <*_ 0-a

But from Eqs. (1) and (2) , /n: f (x^) and M : (x*), and so we have f fb

I f(x) dx =f(x*) f(r^)=eVo

(3)

From inequalities (3) and the intermediate-value theorem (7.5.7),there is some number X in a closed interval containin g x* and ron such that

/(x): or t@)

fb

| fk)dx:f(x)(b-a)

Jo"

, Figure7.5.5

a
I

. rLLUSrRArroN3: To interpret Theorem 7.5.2 geometrically, we consider f(x) > 0 for all values of x in la, bl. Then tl fG) dx is the measure of the area of the region bounded by the curve whose equation is y : (x), the f x axis, and the lines x: a and x: b (see Fig.7.s.il. Theorem7.s.2 states that there is a number X in la, b] such that the area of the recta ngle AEFB of height /(X) units and width (b - a) units is equal to the area of the r€Bion ADCB. . The value of X is not necessarily unique. The theorem does not provide a method for finding X, but it states that a value of X exists, and

7.5 THE MEAN-VALUETHEOREMFOR INTEGRALS 309

this is used to prove other theorems. In some particular cases we can find the value of X, as is illustrated in the following example.

2: Find the value of X EXAMPLE such that f3

I 'f ( * ) d x - l ( x ) ( 3 - 1 ) Jr if f (x) : )c2.Use the result of Example 2 in Sec.7.3.

solurroN: In Example2 of Sec.7.3, we obtained r3 I x 2d x : 8 3

JL

Therefore,we wish to find X such that ' (2) :zfL /(x) that is, X2: # Therefore, X: _+_ +\/99 We reject -+\E

since it is not in the interval [1, 31,and we have

f3

dx- fG\E)(3 - 1) I f
2fk,) i:l

n To generalizethis definition, consider a regular partition of the closedinterval [a, bl,which is divided into n subintervals of equal length Ar: (b - a) ln Let f, be any point in the ith subinterval. Form the sum: n

/(f,) > i:r

( 4)

n

The quotient (4) corresPonds to the arithmetic Because Lx: (b - a) ln, we have

n:

b-a

Ax

Substituting from (5) into (4), we obtain

mean of. n numbers.

(s)

310


or, equivalently, tt

) /(f,) a'

t:1

b-a Taking the limit as n -> *m

(or Ax -- 0), we have, if the limit exists,

nfb

I /(4')a'

l!r_ e!_ b-a ,_;*

:

| to

1tg)ax = b-a

This leads to the following definition. 7 '5'3 Definition

If the function / is integrable on the closed interval la, bf, the aaerageaalue of / on [a, b] is rb

f{r) a*

-J, -F, EXAMPLE3: Find the average value of the function / defined by f (*) : x2 otr the interval [1,,3].

SoLUrroN: In Example 2, Sec.

we obtained

r3 I x2dx: +q

Jr

So if A.V. is the average value of f on 11,gl, we have _

-236 -

3-1

4^

_rJ

3

An important application of the averagevalue of a function occursin physics and engineering in connection with the conceptof center of mass. This is discussedin the next chaDter. In economics,Definition 2.5.1can be used to find an averagetotal cost or an averagetotal revenue. o rLLusrRArroN4: If the total cost function C is given by C(r): #, where C(r) thousands of dollars is the total cost of producing 100r units of a certain commodity, then the number of thousandsof dollars in the average total cost, when the number of units produced takes on all values from 1O0to 300, is given by the averagevalue of C on [1, 3]. From Example 3, this is rra:4.33. Hence, the averagetotal cost is $4333. o

Exercises7.5 ln Exercises 1 through 8, a function / and a closed interval [a, D] are given. Determine if the intermediate_value theorem holds lor the given value of ft. If the theorem holds, find a number c sucihthat c) : lc. If the theorem does not hold, give the f( reason. Draw a sketch of the curve and the line v : k.

THEOREMOF THE CALCULUS 7.6 THE FUNDAMENTAL

311

1. f(x):2 + x - *t la,bl: [0,3]; k: t 2. f({: f a5r-6;la,bf : l-1,2l; k: a 3. f (x) : tfi=P, lq, b1: l-a.5,37;k: 3 a. fk):-vffi=F; fa,bf : f0,81;k: -B 4

s- f {x\ : ;}1; (1*r

6 .f @ : 1 ; - ; (*-+

?.f@ : lf _;

.

,la, bl : [-3, 1]; k: t it-4 =x =-?1

i-i.,

- 1 l; Io'bl:[-4'r],k=L

it-Z-t
ri r =l = i

'],

[s,bl: l-2' 3];k: -1'

e. t k) : r;-_ y ts,b): [0,7h k : 2 In Exercises 9 through 13, find the value of X satisfying the mean-value theorem for integrals. For the value of the delinite integral, use the resilts of the conesponding Exercisd 9 though 13 in Exercises 7.3 Draw a figure illustrating the application of the theorem.

t. ,,

f

rdx

f Q'*x-6)dx

t

to. dx I_,x4 tt

[:,(r'*

f

(x,* 4x*s)dx

t) dx

14. Find the averagevalue of the function / defined byl(r) : f on the interval [2, 4], and Iind the value of r at which it occurs, Make isketch. Use the rcsult of Exercise 8 of Exercises 7.3 for the value of the definite integral. 15. Suppose a ball is dropped from rest and after t sec its velocity is ? ft/sec. Neglecting air r€sistance, exPressr,'in terms of t as v: f(t) and find the averagevalue ot f on 10,21. (r) : 13.Find the 16. Suppose C(r) hundreds of dollars is the total cost of producing 10r units of a ce*ain commodity, and C 7 in ExerExercise result of from 0 to 20. Use the *hen the number of units produced takes on all values .rr6"ig" tot"l "ort integral. cises7.3 for the value of the definite : V16-7 on the interval [-4, 4]. Diaw a figure. (HrNr: Find t Z. Find the average value of the tunction I defined by /(r) the area of a region endosed by a semicircle.) measure of the value of the definite integral by interpr€ting it as the - f on the interval [0, 7]. Draw a figure. (I{INr: Find the 18. Find the averagevalue of the functionl defined byl(r) = @ it as the of the area of a region enclosed by a quarter-circle.) measure interpreting integral by difinite value of the 19. Show that the intermediate-value theoreqr guarantees that the equation r3 4l * r + 3 : 0 has a rcot between I and 2. 20. Suppose f is a function for whidr 0 = f(x) =l if 0 = r = 1. Pncve that ifl is contrnuous on [0, 1] there is at least one (urNr: If neither 0 nor l qualifies as c, then l(0) > 0 and l(1) < 1- consider the in [0, 1] sudr thatl(c):c. trr.-b"" " : x ar\d apply the intermediate-value theotem to 8 on [0, 1] .) function g for which g@) l(x\ 21..lf f is continuous on [a, b] and Jl lG) dx:o,

7.6 Tp.E FUNDAMENTAL THEOREM OF THE CALCULUS

prove that there is at least one number c in [c, b] such thatl(c):0.

Historically the basic conceptsof the definite integral were used by the ancient Greeks,principally Archimedes (287-212B.c.),more than 2000years ago,which was many yearsbefore the differential calculuswas discovered.

312

INTEGRAL THE DEFINITE

In the seventeenth century, almost simultaneously but working independently, Newton and Leibniz showed how the calculus could be used to find the area of a region bounded by u curve or a set of curves, by evaluating a definite integral by antidifferentiation. The procedure involves what is known as the t'undamentaltheorem of the calculus.Before we state and prove this important theorem, w€ discuss definite integrals having a variable upper limit, and a preliminary theorem. Let f be a function continuous on the closed interval la, bl. Then the value of the definite integral ft f Q) dx depends only on the function f and the numbers a and b, and not on the symbol r, which is used here as the independent variable. In Example 2,Sec.7.3,we found the value of .frtx2dx to be 83. Any other symbol instead of x could have been used; for example,

l'

,' or:

u2du: r2dr- 8a 1," l,'

If / is continuous on the closed interval fn, bl, then by Theorem 7.3.3 and the definition of the definite integrcI, [! f (t) dt exists. We previously stated that if the definite integral exists, it is a unique number. If r is a number rnfa, b], then/is continuous onfa, x] because it is continuous on [a, bl.Consequently, If f ft) df exists and is a unique number whose value depends on x. Therefore, I{ fU) dt defines a function F having as its domain all numbers in the closed interval [a, b] and whose function value at any number x in la, bl is given by

_r"f(t) dt

F(x) _

f(t)

F i g u r e7 . 6 . 1

7.6.1 Theorem

J,

(1)

As a notational observation, if the limits of the definite integral are variables, different symbols are uged for these limits and for the independent variable in the integrand. H8nce, in Eq. (1), because x is the upper limit, we use the letter f as the independent variable in the integ.ur,d. If, in Eq. (1),f (t) > 0 for all values of f in la, bl, then the function value F (x) can be interpreted geometrically as the measure of the area of the region bounded by the curve whose equation i" y: f (t), the f axis, and t h e l i n e s t : a a n d f : x . ( S e e F i g . 7 . 6 . 1 ,N. )o t e t h a t F ( a ) : I t fU) dt,which by Definition 7.3.5 equals 0. We now state and prove an important theorem giving the derivative of a function F defined as a definite integral having a variable upper limit. Let the function / be continuous on the closed interval la, bl and let x be any number in la, bl.If F is the function defined by

F(x):f ru>0, then

F'(x): f(x)

THEOREM OF THECALCULUS 313 7.6THEFUNDAMENTAL (If x: x:b,

a, the derivative in (2) may be a derivative from the right, and if the derivative in (2) may be a derivative from the left.)

pRooF: Consider two numbers 11 and (r1 + Ar) in [a, b]. Then

F(r,):

dt f' fG)

and

F(x,* Ax)-

ft) at f'*^* f

so that

F(x,* Ar)- F(r,):

(t) dtdt f' f U) f'*'" f

(3)

By Theorem7.4.6, f e:r+tt

f Jrr+Lx

f rt

dt: or* J" fG\ J., f(t) J,

fQ)dt

or, equivalently, rrr+ir

f n+ Lr

lrt

fo dt- J, fG)dt: J*, f(t)dt

J,

(4)

Substituting.from Eq. (4) into (3), we get F(xr-f,ax)- F(x,) :

f,'*o"

f(t) dt

(s)

By the mean-value theorem for integrals (7.5.2),there is some number the closed interval bounded by x, and (r, + Ax) such that f xr+L.r

J.,

f (t) dt: 1(X)Ar

From Eqs. (5) and (6), we obtain F ( x 1* A x ) - F ( r ' ) : / ( X ) A r or, if we divide by Lx, F(x'*Ax)-F(x'):f(X) r \--/ Ar Taking the limit as Ax approaches zerol we have ,.-

F(rr * A{) - F(xr) :

,n";;'ide

lim f (X)

(7)

or,* ,r, i, r' (',i"il'u"r"r-ine lim f (x), recallthatX is in

314


the closedinterval bounded by .rr and x, * Lx, and because

: ll.T r,: r, and llq (r, * Ar) r, it follows from the squeeze theorem (4.3.3) that continuous at .x1,w€ have lim /(X) :

l:l*:xt.Because/is

lim f (X) : f (xt); thus, from Eq.

(7) we get F'(xr):f(xr)

(8)

If the function/ is not defined for values of x less than abut is continuous from the right at a, then in the above argumenl, if. xr: a rn Eq. (T), A,x must approach 0 from the right. Hence, the left side of Eq. (g) will be Fi(rr). Similarly, if f is not defined for values of x greater than b but is continuous from the left at b, then if xr: b inEq. (7), A,x must approach 0 from the left. Hence, we have F'_(xr) on the left side of Eq. (8). Because r, is any number in la, bl, Eq.(8) states what we wished to Prove. I Theorem 7.6.L states that the definite integral t{ f(t) dt,withvariable upper limit r, is an antiderivative of f.

7.6.2 Theorem F t r n d n m e n t aT l heorem of the Cnlutlus

Let the function / be continuous on the closed interval la, bl and let g be a function such that

g'Q):f(x)

o)

for all x in fa, bl. Then fb

I f U )d t : s f t ) - s ( a ) Ja' (If x: x:b,

a, the derivative in (9) may be a derivative from the right , and if the derivative in (9) may be a derivative from the left.)

pRooF: If is continuous at all numbers in / [a, bf, we know from Theorem 7.6.t that the definite integral Ii f (t) dt, with variable upper limit r, defines a function F whose derivative on la, bl is Becauseby hypothesis /. 8'(x) : f (x), it follows from Theorem 6.3.3 that fr

8 ( r ) - l fJ < tldt+k a

(10)

where k is some constant. Letting x: b and r : a, successively, in Eq. (10), we get

:T:

8@)

f(t) dt + k

(11)

THEOREMOF THE CALCULUS 315 7.6 THE FUNDAMENTAL

and (12)

s(a)-f,urdt+k From Eqs. (11) and (L2), we obtain

8(b)-s@\:["fUl

ot-

dt f fu)

But, by DefinitionT.3.5, f Al dt:0, and so we have [" fb s(b)- s@): J, f U)at

which is what we wished we Prove. If f is not defined for values of x greater than b but is continuous from the left at b, the derivative in (9) is a derivative from the left, and we have g-@) : F'-(b),from which (11) fotlows. Similarly, if f is not defined for values of x less than a but is continuous from the right at a, then the derivative in (9) is a derivative from the right, and we have 7n(a):F'*(a), from I which (12) follows. We are now in a position to find the exact value of a definite integral by applying Theorcm 7.6.2.In applying the theorem, we denote lD ts(b)- stu)l by s(r)l,

o rLLUSrRArroN 1: We apply the fundamental theorem of the calculus to evaluate r3 I x2dx Jr Here, f(x):

x2. An antiderivative of.x2 is ixt. From this we choose .Y3

g(x): ?

Therefore,from Theorem 7.6.2,we get

["*'dr:+-l':t-* 3l'

Jr*

3

:83 Compare this result with that of Example 2, Sec. 7.3. Because of the connection between definite integrals and antiderivatives, we used the integral sign J for the notation J/(r) dx for an an-

316


tiderivative. We now dispense with the terminology of antiderivatives and antidifferentiation and begin to call I f Q) dx the indefiniteintegralof "t' of x, dx." The processof evaluating an indefinite integral or a definite integral is called integration. The difference between an indefinite integral and a definite integral should be emphasized. The indefinite integral Ifk) dx is defined as a function g such that its derivative D*[g(x)]: f (x). However, the definite integral Il f Q) dx is a number whose value dependson the function f and, the numbers a andb, and it is defined as the limit of a Riemann sum. The definition of the definite integral makes no reference to differentiation. The general indefinite integral involves an arbitrary constant; for instance, f*3

I x2dx:++C JJ This arbitrary constantC is calleda constantof integration.lnapplying the fundamental theorem to evaluate a definite integral, we do not need.to include the arbitrary constantC in the expressionforg(r) becausethe fundamental theorem permits us to select any antiderivative, including the one for which C:0. Evaluate

SOLUTION:

- 6xz* ex+ 1,)itx Ii, o'

(xt -

6x2* 9x * 1,) dx:

[:,,

xsdx-6[^rxzdx

*t

lnn,

:t-6.{*, +*,1:, : ( 6 4 - 1 , 2 8 + 7 2 + 4 -) ( # - * + g + * )

Evaluate

SOLUTION:

(r4l3 + Axrts)

dx: *xzrl, + 4 . t*n,tlt_,

:++3- (-++3)

(x6ars * n*rrs) dx

-_ 76

sor,urroN t

:)\

f2

J"

Zx't/xt + '1,dx : 3

rz

J

lFn

(ixz) dx

7.6 THE FUNDAMENTAL THEOREMOF THE CALCULUS

317

: 6 ( 8 * L 1 t r z- # ( 0 * L } s r z

: + ( 2 7- L ) _

104 I

f_

ExavrprE 4: r3

Evaluate

I xt/t+x dx

Jo

solurroN: u:

To evaluate the indefinite integral l.f,+x

u2:'!,*x

J

xVt I x dx we let dx-2udu

x:tt2-"!.

Substituting, we have r_l

I xt/t + x dx: | (u' t)u(2udu) JJ -2

r | (un uz) du

J

: E u ' - f i u Bt C : ? ( 1 * x ) s t z- ? ( 1 * x ) s t za . , Therefore, the definite integral f3 13 I x \ / 1 ,* x d x : 9 ( 1 * x ) s r z 3 ( 1 * x ) s r zI

Jo

lo

: ?@);tz- z(4Y, - !"(1)5t2 + ?(t1r,,

:+!-+-E+3 -_

116 -T5-

Another method for evaluating the definite integral in Example 4 involves changing the limits of the definite integral to values of a. The procedure is shown in the following illustration. Often this second method is shorter and its justification follows immediately from Theorems 6.3.10 and 7.6.2. o TLLUSTRATToN 2: Because u: l.'ffi-x, : and when .r 3, u :2. Thus, we have f3

I xt/t + x dx:2

Jo

12

| (un u2)du

Jr

12

:?u5 - &ut I

It

:91-+q-?+e _

116 15

we see that when ff:

0, u:

L;

318


EXAMPLE 5'

Evaluate

l x + 2 1d x

soLUrIoN: If we let f (x): lr * 21,insteadof finding an antiderivativeof / directly, we write f(x) as ifx>-2 ifx<-2

f(x):l::;

Applying Theorem7.4.6,we have

r_,

2

lx+ 2lO*: f,-3 (.(- ) c - 2 ) a,+ ( x + 2 ) d x [^, T - r,]-_" : [ - _ x2 '*]^-,

*lt*

:++ 1.8 :31ExavrprE6: Find the area of the region in the first quadrant bounded by the curve whose equation is y : xfp + 5, the x axis,and the line r:2. Make a sketch.

solurroN: see Fig. 7.6.2:The region is shown together with one of the rectangular elements of area. We take a regular partition of the interval [0,2l.The width of each rectangle is Ar units, and the altitude of the ith rectangle is &\mT units, where fi is any number in the ith subinterval. Therefore, the measureof the area of the rectangular element is (1@ + s Ar. The sum of the measuresof the areasof n such rectanglesis __ia

2ttY1iz+5 Ar *ht;

y:

xy/izT s

is a Riemannsum.The limit of this sum asAx approaches zero(or

n --> *@) gives the measure of the desired area. The limit of the Riemann sum is a definite integral which we evaluate by the fundamental theorem of the calculus. Let A: the number of square units in the area of the region. Then, n

A: lim ) f,{F+s Ar-0 i:t -

ilx

r2 | x\/x21-S dx

JO

:+l

f2

t/x2*S (2xdx)

:+. & (xx2 *_' _ -

1r 3'L

,]: s;srz

3t2l (e 312.- (s)'

: i ( 27 " 9 )-s5t t./5) Figure7.6.2

Hence, the area is *(27 - 5 \6) squareunits.

7.6 THEFUNDAMENTAL THEOREM OF THECALCULUS 319

ExRtvtptE 7:

solurroN: Let A.V. : the averagevalue of / on [5, 8]. We have,from Definition 7.5.3,

Given

f(x): xtE- E find the average value of /on the interval [5, 81.

118

A.V.:+

| xlx-A

d-5

Letu: lxw h e n x :8 ,

Jt

dx

4 ; t h e n u 2 - *x - 4 ; x : t t z * 4 ; d x : Z u d u . W h e n x : 5 , u : tt:2. Therefore,

A.V.:*l

r2

-: r f6"

J,

:- " f6r

(urt4)u(2udu)

Jr

(un* 4u2)du 12

+ 6u" lius ),

:?l+++-+-*l _ - 4 5466

Exercises7.6 ln Exercises L through 18, evaluate the definite integral by using the fundamental theorem of the calculus.

,':

,.

(x'-

[ f4

+ 1'\dY J, @'- Y2

.ro

x dx

.r5

''J,WWA

f.

,r.

rr

tn.J,ffa*

f, x"*2x2*x*2

@+1y

dx

r3

- 3l dx J_,lx

Y'J,

dx

*.J_,(r*1)t/x*3

V3+ lxl dx J_" f4 l,5-r

,t.J_,t/lxl-xdx ,n-

J^x2!x-4

,fr

o" l t i w d w Jo (1+ 61s t o q15

n ' 1s du J -@ , ;rY .-'' f3 o' J, 1er'- 1y'

M - w2 dut 5./ J_rzw f' (y'* 2V)dy

,{i

r3 (3x2I 5x -'t) dx J_,

,.

2x) dx

a, x

'o'

f'3x"-24x2*48x*5 *z - g* a 16 J_,

(nrxr: Divide the numerator by the denominator.)

rl l,*('n-+.w) at

tt.

fr

J.

{i lT +i{x ax

7 ax

L;

THEDEFINITE INTEGRAL

320

In Exercises1.9through 22, use Theorem 7.6.1 to find the indicated derivative. fr

L9. Dr I

t/++ t6 dt

r5 20.D*l vt+t4 dt

,;

i

Joi

Jr

zr. ur r'x g dt J_. + t,

frz

22.D*l Jr

V1,+t4dt

In Exercises 23 throuth 28, find the area of the region bounded by the given curve and lines. Draw a figure showing the retion and a rectangular element of area. Express the measure of the area as the limit of a Riemann sum and tlen as a definite integral. Evaluate the definite integral by the fundanental theorem of the calculus. "N.

y :

4tc- x2; x axis; tc:']",, x:

3

: F. v \/i + l; x axis,y axis;r: 8 x a x i sx; : - 1 , x : 4 2 7 .y : x f f i ;

r-!';

24.y:x2-2x*3;x axis;r--2,x:l 26' Y : x\/xz * 9 ; x axis' Y axis;x: 4 1 2 8 .-Vx: ' + - x ; x a x i s x; : 2 , x : 3

In Exercises 29 through 32, find the average value of the flnction f on the tiven interval [c, b]. In Exercises 29 and 30, find the value of r at lvhich the average value of f occurs and make a sketch, 2 9 .f ( x ) : 8 r - x 2 ;l a , b f : 1 0 , 4 1 30. /(r) - Q- x2; l a, bl : [0,3] 31.f(x) : az1/a g; la, bl: 17,l2l 32.f(x) :3xt/F - 16; l a, bf: 14,5f 33. A ball is dropped from rest and after t secits velocity is z ft/sec. Neglecting air resistance,show that the averaBevelocity during the first +T sec is one-third of the avenge velocity during the next +T sec. 34. A stone is thrown downward with an initial vqlggqgll"o ftlsec. Neglect air rcsbtance. (a) Show that if ftlsec is the velocity of the stone after falling s ft, then o : lpoz I 2gs . (b) Find the average velocity during the first 100"ft of fall if the initial velocity is 60 ftlsec. Clake g: 32 and downward as the positive direction.) 35. The demand equation for a certain commodity is a2+ pt : 100, where p dollars is the price of one unit when 100r units of the commodity are demanded. Find the average total revenue when the number of units demanded takes on all values ftom 600 to 800. 36. Let/be a function whose derivative f is continuous on [a, b]. Find the averagevalue of the slope of the tangent line of the graph of / on [c, b] and give a geometric interpretation of the result

ReviewExercises (Chapter7) In Exercises 1 and 2, find the exact value of the definite integral by using the definition of the definite integraL do not use the fundamental theorem of the calculus. fr

1,. |

J-z

(r' * 2x) dx

,

[,

(x, - 1.)zdx

In Exercises 3 and 4, find the sum. r00

41

4. > (\?trI_l- V3l+z)

3. > 2i(i3-1.) i:1

i:r

t

5. Provethat)

/^

O: (I

\2

,

and verify the formulalorn:1,2,

and,3.

6. Express_asa definite integal and evaluate th€ definite integral:"lim f which f(t) : Vr.)

@^[1n"',y (nwr: Considet the function f for

REVIEWEXERCISES In Exercises 7 throu gh l6,evaluate the definite integrat by using the fundamental theorem of the calculus.

,- .

f3

r. .

dt J_"U6-3t) s.f -L Jt \/3x-

f7

1 0 .f ' r * f l F + z a x

1

^-' 11.l'z)cygx+ 4 Jo

J-s

12.f =+ Jt \/5

dx

û2 tlw^

fs

f2

dx ts. I lx - 213 J _s

x

fr

' - ' J| - , \ / i + T t n ' J r ( x ' * + / ] ) 1 \ / F + o 7 + t A x

1a

la4dx

L#

16. |

J-z

*l* - 3l dx

L7. Show that each of the following inequalities holds:

r-'dx=l-'!;@1,'+=l: h;t')f h=f+

(u) x_g J_, 18. rf

[0

f (x): lk

[1

ifa
if x: c ifc
prove that f is integrable on la, bf urra fu f k) dx b - c regardlessof the value of k. Ja In Exercises L9 and.2A, find a smallest and a largest possible value of the given integralfl

f4

19. I V5 *4x-x2

dx

Jr

20.I

J-z

Y2x3-3x2+l dx

In Exercises 21 through 24, find the area of the region bounded by the given curve and lines. DIaw a fiturc showing the region and a rectangularelement of area.Expressthe measureof the area as the limit of a Riemann sum and then as a definite integraL Evaluate the definite integral by the fundamenta.l theorem of the calculus. 22. y : 16- x2;x axis;x: -4, x: 4 x: 3 2L. y

9

x2; x axis; y axis;

23. y : 21, - 1 ; x axis;x : 5, x : 17

24.y:

4

rr-x;xaxis;x:

25. It f (x) : 2gz\[x=T, find the average value of / on 17,1,2] 26. Interpret the mean-value theorem for integrats (7.5.2) in terms of an average function value. 27. A body falls from rest and travels a distance of s ft before striking the ground. If the only force acting is that of gravity, which gives the body an acceleration of g ft/sec2 toward the ground, show that the average value of the velocity, expressed as a function of distance, while traveling this distance is ?\/lgs ft/sec, and that this average velocity is twothirds of the final velocity.

28. Suppose a ball is dropped from rest and after f sec its directed distance from the starting point is s ft and its velocity is tr and find the (a) Express ? as a function of. t as a:f(t) ft/sec. Neglect air resistance. When f : tr, s: s1, ?fld o:uy average value of f on [0, t,]. (b) Express z, as a function of s as 7r: 8(s) and find the average velocity of g on [0, st]. (c) Write the results of parts (a) and (b) in terms of. ,, and determine which average velocity is larger. r - 1,0: 0, where p dollars is the price of one unit when 100x units of the commodity are demanded. Find the average total revenue when the number of units demanded takes on all values from 600 to 800.

29. The demand equation of a certain commodity isp'*

322


30. (a) Find the average value of the function / defined byl(r) found in part (a), find lirn A. 31. Given F(x):

tn1

| Jt

: t/ri on the interval tt, rl. (b) If.r{ is the average value

;t dt and r > 0. Prove that F is a constant function by showint that F,(.r):0.

(nrnr: Use Theorem

7.6.1afterwriting the given integlal as the difference of two integrals.) 32' Make up an examPleof a discontinuous function fo{ which the mean-valuetheorem for integrals (a) does not hold, and (b) does hold. 33. Let/be continuous on [a,D] ^na f{t) di * 0. Show that for any numbert in (0,1) there is a nurnber cin (a, b) such Jfu o' that f.

tb

I f(t) dt:k I t'() dt 1"r*i",cor,"ia".*" ,""or." F for whichF(x): [" f(t) d, / [' ruldr andapplythe intermediate-varue theorem.) J"' / J"'

Applications of the definite integral

APPLICATIONS OF THE DEFINITEINTEGRAL

8.1 AREA OF A REGION IN A PLANE

From Definition 7.3.6, if f is a function continuous on the closed interval lo, bl and if f (x) =- 0 for all r in la, bl, then the number of square units in the area of the region bounded by the curve y : f (x), the r axis, and the l i n e s x : a and x: b i s n

lim -) - f(f,) Ax -0 t:l

llAll

which is equal to the definite integral ft f Q) dx. suppose that f (x) < 0 for all x in la, bl. Then eachf (il is a negative number, and so we define the number of square units in the area of the regionboundedby y : f(x), the r axis,and the lines)c: aand x : b to be n

lim ->. t-f(f') I Ax

llAll'-0 i=l

which equals

- |fb f(x) dx Ja'

nxavrprE 1: Find the areaof the region bounded by the cuwe y : f - 4x, the r axis, and the lines x:Landr:3.

soLUrIoN: The region, together with a rectangular element of area, is s h o w n i n Fi g. 8.1.1. If one takes a regular partition of the interval [1, 3], the width of each rectangle is Ar. Because x2- 4x { 0 on [1,3], the altitude of the lth rectangle is - (€r' - a€r) : 4€i - trt.Hence, the sum of the measures of the areas of n rectangles is given by 7Z

i=1

The measure of the desired area is given by the limit of this sum as Al approaches 0; so if A square units is the area of the region, we have n

hm ) Lx:-0

(afi - t,') Lx

i:l

(4x - xz) dx :2)c2 (€i, €i'- atr)

*rt

- T -,.)

F i g u r e8 . 1 . 1

Thus, the area of the region is ]2 square units. rxavrprn 2: Find the area of the region bounded by the curve y : xs -2x2 -5x * 5, the r axis, and the lines x: -L andx:2.

solurroN: The region is shown in Fig. g.l.z.Let (x) : )cB- 2x2 -sx * 6. f Because f (x) > 0 when x is in the closed interval [-1, 1] and (r) = o f when x is in the closed interval [1, 21, we separate the region into two parts. Let At be the number of square units in the area of the region when

8 . 1 A R E A O F A R E G I O NI N A P L A N E

t-

of square units in the area of the t ,,'l,f risin J t, dtlnd let A, be the number region w he e n l x: i si i n lr, zl. Then n

( f Ar l'"1)i =l1 f ( 'f,)

A1

litl :

( € t, f ( t i ) )

rII'r If_ l' -J' l

rr 1( 't -' l

(r)t ddx a x3_ 2x2-5x*6)

dx

F(r,

f(x)

- 2x2 -5x * 6) dx

If A square units is the area of the entire region, then Ar* A,

A:

:

F i g u r e8 . 1 . 2

12 (x'- 2x2-5x * 5) dx G' - 2xz- 5x -f 6) dxJ, J_,

, lr l-_ - 3x'- Ex'* U*1 2 - 3x"- Ex' * U* ), L*"' liro l_r: [ ( *- z ' - E + o -) ( ++ ? - B - 5 ) ] - l ( 4 - a f - 1 0+ 1 2 -) ( +- ? - B + a ) l

: Gi, fGt))

rt

t-

(-?9) The area of the region is thereforc lft

square units.

Now consider two functions/ and g continuous on the closed interval la, bf and such that f (x) > 8(x) for all x in la, bl. We wish to find the area of the region bounded by the two curvesy : f (x) and! :8Q) and the two lines x: a and x : b. Such a situation is shown in Fig. 8.1'3. Take a regular partition of the interval la, bl, with each subinterval having a length of Ar. In each subinterval choose a point f1. Consider the rectangle having altitude lf (€,) - S(6;) ] units and width Ar units. Such a rectangle is shown in Fig. 8.1.3. There are n such rectangles, one associated with each subinterval. The sum of the measures of the areas of these n rectangles is given by the following Riemann sum: (f i, g(ti))

F i g u r e8 . 1 . 3

n -s(fi)l Ax > t/(f')

fni":tiiu*ann

sum is an approximation to what we intuitively think of as

A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL

the number representing the "measure of the area" of the region. The larger the value of n-or, equivalently, the smaller the value of Ax-the better is this approximation. If A square units is the area of the region, we define n,

A : l i m > t / ( f , )* s ( f , ) l A x

(1)

Because f andg are continuouson la, bl, so alsois (f - g); therefore, the limit in Eq. (1) exists and is equal to the definite integral rb

J, tf G) nxavrru 3: Find the area of the region bounded by the curyes A:x2 andy:-x2*4x.

y:

8@)

(2,4)

G' , f ( ti )

- g(x)ldx

soLUrIoN:

To find the points of intersection of the two curves, we solve the equations simultaneously and obtain the points (0, 0) and (2,4). The region is shown in Fig. 8.I.4. Let /(r) : -)c2 * 4x, and g (x) : 12. Therefore, in the interval l0, Zl the curve y : f (x) is above the curve y : g(x). We draw a vertical rectangular element of area,having altitude lf (€r) - s((o)] units and width Ar units. The measure of the area of this rectangle then is given by Ax. The sum of the measures of the areas of n such lf((r) -S({i)] rectangles is given by the Riemann sum n

lf (€,)- s(f,)I Ax

y : f(x) i:1

If A square units is the area of the region, then n

lim ./-/ \

(f;,9(fr))

Ar-0

f:1

lf G,)- s(fi)I Ar

and the limit of the Riemann sum is a definite integral. Hence,

lf(x) - s(x)l dx F i g u r e8 . 1 . 4

*t' -Jo

l(-x'+ 4x)- xzldx

:I:

(-2x'*

:

4x) dx

[- l,x'.-u'f',

:_++g_0 -_ 58

The area of the region is $ square units.

8 . 1 A R E A O F A R E G I O NI N A P L A N E

EXAMPLE4r Find the area of the region bounded bY the Parabola : x - 5. : Az 2x 2 andthe line y

-Z) and (9, 4) . The The two curves intersect at the points (3, solurroN: region is shown in Fig. 8.1.5. Theequationy2:2x-2isequivalenttothetwoequations

and y--\Ei=T

y:\E-z :

, : (9,41

, :

g(xl

ltkl

fr\x)

F i g u r e8 . 1 . 5

u : g(r) v :frkl \e, 4,

Gi, f Jtil) Gt, f r(Eil),

with the first equation giving the upper half of the parabola ald thr sectLx- 2 and ond equation ginittg the bottom half. If we let /r(x): : - \Ex - 2, the equation of the top half of the parabola is.A : f t(x) , f ,(x) and the equation of the bottom half of the parabola is U: fr(x)' If we let gG) : x - 5, the equation of the line is y : S (r) . In Fig. 8.1.6 we see two vertical rectangular elements of area. Each rectangleLas the upper base on the curve A : f r(r). Becausethe base of the frG)] first rectangle is on the curve y: fr(x), the altitude is [/t({,) : y its curve is on the rectangle second the base of 8(r), the units. Because using by problem this solve to wish If we units. gG)l altitude is l/,(f,) vertical rectangular elements of area, we must divide the region into two separate regions, for instance R1 and R2,where R, is the region bounded by and u:fr(x) and the l i ne x:3, and w here R 2 i s th e th e c u rv e s y :fi (r) region bounded by the curyes A : f r(x) and y : 8@) and the line r: 3 (see Fig. 8.1'.7). If A, it the number of square units in the area of region R1, w€ have

Ar: rim ) lf'Go) - frG,)l Ar i:l

(€i, g({i ))

Ae-O

_t' lf ,(x) _ J, -_ f ' l \ E - z J'

(€', /r(€i ))

F i g u r e8 . 1 . 6

fz@)l dx

:, It t E x - Z

+t/zx-zlax ax

l3 : I (Zx- Z)srzI

lr

:+L y : g(x) y : f ,(x')

It Az is the number of square units in the area of region R2, w€ have n

Az: lim )

:J; :T:

l f ' ( € , )- s ( f , ) l A x

[/'(r) - g(x)l dx

Y: f 2Gl

t\E-2

F i g u r e8 . 1 . 7

- (x -s)) dx

A P P L I C A T I O NOSF T H E D E F I N I T E INTEGRAL

:

- z)'z- +x'+ sxf'" [+Qx

:[g-++451-t*-8+ts1 Hence At * Az: # + # : 18. Therefore, the area of the entire region is 1g square units. nxevrprE 5: Find the area of the region in Example 4by taking horizontal rectangular elements of area.

(3, - 2l

soLUrIoN: Figure 8.1.8 illustrates the region with a horizontal rectangular element of areaIf in the equations of the parabola and the line we solve for x,we have x:iQ'+2)

* 2) and r(y) : y + s, the equation of the parabola r : tr(y) Letting QQ) :t(y' : QQ) and the equation of the line as tr: L(y). maybe written as x r*e \9, 4l consider the closed interval l-2,4] on the y axis and take a regullr partition of this interval, each subinterval will have a length of Ly-.In the ith t r ( { i ) ,{ i ) subinterval lyi-r, ai], choose a point f;. Then the lenglh of the ith rectangular element is [r(1,) -d(fr)] units and the width is ay units. The measure of the area of the region can be approximated by ihe Riemann sum: n

F i g u r e8 . 1 . 8

and x:y*S

i:r

[r(f,) - 6G)l Lv

If A square units is the area of the region, then n

A: lim > tl(f,) - d(f,)l Ly A.tl*0 i:1 Because). and Q arc continuouson l-2,4], so arsois (r-d), limit of the Riemann sum is a definite integral:

o:I^,[r(y)- 6U)] dv

: I:,l ( y+ 5 )- i ( y ' + 2 ) l d v

-t -,

:+

l n (-y' + 2y + 8) dy

J_,

[- *y"* v'+ 8yf_,

: + [ - g s + 1 6 + 3 2 )- ( 8+ 4 - 1 , 6 ) l :L8

and the

8.1A R E AOF A R E GION IN A P LA N E 32 9 Comparing the solutions in Examples 4 and 5, we see that in the first casewe have two definite integrals to evaluate, whereas in the second case we have only one. In general, if possible, the rectangular elements of area should be constructed so that a single definite integral is obtained. The following example illustrates a situation where two definite integrals are necessary. EXAMPLE6: Find the area of the region bounded by the two curves A: xB- 6X2* 8r and !: x2 4x.

€i , f(ti))

solurroN: The points of intersection of the two curves are (0, 0), (3, and (4,0). The region is shown in Fig. 8.1.9.

Let f (r) : xB- 6x2* 8x and g(x) : )c2- 4x. In the interval [0, 3] the curvey: f (r) is abovethe curve y: g(r), and in the interval [3, 4] the curvey: g(x) is abovethe curvey: f (r). So the region must be divided into two separateregionsRt and R2,where Rt is the region bounded by the two curves in the interval [0, 3] and R2is the region bounded by the two curvesin the interval 13, 41. Letting A, squareunits be the areaof R1and A, squareunits be the area of Rr, we have n

y:8@) y : f(x)

Ar:

lim

n

Az:lim Ar-0

R2

( f ; , g ( f; ) )

€i,f(ti))

3, -3)

(i,s((;))

F i g u r e8 . 1 . 9

lf (€,)- s(fr)I Ax

AJr- 0 i : 1

)

[ S ( f , )- f ( € , ) ] A x

i:l

so that

A, * Or:

6xzI 8x) - (x'- ax)l dx It [(x'+ [^ l@' - 4x) - (r' - 6x2+ 8x)l dx

:f

Js

(x, - Txz* I2x) dx + f^ (-x, t Txz- I2x) dx J3

- u*]^, :lr*^ - &x,+ 6r]',+ * Tx" f+r-- -4T5--rs7-27 -_ 6

7L

Therefore, the required area is ff square units.

8.1 Exercises h Exercfues 1 though 20, find the area of the region bounded by the given curves. In each Problem do the following: (a) Draw a figure showing the region and a rectangular element of area; (b) expressthe area of the region as the limit of a Rigmann sum; (c) find the limit in part (b) by evaluating a definite ihtegral by the fundamental theorem of the calculus. t J .x. 2 : - y ; y - - 4 4. yt - -x; x: -2; x: -4

330

APPLICATIONS OF THE DEFINITE INTEGRAL

3. tt + y I 4: 0; / : -8. Take the elemmts of area perpendicular to the y axis. 4, The same region as in Exercise 3. Take the elements of area parallel to the y axis. 'l 5. ir:2y2.x=O,y:-2 d.y":4t;r:O;y:-2 ,).y:2-i".y:-7 8.y:f;y:r' 9. y2:y-1;a:g 1 0 -y : y z ; f : 1 6 - , 1 1 .V : { r ; y : *

t 2 .x : 4 - y 2 ; x : 4 - 4 y l\. ys: f;|-gy 4 4:g 1 4 .r y z : y z 7 ; x : 7 t y = l ; y : 4 1 5 .x : y 2 - ) ; y : g - y z 1 6 .t 6 : y z - y ; x : y - y z : : lV. y 2rt 3g 9x; y S 23 tr tB. 3y : as- 2rz - t1x; y : I - ng _ ,r, * *

b. y:lxl,y:S-r,x=-r,x:t

2 0y. : l x + 1 1 * l x l , y : 0 , x : - 2 , x - _ 3

Find by integration the area of the triangle having vertices at (S, l), (1, 3), and (-1,_2). T. 22. Find by integation the area of the triangle having vertice8 at (3,4), (2,0), and (0, 1). 23. Findthe areaof the region bounded by the curvc r1- f +2ry -y2:Oandthelirrer:4. (HrNr: Solvethe cubic equation for y in terms of r, and express y as two functions of r,) 24. Find the area of the region bounded by the three curves y:

f , x: yB, and x + y:2. 2 6 . F i n d t h e a r e ao f t h e r e g i o n b o u n d e d b y t h e t h r e e c u r v e s y : * , y : g - y r , a n d . 4 r - y + 1 2 : 0 . 26. Find the_areaof the region above the panbola * : 4py and inside the triangle formed by the r axis and the lines y : x * 8P and Y : -a 1gp. 27. (a) Find the area of the region bounded by the curves y2 : Apx and f: of the area in part (a) with respect to p when p :3.

4py. (b) Find the rate of change of the measure

28' Find the rate of drange of the measurc of the area of Exercise 25 with respect to p when p: *. 29. If d square unit6 i8 the area of the region bounded by the parabola y, : 4r and the line : y flr (z > 0), find the rate of change of A with respect to t||, 30. Detersdne n so that the region above the line y : 1nx and below the parabola y = 2x _ .f has an area of 36 square units.

8.2 VOLUME OF A SOLID OF REVOLUTION: CIRCULAR-DISK AND CIRCUTAR.RING METHODS

F i g u r e8 . 2 . 1

We must first define what is meant by the "volume" of asolid of revolution; that is, we wish to assign a number,for example V,to what we intuitively think of as the measure of the volume of such a solid. We define the measure of the volume of a right-circular cylinder by mzh, where r and h are, respectively, the number of units in the base radius and altitude. Consider first the case where the axis of revolution is a boundary of

C:I R C U L A R - D I SAKN D C I R C U L A R - R I NM GE T H O D S 3 3 1 8 . 2 V O L U M EO F A S O L I DO F R E V O L U T I O N

the region that is revolved. Let the function / be continuous on the closed i n te rv a l l a ,b l a nd assume that f (x) > 0 foral l x i n l a,bl . Let R be the region bounded bythe curve y: f (r), the x axis, and the lines x: A and x: b. Let S be the solid of revolution obtained by revolving the region R about the r axis. We proceed to find a suitable definition for the number V which gives the measure of the volume of S. Let A be a partition of the closed interval la, bl given by a:xo1xtlxr(

1 xn-t I

xr:

b

. T h e n w e h a v e n subi nterval sof the form l * r-r,r;], w herei :7,2, ,rt , any Lrx: xi-xi-y Choose being length of the ith subinterval with the point fi, with xi-r s ti s xi, in each subinterval and draw the rectangles having widths A1r units and altitudes f (€r) units. Figure 8.2.3 shows the region R together with the ith rectangle. When the ith rectangle is revolved about the x axis, we obtain a circular disk in the form of a right-circular cylinder whose base radius is given by f ((r)units and whose altitude is given by A'ixunits, as shown in Fig.8.2.4. The measure of the volurne of this circular disk, which we denote by L,1V,is given by

(t; 'f(tt))

L,V-- nlf (il)' L,x

F i g u r e8 . 2 . 3

(1)

Because there are n rectangles, n curclrlardisks are obtained in this way, and the sum of the measures of the volumes of these n circular disks is given by

fi:r o,r: fi:r rrlf(€,)l'L,x

(2)

which is a Riemann sum. The Riemann sum given in Eq. (2) is an approximation to what we intuitively think of as the number of cubic units in the volume of the solid of revolution. The smaller we take the norm llAllof the partition, the larger will be n, and the closer this approximation will be to the number V we wish to assign to the measure of the volume. We therefore define V to be the limit of the Riemann sum in Eq. (2) as llAllapproaches zero. This limit exists because f i" continuous on la, bf, which is true because / is continuous there. We have, then, the following definition.

F i g u r e8 . 2 . 4

8.2.1,Definition

Let the function / be continuous on the closed interval lo, bl, and assume that f (r) > 0 for all r in la, bl. If S is the solid of revolution obtained by revolving about the r axis the region bounded by the curve y : f (x), the x axis, and the lines x : a and x : b, and if V is the number of cubic units in the volume of S, then

v-

: ' u|)f' dx f. ,tfG,)l'A,x f ,,t;il,

(3)


v

A similar definition applies when both the axis of revolution and a boundary of the revolved region is the y axis or any line parallel to either the r axis or the y axis.

(tr,€i'

o TLLUSTRATIoN 1: We find the volume of the solid of revolution generated when the region bounded by the curve U : xr, the x axis, and the lines x:"1. and x:2 is revolved about the r axis. Refer to Fig. B.z.s, showing the region and a rectangular element of area. Figure 8.2.6 shows an element of volume and the solid of revolution. The measure of the volume of the circular disk is given by L1V: n(€r')'

Lix:

rr(f A,ix

Then Figure8.2.5

v: .Jimf ntf t* llAll-oi:t :,

r2 I x4dx

Jr

: rr(*rt)l' lr

-_

37_ 5ta

Therefore, the volume of the solid of revolution is #a

cubic units.

o

Now suPPose that the axis of revolution is not a boundary of the region being revolved. Let f and 8 be two continuous functions on the

we wish to find a value for V. Let A be a partition of the interval la, bl, given by A:XolXrlXrl

Figure8.2.6

1 xn_, I xr:

fu

and the lth subinterval fr,-r, ri] has length Lix: xr- xr_.tIn the lth subinterval, choose any €i, with xi-r s €i = xi.Consider the n rectangular elements of area for the region R. See Fig.8.2.7 illustrating the region and the ith rectangle, and Fig. 8.2.8 showing the solid of revolution. When the ith rectangle is revolved about the x axis, a circular ring (or "washe{'), as shown in Fig. 8.2.9, is obtained. The number giving the difference of the measures of the areas of the two circular regions is (rlf ((,)l'rlg(f,)1,) and the thickness is A;x units. Therefore, the measure of the volume of the circular ring is given by

L v : r ( l f ( f , ) l ' - [ s ( { , ) ] 'A ) ir The sum of the measures of the volumes of the n circular rings formed by

METHODS AND CIRCULAR-RING CIRCULAR-DISK 8.2 VOLUMEOF A SOLID OF REVOLUTION:

f(€t) ti, f(t;))

f(x)

s(x)

F i g u r e8 , 2 . 9

Figure8.2.8

Figure8.2.7

revolving the n rectangular elements of area about the r axis is

)

i=L

(4)

l,Y i:l

The number of cubic units in the volume, then, is defined to be the limit of the Riemann sum in Eq. (4) as llAllapproacheszero.The limit exists since , f and g2 are continuous on la, b) because f and I are continuous there. 8.2.2 Definition

Let the functions f and I be continuous on the closedinterval la, bl, and assumethatf (x) > g(x) > 0 forall x in la,bf. Then if Vcubic units isthe volume of the solid of revolution generatedby revolving about the r axis and A: gk) and the lines the region bounded by the curves y:/(r) x : n a n dx : b ,

As before, a similar definition applies when the axis of revolution the y axis or any line parallel to either the r axis or the y axis.

Exavrpr.r l.: Find the volume of the solid generated by revolving about the r axis the region bounded by the parabola : x * 3. A : xcz* L and the line y

solurroN: The points of intersection are (-1., 2) and (2,5). Figure 8.2.10 shows the region and a rectangular element of area. An element of volume and the solid of revolution are shown in Fig. 8.2.1'L.

andg(r) T a k i n gf ( x ) : x * 3 the volume of the circular ring is

n(lf G)l'-

[ s ( 1 ' ) ] ' )A i x

* L, we find that the measure of

334


ti , fGi)

Gi , s ft))

Fi g ur e 8. 2. 10

F i g u r e8 . 2 . 1 1

If V cubic units is the volume of the solid, then n

v : lim ) '( lf G)l' - [s(f )],) Aix iIr llall-o

:7r If2 (lf(x)l'J-t f2

: T I

J-r

ls@)fz)dx

[ ( r ' + 3 ) ' - ( x ' + t ) ' z 1d x

f2

:rT I l-xn-f J-r

+6x*sldx

: nl-Ex'- *x3* gr'+ grlt L

l-r

: rl(-+-

S + L z+ 1 6 )- ( t + + + 3 - 8 ) l

_rL7_ 5tl

Therefore, the volume of the solid of revolution

is LFr

cubic units.

AND CIRCULAR-RING METHODS CIRCULAR-DISK 8.2 VOLUMEOF A SOLID OF REVOLUTION:

ExAMPLE 2r Find the volume of the solid generatedby revolving about the line x: -4 the region bounded by the two parabolas )c:y-yz andx:A'-3.

solurroN: The curyes intersectat the points (-2, -1.) and (-+,il. The region and a rectangular element of area are shown in Fig. P.2.12.Figure 8.2.13shows the solid of revolution as well as an elementof volume, which is a circular ring.

F i g u r e8 . 2 . 1 3

Let F(y) - y - y2 and G(y) = y' - 3. The number of cubic units in the volume of the circular ring is L1V: n(14 + F(f')1, - 14+ G(f')l') Lua

(G(6r),fi)

Thus,

v-

F i g u r e8 . 2 . 1 2

n

+ F(€,)f'- 14+ G(€)]') L,y lim -). l l A l l - 0 i : 1 "-([4

-,

f3l2 -y')' - (4* y'-g)'l dY |J - r [ ( 4 + y

-n

I ?2y'-9y'*8y+L5) dy J-r

l3l2

13tz - 3Yt* 4Y'* tu,l-, : n ll+rn _ -

875* 32,.

The volume of the solid of revolution is then Y*n

cubic units.

8.2 Exercises In Exercises 1. through 8, find the volume of the solid of revolution when the given region of Fig. 8.2.1,4is revolved about the indicated line. An equation of the curve in the figure is y2 : *2.

2. OAC about the line AC

l. OAC about the r axis 3. OAC about the line BC

'

4. OAC about the y axis

5. OBC about the y axis

6. OBC about the line BC

7. OBC about the line AC

8. OBC about the x axis

F i g u r e8 . 2 . 1 4

336


9 . Find

the volume of the sphere generated by revolving the circle whose equation is 12 * A2: 12 about a diameter. 10.Find by integration the volurne of a right-circular cone of altitude /z units and base radius a units.

1 1 .Find

the volume of the solid generated by revolving about the x axis the region bounded by the curve : x3 and the A linesy:0andx:2.

12. Find the volume of the solid generated by revolving the region in Exercise L1,about the y axis. 13. Find the volume of the solid generated by revolving about the line x:

-4

the region bounded by that line and the

p a r a b o l ax : 4 i 6 y - 2 y 2 .

14. Find the volume of the solid generated by revolving the region bounded by the curves U2: 4x andy: r about the x axis. 1 5. Find the volume of the solid generated by revolving the region of Exercise 14 about the line x: 4. 16. An oil tank in the shape of a sphere has a diameter of 50 ft. How much oil does the tank contain if the depth of the oil is 25 ft?

1 7. A paraboloid of revolution is obtained by revolving the parab ola y2 : 4px about the r axis. Find the volume bounded by a paraboloid of revolution and a plane perpendicular to its axis if the plane is 10 in. from the vertex, and if the plane section of intersection is a circle having a radius of 5 in.

18. Find the volume of the solid generated by revolving about the x axis the region bounded by the loop of the curve whose equation is 2y2 : x(* - 4).

19. Find the volume of the solid generated when the region bounded by one loop of the curve whose equation is xzyz: (x'-

9) (1 - x2) is revolved about the x axis.

20. The region bounded by a pentagon having vertices at (-4,4), (-2,0), (0, 8), (2,0), and (4,4) is revolved about the x axis. Find the volume of the solid generated.

8.3 VOLUME OF A SOLID OF REVOLUTION: CYLINDRICAL-SHELL METHOD

Figure8.3.'l

In the preceding section we found the volume of a solid of revolution by taking the rectangular elements of area perpendicular to the axis of revolution, and the element of volume was either a circular disk or a circular ring. If a rectangular element of area is parallel to the axis of revolution, then when this element of area is revolved about the axis of revolution, a cylindrical shell is obtained. A cylindrical shell is a solid contained between two rylinders having the same center and axis. Such a cylindrical shell is shown in Fig. 8.3.1. If the cylindrical shell has an inner radius r, units, outer radius r, units, and altitude h units, then its volume V cubic units is given by V: nrzzh- Trrzh (l) Let R be the region bounded by the curye y : f (x), the r axis, and the lines x: a and r : b, where / is continuous on the closed interval la, bl and f(x) > 0 for all x in la, b]; furtherrnore, assume that a > 0. such a region is shown in Fig. 8.3.2. If R is revolved about the y axis, a solid of revolution S is generated. Such a solid is shown in Fig. 8.3.3. To find the volume of S when the rectangular elements of area are taken parallel to the y axis, we proceed in the following manner. Let A be a partition of the closed interval [a, b] given by A:Xo
1 xn-t 1 xn:

b

METHOD 337 CYLTNDRICAL-SHELL OF A SOLIDOF REVOLUTION: 8.3 VOLUME

tt^.\ trx)

Let mi be the midpoint of the lth subinterval lxr-r, ri]. Then ttti: Lr(xo-,* xi). Consider the rectangle having altitude f (mr) units and width tnlr rectangle is revolved about the y axis, a cylindrical shell is A,* ""itt."if Figure 8.3.3 shows the cylindrical shell generated by the recobtained. tangular element of area. If LiV gives the measure of the volume of this cylindrical shell, we : have, from lormula ( 1.) , whe tE T1: xi-1, 12: xi, and h f (mr) '

o'':i.lr'!',,-;'iir-)',!r'*'

F i g u r e8 . 3 . 2

L rV: rr(x i - )ci -)(xr* xr-t:

Becausexi-

L1V:2nmif

Lrx and becaus€ Ii I

(m)

(2 )

xr-r)f(m) Xi-r:2*r,

We have from Eq. (2) (3)

Lix

lf n rectangular elements of area are revolved about the y axLS,n cylindrical shells are obtained. The sum of the measures of their volumes is given by nn

ii : r o,u: ti : l 2nmif(m,) Lix

(4)

which is a Riemann sum. Then we define the measure of the volume of the solid of revolution to be the limit of the Riemann sum in Eq. (4) as llAll approaches zero' The limit exists because it f is continuous on la, bl, so is the function having function values Znxf (x).

Figure8.3.3

8. 3. 1 D e fi n i ti o n

the volume of S, then V-

n

l i m )Ll , Z r r m j ( m )

llall-o i:1

Arr:,, I: xf(x) dx

(5)

Definition 8.3.1 is consistent with Definitions 8.2.1 and 8.2.2 for the measure of the volume of a solid of revolution for which the definitions apPly. The formula for the measure of the volume of the shell is easily remembered by noticing that 2nmr, f (m), and Lp ate, respectively, the numbers giving the circumference of the circle having as radius the mean of the inner and outer radii of the shell, the altitude of the shell, and the thickness.


ExAMPLEL: The region bounded by the curye A : x2, the r axis, and the line x:2 is revolved about they axis. Find the volume of the solid generated. Take the elements of area parallel to the axis of revolution.

solurroN: Figure 8.3.4shows the region and a rectangular element of area. Figure 8.3.5 shows the solid of revolution and the cylindrical shell obtained by revolving the rectangular element of area about the y axis.

a:

xz

(m r, m i')

'i-,

L i*

Figure8.3.4

Figure8.3.5

The element of volume is a cylindrical shell the measure of whose volume is given by LiV : 2nmi(mr') Aix: 2rmi3 A,ix

Thus,

v-

li* > 2nm13A,6

l l A l l - oi : t

2rl

f2

x 3dx

JO

: 2'(if) f' Jo :8rt Therefore, the volume of the solid of revolution is gzr cubic units.

rxavrpr,E 2: The region bounded by the curye A : xz and the lines y : l and r :2is revolved about the line A : -3. Find the volume of the solid generated by taking the rectangular elements of area parallel to the axis of revolution.

soLUTroN: The region and a rectangular element of area are shown in Fig. 8 .3 .6 . The equation of the curve is y : 12. Solving for x, we obtain x : !fr. Because x ) 0 for the given region, we have ,: fr. The solid of revolution as well as a cylindrical shell element of volume is shown in Fig. 8.3.7. The outer radius of the cylindrical shell is (y1+ 3) units and the inner radius is (yr-, * 3) units. Hence, the mean of the inner

METHOD CYLINDRICAL.SHELL 8.3 VOLUMEOF A SOLID OF REVOLUTION:

y:7

(tffi, m;

__\_

!g-)

Figure8'3'7

Figure8.3.6

and outer radii is (mi* 3) units. Because the altitude and thickness of - {nx) units and Liy units, the cylindrical shell are, resPectively, Q L,V:2n(mi+

3) (2 - l*)

Lry

Hence, if V cubic units is the volume of the solid of revolution, n_

Y:

|i : l 2 n ( m 1 +3 )( 2 ! m , ) L i l J i m ilail-0

: l -r4 zrr(!+3)(2-frlaV Jr r4 : 2rr I ?y"''

+ 2y - Sytrz+ 6) dY

Jr

f14 : 2n | 4y,,' + 6y I * y' - 2yzrz lr L

_ ss_* 5t, Therefore, the volume is Tzi cubic units.

340


8.3 Exercises 1-8. Solve Exercises1 through 8 in Sec.8.2 by taking the rectangular elementsparallel to the axis of revolution. In Fig. 8.3.8the region bounded by the r axis, the line r : 1, and the cuwe y : I is denoted by R1;the region bounded bythetwocurvesy:landy,:risdenotedbyRttheregionboundedbytheyaxis,theliney:l,andthecuweyr:ris denoted by Rs.In Exercises9 through 15,find the volume of the solid generatedwhen the indicated region is revolved abodt the given line.

Figure8.3.8 9. R1 is revolved about the y axisi the rectangular elementsare parallel to the axis of revolution. 10. Sameas Exercise9; but the rectangularelementsare perpendicular to the axis of revolution, 11. R, is revolved about the r axi6; the rectangular elements are parallel to the axis of revolution. 12. Same as Exercise 11; but the rectantular elements are perpendicular to the axis of revolution. 13. & i8 revolved about the line y:2; the rectanguLarelerrents are paiallel to the axis of revolution. 14. same as Exercise 13; but the rectantular elements are perpendicular to the axis of revolution. 15. R, is revolved about the line r : -2; the rectangular elements are parallel to the axis of revolution. 16. Sameas Exercise15; but the rectangularelements are perpendicular to the axis of revolution. 17. Find the volume of the solid generated if the rcgion bounded by the parabola A, : 4ax (a > 0) and the line r : a is revolved about r: a, 18. Find the volume of the solid genented by revolving the region bounded by the curve li3 * yrl": a2i' about the y axis. 19. Find the volume of the solid generatedby revolving about the y axis the region outside the cuwe y : f and between the Lines y : 2a - 1 ?n\d.y : x + 2. 20. Through a spherical shaped solid of radius 6 in., a hole of radius 2 in. is bored, and the axis of the hole is a diameter of the sphere. Find the volume of the part of the solid that remains. 21. A hole of radius 2\6 in. is bored through the center of a spherical shaped solid of radius 4 in. Find the volume of the portion of the solid cut out. 22. Find the volume of the solid generatedif the region bounded by the parabola y, :4pr about the y axis.

and the line r:

p is revolved

PLANESECTIONS 341 8.4 VOLUME OF A SOLIDHAVINGKNOWNPARALLEL

8.4 VOLUME OF A SOLID Let S be a solid. By a plane sectionof S is meant a plane region formed by HAVING KNOWN PARALLEL the intersection of a plane with S. In Sec.8.2 we learned how to find the PLANE SECTIONS volume of a solid of revolution for which all plane sections perpendicular to the axis of revolution are circular. We now generalize this method to find the volume of a solid for which it is possible to express the area of any plane section perpendicular to a fixed line in terms of the perpendicular distance of the plane section from a fixed point. We first define what we mean when we say that a solid is a "cylinder." 8.4.1. Definition

A solid is a right cylinder if it is bounded by two congruent plane regions R1 and R2 lying in parallel planes and by a lateral surface generated by u line segment, having its endpoints on the boundaries of R1 and Rz, which moves so that it is always PerPendicular to the planes of Rt and R2.

F i g u r e8 . 4 . 1

Figure8.4.2

Figure8.4.3

Figure 8.4.1.illustrates a right cylinder. The heighf of the cylinder is the perpendicular distance between the planes of Rr and R2, and the base is either R1 or R2. If the base of the right cylinder is a region enclosed by a circle, we have a right-circular cylinder (see Fig. 8.a.4; if the base is a region enclosed by u rectangle, we have a rectangular parallelepiped (see F i g . 8 .a .3 ). If the area of the base of a right cylinder is A square units and the height is h units, then by definition we let the volume of the right cylinder be measured by the product of A and h. As stated above, we are considering solids for which the area of any plane section that is perpendicular to a fixed line is a function of the perpendicular distance of the plane section from a fixed point. The solid 5 of Fig. 8.4.4 is such a solid, and it lies between the planes perpendicular to the r axis at a and b. We represent the number of square units in the area of the plane section of S in the plane perpendicular to the r axis at x by A(x), where A is continuous on la, bl. Let A be a partition of the closed interval la, bl given by a:xo/-xr1xzl A(€ t) Figure 8.4.4

'1xn:6

. , Wehave, then, n subintervals of the form lxr-r, ri], wherei:1,2, any Choose x^. xiLrx: being n,wittr the length of the ith subinterval


number f1, with xi-t s €i a xi, in each subinterval and construct the right cylin-dersof heights A;x units and plane section areasA(tr) square units. The volume of the lth cylinder is A(f,) A;x cubic units. We obtain n right cylinders,and the sum of the measuresof the volumes of thesen cvlinders is given by the Riemann sum n

A(() ap

(1)

i:r

This Riemann sum is an approximation of what we intuitively think of as the measure of the volume of S, and the smaller we take the norm llAllof the partition A, the larger will be n, and the closer this approximation will be to the number we wish to assign to the measure of the volume. We have, then, the following definition. 8.4.2 Definition

Let S be a solid such that S lies between planes drawn perpendicular to the x axis at a and b. If the measure of the area of the plane section of S drawn perpendicular to the x axis at r is given by A(x) , wherc A is continuous on la, bl, then the measure of the volume of S is given by

lim 2 a(f,) Aix: [' A ( x ) d x .0 i=l J(l

(2)

lllll

Note that Definitions 8.2.7 and 8.2.2 are special cases of Definition 8.4.2. If in Eq,. (2) we take A (x): nl f (x)12,w e have E q. (3) of S e c.g. Z. lf w e t a k e A ( x ) : n ( l f ( x ) l ' - [S(x)]'), *" have Eq. (S) of Sec. 8.2.

nxervrpr.r 1: If the base of a solid is a circle with a radius of r units and if all plane sections perpendicular to a fixed diameter of the base are squares, find the volume of the solid.

solurroN: Take the circle in the xy plane, the center at the origin, and the fixed diameter along the r axis. Therefore, an equation of the circle is 12 I y' : r2. Figure 8.4.5 shows the solid and an element of volume, which is a right cylinder of altitude A1r units and with an area of the base given by l2f (il]2 square units, where /(r) is obtained by solving the equation of thpirc,fe j?, v and setting y : f (x). This computation give f (x) : Yr" - xz. Therefore, if V cubic units is the volume of the solid, "we have

lim -tt llall-0

n

lzf((,)l' Lix

i:r

(r'-

: +l'"*

x 2 )d x

8.4 VOLUMEOF A SOLID HAVINGKNOWN PARALLELPLANESECTIONS

F i g u r e8 . 4 . 5

rxeuPrn 2:

A wedge is cut from a right-circular cylinder with a radius of r in. bY two Planes, one perpendicular to the axis of the cylinder and the other intersecting the first at an angle of measurement 60oalong a diameter of the circular plane section. Find the volume of the wedge.

solurroN: The wedge is shown in Fig. 8.4.6. The xy plane is taken as the plane perpendicular to the axis of the cylinder, and the origin is at the An equation of the circular plane section is then point of plrp"ndicularity. : 12.Every plane section of the wedge perpendicular to the r axis is * * y, a right triangle. An element of volume is,a right cylinder having altitude (€r)12 in.2, where /(r) is obA,r-in., and irea of the base given bV i$lf for y and setting y : f (x), circle tained by solving the equation of the in.3 is the volume of the if V Therefore, \fr'=T. thereby giving f (x) wedge, n_

V:

U, - €rr)L,x

lim )iVe .0,:r

llall

_

:+\/g

fr

|

J-r

(r, - x2)dx I

- i*"1'_, : +\/5lr'* : fi\/3r3 Hence, the volume of the wedge is 3 tEr3 in'3

Figure8.4.6

U4

APPLICATIONS OF THEDEFINITE INTEGBAL

Exercises 8.4 l

The baseof a solid is a circlehawing a radius of / units. Find the volume of the solid if all plane sectionsperpendicular to a fixed diameter of the base are equilateral triangles,

The base of a solid is a circle with a radius of r units, and all pLanesections perpendicular to a fixed diameter of the base are isoscelesright triangles haYing the hypotenuse in the plane of the base. iind the volume of the solid. 3 . Solve Exercise2 if the isoscelesdght t angles have one leg in the plane of the base. 4. Find the volume of a right pyramid having a height of t units and a square base of side n units. 5 . Find the volume of the tebahedmn havint 3 mutually perpendicular faces and three mutually perpendicular edges whose lengths have measuresa, b, and c. 6 . The base of a solid is a cirde with a radius of 4 in. , and each plane section perpendicular to a fixed diameter of the base is an isosceles triangle having an altitude of 10 in. and a chord of the cirde as a base. Find the volume of the solid. 7. The base of a solid is a circle with a radius of 9 in., and each plane section perpendicular to a fixed diameter of the base is a square having a chord of the circle as a diagonal. Find the volume of the solid. 8. Two right-circular cylinde{s, each having a radius of / units, have axes that intersect at right angles. Find the volume of the solid common to the two cylinders. 9 . A wedge is cut ftom a solid in the shapeof a right-circular cylinder with a radius of r in. by a -the plane through a diameter of the base and inclined to the Plane of the base at an angle of measurement 45". Fi;d volume" of the wedge. 10. A wedge is cut from a solid in the shaPeof a right-circular cone having a base radius of 5 ft and an altitude of 20 ft by two half planes through the axis of the cone, The argle between the two pl-aneshas a measurement of 30.. Find the volume of the wedge cut out.

8'5 WORK

The "work" done by u force acting on an object is defined in physics as "force times displacement." For example, suppose that an otiect is moving to the right along the r axis from a point a to apoint b, and.a constant force of F lb is acting on the object in the direction of motion. Then if the displacement is measured in feet, (b - a) is the number of feet in the displacement. And if W is the number of foot-pounds of work done by the force, W is defiped by (1) o tttustRArroN '1-.: lI W is the number of foot-pounds in the work necessary to lift a 70-lb weight to a height of 3 ft, then W:70

' 3:270

.

In this section we consider the work done by a variable force, which is a function of the position of the object on which the force is acting. We wish to define what is meant by the term "work" in such a case. suppose that /(x) , where /is continuous on la, bl, is the number of units in the force acting in the direction of motion on an object as it moves to the right along the x axis from point a to pointb. Let A be a partition of

8.5 WORK

the closedinterval la, bf : 1 xn-r I xr:

a:xolxrlxzl

b

The ith subinterval is [x1-1, x]; and if xi-, is close to xi, the force is almost constant in this subinterval. If we assume that the force is constant in the ith subinterval and if fi is any point such that ri-, s fi s ri, then if AiW is the number of units of work done on the object as it moves from the point xi-1 Io the point xi, from formula (1) we have A ,W: f (€ ,) (x,- xi -r) Replacin E )ci- xi-r by A;x, w€ have

A1W:f (t) L'x and n

AiW:2fG,) L,x

i:1

(2)

i:r

The smaller we take the norrn of the partition A, the larger n will be and the closer the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the total work done. We therefore define the measure of the total work as the limit of the Riemann sum in Eq. (2). 8.5.1 Definition

Let the function / be continuous on the closed interval la, bl and /(r) be the number of units in the force acting on an object at the point / on the r axis. Then if W units is the work done by the force as the obfect moves from a to b, W is given by (3)

In the following examplewe use Hooke'slaw, which statesthat if a spring is stretchedx in. beyond its natural length, it is pulled back with a force equal to kx lb, where k is a constant depending on the wire used. nxalvrpn L: A sPring has a natural length of 14 in. If a force of 5 lb is required to keep the sPring stretched 2rrt., how much work is done in stretching the sPring from its natural length to a length of 18 in.?

solurroN: Place the spring along the r axis with the origin at the point where the stretching starts (see Fig. 8.5.1). 14in.# T

8 .5 .1 F i g u re Let r: the number of inches the spring is stretched; : the number of pounds in the force acting on the spring r in. f (x) beyond its natural length.


Then, by Hooke's law, f (x) : kr. Because/(2) : 5, we have . 5:k'2

k:E Therefore,

f(x):Ex If W: the number of inch-pounds of work done in stretching the spring from its natural length of 'J,4in. to a length of 1g in., we have

w: li3 f. f G,)o,* llall-0 t:I

f(x) dx Exdx

: T x z 1I4

JO

:20 Therefore, the work done in stretching the spring is 20 in.-lb. EXAMPLE2: A water tank in the form of an inverted right-circular cone is 20 ft across the top and 15 ft deep. If the surface of the water is 5 ft below the top of the tank, find the work done in pumping the water to the top of the tank.

solurroN: Referto Fig. 8.5.2.The positive r axis is chosenin the downward direction becausethe motion is vertical. Take the origin at the top of the tank. We considera partition of the closedinterval [5, 15] on the r ixis and let fi be any point in the ith subinterval l*r-r, r;]. An elementof volume is a circular disk having thickness A,ixft and radiu s (il ft, where the f function / is determined by an equation of the line through the points (0, 10) and (L5, 0) in the form y - f (r). The number of cubic feet in the volume of this elementis given by AîV: nlf (&)1, Lrx.If w is the number of pounds in the weight of L fts of water, then the number of pounds in the weight of this element is wnlf(t)1, A;r, which is the force required to pump the element to the top of the tank. If xi-, is closeto xi, then the distance through which this element moves is approximately& ft.Thus, if A'rWft-lb is the work done in pumping the element to the iop of the tank, AiW is approximately(wrrlf G)l' Lrx) . fii so if w is the number of footpounds of work done,

w:

. Ji3 f .df G,)1, €i Lix

llall-0 f:t

: *n Figure8.5.2

fr5

(x)lzxdx J, lf

To determine f (x), we find an equation of the line through the points

8.5 WORK

(15, 0) and (0, 10) by using the interceptform:

*15 + + : 10 1

or y---Bx*Lo

T h e r e f o r e , f ( x ) : - & x * L 0 ,a n d

w - *n f" ?3x * \o)zxdx Js rr5 : rpir | (6rt - #x't Js

toox) dx

|:'u)n lt*n - Xqr3* 5012 1ts rJt : + (10,000nar) Therefore,the work doneis 1,0,000rrw19 ft-lb.

8.5 Exercises 1. A spring has a natural length of 8 in. If a force of 20lb stretchesthe spring i in., find the work done in stretching the spring from 8 in. to 11 in. 2. A spring has a natural length of 10 in.. and a 30-lb force stretches it to 11iin. Find the work done in stretching the sprint from 10 in. to 12 in. Then find the work done in stretchint the sPdng from 12 in. to 14 in. 3. A spring has a natural length of 6 in. A 12,000-lb force compressesthe spring to 5* in Find the work done in compressing it from 6 in. to 5 in. Hooke's law holds for compression as well as for extension. 4. A spring has a natural length o{ 6 in. A 12001b force compresses it to 5} in. Find the wotk done in compressing it from 6 in. to 4* in. 5. A swimming pool full of water is in the form of a rectangularparallelepiped5 ft deep, 15 ft wide, and 25 ft long. Find the work required to pump the water in the pool up to a level I ft above the surface of the pool. 6. A trough full of water is 10 ft long, and its cross section is in the shape of an isosceles triangle 2 ft wide acrossthe top and 2 ft high. How much work is done in pumping all the water out of the trcugh over the top? 7. A hemispherical tank with a radius of 5 ft is filed with water to a depth of 4 ft. Find the work done in pumping the water to the top of the tank. 8. A right-circular cylindrical tank with a depth of 12 ft and a radius of 4 ft is half tull of oit weighing 60 lb/ft3. Find the wotk done in pumping the oil to a height 6 ft above the tank. 9. A cable 200 ft lont and weighing 4 lb/ft is hanging vertically down a well. If a weight of 100Ib is susp€nded from the lower end of the cable, find the work done in pulling the cable and weight to the top of the well, 10. A bucket weighing 20lb containing 60 lb of sand is attached to the lower end of a 100 ft long chain that weighs 10lb and is hanging in a deep well. Find the work done in raising the bucket to the top of the well, 11. SolveExercise10 if the sand is leaking out of the bucket at a constantrate and has all leakedout iust as soon as the bucket i8 at the toD of the well.

348


12. As a water tank is being raised, water spills out at a constant rate of 2 fC per foot of rise, If the tank originally contained 1000 fe of water, find the work done in raising the tank 20 ft. 13. A tank in the form of a rectangular parallelepiped 6 ft deep,4 ft wide, and 12 ft long is fuJl of oil weighing 50lb/fi3. When one-third of the work necessary to pump the oil to the top of the tank has been done, find by how much the sudace of the oil is lowered. 14. A cylindrical tank 10 ft high and 5 ft in radius is standing on a platform 50 ft high. Find the depth of the water whm onehalf of the work required to fill the tank faom the Found level through a pipe in the bottom has been done. 15. A one horsepower motor can do 550 ft-lb of work per second. If a 0.1.hp motor is used to pump water from a full tank in theshapeofarectangularparallelepiped2ftdeep,2ftwide,and6ftlonttoapoint5ftabovethetopofthetank,how long will it take? 16. A meteorite is a miles from the center of the earth and falls to the surface of the earth. The folce of sraviw is inverselv proportional to the square of the distance of a body from the center of the earth. Find the work doie by gravity if the weight of the meteorite is r, lb at the sudace of the earth. Let R rriles be the Edius of the earth,

8'6 LIQUID PRESSURE Another application of the definite integral in physics is to find the force causedby liquid pressureon a plate submergedin the liquid or on a side of a containerholding the liquid. First of all, supposethat a flat plate is insefted horizontally into a liquid in a container. The weight of the liquid exertsa force on the plate. The force per square unit of area exerted by the liquid on the plate is called the pressureof the liquid. Let w be the number of pounds in the weight of one cubic foot of the liquid and h be the number of feet in the depth of a point below the surfaceof the liqui d. If p is the number of pounds per square foot of pressure exertedby the liquid at the point, then 'u:oi"'rne

(1)

number of square feet in the area of a flat plate that is submerged horizontally in the liquid, and F is the number of pounds in the force causedby liquid pressure acting on the upper face of the plate, then F:PA

( 2)

Substituting from formula (1) into (2) gives us F : whA

(3)

Note that formula (1) states that the size of the container is immaterial so far as liquid pressure is concerned. For example, at a depth of 5 ft in a swimming pool filled with salt water the pressure is the same as at a depth of 5 ft in the Pacific Ocean, assuming the density of the water is the same. Now suPPose that the plate is submerged vertically in the liquid. Then at points on the plate at different depths the pressure, computed from formula (1), will be different and will be greater at the bottom of the plate than at the top. We now proceed to define the force caused by liquid Pressure when the plate is submerged vertically in the liquid. We use

8.6 LIQUIDPRESSUBE 349 Pascal's principle: At any point in a liquid, the pressure is the same in all directions. In Fig. 8.6.1 let ABCD be the region bounded by the x axis, the lines x: a and r : b, andthe curve y: f (x), where the function /is continuous and/(r) > 0 on the closed interval fa, bl. Choose the coordinate axes so they axis lies along the line of the surface of the liquid. Take the r axis vertical with the positive direction downward. The length of the plate at a depth r ft is given by f (x) ft. Let A be a partition of the closed interval la, bl which divides the interval into n subintervals. Choose a point f; in the ith subinterval, with xrr 3 €i = xr. Draw n hofizontal rectangles. The ith rectangle has a length of f (€r) ft and a width of L,ix ft (see Fig. 8.6.1). If we rotate each rectangular element through an angle of 90o,each element becomes a plate submerged in the liquid at a depth of f6 ft below the surface of the liquid and perpendicular to the region ABCD. Then the force on the rth rectangular element is given by wtrf (t) L# lb. An approximation to F, the number of pounds in the total force on the vertical plate, is given by

F i g u r e8 . 6 . 1

n

wt,fGi) Air 2 i:r

(4)

which is a Riemann sum. The smaller we take llAll,the larger n willbe and the closer the approximation given by (4) will be to what we wish to be the measure of the total force. We have, then, the following definition. 8'6'1 Definition

Suppose that a flat plate is submerged vertically in a liquid of weight ut pounds per cubic unit. The length of the plate at a depth of r units below the surface of the liquid is /(x) units, where / is continuous on the closed interval la, bl and f(x) > 0 on la, bl. Then F, the number of pounds of force causedby liquid pressureon the plate, is given by

t:

Air: wxf(x)dx [: 2*rOrri) ,,Til,

(s)

Exevpr.E 1: A trough having a soLUrIoN: Figure 8.5.2 illustrates one end of the trough together with a trapezoidal cross section is full of rectangularelement of area. An equation of line AB it y :g - *r. Let water. If the trapezoid is 3 ft wide f (x) : B *x.If we rotate the rectangularelement through 90', th" force on at the top,2 ft wide at the bottom, the element is given by 2w(1fG) Lfi lb. If F is the number of pounds in and 2 ft deep, find the total force the total force on the side of the trough, owing to liquid pressure on one n end of the trough. F - lim zwt,f (t) L#

2

H

llAll-0 f:t


:2us :Ztll

[, oo) d x

I,.*

- ix) dx

lrr'-# ' ] , 2 , O )A ( 2 , l ) Figure8.6.2

Taking w : 62.5, we find that the total force is 291,.2lb.

Exavprr 2: The ends of a trough are semicircular regions, each with a radius of 2 ft. Find the force caused by liquid pressure on one end if the trough is full of water.

soLUrIoN: Figure 8.5.3 shows one end of the trough together with a rectangular element of area. An equation of the semicircle is xz * yr:4. Solving for y gives y : t[=7, and so let f (x) : {E -7. The force on the rectangular element is given by 2w(j(fr) Arr. so if F pounds is the total force on the side of the trough,

F: ,,limf ,*g,f(f,) Air llall.0 t:l :2ut Ir2 xf(x) dx Jo

r2 :2'ut I xt/E=P

dx

JO

: -&w(4- xr)r,rl' JO

Figure8.6.3

Therefore, the total force is ffut lb.

8.6 Exercises 1. A platein the shaPeof a rectangle is submerged vertically in a tank of water, with the upper edge lying in the surface. If -"ia" the width of the plate is 10 ft and the depth is 8 ft, find the {orce due to liquid presiure ott otr" of the plate. 2. A square plate of side 4 ft is submerged ve ically in a tank of water and its center is 2 ft below the surface.Find the force due to liquid ptessure on one side of the plate. 3. Solve Exercise 2 if the center of the plate is 4 ft below the surface. 4. A Platein the shapeof an isoscelesright triantle is submergedvertically in a tank of water, with one let lying in the surface.The legs are each 6 ft lont, Find the force due to liquid pressureon one side of the plate. 5. A rectantular tank fuIl of water is 2 ft wide and 18 in. deep, Find the force due to liquid pressure on one end of the tank. 6. The,endsof a trough are equilateraltriangles having sides with lengtls of 2 ft. If the water in the trough is 1 ft deep, find the force due to liquid prrssure on one end.

8.7 CENTEROF MASSOF A ROD

351

7. The face of a dam adiacent to the water is vertical, and its shape is in the form of an isosceles triangle 250 ft wide across the top and 100 ft high in the center. If the water is 10 ft deep in the center, find the total force on the dam due to liquid Pressure. 8. An oil tank is in the shape of a right-circular cylinder 4 ft in diameter, and its axis is horizontal. If the tank is half full of oil weighing 50 lb/fe, find the total force on one end due to liquid Pressure. 9. The faceof the gate of a dam is in the shape of an isoscelestriangle 4 ft wide at the top and 3 ft high. If the upper edge of the face of the gate is 15 ft below the surface of the water, find the total force due to liquid pressure on the gate. 10. The face of a gate of a dam is vertical and in the shape of an isoscelestrapezoid 3 ft wide at the top, 4 ft wide at the bottom, and 3 ft high. If the upper base is 20 ft below the surface of the water, find the total force due to liquid pressure on the gate. 11. The face of a dam adjacentto the water is inclined at an angle of 30ofrom the vertical. Tlie shape of the face is a rectangle of width 50 ft and slant height 30 ft. If the dam is full of water, find the total force due to liquid pressure on the face. 12. Solve Exercise11 if the face of the dam is an isoscelestrapezoid 120ft wide at the top, 80 ft wide at the bottom, and with a slant height of.40 ft13. The bottom of a swimming pool is an inclined plane. The pool is 2 ft deep at one end and 8 ft deep at the other. If the width of the pool is 25 ft and the length is 40 ft, find the total force due to liquid pressure on the bottom. 14. If the end of a water tank is in the shapeof a rectangle and the tank is full, show that the measure of the force due to liq' uid pressureon the end is the product of the measureof the areaof the end and the measureof the force at the geometrical center.

8.7 CENTER OF MASS OF A ROD

In Sec. 7.5 we leamed that if the function / is continuous on the closed interval l-a,bf', the averagevalue of f on fa, b] is given by

I rat dk b:- a

An important application of the average value of a function occurs in physics in connection with the concept of centerof mass. To arrive at a definition of "mass," consider a particle that is set into motion along an axis by a force of F lb exerted on the particle. So long as the force is acting on the particle, the velocity of the particle is increasing; that is, the particle has an acceleration.The ratio of the.force to the acceleration is.constant regardless of the nagnitude of the force, and this constant ratio is called the massof the particle. . rLLUsrRArror L: If the acceleration of a certain particle is 10 ft/sec2 when the force is 30Ib, the mass of the particle is 30 lb : - 3lb 10 ftlsecz 1 ftlsec2 Thus, for every 1 ft/secz of acceleration,a force of 3 lb must be exerted on . the particle. When the unit of force is 1 lb and the unit of acceleration is7 ftlsecz,

,.-w

352


the unit of mass is called one slug. That is, L slug is the mass of a particle whose acceleration is 1. ftlsec2 when the magnitude of the force on the particle is 1 lb. Hence, the particle of Illustration 1 that has an acceleration of 10 ftlsec2 when the force is 30 lb has a mass of 3 slugs. From physics, if W lb is the weight of an object having a mass of m slugs, and g ftlsec2 is the constant of acceleration due to gravity, then W: mB Consider now a horizontal rod, of negligible weight and thickness, placed on the r axis. On the rod is a system of n particles located at points x 1 , x 2 , .. . , l n . T h e i t h p a r t i d e ( i : 1 , 2 , . ,n) is atadirecteddistance 4 ft from the origin and its mass is mi slugs. See Fig.8.7.1. The n

number of slugs in the total mass of the system ir )

/ni. We define the

i :l

F m2

lTl s

xs

O

x4

x,r

{

t-

x,2

x

F i g ur e 8 . 7. 1

momentof massof the ith particle with respect to the origin as mixi slug-ft. The moment of mass for the system is defined as the sum of the moments .all of mass of the particles. Hence, If M, slug-ft is the moment of mass of the system with respect to the origin, then

Now we wish to find a point f such that if the total mass of the system were concentrated there, its moment of mass with respect to the origin would be equal to the moment of mass of the system with respect to the origin. Then f must satisfy the equation nn

t> mi:2 i:l

flltXt

i :l

and so

(1)

The point r is called the centerof massof the system, and it is the point where the system will balance. The position of the center of mass is indepehd'entof the position of the origin; that is, the location of ttre center of mass relative to the positions of the particles does not change when the origin is changed.

8,7 CENTEROF MASS OF A ROD

ExAMPLE L: Given four particles of masses 2, 3,1,, and 5 slugs located on the r axis at the Points having coordinates 5, 2, -3, and -4, respectively, where distance measurement is in feet, find the center of mass of this system.

solurroN: lf 7 is the coordinate of the center of mass, we have from formula (1) 7

fi, Thus, the center of mass is # tt to the left of the origin.

The preceding discussion is now extended to a rigid horizontal rod having a continuously distributed mass.The rod is said tobehomogeneous if its mass is directly proportional to its length. In other words, if the segment of the rod whose length is Arr ft has a mass of L1mslugs, and'A,1m: k Air, then the rod is homogeneous. The number k is a constant and k slugs/ft is called the.linear dcnsity of the rod. Lim = p((i)Lir

il*],

,

,,_r€i

-;.

O

tfi

,,

L---

Figwe 8.7.2

Suppose that we have a nonhomogeneous rod, in which case the linear density varies along the rod. Let L ft be the length of the rod, and place the rod on the r axis so the left endpoint of the rod is at the origin and the right endpoint is at L. See Fi9.8.7.2. The linear density at any point r on the rod is p(r) slugs/ft, where p is continuous on [0, L]. To find the total mass of the rod we consider a partition A of the closed interval [0, L] into z subintervals. The fth subinterval is [ra-t, ra], and its length is A4x ft. If f6 is any point in lxp1, xrf, ffir approximation to the mass of the part of the rod contained in the ith subinterval is Lam slags, where Arnt: p(t) L* The number of slugs in the total mass of the rod is approximated by nn

s A ; t n - t pG) Lfc

LJ"Ll i:l

(2)

i:l

The smaller we take the norm of the partition A, the doser the Riemann sum in Eq. (2) will be to what we intuitively think of as the measure of the mass of the rod, and so we define the measure of the mass as the limit of the Riemann sum in Eq. (2). 8.7.1 Definition

A rod of length L ft has its left endpoint at the origin. If the number of slugs per foot in the linear density at a point x ft from the origin is p(x), where p is continuous on 10,L), then the total massof the rod is M slugs,


where

p(x) dx

ExAMPrn 2: The density at any point of a rod 4 ftlong varies directly as the distance from the point to an external point in the line of the rod and 2 ft from an end, where the density is 5 slugs/ft. Find the total mass of the rod.

solurroN: Figure 8.7.3shows the rod placed on the r axis. If p(r) is the number of slugs per foot in the density of the rod at the point r ft from the end having the greater density, then P(x):c(6-x) where c is the constant of proportionality. Becausep(4):5, we have 5:2c or c:*. Hence,p(x):E(6x). Therefore,if M slugs is the total mass of the rod, we have from Definition 8.7.1.

M : , l i p f Z < r -f , )A , x llAll-0 i=l E(6- x) dx

-40 The total mass of the rod is therefore 40 slugs. 6-'* ti x

ox.,(E>,46 F i g u r e8 . 7 . 3

we now proceed to define the center of mass of the rod of Definition 8.7.L. However, first we must define the moment of mass of the rod with respect to the origin. ti

L rm : p(t;)Aix

t

,,

,r_, \; Figure8.7.4

As before, place the rod on the r axis with the left endpoint at the origin and the right endpoint at L. see Fig. 8.7.4.Let A be a partition of [0, L] into n subintervals, with the lth subinterval lxo_r, xolhaving length Aix ft. If f' is any point in lx;r, ri], an approximation to the moment of mass

I

8.7 CENTEROF MASS OF A ROD

355

with respect to the origin of the part of the rod contained in the ith subinterval is & Lim slug-ft, where Liftr: pG) L*. The number of slug-feet in the moment of mass of the entire rod is approximated by n

n

i:l

i:l

(4)

PGt) Lix

The smaller we take the norm of the partition A, the closer the Riemann sum in Eq. (4) will be to what we intuitively think of as the measure of the moment of mass of the rod with respect to the origin. We have, then, the followi.g definition. 8.7.2 Definition

A rod of length L ft has its left endpoint at the origin and the number of slugs per foot in the linear density at a point x ft from the origin is p (x) , where p is continuous on [0, L]. The moment of mass of the rod with respect to the origin rs Mo slug-ft, where

(s) The center of massof the rod is at the point r such that if M slugs is the total mass of the rod, iM : Mo. Thus, from Eqs. (3) and (5) we get

(6)

EXAMPLE 3:

Find the center of mass for the rod in Example 2.

soLUrIoN: In Example 2, we found M:4A. E(5 - x), we have

r

Using Eq. (6) with p(r) :

Ex(6- x) dx 40

Theref.ore, the center of mass is at I ft from the end having the greater density. 2: If a rod is of uniform density k slugs/ft, where k is a constant, then from formula (6) we have

O ILLUSTRATION

kLz 2L

kx2 2 L

kx

A T

kL2

0

Thus, the center of mass is at the center of the rod, as is to be expected.o

356


8.7 Exercises In Exercises1 through 4, a system of particles is located on the r axis. The number of slugs in the mass of each particle and the coordinate of its position are given. Distance is measured in feet. Find the center of mass of each svstem. l. mr:5 at 2; m2: 6 at 3; ms: 4 at 5; mq: 3 at 8. 2. mr:2

a t - 4 ; m z : 8 a l - 7 ; m s : 4 a t 2 ;m q , : 2 a t 3 . 3. mt: 2 at -3; frrz: 4 at -2; ms: 20 at 4; mn: 70 at 6; ms: 30 at 9. 4. m, : 5 at -7; ttrz: 3 al -2; m": 5 at 0i ffiE: ! at 2; m5: 8 at 10. In Exercises5 through 9, find the total mass of the given rod and the center of mass. 5 . The length of a rod is 9 in. and the linear density of the rod at a point r in. from one end is (4r * 1) slugs/in. 6 . The length of a rod is 3 ft, and the linear density of the rod at a point r ft from one end is (5 * 2x) slugslft. 7. The length of a rod is 10 ft and the measure of the linear density at a point is a linear function of the measureof the distance of the point from the left end of the rod. The linear density at the left end is 2 slugs/ft and at the right end is 3 slugs/ft. 8 . A rod is 10 ft long, and the measure of the linear density at a point is a linear function of the measureof the distance from the center of the rod. The linear density at each end of the rod is 5 slugs/ft and at the center the linear density is 3i slugs/ft. 9 . The measureof the linear density at a point of a rod varies directly as the third power of the measureof the distance of the point from one end. The length of the rod is 4 ft and the linear density is 2 slugs/ft at the center.

10.A rod is 6 ft long and its mass is 24 slugs. If the measure of the linear density at any point of the rod varies directly as the square of the distance of the point from one end, find the largest value of the linear density. L L . The length of a rod is L ft and the center of mass of the rod is at the point *L ft from the left end. If the measureof the linear density at a point is proportional to a power of the measure of the distance of the point from the left end and the linear density at the right end is 20 slugs/ft, find the linear density at a point x ft from the left end. Assume the massis measured in slugs.

t2. The total mass of a rod of length L ft is M slugs and the measure of the linear density at a point r ft from the left end is

proportional to the measureof the distance of the point from the right end. Show that the linear density at a point on the rod r ft from the left end is 2M(L - x) lLz slugs/ft.

8.8 CENTER OF MASS OF A PLANE REGION

Let the massesof z particleslocated at the points (xr,yr), (x",yr) , . . . , (xn, An)in the ry pline be measuredin slugs by *r,;r, . . . , mn, and consider the problem of finding the center of mass of this system. we may imagine the partides being supported by a sheet of negligible weight and negligible thickness and may assumethat each particle has its position at exactly one point. The center of mass is the point where the sheet will balance. To determine the center of mass, we must find two averages:f, which is the average value for the abscissasof the n points, and y', the averagevalue for the ordinates of the z points. we first define the moment of mass of a system of particles with respect to an axis. ' If a particle at a distance d ft from an axis has a mass of.m slugs,then if

8.8 CENTEROF MASS OF A PLANEREGION

M, slug-ft is the moment of mass of the particle with respect to the axis, Mt:

(1)

md

If the ith particle, having trraissftti slugs, is located at the point (xi, y) ,Its distance from the y axis is xrft; thus, from formula (1), the moment of massof this particle with respectto the y axis ismixl slug-ft. Similarly, the moment of mass of the particle with respectto the r axis is miylslug-ft-The moment of the system of n particles with respectto the y axis is Mu slug-ft, where (2) and the moment of the system with respect to the x axis Ls Mr slug-ft, where (3) The total mass of the system Ls M slugs, where (4)

The center of mass of the system is at the point (f , Y), where

i,:i'.u.llilliillffilr::,i

:'....it.*.. ...':,..

(s)

(6)

The point (i, !) canbe interpreted as the point such that, if the total mass M slugs of the system were concentrated there, its moment of mass with respectto the y axis,Mn slug-ft, would be determin edby Mu : Mi , and its moment of mass with respect to the I axis, M" slug-ft, would be determined by Mr: My. SOLUTION: ExAMPLEL: Find the centerof massof the four particleshavi^g Mo: masses 2, 6, 4, and 1.slugslocated at the points (5, -2) , (-2, L) , (0,3), and (4,-L), respectively. M, '. -

M

4

t f t i x r - 2 ( 5 ) + 6 ( - 2 ) + 4 ( 0 ) + L( 4 ) : 2

i:7 4

\l

.at i:t

- s4

Z,r i:t

rniAt-2(-2) + 6(1)+ 4(3)+ L(-1) :13 lfii:2 + 6 + 4 + L : 13

358

A P P L I C A T I O NO SF T H E D E F I N I T EI N T E G R A L

Therefore,

W:a M13JM13

andy-W-

1 3-

1

The center of mass is at (#, L ).

0i, f0))

xi-rTi

\/

L;x

F i g u r e8 . 8 . 1

f(r)

consider now a thin sheet of continuously distributed mass, for example, a piece of paper or a flat strip of tin. we regard such sheetsas being two dimensional and call such a plane region a lamina.In this sectionw-e confine our discussion to homogeneous laminae, that is, laminae having constant area density. Laminae of variable area density are considered in Chapter 2L in connection with applications of multiple integrals. Let a homogeneous lamina of areaA ff have a mass of M slugs. Then if the constant area density is k slugs/ff , M: kA. rf the homogeneous lamina is a rectangle,its center of mass is defined to be at the center of the rectangle. we use this definition to define the center of mass of a more general homogeneous lamina. Let L be the homogeneous lamina whose constant area density is k slugs/ff, and which is bounded by the curve y : (x), the r axis, and the f lines x : a and x : b. The function / is continuous on the closed interval l!' bl, andf (x) > 0 for all r in f-a,bf . see Fig. 8.8.1.Let A be a partition of the interval la, bl inton subintervals.The ith subinterval is [4_r, 4] and A6x: xi - xi-r The midpoint of [r1-1, xi] is yi.Associatedwith eachsubinterval is a rectangular lamina whose width, altitude, and area densitv are given by Luxft, f (yt) tt, and k slugs/ft2, respectively, and whose centlr of mass is at the point (7,, if 0)). The area of the rectangularlamina is f (yr) L,,xft"; hence, kf (y) Ag slugs is its mass. Consequently,if L,iMo slug-ft is the moment of mass of this rectangularelemeniwith respeciti the y axis, LtMo: yftf(y)

A1x

The sum of the measures of the moments of mass of n such rectangular laminae with respect to the y axis is given by the Riemann sum n

kv,f(y,) aic i:l

If M, slug-ft is the moment of mass of the lamina L with respectto the y axis, we define

Similarly, if AiM, slug-ft is the moment of mass of the ith rectangular lamina with respect to the x axrs,

LoM":if(y)kf(y)

Aux

andthe sumof the measuresof the momentsof massof tt suchrectangular laminaewith respectto the r axis is given by the Riemannsum

> +klf(v,)f'L,x

(8)

Thus, if M* slug-ft is the moment of massof the laminaL with resPect to the r axis, w€ define

' ... . ., ;,.

+*ur+*+il*#*t* i+ffi

r,,

(e)

The mass of the lth rectangular lamina is kf (y) Alx slugs, and so the sum of the measuresof the massesof n rectangular laminae is given by n

)

kf (y) L1x

(10)

the point (x, y) , we detine

which by using formulas (7) , (9) , and (10) gives b

b

k furr*l dx Ja

and

y-

fb

k I f(x) dx Ja

Dividing both the nu merator and denominator by k, we get

*-

rb | *f(*) dx

ttl

rb IJ a f@) dx

(11)

and

lf (x)l'

L

f(x) dx

(12)

360


In formulas (11) and (12) the denominator is the number of square units in the area of the region, and so we have expressed a physical problem in terms of a geometric one. That is, r and ! can be considered as the averageabscissaand the averageordinate, respectively, of a geometric region. In such a case,x and y depend only on the region, not on the mass of the lamina. so we refer to the center of mass of a plane region instead of to the center of mass of a homogeneous lamina. In such a case,we call the center of mass the centroidof the region. Instead of moments of mass, we consider moments of the region. we define the moments of the plane region in the above discussion with respect to the r axis and the y axis by the following:

If. (I, y) is the centroid of the plane region and tf.M, andMu are defined as above,

ExAMPLE2: Find the centroid of the first quadrant region bounded by the curye y' : 4x, the x axis, and the lines N - L and x: 4.

soLUrIoN: Let f (x) : 2x1t2. The equation of the curve is then y : /(x). In Fig. 8.8.2, the region is shown together with the lth rectangular element. The centroid of the rectangle is at (7;, *f @).). The areaA square units of the region is given by

A-

n

lim 2 f0) l l a l l o- i : r

Lic

l,^

f(x) dx

r

0i, f 0))

2xrt2dx

Ti Lix

F i g u r e8 . 8 . 2

We now computeMo and Mr. Mo:

lim f,r,f' (yr) Aix

l l a l l -o , ? : r

xf(x) dx

re;:

8.8 CENTEROF MASS OF A PLANEREGION

361

x(2xrrz1dx

t24 5

-

M*

- li* Zif1,) f(v,) Ln llall-oi:1 : * :t

f4

J, lf(x)f' f4

Jr

dx

4xdx

- xrln lr

:L5 Hence,

+_Ma rw A

_+

Zsg

_93 35

and M-

r, rA: r 2 84

15

:E:-

45

Therefore,the centroid is at the point (89,#).

ExAMPr,n3: Find the centroid of the region bounded bY the curyes y:x2 andy--2x*3.

solurroN: The points of intersection of the two curves are (-1,1) and (3, 9). The region is shown in Fig. 8.8.3, together with the ith rectangular element. Let f (x) : f and g(r) : 2x * 3. The centroid of the ith rectangular elementis at the point (71,ilf 0) + S(yr)l ) where 7; is the midpoint of the ith subinterval l-xr-r, xr]. The measure of the area of the region is given by : tim [s(y,)-f(y,)] Lrx

i

llall-oi:l

:f

f3 J-t

ts(r)-f(x)ldx

f3

- |

J-r

-u

3

lZx+3-xzldx


v : f@lt y : s(*) (3,9)

(v t, g(y ,)

Q,,tlfrl + s(y,)l) 0i, f0))

(-1,1)

Figure 8.8.3

We now compute Mu and M*. n

lim ) y,lSQ) - f(y,)l Lix

Mo:

l l a l l -s , 7 : 1

: f xkk) - f(x)l dx J-r : f x l 2 x + 3- x r l d x :t;'3

M,: lim >+ls4) + f (y)I tsQo)- f (y)l 4x s ,-:1 llallfg

:u

+ (x)]ts(r)-, f (x)l dx J_,[s(r) f

-1 -i

I lZ, + 3 * x'flzx + 3 - xrf dx J-r

-l

-r ,:E l

t/^-\-tr^/

fB

fg J-t

:#

t

l+xr*I2x+9-xnf dx

8 , 8 C E N T E RO F M A S S O F A P L A N ER E G I O N

363

Therefore,

+:#-1, -3 .n

, : 5i # andy : TM: E

_L7

Hence,the centroidis at the point (1, +). If a plane region has an axis of symmetry, the centroid of the region lies on the axis of symmetry. This is now stated and proved as a theorem. 8.8.1 Theorem

f(-v,)) f0t))

Lix

Aix

If the plane region R has the line L as an axis of symmetry, the centroid of R lies on L. pRooF: Choose the coordinate axes so that L is on the y axis and the origin is in the region R. Figure 8.8.4 illustrates an example of this situation. In the figure, R is the region CDE, C is the point (-a,0), E is the point (a,0) , and an equationof the curve CDE is y : f (x). Consider a partition of the interval 10, af .Let y5be the midpoint of the ith subinterval lxo-r, xrl. The moment with resPectto the y axis of the rectangular element having an altitude /(7,) and a width Air is yo[f1) Alr]. Becauseof symmetry, for a similar partition of the interval l-a, 0f there is a corresPondingelement having as its moment with respect to the y axis -ytf 0) A1r.The sum of thesetwo moments is 0; theref.ore,Mo: 0. Because i : Mul A, we concludethat f : 0. Thus, the centroid I of the region R lies on the y axis, which is what was to be proved.

F i g u r e8 . 8 . 4

By applying the preceding theorem, we can simplify the problem of finding the centroid of a plane region that can be divided into regions having axes of symmetry.

ExAMPLE4: Find the centroid of the region bounded bY the semiand the r axis. circle y - \tr4

soLUTIoN: The region is shown in Fig. 8.8.5. Because the y axLSis an axis of symmetry, w€ conclude that the centroid lies on the y axLS;so .r : 0. The moment of the region with respect to the x axis is given by

M*: lim >+l\fy|f'L,x i:1 ;;all-s

- 2 . + f G - x ' )d x JO

F i g u r e8 . 8 . 5

The area of the region .ts 2n square units; so

v-

+8 2rr

3rr

364

A P P L I C A T I O NO SF T H E D E F I N I T EINTEGRAL

There is a useful relation between the force causedby liquid pressure on a plane region and the location of the centroid of the region. As in sec. 8.6, let ABCD be the region bounded by the r axis, the lines x: a artd x:b, and the cur''ey:f(x), where/is continuousand f(x) > 0 on the closed interval la,bl.The region ABCD can be consideredas a vertical plate immersed in a liquid having weight ar pounds per cubic unit of the liquid (see Fig. 8.8.6).If F lb is the force owing to liquid pressureon the vertical plate,

r:,li3

f w{f((1)a1x

llAll-0 i:t

or, equivalently, Figure 8.8.6

xf(x) dx

F_

(13)

If i is the abscissaof the centroid of the region ABCD, then i : Mo lA. BecauseMn: Il xf(x) dx, we have

I:

xf(x) dx A

and so

I:

xf (x) dx - iA

(14)

Substituting from Eq. (14) into (13), *e obtain F- wIA

(ls)

Formula (15) states that the total force owing to liquid pressure against a vertical plane region is the sameas it would be if the region were horizontal at a depth f units below the surface of a liquid. . ILLUsrRArroN1: Consider a trough full of water having asends semicircular regions each with a radius of.2 ft. Using the result of Example4, we find that the centroid of the region is at a depth of Sljn ft. Therefore,using formula (15), we see that if F lb is the force on one end of the trough, _8t6 F:tu.*.2r:;. This agreeswith the result found in Example2 of Sec.8.6.

.

For various simple plane regions, the centroid may be found in a table. when both the areaof the region and the centroid of the region may be obtained directly, formqla (15) is easy to apply and is used in such casesby engineers to find the force caused by liquid pressure.

K 8.8 CENTEROF MASS OF A PLANEREGION

Exercises 8.8 l,2,andSslugsandlocatedatthepoints (-1,3),(2'1), 1. Findthecenterof massof thethreeparticleshavingmassesof -1), (3, respectivelY' and -2) ' 2. Find the center of mass of the four particles having masses of.2, 3,3, and 4 slugs and located at the points (-1, (1, 3), (0, 5), and (2,l) , respectivelY. prove that the centroid of three particles, having equal masses,in a plane lies at the point of intersection of the medians 3. of the triangle having as vertices the points at which the partides are located. In Exercises4 through 11, find the centroid of the region with the indicated boundaries the Y axis. 4. The parabolax:2y - y' aurrd rP and the r axis.

5. The parabolay:4-

6. The parabola A2: 4x, the y axis, and the line y:

4.

and the line Y:4. 8. The lines y : 2x * l, x * Y : 7, and x: 8. 9. The curyes A : f and A : 4x in the first quadrant. 7. The parabola!:

f

10. The curves A: * and'A: x3. -4artd.y:2x-f. 1 1 . T h e c u r v e sy : f 12. prove that the distance from the centroid of a triangle to any side of the triangle is equal to one-third the length of the altitude to that side. 13. If the centroid of the region bounded by the parabola!':4px value of a.

and the line r:

a is to be at the point (p, 0), find the

14. Solve Exercise4 of Sec.8.6 by using formula (15) of this section. 15. Solve Exercise5 of Sec.8.5 by using formula (15) of this section. 16. The face of a dam adjacent to the water is vertical and is in the shapeof an isoscelestrapezoid 90 ft wide at the top, 50 ft wide at the bottom, and 20 ft high. Use forrrula (15) of this section to find the total force due to liquid Pressureon the face of the dam. 17. Find the moment about the lower base of the trapezoid of the force in Exercise L5. L8. Solve Exercise 6 of Sec. 8.5 by using formula (15) of this section. L9. Find the center of mass of the lamina bounded by the parabola 2y' - 18 - 3r and the y axis if the area density at any point (x,y) is \ffi slugs/ftz.

fr(x)

20. Solve ExerciseL9 if the area density at any point (x, y) is r slugs/fP. 2 L . Let R be the regionboundedby the curvesy: fr(r) and y: fr(x) (seeFig. 8.8.7).lf A is the measureof the areaof R and if y is the ordinate of the centroid of R, prove that the measureof the volume,V, of the solid of revolution obtained by revolving R about the r axis is given by V - 2nyA

fr(x)

F i g u r e8 . 8 . 7

Stating this formula in words we have: If a plane regionis revoloeil abouta line in its plane that doesnot cut the region,then the measureof the aolumeof the solid

366

OF THE DEFINITE INTEGRAL APPLICATIONS of rettolutiongenetateilis equalto the productof the measureof the areaof the regionand the measureof the distancetra,eled by the centroidof the region. The above statement is known as the theoremot' Pappusfor volumes of solids of revolution.

22. Us_ethe theorem of Pappus to find the volume of the torus (doughnut-shaped) generatedby revolving a circle with a radius of r units about a line in its plane at a distance of b units from its center, whereb ) t.

23. Use the theorem of Pappus to find the centroid of the region bounded by a semicircle and its diameter. 24. Use the theorem of Pappus to find the volume of a sphere with a radius of r units. 25. LetR be the region bounded by the semicircle V: moment of R with respect to the line y : -4.

t/rt=

land

the x axis. Use the theorem of pappus to find the

26. If R is the region of Exercise25, use the theorem of Pappus to find the volume of the solid of revolution generated by revolving R about the line x - y : r. (nrrrn: use the result of Exercise24 in sec. 5.3.) l'

8.9 CENTER OF MASS OF A SOLID OF REVOLUTION

v:f@) 0i,f0))

xi-tTi)cib Lix

Figure8.9.1

To tind the center of mass of a solid, in general we must make use of multiple integration. This procedure is taken up in chapter 21 as an application of multiple integrals. However, if the shape of the solid is thai of a solid of revolution, and its volume density is constant, we find the center of mass by a method similar to the one used to obtain the center of massof a homogeneous lamina. Following is the procedure for finding the center of mass of a homogeneous solid of revolution, with the assumption that the center of mass is on the axis of revolution. We first set up a three-dimensional coordinate system. The r and.y axesare taken as in two dimensions, and the third axis, the z axis, is taken perpendicular to them at the origin. A point in three dimensions is then given by (x, y , z) . The plane containing the r and y axesis called the ry plane, and the xz plane and the yz plane are defined similarly. suppose that the r axis is the axis of revolution. Then under the assumption that the center of mass lies on the axis of revolution, the y andz coordinates of the center of mass are each zero, and so it is only necessary to find the r coordinate, which we call i. To find r we make use of thb moment of the solid of revolution with respect to the yz plane. Let / be a function that is continuous on the closed interval la , bl , and assumethat/(r) > 0 for all r in la, bf. R is the region bounded by the cufite y : f(x), the r axis, and the lines r : a and x : b;S is the homogeneous solid of revolution whose volume density is k slugs/ft', where k is a constant, and which is generated by revolving the region R about the r axis. Take a partition A of the closed interval la, bf , and denote the ith subintervalby lxn-r,4] (with i: 7,2, . . . ,n ). Let 7i be the midpoint of Lxrr, xtf.Form n rectangleshaving altitudes of f (y) ft and baseswhose width is A;r ft. Refer to Fig. 8.9.1, showing the region R and the ith rectangle.If each of the n rectanglesis revolved about the r axis, n circular disks are generated. The ith rectangle generates a circular disk having a radius of f(y) ft and a thicknessof Lp ft; its volume is rrff(y)fz Aaxff ,

OF MASSOF A SOLIDOF REVOLUTION 8.9 CENTER

and its mass is knlf (y)1, Ap slugs.Figure8.9.2shows the solid of revolution S and the ith circular disk' The center of mass of the circular disk lies on the axis of revolution at the center of the disk (yi, 0, 0). The moment of mass of the disk with respect to the yz plarte is then LMo" slug-ft where LrMo,: Tt(knlf (yr)l' Lrx) The sum of the measuresof the moments of mass of the n circular disks with resPectto the yz plane is given by the Riemann sum

Zy,t

nlf(y)fz L$

(1)

The number of slugs-feet in the moment of massof s with respectto theyz plane, denoted by Mu", is then defined to be the limit of the Riemann sum in (1.)as llAllapproacheszero; so we have

Figure8.9.2

The volume , V ft3, of S was, defined in Sec.8.2 b v fb

v - lim t nlf (yi)12A i r - r T I t f @ ) f ' d x Ja Fr llall-o

The mass,M slugs, of solid S is defined by

We define the center of mass of S as the point (r, 0, 0) such that

+ffi

+i=_qi#+++$ffffi Substituting from Eqs. (2) and (4) into (5), we get

xlf(x)l' dx lf(x)l' dx

368


From Eq. (6), we see that the center of mass of a homogeneous solid of revolution depends only on the shape of the solid, not on-its substance. Therefore, as for a homogeneouslamina, we refer to the center of mass as the centroid of the solid of revolution. when we have a homogeneous solid of revolution, instead of the moment of mass we consider the moment of the solid. The moment, Maz, of the solid S with respectto theyz plane is given by (7)

Thus, if (I, 0, 0) is the centroid of S, from Eqs. (3), (d) , and,(7) we have

ffi ExAMPLE1: Find the centroid of the solid of revolution generated by revolving about the r axis the region bounded by the curve y : x2, the x axis,and the line x=3.

SOLUTION:

The region and a rectangular element are shown in Fig. 8.9.3.

The solid of revolution and an element of volume are shown in Fig . 8.g.4.

f(x) :

x2

Mor:

lim 2 y,nlf (y,)f' L,x

n

Ilall-o f-:r

xlf(x)f' dx

v=f@)

x5 dx

_ryn f(v))

v -,,Til, rilf(vt)f'Nx } f3

I tf?)f'dx Jo -_1r[t*nA* Jo Figure8.9.3

:#n Therefore,

, =T n - 5 x - TM : 4, ,q_ 2 Therefore, the centroid is at the point (8, O, 0 ) .

F i g u r e8 . 9 . 4

8.9 CENTEROF MASS OF A SOLID OF REVOLUTION

0i, f0))

Figure8.9.6

Figure8.9.5

The cenkoid of a solid of revolution also can be found by the cylindrical-shell method. Let R be the region_bou_nded by the curve > 0 on la, bl, the r axis, and the i : f @),where / is continuous and /(r) of revolution generated by solid the be Let s and, x:b. r:a iirr", of S is then at the point centroid y axis. The the about R revolving (0, , , O).lf the rectangular elements are taken parallel to the y axis, then the Llement of volume is a cylindrical shell. Let the ith rectangle have_a width of L1x- xi - xi_1,u.'.dlit 7abe the midpoint of the interval lxi-r, xtf. The centroid of the rylindrical shell obtained by revolving this rectangle about the y axis is at the center of the cylindrical shell, which is the point (0,tf Q), 0). Figure 8.9.5shows the region R and a rectangular element of area,and Fig.8.9.5 shows the cylindrical shell. The moment, Mrz, of s with respect to the xz Plarreis given bY

lf V cubic units is the volume of S,

v-

lim f ,nr,f(y,) \x

Ilall- o t--r

xf(x) dx

i::;

APPLICATIONS OF THE DEFINITEINTEGRAL I

EXAMPLE 2: Use the cylindricalshell method to find the centroid of the solid of revolution generated by revolving the region in Example L about the y axrs.

solurroN: Figure 8.9.7 shows th e region and a rectangular element of area. The solid of revolution and a cylindrical-shell element of volume are shown in Fig. 8.9.8.

M,,: ,lip 2 +ffvt)2nyi(yt)A$ llAll-0i-t f3

v:f@) (3,9)

=rr I xlf (x)f' dx JO -7Tft"d* JO

=Tn V - lim f, rorrf (y,) Lix llall-oFi r3 ,2n I xf(x) dx

0 i, f0 i))

JO

:2tr ft *, d* JO

:#n Q,,tftv,l)

Therefore, 2!3n

M y_f:#_3 aix

Figure8.9.7

Hence, the centroid is at the point

( 0 ,3 , 0 ) .

Figure8.9.8

8.9 CENTEROF MASS OF A SOLID OF REVOLUTION 371

ExAMPLE3: Solve Example L by the cylindrical-shell method.

Figure 8.9.9 shows the region and a rectangular element of solurroN: area, and Fig. 8.9.10 shows the cylindrical-shell element of volume obtained by reiolving the rectangle about the r axis. The centroid of the

,7.'n),0,0)' The cylindricalshellis it its center,i"hi.h is the point (+(3 + (f, 0,0)' is at centroidof the solid of revolution

v (3,9)

M,,

- 6l -- lim i +(3+ 6,1 2nyo(3

Lil

l l a l l *o 7 : t

fs : n l" y(3+ frl(3 - {vl dy JO fs

:n

| (gv-yz) dy

JO

(tE , v,)

(*tt + tfr) ,y)

Ai

v

:#n

AiA

(3, Yi )

Finding V by the cylindrical-shell method, we have

v-

A i-t

lim $ 2nyo(3- ,6)

Lra

l l a l l -o ? t fs

2n I V(3 - y't') dY JO

#n F i g u r e8 . 9 . 9

_

M.,"

*v2 t:-:-

x

5

and the centroid is at (8,0,0). The result agreeswith that of Examplet.

t

8.9 Exercises revolving the given region about the inIn Exercises1 through 16, findthe centroid of the solid of revolution generatedby dicated line. elements perpendicular to the 1. The region bounded by y : 4x f and the x axis, about the y axis. Take the rectangular axis of revolution. 2. Same as Exercise 1, but take the rectangular elements parallel to the axis of revolution.

972


3. The region bounded by x * 2y :2, the x axis, and the y axis, about the r axis. Take the rectangular elementsperpendicular to the axis of revolution.

Same as Exercise3, but take the rectangular elements parallel to the axis of revolution. The region bounded by y' : r3 and x - 4, about the ff axis. Take the rectangular elements perpendicular to the axis of revolution.

Same as Exercise5, but take the rectangularelementsparallel to the axis of revolution. The region bounded by y : x3,x - z, and the r axis, about the lin e x: 2. Take the rectangular elements perpendicular to the axis of revolution. 8. Same as Exercise7, but take the rectangular elements parallel to the axis of revolution. 9. The region bounded by xny: l, y :1, and y :4, about the y axis. 10. The region in Exercise9, about the r axis. 11. The region bounded by the lines y : x, ! : 2x, and.x * y : 6, aboutthe y axis. 12' The region bounded by the portion of the circle r3 -t y' :4 in the first quadrant, the portion of the line 2r - y : 4 in the fourth quadrant, and the y axis, about the y axis. L3. The region bounded by y: rP and !: x + 2, aboutthe line y:4. 1.4.The region bounded by y": 4x and :16 - 4x, about the r axis. !2 L5. The region bounded by y : \/W , the x axis, and the line r: p, aboul the line r: p. 15. The region of Exercise 15, about the line y:2p. L7. Find the centroid of the right-circular cone of altitude ft units and radius r units. L8. Find the center of mass of the solid of revolution of Exercise3 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from tn" y, pi*". 19. Find the center of mass of the solid of revolution of Exercise5 if the measure of the volume density of the solid at any point is equal to a constant k times the measure of the distance of the point from the yz p?ane. 20. that a cylindrical hole with a radius of r units is bored through a solid wooden hemisphere luppose of radius 2r units, so that the axis of the cylinder is the same as the axis of ttre hemifrhere. Find the centroii of the solid remaining.

8.L0 LENGTH oF ARc OF A PLANE CURVE

Let the function f be continuous on the closed interval la, bl. Consider the graph of the equati on y : f (x) of which a sketch is shown in Fig. 9 . 1 0L. .

b

: f(x)

F i g u r e8 . 1 0 . 1

8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E

373

The portion of the curve from the point A(a, f (a)) to the point B(b, f(b)) is called an arc. We wish to assigna number to what we intuitively think of as the length of such an arc. If the arc is a line segmentfrom the point (xr, yr) to the point (rr, Ur), we know from the formula for the distance between two points that its length is given We use this formula for defining the length of av ffi. an arc in general. Recallfrom geometry that the circumference of a circle is defined as the limit of the perimeters of regular polygons inscribed in the circle. For other curves we proceed in a similar way. Let A be a partition of the closed interval la, b) fotmed by dividing the interval into r subintervals by choosing any (n - 1) intermediate numbers between a and b. Let X0: a, and xn:b, and let xt , Xz, Xs, 1 xn-r I xr. Xn-r be the intermediate numbers so that Xo I Xr I xz I

Then the fth subinterval is [ri-1, x1]; and its length, denoted by Air, is . . . ,2. Thenif llAllisthenormofthepartition x1- xi-lrwherei:!,2,3, = A, each Ap llAll. u

8.10.2 Figure Associatedwith eachpoint (4, 0) on the r axis is a point Prktu f k)) on the curve. Draw a line segmentfrom eachpoint Pi-l to the next point P, as shown in Fig.8.10.2 The length of the line segmentfrom Pi-l to Pi is denoted UV lFr-Fr I and is given by the distance formula lPt-t.tr,l:

xr-r)' +

(1)

A*)'

The sum of the lengths of the line segments is

l p a l + l p - l p+r l p r 4 l+ . . . + l n , - p , l + " '

+ lPr-rPnl

which can be written in sigma notation as

lP,-rPil

(2)

i:l

It seems plausible that if n is sufficiently large, the sum in (2) will be "close to" what we would intuitively think of as the length of the arc AB. So we define the length of the arc as the limit of the sum in (2) as the norrn

374

APPLICATIONS OF THEDEFINITE INTEGRAL

of A approacheszero, in which casen increaseswithout bound. we have, then, the following definition. 8.L0.L Definition

If the function / is continuous on the closed interval la, bl and,if there exists a number L having the following property: for any e ) 0 there is a 6>0suchthat l-r. _ | l ) l l P , _ , P -, | L l < e

l - " ' ' I

li:l

I

for every partition A of the interual la, bf forwhich llAll< 6, then we write n

t: ,lip > lF;El llAll-0 i:r

(3)

and L is called the length of the arc of fhe curve y : (x) from the point f A(a, f(a)) to the point B(b, f(U11. If the limit in Eq. (3) exists,the arc is said tobe rectifiable. we derive a formula for finding the length L of an arc that is rectifiable. The derivation requires that the derivative of be continuous on f la,bf; such a function is said to be "smooth,, on lo,bl. 8.1.0.2Definition

A function / is said tobe smootfton an interval I if f , is continuous on I. (xi,yi)

Ui-t

:

LiU

(xi-r,Ai-t)

Figure8.10.3

Refer now to Figure 8.10.3. If. pi_r has coordinates (4_1, yi_1) and.p, has coordinates (xv y), then the length of the chord n*rr, is given by formula (1). Lettin g h - xrr

:

A# and Ui - Ai_r -

lPr-rPrl:

AtA, we have

(4)

or/ equivalently,becauseAg # 0, lPr-tP,l:

(Air)

(s)

CURVE 375 OFARCOFA PLANE 8.10LENGTH Becausewe required that f ' be continuous on [4, b], the hypothesis of the mean-value theorem (4.7.2) is satisfied by f , and so there is a number z1in the open intervd (xr-r, r1) such that ' (6) f (xt) f (xt-') : f (z) (x,- xr-r) BecauseL$: f(x) * f (xr-t) and A6r- x1 ri-1, from Eq. (6) we have

+r!_:f, (zr) Lrx

(7)

Substituting from Eq. (7) into (5), we get (8)

lml--ffiLix where x*r 1 zi 1 xi. For each i from L to n there is an equation of the form (8). nn

(e) i :l

i:l

Taking the limit on both sides of Eq. (9) as llAllapproadres zeto,we obtain n-n-

lim ). lP;F;l : lim ) Vt * lf'(z)12 LP llAll-0 i:l

(10)

llAll-0 t=l

if this limit exists. To show that the limit on the right side of Eq. (10) exists, let g be the function defined by

s@): \ET(@F Because/' is continuous on [4, bl, g is continuous on la, bf. Therefore, . ,n, b e c a u sxe* r 1 ? , 1 l x y f o t i : L , 2 , n-n

\ s@)t* li- ) \trTTfGJP Ap: ,lim llAll-0 l=1

llAll-0 i=l

or, equivalently, (11) ll

B e c a u sge( x ) : f f i ,

f r o mE q . ( 1 1 )w e h a v e

rimtffiLrx:f'rydx \ o'J 116-11*of:r

(12) Jo

Substituting from Eq. (L2) into (10),we get

rim>Finl : ['ry Jo

'=r llall-o

dx

(13)

376


Then from Eqs. (3) and (13) we obtain rb

L-l ffidx Ja In this way, we have proved the followirg 8.L0.3 Theorem

(14) theorem.

If the functi on f and its derivativ e ' are continuous on the closed interval f fa, bf , then the length of arc of the curve y: f (x) from the point (a, (a)) f to the point (b, f (b)) is given by

we also have the following theorem, which gives the length of the arc of a curve when r is expressedas a function of.y. g.10.4 Theorem

If the function F and its derivative F' are continuous on the closed interval lr'lf ,thenthelglgat of arcof thecurvex:F(y) fromthepoint (F(c),c) to the point (F(d) , d) is given by

The proof of rheorem 8.10.4is identical with that of rheorem g.10.3; here we interchange x and y as well as the functions f and F, The definite integral obtained when applying Theorem g.10.3 or Theorem 8.10.4is often difficult to evaluate.Becauseour techniques of integration have so far been limited to integration of powers, we are further restricted in finding equations of curves for which we can evaluate the resulting definite integrals to find the length of an arc. EXAMPLE1.:Find the length of the arc of the curye y : x2t3from the p o i n t ( 1 , 1 ) t o (8, 4) by using Theorem 8.10.3.

'(x): solurroN: See Fig.8.10.4. Because f ( x ) : ) c 2 t 3f, Theorem 8.L0.3we have

&X-rtt. From

v v: To evalu ate this definite integral, let u - gxztT* 4; then du:6x-1t| W h e n x : l , u - L 3 ; when x - 8, Lt- 40. Therefore,

F i g u r e8 , 1 0 . 4

uLtz du

dx.

8 . 1 0 L E N G T HO F A R C O F A P L A N EC U R V E

ExAMPrn 2: Find the length of the arc in Example L by using Theorem 8.L0.4.

soLUTroN: Because y - x2t3and r Lettin g F (y) - At'', we have

377

0, we solve for x and obtain x - y3t2

F ' ( Y )- E Y u z Then, from Theorem 8.'1.0.4,we have fq

L- | Vr+*y dy JT

-rftMdv -,

v

J,

-- l f

14

rg | 3 (a + 9y)3/2|

lr

L

- ;7 (40tr' - 133t2)

: 7.6

is continuous on fa,bl, then the definite integral If f dt is a function of r; and it gives the length of the arc tf \trTV@P of the cuwey:f(x) fromthe point(a,f(a)) to thepoint (x,f(x)),where r is any number in the closed interval la, bf .Let s denote the length of this arc; thus, s is a function of x, and we have ar

s(x):f

Vl+tf'G)12dt

Ja

we have From Theorem7.6.'1.,

s'(r): ttt+TfTdP or, becauses'1r;: dsldxandf'(x): dyldx,

Multiplying by dx, we obtain dsSimilarly, if we are talking about the length of arc of the curve r:80)

(15)

378


f r o m ( S ( c ), c ) t o ( S Q ) , y ) , w e h a v e

ds-

dy

(16)

Squaring on both sides of Eq. (15) gives

ffiffi

ffilF $#rl

(r7)

From Eq. (r7) we get the geometric interpretation of ds, which is shown in Fig. 8.10.5.

Q@ * Ar, V * tV)

Figure8.10.5

In Fig. 8.10.5, line T is tangent to the cur.rtey: f (x) at the

pointP. lFMl= Ax: dx; lMSl: Ay;lMRl: dV;lpRl:}r; ih" l"ngthof a r cP Q : 4 r .

Exercises 8.10 1. Find the length of the arc of the curve 9yz : 4rs from the origin to the point (g, 2\/t) . 2. Find the length of the arc of the curve *: (2V + 3)s from (1, -t) ro (Z\/V, D. 3. Find the length of the arc of the curve 8y : * * 2x-2 fuom the point where r: 1 to the point where r: 2. 4. Use Theorem 8.10.3to find the length of the arc of the curve ys:gf from the point (1,2) to the point (27,7g). 5. Solve Exercise4 by using Theorem 8.10.4. $ Find the entire length of the arc of the curve x2ts+ y2t3: 1 from the point where x : * to the point where r: 1. E Find the length of the arc of the curve y: *(* *21srz from the point where r:0 to the pointwhere r:3. 8. Find the length of the curve 6ry: yo * 3 from the point where y: I to the point where y: !.

BEVIEWEXERCISES 379 1 in the first quadrant from r : [a to x: a. 10. Find the length of the curve 9yz : 4(t * rP)a in the first quadrant from x : 0 to r :2\/2. 11. Find the length of the curve 9y' : x(x - 3)' in the first quadrant from x: L to x:3. 12. Find the length of the curve 9y': x'(2x * 3) in the secondquadrant from r: -1 to r: 0. 9. Find the length of the cuwe (xla)2t} t (ylb)'t":

(Chapter8) ReoiewExercises 1. Find the area of the region bounded by the loop of the curve y2: *(4-

x).

2. The region in the first quadrant bounded by the curyes r : y2 and,x : ya is revolved about'the y axis Find the volume of the solid generated. 3. The base of a solid is the region bounded by the parabola !2:8x and the line r: every plane section perpendicular to the axis of the base is a square.

8. Find the volume of the solid if

4. A container has the sameshape and dimensions as a solid of revolution formed by revolving about the y axis the region in the first quadrant bounded by the parabola *: 4py, the y axis, and the line y : p. If the container is full of water, find the work done in pumping all the water up to a point 3p ft above the top of the container. 5. The sur{aceof a tank is the sameas that of a paraboloid of revolution which is obtained by revolving the parabola y : f about the r axis. The vertex of the parabola is at the bottom of the tank and the tank is 35 ft high. If the tank is filled with water to a depth of.20 ft, find the work done in pumping all of the water out over the top. 6. A plate in the shape of a region bounded by the parabola * : 6y and the line 2y: 3 is Placed in a water tank with its vertex downward and the line in the surfaceof the water. Find the force due to liquid Pressureon one side of the plate' 7. Find the area of the region bounded by the curves y: lx- 11,V: f - 2x, the y axis, and the line r: 2. 8. Find the area of the region bounded by the curves y: lxl + lr 1l and y: x + t, : lx - 21, the r axis, and the 9. Find the volume of the solid generated by revolving the region bounded by the curve y lines l: I and r: 4 about the r axis. 10. Find the length of arc of the curve ayz: f from the origin to (4a,8a). 11. Find the length of the curve 6y': x(x - 2)2 fuom (2, 0) to (8' A\tr) ' (2,2),(2'4)'and(-1,2). 12. Findthecenterofmassofthefourparticleshavingequalmasseslocatedatthepoints(3,0), -l) ' and (5, 2). Find 13. Tfuee particles having masses5, 2, andS slugs are located, respectively, at the points (-1, 3), (2, mass. the center of 14. The length of a rod is 8 in. and the linear density of the rod at a point r in. from the left end is 216 + 1 slugs/in. Find the total mass of the rod and the center of mass. 15. Find the centroid of the region bounded by the parabola !2: x and the line y : x - 2. 16. Find the centroid of the region bounded by the loop of the curve !2: * - x3. 12. The length of a rod is 4 ft and the linear density of the rod at a point r ft from the left end is (3r * 1) slugs/ft. Find the total mass of the rod and the center of mass. 18. Give an example to show that the centroid of a plane region is not necessarily a point within the region. 19. Find the volume of the solid generatedby revolving about the lin e y : -'I..the region bounded by the line 2y : r * 3 and outside the curves!2 I x:0 and.y2- 4x:0. 20. Use integration to find the volume of a segment of a sphere if the sphere has a radius of r units and the altitude of the segment is h units.

380


2 L . A church steepleis 30 ft high and every horizontal plane section is a square having sides of length one-tenth of the distance of the plane section from the top of the steeple. Find the volume of the steeple. 22. A trough full of water is 5 ft long, and its cross section is in the shape of a semicircle with a diameter oI 2It at the top. How much work is required to pump the water out over the top? 23. A water tank is in the shapeof a hemisphere surmounted by a right-circular cylinder. The radius of both the hemisphere and the cylinder is 4 ft and the altitude of the cylinder is 8 ft. If the tank is full of water, find the work necessaryto empty the tank by pumping it through an outlet at the top of the tank. 24. A force of 500lb is required to compress a spring whose natural length is 10 in. to a length of 9 in. Find the work done to compress the spring to a length of 8 in. 25. A cable is 20 ft long and weighs 2lblft, and is hanging vertically from the top of a pole. Find the work done in raising the entire cable to the top of the pole. 26. The work necessaryto stretch a spring from 9 in. to L0 in. is I times the work necessaryto stretch it from 8 in. to 9 in. What is the natural length of the spring? 27. Find the length of the curve 9x2tt+ Ayzrz- 36 in the second quadrant from r: -1 to r: -*. 28. A semicircular plate with a radius of 3 ft is submerged vertically in a tank of water, with its diameter lying in the surface. Use formula (15) of Sec.8.8 to find the force due to liquid pressure on one side of the plate. 29. A cylindrical tank is half full of gasoline weighing 42lblfts.If the axis is horizontal and the diameter is 6 ft, find the force on an end due to liquid pressure. 30. Find the centroid of the region bounded above by the parabola4*:36-9y

andbelow by the r axis. 3 1 . Use the theorem of Pappus to find the volume of a right-circular cone with a base radius of r units and an altitude of h units.

32. Theregionboundedbythe parabolal: of the solid of revolution formed.

4y,thex axis,andthe liner:4

is revolvedaboutthey axis.Findthe centroid

33. Find the center of mass of the region of Exercise30 if the measure of the area density at any point (x, y) is l[ - V 34. Find the center of mass of the solid of revolution of Exercise32 if the measureof the volume density of the solid at any point is equal to the measure of the distance of the point from the rz plane.

Logarithmlc and rl functions

FUNCTIONS AND EXPONENTIAL LOGARITHMIC

9.I THE NATURAL LOGARITHMIC FUNCTION

The definition of the logarithmic function that you encountered in algebra is based on exponents. The laws of logarithms are then proved from corresponding laws of exponents. One such law of exponents is Ac . Aa:

gc*u

(1)

If the exponents, r and ! t altl positive integers and a is any real number, (1) follows from the definition of a positive integer exponent and mathematical induction. If the exponents are allowed to be any integers, either positive, negative, ot zeto, and,a # 0, then (1) will hold if a zero exponent and a negative integer exponent are defined by ao: I and

n>0 If the exponents are rational numbers and a > 0, then Eq. (1) holds when a^tn is defined bv antn: (%)* It is not quite so simple to define a'when r is an ilrational number. For example, what is meant by 4fi? The definition of the logarithmic function, as given in elementary algebra, is based on the assumption that aa exists if a is any positive number and x is any real number. This definition states that the equation 4t:

N

where a is any positive number except 1 and N is any positive number, can be solved for r, and r is uniquely determined by x: loio N In elementary algebra,logarithms are used mainly as an aid to computation, and for such purposes the number a (called the base)is taken as 10. The following laws of logarithms are proved from the laws of exponents: Law 1 logoMN:logo M * logoN .M : loga M- log" N Law 2 tog, f Law 3 logo1:0 Law 4 logoM:nlogoM Law 5 logoa:l In this chapter we define the logarithmic function by using calculus and prove the laws of logarithms by means of this definition. Then the exponential function is defined in terms of the logarithmic function. This definition enables us to define a' when r is any real number and a * 0.

LOGARITHMIC 9.1THENATURAL FUNCTION383 The laws of exponents will then be proved if the exponent is any real number. Let us recall the formula |

+n+l

*C I t"dt:*= n+ L

n#-t

J

Consider the function This formula does not hold when n:-1. defined by the equation A: t-r, where f is positive. A sketch of the graph of this equation is shown in Fig. 9.1.1. LetR be the region bounded above by the curve y : Uf, below by the f axis, on the left by the line t : l, and on the right by the line f : r, where r is greater than 1. This region R is shown in Fig. 9.1.1.The measureof the areaof R is a function of r; call it A(x) and define it as a definite integral by

F i g u r e9 . 1 . 1

(2)

A(x)

Now consider the integral in (2) if 0 < r ( 1. From Definition 7.3.4, we have

f +dt:-t:|0, Then the integral I: Glt) df represents the measure of the area of the region bounded above by the curve y: Ut, below by the t axis, on the left by the line t:x, and on the right by the line f :1. So the integral li $lt) df is then the negative of the measure of the area of the region shown in Fig.9.1.2. rf.x:L, the integral li Glt) df becomesIl Olt) dt, which equals 0 by Definition7.3.5.In this casethe left and right boundariesof the region are the same and the measure of the area is 0. Thus, we see that the integral t{ Glt) dt for x } 0 can be interpreted s in terms of the measure of the area of a region. Its value depends on r and is therefore a function of r. This integral is used to define the "nafural logarithmic function."

Figure 9.1.2

9.1.1 Definition

The natural logarithmicfunction is the function defined by r>0 The domain of the natural logarithmic function is the set of all tive numbers. We read ln x as "the natural logarithm of x."

The natural logarithmic function is differentiable becauseby applying Theorem 7.5.1we have D"(ln r)

384

FUNCTIONS ANDEXPONENTIAL LOGARITHMIC Therefore,

D"(ln i:1

(3)

From formula (3) and the chain rule,7f.u is a differentiable function of x, and u(x) > 0, then

*#*g##;r{f*'-, .,

(4)

ffi'D.ua EXAMPLE 1:

Given

y-ln(3r2-5x+8)

fu__ - r dx

tind dyldx. ExAMPw 2:

solurroN: Applying formula (4), we get

Given

y : l n [ ( 4 x '+ 3 ) ( 2 x- 1 ) ]

SOLUTION:

D"U

find D *U.

3x2 6x + 8

'(6x-6)-

6x-5

3x2- 6x + 8

Applying formula (4) , we get 7

(4x' + 3) (2x- 1)

[8r (2x - 1) + 2(4xz+ 3) ]

24x2-8x+6 (4x'+ 3) (2x- 1) SOLUTION:

dy

From formula (4) , we have 1

(x+1) -)c

It should be emphasizedthat when using formula (4), u(x) must be positive; that is, a number in the domain of the derivative must be in the domain of the given natural logarithmic function. o rLLUsrRArroxL: In ExampleL, the domain of the given natural logarithmic function is the set of all real numbers because3x2- 6x * 8 > 0 for all r. This can be seen from the fact that the parabola having equation - 6x * 8 has its vertex at (1,5) and opensupward. Hence,(6x - 6)l A :3f (3x' 6r * 8) is the derivative for all values of x. In Example 2, because (4x2t 3)(2x- 1) > 0 only when x > i, the domain of the given natural logarithmic function is the interval (*, +-).

FUNCTION 9.1 THE NATURALLOGARITHMIC

Therefore, it is understood that fraction (5) x>i. Becausexl(x * 1) > 0 when either x 1-1 the natural logarithmic function in Example and so 1.lx(x-t 1) is the derivative if either x <

is the derivative only if or r > 0, the domain of 3 is (-oo, -1) U (0, **), -1 or r ) 0. o

We show that the natural logarithmic function obeys the laws of logarithms as stated earlier. However, first we need a preliminary theorem, which we state and prove. positive numbers, then

lf a

9.1.2 Theorem

1

idt PRooF: In the integral ["u (Llt) dt, make the substitution t: aur then u : 1, and when t - ab, u - b. Therefore, we dt-adu. When t:a, have

f'lo':I:# @

du)

: ['Lau u Jr

-1,+"

,

I

F i g u r e9 . 1. 3

o TLLUsTRATIoN2: In terms of geometrY, Theorem 9."1..2states that the measure of the area of the region shown in Fig. 9.1.3 is the same as the measure of the area of the region shown in Fig.9."1..4. o If we take x -'/-. in Definition 9.1.1.,we have

l n 1 : [ ' I ta , Jr

The right side of the above is zero by Definition t

7.3.5. So we have (6)

lnL-0

Equation (5) corresponds to Law 3 of logarithms, given earlier. The following three theorems give some properties of the natural logarithmic function.

F i g u r e9 . 1. 4

9.L.3 Theorem

If a and b arc any Positive numbers, then ln (ab):ln

a+ lnb

386

FUNCTIONS LOGARITHMICANDEXPONENTIAL PRooF:From Definition 9.1.1,

rn(ab):f"oIL at Jr

which, from Theorem7.4.6, gives

tn(ab): I:+ at+l"' I at Applying Theorem 9.1.2 to the second integral on the right side of the above, we obtain

tn(ab):f+at+lolat t t Jr

Jt

and so from Definition 9.1.1 we have ln(ab\ :ltt 9.1.4 Theorem

n * ln b

I

lL a and b are any positive numbers, then lnl:ln

a -lnb

pRooF: Becauseo:

(alb) . b, we have

ln4:hf+.b) \al

Applying Theorem 9.1.3 to the right side of the above equation, we get

l na : t n l + n a ubtractingln b on both sidesof the aboveequation,w e obtain ln4:ha-lnb b

9.1.5 Theorem

rf. n is any positive number and r is any rational number, then lna':rlna pRooF:

From formula (4), if. r is any rational number and r ) 0, we have 1

D"(lnr):*.rrr-r-!* and D,(rlnr):!*

9.1 THE NATURALLOGARITHMIC FUNCTION

Therefore, D'(ln x') : D,(r ln x) From the above equation the derivatives of ln x'and rln x are equal, and so it follows from Theorem 5.3.3 that there is a constant K such that lnx":rlnx*K

forallr>O

(7\

To determine K, substitute 1 for r in Eq. (7) and we have ln 1": r ln 1* K But, by Eq. (5), ln 1 : 0; hence,K : 0. ReplacingK by0 in Eq. (7),we get lrtx':rlnx

forallr>0

from which it follows that if x:

fl, where a is any positive number, then

lna':rlna

I

Note that Theorems 9.1.3,9.1.4,and 9.1.5are the same properties as the laws of logarithms 1.,2, and 4, respectively, given earlier. . rLLUsrRArroN3: In Example 2, if Theorem 9.1.3is applied before finding the derivative, we have y : ln(Axz + 3) * ln (2x - 1)

(8)

Now the domain of the function defined by the above equation is the interval (4, aoo), which is the same as the domain of the given function. From Eq. (8) we get L'tU :

8x2 Ef+ gT E=T

and combining the fractions gives 8x(2x- 1,)* 2(4x2* 3) un t A_.:_@ which is the same as the first line of the solution of Example 2.

.

o rLLUSrnerroN 4: If we apply Theorem 9.1,.4before finding the derivative in Example 3, we have

lnY:ln

x-

ln(r+1)

(e)

ln r is defined only when x ) 0, and ln(x * 1) is defined only when x > -1. Therefore, the domain of the function defined by Eq. (9) is the interval (0, 1o). But the domain of the function given in Example 3 consists of the two intervals (--, -1) and (0, +-;. Finding D*y from Eq. (9), we have

x+"1,

388

FUNCTIONS LOGARITHMICAND EXPONENTIAL

but remember here that r must be greater than 0, whereas in the solution of Example 3 values of r less than -1 are also included. o Illustration 4 shows the care that must be taken when applying Theorems 9.1..3,9.L.4,and 9.1.5to natural logarithmic functions. ExAMPLE4: y:ln(2x find D *A.

Given - l)t

solurroN:

From Theorem 9.1.5we have y :ln(2x - l)t: 3 ln(2x - 1)

Note that ln(2x - 1)3 and 3ln(2x - 1) both have the samedomain, x ) E. Applying formula (4) gives DrA:3 .

1

,rr1

. '_ r-

6

2x_ |

In the di scutssron ror ) n tth ] Iflt tr follo,1 find this by.Utsin S rg fo: rla ((4), )ffn nu fon (l l' D,(ln lrll))- ' D)*"'(ln 1

ve need to make use of D,(ln lrl). $o is substitutedfor l*l,and so we ha{re

tfr,2)

w* r ' D " (({x' , ) .f

1 {x' x x2

Hence, D"(ln lxl):,

1

(10)

From formula (10) and the chain rule, if u is a differentiable function of x, we get (11) The followirg example logarithmic function, given plify the work involved in volving products, quotients, soLUrIoN:

illustrates how the properties oT the natural in Theorems 9.'1,.3,9."1,.4, and 9.'1..5,can simdifferentiating complicated expressions inand powers.

From the given equation

' f?E' = l:, lf'FTl lvl:lr(r*D\mllr+|[ml

FUNCTION 9.1 THE NATURALLOGARITHMIC

Taking the natural logarithm and applying the properties of logarithms, we obtain

l r ,l y l: * l n l r + 1 l- l n l r * 2 l - + l n l r + 3 1 Differentiating on both sides implicitly with respect to r and applying formula (11), we get 1, ^

I-'rY :

I

11

1

2 / -1 rr + 1 \- T Ix 5 - +2-m

Multiplying on both sides by Ur we obtain DrA: y

2 ( x+ 2 ) ( r + 3 ) - 6 ( r * 1 ) ( r + 3 ) - 3 ( r * 1 ) ( x+ 2 )

5(x+1)(r+2)(r+3)

Replacing y by its given value, we obtain D rA equal to Z x z* 1 0 r + 1 2 - 6 x z - 2 4 x - 1 8

-3x2-9x-6

6(x+1)(r+2)(r+3) -7x2*23x-12 Dra:

6 ( x + 1 ) 2 t 3 ( *x 2 ) ' ( r * 3 ) s r z

(r2) we have, for n any real number,

(13)

Evaluate

SOLUTION:

f

xzdx

I f 3x2dx

Jm:3Jffi:3'^ EXAMprn 7: Evaluate

I:#'.

1r

+C lr3+11

sor,urroN: Because(x' * 2) l(x + 1) is an improper fraction, we divide the numerator by the denominator and obtain x2*2

4,

'+1I-x-I+I+1

3

390

LOGARITHMICAND EXPONENTIAL FUNCTIONS

Therefore,

Q-1.#) dx

I#dx:l:

l:

:txr-x*3lnlr+11 -2-2+3ln3-3ln -3In3-3

L

0

-3In3

The answer in the above example also can be written asln27 because, by Theorem 9.'1,.5,3In 3 : ln 33. ExAMPr,r 8: Evaluate

SOLUTION:

f hr , | -ax Jx

LCt LI :

then du : dxlx; so we have

ln

I+dx-1"d t-t I " r +

C: + ( h r )z+ C

Exercises 9.1 In Exercises1 through 10, differentiate the given function and simplify the result. L f (x): ln(l + 4x') 2 .f ( x ) : l n \ M \r '\$; \E3. f(x): ln 4. S(r) : ln(ln r) 5. h(x) : ln (x2ln x) 7.f(x):ln

6.f(x):ln|/# 8. f (x): f4nT --xln(r* \ffi)U.G(r)

lr3+11

O F ( r )- \ E T

-h(1 + lm)

In Exercises11 through 16, findD"y by logarithmic differentiation. xlffi

4A

LZ.y:ffi

ffi 14.1t: \' (x * l)zrs L6. y - (5x - 4) (x, + 3) (3rt - 5) In Exercises L7 and L8, fin d D ry by implicit differentiation.

(b. m!*xy-l

W, lnxy*x*y-z In Exercises 19 through

x

26, evaluatethe indefinite

integral.

\R

9.2 THEGRAPHOF THENATURALLOGARITHMICFUNCTION

21Jt+x L n x

fdx

J s-zx

\q'

ntlffi

5 - nt z,\A + ' f g + 2 * a* J

In Exercises27 through 30, evaluatethe definite integral. 2l dx 27. [5 Jt i2-5

29. 31.P r o v e t h a t l n r n : n ' l n x b y f i r s t s h o w i n g t h a t l n t n a n d n ' l n x d i f f e r b y a c o n s t a n t a n d t h e n f i n d t h e c o n s t a n t b y taking r:

1. ln r ) 0 and 1- ln r-

32. Prove that x-7-

Ux < 0 for all r > 0 and x * l,thus establishingthe inequality

r-1.hr(x-l x 1. (nrnr: Letf(x):x-1-lnrandg(x):t-lnx-llxanddeterminethesigns forallr>0andr# g'(r) on the intervals (0, 1) and (1, +o;.;

off'(x) and

33. Use the result of Exercise32 to prove lt* t-o

h(l*

r) : t

1C

which 34. Establish the limit of Exercise33 by using the definition of the derivative to find F'(0) for the function F for F(r) : ln(l + r). rt rc (Jlt) dt by using Theorem 7.4.8.) (ln r)/r : 0. (mNr: First prove ttrat ol \/i) o, = 35. prove J. J. "lif_ J 1

J l

g.Z TlflIE GRAPH OF THE NATURAL LOGARITHMIC FUNCTION

To draw a sketch of the graph of the natural logarithmic function, we must first consider some Properties of this function' Let / be the function defined by

f(r):lnt:

J[ r' It o ,

r>o

The domain of / is the set of all positive numbers. The function / is differentiable for all r in its domain, and

f'(x):+

(1)

Because/ is differentiable for all x > 0, / is continuous for all r > 0. From (1) weconclude thatf'(r) > 0 forall x) 0, andtherefore/is anincreasing function.

392

FUNCTIONS AND EXPONENTIAL LOGARITHMIC 1

f"(*):-*

Q)

From (2) it follows that f" (r) < 0 when x > 0. Therefore, the graph of y : f (x) is concavedownward at every point. Because/(1) : ln L : 0, the r intercept of the graph is 1. f(2):ln

L t

To determine an approximate numerical value for ln 2, the definite integral |l[lt) df is interpreted as the number of square units in the area of the region shown in Fig.9.2.1. From Fig. 9.2.1 we see that ln 2 is between the measuresof the areas of the rectangles, each having a base of length 1 unit and altitudes of lengths * unit and 1 unit; that is, 0.5< ln2 < 1

F i g ur e 9 . 2 . 1

(3)

An inequality can be obtained analytically, using Theorem 7.4.g,by proceedingas follows.Letf (t) :Llt andg(t) : *. Then/(t) > g(t) for all f in [1,2]. Becausef and g are continuous on 11.,21,they are integrable on f7,2f, and we have from Theorem 7.4.8

f+

dt>

or, equivalently,

l n2 = +

(4)

Similarly, it f (t) : l,lt and h(t'S: t, then h(t) = f (t) for all t in [1, 2]. Becauseh and / are continuous on l'1,,21,they are integrable there; and again using Theorem 7.4.8, we obtain

or/ equivalently, 1 > ln?

(s)

Combining (4) and (5), we get 0.5
< 1

(5)

The number 0.5 is a lower bound of.ln 2 and 1 is an upper bound of ln 2. In a similar manner we obtain a lower and upper bound for the natural logarithm of any positive real number. Later we learn, by applying infinite series, how to compute the natural logarithm of any positive real number to any desired number of decimal places. The value of ln 2 to five decimal places is given by ln 2 - 0.69315

l 9.2 THE GRAPH OF THE NATURALLOGARITHMICFUNCTION

393

Using this value of.ln 2 and applying Theorem 9.1.5,we can find an approximate value for the natural logarithm of any power of 2. In particular, we have :

ln4:tn22

2ln2-

L.38630

:

2.07945 3ln2= ln f : ln 2-r : -'1. ln 2 - -0.69315 ln | : ln 2-z : -2 ln 2 - -1.38630 ln8:1n23

In the appendix there is a table of natural logarithms, giving the values to four decimal places. We now determine the behavior of the natural logarithmic function tn x. for large values of r by considering lig Becausethe natural logarithmic function is an increasing function, if we take p as any positive number, ln x ) ln 2P

x ) 2P

whenever

V\

From Theorem9.1.5 we have rn2p:pk

(8)

2

Substituting from (8) into (7) , we get lnr)

pln2

whenever

x>2e

Because from (6) ,ln 2 > +, we have from the above ln x > +p

whenever

x

Letting p :ZTt, where n ) 0, we have whenever

lnr)n

x

(e)

Itfol1owsfrom(9),.bytakingN-2,n,thatforanyn>

lnr)n

whenever

r>N

So we may conclude that (10) l" r: *oo "li11 To determine the behavior of the natural logarithmic function for postn r. itive values of r near zero, we investigate ]1p Becauseln r: ln(r-t;-t, we have ln r:

-trr 1 x

(11)

The expression "x + 0+" is equivalent to "Ux -->+@," and so from

LOGARITHMIC AND EXPONENTIAL FUNCTIONS

(11) we write lim ln r--

lim

I-o+

llT-*a

tr, 1)c

(r2)

From (10) we have lim tr, 1: X

ll,f-*a

*o

(13)

Therefore,from (L2) and (13) *e get limlnr--@

(ta1

Jl- 0+

From (14) and (10) and the intermediate-value theorem (2.5.1) we conclude that the range of the natural logarithmic function is the set of all real numbers. From (14) we conclude that the graph of the natural logarithmic function is asymptotic to the negative side of the y axis through the fourth quadrant. The properties of the natural logarithmic function are summarized as follows: (i) ( ii) (iii) (iv) (v) (vi)

The domain is the set of all positive numbers. The range is the set of all real numbers. The function is increasing on its entire domain. The function is continuous at all numbers in its domain. The graph of the function is concave downward at all points. The graph of the function is asymptotic to the negative side of the y ax;isthrough the fourth quadrant.

By using these properties and plotting a few points with a piece of the tangent line at the points, we can draw a sketch of the graph of ttre natural logarithmic function. In Fig.9.2.2 we have plotted the points having abscissasof i, t,1,2,4, and 8. The slope of the tangent line is foundlrom D,(ln x) : Ux. slope: I

slope: + slope - 'l,,

fI

rL2 slope - 2 slope - 4

Figure 9.2.2

1, 2

9 . 3 T H E I N V E R S EO F A F U N C T I O N

9.2 Exercises 1. Draw a sketch of the graph of y : 1tr x by plotting the points having the abscissas*, *, 7,3, and 9,.and use tr 3 1.09861.At each of the five points, find the slope of the tangent line and draw a piece of the tangent line. In Exercises2 through 9, draw a sketch of the graph of the curve having the given equation.

S.y:ln(r+l)

!. y:rn(-x) 6.Y:ln(x-l)

8' Y:xlnx

4. vux -tr, !

d.x:lny

t y:rnlx 7.Y:x-lnx

10. Show the geometric interpretation of the inequality in Exercise32 of Exercises9.1 by drawing on the same pair of axes - Ux, g(r) : In x, and'h(x) : x - 7. sketchesof the graphs of the functions f , g, and lr defined by the equations /(r) : t f { firra an equation of the tangent line to the curve A : ln x at the point whose abscissais 2. f.d pina an equation of the normal line to the curve A:lnx that is parallel to the line x*2y - 1:0. In Exercises13 through 20, find the exactvalue of the number to be found and then give an approximation of this number to three decimal places by using the table of natural logarithms in,the appendix. ff. fina the area of the region bounded by the curve y : 2l(x-

3), the r axis, and the lines r : 4 and

la. ft f(x) : Ux, find the averagevalue of f on the interval [1, 5]. 15. Using Boyle's law for the expansion of a gas(seeExercise10 in Exercises3.9); find the averagePressurewith respectto the volume as the volume increasesfrom 4 ff to 8 fts and the pressure is 2000 lb/ft2 when the volume is 4 ff. 'l' 3lx, the r axis, and fe{ fina the volume of the solid of revolution generated when the region bounded by the curve y: the line r: L is revolved about the r axis' tl. lna telegraph cable, the measure of the speed of the signal is proportionalto f ln(Ux), where r is the ratio of the measu.e of the radius of the core of the cable to the measure of the thickness of the cable's winding. Find the value of r for which the speed of the signal is greatest. 18./A particle is moving on a straight line according to the equation of motion s : (t + 1)'zln(f t 1) where s ft is the directed distance of the partide from the starting point at f sec. Find the velocity and accelerationwhen f : 3. 19. The linear density of a rod at a point r ft from one end is 2/(1 * x) slugs/ft. If the rod is 3 ft long, find the mass and center of mass of the rod. 20. A manufacturer of electric generatorsbegan operations on lan. l, 1966.During the first year there were no salesbecause the company concentrated on product development and research.After the first year, the saleshave increased steadily according to the equation y : r ln r, where x is the number of years during which the company has been operating and y is the number of millions of dollars in the salesvolume. (a) Draw a sketch of the graph of the equation. Determine the rate at which the saleswere increasing (b) on Jan. l, L970,and (c) on Jan. 7, 1976.

9.3 THE INVERSE Let us consider the equation OF A FUNCTION x2_u2:4

If this equation is solved for y in Jerms of x, we obtain

Y:+\84

(1)

LOGARITHMICAND EXPONENTIAL F UNCTIONS

Therefore, Eg. (1) does not define y as a function of r becausefor each value of r greater than 2 or less than -2 there are two values of y, and to have a functional relationship, there must be a unique value of the dependent variable for a value of the independent variable. If Eq. (1) is solved for r in terms of y, we obtain

x:+{N and so x is not a function of y. Therefore, Eq. (1) neither defines y as a function of r nor x as a function of y. The equation u:x2-t F i g u r e9 . 3 . 1

expressesy as afunction of r becausefor each value of x a unique value of y is determined. As stated in Sec.1.7, no vertical line can intersect the graph of a function in more than one point. In Fig. 9.3.1 we have a sketch of the graph of Eq. (1). We see.thatif a > 2 or a 1-2, the straight line ;r: a intersects the graph in two points. In Fig. 9.8.2 we have a sketch of the graph of Eq. (2). Here we see that any vertical line intersects the graph in only one point. Equation (2) does not define x as a function of y because for each value of y gteatet than -r., two values of r are determined. In some cases,an equation involving x and y will define both y as a function of r and x as a function oly, as shown in the following illustration. o TLLUSTRATToN1.: Consider

Figure9.3.2

Q)

the equation

Y : )c3+'l'

(3)

If we let f be the function defined by

f(x):f+r

(4)

then Eq. (3) can be written as y : f(x),and y is a function of.x. lf.we solve Eq. (3) for r, we obtain

x-w

(5)

and if we let g be the function defined by

sQ)-w then Eq. (5) may be written as x:

(6) g Q) , and r is a function of y.

o

The functions / and g defined by Eqs. (4) and (6) are called inuerse functions.-wealso say that the function g is the inoerseof the function/and that/ is the inaerseof g. The notation f -1 denotes the inverse of function /. Note that in using -1. to denote the inverse of a function, it should not be confused with the exponent -1. we have the following formal definition of the inverse of a function.

9.3 THE INVERSEOF A FUNCTION

9.3.1 Definition

If the function / is the set of ordered pairs (x, y), and if there is a function f-r such that : ::.::...::j::l::::.:::i::i.r':....r......:..:,.:,,::::r,:::i::ii:..:r': ,: ': . . . . : : .r.i: .r i::::..::..:1..: .1 . :....!..r. ,:. , . :. .,:.::: i: ::. .:i ir

i:::::::!:l:l:ii::::::i::i::i::rli::iil:li;:l;::tji:::::irj::!:::iil]j':::::ii::i::::S:::ii::::

.:.... ::::..:.:

i......:::.......

:. i

:::::i:::;:::i:::i

(7)

then/-1, which is the set of ordered pairs (y, r), is called the inaerseof the function f , and /and f-r arecalled inaersefunctions. The domain of /-1 is the range of f , and the range of /-1 is the domain of /. Eliminating y between the equations in (7) , we obtain

x -- f -'(f (x)),

(8)

and eliminating r between the same pair of equations, w€ get

v:f(f-'(v))

(e)

So from Eqs. (8) and (9) we seethat if the inverse of the function/ is the function /-t, then the inverse of the function /-1 is the function /. Becausethe functions / and g defined by Eqs. (4) and (5) are inverse functions, f-r can be written in place of g. We have from (a) and (6) (10) and f-'(x):9i-1 f(x):f+l Sketches of the graphs of functions / and f-| defined in (10) are shown in Figs. 9.3.3and 9.3.4,respectively. We observe that in Fig. 9.3.3the function / is continuous and increasing on its entire domain. The definition of an increasing function (Definition 5.1.1)is satisfiedby f for all valuesof r in its domain. We alsoobserve in Fig. 9.3.3 that a horizontal line intersects the graph of / in only one point. We intuitively suspect that if a function is continuous and increasing, then a horizontal line will intersect the graph of the function in only one point, and so the function will have an inverse. The following theorem verifies this fact.

Figure9.3.3

Figure9.3.4

9.3.2 Theorem

Supposethat the function / is continuous and increasing on the closedintewalla, bl. Then (i) / has an inverse /-1, which is defined on If @), f (b)l; (ii) /-' is increasineon If @), f (b)l; (iii) /-1 is continuoason lf(a), f(b)|. PRooFor (i): Since/ is continuous on Ia, bl, then if k is any number such that f(a) < k < f(b), by the intermediate-value theorem (7.5.1), there exists a number c in (a, b) such that /(c) : k. So if y is any number in the closed interval lf (a) , f (b)1, there is at least one number x in la, bl such that y : f (x). We wish to show that for each number y there corresponds only one number x. Supposeto a number y, in lf (a),l(b)l there correspond

398

LOGARITHMICAND EXPONENTIAL FUNCTIONS

two numbers x, and x, (x, # 12) in [a, b] such that !r: /(rr). Then, we must have f(x,):

f @r) and y, : (11)

f(xr)

Becausewe have assumedx1 * x2, either xr I xz ot x2 < xr. lt q 1 xr, because/ is increasing on la, bl, it follows that /(r1) < f(xr); this contradicts (11). lf. xr1r1, then f(x") < f(x), and,this also contradicts (11). ThelefoJe,our assumption that x, # xris false, and so to eachvalue of y in Lf(a), f (b)l there corresponds exactly one number x in [a, b] such ihat y : f (x). Therefore,/ has an inverse /-r, which is defined for all numbers in ff(a), f(b)|. pRooFor (ii): To prove that /-1 is increasing on lf(a), /(b)], we must show that y, and Az are two numbers in [/(a), /(b)] such that y1 I y2, if then f-l(yr) < f-,(yr). Because/-1 is defined on lf(a), f(b)|, theie exist numbers r, and x2 in fa, bl such that y, : f (xr) ^dyr: f (xr). Therefore,

f

-tQ) f-'(f (x')) : xr

(12)

f

-t(yr) -

(13)

and f

-'(f

(xr)) : x,

If x2 < 11, then because / is increasing on fa,b), f (x) < (x1) or, f 'th"n equivalently, azl yt. But ur 0 small enough so that f-r(r) - e and f-r(r) * e are both in la, bf, there exists a 6 > 0 such that whenever lV ,l -r(r) : s. Then /(s) : r. Becausefrom (ii), f-t is increasingon ,, .Let f [f(a), f(b)], we concludethat a < s < D. Therefore, a

Let E be the smaller of the two numberc r - f(s- e) and f(s * e) - r; so 6 < t - f(s- e) and 6 = /(s * e) - r or, equivalently, /(s-.)

(15)

and

r+D =f(s*e) WheneverlV- rl

(16)

(17)

r-6
From (14), (15), (15), and (I7), w€ have, whenever ly - ,l

f ( a )= f ( s - € ) < y < f ( s * e ) = f ( b ) Because/-t is increasingon If @), f (b)l,it follows from the above that whenever ly ,l f'(f(s - € )) < f' Q) < f' ( fG + e))

or, equivalently, s- € < f'(Y) < s * e

whenever

lV-rl

or, equivalently, -€ < f -'(Y) s or, equivalently, whenever lY-rl
',l$ffi'ill ilTfff:;,:-#".?H:rilT#::,"1?i1"f r.,l$if,fi tewal la, bl.

9.3.3 Theorem

Supposethat the function / is continuous and decreasingon the closedintewalla, bl. Then -',which is defined o n L f( b ) ,f ( a ) l ; (i) / has an inverse f -r (ii) f is decreasingon lf (b), f (a)l; (iii) f-r is continuouson l,f(b), f (a)1. The proof is similar to the Proof of Theorem 9.3.2 and is left as an exercise (see Exercises 25 to 27). o ILLUsrnATroN 2: We reconsider Eq. (2), y : x2 - "1..As stated earlier in this section, this equation doeS not define .r as a function of y because if we solve for x we obtain x:\F

and

x:-11y11

(18)

t

I


So actually Eq. (2) defines r as two distinct functions of y. If in Eq. (2) we restrict r to the closedinterval [0, c] and lety : fr(x) if r is in [0, c], we have f'(x):

x2- t

and r in [0, c]

(19)

If in Eq. (2) we restrict r to the closed interval [-c, 0] and let y : fr(x) if.x is in [-c,0], we have fr(x) : x2 I'

and r in [-c, 0]

(20)

The functions f, andf, defined by (19) and (20) are distinct functions becausetheir domains .ue different. If we find the derivatives of f and f2, we obtain 'r(x) : zx and r in [0, c] f and f Lk) : Zx

and r in[-c, 0]

Because/,is continuous on [0, c] and f'r(x) > 0 for all r in (0, c),by Theorem5.l.3 (i), f is increasingon [0, c]; and so by Theorep !.1.!, /, has an inverse on [-L, c2-ll. The inverse function is given by

(2r) Similarly, because/2 is continuous on [-c, 0] and becausef L@) < 0 for all x in (-c,0), by Theorem5.1.3(ii), /, is decreasihgon [-c, 0]. Hence, by Theorem 9.3.3,fzhas an inverse on [-1, ,, - tl, and the inverse function is given by

fr-'(y):-\m Figure 9.3.5

(22)

Becausethe letter used to represent the independent variable is irrelevant as far as the function is concerned, r could be used instead of u in (21) and (22), and we write fr-'(r) - \m

and r in t-l , c2- 1]

(23)

and

Qfu,a)

-\ffi fr-t(r) -

o Rb, u)

and r in t-l , cz- 1]

(24) o

In Fig. 9.3.5 there are sketchesof the graphs of /, and its inverse fi-l, as defined in (19) and.(23), plotted on the same set of axes. In Fig. 9.3.6 there are sketchesof the graphs of f2 andits inverse/2-1, as defined in (20) and (24), plotted on the same set of axes. It appears intuitively true from Fig.9.3.5 that if the point Q@, o) is on the graph of fi, then the point R(o, u) is on the graph of fr-'. From Fig. 9.3.6 it appears to be true that if the point Q@, a) is on the graph of fz, then the point R(o, u) is on the graph of fr-'. In general, if Q is the point (u, a) and R is the point (2, z), the line


segment QR is perpendicular to the line y: t and is bisected by it. We state that the point Q is a reflectionof the point R with respect to the line U : x, and the point R is a reflectionof the point Q with respect to the line : A : x.If r and y are interchanged in the equation y f(x), we obtain the x: (y) equation graph of the the and x: equation , f (y) is said to be a f refleciionof thegraph of the equati ony : f (x) with respectto the line y : 7 Becausethe equation x: f (y) is equivalent to the equation y : fr (x) , we conclude that the graph of the equation y: f-t(x) is a reflection of the graph of the equationy : f (x) with respectto the line y : r. Therefore,if a function / haq an inverse /-1, their graphs are reflections of eachother with respectto the line y: v. EXAMPLEL: Given the function / is defined by

f(x):# -t find the inverse f if it exists. Draw a sketchof the graPh of f , and if 1-r exists,draw a sketchof the graph of 1-t on the sameset of axes as the graPh of f . v y -

f-'(x) : /'U

v:fk)

so1urroN: f '(x) : -Sl@- L)'. Because/is continuous for allvalues of r except7, andf' (r) < 0 if x *'l', itfollows fromTheorem9.3.3that/has-an inveise if r is any real number excePt1' To find/-r,lety : f (x), and solve for x, thus giving x: f-t(y).So we have 2x*3 u : -----T " x- L

rY-Y:2x*3 x(y-2):y*3 v*3

i:

----='

a-2

Thus,

f-'(v):

y+3 y-2

or, equivalently, v

-

f-'(xl )

f-'(x) The domain of f-r is the set of all real numbers excePt2. Sketchesof the graphs of f and f-L are shown in Fig. 9.3.7.

Figure 9.3.7

The following theorem bxpressesa relationship between the derivative of a function and the derivative of the inverse of the function if the function has an inverse. 9.3.4 Theorem

closedinSupposethat the function / is continuous and monotonic on the ' (r) * Ofor :f bltr'df on!a, (x).If differentiable tew-al1a,bT,andlety /is any r in la, bl, then the derivative of the inverse function /-1, defined by *:1-r(A), is given by

402


pRooF: Because is continuous and monotonic on / la, b], then by Theorems 9.3.2 and 9.3.3,/ has an inverse which is continuous and monot o n i co n [ f ( a ) , f ( b ) ] ( o r l f ( b ) , f ( a ) l i f f ( b ) < f ( a ) ) . If r is a number in la, bf, LetAr be an increment of.x, Ax # 0, such that x * Lx is also in [a, b]. Then the corresponding increment of.y is given by

Ly:f(x*Ar) -f(x)

(25)

Ly * 0 sinceAr * 0 and/is monotonicon la,bf; that is, either onla,bl f(x+ Lx) < f(x) or f(x+ Lx) > f(x) If r is in [a,b] andy_: f (x),theny is in [/(a) , f (b)] (or lf (b),f (a)l). Also,if x t Ax is in [a, b], theny + Ly is in [/(a) ,f (b)] (or lf (D; @))) f because y + Ly: f(x* Ax)by (25).So x: f-r(y)

e6\

)c+ Ar: f_r(y + Ly) It followsfrom (26) and (27) that Ar: f-t(y + Ly)- f-10)

(27)

and

(28)

From the definition of a derivative,

+ Lv) - f -'(y) Dox: lim f-'Q substit"tiJ;"*

es) ^::(2s) into the aboveequatior,we set A,x

D o x : lim

;;-; f(x*Ar) -f(x)

and becauseAr + 0, Dox:

lim Ag- 0

f(x*Ar) -f(x)

(2e)

Lx Beforewe find the limit in (29) we show that under the hypothesis of this theorem "Lx---> 0" is equivalent to ,'Ly -+ 0.,, First we-show that lim Ar: 0. From (28) we have As-0

hm lf-'!+ In at: As-0

Ar-O

ty) -f-,(y)l

Because/-r is continuous o n l f ( a ) , f ( b ) l ( o r l f ( b ) , f ( a ) l ) , t h el i m i to n the right side of the above equation is zero. So

ln Ar-0

(30)


that li$ AV:0. From (25) we have Now we demonstrate lim AY: lim lf (x + Lx) - f(x)l

Ar-0

At-o

Because/ is continuous on Ia, bl, the limit on the right side of the above equation is zeto, and therefore we have

-o Jn LY

(31)

From (30) and (31) it follows that Ar+0

ifandonlyif

(32)

Ly-0

Thus, applying the limit theorem (regarding the limit of a quotient) to (29) and using (32), we have 1,

Dox:

Because/ is differentiable on la, bl, the,limit in the denominator of the above is-/'(r) or, equivalently,DrA, and so we have

EXAMPLEZ: Show that Theorem g.3.Aholds for the function of Example L.

solurroN: lf y: f (x) : (2x + g)/(r - 1), then

D,A:f'(*):-C+ at all Because/,(r) exists if.x * 1,/is differentiable and hence continuous # l. x forall is_decreasing L,andsof * < ir, b x # l.F.irth"r*o"",f,(x) b] interval any for holds [a, 9.3.4 Theorem of esis Therefore, tt t ypott " 1. number the which does not contain In the solution of Example 1 we showed that x : f - r ( v\ )r :' Y + 3 y-2 Computing Dux ftom this equation, we get

Dox:(f-\'Q):-G+ If in the above equation we let ,: 5

(2x + 3)/(r - 1), we obtain

404

AND EXPONENTIAL FUNCTIONS LOGARITHMIC

EXAMPLE3: Given f is the function defined by f (x) : xs * x, determine if / has an inverse. If it does, find the derivative of the inverse function.

'(x) :3xz + 1. Therefore, ' (x) ) soLUrIoN: Because 0 for f (x) : x3* x, f f all real numbers, and so f is increasing on its entire domain. Because/ is also continuous on its entire domain, it follows from Theorem 9.3.2 that f has an inverse f-r

L e t y - f ( x ) a n d t h e nx - f - ' U ) .

S o by Theorem9.3.4,

r\17

uax:W-

g x z+ l

9.3 Exercises In ExercisesL through 8, find the inverse of the given function, if it exists, and determine its domain. Draw a sketch of the graph of the given function, and if the given function does not have an inverse, show that a horizontal line intersects the graph in more than one point. If the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes as the graph of the given function.

1. f(x) : x3

2. f (x) : x2+ 5

3. f (x)

4 . f ( x )- ( x + 2 ) '

5. f (x)

6. f(x)

7. f (x) - 2lxl+ x

8. f(x)

In Exercises9 through 14, perform each of the following steps: (a) Solve the equation for y interms of r and expressy as one or more functions of r; (b) for each of the functions obtained in (a) determine if the function has an inverse, and if it does, determine the domain of the inverse function; (c) use implicit differentiation to find D,y and.Duxand determine the values of r and y for which D,y and.Dux arereciprocals. 9. x2+A,:9 12.9y'- 813-0

10.x 2 - 4 Y ' - t 6

11.

13.2x2-3xy+L-0

1 , 4 . 2 x 2 * 2 y *L : 0

In Exercises 15 through 20, determine if the given function has an inverse, and if it does, determine the domain and range of the inverse function.

L 6 .f ( x ) - ( x + 3 ) '

1 7f.( x )- x 2- + , r ) 0 x

19. f(x) : x5 * xB

20.f(x)

*x

FUNCTION 9.4 THE EXPONENTIAL

21. Let the function / be defined by f (x):

I

ttr=V

Jo

df. Determine if / has an inverse, and if it does, find the derivative of

the inverse function. : (r * 5)/(r + k) will be its own inverse' 22. Determine the value of the constant k so that the function d.efined by /(r) rfx < L fx it L < x < 9 23.Givenf (x) -- l*' ifx>9 lzrtfr -t Provethat / has an inversefunctionand find f (x) . 24. Given that the function / is continuous and increasing on the closed intervalla,bl.AssumingTheorem9'3'1(i) prove/-1 is continuour i.o* the right atf(a) and,continuous from the left atl(b). 25. Prove Theorem 9.3.3 (ii).

Prove Theorem 9.3.3(i).

and (ii),

27. Prove Theorem 9.3.3 (iii).

Show that the formula of Theorem 9.3.4 can be written as

( f - ' ) 'v(' r ) :' (#f - t ( r ) ) f

29.use

the formula of Exercise 28 to show that

(f-')"(r):-ffi 9.4 THE EXPONENTIAL FUNCTION

9.A.1Definition

Becausethe natural logarithmic function is increasing on its entire domain, then by Theorem 9.3.2 it has an inverse which is also an increasing function. The inverse of the natural logarithmic function is called the "exponential function," which we now define' The exponentialfunction is the inverse of the natural logarithmic function' and it is defined by (1)

The exponential function "exp(r)" is read as "the value of the exponential function at r." 'Becausethe range of the natural logarithmic function is the set of all real numbers, the domain of the exponential function is the set of all real numbers. The range of the exponential function is the set of positive numbers because this is the domain of the natural logarithmic function' Becausethe natural logarithmic function and the exponential function are inverses of each other, it follows from Eqs. (8) and (9) of sec.9.3 that ln(exP(x)) : x

(2)

-- x exP (ln r)

(3)

and Because 0 : ln L, we have exp 0 - exP (ln 1)

406


which from (3) gives us exP0:L We now theorems.

9.4.2 Theorem

(4) state some properties

of the exponential

function

AS

lf. a and b are any real numbers, then exP(a* b) : exp(a)' exp(b) PRooF: Let A:

exp(a), and so from Definition 9.4.1 we have

a:ln A Let B: exp(b), and from Definition 9.4.1.it follows that b:lnB From Theorem9.1.3 we have lnA+lnB:lnAB

(Z\

Substituting from (5) and (5) into (Z), we obtain a * b:ln AB So exp(a * b) : exp(ln AB)

(s)

From Eq. (3) it follows that the right side of Eq. (8) is AB, and so we have exp(a*b):An Replacing A and B by their values, we get exp(a* b): exp(a) . exp(b) 9.4.3 Theorem

r

lI a and b arc any real numbers, exP(a- b) : exp(a) -:- exp(b) The proof is analogousto the proof of Theorem 9.4.2,where Theorem 9.1.3is replacedby Theorem 9.1,.4.rt is left as an exercise(seeExercise1).

9.4.4 Theorem

If a is any real number and r is any rational number, then exp(ra): [exp(a)]" pRooF: If in Eq. (g) x : lexp(a)1,,we have lexp (a)] " : exp{lnfexp(a)1,] Applying Theorem 9.1.5to the right side of the above equation, we obtain

FUNCTION 9.4 THE EXPONENTIAL

[exp@)]r : exP{rln[exP(a)] ] But from Eq. (2),ln[exp @)] - n, and therefore [exp(a)]':

T

exp(ra)

We now wish to define what is meant by a', where a is a positive number and r is an irrational number. To arrive at a reasonabledefinition, consider the casea,, where a ) 0 and r is a rational number. we have from Eq. (3)

(e)

sr : exp[ln (a')] But by Theorem9."!,.5,lna'-

rln a; so from Eq. (9)

61r: exp(r ln a)

(10)

Becausethe right side of Eq. (10) has a meaning if r is any real number, we use it for our definition. 9.4.5 Definition

g.4.6 Theorem

lf a is any positive number and r is any real number, we define

lf. a is any positive number and r is any real number, lna':xlr':.a PRooF: From Definition 9.4.5, a' : exp(x ln a) Hence, from Definition 9.4.1, we have ln n": xln a

I

Following is the definition of one of the most important numbers in mathematics. 9.4.7 Definition

The number e is defined by the forimula e: exp L The letter "e,, was chosen because of the swiss mathematician and physicist Leonhard Euler (7707-1783)' number; that is, it cannot be exThe number e is a transcend.ental pressed as the root of any polynomial with integer coefficients. The number zr is another example of a transcendentalnumber. The proof that e is transcendentalwas first given inl873,by Charles Hermite, and its value


can be expressed to any required degree of accuracy.In Chapter 16 we leam a method for doing this. The value of e to seven decimal places is 2.71,8281,8. 9.4.8 Theorem

ln e: L. PRooF' By Definition 9.4.7, e-expL Hence, ',

ln e - ln(exp 1)

(11)

Becausethe natural logarithmic function and the exponential function are inverse functions, it follows that ln(exP 1) : L

(r2)

Substituting from (L2) into (11), w€ have lne-L 9.4.9 Theorem

exp (x) : st , for all values of x. PRooF' By Definition 9.4.5, st : exp (r ln e)

(13)

But by Theorem 9.4.8, ln e - 1. Substituting this in (13), we obtain sr : exp (r)

From now on, we write e'in place of exp(r), and so from Definition 9.4.1we have

(r4) We now derive the formula for the derivative of the exponential function. Let

a:e' Then from (14) we have x-lny

(ls)

Differentiatirg get

1:1

v

D"U

on both sides of (15) implicitly with respect to x, we

FUNCTION 9,4 THE EXPONENTIAL

409

So,

D"a- y Replacing y by e*, we obtain ( 16,)

D r(e") : sr

If.u is adifferentiable function of.x,7t follows from (L6) and the chain rule that (17)

It follows that the derivative of the function defined by f(x):ke', where k is a constant, is itself. The only other function we have previously encountered that has this property is the constant function zero; actaally, this is the specialcaseof f(r) : ke' when k: 0. It can be proved that the : ke' most general function which is its own derivative is given by f(x) (seeExercise35). EXAMPLE

Given Y : srtrz find

+-

dyI dx.

solurroN;

From (17)

: H:"ttz (-i)

/grlrz

x3

From (I7) we obtain the following

indefinite

integration formula: (18)

EXAMptn 2:

l4a*

J\E

Find

Let

u - t E ; then du

r

2Jeudu 2e'+C 2en+C : Becausefrom (14), e' : A if and only if x : ln y, the graph of y t' i" : identical with the graph of tt: ln y. So we can obtain the graph of y et by interchanging the r and y axesin Fig' 9.2-2(seeFig' 9'a'1)' The gruph ol y : et carrbe obtained without referring to the graph of iogarithmic function. Because the range of the exponential natuial the set of all positive numbers, it follows that e' > 0 for all valis the function graph lies entirely above the r axis' D,A : e" ) 0 for all r; the r. So ues of so the function is increasing for all x. D r'y : e' ) 0 for all r; so the graph is concave upward at all Points.

410


We have the followi^g

two limits:

(le)

lim st--fm X-

*@

and

e*: o

"tlq

(20)

The proofs of (19) and (20) are left as exercises(see Exercises44 and 45). To plot somespecificpoints use the table in the Appendix giving powers of e.

F i g u r e9 . 4 . 1

ExAMPr,u3: Find the area of the region bounded by the curye y : e', thecoordinate axes,and the line x: 2. Y:

The region is shown in Fig. 9.4.2.

solurroN:

tip

l lAal l -o f , 1

T

er dx

e, (2, ezl

et

e2

e0

: e2- 1 From a table of powers of e, we see that the value of e2to two decimal places is 7 .39. Therefore, the area of the region is approximately G.gg square units.

Ei Air

Figure 9.4.2

The conclusionsof Theorems9.4.2,9.4.3,and9.4.4are now restatedby using e" in place of exp(r). If. a and b are any real numbers, then ea+b: eo , eb eo-b: €ra:

ea + eb (sa)r

where r is any rational

number.

9,4 THE EXPONENTIAL FUNCTION

Writing Eqs. (2) , (3), and (4) by substituting have

411

er in place of exp (x) , we

ln(e") : )c -

,lnx

X

and eo:t

ExAMPrn 4: y :

t2r*ln

find D *A.

Given n

t2r*ln r :

soLUTroN '

t2r trnr :

e'" (x). So

y - xez' Therefore, D rU : s2r* 2xe2*

In Definition9.A.7, the number e was defined as the value of the exponential function at L; that is, e: exp L. To arrive at another way of defining e, we consider the natural logarithmic function f(x) :ln x '(x) :1/r; hence, We know that the derivative of / is given by f L. However, let us apply the definition of the derivative to find f,(t): /'(1). We have ' ): !f ' (\ L

:

r(1 l i m E !_L*)._lG)_ Lx

Ar-o

lim Ar-0

_ lim A.r- 0

Therefore, 1

lim ;* ln(1+ Ar) : 1 ar-o AI Replacing Ar by h, we have from the above equation and Theorem9'4.6 (27\ limln(l *h)un-, ft-0

Now, because the exponential function and the natural logarithmic function are inverse functions, we have lim (L * hlun: lim exp[ln(l * h)trn1 h-0

h-O

Q2)


Becausethe exponential function is continuous and lim ln(l + h)unexists and equals L as shown in Eq. (21))rwe can apply H"orem right side of Eq. (22) and get

lg

2.5.5to the

(1+ h)trn-exp h(l + h)',of: exp1 fm

Hence,

(23) Equation (23) is sometimes used to define e; however, to use this as a definition it is necessaryto prove that the limit exists. Let us consider the function F defined by F(h)-(1 +h)un

(24)

and determine the function values for some values of h closeto zero. These values are given in Table 9.4.1. Table9.4.1 h

1 0.5 0.05 0.0L 0.001 0.001-0.0L -0.05 -0.5

F(h):(1 +hyrn

2 2.25 2.65 2.702.71,692.7196 2.73 2.79 4

Table9.4.1leadsus to suspectthat lim (1 + lr; ru ir probably a number that lies between 2.7169 and2.T191.As previously mentioned, in Chapter 16 we leam a method for finding the value of e to any desired number of decimal places.

Exercises 9,4 1. Prove Theorem9.4.g. A

1. 1

2. Draw a sketch of the graph of y :

3. Draw a sketch of the graph of A - el"l

e-s".

In Exercises4 throu gh 14, find dyl dx. 4. a:e5'

5. y :

g-9tz

8' a:e"* \.trt. y: 1'4. y :

*lr-lrna' gnlt/A+rz

I 9.4 THE EXPONENTIAL FUNCTION

413

In Exercises1.5through 1.8,find D*y by implicit differentiation.

r:.

: [F. e' * eu et+u

16. ye'" * xezu- 1

tt[. y'e'* * xyt: I \/ 1.9through26, In Exercises

18. €u: ln(r3 + gy)

r

evaluate the indefinite

rl4't

19. 1 ez-sr dx

20. l tzr+r dx

22. lr"r" 4*

23' J ,-ks"Ydx

J

J f

,]-t"'

integral.

J f

plr

?26. I#.

oZc

-EY.J, Jl # d x ex+J

In Exercises27 through 30, evaluate the definite integral.

27. 29.

fr

J. f2

J,

-?8. e2dx xea-" dx

30.

In Exercises 31 and 32, find. the relative extrema of f , the points of inllection of the graph of.f , the intervals on which / is increasing, the intervals on which / is decreasing, where the gfaph is concave upward, where the graph is concave downward, and the slope of any inflectional tangent. Draw a sketch of the graph of /. 3 1 .f ( x ) : x e - c

&}f(*)-e-*'

d. fitta an equation of the tangent line to the curve !: e-" that is perpendicular to the line 2r !:5. (0, 1) and (1, e). Cl fitra the area of the region bounded by the curve y: e' and the line through the points : 1. If every plane section per35. A solid has as its basethe region bounded by the curves ! : e' and A : e-" and the line r pendicular to the r axis is a square find the volume of the solid. : 35. Prove that the most generalfunction that is equal to its derivative is given by /(r) : ke". (rrrNr: Let y /(x) and solve the differential equation dVldx: y.) id|* lf p lblftz is the atmospheric pressure at a height of h ft above sea level, then p :2ll6e-o'owo31t". Find the time rate of change of the atmospheric press.rr" outside of an airplane that is 5000 ft high and rising at the rate of 160 ftlsec. 3g. At a certain height the gaugeon an airplane indicates that the atmospheric pressrre is 1500lb/ff. Applying the formula of Exercise37, ipproximat- by differentials how much higher the airplane must rise so that the pressure will be 1480 lblft2. Use differentials to find the aPProxg9. lllft is the length of an iron rod when f degreesis its temperature, then I : 60e0'00001'. l'0. imate increase in I when f increasesfrom 0 to (F.

A simple electric circuit containing no condensers,a resistanceof R ohms, and an inductance of L henrys has the electromotive force cut off when the current is 16amperes. The current dies down so that at f secthe current is i amperes and i:

loe-
Show that the rate of dtange of the curent is proportional to the current. 41. A body is moving along a straight line and at f secthe velocity is u ftlsecwhere p: the particle while zr ) 0 after f : 0.

es- e2t.Find the distance traveled by

414


42. An advertising agency determined statistically that if a breakfast food manufacturer increasesits budget for television commercialsby r thousand dollars there will be an increasein the total profit of 25i?e-0.2s hundred dollirs. What should be the advertising budget increase in order for the manufacturer to hive the greatest profit? What will be the coresponding increase in the company's profit? 43. A tank in the shape of the solid of revolution formed by rotating about the r axis the region bounded by the curve yzx: e and the lines r : 1 and x : 4. If the tank is full of water, find the work done in pumplng all the water to a point 1 ft above the top of the tank. Distance is measured in feet. 44. Prove: lim e":*o,by X-*o

showing that for any N>0

there exists an M>0

such that e")N

whenever x>M.

45. P r o v e : l i T ; ' e c : O , b Y s h o w i n g t h a t f o r a n y € > 0 t h e r e e x i s t s a n N < 0 s u c h t h a t e s < e w h e n e v e r r ( N . 46. Draw a sketch of the graph of F if F(h) : (7 t h)tth.

9 q NTHFR FXPONFNTIAL AND I OGARITHMIC FTINCTIONS

From Definition 9.4.5, we have qx :

,"rln

u

(1)

The expression on the left side of Eq. (1) is called the " exponential function to the base a."

Definition

If a is any positive number and r is any real number; then the function / defined by f(x) : 6t is called the exponentialfunction to the basea. The exponential function to the base a satisfies the same Properties as the exponential function to tlE-base e. . TLLUSTRATTON 1.: If x and y are any real numbers and n is positive, then from Eq. (1) we have frr7a

-

gtlnarulna

_

grlna*ulna

:

g(t*u)ln

_

a

Or*U

From Illustration 7 we have the property ar aa -

,r*a

We also have the followirg At+fr.u:ar-a

(2) properties: (3)

9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 415 (a")u -

ata

(4)

(ab)' :

arbt

(s) (6)

ao: I

The proofs of (3) through (5) are left as exercises (see Exercises 1 to 4). To find the derivative of the exponential function to the base a, we set ac - ecrnaand apply the chain rule. We have At:

exlna

s c r n aD r ( x l n a )

Dr(a'):

: errno(lna) :4'ln

a

Hence, itu is a differentiable

function of r, (7)

o TLLUSTRATToN2: If y :

D*A:3"

Jr2, then from formula

ln 3(2x)- z(ln 3)x3"'

(7) we have

o

From (7) we obtain the followi.g indefinite integration formula: (8)

Find dx

SOLUTION: I \4OT dX: I 103't2dx. Let Lt: Ex; then du :8 dx; thus JJ ? du: dx. We have then ff

dx: I to & du | 1or"rz JJ :5 2 1' 0 " r - C lt:r L0

:

ExAMPrn 2: Draw sketches of the graPhs of y -- 2" andY :2-* on the same set of axes. Find the area of the region bounded bY these two graphs and the line x: 2.

2 . t03Jft2

31,,L0

-, C

The region is soLUrIoN: The required sketches are shown in Fig. 9.5.'1.. shaded in the figure. If A square units is the desired atea, we have A-

lim i:r llall-o

416


(2 - 2-") dx (2, 4)

(t i , zE')

2t

ln 2

2-r

12

ln? Jo

\ ((i,z-ti'1

F i g u r e9 . 5 . 1

We can now define the "logarithmic function to the base a', if a is any positive number other than 1. 9.5.2 Definition

'1., If a is any positive number except the logarithmicfunction to the basea is the inverse of the exponential function to the base a; and we write

(e) The above is the definition of the logarithmic function to the base a usually given in elementary algebra; however, (9) has meaning for y any real number becauseaa hasbeen precisely defined. It should be noted,that 7fa: e, we have the logarithmic function to the base e, which is the natural logarithmic function. log" x is read as "the logarithm of.x to the base a.', The logarithmic function to the base a obeys the samelaws as the natural logarithmic function. We list them. logo Qy) - logo x + Iogo y logoQ:U):logox-logoy logoL :0 logo xa : y logo x

(10) (11) (12) (13)

The proofs of (10) through (13) are left as exercises (see Exercises

5 to 8). A relationship between logarithms to the base a and natural logarithms follows easily. Let Y :logo x

9.5 OTHEREXPONENTIAL AND LOGARITHMIC FUNCTIONS 417

Then aa:)c ln aa: ln r ylna--lnr Y-

lnr l"a

Replacing y by logo x, we obtain

ll::r,,., ::rili.iiuiiiniii:ii.:i:i...iiirii:iitixi:liiiii,iiirli.,..ilillir::iiiffiiri ,,*fiffitff't :

r ' : : 1 r : : : : : r : : : r : : : t : : : : : . . : : r'

,tj : : : : . : : : : : : : : . : : :i :. :i : .: j: : : : . : : : :. . . :

(14)

.

Equation (14) sometimes is used as the definition of the logarithmic function to the base a. Becausethe natural logarithmic function is continuous at all x > 0, it follows from (14) that the logarithmic function to the base a is continuous at all r ) 0. If in (14) we take x: e,wehave

ot,

(ls) We now find the derivative of the logarithmic function to the base a. We differentiate on both sides of Eq. (14) with respect to r, and we obtain

D , ( l o g o x ) : + D , ( l nw lna

r)

D*(logox):#'+

(16)

Substituting from (15) into (16), we get e D r(logox) :lo8o x

(rz7

If u is a differentiable function of x, we have (18) Note that if in (18) we take n - e, we get

D*(rog,u) :

lo? t D*u u

FUNCTIONS LOGARITHMICAND EXPONENTIAL

418

or, equivalently, 1

D"(lnu):!D,u which is the formula we had previously for the derivative of the natural logarithmic function. EXAMPTT 3:

Y

Given

soLUrIoN: Using (11), w€ write

: logto x * L

y:

xz +'/',

l o g t o ( r + 1 ) - l o g r o ( x '+ 1 )

From (18) we have

find dy ldx.

dy

e _Iogro e . )y _logn x+1,dx )c2+L'zL

Becauser" has been defined for any real number n, we can now prove the formula for finding the derivative of the power function if the exponent is any real number. 9.5.3 Theorem

rf n is any real number and the function / is defined by wherer)0 f(x):x" then f'(x) : ny"-r PRooF: Lety: Y

rn. Then from (1) :

,nrnr

Thus, DrU -

,nrnt ,nln

ln r)

t

-vn'L x - nxn-r

I

Theorem 9.5.3 enablesus to find the derivative of a variable to a constant power. Previously in this section we leamed how to differentiate a

9.5 OTHER EXPONENTIALAND LOGARITHMICFUNCTIONS

4I9

constant to a variable power. We now consider the derivative of a function whose function value is a variable to a variable power. The method of Iogarithmic differentiationis used and is illustrated in the following example. EXAMPTr 4:

Given

y - x', find

dyldx.

solurroN: BecauseA: f , then lyl : lr'1. We take the natural logarithm on both sides of the equation, and we have ln lYl : ltt lr"l or, equivalent$,

ltt lYl: r ln lrl Differentiating on both sides of the above equation with respect to r,

+ * Lx

H:a}nl'l

+ 1)

- r'(ln lrl + 1)

9,5 Exercises In Exercises1 through 4, prove the given ProPerty if a is any positive. number and x and y are any real numbers. L.

'A*

+ Aa-

3. (ab)t :

2. (a")u - ata

As-a

4 . a o: ' L

arbt 8, prove

5 through

In Exercises

the given

ProPerty

if a \s any positive number

and x and y arc any positive numbers.

5. logok : !) : logo x - logo y

5. logo(rV) - logo )c+ log" y 7. Iogo xu -- y Log" x

8 . l o g oL : 0

'(x). In Exercises9 through24, find f , ' 9 rf.k \ - 3 5 '

i 1 0 .f ( x ) : $ - B t logrox 1 ' 3 f. ( x ) : ' x

;r'

I?. f(x)

- (x3+3)2-'*

l s .f ( x ) : @ 18. f (x) -

vtnx

21.f(x\:1ge' 24. f (x) - (ln x)tn r

i i

t"t. f(x)-25t34cz

14. f ( x ) : l o g r o

#

r) ] 16. f (x) - logo[1o8o(log,

17. f (x) : logro[logto(r+ 1) ]

1 9 .f ( x ) : x 6

20. f ( x ) :

22.f(x):(x)""

23. f (x) : (4e"'1r"

xstz

FUNCTIONS AND EXPONENTIAL LOGARITHMIC In Exercises 25 through 32, evaluate the indefinite integral.

2s. *" a* [

27.

28. Iu**,(2xr+L)dx

'30.

I

a"e"dx

Ir"rnr(lnr+1)dx

31. *r,'3," dx I 33. Find an equation of the tangent line to the curve y24v- /2x at (4,2). 34. A partide is moving along a straight line according to the equation of motion s - A2r' + B2-ht,where s ft is the directed distance of the particle from the starting point at f sec. Prove that if a ft/sec2is the accelerationat f sec, then a is proportional to s. A company has leamed that when it initiates a new salescampaign the number of salesper day increases.However, the number of extra daily salesper day decreasesas the impact of the campaign wears off. For a specific campaign the company has determined that if S is the number of extra daily salesas a result of the campaign and r is the number of days that have elapsed since the campaign ended, then

s: 1000. 3-'/2 Find the rate at which the extradaily salesis decreasingwhen (a) x: 4 and,(b) r:

10.

In Exercises 36 and 37,usedifferentialsto find an approximatevalueof the givenlogarithmandexpressthe answerto three decimal places. 36. logro 1..0L5

37. log,o 997

38. Draw sketchesof the graphs of.y : lo916r and y :ln

r on the same set of axes.

39. Given:f(x) :t@" * a-"). Prove that f(b * c) * f(b - c) :zf(b)f(c). tl(). A particle moves along a straight line according to the equation of motion s : fll, where s ft is the directed distance of the partide from the starting point at f sec. Find the velocity and accelerationat 2 sec.

9.5 LAWS OF GROWTH AND DECAY

The laws of growth and decay provide applications of the exponential function in chemistry, physics, biology, and business. Such a situation would arise when the rate of change of the amount of a substancewith respect to time is proportional to the amount of the substancepresent at a given instant. This would occur in biology, where under certain circumstances the rate of grovWh of a culture of bacteria is proportional to the amount of bacteria present at any given instant. In a chemical reaction, it is often the casethat the rate of the reaction is proportional to the quantity of the substancepresen! for instance, it is known from experiments that the rate of decay of radium is proportional to the amount of radium present at a given instant. An application in business occurswhen interest is compounded continuously. In such cases,if the time is represented by f units, and A units represents the amount of the substancepresent at any time, then dA 7l:

Ktt

9.6 LAWS OF GROWTHAND DECAY

where k is a constant. If A increasesasf increases,then k > O,and we have lhe law of naturalgrowth.If A decreasesas f increases,then k ( 0, and we have the law of natural decny.In problems involving the law of natural decay, thehalf life of a substanceis the time required for half of it to decay. L: The rate of decay of. EXAMPLE is radium proportional to the amount present at any time. If 50 mg of radium are Presentnow, and its half life is 1.690years,how much radium will be Present100 years from now?

Let f : the number of years in the time from now; A : the number of milligrams of radium Present at f years. we have the initial conditions given in Table 9.6.1.The differential equation is

solurroN:

Separatingthe variables, we obtain

9.6.1 Table t

0

1590

100

A

60

30

Aro,

#-kdt Integrating, we have

lnlAl:H+e lAl:ekt+e-F'ekt Letting eE: C, we have lAl : Cro', and becauseA is nonnegative we can omit the absolute-value bars around A, thereby giving us A:

Cekt

BecauseA:

50 when f : 0, we obtain 50 : C. So (1)

A - 50ekt - 'J..690, or we get 30 : 6l0e16e0k Because A - 30 when t 0.5 :

gr6e0k

So ln 0.5 : 1,590k and

llr9:5 : -0.0004L0 k - 1690 :-o='929r 1690 Substituting this value of k into Eq. (1), we obtain A

-

50e-o.o00410t

When t - L00, A

Aroo,and we have

Aroo:50e-o'0410:57.5 Therefore, there will be 57.6mg of radium present 100 years from now.

422


ExAMPtn 2: In a certain culture, the rate of growth of bacteria is proportional to the amount present. If there are 1000 bacteria present initially, and the amount doubles in L2 min, how long will it take before there will be 1.,000,000bacteria present?

The differential equation is the same as we had in Example above, the general solution is

Table9.6.2

t A

0

soLUrIoN: Let f : the number of minutes in the time from now; A: the number of bacteria present at f min. Even though by definition A is a nonnegative integer, we assumethat A can be any nonnegativenumber for A to be a continuous function of f. Table9.6.2 gives the initial conditions. The differential equation is

L2

T

L,000 2 ,000 L,000,000

hence, ds

A - Cekt

When t - 0, A:

L000;hence,

1,000,which gives

A:1000ekt From the condition that A: erzk- 2

2000when f :12, we obtain

and so k: #ln2:0.05776 Hence, we have A-

1000eo.o5776t

Replacing t by T and A by 1,000,000,we get 1,000,000- 1000e0'05776r e0.05776?:1000

0.057767: ln 1000 ln 1( ,: O ffi:179.6 Therefore, there will be 1,000,000bacteria present in t hr, 59 min, 36 sec. ExAMPrn 3:

Newton,s law of cooling states that the rate at which a body changes temperature is proportional to the difference between its temperature and that of the surrounding medium. If a body is in air of temperature 35oand the body cools

solurroN: Let f : the number of minutes in the time that has elapsed since the body started to cool; r: the number of degreesin the temperature of the body at f minutes. Table 9.6.3 gives the initial conditions. From Newton's law of cooling, we have dx

7t-

k(x - 35)

DECAY

from 120"to 60oin 40 min, find the temperatureof the body after 100 min.

Separating the variables, we obtain dx

ffi-kdt

T n b l e9 . 6 . 3

t

0

40

x

I20

60

kl dt

100 '

rroo

kt+ c x: cekt+ 35 Therefore,

When x-85ekt+35

When t - 40, x - 60; and we obtain + 35 85e4ok

60:

40k- ln #

n:l[il;'::'Lz) x

-_ g5e-0.0306t+ 35

:r f f r o o- 8 5 e - 3 ' o 6 1 3 5 : 3 9 Therefore,

ExAMPrn4: There are L00 gal of brine in a tank and the brine contains 70lb of dissolved salt. Fresh water runs into the tank at the rate of 3 gal/min, and the mixture, kePt uniform by stirring, runs out at the same rate. How many Pounds of salt are there in the tank at the end of 1.hr?

the temperature

of the body

is 39o after 100 min.

Let f : the number of minutes that have elapsed since the water started flowing into the tank; t: the number of pounds of salt in the tank at t min' Because1,00gal of brine are in the tank at all times, at f minutes the number of pounds of salt per gallon is r/100. Three gallons of the mixture run out of the tank each minute, and so the tank loses3(r/100) pounds of salt per minute. BecauseDp is the rate of change of x with respect to f, and r is decreasing as f increases,we have the differential equation

solurroN:

dt

100

we also have the initial conditions given in Table9.6.4. Separating the variables and integratingr w€ have

LOGARITHMIC AND EXPONENTIAL FUNCTIONS T a b l e9 . 6 . 4

t

0

60

x

70

reo

I+--o.otIat ln lrl - -0.03f + e x: Ce When

x:70, and so C:70. Letting

60 and x - xeo,we have

x6o: 70e-1'8

:70(0.L553) - 1'/-,.57 So there are 11.57lb of salt in the tank after L hr.

The calculus is often very useful to the economist for evaluating certain business decisions.However, to use the calcuruswe must beconcemed with continuous functions. consider, for example, the following formula which givesA, the number of dollars in the amount after f years, ii P dollars is invested at a rate of 100i percent, compound.edz times per year: A : p (- t\ +- 'Jm- \l ^ '

(2)

Let us conceive of a situation in which the interest is continuously compounding; that is, consider formula (2), where we let the number of interest qerlo$s per year increasewithout bound. Then going to the limit in formula (2), we have

A: p mtim /r + a)'' _*@ m/ \

which can be written as

1)mtil* A : p t nl it -m *@ l(t+ L\

m/

J

(3)

Now consider

tim /r + -!-\''n

z-+o \

ml

Letting h : il m, we have mli : 1lh; and because',m -->*o,, is equivalent tO ,rh + 0+r, we have

== tim /r + l-\'" (1 * hlttn- ' m/ nli$

n-+o \

9.6 LAWS OF GROWTHAND DECAY

425

Hence, using Theorem 2.6.5, we have

.ri* [(t +h)*'')":

:

git

and so Eq. (3) becomes A - Peit

(4)

By letting t vary through the set of nonnegative real numbers, we seethat Eq. (4) expressesA as a continuous function of f. Anotherway of looking at the samesituation is to consider an investment of P dollars, which increasesat a rate proportional to its size. This is the law of natural growth. Then if A dollars is the amount at f years, we have

lnlAl:H+e A- Cekt When t:

0, A:

P, and so C : P. Therefore,we have

A: pekt

(5)

Comparing Eq. (5) withEq. (4), we seethat they are the same if.k: i. So if an investment increasesat a rate proportional to its size, we say that the interest is compoundedcontinuously, and the interest rate is the constant of proportionality. . rLLUSrRArroN1: If P dollars is invested at a rate of 8vo Per year compounded continuously, and A dollars is the amount of the investment at f years, dA

ff: o'oee and A- Peo.ost

o

If in Eq. (4) we take P : L,i:1, and t: l,we getA: e, which gives a justification for the economist's interpretation of the number "e" as the yield on an investment of one dollar for a year at an interest rate of 700Vo compounded continuously.

ii't .'ii


EXAMPIE5: If $5000 is borrowed at an interest rate of L2Voper year, compounded continuously, and the loan is to be repaid in one payment at the end of a year, how much must the borrower repay? Also, find the effective rate of interest which is the rate that gives the same amount of interest compounded once a yeat.

solurroN: Letting A dollars be the amount to be repaid, and becauseP : 5000,i :0.12, and f : 1, we have from Eq. (4) d:

5gggro'tz : s000(1.1275) :5637.50

Hence, the borrower must rcpay $5637.50.Lettingl be the effective rate of interest, we have 5000(1*i) :5000e0'12 1*i:to'rz i:1.1275 - L : 0.1275 - 12,7570

9;6 Exercises 1. Bacteriagrown in a certain culture increaseat a rate proportional to the amount present. If there are 1000bacteria present initially and the amount doubles in L hr, how many bacteria will there be in 3* hr? 2. In a certain culture where_the rate of growth of bacteria is proportional to the amount present, the number triples in 3 hr, and at the end of 12 hr there were 10 million bacteria. How many bacteria were present initially? 3' In a certain chemical reaction the rate of conversion of a substanceis proportional to the amount of the substancestill untransformed at that time. After 10 min one-third of the original amount of the substancehas been converted, and 20 g has been converted after 15 min. What was the original amount of the substance? 4. Sugar decomPosesin water at a rate proportional to the amount still unchanged. If there were 50 lb of sugar present initiallyandattheendof 5hrthisisreduced ro20 lb,howlongwillittaie u n t i l 9 0 V oo f t h e s u g a r i J d e l o m p o s e d l 5. The rate of natural increase of the population of a certain city is proportional to the population. If the population increasesfrom 40,000to 50,000in 40 years, when will the population be 80,000? 6. Using Newton's law of cooling (see Example 3), if a body in air at a temperature of 0ocools from 200oto 100' in 40 min; how many more minutes will it take for the body to cool to 50.? 7. Under the conditions of Exarnple3, alter how many minutes will the temperature of the body be 45'? 8. When a simple electric circuit, containing no condensersbut having inductance and resistance,has the electromotive force removed, the rate of decreaseof the current is proportional to the current. The current is i amperesf sec after the. cutoff, and i: 4Owhen t: 0. If the current dies down to L5 amperes in 0.01 sec, find i in terms of f. 9. If a thermometer is taken from a room in which the temperature is 75ointo the open, where the temperature is 35", and the reading of the thermometer is 55oafter 30 sec, (a) how long after the removal will the reading be Sb.f 1U)What is the thermometer reading 3 min after the removal? Use Newton's law of cooling (see Example 3). 10. Thirty Percent of a radioactive substancedisappears in 15 years. Find the half life of the substance.

REVIEWEXERCISES 427

11. If the half life of radium is 1690years, what percent of the amount present now will be remaining after (a) 100years and (b) 1000years? 12. A tank contains 200 gal of brine in which there are 3 lb of salt per gallon. It is desired to dilute this solution by adding brine containing t lb of salt per gallon, which flows into the tank at the rate of 4 gaUmin and runs out at the same rate. When will the tank contain 1+ lb of salt per gallon? 13. A tank contains 100 gal of fresh water and brine containing2lb of salt per gallon flows into the tank at the rate of 3 gal/min. If the mixture, kept uniform by stirring, flows out at the same rate, how many pounds of salt are there in the tank at the end of 30 min? 14. A loan of $100is repaid in one payment at the end of a year. If the interest rate is 8% compounded continuously, determine (a) the total amount repaid and (b) the effective rate of interest. s 15. If an amount of money invested doubles itself in L0 years at interest compounded continuously, how long will it take for the original amount to triple itself? 15. If the purchasing power of a dollar is decreasingat the rate of 8% annually, cornpounded continuously, how long will it take for the purchasing power to be 50 cents? Professor Willard Libby of University of Califomia at Los Angeles was awarded the Nobel prize in chemistry for discovering a method of determining the date of death of a once-living object. Professor Libby made use of the fact that the tissue of a living organism is composed of two kinds of carbons, a radioactive carbon A and a stable carbon B, in which the ratio of the amount of A to the amount of B is approximately constant. When the organism dies, the law of natural decay applies to A If it is determined that the amount of A in a piece of charcoal is only 15Voof its original amount and the half life of A is 5500 years, when did the tree from which the charcoal came die?

(Chapter9) ReaiewExercises In Exercises1 through 8, differentiate the given function.

1 .f ( x ) : ( l nf ) 2

2 .f ( x ) : @ h

3 .f ( x ) :

a. f @) - 10-5"

5. /(r) - ifttrr

6. f(x)-ttt(

8. f(x):3"'" (7)t"' integral. the indefinite 14, evaluate 9 through Exercises In r aozc dx lo. I eu'-*1x-l)

logro

L*x t-x

+xz)

z. f(x):

s. lfi*ax

r

nlffa,

BI#

(e" * at*) dx

In Exercises15 through L8, evaluate the definite integral.

ts.

fzfr-

ln

17. J,fua* 18

16.

x2e8dx "l l 3

,

19. Find D*A if ye' * xea* x * y - 0. '(x)2 0 . l f f ( x ) : l o g ( " ' y (+x 1 ) , f i n d f

Jo

kb *7)z dx

428


2 1 . T h e l i n e a r d e n s i t y o fa r o d a t a p o i n t x f t f r o m o n e e n d i s U ( x * 1 ) center of mass of the rod. 22. Find an equation of the tangent line to the curve y : xs-r af (2,2).

s l u g s / f tI.f t h e r o d i s 4 f t l o n g , f i n d t h e m a s s a n d

23. A particle is moving on a straight line wheres ft is the directed distance of the particle from the origin, o ftlsec is the velocityof the particle, and aftlse& is theaccelerationof the particle at f sec.If. a: et * e-tand o:1and s:2 when t: 0, find o and s in terms of l. 24.T'heareaoftheregign,boundedbythecurvey:e-s,thecoordinateaxes,andtheliner:b(b>0),isafunctionofb.lf / is this tunction, find /(b). Also find 1im /(b) . 25. The volume of the solid of revolution obtained by revolving the region in Exercise24 about the r axis is a function of b. If g is this function, find g(b). Also find la S(A). 25. The rate of natural increaseof the population of a certain city is proportional to the population. If the population doubles in 60 years and if the population in 1950was 60,000,estimate the population in the year 2000. 27. The rate of decay of a radioactive substanceis proportional to the amount present. If half of a given deposit of the substance disappears in 1900years, how long will it take for 95Voof the deposit to disappear? 28. Prove that if a rectangle is to have its base on the r axis and two of its vertices on the curve y : s-cz, then the rectangle will have the largest possible area if the two vertices are at the points of inflection of the graph. 29. Givenf (x) : ln lrl and r < 0. Show that / has an inverse function. If g is the inverse functioru find g(r) and the domain of g. In Exercises30 and 31, find the inverse of the given function if there is one, and determine its domain. Draw a sketch of the given function, and if the given function has an inverse, draw a sketch of the graph of the inverse function on the same set of axes.

3 0 .Jf '(\ x ) - ! + 4- 3 2x

3 2 . P r o v e t h a t i fr < 1 , L r r ( r .

(nrur: Letf(x):r-lnrandshowthat/isdecreasingon(0,1)andfind/(l).)

33. When a gas undergoesan adiabatic (no gain or loss of heat) expansionor compression,then the rate of changeof the Pressurewith respect to the volume varies directly as the pressure and inversely as the volume. If the pressure is p lb/in.'zwhen the volume is u in.3, and the initial pressure and volume are polblin.z and a6 in.s, show thai pztk: p6ooi. U. ft W in.lb is the work done by a gas expanding against a piston in a cylinder and P lb/in., is the pressure of the gas when the volume of the gas is V in.3, show that if V, in.3 and.V"in.s are the initial and final volumes, respectively, then

W:I

fvz

PN

Jvr

35. Supposea piston comPressesa gas in a cylinder from an initial volume of 50 in.s to a volume of 40 in.3. If Boyle's law (Exercise L0 in Exercises3.9) holds, and the initial pressure is 50 lblin.z, find the work done by the piston. (Use the result of Exercise34.) 36. The charge of electricity on a spherical surfaceleaks off at a rate proportional to the charge. Initially, the charge of electricity was 8 coulombs and one-fourth leaks off in 15 min. When will there be only 2 coulombs remaining? 37. How long will it take for an investment to double itself if interest is paid at the rate of 8% compounded continuously? 38. A tank contains 50 gal of salt water with 120lb of dissolved salt. Salt water with 3 lb of salt per gallon flows into the tank at the rate of 2 gallmin and the mixture, kept uniform by stirring, flows out at the same rate. How long will it be before there are 200 lb of salt in the tank?

REVIEWEXERCISES rt

39.Find I e-t"t dx if f is any real number. Jo

40. Provethat rf.x> 0, and f" ,o-t dt:L,

(1 + h)rm.

then ti* x:lim

Jt

h_O

h_0

4 L . Prove that lim

1og,(l * r) : Logoe X

t-O

(Norr:

Comparewith Exercise33 in Exercises9.1.)

42. Prove that li*a'-1:loo c-O

X

(nrnr: Lety:a'-Tandexpress lim s(y).)

(a"-l)lxas

a f u n c t i o no f y , s a y g ( y ) ' T h e n s h o w t h a t ! + 0 a s x + 0 '

andfind

43. Use the results of Exercises41 and 42 to Prove that

r i mU _ _Lb x'l

(HrNr:

X,-

Write

xb-L -_to'n'-L blnx )c-L

.bLnx x-l

Then let s b ln r and t:

x - L.)

44. Prove that

\

ean- "l' : fl lim X

t-0

eo'and find /'(0) by two methods.) cf(x) for allr where c is a constant, Prove that there is a con45. If the domain of /is the set of allreal numbers and f'(x): function 8 for which 8(x) : f(x)e-co and find 8'(x)') stant k for which f(x) - ke"* for all r. (HrNr: Consider the (nrNr:

Let f (x) :

46. Prove that

Drn(lnx)-(-l)n-r+ (nrxr:

Use mathematical induction' )

each integral' 47. DoExercise 17 in the Review Exercisesof chapter 7 by evaluating

Iiigonometric functions

1 0 . 1T H E S I N E A N D C O S I N EF U N C T I O N S 431

10.1.THE SINE AND COSINE FUNCTIONS

F i g u r e1 0 . 1 . 1

L0.L.LDefinition

In geometry an angleis defined as the union of two rays called the sides, having a cornmon endpoint called the vertex. Any angle is congruent to some angle having its vertex at the origin and one side, called the initial side,lyingon the positive side of the r axis. Such an angle is said to be in standardposition.Figure 10.1.1shows an angle AOB in standard position with AO as the initial side. The other side, OB, is called the terminal side. The angle AAB canbe formed by rotating the side OA to the side OB, arrd under such a rotation the point A moves along the circumference of a circle, having its center at O and radius IOA I to the point B. In the study of trigonometry dealing with problems involving angles of triangles, the measurement of an angle is usually given in degtees. However, in the calculuswe are concernedwith trigonometric functions of real numbers, and we use radian tneasureto define these functions. To define the radian measure of an angle we use the length of an arc of a circle. If such an arc is smaller than a semicircle, it can be considered the by Theorem $aph of a function having a continuous derivative; and so g.r0.z it has length. If the arc is a semicircle or larger, it has a length that is the sum of the lengths of arcs which are smaller than semicircles. Let AOB be an angle in standard position and 16Z'l :1. If s units is the length of the arc oi the circle traveled by point A as the initial side OA is rotated to the terminal side OB, the radianmeasure,f, of angle AOB is given by t:

s

if the rotation is counterclockwise

and t: -s

if the rotation is clockwise

o rr.r,usrnerrow 1: By using the fact that the measure of the length of the unit circle's circumference is 2tr, we determine the radian measuresof the -+T , ,\r , - n%r , and angles in Fig. 10.1.2a,b, c, d, e, and f. They aretzt , *n , tn, respectively.

Ln 2

(d) 10.1.2 Figure In Definition 10.1.1 it is possible that there may be more than one complete revolution in the rotation of. OA'

TRIGONOMETRIC FUNCTIONS o ILLUSTRATIoN 2: Figure

10.1.3a shows such an angle whose radian

measure is +7r,and Fig. 10.1.3bshows one whose radian measure is -*r. An angle formed by one complete revolution so that oA is coincident with oB has degree measure of 360 and radian measure of 2rr. Hence, there is the following correspondencebetween degreemeasureand radian measure (where the symbol - indicates that the given measurementsare for the same or congruent angles): 350" - 2zr rad or, equivalently, 180'- zr rad From this it follows that 1' - r#ozr rad and 180' 1 rad -

57"78'

From this correspondencethe measurement of an angre can be converted from one system of units to the other.

F i g u r e1 0 . 1 . 3

o ILLUSTRATToN 3:

L62" - 162 . tl1-n rad: r%.zrrad - T n5' -

1m" _ 75o n

Table 10.1.1gives the corresponding degreeand radian measuresof certain angles. Table10.1.1

' t

we now define ,n" ,r"" 10'1'2 Definition

; *n

;; tr

""0 "r;""

', &n

"-; tr

;r;r'.;

Ezr

",jr

"""

*o

,5;

Eo

2n

,""

""-0"".Suppose that f is a real number. Placean angle, having radian measure f, in standard position and let point P be at the interseciion of the terminal side of the.angle with the unit circle having its center at the origin. If p is the point (*, y),then the cosinefunctionlJaefinea Uy c o sf : r

10.1 THE SINE AND COSINEFUNCTIONS

and the sinefunction is defined by 0, 1)

ir/z)

F i g u r e1 0 . 1 . 4

sinf:/From the above definition it is seen that sin f and cos f are defined for any value of f. Hence, the domain of the sine and cosine functions is the set of all real numbers. The largest value either function may have is 1, and the smallest value is -1. It will be shown later that the sine and cosine functions are continuous everywhere, and from this it follows that the range of the two functions is [-L, 1]' For certain values of f, the cosine and sine are easily obtained from a figure. From Fig. 10.1.4we see that cos 0: 1 and sin 0: 0, cos*zr: L\/, 0 and sin *z : L, cos rr:-7 and sin 7r:0, and sin 1n: l!2, cosir: Table 10.1'.2gives these values and some cos *zr:0 and sin *rr:-1. others that are frequently used.

+\n +\E 1 +\E +\n -+\n -+\E -1 1 +\E t{z z u - 2 , & 1

-1

n

r;-

An equation of the unit circle having its center at the origin is x2 - cos f and y : sin t, it follows that * y' - L. Because x (1) F i g u r e1 0 . 1 . 5

Note that coszf and sin2 f stand for (cos t)2 and (sin t)2. Equation (1) is called anidentity becauseit is valid for any real number f. Figures L0.1.5and 10.l..6show angleshaving a negative radian measure of lf and corresponding angles having a positive radian measure of t' From these figures we see that . '.i.anii,..., ..,'ffib{.Itli.''l#..'cod.,.*. , sin{*f } +-',,usin:,f

(2)

These equations hold for any real number f becausethe points where the -t) intersect terminal sides of the angles (having radian measures f and the unit circle have equal abscissasand ordinates that differ only in sign. Hence, Eqs. (2) are identities. From these equations it follows that the cosine function is even and the sine function is odd. From Definition 1.0.1.2the following identities are obtained. t

F i g u r e1 0 . 1 . 6

and

Bin:(f,*,2rrr) '= ein f

(3)

The property of cosineand sine statedby Eqs. (3) is called periodicity, which is now defined.

TRIGONOMETRIC FUNCTIONS

L0.L.3Definition

A function f ,t said to be periodicwith periodp * 0 if whenever x is in the domain of /, then x * p is also in the domain of f and

From the above definition and Eqs. (3) it is seen that the sine and cosine functions are periodic with period 2n; that is, whenever the value of the independent variable f is increasedby 2n, the value of each of the functions is repeated.It is becauseof the periodicity of the sine and cosine that these functions have important applications in physics and engineering in connection with periodically repetitive phenomena such as wave motion and vibrations. , we now proceed to derive a useful formula becauseother important identities can be obtained from it. The formula we derive is as fblows: (4)

P1 (cos a, sin a)

x

where a and b are any real numbers. Refer to Fig. 10.1.7 showing a unit circle with the points Q ( 1 , o ) , P r ( c o sa, sin a), Pr(cosb, srn b), Pr(cos(a * b), sin (a * b)), and Po(cosa, -sin a) - Using the notatiot 8 to denote the measure of the length of arc from R to S, we have @, O t r r :n , @ r : b , f jP rPr: a, and

ffi:

l-al: o.Because Gr:@+ffir,

w€have

C4-b*a F i g u r e1 0 . 1 . 7

and because PnP,- a

frr:&+OFr, +b

it fotlowsthat (5)

From Eqs. (5) and (5) we seethat @, :'pnFr;therefore, the length of the joining the p_oints-Qand Ps is the same as the length of the chord :hg1d joining the points Pn and Pr. squaring the measuresof tfiese lengths, we have

loP;|,:l4J'l

(7)

Using the distance formula, we get

IOF;!,: [cos(a + b) - 1], + [sin(a * b) - 0], : cosz(a f b) - 2 cos(a+ b) + L * sin2(a+ b) and because cosz(a * b) + sinz (a * b) : 1,, the above may be written as

l84l':2-2cos(a+b) Again applying the distance formula, we have l@rp:

(cosb-cos a)2+ (sin b+sin a),

(8)

I O . 1T H E S I N EA N D C O S I N EF U N C T I O N S 435 :

cosz b - 2 cos a cos b + cosz a

* sin2 b * 2 sin a sinb*

sin2 a

and because coszb * sin2 b : 1.and cos2a * sinz a: l, the above becomes (9) cosa cosb * 2 sir.a sinb lF/,l':l-2 Substituting from Eqs. (8) and (9) into (7),we obtain 2 - 2 cos(a+ b) : 2 - 2 cosn cos b * 2 sin a sin b from which follows cos(a * b) : cos a cosb sin a sin b which is formula (4). (a) by substiA formula for cos(a - b) can be obtained from formula tuting -b

for b. Doing this, we have cos4 cos(-b) - sin a sin(-b) c o s( a * ( - b ) ) : - -sin b, we get Because cos(-b) - cos b and sin (-b) cssta *',b}"*

s'.n Gosl'li'* sin at'lbin h'

(10)

for all real numberc a NtdbLetting a:Lrr in formula (1'0),we have the equation,cos(Ln-b): : sin *zr : L' it .or J"'-"ol a + ,it Lrr sin b; and'becausecos *rz 0 and - b: c' b".orr,"s cos(*zr- b) : sin b. Now if in this equation we-let $n c and we obtain cos c: sin(*n- c)' We have therefore then b :irrproved the following two identities.

-*$r

(11)

Formulassimilarto(4)and(10)forthesineofthesumanddifference from those we have. A formula for the sine of of two numbers follow ""rity from (11) the sum of two numbers follo*, from (10) and (11)' We have -b) s i n ( a* b ) : c o s ( i z - ( a * b ) ) : c o s ( ( * z t - a ) With (10) it follows that sin(a * b) : cos(tz a) cos b * sin(tzr a) sin b and from (L1) we get

$D' for all real numbers a and b. we To obtain a formula for the sine of the difference of two numbers have we write a -b as a+ (-b) and apply formula (12)' Doing this' sin(a * (-b)) : sin a cos(-b) * cos a sin(-b) :-sin b' we obtain and becausecos(-b) : cos b and sin(-b)

436


(13) :h is valid for all real numb ers a and b. Bylettinga-b- f informulas (L2) and (4) , respectively,we get the (14)

(ls) Using the identity sin2 t + cos2 t -

I we may rewrite formula (15) as

cos2t:2coszt-1,

(16)

or as c o s2 t : 1 , - 2

s i n 2f

(r7)

Replacingt by tt in formulas (15) and (lr) and performingsome algebraicmanipulations, we get, respectively, (18) and

(1e) (19) are identities because they hold for all

real numbers t. By subhacting the terms of formula (4) from the corresponding terms -of formula (10) the following is obtained: sin a sin g:{[-cos(a

* b) * cos(a- b)]

e0)

By adding corresponding terms of formulas (4) and (10) we get cosd cos 6:glcos(a + b) + cos(a- b)l

er) and by adding corresponding terms of formulas (12) and (13) we have sin a cos6:f fsin(a + b) + sin(a _ b)] ez) Bylettingc:a * b and d:ab informulas (22), (21),and (20),we obtain, respectively,

sinc+ sind-2 sin rycos +

(zt1

c o sc + c o s d : 2 cos c + d cos c - d 2 T

(24)

and c o s c - c o s d : - n -z

sln

c*4^r^c-d 2

Sln

2

(2s)

10.1 THE SINE AND COSINEFUNCTIONS

If the terms of formula (13) are subtracted from the corresponding terms of formula (L2) and a * b and a - b arereplaced by c and il, tespectively, we have

P(x,v) Pt (rt , at)

s i n c - s i n d - 2 c o s+ sz i2n

F i g u r e1 0 . 1. 8

v 7

P(

o

i

1

/ \

_L

4

(26)

You are asked to perform the indicated operations in the derivation of some of the above formulas in Exercises7 to 6. When referring to the trigonometric functions with a domain of angle measurements we use the notation f, to denote the measurement of an angle if its degree measure is 0. For example, 45ois the measurementof an angle whose degree measure is 45 or, equivalently, radian measure is *zr' Consider an angle of d|oin standard position on a rectangular cartesian-coordinate system. Chooseany point P, excluding the vertex, on the terminal side of the angle and let its abscissabe r, its ordinate be y, and lOf l : r. Referto Fig. 10.1.8.The ratios xly andy/r ateindependentof the choice of P becauseif the point Pr is chosen instead of P, we see by Fig. 10.1.8that xlr: xlrl and ylr: alrr Becausethe position of the terminal side depends on the angle, these two ratios are functions of the measurement of the angle, and we define cos 0o:x

rr

and

sin 0o:Y

(27)

Becauseany point P (other than the origin) may be chosen on the terminal side, we could choose the point for which / : 1, and this is the point where the terminal side intersects the unit circle * * y': 1 (see Fig. 10.1.9).Then cos gois the abscissaof the point and sin 0ois the ordinate of the point. This gives the analogy between the sine and cosine of real numbers and those of angle measurements.We have the next definition.

.9 Figure10.1

10.1.4 Definition

If a de'greesand r radians are measurements for the same angle, then cos co : cos r and sin ao: sin t

1-0.1. Exercises (10). 1. Derive formula (20) by subtracting the terms of formula (4) from the corresponding terms of formula 2. Derive formula (21) by a method similar to that suggestedin Exercise 1. 3. Derive formula (22) by a method similar to that suggestedin Exercise 1. 4. Derive forrrula (23) by using formula (22). 5. Derive formula QD by using formula (21). 6. Derive formula (25) by using formula (20). I. Derive a formula for sin 3t in terms of sin f by using formulas (72), (74), (15), and (1).

438


8 . Derive a formula for cos 3f in terms of cos f. Use a method similar to that suggested in Exercise7. 9 . Without using tables, find the value of (a) sin #r

and.(b) cos gz.

10. Without using tables, find the value of (a) sin *z and (b) cos *rr. 1 1 .Without using tables, find the value of (a) sin ttn and.(b) cos tzr. T2, Express each of the following in terms of sin f or sin(iz.-t): (a) sin(Ezr-t); (d) cos(Bzrf f).

(b) cos(gzr-

(c) sin(Bzr+ t);

13. Express the function values of Exercise12 in terms of cos f or cos($z - f). t4. Expresseachof the following in terms of sin f or sin (tn - t): (a) sin(z - t); (b) cos(z - f); (c) sin(z + l); (d) cos(z * f). 15. Express the function values of Exercise 14 in terms of cos f or cos(*a. - f).

rc.Find all values of.t

for which (a) sin f :0,

and (b) cos f :0.

17. Find all values of f for which (a) sin f : 1, and (b) cos f : 1. 18. Find all values of f for which (a) sin f : -1, and (b) cost: -1. 19. Find all values of t for which (a) sin f : *, and (b) cos t: i. 20. Find all values of t for which (a) sin t: -+\/r, and (b) cos f : -Lt/-2. 21. Suppose/ is a function which is periodic with period 2r , and,whose domain is the set of all real numbers prove that / is also periodic with period -22. 22. Prove that the function of Exercise 21 is periodic with period 2nn f.orevery integer n. (Hrr.rr: Use mathematical induction. ) 23. Prove that if f is defined by f (x) - )c- [r]] , then ,t periodic. What is the smallest positive period of f? f

1.O.2DERIVATIVES OF THE SINE AND COSINE FUNCTIONS

Before the formula for the derivative of the sine function can be derived we need to know the value of ,.

sin f

1iT t

Letting f(t): (sin t)lt, we see that /(0) is not defined. However, we prove that lim /(f) exists and is equal to L. 10.2.LTheorem

lim t-O

PRooF: We first assume that 0 < t < tn. Refel to Fig. 10.2.7,which shows the unit circle 12*yr:1 and the shaded sector BOP, where B is the point (1, 0) and p is the point (cos f, sin f). The area of a circular sector of radius r and central angle of radian measure f is determined by trzt; and so if s square units is the area of sector BOP,

s -tt

OF THE SINE AND COSINE FUNCTIONS 10.2 DERIVATIVES

439

consider now the triangle BoP, and let K1 square units be the area of this triangle. Hence,

K r:t2l F l ' IOB- I:j( sint) ' ( 1) :f sint The line through the points o(0,0) sin f/cos f; therefore, its equation is

( 2)

and P(cos f, sin f) has slope

sin f x u:----cost " B (L, o)

F i g u r e1 0 . 2 . 1

This line intersectsthe line x:1 at the point (1, sin tlcos t), which is the point T in the figure. Letting K, square units be the area of right triangle BOT, we have

K,:+lBr |. 16|:+'#l' ! : +.#+

(3)

From Fig. 1'0.2.1'we see that K1

(4) , we get Substituting from Eqs. (1), (2), and (3) in inequality L

L

T s i nt < ; t . ; ' f f i

sinf

pos: Multiplying eachmember of the above inequality by Zlsin f , which is itive because0 < f < *n, we obtain r1

1'

"inT';o;7 Taking the reciProcal of each member of the above inequality and reversing the direction of the inequality signs, we get

(s)

si+f .1 c o st <\ t \ From the right-hand inequality in the above we have

(5)

sint
1', - s ol n e ,'

Replacing t by *t in inequality

(7) (5) and squarifr1, we obtain

sinztt < It' Thus, from (7) and (8) it follows that 1-cosf

2

,.-Sz -z

(8)

440


which is equivalent to 1 , - t t ' < c o sf From (5) and (9) and because0

1- !*.#.1

iro

(10)

< t < 0, then 0

If 4n

1 - + e 1 r . s ifn<(1r-) But sin(-f)

i f - i n1. t < 0

sin f; thus, the above can be written as

1-|,'.ff

(11)

From (10) and (11) we conclude that

1-I*.#.

1 ir-+,

t

1 -

t.

Er

r.\,

0

(r2)

2

Becauselim (1 - +t') - 1 and lim 1 - 1,, it follows from (I2) and the t-O t-O

squeeze theorem r.

llmL4 t-O

(4.3.5) that

sin f _1 L

We wish to write the quotient sin 3xlstn 5x in such a way solurroN: '/-.0.2.1. that we can apply Theorem We have, rf x # 0,

As r approaches zeto, so do 3x and Sx. Hence, w€ have 1. sin 3r ltm#:lim Jx i-o

sin 3r - 1 ,. sr-o 3x

and r. 1. sin 5r hm#:lim--1 r-0

5X

sin 5r

b.llr o

5X

Therefore,

singx:'l$ ff) rim ;-:o sin 51

5r) 5 hm fg" r-o 5X \

/

:1.1 . 1 :g 5

5

FUNCTIONS441 OFTHESINEANDCOSINE 10,2DERIVATIVES From Theorem 10.2.1 we can prove that the sine function and cosine function are continuous at 0. 10.2.2 Theorem

the

The sine function is continuous at 0. pRooF: We show that the three conditions necessaryfor continuity at a number are satisfied. (i) sin0:0.

( i i ) l i m s i n f : l i m 4 l t' f

:lim+

,-0

r-o

t-o

t

l i m f: 1 ' 0 : 0 . t-o

(iii) lim sin.f : sin 0. Therefore, the sine function is continuous at 0' 10.2.3 Therirem

I

The cosine function is continuous at 0. PROOF:

( i ) c o s0 : 1 . (ii) lim cosf : lim VT- sinu t-o

t-o

: Vii* (r -i.t)

: vIlI-:1.

sin4 since cos f ) 0 when Nore: We can replace cos f by \G <0. < t
I

The limit in the following theorem is also important. It follows from previous three theorems and limit theorems. the

10.2.4Theorem

ltS T{:

o

PROOF:

(1 - cosf)(1 * cost) ,,_ 1 cosf _,,_ :nm@

111. ,

-: ,'- k-gos2l)-

lS 16+ cosrt

:U*''.sin,f_ r--o-f(l * cos f)


:+$,t* l'-t@ By Theorem 10.2.'l.,lLTtri" tlt):1,

and becausethe sine and cosine

functions are continuous at 0, it follows that Tgor[sin tl(L+ cos f)]: o/(1 + 1) :0. Therefore, L-cosf

l'-TT:o

^

I

/

We now show that the sine function has a derivative. Let f be the function defined by /(r) : sin r From the definition of a derivative, we have f

'

'(x) - lim Lt-O

-

f(x*Ar) -f(x) A,x

r:_ sin (x * Ax) - sin r

Ilm Atr-o

Af

- sin r - lim sin r cos(Ar) * cos r sin(Ar) Lx

A.r- 0

,:_ :ttm A.r-o

sin rfcos(Ax) - 1] , r:- cosx sin(Ar) rufrrAJr*o

LX

LX

(riq sin')* (*+ cos')g* # 1, (by Theorems10.2.4and 1,0.2.1)

Therefore, we have the formula x) : cos r (13) "(sin If u is a differentiable function of x, we have from (13) and the chain

D

rule (14) The derivative of the cosine function is obtained by making use of (13) and the following identities: cos x - sin (in - x) D

"(cos

and

sin x:

r) : D *[sin (in - r) ]

cos(tn - x)

10.2 DERIVATIVES OF THE SINE AND COSINEFUNCTIONS 443

: cos(+zr- x) ' D,(*r - x) : sin r(-1.) Therefore, D'(cos r) : -sin r

(15)

So if. u is a differentiable function of x, it follows from (L5) and the chain rule that (16)

ExAMPLE 2:

Given

SOLUTION:

( 1 - 2 c o sr ) D " ( s i n r ) - s i n r ' D ' ( 1 - 2 c o sr ) -2cosx)' f'(x) (1

(L - 2 cosr) (cosx) - sin r(2 sin r) ( 1 - 2 c o sx ) ' cosr-2(cos2r*sin2r) ( 1 - 2 c o sx ) ' c o sx - 2 ( 1- 2 c o s x Y

ExAMPrn3: Given y : (1 + cos3x,)n find dyldx.

EXAMPLE

du -dx- 4 ( 1+ cos 3x2)t (- sin 3x2)(6r) - -24x sin 3x2(1 + cos 3x')t

Given

r cos f y c o s x : t find D,A.

SOLUTION:

solurloN:

Differentiating implicitly with respect to )c,we get L ' c o sy * x ( - s i n y ) D * A * D r y ( c o s x ) * y ( - s i n r ) : 0

D * A ( c o s) c - r s i n y ) - y

sinx- cosy

srnx-cosY Dw * t"r - Y c o sx - x s l n y

Geometric problems involving absolute extrema occasionally are more easily solved by using trigonometric functions. The following example illustrates this fact.


EXAMPLE5:

A right-circular cylinder is to be inscribed in a sphere of given radius. Find the ratio of the altitude to the base radius of the cylinder having the largest lateral surface area.

soLUrIoN: Refer to Fig. 10.2.2.The measure of the constant radius of the sphere is taken as a. Let 0: the number of radians in the angle at the center of the sphere subtended by the radius of the cylinder; r: the number of inches in the radils of the cylinder; h: the number of inches in the altitude of the cylinder; S : the number of square inches in the lateral surface area of the cylinder. '1,0.2.2, From Fig. we seethat r:Asin0

and h-2acos0

Because S : 2rrrh, we have S - 2n(a sin e)Qa cos 0) :2rA2(2

sin g cos 0) :Znaz sin 20

DsS:4na2 cos20 and DszS:-8na2 sin20 Setting Dfi:0,

we get

c o s2 d : 0 Figure10.2.2

Because 0 < 01trr, 0: i,n We apply the second derivative test for relative extrema and summarize the results in Table 10.2.1,. Table10.2.1 DeS

0:In

0

Dr'S

Conclusion

S has a relative maximum value

Becausethe domain of 0 is the open interval (0,tn),the relative maximum value of S is the absolute maximum value of S. When 0 : *r, r: t\Da, and h : l/-2a.So for the cylinder having the largestlateral surface area,hlr:2. BecauseDr(sin r) : cos r and cos x exists for all values of r, the sine function is differentiable everywhere and therefore continuous everywhere. Similarly, the cosine function is differentiable and continuous everywhere. We now discuss the graph of the sine function. Let f(x):

sinr

OF THE SINE AND COSINEFUNCTIONS 445 10.2 DERIVATIVES

Thenf'(r) : cos xandf" (r) :-sin r. To determinethe relativeextrema, -r2, . . . . If. n is we set f' (x) :0 and get x:$rr * ntr, where rr:0, *7, and [tr:-L; an even integer, then /" (I2n* nn) =-sin(iz * nn\:-sin *nr):-sin *n-l. if n is an odd integer, f "(*o*nn):-sin(*zr Therefore,/ has relative extrema when x : lr * nrr; and if n is an even integer,f has a relative maximum value; and if n is an odd integer, / has a relative minimum value. To determine the points of inflection of the graPh, we set f " (x) :0 . Becausef" (x) changes and obtain x: nnr, where n:0, !L, -r2, . sign at each of these values of x, the graph has a point of inflection at each '(x) : point having these abscissas. At each point of inflection f co's nn:-f].. Therefore, the slopes of the inflectional tangents are either*1 or-1. Furthermore,becausesin(r * 2rr): sin x, the sine function is periodic and has the period 2rr. The absolute maximum value of the sine function is 1, and the absolute minimum value is -1. The graph intersects the r axis at those points where sin r: 0, that is, at the points where x: nzr(n is any integer). From the information obtained above, we draw a sketch of the graph of the sine function; it is shown in Fig. 1.0.2.3'

y:sinr

Figure 10.2.3 To obtain the graph of the cosine function, wp use the identity cosr:

sin(*z * r)

Hence,the graph of the cosine which follows from formula (12)of Sec.1.0.1.. function is obtained from the graph of the sine function by translating the y axis*z units to the right (seeFig. 10.2.a).

q -lf

2

--

7r

y:

Figurc 10.2.4

,;-$bi'

-

cos.r

446


Exercises 10.2 In Exercises 1.through 8, evaluate the limit, if it exists. sin 4r 4 ,. I. lrm

2. lim

X

3.-O

iin - x !__

4. lim

1-cosr F ,. J. rlfrr r-o x2

COS I

s-nl 2

T.lim = r-o

t*'==

1-cos2r ., r. O. uFr 2x2 r-o

1-cos2*r

f-A

x

,r-0 COS.f

In Exercises9 through 24, find the derivative of the given function. 9. f (x) - 3 sin 2r 10.f (x) : cos(3x'+ 1) 1 2 .f ( t ) : c o sP

13.h(t) : 2 cos\tr

15. f (x) : ln sin 5r

16.F(x): sin(lnr)

1 1 .S @ ) : s i n3 x c o s3 r f..,, l + . 8 ( x ): \ f f i r 1 7 .h ( x ) : - ? t nt r 2 + c o st r

18.g(t) : sin(cosf) 20. h(y) : y'22. S(x) :

L9. S@) - 2 cos r sin 2r - sin x cos2x

yz cosy * 2y sin y + 2 cos y

tl

23. h(x) - (cosr)sinr

asinc

21,.f ( x ) :

(sinxz)a*

24. f ( x ) :

(ln sin r)"'

In Exercises 25 and 26, use logarithmic differentiation to find dyldx. .rF zc. y:

sinrffi

2 a"' - : g

V1 -3cosx

In Exercises27 artd 28, find D,y by implicit differentiation. *4.

V: cos(r- y)

28. cos(r + il : y sin x

In Exercises29 through 34, draw a sketch of the graph of the function defined by the given equations. 2 9 .f ( x ) : s i n 2 r 30. 3L. f (x) : lcos3rl rsinr

3 3 .f ( x ) : l

lsin(r-tn\

f(x)-zsin3r 32.f (x) - +(1- cosr)

if0
.

3 4 .f ( x' ) : l

lcosr

if-ir
35. If a ladder of length 30 ft which is leaning against a wall has its upper end sliding down the wall at the rate of tftlsec, what is the rate of change of the measure of the acute angle maae by ttre ladder wiih the ground when the uppei end is 18 ft above theground? 36. The cross section of a trough has the shape of an inverted isoscelestriangle If the lengths of the equal sides are 15 in., find the size of the vertex angle that will give maximum capacity for the trough. ;p. tf a }oay of weight W lb is dragged along a horizontal floor by means of a force of magnitude F lb and directed at ai angle of 0 radians with the plane of the floor, then F is given by the equation -.:

o* ksin0*cos0

where k is a constant and is called the coefficient of friction. Find cos 0 when F is least.

10.3 INTEGRALSINVOLVINGPOWERSOF SINE AND COSINE

447

38. Find the altitude of the right-circular cone of largest possible volume that can be inscribed in a sphere of radius a units. Let 20 be the radian measure of the vertical angle of the cone. A partide moving in a straight line is said to have simple harmonicmotion if the measure of its accelerationis always proportional to the measure of its displacement from a fixed point on the line and its accelerationand displacement are oppositelydirected.Showthatthestraight-linemotionof aparticledescribedbys: Asin2r.ktlBcos2rkt,wheresft is the directed distance of the particle from the origin at f sec, and A, B, and k are constants, is a simple harmonic motion. Show that if a partide is moving along a straight line according to the equation of motion s : a cos(kf * d) (whqre a, k, and 0 are constants) and s ft is the directed distance of the particle from the origin at f sec,then the motion is simple harmonic. (See Exercise39.) 4 L . Given /(r) : x sin(nlx), prove that I has an infinite number of relative extrema. 42. If the domain of the function of Exercise41 is the interval (0, 1], how should/(0) be defined so that / is continuous on the closedinterval [0,1]?

1.0.3INTEGRALS INVOLVING POWERS OF SINE AND COSINE

The formulas for the indefinite integral of the sine and cosine functions follow immediately from the corresponding formulas for differentiation. (1) PRooF: Du(-cos u) : sin z.

(2) PRooF:

EXAMPLE

Find

Du(sin u) :

solurroN:

cos y.

Let u: ln r. Then du: dxlx; so we have

f f cos(lnr) , l--ffrdx:fcosudu JXJ :sinu+C : sin(ln r) + C

EXAMPLE 2: Find

solurroN:

Let u-

f sinxdx

L - cos r. Then du:

f du

J1-cosr: J u

l " l+ c :ln11 -cosrl +C

sin x dx; so

448


EXAMPIn

3:

Find

fn

| ( 1+ sin r) dx

* sin x) dx - x - cos

JO

:Tr-cos

(0 - cos0)

:rr+L-

1)

:Tr+2 We now consider four casesof integrals involving powers of sine and cosine.The method used in each caseis shown by an illustration. Case1: J sinn u du or J cos, u du, whercn is an odd integer. . rLLUsrRATrow L: ff

I cos3r dx:

JJ

I cos2r (cos x dx)

I = | (f - sin2x)(cos x dx) J

So

.or' r ar: I cos x dx- f ,rr,,x cosx dx

Jf J J

(3)

To evaluatethe second integral on the right side of (3), we let a: sin r; then du: cos r dr. This gives

I

ff

I sin2r (cos r d.x): I u2 ilu

JJ

=tu3*C1 =$sin3x*C1 Becausethe first integral on the right side of (3) is sin r * Cr, we have f

J EXAMPTN 4:

/

cos3xdx:sinx-*sin3r*C

Find

sin' x d x

(sin2 x)' sin x dx (1 - cos2x)' sin x dx (1-2coszx+

cos4r) sin xdx

.

10.3 INTEGRALSINVOLVINGPOWERSOF SINE AND COSINE

Therefore, ffrf I sinsr dx: I sinxdx-2 JJJJ

449

| cos2xsinxdx+ | cosar(sinrdr) (4)

To evaluate the secondand third integrals in (4), we let a : cos r and du: -sin x dx; and we obtain Isintr dx:-cosx+Z Iu2du- [unau JJJ :-cosx*&u3-!u5IC r * fi cossr - $ cossr * C

:-cos

Case2: J sinn u du and J cosnu du,wheren is an even integer. The method used in CaseL does not work in this case.We use the following formulas which are obtained from formulas (19) and (18) of Sec. L0.1by substituting 2x for t. . o L-coszx sln- x: 2 s, L + cosz)c c o s -x : 2 O ILLUSTNATION

2:

f

f r

sin'xdx- l1-cos2x rf Jfdx tx - * sin 2x * C EXAMPTN 5:

Find

SOLUTION:

f

/

f

q, \o , : f /1+cos2x\2 , xdx: (cos2 dx x)2dx lcos4 J J (T/

.oro x d x

-1

t

4J

dx*;

1 r

n c ?2vx; vd+xl * i

lcos

: i x1+ ] s t n1 2 x

f

2xdx Jcos2

+1f_t-cos4.x, 4J fax

- t x + + s i n2 x * t r + # s i n 4 x * C - 8x + * sin 2x * # sin 4x * C Case3: I sinn r cosmx dx, where at least one of the exponents is odd. The solution of this case is similar to the method used for Case '/-.. .

ILLUSTNAUON

I

3:

r sin3 x cos4

sin2 r cos4r(sin x dx)

T R I G O N O M E T R IFCU N C T I O N S

:

| (1 cos2r) cosax(sin x dx)

J

ff

I cosar sin xdxJJ

| cos6r sin xdx

:-*cos5x**cos7x*C

.

Case4: t sinn x cos' x dx, where both m and n are even. The solution of this caseis similar to the method used in Case 2. O ILLUSTRATI f

I sin2r J

oN 4:

cosaxdx:f(ry) 1f

(=*)'o*

1f

aO J I d* +* O J | cosZxdx

- i JL r c o s22x d x -1i Jr c o s2sx d x

:* x*+sinzx-+/#0. -

1f

(1 - sin22x) cos2x dx

i J

sin4r : xg - ,r sin2x - x 1,G- 64 L6 - L r cos2x dx+ l f sin22x cos2x dx ; J i J _

x *

sin 2x

G+T

x , sin32x 164g

ExAMPtn6:

Find

I sinn t cosax dx J

sin 4r

sin 2x__ sin3 2r

+6+c

sin4r . A Tt64

o

soLUTroN: If we make use of the formula sin r cos r : { sin 2x, we have, 1 f I sinnr cosaxdx : i

J

lsinnzxax

: t 6LJ \f

/1,- cos 4r\2 ,

z 1o' 1f :A 7 r dr-i r f cos4xdx+fr cos2 4xdx J J J _tc _sin4r * 1 [ 1*cosStr_ *n 64 7zg 64J 2

10.3INTEGRALS POWERS INVOLVING OF SINEAND COSINE :- x 64

sin4r L28

,

sin 4r r28

: - 3x L28

x , sinSr , A '. ' v L28 1024 sin 8r . r. I \/ 1024

The following example illustrates another type of integral involving a product of a sine and a cosine. EXAMPLE7:

solurroN: We use formula (22)of Sec.10.1,which is

Find

r I sin 3x cos 2x dx J

s i na c o s b - * s i n ( a - b ) + + s i n ( a + b ) So ff

I sin 3r cos2x dx - I t+ sin x +t sin 5rl dx

J

10.3 Exercises

ir. ,,r. r. ':* ::t',)))!l:J,::

f_ I J

In Exercises L throu gh 22, evaluate the indefinite integral.

1. 5.

I(3sin /

g. /

x*2cos x)dx

2.

f

I

(sin 3x * cos2x) dx

4

sinxdx

10.

/.oru

7.

1,4. ,ir,' 2t cosa2t dt /

L5.

,r Ii##d.

18.

L9.

2!.

I

(sin 3x - sin 2r)2dx

/

11.

x dx

13. f sint xcoszxdx J

/

,i"

4.

x eeos'dx

/

f cos366* *Jrc

,i" r . sin(cos x) dx

3.

,i" 2x cos4x dx

/ /

/

8.

,ir,' x dx .or' tx dx sin' 3f cosz3t dt

.o, 4r cos 3x dx

22. [ ,ir, r sin 3r sin 5x dx J

In Exercises 23 through 28, evaluate the definite integral.

24.

la;it,, _

lr'^,e'

sine' dx

cossx dx

f

cosr

l2ffidx

r cossx dx I

L2. I sint x cos' x dx J f 16. {cos z sin3 z dz J


26.

fr

|

,ir,- inx dx

27. I sin2 zrx coszrrx dx Jo

If z is any positive integer, prove that f"

"rn,

Jo

If n is a positive odd integer, prove that

l

28.

f rl6

J,

sin 2x cos4x dx

nx dx : Ln.

"o"^*

dx:0.

In Exercises31 through 33, m and rr are arly integers except zero; show that the given formula is true. ft , [ o i f m #' n 3 1 . I cos nTTxcos mTTxdx - { : :-, "' J_r

Lr

fl f0 ifm*n 33. I Srnnix srn mTx.dx: 1J-t

LI

fr 32. I cos nTTx srn mTTxdx - 0 J-r

fim-n

lIm:

n

34. If q coulombs is the chargeof electricity received by a condenserfrom an electric circuit of i amperesat f sec,then 1: Dt1. Supposei:5 sin 50t and 4:0 when t:tr, find the greatestpositive chargeon the condenser. 35. In an electric circuit supPosethat E volts is the electromotive force at f sec and E:2 from f: 0 to f: *zr.

sin 3f. Find the averagevalue of E

For the electric circuit of Exercise35, find the square root of the averagevalue of E2from f:

0 to t:

*r.

Find the area of the region bounded by one arch of the sine curve. Find the volume of the solid of revolution generated if the region of Exercise 37 is revolved about the r axis. Find the volume of the solid generatedif the region bounded by the curve y : sin2 r and the r axis from x: 0 to x: r is revolved about the line y: 1. 40. Find the area of the region bounded by the two curves y : sin x andy : cos x between two consecutivepoints of intersection.

10.4 THE TANGENT, COTANGENT, SECANT, AND COSECANT FUNCTIONS 10.4.1 Definition

The other trigonometric functions, tangent, cotangent, secant, and cosecant,are defined in terms of the sine ind cosine.

The tangenf and secantfunctions are defined by : : : : :i :, , ' ' . . i .f ' +, , S' G :: C,, :#: CSS:ff,

for all real numbers r for which cos r # 0. The cotangentartd cosecantfunctions are defined by

*"*'**;ffi for all real numbers x fot which sin x * 0. The tangent and secant functions are not defined when cos x:0, which occurls when r isir,tr, orin * nrr,wheren is any positive or nega-

10.4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS

tive integer, or zeto.Therefore, the domain of the tangent and secantfunctions is the set of all real numbers except numbers of the form!-r * nrr, where z is any integer. Similarly, becausecot x and csc r are not defined when sin x: 0, the domain of the cotangent and cosecantfunctions is the set of all real numbers except numbers of the form ntr, where n is any integer. By Theorem 2.6.1.(iv),if the functions / and g are continuous at the number a, then f lg is continuous at a, provided that g(a) + 0. Becausethe sine and cosine functions are continuous at all real numbers, it follows that the tangent, cotangent,secant,and cosecantfunctions are continuous at all numbers in their domain. By using the identity (1)

cos2r*sinzr:l

and Definition 10.4.1,we obtain two other important identities. If on both sides of (1) we divide by cos2r, when cos x + 0, we get cos2r, sinzt 1 ;oF;"os';:;;t; and becausesin x/cos r: tan r and 1/cos r:

secr, we have the identiW (2)

By dividing on both sides of (1) by sinz x when sin x + 0, we obtain in a similar way the identity (3) Three other important formulas follow immediately from Definition 10.4.1..They are (4) (5) and (6)

Formulas (4), (5), and (6) are valid for all values of x for which the functions are defined, and therefore they are identities. Another formula that we need is one that expressesthe tangent of the difference of two numbers in terms of the tangents of the two numbers. By using formulas (13) and (10)from Sec.1.0.L,we have tan (a - b) : Dividing

sin(a - Ut) cos(a - b)

sinacosb-cosAsinb

c o sa c o s b + s i n a s i n b

both the numerator and denominator by cos a cos b, we get


tan(a - b) :

sin a cos b cos a cos ? cos a cos b cos a cosb

cos a sin b cgs a cgs -b smj sin b cos a cos b

sin a

sin b

cos b cos a ,, , sin a sin b rT 'cos4cosb

Therefore, we have the identity tan a-tanb

(7)

L + tan a tanb Taking a - 0 in formula (7) gives tan(0- b) -

tan0-tanb 1+tan0tanb

Because tan 0 - 0, we have from the above equation . , , : . : ; ; . . . . , . '.....,:- : : , , .

,,,..,.

.,1 ...,,

lfffi{*;,b),,*:,'- tan'b

(8)

Therefore, the tangent is an odd function. Takingb_-b tn formula (7) gives tan a-

tan (a - (-b) ) :

tan(-b)

L+tan atan(-b) and replacing tan(-b) by -tan b, we get the identity

(e) The derivatives of the tangent, cotangent, secant, and cosecantfunctions are obtained from those of the sine and cosine functions and differentiation formulas. D

"(tan

cos r ' D, (sin r) - sin r . Dr(cos r)

r)

C OS2 T

C OS2 T

c o s zx + s i n 2 x

coszr

-

L

cos,t

: sec2r So we have Dr(tan r) : sec2r

(10)

It u is a differentiable function of r, then from (10) and the chain rule

10,4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS 455

we obtain (11)

ExAMPr.rL:

Find f '(x) if

f(x)-2tantx-x

solurroN: f'(x) -2 secztx(t) - 1 : s e c 2 t x - 1 . If we use identity (2) , this simplifies to ' f (x) - tanztx

The formula for the derivative of the cotangent function is obtained in a manner analogous to that for the tangent function. The'result is D"(cot x) : -csc2 x

(r2)

The derivation of (1,2)is left as an exercise (see Exercise L).

From (12) and the chain rule, 7f u is a differentiable function of x, we have (13)

D

x) : D

"(sec

-t] r) "l(cos -'(sin r) L ( cosx)

- - - - ' SI r n f

cos- r _

1 . sin.r cos I cos I

:secxtanx We have, then, the following formula: (14) x) : secx tan x "(sec lf u is a differentiable function of x, then from (14) and the chain rule

D

(ls)

EXAMPLE 2:

f(x):

Find f

s e c a3 r

'(x)

if

SOLUTION:

' ' f (x) 4 sec33x D"(sec 3r) - 4 sec83r(sec 3r tan 3r) (3) : 12 seca3x tan 3x

456


In a manner similar to that for the secant,the formula for the derivative of the cosecantfunction may be derived, and we obtain (15) D"(csc r) : -csc x cot x So by applying (16) and the chain rule, if u is a differentiable function of r, we have D"(csc a) + *-csc u cot u D*u

(r7)

The derivation of (16)is left as an exercise(seeExercise2).

EXAMPLE

Find f '(x) if

f (x) cot r csc r

SOLUTION:

f'(x) .. : :": r,! ;:: ; ;:; ",: =-csc x cotz x

' D"(cot r) x(-csc2 x)

cscsr

In Definition 10.1.4 we gave the analogy between the sine and cosine of real numbers and those of angle measurements. The following definition gives a similar analogy for the other four trigonometric functions.

10.4.2 Definition

If a degrees and r radians are measurements for the same angle, then tan a": tan r

EXAMpLE4: An airplane is flying west at 500 ftlsec at an altitude of 4000 ft. The airplane is in a vertical plane with a searchlight on the ground. If the light is to be kept on the plane, how fast is the searchlight revolving when the airplane is due east of the searchlight at an airline distance of 2000 ft?

cot a": cot r

secao: sec.x

csc ao: cscr

sor,urroN: Refer to Fig. 10.4.1.The searchlightis at point L, and at a particular instant of time the airplane is at point P. Let f : the number of secondsin the time; r: the number of feet due east in the airline distance of the airplane from the searchlight at time f sec; 0 : the number of radians in the angle of elevation of the airplane at the searchlight at time f sec. given We are Dg: -500. We wish to find D10when x: 2000.

tan o -

4ooo x

Differentiating

on both sides of Eq. (18) with respect to t, we obtain

s e c02D , e - - r y D t x Substituting D'rx: -500 in the above and dividing by secz0 D10:

(18)

2,000,000 x2 secz 0

10.4 THE TANGENT,COTANGENT, SECANT,AND COSECANTFUNCTIONS

Whenx:2000, tan 0:2. Therefore, secz0: L * tan20:5. Substituting thesevaluesinto Eq. (19),we have,when x:2000, Irta:

2,000,000 : 7 4roopoo(s) 10

We concludethat at the given instantthe measurement of the angleis increasing at the rate of rb rad/sec, and this is how fast the searchlight is revolving.

L

xft

Figure 10.4.1

We now consider the graph of the tangent function. Becausefrom Eq. (8) tan(-r) : -tan r, the graph is symmetric with respect to the origin. Furthermore, L__r-. , _.,_ sin(r* t€tnlxfTr):-

z)

cos(r * zr) sin r cos rr + cos x sin n cos x cos zr - sin r sin zr

_ sin r(-1) * cos r(0) cos x(-1) - sin r(0) -sin r _ -cos .r :tanr and so by Definition 10.1.3,the tangent function is periodic with period zr. The tangent function is continuous at all numbers in its domain which is the set of all real numbers except those of the form tn * nzr, where n is any integer. Ffowever, tan r: -roo,and so the lines ,_EEr,, having equations x:Lr-lnn are vertical asymptotes of the graph. If n is any integer, sin nn: 0 and cosnT is either *1 or -L, and so tan nn: 0. Therefore, the graph intersects the r axis at the points (nr, 0). To find the relative extrema of the tangent function and the points of inflection of its graph, we consider the first and second derivatives. So if f(x) : tan x, then f'(x) : sec2r and f" (x) :2 seczr tan x. gives seczr:0. Becauseseczx- 1 for all x, we Setting f'(x):0 concludethat there are no relative extrema.Setting f" (x):0, we obtain

4s8


2seczxtan r:0, from which we get tanr:0 becauseseczx # 0. Therefore, f" (r):0 if. x:nn, wheren is any integer. At these values of x, (x) changessign, and so the points of inflection are the points (nn,0), f" which are the points where the graph intersects the r axis. Because '(no) : sec2nnr:'J., the slopesof the inflectional tangentsare L. f Consider now the open interval (-tn, tn) on which the tangent function is defined everywhere. Because/' (r) : sec, r ) 0 for all values of.x, it follows that the tangent is an increasing function on this interval. When -ir < x( 0, x tanr ( 0; hence,the graph is concave f"(*):2secz downward on the open interval (-tn,O). When 0 < x < +7, f" (x) > 0, from which it follows that the graph is concave upward on the open interval (0, *z). T a b l e1 0 . 4 . 1

x tan r

0

*\R - 0.58

1,

tB - l.Tg

In Table 10.4.1there are some correspondingvalues of r and y satisfying the equation y : tan r. By plotting the points having as coordinates the number pairs (r, y) and using the above information, a sketch of the graph of the tangent function may be drawn, and it is shown in Fig. 1,0.4.2.

'1, 2

y:

tanr

Figure 10.4.2

The graph of the cotangent function can be obtained from the graph of the tangent function by using an identity which will now be proved. sin(tz + r) tan(Ln* r) : cos(*z * r)

10.4 THE TANGENT,COTANGENT,SECANT,AND COSECANTFUNCTIONS

sin *zr cos x * cos *n stn x

cos*z cosr - sin tzr sin r -_ (1) cosr * (0) sin r (0) cosx- (1) sin r

-# : -cot r Therefore, cot x: -tan(*zr * r) From the above identity it follows that the graph of the cotangent function is obtained from the graph of the tangent function by translating the y axis *z units to the right and then taking a reflection of the graph with respect to the r axis. A sketch of the graph of the cotangent function is in Fig. 1,0.4.3.

y:

cotr

Figure 10.4.3 The secant function is periodic with period 2zr because sec(r*2zrl:

coslx.+ z7r)

L :secr cos .r

The domain of the secant function is the set of all real numbers except those of the form*rr * nn, and the function is continuous on its domain. Because lim secr:1o, the graph of the secantfunction has the lines I -fr12+nfr

x: tn * nn as vertical asymptotes. There is no intersection of the graph with the r axis because sec r is never zero. The secant function is even because 11 : -----:-: secI sec(-x) cos(-r, cos r

460


and so the graph of the secantfunction is symmetric with respect to the y axis. lf f(x): sec r/ it follows that f' (r): sec x tan r and f,,(x): sec r(2 tan2 x * 1.).Setting f '(x):0 gives sec x tan x:0. Because secx * 0,f '(t) : 0 when tan r: 0, which is when x: nzr,wheren is any integer. Also, f"(ntr1: secnrr[2tanf nzr+1]. If n is an even integer, a n d i f . n i s a n o d d i n t e g e r ,f " ( n t ) : f"(nr):1'(0*1):1' (-1)(0+ 1):-1. Therefore,when x:nn and n is an even integer,/ has a relative minimum value; and when x: nnTand z is an odd integer, / has a relative maximum value. There are no points of inflection because for all x, f " (x) + 0. Using the above information and plotting a few points, we get a sketch of the graph of the secantfunction shown in Fig. 10.4.4. Because sec(r * tn)

:-:-:-uDL,r

1,

cos(r * in)

1, -sin

r

vv

then cscx: -sec(r * $r) and so the graph of the cosecant function is obtained from that of the secantfunction by translating the y axis$r units to the right and taking a reflection of the graph with respect to the r axis. A sketch of the graph of the cosecantfunction is in Fig. 10.4.5.

y:

secr

Figure 10.4.4

v F i g u r e1 0 . 4 . 5

Exercises 1-0.4 1. Derivei Dr(cot x) : -csC r.

2. Derive: Dr(csc x) : -csc r cot x.

TO THESLOPEOF A LINE OF THE TANGENTFUNCTION 10.5 AN APPLICATION

461

In Exercises 3 through 20, find the derivative of the given function. 3. f(x):

4,a

s(r) : ln .rd?* 7. G(r): \ffi

sec12

6 . h ( x ) : l n l c o tt r l 9. sQ)-

5. F(r) : ln lsec2rl 8. h(t): sec22t - tanz2t

10. f (x) : sin x tan x

secx tan x

13. H(t):

1 2 . 8 ( f ) : 2 s e c\ F ln lsec5r * tan 5rl 18. G(r) logrslcscx - cot rl 15. f(x):

cotat - cscat

L5. f (x) : + seca2x - sec2x 'Ig. F(r) - (sin r)tanc

1 1 .f ( t ) : c s c ( f + a 1) 2: 14.F(r) : ":': l*xz L 7 .S ( x ) : 3 '

s e cx

20. G(r) - (tan r)"

In Exercises2l and 22, use logarithmic differentiation to find D ,y . r

sec rVl=6f zly::ft-

22.'y:ffitarf x

In Exercises23 through 26, hnd D"y by implicit differentiation.

: S, , tan(x-rA) '@ t".' x* cs9y:4

24.cotxy t xy:O 26-csc(x-0 + sec(r* !) : a

In Exercises27 through 30, draw a sketch of the graph of the given function.

2 8 .f ( x ) - + s e c 2 x 30.f (x) - Jcsc $rl

27. f (x) : tan 2x

2e.f (x) -- + lcotrJ

fi"a an equation of the tangent line to the curve y: sec r at the point (In, t/-Z). Q 32. A radar antenna is located on a ship that is 4 miles from a straight shore and it is rotating at 32 rpm. How fast is the radar beam moving along the shoreline when the beam makes an angle of 45" with the shore? (g3l, A steel girder 27 frlongis moved horizontally along a passageway8 ft wide and into a corridor at right anglesto the pas-' sageway.How wide must the corridor be in order for the girder to go around the comer? Neglect the horizontal width of the girder. 34. If two corridors at right angles to each other are 10 ft and 15 ft wide, respectively, what is the length of the longest steel girder that can be moved horizontally around the comer? Neglect the horizontal width of the girder.

10.5 AN APPLICATION OF THE TANGENT FUNCTION TO THE SLOPE OF A LINE 10.5.1Definition

The tangent function can be used in connection with the slope of a straight line. We first define the "angle of inclination" of a line.

Tthe angle of inclination oI a line not parallel to the r axis is the smallest angle measured counterdockwise from the positive direction of the r axis to the line. The inclination of a line parallel to the r axis is defined to have measure zero. If a is the degree measure of the angle of inclination of a line, a may be any number in the interval 0 = c ( 180. Figure 10.5.1shows a line L for which 0 ( a ( 90, and Fig. 10.5.2shows one forwhich90 < a < 180.

462


F i g u r e1 0 . 5 . 1

10.5.2 Theorem

F i g u r e1 0 . 5 . 2

If a is the degree measure of the angle of inclination of line L, not parallel to the y axis, then the slope m of L is given by m:

tatt q.o

PRooF: Refer to Figs. 10.5.1 and'1.0.5.2,which show the given line L whose angle of inclination has degree measure a and whose slope is rn. The line L' that passesthrough the origin and is parallel to L also has slope m and an angle of inclination whose degree measure is c. The point P(cos ao, sin ao), at the intersection of L and the unit circle I * yr:1, lies on L'. And becausethe point (0, 0) also lies on L', it follows from Definition 1.5.1that sin co - 0 *:6;A14:;0;;.

sin co : tan 6r-

I

If the line L is parallel to the y axis, the degree measure of the angle of inclination of L is 90 and tan 90odoes not exist. This is consistent with the fact that a vertical line has no slope. Theorem 10.5.2is used to obtain a formula for finding the angle between two nonvertical intersecting lines. If two lines intersect, two supplementary angles are formed at their point of intersection. To distinguish these two angles, let L2 be the line with the greater angle of inclination of degree measure a, and let L1 be the other line for which the degree measure of its angle of inclination is or. If 0 is the degree measure of the angle between the two lines, then we define 0:

e.2- a1

(1)

If L, and L, are parallel, then c1 : c, and the angle between the two lines

10.5 AN APPLICATIONOF THE TANGENTFUNCTIONTO THE SLOPE OF A LINE

463

has degreemeasure0. Thus, if Lt and L, are two distinct lines,0 < 0 < 180. Refer to Figs. 10.5.3and 10.5.4. v

F i g u r e1 0 . 5 . 3

10.5.4 Figure ./ The following theorem enables us to find 0 when the slopes of L1and Lo ate known. 10.5.3 Theorem

Let L, and L, be two nonvertical lines, which intersect and are not perpendicular, and let L, be the line having the greater angle of inclination. Then if. m1is the slope of Lr, m, is the slope of.Lz, and 0 is the degree measure of the angle between Lt and Lr, (2)

pRooF: If a1 and d.2are degree measuresof the angles of inclination of Lt and L", respectively, from Eq. (1) we have 0:

a2-

e.1

and so tan 0o: tan(azo- oro) Applying formula (7) of Sec. 10.4 to the right side of the above, we get -

L-- Ao_ tan a2o tan d-:1H"fr;5;7tan

c1o

I


EXAMPLEL: Find to the nearest degree the measurements of the interior angles of the triangle having vertices B(-2, l) , C(2, 2) , D (-3, 4).

solurroN: Let B, y, and 6 be the degreemeasuresof the interior anglesof the triangle at the vertic es B, c, and D , respectively. Let u denote ttre line through B and C,let a denote the line through C and D,let w denote the line through B and D, and let m, m, and mrbe their respectiveslopes. Refer to Fig. 10.5.5. Using the formula for the slope of a line through fwo given points, we get ttlu: i

ffio:-?

ffi.:-3

To determine B, we observe that line ar has a greater angle of inclination than line u; so in formula (2), tnz: fitw:-3 and trh- ntu: {. The degree measure of the angle between lines u and. a is B. Thus, from formula (2) we have -e-_L

tanBo:1ffi:-13 F i g u r e1 0 . 5 . 5

Therefore, F:94 To determine 7, becauseline a has a greater angle of inclination than line u, ffiz: ttr: { and rnr: nru: }. The degree measure of the angle between lines u and a is, by definition, 180- 7. Applying formula (2), then, we get

- yo) :#+I tan(180"

: -+g

Hence, 180- Y:144 so that y:35 To determine 6, becauseline o has a greater angle of inclination than line w, friz: frtr: { and rltr: tnw: -3. The degreemeasureof the angle between lines o and w is then 6. Applying formula (2), we obtain 2 -

(-a\

tan 6o: =-'-! ,\ ,el:: 1 + (-f )(-3)

4{ rr

Therefore,

6: 50 The sum of the degree measuresof the angles of a triangle must be r.80.we use this to check the results, and we have

F+y+6:94+35+50:180

OF THE TANGENTFUNCTIONTO THE SLOPEOF A LINE 10.5 AN APPLICATION

10.5.4 Definition

ExAMPLEZ: Find to the nearest degree the measurement of the angle between the cun/es y : f - 2 and y : -2x2 + 10 at their points of intersection.

v

The anglebetweentwo curoesat a point of intersection is the angle between the tangent lines to the curves at the point.

so urroN: Solving the two equations simultaneously,we obtain (2,2) and (-2,2) as the points of intersection.In Fig. 10.5.5sketchesof the two curves are shown together with the tangent lines at one point of intersection. Becauseof the symmetry of the two curves with respect to the y axis, the angle between the curvet at (2,2) has the szunemeasure as the angle between them at (-2 , 2) . Let P be the point (2, 2) , Tt be the tangent line --2x2 * L0 atP. Theline toy: -hasx2 2atP, andTzbe the tangent line to a a greater angle of inclination than Tt. so if 0 is the degreemeasureof z; the angle between T1 arrd T2, -

lllz . ^6 tan fl-:TTffi

tflr

: where mristhe slope of T, and rz, is the slope of Tr. From the equation y :-2x2 L0' + x'- 2,wl get D,A-:2x, and so rttl:4. From the equationy we get DrY:-4x; tlrrtts,ffiz:-8. Hence, P(2,2)

tan 0o:

-R -j-=:*:0.387 (4X-6, Jr l-----:-+

Therefore,to the nearestdegree,0":21".

F i g u r e1 0 . 5 . 6

1-0.5 Exercises (a) I and *; (b) 5 and -*; (c) - 3 and 1. Find tan 0. if 0" isthe measurementof the angle between the lines whose slopesare -rb; (e) -S and -8' *; (d) - * and (a) 5 and -6i (b) -3 and 2. Find, to the nearest degree, the measurementof the angle between the lines whose slopes are -*. -* and 2; (c) t and *; (d) lines that have equa3. Find, to the nearest degree, the measurementsof the interior angles of the triangle formed by the : 8 0. 4y * tions 2r * y 6: 0, 3x y 4: 0, and 3x * (1, 0) , (-3, 2) ' and 4. Find, to the nearestdegree, the measurementsof the interior anglesof the triangle having vertices at (2,3). line having equa5. Find an equation of a line through the point (-1,4) making an angle of radian measure*z with the tion 2r * y 5: 0 (two solutions). -2) making an angle of radian measure tzr with the line having equa6. Find an equation of a line through the poin t (-3 , tion3r-2Y-7:0.


7' Using Theorem 10.5.3,prove that the hiangle having vertices at (2, 3), (6, 2), and (3, -1) is isosceles. 8' Using Theorem 10.5.3,prove that the triangle having verticesat (-3, 0), (-1, 0), and (-2, \/g)is equilateral. 9' Find the slope of the bisector of the angle at A in the triangle having the verticesA(4, 1); 8(6, S);and C(-1, g). 10' Find the slope of the bisector of the angle at A in the triangle having the vertices A(-3,5); B(2, -4); and C(-1,7). 17' Let A' B' and c be the.vertices of an oblique triangle (no right angle), and let a, B, and.7be the radian measures of the interior angles at vertices A, B, and c, respectiveiy. prove-that ti'n a * tan B * tan 7 : tan a tan B tan y. (Hrwr: First show that tan(a * F) : -tan y.) 12' Find the tangents of the measurementsof the interior angles of the triangle having vertices at (-3, -2), Fe ,3), and (5,1) and checkby applyrng the result of Exercise11. 13' Find the points of intersection of the graphs of the sine and cosine functions and find the measurement of the angle between their tangent lines at their poiints of intersection. 14' Find, to the nearest10 minutes, the measurementof the angle between the curvesy:2f11 and !: * -2 at their points of intersecfion.

10.6 INTEGRALS INVOLVING A formula for the indefinite integral of the tangent function is derived as THE TANGENT, follows. Because COTANGENT, SECANT, t'n u AND COSECANT du f ,ur,u du: I J

we let a:

J cosu cos u, da: -sin

u du, and obtain

[ ,ur,u du: - JtU+: -ln lrl+ c

J

Hence, f

u d u - - l n l c o sz l + C Jtan B e c a u s-el n l c o su l : l n l ( c o su ) - t l : l n l s e cu l , w e can also write , '' p : " "

"' '

'

'

t,,. ,

,"j",' ,x4s, 14,t,il''trn |$ecal + c ';j'

' : ,t,1... ri ,':

(1)

..r' :;

=--

(2)

O ILLUSTRATION "1.: f

I

tan 3x dx:

"tf

tan3x (s dx) i J :glnlsecSrl+C

o

The following formula is derived in a way similar to the derivation of (2). (See Exercise 29.)

/

.ot u du- tn lsinzl + C

(3)

ANDCOSECANT 467 SECANT, THETANGENT, COTANGENT, INVOLVING 10.6INTEGRALS To integrate ! sec u ilu, we multiply the numerator and denominator of the integrand by (sec u * tan u), and we have

I r"" u au:= It"t l,(i!f !l,.tT') J J@tanu Letu:

itu: [ (secz * secrl tanil) ' secuTGT- au J

secu * tant. Then 471: (secutarru *seCu) du; andsowehave

["""uAu:[@:;o-lol +c

J

JO

Therefore, (4)

The formula for J csc u ilu is derived by multiplying the numerator and denominator of the integran{ by (csc u - cot z). and proceeding as above. This derivation is left as an exercise(seeExercise30).The formula is

(s)

r t u _ I f ^csc ^ 2x (zdx'l I d * == csc2x dx:; J sin2x J J - f l n l c s c2 x - c o t 2 x l + C The following two indefinite integral formulas follow immediately from the corresponding differentiation formulas. (5)

(7)

Many integrals involving powers of tangent and secant can be evaluated by applying these two forrnulas and the identity given by formula (2) in Sec.10.4,which is (8) 1 * tanz u: se* u There are similar formulas involving the cotangent and cosecant.

(e)


468

PRooF: Du(-cot u) : - (- csczu):

csc2u. (10)

pRooF: Du(-csc n) : - (-csc u cot u) : csc u cot u. . rLLUsrRArrou3:

' . + d u * 3f + [ 2 t 9Sms ous u d u : 2 1 Sln'4 sm

J

J

:2

J

tl

. " ? " ud , sln rl

ff | c s c 2u d u * 3 | c s cu c o t u d u

JJ

:-2cotz-3cscu*C

.

From formula (3) in Sec. 10.4we have the identity 1 * cof z:

(l1.\

csC u

Formulas (6) , (7) , and (8) are used to evaluate integrals of the form f I tan^ u secnu ilu

J

and formulas (9), (10), and (11) are used to evaluateintegrals of the form I I cot' u cscnu ilu

J

where m andn are positive integers. We distinguish various cases. Case1: t tan" u du or ! cot" u du, wheren is a positive integer. We write tattn u:

tanz-2 u tanz u

: fala-z z(seC u- 1) and cot' ll :

ExAMPrn 1:

Find

cotn-2 u cotz u cotn-z u(cscz u -

l)

SOLUTION:

f

I tant x dx J

f

f

I tant r d x J

| t a n r ( s e c 2x J

-+tanz r*ln

lcos

tan x sec2

tan x dx

10.6 INTEGRALSINVOLVINGTHE TANGENT,COTANGENT,SECANT,AND COSECANT

ExAMPrn 2:

Find

SOLUTION:

r

f

f

I cota3x dx - I cof 3r(csc23x l) dx

I cota3x dx J

J

J

I cot' 3r csC 3x dx | co1 lx dx JJ + (-cot3 3x)-

(csc23x - 1) dx I -+cot33x*.*cot3x*x*C

Case2: J sec" u du or J csc" u du, where n is a positive even integer. We write secnU:

gggn-2u seczy:

cscnIt:

sgsn-2u csc2u :

(tanz u * l)@-z)tz seczu

and

EXAMPT,N 3: f

I

Find

"""2

u

SOLUTION: ff

csc6x d x

(cof u * l)@-z)tz

I cscux dx:

J

-f

| (cotz x * L)2 csc2x dx

t,

r c o t a xc s C x d x + 2 | cof x csc2 x JJ lcot5 x-

S c o t 3x -

csC x dx

c o t ) c +c

To integrate J secnu du or ! csc"u dz when n is a positive odd integer, we must use integration by parts, which is discussedin Sec.11.2. Case3: t tar.^ u secnu du or J cot' u cscnu du, wheren is a positive even integer. This caseis illustrated by the following example. EXAMPI.,N 4:

/

Find

tur,' r sec4x dx

SOLUTION:

/

tur,t.r sec4x dx:

I

t a n s x( t a n z x + 1 ) s e c 2x d x

:T

t a n T r s e c x2 d x * f t u r , t r s e c 2x d x

-_ B1

J

t a n sx + * t a n 6 x + C

Case4: t tan* u secnu du or cotmu cscnu du, where m Ls a positive odd "[ integer. The following example illustrates this case.

470


EXAMprr 5: f

I

Find

SOLUTION:

tan5r secTx dx

I

tun'ffseczx dx:

t a n ax s e c 6 r s e c x t a n x d x

I

:I

( s e c 2x -

L)2sec6r (secxtanxdx)

- - [ seclo.r ( secr tan x dx) fr 2 | s e c s r ( s e cx t a n x d x ) + | s e c 6 r ( s e cx t a n x d x ) JJ _ T r1 s e c l l x - 3 s e c e x * * s e c z x * C

Integration by parts (sec. 11.2)is used to integrate J tan- usecou ilu and I cot^ u cscnu du whenz is a positive even integer and n is a positive odd integer.

Exercises10.6 In ExercisesL throu gh 22, evaluate

r

1. I tan 2x dx

{*,

J

the indefinite

integral.

f

2. I cot(3r * t) dx J

3.

I

- cscSxzdx

4.

f\

5, I cotsxdx J

ft^,[tan6rseca xdx 13.

ry")/

f

I

,/

10.

/

t"n x dx

,ur,t r sec' x dx

(tan 3x * cot 3r) z dx

cot'2t dt

7.

/

.r.' 6xdx

11.

cot'3r csca3x dx

15.

,un^tx dx

I

[

f

secr tan r tan(secx) dx I r 8. 12sec2xBdx I r 12. (sec5r * csc5r) 2dx J L6. I t"r,t 3xdx J

fdu

20.[ ""i, ,* J cotz x

J@

In Exercises 23 through 28, ev,aluatethe definite integral. f rl4

25. I

J -nt+ frls

28. I

Jo

sec6xdx -lun} dx

*

,

SeC .f

30. Derive the formula J csc u du - ln lcscu,Z?.Derive the formula J cot u du: ln lsin ul + C. Find the length of the arc of the curve y : ln c,scr from x - *n' to x: 4, in. 32. Find the area of the region bounded by the curye y : tan2x, the r axis, and the lin e x : ir.

cotul + c.

T R I G O N O M E T R IFCU N C T I O N S 471 1 0 . 7I N V E R S E

33. Find the volume of the solid of revolution generated if the rigion bounded by the ctrrvey: seC r, the axes,and the line x: Ir is revolved about the x axis. ' if n*o' 34. Prove: cot r cscn, d x : - c s c " x * r , n I'r - " 6t-tann-r-x - [ ann-'xdx - - J I tannr 35. Prove: n-7 J

if.nisapositiveintegergreaterthanl.

36. Derive a formula similar to that in Exercise34 fot t tan I secnx dx iI n * 0. 32. Derive a formula similar to that in Exercise35 for J col" xdx,if nis a positive integer Sreaterthan 1. and #tW"canintegrate I seCxtanxilx intwoways asfollowi: ! se9 xtanxilx: J tanr(seCx dx):ttan2x*C ll7\ .u:Iwvers. the two of :tseC appearance the in the difference Explain r * C. sec2x tani dx: ! sec r(sec 2ctan x dx) -ln 39. Prove that J csc u du: lcsc z * cot zl * C'

IO.7 INVERSE Consider the equation r: sin y. A sketchof the graph of this equation is the figure, the vertical line r : * is shown; this line interTRIGONOMETRIC in Fig. 1.0.7.1,.In and so on' FUNCTIONS sectsthe graph at the points for which y : +7,8r,,-&zr,-#n, of y as a function not define y does we see,then, that the equation x: sin at in function a graph of the r because every vertical line must intersect have not does function the sine most one point. Hence, we conclude that an inverse function. Refer now to the graph of the sine function (Fig. 10.2.3).We seefrom the figure that the sine is an increasing function on the interval l-Lo,En). This follows from Theorem 5'1.3 because if /(r) : sin r, then /'(r) : cos r > 0 for atl x in (-in, *zr). Thus, even though the sine function does not have an inverse function, it follows from Theorem 9.3'2that the function F for which F(r) : sin t

and -tn

= x -Lrr

(1)

does have an inverse function. The domain of F is l-in,Enl and its range The inverseof is [-1, I ]. A sketchof the graph of F is shown in Fig. 1'0;7.2. and is function" sine the "inverse is called (1) by the function defined definition. formal is the Following sin-l. the symbol denoted by

-z") - TI\ 6r)

Figure

F i g u r e1 0 . 7 . 2

472


10.7.1 Definition

The

denoted by sin-l, is defined as follows: if and only if

The domain of sin-l is the closedinterval [- "!.,'].], and the range is the closedinterval l-+n, +rrl. A sketch of the graph is shown in Fig. 1,0.7.3. It follows from Detinition I0.7.Lthat sin(sin-t x) - x

for r in [-1 , Lf

sin-r(sin y) : y

for y rn l-in,

and

(2)

I

tnf

(3)

In (1) the domain of F was restrictedto the closedinterval l-h,!v1 so the function would be monotonic on its domain and therefore have an inverse function. However, the sine function has period 2n andis increasing on other intervals, for instance, l-Err, -Bz'] and liln, Enl. Also, the function is decreasingon certain closedintervals, in particular the intervals l-Er, -$z] and ltn, Errf. Any one of these intervals could just as well have been chosen for the domain of the inverse sine function. The choice of the interval l-l:r, ]zr], however, is customary becauseit is the largest interval containing the number 0 on which the function is monotonic. The cosine function does not have an inverse function for the same reason that the sine function doesn't. To define the inverse cosine function, we restrict the cosine to an interval on which the function is monotonic. we choosethe interval 10, rrl on which the cosine is decreasing.so consider the function G defined bv

F i g u r e1 0 . 7 . 3

G(r)-cos.r

and 0
jr

(4)

The range of this function G is the closed interval l-1,11, and its domain is the closed interval [0, ,f . A sketch of the graph of G is shown in Fig. 1'0.7.4. Because this function is continuous and decreasing on its domain, it has an inverse function which is called the ,,inverse cosine function" and is denoted by cos-l.

1.0.7.2Definition

The

The domain of cos-l is the closed interval l-1,1], and the range is the closed interval 10, lrl. A sketch of its graph is shown in Fig. 7O.Z.S. From Definition 1.0.7.2it follows that cos(cos-l x) : x

for r in [-1 , lf

(s)

cos-l(cos y) : y

for y rn 10,nl

(6)

and

There is an interesting equation relating sin-r and cos-l that is giv in the following example.

10.7

FUNCTIONS TRIGONOMETRIC

Y

Figure10.7.4

:

479

c o s- 1 r

Figure 10.7.5

Let x be in t-l , Il and Tt): tn - sin-l

ExAMpr,nL:' Prove cos-1 x: irr _ sin-l r

sorurroN:

for lrl < 1,.

and so . sin(sin-1x) : sin (En - w)

Then

s i n _ 1x _ L n _ w

and using (2) , we get x-sin(*n-w) Applying (11)in Sec.10.1to the right side of the above equatio(I, we get (7)

x-cosa)

Because-En = sin-1 r = in, by adding -tn to eachmember we have -rT =-in * sin-l r < 0 Multiplying each mernber of the above inequality by-l direction of the inequality signs, we have 0 =tn

and reversing the

- sin-l r < Tr

Replacing tn - sin-l x by rD, we obtain 0=w<7r From (7), (8), and Definition

10.7.2 it follows that

u) - cos-l r

for lrl < 1. and replacing w by *n - sin-l x,, we get tn-sin-1

x-

c o s - 1r

forltl

(e)

474


Note in the above solution that the relation given by (9) depends on choosing the range of the inverse cosine function to be [0, zr]. To obtain the inverse tangent function we first restrict the tangent function to the open interval (-izr, tn) on which the function is continuous and increasing. We let H be the function defined by H(x)-tanff

and -in
(10)

The domain of His the open interval (-Lzr,ir), and the range is the set of all real numbers. A sketch of its graph is shown in Fig. 1.0.7.6.B,ecause the function H is continuous and increasing on its domain, it has an inverse function, which is called the "inverse tangent function" and is denoted bv tan-r.

F i g u r e1 0 . 7 . 6

10.7.3 Definition

Figure 10.7 .7

The inuersetangentfunction, denoted by tan-l, is defined as follows: Y : tan-1 x i f

and only if

)c-

tan y

and

-tn

< y < *n

The domain of tan-l is the set of all real numbers, and the rangeis the open interval (-hr, ir). A sketch of its graph is shown in Fig. 10.7.7. Before defining the inverse cotangent function, we refer back to Example 1, in which we proved the relationship between cos-1 and sin-l given by Eq. (9). This equation can be used to define the inverse cosine function, and then it can be proved that the range of cos-l is [0, zr].We use this kind of procedure in discussing the inverse cotangent function. 10.7.4 Definition

The inaersecotangentfunction, denoted by cot-l, is defined by cot-r x:

tn - tan-l r

where x is any real number

(11)

It follows from the definition that the domain of cot-l is the set of all

FUNCTIONS 10.7 INVERSETRIGONOMETRIC

475

real numbers. To obtain the range we write Eq. (11) as tan-l x-tn-

cot-l x

Because

-tn
in

by substituting from Eq. (12) into this inequality, we get -in
:

cot

0
-1r

Therefore, the range of the inverse cotangent function is the open interval (0, zr). A sketch of its graph is in Fig. 1'0.7.8' The inverse secant and the inverse cosecantfunctions are defined in terms of cos-l and sin-r, respectively.

Figure 10.7.8

denoted by sec-l, is defined by

10.7.5Definition

The

L0.7.6Definition

The inaersecosecantfunction, denoted by csc-1,is defined by

it follows that the domain of sec-l is (- n,-!f U From Definition "!.0.7,5 Lzn) U (tn, nf . Note that the numb er +rr is [1, +*), and the range is 10, not in the range because sec trr is not defined. A sketch of the graph of sec-r is shown in Fig. 10.7.9.

Y

:

sec-l r

F i g u r e1 0 . 7 . 9

476


It follows from Defini tlon 1.0.7.6 that the domain of csc-l is (-o, - 11 g 11,+*), and the range is L-*r,0) U (0, *z]. The number 0 is not in the range becausecsc 0 is not defined. A sketch of the graph of csc-r is shown in Fig. 1,0.7.10.

Y :

ExAMPtn 2: Find the exact value of sec[sin-l (-t) ].

csc-l r

F i g u r e1 0 . 7 . 1 0

solurroN: Let t : sin 1(-*). Then sin t: -1. Becausethe range of sin-l is l-in, !nl, and becausesin f is negative, it follows that-tzr < f < 0. Figure 10.7.1,1, shows an angle having radian measuref that satisfiesthese requirements. From the figure we see that sec t:#17, and so we conclude that seclsin-l(-il1: +\/7.

Fig ure

EXAMPT,T 3:

Find the exact value

of sin[2 cos- 1 ( - g ) l .

solurroN: Let f : cos-l(-$). so we wish to find the exact value of sin 2f. From formula (14) of Sec. 10.1, we have sin 2t:2

sin f cos t

Becausef _ cos-l(-g), it follows that cos t - -B andtn < f < identity, sin2 t+ cos2t:1., and because sin t > 0, since f is we obtain

sint-1 m-ffi:+ Therefore, from formula (13) we get

sin 2t : z(il (- g) from which we conclude that sin[2 cos-l(-g) ]

(13) From the (*r, n) ,

10.8DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS 477

ExC T C I S C S1 0 . 7 /l).Find the value of each of the following: (p)

V

sin-r(-t);

(h) cos-r(-l);

(f) csc-1(-1)

e) tan-'(-l);

(d) cot-1(-1); (e) sec-1(-1);

(t (S);cos-l(!); (.d') cos-l(-*);

(; Find the value of each of the following: (a) sin-'(i); (b) sin-l(-*);

(e) sec-1(2);

(f) csc-l(-2)

3. Given: 3t: sin-t *. Find the exact value of each of the following: (a) cos y; (b) tan y; (c) cot y; (d) sec y; (e\ cscy. 4. Given: y : cos-t(-I). Find the exact value of each of the following: (a) sin y; (b) tan y; (c) cot y; (d) secy; (e) csc y. 5. Given: y : tan-r(-2). Find the exact value of each of the following: (a) sin y; (b) cos y; (c) cot y; (d) sec y; (e) cscy. 6. Given: y : cot-'(-t). Find the exactvalue of each of the following: (a) sin y; (b) cos y; (c) tan y; (d) secy; (e) csc y. 7. Find the exactvalue of each of the following: (a) tan[sin-'(*16;1,

(b) sin[tan-l(] \6)1.

8. Find the exactvalue of each of the following: (a) cos[tan-t(-3)]; (b) tan[sec-1(-3)]. 9. Find the exact value of each of the following: (a) sin-lltan 1z]; (b) tan-1[sin(-fu)l; (d) cot-l[csc(-*n)1. (b) sin-r[cos(-*zr)];

10. Find the exact value of each of the following: (a) cos-1[sin(-fz)]; (d) sin-1[cos3zr].

(c) sec-r[tan(-*n)]; (c) cos-l[sin 3zr];

In Exercises11 through 18, find the exact value of the given quantity' 12' tartl2 sec-l(-t)l 11. cos[2 sin-'(-t)] 13. sinfsin-l 3* cos-l i] '' L5. tan(tan-l *- sin-'

14' cos[sin-1(-+)+ sin-lrl

17. cos(sin-l *- tan-'*)

18. sin[cos-1(-3) + 2 sin-l(-])l 20. Prove:2 tan-1 f - tan-r(-f):in.

15. tanfsec-l** csc-1(-i8)]

19. Prove:cos-1(3/V10)* cos-1(2/r6):*r.

In Exercises2L through 28, draw a sketch of the graph of the given equation. 2l.y:isin-l

22'Y:sin-rix

7

A. Y :2

23. y : Ian-r 2x 25.y:

26.y:tsec-t2x

c o s - l3 1

28.Y:*csc-tlr

2 7 .y : 2 c o t - r i x

1.0.8DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS

tan-t x

Becausethe inverse trigonometric functions are continuous and monotonic on their domains, it follows from Theorem 9.3.4 that they have derivatives. To derive the formula for the derivative of the inverse sine function, let Y

: sin-l r

which is equivalent to x-

sin y

and

y in l-to,

Lnl

(1)

478


Differentiating on both sides of (1) with respect to y, we obtain Dux:cosy

(2)

and y inl-tn,irf

From the identity sinz y * coszy:

1, and replacing sin y by & we obtain

coszy:l-x2

(3)

If. y is in [-*zr, *z'], cos y is nonnegative, and so from (3) it follows that

cosy: \R

rf y is in l-tn,inl

(4)

Substitutingfrom (4) into (2), we get Dor:\R FromTheorem9.3.4, DrA: llDox; hence, D ' ( s i n - r{ : !

(5)

vI - xz

The domain of the derivative of the inverse sine function is the open interval (-1,1). rf.u is a differentiable function of x,we obtain from (5) and the chain rule

D * ( s i n - ul ) : # D , u o ILLUSTRATTOTV 1: If y:

(6) sin-l

dvl)r dx Vt-(f)'

12, then

.._:_\..1:_

vr-r

a

To derive the formula for the derivative of the inverse cosine function, we use Eq. (9) of Sec.10.2,which is cos-r t:tr

- sin-l r

Differentiating with respect to r, we have D"(cos-l x):Dr(Ln - sin-l r) and so D'(cos-lx):-

7

ffi

(7)

where r is in (-1, 1). If u is a differentiable function of x,we obtain from (7) and the chain rule 1, (8)

479 FUNCTIONS TRIGONOMETRIC 10.8DERIVATIVES OFTHEINVERSE we now derive the formula for the derivative of the inverse tangent function. If U: tan-l x then x:

tan y

and y is in (-in,

irr) Differentiating on both sides of the above equation with respect to y, we obtain D o x : s e c z A a n d y L Sin (-in, tr)

(e)

From the identity sec' y - | * tanzy, and replacing tan y by x, we have (10) sec2Y:t*x2 Substituting from (10) into (9) , we get Dox-1i-x2 So from Theorem 9.3.4 we obtain (11)

x):# D r(tan -'

The domain of the derivative of the inverse tangent function is the set of all real numbers. lf uis a differentiable function of x,we obtain from (11) and the chain rule (12)

o rt,r,usrnerroN2: lf f (x): tan-t #,

1

then

f'(x)

From Detinition

']..0.7.4 we have

cot_l )c: in _ tan_l r Differentiating

with resPect to x, we obtain the formula

D *(cot-l x) : -

L L*xz


From this formula and the chain rule, it follows that if u is a differentiable

(13) EXAMPLE1:

Given

SOLUTION:

du ! -

y : rs cot-t *x

ffic

3x2 cot-l

find dy ldx. :

EXAMPIr 2:

Given

ln(r*y)-tan-r

3x2 cot-l

solurroN: Differentiating implicitly on both sides of the given equation with respect to r, we get (;)

find D *A.

L+Dr!/ w9,

:

-y-7sD*U _

x+y

y'* x'

y'+ )c2 + (y',* x',)D*a: xy * y',- (x',* xy) Dra

D*A:ffi Definition 1,0.7.5and formula (8) of this section are used to derive the formula for the derivative of the inverse secantfunction. From Definition "1.0.7.5 we have

sec-lx - cos-l (+)

for lrl !

vc/

So D

1

"(sec-r

I

r)

1,

(+(-+) )'

1 *21ffi

\E l*l *21ffi Therefore, we have

D * ( s e cx- r) : +

pltF

(r4)

FUNCTIONS 481 10.8 DERIVATIVES OF THE INVERSETRIGONOMETRIC

wherelxl > 1. lf.u is a differentiablefunction of.x, it follows from (14) and the chain rule that

u):ffiD*u D*(sec-l

(ls)

o rLLUSrRArroN3: If f (x) - sec-' 3x, then

f ' ( x )_

1

l3rlffi

From (9) of Sec.10.7we have for lxl = 1

cos-lr

sin-lr:tn-

Replacing xby llx in the above equation, we obtain

1 s'm ' /1\ \-/:Zn

-

(1) "or-' \ r /

ror lrl > 1

From the above equation and Definitions 10.7.5 and L0.7.6 it follows that csc-l x:in

- sec-1r

for lrl > 1

Equations (16) and (L4) are used to find the derivative of the inverse cosecantfunction. Differentiating on both sides of (15) with respectto x, we get D ' ( c s c - rx ) : -

t

Wffi

G7)

where lrl > 1. From (17) and the chain rule, Lf.u is a differentiablefunction of r, we have

u) ffiD*u

ExAMPLE3: A picture 7 ft high is placed on a wall with its bas e 9 ft above the level of the eye of an observer. How far fuom the wall should the obsenrer stand in order for the angle subtended at his eye by the picture to be the greatest?

( 1 8)

sol,urroN: Let x ft be the distance of the observer from the wall,0 be the radian measure of the angle subtended at the observer's eye by the picture, o be the radian measureof the angle subtended at the observer's eye by the portion of the wall above his eye level and below the picture, and F: o + 0. Referto Fig. 10.8.1. We wish to find the value of r that will make d an absolute maximum. Becauser is in the interval (0, *-), the absolute maximum value of 0 will

--]


be a relative maximum value. We see from the figure that cotB-+ and c o.t)oc: o B e c a u s e<0B < t n a n d 0 f' "1,6 B-cot-r

a - cot-t {

|and

9 Substituting these values of p and u in the equation 0 - B - q . , w e g e t x )c e - cot-' : - G - c o t "-r g Differentiating with respect to r gives 1L

D*o

tf

9r*

,rr*

t * (G/

1+ (;/

+rft

L5Lg '1,62*x2'

F i g u r e1 0 . 8 . 1

92*x2 Setting D r0 0, we obtain 9(L62* x') - l5(9' + ,c') :0 -7f+9'16(15-9):0 x2:9.75 x: 12 -12 The is rejectedbecauseit is not in the interval (0, +o;. The resultsof the first-derivative test are shown in Table 10.9.1.Becausethe relative maximum value of 0 is an absolute maximum value, we conclude that the observer should stand 12 ft from the wall Table10.8.1 Dr|

0< x<12

+

X:'/',2

0

Conclusion

d has a relative maximum value

12
Exercises10.8 L. Derive the formula D"(cot-l

-1

x)

2. Derive the formula D*(csc-rx):4

lxl"ffi'

TRIGONOMETRIC FUNCTIONS OF THE INVERSE 10.8DERIVATIVES In Exercises3 through 22, find' the derivative of the given function. (3,, (x) : sin-t tx ,4,-8(r) - tan-r 2x

f

5. 8(t) : sec-r5t 8. f (x) - csc-r2x

ii.:,',1, \{ ' : y sin-t 2y \*---' (y)

7. F ( r ) : c o t - t ? + t " r , - t I x2 f0i f (t) : t2 cos-' f

1 1 .S ( r ) : x . r . - ' I

1 2 ;h ( x ) : t a n - ' , 2 ' , ' , .' L-X-

1 3 .f ( x ) - 4 s i n - 1 i x * x t R

l 4 . S ( s ) c: o s' r-: i +

' 1',1",' 16'. f (x) : }n(tan-l 3x)

17. S(r) : cos-l(sin r)

19. h(x) - sec-l r * csc-l r

20. F(r)-

6. F(r) - cos-t{x

i

I

t

15.f(t) - a sin-r ;+ 18.f(x):sin-l

\m

x*cos-l t

sec-r1ffi

22.f (t) - ,',,-' '* I4 Exercises23 through 26, find D,y by implicit differentiation. 21.G(x):

I cot-' r + hv{T7-

(%'e"*Y:cos-tr 24. ln(sin2 3x) : t" * cot-r y

:ta -'y &:rsiny* f 26.sin-r(ry): cos-t(r* Y) iei. l1ight is 3 mi from a straight beach. If the light revolves and makes 2 rpm, find the speed of the spot of light along the beach when the spot is 2 mi from the point on the beach nearest the light. 2g. A ladder 25 ft long is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away fromthe wall so that the top is s-liding aownit g ftlsec, how fast is the measure of the angle between the ladder and the ground changing when the bottom of the ladder is 15 ft from the wall? 29. A picture 4 ft high is placed on a wall with its base 3 ft above the level of the eye of an observer. If the observer is approactring ttre watt it the rate of.4 fllsec, how fast is the measure of the angle subtended at his eye by the picture changing when the observer is 10 ft from the wall? 30. A man on a dock is pulling in at the rate of 2ftlsec a rowboat by means of a rope. The man's handq are 20 ft above the level of the point where thi rope is attached to the boat. How fast is the measure of the angle of depression of the rope changing when there are 52 ft of rope out? A man is walking at the rate of 5 ftlsec along the diameter of a circular courtyard. A light at one end of a diameter perpendicular to his path castsa shadow on the circular wall. How fast is the shadow moving along the wall when the disiance from the min to the center of the courty atd,is tr , where r ft is the radius of the courtyard?

32, In Exercise31, how far is the man from the center of the courtyard when the speed of his shadow along the wall is 9 ftlsec? 33. A rope is attachedto a weight and it passesover a hook that is 8 ft above the ground. The rope is pulled over the hook at the rate of * ftlsec and drags the weight along level ground. If the length of the rope between the weight and the hook is r ft when the radian measure of the angle between the rope and the floor is 0, find the time rate of change of d in terms ol x. 34. Given: l(r) : tan-'(l/r) - cot-r x. (a) Show that /'(r) : 0 for all r in the domain of /. (b) Prove that there is no constant C for which f (x) : C for all r in its domain. (c) Why doesn't the statement in part (b) contradict Theorem 5.3.2?

(Z

-"t

"

484


10.9 INTEGRALS YIELDING INVERSE TRIGONOMETRIC FUNCTIONS

From the formulas for the derivatives of the inverse trigonometric functions, we obtain the following indefinite integral formulas: (1)

(2) (3)

Formulas(1) and (2) follow directlyfrom the formulasfor the derivatives of sin-r z and tan-l u. Formula(3) needsan explanation. If u > 0, then z : lul, andwe have fdufdu

Jnffi:

J 1"6:

s e c - ' uc* : s e c -l zr l* c

II u < 0, then u: -lul, and we have f d- u - fdu

J u\/u2-7

J -l"lttd= _ f : -ilu J lul\/u2 _ 1, f

dFu\

J |ufftrdt= : sec-1(-z) * C : sec-l lul + c Therefore, if u > 0 or u < 0, we obtain (3). we also have the following formulas. (4)

(s) (5)

These formulas can be proved by finding the derivative of the right side and obtaining the integrand. This is illustrated by proving (Z).

D,(sin-'t):#""e)

v,- \;/

\+.

10.9INTEGRALY S I E L D I N GI N V E R S E T R I G O N O M E T R IFCU N C T I O N S

\tr tfa>0

tfa>0

The proofs of (5) and (6) are left as exercises (see Exercises L and 2).

ExAMPLE l,:

Evaluate

fdx J t/l - 9xz

:+ sin-r ++ c EXAMPug2:

fdx

Jffi

Evaluate

SOLUTION:

df d, I :[ J 3 x ' - 2 x * 5 J 3 ( x -' f x ) + 5 To complete the square of.xz - &x we add *, and because* is multiplied by 3, we actually add * to the denominator, and so we also subtract * from the denominator. Therefore, we have

d* d* _f I J ixz-2x+5- J 3(x,-?x++)+5-+ dx dx 1f - r 3 ( x + ) ' + # s J @- +)r+# J

:l-

solurroN: Because d(xz* 2x * 5) : (2x I 2) dx,we write the numerator as (2x + 2) dx * 5 dx, and expressthe original integral asthe sum of two integrals. d.x I Qx*7\ dx I Qx+2\ dx , - | 'J x'*2x*5 J x'*2x*5 J x'*2x*5

.d 486


dx : l n l x ' l . 2 x + 5 1*u/ (x*I)'+4

+]t an-l tl1+

:ln(x'*2x*5)

2

NorE: Becausexz * 2x + 5 > 0, for all x, Evaluate

:I I

3dx

3dx

(x*z)ffi

(r*Dffi

C

l x ' 1 - 2 x + 5: 1x2* 2x + 5.

3dx

(

-3

sec-1lx + 2l + C ExAMPrr 5: Find the area of the region in the first quadrant bounded by the curue

sot,urroN: Figure 1.0.9. 1, shows the region and a rectangular element of area. If A square units i s the area of the region, w€ have

Athe x axis, the y axis, and the line x: l.

:

lim $= '1

ttiii-o?:'r*€"

Aix

fdx Jo L*x2

: tan-t r It

Jo

:tan-l

1-tan-1

:in-0 :in

The result of Example 5 states that the number z is four times t6'e number of square units in the area of the region shown in Fig. 10.9.1.

Exercises 10.9 In Exercises1 through 5, prove the given formula by showing that the derivative of the right side is equal to the integrand - f --;--:----;: du -7 t"rr-'. u1+ C l. | J A'tU'

A

A

10.9INTEGRALS YIELDING INVERSE TRIGONOMETRIC FUNCTIONS '-'' du 1 ^ f z. t--------::-sec-rlal +C a J u!ut-a' lal

if a>0

Take two cases:a > 0 and z < 0. du ^f 3. | -:-cos-'z*C J \/l- u, Is this formula equivalent to formula (1)? Why?

4'

fdu

Jr;7:

-cot-ru * C

Is this formula equivalent to formula (2)? Why? du -f 5. l;ffi:-csc-l

lul+c

Take two cases:z > 0 and z < 0. Is this formula equivalent to formula (3)? Why? In Exercises5 through 25, evaluate the indefinite integral.

7I#

i

,101 .L

13I#

14t4 J Iz,''.t f

L 6 .l L J\ffi

dx

^fdx

'

t

,89.Jffi a

-,---

22. [ = !d* J x2+x*L f

c o s zx

L7:l " J x2-x*2

i&lJ,'ffi, 6

xdx

1 1Jt LV r 6 - e t '

J xtF

\''--"

i

S I x4+r6

. ,- . , .f - Q + x ) d x J V4-2x-x2

3dx

z) a+l ' J G + D f f i -

In Exercis es 26 through 33, evaluate the definite integral. . . : 'f r 1 * x

27.' J| o #d*

,

LtX'

30. f d x

J, x'z-4x+13

4". Q,

r[L + (ln r)2]

\ 34.Find the area of the region bounded by the curve y : 8l (x2 * 4) , the x axis, the y axis, and the line x: 2. :i5. Find the abscissaof the centroid of the region of Exercise34. 36. Find the area of the region bounded by the curves x2: 4ay and y :8a31 (x2 * 4a2), 37. Find the circumference of the circle rP * y' : I by integration.

*/

t'i


488

38. A particle moving in a straight line is said to have simple harmonicmotion if the measure of its accelerationis always proportional to the measure of its displacement from a fixed point on the line and its accelerationand displacement are oppositely directed. So if at f sec,s ft is the directed distance of the particle from the origin and o ftlsec is the velocity of the particle, then a differential equation for simple harmonic motion is da -- -kzs dt

(7)

where k2 is the constant of proportionality and the minus sign indicates that the acceleration is opposite in direction from the displacement.Becausedoldt : (dulds)(dsldt) : o(dolds), Eq. (7) may be written as I'j

da E:

-kzs

(8)

(a) Solve Eq. (8) for a to get Tr- *k\m. Note: Take a2k2as the arbitrary constant of integration and justify this choice. (b) Letting a - dsldt in the solution of part (u), we obtain the differential equation i<

fi:tktto'-

t"

Taking f : 0 at the instant when o : 0 (and hence s : a), solve Eq. (9) to obtain s: a coskt

(10) (c) Show that the largest value for lsl is c. The number a is called the amplitudeof the motion. (d) The particle will oscillatebetweenthepointswheres:aands:-a.If Tsecisthetimefortheparticletogofrom ato-aandreturn,show that 7: Znlk. The number T is called Ihe periodof the motion. 39. A particle is moving in a straight line according to the equation of motion s:5 - L0 sin2 2t, wherc s ft is the directed distance of the particle from the origin at f sec.Use the result of part (b) of Exercise38 to show that the motion is simple harmonic. Find the amplitude and period of this motion. 40. Show that the motion of Exercise 39 is simple harmonic by showing that differential equation (7) is satisfied.

Reaiew Exercises(Chapter 10) In Exercises L throu gh 4, evaluate the given limit, if it exists. I

l.

.r

r'

r.

llm r-o

2Ljsry

sin2 r

f.

L2 CJL

cos 5r 4. lim -

tan a 2 x

4*^

lt$

5. show tlnat

r-7r12 COS lX

d(Lnsin r) x cot x J(ln rf

6. Prove that lim

sin x2 0.

r-o

X

'1.4, In Exercises7 through find the derivative of the function defined by the given equation.

7.f(x)--ffi g.

S@)

-

xcos!

8. f(x) - \63r . lnf 1 0 ..f ,( \t ') - : secf

1 1 . F ( r ) : t a n -' ; '

12. s(r) : cot SxtTsin2x

13. f(x) - (tanx)rt*'

1 4 . h ( x ) - ( c o sx ) e "

REVIEWEXERCISES 489 In Exercises L5 and L6, find D *A by implicit differentiation. y2

t6. sin(r + V) * sin(x - A) - 1

L 5 . c o t - l| + x A z : 0 5y In ExercisesL7 through 24, find the indefinite integral. r 17. I sinaix cosztx dx

f

18.

J

20. 23.

I

J

IUl-cosxdx

(csc 2x - cot 2x) dx

2t. I#

sin 3r cos5x dx

24. f sint 2x coss2x dx J

r J

dx

ffi

In Exercises25 through 28, evaluate the definite integral. lils

25. I Jo

lal6

secs x tans x ilx (x -t 2) dx

20. I Jo

tana 2x dx

fntz

2 7 . f2 Jrffi

,t.J,,n(sina2x-2sin4x)dx

29. Find the length of the arc of the curve y : ln cos r from the origin to the point (in, -ln Z). 30. Find the averagevalue of the cosine function on the closed interval fa, a * 2nl. 31. Find to the nearest degree the measurementsof the four interior angles of the quadrilateral having vertices at (5, 6) , (-2, 4) , (-2,7) and (3, 1) and verify that the sum is 360'. 32. Findthevolumeof the solidgeneratedifthe regionboundedbythecurve!: revolved about the x axis.

sinzrandther axisfromr:0to

x: zris

33. Two points A and B are diametrically opposite each other on the shoresof a circularlake 1 mi in diameter. A man desires to go from point A to point B. He can row at the rate of 1* mi/hr and walk at the rate of 5 mi/hr. Find the least amount of time it can take for him to get from point A to point B. 34. Solve Exercise33 if the rates of rowing and walking are, respectively,2 milhr and 4 mi/hr. 35. A partide is moving along a straight line and s: sin(4f * tn') + sin(4f * *r) where s ft is the directed distance of the particle from the origin at t sec. Show that the motion is simple harmonic and find the amplitude of the motion. (SeeExercise 38 in Exercises10.9.) 35. If an equation of motion is s: cos 2f * cos f , prove that the motion is not simple harmonic. fn

37. Evaluate: I lcosx+t\tu. Jo

fnl2

38. Evaluate:I

Jo

lcosx - sin xl dx.

39. Find the area of the region in the first quadrant boundedby the y axis and the curves A: seczx and y :2 tanzx. 40. Find an equation of the normal line to the curve /: cos r at the point (32, -*). 41,.A p a r t i c l e i s m o v i n g a l o n g a s t r a i g h t l i n e a c c o r d i n g t o t h e e q u a t i o n o f m o t i o n s : 5 - 2 c o s z t w h e r e s f t i s t h e d i r e c t e d

490

TRIGONOMETRICFUNCTIONS

distance of the particle from the origin at f sec. lf. a fitlsec anda fitlseczare, respectively, the velocity and accelerationof the particle at f sec, find o and a in terms of s. A. ff y ft is the range of a projectile, theS

"c

onzsin 20

where o6ftlsec is the initial velocity, g ftlse3 is the accelerationdue to gravity, and 0 is the radian measureof the angle that the gun makes with the horizontal. Find the value of 0 that makes the range a maximum. 43. A searchlight is * mi from a straight road and it keeps a light trained on an automobile that is traveling at the constant speedof 50 mi/hr. Find the rate at which the light beam is changing direction (a) when the car is at the point on the road nearest the searchlight and (b) when the car is * mi down the road from this point. 44. A helicopter leaves the ground at a point 800 ft from an observer and rises vertically at 25 fllsec. Find the time rate of change of the measure of the observer's angle of elevation of the helicopter when the helicopter is 500 ft above the ground. 4s. In an electriccircuit let E : f (f) and i - g(f) where E volts and i amperes are, respectively, the electromotive force and current at t sec.If P watts is the averagepower in the interval [0 , T), where T sec is the common period of andg, then f

P-+ f fG)su) dt If E - L00sin f and i: A(sin f - *zr), first determine T and then find P. 46. Find the area of the region bounded by the curve y:glt6=7,

x - 2\n.

47. Find.the absolute maximum value attained by the function / if (i / constants.

the two coordinate axes, and the line

: a sin kr * B coskx, where A, B, and k are positive

48. Suppose the function / is defined on the open interval (0, 1) and

f (x): x^:?:'l (x - 1) Define f at 0 and L so that / is continuous on the closed interval 49. Prove: Dru (sin x) : sin(x + tnn). (urNr: Use mathematical cos(x + tn) sin r after each differentiation.)

L O1, J . induction

and the formulas sin(x + trr)_ cos x or

l

Techniquesof integration

492

TECHNIQUEO S F INTEGRATION

11.1 INTRODUCTION

In Chapter 7 the definite integral of a function / from a to b was defined as the limit of a Riemann sum as follows: fbn

dx:,,tiil,> f (€,)Lix J"f k)

(1)

if the limit exists. We stated a theorem, proved in advanced calculus, that the limit on the right side of (1) exists if / is continuous on the dosed intewal fa, bl. The exact value of a definite integral may be calculated by the fundamental theorem of the calculus,provided that we can find an antiderivative of the integrand. The process of finding the most general antiderivative of a given integrand is called indefiniteintegration.We use the term indefinite integral to mean the most general antiderivative of a given integrand. In practice, it is not always possible to find the indefinite integral. That is, we may have a definite integral that exists, but the integrand has no antiderivative that can be expressedin terms of elementary functions. An example of this is [;tz t-t' df (a method for computing the value of this definite integral to any required degreeof accuracyby using infinite series is given in Sec. 1.6.9).However, many of the definite integrals that occur in practice can be evaluated by finding an antiderivative of the integrand. Some methods for doing this were given previously, and additional ones are presented in this chapter. Often you may find it desirable to resort to a tableof integralsinstead of performing a complicatedintegration. (A short table of integrals can be found on the front and back endpapers.)However, it may be necessaryto employ some of the techniquesof integration in order to express the integrand in a form that is found in a table. Therefore,you should acquire proficiency in recognizing which technique to apply to a given integral. Furthermore, development of computational skills is important in all branches of mathematics, and the exercisesin this chapter provide a good training ground. For thesereasonsyou are advised to use a table of integrals only after you have mastered integration. The standard indefinite integration formulas, which we leamed in previous chapters and which are used frequently, are listed below.

1,"du-#+c

u+c :

ALI + C

I l f ( u )+ s @ ) l I un du

f du

lT:

lr n l " l+ c

where a is any constant

:gu+c

f(u)du*

udu--cosu+C

n+-L

u du: sin u + C u du: tan u + C

I

BY PARTS 11.2 INTEGRATION

[ ","' /,".

udu:-cotu*C

/,".

utanudu:secu*C

/.,.

l s e cu * t a n u l + C

udu:lnlcscu-cotul+C d'

I

cot u du

udu:ln

- -:--tu-+c

Jffi:sln';

wherea>o

L+c | =dY ^ -'l' tan-1 l'rr

u du: ln lseczl + C

J

a2+uz

t4

u d u : l n l s i nu l + C

a

fl

J uluz - n2

1 sec-r v Y - lYl+ c

a

lal

wherea > o

parts. It II.2 INTEGRATION A method of integration that is quite useful is integrationby a product: of BY PARTS depends on the formula for the differential d(ua)-uda*adu or, equivalently, udo-d(ua)-adu

\

(1)

Integrating on both sides of (1), we have (2)

Formula (2) is called the formula for integrntionby pntts. This formula expressesthe integral [u do in terms of another integral, lo du. By a suitable choice of u and do, it may be easier to integrate the second integral than the first. When choosing the substitutions for u and do,we usually want da to be the most complicated factor of the integrand that can be integrated directly and u to be a function whose derivative is a simpler funciion. The method is shown by the following illustrations and examples. o ILLUSTRATTON

1.:

We wish to evaluate

r

I tan-r x dx

J

Let u:

t a n - l r and da - dx. Then

TECHNIQUESOF INTEGRATION

494

So from (2) we have f

J

tan-l x dx :

( t a n x- ,) ( x *C , )-

I(r*

-xtan-l x*crtan-rx-[+d*

C , )# n

rR-''JR

r

dx

: x t a n - l x * C , t a n - r x - * l n 1 1+ x r l - C , t a n - r x * C , :xtan-rr-*ln (1,+x2)*C, I In Illustration L observe that the first constant of integration c1 does not appear in the final result. This is true in general, and we prove it as follows: Bv writing u * C, in place of a in formula 2, we have ff

J

u du: u(a* C,)-

+ Cr)du J kt

:uo*cru-[adu-crIau JJ

:uu*cfl-[oa"-cfl :nn-

[ o au J Therefore,it is not necessaryto write C1when finding a from do. EXAMPLE 1:

Find

f

I rln xdx

J

SOLUTION:Let Lt-- ln r and da : x dx. Then ,dxxz au: x2 ,v2

. h r d x - i t n xt I .vz

It+ _ 1;I^

'_. d( ,x

2

|tnx

txz ln lx EXAMPLE 2: f

Find

f rsinxdx

J

xdx

tx' '+ C

SOLUTTON: Let Lt- r andl d v : s i n r : d x Then du-dx

a n d 7 ) - - cos S )c

Therefore, we have f

I r sin x dx:-x J

f

c o sx )+ f c o sx;ddx :

--ficos.

J

x + sinx+. C

BY PARTS 495 11.2INTEGRATION 2: IN

o ILLUSTRATION

Example 2, if instead of our choice s of u and du as

above, we let and

Lt: sin r

da:xdx

we get

and o:tf

du:cosxdx So we have f

J

f r sinx dx:i sin--;

1f

J

f cosx dx

The integral on the right is more complicated than the integral with which we started, thereby indicating that these are not desirable choices for z o and ilo. It may happen that a particular integral may require repeated applications of integration by parts. This is illustrated in the following example.

EXAMPTn 3:

Find

r I xze*dx J

- x2 and da - et dx. Then soLUTIoN: Let u and 7):er

du-Zxdx We have, then, ff

I xre" dx: JJ

xzet - 2 | xe' dx

We now apply integration r,t-x

by Parts to the integral on the right. Let

and d0:etdx

Then and u:e*

dil-dx So we obtain

ff | *t" dx-xex-

JJ

| e'dx

:xet-er+C Therefore, f I x"" dx: )cze" zlxe" er + C]

J

: Nzen- Zxe' * 2e'+ C

Another situation that sometimes occurs when using integration by parts is shown in ExamPle4.

I

TECHNIQUES OF INTEGRATION

Find

soLUTroN: Let u - e, and du - sin x dx. Then du:

sin x dx

e* dx

and

T)-- -cos x

Therefore,

r I e" sin xdx

cos x dx

JJ

The integral on the right is similar to the first integral, exceptit has cosr in place of sin r. We apply integration by purts again by letting fr:et

and d0-cosxdx

So

dil-e'dx

and o-sinx

Thus, we have ff

t * s i nx d x JI J

e'cosx* [e'sin x-

| e* sinxdx]

Now we have on the right the same integral that we have on the left. so we add I e" sin x dx to both sides of the equation, thus giving us

r

2 l e ' s i nx d x J

Dividirg

e ' c o sx * e " s i n x * e

on both sides by 2, we obtain

r

I t " s i n x d x - + e " ( s i n x - c o sr ) + C J In applying integration by parts to a specific integral one pair of / chrrices fot u and du may work while another pair may nofl We saw this in Illustration 2, and another case occurs in Illustration 3. OI

LLUSrRarroN 3: In Example 4, rn the step where we have

fr

l t " s i nx d x : - e * c o s x *

JJ

l t * c o sx d x

i f v ve evaluate the integral on the right by letting

u-cosff di: we get

and d0:e*dx

-sin x dx and u : er

fa-r

t" sin x dx Jl J ff

I t" sin x dx: JJ

e* cosx * (e" cosr * l t" sin x dx) I e* sin x dx

BYPARTS 11.2INTEGRATION

,497""

In Sec. 10.8 we noted that in order to integrate odd powers of the secantand cosecantwe use integration by parts. This processis illustrated in Example 5.

EXAMPLE 5:

soLUTroN: Let Lt- secr and da : sec2x dx. Then du : sec r tan x dx and a - tan x

Find

f

I sect x dx J

Therefore, fr xdx=sec xtanxI sec3 JJ rf I secsx dx: secr tan xJJ

ffr I s e c tx d x : s e c JJJ

xtanx-

f s e cx t a n z x d x | secr(seczx- I) dx | s e c tx d x *

| s e cx d x

Adding I secsx dx to both sidet, yu get

r 2 | t".t xdx:secrtanr*ln J

l s e cx * t a n x l+ 2 C

f

x d x - + s e c r t a n) c + * l n l s e cx + t a n r l + C lsecs

Integration by parts often is used when the integrand involves logarithms, inverse trigonometric functions, and products.

11..2 Exercises In Exercises L through t6, evaluate the indefinite

1.

f

Jrnxdx

u..! x3'dx I

t a n - lx d x lx r L0. sin(ln r) dx J 7.

integral.

, UJ

,6I#"

tA

? I-cos2xdx g,/ 'i"-'\x dx

3.

8. *' stn3x dx I

,.

''ir. I#. 14. /.r.'

6.

xdx I-secz I.'tnxdx I

e* cosx dx

L2. x'e" dx I x dx

1s.

/r".'xdx


In Exercises17 through 24, evaluatethe definite integral. t . [:'" sin 3r cosx dr Jo

F\. Jo ,-j L'^

e*asin4xdx

cos2x itx {9 '-' f _x2 J-n

'-') f ,r,, a, $ Jo

,r. 1""'o-x surr cotxcsc cscx r 4tr itx Jtu

zz. fn '"' ,.,

sec-,t/i dx

,0. [""'' cos\/2x dx Jo

e. ,4'

r Jo

x sin-r x ilx

/,$. fi"a the area of the region bounded by the cule y : In x, the r axis, and the line x: ez. ?$. fi"a the volume of the solid generated by revolving the region in Exercise25 about the r axis. {}. ni"a the volume of the solid generated by revolving the region in Exercise25 about the y axis. 28. Find the centroid of the region bounded by the cule y : er, the coordinate axes, and the line : r 3. 29' The linear density of a rod at a point r ft from one end is 2e-t slugs/ft. If the rod is 6 ft long, find the mass and center of mass of the rod. 30' Find the centroid of the solid of revolution obtained by revolving about the x axis the region bounded by the curve y : sin x, the x axis, and the line x: tr. 31' A board is in the shape of a region bounded by a straight line and one arch of the sine curve. If the board is submerged vertically in water so that the straight line is the lowei bound ary 2 ft below the surfaceof the water, find the force on the board due to liquid pressure. 32' A particle is rnoving along a straight line and s ft is the directed distance of the particle from the origin at t sec.rf o f t / s e ci s t h e v e l o c i t ya t f s e c , s : 0 w h e n f : 0 , a n d o . s : t s i n t , f i n d s i n t e r m so f f a n d a l s os w h e n t : t n . 33' A water tank full of water is inthe shape of the solid of revolution formed by rotating about the r axis the region bounded by the cuwe y: e-", the coordinate axes, and the line r: 4. Find the work d6ne in pumping all the water to the top of the tank. Distance is measured in feet. Take the positive r axis vertically downward. 34. The face of a dam is in the shape of one arch of the curve y :100 cos z&ozrrand the surface of the water is at the top of the dam' Find the force due to liquid pressure on the face of the dam. Distance is measured in feet. 35' A manufacturer has discovered that if x hundreds_of units of a particular commodity are produced per week, the marginal iost is determined by x2'r' and the marginal revenue is d'etermined by s . z-ttz,wliere both the production cost thousands of dollars' If the weekly fixed costs amount to $2000,find the *a"i-.r* *"etaf frorit ili:::T:ftTarein

11.3 INTEGRATION By TRIGONOMETRIC SUBSTITUTION

If the integrand contains an expression of the form \/ar=, \/FTE, or {F=V, where a } O,iiis often possible to perform the integration by making a trigonometric substitution which reiults in an inte_ gral involving trigonometric functions. we consider each form as a separate case. Case 1: The integrand contains an expression of the form \/F7A, a > 0. we introduce a new variable 0 by letting u: a sir, g, where 0 < 0 =+rif u > 0 and -*rr < g < 0 iIu < 0. Ttrendu: a cos0 d0,and,

\/F=7 : \/|r=}Gfr

0:{F6=R6

: a@0

Beggge'-*n= 0 -in,cos 0 - 0; hence,\re 0:cos 0, andwe have \/F=7: a cos0. Because sin 0: ula and.-trr - 0 = t,,it followsthat 0 : sin-r(ula).

SUBSTITUTION 499 BY TRIGONOMETRIC 11.3 INTEGRATION

soLUrIoN:'Letr:3 sin d, where0 < 0 < tnif. x > 0 and-*rr <0
Iryo,:l+#'? (scosodo) :Icop0d0 \lfr F i g u r e11 . 3 . 1

: [ 1"r., 0-1) d0

r>0

+ \/ffi

:1.o, o-o+C Becausesin g: *r and -Lr = 0 =8n,0: sin-l(*r)' To find cot 0 refer to Figs.11.3.1 (for r > 0) and 11'3.2 (for x < 0). We see that cot9: V9 - *lr. Therefore, t/g =7 f ..F--

I

, --t dx

\5=7

_,,__,x c - sin-ri*

F i g u r e1 1 . 3 . 2

Case 2: The integrand contains an expression of the form \817, a > 0. we introduce a new varidble 0 by letting u: a tan 0, where 0 < 0 < tn lf.u > 0 and -Lo < e < O if u < }'Then du: a secz0 il9'and

\/F 1fr : GTV@o

: a{I+ffin:

a{@T

:sec0' andwehave B e c a u s-ei n < 0 < t n , s e c 0> 1 ; t h u s\ / s @ e -in g: < izr, it followsthat < 0 ula and tan \/7 + fr: sec0. Because 0: tan-r(ula).

EXAMPLE

2:

r | !x'+ JI

EVAIUATC

5 dx

solurroN: Letr: tEtan0,where0 < 0 < irif.x = 0and-izr < 0 < 0if r < 0. Then dr: \6 se& 0 d0 and

\ffi

_m-\6\@_\E

sec0

Therefore,

sec'ade 0 d0) dx: | \re secO(fi secz | \ffi / JJ Using the resultof Example5 of Sec.1L.2,we have | \m

J

dx:f

r " . 0 t a n0 + * t " P e ce + t a n g l + C

: TECHNIQUES OF INTEGRATION

r>0 F i g u r e11 . 3 . 3

F i g u r e11 . 3 . 4

we find sec0 from Figs.11.3.3(for x - 0) and 1,1.3.4 (for r < 0), where tarr 0 : xlt6. We see that sec e: \/FTEIVE. Hence, ,5 ax: 2

-tx \ffi -tx \ffi

*Bh l\ffi *Bh (lffi

*xl -Brr,\6 +c *r) +cl

Because\ffi*x)0,wedroptheabsolute-valuebars. case 3:_ The integrand contains an expression of the form t/F1v, a > 0. we introduce a new variable d by letting u: a sec0, where < 0 0 <+r7f.u > a andr < 0 <8rrif u = -r. Th"rrTu: osec0ian 0 it| and

\E=7

: \Gr;;A4-

dt:1/p@[=j

: a{R

o

Becauseeither 0 < 0 < Er or r, < 0 < tr, tan 0 - Q;thus, VTan??: tan 0, and we have l/FjE : a tan 0. Note in Fig. ft.g.S ifrut > u a, sec d: ula >'1., and 0 < 0 < ia. Hence, 6: ssg-"r(ula) if. u > a. ln Fig.11.3.5,u --flt sec d: ula =-1, and z < 0
\/u'z-A

u\a F i g u r e 11 . 3 . s

u1-a

F i g u r e11 . 3 . 0

SUBSTITUTION BY TRIGONOMETRIC 11.3 INTEGRATION

EXAMPLE 3:

sec0,where 0 < d < tn 1f x ) 3 and r < 0
Evaluate

fdx

0

\/F4:ffi64:3ffi0:3tan

J *\ffi

Hence, dx

f

3secgtan0d0

f

J /? r""'o'g t"r'ro

Jffi: :

1f

*

I

(1 * cos 20) d0:*(0

** sin 2e) + C

:#(e*sin0cos0)*C When x) 3,0 < d < Ln, and we obtain sin 0 and cos 0from Fig.1'1.3.7. When x 1 -3, r < 0 < 8zr,andwe obtain sin 0 and cos 0 from Fig' 11.3.8. In either case,sin 0: {FaT lxand cos0: 3lx.To exPress0 in terms of r -3. When we must distinguish between the two cases: , > 3 and x < - sec-r *x (referto x ) 3, d: sec-l *r. However, when x 1-3, 0:2zr Fig.11.3.8).We have, then,

F i g ur e 1 1. 3 . 7

\ffi.3\+c x/ x

x 3

{ffi

+

tfx>3

\ffi

,x

rfx<-3

3)c and letting C Cr * *n, x(

we get

_tx*F 3

-3

Lf+c

ifx>3

F i g u r e1 1 . 3 . 8

_tx*F rf+c 3

EXAMPLE 4:

Evaluate

fdx J t/x2 - 25

tf x<-3

n <0 <*n if solurroN: Letr:5 sec0, where 0 < 0 < Ln lf x > 5 and -5. and Then dx: 5 sec0 tan 0 d0 x <

{F =Z r:{25@

0- 25 :5@6:5tan

0

Therefore,

dx -l 5secltanoitl:I sec'de I Stano J ItlP=6-t : l n l s e c d * t a n 0 l+ C To find sec 0 and tan 0 refer to Fig. 11.3.9(for x > 5) and Fig. tt'g.to (for


< -5). In either case,sec0 - *r and tan

tL*\ffi| l4 Jffi--rnl;--ln lr+ \ffi-l -ln lr+ \ml

I

tlffi.

Wehave,then,

*. -ln5+C +C,

t

r)s F i g u r e11 . 3 . 9

ExAMPr.n 5: Evaluate fz

J,@

dx

F i g u r e1 1 . 3 . 1 0

solurroN: To evaluatethe indefinite integral I dxl(6- *)ttz,we make the substitution r: \6- sin d. In this casewe can restrict 0 to the interval 0 < 0 < *z becausewe are evaluating a definite integrar for which > r 0 berauser-is in [1,2]. So we have i: {d sin 0, 0 2 e
coss 0

Hence, f

dx

l{6cos0do

JG-fft'z:J sG"""'e 7l d0 6J cos20

:i'1 ' f J""coae :$tan0*C Wefind tan d from Fig.L1..3Jt,in which sin 0: xltG and0 < 0 < tr.We

NY T R I G O N O M E T R ISCU B S T I T U T I O N 1 1 . 3 I N T E G R A T I OB

see that tan e-

and so we have

xl\ffi,

fdx)c,

l - - : t f

J G

-

x2)3t2

6t75 - x2

I

Therefore, f2dxx12

|

- - - l

J, rc- x2)3i2-6\ffi

F i g u r e1 1 . 3 . 1 1

),

L

11.3 Exercises In Exercises L through L8, evaluate the indefinite

integral.

3I 7I 11I lsI

14.

(5 - 4x-

x2)3t2

h In Exercises L9 through 26, evaluate the definite integral.

22.

/,

I

re. : \/P -9lx', 27. Find the area of the region bounded by the cufve y 28. Find the centroid of the region of Exercise27. 29. Find the length of the arc of the curve A : ln x from r:

1 to x:

the x axis, and the line r:5.

3.

30. Find the volume of the solid generated by revolving the region of Exercise27 about the y axis. 31. Find the centroid of the solid of revolution of Exercise 30. 32. The linear density of a rod at a point r ft from one end is li17 ter of mass of the rod.

slugs/ft. If the rod is 3 ft long, find the mass and cen-


33. If an gbject is moving vertically due to the forces produced by a spring and gravity, then under certain conditions v2: & - k2s2,where at f sec,s ft is the directed distance of the o-bjectirorith" stlartinjpoint and o ftlsec is the velocity of the object. Find s in terms of f where o is positive. 34. A gate in an irrigation ditch is in the shape of a segment of a circle of radius 4 ft. The top of the gate is horizontal and 3 ft above the lowest point on the gate. If the water level is 2 ft above the top of the gate, find the force on the gate due to the water pressure. 35. A cylindrical pipe is 4 ft in diameter and is.closed at one end by a circular gate which just fits over the pipe. If the pipe contains water to a depth of.3 ft, find the force on the gate due to the water pressua 35. An automobile's gasoline tank is in the shape of a right-circular cylinder of radius 8 in. with a horizontal axis. Find the total force on one end when the gasoline is 12 in. deep and w oz. is the weight of 1 in.B of gasoline.

11.4 INTEGRATION oF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS. CASES 1 AND 2: THE DENOMINATOR HAS ONIY LINEAR FACTORS

In Sec.1.8 a rational function was defined as one which can be expressed as the quotient of two polynomial functions. That is, the function H is a rational function if.H(x): P(x)le(x), where p(x) and e(r) are polyno_ mials. We saw previously that if the degree of the numerator is not less than the degree of the denominator, *" liurr" an improper fraction, and in that casewe divide the numerator by the denominator until we obtain a ProPer fraction, one in which the degree of the numerator is less than the degree of the denominator. For example, x4- 10x2* 3x + 1

-)c2-6*3x,-?3

x2-4

x2-4

So if we wish to integrate

the problem is reduced to integrating

I e'-6) dx* In general, then, we are concerned with the integration of expressions , of the form

where the degreeof p(r) is less than the degreeof e(r). To do this it is often necessaryto write p (r) | e @) as the sum of p artial fractions' The denominators of the partial fractions are obtained by factoring Q(r) into a product of linear and quadratic factors. Sometimes it may be difficult to find these factors of Q (x); however, a theorem from advanced algebra states that theoretically this can always be d.one.We state this theorem without proof. 11'.4.1Theorem

Any polynomial with real coefficients can be expressed as a product of

FRACTIONS. CASES1 AND 2 BY PARTIAL FUNCTIONS 11.4INTEGRATION OF RATIONAL

linear and quadratic factors in such a way that each of the factors has real coefficients. After Q(r) has been factored into products of linear and quadratic factors, the method of determining the Partial fractions depends on the nature of these factors. We consider various casesseParately.The results of advanced algebra, which we do not prove here, provide us with the form of the partial fractions in each case. We may assume,without loss of generality, that if Q(r) is a polynomial of the nth degree, then the coefficient of x" is 1 becauseif * Crxn-r+''' I Cn-6* Cn Q(r) : Csxn '!., we divide the numerator and the denominator of the fracthen if Co + tion P(r)/Q(r) by Co. Case1: The factors of Q(r) are all linear and none is repeated. That is, ' ' ' (x- a,) Q(r) : (x- ar)(x- ar) where no two of the a1are identical. In this casewe write

lql

x-ar'

Q(r)

' + . . . + .'h = + x -A x-fln a2

(1)

where Ar, Ar, . , An are constants to be determined' Note that we used "=" (tead as "identically equal") instead of.":" ln (1). This is because(1) is an identity for the value of each4,. The following illustration shows how the values of A1 ate found' o TLLUSTRATTON1.: We wish

f (r-1)

Jm

to evaluate

dx

We factor the denominator, and we have x-L x3-x2-2x

x-l

So we write x-1

x ( x - 2 )( r + 1 ) Equation (2) is an identity for all x (except x:0,2,

(2) -1). From (2) we

get x-7:

A(x-2)(x+1) tBx(x + 1)+ Cx(x-2)

(3)

Equation(3) is an identity which is true for all valuesof r including0,2,

TECHNIQUES

INTEGRATION

and -1. We wish to find the constants A, B, and C. Substituting 0 for r in (3), we obtain -7:

-2A

or A: t Substituting 2 for x in (3), we get 1,:68

or B: t Substituting -'J. for r in (3), we obtain -2:3C or C:-E There is another method for finding the values of A, B,and C. If on the right side of (3) we combine terms, we have

x-] : (A+B+C)* + (-A+B- 2C)x-2A

(4\

For (4) to be an identity, the coefficientson the left must equal the correspondingcoefficientson the right. Hence, A+B* C:0 -A+B-2C:1 -24--1 Solving these equations simultaneously, we get A:i, -3. Substituting these values in (2), we get

x-t x(x-2)(x*1)-

_+ , x'

+

x-2'

B:

t and C:

-3 x*1.

So our given integral can be expressedas follows:

[ ^' .t -ar:![@ *71 2

J xt-x2-2x-*-

ax -2t

6 J x-2J i:f ln lrl+ *ln lr- 2l: * ( 3 l n l r l + l n l r - 2 l-

:f,mlwl

dx

3J,t+1

Bl n l r * 1 l + g t n C A l n l r * 1 l+ l n c )

.

Case2: The factors of Q(x) are all linear, and some are repeated. Suppose that (r - a) is a p-fold factor. Then, corresponding to this factor, there will be the sum of p paftial fractions A, Ar, .+. , --+, (x - a,)o ' (x - a,)o-r '

*-T A o - , G-F

-r -

Ao i-

where Ar, A", . , Ao are constants to be determined. Example 1 following illustrates this caseand the method of determining each 46.

11.4 INTEGRATIONOF RATIONALFUNCTIONSBY PARTIALFRACTIONS.CASES 1 AND 2

507

solurroN: The fraction in the integrand is written as a sum of partial fractions as follows: x}-L

_ A

Wry:V-

, B ,

C

J-

D

=L

rTGTF-1;-2)''

E

x-2

(5)

Multiplying on both The above is an identity for all r (exceptx:0,2). sides of (5) by the lowest common denominator, we get x3-l = A(x-2)t +Bx(x-2)s * Cxz+D*(x-2) + E*(x-2)z xs- L = A(xs - 5f + l2x - 8) * Bx(xs- 5* + l2x - 8) + C* + Dxs- 2Df + Ef(* - 4x * 4) xs-L = (B* E)xa* (A-58*D-AE)xz

+ (-6A + r2B+ C - 2D+ 4E)* + (L2A- 8B)r- 8,4 Equatingthe coefficientsof like powersof x, we obtain Bi

E:0

A-68+D-4E:L -6A+728+C-zD *4E:0 t2A-88 -0 -84--1 Solvingr we get

A:i

B:*

C:I

Therefore, from (5) we have

D:1

t r -- _ 3

lJ

16


SOLUTION:

1 A _ LB u2-a2 u-a' u*a Multiplying by (u - a) (u * a), *e get

1 - A ( u * a ) + B ( u- a ) L: (A+B)u*Aa-Ba Equating coefficients, we have

A+B_O Aa-Ba-l Solving simultaneously, we get ^1rh1

A-+

za

and B-

2a ' 2Ln

:*rnlu-

ol

I 2a

lnlu+al+C

(6)

-*k

2a

^ ^ 1l u-+ a l '* c lu ol

This is also listed as a formula. (7)

FRACTIONS. CASES1 AND 2 FUNCTIONS BY PARTIAL 11.4INTEGRATION OF RATIONAL

In chemistry, the law of massaction alfords us an application of integration that leads to the use of partial fractions. Under certain conditions it is found that a substance A reacts with a substance B to form a third substance G in such a way that the rate of change of the amount of C is proportional to the product of the amounts of A and B remaining at any given time. Supposethat initially there are a grams of A and B grams of B and that r grams of A combine with s grams of B to form (r * s) grams of C. If we let r be the number of grams of substanceC present at f units of time, then C contains rxl(r * s) grams of A and sxl(r * s) grams of B' The number of grams of substanceA remaining is then q - rxl(r * s), and the number of grams of substAnce B remaining is F- sxl(t * s). Therefore, the law of mass action gives us @ - nu ( ^. - -tt_\/" - --q{-\ A: \d- /+s/\p--l+s/ where K is the constant of proportionality. This equation can be written as

dx ,4rt ,^(+ r A:@f

(8)

a - . * \ ( ' tst - B' - r ) / /\

Letting k__Krs trff

a-

r*s '

ot

b -: * p a

s

r

Eq. (8) becomes dx

ii-k(a-r)(b-x)

(e)

We can seParate the variables in Eq. (9) and get

dx (a-r)(b-x)

:kdt

lf a : b, then the left side of the above equation can be integrated by the power formula. If a + b, partial fractions can be used for the integration.

ExAMPr.n3: A chemical reaction causes a substance Ato combine with a substance B to form a substance C so that the law of mass action is obeyed. If in Eq. (9) a- 8 and b - 6, and 2 g of substance C are formed in L0 min, how many grams of C are formed in 15 min?

soLUrIoN: Lettin g x grams be the amount of substance C present at t Equation minutes, we have the initial conditions shown in Table 11,.4.'1,. (9) becomes dx fr- k(8 x)(6 x) Separating the variables we have

dx (8- x)(6-x)

(10)

510


Writing the integrand as the sum of partial fractions gives

Table11.4.1

t

0

10

15

x

0

2

Xs

1LB @=3_r+

6_x

from which we obtain 1,: A(6-x) +B(8-r) Substituting 6 for x gives B : t, and substituting 8 for x givesA - -+. Hence,Eg. (10) is written as

--rJ8--lld*

f

+lldr_

zJG:k

Jdt

Integrating, we have

*ln 18-xl-*ln 16-xl+*ln lCl:tt

, | 5 - n l ::-_ 2 k t [email protected])-l 5-x ffi:Qs-21;t substituting x:0, 6-x 8-r

t - 0 in Eq. (11) we get c - *. Hence,

_ 3 o_zkt 4v

Substitutingx : 2, t:

(r2) L0 in Eq. (12), we have

6 : te-zort e- 20k:

9€.

Substituting )c- xrs,t - L5 into Eq. (lZ), we get 6-fu-

3 n-sok

3{:4, 4(6-

x.'")-

(11)

3(e-2ok)stz(g- xrr)

24- 4xrs- 3(8)t/,(8- xG)

24- Axrr:ry (8- )crs) 54 - 32\D xrs:W - 2.6 Therefote, there will be 2.6 g of substance C formed in L5 min.

-INTEGRATIONRATIONAL CASES 1 AND FUNCTIONSBY PARTIALFRACTIONS. 11.4 OF

511

L1,,4 Exercises In ExercisesL through "l'6,evaluatethe indefinite integral.

l"v

,Î' d x

J xz_4

(!x-? ax

44 ' [ x 3 - x 2 - 2 x J

.t ,' Jr f f i *dzx

10'r d t J

* z . [ =x 2**ix d -6

4'

J

5I 8I LLI(2x*

3) (x + 1)'

dx

1 4 |. " ' =2 x=5 -' x*nI t a * J

.,F | 44x3*30x2*52x+17 rD.J 9xa-5x3-LLx2*4x+4

, dx

In Exercises17 through 24, evaluatethe definite integral.

(x' - 3) dx

/Fr.. 5 ' Jf' r f f i

24.

t

x2 dx

o 2x3*9x2+t2x+4

& . F i n d t h e a r e a o ft h e r e g i o n b o u n d e d b y t h e c u r v e y : ( x - l ) l e - S x * 5 ) , t h e r a x i s , a n d t h e l i n e s x : 4 a r t d ' x : 6 ' 25. Find the centroid of the region in Exercise25. y axis. 27. Find the volume of the solid of revolution generated by revolving the region in Exercise25 about the : 4 x' 28. Find the area of the regfrn in the first quadrant bounded by the curve (r + z)'y 29. Find the centroid of the region of Exercise28. 30. Find the volume of the solid of revolution generated if the region in Exercise28 is revolved about the r axis. 31. Suppose in Example 3 that a : 5 and b : 4 and I g of substanceC is formed in 5 min. How many grams of C are formed in 10 min? substanceCisformedin4min'Howlongwillittake2gof subSflSoppor"inExample3 thata:5and,b:3and1gof stance C to be formed? of the amount of the substance 33. At any instant the rate at whictr a substancedissolves is proportional to the product in solution at that instant and substance pr"r"rrt at that instant and the difference between the concentration of the is mixed with 10 lb of salt material insoluble quantity of A the concentration of the substance in a saturated solution. in 10 rnin and the condissolves salt 5 lb of If gal water. bf 20 containing in a tank is dissolving initially, and the salt min? in 20 centration of salt in a saturated rolotio.. is 3 lb/gal, how much salt will dissolve then 34. A particle is moving along a straight line so that if u ftlsec is the velocity of the particle at f sec, t+3

a:F+3,tTr. Find the distance traveled by the particle from.the time when t:

0 to the time when f : 2'

OFINTEGRATION 512 TECHNIQUES INTEGRATION OF Case3: The factors of Q (x) are linear and quadratic, and none of the qua1.1..5 RATIONAL FUNCTIONS dratic factors is repeated. BY PARTIAL FRACTIONS. Corresponding to the quadratic f.actor* * px * 4 in the denominator CASES 3 AND 4z is the partial fraction of the form THE DENOMINATOR Ax-t B CONTAINS **px*q QUADRATIC FACTORS

soLUrIoN; The fraction in the integrand is written as a sum of partial fractions as follows: x2-2x-3

:

( x - 1 ) ( x z+ 2 x * 2 )

Ax*B , C x+2x+2T x-"1

(1)

Multiplying on both sides of (1) by the lowest common denominator, we have * - 2x- 3 : *-2x-3:

(Ax * B) (x - 1,)+ C(* * 2x * 2) (A+C)xz+ (B- A-r2C)x+ (zC-B)

Equating coefficients of like powers of r gives Al

C:7

B-Al2C:-2 2C-B:-3 Solving for A, B, and C, we obtain

A:E

B:!

C:_t

Substituting thesevalues into (1.),we get

:

x2-2x-3

(x-l)(x'*2x+2)

I

x2-2x-3

(x-l)(x'*2x+2)

tx+E * -+

x2+2x*2',

Jc-1

dx

xdx 9f dx 4f dx ,7f '5 (2) 5 J x2+2x*2 5J x-t J )c2+2x*2 To integratu I@ dx)l(* -l2x*2), we see that the differential of the denominator is 2(x t 7) dx; so we add and subtract 1 in the numerator, thereby giving us

9f

xdx

9f

(r+1)dx

5Jm:EJW-TJm

9f

dx

Substituting from (3) into (2) and combining terms, we get

(3)

FUNCTIONS FRACTIONS, CASES3 AND 4 11.5INTEGRATION OF RATIONAL BY PARTIAL

I

513

x2-2x-3 (x - l ) ( x ' * 2 x + 2 9. L f 2@+1) dx 5

dx x-"1,

9 ln l*'+2x+21 10

lr- 1l

zJm

l*' + 2x* 2l -?tan-1(r* 1,)- # ln lt 1l + + ln C l c ( x z + z x + z ) n l_ ) (x-1)' I f,tan-l(r*L) | a ILLUSTRATION L: In Example 1, we would have saved some steps if instead of (1) we had expressed the original fraction as

x2-2x-3

D ( 2 x + 2 )+ E , ' \-rw

(x-I)(x'*2x+2)

t

.

-/

F -

-

x2*2x*2

+

x-1

NorE: We write D(2x * 2) + E instead of Ax * B because 2x*2:D,(**2x*2) Then, solving for D, E, and F, we obtain

D: *

E=-3

F:-t

giving us (4) directly.

.

Case4: The factors of Q(r) are linear and quadratic, and some of the quadratic factors are repeated. If. x2* px * q is an n-fold quadratic factor of QG) ' then, corresponding to this factor (f + px * q)", we have the sum of the following n partial fractions: Arxi-B,

Azx*Bz

-

+

@TV|TT'(x'+px*q)"-'

. . -

Anx*Bn

f*px*q

-5x*2)3, o rLLUSrRArroN2: If the denominator contains the factor (f we have, corresponding to this factor, Ax* B

,

Cx*D

@r@ or, more conveniently, A(2x-5) +B -C(Zx-5) *D * E(2r-5) *F ( i , - 5 r * 2 ) t ' ( x ,- 5 x + 2 ) ' ' x 2- 5 x + 2 SOLUTION:

x-2

x(x'-4x+5)'

A , B(2x-4)+C =;+ffi+

D(Zx-4)+E x:2-4x*5


Multiplying on both sidesof (5) by the lowestcommondenominator, we have x - 2 : A(f - 4x * S)2* x(2Bx- 48 + C) + x(xz- 4x * 5) (2Dx- 4D + E) x - 2 = Af + 16A* + 25A- 9Af + tlA* - 40Ax* 28* - 4Bx* Cx * 2Dxa- l2Dxs* Ers* 26D* - 4Ef - 20Dx* SEx x - 2 = ( A + 2 D ) f + ( - 8 A - 7 2 D * E ) x a+ ( 2 6 A + z B+ z 6 D - 4 E ) * + (-40A- 48 + C - 20D* SE)r + 25A Equatingcoefficientsand solving simultaneously,we obtain A:-&

B:*

C:*

D:*

E:-*

Therefore, (x-2\ dx

|

I ,@=a*g

_ ![ , ( ? x - + )a x = + L [ _ _ _ _ _ d r _ _ - 1 f L 2 x - 4 )d x 2 sJ[ @x*' 5 2 sJ F - + x + s J {P=E+W-E J At=7;*5y' dx - E4 f

:-z

JE=E+s

:-E

)

r nlr l-

1

,lr +i

1

flr

Jl1g= .g;iqTp

+litn lx2- 4x +51

_+l

we evaluate separatelythe integrals in the third and right side of (5). tdxfdx

lwffir:Jffi - 2 - tan 0, where 0

dx:

s e c z0 d 0 a n d ( x dx

[(r- 2)'+ L]'

2 ) ' + r _ t a n 2 e + 1 , : s e c 20 . H e n c e ,

515

FRACTIONS. CASES3 AND 4 FUNCTIONS BY PARTIAL OF RATIONAL 11.5INTEGRATION

A1

:;+|sinecos0*C1 - 2)' We find sin 0 Becausetan 0 : x - 2 and -tr < 0 < tr, 0 : tan-l(r and cos g from Figs.11.5.1(if. x > 2) and 11.5.2(if x < 2).In either case x)2

v -+ s i n0 vr-r

F i g u r e11 . 5 . 1

l/xz

-

4x + 5

a n d c o s0 -

ffi

So we have 1 1 d, tan-'tr- 2)'r'ffi rlz:i + 2)z J I(r I

x-2

'

L

7Et=fi77

+ cr

Thus, '

dx

I

lT6zr+W:|tan-'(x-2)

x-2 *'" ', +Z@-4r*5)

V)

Now, consideringthe other integralon the right side of.(6),we have fdxldx

Jffi:tan-l(r-2)

I @ffiT|:

(8)

+cz

Substitutingfrom (7) and (8) into (5), we get f

x
(x-2) dx

JWTW

F i g ur e 1 1. 5 . 2

5(f

r - ll *t'l- 4 x * 5 1 : ; i r1 n

Exercises71.5 In Exercises L through L6, evaluate the ',[dx r' J 2x3*x

2.

c.Jffi

6.

1 - x - '22 2)+ tan-'(x fu ffi*rtat + f r" l*'- 4x*sl - +tan-l(x- z)+ c

- 5 36 taaan- - -r((lxy --r2\ +) + f f i ; ax s- 4f

.-+r?\ -


516

9. 13.

[ (2*-x+2) dx J x'+zx'+x

IJ (4ftsdx + e),

(2t!_9*) ,r 0u,' J[ (,l +. 3 x l - 2 x + d3 *)

,trL. ' J[

" Iffi##

" /ee;f$f;:e

In Exercises17 through 24, evaluate the definite integral. 17.

fg+#4

20. ft

9 dx

(xr+zx-t) dx zTxs-t

f dx

(* + 3x + 3) dx 23. ft J" r"+*+x+l

" I#h

n.l"fffla-

"' rol',,*rIo1,*,

(x -t l) dx

,n

['t, "' Jo r - |

"' J-,(2r -r2x* t)z

J, 8'+l

d* 1 L 2z ' [l 1 t z * r 1 t

fnt2 cosx dx za'J,,usinr+sinsr

25. Find the areaof the region bounded by the curve y:41(xs * 8), the r axis, the y axis, and the line x:2. 26. Find the abscissaof the centroid of the region in Exercise 25. 27. Find the mass of a lamina in the shape of the region of Exercise25 if the area density at any point (x,y) iskf slugslfP. 28. Find the volume of the solid of revolution generated by revolving the region of Exercise25 about the y axis. 29. A particle is moving along a straight line so that if v ftlsec is the velocity of the particle at I sec, then a-

t2-t+l (f+2)r(t2+L)

Find a formula for the distance traveled by the particle from the time when f : 0 to the time when f : fr.

11..6INTEGRATION OF If an integrand is a rational function of sin x and cos x, Ttcan be reduced to RATIONAL FUNCTIONS a rational function of z by the substitution oF SINE AND COSINE z_ hntx (1) c o s zV - 1 and letting y - +x, we have

Using the identity cos 2y : 2 c o sx - 2

c o s 2 L r x -1 -

2.24

-1 sec2+r

1+tanzLx

r

I*zz

I

So .

(2)

In a similar manner, from the identity sin 2y: 2 sin y cosy, we have s i n ) c - 2 s i n t r c o sl x - 2 -2

sin *r cos'tx cos+x ,

r-

tan - --- tt z-

1 sec, tx

--- - Z t a n t x 1 + tanz tx

\ OF RATIONALFUNCTIONSOF SINE AND COSINE 11.6 INTEGRATION

517

Thus, (3) Becausez-tani,x, dz- * sec2tx dx - +(1 + tan2tx) dx Hence, (4)

Find

soLurroN:

Letting

dx l-sin

x*cos )c

I

dx 1-sinxlcosr

tan ix and using (2) , (3), and (4) , we obtain

:T

1-RT

-n

I I

2dz t*22

2z

,1-z' L+22 dz

r *+-z2 '2 ) - 2 2 * ( I - z ' ) / (1 - 2 l d zdz 2 I 22 - t22z

--l

lz dz L 1 --zz

1n l n l L1-- r lz l + C tan Lxl+ C ln ln llr1-; te

ExAMPtn 2:

Find

r secxdx

I

J

by using the substitution of this section.

s o , , u r l o Nl,s e cx d x : l + c o s x

J J using (2) and (4), we get and tan z: Letting tx

I A*:l+.!+,i_2t 1'-z'-'J

J c o si - J L + 7

+

t-22

"!'"1'.4, we have Using formula (7) from Sec. f dz ,11*zl *' 2Jft-rnlffil Therefore,

r r d x - r n1l + t a n ? r*l. Jsec' tt-tan#l Equation (5) can be written in another form by noting that 1' tan *n

7 TECHNIQUEO S F INTEGRATION

andusing the identity tan(a * B) : (tan a * tan illG we have

-tan d tan F). So

ta4 *"t,* tat *r,-l +' c f ,"" r dx: ln | = - tan r tl. *n . tan|xl or, equivalently,

r IJ q*qr dr - ln lt*(*o + *r)l + C

(6)

Formula (U) ,, un alternative form of the formula I

I

secr dx:ln

lsecr * tan rl * C

*nr"n is obtained by the trick of multiplying the numerator and denominator of the integrand by sec r * tan r. It is worth noting that still another form for J sec x dx canbe obtained as follows: f r l secrdx: l

J

dx

J "o--:J

Substituting u : sin x, du:

f c-costt:J osxdx f

cosxdx

l=;fit;

cosx dx, we have

' lt+ul *a:hrll+ti"tl"'*c A r : [ -,d- uu r : t t t1l T -l JI r " . r J tr_sm.rl Because-l. < sin r < l for allx,1.*sin rand 1- sin rarenonnegative; hence, the absolute-value bars can be removed and we have |

/1*sinr\t/r,

secx dx: tn (i:-ffi/ J

/i+sinr * c- _:1 _ln 11=16;

+ c^

(7)

Exercises11.6 In Exercises 1 through'1,2, evaluate the indefinite integral.

,r ' [--4Js+4costr J.

dx ,..' r 2 l3-fu

dx costr-sinx*L

o.

8dx 3 c o s2 x * L

fdxldx l-

J sinr*tanx

s.[-Jro*

J 6+4secx

J tanx-l

1 0 .I

d*

dx

J sinr-tanr

2 s i n x * 2 c o sx * 3

In Exercises L3 through L8, evaluate the definite integral. frtz

1 3 .1

Jo fntz

1 , 6 .|

Jo

dx

5. *s i n x * 3 dx

3 " _+ c o s 2 x

fnta - 8 dx

1 4 .|

ro

tanx+j.

frts 3 dx IT. I Jnt62sin2x*"1,

4I 8I 12I

dx

cot 2x(1 - cos 2r)

1 1 . 7 M I S C E L L A N E O USSU B S T I T U T I O N S 5 1 9

19. Show that formula (7) of this section is equivalent to the formula J sec r dx: lnlsec r f tan rl * C. (xrNt Multiply the numerator and denominator under the radical sign by (1 + sin r).) 20. By using the substitution z : tam*r, prove that f / 1- c o s r

J

c s cr d x : l n V 1 a . * ;

+c

lnlcsc r - cot xl + C. (nrr.rr:Use a method

21. Show that the result in Exercise20 is equivalent to the formula J cscx dr: similar to that suggestedin the hint for Exercise 19.)

IT.7 MISCELLANEOUS If an integrand involves fractional powers of a variable r, the integrand SUBSTITUTIONS can be simplified by the substitution x: zn where n is the lowest common denominator of the denominators of the exponents. This is illustrated by the following example. z 5 tdz. S soLUTroN: we let r : z6; then dxt c : 625

f )crtzdx

J].*,.:J Dividing

- =oL$, l t zt(6zt dz) -

t

28

3-,

;dz

+1 J z 2+

inator, IO mrin the numerator by th ' d €en( z 2-_1 1 +

iz'.3 -

-

1 z2+L

nave

la,

z * tan- ,)

+C

6 ry7t6 - grst6 + 2XU2- 5Xrt6+ 5r tar fr 7* 5

xrt6+ C

No general rule can be given to determine a substitution that will result in a simpler integrand. Sometimes a substitution which does not rationalize the given integrand may still result in a simpler integrand. EXAMPLI. 2:

EVAIUATE

f

I xit/xz+4 dx J

SOL

Let z- \m.

Then

x2+ 4, and 2z dz - 2x dx. So we

hav

rrtffidx

fr

| (*1r\Eq

JJ

x dx- I e, - 4)zz(zdz)

f -!.622) _ _ dz izz - Ez, + #zt + C I ku gza*

J

+ 5601+ C #ztll1za -'1,6822 * 4) + 5601+ C + 4)2 - 1,68(x2 T+5(x, + 4)srz715(x2 - 48x2+ L28l + C r+E(x' * 4)srzSLlxa

520


ExAMPLE3:

Evaluate

ld* Jxzm

sot,urloN: If x - tlz, then dx: -dzlzz. We have, then, -dz

d*

I

-[

J*2ffi

by using the reciprocalsubstitution x - Uz.

,'

J/_1 \ tre; r \;r) \,17*Z-

ifz>0 if z < 0

f

l:

1 f f : + . {?l - 2 2 + 6 ) d z , ^ f

zdz

2JV27*62-22

J\/27*62-22

dz J\/27*62-z2

:-z-J

1ffifdz

f -dz

ffi+"

"J

ffi

+ 3 s i n - , +b + C 1-g

l__

5

1,

\ 1 2+7| - t + 3 ffiLF)

-

s i n -+r + TJ

ffi..r

c

SIN

+JSln

Substituting from (2) into (1) we get

Id*

Jyzffi

--r' ^" 1 , - 3 x .

6i*+c

if x
11.8 THE TRAPEZOIDALRULE

ffi

521

lfx>0

rfx<0

11,7 Exercises In Exercises L through t6, evaluate the indefinite

integral.

r4 .f J f f i ; d x ,l'Jffi o

dx

,- f. J f f i

dx

fdx

72-Jffi Use the reciprocal substitution x: Ilz. t3.DoExercise1,2byusingthesubstitutionffi:z-x.

r o .f Jd xf f i In Exercises 17 through 22, evaluate the definite integral. d* 1 Lnl . Jl no f f i f18dx

20' J,uffi 1.1..8THE TRAPEZOTDAL RULE

1 9 .f2J , , , @dx f2 dx /L'L' J, xlx2 * 4x 4 ,.tt

|

-

We have seen previously how many problems can be solved by evaluating definite integrals. In evaluating a definite integral by the fundamental theorem of the calculus,it is necessaryto find an indefinite integral (or antiderivative). There are many functions for which there is no known method for finding an indefinite integral. However, if a function f is.continuous on a closed interval fa,bl, the definite integral I'"f(x) dr exists and is a unique number. In this section and the next we leam two methods value of a definite integral. These methods for computit g "tt "pptoximate can often give us fairly good accuracyand can be used for evaluating a definite integral by electronic computers. The first method is known as the trapezoidalrule.


We know that if f is continuous on the closed interval [a, b], the definite integral Io"f(x) dr is the limit of a Riemann sum fbn

I f (x) dx: lim ). fG) Lix Ja l l a l l *o 7 : r

The geometric interpretation of the Riemann sum n

fG)

Lix

i:l

is that it is equal to the sum of the measuresof the areasof the rectangles lying above the r axis plus the negative of the measuresof the areasofthe rectangleslying below the.r axis (seeFig.7.3.3). To approximate the measure of the area of a region let us use trapezoids instead of rectangles. Let us also use regular partitions and function values at equally spaced points. Thus, if we are considering the definite integral Io,f (x) ilx,we divide the interval la, bl into n subintervals, each of length Ax: (b - a) ln. This gives the following (n + 1.) points: xo: a, xr: a * Lx, xr: a * 2 Ax, . , xt: a* i Lx, . . r xn-r: a* (n-l) Ax,xn:b. Then the definite integral I2 f @) dx may be expressedas the sum of z definite integrals as follows:

I:

f(x) dx

f(x)

f(x) dx

+ . . . + [:' f@) dx*. . . + f f@ ) dx ( 1) Jrr_r Jrn_,

To interpret (1) geometrically, refer to Fig. 11..8.i.,in which we have takenf(x) > 0forall r in la, bh however, (1) holds for anyfunction which is continuous on la, bf.

f(x)

Figur1 e1.8.1 Then the integral I t' f @) dx is the measure of the area of the region bounded by the x axis, the lines x : a a n d x : x 1 r a n d the portion of the

11 . 8 T H E T R A P E Z O I D ARL U L E

curve from P0to P1.This integral may be approximated by the measure of the area of the trapezoid formed by the lines r : a, x: x1, P6P1,and the x axis. By a formula from geometry, the measureof the areaof 'this trapezoid is

iLf@,)+ f(x)l Lx

Similarly, the other integrals on the right side of (1) may be approximated by the measure of the area of a trapezoid. Using the symbol "-" fot "is approximately equal to," we have then for the ith integral ftt

(*) ilx - tlf (x;,) + f (x,)l Lx J",_,f

(2)

So, using (2) for each of the integrals on the right side of (1) , we have fb

J,f{*) Thus,

d x = E l f ( x )+ / ( r ' ) l L x + i [ f @ ' ) + / ( r , )A] r * ' ' + tlf(x"-,) + f (x"-')fLx+ tlf Q"-r)* /(r")l Ar ,r,

Formula (3) is known as the trapezoidalrule.

ExAMPLE 1:

Compute

Pdx Joffi by using the trapezoidal rule with n: 6. Expressthe result to three decimal places.Check by finding the exactvalue of the definite integral.

Because la, bl - [0, 3] and

solurroN:

6, we have

,o:1:0.5 n6

Lx:b Therefore,

f'

d*

Jo LG*x2

-

+

+ 2f(xr)+ f (x)l [/(ro)+zf(x,)* 2f(xr)+ Zf(xr)+ 2f(xn)

where f (x) - Il(16 + x').The computationof the sum in brackets in the aboveis shown in Table11.8.1.So, [' j'

Jo 6fu:

0'25(0'5427)

I'J*

Q'l.607

Jorcfu:

Rounding the result off to three decimal places, we get

I'J'

Jorcfu:

Q"l'6!

We check by finding the exactvalue. We have It

d*

1 . _^-1rlt

J of f i : a t a n ' Z l o


: + tan-l *-

* tan-l 0

- +(0) : +(0.6435) - Q.1,6A9 to four decimal places Table11.8.1

f (x,)

X1

0 1 2 3 4 5 6

0 0.5 L L.5 2 2.5 3

0.0625 0.0615 0.0588 0.0548 0.0500 0.0450 0.0400

1, 2 2 2 2 2 1,

I

o.o62si

0.L230 I 0.LL76 I 0."1.096 0.1000 0.0900 0.0400 I

- 0.6427 io'r@')

i :0

I

To consider the accuracyof the approximation of a definite integral by the trapezoidal rule, we prove first that as Ar approaches zero and.n increases without bound, the limit of the approximation by the trapezoidal rule is the exact value of the definite integral. Let

T: i Lx[f(x) + zf(x) + . . . -r 2f(x,_,)+ f(x")] Then

T :l f( xr ) + f( x) + . . . + f( x") l A^x+ilf( x)- f( x,) l Lx or, equivalently, n

r: ) f(x) ax + ilf@) - f(b)l Ax i=1 Therefore, if. n -->*o and At -+ 0, we have

lim ttf@) - f(b)) Lx ["] r: aa-o !* i 'f@1)ax* a*_o ?r

Aa-o

fb

: I f @ )d x + 0 Ja'

Thus, we can make the difference between Tand the value of the definite integral as small as we please by taking n sufficiently large (and consequently Ar sufficiently small). The following theorem, which is proved in advanced calculus, gives us a method for estimating the error obtained when using the trapez6idal rule. The error is denoted by e7.

11.8 THE TRAPEZOIDAL RULE

1,L.8.LTheorem

Let thre funct: Iu rcti10n f 1 be continuous on the closedinterval fa, bl, and f ' andf " )n both exisl ist or 1 [ a , bb l . If on fbb

tl f(' ( x )

€yT -

d: dl xx'- T

,J/a

where3 TI'i is tl he aP rro ximate value of [f f (x) dx found by the trapezoidal th PPI rule, th en threre is3 Ssor then ( me number 4 in la, bl such that €y

EXAMPug 2: Find the bounds for the error in the result of Example L.

__

1 t2

,(b

at ) f '" (q) (Ar)'

(4)

soLUrIoN: We first find the absolute minimum and absolute maximum values of f" (x) on [0, 3]. f ( x ) : ( 1 0+ l ) - t '(x) : -2x(16 * f1-z f - 2(LG+ f):2: 10* - g2)(L6-t *1-t f,,(x):B*(1.6 * f;-s -6x(6xz - 32)(1.6* *1-+ * L2x(1'6* l; -r f "' (x) : :24x(L5-

xz)(L6* *1-+

Becausef"'(x) > 0 for all x in the open interval (0, 3), then /"'is increasing on the open interval (0,3). Therefore, the absolute minimum value of f " on [0, 3] is f " (0), and the absolute maximum value of f " on [0, 3] is f " (3).

f"(il --h

andf"\3)-&

T a k i n g n - 0 o n t h e right side of (4), w€ get

3 ( r\L _1 -:r,.\-ru)z

2048 Taking n 3 on the right side of (4), w€ have \L - 3122 L2 \E2s )Z

11 45,000

Therefore, if e1 is the error in the result of Example L, we conclude

1L.8 Exercises In Exercises 1 through 14, compute the approximate value of the given definite integral by the trapezoidal rule for the indicated value of z. Express the result to three decimal places.In ExercisesI through 8, find the exact value of the definite integral and compare the result with the approximation.


526

1. ['4,n:S Jr )c'

4. 7.

r2

Jrx!4-f fr

J,

z . f " Jt *lx ' , n : 8

3.

?rrdx 5 . J,ffi;n:5

6.

J,

dx;n-8

sinr dx;n-5

8.

fn

J,

ln(l * x') d x ;n - 4 \R

dx;n-6

tr cos x2 dx; n:

LL. [' ," dx; nJo

14.

fr

J.

4

5

g.

r2

Jo*"dx;n-4 f3

J,

V l + x 2d x ; n - 6

sin r dx; n [37rtz x Jnn

6

1 . 2f." + i " d x ; n - 6 L*x Jo

VI + xsdx;n: 4

In Exercises15 through 20, find. the bounds for the error in the approximation of the indicated exercises. 15. Exercise1 15. Exercise2 12. Exercise3 18. Exercise6

19. Exercise 1.1

20. Exercise L0

21. The integral Io' e-" dx is very important in mathematical statistics. It is called a "probability integral" and it cannot be evaluated exactly in terms of elementary functions. Use the trapezoidal rule with n :5 to find an approximate value and express the result to three decimal places. 22. The region bounded by the curve whose equation is y : g-ttz, the r axis, the y axis, and the line r: 2 is revolved about. the r axis. Find the volume of the solid of revolution generated. Approdmaie the definite integral by the trapezoidal rule to three decimal places, with z :5. 23' Show that the exact value of l& t/+ - f dx ig z. Approximate the definite integral by the trapezoidal rule to three decimal places,with z:8, and compare the value so obtained with the exact value. 24. Show that the exactvalue of.t losdxl (x * 1) is ln 2. Approximate the definite integral by the trapezoidal rule with n : 6 to three decimal places, and compare the value so obtained with the exact value of ln2 as given in a table.

11'9 SIMPSoN'S RULE

1L.9.LTheorem

Another method for approximating the value of a definite integral is provided by Simpson'srule (sometimesreferred to as the parabotiirute). For a given partition of the dosed interval [a, bl, simpson,s rule rr,r"liy gives a better approximation than the trapezoidal rule. Howevei, simpson's rule requires more effort to apply. tn the trapezoidal rule, successive points on the graph of y: /(r) are connected by segments of straight lines, whereas in simpson's rule the points are conhectJaby ,"gments of parabolas. Before simpson's rule is developed, we statl an--d pnove a theorem which will be needed. If.Po(xs,Ao),Pr(xr, U), and pr(xr, Ar\ are three noncollinearpoints on the parabola having the equation y: A* + Bx * C, where Ao2 0, Ar > 0, Az > 0, xr : xo * h, and xz : xo * 2h, then the measure of the area of the region bounded by the parabola, the r axis, and the lines l: xo arrdx: x, is given by *h(yo*4yr*yr)

(1)

11 . 9 S I M P S O N 'R SU L E

( x o ,y o )

527

pRooF: The parabola whose equation is y : Ax2 * Bx * C has a vertical 'axis. Refer to Fig. 1'1..9.'1., which shows the region bounded by the parabola, the x axis,and the lines r : ro afld x: xz. BecausePo, Pr, and P, are points on the parabola, their coordinates satisfy the equation of the parabola. So when we replace xlby xs * h, and xrby xs*2h,wehave yr: Axoz* Bxs-l C lr : A(xo + h)2 + B (16* h) + C : A(xs2I 2hxs* h2) + B (xo+ h) + C U z : A ( x o + 2 h ) z* B ( x s + 2 h ) + C : A ( x n z* 4 h x s +4 h 2 )* B ( x e+ 2 h ) + C Therefore, Ao* 4yr* yr-

F i g u r e11. 9 . 1

A ( 6 x o ' * 1 2 h x 0+ t h ' ) + B ( 6 x o + 6 h ) + 6 C

(2)

Now if K square units is the area of the region, then K can be computed by the limit of a Riemann sum, and we have

K- lim Atr-0

n sl Ll i:l 2h

(At,'* Bt,+ c) L,x (Ax' * Bx + C) dx

- $Ax3+ LBx' +

- +A(ro* 2h)'

2h)2+ C(ro + 2h) - (*Axr' + +Bxo'* Cxs)

- +hlA(6xo'* L2hx0+ th2) + B(6xo+ 6h) + 6Cl

(3)

Substituting from (2) in (3), we get K- *hlyot 4y, * yrl

I

Let the function f be continuous on the closed interval [4, b]. Consider a regular partition of the interval la, bl of 2n subintervals (2n is used instead of z becausewe want an even number of subintervals). The length of each subinterval is given by Ar : (b - a) l2n. Let the points on the curve y: f(x) having these partitioning points as abscissasbe denoted by Po(xo,yr), Pt(xt, yr), , Pro(xzn,!zn)) see Fig. 77.9.2,where /(r) = O for all x in fa, bl. We approximate the segmentof the ctfivey : f (x) from Psto P2by the segment of the parabola with its vertical axis through Pr, Pr, and Pr. Then by Theorem 11.9.1the measureof the area of the region bounded by this parabola,the r axis, and the lines r:ro and x: xz,withh: Lx, is given by $ Ar(yo * 4y, * yr) or

* Ar[/(ro) + af k)

+ f (xr)]

In a similar manner, we approximate the segment of the curve

528

T E C H N I Q U EO S F INTEGRATION

N

I N

n-

A:Xo.Xr

X2

i

Xs

X+

X5

X6

N F 1

I

\ N

HH

F i g ur e 1 1. 9 . 2

,

+

f1*2n L N

y : f (x) from P" to Paby the segment of the parabola with its vertical axis through P2,Ps,and Pn.The measure of the area of the region bounded by this parabola, the x axis, and the lines r: xs and x: xt is given by

i Lx(yr* 4ys* y,) or t Axff(xr)+ 4f(xr)+ f(x)l This processis continued until we have n such regions, and the measureof the area of the last region is given by

* Axlyro-r*4ar,-'* ar) or t Lx lf(xr^) + +1@r*_r) + f(xr*)l The sum of the measuresof the areas of these regions approximates the measure of the area of the region bounded by thelurv" *io"u uq,rution is y : f(x), the r axis, and the lines r : a andx: b. The measureof the area of this region is given by the definite integral ft fG) dx. so we have as an approximation to the definite integral

t Ax[f(x)+af@)+f(x,)l+ t Axlf(x,) +4f(x")+f(x)]+ . . . + * AxLf@zn-) t 4f(x,,-r)+ f(xr^-r)l+g axlf(xr^_r) + 4f(x2,) + f(xr^)l dx -: * Arlf (xo) + a f Q , ) + z f ( x r ) + a f f t s ) + Z f (+ x .)

+ 2f(xzn_z) + 4f(xrn_J+ f Gill where Lx - (b - a)l\n. Formula (4) is known as Simpson,srule.

ExAMPLEL: Use Simpson,s rule to approximate the value of

soLUrIoN' *, and

Applying Simpson's rule with 2n: 4, we have Ar : +(1 -

1, 12

lf (xi + af@r)+ 2f(xr)+ af@s)+ f (x)l

. . G)

RULE 1 1 . 9S I M P S O N ' S

with 2n: 4. Give the result to four decimal places.

The computation of the expression in brackets on the right side of (5) where f(x): U(x + l)' is shown in Table 1'!'.9.'l', Substituting the sum from Table 11'9'1in (5), we get

:

+

: Q.69325+ (8.31905)

T a b l e 1 7 . 9. 1

xi 0 t 2 3 41

kt

f@)

0 0.25 0.5 0.75

L.00000 0.80000 0.66567 0.57143 0.50000 4

kr'f@i)

| L.00000 3.20000 4 L-33334 2 2.28572 4 10.50000

k,f(x) - 8.31e06

i :0

Rounding off the result to four decimal places gives us F dy: J, x+t

Q v '.vG' vg g g

The exactvalue of y; dxl(x + 1) is found as follows:

I:h:'n

l r + t l ] , : r n 2 - l n 1: r n 2

From a table of natural logarithms, the value of ln 2 to four decimal places is 0.6931,which agreeswith our aPProximllion in the first three -0.0002' plu."t. And the elror in our aPProximationis In applying simpson,s rule, the larger we take the value of.Zn, the smaller will be the vaiue of. Lx,and so geometrically it seemsevident that the greater will be the accuracyof the approximation, becausea parabola p"rJttg through three points of a curve that are close to each other will be close to the curve throughout the subinterval of width 2 Ar' The following theorem, which is proved in advanced calculus, gives a method for determining the error in applying Simpson's rule' The error is denoted by es. 11.9.2Theorem

' Let the function f be continuous on the closed interval la, bl, and f , f " , ftt' and fcv)all exist on la, bl- lf It fb

€s: I f@)dx-S Ja

530


lvhere s is the approximatevalue of [! f (x) dr found by simpson'srule, then there is somenumber 11in fa, bl such that €,s: -rto (b - a)f Gv\ (il (Ax)n EXAMPLN2:

Find the bounds for

(6)

SOLUTION:

the effor in Example1.

f(x): (x* 1)-' f'(x) f " ( x ) 2 ( x* 1 ) - ' f "' (r) : -5(x * t;-+ f ( i v )( r ) : 2 4 ( x * t ; - s (x) : - 120(x * 1) -u 1
_ BecausefF)(x) < 0 for all r in [0,11, fiut is decreasingon [0, 1]. Thus, the absolute minimum value of /(tu) is at the right endpoini 1, and the absolute maximum value of .f(tu)on [0, 1] is at the left endpoint 0. f(iv)(o):24

and /(iv)(l) : *

Substituting 0 for q in the right side of.(6), we get -rto (b - a)/rivl(O)(Ar)a :-t+u Q4) (+)^: -r.#o : -0.00052 Substituting 1. for rl in the right side of (6), we have -

1

( . - -- - \ . r ( i v ) l r \ / ^ - - \ 4

180@

a ) f t v ) ( 1 )( A r ) 4 - -

So we concludethat -0.00052=€s <-0.00002

1'

m

_11_l\

a\z/

(7)

The inequ ality (7) agrees with the discussion in Example L regarding the error in the approximation of [or dxl (r + 1) by Simpson's rule because -0.00052

If f (x) is a polynomial of degree three or less, then ltiv)(r) = 0 and therefore€s: 0. In other words, simpson's rule gives an exactresult for a polynomial of the third degree or lower. This statement is geometrically obvious if f(x) is of the second or first degree becausein the first casethe graph of. y : f(x) is a parabola, and in the second case the graph is a straight line (a degenerateparabola). _ We apply Simpson's rule to the definite integral Il fe) dx, where f(-x) is a third-degree polynomial, and take 2n:2. Here'simpson,s rule gives an exactvalue for the definite integral. Because2n:2,-xs: a, x1:

SULE 11. 9 S I M P S O N ' R

531

i @ + b), and x2: b; and A,x- + (b - a) . So we have

I:f(x)dx-+

lro +4f(+) * f@f

Formula (8) is known as the prismoidalformula. In Sec.8.4, we leamed that Il f Q) dx yields the measure of the volume of a solid for which /(r) representsthe measure of the area of a plane section formed by a plane perpendicular to the r axis at a distance r units from the origin. Therefore, lf f (x) is a polynomial of the third degree or less,the prismoidal formula gives the exactmeasureof the volume of such a solid. EXAMPLE3: Find the volunie of a right-circular cone of height h and base radius r by the prismoidal formula.

soLUrIoN: Figure 11.9.3illustrates a right-circular cone having its vertex at the origin and its axis along the positive side of the r axis. The area of the plane section fi units from the origin is zr[g(fr)]'zsquareunits, where 8(x) : (rlh)x, which is obtained from an equation of line OA in the xy plane. So if. V cubic units is the volume of the right-circular cone, then lim llallo - 7:r

(fi,8(f i ))

nlg(r) I2 dx

\_/ L iX

\_---_.

,, \r - \i-l

,,

B

nf h2

F i g u r e1 1 . 9 . 3

dx

Evaluating t! x2 dx by the prismoidal formula (8) with f(x): a:0, andi@ + b) : th, we have

f , b:h,

:t,r"h v: $ .I o * n'T * h'): # (2h') 71.9 Exercises In Exercises 1 through 6, approximate the definite integral by Simpson's rule, using the indicated value of 22. Express the answers to three decimal places. Also, find the exact value of the definite integral and compare the results.

2

2n

t*.t2n-8 2n-4

6.

|

,ir, '

2n: 6

TECHNIQUES OF INTEGRATION In Exercises 7 throu gh 9, find bounds for the error in the indicated exercise.

7. ExerciseL

8. Exercise2

9. Exercise 5

Each of the definite integrals in Exercises 10 through 15 cannot be evaluated exactly in terms of elementary functions. use Simpson's rule, with the indicated value of 2n, to find an aPProximate value of the given definite integral. Express results to three decimal places. f nl2

Vsin r dx; 2n - G

10. I

Jo f2

13. I e-" dx; 2n - 4 Jo

In Exercises15 through 19, use the prismoidal formula to find the exact volume of the given solid. 16. A sphere of radius r. 17. A right-circular cylinder of height h and base radius r. 18. A right pyramid of height ft and base a square having a side of length s. 19. A frustrum of a right-circular cone of height ft, and base radii r, and.rr. 20' Show that the exact value of 4 lor \/T=7 dx is n. Then use Simpson's rule with 2n:6to get an approximate value of n tott/l=E dr to three decimii places. Compare the results. 21. Find an approximate value to four decimal places of the definite integral ,[f,/r logro cosx dx, (a) by the prismoidal for_ mula; (b) by simpson's rule, taking ax: #n; (c) by the trapezoidal ,-"r", iin"fi, : io. 22- Find' the area of the region enclosed by the loop of the curve whose equation is y2 : 8x2- x5.Evaluate the definite integral by Simpson,s rule, with 2n:9. ExpresJthe result to three decimal places.

ReaiewExercises (Chapter11) In Exercises I through 52, evaluate the indefinite integral.

L.

I

tun'4x cosa 4x dx

4t4

J x2\/a2* x2

T.

[ "or'

10t! J

ly

tx dx

+1

13I# 16.I

2. fsL.- 3 a* xs_x 5.

r I

8. [ffi

J Yx*L

11.

r I

+t -L

,sI#*

a-

sin r sin 3x dx

f

d*

\/4-

tan-l tfr Ax

14. tVzt - szdt J

J \/e, _ 1

314 J

J

tr[ffib, 2 0J X[ 'Y *

e"

6Ih sI#0. 12. .o, g cos20d0 / 15. (sec3x * csc3r)2dx I 18. x"er,dx I 2t.

/

sir,t 3r cosz3x dx

R E V I E WE X E R C I S E S

23I#

x3 cos x2 dx

26Iffi^

28tM du 31I sinb nx dx 34I 5 + 4dxs e c x sin x dx 37I T:r-ffi 40I ffi dx

43I 46I

4x2*x-2

f sin2 2t dt

e* dx I+ - 9e2' cot x dx 3 + 2 sin'x

4sIs i n x -dx 2 c s cx dx

s5t s8I 2 * 2 s dx * c o s x 61,. t x n l n x dx 1nI T

27.

2s|ffi* 32I# 35I

30. 33.

*

38I# 4rIh 44I# 47. l*

dx

x3-5x2*8x-4

s i n - 1x d x

r ln(sinx) dx 50. /.o,

s3IY*

36.

I I I I I I I I I

ett2cos2t dt \6-a

,

7t)

cscs x dx x5 dx (x' - a.')t

39.

tffiat

42.

-dt

45.

cotz 3x csca3x dx

48.

51.

\tr-1 a. yt +L

dx

xffi cos3t dt

sin3fffi

54.

dx s6. /r"Q2tL)

57.

se. ItffiAx

60.

xdx,n>0 52. Irur,"rseca

In Exercises 53 through 94, evaluate the definite integral.

rr

tT-

64'Jr',rl#o* GT f+ Jo \/4+t2

70Iffi 73'

7nl8

2YdY J,,rrcots

5sr@2 xa2x* x * 4 58r 71,. r 74r

,

l,

sins t cos3tdt

l4

seca x dx

(2" * x2) dx


75.["''wry {ax Jo

xB dx

\ffi

7-

76. 79.

t;

(ln r) 2 dx

77.

logro tE

80.

Ax

f!n12

xe" cos x2 dx

83.

fuz 2x dx S5.J,ffi gg.

85.

dx [7'tr2 cosa 3r Jo

g1,.l"''

I;

x3lx

B

l,*

82. I Jo

(2x' - 2x + 1) dx

89.

d,

92.

J o 1 2 + 1 3 c o sf fn

r

Isin x- cosrl dx

T

lcossrl dx

I,', l,'@ay *s1fg

4*

I (rrD\m dr

,. ""i n * f 94. | /'=' at Jrn,, L + cos Lzt

In Exercises95 and 95, find an approximate value-for the given integral by using (a) the trapezoidal rule, and (b) simpson,s rule. Take four subintervals and express the results to thiee decimir places. fsts

96. I

Vl*x7dx

JT

97. Find the length of the arc of the parabola yz :6r

from x:

6 Io x: 12. 98. Find the area of the region bounded by the curve y: sin-l 2x, theline x: *lVI,, andthe r axis. 99. Find the centroid of the solid of revolution obtained by revolving about the r axis the region bounded by the curve y : e-t, the coordinate axes, and the line r : 1. 100. Find the centroid of the solid obtained by revolving the region of Exercise99 about the y axis. 1 0 1 .The vertical end of a water trough is 3 ft wide_atthe top, 2 ft deep,and has the form of the region bounded by the one arch of the curve-y: 2 sin tm. rfthe trough is fulli water, find rhe force due to liquid pressure on ffi:il:t

1,02.The linear density of a rod 3 ft long at a point x ft fuomone end is ke-s, slugs/ft. Find the mass and center of mass of the rod.

103. Find the centroid of the region bounded by the curve y : x ln x, the x axis, and the line x: e. 104. Find the area of the region enclosedby one loop of the curve * : yo(l _ yr). 105' Two chemicals A and B react to form a chemical C and the rate of change of the amount of c is proportional to the product of the amounts of A and B remaining at any given time. Initialiy there are 50 lb of chemical A and 50 lb of dremical B, and to form 5lb of c,3lb of A and zb oin are required. Aiter t hr, 1.5lbof careformed. (a)If xlbof C are formed at f hr, find an expression for r in terms of t. (b) Fini the amount of C after 3 hr. 106' A tank is in the shape of the solid of revolution f:*d by rotating about the x axis the region bounded by the curve ! : h x, the r axis, and the lines r : e and x: e2.rf.the tank is ruu or water, find the work done in pumping all the water to the top of the tank. Distance is measured in feet. Take the positive x axis vertically downward.

Hyperbolicfunctions

536

HYPERBOLIC FUNCTIONS

I2.I THE HYPERBOLIC FUNCTIONS

l2.l.l

Definition

certain combinations of e" and e-'appear so frequently in applications of mathematics that they are given special names. Two of these functions are the "hyperbolic sine function" and the "hyperbolic cosine function.,, The function values are related to the coordinates of the points of an equilateral hyperbola in a manner similar to that in which the values of the corresponding trigonometric functions are related to the coordinates of points of a circle (see sec. 1.7.4)Following are the definitions of the hyperbolic sine function and the hyperbolic cosine function. The hyperbolicsinefunction is defined by

The domain and range are the set of all real numbers. 12.1.2 Definition

The hyperboliccosinefunction LSdefined by

The domain is the set of all real numbers and the range is the set of all numbers in the interval [L, + *;.

y : sinhr

Figure 12.1.1

12.1.3 Definition

Sketches of the graphs of the hyperbolic sine function and the hyperbolic cosine function are shown in Figs. 12.1.1and 12.'1,.2, r"rp".tively. The graphs are easily sketched by using values in the table of hyperbolic functions in the Appendix. Note that these functions are not periodic. The remaining four hyperbolic functions may be defined in terms of the hyperbolic sine and hyperbolic cosine functions. you should note that eachsatisfies a relation analogous to one satisfied by corresponding trigonometric functions. The hyperbolictangentfunction, hyperboliccotangent function, hyperbolic secantfunction, and hyperboliccosecant function are defined, resplctively, as follows: (1) (2) (3)

y:coshr Figure 12.1.2

(4)

12.1THE HYPERBOLIC FUNCTIONS

y:tanhr

F i g u r e1 2 . 1 . 3

Lr)].f urt "L"u'

A sketch of the graph of the hyperbolic tangent function is shown in Fi9.12.1.3, There are identities that are satisfied by the hyperbolic functions which are similar to those satisfied by the trigonometric functions. Four of the fundamental identities are given in Definition 12.1,.3.The other four fundamental identities are as follows: tanh x --

(5)

coth r

cosh2 x - sinh2 )c: t

(6)

1 - tanhz x:

sech2x

(7)

| - coth2 x:

-csch2 r

(8)

Equation (5) follows immediately from (1) and (2).Following is the proof of (6).

cosh2 r- sinh2-: (*)'-

(-ti.:l

: t(er, * 2eo* e-2, - ezr+ zeo- e-rr) :l Equation (7) can be proved by using the definitions of tanh r and sech r in terms of e' and e-' as in the proof above, or an alternateproof is obtained by using other identities as follows: '/., sinh2r cosh2r-sinh2x I - tann. N: | secn-r c6sh7;-: 6;hr;: Ail2f: The proof of (8) is left as an exercise(seeExercise1). From the eight fundamental identities, it is possible to prove others. Some of these are given in the exercisesthat follow. Other identities, such as the hyperbolic functions of the sum or difference of two numbers and the hyperbolic functions of twice a number or one-half a number, are similar to the coresponding trigonometric identities. Sometimes it is helpful to make use of the following two relations, which follow from Definitions 12.1.1 and 12."1,.2. cosh x + sinh x:

er

cosh x-

e-ff

sinh x-

We use them in proving the following

(e) (10) identity:

y sinh(r * A) : sinh r cosh y + cosh x sLnln. From Definition 12.1,.1we have

sinh(r*a):ry:y

( 1 1)

538

FUNCTIONS HYPERBOLIC

Applying (9) and (10) to the right side of the above equation, we obtain sinh(r * y) : *[(cosh r * sinh x) (coshy * sinh y) - (cosh r - sinh r) (cosh y - sinh y)l Expanding the right side of the above equation and combining terms, formula (11) is obtained. In a similar manner we can prove cosh(x + V) : cosh x cosh y + sinh x sinh y

(r2)

If in (11) and (LZ) y i" replacedbyx, the followirg two formulas are obtained: sinh 2x - 2 sinh r cosh x

(13)

cosh 2x - cosh2 x + sinh2 r

(14)

and

Formula (14) combined with the identity (5) gives us two alternate formulas for cosh 2x, which are cosh 2x :2

sinhz x *

'/.,

(15)

and cosh2r:2coslfr-1

(16)

Solving (15) and (15) for sinh x and cosh r, respectively, and replacing xby Lx, we get .lx ffi Srnn =V 2:

(17)

2

and

cosh;:tr

(18)

We do not have a -+ sign on the right side of (18) since the range of the hyperbolic cosinefunction is [ 1,+o;. The detailsof the proofs of Eqs. (12) through (18) are left as exercises(seeExercises2,4, and 5). The formulas for the derivatives of the hyperbolic sine and hyperbolic cosine functions may be obtained by applying Definitions 12.1.1and '1,2.'1,.2 and differentiating the resulting expressionsinvolving exponential functions. For example,

D"(cosh x): D, (ry)

:

=t

--sinhr

In a similar manner, we find D."(sinh r) : cosh r (see Exercise15).

12.1THE HYPERBOLIC FUNCTIONS

To find the derivative of the hyperbolic tangent function, we use some of the identities we have previously proved. D,(tanh x):

- cosh2r--si-nh2 x : D, /g+a) -+: cosh'l cosh'r \cosh r/

sech2x

The formulas for the derivatives of the remaining hyperbolic functions are as follows: Dr(coth r) : -csch2 r; D"(sech r) : -sech r tanh r; D"(csch r) : -csch x coth r. The proofs of these formulas are left as exercises(seeExercises15 and 16). From the above formulas and the chain rule, if z is a differentiable function of.x, we have the following more general formulas: D D

D

"(sinh "(cosh

"(tanh

u) : cosh u D ru

(1e)

u) : sinh u D *u

(20)

u) : sech2u D ru

(2r)

D"(coth u) : -cschz u D*u

(22)

D"(sech u) : -sech a tanh u D*u

(23)

D"(csch u) : -csch u coth u D*u

(24')

You should notice that the formulas for the derivatives of the hyperbolic sine, cosine, and tangent all have a plus sign, whereas the formulas for the derivatives of the hyperbolic cotangent, secant, and cosecantall have a minus sign. Otherwise, the formulas are similar to the corresponding formulas for the derivatives of the trigonometric functions. EXAMPLE 1:

Find D ry tf

SOLUTION'

D rA : -2x

SOLUTION:

+:_

sech2 (L

y - tanh (1 - x') EXAMPLE 2:

Find dyldx if

du

1,

ax

slnn x

cosh x , cosh x - .-: slnn ff

coth .r.

y - ln sinh r Formulas (19) through (24) give rise to the following indefinite integration formulas:

| ,inh u du- coshu + C J

(2s)


(28) (2e) udu

EXAMPTB 3:

Evaluate

f

I sinh3 x coshz x dx

J

(30)

SOLUTION:

r I sinh3r coshz x dx J

I I I 1 b

EXAMPLE

4:

Evaluate

r I sechax. dx J

sinh2 r coshz r (sinh x dx) (cosh2 x - 1) cosh2 r (sinh x dx) cosh4x(sinh xdx)

f

I cosh2 r (sinh x dx)

J

c o s h sx -

*coshsx* C

SOLUTION:

r

I sechax dx

J

I I I

sech2r (sech2x dx) (1 - tanh2 x ) (sechz x dx) sechz x dx -

tan h r - * t a n

f

I

tanh2x (sech2x dx)

h3x+C

A catenary is the curve formed by a homogeneous flexible cable hanging from two points under its own weight. If the lowest point of the catenary is the point (0, a), it can be shown that an equation of it is

y -a cosh(+) \a/

Figure 12.1.4

a> o

A sketch of the graph of Eq. (31) is shown in Fig. 12.1.4.

(31)

12.1THE HYPERBOLIC FUNCTIONS 541

ExAMPLE5: Find the length of arc of the catenary having Eq. (31)from the point (0, a) to the point (xr, Ar) wh€r€ ff1 > 0.

soLUTroN: If.

f(x):, *rn (#) then

f '(x):a sinh(tJ (tJ: 'i"r,(f) and using Theorem 8.10.3, if L units is the length of the given arc, we have fcr

L- | Vl +lf'(x)l'dx JO (32) From formula (6) , L + sinh2 ,c- coshzr, and because cosh(xla)> L, rt follows that

@

: cosh

and substituting fromn t h

to (32)

L- I:.o,n(f) (dx - asinh (il - asinh(+)

l; tr

0

A I sinh

0

- asinh(+)

Exercises12.L In Exercises L through 10/ prove the identities. L. 1-

cothzx:

-csch2 r

2. cosh(r * U) : cosh r cosh y * sinh x sinh y

tanhr*tanhy 3 . tannrr -r v t : -----:--------:1*tanhxtanhy 4. (a) sinh 2r : 2 sinh r cosh r; (b) cosh 2r : cosh2r * sinh2 x : 2 sinhz x * 7 : 2 cosrf x - | 5.

(a)cosh#r: 1$;

6 . csch2r:$

sechr cschr

(b)sinhlr:=,F

542


7. (a) tanh(-r) : -tanh x; (b) sech(-x) : sechr 8. sinh 3x -3 sinh x * 4 sinhsr 9. cosh3x :4 coshsx - 3 coshx 10. sinhz x - srnhzy: sinh (x + y) sinh (x - y) 11. Prove:(sinh x * cosh x)n: cosh nx * sinh nx, tf n is any positive integer. (ntNr: Use formula (9).) 12. Prove that the hyperbolic sine function is an odd function and the hyperbolic cosine function is an even function. tanF x : s,x L4.prove: 1* I-tannx

*'--- L 13. Prove:tanh(ln r) : x2 * "1.

15. Prove: (a) D"(sinh r) : cosh r; (b) D"(coth x) :-csctri r. 15. Prove: (a) D,(sedr x) : -sedr r tanh r; (b) D"(csch r) : -csch r coth r. In ExercisesL7 through 25, find the derivative of the given function. 4v*1

tZ. f (x): tanh aj 1

19. f (x) - sr cosh x

18.g(r) : ln(sinh13)

20. h(x):

coth I

21..h(t):

23. G(x):

sin-r(tanh ri)

2a. t'Q): ln(coth 3r - csch3r)

2 6 .g ( x ) :

( 6 o s hx ) '

22. f (r) : tan-l(sinh 12)

ln(tanh f)

25. f(x) -tsinhr

2 7 . P r o v e :( a ) J t a n h u d u : l n l c o s hu l + C ; ( b ) J c o t h u d u : l n l s i n h u l + C . 2 8 . P r o v e :J s e c hu d u : 2 t a n - i e u l C . 2 9 . P r o v e :J c s c ha d u : l n

ltanh+ul+ C.

In Exercises30 through37, evaluatethe indefinite integral. SO.f sintrnx coshs xdx JJJJ

Sf. I t".rn r In(coshx) dr

tz. I tanhzZxdx

Zl. I coshnzxdx

u.

3s. I sechnSxdx J

so. f s".f, xtanhsxdx J

,r. - " J ["t"h! t/x

| *sinh2 f dx J

a,

In Exercises38 and 39, evaluate the definite integral. 38. [' cosh'rdr J-r

39. [' sinhtxdx Jo

40. Draw a sketch of the graph of the hyperbolic cotangent function. Find equations of the asymptotes. 41.. Draw a sketch of the graph of the hyperbolic secant function. Find the relative extrema of the function, the points of inflection of the graph, the intervals on which the graph is concaveupward, and the intervals on which the graph is concave downward. 42. Prove that a catenary is concave upward at each point. tl3. Find the area of the region bounded by the catenary of Example 5, the y axis, the r axis, and the line r : 11,where 11)0.

tl4. Find the volume of the solid of revolution if the region of Exercise43 is revolved about the r axis.

12,2THE INVERSEHYPERBOLIC FUNCTIONS

45. A particle is moving along a straight line according to the equation of rnotion s: e-ctt2(Asinh f * B cosh f) where s ft is the directed distance of the particle from the origin at f sec.If o ftlsec and a ftlsec2are the velocity and acceleration, respectivelp of the particle at f sec, find o and a. Also show that a is the sum of two numbers, one of which is proportional to s and the other is proportional to p. 45. Prove that the hyperbolic sine function is continuous and increasing on its entire domain. 47. Prove that the hyperbolic tangent function is continuous and increasing on its entire domain. 48. Prove that the hyperbolic cosine function is continuous on its entire domain but is not monotonic on its entire domain. Find the intewals on which the function is increasing and the intervals on which the function is decreasing.

12.2 THE INVERSE HYPERBOLIC FUNCTIONS '

12.2.1 Definition

r

we see that a From the graph of the hyperbolic sine function (Fig. 12.'1..1), point; line the graph in one and only one therefore, to horizontal intersects number in the range of the function there corresponds one and only ""ch one number in the domain. The hyperbolic sine function is continuous and increasingon its domain (seeExercise46 in Exercises12.1),and so by Theorem 9.3.2 the function has an inverse function. The inaerse hyperbolic sine function, denoted by sinh-l, is defined as follows:

A sketch of the graph of the inverse hyperbolic sine function is shown in Fig. 12.2.1'.Itsdomain is the set of all real numbers, and its range is the set of all real numbers. that It follows from Definition 12.2.'1, sinh(sinh-t r) : x

(1)

sinh-r(sinhy\:y

(2)

and

sinh-l r

Figure 12.2.1

we notice that a horizontal line y: k, where k > 1, , From Fig. 1.2.1..2, intersects the graph of the hlryerbolic cosine function in two points. Therefore, for each number greater than L in the range of this function, there correspond two numbers in the domain. So the hyperbolic cosine function does not have an inverse function. However, we define a function F as follows: F(r) : cosh r Figure12.2.2 t '

forx>0

(3)

The domain of F is the interval [0, +-;, and the range is the interval A sketch of the graph of F is shown in Fig. 12.2.2.The function F +-;. [1, is continuous and increasing on its entire domain (seeExerciseL), and so by Theorem 9.9.2, F has an inverse function, which we call the "inverse hyperbolic cosine function."


12.2.2Definition

The inaersehyperboliccosinefunction, denoted by cosh-l, is defined as follows: l:

cosh-t r

if and only if x: cosh y

and y z 0

A sketch of the graph of the inverse hyperbolic cosine function is shown in Fig. 12.2.3.The domain of this function is the interval [1, +oo), and the range is the interval [0, +*;. From Definition12.2.2 we conclude that cosh(cosh-t x) - x

tf x

(4)

cosh-l(coshy):y

ify>0

(5)

and

As with the hyperbolic sine function, a horizontal line intersects each of the graphs of the hyperbolic tangent function, the hyperbolic cotangent function, and the hyperbolic cosecantfunction at one and only one point. For the hyperbolic tangent function, this may be seen in Fig. 12.1.3.Each of the above three functions is continuous and monotonic on its domain; hence, each has an inverse function.

Figure12.2,3

12.2.3 Definition

The inaersehyperbolictangentfunction, the inverse hyperboliccotangent function,and the inaersehyperboliccosecant function,denoted respectively by tanh-r, coth-l, and csch-r, are defined as follows:

A sketch of the graph of the inverse hyperbolic tangent function is in Fi9.12.2.4. The hyperbolic secant function does not have an inverse function. However, we proceed as we did with the hyperbolic cosine function and define a new function which has an inverse function. Let the function G be defined by G (r) : sechr Y

:

tanh-l x

Figure12.2.4

12.2.4 Definition

(5)

forx>0

It can be shown that G is continuous and monotonic on its entire domain (see Exercise 2); therefore, G has an inverse function, which we call the "inverse hyperbolic secantfunction." The inaersehyperbolicsecantfunction, denoted by sech-l, is defined as follows: y : sech-l r

if and only if

x - sechy

/

and

y >0

(7)

12.2THEINVERSE HYPERBOLIC FUNCTIONS545 The inverse hyperbolic functions can be expressedin terms of natural logarithms. This should not be surprising because the hyperbolic functions were defined in terms of the exponential function, and the natural logarithmic function is the inverse of the exponential function. Following are these expressions for sinh-l, cosh-l, tanh-r, and coth-l. sinh-l x:ln(x

+ \F +l)

cosh-1r:ln(r+\/P-)

1

vI1

(s)

x>I

(9)

lrl < 1

(10)

lrl > 1

(11)

tanh-r *:tnffi coth-lx:|lnffi

r any real number

We prove (9) and leave the proofs of the other three formulas as exercises(see Exercises3 through 5). To prove (9), let V: cosh-r x, x > 1.. Then, by Definition 12.2.2, r : cosh U, A > 0. Applying Definition 12.7.2to cosh y, we get ea + e-u

y>0

from which we obtain ,y >0

ezo-2xea+1-0

Solving this equation for e' by the quadratic formula, we obtain

€ u : 2 * r W2: x t \ F

y>o

(r2)

We know that y >O and r > 1. Therefore,eu> 7. W.hen x:1, : 1. Furthermore,when x )']-., 0 < r- 1 < x+ lF - t : x- \/71 x * l, and so \/i=T < t/x + t . Therefore,

\81

\81

< \81

\tr+1

giving us x-'!.1\/r'z--. Hence, whdn x)1,, y-1/p=1 ay Consequently, we can safely reject the minus sign in (12). And since y: cosh-l .r, we have cosh-lr:

ln(x + t/P - t)

x > !

which is (9). ExAMPLE L: Express each of the following in terms of a natural logarithm: (a) tanh-l(- *); (b) sinh-l 2.

SOLUTION: tanh-l

(a) From (10), we obtain

3 (-f):+t"i: +'"i:rn+:-h

(b) From (8), we get sinh-l 2-

ln(2 * \6)

HYPERBOLIC FUNCTIONS To obtain a formula for the derivative of the inverse hyperbolic sine functiofl, let y : sinh-l r, and then x - sinh y. Thus, becauseDox: cosh y and DrU : LlDdc, we have

Dra:

(13)

cosh y

From the identity cosh2y-sinh2 U:1, it follows that cosh y: VA;nt!=. (Norr: When taking the iquare root, the minus sign is rejectedbecausecosh y > 1.) Hence, becauser: sinh /, cosh y: lF +1. Substituting this in (13), we obtain

D *(sinh-x) L :#TT

(14)

If. u is a differentiable function of r, from (14) and the chain rule it follows that (15) Formula (14) can also be derived by using (8), as follows: -r D x) : D r l n ( r * \m) "(sinh 1 r 'r

:

x

1ffi

x,+\m \m ==:

\ffi( r *tffil

-:-

- *x

1 Yxz + 1

The followitg differentiation formulas can be obtained in analogous ways. Their proofs are left as exercises(see Exercises6 throu each formula u is a differentiable function of x.

FUNCTIONS 547 12.2THE INVERSEHYPERBOLIC

(le) (20)

Find f '(x) if

EXAMPLE

EXAMPTE 3: "(r

Verify

Applying formula (L7), w€ get

r, / r : -2 I (x, 1,-

tanh-l(cos 2x)

f(x)

D

solurroN:

stn 2x 2*: "orz

-2

stn 2x : - ., ., z csc zx ,inz ,,

SOLUTION:

cosh-L x -

Dr(r cosh-Lx - tffil

- cosh-t x * x - cosh -r x

The main use of the inverse hyperbolic functions is in connection with integration. This topic is discussedin the next section.

Exercises L2.2 1. If F(r) : cosh x and.x = 0, prove that F is continuous and increasing on its entire domain. 2. lt G(x):

sechx and x = 0, prove that G is continuous and monotonic on its entire domain.

In Exercises3 through 10, prove the indicated formula of this section. 3. Formula (8)

4. Formula (10)

5. Formula (11)

6 . Formula (15)

7. Formula (L7)

8. Formula (18)

9. Formula (19)

1 0 . Formula (20)

ln Exercises 11.through L4, express the given quantity in terms of a natural logarithm. n.

sinh-r *

1,4. coth-l (-2)

13. tanh-l *

12. cosh-r 3

In Exercises L5 through 28, find the derivative of the given function. 15. f (x) : sinh-r x2 18. h(w) - tanh-t Tt)B

1'5.G(x) : cosh-t tx

21. f (x) : x2 cosh-l x2 24. S(r) : tanh-l(sin 3r) 27. G(x) - x sinh-r x- \ffi

22. h(x) - (sech-lr)2

19. S(x) - coth-t(3r + 1) 25. h(x) - cosh-l(cscx) - x t a n h - rx 2 8 .H ( x ) : l n \ F

1 7 .F ( r ) -

tanh-r 4x

- csch-r +r2

20.f(r) 23.f(x) :

sinh-l(tan x)

26.F(x) - c o t h - l ( c o s h x )

In Exercises29 through 32, draw a sketch of the graph of the indicated function. 29. The inverse hyperbolic tangent function.

30. The inverse hyperbolic cotangent function.

31. The inverse hyperbolic secant function.

32. The inverse hyperbolic cosecant function.

HYPERBOLIC FUNCTIONS 1'2.3INTEGRATS YIELDING INVERSE HYPERBOLIC FUNCTIONS

The inverse hyperbolic functions can be applied in integration, and sometimes their use shortens the computation considerably. Il should be noted, however, that no new types of integrals are evaluated by their use. Only new forms of the results are obtained. From formula (15) in 9ec.12.2we have .

D,(sinh-l u):

U;ft7o,u

from which we obtain the integration formula

Idu I ffi:

sinh-ra * C

Expressing sinh-r u as a natural logarithm by using formula (g) of sec. '1,2.2, we get (1) Formula (16) in Sec.12.2 is

D * ( c o s h -zr) - L n

where u

ffiu,u

from which it follows that

[ -!" Jffi:cosh-tu

+C

u

and combining this with formula (9) in Sec.12.2,we obtain (2) Formulas (17) and (18) of Sec.12.2 are, respectively,

D*(tanh-lu):#D,u

wherelul

D,(coth-lD: firD,u

wherelul >t

and

From the above two formulas we get I

d,

_ftanh-l z*C

i f l r l l< 1

J T=7: l"orh-'u * C if lui > t and combining this with formulas (10) and (11) of Sec.12.2 we have (3)

I2,g INTEGRALSYIELDINGINVERSEHYPERBOLICFUNCTIONS 84g; We also have the following

three formulas.

(4)

(s)

(5)

These formulas can be proved by finding the derivative of the right side and obtaining the integrand, or else more directly by using a h1ryerbolic function substitution. For example, the proof of (a) by differentiating the right side is as follows: / r \ 1 1 \ G=o): l D,(sinh-: ffi

VF/*'

' -'':

Uffi7' i

and becausea ) 0, {-a': a, and,we get / .,\ 1 =o):

Du(sinh-'

,6ryt

To obtain the natural logarithm representation we use formula (8) in Sec.12.2and we have

, lu , \/FTF\

:tl-tt--T--l

\4a/

:ln(n+tfft)_lna Therefore, ,rl

sinh-l i + C: ln(z + t/NV) a

- h a* C

: ln(z+ I/FTE) + c,

whereCl=C-lna. Ttre proof of (5) by a hyperbolic function substitution makes use of the identity cosh2r - sinh2r: L. Lettingu: a cosh r, where r ) 0, then


du: a sinh r dx, and r: get f

du

J '\fr'=A-

cosh-l(ula).Substitutingin the integrand,we asinhrdr

f

J \//frcoffi=@ asinhrdr

f

I \/F Vcoiltt=

-J f

asinhxdx

\/F \ffiE;

a. Furtherrnore, Because a ) 0, lF: becausex ) 0, sinh r ) 0, and so \6in6 t: sinh r. We have,then, du I J !u2-a2

f a sinhx dx J asinhx

:Ia, J :r*C

: cosh_r I*, The natural logarithm representation may be obtained in a way similar to the way we obtained the one for formula (a). The proof of formula (5) is left as an exercise(seeExercise17). Note that formulas (1) through (5) give altemate representations of the integral in question. In evaluating a definite integral in which one of these forms occurs and a table of hyperbolic functions is available, the inverse hyperbolic function representation may be easier to use. Also the inverse hy?erbolic function representation is a less cumbersome one to write for the indefinite integral. Also note that the nafural logarithm form of formula (5) was obtained in Sec. 1,1,.4 by using partial fractions.

EXAMPLE

ld* Jffi

Evaluate

sol,urroN: We apply formula (4) after rewriting the denominator. dx

12.3 INTEGRALS YIELDINGINVERSEHYPERBOLIC FUNCTIONS 551

EXAMPTg2: Evaluate dx fto J, !x2 - 25

solurroN:

Using formula (5), we get

fro dx I -Ja vx4

:'.lro cosh-r * | - cosh-l 2c Jo

cosh-l L.2

Using Table 4 in the Appendix, we find cosh-l 2-

c o s h - l( L . 2 ) - ' 1 , . 3 2 - 0 . 6 2 : 0 . 7 0

Instead of applying the formulas, integrals of the forms of those in formulas (1) through (6) can be obtained by using a hyperbolic function substitution and proceeding in a manner similar to that usin g a trigonometric substitution. EXAMPTT 3: Evaluatethe integral of ExampleL without using a formula but by using a hlperbolic function substitution.

solurroN: as

In the solution of Example L, the given integral was rewritten

ldx

J tk=ffiT If we let x-3:2 sinh u,then d.x:2 cosha du and z - sinh-ltk-g). We have, therefore, dx

which agreeswith the result of Example L.

12.3 Exercises In Exercises 1 through 10, express the indefinite integral in terms of an inverse hyperbolic function and as a natural logarithm.


1 r ' J| @ d * =

?f

J

F, f

f

?xdx

r

^ f o .Jffi

,-J@

".Jffi

=

5 . lf ,dgxx 2 - ! 5

. t =2=5 d -x2 cosxdx

d!

rffi

dx

In Exercises LL through 16, evaluate the definite integral and express the answer in terms of a natural logarithm.

13I::'& dx fz l.o ' J - , f f i 17. Prove formula (6) by using a hlperbolic function substitution. L8. Acurvegoesthroughthepoint (0,a),a > 0,andtheslopeatanypointist/f,ia2-l.Provethatthecurveisacatenary. 19. A man wearing a parachute falls out of an airplane and when the parachute opens his velocity is 200 ftlsec. If t, ftlsec is his velocity f sec after the parachute opens, 32ad!.:32484t

az

Solve this differential equation to obtain

t:E (.otn-, #- coth-'ry) ReaiewExercises (Chapter12) In Exercises1.through 3,prove the identities. 1,.(a) coth(-x)- -coth x; (b) csch(-x) : -csch r 2. sinh x * sinh y :2

sinh +(x+ y) cosh t(x - V)

? cosh 2r * cosh 4y _ coth(x + 2y) r. 4. Prove,:f"l j111 coth x - 1; (b)

csch x : 0 "t1i In Exercises5 throu gh 9, find the derivative of the given function. 5. f (x) : coshl 2x

6. 8U) : ln sech f2

8. h(x):

9. f (w) : wz sin1.'1-r 2w

e"(cosh r * sinh r)

T . rf (' \i l' -

cosh tz l*sgcha

10. The gudermannian is a function that occurs frequently in mathematics, and.it is defined by gd x: tan-r(sinh r). Show that D'(gd r) : sech r.

E.i'"

REVIEWEXERCISES 553 In ExercisesL1.through 14, evaluatethe indefinite integral.

11. .tanh tf itx I 13.

/sinnxsin

xdx

In Exercises15 and 15, evaluatethe definite integral. fr 1 . 5 .I s e c h z t d t Jo

In Exercisest7 and 18,obtdn the given result by a hlryerbolic function substitution. 1. dx coth /rir,r,-'f) t c. a > o rt . f al \ ljm q i:- A t "/r*rr-'r)+c.a>o dx I 18. al \ l@:=srnn 19. Find the length of the catenary y:

cosh x from (ln 2, *) to (In 3, t).

20. The graph of the equation

t7-r:cginh-l 'l\-l-'VAry' is called a tractrix. Provethat the slope of the curve at any point (x, y) is -Vl{FT.

Polarcoordinates

13.1 THE POLARCOORDINATE SYSTEM

13.1.THE POLAR COORDINATE SYSTEM P(r, 0) Figure

F i g u r e1 3 . 1 . 2

-

7 61r

F i g u r e1 3 . 1 . 3

F i g u r e1 3 . 1 . 4

F i g u r e1 3 . 1 . 5

P(-n,t,)

a.-A-.

1.1 70 F i g u r e1 3 . 1 . 6

Until now we have located a.point in a plane by its rectangular cartesian coordinates. There are other coordinate systems that can be used. Probably the next in importance to the cartesian coordinate system is the polar coordinatesystem.In the cartesian coordinate system, the coordinates are numbers called the abscissa and the ordinate, and these numbers are directed distances from two fixed lines. In the polar coordinate system, the coordinates consist of a distance and the measure of an angle relative to a fixed point and a fixed ray (or half line). The fixed point is called the pole (or origin), and it is designated by the letter "O." The fixed ray is called the polar axis (or polar line), which we label OA. The ny OA is usually drawn horizontally and to the right, and it extendsindefinitely (seeFig. 13.1.1). Let P be any point in the plane distinct from O. Let 0 be the radian measure of the directed angle AOP, positive when measured counters its initial clockwise and negative when measured clockwise, having aas side the ny OA and as its terminal side the ny OP. Then ift r is the undirected distance from O to P (i.e., ,: lOPl\, one set of polar coordinates of P is given by r and 0, and we write these coordinates as (r' 0)' o rLLUsrRArroN1: The point P(4, ta) is determined by first drawing the angle having radian measure $2, having its vertex at the pole and its initial side along the polar axis. Then the point on the terminal side, which is four uriits from the pole, is the point P (see Fig. 13.1.2).Another set of polar coordinatesfor this samepoint is (4,-&rr); see Fig. 13.1.3.Furthermore, the polar coordinates (4, #rr) would also yield the same point, as ' shown in Fig. 13.L.4. Actually the coordinates (4, $n *2nn), where n is any integer, give the same point as (4,trr). So a given point has an unlimited number of sets of polar coordinates. This is unlike the rectangular cartesian coordinate system becausethere is a one-to-one conespondence between the rectangular cartesiancoordinates and the position of points in the plane, whereas there is no such one-to-one correspondence between the polar coordinates and the position of points in the plane. A further example is obtained by considering sets of polar coordinates for the pole. If r: 0 and 0 is any real number, we have the pole, which is designated by (0, 0). We consider polar coordinates for which r is negative. In this case, instead of the point being on the terminal side of the angle, it is on the extension of the terminal side, which is the ray from the pole extending in the direction opposite to the terminal side. So if P is on the extension of the terminal side of the angle of radian measure 0, a set of polar coordinates of P is (r, 0), where r: -lOPl. is the same . rLLUSrRArroN2: The point (-4, -ttr) shown in Fig. 1"3.1'.5 -&r), another set 1. Still and (4, fzr) in Illustration point as (4, *n), (4, of polar coordinatesfor this samepoint is (-4, l*zr); seeFig. 13.1'6. . The angle is usually meastlred in radians; thus, a set of polar coordi-

556

POLARCOORDINATES

nates of a point is an ordered pair of real numbers. For eachordered pair of real numbers there is a unique point having this set of polar coordinates. However, we have seen that a particular point can be given by an unlimited number of ordered pairs of real numbers. If the point p is not the pole, and r and0 arerestricted so thatr > 0 and0 < 0 < 2rr,then there is a unique set of polar coordinates for p. EXAMPLE1: (a) plot the point having polar coordinates (3, -3r). Find another set of polar coordinates of this point for which (b) r is negative and 0 (c) r rs positive and 0 (d) r is negative and -2n < 0 < 0.

solurroN: (a) The point is plotted by drawing the angle of radian measure -3a in a clockwisedirection from the polar axis. Becauser)0,p is on the terminal side of the angle, three units from the pole; see Fig. 1,3.1,.7a. The answersto (b), (c), and (d) are, respectively,(-3,$r), (3,trl), and (-3, -En).They are illustrated in Fig. 1,3.1.2b, c, and d. 4_

5"

o/ / x

-+" PQ,_3I

7r

* { p(- 3,

\

+c

4 o(_r,

(b)

(a)

-+c

(d)

F i g u r e1 3 . 1. 7

(,, (*,

often we wish to refer to both the rectangular cartesian coordinates and the polar coordinates of a point. To do this, we take the origin of the first system and the pole of the second system coincident, the polar axis as the positive side of the r axis, and the ray for which 0 : in as the positive side of the y axis. suppose that P is a point whose representation in the rectangular cartesian coordinate system is (x, y) and (r, g) is a polar-coordinate representation of P. we distinguish two cases:r > 0 and r < 0.In the first case, if r ) 0, then the point P is on the terminal side of the angle of 0 radians,and r: loPl. Such a caseis shown in Fig. 13.1.g.Then cos d: xl lOPl : xlr and sin 0 : Vllo-Pl : Alr; and so x-rcose

F i g u r e1 3 . 1 . 8

and y-rsrn0

(1)

In the second case, tf r I 0, then the point P is on the extension of the terminal side and r--lOTl (see Fig. 13.1.9).Then if a is the point (- x , -y), we have -x -X -X x cosg=

=l@T

So x- rcos0

;Fl

:-,:

r

(2)

13.1 THE POLARCOORDINATE SYSTEM

557

Also,

-y) Q(- x,

-a-a-va

sino:16T:l6FT:=:; Hence,

P

F i g u r e1 3 . 1. 9

y:rsin9 (3) Formulas (2) and (3) are the same as the formulas in (L); thus, the formulas (1) hold in all cases. From formulas (1) we can obtain the rectangular cartesiancoordinates of a point when its polar coordinates are known. Also, from the formulas we can obtain a polar equation of a curve if a rectangular cartesian equation is known. To obtain formulas which give a set of polar coordinates of a point when its rectangular cartesian coordinates are known, we square on both sides of each equation in (1) and obtain 'f: rz cosz0 and A2: 12 sinz0 Equating the sum of the left members of the above to the sum of the right members, we have **y':12

cos2 o * r z s i n zo

or, equivalently, * * Y': rz(sin2o * coszo) which gives us *+Yz:rz and so (4)

Fromthe equationsin (1) and dividinS,we have r s i ne - _ y ,'cosg i or, equivalently, :':: if lt*i+fi i$il1ii.::':::i:i'iiiiiiirii:i;i::i:li':'ii::i

(5)

ri;1i1r-1i fffitl++i+++

. rr,r.usrRArrorv3: The point whose polar coordinates are (-6, Znt) is plotted in Fig. 13.1.10.We find its rectangular cartesian coordinates. From (L) we have

.10 F i g u r e1 3 . 1

x - r c o s0 - -5 cosZn

y-rsin0 - -6 stn tn

POLAR COORDINATES

--3\n Sothe point is (-3 \n,3\n). The graph of an equation in polar coordinates r and 0 consists of all those points and only those points P having at least one pair of coordinates which satisfy the equation. If an equation of a graph is given in polar coordinates, it is called a polar equation to distinguish if from a cartesianequation,which is the term used when an equation is given in rectangular cartesian coordinates. EXAMPLr'2: Given a polar equation of a graph is 12:4stn20 find a cartesian equation.

SoLUTIoN:

Because y2 - x, + y, and

sin 20:2 sin 0 cose - zfulr)(xlr),

from (1) we have, upon substituting in the given polar equation,

x 2+ Y ' : 4 ( D + ' ; x2* y',:w r2 xz*Y': !!V = x'+y, (x'*y')'-8xy

EXAMPTn3: Find (r, 0) tf r > 0 and0 <0<-2n forthepoint whose rectangular cartesian coordinate representation is

soLUrIoN: The point e{3, cause r ) 0, we have

t\8,-1).

From (5), tan 0:-U(-fi),

-1) is plotted in Fig. 1g.1.1,1. From (4), be_

r: {lT3:2 o:trr So the point is Q, &r).

F i g u r e1 3 . 1 .11

and since r I 0 < $r, wehave

SYSTEM 13.1 THE POLARCOORDINATE

ExAMPtn 4: Given a cartesian equation of a graPh is x2+y2-4x-0 find a polar equation.

soLUrIoN: Substituting x-

/cos g and y:

rsin 0 in

x2+y2-4x-o we have rz cosz 0 + 12 sinz0 - 4r cos0:0

12-4rcos0:0 r ( r - 4 c o sg ) : 0 Therefore, o r r - 4 c o s0 : 0 The graph of r - 0 is the pole. However, the pole is a Point on the g r a p h o f r - 4 cos 0: O becauser:0 when e Lr. Therefore,a polar equation of the graPh is r-0

r-4cos0

The graph of.* * y'- 4r: 0 is a circle.The equationmay be written in the form (x-2)'*y':4 which is an equationof the circlewith centerat (2, 0) and radius 2.

L3.1Exercises then find another set of polar coordiIn Exercises1 through 6, plot the point having the given-set of polar coordinatesi natesforthesamepointforwhich(a)roand-2n<0'0;(c)r(0and-2zr<0<0' 3. (2, Ln) 2. (3, 8zr) 1. (4, in) 6. (2, tn) 5. ?n,fu) 4. (3, szr) give two other sets of polar coordiIn Exercises 7 through 12, plot the point having the given set of polar coordinates; then sign' opposite having r nates of the same point, one with the same Value of r and one with an

,. @,-&n) 10. (-2, tn)

9. (-4, &n) 12. (-3 , -n)

8. ({2, -*n) 1.L.(-2, -*rr)

given: (a) (3' r); 13. Find the rectangular cartesiancoordinates of each of the following points whose polar coordinates are

-tn); (e)1-z,Irr);(f) (-1, -Sa'). b) ({r,-*"');-(") ?a,Ir); (d)(-2,

qartgsil coordinates are given' Take 14. Find a set of polar coordinates for each of the following points whose rectangular -2);

r > 0 and0 = 0 < ir. til 0,-r); (u) (-rfs, r); (c)d',-z);(d) (-s,0); (e)(b,

(r) (-z'-zt/i)'

In Exercises15 through 22, find. a polar equation of the graph having the given cartesian equation. L5. x2+A2:az

L6. xB : 4y2

18.x2-A2:15

L9. (x' * y')'-

4(x' - y')

17.Y' : 4(x + 1) 20. 2xy-- a2

I

I

POLAR COORDINATES

2I. rf+y3-3axy-0

2x 22.Y : x'z+r

In Exercises23 through 30, find a cartesian equation of the graph having the given polar equation. 23. 12: 2 sin 20

24. r2 cos 20:

2 6 .r 2 : 0

27. 12 : 4 cos 20

13.2 GRAPHS OF EQUATIONS IN POLAR COORDINATES

Let r: f (0) be a polar equation of a curve. We first derive a formula for finding the slope of a tangent line to a polar curve at a point (r, g) on the curve. Consider a rectangular cartesian coordinate system and a polar coordinate system in the same plane and having the positive side of the x axis coincident with the polar axis. In Sec.13.1 we saw that the two sets of coordinates are related by the equations r:

25. 12: cos 0

t0

28.r:2sin30

/ cos 0 and tt: r sin 0

r and y can be considered as functions of 0 becauser:f(0).If we differentiate with respect to 0 on both sides of these equations we get, by applying the chain rule, dx Ag:

Adr cos 0 d0-

r sin 0

and

du ffi:sin0

dr cos0 de*r

(2)

If a is the radian measure of the inclination of the tangent line, then

tan o:

du f*

and if dxlde # 0, we have -dv :dx

dy de dx de

substituting from (1) and (2) into (g),we obtain

s i no # * r c o s g c o s0 # - r s i n o If cos 0 * 0, we divide the numerator and the denominator

of the

13,2 GRAPHSOF EQUATIONSIN POLARCOORDINATES

right side of ( ) by cos 0 and replace dyI dx by tan a, thereby giving us (5)

ExAMPLEL: Find the sloPeof the tangent line to the curue whose equation is r 1'- cos A at the point (1 +\n, Ln).

Pt(rt,01)

F i g u r e1 3 . 2 . 1

sor,urrow: Sincer: 1 - cos0,drld0: sin 0.Soat thepoint (1 - i!2, in), d.rlil|: it/2 ana tan0:1; so from (5)we obtain

(1,)+\/, (1,- i\/r\: _-!_ r-tanc^.:_ i!?g2_+

11_+\/2)(L) vz- t:

t/2 + L

If we are to draw a sketch of the graph of a polar equation, it will be helpful to consider properties of symmetry of the graph. In Sec.1..3(Definition L.3.4) we learned that two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ, and that two points P and Q are said to be symmetric with respect to a third point if and only if the third point is the midpoint of the line segment PQ. Therefore, the points 1Z'tr) and (2, $z) are symmetric with respect to the Lzr axis and the points 12,tzr) and (2, -lrr) are symmetric with respect to the pole. We also leamed (Definition 1.3.5)that the graph of an equation is symmetric with respect to a line I if and only' if foi every point F on the graph there is a point Q, also on the graph, such that P and Q are symmetric with resPect to l. Similarly, the graph of an equation is symmetric with respect to a point R if and only if for every poit t f on the graph there is a point S, also on the graph, such that P and s aie symmetric with respect to R. We have the following rules of symmetry for graphs of polar equations. RuIe1. If for an equation in polar coordinatesan equivalent equation is obtained when (r, 0) is replacedby either (r, - 0 * 2nn) or (- r, n 0 * Znt)' where n is any integer, the graph of the equation is symmetric with lespect to the polar axis. RuIe2. If for an equation in polar coordinatesan equivalent equation is obtained wh en (r, 0) is replacedby either (r, o 0 4- 2nn) or (- r, 0 * 2nn), integer, the graph of the equation is symmetric with respect where nis aurty axis. the to tqr RuIe3. If for an equation in polar coordinatesan equivalent equation is obtained when (r, 0) is replacedby either (-t, 0 I 2nzr)ot (r, t * 0 * 2nzr)' where n is any integer, the graph of the equation is symmetric with respect to the pole. The proof of Rule 1 is as follows: If the point P(r, d) is a point on the graph of an equation, then the graph is symmetric with respect to the polar axis if there is a point Pr(rr, 01)on the graph so that the polar axis is the perpendicular bisector of the line segment PtP (see Fig. 13.2.1).So

562

POLARCOORDINATES

if r, : r, then d, must equal - d * 2nn,where n is an integer. And if.r, : - 7, then d1must be n - 0 * 2nrr. The proofs of Rules 2 and 3 are similar and are omitted. . rr,r,usrRArror,l1: We test for symmetry with respect to the polar axis, the in axis, and the pole, the graph of the equation r:4 cos 20. Using Rule 1 to test for symmetry with respect to the polar axis, we replace (r, 0) by (r, -0) and obtain r : 4 cos(-20), which is equivalent to r: 4 cos20. So the graph is symmetric with respect to the polar axis. Using Rule 2 to test for symmetry with respect to the ir axis,we replace (r, 0) by (r, T-0) and get r:4 cos(2(?r-l)) or, equivalently, r: 4 cos(2n- 2e), which is equivalentto the equation r - 4cos20. Therefore, the graph is symmetric with respect to the lr axis. To test for symmetry with respect to the pole, we replace (r, 0) by (-r, 0) and obtain the equation -/: 4 cos 20, which is not equivalent to the equation. But we must also determine if the other set of coordinates works. We replace(r, 0) by (r, n * 0) and obtain r:4 cos2(tr+ 0)or, equivalently, r : 4 cos(2rr+ 20),which is equivalent to the equation r: 4 cos20. Therefore, the graph is symmetric with respect to the pole. . when drawing a sketch of a graph, it is desirable to determine if is on the graph. This is done by substituting 0 for r and solving lhe nofg for'0. Also, it is advantageousto plot the points for which r has a relative maximum or relative minimum value. As a further aid in plotting, if a curve contains the pole, it is sometimes helpful to find equations of the tangent lines to the graph at the pole. To find the slope of the tangent line at the pole we use formula (5) with r: 0. If drldO * 0, we have tan a:

tan 0

(6)

From (6) we conclude that the values of 0, where 0 < 0 < zr, which satisfy a polar equation of the curve when /: 0 are the inclinations of the tangent lines to the curve at the pole. So if.er,02, . , 0* are these values of 0, then equations of the tangent lines to the curve at the pole are 0 : 0 r .0, : e r ,

EXAMpLE2: Draw a sketch of the graph of r:1,-2cos0

, 0: 0*

soLUrIoN: Replacing (r , 0) by (r , -0), we obtain an equivalent equation. Thus, the graph is symmetric with respect to the polai axis. - Table 13.2.1gives the coordinates of some points on the graph. From these points we draw half of the graph and the remainder is-drawn from its symmetry with respect to the polar axis. lf r:0, we obtaincos 0:*, and if 0 < 01r, 0:*r.Therefore, the point (0' *n) is on the graph,and an equation of the tangent line there is 0 : *zr. A sketch of the graph is shown in Fig. 13.2.2.rt is called a limagon.

13,2 GRAPHSOF EQUATIONSIN POLARCOORDINATES 563 Tsble

0

0

tn

tn

tn

3n

r

-1

L-\tr

0

t

2

Err

1+\E

7r

3

Fig ure

The graph of an equation of the form r-s+bcos0 or r-s+bsin0 is a limaqon. r f b If fl:b, the limaqonisacardioid,whichisheart-shaped.Ifa> limaEon has a shape similar to the one in Example 3.

EXAMPIn 3:

Draw

the graph of

3+2stn0 1_

7T

a sketch of

solurroN: The graph is symmetric with respect to the in axis because if (r,0) is replacedby (r, r - 0), an equivalent equation is obtained. Table ti.2.2 gives the coordinates of some of the points on the graph. is drawn by plotting the A sketch of the graph is shown in Fig. 13.2.3.It '1,3.2.2 and using the symin Table points whose coordinates are given metry property. T a b l eL 3. 2 . 2

F i g u r e1 3 . 2 . 3

0

0tntnLniT

r

3

4

3+\E

5

3

&n

*n

En

2

3-\tr

L

564

POLARCOORDINATES

The graph of an equation of the form 'r-acosn0

or r-astnn?

is a rose,having n leavestf n is odd and 2nLeavesrf n is even.

EXAMPur4: Draw a sketch of the four-leafed rose r-4cos20 1

7r

solurroN: In Illustration L we proved that the graph is symmetric with respect to the polar axis, the trr axis, and the pole. Substituting 0 for r in the given equation, we get c o s2 0 : 0 from which we obtain, for 0 < 0 12rr,0:*rr,Zrr,Zr, and #2. Table 13.2.3gives values of r for some values of d from 0 to jz. From these values and the symmetry properties, we draw a sketch of the graph as shown in Fig. 13.2.4. TableL3.2.3 0

0

#n

r

4

2\tr

tn

L+n

tn

#n

tn

2

0

-2

-z\tr

-4

Figure 13.2.4

The graph of the equation 0:C where c is any constant, is a straight line through the pole and makes an angle of C radians with the polar axis. The same line is given by the equation 0:C-rnn where n is any integer. general, the polar form of an equation of a line is not so simple -In as the cartesian form. However, if. the line is parallel to either the poiar axis or the *zr axis, the equation is fairly simple. If a line is parallel to the polar axis and contains the point B whose cartesian coordinates are (0, b) and polar coordinates ^t" 1b,fz'), then a cartesian equation is y : b. If we replace y by r sin d, we have rsin0:b which is a polar equation of any line paraller to the polar axis. If b is posi-

13.2 GRAPHSOF EQUATIONSIN POLARCOORDINATES 565

tive, the line is above the polar

It b is negative, it is below the polar

we have a sketchof the graph of the equa. rLLusrRArroN2: In Fig. 1,3.2.5 ']..3.2.6 g: we have a sketch of the graph of the 3, and in Fig. tion r sin -3. . equation r sin 0: If a line is parallel to the *zr axis or, equivalently, Perpendicular to the polar axis, and goes through the point A whose cartesian coordinates are (a, 0) and polar coordinates are (a, 0), a cartesian equation is r : a. Replacing xby r cos 0, we obtain

F i g u r e1 3 . 2 . 5

rcos0:a

rsin0: Figure cO

tl
which is an equation of any line perpendicular to the polar axis. If. a is positive, the line is to the right of the *zr axis. If a is negative, the line is to the left of the *zr axis. 3: Figure 13.2.7shows a sketch of the graPh of the equao rLLusrRATroN and Fig. \3.2.8shows a sketch of the graPh of the equa0:3, r cos tion 3 . t i o n r c o s0 The graPh of the equation

r:c Figure 13.2.7

where C is any constant, is a circle whose center is at the pole and radius is lcl. The samecircle is given by the equation

r--c

F i g u r e1 3 . 2 . 8

As was the case with the straight line, the general polar equation of a circle is not so simple as the cartesian form. However, there are special casesof an equation of a circle which are worth considering in polar form' If a circle has its center on the polar axis at the point having.polar coord.inates(a, 0) and is tangent to the tr axis, then its radius is lal. The points of iniersection of s.tlh u circle with the polar axis are the pole if P(r, 0) is any point on iO, O; ana the point (2a,0) (seeFig. 1'3.2.9)'Then a right angle. Therefore, is circle the in P inscribed at angle the ihe circle, : equivalentlY, rl2a ot, cos 0 r:2a

c o s0

which is a polar equation of the circle having its center on the polar axis or its extension, and tangent to the Ezt axis. lf. a is a positive number, the circle is to the right of the pole. lf. a is a negative number, the circle is to the left of the pole. In a similar manner, it may be shown that r:2b Fig ure

sin0

is a polar equation of the circle having its center on the in axis or its ex-

566

POLAR COORDINATES

tension, and tangent to the polar axis. If b is positive, the circle is above the pole. If b is negative, the circle is below the pole.

ExAMPrn 5:

Draw a sketch of

the graph of

soLUrIoN: When 0: nn, where n is any integer, the graph intersects the polar axis or its extension, and when 0:inrr, where n is any odd integer, the graph intersects the izr axis or its extension. When r: 0, 0: 0; thus, the tangent line to the curve at the pole is the polar axis. A sketch of the graph is shown in Fig. 13.2.1,0. The curve is calleda spiral of Archimedes.

11 ='lf o

F i g u r e1 3 . 2 . 1 0

ExercisesL3,2 In ExercisesL through 40, draw a sketch of the graph of the given equation. 1 , .r c o s0 : 4 2. r sin 3. r-4

c o s0

4.

2sin0 -4

5. 0:5

6.

7. r:5

8. r : - 4

9. r sin 0: -4 1 1 .r : - 4 13. r:

sin0

eo(ogarithmic spiral)

10. r c o s 0 : - 5 L2. r : - 5 c o sA 1,4.r - eotg(logarithmic spiral)

15. r - ll0 (reciprocalspiral)

L6. r:

L7. r : 2 sin 30 (three-leafedrose)

1 8 . r : 3 cos20 (four-leafed rose)

19. r : 2 cos 40 (eight-leafedrose)

20.

21,.r : 4 sin 20 (four-leafed rose)

22. r : 3 cos 30 (three-leafed rose)

23. r : 2 + 2 sin 0 (cardioid) 25. r : 4 - 4 cos 0 (cardioid)

24.

20 (spiral of Archimedes)

4 sin 50 (five-leafed rose)

3+3cos

0(cardioid)

26. r : 3 - 3 s i n 0 ( c a r d i o i d )

13.3 INTERSECTION OF GRAPHSIN POLARCOORDINATES

2 7 .r : 4 - 3 c o s A ( h m a g o n ) 29.r:4+2

28. r:3-4

sin 0 (limagon)

30. r:2

31. rz :9

32. 12 : -4 sin 20 (lemniscate)

sin 20 (lemniscate) -25 : 33. 12 cos2e (lemniscate) -35. r 2 sin 0 tan 0 (cissoid) 37. r : 2 sec 0 - 1 (conchoid of Nicomedes) 99. r:

cos 0 (limagon) - 3 sin 0 (limagon)

34. 12 : L6 cos 20 (lemniscate) 35. (r - 2)' :80 38. r:

2 csc 0 + 3 (conchoid of Nicomedes)

40.r:2lcos

lsin 201

(parabolic spiral)

0l

41 through 44,find an equationof eachof the tangentlines to the given curveat the pole. In Exercises 4 2 .r : 2 s i n 3 0 4 7 .r : 4 c o s 0 l 2 44.rz :9 sin20 43.rz :4 cos20 r15.Find the slopeof the tangentline to the curver: 4 at (4, *n). rt5. Find a polar equationof the tangentline to the curver = -6 sin g at the point (6, *r) .

13.3 INTERSECTION OF GRAPHS IN POLAR COORDINATES

EXAMPLE "1.: DrAW

sketchesof 20 and r: 2 sin graphs of the r : L, and find the points of intersection.

To find the points of intersection of two curves whose equations are in caftesian coordinates, we solve the two equations simultaneously. The common solutions give all the points of intersection. However, because a point has an unlimited number of sets of polar coordinates, it is possible to have as the intersection of two curves a point for which no single pair of polar coordinates satisfies both equations. This is illustrated in the following example. solurroN: The graph of r:2 sin 20 is a four-leafedrose. The graph of r: 1 is the circle with its center at the pole and radius 1..Sketchesof the graphs are shown in Fig.13.3.1.Solving the two equations simultaneously, we have 1 : 2 sin 20 or sin 20 : i. Therefore, we get 20 : tn, *n , En , and *rr 0:#r,&n,tlr,

andlln

Hence, we obtain the points of intersection (1, $n), 11.,fzr), (1, l$rr) , and that eight points of intersectionare shown. (1,,I12r).Wenotice in Fig. 1.3.3.1 The other four points are obtained if we take another form of the equa-L, which is the tion of the circle r: 1; that is, consider the equation r: same circle. Solving this equation simultaneously with the equation of the four-leafedrose,we have sin 20 : -+.Then we get 20:&n,#n,

#n, andTn

0 - #n, #n,1&n, and #n F i g u r e1 3 . 3 . 1

Thus, we have the four points (-L,#n),1-L,#r),

(-7,1&n), and (-t,1?r).

568

POLAR COORDINATES

Incidentally, (-1 , #n) alsocan be written as (1,,#n), (-1, #n) can be written as (L, # r ) , ( - 1 , # n ) can be written as (1,#n), and (-t ,Hzr) can be written as (1,#n). Because(0, 0) represents the pole for any 0, we determine if the pole is a point of intersection by setting r: 0 in each equation and solving for 0. Often the coordinatesof the points of intersection of two curuescan be found directly from their graphs. However, the following is a general method. If an equation of a curye in polar coordinatesis given by (1) r - f(0) then the same curye is given by (-l)nr:f@*nn) where n is any integer. o rLLUSrRArroN 1: Consider the curyes of Example 1. The graph of r - 2 sin 20 also has the equation (by taking n:'1. in Eq. (2)) (-t)r-

2 sin 2(0irr)

or, equivalently, -r:2 sin 20 If we take n : 2 in Eq. Q), the graph of r : 2 sin 20 also has the equation (-7),r:

2 sin 2(0 + 2n)

or, equivalently, r:2

s i n2 0

which is the same as the original equation. Taking n aurry other integer, we get either r:2 sin20 or r: -2 sin 20.The graph of the equation r: 1 also has the equation (by taking n: I in Eq. (Z)) (-1;r:

1 or, equivalently, r: -7

Other integer values of n in (z) applied to the equation 1 give either r: I or r - -1. o If we are given the two equations r - f(0) and r - g(0), we obtain all the points of intersection of the graphs of the equations by doing the following. (a) Use (2) to determine all the distinct equations of the two cur:ves:

r - f r ( O ,) r : f r ( 0 ), r : f r ( O ,)

(3)

13.3 INTERSECTION OF GRAPHSIN POLARCOORDINATES

r:8t(0) , r:

S r @ ), r :

k @ ) , 'r;;ttaneously

(4)

(b) Solve each equation in (3) with each equation in (a). (c) Check to see if the pole is a point of intersection by setting r: 0 in each equation, thereby giving

and s@)-0

f@l-0

(5)

If Eqs. (5) each have a solution for 0, not necessarily the same, then the pole lies on both cunres.

EXAMPLE2: Find the points of intersection of the two curyes r -- 2 - 2 cos 0 and r - 2 cos 0. Draw sketches of their graphs.

sot-urroN: To find other equations of the cur:ve represented by r 2 2 cos 0, we have (-1)r:2-

2 cos(0+n)

or, equivalently,

-r:2+2cos0 and (-1)'r:2-

2 cos(p+2n)

or, equivalently, r-2-2cos0

which is the same as the original equation. In a similar manner, we find other equations of the curve given by r : 2 c o s0 : (-L)r:2 -r:

cos(O* r) -2 cos0

r:2

c o s0

which is the same as the original equation. So there are two possible equations for the first curve, r: 2 - 2 cos 0 -r :2 * 2 cos0, and one equation for the secondcurve, r:2 cos0. and Solving simultaneously r: 2 - 2 cos 0 and r: 2 cos 0 gives 2cos0-1-2cos0 4 c o s0 : 2 c o s0 : * Thus, e - En and En, giving the points (1,+zr)and (I,8n). Solving simult a n e o u s l y - r - 2 + 2 cos g and r :2 cos 0 yields 2 + 2 c o s0 - - 2

cos 0

570

POLARCOORDINATES

4 cos0: -2 c o sd : - * Hence, 0 : *rr and 0 : *r, giving the points (-1,, ?ir) and (-1, fz). However, (-1, 3z) is the samepoint as (1, *rr), and f 1, *z) is the samepoint as (1, $r). Checking to see if the pole is on the first cuwe by substituting r: 0 in the equationr:2 - 2 cos d, we have 0 : 2 - 2 c o s0 cos 0 : 1

0:0 Figure 13.3.2

Therefore, the pole lies on the first currre. In a similar fashion, by substifuting r: 0 in r:2 cos0, we get 0:2

c o s0

c o s0 : 0 0:ir

or tn

So the pole lies on the second curve. Therefore, the points of intersection of the two curves ate (7, |sr), (1, *n\, and the pole. Sketchesof the two curves are shown in Fig. 13.3.2.

L3.3 Exercises In Exercises1 through L4, find the points of intersection of the graphs of the given pair of equations. Draw a sketch of each pair of graphs with the same pole and polar axis. .' ' [ 2 r : 3 l t:3sin0

^t ' l r(:21r : 3

4- '. [ r : 2 c o s 2 0 [r: 2 sin0

u - '. { r : 4 0 Lr: ir

6 :!n " ' . [[ e r: 2

," [ r : c o s 0 - l lr: cos20

R Jz:l-sind "' [r: cos20

o [r:sin2l " lr: cos20

*cos 0

n.[r':2cos| I r:1

t- .- [' r : t a n ? Lr:4sind

,-^- ' [ r : 4 ( 1 * s i n 0 ) [r(f - sin 0) :3

,ô^ ' [ r ' s i n 2 0 : 8 I r c o s0 : 2

^ f t : 2 c o s0 c'lr:2sinO

p - -.-{ r : 4 t a n 0 s i n 0 [r:4cosO

In Exercises15 and 15, the graph of the given equation intersects itself. Find the points at which this occurs. 'l..6. 15. r: sin *0 r: ! * 2 cos20

13.4TANGENTLINESOF POLARCURVES

13.4 TANGENT LINES In Sec. 13.2. we derived a formula for finding the slope of a tangent line OF POLAR CURVES to a polar curve whose equation is r: f (0).If a radians is the measureof the inclination of the tangent line to the curve at (r,0)' we have

(1)

Formula (1) is generally complicated to aPply. A simpler formula is obtained by considering the angle behmeenthe line OP and the tangent line. This angle will have radian measure 1 and is measured from the tine OP counterclockwise to the tangent line, 0 < X < 7t. There are two possible cases:o > I and ot I 0. These two casesare and 13.4.2.InFig. 13.4.L,a) 0 and 1:a-0' illustratedin Figs. 1.3.4.1 In Fig. 13.4.2,a 1 0 and 1: T - (0 - a). In each case, tan 1: tan(a - 0)

o F i g u r e1 3 . 4 . 1

or, equivalently, tanX:

tan a-

tan 0

(2)

L+tan o-tan0

Substituting the value of tan a from (1,)in (2), we get

(?i' # t,G\;'""') : =^"u : tan x t.K" :') tant #*

r -tan e

#*

r tanz0

o#*r-no

Hence, Figurc 13.4.2

iiit+il++$fi+f+1iffi

(3)

more Comparing formula (3) with formula (1), you can see why it is coordinates. desirable to consider 1 instead of a when working with Polar ExAMPLEL: Find 1 for the cardiotdr- a* a sin 0 at the point (Ea, En).

soLUTIoN: See Fig.

acose

'1.3.4.3. Because

a * a sin 0, we have

572

POLARCOORDINATES

Applying formula (3), we get tan

a*asin0 acos0

1+sin0 cos 0

Therefore, at the point (ra, tn),

tt/s ft-- v3

t a nX - # : = 3 So X: *n

F i g u r e1 3 . 4 , 3

EXAMPLE2: Find to the nearest L0 min the measurement of the smaller angle between the tangent lines to the curyes r - Z cos 20 and r - 3 sin 20 at the point

Pefr,t.o).

If B is the radian measureof the anglebetweenthe tangentlines at P?\/2,grr),then F: xt- x, So tan B: tan(11- 12) or, equivalently, tan '6 - =tT xt tan x' 1 + tan Xr tany"

(4)

and tan X2 arcfound from formula (3). For tan yr, r: _ and, Tul,I, drld0: -6 sin 20. So Figure 13.4.4

1 f-A - -n-Awr_ : _3 6c o s 2 0 sin n:-)cot20 When 0 : tzr, tan 1, : -* cot izr : -*. For tan Xz,r:3 sin 20 and,drlilT: 6cos 20. So

t a n ; i:,m : | t a n z o When e - *n, tan Xz: t tan LEzr : t. Substituting tan Xr: -* and tan Xz:* in (4), we obtain

: 1l + = t a n B--@ --+

-1

4

j

cos20

13.5 AREA OF A REGIONIN POLARCOORDINATES

The angle of radian measure B has a measurementof 125'50'to the nearest L0 min. So the measurement of the smaller angle between the two tangent lines is 180o- 125'50': 53o10'.

1-3.4 Exercises In Exercises1 through8, find 1 at the point indicated. 2. r : a0; (tra, Ezr) l. r0 : a; (a, l) 4. r:asin*g; (tra,*r) 7. t2 : a2 cos20;

la

1

3' r : a sec20;(t/-za,-+r) 6. r:acoslli(iuIa,Jzn)

5. r:02;(*n',tr) \

\U,i")

8' t:

a(l-

sin 0); (a' r)

at the In Exercises9 through 12, hnd a measurement of the angle between the tangent lines of the given pair of curves indicated point of intersection.

t {i=i:trt,(*r2a,rn)

'n {;=Lo"u.e;@,*d

y: 4 cosô. --2, - ' o&n) "/ n. " ' . t r : 4 c o s 2 0 - 3 -: '\

p. {:=;'"::\:t the pole U:l pair of curvesat all In Exercises13 through 15,find a measurementof the anglebetweenthe tangentlines of the given points of intersection.

: 2(1- cos 0) ' 17. Prove that tan 1 : tan *0 at all points of the cardioidt 18. Prove that at each point of the logarithmic spiral r: beoe,1is the same' L9. prove that at the points of intersection of the cardioids r: pe{pendicular for all values of.a arrd b.

a(l * sin 0) and r:

20. Prove that at the points of intersection of the fwo curyes r: Pe{Pendicular.

1.3.5AREA OF A REGION IN POLAR COORDINATES

a sec2te and r:

b(l - sin 0) their tangent lines are b csc2te their tangent lines are

In this section a method is developed for finding the area of a region bounded by a curve whose equation is given in polar coordinates and by two lines through the pole. Let the function / be continuous and nonnegative on the closed interval [a, B]. Let R be the region bounded by the curve whose equation is /: /(0) and by the lines 0 : a and 0: B. Then the region R is the region AOB shown in Fig. 13.5.1' ConsiderapartitionAof[a,B]definedbyo:0o<0,<02<..<

574

POLAR COORDINATES

. 10n_, 19n:B. Therefore,we have z subintervalsof 0rr1&1. the form f0i-1,01f, where i:'1 ,2, . . . ,n.Let $ be a value of g in the ith subinterval f0i-1,061. SeeFig.13.5.2,wherethe ith subinterval is shown together with 0: f;. The radian measure of the angle between the lines 0: 0*r and 0: 0r is denoted by Lg. The number of square units in the area of the circular sector of radius /(f1) units and central angle of radian measure AaOis given by

ElfG)],L,0 There is such a circular sector for each of the n subintervals. The sum of the measuresof the areas of these n circular sectors is F i g u r e1 3 . 5 . 1

+lf(f,)l' a,o+ +lf(t,)12Lzo +

+ +lf(f,)I2 Lio+

+ +lf(€")1,Ln?

which can be written, using sigma notation, as n

> +tfc)f, L,0 Let llAllbe the norm of the partition A; that is, IlAllis the measureof the largest Lr?.Then if we let A square units be the area of the region R, we define

#*ffi .......#=+U a,o I+r, ft$, Figure13.5.2

The limit in (t) is a definite inte gral, and we have

#$

EXAMPLE1: Find the area of the region bounded by the graph of

r:2+2cos0.

(2)

The region together with an element of. areais shown in Fig. ::l]t"_"' 13.5.3.Becausethe curve is symmetric with respectto the polar axis, we take the 0 limits from 0 to z which determint the area or tn" region bounded by the curve above the polar axis. Then the area of the entire region is determined by multiplying that area by 2. Thus, if A square units is the measure of the required area,

A :2 ,!ip i+ fr *2 l l a l l - o- : -n 1

-4

costi) , Aio

r

+(2 + 2 cos 0)2 d0

I F i g u r e1 3 . 5 . 3

(1)

( 1 + 2 c o s 0 + c o s z 0 )d 0

sin 0 + te + * sin *]:

13.5 AREA OF A REGIONIN POLARCOORDINATES

- 4(n* 0 + in * 0 - 0) :6n Therefore, the area ts 6n square units.

ExAMPLE2: Find the area of the region inside the circle r:3

sin0

3 sin0-2-

sin0

sin0:Lz

and outside the limagon r:2-

To find the points of intersection,we set

solurroN:

So 0:tr

sin0 r :, f (0)

Li1 ,0:Ei

' t,) $(Eil (g(€;),tr)

and 0:8n

The curves are sketched and the region is shown together with an element of area in Fig. 13.5.4. - - sin 0, then the equation of the If we let f @):3 sin 0 and g(0) I circle is r: f (0), and the equation of the limagon is r: 8(0). The measure of the area of the element of area is the difference of the measuresof the areas of two circular sectors.

tlf G)l' toe- ils?)l' 7 :, g(0)

: +([/(fn)]' - [g(f') ]') aoo Ln?

The sum of the measuresof the areas of r such elements is given by n

A'o > +(t/(f')l'-[g(f,)]')

C=1

Hence, if A square units is the area of the region desired, we have F i g u r e1 3 . 5 . 4

A: tisr i +ttrtE lt'- [g(f,)]')Aoo llall-oEi This limit is a definite integral. Instead of taking the limits tr to *rr, we or" thu property of symmetry with respect to the ttr axis and take the limits from Szrto ir and multiply by 2.We have, then, fn12

A:2. 1|

( t / ( 0 ) l ' - l s @ ) l ' zd)e

J t16

:

f::re

sin

0)'l d0

-8

e d0+4 s i n0 d e - 4 f: f:sinz -8 - c o s 2 0 )d o + f:(1 :40-2sin20-4

cos o - Ae'n'' lnrc

576

POLAR COORDINATES

sin 20- 4 cos _ (-2 sin 7T- 4 cos

(-2 sin tn - 4 cos En)

-2.+\E+4.+\n - 3{g Therefore,the area is 3 \n square

13.5 Exercises In Exercises L throu gh 6,find

the area of the region enclo sed by the graph of the given equation.

1. r:3cos0

2.r-Z-sin0

3. r-4

c o s3 0

4.r:4sin2t0

5. r2:4sin20

6. r:4

s i n 20 c o s 0

In Exercises 7 through 10, find the area of the region encl sed by one loop of the graph of the given equation. 7. r:

3 cos20

9. r-"1,+3sin0

8. r:a 10. r:

sin30 a ( l - 2 c o s0 )

In Exercises LL through "1,4,find the area of the intersection of the regionsenclosedby the graphsof the two given equations. 1 1 [r-2 ^^' lr - 3 - 2 cos 0 sin20 4 rr'^ l r --3 3 cos 2o lr

l r' : 4 : s i n 0 12..t lr-4cos0 (-2 - 2 cos20 1,4.{ ' I r- 1

In ExercisesL5 through 18,find the areaof the region which is inside the graph of the first equation and outside the graph of the secondequation. 15. [ r --o l r a ( l - c o s0 )

-

1 7 . ft 2 sin o Lr-sin0*cosg

- 2a srn o L6. {r lr-a - 4 stn 20 j.g. r \ " r [r' L t- \n

19. The face of a bow tie is the region enclosed by the graph of the equation 12: 4 cos 20. How much material is necessary to cover the f.aceof the tie? 20. Find the area of the region swept out by the radius vector of the spiral r: not swept out during its first revolution.

a0 duing its second revolution which was

21- Find the area of the region swept out by the radius vector of the curve of Exercise20 during its third revolution which was not swept out during its second revolution.

ReaiewExercises (Chapter13) In Exercises7 and'2, find a polar equation of the graph having the given cartesian equation. l. x2*y'-9x+8y:0

Z.yn:*(ar-yr)

REVIEWEXERCISES

In Exercises3 and4, find a cartesian equation of the graph having the given polar equation. 4. r:

3. r-9sinzi0

a tanz 0

3/d (reciprocal spiral) and (b) r: 013(spiral of Archimedes). 5. Show that the equations r: L I sin 0 and /: sin 0 - t have the same graph. y'lios 0f. 7. Dtaw a sketch of the gaph o1 7 : 5. Draw a sketch of the graph of (a) r:

8. Draw a sketctr of the graph of r:

{@{E|

9. Find an equation of each of the tangent lines at the pole to the cardioid' r: 3 3 sin 0' 10. Find an equation of each of the tangent lines at the pole to the four-leafed rose r:4 cos20. 11. Find the area of the region inside the graph of r:2a

sin 0 and outside the graph of r:

a'

a cos 0 and 12. Find the area of the intersection of the regions enclosed by the graphs of the two equations /: > 0. a 0), cos r: a(l r: stco(k > 0) as 0 varies from 13. Find the area of the region swept out by the radius vector of the logarithmic spiral 0 to 2tr. :2 sin 20 and outside the graph of the circle r: 1' 14. Find the area of the region inside the graph of the lemniscate 12 : (o , n) . 15. Find a polar equation of the tangent line to the curve r 0 at the point the two given equations' In Exercises16 and 17, find all the points of intersection of the graphs of

L6{;:l i;lo,e

aF, | ,-2(l+ Lt' lrr-2cos0

cos 0)

of points of intersection. Also prove that the 1g. prove that the graphs of r: a0 and.r0: a have an unlimited number and find these points' intersection, tangent lines are perpendicular at only two of these points of each point of intersection of the graphs of the equa19. Find the radian measure of the angle between the tangent lines at -6 cos 0)' cos 0 and,t:2(1 tions r: : point (1, *n)' 20. Find the slope of the tangent line to the curve r 6 cos0 2 atthe * 2 cos 0) and also the area of the region en21. Find the area of the region enclosed by the loop of the limagon r:4(7 closed by the outer part of the limagon' n is a positive integer' 22. Find.the area enclosed by one loop of the curve r: a sin n0, wherc is 23. Prove that the distance between the two points PJh, 0) and'PzQz' 0)

@

24. Find the points of intersection of the graphs of the equations /: tan 0 and r: col 0; (2, tr,) and the line 0 : 0 25. (a) Use the forrrula of Exercise23 to find a polar equation of the parabola having its focus at of the paraboh hJving its focus at (0,2) and the r axis as its directrix' as its directrix. (b) Write a cartesian "qtr"tiotr Compare the results of parts (a) and (b). law of cosinesto 25. Find a polar equation of the circle having its center at .(rs,!6) and a radius of a units. (rur'rr: Apply the (r, (ro, and 0)') 0), pole, at the the triangle hiving vertices

14.1 THE PARABOLA

14.I THE PARABOLA

A conic section is a curve of intersection of a plane with a right cirorlar cone of two nappes. There are three types of curr/esthat occur in this way: the parabola, the ellipse (including the circle as a special case), and the h;rperbola. The resulting cuwe depends on the inclination of the axis of the cone to the cutting plane. In this section we study the parabola. Following is the analytic definition of a parabola.

1,4.1.1Definition

A parabolais the set of all points in a plane equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix. We now derive an equation of a parabola from the definition. In order for this equation to be as simple as possible, we choose the r axis as perpendicular to the directrix and containing the focus. The origin is taken as the point on the r axis midway between the focus and the directrix. It should be stressedthat we are choosing the axes (nof the parabola) in a specialway. SeeFig.1'4.1"1" x:

-l

a(- p,v)

.1 14.1 Figure

then P is on the parabola if and only if

lFTl:lOPl Because and

lilPl: t/(.T\YTTW P is on the parabola if and onlY if

\/G=pYTT: {GTff

580

THE CONIC SECTIONS

By squaring on both sides of the equation, we obtain xz - 2px * p' * y' : )c2* 2px * p' Y' : 4Px This result is stated as a theorem.

14.1.2Theorem

An equation of the panbola having its focus at (p,0) and as its directrix the line y:-p is (1)

F(p,o)

F i gur e 14 . 1. 2

14.1,.3Theorem

'!.4.'1..'!,, In Fig. p is positive; p may be negative,however, becauseit is the directed distance Of. Figure 14.1.2shows a parabola for p < 0. From Figs. 14.1.1and 1,4.'1,.2, we seethat for the equation yz :4px the parabola opens to the right if p > 0 and to the leftif p < 0. The point midway between the focus and the directrix on the parabola is called the ztertex.The vertex of the parabolasin Figs. 14.1.1and 14.1.2is the origin. The line through the vertex and the focus is called the axisof the parabola. The axis of the parabolasin Figs. 14.1.1and 1.4.1.2isthe r axis. In the above derivation, if the r axis and the y axis are interchanged, then the focus is at the point F(0, p), and the directrix is the line having the equation y:-p. An equation of this parabola is *:4py, and,thi result is stated as a theorem. An equation of the parabola having its focus at (0, p) and as its directrix the line y : -p is (2)

If p ; 0, the parabola opens upward as shown in Fig. 74.1.2; and ifp as shown in Fig. 14."1,.4. In each case the vertex is at the origin, and the y axis is the axis of the parabola.

F(0,p)

F i g u r e1 4 . 1. 3

Figure 14.1.4

14.1THE PARABOLA

581

When we draw a sketch of the graph of a parabola, it is helpful to draw the chord through the focus, perpendicular to the axis of the parabola. This chord is called the latus rectum of the parabola. The length of the latus rectum is lapl. (SeeExercise1.7.) EXAMPLE1: Find an equation of the parabola having its focus at (0, -3) and as its directrix the line y - 3. Draw a sketch of the graPh.

sor,urroN: Since the focus is on the y axis and is also below the directrix, the parabola opens downward, and p: -3. Hence, an equation of the parabola is *:-12y The length of the latus rectum is l4(-3) | : 12

Q,

Qt

Qt

u

-c

o

A sketch of the graph is shown in Fig. 1'4.1.5. Ary point on the parabola is equidistant from the focus and the directrix. In Fig. 74.L5, three such points (Pr, Pr, an'd Pg) are shown, and we have

- lre-,| I : lPre-,| tFP,l-lk-a,l lFtr,l IFP-, F i g u r e1 4 . 1 . 5

EXAMPLE 2. Given the Parabola having the equation

solurroN:

The given equation is of the form of Eq. (1); so

4p-7

u ': 7 x find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the graPh.

opens to the right. The focus is at the point Becausep - -2. The length of the latus F(+, 0). An equation of the directrix is x rectum rs 7. A sketch of the graph is shown in Fig. 1'4.1.6.

Directrix

F i g u r e1 4 . 1 , 6

THE CONIC SECTIONS

L4.1 Exercises For each of the parabolas in ExercisesI thnrugh 8, find the coordinates of the focus, an equation of the directrix, and the length of the latus rectum. Draw a sketch of the curve. 1. x2:4A 4. xz- -I5y

2' a2:6x

3. y' : -8x

5.f +A:0

6.y2- 5r:0

7.2y'-9x:0

8.3x2*4y-0

In Exercises 9 through 16, find an equation of the parabola having the given properties. 9. Focus, (5, 0); directrix, y: -5. 10. Focus, (0, 4); directrix, A: -4. 11. Focus, (0, -2); directrix, y - 2:0. 12. Focus, (-8, O); directrix, 5 - 3r: 0. 13. Focus,(*, 0); directrrx,2xl l:0. 14. Focus, (0, i); directrix,Sy t2:0. 15. Vertex, (0, 0); opens to the leff length of latus rectum: 6. 15. Vertex, (0, 0); opens upward; length of latus rectum: 3. 17. Prove that the length of the latus rectum of a parabola is l4pl. 18' Find an equation of the parabola having its vertex at the origin, the r axis as its axis, and passing through the point (2, -4\. 19' Find an equation of the parabola having its vertex at the origin , the y axis as its axis, and passing through the point (-2, -4). 20' A parabolic arch has a height of 20 ft and a width of 36 ft at the base. If the vertex of the parabola is at the top of the arch, at what height above the base is it 1g ft wide? 21' The cable of a suspension bridge hangs in th9 form of a parabola when the load is uniformly distributed horizontally. The distance between two towers is rsoo ft, the points ;i;;;6; of the cable on the to*"r, arc 220fr above the road-

il1til*:T#ffiit:HfiJ?ff31""T"70rt

aloveth",oa'd*"v. Findthevertical distance tothecabre rromapoint

22' Assume that water issuing from the end of a horizontal pipe,25 t lb9",u the ground, describesa parabolic curve, the vertex of the parabola being at the end of the pip-e.r, ui u'poit t 8 ft below trr"etine of the pipe, the flow of water has curved outward 10 ft beyond a vertical line throuih th; ;"e iiir" pip", how far beyond this vertical line will the water strike the ground? 23' A reflecting telescopehas a parabolic mirror for which the distance from the vertex to the focus is 30 ft. If the distance acrossthe top of the mirror is il in , how deep is the mirroiai ttre centerz 24' Using Definition 14'1.1,find an equation of the parabola having as its directrix the line a :4 and.asits focus the point (-3, 8). 25' using Definition t4'l'1, hnd' an equation of the parabola having as its directrix the line r:-3 point (2,5).

and as its focus the

26' Find all points on the parabola y2: 8r such that the foot of the perpendicular drawn from the point to the directrix, the focus, and the point itself are verrices of an equirateiJil;gil,' 27' Find' an equation of the circle passing through the vertex and the endpoints of the latus rectum of the parabola ,, - -gy.

14,2 TRANSLATIONOF AXES

2g. prove analytically that the circle having as its diameter the latus rectum of a parabola is tangent to the directrix of the parabola. 29. A focal chord of a parabola is a line segment through the focus and with its endpoints on the parabola. If A and B are the endpoints of ifocal chord of a parabola, and ii C is the point of intersection of the directrix with a line thrcugh the vertex and point A, prove that the line through C and B ii parallel to the axis of the parabola. is half the fl). prove that the distance from the midpoint of a focal chord (see Exercise 29\ of. a parabola to the directrix length of the focal chord.

14.2 TRANSLATION OF AXES

The shape of a curve is not affected by the position of the coordinate axes; however, an equation of the curve is affected' . rllusrRArrorv 1: If a circle with a radius of 3 has its center at the point (4, -1), then an equation of this circle is

(x-4)'+@+1)':e **Y'-8r*2Y*8:0 Howevef, if the origin is at the center, the same circle has a simpler equation, namelY, ' f+Yz:g Ifwemaytakethecoordinateaxesasweplease,theyaregenerally chosen in such a $/ay that the equations will be as simple as possible. Iftheaxesaregiven,however,weoftenwishtofindasimplerequation of a given curve referred to another set of axes' Ingeneral,ifintheplanewithgivenrandyaxes,Itewcoordinate a chosen parallel to the given ones, we say that there has been "re" """s ttanslation of axesin the Plane' y' In particular' let the gi*tett x and'y axes be translated to the x' and that assume Also, axes. axes,hiving origin (h, k)-withrespect to the givgn x' and y' the positive no*b"rs are on the iame side of the origin on the axes as they are on the x and'y axes (see Fig' 1'a'2'1)'

(h,k)

A, l

N,

I I I I

Figure14.2.1

A point P in the plane, having coordinates (x, y) wilh resPectto the the given coordinate axes; wiil have coordinates (x' , y') with resPectto

584

THE CONIC SECTIONS

new axes.To obtain relationships between these two sets of coordinates, we draw a line through P parallel to the y axis and the y' axis, and also a line through P parallel to the r axis and the I' axis. Let the first line intersect the r axis at the point A and the r' axis at the point A', and the second line intersect the y axis at the point B and the y' axisat the point B'. With respect to the r and y axes,the coordinates of P ate (x, y)' the coordinates of A are (r, 0), and the coordinates of A' are (r, k). Because A,p:n-M,wehave y':Y-k or, equivalently, y:y'*k With respect to the x and y axes, the coordinates of B are (0, y), and the coordinates of B' are (h, y). BecauseB'P : BP - BB' , we have X,:X-h

or, equivalently, x:x'*h These results are stated as a theorem. 14.2.1Theorem lf (x, y) representsa point P with respectto a given set of axes,and (x' ,y') is a representation of P after the axesare translated to a new origin having coordinates (ft, k) with respect to the given axes, then

y-v' +k

(1)

y'-y -k

(2)

ot,

Equations (1) or (2) are called the equationsof translating the axes. If an equation of a curve is given in x andy, then an equation in r' and y' is obtained by replacing x by (x' -t h) and y by (y' -f k).The graph of the equation in r and y, with respect to the x and y axes,is exactly the same set of points as the graph of the corresponding equation in r' and y' with respect to the x' and y' axes. ExAMPLE1: Given the equation x2 + Llx * 6y + 19: 0, find an equation of the graph with respect to the x' and y' axes after a translation of axes to the new origin (-5, 1).

soLUrIoN: A point P, represented by (x, y) with respect to the old axes, has the representation(r', y') with respectto the new axes.Then by Eqs. (1), with h: -5 and k : 1, we have x'-5

and y-y'

+1

Substituting these values of r and y into the given equation, we obtain

( x ' - 5 ) , + 1 0 ( r ' , -5 ) + 6 ( y ' , +1 ) + 1 g : 0

14.2 TRANSLATIONOF AXES

x'2 - "1.0x' + 25 * l}x'-

v--+ x'

Figure 14.2.2

50 * 6y' + 6 + 19 - 0 )c'2: -6Y'

The graph of this equation with respect to the r' and y' axesis a parabola with its vertex at the origin, opening downward, andwith 4p :-6. The graph with respect to the x and y axes is, then, a parabola having its vertex at (-5, 1), its focus at (-5, -*), and as its directrix the line y : $ (seeFig. 1,4.2.2).

The above example illustrates how an equation can be reduced to a simpler form by a suitable translation of axes. In general, equations of the second degree which contain no term involving xy can be simplified by a translation of axes. This is illustrated in the following examPle.

EXAMPLE2: Given the equation * 32y + 37: 0, 9xz * 4y' - "1,8x translate the axes so the equation of the graph with respect to the )c' and y' axes contains no first-degree terms.

vy'

solurroN: We rewrite the given equation: 9(* - 2x) + 4(yz * 8y) :-37 ' Completing the squares of the terms in parenthesesby adding 9 I' and 4 ' 16 on both sides of the equation, we have

9(* - 2x* 1) + +(y' * 8y-r 1'6):-37 + 9 + 64 9 ( x - L ) ' + 4 ( y + 4 ) 2: 3 6 Then, if we let r'

1 and y' : y + 4, we obtain

9x'2* 4y'' :35 : yI 4 From Eq. (2), wesee that the substitutions of r' : x - L andy''l'4.2.3' -4). we In Fig. result in a translation of axes to a new origin of.(1, have a sketch of the graph of the equation in r' and y' with resPect to the r' and v' axes. Figure14.2.3

:# Figure 14.2.4

We shall now apply the translation of axes to finding the general equation of a parabola having its directrix parallel to a coordinate axis and its vertex at the point (h, k). In particular, let the directrix be parallel to the y axis. If the vertex is at point V(h, k), then the directrix and the focus is at the point F(h *p, k). x, has the equation x:h-p, Let the .r' and y' axes be such that the origin O' is at V(h, k) (see Fig.1,4.2.4). An equation of the parabolain Fig. 1'4.2.4with respectto the r' and *u' axesis y'':

4px'

THE CONIC SECTIONS

To obtain an equation of this parabola with respect to the x and y axes,we replacex' by (x - h) andy' by (y - k) from Eq,.(2),whichgives us (Y-k)':4P(x-h) The axis of this parabola is parallel to the I axis. Similarly, if the directrix of a parabola is parallel to the r axis and the vertex is at V(h, k), then its focus is at F(h, k + p) and the directrix has the equation A : k - p, and an equation of the parabola with respect to the x and y axes is ( x - 7 1 2 : 4 P ( Y- k ) The axis of this parabola is parallel to the y axis. We have proved, then, the following theorem. 1.4.2.2Theorem

If p is the directed distance from the vertex to the focus, an equation of the parabola with its vertex at (h, k) and with its axis parallel to the x axis is

'i{.p'.H. ; Aik..ru}*.

fr)

(3)

A parabola with the same vertex and with its axis parallel to the y axis has for an equation (r - h)' : 4P{V- k)

EXAMPLE 3: Find an equation of the parcbola having as its directrix the line U : 1,and as its focus the point F(- 3, 7), F(- 3,7)

v(- t, +)l

(4)

soLUrIoN: Since the directrix is parallel to the x axis, the axis will be parallel to the y axis, and the equation will have the form (4). since the vertex v is halfway between the directrix and the focus, v has coordinates (-3, 4).The directed distance from the vertex to the focus is p, and so P:7-4:3 Therefore, an equation is (r * 3)'z:12(y - 4) Squaring and simplifying, we have * * 6 x - ' l - 2 y* 5 7 : 0

Figure14.2.5

ExAMPrn 4: Given the parabola having the equation

y'+6x*8y*1:0

A sketch of the graph of this parabolais shown in Fig. 1.4.2.5.

sol.urroN:

Rewrite the given equation as

y ' + 8 y- - 6 x - 1 Completingthe squareof the terms involving y on the left side of this

OF AXES 14.2TRANSLATION find the vertex, the focus, an equation of the directrix, an equation of the axis, and the length of the latus rectum, and draw a sketch of the graph.

v

equation by adding L5 on both sides/ we obtain

y'+8y*'1.6:-6x+L5 (y + 4)' 6(x- B) Comparing this equation with (3), we let

Directrix

h:E

k--4 and 4P--6

or

P:-g

Therefore,the vertex is at ($, -4); an equation of the axis is y:-4;the focus is at (7,-4); an equation of the directrix is x:4; and the length of the latus rectum is 6. A sketch of the graph is shown in Fig. 14'2.6.

Figure 14.2.6

1-4.2 Exercises In Exercises1 through 8, find a new equation of the graph of the given equation after a translation of axesto the new origin as indicated. Draw the original and the new axes and a sketch of the graph. 2. x2+ y2- L}x * 4y + 13- 0; (5,-2) L . x 2+ y ' + 6 x * 4 Y : 0 ; ( - 3 , - 2 )

3.y2-6x*9:0; (8,01 5. xz+ 4y' * 4x * 8y + 4 : 0; (-2, -l) 7 . y - 4 - 2 ( x- 1 ) t ;( 1 ,4 )

4. y' + 3x - 2y * 7 - 0; (-2, t) 6 . 2 5 x z* y ' - 5 0 r * 2 0 y - 5 0 0 : 0 ; ( 1 , - 1 0 ) 8. (Y+ l)' : 4(x - 2)'; (2' -I)

contain no In Exercises9 through 12, translate the axes so that an equation of the graph with respect to the new axeswill graph. the of and a sketch axes new the first-degree terms. Draw the original and 10. l5x2 * 25y2- 32x - l00y - 284 0 9. x2 + 4y' - l5x * 24y + 84- 0 L2. x:2- y' + L4x- 8y- 35 - 0 L1..3x2- 2y' * 6x- 8Y- LL : 0 In Exercises13 through 18, find the vertex, the focus, an equation of the axis, and an equation of the directrix of the given parabola. Draw a sketch of the graph. 1 5 .y ' + 6 x * 1 0 y + 1 9 - 0 1 , 4 .A x z- 8 r * 3 Y- 2 : 0 L 3 .x 2 * 5 x * 4 y * 8 : 0 1 8 .Y : 3 x 2 - 3 x * 3 17. 2y' 4y 3x 16.3y'-8x-l2Y-4-0 In Exercises19 through 28, find,an equation of the parabola having the given properties. Draw a sketch of the graph. 19. Vertexat (2,4); focusat (-3,4). 20. Vertex at (1, -3); directrix, Y: 1. 21. Focusat (-7,7); directtu, Y:3. 22. Focusat (-*,4);

directrix, x:-*.

588

THE CONIC SECTIONS

23. Vertex at (3 , -2); axis, r : 3; length of the latus rectum is 5. 24. Axis parallel to the r axis; through the points (I,Z), (S,Z), and (11,4). 25. Vertex at (-4,2); acis,y :2; through the point (0, 6). 25. Directrix, x: -2; a
4; length of the latus rectum is 8.

: 27. Directrix , x : 4) aocis, .y 4; through the point (9, Z) . 28. Endpoints of the latus rectum are (1, 3) and (7,3). 29. Given the parabola having the equation V : af I bx -f c, with a * 0, frnd.the coordinates of the vertex. 30. Find the coordinates of the focus of the parabola in Exercise29. 31. Find an equation of every parabola containing the points A(-9, -4) and B (5, -4), such that points A and,B are each 5 units from the focus. 32. Given the equation 4xs- l2xz * l2x - 3y - l0: 0, translate the axes so that the equation of the graph with respect to the new axeswill confail no second-degreeterm and no constant term. Draw a sketch of ttre grap-hand the two sets of axes.(rrrNr:Let 2e:x'Ih andy:y,+ k in the given equation.) 33' Given the equation. xs + 3* - y'-+ 3x * 4y - 3 : 0, translate the axes so that the equation of the graph with respect to the new axes will contain_no first-degree term and no constant term. Draw a sketch of the graph irrd tfru two sets of axes (seehint for Exercise32). 34' lf a parabola has its focus at the origin and the x axis is its axis, prove that it must have an equation of the form !2:4kx+4P,k+0.

1.4.3SOME PROPERTIES In Sec' 14'1we stated that a conicsection(or conic)is a curve of intersection OF CONICS of a plane with a right-circular cone of two nappes, and three types of curves of intersection that occur are the paraboli, the ellipse, and,Tie hyperbola. The Greek mathematician Apollonius studied conic sectiohs, in terms of geometry, by using this concept. we studied the parabola in sec. 14.1,where an analytic definition (ri.l.r) of a parabolawas given. In this section, an analytic definition of a conic r""iior, is given and the three types of curves are obtained as special casesof this d-efinition. generator In a consideration of the geometry of conic sections,a cone is regarded as having two nappes, extending indefinitery far in both directions. A portion of a right-circular cone of two nappes is shown in Fig. r4.g.r. upPer naPPe A generator (or element) of the cone is a line lying in the cone, and all -the the generators of a cone contain the point v, calied aertexof the cone. In Fig. 1'4.3.2we have a cone and a cutting plane which is parallel to one-and only one generator of the cone. Thii conic is a parabola.rf the cutting plane is parallel to two generators, it intersects both nappes of the cone and we have a hyperbola(Fig. 1a.3.3).An ellipseis obtained if lower nappe the cutting plane is parallel to no generator, in which case the cutting plane intersects each generator, as in Fig. 1,4.g.4. generator , A special caseof the ellipse is a cirire, which is formed if the cutting plane, which intersects each generator, is also perpendicular to the axis F i g ur e 14 . 3 . 1 of the cone. Degenerate cases of the conic seitions include a point, a

SF C O N I C S 1 4 , 3S O M E P R O P E R T I EO

F i g ur e 1 4 . 3 . 2

F i g ur e 14 . 3 . 3

F i g u r e1 4 . 3 . 4

straight line, and two intersecting straight lines. A point is obtained if Ure cutting plane contains the vertex of the cone but does not contain a generator. This is a degenerate ellipse. If the cutting plane contains the vertex of the cone and ot ly orr" generator, then a straight line is obtained, and this is a degenerateparabola. A degeneratehlryerbola is formed when the cutting plane contains the vertex of the cone and two generators, thereby giving two intersecting straight lines' There are many applications of conic sections to both pure and applied mathematics. We shall mention a few of them. The orbits of planets and satellites are ellipses. Ellipses are used in making machine gears' Arches of bridges areiometimes ellipticat or parabolic in shape' The path of a projectite Is a parabola if motion is considered to be in a plane and air resisiance is ne-glected.Parabolas are used in the design of parabolic mirrors, searchlighis, and automobile headlights. Hyperbolas are used in combat in "sound tanging" to locate the position of enemy guns by the sound of the firing of ihoi" g,trrr. If a quantity varies inversely as anpt"ssote and volume in Boyle's law for a perfect other quantity, such "s the graph is a hyperbola' temperature, gas at a constant To discuss conics analytically as plane curves, we first state a defini' tion which gives a property common to all conics' L4.3.1 Definition

A conic is the set of all points P in a plane such that the undirected distance of P from a fixed point is in a corrstantratio to the undirected distance of P from a fixed line which does not contain the fixed point. The constant ratio in the above definition is called t}re eccentricityof the conic and is denoted by e. The eccentricity e is a nonnegative number becauseit is the ratio of two undirected distances.Actually for nondegenIf erate conics, e ) 0. (Later we see that when e:0, we have a point') is the conic 14.1.1 that and Definitions'1,4.3.1' e:L,we Seeby comparing

THE CONIC SECTIONS

a parabola. If e < 1, the conic is an ellipse, and if e t L, the conic is a hyperbola. Tlre fixed point, referred to in Definition 14.3.1,is called a focus of the conic, and the fixed line is called the corresponding directrix. We leamed in sec. 14.1that a parabolahas one focus and one directrix. Later we see that an ellipse and a hyperbola each have two foci and two direcfrices, with each focus corresponding to a particular directrix. The line through a focus of a conic perpendicular to its directrix is called the principal axis of. the conic. The points of intersection of the conic and its principal axis are called the oerticesof the conic. From our study of the parabola, we know that a parabola has one vertex. However, both the ellipse and the hyperbola have two vertices. This is proved in the following theorem. 14.3.2 Theorem

If e is the eccentricity of a nondegenerate conic, then if e # 'J.,the conic has two vertices; 7f e: 1, the conic has only one vertex. PRooF: Let F denote the focus of the conic and D denote the point of intersection of the directrix and the principal axis. Let d denote the undirected distance between the focus and iti directrix. In Fig. 14.3.5,F is to the right of the directrix, and in Fig.1.4.3.6,F is to the left of-the directrix. Let v denote a vertex of the conic. we wish to show that if e # L, there are two possible vertices V, and if e: L, there is only one vertex V. From Definition 14.3.1,we have

F i g u r e1 4 . 3 . 5

l-wl:elfil Removing the absolute-value bars, we obtain

Fn:*e(N)

(1)

Because D, F, and v are all on the principal axis,

If F is to the right of D, DF : d. Thus, we have from the above

Tv- d+N

(2)

Substituting from (1) into (2) gives Figure14.3.6

frl- d+ s(fr/) from which we get ,l

Dv:fta

(3)

rf e:1, the minus sign in the above equations must be rejectedbecause we would be dividing by zero. so rf e * !, friro points v are obtained; if e : L, we obtain only one point V.

14.3SOMEPROPERTIES OFCONICS 591 If F is to the left of D, DF: -d. Instead of (3), we get _A

ov:1fi from which the same conclusion follows. So the theorem is proved.

I

The point on the principal axis of an ellipse or a hyperbola that lies halfway between the two vertices is called the centerof the conic. Hence, the ellipse and the hyperbola are called central conicsin contrast to the parabola that has no center becauseit has only one vertex. 14.3.3 Theorem

A conic is symmehic with respect to its principal axis. The proof of this theorem is left as an exercise (see Exercise 7).

EXAMPLE1: Use Definition "1.4.3.1 to find an equation of the conic whose eccentricity is # and havin g a focus at the point F(1 , -2) with the line 4y * 17 :0 as the corresponding directrix.

soLUTroN: Figure 14.3.7shows the focus, the directrix, and the point P (x, y) representing any point on the conic. Letting the point D be the foot of the pe{pendicular from P to the directrix, we have from Defini'1,4.3.'1, tion

lgt:4

lPDl 5

and so

:+l Tol lPTl Using the distance formula to find the above equatiolt, we get

and lfrl

and substituting into

_ +yl + + l P(*,y)

Squaring on both sides of this equation and simplifying, we get

x 2 - 2 x * 1 . +y ' + 4 y * 4 - i # 0 ' + + y

+i#)

25x2- 50x * 25yz* I00y + L25- L6yza B6y + 289 25x2t 9y'- 50r - 35y:164

Figure14.3.7

Becausee - + < L, the equation is that of an ellipse.

1,4.3 Exercises In Exercises1 through 5, use Definition 14.3.1and the formula of Exercise 24 of Exercises5.3 to find an equation of the conic having the given properties. I . F o c u sa t ( 2 , 0 ) ; 2. Focusat (0,

e: i.

e:t.

592

THECONICSECTIONS

3. Focusat (-3, 2); directrix:x:

li e:3.

4. Focusat (1, 3); directrix:!:8i e: *. 5. FocusaI (4, -3); directrix: 2x - y - 2:0; 6. Focusat (-1, 4); directrix:2x - y i 3:0;

e : i. e:2.

7. Prove Theorem 14.3.3. 8. Find an equation whose graph consists of all points P in a plane such that the undirected distance of P from the point (ae' 0) is in a constant ratio e to the undirected distance of P from the line x : al e. Let a2(l - er) :.r b2 and .orr-ridu. the three cases:e : l, e < 7, and,e > 7. 9. Solve Exercise8 if the point is (-ae , 0\ and the line is r : -ale.

14.4 POLAR EQUATIONS OF THE CONICS

D x

lTPl: 'lRPl

F{

u o

I-r

'o

F i g u r e1 4 . 4 . 1

lOFl

Because P is to the right of the directrix, - r because > r 0. So from (1) we have r - e(RP)

(1)

RP > 0; thus, lRPl: RF.

(2) However, RP - m: DO + OQ d * r cos 0. Substituting this expression for RT in e), we get r-e(d*rcosg)

(3) In a similar manner, we can derive an equation of a conic if the direc-

trix coffesponding to the focus at the pole is to the right of the focus, and (4) The derivation of ( ) is left as an exercise (see Exercise L). If a focus of a conic is at the pole and the polar axis is parallel to the

14.4 POLAR EQUATIONSOF THE CONICS

corresponding directrix, then the tn axis and its extension are along the principal axis. We obtain the following equation: (5)

where e and d are, respectively, the eccentricity and the undirected distance between the focus and the coresponding directrix. The plus sign is taken when the directrix corresponding to the focus at the pole is above the focus, and the minus sign is taken when it is below the focus. The derivations of Eqs. (5) are left as exercises(see Exercises2 and 3). We now discuss the graph of Eq. (3) for each of the three cases:e: 1, ( e 1, and e > 1. Similar discussions apply to the graphs of Eqs. (a) and (5)' Case1: e: 1. The conic is a parabola. Equation (3) becomes

to axis

(6)

1 , - c o s0 P(r,0)

Figure 14.4.2

EXAMPLEL:

r is not defined when 0: 0; however, r is defined for all other values of 0 for which 0 < 0 < 2zr. Differentiating in (6), we obtain -dsin0 . . - __ "e, (l _ cos 0)2 polaraxis Setting Der:0, for values of 0 in the interval (0,2n) we obtain 0-n' BecauJewhen 0 < 0 < r,Det < 0, and when zr < e <2n'Dsr ) 0' r has an absolute minimum value at 0: n. when 0: n, r: Ld, and the point distance from the focus 1td, rr) is the vertex. Therefore, the undirected point on the parabola is the when is shortest parabola io a point on the the vertex. By Theorem 14.3.9, the parabola is symmetric with respect to its principal axis; by Theorem 14.3.2there is one vertex. Note that the curve does not contain the pole because there is no value of d which will give a value of 0 for r. A sketch of the parabola is shown in Fig. 1.4.4.2.

A parabola has its focus at the pole and its vertex at (4,n). Find an equation of the parabola and an equation of the directrix. Draw a sketch of the parabola and the directrix.

soLUrIoN: Becausethe focus is at the pole and the vertex is at (4, t), the polar axis and its extension are along the principal axis of the parabola. Furfhermore, the vertex is to the left of the focus, and so the directrix is also to the left of the focus. Hence, an equation of the parabola is of the form of Eq. (3). Becausethe vertex is at (4, n), Ld:4; thus, d:8' The eccentricity e:1, anrdtherefore we obtain the equation r-1-*,0 - -d, and becaused: 8, An equation of the directrix is given by r cos 0

594

THE CONIC SECTIONS

we have r cos 0- -8. Figure 14.4.3 shows a sketch of the parabola and the directrix. |n

axis

a,tr)

polar axis fl_ rocus

F i g u r e1 4 . 4 . 3

EXAMPLE2: Follow the instructions of Example 1 if the parabola has its focus at the pole and its vertex at (3, *o).

polar axis

solurroN: The vertex of the parabola is below the focus, and the lrr axis and its extension are along the principal axis. Therefore, the directrix is also below the focus, and an equation of the parabola is of the form of Eq. (5) with the minus sign. The undirected distance from the focus to the vertex is id:3; thus d:6. Therefore,from Eqs. (5) an equation of the parabola is r' -

6 L-sind

An equation of the directrix is given by r sin 0:-d. Becaused:6,we have r sin d: -6. sketches of the parabola and the directrix are shown in Fig. 1,4.4.4. Figure 14.4.4

Case2: e When e < L, the denominator of the fractiom in Eq. (3) is never zero, and so r exists for all values of 0. To find the absolute extrema of r on the interval [0,2n), w€ find Der and obtain D6r:

-ezd sin 0 (1 - ecos0)2

(7)

For 0 in the interval [0, 2n), D er - 0 when 0 - 0 and zr. When -+ 7r < 0<0,Dsr) 0; and when 0 an absolute maximum value of edl (1,- q. When trr < 0 < n, D6r <--0; and when n

14,4 POLAREQUATIONSOF THE CONICS L

zT ax's

P(r,o

7\

v1

\

of.edl(l* e) when 0: r. The points (edl(l- e), 0) and (edl(1'+ e), n) ate the vertices of the ellipse. We denote them by % and V2, respectively. It follows, then, that the undirected distance from the focus (0,0) to a point on the ellipse is greatest when the point on the ellipse is V1 and smallest when it is Vz. By Theorem L4.3.3,the ellipse is symmetric with respect to its principal axis. There is no value of 0 that will give a value of 0 for r, and so the curve does not contain the pole. A sketch of the ellipse is shown in Fi9.1,4.4.5.

Figure 14.4.5

3: Find an equation of EXAMPLE the ellipse having a focus at the pole and vertices at (5, 0) and (2, n). Write an equation of the directrix corresPondingto the focus at the pole. Draw a sketch of the ellipse.

solurroN: Becausethe vertices are at (5, 0) and Q,tr), the polar axis and its extension are along the principal axis of the ellipse. The directrix corresponding to the focus at the pole is to the left of the focus becausethe ,r"ti"* closest to this focus is at (2, rr). The required equation is therefore of the form of Eq. (3).The vertex vr is at (5,0) and v2is at Q, n).Tt follows from the discussion of Case 2 that and

Solving these two equations simultaneously, we obtain s: $ and d: #' Hence, from Eq. (3) an equation of the ellipse is Xf l.{

U o

\

tzo 1

\7, ;

n.29 -' --1 ---J-----9.*cosO

I2 zraxis

l'{

t

v1

tVz

20 7,

(5, 0)

Ir

F i g u r e1 4 . 4 . 6

EXAMPLE

An equation of a

conic is

r-# Find the eccentricity, identify the conic, write an equation of the directrix corresPonding to the focus at the pole, find the vertices/ and draw a sketch of the culve.

or, equivalently, .20 ' 7-3cos0 f:-

Becaused: T, an equation of the directrix corresponding to the focus - -20. A sketch of at the pole is r cos 0 -+ or, equivalently, 3r cos 0 the ellipse is shown in Fig. L4.4-6.

sor,urroN: Dividing the numeratorand denominatorof the fraction in the given equationby 3, we obtain '-L+?sin0

THE CONIC SECTIONS

are therefore at (1, in) and (5, *o). A sketch of the ellipse is shown in Fig. 14.4.7. I

o axis

directrix

polar axis

t (r, 3i ")l v2

F i g ur e 1 4 . 4 . 7

'1,. The conic is a hyperbola. Case3' e > In the derivation of Eq. (3) we assumed that the point p on the conic is to the right of the given directrix and that p is on the terminal side of the angle of 0 radians, thus making r positive. In the previous two cases, when the conic is an ellipse or a parabora,we obtain tlie entire curve with these assumptions. However, whene ) 1, we shall seethat we obtain only one of two branches of the hyperbola when r ) 0, and this branch is to the right of the directrix. There is also a branch to the left of the given directrix; this is obtained by letting / assume negative as well as positive values as d takes on values in the interval l0,2rrl, lf e> 1 in Eq. (3), r is undefined when cos 0:1.1e. There are two values of 0 in [0, 2r) that satisfy this equation. These varues are 0r: cos-l -1 and p

0z: 2n - cos-l I e

we investigatethe values of r as d increasesfrom 0 to 91.when 0:0, edl(7 - e), and becausee ) l, r < 0. So the point (edl(1.- e), 0) is the vp{ex the hyperbola. The cosine function is decreasing in the infeft 9f terval (0,0r); therefore,when 0 < 0 < 0r, cos d ) cos 0r:Uior, equiv_ alently, 1 - e cos 0 < 0. Hence, for vJues ot 0 in the intervat (0', 0r), r: edl(7 - e cos 0) < 0. When 0 : er, r is undefined; however, r:

P(r,0)

polar axis

\ Figure 14.4.8

lim r: 0_0r

1i^ --3! 0_0rl-eCOS0

^:-*

we conclude, then, that as 0 increasesfrom 0 to 01,r < 0 and lrl increases

.I4.4POLAR EQUATIONS OFTHECONICS 597 without bound. From these values of 0 we obtain the lower half of the left branch of the curye. See Fig. 1,4.4.8.We show later that a hyperbola has two asymptotes. For this hyperbola one of the asymptotes is parallel to the line 0: 01. Now we consider the points on the hyperbola as g increasesthrough valuesbetween 01and 02.When 0, I 0 ( dr)cosg ( cosfi: lleor, equivalently, L - e cos 0 > 0. Therefore, for values of g in the intenral (0r,0r), r)0. lim r:

o-of

Hm3

o-oiL-ecosa

^:-l*

When 0: r, r: edl(L* e); so the point (edl(L+ e), r) is the right vertex of the hyperbola. Finding Dsr from Eq. (3), we get Der: -

ezd srn 0 (1 - e cos 0)2

(8)

When 0, I 0 ( z, sin 0 > 0; hence, Der 10, from which we conclude that r is decreasing when 0 is in the interval (0v r). Therefore, for values of 0 in the interval (0u rr) we obtain the top half of the right branch of the hyperbola shown in Fig.1.4.4.8. When n < 0 ( d2,sin 0 < 0; so from Eq. (8), D6r ) 0. Hence, r is increasing when O'is in the interval (n, 0r).Furthermore, limr: 0-02-

lim=

0-02- L-

ed

;:**

e COS A

Consequently, for values of 0 in the interval (rr, 0r) we have the lower half of the right branch of the hyperbola. The other asymptote of the hyryerbola is parallel to the line 0: 02. The cosine function is increasing in the interval (02, 2n\. So when 0, 1 0 I 2n, cos 0 ) cos 02: 7le or, equivalently, 1 - e cos 0 < 0. Thus, we see from Eq. (3) that for values of 0 in the interval (0r, 2rr), r < 0. Also, €d lim r: lim = =:-* e cos0 ;:;;L ;:;;' It follows from Eq. (8) that Dsr ) 0; hence, r is increasing when 0, < 0 < 2zr. When r < 0 and r is increasing, lrl is decreasing.Thus, for values of 0 in the interval (02,2r) we obtain the upper half of the left branch of the hyperbola, and so the curve of Fig.14.4.8is complete. EXAMPLE5: The polar axis and its extension are along the principal axis of a hyperbola havin I a focus at the pole. The corresponding directrix is to the left of the focus. If the hyperbola contains the point 1L, ?n) and

solurroN: An equation of the hyperbolais of the form of Eq. (3) with e:2.We have,then, '

t:-

2d L - 2 c o s0

Because the point (1, &rr) lies on the hyperbola, its coordinates satisfy the equation. Therefore,

THE CONIC SECTIONS

e : 2, find (a) an equation of the hlperbola; (b) the vertices; (c) the center; and (d) an equation of the directrix corresponding to the focus at the pole. Draw a sketch of the hyperbola.

2d

1- 2(-+)

from which we obtain d:1..

r-

Hence, arrrequation of the hyperbola is

2

1-2cos0

(e)

The vertices are the points on the hyperbola for which 0 :0 and Q'- z'. From Eq. (9) we see that when e - 0, r - -Z;and when 0 - 7r,r : 3. Consequently, the left vertex Vr is at the point (-2, 0), and the right vertex V2 is at the point (?, n).

*ou X l.{ I o F.

'o

(- 2,o)

1-

polar axis

Q,*4 Figure 14.4.9

ExAMPLE 6: Find a polar equation of the hyperboli having tn" line r cos e - 4 as the directrix corresponding to a focus at the pole, and e: 8.

The center of the hyperbola is the point on the principal axis halfway between the two vertices. This is thi point (t, zr). An equation of the directrix corresponding to the focus at the pole . is given by r cos 0: -d. Becaused: l,ihis equation is r cos 0: _1. As an aid in drawing a sketch of the hyperbola, we first draw the two asymptotes. In our discussion of case 3 we stated that these are lines through the center of the hyperbola that are parallel to the lines g: 0r and d : 02,where g, and 02 arethe values of C in the interval [0, 2n) foi which r is not defined. From Eq. (9), r is not defined when 1 - 2 cos 0: 0. Therefore, 0r:$r and 02:$zr. Figure 14.4.9shows a sketch of the hyperbola as well th_" two asymptotes, and the directrix corresponding "r pole. to the focus at the

solurroN. The given directrix is pe{pendicular to the polar axis and is four units to the right of the focus at the pole. Therefore, an equation of the hyperbola is of the form of Eq. (A),which is r' :

ed

1 , * e c o s0

Because d - 4 and e - g, tne equation becomes y' :

E'4

1.+9cos0

or/ equivalently, t' :

12

2 + 3 cos 0

Exercises 14.4 1' show that an equation oJ a conic having its principal axis along the polar axis and its extension, a focus at the pole, and the corresponding directrix to the right o? *r" fo*, is r: e1l o i e cos e). 2' show that an equation of.a conic having its principal.axis along the lr axis and its extension, a focus at the pole, and the coresponding directrix above the fJcus ii ,:iAl(t+ e sin?).

14,5 CARTESIANEQUATIONSOF THE CONICS

3. Show that an equation of a conic having its principal axis along the +zr axis and its extension, a focus at the pole, and the corresponding directrix below the focus is r : edl(l - e sin d) . 4. Show that the equation r:

k cs€ *0, where k is a constant, is a polar equation of'a parabola.

In Exercises5 through 14, the equation is that of a conic having a focus at the pole. In each Exercise, (a) find the eccentricity; (b) identify the conic; (c) write an equation of the directrix whidr corresponds to the focus at the pole; (d) draw a sketch of the curve. v"

-:-

a -v"

11.,:

L4. f' : -

2 t-cos0 4 "1,- 3 cos 0 L 5-5sin0

7 3+4cos0

In Exercises15 through 18, find an equation of the conic having a focus at the pole and satisfying the given conditions. L5. Parabola;vertex at (4,*zr). 15. Ellipse; e: * a vertex at (4' t). 1 7 . Hyryerbola;vertices at (1, tn) and (3 ' trn) . is the directrix corresponding to the focus at the pole. g) and above the parabola r : 3l (7 * sin 0) ' 19. Find the area of the region inside the ellipse r : 6l (2- sin the parabola' 20. For the ellipse and parabola of Exercise 19, find the area of the region inside the ellipse and below

18. Hyperbola; e:*r

cos 0:9

is 80,000,000miles 21. A comet is moving in a parabolic orbit around the sun at the focus of the parabola.Wh-enthe comet of the orbit. (a) Find axis with the from the sun the line segrnent from the sun to the comet makes an angle oi *z radians sun? to the come an equation of the comit's orbit. (b) How close does the comet the pole is at the sun' 22. T\e orbit of a planet is in the form of an ellipse having the equation r: pl(l t e cos-0) where to 0' respect with from the sun Find the a.rerage measune of the distance of the planet pole, the principal axis 23. Using Definition 14.3.1 find a polar equation of a central conic for which the center is at the a/e. is a directrix pole to the from distance and the its extension, is aloing the polar axis and - cos 0) are 24. Show that the tangent lines at the points of intersection of the parabolas r: al(l * cos 0) and r: b/(1 perpendicular.

14.5 CARTESIAN EQUATIONS In Sec.14.4we learned that a polar equation of a conic is OF THE CONICS ed

r:7-ffi7

or, equivalently, r:

e(r cos0 * d)

(1)

A cartesian representation of Eq. (1) can be obtained by replacing r *{7TT and r cos 0 by x. Making these substitutions' we get by

*t/NT:

e(x* d)

600

THECONICSECTIONS Squaring on both sides of the above equation gives x2 + y' : e2x2* 2e2dx* e2d2 or, equivalently, y'+ x'(L-

e 2 )- Z e z d x * e z d z

(2)

If e - L, Eg. (2) becomes

y' :2d(x + +d)

(3)

If we translate the origin to the point Fta,O) by replacirg xby I - +d and y by y,Eq,. (3) becomes

Y' :2dr Equation(4)resembles Eq. (1)of Sec.1,4.1 Y' : 4Px

(4)

1,4.2. '1., If. e * we divide on both sides of Eq. (2) by 1- ez and,obtain z ztexd+17" ]eyd.z: 7 _ r^-. R

r,

Compleling the square for the terms involving r by adding ead2l(l _ e), on both sides of the above equation, we get ( , _-

\r

e d \ r - -r 1 " . ":_ e z i l z t - e) 1--@ Y- O=W

(s)

If the origin is translatedto the point (ezdlG - er),0), Eq. (5)becomes

P 1]-, . t _ e z r i':=4(l_4yz or/ equivalently,

x2-+-L-l

wTafit T1?F:--

Now let ezd'lG -

r

o-1

lft ed

tFI

(6) e2)': a2, whete a > 0. Then

iro
(7\

14.5 CARTESIANEQUATIONSOF THE CONICS

Then Eq. (5) can be written as *

lt2

_--:l

oz, sz(l_d) Replacing x arrd y by r and y, we get

f . --=--'----v' - 1 - -r, oz sz(l_ g)

(8)

Equation (8) is a standard form of a cartesian equation of a central conic having its principal axis on the x axis. Because(5) is an equation of a conic having a focus at the origin (pole), and (8) is obtained from (5) it follows that the by translating the origin to the point(*dl(l-ez),O), point at the central conic having Eq. (8) has a focus eezdl(l-e2),0). However, from (7) it follows that -ezd [-ae ifOL l-rz:l Therefore, we conclude that if (8) is an equation of an ellipse (0 < e < 1), there is a focus at the point (-ae,O). If (8) is an equation of a hyperbola (e > 7), there is a focus at (ae,O). In each case the coresponding directrix is d units to the left of the focus. So if the graPh of (8) is an ellipse, the directrix corresponding to the focus at (-ae,0) has as an equation x:-ae-

d d - a.(I -

Because when 0 x:-ae-a(L-

ez)le, this equation becomes

e2) e

a e

Q r -

Similarly, if the graph of (8) is a hyperbola, the directrix corresponding to the focus at (ae,0) has as an equation x: ae- d When e ) l, d: a(E - L) le; tttus, ttte above equation of the directrix can be written as a e

Hence, we have shown that if (8) is an equation of an ellipse, a focus and and if (8) is an its corresponding directrix are (-ae,0) and x:-ale; equation of a hyperbola, a focus and its coresponding directrix ate (ae,O) and x: a/e. BecauseEq. (8) contains only even Powers of r and y, its graph is symmetric with respect to both the x and y axes. Therefore, if there is syma focus at (-ae,0) having a conesponding directrixof x:-ale,by

THE CONIC SECTIONS

metry there is also a focus at (ae,0) having a conesponding directrix of x: ale. Similarly, for a focus at (ae,0) and a coresponding directrix of x : al e, thete is also a focus at (- ae, 0) and a coffesponding directrix of x:-a/e.These results Ere summarized in the following theorem. 14.5.1 Theorem

The central conic having as an equation x2

1f

a2'

(e)

-r

J

- I

- 4

a2(I-er)

where a ) 0, has a focus at (-ae,0), whose coresponding directrix is x:-a/e, and a focus at (ae,0), whose correspondingdirectrix is x: afe. Figures 14.5.1and 14.5.2show sketchesof the graph of Eq. (9) together with the foci and directrices in the respective casesof an ellipse and a hyperbola. v *v e: -

a

a *:,Q rre

Bb

A

(

)

A

ae, a Xr: L{ii .|.4i: (Jii q.ti t+'li

/

'ui

B, - b

F i g ur e 1 4 . 5 . 1

x

E

(J (U,r tsr

-('

Figure14.5.2

Suppose that in Eq. (9) 0 < e causein this caseL - e2) 0, we conclude that \R Therefore,we let b-a\R

is a real number.

( 10)

Because a ; 0, it follows that b > 0. Substituting from (10) into (9), we obtain (11)

Equation (11) is an equation of an ellipse having its principal axis on the r axis. Becausethe vertices are the points of interiection of the ellipse with its principal axis, they are at (-a,0) and (a, 0). The centerof the ellipse is at the origin because it is the point midway between the vertices. Figure 14.5.1shows a sketch of this ellipse. Refer to this figure as you read the next two paragraphs. If we denote the vertic es (- a, 0) and (a, 0) by A' and A, respectively,


the segmentA' A of the principal axis is called the maior axis of the ellipse, and its length is 2a units. Then we state that a is the number of units in the length of the semimajor axis of the ellipse. The ellipse having Eq. (11) intersectsthe y axis at the points (-b,0) and (b,0), which we denote by B' and B, respectively. The segment B'B is called the minor axis of. the ellipse. Its length is 2b units. Hence, b is the number of units in the length of the semiminor axis of the ellipse. Becausea and b are positive numbers and 0 < e 1L, it follows from Eq. (10)thatb
Given the ellipse having the equation -x:2 -

25

1t2 r' L : 1

"1,6 *

find the vertices, foci, directrices, eccentricity, and extremities of the minor axis. Draw a sketch of the ellipse and show the foci and directrices.

SoLUTIoN: From the equation of the ellipse,we have az:25 andf :16i thus, a : 5 and b : 4. Therefore, the vertices are at the points A'(-5, 0) and A(5, 0), and the extremities of the minor axis are at the points B'(0,-4) from which it follows that and B(0, 4). From (L0) we have 4 :S{T=V, r: $. The foci and the directrices are obtained by applying Theorem 1.4.5.1.Becauseae:3 and ale:ff, we conclude that one focus F is at (3, 0) and its corresponding directrix has the equation x: T; the other focus F' is at (-3, 0) and its corresponding directrix has the equation x: -4. A sketch of the ellipse, the foci F and F' , and the directrices are in Fig. 14.5.3.From the definition of a conic (1'4.3.1)it follows that if P is any point on this ellipse, the ratio of the undirected distance of P from a focus to the undirected distance of P from the corresponding directrix is equal to the eccentricity 9.

F i g u r e1 4 . 5 . 3

ExAMPLE2:

An arch is in the form of a semiellipse. It is 48 ft wide at the base and has a height of 20 ft. How wide is the arch at a height of L0 ft above the base?

soLUrIoN: Figure 14.5.4 shows a sketch of the arch and the coordinate axes which are chosen so that the r axis is along the base and the origin is at the midpoint of the base. Then the ellipse has its principal axis on the r axis, its center at the origin, a: 24, and b :20. An equation of the ellipse is of the form of Eq. (11):

* * a' -1 s76 ' 4 0 0

v

(t, L0)i

Let 27 be the number of feet in the width of the arch at a height of 10 ft above the base. Therefore, the point (t, 10) lies on the ellipse. Thus, *,100 576' 400

-+

-24|+-Zxft+24 *asft# Figure14.5.4

604

THE CONIC SECTIONS

and so

#:432 i:7216 Hence, at a height of L0 ft above the base the width of the arch is 24\/j tt. Consider now the central conic of Theorem 14.s.1.when e _e2 is, when the conic is a hyperbola. Because e e - L > 0. Therefore, for a hyperbola we let

fo-a\F

(r2) is a positive number. Also, from (I2), if b: a; and rf e

( 13) _Equation (13) is an equation of a hyperbola having its principal axis on the x axis,its verticesat the points (-a,0) and(a,0;, ind iti centerat the origin. A sketch of this hyperbola is in Fig. r4.s.2.If we denote the vertices (-4, 0) and (a, 0) by A' and A, the segment A'A is called the transaerseaxis of the hyperbola. The length of the transverse axis is 2a units, and so a is the number of units in the length of the semitransverseaxis. Substituting 0 for r in Eq. (13),we obtain y, : -b", which has no real roots. consequenfly, the hyperbola does not intersect the y axis. However, the line segment having extremities at the points (0,-b) and (0, b) is called the conjugnteaxis of the hyperbola, and its length is 2b units. Hence, b is the number of units in the length of the semiconjugate axis. Solving Eq. (13) fot y in terms of r, we obtain

v:4\m

(r4)

we conclude from (14) that if lxl < a, there is no real value of y. Therefore, there are no points (x, y) on the hyperbola for which -a < x < a. We-also see from (14)that if lxl = a, then y has two real values. As previously learned, the hyperbola has two branches. one branch contains the vertex A(a, 0) and extends indefinitely far to the right of A. The other branch contains the vertex A'(-o,0) and extends indefinitelv far to the left of A'. EXAMPLE

Given the hyperbola

x 2_ y ' 1 g 1,6

SOLUTION:

The given equation is of the form of Eq. (I3); thus, a - 3 and

b : 4 . T h e vertices are therefore at the points A' ?9, 0) and A(3, 0). The number of units in the length of the transverse axis is 2a : 6, and the number of units in the length of the conjugate axis is 2b : 8. Because


find the vertices,foci, directrices, eccentricity, and lengths of the transverseand conjugate axes. Draw a sketch of the hyperbola and show the foci and the directrices.

from (12), b: a171, we have !': l{/i, and so e: t. Therefore, Ae:S and ale:$. Hence, the foci are at F'(-5,0) and F(5,0). The corresponding directrices have, respectively, the equations r: -9 and 1: g. A sketch of the hyperbola and the foci and directrices are in Fig. 14.5.5. From Definition 14.3.1it follows that if P is any point on this hyperbola, the ratio of the undirected distance of P from a focus to the undirected distance of P from the corresponding directrix is equal to the eccentricity*.

F i g u r e1 4 . 5 . 5

EXAMPLE4: Find an equation of the hyperbola having the ends of its conjugate axis at (0,-2) and (0, 2) and an eccentricity equal

to *\6.

solurroN: Becausethe ends of the conjugate axis are at (0, -2) and (0,2), it follows that b : 2, the transverse axis is along the r axis, and the center is at the origin. Hence, an equation is of the form x2

v' - 1

a2

b2

To find a we use the equation b - a\FT. and letb-2 and e-g\8, thereby yielding 2 - a\F. Therefore, a - \8, and an equation of the hyperbola is x 2_ a ' - , , 5

4-r

L4.5 Exercises In Exercises1 through 4, hnd the vertices, foci, directrices, eccentricif, and.ends of the minor axis of the given ellipse. Draw a sketch of the curve and show the foci and the directnces. 2.4x2*9y'-4 4. 3x2* 4y' :9 L. 4xz* 9y' - 36 3. 2x2* 3y' - Lg In Eiercises 5 through 9, find the vertices, foci, directrices, eccentricity, and lengths of the transverseand conjugateaxes of the given hlryerbola. Draw a sketch of the curve and show the foci and the directrices.

5. 9x2- 4y' - 36

6. x2- 9y' :9

7. 9x2- L6y'- 1

8. 25x2- 25yz- 1

THE CONIC SECTIONS

In Exercises9 through 14,find an equation of the given conic satisfying the given conditions and draw a sketch of the graph. 9. Hyperbola having verticesat (-2,0) and (2,0) and a conjugateaxis of length 5. 10. Ellipse having foci at (-5,0)

and (5,0) and one directrix the line x:-20.

11. Ellipse having verticesat (-9,0) and (*,0) and one focus (8,0). 12. Hyperbola having the ends of its conjulate axis at (0, -3) and (0, 3) and e:2. 13. Hyperbola having its center at the origin, its foci on the x axis, and passing through the points (4, -2) and (7, -6). 14. Ellipse having its center at the origin, its foci on the r axis, the length of the major axis equal to three times the length of the minor axis, and passing through the point (3,3). 15. Find an equation of the tangent line to the ellipse 4f -f 9y2:72 at the point (3, 2). 16. Find an equation of the normal line to the hyperbol a 4x2- 3y2: 24 at the point (3, 2). 17. Find an equation of the hyperbolawhose foci are thevertices of the ellipse 7f *17y2:77 the foci of this ellipse.

and whose vertices are

18. Find an equation of the ellipse whose foci are the vertices of the hyperbola l'l,f - 7y, : 77 and whose vertices are. the foci of this hyperbola. 19. The ceiling in a hallway 20 ft wide is in the shape of a semiellipse and is 18 ft high in the center and 12 fthigh at the side walls. Find the height of the ceiling 4 ft from either wall. 20. A football is 12 in. long, and a plane section containing a seam is an ellipse, of which the length of the minor axis is 7 in. Find the volume of the football if the leather is so stiff that every cross section is a square. 21. Solve Exercise20 if every cross section is a circle. 22. The orbit of the earth around the sun is elliptical in shape with the sun at one focus, a semimajor axis of length 92.9 million miles, and an eccentricity of 0.017.Find (a) how close the earth gets to the sun and @f the greatestp'ossible distance between the earth and the sun. 23. The cost of production.of a commodity is $12 less per unit at a point A than it is at a point B and the distance between A and B is 100 miles. Assuming that the route of delivery of the commodity is alonf a straight line, and that the deIivery cost is 20 cents per unit per mile, find the curve at any point of which the commodity can be supplied from either A or B at the same total cost. (nrur: Take points A and B at (-50, 0) and (50, 0), respeciively.) 24. Prove that there is no tangent line to the hyperbol a * - yr: 1 that passesthrough the origin. 25. Find the volume of the-solidof revolution generatedby revolving the region bounded by the hyperbola xzlaz- yzlbz: 1.and the line r: 2a abott the r axis. 25. Find the centroid of the solid of revolution of Exercise25. 27' A rar& has a horizontal axis of length 20 ft and its ends are semiellipses. The width acrossthe top of the tank is 10 ft and the depth is 5 ft. If the tank is full of water, how much work is necessaryto pump all the water to th" top of the tank?

14.6 THE ELLIPSE

In Sec. 14.3 we considered ellipses for which the center is at the origin and the principal axis is on the x axis. Now more general equationJ of ellipses will be discussed. If an ellipse has its center at the origin and its principal axis on the y ucis, then an equation of the ellipse is of the form

14.6THE ELLIPSE

(1)

which is obtained from Eq. (11) of Sec. 14.5 by interchanging x and y. . rLLUsrRArroNL: Becausefor an ellipse a ) b, it follows that the ellipse having the equation

* *t-:t

1,6 25

has its foci on the y axis. This ellipse has the same shape as the ellipse of Example f. in Sec.14.5.The vertices are at (0, -5) and (0, 5), and the foci are at (0, -3) and (0, 3), and their corresponding directrices have the . equations y : -+ and y :4f, respectively. If the center of an ellipse is at the point (h, k) rather than at the origin, and if the principal axis is parallel to one of the coordinate axes, then by a translation of axes so that the point (h, k) is the new origin, an equation -*la2 * y'lb': 1 if the principal axis is horizontal, and of the ellipse is and y'la"+-*1bz:1 if the principal axis is vertical. Becausei:x-h y: y - k, these equations becomethe following in x and y: (2) if the

(3) if the principal axis is vertical. In Sec. 1.6 we discussed the general equation of the second degree in two variables: Axz * Bxy * Cy' * Dx * Ey + F- 0

(4)

where B:0 and A: C. In such a case, the graph of (4) is either a circle a degenerate case of a circle, which is either a point-circle or no real or locus. If we eliminate the fractions and combine terms in Eqs. (2) and (3), we obtain an equation of the form

Axz*Cy'*Dx*Ey+F-0

(5)

where A+ C 1f.a*b and AC>0.It can be shown bycompleting the squares in r and y that an equation of the form (5) can be put in the form

( x- h ) ' 11

(v- k)'

A

C

(6)

CONIC SECTIONS

If. AC > 0, then A and C have the same sign. If G has the same sign as A and C, then Eq. (5) can be written in the form of (2) or (3). Thus, the graph of (5) is an ellipse. 2: Suppose we have the equation o ILLUSTRATToT.T 6**9y2-24x-54y*51:0 which can be written as 6(f-4x)+9(y2-6y):-5L Completing the squares in r and y, we get 6(f -4x* 4) +9(y'-6y t 9):-51 +24+81 or, equivalently, 6(x-2)2+9(V-3)2:54 which can be put in the form (x-2)'

--_-r

(y -3\' , --]-:

J?

This is lr, un,r",roi of the form of Eq. (6).By dividing on both sidesby 54 we have ( x - 2 ) ' _ ( y---z-: -3)'_.,

9 -

'

which has the form of Eq. (2).

o

If in Eq. (5) G has a sign opposite to that of A and C, then (5) is not satisfied by any real values of r and y. Hence, the graph of (5) is the empty set. . rLLUsrRArroN3: Supposethat Eq. (5) is 6**9y2-24x-54y*1L5:0 Then, upon completing the squares in r and y, we get 6(x-2)z a9(! -3)':-115 + 24+8L which can be written as (x-2\' ------i-

(y -3)' T, -=___i_:

_10

(7)

T h i s i s l r , n " f o r - t E q , . ( 6 ) , w h e rGe : - 1 0 , A : 6 , a n d C : 9 . F o r a l l values of r and y the left side of Eq. (7) is nonnegative; hence, the graph of (7) is the empty set. . If G:0

in (6), then the equation is satisfiedby only the point (h, k).

14.6 THE ELLIPSE

609

Therefore, the graph of (5) is a single point, which we call a point-ellipse. o rLLUsrRATroN4: Becausethe equation 6f*9y2-24x-54Y*105:0 can be written as

(*

,2)'*(y *+

_ . 9 )_' o '

its graph is the point (2, 3).

If the graph of Eq. (5) is a point-ellipse or the emPty set, the graph is said to be degenerate. lf. A: C in (5), we have either a circle or a degeneratecircle, as mentioned above. A circle is a limiting form of an ellipse. This can be shown by considering the formula relating a, b, and.e fot an ellipse: bz: az('[,_d) From this formula, it is seent$at as e approacheszero, b2approaches 42. lf b2: az,Eqs. (2) and (3) become (x- h)'+ (y - k)': az which is an equation of a circle having its center at (h, k) and radius a. We see that the results of Sec.1.6 for a circle are the sameas those obtained for Eq. (5) applied to an elliPse. The results of the above discussion are summarized in the following theorem. 14.5.1Theorem

If in the generalsecond-degreeEq. (4) n:0 and AC > 0, then the graph is either an ellipse, a point-ellipse, or the empty set. In addition, if A: C, then the graph is either a circle, a point-circle, or the empty set'

Determine the graph of the equati on 25x2* I6Y2 * 150r - L28y - LIL9: Q.

soLUrIoN: From Theorem'1,4.5.7, becauseB:0 and ng:(25)(t0): 4O0> 0, the graph is either an ellipse or is degenerate. Completing the squares in r and y, we have 25(x2* 6x + 9) + l6(y', - 8y * 16) - 1119+ 225+ 256

EXAMPLE1:

2 5 ( x+ 3 ) ' + l 6 ( y - 4 ) ' : (x*3)2 -(y-+)':1 100 54

L500 *

(8)

Equation (8) is of the form of Eq. (3), and so the graPh is an ellipse having its principal axis parallel to the y axis and its center at (-3' 4).

610

T H E C O N I CS E C T I O N S

ExAMPLE2: For the ellipse of Example 1.,find the vertices, foci, directrices, eccentricity, and extremities of the minor axis. Draw a sketch of the ellipse and show the foci and the directrices.

solurroN: From Eq. (8) it follows that a:'1.0 and b : 8. Becausethe center of the ellipse is at (-3, 4) and the principal axis is vertical, the vertices are at the points V'(-3, -5) and V(-3,l4). The extremitiesof the minor axis are at the points B'(_1t'1.,4)and B(5,4). Becauseb: a{1V, we have

8:101T=7 and, solving for e,weget e: $. Consequently,ae:6 and ale:ry. Therefore, the foci are at the points F'(-3, -2) and F(-3, L0). The corresponding directrices have, respectively, the equations y : -+ and y: 92. A sketch of the ellipse, the foci, and the directricesare in Fig. 14.6.1.

v V:

directrix

3

(- z,t+) v

\

(-3, L0) F

( - 1 1 , 4)

B',

?3,4)

\t | | r I I r

B (s,4) tl

r, r,J

F,

-/ .J

(-3, -6) V',

y:

- T38

directrix

F i g u r e1 4 . 6 . 1

EXAMPLE3: Find an equation of the ellipse for which the foci are at (- 8, 2) and (4,2) and the eccentricity is 3. Draw a sketch of the ellipse.

soLUrIoN: The center of the ellipse which is halfway between the foci is the point (-2, 2). The distance between the foci of any ellipse is 2ae, and the distance between (-8,2) and (4, 2) is L2. Therefore, we have the equation Zae: 12. Replacing e by *, we get 2a(|) : 12, and so a:9. Because b : a{177, we have

b:9lT=@:9lT=4:3rE The principal axis is parallel to the x axis; hence, an equation of the ellipse is of the form of Eq. (2).Because(h, k): (-2,2), a:9, and b : 3V5, the

-

14,6THE ELLIPSE

611

required equation is

( x! 2 ) ' 8L

*

(y _z)'_ 45

1

A sketchof this ellipse is shown in Fig. "1.4.6.2.

(- 2,2+ 3\/s)

\ ,2)

( - 1 1 , 2)

lrrrrltl \

- ___L!.?)____ 7, 2)

lttlll/

o ./

(-2,2 - 3\/-s) F i g ur e 1 4

We conclude this section with a theorem that gives an altemative definition of an ellipse. The theorem is based on a characteristic property of the ellipse. 14.6.2 Theorem

d'

/(-+,vp

r

-/h o .,,/

,p,

An ellipse can be defined as the set of points such that the sum of the distances from any point of the set to two given points (the foci) is a constant. pRooF: The proof consists of two parts. In the first part we show that the set of points defined in the theorem is an ellipse. In the second part we prove that any ellipse is such a set of points. Refer to Fig 14.6.3 in both parts of the proof. Let the point Pt(rt, Ar) be any point in the given set. Let the foci be the points F'(-ae,0) and F(ae,0), and let the constant sum of the distances be 24. Then

lrEl + 1fi1:2o Using the distance formula, we get

Y - - A ' L e

F i g u r e1 4 . 6 . 3

{G=7*Tt7

* {Q,*i{tT y}:26 \/@=1d"TT : 2a- \/Cn ad"W

Squaring on both sides of the above equation, we obtain x( - 2Ae4 * azE * yl : 4a2- Aa\/(it + aAY + y7 * xr' * 2aex1+ a2e2* yr'

7 612

THE CONIC SECTIONS

or, equivalently,

\/GfidrTt!:

a+ exl

Squaring on both sides again, we get xr2+ 2aeh I azez* yr' : a2+ 2nex,* dxr2 ('l'- E)x12* Yt': a2(1'- e2) x,2

-:

71,2

-J- ---------!-:-:1

a2(l - 8) Becauseb2: az(l - e), and replacing 11and yrby x and y, respectively, we obtain a2

f

1t2

7+ fr:1 which is the required form of an equation of an ellipse. Now consider the ellipse in Fig. 1.4.6.3;Pr(xr,yr) is any point on the ellipse. Through P, draw a line that is parallel to the r axis and that intersects the directrices d and d' at the points Q and R, respectively. From Definition 14.3.1it follows that

lF-ql: elR4l and lFEl: 'ltr'al Therefore,

lrEl + lFEl:'(lR4l+ lFrel)

(e)

RPt and PrQ are both positive because an ellipse lies between its directrices. Hence,

:2- (-3):'+ Ro

(10)

Substituting from (10)into (9),we get

l F E l+ l E l : 2 a which proves that an ellipse is a set of points as described in the theorem. I

o rLLUsrRArrow 5: The ellipse of Example L has foci at (-3, -2) and (-3, 10), and 2a: 20. Therefore, by Theorem '1.4.6.2 this ellipse can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (-3, -2) and (-3, 10) is equal to 20. Similarly, the ellipse of Example3 can be defined as the set of points such that the sum of the distancesfrom any point of the set to the points (-8,2) and (4,2) is equal to 18.

14.7THE HYPERBOLA 013

Exercises L4.6 In Exercises1 through 5, find the eccentricity, center, foci, and directrices of each of the given ellipses and draw a sketch of the graph. 3.5x2* 3y'-3y-L2:0 2. 9x2* 4y' - l8x * l6y - 1L : 0 1. 5x2+ 9y'- 24x- 54y+ 51 - 0 5. 4x2* 4y' * 20x 32y + 89 0 6. 3x2* 4y'- 30r * L6y + 100- 0 4. Zxz* 2y' - 2x * L8y + 33 0 In Exercises7 through 12, find. an equation of the ellipse satisfying the given conditions and draw a sketch of the graph. 7. Foci at (5,0) and (-5,0);

one directrix is the line x:-20.

8. Vertices at (0, 5) and (0, -5)

and passing through the point (2,-tr6l

9. Center at (4, -2) , a vertex at (9 , -2), 10. Foci at (2,3) and (2,-7) LL. Foci at (-L,-l)

.

and one focus at (0 , -2).

and eccentricity of &.

and (-1 ,7) and the semimajor axis of length 8 units.

L2. Directrices the lin es y - 3 + +&s and a focus at (0 , -2) .

13. The following graphical method for sketching the graph of an ellipse is based on Theorem 14.6.2.Todraw a sketch of the graph of the ellipse 4* * y2: 15, first locate the points of intersection with the axes and then locate the foci on the y axis by use of compassesset with center at one point of intersection with the r axis and with radius of 4. Then fasten thumbtacks at each focus and tie one end of a string at one thumbtack and the other end of the string at the second thumbtack in such a way that the length of the string between the tacks is2a:8. Place a pencil against the string, drawing it taut, and describe a curve with the point of the pencil by moving it against the taut string. When the curve is completed, it will necessarilybe an ellipse, becausethe pencil point describes a locus of points whose sum of distances from. the two tacks is a constant. 14. Use Theorem 14.6.2to find an equation of the ellipse for which the sum of the distancesfrom any point on the ellipse to (4, -1) and (4, 7) is equal to 12. 15. Solve Exercise 14 if the sum of the distances from (-4, -5) and (5, -5) is equal to 16. 15. A plate is in the shape of the region bounded by the ellipse having a semimajor axis of length 3 ft and a semiminor axis of length 2 ft. If the plate is lowered vertically in a tank of water until the minor axis lies in the surface of the water, find the force due to liquid pressure on one side of the submerged portion of the plate. 77. lI fhe plate of Exercise 15 is lowered until the center is 3 ft below the surface of the water, find the force due to liquid pressure on one side of the plate. The minor axis is still horizontal.

74.7 TII'E HYPERBOLA

In Sec.14.5 we leamed that an equation of a hyperbola having its center at the origin and its principal axis on the r axis is of the fotm *laz - y'lb': 1..Interchanging x and y in this equation, we obtain

(1) which is an equation of a hyperbola having its center at the origin and its principal axis on the y axis.

614

THE CONIC SECTIONS

o ILLUsrnnrroN L: The hyperbola with an equation

v' 9

x2 - 1 4 1,6

has its foci on the y axis because the equation is of the form (1).

o

Note that there is no general inequality involving a andb corresponding to the inequality a > b f.or art ellipse. That is, for a hyperbola, it is possible to have a 1b, as in Illustration 1.,where a:3 and b:4; or it is possible to have a ) b, as for the hyperbola having the equation y'125-*lt6:1, where a:5 and b:4. If.,for a hyperbola,a:b, then the hyperbola is said to be equilateral We stated in Sec.1.4.4that a hyperbola has asymptotes, and we now show how to obtain equations of these asymptotes. In Sec.A.2hofizontal and vertical asymptotes of the graph of a function were defined. What follows is a more general definition, of which the definitions in Sec.4.2 are special cases. 14.7.1Definition Thegraphof theequation y:f(x) hastheliney:mx*basan tote if either of the following statements is true:

asymp_

(i) lim

lf(x)-(mx+b)l:0 (ii) lim l f ( x ) ( m x + b ) l : 0 t-*@

t--@

Statement (i) indicates that for any € > 0 there exists a number N > 0 such that

lf(x)

( m x+ b ) l < e

whenever r > N

that is, we can make the function value /(r) as close to the value of mx * b as we please by taking r large enough. This is consistent with our intuitive notion of an asymptote of a graph. A similar statement may be made for part (ii) of Definitron 1,4.7.1.. For the hyperbola f la' - y'lV : 1.,upon solving for y, we get

\m So if

f(x)

b a

\tr4

we have lim ff-*@

r i mP \ f f i - l r l

r-*6

LA

A

- ffi-i(ffi*r) :_- b , l:r_ m O \m* 6r r-+a

J

x

14.7THEHYPERBOLA615 b ,.

-oP

;"11ffi+x :Q

Therefore, by Definition'1.4.7.1, we conclude that the line A: bxla is an asymptote of the graph of. y:bt@:Fla. Similarly, it can U:_glovvn that the line y:bxla is an asymptote of the graph of y:-bV*azla. Consequently, the line y:bx/a is an asymptote of the hyperbola f laz -y'lV:1.. [n an analogousmanner, we can demonstrate that the line y:-bxla is an asymptote of this same hyperbola. We have, then, the following theorem. 14.7.2Theorem The lines b Y: o*

and

b y:--x,

are asymptotes of the hyperbola

f _u'-t a2W

Figure 14.7.1shows a sketch of the hyperbola of Theorem'1.4.7.2together with its asymptotes. In the figure note that the diagonals of the rectangle having vertices at (a, b), (a, -b) , (- a, b), and (- a, -b) are on the asymptotes of the hyperbola. This rectangle is called the auxiliary rectangle of the hyperbola. The vertices of the hyperbola are the points of intersection of the principal axis and the auxiliary rectangle.A fairly good sketch of a hyperbola can be made by first drawing the auxiliary rectangle and then drawing the branch of the hyperbola through each vertex tangent to the side of the auxiliary rectangle there and approaching asymptotically the lines on which the diagonals of the rectangle lie. There is a mnemonic device for obtaining equations of the asymptotes of a hyperbola. For example, for the hyperbola having the equation 1, if the right side is replaced by zeto, we obtain flaz *la'-y'lbz- y2lb':0. Upon factoring,this equationbecomes(xla Vlb) (xla * ylb) : 0, which is equivalentto the two equationsx/a ylb:0 and xla * ylb:0, which, by Theorem 74.7.2,are equations of the asymptotes of the given hyperbola. Using this device for the hyperbola having Eq. (1), we see and that the asymptotes are the lines having equations yla- xlb:0 yla* xlb:0, which are the same lines as the asymptotesof the hyperbola with the equation *lbz - f lo': 1. These fwo hyperbolas are called conjugatehyperbolas. The asymptotes of an equilateral hyperbola (a: b) are perpendicular to each other. The auxiliary rectangle for such a hyperbola is a square, and the transverse and conjugate axes have equal lengths.

616

THE CONIC SECTIONS

If the center of a hyperbola is at (h, k) and its principal axis is parallel to the x axis, then if the axes are translated so that the point (h, k) is the new origin, an equation of the hyperbola relative to this new coordinate system is *u2

7-F:

t

If we replace 7 by x - h and ! by V - k, this equation becomes (2) Similarly, an equation of a hyperbola having its center at (h, k) and its principal axis parallel to the y axis is (3)

EXAMPLE1: The vertices of a hyperbola are at (-5 , -3) and (-5 , -I), and the eccentricity is \8. Find an equation of the hyperbola and equations of the asymptotes. Draw a sketch of the hyperbola and the asymptotes.

solurroN: The distance between the vertices is 2a, and so ,, : L. For a hyperbola, b: alV - 7, and therefore b: t{5 - 7: 2. Becausethe principal axis is parallel to the y axis, an equation of the hyperbola is of the form (3). The center (h, k) is halfway between the vertices and is therefore at the point (-5,-2). We have, then, as an equation of the hyperbola (y+2)'_(r*5)2_,, 1, 4-:r Using the mnemonic device to obtain equations of the asymptotes, we have lu*2

x*5\ly*2*r*5\:n

\=----u-/t

1

2 t

which gives y*2:*(r+5) Figure 14,7.2

and y *2:-*(x+5)

A sketch of the hyperbola and the asymptotes are in Fig. 1,4.7.2. If in Eqs. (2) and (3) we eliminate fractions and combine terms, the resulting equations are of the form A* + Cyz* Dx * Ey -l F:0 (4) where A and C have different signs; that is, AC < 0. We now wish to show that the graph of an equation of the form (4), where AC < 0, is either a hyperbola or it degenerates.Completing the squares in x and y in Eq. (4), where AC < 0, the resulting equation has the form a2(x-h)2-

B'(y-k)':H

(s)

14.7 THE HYPERBOLA 617

lf H ) 0, Eq. (5) can be written as (x-h)'

(V-k)'_.

T_T:I q:2

B'

which has the form of Eq. (2). o rr,r,usTnarron2: The equation 4xz- l2yz I 24x * 95y - 18L- 0 can be written as 4(* * 6x) - lz(y' - 8y) : L8L and upon completing the squaresin r and Ar wE have , 4(x' * 5x * 9) - tZ(y' - 8y + 16) : 181+ 36 - 192 which is equivalent to 4(x+3)2-12(y-4)2:25 This has the form of Eq. (5), where H: (x*3)'

25 > 0. It may be written as

(y-4)'_.,

TT:I which has the form of Eq. (2). If in Eq. (5), H < 0, then (5) may be written as (x-h\' U-k)' -lE[---lE[--r q2 which

.1

B',

has the form of Eq. (3).

o rLLUsrRArron3: Suppose.thatEq. (4) is 4*-12f*24xI96y-131:0 Upon completing the squares in r and y,we get 4(x*3)2-tZ(Y-4)2:-25 This has the form of Eq. (5), where H:-25 (y-4)'_(x*3)2 2n at

26 -T

( 0; and it may be written as

_,

which has the form of Eq. (3). It H = Q in Eq. (5), then (5) is equivalent to the two equations a(x - h) - BQ - k) : 0 and u(x- h) + P(y - k) : 0

618

THE CONIC SECTIONS

which are equations of two shaight lines through the point (h, k). This is the degeneratecaseof the hyperbola. The following theorem summarizes the results of the above discussion. 14.7.3 Theorem

If in the general second-degreeequation Af + Bxy t Cyz* Dxt Ey * F:0 B : 0 and AC < 0, then the graph is either a hyperbola or two intersecting straight lines.

ExAMPLE2: Determine the graph of the equation 9x2- 4y'- 18r - l6y + 29- 0

soLUTroN: From Theorem 14.7.3, because B:0 and AC: (9) (_4) _ -36 ( 0, the graph is either a hyperbola or two intersecting straight lines. Completing the squares in r and A, we get

9 ( x ' - 2 x + 1 ) - + ( y ' * 4 y+ 4 ) - - 2 9 + 9 - 1 , 5

(6)

Equation (6) has the form of Eq. (3), and so the graph is a hyperbola whose principal axis is parallel to the y axis and whose center is at (1, -2). ExAMPLE3: For the hyperbola of Example 2, find the eccentricity, the vertices, the foci, and the directrices. Draw a sketch of the hlryerbola and show the foci and the directrices.

v(L,1)

//

-/ /a v :

-,2 *, o

nt/t3

x,

t..! F' d

I I \ I V'(L, - 5 ) I I I I

-2 -

O

-

nVts

Figure14.7.3

solurrorT: From Eq. (5) we see that a : 3 and b:2. b: a{e2 - 1; thus,

For a hyperbola,

2:31F1 and solving for e, we obtain , : 4l/I3. Becausethe center is at (1, -2), the principal axis is vertical, and a:3, it follows that the vertices are at the points V'(1,-p andV(1,1). Becauseae: {8, the foci are at the points F' (1, -2 - Vi3) and F(1, -Z + tfr). Furthermore, a/e: #ffi; hun.", the directrix corresponding to the focus at F' has as an equationy:-2 - rtv13, and the directrix corresponding to the focus at F has as an equation y : -2 * r+Vi3. Figure 14.7.3 shows a sketch of the hyperbola and the foci and directrices.

14.7 THE HYPERBOLA 6 1 9

Just as Theorem 1.4.6.2 gives an altemate definition of an ellipse, the following theorem gives an alternate definition of a hyperbola. 1'4.7.4Theorem

A hyperbola can be defined as the set of points such that the absolute value of the difference of the distances from any point of the set to two given points (ihe foci) is a constant. The proof of this theorem is similar to the proof of Theorem 14.5.2 and is left as an exercise (see Exercises13 and 14). In the proof, take the foci at the points F'(-ae,0) and F(ae,O),and let the constant absolute value of the difference of the distances be 2a. See Fig. 14,7.4.

Pt(x1, y)

o rlr.usrnarron 4: Because the hyperbola of Example 2 has foci at (1,-2- Vtgl and (1, -2+ \ffi) and2a:6,itfollowsfromTheoremll7.4 that this hyperbola can be defined as the set of points such that the absolute value of the difference of the distancesfrom any point of the set to the points (1,-2 - vEZTand ('1,,-2 + rffil is 6. The graph of the general second-degreeequation

F'(- ae, o)

F(ae, O)

Af*Bxy*Cy'*Dx*Ey+F-0

(7)

'1.4.6. In this section we where B : 0 and AC > 0, was discussed in Sec. have so far considered the equation when B : 0 and AC < 0. Now if for Eq. (7), B : 0 and AC:0, then either A: 0 or C: 0 (we do not consider A: B: C:0 becausethen Eq. (7) would not be a quadratic equation). Supposethat in Eq. (7), B:0, A:0, and C # 0. The equation becomes Figure 14.7.4

Cy'*Dx*Ey+F-0

(8)

If D + 0, (8) is an equation of a parabola. lf D : 0, then the graph of (8) may be two parallel lines, one line, or the empty set. These are the degenerate cases of the parabola.

o rllusrRArrow 5: The graph of the equhtion 4y' - 9: 0 is two parallel lines; 9y2* 6y * 1 : 0 is an equation of one line; and 2y' * y * 1 : 0 is . satisfied bv no real values of u. R ,irrritu, discussionf,ofi if B :0,C summarized in the following theorem. 14.7.5Theorem

: 0, and A # 0. The resultsare

If in the generalsecond-degreeEq. (T),B:0 and either A:0 and C # 0 or C: 0 and A + 0, then the graph is one of the following: a parabola, two parallel lines, one line, or the empty set. From Theorems 1.4.6.'1., 14.7.3, and 1.4.7.5,it may be concluded that the graph of the general quadratic equation in two unknowns when B:0 is either a conic or a degenerate conic. The type of conic can be deterrrined from the product of A and C. We have the following theorem.

620

THECONICSECTIONS 14.7.6 Theorem

The graph of the equation Af * Cy2* Dx t Ey * F: 0, where A and C are not both zero, is either a conic or a degenerateconic; if it is a conic, then the graph is (i) a parabolaif either A: 0 or C: 0, that is, iL AC: 0; (ii) an ellipseif A and C have the same sign, that is, if. AC > 0; (iii) a hyperbolaif A and C have opposite signs, that is, if AC < 0. A discussion of the graph of the general qu4dratic equation, where B + 0, is given in Sec.14.8.

Exercises 1-4.7 In Exercises1 through 6, tind the eccentricity, center, foci, directrices, and equations of the asyrnptotesof the given hyperbolas and draw a sketch of the graph. 1. 9xz- l8y' * 54x- 36y + 79- 0 2. xz - y' + 6x * l}y - 4- 0 3. 3y'- 4x2- 8x - 24y - 40 - 0 4 . 4 x 2- y ' + 5 6 x * 2 y + 1 9 5- 0

5 . 4 y ' - 9 x 2 * I 5 y * 1 8 r: 2 9

6. y 2 - x 2 + 2 y - 2 x - L : 0

In Exercises7 through 12, find an equation of the hyperbola satisfying the given conditions and draw a sketch of the gruph. 7. One focus af (26, 0) and asymptotes the lines l2y : -+5r. 8 . Center at (3, -5) , a vertex at (7, -5) , and a focus at (8, -5). 9. Center at (-2, -l), a focus at (-2, 14), and a directrix the line 5y : -SS. 10. Foci at (3,5) and (3,0) and passingthrough the point (5,3 + 6l\/5). 1 L . One focus at (-3 - Zt/B,1), asymptotesintersectingat (-3, 1), and one asymptotepassingthrough the point (1,7). 12. Foci at (-1,1)

and (7,4) and eccentricityof 3.

13. Prove that the set of points, such that the absolute value of the difference of the distancesfrom any point of the set to two given points (the foci) is a constant, is a hyperbola. 14. Prove that any hyperbola is a set of points such that the absolute value of the difference of the distancesfrom any point of the set to two given points (the foci) is a constant. 15. Use Theorem 14.7.4Io find an equation of the hyperbola for which the difference of the distances from any point on the hyperbolato (2,1) and (2,9) is equal to 4. 16. Solve Exercise15 if the difference of the distances from (-8,-4)

and.(2,-4)

is equal to 5.

17, Three listening posts are locatedat the points A(0, 0), B(0, +) , and C(T, 0), the unit being 1 mile. Microphones located at these points show that a gun is * mi closer to A than to C, and I mi closer to B than to A. Locatethe position of the gun by use of Theorem 14.7.4. 18. Prove that the eccentricity of an equilateral hyperbola is equal to \n.

14.8 ROTATION OF AXES

We have previously shown how a translation of coordinate axescan simplify the form of certain equations. A translation of axes gives a new coordinate system whose axes are parallel to the original axes. We now

14.8 ROTATIONOF AXES

F i g u r e1 4 . 8 . 1

621

consider a rotation of coordinate axes which enables us to transform a second-degreeequation having an ry term into one having no such term. Suppose that we have two rectangular cartesian coordinate systems with the same origin. Let one system be the ry system and the other the ry system. Suppose further that the i axis makes an angle of radian measure c with the r axis. Then of course the y axis makes an angle of radian measure a with the 3i axis. In such a case,we state that the xy system of coordinates is rotateil through an angle of radian measure a to form the iy system of coordinates. A point P having coordinates (r, y) with respect to the original coordinate system will have coordinates (x' y) wlth respect to the new one. We now obtain relationships between these two sets of coordinates. To do this, we introduce two polar coordinate systems, each system having the pole at the origin. In the first polar coordinate system the positive side of the r axis is taken as the polar axis; in the second polar coordinate system the positive side of the f axis is taken as the polar axis (see Fig. 1a.8.1).Point P has two sets of polar coordinates, (r, 0) and (7, 0), wherc 7:r

(1)

and e-0-a

The following equations hold: x-7cos0

(2)

and y-lsin0

Substituting from (1) into (2), we get x - r cos(e- q and y - r sin(O- q.)

(3)

Using the trigonometric identities for the sine and cosine of the difference of two numbers, Eqs. (3) become 2: r cosOcos a * r sin 0 sin a and ! : r sin0 cos a- / cos 0 sin a Becauser cos0 : x and r sin 0 : y, we obtain from the above two equations x-rcosa+ysLna

(4)

y--xsina+ycosa

(s)

and Solving Eqs. (a) and (5) simultaneously for x and y in terms of i and y, we obtain (5) '

(7)

It is left as an exercise to fill in the steps in going from (4) and (5) to (6) and (7) (see Exercise L).

T H E C O N I CS E C T I O N S

ExAMPLE7: Given the equation xy -- L, find an equation of the graph with respect to the x and y axes after a rotation of axes through an angle of radian measure *r.

sol,urroN: Taking a: *r in Eqs. (6) and (7), we obtain

* : $ x1-_f i Y

1_

a n a A : n q1 x * f i . 1Y_

Substituting these expressionsfor r and y in the equation rU : l, we get

(#-+,)(#c+#,) -1

or/ equivalently,

This is an equation of an equilateral hyperbola whose asymptotes are the bisectors of the quadrants in the ry system. Hence, we conclude that the graph of the equation ry:1is an equilateral hyperbola lying in the first and third quadrants whose asymptotes are the x anil y axes (see Fig. 1a.8.2).

Figure

'1,4.7 In Sec. we showed that when B: 0 and A and C are not both zero, the graph of the general second-degreeequation in two unknowns,

Af*Bxy*Cy'*Dx*Ey+F-0

(8)

is either a conic or a degenerate conic. We now show that if B * 0, then any equation of the form (8) can be transformed by a suitable rotation of axes into an equation of the form

At, + ey,+ Dr+ Ev+ F- o

(e)

where A and C are not both zero.

If the ry system is rotated through an angle of radian measule a, then to obtain an equation of the graph of (8) with respect to the xy system, we replace x by x cos c - ! sin a *d y by r sin c * ! cos a. We get

Af+Bxy+Cy'+Dx+Ey+F-0

(10)

where A-

A cosza + B sin a cos a + C stnz a

B--ZA

sinacos a+ B(cosa z-

C-Astnza-B

sinza)+2Csinacosa

(11)

s i n C I c o sa + C c o s z a

We wish to find an a so that the rotation transforms Eq. (8) into an

14.8ROTATION OF AXES equation of the form (9). SettingB from (11) equal to zero, we have B(cosz a - sinz a) + (C - A)(2 sin d cos a) : 0 or, equivalently, with trigonometric identities, B cos2a * (C - A) sin 2a:0 Because B * 0, this gives

#"Eil$#$F...'$js+++++li $

(r2)

We have shown, then, that an equation of the form (8), where B + 0, can be transformed to an equation of the form (9) by a rotation of axes through an angle of radian measure a satisfying (12). We wish to show that A and C in (9) are not both zero. To Prove this, notice that Eq. (1'0) is obtained from (8) by rotating the axes through the angle of radian measure c. AIso, Eq. (8) can be obtained from (10) by rotating the axes back through the angle of radian measure (-a). If A and C in (10) are both zero, then the substitutions r:r

c o sa * y s i n a a n d ! : - x s i n

o * y c o sa

in (10) would result in the equation D1r cosa* y sin Q + E(xsin a * y cose) * F: 0 which is an equation of the first degree and hence different from (8) because we have assumed that at least B # O. The following theorem has, therefore, beBn proved. 14.8.1 Theorem

If B + 0, the equation Af * Bxy_* Cyz_* Dx_* Ey_* F: 0 can b_etrans; formed into the equation A* + Cyz + Di + E, + F :0, where A and C are not both zero, by a rotation of axes through an angle of radian measure d for which cot2a: (e_ C)IB. By Theorems L4.8.1and'14.7.6,it follows that the graph of an equation of the form (8) is either a conic or a degenerateconic. To determine which type of conic is the graph of a particular equation, we use the fact that A, B, and C of Eq. (8) and A, B, and C of Eq. (10) satisfy the relation Bz - 4AC:

82 - 4AC

(13)

which can be proved by substituting the expressionsfor A, B, and C given in Eqs. (11) in the right side of (13). This is left as an exercise(see Exercise 15).

The expression 82 - 4AC is called the discriminantof Eq..(8). Equation (L3) states that the discriminant of the general quadratic equation in two

F-

I

THECONICSECTIONS

variables is inaariant under a rotation of axes. If the angle of rotation is chosen so that B:0, then (13)becomes Bz- MC:-4AC (14) From Theorem 14.7.6it follows that if the graph of (9) is not degenerate, then it is a parabolaif. Ae : 0, an ellipse it Ae > 0, and a hypeibola if AC < 0. So we conclude that the grapLof (9) is a parabola, an ellipse, or a hyperbola depending on whether-4AC is zero, negative, orpositive. Becausethe graph of (8) is the same as the graph of. (9), it follows from (14) that if the graph of (8) is not degenerate, then it is a parabola, an ellipse, or a hyperbola depending on whether the discriminant 82 - 4AC is zero, negative, or positive. We have proved the following theorem. 14.8.2 Theorem

The graph of the equation A * * B x y* C y ' * D x * E y * F : 0 is either a conic or a degenerateconic. If it is a conic, then it is (i) a parabolaif.82- MC:0; (ii) an ellipseif 82 - MC < 0; (iii) a hyperbolaif 82 - 4AC > 0.

EXAMPLE2: Given the equation lTxz - l2xy * 8y'- 80 : 0, simplify the equation by a rotation of axes. Draw a sketch of the graph of the equation showing both sets of axes.

solurroN: 82- MC: (-12), - 4(17) (8) : -400 < 0. Therefore,by Theorem 14.8.2,the graph is an ellipse or else it is degenerate.To eliminate the xy term by a rotation of axes,we must choose an a such that A-C .F' cotzd.: B

: L7-8 _12:-4

3

There is a 2o. in the interval (0, zr) for which cot2q.: -$. Therefore, o is in the interval (0, trr).To apply (6) and (7) it is not necessaryto find a so long as we find cos a and sin a. These functions can be found from the value of cot 2a by the trigonometric identities cos Because cot

1+ cos2a

2 and 0

and

sin ot:

So

L

\tr sln q. -

2

\E

Substitutingx : il {5 - 2V| \tr andy : 2il \fS + yI \B in the given equa-

REVIEWEXERCISES 625

tion, we obtain

/#-AIy+4Y,'\_rz 17[. s /

)cy- 2y' 5

+8(ry-80-o

Upon simplification, this equation becomes x2+4Y':t6 or, equivalently,

r' *!' - 1 ' 1,5 E So the graph is an ellipse whose maior axis is 8 units long and whose minor axis is 4 units long. A sketch of the ellipse with both sets of axes is shown in Fig. 14.8.3.

Figure 14.8.3

74.8 Exercises y terms of I and'y' 1. Derive Eqs. (6) and (7) of this section by solving Eqs. ( ) and (5) for x and in axes. Draw a sketch of the graph In Exercises 2 through 8, remove the xy term from the given equation by a rotation of and show both sets of axes. 2. x2*xY+A2:3

3.24xy-7y'+36-0

4. 4xy * 3x2: 4

5. xy -8

6. 5x2* 6xy * 5y'- 9

7. 3Lx2+ 106*Y * 2tY2:'!'M

8. 5x2* Z}l.6xy * 25y' :324 Draw a sketch of the graph In Exercises9 throu gh t4, simplify the given equation by a rotation and translation of axes. and show the three sets of axes. 10. x 2 - L } x y * y ' * x * y + 1 - 0 9. x2+xy+y2-3y-6:0 L2. 3 x 2 - 4 x y * 8 r - 1 - 0 11. 17x2- L2xy * 8y' - 68x* 24y - 12 0 14. 19xz* 6xy * t|yz - 25x + 38Y+ 31 - 0 13. Llx2- 24xy * 4y'* 30r + 40y- 45 0 - 4AC is invariant under a 15. Show that for the general second-degreeequation in two variables, the discrimin ant 82 rotation of axes.

Reaiew Exercises(ChaPter14) In Exercises L through of the graph.

6, find a cartesian equation of the conic satisfying the given conditions

1. Vertices at (1, 8) and (1,-4);

e:

?.

2. Foci.at (-5 , L) and (1, !); one vertex at (-4, t). 3. Center at the origin; foci on the x axis; e -2; containing the point (2,3).

and draw a sketch

THE CONIC SECTIONS

4. Vertexat (4,2); focusat (4,-3); e:1. 5. A focus at (-5, 3); directrix: x: -3; e: *. 5. A focus ar (4, -2); directrix: ! :2; e: *. In Exercises7 through 10, find a polar equation of the conic satisfying the given conditions and draw a sketch of the graph. 7. A focus at the pole; vertices at (2, rr) and (4, r). 8. A focus at the pole; a vertex at (6, ln); e : I. 9. A focus at the pole; a vertex at (3, En); e: l. 10. The line r sin d: 5 is the directrix corresponding to the focus at the pole and e: *. In Exercises11 through 14, the equation is that of a conic having a focus at the pole. In each Exercise,(a) find the eccentricity; (b) identify the conic,' (c) write an equation of the directrix which coresfonds to the focus at tfre pole; (d) draw a sketch of the curve.

In Exercises15 through 18, the equation is that of either an ellipse or a hyperbola. Find the eccentricit5z,center, foci, and directrices, and draw a sketch of the graph. If it is a hyperbola, also find Lquations of the asymptotes. 1.5.4f + y' + 24x- l6y + 84 - 0 17. 25x2- y2 * 50r + 5y - 9- 0

16. 3x2- 2y, + 6x - gy + 1L : 0 18. 4x2* gy' + 32x- lgy + 37 - 0

In Exercises19 and 20, simplify the given equation by a rotation and translation of axes.Draw a sketch of the graph and show the three sets of axes. 19.3f -3xy-y2-5y

-0

20. 4x2* 3xy * y, - Gx* 12y- 0

21. Find a polar equation of the parabola containing the point (2,*rr), whose focus is at the pole and whose vertex is on the extension of the polar axis.

22. lf lhe distance between the two directrices of an ellipse is three times the distance between the foci, find the eccentricity. 23. Find the volume of the solid of revolution generated if the region bounded by the hyperbola f la" - yrlbr:1. and the line r: 2a is revolved about the y axis. 24. Show that the hyperbola f - y': 4 has the same foci as the ellipse x2 + 9yz :9. 25. Find an equation of the parabola having vertex at (5, 1), axis parallel to the y axis, and through the point (9,3). 25. Show that any equationof theform xy I ax*by -f c:0canalwaysbewrittenintheform x,y,:kbyatranslationof the axes, and determine the value of k. 27. Any section of a parabolic mirror made by passing a plane through the axis of the mirror is a segment of a parabola. The altitude of the segment is L2 in. and the length of the base iJ ts in. A section of the mirror made by pi"^" p"rpendicular to its axis is a circle. Find the circumference of the circular plane section if the plane perpendicular " to the axis is 3 in. from the vertex. 28. The directrix of the parabola y2 :4pr is tangent to a circle having the focus of the parabola as its center. Find an equation of the circle and the points of intersection of the two curves. 29. A satellite is traveling around the earth in an elliptical orbit having the earth at one focus and an eccentricity of The *.

REVIEWEXERCISES

closest distance that the satellite gets to the earth is 300 mi. Find the farthest distance that the satellite gets from the earth. 30. The orbit of the planet Mercury around the sun is elliptical in shape with the sun at one focus, a semimajor axis of length 35 million miles, and an eccentricity of 0.206.Find (a) how close Mercury gets to the sun and (b) the greatest possible distance between Mercury and the sun. 3 1 . A comet is moving in a parabolic orbit around the sun at the focus F of the parabola. An observation of the comet is made when it is at point Pr, 15 million miles from the sun, and a second observation is made when it is at point Pr, 5 million miles from the sun. The line segments FP, and FP2are perpendicular. With.this information there are two possible orbits for the comet. Find how close the comet comes to the sun for each orbit.

32. The ardr of a bridge is in the shape of a semiellipse having a horizontal span of 40 ft and a height of 15 ft at its center. How high is the arch 9 ft to the right or left of the center? 33. Points A and B are 1000ft apart and it is determined from the sound of an explosion heard at these points at different times that the location of the explosion is 500 ft closer to A than to B. Show that the location of the explosion is restricted to a particular curve and find an equation of it. 3 4 . F i n d t h e a r e a o f t h e r e g i o n b o u n d e d b y t h e t w o p a r a b o l a s : r : 2 1( l - c o s 0 )

andr:21(J+

cosd).

35. A focal chord of a conic is a line segment passing through a focus and having its endpoints on the conic. Prove that if two focal chords of a parabola ar- perpJndicul,ar, the ium of the reciprocals of the measures of their lengths is a constant. (HrNr: Use polar coordinates.) 35. A focal chord of a conic is divided into two segments by the focus. Prove that the sum of the reciprocals of the measures of the lengths of the two segments is the same, regardlessof what chord is taken. (nrNr: Use polar coordinates.) 37. Prove that the midpoints of all chords parallel to a fixed chord of a parabola lie on a line which is parallel to the axis of the parabola.

Indeterminateforms, improperintegrals, and Taylorbformula

FORMO/O 629 15..1 THEINDETERMINATE 15.1 THE INDETERMINATE Limit theoremg (2.2.9)statesthat if lim /(r) and lim g(r) both exist, then FORM O/O

'

f(x) I'qrt'l

i'gm:ffi'(') provided.nu,rri. r,t ) + 0. There are various situations for which this theorem cannot be used. In particular, if lirn g(r) : 0 and lim /(r) : k, where k is a constant not equal to 0, then Limit theorem 12 (2.4.4) can be applied. Consider the and lim g(r) :0. Some limits of this type case when both lim f(x):0

have previously been considered. 1 TLLUSTRATTONL: We wish

to find

f-x-t2 llm -t---=--r-tX1-JX-+ Here, lim (x' - x - t2) :0 r-4

and lim (f - 3x - 4) :- 0. However, the nur-4

merator and denominator

can be factored, which gives

,. f-x-L2 ,um . k-4)(x*3) n\ ir lx _ 4: G_ 4Xx,+ii Because r + 4, the numerator and denominator can be divided b y ( x - 4 ) , and we obtain 1.

x2-x-12

x*3

,.

fim:lTr+1:B

7

- 0 and lim x - 0; and by Theorem "1,0.2."1,, o ILLUSrnerroN 2: lim sin x Jfr0

l-0

lim

o

fr-O

L5.L.L Definition

(x) - 0 and lim S@): 0, then II f and g are two functions such that ftg f the function f lg has the indeterminate form 0/0 a t a . I-A

15.1.1, (x' - x - l2)l(x2 - 3x - 4) ; ILLUSTRATToN3: From Definition has the indeterminate form 0/0 at 4; however, we saw in Illustration 1 that ,. x2-x-!2 ;--;xz-3x-4 rrFr-:-

7 5

Also, (sin xl x) has the indeterminate form 0/0 at 0, while lim (sin x/ x) : 1', r-o as shown in Illustration 2. o We now consider a general method for finding the limit, if it exists,

INDETERMINATE FORMS,IMPROPERINTEGRALS, AND TAYLOR'SFORMULA

of a function at a number where it has the indeterminate form 0/0. This method is attributed to the French mathematician Guillaume Frangois de L H6pital (7561-L707),,,vho wrote the first calculus textbook, published in 7695.It is known as L'H6pital'srule. 15.1.2 Theorem L'Hopital's Rule

Let / and g be functions which are differentiable on an open interval I, except possibly at the numb er a in I. Supposethat for all x # a in I, g, (x) + 0. Then if lim /(r) :0 and lim g(r) :0, and if

,. f' ( x)

lllll*:

L

"-" 8'\x) it follows that llm Ya

f(x)

-;--i-:

L

8\X)

The theorem is valid if all the limits are right-hand limits or all the limits are left-hand limits. Before we prove Theorem 15.1.2,we show the use of the theorem bv the following illustration and examples. . rLLUsrRArror 4: We use L'H6pital's rule to evaluate the limits in - x-12):0 Illustrations 1 and 2. In Illustration 1, because and l2@, lim (l 3x 4): 0, we can apply L'H6pital's rule and obtain r.

x2-x-12

2x-l

7

;:;2x-3

5

r.

riffr-:rllft-:-

;:;xz-3x-4

We can apply L'Hdpital's rule in Illustration 2 because lim sin x - 0 I-O

and lim x: 0. We have, then, Jf-O

r.

llm t-0

EXAMPLE

Find

sin r X

soLUrIoN:

r. - - I -llm .Z*0

cos .r I

Because lim )c: 0 and lim (l - e") ff-O

I-O

be applied, and we have

l i m ; x ; - l i m + : a - - 1- L ;;t-e*

;;-er

SOLUTION:

l i m ( 1 - x + l n r ) - 1 ,- L + 0 - 0 fi-l

if it exists.

L'Hopital's rule can

FORMO/O 631 15,1THEINDETERMINATE and lim (r3-3x*2):l-3*2:0 a-l

Therefore, applylng L'H6pital's rule, we have r - -x +x + l nnxr 1 r. r'il .liii l r - : r l m - -

-1 +1 x ' ili3*- 3

,r ..

x" -3 x*z Now,becaur" (-t + llx) :0 andlim (3f - 3) : 0,weapplyL'H6piItT tal's rule again giving

ft-r 3f

-

3

Therefore,we conclude r. L-x+lnr um;; x'd-3x*2

L 5

To prove Theorem 15.1.2 we need to use the theorem known as Cauchy's mean-ualuetheorem,which extends to two functions the meanvalue theorem (4.7.2)for a single function. This theorem is attributed to the French mathematician Augustin L. Cauchy (1789-1857),who was a pioneer in putting the calculus on a sound logical basis. L5.L.3 Theorem Cauchu's Mean-V alue Theorem

lf. f and I are two functions such that (i) f andgare continuous on the closedinterval la, bl; ( i i ) / a n d g a r e differentiable on the open interval (a, b); (iii) for all r in the open interval (a, b), g'@) * 0 then there exists a number z in the open interval (a, b) such that .

'(z) f(u)- f(a):f s@)-g(o)-g'(z) pRooF: We first show that g(b) + g(a). Assume g(b) : g(a). Becauseg satisfies the two conditions in the hypothesis of the mean-value theorem (4,7.2), there is some number c in (a, b) such that g'(c) : IS(b)- S@)ll(b- a). But 1f g(b) : g(a), then there is some number c in (a, b) such that g' (c) : 0. But condition (iii) of the hypothesis of this theorem statesthat for all r in (a,b), g'@) * 0. Therefore, we have a contradiction. Hence, our assumption that BQI: g(a) is false. So, g(b) + g(a), and consequent$ g(b) g(a) * 0. Now let us consider the function h defined by

: f(x)- f(a)-ffi1 h(x)

frt,l- s@)l

632


Then,

h'(x)-f'(x)-lffils,(,1

(1)

Therefore, ft is differentiable on (a, b) because/ and g are differentiable there, and ft is continuous onfn, b] becausef and g are continuous there.

- f(a) -l#=ffi1 rrr,l- 8(a)l : f(a) :o h(a) - f(a) -U%=#l urrr- g(a)l : f(b) :o h(b) Hence, the three conditions of the hypothesis of Rolle's theorem are satisfied by the function ft. So there exists a number z in the open interval (a, b) such that h'(z) : 0. Thus, from (1) we have

f '(,)-

'\?!- f (!), : o s @ )- s \ at s'Q)

(2)

Becauseg' (z) + 0 on (a, b), we have from (2)

f(b)

-

f(q) _ f' (r)

M:76

where z is somenumber in (a, b).This provesthe theorem.

r

We should note that if g is the function such that g(r) : r, then the conclusion of Cauchy's mean-value theorem becomes the conclusion of the former mean-value theorem because then g'(z) : t. So the former mean-value theorem is a special caseof Cauchy's mean-value theorem. A geometric interpretation of Cauchy's mean-value theorem is given in Sec. 77.4. It is postponed until then becauseparametric equations are needed. o rLLUsrRArroN 5: Supposef(x):3f *3x - L andg(r): f - 4x * 2. We find a number z in (0, L), predicted by Theorem 15.1.3.

f

' ( x ) - 6 x+ 3

g'(x)-3x2-4

Thus, / and g are differentiable and continuous eve4rwhere, and for all r in (0, 1), g'@) # 0. Hence,by Theorem 15.1.3,there existsa z in (0,1) such that /(1) /(0) _ 6 z * 3 - 4 s(1) g(0) 3* Substitutingf(1):5, g(1) :-1", /(0):-1, and g(0):2, and solving for z, we have 5- (-1) _62*3 -l-2 322-4

FORM O/O 15.1 THE INDETERMINATE

633

5 2 2* 5 z - 5 = 0

t2 _-5+2\B T2 - 6- 3 + \ B Only one of these numbers is in

(0, 1), namely,z - +(-3 + t@l.

o

We distinguish We are now in a position to Prove Theorem 1'5.1'.2. three cases:(i) r + a+; (i1) x-+ a-; (iii) r -+ a. P R o o Fo F T H E o R E M1 5 . 1 . 2 ( 1 ) : Because in the hypothesis it is not assumed that / and I are define d at a, we consider two new functions F and G for which F(r):f(x)

1 fx #

G(r):g(r)

tfx#a

a

a n d F ( a ): 0 and G(a) : 0

(3)

Let b be the right endpoint of the open interval I given in the hypothesis. Becausef and g are both differentiable on 1, except possibly at d, we conclude that F and G are both differentiable on the interval (4, rl, where a < x < b. Therefore, F and G are both continuous on (a, rl. The functions F and G are also both continuous from the right at a because G(a). F(a), and lim G(r) : lim 8(r) :0: lim F(r) : lim f(x):0: I

t -A+

-A+

Therefore, F and G are continuous on the closed interual la, xf. So Fand G satisfy the three conditions of the hypothesis of Cauchy's mean-value theorem (Theorem L5.L.3) on the interval la, xf. Hence, F(x) - F(a) _ F' (z) G(r) - G(a) G'(z)

(4)

where z is some number such that a < z < r. From (3) and (4) we have

f@--f'(z)

(s)

ilD-m

Because a < z I x, it follows that as x + a+, z + a+; therefore,

lim f9=-hm f'Q)

t - e , +8 \ x )

'

fg

r-a+ 86--:'I.M

(5)

But by hypothesis,the limit on the right side of (6)is L. Therefore, lim f]

r_a+ g\X )

L

I


The proof of case (ii) is similar to the proof of case (i) and is left as an exercise (see Exercise 27). The proof of case(iii) is based on the results of cases(i) and (ii) and is also left as an exercise (see Exercise28). L'H6pital's rule also holds if either r increases without bound or r decreaseswithout bound, as given in the next theorem. 15.1.4 Theorem Let f and g be functions which are differentiable for all x > N, where N L'H6pital'sRule is a positive constant, and suppose that for all x > N, g'(r) + 0. Then if

- o and - o andif Jl* f@) Jn s(x) f' (x\

r.

J1tre-L it follows that

,. f@) lim --L r-*@ 8\X)

The theorem is also valid if "x -->*q"

is replaced by "x --'s'-6."

pRooF: We prove the theorem for x -+ +6. The proof for r -+ -oo is left as an exercise(seeExercise29). For all r > N, let r: Uf; then t:Llx. Let F and G be the functions defined by F(f) :f(Ut) and G(f) : g(Llt),if.t + 0. Then/(r): F(t) and g(r):G(f), where r )N and 0 < t < UN. From Definitions 4.1.1and 2.3.1,it may be shown that the statements

/tt) : rra and lim F(f) :14

lT

have the same meaning. It is left as an exercise to prove this (see Exercise 30). Because,by hypothesis, lim f (x):0 and lim g(r) : 0, we can r-io t-+@ conclude tJrat lim F(f) : $ and lim G(1) : g

(7)

Considering the quotient F'(t)lG' (t), we have, using the chain rule,

-1r,/1\

F'Gl

Ft \tl

\ t, ' E \t)

P6 \tl

\t/

eit:m:m:N o

f'(x)

Becauseby hypothesis lim f'(x)lg'(x):L, r-*a

it follows from the above

that

lim* t-o+

tr'(t) _ t P

\,

(f)

Becausefor all x ) N, g'Q) # 0,

(8)

FORMO/O 635 15.1THE INDETERMINATE

f o r a l l0 < t . L

G'(t)#0

N

that From (n, (8), and (9), it follows from Theorem 1.5.1..2 r:-

F(f) :

I

I1T@I-L But becauseF(t)lc(t) : f(x)lS@) for all x > N and t + 0, we have

.. f(x) llm -'i:

a-+- 8\r)

L

and so the theorem is proved.

I

Theorems 15.1.2and 15.1.4also hold if L is replacedby *o or -@. The proofs for these casesare omitted. EXAMPIn 3:

Find

solurroN:

lim sin (Ilx) - 0 and lim tan-l (Ux) : Q.So applying L'H6I-*a

, ff-*o

.L-

sln

pital's ryle.we get 1.

lim t-*@

cos-' sin 1 x 4,zi: lim tan-l

(l)

$-*@

\r/

Because lim cos 1= t .?-*@

ryz

and

r.

1

lltfu:,ritfi-l, - 7

.r

it follows that 1, coslim =* -1 t-*@

rya *

f+r Therefore, the given limit is L.

t *E

L5,1 Exercises In Exercises1 through 4, find all values of.z in the given interval (a, b) satrsSng the conclusion of Theorem 15.1.3for the given pair of functions. 1. f (x) : v3S , ( r ) : x 2 ;( a , b ) : ( 0 , 2 )


636

s i n x , g ( r ) : c o st r ; ( a , b ) : ( 0 , n )

3. f(x):

4 . f ( x ) : c o s2 x , g ( r ) - s i n x ; ( a , b ) - ( 0 , t n )

In Exercises5 throu gh 24, evaluatethe limit, if it exists. v5".; l; it m a n,x*

,

.v ( ' l P

O.

,ir, 2

,.

tanr-x

;;

x- sinr

riffr

7. lim I-*@

I

x )r

8. lim=

r-2 z-

9. lim

x

fi

r)11. lim ln(sin @ - 2x)' "-"i, e2" -

1

14. lim*sinz r ;;

20. lim r-o

llII'

uFr

;:; ,.

ufi

;;

10. lim r-o

€t

er-cosr

"1,0t

^{ ,.

cosh r

,\, r. L=..

X2

t\ f7. . -lli _m;t-2

tan3 0

19. lim I-O

L+cos2x

sinh x-

X

+n_2n

0-sin0

zL. Ilmr_7t r-slnI

I

sin-1x

. -2 . r _ 3 r 13. lim X r-o

r sin r

tan\ 2x 1g. lim ;:; tanh x

)C

r-o

(9.

l .

Lz..

rJ.

cos .r -

.r.r r. 1wr.

d n

iF

o.hm+ sin f2 ;; vet -

--q

( t + x ) r r s- ( 1 - x ) r r s (1+x)rrs-(1 -x)trs

1, 1 +-2tan-r*x2 | 22. lim = 1 x I-*a

sinr

llrrl

n -o

sin3 x

A. electrical circuit has a resistance of R ohms, an inductance of L henries, and an electromotive force of E volts, where R, L, and E are positive. If I amperes is the current flowing in the circuit f sec after a switch is tumed on, then

I:fr(1- e-H,) For specific values of t, E, and L, find lim I E-0+

25. In a geometric progression, if a is the first term, r is the common ratio of two successiveterms, and S is the sum of the first n terms, then tf r # l, - 1) o _ a(r" r-L

p ) -

Find lim S. Is the result consistent with the sum of the first n terms if r : L? T-l

27. Prove Theorem 15.1.2(ii). 28. Prove Theorem 15.1.2(iii). 29. Prove Theorem 15."1,.4 for x + -oo. 30. Suppose that f tt a function defined for all r ) N, where N is a positive constant. If t: Llx and F(f) -- f(Llt), where t + 0, prove that the statements lirh f (x) - M and lim F(f ) - M have the same meaning. .T-

I5.2 OTHER INDETERMINATE FORMS

*@

f-0+

Suppose that we wish to determine if cannot appty the theorem involvi.g

(sec2xlsecz3x) exists. We ):_T,, the limit of a quotient because

FORMS 15.2 OTHERINDETERMINATE

lim seC r:

637

*o and lim seC 3x: *oo. In this casewe say that,the funcr-nl2

c-nl2

tion defined by seC xlsef 3x has the indeterminate form (*oo/*o) at x:+,T. t'H6pital's rule also applies to an indeterminate form of this and (*o/-o). This is given by the type as well as to (-ol-o;, (--l**) following theorems, for which the proofs are omitted because they are beyond the scopeof this book. 15.2.1 Theorem Let / and g be functions which are differentiable on an open interval I, L'H6pital'sRuleexcept posiibty at the number a inl, and supposethat forall x * a inl, --, and lim g(r): *o or -oo, and if 8'@) * 0' Then if lim f (x): *o or

f'(x)

12 s'TJ:' it follows that (t -\

lim r*:

L

e-a $\X)

The theorem is valid if all the limits are right-hand limits or if all the limits are left-hand limits. EXAMPLE

SOLUTION: Because lim ln r "'o+ tal's rule and get

Find

@ and lim (Ux) - +@,w€ apply L'H6piJr.-O+

L

limry- hm+7 -0+I7

15.2.2Theorem L'H6pital'sRule

_7,

hm(-r) -0

-0+I3.-0+

,

Let f and I be functions which are differentiable for all x ) N, where N is a positive constant, and suppose that for all r > N, g' @) + 0. Then if lim f (x) - +co or -@, and lim S@)- +m or -@, and if t-*a

The theorem is also valid 7f "x + +m" is replaced by '1.5.2.2 Theorems "1.5.2.'/-. also hold if L is replaced by f m or and and the proofs for these cases are also omitted.

638


ExAMPrn 2:

Find

soLUrIoN: Becauselim ln(2 * e") : *oo and lim 3r :]-6, f-*o

by applying

.T-+@

L'H6pital's rule we obtain 7

..

rlm

ln(2 * e'\ ------;:-:

..

o-*o

5X

r-*a

rlm

2 * e" -

.o*

J

:"t11Tf* ec: la *d,1*

Now becaur" Jit

(6 * 3e") : *@, we applyL'H6pital's

rule again and get

gr":

: :5 "11 &" "1T"3

,.e',.er,.11,

"11 o+ Therefore,

ln(2 -t e')

..

1

--------;-:

ulr-l

-

3x

;-;;

solurroN:

3

lim seC r:

*o and lim seC3x: *o. So by applying L'H6-

I-tlz-

I-Tl2-

pital's rule we get seczx a;-:

r. Ilm

r. Irm

,ili2-secz 3x

;:;;r-6

lim 2 sec2r tan x:

2 secz x tan x seC 3r tan 3r and

*a

c-nlz-

lim

6 sec23r tan 3r:

*o

r-nl2-

It may be seen that further applications of L'H6pital's rule will not help us. However, the original quotient may be rewritten and we have seC x

,.

,.

llm ----: a-ntz-S€.C-3X

cos23x

llm a-v12-COS' X

Now, because lim cos23r:0

and lim cos2x:0,

r-nl2-

we may apply

I-il2-

L'H6pital's rule giving r. Ilm lIIl

|.

"-"ir-

cosz 31 --:

1. llm

e,

cosz)c

i-rir-

-6 cos 3r sin 3r -2 cos x sin x

r:_ 3(2 cos 3r sin 3x) :um r-7ft2- (z cos )c st'a x) r. - 3 sin 6x :rtm t-rrl2-

Sln

ZX

Becauselim 3 sin 6x - 0 and lim sin 2x :0, we apply L'H6pital's I -rrl2-

I -7rl2-

FORMS 15,2 OTHERINDETERMINATE

639

rule again and we have L8 cos5x - 18(-1) - o r:- -3: sin 6r - r:-UIrl T:7 cos2x 2(-Ll sin 2x r-nt2"i"ir-Z Therefore, rlm

1:-__ t""" : llm ,,A 5)C r-zrl2-Se9-

g r

The limit in Example 3 can be evaluated without L'H6pital's rule by using Eqs. (11) in Sec.10.1and Theorem 10.2.1.This is left as an exercise (seeExercise28). In addition to 0/0 and +aF@, other indeterminate forms are 0'(*o), {o - (*o), 0, 1t-;0, and 11-. These indeterminate forms are defined analogously to the other two. For instancei if lim /(r) : *o and lim g(r) : 0, then the function defined by f (xftot has the indeterminate form (*o)o at a. To find the limit of a function having one of these indeterminate forms, it must be changed to either the form 0/0 or 1ol-+o before L'H6pital's rule can be applied. The following examples illustrate the method.

EXAMPLE

Find

solurroN:

Becauselim sin-r r : 0 and lim csc x: *a, the function de.f-0+

lim sin-l r csc tr t-

O+

if it exists.

f-Or

fined by sin-l l csc r has the indeterminate form 0 . (+"o; at 0. Before we can apply L'H6pital's rule we rewrite sin-l r cscr as sin-l r/sin r, and consider lim (sin-1 r/sin r). Now lim sin-l r : 0 and lim sin r: 0, and so I -O+

.?-0+

I -O+

we have the indeterminate form 0/0. Therefore, we apply L'H6pital's rule and obtain l -1,

EXAMPLE

Find

SOLUTION: 1.

BCCAU SC L

lim +Jf;-O

'lr3 *

and

fm

I

,.

rlli:l

r-0 f

"

:TJC

SeC X

we have the indeterminate form {owe have .. ll 1 \ secr-l rrm c-o

l--' \I-

r

SeC Xl

lim ( s e cx - 1 ) : 0

rrifi. a-q

l:

and

t-0

Ltg

f

(fo).

Rewriting the expression,

S€C I

(x' secx)

so we apply L'H6pital's rule

and obtain ,.

s e cx - l

llm-:llm g-O f SeC )C

r. 7-O

sec x tan x 2x secx + rP sec x tan x


640

tan r

!.

:rliFr-

;;2x*x2tanx

Ltg

tanr: 0

and

lim (2x * x2 tan x) I-0

and so we apply the rule again and obtain seczx

:hm lim-talr ;; 2x * xz tan x r-o 2 + 2 x t a n x * x 2 s e c 2 x Therefore,

l i m e - , 1xz secxf\ : 12 "-o

\r'

If we have one of the indeterminate forms, 00, (t*)0, or L!*, the procedure for evaluating the limit is illustrated in Example 6. EXAMPLE 6:

llt

Find

(x * 1)cotr

soLUrIoNl Becauselim (r * 1) : L and lim cot x - *@, we have the t-o+ determinate form r*J.-ilut y:

if it exists.

(1)

(r * t)cotr

Then l n Y : c o tx l n ( r + 1 ) 'lnY- ln(r * 1) tan r So lim ln Y: lim -ot

t -o+

t

ln(r * 1) tan

(2)

X

Because lim ln(r + L) : 0 and lim tan x : 0, we may apply L'H6pi-

tal'sruleto tfJ?gnt sideot (2)ur,a"Jiltuir, ln(r+L)

r.

riifi

-:

.?-o+

tan X

,r l i.f i

1 x + 1 -o-:

n-o+ SeC' X

1

Therefore,substituting L on the right side of (2), we have lim lny - 1

(3)

I-V

Becausethe natural logarithmic function is continuous on its entire domain, which'is the set of all positive numbers, we may apply Theorem 2.6.5and we have from (3) ln lim U:1 -0*

(4)

t

Therefore (because the equation ln a:

b is equivalent

to the equation

LIMITSOF INTEGRATION641 INTEGRALS WITHINFINITE 15.3IMPROPER n:

eb),it follows from Eq. (4) that lim U: er

I -O*'

But from (1), y - (x * 1)cot', and so we have lim (r*1)cot''-s Jf-0*'

L5.2 Exercises In Exercises L throu gh 27, evaluate the limit, if it exists. ,v2

1. lim :, t-+- 0*

r) z. lim 1"(."t

l'1

t_nt2-ln ( tan x)

4.,tlt,ti'

5. lim x csc x f *0+

7 . r i m , J _ _xL- L) l

8.

;:;\lnr

lim

10) t;n (r + eza)ttn

L1.. l i m ( x ' - f f i ) t-t@

w J1t *'tt 16. lim

9. lim tsin r

(sinh x)tanr

f -0+

L4. l i m ( 1 + x ) t n t

/ s 12. l i m i * , , ;-;\r'*x-6 @,

J/-0*

*rtrn r

17. lim (1 + ax)rt";a + 0

f -0+

&

.T-0

/

;r-*@ \

hm Q-2) taninx

21,.

t-2

23. lim [ (ru * 3x5+ 4)rt6- xf

22. lim (cos x)rrt'

k" * x)zrt 1 \r2

18. l i m ( 1 + ; l

t-0

19. lim (1 + sinh x)zrt

J1*

L\ x-2t

ft*

2x/

[(cos x)et2tzfuue

24. l i m x " "

t -*a

W

lim t -*o

ln(r * e") 3;x

u"t1i#

Evaluatethe limit in Example3 without using L'H6pital's rule, but by using Eqs. (11)in Sec.10.1and Theorem10.2.1. (a) Prove that lim (e-rtc2lxn): 0 for any positive integer n. (b)lI f (x): e-rto',use the result of (a) to prove that the limit of f and all of its derivatives, as r approaches0, is 0. - - lim P-+\',:9,frnd,n. 30. v v ' If - rl ;;;

\nx

31.Suppose f(x):

f

:r,hndn. e$t/iF+tittand,g(r): xnes,.rf,t* tffi]

15.3 IMPROPER INTEGRALS WITH INFINITE LIMITS OF INTEGRATION

In defining the definite integral Il f @) dx,the function/was assumedto be defined on the dosed interval [a,bl.We now extend the definition of the definite integral to consider an infinite interval of integration and


call such an integral an improper integral.In sec. 15.4 we discuss another kind of improper integral. o rLLUsrRArroN1: Consider the problem of finding the area of the region bounded by the curve y: s-r, the r axis, the y axis, and the line x:b, where b > 0. This region is shown in Fig. 15.3.1.Letting A square units be the area of this region, we have A_

-lD

: -e-, I

lo :7 - e-b If we let b increase without bound, then f-

[o e-' ilx: lim (1-

D-+6 Jo

s-o1

b-*o

rb

lim I e-" dx:l

L

(1)

b-+q Jo

We conclude from Eq. (1) that no matter how large a value we take for b, the area of the region shown in Fig. 15.3.1will always be less than L square unit. . Equation (L) states that if b > 0, for any e ) 0 there exists an N > 0 such that ltb I lf e-cilx-11
tJo

wheneverblN

I

In place of (1) we write f+*

I

e-, dt:L

JO

In general, we have the following definition. L5.3.L Definition

It f is continuous for all x f+*fb

I

Ja

I f(x) dx f ( x ) d x - bl i- m +* Ja

if this limit exists. If the lower limit definition.

of integration is infinite,

we have the following

15.3 IMPROPERINTEGRALSWITH INFINITELIMITS OF INTEGRATION 643

15.3.2Definition

If / is continuous for all r < b, then rb

I

J-*'

fb

f(x) dx: rim I f(x) dx a--@ Ja'

if this limit exists. Finally, we have the casewhen both limits of integration are infinite. 15.3.3Definition

If / is continuous for all values of r, then f+a

fo

fb

I f<*l dx: arim I ftrl dx-t brim I f(x) dx -+@JO --a Ja' J-a' if these limits exist. '1,5.3.2, and 15.3.3,if the limits exist, we say In Definitions 15.3.1, that the improper integral is conaergenf.If the limits do not exist, we say that the improper integral is diaergent. EXAMPLEl,:

Evaluate

dx

f2

dx

f2

SOLUTION:

J-*ffi:J1a

J-*@ if it converges.

_ lim A--Q

-

lim Q,--@

-+-o 1 -- 2

/.i

EXAMPLE

Evaluate if they

exist:

solurroN:

(a) Using Definition 15.3.3,we have

f+nf0fb

I

r+-

J -a

(a) |

:

J-a

and (b) lim T- t@

xdx: lim I xdx* lim I xdx a--a J a D-+m J0

xdx

:

rimL+*J, Ia L+*)"* fl0t-lD

Ia ?ta')* I** *b'

Because neither of these two limits exists, the improper integral diverges. frflr (b)ln J_,x dx: P*,1_, "l1t : (tr' *r')

"11T"


:lim0 ?n*-t-@

-0

Example 2 illustrates why we cannot use the limit in (b) as the definition of an improper integral where both limits of integration are infinite. That is, the improper integral in (a) is divergent, but the limit in (b) exists and is zero.

Evaluate dx x2+6x*L2

dx x2+6x*12

dx 1. fo , r. fb -:]3-J"m*J1tJ,m

:

if it converges.

#]: .,It t+ tan-l #], .t1it+ tan-r

:Ii

_

ExAMPLE'4: Is it possible to assign a finite number to represent the measure of the area of the region bounded by the graphs of the equations y : Llx, y - 0, and x - L?

dx

\tr-+tan-l#) Gtan-r

7r_

\E

Let L be the number we solurroN: The region is shown in Fig. 1,5.3.2. wish to assign to the measure of the area,7f.possible. Let A be the measure of the area of the region bounded by the graphs of the equationsU :11x, y : 0 , x : 7 , a n d r : b w h e r eb > 1 . T h e n

A-,,'iil, f la. 2+,Ln: So we shall let L J:tA if this limit exists.But lim

lim AD-*oo

-

Jr x

lim

-ln1l lln b

b-*a -

r,Ldx

b-*a

T

l m

\-.\'

15.3IMPROPER INTEGRALW S I T H I N F I N I T EL I M I T SO F I N T E G R A T I O N

Therefore, it is not possible to assign a finite number to represent the measure of the area of the region.

Figure 15.3.2

ExAMPLE5: Is it possible to assign a finite number to represent the measure of the volume of the solid formed by revolving the region in Example 4 about the x axis?

solurroN: The element of volume is a circular disk having a thicl,ness of. A1xand a base radius of. U€r.Let L be the number we wish to assi'gn to the measure of the volume, and let v be the measure of the volume of the solid formed by revolving about the r axis the region bounded by the graphs of the equations A : 1lx, ! : 0, x : 1, and r : b where b > 1,. Then V-

7T(*)'

lim

$ lltll-o --r

Lix:n * Ib

We shall let L - lim V, if the limit exists. D* *o

lim

lim

D- *o

D-*o

7T

li1mn b -- * o- a :Tf

ti: imn

1)

b -- * o- @ -77

Therefore, we assign z to represent the measureof the volume of the solid.

EXAMPLE 6:

Determine

if

sin x dx is convergent or divergent.

SOLUTION: f+*

I

sin x dx: lim

JO

b-*o

:

lim b-*oo

646


: lim (-cos b + L) For any integet fi, as b takes on all values from nn to 2nn, cos b takes on all values from - L to L. Hence, lim cos b does not exist. Therefore, the b- *o

improper integral is divergent.

Example 6 illustrates the case for which an improper integral is divergent where the limit is not infinite.

15.3 Exercises In ExercisesL through 14,determine whether the improper integral is convergent or divergent. If it is convergent,evaluate it. 3.

/:."

x 5-*' dx

8.

*'rn dx

[o_* a)

4.

I:-

s.

2-0dx

I-s-ttttuc

1- ( 1 13. 1 5 . Evaluate if they exist: (a) JJJ sin x dx; (b)

IT

tn x dx Ir**

74. f-

e-" cosx dx

Ja, sin x ilx.

15. Prove that it !!- f (x) dr is convergent, then I:f f(-x) dr is also convergent and has the same value. L7. Show that the improper integral ES x(l -l *1-z dx is convergent and the improper integral IJX x(7 t xz)-r dx is divergent. 18. Prove that the improper integral !{- dxlx" is convergent if and only if n > 1' L9. Determine if it is possible to assign a finite number to represent the measure of the area of the region bounded by the curve whose equation is y : 1-1 1r" * e-") and the x axis. If a finite number can be assigned, find it. 20. Determine if it is possible to assign a finite number to represent the measure of the area of the region bounded by the x axis, the line x:2, and.the curve whose equation is y:11 1rz 1). If a finite number can be assigned,find it.

a

Determine if it is possible to assign a finite number to represent the measure of the volume of the solid formed by revolving about the r axis the region to the right of the line r: 1 and bounded by the curve whose equation is ! : Tlxt/i and the r axis. If a finite number can be assigned, find it.

22. Determine f+*/

a value of n for which nxz

J, \ffi

-

the improper

integral

1. \ j sx+r)o*

is convergent and evaluate the integral for this value of n. 23. Determine a value of. i for which the improper integraL

l.*(T=-=?)4* 2r+n1 \x*1

J'

is convergent and evaluate the integral for this value of n.

15.4 OTHER IMPROPERINTEGRALS

647

Exercises24,25, and.25 show an interesting application of improper integrals in the field of economics. Suppose there is a continuous flow of income for which interest is compounded continuously at the annual rate of 100l percent and /(f) dollars is the income per year at any time f years. If the income continues indefinitely, the Present value, V dollars, of all future income is defined by l+@

V:

I

Jo

f ( t ) e - i ld t

2/. A eontinuous flow of income is decreasing with time and at f years the number of dollars in the annual income is 1000 . 2-t. Find the present value of this income, if it continues indefinitely using an interest rate of 8 percent compounded continuously 6

fA. continuous flow of profit for a company is increasing with time, and at f years the number of dollars in the profit per year is proportional to t. Show that the present value of the company is inversely proportional to i2, where 1.001 percent is the interest rate compounded continuously.

8. fn British Consol is a bond with no maturity (i.e., it never comes due), and it affords the holder an annual lump-sum payment. By finding the present value of a flow of payments of R dollars annually and using the current interest rate of 1001percent, compounded continuously, show that the fair selling price of a British Consol is R/i dollars.

15.4 OTHER IMPROPER INTEGRALS

Figure 15.4.1shows the region bounded by the curve whose equation is y : U!x, the r axis, the y axis, and the line x: 4.lf it is possibleto assign a finite number to represent the measure of the area of this region, it would be given by lim llall-o If this limit exists, it is the definite integral denoted by

t

dx

(r)

\fr

However, the integrand is discontinuous at the lower limit zero. Furthermore/ lim UVx : *@, and so we state that the integrand has an infinite discontinuity at the lower limit. Such an integral is improper, and its existencecan be determined from the following definition.

F i g u r e1 5 . 4 . 1

L5.4.1 Definition

If / is continuous at all r in ttie interval half open on the left (a, b], and if lim /(r) : t@/ then

L

f(x) dx:

f(x) dx

if this limit exists. . rLLUSTRArroul.: We determine if it is possible to assign a finite number to represent the measure of the area of the region shown in Fig. 15.4.1. From the discussion preceding Definition 15.4.1 the measure of the area of the given region will be the improper integral (1) if it exists. By Defini-

648


tion 1,5.4.L, we have fn dx

f4 dx

1.

Jo&::'_TJ. &

: tt-T'*t''1 4 f, - lim 14- 2*l :4-0 -t Therefore, we assign 4 to the measure of the area of the given region.

o

If the integrand has an infinite discontinuity at the upper limit of integration, we use the following definition to determine the existence of the improper integral.

15.4.2Definition

If f rs continuous at all x in the interval half open on the right la, b) and if lim f (x) - -f m, then I - b-

f b-e

fb

f(x) dx I ff*l dx- lim e-o*Ja Ja' if this limit exists.

If there is an infinite discontinuity at an interior point of the interval of integration, the existence of the improper integral is determined from the following definition. 15.4.3 Definition

If / is continuous at all r in the interval fa,bl exceptc when a I c 1 b and if lim l/(r) | : *o, then fb

fc-e

fb

I f(x) dx:rim I f@) dx*rim I f(x) dx D-o*Jr+6' e-o+Ja Ja'' if these limits exist. If Ilf@) dx is an.improper integral, it is convergent if the corresponding limit exists; otherwise, it is divergent. EXAMPLE1.: Evaluate fz

t -

dx

J o@ - r ) ' if it is convergent.

SOLUTION: ThC iNtcg 15.4.3, we

Definition f2

dx

- I lr.m

Jo1, -lim €-0+

15,4 OTHER IMPROPERINTEGRALS 649

:tl1

|}-

t] * t'_T l-t.*]

Because neither of these limits exist, the improper integral is divergent.

. rLLUsrRArrou 2: Suppose that in evaluating the integral in Example 1 we had failed to note the infinite discontinuity of the integrand at 1. We would have obtained L12 r-1lo

1 ,- 1- : - '-1 L

^

This is obviously an incorrect result. Becausethe integrand 1.1(x-7)2 is never negative, the integral from 0 to 2 could not possibly be a negative number.

EXAMP

fr

rn 2: Evaluate

I rln JO

SOLUTION:

irg

xdx

if it is convergent.

Definiti

Th ne integrand has a discontinuity at the lower limit. Applyion n 15 . 4 . 1 w , e h ave fr

fr

: - l i m I x l n r dx I x l n ,x ddx e-0+ J e Jo t-

- lim ltr' €-0+

L

€ - tr'l

: tt5 [] ln Hence, fr

I xlnr dx:0-+-tlime2

Jo

ln€-0

€*o+

To evaluate lim ezln € - lim I!-s €-0+

€-0+ L e2

rule because lim ln

we apply L'H6pital's

€-0+

have

T lne r. llm T: €-0+ r e2

1. llm €-0+

€

-lim 2

€- 0-

€3

Therefore, from (2) we have f1

I x lnx dx--*

Jo

-0

and l i m l l e ' : €-0+

+@. We

AND TAYLOR'SFORMULA FORMS,IMPROPERINTEGRALS, INDETERMINATE

650

EXAMPIn 3:

Evaluate

dx

,f** -

J' )*FT.

SOLUTII Of N : Fr f,or: this integral, there is both an infinite upper limit and an :onltinuity of the integrand at the lower limit. We proceed infinit,te: d iisco as foll< OIws. +€ r+ @

I Jr

if it is convergent.

dx -:rilrl x!x2 -

r.

f2 t a-:-Ja

1

::

dx

, r.

fb -

l l ]l r

dx

|

u-**J, xlFT.

xlF'

t. lD - lim |l sec-t r 12 l + lim l sec-l r r a-r* L

)"

D-*m L

)z

(sec-l 2 - sec-l a) * lim

:

lt_T. - -lim sec-l a * lim sec-l b a-!+

(sec-l b - sec-l 2)

b-*a

-0 **n -- l * 2 u

Exercises L5.4 In ExercisesI through 20,determine whether the improper integral is convergent or divergent. If it is convergent,evaluateit. 1. f

d-

2'

J o! t - x

4r+

Jo \/L5-

x2

7 ' Jlntz, l - sdti n r

f2

dx

Jrffi

5I:# 8 ' Jfz , f f dx i f+-

3' 6. (1

fnl2

J,,n frl2

J,

secx dx tan0 d0

f4dx

t-

Jo x2-2x-3

dX 10.f+J, ffi

L1' rnxdx J,

12.f' 4x ' J-r

l 3 ' rJo- r f fdit

H.fL

4r

Jo\m

16. J, fta*

17. J,,,ffi

L L/"s f L x\FT.

z oJo f LxlR

7+* o-tlt

J,

f'

dt

rc'

f2

dx

J-, *'

18I*

2L. Evaluate, if they exist:

4+ (a) l'-T [/-, l,+) I:,*,(b)

by the 22. Show that it is possible to assign_afinite number to represent the measure of the area of the region bounded a finite curve whose equation is y : tit/i, the line x : 1, and ihe r and y axes,but that it is not possible to assign about number to represent the ileasure of the volume of the solid of revolution generated if this region is revolved the r axis.

15.5 TAYLOR'SFORMULA

651

In Exercises23 through 25, find. the values of.n lor which the improper integral converges and evaluate the integral for these values of n. 23.

J:

24.

xn dx

15.5 TAYLOR'S FORMULA

L5.5.1 Theorem

['

xn lnz x dx

*" Ln x dx

Values of polynomial functions can be found by performing a finite number of additions and multiplications. Howevet, there are other functions, such as the logarithmic, exponential, and trigonometric functions, that cannot be evaluated as easily. We show in this section that many functions can be approximated by polynomials and that the polynomial, instead of the original function, can be used for computations when the difference between the actual function value and the polynomial approximation is sufficiently small. There are various methods of approximating a given function by polynomials. One of the most widely used is that involving Taylor's formula named in honor of the English mathematician Brook Taylor (16851731).The following theorem, which can be considered as a generalization of the mean-value theorem (4.7.2), gives us Taylor's formula. Let / be a function such that / and its hrst n derivatives are continuous on the closed interval fa, bl. Furtherrnore, let f{n+r)(x) exist for all r in the open interval (a, b). Then there is a number f in the open interval (a, b) such that

fO) - f(a)

(b- a)

(b-a)'+ i(n+l\( t'\

. f@(a) ,, (b- a)"+ffi1i *';

(b- 61"+r (1)

Equation (1) also holds if b < a; in such a case/la,blis replacedby fb, al' and (a, b) is replacedby (b, a). Beforeproving this theorem, note that when n:0,

(1) becomes

f(b):f(a)+f'G)@-a) where f is between a and b. This is the mean-valuetheorem (4.7.2). There are several known proofs of Theorem 15.5.1,although none is very well motivated. The one following makes use of Cauchy's meanvalue theorem (15.1.3). pRooFoF THEoREM15.5.1: Let F and G be two functions defined by

F(r): f@l- f@)- f' (x)(b-x)-'+

-ffi

(b- r)'

-ry (b- x)*(2) (b- v1n-r


and

G(r):ffi

(3)

It follows that F(D) :0

and G(b):0.

Differentiating in e), we get

F ' ( r ) : - f ' ( x ) * f ' ( x ) - f " ( x ) ( b - x ) * 2 f " ( x ) - ( b -x ) 2! _f"' (x)(b x)2, 3f"' (x)(b x)2 (n -

(b - x)"-z

7)f@-\(x) rE____6=l[l_ r-nf@(x)(b n!

frtv)(x)(b x)g

f(n)(x)(b v1"-r

- x)n-1 _f("+r)(x)(b x)n n!

Combining terms, we see that the sum of every odd-numbered term with the following even-numbered term is zero, and so only the last term remains. Therefore, F,(x):

- f("*D.(x) - )c)n @ n!

Differentiating

(4)

in (3), w€ obtain

G'(x):-+,@-x),

(s)

Checking the hypothesis of Theorem 15.1.3, we see that (i) F and G are continuous on Ia,bl; (ii) F and G are differentiable on (a,b); (iii) for all r in (a,b),G'(x) + 0. So by the conclusion of Theorem 1.5.1..3it follows that F(b) - F(a) _ F'({)

G(b)-G(a)-c'(f)

where f is tn (a, b). But F(b) - 0 and G(b) - 0. So

F(il- GryG(a) \J* '(f)

(5)

for somef in (a,b). Letting tc: a irt (3), x : ( in (4), and r : f in (5) and substituting into (6), we obtain

F(a):

f(n+l)(t\

or, equivalently,

(b - €)"

|

nt

l- €€)"j

-]

(b_

a)"*,

(n + 1)!


' - a)n+r F(il - - f,'*:'(i)= (n+ 1)! @ Letting x - a in (2), w€ obtain

(7)

F ( i l- f ( u ) - f ( a ) - f ' ( o ) ( b - a ) - ' + @ - a ) ' -

" '

_,ffi@_a)n-=_,#@_a)n (8) Substituting from (7) into (8), we get

f ( u )- f ( o )+ f ' ( a ) ( b -a )* t : #

'

@- a)'+. . . f(',+l)(f)

,7 , i - @ - a ) "+ f f i

*f{")(a)

(, 7b - a ) ' + r

which is the desired result. The theorem holds if b < a becausethe conI clusion of Theorem 15.1.3 is unaffected if. a and b are interchanged. If in (1) b is replaced by x, Taylor's formula is obtained. It is

(e) where f is between a and x. The condition under which (9) holds is that / and its first n derivatives must be continuous on a closed interval containing a and x, and the (tc* 1)st derivative of / must exist at all points of the corresponding open interval. Formula (9) may be written as f (x) : Pn(x) * R,(r)

(10)

where f" (a)

( x - a ) * ' t @ - n ) r, + . . .

P"(x) : f (a)

+ - f(")(a)Q - a ) " n!

and

R,(r) :m

Q-

a)n+t

where f is between aand r

( 1 1)

(12)

P,(r) is called the zth-degree Taylor polynomialof the function / at the number a, and Rr(r) is called the remainder.The term R,(r) as given in (12) is called the Lagrangeform of the remainder, named in honor of the French mathematician Joseph L. Lagrange (1736-1813).

654


ExAMPLE 1:

Find the thirddegree Taylor polynomial of the cosine function at L+rratrd the Lagrange form of the remainder.

soLUrIoN: Letting f (x) : cos r, we have from (11.)

p,(r) -i).LP (.-o)'*r+ (,-o)" : re). r'G)Q Because/(x) : cos x, f(*rr) : +{Z; f' (x) :-sin r, f' (*zr): -+t/2; -rlf} :-cos : : (x) (x) (*d:'4r2.Therefore, x, " x, sin f "' f f " Gd f "' : P,(r) +\/2 *{2(x Ln) i{2(x *o), + jrt/2(x - *o)' Because /(to)(r) : cosr, we obtain from (12) wheref is between*r andx &(r) : 2l(cos€) (x - In)n Because lcosf | = 1, we may concludethat for all x, lRr(r) | < *(x - *rr)n. Taylor's formula may be used to approximate a function by means of a Taylor polynomial. From (10) we obtain

lR"(r)| - lf(x) - P,(r)I

(13)

If. P"(x) is used to approximate f(x), we can obtain an upper bound for the error of this approximation if we can find a number E ) 0 such that lR"(r) | - E or, because of (13), such that lf (x) - P"(x)l < E or, equivalently, P"(x)-f-fQ)=P"(r)+E ExAMPLE2: Use the result of Example L to compute an approximate value of cos 47" and determine the accuracy of the result.

sor.urroN: 4T - f{or radians. Thus, in the solution of Example L, we takex:f&r and r- in:#n, and we have

cos4T : *12L' - #rr - L(#d' + 6(#zr)'l + &(rttzr) where Rr(ttrr) -;r

cosf (#n)n

within<€<#n

Because0 0 Takin g noLn: 0.0349056,we obtain cos 47o - 0.58L998 which is accurate to six decimal places.

EXAMPLE3: Use a Taylor polynomial at zero to find the value of \fr, accurate to four decimal places.

soLUrIoN: If we let f (x) : sr, all the derivatives of f at x are et and all the derivatives evaluated at zero are 1. Therefore, from (11) we have

P,(x)-1.*x+**#*


and from

R"(r)

where f is between 0 and r

We want lR,(+)I to be less than and becauseertz1 2, we get

: t '-.< +\ ' l R , ( +.-) l 2 n,+, 9 r (+ n 1)I zn+r(n r

If we take x: t in the above

1)I

-

1 2(n * 1,)I

When ft: 5, lR"(+)l will be lessthan 0.00005if Llz(n * l,Y < 0.00005. 11-

ffi:ffi-o.oooo4

\fr:ps(*):1+ t+*+t+E+4+88% from which we obtain

fr:'!,.64870.

A special caseof Taylor's formula is obtained by taking a:0 and we get

*+ f(x):/(o)

xt'+xz+

*f#xn*ffi*n*,

in (9)

(14)

where f is between 0 and r. Formula (1a) is called Maclaurin's formula, named in honor of the Scottish mathematician Colin Maclaurin (15981746). However, the formula was obtained earlier by Taylor and by another British mathematician, f ames Stirling (1692-1770). There are other forms of the remainder in Taylor's formula. Depending on the function, one form of the remainder may be more desirable to use than another. The following theorem expressesthe remainder as an integral, and it is known as Tayloy'sformula with integral form of the remainder. L5.5.2 Theorem

If.f is a function whose first n * 1 derivatives are continuous on a closed interval containing a and r, then f (x): P"(x) -l R,(r), where Pn(r) is the zth-degree Taylor polynomial of.f at a and R,(r) is the remainder given by Rr(r)

(x-

t ) " 1 t " + t ) ( f )d t

The proof of this theorem is left as an exercise (see Exercise 22).

(15)

656


Exercises L5.5 In-Exetcises 1 through 10, find the Taylor polynomial of degree n with the Lagrange form of the remainder at the number a for the function defined by the given equation. l. f(x):sin x;a-tn;n:3 4.f(x) -cosh x;a-0;n:4 7. f(x) : ln costr; a- *n; n- 3 1 0 .f ( x ) - ( 1 - x ) - u z 'a - 0 ; n - 3

2.f(x):tanx;a-0;n-3

3 . f ( x ) : s i n hx ; a - 0 ; n : 4

5.f(x):ln x;a-1;n:3

6.f(x)--{x;a-4;n-4

8.f(x):e-*';a-0;n-3

9.f(x) - (t + x)'tr;a-0;n-3

11. Compute the value of e cbrrect to five decimal places, and prove that your answer has the required accuracy. 12. Use the Taylor polynomial in Exercise5 to compute \6 accurateto as many places as is justified when Rais neglected. 13. Compute sin 31oaccurateto three decimal placesby using the Taylor polynomial in Exerciset at tr. (Use the approximation rluz - 0.0175.) 14. Use a Taylor polynomial at 0 for the function defined by f(x): four decimal places.

ln(l * r) to compute the value of.lnl.2, accurateto

15. Use a Taylor polynomial at 0 for the function defined by 1-L+

f(x): tnffi

to compute the value of ln 1.2 accurateto four decimal places. Compare the computation with that of Exercise 14. 16. Show that the formula ( l + x g 1 t r '- l 1

*x is accurateto three decimal places if -0.03 < x < 0. 17. Show that the formula (1+l;-vz-l-tx is accurateto two decimal places if -0.1 < r < 0. 1 8 .S h o w t h a t i f 0 - x - t , Y3

sinr:r-i+R(r) 5I

where lR(x)l <

"t". 19. Use the result of Exercise 18 to find an approximate value of 6rc sin 12 dr and e 20. Draw sketches of the graphs of y: tir, x and y : mx on the same set of axes.N zero, then the graphs intersect at a point whose abscissais close to z. By finding t at r fior the function / defined by f(x): sin r - mx, show that an approximate when rz is positive and close to zero, is given by i : zrl0 + m). 2 L . Use the method described in Exercise20 to find an approximate solution of the r tive and close to zero. 22. Prove Theorem 15.5.2 (nrwr: Let I f'(t) dt: f(x) f(a)

Ja'

REVIEWEXERCISES Solve forl(r)

and integrate

ft

at )" f'{t) by parts by letting u: f ' (t) and.do: df. Repeatthis process,and the desired result follows by mathematical induction.)

Reaiew Exercises(Chapter15) In ExercisesL throu g;h 14, evaluate the limit, if it exists. 4 ,. tanh 2x I. lrm;-:;l sinh2 r

L.

4. lime-(l+x)rr"

F J.

X

fr-O

7 . lim

1

,! _ ln(1 II.IJ:I t-*cn

2 \

Sln I *

eztlt)

Stlz

1100

( sinz x)tan r

8. lim ? t-+ q

I-rr12

x - ta!-t x L0. lim 4xg .r-o 13. lim

/

rJ.rrll-r-nt2 \I

e'

tan 0 * 3 LL. lim i-.'/rsec0-l

(e" - x)rtx

-l COS- X/

3. lim (csc2 x - x-2) t-O

6. lim rln + X;z-*@

L

In(ln r) 9. lim-- ln r) ln(r ;:;;

L2.tim rti" r-o \

x

t)"" /

14. lim /1\'"'" l,o* \x/

f-*o

In ExercisesL5 through 24, determine whether the improper integral is convergent or divergent. If it is convergent, evaluate it.

1s. f'

dx

L5.

J - z 2 x+ 3

1 8 . [ ^\ /4x - 2

L9.

2r'J_-ffi

22.

Jz f+*dx fo

2+.J_sffi

dx

In Exercises25 and 26, find, the Taylor polynomial of degree n with the Lagrange form of the remainder at the number a for the function defined by the given equation. % . f ( x ) : c o sr ; a : 0 ; n : 6 27. Given:

[4J f(x): I x u

2 6 .f ( x ) : ( 1 + * ) - ' ; a : l ; n : 3

ifx*o ifr:o

(a) Prove that / is continuous at 0. (b) Prove that / is differentiable at 0 and find /'(0). 28. For the tunction of Exercise27, find, if they exist (a) lim f (x); (b) lim /(r).

658


29. Given: foc-o-r

f(x):1t""-, u

ifx*O

ifr:o

(a) Is / continuous at 0? (b) Find 1im /(r), if it exists. 30' If the normal line to the curve y:ln a, prove that lim (a - rr; : g. tr-

x at the point (x1, ln 11) intersects the r axis at the point having an abscissaof

+6

31. If the first four terms of the Taylor polynomial at 0 are used to approximate the value of {i,how 32. Express both e"'and cos r as a Taylor polynomial of degree 4 at 0 and use them to evaluate ,.

accurateis the result?

e"-cosa

llm----'t-0

X=

33. Evaluatethe limit in Exercise32 by L'H6pital's rule. 34. Apply Taylor's formula to expressthe polynomial P(x) : 4)c3* 5x2- 2x + 1 as a polynomial in powers of (x # 2). 35. Determine if it is possible to assign a finite numhêrto_represent the measure of the area of the region to the right of the y axis bounded by the clowe 4y2- xy'- r3:0'and iti asymptote. If a finite number can be as-signed, find ii. 35' Determine if it is possible to assign a finite number to represent the measure of the area of the region in the first quadrantandbelowthecurvehavingtheequationa:e_".Iiafinitenumbercanbeassigned,findit. 37' Assuming a continuous flow of income for a particular business and that at f years from now the number of dollars in the income per year is_given by 1000f- 300, what is the present value of all iuture income if gVo isthe interest rate compounded continuously? (nrwr: See the paragraph precedlng Exercise24 in Sec.15.3.) 38' Find the values of z for which the improper integral dx convergesand evaluate the integral for those values of z. Jf-r * 39. (a) Prove that for n any positive integer. (b) Find lim (e-tr"lx"), where r > 0 and n is any positive ]il-G"le'):0, integer, by letting x: Ut and using the result of part (a).

I N F I N I T ES E R I E S

15.1 SEQUENCES You have undoubtedly encountered sequencesof numbers in your previous study of mathematics.For example,the numbers5,7,9,l'J..,13, 15 define a sequence. This sequence is said to be finite because there is a first and last number. If the set of numbers which defines a sequencedoes not have both a first and last number, the sequenceis said to be infinite. For example, the sequencedefined by t

2

3 A

(1)

31 51 7 r 9r

is infinite because the three dots with no number following indicate that, there is no last number. We are concerned here with infinite sequences, and when we use the word "seguence" it is understood that we are referring to an infinite sequence.We define a sequenceas a particular kind of function. 16.1.1 Definition

A sequenceis a function whose domain is the set of positive integers. The numbers in the range of the sequence,which are called the elements of.the sequence,at'e restricted to real numbers in this book. If the nth element is given by f (n), then the sequence is the set of ordered pairs of the form (n, f (n)), where n is a positive integer. o rllusrRArrorv 1: If.f 1n1: nl(2n * 1), then

f(L):t

fQ):t

f@):+

f(4):#

and so on. We seethat the rangeof / consistsof the elementsof sequence (1). Someof the orderedpairs in the sequencef are (1, +), (2, g), (3, +), (4, *), and (5, +). A sketchof the graph of this sequenceis shown in Fig. 16.1.1.

(t,3) Q'.?) (r,+) (n,3)

Figure'16.1.1

Usually the nth element f(n) of the sequenceis stated when the elements are listed in order. Thus, for the elements of sequence(1.)we would write n 1.234 ''' 5'6'V'g' 'r;!i' Becausethe domain of every sequence is the same, we can use the notation U@)I to denote a sequence.So the sequence(L) can be denoted

1 6 . 1S E Q U E N C E S 661

by {nl(2n + 1)}. We also use the subscript notation {a"} to denote the sequencefor which f(n): a,. You should distinguish between the elements of a sequenceand the sequenceitself, as shown in the following illustration. . rLLUsrRArror2: The sequence{Un} has as its elementsthe reciprocals of the positive integers L ., 111 ,r7 gr4, ' ' ' ,-n,' ' '

'

(2)

The sequencefor which 11 f(n):.| 2 V+2

if n is odd :L(r n l. s e v e n

has as its elements

l , + , 1+ , , 1 i,, . . .

(3)

The elements of sequences (2) and (3) are the same; however, the sequences are different. Sketches of the graphs of sequences (2) and (3) are shown in Figs. 16.1.2 and 1,6.1..3, respectively. f(n)

(n,i)

(u,t)

F i g u r e1 6 . 1 . 2 f (")

G'.+)

(u'.*)

16.1.3 f(n) Figure

F i g u r e1 6 . 1 . 4

We now plot on a horizontal axis the points corresponding to successive elements of a sequence. This is done in Fig. L6.L.4 f.or sequence (1) which is {nl (2n + t ) ). We see that the successive elements of the se-


quence get closer and closer to *, even though no element in the sequence has the value *. Intuitively we see that the element will be as close to * as we please by taking the number of the element sufficiently large. Or stating this anothet way, we can make lnl(2n*l) -fl less thair any given e by taking n large enough. Becauseof this we state that the limit of the sequence {nl (Zn + 1) } is i. In general, if there is a number L such that lan- Ll is arbitrarily small for n sufficiently large, we say the sequence {an} has the limit i. Following is the precise definition of the limit of a sequence. 16.1.2 Definition

A sequence {a"} is said to have the limit L if for every e ) 0 there exists a number N > 0 such that la^- Ll ( e for every integer n >N; and we write lim an: L n-*@

E1AMpLE1: Use Definition "1,6.1.2 to prove that the sequence

solurroN: such that

We must show that for any e ) 0 there exists a numberN ) 0

[-u-I

for every integer n > N

l2n+U

- 1 l: l z n - Z n - L l : | - r I -t + l2n rl lW | 14"'r11 4 n * 2

has the limit #.

I n

Hence, w€ must find a number N > 0 such that for every integer n > N But

i s e q u i v a l et o nt 2n*l r+

#.€

which is equivalent to nrL-2e 4e So it follows that

1l I n + 1 r l l2n

for every integer n u'l'

- 2e 4e

Therefore, if N: (1 - 2e)14e,Definition 16.'!,.2holds. In particular, if e: t, N - (1 - +)l+: g. So

.*

for every integer n , ,

1 6 . 1S E Q U E N C E S

For instance,

o rLLUsrRArron3: Consider the sequence{(-1)*'lnl,. Note that the nth element of this sequence is (-l)+rln and 1-1;"+t is equal to *L when n is odd and to -1 when z is even. Hence, the elements of the sequencecan be written L 1 1L "t , - 2 ' i ' - Z ' E ' '

' -1---1' 1 n + t

Itr Fig. 16.1.5 are plotted points corresponding to successive elements of this sequence.In the figure, ar:1, az:-t, as:t, a+:-*, aa:*, ae:-t, az:*, ae:-t, as:*,4r0:-*. The limit of the sequenceis 0 and the elements oscillate about 0. 1. 7 rf, 3 - 5

t

1 dS:

As

:

L 9

t2 5

Figure 16.1.5 Compare Definition 16.1.2with Definition 4.7.'1,of the limit of.f(x) as x increases without bound. The two definitions are almost identical; however, when we state that /tr) : L, the function / is defined for "l1a all real numbers greater than some real number R, while when we consider an, n is restricted to positive integers. We have, however, the nlim following theorem which follows immediately from Definition 4.1.1. 15.1.3 Theorem . If lim /(x) : L, and / is defined for every positive integer, then also L when n is any positive integer. yg-iAl: The proof is left as an exercise (see Exercise20). o rLLUsrRATroI.t4: We verify Theorem 16.1.3for t'he sequence of Example L. For that sequencef (n) : nl(2n * 1). Hence,f (x) : xl(2x * 1) and


It follows then from Theorem 16.1.3 that

when z is any ]!!_f(n):* positive integer. This agreeswith the solution of Example 1. o 16.1.4 Definition

ExAMPLE2: sequence

Determine if the

[4n'I l2n'+ L) is convergent or divergent.

If a sequence {a,} has a limit, the sequenceis said tobe conaergent,and, we say that an conoergesto that limit. If the sequenceis not convergent, it is said to be diaergent. solurroN:

We wish to determin" if

4*lQf

* 1) and investigate "tl1/(r). ,.4*..4 Irm;];: lim -:2 I I I)+@-e ,-*-2**

,It

4n2l(2nz* 1) exists.Let /(r) :

Therefore, by Theorem 16.1,3, Ol: 2. We conclude that the given ]1g_f sequenceis convergent and that4n2l(2n2 * 1) converges to 2. ExAMPu 3: Prove that if lrl the sequence{r"} is convergent and that rn convergesto zeto.

O: 0. Hence, Jl* the sequenceis convergent and the nth element converges to zero. If 0 < Irl < t, we consider the function / defined by f (x): f, where r is any positive number, and show that lim fc :0. Then from Theorem solurroN: First of all, if.r:

0, the sequenceis {0} and

1.6.L.3itwill follow that lim r^: O wneii.tl

ury positive integer.

n-+@

To prove that

(0 < lrl < L), we shall show that for any r,:0 "ljt e > 0 there exists a numberN > 0 such that lf-Ol

wheneverr)N

(4)

Statement (4) is equivalent to wheneverr>N

lrl'<,

which is true if and only if ln lrl'<

ln e

wheneverr > N

or, equivalently, rlnlrl
wheneverr>N

(5)

Because0 < ltl ( L, ln lrl < 0. Thus, (5) is equivalent to - tre x># ln l/l

wheneverx>N

Therefore, if we take N: ln elln lrl, we may conclude (4). Consequently, lim rr : 0, and so r" : 0 if n is any positive integer. Hence, by Defi,liT" '/..5.1.2 nitions and'I-.6.1.4, {l} is convergent and f, convergesto zero.

1 6 . 1S E Q U E N C E S

ExAMPtn4: Determine if the sequencet (- I)" + 1) is convergent or divergent.

soLUTroN: The elements of this sequence are 0, 2, 0, 2, 0, 2, , (-t)"+'!., . .. Becausean:0 if. n is odd, and an:2 if. n is even, it appears that the seQuenceis divergent. To prove this, let us assume that the sequenceis convergent and show that this assumption leads to a contradiction. If the sequence has the limit L, then by Definition 16.1.2,for every € > 0 there exists a numberN > 0 such that ln"- Ll ( e for every integer n ) N. In particular, when 6: f, there exists a number N > 0 such that for every Lntegern > N

lo"- Ll < + or, equivalently, ,-+< nn-L<+

f o r e v e r y i n t e g e r> nN

(6)

Becausean: 0 if n is odd and an: 2 if. n is even, it follows from (5) that

-+<-L<+

and -t
But if -L > -+, then 2 - L > !z;hence, 2 - L cannot be less than *. So we have a contradiction, and therefore the given sequence is divergent. EXAMPLE

Determine if the

solurroN:

We wish to determine if lim n sin(nln) exists. Let f(x): n-+@

sequence

x sin(nlx) and investigate lim f(x). Becausef(x) can be written as - 0 and lim (Ux): 0, we can apply [sin(nlx)]l (Llx) and lim ti"(;;t)

is convergent or divergent.

L'H6pital's rule and obtain -

7T 7T -c, cos -

'-

Jlt f@):l i m

.T-*@

-

,

*--

1

lim ?r cos !: X

r-4,@

,

x2

Therefore, lim f ( n ) : n

w h e n n is a positive integer. So the given se-

quence is convergent and n sin(nln) convergesto rr. We have limit theorems for sequences,which are analogous to limit theorems for functions given in Chapter 2. We state these theorems by using the terminology of sequences.The proofs are omitted becausethey are almost identical to the proofs of the corresponding theorems given in Chapter 2. L5.L.5Theorem

If {a"} and {b"} are convergentsequencesand c is a constant, then (i) the constant sequence{c} has c as its limiU (ii) lim can: c Itm an; n-*@

(iii)

lim fl-I@

(an !

br) :

lim h-*@

an !

lim ll-*@

bn)

INFINITESERIES

(iv) lim nnbn: ( lim a") ( lim ll-*@

fl-*@

fl-*@

(v) lim

if lim b n + 0 .

ll,-*@

fl-t@

EXAMPTg5: Use Theorem "1,6."1,.s SOLUTION: to prove that the sequence n2.7Tn (

n2

W+I

'

2 n * 1 s mi : r f i 1

s r nerl- J

is convergent and find the limit of the sequence.

.Tr nslnn

In Example 1 we showed that the sequence {nl (2n + 1) } is convergent and lim lnl Qn + 1)l :*. In Example 5 we showed that the sequence ll-*@

{n sin(nln)} is convergent and lim [n sin (zr/n)] : zr. Hence, by Theorem 16.1.S(iv), r.

l-

n

.

T1

lim n sin T:! n2

ItLffi'ns'\n;l:Hh

' rT

Thus, the given seguence is convergent, and its limit is in.

Exercises L6.1. In Exercises1 through 4, use Definition 16.1.2to prove that the given sequencehas the limit L. L.

\) t/\ 1

3.

4.

In Exercises5 through 19, determine if the sequenceis convergent or divergent. If the sequenceconverges, find its limit.

s{#} 8{ffi}

7.

10.

11. {tanh n}

14.{et$i} LSlNNJ

'7 {('.#)"} (nrrrrr: Use lim (1 + x)rtt : e.)

See Hint for Exercise L7.

19. {rrr"y and r } 0. (HrNr: Consider two cases:r < 1 and r } l)

16.2 MONOTONICAND BOUNDEDSEQUENCES

20. Prove Theorem 16.1.3. 21. Prove that if lrl < 1, the sequence {nt'I is convergent artd nr" converges to zero. 22. Prove that if the sequence {a,} is convergent and lim dn:L, then the sequence {la"l} is also convergent and lim lo"l- lLl. fl-

*6

then the sequence {a"'}

23. Prove that if the sequence {a"} is convergent and lim an tL-*@ an2: L2' In

is also convergent and

24. Prove that if the sequence{ao} converges,then lim ao is unique. (nrrt: Assume that lim an has two different values, L and,M, and show that this is impossible by taking e:ilL - Ml in Definition 76.1.2') fl-*@

16.2 MONOTONIC AND We shall be concerned with certain kinds of sequenceswhich are given BOUNDED SEQUENCES sPecialnames. 15.2.LDefinition

A sequence{a"} is said to be (i) increasingif. an = an*t for all n; (i1) decreasing7f ao 2 an+rfor all n. If a sequenceis increasing or if it is decreasing, it is called monotonic. Note that if. an< aa+r(a special caseof an< anal),we have astictly lN€ have a strictly decreasingsequence. increasingsequence; if. an > anq11

ExAMPLEL: For each of the following sequences determine if it is increasing, decreasing, or not monotonic:

(a) {nlQn + 1)} (b) {Un} (c) t (- L)"*'ln\

solurroN:

(a) The elements of the sequence can be written

12 3 4

n

3'E'r'9''"761'

n+!

zn+g''''

Note that no*, is obtained from an by replacing n by n * l. Therefore, becausa eo:nl(2n*L), n+l an+r:ffiffi:m

n+l

Looking at the first four elements of the sequence,we see that the elements increase as n incteases. Thus, we suspect in general that

n n+l zn+r= h+g

(1)

Inequality (1) can be verified if we find an equivalent inequality which we know is valid. Multiplying each member of inequality (1) by (2n + 1) (2n + 3), w€ obtain n ( Z n+ 3 )

(2)

668


Inequality (2) is equivalent to inequality (1) because (1) may be obtained from (2) by dividing each member by (2n + t)(Zn * 3). Inlquatity (2) is equivalent to 2n2*3n=2n2*3n+L

(3)

Inequality (3) obviously holds becausethe right member is L greater than the left member. Therefore, inequality (1) holds and so the given sequence is increasing. (b) The elements of the sequencecan be written

for all n, the sequenceis decreasing. (c) The elements of the sequencecan be written

-( ;- 1, )R' ' +,r. . .( - 1 ) t t + 2

, , - 21'1i , - 4 '1

,

ar:l and az:-*, and so ar) az.But ar:t, and so az< as.In a more generalsense,consider three consecutiveelementsan: (-l)n+rln, an+r: (-1)"*'l@ * 1), and anrz: FL)"*"|(n+2).If. n is odd, an) an*, and an+r1 aaa2,?trd, if. n is evett, an 1 an*, attd anal) ar+z,Hence, the sequence is neither increasing nor decreasing and so is not monotonic.

16.2.2 Definition

The number C is called a lower bound of the sequence {a"} if. C < an for all positive integers n, and.the number D is called arr upper boundof the sequence {a"} if. an = D for all positive integers n. o ILLusrRATrouL: The number zero is a lower bound of the sequence {nl(2n * 1)} whose elementsare 12

3 4

g's'7't"

"

r f f inr

Another lower bound of this sequence is *. Actually any number which is less than or equal to $ is a lower bound of this sequence. . O ILLUSTRATION

"' ,L ' t

1

L 'g ' 4 "

For the sequence {1,1n} whose elements are

'L " A"

1 is an upper bound; 26 is also an upper bound. Any number which is greater than or equal to f. is an upper bound of this sequence,and any nonpositive number will serve as a lower bound. o

16.2 MONOTONICAND BOUNDEDSEQUENCES 669

From Illustrations 1 and 2, we see that a sequence may have many upper and lower bounds. 16.2.3 Definition

If A is a lower bound of a sequence {an} and if A has the property that for every lower bound Cof {a"}, C - A, then A is called the greatestlower bound of the sequence. Similarly, if B is an upper bound of a sequence {an} and if B has the property that for every uPPer bound D of.{an}, B = D , then B is called the least upper boundof the sequence. . rLLUsrRArroN3: For the sequence {nl(2n + 1)} of Illustration 1, the greatest lower bound is * because every lower bound of the sequenceis less than or equal to *. Furthermore,

2n*L

z+!

.t

for all n, and * is the least upper bound of the sequence. In Illustration 2, we have the sequence{Uz} whose leastupper bound is 1 because every upper bound of the sequenceis greater than or equal o to 1. The greatest lower bound of this sequenceis 0. 1,6.2.4Definition

A sequence{a"} is said to beboundedif and only if it has an upper bound and a lower bound. Becausethe sequence {1,1n}has an upper bound and a lower bound, it is bounded. This sequence is also a decreasing sequenceand hence is a bounded monotonic sequence.There is a theorem (1,6.2.6)that guarantees that a bounded monotonic sequence is convergent. In particular, the sequence {Un } is convergent because lim (1/n) :0. The sequence {n} whose elements are L,

n,

monotonic (because it is increasing) but is not bounded (because there no upper bound). It is not convergent because lim n - f o.

For the proof of Theorem 16.2.6we need a very important property of the real-number svstem which we now state. 16.2.5 The Axiom of Every nonempty set of real numbers which has a lower bound has a greatCompleteness est lower bound. Also, every set of real numbers which has an upper bound has a least upper bound. The axiom of completenesstogether with the field axioms (1.1.5 through 1.1.11)and the axiom of order (1.1.14)completely describe the real-number system. Actually, the second sentence in our statement of the axiom of completenessis unnecessarybecauseit can be proved from


Ar flz

As da

fl5&5Tz

Aa As

flto

D

F i g u r e1 6 . 2 . 1

16.2.5Theorem

the first sentence. It is included in the axiom here to expedite the discussion. Suppose that {ar} is an increasing sequence that is bounded. Let D be an upper bound of the sequence.Then if the points corresponding to successive elements of the sequence are plotted on a horizontal axis, these points will all lie to the left of the point corresponding to the number D. Furthermore, because the sequence is increasing, each point will be either to the right of or coincide with the preceding point. See Fig. 1,6.2.1,. Hence, as n increases,the elements increase toward D. Intuitively it appears that the sequence {a"} has a limit which is either D or some number less than D. This is indeed the case and is proved in the following theorem. A bounded monotonic sequenceis convergent. PRooF: We prove the theorem for the casewhen the monotonic sequence is increasing. Let the sequencebe {a"}. Because{a"} is bounded, there is an upper bound for the sequence. By the axiom of completeness(15.2.5), {a,} has a least upper bound, which we call B. Then if e is a positive number, B - e cannot be an upper bound of the sequencebecauseB - e ( B and B is the least upper bound of the sequence.So for some positive integer N B- € < a1s

(4)

BecauseB is the least upper bound of {a"}, by Definition 1,6.2.2it follows that an N

(5)

From (4),(5), and (6), it follows that B - € < ay from which we get B- € < an

whenever n > N

or, equivalently, -€ < fln- B which can be written as

l a "- B l

(7)

16,2 MONOTONICAND BOUNDEDSEQUENCES

But by Definition 1.6.1.2,(7) is the condition that lim an:B. Therefore, the sequence {a"} is convergent. To prove the theorem when {a"} is a decreasing sequence, we consider the sequence {- a"}, which will be increasing, and apply the above results. We leave it as an exercise to fill in the steps (see Exercise 13). t '!,5.2.6 states that tf {a"} is a bounded monotonic sequence, Theorem there exists a number L such that lim an: L, but it does not state how fl-*@

to find L. For this reason, Theorem 16.2.6 is called an existencetheorem. Many important concepts in mathematics are based on existence theorems. In particular, there are many sequences for which we cannot find the limit by direct use of the definition or by using limit theorems, but the knowledge that such a limit exists can be of great value to a mathematician. In the proof of Theorem 76.2.6we saw that the limit of the bounded increasing sequenceris the least upper bound B of the sequence.Hence, if D is an upper bound of the sequence, lim an: I < D. We have, then, the following theorem. 15.2.7 Theorem

Let {aJ be an increasing sequence,and suppose that D is an upper bound of this sequence.Then {a"} is convergent and lim an
In proving Theorem'1.5.2.5fot the casewhen the bounded monotonic sequence is decreasing, the limit of the sequence is the greatest lower bound. The following theorem follows in a way similar to that of Theorem'16.2.7. 16.2.8 Theorem

Let {a"} be a decreasing sequence,and suppose that C is a lower bound of this sequence.Then {an} is convergent and lim an> C ll-*@

EXAMPLE2: Use Theorem "1,5.2.6 solurroN: to prove that the sequence

The elements of the given sequence are

21 22 23 24

l2)

WJ is convergent.

u'21 3!'4!' :1 . 2.3:6,4!:L.2.3. lt:L, 2l:1.2:2,3! elements of the sequencecan be written as ^ ^' z4' 2j ' 2 g'''

"

4 : 2 4 . H e n c e t, h e

2 2n+r il.' (!r+l)t''''

We see, then, that ar:

az) a"t

an, and so the given sequencemay be


decreasing.we must check to seeif.an 2 an+i that is, we must determine if 2n = -

n!

2n*'

(8)

(n+I)I

which is equivalent to 2(n+1)! =2n+rnl which is equivalent to 2 n n \ ( n + L )> 2 . Z n n ! which is equivalent to n+"1,>2

(e)

When n:1, inequality (9) becomes2:2, and (9) obviously holds when n ) 2. Because inequality (8) is equivalent to (9), it follows that the given sequenceis decreasing and hence monotonic. An upper bound for the given sequence is 2, artd.a lower bound is 0. Therefore, the sequence is bounded. The sequence {2ln!} is therefore a bounded monotonic sequence, and by Theorem 1,6.2.6it is convergent. Theorem 16.2.5states that a sufficient condition for a monotonic sequence to be convergent is that it be bounded. This is also a necessary condition and is given in the following theorem. 15.2.9 Theorem

A convergent monotonic sequenceis bounded. pRooF: We prove the theorem for the casewhen the monotonic sequence is increasing.Let the sequencebe {an}. To prove that {a"} is bounded we must show that it has a lower bound and an upper bound. Because{ar} is an increasing sequence,its first element serves as a lower bound. We must now find an upper bound. Because{an} is convergent, the sequencehas a limi! call this limit L. Therefore, lim an: L, and so by Definition 76.1.2,for any e > 0 there exists a number N > 0 such that

l a "- Ll

N

or, equivalently, -€<

an-L
whenevetn >N

or, equivalently, L-€<

anlL*e

whenevern>N

Because {an} is increasing, w€ conclude that an

16.3 INFINITESERIESOF CONSTANTTERMS

Therefore, L * e will serve as an upper bound of the sequence {ar}. To prove the theorem when {a,,} is a decreasing sequencewe do as '1.6.2.6:. Consider the sequence {-a,'}, suggested in the proof of Theorem which will be increasing, and apply the above results. You are asked to do this proof in Exercise L4.

16.2 Exercises L through In Exercises

12, determine if the given sequence is increasing, decreasing, or not monotonic.

,4' [ 3 n - I l

3.

2. {sin nn}

lan * 5)

z^ [' 2 " I lt+zJ

6.

7 ' . ["tI L3'J

9. L2.

10. {n' * (-l)"nI

nn

n!

r.3.5

13. Use the fact that Theorem 15.2.5holds for an increasing sequenceto prove that the theorem holds when {co} is a decreasing sequence.(nrrt: Consider the sequence {-a"1.) 14. ProvglTheorem 15.2.9when {an} is a decreasing sequenceby a method similar to that used in Exercise 13. In Exercises15 through 21, prove that the given sequenceis convergent by using Theorem76.2.6, 15. The sequenceof Exercise 1.

16. The sequenceof Exercise4'

17. The sequenceof Exercise 5.

18. The sequenceof Exercise8.

19. The sequenceof Exercise 11.

20. The sequenceof Exercise 12'

,1 ' - ' 1ttl 2)

16.3 INFINITE SERIES OF CONSTANT TERMS

L5.3.L Definition

The familiar operation of addition applies only to a finite set of numbers. We now wish to extend addition to infinitely many numbers and to define what we mean by such a sum. To carry this out we deal with a limiting process by considering sequences. If t un\ is a sequence and n

sn:

). ur:

L l r* u , * u g *

' ' '*lln

i:l

then the sequence{s"} is called an infinite series. Thenumbetsu1,u2rxs, . . . arecalled thetermsof theinfiniteseries. We use the following symbolism to denote an infinite series:

674


)

un: ut* uz* us* . . ' + un* . . .

(1)

Given the infinite series denoted by (1), sr: Llr, sz: tt, * ttr, ss: ur * uz * ^nd in general :r,

s k:

Ltr* ur* us* . . . * urc

Z"r:

(2)

where sa, defined by (2),is called the kth partial sum of the given series, and the sequence {sr} is a sequence of partial sums. . . + iln_t Because sn-t: Llr* u, * . . . + lln_r and sn: Ltr* u r * . * ,r, we have the formula Sn:

EXAMPLE

Given the infinite

Sn-t *

sot,urroN: Applying formula (3), we get

series

f ":,z#n find the first four elements of the sequence of partial sums {sn]t, and find a formula for sn in terms of n.

(3)

Un

St: ut:

32:

Srt

L1 L 2:,

,L12

U2:;f

z

2'3

r

3'4

,213

SB:SztUs:;t

:-

:-

3 4

314 S + : S s* L l 4 : ; + f t : ; By partial fractions,we see that 1 1 wrk, _- k ( k + 1 ) - k_ t _

1

k+L

Therefore, .1 Ltr: 1 -

i.,

ur:

11 1,- 4 1, ' Us: ;- ;, 3

n*l

: Llr * u, *+ . . . + Lln-r* un,we have Thus, because s?z

,.:('-il*(,t-*)*(}-r,). .(-,-,t).e-h) Upon removing parenthesesand combining terms, we obtain -1,n sn:l-n1-1:rt7 Taking n:

1

l, 2, 3, and 4, we see that our previous results agree.

Note that the method of solution of the above example applies only

16.3 INFINITESERIESOF CONSTANTTERMS

to a special case. In general, it is not possible to obtain such an expression forsr. We now define what we mean by the "sum" of an infinite series. 16.3.2 Definition

f"t j

unbe agiven infinite series, and let {s,} be the sequenceof partial

n=l

t" exists and is equal to S, ,lit we say that the given series is conuergentand that S is the sum of. the sn does not exist, the series is said to be digiven infinite series. If ,lin oergentand the series does not have a sum. sums defining this infinite series. Then if

Essentially Definition 16.3.2statesthat an infinite seriesis convergent if and only if the corresponding sequenceof partial sums is convergent. If an infinite series has a sum S, we also say that the series converges to S. Observe that the sum of a convergent series is the limit of a sequence of partial sums and is not obtained by ordinary addition. For a convergent series we use the symbolism +@

2o'. n=t to denote both the series and the sum of the series. The use of the same symbol should not be confusing because the correct interpretation is apparent from the context in which it is used. EXAMpLE2: Determine if the infinite series of Example L has a sum.

SOLUTION: IN thc

solution of Example L, we showed that the sequence

of partial sums for the given seriesis {s"} : {nl (n * 1)}. Therefore, lim sn: n-

*@

lim ll-

*@

nr.L L

::uur:r

ni

h_.re 1+l

n

So we conclude that the infinite series has a sum equal to 1..We can write

. . . +_+:* ' s_L:1+!+L+l+ ' n(n-tL) 2' 6 12' 20

f_,n(n*l)

. :1

As we mentioned above, in most cases it is not possible to obtain an expression for sn in terms of. n, and so we must have other methods for determining whether or not a given infinite serieshas a sum or, equivalently, whether a given infinite series is convergent or divergent. 16.3.3Theorem If the infinite series !

a, is convergent,then lim un:0.

676


PRooF: Letting {s,} be the sequenceof partial sums for the given series, and denoting the sum of the series by S, we have, from Definition 16.3.2, Therefore,for any € >0 there exists a number N>0 such "lit*:S. that lS snl ( *e for every integer n > N. Also, for these integers n > N we know that lS - sn+rl< te. We have, then, lun*rl: ls,*,

- s,l : lS sn*sn*l Sl s lS- r,l * ls"*r- Sl

So

l ' n * r l< t e * l e Therefore, lim ur:

e

for every integer n > N

0.

I

The following theorem is a corollary of the preceding one.

1.6.9.4Theorem

If

lln

unLsdivergent.

un * 0, then the series 5

n:l

PRooF: Assume that i

un is convergent. Then by Theorem 16.3.3

n:l

lim un:0.

But this contradicts the hypothesis. Therefore, the series is

lI-*@

I

divergent.

EXAMPLE3: Prove that the followitrg two series are divergent.

(a)yry -z+Z+# *#* +@

(b) ':t3-3+3-3+

soLUrIoN:

(a) lim un: tt- t@

lim It- *@

n2 +'l' nz

_ lim n-*@

+0 Therefore,by Theorem15.3.4the seriesis divergent. (-1)n*t3, which does not exist. Therefore,by (b) JiT" "": l*_ Theorem15.3.4the seriesis divergent. Note that the converseof TheoremL6.3.3is false.That is, if

un: 0, olim it does not follow that the series is necessarilyconvergent. In other words, it is possible to have a divergent series for which lim zn: 0. An example of such a series is the one known as the lror*onit'rln1es, which is (4)

16,3 INFINITESERIESOF CONSTANTTERMS

gI7

0. In Illustration 1 we prove that the harmonic series ):*tl": diverges and we use the following theorem, which states that the difference between two partial surls sa and s7of a convergent series can be made as small as we please by taking R and T sufficiently large. Clearly,

15.3.5 Theorem

Let {sn} be the sequence of partial sums for a given convergent series +@

2 u". Then for any € > 0 there exists a number N such that lsa-s7l< e

w h e n e v e r R> N a n d T > N

pRooF: Because the series i

unis convergent, call its sum S. Then for

any e ) 0 there exists an N = O such that lS - rrl < te whenevet n > N.

Therefore, if R > N and T > N, l s r - r . l : l s r - S + S - s r l < l s r - S l+ l S - s r l NandT>N

. rLLUsrRArroN L: We prove that the harmonic series (4) is divergent. For this series, sn:L+|+.. and szn:l +

|+

.

So szn- sr:

L,L,L

"*l-

r+Z-

(s)

n+3-

lfn

11,1L1,111 n+lntrr- n+3-

' T

2n>mr

hTmr

' ' ' Tm

(oi

There are n terms on each side of the inequality sign in (6); so the right side Lsn(ULn) : *. Therefore,from (5) and (5) we have szn-snlt

whenevetn

(7)

But Theorem 1.6.3.5states that if the given series is convergent, then szn- sn may be made as small as we please by taking n large enough; that is, if we take e - *, there exists an N such that sro- sn 1* whenever 2n > N and n ) N. But this would contradict (7). Therefore, we conclude that the harmonic series is divergent even though lim Lln:0. ll-*@

678

INFINITE SERIES

A geometricseriesrs a series of the form +co

'

.LJ n:L

orn-r- a* ar* arz*.,.+arn-L*

(8)

The nth partial sum of this series is given by sn:a(l

*r*r2*

' ''+r"-t)

(e)

From the identity 1 - vn:

(1 - r) (1 * r * r2 I

. . . + r"-t)

we can write (9) as a(r-/)_ sn: _1 _ r

16.3.6 Theorem

ifr#L

The geometric series converges to the geometric series diverges if lrl

(10)

sum al(1 - r) if lrl

PRooF: In Example 3, Sec. 1,6.1,, w€ showed that lim r' : 0 if lrl Therefore, from (10) we can conclude that if lrl ,.4

,t]1t':-1-, So if lrl ( L, the geometricseriesconverges,and its sum is al(1.- r). lf. r:1, sn: na.Then lim s, : fo if.a > 0, and lim s,r: -@ if a < 0. n- *@ lf. r:-l,the geometricseriesbecomesa- a * a - . . - * (-Ll"-to +' ' ' ; so sz:0if ziseven,andsr:aif.n isodd.Therefore, rll1t does not exist. Hence, the geometric series diverges when lrl : 1. If lrl > 1,,li1 ay't-t:, Clearly, rn-t * 0 becausewe )!T_r"-1. ,li1 can make lro-11as large as we pleaseby taking nlarge enough. Therefore, I by Theorem 16.3.4 the series is divergent. This completes the proof. h-*@

* +++1** which is the geometric series, with a : 1 and r: !. Therefore, by Theorem 15.3.5the series is convergent. Becauseal(L - r) : Ll(l - L) :2, . the sum of the series is 2. The following example illustrates how Theorem 15.3.6 can be used to expressa nonterminating repeating decimal as a common fraction.

16.3INFINITE S E R I E SO F C O N S T A N TT E R M S

EXAMPw 4: Express the decimal 0.3333. . as a common fraction.

SOLUTION:

3333 , 10 100 L,000 10,000 We have a geometricseriesin which a: fo and r : $. Becauselrl < 1,,It follows from Theorem 1.6.3.6that the series converges and its sum is al (L - r). Therefore,

The next theorem states that the convergence or divergence of an infinite series is not affected by changing a finite number of terms. 1,6.3.7Theorem

two infinite series, differing only in their ffust m

If

,.;,

(i.e. , oi! rrif k

series diverge. PRooF: Let {sr} and {t"} be the sequencesof partial sums of the series +@

+@

n:l

n:1

Then Sn:

At*

Az*

*

fl* *

a**,

*

*an

A*'r2 *

and ' ' ' + b**b**t+b*+z+

tn:bt*br*

' '

*bn

Becausl ak: brcrf k > m, then rf n sn- tn: @r* az* . . . + a*)

(br*br*

' ' +b*)

So Sn-

tn:

t*

S*-

-

whenevet n

We wish to show that either both lim sz and jlt

(11)

f, existor do not

exist. Supposethat lim fn exists.Then from Eq. (11)we have ll-*@

sn: tn* (s*- t*)

whenever n > m

and so lim sn: n-*@

lim tn * (s* - t*) n-*@

Hence, when lim tn exists, lim sn also exists and both series converge. ll-

*@

fl,-*@

SERIES INFINITE

Now suppose that lim f, does not exist and lim s, exists. From Eq. (11.)we have tn: sn* G--

s^)

whenevern > m

Because,lit * exists,it follows that lim fn: lim sn* (t^- s*) n-+6

n-+@

fo does to exist, which is a contradiction.Hence,if ,111 )!I_t"has I not exist, then lim s, does not exist, and both series diverge. and so

glven serles ls

EXAMPLE5: Determine whether the infinite series

s1 n ? : r f l+ 4 is convergent or divergent.

which can be written as

o+o+o+o+1*!*1 * 5'6'7'

+ !n+

(12)

Now the harmonic series which is known to be divergent is

1+l+|+I+f+l+|+

+ !n+

(13)

Series (L2) differs from series (13) only in the first four terms. Hence,by '1,6.3.7, Theorem series (12)is also divergent.

EXAMPLE6: Determine whether the following infinite series is convergent or divergent:

solurroN:

The given series can be written as

ry*%*W*%

+@ n:l

_ - 31 L' 2 r gr'

L r t + 1 L2 L2 -2 _' ' ' ' ' 3t gn' 3t 3u gr' 38

Consider the geometric series with a - 3 and 2,2,2,2,2,2,2,2,

5- y' F- y' F-F*

y- F- ' '

(r4) l. - _ r-3.

(15)

This series is convergent by Theorem 16.3.6.Becauseseries (L4) differs from series (15) only in the first five terms, it follows from Theorem16.3.7 that series (14) is also convergent.

1 6 . 3 I N F I N I T ES E R I E SO F C O N S T A N TT E R M S

681

As a consequenceof Theorem 1,6.3.7,for a given infinite series, we can add or subtract a finite number of terms without affecting its convergence or divergence. For instance, in Example 5 the given series may be thought of as being obtained from the harmonic series by subtracting the first four terms. And because the harmonic series is divergent, the given series is divergent. In Example 6, we could consider the convergent geometric series )))

(15)

L t L ) L I

36'37'38r

and obtain the given series (1a) by adding five terms. Becauseseries (L6) is convergent, it follows that series (14) is convergent. The following theorem states that if an infinite series is multiplied term by term by a nonzero constant, its convergence or divergence is not affected. 16.3.8 Theorem

Let c be any nonzero constant. +@

(i) If the series n- - l

+m

(ii) li'.nu seriesX Ltnisdivergent,then the seriesi cu,is also rt:r n:r divergent.

F@

be sn. Therefore, 7rr" ' * un. The nth partial sum of the series 5 cun is

PRooF: Let the ntt:- partial sum of the series ' sn:ur*uz* ' ' ' * u) -- csn. c(urlur* +co

nenr (i): If the series Therefore,

h'*a

ft:l

lim csn- c l i m s n : c ' S n- *@

lL-l@

+@

Hence, the series 2 ,un is convergent and its sum is c ' S. n=l

ranr (ii):

If the series )

z, is divergent, then lim s, does not exist' Now n-+@

n:l +@

suppose that the series n:I sn: csrlc, and so lim sn:

lim

1 L ( c s " )- - ; c ):\""

ll'*@


Thus, lim sn must

which is a contradiction. Therefore, the series

fl-t'a

+@

I n:1

ExAMPLE7: Determine whether the infinite series

$1

solurroN'h:l+$+ ,!r+ft+ ,i Because 2 tl" is the harmonic series which is divergent, then by The-

4n

"4r is convergent or divergent.

orem 1,5.3.8(ii)with c - *, the given series is divergent.

Note that Theorem 15.3.8(i) is an extension to convergent infinite series of the following propefiy of finite sums: nn

2 , o * : , 2k : l o o

k:l

Another property of finite sums is nnn k:l

k:l

k:7

and its extension to convergent infinite series is given by the following theorem. +@

L5.3.9 Theorem

If )

n:l

+co

a" and )

bn are convergent infinite series whose sums are S and R,

respectively,then {m

(i)

is a convergentseriesand its sum is S * R; \::

(ii) @,-bn)is a convergentseriesand its sum is S u

R.

The proof of this theorem is left as an exercise(seeExercise11). The next theorem is a corollary of the above theorem and is sometimes used to prove that a series is divergent.

L6.3.L0 Theorem

If the series 5 an is convergentand the series n:l

the series X @, * b,) is divergent. n:l

+@ n:l

1 6 . 3 I N F I N I T ES E R I E SO F C O N S T A N TT E R M S

pRooF: Assume that i

683

@** b,) is convergentand its sum is S. Let the

n=l

sum of the series )

a"be R. Then because

n=l

+@

+6 A " : 2 ) l( a"* b") a"l n:l n=7

+@

bo is convergent and its sum is

it follows from Theorem 16.3.9(ii)that \ n=l

+m

S - R. But this is a contradiction to the hypothesis that )

b, is divergent.

n-l

+@

Hence

I

+m

8: Determine whether EXAMPTE the infinite series

s(t

1\

A\n*n)

solurroN:

Ll4n is divergent.

In Example 7 we Proved that the series n:l

Because the series i

U4' is a geometric series with

lrl -

+ < 1.,it is con-

vergent. Hence, by Theorem 16.3.10the given series is divergent.


Note that if both series5 an and i i

b, are divergent,the series

!e, tf an- Un and (ant bn)mayor may not ol'.or,""rr;r1 For examp

T,t: Lln,then a, * bn: 2lnand b n : - U n , t h e n a nl b n : 0

divergent.But tf a, : Lln and

2rtris

+m

and

L6.3 Exercises In Exercises1 through 6, find the first four elements of the sequenceof partial sums {s,} and find a formula for snin terms of n; also,determine if the infinite series is convergent or divergent, and if it is convergent, find its sum' r'

o 5t"h

a S l

?:, (2n 1) (2n + 1)

6'>,7 +oo

In Exercises 7 through 10, find the infinite series which is the given r::;:"ce nite series is convergent or divergent, and if it is convergent, find its

8.{s,}:{h}

,n_l

of partial sums; also determine if the infi-

s. {s,} : {1}

lz)

10. {s,} - {3n}

684

I N F I N T ES E R I E S

11. Prove Theorem 15.3.9. Iir Exercises12 through 25, write the first four terms of the given infinite series and determine if the series is convergent or divergent. If the series is convergent, find its sum.

12. L L ' $ --l . i !--rn+

-,?'

t?

17. (-1)"*'* ,;, "z .f !

22.$

i!-t

sinhn n

+@r+@

2tz+ 1 1? Lv' $ A3n+2

L4.s+ A3n

18.$ r, n1

+00

l r t5 . ! 1

L6.

?:,

n=l +@

fm+co

20. .L/ '

L9. Z.J S e-n

zs.i (2-n+3-,)

sln rrn

n:l

n:l

"?t

2 3n-r

+@

21,.

cos nTn n:l

+@

24. n:l

In:l

In Exercises25 through 29, express the given nonterminating repeating decimal as a common fraction. 26. 0.2727 27 . . .

27. 1.234234234 . . .

28. 2.04545 45 . . .

29. 0.4653 46534553 . . .

E\- A ball is dropped from a height of.12 ft. Each time it strikes the ground, it bounces back to a height of three-fourths the distance from which it fell. Find the total distance traveled by the ball before it comes to rest.

16.4 INFINITE SERIES If all the terms of an infinite series are positive, the sequence of partial OF POSITM TERMS sums is increasing. Thus, the following theorem follows immediately from Theorems15.2.5and'1,6.2.9. 16.4.1 Theorem

An infinite series of positive terms is convergent if and only if its sequence of partial sums has an upper bound. pRooF: For an infinite series of positive terms, the sequenceof partial sums has a lower bound of 0. If the sequenceof partial sums also has an upper bound, it is bounded. Furthermore, the sequenceof partial sums of an infinite series of positive terms is increasing. It follows then from '1,6.2.6 Theorem that the sequence of partial sums is convergent, and therefore the infinite series is convergent. Supposenow that an infinite series of positive terms is convergent. Then the sequence of partial sums is also convergent. It follows from 'l'6.2.9 Theorem that the sequenceof partial sums is bounded and so it has an upper bound. r

EXAMPLE 1: Prove that the series \{1 n! -,.4:, is convergent by using Theorem 15.4.'1,,

SOLUTION:

WC

MUSI

find an upper bound for the sequence of partial

-f@

sums of the series LIS U n ! .

s r : ! , s z -t + # ,

16.4 INFINITESERIESOF POSITIVE TERMS

sn:r+h+#+ffi+ Now consider the first n and r-t:

*' 1 L'2 '3'

. . . 'n

geometric series with A : " 1 ,

1

$ l-

(1)

1 + ; * 8-.l - * 1 * 2r* #_r2*4-

(2)

By Theorem 16.3.6 the geometric series with a : 1 and 1: $ has the sum a/(1 - r) : U G - t) :2. Hence, sumrnation (2) is less than 2. Observe that each term of summation (1) is less than or equal to the corresponding term of summation (2); that is, 11 kI zk-r This is true becausek! - | ' 2 ' tor L contains k- 1factors each

' k, which in addition to the facthan or equ al to 2. Hence,

n Zr

From the above we see that sn has an upper bound of 2. Therefore, by '1.6.4."1. Theorem the given series is convergent.

In the above example the tenrrs of the given series were compared with those of a known convergent series. This is a particular caseof the following theorem known as the comparisontest. 16.4.2 Theorem Comparison Test

Let the series5 un be a series of positive tenns. n:l

+o

(i) rf n:l

+m

vergent,andllnSanforallpositiveintegersn,then> n:r convergent. {o

(ii) If

is a series of positive terms which is known to be diA*" +@ vergent, and un 2 vonfor all positive integers n, then n:l divergent.

PRooF oF (i): +@

Let {sr} be the sequence of partial sums for the series

sums for the series 5 T)n.Beh:t

I N F I N I T ES E R I E S +@

cause

a, is a series of positive terms which is convergent, it follows ,) from Theorem 15.4.1that the sequence {t } has an upper bound; call it B. Becauseun < onfor all positive integers n, we can conclude that sn - tn < B for all positive integers n. Therefore, B is an upper bound of the sequence {sr}. And becausethe terms of the series !

un areall positive, it follows

n:l

*oo

from Theorem 16.4.7that

convergent. n:l

PRooFoF (ii): Assume that i + co

+@

n:l

n:1

r,tnis convergent. Then because both

n:l

of positive terms and znn

positive integers n, it follows from part (i) that i ,,is convergent. However, this contradicts the hypothesis, and assumption is false. "J-irr" Therefore,

I

As we stated in Sec. '1.6.3,as a result of Theorem 76.3.7, the convergence or divergence of an infinite series is not affected by discarding a finite number of terms. Therefore, when applying the comparison test, if ui s ut or tti 2 rDi when i > m, the test is valid regardless of how the first m tenrts of the two series compare.

ExAMPLE2: Determine whether the infinite series \'4 n?:r3" + 1 is convergent or divergent.

soturroN: The given serles rs a

a

4,4,4,4 n+ l,o 2g- g2+ comparing the nth term of this series with the nth term of the convergent geometric series 4+4+4 +4 3'.9',27',9L'

L

r-*.

1

we have 44 -3n 3n+1 for every positive integer z. Therefore, by the comparison test, Theorem 1,6.4.2(i),the given series is convergent.

1 6 . 4 I N F I N I T ES E R I E SO F P O S I T I V E TERMS

ExAMPLE 3: Determine whether the infinite series

sol,urroN:

The given series is

L+ t:+

L

$-L

rtr

AW,:4+V*-.6*

+@1

s^: r.?:t \/n


687

1

Yn

Comparing the nt}:.term of this series with the nth term of the divergent harmonic series,we have

+YLnLn, +@

So by Theorem \6.4.2(ii) the given series l:I

The following theorem, known as the limit comparisontest,is a conand is often easier to aPply. sequenceof Theorem "1.6.4.2 +@

15.4.3Theorem Let )

Limit ComparisonTest

n=r

un and \

a"be two seriesof positive terms.

n=r

(i) If lim (unla*): c ) 0, then the two serieseitherboth converge fl-*'@

or both diverge. (ii) If lim (unla) - 0, and if

+@

+@

n:l

n:l

ll-tcp

un converges.

un converges, then )

+oo

+oo

(iii) If lim (unla) - f m, and if n:l

fl-*@

an diverges, then )

un dlerges.

n:\

pRooFoF (i)t Because lim (u*lu) : c, it follows that there exists an N>0suchthat

ll a* n- l z d . *

f o ra r r n >N

or, equivalently, - t
un-c.* vn

z

foralln>N

or, equivalently,

c 'u"<+ z ,-;,

foralln>N

(3)

From the right-hand inequality (3) we get u, l Ecan

(4)

+6

lf \

+6

a" is convergent, so is \

{ca,. It follows from inequalif

(a) and

the comparison test that 5 un LSconvergent. From the left-hana ifi:"tquatity(3) we get

(s) +@

tt E

un rs convergent,so is

son test it follows that )

)

) ,n.From inequality (5) and the comparii

u, is convergent.

n=l

+@

It \

n:t

+@

a" is divergent, we can show that )

un is divergent by assum-

n=7

*-

ing that \ u^ is convergentand getting a contradictionby applying inequality (5) and the comparisontest. In a similar manner,tf

u,is divergent,it follows that j

oois di-

vergent becausea contradilltr, i, obtained from inequ"firy 17')and the +@

comparison test if )

u, is assumed to be convergent.

we have trru"JrL proved parr (i). The proofs of parts (ii) and (iii) are left as exercises(see Exercises19 and 20). I A word of caution is in order regarding part (ii) of rheorem 1,6.4.g. Note that when

nlim

(uJa,):0,

the divergenceof the series

f

,, do",

ll:l

not imply that the series 5 un diverges. ExAMPLE 4: Solve Example 2 by using the limit comparison test.

soLUrIoN: Let un be the zth tenn of the given series )

4/(3, + 1) and

n=l

c,nbe the nth term of the convergentgeometricseries +rc".Therefore, f n=t 4

" lim 3:

lim i-i*a" ;;;

en+L e

&

7n : lim 3" * L ,-+-

TERMS 16.4 INFINITESERIESOF POSITIVE

Hence, by part (i) of the limit comparison test, it follows that the given series is convergent.

+m

EXAMPLE5: Solve ExamPle 3 bY using the limit comParison test.

Let un be the nth term of the given series

sorurroN:

Ll \fr and anbe

n:l

the nth term of the divergent harmonic series. Then 1 lim n-a*

un:.lim 7)v

n-*@

V: I

hm \n:*m n-*cn

n Therefore, by part (iii) of the limit comparison test, we conclude that the given series is divergent.

+6

6: Determine whether EXAMPLE the series

In Example 1, we proved that the series )

solurroN:

Using the limit comparison test with un: rf lnt

$ng

: Unt',we have

ns

3'n! is convergent or divergent.

",,a5l

llnl is convergent.

tr

nl

1im :2 : lim T: n-|a

An

n-+o

I

lim ns : *a 't-*@

nl part (iii) of the limit comparison test is not applicable becaur" 5 o,,.or,verges. However, there is a way that we can use the limit "3:rrlpu"iror, test. The given series can be written as

l" *2"*3t*4 t * I'+ . ' ' 4 ! 5! T-t-3!

. *4*.' nt

. .

Because Theorem 15.3.7 allows us to subtract a finite number of temrs without affecting the behavior (convergence or divergence) of a series, we discard the first three terms and obtain

4'*5'*{* 4!-F-6!'

+' ( (n r* t3?) !) l + .

Now lettin E un: @ + 3)sl@ + 3) ! and, as before, letting an: llnl' we have u, r. Irm J-

n-1 * An

INFINITESERIES

(n+3)ln! : lim i-+- n!(n+ 1)(n + 2) (z + 3) : rim - @*-3)2 . il-i:-@+L)(n+2)

: , ',1. T n" 2F* 6+ng* 9n + 2 :rrl

r+9+Z n n'

n - +L- + t + 4

:l It follows from part (i) of the Limit comparison Test that the given series is convergent.

o rLLUsrRATron 1.: Consider the geometric series

1+tr+Ln*|* ft*$*. . .+f,*. . .

(5)

which convergesto 2 as shown in Illustration 2 of Sec.16.3.Regrouping the terms of this series, we have

+(#*+;+ (,.;)*(l*rJ.(#*#)* which is the series 3 9*9*3*. .r 2-E-92-r--'Tr4*rr.

- L . .. .

(7)

S"1": fi) is the geometricserieswith o: g andr: *. Thus,by Theorem 1,6.3.6 it is convergent,and its sum is a : I ,:" '1,r t- + We see,then, that series (7), which is obtained from the convergent series (6) by regrouping the terms, is also convergent. Its sum is the sameas that of series (5). . Illustration 1 gives a particular caseof the following theorem. +@

16.4.4 Theorem

If

is a given convergent series of positive terms, its terms can be )u" grouped in any manner, and the resulting series also will be convergent and will have the same sum as the given series.

16.4 INFINITESERIESOF POSITIVETERMS

pRooF: Let {s"} be the sequenceof partial sums for the given convergent series of positive terms. Then lim sn exists; let this limit be S. Consider +@

are obtained by grouping the terms of > ttn

a series

n:t

n:l

+oo

may be the series

in some manner. For examPle,

u r * ( u ,* u r ) + ( u n* u s* " " r +

(ur*us*ug*uro) + ' ' '

or it may be the series (ur*ur)*(ue*u)+(ur*u)+(uz*u")+'''

+@ and so forth. Let {t*} be the sequenceof partial sums for the series 2 o". Each partial sum of the sequence {f,,} is also a partial sum of a["t r"qrtettc" {sn}. Therefote, as m increaseswithout bound, so does n' Because I t, - S, *e conclude that lim t*: S' This proves the theorem' ,11 Theorem 1.6.4.4and,the next theorem state properties of the sum of a convergent series of positive terms that are similar to properties that hold for the sum of a finite number of terms' +@

15.4.5 Theorem

If

given convergent series of positive terms, the order of the

n:l

' terms can be rearranged, and the resulting series also will be convergent and will have the same sum as the given series.

pRooF: Let {sr} be the sequenceof partial sumsfor the given convergent seriesof positive tems, and let lim s,: S. Let i

o, U" a seriesforrred

n=1

+oo

by rearranging the order of the terms of.)

n:l

+@

un' For example, 2 ao may n:l

be the series

u+*us*uz*ut*ug*us*

' +@

Let {t"I be the sequenceof Partial sums for the series n:L sum of the sequence{t"}

T]l terms of the infinite series

O" less than S becauseit is the sum of.n Therefore, S is an uPPer bound of the

sequence {t"}. Furthermore, because all the terms of the series

n:l

positiV€, {t"} is a monotonic increasing sequence. Hence, by Theorem < S. Now because 16.2.7 the sequence {t"} is convergent, and lim tn: T

SERIES

j

the given ,"ri",

Ltn can be obtained from the series

)

ing the order of?u terms, we can use the same urg.r;:"lt that S and S that S - T. This proves the theorem.

anby rearrangand conclude I

A series which is often used in the comparison test is the one known as the p series,or the hyperharmonicseries.Itis

$+$+$+

where p is a constant

(8)

In the following illustration we prove that the p series diverges if p <1. and convergesit p > 1.. . rLLUsrRATrow2: rf p : 1., the p series is the harmonic series, which diverges. If p < 1.,then no - n, and so 11 > n" i

tor every positive integer z

Hence, by Theorem 1,6.4.2(ii)the p series is divergent if p < 1.. lI p > 1.,we group the terms as follows:

*#* . ++,. (e (h.+*h*+).(# +.(+*#)* Consider the series

1*Z+4+8+

(10)

Lp'2p,4o'gor

This is a geometric series whose ratio is 2l2o - tl2?-1, which is a positive number less than 1,.Hence, series (10) is convergent. Rewriting the terms of series (10),we get

1 , lr , 1\ , /1 *L+l*1\*/-1-*a*...

1'-\F-2,)-

\F-F-@-F/*\a"

8p

_, s 1o /\ -,

(11)

comparing series (9) and series (1r.), we see that the group of terms in each set of parentheses_ after the first group is less in Jum ior (9) than it is for (11). Therefore, by the comparison test, series (9) is convergent. Because(9) is merely a regrouping of the terms of the p rlri", whenp-> 1, we conclude from Theorem '1,6.4.4that the p series is convergent ifp>l,. . .1.; Note that the series in Example 3 is the p series where p: t < therefore, it is divergent.

TERMS 1 6 . 4 I N F I N I T ES E R I E SO F P O S I T I V E

7: Determine whether EXAMPLE the infinite series {o1 sr

?:, (n' + 2)rrs is convergent or divergent.

solurroN: Becausefor large values of n the number nz + 2 is closeto the numb et n2, so is the number U (n' + 2) 1/3closeto the numb er tlnzl3.The +@ because it is the p series with p -- 3 < L. series Using ;;" hmit comparison test with un: Ll(n' + 2)1/3 and an: Llnzrs we have

-T Therefore, the given seriesis divergent.

16.+ Exercises In ExercisesL through 18, determine if the given series is convergent or divergent. +oo 1 1 . Lst n )^n n:l

!S

n?:rt/2n + L

'"-

+@

1

rs

n-l +@

nz

,?:r4nt + L

,?:r!nz + 4n

+oo

+@1

nl

7 " .tfr(n+Z)l

6. +@

f-, ln3 + 1.

n

n:l

j3 $1ryl ln

1 r^'1 $ 4 (zn)I f-, r\f

L

il4t . \7' 3_r+ \fr

13# r ( n + 2 ) ( n + 4 )

L7sg_ 2-n'+2

19. Prove Theorem "l'5.4.3(ii).

20. Prove Theorem t6.4.3(iii).

n:l

1 nn

lsinn I nz

n:l

9. n:l +@

+@

1_ 10. ).Lt sin -

n?:,

+@

3.

12. n:l +co

L5. n:2 .

t)

n 5n2*3 2n nl.

1 n\/nz - 1.

+co

n:l

3n -

cos t4

694

INFINITESERIES +@

+6

21. If > an and )

b, are two convergent series of positive terms, use the Limit Comparison Test to prove that the series

n=l *.n=t

is alsoconvergent' Zo"o" 22. Suppose/ is a function such that f(n) > 0 for n any positive integer.Furthermore,supposethatif p is any positive numbernlimnpf(n)existsandispositive.Provethattheseries) if p>landdivergentif.o

16.5 THE INTEGRAL TEST

The theorem known as the integral fesf makes use of the theory of improPer integrals to test an infinite seriesof positive terms for convergence.

16.5.1Theorem Let / be a function which is continuous, decreasing,and positive valued lntegralTestfor all x >'!.. Then the infinite series +@

2 f t " > : / ( l ) + f ( 2 ) + / ( 3+) . . . + f ( n ) +. . . is convergent if the improper integral f+*

I f(x) dx Jr exists, and it is divergent if the improper integral increases without bound. pRooF: If i is a positive integer, by the mean-value theorem for integrals (7.5.2)thereexists a number X such that i - 1 < X /( x) > f( i) and so from (1) we have

f ( i - D = Jfit- r ' f @ )d x > f ( i ) Therefore, if. n is any positive integer, nnfin

> Jit| f < * ld x >i:2\ f ( i ) 2 f U - 1 ) > 7=z Fz or, equivalently, n-l

fn

n

) f t ; l = ; f (JxI ) d x = ) / ( ; i:l) - / ( 1 ) i-l

e)

16.5 THE INTEGRALTEST

F i g u r e1 6 . 5 . 1

Figures 16.5.1.and 16.5.2show the geometric interpretation of the we have a sketch of the graph above discussion for n:5. In Fig. 1.6.5.'1., of a function / satisfying the hypothesis. The sum of the measures of, the areas of the shaded rectanglesis /(1) +f(2) +/(3) +f(4)+f(5), which is the left member of inequality (2) when n:6. Clearly, the sum of the measuresof the areasof these rectanglesis greater than the measure of the area given by the definite integral when n:5. In Fig. 15.5.2the sum of the measures of the areas of the shaded rectangles is f(2) + f (3) +f(4) +/(5) +f(6), which is the right member of the inequality (2) when n : 6. This sum is less than the value of the definite integral when n:6, If the given improper integral exists, let L be its value. Then

t

v

f(x)'a*< L

(3)

From the second and third members of the inequality (2) and from (3) we obtain

ifr,l = f:-). f, f@)dx= f(t) + L +00

Consider now the infinite series

(4)

f ( n ) . L e t t h e sequence of partial

n:l

F i g u r e1 6 . 5 . 2

sums of this series be {sn}, where sn: )

has an upper bound of f (L) + +m

that

/(i). From (4) we see that {s"}

L.H.r,..,:ty rneorem15.4.'1.we

conclude

f (n) is convergent.

n:l

Suppose that the given improper integral increaseswithout bound. From (2) we have n-l i:l

f (il

t,

f(x) dx

for all positive integers n. Thus,5 f (i) must also increasewithout bound as n +f m. Therefore, lim ,r= divergent.

lim iftL) :fm . i:l

ll-*e

Hence,t f@) n:\

1S I

L: Use the integral test solurroN: The p series is ) tlnp. The function /, defined by f (x) : Uxo, EXAMPLE to show that the p series diverges satisfies the hypothesis of Theorem 1.6.5.L. Thus, considering the Lf p < J.and convergestf P have proper integral, we

- lim lu dx b- *@

J, x'


If p - 1, the above integral gives

J:t

r" '],,:Jit tnb: *m

If p + L, the integral gives

h m bL:--p' - ,

lim #10:

b-*q I-PJt

b-*q

This limit is *- when p is -U (1 - p) if p integral test, it follows that the p series convergesfor p for p ExAMPLE2: Determine if the infinite series +6

solrrrroN: Letting f(x) : xe-r, we see that / is continuous, decreasing, and positive valued flor x > 1.;thus, the hypothesis of the integral test is satisfied. Using integration byparts, we have I xe-" dx : - s-" 1i + 1) + C.

Hence, n:l


f**

f1b

f : * l _ b + 1 ,*' ? e )1 I *'-' dx: l i m l - e - " ( r + 1')l r| - 11m eb JT Dr*o ;;;L

L

Becauselim (b + 1 ) - + m a n d l i m e b : *@, we can use L'H6pital's rule D- *oo

and obtain linm

b+1

lim

eb

b - *+ @

b-fm

Threrefo :ore, f * F"@

I Jr

xe

-& a.r2 AX:-

e

an(d sSO o l the given series is convergent. EXAMPLE

Determine if the

series +@

1,

S O) L (UTIO ITIO N: The function f defined by f (x) - ' / . , 1 ( r *1 ) @ i s c o n tir:tulous, un ous, decreasing, and positive valued for x > "1.;hence, the integral tes st can can be applied. f *+mm


dx

I u l

:

lim D- *m

:

lim D- *m

:*rc

We conclude that the given series

16.6INFINITE SEBIES OF POSITIVEAND NEGATIVETERMS

L6.5 Exercises In Exercises1 through 12, determine if the given series is convergent or divergent.

2.S I

1 '' . S b a n

3.

f,:rnlnn

3,

+6

+@

6.

5. \nze-"

4. \ne-"'

n=l

n=l

+@ ^|il'r

+@

8.\F+1

7.)cschn

n

e.

n=L

n:l

*-. /n+j\ 11. ) ln \T) n=l

+€ 10. ) sech2n n:L

13. Prove that the ,"ri"

t

1

*

>-ffiW

L4. Prove that the series )

12.

is convergent if andonlyif p > 1' is convergent if and only if p > 1.

t6ilh.,,--

15. If s;,is the kth partial sum of the harmonic series, prove that ln(k+1)
ifo < m < x
Integrate eaclr member of the inequality fuom m to m I l; let m take on successivelythe values 1, 2, add the results.) vE--- S 16.Use the result of ExerciseL5 to estimate thev sum

,n-Lrand

I : *5 0 *' *5 L *

#^rm

16.6 INFINITE SERIES OF In this section we consider infinite series having both positive and negaPOSITIVE AND tive terms. The first type of such a series which we discuss is one whose NEGATIVE TERMS terms are alternately positive and negative-an "altemating series." 15.6.1Definition

If an > 0 for all positive integers n, then the series +@

n:l

and the series +@

-) .

t r \

ur" 5"a

(-l)nan:

- ar*

fl2-

as* fla-

' ' ' + (-l)"an*

ahernatingseies.

The following theorem gives a test for the convergence of an alternating series.

INFINITESERIES

16.5.2Theorem If thenumb€tst\tu2tus, . . . ,un, . ..arealternatelypositiveandnegaTest 'ive, Alternating-series lun*rl < lrrl for all positive integers n, and lim un: 0, then the altemating r"ri", j

uo is convergent.

PRooF: *" that the odd-numbered terms of the given series are positive "rr;;" and the even-numbered terms are negative. This assumption is not a loss of generality because if this is not the case, then we consider the series whose first term is z, because discarding a finite number of terms does not affect the convergenceof the series.So urn-, ) 0 antdu2n( 0 for every positive integer r. Consider the partial sum szn: (u, * u) + (u, * u) +

+ (urn-, * urr)

(1)

The first term of each quantity in parentheses in (1) is positive and the second term is negative. Because by hypothesis lun*rl < lu"l, w€ conclude that each quantity in parentheses is positive. Therefore,

0 We also can write s2nds szn: ut *

(ur*ur) + (ua*u) +

+ (urn-, * urr-r) * urn

(3)

Becauselrn*rl so also is u2n.Therefore, Szn 1

l,lt


(4)

From (2) and (a) we have 0 < s2o< u,


!o the sequence {srn} is bounded. Furthermore, from (2) the sequence {sz,} is monotonic. Therefore, by Theorem 1,6.2.6the sequence{s2n}is convergent. Let szo: S, and from Theorem '1,6.2.7 we know that S = ur. ,lim Becauseszn+r: s2n]-u2n411 we have olim

s2a1:"li1

srn*

IT_urn*,

but,by hypothesis, lim u2o*r:0, and ro

szn+r:,lt

rrr.Therefore, "li1 the sequence of partial sums of the even-numbered terms and the sequence of partial sums of the odd-numbered terms have the same limit s. We now show that lim sn: S. Because lim szn: S, then for any

€ > 0 t h e r e e x i s tan s integer Nl > 0 such that whenever 2n > N1 ltrn- sl < € And becaur" szn+r: S, there exists an integer N2 > 0 such that Jlt whenever 2n * 1, > N, ltrn*r- sl < e

16.6 INFINITESERIESOF POSITIVEAND NEGATIVETERMS

699

If N is the larger of the two integers Nt and N2, it follows that if n is any integer, either odd or even, then lsr-Sl
whenevern>N +@

Therefore, lim sn: S, and so the series )

EXAMPLEL: Prove that the alternating series + o (_ 1 ) n + 1 srr Ltn n:l

r

The given series is

solurroN:

1-|+i- !+ BecauseU(n + 1.) < Un for all positive integers n, at:rdlim (Un): 0, it follows fromTheorem1,6.6.2that the given altemating seriesis convergent.

is convergent.

EXAMPLE

z, is convergent.

n:l

n-+@

Determine if the

series

)"ffi :l(-1 is convergent or divergent.

sor,urroN: The given series is an alternating series, with

un: (-1)"

n+3 a n d u n + r : ( - 1 ) n + l( n * I ) ( n + 2 )

ffi

L*2

n+) r. rrm

#:limry-o L) n-**

n-*@n\nf

t*;

Before we can apply the altemating-series test we must also show that lun*rl < lunl or, equivalently, lun*tlllunl < 1'.

l r n * r_l

l'"1

n+3 (n*1)(n+2)'

. n ( nn + * LL) ) - n ( n+ 3 ) n+2

(n*2)'

Then it follows from Theorem1,6.6.2that the given series is convergent.

15.G.3Definition

If an infinite series j

,, is convergent and its sum is S, then the re-

mainderobtainedUy l]tpro*imating the sum of the seriesby the kth partial sum s7,is denoted bY Ra and R*:S-s* 15.5.4Theorem Suppose 2-u^ i" an altemating series, lrn*rl < lunl, and ]l}rn:O. the sum of the approximating by obtained Then, U r{-ttr the remainder < first k terms, seriesby the sum of the lRol luo*tl' pRooF: Assume that the odd-numbered terms of the given series are positive and the even-numbered terms are negative. Then, from (2) in

700


'J-.6.6.2, we see that the sequence {s2,} is increasing. the proof of Theorem given of the series, we have if S is the sum So

(s)

Szrc1 Szrc+z To show that the sequence {srn-t} is decreasingr w€ write szn-r:ur*

(u2*ur) + (un*ur) + ' ' ' + (urn-r*urn-r)

(6)

The first term of each quantity in parentheses in (5) is negative and the second term is positive, and becaus" lun*rl < lu,l, it follows that each quantity in parentheses is negative. Therefore, because u, ) 0, we conclude that s1 ) ss ) s5 ) {srn-t} is decreasing.Thus, (7)

S From (7) we have S - srrc S - sn .Therefore,

and from (5) we have 0

0

(8)

From (5) we have -S 1-szrc; hence, szk-t- S cause from (7) it follows that 0 1 sn -r - S, we have

-

Sac

Llzk. Be-

(e)

0

Becausefrom Definition 16.6.3,Rrc: S - to, (8) can be written as 0 we lRr,,l< luro*rl and (9) can be written as 0 have lRol I

EXAMPLE 3:

A series for com-

puting ln ( I - x) rf x is in the open interval (-1 ,I) is

l n (I - x ) : S - 4 f,:r n Find an estimate of the error when the first three terms of this series are used to approximate the value of ln L.L.

SoLUTION: Using the given series with x-

-0.1-ry*ql)lln1.1

-0.L, we get

(oil)*+

This series satisfies the conditions of Theorem'15.6.4; so if R, is the difference between the actual value of ln 1.1 and the sum of the first three terms, then lR'l < lunl:O.OOOOZS Thus, we know that the sum of the first three terms will yield a value of ln L.l. accurateat least to four decimal places. Using the first three terms, we get ln l.L : 0.0953

Associated with each infinite series is the series whose terms are the absolute values of the terms of that series.


16.5.5 Definition

Tlie infinite ,"rie, !

z, is said to be absolutelyconoergent if. the series

n:l

t@

)

701

lut is convergent.

.: ,r"arSTRATroN1: Consider the series +@

(10)

3n'

reries will be absolutely convergent if the series

T#

$3

2'2'Z+Z+...+Z*... -T3.-

Ar:5*ytF-Fis convergent. Becausethis is the geometric serieswith r: $ < 1, it is convergent. Therefore, series (10) is absolutely convergent.

o

. rLLUsrRArroN2: A convergent serieswhich is not absolutely convergent is the series €

(-l;',*'

A" In Example 1 we proved that this series is convergent. The series is not absolutely convergent because the series of absolute values is the har. monic series which is divergent. The series of Illustration 2 is an example of a "conditionally convergent" series. 16.6.6 Definition

A series which is convergent, but not absolutely convergent, is said to ergent. be conditionallyconz: It is possible, then, for a series to be convergent but not absolutely convergent. If a series is absolutely convergent, it must be convergent, however, and this is given by the next theorem.

16.6.7 Theorem

If the infinite series j

u, is absolutely convergent, it is convergent and

'f v*1"=' 1,5,',1 pRooF: Considerthethreeinfinite s"ri", j

,^,f,lu^l,"na

j

@,* lunl),

of partialsumsbe iti, tT"i, unaU,Iill.spectively. and let their sequences For everypositiveintegern, un* lurl is either0 or 2lunl;so we havethe inequality 0 =un+lr"l <2lr"l

(11)

INFINITESERIES +6

Because)

lu"l is convergent, it has a sum, which we denote by T.

{fr} is an increasing sequence of positive numbers, and so tn < T for all positive integers n. From (11), it follows that O=rn<2t;<27 Therefore, the sequence {rn} has an upper bound oL 2T. Thus, by The(r, + lu*l) is convergent,and we call its sum R.

orem 16.4.1the series i

Becausefrom (11), {r|t* an increasing sequence,it may be concluded from Theorem 16.2.7that R < 27.

Eachof the seriesf

n:l

@,+ lu*l) ur,ai

from Theorem"1,6.3.9 the series +m

+oo

n:l

n:l

hence, lu,l is convergent;

n:l

is also convergent. '1,6.g.9, Let the sum of the series I ," U" S. Then, also from Theorem S : R - T. And becauseo = r:i, S < zT - T : T. +o

Because)

u,is convergent and has the sum S, it follows from The-

n=l +@

orem 16.3.8 that

(-u") ,)

tg

I

n:l

is convergent and has the sum -S. Because

+@

+@

+@

l-z'l : ) lu"l: T, we can replace2 ,, by I tD n=l n=l n:t

< T. BecauseS < Tand -S =7, we have

discussion and show that-S .

l+@

|

in the above

+e

lsl s T; therefore,l) ,,1 - \ Iu,,l,and the theoremis proveci. tn-l I n=l

ExAMPrn4: Determine if the series € cos tnrr

z7

n:l

I

+m

soLUTroN: Denote the given series byJ t .LI

uo. Therefore,

n:I

#*frZ,un:i-fi-i--i*+* -+-+-+-#+#+++eb-.


..

we have a series of positive and negative terms. we can prove this series is convergent if we can show that it is absolutely convergent.

$ lr,l : $ lcosizzl n:l

n=l


Because lcostnnl < L lcos*nrrl = 1 . n2

*

for alln for all positive integerc n

Theseriest Un'is the p series,with p:2, So by the comparison test )

and is thereforeconvergent.

lznl is convergent. The given series is there-

fore absolutely convergen! hence, by Theorem 1'6.6.7it is convergent. You will note that the terms of the series )

lu"l neither increase

n=l

monotonically nor decreasemonotonically. For example, lrnl: and so larl < lunl,but lu5l > lrrl. it,luul: "f;

ir, lutl:

Ttre ratio test, given in the next theorem, is used frequently to determine whether a given series is absolutely convergent. +@

15.6.8 Theorem Ratio Test

Let ) urbe a given infinite series for which every un is nonzero. Then "-lU o,olun*rlunl L <1, the given seriesis absolutd convergent; the seriesis (ii) if lun*rluol- L> L or if )Y*lu"*tlunl:**, "lim divergent; no conclusion regarding convergencemay (iiil if )lnlu"*tlunl:1, be made. pRooF or (i): We are given that L < L, and let R be a number such that ( 1. Becauselim lu,alunl:L, there existsan L < R < 1. Let R-L:e ?r-*o integerN> o suchthat whenevet n > N

(12)

0

lunl Letting n take on the successivevalues N, N + L, N + 2, so forth, we obtain from (12)

lu**rl< Rlu"l lu**rl< Rlz"*,| < r{3lr"l

t

and


l,**,1<-],::l
lu**ol< Rnlr"l

for every positive integer k

The series +@

+ lu*lR"+

is convergent because it is a geometric series whose ratio is less than L. So from (13) and the comparison test, it follows that the ,"ri",

j

lu"**l

+@

is convergent. The series )

k :l

lu"**l differs from the series )

lu"l in only

the first N terms. Therefore, ) lurl is convergent, and so the given series n=r is absolutely convergent. pRooFor (ii): If lim lun*tlunl:L>

1 or lim lun*rlunl:*@, then in either casethere is an integer N ) 0 such that lun*rluol > t for all n > N. Lettingz take on the successivevaluesN,N+t,N+ 2, . . .,andsoon, we obtain fl-*@

ll-*6

lz"*tl> larl lr**rl> lz"*rl> lurl lr"*rl > lu**rl>lu*l So we may conclude that lu"l

for all n > N. Hence, lim un * 0, ?L-t@

and so the given series is divergent. PRooF oF (iii):

If we apply the ratio test to the p series, we have

:rT.l(#.J'l :' Becausethe p series diverges if p - 1 and converges if p > 1.,we have shown that it is possible to have both convergent and divergent series for which we have lim lun*rlunl : 1. This proves part (iii). I


ExAMPLE 5: series

Determine if the

iln:

+ € (_ I ) n + t n sr\ ./J n:l

( - 1 ) " * ' n l 2 ' a n d u n + r : ( - l ) n + z ( n * 1 )l z * t

nIL

2n

)n

2n


Therefore, by the ratio test, the given series is absolutely convergent '1.6.6.7, it is convergent. and hence, by Theorem

EXAMPLE

In Example 2, the

series

;(-1)"ffi

solurroN: To test for absolute convergencewe apply the ratio test. In the solution of Example 2 we showed that lu"*rlllu"l: (nz * 3n)I @2t 4n * 4) . Hence,

1 + fl3 t-., , z :l n-a_|unIn_+_t+j+ft 1-

was shown to be convergent. Is this series absolutely convergent or conditionutly convergent?

lun+rl: fim

So the ratio test fails. Because n*2 , , lu"l: n@:41.

n*2

1.-1 n- n

we can apply the comparison test. And because the series i

tt" is the

harmonic series, which diverges, we conclude that ,h" rJ;,

5 l,'l t,

+d

divergent and hence )

z" is not absolutd

convergent. Therefore, the

n:l

series is conditionally convergent.

It should be noted that the ratio test does not include all possibilities lun*rlu,l becauseit is possiblethat the limit does not exist and is ,li1 not *oo. The discussion of such casesis beyond the scope of this book. To conclude the discussion of infinite series of constant tetms, we suggest a possible procedure to follow for determining the convergence un * 0, we may conor divergence of a given series. First of all, if "1111 clude that the series is divergent. If lim un:0 and the series is an alterfot

INFINITESERIES

706

nating series, then try the alternating-series test. If this test is not applicable, try the ratio test. If in applying the ratio test, L:1, then another test must be used. The integral test may work when the ratio test does not; this was shown for the p series.Also, the comparison test can be tried.

Exercises 16,6 In ExercisesL through 8, determine if the given alternating serles is convergent or divergent.

I ir-D"#

+@

4.

nz

3.

n

n:2

+@

+oo

7T

2.

n:l +@

ln n sn+r

on

5. .LJ ) (-t)" -n n:l

n:l +a

7.t (-1)"# n:l

In Exercises9 throu gh L2, find the error if the sum of the first four terms is used as an approximation to the sum of the given infinite series.

s.i (-1 )*'I

10 (-1)"*. >

n:l

+@1

1 1 . 5 ( - 1) n + , # D ,

12. .LJ ).

(-11n+r \ /

vllr

n:l

In Exercises L3 through 15, find the sum of the given

infinit" ,Jl"r, accurut! tothreedecimalplaces. L (2n)I

13. (-1)nr1 > # "::

1,6 5(-1)"ffi1,

ls.,e (-1)n+1 #,

In Exercises17 through 28, determine if the given series is absolutely convergent, conditionally convergent, or divergent. Prove vour answer. +@

18. n:l

1. ( 2 n- r ) I

+@

L9. n:l

n2 nl,

+co

22. n:l +@

23.

+a

3n

n!

25.

26. 2 S ' n 2e i l n c o s = n

28.

n:2

+@

+@

n:l

n:l

29. Prove by mathematicalinduction that Un! = 1l2n-t +@

30. Prove that if i n:l

un is absolutely convergent and,un # 0 for all n, then

Lllu"l is divergent. n:\

16.7 POWERSERIES

3 1 . Prove that if > un is absolutely convergent,then )

unzis convergent.

32. Show by means of an example that the converseof Exercise3L is not true.

16.7 POWER SERIES We now study an important type of series of variable terms called "power series." L5.7.1Definition

A power seies in (x - a) is a seriesof the form co* cr(x- a) * cz(x- a)'+''

I cn(x- a)"+'''

(1)

We use the notation 5 ,"(t - a)n to representseries (1). (Note that x : a, for conveniencein writing the genwe take (x - a)o: 1, "lr"rr"#hen eral term.) If r is a particular number, the power series (1) becomes an infinite series of constant terms, as was discussed in previous sections' A special case of (1) is obtained when a: 0 and the series becomes a power series in x, which is +@

(2)

n:o

In addition to power series in (r = a) and t, there are Power series the form of +@

c"l6@)1"+''' ) t " t 0 ( t ) l ' : c o *c r 6 @+) c r l f @ ) 1 z + ' ' ' + n:0 where d is a function of r. such a series is called a power series in {(r). In this book, we are concerned exclusively with Power series of the forms (1) and (2), and when we use the term "power series," we mean either of these forms. In discussing the theory of power series, we confine ourselves to series (2). The more general power series (1) can be obtained from (2) by the translation x: i - a; therefore, our results can be applied to series (1) as well. In dealing with an infinite series of constant terms, we were concemed with the question of convergence or divergence of the series. In considering a power series, we ask, For what values of x, if any, does the power series converge? For each value of r for which the power series conve"ges, the series represents the number which is the sum of the series. Therefore, we can think of a Power series as defining a function. The function f, with function values

f(x):

(3) n:o

INFINITESERIES

has as its domain all values of.x for which the power series in (3) converges. It is apparent that every power series (2) is convergent for r: 0. There are some series (seeExample 3) which are convergent for no other value of r, and there are also series which converge for every value of r (seeExample 2). The following three examples illustrate how the ratio test can be used to determine the values of.x f.orwhich a power series is convergent. Note that when n! is used in representing the zth term of a power series (as in Example 2), we take 0!: L so that the expression for the nth term will hold when z:0.

EXAMPLE1: Find the values of x for which the power series +@

' S el) ' n+l !,, zr n3n n:l )flyn

is convergent.

solurroN: un :

For the given

-, (- 1 )n+r

rio, lful: ttn

Znxn

W

rnd. ( - 1t )\n+2 n + 2 2n+rx6n+t and u Ltn+.t n +:r : (-

TffiTI

| 2n+rxn+t. n3 l - ,,- 2 ,-., n

hm n-p*l I ,_*l lffiF'Nl:,li1ilxlfr_l:i

2

lxl

Therefore, the power series is absolutely convergent when $lrl < 1 or, eguivalently, when lrl < *, The series is divergent when flrl )'1 or, equivalently,when ]rl = *. When Slrl:1 (i.e., when x:!8), the ratio test fails. When x: $, the given power series becomes

+-I+*- I+. . . + (-1)n+, 1+ which- is convergent, as was shown in Example 1 of Sec. 16.6. When x:-t, we have

which by Theorem 15.3.8is divergent. we conclude, then, that the given power series is convergent when -* < x < 9. The series is absolutely convergent when -* < .r < * and is conditionally convergent when r: g. If r < -* or x > 8, the series is divergent.

EXAMPLE 2: Find the values of x for which the power series +m

yftt

\* ,?on! rs convergent.

solurroN: For the given series, ttn: tnln! and un*r: xn+tl(n * 1) !. So by applpng the ratio test, we have

| ,"*' .ntl-,-, 'Fl: l'l rio,lgul: un 1im

n-+* | I z;; l1;Tln

1

Jii;ft-:

o< r

we conclude that the given power series is absolutely convergent for all values of r.

16.7 POWERSERIES

ExAMPLE3: Find the values of x for which the power series +@

S nlxn

LJ n:0

(n * l)lx"+r. Apply-

soLUrIoN: For the given seriesun: nlN,nandur*r: ing the ratio test, we have t" I fim l'n+11: lim l ( n * ' l ' ) l x " + 1 1 n-+-lunl n-+*l--M-l

:

is convergent.

:.1

709

+ 1)rl ,111l(n f 0 if r:0

t**

if.x*0

It follows that the seriesis divergentfor all valuesof r except0. 16.7.2 Theorem

If the power series f

,r*" is convergent for r : xr (xr + 0), then it is ab-

n=O

solutely cohvergent for all values of r for which lrl < pRooF: ff. f

cnxr" is convergent, then

l*-r,rr":0.

l r 'l . Therefore, if we

take e : L in Definition 4.1.1,there exists an integer N > 0 such that Itnxr"l<1

w h e n e v e r n> N

Now if x is any number such that lrl < lrtl, we have |

-nl

lrln

ltln

. lc,x"l: lc"x,^ ful: lt"tr'l lil El

n> N whenever

(4) (s)

is convergent becauseit is a geometric series with r : lxlxl < 1 (because lrl < lrtl).Comparing the series )

lc^x"|,where ltl < lr'1, with series

n:N

+@

lcnx"l is convergent ") for lrl < ltrl. So the given power series is absolutely convergent for all r values of.xfor which ltl < lrtl.

(5), we see from (4) and the comparison test *",

o rLLUsrRArrou1: An illustration of Theorem'l'6.7.2is given in Example 1. The power seriesis convergentfor r:8 and is absolutely convergentfor all values of.x for which lrl < g. The following theorem is a corollary of Theorem 16.7.2. L5.7.3 Theorem

'

If the power series )

cnxnis divergent for r:

values of x for*r,t"fr=itt> lxrl'

xr, it is divergent for all

710


pRooF: Suppose that the given power series is convergent for some number r for which lrl > lrrl. Then by Theorem 16.7.2the series must converge when r: rz. However, this contradicts the hypothesis. Therefore, the given power series is divergent for all values of x for which

.

lrl > ltrl.

I

o rLLUsrRArror 2: To illustrate Theorem 16.7.3 we consider again the power seriesof Example 1. It is divergent for x: -B and is also divergent . for all values of x for which |tl > l-91. '1.6.7.2 From Theorems and t6.7.3, we can prove the following important theorem.

,

15.7.4 Theorem

fet i

coxnbe a given power series. Then exactly one of the following

"o.iioror,,

holds:

(i) the seriesconvergesonly when x: 0; (ii) the series is absolutely convergent for all values of x; (iii) there exists a number R > 0 such that the series is absolutely convergent for all values of x for which lrl for all values of.x for which lrl

pRooF: If we replacex by zero in the given power series,we havecs* 0 + 0+ ' , which is obviously convergent.Therefore,every power +@

series of the form )

c;cn is convergent when x:

0. If this is the only

value of r for whicil1l" ,"ri", converges, then condition (i) holds. Suppose that the given series is convergent for r: rr where x1 * 0. Then it follows from Theorem1,6.7.2 that the series is absolutely convergent for all values of r for which lrl < lrrl. Now if in addition there is no value of x f.ot which the given series is divergent, we can conclude that the series is absolutely convergent for all values of r. This is condition (ii). If the given series is convergent fot x: xr, where xr* 0, and is divergent for x:.rr, where lx"l> l*rl, it follows from Theorem 15.7.3that the series is divergent for allvalues of xfor which lrl > lrrl.Hence, lrrl is an upper bound of the set of values of lrl for which the series is absolutely convergent. Therefore, by the axiom of completeness (1.6.2.5),this set of numbers has a least upper bound, which is the number R of condition (iii). This proves that exactly one of the three conditions holds. I

serresconvergent

for lxl < R

Theorem 1,6.7.4(iii) can be illustrated on the number line. See Fig. 1,6.7.1. seriesdivergent for lrl )

Figure 16.7.1

R

+@

If instead of the power series )

+@

cnxnwe have the series )

cn(x - n)n,

16.7 POWERSERIES

711

in conditions (i) and (iii) of Theorem'1.5.7.4,r is replacedby x - a. The conditions become a-R

n

a+R

seriesdivergentfor lx - ol > R Figure16.7.2

(i) the series converges only when x: ai (iii) there exists a number R > 0 such that the series is absolutely convergent for all values of. x for which l* - ol < R and is divergent for all values of.x for which lrol > R. (SeeFig. 16.7.2 for an illustration of this on the number line.) The set of all values of.x for which a given Power series is convergent is called the interztalof conoergenceof the Power series. The number R of 'J,6.7.4 is called the radius of conaergenceof the condition (iii) of Theorem power series. If condition (i) holds, we take R: 0; if condition (ii) holds, we write R: *o. 3: For the power seriesof Example 1, R: $ and the interval o TLLUSTRATTOTI is of convergence (-8, g7.In Example2, R: ao, and we write the interval o of convergenceas (-o, *oo). +@ If R is the radius of convergenceof the power series ) cnro, the irrn=0 terval of convergenceis one of the following intervals: (-Oi}), [-R, R], (-R, R], or [-R, R). For the more general Ponrer series

Ar"@-a)n, (a-R, a*R), following: of the one the interval of convergence is R ) . R , a * la- R, a * R], (a- R, a* R], or laA given power series defines a function having the intenral of con.r"rg"rrJ" as iis domain. The most useful method at our disposal for deteriining the interval of convergence of a Power series is the ratio test' However, the ratio test will not reveal anything about the convergence

which it converges (see Exercise 22). There are casesfor which the convergence or divirgence of a power series at the endpoints cannot be detirmined by the methods of elementary calculus. ExAMPLE 4: Determine the interval of convergence of the Power series +6 n:l

n(x- 2)"

solurloN: The given seriesis ( x - 2 ) + 2 ( x - 2 ) , + . . . + n ( x - 2 ) " + ( n * 1 ) ( x 2 ) n +*r ' ' ' Applying the ratio test, we have

+ 1 ) ( r = ? ) " . ':l limlwl-- hml(n n(x-z)" W - z l t i m# :nl x - 2 1 |

;;;lu,

l

"-+;l

"-;;

712


The given series then will be absolutely convergent if lx - 2l equivalently,1 1x (-3. equivalently,-L < x-2 When x : l, the series is

jsu"

+oo

# 0. When x:3, the seriesis 2 n, which is alsodivergentben=l

cause lim un * 0. Therefore, the interval of convergence is (1, 3). So the ll-*'@

given power series defines a function having the interval (L, 3) as its domain.

EXAMPLE5: Determine the interval of convergence of the power series +6 Sl

yn J+

t,-ffi

n:l-

SoLUTIoN: The given series is x,x2,x3l

Applying

|

'

2t, 1zT 2+22-

2+y-

ry,?t

'

l * , - ,

2*nz'

Xn+L

2+ (n*'1,)2

I

the ratio test, we have

2*nz ,., r. --Tl: ltl ri- l1*l : fim l-, T'*l ^. .2*-n2l n-+-tun | ,-*-lz+ @+ty ,li* 2ffi:lxl So the given series will be absolutely convergent if lrl < L or, equivalently, -l < x < 1. When x:1, we have the series L,L,L ' 2+12' 2+22' 2+92 __I

BecauseU (2 + n2)< 7ln2for all positive integersn, and,because ,fu, f is a convergent p series, it follows from the comparison test that the given power series is convergent when r : L. When x: -1, we have the series +@

>

(-t)"1(2*

n), which is convergent becausewe have just seen that

liti, convergent. Hence, the interval of convergence of the given"Urot.rtely power seriesis [-1, 1].

Exercises L6,7 In ExercisesL through20, find the interval of convergence of the given power series. +@

1.P(-1)n+'fu1, *2n_l

4.X(-1)"# n:l

r.)(-lvffi*_, +6

1 6 . 8D I F F E R E N T I A T I O ON F P O W E RS E R I E S 7 1 3 +@

9.

t*-11n (-1)n-r-tT

+oo

10.

n:l

n:l $co

*co

11.

12. LlI

( sinh 2n) x" n:0

n:l

+@

( -1\^ ' z+1

L3.

\

n:2 +@

15. n:l +co

17. n:l

+@

xn

1,4.

n(ln n)z

n:l

xn

ln(n + 1) (x * 5)z-t n2

n:l +@

nlxn nn

1,6. n:l

ln n(x - 5)"

xn nn

+@

18.L/s n"(x - 3)

n+l

lI:1 +@

+@

19.

(x* 2)" (n * 1)2

2'4'6'.

. .'2n

20.

xzn+l

n:1

( - 1 ) n + r 1' 3 ' 5 ' 2'4'5'.

. . . ' (zn-l) . '2n

xn

21. Prove that if a power series convergesabsolutely at one endpoint of its interval of convergence,then the power series is absolutely convergent at each endpoint. 22. Prcve that if a power series converges at one endpoint of its interval of convergenceand diverges at the other endpoint, then the power series is conditionally convergent at the endpoint at which it converges. 23. prove that if the radius of convergence of the powercerie, j

is {r. 2ru,*2n 24.Provethat if lim +M - L (L + 0), thenthe radiusof

unxnisr, then the radius of convergenceof the series

+!p

convergence of the power series

unx'is LlL. n:l

A power series 2 cnx" defines a function whose domain is the interval 16.8 DIFFERENTIATION N:O OF POWER SERIES . of convergence of the series. o rLLUsrRArrow1: Consider the geometric series with a:1 +o

which is )

and r:x,

r". By Theorem 16.3.6 this series converges to the sum

- )c)if ltl 1.1(1,

)cndefines the func-

tion f forwhich f (x)- 1l(1 - x)and ltl

we can write

1+ x+x2+),3+ ' ' '+ xn+ ' ' ':

1 1- x

if lrl

The series in (1) can be used to form other power series whose sums can be determined. o rLLUSrRArroN2: If in (1) we rePlace x by -*,

1-x+x2-x3+

+ (-1)"x"* ' ' '

we have 1

L*x

if lrl

714


Letting )c: x2 in (1), w€ get 1 + )cz+ xa * xG+ . . . + x2' *

. . .

if lrl

(3)

lf x is replaced by - x2 in (1) we obtain 1- x2+x4-x6+

' ' ' + (-1)'x'"+

if lrl

In this section and the next we learn that other interesting series can be obtained from those like the above by differentiation and integration. We prove that if R # 0 is the radius of convergence of a power series which defines a function /, then / is differentiable on the open interval (-R, R) and the derivative of / can be obtained by differentiating the power series term by term. Furthermore, we show that / is integrable on every closed subinterval of (-R, R), and the integral of f is evaluated by integrating the power series term by term. We first need some preliminary theorems. +@

L5.8.1 Theorem

If ) cnx, i s a power series having a radius of convergence of R ) 0, then Eo +co the series LI ncnxn-l also has R as its radius of convergence. n:l

This theorem states that the series, obtained by differentiating term by term each term of a given power series, will have the same radius of convergenceas the given series. pRooF: Let r be any number in the open intewal (-R, R). Then lrl < R. Choose a number r, so that lrl < lrrl < R. Becauselxrl < R, j

corrois

convergent.Hence,

rn:0. So if we take e : L in Oefiniiln A.t.t, ]!I_rn there exists a number N > 0 such that lcn*rnl< 1

whenevern >N

Let M be the largestof the numbers Icfirl , lcrxrrl, lrr*rtl, Then lrnrrnl = M

for all positive int egerc n

, lcNxrNl,1. (5)

Now

: lnc,x'-' |: lrr,.# . xrnl " W l;l*' From (5) and the above equatiorr, we get

="#l;l' lncnx,-,|

(5)

16.8 DIFFERENTIATION OF POWERSERIES 715

Applytng the ratio test to the series M t{ lr lo-1 ri )l n l:l lxrl #-r lrrl

(7)

we have

,,_ l z,*rl:_ llm ,:_ l( n+ 1) lxl'. lr ,l.' - tl llm l-l

,il'i!*| u" | ;].i; I :

lr'1"

nlxl"-r1

-n * L

l r l llrl ,. l-t

lrrl n_+-

n

: l _l =r ll ( 1 lrrl Therefore, series (7) is absolutely convergenf so from (6) and the comparison test, the series )

ncnxn-ris also absolutely convergent. Because

r is any number i" t-nl-ntl, it follows that if the radius of convergenceof +@

> R. \ ncnx"-ris R', then R' n=t To complete the proof we must show that R' cannot be greater than R. Assume lhat R' > R and let 12 be a number such that R < lr2l < R'. Becauselrrl > R, it follows that +@

(8) n'-o

Becausel*rl ( R', it follows that i

ltcnx2n-Lis absolutely convergent.

Furtherrnore, +@

*rc

l*rl and so from Theorem 16.3.8 we may conclude that +@

(e)

positiveinteger "", (10) Itn*rnl= nlcnxr"l- lncnxr"l From statement (9), inequality (L0),and the comparison test it follows that

If n,

+@

)

+@

lc,xr"l is convergent.Therefore, the series 2 ,n r" is convergent, which

that R' > R is false. Iot,o"Ur",, statement (8). Hence, ttre assffition Therefore, R' cannot be greater than R; and because we showed that I R' ) R, it follows that R' : R, which proves the theorem.

716


. rLLUsrRArroN 3: We verify Theorem 16.8.1.for the power series S

tn*t

_-,f

?^@*:'*T*e

+d+

*

.

xn+r ,n*, r_. +@T+G+'/r+'

_L

. .

To find the radius of convergence,we apply the ratio test.

,- ln'* 2n* 1,1 - r"r l : Hml@+ t!?*+:.1 lxl ::T-17;+ 4n+ 4l : lrl ,-*l l@lWl

rio,l

n-+*| iln |

Hence, the power series is convergent when lxl < 1., and so its radius of convergenceR: L. The power series obtained from the given series by differentiating term by term is $(n+1)r':t{_j'

.

x.f

*4Try:,17"+t:r+r+T*

.fr...r

++''

n*t + nx+nt +, x; + 2+'''

Applyirg the ratio test for this power series,we have

*,L) !n+tt

:- - ln+ r l hm l( F l'l n-J; ;-i;l @+z)*l: I* lffil: l'l l |

l i m lwl: un

This power series is convergent when lrl vergenceR'- 1. BecauseR - R', Theorem 16.g.lis verified. 1.6.8.2Theorem

If the radius of convergence of the Power series

of cono

+@ n:O

is also the radius of convergence of the series

X n(n- 1)cnx,-r.

n:2

+co

pRooF: If we apply Theorem 16.8.1to the series ncnxn-l, w€ have the desired result. n:l I We are now in a Position to prove the theorem regarding term-byterm differentiation of a power series.

1'6.8.3Theorem

Let ) coxnbe a power series whose rad.iusof convergenceis R > 0. Then tr 1ir-irr" function defined by +€

f(x)

( 1 1)

' f (x) exists";revery r in the open interval (-R, R), and it is given by +@

f'(x)

SERIES 717 OFPOWER 16,8DIFFERENTIATION pRooF: Let r and a be two distinct numbers in the open interval (-R, R). Taylor's formula (formula (9) in Sec.15.5),with n: L, is (" (l\

('(o\

f(x): f(a\+Lff (x- a)+'-f,a k- a)' Using this formula with f (x) : xn, we have for every positive integer z (12) { : nn* non-r(x- a) + in(n - l) ((")"-2(x a)z where f, is between a and r for every positive integer z. From (11)we have

f(x) - f(a) :2

+@

+00

2 cna" n=o "n*" n-0

: co*5 ,-*- ce-5 ,,o" n:l n-l +@

:)cn(x"-a") n=l

(becausex*

Dividing by x-a

+@

a) and using (12)' we have from the

n - z ( x- a ) ' f c n l n a n - t ( -r a ) + + n ( n 1 ) ( f , )

n:l

+@

+@

ncnan-r++(x-a) n:l

n(n - 1)c"(tn)"-'

(13)

n:2

Because a is in (-R, R), it follows from Theorem 16.8.1that )

ncnan-r

is absolutely convergent. K> 0 Becauseboth a and x are in (-R, R), there is some number "l'6.8'2 that Theorem suchthat lal < K < R and lrl < K < R. It follows from +@

is absolutely convergent. Then because ln(n-

L)c,(€*)"-'l < ln(n L)cnK-'l

for each €n, we can conclude from the comparison test that +@

,,

";;utely

conversent.

(14)

SERIES

It follows from (13) that lf(x)-f(a\ t# x-a I

ro

.l I

n:r

l. I

+@ . n:2'\'?

However, from Theorem 1-,6.6.T we know that if vergent, then Itg

.

L,vna

I I

*o

un is absolutely conn:l

+-

|

l',1 l},""l=p,

Applying this to the right side of (15),w€ obtain

-f@) ll M ' * -J c - a

I

{ { n c r a ".-l r=l + l * - - ' t + @ > 'trv ol

#,

I

(16)

"1:,

From (14) and (16) we get

(x)- f-(n)lf t A , n(n- 1)lc,lKn-, +lx - watj ; ncnan-,1= I x-a | A "n

(r7)

where 0 < K < R. Becausethe series on the right side of (L7)is absorutely convergent, the limit of the right side, as r approaches a, is zero. Therefore, from (17) and Theorem 4.3.3 it follows thit

y ^ f @ ) - f ( a ) - *' \-' n c n a n - r c'a

x-A

or, equivalently, +@

f'(a):

\

ncna"-r

and becausea may be any number in the open interval (-R, R), the theorem is proved. I

ExAMrLE 1: Let b" the function f defined by the power series of Illustration 3. (a) Find the domain of f; (b) write the power series which defines the functi on , and f find the domain of f ,.

The domain of f is the intervar of convergenceof the power series. In Illustration 3 we showed that the radius ofof *re pou/er series is 1; that is, the series converges when "or,rr"rg"n"e lrl < rl we now consider the power serieswhen lrl : 1. When-r: 1, the seriesis

1+ l+!r*

719

OF POWERSERIES 16.8 DIFFERENTIATION

which is convergentbecauseit is the p serieswithp:2. +@

we have the series > (-1)'*Y(n*2)2,

When x:-L,

which is convergentbecauseit

is absolutety converg:.1',t.Hence, the domain of f is the interual [-1, 1]. ' (b) From Theorem15.8.3it follows that f is defined by

f'(x)-$+ +1 *?:^n

(1s)

and that f '(x) exists for every x in the open interval (-1, L). In Illustration 3 we showed that the radius of convergenceof the Power series in (18) is L. We now consider the power series in (18) when x +1. When x: !, the series is

' . '+' n * \1=*"' 1+****1* z 3 4 which is the harmonic series and hence is divergent' When t: series is 1,-

r , L- 1 , i;rs

-1, the

+ (-1vftr+

which is a convergent alternating series. Therefore, the domain of f the interval [ - 1 , 1 ) .

is

Example 1 illustrates the fact that if a function / is defined_by a power series and this Power series is differentiated term by term, the resulting ',has the same radius of convergence but power series, which defines f not necessarily the same interval of convergence. ExAMPrn2: obtain a Powerseries representation of

solurroN:

if ltl

+'/,,-1*x*x2*13*...+)cn+ x

L

@

From (1) we have

using Theorern 1.5.8.3,differentiating on both sides of the above, we get L

ffi-1*2x*3x2* EXAMPT,n 3:

Show that

solurroN: +6

+@

vsr--

q2fl

S + n!

#-,

:L* for all real values of x.

I

* nxn-r +

if ltl

In Example 2 of Sec. \6.7 we showed that the Power series

x"lnl is absolutely convergent for all real values of r. Thereforc, 7f.f is

lil rrrr,.tior,definedby (1e)


the domain of / is the set of all real numbers; that is, the interval of convergence is (-rc , +m). It follows from Theorem 16.8.3 that for all real values of x

f'(x):S+n! "1

Becausenln!:

(20)

- l)1, (20) can be written as U (n

,*'-'. f'(*):S #, (n L)I or, equivalently,

f'(x):fSo n+!

(21)

Comparing(19)and (27),we seethat f ' (x) : f (x) tot all real valuesof x. Therefore,the function / satisfiesthe differentialequation du

fr:y for which the general solution is y: Cec. Hence, for some constant C, f(x) : Ce'. From (19)we seethat/(0) : 1. (Rememberthat we take r0: L even when r: 0 for convenience in writing the general term.) Therefore, C : 7, and so f (x) : ,','and we have the desired result. ExAMPtn 4: Use the result of Example 3 to find a power-series representation of s-t.

SOLUTION: e &: - 4

If we replac e x by -x in the series for er, it follows that n2

ng

1- )c+n-fr+

vTI

+ (-r)"\n ! +

for all real values of )c. EXAMPLE5: Use the series of Example 4 to find the value of e-t correct to five decimal places.

solurroN:

Taking x - 1 in the series for e-r, w€ have

,1 -3.r'f8ff - | - 1 + 0.5- 0.1.66667 - 0.008333 + 0.041667 + 0.001389 - 0.000198 - 0.000003 + 0.000025 + 0.0000003 We have a convergentaltemating seriesfor which lur*rl < lnrl. So if we usethe first ten termsto approximatethe sum,by Theorem16.5.4the

OF POWERSERIES 16.8 DIFFERENTIATION

errof is less than the absolute value of the eleventh term. Adding the first ten terms, we obtain 0.357880.Rounding off to five decimal places gives e-r - 0.36788 In computation with infinite series two kinds of errors occur. One is the error given by the remainder after the first n terms. The other is the round-off error which occurs when each term of the series is approximated by a decimal with a finite number of places.In particular, in Example 5 wewanted the result accurateto five decimal places;so we rounded off each term to six decimal places.After computing the sum, we rounded off this result to five decimal places. Of course, the error given by the remainder can be reduced by consideriirg additional terms of the series, whereas the round-off error can be reduced by using more decimal places.

L6.8 Exercises exerciie do the following: (a) Find the radius of In Exercises1 through 8, a function / is defined by a power series.In each series which defines the function f' and the (b) write of domain the and Power power series f; convergence of the given Theorem 15.8.1);(c) find the domain of /'' find its radius of .orr',rerg"nceby using methods of Sec. 16.7 (thus verifying +@

+co

2' f(x)

L.f(x) n:l +co

s. f (x)

4. f(x)

yn

3. f (x) --

n

,:| +@

2#"

XZn-t

l\-

n:2

+co (x _

7.f(x):ZT

+@

8. f(x)

1)"

n:2

- x)3' 9. Use the result of Example 2 to find a power-series representation of U(1 of e€. 1.0.Use the result of Example 3 to find a power-series representation series (2) term by term' 11. Obtain a power-series representation of U(l l- x)2 if lrl < r by differentiating series (4) term by term' 12. Obtain a power-series representation of fl\ + *)2 it lxl < 1 by differentiating decimal places' 13. Use the result of Example 4 to find the value of u {i correct to five M. rt f (x):

j n=u

{-r)'

$,

ma/'(*)

correctto four decimal places'

of (a) sinh x and (b) cosh r' 15. Use the results of Examples 3 and 4 to find a power-series representation 15 can be obtained from the other by term-by-term 15. show that each of the power series in parts (a) and (b) of Exercise differentiation. 17. Use the result of ExampIe Zto find the sum of the series ;

n:l

#. -

- 1)lx' 18. (a) Find a power-series rePresentation for (e" (b) By differentiating

term by term the power series in part (a), show that


19. (a) Find a power-series representation for fe-c. (b) By differentiating term by term the power series in part (a), show t h a t S ( _ 2 ) r + 1" \ 2 : n! f=,

n. +@

20. Assume that the constant 0 has a power-series representation

cnxn,where the radius of convergence R > 0. Prove n:o

that cn: 0 for all n.

21..Supposeafunctionfhasthepower-seriesrepresentation!

cnro,wheretheradiusof convergenceR) O.lf f,(x):f(x)

and/(O) : 1, find the power seriesby using only propertrlitf

po*"r reries and nothing about the exponential function. 22. (a) Using only properties of power series, find a power-series representation of the function / for which /(r) > 0 and '(x) :2xf (x) for all r, and f /(0) : 1. (b) Verify your result in part (a) by solving the differential equation D,y :2xy having the boundary condition y: 1 when r: 0.

15.9 INTEGRATION The theorem regarding the term-by-term integration of a power series OF POWER SERIES is a consequenceof Theorem 76.8.9. 16.9.1 Theorem

fet i

coxnbea power series whose radius of convergenceis R > 0. Then

if 1 ir=tU""function defined by +@

f(x):2

cnx"

n=0

/ is integrable on every closed subinterval of (-R, R), and we evaluate the integral of f by integrating the given power series term by term; that is, if r is in (-R, R), then +@ fc -dt: ' S cn ,ni-r l- 'ro \ ' ' Jo fton+l*

Furthermore, R is the radius of convergenceof the resulting series. pRooF: Let g be the function defined by tr,rn*, .s(r):$ flon + l Becausethe terms of the power-series representation of f(x) are the derivatives of the terms of the power-series representation of g(r), the two series have, by Theorem 16.8.']-., the same radius of convergence.By Theorem 16.8.3it follows that for every r in (-R, R) f (x) '1,6.8.2, By Theorem it follows that /'(x): g,, (r) for every r in (-R, R). Because/ is differentiable on (-R, R) , / is continuous there; consequently, / is continuous on every closed subinterval of (-R, R). From Theorem g'(x):

SERIES 723 OF POWER 16.9INTEGRATION 7.6.2ilt follows that if r is in (:R, R), then fr IJ o -f ft) itt: s@)- s(o): s(x)

or, equivalently, ' [' nl $ -h- ,n*r t t , t *dr: *Lon*L-

r

Jo

Theorem 15.9.1 often is used to comPute a definite integral which cannot be evaluated directly by finding an antiderivative of the integrand. Examples 1 and 2 illustrate the technique. The definite integral j{ t-o dt appeaing in these two examples rePresentsthe measure of the area of a tegiot under the "notmal probability curve'" EXAMPLE1: Find a Power-series representation of fr

I e-t' dt

Jo

SoLUTIoN: In Example 4of Sec.'1.6.8we showed that H

€-0:

)'-

(-l)nxn n!

"A

for all values of )c.Replacing x by t2, we get

+#- fi*''' e-t'-1- 1z

+ (-1Y#*

for all values of. t

we integrate term by term and obtain Applying Theorem 1,6.9.1, fx

J,

e-t'dt:

f

lzn

ft

x3,

x5

'2! '5 3

-LJ-

x7

3!'7

nZn*L

'

+(-l)"ffi+

The power series represents the integral for all values of x. ExAMPLE2: Using the result of Example L, comPute accurate to three decimal Places the value of frl2

I e-t' dt Jo

SOLUTION:

Replacing x by t in the power series obtained in Example L,

we have f rt2 ' ' ' IJ O t-" d t = + - # + # - ; = b a +

: 0 . 5 - 0 . 0 4 1 . 7 + 0 . 0 0 3-10 . 0 0 0 2 + ' ' ' We have a convergent alternating series with lao*tl < lu"l' Thus' if we use the first three terms to approximate the sum, by Theorem 15.6.4 the error is less than the absolute value of the fourth term. From the first three terms we get [t'' ,-o dt - 0.461 Jo


ExAMPLE 3: Obtain a powerseries representation of ln(1 * r).

solurroN: Consider the function / defined by f (t) : U (l + t). A powerseries representation of this function is given by series (2) in sec. 16.g, which is

+ (-1)"t"+

1z-t3+

#-1-t+

if lr l

Applying Theorem 16.9.1,we integrate term by term and obtain

f' .d, _i{

r

Jo'].+t f,u.,

(- 1)"t" dt

and so

rn(r +r): x-$*t-f;*

iflrl <1

or, €guivalently,

(-1)"-'#

+@

ln(l +r):) n:I

if

(1)

Note that because l * l bars are not needed when writing ln(L * r).

)

(-t;'-'7n isln2.

O ILLUSTRATION

For the infinite

fo

(-1)"-tln,

series

sum is

the nth partial

n:t

1-

|+*-

|+

It follows from Definition

1 + (- 1)"-t n '1,6.9.2 that if we show

(2)

Jlt

sn: ln 2, we will

have proved that the sum of the series is ln 2. From algebra we have the following formula for the sum of a geometric progression:

16.9INTEGRATION OF POWERSERIES 7i28

a * ar * atz * ars+ . . ' * a4l:# Using this formulawith a : 1 dnd r: -t, we have . . .+ (-t)n-r-1: (-r)' t-t+p-tt+ 7+t which we can write as . .$(-110-rp-r=:1 :+(,l)n*trll , l+t'''' l+t

1-t+P-ts+.

Integrating frcm 0 to 1, we get fr

f' +n Jrt+rat

. . .**((-_11; ,;-, r_vr-rrnl d _ tr :l [d' t4 - * 1 - - 1 1, .n * ' Jo1+f

I tr- t+F-tt+. ro which gives

+n

1ft

',3 r-1+1-1+ 4 2

+ (-1)"-' ;: ln 2 + (-t)'*' Jo1t

tdt

(3)

Referringto Eq. (2),we see that the left side of Eq. (3) is s,,.Letting R n : ( _ 1) n + '[ : h " JO J

Eq. (3) may be written as (4)

Sn:ln2 * R,

Becausef l(1 + f) < f for all t in [0, 1] it follows from Theorem 7.4.8that

f r = JFo t " d t 1 , + t d t J, Hence,

0

.nt

J or + t d t

Because lim U(n + 1) :0,

=

J,

t"dt:;

it follows from the above inequality and the

squeezetheorem (4.3.3)that lim Rn:0' So from Eq. (4) we get ll-*@

lirn sn: ln 2 *

lim R,, fl-*a

fl,-*6

-ln2 Therefore,

' (-1)n-'I:1 |+ |+i5

:ln2

(5) o

The solution of Example 3 showed that the power series in (1) represents ln(r + 1) if lrl < 1. Hence, with the result of Illustration 1 we can condude that the power series in (1) represents ln(r * 1) for all r in its interval of convergence (-1, 1]. Although it is interesting that the sum of the series in (5) is ln 2, this series converges too slowly to use it to calculate ln 2. We now proceed to obtain a power series for computation of natural logarithms. From (1) we have

' ' ' + (-1 )'-'* + . . 23 n Replacingx by -x in this series,we get ln(l+ x):)c-4*t--

ln(1,x):-)c-t-t-t

xn

for r in (-1 ,1,1 (6)

for r in [-1 , I)

n

(T)

Subtracting term by term (7) fuom (6),we obtain

(8)

The series in (8) can be used to compute the value of the natural logarithm of. any positive number. . rLLUsTRArroN2: lf y is any positive number, let l*x

andthen ,:ffia - l

Y: ft

andlrl < 1

For instance,if y :2, then r : *. We have from (g)

rn2:,(3*+.# +/v+qlF*f3".

)

1 1 : r ( ! r 1 -- r j = -- , = l . . , , \ E f L z r s - t g o g - 1 7 7 , . t 4 7- t'g 4 g , 6 1 l , ' ' ) - 2(0.333333 + 0.012345 + 0.000823 + 0.000065 + 0.000006 + 0 . 0 0 0 0 0+1.

. .)

Using the first six terms in parentheses,multiplying by 2, and roundirg off to five decimal places,w€ get ln 2: ExAMPyn 4: Obtain series representation

a Powerof tan-r x.

solurroN:

Q.693Ls From series (a) in Sec.1.6.8we have

1, 1+ xr:l-

x 2 + x 4 - 1 6 + . . ' + (-l)nxzn a

itlxl < 1

OF POWERSERIES 16.9 INTEGRATION

APrplp Pl mg 8 rTheorem

integrating term by term, w€ get I integr

l" 1L +-+Lt 2

+(-lvffi+

Jo

dt:

lE' :ore Th,erefo {m

if ltl

tan-1 n:o

'1.6.9.'j-. allows us to conclude that the Power series Although Theorem values of r such that lrl < 1, it can be for r only in (9) repreients tan-r of the power series is [-L, 1] and convergence of shown that the interval that the power series is a representation of tan-l r for all r in its interval of conveigence. (You are asked to do this in Exercise 15.) We therefore have (10)

- f. in (10), we get o rLLUsrnArroN 3: Taking x

T:1 ]+*- |+

+ (-1)"V!*r+

The series in Illustration 3 is not suitable for computing zr becauseit converges too slowly. The following example gives a better method. ExAMPLE5: in:

Prove that

tan-l i+

tan-l *

solurroN:

Let a - tan-l * and B - tan-1 *. Then

tan(o*F):ffi

Use this formula and the Power series for tan-r x of ExamPle 4 to comPute the value of. n accurate to four decimal Places.

1++ -1-+.+ _2

_3+2 6-1, -L - tan In Therefore, 1 4 '7 T - a * F : t a t

Firom formula -1

n-1 b+ tan (LL0)with

(1 1 )

B

r -+ we get

*1,(i)' +I (i)'-+(i)' 1, 11

1 /L\15 . c\r/ +

1 + 1 _ 1 _L 1 1 ' -r- ' ' _1_!*l__ _ -t 24- 1,60 896- 4s0s- n,nto6Ag2- 491'zo - 0.00L12 - 0.00004 = 0.5- 0.041,67 + 0.00625 + 0.00022 -0.000002+. . . +0.00001

Because the seriesis altematingand lrn*rl < lunl,we know by Theorem 16.6.4that if the first seventerms are used to approximatethe sum of the series,the error is less than the absolutevalue of the eighth term. Therefore, tan-r + : 0.46365

(12)

Using formula (10)with x: t, we have

+;(i)'-# * (i)" +G)'

1_1_1/1\'+1e\'_ tan-ri:i-i(,al 5\3/ L _ '1,,2L5

1 _ 1 1 _ _ '. 15,309 L77,L47 r,949,617' - 0.0L235+ 0.00092- 0.00007+ 0.0000L : Q.33333

_1_1_ 3 g1

- 0.000001 + Using the first five terms to approximate the sum, we get tan-l + - 0.32174

(ta1

Substitutingfrom (12)and (13)into (11),we get irr - 0.45365 + 0.32174:0.78539 Multiplying by4 and rounding off to four decimalplacesgivesz' - j.'1.4'1.6.

Exercises 76.9 In Exercises 1.throu gh T,compute the value of the given integral, accurate to four decimal places, by using series. f rts rlx rl 3. I e-r' dx

1|:',#

zI:',"h^

Jo

f l n ( l* r ) fuz , 4 . I f t * ) dt _x ., w,h e r e f ( x ) : { x Jo Lr ft . where , , , , - .l| - 4 4 i 6. I h(r) dx, h(r) r ro Lt

ifx+o tfx-o ifx+o rfx-o

5.

7.

t

g(r) s(r) dx, where

dx, whercf (x) f, f fO

1 6 . 1 0T A Y L O RS E R I E S

8 . Use the power series in Eq. (8) to compute ln 3 accurateto four decimal places. 9 . Use the power series in Eq. (9) to compute tan-l,i accurateto four decimal places. 10. Integrate term by term from 0 to r a power-series representation for (1 - P)-' to obtain the power series in Eq. (8) for ln[(1 + r)/(1 - r)]. 11. Find a power-series representation for tanh-l r by integrating term by term from 0 to r a power-series representation for (1 - P)-1. 12. Find a power series for xet by multiplying the series for e' by x, and then integrate the resulting series term by term '* that ttut $ .-!--.: fa' from 0 to 1 and show o2_nrln + 2y:

13.By integrating terrr by term from 0 to r a power-series representation for ln(1 - f), show that +@

)

+n

;(n - - -: ^

f,:"

r)n

:r*

(1-r)

ln(1-r)

L4. By integrating term by term from 0 to r a power-series representation for f tan-r f, show that +6

j2n+1

(-t)'*t

: +te * 1) tan-l r - rl

n) 15. Find the power series in x of f (x) if. f " (x) : -f (x) , /(0) : 0, and /'(0) : 1..Also, find the radius of convergenceof the resulting series. 16. Show that the interval of convergenceof the power series in Eq. (9) is [-1, 1] and that the Power series is a rePresentation of tan-r I for all r in its interval of convergence.

16.L0 TAYLOR SERIES If / is the function defined by crx':co*crx*crxz*car3*

f(x):j

' ' ' + c n x n +' ' '

(1)

whose radius of convergence is R > 0, it follows from successiveapplications of Theorem 15.8.3that / has derivatives of all orders on (-R, R). We say that such a function is infinitely differentiableon (-R, R). Successive differentiations of the function in (1) give f'(x):cr*2crx*3c"f*4cax3+''' : Z c r * 2 ' 3 c s *x 3 ' 4 c a f + ' ' ' f" (*) f"'(x)

:2'

l t i v r ( r ): 2 '

3cs+2' 3' 4cax*''' 3' Aca*'''+

etc. Letting r:

f'(o): c'

(2)

* (n- L)ncnxn-z+'''

(3)

+ (n-z)(n-7)ncnxn-g+'''

(n-3)(n-2)(n-L)nc"xn-4+'

0 in (1.),we get

f(o): co For t: 0 in (2) we see that

*ncnxn-r+'''

"

(4) (5)


If in (3) we take x:0, f " ( 0 ): 2 c 2

we have

a n ds o q : #

From (4), taking x:0,

we get

so ^:L"rl.o) f "'(0):2 ' 3csand. In a similar manner from (5), if x:0,

/(iv)(g):2

we obtain

/(i'if0)-

. 3 . 4cE and so ,.:

In general, we have f(n)(0) c":'-T-


(6)

(5) also holds when n - 0 if we take /(0)(0) to be /(0) and 0! : 1. (1) and (6) we can write the power seriesof f in x as (7) a more general sense, consider the functi on f as a power series in that is, f(x):)o

cn(x-a)"

: c 0* c r ( x - a ) * c r ( x - a . ) ,+ . . . + c n ( x- a ) " + . . .

(8)

If the radius of convergenceof this series is R, then is infinitely differ/ entiable on (a-R, a*R). successivedifferentiationsof the function in (8) give f' (x) : cr* 2cr(x- n) + 3c"(x- a)z+ 4cn(x- a)t + . *nco(x-a)n-r + " : . f " ( x ) 2 c r * 2 . ' 3 c " ( x a ) + 3 4 c a ( x- a ) , + .

'

* (n-l)ncn@3 c z + 2 ' 3 ' 4 c o ( x -a ) + . . .

'

f"' (x):2'

a ) n - z+ "

-t (n - 2)(n - 1.)ncn@ - a)n-t+ . . . etc. Letting x: atives, we get co: f @)

a in the power-series representations of f and its derivcr: f '(a)

v( n 2 -: f g

2I

Lc3B - :f

"'_(a) 3l

and in general Crr:

lrnt(a) "I

(e)

( 10)

case

EXAMPLEL: Find the Maclaurin series for e'.

soLUrIoN: It f (x) : st, f{n)(x) : e' for all r; therefore,f {n)(0): 7 for all n. So from (4 we have the Maclaurin series for e': *2

y3

7*x*z*g.+.

yn

. . +h+. . .

(11)

Note that series (1.1)is the same as the series for e' obtained in Example 3 of Sec.16.8. ExAMPLE2: Find the Taylor series for sin x at a.

sor,uriox: ltf(x): sinr, f'(*):cos x,f" (r):-sin x,f"'(x):-cos rr f(tu)(x): sin .r, and so forth. Thus, from formula (9) we have cs: sirl.A, cr:cos a, c2: ?sin a)121,cr: (-cos a)13!,cn: (sin a)141,and so on. The required Taylor series is obtained from (10), and it is sina+cosa(x-n)-srna

(x-

a)' _

zl

(x-

cAoAsâ

a)t

a

*sinrw*

(r2)

We can deduce that a power-series representation of a function is unique. That is, if two functions have the same function values in some interval containing the number n, and if both functions have a Powerseries representation in (x - a), then these series must be the same because the coefficients in the series are obtained from the values of the functions and their derivatives at a. Therefore, if a function has a powerseries representation in (r - a), this series must be its Taylor series at a. Hence, the Taylor series for a given function does not have to be obtained by using formula (10). Any method that gives a power series in (x- a) representing the function will be the Taylor series of the function at a. o rLLUsrRArroN L: To tind the Taylor series for et at a, we can write sr : eaer-a and then use series (11), where x is replaced by (x a). We then have vs r

- vs a f t *| ( x - va' )/ *, k : . a ) ' *, ( x - . a ) ' * \.^' L^

2!

3!

732

INFINITESERIES

. rLLUsrRArroN2: We can use the seriesfor ln(1 + r) found in Example3 '1.6.9 of Sec. to find the Taylor series for ln r at a (a > 0) by writing

l n r : l n [ a* ( x - a ) l : l n a * r " ( r . T ) Then because ln(1 + x) : n:l

we have

n - (*

(x!)t 3aB and the series represents ln x if. -L < (x- a)la - 1 or, equivalently, 0
ln a +x-

n

= ! ) ' *' 2a2

A natural question that arises is: If a function has a Taylor series in (x- a) having radius of convergence R > 0, does this series represent the function for all values of r in the interval (a - R, a * R) ? For most elementary functions the answer is yes. However, there are functions for which the answer is no. The following example shows this. ExAMPLE3: defined by ,/,

Let f bu the function

solurroN:

To find /'(0) we use the definition of a derivative. we have

(e-rttz if x*0

Ivc):tO

f ' ( u - l i m t - ' : :r--o

ifr:0

Find the Maclaurin series for f and show that it converges for all values of x but that it represents f (x) only when r: 0.

r-o

Becauselim (Ux): I-0

ger.

o

: ' j1T'

1, x

P

+@ and lim srttz: +oo we use L'Hopital's rule and 71-0

1 f

'(il - lim -rJ

1'-

x

r

r - op u r z l - \ : t " T f u - o \

x3/

By a similar method, using the definition of a derivative and L,H6pital,s rule, we get 0 for every derivative. So f(o(O):0 forall n. Therefore,the Maclaurin series for the given function is 0 + 0 + 0 + . . . +0 + This seriesconvergesto 0 for all r; however, if x + A, f @)- e-un # 0. A theorem that gives a test for determining whether a function is represented by its Taylor series is the following. L5.L0.L Theorem

Let f be a function such that / and all of its derivatives exist in some interval (a - r, a * r). Then the function is represented by its Taylor series *_ lot(a) (x '-:j.a)" ): n! "1"

16.10TAYLORSERIES 7gg

for all r such that l* - ol

: J1tR,(r) ;11'ffi

o @- a)n+,-

where each f, is between x and a. PRooF: In the intenral (a - r, A * r) , the functi on f satisfies the hypothesis of Theorem 15.5.1 for which : Pn(x) * R"(r) f (x)

(13)

where P is the nth-degree Taylor polynomial of f at a and R"(r) is "(x) the remainder, given by

R,(r) :f:*')(!!), ( n + 1 ) ! k-

(14)

a)n+r

where each f, is between x arrd a. Now Pn(r) is the nth partial sum of the Taylor series of / at a. So if we show that lim Pn(r) exists and equalsf (x) if and only if lim Rn(r) : 0, n-*@

i

the theorem will be proved. From Eq. (13) we have

P"(x): f(x) - R,(r) R"(r) :0, it followsfrom Eq. (15)that tf JIT

H

(1s)

R,(r) P"(x):f(x)Jlt :f(x)-o : f(x)

: f(x) we wish to show that Now under the hlryothesisthat ,111t,(t) lim R,(r) :0. FromEq. (13)we have ll-*'@

R,(x):f(x)-P"(x) and so

,tit

o,(t) : f(x)-,ti1" o,(t) : f(x)_ f(x) :Q

This proves the th6orem.

t

Theorem 15.10.1 also holds for other forms of the remainderR'(r) besides the Lagrange form. It is often difficult to apply Theorem 16.10.1in practice becausethe

794


values of {n ate arbitrary. However, sometimes an upper bound for Rn(r) can be found, and we may be able to prove that the limit of the upper bound is zero as n -+ *oo. The following limit is helpful in some cases: ryll

lim 1: n-+a

for all r

0

Tl!

(16)

This follows from Example 2 of Sec.76.7,wherc we showed that the power +@

series )

x"ln! is convergent for all values of r and hence the limit of its +@

nth term must be zeto. In a similar manner, because convergent for all values of x, we have (x- a)" lim forallr T-0

h:O

(17)

EXAMPLE 4: Use Theorem 1.5.10.l" solurroN: The Maclaurin series for e'is series (11) and to show that the Maclaurin series e€n for e*, found in Example L, represents the function for all values of x. where each fo is between 0 and r. We must show that lim Rn: 0 for all x. We distinguish three cases:

R"(r): !ffitrr"*'

tt'*@

x > 0 , x < 0 , a n d .r : lfx)0,then0

0.

t,r'=r, o v a\ , *n*r \ v (n* 1)!*

(18)

(n+l)I

From (16) it follows that lim )cn+rl(n* 1)l:0,

and so

rn+l

lim et zj;--:-:0 r-*e ln+ I)l Therefore, from (18) and the squeeze theorem (4.3.3) we can conclude that Rz(r) : 0. nlim If r < 0, then x 1 tn< p4,

o.6i1;1

x n +
0 and 0 < ero < 1..Therefote, if xn+r ) 0, yn+l

GT1;1

|

and if *n+r 10, *:.*:a - , , xo*t
,It

x"*tl(n + l) ! : 0, it follows that lim R, : 0.

Finally, if x:0, the serieshas the sum of 1 which i, ,{lllr"", can concludethat the series(1L)representseafor all valuesof r.

*"

From the results of the above example

(1e) and this agrees with Example 3 of Sec. "1.6.8.

ExAMPLE5: Show that the Taylor series for sin x at a, found in Example 2, rcpresents the function for all values of x.

solurroN:

so we must show that We use Theorem'1,6.'J",0.1;

g1n,(r):,1* 'ffi

(x-a)n+'l:s

t, f("*D(tn) will be one of the following numbers: Becausef(x):sin -sin fn. In any case,l/tz+tl1f")| s 1. Hence, -cos cos f,, sin fn, fn, or 0

l-

(20)

(n+l')I

From $n we know that lim lx - al"*tl(n * 1) ! : 0. Thus, by the squeeze n-*@

theorem (4.3.3)and (20) it follows

ExAMPLE 6: ComPute the value of sin 47" accurate to four decimal places.

that,ln R"(r)- 0.

solurroN: In Examples 2 and 5 we obtained the Taylor series for sin r at a which represents the function for all values of r. It is

s i n x - s i n a + c o sa ( x - a ) - s i n r r y - c o s

o(x-a)t* ' 3!

To make (x - a) small we choosea value of a near the value of r for which we are computing the function value. we also need to know the sine and cosine of a. We therefore choose a: Irr and have @ --in)' sin x - sin In *cos in(x - tn) - sin tn 2l (21)

Because4/ is equivalent to f{on radians: (*rr + erot) radians, from Eq. (21) we have

'*(#z)s+ ' ' ' sinrfr%.rr:trt/-z++{1.#rr-i{2.tGtr)'-+\/r : tl2(t+

+ 2' ' ' ) 0 . 0 3 4 9- 00 . 0 0 0 6- 10 . 0 0 0 0 0

Taking 11:1.4L421 and using the first three terms of the series,we get :0.73136 (1.03429) sin rttzr - lO.ZOZtt) Rounding off to four decimalplacesgives sin 4T - 0.7374.The error in-


troduced by using the first three terms is Rr(rttrr), and from (20)we have

Our result, then, is accurate to four decimal places.

The following Maclaurin series represent the function for all values of r: (22)

(23)

(z+1 (25) Series (22) is a direct result of Examples 2 and 5 with A : 0. You are asked to verify series (23),(24), and (25) in Exercises 1,2, and 3.

EXAMPLE 7:

f'

sin x

Jtiz x

Evaluate

O*

accurate to six decimal places.

solurroN: We cannot find an antiderivative of the integrand in terms of elementary functions. However, using series (22) we have sin r : - '1, sin.r )cx :- l/ r\^

x3,x5 3! 5!

xz xs 7!',9!

-14 -x2+t-x6+x8-... 3!'5! 7l',g! which is true for all x * 0. Using term-by-term f' sin r a,-.x : ) c - 3 ' 3 1x3 -,5 ' 5xi1 Jr,, x -

)cT ,

xe

integration we get ll

(1 - 0.0555555+ 0 .00L6667- 0.0000293+ 0.0000003- (0.5 - 0.0069444+ 0.00005210.0000002+ . . .)

.)

In each set of parentheses we have a convergent alternating series with lun*rl 1 lu,l' tn the first set of parentheseswe use the first four terms because the eror obtained is less than 0.0000003.In the second set of parentheses we use the first three terms where the error obtained is less

1 6 . 1 0T A Y L O RS E R I E S

than 0.0000002. Doing the arithmetic and rounding off to six decimal places,we get sin r ih - 0.4s297s [1 Juz x

16.10 Exercises 1. Prove that the series S

2. Prove that the series +@

(-l)nxtzn

sr*.

AW

3. Prove that the series

n2n+l

+@ \t

Qn + L)I

representscos x for all values of x.

"?o representssinh r for all values of x.

y2tt .v

Qn)I "?o representscosh x for all values of.x.

4. Obtain the Maclaurin series for the cosine function by differentiating the Maclaurin series for the sine function. Also obtain the Maclaurin series for the sine function by differentiating the one for the cosine function. 5. Obtain the Maclaurin series for the hyperbolic sine function by differentiating the Maclaurin seriesfor the hyperbolic cosine function. Also differentiate the Maclaurin series for the hyperbolic sine function to obtain the one for the hyperbolic cosine function. 6. Find the Taylor series for e" aI 3 by using the Maclaurin series for e". 7. Use the Maclaurin series for ln(l * r) to find the Taylor series for lnx at 2. 8. Given ln2-0.6937,

use the series obtained in ExerciseT to find ln 3 accurateto four decimal places.

In Exercises9 through 14, find a power-series representation for the given function at the number a and determine its radius of convergence. 9.f(x):ln(l

+x);a-L

1 2 .f ( x ) : 2 ' ; a : 0

1 1 .f ( x ) : t f r ; a : 4

10.f(x) : ! ,x a : L

U' f(x):ln

l3.f(x):cosr; a:tzr

15. Find the Maclaurin series for sin2 r. (nrNt: Use sinz x:

l r l ;a : - T

i0 - cos 2r).)

16. Find the Maclaurin seriesfor cos2r. (nrrvr: Use cos2x:*(7

* cos 2r).)

77. (a) Find the first three nonzeno terms of the Madaurin series for tan x. (b) Use the result of part (a) and term-by-term differentiation to find the first three nonzero tenns of the Maclaurin series for seC r. (c) Use the result of part (a) and term-by-term integration to find the first three nonzero terms of the Maclaurin series for ln cos r. In Exercises18 through 23, use a power series to compute the value of the given quantity to the indicated accuracy. L8. sinh i; five decimal places.

19. cos 58o; four decimal places.

2L. fFgO;five decimal places.

22. W;

three decimal places.

20. {r;

four decimal places.

23. ln(0.S); four decimal places.

24. Compute the value of e correct to seven decimal places, and prove that your answer has the required accuracy. In Exercises25 through 29, compute the value of the definite integral accurate to four decimal places. ru2 25. I sin x2 dx J0

26. .o, tE Ax |

r0.r

27. I

Jo

lr,(1+ sin x) dx


28.

rrB(*) , ,, \ tgry if x+o dx,where f f (x) : I x

JI o

fL i r , \ F-lgry 29. I S(x) dx, whereg(x) - I r Jo Lo

ifx-o

Lt

if x+o ifx-o

30. The function E defined by

E(x): +

[" e-t'dt

Yn Jo

is called the error function, anditis important in mathematical statistics.Find the Maclaurin seriesfor the error function. 31. Determine ao (n : 0, l, 2, 3, 4) so that the polynomial - 77xs+ 35f - 32xt 77 f(x) :3f is written in the form f(x): ar(x-7)a-t a"(x 1)t+ az(x-t)2* ar(x-t) * an

16.11'THE BINOMIAL

SERIES In elementary algebra you learned that the binomial theorem expresses (a * b)^ as a sum of powers of.a arrd b, where m is a positive integer, as follows: (a * b)*:

am* ma^-rb*m(m='2!9.v' rm(m-l).

l)

am-2b2 + .

. ._,.(m-k*7) kt

a n _ k b k + .. . + b n

we now take a : 1 and b: x, and apply the binomial theorem to the expression ('l' * x)^, where m is not a positive integer. We obtain the power series

L* mx*ry

*z*m(m 1)-(mz)f + (1)

-m\m-l)lm-2). .

. . ..(m-n*7\ n!

Series(1) is the Maclaurin seriesfor (l * x)*.It is calleda binomialseries. To find the radius of convergence of series (1), we apply the ratio test and get

1 6 . 1 1T H E B I N O M I A LS E R I E S

l'l : lrl So the series is convergent if lrl < 1. We now prove that series (1) represents (l * x)* for all real numberc m if.r is in the open interval (-1, 1). We do not do this by calculating Ro(r) and showing that its limit is zero becausethis is quite difficult, as you will soon see if you attempt to do so. Instead, we use the following method. Let

'(m-n+1)

m(m-1)'

xn

n!

n:l

We wish to show that f (x) -(1 * x)* , wherelrl we have +@

f'(x):

m(m-'|,)'

n:t

'(m-n*'L,) (n r)I

l'l Theorem L5.8.3

xn_r

lrl

(3)

Multiplying on both sides of Eq. (3) by x, we get from Theorem 1.6.3.8'

$

m(m- l)',

n:L

. . _ ' - t m -n * t ) n

(n- 1,)I

(4)

Rewriting the right side of (3), we have

f , ( x ): m * f m ( m - ' l ' ) ' ' '

( m- !-- n i n+ -r1) (n-lq

(s)

in (5) by decreasing the lower limit by 1 and Rewriting the ;;ation replacing n by n * L, we get

m(m-l)

f'(x):m*5@-11 If in (4)we *"n;; xf,(x)-y

',' n!

(m-n*l)rn

(6)

the numeratorand the denominatorby ,,we get

nm(m-L)'

(m-n*L) n!

*n

(7)

Becausethe series in (5) and (7) areabsolutely convergent for lrl < 1, then by Theorem 16.3.9we can add them term by term, and the resulting series will be absolutely convergent for lrl < 1.When we add we get

( 1+ x ) f(' r )- m f t * i m ( m r ) ' I

L

n:l

Becauseby (2) the expressionin bracketsis /(r), w€ have (1 + x)f '(r) - mf(x)

740


or, equivalently,

f ' ( x ): * f(x) L*x The left side of the above equation is Drfln f (x)]; so we can write

D , [ l nf ( x ) ] : h However, we also know that

D,[ln(l + x)*f _ m

L*x

Becauseln f (x) and 1n(1 * x)' have the same derivative, they differ by u constant.Hence, ln /(r) - ln(l + x)* + C We seefrom (2) that /(O) : L. Therefore,C - 0 and we get f(x):(1+x)* We have proved the generalbinomial theorem, which we now state. 16.ll.l

Theorem

lf m is any real number, then

Binomial Theorem

for all values of x such that lrl If m is a positive integer, the binomial series will terminate after a finite number of terms.

EXAMPLE

Express

sor,urroN: From Theorem 16.11.1we have, when lrl < 1,

(t + yy-',-1- L, * CA;E:!i * * W (-+)(-B)(-s) + +

1

\m as a power series in r.

*" ( - + - n+ 1 )

n!

-: Tr ^

1-

of

SOLUTION:

l'l

L '3

z*'22-2!*

+ (-L)" EXAMPLE 2: From the result Example L obtain a binomial

,

q,

'/."

3'5

-,,

23.3! 2n!

Replacing x by -x2 in the series for (1 + x)-rrz, we get for

1 6 . 1 1T H E B I N O M I A LS E R I E S 7 4 1

series for (1 - xz)-Ltzand use it to find a power series for sin-l .tr.

( 1- x z ) - u z - 1 + * x 2 '+2!2=- '2?! *= x 4' + 1 =' ;33! ; 9x 6 * . . . 23 z

( 2 n- 1 )

L.3.5 Zntt

x2n *

we integrate term by term and obtain Applying Theorem "1.6.9.L, il fr Jo \R-*'Z

.L

x3, L'3 '2r.2! 3

xr, I T

)r.5,L'3'5 ' 2t.3! 5

@_t) 2nl and so +00

sin-r x-

x*

1,.3.5

2"n!

n:l

ExAMPLE3: ComPute the value of % accurateto three decimal placesby using the binomial seriesfor (1 + x)ttt.

. . - (2n-1)

*2n*r

2n*1

for ltl

soLuTIoN: From Eq. (8) we have (1 +x)us-L

+1x*(il(-',J $* X*('J(-A(-:)

(e)

if ltl We can write (10)

: l - 0 . 0 2 4 7 - 0 . 0 0 0 5 - 0 . 0 0 0 0 3'- ' '

(11)

If the first three terms of the above series are used, we see from Eq. (14) of Sec.16.10 that the remainder is

/ 2\ _f"'(t) 1_e\' 2?) 3r-

Rz rrt \-

V

: (i)(-z)(-!)(il (1+€z)G*,-'(?)' where-+ < €,

:(ffi)t+#&)" lo,(?)l

Because -h

(12)

< t, ( 0, it follows that

(t + €r)'t'

(13)


Furtherrnore, 2.5

2.5 5t-.3i:

: 3 2' 6 ' 5 1 , 1 9=9

(14)

and so using inequalities (13) and (1a) in (L2)we see that |

/

2\l 1 | . ( 2 \ ' :ffi
2 \1/3

(1-+l \

:Q-9747

ztl

with an error less than 0.00005.From (10)we obtain %

- J(0.9747):2.9241

with an effor less than 3(0.00006)- 0.000L8.Rounding off to three mal placesgives % - /.924.

Exercises 1.6.11. In ExercisesI through 5, use a binomial series to find the Madaurin series for the given function. Determine the radius of convergence of the resulting series.

1 .f ( x ) : \ m

2. f(x):

4. f(x) - ff; 1t

s.f(x):A+-

3.f(x): (4+ x')-'

(9+ x4)-ttz

6.f(x):g+7

7. Intetrate term by term from 0 to r the binomial series for (l + 5"1-tr"to obtain the Maclaurin series for sinh-l r. Determine the radius of convergence, 8. Use a method similar to that of Exercise7 to find.the Maclaurin seriesfor tanh-l r. Determine the radius of convergence. In Exercises9 through 12, compute the value of the given quantity accurateto three decimal places by using a binomial series.

g. t/u

to.flu

rr. Veso

n. L

v3l

In Exercises 13 through 15, compute the value of the definite integral accurate to four decimal places. d, 1a. ft't 13' J. ffi

u:

Ittz

t/T=F ax J"'"

,,.

frt2

dx

Jo f,1- *t

,,

f1t2 . .

fsin-r r

(x)dx,where 'f(r) : l-r J:'"f Lt

'. ir x* o ifr:o

Reuiew Exercises(Chapter 16) In Exercises L throu gh 6, write the first four numbers of the sequence and find the limit of the sequence, if it exists.

s. {2+(-1)"}

6.{+--"' n - 2 )-Î ln*2

REVIEW EXERCISES743 through 13, determine if the series is convergent or divergent. If the series is convergent,find its sum. f n-"1, s \t" .?:,fl+ L

+6

S e-2' 8. LJ n:l

+@

+@

12.

11.

n:o

n:o

(nrNr: To find the sum, first find the sequence of partial sums.)

In ExercisesL4 through 25, determine if the seriesis convergent or divergent. 1,4

+@)

+@1 \r4

15.t " ,?- n' * 5n

?:, (2n * l)B

+@

+@/2,\

18.s+

17. ) c o s \ p 1 f

2 0 t. + "?,

f,:r \/3n + 2 +@1

e1)?'+1

2T,

1 + \/n

+@1 'r?\^ tLV' n(ln ?:,

n

n:l

-)a, ''

n)2

+co

nl s "'

*2oL0"

or divergent. In Exercises26 through 33, determine if the given series is absolutely convergent, conditionally convergent, Prove your answer. +@

q?n+l

+@

28.

26. Z(-l)"ffi.

n:l

n:0

+@

+@

29.

,n-l

3L.

1

n:l

n:l

t -+

cn:l 32.f ,*,where

i

t

7

(-+

33. frr*,wherecn:l

L

t

E

tf *n is an integer

,*, tf in is not an integer tf nis a perfectsquare

ii( n ii is not a Perfect square if n

In Exercises 34 through 42, find the interval of convergence of the given Power series. +@

34. .L/ s

+@

yn 'v

3s.t+

f,:r Yn

1n n:O ' +@

afl

yn

37').'ffi +@ (_I)n-txzn-r

40',,>-ffi

- L)" 38. ,,f;^"(2x

41. Z,#@*r)"

744

INFINITE SERIES

In Exercises43 through 50, use a power series to compute the value of the given quantity accurateto four decimal places. 43. tan-l + 47. cosio

44. sin 0.3

45. \7130

48. V-e

nt. [''n {i Jo

46. sin-r L sin x dx

so. ft cos f dr Jo

In Exercises51 through 53, find the Maclaurin series for the given function and find its interval of convergence. 5r. f(x):

a"(a > 0)

s2. f(x) -*

$. f(x):

s i n sr

In Exercises54 through 55, find the Taylor series for the given function at the given number. 5a. f @) : st-z; at 2

55. f (x):

sin 3r; at -tzr

56.f (x) :!;

at 2

Exercises57 through 61 pertain to the functions /s and /, defined by power series as follows:

: f, r-tl"#, /o(r)

I,k):;[,-t,' a#

The functions /6 and /, are called Besselfunctions of the first kind of orders zero and one, respectively. 57. Show that both Io aurrd converge for all real values of r. ]r 58. Show that /s'(r) : -l (x). 59. Show that D,(xlr(x)) : xlo@). 60. Show that y: Jo@) is a solution of the differential equation fru

ht

,#+ELy:o 51. Show that y : /, (r) is a solution of the differential equation ila ^ *u

f#+r&+(*-ryy:s

Vectors in the plane and parametric equations

VECTORSIN THE PI-,ANEAND PARAMETRICEQUATIONS

I7.I VECTORS IN THE PLANE

ft: RB F i g u r e1 7 . 1 . 1

In the application of mathematics to physics and engineering, we are often concerned with quantities that possessboth magnitudeand direction; examples of these are force, uelocity, acceleration,and displacement.Such quantities may be represented geometrically by a directedline segment. Physicists and engineers refer to a directed line segment as aoector, and, the quantities that have both magnitude and direction are called aector quantities.The study of vectors is called oector analysis. The approach to vector analysis can be on either a geometric or an analytic basis. If the geometric approach is taken, we first define a directed line segment as a line segmentfrom a point P to a point Q and denote this directed line segmentby PA,.The point P is called the initial point, and,the point Q is called the terminal point. Then two directed line segments PQ and R3 are said to be equal if they have the sane length and.direction, and, we write p? : nt (see Fig. 17.1,.1,). The directed line segmentPQ is called the aector fuomP to Q. A vector is denoted by a single letter, set in boldface type, such as A. In some books, a letter in lightface type, with an arrow above it, is used to indicate a vector, for example, Zl Continuing with the geometric approach to vector analysis, we note that if the directed line segment PQ is the vector A, and p?: R3, th" directed line segment n3 is also the vector A. Then a vector is considered to remain unchanged if it is moved parallel to itself. With this interpretation of a vector, we can assume for convenience that every vector has its initial point at some fixed reference point. By taking this point as the origin of a rectangular cartesian-coordinate system, a vector can be defined analytically in terms of real numbers. such a definition enablesus to study vector analysis from a purely mathematical viewpoint. In this book, we use the analytic approach; however, the geometric interpretation is used for illustrative purposes. we denote a vector in the plane by an ordered pair of real numbers and use the notation (x, y) instead of (x' y) to avoid confusing the notation for a vector with the notation for a point. we let vr be the set of all such ordered pairs. Following is the formal definition.

17.l.l Definition

A aector in the plane is an ordered pair of real numbers (x, y). T}r" numbers r and y are called the componenfsof the vector (x, yl. There is a one-to-one correspondencebetween the vectors (x, yl in the plane and the points (x,y) in the plane. Let the vector A be the ordered pair of real numbers (a1, a2l. lf we let A denote the point (ar, ar), then the vector A may be represented geometrically by the directed line segment 07. Such a-directe-dline segirent is called'a reprcsentationof. vector A. Any directed line segment which is equal to O-Ais also a representation of vector A. The particular representation of a vector which has its initial point at the origin is called the position representationof the vector.

17.1VECTORS IN THEPLANE 747

+z,k+g)

o rLLUsrRArrow1: The vector (2, 3) has as its position representationthe directed line segment from the origin to the point (2,3). Therepresentation of the vector (2, 3) whose initial point is (h, k) has as its terminal point (ft + 2,3 * k); refer to Fig. 17.1.2. o The vector (0, 0) is called the zero aector, and we denote it by 0; that is,

o-(o,o) Any point is a representation of the zero vector. F i g ur e 1 7. 1. 2

17.1..2Definition

Thremagnitudeof a vector is the length of any of its representations,and the directionof a nonzero vector is the direction of any of its representations. The magnitude of the vector A is denoted by lAl.

17.1.3 Theorem

If A is the vector (ar, arl, then lAl :!i718 pRooF: Becauseby Definition L7.1.2,lAl is the length of any of the representations of A, then lAl will be the length of the position representation of A, which is the distance from the origin to the point (ay az).So from the formula for the distance between two points, we have

r

l\l:ffi:Far4ar,

It should be noted that lAl is a nonnegative number and not a vector. it follows that l0l : 0. From Theorem 17.'1..3 o rLLusrRArroN2: If A:

(-3, 5), then

lAl: V(=5FTF :{fi

o

The direction of any nonzero vector is given by the radian measure 0 of the angle from the positive r axis counterclockwise to the position representationof the vector:0 < 0 < 2r. So if A: (at, arl, then tan 0: a2la1 if. a1# 0. If. ar:0 and a2) 0, then 0:tT. If. ar:O and a, ( 0, then and 17.1.5show the angleof radian measure e: fu. Figures 77.'1..3,17.'I...4, 0 for specific vectors (at, a"l whose position representations are drawn. (ar, ar) (ar, ar)

o Figure17.1.3

F i g ur e 1 7. 1, 4

Figure

748

VECTORSIN THE PLANEAND PARAMETRICEQUATIONS

ExAMPLEL: Find the radian measure of the angle giving the direction of each of the following v e c t o r s :( a ) ( - 1 , L > ; ( b ) ( 0 , - 5 ) ;

(c) (1,-2>.

soLUrIoN: The position representation of each of the vectors in (a), (b), '1.7 'J.7 and (c) is shown in Figures 17.7.6, ."1. .7, and .'J..8,respectively. (a) tan 0 -L so 0 - tn; (b) tan 0 does not exist and a2 (c) tan e - -2; 0 - tan-l (-2) * 2n.

0:

tan-t (- z)

F i g ur e 1 7 . 1. 7 Figure 17.1.6 -2) (0, - 5)

F i g u r e1 7 . 1 . 8

If the vector A: (ar, ar), then the representation of A whose initial point is (x, y) has as its endpoint (x * ar,A * ar).Inthis way a vectormay be thought of as a translation of the plane into itself. Figure 17.1.9illustrates five representations of the vector A : (nr, a2). In each case, A translates the point (xi, yr) into the point (xi I a1,yi * ar). v (at,,az)

(tt

*

(xn *

at,Ar._* or)

frt,U+ 1, or)

(*n, y n)

(xr,y,)

(xt, y t)

(xr, yr)

F i g ur e 1 7 . 1. 9

The following definition gives the method for adding two vectors. 17.1..4Definition

The sum of two vectors A: defined by

(nr, arl and B:

(br, br) is the vector A + B,

A + B : ( a r l b 1 ,a 2 * b 2 l . rLLUsrRArroN 3: If A:

A+B:(3+

(3, -1) and B : (-4,5), then

(-4),-1+5): <-I,4,

.

17.1VECTORS IN THEPLANE 749

(a2* b))

flt,A +

az)

P(x'y) x

F i g u r e1 7 . 1 . 1 0

17.1.5 Definition

The geometric interpretation of the sum of two vectors is shown in Fig. 17.1.10.Let A : (at, ar) and B : (br, bzl and let P be the point (r, y). Then A translatesthe point P into the point (x * at, y * a): Q. The vector B translatesthe point Q into the point ((r * a1)* br, (! * a) * b) or, equivalently, (x* (a1*b), A * @r* br)):R. Furthermore,A*B: (ar*br, arlbrl. Therefore,the vector A+B translatesthe point P PQ into the point (x * (at * bt), y * (ar+ br)): R. Thus, in Fig. 17.1'.10 is a representation of the vector A, QR is a representationof the vector B, and PR is a representation of the vector A + B. The representations of the vectors A and B are adjacent sides of a parallelogram, and the rePresentation of the vector A + B is a diagonal of the parallelogram. Thus, the rule for the addition of vectors is sometimes refened to as the parallelogramlaw. If A: \ar, az),thenthevector (-ar,-arl A, denoted by -A.

of is defined to be the negatitse

If the directed line segment p? is a representation'of the vector A, -A. Any directed then the directed line segment Q? is a representation of line segment which is parallel toPA,has the same length u" fp,and has a -A (see Fig. direction opposite to that of PA is also a representation of We now define subtraction of two vectors. 17.7.11').

F i g ur e 1 7. 1. 11

17.1.5Definition

The differenceof the two vectors A and B, denoted by A- B, is the vector obtained by adding A to the negative of B; that is, A-B:A*

(-B)

So if A: (ar, a2l andB: A - B : ( a r - b r ,a z - b r l

(br, br), then -B:

(-br, -br), and so

o rLLUsrRArroN4: If A : (4, -2> and B : ($, -3), then

A-B:<4,-2) -(6,-3) - (4,-2) + (-6, 3> - (- 2, L,

.12 Figure 1 7 . 1

To interpret the difference of two vectors geometrically, let the rePresentations oi the vectors A and B have the same initial point. Then the directed line segment from the endpoint of the representation of B to the endpoint of the iepresentation of A is a representation of the vector A B. This obeys the parallelogramlaw B + (A - B) : A (seeFig.17'1"72)' Another operation with vectors is scalarmultiplication.A scalaris a real number. Following is the definition of the multiPlication of a vector bv a scalar.

750

VECTORSIN THE PLANE AND PARAMETRICEQUATIONS

17.1.7 Definition

If c is a scalar and A is the vector (ar, ar), then the product of c and A, denoted by cA, is a vector and is given by cA: c(ar, a2l: (car, ca2l o rLLUsrRArroN5: If A:

(4, -5), then

3,4': 3(4, -5) - (12, -15) ExAMprn 2: If A is any vector and c is any scalar, show that 0(A)-0andc(0):0.

solurroN:

From Definition 77.1,.7it follows that

0 ( A ) - O ( a r ,n z ) - ( 0 , 0 ) : 0 and c(0) : c(0, 0) - (0, 0) : 0

We computethe magnitudeof the vectorcA as follows. cL:lGir)@

: lFG,4i7l :l? \/ii-+ a,, : lcllAl Therefore, the magnitude of cA is the absorute varue of c times the magnitude of A. The geometric interpretation of the vector cA is given in Figs. 17.1,.1g and 17.1.14.If c ) 0, then cA is a vector whose representationhai a length c times the magnitude of A and the same direction as A; an example of tlis is shown in Fig. 17.1.13,where c:3. If c ( 0, then cA is a vector whose representation has a length which is lcl times the magnitude of A and a direction opposite to that of A. This is shown in Fig. 1r.1.'t4,where

c:-t.

F i g ur e 1 7. 1. 1 3

F i g u r e1 7 . 1 ,14

Exercises L7.1 In Exercises I through 6, draw the position representation of the given vector A and also the particular representation through the given point P; find the magnitude of A.

17,2PROPERTIES OF VECTOR ADDITION AND SCALAR MULTIPLICATION 1.A- (3,41;P- (2,1) 4 . A - ( 4 , 0 > ;P : ( 2 , 6 )

2. A'-(-2,5);P:(3,-4)

751

<0,-2); P -- (-3 , 4) 6. A- (e,-t);P: (-2,-e)

3. A-

5 . A - ( 3, { Z > ; P : ( 4 , - \ n )

In Exercises7 through 12,find the vectorA havingPQ asa representation. Draw PQ and the position representation of A. 7.P: (3,7);Q: $,4) 1 0 .p : @ , 6 ) ; g : 1 2 , 3 6 )

8 .P : ( 5 , 0 ; Q : Q , 7 ) r t . p : ( - 5 , - 3 ) ; p : ( 0 ,3 )

9 .P : ( - 3 , s ) ; Q : e 5 , - 2 ) 12p . : t-t/1,0); e: (0,0)

R3 ur" eachrepresentations of the samevector. In Exercises13through 15,find the point S so that pi ""a 14.P:(-r,D;Q:e,-3);R:(-s,-2) 73.P:(2,5);Q:0,6);R:(-3,2) ts.P:(0,3);Q:(s,-z);R:(7,0)

15.P:(-2,0);Q:e3,-4);R:(4,2)

In Exercises17 through22, frnd.the sum of the given pairs of vectorsand illustrategeometrically. 1 9 .( - 3 , 0 ); ( 4 , - 5 )

1 8 .( 0 ,3 ) ;( - 2 , 3 > 2 r .( 0 ,0 ) ;( - 2 , 2 >

1 7 .( 2 , 4 1(;- 3 , 5 ) 2 0 .( 2 , 3 ) ;( - \ n , - 1 )

23 through28,subtractthe secondvectorfrom the first In Exercises 2 4 .( 0 ,5 ) ; ( 2 , 8 > 2 3 .( 4 , 5 ) ;( - 3 , 2 >

2 2 .( 2 , s ) (; 2 , 5 > and illustrate geometrically.

2 7 .( 0 , \ 6 ) ; ( - t D - , 0 >

2 5 .( l , e ) ; ( - 3 , 2 e 1

25.(-3, -4>; (6, 0) 2 8 .( 3 , 7 > (; 3, 7 >

29. Given A: (2, -5); B: (3, 1); C: (-4,2). (a) Find A + (B * C) and illustrate geometrically.(b) Find (A + B) + C and illustrate geometricallY. In Exercises30 through 35, let A - <2,4>, B (4, -3) and C (-3, 2>. 3 0 .F i n d A + B 33. Find lc - Bl 3 6 .G i v e nA = ( 9 , 2 > ;C : 37. Lett?

3 1 .F i n d A - B

32.Find lCl

34. Find 2A + 38

35. Find 17A- Bl

(8,8); A + B: C; find lBl.

U" a representation of vector A, Q? be a representation of vector B, and RS b" u representation of vector C.

A=n,and RT are sides of a triangle, then A * B * C:0' 38. Prove analytically the triangle inequality for vectors lA + Bl < lAl + lBl' Prove that t&,

17.2 PROPERTIES OF VECTOR ADDITION AND SCALAR MULTIPLICATION

The following theorem gives laws satisfied by the operations of vector addition and scalar multiplication of any vectors in V2.

L7.2.1 Theorem

If A, B, and C are any vectors inVr, and c and d are any scalars,then vector addition and scalar multiplication satisfy the following properties: (commutative law) (i) A + B : B * A (associativelaw) (ii) A + (B + C) : (A + B) + C (iii) There is a vector 0 inV, for which (existence of additive identity) A * 0: A (iv) There is a vector -A in V, such that (existenceof negative) A + (-A) :0

EQUATIONS VECTORSIN THE PLANEAND PARAMETRIC

(v) (vi) (vii) (viii)

(associativelaw) (cd)A - c(dL) (distributive law) c(A + B) - cA * cB (distributive law) (c * d)A: cA + dA (existenceof scalarmultiplicative identity) 1(A) - A

pRooF: We give the proofs of (i), (iv), and (vi) and leave the others as exercises (see Exercises 16 through 19). Let A : (at, a"l and B : (bt, b"l. PROOFOF ( i ) : By the commutative law for real numbers, ar * br: and flz * b z : b, * az, and so we have

A+ B-

b1* a,

(ar, arl + (br,br) (at * br, a, * br) (b, * flr, bz * azl (bt, brl + (ar, qrl

B+A pRooF or (iv): The vector -A A + (-A)

is given by Definition 17.1.5 and we have

:

(ar, azl * (-au-azl

:

(at *

(-at) , a, *

(-ar)l

- (0,0) -0 PRooF oF (vi):

c ( A + B ) - c ( ( a r , n z l+ ( b r , b r l ) c ( (a t * b r ,a r + b r ) ) ( c ( a r * b , , ) ,c ( a r * b r ) l (ca, * cbr,caz* cbz) (car, caz)+ kbr, cbrl c(ar, nzl* c(br, brl cA* cB '1.: Verify (ii), (iii), (v), EXAMPLE (vii), and (viii) of Theorem 17.2.1

SOLUTION:

A+(B+C):

if A: <3,4),8: (-Z,L>, c - ( 5 ,- 3 ) , c : 2 , a n dd - - 6 . (A+B)*C:

(3,4> + ( ( - 2 ,1 )+ ( 5 ,- 3 ) ) (3,4) + <3,-2> < 62, > ( ( 3, 4 >+ ( - 2 ,1 ) )+ ( s ,- 3 )

OF VECTORADDITIONAND SCALARMULTIPLICATION 753 17,2 PROPERTIES

: (1,5)+ (5,_3)

: (6'2> A + (B + C) : (A+ B) * C, and so (ii) holds. Therefore, A + 0 : \ 3 , 4 l + ( 0 , 0 ) : ( 3 ,4 ) : A Hence,(iii) holds. (cd)A: [(2)(-6)](3,4)

: (_12)(3,4> : (_96,_49) c(dl ) :2(-6< 3' 4> ) : 2(-18, -24) : (_36, _49, Thus, (cd)A:

c(ilA), and so (v) holds.

(c+ d)A : 12+( - 5) l( 3,4): ( - 4) ( 3,4;: ( - 12' - 16l cA * dA: Z(9,4)+ (-6) (3,4l : (5,8) + (-78, -24>: \-12, -15> Therefore, (c + d)A:

cA * ilA, and (vii) holds.

1 A : 1 ( 3 ,4 ) : ( ( 1 )( 3 ) , ( 1 )( 4 )) : ( 3 , 4 ) : A and so (viii) holds.

Theorem 17.2.':.isvery important becauseevery algebraic law for the operations of vector addition and scalar multiplication of vectors in Vr can be derived from the eight properties stated in the theorem. These laws are similar to the laws of lritirmetic of real numbers. Furtheruror€r ill linear algebra, a "real vector space" is defined as a set of vectors together with a sel of real numbers (scalars)and the two operations of vector addition and scalar multiplication which satisfy the eight properties given in Theorem 17.2.1.Following is the formal definition' 17.2.2 Defrnition

A real oectorspaceV is a set of elements, called z)ectors'together with a set of real numbers, called scalars,with two operations called aectot addition and.scalarmultiplication such that for every pair of vectors A and B in V and for every scalar c, a vector A + B and a vector cA are'defined so that properties (i)-(viii) of Theorem 17'2"1'are satisfied' From Definttion 77.2.2 and Theorem 17.2.1 it follows that v, is a real vector space.

754

VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS

We now take an arbitrary vector in V, and write it in a special fbrm. A:

\ a r , a " l : ( a r , 0 ) + ( 0 , a r l : a r ( L , 0 )* a r ( 0 ,L )

Becausethe magnitude of each of the two vectors (1, 0) and (0, 1) is one unit, they are called unit aectors.we introduce the following notations for these two unit vectors: ( 1 ,0 ) a n d i : ( 0 , t ) The position representation of each of these unit vectors is shown in Fig. Because 17.2.1.. i:

F i g u r e1 7, 2 . 1

: arii- ari (ar, ar): a1(1,01* ar(O,'!,) it follows that any vector in v, can be written as a linear combination of the two vectors i and j. Becauseof this the vectors i and j are said to form a basis for the vector space vr. The number of elemenis in a basis of a vector spaceis called the dimensionof the vector space.Hence, v, is a twodimensional vector space. . rLLUsrRATrow L: We expressthe vector <9,-4, in terms of i and i. (3, -41 : 3(1, 0),+ (-4) (0, 1; : 3i - 4i

(ar, ar,)

.

Let A be the vector \ar., arl and 0 be the radian measure of the angle giving the direction of A (see Fig. 17.2.2where (ar, az) is in the se"ottd quadrant). ar : lAl cos g and ar: lAl sin 0. BecauseA: ari * ari,we can write

A - lAl cos0i + lAl sin 0j

Figure 17.2.2

or/ equivalently, A - lAl(cos 0i + sin gj)

(1)

Equation (1.) expressesthe vector A in terms of its magnitude, the cosine and sine of the radian measure of the angle giving the direction of A, and the unit vectors i and i. EXAMPLE

(-5 , -2),

Expressthe vector the form o f E q . ( 1 ) .

SOLUTION:

RCfCr

to Fig. 17.2.3,which shows the position representation

of the vector (-5,

.2>.

l(-s,-2)l: s i ne - - +

\E

and cos

So from (1) we have

(-5,-2) - \E Figure 1 7 . 2 . 3

/5.2.\

\-aEr-@t)

\E

AND SCALARMULTIPLICATION755 ADDITION 17,2PROPERTIES OF VECTOR

17.2.3 Theorem

If the nonzero vector A: ari * ari, then the unit vector U having the same direction as A is given by (2)

PRooF:

(2),

From 11

U: ri

lAl

@j* a2j):;T

(A)

lAl

Therefore,U is a positive scalartimes the vector A, and so the direction of U is the stune as the direction of A. Furthermore,

:,t(ft)'*(ft)' tur _\ffi lAl _ lAl lAl -1 Therefore,U is the unit vector having the same direction asA, and the theorem is proved. ExAMPrn3: Given A - (3, 1) and B - <-2, 4>, find the unit vector having the same direction as A-B.

solurroN: A - B - (3, 1) - (-2, 4) :

So we may write

A-B-5i-3i Then

lA-Bl :ffi-\E By Theorem 17.2.3,the desired unit vector is given by

u-#i-fti Exercises1-7.2 In Exercises1 through 8, let A -2i + 3i and B 4i- i.

L.FindA+B

2. FindA-B

4 .F i n dl A l l B l 7. Find l3A - 2Bl

5. Find lA + Bl 8. Find l3Al- l2Bl

9. Given A:

8i * 5j and B :3i

-

3. Find 5A - 68 6. Find lAl + lBl

i; find a unit vector having the same direction as A * B.

I


10. Given A :-8i + 7i; B : 5i- 9i; C: -i- i;find l2A - 38 - Cl. 11. Given A : -2i+ j; B : 3i - 2i; C : 5i - 4j; hnd scalarsh and,k such that C : hA * kB. In Exercises12 through 15, (a) write the given vector in the form r(cos 0i * sin 0i), where r is the magnitude of the vector and 0 is the radian measure of the angle giving the direction of the vector; and (b) find a unit vector having the same direction. 12. 3i-3i 13. -4i+ 4\/3i 14. -r6i rs. 2i 16. Prove Theorem 17.2.1(ii).

17. Prove Theorem 17.2.1(v).

18. Prove Theorem 17.2.1(vii).

19. Prove Theorem 17.2.7(iii)and (viii).

20. Two vectors are said to be independentif. and only if their position representations are not collinear. Furthermore, two vectors A and B are said to form a basisfor the vector spaceV, if and only if any vector in V, can be written as a linear combination of A and B. A theorem can be proved which statesthat two vectors form a basis for the vector spaceV, if they are independent. Show that this theorem holds for the two vectors (2, 5) and (3, -1) by doing the following: (a) Verify that the vectors are independent by showing that their position representations are not collinear; (b) verify that the vectors form a basis by showing that any vector ari l a2ican be written as c(2i + 5i) + d(31- il, where c and . d are scalars.(nrwr: Find c and il in terms of a, and ar.\ 21' Refer to the first two sentenies of Exercise 20.A theorem can be proved which statesthat two vectors form a basis for the vector spaceV, only if they are independent. Show that this theorem holds for the two vectors (9,-2) and (-6,4) by doing the following: (a) Verify that the vectors are dependent (not independent) by showing that their position representations are collinear; (b) verify that the vectors do not form a basis by taking a particular vector and ihowing that it cannot be written as c(3i + 2j) + d(-6i + 4i), where c and d are scalars.

17'3 DOT PRODUCT

17.3.1 Definition

In Sec.17.1 we defined addition and subtraction of vectors and multiplication of a vector by a scalar.However, we did not consider the multiplication of two vectors. we now define a multiplication operation on two vectors which gives what is called the ,,dot product.,, If A : (at, azl and B : (bt , brl are two vectors in Vr, then the dot product of. A and B, denoted by A . B, is given by A . B:

( a r , a r l . ( b r , b r l : a r b r* a " b ,

The dot product of two vectors is a real number (or scalar) and not a vector. It is sometimes called tlrrescalarproduct or inner product. o rLLUsrRArroN 1: If A:

(2,-9) and B : (-+,4), then

A . B : (2,-3) . (-i, +>: (2)(-+) + (_3)(4): _13

.

The following dot products are useful and are easily verified (see Exercise 5). i.i:L

(1)

i .i - L i'i:o

(2) (3)

17.3 DOT PRODUCT

757

The following theorem states that dot multiplication is commutative and distributive with respect to vector addition. 17.3.2 Theorem

If A, B, and C are any vectors in V2, then (i) A. B: B . A (ii) A' (B+C) :A.

(commutativelaw) (distributivelaw) B+A.C

The proofs of (i) and (ii) are left as exercises(see Exercises5 and 7). Note that becauseA. B is a scalar, the expression (A . B) ' C is meaningless. Hence, we do not consider associativity of dot multiplication. Some other laws of dot multiplication are given in the following theorem. 17.3.3 Theorem

If A and B are any vectors in V2 and c is any scalar, then (i) c(A. B) : (cA)'B; (ii) 0'A:0i (iii) A' A: lAl'. The proofs are left as exercises(seeExercises8 through 10). We now consider what is meant by the angle between two vectors, and this leads to another expression for the dot product of two vectors.

17.3.4 Definition

Let A and B be two nonzero vectors such that A is not a scalarmultiple of B. If O-Fis the position representation of A and O? is the position rePresentation of B, then the angle betweenthe aectorsA and B is defined to be the angle of positive measure between O? and OQ imerior to the triangle POQ.If A: cB, where c is a scalar,then if c > 0 the angle between the vectors has radian measure 0; if. c < 0, the angle between the vectors has radian measure ?r. It follows from DefinitionL7.3.4 that if a is the radian measure of the angle between two vectors, then 0 < d. < z. Figure 17.3'1 shows the angle between two vectors if A is not a scalar multiple of B. The following theorem is perhaps the most important fact about the dot product of two vectors.

F i g u r e1 7 . 3 . 1

17.3.5Theorem

If a is the radian measure of the angle between the two nonzero vectors A and B, then A . B:

l A l l n l c o sa

(4)

thepositionrePreari* ariandB :bi*b2i'LetOFUe sentation of A and oQ be the position representation of B. Then the angle

pRooF: LetA-

THE PLANEAND PARAMETRIC EQUATIONS

Q(br, bz) P(or, a z)

between the vectors A and B is the angle at the origin in triangle POQ (see Fi9.77.3.2); P is the point (ar, ar) and Q is the point (01,b2).In triangle OPQ, lAl is the length of the side OP and Inl is the length of the side Oe. So from the law of cosines we obtain lvog-------=_-

l A l ' + l B l -, l P Q l , 2lAllBl

_ (arz+ ar2)I (brz* br2).-. .l(a, - br)z* (a, - br)zf

2lAllBl

Figure17.3.2

_ 2arb,* 2arbz

2lAllBl

_ lrbr * arb,

lAllBl

Hence, cosa:

,A;, B,

lAllBl


A . B : l A l l n lc o sa

r

Theorem L7.3.5states-thatthe dot product of two vectors is the product of the magnitudes of the vectors and the cosine of the radian measure of the angle between them. o rLLUsrRArroN2: If A : 3i - 2i, B :2i * i, and a is the radian measure of the angle between A and B, then from iheorem 17.3.5,we have

cosc: I 'l B, lAllBl GX2) + (-2)(1)

\M\M

6-2 ____,V13 V5

:+y65

0

we leamed in sec. 17.2 that if two nonzero vectors are scararmultiples of each other, then they have either the same or opposite directions. We have then the following definition. 17.3.6 Definition

Two vectors are said tobe parallel if. and only if one of the vectors is a scalar multiple of the other. . rLLUsrRArrow3: The vectors <9, _4, and (*, _l) '(3' -4) : 4(1' -11'

are parallel because o

17,3 DOT PRODUCT

759

If A is any vector, 0 : 0A; thus, from Definition 17.9.6it follows that the zero vector is parallel to any vector. It is left as an exercisefor you to show that two nonzero vectors are parallel if and only if the radian measure of the angle between them is 0 or zr (seeExercise33). If A and B are nonzero vectors, then from Eq. (4) it follows that c o sd : 0

if andonlyif A . B:0

Because0 < a < fi, it follows from this statement that a:

trr if and only if

A . B :0

We have, then, the following definition. 17.3.7Definition

Two vectorsA and B are said tobe orthogonal(perpendicular) if and only if A ' B:0. . ILLUsrRArrou4: The vectors <-4,5) and (10, 8) are orthogonalbecause

(-4, s, . (10,8) (10)+ (s)(8) : F*,

o

If A is any vector, 0 . A:0, and therefore it follows from Definition 17.3.7 that the zero vector is orthogonal to any vector.

EXAMPLE1: GivCN A _ 3i + 2i and B - 2i + ki, where k is a scalar, find (a) k so that A and B are orthogonal; (b) /c so that A and B are parallel.

solurroN: (a)By Definition17.3.7, A artdB areorthogonalif and only if A . B : 0 ; t h a ti s , (3)(2) + 2(k) : s Hence,

k: -3 (b) From Definition 17.3.6,A and B are parallel if and only if there is some scalarc such that \3,21 : c(2, kl; that is, 3-2c

and 2-ck

Solving these two equations simultaneously, we obtain

4 3.

A geometric interpretation of the dot product is obtained by considering the projectionof a vector onto another vector. Let O-?and O! be representations of the vectors A and B, respectively. See Fig. 17.3.3.The ++ projection of.OQ in the direction of.OP is the directed line segment OR, wherb R is the foot of the perpendicular from Q to the line containing O?. Then the vector for which O? is a representation is called the oectorprojec-


tion of the vector B onto the vector A. The scalarprojectionof B onto A is defined to be lBl cos d, where a is the radian measure of the angle between A and B. Note that lBl cos cr may be either positive or negative depending on a. BecauseA . 3: lAllBl cos a, it follows that A . B : l A l ( l B l c o sa )

F i g u r e1 7 . 3 . 3

F)

Hence, the dot product of A and B is the magnitude of A multiplied by the scalar projection of B onto A. See Fig. 17.3.4aand b. Becausedot multiplication is commutative, A . B is also the magnitude of B multiplied by the scalarprojection of A onto B.

l B lc o s o ) 0 (a) Figure 17.3.4

EXAMPLE 2: Given A: -5i + i and B - 4i + 2i, find the vector projection of B onto A.

solurroN: Figure 17.3.5shows the position representations of vectors A and B as well as that of c, which is the vector projection of B onto A. From (5) we get

(-s)(a)+ (t) (z) \B

l B lc o sa : t ; P : lAl

Therefore,lCl : 181\8. Becausecos a 1 0,tn ( a ( n, and so the direction of c is opposite that of A. Hence, c: cA, and c < 0. Then we have C--Sci + ci Figure17.3.5

BecauselCl : l8l \8,

w€ get

\DW and so c = -+,

If A:

ai*

A.i:ar


azi,then and A.i:a,

Hence, the dot product of A and i gives the component of A in the direction of i and the dot product of A and i gives the component of A in the direction of j. To generalize this result, let u be any unit vector. Then

17,3DOTPRODUCT 761

from (5) A.

lAl(lul cosa) : lAl cosa

and so A . U is the scalarprojection of A onto U, which is called the componentof the vector A in the direction of U. More generally, the component of a vector A in the direction of a vector B is the scalarprojection of A onto a unit vector in the direction of B. In Sec.8.5 we stated that if a constant force of F pounds moves an object a distance d feet along a straight line and the force is acting in the direction of motion, then if W is the number of foot-pounds in the work done by the force, W: Fd. Suppose, however, that the constant force is not directed along the line of motion. In this casethe physicist defines the work done as the productof the componentof theforce alongthe line of motion timesthe displacemenf.If the object moves from the point A to the point B, we call the vector, having aE ^s a representation, the displacementaector and denote it by V(Al). So if the magnitude of a constant force vector F is expressedin pounds and the distance from A to B is expressedin feet, and a is the radian measure of the angle between the vectors F and V(fr), then if W is the number of foot-pounds in the work done by the force F in moving an object from A to B,

w: (lFlcoso) lv(AT)| : lFllv(AE)lca os : F . V(A?) EXAMPLE3: Suppose that a force F has a magnitude of 6lb and trr is the radian measure of the angle giving its direction. Find the work done by F in moving an object along a straight line from the origin to the point P (7 , t) , where distance is measured in feet.

solurroN: Figure 17.3.5shows the position representationsof F and V(O?). If W ft-lb is the work done,then W: F .V(O?) F: (6 costzr, 6 sin $zr): 13t6, 3), andV(Oi') : (7, 1l. So YJ: (313,3, . (7, t) :27l.8 + g - 39.37 Therefore,the work done is 39.37ft-lb. v

P(T ,

-l.,)

x

Figure 17.3.6

Vectors have geometric representationswhich are independent of the coordinate svstem used. Becauseof this, vector analysis can be used to


prove certain theorems of plane geometry. This is illustrated in the following example. ExAMPLE 4: Proveby vector analysis that the altitudes of a triangle meet in a point.

sot,urroN: Let ABC be a triangle having altitudes AP and BQ intersecting at point S. Draw a line through C and S intersecting AB at point R. We wish to prove that RC is perpendicular to AB (see Fig.17.3.7).

Letfr, Bt., eA.,fS, n3,CJ U" representations of vectors. ret v(A-b)

C

be the vector having directed line segment Ai ur a representation. In a

similarmanner,let V(Bt), v(a?), v(I3), v(B), and v(G) be the vectors having the directed line segment in parentheses as a representation. BecauseAP is an altitude of the triangle,

v(,B).vtrtl-o

(5)

Also, because BQ is an altitude of the triangle,

-o v(BT)-vtfrl

A Figure17.3.7

(7)

To prove that RC is perpendicular to AB, we shall show that

v(c3).v(fr):o v(G).vtfrl -vtGl . [vtfrl +v(eB)] -v(G) .vtfrl +vtdl .vrcil ->+->

: L V ( C B+) V ( B S ) l . V ( A C )

+ tv(Aal+ v(AJ)l. vtdEl

- v(ci).v tfrl + v(r3).v(fr) +v€al.vteEl+v(i3).v(eB) Replacirgv(C7) by -v tfrl and using (6) and (T),we obtain

v(Gl 'v rfiu '- v(eB)" v (fr) + 0+ t-v(At)l.v(eB)+ 0 -0 Therefore, dtitudes AP, BQ, and RC meet in a point.

Exercises17.3 In ExercisesL through 4, find A . B.

1 .A - ( - 1 , 2 > ; B : ( - 4 , 3 > 3 .A - 2 i - i ; B : i * 3 i 5 . S h o wt h a ti . i : 1 . ; i , i : l ; i 8. Prove Theorem 17.3.3(i).

2 .A - ( + , - + ) ; B(:* , * > 4.4--2i;B--i+i .

i:0.

6. ProveTheorem17.3.2(i).

7. Prove Theorem 17.3.2(ii).

9. ProveTheorem17.3.3(ii).


17.4 VECTOR-VALUED FUNCTIONSAND PARAMETRICEQUATIONS

In Exercises11 through 14, if. a is the radian measure of the angle between A and B, find cos a. 1 1 .A :

(4,31;8: (1,-1)

1 2 .A :

1 3 .A : 5 i - 1 2 j ; B : 4 i * 3 i

(-2,-31;8:

(3,2>

1 4 .A : 2 i + 4 j ; B : - 5 i

15. Find k so that the radian measure of the angle between the vectors in Example 1 of this section is *a. 16. Given A:

ki-2i

and B:

17. Given A:5i-ki;B:ki+5j, are parallel.

ki* 6i, where k is a scalar.Find k so that A and B are orthogonal. where k is a scalar.Find (a) k so that A andB are orthogonal;(b) k so thatAandB

18. Find k so that the vectors given in Exercise15 have opposite directions. 19.GivenA:5i+12j;B:i*kj,wherekisascalar.Findksothattheradianmeasureof

theanglebetweenAandBislzr.

20. Find two unit vectors each having a representation whose initial point is (2,4) and which is tangent to the parabola Y : 12 there' 21,.Find two unit vectors each having a representation whose initial point is (2, 4) and which is normal to the parabola Y - 12 there' 22. If A is the vector ari.| arj, find the unit vectors that are orthogonal to A. 6i, frnd the vector projection of A onto B. 23. If A: -8i * 4i and B:7i24. Find the vector projection of B onto A for the vectors of Exercise23. 25. Find the component of the vector A : 5i - 6i in the direction of the vector B:7i

+ i. direction of vector A. 25. For the vectors A and B of Exercise 25, find the component of the vector B in the

27. A vector F represents a force which has a magnitude of 8 lb and *z as the radian measure of the angle giving its direction. Find the work done by the force in moving an object (a) along the r axis from the origin to the point (5, 0) and (b) along the y axis from the origin to the point (0, 6). Distance is measured in feet. 28. Two forces represented by the vectors F1 and F2 act on a particle and cause it to move along a straight line from the point (2, 5) to the point (7, 3). If Fl : 3i - i and F2: -4i + 5j, the magnitudes of the forces are measured in pounds, and distance is measured in feet, find the work done by the two forces acting together. 29. If A and B are vectors, prove that

( A + B ) ' ( A + B ): l ' '

A+2A' B+B'B

30. Prove by vector analysis that the medians of a triangle meet in a point. 31..Prove by vector analysis that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and its length is one-half the length of the third side. 32. Prove by vector analysis that the line segment joining the midpoints of the nonparallel sides of a trapezoid is parallel to the parallel sides and its length is one-half the sum of the lengths of the parallel sides. 33. Prove that two nonzero vectors are parallel if and only if the radian measure of the angle between them is 0 or z.

17.4 VECTOR-VALUED FUNCTIONS AND PARAMETRIC EQUATIONS

We now consider a function whose domain is a set of real numbers and whose range is a set of vectors. Such a function is called a "vector-valued fimction." Following is the precise definition.

17.4.1 Definition

Letl and I be two real-valued functions of a real variable f. Then for every

764

VECTORS IN THE PLANEAND PARAMETRIC EQUATIONS number f in the domain common to / and g, there is a vector R defined by

R(f):f(f)i+s(f)i

(1)

and R is called a aector-aalued function. o ILLUSTRATToNL: Suppose

R(f): \Fi+

(t- 3)-'i

Let a n dg ( f ) : ( t - 3 ) - '

f (t): \F

The domain of R is the set of values of f for which both /(t) and g(f) are defined. The function value /(t) is defined for f * 2 and g(f) is defined for all real numbers except 3. Therefore, the domain of R is

{tlt-z,t+S}.

.

If R is the vector-valuedfunction defined by (1), as f assumesall values in the domain of R, the endpoint of the position representation of the vector R(f) haces a curye C. For each such value of f, we obtain a point (x, y) on C for which

x-f(t)

and y-sG)

(2)

The curve c may be defined by Eq. (1) orEqs. (2). Equation (1) is calleda aectorequationof c, and Eqs. (2) are called parametricequationsof C. The variable t is aparameter.The curve C is also called a graph; that is, the set of all points (x, y) satisfying Eqs. (2) is the graph of the vector-varued function R. A vector equation of a curve, as well as parametric equations of a curve, gives the curve a direction at each point. That is, if we think of the curye as being traced by a particle, we can consider the positive direction along a curve as the direction in which the particle moves as the parameter f increases.In such a caseas this, f may be taken to be the measure of the time, and the vector R(t) is called the positionaector.sometimes R(f) is referred to as the radius aector. If the parameter f is eliminated from the pair of Eqs. (2), we obtain one equation in r and y, which is called a cartesianequationof c. It may happen that elimination of the parameter leads to a cartesian equation whose graph contains more points than the graph defined by either the vector equation or the parametric equations. This situation occurs in Example 4. EXAMPLE

Given the vector

equation R(f)

cos ti + 2 srn ti

SOLUTION: The domain of R is the set of all real numbers. We could tabulate values of x and y for particular values of t. However, if we find the magnitude of the position vector, we have for ev ery t

l R ( f )-l t / 4 c o s z i + Z ; i p i _ 2 m _ z r

(a) draw a sketchof the graph of

17.4 VECTOR-VALUED FUNCTIONSAND PARAMETRIC EQUATIONS

this equation,and (b) find a cartesian equation of the graph. v P(2 cost, 2 sin t)

Therefore, the endpoint of the position representation of each vector R(f) is two units from the origin. By letting f take on all numbers in the closed interval 10,2nl, we obtain a circle having its center at the origin and radius 2. This is the entire graph becauseany value of f will give a point on this parametric equations of circle. A sketch of the circle is shown in Fig. 17.4.1,. the graph are x,:2cosf

and y:2sint

A cartesian equation of the graph can be found by eliminating f from the two parametric equations, which when squaring on both sides of each equation and adding gives 17.4.1

* + Yz:4 As previously stated, upon eliminating f from parametric equations (2) we obtain a cartesian equation. The cartesian equation either implicitly or explicitly defines y as one or more functions of r. That is, if x: f (t) andy : g(f), then y : h(x).lf.h is a differentiable function of r and / is a differentiable function of.t, it follows from the chain rule that DtA: (D"y)(Dtx) or

g,G): (h,(x))(f,(t)) or, by using differential notation, dy :dy dx dt dx dt If. dxldt # 0,we can divide on both sides of the above equation by dxlilt and obtain

(3)

Equation (3) enables us to find the derivative of y with respect t o x directly from the parametric equations.

ExAMPun2: Given x:3t2 and y -- 4tB, find dy ldx and dzyldxz without eliminating t.

solurroN:

APplying (3), we have

EQUATIONS VECTORSIN THE PLANEAND PARAMETRIC

d(y') dt dx dt Becausey' :2t,

d(y')ldt:2;

thus, we have from the above equation

dry dx2

ExAMPrn 3: (a) Draw a sketch of the graph of the curye defined by the parametric equations of Example 2, and (b) find a cartesian equation of the graph in (a).

solurroN: Becauser:3P, we conclude that r is never negative. Table gives values of r and y for particular values of f. BecauseDry:2t, 17.4.1, we see that when t:0, Dry: 0; hence, the tangent line is horizontal at the point (0,0). A sketch of the graph is shown in Fig. 17.4.2.From the two parametricequationsr:3f2 and y:4t3, we getf :27F andy2:'),6f . Therefore,

*"-a'

6 5 4

27

or, equivalently, 15xs- 27y'

3 2

16

which is the cartesianequation desired.

T

o

T a b l e7 7 . 4 . 1

-1_

-2 -3

000

-4

24'

-5

134 21232

-6 Figure17.4.2

131

+*-+

-L -2

-4 -32

3 12

o rr,lusrru,rrox 2: If in Eq. (4) we differentiate implicrtly, we have 4gx2 -

dlt

s4y -d*

and solving for dyldx, we get dV _ 8*' dx 9y

17.4VECTOR-VALUED FUNCTIONS ANDPARAMETRIC EQUATIONS767 Substituting for r and y in terms of t fuom the given parametric equations, we obtain -dy :-: dx

8(3t2)2 9(4ts)

'''

which agreeswith the value of dyldx found in Example 2. ExAMPrn 4: Draw a sketch of the graph of the curue defined by the parametric equations x-coshf

and y:sinhf

(5)

Also find a cartesian equation of the graph.

.

solurroN: Squaring on both sides of the given equations and subtracting, we have * - y': cosh2f - sinlf f From the identity cosh2t - sinh2 f : 1, this equation becomes (5)

f-y':l

Equation (5) is an equation of an equilateral hyperbola. Note that for f any real number, cosh f is never less than 1. Thus, the curve defined by parametric equations (5) consists of only the points on the right branch of the h;nperbola.A sketdr of this curve is shown in Fig. 17.4.3.A cartesian equationis * - A2:1, where x >'1..

Figure17.4.3

The results of Example 4 can be used to show how the function values of the hyperbolic sine and hyperbolic cosine functions have the samerelationship to the equilateral hyperbola as the trigonometric sine and cosine functions have to the circle. The equations x-cosf

and y-sinf

(7)

are a set of parametric equations of the unit circle becauseif f is eliminated from them by squaring on both sides of each and adding, we obtain

Figure17.4.4

f * Y': cos2f * sin2f: 1 The parameter f in Eqs. (7) can be interpreted as the number of radians in the measure of the angle between the r axis and a line from the origin to P(cos f, sin f) on the unit circle. See Fig. 17.4.4 Becausethe area of a circular sector of radius r units and a central angle of radian measure f is

768


cosh f, sinh f

given by *r2f square units, the area of the circular sector in Fig. 17.4.4 is +t square units becauser: 1. In Example 4 we showed that parametric equations (5) are a set of parametric equations of the right branch of the equilateral hyperbola f -U2:1. This hyperbola is calledtheunit hyperbola.LetP(cosh f, sinh l) be a point on this curve, and let us calculatethe areaof the sectorAOP shown in Fig. 17.4.5.The sector AOP is the region bounded by the x a>
(8)

Using the formula for determining the area of a triangle, we get Figure17,4.5

(e)

Az: * cosh f sinh f We find A, by integration:

sinh u d(coshz) sinhz u du (cosh2u - L) du

- + sinh2u- +r1' lo

and so

As:*coshf sinh t-+t

(10)

substituting from (9) and (10) into (8), we have A r : * c o s h f s i n ht - ( * c o s h f s i n h t - l , / ) -- 2 0 l g

so we see that the number of square units in the areaof circular sector AoP of.Fig.17.4.4 and the number of square units in the area of sector AoP of Fig. 17.4.5is, in each case,one-half the value of the parameter associated with the point P. For the unit circle, the parameter iis the radian measure of the angle AoP. The parameter f for the unit hyperbola is not interpreted as the measure of an angle; however, sometimes the term hypeftolic radian is used in connection with f. In sec. 1.5.L,where cauchy's mean-value theorem (15.1.3)was stated and proved, we indicated that a geometric interpretation would be given in this section because parametric equations are needed. Recall thal the theorem states that if f and g are two functions such that (i) and g are / continuoirs on fa, bl, (ii) f and g are differentiable on (a, bj,' and.(iiiJ for

FUNCTIONSAND PARAMETRICEQUATIONS 17.4 VECTOR-VALUED

all r in (a, b) g' @) * 0, then there exists a number z in the open intenral (a, b) such that

f(b)- f(a) _f'(r)_ s@)- s@) s'k) v

f(b) -

(a),f(a))A

17.4.6 Figure Figure 17..4.6shows a curve having the parametric equations 1: g(f) and < < b. The slope of the curve in the figure at aparticu' V = fG),where a t lar point is given by

and the slope of the line segment through the points A(g(a)'/(a)) B(g(b), /(b) ) is given bY

and

f(b) - f(a)

s@)- s@)

Figure17.4.7

Cauchy's mean-value theorem states that the slopes are eq-ualfor at least one vai.ueof f between a and b. For the curve shown in Fig. 17.4'6,there are four values of f satisfying the conclusion of the theorem: t: "r, t: 'r, t: zs, and t: z+. w" tro* show how parametric equations can be used to define a curve which is described by a physical motion. The curve we consider is a cycloiil, which is the curve triced by a point on the circumference of a circle as the circle rolls along a straight line. Suppose the circle has radius a. Let the fixed straight finJon which the circle rolls be the r axis, and let the origin be one of the points at which the given point P comesin contact with the r axis. See Fig. 17.4.7,which shows the circle after it has rolled through an angle of f radians.

770


We seefrom Fig. 17.4.7that

v(o?)+ v(A) + v(AT)-vtoFl

(11)

V(Of) : ati

(r2)

. lV( Of) | : length of the arc PT : at. Becausethe direction of V(O?1 is along the positive )caxis, we concludethat

- Also, lVtf,l l: o a cost. And becausethe directionof V(f,) the sameas the directionof i, we have vtfil : a('!.- cosr)i

i,

1ra) : lv(riF) | a sin f, and the directionof V(a?) is the sameasthe direction of -i; thus, V(A?): -a sin fi

(14)

substituting from (rz), (13), and (14) into (11), w€ obtain ati + a(l - cost)i - a sin ti: V(O?) or, equivalently, V ( O ? ) : a ( t - s i n f ) i + a ( I - c o sf ) i

(1s)

Equation (15) is a vector equation of the cycloid. so parametric equations of the cycloid are x-

a(t-

sin f)

and y : a(I-

cosf)

(16)

ynq"_f il any realnumber.A sketchof a portion of the cycroidis shownin Fig.17.4.8.

Figure 17.4.8

Exercises L7.4 In Exercises1.through 6,find the domain of the vector-valued function R. 1. R(f) - (Ut)i+ l.Fti 2 . R ( f ) ( t 2+ 3 ) i + ( f 1 ) i 4 . R ( f ) : l n ( t + 1 ) i + ( t a n - rf ) i s.R(t;:tffii+ffii

3. R(f)

6 .R ( f )

In Exercises7 throu gh L2, find dyldx and fry ldxz without eliminating the parameter. 7. x:3t,y:2tz 8. x- L-t2,y:L*f 9. )c-t'et,y:tlnt 1 0 .x : s z t , A - L * c o s f 1L.x- acost,y-b sinf L 2 .x - a c o s h t , y - b s i n h f

EQUATIONS FUNCTIONSAND PARAMETRIC 17.4VECTOR-VALUED In Exercises13 through 16, draw a sketch of the graph of the given vector equation and find a cartesian equation of the graph. 4.,4. 14.R(f) : Prt'll

13.R(f) : tzi+ (f + l)i

L5. R(t) - 3 cosh ti + 5 sinh fi

16. R(f) - cos fi + cos fi; f in 10, inf and In ExercisesL7 and 18, find equations of the horizontal tangent lines by finding the values of f for which dy6t:o, graph of the a sketch draw find equations of the vertical tingent lines by finding the values of f for which dxldt: 0. Then of the given pair of parameteric equations'

fr,y:ffi r u .x - T 3at

a

1 7 .x - - 4 t 2 - 4 t , y - 1 - 4 t 2

3at2

cos 0, x:2 sin d, at the point where 0: *z' 19. Find an equation of the tangent line to the curve, /:5 given by the equations x:60t and 20. A projectile moves so that the coordinates of its position at any time I are projectile' path of the the of sketch a V: SOi 16t2.Draw which y has its largest value when r is 21. Find dyldx, fryld*, md tryldf at the point on the cycloid having Eqs. (16) for in the closed interval l0,2tral. that the tangent 22. Show that the slope of the tangent line at t : tr to the cycloid having Eqs. (16) is cot *tr. Deduce then line is vertical when t:2nn, where n is any integer. circle of ndius a, 23. A hypocycloid is the curve traced by a point P on a circle of radius b which is rolling inside a fixed P comesin conpoint the points at which the one of A(a,0) is o >-b.lf-th" origin is at the center of tne fixea circle, f is the number of parameter and the circles, two the of of tangency point moving is the tact with the fiied circle, B are hypocycloid of the equations parametric that ptove AOB, angle radians in the

x:(a-b)cosf*bcos+, and

y:(a-b)sinf -bsn+t of this cuwe are x: 24. If a : 4b in Exercise 23, we have a hypocycloidof f our cusps.Show that parametric equations : t. y a sirts 4 cossf and of four cusps, and draw a 25. Use the parametric equations of Exercise24 to find a cartesian equation of the hypocycloid equation. resulting of the graph sketch of the

26. Parametric equations for the ttactfix arc

x : t - a t a n hi

,:o"""hl

Draw a sketchof the curve for a:4' intercept of the tangent line. 27. provethat the parameter t in the parametric equations of a tractrix (seeExercise26) is the r line from the point 2g. Show that the tractrix of Exercise26 is a curve such that the length of the segment of every tangent to a' and equal constant is r axis of tangenry to the point of intersection with the (16). 29. Find the area of the region bounded by the r axis and one arch of the cycloid, having Eqs. 30. Find the centroid of the region of Exercise29.

772

VECTORS IN THEPLANEANDPARAMETRIC EQUATIONS

17.5 CALCUTUS OF VECTORVALUED FUNCTIONS !7.5.1 Definition

We now discuss limits, continuity, functions.

and derivatives of vector-valued

Let R be a vector-valued function whose function values are given by

R(f) -- f(f)i + s?)i Then the limit o/ R(t) as t approaches tr is defined by

rimR(f): tt

+ [ln fuli [n,

(f)]t

and g@ bothexist. 1tTf@ 1T

o rr.r.usfRArroN L: If R(f) : cos fi + 2t j, then

lim R(r;: (ti3 cosf)i + (liT 2e,)i: i+ zi 17'5'2 Definition

The vector-valued function R is continuousat trifand only if the following three conditions are satisfied: (i) R(fr) exists; (ii) lim R(f) exists; t-tr

(iii) lim R(f) : R(r,). 2-tr

From Definitions 17.5.'Land']-,7.s.2,it fonows that the vector-valued functio-nR, defined uy n(t-) : f (t)i + g(t)i, is continuous at f, if and onry if / and g are continuous there. In the following definition the expression R(f+Af)-R(r) At is used' This is the division of a vector by a scalarwhich has not yet been defined. By this expression we mean 1 6 tntt+ ar) R(r)l

L7'5'3 Definition

If R is a vector-valued function, then the deivatiae of R is another vectorvalued function, denoted by R, and defined by R ' (t ) : l i m

R(t + Af) - R(f)

^t

ir thistimit I:,,. The notation DrR(t) is sometimes used in place of R'(f).

FUNCTIONS773 17.5CALCULUS OFVECTOR-VALUED The following theorem follows from Definition 17.5.3and the definition of the derivative of a real-valued function. 17.5.4 Theorem

If R is a vector-valued function defined by (1)

R(f):f(f)i+s(f)i then

R ' ( f )- f ' ( t ) i +

g'G)i

if f ' (t) andg' G) exist. PRooF: From Definition

17.5.3

R(f+af)-R(f) R ' ( t ) : lim Lt A/-0

-lim

lf ( t + af) i+ s( t + At) il [/( t) i + s( t) i] t

A/-0

:lim

i+lim Af -O

Af - 0

: f,(t)i+g,(f)i

I

The directionof R'(f) is given by d (0 < e < 2n), where dy dt _ dY e'(t) tan d: dx dx f'(t)

E

From the above equation, we see that the direction of R'(f) is along point the tangent line to the curve having vector equation (1) at the

tation of R(f) as f assumesall values in the domain of R. Let OP be the position representation of R(f) and OQ be the position replesentation of

F i g u r e1 7 . 5 . 1

sentations tangent to the curve C at the point P.


o rt,t,usrnerron 2: If R(f) - (2 * sin f)i + cos fj, then R'( t) : cos tr - sin fj Higher-order derivatives of vector-valued functions are defined as for higher-order derivatives of real-valued functions. so if R is a vectorvalued tunction defined by R(f) : f (t)i + g(t)j, the secondderivative of R, denoted by R"(t), is given by R"(f) : Dr[R'(f)] We also have the notation DrrR(t) in place of R,,(t).By applying Theorem 17.5.4to R'(t), we obtain

R " (f) : f" ( t) i* g" G) i if f" (t) andg"(f) exist. o rLLUsrRArroN3: rf R(r): (ln f)i + (]) i, tr,""

R',(t):lt-ii and

- -+i+ ii R"(t) 17.5.5 Definition

A vector-valued function R is said to be differentiableon an interval if R'(f ) exists for all values of f in the interval.

The following theorems give differentiation formulas for vectorvalued functions. The proofs are based on Theorem 17.5.4 and theorems on differentiation of real-valued functions. 17'5'6 Theorem

If R and Q are differentiable vector-valued functions on an interval, then R + Q is differentiable on the interval, and

D,[R(r)+ Q(r)] : D,R(r)+ D,Q(f) The proof of this theorem is left as an exercise (see Exercise 16). EXAMPLE 1:

If

R(f) -tzi+ (t-L)i and Q(f) : sin fi + cos fi verify Theorem 77.5.6.

SOLUTION:

D'[R(t) + e(r)] :Dt(Jtzi+ (t- 1)i] + [sin ti+cos ti]) : DtlG, * sin f)i + (t - 1 + cost)il - (2t+ cosf)i + (1 - sin f)j D'R(r) + DIQG): Dtltzi+ (r - 1)il * D,(sinfi + cosri)

FUNCTIONS 17.5 CALCULUSOF VECTOR-VALUED

: (zti + i) + (costi - sin fi) - ( 2 t + c o s f ) i + ( 1- s i n f ) i D'[R(f)+ Q(f)l- DrR(f)* D'Q(t). Hence, 17.5.7 Theorem

If R and Q are differentiable vector-valued functions on an intervd, then R' Q is differentiable on the interval, and

D , [ R ( r ) .Q ( r ) ] : [ o , n ( t ) ] ' Q 0 ) + R ( t ) ' [ D ' Q ( t ) ] and Q(f) :f2!)i+8r(t)i' Then by PRooF:Let R(f) :flt)i+8,(t)i Theorem17.5.4 DE(f) : fl (t)i+8,'(t)i and D'Q(f): fz'G)i+ 82'G)i

R(f). Q(f): 11,(t)l lf,(t)l + [s'(t)][s'(t)] So

.Q(r)] D,[R(r) + [s'(f)][g,'(t)] : lfr'(r)lt/r(f)l + t/,(r)llf,'(t)l* [8,'(t)]lsr(t)] - {lf,'(r)lt/2(r)l + {[/,(t)]lf,'(t)l+ [s'(t)][g"(t)]] * [8,'(r)]ts,(r)l] I : [P'n(t)]' Q(t)+ R(t)' tD'Q(f)l EXAMP:n2: Verify Theorem 17.5.7 for the vectors given in Example 1.

SOLUTION:

R ( r ) ' Q ( t ) : f 2s i n t + ( t - 1 ) c o sf '

Therefore,

Q ( f ) l - 2 f s i n t + f 2 c o st + c o s t + ( f t ) ( - s i n f ) D,[R(r) Q ( f ) l - ( f + 1 ) s i n t + ( f 2 + 1 ) c o sf

D,[R(f)

Because

D,R(r)- D,ltzi+ (t - 1)il - zti+ i [D,R(f)] . Q(t) : (2ti+ i)' (sin fi + cosfi) - 2 t s i n t + c o sf Because D,Q(t) : D,lsin fi + cos til:

cos ti- sin fi

R ( f ) . [ D , Q ( f ) ] : [ t z i + ( t - 1 ) i ] ' ( c o st i - s i n f i ) : t2 cos t - (f - 1) sin f Therefore, : (2t sin t + cosf) tD,R(f)l . Q(f) + R(f) ' [D'Q(f)] + [f2 cost-

(f - 1) sin f]


Thus, [D,R(f)]. Q(f) + R(r) . [ D , Q ( f ) ] : ( f + 1 ) s i n f + ( t 2 + L ) c o s f (3) ComparingEqs. (Z) and (3), we see that Theorem 17.5.T holds. 1'7.5.8Theorem

If R is a differentiable vector-valued function on an interval and f is a differentiable real-valued function on the intenral, then

Dilf@ltR(f)ll:

[D,/(f)]R(t)+ f(t) D,R(f)

The proof is left as an exercise(seeExercise1Z). The following theorem is the chain rule for vector-valued functions. The proof which is left as an exercise (see Exercise 1g) is based on Theorems 2.6.6 and 3.6.1, which involve the analogous conclusions for realvalued functions. 17'5'9 Theorem

Suppose that F is a vector-valued function, h is a real-valued function such that 6: h(t), and G(f) : F(/r(O).If ft is continuous at f and F is continuous at h(t), then G is continuous at f. Furthermore, D1s if and DoG(f) exist, then D$(t) exists and is given by D$(t):

lDoG(t))Dt6

we now define an indefinite integral (or antiderivative) of a vectorvalued function. 17.5,10 Definition

If e is the vector-valued function given by

Q(f):f(t)i+s@i then the indefinite integral of eG) is defined by f.

lr

d t : i f G )d t + i s @a t J a(r) J J

(4)

This definition is consistent with the definition of an indefinite integral of a real-valued function because if we take the derivative on both sides of (4) with respect to f, we have rff

D , Jq ( t )d t : i D t f @ d t + i D t s @ a t J J which gives us f

D, Q@ itt:if(t) + wG) J For eachof the indefinite integrals on the right side of (4) there occurs an-arbitrary scalar constant. when each of theJe scalarsis muttiplied by either i or i, there occurs an arbitrary constant vector in the sum. so we

17,5 CALCULUSOF VECTOR-VALUED FUNCTIONS

have f

I Q(f) dt _ R(f)

+c

J

where DrR(t) : Q(f) and C is an arbitrary constant vector.

ExAMPrr3: Find the most general vector-valued function whose derivative is

If D'R(t) - Q(t), then R(t; : ,l"Q(f)dt, or

SOLUTION:

ff

R ( r ; : i I sin t dt - 3i' J I cosf dt J

Q(f) : sin ti- 3 cos fj

i(-cos t+ C') -3i(sin t+cr) -cos ti - 3 sin ti + (C,i - sCri) -cos ti- 3 sin ti+C

EXAMPTu 4: Find the vector R(0 for which DB(t):

solurroN:

-t dt+i a a, I

R(f)

So

e-ti* e'i

and for which R(0) : i + i.

R(f) - i (-e-' * C') + i k' + Cr) Because R(0) : i + i, we have i + i: i(-1 + C') + i(1 + Cr) So Cl -L:1

and

Cr*1.:1.

Therefore, Cr: 2

and

Cz: 0

So

R(t;: (-e-,+2)i+eti

The following

l7.5.ll

Theorem

theorem will be useful later.

If R is a differentiable vector-valued function on an interval and lR(t)l is constant for all t in the interval, then the vectors R(t) and DrR(f) are orthogonal. pRooF: Let lR(01: k. Then by Theorem 17.3.3(iii),

R (t)' n1 4:12 Differentiating on both sides with respect to f and using Theorcm17.5.7

VECTORSIN THE PLANEAND

PARAMETRIC EQUATIONS we obtain

. R(r)+ R(r). [D,n1r;1 :s [D,R(r)] Hence, 2R(f)'D'R(t):0 Becausethe dot product of R(f) and DlR(f) is zero, it follows from Definition 77.3.7that R(f) and D1R(f)are orthogonal. I Df R(f)

The geometric interpretation of Theorem 17.5.1.7is evident. If the vector R(t) has constant magnitude, then the position representatiotr OF of R(t) has its terminal point P on the circle with its center at the origin and radius k. So the graph of R is this circle. BecauseDrR(t) and R(f) are +

orthogonal, OP is perpendicular to a representation of DE(t).

Figrre

17.5.2shows a sketch of a quarter circle, the position representation OP of R(t), and the representationPB of DE(f).

Figure 17.5.2

Exercises 77.5 In Exercises L through 4, find the indicated limit, if it exists.

1 . R ( f )- ( t - 2 ) i +

2. R ( f ) - e t + t i+ l f + 1 . l i ; t i m R ( f )

=i,rr,gR(r)

3. R(f) - 2 sin fi * cos,t,

4. R(f):

) l:nR(f In Exercises 5 throughL0,find R'(f ) and R" (f ). t-L.

t-2.

s. R(f) : (t2- 3)i + (2t+ r)i

6 .R ( f ) : f i i + i i,

8. R(f) : coszti + tan fi

9. R(f) : tan-r ti + zti

In Exercises11 and 12, find.D,lR(f) l. (2-t)i 1 1 .R ( t ) : ( f - l ) i +

12-2t-3.,t2-5t+6.1

fi:i

+

timR(f) f _ 3i i;

7. R(f) : szti+ ln fi

1 0 .R ( t ; : \ t r t + r i + ( t - l ) ' i

1 2 .R ( f ) : ( e t * l ) i + ( e ' - l ) j

In Exercises13 through 15, find R'(t) . R"(t). 1 3 .R ( t ) :

(2P- l)i+ (f +3)i


1 4 .R ( t ) : - c o s

2ti*sin2ti


ls. R(t):ezti+e-zti 18. Prove Theorem 17.5.9.

In Exercises19 through 22, find, the most general vector whose derivative has the given function value.

t s .t a n r- i+ i

20.hi-+t,i

sint fi + 2 coszti, andR("t) :0, find R(f). 24. If R'(f) : et sin ti+ et cos ti, and R(0) - i - i, find R(f). 23. If R'(t):

2t. ln fi + Pi

22. }ti - zti

17.6 LENGTHOF ARC

25. Given the vector equation R(t) : cos fi * sin fj. Find a cartesian equation of the curve which is traced by the endpoint of the position representationof R'(f). Find R(t) . R'(t). Interpret the result geometrically. 26. Given R(t):zti+ find Dp(t).

(tz -1)j

and Q(t):3ti.

lf. a(t) is the radian measureof the angle between R(t) and Q(f),

27. Suppose R and R' are vector-valued functions defined on an interval and R' is differentiable on the interval. Prove D ' [ R ' ( t ) . R ( t ) ] : l R ' ( t ) | ' z+ R ( t ) ' R " ( t ) 28. If lR(t) | : h(t) , prove that R(t) . R'(t) : lh(t)llh' (t)1. 29. If the vector-valued function R and the real-valued function / are both differentiable on an interval and /(t) * 0 on the interval, prove that Rff is also differentiable on the interval and

(r)- /'(f)R(f) _ /(r)R', l] fRC)..l -'LfG) lf@l' J 30. Prove that if A and B are constant vectors and / and g are integrable functions, then fff

+ Bs(t)litt: A f@ at+n s@ d.t J J tA/(t) J (nrNr: Express A and B in terms of i and i.)

17.5 LENGTH OF ARC

In Sec.8.10 we obtained a formula for finding the length of arc of a cunre having an equation of the form y : f (x).This is a special kind of cunre becausethe graph of a function /cannot be intersected by a vertical line in more than one point. We now develop a method for finding the length of arc of some other kinds of curves. Let C be the curve having parametric equations x-f(t)

and y:g(f)

(1)

and suppose that / and g are continuous on the closed interval [a, b]. We wish to assign a number L to represent the number of units in the length of arc of C from t: a to t: b. We proceed as in Sec.8.10. Let A be a partition of the closedintenral [a, b] formed by dividing the interval into n subintewals by choosing n - L numbers between a and,b. L'et fo: a and tn: b, and let tr, tz, . . . , tn-r be intermediate numbers: to1t, The dth subinterval is [/i-1, t1] and the number of units in its length, denoted by Llt,is \- f6-r,where i: L,2, . . . , n. Let llAllbe the norm of the partition; so each A/ = llAll. Associatedwith each number f1is a point P{f (t),9(4)) on C. From each point P1-, draw a line segment to the next point P1 (see Fig. 77.6.7). The number of units in the length of the line segment from P1-, to P; is denoted by lPprPTl. From the distance formula we have

IFFIP{l :

e)

780


v Pi(f(t),s(f ;))

P"(f(b),s(b))

Pi-t(f (t i-t), s$ r-r))

F i g u r e1 7 . 6 . 1

The sum of the numbers of units of lengths of the n hne segments is n

(3) t-l

Our intuitive notion of the length of the arc from f : a to f : b leads us to define the number of units of the length of arc as the limit of the sum in (3) as llAllapproacheszero. 17.5.1 Definition

Let the curve C have parametric equations r : /(t) and y : g!).Then if there exists a number L having the property that for any € > 0 there is a E>0suchthat ln

I

l) lP,_rPiI-Ll l?_r'| for everypartition A of the intenral la, bl for which llAll

L- lim i lml s f:1

(4)

llall-

and L units is called the length of arc of the curue C from the point (f (a), g@)) to the point (f (b), g(b) ).

The arc of the curve is rectifiable if the limit in (4) exists. If f ' and g' are continuous on la, bl, we can find a formula for evaluating the limit in (a). We proceed as follows. Because/' andg' are continuous on [a, bl,lhey are continuouson each subinterval of the partition A. So the hypothesis of the mean-value theorem (Theorem 4.7.2) is satisfied by f andg on each lti,_r,tl; therefore, there are numbers zi and ari in the open interval (tr_r, f1) such that

' f (t,) - f (tr) - f (2,) Aot

(s)

gG) - s1r-) - g'(w) L,t

(6)

and

17.6LENGTHOF ARC

781

Substituting from (5) and (5) into (2) , we obtain

tmt: or, equivalently, (7)

Lit

lffil:

inthe openintervil (to-r,tu).Thenfrom (4) and (7),if where z.land,wieltE the limit exists, n

L-

(8)

lim ' l l a l l - of i

The sum in (8) is not a Riemann sum becausezi and wl N€ not necessarily the same numbers. So we cannot apply the definition of a definite integral to evaluatethe limit in (8). Howevet, there is a theorem which we can apply to evaluate this limit. We state the theorem, but a proof is not given becauseit is beyond the scope of this book. You can find a proof in an advanced calculus text. 17.6.2 Theorem

If the functions F and G are continuous on the closed interval la, bl, then the function \/FtTtr, is also continuous on la,bf , and'if A is a partition I tr, < ti < o f t h e i n t e r v a fl a , b l ( A , a: : t o I t , I ' ' ' (f1-1, f1), then in and z1 artd wl are any numbers Lit:

['

dt

(e)

' Applytng (9) to (8), where F is f and G is g', we have

t:

fb

J"

dt !lf,(t)1,+ 18,G)12

We state this result as a theorem. 17.5.3 Theorem

Let the curve C have parametric equations r: l(t) and y : 8G), and suppose that f' andg' are continuous on the closed interval [4, D]. Then the length of arc L units of the curve C from the point (/(a), g(a) ) to the point (f (b), S(b) ) is determinedby (10)

EXAMPLE1: Find the length of the arc of the curye havingParametric equations x: f 3 and y : 2t2 in each of the following cases:(a) from t - 0 to t - 1; (b) from t--2tot-0.

sor,urroN: A sketch of the curve is shown in Fig. L7.6'2' (a) Letting a n d l e t t i n gy : 8 G ) , 8 ' ( t ) : D , Y : 4 f ' s o f r o m x: f(t), f'(t):Dp:3P; Theorem L7.6.3,if L units is the length of the arc of the curve from f : 0 to t:1,

VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS fr

:l

t\ffidt

JO l1

: + . ? ( g t r + 1 5 ) 3 / 21 lo

-;r[(25)Btz - (16)3/2] -L0-8-6-4-Z

F i g ur e 1 7. 6 . 2

-+(12s-64) -n61 -

(b) If L units is the length of the arc of the curye from we have from Theorem 17.6.3

L-l

fo

f0

I \F\ffi J-z

ffidt:

J-z Because -2 < t S 0, \F ro

L- | -t\ffi J-z

-2tot:0

dt

- -t. So we have

dt

- -;?

lo (9t' + I6)st2 | J-z

- -;r

[ (16)3t2

(50) 3/2]

-;r(2s0 \D - 64) : 10.7

The curve C has parametric equations (1). Lets units be the length of arc of C from the point (f(t),g(tr)) to the point (/(r), gG)), and let s increase as f increases.Then s is a function of f and is given by ft

s- l Jto

du

(11)

From Theorem 7.6.1, we have

A vector equation of C is

R(t):f(f)i+ s?)i

(13)

Because

R'(t):f'(t)i+ g'U)i we have (14)

17.6 LENGTHOF ARC

783

Substituting from (14) into (12),we obtain

(1s)

lR'(f)l

From (15) we conclude that if s units is the length of arc of curve C having vector equation (13) measured from some fixed point to the point (f (t), SG)) where s increasesas f increases,then the derivative of s with respect to f is the magnitude of the derivative of the position vector at the point (/(f),9(f)). If we substitutefrom (14) into (10), we obtain f : "ljlR'(t)ldf. So Theorem 17.5.3 can be stated in terms of vectors in the following way. 17.6.4 Theorem

Let the curve C have the vector equation R(t) : f (t)i + SG)i, and suppose thatf'andg' are continuous on the closed interval [a, b]. Then the length of arc of C, traced by the terminal point of the position representation of R(f) as f increases fuom a to b, is determined by (15)

EXAMPrn 2: Find the length of the arc traced by the terminal point of the position rePresentation of R(f) as f increases from L toAtf R(f) : et sin fi + et cos ti

sorurroN: R'(t) : (e'sin t + et cost)i + (et cos t et sin t)i. Therefore, -2

l R ' (t ) l : e '

sin f cos t*

stnz t

-- et{z From (16) we have

f,et dt: nr)^r-- \D,(en- e)

An altemate form of formula (10) for the length of an arc of a curve C, is obtained by reand y:8(t), having parametric equations x:f(t) placing f'(t) by dxldt and g'U) by dyldt.Wehave dt

(r7)

Now supposethat we wish to find the length of arc of a curve C whose (x, y) is the cartesianrepresentationof a polar equation is r:F(0).If. point P on C and (r, 0) is a polar rePresentation of P, then x-rcose

and y-rsrn0

(18)

Replacirg r by F(0) in Eqs. (18), we have x - F(0) cos 0

and

F(0) sin 0

(le)


Equations (19) can be considered as parametric equations of c where 0 is the parameter instead of f. Therefore, if.E' is continuous on the closed interval fo-,Ff , the formula for the length of arc of the curve c whose polar equation is r:F(d) is obtained from (12) by taking f :9. So we harre d0

(20)

From (18) we have dx dr UE-r d0_cos

sin0 and

du dr d0:sin0 dn*rcos0

Therefore,

ffi: * 2r sin 0 cosd

EXAMPLE3: Find the length of the cardioid r - 2(1 + cos A).

fi*

r2 cosz 0

solurroN: A sketch of the curve is shown in Fig. 17.6.9.To obtain the length of the entire curve, we can let d take on values from 0 to 2zr or we can make use of the symmetry of the curve and find half the length by letting 0 take on values from 0 to z. Becauser:2(l * cos d), drld|: -2 sin 0. Substitutinginto (21), integrating from 0 to z, and multiplying by 2, we have frr

L-'J,

dE de

: 4 ! 7- f n | t / t + c o s e a e JO

Figure 17.6.3

To evaluatethis integral, we use the identity cos2*g:+(1+ cos 0), which gives vT +cos 0 : V2 lcosggl.Becauseb = 0 = 7T,0
17.7 PLANE MOTION

17.6 Exercises

a appears' a > 0' Find the length of arc in each of the following exercises'When : 7. x:tP * t, y : tP t;from t: 0 to f 1' -2 to t : 0. 2. x: P, ! : 3P;from f = 3 . R ( f) : 2 P i * 2 F i ; f t o m t : 7 t o t : 2 ' :0 to t:*n 4 . R ( f ) : a ( c o sf * t s i n f ) i * a ( s i n f f c o st ) i ; f r o m f a sins t' 5. The entire hypocycloid of four cusps: r: 4 cos3t' A: : : 5. x : e-t cost, y: e-t sin f; from t 0 to t n' 7. The tractrix x:

t -a tanh

: I, r

o'"cl' 1 f"o* t : -a lo t : 2a'

- cos f) ' a(t - sin f) , y : a(l - ct cos fi * a sin fi. 9 , The circumference of the circle: R(f)

8. One arch of the cycloid i x:

10. The circumference of the circle: r:

a sin 0'

L L . The circumference of the circle: r:

a'

L3. The entire curve: r:3

cos2+0.

L5. r -- a sin3 Ls|; from 0 0 to 0 - 0r

I7.7 PLANE MOTION

L2. The entire curye: r : | - sin 0. 1 4 .r : a 0 ; f r o m 0 - 0 t o 0 : 2 z r L 6 .r : A 0 2 ; f r o m 0 - 0 t o 0 : r '

to The previous discussion of the motion of a particle was confined acceland velocity the straight-line motion. In this connection, we defined the moeratiin of a particle moving along a straight line. We now consider tion of a particle along a curve in the plane' Suppose that C ls the plane curve having parametric equations x: f Gi, V:8(t), where f units denotestime' Then R(t;:/(t)i+g(t)i is a vector equation of c. As f varies,the endpointP(f(t),8(t)) of oP moves along the cunre C. The position at time f units of a particle moving along C is tie point P ( f (t),8(f ) ). The oelocityoectorof the particle at time t unils is defined to be'R' (f). W" denote the velocity vector by the symbol V(f) insteadof R'(f).

17.7.L Definition

: Let C be the curve having parametric equations x: f (t) and y g(t) ' tf a is the units f particle is moving along C so that its position at any time units is f time at point (r, y), then the initantaneousaelocity of the particle determined by the velocitY vector

if f ' (t) and g'(f) exist.


In sec. 17'5 we saw that the diiection of R'(f) at the point PV(t)' 8(t) ) is along the tangent line to the curve c at p. Therefore, the velocity vector V(f) has this direction at p. The magnitude of the velocity vector is a measure of the speed of.the partide at time f and is given by

lv(r)l:

(1)

Note that the velocity is a vector and the speed is a scalar.As shown in sec. 17..6,the expressionon the right side of (1) is dsldt.sothe speed is the rate of change of s with respect to f, and we write (2) T'he accelerationuector of the particle at time f units is defined to be the derivative of the velocity vector or, equivalenfly, the second.derivative of the position vector. The accelerationvector is denoted by A(l). 17'7'2 Definition

The instantaneousaccelerationat time f units of a particle moving along a curv€ c, having parametricequationsx: (t) undy :g(t), is delrmined f by the accelerationvector

and R" (f) exists.

ligure 17.7.1showsthe representations of the velocity vector and the accelerationvector whose initial point is the point p on i.

Figure 17.7.1

EXAMPLE1: A particle is moving along the curye havin gparametric equations x- 4 cos tt andy: 4 sin tt. If t is the number of seconds in the time and r and y represent number of feet, find the speed and magnitude of the particle's acceleration vector at time f sec. Draw a sketch of the particle,s path, and also draw the representations of the velocity and acceleration vectors having initial pointwhere t-fn.

solurroN:

A vector equation of C is

costti + 4 sin tti V(f) R' (t) - -2 sin tti + 2 costti A(f ) : V'(t) : -cos tti - sin*fj R(f)

l v ( r ) l : (-zsin}:t)W -2

lA(r)l:

:

1,

Therefore, the speed of the particle is constant and is 2 ftlsec. The magnifude of the acceleration vector is also constant and is 1 ft/sec2.

17.7 PLANE MOTION

P(z{3,2)

Eliminating f between the parametric equations-o! C' we obtain the I * Y':16, which is a circle with its center at 0 and "q""tlor, ""tt"ti* radius 4. When t: In, the direction of V(t) is given by tn < 0 < n and' tan 0:-

cos *z-:-cot sir, E-r.

frr:-6

and the direction of A(f) is given by n < 0 < tr

tan0:g: cos t1r Figure17.7.2

ExAMPrn 2: The Position of a moving Particle at time f units is given by the vector equation R(f) - s-2tf*Seti

F i n dV ( r ) , A ( t ) , l V (t ) 1 ,l A ( r )l . Draw a sketch of the Path of the particle and the rePresentations of the velocitY and acceleration vectors having initial point where t: *.

v(+)

and

tan*zr:+

acceleration Figure 17.7.2 shows the representations of the velocity and t: tt' ,ru-.torchaving initial point where

SOLUTION:

-2e-2ti * 3eti V (f) - R' (f) : A(f) : V' (t) : 4e-2ti * Seti

lv(t)l:ffi l A (t ) l : ffi

lv(+)l: {Et+ei - 5.01 lA(+)l:\re7T6 - 5.1'5 Parametric equations of the path of the particle Tt x: :3et. Etiminating f between these two equations' we obtain A

e-a and'

ryz:9 Becauser ) 0 and a ) 0, the path of the particle is the portion of the parcrrweryz: 9 in the firsi quadrant. Figure L7.7.3shows the path of the (l) : V of slope The I when vectors f. ticle and the velocity and icceleration -6], 3.4. is of A(+) lstrz and the slope ,"-gtt,"

F i g u r e1 7. 7 . 3

we now derive the equations of motion of a projectile by assuming the that the projectile is moving in a vertical plane. we also-assume that a downward has whlch weight, is its only force acting on the pr6jectile d#ction and a magnituae of mg lb, where z slugs is its mass and g ft/sec2

788

VECTORSIN THE PLANEAND PARAMETRIC EQUATIONS (oo cos a, oosin a)

r Figure17.7.4

is the constant of acceleration caused by gravity. we are neglecting the force attributed to air resistance (which flrleavybodies traveling at Imail speeds has no noticeable effect). The positive direction is taken as vertically upward and horizontally to the iight. 'suppose, then, that a projectile is shot from a gun having an angre of elevation of radian measure a. Let the number of"feet per second in the initial speed, or muzzlespeed,be denoted by ao.we set up the coordinate axesso that the gun is located at the origin. Refer to Fig. iz.z.+. The initial velocity vector, Vs, of the projectile is given by Vo : ?6 CoSqi + as sin aj

(3) : Let f the number of secondsin the time that has erapsed since the gun was fired; r: the number of feet in the horizontar distance of the projectile from the starting point at f sec; /: the number of feet in the verticar distance of the projectile from the starting point at f sec; : the position vector of the projectile at f sec; I_!tl V(f) : the velocity vector of the p*p"titu at f sec; A(t; : the accelerationvector of theprojectile at f sec. r is a function of t; so we write r(f). similarry, y is a function of t; so we write y(t). Then

R(r)-)c(r)i*y(r)i v(f) - R'(t) A(t):

(4)

(s)

Y'1t;

(6) Becausethe onlv force acting on the project'e has a magn itude of mg lb and is in the downward direciion, *re' ir F denotes this force we have

F _ _mTi

(7) Newton's second law of motion states that the net force which is acting on a body is its "mass times acceleration.,, so F- mA

(8)

From (7) and (8) we obtain

mA Dividing

mgi by *, we have

A__8i BecauseA(t) : V'( t), we have from (9)

V'(f) : -gi Integrating on both sides of Eq. (10) with respect to t, weobtain V(f) :-gti * C,

(10)

(11)

17.7 PLANE MOTION

where C1 is a vector constant of integration. when t-

0, v:

v(f) --gti

v0. so cr:

v0.Therefore, from (11) we have

* Vo

or, becauseV(f ) : R' (t), we have R'(f) : -gti * Vo

(12)

Integrating on both sides of the vector equation (I2) with resPect to t, we obtain R(f):-{gt'i+Vof+C2 where C2 is a vector constant of integration. when f : 0, R : 0 because the projectile is at the origin at the start. so Cz:0. Therefore, (re) R(t): -tgt2i*Yrt Substituting the value of V6 from Eq. (3) into (13), we obtain R(t) : -4ilt'i + (uo cos oli* ag sin oi)f or, equivalently, R(f) : tuocos CIi+ (taosin a-trt')i

(14)

Equation (14) gives the position vector of the proiectile at time f sec. From tiis equation, we can discuss the motion of the projectile. We are usually concerned with the following questions: 1. What is the range of the projectile? The range is the distance lO-Al along the r axis (see Fig. 17.7.4). 2. Whai is the total time of flight, that is, the time it takes the projectile to go from O to A? 3. What is the maximum height of the projectile? 4. what is a cartesian equation of the curve traveled by the projectile? 5. What is the velocity vector of the proiectile at impact? These questions are answered in the following example' ExAMpln 3: A projectile is shot from a gun at an angle of elevation of radian measure tzr. Its muzzle speed is 480 ftlsec. Find (a) the position vector of the Proiectile at any time; (b) the time of flight; (c) the range; (d) the maximum height; (e) the velocitY vector of the proiectile at imP act; (f) the position vector and the

and c: $zr.So "o:480 Vo:480 cos *zri * 480 sin $zi

soLUTroN:

:24a16i+ 24oi (a) The position vector at f sec is given by Eq' (L4)' which in this caseis (15) R(r) : 240\/5ti+ (240t-tlt')i

790

VECTORSlN THE PLANE AND PARAf\4ETRIC EQUATTONS

velocity vector at 2 sec; (g) the speed at 2 sec; (h) a cartesian equation of the cun/e traveled by the projectile.

if (1, y) is the position of the projectile at f sec,x:240 V3t and _So_ y :240t - tSP. (b) To find the time of flight, we find f when : 0. From part (a), we A gety:0 and have 240t-l2gf2:g

t( 240- lg1) :s f:0

and

The value f : 0 is wh91 tle projectile is fired because then y : 0. If we tlkeg: 32, the time of flight is determined by t: $O1g: ieilgZ: 15.So the time of flight is 15 sec. (c) To find the range, we must find r when f : 15. From part (a), x:240rfgt. so when f J rs, ,: 3500\6. Therefore,tlre ,aige is 6235ft (approximately). (d) The maximum height is attained whenDg: 0, that is, when the vertical component of the velocity vector is 0. Beciuse y:240t-tgt, DtA:240 - gt So Dg: 0 when t:240lg.If we take g: j2, the maximum height is attained when f : 7*, which is half the toial time of flight. when t-:T+, y :90f. So the maximum height attained is 900 ft. (e) Becausethe time of flight is L5 sec,the velocity vector at impact is V(15). Finding V(f) by using (15), we have

v(t; : D,R(r): Z4ot6i+ e4o- gt)i Takingg:32,

we get

V(ls) :240\/-3i(f) Taking t:2

240i in Eq. (15), we obtain

R(2) : 480\/3i+ 416i BecauseV(t) : R'(f), we have

V(f) : 240\Ei+ e4O- gt)i v (2) - 24ot6i + r76i (g) The speedwhen t - 2 is determinedby

lv(z)| :

(24ovS)t+-O=W

:32\M So at 2 sec the speed is approximately 4s1,.4ft/sec.

17,7 PLANE MOTION

791

(h)Tofindacartesianequation,weeliminatefbetweentheparametric equations x-- 2406t y :2tL0t - tSF equasolving the first equation forf and substituting into the second tion, we obtain 1-.

Y:4t

-

L_nz

LoSoot-

which is an equation of a Parabola.

L7.7 Exercises is the f the curve having the given parametric equations' where sec ln Exercises 1 thrcugh 6, a particle is moving along the of (d) the magnitude t'; at ipeed the A(t); vector 1cJ !: time. Find: (a) the velocity vector V(f); (b) the acclleration vector and velocity of the in" representations *d particle of the p"trt acceleration vector at f : fr. Draw a sketch or trt" the accelerationvector at t: tv t1:*n 3. x:t,y:lnsecf; 2 . x - 2 c o st , y - 3 s i n f ; t r : * n 1. x-t2+4,Y-t-2;tt-3 4. x:Zlt,y:-!t;tr:4

5' r:sin

t,y:tant;h=tn

6' x:ezt'A:estitt:0

is determined from a vector equation' Find: (a) v(fr); In Exercisesz through 12, the position of a moving-aparticle at,f sec path of the particle containing the position of the par(b) A(tr); (c) lv(rr)l; tOl.ltrril. Draw a sketch of portion of the where f : fr' and draw'tire representationsof v(fr) and A(tr) having initial point ii.t" "iJ:'rr, 8 . R ( t ; : ( t ' + 3 f ) i + ( 1 - 3 t ' ) i ;h - t 7 . R ( f ) - ( 2 t - 1 ' ) i + ( t ' + 1 ) i ;t , , : 3 9. R(f) : cos zti- 3 sin ti; h: n -sinf)i;h-En 1 1 .R ( f ) - 2 ( 1 - c o s f ) i + 2 ( 1

10.R(f) : e-ti+ ezti;tr:ln2 12.R(f) : ln(f + 2)i + tt'i; tr:1

L3. Find the position vector R(f ) if the velocity vector

v(t) :

1

d- ryi

- (r+ l)i and R(o): 3i+ 2i

14. Find the position vector R(f) if the accelerationvector s i n 2 f i a n d V ( 0 ) : i * i , a n dR ( 0 ) : + r - + i A(t):2cos2ti*2 of 45" with a mtzzle speed of 2500ftlsec' Find (a) the range of 15. A projectile is shot from a gun at an angle of elevation (c) the velocity at impact' th; projectile, (b) the maxiirum height reached' and of 50". The muzzle speed is 160 ftlsec' Find (a) the position vec16. A projectile is shot from a gun at an angle of elevation (d) the maximum height reached; (e) the velocity at tor of the projectile at f sec; (b) the time of flight; (c) the range; impact; (f) the sPeedat 4 sec' gun at an angle of 30'w-ith the horizontal' If the muzde 17. A projectile is shot from the top of a building s,aft niqrr from-a the base of the building to the point where the profrom distance the speed is 1500ftlsec, find the tirne of flight id jectile lands'


18. A child throws a ball horizontally from the top of a cliff 256 ft high with an initial speed of 50 ftlsec. Find the time of flight of the ball and the distance from the base of the cliff to the point where the bill lands.

19' A child throws a ball with an initial speed of 60-Itlsecat an angle of elevation of 50otoward a tall building which is 25 ft from the child' If the child's hand is-5 ft from the ground, stri,r, that the ball hits the building, and find the direction of the ball when it hits the building. 20' The muzzle speed of a gun is 150 ftlsec. At what angle of elevation should the gun be fired so that a projectile will hit an object on the same level as the gun and a distance of 400 ft from it? 21' what is the muzzle speed of a gun if a projectile ffued from it has a range of 2000ft and reachesa madmum height of 1000ft?

I7.8 THE UNIT TANGENT AND UNIT NORMAT VECTORS AND ARC LENGTH AS A PARAMETER 17.8.1Definition

we prwiously noted that a unit vector is a vector having a magnitude of 1, examplesof which are the two unit vectors i and wiit point on a i. curve in the plane we now associate two other unit vectors, "".t the ,,unit tangent vector" and the ,,unit normal vector.,, R(f) is the position vector of curve c at a point p on c, then the unit If tangentoectorof c at p, denoted by T (t) , is the unit vector in the direction of D1R(f) if D,R(r) # 0. The unit vector in the direction of D,R(f) so we may write

is givenby D,R(t)l lD,n(r)f; (1)

_ BecauseT(f) is a unit vector, it follows from Theorem 12.5.11that D;t (t) must be orthogonal T ( r) . D (t) is not necessarily a unit vector. 1o lT However, the vector Dilg)llDl(t)l is of unit magnitude and has the samedirection asDlT ( f ) . Therefore, D il U) I IDtT ( f ) | ir a unit vector which is orthogonal to T(f), and it is called the "unit normal vector.,, 17'8'2 Definition

If T(f)-is the unit tangent vector of curve C at a pointp on C, then the nzif normal uector, denoted by N(f), is the unit vector in the direction of DtT(t). 17.8.2 and the previous discussion, we conclude that

(2)

EXAMPLEL: Given the cur:ve havin g p arametric equations r c - S t- 3 f a n d y : 3 t 2 , f i n d T ( f ) and N(f). Draw a sketch of a portion of the curye at t - 2 and

solurroN:

A vector equation of the curye is

R(r; : (f'- 3r)i + 3t2i So D , R ( t ) : ( g t ,- 3 ) i + G t i

17.8 THE UNIT TANGENTAND UNIT NORMALVECTORSAND ARC LENGTHAS A PARAMETER

draw the representations of T(2) and N(2) having their initial point at t - /.

Then

-3(t2+D

:

lD,n(r)l: Applying (1), we get

r(r):#i+hi L3 l2 L1

r(2)

Differentiating N(2)

T(f) with resPect to t, we obtain

D;r(r):dryi+ffit Therefore,

: lD,T(r)l *ry^*ffi W Y@

tWry VT'Try :g -2 -L -I Figure17.8.1

2

Applying (2) , we have

N(r):#ri+#i Finding R (t), T(f), and N(f), when t 2, we obtain

R ( 2 ): 2 i + r z i

T ( 2 ): g i + * i

N ( 2 ): * i - g i

The required sketch is shown in Fig.17.8.'1..

FromEq. (1) we obtain D'R(f): lD'R(t)lT(f)

(3)

The right side of (3) is the product of a scalar and a vector. To differentiate this product, we apply Theorem L7.5.8, and we have

D,rR(r): lD,lD,R(f)llT(f) + lD,R(f)| [D,T(r)]

(4)

From (2) we obtain D;r(t): lD'r(f)lN(f)

(s)

Substitutingfrom (5) into (4), we get D,'R(t): lD,lD'R(f)llr(f) + lD'n(f)| lD'r(f) lN(f)

(5)


lDtlDtR(f)llr(f) D,2R1t;

llD,r(f)lN(r lD,R(f)

Df R(f)

r(f) N(f)

R(f)

Figure17.8.2

Equation (3) expressesthe vector D,R(t) as a scalartimes the unit tangent vector, and Eq. (5) expressesthe vector DrrR(f) as a scalar times the unit tangent vector plus a scalartimes the unit normal vector. The coefficient of T(f) on the right side of (6) is the component of the vector Dr'zR(t) in the direction of the unit tangent vector. The coefficient of N(f) on the right side of (6) is the component of D12R(f)in the direction of the unit normal vector. Figure 17.8.2 shows a portion of a curve C with the position representation of R(f) and the representationsof T(t), N(f), D,R(r), D,'R(f), [D,|D,R(r)l]T(r), and lD,R(r)llD,T(r)lN(r), all of whose initial points are at the point P on C. Note that the representation of the unit normal vector N is on the concaveside of the curve. This is proved in generalin Sec.17.9. Sometimesinstead of a parameter f, we wish to use as a parameter the number of units of arc length s from an arbitrarily chosen point Ps(xe,yo) on curve C to the point P (r, y) on C. Let s increaseas f increasesso that s is positive if the length of arc is measuredin the direction of increasing f and s is negative if the length of arc is measured in the opposite direction. Therefore, s units is a directed distance. Also, D1s> 0. To each value of s there corresponds a unique point P on the curve C. Consequently, the coordinates of P are functions of s, and s is a function of f. In Sec.17.6we showed that lD'R(f) | : Dts

(7)

Substituting from (7) into (1), we get

T(f):ry D,R(t) : DrsT(f) If the parameter is s instead of f, we have from the above equation, by taking f : s and noting that Dp : L, D-R(s) : T(s) This result is stated as a theorem. 17.8.3Theorem

If the vector equation of a curve c is R(s) : /(s)i + g(s)i, where s units is the length of arc measured from a particular point Poon C to the point P, then the unit tangent vector of C at P is given by

Now suppose that the parametric equations of a curve C involve a parameter f, and we wish to find parametric equations of C, with s, the number of units of arc length measured from some fixed point, as the

AND ARC LENGTHAS A PABAMETER AND UNITNORMALVECTORS 17.8THEUNITTANGENT

parameter.Often the operations involved are quite complicated. However, the method used is illustrated in the following example. ExAMPLE2: Suppose that parametric equations of the cunre C a r ex - f 3 a n d y : t 2 , w h e r e f > 0 . Find parametric equations of C having s as aparameter, where s is the number of units of arc length measured from the point where t:0.

solurroN: If Peis the point where t:O,Po is the origin. The vector equation of C is R(f) : tsi+ t2i BecauseD6s: lDlR(f)1, we differentiate the above vector and obtain DrR(t):3t2i+2ti So

lD,R(r)l: y9FT@ : {F 6FTT Becauset > 0, lF

: f. Thus, we have

l D ' R (lr:) t @ + + Therefore, Dos,:115P14 and so ft

s: I u\/9uz*4 du JO

- * J, fot t8u\m

du

: iv (9u,+ 4rt,r]. We obtain (S)

t:419t2*47ttz-* Solving Eq. (8) for f in terms of s, we have (9f+.41"r':27s*8 9p + 4: (Z7s* 8;zra Becauset > 0, we get

t:t{@iT?ryEZ Substituting this value of f into the given parametric equations for C, we obtain ,:fi1(27s*8)za-4ltrz

and y:*l(27s*8)ztr-n,

(9)

Now becauseD"R(s) : T(s), it follows that if R(s) : x(s)i + y(s)i, then T(s) : (Drr)i + (Dry)i. And becauseT(s) is a unit vector, we have (10) ( D , r )z 1 ( D ,y ) ' - 1

796


Equation (10) can be used to check Eqs. (9). This check is left as an exercise (see Exercise 11).

Exercises 17.8 In Exercises1 through 8, for the given curve, find T(f) and N(f ), and at f : fr draw a sketch of a portion of the curve and draw the representationsof T(tr) and N(fr) having initial point at t: h. 1. x:*t3-t,y:F;tr:2 2. x:itz,y:+f;tt:l 3. R(f) : etiI e ti;

\:0 4 . R ( t ) : 3 c o sf i * 3 s i n 4 ; h : t r 5. r : cos kt, y : sin kt, k > 0; tr: nlk

6 . x : t - s i n f , y : | - c o sf i \ : n 7. R(t) : ln cos fi * ln sin ti,0 < t < tn; tt: ir 8. R(t) : f cos fi * f sin ti; \:0 9 . I f t h e v e c t o r e q u a t i o n o fc u r v e C i s R ( f ) : 3 t 2 i * ( t t - 3 t ) i , vectorsR(2) and T(2).

f i n d t h e c o s i n e o f t h e m e a s u r e o ft h e a n g l e b e t w e e n t h e

10. If the vector equation of curve C is R(f ) : (4 - 3P)i + (ts - 3t)i, find the radian measure of the angle between the vectorsN(1) and D,'?R(l). 11. Check Eqs. (9) of the solution of Example2 by using Eq. (10). In Exercises12 through 15, find parametric equations of the curve having arc length s as a parameter, where s is measured from the point where f : 0. Check your result by using Eq. (10). 72. x: e cost, !:

a sin t

1 3 .x : 2 * c o st , y : S * s i n f 1 4 . x : 2 ( c o s f * f s i n t ) , V : 2 ( s i n t - f c o sf ) l.5. One cusp of the hypocycloid of four cusps:R(t) : a cossfi * a sin3 ti, 0 - t =tr. 16. Given the cycloid x:2(t-sin from the point where f : 0.

f), l:2(1-cos

t), expressthe arc length s as a functionof t, where sismeasured

L7. Prove that parametric equations of the catenary y : a cosh (xl a) where the parameter s is the number of units in the lengthofthearcfromthepoint(0,a)tothepoint(x,y)and.s=0whenx=0ands(0whenx(0,are )c-asinh-r!

and y-\m

17.9 CURVATURE

Let @ be the radian measure of the angle giving the direction of the unit tangent vector associated with a curve C. Therefore, S is the radian measure of the angle from the direction of the positive r axis counterclockwise to the direction of the unit tangent vector T(f ). See Fig. 17.9.1,.

17.9CURVATURE BecauselT(t)l:

L, it follows from Eq. (1) in Sec.l7.2 that

T(f) : cos 0i + sin rfi

(1)

Differentiating with respect to Q, we obtain DoT(f)

(2)

0i + cos,fi

- 1 , D a T ( t ) i s a unit vector. B e c a u s leD r T ( t ) l : that BecauseT ( f ) has constantmagnitude it follows from Theorem 17.5.11 DoT(t) is orthogonalto T(f). Replacing-sin f by cos(*n*d) and cos f by sin(*zr*f), we write (2) as F i g u r e1 7 . 9 . 1

,

(3)

DoT(f)- cos(ln + 0)i + sin(tn + 0)i

From (3) and the previous discussion, the vector DoT(f) is a unit vector orthogonal to T(t) in the direction*zr counterclockwise from the direction of T(t). The unit normal vector N(t) is also orthogonal to T(f). By the chain rule (Theorem 17.5.9),we have D;I(t):

(4)

[Dor(f)]DtQ

Becausethe direction of N (f ) is the same as the direction of.D it (t), we see from (4) that the direction of N(t) is the sameas the direction of DoT(f) if D# > 0 (i.e., if T(f) turns counterclockwiseas f increases),and the direction of N(t) is opposite that of D6T(t) if D# < 0 (i.e., if T(f) tums clockwise as f increases). Becauseboth D6T(f) and N(t) are unit vectors, we conclude that

Dar(r): t_N[t] D,6>o

ifD#>0 ifD$<0

(s)

. rLLUsrRArroN1: In Fig. 17.9.2a,b, c, and d various casesare shown; in a and b, Dtb > 0, and in c and d,DtO < 0. The positive direction along the curve C is indicated by the tip of the arrow on C. In each figure are shown the angle of radian measure { and representations of the vectors T(f),

D,6( o F i g u r e1 7 . 9 . 2


DoT(t), and N(f). The representation of the unit normalaectorN(f) is always on the concaaesideof the curT)e. o Consider now DrT (t) , where s units is the arc length measuredfrom an arbitrarily chosen point on C to point P and s increasesas f increases. By the chain rule D,T(r) - DaT(t)D,o Hence,

: lDrT(f)|lD,dl lD,T(r) | : lDrT(f)D,dl But becauseD6T(f) is a unit vector, lDrT(f) | :

thus, we have

lD,T(r)l:lD,4l

(5)

The number lD-dl is the absolute value of the rate of change of the measureof the angle giving the direction of the unit tangent vector T(f) at a point on a curve with respect to the measure of arc length along the curve. This number is called the curaatureof the curve at thJpoint. Befo.e giving the formal definition of curvature, let us see that taking it as this number is consistent with what we intuitively think of as theturvature. For example, at point P on C, { is the radian measure of the angle giving the direction of the vector T(f ), and s units is the arc length from a poi"t po on c to P. Let Q be a point on C for which the radian measureof the angle giving the direction of T(f * Af) at e is d * A@and s * As units is the arc length from Poto Q. Then the arc length from p to e is As units, and the ratio AdlAs seemslike a good measure of what we would intuitively think of as the aaeragecurvature along arc PQ. o ILLUsrRArroN2: See Fig. 17.9.3a,b, c, and d: In a, A0 > 0 and As > 0; in b, Af ) 0 and As ( 0; in c, A{ ( 0 and As > 0; and in d, 46 ( 0 and As<0. aC)o

As)O

T

6 +^ +

As(0

Figure17.9.3

ad<0

As)0

ad(o

As(0

17.9CURVATURE

L7.9.LDefinition

IfT(f)istheunittangentvectortoacurveCatapointP,sisthearclength measured from an arbitrarily chosen point on C to P, and s increasesas f increases,then the curaatureaectorof C at P, denoted by K(f), is given by

ffiQ) The curvature ol C at P, denoted by K(f), is the magnitude of the cunrature vector. To find the curvature vector and the curvature for a particular curve it is convenient to have a formula expressingthe curuature vector in temrs of derivatives with respect to f. By the chain rule D'T(f):

IDJ(f)]D's

Replacing Dp by IDE(t)l and dividing by this, we obtain DJ(f) DrT(t)/lDrR(t)1. So from (7) we have for the curvature vector

BecauseK(f ) : lK( t)1, the curyatureis given by

=llDA?i)Tl LP+(tl ,l K(r;: EXAMPLE L:

Given

the circle

x:acos1,!:osinf

fl>0

find the curvature vector and the curuature at any t.

soLUrIoN: The vector equation of the circle is R(f) - a cos ti* a sin fi So DrR(t) : -a sin ti + a costi

l D , R ( r ) l : (-

a sin t)' + (a cost)2 - a

, r(f):#!tl lD,R(f)l D;T(t) : -cos ti - sin fi QTJil lpRbl

cosf , sinf , ra ol

So the curyature vector

1 ' K ( f ) : - l c o s t i t - A s m t. t. a and the curyature

K(r;:lK(ill:+

:

VECTORS IN THEPLANE ANDPARAMETRIC EQUATIONS The result of Example L statesthat the cunrature of a circle is constant and is the reciprocal of the radius. Suppose that we are given a curve C and at a particular point p the curyature exists and is K(f), where K(t) + 0. consider the circle which is tangent to curve C at P and has curvature K(f) at P. From Example 1.,we know that the radius of this circle is UK(f) and that its center is on a line perpendicular to the tangent line in the direction of N(f). This circle is called the circle of curaature,and its radius is the radiusof curoatureof. C at P. The circle of curvature is sometimes refered to as the osculating circle. 17.9.2 Definition

If I((t) is the curvature of a curve C at point P and K(t) + 0, then the radiusof curlature of C at P, denoted by p(t), is defined by

p(r): fri, 1

ExAMPrn 2: Given that a vector equation of a curye C is R(f) :2ti * (t'-

SOLUTION:

R ( f ): 2 t i * ( t ' - L ) i D , R ( f )- 2 i + z t i

l)i

find the cunrature and the radius of curvature at t - 1. Draw a sketch of a portion of the curye, the unit tangent vector, and the circle of cunrature at t : t.

lD,n(f)l-21ffi2

1-=i+ t ( f ') : , ? ' l l t ) -rt \ m lD,R(fl- irffii

'

D;r(r)-G*fui+&i DTG)lD,n(f)l

2(L+t), r'2(1 +t)rt | I sz ,- l.o\ K (t): r )- l g - _ " ) : tlVg-, r -4(r*tzy+

hffitfiTl

K ( t') r) -

x(t):

1

+b

L

20

+ t2)3t2

pill) :4t/I

and 11

r(1):1ri+Vri Figure17.9.4

Figure

17.9.4 shows the sketch that is required.

The accompanying

17.9CURVATUREE01 T able 17.9.'1.gives the corresponding values of r and ! for t': -2, -'L , 0, 1, and2. Table L7.9.7

-2 -1.

-43 -20

0 L 2

0 20 43

-L

We now find a formula for computing the cunrature directly from parametric equations of the cun/e, x- f(t) and y: g(f). From Eq. (5)

IDJ(f) | : lD,,$l,and so

K(r;- lD,ol Assuming that s and f increase together, we have

So

do Dr6:

dt

ffi

(10)

To find dfildt, we observe that because@is the radian measureof the angle giving the direction of the unit tangent vector,

(11)

Differentiating implicitly with respect to f the left and right members of (11), we obtain

802


So d0_ dt

(#)(#)-(HGry)

But

Substituting this expression for sec2

(#)(#)-(#)(#)

(14)

ExAMPr,n 3: Find the cunratureof the curuein Example2by using formula (14).

soLUTroN: Parametric equations of C are x - 2t and y : t2 -'j... Hence,

dx - " ,

d2x

dt-L

dt2

-0

du +:2t dt

d2u a/

_n

dtz-L

Suppose that we are given a cartesian equation of a curye, either in the form A : F(x) or x: G(y). Specialcasesof formula (L4) can be used to find the cunrature of a curve in such situations. If y : p 1r1 is an equation of a curve C, a set of parametric equations of C is r: f and y:F(t). Then dxldt:1, dzxldt2:0, dylilt: dyldx, and dzyldt2: dzyld*. Substituting into (14), we obtain

(ls)

17,9CURVATURE 803

C is r-

Similarly,

G(y), we obtain

(15)

EXAMPTS4: If the curye C has an equati on )cy - 1, find the radius of cunrature of C at the point (1, 1) and draw a sketch of the curye and the circle of cunrature at (1, 1).

solurroN: Solving for !, we obtain A:Ux. dzyldf :21x3. Applying formula (15), we have

K-

l*l

2xu:ry:ry

So dylik--11f

and

Zxa

v Becausep:

LIK we have

lrl(# + 1)3/2 p:----ESo at (1, l), p:tE.fne

required sketch is shown in Fig.77.9.5.

ure17.9.5

Exercises 17.9 In Exercises L through 4, find the curvature K and the radius of curvature p at the point where f - fr. Use formula (9) to find K. Draw a sketch showing a portion of the curve, the unit tangent vector, and the circle of curvature at l: fi. 2. R(f) - (sz- zt)i+ (f3- t)i; tr- 1 1. R(f) : 2ti* (t' - r)i; h: L 3. R(f ) - Zeti+ Ze-ti)h:

4. R(f ) : sin fi + sin 2fi; h:

0

Ln

In Exercises5 and 6, find the curvature K by using formula (14).Then find K and p at the point where f : fr and draw a sketch showing a portion of the curve, the unit tangent vector, and the circle of curvature at t: ty 5. )c--+

{y:

t_ t;

6. X: et* e-r, y - et_ e-t. tr- 0

fr: o

In Exercises7 through t2, find the curvature K and the radius of curvature p at the given point. Draw a sketch showing a portion of the curye, a piece of the tangent line, and the circle of curvature at the given point. (+,t) 9. a : e'; (0, 1) 8. 7. y : 2{x; (0, o)

a':f;

10. 4x2* 9y' : 36; (0, 2)

11. x : sin y; (*, tn)

12. x:

tan y; (l, Ln)


In ExercisesL3 through L8, find the radius of cunrature at any point on the given curue. 1 3 .y - s i n - r x 15. xuz+ yuz- auz

ln secx L5. R(f) -- et sin fi * et cos fj

17. The cycloid x -- a(t - sin f), A : a(l - cos f)

L8. The tractrixx- t - a tanh

14. y:

1,,

- a r".h 1

19. Show that the cuwature of the catenary y: a cosh (xla) at any point (x, y) on the curve is aly2. Draw the circle of curvature aI {0 , a) . Show that the curvature K is an absolute maximum at the point (0, a ) without referring to K' (r) . In Exercises20 through 23, find a point on the given curve at which the curvature is an absolute maximum. 2 0 .y : s t

2 1 .y : 6 x - f

2. y:

2 3 . R ( r ) : ( 2 t- 3 ) i + ( p - l ) i

sin x

24' lf a polar equation of a curve is r:

F(0), prove that the curvature K is given by the formula

K _ lr' + ?@tae), _r\g!do,)l (drlde)zf'n lr' *

In Exercises25 through 28, find the curvature K and the radius of curyature p at the indicated point. Use the formula of Exercise 24 to find K. 25. r: 4 cos 20; e - #rr 2 6 . r : ' / . .- s i n 0 ; 0 - 0 2 7 .r : A s e c 2 t 0 ; 0 : 3 n

2 8 .r - a 0 ; 0 : l

29. The center of the circle of curvature of a curve C at a point P is called the centerof curcature at P. Prove that the coordinates of the center of curvature of a curve at P(x, y) are given by

xc: x - @vld*)lf,t=7vldx)'l d2yI dxz

, (dyldx)z+ L A":Y+@

In Exercises30 through 32, find the curvature K, the radius of curvature p, and the center of curvature at the given point. Draw a sketch of the,curve and the cirde of curvature.

30. y : cosr; (tn, t)

3 1 .y - f

-x'; (0,0)

3 2 .y : l n

x; (1,0)

In Exercises33 through 35, find the coordinates of the center of curvature at any point. 33. y':

4px 35. R(f) - a cos fi +b sin fi

I7.IO TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION

34. At : a'x 36. R(f) - a coss fi + a sins ti

If a particle is moving along a curve C having the vector equation

R ( f ): f ( t ) i + s @ i from Definition

(1)

17.7.1, the velocity vector at P is given by

v(f) - D/R(f)

(2)

From Sec.17.8,if T(t) is the unit tangent vector atP, s is the length of arc of C from a fixed point P0 to P, and s increasesas f increases,we have D,R(t) - DtstT(f)l

(3)

OF ACCELERATION 805 AND NORMALCOMPONENTS 17.10TANGENTIAL

Substituting from (3) into (2) , we have

v(f) - Dtstr(f)l

(4)

Equation (4) expressesthe velocity vector at a point as a scalartimes the unit tangent vector at the point. We now proceed to expressthe acceleration vector at a point in terms of the unit tangent and unit normal vectors at the point. From Definition17.7.2 the accelerationvector at P is given by A(f) : Dr'zR(f)

(s)

From Eq. (6) in Sec.17.8we have

+ lD' R( t) llD' r ( t) lN( f) D ,'zR (t):[D' lD' R( t) l]r ( f)

(6)

Because D t s : l D r R ( f )|

(7)

if we differentiate with resPect to t, we obtain

Dr,s: D|D,n(t)|

(8)

Furtherrnore,

lD,R(r)P tD,R(r)llD,r(r)l: l#fr&l

t'l

Applying (7) above and Eq. (9) of Sec. 17.9 to the right side of (9), we have lD,n(f) | lD,T(f)| : (Dts)'K(t)

(10)

Substitutingfrom (5), (8), and (10) into (6), we obtain A(r; - (D,rs)T(f)+ (D,s),K(f)N(f)

(11)

Equation (1L) expressesthe accelerationvector as the sum of a scalar times the unit tangent vector and a scalartimes the unit normal vector. The coefficient of T(f) is called the tangentialcomponenfof the acceleration vector and is denoted by Ar(t), whereas the coefficient of N(t) is called the normal componentof the accelerationvector and is denoted by A1s(t). So

(r2) and (13)

(14)

806

VECTORS IN THEPLANE ANDPARAMETBIC EQUATIONS Becausethe number of units in the speedof the particle at time f units is lv(f ) | : DF, ArG) is the derivative of the measureof the speed of the particle and,4"(f) is the square of the measureof the speed divided by the radius of curvature. From Newton's second law of motion F:mA where F is the force vector applied to the moving object, m is the measure of the mass of the object, and A is the accelerationvector of the object. In curvilinear motion, the normal component of F is the force t or11121 to the curve necessaryto keep the object on the curve. For example, if an automobile is going around a curye at a high speed, then the normal force must have a large magnitude to keep the car on the road. Also, if the curve is sharp, the radius of curvature is a small number, and so the magnitude of the normal force must be a large number. Equation (4) indicates that the tangential component of the velocity vector is Dp and that the normal component of the velocity vector is zero. Substituting from (12) and (13) into (11), we have A(f) : Ar(f)T(f) * Ar(f)N(f)

(1s)


l A ( r ) l : lAr(f)lz a lA;wP Solvin g for tAr(f ), noting from (13) that A*(t) is nonnegative, w€ have iiiii

ExAMPrn L: A particle is moving along the curye having the vector equation

R ( r ;- ( t ' - l ) i + ( * r ' - t ) i Find each of the following vectors: V(f ), A( t) ,T (t), andN(f ). Also, find the following scalars: lV(f) l, Ar(t) , A*(t) , and K(f ). Find the particular values when t: 2. Draw a sketch showing a portion of the curye at the point where t - 2, and representations of v (2), A (2), Ar(2)T (2), and Ar(2)N(2), having their initial point at t: 2.

(15)

SOLUTION:

v(r):DrR(t) - zti+ (t'-L)i A(f) -DN(r) - 2i+zti

lv(r)l: lA(t)l:\M

-m:t2+1 -zt/L+t2

Dts:lv(t)l:t2+L Ar(t):D,'s-2t From (16) A*(t)=

-@-2

r(r):ffi:hi+#i To find N(f), we use the following formula which comes from (11):

(r7)

EXERCISES REVIEW

r (t) = 2i+ zti- r, (#i A(r)- (D,'s) -ht(1 A(f)- (D,'s)r(t)

+#

-1\i+ ztil

t) 1ra)

From (17) we see that N(t) is a scalar times the vector in (18). BecauseN(f) is a unit vector, N(f) can be obtained by dividing the vector in (18) by its magnitude. Thus, we have N(f): =

6 5

A(2)

4 3 2 -1

v(2)

rvN(z),"

s,3)

1 3

4

5

6

7I

2t ! ': t - S z r' +' | l+t'l r+tz

K(f ) is found from (13), and we have

r((f)=ffi:& When t: 2, we obtain V(2) : 4i + 3i, A(2) : 2i + 4i, lVlZ;; : 5, Ar(2\: 4, AN(2):2, T(2) : *i + gi, N(2) : -ti + €i, K(2): 4z The required sketchis shown in Fig. 77.10'1.

F i g u r e1 7 . 1 0 . 1

L7.10 Exercises In Exercises1 through 5, a particle is moving along the curve having the given vector equation. In each problem, find the vectorsV(t), A(f), T(t), and N(f), and the following scalarsfor an arbitrary value of f: lV(f) l, Ar, A*, K(f). Also find the particular values when f : fr. At f : fr, draw a sketch of a portion of the curve and representations of the vectors V(ft), A(tr), A"T(f'), and ArN(h). 3 . R ( f ) - 5 c o s3 f i + 5 s i n 3 f i ; h : t n 2 . R ( t ; : ( t - l ) i + t z i ;h : f I . R ( f ) : ( 2 t+ 3 ) i + ( t ' - L ) i ; t r : 2 5. R(f ) : 3tzi+ 2Fi; tt: I 5 . R ( f ) : e t i * e - t i ;f r : 0 4. R(f) : cos tzi+ sintzi;tr: +\G 7. A particle is moving along the parabola A":8x and its speed is constant. Find each of the following when the particle is at (2,4): the poJ-itiotr vector, the velocity vector, the acceleration vector, the unit tangent vector, the unit normal vector, A7, and A11. - * : 9, such that D6, is a positive constant. Find each 8. A particle is moving along the top branch of the hyperb ola y' (4, position vector, the velocity vector, the acceleration vector, the unit 5): the of ihe fo[owing when the partid- is at tangent vector, the unit normal vector, A7, arrd Ap

(ChapterL7) ReaiewExercises In Exercises1 through 12,let A:4i1. Find 38 - 7A 4. Find lsBl - l3cl

and C:9i - 5i. 2. Find 4A + B - 6C 5. Find (A - B) ' c

5i,B:i*7i,

7. Find a unit vector having the same direction as 2A + B. 8. Find the unit vectors that afe orthogonal to B.

3. Find 5B - 3C 5. Find (A . B)C

808


9. Find scalarsh artd k such that A:

hB + kC.

10. Find the vector projection of C onto A. 11. Find the component of A in the direction of B. 12. Find cos a if a is the radian measure of the angle between A and C. In Exercises13 and 14, f.or the vector-valued function, find (a) the domain of R; (b)

i'l 1 4 .R ( r )- l t - 1 l i + l n ti

R(t);(c)DyR(t).

In Exercises15 and 16, find, equations of the horizontal and vertical tangent lines, and then draw a sketch of the graph of the given pair of parametric equations. 1 5 .r : 1 2 - t 2 , y : l z t - t 3 t6.

x : T + t z , Y : T 2aF R,a 2at2

> 0

(the cissoid of Diocles)

17. Find the length of the arc of the curve R(f) : (2 - t)i* pi from t: 0 to t: 3. 18. Find the length of the arc of the curve r: 3 sec d from 0 : 0 to 0: *r. t9. (a) Show that the curve-defined by the parametric equations, x: a sin t and.y: measure of the length of arc of the ellipse of part (a), ihow that

t=4

Frl2

Jo

at/t=FRl

b cos f, is an ellipse. (b) If s is the

At

where k2: (a2- b') I a' < 1. This integral is called an elliptic integral and cannot be evaluated exactly in terms of elementary functions. There are tables available that give the value of the integral in terms of k. 20. Draw a sketdr of the graph of the vector equation R(t) : eti* e-ti and find a cartesian equation of the graph. 2 1 . S h o w t h a t t h e c u r v a t u r e otfh e c u r v e ! : l n x a t a n y p o i n t (x,y)isxl (f*l;arz.Alsoshowthattheabsolutemaximum curvature ils2lry'i which occursat the point (+{r,^-+ln2)."' 22. Find the curvature at any point of the branch of the hyperbola defined by x:a the curvature is an absolute maximum at the vertex. 23. Find the radius of curvature at any point on the curve r:

cosh t,y:b

sinh f. Also show that

a(sin f - f cos f). 24. Find the curvature, the radius of curyature, and the center of curvature of the curve : e-" at the point (0, 1). A 25. Forthehypocycloidof fourcusps,r:4cosstandy:asinat,hnd,dyldxand,&yldfwithouteliminatingtheparameter. a(cos t * f sin t),y:

.26. A particle is moving along-a : 3ti + (4t - p)i. Find a cartesian equation of the -curvehaving the vector equation R(t) path of the particle. Also find the velocity vector and the accelerationvector at t: l'. Find the tangential and nonnal components of the accelerationvector for the particle of Exercise25. If a particle is moving along a curve, under what conditions will the accelerationvector and the unit tangent vector have the same or opposite directions? In Exercises 29 and. 30, for the given curve find T(f) and N (f), and at t : h draw a sketch of a portion of the curve and draw the representations of T(fr) and N(fr) having initial point at t: tr.

29. R(t) - ( et+ e-t)i+ Zti; h:2 3 0 . R ( f ) - 3 ( c o st + t s i n f ) i + 3 ( s i n t - t c o s f ) i , t >

tr: tn

REVIEWEXERCISES f = 0, find parametric equations having the 31. Given the curve having parametric equations x:4t, y:*(2t*l1srz, measure of arc length s-al a parameter, where arc length is measured from the point where f : 0. Check your result by using Eq. (10)of Sec.17.8 32. Find the radian measure of the angle of elevation at which a gun should be fired in order to obtain the maximum range for a given muzzle speed. 33. Find a formula for obtaining the maximum height reached by a projectile fired from a gun having a given mazAe speed of vn ftlsec and an angle of elevation of radian measure tr. 34. Find the position vector R(f) if the accelerationvector 1

A(r): fi- ui andV(1) : i, and R(1) : -*i + +i. 35. Proveby vectoranalysisthat the diagonalsof a parallelogrambisecteachother. : *V(AT); V(Ei) : 36. Given trianglez{BC,points D, E, and.F are on sidesAB, BC, andAC, respectively,and V(.D)

v(Al) + v(ti) + v(cE): o. *v(rZ);v(ai) :*vtc?l. Prove In Exercises37 and 38, find the velocity and acceleration vectorq, the speed, and the tangential and normal components of acceleration. 37. R(f ) : cosh zti + sinh 2fi

38. R(t; : (2 tan-r s - f)i + ln(1 + t')i

39. An epicyctoidis the curve traced by a point P on the circumference of a cirde of radius b which is rolling extemally on a fixed circle of radius a. If the origin is at the center of the fixed circle,A(a,O) is one of the points at which the given point p comes in contact with the iixed circle, B is the moving point of tangency of the two circles, and the Parameter 1 ir the radian measure of the angle AOB, prove that parametric equations of the epicycloid are

x-(a*b)cosf-bcos+, and

y : ( a t b )s i nt - b s i n f f t

Vectors in three-dimensionalspace and solid analytic geometr

NUMBER SPACE 811 18.1tr, THETHREE-DIMENSIONAL 19.1 Rt, THE In Chapter 1 we discussed the number line Rl (the one-dimensional THREE-DIMENSIONAL number space) and the number plane R2 (the two-dimensional number NUMBER SPACE space).we identified the real numbers in Rl with points on a horizontal axis and the real number pairs in R2with points in a geometric plane. In an analogous fashion, we now introduce the set of all ordered triples of real numbers. L8.1.L Definition

The set of all ordered triples of real numbers is called the three-dimensional numberspace and is denoted by Rt. Each ordered triple (x, y, z) is called a point in the three-dimensionalnumber sPace. To represent R3 in a geometric three-dimensional spacewe consider the directed distances of a point from three mutually perpendicular planes.The planes are formed by first considering three mutually perPe,ndicular lineJwhich intersect at a point that we call the origin and denote by the letter O. These lines, called the coordinate axes,are designated as the r axis, the y axis,and the z axis.Usually the r axis and the y axis arc taken in a horizontal plane, and the z axis is vertical. A positive direction is selected on eachaxis. If the positive directions are chosen as in Fig. 18.1.1,the coordinate system is called a right-handedsystem.This terminology follows from the fact that if the right hand is placed so the thumb is pointed in the positive direction of the r axis and the index finger is pointed in the posiiive direction of the y axis, then the middle finger is pointed in the positive direction of the z axis.If the middle finger is pointed in the negative direction ofthe z axis, then the coordinate system is called lefthanded. A left-handed system is shown in Fig. 78.1.2. In general, we use a righthanded system. The three axes determine three coordinate planes: tl]rexy plane containing the x and y axes,the xz plane containing the r and z ixes, and the yz plane containing the y and z axes.

F i g u r e1 8 . 1 . 1

Figure 18J.2

An ordered triple of real numbers (x, y, z) is associated with each point P in a geometricthree-dimensional space.The directed distance of P

812

VECTORSIN THREE-DIMENSIONAL SPACE AND SOLID ANALYTICGEOMETRY

v (o,2, o)

(

G , o ,o )

from the yz plane is called the x coordinate,the directed distance of p from the xz plane is called the y coordinate,and the z coordinateis the directed distance of P from the xy plane. Thesethree coordinates are called the rectangularcartesiancoordinates of the point, and there is a one-to-onecorrespondence (called a rectangularcartesiancoordinatesystem)between all such ordered triples of real numbers and the points in a geometric threedimensional space. Hence, we identify R3 with the geometric threedimensional space, and we call an ordered triple (r, A, z) a point. The point (3, 2,4) is shown in Fig. 18.1.3,and the point (4, -2,-S) is shown in Fig. 18.1,.4.The three coordinate planes divide the space into eight parts, called octants.The first octant is the one in which all three coordinates are positive. z A

F i g u r 1e 8 . 1 . 3 (4, 0, 0)

-5) ;*tl(o,0,

(4,

F i g u r 1e 8 . 1 . 4 A line is parallel to a plane if and only if the distance from any point on the line to the plane i s the same. O ILLUSTRATION

1.: A line parallel to the yz plane, one parallel to the xz plane, and one parallel to the xy plane are shown in Fig. 18.1.5a,b, and c, respectively.

at'

aa'

Figure 18.'1 .5

AU L M B E RS P A C E 1 8 . 1 R 3 ,T H E T H R E E - D I M E N S I O NN

813

We consider all lines lying in a given plane as being parallel to the plane, in which casethe distance from any point on the line to the plane is zero. The following theorem follows immediately. 18.1.2 Theorem

,

18.1.3 Theorem

( i ) A line is parallel to the yz plane if and only if all points on the line have equal x coordinates. ( i i ) A line is parallel to the xz plane if and only if all points on the line have equal y coordinates. (iii) A line is parallel to the xy plane if and only if all points on the line have equal z coordinates.

In three-dimensional space,if a line is parallel to each of two intersecting planes, it is parallel to the line of intersection of the two planes. Also, if a given line is parallel to a secondline, then the given line is parallel to any plane containing the second line. Theorem 18.1.3follows from these two facts from solid geometry and from Theorem 18.1.2. (i) A line is parallel to the r axis if and only if all points on the line have equal y coordinates and equal z coordinates. (ii) A line is parallel to they axis if and only if all points on the line have equal r coordinates and equal z coordinates. (iii) A line is parallel to the z axis if and only if all points on the line have equal r coordinates and equal y coordinates. . rLLUsrRArroN2: A line parallel to the x axis, a line parallel to the y axis, 'l'8.1.6a, b, and c, reand a line parallel to the z axis are shown in Fig. ' sPectivelY.

18.1.6 Figure The formulas for finding the directed distance from one point to another on a line parallel to a coordinate axis follow from the definition of directed distance given in Sec.1.4and are stated in the following theorem.

lII

VECTORS IN THREE-DIMENSIONAL SPACEAND SOLIDANALYTIC GEOMETRY (i) If A(xr,y, z) and B(xr,y,z) are two points on a line parallelto

18.1.4 Theorem

the r axis, then the directed distancefrom Ato B, denoted by B, is given by AB:xz-x1 (ii) If C(x, At, z) and D(x, Az,z) are two points on a line parallelto W V axis, then the directed distance from C to D, denoted by CD, is given by CD: Uz- Ar (iii) If E(x, y, z) and F (x, y, zr) aretwo points on a line parallel to the

z axis, then the directed distance from E to F, denoted by EF, is given by EF: zz- zr o rLLUsrRAtroN3: The directed distancePQ from the point P(2,-5,-4) to the point Q(2, -3, -4) is given by Theorem 1S.1.4(ii).We have

The following theorem gives a formula for finding the undirected distance between any two points in three-dimensional space. 18.1.5 Theorem

B(xr,Ar,zrl Pt (xt , At, Zr) i-..

A(x2, Uz, zt)

The undirected distance between the two points Pr(xr, yr, zr) and Pr(xr, Az,z) is given by

pRooF: Construct a rectangular parallelepiped having Pr and P, as opposite vertices and faces parallel to the coordinate planes ( see Fig. '/-,8.'J-,.7). By the Pythagorean theorem we have

INF;P:ITEI'+ I7F,I'

(1)

Because

lmP: lF,fl,+l$l, ,)-----/-,/

F i g u r e1 8 . 1 . 7

Pz(x2,yr,zr)

----) / ,/

(2)

we obtain, by substituting from (2) into (l) , Applying Theorem 18.1.4(i),(ii), and (iii) to the right side of (3), we obtain ffpr-;z : (xr- xr), I (Ar- yr), * (zz- zr)z So and the theorem is proved.

I

AU L M B E RS P A C E 1 8 . 1 F ' , T H E T H R E E - D I M E N S I O NN

l.: Find the undirected distance between the points P (-3, 4, - 1 ) a n d Q ( 2 , 5 , - 4 ) . EXAMPLE

815

soLUTroN: From Theorem 18.L.5, we have

(5 - 4)'

:m-{gs

Note that the formula for the distance between two points in R3 is merely an extension of the corresponding formula for the distance between two points in R2 given in Theorem 1-4.I- It is also noteworthy that the undirected distance between two points x2 and x1 in Rl is given by

lxr-rrl:@ The formulas for the coordinates of the midpoint of a line segmentare derived by forming congruent triangles and proceeding in a manner analogous to the two-dimensional case.These formulas are given in Theorem and the proof is left as an exercise(seeExercise15). 18.'/-..6, 18.1.6 Theorem The coordinates of the midpoint of the line segment having endpoints Pr(xr,Ur,zr) and Pr(xr,Az,z2)aregiven bY

18.1.7 Definition

The graph of an equationin RBis the set of all points (x, y, z) whose coordinates are numbers satisfying the equation. The graph of an equation in RBis called a surface.One particular surface is the sphere, which is now defined.

18.L.8 Definition

A sphere is the set of all points in three-dimensional space equidistant from a fixed point. The fixed point is called the centerof the sphere and the measure of the constant distance is called t}:retadius of the sphere.

18.1.9Theorem An equation of the sphere of radius r and centet at (h, k,l) is (4)

pRooF: Let the point (h, k,l) be denotedby C (seeFig. 18.1.8).The point P(x, y, z) is a point on the sphere if and only if

lcnl: t or, equivalently, @:7 Squaring on both sides of the above equation, we obtain the desired I result.

816

VECTORS IN THREE-DIMENSIONAL SPACEAND SOLIDANALYTIC GEOMETRY

c(h,k,l)

If the centerof the sphereis at the origin, then L :k:l: equation of this sphere is * + y'l

0, and so an

22: 12

If we expand the terms of Eq. (a) and regroup the terms, we have * * y' * z2- Zhx - 2ky - 2lz * (h2+ P + 12- rz) : 0 This equation is of the form x2 + y',+ 22 + Gx * Hy * lz + I - 0 P(*, v, z)

x F i g u r e1 8 . 1 . 8

(5)

where G,H,l, and/ are constants.Equation (5) is called the general formof an equation of a sphere, whereas Eq. (4) is called the center-radiusform. Becauseevery sphere has a center and a radius, its equation can be put in the center-radius form and hence the general form. It can be shown that any equation of the form (5) can be put in the form

( x - h ) , + ( y - k ) , + ( z- t ) , : K

(6)

where

h--*C

k--tU

l:-iI

K-+(Gr+H2+tz-41)

It is left as an exerciseto show this (see Exercise 16). If K > 0, then Eq. (5) is of the form of Eq. (4), and so the graph of the equation is a spherehaving its center at (h, k,l) and radius l'R.lt K: 0, the graph of the equation is the point (ft, k, l). This is calledapoint-sphere. If K < 0, the graph is the empty set becausethe sum of the squaresof three real numbers is nonnegative. We state this result as a theorem. 18.1.10 Theorem

The graph of any second-degree equation in r, y, and z, of the form *+y,+22+Gx*Hy*tz*J:g is either a sphere,a point-sphere, or the empty set.

EXAMPLE2: Draw a sketch of the graph of the equation x2+y'+22-6x-4y+22-2

solurroNt Regrouping terms and completing the squares,we have f - 6x -t 9 -f y2 - 4y + 4 * z2* 2z * l: 2 * 9 + 4 + l ( r - 3 ) ' + 0 - 2 ) 2* ( z * 1 ) 2: 1 5 So the graph is a spherehaving its centerat (3,2,-1) and radius 4. A sketch of the graph is shown in Fig. 18.1.9.

F i g u r e1 8 . 1 . 9

SPACE 817 NUMtsER 18.1 R',THETHREE-DIMENSIONAL

3: Find an equation of EXAMPT,n the sphere having the Points A ( - 5 , 5 , - 2 ) a n d B ( 9, - 4 , 0 ) a s endpoints of a diameter.

solurroN: The center of the sphere will be the midpoint of the line segment AB. Let this point be C(7,y,2).By Theorem 78.'l'.6'we get -4+6-a 9-5 ^ z=:--0- -: 2 - L- - . . L i:-::z t:T: So C is the point (2,1,-l).

The radius of the sphere is given by

rTherefore, from Theorem 18.L.9,an equation of the sphere is ( x - 2 ) ' + Q - r ) ' ,+ ( z * r ) 2: 7 5 or, equivalently, x2 + y', + z2- 4x - 2y * 2z - 69 : 0

18.1 Exercises parallelepiped, hawing its faces In Exercises 1 through 4, the given points A and B are opposite v,ertices of a rectangular the coordinates of the other sii find p"r*"r to the coordin"t" ptunlr. In each problem 1a)drari a sketch of the figure, 1b) vertices, (c) find the length of the diagonal AB. 2 . A ( I , 1, l ) ; B ( 3 ,4 , 2 ) 1 . A ( 0 ,0 , 0 ); B ( 7, 2 , 3 ) 4. A(2, -1, -g); B(4, 0, -l)

3 . A ( - 1 , l , 2 ) ; B ( 2 ,3 , 5 )

th9 first comer. (a) Draw a sketch 5. The vertex opposite one comer of a room is 18 ft east, 15 ft south, and 12 ft up from (c) find the coordinates of all eight vertices, joining opposite two diagonal of the length of the figure, (b) determine the vertices of the room. In Exercises 6 throu gh g,find segment joining A and B. 6. A(3, 4, 2); B(1,6,3) -5) 8. A(4, -3, 2); B(-2, 3,

(a) the undirected distance between the points A and B, and (b) the midpoint

of the line

7 . A ( 2 , - 4 , 1 ) ;B ( t , 2 , 3 1 9. A(-2, -L,5); B(5,1, -4)

-1,3) , (2, L,7) , and,(4,2,6) are the verticesof a right triangle, and find its area' 10. prove that the three points (7, of the points on this 11. A line is drawn through the point (6,4,2) perpendicular to the yz plane. Find the coordinates (0,4,0)' point the from 10 units of line at a distance 12. Solve Exercise11 if the line is drawn perpendicular to the ry plane. distancefonnula' 13. Prove that the three points (-3,2,4), (6,1,2), and (-12,9,5) are collinear by using the (0,3,4)' 14. Find the verticesof the triangle whose sides have midpoints at (3,2,3), (-1,1,5), and 15. Prove Theorem 18.1.6' 1 5 . S h o w t h a t a n y e q u a t i o no f t h e f o r m * * y " * z 2 l G x ' l H y * l z + 1 : 0 * (z- I)': X.

can be putin theform (x-h\"+

In Exercises17 through 21, determine the graph of the given equation. L8. x2+ y', + 22- 8y * 5z - 25 : 0 17. x2+ y' + 22-8r * 4y * 2z -4 : 0

(y-k)'

1 9 .x 2 + y ' + z , 2 - 6 2 * 9 : 0

818

VECTORSIN THREE-DIMENSIONAL SPACE AND SOLID ANALYTICGEOMETRY

2L. x2+ y' + zz- 6x * 2y - 4z *L9 : 0

20. x 2 + y ' + 2 2 - x - y - 3 2 * Z - - 0

In Exercises22 through 24, find, an equation of the sphere satisfying the given conditions. 22. A diameter is the line segmenthaving endpoints at (6,2,-5) and (-4,0,2). 23. It is concentric with the sphere having equation f * y, * z2- 2y I 8;z- ) : g. 24. It containsthe points (0,0,4)', (2,1,2),and (0, 2,6) and,hasits centerin the yz plane. 25. Prove analytically that the four diagonals joining opposite vertices of a rectangular parallelepiped

bisect each other. 26. I!!: Q'4, S are four points in three-dimensional space and A, B, C, and D are the midpoints of pe, T{ eR, RS,and SP, respectively, prove analytically that ABCD is a parailelogram.

18'2 VECTO_RSIN The presentation of topics in solid analytic geometry is simplified by the THREE-DIMENSIoNAL SPACE use of vectors in three-dimensional ,p".". irr" a"fir,itions and theolms given in Secs.17.1 and17.2 for vectors in the plane are easily extended. 18.2.1Definition

A aector in three-dimensional spaceis an ord.eredtriple of real numbers (x, y, zl . The numbers tt, y, and z are called the componezlfs of the vector

(x,y, zl.

(r * ar,y * or,z *

as)

(x, v, z)

x F i g ur e 1 8 . 2 . 1

We let Vs be the set of all ordered triples (x, y, zl for which x, y, and.z are real numbers. In this chapter, a vector is always in v3 unless otfrerwise stated. Just as forvectors inv2, a vector in v, can be representedby a directed line segment. If A : (ar, a2, asl, then the directed line segmeni having its initial point at the origin and its terminal point at the polnt (ar, ar, af,)is the position representationof A. A directed line segment travinf its :{tdinitial point at (x, y, z) and its terminal point at (x * ar, y * a2,z+ ar) is also a representationof the vector A. SeeFig. 1g.2.1. The zeroaectoris the vector (0, 0,0) and is denotedby 0. Any point is a representation of the zero vector. - The magnitudeof a vector is the length of any of its representations.If the vector A: (ar, ez, fls), the magnitude of A is denotei by lAl, and it follows that

Wl: t/ar'177 t, og The directionof anonzero vector in v, is given by three angles, called the direction anglesof the vector. 18.2.2 Definition

The direction anglesof a nonzero vector are the three angles that have the smallestnonnegative radian measuresa, B, andT measuied from the positive x, A, and z axes, respectively, to the position representatio' o] the vector. The radian measure of each direction angle of a vector is greater than

SPACE 18.2VECTORSIN THREE.DIMENSIONAL

819

or equal to 0 and less than or equal to zr.The direction angleshaving radian meaiu"es ot,F, andT of the vector A: (ar, az, asl are shown in Fig' 18'2'2' In this figure the components of A are all positive numbers, and the direction angles of this .r"itot all have positive radian measure less than *2. From thl figure we see that triangle POR is a right triangle and

a1 lml loPl lAl

COSC:__:_

It can be shown that the same formula holds if. Ln = q. < 7r.Similar formulas can be found for cos B and cos 7, and we have (1) Figure18.2.2

The three numbers cos a, cos B, and cos 7 are called the directioncosinesof vector A. The zero vector has no direction angles and hence no direction cosines. . rLLUsrRArroN1: We find the magnitude and direction cosines of the vector A : (3, 2, -6>.

:!9T4+35:t/4s -7

lAl: m From Eqs. (1) we get

cosa-+

cosy:-+

cosF:?

If we are given the magnitude of a vector and its direction cosines,the vector is uniqud determined becausefrom (1) it follows that (2) n s : l A l c o s7 n z : l A l c o sB n r : l A l c o sa The three direction cosines of a vector are not independent of each other, as we see by the following theorem' 1'8.2.3Theorem

If cos a, cos B, and cos 7 are the direction cosines of a vector, then coszc*cos2F*cosz7:1 pRooF: If A : (a1,az,a.), then the direction cosinesof A are given by (1) and we have

y:ffi*#i.ffi d+cosz cos2 P+cos2

-1

I

VECTORSIN THREE-DIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY

o rLLUsrRArroN2: we verify Theorem1.8.2.3 f.orthe vector of Illustration 1. We have 'cos2 a+ coszB+ aG i:

The vector A: (ar, ar, a"l is a unit vector if lAl : t, and from Eqs. (1) we see that the comPonents of a unit vector are its direction cosines. The operations of addition, subtraction, and scalar multiplication of vectors in Vs are given definitions analogous to the corresponding definitions for vectors in V2. 18.2.4Definition IfA: bv

(ar,nz,ar)andB:(br,br,brl,thenthesumof thesevectorsisgiven

A+ EXAMPLE1: Given A - (5, -2, 6> and B - (8, -5, -4), find A + B.

(a, * br, flz * br, fls * br)

solurroN: A * B:

(5 + 8, (-2) + (-S),5 + (-4)) : (13,-7,2>.

The geometric interpretation of the sum of two vectors in v3 is similar to that for vectors in Vr. See Fig. 18.2.3.If p is the point (x, y, z), A: (a1, a2, ar) and tr a representation of A, then e is the point "? (x * ar, A * az,z * ar).Let B : (br, br,b3) and let e? be a representation of B. Then R is the point (r * (a1*br), A-f @r*br), z* (ar*b")). Therefore, p? is a representation of the vector A + B, and the parallelogram law holds.

x

L8.2.5 Definition

Figure18.2.3

lf A: (ar, ar,ar), then the vector(-ar,-az,-ar) tioe of.A, denoted by -A.

is definedto be thenega-

SPACE 18.2VECTORSIN THREE-DIMENSIONAL

18.2.6 Definition

821

The difierenceof the two vectors A and B, denoted by A B, is defined by A-B:A*

(-B)

: From Definitions 18.2.5and 1'8.2.6it follows that if A (ar, ar, asl and B : ( b r , b r , b " l ,t h e n - B : ( - b r , - b 2 , - b " l a n d A - B : \ a r - b 1 ,a 2 - b 2 ,a s - b s l

ExAMPrn 2: For the vectors A and - B. B of Example 1, find A

SOLUTION:

A-B-(5,

-4>

6 >- ( 8 ,

5 , 4>

- ( 5 , - 2 , 6 >+ : (-3 , 3, 10)

The difference of two vectors in Vais also interpreted geometrically as it is in vz. see Fig. 1,8.2.4.4representation of the vector A B is obtained point. by choosing representations of A and B having the same -initial segment line Tilen a ,"pi"r"-t t"tion of the vector A B is the directed point of from the terminal point of the representation of B to the terminal the representation of A. o rLLUsrRAuoN3: Givgl the points P(L,3,5) and Q(2,-L, a)' Fig- 19"2'5 shorys p-i as well as O-Pand O0. We see from the figure that V(PQ): V(OQ) - v(OP). Hence,

x

5): (L,-4, -L> v(t[) : (2,-1, 4>- <7,3,

F i g ur e 1 8 . 2 . 4

'

P(L,3, 5) QQ,

-L,4)

Lo__-i

Figure 18.2.5 18.2.7 Definition

If c is a scalar and A is the vector (ar, fl2, as), then the product of c and A, denoted by cA, is a vector and is given by cA -

c(a1, fl2, flB) -

(car, cflz, cfls)


EXAMPrc 3: Given A (-4,7 , -2), find 3A and -5A.

SOLUTION:

3A : 3(-4,7 , -2) - (-12, 2l , -5>

and -5A : (-S) (-4 ,7 , -2> - (20, -gS, 10) Suppose that A - (ar, flz, azl is a nonzero vector having direction cosinescos 0, cos P, and cos y, and c is any nonzero scalar. Then cA : ( c a r , c a z , c f l s ) ; a n d if cos o1, cos Fr, and cos yr are the direction cosines of cA, we have from Eqs. (1) cos 01

:

cat

TrAJ

or/ equivalently, cos cl

-FIc a 1 IAT

from which we get

cosot :

c

IZI

cos a

cos

cos B

c

cos Tr:1;1 cos7

(3)

So if c > 0, it follows from Eqs. (3) that the direction cosines of vector cA are the same as the direction cosines of A. And if c < 0, the direction cosinesof cA are the negatives of the direction cosines of A. Therefore, we conclude that if c is a nonzero scalar,then the vector cA is a vector whose magnitude is lcl times the magnitude of A. If c > 0, cA has the samedirection as A, whereas if c < 0, the direction of cA is opposite that of A. The operations of vector addition and scalar multlplication of any vectors in Vr satisfy properties identical with those given in Theorem 17.2.7.These are given in the following theorem, and the proofs are left as exercises(see Exercises1 through 6). 18.2.8 Theorem

If A,B, and C are any vectors in V3and c and d are any scalars,then vector addition and scalar multiplication satisfy the following properties: (i) A + B - B + A (commutative law) (ii) A + (B + c) : (A + B) + c (associativelaw) (iii) There is a vector 0 in Vs for which (existenceof additive identity) (iv) There is a vector -A in V, such that A+0:A

A+(-A)-o (v) (vi) (vii) (viii)

(existence of negative)

(cd)A - c(dl) (associative law) c(A + B) - cA * cB (distributive law) (c * d)A: cA + dA (distributive law) (existence of scalar multiplicative identity) 1(A) - A

SPACE 18,2VECTORSIN THREE-DIMENSIONAL

From Definition 17.2.2 and Theorem real vector space. The three unit vectors

i:

( 1 ,0 , 0 )

i : (0,1,0)

1:

823

'1.8.2.8 it follows that Vs is

( 0 ,0 , 1 )

form a basis for the vector spaceV, becauseany vector (ar, az, at) can be written in terms of them as follows: ( a r , a r , a s l: a t ( L , 0 , 0 ) + a r ( o , 1 , 0 )* c r ( O ,0 , 1 ) Hence, if A : (ar, az, as), we also can write A:

ai* a"i * ask

(4)

Becausethere are three elements in a basis, Vs is a three-dimensional vector sPace. Substituting from Eqs. (2) into Eq. (4), we have

A: lAl cosci + ll,l cosBi + lAl cos7k or, equivalently, A:

lAl(cos ai * cos Bi * cos7k)

(s)

Equation (5) enablesus to expressany nonzero vector in terms of its magnitude and direction cosines.

rn 4: Expressthe vector of EXAMP Illustration L in terms of its magnitude and direction cosines.

18.2.9 Theorem

-6> , lAl solurroN: In Illustration!, A: ( 3, 2, -+,Hence, : have we (5) from Eq. and cos y

c o sa - ? , c o sF : ? ,

A-7(?i++i-+k) If the nonzero vector A: ai* a2i* ark, then the unit vector U having the same direction as A is given bY

"l'8.2-9is analogous to the The proof of Theorem Proof of Theorem 30). Exercise (see exercise an 17.2.3for a vector in V2 and is left as

EXAMPLE5: Given the Points R ( 2, - 1 , 3 ) a n d S ( 3 , 4 , 6 ) , f i n d t h e unit vector having the same direc-

tion asV(ffi).

SOLUTION:

v(ff) - <3,4, 5>- ( 2,- r , 3) - ( 1,5, 3> -i+5i+3k So

-->

lvtns)l-ffi-\B

VECTORSIN THREE.DIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY

Therefore, by Theorem L8.2.9 the desired unit vector is

u__1 . , 5 . , .Fir@i1155k

3

1-8.2 Exercises 1..Prove Theorem 18.2.8(i).

2. prove Theorem 1S.2.S(ii).

3. Prove Theorem 18.2.8(iii),(iv), and (viii).

4. prove Theorem 19.2.g(v).

5. Prove Theorem 18.2.8(vi).

5. prove Theorem lg.2.g(vii).

I n E x e r c i s e s T t h r o u g h l S , l e t A : ( 1 . , 2 , 3 ) , 8 :( 4 , - 3 , - 1 ) , C : ( - 5 , - g , S ) , D : < - 2 , 1 , 6 > . 10. Find 48 + 6C - 2D

8. Find 2A - C 1,1,. Find lTCl- lsDl

1 3 .F i n d C + 3 D - 8 A

14. Find 3A - 28 + C - tZD

7. Find A + 58

9. Find 7C- 5D 12.Find l4Bl+ l6cl - l2Dl l s . F i n dl A l l n l ( c - D )

16.Find lAlc - lBlD 17. Find scalarca and.b such that a(A + B) + b(C + D) : 0. 18. Find scalarsa, b, artd c such that aA + bB + cC: D. In ExercisesL9 through 22, find. the direction cosines of the vector Vtfrl sum of their squaresis L. L9. P,(3, -L , -4) ) Pz(7, 2, 4) 21. Pr(4, -3, -L); Pr(-z, -4, -8)

and check the answers by verifying that the

2 0 . P ,( 1, 3 , 5 ) ; P r ( z ,- 1 , 4 ) 2 2 . P , ( - 2 , 5 , 5 ) ; P r ( 2 ,4 , ! )

23. Using the points P, and P, of Exercise19, find the point e such trrat v(p?r) :3v(&a). 24. Using the points P, and P, of Exercise20, find the point R such ttrat v(fr) : -2v(p?). In Exercises25 through 28, express the given vector in terms of its magnitude and direction cosines. E.-6i+2j+3k 25.2i_2i+k 2 i + i 3 k 27. 28.3i+ 4i_5k 29. lf the radian measure of each direction angle of a vector is the same, what is it? 30. Prove Theorem 18.2.9. In Exercises31 and 32, find the unit vector having the same direction as v1p,pr;. 3 1 .P { 4 , - 1 , - 6 ) ; P , ( 5 , 7 , - 2 )

22.pJ-8,-5,2);p,(-3,-s,4)

33. Three vectortsin Vr are said to be independentif and.only if their position representations do not lie in a plane, and three vectors Et, Ez, ar.d'E3 are said to form a basisfor the vector spaceVr if and only if any vector in V, can be written as a linear combination of E1, E2, and Eg.A theorem can be prorred which statesthai three vectors form a basis for the vector sPaceVr if they are independent. Show that this theorem holds for the three vectors (1, O, O>, (1, 1, 0), and (1, 1, 1) by doing the following: (a) Verify that the vectors are independent by showing that their position representations are not coplanar; (b) verify that the vectors form a basis by showing that any vector A can be written

1 8 . 3T H E D O T P R O D U C Tt N V 3

825

(6)

r ( 1, 0 , 0 ) * s ( 1, 1 , 0 ) + t ( ' l . , l , l > where r, s, and t are scalars.(c) If A : <6,-2,5),

find the particular values of.r, s, and f so that Eq. (6) holds.

34. Refer to the first sentenceof Exercise33. A theorem can be proved which statesthat three vectors form a basis for the vector spaceV3 only if they are independent. Show that this theorem holds for the three vectorsFt: (1,0,7>,Fz: <1,l,lr, and Fg: (2,l,2) by doing the following: (a) Verify that Fr, F2,and F3are not independent by showing that their position representations are coplanar; (b) verify that the vectors do not form a basis by showing that every vector in V3 cannot be written as a linear combination of Fr, Fz, and F3'

1.8.3THE DOT The definition of the dot product of two vectors in V3is an extension of the PRODUCT IN V3 definition for vectors in V2. L8.3.1Definition

(br, br, brl , then the dot product of A and B,

fl2,

by

A tiiiiri+

1 . :I f A : o rLLUSrRArroN

(4,2,-6)l and B - (-5, 3,-2), then

A . [ - ( 4 , 2 , - 6 >. ( - 5 , 3 , - 2 > - 4(-5) + 2(3) + (-6) (-2) --20+6+12 --n

t

For the unit vectorsi, j, and k, we have i'i:j'i:k'k:1 and i'i:i'k:j'k:0 Laws of dot multiplication that are given in Theorem 18.3'2are the same as those in Theorems 17.3.2and17.3.3f.orvectors in Vr. The proofs are left as exercises(seeExercises1 through 4). 18.3.2Theorem

If A,B, and C are any vectors in V3 and c is a scalar,then (commutative law) (i) A'B: B'A ' (distributive law) (ii) A'(B+C):A B + A' C (iii) c(A ' B) : (cA) . B (iv) 0'A:0 (") A'A: lAl' Beforegiving a geometric representation of the dot product for vectors in Vr, we do as we did with vectors in Vz.We define the angle between two vectors and then express the dot product in terms of the cosine of the radian measure of this angle.

SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREE-DIMENSIONAL

18.3.3Definition

Let A and B be two nonzero vectors in V3 such that A is not a scalarmultiple of B. If O? is the position representation of A and O? is the position representation of B, then the angle between the vectors A and B is defined to be the angle of positive measurebetween O? and Oi ir,t""io, to the triangle POQ. If A: cB, where c is a scalar,then if c ) 0, the angle between the vectors has radian measure 0, and if. c < 0, the angle between the vectors has radian measure n'. Figure 18.3.1shows the angle of radian measure 0 between the two vectors if A is not a scalar multiple of B.

x Figure 18.3.1

18.3.4 Theorem

If 0 is the radian measure of the angle between the two nonzero vectors A and B in V3, then (1) The proof of Theorem 1,8.3.4is analogous to the proof of Theorem 17.3.5 for vectors in V2 and is left as an exercise (see Exercise 15). If U is a unit vector in the direction of A, we have from (1)

U . B : lUllr l cos6: lBlcos0 As with vectors in Vr, lBl cos d is the scalarprojectionof B onto A and the componentof B in the direction of A. It follows from Theorem 18.3.4that the dot product of two vectors A and B is the product of the magnitude, lAl, of one vector by the scalarprojection, lBl cos 0, of the secondvector onto the first. EXAMPLE1:

Given the vectors

A - 6 i - 3 i + 2 k a n dB : 2 i + i - 3k, find (a) the component of B in the direction of A, (b) the vector proiection of B onto A; and (c) cos 0 if g is the radian measure of the angle between A and B.

solurroN: lAl :

:m

- 7. Hence a unit ,

vector in the direction of A is

u- +i-+i+?k BecauseU . f, - lBl cos0, the component of B in the direction of A is

+k)'(2i+i - 3k)

u' t S'_11; - 3T _

The vector proiection of B onto A is therefore

tU: +8i- zti + abk From Eq. (1) A'B tI'B c o s o ==lAIlBT:lBT-

18.3 THE DOT PRODUCTtN V3

B e c a u s e U .:ft,a n d l n l : 3 cosQ-=ffi:

ExAMPtn2: Find the distance from the point P (4,1,6) to the line through the points A (8,3 ,2) a n dB ( 2 , - 3 , 5 ) .

:ffi

_ 3\m

gg

soLurroN: Figure 18.3.2 shows the point P and a sketch of the line through A and B. The point M is the foot of the perpendicular line from P to the line through A and B. Let d units be the distance lF[1. fo find d we find lffil and lZFl an{use the Pythagorean theore*. lIFl is the magnitude of the vector V(AP).

v(A?):v(&) -v(A) : ( 4 , 1 , 6-> < 8 , 9 , 2 > : (_4,_2,4>

B ( 2 ,- 3 , 5 )

a

P(4,L,6)

Hence,

lZPl:@TTR:ffi:5 we find the scalarprojectionof V(;F) onto V(A-i).

To find lMl

v(AT):v(ob)-v(o7) : ( 2 , - 1 , 5 )- ( 8 , 3 , 2 ) : (-6, -6 ' 3> The scalarprojection of v(a-F) onto V(Ai)

is then

v(,4Ft. v(aEr_ (-4,-2, 4) . (-6,-6, g') v36+ 36+e V(;;) | 24+12+ 12

x

Figure 18.3.2

\ET

_+1 Thus,r^r): :ffi -6\ffi :3\ffi The definition of parallel vectors in V3is analogousto Definition17.3.6 for vectors inV2, and we state it formally. 18.3.5 Definition

Two vectors in V3 are said to be parallel if. and only if one of the vectors is a scalar multiple of the other.

828


The following theorem follows from Definition 18.3.5and Theorem 18.3.4.Its proof is left as an exercise(seeExercise16). 18.3.6 Theorem

Two nonzero vectors in Vs are parallel if and only if the radian measure of the angle between them is 0 or zr. The following definition of orthogonal vectors in Vr corresponds to Definition 17.3.7f.orvectors in Vr.

18.3.7 Definition

If A and B are two vectors in Vs, A and B are said to be orthogonal if and onlvifA.B:0.

ExAMPln3: Prove by using vectorsthat the pointsA(4,9 , L) , B ( - 2 , 5 , 3 ) , a n d C ( 6, 3 , - 2 ) a r e the vertices of a right triangle.

solurroN: Triangle CAB is shown in Fig. 18.3.3.From the figure it looks as if the angle at A is the one that may be a right angle. We shall find

B(- 2,6,9)

V(AB) and V(AC), and if the dot product of thesetwo vectorsis zero,the angle is a right angle. -+

-+

v(AB)-v(oB)-v(oA) - ( - 2 , 6 , 3 )- ( 4 , 9 , ' ] . , > : (_5,_3, 2>

v(A?)-v(oa)-v(o?) A(4, 9, L)

- ( 6 , 3 , - 2 )- < 4 , 9 , ' j . , > : (2,-5, -3)

C(6, g, - z) F i g u r e1 8 . 3 . 3

v ( f r ) . v ( f r ) - F 6 , - g ,2 >. < 2 , - 6 , - g-) - 1 2+ Therefore;V(AB) and V(;E) angle CAB is a right angle.

L8- 6 - 0 are orthogonal,and so the angle at A in tri-

1-8.3 Exercises 1. Prove Theorem 18.3.2(i).



4. Prove Theorem 18.3.2(iv)and (v).

InExercises5through14,letA:(-4,-2,4r,8:<2,7,-lt,C:<5,-3,0),andD: 5. FindA'(B+C) 8. FindA'D-B'C

6.FindA'B+A'C 9. Find (B ' D)A - (D ' A)B

<5,4,-3>. 7. Find (A ' B) (C . D )

10. Find (2A + 3B) . ( 4 C - D )

11. Find the cosine of the measureof the angle between A and B. L2. Find the cosine of the measureof the angle between C and D. 13. Find (a) the component of C in the direction of A and (b) the vector proiection of C onto A.

18.4 PLANES

14. Find (a) the component of B in the direction of D and (b) the vector projection of B onto D. 15. Prove Theorem 18.3.5.

1 5 . Prove Theorem 18.3.4.

17. P r o v e b y u s i n g v e c t o r s t h a t t h e p o i n t s( 2 , 2 , 2 ) , ( 2 , 0 , 7 ) , ( 4 , 1 , - l ) , a n d

( 4 , 3 , 0 ) a r e t h e v e r t i c e s o fa r e c t a n g l e .

18. Proveby using vectorsthat the points (2,2,2), (0,1,2), (-1,3,3), and (3,0, L) are the verticesof aparallelogram. to the line through the points (3,-2,2) and (-9,-5,6). 19. Find the distancefrom the point (2, -t,-4) 20. Find the distancefrom the point (3,2,1) to the line through the points (1,2,9) and (-3, -5,-3). 2 L . Find the areaof the triangle having vertices at (-2, 3, l) , (l, 2, 3) ,,md (3, -l, 2). 22. If a force has the vector representation F : 5i - 3k, find the work done by the force in moving an object from the point P J4, | ,3) along a straight line to the point Pr(-S , 6 , 2) . The magnitude of the force is measured in pounds and distance is measured in feet. (nrxr: Review Sec. 17.3.) 23. A force is represented by the vector F, has a magnitude of 10 lb, and direction cosines of F are cos d : tV6 and cos B : +\/6. lf the force moves an object from the origin along a straight line to the point (7, -4,2) , find. the work done. Distance is measured in feet. (Seehint for Exercise22.) 24.lfAandBarenonzerovectors,provethatthevectorA-cBisorthogonaltoBifc:A'B/lBl'z. 9i - 5k and B:4i + 3i - 5k, use the result of Exercise 24 to find the value of the scalarcso that the vec25. If A :l2iI tor B cA is orthogond to A. 26. For the vectors of Exercise 25, use the result of Exercise24 to find the value of the scalard so that the vector A dB is orthogonal to B. 27. Prcve that if A and B are any vectors, then the vectors lBlA + lAlB and lBlA - lAlB are orthogonal. 28. Prove that if A and B are any nonzero vectors and C: measure as the angle between B and C.

18.4 PLANES '

1g.4.1 Definition

lBlA + lAlB, then the angle between A and C has the same

The graph of an equation in two variables, r and !, is a curve in the xy plane. The simplest kind of curve in two-dimensional space is a straight line, and the general equation of a straight line is of the form Ax * By * C:0, which is an equation of the first degree. In three-dimensional space,the graph of an equation in three variables, x, y, and z, is a surface. The simplest kind of surface is aplane, and we shall seethat an equation of a plane is an equation of the first degree in three variables. If N is a given nonzero vector and Pe is a given point, then the set of all points P for which V(P'P) and N are orthogonal is defined to be a plane through Ps having N as a normal aector. Figure 18.4.1 shows a portion of a plane through the point Po(ro, Ao, zo) and the representation of the normal vector N having its initid point at Ps. In plane analytic geometry we c€rnobtain an equation of a line if we are given a point on the line and its direction (slope). In an analogous manner, in solid analytic geometry an equation of a plane can be deter-

830


mined by knowing a point in the plane and the direction of a normal vector. L8.4.2Theorem

If.Ps(xs,yo, zo)isapoint in a plane and a normalvector to the plane isN : (a, b, c) , then an equation of the plane is (1)

+y

pRooF: Refer to Fig. '1,8.4.'1,. Let P(x, !, z) be any point in the plane. v(4?) is the vector havin g P;P as a representation,and so V(P;P) - (x - xo,a - Uo,z- zo) (2) From Detinitions 1,8.4.1 and 18.3.7,it follows that V ( P g P ). N : 0 BecauseN : (a, b, c>, from (2) and the above equation we obtain a ( x - x s )+ a Q - A ) * c ( z - z o ): 0 which is the desired equation.

F i g u r e1 8 . 4 . 1

ExAMPLEL: Find an equation of the plane containing the point (2, 1, 3)and having 3i - 4i + k as a normal vector.

(3) I

sot,urroN: Using (1) with the point (xo, Uo,zo)- (2,I, 3) and the vector (a, b, cl : (3,-4,']-.>,we have as an equation of the required plane

3(x-2)-4(v-1)+

(z- 3):0

or/ equivalently, 3x-4y*z-5:0 L8.4.3 Theorem

If. a, b, and c are not all zero, the graph of an equation of the form ax*by * cz* d:0

(4)

is a plane and (a, b, cl is a normal vector to the plane. PRooF: Supposethatb * 0. Then the point (0,-dlb,0) is on the graph of the equation becauseits coordinates satisfy the equation. The given equation can be written as /s\ a(x- o)+ b \a + t) + c(z- o): 6 which from Theorem 18.4.2is an equation of a plane through the point (0, -dlb,0) and for which (a, b, cl is a normal vector.This provesthe theorem if b # 0. A similar argument holds if b:0 and either a * 0 or c#0. r Equations ( 1) and (4) are calledcartesianequationsof a plane.Equa-

18.4 PLANES

831

tion (1) is analogousto the point-slope form of an equation of a line in two dimensions. Equation (4) is the general first-degree equation in three variables and it is called a linear equation. A plane is determined by three noncollinear points, by a line and a point not on the line, by two intersecting lines, or by two parallel lines. To draw a sketch of a plane from its equation, it is convenient to find the points at which the plane intersects each of the coordinate axes. The r coordinate of the point at which the plane intersectsthe r axis is called the x interceptof the plane; the y coordinate of the point at which the plane intersectsthe y axis is called the y interceptof the plane; and the z interceptof the plane is the z coordinate of the point at which the plane intersectsthe z axis. . rLLUsrRArroN 1: We wish to draw a sketch of the plane having the equation 2 x- l 4 Y * 3 z : 8 Figure 18.4.2

By substituting zero for y and z, we obtain .x: 4; so the r intercept of the plane is 4. In a similar manner we obtain the y intercept and the z intercept, which are 2 and $, respectively. Plotting the points corresponding to theseinterceptsand connectingthem with lines, we have the sketch of the plane shown in Fig. 18.4.2.Note that only a portion of the . plane is shown in the figure. I rLLUSrRArrox2: To draw a sketch of the plane having the equation 3x*2Y-62:0

F i g u r e1 8 . 4 . 3

Figure 18.4.4

we first notice that because the equation is satisfied when r, y, and z are all zero, the plane intersects each of the axes at the origin. If we set r: 0 in the given equation, we obtain y - 32: 0, which is a line in the yz plane; this is the line of intersection of the yz plane with the given plane. Similarly, the line of intersection of the xz plane with the given Drawing a plane is obtained by setting A:0, and we get r-22:0. sketch of each of these two lines and drawing a line segment from a point on one of the lines to a point on the other line, we obtain Fig. 18.4.3. . In Illustration 2 the line in the yz plane and the line in the xz plane used to draw the sketch of the plane are called the tracesof the given plane in the yz plane and the xz plane, respectively. The equation r : 0 is an equation of the yz plane becausethe point (x, y , z) is in the yz plane if and only if r : 0. Similarly, the equations y : 0 and z:0 areequationsof the xz plane and the xy plane, respectively. A plane parallel to the yz plane has an equation of the for:n x: k, where k is a constant. Figure 18.4.4shows a sketch of the plane having the equation x: 3. A plane parallel to the xzplarrehas an equation of the form A : k, and a plane parallel to the xy plane has an equation of the form

832


z: k. Figures 1.8.4.5and 18.4.6show sketchesof the planes having the equationsy: -5 and z- 6, respectively.

F i g u r e1 8 . 4 . 5

18'4'4 Definition

F i g u r e1 8 . 4 . 6

The anglebetweentwo planesis defined to be the angle between the nonhal

vectors of the two planes.

EXAMPTB 2: Find the radian measure of the angle between the two planes5x- 2y * 5z- l2:0 and 2x*y-72* 11:0.

solurroN: Let N1 be a normal vector to the first plane and N1 : Si-2i * 5k. Let N, be a normal vector to the second plane and N2 : 2i + - 7k. i By Definition 78.4.4the angle between the two planes is the angle between N1 and N2, and so by Theorem 18.3.4if d is the radian measureof this angle, c o so -

N' 'N' :

lN,llN,l ffiffi

-27

1

542

Therefore,

o- &n 18'4'5 Definition

Two planes are parallel if and only if their normal vectors are parallel. From Definitions 18.4.5 and 18.3.5, it follows that if we have two planes with equations arx*bry+cfi+dt:0

nzx+ bzA* c2z* dr- 0

18.4 PLANES

and normal vectorsNr : (4r, bt, ct) and Nz: (az, br, crl, respectively,then the two planes are parallel if and only if N1 : kN2

where k is a constant

Figure 18.4.7shows sketchesof two parallel planes and representations of some of their normal vectors.

L8.4.6 Definition

Two planes are perpendicularif and only if their normal vectors are orthogonal. From Definitions L8.4.6and 18.3.7it follows that two planes having normal vectors Nr and N, are perpendicular if and only if N1'N2:0

3: Find an equation of ExAMPLE the plane pe{pendicular to eachof the planesx- y + z - 0 and 2x* Y - 4z- 5 - 0 and containing the point (4, 0 , -2) .

Q)

soLUrIoN: Let Mbe the required plane and (a,b, cl be a normal vectorof M.Let Mr be the plane having equation te- y + z: 0.By Theorem18.4.3a normal vector of.Mr is <1,-l ,1). BecauseM and M, areperpendicular,it follows from Eq. (7) that (a,b,cl ' <1,-7,1):0 or, equivalently, a-b+c:0

(8)

- 4z - 5: Q.A normal Let M2 be the plane having the equation 2x * y -$. BecauseM and M2 are PerPendiculat,we have vector of Mz is (2 , l,

( a , b , c >' ( 2 ,L , - 4 ) : o or, equivalently, 2a*b-4c:0

SPACEAND SOLID ANALYTICGEOMETRY VECTORSIN THREE.DIMENSIONAL

Solving Eqs. (8) and (9) simultaneouslyfor b and c in terms of a, we get b - 2a and c: a. Theref.ote,a normal vectot of M is (a, 2a, a>. Because (4, 0 , -2) is a point in M, it follows from Theorem "1,8.4.2 that an equation ofMis

a ( x- 4 ) + z a ( y- 0 ) * a ( z+ 2 ) or, equivalently, x*2Y*z-2:0 Consider now the plane having the equation ax * by + d: 0 and the xy plane whose equation is z:0. Normal vectors to these planes are (a, b,0) and (0, O, 1.>,respectively. Because(a, b,0) ' (0, 0, 1):0, the two planes are perpendicular. This means that a plane having an equation with no z term is perpendicular to the xy plane. Figure 18.4.8 illustrates this. In a similar manner, we can conclude that a plane having an equation with no r term is perpendicular to the yz plane (see Fig. 1'8.4.9),and a plane having an equation with no y term is perpendicular to the xz plane (see Fig. 18.4.10). An important application of the use of vectors is in finding the undirected distance from a plane to a point. The following example illustrates this.

F i g u r e1 8 . 4 . 8

Figure 18.4.9 EXAMPI^.S4:

Find the distance

from the plane 2x - y + 2z * I0 - 0 to the point (l , 4, 5) .

F i g u r e1 8 . 4 . 1 0

soLUrIoN: Let P be the point ('1,,4,5) and choose any point Q in the plane. For simplicity, choosethe point Q as the point where the plane intersectsthe r axis, that is, the point (-5, 0, 0). The vector having QP as a representation is given by

v(a?):6ir 4i+ 6k A normal vector to the given plane is

N:2i-i+zk

18.4 PLANES P ( 1 ,4 , 6 )

The negative of N is also a normal vector to the given plane and -N: -2i+ i- 2k We are not certain which of the two vectors, N or-N, makes the smaller angle with vector V(QIF). Let N' be the one of the two vectors N or -N

-5r0,0)

which makes an angle of radian measure 0
(10)

coso d: lv(e=-P)l

Becauseg is the radian measute of the angle between N' and V(Q?), we have (11)

Substituting from (1,1)into (10) and replacing lN'l by lNl, we obtain

l v ( d l l ( N '. v t O F l l lNllv(aF)l Becaused is an undirected distance, it is nonnegative; hence, we can replacethe numerator in the above expressionby the absolute value of the dot product of N and v(QF). Therefore,

I w. v r & i l

l ( z i - i + 2k) . (5i + 4i + 6,k)| _ 20 ffi

78.4 Exercises In Exercises1 through 4, find an equation of the plane containing the given point P and having the given vector N as a normal vector.

L . P ( 3, L , 2 ) ; N : ( L ,2 ,- 3 ) 3 . P ( 2 , t ,- 1 ) ; N - - i + 3 i + 4 k

2 . P ( - ! , 8 , 3 ) ;N - ( - 7 , - t , l > 4 . P ( l, 0 , 0 ) ; N - i + k

In Exercises 5 and 6, find an equation of the plane containing the given three points. 6 . ( 0 , o, 2 ) , ( 2 ,4 , L ), ( - 2 , 3 , 3 ) 5. (3, 4, L), (L,7 , L), (-1 , -2, 5)


In Exercises'7 through12, draw a sketch of the given plane and find two unit vectors which are normal to the plane. 7.2x-y+22-5:0 8. 4x-ay-22-9:0 9.4x*3y-l2z-0 1 0 .y + 2 2 - 4 : 0 1 . 1 . . 3 x * 2 2 -5 : 0 L 2 .z : 5 In Exercises13 through 17, find an equation of the plane satisfying the given conditions. 13. Perpendicularto the line through the points (2,2,-4) 14. Parallelto the plane 4x-2y

t z- l:0

and,(7,-1,3)

and containing the point (-5, 1,2).

and containing the point (2,6,-l).

15. Perpendicularto the plane x*3y - z-7:0 16. Perpendiculartoeachof the planesx-y I

and containingthe points (2,0,s) and (0, 2,-l). + z:0 and 2x*y -42- 5:0 and containingthe point (4,0,-2).

17. Perpendicular to the yz plane, containing the point (2,l,l\ plane2x-y+22-3:0.

, and making an angle of radian measure cos-r(i) with the

18. Find the cosineof the measureof the anglebetweenthe planes2x-y-22-

5:0

and5x -2y -l3zt8:0.

19. Find the cosine of the measure of the angle between the planes 3x -l 4y: 0 and 4x - 7y I 4z - 6: 20. Find the distancefrom the plane 2r + 2y - z - 6:0 to the point (2, 2, -4).

0.

21. Find the distance from the plane 5r -l l7y * 2z - 30: 0 to the poinr (-2, 6, 3). 22. Find the perpendicular distance between the parallel planes and 4x-By - z-6:0

k-8y-z*9:0

23. Find the perpendicular distance between the parallel planes 4y-32-6:0

and 8y-62-22:0

24. Prove that the undirected distance from the plane ax -f by t cz * d:0

to the point (16, yo, zs) is given by

laxn-l byn* czs-t dl

\/F 175+7 25. Prove that the perpendicular distance between the two parallelplanesax + by * cz * dr- 0 and ffic+ by + cz + dz:0

26. If a' b, and.c are nonzero and are the r intercept, y intercept, and z intercept, respectively, of a plane, prove that an equation of the plane is -r *, a i *z- : 1 aoc This is called the interceptform of an equation of a plane.

18'5 LINES IN R3 Let L be a line in R3such that it contains a given point Po(ro, Ao, zo)and.is parallel to the representationsof a given vectorR : (a, b, c). iigure 1g.5.1 shows a sketch of line L and the position representation of vector R. Line L is the set of points P (x, y ,z) such that v (p;F) is parallelto the vector R. so

1 8 . 5 L I N E SI N R 3

P is on the line L ( x o ,Y o , z o ) (a, b, c)

and only if there is a nonzero scalar t such that (1)

V(EB): fR BecauseV(fl'p) - (x- xo,A- Ao,z- zolwe obtain from (1) ( x - x o ,A - U o ,z - z o l: t ( a , b , c l from which it follows that x-Xs:ta

Y-Ao:

tb

z- zs: tc

or, equivalently, Q) Letting the parameter t be any real number (i.e., f takes on all values in the interval (-o, fm)), the point P may be any point on the line L. Therefore, Eqs. (2) represent the line L, and we call these equations parametricequationsof the line'

F i g u r e1 8 . 5 . 1

. rLLUsrRArron 1: From Eqs. (2), parametric equations of the line that is parallel to the representationsof the vector R: (11, 8, 10) and that contains the point (8,12,6) arc r:8*11t

y:L2+8t

z:6*10t

Figure 18.5.2shows a sketch of the line and the position representation o ofR. If none of the numbers A,b, or c is zero,we can eliminate f from Eqs. (2) and obtain (11,8, 10)

(3) Po(8, L2, 6

F i g u r e1 8 . 5 . 2

These equations are called symmetricequationsof the line. The vector R: (a, b, c) determines the direction of the line, and the numbers a, b, and c are called directionnumbersof the line. Any vector parallel to R has either the same or the opposite direction as R; hence, such a vector can be used in place of R in the above discussion. Becausethe cornponents of any vectoiparallel to R are proportional to the components of R, *e cur, conclude that any set of three numbers proportional to a, b, arrd c also can serve as a set of direction numbers of the line. So a line has an unlimited number of setsof direction numbers. We write a set of direction numbers of a line in brackets as la,b, cf . . rLLUsrRArroN2: If. 12,3, -4] represents a set of direction numbers of a line, other sets of direction numbers of the same line can be represented o a s 1 4 , 6 , - 8 1 ,[ 1 , t , - 2 ] , a n d l 2 l 1 ,U\f29,-41\/Dl. . ILLusrRArroN 3: A set of direction numbers of the line of Illustration 1

838

V E C T O R SI N T H R E E - D I M E N S I O NSAPLA C EA N D S O L I DA N A L Y T I CG E O M E T R Y

is [11,8, 10], and the line containsthe point (8,12,6). Thus, from (3)we have as symmetric equations of this line x - 8 _ y - 1 2_ z - 6 g 1,1, 10 EXAMPLE1: Find two sets of symmetric equations of the line through the two points (-3, 2, 4) and (5, 1, 2) .

sor.urroN: Let P1 be the point (-3, 2, 4) and P2 be the point (6,'1,,2). Then the required line is parallel to the representations of the vector V (P.,P) , and so the componentsof this vector constitute a set of direction numbers of the line. V(FE): \9,-L,-2>. (-3,2,4), we have from (3) the equations x*3:y-2 9-L-2

Taking Po as the point

_"-4

Another set of symmetric equations of this line is obtained by taking P6as the point (6,'/.,,2),and we have x-6 9

z 2 -2

_y-L -1

Equations (3) are equivalent to the system of three equations b ( x - x s ): a ( y - y )

c ( x - x s ): a ( z - z )

c(y-y)

:b(z-z)

(4)

Actually, the three equations in (a) are not independent becauseany one of them can be derived from the other two. Each of the equations in (4) is an equation of a plane containing the line L represented by Eqs. (3). Any two of these planes have as their intersection the line L; hence, any two of the Eqs. (4) define the line. However, there is an unlimited number of planes which contain a given line and becauseany two of them will determine the line, we conclude that there is an unlimited number of pairs of equations which represent a line. , If one of the numbers a, b, or c is zero, we do not use symmetric equations (3). However, supposefor examplethat b:0 and neither a nor c is zero. Then we can write as equations of the line X-Xo

:TZ-Zo

a F i g u r e1 8 . 5 . 3

ExAMPtn 2: Given the two planes

x+3y-z-9:0

1

anq A:Uo

A line having symmetric equations (5) lies in the plane y: hence is parallel to the xz plane. Figure 18.5.3 shows such a line.

ys and

If we solve the two given equations for x and y Ln terms of z,

SOLUTION:

we obtain y : - , > * -

(5)

k

+2

y-?z+&

1 8 . 5 L I N E SI N R 3

from which we get 2x-3Y*42*3:0

x-2

-1 For the line of intersectionof these two planes, find (a) a set of sym- or, equivalently, metric equations and (b) a set of y-+ x-2 parametric equations. -3 2

z-0

which is a set of symmetric equations of the line. A set of parametric equations can be obtained by setting eachof the above ratios equal to f, and we have

y:&+2t

x:2:3t ExAMPrr 3: Find the direction cosines of a vector whose rePresentations are parallel to the line of Example 2.

4; Find equations of the ExAMPLE line through the Point (L, t , !) , perpendicular to the line

solurroN: From the symmetric equations of the line in Example 2, we see that a set of direction numbers of the line is [-3,2,3]. Therefore, the vector (-g,2,3) is a vector whose representationsare parallel to the line. The direction cosines of this vector are as follows: cos a:-3llD'

cosY : 3l\8.

cosF:21 \8,

solurroN: Let la, b, cl be a set of direction numbers of the required line. The equations 3r : 2y : z can be written as

x-0

y-0

x*Y-z:0

z-0

1 2

\x:Zy-z and parallel to the Plane

z:3t

which ares;rmmetricequationsof a line. A setof directionnumbersof this line is lt,t, tl. Becaus6the required line is perpendicularto this line, it follows that the vectors(a, b, cl and (*, *, 1) are orthogonal'So

( a , b , c > ' ( + ,1+),: 0 or, equivalently,

ta**b+c:0

(5)

Anormal vector to the plane r * y z:0 is <1,1,-tr' Becausethe required line is parallel to thls plane, it is perpendicular to representations of the normal vector. Hence, the vectors (a,b,cl and (L,1',-1) are orthogonal, and so ( a , b , c >' ( L , l , - 1 ) - 0 or, equivalently, a+b-c-0

(7)

Solving Eqs. (6) and (7) simultaneously for a and b in terms of- c we get fl : gc and b - -8c. The required line then has the set of direction


numbers l9c, -8c , cl and contains the point (1, metric equations of the line are x - " 1 ._ y + I %-8cc

1). Therefore,sym-

_r-L

or, equivalently, x-l 9-8

- . t -

ExAMPrn 5: If I, is the line through A(L, 2,7) and B(-2,3, -4) and lrts the line through C (2, -'J,, 4) and D (5 , 7 , -3) , prove that I, and 12 are skew lines (i.e., they do not lie in one plane).

u*l

z-L

sor-urroN: To show that two lines do not lie in one plane we demonstrate that they do not intersect and are not parallel. Parametric equations of a line are x-

xs* ta

y-yo+tb

z- zs* tc

where.[a,b, cf is a setof directionnumbersof theline and (16,Ao, zo)is any point on the line. BecauseVta7l: (-3, l, -!'!,>, a set of direction numbersof It is l-3, 1, -111. TakingA as the point Po,we haveas para-

metric equations of h x-l-3t

y-2+t

z-7-ILt

(8)

Becausev(C-D) - (3,8, -7>, and Iz containsthe point C, we have as parametric equations of lz x-2*3s

y--1

*8s

z-4-7s

(e)

Becausethe setsof direction numbers are not proportional, l, andl2are not parallel. For the lines to intersect, there have to be a value of f and a value of s which give the samepoint (xr, Au zr) in both setsof Eqs. (8) and (9). Therefore, we equate the right sides of the respective equations and obtain l-3t:2*3s 2* t: -1* 7 -llt:4-7s

8s

Solving the first two equations simultaneously, we obtain s: f7 and t: -tl.This set of values does not satisfy the third equation; hence, the two lines do not intersect. Thus, l, andl, are skew lines.

18.5 Exercises In Exercises 1 throu gh 5, find parametric and symmetric equations for the line satisfying the given conditions. L. Through the two points (I,2, t) and (5, - L , l ) .

1 8 . 5 L I N E SI N R 3

841

2. Through the point (5,3,2) with directionnumbers 14,t, -1]. 3 . Through the point (4,-5,20) and perpendicularto the plane x*3y - 5z-8 - 0 . 4. Through the origin and pe{pendicular to the lines having direction numbers 1 4 , 2 , " 1a. n 1 d[ - 3 , - 2 , l f . 5 . Through the origin and perpendicular to the line i@ - L0) - *y - tz at their intersection. 5 . Through the point (2,0,-4)

and parallelto eachof the planes2x* y - z:0

and r * 3y * 5z:0.

7. Show that the lines

ry

- a * = 4: ' = 2 a n d ) c- 2- 35

y+1,4

2-53

z-8 -3

are coincident. 8. Prove that the line r * t:

-t(y - 6) : z lies in the plane 3x * y - z: 3.

The planes through a line which are perpendicular to the coordinate planes are called the projecting planesof the line. In Exercises9 through 72, find equations of the projecting planes of the given line and draw a sketch of the line. 1 0 .x * y - 3 2 * ' 1 . : 0 2x- y -32*'l'4:0

9.3x-2yI5z-30:0 2 x * 3 y - 1 0 2- 6 : 0

72.2x-y-tz-7:0 4x- y *32- 73:0

1 7 .x - 2 y - 3 2 * 6 : 0 x I y * z-l:0

*4, and x:y*7, 13. Find the cosine of the measureof the smallestangle between the two lines r:2yl4,z:-y 2y:y12. 14. Find an equation of the plane containing the point (6,2,4) and the line *(r - 1) : t@ + Z) : i(z- 3)In Exercises15 and 16,find an equation of the plane containing the given intersecting lines.

) c - Z - - 4y: -+ 3

1 L6 v'4-L3

z+2

- (3x+2ylz+2-0 and )c-y+22-L:o t

x u-2 r5.;:T:

x z-"1, and L 1

y-2 -L

z-1 1.

L7. Show that the lines

Fx-Y-z:0 _0 L8x-2y-32*t

and

[x-3y+z+3:0 L3r-y-z+5:0

are parallel and find an equation of the plane determined by these lines. 1 8 . F i n d e q u a t i o n s o ft h e l i n e t h r o u g h t h e p o i n t ( 1 , - 1 , 1 ) , p e r p e n d i c u l a r t o t h e l i n e S x : 2 y : z , a n d . p a r a l l e l t o t h e p l a n e x+y-z:0. 19. Find equationsof thelinethroughthepoint (3,5,4),intersectingthezaxis,andparalleltotheplanex- 3y *52-6:020. Find equations of ttre line through the origin, perpendicular to the line x : A.- 5 , z : 2y - S, u;iintersecting the line !:2xt1'z:x*2. 21,. Find the perpendiculardistancefrom the point (-1,3, -1) to the line x-22:7,Y:1. 22. Find the perpendicular distance from the origin to the line

x:-2-ltt

y:7-it

z:4*trt

23. Prove that the lines

!=:L:Jl: 5-2-or-J2

un4'-2:y+:ry

are skew lines. 24. Find equations of the line through the point (3, -4, -5) which intersects each of the skew lines of Exercise23.

842


Let A and B be two nonparallel vectors. Representations of these two vectors having the same initial point determine a plane as shown in Fig. 18.5.1.We show that a vector whose representations are perpendicular to this plane is given by the vector operation called the "cross product" of the two vectors A and B. The cross product is a vector operation for vectors in Vr that we did not have for vectors in Vr. We first define this operation and then consider its algebraic and geometric properties. F i g u r e1 8 . 6 . 1

L8.6.LDefinition

If A : (ar.,ar, a3) and B : (bt, br, b"l , then the crossproducfof A and B, denoted by A x B, is given by - atbs, fltbz -

arbtl

(1)

Becausethe cross product of two vectors is a vector, the cross product also is called the aectorproduct.The operation of obtaining the crossproduct is called aectormultiplication. . rLLUsrRArroul.: If A : (2, 1,,-3, and B : (J,-1,4), then from Definition 18.6.1we have A X B : (2, 1, -3) X <3,-1., 4l : ( (1) (4) - (-3) (-1), (-3) (3) - (2) (4), (2) (-1) - (1) (3))

: <4-3,-9-8,-2-3) : ( 1 ,_ l T , _ S l :i-17i-5k

o

Thereis a mnenomicdevicefor rememberingformula (1) that makes use of determinantnotation.A second-orderdeterminantis defined bv the equation la

l;

bl

alaa-a'

where a, b, and c are real numbers. For example, I g | ; l-z

Gl ^/F\ . ll: 3 ( 5 ) _ ( G ) ( _ z ) : z z 5

Therefore, formula (1) can

AxB-lZ: f:)iThe right side of the above expression can be written symbolically as

18.6 CROSSPRODUCT

843

which is the notation for a third-order determinant. However, observe that the first row contains vectors and not real numbers as is customary with determinant notation. . rLLUsrRArioN2: We use the mnemonic device employing determinant notation to find the cross product of the vectors of Illustration 1.

-i-L7i-5k

18.5.2 Theorem

If A is any vector LrrVy then

(i)AxA:o (ii) 0 x A:0 (iii) A x 0:0 pRooFor (i): A X A:

(ar, ar, ar), then by Definition 18.6.1'we have

If A: (arnr-

f l s a 2 ta s a r - 4 1 a s ,f , 1 a 2 - A z a r l

: (0,0,0) -0 The proofs of (ii) and (iii) are left as exercises(seeExercise13).

I

By applying Definition 18.6.1to pairs of unit vectors i, i, and k, we obtain the following: iXi:iXi:kxk:0

F i g u r e1 8 . 6 . 2

ixi:k

ixk--i

kxi:i

jxi:-k

kxl:-i

ixk:-i

As an aid in remembering the above cross products, we first notice that the cross product of any one of the unit vectors i, i, or k with itself is the zero vector. The other six cross products can be obtained from Fig. 18.6.2by applying the following rule: The crossproduct of two consecutive vectors, in the clockwise direction,'is the next vector; and the crossproduct of two consecutivevectors, in the counterclockwise direction, is the negative of the next vector. It can be easily seen that cross multiplication of two vectors is not

844

VECTORS IN THREE-DIMENSIONAL SPACE ANDSOLIDANALYTIC GEOMETRY commutativebecausein particulari x I + iX i. However, i X;: Luttd and so ix;:-(ixi). It is true in generalthat iXi:-k; A x B : - (B x A), which we state and prove as a theorem. 18.5.3 Theorem

If A and B are any vectors in Vr,

pRooF: If A : \ar, az,a3)and B : (b,,,bz,bsl,thenbyDefinition 18.6.1we have A X B : (azbs- asbz,f,abr- arbr,nrb2- arb) : -7(asb2- a2bs,atbs- a"b1,arb,- atbz) :_(B

x A)

I

Crossmultiplication of vectors is not associative.This is shown by the followi.g

example:

ix(ixi)-ix

k: -i

(ixi)Xi:ox

i:0

So

ix(ixil+Gxi)xi Cross multiplication of vectors is distributive with reSpectto vector addition, as given by the following theorem. 18.6.4Theorem

If. A,B, and C are any vectors in 7r, then

Ax(B+C):AxB+AxC

(2)

Theorem 18.5.4 can be proved by letting A: (ar, ar, a"), B: (bt, br, bs), and C: (cr, cz, cs), and then showing that the componentsof the vector on the left side of (2) are the same as the components of the vector on the right side of (2). The details are left as an exercise (see Exercise14). L8.5.5 Theorem

If A and B are any two vectors in V, and c is a scalar, then (i) (cA)xB:Ax(cB); (ii) (cA) x B: c(A x B). The proof of Theorem 18.6.5is left as exercises(see Exercises15 and 16). Repeatedapplicationsof Theorems18.6.4and 18.6.5enableus to compute the cross product of two vectors by using laws of algebra, provided we do not change the order of the vectors in crossmultiplication, which is

18.6 CROSSPRODUCT

845

prohibited by Theorem 18.6.3. The following illustration demonstrates this. o rLLUerRArroN3: We find the cross product of the vectors in Illustration 1 by applying Theorems 18.6.4and 18.6.5.

AxB:

(2i+i=3k) x (3i-i+4k) : 5 ( i x i ) - 2 ( i x i ) + 8 ( i x k ) + 3 ( i x i ) - 1 ( ix i ) + 4 ( i x k ) - e ( k x i ) + 3 ( k x i ) - 1 2 ( kx k ) : 6 (o)- 2( k) + 8( - i) + 3( - k) - 1( o) + 4(i) - e(i) + 3(-i) - 12(o) :-2k-8i-3k+4i-9i-gi : i -L7i- Sk

.

The method used in Illustration 3 gives a way of finding the cross product without having to remember formula (1) or to use determinant notation. Actually all the steps shown in the solution need not be included because the various cross products of the unit vectors can be obtained immediately by using Fig.1'8.6'2 and the corresponding rule. 4: We prove that if A and B are any two vectors in v3, then o rLLUSrRArrOw ' l A x B l ' ? : l A l ' z l B l ' ?( A n ; ' LetA:

(a1,a2,a3) and B:

\br,bz,bsl.Tllren

lA x Bl'z: (azbs- a"br)' * (agbr- arb")'* (nrbz- f,rbt)' - Za2arbrbg : azzbsz * albrz - 2a$sb1bg * as2b22 - 2ararb1b,* a22b12 + arzbrz* a12b22 lAl'lBl'

- (A' B\2 : (arz+ a22 + ar')(b" + bzz+ bsz) - (arbr* a2b2* asbs)z = arzbzz + aszbrz * ar2b"2 * a22bt2 * arzb"2 - Zarasbrbs-2arasb2bs2ata2b1b2 + a,zbzz

Comparing the two expressions,we conclude that

- (A ' 3;' lA x Bl'z: lAl'zlBl'z

o

The formula proved in Illustration 4 is useful to us in proving the following theorem, irom which we can obtain a geometric interpretation of the cross product. 18.6.6 Theorem

If A and B are two vectors in v, and g is the radian measure of the angle between A and B, then

ln x sl: lAllBlsino

(3)

GEOMETRY SPACEAND SOLIDANALYTIC VECTORSIN THREE-DIMENSIONAL PRooF: From Illustration 4 we have

l A x B l ' : l A l ' l B l ' -( A . B ) ' From Theorem and B, we have

(4)

'1,8.3.4, if 0 is the radian measure of the angle between A

A.B-lAllBl cos0

(s)

Substitutingfrom (5) into (4) we get l a x B l , : l A l , l B l -, l A l , l B l ,c o soz : l A l r l B l t ( -1 c o s z0 ) So l A x B l r : l A l r l B l 2s i n , 0 Because0 < 0 < z, sin 0 > 0. Therefore, from Eq. (5), we get

I n x n l : l A l l Bsl i n0

r

We considernow a geometricinterpretationof lA x Bl. Let p? be a representationof A and let p? f" a representationof B. Then the angle betweenthe vectorsA and B is the angle at P in triangleRPQ (seeFig. 18.6.3).Let the radianmeasureof this anglebe 0.Therefore,the numberof squareunits in the areaof the parallelogram having pT anaFO ur adjacent sidesis lAllnl sin 0 becausethe altitude of the parallelogramhas iength lBl sin 0 units and the lengthof the baseis lAl units. Sofrom Eq. (g) it follows that lA x Bl squareunits is the areaof this parallelogram.

F i g u r e1 8 . 6 . 3

EXAMrLE 1: Show that the quadrilateral having vertices at

P(1, -2, 3), Q(4,3, -r) , R(2,2, 1,),andS(5, 7,-3) is a

parallelogram and find its area.

solurroN:

Figure 18.6.4shows the quadrilateral peSR.

v ( P ? ): ( 4 - ' t , s - ( - 2 ) , ( - t ) - 3 ) : ( s , s , - 4 ; v ( F f t :) ( 2 - ' ! . , 2 - ( - 2 ), t - 3 ): ( t , 4 , - 2 > v ( n - 3 ) :( 5 - 2 , 7- 2 , - g - 1 ) : ( 3 ,s , - 4 1 v(0-) : (5- 4,7- s,-g - (-1)) : (t, 4, -2l : V(n3) andV(P?) : V(Q*S),itfollows thatPQisparalBecauseV(FQ) lel to R3 and P? is parallel to Q3. Therefore,PQSRis a parallelogram. Let A: v(pT) and B: V(P!), then A x B : (i + 4i- 2k) x (3i + si -4k)

/

b, R(2, z, !)

(4, 3, - L)

Figure18.6.4

(6)

S(s,7, - g)

:3(i x i) +5(i x -4(i x k) +72(ix i) +20(ix i) i) - t i ( i x k ) - 5 ( k x i ) - 1 0 ( kx + 8 ( k x k ) i) : 3(0) + s(k) - 4(-i) + 12(-k) + 20(0)- 16(i) - 5(i) - 10(-i) + 8(0)

18.6 CROSSPRODUCT

847

2i-7k Hence,

lexBl:m-\@ The area of the parallelogram is therefore \@

square units.

The following theorem, which gives a method for determining if two vectors in V3 are parallel, follows from Theorem 18.6.5. 1g.6.7Theorem If A and B aretwovectorsinVg,AandB areparallelifand onlyif AxB:0' pRooF: If either A or B is the zero vector, then from Theorem 1'8.6.2, A x B:0. Becausethe zero vector is parallel to any vector, the theorem holds. If neither A nor B is the zero vector, lAl + O and lBl # 0. Therefore, from Eq. (3), lA x Bl :0 if and only if sin 0:0. BecauselA x Bl : 0 if and only if A X B : 0 and sin 0 : 0 (0' 0 = r) if and only if 0 : 0 ot r, we can conclude that AxB:0

if andonlyif 0:0otr-

However, from Theorem 18.3.6, two nonzero vectors are parallel if and only if the radian measure of the angle between the two vectors is 0 or zr. I Thus, the theorem follows. The product A ' (B x C) is called the triple scalar product of the vectors A, B, and C. Actually, the Parenthesesare not needed because A ' B i s a s c a l a r / a n d therefore A ' B X C can be interpreted only in one way. 18.6.8 Theorem

If A, B, and C are vectors in Vs, then (7) Theorem 18.6.8 can be proved by letting A: (at, az, ag), : (br, br, b"l , and C : (cr, c2, ca) and then by showing that the number B on the left side of (7) is the same as the number on the right. The details are left as an exercise (see Exercise 17). -1, 21, B: 11rLLUSrRArroN 5: We verify Theorem 18.6'8 if A: (1,

( 3 , 4 , 2 > ,a n dC - ( - 5 , 1 , - 4 > . B x c- (3i+ 4i- 2k) x (-5i +i-4k)

- 3k - 12(-i) - zo(-k) - L6i+ loi - 2(-i) 1,4i+22i+23k

IN THREE-DIMENSIONAL VECTORS SPACEAND SOLIDANALYTIC GEOMETRY

A . ( Bx C ): ( L , - 1 , 2 , . ( - L 4 , 2 2 , 2 3 > : - L 4 - 2 2 + M :L0 A x B - ( i - i + 2 k ) x ( 3 i+ 4 i - 2 k ) :4k-2(-i)-3(-k) +2i+6i +8 (-i) :-6i+8i +7k (A x B) . C: (-5,8,7>. (-5,1,-4> :30+8-28 :L0 This verifies the theoremfor thesethreevectors. L8.5.9 Theorem

If A and B are two vectors in 7r, then the vector A x B is orthogonal to both A and B. PRooF: From Theorem 18.6.8we have

A.AXB_AXA.B From Theorem18.5.2(i), AxA:0. we have

Therefore, from the above equation

A'AxB:0.8:0 Becausethe dot product of A and A x B is zero, it follows from Definition 18.3.7that A and A x B are orthogonal. We also have from Theorem 18.6.8that

AxB.B:A.BxB Again applytt g Theorem 1,8.6.2(i),we get B x B : 0, and so from the above equation we have AxB.B:A.0:0 Therefore, since the dot product of A X B and B is zero, Ax B and B are orthogonal and the theorem is proved. I From Theorem 18.5.9we can conclude that if representations of the vectors A, B, and A X B have the same initial point, then the representation of A X B is perpendicular to the plane formed by the representations of A and B. EXAMPTu 2: Given the points P(-1, -2, -3) , Q(-2,1,0), and R(0, 5, L), find a unit vector

solurroN: Let A - V(p0) and g - v(p?). Then A - (-2- (-1),1- (-2), 0 - (-s)) : (-1 ,3,3>

18.6 CROSSPRODUCT

whose representations are Perpendicular to the plane through the points P, Q,,and R.

B- (0- (-1),5- (-2),r- (-3)): (L,7,4> p0 and Pl, The plane through P, Q, and'R is the plane formed by any Therefore, and B. A vectors which are, respectively, representationsof plane. to this x perpendicular representation of the vector A B is

A x B : (-i + 3i + 3k) x ( i+ 7i + 4k): - 9i+7i- 10k The desired vector is a unit vector parallel to A X B. To find this unit vector we apply Theorem 18.2.9and divide A x B by lA x Bl, and we obtain 7 9_rAXB _ ri__10

lffiI:-n5-r+

ExAMPln3: Find an equation of the plane through the Points P ( 1 ,3 , 2 ) , Q ( 3 , - 2 , 2 ) , a n d R(2,L,3).

1\/m-t-16-x

solurroN: v(6) : -i * 3i + k and v(PR) : i- ?i + k. -A normal vectorto the requiredplaneis the crossproductVtp?) x V(p?)' which is (-i+ 3i + k) x (i- 2i+ k) : si + 2i - k Soif Po: (1,3,2) and N: (5, 2,-ll, equation of the required Plane

from Theorem18'4'2we have as an

5 (r- 1 ) + z( V- 3) - ( z- 2) : 0 or, equivalently, 5x*2Y-z-9:0 A geometric interpretation of the triple scalar product is obtained by consideringa parallelepipedhaving edgesPn,Frt.,and fi, and letting A: V(P?), B:V(PR), and C:V(PS). See-Fig'18'6'5'The vectorA X B is pA. fne vector-(A X B) is also a a normal vector to the plane of p0 and normal vector to this plane. we are not certain which of the two vectors,

PA F i g u r e1 8 . 6 . 5

(AxB)or_(AxB),makesthesmalleranglewithC.LetNbetheoneof radian ti\e two vectors (A x B) or (A x B) that makes an angle of their having C N and of representations the Then < with C. measure0 tn FR as shown F0 plane of of the initial points at P are on the same side "",{ in Fig.'18.5.5.The area of the base of the paralletepiped i9 lA X Bl square unitsl If h units is the length of the altitude of the parallelepiped, and if V cubic units is the volume of the parallelepiped,

v-lAxBlh

(8)

"l'8.3.4,N . C _ Considernow the dot product N ' c. By Theorem B e c a u sNe is either l N l l g l c o s0 . B u th - l C lc o s0 , a n ds oN ' C : _ l N l h . -(A Hence,we have x nl. x B), it followsthat lNl lR. tA r b) ot (e) N.e:lAxBlh

VECTORSIN THREE.DIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY

Comparing Eqs. (S) and (9) we have N.C:V It follows that the measure of the volume of the parallelepiped is either (A x B) ' c or - (A x B) ' c; that is, the measure of the volume of the parallelepiped is the absolute value of the triple scalarproduct A X B . c. ExAMPLE4: Find the volume of the parallelepiped having vertices P ( 5, 4 , 5 ) , Q ( 4 , 1 , 0 ,6 ) , R ( ' ! . 9 , ,7) , a n d S ( 2 , 6 , 9 ) a n d e d g e s& , , P ? ,

solurroN: Figure 18.6.6shows the parallelepiped.Let A: V(pQ) : ( - 1 , 6 , 1 ) , B : V t p ? ) : ( - 4 , 4 , 2 ) , a n dC, : V ( p 3 ): ( - 3 , 2 , 4 ) . T h e n A x B : (-i + 6i + k) x (-4i + 4i +2k) : 8i- 2i+ 20k

andu;

Therefore, ( A x B ) . C : ( 8 ,- 2 , 2 0 > . ( - 9 , 2 , 4 l : - 2 4 -

4 * g0:52

Thus,the volume is 52 cubic units. s (2,6,9)

R (1,9,7)

Q @ , T O6, ) P (5, 4, 5)

F i g u r e1 8 . 6 . 6

EXAMPLE5: Find the distance between the two skew lines, /, and 12,of Example 5 in Sec. 18.5. N

solurroN: BecauseIt and /, are skew lines, there are parallel planes p, and Pt containing the lines l, and 12,respectively.see Fig.1,g.6.7. Let d

units be the distancebetweenplanesp, and pr. th" distaice between/, andl, is also d units. A normarvectorto the two planesis N: vterti x v(cD). Let u be a unit normarvector in the direction of N. Then

x v(c-i; u: v(a-B) x v(6) | lv(a-B) Now we taketwo points,onein eachplane(e.g.,BandC). Thenthe scalar projectionof V(CT) on N is V(CTI . U, and

d: lv(c?)- ul: Ittcll . v(AE)x v(cB)| I

lv(AB)x v(cD) | I

Performing the computations required, we have

v(rt):-3i+i-llk F i g u r e1 8 . 6 . 7

v(cB)-3i+Bi-rk

18.6 CROSS PRODUCT

851

kl -rrl: zr(3i- 2i- k) -7 1

-?i-zi-k

u-

\n,

Finally,vtdBl : -4i + 4i- 8k, and so

tI|: # d-lv(c?)'

l-D- 8+8|: h:

+ ffi

1'8.6 Exercises

- 1 ) , C : ( - 5 , - 3 , 5 ) , p - ( - 2 , L , 5 ) , E- < 4 ' O ' - 7 > ' a n d F<-0 ' 2 ' 1 ) ' 1.throu ghl2,let A - (1,2'3)' B - (4'-3' In Exercises 2.FindDxE L.FindAxB 4.Find(CxE)'(DxF) 3.Find(CxD)'(ExF) 5. Verify Theorem18.6.3for vectorsA and B.

6. Verify Theorem 18.6.4 for vectors A, B, and C. 7. Verify Theorem 18.6.5(i) for vectors A and B and c:3' c :3' 8. Verify Theorem 18.6.5(ii) for vectors A and B and 9. Verify Theorem 18.6.8 for vectors A, B, and c.

10.Find (A x B) x C andA x (B x C)' verify theY are equal. L 1 .F i n d( A + B ) x ( c - D ) a n d( D - c ) x ( A + B ) and 12.Find lA x Bllc x Dl. L3. Prove Theorem18.5.2(ii)and (iii)'

14. Prove Theorem L8.6.4.

15. Prove Theorem18.6'5(i)'


17. Prove Theorem 18.6.8. 18. Given the two unit vectors

A-6i+6i -+k and B--&i+3i ++k

ways: (a) by using the crossproduct (formula If 0 is the radian measure of the angle between A and B, find sin 0 in two identity. trigonometric (3) of this section); (b) by using the dot product and a 1g. Follow the instructions of Exercise 18 for the two unit vectofs:

a- 1:1 1u 5t - V r t * + k

1R1

a n dn : 5 f t t * 5 f t i + 3 v g k

is a parallelogram and (1,1 ,3), (-2,'l',-1), (-5 ,4,0), and (-8,4,-4) zo. show that the quadrilateralhaving verticesat find its area. and (4,3, -1) , (2,2,L), and (5,7,*3) is a parallelogram 21. show that the quadrilateralhaving verticesat (1,-2,3) ,

find its area.

8s2

VECTORSIN THREE-DIMENSIONAL SPACEAND SOLID ANALYTICGEOMETRY +

-+

22. Find the areaof the parallelogramPQRSif V(pe) : 3i - 2i and V(pS) - 3i + 4k. 23. Find the areaof the triangle having verticesat (0, 2, z), (g, g, -2) and (g 1 2 ,6 ) . , , 24. Find the area of the triangle having verticesat (4, 5 , 6) , (4, 4, 5) and (3 5 5 ) . , , , 25' Let OP be the position representation of vector A , dQ b" the position representation of vector B, and o-i-- be - - the ---- Iposition representation of vector c. prove that the area of triangle pen is (b - A) x (c - A) d l. 26. Find a unit vector whose representations are Perpendicular to the plane containi.g pQ urrdFR if P0 is a representation of the vector i + 3i Zk and FR i, u representationof the vector 2i - - k.

i 2 7 ' G i v e nt h e p o i n t s P ( 5 , 2 , - D , Q e , 4 , - 2 ) , a n d R ( 1 1 , 1 , 4 ) . F i n d a u n i t v e c t o r w h o s e r e p r e s e n t a t i o n s a r e p e r p e n d i c u l a r to the plane through points p, e, and R. 28. Find the volume of the parallelepiped having edgesPi, Fi, and E (1,3,4), (3,5,3), (2,1,6), and (2,2, 5). 29. Find the volume of the parallelepiped peRS if the vectorsv(P?), 2i+ i - k, and i-2i + k.

ir the points p, e, R, and s are,respectively,

v(p?), and v(pB) are,respectively,i + 3i + zk,

30. If A and B are any two vectors rnvs, prove that (A - B ) x ( A + B ) : 2(A x B). In Exercises31 and 32, use the crossproduct to find an equation of the plane containing the given three points. 3 1 . ( - 2 , 2 , 2 ) , ( - g , l , 6 ), ( 3 , 4 , _ L ) 3 2 . ( a ,b , 0 ) , ( a , 0 , c ) , ( 0 ,b , c ) In Exercises33 and i4, find the perpendicular distance between the two given skew lines. x-"1,_y-2_z+1 x * 2 y + l z _ l a ? ^-J z-l rr'5:3:2andi:T:== .A xt'/., y+Z Jt.::-:-

and

x-l _y-t 532

_z*t

35. Let P, Q, and R be three noncollinear points in R3 and dP' oa, and ol be the position representations of vectors A, B, and C, respectively. Prove that the representations of the vector A x B + B x c + c x A are perpendicular to the plane containing the points p , e, and R.

I8.7 CYLINDERS AND SURFACESOF REVOLUTION

18.7.L Definition

As mentioned previously, the graph of an equation in three variabres is a surface.A sudace is representedby an equation if the coordinates of every point on the surface satisfy the equatiorrand if every point whose coordinates satisfy the equation lies on the surface. we have already discussed two kinds of surfaces,a plane and a sphere.Another kind of surface that is fairly-simple is a cylinder. you are probably familiar with right-circular cyfinfels from previous experiet W" now consid", u -6r" general "e. rylindrical surface., A cylin.deris a surface that is generated by a line moving along a given plane curve in such away that it always remains parallel to"a fixea tne not lying in the plane of the given curve. The moving line is called generator a of the rylinder and the given plane curve is called a directix of the der. Any position of a generator is called a ruling of the cylinder. rylinwe confine ourselves to cylinders having a directrix in a coordinate

18.7 CYLINDERSAND SURFACESOF REVOLUTION 853

F i g u r e1 8 . 7 . 1

plane and rulings perpendicular to that plane. If the rulings of a cylinder are perpendicular to the plane of a directrix, the cylinder is said to be perpendicular to the plane. The familiar right-circular cylinder is one for which a directrix is a circle in a plane pe{Pendicular to the cylinder. 't8.7.'1,, we show a cylinder whose directrix is . rLLUSrRArroN1: In Fig. in the xy plane and whose rulings are parallel to the parabola y':8r the i axis. This cylinder is called a paraboliccylinder.An elliptic cylinder its directrix is the ellipsegxz*16y2:144 i4 the is shown in Fig. 1'8.7.2; ry plane and its rulings are parallel to the z axis. Figure 18.7.3shows a hyp-erboliccylinder having as a directrix the hyperbola 25x2 4A2:700 in the xy plane and rulings Parallel to the z axLS.

Figure 18.7.2

Let us consider the problem of finding an equation of a cylinder having a directrix in a coordinate plane and rulings parallel to the coordinate a-xisnot in that plane. To be specific, we take the directrix in the ry plane and the rulings paraltel to the z axis. Refer to Fig.1'8.7.4' Suppose that an equation of the directrix in the xy plane is y : f (x). If the point (xo,yo,0) in the xy plane satisfiesthis equatibn, any point (ra, Ao,.z)-in where z is any real number, will satisfy the'sdme iftt"!-ai*""sionaispace, equation because z does not appear in the equatiqn. The points having representations (ro, ao,z) all lie on the line parallel to the z axis through ttre point (xo,yo,0). This line is a ruling of the cylinder' Hence, any point whose x and y coordinates satiqf the equation A : f @) lies on t_herylinder. Conversely, if the point P(x, y, z) lies on the cylinder (see Fig' 18.7.5), then the point (r, y, 0) lies on the directrix of the cylinder in the xy plane, and hence the r and y coordinates of P satisfy the equation y: f(x). Therefore, ity: f(x) is consideredas an equation of a $raph in

Figure 18.7.3

(xo, yo, O)

Figure 18.7.4

F i g u r e1 8 . 7 . 5


three-dimensional space,the graph is a cylinder whose rulings are parallel to the z axis and which has as a directrix the curve y: f(x) in the plane z:0. A similar discussionpertains when the directrix is in either of the other coordinate planes. The results are summarized in the following theorem. 18.7.2 Theorem

In three-dimensional space, the graph of an equation in two of the three variables x, y, and z is a cylinder whose rulings are parallel to the axis associated with the missing variable and whose directrix is a curve in the plane associatedwith the two variables appearing in the equation. , . rLLUsrRAtroN2: It follows from Theorem 18.7.2that an equation of the paraboliccylinder of Fig. 18.7.1is A" :8x, consideredas an equation in RB. Similarly, equations of the elliptic cylinder of Fig. 18.7.2and the hyperbolic cylinder of Fig. 1.8.7.3are,respectively, 9x?* I6y2 : 1,44and.25xz4A2:100, both consideredas equationsin RB. o A crosssectionof a surface in a plane is the set of all points of the surface which lie in the given.plane. If a plane is parallel to the plane of the directrix of a cylinder, the cross section of the cylinder is the iame as the directrix. For example, the cross section of the elliptic cylinder of Fig. 18.7.2Lnany plane parallel to the xy plane is an ellipse.

EXAMPLEL: Draw a sketch of the graph of each of the followirg equations: (a) y : In z; (b) z2: )c3.

solurroN: (a) The graph is a cylinder whose directrix in the yz plane is the curve a : ln z and whose rulings are parallel to the r axis. A sketch of the graph is shown in Fig. 1,8.7.6. (b) The graph is a cylinder whose directrix is in the rz plane and whose rulings are parallel to the y axis.An equation of the directrix is the crrrve22: rf in the xz plane. A sketch of the graph is shown in Fig. 1,9.7 .7.

F i g u r e1 8 . 7 . 6

F i g ur e 1 8 . 7 . 7

18.7 CYLINDERSAND SURFACESOF REVOLUTION 855

18.7.3 Definition

F i g u r e1 8 . 7 . 8

If a plane cuwe is revolved about a fixed line lying in the plane of the the sur{acegenerated is called a surfaceof reoolution The fixed line "..r*^", is called the axisof the surface of revolution, and the plane curve is called the generatingcurae. Figure 18.7.8shows a surface of revolution whose generating cunre is the curve c in the yz plane and whose axis is the z axis. A sphere is a particular example of a surface of revolution because a sphere can be generated by revolving a semicircle about a diameter. o rLLUsrRArroN3: Figure 18.7.9shows a sphere which can be generated by revolving the semicfucley2*22:12, 22 0, about the y axis' Anof a surface of revolution is a right-circular cylinder for oih", "*"-ple which the generating curve and the axis are parallel straight lines. If the generating curve 1s the line z: k in the xz plane and the axis is the ' we obtain the right-circular cylinder shown in Fig' 1'8'7'10' r "*ir, we now find an equation of the surfacegenerated by revolving about the y axis the curve in the yzplane having the two-dimensional equation (1) z: f(y) y , z) be any point on the surfaceof revoReferto Fig. 1,8.7.1,1..LetP(x, to the y axis. Denote the perpendicular plane lution. Through P, pass a y axis by Q(0, y,0), and let the with plane point of interiection of this with the generating plane the Po(o,a, z) be the point of intersection of plane through P is the B".urrse the cross section of the surfacewith "rr*". a circle, P is on the surface if and only if (2)

l@12-lOP,l' x F i g u r e1 8 . 7 . 9

Becausel@l : \m

and lOPol: zo,w€ obtain from (2) (3)

x2+22:zo2

The point Po is on the generating cun/e, and so its coordinatesmust satisfy Eq. (1). Therefore, we have (4)

zo: f Q)

P(*,u, z)

P o ( 0 ,U , z o )

Q(o,y,o) F i g u r e1 8 . 7 . 1 0

Y

F i g u r e1 8 . 7 .11

856


From Eqs. (3) and (4), we concludethat the point p is on the surfaceof revolution if and onlv if x 2+ 2 2: l f Q ) 1 , (s) Equation (5) is the desired equation of the surface of revolution. Because (5) is equivalent to

_+t/ptry' :

fe)

we can obtain (5) by replacing z rn (1) In a similar manner we can show that if the curve in the yz plane having the two-dimensional equation

y: s(z)

(6)

is revolved about the z axis, an equation of the surface of revolution generated is obtained by replacing y in (6) by *I/FTT. Analogous remarks hold when a curve in any coordinate plane is revolved about either one of the coordinate axes in that plane. In summary, the graphs of any of the following equations are surfacesof revolution having the indicated axis: ,'+ y' : lp(z)fz-z axis;f + z2 : lF(y)1"-y axis; y'+ z': [f(x)],-r axis. In each case,cross sections of the surface in planes perpendicular to the axis are circles having centers on the axis. ExAMPLE2: Find an equation of the surface of revolution generated by revolving the parabola y' : 4x in the xy plane about the r axis. Draw a sketch of the graph of the surface.

soLUrIoN: In the equation of the parabola, we replace y by tWTz and obtain y2 + z2: 4r A sketch of the graph is shown in Fig. 18.7.1.2. Note that the same surface is generated if the parabola 22:4r in the xz plane is revolved about the r axis.

rrrlrr,

F i g ur e 1 8 . 7 .12

The surfaceobtained in Example2 is calledaparaboloidof reaolution.If an ellipse is revolved about one of its axes, the surface obtained is called an ellipsoidof reaolution.A hyperboloidof reaolutionis obtained when a hyperbola is revolved about an axis.

18.7 CYLINDERS AND SURFACESOF REVOLUTION 857 EXAMPLE surfac e x2

Draw a sketch of the 2 2- 4 y ' : 0 , i f y > 0 .

solurroN: The given equation is of the formrP -f 22: lF(y)lr,and so its graph is a surface of revolution having the y axis as axis. If we solve the given equation for y, we obtain

2Y:!\/7+V Hence, the generating curve can be either the straight line 2y: r in the xy plane or the straight line2y : z in the yz plane. By drawing sketchesof the two possible generating curves and using the fact that cross sectionsof the surface in planes perpendicular to the y axis are circles having centers on the y axis, we obtain the surfaceshown in Fig. 1,B..Z.lg (note that because y > 0 we have only one nappe of the cone).

F i g u r e1 8 . 7 . 1 3

The surface obtained in Example 3 is called a right-circular cone.

Exercises 18.7 In Exercises1 through 8, draw a sketch of the cylinder having the given equation. t. 4x2* 9y' - 36

2. x:2-22:4

3.y:lzl

4. z-siny

5. z: 2x2

6. x2:A3

7.y:coshr

8. z2-4y2

In Exercises 9 througt:.L4, find an equation of the surface of revolution generated by revolving the given plane curue about the indicated axis. Draw a sketch of the surface. 9. x2 : 4y tn the xy plane, about the y axis. 10. x2 : 4y in the xy plane, about the r axis. 11. x2 + 422: L6 in the xz plane, about the x axis. 12. x2 + 422: L6 in the xz plane, about the z axis. 13. y : sin r in the .ry plane, about the x axis. 14. y2:

d in the yz plane, about the z axis.

In Exercises 15 through 18, find a generating curve and the axis for the given surface of revolution. surface. 1'6. y' + z2 - ezr

15.x2+y'-22:4

17.x2+ z' - lyl

19. The tractrix

x_ t-atanhj

a-a

y_-asechl

from x - -a to x - 2a ts revolved about the x axis. Draw a sketch of the surface of revolution.

Draw a sketch of the

18. 4x2* 9y' + 422- 36


18.8 QUADRIC SURFACES The graph of a second-degreeequation in three variables x, y, and z is called a quadric surface.These surfaces correspond to the conics in the plane. The simplest types of quadric surfaces are the parabolic, elliptic, and hyperbolic cylinders, which were discussed in the preceding section. There are six other types of quadric surfaces,which we now consider. We choosethe coordinate axesso the equations are in their simplest form. In our discussion of each of these surfaces,we refer to the cross sections of the surfaces in planes parallel to the coordinate planes. These cross sections help to visualize the surface. >y

The ellipsoid -1

F i g u r e1 8 . 8 . 1

(1)

where a, b, and c are positive (see Fig. 18.8.1). If in Eq. (1) we replacezby zero,we obtain the crosssection of the ellipsoid in the ry plhne, which is the ellipse *laz * yrlbr: 1..To obtain the cross sections of the surfacewith the planes z : k, we replace z by k in the equation of the ellipsoid and get

f *t-,

-kz

a2b2^c2

If lkl < c, the cross section is an ellipse and the lengths of the semiaxes decreaseto zero as lkl increasesto the value c. Iflkl: c, the intersection of a planez : k with the ellipsoid is the single point (0,0, k). If lkl ) c, there is no intersection. we may have a similar discussion if we consider cross sections formed by planes parallel to either of the other coord.inate planes. The numbers a, b, and c are the lengths of the semiaxesof the ellipsoid. If any two of these three numbers are equal, we have an ellipsoid-of revolution, which is also called a spheroid.If we have a spheroid and the third number is greater than the two equal numbers, the spheroid is said to be prolafe. A prolate spheroid is shaped like a foot6afl. An oblate spheroid is obtained if the third number is less than the two equal numbers. If all three numbers a,b, and c in the equation of an ellipsoid are equal, the ellipsoid is a sphere. The elliptic hyperboloidof one sheet fuzzz

a+E

Figure18.8.2

7

e)

\( where a, b, and,c are positive (see Fig. 18.3.2). The cross sections in the planes z: k are ellipses *laz * Azlbz: l + kzle. when k: 0vthe lengths of the semiaxesof tlie ellipse are smallest,

1 8 . 8 Q U A D R I CS U R F A C E S

and theselengths increaseas lkl increases.The cross sections in the planes x : k arehyperbolas y2I b2- zzI cL: L - kz.ln2.If Ilcl < a, the transverseaxis of the hyperbola is parallel to the y axis, and if lkl > o, the transverse axis the hyperbola degenerates into two is parallel to the z axis. If k:a, straight lines: ylb zlc: Oandylb * zlc :0. In an analogousmanner,the cross sections in the planes y : k are also hyperbolas. The axis of this hyperboloid is the z axis. If a : b, the surface is a.hyperboloid of revolution for which the axis is the line containing the conjugate axis. The elliptic hyperboloidof two sheets _ - xz_y'

_ z' _ - 1r

(3)

,r- w-7

where a,b, and c are positive (see Fig. 18.8.3). Replacing z by k in Eq' (3), we obtain x2la2+ A'lb'= Plcz L' If surface;hence, lkl < c, there is no intersection of the plane z = k with the-c and z: c' lf planes z: the there are no points of the surfacebetween is the single surface wlth the k z: lkl: ,, the iirtersection of the. plane in the plane the surface of section point (0, 0, k). When lkl > t, ihe cross as increase ellipse of the semiaxes 2: k is an ellipse and the lengths of the lkl increases. The cross sectionsof the surfacein the planes x: k arethe hyperbolas ,z1sz- y2lb2: l. * k2| a2whose transverseaxesare parallel to the-z axis' In a : similaifashion, the cross sections in the planes A k arc the hyperbolas axesare also parthe transverse given by z2lc2 f la2: 1 * Plbz forwhich allel to the z axis. If.a : b, the surfaceis a hyperboloid of revolution in which the axis is the line containing the transverse axis of the hyperbola' Each of the above three quadric surfacesis symmetric with respect to each of the coordinate planes and symmetric with respect to the origin' Their graphs are called ienttal quadricsand their center is at the origin. The graph of any equation of the form

Figure 18.8.3

+t*Y'*z' a2-

b2-

-1 c2

where a, b, and c ate Positiv€, is a central quadric.

nls.rvrptE L: Dra* asketch of the graph of the equation

4x2-y'+

soLUTIoN: The given equation can be written as

100

and name the surface.

which is of the form of Eq. (2) with y and z interchanged. Hence, the sur-


face is an elliptic hyperboloid of one sheet whose axis is the y axis. The cross sectionsin the planesy : k arethe ellipses*125 + z2l4: | + k21700. The cioss sectionsin the planes x: k are the hyperbolasz2l4- y'1100: l-k2125, and the cross sections in the planes z:k anethp hyperbolas fl25 - y21700: ! - kzl4. A sketch of the surfaceis shown in Fig. 18.8.4.

F i g u r e1 8 . 8 . 4

ExAMPLE2: Draw a sketch of the graph of the equation 4x2-25y2-22:100 and name the surface.

solurroN:

The given equation can be written as

!--t-,'l_:1. 25 4 100 which is of the form of Eq. (3) with r and z interchanged; thus, the surface is an elliptic hyperboloid of two sheetswhose axis is the r axis. The cross sections in the planes x:k, where lkl >5, are the ellipses yzl4 - 1. The planes x = k, where * 221100:k2125 lkl < 5, do not intersectthe surface. The cross sections in the planes A:k are'the hyperbolas fl2| - 221100:1* k214,and the cross sections in the planes z:k are the hyperbolasf 125- U'14:1 + k,/100.The required sketchis shown in Fig. 18.8.5.

F i g u r e1 8 . 8 . 5

18.8QUADRIC SURFACES The following two quadrics are called noncentral quadrics. The elliptic p araboloid

fuzz

7* w:;

F i g u r e1 8 . 8 . 6

G)

where a and b are positive and c * 0. Figure 18.8.5 shows the surface if c>0. Substituting k for z in Eq. (4), we obtain *laz * y'lb' : k/c. When k: 0, this equation becomes *laz * y2lb2-- 0, which represents a single point, the origin. If. k + 0 and k and c have the same sign, the equation is that of an ellipse. So we conclude that cross sections of the surface in the planes z: k,where k and c have the samesign, are ellipses and the lengths of the semiaxesincreaseas lkl increases.If k and c have opposite signs, the planes z: k do not intersect the surface.The cross sections of the surface with the planes x: k and y : k arc parabolas' When c ) 0, the parabolas open upward, as shown in Fig. 18.8'6; when c < 0, the parabolas open downward. lI a: b, the surface is a paraboloid of revolution. The hyperbolicparaboloid -u2 - - : - )c2 z

(s)

b2n2c

F i g u r e1 8 . 8 . 7

ExAMPLE3: Draw a sketch of the graph of the equation

where a and b are positive and c * 0. The surface is shown in Fig. 18.8.7 forc>0. The cross sections of the surface in the planes 2: ft, where k # 0, ate hyperbolas having their transverse €xes parallel to the y axis if k and c v t the samesign and parallel to the r axis if k and c have opposite signs. ""e The cross section of the surface in the plane z: 0 consists of two straight lines through the origin. The crosssectionsin the planesx: k ateparabolas opening upward if c > 0 and opening downward if c < 0. The cross sectionsin the pl".t"t y: k areparabolas opening downward if c ) 0 and opening upward if c < 0.

solurroN:


3y'*1222:"1.6x and name the surface.

which is of the form of Eq. (4) with r and z interchanged. Hence, the graph of the equation is an elliptic paraboloid whose axis is the r axis. The -f crois sections ir, the planes x: k ) 0 are the ellipses y2lL6 zz| 4: kl3, and the planesx: k 10 do not intersectthe surface.The crosssectionsin the planes y : k are the parabolas L2z2:16x - 3k', and the cross sections in


the planesz : k arethe parabolas3yz: L5x- Ue. A sketchof the elliptic paraboloidis shov,rnin Fig. 18.8.8.

F i g u r e1 8 . 8 . 8

EXAMPLE4: Draw a sketch of the graph of the equation 3y'-

1222: l6x

and name the surface.

solurroN:

Writing the given equation as

-tlz - - : - 22 x 1643 we see it is of the form of Eq. (5) with x artd z interchanged. The surfaceis therefore a hyperbolic paraboloid. The cross sections in the planes x: k * 0 are the hyperbolas y'lL5 - z2l4: k/3. The cross section in the yz plane (r: 0) consistsof the two lines y : h ffid y - -2z.In the planes z: k, the cross sections are the parabolas 3A, : L6x * l2t9; in the planes A : k, the clrosssections are the parabolas l2z2 : 3k2- t6x. Figure 1g.g.9 shows a sketch of the hyperbolic paraboloid.

z

Figure18.8.9

The elliptic cone

5.#-$:o

F i g u r e1 8 . 8 . 1 0

(6)

where a, b, arrdc are positive (see Fig. 18.8.10). The intersection of the plane z: 0 with the surface is a single point, the origin. The cross sections of the surface in the planes z: k, *hete k * 0, are ellipses, and the lengths of the semiaxesincreaseas k increases. Cnr-ssse-ctionsin the planes x: 0 and,y: 0 are pairs of intersecting lines. In the planes x: k arrd U: k, where k * 0, the cross sections a"e hyperbolas.

18.8 QUADRICSURFACES

ExAMPLE5: Draw a sketch of the graph of the equation

soLUrIoN: x2

4xz - yz * 2522- 0 and name the surface.


25 which is of the form of Eq. (6) with y and z interchanged. Therefore, the surface is an elliptic cone having the y axis as its axis. The surface intersectsthe xz plane (y : 0) at the origin only. The intersection of the surface with the yz plane (r : 0) is the pair of intersecting lines U : +52, and the intersection with the xy plarre (z: 0) is the pair of intersecting lines y:-+zx. The cross sections in the planes A:k+ 0 are the ellipses f 125+ z2l4:WlL00.In the planesx:k * 0 and z:k # 0, the crosssections are, respectively, the hyperbolas y2lt00-z'14:l&125 and y21100 - f IZS: k214.A sketch of the surfaceis shown in Fig. 18.8.11.

F i g u r e1 8 . 8 . 1 1

The general equation of the second degree in x, y, and z is of the form af -t byz * czz* dxy * exz + fyz + gx * hy * iz * i : g where a,b, . . . , j areconstants. It can be shown that by translation and rotation of the three-dimensional coordinate axes (the study of which is beyond the scope of this book) this equation can be reduced to one of the following two forms: Axz*By',*Czz+l-0

(7)

A* + Byz* lz:0

(8)

or Graphs of the equations of the seconddegree will either be one of the above six types of quadrics or else will degenerateinto a cylinder, plane, line, point, or the empty set. The nondegenerate cur:vesassociatedwith equations of the form (7) are the central quadrics and the elliptic cone, whereas those associated with equations of the form (8) are the noncentral quadrics. Following are examples of some degeneratecases: f - y ' : 0 ; t w o p l a n e sx, - Y : 0 a n d r * Y = o

ANDSOLIDANALYTIC GEOMETRY IN THREE-DIMENSIONAL SPACE VECTORS

864

z2:0;

one plane, the ry plane

* * Y': 0; one line' the z axis f * y'l z2:0; a single Point, the origin f -t y' I z?* 1 : 0; the empty set

1.8.8 Exercises In ExercisesL through 12, draw a sketch of the graph of the given equation and name the surface. t.4*+9y2lz2:36

2.4*-9yz-"2-*

3.4*+9y2-*:36

4.4*-9y2**:36

s.f:Vz-2z

6.f:y,+2,

,.***:n,

a,.fi+{:+

,"2

9.

,2

1 0 .f : 2 y * 4 2

re-*:gy

1 7 .f + 7 6 2 2 : 4 y 2 - 7 6

13. Find the values of k for which the intersection of the plane x + W: y'- * - z2:1 is (a) an ellipse and (b) a hyperbola.

12.9y2-4z2t t8r:0

1 and the elliptic hyperboloid of two sheets

L4. Find the vertex and focus of the parabola which is the intersection of the plane y :2 y' _x" _1 1649

with the hyperbolic paraboloid

15. Find the area of the plane section formed by the intersection of the plane y : 3 with the solid bounded by the ellipsoid f7f22 *1 9254

:

*_:1

16. Show that the intersection of the hyperbolic phraboloid yzlbz- * I a, : zl c and the plane z : bx I ay consists of two intersecting straight lines. 17' U1e the method of parallel plane sectionsto find the volume of the solid bounded by the ellipsoid rfl a2+ y17 * zzlcz: l. (The measure of the area of the region enclosed by the ellipse having semiaxesi and,b is nab.)

1'8.9 CURVES IN R3 We consider vector-valued functions in three-dimensional space. 18.9.1 Definition

Let f u fzi and1" be three real-valued functions of a real variable f. Then for every number f in the domain common to fr, fz, and fs there is a vector R defined by

R(r): fJt)i+ f,(t)i+ fr(t)u

(1)

and R is called a aector-oalued function. The graph of a vector-valued function in three-dimensional spaceis obtained analogously to the way we obtained the graph of a vector-valued function in two dimensions in sec. 17.4. As f assumes all values in the domain of R, the terminal point of the position representation of the vector R(f) traces a curve C, and this curve is called the graph of (l). A point on the curve C has the cartesian representation (r, a, z) , where

18.9 CURVESIN R3

865

(2) y: fr(t) z: fr(t) Equations (2) are called parametricequationsof C, whereas Eq. (L) is called a aectorequationof.C. By eliminating f from Eqs. (2) we obtain two equations in x, y , and z. Theseequations are called cartesianequationsof.C. Each cartesian equation is an equation of a surface, and curve C is the intersection of the two surfaces. The equations of any two surfaces containing C may be taken as the cartesian equations defining C. x: fr(t)

o rLLUsrRArroN 1: We draw a sketch of the curve having the vector equation R(f) : a cos fi * b sin fi + tk Parametric equations of the given cuwe are ' x:acost z:t A:bsint To eliminate f from the first two equations, we write them as f : a2 coszt

and A2: V sinz t

from which we get x2 ;:

cosz t

and

Adding correspondingmembersof thesetwo equations,we obtain f a2

+a' :1 b2

Therefore, the cuwe lies entirely on the elliptical cylinder whose directrix is an ellipse in the xy plane and whose rulings are parallel to the z a>
Table18.9.1

v

z 0

a

0 b

4

\D.

\D

4

7T

0

b

t

x

0

fl

TT

2 3n 4 7f

3rr 2

-- a t/z -a 0

b

\D 0 -b

ll

7T

2 3n 4 7T

3n 2

GEOMETBY ANDSOLIDANALYTIC SPACE VECTORSIN THREE-DIMENSIONAL The curve of Illustration 1 is called a helix. If a : b, the helix is called a circular helix and.it lies on the right-circular cylinder f + yz: oz. o rLLusrRATrow2: The curve having the vector equation R(f) - fi+ Pi+ f3k is called a twisted cubic.Parametric equations of the twisted cubic are x:t

A:P

z:F

Eliminating f from the first two of these equations yields A : *, which is a cylinder whose directrix in the xy plane is a parabola. The twisted cubic lies on this rylinder. Figure 18.9.2shows a sketch of the cylinder and the o portion of the twisted cubic from t:0 to t:2' Many of the definitions and theorems pertaining to vector-valued functions in two dimensions can be extended to vector-valued functions in three dimensions Fig ure

18.9.2Definition

If R(f) - f,(t)i + hgi

* /3(f)k, then

lim R(f) : lim fi(t)i + lim hg)i + lim /s(f)k t-tr

iflim/r(f), t-tr

18.9.3 Definition

,

t-tr

l- fi

t-tr

lilrl.f2Q),and lim/3(f) all exist. t-tr

t-tr

The vector-valued function R is continuous at ft if and only if (i) R(tl) exists; (ii)

R(f) exists; ,lig (iii) lim R(t) : R(t,). t-h

18.9.4 Definition

The derivative of the vector-valued function R is a vector-valued function, denoted by R' and defined by

R'(t):tt-M{f-E(1I if this limit exists. 18.9.5 Theorem

If R is the vector-valued function defined by

R(f): f(t)i + h@i +/,(r)r. and R'(f) exists,then

R'(t): fr'(t)i + fr'(t)i + fs'(t)k The proof of Theorem 18.9.5is left as an exercise (see Exercise9).

1 8 . 9 C U R V E SI N R 3 R(f + af) - R(f)

z

867

The geometric interpretation for the derivative of R is the sameas that for the derivative of a vector-valued function in R2.Figure 18.9.3shows a portion of the curye c, which is the graph of R. In the figure oF is the position representation of R(f), O? is the position representation of R(t + Af ), and so fQ is a representationof the vectorR(f + At) R(f )' As Af approacheszero,the vector tR(t + af) R(f)l/At has a representation appioaching a directed line segment tangent to the curve C at P' The definition of the unit tangent vector is analogous to Definition 17.8.'1, for vectors in the plane. So if T(t) denotes the unit tangent vector to curve C having vector equation (1), then (3)

F i g u r e1 8 . 9 . 3

. rLLusrRArroN 3: We find the unit tangent vector for the twisted cubic of Illustration 2. BecauseR(f) : ti+ t2i + fsk, D'R(t) : i+ 2ti + 3t2k and

lD'R(t)l: IGW49F From Eq. (3), we have, then,

T(t) :

1

(i+ 2ti + 3f'zk)

Therefore,in particular, 2

r ( 1 )- \ M1 r ' \i rm t ' @ k i r

?

' Figure 18.9.4shows the representation of T(1) at the point (1, 1, 1)' and 17.5.8regarding derivativesof sums and Theorems 17.5.6,77.5.7, products of two-dimensional vector-valued functions also hold for vectors in three dimensions. The following theorem regarding the derivative of the cross product of two vector-valued functions is similar to the colresponding formula for the derivative of the product of real-valued functions; however, it is important to maintain the correct cirder of the vectorvalued functions becausethe cross product is not commutative.

F i g u r e1 8 . 9 . 4

18.9.5 Theorem

If R and Q are vector-valued functions, then D ' [ R ( t ) x Q ( f ) ] : R ( f ) x Q ' , ( t )+ R ' ( f ) x Q ( f ) for all values of f for which R'(f) and Q'(t) exist' The proof of Theorem 18.9.6is left as an exercise(seeExercise10).

868


We can define the length of an arc of a curve C in three-dimensional spacein exactly the same way as we defined the length of an arc of a.curve in the plane (see Definition 77.6.1). If C is the curve having parametric equations (2), fr' , fr' , f"' are continuous on the closed interval fa, bf , and no two values of f give the same point (r, y, z) on C, then we can prove (as we did for the plane) a theorem similar to Theorem 17.6.3, which states that the length of arc, L units, of the curve C from the point (f'(n), fr(a), ft(a)) to the point (/,(b), f"(b), fs(b)) is determinedby (4)

If s is the measureof the length of arc of C from the fixed point (fr(t), fr(t), h!i) to the variablepoint (/,(f),fr(t),f"(t)) and s increasesas f increases,then s is a function of fand is given by du

(5)

As we showed in Sec. 17.6 for plane curves, we can show that if (1) is a vector equation of C, then

D6: lD'R(f)l

(5)

and the length of atc, L units, given by (4), also can be determined by (7)

ExAMPLE1: Given the circular helix R(f) - n cos ti* a sin fj + tk, where a ) 0, find the length of arc from t - 0 to t - 2n.

sol.urroN: DrR(t) : -a sin ti + a cos fi + k. So from (7) we obtain dt

\m

dt-zn\ffi

Thus, the length of arc is 2nTa4

"1,units.

The definitions of the curaature aector K(f) and the curoatureK(t) at a point P on a curye C in Rsare the sameas for plane curyes given in Defininon 17.9.1.Hence, if T(f) is the unit tangent vector to C at P and s is the measureof the arc length from an arbitrarily chosenpoint on C to P, where s increasesas f increases,then K(t) : D".1(t) or, equivalently,

1 8 . 9 C U R V E SI N 8 3

and

K(r): IDJ(')l or, equivalently,

(e)

K(r):lP,I!l1 | l l D ,R( ll f) Taking the dot product of K(t) and T(f) and using (8), we get

(10) Theorem 1,7.5.1,1, statesthat if a vector-valuedfunction in a plane has a constant magnitude, it is orthogonal to its derivative. This theorern arrd its proof also hold for vectors in three dimensions. Therefore, because 'T(t) : 0. And so the curlT(t) | : L, we canconcludefrom (10) that K(f) vature vector and the unit tangent vector of a curve at a point are orthogonal. We define the unit normal aector as the unit vector having the same direction as the curvature vector, provided that the curvature vector is not the zero vector. So if N(t) denotes the unit normal vector to a curve C at a point P, then if K(t) + 0, (11)

From (L1) and the previous discussion, it follows that the unit normal vector and the unit tangent vector are orthogonal. Thus, the angle between these two vectors has a radian measure of lzzr,and we have from Theorem 'J,8.6.6,

l r(r) x N (r)| : lr ( r )llN( r )| sin#z': ( 1)( 1)( 1): 1 Therefore,the crossproduct of T(t) and N(f ) is a unit vector.By Theorem 18.5.9T(f) x N(t) is orthogonalto both T(f) and N(f); hence,the vector B(f), defined by (12)

B(f)-r(f)xN(r)

F i g u r e1 8 . 9 . 5

EXAMPrn 2: Find the movin g trihedral and the curuature at any point of the circular helix of Example 1.

is a unit vector orthogonalto T(f ) and N(t) and is called the unit binormal aector to the curve C at P. The three mutually orthogonalunit vectors,T(t), N(f), and B(f), of a curve C are called the moaingtrihedralof C (seeFig. 18.9.5). solurroN:

A vector equation of the circular helix is

R(f) - a cos ti* SoDrR(f) get

T(f):ffi

a sin ti+ tk

k and lD,R(r)l: \m. 1

(-a sinti+ a cosfi + k)

From(3) we

870


So

D ; r ( r ) :+ ? a c o s

sin fj )

\/az + i

Applying (8), we obtain (-a cos ti-

K(f)

a sin fj)

The cunrature, then, is given by

K ( r ;- l K ( f ) l and so the curvature of the circular helix is constant. From (1r) we get N(f) : -cos fi - sin fi Applyrng (12), we have

B(t) : --

1

ffi

(-a sin fi* a costi + k) x (-cos fi- sin fi)

1 (sin ti- cos ti+ ak) \/a2 + 1

A thorough study of curves and surfaces by means of carculusforms the subject of.differentialgeometry.The use of the calculus of vectors further enhancesthis subject. The previous discussion has been but a short introduction. we now consider briefly the motion of a particle along a curve in three-dimensional space. If the parameter f in the vector e-quation (1) measurestime, then the position at f of a particle moving along the curve C, having vector equation (t ), is the point p(f Jt) , fr(t), /r(r) ). fn e aeloc_ ity aector,V(f), and the accelerationoector,A(f), are defined as in the

plane. The vector R(f) is called the position aector, and, V(f):

D'R(f)

(13)

and A(f) :DtY(t):

DfR(t)

The speed of the particle at t is the magnitude By applying (5) we can write

lv(t)l: Dts EXAMPLE3: A particle is moving along the curve havin gparametric equations x: 3t, y : t2, and z : zrtt.Find the velocity and acceleration vectors and the speed

sol,urroN:

A vector equation of the curye is

R(f)-3fi+tzi+&t'k Therefore, V(f) : DrR(t) - 3i + 2ti + 2t2k

(14) of the velocity vector.

1 8 . 9 C U R V E SI N R 3

of the particle at t - L. Draw a sketch of a portion of the cunre at t - 1,,and draw representations of the velocity and acceleration vectors there.

871

and

A(t; : DNG):2i + 4tk Also,

lv(r)l: \trTW+4 So when t: !, Y: 3i * 2j + 2k, A:2i * 4k, and lv(l) | : \/F. The requiredsketchis shownin Fig. 18.9.6.

v(1)

F i g u r e1 8 . 9 . 6

18.9 Exercises In Exercises1 through 4, find the unit tangent vector for the curve having the given vector equation. 1. R(r): (r+l)i-r3i+

(1-2f)k

3. R(f):etcos ti-t etsinti* etk

2. R(t):sin2fi lcos2ti+2t}t?k 4, R(f) :tzi*

(t++f3)i+ (t-+t3)k

In Exercises5 through 8, find the length of arc of the curve fuom t' to t2. 5. The curve of Exercise2; tt : Q; 1,: 1. 5. The curve of Exercise7; \ : -l; S": 2. 'l'. 8. The curve of Exercise4; h: 0; tz: 7. The curve of Exercise3; tr: Q; 1r: 3. 9. Prove Theorem 18.9.5.


1 1 . W r i t e a v e c t o r e q u a t i o n o ft h e c u r v e o f i n t e r s e c t i o n otfh e s u r f a c e s A : { a n d ' z : x ! . 12. Prove that the unit tangent vector of the circular helix of Example 1 makes an angle of constant radian measure with the unit vector k. 13. Find the moving trihedral and the curvature at the point where f : 1 on the tWisted cubic of Illustration 2. 14. Find the moving trihedral and the curvature at any point of the curve R(t) : cosh fi * sinh ti + tk. In Exercises15 through 18, find the moving trihedral and the curvature of the given curye at t: 15.Thecurveof Exercisel;tr:-1.

15.ThecurveofExercise2;tt:9.

17. The curve of Exercise3; fr: 0.

18. The curve of Exercise4; L:7.

tt if they exist.

In Exercises19 through 22, aparticle is moving along the given curve. Find the velocity vector, the accelerationvector, and the speed at t: ty Draw a sketch of a portion of the curve at t: tr and draw the velocity and accelerationvectors there. 19. The circularhelix of ExampleL; tr: itr.

20. x: t,y:

$t2,z:

2 1 .x : e 2 t , A : r - z t , 2 : t e z t ; t r : 1 .

2 2 .x : L l 2 ( t 2 + t ) , y :

iF; h:2. ln(l * P),2:tan-tt;tr:1.


23. Prove that if the speed of a moving particle is constant, its accelerationvector is always orthogonal to its velocity vector. 24. Prove that for the twisted cubic of Illustration 2, iI t + 0, no two of the vectors R(f), V(f), and A(f) are orthogonal. 25. Prove that if R(t) : f ,(t)i + h@i + /r(t)k is a vector equation of curve C, and K(t) is the curvature of C, then lD,R(f)x D,2R(f)l

K ( 1 1:

lD,R(f)l'

26. Use the formula of Exercise25 to show that the curvature of the circular helix of Example I is al(az + l). In Exercises27 and28, find the curvature of the given curve at the indicated point. 2 7 .x : t , y : t " , 2 : f ; t h e

origin.

2 8 .x : e ' , y : e - t , 2 : t ; t : 0 . 29. Prove that if R(f) : f'(t)i+ fr(t)j + /B(f)k is a vector equation of curve C, K(t) is the curvature of C at a point P, and s units is the arc length measured from an arbitrarily chosen point on C to P, then D,R(t) . D"3R(f): -[K(f)],

18.10 CYLINDRICAL AND SPHERICAL COORDINATES

Cylindrical and spherical coordinatesare generalizations of polar coordinates to three-dimensional space. The cylindrical coordinate representation of a point P is (r, 0, z), where r and 0 are the polar coordinates of the projection of P on a polar plane and z is the directed distance from this polar plane to P. SeeFig. 18.10.1. ( r , 0 ,z )

x Figure 18.10.1

EXAMPLEL: Draw a sketch of the graph of each of the following equations where c is a constant: (a) r:c; (b) 0:c; (c) z:c.

solurroN: (a) For a point P(r,0, z) on the graph of r: c, 0 and z can have any values and r is a constant. The graph is a right-circular cylinder having radius lcl and thez axis as its axis. A sketch of the graph is shown in Fig. 18.10.2. (b) For all points P(r, 0, z) on the graph oL0: c, r andz can assume any value while 0 remains constant. The graph is a plane through the z axis. See Fig. 18.10.3for a sketch of the graph. (c) The graph of z: c is a plane parallel to the polar plane at a directed distance of c units from it. Figure 18.10.4shows a sketch of the graph.

1 8 . 1 0C Y L I N D R I C AAL N D S P H E R I C A C L OORDINATES

F i g u r e1 8 . 1 0 . 4

z

P(*, a, z) (r, o, z)

z

,/

7X

The name "cylindrical coordinates" comesfrom the fact that the graph : r c is a right-circular cylinder as in Example 1(a). Cylindrical coordiof nates are often used in a physical problem when there is an axis of symmetry. Suppose that a cartesian-coordinate system and a cylindrical-coordinate system are placed so the ry plane is the polar plane of the cylindrical-coordinate system and the positive side of the r axis is the polar axis as shown in Fig. 18.10.5.Then the point P has (r, y, z) and (r, 0, z) as two sets of coordinates which are related by the equations

v

x

(1)

(2)

F i g u r e1 8 . 1 0 . 5

ExAMPLEZ: Find an equation in cartesian coordinates of the following surfaces whose equations are given in cylindrical coordinates and identify the surface: (a)r-6sin0; (b) r(3 cos0+ 2 sin 0) *52 -0.

solurroN: (a) Multiplying on both sides of the equation by r, we get t2: 5r sin 0. Becauset2: X, * y, and r sin 0: y, we have X2+ yz: 6r. This equation can be written in the fonrt xz + (V - 3)': 9, which shows that its graph is a right-circular cylinder whose cross section in the ry plane is the circle with its center at (0,3) and radius 3. (b) Replacing'/ cos 0 by x and r sin 0 by y, we get the equation 3r Hence, the graph is a plane through the origin and has *2y*62:0. (3,2, 6> as a normal vector.

EI4

IN THREE-DIMENSIONAL VECTORS SPACEAND SOLID'ANALYTIC GEOMETRY

EXAMPLE3: Find an equation in cylindrical coordinates for each of the following surfaces whose equations are given in cartesian coordinates and identify the surface: (a) xz + !' : z; (b) x2-A':2.

F i g u r e1 8 . 1 0 . 6

ExAMPrn 4: Draw a sketch of the graph of each of the followirrg equations where c is a constant: (a) p:c, and c >0; (b) 0_ c; (c) 6-c,and0

F i g u r e1 8 . 1 0 . 7

solurroN: (a) The equation is similar to Eq. ( ) of Sec. 18.8,and so the graph is an elliptic paraboloid.lf. * * y2 is replaced by r', the equation becomesf : z. (b) Ttre equation is similar to Eq. (5) of Sec.18.8with x and y interchanged. The graph is therefore a hyperbolig paraboloid having the z axis as its axis. When we replacex by r cos 0 and y by t sin 0, the equation becomesrz cosz0 - rz sinz 0 : 4 andbecause cosz0 - sinz 0: cos 20,we can write this as z:12 cos20.

In a spherical-coordinatesystemthere is a polar plane and an axis perpendicular to the polar plane, with the origin of the z axis at the pole of the polar plane. A point is located by three numbers, and the spherical-coordinate representationof a point P is (p, 0,S), where p : lOPl, I is the radian measure of the polar angle of the projection of P on the polar plane, and @is the nonnegative radian measure of the smallest angle measured from the positive side of the z axis to the line OP. SeeFig. 18.10.6. The origin has the spherical-coordinaterepresentation(0,0,6), where 0 and $ may have any values. If the point P (p, 0, @) is not the origin, then p > 0 and 0 < 6 < zr, where Q:0 if P is on the positive side of the z axis and 6: r if P is on the negative side of the z axis.

soLUrIoN: (a) Every point P(p, 0, S) on the graph of p: c has the same value of p, 0 rnay be any number, and 0 = Q = zr.It follows that the graph is a sphere of radius c and has its center at the pole. Figure 18.1,0.2 shows a sketch of the sphere. (b) For any point P (p, 0, Q) on the graph of 0 : c, p may be any nonnegative number, d may be any number in the closed interval 10,nf , and 0 is constant. The graph is a half plane containing the z axis and is obtained by rotating about the z axis through an angle of c radians that half of.the xz plane for which r > 0. Figure 18.10.8shows sketchesof the half planes for 0:ir,0: & r r , 0 : t r r , a n dg : - * 2 .

1 8 . 1 0C Y L I N D R I C AALN D S P H E R I C A C L O O R D I N A T E S 875

(c) The graph of 6: c contains all the points P(p,0, d) for which p is any nonnegative number, 0 is any number, and S is the constant c. The graph is half of a cone having its vertex at the origin and the z axis as its axis. Figure 18.10.9aand b each show a sketch of the half cone for 0 < c < *n and izr < c < zr, respectively.

o: :27 0
F i g u r e1 8 . 1 0 . 8

I

+"
(a)

(b)

F i g u r e1 8 . 1 0 . 9

Becausethe graph of p: c is a sphere as seen in Example 4(a),we have the name "spherical coordinates." In a physical problem when there is a point that is a center of symmetry, spherical coordinates are often used. By placing a spherical-coordinate system and a cartesian-coordinate system together as shown in Fig. 18.10.10,we obtain relationships between the spherical coordinates and the cartesian coordinates of a point P from P(*, a, z) (P,0,6)

*4/'

-

Because

cos0

y:

sin 0

z- l a P l

: p sin ,f and

become

(3) By squaring each of the equations in (3) and addingr w€ have )c2+ y, + 22: p2 sin, + cos20 + p2 sin'0 sin2 0 + p2 cos'0

-- p2 sinz (cosz0 + sin2 g) * p' cosz(0 @ x F i g u r e1 8 . 1 0 . 1 0

5: Find an equation in ExAMPLE cartesiancoordinatesof the following surfaceswhose equations

: pz(sin2Q + cosz0) :p2

solurroN: (a) Becausez: p cos{, the equation becomesz: 4. Hence, the graph is a plane parallel to the xy plane and 4 units above it' (because0< (U) for spherical coordinatesp>0 and sinS>0

876


are given in sphericalcoordinates and identify the surface: ( a ) p c o sQ : 4 ; ( b ) p s i n Q : 4 .

Q = rr); therefore, if we square on both sides of the given equation, we obtain the equivalent equation pz sinz 6:16, which in tum is equivalent to P " ( l - c o s 2{ ) : 1 5 or Pz- P2cos' 0:

L6

Replacing p'by x2+ y' + z2 and p cos 6 by z, we get x2+y'+22-22:L5 or, equivalently, x 2+ A 2: 1 5 Therefore, the graph is the right-circular cylinder having the z axis as its axis and radius 4. ExAMPLE6: Find an equation in spherical coordinates for: (a) the elliptic paraboloid of Example 3(a); and (b) the plane of Example 2(b).

sor,urroN: (a) A cartesian equation of the elliptic paraboloid of Example 3(a) is *'+ y': z. Replacing xby p sin @ cos0, y by p sin @sin 0, and z by p cos Q, we get p2 sin' Q cosz0 * p'sinz ,f sin2

coso

or, equivalently, p2 sin2S(coszd * sinz 0) : p cos S which is equivalent to the two equations P:0

and p sin2S : cosS

The origin is the only point whose coordinates satisfy p : 0. Becausethe origin (0, 0,En) lies on p sinz @: cos 6, we can disregard the equation p:0. Furthermore, sin { # 0 because there is no value of d for which both sin d and cos f are 0. Therefore, the equation p sin2{ : cos f can be written as p: cscz@cos @,or, equivalently, p: csc @cot {. (b) A cartesianequation of the plane of Example 2(b) is 3x * 2y * 6z:0. By using Eqs. (3) this equation becomes 3p sin @cos 0 *2p sin $ sin 0 * 6p cos g : g

Exercises 18.10 1. Find the cartesian coordinates of the point having the given cylindrical

(a) (3, tn, 5)

(b) (7, 3n, -4)

coordinates:

(c) (L,1,'1.)

2. Find a set of cylindrical coordinates of the point having the given cartesian coordinates: (a) (4, 4, -2) (b) (-3 \n, 3, 6) (c) (1,1,'1,)

L OORDINATES 877 1 8 . 1 0C Y L I N D R I C AAL N D S P H E R I C A C

3. Find the cartesian coordinates of the point having the given spherical coordinates:

(c) $re, *n, *n)

(b) (4, tn, *n)

(a) (4, tn, tr)

4. Find a set of spherical coordinates of the point having the given cartesian coordinates: (c) (2, 2, 2) (b) (-1 , tE, 21 (a) (1,-L, -rt) 5. Find a set of cylindrical coordinates of the point having the given spherical coordinates:

(d Q|l,tu,*n)

(b) (\/r,tu,n)

(a)(4,32r,*r)

5. Find a set of spherical coordinates of the point having the given rylindrical coordinates: (c) (2,fu'-4) (b) (3,1n,2) (a) (3,tn,3) In Exercises7 through 10, find an equation in cylindrical coordinates of the given surface and identify the surface. 8' *-y2:9

7. f * V ' 1 4 2 2 : 1 6

l0' *+A2:zz

9. f - Y 2 : 3 2 '

In Exercises11 through 14, find an equation in spherical coordinates of the given surface and identify the surface. L2.x2+A2:z'

LL, x 2 + y ' , + 2 2 - 9 2 - 0 1 3 . )c2+ a',:9

1,4. x2 + U' : 2z

given in cylindrical In Exercises15 through 1g, find an equation in cartesiancoordinates for the surfacewhose equation is surface' the identify and 16, 15 In Exercises coordinates. L 6 . r : 3 c o s0 1 5 .( a ) r - 4 ; ( b ) 0 - i n L8. z2 sin3 0 : rg

L7. r2 cos 20 - zB

is given in spherical In Exercises1g through 22, find,an equation in cartesian coordinates for the surface whose equation coordinates. In Exercises19 and 20, identify the surface' -20' p:9 sec@ 19. (a) p: s; (b) e in; (c) Q: *r 2 2 . p : 6 s i n d s i n0 + 3 c o s t f 2 I . P : 2 t a n0 29. A curve C in R3has the following Parametricequations in cylindrical coordinates: r:FJt)

0-Fr(t)

z-F(f)

arc of C from the Use formula (4) of Sec. 18.9 and formulas (1) of this section to Prove that if L units is the length of : point where t: a to the point where f b, then lb

r:l-661 Ja

24. A curve C in Rr has the following parametric equations in spherical coordinates: o: GzQ) p: G,(t) d : G'(f) L units is the length of arc of C from the Use formula (4) of Sec. 18.9 and formulas (3) of this section to prove that if : : then t b, where point a to the point where t L-

rb l

Ja

fi)z * p'(D,Q)z dt

878


25. (a) Show that parametric equations for the circular helix of Example a

0-t

18.9, are

z-t

(b) Use the formula of Exercise 23 to find the length of arc of the circular helix of part (a) from f : 0 to f : 2zr. Check your result with that of Example 1, Sec.18.9. 26. A conical helix winds around a cone in a way similar to that in which a circular helix winds around a cylinder. Use the formula of Exercise 24 to find the length of arc from f : 0 to f :2r oI the conical helix having parametric equations p:t

0:t

Q:tn

(Chaqter18) ReaiewExercises 1. Draw a sketch of the graph of x-

3 in Rt, R', and R3.

2. Draw a sketch of the set of points satisfying the simultaneous equations x:

6 and,y:

3 in Rz and R3.

In Exercises3 through LL, describe in words the set of points in Rs satisfying the given equation or the given pair of equations. Draw a sketch of the graph. v?'

fv:o [z-o

5. x2+22:4 9 . x 2 + A 2: 9 2 In Exercises 12 through 17, let A - - i indicated vector or scalar.

12.3A-zts_+C

4. {*_, ly:z

5.

7. x - y

8.

10. )c2+ A, : z, + 3i + 2k, B - 2i+ i - 4k, 13.6C+4D-E

11. i+2i-2k,

E - 5i - 2i, and find the

T4.28.C+3D.E

1 5 .D . B x C

16.(AxC)-(DxE) 1 7 .l A x B l l Dx E l In Exercises18 throrrgh 23, there is only one way that a meaningful expression can be obtained by inserting parentheses. Insert the parenthesesand find the indicated vector or scalar if A - ( 3 , - 2 , 4 > , B : ( - 5 , 7 , 2 ) , a n d C - ( 4 , 6 , - 1 ). 1 9 .B . A - C 1,9.A+B.C 20.A.BC 21,.A8.C 24. If A is any vector, prove that A -

22.AXB.CXA (A. i)i+ (A . i)i + (A . k)k.

23.AxB.A+B-C

25. Find an equation of the sphere concentric with the sphere * * y, + zz+ 4x + 2y - 6z * 10 : 0 and containing the point (-4, 2,5). 26. Find an equation of the surface of revolution generated by revolving the ellipse 9* * 422: 36 in the xz plane about the r axis. Draw a sketch of the surface. Determine the value of c so that the vectors 3i * ci - 3k and 5i - 4i * k are orthogonal. ShowthattherearerepresentationsofthethreevectorsA:5i+i-3k,B:i*3i-2k,andC:-4i+2i*kwhich form a triangle. 29. Find the distance from the origin to the plane through the point (-6,3,-2) and having 5i - 3j + 4k as a normal vector. 30. Find an equation of the plane containing the points (1,7 , -3) and (3, 1, 2) and,which does not intersect the r axis. 3r. Find an equation of the plane through the three points (-1,2,l) , (1,4, 0), and (1,-1,3) by two methods:(a) using the cross product; (b) without using the cross product.

REVIEW EXERCISES 879

32. Find two unit vectors orthogonal to i - 3i * 4k and whose representations are parallel to the yz plane. 33. Find the distancefrom the point P(4, 6,-4) to the line through the two points A(2,2,1) and B(4,3,-1'). 34. Find the distancefrom the plane 9r - 2y 'f5z * 44: 0 to the point (-3, 2, 0). * 5kfindcos 0 intwo ways: i+ kandB:4i-3i identity. and a trigonometric product cross producf (b) using the by dot the (a) by using

35. I f 0 i s t h e r a d i a n m e a s u roef t h e a n g l e b e t w e e nt h e v e c t o r sA : 2 i + 36. Prove that the lines +2 _x_- l_ : 2 : a2

z-2

. x-2 Y and-n-:3:---

-S

z-5

are skew lines and find the distance between them' to each of the lines of 32. Find symmetric and parametric equations of the line through the origin and perpendicular Exercise36. - 3k and 5i * 4k' 38. Find the area of the parallelogram two of whose sides are the position representationsof the vectors2i -1, 3) , (-2,2' -1), and (-1,l' 2) ' 39.Find the volume of the parallelepipedhaving verticesat (1, 3, O), (2,

40. Find the length of the arc of the curve y:tsint

l:fcost from f: 0 to t:

z:l

itr.

vector, and the speed at 4r. Aj-particle is moving along the curve of Exercise40: Find the velocity vector, the acceleration and acceleration t tn. Draw a sketch of iportion vectors there.

of the curve at t:

Ln and draw the representationsof the velocity

curve having the vector equation r12. Find the unit tangent vector and the curvature at any point on the R(t):

eti* e-tiI2tk

: equations: (a) (r * !)2 + t : z; (b')25* r +yz 1gg' 43. Find an equation in cylindrical coordinates of the graph of eachof the - 4y2 of the equations: (a) r' i Y'+ 422:4; (b) 4f 44. Find an equation in spherical coordinates of the graph of each l9z2:36' with respect to f exist, prove that 415.If R, e, and W arethree vector-valued functions whose derivatives D,[R(r).e(r)xw(f)]:D,R(f).Q(f)xw(t)+R(t)'DrQ(t)xw(t)+R(t)'Q(t)xD'w(t)

Differentialcalculus of functions of severalvariables

19.1 FUNCTIONSOF MORE THAN ONE VARIABLE

I9.I FUNCTIONS OF MORE THAN ONE VARIABLE

1r9.L.1Definition

19.1.2 Definition

In this section we extend the concept of a function to functions of n variables,and in succeedingsectionswe extend to functions of n variables the concepts of the timit of a function, continuity of a function, and derioative of a function. A thorough treatment of these topics belongs to a course in advanced calculus. In this book we confine most of our discussion of functions of more than one variable to those of two and three variables; however, we make the definitions for functions of n variables and then show the applications of these definitions to functions of two and three variables. W" utto show that when each of these definitions is applied to a function of one variable we have the definition previous$ given. To extend the concept of a function to functions of any number of vari' ables, we must first consider points in n-dimensional number space.Just as we denoted a point in Rl by a real numb et x, apoint in R2by an ordered pair of real numterc (x, y), and a point in R3 by an ordered triple of real numbers (x, y, z), we rePresent a point in n-dimensional number sPace' : R", by an'oriered n-taple of real numbers customarily denoted by P ( x ' y ) ; n:2'P: ( r r , * r , . . . , r n ) . I n p a r . t i c u l a irf,. n : 7 , w e l e t P : x i i f if.n: g, P : (x, y, z)t if.n: 6, P : (xr.,x2,xg,xa,x5,x6). The set of all ordered z-tuples of real numbers is called the n'dimensional xr, . ' , xo) numberspace andis denoted by R'. Eachordered n-tuple (1,r, space' number is called a point in the n-dimensional Afunctionofnaariablesisasetoforderedpairsoftheform(P-.al)inwhich P is a point in ,ro t"o distinct ordered pairs -rpu." have the same fiist element. set of all The number' real and w is a n-dimensional numbe, the set and function, the p"rriUi" values of P is called the domain of function' ff ail possible values of w is called the rnnge of the of.n variFrom this definition, we see that the domain of a function or, numbers feal of a set ables is a set of points in R" and that the range is of one function a have we of points in Rl' When rt: L' equivalently, " ""t -of variable;thus,thedomainisasetofpointsinRlor'equivalently'asetof real numbers' Hence' we see that a set is range the and real numbers, .1..7.L d specia-icaseof Definition 19.1..2.lf.n:2, wehave a is Definition and the domain is a set of points in R3or, equivvariables, two of function is a set of a set of ordered pairs of real numbets (x' y)' The range ;iliit, real numbers. variables r and y be the set . rLLUsrRArroN1: Let the function f of two that such (P, z) of all ordered pairs of the form

z: {zs777l

y) for which 25 - * The domain of I is the set of all ordered pairs (x' I * Az:25 y, > o.This is tire set of all points in_the xy plane on the circle and in the interior region bounded by the circle'

CALCULUSOF FUNCTIONSOF SEVERALVARIABLES DIFFERENTIAL

Becausez: ry'F=@Tfr, we see that 0 < z < 5; therefore, the range of / is the set of all real numbers in the closed interval [0, 5]. In Fig. 19.1.1we have a sketch showing as a shaded region in R2 the set of points in the domain of /. o . rLLUsrRArron2: The function g of two variables r and y is the set of all ordered pairs of the form (P, z) such that \/F t 7 -,5 F i g u r e19 . 1. 1

v The domain of g consists of all ordered pairs (r, y) for which * + yz - 25 and y # 0. This is the set of points, not on the r axis, which are either on the circle * * y, :25 or in the exterior region bounded by the circle. Figure 19.7.2is a sketch showing as a shaded region in R2the iet of points in the domain of g. o

F i g u r e1 9 . 1. 2

F i g u r e1 9 . 1. 3

o rllusrRArror.r 3: The function F of two variables x au';rd, y is the set of all ordered pairs of the form (P, z) such that

z:yG+f,-2s lf y:0, then z -- 0 regardlessof the value of r. Howeveg iI y # 0, then f * y'- 25 must be nonnegative in order for z to be defined] Therefore, the domain of F consists of all ordered pairs (r, y) for which either y = g or rP * y' - E > 0. This is the set of all points on the circle rp * yr-:25, all points in the exterior region bounded by the circle, and all poitrts on the r axis for which -5 . r < 5 In Fig. 19.1.3we have a sketclishowing as a shaded region in R2 the set of points in the domain of F. o o rr,r,usrRArror.r4: The function G of two variables r and y is the set of all

19.1FUNCTIONS OF MORETHANONEVARIABLE ordered pairs of the form (P, z) such that z-

'

ffi

v

lf y:0, then z:0 provided that * * y'- 25 + 0. lf. y # 0, then 12* y' - 25 must be positive in order f.orz to be defined. Hence, the domain of G consists of all ordered pairs (x, y) for which x' * y' - 25 > 0 and those for which 3r: 0 and x # +-5. These are all the points in the exterior region bounded by the circle 12* y' :25 and the points on the r axis for which -5 ( r < 5. Figure 19.1.4is a sketch showing as a shaded region in R2 o the set of points in the domain of G. If f is a function of n variables, then according to Definition 19.'1"2' / i s a s e t o f o r d e r e d p a i r s otfh e f o r m ( P , w ) , w h e r e P : ( x r , x 2 , . - . , x n ) is a point in R" and ar is a real number. We denote the particular value of ar, which corresponds to a point P, by the symbol f (P) or f (xr, h, . . , xo). In particular, 7f.n:2 and we let P : (x, A) , we can denote the function value by either /(P) or f(x,y). Similarly, if.n:3 and P: (*, y, z), we denote lhe function value by either /(P) or f(x, y, z). Note that if n: l, P : xi hence, if f is afunction of one variable, f (P) : /(r). Therefore, this notation is consistent with our notation for function values of one variable. A function f of.n vaiables can be defined by the equation

F i g u r e1 9 . 1. 4

w:f(\,x2,.

.

,xn)

The variablES X1,Xz, . . ., xn ate called the independentaariables,arrd w is called tl:.edependentaariable. o ILLUsrRArron 5: Let / be the function of Illustration 1; that is,

f(x' Y): \/8=7f (9,

-4) -

f (-2, r) : f (u, 3a) : EXAMPLE1: The domain of a function I is the set of all ordered triples of real numbers (x, Y, z) such that

g(x, y, z)

Sxz* yz'

Find (a) S(1, 4, -2); (b) S(2a, -b, 3c); (c) 8(x', y', z'); (d) Sg, z,-x).

:ffi-g :ffi-2\E _ffi

SOLUTION:

( a )8 ( 1, 4 , - 2 ) (b) B(2a,

: 1 2 - 5 ( 1 ) ( - 2 ) + 4 ( - Z ) ' : 1 ' + 1 0+ 1 6 : 2 7 3c) - (2a)2- 5(2a)(3c) + (-b) (3c)' : 4a2- 30ac- 9bc2

rc4- 5x222* y'zn (c) g@r, Ar, zr) : (x')2 - 5(x')(z') + (y')(z')2: : yz - Sy(-r) * z(-r)2 : y' + Sxy * xzz (d) SQ, z, -)c)

DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES

19.1.3 Definition

If / is a function of a single variable and g is a function of two variables, then the compositefunction f ' I is the function of two variables defined by

(f ' s)(x,y) : fQ@'y)) and the domain of f " I is the set of all points (x, y) in the domain of g such that g@, y) is in the domain of /. EXAMPTr2: Given f (t) - ln f a n d g @ , y ) - x 2* y , f i n d h ( x , y ) if h - f o S, and find the domain of. h.

SOLUTION]

h(x,y) : ( f " il@, y) : f k@' y) )

: 1 1 *+ y ) :ln(*

I y)

The domain of g is the set of all points in R2,and the domain of / is (0, +.;. Therefore, the domain of h is the set of all points (r, y) for which f*y>0. Definition 19.1.3 can be extended to a composite function of n vaiables as follows. 19.1.4Definition

If / is a function of a single variable andg is a functionof.n variables, then the compositefunction f " g is the function of.n vartables defined by U"il(xr,xr,.

. . ,x):f(g@r,x2,.

. , ,xn))

and the domain of f " g is the set of all points (x1, x2, . . .,r,) domain of g such thatg(rr, x2, . . . , x,\ is in the domain of f. EXAMPLE 3:

Given F(r) - sin-l r

and G(x, A, z) find the function F o Q and its domain.

,

in the

(F.c)(x, y,z) :ii:##_r,

SOLUTION:

: sin-l tp+f+Za The domain of G is the set of all points (x, y, z) in Rs such that I * y, * zz - 4 > 0, and the domain of F is [-1, 1]. So the domain of F. G is the set of all points (x, A, z) in R3 such that 0
A.polynomial function of two variables r and y is a function such / that f (x, y) is the sum of terms of the form cxoy^, where c is a real number and n and rfl are nonnegative integers. The degreeof the porynomial function is determined by the largest sum of the exponents of x nd y appearing in any one terrn. Hence, the function / defined by f (x, y) : 6ruy' Sry" r 7*y - 2f + y is a polynomial function of degree 5.


A rational function of two variables is a function h such that h(x, y) : f(x,y)lg@, y), where f and,garetwopolynomialfunctions. Forexample, the function / defined by ,

x2ll2

l \ x , y ) : f f^ i' !E

is a rational function. The graph of a function / of a single variable consists of the set of points (x, y) in R2 for which y : f (x). Similarly, the graph of a function of two variables is a set of points in R3. 19.1.5Definition

If / is a function of two variables, then the graph of f is the set of all points. (x, y, z) in Rs for which (x, y) is a point in the domain of / and z : f(x, Y). Hence, the graph of a function f of.two variables is a surface which is the set of all points in three-dimensional spacewhose cartesiancoordinates are given by the ordered triples of real numbers (x, y , z). Because the domain of.f is a set of points in the xy plane, and because for each ordered pafu (x, y) in the domain of / there corresponds a unique value of.z, no line perpendicular to the xy plane can intersect the graph of / in more than one point. o rLLUsrRArroN6: The function of Illustration 1 is the function f which is the set of all ordered pairs of the form (P, z) such that

z:VFE=T So the graph of / is the hemisphere on and above the xy plane having a radius of 5 and its center at the origin. A sketch of the graph of this hemio sphere is shown in Fig. 19.1.5.

F i g u r e1 9 . 1. 5

4; Draw a sketch of EXAMPLE the graph of the function / having function valuesf (x, A) : x' + y'.

F i g u r e1 9 . 1

soLUrIoN: The graph of / is the surface having the equation z: f * y'. The trace of the surface in the xy plane is found by using the equation z = 0 simultaneously with the equation of the surface. We obtain f I y' :0, which is the origin. The traces in the xz and yz planes are found by using the equations y:0 and x:0, respectively,with the equation z: * * yz.We obtain the parabolasz: f andz: A2.The crosssectionof the surface in a plane z:k,Parallel to the xy plane, is a circle with its center on the z axis and radius Vt. wittr this information we have the required sketch shown in Fig. 19.1..6.

886


z:2 z-- 3 z-- 4 z:5 z:6

Figure 19.1,7

EXAMPLE5: Let f be the function for which f(x, y) - 8 - x2 - 2y. Draw a sketch of the graph of f and a contour map of f showing the level curyes of f at L0,g, 6, 4, 2, 0, -2, -4, -6, and -8.

There is another useful method of representing a function of two variables geometrically. It is a method similar to that of representing a threedimensional landscapeby a two-dimensional topographical map. Suppose that the surfacez: f (x, y) is intersected by the plane z: k, and the curve of intersection is projected onto the xy plane. This projected curve has f (x, y): k as an equation, and the curve is called the leaelcurae (ot contour curce) of the function / at k. Each point on the level curve corresponds to the unique point on the surfacewhich is k units above it if k is positive, or k units below it if k is negative. By considering different values for the constant k, we obtain a set of level curves. This set of curves is called a contour map. The set of all possible values of k is the range of the function /, and each level curve, f (x, V) : k, in the contour map consists of the points (x, y) in the domain of / having equal function values of k. For example, for the function f of Example 4, the level curves are circles with the center at the origin. The particular level curyes for z:7,2,9, 4,5, and6 are shown in Fig. 19.1,.7. A contour map shows the variation of z with x and y. The level curves are usually shown for values of z at constant intervals, and the values of z

solurroN: A sketch of the graph of is shown in Fig. 19.1.9.This is the f surfacez:8f -2y. The trace in the xy plane is obtained by setting z : 0, which gives the parabolax2: -2(V - 4). Settingy : 0 and r : 0, w6 obtain the traces in the xz and yz plarres,which are, respectively, the parabola rf : - (z - 8) and the line 2y * z: 8. The crosssection of the surface made by the plane z : k is a parabola having its vertex on the line 2y * z : 8 in the yz plane and opening to the left. The crosssectionsfor z : g, 6, 4,2, -2, -4, -6, and -8 are shown in the figure. The level curves of.f arc the parabolasf : -Z(y - 4++k). The con-


887

tour map of / with sketchesof the required level curves is shown in Fig. 1,9.1,.9.

x 8 1.0

F i g u r e1 9 . 1. 9 F i g u r e1 9 . 1. 8

Extending the notion of the graph of a function to a function of nvariables, we have the following definition. L9.1.6 Definition

If / is a function of n variables, then the graph of.f is the set of all points , xn)is a point in the (xr, xr, . . . , xn, ar) in Ro+l for which (xr, xr, domain of.f and w : f(xr, xz,' . ., xn). Analogous to level curves for a function of two variables is a similar situation for functions of three variables. If / is a function whose domain is a set of points in R3,then if k is a number in the range of /, the graph of the equation /(r , y , z) : k is a surface.This surfaceis called the leoelsurfaceof as a 1 it tc.Every sur{ace in three-dirnensional space can be considered level surface of some function of three variables. For example, if the function g is defined by the equation g(x, y, z) : f * y' z, then the surface shown in Fig. 19.1,.6is the level surface of g at 0. Similarly, the surface having equationz- f -y'+ 5:0 is the level surfaceof g at 5'

19.L Exercises (r* y)l(x- V)' l. Let thefunctionf of twovariables xandybethesetof allorderedpairsofthe form(P,z) suchthatz: t h e r a n g eo f f ' D t a w a (b)f(f ,y'); (c) lf(x,V)l'; (d)f(-x,y)-f(x,-V); G) the domain of.f;(f) F i n d : ( a )f ( 4 , $ ; sketch showing as a shaded region in R2 the set of points not in the domain of /. (P,w) suchthatar: 2 . L e t t h e f u n c t i o n g o f t h r e e v a r i a b l e sx , y , a n d z , b e t h e s e t o f a l l o r d e r e d p a i r s o ft h e f o r m

888


-x, -y); (d) the domain of g; (e) the range of g; w.Find:(a)s(l',-t,_t);(b)s?a,2b,*c);(c)g(y, (f) lgQ, U,z)l'- ISQ + 2, y * 2, z)l'. Drawa sketchshowingas a shaded solid in RBthe set of points in the domain of g. In Exercises3 through L1, find the domain and range of the function / and draw a sketch showing as a shaded region in R2 the set of points in the domain of./. 3.

f(x,a):ffi

5.f(x,a):rffi s .f ( x ,D : 1

lvl

4 .f ( * , A ) : x \ f r - x 2 - Y z

5 .f ( x ' a ) :

7 . ,f ( x , t t \ : U ' x-y

8 . f ( x ,a ) : ) c

ffi +Y xy

ll. f (x, y) : ln (xy- I)

10.f (x, A) : sin-l(x + Y)

In Exercises12 through 14, find the domain and range of the function f . 12. f (x, y, z) - (x * y) tF. 13. f (x, y, z): 1,4.f (x, y, z) - lxlsur'

sin-r x * cos-l y + tan-r z

In ExercisesL5 through 20, frnd the domain and range of the function / and draw a sketch of the graph. 15. f (x, A) : 4x2* 9yz 17.f (x, A) : 16- x'- A'

20. ::. * v f (r,n : [7 f.0 if.x:y

1 8 .f ( * , A ) :

In Exercises21 through 26, draw a sketch of a contour map of the function showing the level curves of at the given / / numbers. 21. The function of Exercise15 at 36, 25, 16,9, 4, l, and,0. 23. The function of Exercise 17 aL8, 4, 0, -4, and -8.

22. T]nefunction of Exercise16 at 5, 4, g,2,1, and 0. 24. The function of Exercise18 at 10, g, 5, 5, and 0.

25. The tunction / for which f(r, V) : t(* + y2) at g, 6, 4, 2, and 0. 2 6 . T h e f u n c t i o n / f o r w h i c hf ( x , y ) : ( r - 3 ) l ( V + 2 ) a t 4 , 2 , 1 , i , + , 0 , - + , - + ; , - t , - 2 , a n d . - 4 . In Exercises27 and 28, a function / and a function g are defined. Find, h(x, y) it h: f . g, and also find the domain of Iz. 2 7 .f ( t ) : s i n - r t ; S ( x , y ) - @ 28.f (t): tan-rt; S(x,y) - tffiz

2 e .G i v e n l ( x , ! ) : x - y , s G ) : t / l , n ? ) : s 2 . F i n d ( a ) ( 8f.) ( s , r ) ; ( b ) f ( h ( s ) , 5 ( q ) ; G ) f @ ( x ) , h ( y ) ) ; ( d ) S ( ( r , . f ) ( x ( e )( s . h ) ( f ( x , v ) ) . 30. G i v e n f ( x , y ) : x l y ' , s k ) : * , h ( x ) : 1 6 . r i n d ( a ) ( h " f ) ( z , r ) ; ( b ) f ( s ( z ) , h G ) ) ; ( c ) f ( s 6 / i ) , h ( r , ) ) ; ( d ) h ( ( S "f)(x,il\; ( e )( / t .i l ( f @ , v ) ) . 3 L . The electric potential at a point (x, y) of the ry plane is V volts and V: for V : 75, 12, 8, 4, l , t, and *.

+t {i=7=7.

Draw the equipotential curves

32. The temperatureat a point (x,y) of a flat metal plate is f degreesand f : 4* *2y2. Draw the isothermals for t:12, 8, 4, 1, and,0. 33. For the production of a certain commodity, if r is the number of machines used and y is the number of man-hours, the number of units of the commodity produced.is f (x, y) and f (x, y): 5ry. Such a function is called a production / function and the level curves oI f are called constant product curyes. Draw the constant product curves foi this function / at 30,24, 18,12,5, and0.

19.2 LIMITSOF FUNCTIONSOF MORE THAN ONE VARIABLE

889

In Exercises34 and 35, draw sketchesof the level surfacesof the function f at the given numbers. 3 5 .f ( x , ! , 2 ) : f * y " - 4 z a t 8 , 4 , 0 , - 4 , a n d - 8 . i l . f ( * , y , 2 ) : * * y ' I z 2a t 9 , 4 , 7 , a n d 0 .

I9.2 LIMITS OF FUNCTIONS OF MORE THAN ONE VARIABLE

In Rr the distance between two points is the absolute value of the difference of two real numbers. That is, lx - al is the distance between the points r and a. In R2 the distance between the two points P(x, y) and In R3 the distance between Po(xo,y6) is given av ffi. the two points P(x, y, z) and Po(xo,!o, zo) is given by In R" we define the distance between two points analogously as follows.

19.2.1Definition

lf.P(xr,x2t . . . ,xn) andA(a1,a2, . . ,an) aretwopointsinR',then the distance between P and A, denoted by llp All, is given by (1)

If in Rl we take P : x and A:

llr- nll: {G=W:

a, (L) becomes

Q)

lx- al

If in R?we take P : (x, y) and A:

ll(x,y)-(xo,yo)ll:ffi

(xo,A), $) becomes

(3)

And, if in Rs we take P : (x, A, z) and A: (xr, !0, zo),(1) becomes (4) - (xo, Ao, zr)ll: ll(x, y, z) llP All is read as "the distance between P and A-" lt is a nonnegative number. 19.2.2 Definition

lf A ]s a point in R" and r is a positive number, then the openball B(A; r) is defined to be the set of all points P in R' such that llP All < t'

19.2.3 Definition

If A is a point in R, and r is a positive number, then the closedball BIA; rl is defined to be the set of all points P in Rn such that lp a1; = '1'

a

f,- r

oPen ballB(a; r) in Rl F i g ur e 1 9 . 2 . 1

a*r

To illustrate these definitions, we show what they mean in R', R', and R3.First of all, if a is a point in Rl, then the open ballB(a; r) is the set of all points r in Rr such that (5) lx-al

a-r

a

a*r

closedball Bla; rl in Rl Figure19,2.2

a - r and,a * r. The closed ball Bla; rl in Rr (Fig.19.2.2) is the closed interval [a - r, a * rf . If (ro, ys) is a point in R2,then the open ball B( (xo, yr); r) is the set of all points (x, y) in R' such that

ll(*'y)

(xr,yo)ll

(5)

From (3) we see that (5) is equivalent to )'

So lhe open ball n(@p, yi; r) in R2 (Fig. tg.Z.g) consists of all points in the interior region bounded by the circle having its center at (xs, ys) and radius r. An open ball in R2 is sometimes called an open a*i*. rn6 closed ball, or closed disk, B[(re, yo); r] in R, (Fig. 19.2.4)is the set of all points. in the open ball B ( (ro, yo); r) and on the circle having its center at (xr, yi and radius r. If (xo, !0, zo) is a point in R3, then the open ball B((xo, Ao, zo);r) is the set of all points (x, y , z) in Rs such that

Figure

ll(x, y, z)

(xo,Ao,zo)ll< t

(7)

closed ball B [(ro ,Uo); r ] in R2 Figure 19,2.4

19.2.4Definition

open ball B((r o, Ao, zo); r) in R 3

closedball B[(r o,ao, zo); r] in R3

Figure 19.2,5

Figure 19.2.6

Let / be a function of. n vaiables which is defined on some open ball B(A; r),,except possibly at the point A itself. Then the limit o1 f @) as e approachesA is L, written as limf(P)-L P-A

(8)


891

if for any € ) 0, however small, there exists a 6 > 0 such that

lf(P)- Ll

(e)

If f is a function of one variable and if in the above definition we take A: a in Rr and P: r, then (8) becomes

: f'T/t') r and (9) becomes <6 l f ( x ) - L l < " w h e n e v e r0 < l t - r l So we see that the definition (2.1.1) of the limit of a function of one variable is a special caseof Definition 19'2'4' we now state the definition of the limit of a function of two variables. (re, ys) and It is the special caseof Definition 7g.z.4,whereA is the point P is the Point (x' Y). 19.2.5 Definition

Let f be a function of two variables which is defined on some open disk B((n, Yr); r), exceptpossibly at the point (16' y6) itself' Then

f (x' Y): L ,".,1t-T".r", ifforanye)0,howeversmall,thereexistsa6>0suchthat ( x o ,y o, f ( x o , v o ) )

( x o ,U o ,o )

Figure 19.2.7

lf (x,Y)

- Ll < € whenever 0

(10)

In words, Definition 19.2.5 states that the function values f(x, v) (rq' yq) if the approach a limit L as the point (r, y) approachesthe point arbibemade I, can and a'bsolutevalue of the diffeience between f(x, V) (xo' yo) but to close (!,y) sufficiently tt*ify small by taking the point about said is nothing 19.2.5 Definition (io, yJ. frote that in "lf"J," ""i valul at the point (xo,yJ; that is, it is not necessarythat the the fulnction to exist' function be defined at (xo, /o) in order for ,,,rli1.,*, f(x,y) in Fig' A geometric interpretation of Definition 19.2.5 is illustrated having surface ( (ro, of the B Ao\;r) 1g.2.7.1heportion above the open disk lie will axis z the on that f(x, U). ug";,io" ,': f(*, y) is shown] W" t"" in plane is (r, ry the y) in point the 2 and L + e whenever between f tnatf(x'A) this.is stating of way tfr" op"r, disk 8((16, /o); S). Another on th" z axiscan be."ttti.t"d toliebetweenL-e and L* e byrestrictiig ifr" point (r, y) in the xy planeto be in the open disk B((ro' /o);6)' that . rLLUsrRArroN1: We apply Definition l'9'2'5 to prove lim

(c,rr)-(r,s)

(2x * 3Y) = ll

> 0 such that We wish to show that for any € > 0 there exists a 6

VARIABLES OF FUNCTIONS OF SEVERAL DIFFERENTIAL CALCULUS

l ( 2 x + 3 1 ) - 1 1< 1 e w h e n e v eOr < V G = I I + f y - g ) z < 6 Applying the triangle inequality, we get

l 2 x + 3 -y 1 1 1 :l 2 x - 2 t 3 y - e l = z l x- 1 l+ s l V- 3 1 Because

and ly-gl - I/G=TTQ=T

lx- 1l - t/@ffiQ$ we cnn conclude that

<26+36 2lx-11+}lv-31 whenever

o < V{FT)ITIY=

a)z< a

So if we take 6 : *e, we have

l 2 x + 3 y- 1 1 1= z l x - L l + 3 l V - 3 1 < 5 6 : e whenever

o < V(I= r)zTly= s)z< a (2x+ 3y): tt. Thisproves,hu, ,",llT,r, ExAMPTE 1.: Prove that lim

(r,a)-(r,z)'

by

"pplying

(3x' * rv)' : 5 \ Definition 79.2.5.

o

sor,urroN: Wewish to show that foranye ) 0 thereexistsa6 > 0such that

l(3* +y) -51 < e wheneverO< V(7=lzTly-Z)t

Applyirg the triangle inequality, w€ get

-t *y -2]t= slx- lllr+ 1l+ lV-zl. l3* + y -51: lzxz

(11)

If we require.the 6, for which we are looking, to be lessthan or equalto 1.,thenlr- 1l < 6 < 7 andly- 2l
)c+'/., < 3. Hence,

3 l x - r l l r + 1 l+ l y - 2 | ,< 3 . 6 . 3 + 6 - 1 0 6

(12)

whenever

0<

(x- 1)

So if for any € and (12)

+ Q -z)', 0 we take E - m i n ( L , #e),

l3*'*y -51 < 100< € whenever 0 This proves that (Jx, * y) : 5. r;,lift,zt

then we have from (11)


893

We now introduce the concept of an "accumulation point," which we need in order to continue the discussion of limits of functions of two variables.

19.2.5Definition

(*,, + 1) (m-T,n*T)

I

| (*+T,n*I)

(m'' - 1)

F i g u r e1 9 . 2 . 8

A point Pois said to be an accumulationpoint of a set S of points in R' if every open ball B(P6;r) contains infinitely many points of S. o rLLUsrRArron2: If S is the set of all points in R2on the positive side of the r axis, the origin will be an accumulation point of S becauseno matter how small we take the value of r, every open disk having its center at the origin and radius r will contain infinitely'many points of S. This is an example of a set having an accumulation point for which the accumulation point is not a point of the set. Any point of this set S also will be an accumulation point of S. . I rLLUSrRArror3: If S is the set of all points in R2for which the cartesian coordinatesare positive integers,then this set has no accumulationpoint. This can be seen by considering the point (m, n), where m and n are positive integers. Then an open disk having its center at (m, n) and radius less than 1 will contain no points of S other than (m, n); therefore, Definition 19.2.6will not be satisfied.(seeFig. 19.2.8). r We now consider the limit of a function of two variables as a point (x, y) approaches a point (xo, Ui, where (x, il is restricted to a specific set of points.

19.2.7Definition

Let f b" a function defined on a set of points S in R2, and let ( x o , y ) b e an accumulationpoint of S. Then the limit of f(x, V) as (x, y) approaches (xr, U) in S is L, written as lim

(r,a)-(l,o,ud (p in s)

(13)

f(x,v)-L

if for any € lf (x,y)

- Ll < €

whenever 0

and (x, y) is in S.

A specialcaseof (13)occurswhen S is a setof points on a curvecontaining (xo, yo).In such casesthe limit in (1.3)becomesthe limit of a function of one variable.For example,consider,,,llTr,,f(x' a). Then if S, is the set of all points on the positive side of the r axis,we have l i m f ( x ,y ) : l i m f ( x , 0 ) r-o'

(r,r,)-(oo) (P inSr)


If 52 is the set of all points on the negative side of the y axis,

(x' v) : f (o'Y) ,",li,To ,orf lt-T (p in sz) If Ss is the set of all points on the x axis, lim

, (r,a)-(o,o) (p in ss)

f (x, A) : lim f (x, 0) r-o

If 54 is the set of all points on the parabola A : x2,

(*'a): f (x'x') ,,,1,1T, ,orf lg @ in s+) 79.2.8Theorem

Suppose that the function f is defined for all points on an open disk having its center at (xo, A) , exceptpossibly at (xo,y) itself, and

(*'Y) - L r",rliTo ,uorf Thenif Sis anysetof pointsin R2having (xo,yo)as an accumulationPoint, lim (r,g)-(co,go)

f(x,y)

exists and always has the value L.

PRooF: Because

li+

: L, then by Definition 19.2.5,for any . f (x, y)

U,i-bo,ur)'

e ) 0 there exists a 6 > 0 such that < e whenever 0 ( ll(x,y) - (xo,/o)ll< S The above will be true if we further restrict (x, y) by the reguirement that (x, y) be in a set S, where S is any set of points having (io, yo) as an accumulation point. Therefore, by Definition 1.9.2.7, lf(x,y)-Ll

lim

(t,u)-(to,uo) (p in s)

f(x,A):L

and L does not depend on the set S through which (x, y) is approaching (xo, Ao).This proves the theorem. I The following is an immediate consequenceof Theorem 19.2.g. 19.2.9 Theorem

If the function / has different limits as (x,y) approaches (ls, yo) through two distinct sets of points having (xr, yJ as an accumulation point, thett does not exist' t"''li["'" f(x'Y) PRooF: Let 5r and Sz be two distinct sets of points in Rz havin g (xr, yo)

19.2 LIMITSOF FUNCTIONSO F M O R ET H A N O N E V A R I A B L E

895

as an accumulationpoint and let lim

(r,u)-(ro,uo) (p in sr)

f (*, a) : Lt and

lim ( r ,d- ( to,uo) (p in sz)

f (x, y) : Lz

f (x,y) exists.Then by Theorem 19.2'8Lt must equal L", but by hypothesis Lr * L2, and so we have a contradiction. (x'y) does not exist' I Therefore' ,r,rl1T,,rorf

Now assumethat

ExAMPtn 2:

Given

xY f ( x , a' ) : x 2 * y ' lim

find

f (x, y) if it exists.

(r,a)-(o'o)-

lim

(r,y)-(ro,yo)

sol.urroN: Let S, be the set of all points on the r axis. Then lim

(r,u)-(o,o) (p in sr)

m f (x,0) : f ( x , y ) - lr i-o

0 t;9ffi-o r.

Let 52be the set of all points on the line y - r. Then lim (r,g) - (o'o) @ in sz)

f (x,y) :

Because

lim f(x,y) +

(r,s)*(o,o) (p in sr)

li*

(r'u)*(o'o)

f Q,y)

(p in sz)

it fotlows from Theorem1,9.2.9that ( r , ulim f (x , y) does not exist. )-(o,o)In the solution of Example 2, instead of taking s1 to be the set of all the points on the x axis, we could just as well have restricted the points of S, to be on the positive side of the r axis becausethe origin is an accumulation point of this set. EXAMPLE 3:

Given

f(x,Y):ffi find

lim ( r ' g ) - ( o , o)

f (x,y) if it exists-

solurroN:

Let 51 be the set of all points on the x axis. Then

f (*'Y): t"T',h - o ,,,11%,or (p in sr) Let Sz be the set of all points on any line through the any point (x, y) in Sz, A : mx. :lim lim f (x, a) x' + m'x': (r,s)-(o,o) '--o:{'}:^

r' tJT

that

for

3mx : o u Jt, ntz:

(p in sz)

Even though we obtain the same limit of 0,if. (x, y) approaches(0,0) through a set of points on any line through the origin, we..cannot conHowever' cludelhat f(*, y) exists and is zero (see Example4)' *,ll%., let us attempt to prove an",,,,llTo,orfG,V):0' FromDefinition 19'2'5'


if we can show that for any e ) 0 there exists a E > 0 such that

l%l

lx'*y'l

.\€\

' Yh r eneve w or< t W < D

then we have proved that

lim

f (x, A) : g.

Because w€have rp- xz* yr"fi'ilj''! W, 3(x'+U\W:_g | 3 * ' y^ 1: g l l v l W lry,l:m So if 6 - *e, we can conclude that

l+l

lf+vl

. € w h e n e voe
which is (1a).Hence,we have proved that.r,ll%

EXAMPTr' 4:

U) : g.

solurroN: Let Sr be the set of all points on either the r axis or the y axis. So if (r, y) is in Sr, rA:0. Therefore,

Given

' : f *:'! f (*,y\ y, , find

,orf(r,

(x'v) if it exists' ,,,liT,,orf

f(x' Y): o ,",}}1r, (P tnSr) Let Sebe the set of all points on any line through the origin; so if. (x, y) is a point in 52,y : mx.'We have, then,

f(x,y):L,T #: ,,,11T.,, (P in s2)

fi ffi : o

Let Ss be the set of all points on the parabola y:

liry

.fk, (t,a)-(o,o)' (p in ss)

f. Then

x4

A):lim =x: :e: lim*:* ;;x4+x4

;*o2-2

Because

(x'u)* , (r,s)-(o,o)' liq _f(x,y) ,,,llT, ,r,f (p in ss) (p in sr) it follols tha, (x,U) doesnot exist. ,,,1111 , rf soLUrIoN:

EXAMPLE

tf.x+A

f(x,a): find

if x - 0

(x,u) if it exists. ,",11% ,orf

Let Sr be the set of all points on the y axrs. Then

lim ,f(x,A):lim0:0

(a,g)-(o,o) (p in sr)

I-O

Let s2 be the set of all points on any line through the origin except points on the y axis; that is, it (x, y) is a poinf in 52,y: kt, where r * 0. Ttren


(r * kx) rir, 1 , liq, ,f (x, A) :lim X 1^- o

( a,g) - ( o,o) ' (P in ^gz)

To find the abovelimit we makeuseof the fact that lim (r * kr) : g. Becauselim sin(Ur) doesnot exist, we cannotupprytri"-lrreoremabout c-0

the limit of a product.However,because (r * kx) :0, it follows that LtT for any e ) 0 there existsa 6 > 0 suchthat l x * k x l < e w h e n e v e r0 < l r l < 6 The 6 is, in fact, elll + kl. But I

rl

I

rl

+ kr) sin:l : lx + kxl lsinjl = lr * kxl. L l(r lxllxl Hence, for any e ) 0 there exists a 6 > 0 such that |

1l

l(x+kx)sinjl
w h e n e v e r0 < l r l < 6

l4l

Thus, . l i - . f ( x ,A ) : 0

(ls)

(r,g)-(o,o) (p in sz)

Let 53 be the set of all points (x, y) for which U:k*, where n is any positive integer and,x * 0. Then by an argument similar to that used for proving (15) it follows that

f(x,il:lt1 ,",5T0," (P

(t *kxn)sin]:o

in,ca)

We now attempt to find a E > 0 for any € > 0 such that

lf(x,v) - ol which will prove that

(15)

lim

(c,a)-(o'o)

f (x, A) :0

We distinguish two cases:

a n dx + 0 .

r:0

CasL e : I f x - 0 , l f ( * , V ) - 0 l = l 0- 0 l

which is less than e for any

6>0. Cqse2: lf x

+ 0 , l f( * , y )- 0 l: l ( r + V) sin(Ur)l.

: V+vt lI ( x+ v ) sin*l 1""11 I


:zvf

+ y2

Then |

1l

+ v) sinjlx l < z .*e whenevero < lTf l@ l'

< te

So take 6: te. Therefore, in both caseswe have found a E > 0 for any e ) 0 such that lim f(x, y):0. that (16)holds, and this proves (c,g)-(o,o)'

The limit theorems of Chapter 2 and their proofs, with minor modifications, apply to functions of several variables. We use these theorems without restating them and their proofs. o rLLUsrRArrow4: By applying the limit theorems on sums and products, we have lim G,il-Gz,r)

(r'* 2*y - y' + 2):

(-2)' + 2(-2)'(1,) - (1)' * 2:7

o

Analogous to Theorem 2.6.5 fot functions of a single variable is the following theorem regarding the limit of a composite function of two variables. l9.2.l0 Theorem

If g is a function of two variables and

!i+

.g@, y):b,

and f is a

tunctionof a singlevariable continuous "gi?;tfi:"rl . l i + . ( f "d @ , y ) : f ( b ) (;rll)-(co,aol

or, equivalently,

fQ@, D) :/(",,1iT".,", s@,v)) ,",,IiT".,u The proof of this theorem is similar to the proof of Theorem 2.6.5 and is left as an exercise (see Exercise20). ExAMPtn6: Use Theorem to find lim ln (xy - 1). 19.2.10 (t,u) - ( 2,1)

soLUrIoN: Let g be the function such that g@, y) - ry - 1, and / be the function such that f(t):ln t. lim (w L\ :'t G,i-(z,i

-

and because/ is continuous at 1, we use Theorem 19.2.10and get

- t) : t"(,,.11T.,, (v - t)) rn(xv ",ll%,,, -lnL -0

19,2 LIMITSOF FUNCTIONSOF MORE THAN ONE VARIABLE

Exercises19.2 In ExercisesL through 6,establish the limit by findi.g a 6 > 0 for any e ) 0 so that Definition 1,9.2.5holds. 1. lim (3x - 4V) - 1 2. lim (5x- 3y) - -z 3. lim (x' * A') :2 (c,u)-(g,z)

4.

lim

(c,u)-(z.g)

(x,u)-(2,4\

(2x' - yz) : -L

5. li-

(r,g)-(1,1)

(x'*2x-y):4

5'

(t,g)-(2,4)

(x' + yz- 4x * 2y) : -+ ,r,jt-T,-r,

In Exercises7 through 12, prove that for the given function f ,,r,llTo,o,f (x,y) doesnot exist.

V1 7.f (x,A):.?x'+y'

8 .f ( x , a ) :xJ' + y '

xa*3*'y'*Zxyg

L 0 .f ( x ,A ) :

11.f (x,A): , =t'n * (x' + yn)t

(x' * y')'

L3through15,provethat,",llTo, @,U) exists. In Exercises orf * Yt :*t u. f(x,y) x'+ y'

13. f(x,a):ffi (,

1 5 .f ( x , a ) : l

L

L

( r +. y )\ s i n 2 s r n ' =r f x + 0 a n d y + 0 if eitherx:0 or y - 0

L0

r c . f @ , y ) : l x[f

i" (xy) \ '.

lv

if x+0 rf.x-0

In Exercises17 through 19, evaluatethe given limit by the use of limit theorems. 17.

lim (c,u)-(2$)

(3x' * xy - 2y')

18. lim v+m (c,a)-(-2,E)

lg.

e* * eu Iim (a,s)-(o,o) COSX * Sin y

20. Prove Theorem 19.2.10 In Exercises2L through 23, show the application of Theorem 19.2.L0to find the indicated limit. 21.

lim

u

tan-l 1

22.

X

(c,y)-(2,2'1

lim (t,s)-(-2,g)

[5r + ty'\

23.

tl r lim \'l (x,u\-(t,2, g* - +V

In Exercises24 through 29, determine if the indicated limit exists.

24. lim

(c,s)*(o,o>f

!'a' =

25. lim

!'a' , * Yn

.L sln x

i f x + 0-1h

ti" zs.f (x,a):{t i* '

.1 sln -

t f x + L a n d ' la * 0

L0

lim

sin(r + y)

(c,u)-(2,-2)

X+ y

22. tim

*: + Y=

(c,s)-(o,o)X' + y2

-l

f-*v- + 28f(x,a):wtv .

x

26.

lim

f(x,y) (e's)-(o'o) if x - 0J

if either x:0

I lim f(x,y) I (c'a)'(o'o) : or y o)

30. (a) Give a definition, similar to Definition 19.2.5, of. the limit of a function of three variables as a point (x, y, z) approaches a point (xo, yo, zo). (b) Give a definition, similar to Definition 19.2.7,of the limit of a function of three variables as a point (x, y, z) approachesa point (xo, yo, zs) in a specific set of points S in R3. 31. (a) State and prlove a theorem similar to Theorem 19.2.8f.or a function / of three variables. (b) State and prove a theorem similar to Theorem 19.2.9for a function f of three variables.

VARIABLES CALCULUS OFSEVERAL OF FUNCTIONS 9OO DIFFERENTIAL 30 and 31 to proveanu,,",r,ltTo,o @,A, z) does In Exercises32 through 35, use the definitions and theoremsof Exercises ,orf not exist. g2. f (x, y, z) '/ -f

* Yx,'* z2x2 f + Yn* zn

*+y2-22

33.f(x,v,"):FTFr,

g5. f(x,y,z):Fffi

u . f ( x , A z,' ') : , O . *! " f + Y'+zn

In Exercises36 and 37, use the definition in Exercise30(a)to provethat

f @, y, z) exists. ,,.r.1l1o.o.o,

96.f (x,A,z) : *r!l;r-( ,, ;): {e*y*z) L0

r i r ,1 r i r ,1 i f x + o a n d , y + o if eitherx :0 or y : 0

19.3 CONTINUITY OF FUNCTIONS OF MORE THAN ONE VARIABLE 19.3.1 Definition

We define continuity of a function of.n variables at a point in Rz

Suppose that / is a function of n variables and A is a point in R". Ttren / is said to be continuousat the point A if and only if the following three conditions are satisfied: (i) f(A) exists; (ii) lim /(P) exists; P.A (iii) lim f(P):f(A). If one or more of these three conditions fails to hold at the point A, then / is said to be discontinuousat A. Definition 2.5.1 of continuity of a function of one variable at a number a is a special caseof Definition 19.3.1. If / is a function of two variables, A is the point (re, ys), and.P is a point (r, y), then Definition 19.3.1becomes the following.

19.3.2 Definition

The function f of. two variables r and y is said to be continuousat the point (rs, ye) if and only if the following three conditions are satisfied: (1) f(xo, yo) exists; (ii) y) exists; lim f(x, (t,il-(so,ud

(iii) _ !i+ . f(x, y) : f (xo,y). (a,i-(h,sd' soLUrIoN: We check the three conditions of Definition 19.3.2 at the point (0, 0).

+ (0,0) - (0,0)

(i) f (0, 0) : Q.Therefore,condition (i) holds.

19.3 CONTINUITY OF FUNCTIONSOF MORE THAN ONE VARIABLE

determine If f is continuous at ( 0 ,0 ) .

(tt) f(r, v):,",llTo,o, ,,.11%,, ffi:0,

901

whichwasproved in Ex-

ample 3, Sec.19.2.

(ttt),".ll1, u): /(0,o). rf(x' So we conclude that / is continuous at (0, 0). EXAMPLE 2:

LCt

the function f

be defined by

f*Y f(*,a):|Tf Lo

solurroN: following.

Checking the conditions of Definition "1.9.3.2,we have the

(i) f (0, 0) - 0 and so condition (i) holds. ( i i ) W h e n ( x , y ) + ( 0 , 0 ) , f ( x , U ) : x y l ( x ' + y ' ) . I n Example 2, Sec. '!.9.2, + y') does not exist and so we showed that ,r,llT, ,rr*Al@' lim f (x,y) doesnot exist.Therefore,condition (ii) fails to hold.

if (x, y) + (0, 0) i f ( x , y ) - ( 0 ,0 )

Determine tf.f ts continuous at ( 0 ,0 ) .

(t,a)-(o,o)

We concludethat / is discontinuousat (0, 0).

If a function f of.two variables is discontinuous at the point (ro, yo) tim f(x, -y) exists, then / is said to have a remoaabledisconti(r,u)-(co,ui' nuity at (xs, !) becauseif f is redefined at (xo, /o) so that f(xo, y): . li* f (x, y), then / becomescontinuous at (xs, y). It the discontibut

(t,il-(co,uo)

nuity is not removable, it is called art essentialdiscontinuity. * yz), then g is discontinuous o rLLUsrRArron1: (a) If g(r, y):3x'yl(f at the origin becauseg(0, 0) is not defined. Howevet, in Example 3, Sec. * y'):0. Therefore, the discon19.2, we showed an"a or3*yl(xz ,,,11T0, tinuity is removable if g(0, 0) is defined to be 0. (Refer to Example 1.) (b) Let h(x, y):xyl(xz *y2). Then ft is discontinuous at the origin because h(0, 0) is not defined. In Example 2, Sec. 79.2, we showed that lim xyl(x2 * y) does not exist. Therefore, the discontinuity is essential. k,u)-(0,0)

(Refer to Example 2.)

o

The theorems about continuity for functions of a single variable can be extended to functions of two variables. 19.3.3 Theorem

If f andg are two functions which are continuous at the point (16,yo), then (i) / + g is continuous at (xo, Y); (ii) / - g is continuous at (xo, Y); (iir) /S is continuous at (ro, yo); (iv) fl7 is continuous at (ro, /o), provided that g(xo, yr) * 0. The proof of this theorem is analogous to the proof of the corresponding theorem (2.6.1)for functions of one variable, and hence it is omitted.


19.3.4 Theorem

A polynomial function of two variables is continuous at every point in R2. PRooF: Every polynomial function is the sum of products of the functions defined by f (x, y) : x, g@, y) : y, and h(x, y): c, where c is a real number. Becausef , g, and ft are continuous at every point in R2,the theorem follows by repeated applications of Theorem 19.3.3, parts (i) and (iii). I

19.3.5 Theorem

A rational function of two variables is continuous at every point in its domain. pRooF: A rational function is the quotient of two polynomial functions f and g which are continuous at every point in R2 by Theorem 1,9.9.4.If. (xo, y) is any point in the domain of flg, then g(ro, y) # 0; so by Theorem 19.3.3(iv)f lS is continuous there. I

EXAMPTT3: Let the functi on f be defined by

f (x, y) : [ r ' * y ' L0

ifx2*y, rfxz*y,

Discussthe continuity of f .What is the region of continuity of f?

solurroN: The function / is defined at all points in R2. Therefore, con_ dition (i) of Definition 19.3.2holds for every point (rs, y6). Consider the points (xr, yi if.xszI Ao2+ 1.If. xoz+ yoz( 1, then

(f + y') : xo2 * yo': f (xr,yo) f k' y): (",,liTo.,o) ,",,f1T,,u lf xoz* yo2) 1, then

f@'y) :,".rli2",,u0: 0: f(xo'yo) ,,.,11T".,u Thus, / is continuous at all points (xo, y) for which xoz+ yo2+ L. To determine the continaity of.f atpoints (re, yo) for which xoz+ yoz: L, we determttu tt,",rrti2" (*, y) exists and equals 1. ,uorf Let 51 be the set of all points (r, y) such that f * y, = 1. Then (x, a) :,r,rfi2o,ro,Q' + y') - xoz* yo'- 1 lim 'f ( r ,u) - ( to ,uo) (p in sr)

@ in sr)

Let s2 be the set of all points (x, y) such that x, + y, lim lim 0:0 f (x, A ) :

( t,a) - ( r o ,go) ' (p in sz)

(t,a)-(ro,go) (p in sz)

Because lim

( t,u\- ( co,ao) ' (p in sr)

f (x, y )

*.

liry ,f @,y)

( r ,a) - ( to,uo) ' @ in sz)

we conclude that

li+ . f (x, y) does not exist. Hence, / is discontinuous at-allpoints (xo,yi fo-rwhic! xo2+ yo2: L. Theregionof continuityof /consists of all points in the ry plane exiept those or, tl.r" circle rp * y, ! t. @,i-bo,ud

I

OF FUNCTIONSOF MORE THAN ONE VARIABLE 19.3 CONTINUITY

903

The function f of.n variables is said to be continuouson an open ball if. it is continuous at every point of the open ball.

19.3.6 Definition

As an illustration of the above definition, the function of Example 3 is continuous on every open disk that does not contain a point of the circlef * y':7. The following theorem states that a continuous function of a continuous function is continuous. It is analogous to Theorem 2'6'5' 19.3.7 Theorem

suppose that / is a function of a single variable and g is a function of two .r""ilbles. Suppose further that g is continuous at (ro, ye) and f is continuous at g(xo, yo). Then the composite function.f .8 is continuous at ( x o ,y l The proof of this theorem, which makes use of Theorem 19.2.70'is similar to tt proof of Theorem 2.6.6 and is left as an exetcise (gee Ex" ercise 7).

4 3 2 -5-4-3-Z-LL L 2 -1

3

4

5

x

-2 -3 -4 F i g u r e1 9 . 3 . 1

ExAMPrn 4:

Given that

f(x,u):ffi discuss the continuity of fand draw a sketch showing as a shaded region in R the region of continuity of f .

o rLLUsrRATroN2: Let h(x, Y) : ln (rY - 1) we discuss the continuity of.h.If g is the function defined by 8@,y): xa - 7, gis continuous at all points in R2.The natural logarithmic function numis continuous on its entire domain, which is the set of all positive for all is-continuous t,.f by defined f(t):.ln bers. So if / is the function f>0.Thenthefunctionhisthecompositefunctionf"gand'byTheorcmlg.3.T,iscontinuousatallpoints(x,y)inR2forwhichxy-L)0or' region of equivalently , xy > 1.. The shaded region of Fig. 19.3.1 is the ' continuity of h' soLUrIoN: This is the same function as the function G of lllustration 4 in sec. 19.1. we saw there that the domain of this function is the set of all and the points in the exterior region bounded by the circle rc *y',:25 -5 < < x 5' points on the r axis for which The function / is the quotient of the functions g and h for which The tunction g is a polynomial g@, y) : y and h(x, V): \/FW=E irnction and therefore is continuous everywhere. It follows from Theorem 19.3.7that h is continuous at all points in xP for which * * y2 > 25. Therefore, by Theorem 19.3.3(iv) we conclude that / is continuous at all :25' points in the exterior region bounded by the circle f * y' -5 < r < 5, that is, Now consider the points on the x axis for which -5 I a < 5. If Sr is the set of points on the line the points (a, 0) wherl x: a, (a, 0) is an accumulationpoint of Sr, but

f(x'Y):tu#x *.}T,r, (P tn Sr)

gO4

DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES

which does not exist becauseyl{ffi is not defined if lyl = \E-. Therefore,/ is discontinuousit *re points on the r axis for which -5 ( r < 5. The shadedregion of Fig. 19.3.2is the region of continuity of /.

Figure19.3.2

Exercises19.3 In Exercises 1 through 6, discuss the continuity of /.

t+

1. f (r, V) : ]{Ff t0 (x'y

2 .f ( x , a ) : | f f i f.0

3 .f ( x , a ) : t # L0

ir @,y)+ (o,o) if (x,y) : (0,0)

i. r ( x ' Y *) ( o o' ) if (x, y) : (0, 0)

(nrNr: SeeExample 4, Sec.19.2.)

i f @ , v+) ( oo, )

(x, y) + (0,0) ( x , y ) : ( 0 ,o )

if (x, y) : (0, 0)

(ry

ir (x'Y)+ (o'o)

L0

i f ( x , y ) : ( 0 ,0 )

5.f(x,a)={lffiT

lnrNr: SeeExercise13,Exercises19.2)

if (x, y) + (0, 0) if (x, y) : (0, 0)

7. ProveTheorem19.3.7. In Exercises 8 through 17, determine the region of continuity of I and draw a sketch showing as a shaded regron in R2the region of continuity of /.

L1..f ( x , a ) :

W

L3. f (x, A) : r ln (xV')

19.4PARTIAL DERIVATIVES 905 L4. f (x, A) : sin-l (xy)

15. f (x, A) : tan-r L + sec-r(xV)

(* - y' l-i f".-x' r* y 16.f(x,D:lx-y ifx:y Lx-y

rsin(x* u) ifx+y+O 1 t zj .. ft (_x ,"y.), :_t l, # x - t y ifx*y:g U

v

In Exercises18 through 27,provethat the function is discontinuousat the origin. Then determineif the discontinuityis removableor essential.If the discontinuityis removable,define/(0, 0) so that the discontinuityis removed. 1 9 f. ( x , A ) : ( x + y ) s i n l

v

2 1f.( x , A ) : Y

xu*yn

22. (a) Give a definition of continuity at a point for a function of three variables, similar to Definition 19.3.2.(b) State theorems for functions of three variables similar to Theorems 19.3.3and 79.3.7.(c) Define a polynomial function of three variables and a rational function of three variables. In Exercises23 through 25, use the definitions and theorems of Exercise22 to discuss the continuity of the given function.

23.f (x, A, z) :

24. f(x, y, z) : ln(36 - 4x2- y2 - 9z')

W Sxyz

(

2 5f.( x , y , z ) : ] W L0

if ( x , y , z ) # ( 0 , 0 , 0 ) if ( x , y , z ) : ( 0 , 0 , 0 )

I9.4 PARTIAL DERIVATIVES

19.4.1 Definition

' I xz-v'

25.f (x, A,z) - fr' * y' * z2 L0

if

( x , y , z ) + ( 0 ,0 , 0 )

if

(x,y,z)- (0,0,0)

We now discuss differentiation of real-valued functions of n variables. The discussion is reduced to the one-dimensional caseby treating a function of n variables as a function of one variable at a time and holding the others fixed. This leads to the concept of a "partial derivative." We first define the partial derivative of a function of two variables. Let / be a function of two variables, r and y. The partial deriaatitseof f with respect to x is that function, denoted by Dtf , such that its function value at any point (x, y) in the domain of / is given by (1) if this limit exists.Similarly, the partial deriaatiaeof f u)ith respectt o y r s that function, denoted by Drf , such that its function value at any point (x, y) in the domain of / is given by (2)

The process of finding a partial derivative is called partial differentiation.

FUNCTIONS OFSEVERAL VARIABLES Drf is read as "D sub t of f ," and this denotes the partial-derivative function. Drf (*, y) is read as '? sub L of.f of.x and yi' and this denotes the partial-derivative function value at the point (x, y). Other notations for the partial-derivative functionDrf arefr, f ,, and O/dr. Other notations for the partial-derivative function value D/(r, A) arc fr(x, A), f ,(x, y), and |f(x, y)lax. Similarly, other notations for Drf are fz, fa, and Aflay; other notations for Dd@, y) are fr(x, y), fu(x, y), and \f(x, V)laV.It z : f (x, A) , we can write 0zl0x for D rf (r, y) . A partial derivative cannot be thought of as a ratio of 0z md Er becauseneither of these symbols has a separate meaning. The notation dyldx can be regarded as the quotient of two differentials when y is a function of the single variable r, but there is not a similar interpretation for 0zl0x. SOLUTION:

find Drf (x,y) and Drf (x,y) by applying Definitio n'!.9.4.1.

-Drf(r,/) r/\.-'"' : lim f@

+ ax'Y.)- f(x' Y)

4* * d^r1|'-2(x+ ax)y* y, - (g* - 2ry-!n-

: ;;

Ar

An-o

: lim 3f*6x Ax+3(Ax)2-2ry:2y Lx+y2-3f+zry-yz Ax

Ar-o

: :

- 2u a'x 5xLx+ 3(4t!)2 ^tlll j lim (Gx+ Lx-2y)

Ao-0

:6x - 2A Drf(x,/) : lim f @'v + AY) f(x' v) aa-o

Ly

: tim 3* Zx(y+ AV)+ (y * Ay),- (gf - 2ry * yr) au-o

Ly _ r.^ 3* 2xy 2x A,! * y2'* 2y Ay + (Ay)z- 3* I 2xy - yz Au-o

: lim '

Ly

-2x LY + 2YLY + (LY)z

aa-o

Ly

: lim (-2r + 2y + Ay) Aa-O

:-2x * 2A If (xr, y)

is a particular point in the domain of /, then

(3)

19.4 PARTIALDERIVATIVES

if this limit exists, and (4)

if this limit exists. o rLLUSrRArroNL: We apply formula (3) to find Drf(3,-2) for the function / of Example 1.

D,f(3,-2):ll* At + ^x'-fI- f@'-2\ :,::, + L2+ 4 Lx * 4- 43 : lim 27+ LBAr * 3(Ax)'? Ax

Ar*o

:lim

(18+3 Lxt4)

A.r-0

:22 Altemate formulas to (3) and (4) for Drf (xo, ys) and Ozf(n,/o) given by

o are

(s) and (5)

if this limit exists. -2) for the function o rLLUSrRArrow2: We apply formula (5) to find Drf(3, f of Example 1.

D,f(3,-2):n^t@AFA x- 3 "'t


o rLLUsrRArrON3: In Example 1, we showed that Drf (x, V) :6x - 2y Therefore, Dtf(9,-2) :18 + 4 :22 This result agreeswith those of Illustrations 1 and 2.

o

To distinguish derivatives of functions of more than one variabre from derivatives of functions of one variable, we call the latter derivatives ordinary derioatiaes. comparing Definition 19.4.1with the definition of an ordinary derivative (3.3.1),we see that Drf (x, y) is the ordinary derivative of if is f f considered as a function of one variable x (i.e., y is held constant), and Drf (r, y) is the ordinary derivative of.f if/ is considered as a function of

ExAMPrn2: Given f(x, A) : 3x3 - 4x'y * 3xyz a Tx - By, find D r f( x , y ) a n d D r f( x , y ) .

soLUTroN: Considering f as a function of r and holding y constant, we have Dtf (x, A) :9*

- *ry + 3y2+ 7

Considering / as a function of y and holding r constant, we have Drf (x, A) : -4* * 5ry - 8

solurroN:

+ (0,o) : ( 0 ,0 )

(a)f '(0,

(b)f ,(x, 0).

(a) If y # 0, from (5) we have

f@' Y) - f(o' Y) /,(0, y)- lim r-0

19.4 PARTIALDERIVATIVES 909

If y:

0, we have

f /'(0,0):l-iT

:tTT-0

Because/, (0, y) - -y i f y + 0 a n d / ' ( 0 , 0 ) : 0 , - -y for all y. /t (O, y) (b) lf x * 0, from (5) we have

we can conclude that

' fr ( x ,o ) : t J T - lim A'0

: lim U-0

x3 x:2

If x:

0, we have

f ( 0 , y ) - f ( 0 , 0 )_ lim fr(O,o) - lim y-0 U-0 Becausef ,(x, o ) : r , i f x # 0 a n d fr(O, 0) :0, f r ( x , 0 ) - x f o r all x.

we can conclude that

Geometric interpretations of the partial derivatives of a function of two variables €uesimilar to those of a function of one variable. The graph of a function f of two variables is a surfacehaving equation z:f(x,U). If y is held constant (say,y : yo),then z : f (*,yo) is the equation of the trace of this surface in the plane y: yo. The curve can be represented by the two equations (a) x

a:ao

and z-f(x,y)

(7)

becausethe curve is the intersection of these two surfaces. Then Dtf(rq, /o) is the slope of the tangent line to the curve given by Eqs. (7) at the point P6(16,Ao,f (xo,yo)) in the plane A : Ao.In an analogous fashion, Drf (xs, ye) represents the slope of the tangent line to the curve having equations x:xo (b) x Figure19,4.1

and z:f(x,y)

at the point Po in the plane r:xe. Figure L9.4.Laand b shows the portions of the curves and the tangent lines.

910


EXAMPIE4: Find the slope of the tangent line to the curye of intersection of the surface

SOLUTION:

(2,2, \E).

-x

0z

z-+@withthe planey:2at

The required slope is the value of 0zl0x at the point

thepoint

( 2 , 2 ,\ E ) .

A partial derivative can be interpreted as a rate of change. Actually, every derivative is a measure of a rate of change. lf. is a r,ri.tio'of the f two variables r and y, the partial derivative oi1*iit respect to r at the point P1(ro, /o) gives the instantaneous rate of change, at ps, of f(x, y) per unit change in r (r alone varies and y is held fixed at ye). similariy, the partial derivative of / with respect to'y atpo gives the instantaneous rate of change,at Po,of f(x,y) per unit changein y. EXAMPTn5: According to the ideat gaslaw for a confined gars, if P pounds per square unit is the pressure, V cubic units is the volume, and T degrees is the temperature, w€ have the formula

PV: kT

(8)

where k is a constant of proportionality. Suppose that the volume of gas in a certain container is 100 in.3 and the temperature is 90o,andk:8. (a) Find the instantaneous rate of change of P per unit change in T if V remains fixed at 100. (b) Use the result of part (a) to approximate the change in the pressure if the temperature is increased to 92". (c) Find the instantaneous rate of change of V per unit change in P if T remains fixed at 90. (d) Suppose that the temperature is held constant. Use the result of part (c) to find the approximate change in the vol-

soLUrIoN: SubstitutingV:100, T:90, andk:8 in Eq. (g),we obtain P : 7.2. (a) Solving Eq. (8) for P when k : 8, we get

Therefore, -a: P -8 aTv When T:90 and V: l0o,lPl0T:0.08, which is the answerrequired. (b) From the result of part (a) when T is increased by z 1ind,v rep is 2(0.09) :0.16. We conclude, mains an approximate increase in {ixed) then, that if the temperature is increased from 90" to 92", the increase in the pressure is approximately 0.1,6lblin.z (c) Solving Eq. (8) for V when k: 8, we obtain

Therefore,

_av : AP

8T P2

WhenT-90andP-

t25

19.4 PARTIALDERIVATIVES 9 1 1

ume necessary to produce the same change in the pressure as obtained in part (b).

which is the instantaneous rate of change of V per unit change in P when T: 90 and P :7.2 if T remains fixed at 90. (d) If P is to be increased by 0.L6 and T is held fixed, then from the result of part (c) the change in V should be approximately (0.16) (-435) : -#. Hence, the volume should be decreasedby approximately # in.3 if the pressureis to be increasedfrom 7.2lblin.z to7.36lblin.'z. We now extend the concept of partial derivative variables.

19.4.2Definition

to functions of n

x) be a point in R', and let f be a function of the n Let P(xt, x2, . variables x1r x2t . . . , xn. Then the partial derivative of f with respect to *2t

.

.

o

i

*nl

lv'\-

1

I/Ltrr|'L

IIr

rV

sraY

rvL

/

I

xk is that function, denoted by D rcf, such that its function value at any point P in the domain of f ,t given by Duf@t, x2, lim

.

.

,Xn):

(xr,

(x1, X2, . .

Lxrc

A'ro-6

if this limit exists. In particular, If f is a function of the three variables r, y, and z, then the paftLalderivatives of f are given by D t f( x , U , z ) : l i m

f(x + L x , y , z ) - f ( x , y , z ) Ar

and

if these limits exist. EXAMPtn 6:

Given

-f (x, y, z) x'U * Yz' * zg verify that xf t(x, Y, z) * yfr(x,y, z) t ,ft(x,Y, z) : 3f(*,y,2).

soLUTroN:

Holding y and z constant, we get

f t(x, v, z) :2xY Hotding x artdz constant,we obtain fr(x,Y,z):f+z' Holding x and y constant,we get f"(x, y, z) :2Yz * 3z' Therefore, ) A ( * * z 2 *) z ( 2 y z * 3 z z ) x f { x , y , z )+ y f r ( x , y , z )+ z f r ( x , A , z ) : x ( 2 x 1 +

912


:' 2x2A* r'y * yz' * 2yz2* 3zg -3(x'y*yzr*zt) : 3f (x, y, z)

Exercises19.4 In ExercisesL through 6, apply Definition '!.g.4.'!, to find each of the partial derivatives. 1. f(x, A) :6x * 3y -7; Drf(x,y) Z . f ( x , A ) : 4 x z - 3 x y ;D r f ( x , y ) 3. f (x, A) : 3xy* 6x - Ar;Drf(x, y) 4 . f ( x , A ) : r y z S y * G ; D r f( x , y ) 5.

6.f (x,a):#,

f(x,a): ffi';Drf(x,y)

D,f(x,y)

In Exercises7 through 10, apply Definition 1,9.4.2to find each of the partial derivatives. 7 . f ( x , y , z ) : f y - 3 x y ,* 2 y z ;D r f ( x , y , z ) 8. f (x, y, z) : x2* 4y, * 9zz;Drf (x, y, z) 9. f(x, !, z, r, t) - xyr * yzt * yrt * zrt; Dnf@, z, r, t) !,

70.f (r, s, t, u, o, 7u): 3rzstI spa- 2tup2- tow * 3uutz; Dsfe, s, t, u, o, w) LL. Givenf (x, y) - x! - 9y2.Find Drf (2,1) by (a)apglyrng formula (3);(b) apptnng formula(S);(c)applyrngformula (t) and then replacingx and,y by 2 and 1, respectively. 12' For the tunction in Exercise11,find Drf (2,11by tul applyrr,g*formula (l); (b) applying formula(5);(c) appryingformula (2) and then replacingx and,y by-2'and't,i"#"tii"iy. "

llrtil::'.Hl?i[::?l#

j1fi,1i,'*fted

13.f(x, A) :4yt + ffir;

partialderivatives bvholdinsaltbutoneorthevariables constant andappry-

Drf (x,y)

L 4f.( x , A )=: &@, i ; D r f ( x , Y )

15.f @, Q) : sin 3d cos2g; Dzf@, O) 1 7 .z - s a t r :

0z

t" " v' uv

1,6.f (r, 0) : r2 cos0 - 2r tan 0; Drf (r, 0)

L8.r: e-ocos(0 + e); ! a0

21..f (x, A, z) : 4xyz * ln(2xyz); fs(x, y, z)

29.f (x, y, z) : suuz * tan-t!,

19.u - (x' + y, + rr1-t,r.4

dz

22. f(x, y, z):

2A.u - tan-r(xyzut), #

stu sinh 2z- eru cosh2z;fr(x, A, z)

fr(x,y, z)

2 4 -f ( r , 0 , 0 ) : 4 r z s i n 0 * s e 'c o s0 s i nr f - 2 c o s @ ;f r ( r , 0 ,o ) In Exercises25 and 25, find fr(x,y) and fu(x,y). 25. f(x, A) :

fu

J, l n

s i nt d t

27.Givenu -sin * h'-. v"nfyt +, | ff #:

26.f (x, A) : fu e"*t dt Jr

o.

2 8 . G i v e n r p - * y + y , z I z z x . V e r i f yU + @ , d u )- , "y AxWnE:(r+y*z),

AND THE TOTAL DIFFERENTIAL 913 19.5 DIFFERENTIABILITY

It+ u' if (x'Y)+ (o'o) :1ffi (x, 29.Given f u) rf (x, y) : (0, 0) L0 Find (a\ f'(0, 0); (b)fr(O,0).

^r (\"x,y)+ ( 0,o) , \ :1 l-,' +t-yl if 30. Givenf(x, A) if (x, y) _ (0, 0) L0 Find(a)f'(0,y) if y * 0; (b)/,(0, 0).

31. For the tunction of Exercise30 find (a) fr(x,0) if r + 0; (b) fr(O,0). - 9yz + 4z2* 36: 0 with the plane 32. Find the slope of the tangent line to the curve of intersection of the surface 36f -3). t/tZ, partial (t, as a derivative. this slope Interpret x:1 at the point 33. Find the slope of the tangentline to the curve of intersection of the surfacez: * * y2 with the planey: (2, 1,5). Draw a sketch. Interpret this slope as a partial derivative.

l atthepoint

34. Find equations of the tangent line to the curve of intersection of the surface * * y'* z2:9 with the plane y: 2 at the point (1,2,2). - 4y2. lf. distance is measured in feet, 35. The temperature at any point (r, y) of a flat plate is T degrees and 7 : S+ W along the plate in the directions of the find the rate of change o1 the temperature with respect to the distance moved (3, 1). point positive r and y axes, respectively, at the 36. Use the ideal gas law for a confined gas (seeExample 5) to show that

v-_ .{. e!_:-r aT dP dV g7.lf.V dollars is the present value of an ordinary annuity of equal payments of $100 Per year for f years at an interest rate of 100i percent Per Year, then

11-(1+t)-'l V : 1 0 0l -

til result of part (a) (a) Find the instantaneous rate of change of V per unit change in i if f remains fixed at 8' (b) Use the time remains to 7vo and the 6vo from changes rate interest if the io find the approximate change in the lresent .raloe (d) Use the at 0'06'' fixed remains if I in f change per unit oIV fixed at 8 years. (c) Find the instantaneous rate of change and T.years 8 to from decreased is time the if value present tlie in change result of part (c) to find the approximate the interest rate remains faed at 6Vo.

19.5 DIFFERENTIABILITY AND THE TOTAL DIFFERENTIAL

In Sec.3.6 in the proof of the chain rule we showed that if I is a differentiable function Of the single variable r and y : f (x), then the increment Ay of the dependent variable can be expressedas Ay: f' (r) Ar * q Lx -+ 0' where 4 is a function of Ar and q -+ 0 as Ax From the above it follows that if the function / is differentiable at ro, the increment of.f at ro, denoted by Lf (xs)' is given by A / ( x o )- f ' ( x o ) A r * n L x

where r: O. ligo

For functions of two or more variables we use an equation colTesponding to Eq. (1) to define differentiability of.a function. And from the almition we determine criteria for a function to be differentiable at a

I

914


poinl we give the details for a function of two variables and begin by defining the increment of such a function. 19.5.1Definition If /isafunctionof twovariablesxand !,thentheincrementof fatthe point (ro, ye), denoted by Lf(xs, AJ, is given by (2)

Figure 19.5.1illustratesEq. (2) for a function which is continuous on an open disk containingthe points (xo, yo) and (16* Lx,ys+ Ay). The figure shows a portion of the surface,J f(*, y). Af(xo,iJ : ffn, is (16 the point * M,yo!.Ly, f(xo,'y))-ana'i isihe point lvhere^Q ( x s * L x , A o *N A , f @ o +A x , y o + A y ) ) . z

(xo,yo,f(xo,VJ)

z: f(x,y) \

(ro * ax,uo* Lv,f@o* ax,ys+ av))

+ Ay)

o)

F i g u r e1 9 . 5 . 1

o rLLUsrBArrow1: For the function defined by /

f(x, y) :3x - xaz we find the increment of at any point (xn, ys). / Lf (xo,Ai : f (xr* M, yo+ Ail f Go, yo) :3(ro * Ar) - (ro * Lx)(yo+ Ay), _ (3ro _ xoAoz) : 316* 3 Ar - xeiloz_ AozLx _ 2*oyoAy _ 2yoAx AV - xo(Ay), - Ax(Ay)z _ 3xs * x6ysz : 3 Ax Ao2Ax ZxoyoLy 2ys Ax Ay _ xs(Ay), _ ArlAyy, . 19'5'2 Definition

rf f is afunction of two variables x and y and the increment of f at (xo, yo) can be written as Af (xo, y) : Drf(xo,Ui Ax * Drf (xs, A) Ly * e1Ar + ez Ly.

(3)


where e1and e2are functions of Ar and Ay such that e1-+ 0 and €z+ 0 as (M, Ly) - (0,0), then/is said tobe differentiable at (xs,ys). . rLLUsrRArrou 2: We use Definition 19.5.2 to Prove that the function of Illustration 1 is differentiable at all points in R2. We must show that for all points (xo, yo) in R2 we can find an e1 and an e2 such that

Af(xo,y) - Dtf(xo,U) Lx - D"f(xo,y) Ly: e1Ar I e2Ay and e1-+ 0 and €z+ 0 as (Ar, Ay) -+ (0,0). Because f(x, y):3x- rA2, Drf(xo,y) :3 - voz and D"f(xo,ao): -2xoyo Using thesevaluesand the valueof Lf (xo'yd from IllustrationL, we have Lx-Drf (xo,yi AA:-xo(N)'-2ys Lx LV- Lx(LV)' Lf (xo,yo)-Dtf(xo,Uo) The right sideof the aboveequationcanbe written in the followingways: l-2yo Ay - (AV)'l Ar * (-ro Ay) LV or (-2y, Ly) Ar * (-Ar Ly - xo LY) LV or [-(Ay)'] Lx* (-2ysAr-re Ly) LV or 0'Ax*

f - Z y oA r - L x L y xs Lyl Ly

So we have four possible pairs of values for €1 efld €21 €1: -2yo Ly - (Lv)' and €2: -xo LA or €r : -2yo Ly

and

G2: -Lx Ly - xo LU

€1: - (Ly)'

and

€z: -2Yo Ax - xo LA

€r:0

and

€z: -2Yo Lx - Lx LY - xo LU

or or For each paut,we see that lim

(Lr,au)-(o,o)

€r:0

and

lim

(ar,Lu)-(o,o)

€z:0

It should be noted that it is only necessary to find one pair of values o for e1 and e2.

L9.5.3 Theorem

If a function f of two variables is differentiable at a point, it is continuous at that Point.


pRooF: If / is differentiable at the point (ro, a), it follows from Definition 19.5.2that f(xo+ Ax, yo+ Ay) - f (xo,yi : Drf(xo,y) Lx * Drf(xo,y) Ly *erAri-e2Ly where €r + 0 and e2+ 0 as (Lx, Ly) -t (0,0). Therefore, f (xo+ Ax, Ao+ Ay) : f (xr, ys) * Dlf @s,U) Ax * Drf (xo,y) Ay *erAr*ezLA

Taking the limit on both sides of the above as (Lx, Ly) -

(0, 0), we obtain

l i m . , f ( x o * A x ,A o * A y ) : f ( x o , A o )

(4)

(Ar,ag)-(o,o)'

If we let xo* Ax:, -?n!yoi- AA:y,,'(Ax, Ay) (0,0)" is equivalent to "(x, y) t (xo, yo)." Thus,we have from (4)

y) : ,,,,1\T,*rf(x' f(x" yo) which proves that / is continuous at (xs, ys).

r

Theorem 19.5.3 states that for a function of two variables difierentiability implies continuity. However, the mere existence of the partial deriv.ativr D'! Drf at a point does not imply differentiability at that -ya point. The following example illustrates this. ExAMpLE 1.: Given

f(x,y)

SOLUTION:

-

f - ( 0 ' 0 )- t i m 0 - 0 0

if (x, y) # (0,0)

D r f ( 0 , 0 -) t i m f k ' 0 )

if (x, y) : (0,0)

(0,a'r): D,f ; t*' t, l,l''g: ; &],

prove that Dl(0, 0) and Dd@, 0) exist but that f is not differentiable at (0, 0).

u'0

Y-0

u-0 Y

\

Therefore,both Drl(0, 0) and Drf(0,0) exist. In Example2 of Sec.L9.2we showedthat for this function lim f(x, -a) (.r,g)-(o,o)'

does not exist; hence,/ is not continuous at (0,0). Because/is not con_ tinuous at (0, 0), it follows from Theorem 19.5.3that is not differenti/ able there. ... P"f9f stating a theorem that gives conditions for which a tunction will be differentiable at a point, we Jonsider a theorem needed in its proof. It is the mean:value t*reoremfor a function of a single variaule appli"d to function of two variables. "


19.5.4 Theorem

Let / be a function of two variables defined for all r in the closed interval la, bl and all y in the closedinterval lc, d). (i) If.D1f@,/o) existsfor some yoinlc, d] and for all xinfa, b], then there is a number f, in the open interval (a,b) such that

f(b,y) - f(a,AJ: (b- a)D'.f(t',A)

(s

(ii) lt Drf (xo,y) existsfor somexoin la, bl and for all y in lc , dl , then there is a number f, in the open intewal (c, d) such that (5) : f (xo,d) f (xo,c) (d c)Drf(xo, tr) Beforeproving this theorem, we interpret it geometrically. For part (i) refer to fig. tS.S.i, which shows the portion of the surface z:f(x,y) above the rectangular region in the xy plane bounded by the lines x: a, x: b,y -- c, and y : d: The plane y : yo intersects the surface in the curve represlnted by the two equations y: /o and z: f(x, y). The.slopeof the line through the points A(a, yo,f@, y)) and B(b, Ao,f(b, /o)) is

f (b,v)

- f(a' Yo)

b-a Theorem 19.5.4(i)statesthat there is some point (tt, Ao,fGr, yJ ) on the curve between the points A and B where the tangent line is parallel to the secantline through A and B; that is, there is some number (1in (a, b) such - a), and this is illustratedin that D/(f1, yJ: lf (b, y) - f(a, y)ll(b the figure, for which Drf (€t, /o) < 0. Figure 19.5.3illustrates part (ii) of Theorem 19.5.4.The plane x: xo interseits the surface,: f (x, y) in the curve rePresentedby the two equaslope of the line through the points tions r:xo ;lfld z:f(x,y).The D(*o, d, f (ro, d))

z:

(tr,ao,fGr,Yo))

f(o,y)

-

(ro, tr, f(xo' tr)) -

f(b,vo)

z: f(x,Y B(b,Ao, f (b, yo))

Figurc 19.5.2

f(x,Y)

F i g u r e1 9 . 5 . 3

f ( x o ,c )

918


C(xo,c,f (xo,c)) and D (xs,d,f (xo,d) ) is [f (xo,d) - (xo,c)]tG- c), and f Theorem 19.5.4(ii) states that there is some point (xo, tr, f(A, €")) on the curve between the points c and D where the tangent tine is parallel to the-secantline through c and D; that is, there is som-enumber in {, 1c,d,1 such that Drf (xo, €r) : lf (xo,d) - f (xo,c)llG - c). PRooFoF THEoREM19.5.4(i): Let g be the function of one variable r defined by g(r): f(x,yo) Then 8'(r) : Drf(x, yo) Because-Daf (x, y) exists for arl r in fa, bl, itfoflows thatg'(r) exists for alr x in fa, bf, and therefore g is coniinuous on [a, bl. so u]' ir," mean-varue theorem (4.7.2) for ordinary derivatives there exiits a number (, in (a, b) such that

8'(f') or, equivalently, (b' Yo)- f (o' Yo) Drf (€r,Ao)- f a-a from which we obtain f(b, yr) - f(a, Ao): (b - a)Dl(€r, Ao) The proof of part (ii) is similar to the proof of part (i) and is left as an exercise(seeExercise11). I Equation (5) can be written in the form

f(xo+ h, Ui - fko, A) : yDJ(€r, Ur)

(7)

where f1 is befween xe and xs * h and h is either positive or negative (see Exercise 12). Equation (6) can be written in the form f (xo, yo+ I() - f(xo, yr) : lDrf (xo, €r) (8) is between yo and ao * 11andk is either positive or negative (see Yh"r" f, ExerciseL3). EXAMPIr 2:

f (x, y)

Given

SOLUTION:

By Theorem "l'9-5.4,there is a number f1 in the open interval

(2, 5) such that

f (s'4) - f (2,4) : (s- z)Dl(€,,4)

ANDTHETOTALDIFFERENTIAL9 1 9 19.5DIFFERENTIABILITY find a fr reguired by Theorem 19.5.4if x is in 12,5l a n d y : 4 .

So Fr6,^24 c5:J

(3+fJ

972

5-re-|F (3+ t)':40 Therefore,

3+ fr: +2\m sign and obtain

But because2

€r:z\rc - 3

The following theorem states that a function having continuous partial derivatives at a point is necessarily differentiable at the point. 19.5.5Theorem

Let/be a function of two variablesx andy. SupposethatDtf NrdDrf exist on an open disk B (Po;r), where P6is the point (16,y6). Then if Dtf andD2f are continuous at Ps,f is differentiable at Pq. pRooFr Choose the point (xs* Lx, yo* Ly) so that it is in B(Po; r). Then Lf(xo, y) : f(x, * Lx, ys + LV) - f(xo, yo) Subtractingand adding f (xo+ Lx, lJ to the right side of the above equation, we get Lf (xo,y) : lf (xoi- Lx, ys+ LV) - f (xo* A,x,yr)f

, + l f ( x o* A r, A ) - f ( x o a)l

(e)

Because Drf and Drf exist on B(Po; r) and (ro * Lx, Ao* Ly) is in B (Po; r) , rt follows from (8) that f ( x o * A , x ,A o * A y ) f ( x o * L x , y ) :

( A V ) D r f( x 0+ L x , t r )

(10)

where fr is between lo and Ao* Ly. From (7) it follows that f (xo* Lx, ys) f (xo,Ao) (Ar)D'f (€', Ao)

(11)

where fr is betwe€n .rs and xo * A,x. Substituting from (10) and (11) in (9), w€ obtain L f ( x o , A o )- ( L y ) D r f ( r o * L x , t )

+ (Ar)D'f (€r,Uo)

(12)


Because(ro * Lx,Ao* Ly) is in B(Po;r),€zisbetweenyoandAo* Ly, and Drf is continuous at P0,it follows that lim.-

,Drf (Ar,Au)-(o,o)

(xo* Lx, (r) : Drf (xo, Uo)

(13)

and, becausef, is between xs and xo* Lx and Drf rs continuous at Po, it follows that l i m . , D r-'f ( h , A ) (Ar,As)- (o,o)

: D r f( x o , A o )

(14)

If we let er: Drf(tv y) - D,,f(xo,yr)

1rs)

it follows from Eq. (14)that lim €r -' :0 ta',lir[to,ot

(16)

and if we let e2: Dd(xs * Ax, {r) - Drf (xr, yJ

Gz)

it follows from Eq. (13) that (18)

,. li+.-..G2:0 (AJ',au)-(o,o) Substituting from Eqs. (15) and (12) into (12), we get Lf (n, yd : LylDd@o, yi * erl * AxlDi@s, yo) * erl from which we obtain Lf (xo, yi : Drf (xo, Ao)Lx * D2fks, y) Ly * e1Ar * e2Ay

(19)

From Eqs. (15), (18), and (19) we see that Definition 19.5.2holds; so is differentiable at (xo, Ai. / I A function satisfying the hypothesis of Theorem 19.5.5is said to be continuouslydiferentiable at the point po. ExAMpu 3: Given f :x2u2 | -r

.t ir(x,y)*

f(x,A):1r'*y'-if (x,y): t 0

(0,0) (0,0)

use Theorem 1.9.5.5to prove that f is differentiable at (0, 0).

solurroN: To find Drf we consider two cases: (x, (x,y) * (0,0). If (x,y): (0,0), w€ have Dl(0,0) -lim

f(x,O)-f(0,0)

: t T, .; - 00 - 0

(0, 0) and

A

lf. (x, y) * (0, 0) , f (x, y) : *yrl(* * yr). To find Drf (x, y) we can use the theorem for the ordinary derivative of a quotient and consider y as a constant. We have Drf(*, r, -2rY'z(xz !!'z) :-?x(fY2) (* * y')'


2xyn

:

1x4 11'

The function Dtf is therefore defined by

2*Yn | D r f( x , A ) : l ( * ' * y ' ) ' Lo

+ ( 0,0) if ( x , y) - ( 0,o)

rir^ (x, y)

In the same manner we obtain the function Drf defined by

, t , ry:^, ir(x,y)+ (o,o)

Drf (x, y) : ](x2 - y t

L0

1f.(x, y) : 10,0)

Both D/ and Drf exist on every open disk having its center at the origin. It remains to show that Dtf and D2f are continuous at (0, 0). BecauseDrf(O,0) :0, Drf will be continuousat (0,0) if lim

D'f(x, U):O

(e,s)-(oo)

Therefore, we must show that for any e ) 0 there exists a 6 > 0 such that | |

'r*ol4 ^nJ

I

l(* *'r1'l

. A o ' e whenever

- ^/-

- , l , l y-n, = n W f W ) ^ l, ?*!n (x'+Y')' lry| " r r l @'* y')':

(20)

- 2\W

Therefore, | I

7rt14 tnl

|

l(x'+fuI "u

o whenever

So if we take 6 : !e, we have (20).Hence, Dt;f is continuous at (0,0)' In the same way we show that D2f is continuous at (0' 0). It follows from Theorem 19.5.5that f is differentiable at (0,0). If we refer back to Eq. (3), the expression involving the first two terms on the right side, Dgf(ro, A) Lx * D2f(xs, yr) NA,is called the principal part of.Lf (xs, yq) or ttre "total differential" of the function/ at (n, y). We make this as a formal definition' 19.5.5Definition lf f isafunctionof twovariables xandy,and/isdifferentiableat(x,y), ttren the total differential of f is the function df having function values given by

(2r) Note that df is a function of the four variables x, y, Lx,and Ly.If


z: f(x, A), we sometimes write dz in place oI df(x, y, Lx, Ay), and then Eq. (21) is written as (22) If in particularf (x, y) : x, then z : x, D 1f(x, y) : 1, and D2f(x, y) : 0, and so Eq. Q2) gives dz: Ar. Becausez: /, we have for this function dx: Ax.In a similar fashion,if we take/( x,U) : y, then z: U,Drf (x,y) : 0, and Drf (x,A) : l, and so Eq. Q2) givesdz: Ay. Becausez: y, we have for this function dy : Ay. Hence, we define the differentials of the independent variables as dx: Ar and dy : Ly. Then Eq. e2) can be written as d z - D r f( x , y ) d x * D r f( x , y ) d y and at the point (xo, Uo), we have dz - Drf (xo, A) dx * Drf (xo, U) dy In Eq. (3),letting Lz: Lf (xo,U), dx: Lx, and dy:

(23)

(24)

Ly, we have : Az Drf (xo, Ao)dx + Ddjco, Ao)dy + e1dx * e2dy (25) Comparing Eqs. @a) and (25), we see that when dx (i.e., Ax) and dy (i.e., Ay) are close to zero, and becausethen e, and eralso will be close to zero, we can conclude that dz is an approximation to Az. Becausedz is often easier to calculate than az, we make use of the fact that ilz - Az incertain situations. Before showing this in an example, we write Eq. (23)with the notation 0zl0x and 0zlOyinsteadoLDlf(d, y) and Drf(x, yl, respectively: (26)

ExAMPrn 4: A closed metal can in the shape of a right-circular rylinder is to have an inside height of 5 in., €ln inside radius of 2 in., and a thickness of 0.1.in. If the cost of the metal to be used is L0 cents per in.t, find by differentials the approximate cost of the metal to be used in the manufacture of the can.

solurroN: The formula for the volume of a right-circular cylinder, where the volume is V in.8, the radius is r in., and the height is h in., is V - nrzh

(zz7

The exact volume of metal in the can is the difference between the volumes of two right-circular rylinders for which r : 2.'!.,h: 6.2, and,r : 2, h: 5, respectively. AV would give us the exact volume of metal, but becausewe only want an approximate value, we find dV instead. Using (26), we have

dv: ff a,+{, an

(28)

From Eq. (27) it follows that av _ ^_-.t- _,_r av and ;:2nrh ffi:

q, Trrz

t


Substituting these values into Eq. (28) gives dV :Znrh dr * nrz dh B e c a u s er - 2 , h : 6 , d r 0 . L , a n d d h : 0 . 2 ,

w€ have

dV :2n(2) (5)(0.1)* n(2)'(0.2)

- 3.2Tr Hence, AV : 3.2rr, and so there is approximately 3'2rr in'8 of metal in the ' : can. Becausethe cost of the metal is 10 centsper in.3and 1.0 3.2tt:32tr manufacture the in used 100.53,the approximate cost of the metal to be of the can is $1.

we conclude this section by extending the concepts of differentiabilitv and the total differential to a function of n variables. 19.5.7 Definition

L9.5.g Definition

point If l is a function of the n variables x1'11z" " ' xn' antdP is the by given P is at of (xi, xr, . . . 'r'), then the increment f (29) f(P) Af(P) : f(x'+ Lx1,ir* Lx2,. . - ' ?n* Lx) If f is a functio_n of the n variables x1, x2, as f at the Point P can be written

. . , xn, and the increment of

' . . +D"f(P) Axn Lf (P) : D'f (P) Ar, + Drf(P) Lx2* * er Lxt * e z L x 2 * " ' + e n L x n

w h e r e € 1+

(30)

0r€z-)0, . . ' ,€n--+Q'a;s

(Arr,Lx2,...

,Lxr) +(01 0,

"

,0)

then f ," said to be differentiableat P '

Analogously to Theorem 19.5.5,it can be proved that sufficient conditions foia functio n f of.n variables to be difflrentiab-le at a point P are that Df , Drf , . . . , bnf all exist on ar-ropen ball B(P; r) and that D/' Dzf, .-. . ,-'Dnf are all continuous at P. As was the case for functions of" two variabies, it follows that for functions of n vatiables differentiability implies continuity. Howevet, the existence of the partial derivatives Drf, Drf, . . ., Dof ata point does not imply differentiability of the function at the Point. 1.9.5.9Definition

If f isafunctionof thenvariables xy,x2,. . . 'xnarrd/isdifferentiable values at,P, then the total difierential of / is the function df havtng fu nction. given bY df(P,Lxr,Lx2,""Aro):Dl(P)Lx11-D2f@)Mz*"'+DJ(P)Ar"(31)

92/T


Letting w: f (xr, x2, . . . , xn), definingdx, : Lxt, dxr: L,x2, . dxn: Lx, and using the notation 6w| 0x; instead of D1f(P), w€ can Eq. (31)as (32)

EXAMPTn5: The dimensions of a box are measured to be 1o ln., 12 in., and 15 in., and the measurements are correct to 0.02 in. Find approximately the greatest error if the volume of the box is calculated from the given measurements. Also find the approximate percent error.

solurroN: Letting V in.3 be the volume of a box whose dimensions are tr in., y in., and z in., we have the formula

xyz The exactvalue of the error would be found from LV; however, we use dV as an approximation to LV. Using Eq. (32) for three independent variables,we have

av,,av da*{a, dx dy v d z

dV:+- dxt+ and so

dV:yzdx*xzdy+xy

dz

(33)

From the given information lArl < 0.02,lAyl < 0.02, and lArl = 0.02.To find the greatest error in the volume we take the greatest error in the measurementsof the three dimensions. So taking dx:0.02, dy:0.02, dz: 0.02,and r : 10,! : 12,z:1.5, we have from Eq. (33) 6y: (12)(1s) (0.02)+ (10)(15)(0.02)+ (10)(12)(0.02) -9 so, AV - 9, and therefore the greatest possible error in the calculation of the volume from the given measurements is approximately 9 ins. The relative error is found by dividing the error by the actual value. Hence, the relative error in computing the volume from the given measurementsis LVIV;= ilVlV:r*o:zfo:0.005. So the approximatepercent error is 0.57o.

Exercises 19.5 1'. If f(x, A) : 3x2* 2xy - A', Ar - 0.03,and Ay : -0.02, find (a) the increment of ar ( 1 , 4) and (b) the total differential f off at (1,4). 2. It f(x, y) : xye"u, Ax: -0.1, and Ay : 0.2, find (a) the increment oI at (2,-4) f (2,-4). 3. If f(1, y, z): xy *ln(yz), Ax:0.A2, Ay:0.04, and &:-0.03, total differential of f at (4, 7, S).

and (b) the total differential of f at

find (a) the incrementof / at (4, I , 5) and (b) the


4. If f (x, y, z) - x'y * Zxyz- z.3,Lx: 0;0L, Ay: 0.03, and the total differential of f at (-3 , 0, 2) .

find (a) the increment of f ut (-3 , 0, 2) and (b)

In Exercises5 through 8, prove that / is differentiable at all points in its domain by doing each of the following: (a) Find Lf (xo, y) for the given function; (b) find an e1and an €2so that Eq. (3) holds; (c) show that the et and the e2found in part (b) both approachzero as (Lx, Ay) --, (0,0). 6. ( x ' y ) - 2x2i 3y' 5. (x, : x'Y- 2xY

f

f

U)

8'f ( x , y ) vx

7.f (x,,) :i s. Given f (x,a) :

Y- 2

{;.

;l:}'r;i;t.

,

and,D2f(1,1) exist, but / is not differentiable at (1, 1).

Prove that Dl[,l)

-' (x,d . l3!I- ir \-" r ' + @,0\ 10. Given f (x, y) : 1f + Vn it (x, y\ : (0, 0) l0 Prove that DlQ, 0) and D2f(0,0) exist, but / is not differentiable at (0, 0). 11. Prove Theorem 19.5.4(ii). 12. Show that Eq. (5) may be written in the form (7) where {1 is between 16 and xs I h. 13. Show that Eq. (6) may be written in the form (8) where {2 is between yn and lo * k. In Exercises14 through 77, use Theorem19.5.4 to find either a f1 or a f2, whichever applies.

14.f(*, a) : x2+ 3xY- Y';r is in [1, 3]; Y - 4 16.f(x, A) :h,

y LSin l-2, 2f;x : 4

15.f (x, y) : xs- y'; x is in 12,61;y - 3

tr. f (x,y):'ffi;

y LsLnlo,4l;x- Z

\'-' y) + (0, o) , \ | ?:Y:i ir (x, L8.Givenf (x, A) : 1*' + y' rf (x, y) : (0,0) L0 This function is continuous at (0, 0) (see Example 3, Sec. 19.2, and Illustration L, Sec. L9.3). Prove that Drf (0,0) and Drf (0,0) exist but Dtf and Drf are not continuous at (0, 0).

?ve-n if: (x' Y) + (o'o)

rs.Givenf (x,a) : I # L0

tf (x, y) _ (0, 0)

Prove that f is differentiable at (0, 0) by using Theorem 19.5.5. In Exercises20 and 21, prove that / is differentiable at all points in R3by doing each of the following: (a) Find, Af (xo)ys, zn); G2,and e3,so that Eq. (30) holds; (c) show that the e1,e2,and e3found in (b) all approach zero as (Ax, Ly, Lz) (b) find drr G11 approaches(0, 0, 0). 21,.f (x, y, z) : )cy- xz * z2 20. f (x, A, z) :2x22 - 3yz'

|*"

i I(x'v'z)+ (0'0' 0)

L0

it (x, y, z) : (0, 0, 0)

22 .G ivefn( x , y , z ) : Vi fu

(a) Show that D1l(0, 0, O), D2f@, 0, 0) , and D"f (0 , 0, 0) exist; (b) make use of the fact that differentiability implies continuity to prove that / is not differentiable at (0, 0, 0).

DIFFERENTIAL CALCULUSOF FUNCTIONSOF SEVERALVARIABLES (

xttzz

if (x'v'z) * (0,0,0) if (x,y, z): (0,0, 0)

23. G i v e n / ( r!,, 2 ) : l 7 T 7 I 7

|.O

Prove that / is differentiable at (0, 0, 0). 24. Use the total differential to find approximately the greatest error in calculating the area of a right triangle from the lengths of the legs if they are measured to be 6 in. aid 8 in., respectively, with i possible error of 0.1 in. for each measurement. Also find the approximate percent error. 25. Find approximately, by using the total differential, the greatesterror in calculating the length of the hypotenuse of the right triangle from the measurementsof Exercise24. Alio find the approximate percent error. 26' If the ideal gas law (see Example 5, sec. 19.4) is used to find P when T and,v are given, but there is an error of 0.37o in measuring T and an error of 0.8voin measuring V, find approximately the greatlst percent ertor in p. 27. The specific gravity s of an object is given by the formula A -S : A-W where A is the number of pounds in the weight of the object in air and W is the number of pounds in the weight of the object in water. If the weight of an obiect inair is read is 20 lb with a possible error of 0.0i lb and its weighiin water is read as 12 lb with a possible enor of 0.02.1b,find approximately the ligest possible ertor in calculating ifrom these measurements.Also find the largest possible relative error. 28' A wooden box is to be made of lumber that is 3 in. thick. The inside length is to be 5 ft, the inside width is to be 3 ft, the inside depth is to be 4 ft, and the box is to have no top. use the total differential to find the approximate amount of lumber to be used in the box. 29' A- company has contracted to manufacture 10,000closed wooden crates having dimensions g ft, 4 ft,and 5 ft. The cost of the wood to be used is 5c per square foot. If the machines that are used to cut the pieces of wood have a possible eror of 0'05 ft in each dimension,-find approximately, by using the total differential, the greatest possible error in the estimate of the cost of the wood. In Exercises30 through 33, we show that a function may be differentiable at a point even though it is not continuously differentiable there' Hence, the conditions of Theorem 19.5.5are sufficient but not necessaryfor differentiability. The function / in these exercisesis defined bv

f (x,y)

.1 sin

u76,

if (x, Y) + (0,o) if (x, y) : (0, 0)

30. Find A/(0, 0).

31. Find Drf(x, y) and Drf (x, y).

32. Prove that / is differentiable at (0, 0) by using Definition 19.5.2and the results of Exercises 30 and 31. 33. Prove that D1f and.Drf arc not continuous at (0, 0).

19.6 THE CHAIN RULE

In Sec.3.6 we had the following chain rule (Theorem 3.d.1) for functions of a single variable: rf y is afunction of a, defined by y : (u) , and,Dzl exists; f and z is a function of x, definedby u: g(x), anaoru exists; their y is a function of x, and D"y exists and ii given by Dry : D;y Dru

1 9 . 6T H E C H A I NR U L E

or, equivalently, da "/ dx

dy du du dx

(1)

We now consider the chain rule for a function of two variables, where each of these variables is also a function of two variables. 19.6.1 Theorem The Chain RuIe

and lf. u is a differentiable function of x and y, defined by u:f(x,y)' x: F(r, s) , U : G(r, s), and \xlEr, \xl0s, 0yl0r, and Eyl0sall exist,then u is a function of r and s and

pRooF: We prove (2). The proof of (3) is similar. If s is held fixed and r is changed by an amount Ar, then r is changed by an amount Ar and y is changed by an amount Ay. So we have (4) Lx: F(r * Ar, s) - F(r, s) and L y : G ( r * L r ,s ) - G ( r , s )

(5)

Because/ is differentiable, (6) Lf (x, y) : Drf (x, y) A'x* Drf (x, v) Ay * e1Ar * e2Ly where er and e2both approach zero as (Lx, Ly) approaches(0, 0). Furthermore, we require that e1:0 and €z:0 when Ax: Ly:0' We make this requirement so that e1and er, which are functions of Ax and Ly, will be continuousat (Ax, Ly\: (0,0). If in (6) we replaceLf (x, y) by Lu,Drf (x, V) by \ul6x, andDzf G, U) by \ulay and divide on both sides by Lr (Lr * O), we obtain Lu Lr

-:-

0u Lx 0x Lr

Ly A'x , 0u La - T € r - - : - T 'Lr Cr.Lr 0Y Lr

Taking the limit on both sides of the above as Ar approaches zero, we get

li$fl vt li$# * ,1i*1.,) Hql # +(1iq.,) H # :#rin ^+,.H Becauseu is a function of r and y and both r and y are functions of r and s, a is a function of r and s. Becauses is held fixed and r is changed by an


amount Lr, we have lim

A,r - O

L u _ r : _ u ( r * A r ,s ) - u ( r , i l _ O , 'm Lr or

(8)

Also, Lx _ r:_ F(r * Lr,s) - F(r,s)

Lr: ll$

: 0x a,

l i mr y : hAm r-o

ay

lim

Ar - 0

(e)

and Ar-o Lf

0r

(10)

Because 0xl0r and 0yl6r exist, F and G are each continuous with respect to the variable r. (rorr: The existence of the partial derivatives of a function does not imply continuity with respect to au of the variables simultaneously, as we saw in the preceding section, but as with functions of a single variable it does imply continuity of the function with respectto each variable separately.) Hence, we have from (4) lim tF(r + L,r,s) - F(r, s)]

lim Ar:

Ar-0

Ar-0

: F(r, s) - F(r, s)

-0 and from (5) lim AY: lim lc(r + Lr, s) - G(r, s)f

Ar-0

Ar-0

: G(r, s) - G(r, s) -0 Therefore, as Ar approacheszero, both Ar and Ay approach zero. And becauseboth e1and e2approachzeroas (Lx, Ay) approaches (0,0), we can concludethat €z: 0 li-^ €r : 0 and Alim r-0

( 1 1)

Ar-0

Now it is possible that for certain values of Lr, Lx: Ly:0. Becausewe required in such a case that er:€2:0, the limits in (11,)are still zero. Substituting from (8), (9), (10), and (1.1)into (Z), we obtain

.(,H(y) #:(,H(#)

which proves (2), EXAMPLE

].:

u-ln@

GiVCN

0u

ay

0x

__ a V

dr

p8

19.6THECHAINRULE x : T€r,and y : f o-r, find 6ul0r

and oulos.

929

0x E:"r From (2) we get 0ux

Ar:W From (3) we get 0u:ffi(rer)+ffi(-rt-r)-W x U , ^\ / .. _c\ r(xes ye-r) ds

As mentioned earlier the symbols 0ul0r,0ul0s, Eul\x,0ul0y' and so forth must not be considered as fractions. The symbols 0u, 0x,,and so on have no meaning by themselves. For functions of one variable, the chain rule, given by Uq. (t), is easily remembered by thinking of an ordinary derivative as-the quotient of two differentials, but there is no similar interpretation for partial derivatives. Another trouLlesome notational problem arises when considering u as a function of x and y and then as a function of r and s.lf. u: f (x, y) , x: F(r, s), and y: G(r, s), then u: f(F(r, s), G(r, s)). [Note that it is incorrect to write u : f (r, s).1 1: In Example 1, o rLLUsTRArroN

u: f(x, y) :lnty'p +V x: F(r, s) : re" A : G ( r ,s ) : r e - " and so

L t - f G ? , s ) , G ( r ,s ) ) : l n f f i ff(r, s): 6tffi

+ u'f

If we let f(E(r, s), G(t, s)):h(r, written resPectivelYas

s), then Eqs. (2) and (3) can be

hr(r, s) : f ,(x, y)Er(r, s) + fr(x, Y)Glt, s) and hr(r, s) : fr(x, A)F"(t, s) + 1r@,Y)Gr(r' s) In the statement of Theorem 19.6.1 the independent variables are r antd.s, and,u is the dependent variable. The variables r and y can be :alled the intermediate variables. We now extend the chain rule to z intermediate variables and m independent variables. 19.6.2 Theorem Suppose that u is a differentiable function of the n variables.rr, x2t . . . , ChainRule xn,' ind each of these variables is in turn a function of the m variables TheGeneral


Ar, Uz, . . ., A*. Supposefurther that each of the partial derivatives A x iAl y j ( i : 1 , 2 , . . . , n ; j - 1 , 2 , . , m) exists. Then u is a function

ffi(,H #,:mGH.(#: (,HW) )+

dz\ )xr/

,)

*

:

#--(,Hffi).(,H(,H. (.9 The proof is an extension of the proof of Theorem 19.6.1.. Note that in the general chain rule, there are as many terms on the right side of each equation as there are intermediate varia-bles.

EXAMPLE

Given

solurroN:

u-xy+xz*yz X: r, A - r cos t, and Z: find 0ul0r and 0ul0t.

r stn t,

By applying the chain rule, we obtain

#:e;e).w)(,*).(Hff) : (y * z)(1) + (x I z) (cosf) * (r * y) (sin f) : y * z * r cos t * z cos f * r sin t i- y sin t : / cosf * r sin f * r cosf * (r sin t)(cos t) * rsinf * (rcos f)(sin t) :2r(cos f * sin t) + r(2 sin f cos f) : 2r(cos f * sin t) * r sin 2t

.ffi)(y).ffie,) #:(rJ(#J : (y * z) (O) + (x *-z) (-r sin r) + (s * y) (r cos f) : (r* r sin f)(-r sin t) + (r* r cosf)(r cosf) :-r2 sin f - rz sinzt * 12 cost I rz coszt : r2(cosf - sin t) * rz(coszf - sin2f) : rz(cos I - sin t) + 12 cos2t

Now suppose that z is a differentiable function of the two variables r and y, and both r and y are differentiabre functions of the single variable f. Then z is a function of the single variable t, and so instead of the partial derivative of z with respect to f, we have the ordinary derivative

1 9 . 6T H E C H A I N R U L E

given by

of. u

(12) we call duldt given by Eq. (12) the total deiaatiae of u with respect to f. If u is a differentiable function of the n variables x1, x2, . . ., xn and each ri is a differentiable function of the single variable t, then z is a function of f and the totdl derivative of a with resPectto f is given by

EXAMPLE

Given

u-x2+2xy*y' )c: f cos t, and y : t sin t, find duldt by two methods: (a) using the chain rule; (b) expressing u in terms of f before differentiating.

solurroN: (a) 6uldx: 2x * 2y, \ul0y - 2x * dyldt: sin t + f cos f. So from (12)we have

dxldt: cos t- t sin

( 2 x * 2 y ) ( c o s t - f s i n f ) + ( 2 x * 2 y ) ( s i n t + t c o st )

#:

-2(r*

il(cos t- f sint+ sint+ f cosf) : / ( t c o s t + f s i n f ) ( c o st - f s i n t + s i n t + f c o s f ) : 2 t ( c o s 2 t - f s i n f c o st + s i n f c o st + f c o s 2t + s i n f cos t - t s i n 2t + s i n 2t + f sin f cos f) - 2 t [ 1 + 2 s i n f c o st + f ( c o s 2 t- sin2t)] :2t(1 + sin 2t + t cos2t) (b) u : (f cos t)' + 2(t cos t) (t sin f ) + (t sin t)2 - szcos2t + t'(2 sin f cos f ) + t2 sinz f : t2 + tz stnZt

4:2t dt

+ 2t stn2t + Ztz cos2t

4: lt f is a differEXAMPLE soLUTroN: Let u - tbxz - *ay'. We wish to show that z f (u) satisfies the given equation. By the chain rule we get entiable function and a and b z: that prove are constants, a n d *ay: + Ydu: ay f ' ( u )( - n y ' ) !*a'lE \f Gax'- *ayt) satisfiesthe Partial 0x du 0x J' differential equation

Y:+Y:f'(u)(bx)

Therefore,

' 0 2* b x * : a'Y' dy ax

a y z l f ' ( u ) ( b x )*l b x l f ' ( u ) ( - a y ' ) l: 0

which is what we wished to Prove.

gilI


EXAMPLE5:

Use the ideal gas law (see Example 5, Sec. "1,9.4) with k - 10 to find the rate at which the temperature is changing at the instant when the volume of the gas is L20 in.3 and the gas is under a pressure of 8 lb/in.2 if the volume is increasing at the rate of 2 in.t/sec and the pressure is decreasin g at the rate of 0.1.lb/in.2 per sec.

soLUTroN: Let t - the number of seconds in the time that has elapsed 7Tr

T D-

Tf -

PV:

since the volume of the gas started to increase; the number of degreesin the temperature at t sec; the number of pounds per squareinch in the pressure at t sec; the number of cubic inches in the volume of the gas at t sec. F^ PV

L0T and so

r:1o

At the given instant, P : 8, V - !20, dPldt: -0.1, and dvldt- 2. Using the chain rule, we obtain

-d: T - aTdP.aTdv dt aPift'avdt : y d P _' L u 70 itt t0 dt :1#(_0.1) + fo(Z) :-1.2* 1.6 :0.4 Thereforethe temperatureis increasingat the rate of 0.4 degreeper second at the given instant.

Exercises19.6 In Exercises1 through 4, find-the indicated partial derivative by two methods: (a) Use the chain rule; (b) make the substitutions for r and y beforc differentiating.

1. u- xz- y'; x: 3r- s;y : r * 2s;#ry, 3. u - sutc'x:2r

cost; A : 4r sint, !, U 0r' at

2. u - 3x2 * )cy- 2y,+ 3x- y; x: 2r- 3s;y : r * s;#, 4 . u - x 2+ y , ; x : c o s hr c o st ; A : s i n h r s i n t ; U = O u or' at

In Exercises 5 through 10, find the indicated partial derivative by using the chain rttle.

5. Lt: sin-l(3r* il; x: rler;y : sin rs; #, X

5. u - xe-o; )c- tan-r (rst); y : ln(3rs

7. u -cosh * : 3r2s; y : 5se'; X, #r#

8 . u : , c y + x z * A z ;x :

9. u- xzI y' + 22;x - rsin @ cos0; y :r 10. u - x'yz; )c-!;

ou. ou V : res;z re-'r; 0r' ds

' ott ou ou sin @sin o; z: r cos a;

*; ao,M

r s ;y : r z - s 2 ;

X

19.6 THE CHAIN RULE

(b) make the substitu-

In Exercises L1 through 14, find the total derivative duldtby two methods: (a) Use the chain tions for x and y or fiorx, y, and z before differentiating' 12. u : l n x: cos f; y - sin f LL. u: ye' + )cea; 13.u-

x:

W;

t*

In ExercisesL5 through 18, find the total derivative duldt by using the chain differentiating.

x - rnt; a - et 1s.u -tan-r e), L 7 u. - Hy t,

xy+y';x:et)y: er

1,4.u : - . t ; x : 3 y-e'

t a n t ; y - c o sf ; z - s t n t ; 0 < t < L n

933

e-t

sin t; A:

lnf

do not expres s u as a function of t before

1 , 6u. - x y + x z +

f cos

f sin

18. u- ln(xz* y'+ t2);x - f sin t; U: cos f

x : l n f ;y : t " 1

In Exercises1"9through 22, assumethat the given equation defin es z as a function of x and y. Dlfferentiate implicitly to find 0zl0x and 0zlAy. 19. 3x2+ y'+

z2 - 3xy * 4xz - 15 0

20. z-- (x'*

yz) sin xz

Zg. If f is a differentiable function of the variable u,let u: a(azlAx) + b(azlay) : 0, where a and b ate constants.

21. ye*o" cos 3xz - 5

bx - ay and prove that z:

24. If.f is a differentiable function of two variables u arrd.o,let u: satisfiesthe equation 0zl6x* Azl0y 0.

- ay) satisfies the equation f (b*

x - y and a : ! - r and prove that z: f (x - y ' y ')

ZS.If f rs a differentiablefunction of x and y andu- f (x,A), x- r cos 0, and 0u dx

22. zeazI Zxe*' - 4e'o : 3

r sin 0, show that

0u ^ 6usin0 cost--A0 r 0r----

0u 0u ^ ducos0 srn0+ ae r a": ar 2 6 .| f f a n d g a r e d i f f e r e n t i a b l e f u n c t i o n s oxfa n d y a n d u : f ( x , i l -dvl}x, then if x: r cos0 and'y: r sin 0, show that 'l,. | 0u 0o 6v du

ar:V N

anda: g(x,V),suchthat dulax:|oldyandaul|y:

and ar:-7 ao

27. Supposefis a differentiablefunction ofr and y and.u: f(x,y). Then ifx: 6ul0p and duldw in terms of 0ul0x and 6ul6y.

cosh 7)cosToandy:

sinh o sin at, exPress sin 0, and

r sin d 2 8 . S u p p o s e / i s a d i f f e r e n t i a b l e f u n c t i o nxo, yf , a n d z a n d u : f ( x , V , z ) . T h e n l f x : r s i n @ c o s 0 , and |ul0z' \ul0y, of. |ul6x, terms in arrd aulae 6ul0Q, ,:, {, expressdul0r, "o" the 29. At a given instant, the length of one leg of a right triangle is 10 ft and it is increasing at the rate of 1 ftlmin and lengti of the other leg of tie right triangle is 12 ft and it is decreasing at the rate of 2 ftlmin. Find the rate of change of the measure of the acute angle opposite the leg of length 12 ft at the given instant'

the 30. A vertical wall rnakes an angle of radian measure Sa with the ground. A ladder of length 20 ft is leaning against ladder, by the formed triangle the area of is the How fast ftlsec. rate of 3 at the wall doiyn tne wall and its top is sliding the wall, and ihe g"orttJ changing when the ladder makes an angle of ta'radians with the ground? which 31. A quantity of gas obeys the ideal gas law (see Example 5, Sec.19.4) with k: 12, and the gas is in a container lb/in'2 and is 5 pressure is is being heated at a raie of 3o per second. If at the instant when the temperature is 300o,the instant. that at the volume of decreasing at the rate of 0.1 lb/in.'z per second, find the rate of change

9:II


ZZ. l'tater is flowing into a tank in the form of a right-circular cylinder at the rate of *z fd/min. The tank is stretching in such a way that even though it remains cylindrical, its radius is increasing at the rate of 0.002ftlmin. How fast is the surface of the water rising when the radius is 2 ft and the volume of water in the tank is 20z-fp?

19'7 HIGHER-OR_DER If f is a function of two variables, then in general Drf and.D4f are also PARTIAL DERIVATIVES functions of two variables. And if the pa*iil derivatives of these functions exist, they are called second partial derivatives of. In contrast, f. Dtf and D"f are called first partial derivatives of f. There are four second partial derivatives of a function of two variables. If f is a function of the two variables r and !, the notations

Dr(Dl)

Drrf

f ,,

f *o

Arf 0y 6x

all denote the second partial derivative of /, which we obtain by first partial-differe_ntiating/ with respect to r and then partial-differentiating the result with respect to y. This second partial derivative is defined bi ft(x,U):lim

r

(1)

Ly

Au-O

if this limit exists. The notations

Dr(Drf )

Drrf

fn

f ,,

Arf 0xz

all denotethe secondpartial derivativeof /, which is obtainedby partialdifferentiatingtwice with respectto r. we have the definition r (x * Ax,y) fr(x, y)

fu@,y):lntr

e)

if this limit exists. we define the other two second partial derivatives in an analogous way and obtain frr(x, a) : lim

(3)

f'@' Y * LY)- fr(x' Y) f ,r(x, a) : lim ay

(4)

and As-0

if these limits exist. The definitions of higher-order partial derivatives are similar. Again we have various notations for a specific derivative. For example,

Drrrf

frr"

f rcu

atf av axax

tlay$

all stand for the third partial derivative of which is obtained by partial/, differentiating twice with respect to r and then once with respect to y.

PARTIALDERIVATIVES 935 19,7 HIGHER-ORDER

Note that in the subscript notation, the order of partial differentiation is from left to right; in the notation 03f l0y 6x 0r, the order is from right to left.

EXAMPLE

f (x,u) :

Given e* sin

find: (a) D'f (x, (c) atfli.x oyz.

SOLUTION:

y+lnxy (b) D 'rf (x,

Drf (x, Y)

.,1 s r n '/u t -

)cy

er stn V +*

So (a)Dr,;f(r,A) : e" siny - Ll*; and (b) Drrf (x, y) : s" cos y. (c) To find AsflaxAy2,we partial-differentiate twice with respect to y and then once with respect to x. This gives us .L slnut' - -,

cos

y'

a'f 6x 0y2

sin y

Higher-order partial derivatives of a function of n variables have definitions which are analogous to the definitions of higher-order partial derivatives of a function of two variables. lf f is a function of n variables, there may be n2 second partial derivatives of f at a particular point. That is, for a function of three variables, if all the second-order partial derivatives exist, there are nine of them: /rr, f ,r, fr", frr, frr, fr", f"r, 1"r, and flr.-

EXAMPLE

Given

f (x, A, z) - sin(xy * 2z) findDrtf(ic,y,z).

SOLUTION:

D'f (x, A , z ) - y c o s ( x y * 2 2 ) D"f (x, y , z) - -2y sin (xy * 2z) D t r r f( x , y , z )

EXAMPLE

Given

f ( x , a ) : f Y - y cosh xy find: ( a) Dt, f (x, U); ( b ) D r , , (f * , y ) .

sin (xy * 2z) - 2xy cos(xY * 2z)

SOLUTION:

(a) Drf (x,y) - 3x2U- Y2 srnh xY Drrf(x,Y) :3x2 - 2Y sinh xY - xY' coshxY (b) Drf (x,Y) : x3- cosh )cY- xY sr'.h xY Dr,,f(x,y):3x2 - y sinh xy - y sinh xy xy'cosh xy :3)c2 - 2y sinh xy - xy' cosh ry We see from the above results that for the function of ExamPle 3 the "mixed" partial derivatives Dtrf (x, y) and Drrf(x, V) are equal. So for with this particular functioh, when finding the second partial derivative This immaterial. is differentiation respect to x and then !, the order of examPle following the However, condition holds for many functions' shows that it is not alwaYs true.

93G


ExAMPLE4: Let f be the function defined by t.

\ x2-7rz

f ( x , a ) : l ( xfvf )i Lo

i r( x , Y ) ( 0 ,0 ) if (x,y)

( 0 ,0 )

Find frr(0, 0) and frr(0, 0).

solurroN: In Example3, Sec. 19.4,we showed that for this function - -y for all y f '(0, Y) and for allx fr(x,O) : r From formula (1) we obtain ft (0, o) : lim Ag-0

But from (5),f ,(0, Ly) Ay andf ,(0, 0) and so we have -O{,, O: lim (-1) : -1 frr(0,0): lim Au-o

ay

Au-o

From formula (3), we get

f'(o * Lx' o) - fr(o' o) frr(o,o) : lim A,x From (5),fr(Lx,0) : Ax and fr(0,0) :0. Therefore,

lim#: f r , , ( oo,) : Aal-0 AI

hm L: l Atr_Q

For the function in Example 4 we see that the mixed partial derivatives/tr(r, y) and frr(x, y) are not equal at (0,0). A set of'conditions for yo):fu(xo, ye) is given by Theorem 1g.7.'!., which follows. thi+ ft(xo, The function of Example 4 does not satiify the hypothesis of this theorem becauseboth f' and fn are discontittuou" at (0, 0). It is left as an exercise to show this (see Exercise20). 19.7.1 Theorem

*qp::g that / is a funcrion of rwo variables x and y defined on an open disk B((16, yJ; r) and f*, fu, f,r, ^d fu, also are aefinea on B. Furthermore, suppose that f ,u and fu, are continuous on B. Then f ,u(h, uo): fo,(xo, yo1 pRooF: consider a square hgifg its center at (xs, yo) and the length of its side 2lhl such that 0 < frlhl < r. Then ail the ioi"t, in the interior 9f the square and on the sides of the square are irithe open disk B (see F i g . . l 9 . 7 . l _S)o. t h e p o i n t s ( x r * h , y o * h ) , (xr*h,yo), ind 1xo,yoih1 are in B. Let A be defined by A: f(xo* h,yo+ h) - f@o+ h, y) f ( x o , V o +h ) + f ( x o , y o ) ( z ) Consider the function G defined by

Figure

c(r) - f(x, Ao* h) - f(x, yr)

(8)

PARTIAIDERIVATIVES937 19.7HIGHER.ORDER Then

G(x+h)- f(x+h,Ao+h)- f(x+h,Ar) So (7) can be written as A-G(ro+h)-G(ro) From (8) we obtain (10) G' ( x) : f"(x, Uo+ h) - f"(x, yo) Now, becausefr(x, Ao*h) and fr(x, Ao) are defined on B, G'(r) exists if r is in the closed interval having endpoints at re and xn*h. Hence, G is continuous if r is in this closed interval. By the mean-value theorem (4.7.2)there is a number c, between xq and xs1' h such that

G ( r o+ h ) - G ( r o ) : h G ' ( c ' )

(11)

Substituting from (11) into (9), we get A - hG'(ct)

(r2)

From (12) and (10) we obtain * h) - f,(ct, U)f A - hlf " ( c r ,U o Now Lf I is the function defined bY

s @ )- f " ( c r , Y )

(13)

(14)

we can write (L3) as

A - hls@o+ h) - s(yo)l

(ls)

From (L4) we obtain 8'(Y) - f"o(h,a)

(15)

Becausef,u(cr, y) is defined on B,g'(y) exists if y is in the closed interval having endpoints at yq and AoI h; hence, g is continuous if y is in this closed interval. Therefore, by the mean-value theorem there is a number d1 between Ao and Ao* h such that (J7) g\o + h) - sg) : h7'(d') Substituting from (1,7)into (15),we get A:h2{ lows that L:

hzf,o(ct,dt)

(dr); so from (16) it fol(18)

for some point (c1,d1)in the open disk B. We define a function { by (1e) QQ) : f (xo+ h,y) f ( x o ,V ) + h ) . T h e r e f o r cQ, ) c a n b e and so Q@+h):f(xo*h,y+h\-f(xo,A written as

a - 6 u o + h )- f u o )

(20)

DIFFERENTIAL CALCULUS OF FUNCTIONS OF SEVERAL VARIABLES From (19) we get

Q ' @ ): f o ( r o* h , Y ) - f o 7 o , Y )

(2r)

{'exists if y is in the closed interval having yo and Ao*h as end.points because by hypothesis each term on the right side of (21) exists on B. Therefore, Q is continuous on this closed interval. so by the meanvalue theorem there is a number d2 between ye and * h such that ao

Q Q o +h ) - A Q J = h O ' @ , ) From (20),(2L),and (22)it follows that A - hlfr(ro * h, dr) - fr(xo, dr)f

(22)

(23)

Define the function 1 by

x(r) : fo@,k)

e4)

and write (23)as A : h l y ( x 6 +h ) - x ( r o ) l From (24)we get x'(x) : fo,@, d")

(2s)

e6) and by the mean-valuetheoremwe concludethat there is a number c2 betweenxoand xs* h suchthat x ( r o+ h ) - X @ ) : h x ' ( c r ) e7) From (25),(26),and e7) we obtain A: h2fnr(c2,dr)

(2S)

Equating the right sides of (1g) and (2g), we get hzf,n(cr, ilr) : hrfur(cz, ilz) and becauseh # 0, we can divide by hr, which gives fru(cr, d):

fnr(cr, dr)

eg) where (cr, dr) and (c2,d) arc in B. Becausec, and c, are each between ro and xs * h, we can twite c, : xs*e1h, where 0 ( e, < L, and cz:xo*e2h,wheie0 ( e, < 1. Similarly, becauseboth i\ and d2,arebetween youia!o*h,*".1n iit" d,r:yo ,"h, where 0 ( es <_-1a.and ilr: ! Ao* enh,-where 0 ( ea( 1.. Making these substitutions in (29) gives f ,o(xo I e1h,UoI esh) : f o*(xo * €zh, Uo* e+h)

(30)

Because fry and far are continuous on B, upon taking the limit on both sides of (30) as h approaches zero, we obtain fro(xo,U):

fo*(xo,yo)

I

PARTIALDERIVATIVES 939 19.7 HIGHER-ORDER

As a result of the above theorem, if the function / of two varidbles has continuous partial derivatives on some open disk, then the order of partial differentiation can be changed without affecting the result; that is, Drrrf : Dt rf : Drttf Drrrrf : Dt rrf : Drrrrf : Drr,rrf: D"t tf : Dr"rtf and so forth. In particular, assuming that all of the partial derivatives are continuous on some open disk, we can Prove that Drttf : Dtt"f bY applying Theorem 19.7.7 rcpeatedly. Doing this, we have Drrrf: Dr(Drrf) : Dr(Drrf) : DlD2(D1f)l: DzIDJDrf)] :Dr(Dtf):Dttf

EXAMPLE 5:

Given

that u -

f ( x , y ) , x - F ( r , s ) , a n dY G (r , s) , and assumingthat f ,o: f o*,ProvebY using the chain rule that -62u 6or: ,*(x,Y) [r'?' s)f' f * 2 f* o ( x ,y ) F , ( r , s ) G , ( r ,s ) * f o o ( x y, ) [ G ,( r , s ) l ' +f , y) F,,(r, s) "(x * f o(x, U)G,,(r,s)

sol-urroN: From the chain rule (fheorem 19.6.1\we have

y)G,(r,s) #: f ,(*, y)F,(r,s)+ fu@, Taking the partial derivative again with respect to r, and using the formula for the derivative of a product and the chain rule, we obtain t2n '

ffi:

lf,,(x,y)F,(r,s)* f*o(x,y)G,(r,s)lF,(r,s)'f f,(x, y)F,,(r,s) y)G,(r,s)fG,(r,s)+ f uQ,y)G,,(r,s) * lf o,(x,y)F,(r,s)+ f oo(x,

Multiplying and combiningterms,and using the fact that f ,u(x, A): f ,,(x, y) , we get t#:

f ,,(r, y)lF,(r, s)l' + 2f,o(x,y)F,(r,s)G'(r,s)

I f o u ( xy, ) l G , ( r ,s ) l ' * f , ( x , y ) F * ( r ,s ) + f o @ ,y ) G , , ( r ,s ) which is what we wished to Prove.

19,7 Exercises (x,il; b) find Drrf (x,y); (c)show thatDrrf(x,A) : Dr',f(x,y)In Exercises L through 8, do each of the following: (a)Find Dr,.f

L . f ( x ,y ) -t-L yx2 4.f(x,y)-t-rta*ln 7. f (x, A) :4r

Y-

x

s i n h y + 3y cosh r

2 ' f ( x , a) :2xB - 3x'y * xy'

3. f (x, U) : e2r svt Y

U u2t t r t /(*, * . , \ - -- ((x' * Y') tan-r 1 5. v) f

6. f (x,A):

8. f(x, A) : tr cos y - ye*

In Exercises9 through 14, find the indicated partial derivatives.

"rn-t!

940


9. f (r, y, z) : ye, I zevt ez;(a)f ,"(x, y, ); 1ayyo,Q,y, z) ll. f(r,s) :2rtsl r2* -5rs3; (a) frrr(r,s); @) frrr(r,s) 73. g(r, s, t) : ln(r, * 4s2- SP); (a) grrr(r, s, t); b) gp2e, s, t)

10.g\x, y, z) : sin (xyz); (a) gza(x,y , z); (b) gr,(x, y , z) 12. f(u, o) : ln cos(u - a); (a)f uuo(u,u); (b) f ,uo(u,a)

14. f(x, A, z) : tan-t(3xyz); (u) fu"k, y, z); (b) frrr(x, y, z) In Exercises15 through 18, show that u(x, y) satisfies the equation

#,.#-o which is known as Laplace'sequatidnin R2. 15. u(x, A) : ln(rz * y') 16. u(x, A) : er sin y + ea cosx

18.u(x,A): tan-l ;l! ,,

17. u(x, A) : tan-l

19. Laplace'sequation in R3is

u"*ry+92-o a-r- tvr-M show that u(x, y, z) :

20' For the function of Example tlo* 1 is not satisfiedif (xs,go) : (0, 0).

(x, * y, * zr)-u2 sTlisfies this equation.

that f nis discontinuous at (0, 0) and hence that the hypothesis of Theorem 19.7.1

In Fxercises21 through 23, find frr(O,0) and fzJ},0), if they exist.

( 2xu

2 r f. ( x , a ) : ] f f i L0

) ( 0 o) i r ( x ' Y+ ,

if (x, y) + (0, 0)

if (x, y) - (0,o )

if (x, y) : (0, 0)

-'XY2tan-'L if x + 0 and y + 0 2s.f (x, a) :{t' tu" 1.0 if eitherx :0 or y : 0 24. Given that u -- f (x,y), s2,,

F(t), and y : G(t) , and assuming that ,u: f f ax, ptove by using the chain rule that

ffi:f*tr'oLF'(t)l't2f*(x,y)F'(t)G'(t)*f*(x,y)[G'(t)]rtf,(x,y)F,,(t)+t'"k,y)G,,(t) 2 5 . G i v e nt h a t u - f ( x , U ) , x - F ( r , s ) , a n d

G(r, s), and assumi.g that f*o: fo* prove by using the chain rule that y ) l F , ( r , s ) G , ( r ,s ) + F , ( r , s ) G ,( r , s ) f * f o o ? , y ) G , ( r , s ) G , ( r ,s ) * f r ( x , y ) F , , ( r , s ) * f u @ ,y ) G , " (r , s )

26. Given Lt- ea cos x, x:2t, - tz. Find dzuldtz in three ways: a (a) by first expressing u in terms of t; (b) by using the formul a of Exercis e 24; (c) by using the chain rule.

27' Given u:3xy - 4y', x:2se', ! : re-". Find, }z_urdr2 in three ways: (a) by first expressingz in terms of r and s; (b) by using the formula of Example 5; (c) by using the chain rule. 28. For u, x-, and,y as given in Exercise27, hnd A2ulAs 0r in three ways: (a) by first expressing z in terms of r and s; (b) by using the formula of Exercise 25; (c) by using the chain rule. 29. Given Lt- 9x2* 4y,, x - r cos 0, y: r sin 0. Find dzulyrz in three ways: (a) by first expressing u in terms of r and,0; (b) by using the formula of Example 5; (c) by using the chain rule.

30' For u' x' and'y as given in Exercise 29, find,02ul0tr in three ways: (a) by first expressing z in terms of r and 0; (b) by using the formula of Example 5; (c) by using the chain rule.

REVIEWEXERCISES 941

31. For Lt, x, and y as given in Exercise29,find 62ul0r d0 in three ways: (a) by first expressing utn terms of r and 0;\b)by using the formula of Exercise25; (c) by using the chain rule. 9 2 . I f u : f ( x , y ) a n d a - g ( x , y ) , t h e n the equations 0r-_A! 0u_0a and AY 0x 0x Ay are called rhe Cauchy-Riemannequations.If / and g and their first and second partial derivatives are continuous, Prove that if z and u satisfy the Cauchy-Riemann equations, they also satisfy Laplace'sequation (seeExercises15 through 18)' 33. The one-dimensional heat-conduction partial differential equation is 0u At

-:e-

,, aru " 6xz

Show that if / is a function of r satisfying the equation ,flt

ftrt+x'11x1:o ) : 0, then 11v: and g is a function of f satisfying the equation dgldt * k'zI'Dg(f tion is satisfied. k and )t are constants. 34. The partial differential equation for a vibrating string is

f (x)g(t), the partial diffetential equa-

azu , ozu o' ar' at,: f satisfyingtheequaShowthat if f isatunctionof xsatisfyingtheequationfrftdf+fT(r):Oand{isafunctionof a and )t are constants' is satisfied. equation partial differential the u: lf. 0, then az\zg(t): t Bgldp iion f(i)g(t), of a real variable having continuous secondderivatives andu: f (x + at) prove functions arbitrary two g are if and that 35. / Exercise 34' (ruNr: Let at), theni satisfies the partial differential equation of the vibrating string given in t g@ x and l') of functions in tum are andw anda w, o and of a: xi at andTo: x- at; lhenals a function up to the fourth order are continuous on 35. prove that if / is a function of two variables and all the partial derivatives of.f some open disk, then Dtrrrf : Drt tf

ChaPter19' ReaiewExercises, In ExercisesI through 5, find the indicated partial derivatives' &-tt

7.f (r, D :ff;

D,f(x,y), D"f(x'Y), D'"f(x,Y)

2.F(x,A,z):2xy'43y2'-Sxzs;Dj(x,y,z),Dsf(x,y,z),DpfQ'y,z) r 4'h(x'y):tan-1 3 . g G , f ) : s i n ( s P )* t e s ; D 1 g @ , t ) , D 2 g @ , t ) , D z , l G , t ) fo;o'h(x'y)'Dzh(x'y)'Dth(x'v) s. f(u, u, w) :t#t

Di@, tt, w), Dpt'@,a' u)' D',i(u' tt' w)

s i nu - 2 u a t a n w ; D z f ( u ' t t ' w ) ' D i @ ' a ' w ) ' D ' " ' f \ u ' a ' w ) 6 . f ( u , 7 ) , 7 D ) : wc o s 2 t t * 3 o In Exercises7 and'8 find,6ul6t and aulasby two methods' 8 ' u : e 2 ' + oc o s ( Z yx- ) ' x : 2 s 2 - t ' ' y : s 2 * 2 t 2 7.u:yln(r2+ y'),x:2s-l 3t,y--3t-2s

942

OF SEVERAL VARIABLES CALCULUS OF FUNCTIONS DIFFERENTIAL

9.fru:xylf , r : 4 c o s t , a n d y : 3 s i n f , f i n d t h e v a l u e o f t h e t o t a l d e r i v a t i vdeu l d t a t t : * r n b y t w o m e t h o d s : ( a ) D o not express u in terms of f before differentiating; (b) express z in terms of f before differentiating. l$. If f (x, !, z) :3ry'5xz2- 2ryz, Lx:0.02, Ay: -0.01, and Az : -0.02, find (a) the increment of.f at (-1,3,2) and (b) the total differential of f at (-7 , 3, 2) . In Exercises11 through 14, find the domain and range of the function / and draw a sketch showing as a shaded region in R2 the set of points in the domain of /.

rr. f(x,il: \/@ V 13.f (x,y) :161;r{p=fi

72.f(x,y):sin-'{r=7=T ra.fQ, y) : flrn + [{1,111

In Exercises 15 through 17, hnd. the domain and range of the function /. xv x u l 'x-z- t- r t l zr 7 5 .f ( x , ! , 2 ) : 2 . 1 6 .f ( x , ! , 2 ) : - - r Yz xyz

x -- lyl lzl InExercises18through20,establishthelimitbyfindinga6>0foranye)0sothatDefinitionlg.2.5holds. 18..1i+

(t,r)-(4,-1)

.( x-Sy):zt

19.

!i+

(t,t)-Q,-il

7 7 .f ( x , ! , 2 ) :

( 3 f - 4 y ,- ) : - e

20. lim

(f-yr*2x-4y):t0

(c.s)-(e,r)

In Exercises21 and 22, determine if the indicated limit exists.

2r. lim :I**y'

22. lim

f,- v',

f *yn (c,y)-(o,o)

t",yt
In Exbrcises23 through 27, discuss the continuity ofl. ( f'us if (x' y) ,, 23.f(x, il :|1ri if (x,y) 1.0 (f-ua t-----=- if (x,u) 2a. f@, a) : lf + yt it (x, y) t0

+ (0' 0) : (0,0) + (0,0) : (0, 0)

(nwr: see Exercise21.)

(HrNr: SeeExercise22.)

25,f (x,A):':* - +.Y: xz 4Yz

cos2tnx * cos2try In Exercises 28 and 29, ptove that the function / is differentiable at all points in its domain by showing that Definition 19.5.2holds. 28. f (x, A) :3xy'-

4x2* y'

2 9f.( x ,A ): ' ? - I Y

v'

30. Suppose c is the radian measure of an acute angle of a right triangle and sin a is determined,by alc,where a in. is the length of the side opposite the angle and c in. is the length of the hypotenuse. If by measurement a is found to be 3.52 and c is found to be 7.14, and there is a possible error of 0.01 in eich, find the possible error in the computation of sin c from these measurements. 31. A painting contractor charges l2A pet square foot for painting the four walls and ceiling of a room. If the dimensions of the ceiling are measured to be 12 ft and 15 ft, the height of the room is measured to be10 ft, and these measurements are correct to 0-05 ft, find approximately, by using the total differential, the greatest error in estimating the cost of the job from these measurements. 32. At a given instant, the length of one side of a rectangle is 6 ft and it is increasing at the rate of 1 ftlsec and the length

EXERCISES 94II REVIEW of another side of the rectangle is 10 ft and it is decreasing at the rate of 2 fitlsec.Find the rate of change of the areaof the rectangle at the given instant. gg. If f is a differentiable function of the variable z, let u: * + yz and,prove that z: ry + f(f 6z dz u -x?:1t2-f - ?i|x dy

+ y') satisfies the equation

34. Verify rhat u(x, y) : (sinh r) (sin y) satisfies Laplace's equation in R2: 0 2 u* 0 2 u _ n 0*' aYz 3.5. Verify that nffi ' u\x, t) : srn ,

e(ntnzktllzY

satisfies the one-dimensional heat-Conduction partial differential equation: du at

- : P -

,, dzu '" a*

35. Verify that u(x, t) : A cos(kat) sin(kr), where A and k are arbitrary constantS,satisfies the partial differential equation for a vibrating string: d2u ^ dzu a" aP: a* 97. lf f is a differentiable function of x and y and,u: f(x,V),I:

/ cos 0, andy:

r sin 0, show that

.(#)', .i,G*;': (*J', (#)'

38. Given

,

- - (x, if \ - ' o) \ - - 'rd' + (0,

l_t!_

f(x'V):l#+Yz [0

if. (x, y) : (0, 0)

ProvethatD,/(0,0) andDd@,0) existbutthat/isnotdifferentiableat (0,0). (nnrr:SeeExample4,9ec.19.2,and Exercise 2 in Exercises19.3.) 39..Given t_

fa2z2

f(x,v,O:troffi

i r ( x ' Y ' z+) ( o ' o ' o ) if. (r' Y' z) : (0, 0, 0)

[o

Prove that / is differentiable at (0, 0, 0). t(). Let / be the function defined by I

eatr"u

f(x,il:le-u"Tft L0

ifx*o if r:0

Prove that f is discontinuous at the origin' 41. For the function of Exercise 40, prove that Dl(0,0)

and Drf(0'0) both exist.

graDirectionalderivatives, dients,applicationsof partial derivatives,and lineintegrals

20j DIRECTIONALDERIVATIVES AND GRADIENTS

20.L DIRECTIONAL DERMTIVES AND GRADIENTS

We now generalize the definition of a partial derivative to obtain'the rate of change of a function with respect to any direction. This leads to the concept of a "directional derivative." Let / be a function of the two variables x and y and let P(r, y) be a point in the xy plane. Suppose that U is the unit vector making an angle of radian measure 0 with the positive side of the r axis. Then LI: cos 0i * sin 0i Figure 20.1.1 shows the representation of U having its initial point at the point P(x, y).

P(x,y)

Figure 20.1.1 20.1.1 Definition

Let f b" a function of two variables r and y. If. U is the unit vector cos 0i * sin ei, then the directional deriaatiae of f in the direction of U, denoted by Duf , is given by

if this limit exists. z:

+ h sin d,0)

Figure20.1.2

f(x'Y)

The directional derivative gives the rate of change of the function values f(x, y) with respect to the direction of the unit vector U' This is illustrated in Fig. 20.L.2.The equation of the surface S in the figure is z: f(x, A). Po(xo,Ao,zo)is a point on the surface, and the points R(rs, ys, 0) and Q(ro * h cos 0, yo * h sin 0,0) are points in the xy plane. The plane through R and Q, parallel to the z a>tis,makes an angle of 0 radians with the positive direction on the r axis. This plane intersects the surface S in the curve C. The directional derivative Dol, evaluated at Pe,is the slope of the tangent line to the curve C at Po in the plane of R, Q, and Pe. If U : i, then cos 0 : 1 and sin 0: 0, and we have from Definition 20'1"1' .. f (x + h, y) - f (x, y) : D'f(r,

/)

liq'h-o

h

APPLICATIONS DIRECTIONAL DERIVATIVES, GRADIENTS, DERIVATIVES, OF PARTIAL AND LINEINTEGRALS

which is the partial derivative of / with respect to r. If U : i, then cos 0: 0 and sin 0 : l,and we have \

: lim Dtf -;-; \ ' . u) J, t r G,

f(x,V+h)-f(x,y) h

which is the partial derivative of / with respect to y. So we see that f , and /, are special casesof the directional derivative in the directions of the unit vectors i and i, respectively. o rLLUsrRArrow1: We apply Definition 20.1.1to find Dul if - y2* 4x f(x, Y) :3* and U is the unit vector in the direction *zr. Then LJ: cos *zri * sin f7i : iltfgi+ *i. So from Definition 20.1.1we have

f@+ +\/-gh' Y! +h)- f (x'v) 4,f @,/) : lim h h-o 3(x + +\/3h)2- (y + +h)2+ 4(x + +\/-gh)- (g* - y2+ 4x) _ : [ mr_ + h

- - ) ^ r * * r { r r r * * Z n " - y , - r r-, * t ,* n * r t / - ! , - u l , *r , - n : rrm3\/3hx+ Zh'z-UY +h2+ 2\fgh n

h-o

: lim @{3x + *h - V - +h+ 2\/g) lr-0 :3{3x-y+2fi

.

we now proceed to obtain a formula that will enable us to calculate a directional derivative in a way that is shorter than if we used the definition. We define g as the function of the single variable t, keeping x, y, and d fixed such that g ( t ) : f ( x + t c o sd , y * t s i n ? )

(1)

and let U: cos 0i * sin 0i. Then by the definition of an ordinary derivative we have 8,(0):,.*/(x+ "'

(0 + h) cos0, y * (0 + ft) qind) -/(r*0

, r r - . h oha ccos A 0'

g,(0): rr^,'-' h-o

cos0, y +0 sind)

h t t - . i t r sin c i - A0) \ Y'*-h f(x, Y)

h

Becausethe right side of the above is Dof (x, y), it follows that

8'(o): Duf@,y)

e)

We now find g'(t) by applying the chain rule to the right side of

AND GRADIENTS 947 DERIVATIVES 20.1 DIRECTIONAL

(1), which gives g'(t):fr(x+t

sin0)

c o s0 , y + t

* f'(r * :fr(r*f

d(r * f cos 0)

at f c o so , y + t s i n g )

o(Yttsrno) at

c o s0 , y + t s i n 0 ) cos 0 *fr(r*f

c o s0 , Y + t s i n 0 ) s i n 0

Therefore, y) cos 0 * f oQ, Y) sin 0 g'(0) : f "(x, From (2) and (3) the following theorem is obtained' 20.1.2 Theorem

(3)

If f is a differentiable function of r and u, and LI: cos 0i * sin 0i, then

Illustration 1, o rLLUSrRArrow2: For the function / and the unit vector U of we find Duf by Theorem 20.1.2' - y2 * 4x, f ,(x, y): 6x * 4 ul1 y) - -2y' geca;e fir, y):3f ["\*' BecauseU: costri * sin tzi, we have from Theorem20"l"2

Duf(x,11: (6x+ 4)+\/g+ (-2y)l: gtfgx+ Z{3 y ' which agreeswith the result in Illustration 1' The directional derivative can be written as the dot product of two vectors. Because g: (cos0i + sin 0i)' lf"(x' y)i+ f"(x' y)il f,(x, y)cos 0 * fu@,y) sin it follows from Theorem 20.1,.2that

D u fk , A ) :

( c o s0 i + s i n 0 i ) ' l f * ( x , y ) i + f r ( x , y ) i l

(4)

The second vector on the right side of Eq' (4) is a very important one' use for and it is called the "gradient" oithe function/. The symbolthat we read is and delta capital inverted an is V where is"if, ,t" gt"af""t of f "del" sombtimes the abbreviation grad f is used' 20.1.3 Definition

then the If / is a function of two variables x and y and f* ffid fu exist' gradientoff,denotedbyVf(read:"delf")'isdefinedby

Using Definition 20.!.3, Eq. (4) can be written as

D u f@ , y ) - U , . Y(f x , Y )

(5)

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINEINTEGRALS

Therefore, any directional derivative of a differentiable function can be obtained by dot-multiplying the gradient by a unit vector in the desired direction. ExAMPLE 1:

If

solurroN: Becausef"(x, y):

f (x,a):#.n;

9f (x,y) : txi * &yj

find the gradient of f at the point (4, 3) . Also find the rate of changeof f (x, y) in the direction *n at (4, 3) .

$r and f ,(x, y) : &y, we have

Therefore,

V f ( 4 , 3 ) : t i +? i The rate_of change of f(x, y) in the direction |rr at (4,3) is Dvf(4,3), where U is the unit vector 1.,

1

v.'*v.l This is found by dot-multiplying Vf (4, g) by U. We have, then, t1 1.\ 7 /1,..2.\ Duf(4,3):

(- i*#

t/ 6 t+ii):;

\n

If a is the radian measure of the angle befween the two vectors u and Vf, then

U . Yf: lullv/l cosa

(6)

From Eqr. (5) and (d) it follows that

Duf: lullv/l cosd

v f(4,3)

' rLLUsrRArron 3: In Figure 20.1.3 we have a contour map showing the level curves of the function of Example '1.at'!.,2, and.3.The level curves are The figure also shows thJrepresentation of.Vf(4,3) having its 9U]ns9s. initial point at (4, 3). .

Figure20.1.3

EXAMPTN 2:

(7)

We see from Eq. (7) that Duf will be a maximum when cos a : 1., that is, when U is in the direction of Vf; and in this case,Duf : lV/|. Hence, the gradient of a function is in the direction in which thelunciion has its max_ imum rate-of change. In particurar, on a two-dimensional topographical map- of a landscape where z units is the elevation at a point 1i, yy ana z: f(x, y), the direction in which the rate of change of z is the gr€atest is given byYf (x' y); that is,yf (x, y) points in the direction of steepest ascent. This accounts for the name "gradient" (the grade is steepestir, tr," direction of the gradient).

Given

f (x,y) - 2 x 2 - y 2 * J x - y

sot,urroN: f*(x, U) : 4x * 3 and fo(x, A) : -2yYf (*, A) : @x + 3)i + (-2y - l)i

L. So,

AND GRADIENTS 20.1 DIRECTIONALDERIVATIVES

find the maximum value of Dvf at the point where x - l a n d Y:-2-

Therefore,

v f ( \, - 2 ) - 7 i + 3 i So the maximum value of Du f ut the point (L,-2) is

lv/(r,-2)l: \M: ExAMPLE3: The temperature at any point (x, y) of a rectangular plate lying in the xy plane is deter- xz * mined by T(x, y) Y'. (a) Find the rate of change of the temperature at the Point (3, 4) in the direction making an angle of radian measute +n with the Positive r direction; (b) find the direction for which the rate of change of the temPerature at the point (-3 , "l') is a maximtlm.

solurroN:

\68-

(a) We wish to findDuT(x, y), where

U : cos*zri* sin *ni : Li + +\/3i BecauseT(x, y) : f * y', Tr(x, y) : 2x, and To(x, y) : 2y- Hence, VT(x, Y) : T,(x, Y)i + Tok, Y)i:2xi + 2Yi Therefore, DvT(x, /) : g 'YT(x, Y)

: (ti+ +{'il.Qrt+2vi) : x* $y Hence, DuT(3, 4) : 3 + +tE = 3 * 4(1.732): 9'93

(-3, 1) is a maximum in the direction making an angle of radian measure zr - tan-l * with the positive side of the r axis. we extend the definition of a directional derivative to a function of three variables. In three-dimensional space the direction of a vector is determined byits direction cosines. So we let cos d/ cos F, and cos 7 be the direction cosines of the unit vector U; therefore, IJ: cos oi * cos Bi * cos 7k. 20.1.4Definition

Suppose that / is a function of three variables x, y, atrd z' lf U is the unit .tr""io, cos ai i cos Bi * cos 7k, then the directional derivative of / in the direction of U, denoted by Duf , is given by *+u*ul+$**u'.;i*+i

if this limit exists. The directional derivative of a function of three variables gives the

APPLICATIONS DERIVATIVES, AND LINEINTEGRALS OF PARTIAL GRADIENTS, DERIVATIVES, DIRECTIONAL

rate of change of the function values f (x, y, z) with respect to distance in three-dimensional space measured in the direction of the unit vector U. The following theorem, which gives a method for calculating a directional derivative for a function of three variables, is proved in a manner similar to the proof of Theorem 20.7.2. 20.1.5 Theorem

If. f is a differentiable function of x, y , and z and u: then

cos ai * cos Bi * cos 7k ' (8)

ExAMPrn 4: Given f (x, A, z) : 3x2 * xy - 2y' - yz * 22, find the rate of change of f (x, y, z) at (I , -2, -L) in the direction of the vector 2i - 2i - k.

solurroN:

The unit vector in the direction of 2i - 2i - k is given by

U:3i-gi-+k Also, f(x, U, z) :3*

* rY - 2Y2- Yz * z2

So from (8)

Duf@,y, z) : &(6x+ y) - 3@- 4y- z) - tFy + 2z) Therefore,the rate of change of f(x,y,z) of U is given by

at (!,-2,-1)

in the direction

- +(o) D"f(r,-2, -r) - 3@)- 3(10) 20.1.6 Definition

If / is a function of three variables x, y, and z and the first partial derivalives f ,, fu, and f , exist, then the gradient of /, denoted,byvf , is defined by

Just as for functions of two variables, it follows from Theorem 20.1.5 and Definition 20.1.6that if IJ : cos ai * cos Bi * cos 7k, then

D u fQ , y , z ) - u ' Y f ( * , y , 2 )

(e)

Also, the directional derivative is a maximum when U is in the direction of the gradient, and the maximum directional derivative is the magnitude of the gradient. Applications of the gradient occur in physics in problems in heat conduction and electricify. Suppose that w:f(x, y, z). A level surface of this function f at k is given by f(x, y, z) : k

(10)

If w is the number of degrees in the temperature at point (x, y, z), then

AND GRADIENTS 951 DERIVATIVES 20.1 DIRECTIONAL

all points on the surface of Eq. (10) have the same temperature of k degrees, and the sur{ace is called an isothermalsurface.lf w is the number of volts in the electric potential at point (x, y , z) , then all points on the surface are at the same potential, and the surface is called an equipotential surface.The gradient vector at a point gives the direction of greatest rate of change of w. so if the level surface of Eq. (10) is an isothermal surface,Yf(x, y, z) gives the direction of the greatest rate of change of temperatuie at (x , y , z) . If. Eq. (10) is an equation of an equipotential surface, ihen V/(r, y, z) gives the direction of the greatestrate of change of potential at (x, y , z). 5: If V volts is the ExAMPLE electric potential at anY Point (x, y , z) in three-dimensional

v-L|ffi, rpu..and find: (a) the rate of change of V at the point (2, 2, -L) in the di- 3i + 6k; rection of the vector 2i and (b) the direction of the greatest rate of change of-V at (2,2, -1).

+ 22. solurroN: Let f (x, y , z) -'J,l\m2 2i - 3i * 6k is of direction in the vector (a) A unit

u-?i-+i ++k we wish to find D"f (2,2, -1). V f( x , y , z ) : f * ( x , y , z ) i + f , ( x , A , z ) i + f " ( x , U , z ) k -x (x'*y'+22)3t2

k

(x'*y'+22)st2

Then we have D"f (2,2, -1) : lJ ' Yf(2, 2, -L)

:g;,Jru?hi-+j++k) :T€g

- 0.042 Therefore,at (2, 2,-1) the potentialis increasingat the rateof 0.042volt per unit changein the distancemeasuredin the direction of u. (b)Vf(z,2,-L):-*i-+i+#keunitvectorinthedirectionof v f ( 2 , 2 ,- 1 ) i s

y-\7,?,-:),, --f,Ti - &i + +k : -1i- Ai+ +k -lvf(2,2,-7)l+' -?, -3, and *, which give the diThe direction cosines of this vector are -1). rection of the greatest rate of change of V at (2, 2,

20.1 Exercises given function in the direction of the given unit vector U In Exercises 1 through 4, tind the directional derivative of the verify your result by upplying either Theorem 20.L.2 or by using either pefinition zo.I.L or Definition 20.I.4, and then Theorem 20.1.5, whichever one applies'

952

APPLICATIONS DERIVATIVES, GBADIENTS, OF PARTIAL AND LINEINTEGRALS DERIVATIVES, DIRECTIONAL

l. f(r, V) :21' + 5y\ V: cos*zri* sin *zj

2. gG,v) : '"

1

r, + yr;u

3. h(x' A' z) :3*

: gi- *i

-f y2 - 4z2iU:

cos izri * cos {zi * cos 3zk

4 .f ( x , V , z ) : 6 * - 2 x y t V z ; U : + i + + j + + k In Exercises5 through 1.0,find the value of the directional derivative at the particular point Pofor the given function in the direction of U. 5. g(*, y\ : y, tan2r; U : -i{Si * fi; Po: (*n, 2)

5. f (r, !) : xezu; U: +i + t!'i;

Po: (2, 0)

7 . h ( x , y , z ) : c o s ( r y*)s i n ( y z ) ;U : - + i + A j + 3 k ; f o :

s .f ( r ,a ,z ): t n ( f * y 2+ 2 2 ) ; u : +F + t - + k ; 9. f (r, y) : s-et cos3y; U:

e,0,-3)

p o :0 , s , 2 )

cos(-#z)i f sin(-# n)j; Po: t#n,

r0.g(x,a,z):cos 2r cos3ysinh4z;rt:+

t-

0)

u, : (tn,o,0) ".

*$ # t

In Exercises11 through 14, a function /, a point P, and a unit vector U are given. Find (a) the gradient oI f atP,and (b) the rate of change of the function value in the direction of U at P. 11..f (x, U) : x2- 4y; P : (-2, 2) ; V : cos *zri * sin *zj 1 2 .f ( x , y ) : e z " uP; : ( 2 , 1 ) ; U : € i - g i 13. f (x, !, z) : y' + zz- 4xz; P : (-2, l, 3) ; U : +i - +i + +k M. f(x,!,2):2f * t y z * x z zP ; : (1,1,1); U: +\/-21i-+\nk 1.5.Draw a contour map showing the level curves of the function of .Exercise11 at 8, 4,0,-4, and -8. Also show the representationotVf (-2,2) having its initial point at (-2,2). L6' Draw a contour map showing the level curyes of the function of Exercise12 at eB, ea,7, e-a, and e-s. Also show the representationof Yf(z,1) having its initial point at (2,1). In Exercises17 through 20,IindDvf at the given point P for which U is a unit vector in the direction of f!. Dvf , if. U is a unit vector for which Dul is a maximum. 1 7 .f ( x , A ) : e " t a n - l y ; P ( 0 , 1 ) , Q ( 3 , 5 ) 7 9 .f ( x , A , z ) : x - 2 y * 2 2 ; P ( 3 , 1 , - 2 ) , Q 0 0 , 7 , 4 )

1 8 .f ( x , ! ) : e "

atso atpfind

c o s y * e us i n x ; p ( t , O ) , e ( - 3 , 3 ) -tyr-4xz;p(3,1,-2),ee6,g,4)

2 0 .f ( x , ! , 2 ) : f 21. Find the direction from the point (1, 3) for which the value of does not change if. (x, y) : e2otan-l(yl3x). / f 22. The density is p slugs/fP at any point (x, y) of a rectangular plate in the ry plane and p : U \/F + it-+ j. (a) Find the rate of change of the density at the point (3,2) in the direction of the unit vector cos Szri * sin |r,j. (b) Find the direction and magnitude of the greatest rate of change of p at (3,2). 23. The electric potential is V volts at any point (x, y) in the ry plane and V: e-2" cos2y. Distance is measured in feet. (a) Find the rate of change of the potential at the point (0, *z) in the direction of tlie unit vector cos *zi + sin *zi. (b) Find the direction and magnitude of the greatestrate of change of V at (0, *z). 24. The temperature is T degreesat any point (x, y, z) in three-dimensional spaceand T: 601(* t y, + zz -t 3). Distance is measured in inches. (a) Find the rate of change of the temperature at the point (3,-2,2) in the direction of the vector -2i + 3i - 6k. (b) Find the direction and magnitude of the greatest rate of change of T at (g , -2, 2) .

20,2TANGENTPLANESAND NORMALSTO SURFACES 953

25. An equation of the surface of a mountain is z : 1200- 3f - 2y2, where distance is measured in feet, the r axis points to the east,and the y axis points to the north. A mountain climber is at the point correspondingto (-10,5,850). (a) What is the direction of steepestascent?(b) If the climber moves in the east direction, is he ascending or descending, and what is his rate? (c) If the climber moves in the southwest direction, is he ascending or descending, and what is his rate? (d) In what direction is he traveling a level path?

20.2 TANGENT PLANES AND Let S be the surface having the equation NORMALS TO SURFACES - 0 F(x,y, z)

(1)

and supposethat P6(16,!o, zo)is a point on S. Then F(xo,yo, zo) : 0' Suppose further that C is a curve on 5 through P6and a set of parametric equations of C is x-f(t)

y:8(f)

z-h(t)

(2)

Let the value of the Parametert at the point Psbe fo.A vector equation ofCis (3) + h(t)k + R ( f )-

f(f)i

s(f)i

Becausecurue C is on surface 5, we have, upon substituting from (2) i n ( 1 ), F ( f ( t ) , g G ), h ( t ) ) - 0

(4)

Let G (f ) : F (f (t), 8(t), h(t)) . If F,, Fa,and F" are continuousand not , all zero at Po,and If f (t) , g' (to), and h'(fe) exist, then the total derivative of F with respect to f at Ps is given by ' G'(fo; : F,(xo, ao, zo)f (to) + Fr(ro, !s, zs)g'(to)+ F,(ro, ys, z)h' (tr) which also can be written as G'(fo) : VF(xo,!o, zo)' DrR(to) Because G'1f;:0 for all f under consideration (because of (4)), :0; so it follows from the above that G'1t01 (5) YF(xs,Ao, zo)' D1R(fs): 0 From sec. 18.9, we know that DrR(fo) has the same direction as the unit tangent vector to curve C at P6.Therefore, from (5) we can conclude that the lradient vector of F at P6is orthogonal to the unit tangent vector of C on S through the point Ps.We are led, then, to the following ".i*" ".r"ry definition. 20.2.L Definition

A vector which is orthogond to the unit tangent vector of every curve C through a point Pe on a surface S is called a normal aector to S at Ps. From this definition and the preceding discussion we have the following theorem.

DIRECTIONAL DERIVATIVES, GRADIENTS,'APPLICATIONS OF PARTIAL DERIVATIVES, AND LINEINTEGRALS

20.2.2Theorem

If an equation of a surface5 is F(r, y, z) : 0, and Fa, Fu,and F, are continuous and not all zero at the point Po(xo,yr, zs)on S, then a normal vector to 5 at Po is VF(ro, Ao, zo). We now can define the "tangent plane" to a surface at a point.

20.2.3 Definition , zo)

If an equation of a surfaceS is F(r, y, z) :0, then the tangentplaneof S at a point Po(ro, !o, zi is the plane through Pe having as a normal vector VF(xo,Ao,zo). An equation of the tangent plane of the above definition

is Fr(xo,Uo,zo)Q ro) * Fu(xo,!0, z) (U - A) * Fr(xo,!o, zo)k - zo) : 0

(6)

Refer to Fig. 20.2.1,which shows the tangent plane to the surface S at

Po and the representation of the gradient vector having its initial point at Po. A vector equation of the tangent plane given by (5) is Figure20.2.1

EXAMPLE 1: Find an equation of the tangent plane to the elliptic paraboloid 4x2* y, - l,Gz: 0 at the point (2, 4, 2).

VF(ro,Ao, zo). l(x- xs)i+ (y - A)i + e- zo)kl- 0

(7)

solurroN: Let F(x, y, z) 4xz+ y2- 16z.Then VF(r, A, z) : gxi* 2yi .: -76k, and so VF(2,4, 2):tli+Si - 15k. From ii toUowsthat in 1Z; equationof the tangentplane is 1 . 6 ( x - 2 )+ 8 ( y - 4 ) - 7 6 ( z - 2 ) : 0 2x*y-22-4:0

20.2.4 Definition

The normal line to a surfaces at a point po on s is the rine through pohaving as a set of direction numbers the components of t ot-al vector to s "ny at Ps. , If an equation of a surface S is.F(r, y, z) :0, symmetric equations of the normal line to S at Ps(xs,ys, zs) are (8) The denominators in ( 8 ) are components of YF(xs, Ao, zo), which is a normal vector to S at Po)thus, (8) follows from Definition 20.2.4. The normal line at a point on a surfaceis pe{pendicular to the tangent plane there.

EXAMPLE2: Find symmetric equations of the norrnal line to the surface of Example L at (2,4,2).

BecauseVF(2,4,2) :16i + 8i - '!,6k,it follows that symmetric equations of the required normal line are

SOLUTION:

x-2

y-4 _r-2 1-2

20.2 TANGENTPLANESAND NORMALSTO SURFACES

20.2.5 Definition

Figure 20.2.2

The tangent line to a curve C at a point Pois the line through Ps having as a set of direction numbers the components of the unit tangent vector to C at Po. From Definitions 20.2.3 and 20.2.5we see that all the tangent lines at the point Po to the curves lying on a given surface lie in the tangent plane to the surface at Pe.Refer to Fig. 20.2.2,showing sketchesof a sur?aceand the tangent plane at Po. Some of the curves through Ps and their tangent lines are also sketched in the figure. Consider a curve C which is the intersection of two surfaceshaving equations (e) F(x,y,z) - 0 and

G(x,y,z) - 0

(10)

(9) is given by N1 : VF(x6, !o, zo): Fr(rco,Uo, zo)i* Fo(xs,Ao,z)i I Fr(xo,Ao, zo)k and a normal vector at Ps to the surface having Eq. (10) is given by N2: VG(ro, Uo,zo): G,(xo,Ao,zo)i* Go(xs,Ao,z)i*

G"(xs'Ao' zo)k

of direction numbers of the tangent line. From this set of direction numbers and the coordinates of Pewe can obtain symmetric equations of the required tangent line. This is illustrated in the following example.

ExAMPLE3: Find symmetric equations of the tangent line to the curye of intersection of the surfaces 3r2 * Zy', * z2 - 49 and x2 + y2 - 222- L0 at the point

solurroN:

Let F(x, y, z) :3x2 * 2y2* z2- 49

and G(x, Y, z) : f f- Y' - 2zz- L0 : Then VF(r, y, z) : 5xi + ayj * 2zk and VG(r, y, z) 2xi + Zyi'

( 3 ,- 3 , 2 ) .

Nr: VF(3,

N,

2) :18i - rzi+ 4k:2(9i- 6i + 2k)

Nz : VG(3, -3, 2) - 6i - 6i- 8k 2(3i 3i 4k) X Nz: 4(9i- 6i + 2k) x (3i- 3i - 4k)

4zk' So


:4(30i + 42i-9k) : 1 2 ( 1 0+i 1 4 i : 3 k ) Therefore,a set of direction numbers of the required tangent line is -3]. Symmetricequationsof the line are,t'h"rr, 110,'J-.4, x-3 a *3 z-2 -3 10 14 If fwo surfaces have a common tangent plane at a point, the two surfacesare said lo be tangenfat that point.

Exercises 20.2 In Exercises1 through 72, frnd an equation of the tangent plane and equations of the normal line to the given surface at the indicated point. 1 . x 2+ y , + z 2: l T ; ( 2 , - 2 , 9 ) 4. x2* yz - z2: 6; (3, -L, 2) 7. x2: L2y; (6, g, g) L 0 .z f - x y ' - y z z : 1 8 ;

2. 4x2* y' + 222: 26; (1, -2, g) 5. y : e0 cosz; (1, e, 0) 8. z:

(0,-2,3)

-t,yttr; (1, L, 2) 76rtz

l l . x 2 t s l y z t a l z z 1 s : 1 4 ;( - g , 2 7 , 1 )

3 . x 2+ y 2- 3 z : 2 ; ( - 2 , - 4 , 6 ) 6. z: e3" sin Jy; (0, trr, l) 9. xrtz+ ytt' + zrtz: 4; (4, I , l.) 1 2 .x u z + z l t z : g ; ( 8 , 2 , 9 )

In Exercises13 through 18, if the two given surfacesintersect in a curve, find equations of the tangent line to the curve of intersection at the given point; if the two given surfacesare tangent at the given point, prove it. 13. x2+ y2- z- 8, x - y2 * z2- -2; (2, -2, 0) L 4 . x z + y 2- 2 z * 1 - 0 , x 2* y , - z z - 0 ; ( 0 , 1 , ! ) 1 5 . y : x 2 ,A : L 6- z 2 ;( 4 , L 6 ,0 ) L5. x: 2 + cosrTyz,y- 1 * sin nxz; (9, l, Z) 1 7 .x 2 * 2 2 + 4 y : 0 , * * y " + 2 2 - 6 2 I 7 : 0 ; (0,-1,2) 1 8 .f + y " + z r : g , y z : 4 ; (0,2,2) 19' Prove that every normal line to the sphere * + y" * 22: a2 passes through the center of the sphere.

20.3 EXTREMA OF FUNCTIONS OF TWO VARIABLES

20.3.1Definition

An important application of the derivatives of a function of a single variable is-in the study of extreme values of a function, which leads to a variety of problems involving maximum and minimum. we discussed this thor_ oughly in chapters 4 and 5, where we proved theorems involving the first and second derivatives, which enableilus to determine relative maximum and minimum values of a function of one variable. In extending the theory to functions of two variables, we see that it is similar to the oire-variable - case;however, more complications arise. I The function / of trvo variables is said to have an absolutemaximumaalue on a disk B_in the xy plane if there is some point (re, /o) in B such that f (xo, y) > f (x, y) for all points (x, y) inA. h such .*", f (xs, ys) is the " absolute maximum value of f on B.

OF TWOVARIABLES 957 OF FUNCTIONS 20.3EXTREMA

20.3.2Definition

The function / of two variables is said to have an absoluteminimumaalueon a disk B in the ry plane if there is some point (x6,90) in B such that (xs, ys) is the f (xo,Ai - f (x, y) f.ot all points (x, y) in B. In such a case,f absolute minimum value of f on B.

20.3.3Theorem Let B be a closed disk in t]|texy plane, and let f be a function of two variableswhich is continuous on B. Then there is at least one point in B where an f has an absolute maximum value and at least one point in B where/has value. absolute minimum

The Extreme-Value Theorem for Functions of Two Variables

The proof of Theorem 20.3.3 is omitted because it is beyond the scopeof this book. 20.3.4Definition

The function / of trvo variables is said to have arelatiue maximumaalue at the point (xs, !) if there exists,an open disk B((xo, y); r) such that disk' f(x, y) < f (xo, y) fot all (r, y) in the open

20.3.5Definition

The function / of two variables is said to have a relatiae minimum oalue at the point (ro, yo) if there exists an open disk B((xo, yr); r) such that disk' f(x, y) > f (xo, V) for all (x, y) in the open

20.3.6 Theorem

lf f (x, y) exists at all points in some open disk B((xo, yr); r) and if f has a relative extremum it (xo, yo), then if f,(xo, ys) and fu(xn, ys) exist, : o : f,(xo, Yr) fo(xo,U) pRooF:Weprovethatif/hasarelativemaximumvalueat(xo'yiandif :0. By the definition of a partial deriva1,t o, yr) exists, then f,(xo, /o) tive,

Because/ has a relative maximum value at (xo, ao),by Definition 20.3.4it follows that - Q o , U o )< o f f ( x o* A r , A J L'x whenever A,x is sufficientlY small so that (xo* L,x,Y) is in B. If > therefore/ 0; Ar right, the approacheszeto from Ax < 0' Hence, by Theorem4.3.4,if f *(xo,Ao)exists,f *@o,Ui Simila rly, tf L,x aPProacheszero from the left, Ar < 0; so f (xo* Lx, yo) f (xo,y) > o Lx

DIRECTIONALDERIVATIVES, GRADIENTS,APPLICATIONSOF PARTIALDERIVATIVES, AND LINE INTEGRALS

Therefore, by Theorem 4.3.5, if fr(xo, Ad exists, f ,(xo, Ui > 0. We conclude, then, that becausefr(xo, yi exists, both inequalittes, fr(xs, yo) = 0 and f ,(xo, Ao)> 0, must hold. Consequently,f ,(xo, Ad:0. The proof that f u(xs,Ad: 0, if.f u@n,ye)exists and/ has a relativemaximum value at (xo, yo),is analogous and is left as an exercise(seeExercise 13). The proof of the theorem when /(rs, ys) is a relative minimum value is also left as an exercise (see Exercise 14). I 20.3.7Definition

A point (xo,A) for which both f ,(xo, /o) : 0 andf n(xs,U) :0 is calleda critical point. Theorem 20.3.5states that a necessarycondition for a function of two variables to have a relative extremum at a point, where its first partial derivatives exist, is that this point be a critical point. It is possible fo" function of two variables to have a relative extremum at a point at which" the partial derivatives do not exist, but we do not consider this situation in this book. Furthermore, the vanishing of the first partial derivatives of a function of two variables is not a sufficient condition for the function to have a relative extremum at the point. such a situation occurs at a point called a saddlepoint. . rLLUsrRArroN 1: A simple example of a function which has a saddle point is the one defined by

f(x, y) : y'- * For this function we see that f,(x, and fu@, !):2y. a):-2x Both f*(0'0) and fr(0,0) equal zero. A skeich of the grapiiof tfre function is $own in Fig. 20.3.1,and we see that it is saddle srrapeaat points close to the onsin. It is apparent that this function does not-satisfy either Defini/ tion 20.3.4or 20.3.5when (x6,yo): (0, 0). .

Figure20.3.1

we have a second-derivative test which gives conditions that guarantee a function to have a relative extremum at i point where the first partial derivatives vanish. However, sometimes it is possible to determine rela_ tive extrema of a function by Definitions 20.3.+and 20.3.5,as shown in the following illustration. o rLLUsrRArrow2: Let be the function defined by / f(x,y):6r-4y-*-zyz We determine if f has any relative extrema. Because/ and its first partial derivatives exist at all (x, y) in R2, The_ orem 20.3.6is applicable. Differentiating, we get f,(x,y) :6-2x

a n d f u @ ,y ) : - 4 4y Settingfr(x,y) andfu(x, y) equal to zero,we get r:3

and y:-1.

The

OF TWOVARIABLES 959 OF FUNCTIONS 20.3EXTREMA

graph (seeFig. 20.3.2f.or a sketch) of the equation z:6x-4y-f-2y' is a paraboloid having a vertical axis, with vertex at (3,-L,1.L) and oPening downward. We can concludethat f(x,y)
Figure20.3.2 The basic test for determining relative maxima and minima for functions of two variables is the second-derinative test, which is given in the next theorem. 20.3.8 Theorem Second-Deriaatiae T est

Let f be a function of two variables such that f and its fils!- 1d secondorder partial derivatives are continuous on some oPen disk B((4, b); r\' Suppoie further that f ,(a, b) : f u(a, b) : 0. Then (i) / has a relative minimum value at (a' b) if b) >0 f,,(a,b)fuu@,b)-f'n'@,b) >0 and f,,(a' (ii) f has a relative maximum value at (a ' b) if > 0 and f*,(a'b) <0 f,,(a,b)fuu@,b) f*'(a,b) (1ii) f(n, b) is not a relative extremum if < 0 f,,(a,b)f,u@,b) f*'(a' b)

AND LINEINTEGRALS APPLICATIONS DERIVATIVES, OF PARTIAL GRADIENTS, DERIVATIVES, DIRECTIONAL

(iv) We can make no conclusion if : f ,"(a, b)foo@,b\ f *'(a, b) 0 pRooF or (i):

For simplicity of notation, let us define QQ' y) : f,,(x, a)foo@,Y) f ,o'k' Y)

We are given S(a, b) > 0 and f ,,(a, b) > 0, and we wish to prove that f(a, b) is a relative minimum function value. Becausefr, f ,n, and,fuo are continuous onB((a,b);r), it follows that f is alsocontinuous on B. Hence, there existsan open disk B' ((a, b); r' ), where r' = r,such that Q@,y) > 0 and f",(x, A) > g for every point (r, y) in B'. Let h and k be constants, not both zero, such that the point (a + h, b + k) is in B'. Then the two equations x:a*ht

and y:b+kt

0
define all the points on the line segment from (a, b) to (a * h, b + k) , and all these points are in B'. Let F be the function of one variable defined by

F ( t ) : f ( a + h t , b + kt)

(1)

By Taylor's formula (formula (9), Sec.15.5),we have

F ( t ) : F ( 0+ ) F ' ( O*) # f

n

where f is between 0 and f. lf t - 1 in Eq. (2), we get

F(l) : F(0) + F'(0) * tF" (f)

(3)

where 0 < f < L. BecauseF(0) - f(a, b) and F(1) - f(a + h, b + k), it follows from Eq. (3) that

f(a + h, b + k) : f(a,b) + F'(0)+ +F"(€)

(4)

where0
F" (t) : h2f,, + hkfu" + hkf,a + Pf uu where each second partial derivative is evaluated at (a * ht , b + kf ) . From Theorem lg.7.l,itfollows thatf,u(x,A) : fu,(x,y'| for all (x,y) in B'. So F" (t) : hzfr, + zhkfw * k'fou

(6)

where each second partial derivative is evaluated at (a * ht,b + kt). Substituting 0 for f in Eq. (5) and { for t in Eq. (5), we get F ' ( 0 ) : h f , ( a , b )* k f u @ , b ) : 0

(z)

OF FUNCTIONS OF TWOVARIABLES ,961 20.3EXTREMA and

F'(€):

(8)

h'f*,+2hkf*u*k'fru

where each second partial derivative is evaluated at (a*ht, b+kf)' where 0 < f < 1. Substituting from (7) and (8) into (4), we obtain

(e)

f(a*h,b+k)-f(a,to):$(hzf,,+2hkf"u*k'f*)

The terms in parentheseson the right side of Eq. (9) can be written as

. * hzf ,u* tsfuo: f,, lr,'+ 2hk ,, + 2hkf * (o'#)' (r'#)' "'#l So from (9) we have

f(a* h,b+k)- f(,,D:+lV .tr)' +rJTjld wl (10) Becausef ,,f ,o - f ,o2 evaluatedat (a + ht, b * kf) equals

6@+ht,b+kf) > o it follows that the expression in brackets on the right side 9f Eq. (10) is b+k0 > 0, it follows from Furthermori, b""uttt" fr*(a*ht, positive. êq. provedthat ) Hence,wehave 0. (fO) thatf(a+h,b +k) f(a,b) f(a*h,b+k)

> f(a,b)

I

(iv) is included to cover all possible cases.

ExAMPLEL: Given that / is the function defined bY

f(x,A):2f*Y'-x2-2Y determine the relative extrema of f it there are arry.

SOLUTION:

:2Y - 2 f , ( x , y ) : 8 x B 2 x f u @ ,Y ) x:0, and x:*' Setting Settingfr(x,y):0, we get x:-t, and Therefore y:1. git , f n bolh vanish at the Points f, fok, y):6, *" (-+, L), (0, 1), and (t, 1). To apply the second-derivative test, we find the second partial derivatives of / and get

- Z f,,(x, Y) :24f f, , ( - i , 1 ) : 4 > 0 and

foo(r,Y) :2

f,u(x'Y) :0

:4' 2- o:8 > o f . , ? E , l ) f o o ? t ,t ) f * o ' e + , 1 )

Hence , by Theorem z}.g.g(i) , f has a relative minimum

value at

r)

AND LINEINTEGRALS OF PABTIALDERIVATIVES, APPLICATIONS GRADIENTS, DERIVATIVES, DIRECTIONAL

- : -4 ( 0 : f ,,(0, 1)fou(0,L) f ,u'(0, 1) (-2) (2) 0 and so by Theorem20.3.8(iiil,/ does not have a relative extremumat (0,1). and

>0

f ""(*' l)

t) - f,o'G, t1 - 4 ' 2 - 0 : 8 > 0 f ,*(t, L)foo(L, Therefore,by Theorem 20.3.8(i)/ has a relative minimum value at (+, 1). Hence, w€ concludethat / has a relative minimum value of - I at each of the points (-i, t) and (t, L7.

ExAMP:.;.2. Determine the relative dimensions of a rectangular box, without a top and having a specific voluril€, if the least amount of material is to be used in its manufacture.

Let )c- the number of units in the length of the base of the box; y - the number of units in the width of the base of the box; z - the number of units in the depth of the box; S - the number of square units in the surface area of the box; V - the number of cubic units in the volume of the box (V is constant). x, y, and z are in the interval (0, +o;. Hence, the absolute minimum value of S will be among the relative minimum values of S. We have the equations solurroN:

S:xy*2xz*2yz

and V:ryz

Solving the second equation fot zin terms of the variables r and y and the constant V, we get z : Vlry . And substituting this into the first equation gives us S : x-1y*tx! + ! Differentiating, we get

as E:y-

2v f

azs 4v

M:7

ds 2v ay:r-T azs

ffi_l

A2S

4V

oy'

y'

Setting 6Sl0x:0 and 0Sl0y:0, and solving simultaneously,we get x 2 Y- 2 V : 0 r y 2- 2 V : 0 from which it follows that r : 9fr

, and y : W.

For these values of r

20.3 EXTREMAOF FUNCTIONSOF TWO VARIABLES

and A, we have

azs_ 4v dxz (W)3 a2S 0x2

_ - 4 v : -) '> n/ v 2v

a 2 S _ ( a r s \ r _:@'798ff-r _4V L_l_3>o

@-w)

From Theorem 20.3.8(i), it follows that s has a relative minimum value nd y:{N' and hence an absoluie minimum value when x:flTv From these values of r and Y we get

etrWe therefore conclude that the box should have a square base and a depth which is one-half that of the length of a side of the base' Our discussion of the extrema of functions of two variables can be extended to functions of three or more variables. The definitions of relative extrema and critical point are easily made. For example, if.f is a function : : of the three variablis r, y, alrrdz, arrd f ,(xo, Yr, zo) fo(xo, yo' zo) (rs, ys, zo) is a critical point of /' Such a point is fr(xo, yo,zo):0, then 'oUtui""a by solving simultaneously three equations in-three unknowns. For a functio n of.n variables, the cltical points are found by setting all the n first partial derivatives equal to zero and solving simultaneously the n equati6ns in n unknowns. The extension of Theorem 20.3'8 to functions of tht"u or more variables is given in advanced calculus texts. In the solution of Example 2 we minimized the function having function values ry * 2xz * 2yz, iubject to the condition that r, A, ad:rdz satr;sfy Cbmpare this with Example 1, in w-hich we found the equatioi ryr:V. :2f * y2- *- 2V' the relative extrema of the filnction f for which/(', y) These are essentially two different kinds of problems becausein the first casewe had an additional condition, called a constraint (or sidecondition). Such a problem is called one in constrainedextterna,whereas that of the second type is called a problem in free extrema' Ihe solution of txample 2 involved obtaining a function of the two variables x and y by replacing z in the first equation by-its value from fhe secondequation. Another method that can be used to solve this exampleis due to ]oseph Lagrange, and it is known as the method of.Lagrangemultipliers. The- theory Uifrina this method involves theorems known as implicit function theorems, which are studied in advanced calculus. He^nce,a proof is not given here. The procedure is outlined and illustrated by" examples. Supiose that we wish to find the critical points of a function /of the : three variables x, y , and z, subiect to the constraint g (r, y ' z) 0' We in-

964

AND LINEINTEGRALS DERIVATIVES, OF PARTIAL APPLICATIONS GRADIENTS, DEBIVATIVES, DIRECTIONAL

troduce a new variable, usually denoted by )t, and form the auxiliary function F for which F(x, y, z, I") : f (x, y, z) * I,,g(x,Y, z) The problem then becomes one of finding the critical points of the function F of the four variables x, y , z, and )t. The method is used in the following example. EXAMPT,N 3: Solve Example 2 by the method of Lagrange multipliers.

sor.urroN: Using the variables x, A, and z and the constant V as defined in the solution of Example 2,let S: f(r, A,z) : xy * 2xz* 2yz and g(x, y, z) : ryz - V We wish to minimize the function / subject to the constraint that g(x, y, z) : o Let F(x, y, z, \) : f (r, y, z) + lrg(x, y, z) :xyIZxz*2yz+L(xyz-V) Finding the four partial derivatives F*, Fo, F, and Fr and setting the function values equal to zero, we have Fr(x, A, z, ),) : y * 2z * \yz- 0

(1 1 )

F o ( x ,y , 2 , t r ) : x + 2 z * \ x z - 0 Fr(x, U, 2,,^,) :2x * 2y * \xy - 0 F ^ ( x , y , z ,t r ) : x y z - V - 0

(r2) (1 3 )

(r4)

Subtracting corresponding members of Eq. (L2) from those of Eq. (11)we obtain y-x+)\z(y-)c)-0 from which we get Q-tr)(L+|,2)-0 giving us the two equations y:)c

(15)

r__1z

(15)

and

Substituting from Eq. (15) into Eq. (L2) we get x + 2z - x : 0, giving z : 0, which is impossible becausez is in the interval (0, +m). Substi-

203 EXTREMAOF FUNCTIONSOF TWO VARIABLES 96s

tuting from Eq. (15) into Eq. (13) gives 2x*2x*\x2-0 x(4 * Ir) - 0 and becausex * 0, we get

- -41x, w€ have If in Eq. (L2) we take I

x*zr-I.Qz)-o xi2z-42-0 )c zr_

Fl

(17)

z-

Substituting from Eqs. (15) and (17) into Eq. (14), we get L.x3-V:0, From Eqs. (15) and. (17) It follows from which it follows that r:{2V. agreewith those found in the results These and z: that y:lfl27 ^W. 2. Example solution of Note in the solution that the equation F1(r, A, z, L) 0 is the same as the constraintgiven by the equationV: xyz. If several constraints are imposed, the method used in Example 3

and p. for which F(x, y, z, \, p) : f (*,y, z) + t g(x, V, z) + p'h(x,y, z) The following example illustrates the method. ExAMPLr'4: Find the relative extrerna of the function / if - )cz+ Yz and the Point f (*,A, z) (x, y , z) lies on the intersection of the surfac es )c2* z2 2 and yz:2-

soLUrIoN: We form the function F for which F(x, y, z, I', p) : x,z* yz + \(* * z2- 2) + t'c(yz 2) Finding the five partial derivatives and setting them equal to zero, we have ( 1 8) Fr(x,U,z, tr,p) : z + 2\x - 0

Fo(x,y, z, tr,p) : z +

(1e)

F " ( x ,! , z , t r , p ) - ) c * y * 2 ) r z * p A - 0 -x2*22-2-0 F ^ ( r ,! , z , t r , F r )

(20) (2t)

966

DERIVATIVES, AND LINEINTEGRALS APPLICATIONS OF PARTIAL GRADIENTS, DERIVATIVES, DIRECTIONAL

F * ( x ,A , z , I r ,p ) : y z - 2 : 0

(22)

From Eq. (19) we obtain p - -l and z: 0. We reject this contradicts Eq. (22). From Eq. (18) we obtain

0 because

(23)

-t

Substituting from (23) and p:

into Eq. (20),we get

x* and so (24)

x2:22

Substituting from Eq. (24) into (21.),we have 2* - I- 0, or *: L. This gives two values for r, namely 1 and -1; and for each of these values of r we get, from Eq. (24), the two values 1 and -1 for z. Obtaining the corresponding values for y fuomEq. Q2), we have four sets of solutions for the five Eqs. (18) through (22). These solutions are

2

x-

1

Y:

x-

|

Y:-2

x,:-t

y:

x:-t

y:-Z

z-

1

z:-t 2

z-

|

z:-l

r--+ r- + r- + r--+

p--r p--r p--L p--L

The first and fourth sets'of solutions give f(x, y, z):3, and the second and third sets of solutions give f (x, y , z) : 1. Hence, / has a relative maximum function value of 3 and a relative minimum function value of 1.

Exercises20.3 In Exercises L through6,

determine the relative extrema of f ,if there are any.

1 64, T xy

1. f (x, A) : l8x2- 32y' - 35x- l28y - 110

2 .f ( x ' y ) - -

3. f(x, A) : sin(r * y) * sin x + sin y

4 . f ( x , y ): x 3 * y ' - I 8 x y

5. f (x, A) : 4xY' - 2x'y - x

6 . f ( x , y )_ 2 x * 2 y t L

Xy

x2*y'+t

In Exercises7 through 12,use the method of Lagrangemultipliers to find the critical points of the given function subject to the indicated constraint. 7. f(r, y) : f * 2ry * y' with constraint x - y : 3 8. f (t, V) : vz -t xy * 2y2- 2x with constraint r - 2y -t 1:0 9' f(r, y) : Zs - f - y2 withconstraint * -f y' - 4y : o

TO ECONOMICS DERIVATIVES OF PARTIAL 20.4SOMEAPPLICATIONS * 2y21- 5 with constraint* + y' - 2y :0 L L .f (x, y, z) : * * y' I z2 with constraint 3x - 2y * z - 4 : 0 L2. f(x, y, z) : f * y' * z2 with consilaint a2- * : I L3. Prove that f u(n, y) : 0 if.f o(xn,y6) exists and / has a relative maximum value at (xo, y). 10. f (x, y) :4*

14. Prove Theorem20.3.6 when f(rs, yo) is a relative minimum value. 16. Prove Theorem20.3.8(iii). 15. Proye Theorem 20.3.8(ii). 1 7 . Find three numbers whose sum is N(N > 0) such that their product is as great as possible. L8. Prove that the box having the largest volume that can be placed inside a sphere is in the shape of a cube. 19. Determine the relative dimensions of a rectangular box, without a top, to be made from a given amount of material in order for the box to have the Sreatest possible volume. run, and classB 20. A manufacturing plant has two classifications for its workers, A and B. Class A workers earn $14 Per salaries of the workers, the to in addition that is determined run, it production certain workers eam $15 p", *tt. For a ifrclassAworkersandyclassBworkersareused,thenumberofdollarsinthecostoftherunisys*f-Bry+600' workers of How many workers of each class should be used so that the cost of the run is a minimum if at least three each class are required for a run? :1. If T degrees-is the temperature at any 21. A circular disk is in the shape of the region bounded by the circle I I y' points on the disk' coldest and hottest the yi tind point 1r, y) of the disk and i:2xz + !, distance. 22. Find the points on the surface yu - xz:4 that are closestto the origin and find the minimum and the plane x- 4y - z:0 that are 23. Find the points on the curve of intersection of the ellipsoid I + 4y2* 422:4 closest to the origin and find the minimum distance' the material for the bottom of the box 24. A rectangular box without a top is to be mad,eat a cost of $10 for the material. If find the dimensions of the box of foot, costs l5c per square foot and the material for the sides costs 30g per square greatest volume that can be made. The cost of the material for the top 25. A closed rectangular box to contain 16 ff is to be made of three kinds of material. per !f , and the cost of the material 8c is back and the front for the material and the bottom is 99 per ff, the cost of the of the materials is a minimum' cost the that so box the of dimensions the Find per f8. is 6c for the other two sides f * y" I zz : 4' andT:700xy22'Find 25. Suppose that T degrees is the temPeratufe at any point (x, y, z) o1 the sphere where the temperature is the least' points the also and the points on the sphere where the temperature is ttre greaiest Also find the temperature at these points. * * y'* z2 wilh the two constraintsx-y + z:0 and 27. Find the absolutemaximum function value of f it f(x,y,z\: = multipliers' 25f + 4y2+ 20z,2 100' Use Lagrange 3 y 2 * 2 z 2 w i t h t h e t w o c o n s t r a i n t sx - 2 y - z : 6 28. F i n d t h e a b s o l u t e m i n i m u m f u n c t i o n v a l u e o ff i t f ( x , a , z ) : * - 1 multipliers. Lagrange 4. Use 2z: and x 3y *

20.4 SOME APPLICATIONS OF BARTIAL DERMTMS TO ECONOMICS

In Sec. 5.6 we discussed a demand equation giving the relationship between x andp, where p dollars is the price of one unit of a commodity when r units are demanded. In addition to the price of the given commodity, the demand often will depend on th9 prices of other commodities 1."1ut"dto the given one. In particular, let us consider two related commodities for wiich p dollars is the price per unit of r units of the

DERIVATIVES, AND LINEINTEGRALS APPLICATIONS OF PARTIAL GRADIENTS, DERIVATIVES, DIRECTIONAL

first commodity and 4 dollars is the price per unit of y units of the second commodity. Then the demand equations for these two commodities can be written, respectively, as

o(x,p,q)-0

and FU, p,q):0

ot, solving the first equation for x and the second equation for !, as

x - f(p, q)

(1)

y - s ( p ,q )

(2)

and

Figure 20.4.1

Figure 20.4.2

The functions / and g in Eqs. (1) and (2) are the demand functions and the graphs of these functions are surfaces. Under normal circumstancesx, A, p, and q are nonnegative, and so the surfacesare restricted to the first octant. These surfacesare called demandsurfaces.Recalling that p dollars is the price of one unit of r units of the first commodity, we note that if the variable 4 is held constant, then r decreasesas p increasesand r increasesas p decreases.This is illustrated in Fig. 20.4.1,,which is a sketch of the demand surface for an equation of type (1) under norrnal circumstances.The plane 4 : b ,intersectsthe surface in section RST. For any point on the curve RT, q is the constant b. Referring to the points M(pr, b, rt) andN(pr,b, xr),we seethatr, ) xrif.andonly if p, < pr;that is, r decreasesas p increasesand r increasesas p decreases. When 4 is constant, therefore, as p increases,I decreases;but y may either increase or decrease.If y increases,then a decreasein the demand for one commodity correspondsto an increasein the demand for the other, and the two commodities are said to be substitutes(for example,butter and margarine). Now if, when 4 is constant, y decreasesas p increases,then a decreasein the demand for one commodity corresponds to a decreasein the demand for the other, and the two commodities are said to be comolementary (for example, tires and gasoline). o rLLUsrRArror L: Figures 20.4.2 and 20.4.3 each show a sketch of a demand surfacefor an equation of type (2). In Fig.20.4.2, we seethat when q is constant, y increases as p increases,and so the two commodities are substitutes. In Fig. 20.4.3 the two commodities are complementary because when 4 is constant, y decreasesas p increases. .

Figure 20.4.3

Observe that in Figs.20.4.1 and20.4.2,which show the demand surfaces for equations of types (1) and (2), respectively, the p and q ixes are interchanged and that the vertical axis in Fig. 20.4.1is labeled x and in Fi9.20.4.2 it is labeled y. It is possible that for a fixed value of q, y may increasefor somevalues of.p artd decreasefor other values of p. Fot example,if the demand surface of Eq. (2) is that shown in Fig. 20.4.4,then for q:b,whenp:auy is increasing, and when p : az, y is decreasing. This of course means that

TO ECONOMICS 20,4 SOME APPLICATIONS OF PARTIALDERIVATIVES

-

4

if the price of the second commodity is held constant, then for some prices of the first commodity the two commodities will be substitutes and for other prices of the first commodity the two commodities will be complementary. These relationships between the two commodities that are determined by the demand surface having equation y : 8(p, q) will correspond to similar relationships determined by the demand sur. face having equation x: f(p, q) for the same fixed values of p and q' An economic example might be one in which investors allocate funds between the stock market and real estate. As stock prices climb, they invest in real estate. Yet once stock prices seem to reach a crash level, investors begin to decreasethe amount of real estate purchases with any increase in stock prices in anticipation of a collapse that will affect the real estate market and its values. The demand functions are now used to define the (partial) "marginal demand."

20.4.4 Figure

20.4.1 Definition

Let p dollars be the price of one unit of x units of a first commodity and 4 dollars be the price of one unit of y units of a second commodity. suppose that / and g are the respective demand functions for these two commodities so that x: f(p, q) and y: g(p, q) Then (i) *gives dp"

the (partial) marginal demandof x with respectto p;

(ii) + gives the (partial) marginaldemandof x with respectto q; dqAu (iii) = gives the (partial) marginal demand of y with respect to p; dp

,.,0a (i") i

gives the (partial) marginal demandof y with respectto q'

Becausein normal circumstancesif the variable 4 is held constant, I decreasesas p increasesand r increasesas p decreases,we conclude that AxlAp is negative. Similarly, AylAq is negative in normal circumstances. Two commodities are said to be complementary when a decreasein the demand for one commodity as a result of an increase in its price leads to a decreasein the demand for the other. So when the goods are complementary and. q is held constant, both dr/Ep < 0 and AylAp < 0; and when p is held constant, then Oxl0q< 0 aswell as AyI Aq < 0. Therefore,we can conclude that the two commodities are complementary if and only if both 6xl0q and aylaP are negative'

970

GRADIENTS, APPLICATIONS DERIVATIVES, OF PARTIAL DERIVATIVES, AND LINEINTEGRALS DIRECTIONAL

When a decreasein the demand for one commodity as a result of an increasein its price leads to an increase in the demand for the other commodity, the goods-are substitutes. Hence, when the goods are substitutes, because\xldp is always negative, we conclude that Ayl\p is positive; and becauseAylAqis always negative, it follows that 0xl0q is positive. Consequently, the two commodities are substitutes if and only If. 0xl6q and dyl6p arc both positive. If. Axl\q and AylAp have opposite signs, the commodities are neither complementary nor substitutes. For example,lf. dxl\q < 0 and 0yl0p > 0, and because 0xl6p and AylAq are always negative (in normal circumstances),we have both 0xl0q ( 0 and AylAq < 0. Thus, a decreasein the price of the second commodity causesan increasein the demands of both commodities. Because0xl0p < 0 and AylAp ) 0, a decreasein the price of the first commodity causesan increasein the demand of the first commodity and a decreasein the demand of the second commodity. EXAMPLE1: Suppose that p dollars is the price per unit of x units of one commodity and q dollars is the price per unit of y units of a second commodity. The demand equations are

solurroN: given by

If we use Definition 20.4.1,.the four marginal demands are

_^ a _ _/) dp

Y -3 aq

a Y- 1 op

Because0xl0q > 0 and AylAp > 0, the two cot'nmoditiesare substitutes. A sketch of the demand surface of Eq. (3) is shown in Fig. 20.4.5.To )c--2p*3q+L2 (3) draw this sketch we first determine from both equations the permissible and values of.p and 4. Becausex and y must be positive or zero, p and must 4 y:-aq*p+8 ( 4 ) satisfy the inequalities -2p * 3q * 12 > 0 and-Aq * p * I > 0. Also, p and q are each nonnegative. Hence, the values of p and q are restricted to the Find the four marginal dequadrilateral AOBC. The required demand surface then is the portion in mands and determine if the comthe first octant of the plane defined by Eq. (3) which is.above AOBC. This modities are substitutes or comis the shaded quadrilateral ADEC in the figure. In Fig. 20.4.6we have a plementary. Also draw sketches of the two demand surfaces.

Figure 20.4.5

Figure 20.4.6

TO ECONOMICS OF PARTIALDERIVATIVES 20.4 SOME APPLICATIONS

sketch of the demand surface defined by Eq. (4). This demand surface is the shaded quadrilateral BFGC,that is, the portion in the first octant of the plane defined by Eq. (4) which is above the quadrilatenl AOBC. For nonlinear demand functions of two variables, it is often more convenient to represent them geometrically by means of contour maps than by surfaces.The following example is such a case. ExAMPLE 2: If p dollars is the price of one unit of x units of a first commodity and q dollars is the price of one unit of y units of a second commodity, and the demand equations are given by

8 pq

solurroN: Let the two demand fudctions be / and g so that x: f (p , q) : Slpq and y:g(p, q):l2lpq.sketches of the contour maps of f and g showing the level curves of these functions at the required numbers are shown in Figs. 20.4.7 and20.4.8, respectively. pq2

and

ay ap

L2 p'q

Because\xl 0q < 0 and aylAp ( 0, the commodities are comPlementary. q

q

draw sketches of the contour maPs of the two demand functions showing the level cun/es of each function at 6,4,2,1, and*. Are the commodities substitutes or com-

v:+

plementary? y--6

Figure20.4.8

Figure 20.4.7

EXAMPLE3: The demand equations for two related commodities are *c-Ae-pq and

y:8eo-q

Determine if the commodities are complement dr4, substitutes, or

SOLUTION: 6x - -4Pe-P(t

aq

and

#-

8e--q

Because0xl0q< 0 and aYIAP ) 0, the commodities are neither comPlementary nor substitutes.

neither.

If the cost of producing r units of one commodity alnd y units of another commodity is given by c(x, y), then c is called a ioint'cost function.Thepartial d"rin"ti*'"t of c are called marginal costfunctions.

AND LINEINTEGRALS DERIVATIVES, OF PARTIAL APPLICATIONS GRADIENTS, DERIVATIVES, DIRECTIONAL

suppose that a monopolist produces two related commodities whose demand equations are)c: f (p, q) and y : g(p, q) and the joint-cost function is C. Because the revenue from the two commodities is given by px * qy, then if S is the number of dollars in the profit, we have S : p x * q y - C ( x ,y ) To determine the greatest profit that can be earned, we use the demand equations to express S in terms of either p and q or r and y alone and then apply the methods of the preceding section. The following example illustrates the procedure. EXAMPLE4: A monopolist Produces two commodities which are substitutes having demand equations

x- 8- p + q and

y--9+p-Sq where 1,000runits of the first commodity are demandedif the price is p dollars per unit and 1000y units of the secondcommodity are demandedif the price Lsq dollars per unit. It costs$4 to produce eachunit of the first commodity and $2 to produce eachunit of the secondcommodity. Find the demands and prices of the two commodities in order to have the greatestprofit.

solurroN: When 1000runits of the first commodityand 1000yunits of the secondcommodityareproducedand sold,the numberof dollarsin the and the numberof dollarsin the total cost totalrevenueis 1000pr* L000qy, of production is 4000r*2000y. Hence,if S dollarsis the profit, we have S:L000px+1000qy- (4000r+2000y) - p -r 4) + 10004(9 S : 1000p(8 + p - 5q) - 4000(8- p * q) - 2000(e+ p - 5q) S : 1000(-p' * Zpq- 5q' * l0p + LSq- 50)

$dp:

r o o o l - 2*p2 q + t 0 )

9dq: 1 0 0 0 ( 2 p - 7 0 + q7 5 )

Setting 0Slap:0 and 0Sl6q: 0, we have -2p*2q*70:0 and 2p-10q *15:0 from which it follows that p:ga

and q:T

Hence, (3F, +) is a critical point. Because

-2ooo ?'l : -1o.ooo S-: 2ooo +?: dpdq' dq dp we have

azsazs_ ( a'S ,' : (-2000)(-10'000)- (2000)2 >0 0p, 0q, \aq ap) /\

Also, 02Sl0p'10, and so from Theorem 20.3.8(ii)we concludethat S has a relative maximum value at (S, +5). Becausex, U, p, and q must be nonnegative, we know that 8-p*q>0

Figure 20.4.9

9*p-5q-0

p>0

q>0

From these inequalities we determine that the region of permissible values is that shaded in Fig. 20.4.9.It follows that S has an absolutemaximum


value at the point (6F, ?), and the absolute maximum value of S is '1,4,062.5. From the demand equations we find that when p : $a and q: T, x:3 andy:t. is attained We therefore conclude that the greatestprofit of.$14,052.50 at produced and sold are commodity $8'12*per when'3000units of the first produced and sold at are commodity unit and 1500 units of the second $3.12+per unit. o rl-I.usrnArrox 2: In the preceding example, if the demand equations are solved for p and 4 in terms of r and y, we obtain

p - + ( 4 9- u r - y ) a n d q : + ( L 7- x - Y ) Becauseq and p must be nonnegative as well as x and y, we know that 17 - x- y > O,49 - 5x- y > 0, x = 0, andy > 0. From these four inequalities we determine that the region of permissible values is'that shaded in Fig.20.4.10.The problem can be solved by considering r and y as the in. dependent variables. You are asked to do this in Exercise 12.

Figure 20.4.10

Suppose that in the production of a certain commodity, r is the number of machines, y is the number of man-houts, and z is the number of units of the commodity produced, where z depends on r and y, and z : f (x, y) . Such a function / is called a productionfunction. Other factors of production may be working capital, materials, and land. Let us now consider, in general,a commodity involving two factors of production; that is, we have a production function f of two variables. If the amounts of the inputs are given by r and y, and,z gives the amount of the outlrut, then z : f(x, y). Suppose that the prices of the two inputs are a dollars and b dollars per unit, respectively, and that the price of the output is c dollars (a, b, and c are constants). This situation could occur if there were so many producers in the market that a change in the outPut of any particular producer would not affect the price of the commodity. Such a market is called a perfectly competitiaemarket.Now if P dollars is the total profit, and becausethe total profit is obtained by subtracting the total cost from the total revenue, we have P:cz-(ax+by) and becausez: f(x, y), P: cf(x, y) - ax- bY It is, of course, desired to maximize P. This is illustrated by an example.

ExAMPLE5: Suppose that the production of a certain commodity depends on two inPuts. The

soLUrIoN:

If P dollars is the proftt,

P - e(1002) - 4(100r) - L00y

974

APPLICATIONS DERIVATIVES, AND LINEINTEGRALS GRADIENTS, OF PARTIAL DERIVATIVES, DIRECTIONAI-

amounts of theseare given by 100r and L0Ay,whosepricesper unit are,respectively,$4 and $L.The amount of the output is given by 1002,the price per unit of which is $9. Furthermore,the production functi on f has the function values - Llx- UY.Determine f (x, A) :5 the greatestprofit.

Letting z - f(x, y), we get

- Looy - oox P -eoo(s- +x -+) y/ \ x and y arc both in the interval (0, +o). Hence,

AP 0x

:ry-4ooand g:dy

9oro _ 1oo y'

Also, a2P 0xz

1800 x3

azP

1800

V

T

a2P

n

ayax:u

Setting APlAx - 0 and API 0y - 0, we have

ry-4oo:oandry-loo:o from which we obtain 5 : $ and y : 3 (we reject the negative result becausex and y must be positive). At (t,3) -^^^, azP azP / a2P \2 /1800\ /

;a' ar- \i*) :- (F/

\-T)-

(o)'>o

From the above and the fact that at (*, 3) azPlaf < 0, it follows from Theorem 20.3.8(ii)that P has a relative maximum value'at (8,3). Because x and y are both in the interval (0, +-1, and P is a negative number when x and y are either close to zero or very large, we conclude that the relative maximum value of P is an absolutemaximum value. Becausez: f (x, V), the value of.z at (*,9) is f(2,3) -5 - B - *:4. Hence, from Eq. (5) P , n o: 9 0 0 ' 4 - 4 0 0 ' 8 - f O O ' 3 : 2 7 0 0 The greatest profit then is 92700.

EXAMPLE6: Solve Example 5 by the method of Lagrange multipliers.

solurroN: We wish to maximize the function P defined by Eq. (5) subject to the constraintgiven by the equationz :5Ux- 1/y,which we can write as g(x, y, z) -1+1*z-5-o

xy

F(x,A, z, ),)- P(x, y, z) * Ig(x, y, z) - 9002- 400x- I00y + I We find the four partial derivatives F, Fs, F, and F1 and set them equal to zero.


F*(x, A , z, Ir)

F,(x,

Fo( x, A,z,) ,): - 100

--4oo-):o

r)

h-

975

0

F ^ ( x , A , z ,) ' ) - L + 1 * z - 5 - o lcy

Solving these equations simultaneously, we obtain z:4 x:8, )r: -900 A:3 The values of x, y,and z agreewith those found previously, and P is shown to have an absolute'maximum value in the same way as before. In some applications of functions of several variables to problems in economics, the Lagrange multiplier )t,is related to marginal concepts, in particular marginal cost and marginal utility of money (for details of this you should consult referencesin mathematical economics).

20.4 Exercises In Exercises1 through 9, demand equations for two related commodities are given. In each exercise,determine the four partial marginal demands. Determine if the commodities are complementary, substitutes, or neither. In Exercises1 through 6, draw sketchesof the two demand surfaces. 2. x- 5-2p*q,y-6*p-q 1. x- 14- P-2q, Y:L7 -2P - q 4 . x : 9 - 3 p * q , y : L 0- 2 p - 5 q 3 . x - - 3 p + 5 q + 1 5 ,A : 2 P - 4 q * I 0 6 . x - - p - 3 q* 6 , y - - 2 q - p + 8 5. x-6-3p-2q,y-2*P-2q 8. pqx - 4, p'qy : L5 7. x - P-o'4qos, A Po'4q-rs 9. px: q, qy: p' 10. From the demand equations of Exercise8, find the two demand functions and draw sketchesof the contour maps of these functions showing the level curves of each function at5,4,3,2,1,+,+. 11. Follow the instructions of Exercise 10 for the demand equations of Exercise 9. 12. Solve Example 4 of this section by considering x and y as the independent variables. (Refer to Illustration 2.) 13. The demand equations for two commodities that are produced by a monopolist are

and y:7+p-q

x:6-2p+q

where 1.00r is the quantity of the first commodity demanded iI the price is p dollars per unit and l00y is the quantity of the second commodity demanded if the price is q dollars Per unit. Show that the two commodities are substitutes. If it costs $2 to produce each unit of the first commodity and $3 to Produce each unit of the second commodity, find the quantities demanded and the prices of the two commodities in order to have the greatest profit. Take p ar.d q as

the independent variables. 14. Solve Exercise 13 by considering r and y as the independent variables. 15. The production function f for a certain commodity has function values

f(x,v):4-;

R

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINEINTEGRALS The amounts of the two inputs are given by 100r and 100y, whose prices per unit are, respectively, gl0 and $5, and the amount of the output is given by 7002,whose price per unit is $20.Determine the greatestprofit by two methods: (a) without using Lagrange multipliers; (b) using Lagrange multipliers. 16. Solve Exercise15 if

f(x'v):x**y -*f-

*y'-*

the prices per unit of the two inputs are $4 and $8 (instead of $10 and $5, respectively), and the price per unit of the output is $16.

20.5 OBTAINING A FUNCTION FROM ITS GRADIENT

We consider the problem of how to obtain a function if we are given its gradient; that is, we are given

Yf(*, y) : f ,(x, y)i + f u(*,y)i

(1)

and we wish to find f (x,y). o TLLUSTRATTON 1.: Suppose

Yf(*, A) : (y, * 2x * 4)i + (Zxy* 4y - 5)i

(2)

Then because Eq. (1) must be satisfi€d, it follows that

f"(x,y):y2*2x+4

(3)

and

fok,y) :zxy * 4y 5 By integratingboth membersof Eq. (3) with respect to x, we have f (x,y) : yzx* x2* 4x * g(V)

(4)

(s)

Observe that the "constant" of integration is a function of y and independent of r because vve are integrating with respect to r. If we now differentiate both members of Eq. (5) partially with respectto y,we obtain f oG, y) :2xy + g'(y)

(5)

Equations (4) and (6) give two expressionsfor fu@,y).Hence, 2xy*4y-5:2xy+g'(y) Therefore, - 5 8 ' ( Y ): 4 Y g(Y) :2Y2 - 5Y * c Substituting this value of g(y) into Eq. (5), we have -l4x*2y'-5y + C f(x, y) y'x* * EXAMPLE

'1,:

Find f (x, y) it

vf(x, V) : gu2cos xi + Zyeu'sin xl

SOLUTION:

Because Eq. (1) must hold, w€ have

f "(x, y) : guzcos r

o

FROMITSGRADIENT 977 A FUNCTION 20.5OBTAINING and f o@,y)

- Zye" sin r

Integrating both members of Eq. (8) with respect to A, we obtain -- su2sin x * S(x) f (x, y) where S@) is independent of y. We now partially differentiate both bers of Eq. (9) with respect to r and get f*(x,Y)

: s u 2c o s x * 8 ' Q )

(10)

By equating the right members of Eqs. (7) and (10), w€ have eu' cos x:

ea' cos x * 8' @)

g ' ( r )- o s(r) c We substitute this value of g(r) into Eq' (9) and obtain f (x,y)

: s a 2s i n x * L

gradients' All vectors of the fotm M(x,y)i * N(r' y)i atenot necessarily as shown in the next illustration' such that o rLLUsrRArroN2: We show that there is no function f (11) 2xi

Vf(x, U)

3Yi-

that Assume that there is such a function. Then it follows

f *(x, Y) - 3Y

(12)

-2x f o @ , Y :)

(13)

and and obtain we integrate both members of Eq. (12) with respect to r f (x,y)

- 3 x A* g Q )

Wepartiallydifferentiatebothmembersofthisequationwithrespecttoy and we have

fo@,Y):3x+g'(Y)

(14)

w€ obtain Equating the right members of Eqs. (13) and (14),

- 2 x : 3 x * g 'Q ) _ S x : g ,( a ) with resPect to )c, it If both members of this equation are differentiated must follow that -5:0

978


which, of course, is not true. Thus, our assumption that 3yi - 2xi is a gradient leads to a contradiction. . we now investigate a condition that must be satisfied in order for a vector to be a gradient. Suppose that Mu and N' are continuous on an open disk B in R2. If

M(x, y)i + N(r, y)i

(15)

is a gradient on B, then there is a function / such that and

f "(x, y)

- M(x, y)

(16)

fu?,y) : N(x, y) G7) for all (x,y) in B. Because Mu@,y) existson B, thenfrom Eq.(15)it follows that MaQ, y) : f ,uQ, y)

(18)

Furthermore, becauseNr(x, y) exists on B, it follows from Eq. (17) that N"( x, y) : f *ok, y) (1e) BecauseMn and N, are continuous on B, their equivalents cu and, are f'that f o, also continuous on B. Thus, from Theorem 1.9.7.1 it follows f ,u(i, y1: fo'@' y) at all points in B. Therefore, the left members of Eqs. 1rs) and 119) are equal at all points in B. we have proved that if M, and N, are continuous on an open disk B in R2,a necessarycondition for vector (15)to be a gradient on B is that Mo(x, y) : N,(x, y) This equation is also a sufficient condition for vector (15) to be a gradient on B. However, the proof of the sufficiency of the condition involves concepts that are beyond the scopeof this book, and so it is omitted. we have then the following theorem. 20.5.L Theorem

Suppose that M and N are functions of two variables x and.udefined on an open disk B((xo,yr); r) in R2, andMu and N, are continuorrr o'B. Then the vector M(x, y)i + N(x, y)i is a gradient on B if and only if Mo(x, y) : N,(x, y) at all points in B. . rl-I.usrRArrorrr3: (a) we apply Theorem 20.5.1to the vector in the right member of Eq. (2) in Illustration 1. Let M(x, y): y, * 2x * 4 and N(r, yj:

20.5 OBTAININGA FUNCTIONFROM ITS GRADIENT 979

2xy * 4y - 5. Then Mak, y) - 2Y and N,(x, Y) 2Y Thus, Mo(x,a): Nr(r, a), and thereforethe vector is a gradient. (b) If we apply Theorem 20.5.1to the vector in the right member of : -2x, we obtain Eq. (11)in Illustration2, with M(x,a):3y and N(r, A) -2 Ma(x, y) - 3 and N"( x, Y) o Thus, Mo(x,y) + N*(x,a), and so the vectoris not a gradient. rn 2: Determine if the EXAMP following vector is a gradient Yf (*, A), and if it is, then tind

f@,v)t (e-o - 2x)i - (xe-o * sin V)i

solurroN:

-2x and N(r' y) : We apply Theorem 20.5.1.Let M(x, U): e-u

-xe-a - sin y. Then MaQ,y) : -e-u

and

N " ( x , Y ) : -e-U

Therefore, Mo(x, A): N'(r, Yf (x, y). Furthermore, f

"(x,

Y)

,given Y), and so the

vector

is a gradient

: s-u - 2x

and f o@,y)

x e - a- s i n Y

(20)

(21)

to x, we obtain Integrating both members of Eq' (20)with resPect (V) : f (x, y) xe-a )cz* g differentiate both where S(fi is indePendentof x' We now partially have we bers of Eq. (22)with resPectto A, and : -xe-a + g'(Y) f,(x, y) (21) and get we equate the right members of Eqs. (23) and -xe-a * 8' Q) : -xe-a - sin Y g' (Y) : -sLn Y SQ):cosY+C (22) and we have We substitute this exPression for g(y) into Eq. +C : f (x,Y) xe-a xz* cosY of three variables' we can extend Theorem 20.5.1 to functions

20.5.2Theorem

x' y' and z defjned on an Let M,N, and R be functions of three variables M"' N'' N" R'' and R' are conopen ball B((xo,yo, zo);r) inf,-11d Ma' z)i I ru(r' y' z)i + \!*'v' z)k is tinuous on B' Then iii,el,"cto' M(x, y' M"(x'y'z): y; z):N'(x'y'z)' a e r a d i e n to n B i f u r r ao " i y ' t i u 7 - , : ' z) Ru(x,Y' I.l!, y, z) , and N"(x, Y, z)

980

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PABTIALDERIVATIVES, AND LINEINTEGRALS

The proof of the "only if" part of rheorem 20.5.2is similar to the proof of the "only if" part of rheorem 20.5.1and is left as an.exercise(seeExer-

cise 23).The proof of the 'tif" part is beyond the scope of this book.

ExAMPlr 3: Determine if the following vector is a gradient Yf (x, A, z), and if it is then find f (x, U, z): (e' sin z * Zyz)i + (2xz+ 2V)i + (e* cosz * Zxy * 322)k

solurroN: We apply Theorem 20.5.2.Let M(x, y, z): e, sin z*2y2, N(x, y, z) : 2xz * 2y, and R(x, y, z) : e, cos z * 2iy * 322.Then Mu(x, y, z) :22

Mr(x,y,z)-sr

N"(r, y, z) :22

N r ( x ,y , z ) : 2 x

R*(x, y, z) : e, cosz * 2y

Ro(x,U, z) : 2x

c o sz t 2 y

Therefore,Mo(x, y, z) - At(r , y, z), Mr(x, y, z)- Rr(r, A, z), andNr(r, A, z) : Ro(x,y, z). Thus, the given vector is a gradient Yf (x, A, Z).Furthennore, : y, z) sr sin z * 2yz f *(x,

(24)

f u ( x ,y , z ) : 2 x z * 2 y

(25)

f"(x, y, Z) : e'cos z * 2xy * 3zz

e6) Integrating both members of Eq. e4) with respect to x, we have : f(x, y, Z) e, sin z * 2xyz* Sg, z) ez) *-Ett g(y, z) is independent of r. we partial differentiate both members of Eq. (27) with respect to y and,obtain fu(x, y, z) : 2xz * gu(y, z) (28) Equating the right members of Eqs. (2g) and (25), we have 2xz * gn(y, z) :2xz * 2y S o ( Yz' ) : 2 Y we now integrate both members of this equation with

respectto y, and get

gQ, z) : yz+ h(z)

(2e) where h is independent of tr and y. Substituting from Eq. (zg)into Eq. (27), we obtain f(x, y, z) : sr sin z + 2xyz * y, + h(z) (30) We now Paftial differentiate with respectto z bothmembers of Eq. (30). We get fr(x, U, Z) - er cosz + 2xy + h,(z) Equating the right membersof Eqr. (31)and (26),we have et cosz + 2xy + h, (z) sr cos z + 2xa * jzz h' (z) :

(31)

322

h(z):23+C

(32)

20.6LINEINTEGRALS 981

We substitutefrom Eq. (32) into Eq. (30), and we obtain -f (x, y , z) sr sin z * 2xyz * v' * 2 3 + C

Exercises 20.5 In the following exercises, determine if the vector is a gradient. If it is, find a function having the given gradient

1. axi- 3yi 3. (6x- sy)i - (5r - 6V')i

4. (4y' * 6xy- 2)i + (3x'* 8xY+ l)i

5. (6x'y' - l4xy + 3)i + (4xty '7xz - 8)i

6 . 3(2x2* 6xy)i + 3(3x2+ 8V)i

7. (Zxy* y'+ L)i + (x' * Zxy + r)i

8.

2. y'i + lx'i

-)l +(ry\. s (L*1)r \ yo \7Y'

2x-L. , x-x2. 'y' I y

1 0 . (2x* rn y)i + (y' +i) i yl \L2. (sin 2x - tan y)i - x seczyi

,',

11. (2x cosy - l)i - x2 sin Yi 13. (eu- Z*y)i * (xeo- x')i 15. (2V - Sz)i + (2x * 8z)i - $* - 9Y)U

1,4.4 x 3 i 1 - 9 y ' i - Z z k - Ay)U 1,6.(Zxy * 7zs)i+ ( xz + 2y' - 3z)i + (21'xz2

17. (4xy * 3yz - 2)i + (2x' * 3xz - Szz)i+ Gxy l}yz * 1 ) k L8. (2y + z)i * (2x - 3z * 4yz)i + @ - 3y * 2y')U 19. z tan yi + xz seczyi + x tan Yk 20. e' cosri + z sin yi + (e" sin x- cos y)k ZL. e,(e, - ln ili+ (eoln z- e'A-t)i + (s'+" * tu7-r)k t - x-z ' x+a y+z i- (y+4'zt-ffiK

22.

23. Prove the "only tf" part of Theorem 20.5.2.

20.d LINE INTEGRALS

In Chapter 7 we used the concept of area to rnotivate the definition of the definite integral. To motivate the definition of an integral of a vectorvalued function, we use the physical concept of work. We have seenthat if a constant force F moves an obiect along a straight line, then the work done is the product of the component of F along the line of motion times the displacement.We showed in Section 17.3that if the constant force F moves an object from a point A to a point B, then if W is the measure of the work done,

W:F'V

@3)

(1)

982

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIALDERIVATIVES, AND LINEINTEGRALS

suppose now that the force vector is not constant and instead of the motion being along a straight line, it is along a curye. Let the force that is exerted on the object at the point (x, y) in some open disk B in R2be given by the force vector F(x, y) : M(x, y)i + N(x, y)i where M and N are continuous on B. The vector-valued function F is called a force field on B. Let C be a curve lying in B and having the vector equation R(f):f(t)i+S@i

a
We require that C be smooth, that is, that f, and g, be continuous on [a,bf . we wish to define the work done by the varia6re force F in moving the object along C from the point (f(a), g(a)) to (f(b), S@)). At a poinl (f(t), S(t)) on C, the force vector is

F ( / (t ) , s ( f ) ) - M ( f ( t ) ,s ( f ) ) i + N ( / ( t ) , 8 ( f ) ) i

(2)

Let A be a partitionof the interval fa, bf : ct: fo ( tr

On C let [$: th" point (ri, At): (f(t), g(f)). Refer to Figure 20.6.1.The vector V(Pi-lPi) : R(tr) - R(4-r); therefore,

v (P;7i) : f (f,)i+ s(tt)t- lf (t,-,)i + s (f,-,)i l or, equivalently,

v (+7,) : lf(t,)- f(t*r)li + ls(t,)- s(ro-,) li

R(f;-t)

Figure20.6.1

(3)

Becausef' and g' are continuous on la, bJ, it follows from the mean-vhlue theorem (4.7.2) that there are numbers ci and rdi in the open intewal (t*t t) such that f (t,)

-

f (t,-r) : f' (c) (t, tr-r)

(4)

and g(t) - g(f,-r) : g' @,)(t,- tr_r) (s) Letting \t : t1- t;1 and substituting from Eqs. (4) and (5) into Eq. (3), we obtain

v(PFi)

: lf '(c)i * s,(d)il aot

(5)

For each i, consider the vector

Fr: M(f (ct), Bk) )i + N(/( d,), S@ili

(7)

Each of the vectors Fr (i: 1,2, . , n) is an approximation to the force vector F(f(t), g(f)), given by Eq. (2), along the arc of C from p;1to p1. Observe that even though c1 and di are in general different numbers in the open interval (trr, tr), the values of the vectors F(f (t),g(f) ) are close to

20.6 LINE INTEGRALS

the vector F1.Furtherlnore, we approximate the arc of C from Pi-l to Pi by the line segment Pr-rP,.Thus, we apply formula (1) and obtain an approximation for the work done by the vector F(f(t), g(f)) in moving an object along the arc of C from Po-, to Pi. Denoting this approximation by A1W, we have from formula (1), and Eqs. (7) and (6), ' AiW : lM(f (c), 8(c,))i + N(/(d,), S(do) ) jl ' lf (c)i * I' G)jl Alt or, equivalently, AlW: lM(f (c,), g(c))f '(c)7 Att+ lN(f (d) , s@))S'(4)l A,t An approximation of the work done by F(f (t),g(f)) along C is i A1Wor, o:t equivalently, nn

(c,),ski))f'(c)lair+ > tN(/(4), B(d))s'@)l\t 2 tu(f ,""ri l, these sums is a Rieman" J;. The first summation is a Riemann sum for the function having function values M(f(t), g(t))f'(t) and the second summation is a Riemann sum for the function having function values N(/(0, S(0)S'(0. If we let n-->*a, these two sums approach the definite integral:

I"

l M ( f( t ), s ( t ) ) f ' ( t )+ N ( f ( t ), s ( f )) s ' ( f I) d t

We therefore have the following definition. 20.5.L Definition

Let C be a curve lying in an open disk B in R2for which a vector equation of C is R(0:/(t)i+g(f)i, where f' andg'are continuousonfa, b]. Furthermore, let a force field on B be defined byF(x, y) : M(x, y)i +N(x, y)i, where M and N are continuous on B. Then if W is the measure of the work done by F in moving an object along C from (/(a), g(n)) to

( f ( b')s ( b )')

(8)

or, equivalently, by using vector notation,

( M ( f ( t ) , g ( t ) ) , N ( / ( t ) , g ( f )) ) . ( f ' ( t ) , 9 ' ( f ) ) d t

(e)

or, equivalently,

. R'(f) dt

(10)

DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIAL DERIVATIVES, AND LINEINTEGRALS ExAMPLEL: Suppose an object moves along the parabola y : x2 from the point (-1 , 1) to the point (2, 4).Find the total work done if the motion is caused by the force field F(x,A) : @' * y')i + 3x'yi. Assume the arc is measured in inches and the force is measured in pounds.

sot,urroN:

Parametric equations of the parabola are

x:t

-1
and U:tz

Thus, a vector equation of the parabola is

R(f):ti+t'i Because ,then /( t) - t, gU): t2,and F(x,U): Q' t A,, Sxry> F ( / (t ) , 9 ( f ) ) : F ( t ,

(t, + t4,gt4ll

t)-

If W in.-lb is the work done, then from formula (10) we have f2 w - | n G ,P ) . R ' ( f ) d t

J-r

:

f2

|

J-t

( t 2 + t 4 , g t 4 > (. 1, 2 t > d t

f2

: I Gr+t4+6t) dt J-, 15 12 ts :?+;* ruj_,

:8+s#+64-(-+-++1) _

363 1.

Therefore, the work done is 3€3 in.

i) is calllle d a "line integral." A common notation Th e lintr ral in E'.q, n tegral iq. (8) for the IC lint I] n e: :integ tegral olfrf Eq. 1q. . (8), i S

t AMt((: x: ,,y )) d Nt( x ,, !V : ) l dy a x *1 N

t -_ f t aa

Thi1sS nnol atio (rtati lgg5esteC on is' su8 sui ted d b :v F t he h fact that because parametric equations ',t) o f (CI are ar,E X : =f(t) and an(d 9y 8 ( 8( iu),tltheen dx : f ,(t) dt and dy - g'(t) dt. We have tlloowi nrin thet ffoll rfl. ng forma forrmal al< detin effiniritio)n 20.5.2 Definition

Let M and N be functions of two variables r and y such that they are continuous on an open disk B in R2.Let C be a curve lying in B and having parametric equations x-f(t)

y:gG)

a
such that f ' and g' are continuous on la, bl. Then the line integralof M(*, y) dx * N(x, y) dy over C is given by

20,6 LINE INTEGRALS 985

by using vector

or, eq

. ILLUsrRArror 1: In Example 1, W is given by the line integral f

| (r'+ y') dx + 3fy ilY

Jc

where C is the arc of the parabola A : xz from (-1, 7) to (2, 4)'

o

I f a n e q u a t i o n o fC i s o f t h e f o r m y : F ( r ) , t h e n r m a y b e u s e d a s a parameter in place of f. similarly, if an equation of c is of the form x:G(y), then y may be used as a Parameter in place of f. . rLLUSrRArroN2: In Example 1 and Illustration 1.,the equation of c is : can use r as a Parama : xr, which is of the form y F(x). Therefore, we eter instead of f. Thus, in the integral of Illustration 1 we can replace y by xz and dy by 2x dx, artd we have * :

: Figure 20.6.2

f2

J_, r2

J_,

(x2 * xa) dx * 3x2xz(2x dx)

@' * x4+ 6x5)dx

This integral is the same as the third one appearing in the solution of o Example 1.,except that the variable is r instead of f. If the curve c in the definition of the line integral is the closedinterval bf on the x axis, then y :0 and dy :0. Thus, fa, f

fb

-.

I t v t ( r , y )d x + N ( r , y ) d y : I M ( x , O ) d x

Jc-Js

Figure20.6.3

Therefore, in such a case, the line integral reduces to a definite integral. In the definition of a line integral we required that the functions / and g in the parametric equations defining C be such that /' and g' are continuous. Such a curve C is said to be smooth. If the curve C consists of a finite number of arcs of smooth curves, then C is said to be sectionally


986

smooth. Figures 20.6.2 and 20.6.3 show two sectionally smooth curves. we can extend the concept of a line integral to include curves that are sectionally smooth. 20.6.3Definition

Let th,e curye C consist of integraI of M(x, il dx * N(r, r

I M(x, y) Jc ExAMPrn2: Evaluatethe line integral

r

I 4*y dx * (2x' 3*y) dy

JC

over the curye C consisting of the line segment from (-3, -2) to (1, 0) and the first quadrant arc of the circle x2 * y' : 1,from (1, 0) to (0, 1), traversed in the counterclockwise direction.

Cr, Cr, . , Cn. Then the line is defined by

dx * N (x, y)

M(x, y) dx + N (x, y) dy)

solurroN: Figure 20.6.4shows the curve C composedof arcs Cr and G. The arc Ct is the line segment.An equation of the line through (-9,-2) and (1.,0) is r - 2y : 1. Therefore, C1c?rtbe representedparametricallyby x:7*2t

-2
y:t

The arc Cr, which is the first quadrant arc of the circle I * yr:1, representedparametrically by r:cost

y:sinf

can be

0=t-+n

Applying Definition 20.6.2f.or each of the arcs C1 and C2 we have I

dx+ (2f - 3xy)dy J",n*A

-

r 4(1 + 2t)t(2 dt) + l2(1 + 2t)2-

J-,

ro : J_, (8f +'1.6t2+ 2 + 8f + 8t2-

- ro (r8t2 + J-,

3(1 + 2t)tl dt

3t - 5t2)dt

13t + 2) dt

- 6F+ +t'+ 2t10

Figure20.6.4

-l-z

:_(_49 +26_4) :26

dx t (2x' - 3xy) dy

: l, --L

4 cos f sin f (-sin t dt) + [2 coszs- 3 cos f sin f](cos t dt)

n12

( - 4 c o s f s i n 2t + 2 c o s 3t - 3

c o s 2 fs i n f ) d t

20.6LINEINTEGRALS 987

:

f n12

|

Jo

:

f nl2

|

- 3 c o s 2t s i n t ] d t [ - 4 c o s t s i n z t * 2 c o s t ( 1 - s i n 2f ) ( 2 c o sf - 5 c o s f s i n 2f - 3 c o s z t s i n t ) d t

JO

: 2 sin f - 2 sin3 f + cos"l''' lo :2- 2-'L --1 Therefore, from Definition 20.6.3, f

| + x yd x + ( 2 * - 3 r Y ) d Y : 2 6 + ( - 1 ) Jc

The concept of a line integral can be extended to three dimensions. 20.6.4 Definition

Let M, N, and R be functions of three variables, x, y, and z such that they are continuous on an open disc B in R3.Let C be a curve, lying in B, and having parametric equations x:f(t)

y:gU)

z:h(t)

a
such that f ', S', and h' are continuous on la,bf . Then the line integralof M(x, y, z) dx -t N(x, y, z) dy + R(x, !, z) dz over C is given by

or, equivalently, by using vector notation

We can define the work done by a force field in moving an object along a curve in R3as a line integral of the form of that in Definition20.6.4. ExAMPr,n3: An obiect traverses the twisted cubic

R(f):ti+t'i+Pk

0
solurroN: Becausef (t) : t, S$) : F, h(t) : f , M(x, y, z) : e' , N(x, y ' z) : xe",ar]ldR(x,y,z):r sin ry2,thenM(f(t),g(t),h(t)):et,N(f(t),g(t)'h(t))-tet", and R(f (t), g(t), h(t)) : f sin rf . Thrts, if W in.-lb is the work done, we have from the line integral of Definition20.6.4,

988


Find the total work done if the motion is caused by the force field F(x, y, z): e'i * xe"i + r sin ny'k.Assume that the arc is measured in inches and the force is measured in pounds.

ft w - | t @) Ut) + (tt") (2t dt)+ (t sin ntn)(3t' dt)l

Jo

-

fr

| t et * 2t2et"+ gt} sin ntn) dt Jo

: et+? eF-* J

+7T

.o, ntnl lo

- e+ e- Lnn n - 1- +fi .o, ? I "o, 5 ,3 3"'2zr

:-O+

5 3

Therefore,the work done is

I

tr

- I) +

*]

Exercises 20.5 In Exercises1 through 20, evaluate the line integral over the given curve. ,. I e + ry) dx + (y'Jc

xy) dy;C: the line y:

x fromthe origin to the point (2, 2).

2. The line integral of Exercise 1; C: the parabola * :2y

from the origin to the point (2, 2).

3. The line integral of Exercise 1; C: the x axis from the origin to (2, 0) and then the line r:2 a. I V* dxl- (x+y) dy;C: the line !:-xfrom Jc'

from ( 2 , 0 ) t o ( 2 , 2 ) .

the origin to the point (1,-l).

5. The line integral of Exercise4; C: the curve y : -rs from the origin to the point (1, -1). 6. The line integral of Exercise4; C: the y axis from the origin to (0, -1) and then the line / : -1 from ( 0 , - L ) t o ( 1 , - 1 ) . ,. I V dx* x dy;C: R(t) : Jc-

ti+ fj,O < t < ]..

f

8 . | ( r - 2 y ) d x * x y d y ; CR : ( t ) : 3 c o st i * 2 s i n4 , 0 = t - t r , Jc f

9. | (x-y) JC

ax+ Q+x)

d y ; C : t h e e n t i r ec i r c l e l t y ' : 4 .

rc. [ ,y- d, - y' d.y;C: R(t) : Jc tt. 12.

I

| (xV + f) JC f

JcV

l2i+ fsi, from the point (1, L) to the point (4, -8).

dx * * dy; C: the parabola y :2rz from the origin to the point (1, 2).

sin x dx - cos r dy; C: lhe line segment from (Lr,0) to (2, 1).

20.7LINEINTEGRALS INDEPENDENT OFTHEPATH 989 n. I g t y) dx + (V + z) dy 1- (x + z) dz;C: thelinesegmentfrom the origin to the point (1,2,4). Jc U. I gy - z) dx * e' dy * y dz; C: the line segmentfrom (1, 0, 0) to (g, 4, 8). Jc'"

1 5 . T h e l i n e i n t e g r a l o fE x e r c i s e 1 4 ;C : R ( f ) :

(t+ l)i+tgj + fsk,0 < t <2.

rc. I V' ilx * z2dy + f dz; C: R(t) : (t -1)i + (t + l)i + Fk, 0 < t < 7. Jc' V. I z dx * x dy -t y dz; C: the circular helix R(t) : a cos fi * a sin fi + tk; 0 < t < 2n. Jc r c . I Z x yd x * ( 6 y 2 - x z ) d y * l 0 z d z ; C : t h e t w i s t e d c u b i c R ( t ) : t i + t z i + t s k , 0 < t < 7 . ,Jc 19. The line integral of Exercise18; C: the line segmentfrom the origin to the point (0,0,l); ( 0 , 0 , 1 ) t o ( 0 , 1 , 1 ) ; t h e n t h e l i n e s e g m e n ft r o m ( 0 , 1 , 1 ) t o ( 1 , 1 , 1 ) .

then the line segmentfrom

20. The line integral of ExerciseL8; C: the line segmentfrom the origin to the point (1,l, l). In Exercises21 through 34, find the total work done in moving an object along the given arc C if the motion is causedby the given force field. Assume the arc is measured in inches and the force is measured in pounds. 27. F(x, y) :2xyi + (* + y')i; C: the line segmentfrom the origin to the point (1, 1). 22. The force field of Exercise2\; C: lhe arc of the parabola A' : x fuom the origin to the point ( 1, 1) . 23. F(x, V) : fu - r)i * *yj; C: the line segmentfrom the point (1, 7) to Q, a\. 24. The force field of Exercise23; C: the arc of the parabola ! : f from the point (l , l) to Q, $ . 25. The force field of Exercise23; C: Ihe line segmentfrom (1, l) to (2,2) and then the line segmentfrom (2, 2) to Q,0. 26. F(x, V) :-*Vi * 2yi; C: the line segment lrom (a,0) to (0, a). 27. The forcefield of Exercise25;C:R(t):

a cos fi * a sin 4,0 =t =ir.

28. The force field of Exercise 26; C; the line segment ftom (a,0) to (a, a) and then the line segment fuom (a, a) to (0, a). 29.8(x,l,z):(y+z)i*(r*z)i+Q+y)k;

C : t h e l i n e s e g m e n t f r o m t h e o r i g i n t o t h e p o(iLn,t1 , 7 ) .

30. F(r, A, z) : zzi * yzj l xzk; C: the line segment from the origin to the point (4, 0, 3) . 31. F(r, A, z) : e"i * euiI e"k C: R(t) : ti + tzj + tsk, 0 < t < 2. 32. F(x, I, z\ : (ry" * x)i I (xzz+ V)i + (*y + z)k; C: the arc of Exercise31. 33. The force field of Exercise32; C; lhe line segment from the origin to the point (1,0,0); then the line segmentfrom ( 1 , 0 , 0 ) t o ( 1 , 1 , 0 ) ; t h e n t h e l i n e s e g m e n ft r o m ( 1 , 1 , 0 ) t o ( 1 , 1 , 7 ) . 34. F(x, y, z): xi* yj + (yz-r)k;

C: R(f :211 + t2j + 4tsk,0 < f < 1.

20.7 LINE INTEGRALS INDEPENDENT OF THE PATH

We learned in Sec. 20.6 that the value of a line integral is determined by the integrand and a curve C between two points P, and Pr. However, under certain conditions the value of a line integral depends only on the integrand and the points P1 and P, and not on the path from P, to Pr. Such a line integral is said to be independent of the path.

9OO DIBECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIALDERIVATIVES, AND LINE INTEGRALS

. rLLUsrRAflow 1: Suppose a force field

F(x,y) : (y, * 2x* 4)i + (2ry+ ay- 5)i moves an object from the origin to the point (1, 1). We show that the total work done is the same if the path is along (a) the line segment from the origin to (1, 1); (b) the segment of the parabola a - * from the origin to (1,1); and (c) the segmentof the curve x: ys from the origin to (1, 1). If W in.-lb is the work done, then f

* :

J,

(y' * 2x * 4) dx * (Zxy* 4y - 5) dy

(1)

(a) An equation of C is A : x.We use ras the parameter and let y : x arrd dy : dx in Eq. (1). We have then ft

r:

(x'*2x*4) dx*(2x'*4x-5)dx

Jo

r

- J o ( 3 x ' * 6 x - 1 ) dx -

- r L

"r3

-3 (b) An equation of C is A: *. Again taking x as the parameter and in Eq. (1) letting U : * and dy - 2x dx, we have lr

| (f +2x + 4) dx + (2f + 4* - 5)2xdx JO

w:

fr

:l

(51+8x3-8x+4)dx

.Jo

: r5 + 2f - 4x2. *],

-3 (c) An eqluati,onr o,ff(c i l S T = y3- We take y as the parameter and in Eq. (1) we let x - - y s :rn(d ddxx - - 3yy' t dy. We have

T

w

I( I(

'+ 44 ) 3ly' td y + (Zyn* 4y - s) dy (y y 2 *- 22yt y 3+

(6 6y't +-5Syor + 1 2 y '* 4 y : 5 ) dy

: yG* yu * 4yt* 2yr- ty], -3

INDEPENDENT OF THE PATH 20.7 LINE INTEGRALS

Thus, in parts (a), (b), and (c) the work done is 3

991

-lb.

In Illustration 1.,we see that the value of the line integral is the same over three different paths from (0, 0) to (1, 1). Actually the value of the line integral is the same over any sectionally smooth curve from the origin to (1, 1), and so this line integral is independent of the path. (This fact is proved in Illustration 2, whibh follows Theorem 20.7.1'.) We now state and prove Theorem 20.7.'1,,which not only gives conditions for which the value of a line integral is independent of the path but also gives a formula for finding the value of such a line integral.

20.7.1 Theorem

Suppose that M and N are functions of two variables x andy defined on an open disk B((xo, yo);r) in R2,Mu and N' are continuous on B, and VQQ, y) : M(x, y)i + N(x, y)j Suppose that C is any sectionally smooth curve in B from the point (xr, yr) to the point (r2, y2). Then the line integral

I,

M(x, y) dx * N (x, y) dy

is independent of the path C and

pRooF: We give the proof if C is smooth. If C is only sectionallysmooth, then we consider each piece separately and the following proof applies to each smooth piece. Let a vector equation of C be R(t) : f(t)i+ S@j

t1< t < t2

where (h, At):(f(tt), g(tJ) and (x2, yz):(f(t"),8(fr)). definition of a line integral,

I,

M(x, y) dx* N (x, y)

Then from the

v $ ( f @ , g G ) ) . R ' ( f )d t

(2)

We proceed to evaluate the definite integral in the right member of Eq. (2). BecauseYQ@, y): M(x, y)i * N(r, y)i, then 6"@, y): M(x, y) and 6',(x.,y'S: N(r, Y).Therefore, d6@, y) : M(x, y) dx + N(x, y) dy for all points (x, y) in B. Then If (f (t), g(t) ) is a point on C,

d6(fG),8(f)) : tM(f(t), s?))f'(t) + N(/(f) , g(t))g'(t)ldt or, equivalently,


df(f(t), 8(f)) : tvd (f@, 1GD . R' (f) I dt

(3)

for t1 < t 3 tz.Consider the function G of the single variable t for which

0 ( f ( t ), s ( f )) : G ( f ) Then

h

d+(f(t) , 8G)) : G'(t) dt From Eqs.(3) and (5) it follows that v6(ff),8(f)) . R' (t) dt : G'(t) dt

(4)

(s)

Hence, ft"

. R'(t)dt: ftz G'{t)dt S(t)) J,,VOUG), J,,,But from the fundamental theorem of the calculus (Theorem 7.6.2)the nght member of this equation is G(fr) - G(D. Furthermore, from Eq. (4), G(t,) - G(D : 6(f $r), S(t)) - 6(f (t,), S(r,)). Therefore,we have ftz IJ t , vf(f@, s(t)).R'(t)dt: Q(fG),sG)) - eUG),s(f,))

(6)

From Eqs. (2) and (5) and because(f (D, S(tr)): (xr, yr) and (f (t1),g(h)): (xr, yr), we obtain J[c

,k, fi dx* N(x,y) 4y: e@r,y,) - 6@r,y,)


I

Because of the resemblance of Theorem 20.7.1,to the fundamental theorem of the calculus, it is sometimes called the fundamental theorem for line integrals. . rLLUsrRATrow2: We use Theorem 20.7.1 to evaluate the line integral in Illustration 1. The line integral is f

I ( V ' + 2 x * 4 ) d x * ( 2 x y + a y - 5 )d y JC From Illustration 1, Sec. 20.5, it follows that if 6@, y) - yrx * x2 + 4x + 2y2- 5y, then Ybk, y) : (y, * 2x * 4)i + (2xy + ay - 5)i

Thus, the value of the line integral is independent of the path, and C can be any sectionally smooth curve from (0, 0) to (1, 1). From Theorem 20.7.'J.., we have, then,

OF THE PATH 20.7 LINE INTEGRALSINDEPENDENT

993

I,

(y' * 2x* $ dx* (Zxy* 4y - s) dy - 0(1, t) - 0(0, 0) -J-0 -3

This result agreeswith that of Illustration L. If F is the vector-valued function defined by F ( x , Y ) : M ( x , Y )i + N ( r , V ) i

(7)

F (x, y) : V0@, Y)

(8)

and then F is called a gradientfield, $ is called a potentialfunction, and $(x, y) is the potential of F at (r, y). Furthermore, if F is a force field satisfying Uq. (g), then F is said to be a conseroatiaeforce field. Also, when a function F,- defined by Eq. (7) has a potential function {, then the expression M(x, y) dx * N(x, y) dy is called an exact differential;this ter'minology is reasonablebecauseM(x, y) dx * N(x, y) dy : d6. . rLLUsrRArrow3: In Illustration 1, we have the force field F(x, y) : (y' * 2x * 4)i t (2xY+ aY - 5)i In Illustration 2, we have the function { such that -f 6k, y) : yLx* xz 4x * 2yz 5y Because 96G, y) -F(x, y) it follows that F is a conservative force field, and q| is a potential function of F. In particular, becauseOe, D:11, the potential of F at (2, 1) is 11' Furthermore, (y' * 2x * 4) dx * (2xY* 4Y - 5) dY is an exact differential and is equal to d0. .ov edI t h at PIE 2, Sec. 20 5, we sh, Pl '.xe- a : l= ( e--aa Zx)i - ( 'e-a * s in S y' ) = c Is : *l cco Y (xe-a - x2

SOLUTION:

I Nr E x

v)i

t."1, 'em 2( '(x, with rQ( 20. 1.7 )rel lv Th.eo , a/ ) =: X e - a - X 2 + )Ply dy Therefore, we apl r

I te-a 2x)) d x JC I

t

-4

(xte- a +- Ssir in, y ) d y: a-U

=+ 0 (00, ?r) -

:cos :

O ( n ,0 )

7r-(n-n2*I)

T r 2- T r - 2

cos y.


If the value of a line integral is independent of the path, it is not necessary to find a potential function 0. we show the procedure in the next example.

ExAMpw 2: Show that the value of the followirg line integral is independent of the path and evaluate it:

C is any sectionally smooth curye from the point (5, - 1) to the point (9,-3).

solurroN:

Let M(x, A):

Ma(x, y) -

-xly'. Then

Uy and N(x,

and

N

"(x,

y)

1,

v'

BecauseMo(x, y):N,(r, y), it follows from Theorem 20.5.1that (Uy)i(xlV')i is a gradient, and therefore the value of the integral is independent of the path. We take for the path the line segmentfrom (5,-1) to (9,-3). An equation of the line is x * 2y: 3. Thus, parametric equations of the line are x:3*2t

y:-t

l.
We compute the value of the line integral by applying Definition 20.6.2.

(M( (3 + 2t),

N( (3 + 2t), -f) ) . (2, -1) dt

The results of this section can be extended to functions of three variables. The statement of the following theorem and its proof are analogous to Theorem 20.7.1,.The proof is left as an exercise (see Exercise 31). 20.7.2 Theorem

.

Suppose that M, N, and R are functions of three variables x, y, and,z defined on an open ball B((xr, !o, zo))r) in R3;Ms, M", N' N, R' and R, are continuous on B; and Vd (x, y, z) : M(x, y , z)i * N(x, y, z)i + R(x, y, z)k Suppose that C is any sectionally smooth curve in B from the point (rr, yr, z) to the point (r2, Az, z2).Then the line integral

OF THE PATH INDEPENDENT 20,7 LINE INTEGRALS

995

f

(x, ly, z) dz I tvt(x,y, z) dx * N(r, y, z) dy * R(r,

Jc

is independent of the path C and .........'...11.1111..1..l.i.ill..liir.,.il.,.i.riiiii.l.liiiii itt'i',,,'.l..,.iin..tt.t'"t*l+i'i'li+Hil+i+l"ilit+ii#-iffi

+#*+l:ffi il+ii:iuilil+: il#*+'ifi lifflffifti.ll..$+#.ru'i#J.iri#.l.il.i'*flEi#ft#.lii i*:i:iiiiit+:+lii ii i$*+ii .*H#..'

it r€.'"i++f+'t"+

.1i+.lrti!;ffir

r'il

itt rijr,i"r,,r;riiir+.itl;.;l.,r.ti.rt.=ffiii.

ttl'nt'

ffi ffi Lir'#titilintlii-ittiii'iffi

ir'tii+tnitirllli

we showed that the vector ,5, rn 4: In Example3, Sec.20.5, o TLLUSTn^lrroN c o sz * 2 x y * 3 2 2 ) k (e' sin z * 2yz)i + (Zxz* 2y)i + (e' cos

is a gradient V/(x, U, z) and f(x,A,z) - e'sinz*2xyz*y'*zB

+C C

cur from (0, 0, 0) to (1, -2, n) Therefore,if C is any sectionallysmooth curue then it follows from Theorem 20.7.2that the line integral f

ly + (e' cos z * 2xy * 322)dz I t e' sin z * 2yr) dx * (2xz * 2y) dY

Jc

is is independent of the path and its value is f (t,-2,

n) - f (0,0, 0) :

( e s i n n - 4 nn + 4 + 7 r 3 ) - 0

-rf-4r*4

o

As with functions of two variables, If F(x, y, z) : Y6(x, y,z), then F is a gradient field, { is a potential function, and f(r, y, z) is the potential of F at (x, y, z).If F is a force field having a potential function, then F is a conservativeforce field. Also, if the function F, defined by F(x, !, z): M(x, y, z)i * N(r, y, z)i * R(x, y, z)k, has a potential function, then the expression M(x, y, z) dx * N(x, y, z) dy + R(x, y, z) dz is an exact differential. 3: If F is the force field EXAMPLE definedby F(x, y, z) - (z'+ 1)i + 2yzi * (Zxz* y')U Prove that F is conseryativeand find a Potential function.

solurroN: To determine if F is conservative, we apply Theorcm 20.5'2 to find out if F is a gradient.Let M(x, y, z): z2+'l', N(x, y, z):2y2, and R(x, y, z):2xz * y2.Then Nr(x, Y,z) :0 M,(x, !, z) :22 M u ( x ,y , z ) : 0 N"(x, y, z) :2y

R,(x, y, z) - 2z

Ru(x,Y, z) :2Y

Therefore, Mu(x, y, z) : N r(x, y, z), M"(x, a, z) : Rr(x, y, z), and N r(x; y, z) : Ru(x,y, z). Thus, by Theorem 20'5.2,F is a gradient and hence a conservative force field. We now find a potential function f such that F(r, y, z\: YQ(x,y, z); hence, :2yz (9) Q " ( x , y , z ): 2 x z * y 2 Q r ( x , y ,z ) : z ' + ' ] - , 6 u ( x , A , z ) Integrating with respect to r both members of the first of Eqs. (9), we have 6(x, y, z) : xzz* x * g(Y, z)

(10)

g9O ] DIRECTIONIAL DERIVAT]VES, GRADIENTS,'APPLICATIONS OF PARTIALDERIVATIVES, AND LINE INTEGRALS

we now partial differentiate with respect to y both members of Eq. (10) and we obtain 6o(x, y, z) : go(y, z)

(tt)

Equating the right membersof Eq. (11)and the secondof Eqs. (9),we have

go(Y, zl :zvz We now integrate with respect to y both members of this equation, and we get Sg, z) : yzz+ h(z)

(12)

Substituting from Eq, (I2) into Eq. (10),we obtain Q@' Y, z) :1cz2* x * Y'z + h(z) (13) We partial differentiate with respect to z both members of Eq. (13)and get Q"(x, y, z) :2xz * yz t h'(z)

(14)

We equate the right members of Eq. (1a) and the third of Eqs.(9) and we have 2xz * y' + h' (z) :2xz * y' h ' ( z ): o h(z): C

(1s)

Substituting from Eq. (15) into Eq. (13), we obtain the required potential function 0'such that ' O ( x , Y ,z ) : J c z+z ) c + Y ' z+ C

Exercises20.7 In Exercises L through 10, prove that the given force field

1. F (x, A) : yi-+ xi

1 S conservative

and tind a potential ftrnction.

2. F(x, A) : xi + yi

3. F(x, A) : e' sin W + e' cosyi 4. F(x, A) : (sin y sinh x * cos y cosh r)i * (cos y cosh x - sin y sinh r)i 5u F (x, !) : (2xy' - yr)i + (2xzy- 3xy, + 2)i 6. F(x, A) : (3x'* 2y - y'e")i* (2x- 2ye")i 7. F(x, y, z) - (x2- V)i - (r - 3z)i + k + 3y)k 8. F(x, y,z) : (2y'- 8xi2)i * (5xy, + l)i - (8xzz* lzz)k 9 . F ( x , y , z ) - ( 2 x c o sy - 3 ) i - ( x , s i n y * z r ) i - Q V z - 2 ) k L0. F(x, y, z): (tan y + 2xy secz)i * (r seczy * x2 sec + secz(x,y t a n z - s e c z ) k ")i

INDEPENDENT OF THE PATH 20.7LINEINTEGRALS

997

In Exercises11 through 20, use the results of the indicated exerciseto prove that the value of the given line integral is independent of the path.-Then evaluate the line integral by applying either Theorem20.7.7 or Theorem 20'7.2 and using the potential function found in the indicated exercise.In each exercise, C is any sectionally smooth curve from the point A to the point B. f

11. I y dx * x dy; A ts (L, 4) and B is (3, 2); Exercisel JC

12. 13. L4. L5.

L I, I, I, I, I, I, L

x dx * y dy; A is (-5 , 2) and B is (L, 3); Exercise2 e" siny ilxl e" cosy dy; A is (0,0) and B is (2, iz); Exercise3 ( s i n y s i n h r * c o s y c o s hx ) d x * ( c o s y c o s h x - s i n y s i n h x ) d y ; A i s ( 1 , 0 ) a n d B i s ( 2 , n \ ; E x e r c i s e 4

(2xy'- A\ dx + (zfv - 3xy2* 2\ dy; A is (-3, -1) and B is (1, 2); Exercise5 dx-f(2x-2ye') dy;Ais (0,2) andBis (1,-3);Exercise6

1,6.

(3f +2y-y'e")

17.

@ - il dx (x 3z) dy * (z I 3y) dz; A is (-3, l, 2) and'B is (3, 0, 4); Exercise7

18.

(2y3- 8xz2)dx -t (6xy2+ l) dy - (Sfz 1 322)dz; A is (2,0, 0) and B is (3, 2, 1); Exercise8

1,9.

(2xcosy-3)

9 d x - ( x 2s i n y t z r ) d y - ( 2 y z - 2 ) d z ;A i s ( - 1 , 0 , 3 ) a n d B i s ( 1 , z r , 0 ) i E x e r c i s e

ZO.[ (tanv-l2xusecz)dx*(xseczytx2secz)dy*secz(x2ytanz-secz)dz;Ais(2,*n,0)andBis(3,n',2);Exercisel0 Jc' In Exercises21 through 30, show that the value of the line integral is independent of the path and compute the value in any convenient manner. In each exercise,C is any sectionally smooth curve from the point A to the point B. I

21. | (2v - x) dx + (y' t 2x) dy; A is (0, -1) and B is (L' 2) Jc'zz. I Onx*Zti Jc'

d x * ( e u+ 2 x ) d y ; A i s ( 3 , 1 ) a n d B i s ( 1 , 3 )

z . I an y dx'r x sei y dy; Ais (-2, 0) andBis G,tn) Jc

dx* (siny* 2a. ["siny

,t.

l2a'+

1"67r*dx

25.[ *JJcrry-tz-

+

xcosy) dy; Ais (-2,0) andB is (2,*rr)

A is (0,2) andB is (1,0) 1rft g,dv;

av + --J* a, + --=!*-ly'!22*v' f+y'a""

dz;A is(1, 0, 0) and B is (1,2, g)

z z . I f u 1 2 ) d x * ( r * z ) d y + ( x + y ) d z ; A i s( 0 , 0 , 0 )a n d B i s( l , r , r ) Jc.-

DIRECTIONALDERIVATIVES, GRADIENTS, APPLICATIONS O F PARTIALDERIVATIVES, AND LINE INTEGRALS

998

28. 29.

f

J" Qz

+ x) dx I (xzi y) dy + (xy + z) dz;A is (0, 0, o)an B i s ( 1 , 1 , L )

r Jr(e'sin

xy- cosy) y + y z ) d x * ( e ' c o sy * z s i n y * x z ) d y + ( x!

A i s ( 2 , 0 , ' 1 . )a n d B i s ( 0 , n , 3 )

f

30. I Qx ln yz - Sye") dx - (5e" - x'y-t) dy + (x27-r* 2z1 ) dz; A i s ( 2 , 1 , 1 ) a n d B i s ( 3 , 1 , e ) JC


ReaiewExercises (Chapter20) 'J. In Exercises and,2, find the value of the directional derivative at the particular point Po for the given function in the direction of U. 7. f(r, y, z) : xy"z - 3xyz * 2x*; U : -*i + 3i - *k; Po: (2, l, 1)

Y,v:;dz.f(r,y): tan-1

fri,po:

(4,-4)

In Exercises3 and 4, find the rate of change of the function value in the direction of U at P.

3.f(x,a):f,r" (x'* u\;u- l, * + i;P- (r,L) 4. f(x, y, z) : yz - y2- xz;U - 9i + +i + ?k;P - (1, 2, g) In Exercises 5 and 6, determine if the vector is a gradient. If it is, then find a function having the given gradient. 5. y(cos)c- zsinr)i*z(cos

)c+siny)i-

(cosy-y

cosr)k

6. (e" tan y - sec y)i - sec AQ tan y - er sec y)j

In Exercises7 and8, find an equation of the tangent plane and equations of the normal line to the given sur{aceat the indicated point. 7. z2+2y+z--8;

(2,1,2)

8. z:x2+Zxy; (L,3,7)

In Exercises 9 and 1.0,determine the relative extrema of f , if there are any.

9.f(r,Y):P-tY3t3xY 10. f(x, V) : Z* - 3xy -f 2y2* l}x - lly In Exercises 11 and 12, evaluate the line integral over the given curve.

t t . I p * * 3 y ) d x * x y d . y ; CR: ( t ) : 4 s i n f i - c o s t i , o < t - * r Jc. I

1 2 . I x e ai l x - x e " d y * e " d z ; C : R ( t ) : t i +

P i + t 3 k , 0< f < L

JC

In Exercises13 and 14, find the total work done in moving an object along the given arc C if the motion is caused by the given force field. Assume the arc is measured in inches and the force is measured in pounds. 13.F(r,y,z):(xy-z)i+yi+zlqC:thelinesegmentfromtheorigintothepoint(4,7,2)

REVIEWEXERCISES 999

14. F(x, y) : xyzi - x'yj; C: the arc of the circle x' + y':

4 from (2, 0) to (0, 2)

In Exercises15 and L6, prove that the value of the given line integral is independent of the path, and compute the value in any convenient manner. In each exerciseC is any sectionally smooth curve from the point A to the point B. I

1 5 . | 2 x e ud x l f e u d y ; A i s ( 1 , 0 ) a n d B i s ( 3 , 2 ) JC

rc. I z siny dx I xz cosy dy + x siny dz;A is(0, 0, 0) and B is (2, g, +n) Jc

17. lf f (x, !, z) : 3xy' - 5xz2- 2xyz, find the gradient of f at (-7, 3, 2). 1 8 .l t f ( x , l , z ) : s i n h ( r * z ) c o s h A , h n d t h e r a t e o f c h a n g e off( x , y , z ) w i t h r e s p e c t t o d i s t a n c e i n R s a t t h e p o i n t P ( 1 , 1 , 0 ) in the direction PQ if Q is the point (-L , O, 2) . 19. Find the dimensions of the rectangular parallelepiped of greatest volume that can be inscribed in the ellipsoid rf + 9y2-f z2: 9. Assume that the edges are parallel to the coordinate axes. 20. Find equations of the tangent line to the curve of intersection of the surfacez:3* the point (2,-1,74).

f' y' * L with the plane x:2

at

21. The temperature is T degrees at any point on a heated circular plate and t:

M f +a,+9

where distance is measured in inches from the origin at the center of the plate. (a) Find the rate of change of the temperatureat the point (3, 2) in the direction of the vector cos6zi * sinfzri. @) Find the direction and magnitude of the greatestrate of changeof T at the point (3,2). 22. The temperatureis T degreesat any point (x, y) of the curve 4x2* L2y2: L and T : 4f * 24y2 2x. Find the points on the curve where the temperature is the greatest and where it is the least. Also find the temPerature at these points. 23. A monopolist produces two commodities whose demand equations are x : 1 6 - 3 p - 2 q a n d A : 1 l - 2 p- 2 q where 100r is the quantity of the first commodity demanded if the price is p dollars per unit and 100yis the quantity of the second commodity demanded if the price is 4 dollars per unit. Show that the two commodities are complementary. If the cost of production of each unit of the first commodity is $1 and the cost of production of each unit of the second commodify is $3, find the quantities demanded and the price of each commodity in order to have the greatest profit' 24. Find the value of the line integral

f

-Y

)^-,

x

J ,x ' +v ' d x * f i a v if C is the arc of the cir c\exz * y' - 4 from (rt,

\D

to

o).

25. Given the force field F such that F (x, y ,' z) : 22 sec2xi r 2ye3J* (3y2e3"4 2z tan x)k prove that F is conservative and find a potential function. 26. A piece of wire L ft long is cut into three pieces. One piece is bent into the shape of a circle, a second piece is bent into the shape of a square, and the third piece is bent into the shape of an equilateral triangle. How should the wire be cut so that (a) the combined area of the three figures is as small as possible and (b) the combined areaof the three figures is as large as possible?

1OOO DIRECTIONAL DERIVATIVES, GRADIENTS, APPLICATIONS OF PARTIALDERIVATIVES, AND LINEINTEGRALS 27- Use the method of Lagrange multipliers to find the critical points of the function for which / /(r, A, z) : y * xz 2f - yz - z2 subject to the constrainl z:35 - r: y. Determine if the function has a relative maximum or a relative minimum value at any critical point. 28. In parts (a), (b), and (c) demand equations of two related commodities are given. In eachpart, determine if the commodi t i e s a r e c o m p l e m e n t a r ys,u b s t i t u t e so, r n e i t h e r :( a ) r - - 4 p - l 2 q * 6 , + 1 0 ;( b ) x : 6 - 3 p - 2 q , A:5p-q A:4+ 2p - q; (c) r : -7q - p * 7, y : 18 - 3q - 9p. 29. Determine the relative dimensions of a rectangular box, without a top and having a specific surfacearea,if the volume is to be a maximum.

Multiple integration

MULTIPLEINTEGRATION

2l.l THE DOUBLE INTEGRAL

(bt, b2) (ti,yt) Liy

(a,, ar)

F i g ur e 2 1. 1. 1

The definite integral of a function of a single variable can be extended to a function of several variables. We call an integral of a function of a single variable a single integral to distinguish it from a multiple integral, which involves a function of several variables. The physical and geometric applications of multiple integrals are analogous to those given in Chapter 8 for single integrals. In the discussion of a single integral we required that the function be defined on a closed interval in Rl.}or the double integral of a function of two variables, we require that the function be defined on a closed region in R2. By a closed region we mean that the region includes its boundary. In this chapter, when we refer to a region, it is assumed to be closed. The simplest kind of closed region in R2 is a closed rectangle, which we now proceed to define. Consider two distinct points A(ar, ar) arrd B(bt, bz) such that a1 = b1 and az - bz.These two points determine a rectangle having sides parallel to the coordinate axes. Refer to Fig. 2L.1,.1,. The two points, together with the points (br, ar) and (ar, bz), arc called the aerticesof the rectangle. The line segmentsjoining consecutive vertices are called the edgesof the rectangle. The set of all points interior to the rectangle is called an open rectangle,and the set of all points in the open rectangle, together with the points on the edges of the rectangle, is called a closedrectangle. Let the closed rectangular region of Fig. 21,.1.1be denoted by R and let / be a function defined on R. The region R can be considered as aregion of integration. Our first step is to define a partition, A, of R. We draw lines parallel to the coordinate axes and obtain a network of rectangular subregions which cover R. The norm of this partition, denoted by llAll,is determined by the length of the longest diagonal of a rectangular subregion of the partition. We number the subregions in some arbitrary way and let the total be n. We denote the width of the ith subregionby A4xunits and its height by LtV units. Then it LA square units is the area of the lth rectangular subregion, AA: L'P A'g ((1, yr) be an arbitrary point in the ith subregion and f({i,7i) be the Let function value there. Consider the product f(€u y) A/. Associated with each of the n subregions is such a product, and their sum is n

) ff €i,ri) A,A

(1)

There are many sums of the form (1) because the norm of the partition can be any positive number and each point (f;, 7;) can be any point in the ith subregion. If all such sums can be made arbitrarily close to one number L by taking partitions with sufficiently small norrns, then L is defined to be the limit of these sums as the norm of the partition of R approacheszero. We have the following definition. 21.1*1Definition

Let / be a function defined on a closed rectangular region R. The number

2 1. 1 T H E D O U B L EI N T E G R A L

L is said to be the limit of sums of the form

i

1OO3

f (€,, Tr) L,A if L satisfies

i:l

the property that for any € ) 0, there exists a 6 > 0 such that ln

I

ti,ri)^,A- Ll' € l) /( for every partition A for which llAll < 6 and for all possible selections of . . . , n. If such a number the point (tu y) in the ith rectangle,i:7,2, L exists, we write n

lim ). f(ti,y)

L1A:L

llAll-o {:1

it can be shown If there is a number L satisfying Definition 21.1.'1., that it is unique. The proof is similar to the proof of the theorem (2.1,.2) regarding the uniqueness of the limit of a function. 21.L.2 Definition

A function / of trvo variables is said tobe integrableon a rectangular region R if / is defined on R and the number L of Definition 21.1.1exists.This number L is called the doubleintegral of / on R, and we write

ffiju'dffi* #ffi

. : : . . ,

. :

: . : r . . : . : : . : . : t .

, . .

:

.

: r : :

. :

: .

t .

.

:

e,

:

Other symbols for the double integral in (2) are rr | | f ( x , y ) d xd y a n d

JnJ

f f | | f (x,y) dydx

JnJ

The following theorem, which is stated without proof, gives us a condition under which a function of two variables is integrable. 21..1.3Theorem

If a function f of two variables is continuous on a closedrectangular region R, then / is integrable on R. The approximation of the value of a double integral by using Definition 21.1.3is shown in the following example.

(AMFP L EE I EXA

Find an approxi-

mat ate rvalrlue of the double inte gral

Î I

Jn

(2 (2x' - 3y) dA

s th regi10n :tanng wht here3 R( iis tne rect ' 5ular v erticr hav lvinlr g ver ficerS (-:- L , Lt) and ( 2t , 3 )t ..Ttak nofR titic.or rkee a l P)arti -0tX:-' esx r= y thr the Iiner ; bv rrne( forr:

t,

solurroN: Refer to Fig. 21..1.2,which shows the region R partitioned into six subregions which are squares having sides one unit in length. So for each i, LoA:l. In each of the subregions the point ({i,y) is at the center of the square. Therefore, an approximation to the given double integral is given by ff

| | t z * - 3 y )a e - f e + , * ) . 1 +f ( + , t ) . 1 + f ( 9 , * ) . 1 JnJ

+ f G ,t ) . 1 + f ( + , i l . t + f ( - + , i l . L

1OO4

MULTIPLEINTEGRATION

and y - 2, and take (€u, yr) center of the ith subregion.

3)

(2,

(-+,*) (+,+) G,*) 1)

(-+,+) G,i) G,*)

(2,

o

r F i g ur e 2 1. 1. 3

F i g ur e 2 1. 1. 4

Figure 21.1.2

The exact value of the double integral in Example 't is-24, as will be '1. shown in Example of Sec. 27.2. We now consider the double integral of a function over a more general region. In Definition 8.10.2we defined a smooth function as one that has a continuous derivative, and a smooth curve is the graph of a smooth function. Let R be a closed region whose boundary consists of a finite number of arcs of smooth curves that are joined together to form a closed curve. As we did with a rectangular region, we draw lines parallel to the coordinate axes, which gives us a rectangular partition of the region R. We discard the subregions which contain points not in R and consider only those which lie entirely in R (these are shaded in Fig. 21.1,.3). Letting the number of these shaded subregions be n, we proceed in a manner analogous to the procedure we used for a rectangular region. Definitions 21.1.1and 21.1,.2apply when the region R is the more general one described above. You should intuitively realize that as the norm of the partition approaches zero, n increases without bound, and the area of the region omitted (i.e., the discarded rectangles)approacheszero. Actually, in advanced calculus it can be proved that if a function is integrable on a region R, the limit of the approximating sums of the form (L) is the same no matter how we subdivide R, so long as each subregion has a shape to which an area can be assigned. Just as the integral of a functior geometrically as the measure of the a: tegral can be interpreted geometricall three-dimensional solid. Suppose tha closed region R in R2. Furthermore, I sume that/( x,y) is nonnegativeon R.' is a surface lying above the ry plane

21.1 THE DOUBLEINTEGRAL

1OO5

shows a particular rectangular subregion of R, having dimensions of measures Aal and A1y.The figure also shows a rectangular solid having this subregion as a base and f((p yr) as the measure of the dtitude where (€t, y) is a point in the ith subregion. The volume of the rectangular solid is determined by LrV:f(€o,y)

L{:f(tr,y)

L,p A,g

(3)

The number given in (3) is the measure of the volume of the thin rectangular solid shown in Fig. 2L.1.4;thus, the sum given in (L) is the sum of the measuresof the volumes of n such solids. This sum approximates the measure of the volume of the three-dimensional solid shown inFi9.21..1..4 which is bounded above by the graph of / and below by the region R in the xy plane. The sum in (1) also approximates the number given by the double integral

v) dA It can be proved that the volume of the three-dimensional solid of Fig. 21.L.4is the value of the double integral. We state this in the following theorem for which a formal proof is not given. 21.1..4Theorem

ExAMPrn 2: Approximate the volume of the solid bounded by the surface

f(x,a):4-*x'-#y' the planes x- 3 and y -2, and the three coordinate Planes. To find an approximate value of the double integral take a Partition of the region in the xy Plane bY - L, x : 2, drawing the lines x L, and take (tt, yr) at the and y center of the ith subregion.

Let / be a function of two variables that is continuous on a closed region R in the xy plane andf (x,A) > Ofor all (x,y) inR.If y(S) is the measure of the volume of the solid S having the region R as its base and having an dtitude of measuref (x, y) at the point (x, y) in R, then

The rectangular region R solurroN: The solid is shown in Fig. 21..1..5. is the rectangle in the xy plane bounded by the coordinate axes and the if V cubic units is the volume lines r : 3 and A : 2. From Theorem2'1,.1..4, of the solid,

(4 - tx' - #y') dA Figure 21:1.5 shows R partitioned into six subregions which are squares having sides of length one unit. Therefore, for each i, LA:1.. The point (tu y) in,each subregion is at the center of the square. Then an approximation of V is given by an approximation of the double integral, and we have

t/ : f(+,+) . 1+ f(9,+) . L+ f(8,il . 1 + f(+,g) . L + f(9,+) 1 * f(8,"r). L : (4-ffi) + (4-#) + (4-ffi) + (4-ffi) + (4-H) + @- +++)

1006

MULTIPLE INTEGRATION

: 24* 88t : 21.59 Thus, the volume is approximately 2L59 cubic units.

Figure21.1.5

The exact value of the volume in Example 2 is shown to be 21.5 cubic units in Example 2 of Sec. 21.2. Analogous to properties of the definite integral of a function of a single variable are several properties of the double integral, and the most important ones.ue given in the following theorems. 21.1.5 Theorem

If c is a constant and the function / is integrable on a closed region R, then c/ is integrable on R and lf

rf ae: , 4@,v) JJ J*Jf@,y)dA 21.1.6 Theorem

If the functions / and g are integrable on a dosed region R, then the func-

tion f + I is integrable on R and

rf (x,y)ldA: f f f @,y)dA+ s@,y) dA J.Jlf@,y)* s J.J I.l

The result of Theorem 21.1.5can be extended to any finite number of functions which are integrable. The proofs of Theorems 21.1.5 and,2']-..1.6 follow directly from the definition of a double integral. These proofs are left as exercises(see Exercises13 and 14). 21.1.7 Theorem

If the functigls / and g are integrable on the closed region R andf (x,y) > gk, y) for all (x, y) in R, then f I

tr dAf@,y) J.J JJ t@,v)ae Theorem 21.7.7is analogous to Theorem 7.4.8 for the definite inte-

21.1THEDOUBLE INTEGRAL1OO7 gral of a function of a single variable. The proof is similar and is left as an exercise(seeExercise15). 21.1.8 Theorem

Let the function / be integrable on a closed region R, and suppose that m and M arc two numbers such that m = f ( x , y ) < M f o r a l l ( x , y ) i n R . Then if A is the measure of the area of region R, we have

mA=

dA< MA I.l f @,y)

The proof of Theorem21.1.8is left asan exercise(seeExercise16).The proof is similar to that of Theorem 7.5.2 and is based on Theorem 2L.1..7. 21.1.9 Theorem

Suppose that the function / is continuous on the closed region R and that region R is composed of the two subregions R1 and R, which have no points in common except for points on parts of their boundaries. Then f I

f

f

f

f

| | 'f ( x ,-v )d e : 1J n , |J f k , y ) d A + J|n J| f ( x , y ) d A JnJ The proof of Theorem 21.1..9is also left as an exercise and depends on the definition of a double integral and limit theorems (seeExercise17).

2L.1Exercises In Exercises1 through 6, find an approximate value of the given double integral where R is the rectangular region having the vertices P and Q, A is a regular partition of R, and (€t, y) is the midpoint of each subregion. fl

; G , 2 ) ;A : t , : 0 , x 2 : 7 , x s : 2 , x r : 3 , A r : 0 , A " : 1 , . l . | | ( f + y ) d A ;P ( 0 , 0 ) Q JRJ ft z. I I Q-, JEJ

- y) dA; P(0,0); QG, 4); A; x,:

0, x2: 2, xt: 4, lr:0,

yz: 2.

rf l. | | (xV * 3y2) dA; P (-2, 0) ; QG, 6) ; A: x, : -2, xz: 0, x": 2, y, : 0, lz : 2, Ar: 4. JRJ ff

-f 3y2) dA; P(0, -2) +. | | (xy ; Q(6, D ; A: r, : 0, xz: 2, x": JnJ

4, Ar: -2, y":

0, ys: 2.

tf - 2xy') M; P(-3, -2) ; Q0, 6); A: \ : -3, xz: -1, : -2, s' | | (x'za lt !z: 0, !t: JRJ tf

2, yo: 4.

6. | | GV-2xy') dA;P(-3,-2);QQ,5); L:xt:-3,x2:-2,h:-l,xa:0,y':-2,A2:-1,As:0,yn:1,As72, JnJ Ve: 3' lz: 4' !e: 5Exercises 7 through 10, find an approximate value of the given double integral where R is the rectangular region having In the vertices P and Q, A is a regular partition of R, and (€u y) is an arbitrary point in each subregion. 7. The double integral, P, Q, and A are the same as in Exercisel; (€', y) : (i, i); (€2, yr) : (1,0); (fa, ya) : G, D, G n , y ) : G , r ) ; ( & , y " ) : G , D ; ( f o ,y u ): G , i l ; G , , r t ) : G , z 1 ; ( f a ,y e ) : ( 3 , 1 ) .

1OO8

MULTIPLEINTEGRATION

S.Th"doubleintegral,P,Q,andAarethesameasinExercise2;(€r,y):(t,t);(tr,l):(3,1);Gs,ys):(t,t); G r , y ) : ( 2 , 2 ) ; ( f s , % ) : ( 2 , 2 ) ; ( f e , y e) - ( 5 , 3 ) . 9.T"doubleintegral,P,Q,andAarethesameasinExercise3;(€r,yr):(-t,t);(€r,y):(l,t);(€",f):(g,2);

G+,y):

: (0,3); (f., ye): (4,4);(€r,rr1: 1-i, il; Gr,y"): (-*, t); (€0,7s) |i,.ElitA, yrl : ts,*1.'"'

10. The double integral,P, Q, and.A are the same.asin Exercise3; (tr,yr):

(-2,0); (€r,^yr): (0,0); (6r, ys): (2,0);

(tu y) : (-2,2); ({0,7s): (0,2); (t", y) : (2,2);(tr, rr1: 1-2,4); (fs,ye): 10,+li (fs,ys): lii +1.

11. Approximate the volume of the solid in the first octant bounded by the sphere * * y" I z2:64, the planes r: 3 and ! : 3, and the threecoordinate planes. To find an approximate valui of the double intbgral take a partition of the region inthexy planebydrawingthelines x-l,x:2,y:l,andy:2,arrd.take ( f 1 , n ; a i t t r e c e n t e i o ft h e i t h s u b r e g i o n . 12. Approximatethevolumeof thesolidboundedbythesurfacelfi)z:3OO-2Sf -4y2,theplanesr: -7,x:3,y:-g, and y :5, and the ry plane. To find an approximate value of the double integral i"t" ia"tition of the re6oi in the " x y p l a n eb y d r a w i n g t h e l i n e s x : 1 , y - - l , y : 1 , a n d A : 3 , a n d , t a k e ( f 1 , y , ; u t t t r " c e - n t e r o tf h e i t h s u b - r e g i o n .

13. ProveTheorem2L1.5.

L4. Prove Theorem 2L.L.6.

L6. ProveTheorem2L1.8.

17. Prove Theorem 21.'1..9.

2I.2 EVALUATION OF DOUBLE INTEGRATS AND ITERATED INTEGRALS

z : f(x, Y)

x, F i g ur e 2 1 . 2 . 1

15. Prove Theorem 2L.1,.7.

For functions of a single variable, the fundamental theorem of calculus

one, and we use the geometric interpretation of the double integral as the measure of a volume. we first develop the method for the doubie integral on a rectangular region.

is the measure of the volume of the solid between the surface and the region R. we find this number by the method of parallel plane sections, which we discussedin Sec.8.4. Let y be a number in la2, br]. consider the plane which is parallel to the xz plane through the point (0, y,0). Let A(y) be the measureof the area of the plane region of intersection of this plane with the solid. By the method of parallel plane sections, as discussedin sec. g.4, we express the measure of the volume of the solid bv

f", etvldv Becausethe volume of the solid also is determined by the double integral,

INTEGRALS 1OO9 AND ITERATED OF DOUBLEINTEGBALS 21.2EVALUATION we have fbz

(1)

dA: I efvl dy v) Jaz

By using Eq. (1) we can find the value of the double integral of the function f on R by evaluating a single integral of A(y). We now must find A(y) when y is given. BecauseA(y) is the measure of the area of a plane region, we can find it by integration' In Fig.21'.2.L,notice that the uPPer bolndary of the plane region is the graph of the equation z = f (x' y) when r is in lar, brl. Therefore, A(y) : [l: f k, !) dx. If we substitute from this equation into Eq. (1), we obtain (2) The integral on the right side of Eq. (2) is called an iteratedintegral.Usuutly the brackets are omitted when writing an iterated integral. So we write Eq. (2) as (3) When evaluating the "inner integral" in Eq. (3), remember that r is the variable of integration and y is considered a constant. This is analogous to considering y as a constant when finding the partial derivative of

f (x, y) with respectto x. By considering plane sections parallel to the yz plane, we can develop the following formula, which interchangesthe order of integration. dy dx dA:f:,f:,r@,v) H r@,v)

(4)

A sufficient condition for formulas (3) and (4) to be valid is that the function be continuous on the rectangular region R. Evaluate the double

EXAMPLE

integral

LI

(2x' - 3y) dA

soLUTIoN: ar: -l , br = 2, az: 1, andbz: 3' So we have from formula (3)

LI

(2x' - 3y) dx dU

(2f - 3y) dA

(2x'- 3il d.f dy

if R is the region consisting of all points (x,y) for which -L x<2and1=y<3.

: -24

1O1O

MULTIPLEINTEGRATION

In Example 1 of sec. 27.1wefound an approximate value of the double integral in the above example to be -25. EXAMPLE2: Find the volume of the solid bounded by the surface

f(x,y) -4-ix'-#y, the planes x:3 and y : 2, and the three coordinate planes.

soLUTroNr Figure 21.2.2shows the graph of the equation z: f(x,y) in the first octant and the given solid. rf y cubic unitjis the volume or tn" solid, we have from Theorem 21..1.4

u:

,,lit.i fc,, vt)aA:LI f@,v)itA

:

f3 f2

:

Ff

:

@- +* - /rY2)dYdx JoJo 12

J"Lor-t*v-tv"]odx f3

(#-&*) dx Jo : # x _ t x " ll3 Jo

:2L.5

x

Figure 21.2.2

The volume is therefore2!.5 cubic

x i-r ti xi V Ltx

Figure 21.2.3

In Example 2 of.sec. 21.1 an approximate value of the volume in Example 2 was found to be 2L.59cubic units. Y:6r@) suppose now that R is a region in the xy plane which is bounded by t h e l i n e sx : a a r r d x : b , w h e r ea 1 b , a n d b y t h e c u r v e sy : e 1 @ ) a n i y : 6r(x), where 0r and $, arc two functions which are continuous on 4r@) the closedinterval la,bl;furthermore,6r@) = 6r@) whenevera < x < b (seeFig. 21.2.3).Let A be a partition of the interval fa,bf defined byA: a:.xo <.r1 < divided into vertical strips with widths of meisure Alx. A particular strip is shown in the figure. The intersection of the surface ,J f(*, y) and a plane r: f1, where xr-t s €t = xr.,is a curve. A segment of this curve is over the ith vertical strip. The region under this curve segment and above the xy plane is shown in Fig. 2't.2.4, and the measure o1 the area of this region is given by

r::,:

f G,,a)dy

The measure of the volume of the solid bounded above by the surface z - f ( x , y ) a n d below by the ith vertical strip is approximately equal to | fazgil

f(€" y) LJo,tE,r

I

dV I Lix J

INTEGRALS 1011 AND ITERATED OF DOUBLEINTEGRALS 21.2EVALUATION

v :4t(x)

Y : dt(x) Y +r(x)

Y : Qr@)

F i g ur e 2 1. 2 . 5

Figure 21.2.4

If we take the limit, as the norm of A approacheszero, of the sum of these measuresof volume for n vertical strips of R from x: a to x:b, we obtain the measure of the volume of the solid bounded above by the surface z: f (x, y) and below by the region R in the xy plane. (SeeFig. 21'.2.5.) This is the double integral of / on R; that is,

(s)

Sufficient conditions for formula (5) to be valid are that f b" continuous on the closedregion R and that rf1 and Q, be smooth functions. ExAMPLE3: Express as both a double integral and an iterated integral the measure of the volume of the solid above the xY plane bounded by the elliPtic Paraboloid z x2 + 4y', and the - 4. Evaluate cylind er xz * 4y' the iterated inte gral to find the volume of the solid.

soLUrIoN: The solid is shown in Fig. 21'.2.6.Using properties of symmetry, we find the volume of the portion of the solid in the first octant which is one-fourth of the required volume. Then the region R in the : ry plane is that bounded by the x andy axesand the ellipse * + 4yz 4. This region is shown in Fig. 27.2.7,which also shows the fth subregion of a rectangular partition of R, where (tu yi) is any point in this ith subregion. If v cubic units is the volume of the given solid, then by Theorem 21.1.4we have lim l l a l l - ot

A | | (x'+ 4y') dA ( € o ' * A v o \L A - =

J*J

1012

MULTIPLEINTEGRATION

(ti,Yi)

Figure 21.2.7

F i g ur e 2 1. 2 . 8

To express the measure of the volume as an iterated integral we divide the region R into n vertical strips. Figure 21.2.g shows the region R and the ith vertical strip having width of L6xunits and length of units, where x*r s €i = xt. Using formula (5) we have +IT=E f f\rytz

V:4 Figure21.2.6

LJ,

(€; * 4v') dyI Lix )

t2

-4

@, + 4yr) dy dx

Evaluating the iterated inte gral, we have

v:4 t:l*,,++v'])o*'' o* :n

r2

Jolt*I+-P (f +z)tR

+ + ( 4- x : 2 ) 3 d t 2x1 dx

- -*r (4 - xz)3t2 * 2x\tr4 :4r

+ 8 sin

-'*'],

Therefore, the volume is 4n cubic units.

If the region R is bounded by the curvesr: trr(y) and.l: ),r(y) and the lines y: c andA: d, where c < d, and ),, and )r, are two functions which are continuous on the closedinterval lc, d] forwhich Ir(y) < Ir(y) whenever c < y - d, then consider a partition A of the intervJ lc, dl-ia divide the region into horizontal strips, the measures of whose widths are Ag. see Fig. 27.2.9,which shows the lth horizontal strip. The intersection of the surface z: f (x, y) and a plane A = lt, where Ai_r = Tt = Ut, is- a curve, and a segment of this curve is over the ith horizontal strip. Then, as in the derivation of formula (5), the measure of the volume of the solid bounded above by the surface z- f(x, y) and,below by the ith

21.2 EVALUATIONOF DOUBLE INTEGRALSAND ITERATEDINTEGRALS

1013

vertical strip is approximately equal to x:

|rz(U)

d'lL,a t/ff1r@,v,) Taking the limit, as llAll approaches zero, of the sum of these measures of volume f.or n honzontal strips of R from y: c to U: d, we obtain the measure of the volume of the solid bounded above by the surface z: f (x, y) and below by the region R in the xy plane. This measureof volume is the double integral of / on R. Hence, we have

(6) Figurc 21.2.9

Sufficient conditions for formula (5) to be valid are that trt and )t2 be smooth functions and / be continuous on R. In applying both formula (5) and formula (6) sometimes it may be necessaryto subdivide a region R into subregions on which these sufficient conditions hold. ExAMPLE4: Express the volume of the solid of Example 3 by an iterated integral in which the order of integration is the reverse of that of Example 3. Compute the volume.

solurroN: Again we find the volume of the solid in the first octant and multiply the result by 4. Figure 2L.2.70shows the region R in the xy plane and the ith horizontal strip whose width has a measure of.Ag, and whose length has a measure of 2!1- y;2. Then by formula (5) we have

V:4

lim $ l l a l l -eo

d.f Liu @'*4vr)

U:*

(x'* 4y') dx dy

\v ?,

-4

ft

t-

12\n42

| l * r ' * 4 yez x l

Jo L

dy

Jo

F i g ur e 2 1. 2 . 1 0

- -+r y(1 - yz)s/z 1 sytffi+ :4n Hence, the volume is 4n cubic Example 3.

8 sin-l y]', which agrees with the answer of

1014

MULTIPLEINTEGRATION

From the solutions of Examples 3 and 4 we see that the double inteff (* + 4y2)ilA can be evaluated by either of the iterated integrals gril J*J f2 f\Etrl2

(f + +y2)dyitx or

J,J,

-

fr fzt/r-az

J,J,

(* + 4yz)dxity

If in either formula (5) or (6),/(r ,A) :1 for all x andy,then we obtain a formula that expressesthe measureA of the areaof a-regionR as a double integral. We have

AExAMPLE5: Find by double integration the area of the region in the xy plane bounded by the cun/es y -- x2 and y : 4x - x2.

I^l

solurroN: we have

itudx:

[.]

dxdy

(7)

The region is shown in Fig. 21..2.1,1. Applying formula (Z),

A- I.l dydx: I:l;-" dyd* :

I:

@x- x2- x2) dx - 2x2- 3*]',

-_38

Hence, the area of the region is I square units.

F i g u r e2 1 . 2 . 1 1

Exercises21.2 In Exercises L through 8, evaluate the given iterated integral.

3. 7.

TI:'

ttta dx dy

[[

l * - y l d yd x

4. 8.

AND ITERATEDINTEGRALS 1015 OF DOUBLEINTEGRALS 21.2 EVALUATION

In Exercises9 through 14, find the exact value of the double integral. 9. The double integral is the same as in Exercise 1 in Exercises2l'.1. 10. The double integral is the same as in Exercise4 in Exercises21.1. I.

:'r' : : [ [ sin x dA;R is the region bounded by the lines y 2x, y !x, andx JaJ fl

lZ. | | cos(r * y) dA; R is the region bounded by the lines y:

x and x:

r, artd the x axis.

JRJ tf

d.A;R is the region bounded by the cirde I t y':9.

13. | | f \/rJnJ t lu2

:1. dA; R is the regionboundedby the linesy : r and A :2, andthe hyperbolaxy J.J b 1 5 . Find the volume of the solid under the planez : 4x andabovethe circle12I y':15 in the ry plane.Draw a sketch 14.

of the solid. 1 6 .F i n d t h e v o l u m e o ft h e s o l i d b o u n d e d b y t h e p l a n ex s: y + 2 2 * 1 , x : 0 , ! : 0 , 2 : 0 , a n d 3 y * z - 3 : 0 ' D r a w a sketch of the solid. 17. Find the volume of the solid in the first octant bounded by the two cylinders * * y2: 4 and x2 + z2: 4. Draw a sketch of the solid. - f - 3y'' Draw a sketch of the solid. 18. Find the volume of the solid in the first octant bounded by the paraboloid z:9 19. Find the volume of the solid in the first octant bounded by the surfacesxl z2:1, x:y, and x: y2.Dtaw a sketch of the solid. 'l :16 which lies 20. Find by double integration the volume of the portion of the solid bounded by the sphere * I y' z2 in the first octant. Draw a sketch of the solid' In Exercises21 through 24,use double integrals to find the area of the region bounded by the given curves in the xy plane. Draw a sketch of the region. 2 4 .x 2 + y ' : L 5 a n d A ' : 6 x 2 3 . y : x 2- 9 a n d y - 9 - x 2 2 2 . y ' : 4 x a n dx 2 : 4 a 2L.y--rBandA:x2 25. Express as an iterated integral the measure of the volume of the solid bounded by the ellipsoid *

* a ' *' Z ' : 1

a2'

b2

c2

25. Given the iterated integral fq

lf

fr

!a2-*dydx

Jo Jo

(a) Draw a sketLh of the solid the measure of whose volume is representedby the given iterated integral; (b) evaluate the iterated integral; (c) write the iterated integral which gives the measure of the volume of the same solid with the order of integration reversed. 27. Given the iterated integral 1

1a 1t/F=it

iJ | |

(2x-r y) dy dx

JoJo

The instructions are the same as for Exercise26.

1016

MULTIPLEINTEGRATION

28. Use double integration to find the area of the region in the first quadrant bounded by.the parabola x,the circle lz: * -t y': 5, and the r axis by two methods: (a) Integrate first with respectto x; (b) integrate first with iespect to y. Compare the two methods of solution. 29. Find" by two methods, the volume of the solid below the plane 3r -l 8y -l 5z:24 and above the region in the ry plane bounded by the parabola A2:2rc, the line 2x -l 3y: 10, and the r axis: (a) Integrate first with respe-ctto r; (b) lnt"grute first with respect to y. Compare the two methods of solution. In Exercises 30 and 3L, the iterated integral cannot be evaluated exactly in terms of elementary functions by the given order of integration. Reverse the order of integration and perform the computation. rl

rl

30. I I e"' dxdy JoJa 32. Use double integration to find the volume of the solid conunon to two right-circular cylinders of radius r units, whose axes intersect at right angles. (SeeExercise 8 in Exercises g.4.)

21.3 CENTER OF MASS AND MOMENTS OF INERTIA

In Chapter g we used single integrals to find the center of massof a homo_ geneous lamina. In using single integrals we can consider only laminae of constant area density (except in special cases);however, with double integrals we can find the center of mass of either a homogeneous or a nonhomogeneous lamina. Suppose that we are given a lamina which has the shape of a closed region R in the xy plarre. Let p(x, y) be the measure of the area density of the lamina at any point (r, y) of R where p is continuous on R. To find the total mass of the lamina we proceed as follows. Let A be a partition of R into r rectangles. lf (t0,1) is any point in the ith rectangle having an area of measure Arr4,then an approximation to the measureof the mass of the lth rectangle is given by p(tr, y) AtA, and the measure of the total mass of the lamina is approximated by n

PGt' Yi) LoA i:l

Taking the limit of the above sum as the norm of A approaches zero, we express the measure M of the mass of the lamina by (1)

The measure of the moment of mass of the lth rectangle with respect to the r axis is approximated by fp(€u y) AA.The sum of the measuresof the moments of mass of the n rectangleswith respect to the r axis is then approximated by the sum of n such terms. The measureM" of the moment of mass with respect to the r axis of the entire lamina is given by (2)

2 1, 3 C E N T E RO F M A S SA N D M O M E N T SO F I N E R T I A

1017

measureMa of its moment of mass with respectto the

(3) The center of mass

'/..: ExAMPLE A lamina in the shape of an isosceles right triangle has an area density which varies as the square of the distance from the vertex of the right angle. If mass is measured in slugs and distance is measured in feet, find the mass and the center of mass of the lamina.

the lamina is denoted by the point (I, y) and

solurroN: Choose the coordinate axes so that the vertex of the right triangle is at the origin and the sides of length a ft of.the triangle are along the coordinateaxes(seeFig. 21,.3.1).Let p(x, y) be the number of slugs/ff in the area density of the lamina at the point (r, y). Then p(x, y): k(xz * y2) where k is a constant. Thereforc, if M slugs is the mass of the lamina, we have from formula (1.)

M : lisr L o$f + yf) LtA llall-o f!

(x' * y') dA * y') dy dx

lr*'++u'] i-"o* (*at - azx * 2ax2- txt) dx

: k(taa - laa I &aa- *aa) : gkaa F i g ur e 2 1. 3 . 1

To find the center of mass, we observe that because of symmetry it must lie on the line y: r. Therefore, if we find x, we also have y. Using formula (3), we have n

M,:

.!iln ). kf,(f,, + :|12)LtA l l A l l - oi = i

*'* y') dA x(x' * y') dy dx 1a-r

txvt v J ol

ax

1018

M U L T I P L EI N T E G R A T I O N

(*atx - a2)c2 * 2ax3- *xn) dx

- k(tat - tas * ta; - #a')

- +ka, Because MI: My, we have MI - +ka'; and becauseM - tkan, we get I ?a. Therefore, the center of mass is at the p o i n t ( ? a , ? a ) . 21.3.1 Definition

The moment of inertia of a particle, whose mass Ls m slugs, about an axis is defined to be mrz slug-ft?, where r ft is the perpendicular distance from the particle to the axis.

If we have a system of n particles, the moment of inertia of the system is defined as the sum of the moments of inertia of all the particles. That is, if the ith particle has a mass oI m7 slugs and is at a distance of ri ft from the axis, then I slug-ft2 is the moment of inertia of the system where n

I_

ttltT12

(4)

i:1

Extending this concept of moment of inertia to a continuous distribution of mass in a plane such as rods or laminae by processessimilar to those previously used, we have the following definition. 21..3.2Definition

Suppose that we are given a continuous distribution of mass occupying a region R in the xy plane, and suppose that the measure of area density of this distribution at the point (x, y) is p(x, y) slugslfP,where p is continuous on R. Then the moment of inertia I" slug-ff about the r axis of this distribution of mass is determined by

(s) SimilarLy, the measure la of the moment of inertia about the y axis is given by (6) and the measure /0 of the moment of inertia about the origin, or the z axis, is given by (7)

The number Ie of formula (7) is the measure of what is called the polar momentof inertia.

21.3 CENTEROF MASS AND MOMENTSOF INERTIA

EXAMPLE2: A homogeneous straight wire has a constant linear density of p slugslft. Find the moment of inertia of the wire about an axis perpendicular to the wire and passing through one end.

1019

solurroN: Let the wire be of length a ft, and suppose that it extends along the x axis from the origin. we find its moment of inertia about the y axis. Divide the wire into n segments; the length of the ith segment is L,ix ft. The mass of the ith segment is then p Aaxslugs. Assume that the mass of the ith segment is concentrated at a single point fi, where xa-,rs €i = x* The moment of inertia of the ith segment about the y axis lies between pxi-12Lg slug-ff and pxf Ap slug-ff and is approximated by ptt" A& slug-ff, where x*r s & = xr.If the moment of inertia of the wire about the y axis is I, slug-fP, then

lo - :,lip

l l a l l - oi): r

ot,'

pxz dx - *pot

Therefore, the moment of inertia ts ! pa3 slug-ft2. EXAMPLE3: A homogeneous rectangular lamina has constant area density of p slugslfP. Find the moment of inertia of the lamina about one corner.

solurroN: Supposethat the lamina is bounded by the lines r : d, A : b, the r axis, and the y axis. See Fig. 21.3.2.If 16slug-ff is the moment of inertia about the origin, then n

Io: ,liF ) p(f,, + y?) ArA llAll-o;=

((i,yi)

p(x' * y') dA

(x' + y') dx dy

Figure21.3.2

- +pab(a' + br) The moment of inertia is therefore * pab (a2 r bz) slug-ftz.

It is possible to find the distance from any axis L at which the massof a lamina can be concentratedwithout affecting the moment of inertia of the lamina about L. The measure of this distance, denoted by r, is called the "radius of gyration" of the lamina about L. That is, if the massM slugs of a lamina is concentrated at a point r ft from L, the moment of inertia of the lamina about L is the same as that of a particle of mass M slugs at a distance of r ft from L; this moment of inertia is Mrz slug-ft2.Thus, we have the following definition. 21.3.3Definition

If 1 is the measure of the moment of inertia about an axis L of a distribution of mass in a plane and M is the measure of the total mass of the

1O'IO

MULTIPLEINTEGRATION

distribution, then the radius of gyration of the distribution about L has measute r, where

EXAMPLE4: Suppose that a lamina is in the shape of a semicircle and the measure of the area density of the lamina at any point is proportional to the measure of the distance of the point from the diameter. If mass is measured in slugs and distance is measured in feet, find the radius of gyration of the lamina about the r axis.

solurroN: Choose the r and y axes so that the semicircle is the top half of the cirde f * y': a2. See Fig. 21.3.3. The area density of the lamina at the point (r, y) is then ky slugslfP. So if M slugs is the mass of the lamina, we have

M:

lim $. t", Aa

l l a l l - of i rf

:llkuM JnJ

:l

fa f{E--V

| _kydxdy

J o J _\ta2_a2

: k Ifa Il v ,lroz=F l -dY Jo L )-\,td-Yz f@

:zk I y!a2-yz dy Jo-

: -?k(a,- ,r'rt,rfi : &kas Figure21.3.3

If I, slug-fPis the momentof inertia of the lamina about the r axis,then lim i.yrt(h'y,) M

I":

llall-o i=i

ft :J)W"ava'

:f

fd NE=F

|

.kV"dUdx

J-aJ0

:k :*k

fa

t

1{Fst

J_"1+r.1,dx la

|

(a+-2azf +f)

J-a

: *k(2ai - $aa * ?a5)

_ftkas

dx

21.3CENTER OF MASSAND MOMENTS OF INERTIA 1021 Therefore, tf r ft is the radius of gyration ,Lka' 2 q, 12__ffi_to,

and so r - t{t}a.

The radius of gyration is therefore *ffia

tt.

2L.3 Exercises In Exercises1 through 10, find the mass and center of mass of the given lamina if the area density is as indicated. Mass is measured in slugs and distance is measured in feet. 1. A lamina in the shape of the rectahgular region bounded by the lines r : 3 and y : 2, arrd the coordinate axes.The area density at any point is ry2 slugslfP. 2. A lamina in the shape of the region in the first quadrant bounded by the parabola A : * , the line y : 1., and the y axis. The area density at any point is (x * y) slugs/ff. 3. A lamina in the shape of the region bounded by the parabola f :8y, varies as the distance from the line y : -1.

the line y:2,

4. A lamina in the shape of the region bounded by the curve y : e', the line l: density varies as the distance from the r axis.

and the y axis. The area density

L, and the coordinate axes. The area

5 . A lamina in the shape of the region in the first quadrant bounded by the circle I * y': The area density varies as the sum of the distances from the two straight edges.

a2 arrd,the coordinate axes.

6 . A lamina in the shape of the region bounded by the triangle whose sides are segmentsof the coordinate axesand the line 3r I 2y :18. The area density varies as the product of the distances from the coordinate axes. 7. A lamina in the shape of the region bounded by the curve y: sity varies as the distance from the r axis. 8 . A lamina in the shape of the region bounded by the curve y : distance from the y axis.

sin r and the r axis from r : 0 to x:

r. The area den-

fE and the line t : x. The area density varies as the

9. A lamina in the shape of the region in the first quadrant bounded by the circle f * y' :4 area density at any point is xy slugslfP.

10.A l a m i n a i n t h e s h a p e o f t h e r e g i o n b o u n d e d b y t h e c i r c l e f l y ' : 1

and the line r * y :2.The

andthelinesr:landy:l.Theareadensity

at any point is xy slugslfP. In Exercises11 through 16, find the moment of inertia of the given homogeneous lamina about the indicated axis if the area density is p slugs/ff, and distance is measured in feet. 11. A lamina in the shapeof the region bounded by 4y : 3x, x: 4, and the x axis; about the r axis. 12. The lamina of Exercise 11; about the line r:

4.

L3. A lamina in the shape of the region bounded by a circle of radius a units; about its center. 14. A lamina in the shape of the region bounded by the parabola f :4-

4y and,the r axis; about the x axis.

15. The lamina of Exercise 1.4;about the origin. 16. A lamina in the shape of the region bounded by a triangle of sides of lengths a ft, b ft., and c ft; about the side of length a ft.

1022

MULTIPLEINTEGRATION

In Exercises17 through 20, find for the given lamina each of the following: (a) the moment of inertia about the r axis; (b) the moment of inertia about the y axis; (c) the radius of gyration about the r axis; (d) the polar moment of inertia. 17. The lamina of Exercise 1.. 18. The lamina of Exercise2. 19. The lamina of Exercise 7. 20. The lamina of Exercise8. 21. A homogeneous lamina of area density p slugs/ff is in the shape of the region bounded by an isosceles triangle having a base of length b ft and,an altitude of length ft ft. Find the radius of gyration of the lamina about its line of symmetry. 22. A lamina is in the shape of the region enclosedby the parabola ! :2x - f and the r axis. Find the moment of inertia of the lanina about the line y: 4 if the area density varies as its distance from the line y : 4. Mass is measured in slugs and distance is measured in feet.

2I.4 THE DOUBLE INTEGRAL We now show how the double integral of a function on a closed region IN POLAR COORDINATES in the polar coordinate plane can be defined. We begin by considering the simplest kind of region. Let R be the region bounded by the rays 0 : c and g : B and by the circles r : a and r : b. Then let A be a partition of this region which is obtained'by drawing rays through the pole and circles having centers at the pole. This is shown in Fig. 2'1..4.1.. We obtain a network of subregions which we call "crJrved" rectangles.The norm llAllof the partition is the length of the longest of the diagonals of the "curyed" rectangles. Let the number of subregions be n, and let {r4 (ri,or) be the measure of area of the ith "curved" rectangle. Becausethe area of the ith subregion is the difference of the areas of two circular sectors, we have AA - +riz(0i - 0r-r) - trr-rz (0t :t(rt-

ri-)(rr*

Letting 7i: t(k l F i g ur e 2 1. 4 . 1

LtA:

ri-)(0i-

0r-r) fli-r)

rr-) , Ltr : rt - ri-r, and A,0 - 0i - 0a-1,we have

11 Llr L10

Take the point (71,0i) in the ith subregion, where 0*r =di the sum

|rr',,E,)

LtA: f G,,a,)Fi a^sL,o ,i

= 01,and form

(1)

It can be shown that'it 1iscontinuous on the region R, then the limit of the sum in (1), as llAllapproacheszero, exists and that this limit will be the double integral of f on R. We write either

Observe that in polar coordinates, dA-

r dr d0.

21,4 THE DOUBLEINTEGRALIN POLARCOORDINATES 1023

r:

The double integral can be shown to be equal to an iterated integral having one of two possible forms:

6z(0)

r: (r i , 0 i ) r:

6 ,( o )

o: oi-t

r 0:ot

F i g ur e 2 1. 4 . 2

0 - xr(r)

f(r, 0) dA:

(ri,oi) o: oi-.t 0 : X{r)

r t.- L

rT

f(r, 0)r dr d0

f ( r , 0 ) r d 0d r

(4)

We can define the double integral of a continuous function / of two variables on closed regions of the polar coordinate plane other than the one previously considered. For example, consider the region R bounded by r: dr(O) and r:6"(0), where @rand S2 are smooth functions, and In the figure, 6t@) - 6r@) by the lines d : a and e: B. SeeFig. 21..4.2. for all d in the closed interval lo, Fl. Then it can be shown that the double integral of / on R exists and equals an iterated integral, and we have

LI

rr;::

(5)

LI

LT::

(6)

f (r,0)dA:

ri r:

LI

f(r, 0)r dr d0

If the region R is bounded by the cun/es 0 Xt?) and 0 - Xr?), where Xr and Xz are smooth functions, and by the circles r: a and r- b, as shown in Fig. 21.4.3, where Xt?) = Xr!) for all r rn, the closed interval la, bf, then

f (r, 0) dA:

f(r, 0)r d0dr

We can interpret the double integral of a function on a closed region in the polar coordinate plane as the measure of the volume of a solid by using cylindrical coordinates. Figure 2'1..4.4shows a solid having as its base a region R in the polar coordinate plane and bounded above by the surfacez: f (r, 0) where / is continuous on R and f (x, y) > 0 on R. Take a partition of R giving a network of n "crlrued" rectangles.Construct the n solids for which the ith solid has as its base the lth "cvwed" rectangle and as the measure of its altitude f (fo, 0) where (7r, 0r) is in the ith subregion. Figure 21..4.4shows the ith solid. The measure of the volume of the ith solid is

o Figure 21.4.3

z : f ( r ,o )

fju,0) AoA: f(li,0)n AlrLg The sum of the measuresof the volumes of the n solids is n

0:*o

s

LJ i:l

f (7r, |r)Fi Air Ai?

I f V i s the measure of the volume of the given solid, then

EXAMPLE L:

Find the volume the solid in the first octant

solurroN: The solid and the ith element are shown in Fig. 21.4.5.Using formula (7) with f (r, 0) : r, we have, where V cubic units is the volume

1024

MULTIPLEINTEGRATION

bounded by the cone z - r and thecylinderr-3sin0.

of the grve en so )li,d, n lilim \ , ffii l laal l--ooifE :7

v

z

7i L,ir Lrfl

L-I r,r, t,f,

rn 22 dr d0

'zrl2

:

rz dr d0

03

t,[+ 'nl2 7rlz 0

t:L,

Jo

de

ft7t

9 0:O

--9

Figure21.4.5

l 3 sln 0

r3t t

sins 0 d0

Jo

c o s0 + 3 c o s s

-5 The volume is thereforc 6 cubic units. EXAMPLE 2: Find the mass of the lamina in the shape of the region inside the semicircle r a cos 0, 0 < 0 = tn, and whose measure of area density at any point is proportional to the measure of its distance from the pole. The mass is measured in slugs and distance is measured in feet.

solurroN: Figure 21.4.5 shows a sketch of the lamina and the I'th "crrwed" rectangle. The area density at the point (r, 0) is kr slugslff , where k is a constant.lf M slugs is the mass of the lamina, then M_

>t(

')Ii A,ir A,$ (k7 1i Im l lAa l l -o' ii : =l l 'cr2 2d ' f,rrr dle kr Jn

-I,t

t

7f

'12 'a co lsd t cos

r

n12

kag it<

r2 dr d0

0

( s30d0 col

f : +kas iin e--+ s i n 3 01nl ['i,

Jo

: &kag Therefore, the mass is 3 kasslugs. Figure21.4.6

EXAMPLE 3:

FiNd thc center of

mass of the lamina in Example 2.

Let the cartesian coordinates of the center of ma lamina be i and y, where, ds is customdr4, the r axis is along axis and the y axis along the Ln axis. Let the cartesian coordi

SOLUTIoN:

21.4THE DOUBLEINTEGRAL IN POljR COORDINATES 1025

resentation of the point (r1, 0r) be (ir, !).Then if M" slug-ft is the moment of mass of the lamina with respect to the x axis,

M":,fffilo iv,ttnlna,ya,g Replacing |rby it sin 0r, we get M":'

.ti.r.n i. *rlt sin o1A1rA1o Iltll'ofi :[

[ kfsin oitrito JnJ lTl2 faco60

:f

Jo Jo

tBsin9dril|

ft12

- lkaa I

cosa0 sin 0 dA

JO

- -fikaa.or'

0lo" Jo

:,itka4 If M, slug-ft is the moment of mass of the lamina with respect to the y axis, then

t,:

a,rr L$ ,,Til'2r,
Replacing fuby rlcos Oilwe have Mr: "

tim i. tr,t cos 0r Arr Ar0 llall-o;!!

: I lkf coso itr do JnJ

:k

filz

lacoA0

Jo Jo

:*kaa

It12

Jo

f cos0 ilr il|

cosi0 il9

: lkaafsine - &sinso + + rir,t o];" : &ka4 Therefore, + _Mu _ {tka4 _ 3 *:fr:w:5o

MULTIPLEINTEGRATION

1026

Ms _#kao 9 Y: M: &kas:4oo Hence,the centerof mass is at the point (Ea,#a). The area of a region in the polar plane can be found by double integration. The following example illustrates the method. EXAMPLE4: Find by double integration the area of the region enclosed by one leaf of the rose r - sin 30.

SOLUTION:Figure 2'1,.4.7 shows the region and the ith "cufrred" tangle. If A square units is the area of the region, then A-

lim $

lltll-o

rec-

aoA

"?:t

: lim 9. r, AirL,g l l a l l -o , ? : t

0:

0i

(,,, 0 , ) 0-

:

0 i-t

:l

rr

| | rdrd| JnJ fnlS fsin g0

I

,Jo Jo

0:O

o Figure 21.4.7

rdrdT

- + sin2sodo f'' - +o-# sin 6of"'' Jo

: frni square units.

Hence, the area is #n

Sometimes it is easier to evaluate a double integral by using polar coordinates instead of cartesian coordinates, Such a situation is shown in the following example. PLE E XAMP X

Evaluate the double

r'*l'

inntegri rtegr€ ral

,-1rz*U2) g-(tz+

JJn, n

dA

rheree t hlee regio reg ron r R 1 S in the first wrhere )unde d by the qrluadri rantt and bbor uadre ircle ircle x:2 :r -* r7 y ci: t 2'=: ,= a 2 nd the co,rdina OI rdinaate axes. 2 A <

soLUTroN:

Because Jcz* y':

rf | | s-(tz*u,) dAJnJ

y2, and dA:

f f | | e-r, r dr d0 JnJ

--

f'' J["O ,-" r dr do

JO

- -+- J r"'' o lr-,'f"d0 L

f rrl2

=-+ |

Jo

(e-o'-1) d0

JO

: +n(l - e-o')

r dr d0, we have

IN POLARCOORDINATES 1027 21.4IHE DOUBLEINTEGFAL

21.4 Exercises In Exercises1 through 5, use double integrals to find the area of the given region. 1. The region inside the cardioid r:2(l 3. The region inside the cardioid r:

* sin 0).

2. One leaf of the rose r:

a(l * cos 0) and outside the circle r:

a cos20.

a.

1 and outside the lemniscate 12: cos 20. 5. The region inside the large loop of the limagon r:2 - 4 sin d and outside the small loop. 6. The region inside the limagon t:3 - cos 0 and outside the circle r: 5 cos d.

4. The region inside the circle r:

In Exercises 7 through 12, find the volume of the given solid. 7. The solid bounded by the ellipsoid z219r2:9' 8. The solid cut out of the sphere z2* 12: 4by the cylinder r: 1. 9. The solid cut out of the sphere z2I t2 : 16 by the cylinder r: 4 cos 0. - cos 0. 10. The solid above the polar plane bounded by the cone z:2r artd.the cylinder r:1 11. The solid bounded by the paraboloid z: 4 12, Ihiecylinder r: L, and the polar plane. 12. The solid above the paraboloid z: r2 and below the plane z:2r

sin 0-

In Exercises13 through 19, find the mass and center of mass of the given lamina if the area density is as indicated. Mass is measured in slugs and distance is measured in feet. 13. A lamina in the shape of the region of Exercise 1. The area density varies as the distance from the pole. 14. A lamina in the shape of the region of Exercise2. The area density varies as the distance from the pole. cos 0. The area density varies as the distance from 15. A lamina in the shape of the region inside the limagon r:2the pole. cos 0, 0 - 0 < tr, and the polar axis. The area 15. A lamina in the shape of the region bounded by the limagon r:2* density at any point is k sin 0 slugs/ff. 17. The lamina of Exercise 15. The area density at any point is kr sin 0 slugsiff' 18. A lamina in the shape of the region of Exercise5. The area density varies as the distance from the pole. 19. A lamina in the shape of the region of Exercise5. The area density varies as the distance from the pole. In Exercises20 through 24, frnd.the moment of inertia of the given lamina about the indicated axis or point if the area density is as indicated. Mass is measured in slugs and distance is measured in feet. 20. A lamina in the shape of the region enclosedby the cirde r: k slugs/ff.

sin 0; about the *z axis. The area density at any point is

21. The lamina of Exercise20; about the polar axis. The area density at any point is k slugs/fP. - cos 0); about the pole. The area density at any 22. A lamina in the shape of the region bounded by the cardioid r: a(l point is k slugs/ff. 23. A lamina in the shape of the region bounded by the cardioid t : a(7 f cos 0) and the circle r: 2a cos 0; about the pole. The area density at any point is k slugs/ff. 24. A lamina in the shape of the region enclosed by the lemniscate 12: a2 cos 20; about the polar axis. The area density varies as the distance from the pole. 25. A homogeneous lamina is in the shape of the region enclosedby one loop of the lemniscate12: cos20. Find the radius of gyration of the lamina about an axis perpendicular to the polar plane at the pole.

1028


25. A lamina is in the shape of the region enclosedby the circle r: 4, and the area density varies as the diptance from the pole. Find the radius of gyration of the lamina about an axis perpendicular to the polar plane at the pole. 27. Evaluate by polar coordinates the double integral

[ [ ,**",4a

JnJ where R is the region bounded by the circlesI * y, :7

and f -l y\:9.

28. Evaluate by polar coordinates the double integral

[[-Laa l*l 1wt-77-' where R is the region in the first quadrant bounded by the circle rP * y":1 29. In advanced calculus, improper double integrals are discussed, fhfh

".ra

and the coordinate axes.

[.- f*' f(x, y) dx dv isdefined to be

Jo

Jo

lim | | f(x,y)dxdy h-+@ JoJo Use this definition to prove tnut f*- e-" ilx: JO

+\/; by doing the following: (a) Show that the double integral in Ex-

ample 5 can be expressed^l[

because|[*- ,-"'dt]': LJo J

r-"'a*]';b)' LJo I

5* |['.""drl',rr" o-+-LJo I

theresultof Example5

to obtain the desired result.

2I.5 AREA OF A SURFACE The double integral can be used to determine the area of the portion of the surface z: f(x, y) that lies over a closed region R in the ry plane. To show this we must first define what we mean by the measure of this area and then obtain a formula for computing it. We assumethat / and its first Q(f;,ri,f(ti,y)) partial derivatives are continuous on R and supposefurtherthatf (x,y) > 0 on R. Let A be a partition of R into n rectangular subregions. The ith rectangle has dimensions of measures Ag and {y and an area of measure Ay'. Let (€i, y) be any point in the ith rectangle, and at the point z: f(x,y) Q(ti, fi, fGo, y)) on the surface consider the tangent plane to the surface. Project vertically upward the ith rectangle onto the tangent plane and let {o be the measure of the area of this projection. Figure 21.5.1 shows the region R, the portion of the surface above R, the lth rectangular subregion of R, and the projection of the fth rectangle onto the tangent plane to the surface at Q. The number A;cris an approximation to the mea(ti,Yi) sure of the area of the piece of the surface that lies above the ith rectangle. F i g ur e 2 1. 5 . 1 Becausewe have z such pieces, the summation n

)

A,o

i:l

is an approximation to the measure a of the area of the portion of the surface that lies above R. This leads to defining o as follows: n

lim ' A;0 l l a l lo- e

21.5AREAOFA SURFACE1029 We now need to obtain a formula for computing the limit in Eq. (1). To do this we find a formula for computing Aio as the measure of the area of a parallelogram. For simplicity in computation we take the point (ily) in the lth rectangle at the comer (xi-r, yi-). Let A and B be vectors having as representations the directed line segments having initial points at Q and forming the two adjacent sides of the parallelogram whose area has measureA;c. SeeFig.21.5.2.Then A;o: lA x Bl. Because , A-A;ri*f"({r,y)A,6k and

n: LAi* fu(t,,y) Ailk it follows that

ikl

0 L'il

Figure 21.5.2

f"(t,,f i) A#l f"Gr,Tr) Lral

- Lix Lilf - - Aix LiUf T)i o(€r,Di * A,p LiAk "(tr, Therefore, Licr: lA x Bl :

(2)

Substituting from Eq. (2) into Eq. (1),we get cr -

lim llal;o - l:r

* f o'(tr, Yt)

Lfc Lil

This limit is a double integral which exists on R becauseof the continuity of f , and fu on R. We have, then, the following theorem. 21.5.1 Theorem

derivatives are continuous on the closed is the measure of the area of the sur-

(3)

EXAMPLEL: Find the area of the surface that is cut from the cylinder x2 + 22: L5 by the planes x-0,x-2,A:0,andy:3.

solurroN: The given surface is shown in Fig. 21..5.3.The region R is the rectangle in the first quadrant of the xy plane bounded by the lines r: 2 and y :3. The surfacehas the equation * * z2: 15. Solvingfor z, we get z: {G77. Hence, f(x, y): t/(-p. So if o is the measure of the area of the surface, we have from Theorem 21..5.L * fo'(x'

1 dxdy dx dy

1030

MULTIPLEINTEGRATION

-:

ff

+dxdy JoJotfio - xz

The surface area is therefore 2n square Figure 21.5.3

ExAMPLE 2:

Find the area of the

paraboloid z -- x' + y' below the plane z -- 4.

solurroN: Figure 2L.5.4 shows the given surface. From the equation of the paraboloid we see that f (r, y) : * * y2. The closed region in the xy plane bounded by the circle * * y': 4 is the region R. If o square units is the required surface area, we have from Theorem 21.5.L tf

cr: | | lf ,'(x, y) * fu'@,y) + | dx dy JRJ

:l

t!

I V4(**yr)+7dxdy

JNJ

*V,

Becausethe integrand contains the terms 4(f + y'), the evaluation of the double integral is simplified by using polar coordinates. Ttren I * y' : 12, and dx ily : dA: r dr d0. Furtherlnore, the limits for r are from 0 to 2 and the limits for 0 are from 0 to 2rr. We have then tf

| | \/4r2 l'1, r dr ilo JnJ

o:

:f

Figure21.5.4

f2n f2

| !4r2*\rdrd0

JO JO f2n f

: I

Jo

-12

l # ( + r r + t ) B t z ld 0 L

lo

:tr(tzt/a -D Hence,the areaof the paraboloid below the given plane is*zr(tZtE square units. EXAMPLE 3:

Find the area of the

top half of the sphere x' * y' *

z2

-- g2.

- t)

solurroN: The hemisphere is shown in Fig. 21..5.5.Solving the equation of the sphere for z and setting this equal to f(x, y), we get

y' f (x, y) : \F7= Because f,(x, y):-rl{7=7=7,

flrd fr(x, y):-y1lFF=!,

21.5AREAOF A SURFACE 1031 we n10te t e lthrat If"r o f lnd ,d fo are not defined on the circle x2 + y' : a2 which is y oof thr fl rdiary the the re I egic t. ;1on R in the xy plane. Furtherrnore, the double int orbtair m lEq, q. (3) is inee d ffrom I

w

a

d x tdy

which is improper because the integrand has an infinite discontinuity at each point of the boundary of R. We can take care of this situation by considering the region R' as that bounded by the circle rP I y' : b2,where b 1a, and then take the limit as b--->a-. Furthermore, the computation is simplified if the double integral is evaluated by an iterated integral using polar coordinates. Then we have

F i g u r e2 1 . 5 . 5

a :-rd|dr

l/a2-12'

:Zrra

!':-I

:2rnlim l-"rp)' b-a- L

:2na

F\ffi lt_T_

lo

* al

:2rra2 The area of the hemisphere is therefore 2na2 square units.

Consider now the curve y: F(r) with a < r < b, F positive onla,bl and F' continuous on fa, b]. If this curve is rotated about the r axis, we obtain a surfaceof revolution. From Sec.18.7an equation ofthis surfaceis y'+ z": [f(x)]'z

r

Figure 21.5.6

(4)

Figure 21.5.5shows the surface of revolution. In the figure we have taken the xy plane in the plane of the pape\ however, we still have a righthanded system. We wish to obtain a formula for finding the measure of the area of this surface of revolution by using Theorem 21..5.1From properties of symmetry, the measure of the area of the surface above the xz plane and in front of the xy plane is one-fourth of the measure of the area of the entire surface. Solving Eq. (4) for z (neglecting the negative square root because z = 0), we get f(x, y) : !E6JP=V. The region R in the xy plane is that bounded by the x axis, the curve y : F(x), and the lines x: a and r : b. Computing the partial derivatives of f , we obtain

f,(x,y):ffi

ful,y):a@fu

We see that f,(x, y) and fu@, y) do not exist on part of the boundary of

1032

MULTIPLEINTEGRATION

y : -F(r) (3) is

and when y : F(r) ). The double integral obtained

^

!ffi+ffi+ldYdx : t I F ( x ) \ / l F ' ( i P + rdy dx J^JW

This double integral is improper because the integrand has an infinite discontinuity at each point of the boundary of R where /: -F(r) and y: F(x). Hence, we evaluate the double integral by an iterated integral for which the inner integral is improper.

c,-^l:[t,',ffif"'ffi]dx

(s)

where

I:",ffi:1,_Tf",_,ffi : t,_T lrir,-'h]',(r)-e : t t _ T s i n - , (-16 ) :Lr Therefore, from Eq. (5) we have

q:2rr [' ,G){FTil]' Ja

* L itx

We state this result as a theorem, where F is replacedby f. 21.5.2 Theorem

ExAMPLE 4: Find the area of the paraboloid of revolution generated by revolving the top half of the parabola y' : 4px, with 0 < .r < h, about the x axis.

Suppose that the function / is positive on fa, b] and /' is continuous on la, bf .Then if o is the measure of the area of the surface of revolution obtainedbyrevolving the curvey: f(x),witha < x < b,abouttheraxis,

solurroN: The paraboloid of revolution is shown in Fig. 2l.S.T.Solving the equation of the parabola for y (y = 0), we obtain !-2prnyrn. So if o square units is the area of the surface,from Theorem2'1..5.2,with f(r) : 2prtzyrtz,we have 2pttzTtttz

dx

21.5 AREA OF A SURFACE

- AnPtt'

1033

t - p')

The area of the paraboloid of revolution is therefore &n(@ square units.

F i g ur e 2 1. 5 . 7

2L.5 Exercises 1. Find the area of the surfacewhich is cut from the plane 2x * y * z: 4 by the planes x: 0, x: t, y :0, and y : 1.. 2. Find the area of the portion of the surface of the plane 35x -l 1.6yl- 92:144 which is cut by the coordinate planes. 3. Find the area of the surface in the first octant which is cut from the cylinder f * y':

g by the plane r:

z.

the plane x*y:4. 4. Find the areaof the surfaceinthefirst octantwhich is cutfromthe conef *y':z2by 5. Find the area of the portion of the surface of the sphere f * y' * zz: 4x which is cut out by one nappe of the cone y2**:*. 6 . F i n d t h e a r e a o f t h e p o r t i o n o f t h e s u r f a c e o ft h e s p h e r e f + y 2 * 2 2 : 3 5 w h i c h l i e s w i t h i n t h e c y l i n d e r f + y z : 9 . 7. Find the area of the portion of the surfaceof the sphere x' + y' * z2: 4z which lies within the paraboloid f + !2 :32. 8. For the sphere and paraboloid of Exercise 7, find the area of the portion of the surface of the paraboloid which lies within the sphere. 9. The line segment from the origin to the point (a, b\ is revolved about the x axis. Find the area of the surface of the cone generated. 10. Derive the formula for the area of the surface of a sphere by revolving a semicircle about its diameter. 11. Find the area of the surface of revolution obtained by revolving the arc of the catenary y : a cosh(r/a) from r: x: a about the y axis.

0 to

12. Find the area of the surface of revolution obtained by revolving the catenary of Exercise 11 about the x axis. 13. The loop of the curve lSyz : x(6 - x)2 is revolved about the r axis. Find the area of the surface of revolution generated. 14. Find the area of the surfaceof revolution generated by revolving the arc of the curve A : ln x from r :1 lo x: the y axis.

2 about

15. Find the areaof theportionof theplaner: zwhichliesbetweentheplanes/:0and A: Sand,withinthehyperboloid 9f-4y2*1622:144. 15. Find the area of the surface cut from the hlryerbolic paraboloid y' - * : 6z by the rylinder f + yz : 35.

'103'f MULTIPLE INTEGRATION 21.5 THE TRIPLE INTEGRAL

The extension of the double integral to the triple integral is analogous to the extension of the single integral to the double integral. The simplest type of region in R3 is a rectangular parallelepiped which is bounded by six planes:x:

a 1 tx :

= fal z2, t A:br,

!:bz,

Z:

c1t and z:

cr, wlth a1 1 a2,

br 1br, and c1 ( cr. Let f be a function of three variables and suppose that / is continuous on such a region S. A partition of this region is formed by dividing S into rectangular boxes by drawing planes parallel to the coordinate planes. Denote such a partition by A and suppose that n is the number of boxes. Let AlV be the measure of the volume of the rth box. Choose an arbitrary point (fi, yt, p) in the ith box. Form the sum

>-f

(€uru Pt)LoV

(1)

Refer to Fig. 2'1,.6.L,which shows the rectangular parallelepiped together with the ith box. The norm llAll of the partition is the length of the longest diagonal of the boxes. The sums of the form (1) will upproach a limit as (ot,bt,ct)

the norm of the partition approaches zero for any choices of the points (€r, ^yr,lr) if f is continuous on S. Then we call this limit the triple integral of f on R and write n

lim Il a Il ' o i : l

x F i g ur e 2 1 . 6 . 1

EXAMPLE

Ai if(€,, Tt, 1-ci')) L,V

ft\ X,,Y , z ) d v T f(

-ite: Analogous rs; ttoo a d orubl urble in l l beiing equal to a twice :-) eratted jin rntegral, the tri Ple l int int egral i s e qluat thri ice-iterated integral. v tl Whhenr S i rs; lthe re rectangular pa arall elepilped d a rbove, and f is con'titinu e d (des scribed uouus u s o n SS, , we have

Evaluate the triple

SOLUTION:

integral sl^n yz

sin yz dV

dv:rff r'' *,sin yz dz dy dx ava* [-'cosv"]i'

if S is the rectangularparallelepiped bounded by the planes ,c-- zr,'U : tr , z -- *o, and the coordinate planes.

x(l - cos*nV) dy dx

:

o* f ,(r-*sin t*)lno''

: .(T-""sin T) * f

21.6 THE TRIPLEINTEGRAL

1035

x2

2

- 47r

z: Fz@'Y)

\

9 l

x,0

F i g ur e 2 1. 6 . 2

Y : 6z@)

We now discuss how to define the triple integral of a continuous function of three variables on a region in R3 other than a rectangular parallelepiped. Let S be the closed three-dimensional region which is and x:b, the cylindets y: dr(r) and bounded by the planes x:a : and the surfaces z F = r(x, y) and z : Fr(x, /), where the funcA 6r(x), tions @1,6r, Ft, and F2 are smooth (i.e., they have continuous derivatives or partial derivatives). See Fig. 21,.6.2.Construct planes parallel to the coordinate planes, thereby forming a set of rectangular parallelepipeds that completely cover S. The parallelepipeds which are entirely inside S or on the boundary of S form a partition A of S. Choose some system of numbering so that they are numbered from L to n. The norm llAllof this partition of S is the length of the longest diagonal of any parallelepiped belonging to the partition. Let the measureof the volume of the ith parallelepiped be {V. Let f be a function of three variables which is continuous on S and let (fi, Ti, F) be an arbitrary point in the ith parallelepiped. Form the sum

2 f rci,Tr,F) A,V

(2)

i:l

If the sums of form (2) have a limit as llAllapproacheszero, and if this limit is independent of the choice of the partitioning planes and the choices ofthe arbihary points ((u yu pi) in each parallelepiped, then this limit is called the triple integral of / on S, and we write (3)

It can be proved in advanced calculus that a sufficient condition for the limit in (3) to exist is that / be continuous on S. Furthermore, under the condition imposed upon the functions 6r, 6r, Fr, and F2 that they be smooth, it can also be proved that the triple integral can be evaluated by the iterated integral

H,::I:::

f(x, y, z) dz dy dx

just as the double integral can be interpreted as the measure of the area of a plane region when /(r, U): L on Rr, the triple integral can be interpreted as the measure of the volume of a three-dimensional region. If

r1036

MULTIPLEINTEGRATION

f (x, y , z) :, 1 on S, then Eq. (3) becomes lim llall-o and the triple integral is the measure of the volume of the region S.

EXAMPTN 2:

Find by triple inte-

If V cubic units is the volume of the solid, then

solurroN,

gration the volume of the solid of Example 3 in Sec.2L.2.

lim llall-o

where S is the region bounded by the solid. The z limits are from 0 (the value of z on the ry plane) to f * 4yz (the value of z on the elliptic paraboloid). Th" y limits for one-fourth of the volume are from 0 (the value of y on the xzplane) tot:y'T=P (the value of y on the cylinder). The rlimits for the first octant are from 0 to 2. We evaluate the triple integral in (4) by an iterated integral and obtain

Hence,

(x' * 4y') dy dx

V:4

(s)

The right side of Eq. (5) is the same twice-iterated integral that we obtained in Example 3 in Sec. 2'1..2,and so the remainder of the solution is the same.

E1AMpLE3: Find the volume of the solid bounded by the cylinder x2+ y' :25, the plane x + y + z - 8 and the xy plane.

solurroN: The solid is shown in Fig. 21,.6.3.The z limits for the iterated integral are from 0 to 8 - x- y (the value of z on the plane). The y limits are obtained from the boundary region in the xy plane which is the circle * + y':25. Hence, the y limits are from -\/25=7 to t/x=E The r limits are from -5 to 5. If V cubic units is the required volume, we have

v- l rim a,u:Jfffl * l a l l - offi

:f,ml:-"-"dzdvdx fs

f\ffi

:

- -'*-,, (8 x Y)dYdx J-rJ

:

l rE"e - x)y- tr,)_*ud* J_,L(r fb

f

21.6 THE TRIPLEINTEGRAL

(8- itffi

:16 f

J-s

:

1037

dx

\E4dx*

f

J-s

\84(_2x)

dx

+ z*sin-r Ex) + |eS - xzr',r]:,

L6(ix\84

:200n The volume is therefore 200n cubic units.

F i g u r e2 1 . 6 . 3

EXAMPLE4: Find the mass of the solid above the xY Plane bounded by the cone 9xz * z2: Y' and the Plane Y : 9 if the measure of the volume densitY at any point (x, y , z) in the solid is proportional to the measure of the distance of the Point from the xy plane.

((i,yi,

solurroN: Figure 2']-..5.4shows the solid. Let M slugs be the mass of the solid, and let the distance be measured in feet. Then the volume density at any point (r, y, z) in the solid is kz slagslfts,where k is a constant. Then if (tr, yt, p1) is any point in the ith rectangular parallelepiped of the partition, we have

M-

lim Zkp

l l a l l - oi : 1

fs fals h@ifr

:ZkJ,J,

J,

zdzdxdY

pi)

: z ek [ n v ' d y - T k Jo -

The mass is therefore ryk slugs.

Figure 21.6.4

1038

MULTIPLEINTEGRATION

Exercises 21,.6 In Exercises1 through 4,evaluate the iterated integral. -

1'

f

fF*f+uz

JrJ, Jr"

r x dzdYdx

a 2'

12fu,frnr

dzdxdY JrJ"J, Ye'

"or!

dy dx dz

4 fff'#dxdzdy

In Exercises 5 through 10, evaluate the triple integral.

5.

rrf JJJ

y dV if S is the region bounded by the tetrahedron formed by the plane t2x + 20y * l5z:

50 and the coordinate

oiur,ur.

6. @z+22) dvirsis the Iil

same region as in Exercise 5.

ltl 7. | | | z dV if S is the region bounded by the tetrahedronhaving vertices (0, 0, 0), (7,1, O), (1, 0, 0), and (1, 0, 1). JJJ

i,, i,,

8. | | | -yz dV if S is the same region as in Exercise7. JJJ 9. llt JJJ

( n + 3 2 ) d V i t S i s t h e r e g i o nb o u n d e db y t h e c y l i n d e rf l z 2 : 9

a n d t h e p l a n e sx * y : 3 ,

z:0,

and y:0,

s above the ry plane.

rft 10. lf f xyzi[V, if S istheregionboundedbythecylinders JJJ

tr, Uir.ire,

**y":4

and f Iz2:4.

11 through 21.,use triple integration.

11. Find the volume of the solid in the first octant bounded below by the xy plane, above by the plane z: y, andlaterally by the rylinder y': r and.the plane x: 1.. 12. Find the volume of the solid in the first octant bounded by the rylinder f + z2: 16, the plane r * y:2, coordinate planes. 13. Find the volume of the solid in the first octant bounded by the cylinders * * yr: ordinate planes. 14. Find the volume of the solid bounded by the elliptic cone 4* I9y" - 3622:0

and the three

4 and x2 + 2z: 4, and.the three co-

and the plane z : 1.

15. Find the volume of the solid above the elliptic paraboloid 3* * y2: z and below the cylinder f * z:4. 15. Find the volume of the solid enclosed by the sphere * * y, * 22: a2. 17. Find the volume of the solid enclosed by the ellipsoid *u222 _l__: _ +L a2 b2 c2

1

18. Find the mass of the solid enclosedby the tetrahedron formed by the plane 100r + 25y * 762: 400and the coordinate planes if the volume density varies as the distance from the yz plane. The volume density is measured in slugs/ftr. L9, Find the mass of the solid bounded by the rylinders x: z2 andA: *, and the planesx:l,y:0, andz:0. The volume density varies as the product of the distances from the three coordinate planes, and it is measured in slugslfd.

COORDINATES 1039 AND SPHERICAL IN CYLINDRICAL 21.7IHE TRIPLEINTEGRAL 20. Find the mass of the solid bounded by the surface z: 4 4xz !2 and the xy plane. The volume density at any point of the solid is p slugs/ff and p:3"111' 2L. Find the mass of the solid bounded by the sutf.acez - xy, and the planes r: L, y 1,,and z 0. The volume density y2. p: 3V* + at any point of the solid is p slugs/ff and

2I.7 THE TRIPTE INTEGRAL If a region S in R3has an axis of symmetry, triple integrals on S are easier IN CYLINDRICAL AND to evaluate if cylindrical coordinates are used. If there is symmetry with SPHERICAL COORDINATES respect to a point, it is often convenient to choose that point as the origin and to use spherical coordinates. In this section we discuss the triple integral in these coordinates and apply them to physical problems. To define the triple integral in cylindrical coordinates we construct ( , , , 0 ,, , , ) a partition of the region s by drawing planes through the z axis, planes perpendicular to the z axis, and right-circular cylinders having the z axis ls axis. A typical subregion is shown in Fig. 2L.7.1.The elements of the constructed partition lie entirely in S. We call this partition a cylindrical partition. The measure of the length of the longest "diagonal" of any of the subregions is t:nenorm of the partition. Let n be the number of subI 1 I regions of the partition and A1v be the measure of the volume of the ith 0 zn subregion. The measure of the area of the base is li L1rA,60,whete 7i: t?r * ri-). Hence, if.A,Ais the measureof the altitude of the lth subregion, x F i g ur e 2 1 . 7. 1

L1V-

L4r Aî0 Liz

Let f b" a function of r, 0, and z, and suppose that / is continuous on S. Choosea point (Vr,0r, Zi) in the i th subregion such that 0r-r, = 0i = 0i and zr-t < Zi s zr Form the sum n i:l

f Gr, 0r, 2r) LrV

n

f (7r, 0i, 2r)7, A^1rLi? Aiz

(1)

i:l

As the norm of A approaches zeto, it can be shown, under suitable conditions on S, that the limit of the sums of form (L) exist. This limit is called the triple integralin cylindricalcoordinatesof the function f on s, and we write (2)

(3)

Note that in cylindrical coordinates, dV - r dr d0 dz. We can evaluate the triple integral in (2) and (3) by an iterated integral. For instance, : suppose that the region S in R3 is bounded by the planes 0 a and

1O4O

MULTIPLEINTEGRATION

0:B, with a
I sil ,r,, 0,z)rdrd0dz: ff::[:::

r(r,0,z)ritzdrit.

(4)

There are five other iterated integrals that can be used to evaluate the triple integral in (4) because there are six possible permutations of the three variables t, 0, and z. - -Triple integrals and cylindrical coordinates are especially useful in finding the moment of inertia of a solid with respect to ihe z uxis because the distance from the z axis to a point in the solid is determined by the coordinate r. ExAMPLE 1: A homogeneous solid in the shape of a rightcircular cylinder has a radius of 2 ft and an altitude of 4 ft. Find the moment of inertia of the solid with respect to its axis.

solurroN; choose the coordinate planes so that the xy plane is the plane of the base of the solid and the z axis is the axis of the-solid. Fig:uu;e)l.z.z shows the portion of the solid in the first octant together with the ith subregion of a rylindrical partition. using cylindrical coordinates and taking the point (Fo,0r, z1) in the ith subregion with k slugs/fp as the volume density at any point, then if I, slug-ff is the tt o*urrt of inertia of the solid with respect to the z axis, we hlve

I':rr1t, a,v:III krz dv 2rru There are six different possible orders of integration. Figure 21.7.2shows the orderdz dr d0.Using this order, we have

or,itzritrdo:4k L: Jff[ f''f ['*itzdrdo JJ Jo JoJo

,s 11 the first integration,the blocks are summed from z:0toz:4;the blocks become a column. In the second integration, the columns are summed from r: 0 to r:2; the columns become a wedge-shapedslice of the cylinder. In the third integration, the wedge-shaped-sliceis rotated from d : 0 to 0: Ini this sweepsthe wedge about the entire three-dimensional region in the first octant. we multiply by 4 to obtain the entire volume. Performing the integration, we obtain Figure21.7.2

t": tilk ["'' [' * a, ao:54k f'' ar: 32kr Jo Jo Jo

Hence, the moment of inertia is 32kn slug-ff.

COORDINATES 1041 AND SPHEBICAL IN CYLINDRICAL 21.7THETRIPLEINTEGRAL ExAMPLE 2: Solve Example L by taking the order of integration as (a) dr dz d0; (b) d0 dr dz.

soLUrIoN: (a) Figure 21.7.3 reptesents the otder drdzd0' It shows the block summed from r: 0 to r: 2 to give a wedge-shaped sector' We then sum from z : 0 to z: 4 to give a wedge-shaped slice. The slice is rotated from 0 : 0 to 0 = *n to cover the first octant. We have, then,

t,: nnf'' ffi: r drdzdo:32kn (b) Figure 2']...7.4represents the oider d|ilrilz. It shows the blocks summed from 0 : 0 to 0: tr to give a hollow ring inside the cylinder. These hollow rings are summed from r: 0 to r: 2 to give a horizontal slice of the cylinder. The horizontal slices are summed trom z: 0 to z: 4. Therefore, we have f d0 dr dz: 32kn

Figurc 21.7.3

ExAMPLE3: Find the mass of a solid hemisphere of radius a ft if the volume densitY at any Point is proportional to the distance of the point from the axis of the solid.

Figure 21.7.4

soLUrIoN: If we choose the coordinate planes so that the origin is at the center of the sphere and the z axis is the axis of the solid, then an equation of the hemispherical surface above the xy plane is z:{72777Figure 21.7.5 shows this surface and the solid together wit! the ith subt"liot of a cylindrical partition. An equation of the hemisphere in rylin(ti, ei, zl) is a point in the ith subdrical coordinates is z: \/F7.lf region, the volume density at this point is kr6 slugs/ft3, where k is a constant; and if M slugs is the mass of the solid, then

M:

lim $ n, a,Y

l l a l l - of i

1O:42

MULTIPLEINTEGRATION

: lil krdv -k : o (vi,oi,o) Figure21.7.5

r2dzdrdo

I:l:13

: O

fzn fa

J oJ ,

,21ffi'

drdo

fzn I

- rz)Btz* tazrlprz Jo l-Lnr(az f2n

-#kaan I

* taosin

d0

JO

- +kAan' The mass of the solid hemisphere is thereforetkann2 slugs. ExAMPLE 4: Find the center of mass of the solid of Example 3.

solurroN: Let the cartesian-coordinate representation of the center of mass be (i , , Z). Becauseof symmetry , i: !: 0. We need to calculate " Z.,rt M,u slug-ft is the moment of mass of the solid with respect to the ry plane, we have M,u:

lig, i

ilail_oGi ffr

zt(ktt) Atv

: lll kzrilV JJJ s

:kl

fzT fa f\/a'=Fz

, I Jo JoJo fzn fa

: tk I

|

Jo Jo

zrzilzdril|

(a2- r2)r2ilr do

- J3ka5l2n de I Jo

: frkaizr BecauseM2:

Mrn, we get

#ka,rr .; -_TMri : W_: G o

L6

The center of mass is therefore on the axis of the solid.at a distance of 'l,6al75zr ft from the plane of the base.

We now Proceed to define the triple integral in spherical coordinates.

ANDSPHERICAL COORDINATES1043 IN CYLINDRICAL 21.7THETRIPLEINTEGRAL A spherical partition of the three-dimensional region S is formed by planes containing the z axis, spheres with centers at the origin, and circular cones having vertices at the origin and the z axis as the axis. A tyPical subregion of the partition is shown in Fig. 27.7.6.If A1Vis the measure of the volume of the l'th subregion, and (Fo,0r, dr) is a point in it, we can get an approximation to A1Vby considering the region as if it were a rectangular parallelepiped and taking the product of the measures of the three dimensions. These measures are p; sin 6, Lol, Fi LrQ,and Ap. Figures 2'1,.7.7and 21..7.8show how the first two measuresare obtained, and Figure 21.7.6 shows the dimension of measure dp. Hence, LtV: ,'z sin {; Aip A$ A1Q

Vi Li6

p; sin$,

(i,,0,,6,)

\---\.. --.. \ t \ \

pi sin$i

o

F i g ur e 2 1. 7. 6

\\

F i g ur e 2 1. 7. 8

The triple integral in sphericalcoordinatesof a function / on S is given by (5)

Observe that in spherical coordinates,N: p2 sin $ dp d0 dS. The triple integrals in (5) or (6) can be evaluated by an iterated integral. Spherical coordinates are especially useful in some problems involving spheres, as illustrated in the following example.

1044

MULTIPLEINTEGRATION

ExAMPTE5: Find the mass of the solid hemisphere of Example 3 if the volume density at any point is proportional to the distance of the point from the center of the base.

solurroN: rf (po, at, 6) is a point in the ith *bt"gior of u spherical partition, the volume density at this point is kpl slugslfts, where k is a constant. lf M slugs is the mass of the solid, then

M-

lim

l l a l l - o Zkp, i:-r

L,V

p3 sin 6 dp dO d0 sin O d0 d0

- $aakn Hence, the mass of the solid hemisphere is taakzr slugs.

It- is- interesting to compare the solution of Example 5 which uses spherical coordinates with what is entailed when using^cartesiancoordinates. By the latter method, a partition of s is formed by dividing s into rectangular boxes by drawing planes parallel to the coordinate planes. If (!u Ti, g.) is any point in the lth subregion, and becausep : {frV + i, then

M: ,trs, 9rt/€7 + yFTE n,v lltll-o "?=t

: [[ [ rtF+j JJJ

+ zzav

,s

: no fa f\/F-zz fla=uz-F JoJ" J,

l.F+T+z dxdydz

The computation involved in evaluating this integral is obviously much more complicated than that using spherical coordinates. EXAMPLE6: A homogeneous solid is bounded above by the sphere p: a and below by the cone Q,- a, where 0

AND SPHERICALCOORDINATES 1045 21.7 THE TRIPLEINTEGRALIN CYLINDRICAL

Find the moment of inertia of the solid about the z axis.

if I, slug-ff is the moment of inertia of the given solid about the z axis, then (pr sin 6)'k L,V

;

kp'sin2 + dV

- k

f

fl:

(p2sin'Q)p2 sinQdpdodO

fa fzn

-+ka' ll sin30d0d0 Jo Jo fa

- ?ka'n sinsOdO Jo l.1" - ?ka\zr l-cos O + * coss0 |

F i g ur e 2 1. 7. 9

lo

L

-+kasn(cos3 a-

3 c o sa * 2 )

The moment of inertia of the solid about the z axis is therefore -3cos a*2) #ka'n(cos3 o

slug-ff

21.,7 Exercises In Exercises L throu gh 4,evaluate the iterated integral.

t t.

lit4 fa frcosl

J; JJ. lnl4

lzacoso fzt

f:^ I:'""'[""

^

J_ dr )- d0 )a o dz t secs

o' "in S dedpdg

'' n'

ltl4 fzcos9 ['"'"t

Jo J"",^,Jo lnt2 16

J,^J),

facsco

,,

cos 0 dz dr d0

pssin20 sin @dpd0dS

: using (a) cylindrical coordinates and (b) spheri5. Find the volume of the solid enclosedby the sphere x' * y' * z2 a2by cal coordinates. : 16 and the coordinate planes, evaluate the triple 6. If s is the solid in the first octant bounded by the sphere f * y" I z2 coordinates; (b) using rectangular coordinates; (c) using *"dv by thtee methods: (a) using spherical III s cylindrical coordinates.

integral

In Exercises7 through 10, use cylindrical coordinates' density varies as the square of the distance 7. Find the mass of the solid bounded by a sphere of radius fl ft if the volume in slugs/ff' from the center. The volume density is measured : y' -f z2:76' g. Find the mass of the solid in the first octant inside the cylinder * l- y' 4x and un-der the sphere f t measured in slugs/ft3' The volume density varies as the distance from the xy plane, and it is

1046


9. Find the moment of inertia with respect.to the z axis of the homogeneous solid bounded by the rylinder r: cone z: r, and the ry plane. The volume density at any point is k slugs/fts.

5, the

10. Find the moment of inerlia of the solid bounded by a right-circular cylinder of altitude h It and,radius a ft, with respect to the axis of the cylinder. The volume density varies as the distance from the axis of the rylinder, and it is measured in slugs/ff. , In Exercises11 through L4, use spherical coordinates. 11' Find the center of mass of the solid bounded by the hemisphere of Example 5. The volume density is the same as that in Example 5. 12. Find the moment of inertia with respect to the.z axis of the homogeneous solid bounded by the sphere f * y, + z2: 4. The volume density at any point is k slugs/fts 13' Find the moment of inertia with respect to the z axis of the homogeneous solid inside the cylinder f I yt - 2x: 0, below the cone t2 * y":22, and above the xy plane. The volume d-nsity at any point is k slugs/ff. 14' Find the mass of a spherical solid of radius n It if the volume density at each point is proportional to the distance of the point from the center of the sphere. The volume density is measured in slugs/ff. In Exercises15 through 18, use the coordinate system that you decide is best for the problem. 15. Find the center of mass of the solid inside the paraboloid f * yr: z artd.outside the cone * * y": volume density is k slugs/fts.

22.The constant

16. Find the moment of inertia with respect to the z axis of the homogeneous solid of Exercise 15. 17' Find the moment of inertia about a diameter of the solid between two concentric spheres having trdii a ft and,2a ft. The volume density varies inversely as the square of the distance from the center, and it i, *"urore-d in slugs/ft3. 18' Find the mass of the solid of Exercise 17. The volume density is the same as that in Exercise12. In Exercises19 through 22, evaluatethe iterated integral by using either cylindrical or spherical coordinates.

ReaiewExercises (Chapter21) In Exercises1 through 8, evaluate the given iterated integral.

r. ['l* *a ava,

, IJZ_xy itzc dy

3.

n.

u. dxitydz I:f"f" eceyez

6.

,

f I"*"^'r, sin0dr ttl

f'f,:Ip3

sin{ cos e dpded0 t f f'["*

zre-ndrd0dz

In Exercises9 through 12, evaluate the multiple integral. t

LI

ru dA;R is the region in the first quadrant bounded by the circle 12* yr:

L and the coordinate axes.

REVIEW EXERCISES 1UI7 tf 10. | | (x + y) dA; R isthe region bounded by the curve y : cos r and the r axis from x:-in

to x:

JRJ ttl

11. f f f z2dV; S is the region boundedbythe cylindersx2*z:l

in.

and the ry plane.

andy2*z:1,

JJJ

t

jffycos(x t z ) d V ; S i s t h e r e g i o n b o u n d e d b y t h e r y l ixn:dAe' ,r a n d t h e p l a n exsI z : i z r , V : 0 , a n d z : 0 . JJJ "

rS. f:raluateby polar coordinatesthe doubleintegral tt

1

JrJ7i v'dA where R is the region in the first quadrant bounded by the two circles f * y': 14. Evaluate by polar coordinates the iterated integral f [*

1 and x2 + yz : 4.

h(rz * yz) dx dy.

JoJ1l}r

In Exercises15 and L6, evaluate the iterated integral by reversing the order of integration. 15

I:Isinyz

dydx

L5.

ft

fcos tu

JrJ,

rsinrdxdy

In Exercises17 and 18, evaluate the iterated integral by changing to either cylindrical or spherical coordinates. f2 6m

17.

tFI#dzdvdx

1m

18' J,J. J,

z@

dzdvdx

19. Use double integraticin to find the area of the region in the first quadrant bounded by the parabolas*: 8 - 4y by two methods: (a) Integrate first with respect to x; (b) integrate first with respect to y.

4y and I :

20. Use double integration to find the volume of the solid above the xy plane bounded by the cylinder f -t y': the plane z:2y by two methods: (a) Integrate first with respect to r; (b) integrate first with resPectto y. 21,. Find the volume of the solid bounded by the surfacesx2:4A, !2:4x,

and.*:

16 and

z- !.

22. Find the mass of the lamina in the shape of the region bounded by the parabolaA: x2 and the line x- y + 2:0 the area density at any point is fy2 slugslfP. 23. Find the area of the surfaceof the cylinder f * yz :9lying

in the first octant and between the planes x:

if

z and3x: z.

24. Find the area of the surface of the part of the cylinder * t y' : a2 that lies inside the cylinder yz * zz : az. 25. Usedoubleintegrationtofindtheareaoftheregioninsidethecircler:landtotherightoftheparabolar(1 *cos0):1. 26. Find the mass of the lamina in the shape of the region exterior to the limagon r: 3 - cos d and interior to the circle r: 5 cos 0 if the area density at any point is 2lsin 0l slugs/ff. 27. Find the center of mass of the rectangular lamina bounded by the lines r : 3 and A : 2 and the coordinate axesif the area density at any point is xyz slugslfP. 28. Find the centerof massof thelaminaintheshape of theregionboundedbytheparabolasf :4*4A if the area density at any point is kf slugslfP.

andl:4-8y

29. Find the mass of the lamina in the shape of the region bounded by the polar axis and the curve / : cos 20, where 0 < 0 < *zr. The area density at any point is r0 slugs/ft2. 30. Find the moment of inertia about the r axis of the lamina in the shape of the region bounded by the circle r3 * y': slugs/fP. if the area density at any point is klFl!

a'

10.18

MULTIPLEINTEGRATION

31. Find the moment of inertia about the r axis of the Lamina in the shape of the region bounded by the curve y : line r: 2, and,the coordinate axes if the area density at any point is xy slugslff .

ea,the

32. Find the moment of inertia of the lamina of Exercise31 about the y axis. 33.Find the moment of inertia with respect to the *z nxis of the homogeneouslamina in the shape of the region bounded by the curye 12:4 cos 2d if the area density at any point is k slugi/ff. Find the mass of the lamina of Exercise33.

Find the polar moment of inertia and the corresponding radius of gyration of the lamina of Exercise 33. Find the moment of inertia about the y axis of the lamina in the shape of the region bounded by the paraboluy : , - * and the line r * I :0 rt the area density at any point is (r * y) slugs/fp. Find the mass of the solid borrnded by the spheres f * y, * ?r:4 point is kll*W shgs/fF.

and.f * y, * zz:9

if.the volume density at any

Find the moment of inertia about the z axis of the solid of Exercise 32.

The homogeneous solid bounded by the conez2:4* * 4y2 between the planes z: 0 and z: 4hasa volume density at any point of k slugs/ff. Find the moment of inertia about the z axis forthis solid. Find the center of mass of the solid bounded by th9 sphele * * y, * z2- 6z:0 and the cone * * yr: zz, and above the cone, if the volume density at any point is *z slugi/ft

App endlx

Table7 n L 2 3 4 5 5 ,7 8 9 10 L1 12 L3 l4 L5 L5 17 18 t9 20 21 22 23 24 25 25 27 28 29 30 31 32 33 34 35 35 37 38 39 40 4L 42 43 44 45 46 47 48 49 50

r

Powersand roots n2

\E

nB

w

1 L.000 1 1.000 g 1.260 4 1.414 9' 1.732 27 1.442 L6 2.000 64 1.587 25 2.236 I25 L.7L0 35 2.449 216 1.817 49 2.546 343 1.9L3 64 2.929 5L2 2.000 81 3.000 729 2.080 100 3.',1.62 L,000 2.154 LzL 3.317 L,33L 2.224 '1,,729 144 3.464 2.289 L69 3.606 2,L97 2.35L 196 3.742 2,744 2.4L0 225 3.973 3,375 2.466 255 4.000 4,095 2.520 289 4.t23 4,913 2.571 324 4.243 5,932 2.621 351, 4.359 5,959 2.568 400 4.472 9,000 2.714 441 4.593 9,25L 2.759 484 4.690 L0,649 2.802 529 4.796 12,L57 2.844 576 4.ggg 13,924 2.884 525 5.000 15,625 2.924 676 5.099 t7 ,576 2.962 729 5.L96 Lg,6g3 3.000 784 5.291 2L,952 3.037 841 5.395 24,389 3.072 900 5.477 27,000 3.107 96L '/',,024 5.559 29,791 3.141 5.657 32,768 3.175 1,,089 5.745 35,937 3.208 L,L55 5.931 39,304 3.240 "1,,2255.9"1,6 42,975 3.27r 1,296 6.000 46,655 3,302 1,369 5.093 50,553 3.332 L,444 6.154 54,972 3.362 L,52L 6.245 5g,3Lg 3.391 L,600 6.325 64,000 3.420 l,6g'j, 6.403 6g,92l 3.448 L,764 6.481 74,099 3.476 L,849 5.557 79,507 3.503 "1,,9355.533 g5,lg4 3.530 2,025 6.708 91,L25 3.557 2,115 5.782 97,336 3.583 2,209 6.855 103,823 3.609 2,304 6.928 Lt},592 3.634 2,401, 7.000 tL7,549 3.659 2,50Q 7.07L L25,000 3.584

n

5L 52 53 54 55 56 57 5g 59 60 61' 62 63 64 55 66 67 5g 69 70 7r 72 73 74 75 76 77 78 79 80 81 82 83 84 85 85 87 88 89 90 9L 92 93 94 95 95 97 98 99 100

n2

\E

n3

%

2,60L 7.14L L32,65'1.3.709 '1,40,509 2,704 7.21r 3.732 2,809 7.280 148,977 3.755 2,9'1,6 7.348 t57,464 3.790 3,025 7.4L6 L66,375 3.903 3,',1,36 7.483 175,61,63.926 3,249 7.550 185,L93 3.949 3,354 7.5L6 Lgs,Ll2 3.971 3,48L 7.68"1, 205,379 3.993 3,500 7.746 21,6,000 3.91,5 3,72L 7.810 226,99'j. 3.936 3,844 7.874 238,329 3,959 3,969 7.937 250,047 3.979 4,095 8.000 262,"1.444.000 4,225 8.062 274,625 4.021 4,356 8.L24 297,496 4.041 4,489 8.1.85 300,763 4.052 4,624 8.245 3'1,4,4324.092 4,761 8.307 329,509 4.1.02 4,900 8.357 343,000 4.12L 5,04/., 8.425 357,9'1,"1, 4.L4L 5,',1,84 8.485 373,249 4.160 5,329 8.544 389,0L7 4.179 5,475 8.602 405,224 4.lgg 5,525 9.550 421,875 4.217 5,776 g.7Ig 439,976 4.236 5,929 8.775 456,533 4.254 5,094 9.932 474,552 4.273 6,241 g.ggg 493,039 4.29'1, 6,400 9.944 5L2,000 4.309 6,55L 9.000 53'1,,441.4.327 6,724 9.055 551,358 4.344 6 ,ggg g.1L0 57L,797 4.352 7,055 9.165 592,704 4.390 7,225 9.220 61.4,L25 4.397 7,396 9.274 636,055 4.4L4 7,569 9.327 658,503 4.43I 7,744 g.3gt 68'1,,4724.449 7,921 9.434 704,969 4.455 9,100 9.497 729,000 4.491 g,28L 9.539 753,57'1,4.499 778,688 4.5L4 8,464 9.592 8,,549 9.543 904,357 4.53L 8,835 9.695 830,584 4.547 9 ,025 9.747 .857,375 4.563 9 ,2L6 9.798 884,735 4.579 9 ,4W 9.849 912,673 4.595 9 ,604 9.899 94L,L92 4.6L0 9,801 9.950 970,299 4.626 L0,000 L0.000 L,000,000 4.542

Table 2 N

Natural logarithms

0

I

2

3 0296 L222 2070 2852 3577

L.0 1.L 1.2 1.3 '1,.4

0000 0953 L823 2624 336s

0100 L044 1906 2704 3436

0198 1133 1989 2776 3507

1.5

L.8 1.9

4055 4700 5306 5878 6419

4L2L 4752 5365 s933 647L

4t87 4253 4824 4886 5423 5481 5988 6043 6523 6575

2.0 2.I 2.2 2.3 2.4

693L 74L9 7885 8329 8755

698L 7467 7930 8372 8796

7031, 75L4 7975 841,6 8838

r.6 t.7

9L63 2.5 955s 2.6 2.7 9933 2 . 8 r.0296 2.9 0647

7080 756L 8020 8459 8879

4

5

5

7

8

I

0392 0488 0583 0577 1310 L398 L484 L570 2I5I 223L 23rL 2390 2927 3001 3075 31,48 3546 37L6 3784 38s3

0770 L555 2469 3221, 3920

0862 1740 2546 3293 3988

45LL 5L28 5710 6259 6780

4574 5188 5756 6313 5831

4637 5247 5822 6366 5881

7227 7275 7324 770L 7747 7793 8L54 8198 8242 8587 8629 8571, 9002 9042 9083

7372 7839 8286 8713 9L23

43L8 4383 4447 4947 5008 5058 5539 s596 5653 6098 6L52 6206 6627 6678 5729 7L29 7508 8065 8502 8920

7L78 7655 8L09 8544 8961,

9203 9243 9282 9322 936L 9400 9594 9632 9570 9708 9746 9783 996e *0006 *0043 *0080 *0LL6 *0152 0332 0367 0403 0438 0473 0508 0682 07L5 0750 0784 0818 0852

9439 9478 9517 9821 9858 9895 *018g *0225 *0260 0543 0886

0578 09L9

0613 0953

3 . 0 1..0985 L019 L053 3.1. L3L4 1346 L378 3.2 1632 L663 L594 3.3 1939 1969 2000 3.4 2238 2257 2296

L085 L4r0 1725 2030 2326

LLL9 L442 1756 2060 2355

1151. L474 1787 2090 2384

rL84 L505 1,817 2Lr9 24L3

L2L7 1537 1848 21,49 2442

L249 L569 1878 2179 2470

L282 1500 DAg 2208 2499

3 . 5 L.2528 2556 3.6 2809 2837 3.7 3083 3110 3.8 3350 3376 3.9 3510 363s

2613 2892 3164 3429 3686

254L 2920 3191 3455 3712

2569 2947 32L8 3481 3737

2598 2975 3244 3507 3762

2726 3002 3271, 3533 3788

2754 3029 3297 3558 38L3

2782 3056 3324 3584 3838

4 . 0 L.3863 3888 39L3 3938 4.L 4LL0 4L34 4L59 4183 4.2 4351, 4375 4398 4422 4.3 4586 4609 4633 4656 4.4 48L6 4839 486L 4884

3962 4207 4446 4679 4907

3987 423L 4469 4702 4929

40L2 4036 4255 4279 4493 4516 4725 4748 495L 4974

51.51 5369 5581 5790 5994

5173 5390 5602 5810 601,4

6194 6390 5582 6771 6956

62L4 6233 6409 6429 5601 5620 6790 5808 6974 6993

2585 2865 3137 3403 356L

4 . 5 1.504L 5063 4.6 5261 5282 4.7 5476 5497 4.8 5686 5707 4.9 5892 59L3

5085 5L07 5L29 5304 5326 5347 5518 5539 5550 5728 5748 5769 5933 5953 5974

5 . 0 L.5094 6LL4 5.L 6292 63L2 5.2 5487 5506 5.3 6677 6696 5.4 6864 5882

6134 5L54 6332 635L 5525 5544 6715 6734 5901 69L9

6L74 537r 5563 6752 6938

4061 4085 4303 4327 4540 4563 4770 4793 4996 5019

5L95 5217 54L2 5433 5623 5644 5831 585L 6034 6054 6253 6448 5539 6827 70Lr

5239 5454 5565 5872 6074 6273 6467 6658 5845 7029

Table 2

(Continued)

0123456789 5 . 5 1.7047 7056 5,6 7228 7245 5.7 7405 7422 5.8 7579 7596 5.9 7750 7766 6 . 0 "l..7gLg 7934 6.1 6.2 6.3 6.4

8083 8245 8405 8563

6 . 5 l.g7lg 6.6 8871 5.7 902L 5.8 9t69 6.9 9315

8099 8252 8421 8s79

7084 7263 7440 76L3 7783

7102 7281 7457 7630 7800

7120 7299 7475 7647 7817

7L38 7317 7492 7664 7834

7155 7334 7509 7591, 7851

7174 7352 7527 7699 7857

795L 7967 7984 8Lt6 8L32 8148 8278 8294 8310 8437 8453 8469 8594 8610 8625

8001 8165 8326 8485 8641,

80L7 8181 8342 8500 9656

8034 80s0 8066 8197 82L3 8229 8358 8374 8390 8516 8532 8547 9672 8587 8703

9081 9228 9373

8795 8946 9095 9242 9387

8810 896L 9110 9257 9402

8825 8976 9125 9272 94L5

9733 8749 8886 8901 9036 9051 9L84 9t99 9330 9344

8764 991,5 9065 9213 9359

8779 gg31

7L92 7370 7544 77L6 7884

7210 7387 7561 7733 7901,

8840 8856 8991 9006 9140 9155 9285 9301 9430 9445

7 . 0 L.9459 9473 7.L 9601 9615 7.2 9741, 9755 7.3 9879 9892 7 . 4 2.0015 0029 7 . 5 2.0L49 0152 7.6 0281 0295 7.7 04t2 0425 7.9 0541 0554 7.9 0669 0681

9488 9502 9629 9643 9769 9782 9905 9920 0042 - 0055

9516 9657 9796 9933 0069

9530 9671 9810 9947 0082

0175 0308 0438 0567 0594

0189 0321 0451 0580 0707

0202 0334 0464 0592 0719

02L5 0347 0477 0605 0732

8 . 0 2.0794 8.L 09L9 9.2 1041 8.3 1,163 8.4 1282

0819 0943 1056 1,187 1305

0832 0844 0955 0958 L078 1090 llgg 121,1 1318 1330

0857 0980 L102 1223 1342

9544 9559 9573 9587 9685 9699 9713 9727 9824 9838 9851 9865 995L 9974 9988 *000L 0096 0109 0122 0136 0229 0242 0255 0268 0360 0373 0386 0399 0490 0503 0516 0528 0518 0530 0543 0655 0744 0757 0759 0782 0859 0882 0894 0905 0992 1005 1017 L029 tLL4 L126 L138 1L50 1235 1,247 1258 1,270 1353 1365 1377 1389

0907 0931 1054 1,175 L294

8 . 5 2.140L t4l2 8.5 1518 1529 8.7 L633 1645 8.9 L748 1759 8.9 1861 1872 9 . 0 2.L972 1gg3 9.'/', 2083 2094 9.2 2L92 2203 9.3 2300 23tl 9.4 2407 24t8

1424 L54L L656 L770 1883

1435 L552 L668 L782 L894

L448 1554 1679 1793 L905

L459 Ls76 L69L 1804 t9t7

L47L 1483 1587 1599 1702 1713 18L5 L827 1928 1939

1494 L61,0 1725 1838 1950

1505 1,622 1736 1,949 t96L

1994 2105 22L4 2322 2428

2005 21,1,5 2225 2332 2439

20L7 2127 2235 2343 2450

2028 2138 2246 2354 2460

2039 2148 2257 2364 2471,

20s0 2159 2268 2375 24gl

2061 2170 2279 2386 2492

2072 2191, 2289 2395 2502

9 . 5 2.2513 9.6 26L8 9.7 272L 9.9 2824 2925

2534 2638 2742 2844 2946

2544 2649 2752 2854 2956

2555 2659 2762 286s 2955

2565 2670 2773 2875 2976

2576 2680 2783 2885 2986

2586 2690 2793 2895 2996

2597 270L 2803 2905 3005

2607 27Ll 28L4 29L5 3015

2523 2628 2732 2834 2935

use ln 10 :2.30259 to find logarithms of numbers greater than 10 or less than l. Example: ln 220: ln 2.2 * 2 ln 10 : 0.7885+ 2(2.90259): 5.3937.

Table3

Expoientialfunctions et

0.00 0.0L 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1.0 0.11 0.L2 0.L3 0.14 0.15 0.L5 0.L7 0.L8 0.1g 0.20 0.2r 0.22 0.23 0.24 0.25 0.25 0.27 0.28 0.29 0.30 0.31 0.32 0.33 0.34 0.35 0.35 0.37 0.38 0.39 0.40 0.41 0.42 0.43 0.44 0.45 0.45 0.47 0.48 0.49 0.50

1.0000 L.0L01. L.0202 1.030s 1.0408 1.0513 1.0618 L.0725 1.0833 r.0942 L.1052 1.1163 L.L275 1.1388 1.1503 '1,.1618

logro(e")

0.00000 .00434 .00859 .01303 .01737 0.0217"t .02606 .03040 .03474 .03909 0.04343 .04777 .05212 .05646 .06090 0.06574 L.L735 .05949 1 . 1 8 5 3 .07383 1.1972 .07817 L.2092 .08252 L.22r4 0.08685 1.2337 .09120 L.246L .09554 L.2586 .09989 r.2712 .1,0423 1.2840 0.10857 L.2959 .L1292 L.3100 .LL726 7.323r .12160 1,.3364 .12595 1.3499 0.13029 l.3634 .L3453 L.377L .1,3897 1 . 3 9 1 0 .L4332 r.4049 .L4766 '1,.4191. 0.15200 1.4333 .15635 r.4477 .16069 1.4623 .15503 L.4770 .L6937 L.49L8 0.17372 L.5058 .17806 r.5220 .L8240 1,.5373 .18675 1,.5527 .19109 1,.5583 0.19543 1.5841 .19978 1.6000 .20412 L"6L6L .20846 L.6323 .21290 L.6487 0.2L7L5

e-"

1.000000 0.990050 .gg}Lgg .970446 .950789 0.951229 .94L765 .932394 .9231.1.6 .gL3g3L 0.904837 .895834 .886920 .978095 .g6935g 0.860708 .852744 .843655 .835270 .826959 0.818731 .810584 .802519 .794534 .786;,528 0.778801, .77L052 .763379 .755784 .748264 0.74081 8 .733447 .7261,49 .7L8924 .71,L770 0.704688 .597576 .690734 .683851, .677057 0.670320 .663650 ,557047 .550509 .544036 0.637628 .631.294 .625002 .618783 .6L2626 0.605531

e'

0.50 0.51 0.52 0.53 0.54 0.55 0.56 0.57 0.58 0.59 0.60 0.5'1. 0.62 0.53 0.64 0.65 0.55 0.57 0.58 0.59 0.70 0.71, 0.72 0.73 0.74 0.75 0.76 0.77 0.78 0.79 0.80 0.81 0.82 0.83 0.84 0.85 0.86 0.87 0.88 0.89 0.90 0.91 0.92 0.93 0.94 0.95 0.96 0.97 0.98 0.9e 1.00

L.6487 1.6653 r.6820 1,.5999 1,.7160 1.7333 L.7507 L.7693 'j..7960

logto(e")

0.2L7L5 .22149 .22583 .230L8 .23452 0.23886 .24320 .24755 .25't89 1.9040 .35623 L.8221 0.26058 1,.8404 .26492 1.8589 .26926 L.8776 .2736r 1.8965 .27795 1.9155 0.28229 L.g34g .28653 L.9542 .29098 1.9739 .29532 L.9937 .29966 2.0138 0.30401 2.0340 .30835 2.0544 .3L269 2.075L .37703 2.0959 .321,38 2.1170 0.32572 2.L383 .33006 2.L5gg .3344L 2.19"1,5 .33875 2.2034 .34309 2.2255 0.34744 2.2479 .35178 2.2705 .35612 2.2933 .36046 2.31,54 .36481, 2.3395 0.359L5 2.3632 .37349 2.3869 .37784 2.4'1,09 .38218 2.435L .38652 2.4596 0.39087 2.4843 .39521, 2.5093 .39955 2.5345 .40389 2.5600 .40824 2.5957 0.4L258 2.61,17 .41,692 2.6379 .42127 2.6645 .4256L 2.6912 .42995 2.7L83 0.43429

e-"

0.506531 .600496 .594521, .588505 .582748 0.576950 .57L209 .565525 .559899 .554327 0.548812 .543351. .537944 .532592 .527292 0.522046 .51685L .51L709 .506617 .50L576 0.495585 .491644 .486752 .481909 .4771,1,4 0.472367 .467666 .463013 .458406 .453845 0.449329 .444858 .440432 .436049 .43171,1 0.4274L5 .4231,62 .418952 .414783 .4L0656 0.405570 .402524 .39851,9 .394ss4 .390528 0.38674L .382893 .979083 .3753LL .371,577 0.367879

Table3 (Continued) x

1.00 L.0L 1.02 1..03 L.04 1.05 1.05 L.07 L.08 1.0g 1.10 1.1L L.lz L.L3 t.t4 1.15 l.l5 L.L7 L.L8 l.l9 1.20 t.2l 1.22 "1,.23 1.24 1.25 1.26 1.27 1.28 '/.,.29

e'

2.7t93 2.7456 2.7732 2.90LL 2.9292 2.9577 2.9964 2.9154 2.9447 2.9743 3.0042 3.0344 3.0649 3.0957 3.L269 3.1592 3.Lg99 3.2220 3.2544 3.2971 3.3201 3.3535 3.3972 3.4212 3.4556 3.4903 3.5254 3.5609 3.5966 3.6329 1.30 3.6693 L.3L 3.7062 1..32 3.7434 L.33 3.7810 1.34 3.81.90 1.35 3.9574 t.35 3.9952 1.37 3.9354 1.38 3.9749 L.39 4.0149 1.40 4.0552 '/.,.4'1, 4.0960 1.42 4.137L "1..43 4.L797 1.44 4.2207 4.2531 L.45 L.46 4.3060 '/.,.47 4.3492 l.4g 4.3929 r.49 4.437t 1.50 4.4817

logro(e")

0.43429 .43864 .44298 .44732 .45L57 0.4560L .46035 .46470 .45904 .47338 0.47772 .48207 .4864L .49075 .495L0 0.49944 .50378 .508L2 .51247 .5158t 0.52115 .52550 .52984 .53418 .53853 0.54287 .54721 .55L55 .55590 .56024 0.56459 .55993 .57327 .5776t .58L95 0.58630 .59054 .59498 .59933 .60367 0.60801 .61236 .6L570 .62L04 .62539 0.62973 .63407 .6384r .64276 .647t0 0.651,44

e-"

0.367879 .3542L9 .350595 .357007 .353455 0.349938 .346456 .343009 .339595 .3362L5 0.332871 .329559 .326290 .323033 .3r9819 0.3t6637 .3r3486 .3L0367 .307279 .30422L 0.30L194 .298197 .295230 .292293 .289384 0.286505 .283554 .280832 .278037 .275271 0.272532 .269920 .257135 .26M77 .25L946 0.259240 .25566L .254L07 .251579 .249075 0.246s97 .244'1,43 .2417'1,4 .239309 .236929 0.234570 .232236 .229925 .227639 .225373 0.2231,30

x

e'

logre(e")

e-"

1.50 4.48L7 0.65L44 0.2231.30 L.5L 4.5257 .65578 .2209L0 '1..52 .660L3 4.5722 .2lg7r2 L.53 4.6192 .66447 .215535 L.54 4.6646 .66881 .21439l 1.55 4.7L15 0.673L5 0.212249 L.56 4.7588 .67750 .210135 L.57 4.9066 .58194 .209045 L.58 4.9550 .6g6tg .205975 1.59 4.9037 .69053 .203926 1.60 4.9530 0.59487 0.20L997 L.5L 5.0029 .5992'1. .lggggg 1,.62 5.0531 .70356 .197899I L.63 s.1039 .70790 .19s930 I L.64 5.1552 .7L224 .1g3gg0 1.65 5.2070 0.71659 0.192050 1.56 5.2593 .72093 .1901.39 1.67 5.3122 .72527 .lgg247 1.68 5.3555 .72961 .196374 L.69 5.4195 .73395 .194520 1.70 5.4739 0.73930 0.1,92594 L.7l 5.5290 .74264 .190955 1.72 5.5945 .74699 .L79066 "1,.73 5.6407 .75L33 .177294 1.74 5.5973 .75567 .175520 1.75 5.7545 0.75002 0.173774 '/.,.76 5.8124 .76435 .172045 r . 7 7 5.8709 .76870 .L70333 L . 7 8 5.9299 .77304 .1.68639 "1,.79 5.9895 .77739 .156960 1.80 6.0495 0.79173 0.165299 1.81 6.Lr04 .79607 .153554 '1,.82 5.1719 .79042 .'1,62026 L.83 5.2339 .79476 .L604L4 1.84 6.2965 .7ggl0 .159917 1.85 6.3599 0.80344 0.157237 L.85 6.4237 .90779 .L55673 1.87 6.4493 .81213 .154L24 1.88 6.5535 .91647 .152590 1.8g 6.6194 .82082 .r5L072 1.90 6.6959 0.92515 0.L49569 '1,.9"1, 5.7531 .92950 .1490g0 L.92 6.92L0 .93395 .L46507 L.93 6.8895 .83819 .L45L4g 1.94 6.95gg .84253 .143704 1.95 7.0297 0.84697 0.142274 L.96 .85t22 .14095g '1..97 7.0993 7.1707 .85555 .139457 L.98 7.2427 .85990 .139069 t.9g 7.3155 .86425 .L35595 2.00 7.399'j, 0.86959 0.135335

(Continued) e'

2.00 2.01 2.02 2.03 2.04 2.05 2.05 2.07 2.08 2.09 2.lo 2.lt 2.12 2.13 2.1,4 2.15 2.16 2.L7 2.18 2.L9 2.20 2.21 2.22 2.23 2.24 2.25 2.26 2,27 2.28 2.29 2.30 2.31 2.32 2.33 2.34 2.35 2.35 2.37 2.38 2.39 2.40 2.4L 2.42 2.43 2.44 2.45 2.46 2.47 2.48 2.49 2.50

log1s(e"')

e-"

x

7.3ggL 0.96959 0.135335 2.50 .1,33989 2.51 7.4533 .97293 .132655 2.52 7.5393 .87727 7.614'1, .88152 .131336 2.53 .88595 .L30029 2.54 7.6905 7.7679 0.89030 0.L28735 2.55 .89465 .L27454 2.56 7.8450 .89899 7.9249 .L26L86 2.57 .124930 2.58 .90333 9.0045 g.0g4g .123587 2.59 .90755 9.t662 0.91202 0.122455 2.60 .t2L238 2.6L .9L636 9.2492 9 . 3 3 1 1 .92070 .L20032 2.62 g.4l4g 2.63 .92505 .Lt8837 g.4gg4 .92939 .1,17655 2.64 8.5849 0.93373 0.1,16494 2.65 .1.L5325 2.66 8.571'1, .93808 2.67 .tL4l78 .94242 8.7583 .'1,13042 2.68 .94675 8.8463 2.69 .lllgl7 .95LL0 8.9352 2.70 0. L 10803 9.0250 0.95545 .10970L 2.71 .95979 9.1157 .L08509 2.72 .96413 9.2073 .107528 2.73 .95848 9.2999 .106459 2.74 .97282 9.3933 9.4877 0.9771,6 0.L05399 2.75 . L04350 2.76 .98L5L 9.583L .L033L2 2.77 .98585 9.6794 .L02284 2.78 .990L9 9.7757 ."1,01266 2.79 .99453 9.8749 L00259 2.80 0. 0.99888 9.9742 .09926L 2.8L L0.074 1.00322 10.175 1.00756 .098274 2.82 .097296 2.83 1.01191 10.278 10.381 1,.0L625 .096328 2.84 "1.0.485 L.02059 0.095369 2.85 .094420 2.85 L0.59L L.02493 L0.597 1,.02928 .09348'1, 2.87 .09255L 2.88 L0.805 I.03362 L0.91.3 L.03796 .09L530 2.89 1,L.023 L.0423L 0.090718 2.90 L.04665 .089815 2.9L Lt.L34 .088922 2.92 1.05099 L1.246 .088037 2.93 11.359 1.05534 .08716t 2.94 1.05968 11.473 11.588 1.06402 0.086294 2.95 .085435 2.96 L.06836 11.705 .084585 2.97 L.07271 1L.822 LL.94L L.07705 .083743 2.98 L2.06L 1.0g139 .0829L0 2.99 L.09574 0.082085 3.00 L2.r82

e'

t2.L82 L2.305 12.429 12.554 12.580 t2.807 12.936 13.056 13.197 L3.330 L3.464 13.599 L3.736 '1,3.874 L4.0L3 L4.I54 14.296 L4.440 14.585 14.732 14.880 L5.029 L5.L80 L5.333 L5.487 15.643 15.800 L5.959 r6.tL9 '1,6.281 16.445 16.510 '1,6.777 16.945 17.116 17.288 17.462 17.637 17.81.4 17.993 L8.r74 L8.357 L8.541 18.728 1g.9L5 L9.r06 19.298 19.492 19.688 19.886 20.085

logrs(e")

g- r

t.09574 0.082085 1.09009 .081269 L.09442 .080460 1.09877 .079659 1.L03LL .078966 t.t0745 0.078082 L.LLlTg .077305 1.1,161,4 .076536 1,.1,2049 .075774 L.12482 .075020 ',1,.12917 0.074274 1,.L335L .073535 L.13785 .072803 l.l421g .072078 1.1,4654 .071351 1.15088 0.070651 .069948 ],.15522 .069252 L.L5957 'l.,."1.6391 .058553 .067881, L.L6825 1.17260 0.067206 1,.17594 .056537 1.18128 .065875 .0552L9 r.L8562 .064570 L.18997 1,.L9431 0.063928 L.19865 .053292 L.20300 .052652 .a62039 L.20734 '1,.2L168 .051,421, 1.21602 0.050810 '1,.22037 .060205 1.22471 .059606 '1.22905 .059013 .05u26 1.23340 1.23774 0.057u4 r.24208 .057259 1.24543 .056699 r.25077 .055L35 L.255LL .055575 L.25945 0.0s5023 1,.26380 .054476 1,.2681,4 .053934 .053397 r.27249 1.27693 .052866 L.28Lr7 0.052340 1.2955L .051819 1.2898s .05L303 1,.29420 .050793 1.29854 .050287 L.30288 0.049787

Table 3 x

3.00 3.01. 3.02 3.03 3.04 3.05 3.05 3.07 3.08 3.09 3.10 3.L1. 3.r2 3.L3 3.I4 3.15 3.15 3.L7 3.L8 3.79 3.20 3.21 3.22 3.23 3.24

3.25 3.26 3.27 3.28 3.29 3.30 3.31 3.32 3.33 3.34 3.35 3.36 3.37 3.38 3.39 3.40 3.41 3.42 3.43 3.44 3.45 3.46 3.47 3.48 ,3.49 3.50

(Continued) e'

20.085 20.287 20.491 20.597 20.905 21.'/',15 21.329 21.542 21.759 21,.977 22.Lgg 22.421 22.645 22.974 23.L04 23.336 23.571 23.907 24.047 24.299 24.533 24.779 25.029 25.290 25.534 25,790 26.05A 26.3L1 26.576 26.843 27.113 27.3g5 27.660 27.939 2g.2lg 29.503 2g.7gg 29.079 29.371 29.655 29.964 30.265 30.559 30.977 3l.lg7 31.500 3'l,.gl7 32.137 32.450 32.796 33.L15

logro(e")

e-"

L.30288 0.049787 L.30723 .049292 1.3Lr57 .04880L .048316 r.3L59L .047835 L.32026 1.32460 0.047359 1.32894 .046988 1.33328 .046421 1.33763 .045959 1,.34197 .045502 r.34631 0.045049 1.35066 .04460L 1..35500 .044157 t.35934 .043719 L.36368 .043293 L.36803 0.042952 1.37237 .042426 L.35571 .042004 L.38L06 .0415g5 1.38540 .041172 1.38974 0.040764 l.3g40g .040357 "1,.39843 .039955 1.40277 .039557 l.407ll .039164 l.41146 0.038774 1..4L580 .038388 1.42014 .039005 1.42449 .037629 '/.,.42883 L.44317 L.4375L l.44lg6 1.44520 L.45054 1.45489 1.45923 1.45357 '1,.46792 1.47225 1.47650 1.48094 1.48529 1.48963 L.49397 1,A9932 1.50256 L.50700 1.5L134 l.51569 1.52003

x

3.50 3.5L 3.52 3.53 3.54 3.55 3.56 3.57 3.59 3.59 3.60 3.6L 3.62 3.63 3.64 3.55 3.56 3.67 3.58 3.69 3.70 3.71 3.72 3.73 3.74 3.75 3.75 3.77 3.78 .037254 3.79 0.036993 3.90 .036516 3.91 .035L53 3.92 .035793 3.93 .035437 3.94 0.035094 3.95 .034735 3.96 .034390 3.97 .034047 3.gg .033709 3.gg 0.033373 3.90 .033041 3.gt .0327L2 3.92 .032397 3.93 .032065 3.94 0.031746 3.95 .031430 3.96 .031. Ll7 3.97 .030907 3.gg .030501 3.gg 0.030197 4.00

e'

33.1.L5 33.448 33.784 34.r24 34.467 34.8L3 35.1.63 35.517 35.874 35.234 35.599 36.966 37.339 37.713 39.092 38.475 38.861 39.252 39.646 40.045 40.447 40.854 41.264 41.679 42.098 42.521 42.948 43.380 43.81,5 M.256 44.70L 45.L50 45.604 45.063 46.525 46.993 47.465 47.942 49.424 4g.gt1, 49.402 4g.ggg 50.400 50.907 5t.4tg 51.935 52.457 52.995 53.517 54.055 54.599

logt6(e")

1.52003 L.52437 L.52872 1.53305 1.53740 1,.54175 7.54609 L.55043 L.554n 1.559L2 1,.55346 L.56780 t.572t5 L.57649 1.58083 L.58517 '/...58952 1.59385 1.59820 t.60255 1,.50699 '1..6LL23

e-"

0.030L97 .029897 .029599 .029305 .0290t3 0.028725 .028439 .028L56 .027876 .027598 0.027324 .027052 .026793 .026516 .026252 0.025991 .025733 .025476 .025223 .024972 0.024724 .024479 .024234 .023993 .023754

1.5155g t.6lgg2 L.52426 1.62960 0.02351g 1,.63295 .023294 '/.,.63729 .023052 '/.,.64L53

.022923

1.54598 .422595 1.55032 0.022371 1,.65456 .022L49 1.55900 'j,.56335 .02lg2g .021710 1.65769 .021494 r.57203 0.0212g0 r.67639 .021069 r.68072 .020858 1.68505 .020651. 1.68941 .020445 1.69375 0.020242 t.69g0g .020041 1.70243 .019940 1.70679 .0tg6$ |.7LLLT .0lg44g 1.71546 0.019255 l.7lg8.l, '/.,.724L5 ,019063 '1,.72849 .01.9973 .019696 1.73293 .01g500 '/,,.73718 0.01.9316

Table3 (Continued) x

4.00 4.01 4.02 4.03 4.04 4.05 4.05 4.07 4.08 4.09 4.10 4.t1 4.12 4.13 4.L4 4.15 4.16 4.17 4.18 4.19 4.20 4.21 4.22 4.23 4.24 4,25 4.26 4.27 4.28 4.29 4.30 4.3'1, 4.32 4.33 4.U 4.35 4.36 4.37 4.38 4.39 4.40 4.41 4.42 4.43 4.44 4.45 4.46 4.47 4.48 4.49 4.50

e,

54.598 55.147 55.701 56.261 56,825 57.397 57.974 58.577 59.145 59.740 60.340 60.947 6L.559 62.178 52.803 63.434 54.072 64.715 55.366 56.023 66.686 67.357 58.033 58.717 69.408 70.1.05 70.810 7t.522 72.240 72.956 73.700 74.440 75.tgg 75.944 76.709 77.479 79.257 79.044 7g.g3g 90.540 91.451 92.269 93.095 93.931 94.775 95.627 g6.4gg 97.357 99.235 99.12'j, 90.017

logro(e")

e-"

x

1.73718 0.018315 4.50 1.74152 .0L8133 4.51 L.74596 .0L7953 4.52 L.75021 4.53 .0t7774 "1..75455 .017597 4.54 t.7sggg 0.017422 4.55 1.76324 .017249 4.56 1.75758 .0L7077 4.57 l.77rg2 .016907 4.58 t.77625 .0L6739 4.59 L.78061 0.0L6573 4.60 1.78495 .01,5408 4.61 '1,.78929 .016245 4.52 1.79354 .0L5083 4.63 t.79798 .015923 4.64 1.80232 0.0L5764 4.55 t.80667 .01.5608 4.66 1.81.101 .015452 4.67 4.68 1.8L535 .0t5299 1.81969 .0L5L45 4.69 1.82404 0.014995 4.70 1.82838 .014846 4.7L 1.83272 .0"1,4699 4.72 1.83707 .01.4552 4.73 '1,.84141 .014408 4.74 1.84575 0.014264 4.75 1.85009 .014122 4.76 1.95444 .01.3982 4.77 1.g5g7g .013843 4.78 1,.8531.2 .013705 4.79 t.85747 0.0L3569 4.80 l.87l8l .013434 4.81 l.g76L5 .013300 4.82 1.9g050 .013158 4.83 1.884g4 .01.3037 4.84 L.gg91g 0.012907 4.95 l.gg352 .0L2778 4.86 L.gg7g7 .0L265L 4.87 l.g022l .012525 4.88 1.90555 .0L240L 4.89 l.glOga 0.412277 4.90 l.gl524 .0L2L55 4 . 9 L 1.g1g5g .012034 4.92 L.g23g2 .0119'/.,4 4.93 l.g2g27 4.94 .0LI795 l.g326L 0.01L679 4.95 l.93695 .011552 4.96 l.g4l30 .011447 4.97 L.94564 .01L333 4.98 l.949gg .0"1,"1,2214.99 L.95433 0.01.1109 5.00

ea

logrs(e")

e-"

"1,.954330.01.110g L.95867 .0L099g 1.95301 .010999 L.96735 .010781 1.97L70 .0L0673 1.97604 0.0L0567 1.98038 .010462 L.98473 .0L0358 1.98907 .0L0255 L.99341 .010L53 L.99775 0.0L0052 2.00270 .009952 2.00544 .009853 2.01078 .009755 2.01513 .009658 2.01947 0.009562 2.0239L .009466 L06.70 2.02916 .009372 L07.77 2.03250 .009279 L08.85 2.03684 .009L87 109.95 2.04'1,18 0.009095 L11.05 2.04553 .009005 '1,'1,2.17 2.04987 .008915 LL3.30 2.05421 .008826 1"1,4.43 2.05956 .008739 2.05290 0.008652 115.58 2.06724 LL6.75 .008556 2.07L58 .008480 tl7.g2 119.10 2.07593 .008395 1.20.30 2.08027 .0083L2 2.08461 0.008230 !21.5L 2.08896 122.73 .008L48 2.09330 123.97 .008057 125.21 2.09764 .007987 2.t0199 L26.47 .007907 2.10633 0.007928 127.74 L29.02 2.LL067 .007750 130.32 2.71,501, .007673 1,31,.63 2.71936 .007597 2.L2370 t32.95 .007521 2.12804 0.007477 134.29 2.13239 L35.64 .007372 "t37.00 2.13673 .007299 L38.38 2.L4L07 .007227 L39.77 2.L4541 .007L55 2.1,4976 0.007093 L4l.L7 1,42.59 2.t54L0 .007013 144.03 2.'1,5844 .006943 L45.47 2.16279 .006974 146.94 2.16713 .006905 148.41 2.L7147 0.006739 90.017 90.922 gL.g36 92.759 93.691, 94.632 95.593 96.544 97.5'/.,4 98.494 99.484 L00.48 10L.49 102.51 103.54 104.58 '1,05.64

Table 3

(Continued) e*

'/.,49.41 5.00 5.01 L49.90 5.02 L5t.4L 5.03 L52.93 5.04 154.47 5.05 L55.02 5.05 157.59 5.07 159.17 5.08 L60.77 '1,52.39 5.09 '1,64.02 5.10 5.11 L65.67 5.L2 L67.34 5.13 159.02 5.L4 L70.72 5.L5 172.43 5.L5 174.1,6 5.L7 L75.91 5.1.8 L77.68 5.L9 t79.47 5.20 L81.27 5.21 193.09 5.22 Lg4.g3 5.23 Lg6.7g 5.24 L88.57 5.25 190.57 5.26 Lg2.4g 5.27 194.42 5.28 196.37 5.29 rgg.34 5.30 200.34 5.3L 202.35 5.32 204.39 5.33 205.44 5.34 209.5L 5.35 2L0.61 5.35 212.72 5.37 2L4.86 5.38 217.02 5.39 2L9.20 5.40 22L.4L 5.41 223.63 5.42 225.99 5.43 229.15 5.44 230.44 5.45 232.75 5.46 235.L0 5.47 237.45 5.48 239.95 5.49 242.26 5.50 2M.59

logls(e')

2.17147 2.L7582 2.18015 2.18450 2.18884 2.L9319 2.19753 2.20L87 2.20522 2.2L056 2.2L490 2.2L924 2.22359 2.22793 2.23227 2.23562 2.24095 2.24530 2.24955 2.25399 2.25833 2.26267 2.26702 2.27L36 2.27570 2.28005 2.28439 2.28873 2.29307 2.29742 2.30L76 2.30610 2.31045 2.31479 2.31913 2.32348 2.32782 2.33215 2.33550 2.34085 2.345L9 2.34953 2.35388 2.35822 2.35256 2.36690 2.37L25 2.37559 2.37993 2.38428 2.38862

e'

e-r

0.006738 .006671 .005605 .005539 .005474 0.005409 .005346 .006282 .005220 .006159 0.006097 .005035 .005976 .0059L7 .005959 0.005799 .005742 .005695 .005529 .005572 0.005s17 .005462 .005407 .005354 .005300 0.005249 .005195 .005L44 .005092 .005042 0.004992 .004942 .004993 .004944 .004796 0.004749 .00470r .004554 .004509 .004562 0.004517 .004472 .004427 .004393 .004339 0.004295 .004254 .0042L1 .004L69 .0041,29 0.004087

5.50 5.55 5.50 5.55 5.70 5.75 5.80 5.85 5.90 5.95 6.00 6.05 5.10 5.15 6.20 5.25 6.30 6.35 6.40 5.45 6.50 6.55 6.60 6.55 6.70 5.75 6.80 5.85 6.90 6.95 7.00 7.05 7.10 7.L5 7.20 7.25 7.30 7.35 7.40 7.45 7.50 7.55 7.60 7.55 7.70 7.75 7.80 7.85 7.90 7.95 8.00

244.69 257.24 270.43 284.29 298.87 3L4.Ig 330.30 347.23 365.04 393.75 403.43 424.L1 445.96 469.72 492.75 519.01 544.57 572.49 601.85 532.70 665.14 699.24 735.10 772.79 8L2.4L 954.05 gg7.g5 g43.gg 992.27 1043.1 L096.6 1,L52.9 12L2.0 L274.L 1339.4 1409.1 14g0.3 L556.2 L636.0 L719.9 1909.0 1900.7 lggg.2 2L00.6 2209.3 232L.5 2440.6 2565.7 2597.3 2935.5 2ggL.0

log16(e")

2.39962 2.41033 2.43205 2.45375 2.47548 2.4g7Lg 2.51,99L 2.54052 2.56234 2.59405 2.60577 2.62749 2.64920 2.67091 2.59263 2.71434 2.73506 2.75777 2.77948 2.80120 2.92291, 2.94463 2.96634 2.ggg06 2.90977 2.g3l4g 2.95320 2.97492 2.99653 3.01835 3.04006 3.06L78 3.08349 3.',1,0521 3.12692 3.1.4963 3.17035 3.L9206 3.2L378 3.23549 3.2572L 3.27992 3.30064 3.32235 3.34407 3.36579 3.38750 3.40921 3.43093 3.45264 3.47436

e-t

0.0040868 .0038875 .0035979 .0035175 .0033460 0.0031828 .0030276 .0028799 .0027394 .0026058 0.0024788 .0023579 .0022429 .0021335 .0020294 0.0019305 .0018353 .0017467 .00L56L5 .001.5805 0.0015034 .0014301 .0013604 .0012940 .0012309 0.001L709 .0011138 .0010595 .0010078 .0009s85 0.0009119 .0008674 .000825L .0007849 .0007465 0.0007L02 .00057s5 .0006426 L3 .00061 .0005814 0.0005531 .0005261 .000s005 .0004760 .0004528 0.0004307 .0004097 .0003898 .0003707 .0003527 0.0003355

Table 3

(Continued) e'

8.00 9.05 8.L0 9.1.5 8.20 9.25 9.30 8.35 9.40 8.45 8.50 9.55 8.50 8.55 9.70 9.75 8.80 8.85 9.90 8.95 9.00

298L.0 3L33.8 3294.5 3453.4 354L.0 3827.5 4023.9 4230.2 4447.1 4675.1 49L4.8 5L56.8 543L.7 57L0.0 5002,9 5310.7 6634.2 6974.4 7332.0 7707.9 8L03.1

logrc(e")

3.47436 3.49607 3.51779 3.53950 3.55121 3.58293 3.60464 3.62636 3.64807 3.56979 3.69L50 3.7L322 3.73493 3.75565 3.77836 3.80008 3.82179 3.8435L 3.86522 3.88694 3.90865

e-"

x

0.0003355 9.00 .0003L9L 9.05 .0003035 9.10 .0002887 9.15 .0002747 9.20 0.0002513 9.25 .0002495 9.30 .0002364 9.35 .0002249 9.40 .0002L39 9.45 0.0002036 9.50 .0001935 9.55 .0001.841. 9.60 .000L751. 9.55 .0001.656 9.70 0.00015g5 9.75 .000L507 9.80 .0001434 9.85 .000L364 9.90 .0001297 9.95 0.000L234 10.00

e*

8L03.L 85L8.5 8955.3 94L4.4 9897.1 10405 10938 L1499 12088 L2708 13360 14045 t4765 15522 L6318 L7154 18034 18958 19930 20952 22026

logro(e")

3.90855 3.93037 3.95208 3.97379 3.9955L 4.01722 4.03894 4.06065 4.08237 4.L0408 4.12580 4.t475I 4.t6923 4.L9094 4.2L255 4.23437 4.25609 4.27780 4.29952 4.32L23 4.34294

e-c

0.0001234 .0001174 .0001r17 .0001062 .000L010 0.000095L .0000914 .0000870 .0000827 .0000787 0.0000749 .00007t2 .0000577 .0000544 .00005L3 0.0000583 .0000555 .0000527 .0000502 0.0000477 0,0000454

Table4 )c

0 0.1. 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.1 1,.2 L.3 L.4 1..5 '/-,.6 L.7 L.8 t.9 2.0 2.1 2.2 2.3 2.4 2.5 2.5 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4.0 4.L 4.2 4,3 4.4 4.5 4.5 4.7 4.8 4.9 5.0

Hyperbolicfunctions sinh r .00000 .100L7 .20t34 .30452 .4L075 .52Lr0 .63565 .75858 L .888L '/.,.0265 'j.,,1752 L.3356 1.5095 L.6984 7.9043 2.1293 2.3755 2.6456 2.9422 3.2682 3.6269 4.02L9 4.457r 4.9370 5.4652 5.0502 6.6947 7.4053 8.19L9 9.0596 10.018 rL.076 L2.246 13.s38 t4.965 16.543 L8.285 20.zLL 22.339 24.691 27.290 30.L62 33.335 36.843 40.719 45.003 49.737 54.959 60.75t 67.1,41 74.203

cosh r 1.0000 1.0050 1.020L 1.0453 1.0811 1.1276 1.1855 1,.2552 1.3374 "1,.433"1, 1.5431 1.6685 'l.,.8107 1,.9709 2.L509 2.3524 2.5775 2.8283 3.1075 3.4L77 3.7622 4.1,443 4.5579 5.0372 5.5559 6.1323 6.7590 7.4735 8.2527 9.L146 10.058 Ll.l22 12.287 1,3.575 t4.999 16.573 L8.3L3 20.236 22.362 24.7LL 27.308 30.178 33.351 36.857 40.732 45.0L4 49.747 54.978 60.759 57.149 74.210

tanh r .00000 .09967 .r9738 .29L31 .37995 .462L2 .5370s .50437 .66404 .7L630 .76159 .80050 .83365 .86172 .88535 .905L5 .92L67 .93541 .94681, .95624 .96403 .97045 .97574 .980L0 .98367 .9865'1. .98903 .99"1,0'/., .99253 .99396 .99505 .99595 .99568 .99728 .99777 .99818 .9985L .99878 .99900 .999L8 .99933 .99945 .99955 .99963 .99970 .99975 .99980 .99983 .99986 .99989 .99991

Table5

Trigonometric functions

Degrees Radians

Sin

Cos

Tan

Cot

0 L 2 3 4

0.0000 0.0L75 4.0349 0.0524 0.0598

0.0000 0.0L75 0.0349 0.0523 0.0598

1.0000 0.9998 0.9994 0.9985 0.9975

0.0000 0.0175 0.0349 0.0524 0.0699

57.290 28.535 1.9.081 14.30L

5 6 7 8 9

0.0873 4.t047 0.L222 0.1395 0.L57'/',

0.0872 0.L045 0.t219 0.1392 0.1554

0.9962 0.9945 0.9925 0.9903 0.9877

0.0875 0.1051 0.L228 0.1405 0.1584

10 11 12 13 L4

0.t745 0.1920 0.2094 0.2269 0.2443

0.1735 0.1908 0.2079 0.2250 0.24L9

0.9848 0.981.6 0.978L 0.9744 0.9703

15 16 t7 18 L9

0.26L8 0.2793 0.2967 0.31.42 0.3315

0.2588 0.2756 0.2924 0.3090 0.3255

0.9659 0.9613 0.9563 0.9511 0.9455

20 21, 22 23 24

0.349L 0.3565 0.3840 0.40L4 0.4lgg

25 26 27 28 29

1.5708 L.5533 1.5359 1.5L84 1.501.0

90 89 88 87 86

11.430 9.5'1,44 8.L443 7.1154 5.3138

L.4835 L.465'j, L.4485 1.431,2 '1,.4137

85 84 83 82 81

0.t753 0.1944 0.2125 0.2309 0.2493

5.6713 5.1446 4.7045 4.33L5 4.01.08

r.3963 1.3788 L.36r4 L.3439 1.3255

80 79 78 77 76

0.2679 0.2867 0.3057 0.3249 0.3443

3.7321 3.4874 3.2709 3.0777 2.9042

L.3090 L.29L5 '1..274'1,

75 74 73 72 7L

0.3420 0.3584 0.3746 0.3907 0.4067

0.9397 0.3640 0.9336 0.3839 0.9272 0.4040 0.9205 0.4245 0.9L35 0.4452

2.7475 2.505L 2.4751 2.3559 2.2460

1.2217 L.2043 1.L868 L.L594 '/.,.1519

70 59 68 57 66

0.4353 0.4538 0.47'1.2 0.4887 0.5061

0.4226 0.4384 0.4540 0.4695 0.4848

0.9053 0.8988 0.8910 0.8829 0.8746

0.4563 0.4877 0.5095 0.531.7 0.s543

2.L445 2.0503 1.9626 L.8807 1.8040

1,.1345 t.tt70 "1,.0996

65 64 63 62 6L

30 3L 32 33 34

0.5235 0.5411 0.5585 0.5750 0.5934

0.5000 0.5150 0.5299 0.5446 0.5592

0.8660 0.8572 0.8480 0.8387 0.8290

0.5774 0.5009 0.5249 0.6494 0.6745

L.732L 1.0472 1.6543 L.0297 L.5003 L.0123 r.5399 0.9948 L.4825 0.9774

35 36 37 38 39

0.6109 0.6283 0.5458 0.6632 0.5807

0.5736 0.5878 0.5019 0.6L57 0.5293

0.8192 0.8090 0.7986 0.7880 0.777L

0.7002 0.7265 0.7536 0.7813 0.8098

I.4281 t.3764 L.3270 L.2799 1.2349

0.9599 0.9425 0.9250 0.9076 0.8901.

55 54 53 52 51

40 41 42 43 44

0.5981. 0.7L56 0.7330 0.7505 0.7679

0.6429 0.6561. 0.6691, 0.6820 0.5947

0.7650 0.7547 0.743"1. 0.73L4 0.7L93

0.8391 0.8693 0.9004 0.9325 0.9657

L.1918 1..L504 1..71.05 L.0724 L.0355

0.8727 0.8552 0.9378 0.9203 0.8029

50 49 48 47 46

45

0.7854

0.7071, 0.7071

1.0000

1..0000 0.7854

45

Cos

Sin

Cot

Tan

1,.2566 L.2392

1..0821 r.0647

Radia

60 59 58 )-

l,

c/

56

Degrees A-13

Table6 N

Commonlogarithms 0123456789

t0 11 L2 13 t4

0000 04t4 0792 1139 1461

0043 04s3 0828 IL73 1492

0085 0492 0864 1206 1523

0128 0531 0899 L239 1553

0170 0569 0934 I27L 1584

0212 0607 0959 1303 1,514

0253 0645 1004 1335 L544

0294 0682 1038 L367 1,673

0334 0719 1072 L399 L703

0374 0755 1106 1430 1732

L5 L5 t7 1,8 L9

L76L 2041, 2304 2553 2788

1790 2068 2330 2577 28L0

1818 2095 2355 260L 2833

L847 2t22 2380 2625 2856

1875 2L48 2405 2648 2878

1903 2L75 2430 2672 2900

t93r 2201 2455 2595 2923

1959 2227 2480 2718 2945

1987 2253 2504 2742 2967

201,4 2279 2529 2765 2989

20 21, 22 23 24

3010 3222 3424 3617 3802

3032 3243 3444 3536 3820

3054 3263 3464 3555 3838

3075 3284 3483 3674 3856

3096 3118 3304 3324 3502 3522 3692 37LL 3874 3892

3L39 3345 354L 3729 3909

3150 3355 3550 3747 3927

3181 3385 3579 3755 3945

3201, 3404 3598 3784 3952

25 26 27 28 29

3979 4150 431,4 4472 4624

3997 4L56 4330 4487 4639

40L4 4183 4346 4502 4654

403L 4200 4362 4518 4669

4048 42L6 4378 4533 4583

4055 4232 4393 4548 4698

4082 4249 4409 4564 4713

4099 4255 4425 4579 4728

4L1,5 428L 4440 4594 4742

4133 4298 4455 4609 4757

30 31 32 33 34

477L 49L4 5051 s185 5315

4786 4800 48L4 4928 4942 4955 505s 5079 s092 5198 521,1 5224 5328 5340 5353

4829 4959 5105 5237 s356

4843 4857 4983 4997 5 1 1 9 5L32 5250 5263 5378 5391.

4871 5011 5145 5276 5403

4885 5024 5159 5289 5415

4900 5038 5L72 5302 5428

35 36 37 38 39

5441 s553 5582 5798 5911

5453 5575 5594 5809 5922

5465 5587 5705 582L 5933

5478 5599 57L7 5832 5944

5490 561,1. 5729 5843 5955

5502 5523 5740 5855 5966

55L4 5635 5752 5866 5977

5527 5647 5753 5877 5988

5539 5558 5775 5888 5999

555L 5570 5786 s899 5010

40 4L 42 43 44

5021 6L28 5232 533s 6435

5031 5138 5243 5345 64/,4

6042 6L49 5253 6355 6454

60s3 6L60 6253 6365 5454

6064 6L70 6274 5375 5474

607s 6180 6284 5385 6484

5085 6L9r 6294 6395 5493

5095 6107 620L 6212 6304 6314 5405 6415 6503 5513

5lL7 6222 6325 5425 6522

45 46 47 48 49

6532 6528 672L 6812 5902

5542 5551 656L 5637 5545 6656 6730 5739 5749 682r 5830 5839 59rr 692A 6928

657L 5565 5758 6848 6937

5580 6675 6767 6857 6946

5590 6684 5775 6866 6955

6599 6609 6693 5702 6785 6794 6875 6884 6964 6972

6618 6712 5803 6893 6981,

50 51 52 53 54

6990 7076 7150 7243 7324

5998 7084 7158 725r 7332

70L5 7I0L 7185 7257 7348

7024 7L10 7193 7275 7356

7033 7LL8 7202 7284 7364

7042 7L26 7210 7292 7372

7050 7135 721,8 7300 7380

7067 7L52 7235 7316 7395

7007 7093 7177 7259 7340

7059 7L43 7226 7308 7388

Table 6 N

(Continued)

0r234s6789

55 55 57 58 59

7404 7482 7559 7634 7709

7412 7490 7565 7642 7715

74L9 7497 7574 7649 7723

7427 7505 7582 7657 7731

7435 751,3 7589 7664 7738

7443 7520 7597 7672 7745

745L 7528 7504 7679 7752

7459 7536 7512 7686 7760

7466 7543 76L9 7694 7767

7474 7551, 7627 7701, 7774

50 62 53 64

7782 7853 7924 7993 8062

7789 7860 793L 8000 8069

7796 7868 7938 8007 8075

7803 7875 7945 8014 8082

7810 7882 7952 802L 8089

78t8 7889 7959 8028 8096

7825 7896 7966 803s 8102

7832 7903 7973 8041 81.09

7839 7910 7980 8048 8115

7845 7917 7987 8055 8L22

65 56 67 68 59

8129 8195 826L 8325 8388

91,36 8202 8257 8331 8395

81.42 8209 8274 8338 8401

81.49 8215 8280 8344 8407

9155 8222 8287 8351 84L4

81,62 8228 8293 8357 8420

81,59 8235 8299 8363 8426

8176 8241, 8305 8370 8432

81,82 8248 83L2 8376 8439

8189 8254 8319 8382 8445

70 7T 72 73 74

845L 8513 8573 8633 8592

8457 8519 8579 8639 8598

8463 8470 8525 8531 8585 8591 864s 8651 8704 .8710

8476 8537 8597 8657 871,5

8482 8543 8503 8663 8722

8488 8549 8509 8669 8727

8494 85s5 851.5 8675 8733

8500 8561 862L 868L 8739

8506 8567 8627 8686 8745

75 76 77 78 79

875L 8756 8808 8814 8865 8871 892L 8927 8976 8982

8762 8820 8876 8932 8987

8758 8825 8882 8e38 8993

8774 8831 8887 8943 8998

8779 8837 8893 8949 9004

8785 8842 8899 8954 9009

879L 8848 8904 8960 9015

8797 8854 8910 8965 9020

8802 8859 8915 8971 9025

80 81 82 83 84

9031 9085 9138 9I9I 9243

9036 9090 91,43 9196 9248

9042 9096 9L49 920L 9253

9047 9101, 9154 9206 9258

9053 91,06 9L59 9212 9263

9058 91,1,2 9L55 9217 9269

9053 91,17 9170 9222 9274

9059 9122 9175 9227 9279

9074 91,28 9L80 9232 9284

9079 9L33 9186 9238 9289

85 86 87 88 89

9294 9345 9395 9445 9494

9299 9350 9400 9450 9499

9304 9355 9405 9455 9504

9309 9360 94L0 9460 9509

93L5 9365 9415 9465 95L3

9320 9370 9420 9469 9518

9325 9375 9425 9474 9523

9330 9380 9430 9479 9528

9335 9385 9435 9484 9533

9340 9390 9440 9489 9538

90 9L 92 93 94

9542 9590 9638 9685 9731

9547 9595 9643 9689 9736

9552 9600 9647 9694 974t

9557 9605 9652 9699 9745

9562 9609 9657 9703 9750

9566 96L4 9661 9708 9754

957L 95L9 9555 971,3 9759

9575 9624 9671 9717 9753

958L 9628 9575 9722 9758

95 96 97 98 99

9777 9823 9868 99L2 9956

9782 9827 9872 99L7 995L

9786 9832 9877 992L 9955

9791, 9836 9881 9926 9969

9795 9841 9886 9930 9974

9800 9845 9890 9934 9978

9805 9850 9894 9939 9983

9809 9854 9899 9943 9987

6r

9585 9533 9580 9727 9773 991,4 gglg 9859 9863 9903 9908 9948 9952 999L 9996

Table 7

a

The Greek alphabet

alpha

B beta y gamma 6 delta € epsilon ( zeta' q eta 0 theta c iota K kappa I lambda pmu

vnu €xi o omicron tr pi p rho (, sigma r tau u upsilon 0 X ,lt @

Phi chi psi omega

Answers to odd-humberedexercises

ExercisesL.1 (Page14) 1. {0.,1,,2,4,6,8} 3. {2,4,8} 5. {0,1,2,3,4,6,8,e} z. A 17. (-*, Sl 1e. (-e,-t) u (0,+-) 2r. (-a,-2) U (2,+q) 2e. (-a, -1) u (+,3) 31. (t +il u (3,+-)

e . { 0 , 1 . , 6 } 7 r . ( - 2 , + @ ) 1 3 .( - o , { ; ts. [4,8] 23. (-o,-s) U (3,+o) zs. t-t,Ll zz. (-s,i)

ExercisesL.2 (Page20) 1. 1,-E 3. -3,8 5. -*,4 1 9 .( - a , 1 ) u ( 4 , + @ ) 35. lcl < a- b

7.-&,+ 9.+,3 1 1 .[ 9 , + o ) 1 3 .( - a , 1 ] u [ a , + o ) ts. (-11,3) 1 7 .I g , 2 ] 2r. (-o,31 u [ro,+-) 23.[-g,g] 2 5 .( - a , * ) u ( 2 , + * ) zs.lxl>lal sr. lx-2l>2

Exercises'1..3(Page27) 1. (a) (1,2); (b) (-1, -2); (c) (-7, 2); (d) (-2, 1) (b) (t) -3); (c) (1,3); (d) (-3, -l)

3. (a) (2,-2); (b) (-2, 2); (c) (-2, -z); (d) doesnot appty

5. (a) (-1, 3);

Exercises1.4 (Page32) s. *163;Ll-$et\/B .z.X tt.-2,8 1 3 .( - 2 + t \ / s , t + 2 \ / 1 ) a n d ( - 2 - t \ / 5 , t - 2 \ f r ) l s . 7 7 ; l* e 1 x y * 2 r y z - r 2 2 x -66y-55:0 1 7 .( - 2 , * ) ; $ , * ) ; ( t * , T ) le. (1,+) 2r. (-t,D 2 s .( s , - 7 ) 2 s .( a ) 2 x + y : s Exercises1.5 (Page41) 1.-1

3.-+

rs.f +$:1

5.4x-y-t7:0

7 .x : - 3

9.4x-gy+12:o

le. (a)yes;(b)yes;(c)no;(d)n2 ot.2x*3y+z:0;lE

tt. t6l-y+(2\6-s):o

1 3 .x r y : g

z z .@ - $ ; 1 u ) .

(d)Bx-Ay:s tg-^g; 2s.y:'t;9x-4y-11:0;9x*4y-19:0 27.2x*3y-t2:0;(2+zt/Z)x+(S-tt41y-12:0;(Z-zt/Z)x+(g+S{|)V - 72: 0 29. (a) x : t; (b) t : l; (c)x : l; (dl 2x + y - 3 : 0; (e)r - 2y + 1 : 0; (I) x + y - 2 : O ExercisesL.6 (Page47) L.(N-4)'+(y +.3)':2f;*+y,-8x+6y:g 3. (r+5)2+(y+t272:g;*+f*10r*2ay+1,1;0:O S .* * y z - 2 j - 4 y -8:0 7.**y2+6x*lOy +9:0 9.f *y2-U-4y-2:0 tL. @,D;4 1 3 .( 0 , - i ) ; * 1 5 .c i r c l e L Z .r h e e m p t y s e t 1 9 .c i r c l e 2 1 .x * y + 5 : 0 2 3 .3 x 1 - 4 y - 1 9 : 0 A-17

A.18

ANSWERSTO ODD-NUMBEREDEXERCISES

Exercises7.7

(Page 55)

5. domain: [*, +*); range: [0, +o) 1. domain: (--, *-); range: (-o,4o; 3. domain: (-., *-); range: [-5, *o) 9. domain: (--, **); range: [0, +o) 11. domain: (-*, *-); range:-2 and 2 7. domain: all r not in (-2,2); range: [0, +-; 17. domain: all real numbers range:all real numbers except3 13. domain: (--,4-); 15. domain: (--, *-); range: [-4, +-) except-5and-1;range:allrealnumbersexcept-7and-3 19. domain:allrnotin(-1,4);range:[0,+-; 21.domain:allreal 25. domain: (--, *.);range: [1, +*) numbersexcept2;iange: [0, +*; 23. domain:allrealnumbersexcept-5;range: [-5, +o) 29. domain: (--, **); range: the nonnegative (see Fig. EXl.7-251 27. domain: (-*, +-); [0, 1) (seeFig. EXl.7-27) integers (see Fig. EX1..7-29) 31. domain: all real numbers except 0; range: (--, -11 U {0} U (}, 1l (see Fig. EX1.7-31) 33. domain: (--, +*), range: {1, 3} (seeFig. EX1.7-33)

il I 2l 'E lt ,E

Figure EXl .7-27 Figure EXl .7-25

Figure EX1.7-29 Figure EXl .7-33

1.,8 (Page61) Exercises 1. (a)-5; (b)-6; (c)-3; (d)

(f) 8f * 10x2- 3; (d zt' - 7x2;(h) 2 x 2 * 4 h x * 5 x * 2 h 2 + 5 h 2

(e) 2h2 + th

+5h-6;$)ax+2h+s

(c)2; (d) 3rt; @)

(b)

(a)

(f)

(i) Zxz* 5r + 2h2

s. (a)

(ffi+tffi)

( d )- 1 ;

(b) -1; (c)

(e)-l if x > 0, t if x = 0; (f) 1 if x > -"1.,-1, if x <-1; (g) 1; (h) -1 7. (a) x'* x- 6, domain: (-*, +co); (b) -xz * x- 4, domain: ( - m , + @ ) ; ( c )f - 5 x 2- x * 5 , d o m a i n :( - - , + - ) ; ( d ) ( r - 5 ) l ( x , - L ) , domain: all real numbers except -1 and 1; Id (x'- 1)/( x - 5), domain: all real numbers except real numbers except0 and

(b)

domain: (-*,

(f)

*o);

(g)x,-1.0r+24,domain:(-*,+.o)9.@iw,domain:all

-2L1

ffi,

domain: all real numbers except0 arid

-2 J-. -

(d)

}f,

x*'/', domain: all real numbers except 0 and fi,

all real numbers except -L, 0, and l; (f)#,domain:

domain: all real numbers except 0 and

numbers except 0 and

(r)

domain: all real numbers except-l

and 1

11. 1a)lF

- | + t/x-

t, domain: [t, +-;,

D vETT, domain:[t, +-;, Gl tlx + t, domain:(1, +-;, C) t#T, \tr) andl\8, +@) (--, 13. (a) even; (b)

(b)t-11+ [r']; (c) [l

(c) neither; (d) even; (e)

+ 3] * [x' - xl; (d) t0]

(f) even; (g) odd; (h) neither

(")[fil].1#l;

all real

domain:(1, +*;,

17. (a) [r' + 2] + [0];

tr) tttzxt+ lx- 1l+ lr +

++ilx-11-lr+1ll

(

ANSWERSTO ODD-NUMBERED EXERCISES A-19 L9. (a) even; (b) odd; (c) even; (d) even

21,. domain: (-@, +oo); range {0, 1} (Sketchesof the graphs for Exercises23-33

appear in Figs. EXt .8-23 through EX1.8-33.)

o Figure EXl .8-29

Figure EXl .8-23 Figure EX1 .8-27

Figure EXl .8-25

v

Figure EX1.8-31

Figure EX1.8-33

3s. (f " S)(x) :

37. (f " S) (r) -

ifr<0 if0'/.,

[l

rf. x < 0 if0 - x =i ifi- ( - x < t

ll

3 9 . S ( r ) - x - 3 ; S ( r ) : ' / . .- x

ifx >L

Reoieat Exercises for Chapter 1. (Page 63) 3. (-+,1) s. t-#,*l 7. (a)(x-3)"+ (y+s)'z:0;(b) (r-3)'+(y +s)'< 76;(c)(x-3)"+(y+5)"-25 1. (*,gl 9. x:0,3x*4A:0 1 1 .k : - + r h : + 7 3 .( 3 , - 6 ) ; ( - 2 , - 5 ) 7 5 .0 < k < 2 . 17.7x2*7yr*llx-1,9y-6:0 1 9 .l 2 x * 3 y - 2 : 0 21. domain: [-8, +*); range: [0, 4o) 23. domain: all real numbers except -B and 2; range: all real numbers except] and 5 25.(a)f * 4x-7,domain: (--,**); 2,

Ar-a

all real numbersexcept @ i; ffi, domain:(--, *-)

x2-3x ,, + I

1.,domain: (-o,4o;.

(c) (rP- a)Gx-3),

domain: (--,**);

@)#,domain:

domain:all realnumbersexcept-2 and 2; (f) L6xz- 24x* 5, domain:(--, *-); G) a* - 19, domain:all realnumbersexcept-1 and 3; (b)

27. (a) #h,

except-1 and 3; (c) G=;T;TT; (e)

(b) l-4r-

G4+#$,

domain:all realnumbersexcept-1 and Z; @) #*,domain:

, domain: all real numbers except -1 and 3; (f) -

#,

domain:all realnumbers all realnumbersexcept0, -1, and 3;

domain:all realnumbersexcep,-Xand-1;

(g)

#,

domain:all real

35. (+s, +a); (9,7); (2,3)

numbers except 2 and 3

Exercise2.1. (Page73) 1. 0.005

3. 0.005

79. E: min(l, *e)'

5. rtr

7. 0.01

2L E - min(l, *e)

g. a: i 5

1 1 .D : e

1 3 .E : m i n ( l , * e )

2 3 .D : m i n ( 1 , ( 2 t E + 3 ) e )

15. 6 - min(L, te)

17.0-min(*,*e)

EXERCISES ANSWERSTO ODD.NUMBERED

A-20

Exercises 2.2 (Page83) 3.-#

1.7

s. 12

e. 1, ir. +\r0

'7. +

1s.+

13.*{z

17.++

23. (a) 0

2.3 (Page87) Exercises e 1. (a)-3; (b) 2; (c)doesnot exist

3. (a) 7; (b) 7; (c) 7 5. (a) 5; (b) s; (c) s 11. (a)-2; (b) 2; (c) does not exist L3. (a) 2; (b) 1; (c) does not exist

exist

Exercises2.4 1. frc

e. (a) 1; (b) -1; (c) doesnot

7. (a) a) 0; (b) 0; (c) 0

(Page 96) 5. -@

3. *o

7. -@

9. *m

11. -@

13. -@

15. *rc

2.5 (Page100) Exercises l. 4; lim f (x) does not exist.

3. -3; F(-3) does not exist.

7. -3,2; h(-3) and h(2) do not exist.

5. -r, jt11;g(r) + Sei- 3 )

t-4

9. 0; lim f (x) does not exist. r-0

11. continuous everywhere 1,3.3; f (3) does not exist. 15. 2; lim Sk) does not exist. 17. all integers; r-z

any integer.

t9. O,t"tT f G) does not exist.

25. removable;0

. all real numbers . all x not rn l-4,4]

27. removable;*

2L -1, O, 0 jli

29. essential

f Q) does not exist, lim f (x) does not

31.. / is continuous everywhere.

-3, and -4 13. all real numbers except2 and -22 15. all real numbers except "1., 21,.all real numbers 23. all x ir in (0, +@) 25. all x in (-1,0) and (0, 1)

ReaipwExercises for Chapter2 (Page1-08) 1. 9

3. -+

-9] and [8, * *)

5. tontinuous on (-o

13. 6 - min (1, 5e)

17. +

19. -@

7. continuous on 13,4]

23. -+

25. (a) yes; (b) no

11. E- min (1, e)

9' 6 - *e

27. f (x)

29. (b) all real numbers;

(c) any number that is not an integer

Exercises 3.1 (Page 114) 1. -2xt 3. -6 - 2x, 5. 3xr2- 3 7. 12xr2- 26x, I 4 9.8r* y+9 -0; x-8y *58:0 1 1 .6 x - y - 1 6 - 0 ; 5 2 : 0 1 7 2 y * 3 0 : 0 : 0 ; x*6y 1 3 .2 x * 5 y t7. x - 1 2 y + 1 6 - 0 ; 5r 1 5 .2 x * 3 y - L 2 : 0 ; 3 x - 2 y - 5 : 0 -30 +4{30:0; (12+z@lx-y-30 lZx*y-98-0 1 9 .8 x - y - 5 - 0 2 L .2 x - y - 2 - 0 2 3 .( 1 2 - 2 \ m ) x - y

- 4\F0: 0 Exercises3.2 (Page1.20) 1. 6tr'- gtr+ 2

3. 6fi L8

5.

1

.L

2tffi1'n

.t ,'

-5 ._ 5 ( 5 f 1+ 5 ) t t z ' 6 4

and t:l moving to right; changesdirection w h e n t : - 3 1 , 1 .t < - L right; t > -1 + \6, moving to lefq changes direction when | - -l + \R 15. (a) (20h + 24) ftlsec; (b) $ sec

Exercises3.3 1.8r*5

2r. -+

< t 1 l, moving to left; t > l,

\ 6 , moving to left; -1. - tE < t 13. (a)-32 ftlsec; (b) -64 ftlsec; (c) 4 sec; (d) -128 ftlsec

(Page 125)

7 - 4f, - '

3.

23.

Exercises3.4

q t9<\ . -3, moving to right;-3 "

1 1 .- 6

13. -#-6

15. z

17. 3

19. -12

s@) (Page 129)

1. (b) yes; (c)+1,-1; (d) no

3. (b) yes; (c)-1, +1; (d) no

5. (b) yes; (c) 0, 1; (d) no

7. (b)

(d) yes

e. (b)

A-21

EXERCISES ANSWERSTO ODD-NUMBERED 13. (b) yes; (c)-6,

11. (b) yes; (c) neither exisU (d) no (c) does not exist, 0; (d) no 21. (a) 0; (b) 1; (c) does not exist b : -t

'1.9. a:2

17. /i (0)

(d) yes

and

(Page 137)

Exercises3,5

(x -

l)t

3 L . 3 ( 2 x 2* r * x*8Y-2:0

L ) 2 ( 4 x+ 1 )

Exercises3.6

(Page 141)

6 (r2.+ 10r-f L) (r * 5)2

t s . - ,( x - t =1 ),, = 32L '1 ,( 1. 1+ ! 21x,' )-2= 2 f i-z] i . 33.x * (4\re- I})y - 8\re+ 2L: 0;x -

+ t)

-t@ L L "7 .

- t s x z- 6

13. 3 6sz - 2 6s

7.4rrrz g.2x*3-?,

5. Sz-t

3.x7-4f

1.3x2-6x*5

15. 70x6+ 60/. 29.2(ix + 2) (6x, * 2x- 3)

3 5 .x * 8 y * 2 - 0 ;

8\re*2L:0

7.6(3u-l)(3u2+5)2(L2u2-3u+5) 5.-2(xl-4)-3 3 . 2 ( 8 1 3 - 2 1 , t 2 + 2( )2 t 4 - 7 t s + 2 t - L ) 2 ( l 4 x + 3 ) ( 7 f 7 ) ( y + 2 ) 3 + 3 x -L)-2 13. 11.18(y s.2(-t2x+17)(2x-5)-'z(4r+l)-a ! 1 7 . 2 z ( z z 5 ) ' ( z ' 1 2 2 ) ( 2 2* 4 ; - e 1 5 . i ( r , + 1 ) ' ( 2 r + 5 ) ( 8 r ' * 1 5 r+ 2 )

1.6(r+z) (xz+4x-5yz

ftlsec;ffiftlse.

+5)-3 + 49x2-4x+30)(31 19.4(4x-1.)2(*+2)3(2txa-3xf

x*l6y-35:0

".ffiftlsec;0

25. (a) 3xa;(b) 615 (Page 145)

Exercises3.7

3.

1,.2(3x * S;-tra

LJ'

L7

r*5

^

- A

Lv'

|

r

v

x ( 5 x 2- 1 )

1 E

6\tr-1\7(fry \i

I

5. Z,-rtz - EX-t,,

ztTzx-s\@Tl)T

r/

t7.

yffi\?reW /

Exercises 3.8

1. -;

x

^

3.

6x2-10x*1

Lg.4x-5Y+9:0 +t/g + \ffitFx

v2-4\

27. sxlxl 2s.z,lffi) -r3(31 (t2+ f + 5) t4+ 29t3+ ]i09P 18t- 88) 35. +(t3- 2t + 1)tt2 23. (a)0; (b)4 (c)no valueof t

7.

sg. *(3r + 2)3(f - 1)-trs(L2f+ 2x - e)

(P age 1.50) gy_ lxz

3yz_8x

-by , tr _ i.

1rtz.t_rz 7 7. _-yttzx-ttz

9

x-xttz

Lr

_,1u

3x2-4u

7;-_r,

xs + 8tr3 - -z; 1 g . _ # 2 1 -' . 2 x i a : 4 2 5 . ( a ) f , ( x ) _ 2 \ f f i . , d o m a i n : . t r > _ 2 ; f , ( x ) 4x" !x'y -!0 ;x + y - L : 0 d o m a i nx: ) 2 ; f L ( x -) - ( x - 2 ) - t t 2 , d o m a i n :x ) 2 ; ( e ) l ; 0 x - y v domain: lxl > S, (d) /i(r) : x(f - 9)-1/2,domain: lrl > 3; fLG):-x(x" fr(x):-{V4, (f)5x*4y*9:0;5x_ay*9:02g.(u)f,(x):\mi+2,domain:-2< -2 < x < 4; (d)(l-

3x2-v3

y+4@

i 3 .v * 2 x ( x - i l '

;7.&

i5.ffi

domain i )c > 2; (d) f

27. (a)f ,(r) - \E=

'r(x)-

9, domain:

-9)-trz, domain: lxl > 3; (e)

x = 4; fr(x)_

(x - z)-ttz,

- \FPT|

i;

* 2, domain:

-2 I x 1a; (e) x ) ( 8 - x 2+ Z x ) - t n , d o m a i n ; - 2 1 x < - 4 ; ( x - 1 ) ( 8 - x 2* ' 2 x ) - t r zd, o m a i n: =; v

( f) y - 5 :

y +7:0 Exercises3.9 (Page153) 1. (a)8.6;1b)8.3;(c)8.1;(d) 8 (b) not profitable

s. oa:-E

3. (a)-'t, -*; (b) -rlr, -*

5. (a) 18,750gallmin; (b) 17,s00gallmin

7. (a) profitable;

11.2.7mi/min

3.10 (Page1,56) Exercises T. f r,lr". ftlmin 5. ftlsec a. * *ftlmin 3 1.5.rta(3\E + 97) ftlsec- 0.65ftlsec L7. 22 ff lmtn L

s.

#ftlmin

11. 1800 lblft2 per min

L3.

'1.4 ftlsec

0,

A.22

EXERCISES TO ODD.NUMBERED ANSWERS

Exercises 3.L7 (Page160) 3 . 9 ' ( s ) : 8 s 3 - 1 2 s 2 + 7 ; S " ( s ) : 2 4 s 2 - 2 4ss. f ' ( x ) : x ( i ? + 1 ) - t t 2 ' t. f'(tc):5f-6**7;f"(r):20f-12x 7 . F ' ( x ) : W 2 5 . F " ( x ):tlarrz 9 . G ' , ( x ) : - 2 x ( 3 + 2 x 2 ) - 3 t 2 . G " ( (x8) r: r - 5 ) ( 3 + 2 x 2 ) - 5 t 2 + 1 1 t r , f,,(x):(f 15.D,ay: *(+trt2 | lsstz - tx-ttz1 19.D,2y: -lstxly-z 21,.-t;5 L3.D,8f(x): 6(3+ r) (t - a;-s 11.D,1y: ll,y - L z t + 8 ; a : 0 w h e nf : 0 a n dt : 4 i t o w a r dt h eo r i g i n : 2 < t < 4 ; a w a yf r o mt h eo r i g i n : 0< t < 2 2 3 .v : t 3 - 6 P + 8 t ; a : 3 f and t > 4

25.t; #; -+

27. sr;tfi;

31. f '(r) - 4l.fl, domain: (-*, +*) ) f " (x) + (f ' " g) (x)g" (x)

ReviewExercises for Chapter3

domain:(-*, +*) ; f " (x) : r4rdomain: (-*, o) u (0,+*; x'

2 9 .f ' ( x )

?\re

(--, + o )

domain:

3 3 .f " ' ( x ) : 2 4 l x l

(Page161)

str(**+)'(#) (xn* x)'l'l3x' * 2(xa* r) (4xt + 1) I lefq when t:

3 5 .h " ( x ) - (f" " s)(r) (g'Q))'

"1,-2x3 n o'' " gxzts(xt* '!.)+rt \F@15. (a) t : 3 a n d f : 8 ; ( b ) w h e n t : 3 ,

8, o: 40, and particle is moving to the right

z|x+1l- l'l, (ffi-fr) 21,.

zs.(b)0;(c)0

3 1 .a : - L ; b : t

445-1,80\/2, lZ. jffiknots

#t lblft' per min

a 1 .( f . s ) ' ( 0 ) : 2

11 x(4x2 13)\F \F> \E__4 7 )- - 1 5 , a n d particle is moving to the 1,

17. 5x - 4y - 6: 0; 4x + 5y - L3 : 0

2s.-1,

27.

0)

79.

29. (a) 32 ftlsec; (b) 255

E1t 37. -?units/sec frU.in./sec 4 3 .f ( x ) : l x l , s k ) : x , M. p'(r) - 10.6knots

l

35.

l

2

\Fg (c) 7 sec;(d) -128 ftlsec

39. decreasing at a rateof

4.1 (Page170) Exercises 1.?

3.0

s.1

7.+

9. 0

11. 0

13. f o

23. f is discontinuousat 0

4.2 (Page173) Exercises 1 .x : 5 ; y - 0 1 1 .y - - 3 , A : 3

3.x--2;y-0 5 .x : - 6 , x : 1 ; y : 0 1 3 .x - - \ n , x - f i . 1 5 .x - t i y - &

7.x=3,x:-3;y-4 17.y--1,y-1

9. x:-2,x:2;y:0 t9.x-3;y--L,y:1

27.

4.4 (Page180) Exercises 1. continuous; discontinuous; discontinuous; continuous; discontinuous; continuous 3. continuousi continuous; continuous; 5. continuous; discontinuous; discontinuous; discontinuous; continuous 7. continuous; continuous; continuous; discontinuous continuous; continuous; discontinuous; discontinuous 9. continuous; discontinuous; discontinuous; discontinuous 11. discontinuous; continuous; continuous; continuous; discontinuous; discontinuous; discontinuous; continuous 73. (-a,-21' (-2,2); (2,+a1 (-1,1); (1,+o) 15. (-o,-31; [4,+oo) L7. (-a,-l); 19. continuouson (0, 1) and all intervalsfn,n+']..) wheren is anyintegerexceptzero

ZZ.-!.continuousatalxin vx- z at all r in (0, L) and (1, +o1 29. 5

Zt. t/F;continuousatalltin (0,+-;

continuous at all r in (0,4) and (4, +o1

ZZ. !!],continuous lx- l' 33. no; / will be continuouson la, cf iflim g(r) existsand is equal to h(b). Exercises 4.5

(2,+o) 31.

25.::-, !x-2' and k:

4

(Page 188)

13. no absolute extrema 1. -5, * 3. -3, -L,1 5. 0, 2 7. -2,0,2 9. no critical numbers 1.1.abs min: l(2) : -2 : 2 : 1 1 5 . a b sm i n : / ( - 3 ) : 0 1 7 . a b sm i n : f ( S ) : L 1 9 . a b sm i n : f ( + ) 2 L . a b s m a x :f ( 5 ) 23. abs mint f (2) - 0 -6; -1p : : : : 25. abs min: l(-3) 29. abs min: f (21'.:0; abs max: abs max: /(-1) 27. hbs min: /(-2) 0; abs maxt f (-4) LM 3 1 .a b s m i n : / ( - 1 ) : - 1 ; a b s m a x : f ( 2 ) : l 35. abs min: /(-3| -- -13; 3 3 . a b s m i n : / ( - l ) : 0 ; a b s m a x : / ( 1 ) - VE f(3):25 :7 abs max:f (3)

TO ODD-NUMBERED EXERCISES A.23 ANSWERS 4.6 (Page194) Exercises 5. From Ato P to C, where P is 4 miles down the river from B.

3. * in.

1. 2500 ftz

height - GtQ.in.

11. (a) 3.a units; (b) 9.4 units

10

.

t

5

..

13. 225

15. 400

T. g mi

r s17. (a) radius of circle

g. radius _ 3tD.in., 5 ft and length of side of G+a)

r.r

square:#fU(b)radiusofcircle-5ftandthereisnoSquare. 7T

i

4.7 (Page200) Exercises (a,t \fr) or *(2- rt) 1. 2 3:".. 11.+(3+ 4\E) 7. + s. #

5. 1-t,-*),c:-t-t.VD;.t8,+f I:-r1+t/6 B. r. 1s. (0, (ii), (iii) satisiied; (c) (*, -*v6)

satisfied

21. (b) (ii) not satisfied

satisfied

33. (ii) not satisfied

23. (b) (i), (ii), (iii) satisfied; (c) (0, 9)

25. 4

orc:-1 -trt/w;(-+,+),':-r+*t/e 19. (b) (i) not 17. (b) (ii) not satisfied 'Zz. 31. (i) not zs. *+\/39 s + '\\/,

(Page 202)

ReaiewExercises f or Chapter4

,/x=1'

continuous at all r in (1, +o;; continuous on the closedinterval 13.absmax: 11.absmin:l(-5):6 9.(f "S)(r):sgn(f-1);continuousatallrealnumbersexcept-1 andl [1,+-1 l e n g t h : 1 8 o i n . ; w i dth:depth: 1 9 . L 7 . 6 a n d 6 1 5 .a b s m a x : / ( 5 ) : 3 5 1 ; a b s m i n : f ( V 6 ) : 0 f(O):9;absmin:/(3):o ( a ) ( x ) : s g n r ; 2 9 ' n e a r e s t A 2 7 . 1 2 l o r fr ri o m p o i n t o n b a n k x:-L f 2 5 .y : 2 , y : - z ; x : 7 , *in.

L. 4

Z. -i

( b )/ ( x ) :

5. y:5;

x:2,

x:-2

7. (f . ilQ):

( x r f x< L [ x i f r < 0 <1; ( c )f ( x ) : J r + r i f 1 < x if 0 < x l+ tr rf!
Lr

5.L (Page21'0) Exercises 5. (a) and 3. (a) and (b) no relative extrema; (c) (-*, +*); (d) nowhere 1. (a) and (b) /(2): -5, rel min; (c) [2, +o;; (d) (-*, Z] -21, (0' +o); (c) extrema; relative : (b) no -50, (-*, 7. (a) and +*'1, (c) 46,rcl max; lz, rel min; f(-2) @) l-2,21 (b) l(2) : 11. (a) and (b) /(+) : 3'59, rel max; f (1) : 0, rel min; (d) [2, 3] 9. (a) and (b) /(2) : 4, rcl max; (c) (--,2f; (rd)nowhere :t, 15' (a) and (b) l(4) :9, rel max; (--, +); (d) [4, +o) rer max; (c) 13. (a) and (b) l(4) (c) (--, *1, [r, +-;, (d) t*, 1] : 19' (a) and (b) (-*,-z], 1, rel min; (c) rel max; /(0) [0, +-;, (d) [-2, 0] 17. (a) and (b) f(-2):5, (c) (-*,41; (d) [4, +-) 2 1 .( a ) a n d ( b ) / ( - 1 ) : 2 , r c l m a x ; l ( 0 ) : l , r e l m i n ; f ( 2 ) : 5 , r e l m a x ; ( c ) ( - * ' - t l , f(2):-g,relmin; (c)12,+@);(d)(--,21 f(-7):-4,relmax; 23.(a)and(b)f(-g):-8,relmin;/(-4):-5,relmin;f(2):-7,relmin t0,21;(d)t-1,01,f2,+*1 ( c; ) [ 0 , + - ; ; -71,1-4,01,12,+*1;(d) (--, -sl,l-7,-4l,lO,21 25. (a)and(b)norelativeextrema ( c ) t 9 , f(O):-g,relmax; -11, -94, (a) and (b) : 29. (--, (d) 1l : +*); (c) rel min; I-1, f (4) : +f/4' 11, 27. (a) and,(b) /(-1) Q rel max; /(1) (d) nowhere 4 , l a , + * ) 3 3 .a : - 2 , b : 9 , c : - 1 2 ' d : 7 3r. a:-3,b:7 r e l m a x ;( c ) ( - 4 , 4 1 ; ( d ) ( - - , Exercises 5.2

(Page 213)

3. f(g):T, 1. /(+):3, relmin IL. f (l) :8, rel min rel min Exercises5.3 1. /(0) :0, by 60 yd.

5.f(4)

rel min

7. G(3) :0,

rel min

9. h(-2)

(Page 219)

5. 8(+) :*, abs min 13. \n 1,1. Depth is one-half the length of the base. : R; depth : L7. If R is the radius of the cylinder, breadth

abs min

*(4 + n) .

r e l m a x ;f ( - 1 ) : - 1 1 , r e l m i n 13' F(27):9, rel max

3. no absolute extrema

9 . a5 yd 7 . / ( 0 ) : 0 , a b sm i n ; f ( f r ) : + \ / 5 , a b sm a x 15. Ratio of height of rectangleto radius of semicirclers 21'.R : f rg. 30( \m + 1) ft by 44({m + 1) ft rfgR.

23. 2\/2 5.4 (Page226) Exercises 3. concaveupwardeverywhere 5. concave 1. concavedownwardftorx 10; concaveupward for x ) 0; (0, 0) pt. of infl. ( forx<2; 7. concaveupward ( ( (0,0)pt. of infl. < x>'!.; r 0 and r 1; concaveupwardfor-1 <-1 and0 downwardforr pt. ) (0' 0) of infl. 0; fot x < downward 0; concave for x upward pt. 9. concave ) (2, O), of infl. 2; concavedownward for x ll. a:-l,b:3

EXs.4-2s.)

13. c:2,b:-6,c:O,il:3

( S k e t c h e s o ft h e g r a p h s f o r E x e r c i s e s 1 5 - 2 5 a p p e a r i n F i g s . E X 5 . 4 - l 5 t h r o u g h

A-24

EXERCISES ANSWERSTO ODD-NUMBERED

(c, f(cl) (c, f(c)l

c

c

Figure EX5.4-15 Exercises5.5

Figure 8X5.4-17

Figure EX5.4-19

Figure EX5.4-21

Figure EX5.4-25

(Page 229)

(Sketchesof the graphs appear in Figs. 8X5.5-L through EX5.5-33.) 1. f(-1):5, rel mn<;f(l):-3, rel min; (0, 1), pt. of infl;/increasing on (-o,-11 and Il,+*1;/decreasing on [-1, 1]; graph concavedownward for x ( 0; graph concaveupward for r > 0. rel min; (0, 0), (1, -1), pts. of infl.; f increasing on 3. f(D:-1+, B,+-);/decreasingon (-m,*J; graph concaveupwardforr < 0andx > 1; graphconcavedownwardfor0( r ( 1 5. /(-3) -t rel max; f (-+) : -9, rel min; (-*,,H, pt. of infl.; f increasing on (-o, -31 and [-*, +-); / decreasingon [-3, -*J; graph concave downward for r < -*; graph concaveupward for r > -* 7. f (0) :1, rel min; (t,lil , O,2) , pts. of infl.; / decreasingon (-@, 01. 9. f (-l): / increasing on [0, +o); graph concaveupward for r < * and r > 1; graph concavedownward for t < x < 7 r1, rel min; ( r c , 1 1 r e l m i n ; p t s . o f i n f l . a t r : i ( l = l V ) ; / d e c r e a s i n g o n a n d [ 0 , 2 ] ; l i n c r e a s i n g o n [ 1 ,0] and f(O):t,relmax; f(2):-t, \/7) 12,+*1t graphconcaveupwardforr < +(l- fr) andr > *(1 + .y'7);graphconcavedownwardfor*(1 - 'y'V)
tu. /G4) :'+#,relmax;

f(2):O,relmin;pts.of infl.at (-1,0)andr:rt(s'r3{6);f

increasing

and+(8-3V6)lb(8+3\6) . 17.f(-+):{f,relmax;/(0):0, concave upwardfor-l -1 -*] and [0, +-;; < for r (-o, graph downward decreasing on 0]; concave on increasing [-*, f I 19. f (0): 0, rel min; l(1) : 1, rel max; / decreasingon (-o, 0) and [1, 4a); /increasing on (0, 1]; graph concavedownward forr < 0 21. /(-*) : -8, rel min; (0, 0), (+, *.92), pts. of infl.; f decreasingon (-o, -*l; I increasing on [-L ao); graph concave and r ) 0 23. no relative extrema; (3, 2), pt. of infl.; f increasing on upward for r ( 0 and r > *; graph concavedownward for 0 < r 3; graph concaveupward for r < 3 25. no relative extrema; (3, 2) , pt. of infl. with 27. f(-l):3, horizontal tangent; /increasing on (-o, 1o); graph concavedownward for r( 3; graph concaveupward forr> 3 -11; (-m, (O): 29. f 0, rel min; l(f) :1X\/5, rel min; / decreasingon / increasing on [-L, 1o); graph concaveupward for all r r*J; (-co, graptr upward for increasing on concave 8{6); / decreasingon 0l and [1F,4]; / rel max; pt. of infl. at r: +(48 [0, : :2, < < rel rcl max; 8\6) x 4 31. f(-1) min; /(l) 816); graph concavedownward for *(rtt| r < 'i(4U Q Pts. of infl. at r: 0 -16 -1] graph concave (-o, < < upward for r 0 and r > 16; and [1, +-;, / increasing on [-1, 1]; and x: d5; / decreasingon (-o,-31; graph concavedownwardforr ( -rEand0 < r < tE relmax;/increasingon n.feil:{/6, /decreasingon [-3,0]; graph concave downward for r (

Figure EX5.5-3

Figure EX5.5-1

Figure 8X5.5-7 Figure EX5.5-5

EXERCISES A-25 ANSWERSTO ODD-NUMBERED

Figure EX5.5-9

Figure EX5.5-17

Figure EX5.5-L3

Figure EX5.5-15

Figure EX5.5-19

Figure EX5.5-21

Figure EX5.5-23

Figure EX5.5-31

Figure EX5-5-27

Figure EX5.5-25

Figure EX5.5-29

Figure EX5.5-33

5.6 (Page238) Exercises ( a ) 5 0 c e n t s p e r g a l l o n ; ( b ) 255. ( a ) Q ( r ) : x - I - 4 + i ; ( b ) c ' ( x ) : 2 a 1 4 ; ( c ) Q ' ( r ) : L - 7 , t - - -2 - '9 ' . 9\ 9. ; ( c ) g 2 e . 9 s3 rr r. i\ sa, l we o - 4x 2; , @ \ - , )- $ 7.8 (d) $9.55

_ 4x* 5; (c) decreasingon [0,2] and increasingon [2,.to); (d) graph concavedownward for 7' (a) f2,+a1;(b) C,(r) : f -L 4nn' ,.1-\:2-

;-:;:r,r;";;;:"""upwardforr (b)Q(r)-a*$,(c)C'(r):3;(e)e53

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1 3 '( a ) P ( r ) : t / F a ; & ) R ( r ) : r 0 0 x { f i = 7 ;

A.26

ANSWERS TO ODD-NUMBERED EXERCISES

( c ) R ' ( r ) - 3 5 0 0 - 2 0 0' f ',-r . 1 ) $ 1 8 0 01 5 .( a ) S ( r :) - ? f y36- *

+ t 6 x - 2 ; ( c ) $ 3 0 0 0 ; ( d ) R , :( 1r )4 - 2 x , C , ( x ) : 2 a - 2

(b) R'(r) : 3x2-32x * 64;C' (r) - 18 - 2x; (c)+(15- tffil Reaiew Exercisesfor Chapter S .1.t E 7 F ;

- t.Bg

19. (a) 1625; (b) 67.5 cents; (c) $3Zg.tZ

1 2 .( a ) [ 0 , 8 ] ; 21,. g7.S cents

(Page 240)

5. B sec;velocity of horizontal particle -'!. ftlsec;velocity of vertical particle - 3 ftlsec

f(4):0, relmin;pts.of infl. atx:-2andr:*(g-+S\/6);fincreasingon ( _ o o , T la n d [ a , * * ) ; / d e c r e a s i n g o n [ g , 4 ] ; g r a p h concaveupward lor-2 < r < *(8 -316) and g(8 +lt/6,1 < x < +o; graph concavedownward for_o ( x 1_2and, *(8 - 3V6) < r < *(8 + g\/6) 9. no relative extrema; pt. of infl. at x:3;/ increasing on (-o, 4o); graph concaveupward for3 ( I < +€; graphconcavedownward for-e < r < 3 15. (a) increasingata rate of$10100 peryear; (b) increasingat a rate of $800peryear 77. radius:*rin.; alritude:3ft in. 19. (a) p(r) :TY;15) R(r) -3600t:400x2.

(c)R,(r) : - aoo(r+ 9)(x 3). (d) 12oo (x * 3)2

23. (hut t7 ,zrz)stz f1

21,. 1000, $11

Exercises6.1 (Page248) 1. (a)3x2Ax * 3x(Ar)2+ (Lx)t; (b)3xzAx;(c)3x(Ar),+ (Ar)t

3. 1a) ffi

- fr; b) #;

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(Page 252) 1. 6(3x 1) (3x'- 2x + 1)2dx

3. ""(7x,* 9x)(2x* 3)-zrsn*

5 . ( x - 1 . ) - r r z ( x* l ) - s n O ,

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17. Lgf5_ 55t3+ 3g,

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t e . ( 2 t - 1 ) [ 1- 9 ( y z - t + A ) - u z f

2 1 , .6 t s - 4 t B- 2 t - 9 t ( t 2 + 1 ) r t z+ 6 t ( t z * l ) - u z Exercises 6.3

7.Ert+C

(Page 260)

3.*tt-t2+gt+C

1/ 1 \3/2 1 3 -. i ( l + i r l +c

L 3 5.F 5 x- p - ; +

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(Page 268)

l. o:2I5t-

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5 . * s e c ; 2 0 f t l s e c ; * s e c?;f t

13. r*sec;r#!it

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EXERCISES A.27 TO ODD-NUMBERED ANSWERS

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l. &

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Exercises7.2 (Page287)

2

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ZS. gzn_ +(3-rn) - #n - I

2 3 'n a - I n 3 - 3 n ' - t n

2 1 '+ 8 +

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7' 9 sq units

9' B tq units

1,1,.* tq units

13. tm(b' - a') sq units

(Page 295) 19. 38" sq units

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3. c : -4

tr(1+ \G)

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19.4\fr =

13.o

17.rr

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7. f drscontinuous at L

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(Page 320)

g. ?$m-

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units

9. trr'

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cu units

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11.#n cu units

13. 8zr cu units

15. #rr cu units

L7. #nAB crt units

19. tn cu units

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A-28


Exercises8,4 (Page344) 1. *tfgrg cu units

3. 8r3cu units

5. tabccu units

7. l9M in.8

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3. 12,000 in.-lb

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7. 256nw ft-lb

9. 100,000ft-lb

11. 5500ft-lb

13.Ztfr tt

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3. 64utlb

T. r+r! w lb

5. 2.2\atlb

9. 96w lb

L1,.1,L,250fiw lb

lg. 2SO\@w lb

Exercises 8.7 (Page355) 1. 4 44

20

3. 6 o

5. 171slugs;5.92in. from oneend

7. 25 slugs; 5.33 ft from left end

9. 16 slugs;#ft from one end

r

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1. (2,+)

s. (0,g)

7. (0,+)

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17.100,000p ft_lb

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7. (2,+, O)

9. (0,t, 0)

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25. 400 ft-lb

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(Page 390)

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1 3 .l n 4 s q u n i t s

1 5 .2 0 0 l0n 2 l b l f t z

17.ln x - -+

19.ln 1,6slugs; ;:

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ExercisesI .3 (Page 404) l. f-t(r) -ffx;

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EXERCISES A.29 ANSWERSTO ODD-NUMBERED 19. domain: (-@, +*); range:(-o, +-)

+m); range:(0, +-;

15. domain: [0, +m); range: [4, +-;

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9' 1'02

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(Page 427)

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ANSWEBSTO ODD-NUMBEREDEXERCISES

Exercises 10.3

(Page 45L)

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3 1 . y. : { 2 ( x + 1 - I n ) 33.5VEft Exercises 1.0.5 (Page 465) g.2T,4s",tog" 5.3x-y+z:0;x*3y-11:0 l.(a)*;(b)e,;(c)-i;(d)&t(e)t 9._+_+\Am ( + ( 1 13. *4n)n,*V2),wherezisanyeveninteger; (i0,+4n)r,-+{2),wherezisanyoddinteger;109.30, Exercises L0.5

(Page 470)

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(Page 477)

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s. sin-r2y*& 23.-er -#

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EXERCISES A-31 ANSWERSTO ODD.NUMBERED 31. L06",90",'!.'!.2",52"33. :|.rrhr; he walks 29. ln (2 + v3) 27' ?n * t[g- z 25. +89 23. * cos 2x - t cos 8x * C 4 3 . ( a ) 1 2 0t a d l h r ;( b ) 6 0 r a d / h r 4 1 .7 r 2 : 4 ( 5 s ) ( s 3 ) ; a : a G s ) 39. Ln - 1' 37. tfr - tn all the way 47. \m 4 5 .T : 2 n , P : 2 0 0 Exercises1L.2 (Page 497)

7.itan-tx(x2+1)-ix+C 5 . r s i n - lx + t / t = T + C 3.rtanr*lnlcosrl*C 1 .r ( l n x - 1 ) * C 15'*secsrtanr l z 1 t t ' * C ( 2 x ' ) c o s I +2rsinl*C 1 3 . tt.-*^/t=F-3(1 9.*e"(cosr*sinr)*C ) 2 5 '( e ' z + 1sq 21'$(g3tra*l) ZS'','-15 1 9 . i ( 3 e a + 1 ) t Z . & +g[secrtanr+lnlsecr+tanxl]+C a g ' * ( 1- g s - a h o i f t - l b 31' (4-Ir)wlb ZT.tzr(Je4*1)cuunits 29.2(l-e-6)slugs;ffin fromoneend

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+C

11.-bx(4x'-s)-ttz+ C 13'ln lr +2+ \trRl

1 / 2 ( 8-r f ) + C 9 . 3 2 s i n - l ( * x 1- + ( 1 6- x z )

tary- t) * a

L7. 6ln 2

-+tan-'1(#)

(#)

13*tan-'3x

*.

21,.*n

1 9 .l n r * + g l n i *

27.*k1n2s1ugs2g.*,.ffi_!,,,+2)_|-ft,u^_,,,*h

Exercises1"1-.6 (Page 518)

1 .t t a n - , ( * t a n l x ) - F c3 .- l n 1 1 - t a n l r l * c

e.!,u.,-,(.""14-f r(Sffi:lJ..,'.

s .* l n l t a nL x l - r t a n 2 l x r _ Cz . z t / l l " l f f i l . c

+ Vs) +c t3.trn3 ls.zVSh(r 17.2-..,(2+tan+x)

17.+\R h(l + +\r3) .7 (Page52L) 1.1. Exercises j.. *f,, - 3,x+tg\fr - 54ln(3+ \n) + C

'*{

3'"1ffi| *.

i

-

' '---'lVt*x-!21 5.-2\F*+\6nlffil

- l

*.

23.*ln2*En

A-32

ANSWERSTO ODD-NUMBERED EXERCISES

_3(x-2)rlo*,ln|1+(I_2)1/s|+cg.2\/2x+2\tr+4+4!2|n|W|-, ' x + f f3{x i+C L C 11,.3 tan-r ffi, -L

13 \m1) 1 3 .)2 ttan-r(r a n - l (*x *\f-

Cx

+C

21,.+(54- 2{3)

Le.ln #

1 , 74 . -2ln 3

x*2+z\Gi+It Cx x*2-2V1,*x*x2


l. approx: 0.695;exact:ln 2 : 0.693 3. approx: 4.250; exact:4 5. approx:0.880; exact: ln(1 * !2) : 0.881 T. approx:1.954; exact:2 9. 0.248 11. 1.481, 13. 3.694 15. -0.007 s €r < -0.001 1T. -0.5 = €r < 0 L9. -#e s €r 21..0.882 23. 3.090 ExercisesL1.9 (Page 531) 1. by Simpson's rule: !r; exact:g - 0.6044, 7. e,:Q

*t{5

3. by Simpson,s rule: 0.gg1;exact: ln(l + fD) - O.SS1 t. -

#=

€, = 0

11. 0.248

13. 0.883

5. by Simpson's rule: 0.G045;exact:

15.1,.402

17. nrzh

19. inhr(rr, * rzz* r{z)

21,.(a)-0.0952;(b) -0.0950;(c)-0.0991 Reaiew Exercisesfor Chapter 11. (page 532) g.-2\/T=7+c 1.*r-rzlssinL6r*c 5 . ( r * r ) t a n - r. y ' i - . V v + c z.ir**sin3r*C e . l n l r - 1 l - 2 ( x- 1 ) - ' | -rl3 I - (r-t;-,-t 6 1 1 .* s i n 2 x - i s i n 4 x * C 13 st" +c 1 5 .* t a n 3 r - * c o t 3 x * 3 l n ltan lxl + C l1f;ml 17. 2 t + h # - # + c

L sr.- t a n -x, * * r nl - l

25. t r 2 s i n x 2 + { c o s x 2 + c

+ c 21,.#r - r# sin 12x - r*a sins 6x * C lr * 1l 2 7 . h e t r z ( 4 s i n2 t + c o s 2 f ) + c 2 9 .* tutt-t (* sin2x) + C

trl .2 1 _ \ s r . . f ; \ - " o " n r * j c o s r z r - t c o s 5n x ) + C i f n # 0 [c irn:o 3 5 .Z L nl v - 2 l - 8 ( V- 2 ) - , - t e - z ) - 2+ C 'l-

4t.itn

|

lt

- At l + C

4 3 *. s i n - , ( * e +" 1c

49.tan-r(cos r) + C

s r.# \ / z + f f i O l ; _ t-4) + c

$. -*cscrcotx-*cscrcotx*gln lcscx-cotr+ ] c

3 7 .- t a n - r ( c o sx ) + C

s s . 2s i n - ,( + ) * l U - z ) ( 4 t -

45.-+ cots3x- $ cots3x* C

51.B r".-' lz sin3fl + C

2 3s. i n - r G ) + c

53.-O'e,triT+

C

47.*r3 sin-t x * te, + 2)\R

ss. \Ei_ \Fn

+C

sin-r{n+ C

- -x - 1 ) +, ;\ n t a n -(.1l/M

,, +

t 2 ) u+ 2c

- ,@ r*ci . * '.== + c ti ft n ** - -r ,6. n 1 'f.x ", *+lr , l y 6g. 6 5 .t + z l n g 6 7 .+ o - E \ / ' 6 s .# \ t r - t n I ifn--'!, [*tn'x*C 75.a'(tn - 8\6) 77.* ln B- tn 7s. s 8 1 . t + t n g 8 3 . + 8 s .l , - + l n 3 87.hr 91..* ln I e3. +f es. (a)1.624;(b) 1.s63 e 7 .e \ n - 3 \ 6 * g h Q \ m + 3 \ 6 - 4 \ n - 6 )

\D

+ttan-,(tffi+1) 7 1 ,+.

7 3 .+ _ + t n 2

89. \ E - t l n ( 2 + 99. (#,0,0)

103.7:t(#);a:+(#)105.(a)x:300(m);@)35.g4|b ExercisesL2.1 (PageS4l,) i?'

,r.

4

S

,r4x*l

sec.rt'

5 3 1 . i l n ' c o s hx * C

19. e"(cosh r * sinh r)

21..Z cschZt

33.#zsinh Z8x*#sinh I4x**r*C'

3 9 .* c o s h 3 2 - c o s h z + 3

4 g .a 2s i n h T"runits

23. 2x sechx2

+c

2s. rsinhr-r(x cosh r ln lrl + sinh x)

35. t tanh 3x - $ tanh' 3x * c 37. 2 cosh tfx + c -cttzl(B 4s.7r:e tcA) sinh t + (A - tcB) cosh fl;

\6) 1 o t .s u ) t b

EXERCISES A-33 ANSWERSTO ODD-NUMBERED n : s-ct,rl(A _ cB * IczA) sinh t+ (B - cA + Ic'B) cosh

a : K f * K r a ,w h e r eK r : 1 - + c 2 a n d K z : - c

L2.2 (Page547) Exercises 13.* ln 3

11.tn(* + +\n)

4'^|

1 F l

L''

21..Z* (rorh-r x2+4\

L/'

\

2x*3x2

1,-16x2

\/xa -

23. lsecrl

I/

25.-csc x cotx/lcotrl L2.3 (Page551) Exercises 1. sinh-lLx+C:tn*(r+

3. !cosh-rx2+C:}ln1r'?* t/Ft)+C

{F+1)+C

x)+C x+r'---coi' 7 sinh-' T++c:,n(sin

' l_t::ll-l,gli: I |;|:;]:*!" lffil *.

*J:-1=m +c ntx+2t' rn3 rs.}r.,'# r(*@uu) 13. ', [+tanh-'ffi 1{++?+r,l+c,,. l{o-z-'t' *J'&^ irlx+21' +c ffi l+cotn-' ReoiewExercises for Chapter12(Page552\ 5. 5 cosh,2x sinh2x z.

11.ln coshix'+ c

t 9. 2w sinh-L2w #

ffiffi

1 9 '+ t

1 5 .t a n h l '

1 3 .* c o s h r s i n r - * s i n h r c o s r * C

Exercises13.1 (Page559) s. (a) (-rD,*z); (b) $/1,-in); 3. (a)(-2,*a'); (b) (2,-8tr);(c) (-2,-it) 1. (a)(-4,*a); (b) (4,-t,n);(c) (-a,-*z') 1 3 .( a )( - 3 , 0 ) ; ( b ) ( - 1 , - 1 ) ; 1 1 .( - 2 , i f l ; Q , I z r ) z. (3,tr); (-3, *zr) s. (-4,-&d; @,-tr) G)?11,-Zd Le'r2:4cos20 t5'r--lal ( c ) ( 2 , - 2 { 5 ) ; ( d ) ( 0 , 2 )(;d ? 1 2 , { 1 ) ; $ ) ( + \ / 3 , - + ) "'':r}* -3q stn 20 29.4x2-5y'-36y- 36:0 2g' (x' * y')'-- 4xA 25' (x' * y')t xz 27. (x' * y')' - 4(x' y') 21'.r: fu ExercisesL3.2 (Page 566) 41.0:3n,0-tn

43.0:ir,0-*n

45.-L

13.3 (Page570) Exercises 7. (1',&n);(1,it); (i,tn); (i,3,,); pole; (0.22,2.47); 5. (fu,tzt); (-*n,-ir) 3. pole;(t/7,*d i,. (t,*t); (t,Et) 1 1 .p o l e ;( V T 5 , c o s - ' i l ; ( ! 1 5 , r - cos-r +) ( 0 . 2 2 , g . 8 2 ) 9 . p o l e ;G l r , t Q n + \ ) z r ) , w h e r e n : 0 , 1 , . . , 7 ( i l 2 ' t Q n + 1 ) z r ) , w h e r e z:0,1',2,3,4,5 t z . 1 d , i d ; $ , E z r ) ; ( 2 , 8 ' . ) ; Q , J * r1 ) 5 .p o l e ; 13.4 (Page573) Exercises '!,53" 5. 38"g' 26' (-L\E at 7go 6' , En) G\8, tn);

7. trr

3.

Exercises13.5 t. *r 19. 4

5.4 3.4n '1.6a2n3 21..

c o sd - 8 s i n 0

15. r:ffi

Trz

11' *n

f . irr

13. 0 at pole; tn at (1, n); in at (1, 0)

15. 0oat (0, tn); 79"6' at

(P age 576) 7. &n

Reaiew Exercisesfor Chapter 73 t. r:9

7- En

g. *t-1lsin-'

+-g\/-2

tt. *n-t*15

t3.9zr-9

15.a2(2-i',)

L 7 .i @ + 1 , )

(Page 576)

3. 4xa*8x'y'*Ayn*36x3*36xy2-8'l'y2-g 17. no points of intersection

25. (a) 4r sin 0 - 12 cosz 0 : 4 ; ( b ) x ' : 4 y - 4

19. tn, tn, Ln '

9.0-irr

1 1 .a ' ( + n* + * )

27. 16n- 24{5; 32n + 24{3

13.

g4kr -

4k

1

A.34


Exercises1.4.1 (Page 582) 3. (-2,0); x:2; 8 1 . ( 0 ,1 ) ;Y : - 1 ; 4 19. x2: -A 21. 1,56ft 15. y' : -6x Exercises14.2

s . ( 0 ,- + ) ; y : * ; 1 7 . ( 8 ,O ; x : - * ; g 9 . A z: 2 0 x 1 1 . x 2: - 8 y '1.0y- g 23.*?in. * 20:0 25.yz- 10x- 1,0y 27. xz + y' +

13.

(Page 587)

l . x ' 2l y ' 2 : 1 3 3.y'':6x' 5.x'2*4y'2:4 7.y':27't 9 . x ' 2 * 4 y ' , : 1 6 l ' ! . . 3 x ' 2 - 2 y ' " : 6 7 3 .( - 3 , * ) ; (-3,-*); x:-3;y:l ( 1 5 .( 1 , - 5 ) ; ( - i , - s 1 ; y : - 5 ; x : t 1 7 . t , 1 ) ;( i a , t 7 ; y : 1 ; x : 3 t Ie.y2+20x-By-24:0 2 1 .*.+ 2 x - B y + 4 1 : 0 2 3 .* - 6 x - 6 y - 3 : 0 ; * - 6 x - t 6 y + 2 1 , : 0 25.y"-4x-4y-72:0 27.y"-2x-By+28:0; y2-tlx-By+1,6e:0 *49:0

,t (-*,9#)

3 3 .x ' 3 - A ' ' : 0

Exercises 1,4.3 (Page 591) 2 4 - 2 4 x - y ' + a y- 4 : 0 1 .3 x z x * U ' : 0 3. 8x2 *

1tz

a r +W : 1 ;

f -2x+1.6y

z r .* - 2 x * 4 y * t : 0 ; * - 2 x - r 6 y - z e : 0 ; * - 2 x - 4 y - 3 1 : 0 ;

x2

e > t,

5.

'/'.6x2 * 4xy * lgyz - l52x + t1,6y* 496: 0

9.

ttz

or- W:t

Exercises 14.4 (Page598) 5. (a) 1; (b) parabola; (c) r cos g - -2 (b) hyperbola; (c) 2r sin d - -3

7. (a) i; (b) ellipse; (c) r sin 0 : S

13. (a) 4; (b) ellipse; (c) r sin 0:

9. (a) ?; (b) ellipse; (c) r cos 0:

-j

11. (a)

-S

1g.#6,squnits2L,(a)r:ffi,q@)20,000,000miles

)? -v"

a2(1-ez) l - e2 cosz 0

rz-

Exercises 14.5 (Page505) 1. vertices:(-f3,0);foci: (-r1,f,0);directrices:x:=E{-5;t:trt/-S;endsof minoraxis: (0,-+2) 3. vertices:(-F3,0); toci:(-r1/j, directrices:x:-+313; e: *{3; ends of minor axis: 10,-rV6y 5. vertices:(+2,0); foci: (+y'i}, 0); directrices:x:-rt*.V-l}; t:*\/tZ;2a:4;2b:6 7 . v e r t i c e s(:l * , 0 ) ; f o c i : ( - r 1 t r , 0 ) ; d i r e c t r i c e s : r : : L r ts : f l ; 2 a : | ; 2 b : * g.gf-4y2:96 ll. 16f *25y2:1gg 13.32f -33y2- 380:0 15.2x]-3y-12:0 tZ. Zxz-4yz:29 19.76*ft 2.t. 98r in.r 23. the right branch of the hyperbola l6f -9y2: 111400 25. *rabz 27. 2400w ft-lb Exercises 14.6

(Page 513)

t. e : *16; center (2, 3); foci: (2 t \8, y - + ! +\f170

,u. k 64 -,r)' * Exercises L4.7

directrices: x:2'+313

5. point-circle (-8, +1

(y1-s)'_ | 39

7 .s r + 4 y 2 : s o o, .

17..tlnwtb

3. e:*t/-10; cenrer:(0,*); foci: (0,+*lnt/tm);

" *n''

*QtzY -1

direcrrices:

(tirt)':,

1r# .*

(Page 520)

1 . .e : 1 3 ; c e n t e r (: - 3 , - 1 ) ; f o c i : ( - 3 , - 1 * 3 { 6 ) ; d i r e c t r i c e s :y : - t * & t / - e ; a s y m p t o t e s : - + x + { 2 y + 1 2 * S : O g. s:ffi;center: (-l,a); foci: (-L,-3), (-1, 11); directriceszy:0, y:8; asymptotes2-+2x+ {-Zy*Z-+.y'l:0 5. ,:+t/tg;center: (1, -2);toci: (t,-z'+ Vi3); directrices2 y:-2t+t{B; asymptotes:3x*2y + 1:0, 3x-2y -7:0 7. 25x2 t44y2:14am n

Q+1)' 1,44

_ _( x: * 2 ) z g1

4 r

17. (3, 4)

Exercises14.8 (Page 625) 3. 16y' - 972:36 5. 7z- A2: 16 Reaiew Exercisesfor Chapter 14 1,.9(x - 1)' + 5(y - 2)' : 405

7. 972+ 4A' :36

9. 3i'2

18

1,1,.7'2+ 4y'' : 16

13. 7'z - 4y'' : 15

(Page 625)

3. 3 x 2- y 2- 3 : 0

5. 49(x+ 9a)2+ 76(V- 3)2:576

7. ' l

-

-

8 | - 3 c o s0

ANSWERSTO ODD.NUMBEREDEXERCISES A.35 15. e - +\re; center: (-3 , 8); foci: 13. (a) e: g' (b) hyperbola;(c) 3r cos g - 4 + 1 I zbffi; asymptotes: \m, 3); directrices:r 1 7 . e t B ; center:(*1,3); foci: (- t

11. (a) e - +'(b) ellipse;(c) r sin 0: -2 ( - 3 , 8 + 2 \ 6 ) ; d i r e c t r i c e sA: : 8 1 8 \ 6

7 9 .2 1 ' x -' 2 4 9 y ' z : 7 2

*8_-0

5r * y +2:0,5x-y

31. (3 I *\ft) million miles

29. 600miles

L5.1 (Page 635) Exercises 7.2 5.7 1.+ 3.tn 75.2 (Page641) Exercises 7.+ s.t 3.0 1.0

11.+

9.1

25. (x - 5)' : 8(V - 1)

23. 4tf3na2b

ft-e

27. 9n in.

'l',440,000 33. the left branch of the hyperbola 16xz- 9y' :

1 3 .l n 3

11.-t

9.fo

1

21,.r :

13.1

lg. ez

17. e,,

15. ez

23. +

21,. e-rt\

25-Y

23.-1

21.2

ls.g

17.1

ls.+

L V I. L

31. 2

27. 1

25. 0

15.3 (Page 646) Exercises 1.1

21,.Ln

T.divergent

5.1

a.-ln15

g.2

11.1

19'tr

13.divergent 15.(a)divergent;(b)0 17.(a)0

23.n:g; ? ln #

ExercisesL5.4 (Page 650) 1. 2

2t. (a) divergent; (b) 0

23. n > _t,

#

25' n

\\

-t.

''

1'3. 0

11. divergent

9. divergent

7. divergent

5. divergent

3. divergent

L9. *rr

17.0

L5. divergent

)

-

(, + 1)t

Exercises75.5 (Page656) E @ - t r ) a ,{ b e t w e e n } z a n d r 3 . P n ( r ) : x * t x 3 ; 1 .p 3 ( r ) : t + i t / j ( x - * z ) - i ( r - * z ) ' - # l i ( r - * z ) ' ; & ( r ) : # s i n 1 ) ' + + ( t - l ) 3 ; R r ( r ) : - + t - 4 ( x - l ) n , t b e t w e e1n' a n d x P s ( r ) : x 1 . * ( x 5 . R o ( r ) : r , r o c o s h f rf;b, e t w e e n 0 a n d r *seC (\(x-*t)a' f between*zandr_ p s ( x ) : l n 2 t / i ( x 1 " i l 2 ( x i n ) " + V g ( r * z ) 3 ; R s ( r ) : a r e s e C {tan2f 7. 9.ps(x):t+8x+*x,-*ri;&(r):a*E(1 lerrorf< a+"6\/2

2L.x:

* { ) - s r z y , ( b e t w e e n 0 a n d r 1 ' 1 . 2 . 7 7 8 2 81 3 . 0 . 5 1 5 1 5 . 0 ' 1 8 2 3 t t ' f f i

#6

ReoiewExercises for Chapter75 (Page657) 11. 1 9. 0 Z. | 5. +3. + 1. +o

13.e

25.PuQ):1-ixz-rtaxn-ttrl;&(r):o-#osinfrT 37. $152,500 39. (b) 0 35. # sq units

17.+

15. divergent

19. divergent

23. divergent

27.tr

27.f'(0):+

2 9 ' ( a ) n o ; ( b ) 0 3 1 ' l R 3 ( +<) lr l z - 0 ' 0 0 5

15. divergent

17. erts

ExercisesL6.7 (Page666) 5. +

7. divergent

9. 0

11. 1

13. divergent

'l'9-

1

ExercisesL6.2 (Page673) f . increasing

3. not monotonic

Exercises 16.3

(Page 683)

NL

1. sn-ffi;

,

3. sn: ln

f. increasing

7. not monotonic

+6

s.sn:ffi,t

7. n:l

17.

13. divergent

15. 2

ExercisesL6.4

(Page 693)

1. convergent 15. convergent

5. decreasing

5. divergent 3. convergent 17. convergent

21,.divergent

7. convergent

23. g

11. decreasing

2 ( 3 n - 2 ) ( 3 n+ 1 ) ' 3

25. divergent

9. divergent

27. +++

11. convergent

1+@1

e' ' . )+ - \ZiJ: 0) n ' n:2

29.

-

jt&

13. divergent

33'*o

A-36


Exercises 15.5 (Page597) 1. divergent

3. convergent

5. convergent

7. convergent

9. convergent

11. divergent

(Page 706)

Exercises 16.6

3. convergent

1. convergent

5. convergent

7. convergent

19. absolutely convergent


9.

l R o
21,. divergent

1 1 .l R n
1 3 .0 . 1 1 3

15.0.406 25. absolutely convergent


27. divergent

Exercises L6.7 (Page712) I

1. (--, +€)

3. l-+, +l

5. 0

g.(0,21

7. ( - s , g )

11.(-

13. [-1, 1]

15.

e)

1 7 . 1 4 ,6 )

19. [-1, 1] ExercisesL6.8 (Page 721)

1.(a)r - t; l-r, tl; (b)y, +; r:1; (c)[ - 1 ,

3 . ( a )r - t ; l - 1 , t ) ; ( b ) )

1)

=t

+oo

t/nx"-r; r -'1,; (c) ( - 1 , 1 )

5. (a)r:

(b)\ (-1)"-' lrf=\ilo ,--, +@)

r - 3; (c) (-2, 4)

*m);

-

l)Yn-z

n:2

+@

(b)i#

11.

v 2

n:o

+@

(-1)n

1,9. (a) n:0

r. 0.4854

(--,

+00

e.i)n(n

n:l

Exercises16.9

*oi

n-l

^.2n-2

+@

*u n 1 2 i n!

2 1 ,Lt.t.

yn nl

n:O

(Page 728)

3. 0.7468

7. 1.3179 g. 0.2450

5. 0.2483

+@

ls. s

11,.$ 4

,-2" 2n * |

.Lt n:O

ExercisesL6.10 (Page 737) +co

n:1

R- 4

t

nTn

tr. z+ *(x- 4)+2 t

2.4.6.

n:z

13.+- +\B(x- *n) - *(x- *n),+ +fi(*-

t"r)' + # ( x - * n 1 + -

t7. (a) r * *rt + i"x'; (b) 1 * x2* &xn;(c) tx, * #xn* *ru 2 9 .0 . 2 3 9 8 3 1 . a 4 : 3 ; a s - - 5 ; a z : 2 ; a , l; as: 6

. ' (2n-3)(x- 4)".

( - 1 ) n - r ' t "3 ' 5 '

25. 0.0415

27. 0.0048

Exercises1.6.11. (Page 742) (2n

1 .1 +5 1

+co'-]):f-.2 3' 2t2,n*, 'b

1

5. 1 +

n:o

7. r+ 5 nrrir*

(-1)n

(2n - 1,)

' "1. 3

Xn+z; 1

2"n!

15.0.s082

ExercisesforChapterL6 (Page742)

1' 1' t' t' z; s

3. 0' +, t, i+ 1

97. l-3,g1

39.x:3

7. convergent; g

5. 1.,3, 1,,3; no limit

15' convergent


s 3i.)

5

n=l

)czn+r 9 . 4 . 8 9 8 9 1 1 . 5 . 0 1 0 1 3 . 0.3351

13. convergent; {

17. divergent

29. conditionally

4r. (-7,5)

(-1)"-'ffiv2n,' (-oo +@)

19. convergent

convergent

31. divergent

9, divergent 27. divergent

11. convergent;4 -l 2{! 23. convergent


43.0.1973 45.5.0658 47.O.ss86 49.0.0124 Sr. T +o

55.

("v

\ '7

-''

I

n)zn-r

(2n - 1')l

'

';R=*m 1s.;2W

21. 1,.97435 23. -0.223L

19. 0.5299

(Zn).4n

25. divergent 35. [-1,

(1"1)"

1)

rr; (-o.*o)

TO ODD-NUMBERED EXERCISES A-37 ANSWERS Exercises 1,7,1 (Page7 50) L g . ( 1 ,- 5 ) 1 7. ( - 1 , g ) 1 5 . ( 1 2 ,- 5 ) (5, 6) g. 13. (-4, 3) 1,1,. 9. (-2, -7> 7. (2, -3> 2 5. \n 1. 5 35. \M 33. \m 3 1. ( - 2 , 7 > 27. ({2, V-g) 2e. (a) ( 1 , - z ) ; ( b ) ( 1 , - 2 ) 21.(-2, 2> 23. (7, 3> 25. (-9, -4> 17.2 (Page755) Exercises 3. -1,4i+21.i s. z{fr

1. 6i+2i

(b)-+i + L{zi

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3 . filef|a,

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13. 12

Exercises17.7 (Page 791) 5. (a) cos fi * seczti; (b) -sin fi + 2 sec2t tan ti; 3. (a) i * tan ti; @) seczfj; (c) t/2; G) 2 1. (a)Zti+ i; (b) 2i; (c) tE; (d) 2 (c)2; (d)2 (a) i + {3i; (b) -fsi+i; 1'1.. s. (a) 3i; @)-ai; (c) 3; (d) 4 7. @)2i + 6i; @) 2i; G) 2t/t0; (d) 2 (c) * {%; (d) + {W 3ea,625 ,#r*#i sec; (20,000t/5 + 800\m93) ft 1,7.(zs + {ffi) 12s0\nj (c) 1250{2iru(b) 1s.(a)

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17.9 (Page 803) Exercises

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1 . ( b )( 7 , 2 , 0 ) ,( 0 , 0 , 3 ) ,( 0 , 2 , 0 ) ,( 0 , 2 , 3 ) ,( 7 , 0 , 3 ' )(,7 , 0 , 0 \ ; G )@ . 3 . ( b )( 2 , 1 , 2 ) ,( - L , 3 , 2 ) , ( - r , r , 5 ) , ( 2 , s , 2 ) ,( - 1 , , s , 5 ) , (2,r,5);G){E 5 . b \ 3 f r ; ( c ) ( 0 , 0 , 0 ) ; ( 1 s , t 8 , 1 2 )( ,1 s , 0 , 0 ) , ( 1 5 , 1 8 , 0 ) , ( 0 , 1 8 , 0 ) , ( 0 , 1 8 , 1 2 ) , ( o , o , 1 2 ) , ( t 5 , 0 , L 2 ) 7 . ( a ) t ; ( b )( * , - 1 , 2 ) 9 . ( a ) E ; ( b )( * , + , + ) l r . f t a t f t , a , r ' , 1 7 .s p h e r e w i t h c e n taetr( 4 , - 2 , - l ) a n d r : 5 1 9 .t h e point (0,0,3) 21. the emptyset 23.f + (y - l)'t (z * 4')2: r2,lrl > 0 Exercises18.2 (Page824) z. <21,-rs,-21

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TO ODD.NUMBERED EXERCISES 4.39 ANSWERS Exercises18.3 (Page 828) 5. -44 25. g

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9.8x-y -66:0;l1x-52-702:0;l3y-402* 42:0

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9.(+i,a\,-*3);(-rt,-fr,+3)

13.5r-3y*72+14:015.2x-y-z+L:0 17.4v-32-r:0andz:1

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1 1 . x 2+ 4 y ' * 4 2 2 : " l ' 6

13. y'+ z2: sinzx

15.xz- z2:4; z axrs

17. z:

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y axis

(Page 864)

7. elliptic paraboloid 5. elliptic cone 3. elliptic hyperboloid of one sheet 1. ellipsoid < 17. *tabc 15. *$z < (b) 1. < (a) rD; 13. 1 two sheets of lkl lkl 11. elliptic hyperboloid

9. hyperbolic paraboloid

Exercises 18.9 (Page 871)

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lv(l)l: {trVT@

27.2

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A.4O

ANSWERS TO ODD.NUMBERED EXERCISES

2 L .< 6 0 , - 4 0 , 8 0 ) 2 3 . 2 9 s 2 5 .( r + 2 ) r + ( y + 1 ) r + ( z - 3 ) 2 : 1 7

3s.*\fr sr.:: L:1 ,: 4 _3 I;

4t, y : -3t, z: t

39.24

27.3

2s.+t\/,

3 1 .r - 6 v - t o z + 2 3 : 0

3 3 .3

41.v(tr)-_l2ni+i+k;A(tn):-2i-tni;|v(tn)|:*\ffi

43. (a) z : r2(7 * sin 20) * l; (b) rr(25 cos20 * 4 sine 0) = 1gg Exercises 19.1 (Page 887) '1....f *uz .f+2w+u2 r' \a)-V;folV;:f7;(c)7-AT;;

( d ) 0 ; ( e ) t h e s e t o f a l l p o i n t s ( x , y ) i n R 2 e x c e p t t h o s e o n t h e l i n ex : y ; ( f )

(-.,+-)

3. domain: set of all points (r, y) in R'interior to and on the circumferenceof the circle I + yz:21except those on the line x:0; range: (-rc, 1o) 5. domain: set of all points (r, y) in .* interior to the circle f + yz:25 and all points on the y axis except (0, 5) and (0, -5); range: (--, *-) 7' domain: set of all points (r, y) in R'zexcept those on the line r : y; rantei (-€, +a) 9. domain: set of all points (r, y) in R'zexcept those on the r axis; range: (--, +.) 11. domain: set of all points (r, y) in Rz for which ry ) 1.;range: (--, *-) 13. domain: set of all points (r, y, z) in R3for which lrl - I and lyl = u nnge: (-r,2r) 15. domain: set of all points (x, y) in R ; range: [0, +o) X7. domain: set of all points (x, y) in Rr; range: (-@, ld] 19. domain: set of all points (/, y) in R'zfor which x + y, = 10; range: [0, +-; 27. h(x, /) - sin-r \/f=F=T; domain: set of all points (r, y) in R2interior to and on the circle f + yz:1 29, (a) 2; (b) 6; G) {V - yr; (dl lx - yl; (e\ lx - yl

Exercises 19.2 PageBgg) 1. 6: *e 3 . 6 : m i n ( l ,* e ) 5 . 6 - m i n ( l ,* e ) 1 7 .0 27. limit doesnot exist 29. limit existsand equals0 Exercises 19.3

lg. Z

Zl.In

23. *

25. limit existsand equals0

(Page 904)

1' continuous at every point in Rl 3. continuous at every point (r, il * Q,0) in R' 5. continuous at every point in R2 9. all points (x, y) in R2 which are interior to the circle * + yz:15 11. all points (x, y) inR, which are exterior to the ellipse 4*+9yz:36 L 3 ' a l l p o i n t s( x , y ) i n R 2 w h i c h a r e i n e i t h e r t h e f i r s t o r t h i r d q u a d r a n t 1 5 . a l l p o i n t s( x , y ) i n R z f o r w h i c h 17. all points in P lxyl > 7 19. removable; /(0,0) : O 21. essential 23. continuous at every point (r, y, z) in Rs for which rz * y2 * zz ) 1. 25. continuous at all points in Rl Exercises 19.4

1.6

(Page 912)

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t.#(rt"i-.) (b) 1

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33. 4

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5000ln 1.06- -57 ..

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25.-ln sin x; ln sin y

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sr(")# l##' -'], tu)0,#36 -'] - -24 4; t#e

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ANSWERS TO ODD-NUMBERED EXERCISES A.41 (LY1

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(Page 951)

1. 2{Lx + stfzy

13.(a)(-12,2,'!.4>;(b)+

17.#n * ?;*ffi

25.(a)directior,of nfur-fri;{b)

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11.f(x,y):x2cosy-x+C

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Index Abscissa, 22 Absolute extrema of functions, 185, 956-966 on a closed interval, applications in- . volving, 189-795 problems involving, 213-219 Absolute maximum value, 185 of functions of two variables,956 Absolute minimum value, L85 of functions of two variables,957 Absolute value, t4-21 Absolutely convergent infinite series, 70L Acceleration, instantaneous, 158, 786 normal component of, 805 tangential component of, 805 Acceleration vector, 786, 870 Accumulation point, 893 Addition,4 of vectors, 748, 820 Additive identity, existence of for vectors, 75'1.,822 Additive inverse, 5 Algebraic function, 60 differentiation of, 130-137 Altemating series, 697 Alternating-series test, 698 Analytic geometry, 21,-55 Angle, 431 between two curves, 465 between two lines, 462-463 between two planes, 832 between two vectors, 757, 826 direction of a vector, 818 of inclinatron, 46'1. Antiderivative, 253 253 -260 Antidifferentiation, application of to economics, 269-272 chain rule for, 257-258 and rectilinear motion, 265-268 Apollonius, 588 Arc, of a curye in R3, 868 length of ,779-784 as a parameter,794-795 of a plane curye, 372-378 rectifiable, 374 Archimedes, 31L spiral of , 565 Area, 281-287 of'a region in a plane, 324-329

of a region in polar coordinates, 573-576 of a surface, 1028-1033 Associative laws, for real numbers, 4 for vectors, 751,-752, 822 Asymptotes, of a hyperb oLa, 597, 614-615 horizontal, 171-173 vertical , 172-173 Auxiliary rectangle of a hyperb ola, 615 Average concept of variation, 230 Average cost cuwe,232 Average cost function , 230 Average value of a function, 309-310 Axes, coordinate, 22 rotation of, 620-625 translation of, 583-587 Axiom, 4 of completeness , 9, 669-670 of order, 6 Axis, conjugate, of a hyperbola, 604 major, of an ellipse, 503 minor, of an ellipse, 603 polar, 555 principal, of a conic, 590 of a parabola,579 of revolution, 330 of surface of revolution, 855 of symmetry,363 transverse, of a hyperbola, 604 Bases for Bernoulli, Binomial Binormal Boundary Bounded

vector space,754, 824 Johann, 389 series, 738-742 vector, 869 conditions, 262 sequences/ 667-673

Calculus, fundamental theorem of, 31,1,-319 of vector-valued functions, 772-778 Cardioid,563 Cartesian coordinates, rectangulat, 22 three- dimensio nal, 812 Cartesian equations, 558, 764 of the conics, 599-605 of a plane, 830 in three-dimensional space, 855 Catenary, 540 Cauchy, Augustin L., 631

Cauchy-Riemann equations, 941 Cauchy's mean-value theorem, 631,-632 Center, of a circle, 43 of a conic, 591 of curvature, 804 of mass, 352, 1016-1018 of a plane region, 356-364 of a rod, 351-355 of a solid of revolution, 366-371, of a sphere, 815 Center-radius form, of an equation of a ctrcle, 44 of an equation of a sphere, 816 Central conics, 59L Central quadrics, 859 Centroi d, 360 Chain rule, 138-1,41 for antidifferentiation, 257 -258 general, 926-932 Circle, 43-48, 588 center of, 43 of curvature, 800 degenerate, 607, 609 equation of, center-radius form, 44 general f.orm, 44 osculating, 800 point-, 607 radius of., 43 Circular-disk method for volulr€, 330-332 Circular helix, 866 Circular-ring method for volume, 332-335 Cissoid, 567 Closed ball, 889 Closed disk, 890 Closed interval, 10 absolute extremum on, 1'89-t95 Closed rectangle, "1,002 Closure Law, 4 Commodities, complementary, 968 substitute, 968 Common logarithms, table of , A-14-A-15 Commutative laws, for real numbers, 4 for vectors, 751-752,757, 822, 825 Comparison test, 685-687 Complementary commodities, 968 Completeness, axiom of , 669-670 Components of a vector, 746,76'1.,8'l'8, 826 Composite function, 58, 884 derivative of, 1.38-742 A-45

I

I A-46

INDEX

Compound interest, 424-426 Concavity, 220-225 applications of to drawing a sketch of the graph of a function,227-229 downward,22'l' upward,221. Conchoid of Nicomedes, 567 Conditionally convergent infinite series, 70L Cone, elliptic, 852 generator of, 588 nappe of, 588 right-circular, 857 vertex of, 588 Conic sections, 579-627 Conics, cartesian equations of ,599-505 center of., 591 central, 591 degenerate cases of, 588-589 directrix of.,579 eccentricity of , 589 focus of, 579 polar equations of, 592-598 principal axis of, 590 properties of, 588-591. vertices of, 590 Conjugate axis of a hYPerbola, 504 Conjugate hyperbolas, 615 Conservative force field, 993 Constant, derivative of, 13L Constant function, 59 Constant of integration , 3'l'6 Constant terms, infinite series of , 673-683 Constant times a function, derivative of, L32 Constrained extrema, 963 Constraint,963 Continuity, 97-107 126-130, 9'/..6 and differentiability, of a function , 97 -107 on an interval, 177-180 left-han d, 178 at a number, 97-l0l on an open ball, 903 right-hand, 178 theorems on, L0l-107 of a function of more than one variable, 900-904 of a vector-valued function , 772 920 Continuous differentiability, Continuous function, 97, 88L ' Contour curve of a function, 886 Contour map, 886 Convergence, interval of , 7ll radius of ,71'1. Convergent improper integtal, 643, 648 Convergent infinite series, 575 Convergent sequence, 654, 670-672 Coordinate axes, 22 Coordinates, cylindrical, 872-87 6

and triple integrals, 1039-1042 left-handed, 811 polar, 555-577 and double integrals, 1022-1'026 rectangular cartesian, 22, 812 spherical, 872-876 and triple integrals, L043-1045 Cosecantfuncti on, 452-450 derivative of, 456 integrals involving powers of, 458-470 inverse,475 derivative of, 481 Cosine function, 431,-437 derivative of, 438-445 integrals involving powers of, 447-451' integration of rational function of, 516-518 inverse, 472 derivative of , 478 Cosines,direction, 819 Cost function, average,230 joint,97"l, margin al, 23!, 971 marginal average, 23'1. total, 230 Cotangent function, 452-460 derivative of, 455 integrals involving powers of, 468-470 inverse, 474-475 derivative of, 479-480 Critical number, 184 Critical point, 958 Cross product of vectors, 842-851' Cross section of a surface in a plane, 854 Cubic function, 60 Curyaturc,796-803,858 center of.,804 circle of, 800 radius of, 800 vector, 799, 868 Curve(s),23 angle between two, 455 equipotential, 886 ' generating, 855 plane, length of arc of.,372-378 in R3, 864-87L smooth, sectionally, 985 Cycloid,769 Cylinder, 341.,852-854 directrix of , 852 elliptic, 853 generator of, 852 height of,34l hyperbolic, 853 parabolic, 853 of revolution, 855 right,34'1. right-circular, 341 ruling of,852 as a solid, 341

as a surface,852 Cylindrical coordinates, 872-875 and triple integrals, 1039-1042 Cylindrical partition, 1035 Cylindrical-shell method for volumes, 336-339 Decay, and growth, laws of,420-426 natural, law of, 42'1. Decimals,nonrepeatin9,6 nonterminating, 5 repeatin9, 6 terminatrng, 5 Decreasingfunction, 205 Decreasingsequence,667 Definite integr al, 276-322 applications of, 324-380 definition of, 291,294 properties of, 296-305 Degeneratecasesof the conic sections, 588-589,620,624 Degeneratecircle, 607,509 Degenerateellipse, 588-589,620,624 Degeneratehyperbola, 589, 620,624 Degenerateparabola, 620,624 Del (v),947 Delta (A) notation, 29 Delta (6), 67 Demand, marginal, 969-971, Demand curve,235 Demand equation,234 Demand surface,968 Density, linearr 353 Dependent variable(s) , 49, 883 Derivative(s), L2L-L63 applications of, 204-242 to economics, 230-238 of a composite function, 138-142 of a constant, 131 of a constant times a function, 132 of cos u,438-445 of cos-r u, 478 of cosh u, 539 of cosh-t u,546 of cot u, 455 of cot-l u, 479-480 of coth u, 539 of coth-r u, 546 of cscu,456 of csc-r u, 48'j. of csch u,539 of csch-r u, 547 definition of, l2'/-. directional, 945-951 of exponential functions, 408-4@, 4L5 first through nth, 157 ofafunction,12l'-126 of higher order, 157-1'62introduction of concept of,25l of inverse functions, 401-403

INDEX of inverse hyperbolic functions, 546-547 of inverse trigonometric functions, 477-482 from the left, 128 of logarithmic function , 417 of natural logarithmic function , 384 notation f.or, 250-251, one-sid ed, "1,27 ordinary, 908 partial, 905-91,2 applications of to economrcs, 957 -975 definition of, 905 higher order, 934-939 of power function for rational exponents, 1,42-1'46 of the product of two functions, L35 of the quotient of two functions,'1.36 as a rate of change, 150-L53 from the right, 127 of sec u,455 of sec-r u, 480-48L of sech u,539 of sech-t u, 547 of sin u, 438-445 of sin-l u, 478 of sinh u, 539 of sinh-r z 546 of the sum of two functions, 133 of tan u, 454-455 of tan-l u, 479 of tanh u,539 of tanh-t u, 546 total, 93L of trigonometric functions, 438-445, 454-460 of vector-valued functions, 772, 866 See also Differentiation Descartes, Ren6,, 2L Difference, 5 of functions, 57 of vectors,749, 82'1. Differentiability, 125, 9L5 and continu ity , "1,26-130,9'1.6 continuous,920 of vector-valued functions, 774 Differential(s), 244-249 exact, 993 formula s, 249-252 total, 921.-924 Differential equations, first-ord er, 26'1. partral,94l. secondorder, 262 with variables separable, 261'-264 Differential geometry, 870 Differentiation, of algebraic functions, 130-137 implicit, 146-tSA inverse of ,253-260 logarithmic, 389, 419

partial, 905 of power series, 7'1,3-721 of trigonometric functions, 438-445, 454-460 See also Derivative(s) Dimension of vector space,754 Directed distance, 28, 81,3 Directed line segment, 746 Direction of a vector, 747, 8'1"8 Direction angles, 81.8 Direction cosines, 8L9 Direction numb ers, 837 Directional derivative, 945-951, Directrix, of a conic, 579 of a cylinder, 852 of a parabol a, 579 Discontinuity, 99 essential, 99, 90'!. removable, 99, 90'1, Discontinuous function , 97, 900 Discrimtnant, 623 Displacement vector, 76L Distance, between two points, 28,874 directed , 28, 8j-,3 from a point to a line, 220, 827 from a point to a plane, 834-835 undirected, 29 Distributive law, for real numbers, 4 for vectors, 752, 757, 822, 825, 844 Divergent improper integral , 643, 648 Divergent infinite series, 675 Divergent sequence, 664 Divisiofl, 5 Domain of a function , 49, 58, 881 Dot product of vectors, 756-762, 825-828 Double integral , 1002-1007 definition of, 1003 evaluation of, 1008-1014 in polar coordinates, 1022-1026 Drawing the sketch of a graph of a function, 227-229 e (the base of natural logarithms), 407, 4L6-417 Eccentricity of a conic, 589 Economics, applications of antidifferentiation to, 269-272 applications of derivatives to, 230-238 applications of partial derivatives to, 967 -975 Edges of a rectangle, 1002 Elernents of a sequence, 660 Ellipse, 588, 602-603, 606-672 degenerate,5SS-589 , 620, 624 major axis of, 603 minor axis of, 603 point-, 609 Ellipsoid, 858 of revolution, 856 Elliptic cone, 862

A-47

Elliptic cylinder, 853 Elliptic hyperboloid, of one sheet, 858 of two sheets, 859 Elliptic integral, 808 Elliptic paraboloid, 861 Empty set, 3 Endpoints of an interval, 10 Epicycloid, 809 Epsilon (e), 67 Equation(s), cartesian, 558, 764, 865 of a circle, 44 of the conics, cartesian, 599-605 polar, 592-598 differential, 261,-264 of a graph,25 graph of , 23-27 , 8'1,5 linear, 38, 835 of motion, L15 parametric, 764, 837, 865 of a plane, 830 of rotating the axes/ 627-623 of a sphere, 816 of a straight line in a plane, 33-40 intercept form, 37 point-slope form, 36 slope-intercept form, 36 two-point form, 35 of a straight line in R3, 836-840 parametric, 837 symmetric, 837 of translating the axes, 584 vector, 764, 865 Equilateral hyperb ola, 614 Equipotential curves, 886 Equipotential surfaces, 95I Essential discontinuity , 99, 90'1. Euler, Leonhard, 407 Euler's number (e), 407 Even function, 59 Exact differen tral, 993 Existence theorem, 67'1. Explicit function, 146 Exponential function (s), 405- 41'2 to base a, 4'1.4-419 definition of, 4'1,4 definition of, 405 derivative of, 408-409 , 415 table of, A-5-A-11 Exponents, rational, derivative of power function for, 742-1'46 Extrema, absolute, 1.85,956-966 applications involvin g,'1.89-195, 21,3-219 constrained, 963 free, 963 of functions of two variables,956-966 relative , 182,957 first-derivative test for, 206-209 second-derivative test fot, 2'l'1'-213 Extreme value theorem , 1,87-L88

A-48

INDEX

Extreme value theorem (Continued) for functions of two variables,957 Field(s), force,982 gradient,993 Field axioms, 5 Finite sequence, 660 First derivative, 157 First-derivative test for relative extrema, 206-209 First derived function, 157 First-order differential equations, 267 Fixed cost, 238 Focus, of a conic, 579 Force field, 982 conseryative, 993 Fractions, 5 partial, use of in integration, 504-515 Free extrema, 953 Function(s), 48-60 absolute extrema of ,185, 956-966 applications involvin g, 189-195, 213-219 algebrarc, 60 differentiation of ,'1,30-137 antiderivative of , 253 average values of, 309-310 composite, 58, 884 derivative of, 1,38-142 constant, 59 continuity of ,97-107 on an interval, 177-180 left-han d, 178 with more than one variable, 900-904 at a number,97-101, on an open ball, 903 right-hand, 178 theorems on, l0L-107 continuously differentiable, 920 contour curves of, 885 cosecan t, 452-450, 457 -470 cosine, 431-437, 447 -451 cotange nt, 452- 450, 466-470 cubic, 60 decreasing,20S definition of, 48, 881 derivative of, 121-126 total,931 difference of , 57 differentiability of , 913-920 and continuity,'1.26-130 differentiable, L25 discontinuous, 97, 900 domain of' 48,58, 881, even, 59 explicit, 146 exponential, 405-412 to base a, 4'1.4-419 extrema of, absolute, 185 relative , 182

first derived, 157 gradient of , 947-951, 976-981 graph of,48-55, 885 sketching of,227-229 greatestinteger, 54 hlryerbolic, 535-553 derivatives of, 539 integrals of, 539-540 identity, 60 increasing, 205 integrable, 290-291,1003 inverse,395-404 inverse cosecant,475 inverse cosine,472 inverse cotangent, 474-475 inverse hlryerbolic, 535-553 integrals yielding, 548-551 inverse secant,475 inverse sine, 472 inverse tangent, 474 inverse trigonometric, 471,-488 integrals yieldi ng, 484-488 level curves of , 886 level surface of.,887 limit of, 66-97 involving infinity, 88-97, 165-17'1. left-frand, 85 with more than one variable, 889-898 one-sided,85-88 right-hand, 85 theoremson, 74-83,91-96, 174-177 two-sided, 85 undirected, 85 linear, 60 logarithmic, to base a, 41,4-419 maximum value of, 181-189,956-966 - absolute,L85,955 relative, L82,957 mean value of, 309 minimum value of.,l8L-189, 956-966 absolute,185,957 relative, 182,957 monotonic,206 of more than one variable, 8&1-887 continuity of,900-904 limits of, 889-898 of n variables,881,883, 953-966 natural logarithmic, 382-394 graph of.,39t-394 notation f.or,56-57 obtaining of.,from its gradient, 976-981' odd, 59 one-parameterf.amily, 252 operations on, 57-59 periodic, 434 polynomial, 59, 884 potenti al, 993 power,'1,42 product of'57 derivative of, L35

quadratrc, 60 quotient of , 57 derivative of, 136 range of., 48, 881 rational, 60, 885 integration of, 504-515 , 516-518 relative extrema of , L82, 957 secant, 452-460, 467 -470 of several variables, differential calculus of, 88L-943 sign,72 signum,72 sine, 431-437, 447-451 smooth, 374 sum of,57 derivative of, 133 tangent, 452-460, 468-470 application of to slope of a line, 451,-465 total differential of, 921-924 transcendental, 60 trigonometric, 431-488 inverse, 471-488 two-parameter family, 262 of two variables, absolute maximum value of.,956 absolute miminum value of ,957 extrema of ,956-966 extreme-value theorem for, 957 relative maximum value of ,957 relative minimum value of,957 types of, 58-60 unit step, 72 vector-valued, 763, 778, 8& calculus of , 772-778 Fundamental theorem of the calculus, 311-319 Gas, ideal, law of, 910 Generating curve of a surface of revolution, 855 Generator, of a cone, 588 of a cylinder, 852 Geometric series, 678-679 Geometry, analytic, 2l-55 differential, 870 Gradient field, 993 Gradient of a function, 947-951 obtaining a function from, 976-987 Gradient vector, 947 Graph(s), of equations, 23-27, 815 equations of, 25 in polar coordinates, 560-566 of a function, 48-56, 885 sketching of, 227-229 intersection of, in polar coordinates, 567-570 of natural logarithmic function, 391-394 reflection of, 40L symmetry of,25

INDEX

Gravity, acceleration of.,265 Greatest integer function, 54 Greatestlower bound of a sequence,669 Growth and decay,laws of,420-426 Gyration, radius of, 10L9 Half life, 42"1. Harmonic motion, simple, M7 Harmonic series,676-677 Helix, 866 circular, 866 Hermite, Charles,407 Higher-order derivatives' 157*152 Higher-order partial derivatives, 934-939 Homogeneouslamina, 358 Homogeneousmass,353 Homogeneous rod, 353 Homogeneoussolid of revolution, 366-37r Hooke's law, 345 Horizontal asYmPtotes,l7l-173 Hyperbola, 588, 604,513-620 asymptotes of , 597, 61'4-515 auxiliary rectangle of, 515 coniugate,615 coniugate axis of.,604 degenerate,589, 620,624 equilateral, 614 transverse axis of', 504 unit,768 Hyperbolic cosecantfunction ' 536 derivative of, 539 inverse,SM derivatle of.,547 Hlperbolic cosine function, 536 derivative of, 539 inverse,5M derivative of, 545 Hlperbolic cotangent function, 536 derivative of, 539 inverse, SM derivative of , 546 Hlryerbolic rylinder, 853 Hlperbolic functions, 535-553 derivatives of, 539 integration of, 539-540 inverse, 543-553 derivatives of.,546-547 yielded bY integrals, 548-551 table of , A-12 Hlryerbolic Paraboloid, 861' Hlryerbolic radian,768 Hyperbolic secant function, 536 derivative of, 539 inverse, sM-545 derivative of, 547 Hlryerbolic sine function, 536 derivative of, 539 inverse, 543 derivative of,545

Hyperbolic tangent function, 536 derivative of, 539 inverse, 544 derivative of,546 Hlperboloid, of revolution, 855 Hyperharmonic series,692 Hypocycloid, 77'1. Ideal gas law,9t0 Identity, 433 Identity function, 50 Implicit differentiation, 146-150 Improper integrals, 54L-650 convergent, 543, 548 divergent, 643, 648 with infinite limits of integration, 64L-650 Inclination, angle of , 461 Increasingfunctions, 205 Increasingsequence,667 Increment of function of two variables, 914 3'l'6, 492 Indefinite inte gral, 29'/.,, Indefinite integration, 492 Independent variables, 49, 883 Independent vectors, 756, 824 Indeterminate forms, 629-641 Index of summation,276 Inequalities, 7 nonstrict, T strict, 5 Inertia, moment of, 1018-1021 polar, L018 Infinite sequence,660 Infinite series, 550-744 absolutely convergent,70'l' alternating, 697 alternating-seriestest, 698 binomi al, 738-742 comparison test, 685-687 conditionally convergent, 701 of constant terms, 673-683 convergent, 675 definition of , 673 divergent, 675 geometric,678-679 harmonic, 576-577 hyperharmonic, 692 integral test, 694-696 limit comParison test, 687-690 Maclaurtn, TSl p, 692 partial sum of.,574 bf positive and negative terms, 697-706 of positive terms, 684-693 -712 -power series,707 differentiation of, 713-727 integration of.,722-728 interval of convergenceof.,7ll radius of convergenceof',711

A-49

ratio test, 703-7A6 remainder of, 699 sum of , 575 Taylor, 729-737 terms of., 673 Infinity, limits involvin g, 88-97 , 165-L7'l' '1,0, 90 negative, positive, 10, 90 Inflection, point of , 220-225 applications to drawing a sketch of the graph of a function, 227-229 Inflectional tang ent, 224 Initial conditions, 262 Initial point,746 Initial side of an angle, 43'l' Inner product, 756-762 Instantaneous acceleration, L58, 786 Instantaneous rate of change, LS"l' Instantaneous velocity , L"l'6-117,785 in rectilinear motion, 115-L2A Integers, 6 Integrable function, 290-291, 1003 Integral(s), definite, 276-322 applications of, 324-380 definition of, 291, 294 properties of, 296-305 double, 1'002-L007 definition of, 1003 evaluation of, 1008-1014 in polar coordinates, 1022-1026 elliptic, 808 of exponential functions, 409' 4'l'5 formulas, standard indefinite integration, 492 of hyperbolic functions, 539-540 improp er, 641-650 indefinite, 29'l', 316, 492 involving Powers of sine and cosine, 447-451 involving powers of tangent, cotangent, secant, and cosecant, 466-470 iterated, 1009 line, 981-988 indePendent of the Path, 989-996 mean-value theorem fot, 306-310 multiple, 1002 sign ([),291' single, 1002 table of , 492 test, 694-696 of trigonometric functions, 447 -45'l', 466-470 triple, 1'034-1037 in cytindrical coordinates, 7039-1042 definition of, 1.034 in spherical coordinates, 1043-1-045 yielding inverse hyperbolic functions, 548-551 yielding inverse trigonometric functions, 484-488

A-50

INDEX

Integral(s) (Corztinu ed) See also Integration Integrand,29l lower limit of , 291, upper limit of ,291 Integration, 316 constant of, 316 indefinite, 492 limits of,29l infinite, 64L-646 miscellaneous substitutions, 5t9-521 multiple, 1002-1045 by parts, 493-497 formula f.or, 493 of power series, 722-728 of rational functions by partial fractions, 504-515 denominator contains quadratic factors, 512-515 denominator has only linear factors, 504-s1.0 of rational functions of sine and cosine, 516-518 'J,002 region of, techniques of, 492-534 by trigonometric substitution, 498-503 See also Integral(s) Intercepts, of a line, 35 of a plane, 831 Interest, compound, 424-426 Intermediate-value theorem, 306-907 Intersection, of graphs in polar coordinates, 567-570 'of sets, 4 Interval, 9 closed, 1.0 continuous on, L78 of convergence of power series, T'!.L endpoints of, 10 half-open on the left, 10 half-open on the right, 10 open, 9 partition of , 288 Invarian t, 624 Inverse cosecant function, 475 derivative of, 48L Inverse cosine function, 472 derivative of, 478 Inverse cotangent function, 474-4Ts derivative of, 479-480 Inverse of a function , 395-404 definition of , 397 derivative of, 401,-403 Inverse hyperbolic cosecant function , 544 derivative of, 547 Inverse hyperbolic cosine fun ctioni 544 derivative of, 546 Inverse hyperbolic cotangent function, 544 derivative of., 546

Inverse hyperbolic functions, 543-553 derivatives of, 539 yielded by integrals, 548-551 Inverse hyperbolic secant function, 544 derivative of ,547 Inverse hyperbolic sine function, 543 derivative of , 546 Inverse hyperbolic tangent function , 544 derivative of., 546 Inverse of the logarithmic function , 4'l.L Inverse operations, 253 Inverse secant function , 475 derivative of, 48L Inverse sine function, 472 derivative of, 478 Inverse tangent function , 474 derivative of, 479 Inverse trigonometric function s, 47'1.-476 derivatives of, 477-482 integrals yieldi ng, 484-488 ' Irrational numbers, 6 Isothermal surface, 95L Isothermals, 886 Iterated integrals, L009 Joint-cost functron, 971. Lagrange, Joseph, 250, 653, 963 Lagrange form of the remainder, 553 Lagrange multi plierc, 975 ' Lamina, 358 Laplace's equa tion, 940 Latus rectum of a parabola, 58L Law of mass action, 509 Law of natural decay, 42'1. Law of natural growth, 421 Laws of growth and decay, 420-4ZG Least upper bound of a sequence, 669 Left*hand continuity, 178 Left-hand limit, 85 Left-handed system, 8LL Leibniz, Gottfried Wilhelm, 250, gl2 Lemniscate, 567 Length of arc, of curves,372-378, 779-794, g6g as a parameter,794-795 Level curye of a function, 885 Level surface of a function, 887 L'H6pital, Guillaume FranEois de, 630 L'H6pital's rule , 630-63"1,,534-695, 637-639 LimaEon,562,567 Limit(s), of a functi on, 66-97 theorems on, 74-83, 1V4-177 of functions of more than one variable, 889-898 involving infinity, 88-97, 165-171 theorems on, 9l-96 left-hand, 85 lower, of integrction, 291

of a sum,276 onersided, 85-88 right-hand, 85 of a sequence, 662 of sums, 29"1, trigonometric, 438- 44I two-sided, 85 undirected, 85 upper/ of integratton, 29L of a sum,275 of a vector-valued function ,772, 866 Limit comparison test, 687-688 Line(s), equations of in R2, 33-40 intercept form, 37 point-slope form, 36 slope intercept form, 36 two-point form, 35 equations of in R3, 836-840 normal , 1!3,954 parallel , 38-39 perpendicular, 39-40 slope of , 33 application of tangent function to, 461-465 symmetry with respect to,25 tangent, 1,'!."1.-'1,15, 955 Line integrals, 98L-988 independent of the path, 989-996 Line segment, directed, 746 midpoint of, 30-31, 815 Linear equation, 38, 831. Linear function, 60 Liquid pressure, 348-350 Logarithmic differentiation, 389, 4I9 Logarithmic function, to bas e a, 4'!,4-419 definiti on of , 4'1,6 derivative of., 417 natural, 382-394 definition of, 383 derivative of, 384, 41.8 graph of , 39'1,-394 inverse of., 4'j-,"1, Logarithmic spfual,566 Lower bound of a sequence, 668 greatest,669 Lower limit, of integration, 29"1, of a sum,276 Maclaurin, Colin, 655 Maclaurin seri es, 73'1, Maclaurin's formula, 655 Magnitude of a vector, 747, 818 Major axis of an ellipse, 503 Map, contour, 886 Marginal average cost function , 231 Marginal concept of variation,230 Marginal cost curve, 231 Marginal cost function, 231, 97'1, Marginal demand, 969-971 Marginal revenue curye, 235

INDEX

Marginal revenue function, 235 Mass, centerof, definition of,352 of a lamina, 1016-1018 of a plane region, 356-364 of a rod, 351-355 of a solid of revolution, 366-371' definition of, 351, 1017 homogeneous,353 moment of.,352,366 Mass action, law of, 504 Maximum value of a function, L81-1'89, 956-966 absolute,185 of functions of two variables,956 relative, 182 of functions of two variables,957 Mean value of a function, 309 Mean value theorem, 197-201 Cauchy's, 53L-632 for integrals, 306-310 and Rolle's theorem, 195-20'l' Measure,ZSL Midpoint of a line segment,30-31, 815 Minimum value of a function , 18'l'-189, 956-966 absolute,L85 of functions of two variables,957 relative, 182 of functions of two variables,957 Minor axis of an elliPse,503 Moment, of inertia, 1018-1021 polar, 1018 of mass,352,366 Monopoly,236 Monotonic function, 206 Monotonic sequence,563-673 Motion, equation of, of a particle, Lls plane, 785-791 iectilinear, and antidifferentiation, 265-268 instantaneous velocity in, L15-120 simple harmonic, 447 Moving trihedral,869 Multiple integral, 1002 Multiple integration, 1002-1045 Multiplication, 4 of vectots, 842 Multiplication cross-product, 842-85L dot product, 756-752, 825-828 scalar, 749 -750, 75'l'-755, 821 Multiplicative inverse, 5 Muzzle speed, 788 n-dimensional number sPace, 881' Nappe of a cone, 588 Natural decay, law of'42l Natural growth, law of., 421 Natural logarithmic function s, 382-394 graph of.,39'l'-394 Nitural logarithms, table of, A-3-A-4

Newton, Sir Isaac, 250,312 Nicomedes, conchoi d of.,567 Noncentral quadrics, 861 Norm of partition, 288, 1002, 1022' L035 Normal component of acceleration, 805 Normal line, L13 to a surface, 954 Normal vector, to a plane, 829 to a surface, 953 unit, 792,869 Notation, for functions, 56-57 sigma, 276-280 rth derivative, 157 Number line, 9 Number plane, 21'-22 Number space, n-dimensional, 881 three- dimensional, 8ll -8t7 Numbers, critical, L84 irration al, 6 rational, 6 real, 4-14 transcendental, 407 Numerical tables, A-2-A-15 Oblate spheroid, 858 Octants, Sl2 Odd function, 59 One-parameter family of functrons, 262 One-sided derivatle, 127 One-sided limit, 85-88 One-to-one corresPondenc e, 7 46, 812 Open ball, 889 continuity of a function on, 903 Open disk, 890 Open interval, 9 Open rectangle, 1002 Operations on functions, 57-59 Order, axiom of., 6 Ordered fields, 7 Ordere d n-tuPle, 881 Ordered pair, 21,746, 882 Ordered triple, 81L Ordinary derivative, 908 Ordinate,22 O r i g i n , 9 , 2 ' 1 , 5 5 5 ,8 L L Orthogonal vectors, 759, 828 Osculating circle, 800 Overhead cost, 230 p series, 692 Pair, ordered , 21,746, 882 Pappus, theorem of., 366 Parabol a, 23, 579-581', 588 axes of., 579 degenerate, 620, 624 directrix of , 579 focus of.,579 latus rectum of, 581 vertex of, 580 Parabolic cvlinder, 853

Parabolic rule, 526-53L Parabolic spiral, 567 Paraboloid, elliptic, 861 hyperbolic, 86L of revolution, 855 Parallel lines, 38-39 Parallel plane sections, of a solid, volume of, 34'1.-343 Parallel planes, 832 Parallel vectors , 758-759 , 827 -828 Parallelepiped, rectangular, 34'l' Parallelogram law, 749 Parameter,764 Parametric equations, 7 63-770 of a line, 837 in R3, 865 Partial derivatives, 905-912 applications of to economics, 967-975 higher-order, 934-939 Partial differential equ atrons, 94'l' Partial differentiation, 905 Partial fractions, integration by, 504-515 Partial sum of an infinite senes, 674 Partial sums, sequence of., 674 Partition, cylindrical, 1035 of an interval, 288 norrn of , 288, t002, 1022, 1035 regular, 293 Percent rate of change, 153 Perfectly comPetitive matket, 973 Penod,445 Periodic function, 434 Perpendicular lines, 39-40 Perpendicular planes, 833 Perpendicular vectors, 759 Plane curve, length of arc of , 372-378 Plane motion, 785-79'l' Plane region, center of mass of ,356-364 Plane section, 341 Plane(s), 829-835 angle between,832 cartesian equations of, 830 definition of, 829 equation of., 829 parallel ,832 perpendicular, 833 tangent,953-956 traces of, 831 Point, 2"1,,22 accumulation, 893 critical, 958 in n-dimensional number sPace, 881 reflection of, 401 saddle, 958 symmetry with resPect to, 25 in three-dimensional number space, 811 Point-circle, 45, 607 Point-ellipse, 609 Point of inflection, 220-225

4.52

INDEX

Point of inflection (Continued) applications to drawing a sketch of the graph of a function, 227-ZZ9 Point-slope equation of a line, 35 Point-sphere, 816 Polar axis, 555 Polar coordinates, 555-577 arc length in,784 area of a region in, 573-576 double integrals in, 1022-1026 graphs of equations in, 560-566 intersection of graphs in, 56T -570 Polar curves, tangent lines of, S7'1-57j Polar equations, 558 ,and the conics, 592-598 Polar line, 555 Polar moment of inertia, 1018 Pole, 555 Polynomial function, 59, 884 Position representation , of a vector, 746, 818 Position vector, 764, 870 Potential function, 993 Power function, 142 derivative of, for rational exponents, L42-1,46 for real exponents, 41.8 Power series, 707-712 differentiation of , 713-722 integration of, 7ZZ-728 interval of convergence of , Tl"!, radius of convergence of.,7l'!, Powers and roots, table of, A-2 Pressure, liquid, 348-350 Price function, 294 Principal axis of a conic, 590 Principal part of a function , g2j, Principal square root, 19 Prismoidal formula, 530-531 Product, 4 of functions, 57 of two functions, derivative of, 135 of vector and a scalar, 750, 920 cross/ 842-8Sl dot, 756-762,825-B2B triple scalar, 847 Production function, 9Zg Profit function, 296 Projectile, motion of, T8T-7gl muzzle speed of ,788 Projection, scalar, of.vectors, T59-760, g26 Prolate spheroid, 858 Pythagorean theorem , 29, 40 Quadrants,22 Quadratic factors in rational functions, 512-515 Quadratic function, 60 Quadric surfaces, 858-864 central, 859

noncentral, 86'1. Quotient, 5 of functions, 57 of two functions, derivative

of, LJ6

R2, equation in, 22 graphs of equations in, 23-25 R3 (three-dimensional number space), gtt-8L7 curves in, 864-87'1, graph of an equation in, 815 lines in, 836-840 Radian, hyperb olic, 768 Radian measure, 43'!, Radius, of a circle, 43 of convergence of a power series, 211 of curvature, 800 '1,9 of gyration, of a sphere, 815 Radius vector,764 Range of a function , 49, BB'J-, Rate of change, derivative as, 150-L53 instantaneous, L5L percent, L53 relative , 152-L5g Rates, related, L54-lS7 Ratio test, 703-706 Rational exponents, derivative of power function for, 142-146 Rational function, 60, 885 Rational numbers, 6 Real number line, 9 Real numbers, 4-"1,4 Real vector space, TSg-7Ss b a s i s o f, 7 5 4 dimension ot,7S4 Reciprocal spiral, S7T Reciprocals, 5 Rectangle(s), auxiliary , of a hyperb ola, 6'l.5 closed, 1.002 edges of,'1,002 open, 1002 vertices of, 1002 Rectangular cartesian coordinates, 22 three- dimensio nal, B!2 Rectangular parallelepiped, 341 Rectifiable arc, i74 Rectilinear motion, and antidifferentiation, 265-268 instantaneous velocity in, 115-120 Reflectiofl, of a graph , 401 of a point, 401. Region of integration, 1002 Regular partition, 293 Related rates, 1,54-j,57 Relative extrema, lB2, 957 first-derivative test for , 206-209 second-derivative test for, 2ll-219 Relative maximum value, of a function, 182

of functions of two variable s, 957 Relative minimum value , of a function, 182 of functions of two variables, gST Relative rate of change, 1,52-"1.53 Remainder in infinite series, 699 afternterms,6sg in Taylor's formula, integral form of, 655 Lagrange form of, 653 Removable discontinui ty, 99, 901 Representation of a vector,746 Revolution, axis of, 330 cylinders of, 855 ellipsoid of,856 hyperboloid of, 856 paraboloid of, 856 solid of, 330 center of mass of , 366-3TI volume of , 320-Zi9 surfaces of , 854-857 Riemann, Georg Friedrich Bernhar d, Zgg Riemann sum, 289 Right-circular cone, 857 Right-circular cylinder, 34L Right cylinder,34I Right-hand continu tty, l7g Right-hand limit, g5 Right-handed system, 81L Rod, center of mass of, 351-355 homogeneous, 353 moment of mass of, 355 Rolle, Michel, 195 Rolle's theorem, L}S-Z}I and the mean-value theorem , tg?_Z}t Rose, eight-le afed, 566 five-lea fed, 566 four-lea fed, 566 three-le af.ed, 566 Rotation of axes, 620-625 Ruling of a cylinder, BSz Saddle point, 958 Scalar, 749 Scalar multiplication of vecto rs, 749-T50, 75r-755, 920 Scalar (dot) product, T56-762, g2}-g}g Scalar projection of vectors, TG\, g26 Secant function, 452-460 derivative of, 455 integrals involving powers of, 467-470 inverse , 475 derivative of, 480-481 Secant line, 11L Second derivative,'J,ST Second derivative test for relative extrema, 21.1,-21,9 applications to drawing a sketch of the graph of a function , 227-Z2g for functions of two variables, g5g-963

INDEX A-53

Second order differential equations' 262 Sectionally smooth cunre, 985 Sequence(s), 660-673 bounded,667-673 convergent, 664, 670-672 decreastng, 667 definition of, 660 divergent,664 elements of , 660 finite, 650 greatest lower bound of', 669 increasing, 667 infinite, 660 least upPer bound of-, 659 limit of, 662 lower bound of , 668 monotonic, 667-673 of partial sums, 674 strictly decreas ing, 667 strictly increasi ng, 667 upper bound of , 668 Series. See Inftnite serles Set(s), 2 intersection of., 4 union of, 3 Set-builder notation, 2 Side condition,963 Sides of an angle, 43L Sigma notation, 276-280 Sign function, 52 Signum function, 52 Simple harmonic motion, 447 Simpson's rule, 526-53t Sine function, 431-437 derivative of, 438-445 -451 integrals involving powers of, 447 integration of rational functions of,

516-s18 inverse , 472 derivative of., 478 Single integral , L002 Slope, of. a line, 33 application of tangent function to, 46t-465 of a tangent line to a graPh, 112-113 of a tangent line to a Polar curve/ 560-551 Slope-intercept form of equation of a line, 36 Slug, definition of, 352 Smooth curve, 985 Smooth function, 374 Solid, with known parallel plane sections, volume of ,341-343 of revolution, center of mass of, 366-37L volume of ,330-339 Solution set, 11 Special functions, 58-60 Speed, 117,786, 870

muzzle,788 Sphere, 8L5, 858 center of, 815 equation of, center-radius form of ' 8l'6 general form of , 8-16 point-, 8L6 radius of, 815 Spherical coordinates, 872-87 6 and triple integrals, L039-1042 Spheroid, 858 oblate, 858 prolate, 858 Spiral, of Archimedes, 566, 577 logarithmic, 566 paraboltc, 567 reciproc al, 577 Square root, PrinciP al, 19 Squeeze theorem, 175 Stindard position of an angle, 43'l' Stirling, James, 655 Strictly decreasing sequence, 667 Strictly increasing sequence, 667 Subsets, 2 Substitute commodities, 968 Subtraction, 5 of vecto rs, 749, 82'l' Sum, 4 of functions, 57 of infinite series, 675 limits of , 29L partial, sequence of., 674 Riemann,289 of two functions, derivative of , 133 of vecto rs, 7 48-7 49 , 820 Summation, index of ,276 Surface(s), 8L5, 852 area of, 1.028-1033 demand,968 equipotential, 95L isothermal, 951 level, 887 quadric, 858-864 of revolution, 854-857 Symmetric equations of a line, 837 Symmetry, axis of.,363 of a graph,25 tests f.or,26 Tables, numerical, A-2-A-15 Tangent functi on, 452-460 application of to slope of a line, 461'-465 derivative of, 454-455 inflectional,224 integrals involving powers of, 468-470 inverse , 474 derivative of , 479 Tangent line, 11'1'-1'15,955 to a graph, sloPe of ,'l'1'2-1'1'3 of a polar curve, 57L-573 slope, of, 560-561

Tangent Plane, 953-956 Tan[ent vector, unit, 792, 869 i"t ["tttial component of acceleration' 805 Taylor, Brook, 651 Taylor polYnomial, 653 Taylor series, 729-737 Taylor's formula, 651-657 with integral form of the remainder' 555 with Lagrange form of the remainder' 653 Terminal point,746 Terminal side of an angle, 431 Terms of infinite series, 673 Third derivatle, L57 Three-dimensional rectangular cartesian coordinates, 812 Three-dimensional number space (R3), 8"1.1-817 cartesian equations in, 865 vectors rn, 8L8-824 Total cost cuwe,232 Total cost function, 230 Total differential of a function,92L-924 Total derivative, 931 Total revenue curve, 235 Total revenue function ' 235 Traces of a Plane, 83L Tractrix , 77'1. Transcendental functions, 50 Transcendental number, 407 Translation of axes, 583-587 equations of, 584 Ttansverse axis of a hYPerb ola, 604 Trcp ezoidal rule, 521'-525 Triangle inequ alttY, L9-20 Trigonometric function s, 43'l'-488 derivatives of, 438-445, 454-460 integration of, 447-45L, 466-470 inverse , 471-488 derivatives of, 477 -482 integrals Yieldi ng, 484-488 limits of, 438-441 table of, A-13 Trigonometric substitution, integration by, 498-503 Trihedral, moving,869 Triple integral , 1034-1037 -1042 in cylindrical coordinates, 1'039 in spherical coordinates, L043-1045 Triple scalar product of vectorc, 847 Triple, ordered, 8I'l' Twisted cubic, 866 Two-parameter family of functrons, 262 Two-point form of equation of a line, 35 Two-sided limit, 85 Undirected distanc e, 29 Undirected limit, 85 Union of sets, 3

Uniqueness theorcm,72 Unit binormal vector, 869 Unit hyperbola,768 Unit normal vector,792,869 Unit step function,72 Unit tangentvector,792,869 Unit vector, 754, 820 Upper bound of a sequence,668 Ieast, 669 Upper limit, of integration,29'l' of a sum,276 Variables,2,49 dependent,49,883 independent, 49, 883 Vector(s),746 acceleration,786,870 normal component of, 805 tangential comPonentof, 805 addition of, 7 48-749,75'l'-755,820 analysis of , 746 angle between,757, 826 associativelaws for, 751'-752,822, 844 commutative laws for, 751-752,757, 822,825 componentsof,746,76'l',826 cross product of, 842-851' cunraturc,799,868 differenceof, 749,821 direction of, 747, 818 direction anglesof, 818 direction cosinesof, 81.9 displacement,761 dislributive laws fot,752,757,822,825, 844 dot product of , 756-762,825-828 equation,764,865 existenceof additive identity for,75t' 822

existenceof negative for, 75L, 822 existenceof scalarmultiplicative identity for, 751,,822 gradient,947 independent,756,824 magnitude of,747, 818 multiplication of , 749-750, 75'1.-755, 842 756-762,82-1,, negative of, 749, 820 normal, 829, 953 orthogonal, 759, 828 parallel, 758-759, 827-828 in the plane, 746-750 position,764,870 position representation of.,746, 818 product,750,842 dot, 756-762 projection of onto another vector, 759-760,826 quantitres, T46 radtus,764 scalarmultiplication of, 749-750, 751-755, 820 scalar(dot) product of,756-762, 825-828 scalarproiectionof, 760,826 space,753-755 basis for,754, 824 dimensronof,754 subtraction of, 749, 821 sum of,748-749,820 in three-dimensional sPace/818-824 triple scalarproduct of,847 unit, 754,820 unit binormal,869 unit normal, 792,869 unit tangent,792, 869 velocity,785, 870 zero,747,9tB

Vector-valued function (s), 763-778, 864 calculus of, 772-778 continutty of.,772 derivatives of, 765-766, 773-778, 866-867 differentiability of , 774 graph af , 764 limits of ,772, 866 Velocity, instantaneous , 1,,L6-117,785 in rectilinear motion, 115-120 Velocity vector, 785, 870 Vertex, of an angle, 431 of a cone, 588 of a conic, 590 of a parabola, 580 of a rectangle, 1002 Vertical asymptotes, 172-L73 Volume, of a solid having known parallel plane sections , 341-343 of a solid of revolution, 330-339 circular-disk method, 330-332 circular-ring method, 332-335 cylindrical- shell metho d, 336-339 Work, 344-347,76L,983 x axts,2L r coordinate, 22,812 x interc ept, 36 of a plane, 831 y axis,2l y coordinate, 22,8L2 y intercept, 36 of a plane, 831 z coordinate, 8L2 z intercePt, of a Plane, 83L Zero vector, 747, 818

789

8 7 5 5 4

The Calculus With Analytic Geometry

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