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Calculus BRIAN E. BLANK STEVEN G. KRANTZ Washington University in St. Louis
� WILEY
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This book was set in 10/12 TimesTen-Roman at MPS Limited, a Macmillan Company, and printed and bound by R. R. Donnelley (Jefferson City). The cover was printed by R. R. Donnelley (Jefferson City). Founded in 1807. John Wiley & Sons, Inc. has been a valued source of knowledge and understanding for more than 200 years, helping people around the world meet their needs and fulfill their aspirations. Our company is built on a foundatin of principles that include responsibility to the communities we serve and where we live and work. In 2008, we launched a Corporate Citizenship Initiative, a global effort to address the environmental, social, economic, and ethical challenges we face in our business. Among the issues we are addressing are carbon impact, paper specificatinos and procurement, ethical conduct within our business and among our vendors, and community and charitable support. For more information, please visit our website:
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This book is printed on acid free paper. 8 Copyright © 2011, 2006 by John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 646-8600. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 748-6008. Evaluation copies are provided to qualified academics and professionals for review purposes only, for use in their courses during the next academic year. These copies are licensed and may not be sold or transferred to a third party. Upon completion of the review period, please return the evaluation copy to Wiley. Return instructions and a free of charge return shipping label are available at www.wiley.com/go/returnlabel. Outside of the United States, please contact your local representative. ISBN 13
978-0470-45359-9 (Multivariable)
ISBN 13
978-0470-45360-5 (Single & Multivariable)
ISBN 13
978-0470-60198-3 (Single Variable)
Printed in the United States of America 10 9 8 7 6 5 4 3 2 1
A pebble for Louis.
BEB
This book is for Hypatia, the love of my life.
SGK
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Preface
ix
Supplementary Resources
c H A pT E R 9
Acknowledgments
xiv
About the Authors
xvi
Vectors Preview
721 721
1
Vectors in the Plane
2
Vectors in Three-Dimensional Space
3
The Dot Product and Applications
4
The Cross Product and Triple Product
5
Lines and Planes in Space
722
Summary of Key Topics Review Exercises
782 787
191
791
1
Vector-Valued Functions-Limits, Derivatives, and Continuity
2
Velocity and Acceleration
3
Tangent Vectors and Arc Length
4
Curvature
5
Applications of Vector-Valued Functions to Motion
Review Exercises
813 834
850
853
Genesis & Development 10
856
Functions of Several Variables Preview
792
803
824
Summary of Key Topics
C H A PT E R 1 1
753
766
Vector-Valued Functions Preview
732 741
785
Genesis & Development 9
C H A PT E R 1 o
xiii
861
861
1
Functions of Several Variables
2
Cylinders and Quadric Surfaces
3
Limits and Continuity
4
Partial Derivatives
5
Differentiability and the Chain Rule
6
Gradients and Directional Derivatives
7
Tangent Planes
8
Maximum-Minimum Problems
863 874
882
890 900 913
921 932
vii
viii
Contents 9
Lagrange Multipliers
Summary of Key Topics Review Exercises
946 957
960
Genesis & Development 11
CH APTER 1 2
Multiple Integrals 967 Preview
1 2 3 4 5 6 7 8
967
Double Integrals Over Rectangular Regions Integration Over More General Regions Polar Coordinates
976
990
Integrating in Polar Coordinates
999
1014
Triple Integrals
Physical Applications
1020
Other Coordinate Systems
Review Exercises
968
984
Calculation of Volumes of Solids
Summary of Key Topics
1029
1037
1042
Genesis & Development 12
CH APTER 1 3
963
1045
Vector Calculus 1049 Preview
1 2 3 4 5 6 7 8
1049
Vector Fields Line Integrals
1050 1061
Conservative Vector Fields and Path Independence Divergence, Gradient, and Curl
1085
1094 1104 Stokes's Theorem 1116 Green's Theorem
Surface Integrals
1131
The Divergence Theorem
Summary of Key Topics Review Exercises
1139
1143
Genesis & Development 13
1146
Table of Integrals T-1 Formulas from Calculus: Single Variable T-14 Answers to Selected Exercises A·1 Index 1-1
1071
Calculus is one of the milestones of human thought. In addition to its longstanding role as the gateway to science and engineering, calculus is now found in a diverse array of applications in business, economics, medicine, biology, and the social sciences. In today's technological world, in which more and more ideas are quan tified, knowledge of calculus has become essential to an increasingly broad cross section of the population. Today's students, more than ever, comprise a highly heterogeneous group. Calculus students come from a wide variety of disciplines and backgrounds. Some study the subject because it is required, and others do so because it will widen their career options. Mathematics majors are going into law, medicine, genome research, the technology sector, government agencies, and many other professions. As the teaching and learning of calculus is rethought, we must keep our students' back grounds and futures in mind. In our text, we seek to offer the best in current calculus teaching. Starting in the 1980s, a vigorous discussion began about the approaches to and the methods of teaching calculus. Although we have not abandoned the basic framework of calculus instruction that has resulted from decades of experience, we have incorporated a number of the newer ideas that have been developed in recent years. We have worked hard to address the needs of today's students, bringing together time-tested as well as innovative pedagogy and exposition. Our goal is to enhance the critical thinking skills of the students who use our text so that they may proceed successfully in whatever major or discipline that they ultimately choose to study. Many resources are available to instructors and students today, from Web sites to interactive tutorials. A calculus textbook must be a tool that the instructor can use to augment and bolster his or her lectures, classroom activities, and resources. It must speak compellingly to students and enhance their classroom experience. It must be carefully written in the accepted language of mathematics but at a level that is appropriate for students who are still learning that language. It must be lively and inviting. It must have useful and fascinating applications. It will acquaint students with the history of calculus and with a sense of what mathematics is all about. It will teach its readers technique but also teach them concepts. It will show students how to discover and build their own ideas and viewpoints in a scientific subject. Particularly important in today's world is that it will illustrate ideas using computer modeling and calculation. We have made every effort to insure that ours is such a calculus book. To attain this goal, we have focused on offering our readers the following: •
A writing style that is lucid and readable
•
Motivation for important topics that is crisp and clean
•
Examples that showcase all key ideas
•
Seamless links between theory and applications
•
Applications from diverse disciplines, including biology, economics, physics, and engineering
ix
x
Preface •
Graphical interpretations that reinforce concepts
•
Numerical investigations that make abstract ideas more concrete
Content We present topics in a sequence that is fairly close to what has become a standard order for multivariable calculus. This volume refers to only ten formulas from Calculus: Single Variable. To facilitate the use of this book for students who own a different single variable text, we have collected these equations and presented them on p. T-14. In the outline that follows, we draw attention to several sections that may be regarded as optional. Chapter numbering follows that of our eight chapter single variable text. Chapter 9, the first chapter of multivariable calculus, begins with a section that introduces vectors in the plane. It is followed by a section that covers the same material in space. After sections on the dot and cross product, Chapter 9 concludes with a comprehensive account of lines and planes in space. Chapter 10 is devoted to the differential calculus and geometry of space curves. The final two sections are concerned with curvature and associated concepts, including applications to motion and derivations of Kepler's Laws. Because these two sections are not used later in the text, they may be omitted by instructors who need additional time for other topics. The differential calculus of functions of two and three variables is taken up in Chapter 11. All topics in the standard curriculum are discussed. One less con ventional topic that we treat is the development of order 2 and order 3 Taylor polynomials for functions of two variables. We use these ideas in the discriminant test for saddle points and local extrema but nowhere else. It is therefore feasible to omit the discussion of multivariable Taylor polynomials. Chapter 12 is devoted to multiple integrals and their applications. In a section that precedes cylindrical and spherical coordinates, we develop polar coordinates ab initio. This section may be omitted, of course, if polar coordinates have been introduced earlier in the calculus curriculum. The concluding chapter on vector calculus, Chapter 13, covers vector fields, line and surface integrals, divergence, curl, flux, Green's Theorem, Stokes's The orem, and the Divergence Theorem.
Structural Elements We start each chapter with a preview of the topics that will be covered. This short initial discussion gives an overview and provides motivation for the chapter. Each section of the chapter concludes with three or four Quick Quiz questions located before the exercises. Some of these questions are true/false tests of the theory. Most are quick checks of the basic computations of the section. The final section of every chapter is followed by a summary of the important formulas, theorems, definitions, and concepts that have been learned. This end-of-chapter summary is, in tum, followed by a large collection of review which are similar to the worked examples found in the chapter. Each chapter ends with a section called Genesis &
Preface
xi
Development, in which we discuss the history and evolution of the material of the chapter. We hope that students and instructors will find these supplementary dis cussions to be enlightening. Occasionally within the prose, we remind students of concepts that have been learned earlier in the text. Sometimes we offer previews of material still to come. These discussions are tagged A Look Back or A Look Forward (and sometimes both). Calculus instructors frequently offer their insights at the blackboard. We have included discussions of this nature in our text and have tagged them Insights.
Proofs During the reviewing of our text, and after the first edition, we received every possible opinion concerning the issue of proofs-from the passionate Every proof must be included to the equally fervent No proof should be presented, to the see mingly cynical It does not matter because students will skip over them anyway. In fact, mathematicians reading research articles often do skip over proofs, returning later to those that are necessary for a deeper understanding of the material. Such an approach is often a good idea for the calculus student: Read the statement of a theorem, proceed immediately to the examples that illustrate how the theorem is used, and only then, when you know what the theorem is really saying, turn back to the proof (or sketch of a proof, because, in some cases, we have chosen to omit details that seem more likely to confuse than to enlighten, preferring instead to concentrate on a key, illuminating idea).
Exercises There is a mantra among mathematicians that calculus is learned by doing, not by watching. Exercises therefore constitute a crucial component of a calculus book. We have divided our end-of-section exercises into three types: Problems for Practice, Further Theory and Practice, and Calculator/Computer Exercises. In general, exercises of the first type follow the worked examples of the text fairly closely. We have provided an ample supply, often organized into groups that are linked to particular examples. Instructors may easily choose from these for creating assignments. Students will find plenty of unassigned problems for additional practice, if needed. The Further Theory and Practice exercises are intended as a supplement that the instructor may use as desired. Many of these are thought problems or open ended problems. In our own courses, we often have used them sparingly and sometimes not at all. These exercises are a mixed group. Computational exercises that have been placed in this subsection are not necessarily more difficult than those in the Problems for Practice exercises. They may have been excluded from that group because they do not closely follow a worked example. Or their solutions may involve techniques from earlier sections. Or, on occasion, they may indeed be challenging.
xii
Preface The Calculator/Computer exercises give the students (and the instructor) an opportunity to see how technology can help us to see and to perceive. These are problems for exploration, but they are problems with a point. Each one teaches a lesson.
Notation Throughout our text, we use notation that is consistent with the requirements of technology. Because square brackets, brace brackets, and parentheses mean dif ferent things to computer algebra systems, we do not use them interchangeably. For example, we use only parentheses to group terms. Without exception, we enclose all functional arguments in parentheses. Thus, we write sin(x) and not sin x. With this convention, an expression such as cos(x)2 is unambiguously defined: It must mean the square of cos(x); it cannot mean the cosine of (x)2 because such an interpretation would understand (x)2 to be the argument of the cosine, which is impossible given that (x)2 is not found inside parentheses. Our experience is that students quickly adjust to this notation because it is logical and adheres to strict, exceptionless rules. Occasionally, we use exp(x) to denote the exponential function. That is, we sometimes write exp(x) instead of
because students will need to code the exponential function as exp(x) in Matlab and in Maple, and as Exp[x] in Mathematica, we believe we should introduce them to what has become an important alternative notation. Our classroom experience has been that once we make students aware that exp(x) and c mean the same
thing, there is no confusion.
The Second Edition For this second edition, we have moved our coverage of inverse functions to Chapter 1. Logarithmic functions now closely follow the introduction of expo nential functions in Chapter 2. The remaining standard transcendental functions of the calculus curriculum now appear in Chapter 3. The topic of related rates has been moved to the chapter on applications of the derivative. We deleted a section entirely devoted to applications of the derivative to economics, but some of the material survives in the exercises. We did not discuss the use of tables of integrals in the first edition, but we now give examples and exercises in Chapter 5, The Integral. Accordingly, we now provide an extensive Table of Integrals. In the chapter on techniques of integration, the two sections devoted to partial fractions were separated in the first edition but are now contiguous. In Chapter 7, Applications of the Integral, the discussion of density functions has been expanded. The section devoted to separable differential equations has been moved to this chapter, and a new section on linear differential equations has been added. In that chapter, we also have expanded our discussion of density functions. Our treatment of the Taylor series in the first edition occupied an
Preface
xiii
entire chapter and was much longer than the norm. We shortened this material so that it now makes up two sections of the chapter on infinite series. Nevertheless, with discussions of Newton's binomial series and applications to differential equations, our treatment remains thorough. The second edition of this text retains all of the dynamic features of the first edition, but it builds in new features to make it timely and lively. It is an up-to-the moment book that incorporates the best features of traditional and present-day methodologies. It is a calculus book for today's students and today's calculus classroom. It will teach students calculus and imbue students with a respect for mathematical ideas and an appreciation for mathematical thought. It will show students that mathematics is an essential part of our lives and make them want to learn more.
Supplementary Resources WileyPLUS is an innovative, research-based, online environment for effective teaching and learning. (To learn more about WileyPLUS, visit www.wileyplus.com.) What do students receive with WileyPLUS? •
A research-based design. WileyPLUS provides an online environment that
integrates relevant resources, including the entire digital textbook, in an easy-to navigate framework that helps students study more effectively. •
WileyPLUS adds structure by organizing textbook content into smaller, more
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Innovative features such as calendars, visual progress tracking, and self
manageable "chunks." objectives. evaluation tools improve time management and strengthen areas of weakness. •
One-on-one engagement. With WileyPLUS for Blank/Krantz, Calculus znct edi
tion, students receive 2417 access to resources that promote positive learning outcomes. Students engage with related examples (in various media) and sample practice items, including the following: •
Animations based on key illustrations in each chapter
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Measurable outcomes. Throughout each study session, students can assess their
progress and gain immediate feedback. WileyPLUS provides precise reporting of strengths and weaknesses, as well as individualized quizzes, so that students are confident they are spending their time on the right things. With WileyPLUS, students always know the exact outcome of their efforts. What do instructors receive with WileyPLUS? •
Reliable resources. WileyPLUS provides reliable, customizable resources that
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xiv
Preface
•
Customizable course plan. WileyPLUS comes with a precreated course plan
designed by a subject matter expert uniquely for this course. Simple drag-and drop tools make it easy to assign the course plan as-is or modify it to reflect your course syllabus. •
Course materials and assessment content. The following content is provided. •
Lecture Notes PowerPoint Slides
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Classroom Response System (Clicker) Questions
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Instructor's Solutions Manual
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Gradable Reading Assignment Questions (embedded with online text)
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Question Assignments (end-of-chapter problems coded algorithmically with hints, links to text, whiteboard/show work feature, and instructor-controlled problem solving help)
•
Gradebook. WileyPLUS provides instant access to reports on trends in class
performance, student use of course materials, and progress towards learning objectives, helping inform decisions and drive classroom discussions. Powered by proven technology and built on a foundation of cognitive research, WileyPLUS has enriched the education of millions of students in more than 20 countries around the world. For further information about available resources to accompany Blank/Krantz,
Calculus 2nd edition, see
www.wiley.com/college/blank.
Acknowledgments Over the years of the development of this text, we have profited from the com ments of our colleagues around the country. We would particularly like to thank Chi Keung Chung, Dennis DeTurck, and Steve Desjardins for the insights and suggestions they have shared with us. To Chi Keung Chung, Roger Lipsett, and Don Hartig, we owe sincere thanks for preventing many erroneous answers from finding their way to the back of the book. We would also like to take this opportunity to express our appreciation to all our reviewers for the contributions and advice that they offered. Their duties did not include rooting out all of our mistakes, but they found several. We are grateful to them for every error averted. We alone are responsible for those that remain.
Debut Edition Reviewers
David Calvis, Baldwin-Wallace College Gunnar Carlsson, Stanford University Chi Keung Cheung, Boston College Dennis DeTurck, University of Pennsylvania Bruce Edwards, University of Florida Saber Elyadi, Trinity University David Ellis, San Francisco State University Salvatrice Keating, Eastern Connecticut State University Jerold Marsden, California Institute of Technology Jack Mealy, Austin College Harold Parks, Oregon State University Ronald Taylor, Berry College
Preface
Second Edition Reviewers
Anthony Aidoo, Eastern Connecticut State University Alvin Bayless, Northwestern University Maegan Bos, St. Lawrence University Chi Keung Cheung, Boston College Shai Cohen, University of Toronto Randall Crist, Creighton University Joyati Debnath, Winona State University Steve Desjardins, University of Ottawa Bob Devaney, Boston University Esther Vergara Diaz, Universite de Bretagne Occidentale Allen Donsig, University of Nebraska, Lincoln Hans Engler, Georgetown University Mary Erb, Georgetown University Aurelian Gheondea, Bikent University Klara Grodzinsky, Georgia Institute of Technology Caixing Gu, California Polytechnic State University, San Luis Obispo Mowaffaq Hajja, Yarmouk University Mohammad Hallat, University of South Carolina Aiken Don Hartig, California Polytechnic State University, San Luis Obispo Ayse Gui Isikyer, Gebze Institute of Technology Robert Keller, Loras College M. Paul Latiolais, Portland State University Tim Lucas, Pepperdine University Jie Miao, Arkansas State University Aidan Naughton, University of St. Andrews Zbigniew Nitecki, Tufts University Harold Parks, Oregon State University Elena Pavelescu, Rice University Laura Taalman, James Madison University Daina Taimina, Cornell University Jamal Tartir, Youngstown State University Alex Smith, University of Wisconsin Eau Claire Gerard Watts, Kings College, London
xv
Brian E. Blank and Steve G. Krantz have a combined experience of more than 60 years teaching experience. They are both award-winning teachers and highly respected writers. Their extensive experience in consulting for a variety of pro fessions enables them to bring to this project diverse and motivational applications as well as realistic and practical uses of the computer.
Brian E. Blank
Brian E. Blank was a calculus student of William 0. J. Moser and Kohur Gowri sankaran. He received his B.Sc. degree from McGill University in 1975 and Ph.D. from Cornell University in 1980. He has taught calculus at the University of Texas, the University of Maryland, and Washington University in St. Louis.
Steven G. Krantz
Steven G. Krantz was born in San Francisco, California in 1951. He earned his Bachelor's degree from the University of California at Santa Cruz in 1971 and his Ph.D. from Princeton University in 1974. Krantz has taught at the University of California Los Angeles, Princeton, Penn State, and Washington University in St. Louis. He has served as chair of the latter department. Krantz serves on the edi torial board of six journals. He has directed 17 Ph.D. students and 9 Masters students. Krantz has been awarded the UCLA Alumni Foundation Distinguished Teaching Award, the Chauvenet Prize of the Mathematical Association of America, and the Beckenbach Book Award of the Mathematical Association of America. He is the author of 165 scholarly papers and 55 books.
xvi
Vectors P
R
E
V
E
W
In this chapter, we lay the geometric foundation for all of our coming work with functions of two or more variables. Those studies will require us to have a mathematical model of three-dimensional space. Just as we use a real number as the mathematical model of a point on a line, and an ordered pair of real numbers as the mathematical model of a point in the plane, so we use an ordered triple
(x, y, z)
as the mathematical model of a point in three-dimensional space.
Equations among these space variables describe curves and surfaces in space. In this chapter, we will study the simplest such objects, lines and planes, in detail. As an aid to our investigations, we introduce a new concept, the vector, that will be a vital tool in every chapter to follow. A vector may be thought of as an arrow that has a length and a direction, but whose initial point is of no importance. If we reposition the arrow while preserving its direction, then it still represents the same vector. We will learn how to add vectors and how to perform a number of other algebraic operations with them. The interplay between the geometry and algebra of vectors is the perfect device for understanding the relationship between the geometry of lines and planes and their Cartesian equations. The vector construction is ideal for capturing many concepts in both algebra and geometry. But the use of vectors extends to other sciences as well. Many fundamental quantities in physics are understood by means of their direction and magnitude. Force, velocity, and acceleration are examples. Using the length of a vector to represent magnitude, we can model these physical quantities by means of vectors.
721
722
Chapter 9
Vectors
9.1
Vectors in the Plane For many purposes in calculus and physics, we need a concept that simultaneously contains the notions of direction and magnitude. For instance, force, velocity, and acceleration are all quantities that have both direction and magnitude. In this
vector,
section, we will develop a mathematical tool, the
for handling these
concepts.
l•JM@hi[.]:@
A line segment AB between two points
directed line segment
A
and B is said to be a
if one endpoint is considered to be the initial point of
the line segment and the other endpoint the terminal point. We denote the
directed line segment with initial point A and terminal point B by AB. The same pair of points determines a second directed line segment, BA, which is said to be
opposite in
direction to AB (see Figure
1).
x
x
_. Figure 1 Line segment AB, directed line segment
x
AB, and directed line segment BA
The directed line segment AB may be thought of as a straight path from B. Its direction is indicated by an arrow, as in Figure
1.
A
to
We often find it useful
to parameterize AB so that its direction is "respected." In such a parameterization, the initial point of the interval of paramaterization corresponds to
A,
and the
endpoint of the interval of parameterization corresponds to B. Example
1
demonstrates this technique. � EXAM PL E meterization x
1
Suppose
that
A =(5, 6)
=f(t),y =g(t),0� t�1ofAB
Solution We are to find functions f and the line segment from A to B as
t
and B
=(9, 14).
Find a para
that respects its direction.
g so
that the point
increases from 0 to
to the terminal point B, the x-displacement is
9 - 5,
1.
(f(t), g(t))
traverses
From the initial point A
or 4. The y-displacement is
14 - 6, or 8. We use these displacements together with the coordinates of A to define f(t) = 5 + 4t and g(t) =6 + 8t. Then, at the endpoints, we have (f(O), g(O)) = (5,6) =A and (f(l), g(l)) = (5 + 4, 6 + 8) =B. Thus x =f(t), y =g(t), O�t�1 parameterizes a curve that begins at A and ends at B. To see that this parameterized
=f(t) =5 + 4t to obtain t = (x - 5)/4. =g(t) =6 + 8t =6 + 8(x - 5)/4 =6 + 2(x - 5), or y =2x - 4, which equation of a line. Thus the �uations x = 5 + 4t, y =6 + 8t, 0 �t�1
curve is a line segment, we use the equation x It follows that y is the
parameterize the directed line segment A!J.
..,.
9.1 Vectors in the Plane INSIGHT A
The calculation of Example 1 may be carried = (xo, Yo) and B = (x1, Y1). The result is that the equations
x = xo+ t(x1 - xo),
out with any two points
=Yo+ t(y1 - Yo) (0:::::; t:::::; 1)
y
723
(9.1.1)
-->
parameterize the directed line segment AB.
A directed line segment is determined by its initial point, direction, and length. -
Given a directed line segment AB, it is important to be able to construct directed -
line segments with the same length and direction as AB but with other initial points. The next example shows how this is done.
(5, 6) and B (9, 14) as in Example 1. Let 0 denote (3, -6) and S (0, 10), then find points D, P, and R so that the
� EXAM PL E 2 Let A the origin
(0, 0). If C
=
=
=
=
-----+
;::;;t
-----+
-----+
directed line segments, 0P, CD and R.l have the same length and direction as AB.
Solution The x-displacement
length
9
-
5
and =
4
direction
and
of
-
AB
y-displacement
are
both
14 - 6
=
8.
determined If
we
by
add
its
these
displacements to the coordinates of the initial points 0 and C, then we obtain the terminal points P
=
(0 + 4, 0 + 8)
=
(4, 8) and D
=
(3 + 4, -6 + 8)
=
(7, 2).
If we
subtract these displacements from the coordinates of the terminal point S, then we obtain the initial point R
_. Figure 2 Directed line seg ments with the same length and
direction as AB.
Vectors
=
(0 - 4, 10
-
8)
=
(-4, 2).
-
.
-
-
-
In this way, OP, CD, and RS
have the same x- and y-mcrements as AB and therefore the same length and direction (see Figure
2.)
-11111
Imagine that you are exerting a force of constant magnitude to push a stalled automobile (Figure
3). If the street is straight, then the direction of the force F that
you exert is also constant. An arrow may be used to represent the direction of the force F, as in Figure
3.
We can use the length of the arrow to represent the
magnitude of F. The initial point of the arrow can be used to represent the point of application of F. As the car moves, the position of the arrow changes, but the direction and length do not. We can therefore regard the force F as a directed line segment that (a) has a fixed direction, (b) has a fixed length, and (c) can be applied
_. Figure 3
at any point. These considerations suggest creating a mathematical object that captures the direction and length of a directed line segment but which disregards its initial point. Such an object is called a vector.
Let A and B be points in the plane. The collection of all directed -
line segments having the same length and direction as AB is said to be a vector. Every directed line segment in this collection is said to represent the vector.
_J
To help distinguish between vectors and ordinary numbers, we often refer to real numbers as scalars. In textbooks, boldface type such as vis frequently used to denote a vector. When rendered by hand, the vector vis often denoted by v. Informally, we can think of a vector as a "floating" arrow whose initial point can be chosen in any convenient way. Figure
2
illustrates four representations of a single
vector. In practice, we often refer to a vector by one of the directed line segments that represent it. This can lead to no confusion and is often very helpful. Thus we
724
Chapter 9
Vectors may refer to the vector that appears in Figure 2 as vector -
AB,
-
PQ, vector
vector
ON, or vector RS. Even though these four directed line segments are all different, they all represent the same vector.
�
When a particular directed line segment P0P
is used to represent a vector
v
(as in Figure 4), the x-displacement v1= x1 - x0 and they-displacement v2=y1 -y0 do not depend on the choice of initial point P0• We can therefore use the notation
( v1, v2)
v = ( v1, v2) . The quantities v1 and v2 are ( vi, v2) . Notice that, if P= (x, y) and 0= 0,y 0) or (x,y). Thus every point P = (x,y)
to unambiguously denote v. Thus
said to be the components or entries of -
(0, 0), then vector OP
is equal to (x
-
-
gives rise to the vector (x, y) that is represented by the directed line segment from
(
the origin to P Figure
.A Figure 4 The components v1 and v2 of vector v are the same in each directed line segment representation.
vector of P = (x, y ) .
5).
-
The vector OP= (x, y) is sometimes called the position
� EXAM PL E 3 Let A = (5, 6) and B= (9, 14) as in Examples 1 and 2. Let v be the vector that is represented by AB. Write v in terms of its components. If v is represented by the directed line segment 0 P with the origin 0 as initial point, then what is the terminal point P?
v=(a, b), then the components a and b of v are given by a= 9- 5= 4 b= 14- 6= 8. Thus v= (4, 8). The point P = (4, 8) is the terminal point of the directed line segment that begins at the origin 0 and that represents v = (4, 8).
Solution If and
Ref er again to Figure 2.
Vector Algebra
.,..
In this subsection, we introduce some algebraic operations that can be performed with vectors.
The sum
v+w
of two vectors
v= ( v1, v2)
(
and w= w1, w2
)
is
formed by adding the vectors componentwise:
v + w= ( v1, v2 ) +
( w1, w2) = ( v1 + w1, v2 + wz ) .
We can interpret vector addition geometrically by first drawing a directed line segment
AB that represents v and then drawing a directed line segment BC that (as shown in Figure 6) that the initial point of the second
represents w. Notice
v + w is then equal -=-+ ____,. + BC=AC. Figure 7 shows that the sum v + w is parallelogram determined by v and w. It follows that
directed line segment is the terminal point of the first. The sum ____,.
____,.
to vector AC. In other words, AB the diagonal of the v
y
+ w= w + v,
as is also evident from the algebraic definition. c
p
=
(x,y)
x 0 A
.A Figure 5 The position vector of point P (x, y) =
.A Figure 6
.A Figure 7
9.1 Vectors in the Plane
� EXAMPLE 4 Add the vectors v = (-3, 9) and
725
w = (1,8).
Solution We have v+w = (-3+1, 9+8) = (-2, 17) . <11111
0.
Next we define the vector analogue of the scalar
0.
The zero vector 0 is the vector both of whose components are That is, 0 = (0,
0).
The zero vector is the identity for vector addition. This means that, if v is any vector, then v+0
=
0+ v
(
v.
=
)
If v = v1, v2 is a vector, and >.is a real number, then we define
y
the scalar multiplication of v by >.to be >.v = (>.v1,>.vz ) . ,\v2 x
0
V1-------I >+---- A V1------+1•
I
Geometrically, we think of scalar multiplication as producing a vector that is "parallel" to Figure
_. Figure Sa >..>O
v. One may use the idea of similar triangles to check this idea, as
8 suggests.
If>.> 0, then>.v and v have the same direction. If>.< 0, then >..v
and v have opposite directions. These intuitive notions will be made more precise when a formal definition of direction is given later.
y
� EXAMPLE 5 If v = (2,-1), then calculate 2v and -3v. Solution We have 2v = (2 = (-6, 3).
·
2, 2 (-1)) ·
= (4, -2) and - 3v = ((-3)
·
2,(-3) (-1)) ·
...
Vector addition and scalar multiplication can be used together to define vector
w are given vectors, then the expression v - w is interpreted to 9, which features the parallelogram that is generated by v and w. This same parallelogram has been used in �ure 7 to visualize the sum v+w, which is represented by the directed diagonal AC. In Figure 9, notice that v -w is represented by the other directed diagonal, DB, of the parallelogram. To
subtraction. If v and
mean v+((-l)w). See Figure
_. Figure Sb >.. < 0
-----+
understand why, focus on the three sides of the upper-right triangle, which show -----+
-----+
-----+
that DB =DC +CB = v + ( -w) = v
- w.
Furthermore, by concentrating on the
three sides of the lower-left triangle of Figure C
we add to A
w
9, we see that v - w is the vector that
to obtain v.
simple
calculation
shows
that
subtraction
of
(
vectors
)
is
componentwise, just like addition of vectors. Thus if v = vi. v2 and then
(
v -w = v1, v2
performed
w = (w1,
)
w2 ,
) + ( - l )(w1, w2 ) = (v1, v2 ) +( -w1, -wz ) = ( v1 - w1, vz - wz ) .
� EXAMPLE 6 Calculate the difference v - w of the vectors v = (-6,12) and A _. Figure 9
w = (19,-7). Solution We have v - w = (-6 - 19,12 - (-7)) = (-25, 19).
<11111
726
Chapter 9
Vectors
The Length (or Magnitude) of a Vector
Ifv =
(v1, v ) is a vector, then its length is defined to be 2 llvll = (v1 )2+(v2 )2. This quantity is also called the magnitude ofv. If AB is a directed line segment that representsv, then IIv II is just the distance between the points A and B. The length
V
-
of a nonzero vector is positive. The zero vector 0 is the only vector that has length equal to
0.
=(0,0) denote the ongm. Calculate llPQII, II OP II, and Q P OQ (-9,6) = = (2,1). ll ll, ::;-h PQ= (2 - (-9), 1 - 6)=(11,-5), OP= (-9, 6), and O� = (2, 1), we have llPQll, = 112+(-5)2=JI46, II oPll = (-9)2+62 =y1Il7, and •
� EXAMPLE 7 Let 0 for and Solution Because
-----+
•
J
-----+
V
llOQll, =V22+12 =v's. .... THEOREM 1
A. is a scalar, then 11>.v11 =1>-111v11·
If vis a vector and
In words: The length of
Proof. Equation
(9.1.2)
A.v is the absolute value of A. times the length of v.
(9.1.2) is easily derived from
the definitions of length and scalar
multiplication:
J
V
II .\vII =II ( .\vi, .\v2)II = ( .\v1 )2+(.\ v2)2 = v'>!- (v1 )2+( v2)2 = l>-111vII · � EXAMPLE 8 Verify equation
•
(9.1.2) forv=(4, 12) and A.=-5.
Solution We calculate
l>-1 11v II= l-5IV42+122 = 5v'l60 = 2ov1o and
J
V
II>. vII = II (-5) (4, 12) II = II (-20, -60) II = (-20)2+ (-60)2 = 20 (-1)2+(-3)2 =20JIQ. ....
Unit Vectors and Directions
u = (u1, u ) is 1, that is, if II uII = 1, then u is called a unit vector. u= (u1, u2) is2 a unit vector if and only if the point (u1, u ) lies on the 2 unit circle (Figure 10). A unit vector u therefore has the form u =(cos( a) , sin( a)) ( 9.1.3) for some angle a in [0,27!"). Sometimes we refer to a unit vector u as a direction
If the length of Observe that
v=(vi. v ), the vector 2
vector or simply a direction. For any nonzero vector
u= 1 v w
is a unit vector because, according to equation
llull
(9.1.2),
=I 11!11vi = I 11!11 I llvll = 11!11 llvll = 1.
( 9.1.4)
9.1 Vectors in the Plane
1
y
We refer to the direction vector u defined by equation (9.1.4) as the direction of v and write dir(v) x
-1
=
!
11 11 v
(v-=!= 0).
(9. 1. 5)
We do not define the direction of the zero vector. By rearranging equation (9.1.5), we see that every nonzero vector v can be expressed as the scalar multiplication of the direction vector of v by the magnitude of v:
1
v
=
(9. 1. 6 )
llv ll dir(v).
The right side of equation (9.1.6) is sometimes said to be the polar form of v.
• Figure 10 A unit vector u =
727
(u1, Uz)
Let P=(-1,2) and Q=(2, 1). What is the direction u of the . vector v represented by PQ? What angle a, as given by formula (9. 1.3), does u make with the positive x-axis? � EXAMPLE 9
J
We calculate v=(2- (-1), 1-2)=(3,-1) and llvll= 32+(-1)2= JIO. The unit vector
Solution
y
Q
=
(2, 1)
is the direction of v. Figure 11 illustrates both v and its direction u. The angle a that u =(cos(a) , sin(a)) makes with the positive x-axis is the value of a between 37r /2 and 27r such that cos(a) =3/M and sin(a)= -1/JIO. That is, a=27r arctan ( 113). With the aid of a calculator, we find that a� 5.96 radians.
x
2
-1
-1
• Figure 11
Let v and w be nonzero vectors. We say that v and w have the same direction if dir(v)=dir(w). We say that v and ware opposite in direction if dir(v)= -dir(w). We say that v and ware parallel if either (a) v and whave the same direction or (b) v and w are opposite in direction. Although the zero vector 0 does not have a direction, it is conventional to say that 0 is parallel to every vector. Notice that nonzero vectors v and ware parallel if and only if dir(v)= ±dir(w). Our next theorem provides us with a simple algebraic condition for recognizing parallel vectors. THEOREM 2
Vectors v and w are parallel if and only if at least one of the following two equations holds: (a) v=0 or (b) w=A.v for some scalar A.. Moreover, if v and ware both nonzero and w A.v, then v and whave the same direction if 0 < A. and opposite directions if A. < 0. =
We begin by supposing that equation (a) or equation (b) is true. If v=0, then v and w are parallel by convention. Now suppose that v-=!= 0 and w= .Xv for some scalar .X. If .X=0, then w=0, in which case v and ware parallel. If .X -=!= 0, then
Proof.
728
Chapter 9
Vectors the following three facts hold: neither v nor w is 0, both and both
dir(v)
llvll
and
and dir(w) are defined. Consequently, we have
llwll
are nonzero,
dir(v) = ±dir(v). v= (>.v) = v l I l I ii II This equation tells us that v and w are parallel. Furthermore, because A. I IA.I = + 1 when A.> 0 and A. I IA.I = -1 when A.< 0, we see that v and w have the same direction
dir(w) =
1 �1 w
=
( � ) 1 !1 ( � )
;
if A.> 0 and opposite directions if A.< 0.
To prove the converse, we suppose that v and w are parallel and that v =I= 0. We
must prove that w=A.v for some scalar A.. If w= 0, then w=A.v for A.= 0. We may
w, in addition to v, is nonzero. In this case, both dir(v) and dir(w) are defined. Moreover, because v and ware parallel, we have dir(v) = ±dir(w). Finally, noting that llvll =I= 0 and using equation (9.1.5), we calculate therefore assume that
w= llwlldir(w) = ± llwlldir(v) = ± (O, 100) for >.=
±llwll/llvll·
� EXAMPLE
1/1:111 v
= >.v •
1 0 For what value of
a
are the vectors
(a, -1)
and
(3,4)
parallel? Solution For the two given nonzero vectors to be parallel, it is necessary and (75,0) _. Figure 12
An Application to Physics
(3, 4)=A. (a, -1) for some scalar>.. This vector equation is equivalent to the two scalar equations, A.a=3 and -A.=4. We obtain a=3/A., or a= -3/4.
sufficient for
As discussed earlier,
force
is a quantity that is naturally described by the vector
concept. If a force of magnitude F is applied in direction (cos(o:), sin(o:)), then the
principle of superposit ion says the resultant force is obtained by adding the
force is written as the vector (Fcos(o:), Fsin(o:)). The that, if two forces act on a body, then
vectors corresponding to the two given forces. � EXAMPLE 1 1 Two workers are each pulling on a rope attached to a dead tree stump. One pulls in the northerly direction with a force of
100 pounds and the
other in the easterly direction with a force of 75 pounds. Compute the resultant force that is applied to the tree stump. Solution Look at Figure
12.
The force vector for the first worker is
(0, 100)
and
that for the second worker is (75, 0). The resultant force is the sum of these, or (75, 100). The resultant force vector is shown in Figure line segment has magnitude (length) equal to (752 +
The Special Unit Vectors i and j
It is common to let in Figure 13. If v
follows:
=
12. The associated directed 10a2)112= 125.
i denote the vector (1, 0) and j denote the vector (0, 1), as shown (a, b) is any vector, then we may express v in terms of i and j as
v= (a,b)=a (l , O) + b(O, 1)=ai + bj.
� EXAMPLE 1 2 Suppose that
v + w in terms of i and j.
v= (3, -5)
and
w= (2,4).
Solution We have v=3i- 5j and w=2i + 4j. We can calculate
v + w= (3, -5) + (2,4)= (5, -1)= 5i - j.
Express
v, w,
and
9.1 Vectors in the Plane
Alternatively, we can express
729
v and win terms of i and j, as we have already done,
and add these expressions:
v+ w=(3i- 5j) +( 2i +4j) =(3+ 2) i +(-5+4) j=Si-j. The Triangle Inequality
7 suggests that, if v and w are nonparallel v+ w is never greater than the sum of the lengths of v
A glance back at MBC in Figure vectors, then the length of and w. In other words,
y
llv+wll
1 j
=
"ill
In fact, this inequality, known as the
(0, 1)
:5
llvll + llwll
(9.1.7)
·
Triangle Inequality,
is just a vector inter
pretation of the familiar theorem in Euclidean geometry that states that the sum of x
i
=
(1, 0) 1
the lengths of any two sides of a triangle is greater than the length of the third side. Inequality
(9.1.7)
is also true when
v and w are
parallel, for, in this case, we may
write one of the vectors, say w, as a scalar multiple of the other: use equation
(9.1.2)
v =A. w. We then
together with the Triangle Inequality for scalars to obtain
llv+wll =llv+>.vii = II (1+>. )vii =11+>.i llvll :5 (1+ i>.I) llvll = llvll +i>.1 llvll = llvll +II>.vii = llvll + llwll <411
.A Figure 13
·
� EXAMPLE 13 Verify the Triangle Inequality for the vectors
= (8, 6).
v=(-3, 4)
and w
Solution After calculating find that
llvll + llwll =J(-3) 2+42+ V82+62=5+ 10=15, we
llv+wll =II (5, 10) II= villi � 11.18<15= llvll + llwll
·
<1111
We conclude this section by stating the distributive, associative, and commu tative laws for scalar multiplication and vector addition . If
u, v, ware vectors and>.,
µ are scalars then a. b. c. d. e.
Q UIC K
Q UIZ
1. 2. 3. 4.
v+w=w+v u +(v+ w) = (u +v) + w >.(µv) =(>.µ )v >.(v+ w) = >.v+>.w (>. + µ )v=>.v+ µv P= ( 2, - 1) , then for what pomt Q does PQ represent (-3, - 2)? What is the magnitude of (-3, 4)? What is the direction vector of (12, -5)? If (a, b) is opposite in direction to (-4, 3) and has length 2, then what is a?
If
Answers
1. (-1,-3)
•
-----+
2. 5 3. (12/13,-5/13) 4. 8/5
730
Chapter 9
Vectors
EXERCISES
In each of Exercises 1-8, sketch form (a,b).
1. 2.
P= (5,8) P= (1,1) P= (0,0) P= (1,1)
PQ,
and write it in the
Q=(4,4) Q=(3,9) Q=(-5,1) Q = (2,2) Q=(4,-3) Q=(0,0) Q = (1,0) Q = (2,-2)
3. 4. 5. P= (-1,8)
P= (0,1) 7. P= (-5,6) 8. P= (2,2) 6.
In each of Exercises
12.P=(l,1)
Q=(-7,5)
7r/6 37r/4 7r 57r/3
In each of Exercises 33-36, write the given vector v in the form >.u where >. is a positive scalar, and u is a direction vector.
9-12, calculate the lengths of PQ and
Q = (2,0) Q = (3,1) Q= (- 6,4)
In each of Exercises 29-32, determine the direction vector u that makes the given positive angle a with the positive x-axis.
29. 30. 3L 32.
RS. Also determine whether these vectors are parallel. 9. p = (1,2) 10. p = (1,3) lL P= (3,1)
Fz=-5i + 13j
28. Fi=i-j
Problems for Practice
R= (3,6) R=(l,2) R=(7,-2) R= (1,0)
S= (6,0) S= (6,3) S= (5,5) S= (17,-8)
In each of Exercises 13-16, calculate the indicated vector given that v= (4,2) and w= (1, -3). For each exercise, draw a sketch of v, w, and the vector you have calculated. 13. 2v+w
33. v= (6,-5/2)
34. v = (-Vil, 5) 35. v= 3i-2j 36. v = - i-
� v:. j
In each of Exercises 37-40, write the given vector v in the form (a,b).
37. 2( 5i + 3j) - 3(i - j) 38. 4(i - 2j)+ 5(i+2j) 39. 3j - 2(-3i+j) -2i 40. 5j + 2(-j + 4i) -3(2j + i)
14. (1/2)v + 3w 15. v - 2w
Further Theory and Practice
16. -3v+2w In each of Exercises
17-20, two points P and Q are given.
Determine (a) the vector v that is represented by PQ, (b) the length of v, (c) the vector that has the same length as v but is in the opposite direction of v, (d) the direction vector of v, and (e) a unit vector that is in the opposite direction of v.
17. p= (5,-7) 18. p = (-7,1) 19. p = (-3,7) 20. p = (3,2)
Q=(0,5) Q = (-14) Q = (1,-9)
Q = (l,1)
In Exercises 41-54, determine the value of a from the given information about v = (a, b).
ll
4L a>b=3 and v =7 42. a < b=5 and v =13 43. v is parallel to
(5, 3b)
44. v is parallel to (7, -b/2)
ll v ll = 3 ll w ll = (-3, c) and ll v ll = II II
45. v is opposite in direction to w= (2, c) and 46. v is opposite in direction to
�
47. both a and b are positive, and v is a diagonal of the parallelogram generated by (3,5) and (-1,7)
In each of Exercises 21-24, vectors v and w are given.
48. both a and b are negative, and v is a diagonal of the
Express v, w,-4v, 3v - 2w, and 4v + 7j in terms of the unit vectors i and j.
49. both a and b are positive, and v is a diagonal of the
2L v= (-7, 2) 22.
v= (0,3)
23. v = (6,-2)
24. v = (1/3, 1)
w= (-2, 9) w = (-5 , 0) = (9,-3) w= (-3/2,1/2)
parallelogram generated by (4,7) and (-1,8) parallelogram generated by (-7, 6) and (3,7)
50. both a and b are negative, and v is a diagonal of the
parallelogram generated by (2,9) and (-5, 3) SL dir(v) = dir( )and II v ii = 3 for w = (1, v'3)
25-28, calculate the resultant force F =Fi+ F2• What is the magnitude of F?
52. dir(v) = dir(2w)and ll v ll = 1/2 for w = (2,1) 53. dir(v)= dir(-w) and ll v ll =10 for w= (3, -4) 54. dir(v)=-dir(w) and ll v ll =1/../2 for w= (-v'3, v's)
25.
55. Mr. and Mrs. Woodman are pulling on ropes tied to a
In each of Exercises
Fi= (3,0) Fi= (1,2) 27. Fi=i-2j
26.
Fz= (0,4) Fz= (2,2) Fz=2i + j
heavy wagon. Refer to Figure 14. If Mr. Woodman pulls with a strength of 100 pounds, then how hard must
9.1 Vectors in the Plane
731
60. Let P= (P1. pz) and Q =qi, qz) be distinct points in the plane. Let M be the midpoint of the segment joining P nd Q. Use the relationship
OM=OP+ iPQ
to deter
mine the Cartesian coordinates of M.
Iman
w be
61. Let v and
two given nonzero vectors. Prove that
there is a unique number
such that
v + >..w is as short as
possible. In other words, the function from JR o by
v + >.wll
H
efined
ttains an absolute minimum value.
62. Let m and b be constants. Show that v is parallel to PQ for every pair of points P and Q on the line y = mx + b f and only if
63. Verify that 64. Let v nd
Mrs. Woodman pull so that the wagon moves along the dotted line? What is the magnitude of the resultant force?
i, F2,
120 newtons applied in the direction
must F3 be if the point mass
is to
2
50
has magni
5, 415
. What
remain at rest?
57. A boat maintains a straight course across a river at a rate of
500
m wide
m/min. The current pulls the boat
down river at a rate of carried?
3
1
mile wide and flows south with a current of
miles per hour. What speed and heading should a
motorboat adopt in order to cross the river in
10 minutes
and reach a point on the opposite bank due east of its point of departure?
59. Bjarne, Leif, and Sammy are towing their vessel. The forces that they exert are directed along the tow lines, as indicated in Figure
15,
which also provides the magni
a,
u
be any other vector. Prove that there
re unique scalars>.. andµ such that
u=.Av+ µw. If v= i nd
w=j, then .A andµ are the entries of vector n. think of v and w generating a coordi
In general, we can
nate system with }.. and µ, the entries of vector
n
in this
new coordinate system.)
65 and 66, develop a formula for the distance of a point to a line.
65. Suppose that A and B are not 0. Consider the line l whose Cartesian equation
Po=(.xo,y0)
is Ax+By + D= 0.
does not lie on l. Show that
Suppose that
=A, B is a x, -Ax/B
n
vector that is perpendicular to l. Let Q"' =
DIB) be a point on l. Calculate Po� x is� parallel to n?
�-For what value
66. For the value of� determined in Exercise
Ax +By +D
tudes of their forces. (Note that the actual force vectors are not drawn in Figure
v= v.
w be two nonzero vectors in the plane that are
parallel. Let
20 m/min. When the boat docks on
the other shore, how far down river will it have been
58. A river is
not
and F3 ·re applied to a point mass.
Suppose that F1=lOOj newtons and tha1 r
tude
to
Describe all vectors that are perpendicular to
A Figure 14
56. Three forces
v=>.., 1, m) for some scalar>... w= - , ) is perpendicular
= o llP-'011 JA2 + B2
15.) What is the resultant force?
of
65, show that
.
What does it mean to state that this is the distance of
Po
o l?
67. Let P =(pi,pz) and Q = qi, qz) be distinct points in the plane. Suppose that
0
coordinates of the point distance from Pis
·
P,
1.
What are the Cartesian
on line segment PQ whose
PQ ? Use vectors to find an elegant
solution to this problem. Let
A, B, and C
not collinear. Let
enote any three distinct points that are , and "I denote the midpoints of line
a:,
BC, AC, and AB, respectively. The line segments Aa, B{3, and C"( are called medians of 6.ABC. Exercises segments
68-70 concern these medians and their common point of intersection, the
centroid 4-
of
6.ABC.
4
....+
68. Prove that Ao: + B/3+C = 0. 69. Let M e point of intersection of Aa ndB/3. Show that A Figure 15
AM=
2 3}.A
-
ndBM =
2/3) nJJ.
Symmetrically,
same relationship holds when the medians
Bf3
the
nd C"f are
732
Chapter 9
Vectors
selected. Deduce that all three medians intersect in one
point. It is called the
centroid
of MBC.
70. Express the coordinates of the centroid of MBC in terms of the coordinates of the vertices A, B, and C. 71. Let P, be the point that
P1P1+3
{r, r) for t > 0. Find the value oftsuch
has the same direction as
(1,7).
72. In his first season in Major League Baseball Musial's batting
0.426.
average was
(1941), Stan
That
DiMaggio, already an established star, hit for next year, Musial's average was was
0.305.
0.315
year, Joe
0.357.
The
and DiMaggio's
Because Musial's average exceeded DiMag
gio's in both of those years, it seems obvious that his combined average over the two years must have exceeded
parallelograms. Observe that a diagonal of parallelogram
P2
may have greater slope than the corresponding diag
onal of P1 even if the edges of P1 have greater slopes than
the the corresponding edges of
Calculator/Computer Exercises 73. Plot y = c and y = 2 -x2 for
y
year period. Do the same for DiMaggio-you will find that, contrary to expectation, it is
greater
than Musial's.
The phenomenon at work here is known as
Paradox
Simpson's
in statistics. Explain geometrically how this
and
Q
be
PQ?
x2 + 2x+ 2. Let P" be the point (x, f(x)). Plot II OPx II· What is the point on the graph of f that is
=
75. For
f(t)
=
P
=
closest to the origin?
tis
=
Let
and first quadrants respectively. What is vector
74. Let f(x)
be
Calculate the combined average for Musial over the two
-2 ,s x ,s 1.
the points of intersection of the two curves in the second
that of DiMaggio. Not so fast! Define the
slope of (a, b) to b/a if a =f 0. We can represent Musial's number of at bats in 1941 (47) and hits {20) by the vector m1 =(47,20). The slope of this vector, 20/47, or 0.426, is Musial's bat ting average for 1941. The vector m2 representing his data for 1942 is m2=(467,147). The corresponding vectors for Joe DiMaggio are d1 (541, 193) and d2 (610, 186).
P2•
-2oStoS3 let Q1, llPG II where P
=
f(t) minimized?
(t, e-1+t 2). Let (1,4). Plot f(t). At what value of
denote the point =
76. Suppose that for each point P1= (t, r) on the curve y =x2, a point Q1= (�(t), rJ(t)) is found such that (a) �(t) > t,
{b) ®'
is a unit vector, and (c)
P,Qi
is tangent to the
curve. Plot the curve whose parametric equations are x
= �(t), y = rJ(t).
What point on this parametric curve is
farthest below the y-axis?
77. Suppose that, for each point P1= (t, r) on the curve y = x3, a point Q, =Wt), rJ(t)) is found such that (a) �(t) >
t, {b)PiG' is
a unit vector, and (c)
P,Qi
is tangent to the
curve. Plot the curve whose parametric equations are
e{t), y rJ(t). What are its x-and y-intercepts? To what
paradox can occur by representing the combined averages
x
that you calculated as the slopes of the diagonals of two
values oft do these intercepts correspond?
=
=
9.2 Vectors in Three-Dimensional Space We are familiar with locating points on the line by using one coordinate and points in the plane by using two coordinates. Now we turn to three-dimensional space, where locating points requires three coordinates.
.J-i.�
��.Jf"
_. Figure 1
Figure 1 exhibits three mutually perpendicular axes: the x-axis, the y-axis, and the z-axis. The yz-plane contains the y- and z-axes. It coincides with the plane of this page. The xy- and xz-planes are defined analogously. They emerge perpendi cularly from this page. A point with spatial coordinates (x, y,
0) lies in the xy-plane.
In general, the third coordinate z of a point (x, y, z) indicates the height of the point above or below the xy-plane. Figure 2 exhibits a point (x, y, z) with z >
0:
It lies z
units above the point (x, y, 0) in the xy-plane. Notice how we use dotted lines in our picture to create a sense of perspective so that we can picture a three-dimensional concept on two-dimensional paper. Points (x, y, z) with z < 0 lie below the xy plane. The coordinates x, y, and z of the point P
=
(x, y, z) are called the Cartesian
coordinates of P. An equation involving Cartesian coordinates is said to be a Cartesian equation. The terms rectangular coordinates and rectangular equation are sometimes used instead. The set of all spatial points is denoted by JR3 to signify that three coordinates are needed to describe such points.
9.2
z , ) (xy , z
•I 1 I 1
Vectors in Three-Dimensional Space
733
Just as the coordinate axes in the plane divide the plane into four quadrants, so do the xy-plane, the xz-plane, and the yz-plane divide space into eight octants. The octant that contains points with all three coordinates positive is called the first octant. It is sometimes helpful to think of the xy-plane as the floor of a room. 1bink of the origin 0 (0, 0, 0) as a corner of the floor and of the positive z-axis as the join of two walls(the two walls are the xz-plane and the yz-plane). Look at Figure 3, which depicts the first octant from this point of view. Let P (x0, y0, zo) be a point with all three coordinates positive. Then the first two coordinates of P tell which position P' (x0, y0,0) on the floor the point Plies over. The last coor dinate tells how high above the floor Pis. =
:
Y li-+--��-
=
x .A.
=
Figure 2
Sketch the points(3, 2, 5),(2, 3, -3), and(-1, -2, 1).
� EXAMPLE 1
These points are graphed in Figure 4. Some polygonal paths connecting these points to the origin are also plotted for perspective. For example, to reach the point(-1, -2,1) from 0 (0, 0, 0), we may go 2 units in the negative y-direction, arriving at the point(0, -2, 0), then 1 unit in the negative x-direction, arriving at the point (-1, -2, 0), and finally 1 unit in the positive z-direction, finishing at (-1, -2,1). .... Solution
=
z
z
z 3 ( ,2, 5)• y 1
t
1 23 y
/ 2 i 3 ( 3 , 2 0 , ) x -·
x .A. Figure 3 A point P (xo,yo, zo) in the first octant and the projection of P to the xy-plane =
Distance
.A.
z
( 1-2 , ,1) • 123y (- 1'-2'0) .I 0 ( , 20 , )9 � 3 x
:,,.'
123y
:2
--±(2,3,0)
1
�
(2,3,-3)•
Figure4
Two distinct points in space determine a box with sides parallel to the coordinate axes(see Figure 5). The length of a main diagonal of this box coincides with the di.stance d between these two points. We have labeled length, width, and height of the box as a, �.and 'Y· One main diagonal with length dis drawn while 8 is the length of a minor diagonal that is shown. Notice that, by the Pythagorean Theorem, 82 ff + ry2 and d1- a2 + 82• Substituting the first of these equations into the second yields d2 a2 +ff+ ry2. Translating this calculation into coordinates gives us a formula for the distance between two points in space: =
=
=
THEOREM 1
The distance d(P, Q) between the points P
=
(qi. q2, q3) is given by d (P, Q) .A.
Figure
5
=
(p1, p , p3) and Q 2
V(q1 -P1)2 + (q2 -P2)2 + (q3 - p3)2.
=
(9.2.1)
734
Chapter 9
Vectors � EXAM P L E 2 Calculate the distance between the points ( 4 , -9 , 7) and (2,1,4). Solution
According to Theorem 1, the required distance is
VC2 - 4)2+ (1 - (- 9))2+ (4 -7)2=VC-2)2+ (10)2+ (-3)2=Jill.
....
If Po=(xo, Yo. zo) is a fixed point in space and if r > 0 is a fixed number, then
the collection of all points with distance r from Po is the sphere with center Po and
radius r. From Theorem 1, we see that the set of points that satisfy the equation
or (9.2.2) is the sphere with center P0 =(x0, y0 , z0) and radius r. Notice that the left side of equation (9.2.2) is quadratic in the three space variables x , y , and z. The coefficients of I. and z2 are all 1. There are no mixed products xy, xz, or yz.
x2,
� EXAM PL E 3 Determine what set of points is described by the equation
x2+1+r - 6x + 4y + 6
=
o.
Solution We complete the squares for x and y (because of the presence of the linear terms and 4y): The equation y2 z2 4y 6 =0 is equivalent to (y2 4y) z?= -6 or
-6x x2+ + -6x + + (x2-6x) + + + (x2-6x + (-6/2)2) + (y2+ 4y + (4/2)2) + i1'=-6 + (-6/2)2+ (4/2)2• After simplification, we have (x - 3)2+ (y + 2)2+z2=7, or (x - 3)2+ (y (-2))2+ (z - 0)2=7. We see that the given equation describes a sphere with center (3, -2,0) and radius v'7 (see Figure 6) . ...z
.A. Figure 6 The sphere (x -3)2
+(y - (-2))2
+
(z - 0)2
=
7
9.2 Vectors in Three-Dimensional Space
�·Mll!!l[.Jll
Po� (x0, y0, z0)
Let
be a point in space, and let
r
735
be a positive
number. The set
{PE �3: d (P,Po)
{(x,y, z): (x - xo)2 + (y - Yo)2 + (z - zo)2
is the set of all points inside the sphere (x set is called
This
{PE �3: d (P,Po) < r}, or, equivalently,
{(x,y, z): (x - xo)2 + (y - Yo)2 + (z - zo)2 < r 2}, is the preceding open ball plus its boundary, the sphere. We call this set the closed
ball with center P0 and radius r.
� EX A M P L E
4 Describe the set of points S that satisfy the equation
x2 + y2 + z2 + SO + Sy + 16z <6x. Solution This set can be identified by rearranging the given inequality as
(x2 - 6x)+ (y2 +Sy)+ (z2+ 16z) <-SO
and completing the squares in each set
of parentheses:
( x2 - 6x + ( �) 2 ) + (y2 + Sy + ( �) 2 ) + ( z2 + 16z + ( �6) 2 ) 2 �) 2+ ( 126) 2 < -so + ( �) + (
This inequality simplifies to
(-4))2 + (z - (-S))2 <32
•
(x - 3)2+ (y +4)2 +(z+ s)2 <9
other words, S is the open ball with center
A
vector
(x - 3)2+ (y -
From this last inequality, we see that the given set
consists of those points whose distance from the point
Vectors in Space
or
(3, -4, -S)
(3, -4, -S)
and radius
in three-dimensional space is an ordered triple
is less than
3.
3.
In
<11111
(vi. v , v3). 2
This is the
algebraic definition of vector and is the one that we use in calculations. But there is a geometric interpretation as well. Like a vector in the plane, a vector in space can be represented by a directed line segment that specifies magnitude and direction. We say that a directed line segment
represents satisfy
the vector
v
= (vi . v , v3) 2
----+
PQ
P and terminal point Q P =(pi,p ,p3) and Q =(qi, q , q3) 2 2
with initial point
if the points
(9.2. 3) P= (x,y, z) is a point in space, the position vector of P. If
then the vector
----+
OP= (x,y,z)
is sometimes called
736
Chapter 9
Vectors
= (3, -1, 4), Q = (2, 1, 0), R = (-6, 9, 7), and S= (-7, 11, 3). What vectors are represented by the directed line segments PQ, PR, and @?
� EXAMPLE 5 Let P
Do any of these directed line segments represent the same vector?
PQ is (2-3, 1-(-1), 0- 4) or (-1, 2, -4), . __!?,y.PR 1s (-6-3, 9- (-1), 7- 4) or (-9, 10, � and t� vector represented by QS 1s (-7-2, 11- 1, 3-0) or (-9, 10, 3). Thus PQ and PR represent different vectors. However, PR and QS represent the same vector, (-9, 10, 3). ... Solution The vector represented by -
the vector represented
Vector Operations
z
Addition. If
v (v1, v2, v3 ) =
and
w ( w1, w2, w3) , then we define =
As in the plane, the geometric interpretation of vector addition is obtained by using v
v+ w, indicated by the dotted directed line seg 7, is the diagonal of the parallelogram determined by v and w.
the parallelogram rule. The sum ment in Figure
� EXAMPLE 6 Suppose that
v= (3, 0, 1) and w= (0, 4, 2).
Calculate
v+w,
and sketch the three vectors.
Solution We calculate
_. Figure 7
drawn in Figure z
3
v+
w =
(3, 4, 3)
8.
v+ w= (3+0, 0+ 4, 1+2)= (3, 4, 3). The three vectors are
"ill
Scalar Multiplication. If v= ( v1, v2, v3 ) is a vector and .A is a real number (i.e., a scalar) then we define the scalar multiplication of v by .A to be
Geometrically, we think of scalar multiplication as producing a vector that is parallel to
_. Figure 8
v but
with different length when
I .Al -=/= 1.
An argument with similar
triangles establishes that two nonzero vectors have the same direction if one is a positive scalar multiple of the other. They have opposite directions if one is a negative scalar multiple of the other. For now, these notions are intuitive gen eralizations of facts that we already know for vectors in the plane. They will be made more precise in the discussion that follows later in this section.
v= (2, -1, 1). Calculate 3v and -4v. Solution We see that 3v= ( 6, -3, 3) and -4v= (-8, 4, -4). "ill Subtraction. If v and w are given vectors, then the expression v-w is interpreted to mean v+ ((-l)w). If v = ( vi, v2, v3 ) and w= ( w1, w2, w3 ) , then � EXAMPLE 7 Suppose that
v-w= v+ (-1)w= (V1 - W1' Vz - Wz' V3 - W3 ) . � EXAMPLE 8 Suppose that
v-w and w-3v.
v= (-6, 12, 5) and w= ( 19, -7, 4).
Calculate
that v-w= (-6- 19, 12- (-7), 5- 4)= (-25, 19, 1) w-3v= ( 19-3(-6), -7-3 12, 4-3 ·5)= (37, -43, -11). "ill
Solution We see
·
and
9.2 Vectors in Three-Dimensional Space
The Length of a Vector
737
If v= (v1, v2, v3) is a vector, then the length or magnitude of vis defined to be (9.2.4 ) If PQ is a directed line segment that represents v, then IIvII is just the distance between the points P and Q, as can be seen from formulas (9.2.1 ) and (9.2.3). If vis a vector and >. is a scalar, then the magnitudes of vectors v and >.v are related by (9.2.5)
I i = l>-1 llvll· I>.v Thus the length of A.vis IA.I times the length of v. �
EXAM PL E 9
Let
0
denote the origin . Suppose that P= (-9, 6, -3) and
Q= (2 , 1, 7). Calculate llfiQ II, II oPII, and Solution
We find that
II 0Q II·
2 J11 2+ 5+ 2 (7- (- 3))= 2 102= v'246. llPQ II= (2- (-9 ))2+(1- 6)+ Also Unit Vectors and Directions
V 2 v'I26 and lloPll= V(- 9 )2+ 62+(-3)=
2 72= J54. .,.. llOh II= V2 2+1 +
A vector with length one is called a unit vector. We sometimes refer to a unit vector as a direction vector or a direction. � EXAM PL E 1 0 Is there a value of r for which u= (-1/3, 2/3, r) is a unit vector? Is there a value of sfor which v= (-4/5, 415,s) is a unit vector?
We set II u II=1 and attempt to solve for r. This gives us (-1/3)2+(2/3)2+r= 2 1orr= 2 1- 1/9- 4/9= 4/9. Thus if r= 2/3 or r= -2/3, Solution
is a unit vector . On the other hand, II vII= (-4/5)2+(4/5)2+s= 2 yf32/25+s2 > J1+s2• Because this inequality is strict , we see that llvll > 1 for every value of s. Therefore vector vcannot be a unit vector for any value of s . .,.. then
V
u
If vis any nonzero vector, then we may form the vector dir(v)=
1 !1 v
(v# 0),
(9.2.6)
which , according to equation (9.2.5) with>.= 1/ llvll, has length 1. We call dir(v) the direction of v. If we multiply each side of equation (9.2.6) by llvll, then we obtain v= llvll dir(v)
(v# 0).
(9.2.7)
The right side of equation (9.2.7) is sometimes said to be the polar form of v. Let v and w be nonzero vectors . We say that v and w have the same direction if dir(v)=dir(w). We say that vand ware opposite in direction if dir(v)= -dir(w). We say that vand ware parallel if either (a ) vand whave the same direction or (b ) v and w are opposite in direction . Although the zero vector 0 does not have a direction, it is conventional to say that 0 is parallel to every vector.
738
Chapter 9
Vectors Notice that nonzero vectors v and ware parallel if and only if dir(v) = ±dir(w). Our next theorem provides us with a simple algebraic condition for recognizing parallel vectors.
THEOREM 2
Vectors v and w are parallel if and only if at least one of the
following two equations holds: (a) v = 0 or (b) w= >..v for some scalar >... Moreover, if v and ware both nonzero and w='Av, then v and whave the same direction if >.. > 0 and opposite directions if >.. < 0.
The proof of this theorem is the same as that of its planar analogue (Theorem
� EXAM PLE 1 1 Suppose that v = (4, of
b
and
2
1 ).
of Section
c
3, 1) and w= (2, b, c). Are there values
for which v and w are parallel?
>. then (4, 3, 1) = >.(2, b, c) = (2>., >.b, >.c). It >. = 2. Therefore 3 = >.b = 2b and 1 = >.c = 2c. We conclude w= (2, b, c) are parallel if and only if b = 3/2 and c = 1/2.
Solution If v = >.w for some scalar follows that
4 = 2>.
4, 3, 1)
that v = (
The Special Unit Vectors i, j, and k z 2
1
or
and
Let i denote the vector (1, 0, 0), j the vector
(0, 1, 0), and k the vector (0, 0, 1). These
are the unit vectors that give the positive directions of the coordinate axes, as
9. If v = (v1, v2, v3) is any vector, then we may express v in terms of follows: v = v1i + v2 j + v3k.
shown in Figure
i, j,
and
k
as
� EXAMPLE 1 2 For what values of band care the vectors v and w= k
-3i + 2j +ck
Solution The
vectors
v=
9i +bj -k
and
w= -3i + 2j +ck
direction if and only if v = >.w for some negative scalar
>..
A Figure 9
The Triangle Inequality
-1
9i +bj-k
are
opposite
in
The vector equation
v = >.w is equivalent to the three simultaneous scalar equations and
=
opposite in direction?
9
= -3>.,
b = 2>.,
= c>.. The first scalar equation tells us that >. = -3, which indeed is negative.
b = 2 (-3) = -6 and c = -1/(-3) = 1/3. Substituting these values for b and c in the formulas for v and w, we obtain the vectors 9i - 6j -k and -3i + 2j + (1/3)k, which are indeed opposite in direction.
The second and third scalar equations give us
The Triangle Inequality,
ll v+wll :5 ll v ll + ll wll ,
(9.2.8)
1, holds for spatial vectors as well. 1 for planar vectors also applies in space. In the case that ware nonparallel, we see from Figure 7 earlier that equation (9.2.8) is just a
which we learned for planar vectors in Section The proof given in Section v and
vector interpretation of the theorem in Euclidean geometry that states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side.
� EXAMPLE
(2,-2,-1)
1 3 Verify
and w
=
(3,-6,-2).
the
Triangle
Inequality
for
the
vectors
v=
9.2 Vectors in Three-Dimensional Space Solution After calculating
V
739
llvll + llwll = J22+ (-2)2+ (-1)2+
32+ (-6)2+ (-2)2= 3+7=10,
we find that
llv+wll = II (5, -8, -3) II = V52+ (-8)2+ (-3)2=v'98
·
"ill
We conclude this section by stating the distributive, associative, and commu tative laws for scalar multiplication and vector addition. If
>., µ are scalars, then v+w=w+v +(v+w) = ( +v) + w >.(µv) =(>.µ )v >.(v+w) =>.v+ >.w (>. +µ )v=>.v+µv
u,
v, w are
vectors in
space and a. b. c.
d. e.
u
u
Equations a through e may be routinely verified.
Q UIC K
Q U IZ
1. 2. 3. 4.
What is the distance between the points
(3, 1, 5) and (1, -1, 6)?
= (-2, -1, 2) and Q = (5, 3, 6), then what is the length of the vector represented by the directed line segment PQ?
If P
(8, -1, -4)? 2k - 2(i+ 4j - 7k)+ 3(-i+ 4j - 5k)
What is the direction of Write
in the form
(a, b, c).
Answers
1. 3
2. 9
3. (8/9, -1/9, -4/9)
4. (-5, 4, 1)
EXERCISES Problems for Practice In each of Exercises 1-4, graph the given point graph the point that lies directly above or below
P
P.
Also
in the xy
12. 13. 14.
plane.
In each of Exercises 15-20, explain how to recognize that the given Cartesian equation is not the equation of a sphere.
1. P={3,2,1) 2. p=(2,0,5) 3. P=(-4,1,-2) 4. p=(3, -1,2)
15.
In Exercises 5-8, calculate the distance between the given pair of points.
5. 6. 7. 8.
(2,0,-3) (-2,-3,-5) (4,4,1) (-4,2,1)
(8,3,6) (4,2,0) (1,1,4) (8,0,4)
In each of Exercises 9-14, determine the center and radius of each sphere whose Cartesian equation is given.
9. 10. 11.
-x2-y2-z2 = 6x -Sy+ 12z-4 x2+y2+z2=x -4x2 - 4y2 - 4z2=-2x - 2y - 2z - 6
x2+ y2+ z2+2x+ 4y+ 6z= 8 3x2+ 3y2+ 3z2 - 12x+ 6y= 0 4x2+4y2+4z2+ 4x+4y+ 4z=2
x2+y2+2z2+ 2x+4z = 8
16. 2x2 - 2y2+ 2z2= 1 17. x2+y2+z2+4xy=2 18. x2+y2+z2+2z+2=0 19. (x - y)2+z2=1 20. x2+y2 = 2x+ 4y -7-z2
In each of Exercises 21-26, find an equation or inequality P(x, y, z) such that the given set is described by {(x,y,z) : P(x, y,9z)}.
For
example, the
x2+ y2+ z2 = O}, and
origin
is
the
set
{(x,y,z)
:
the set of points above the xy-plane is
{(x,y,z):O
2L The open ball with center (3, -2,6) and radius 4. 22. The closed ball with center (1,2,9) and radius 1. 23. The set of points whose distance to the point (11', 7!', -7r) is 7!'.
7 40
24. 25. 26.
Chapter 9
Vectors
The set of points whose distance to the point (1, -2, 0) is not greater than 3. The set of points outside the closed ball with center (2, 1, 0) and radius 2. The set of points outside the open ball with center (-3,-1, -5) and radius 5.
Points P, Q, R, and S are given--> in each of Exercises 27-30, --> Calculate the lengths of vectors PQ and RS. Also determine if the two vectors are parallel. 27. 28. 29. 30.
P=(l,2,-1) P=(-1,3, 6) P=(0,1, 4) P=(l, 1, 2)
Q=(2,0,-7) R=(2,6,9) S=(4, 2,-3) Q=(4,1,5) R=(l, 1,0) S=(l, 3, 5) Q=(-1,3,6) R=(6, -3,0) S=(l, 1,2) Q=(-7, 5, 4) R=(l, 0, 6) S=(l9, -12, 0)
In each of Exercises 31-38, let v= (3, 2, 6) and w= (4,-1, 8). Calculate the specified vector. 31. 32. 33. 34. 35. 36.
-3w 3v - 4w -4v+3w (1/2)w - (l/3)v 2dir(v) 21dir(v) - 18dir(w)
37. v+3k 38.
w - 4i+7k
In each of Exercises 39-42, points P and Q are given. State (a) the vector v which is represented by PQ, (b) the vector that has the same length as v and direction opposite to that of v, (c) the length of v, (d) the direction dir(v) of v, and (e) the vector with length 12 that has the same direction as v. 39. 40. 41. 42.
P=(l,2,-1) P=(-2,2, 3) p=(3, 1, -3) p=( 9,8, -1)
Q=(3,-1,5) Q=(4,-4,-4) Q=(4, 0,-5) Q=(5, 12,6)
In each of Exercises 43-48, let v= (3, -4, 1) and w= (-5,2, 0). Express the given vector in the form ai+bj+ck. 43. 44. 45. 46.
-5v 4w w+2v 2v - w
Further Theory and Practice 53.
In each of Exercises 54-61, find an equation or inequality P(x, y,z) such that the given set is described by {(x,y,z): P(x, y, z)}. See the instructions to Exercises 21-26 for examples. 54. 55. 56. 57. 58.
59. 60.
61.
62. 63.
64.
49. 50. 51. 52.
(3, -2, 1), (1, -4,2) (-1, 1,0), (4, 0, 1) -i+ j+k, 2j - k -2i+3k, -j +k
What are the lengths of the diagonals of the parallelo gram determined by the vectors (1, 1,0) and (0, 1,2)? Give a geometric description of the set of all vectors of the form
Consider the vectors u= (1, 3, 4), v= (-2, 1, 6), and w= (0, 1, 5). If mis any other vector in space, then show that there is a unique triple of numbers a, b, c such that m=au+bv + cw .
65.
66.
2w+4j
In each of Exercises 49-52, the two given vectors deter mine a parallelogram P. Calculate the vectors with positive first entries that represent the diagonals of P.
The yz-coordinate plane. The xz-coordinate plane. The plane that is parallel to the xy-plane and that passes through the point (rr, -?, 7r3). The open half space that contains all points that lie on the same side of the xz-coordinate plane as the point (1, 1, 1). The half space that comprises the yz-plane as well as all points that are on the same side of the yz-plane as the point (1, 1, 1). The set of points whose distance to the xy-coordinate plane is greater than 5. The set of points that lie in one or more of the three coordinate planes. (In other words, the union of the three coordinate planes.) The set of points that are farther from the origin than from (2, -1,-2)
(cos(O)sin(¢), sin(O)sin(¢), cos(¢) ) 0 ::5 (} < 27r, 0 ::5 ¢ ::5 7r.
47. v-3i 48.
Let P0 =(x0, y0, zo) be a point in space. Which point in the xy-plane is nearest to P0? Which point in the yz-plane is nearest to P0? How about the xz-plane?
A student walking at the rate of 4 feet per second crosses a 16 foot high pedestrian bridge. A car passes directly underneath traveling at constant speed 40 feet per second. How fast is the distance between the student and the car changing 2 seconds later? Suppose that P, Q, and R are three noncollinear points. Show that there are three points A, B, and C, any one of which together with P, Q, and R forms a parallelogram. Show that �
�
�
�
�
�
OA +OB +O C=OP+OQ +OR. 67. 68.
69.
On which sphere do the four points (5, 2,3), (1,6, -1), (3,-2,5), and (-1,2,-3) lie? What is the minimum value of d(P, Q) if P is a point on the sphere x2 + lOx + y2 -l2y + z2 -2z + 37=0 and Q is a point on the sphere x2 -12x +y2 + Sy + z2 - 6z +45= O? Atomic particles may carry a positive charge (like a proton) or a negative charge (like an electron). A charged
9.3 The Dot Product and Applications
7 41
Suppose that v1, v2, ... , vN are vectors and >..1, >..2 , ... ,>..N are scalars. The vector w = >..1 v1 + >..2v 2 + ... + >..NvN is said to
particle exerts a force on every other charged particle. Suppose that P is the location of a charge p and Q is the
force exerted by the particle at P on the particle at Q is
be a linear combination of the vectors V1, v2 , ,VN· If every vector in R3 is a linear combination of vi, v2, , VN, then we say that the vectors v1, v2, ... ,vN span or generate R3. The
given by
vectors
location of a charge q. Let r be the distance between P
and
where
Q.
Coulomb's Law
states
that
the
-
u
• • •
electric
is the unit vector in the direction of PQ and
co
• • •
i, j, k, for example, span R3• In each of Exercises 73-76, determine whether the three given vectors span R3•
73. 74. 75. 76.
is
a positive constant, the numerical value of which depends on the units
of measurement. Discuss the
physical
meaning of the sign of the coefficient of u in this equation.
i+ j, j+ k, i + k i+ j + k, i+ j, i-j i+ j + k, i+ j, -i-j+ 2 k i+ 2j+ 3k, i- 2j + 3k, i- 3k
In particular, explain how the principle "Like charges
Calculator/Computer Exercises
repel and opposite charges attract" is captured by the
77. For what values of t, to four decimal places, are the
formula for the force.
lengths of (2t, t, t), and
2 + y t4 with
70. Describe the intersections of the surface z = x2 the coordinate planes and with the planes z = h where h is a constant. Sketch the surface.
7L Describe the intersections of the surface 2 z /16
=
x2 + y
2
=
,,/
with the coordinate planes and with the planes z
=
1) equal?
same direction?
!9 +
79. A particle's position P, at time t is the point (t - cos(t),
1 with the coordinate planes. Sketch the surface. z 9x2 + y2
72. Describe the intersections of the surface
(r, 1 - t,
78. For what values, to six decimal places, of s and t in (0, 1) 2 do the vectors (s - t, t2 + s, st+ 1) and (1,2,3) have the
sin(t), t sin(t)). How close to the origin does the particle come?
h
80. At time t 2': 0, a particle's position P, is the point (t, t cos (t),
where h is a constant. Sketch the surface.
t sin(t)). Does the vector P,Pt+� ever have length 2?
9.3 The Dot Product and Applications If two nonzero vectors v and
w
have the same direction, then the angle between
them is 0. If the two vectors are in opposite directions, then the angle between them
is
'Tr.
In all other cases, we may use the two vectors to form a plane P by selecting
directed line segment representatives AB and (see Figure 1). The angle between v and ---+
---+
w
AC that have the same initial point () E (0, 7r)
is understood to be the angle
between AB and AC in plane P. It does not depend on the initial point A that is selected. In this section, we will study an algebraic operation that can be used to
_. Figure 1
calculate the angle between two vectors.
The Algebraic
In Sections 1 and 2, we discussed the addition of vectors and the multiplication of
Definition of the
vectors by scalars. We now introduce a useful way to form the product of two vectors.
Dot Product
•·
[email protected]
The dot product v .
w
of two vectors v and
the products of corresponding entries of v and
w
=
(wi. w2)
are planar vectors, then v ·
w
V·W=V1W1
If v
=
(vi. v2, v3)
and
w
=
(wi. w2, w3)
w.
w
is the sum of
That is, if v = ( vi.
v2
)
and
is defined by
+ V2Wz .
are spatial vectors, then v · w is defined by
V · W=V1W1 + V2W2 + V3W3.
7 42
Chapter 9
Vectors Notice that the dot product of two vectors is a scalar, or real number. For that
scalar product of v and w. This terminology should not be confused with scalar multiplication, which we have discussed in the preceding two sections. When we create a vector >.v by scalar multiplication, we multiply one vector v by a scalar >.. The result of scalar multi plication is a vector. We never use a dot to signify this type of multiplication. By contrast, the scalar product is an operation between two vectors, and the result is a scalar. We always use a dot to signify this type of multiplication. reason, the quantity
v·w is
sometimes called the
� EXAMPLE 1 Let u = (2,3, -1), v= (4,6, -2), and
culate the dot products u ·v, u ·w, and v · w.
w= (-2, -1,-7).
Cal
Solution We have u. v
= (2)(4)+(3)(6)+(-1)(-2) = 28
U·W = (2)(-2)+(3)(-1)+(-1)(-7)=0 V·W
= (4)(-2)+(6)(-1)+(-2)(-7)=0.
In the next theorem, we collect several simple algebraic rules for working with the dot product.
THEOREM 1
Suppose that u, v, and w are vectors and that A. is a scalar. The dot
product satisfies the following elementary properties: a. b. c.
u· (v+w) = u · v+u·w v · w=w·v (A.v) ·w=v· (A.w) = A.(v·w).
INSIGHT
Theorem le tells us that the value of the expression >.v wis unambiguous: ·
Both of the two possible interpretations, ( .Av) · wand >. (v w) , result in the same number. ·
Nevertheless,we will generally insert parentheses for clarity. A similar remark applies to the expression v ·>.w. Because we do not form the dot product of a vector v and a
scalar >.,we cannot interpret v ·>.was (v ·>. )w. Thus we can understand v >..w only as ·
v · (>..w) . By including the parentheses,we make this meaning instantly evident.
� EXAMPLE 2 Let u=(3,2) and v=(4, -5). Calculate (u ·v) u+(v·u)v. Solution Using Theorem lb and then la, we have
(u · v)u+ (v· u)v = (u · v)u+ (u ·v)v = (u · v)(u+v) = ((3)(4)+(2)(-5))(3+4,2 - 5) = 2(7, -3) = (14, -6).
A Geometric Formula for the Dot Product
v · v = (v1)2+ (v2)2+ (v3)2, we of v by rewriting equation (9.2.4)
Because
can relate the dot product
length
as
v· v
....
to the
llvll = VV:V,
(9.3.1)
llvll2=v. v.
(9.3.2)
or
9.3 The Dot Product and Applications
7 43
We will use this formula to develop a method for calculating the angle between two vectors and for calculating projections. To that end, recall that the Law of Cosines says that if the sides of a triangle measure first two sides, as in Figure b
2, then c2 a2 + b2 -2ab cos(O).
(9.3.3)
=
_. Figure 2 Law of Cosines: c2 a2 + b2 - 2abcos(O) =
Supposing that determined by
vandware not parallel, we apply the Law of Cosines to the triangle v,w, and v -win Figure 3. This trigonometric formula tells us that llv-wll2
r� /---:> ,/------2 _. Figure 3
a, b, c and if(} is the angle between the
=
llvll2 + llwll2 -2 llvll llwllcos(O).
(9.3.2) and the rules of Theorem 1, we may (9.3.4), obtaining
Using equation equation
(9.3.4)
expand the left side of
2 llv- wll2 = (v-w) ·(v-w) = v v- v w-w v +w·w= llvll2 - 2v w+ llwll . Substituting this expression for llv- wll2 into the left side of equation (9.3.4), we ·
·
·
·
obtain
Canceling
llvll2 - 2v·w+ llwf = llvll2 + llwll2 - 2 llvll llwllcos(O). like terms from each side results in the identity -2v·w= -2 llvll llwll · ·
cos(O), or
v·w= llvll llwllcos(O). THEOREM 2
Let
(9.3.5)
v andwbe nonzero vectors. Then, the angle(} between v and
wsatisfies the equation cos(O) =
V·W llvll llwll '
(9.3.6)
v andware not parallel, then equation (9.3.5) holds. We obtain equation llvll llwll, which is not 0. If v andware parallel and have the direction, then ()= 0 and w= >..v for some positive scalar >... The left side of
Proof. If
(9.3.6) same
by dividing by
equation
(9.3.6)
is cos(O), or 1. Using Theorem le, we calculate
9 2 9 5 llvll llwll = llvll 11 >..vii ( ;; ) i>..1 llvll llvll = >.. llvll2 ( � ) >..(v v) = v (>..v) = v w, (9.3.7) which shows that the right side of equation (9.3.6) is also 1. Finally, if v and ware parallel and have the opposite direction, then()= 7r and w= >..v for some negative scalar >... The left side of equation (9.3.6) is cos(7r), or -1. Using Theorem le, we ·
·
·
calculate
(9 2 ) (9 5) llvll llwll = llvll 11 >..vii ;; i>..1 llvll llvll = ->.. llvf � ->..(v·v) = -v·(>..v) = -v w, (9.3.8) ·
which shows that the right side of equation
(9.3.6)
� E X A M P L E 3 Calculate the angle between the two vectors w=
(2, -1, 1).
•
is also -1.
v= (2, 2, 4)
and
7 44
Chapter 9
Vectors
J22+22+42 . 6
Therefore () is Equation
7r/3.
J22+(-1)2+12
(9.3.5) gives rise to an important inequality.
Because lcos(O)I :5 l for
every angle(), we have
Iv· wl :5 llvll llwll ·
(9.3.9)
This relationship, called the Cauchy-Schwarz Inequality, can also be deduced from the identity
2 2 llvll llwll -(v· w)2 for
v= (v1, v2, v3)
=
and
(v1 w2 -w1v2 )2+(v1w3 -w1v3) 2+(v w3 -w v3)2 (9.3.10) 2 2
w= (w1, w2, w3).
See Exercise
� EXAMPLE 4 Verify that the two vectors
59.
v= (2,2,4)
and
w= (12,13,24)
satisfy the Cauchy-Schwarz Inequality. Solution We calculate
llvll llwll
=
v · w= (2)(12)+ (2)(13)+ (4)(24) = 146
V22+22+42V122+132+242
which is greater than
Iv· wl,
=
( 2v'6)(v'889)
and
=
146.068 . . .
as the Cauchy-Schwarz Inequality asserts.
Let () be the angle between nonzero vectors then we say that the vectors
v and w are orthogonal
v and w.
,
If ()
= 7r/2,
or mutually perpendicular.
Although the zero vector 0 has no direction, it is conventional to say that 0 is orthogonal to every vector.
Our next theorem serves two important purposes. The definition of ortho gonality of two vectors is geometric, as it should be. However, because the defi nition does not involve the entries of the vectors, it does not allow us to easily verify
orthogonality. Theorem 3 provides us with a simple algebraic criterion for doing so.
By contrast, our treatment of parallel vectors has been, up until now, algebraic and easy to apply. What it has lacked has been a transparent geometric foundation that refers to the angle between the two vectors. Theorem
3
fills in this gap.
9.3 The Dot Product and Applications
v and w be any vectors Then: v and w are orthogonal if and only if v· w = 0. v and w are parallel if and only if Iv· wl = IIv1111wII· If v and w are nonzero, and if(} is the angle between them, then v and w are parallel if and only if(}= 0 or(}= 7r. In this case, v and w have the same
THEOREM 3
a. b. c.
7 45
Let
direction if (}=
0 and opposite directions
Proof. Recall that a vector is
see from equation
(9.3.5)
0
that
if (}= 11.
if and only if its length is
v· w = 0
if and only if
0. With this in mind, we
v = 0, w = 0,
Statement a is an immediate consequence of this observation. To prove statement b, recall from Theorem
2,
Section
9.2
or (}= 7r/2.
v and w are
that
v = 0 or w = A.v for some scalar >.. Thus statement b asserts v = 0 or w = >.v for some scalar >.if and only if Iv· wl = llvll llwll It is evident that v = 0 implies Iv· wl = llvll llwll: Both sides are 0 in this case. That w = >.v implies Iv· wl = llvll llwll has been proved in lines (9.3.7) and (9.3.8). For the last step of the proof of statement b, we must show that if Iv· wl = llvll llwll and v =I= 0, then w = >.v. In this case, we write v = (v1, v2, v3), w = (wi, w2, w3), and use equation parallel if and only if that
·
(9.3.10) to
deduce that
(v1W2 - W1V2)
2
+
(v1W3 - W1V3)
2
+
(v2W3 - W2V3)
2
=
0,
or, equivalently,
(9.3.11) Because
v =I= 0, at least one of the entries v1, v2, v3 is nonzero. Whichever entry is 0,
the argument is the same. Suppose, for example, v2 =I= 0. Let >. = w2/v2. Then w2 = >.v2. Using this equation with the first equation of line (9.3.11), we obtain v1(>.v2) = w1v2, or w1 = >.v1. Similarly, using the last equation of line (9.3.11), we obtain v2w3 = (>.v2)v3, or w3 = >.v3. Thus w = (wi, w2, w3) = (>.vi, >.v2, >.v3) = >.(vi, v2, v3) = >.v.
(9. 3.6), we see that Iv· wl = llvll llwll holds for nonzero vectors v and w if and only if lcos (O) I = 1, which is to say(}= 0 or(}= 7r. Statement c Finally, from formula
•
now follows from statement b. � EXAMPLE
(-2, -1, -7)
5 Consider the vectors
u = (2, 3,-1), v = (4, 6, -2),
and
w=
of Example 1. Are any of these vectors orthogonal? Parallel?
u· w = 0 and v· w = 0 that we calculated in Example 1, we conclude that u and v are both orthogonal to w. Because u· v = 28 =I= 0, we can tell that u is not orthogonal to v. Indeed, the equation Solution From the dot products
2 2 2 2 llull llvll = J2 + 3 + (-1)2 V42 + 6 + (-2) =v'14 56 =J784 =28 =lo· vi tells that u and v are parallel (by Theorem 3b ). Since llull llvll actually equals+ u· v, we may further observe that u and v have the same direction. Notice that the equality v = 2u is another way to reach this conclusion. ..,.. ·
Projection
One of the most powerful constructions in geometry is the projection. Figure 4a shows two nonzero vectors
v and w represented
by directed line segments that
7 46
Chapter 9
Vectors
� £ Figure 4a
£ Figure 4b
£ Figure Sa
£ Figure Sb
share the same initial point. The projection
£ Figure 4c
P w(v) of v onto w
The main thing to notice in Figure 4c is that the projection of
w
and determines, along with
v
and
Q,
is shown in Figure 4b.
v
onto
w
is parallel to
a right triangle. Figure 5 shows two other
projections that will aid your geometric understanding of the concept. The next theorem provides us with an analytic expression for the direction,
dir (w) =(1/llw ll)w, of w.
THEOREM 4
If
v
and
w
P w(v)
as a scalar multiple of
are nonzero vectors, then the projection
onto w is given by
Pw(v)
=
( -V·W ) w W·W
Pw(v)
of
v
(9.3.12)
or, equivalently,
Pw(v) (v dir(w))dir(w). =
The length of
·
P w(v) is given by llPw(v)ll =Iv·dir(w)I,
or, equivalently,
IIPw(v) 11 Proot Let
Q
v w - i;r·
_
l
l
·
denote the vector orthogonal to
(9.3.13)
(9.3.14)
(9.3.15)
w
that is depicted in Figure 4c. We
v.
(9.3.16)
have
Pw(v) + Q
=
9.3 The Dot Product and Applications
747
Because Pw(v) is parallel to w, we may write
Pw(v) =cw
(9.3.17)
for some scalar c. Substituting this expression for Pw(v) into equation (9.3.16), we arrive at cw+ Q = v. Taking the dot product of each side of this equation with w and using the distributive law for the dot product gives us cw· w + Q w= v w, or (w w)c + 0= v w. Thus c= (w w)-1v· w. With this value for c, equation (9.3.17) becomes ·
·
·
·
·
which establishes formulas (9.3.12) and (9.3.13). By equating the lengths of the vectors on each side of formula (9.3.13), using equation (9.2.5), and noting that IIdir(w) II = 1, we see that I I ( ll!ll w) I = lii�� ,
llPw(v) II =Iv· dir(w)l lldir(w) II =Iv· dir(w)I = v· which establishes both (9.3.14) and (9.3.15).
In formulas (9.3.12) and (9.3.13), notice that (a) the expressions
•
( v w ) and ·
W·W
(v dir(w)) are scalars, and (b) each of these scalars multiplies a vector. The result of these scalar multiplications, the projection Pw(v) of v onto w, is a vector. Because equation (9.3.16) decomposes v as a sum of two orthogonal vectors, Pw(v) is also called the orthogonal projection of v in the direction of w. The scalar v· dir(w), ·
which may also be written as
( Ii�jj' ) , is called the component of v in the direction
of w. The absolute value of the component of v in the direction of w is the length of the projection Pw(v). � EXAMPLE 6 Let v= (2,1,-1) and w= (1,-2,2). Calculate the projection of v onto w, the projection of w onto v, and the lengths of these projections. Also calculate the component of v in the direction of w and the component of w in the direction of v.
To efficiently calculate Pw(v) and the other quantities associated with this projection, we determine two scalars: llwll and v·dir(w). We first calculate
Solution
J
llwll = 12+(-2)2 + 22= 3 and dir(w)= 1 ;1 w= (1/3,-2/3,2/3). The component
of v in the direction of w is v·dir(w), or (2)(1/3)+(1)(-2/3)+(-1)(2/3), which simplifies to -2/3. The absolute value of this quantity is the length of the projection. That is, llPw(v) II=l-2/31=2/3. Formula (9.3.13) gives us the projection Pw( v) itself:
Pw(v) =(v·dir(w))dir(w)=For Pv(w), we calculate llvll =
� \ � , - � , � ) = \- � ,� , - � )
·
J22+12+(-1)2 = v'6 and dir(v) = H v =
(2/v'6,1/v'6,-1/v'6). The component of w in the direction of v is
w·dir(v), or
7 48
Chapter 9
Vectors
(1)(2/\/'6)+(-2)(1/\/'6)+(2)(-1/\/'6), which simplifies to -2/\/'6. The absolute value, 2/v'6, of this quantity is the length of ll Pv (w) II· Formula (9.3.13) gives us the projection
Pv(w) itself:
Pv(w) = (w · dir(v)) dir(v) =-
�( �, �, -�J= (-�, -�,�J·
As we could have anticipated from the geometry of projections, the two pro jections
Pw(v)
INSIGHT
Pv(w)
and
are different.
<11111
For theoretical reasons, it is useful to have a variety of formulas. Thus
equation (9.3.15) allows us to easily prove that
As a practical matter, however, Example 6 shows that formula (9.3.13) contains all the information that pertains to the projection of one vector onto another.
Projection onto a unit vector u is simple. Setting
(9.3.14)
w=u in equations (9.3.13) and dir(u) =u, we obtain the simplifications (9.3.18) Pu(v) = (v ·u)u and ll Pu(v) II =Iv· nl.
and noting that
Also, the component of
v in the direction of u is just the dot product v· u.
v= (v'2,8,-2) and u =(1/v'2,1/2,-1/2). Calculate the v onto u. What is the component of v in the direction of u?
� EXAMPLE 7 Let projection
P0(v)
of
Solution To calculate
P u(v),
we proceed as in Example
6
and determine the two
llu ll =V(1/v'2)2+(1/2)2+(-1/2)2=1 and v · u=(v'2,8,-2)(1/../2,1/2,-1/2)=1+4+1=6. Because u is a unit vector, the component of v in the direction of u is v· u, or 6, and llull
scalars
and
v·dir(u).
P0(v) = (v ·u)u =6 Projection and the Standard Basis Vectors
As noted in Section
( }i.�, -� J=( }z,�, -�J= ( 3v'2,3,-3J·
� EXAMPLE express
2, it is often useful to express vectors in terms of the three unit
8 Let
v=(2,- 6,12).
Calculate
Pi(v), Pj(v),
v as a linear combination of these projections.
Solution Because
(9.3.18):
and
Pk(v),
and
i, j, and k are unit vectors, we use the projection formula found
Pi(v) = (v ·i) i=((2,-6,12) (1, 0, 0))i=2i=(2,0, 0), ·
�(v) =(v ·j) j=((2,-6,12) · (0, 1, 0))j=-6j= (0, -6,0), and
.._
i=(1,0, 0), j= (0, 1,0), and k= (0, 0, 1).
vectors
in line
Thus
l\(v)=(v · k) k=((2,-6,12) (0, 0, 1))k=12k= (0, 0, 12). ·
9.3 The Dot Product and Applications
7 49
Observe that
v (2,-6,12) (2,0,0) + (0,-6,0) + (0,0,12) Pi(v) + Pj(v) + Pk(v) =
=
=
INSIGHT
If we repeat the calculations of Example 8 with a general vector v = (a, b, c) , then we find that the projections of v on i, j, and k give the displacements of v in the x-, y-, and z-directions, respectively: Pi(v) = ai, Pj(v) = bj, and Pk(v) =ck. In particular,
v = Pi(v) + Pj(v) + l\(v).
Direction Cosines and Direction Angles
(9.3.19)
that u= (u1, u2, u3) =u1i + uzi + u3k is a unit vector. Because ui + u� + u� = 1, it follows that the numbers u1, u2, u3 all lie between -1 and 1. Thus there are unique numbers a, (3, "Y between 0 and 7f such that u1=cos(a), u2=cos((3), and u3=cos(1). As a result, Suppose
u=cos(a)i + cos((3)j
+
cos("Y)k.
The numbers a, (3, and "Y are called the direction angles for u, and the entries u1, u2, u3 are called the direction cosines of u. Notice that, by Theorem 2: a is the angle that u makes with the positive x-axis. (3 is the angle that u makes with the positive y-axis. "Y is the angle that u makes with the positive z-axis. Refer to Figure 6. z
z
z
x
• Figure 6
u =
cos(a)i + cos(,B)j + cos('y)k.
If is any nonzero vector, then the direction angles and direction cosines for v v are understood to be the direction angles and cosines for the unit vector (1/l l vll )v. The direction angles for v are the angles that v makes with the positive x-, y-, and z-axes. Calculate the direction cosines and direction angles for the vector v=(0,-3v'3,3) � EXA M P L E 9
Solution
The associated unit vector is
1 u=v= l lvl l
J
oz
(
)
r,; r,; v'3 1 1 (0,-3v3,3)=- (0,-3v3,3)= 0,-- ,- . 2 2 6 2 2 + (-3v'3) + 3
1
The direction cosines for v are therefore cos(a)=0,cos((3)=-v'3/2, and cos("!)= 1/2. It follows that a=7r/2, (3=57r/6, and "Y=Jr/3.
750
Chapter 9
Vectors
Applications
If a constant force Fis applied along the line of motion to move an object a distance
d, then, as we learned in Chapter 8, the work performed is W = Fd. However, in many
applications, the force
F is
Imagine a truck towing a
a vector that is not applied in the direction of motion.
(Figure 7). The force is applied in a direction that makes
car II II cos
an angle () with the direction of motion. Let u be a wtlt vector along the direction of
motion. Let g=
F u= F (0) be the component of F in the direction u. Then the work performed in moving the body in the figure a distance d is W = gd. ·
.._ EXAM PL E 1 0 A tow truck pulls a disabled vehicle a total of 20,000 feet.
.A. Rgure 7
To keep the vehicle in motion, the truck must apply a constant force of 3,000
pounds. The hitch is set up so that the force is exerted at an angle of 30° with the horizontal. How much work is performed?
1 2
Solution We use a 30° -60° -90° triangle, as shown in Figure resolve the force vector direction
v=
{3 2
of
8,
panel (a), to
F as a sum of two orthogonal vectors, one of which is in the We find that the direction of F is given by (see panel (b) of Figure 8). Because the force vector F has
motion.
(V3/2)ii (1/2)j +
magnitude 3,000 and direction v, we have
.A. Figure Ba
F = llFllv = 3000 as
(�i �j) +
= 1500J3i + 1500j,
is indicated in Figure 8, panel (c). The component of vector
motion is
g = 1500v'3.
As a result, the work performed is
F in the direction
W=gd=1500¥'3 20, 000=3¥'3·107foot-pounds. ·
F=
3). In
(2, 1, (1, 1,
navigation, it is useful to resolve this force into its component in the direction of motion and its component perpendicular to the direction of motion. If
represents the direction of motion, then perform this resolution. Solution The component of P,..(F). We calculate
F in the direction v=(1, 1, -4) P
...(F) (;�II� ) =
1500{'Ji .A. Rgure 8c F = 1S00./3i +
of
<1111
.._ EXAM PL E 1 1 The force of the wind is given by the vector
.A. Figure 8b
v = (./3/2}i + (1/2}j
i
=
lSOOj =
-9
18
(1, 1,
-4)
is just the projection
v
- 4)
\-�,-�,2)
·
Then we write 1, 3)
(2,
=
F
=
=
...(F) (F- ...(F)) \-�, -�,2)+\�·�·1).
P
+
P
Notice that the two vectors into which we have decomposed
F are orthogonal. The
first is a multiple of v, while the second is perpendicular to v. This is the resolution that we desire. ..._
9.3 The Dot Product and Applications Q UIC K
1. 2. 3. 4.
Q U IZ
751
Calculate (1,2, -1) (3,4,2). Use the arccosine to express the angle between (6,3,2) and (4,-7,4). For what value of a are (1,a,-1) and (3,4,a) perpendicular? Calculate the projection of (3,1,-1) onto (8,4,1). ·
Answers 1. 9 2. arccos (11/63)
3. -1
4. (8/3,4/3,1/3)
EXERCISES Problems for Practice In each of Exercises 1-6, calculate the dot product of the given vectors.
(2,1,6) (-8,1,1) (5,1,0) (-4,2,5) (2,3,-5) (-8,0,2)
(3,-2,4) (-6,-3,-5) (0,4,0) (0,-2,6) (1,4,9) (-2,-10,4)
L 2. 3. 4. 5. 6.
In each of Exercises 29-32, calculate the dot product of the given vectors and their lengths. Verify that the Cauchy Schwarz Inequality holds for the pair.
In each of Exercises 7-12, calculate the angle between the
(1,1,0) (2,-2,2) (3,0,4) (3,4,7) (2,-1,9) (1,1,2)
8. 9. 10. lL 12.
(0,1,1) (-1,1,5) (0,../7, -5) (18,24,5) (-4,1,1) (2,-1,1)
vectors
are
orthogonal.
30.
In each of Exercises 33-36, calculate the projection of the and
(-3,1,5) (0,-6,7) 15. (2,-5,8) 16. (1,6,5) 14.
Give
a
reason for
your
17. 18. 19. 20. 21. 22.
Verify that
w in
in the direction of
v=(-3,7,2) v=(0,0,1) v=(2,4,6) v=(2,1,9) v=(l,0,1) v=(2,4,6)
u u
(3, 2, -1) (s, 5, -12) (3, 1, 1) 40. (s, -5, 1) 37.
s
for which
Pv(w) and Pw(v). State v and the component
(s, 1, -4) (1, s, 2) (-7, s2, 5) (s, s, 6)
the direction of
w.
w=(1,4,-4) w=(-9,7,5) w=(4,8,12) w=(1,0,1) w=(0,1,0) w=(-2, -4,-6)
vector. Then find the direction angles for the vector.
(6,-2v'3,0)
Pu(v)
=(2/3, 2/3, -113) =(3/5, 0, -4/5) u =(1/v'3,1/v'3, -1/v'3) u =(1/9, -4/9, 8/9)
v =(4, 1, -8) v =(4, 2, -3) 35. v = (vU, 0, \!'48) 36. v =(3, 2, -1) 34.
38. 39.
In each of Exercises 23-28, find the direction of the given
23.
u.
the two given vectors are orthogonal.
(4,-2,3) (8,14,12) (-2,4,3) (-8,3, -2)
In each of Exercises 17-22, find
v
direction
In each of Exercises 37-40, find all values of
the component of of
v onto the given v - P0(v) are orthogonal.
33.
answer.
13.
(-2, 1, -2) (1, 3, 2) (4,v'3,1) (4, 1, -8)
given vector
In each of Exercises 13-16, determine whether the two given
(2, -4, 4) (1, 2, 3) 3L (2,0,1) 32. (2, 1, -2) 29.
given pair of vectors.
7.
(0,3v'3,-9) (0, 1, 0) 26. (3v'2,-3,3) 27. (-6, 0, 6) 28. (0, 1, 1) 24. 25.
Further Theory and Practice 4L Find a unit vector
v =i + 2j + k
and
u
that is perpendicular to both
w= i - j.
Show that any other vector
v
that is perpendicular to both
42. Show that if
v
and
w
is parallel to
u.
is any vector, then
v = (v i)i + (v j)j + (v k)k. ·
43. Prove that the vectors b=(0,1/v'z,1/v'z ), and
·
·
a=(v'3/2,1/(2v'2),-1/(2v'2)), =(1/2,-.../6/4, .../6/4) are each
c
perpendicular to the other two and are of unit length.
Chapter 9
752
Vectors
44. If T is any vector and if a, b, and c are the three vectors of
Exercise 43, then prove that
v= (v a)a +(v· b)b +(v · c)c. ·
45. Mrs. Woodman pulls a railroad car along a track using a
rope that makes a 30" angle with the track (Figure 9). Calculate how much work is performed if she exerts a force of 200 pounds and succeeds in pulling the car 1000 feet
54. Suppose that P, Q, and R are vertices of a cube with PQ the diagonal of the cube and PR the diagonal of a face of
the cube. What is the angle between PQ and PR? 55. Suppose that a and b are nonnegative. When the Cauchy Schwarz Inequaliy t is applied to T= v'Oi +Vbj and w= Vbi +vfa.j, what inequality results? 56. Let v1, v2 , and v3 be any numbers. Find a suitable vector" such that the Cauchy-Schwarz Inequality applied
v= (v1, v2, v3) and w results in the inequality 2 (v1 +v2 +v3) :S3(v� +� +�).
to
57. Use the Cauchy-Schwa:rz Inequality to prove the Triangle
Ineq uality:
l v+wll s l vll + llwll· 58. Suppose that a, b, and c are positive numbers. The equation x/a +y/b +z/c = 1 represents a plane. What are the intercepts A, B, and C with the x-axis, y-axis, and
.& Figure 9 46. Show that the direction cosines for any vector v satisfy
2
cos2(a) +cos (fJ) +cos2('Y) = 1. If v lies in the .xy-plane, then does this equation reduce?
to
what familiar identity
z-axis, respectively? Calculate the cosines of the three angles of 6.ABC. What arccosine identity can be deduced &om these angles? 59. Verify the identity
(� +� +vj)(;vi +� +wi) -(v1 w1 +v2w2 +v3w3)2 2
2
2
= (v1w2 -w1v2 ) +(v1w3 -w1v3) +(v2w3 -w2v3) •
47. Prove the Parallelogram Law:
Let v= (v1, vz, v3} and w= (wi. w2, w3). Deduce that
2 2 2 2 llTl llwll -(v· w) =(v1w2 -w1v2) +(v1w3-w1v3) 2 2 +(v2w3 -w2v3) •
Give a geometric interpretation for this equality.
48. Prove the Polarization Formula:
2 llv +wll - llv -w!r� = 4v w. ·
49. In each of the following, find a real number >. such that P11(v) = >.w: a. n = (3,6, 9) v= (1,-2,5} w = (1,2,3) 6 W = (-1,0,2) = (1, -4, ) T b. U = (-2,0,4) c. u= (2,10,-6) w= (l,5,-3) v= (-3,4,8) ... = (1, 1, 3) w= (3, 0, -4) d. u= (12, 0, -16) 50. Prove that for any two vectors vand w, the angle between u
= llwl v + llvllw and vequals the angle between u and"·
oP be the position vector for an arbitrary point P = (.x,y, z) in space. Using vector methods, find an equation involving oP,j, the dot product, and norm to describe the cone with vertex at the origin, forming an angle of 60° at
SL Let
the vertex, and axis of symmetry the y-axis.
oP be the position vector for an arbitrary point P = (x,y, z) in space. Using vector methods, find an equation involving oP, k, the dot product, and norm to describe
52. Let
the cone with vertex at the origin, forming an angle of 45° at the vertex, and axis of symmetry the z-axis. 53. Show that the points (2,1, 4), (5,3, 2), and (7,4,6) are the vertices of a right triangle.
60.
Prove the Cauchy-Schwarz Inequality from this identity. Suppose that v= (v1, vb v3) and w= (w1, w2, w3) are par allel Use Theorem 3b to deduce that
Use the second identity of Exercise S9
to
conclude that
2 2 2 (v1W2 -w1v2) +(v1W3 -Wl.V3) +(v2W3-W2V3) =O, which implies the following three equations:
If one entry v; of vis nonzero, then show that"= (w/v1) T. Deduce that in any event, one of v and w is a scalar mul tiple of the other. 6L Let V denote the set of planar vectors. Let T : V-.. V be defined by
T((a , b} ) = (acos(O) - bsin(O),asin(O) +bcos(O)). Show that the angle between (a, b} and T((a, b)) is 0. Also
show that II T({a, b)) II = II (a, b) II·
9A The Cross Product and Triple Product
64. For what value(s ) of a are the vectors (a2, 2a, -3} and
Calculator/Computer Exercises
(a,3a,1} orthogonal?
62. Calculate the angles of the triangle with vertices (1,1,2),
65. Plot f(t)=(e-',t,1)·(1,2t,3) for
(2,0,3),and (1,2,-5). What is the sum of these numbers? (1,2,-5).
What
is
the
sum
of
-2:5t:52. At what
value of t does f(t) attain a minimum?
63. Calculate the angles of the triangle with vertices (1,1,2), (2,0,3), and
753
66. At time t 2' 0,a particle's position P, is the point (t,tcos(t),
these
tsin(t)). Is the vector
numbers?
P,P,)1 ever perpendicular to (1,2,3}?
9.4 The Cross Product and Triple Product If we are given two vectors v and w in space, then it is frequently useful to find a nonzero vector that is perpendicular to both of them. Of course such a vector is not unique: If
n
is one such vector, then, for any scalar c, the vector
pendicular to both
v
en
is also per
and w. In this section, we present an algebraic method for
producing a vector that is not only perpendicular to both v and w but also has a length with particular geometric significance.
The Cross Product of
In the preceding section, we learned a product of two vectors, the dot product, that
Two Spatial Vectors
results in a scalar. We now define a product of two spatial vectors that results in another vector.
l•l§inhi[.]�I product
v
If v =
(vh v2, v3)
and w
=
(wh w2, w3 ) ,
then we define their cross
X w to be
(9.4.1)
Formula
(9.4.1)
for the cross product is somewhat odd-looking and cannot
be easily committed to memory. We will soon present an alternative approach to calculating cross products, but first we verify that v X w performs an important geometric job.
+ii!g.l;l§HI
If v and w are vectors, then vxw is orthogonal to both v and w.
(vi. v2, (wi. w2, w3) . We calculate the dot product of v with v X w: (v X w) (vi, v2, v3) ((v2w3 -w2v3 ), - (v1w3 -w1v3 ), (v1w2 -w1v2 )) v1 (VzW3 -wzv3 ) -Vz (V1W3 -w1v3 ) + V3 (V1W2 -W1V2 )
Proof. Recall that "orthogonal" is a synonym for "perpendicular." Let v = v3
)
and w v
·
=
·
=
=
=
0.
The verification that w (v X w) ·
� EXAMPLE 1 Let v = that
v
=
0 is similar.
(2,-1,3)
and w
X w is orthogonal to both v and w.
• =
(5,4,-6).
Calculate vxw. Verify
754
Chapter
9
Vectors Solution We calculate
v x w = (2,-1,3) x (5,4, -6) ((-1) . (-6) - 4 . 3)i - (2. (-6) - 5 . 3)j +(2 . 4 - 5. (-1) )k = (-6,27,13). =
We may now use the dot product to verify that and
w:
vXw
is perpendicular to both
v
v (v X w) = (2,-1,3) (-6, 27, 13) = 2 (-6) +(-1) 27+ 3 13 = 0 ·
·
·
·
·
and
w (v x w) = (5,4,-6)· (-6,27, 13) = 5 (-6) + 4 27+(-6)·13 = 0. ·
The Relationship between Cross Products and Determinants
·
.,..
·
There is a nice way to remember the formula for a cross product using the language of determinants. The determinant is a procedure for calculating a number based on the entries of a square array (or square
[� �]
matrix)
of numbers such as
[(); /3 '] a d
or
.._____ 2 x 2array
b
c
.
e f
-----3 x 3array
By "square," we mean that the number of rows in the array equals the number of columns. In the general theory of determinants, the number of rows and columns may be any positive integer. For our purposes, we need only consider
3X3
arrays.
The
det(M)
=ad
_. Figure
1
-
determinant
of a
2X2
array
det be
[� ] b
d
( [ � �] )
=
2X2
and
is defined by
ad - be.
(9.4.2)
Notice that the determinant is the alternating sum of the product of diagonal
1. 3X3
entries indicated in Figure The determinant of a
This definition is said to be an side of equation
(9.4.3)
array is defined by
expansion along the first row
of the array. The right
is an alternating sum of three products, one for each entry
in the first row. The factor of each entry is the determinant of the obtained by crossing out the row and column of the entry. Figure
2
2X2
array
illustrates this
procedure. Notice the signs, +, - , +, that appear on the right side of formula
(9.4.3). This pattern of alternating sums is the reason for describing the determinant as an "alternating sum."
An equivalent scheme for calculating the determinant of a sometimes used. As Figure
3
shows, two copies of the
3X3
3X3
array is
array are placed side
by-side. The products of the six indicated diagonals are then computed. The determinant is the sum of the products that arise from the diagonals that slope down to the right minus the sum of the products that arise from the diagonals that slope down to the left.
9.4 The Cross Product and Triple Product
� �
�� ifett
@det
itttT
[ *]]- [ *] @det
755
+
G)det
• Figure 2
[ *]]
'---v----1 '---v----1 M M
• Figure 3 det(M) a.bf+ f3cd + 'Yae - ace - f3af - "fbd =
If v = (v1, v2, v3) and w= (w1, w2, w3), then a simple calculation shows that vxw= det
([
i V1 W1
j Vz Wz
k V3 W3
l)
.
(9.4.4)
Here the determinant is expanded just as if i,j, k were scalars. When we do so, we obtain vxw = det
( [:: :: ] }- ( [ :� ::J ) ( [:� ::J ) j+ det
det
= (v2w3 - w2v3)i - (v1w3 - w1v3)j
+
k
(v1w2 - w1v2)k,
which is the formula for the cross product, as given by equation (9.4.1). E X A M P L E 2 Use a determinant to calculate the cross product of v = (2,-1, 6) and w= (-3,4,1).
�
Solution
We write
vxw
�
det
([ i � m 1
3 = i((-1) 1 -4 . 6) -j(2. 1 -(-3) . 6) = -25i- 20j + 5k. ·
+
k(2 . 4 -(-3) . (-1))
Thus vxw= (-25, -20, 5). ..,.. INSIGHT
Recall that the dot product v w of two vectors is a scalar; however, the cross product of two vectors is another vector. Texts that call v w the scalar product often call v X w the vector product. We will not use this alternative terminology. ·
·
Algebraic Properties of the Cross Product
If u, v, and ware vectors and >.and µare scalars, then 0x v uX 0 u x(v+ w) (v + w)Xu vxw (>.v )x(µw)
=
0, 0, (u xv)+ (u xw), (vXu)+ (wXu), -wxv, (>.µ)(vxw).
756
Chapter 9
Vectors These algebraic properties for the cross product can be verified by routine calcula tion. The properties of distribution are as expected. However, pay particular atten tion to the law
v Xw
=
-w Xv,
which shows that the cross product is
not a v Xw anticommutative.
commutative operation. In fact, because reversing the order of the operands of changes the sign of the product, the cross product is said to be THEOREM 2
If vis any vector, then
parallel vectors, then
u Xv
=
0.
v Xv= 0.
More generally, if
u
and
v
are
That is, the cross product of any two parallel
vectors is the zero vector.
Proot The
anticommutative
property
of
the
cross
product
states
that
v Xw= -w Xv for any vectors v and w. It follows that, if w= v, then v x v= -v x v. By adding v x v to each side of this equation, we conclude that 2v Xv= 0, or v Xv= 0. Next, suppose that u and v are parallel. According to Theorem 2 of Section 2, either u= 0 or v= ,\u for some scalar A.. In the first case, u Xv= 0 Xv= 0, as we have already observed. In the second case, we use the equation u Xu= 0, which we proved as the first part of this theorem, and obtain • u Xv= u X(.\u)= ,\ (u Xu)= ,\0= 0. � EXAM P L E 3 Give an example to show that the cross product does not satisfy a cancellation property. In other words, the equality not imply that w=
u.
v Xw= v Xu
does
Give an example to show that the cross product does not satisfy
the associative property. In other words,
(u Xv) Xw =I- u X(v Xw)
in general.
Solution A counterexample to the cancellation property is given by the parallel
v= (1,0,0), w= (2,0,0), and u= (3,0,0). We have v Xw= 0= v Xu yet w =f- u. A counterexample to the associative property is given by u= (1,0,0), v= (1,0,0), and w= (0,1,0). Because u= v, we have (u Xv) Xw= 0 Xw= 0. On the other hand, we calculate v Xw= (0,0,1) and u X(v Xw) = (1, 0,0) X (0,0,1)= (0, - 1 , 0) =I- 0. .... vectors
� EXAM P L E 4 Use Theorem vector of the form Solution z
y �-----.-
x
.& Figure 4
A Geometric Understanding of the Cross Product
v
(a,b,b2)
2
to determine whether there is any nonzero
that is parallel to
u (1, 2,3). 2 2 We calculate u Xv= (2b -3b, 3a -b ,b -2a). Theorem 2 tells us that if =
=
u and
v are parallel, then the entries of this cross product must all be 0. That is, 2b2 -3b= 0, 3a -b2= 0, and b -2a= 0. The last of these three equations gives us b= 2a. If we substitute this value for b into the second equation, then we obtain 3a -4a2= 0, or a(3/4 -a)= 0. Because b= 2a, the solution a= 0 implies b= 0, which results in v being the zero vector. Therefore the only admissible solution is a= 3/4. This leads to b= 3/2. We note that this value of b also satisfies the first 2 equation, 2b -3b= 0. Thus Theorem 2 tells us that v= (3/4, 3/2, 9/4) is the only nonzero vector that can be a solution. Because v= (3/4)(1,2,3)= (3/4) u, we see that v is parallel to u. <1111
Now we would like to develop a geometric way to think of the cross product. If we
v and w (Figure 4), then how can we visualize the cross product v Xw? The first step is to use v and w to determine a plane. If P0 is a point in space, then we may represent v and w by directed line segments that
have a picture of nonparallel vectors
9.4 The Cross Product and Triple Product have
P0 as their common initial point.
that contains
v
and w (see Figure
757
In doing so, we obtain a plane V through
P0
5).
Next, we observe that, for any point P0 on a plane V in space, there is a unique line f, through
v
P0
that is perpendicular to V (see Figure
normal line to V through y
P0•
6).
We say that f, is the
If we do not refer to the point of intersection
P0
of
f, and V, then we use the indefinite article and say that f, is a normal line to V. A nonzero vector n is normal to V if n can be represented by a directed line segment Figure
_. Figure 5
P1P;
where
and
P1
P2
are distinct points on a normal line f, to V (see
7).
If n is normal to a plane V that is determined by
v and w, then we say that n is a
v and w. If n is also a unit vector, then it is said to be v and w. A pair of nonparallel vectors always has two unit one of them, then -n is the other (see Figure 8). With the
normal vector for the pair a unit normal vector for normal vectors: If n is
following right-hand rule, we will associate one of these two unit normal vectors with the ordered pair (v, w) and call it the standard unit normal vector for (v, w) . The negative of this vector is the standard unit normal vector for the ordered pair (w, v).
\
P,
�
P1
v
v
Po
_. Figure 6
\
_. Figure 7
r
n
-�I
\
y
•
� ..
Figure 8
The Right-Hand Rule for Finding the Direction of the Standard Unit Normal Let
(v, w)
be an ordered pair of two nonparallel vectors. Represent
v and
w by directed line segments with a
common initial point. Point the fingers of your right hand along the first vector
v and then curl them toward the
second vector w, as in Figure 9. The direction that the thumb points during this process is the direction of
v, w. (v, w) .
the standard unit normal vector for the vectors standard unit normal for the ordered pair
Look again at Figure
8.
The direction of vector n is the
Standard unit normal
� EXAMPLE 5 Find the standard unit normal for the pairs
(i,j)
and
(k,i). (i,k).
and
(k,j)
and
Find also the standard unit normal for the pairs
Solution If you curl the fingers of your right hand from
i
(j,i)
toward
j,
(j,k) and
then your
thumb will point in the direction of the positive z-axis. Thus the standard unit normal for the pair normal for
(j,i)
is
-k.
unit normal for the pair
j.
is
k.
Similar reasoning shows that the standard unit
(j,k) is i, and the
standard unit normal for the pair
In addition, the standard unit normal for
normal for
_. Figure 9
(i,j)
We leave it as an exercise to check that the standard
(i,k)
is
-j.
(k,j)
is
-i,
(k,i)
is
and the standard unit
758
Chapter 9
Vectors INSIGHT
According to the right-hand rule, the standard unit normal vector for
the ordered pair (v, w) points in the opposite direction to the standard unit normal vector for the ordered pair (w, v). Example 4 provides particular cases of this general fact.
If v x w i- 0, then we may form the unit vector dir(v x w)= ll v x w r1 (v x w). Because v X w is perpendicular to both v and w, the vector dir(v X w) must be the standard unit normal for either the ordered pair (v, w) or the ordered pair (w, v). Our next theorem tells us which. It also specifies the length of v X w and tells us precisely when the equation v X w 0 holds. =
THEOREM 3
a. b. c. d.
Let v and w be vectors. Then: ll v X w ll ll v ll 2 ll w ll 2 - (v w)2 . If v and w are nonzero, then ll v X w ll = ll v ll ll w ll sin(O) where 0 E [O, rr] denotes the angle between v and w. v and w are parallel if and only if v X w=0. If v and w are not parallel, then v x w points in the direction of the standard unit normal for the pair (v, w). In particular, dir(v X w)= ll v X w r1 (v X w) is the standard unit normal vector for the pair (v, w). 2
=
·
Proot Part a is an identity among the entries of v=(v1, v2, v3) and w=(w1, w 2, w3).
We have
ll v X w ll
2
+ (v1w3 - w1v3)2 + (v1w2 - w1v2)2 (Vt+ V� + V�)(wr + W� + W�) - (V1W1 + VzWz + V3W3)2 2 2 2 = ll v ll ll w ll - (v w) . =
(v2w3 - w2v3)
2
=
·
The second equality in this chain is not obvious, but it may be verified by multiplying everything out. If neither v nor w is the zero vector, then the scalars ll v ll and ll w ll are nonzero. In this case, the identity of part a becomes
ll v X w ll
2
= ll v ll
2
ll w f - (v · w)2= ll v ll 2 ll w ll 2
(i
-
2 2 = ll v ll ll w ll (1- cos2 (0)) , \ ) ll w ll
(v w)
�
ll v ll
where the last equality is obtained by using formula (9.3.6). It follows that ll v X w ll 2=ll v ll 2 ll w ll 2sin2(0) and, on taking the square root, ll v X w ll =ll v ll ll w ll lsin(O)I. Because 0 :5 0 :5 rr, it follows that 0 :5 sin(O). Therefore lsin(O)I=sin(O) and ll v X w ll = ll v ll ll w ll sin(O), as asserted in part b. Theorem 3 of Section 3 tells us that vectors v and w are parallel if and only if Iv· wl= ll v ll ll w ll Using the identity of part a, we conclude that v and w are parallel if and only if ll v X w ll =0. Because 0 is the only zero length vector, assertion c follows. For part d, we will limit our verification to a few special cases. Notice that i X j=k, j X k i, and k X i j. Based on our observations in Example 4, we conclude that, for the basic vectors i,j, and k, the operation of cross product produces the standard unit normal. Geometric reasoning can be used to show that the cross product v X w always points in the direction of the standard unit normal for the vectors v,w. • ·
=
=
� EXAMPLE 6 Let v=(l,-3,2) and w=(l,-1,4). What is the standard unit normal vector for (v, w)?
9.4 The Cross Product and Triple Product
759
Solution We calculate the vector
vxw = (( -3) . 4- (-1) . 2) i- ( 1 . 4- 1 . 2)j+ ( 1 . (-1)- 1 . ( -3)) k -10i- 2j+2k =
and its length
llvxwll = Part d of Theorem
V(-10)2+(-2)2+22 = v'T08= J(36)(3)= 6v'3.
2 tells us that
dir(vxw)= llvxwr1(vxw)= 1;;;-(-10i- 2j+2k)=- 5;;;-i- 1;;;-j+ 1;;;-k 3v3 3v3 3v3 6v3 is the vector we seek.
Cross Products and the Calculation of Area
<11111
Now we learn a connection between cross products and areas of triangles and parallelograms. To be specific, if
v
and
represented by directed line segments then we speak of triangle
are two nonparallel vectors that are -
OPQ as "the triangle determined by the vectors v and w."
v and w is 11. Notice that, when we declare OPQ to be the triangle deter mined by v and w, the third side QP represents v-w. Had we positioned v so that its initial point coincided with the terminal point Q of w, then we would have obtained triangle OQR with third side v+w. Notice that the areas of triangles See Figure
10
w
-
OP and OQ with common initial point 0,
for this triangle. The parallelogram determined by
shown in Figure
OPQ and OQR are the same because each is half the area of the parallelogram. Thus had we declared OQR as the triangle determined by the vectors v and w instead of 0PQ, we would have been speaking of a triangle with the same area. R
_. Figure 11
w are nonparallel vectors. The area of the triangle determined by v and w is IIv X w II The area of the parallelogram determined by v and w is IIv X w II·
THEOREM 4
Suppose that
v
and
�
·
10, the altitude from vertex P to base OQ has length llvll sin(O). The area of triangle OPQ, namely half the product of its base and height, is therefore llwll llvllsin(O). This quantity is llvxwll by Theorem 3b. As
Proof. As we see from Figure
�
�
760
Chapter 9
Vectors discussed, the second assertion follows from the first because the area of the parallelogram determined by v and
w is
twice the area of the triangle that the two
•
vectors determine.
� EXA M P L E
v
=
(0,2, 1)
and
7 Find the area of the triangle determined by the vectors
w (3, 1, -1). =
Solution We calculate that
� llvxwll = � ll-3i+3j -6kll = � V54= �v'6.
<1111
� E X A M P L E 8 Calculate the area of the parallelogram determined by the vectors
v
= (-2,1, 3)
and
w
= (1, 0,4).
Solution The required area is
A Physical Application of the Cross Product
llv X wll
= ll4i+ 11j - kll = v'I38.
"ill
Experimental evidence teaches us that if a magnetic field M acts on a charged
particle with charges, and if the charged particle has velocity v, then the resultant
force
F
that is exerted is
F= (sv) X
M.
� E X A M P L E 9 In the picture tube for an oscilloscope, a magnetic field is used to control the path of ions that transmit the image to the screen. If s is the
charge of the particle, v is its velocity, and M is the magnetic field, then the force
F
(sv) X M. Suppose that the velocity vector for the (c, -c, 0), where c is a physical constant. The magnetic field will have the form (a, 1, 1), where the value of a will be varied to force the ions to go in different
exerted on the particle is ions is
F
=
directions. If we want to exert a force on a positively charged ion in the direction
(c/2,c/2, c/2), then
a?
how should we select
>..(c/2,c/2,c/2) s(c, -c, 0) X (a, 1, 1). The >.. on the left side allows us to adjust for length. Calculating the cross product (c, -c, 0) X (a, 1, 1), we find that
Solution We need to solve the equation
=
constant
>.. 2 (c, c, c) We choose
>..
= -2s
= s(-c, -c, c +ca).
so that the first two entries on either side match up. For this
vector equation to hold, the third entries
must therefore choose a so that
The Triple Scalar Product
(-2s) c/2
>..c/2 and s(c +ca)
= s(c +ca).
must also be equal. We
This forces us to set a
= -2.
"ill
We have just seen that the cross product is useful for finding the areas of triangles and parallelograms. We now develop this idea further by introducing a product that involves three vectors.
product
If u, v, and ware given vectors, then we define their to be the number (u
Xv)· w.
triple scalar
Notice that the triple scalar product (u Xv)· w of three vectors u, v, and w is a scalar because it is the dot product of the two vectors u Xv and w. The parentheses in the triple scalar product are not really necessary: The association u X (v · w)
9.4 The Cross Product and Triple Product
makes no sense because the cross product of vector
761
u with scalar v · w is not defined.
u · (vXw) is defined, and, by expanding both it and the (uXv)· win terms of the entries of u, v, and w, we find that (uXv)·w=u·(vXw). (9.4.5) In other words, to compute the triple scalar product of u, v, and w, we write down the vectors in the given order, we insert the operations· and X in either order, and we then associate in the only way possible (see Figure 12).
However, the expression triple scalar product
uvw
u xv·w or u·v xw
(u xv)·w or u·(v xw)
Insert and x
Associate
�
'-------y---'
11rree vectors, in order
'----y---1
'------y--
·
_. Figure 12 The triple scalar product of � EXA M P L E
u, v,
and
w
1 0 Calculate the triple scalar product of
v= (7, 2, 3), and w= (-1, 1, 2)
u= (2, -1, 4),
in two different ways.
Solution We may calculate the triple scalar product as
u x v= (-11, 22, 11).
Therefore
(uXv)· w= (-11, 22, 11) · (-1, 1, 2) = 11+22+22 = 55 or
u · (vXw)= (2, -1, 4) · (1, -17, 9) = 2+17+36 = 55.
.,..
The next theorem shows that we can compute a triple scalar product without first calculating a cross product.
THEOREM 5
The triple scalar product of
u= (u1, u2, u3), v = (v1, v2, v3),
and
w= (w1, w2, w3) is given by the formula (9.4.6)
Proot Because
(uXv)· w= u· (vXw) and
v x w= det _. Figure 13
( [ :� :: ] ) - ( [ :� ::] ) j+ ( [ :� :�]) i
det
det
k,
we have
•
To understand the geometric significance of the triple scalar product, we look
13. The solid region determined by u, v, and w is called a "parallelepiped." By the right-hand rule, uXv points upward in the figure and is perpendicular to the base of the parallelepiped. Moreover, lluXvii is the area of the base. The height of at Figure
762
Chapter 9
Vectors the parallelepiped is in Figure
13).
We
ll wll cos(O), where 0 is the angle between wand u Xv (as shown may always take 0 to be between 0 and 7r/2. We calculate the
volume of the parallelepiped as follows:
(volume of parallelepiped)
= (area of base) (height) llu Xvii llwll cosO = l(u Xv)·wl. ·
=
·
Because the volume of the parallelepiped is obviously independent of the order in which we take the vectors u, v, and w, we deduce that the absolute value of the triple
product, taken in any order, gives the same answer. We record the results of this investigation as a theorem: THEOREM 6
Let
u, v, w be
vectors. The volume of the parallelepiped deter
mined by these vectors is given by any of the expressions
l(uxv)·wl=l(uxw)·vl=l(vxw)·ul= det
([
U1
U2
U3
v1
v2
v3
W1
W2
W3
l)
(9.4.7)
In case uxv and w point to the same side of the plane determined by u and v (the vectors form a right-hand system), then equals the volume of the parallelepiped.
�
(u Xv)·w is already positive and
EXAM PL E 1 1 Use the determinant to calculate the volume of the par
allelepiped determined by the vectors
(- 3,2,5), (1,0, 3), and (3, - 1, -2).
Solution The required volume is
det
-
([ � -� _m
�
l -3 ([- � - m -2 ( [� -m det
det
+Sdet
m -ml
=l-3(3) -2(-11) +5(-1)1=8. ....
Vectors are said to be coplanar when they lie in the same plane. Our next
theorem gives us a simple way to tell when three vectors are coplanar:
u (u1, u2, u3 ) , v ( v1, v2, v3 ) , and w (w1, w2, w3) are coplanar if and only if u· (v Xw)=0. Equivalently, they are coplanar if and
THEOREM 7
Three vectors
=
=
=
only if
Proot According to Theorem 6, this determinant is parallelepiped determined by the three vectors has
that the three vectors are coplanar.
0
0
if and only if the
volume. But this means •
9.4 The Cross Product and Triple Product
� EXAMPLE 1 2 Show that u
coplanar.
=
(3, 1, 1),
v
=
(1,2,0), and w
=
763
(1, -3, 1) are
Solution We calculate
It is reasonable for us to ask, "Is there a triple product of spatial vectors that results
Triple Vector Products
in a vector?" An affirmative answer is contained in the next definition.
u
X (v X
If u,
w) and (u
INSIGHT
v, and w are given spatial vectors, then each of the vectors Xv) X w is said to be a triple vector product of u, v, and w.
u v X w does not require parentheses, u Xv X w is ambiguous because its meaning depends on how the terms are associated. In general, (u Xv) X w =I- u X (v X w), Although the triple scalar product
·
a triple vector product does. The expression as we observed in Example 3.
Our final example, which involves a triple vector product, illustrates several
geometric properties of the cross product.
� EXAMPL E 1 3 Let v and w be perpendicular spatial vectors. Show that
v X (v X w) = -(v v) ·
w
(9.4.8)
and w
X (v X w) = (w
·
w)
v.
(9.4.9)
Solution If either v or wis 0, then the asserted equation holds because each side is
n=vxw
0. Suppose that v and ware both nonzero. Then n =v X wis normal to the plane V
determined by
v and
v Xn. Then mis perpendicular to v and n. Because
w. Let m =
mis perpendicular ton, a normal to plane V, it follows that m lies in V. Because m,
v, and
w
wall lie in V, and mand ware both perpendicular to
v, it follows that m and
ware parallel. Thus there is a scalar A such that m=Aw. If we apply the right-hand
v
n =v X w and a second time to find the v Xn, then we see that m and w have appositive directions-use
rule a first time to find the direction of
direction of m =
_. Figure 14
Figure 14 to help you see this. It follows that A< 0 and A= -IAI. Finally, by applying Theorem 3b, first to the perpendicular pair
perpendicular pair IAI
llwll
II
ll
= Aw =
llmll
=
llv Xnil
=
v,
w, we obtain
llv ll ll nll
=
llv ll llv Xwll
=
llvll llvll llwl
=
v, n
ll vll2 llwl
v·v and A= -A I I = -(v·v). Thus
We conclude that A I I=
v X (v X w) =v Xn =m=Aw= -(v v)w, ·
and then to the
=
(v v) llwl ·
·
764
Chapter 9 Vectors which is equation (9.4.8). We obtain equation (9.4.9) from this equation (with v and
w interchanged) as follows:
(
(
)
wx vxw =-wx wxv
) (9.4.8) ( =
)
w·w v.
Formulas (9.4.8) and (9.4.9) can also be proved by direct calculation with the entries of
Q U IC K
U
Q
IZ
v and w. <11111
1. Calculate (2,1,2)x (1, -2, -1).
ined by (2,1, -2) and (1,1,0). 2. Find the area of the parallelogram determ ordered pair ( (2,1, -2), (1,1,0)). the for 3. Find the standard unit normal vector = (uXv)Xw; (c) (vxw) X u (b) wxv; 4. True or false: (a) vX w=
u·vXw=uXv·w? Answers
1. (3,4, -5)
2. 3
3. (2/3, -2/3, 1/3)
4. (a) False; (b) False; (c) True
EXERCISES Problems for Practice that v and In each of Exercises 1-8, compute v X w. Verify (v X w) and are perpendicular to v X w by showing that v
w w· (vx w)
·
are both
1. v=(3,3,-5) 2. v =(0,1,1) 3. v=(4,-3,6) 4. v=(2,0, -2) 5. v=(-4,-2,3) 6. v=(-9,4,6) 7. v=(1,6,8) 8. v=(2,3,3)
0.
w=(-1,3,-2) w=(1,1,0) w=(2,1,2) w=(2,0,2) w=(7,1,-5) w=(0,0, 1) w=(1,2,8) w=(-2,-2,3)
unit normal In each of Exercises 9-14, find the standard vectors. vector associated to the given ordered pair of
9. (2,1,3) 10. (0, 1, 1) 11. (3,-1,1) 12. (-2,-4,3) 13. (1,2,2) 14. (0,1/2,1)
(-3,-2,-5) (1, 1,0) (1,1,1) (2,2,-1) (2,-1,1) (1/2,0, 1)
triangle In each of Exercises 15-18, find the area of the determined by the two given vectors.
15. (2,1,2) 16. (1/3,0,-1) 17. (-3,4,1) 18. (1,3,4)
(3,2,3) (1,3,3) (-2,0,1) (0,-3,-4)
of the In each of Exercises 19-22, calculate the area . vectors given two the by ned determi parallelogram
19. (2,1,1) 20. (-1,3,2) 2L (1,0,2) 22. (1,3,4) In
(0,-3,1) (-1,1,1) (1,-3,0) (1,1,2)
vector each of Exercises 23-26, calculate the triple u X (v X w) and (u Xv) X w. (In each exercise,
products
notice that
u X (v X w) f (u Xv) X w.)
u=(-4,1,0) 24.u=(0,4,-3) 25.u=(1,1,2) 26.u=(2,-3,-2)
23.
v=(2,-1,-3) v=(3,0,-2) v=(2,1,2) v=(-3,1,-5)
w=(3,-2,-3) w=(-4,-3,-2) w=(-3,4,-3) w=(2,1,0)
In each of Exercises 27-30, calculate u Xv,v X w, and the (u Xv)·w. Verify that (uxv)·w=u· (vx w). w=(-1,-3,2) v=(2,-1,-3) 27.u= (3,0,1) w= (-3, -1, -2) , , v= (2,0,-6) 28.u=(0,4, -3) w=(3,2,3) 29.u=(1,1, -2) v=(2,1,-2) w=(2,1,0) 30.u=(4,3, -2) v=(2,1, -5)
triple scalar product
In each of Exercises 31-34, use a determinant to calculate (uxv)·w.
the triple scalar product
3L u=(1, -2,4) 32.u=(2,1,2) 33.u=(l,1,1) 34.u=(-5, -1,0)
v=(2,0,1) v=(2,2,3) v=(3,0,2) v=(l,0,3)
w=(3,1,1) w=(3,3, -5) w=(-2, -3, 3) w=(0,2,2)
In each of Exercises 35-38, use the triple scalar product to r. verify that the three given vectors are coplana 35.u=(1, -2,4)
v=(2,0,1)
w=(5, -2,6)
9A The Cross Product and Triple Product 36. D= (2,1,2} 37. D=(1,1,1} 38. o=(-5, -1,0}
"= (6,4,7} w=(8,2,6} w=(-6, -1, -3}
= (2,2,3} =(3,0,2} v=(l,0,3}
V
V
In each of Exercises 39-42, vectors
Calculate
determine.
the
volume
of
the
u, v, and w
parallelepiped
v=(1,1,-1} v= (3,1,-1) v=(1,2,3} v=(4,-2,3}
39. D=(2,1,1} 40. D= (2,1,-2} 4L u=(3,2,l} 42. D=(5,2,3}
are given.
that
they
w=(l, -1, -1} w= (1,-1,1} w=( 1,4, -2} w=(3, -1,2)
Further Theory and Practice In each of Exercises 43-46, find scalars s and
uX(vxw)=sv+tw.
43. u=(1,-2,4} 44. D=(2,1,2} 45. D=(1,1,-1} 46. D=(5,-1,O}
v=(2,0,1} v=(2,2,3) v=(3,0,2} v=(1,0,3}
In each of Exercises 47-SO, vectors
t
for which
47. 48. 49. SO.
v
and
w
The cross product can be interpreted physically in terms of
moment of force. Namely , let F be a force (vector)applied at a point Q in space, and let P be another point in space. The moment or torque r of the force F at the point Q about the point P is defined to be ---+
T=PQXF. See Figure
15. Notice that r is a vector. If a certain body that P extends to the point Q, and if Fis applied at Q, then II T II measures the tendency F to make the body rotate about P. The direction of r gives the axis of rotation. A bolt at point P will be driven by a wrench in the direction of r (see Figure 15). Exercises 60-62 concern torque.
can pivot about
�
.,-=PQX F
are given.
lv X wl[2 and verify that this quantity a. ll2llwll2- (v w)2, as asserted by Theorem 3
equals F
·
v= (2,1,2} v=(3,-1,1} v=(2,3,1} v= (4,0,1)
w= (1, -2, -2} w=(1,-1,l} "=(2,3,-1} w= (7,1,2)
In each of Exercises S1-S4, vectors v and ware given. Let()
be the angle between
(c) v
·
w, and
v and w. Calculate (a)v x w, (b)sin(O),
(d) cos(O). Verify that the values of sin(O) and
cos(O)are consistent.
SL S2. S3. S4.
w is a fixed spatial vector. Find all vectors v v X w= v + w.
that satisfy the equation
w=(5,-3,2} "=(1,1,1) w=(4,2,1} "=(-2, -1, 7}
Calculate llv
S9. Suppose that
v=(2,1,1} v= (1,0,-2} v=(2, -2,l} v=(v'2, v'3, 2)
SS. Suppose that
"=(1,2,-1} w= (2,1,-1} "=(8 4 , , l} "=(2v'2, 0, 1)
u,v, ware
spatial vectors. Prove that
u X (v + w)= (uXv) + (uXw). S6. Suppose that
o,v, ware
spatial vectors. Use the result of
the preceding exercise to deduce that
(v + w)Xu= (vXu) + (wXu). S7. Suppose that v and ware spatial vectors and that A and µ are scalars. Prove that
(.Av)X (µw)=(.Aµ)(vX w). SS. Suppose that
v
and
w
are spatial vectors. Prove that
(uxv)Xw=wX (vxu).
765
£. Figure 15
60. A diver who weighs
520
rigid diving board: It is
board is
from
newtons stands at the end of a
not
a spring board. The diving
2 meters long and extends in an upward direction the edge of the pool at a 30 degree angle with the
horizontal. The weight of the diver results in a torque
about the point at which the diving board is attached.
What is the magnitude of this torque?
6L An 8-inch wrench is used to drive a bolt at point
P. A F is applied to the end of the handle (point Q). If II Fil=60 pounds and the angle between and PQ is 90°,
force
then what is the magnitude (in foot-pounds)of the torque that is produced?
62. Repeat Exercise cation is
120",
61
63. Two distinct points
but suppose that the angle of appli
P and Q determine a line i. Suppose i. The distance d(R,i)of Rto i is defined to be the minimum value of II iiT II as T varies over all points of i. Let S be the point on i such that RS and PQ are mutually perpendicular. Show that d(R, i) = II RS II· Use the cross product to calculate d(R,l)in terms of the given points P,Q, and R 64. Prove the converse to equation (9.4.8): If vX (vXw)= -(v · v), w, then v and ware perpendicular. that Ris a point not on
766
Chapter 9
65. Let
s
and
t
Vectors
be scalars. Show that
u, v,
and
su
+ tv
Cartesian equation in x, y, and
are
coplanar.
66. The
Gram determinant of
defined to be
G=det Show that
v
and
w
two spatial vectors
v·v ([W·V
v· w
W·W
and
v
w
is
relationship to the vector v X w? 70. Consider the curve that is parameterized by t r-> (5 + cos(t), 3 + sin(t), 5), 0 =:; t =:; 27!". Find the point P on the curve that is closest to the origin. Find a vector v that is tangent to the
]) ·
are parallel if and only if
G
=
curve at P. Calculate
0. 7L
(v X w)· (p X q) = (v· p)(w· q) - (v· q)(w· p).
f(t)= llv1 X wll
68. Prove the Jacobi Identity:
for -3
�x�xw+�x�xu+�x�xv=Q
(x,y, z), (5,-2,1}. Write
P0
=
(1,2, 3), v
the equation
=:;
and
llvll · llDPll·
What
f(t)
=
-Vt· w
t =:; 3. What is the minimum value off? v, (t, r, 1 - r} and w, (1 - r, t, r}.
72. Suppose that
Calculator/Computer Exercises =
llvxDPll
OP do these calculations have? Suppose that v1 = (1- t2, t, 4- t2} and w= (1,2,3}. Plot significance for v and
67. Prove the Lagrange Identity:
69. Let P
z. Plot the solution set S of
the equation. What geometric figure results? What is its
llv1 X W1 f
=
for
=
-1 =:; t=51.
Plot
What are the local and
global minima of f? =
(2, 1,-4},
and
Fr:P · (v X w)= 0
w as
=
a
9.5 Lines and Planes in Space When we learned to graph with Cartesian coordinates in the two-dimensional plane, we found that the set of points satisfying one linear equation is a line, while the set of points satisfying two linear equations is (usually) a point. The philosophy here is that each equation "removes a degree of freedom" or takes away a dimension. The same philosophy works in three dimensions: The set of points in space satisfying one linear equation will be a plane, and the set of points in space satis fying two linear equations will (usually) be a line. Notice that we had to add the word "usually" because sometimes two equations have no solution or have too many solutions.
Cartesian Equations of
We can describe a line in the plane by giving a point through which the line passes
Planes in Space
and the direction or slope of the line. We want to use the same idea for determining
\
a plane V in space. However, we specify the "direction" of a plane in a rather indirect way. Namely, we give a nonzero vector n that is perpendicular to V. By this, we mean that n is perpendicular to every vector that lies in V. Recall that such a vector n is said to be a normal vector for V. All normal vectors for V are then of the form .An for a nonzero scalar .A. Notice that a normal vector determines the "tilt" of a plane in space, but not its position. Any two parallel planes will have the same normal vectors (see Figure
1). Finally, any given point in space lies in exactly
one of these parallel planes. That is the geometric reasoning that underlies the next theorem.
@Wg.l;l§.tli
Let V be a plane in space. Suppose that n
vector for V and that
P0
=
(x0,y0, z0)
A(x - xo) + B(y - Yo) + C(z - zo) _. Figure 1 Parallel planes have the same normal vectors.
=
(A,B, C)
is a normal
is a point on V. Then, =
0
(9.5.1)
9.5 Lines and Planes in Space
767
is a Cartesian equation for V. This equation may be written in the alternative form
Ax+By+Cz=D where
n =
(A,B,C)
(9.5.2)
D= Ax0+By0+Czo. 2 shows a plane V and a nonzero vector n=(A, B,C) that is normal 2 also shows a fixed point Po= (x0,y0,z0) in V. Now let P= (x,y,z) be a
Proot Figure to V. Figure
general point in the plane. The key geometric fact that we need is that the vector
P0P lies in the plane. This is true precisely when the vector P0P is perpendicular to n. In other words, P lies on V if and only if n P0P= 0. In coordinates, the vector PoP is given by (x - x0,y - y0,z - z0), and the equation n PoP=0 becomes (A,B,C) (x - x0,y - y0,z - zo)=0. By expanding the dot product we obtain A(x - x0)+B( y - y0)+C(z - z0)=0, which is equation (9.5.1). Equation (9.5.2)
v
·
·
·
-y
results from this one when the products on the left are expanded and the constant terms are brought to the right side.
•
� EX A M PL E 1 Determine a Cartesian equation for the plane that has nor
_.Figure 2
n
·&=o
mal vector n=(-2,7,3) and passes through the point
(9.5.2)
Solution Equation
-2x+7y+3z=D
P0
=
(-5,-2,1).
tells us that the required equation has the form
D = -2x0+7y0+3z0 for any point (x0,y0,z0) in the plane. We are given one such point, P0. Using its coordinates, we calculate D = -2(-5)+7(-2)+3(1)= -1. Consequently, -2x+7y+3z= -1 is a Cartesian where
equation for the plane.
The chain of reasoning that we have used to determine a Cartesian equation for a given plane can be reversed. We state this observation as a theorem.
THEOREM 2
A,B,C is nonzero. A(x - x0)+B( y - y0)+C(z - z0) 0 is
Suppose that at least one of the coefficients
Then, the solution set of the equation
=
(A,B,C ) as a normal vector and passes through the point (x0,y0,z0). The solution set of the equation Ax+By+Cz D is a plane that has (A,B,C) as a normal vector. the plane that has
=
� EX A M PL E 2 Find a vector that is normal to, and two points that lie in, the plane V whose Cartesian equation is
z
=
3y - 2x.
Solution By writing the given equation in the standard form may conclude that the vector
Po= (x0,y0,zo)
(2,-3,1)
2x - 3y+z=0,
we
is a normal vector to V. To find a point
in V, we may choose two of the entries of
Po
in an arbitrary way
and use the equation to solve for the third entry. For example, if we arbitrarily
x0=7 and y0=5, then we obtain zo= 3y0 - 2x0= (3)(5) - (2)(7)=1. Thus the (7,5,1) is on V. Similarly, if we (arbitrarily) set y0=3 and zo=1, then we have 2x0 - 3(3)+(1)=0, or x0=4. Therefore the point (4,3,1) is also on V.
point
768
Chapter 9
Vectors INSIGHT
If the three space variables
x,y,z
have no equations or restrictions
imposed on them, then they have three "degrees of freedom" and generate all of three dimensional space. In general, each new equation imposed on
x,y,z
removes one
degree of freedom. We therefore expect the solution set of one Cartesian equation among the variables x,y, z to be a two-dimensional surface. Theorem 2 tells us that, if the p (2, -1, 4)
�
V R (6
.& Figure 3
Q (3, 1, 2)
equation is linear, then the two-dimensional surface is a plane.
It is intuitively clear that any three points that are not on the same straight line determine a plane; imagine that the three points are three fingertips and balance your notebook on the fingertips-that is the plane we seek. To use equation or
(9.5.2), we must be able
(9.5.1)
to obtain a normal vector from the three points. The
next example shows how this is done. � EX A M P L E 3 Find an equation for the plane V passing through the points and R Q
P
= (2, - 1,4),
Solution To
=
( 3, 1,2),
determine
a
= (6,0,5).
normal
vector to V,
we notice that the
vectors
PQ = (3 - 2, 1 - ( -1),2 - 4) = (1,2, -2) and PR= (6 - 2,0 - ( - 1), 5 -4) = (4, 1, 1) lie in the plane (see Figure 3). The cross product of these two vectors will be perpendicular to both of them, hence perpendicular to V. Thus a normal vector will
PQ X PR= 4i -9j - 7k. According to formula (9.5.2), the desired equation has the form 4x -9y - 7z = D. Each point in the plane satisfies this equation, yielding be
the same value for D on the right side. We are given three such points. By choosing one, say R, we obtain D is a Thus
= 4(6) - 9(0) - 7(5) = -11.
4x -9y - 7z = -11
Cartesian equation for V. INSIGHT
In working the type of problem illustrated by Example 3, there is a lot of
choice. We started by choosing two pairs of points: P, Q and P,R. We might have chosen P, Q and Q,R (or P,R and Q,R) equally well. We used each pair of points to determine a vector in the plane, but we could just as well have chosen the opposite vector. For
--->
example, in taking the cross product to find a normal vector, we could have used QP instead of
PQ. After we found a normal vector, we used equation (9.5.2) to obtain a
Cartesian equation for the plane V. We might have used equation (9.5.1) instead. The coordinates of any one of the three given points could then have been used for (x0,y0, z0). To illustrate with just one of these combinations,
-4i + 9j + 7k. (x0,y0,z0), we
RQ X QP results in the normal
Using this normal in equation (9.5.1) and choosing P= (2, -1, 4) for obtain
-4(x -2) + 9( y - (-1)) + 7(z - 4)=0,
or
-4x + 9y + 7z - 11=0,
which is equivalent to the equation obtained in Example 3.
We define the angle between two planes to be the angle () between their normals, as in Figure
.& Figure 4
4. We avoid ambiguity by always selecting the angle() such that 0
:s () <
7f.
� EX A M P L E 4 Find the angle between the plane with Cartesian equation
x - y - z = 7 and the plane with Cartesian equation -x + y - 3z = 6. Solution As we have noticed, (1, -1, -1) is a normal vector for the first plane, and ( -1, 1, -3) is a normal for the second plane. The angle () between the two given planes satisfies
9.5 Lines and Planes in Space
769
(1, -1, -1). (-1,1, -3) 11(1,-1, -1) 11 11(-1,1, -3) 11
cos(O)
1
Thus using a calculator, we see that the angle between the two planes is about radians.
Parametric Equations of Planes in Space
<11111
Suppose that u plane V. Let
P0
=
=
( u1, u2, u3) (x0,y0,z0)
and v
=
( v1, v2, v3)
1.4
are nonparallel vectors that lie in a
P0 as (x,y,z) in the plane can be reached from P o
be any fixed point that lies in V. We can think of
P
an "origin" in the plane. Any point
=
by adding a scalar multiple of u and then a scalar multiple of v, as illustrated in Figure
5.
The position vector
-
OP
can then be obtained by vector addition as
-
OP
-
=
OP0
+ su + tv.
If we write out this vector equation in coordinates, then we get three scalar parametric equations for the plane: --->
_. Figure 5 OP
x
--->
=
OP0 + su + tv.
Y Z
=
=
=
xo + su1
+ tv1
Yo + su2 + tv2 Zo + SU3
+ tv3.
The parameters s and t may be thought of as coordinates in the plane V: Each point of V is determined by specifying the values of s and t. We state these observations as a theorem.
THEOREM 3 v
=
( v1, v2, v3)
If
P0
=
(x0,y0,z0) is a point in a plane V, and if u
=
( u1, u2, u3)
and
are any two nonparallel vectors that are perpendicular to a nor
mal vector for V, then V consists precisely of those points
(x,y,z)
with coordi
nates that satisfy the vector equation
(x,y,z)
=
(xo,yo,zo)
When written coordinatewise, equation
+ s u + tv.
(9.5.3)
(9.5.3)
yields parametric equations for
V:
x Y Z
=
=
=
xo + su1 + t v1 Yo + su2 + t Vz Zo + SU3
+ t V3
It should be noted that parametric equations for planes are not unique. Other choices for �
P0, u
and v will result in a different parameterization.
E X A M P L E 5 Find parametric equations for the plane V whose Cartesian
equation is
3x -y + 2z
=
7.
770
Chapter 9
Vectors Solution We must find one point P0=(x0, y0, z0) and two nonparallel vectors u and
v in V. To find P0, we may select any values for
x0 and y0 and use the Cartesian 3x - y+ 2z=7 to solve for z0• For example, if we (arbitrarily ) set x0=1 y 0=0, then we obtain 3(1)- (0)+ 2z0=7, or zo=2. ThusP0=(1,0,2) is a
equation
and
point on V. The vectors u and v can be any nonparallel vectors that are perpendicular to the normal vector n=(3,-1, 2). For example, we may take
u=(0, 2, 1) and v=(2, 0,-3), as you can verify by calculating u n=0 and v n=0. ·
With
these
choices,
the
y=0+2s+Ot,z=2+s - 3t.
parametric
We
may
equations
(and
should )
·
become
verify
x=1+Os+ 2t,
these
parametric
equations by substituting them into the Cartesian equation and checking that the resulting equation holds for all s and t. Here (1+ 2t)- (2s)+ 2(2+s - 3t) = 7, which we
the substitutions yield the equation
3 s and t by x=1+2t,y=2s,z =2+s - 3t are
see is true for every
simplifying the left side. Having verified that
valid parametric equations for V, we should remark that alternative choices lead to equally valid yet quite different parametric equations. See Exercise INSIGHT
92.
The cross product is useful for producing one vector that is orthogonal to
two given vectors. The solution of Example 5 illustrates a simple procedure for finding two nonparallel vectors that are orthogonal to one given vector
n
=(A, B, C). To do that,
we may replace one component of n by 0, interchange the other two, and change the sign of one of them. Thus
Parametric Equations of Lines in Space
(0, C, -B), (C, 0, -A),
and
(B, -A,O)
are all orthogonal ton.
A line f, in space can be described by a point that it passes through and a vector that is parallel to it. Look at Figure 6. Let P0=(x0,y0,z0) be the point, and let m=(a, b, c)
the vector parallel to L An arbitrary point P=
(x,y,z) lies on f, if and only if the vector
P0P is parallel to m. This happens if and only if P0P is a multiple of m. That is, P0P is parallel to m if and only if, P0P=tm for some scalar t. In coordinates, this last equation becomes
(x - x0,y - y0,z - z0) = t(a, b, c), or (x,y, z) - (x0,y0, z0) = (ta,tb,tc). By adding vector (x0,y0,z0) to each side, we obtain (x,y, z)=(xo+ta, yo+tb, zo+tc).
Notice that tis a scalar and will play the role of a parameter. After we match up coordinates, the parametric equations for a line in space become
x = xo+ta y =Yo+tb z = zo+tc
.A Figure 6
Let us summarize our observations with a theorem.
THEOREM
4
(Parametric
Equations for a
Line)
The line in space that passes
through the point P0=(x 0, y 0, z 0) and is parallel to the vector m=(a,b,c) has
equation
-------+
PoP=tm. Here,P=
(x,y,z)
is a variable point on the line. In coordinates, the equation
may be written as three parametric equations:
9.5 Lines and Planes in Space
771
x = xo+ta y =Yo+tb z =zo+tc.
� EXA M P L E
6 Give three points that lie on the line f, whose parametric
equations are
x = -5+3t y = 7- St z = 1+2t. Does the point (1,2,3) lie on£? How about the point ( -11,23, - 3)? Give para metric equations for the line through the origin that is parallel to£.
Solution Each value assigned to
t
will generate a point on£. For instance,t
=0
givesx=-5,y=7,z=1, or the point (-5, 7,1). The value t=3 gives x=4,y= -17, z=7, or the point (4, -17, 7). The value t=-1 gives x=-S,y=15,z=-1, or the point (-S, 15, -1). To determine whether the point (1,2,3) lies on the line, we need to determine whether some value of
t,
when substituted into the parametric
equations, yields this point. Thus we need to find a single
t
that satisfies all three
equations
3t- 5 = 1 -St+7 = 2 2t+1=3 simultaneously. The first equation gives t = 2, the second gives t = 5/S. We need look no further: There is no single value of t which satisfies all three equations. Thus the point (1,2,3) does not lie on the line. We have better luck with the point (-11,23, - 3). We endeavor to find a (single)
t by
solving
3t- 5 = -11 -St+7 = 23 2t+1= -3. The first equation gives substituting
t=
t=
- 2, and so does the second, and so does the third. Thus
- 2 into our parametric equations yields the point (-11,23, - 3), and
this point therefore lies on the line. Finally, the parametric equations tell us that the
4 with P0 = (x0,y0,z0) = x = 0+3t,y = 0- St,z = 0+2t is
vector (3, -S,2) is parallel to£. We now use Theorem
(0, 0, 0)
to see that the line parameterized by
parallel to f, and passes through the origin. ..,...
Cartesian Equations of Lines in Space
As our next theorem shows, Cartesian equations of a line are obtained by elim inating the parameter from a parameterization.
THEOREMS
Suppose that a line f, passes through the point P0
is parallel to the vector
m
= (a,b,c)
with
a,b,
and
c
= (x0,y0 ,z0) and
nonzero. Then f, is the
solution set of the following system of equations:
x- xo Y- Yo z-zo = = b c a
(9.5.4)
772
Chapter 9
Vectors Proot According to Theorem
4, we may parameterize f by the equations x = x0+ta,y = y0+t b, z= zo+t c. After solving for the parameter t in each of these equations, we havet = (x - x0)!a, t = (y -y0)/b, andt = (z- z0)/c. We equate • these three expressions fort to obtain line (9.5.4). The equations of line
(9.5.4)
are known as
symmetric equations 96.
for£. For the
cases in which one or two of a, b, c are zero, see Exercise �
7 Line f, of Example
EXAM PL E
y=7 - St, z= 1+2t.
6
is parameterized by
Find Cartesian equations for £. Is the point
How about the point
x = -5+3t, (1, -9, 5) on £?
(4, -1, 7)?
Solution We solve fort in each of the parametric equations, obtaining t
= (y - 7)/(-8)
and t
= (z-1)/2.
t = (x+ 5)/3,
We now equate these three expressions for t to
obtain the symmetric form of£:
x+5 y - 7 z-1 -2- . -3(-8) _
A point
P = (x, y, z)
(9.5.5)
_
lies on f if and only if all three of these quantities are equal.
For example, the point (1, -9, 5) lies on the line because each quotient in line (9.5.5) becomes 2 when we substitute x=1,y= -9, and z=5. However, the point (4, -1, 7) is not on the line because when we substitute x=4,y= -1, z=5 into line (9.5.5), the first quantity becomes 3 and the second becomes 1. <11111 Just as two points in the plane determine a straight line, so do two points in space. The next example shows how to find the symmetric form of a line that passes through two given points. �
8 Let
EXAM PL E
P = (2, -3, 5) P
equations for the line f passing through
and
Q
= (-6, 1,12).
Write
Cartesian
and Q.
Solution The symmetric form of a line requires a point on the line and a vector m that is parallel to the line. The piece of information we are initially missing is m, but the vector
PQ = (-6 - 2, 1 - (-3), 12 - 5) = (-8, 4, 7) will certainly do the job. We
can use either choose
P,
P
or Q as the point we take to be
in line
(9.5.4).
then the symmetric form for f is
x - 2 y+3 z- 5 (-8) -4-7- . _
_
�
(x0,y0, z0) ....
EXAM PL E 9 Convert the symmetric equations
x+1 l 2z- 3 = = 6 3 2 to parametric form. Solution We set each of these three equal quantities equal tot. We obtain
x+ 1 =t ' 3
y
x=3t - 1,
y=2t,
2 =t,
2z- 3 =t 6
--
or, equivalently,
These are parametric equations for the given line.
3 z=3t+ 2· <11111
If we
9.5 Lines and Planes in Space
773
� EX A M P L E 1 0 Find symmetric equations for the line f that is perpendi cular to the plane 3x- 7y+ 4z = 2 and that passes through the point Solution Notice that the vector m
= (3,-7,4)
(1,4,5).
is normal to the plane. So the line
we seek will be parallel to m. Because the line passes through
(1,4,5),
its
parametric equations will be
= 1 + 3t = 4-7t = 5 +4t.
x y z
To pass to symmetric equations for f, we solve for equate the resulting expressions for
t.
t
in each equation and then
Thus
y-4 _ z-5 -3- - --=? - -4-,
x-1 _ are symmetric equations for f. INSIGHT
<1111
Symmetric equations for a line f describe f by means of
two
linear
equations. Each of these equations is, taken by itself, the equation of a plane. Therefore symmetric equations of a line permit us to visualize the line as the intersection of two 4y + 7z
=
planes. The Cartesian equations in Example 10 may be rewritten as
51 7x + 3y
=
x-l y-4 = 3 -7
19
and
y-4_z-5 - -· �- 4
and
4y+ 7z=51,
Rewriting these equations as
7x+3y=19 we see that
�
x 1 = .A Figure 7 The line z-5 y-4 = - - as an intersection of � 4 two planes.
our line is the set ofpoints that lie in the intersection of two planes. These ideas
are illustrated in Figure 7 .
� EX A M P L E 1 1 Find parametric equations of the line of intersection of the two planes x- 2y+z =4
and
2x+y-z =3.
Solution To do so, we set x =t and substitute this into the given equations: -2y+ z
=4-t
and
y-z =3- 2 t .
We may solve these equations simultaneously for y and z in terms of y z
t:
= 3t-7 = 5t-10.
If we combine these equations with the equation x =t with which we began our computation, then we see that x y z
=t = 3t-7 = 5t-10 t in each of the t yields x = ( y+7)/3 =
are parametric equations for the line of intersection. Solving for parametric equations and then equating the expressions for (z+ 10)/5 for symmetric equations.
<1111
774
Chapter 9
Vectors INSIGHT
We could have done the last example by setting either y =t or z =t. These
choices would have resulted in different parameterizations for the same line. For clarity, let's set y =
s
(instead oft) and see what happens: We obtain
x+z=4+2s and
2x-z =3-s. Solving these simultaneously, for
x
and
z in
terms oft, yields
s 7 x = -+ 3
3
5 5 z = 3s+ 3" This leads to parametric equations
s 7 x = -+ 3
3
s 5 5 z = s+ 3· 3 y
=
These three equations describe the very same line as in the last example, but they look quite different from the parametric equations that we found there. What is the rela tionship between these two parameterizations? The answer is that the change of variable
s = 3t -7 transforms � EXAM P L E 2x+ y-2z
one parametrization into the other.
1 2 Let
V
be
the
plane
whose
Cartesian
equation
is
10. Let l be the line that is perpendicular to V and that passes through P= (3, -8,3). Find the point R at which l intersects V. =
Solution For the first step, we follow the method of Example
10 to parameterize £.
Because m= (2, 1, -2) is orthogonal to V, l has parameterization x=3+ 2t,
y= -8+ t,z =3 - 2t. We substitute the parametric equations for x,y, and
z into the equation of the plane. This gives us 2(3+ 2t)+ (-8+ t) - 2(3 - 2t)= 10, or -8+9t=10. Solving for t, we obtain t=2. The coordinates of R are therefore x =3+ 2(2) =7,y = -8+ 2= -6, and z =3 - 2(2)= -1. ThusR= (7, -6, -1) is the point of intersection that we seek. INSIGHT
p� €
R
llPQll > llPRll Q
v �
A Figure 8 R is the point on V that is closest to P
In Example
12,
the distance between P and R is
J(7 - 3)2+ (-6 - (-8) ) + (-1 - 3)2, or 6. Using the Theorem of Pythagoras, we see 2
that R is the point on V that is closest to P (see Figure
8).
Later in this section, we will
discuss an efficient method of calculating this distance without first locating the point R of intersection.
� E X A M P L E 1 3 Determine whether the lines parameterized by x
3t-7 = -2t+5 z t+ 1 =
y
=
9.5 Lines and Planes in Space
775
and
x y z
=
=
=
-3t+2 t+ 1 2t-2
intersect. Solution First of all, let us note that we cannot expect that the point of intersection, if there is one, will correspond to the same value of
t for
each line.
(Imagine two cars crossing each other's path at an intersection. When all goes well the two cars pass through the same point but
at different times.)
We overcome this
complication by using a different parameter, says, for the first line. Now we equate the expressions for
x,y, and z .
We have
3s-7 -2s + 5 s+ 1
=
=
=
-3t+2 t+ 1 2t-2.
Notice that there are three equations in only two unknowns. Solving the first two
s 1 and t 2. There will be a point of intersection s and t also satisfy the third equation. That is the case s 1 corresponds to the point (-4, 3,2) on the first line;
equations simultaneously gives
=
=
if and only if these values for in this example. The value the value
t 2 =
corresponds to the same point on the second line. This is the only
point of intersection. INSIGHT
=
<11111
When we determine the (line of) intersection of two planes, we solve two
equations in three unknowns. Such a system generally has a solution (corresponding to
the fact that two nonparallel planes in space intersect) . However, when we determine the
(point
of) intersection of two lines, we solve three equations in two unknowns. Such a
system generally does not have a solution (corresponding to the fact that two lines in
space generally do not intersect) . Much of the best mathematics arises from the elegant
fashion in which algebra reinforces geometry. The situation of intersecting planes and intersecting lines is a simple instance of this principle.
Calculating Distance
If
S1 and S2 are
and
S1
two sets of points in space, then we define the distance between
S1
S2 to be the minimum, if it exists, of all distances d(P1, P2) where P1 belongs to P2 belongs to S2• We now use vector methods to find formulas for the
and
distances between various geometric figures.
(x0,y0,z0) is a point and that Vis a plane. Let n (A, be a normal vector for V and let Q (x1,y1,z1) be any point on V. Then the
THEOREM 6
B, q
Suppose that
P
=
=
=
P and V is the length llPn(PQ) II of the orthogonal projection of PQ in the direction of n. That is, the distance between P and V is equal to distance between
IPQ·nl 11°11 -----+
(9.5.6)
776
Chapter 9
Vectors or, equivalently,
IAxo +Byo + Czo -DI JA2+B2+ C2 where D
=
(9.5.7)
Ax.1 + Byz + Czz.
Proof. Let R be the intersection of V with the line f through P that is normal to V, as in Figure
8.
According to Pythagoras's Theorem,R is the point on V that is
closest to P. The distance of P to V is therefore
llPR II·
direction of n. Formula
(9.3.15). Formula (9.5.7) JA2+B2+ C2 and
Because
n an�R
are
parallel, this length is precisely the length of the orthogonal projection of PQ in the
(9.5.6)
follows from equation
then obtained from (9.5.6) by observing that
llnll
=
is
IPQ · nl = l (x1 -xo,Y1 -yo,z1 -zo) · (A,B, C)I =
IAx1 +By1 + Cz1 - (Axo +Byo + Czo)I
The proof of Theorem
=
•
IAxo +Byo + Czo -DI.
6 demonstrates how powerful the concept of
projection
can be. Elementary geometry tells us that the distance we seek is the length
llPR II,
but this quantity depends on a point R that has not been given. By calculating a
projection, we are able to find the length
llPR II
without first finding the point R.
� EXAM P L E 1 4 Find the distance between the point P = (3, -8, 3) and the plane V whose Cartesian equation is 2x +y - 2z 10. (The point and the plane of =
this example are the same as those of Example
12.)
Solution From the given Cartesian equation for V, we see that n normal for V. If Q
=
According to formula
(2, 1, -2) is a (x1,y1,z1) is any point on V, then D 2x1 +y1 - 2z1 10. (9.5.7), the distance between P and V is 12(3) +(-8) - 2(3) - 101 =
=
=
J22+12+ (-2)2 or
6. This answer agrees with the distance that we calculated in the Insight following Example 12. That computation was considerably more laborious because it required the calculation of the point R at which V and the line normal to V through P intersect.
<1111
The next theorem, which gives a method for calculating the distance of a point
to a line, is a nice application of vector projection, the vector triple product, and the scalar triple product.
THEOREM 7
Suppose thatP
is parallel tom. Let Q
( x0 ,y0 ,z0) is a point and thatf is a line in space that (x1,y1,z1) be any point onf. Let n m X (m X PQ). Then, =
-----+
=
=
II (PQ) II of the orthogonal projection of
the distance between P and f is the length P0
PQ in the direction of n. That is, the distance between P and f is equal to P I Q -----+
· nl
llnll
(9.5.8)
9.5 Lines and Planes in Space
777
Let V be the plane containing both P and £, let f' be the line in V that passes through P and that is perpendicular to f, and let R be the point of intersection of f and f,' (see Figure 9). Reasoning as in Theorem 6, we see that R is the point on f that is clos� to P. The distance of P to f is therefore llPR II· Our next step is to show that PR and n are parallel. To do this, we need to see that n is perpendicular to f and lies in V. Because n m X (m XPQ), we know that n is perpendicular to the first operand of the cross product, m, which is to say n is perpendicular to f. We also know that n is perpendicular to the second operand of the cross product, m XPQ. But m and PQ are nonparallel vectors in V. Therefore m XPQ is orthogonal to V. Because n is perpendicular to a normal for V, it follows that n lies in V. Thus n and PR are p�allel, and the length llPR II is precisely the length of the orthogonal projection of PQ in the direction of n. Formula (9.5.8) therefore follows • from equation (9.3.15). Proot
e v
=
Q • Figure 9
Find the distance between P (5, 3, 3) and the line f with x - 1= (y+8)/ (-4)= (z - 7)/3 Solution Writing the term x - 1 as (x - 1)/1 (to attain the exact form recorded in line (9.5.4) in which each expression is a quotient), we see that Q (1, - 8, 7) is a point on f, and m (1, -4, 3) is parallel to f. We calculate PQ ( -4, -11, 4), m XPQ (17, -16, - 27), n m X (m XPQ) (156, 78, 52) 26(6, 3, 2), 1 n 1 ,3 2) - 1 (6 3 2) and IPQ nl (6 26 , 7 , , , M llnll 26 11(6,3,2) 11 1(-4)( 6) + (-11)(3) + 4(2)1 l -49 1 7. 7 7 � EXAM PL E 1 5
=
symmetric equations
.
=
=
=
=
=
=
=
----+
·
_
_
=
=
=
Thus the distance between P and f is 7. In Exercise 93, you are asked to verify this result by using the methods of Chapter 4. <1111 Our next theorem provides a method for calculating the distance between two lines. Suppose that m1 and m2 are nonparallel vectors, that £1 is a line through P1 (x1,y1, z1) parallel to m1, and that £2 is a line through P2 (x2,y2, z2) parallel to m2. Let n m1 X m2. Then, the distance between £1 and £2 is the length llPn(P1P;) II of the orthogonal projection of P1P; in the direction of n. That is, the distance between f1 and £2 is equal to
THEOREM 8
=
=
=
IP1P2 nl 11°11 ----+
·
(9 .5 9) .
Let R1 and R2 be the points on £1 and £2, respectively, that are closest to each other. Then, R iR2 is perpendicular to both £1 and £2• It follows that RiR2 is parallel to n, and the length llR1R;ll is precisely the length of the orthogonal projection of P1P; in the direction of n. Formula (9.5.9) therefore follows from • equation (9.3.15).
Proot
778
Chapter 9
Vectors �
EX A M P L E 1 6 x = y+ 1 = z
Let £1 and £2 have symmetric equations -3 -4
x
and
-9 2
=
y
-2 -1
+3 = z . -2
respectively. Find the distance between £1 and £2. Solution From the symmetric equations for £1 we see that £1 passes through P1=(0, -1, 3) and is parallel to m1=(1,1, -4). From the symmetric equations for £2 we see that £2 passes through P2=(9,2, -3) and is parallel to m2=(2, -1, -2). We calculateP1P�=(9,3,-6) =3(3,1,-2),n=m1 X m2= (-6, -6, -3)=-3(2,2,1), and 2 IM. nl. = 3 1(3,1,-2) (2 , 2,1)1 =(3)(2)+(l)(2)+(-2)( )=6 l . 2 2 2 3y'2 +2 +1 11°11 ·
Thus the distance between £1 and £2 is 6. With considerably more labor, the points R1 =(1,0, -1) on £1 and R2=(5,4,1) on £2 that are closest can be located. Vector projection allows us to avoid such calculations.
If v and w are nonparallel vectors represented by directed line segments with a common initial point, then they determine a plane V with v Xw as a normal. Because the triple vector product u X (v Xw) is perpendicular to the second operand v Xw, we see that u X (v Xw) lies in V. In our work on the parametric equations of planes, we observed that every vector in V can be written as sv+tw for some scalars s and t. The consequence is that there are scalars s and t such that u X (v Xw)=sv+tw. Our next theorem provides specific formulas for these scalars. A shorter proof is discussed in Exercise 94; the argument uses tedious algebraic identities involving the nine entries of u, v, and w. The triple vector products u X (v Xw) and (u Xv) Xw satisfy
THEOREM 9
u x (v x w)= (u w) v - (u v) w
(9.5.10)
(u Xv) Xw= -(v w) u+ (u w) v.
(9.5.11)
·
·
and ·
·
We prove the first equation and leave (9.5.11) as Exercise 95. If v and w are parallel, then v Xw=0, and either v=0 or w=.Xv. For the left side of (9.5.10), we have u X (v Xw)=u X 0=0. For the right side of (9.5.10), we have either (u w) v - (u v) w = (u w) 0 - (u 0) w = 0 or (u w) v - (u v) w = (u .Xv) v - (u v) .Xv= >.(u v) (v - v) =0. Thus equation (9.5.10) holds when v and w are parallel. Now suppose that v and w are perpendicular. Let V be the plane they deter mine. We may write u=o:v+{3w+ n where n is orthogonal to V and therefore parallel to v Xw. Then (9.5.10) will hold for u if it holds for each summand. Starting with o:v and noting that o:v w= 0, we have Proot
·
·
·
·
·
·
·
·
·
·
o:v X (v Xw) (9;!8) -o:(v v) w= -(o:v v) w= (o:v w) v - (o:v v) w, ·
·
·
·
9.5 Lines and Planes in Space
779
which is (9.5. 10) for the vector av. Next, working with (3w and noting that ( (3w) · v we have
=
0,
(3wX (vXw) (9�9) (3 (w w) v=((3w w) v=((3w w) v - ((3w v)w, ·
·
·
·
which is (9.5. 10) for the vector (3w. Finally, because nX (vXw)=O,n·w=O, and 0, we have
n· v
=
nX (vXw)=0= (0) v - (0)w=(n ·w) v - (n · v)w, which is (9.5. 10) for the vector n. Now we make no assumptions about v. As in equation (9.3.16), we write
v=Pw(v)+ Q where Pw(v) is parallel tow and Q is perpendicular tow. Because equation (9.5.10) has been proved for both summands of v, it holds for v as well. • � EXAMPLE 17 Let u=(2,-1,3),v=(3,2,-2), and w= (-1,0,2).Calcu late uX (vXw), and verify equation (9.5.10). Solution We have
and
( [ ; =! � l) i
u x (v xw) =det
j
k
= 10i+ 8j -4k.
Also,u·w= (2)(-1)+ (-1)(0)+ (3)(2)=4 and u· v= (2)(3)+ (-1)(2)+ (3)(-2)= -2. Therefore (u ·w) v - (u · v)w=4( 3i+ 2j - 2k) - ( -2)( -i+ Oj + 2k) =( 12 - 2)i + (8 - O)j + ( -8+4)k=10i+ 8j -4k, which agrees with our direct calculation of uX (v Xw ). .,..
Q UIC K
Q UIZ
1. If n=(1,(3 ,1) is orthogonal to the plane with Cartesian equation 2x- Sy+ 6z =11, then what is 1? 2. The point (3, 15, -2) is on a plane that has (3, 1,6) as a normal and Cartesian
equation x+By+ Cz =D. What is D? 3. If x= (3 - 2y)/8=1 + 2z are Cartesian equations for a line f, and (3,b, c) is parallel to f, then what is b? 4. What is the distance of the plane x+ y+ 2z = 2 to the origin? Answers
1. 3
2. 4
3. -12
4. 2/../6
780
Chapter 9
Vectors
EXERCISES In Exercises 25-28, find parametric equations for the line
Problems for Practice In each of Exercises 1-4, a Cartesian equation that describes a plane V is given. Using only the given equation,
perpendicular to the given plane and passing through the given point.
describe all normals to V. Then, verify that the given points
25.
P,Q, and R are on V. Calculate n=PQ x PR, n is normal to V.
26.
x- 3y+4z=2 P=(2,0,0) 2. z=5 - 7x+l lyP=(0,0,5) 3. 3x- 4z=1 P=(3,0,2) 4. x=2y P=(0,0,0) L
and verify that
Q=(l,1,1) R=(5,l,O) Q=(l,0, -2) R=(2,1,2) Q=(7 ,1,5) R=(-5,2,-4) Q=(-4, -2,3) R=(2,1,1)
In each of Exercises 5-8, describe all normals to the plane that is parameterized by the given equations. 5. x=3 +2s+t 6.x=l+s+t 7. x=2s 8. x=2s+t
y=-1 +2s-t y=2 +2s+2t y=l -t y=2s
z =2 +s- 3t z=3 +2s+3t z=s- 2t z =-3t
In each of Exercises 9-12, find a Cartesian equation for the plane with the given normal vector n and passing through the given point P.
n=(-3,7,9) n=(9,-3,7) 11. n=(2,2,5) 12. n=(7,1,1) 9.
10.
P=(1,4,6) P= (6,0,-9) P=(-5,9,2) p =(2,0,0)
In each of Exercises 13-16, find a Cartesian equation for the plane determined by the three given points.
27. 28.
(7,-4, 6) (2,3,8) (-1,-1,-8) (0,2,1)
In each of Exercises 29-32, find symmetric equations for the line that passes through the two given points.
(1,0,1) (1,1, -1) 3L (1,2,1) 32. (2,2,1) 29.
30.
( 2,1,0) (-1, -1,1) (3,1,0) (0,0,0)
In each of Exercises 33-36, find, in parametric form, the line of intersection of the two given planes.
x- 3y+5z=2 x- 3y=4 35. 4x+6y-z=1 36. x+2y -z=1
33. 34.
2x+6y+3z=4 y+8z=6 4x+z=0 2x-y-z=2
In each of Exercises 37-40, find symmetric equations for the line of intersection of the two given planes.
x-y+z=2 x- 3y=1 39. 2x+y-z=2 40. x+2y-z=0 37.
38.
x+y+3z=6 y+8z=l 4x-z=3 2x-y-z=l
In each of Exercises 41-44, find the point of intersection
13.
(4,1,6) (3,-9,5) (0,1,3) (1,9, 5) (1, 1,8) (0,1,2) 15. (2,-6,8) (1,-9,-4) (-5,1,4) (-5,4,6) (8,2,5) 16. (6,1,3)
x- 3y+7z=6 3x+Sy+z=9 x-y+z/3 =0 x/2 +y +2z=1
14.
of the given plane and the given line.
that passes through the given point Po and that is parallel to
x- 3y+5z=0 2x+3y- 5z=1 6 43. x+y=4 44. x+y- 3z=5
In Exercises 17-20, find parametric equations for the line the vector
4L 42.
m
Po=(2,1,9) 18. Po= (7,-1 1,3) 19. Po=(0, 1, 0) 20. Po= (2, 2, -5) 17.
z=-t- 3 z=4t+l z=-2t+3 z=2t+5
of the given plane and the given line. 45. 46. 47. 48.
that passes through the given point P0 and that is parallel to the given vector:
Po=(2,1,5) Po=(-4, -2,-1) 23. Po=(0,1, 0) 24. Po= (-5, -3,-9) 21.
y=6t+4 y=-3t+2 y=2t+6 y=2t+6
In each of Exercises 45-48, find the point of intersection
m=(-3,1,7) m= (4,6,-8) m=(2,-1,1) m=(1, -1,-3)
In Exercises 21-24, find symmetric equations for the line
22.
x=2t+6 x=t- 4 x=-t x=4 -t
x- 3y+5z=1 9 2x+3y- 5z=0 x+y=4 x+y- 2z=5
x=y+ 1 =z/2 (x- 1)/2 =(y+1)/4 =2z- 5 (x+1)/2 =y+1 =3z+2 (x- 2)/2 =(y+1)/3 =z- 1
In each of Exercises 49-52, find the cosine of the angle between the two given planes.
(-5,8,2) (-6,-3,-7) (1,1,9) (9,3,-1)
49. 50. SL 52.
4x- 3z =4 2x- 2y+z=1 -5x- 3y- 4z=0 8z-x- 4y=1
2x+y+2z=2 x- 2y+2z=5 x+y+z=l z-x-y=5
9.5 Lines and Planes in Space
In each of Exercises 53-56, find the distance between the
three coordinate axes. Sketch the part of the plane that lies in
given point and the given plane:
the first octant.
53.
77.
54. 55. 56.
(1,0,1) (-2,1,1) (1,1,1) (0,0,0)
2x+y+z=9 2.x- 2y+z=13 2.x+2y+z=14 x- 6y+4z=3
78. 79. 80.
In each of Exercises 57-60, determine whether the given
58. 59. 60.
(-7,6,2) (-4,7,8) (1,S,1) (21,4,-4)
x=3t-1 x=-3t+S x=t x=6t-3
parallel:
y=-t+4 y=t+4 y=S y=t
a.
z=2t+6 z=St- 1 z=-3t+1 z=-t
b. c.
d. e.
In each of Exercises 61-64, determine whether the pairs of
lines
intersect;
if
they
do,
find
the
point(s)
62. 63. 64.
x=3t- S x=t+1 x=-t+6 x=2t-4 x=t-1 x=3t+1 x=t+4 x=t-2
with
of
y=-4t- S y=t- 6 y=t+3 y=t-S y=-2t+14 y=-t+4 y=-2t+1 y=t+1
z=t+1 z=t+S z=4t-6 z=-3t+4 z=2t+3 z=-2t+11 z=-3t-3 z=2t+6
67. 68.
71. 72.
£ and a Cartesian equation for a plane V. How do you tell if £ lies in V? If £ does not lie in V, how do you tell
whether or not £ and V intersect (without attempting to solve the equations simultaneously)? In each of parts a to d, apply your answer to each given line and plane. If the line and plane do intersect, find the point of intersection. a.
b. c.
d.
=(1,-2,4) u =(2,1,2) u =(1,1,1) u =(-S, -1,0) u
v=(2,0,1) v=(2,2,3) v=(3,0,2) v=(l,0,3)
x- Sy+2; z=8; x=-3t+4; y=-t- 6; z=-t+2 3x- 7y+2; z=-22; x=3t- 6; y=t+2; z=-t+S x+y- 2; z=10; x=4t; y=t+6; z=-3t+9 -3x- 2y+6; z=1; x=2t+1; y=3t- 4; z=2t+8
84.
w=(3,1,1) w=(3,3,-S) w=(-2,-3,3) w=(0,2,2)
85. 86. 87.
x=1 - y=z+1 x- 1 =y/2 =z x- 1 =y- 1 =z x- 1 =y+1 =-z
Further Theory and Practice
In each of Exercises 88-91, a point
88.
the plane that is parallel to the given plane and that passes
90.
74. 75. 76.
3x+4y+z=0 z=2.x (x-1) +2(y-2) +3(z -3) =1 z=2.x- 3y In each of Exercises 77-80, a Cartesian equation for a
plane is given. Calculate the intercepts of the plane with the
and a Cartesian
Q that
Q
is perpen
dicular to V; next, find the point R of intersection of this line and V; finally, calculate the distance from
89.
(1, 2, 3).
Q
to V as follows: First find the line through
In each of Exercises 73-76, find a Cartesian equation for through the point
x=y=z x=y=z 2.x=y- l =z- 1 x=y=2z
equation for a plane V are given. Compute the distance of
distance between
73.
and
lines with given symmetric equations.
In each of Exercises 69-72, verify equation (9.5.10) for the
70.
Ax+By+ Cz=D1
In each of Exercises 84-87, find the distance between the
x- 1 =y- 2 =z- 3 x=y=-z x=1 - y=2z x- 2 =y- 2 =z/2
(0,0,0) (0,-1,2) (1,0,-1) (1,0,-1)
triple vector product.
69.
equations
83. Suppose that you are given parametric equations of a line
given point and the line with given symmetric equations.
66.
Cartesian
Ax+By+ Cz=Dz.
In each of Exercises 65-68, find the distance between the
65.
2.x- 4y+8z=S 3x+3y+1Sz=-3 x- y+Sz=8 -2.x- Sy+12z=1 2.x+2y+2z=9
x- 2y+4z=7 x+y+Sz=1 4x- 6y+2z=1 x+4y- 6z= -s x+y+z=0
82. Find a formula for the distance between the two planes
intersection.
61.
2.x+3y+z=6 x+4y+6z=12 3x+Sy+1Sz=lS 2.x+3y+9z=18
81. Determine whether the following pairs of planes are
point lies on the given line.
57.
781
91.
Q and
Q=(-6,-3,13) Q=(8,-2,4) Q=(-1,S,3) Q=(2,-3,6)
92. Verify
that
Q to R to obtain the
V.
2.x+y- 3z=2 x- y+z=4 -3x+y- 4z=2 2.x- Sy+1z=4
x=2 +s+t, y=-1 +s- t, z=-s- 2t
are
equally valid parametric equations for the plane V of Example S even though they are obviously quite different from the ones found in the solution.
93. In Example
lS,
vector methods are used to show that the
distance between P =(S,3,3) and the line £ with sym metric
equations
x- 1 =( y+8)/(-4) =(z- 7)/3
is
7.
782
Chapter 9
Vectors
Obtain this result by parameterizing the line, expressing
and
the distance between P and a point on the line in terms of
y=yo, z=zo
when
m=(a,0,0),
x=xo, z=zo
when
m=(O,b,O),
x=xo, y=Yo
when
m=(0,0,c).
the parameter, and using calculus to minimize the dis tance function.
94. Writing u=(a,b,c),v=(d,e,f), and w=(g,h,p), expli
(9.5.10). Verify
citly calculate each side of vector equation
that corresponding entries of each side are equal.
95. Derive equation (9.5.11) by applying (9.5.10) with the
Calculator/Computer Exercises
vectors appropriately permuted to the right side of
(uxv) Xw= -w x(uxv).
97. Plot the sphere with center (2,1,2) and radius 3. Verify that
96. Let£ be a line whose direction m=(a,b,c) has one or two
P0=
zero entries. Show that £ has the Cartesian equations
x -xo a
_ _
y-yo z-zo b '
_ _
_
_ _
when
that is tangent to this plane at the point has radius
when
m=(a,O,c),
y -Yo z -zo = x=xo b c '
when
m=(0' b c)
--
to your plot the
98. Plot the plane 2x +y +z= 5. Add to this plot a sphere
m=(a,b,O),
x-xo z -zo , y=yo = a c --
(3, 3, 4) is a point on this sphere. Add
plane that passes through P0 and that is tangent to the sphere.
(1,2, 1)
and that
1.
99. The line£ that is parallel to(-1,- 2,3) and passes through the point (1,1,0) intercepts the solution set S of the Car tesian equationz=x4 +y4 in two points. Find them.
'
Summary of Key Topics in Chapter 9 Points in Space (Section 11.2)
(x,y,z). The
A point in space is located with three coordinates:
distance between
two points (ai. bi, c1) and (a2, b2, c2) is
V( ai - az)2 + (bi - bz)2 + (c1 - cz)2 . It follows that the equation of a sphere with center (a, b, c) and radius r > 0 is
(x - a)
2
+
(y - b)
The sets
and
{(x,y,z): (x - a)
2
{(x,y,z): (x - a)
2
2
+
(z - c)
+
(y - b)
2
+
(y- b)
2
2
= r2.
+
(z - c)
+
(z - c)
2
2
< r2}
�
r2}
are called, respectively, the open and closed balls with center (a, b, c) and radius r.
Vectors (Sections 11.1, 11.2)
A vector in two dimensions is an ordered pair dimensions is an ordered triple points in the plane, and if
v
w
If
v
v.
=(vi. v2). A vector in three
=(v1, v2) where v1 = q1 -p1 and v2 = q2 -p2, then we
say that the directed line segment the vector
v
=(wi. w2, w3). If P =(pi,p2) and Q =(qi, q2) are
PQ with initial point P and endpoint Q represents
A similar definition is used in three dimensions.
=(vi. v2, v3) and
w
= (wi. w2, w3) are vectors, then
V + W =(v1 + W i , V2 + Wz, V3 + W 3 ). Geometrically, two vectors are added by following the displacement of the first by the displacement of the second. The zero vector is defined by 0 = (0, 0, 0). If ..\ is a real number, then
Summary of Key Topics
The most important attributes of a vector v =(vi.
and, if v-1 0, its
>.
its
magnitude
(or
length)
direction
dir(v) If
v2, v3) are
783
is positive, then
>.v
! (vi, v2, v3).
=
ll ll
points in the same direction as
v: dir(>.v) = dir(v).
If
>.
is
negative, then >.v points in the direction opposite to v: dir(>.v) = -dir(v). In both cases,
11 >.vll l>-l llvll·
is obtained by dilating the length of llvll by a factor of The Triangle Inequality tells us that the length of a sum of vectors is less than the sum of the lengths of the summands: the length of
11>-11: 11>-vll
=
llv+wll :S llvll + llwll· Dot Product
If v
=
(v1 , v2 , v3) and w (w1, w2, w3) are =
v. w
(Section 11.3)
=
(v·w w·v) and distributes over addition (u· (u +v)·w u·w+v·w). The dot product commutes ((>.v)·w w·(>.v) >.(v·w)). The length of a vector is
The dot product is commutative
(v+w) u · v+u·w =
and
with scalar multiplication
vectors, then their dot product is
V1W1+ V2W2+ V3W3. =
=
=
=
related to the dot product by the equation
v. v
=
2 llvll ·
If v and w are nonzero vectors, then cos(B) where 9 E
[O, 7r]
=
. 11 11 11
: :11 '
is the angle between the two vectors.
Two vectors
v and w are
orthogonal (that is, mutually perpendicular) if and
llvll llwll =Iv· wl, or, equivalently, if and only if v = >.w or w =>.v for some scalar >. By convention, 0 is both orthogonal
only if v·w = 0. They are parallel if and only if
.
and parallel to every vector.
Projection
The projection of a vector
(Section 11.3)
v onto
a vector w is
Pw(v)
=
( ��f ) w.
The component of v in the direction of w is v is
Direction Vectors (Section 11.3)
Iv· wl/llwll·
·
w!llwll· The length of
the projection
If vis any nonzero vector, then the vector (1/llvll )v is a unit vector pointing in the same direction. A unit vector is called a
direction.
The unit vectors
784
Chapter 9
Vectors
j k are called the
u u
=
(0, 1,0) (0, 0,1)
=
=
standard basis vectors.
uniquely in the form
If
= (1, 0,0) Any vector
u
= (ui. u2, u3)
can be written
u1i + uzj+ u3k.
is a unit vector, then we can write u = cos(a)i+ cos(,B)j + cos(r)k.
The coefficients of i,j, and
k are called the direction cosines for u, and the angles
a,
,8, and "fare called the direction angles.
Cross Product (Section 11.4)
The
cross product of two vectors v = (v1,v2,v3)
and w=
(w1,w2,w3)
is defined to be
the vector
vXw= (v2w3-w2v3)i-(v1w3-w1v3)j + (v1w2 -w1v2)k. The geometric interpretation of cross product is that
vXw= llvll llwll sin(9)n, where (} is the angle between the two vectors, and n is the standard unit normal to v and w determined by the right-hand rule. We may use determinant notation to express the cross product as
V
X W = det
([
i
V1 W1
The cross product is anticommutative addition
j Vz Wz
(vXw= -wXv)
the triangle determined by the two vectors v and wis
(Section 11.5)
IIvXwII·
The plane with normal vector N =
Yo,zo)
. and distributes over
(u X (v+ w) =u Xv+ u Xwand (u+ v) Xw =u Xw+ vXw). The area of
paralellogram they determine is
Lines and Planes
l)
k V3 W3
(A,B, C)
IIvXwII /2, and the area of the
and passing through the point P0 =
(x0,
has equation
A(x - xo)+B(y - yo) + C(z - zo) = 0. A line in space that passes through m
= (a, b, c)
P0 = (x0,y0,zo)
and is parallel to a vector
is given parametrically by
x = xo+ta y =Yo+ tb z = zo+ tc. t and equating, we can express the line in z only. This is called the symmetric form for the line. If a, b, and c
By solving each of these equations for terms of x, y, and
are all nonzero, then the symmetric equations are
x-- xo Y-- Y zo -o - z-- b c a _
_
Review Exercises
Triple Scalar Product (Section 11.4)
If
u, v,
and
w
785
are vectors, then the volume of the parallelepiped which they
determine is given by the triple product any order. The vectors
u, v, w
l(u Xv)· w l.
The product may be taken in
are coplanar if and only if
(u Xv)· w
=
0.
Review Exercises for Chapter 9 In Exercises 1-4, calculate the distance between the given
(-1,0,-3) 2. (-3,-4,3) 3. (5,1,-2) 4. (-1,2,1)
27. (1,1,-1) (2,2,-2) 29. (1,1,4) 30. (4,-2,2)
(6,4,1) (4,2,-3) (6,-7,2) (1,6,-3)
1.
the arccosine to express
In each of Exercises 5-8, determine the center and radius
In each of Exercises 31-34, calculate the projection of the given vector
x2 + y2 + z2 - 4x+Sy+ 2z=15 6. x2 + y2 + z2 + 12x - 6y + 4z=0 7.4x2 + 4y2 +4z2 + 4x+4y- 4z=1 2 2 2 8. 2 - x - y - z =v'3x- y+2z In each of Exercises 9-12Joints P and Q are given. Write v represented by PQ in the form ll vlldir(v). Q= (5,8,1)
Q= (0, 1,2)
Q= (12,-1,1)
13.ll vll i +ll wll j + (v w) k 14. (v w) v - (v i) w 15. llw ll v - (v v) w 16. (v k) j + 6dir(w)
In each of Exercises 35-38, calculate the dot product of
35.v=(6,-2,3) 36.v=(7,4,4) 37.v=(10,10,5) 38.v=(3,V2,5) vXw.
·
·
18. What vector of length
(11,10,2)?
3 5
has the same direction as is opposite in direction to
19. For what value of a is (a, 3, 1) perpendicular to (4, 5, 13)? 20.If (a, b, -12) is parallel to (27, 5,9), then what is a? In each of Exercises
21-24, calculate
the dot product of
the given vectors.
2L (2,-3,4) (-3,3,-5) 23. (V2,v'3,-1) 24. 2i + 3j +4k
(2,4,5) (-8,1,6) (3V2,3, 3 + 3v'3) 3i- 4j +5k
22.
25. Find all values of
s
such that
(3,s,s)
the cross product
w=(3,-2,-3) w=(-1,3,2) w=(-2,2,2) w=(2,1,1)
In each of Exercises 43-46, find the area of the triangle determined by the two given vectors.
43. 44. 45. 46.
(4,2,4) (2,2,-1) (2,2,3) (3,1,3)
(1,-1,1) (4,6,2) (2,2,-2) (3,2,1)
In each of Exercises 47-50, calculate the area of the
and
(2,s, -5)
are
perpendicular.
(3,2, -1)?
39.v=(-4,1,0) 40.v=(3,1,-2) 4L v=(2,1,2) 42.v=(3,1,-5)
39-42, calculate
parallelogram determined by the two given vectors.
26. What are the lengths of the parallelograms determined by and
w=(8,-1,4) w=(7,6,6) w=(11,10,2) w=(3,3V'2,13)
In each of Exercises
·
·
of length
w =(2,2,1) w=(9,2,6) w=(7,4,4) w=(2,6,-3)
Schwarz Inequality holds for the pair.
Q= (4,-1,0)
In each of Exercises 13-16, letv= (4,2,4) and w=(2,2,-1 ).
17. What vector (10,10,5)?
w. Verify that P,,.(v) and
the given vectors and their lengths. Verify that the Cauchy
Calculate the specified vector.
·
onto the given vector
31.v =(4,2,6) 32.v=(20,4,9) 33.v=(75,30,21) 34.v=(10,3,-20)
the vector
P= (-1,2,-2) 10.P= (1,-1,4) 11. P= (2,1,2) 12. P= (1,-11,3)
v
v - P,.(v) are mutually perpendicular.
5.
9.
(-1,1,1) (1,-1,-1) (3,0,3) (1,2,1)
28.
of the sphere whose Cartesian equation is given.
(2, -1, -3)
27-30, use
In each of Exercises
the angle between the given pair of vectors.
pair of points.
47. (4,2,1) (5,1,1) 49. (3,1,-1) 50. (4,1,-2) 48.
(1,1,3) (1,1,-1) (1,2,2) (2,0,-3)
786
Chapter 9
Vectors
In each of Exercises 51-54, vectors Calculate the triple scalar product 51. 52. 53. 54.
u,v, and w (u Xv)· w.
v=(1,-1,-1) v=(1,1,-4) v=(1,-1,17) v=(1,1,3)
u=(2,1,1) u=(3,1,2) u=(1,1,-3) u=(2,3,-6)
In each of Exercises 55-58, vectors Calculate
the
volume
of
the
u,v,
are given.
75.
In Exercises 76-79, find parametric equations for the line
w=(1,1,-2)
that passes through the given point
w=(2,1,2) w=(2,1,5) w=(3,2,5)
the vector 76. 77.
and w are given.
parallelepiped
that
they
78. 79.
determine. 55. 56. 57. 58.
x
80. 81. 82.
is
describes a plane V is given. If
83.
(1,B, C)
is normal to V, then
what is B?
63.
86.
87.
x=1+3s+4t x=-1+s - t 66.x=1+2s 67. x=s+2t 65.
y=-1+2s+3t y=2s+t y=2+3t y=2s
89. 90. 91.
and passing through the given point
70. 71.
P.
n
What is A?
P=(l,1,-4) p=(2,-5,5) p=(3,-2,7) p=(-1,-1,0)
In each of Exercises 72-75, find a Cartesian equation for the plane determined by the three given points. 72. 73. 74.
(-3,7,2) (1/2,1,-1) (1,-1,-2/3) (1/3,2/3,-2/3)
(1,-1,1) (1,1, -1) (-1, -2, 7) (1/3, 4/3, 213)
(2,3,0) (-1,0,1) (3,1,4) (1,2,1)
(1,1,1) (2,3,1) (3,1,4) (4,2,1) (5,3,2) (6,1,0) (1,2,1) (2,1,3) (2,2,0)
2x+y+z=6 x+y+2z=4 x+2y - z=0 x+2y - z=2
x+y+2z=6 x - y+z=2 x+z=4 5x+y+z=4
In each of Exercises 92-97, find the point of intersection
In each of Exercises 68-71,Ax+By+ Cz=1 is a Carte
69.
Po=(3,0,-5) Po=(-2,2,-1) Po=(1,6,-1) Po=(-2,-3,-5)
line of intersection of the two given planes. 88.
z=s+t z=3+3s+2t z=s - t z=3t
sian equation for the plane with the given normal vector
=(3,2,-1) n=(10,3,2) n=(6,5, 1) n= (16, -2,-3)
and that is parallel to
In each of Exercises 88-91, find, in parametric form, the
In each of Exercises 64-67, (A, B,1) is normal to the plane
64.
P0
In each of Exercises 84-87, find symmetric equations for
85.
2x - 6y+z=3 z=2 - 5x +7y 3x - 4z=1 x=2y
n
m=(-5,2,4) m=(l,0,3) m=(1,-1,2) m=(2,-2,-3)
the line that passes through the two given points. 84.
described by the given parametric equations. What is A?
68.
and that is parallel to
the given vector:
In each of Exercises 60-63, a Cartesian equation that
62.
Po=(0,1,-1) Po=(1,3,5) Po=(1,-1,2) Po=(0,0,1)
that passes through the given point
w=(2,1,2) w=(1,1,-2) w=(3,2,5) w=(2,1,5)
=(1,1,-4) v=(1,-1,-1) v=(1,1,3) v=(1,-1,17) v
(1,x,x) X ((x,x,1) X (x,1,x))=0?
60.
P0
m.
In Exercises 80-83, find symmetric equations for the line
u=(3,1,2) u=(2,1,1) u=(2,3,-6) u=(1,1,-3)
59. For what values of
61.
(6,1,3) (8,2,3) (7,1,5)
of the given plane and the given line. 92. 93. 94. 95. 96. 97.
x=2t - 6 y=-3t+7 x - y+2z=0 z=-t - 1 x+y - z=2 x=2t - 4 y=4t+2 z=2t+2 x+y+2z=4 (x+3)/2=y+2=(z - 12)/(-3) x=-1+s+3t, y=-10+2s+t,z=-2+s; x=y+1=z/2 (x,y,z)=(1,0,1)+s(O,1,1)+t(l,0,2); (x,y,z)=(6,5,13)+u(l,2,3) x=s+t,y=-s,z=4 - t; (x - 3)/2=y/2=(z+11)/4
98. Find the distance between the point
(1,1,2) and the plane 2x - y+z=5. Find the distance between P= (2,0,0) and the line with symmetric equations x=y=2z. that has Cartesian equation
99.
The roots of vector calculus can be traced to 1799. In
study of languages, a field in which he finally earned
that year, both Carl Friedrich Gauss and a Norwegian
recognition. At the time of his death, and for some time
surveyor, Caspar Wessel (1745-1818), "identified" the complex numbers IC with the plane JR2• They did this by
thereafter, his Sanskrit-to-German translation of the Rig
associating each point
mann's mathematical work did not attract any serious
complex number
attention until nearly fifty years after his death.
(x,y) of the plane with the x + yv'-I. Gauss and Wessel used
Veda appeared to be his primary legacy. In fact, Grass
this identification to transfer the geometry of the plane
One of Grassmann's most original ideas was the
to the field of complex numbers. This geometric
creation an anticommutative algebra of vectors that is
interpretation of complex numbers gave rise to a new
now called "exterior algebra." In the 20th century, this
mathematical subject, the calculus of complex-valued
algebra has played a prominent role in geometry and
functions of a complex variable.
analysis. As Grassmann developed his theory of exterior
In fact, the identification of JR2 with IC is a two-way
algebra, he discovered not only the cross product but
street: We can also use it to transfer the algebraic
also the dot product, the vector norm, orthogonal pro
structure of the complex number system to the plane.
jection, and several sophisticated concepts such as the
For example, the multiplication of complex numbers,
(x + yH)(x' + y'H)
=
(xx' -yy') + (xy' + yx')H,
can be used to define a product of points in the plane:
(x,y) O (x',y')
=
(xx' -yy',xy' + yx').
"Hodge star operator" that were rediscovered and introduced into mainstream mathematics at a much later date. Frustrated by the apathy with which his research was greeted, Grassmann did not waver in his belief of its importance. As he expressed it, I remain completely confident that the labor I have
The search for an analogous algebraic structure that
expended on the science presented here and which
could be imposed on the points of three-dimensional
has demanded a significant part of my life as well as
space led to the creation of vector calculus in 1843. The
the most strenuous application of my powers, will
breakthrough was attained more or less simultaneously
not be lost.... I know and feel obliged to state
by two mathematicians, Hermann Grassmann and Sir
(though I run the risk of seeming arrogant) that even
William Rowan Hamilton, working independently.
if this work should again remain unused for another
Hermann GUnther Grassmann and Vector Calculus
into the actual development of science, still that time
seventeen years or even longer, without entering
Hermann Giinther Grassmann (1809-1877) was born and raised in Stettin, a city that is now known by its
will come when it will be brought forth from the dust of oblivion and when ideas now dormant will bring forth fruit.
Polish name, Szczecin. Grassmann acquired his educa tion in Berlin, where he studied philology and theology but received no mathematical training at the university level. After returning to Stettin, he became a high school teacher and father.
Sir William Rowan Hamilton and Vector Calculus In stark contrast to Grassmann, Sir William Rowan
The demands of rearing eleven children notwith
Hamilton was the best known scientist of his day. Even
standing, Grassmann maintained his interest in languages
before his talents in mathematics and physics became
and also carried out a program of original research in
evident, Hamilton attracted attention for his pre
mathematics and physics. He first published his work on
cocious ability to learn languages. By five years of age,
vector calculus in a book that appeared in 1844. Almost
he was already proficient in Latin, Greek, and Hebrew.
no notice was taken of it during his lifetime. After selling
As a teenager he was versed in the rudiments of thir
only a few copies, his publisher rendered the remaining
teen languages. All this knowledge was the result of
inventory of 600 copies into waste paper. With that fail
self-study under the tutelage of an uncle. Hamilton
ure, Grassmann's hopes for landing a university position
received his only formal schooling when he attended
were dashed. Spumed in mathematics, he returned to the
university.
787
788
Chapter 9
Vectors
Hamilton's interest in mathematics was kindled by
electric current seemed to close; and a spark flashed
the reading of a Latin copy of Euclid when he was ten
forth, the herald . . . of many long years to come of
years old. When he was sixteen, Hamilton detected a
definitely directed thought and work ... Nor could I
hitherto unsuspected error in Laplace's treatise on
resist the impulse to cut with a knife on a stone of
celestial mechanics. He published original mathematical
Brougham Bridge ... the fundamental formula with
research one year later and made his name in physics
the symbols
i, j, k;
over the next ten years by applying the calculus of var
namely
iations to discover fundamental principles of mechanics
i2 = j2 = k2 = ijk = -1,
and optics. The Nobel laureate Erwin Schrodinger
which contains the Solution of the problem, but of
summarized Hamilton's influence as follows
course, as an inscription, has long since mouldered
The central conception of all modem physics is the "'Hamiltonian." If you wish to apply modem theory to any particular problem, you must start with put ting the problem "in Hamiltonian form."
predicted
the
phenomenon
of
conical
refraction. In the words of a prominent scientist of the time, this phenomenon was "unheard of and without analogy." The experimental verification of conical refraction created a sensation and brought fame to Hamilton beyond scientific circles. Hamilton's introduction of vector calculus was a by-product of his creation of the quaternions
Q,
an
algebraic system that may be identified with JR4• A typical quaternion is of the form
xi +yj + zk
may be iden
(x, y, z)
xi+ yj + zk.
-->
At
first, Hamilton referred to these spatial quaternions as triplets. Addition and subtraction of quaternions was no problem. Finding a suitable multiplication proved dif ficult. The stumbling block was Hamilton's initial reluctance to abandon the commutative law:
(p,q E Q). The answer came to 1843. Shortly before his death
pq = qp 16,
him on October in
1865,
Hamilton
described the circumstances in a letter to his son Archibald: Every morning in the early part of [October
1843],
on my coming down from breakfast your (then) little brother William Edwin, and yourself, used to ask me "Well, Papa, can you multiply triplets?" Whereto I was always obliged to reply, with a sad shake of the But on the sixteenth day of the same month ... I walking ...
along
the
(p, q, r
E
Q);
he even coined
relation
ijk = -1
that Hamilton carved in stone. But
Hamilton recognized that he would have to give up the commutative property of multiplication. To see how to multiply two quaternions, imagine the face of a clock
i, j,
with
and
k
12, 4, and 8 o'clock, 1). A clockwise product of any two
positioned at
respectively (Figure
adjacent terms results in the third term; a counter clockwise
product
results
remaining term. Thus
in
the
ij = k = -ji.
negative
of
the
Two general qua
ternions are then multiplied by the ordinary rules of
Royal
Canal .. .
(2 +3j)(5i+6j+7k)= lOi+ 12j+14k+15ji+18j2+21jk = -18+(10+21)i+12j+(14-15)k = -18+31i+12j-k.
(( k
ij=k
' ;
�
ki =j
� Figure 1
By 1846, Hamilton was calling the triplet xi +yj + zk a vector. This terminology was derived from radius vector, a term that was already used in analytic
head: "No. I can only add and subtract them." was
(pq)r = p( qr)
the term. Therefore no bracketing is necessary in the
preserved. For example,
tified with points of three-dimensional space JR3 by the one-to-one correspondence
Hamilton retained the associative law of multi
arithmetic except that the order of factors must be
t +xi +yj + zk. Quaternions of the form
1954, a plaque on Brougham Bridge has
commemorated the vanished engraving.]
plication:
On the basis of his theoretical investigations, Hamilton
away. [Since
An
geometry. At the same time, he wrote that "the algeb raically real part [namely, the quantity tin the quaternion
Genesis & Development
t +xi+ yj + zk] may receive . .. all values contained on
789
The famous astronomer, Sir John Herschel, complained
the one scale of progression of number from negative to
that it would "take any man . .. half a lifetime to digest."
positive infinity; we shall call it therefore the scalar part."
In the United States, the mathematician Thomas Hill,
quaternion
who later become president of Harvard, reviewed the
S.q, the scalar
book in extremely favorable terms. He concluded that
component of q, equal to t and V.q, the vector compo
"in quaternions .. . there is as much real promise of
In q
his
=
notation,
he
wrote
the
t +xi+ yj + zk as -S.q + V.q with
-
benefit to mankind as in any event of [Queen] Victoria's
nent of q, equal to xi+ yj + zk. Hamilton also introduced the scalar product (i.e.,
reign." Hill lamented that "Perhaps not fifty men on this
the dot product) and the vector product, (i.e., the cross
side of the Atlantic have seen [Hamilton's book], cer
product). These products arise simultaneously when
tainly not five have read it." In his review, he admitted
the scalar and vector parts of a quatemionic product of
that he himself was not actually among that handful.
two vectors are separated. In other words, if u and v are vectors (quaternions u and v with S.u uv
=
-S.uv + V.uv with S.uv
=
=
S.v
=
u v and V.uv ·
0),
=
then
u Xv.
After Catherine's death in
1853,
Hamilton's life
was marked by grief, reclusiveness, concentrated work at irregular intervals, and a struggle with alcoholism.
Despite his professional success, Hamilton's per
As one biographer put it, "When thirsty, he would visit
sonal life was a shambles. At the age of twenty,
the locker, and the one blemish in the man's personal
Hamilton was rejected in love by Catherine Disney, the
character is that these latter visits were sometimes paid
sister of a friend. He never really recovered from this
too often." It was during this last phase of his work,
unrequited first love. A second courtship six years later
when he was organizing his final thoughts on vectors
led to a second rejection. Hamilton finally proposed to
and quaternions, that Hamilton discovered the theo
a woman who was too timid to reject him but who was
rem on square arrays (or matrices) that is fundamental
also too timid for very much else. Throughout his
to matrix algebra.
marriage, he remained in love with Catherine Disney
Although the theory of quaternions never assumed
but was tom by guilt over it. Depression and drink
the central importance that Hamilton expected, his
1848, Catherine initiated
influence on mathematics has been as enduring as his
a correspondence with Hamilton, but her conscience
became frequent problems. In
influence on physics. The study of noncommutative
was not up to it: She soon confessed to her husband and
algebraic systems that he initiated was taken up by first
attempted suicide. Hamilton responded to the stress by
rate mathematicians such as Arthur Cayley
drinking more.
and William Kingdon Clifford
By
(1821-1895) (1845-1879). Their work
Hamilton had collected his work on
gave rise to the subject of noncommutative algebra. To
quaternions and vectors into a book which he titled
this day noncommutative algebra continues to be an
Lectures on Quaternions. It ran to more than
important branch of mathematical research.
1853,
800
pages.
This page intentionally left blank
Vector-Valued Functions P
R
E
V
E
W
Until now, all of the functions
x
1-+
f(x)
valued: We evaluate fat a real number
that we have studied have been scalar
x,
and we obtain a real number f(x) as the
result. In this chapter, we will study the calculus of vector-valued functions
t r(t). Here we evaluate a function rat a real number t, and we obtain a vector r(t) as the result. By writing out the vector r(t) in terms of its components (x(t),y(t), z(t)), we see that we can identify the value of r(t) with the point (x(t),y(t), z(t)) in space. Ast varies in the domain of r, the values r(t) trace out a curve in three-dimensional 1-+
space. By developing the calculus of vector-valued functions, we will be able to analyze such space curves. For example, if we differentiate each component of
r(t),
then we obtain a vector that is tangent to the space curve. That much is analogous to the scalar-valued theory that we already know. But the geometry of curves in three-dimensional space is very rich, and there will be much that is new in this chapter. Early in the 16th century, Johannes Kepler deduced the three laws of planetary motion that now bear his name. They were among the first precise quantitative laws known to science. What was lacking, however, was any theory that provided a reason for those laws. Half a century later, Isaac Newton demonstrated the power of calculus, then in its infancy, by deducing Kepler's Laws from basic physical principles. Chapter 10 concludes by applying the calculus of vector-valued functions to derive Kepler's Laws.
791
792
Chapter 10
Vector-Valued Functions
10.1 Vector-Valued Functions-Limits, Derivatives, and Continuity
y
A function whose range consists of vectors is called a vector-valued function. For example, the formula r(t) =cos(t)i+sin(t)j +tk defines a vector-valued function r that depends on a single variable t. Such functions will be the focus of our study in this chapter. One reason for our interest in this matter is that a vector-valued function r can be used to describe a curve C in space. To do so, we draw r(t) as a directed line segment with initial point at the origin. Figure la shows such directed line segments for several values, t0, t1, ...,tN, in the domain of r. The curve C is the collection of terminal points of the vectors r(t) as t runs through all values of the domain of r, as Figure lb illustrates. We say that the curve C is parameterized by r. We refer to the vector-valued function r as a parametric curve. � EX A M P L E 1 Describe the curve that is parameterized by the vector valued function r(t)= (5 - t)i+ (1+2t)j - 3tk.
_. Figure 1a
Solution From Section 11.5, we know that the equations x =5 t, y =1+2t, z = -3t are parametric equations of the line L that (a) is parallel to the vector {-1,2, -3) and (b) passes through the point (5, 1, 0). This straight line is therefore the curve described by r.
y
� EX A M P L E 2 The position of a particle moving through space is given by r(t)=cos(t)i+sin(t)j +tk. What is the position of the body at t=0, t=7r/2, t='Tr, t= 37r/2 and t= 27r? Describe the curve C along which the particle moves.
.. Figure 1b
6
z
Solution
··(3;) •
r(27r)
4
Suppose that a curve C is parameterized by t 1-+ r(t). If we think of the variable t as "time," then we can imagine a particle tracing the curve so that its position at time tis the terminal point of r(t). With this interpretation of the curve, we say that r(t) is the position vector of the particle at time t. If we write the position vector as r(t) =x(t)i +y(t)j + z(t)k, or, equivalently, r(t) = {x(t),y(t),z(t)), then (x(t), y(t), z(t)) is the position of the particle at time t, and x=x(t), y=y(t), z= z(t) are parametric equations for the motion. In this context, we often call C a trajectory.
At time 0, the body has position vector r(O)= cos(O)i+sin(O)j +Ok=i= (1,0, 0).
r(7r) •
This means that, at time t=0, the particle is at the point (1,0, 0 ) in space. At time t=7r/2, the particle has position vector r
G) =cos G)i+sin G)j + � k =(0,1,�)
Thus at t=7r/2, the particle is at the point (0, 1,7r/2) in space. Similarly, we find that the particle is at the points ( l,0,7r), (0, -l,37r/2), and (l, 0, 27r) at times t= 'Tr, t =37r/2, and t =27r, respectively. These five points are plotted in Figure 2; they do not, however, give us a very good idea of the particle's path! -
_. Figure 2
·
10.1 Vector-Valued Functions-Limits, Derivatives, and Continuity
793
To better understand the trajectory, we observe that the values x cos(t), y sin(t), z t satisfy the equation x2 + y2 1. Now the set {(x, y, 0) : x2 +fl = 1} is a circle in the xy-plane ofxyz-space, as is shown in Figure 3. The coordinates of every point above or below that circle also satisfy the equation x2 + r 1 because the equation imposes no requirement on the z-coordinate. Therefore inxyz-space, the graph of x2 +fl 1 is a cylinder, as can also be seen in Figure 3. Because curve Clies on this cylinder, we can picture how Cpasses through the five plotted points. See Figure 4 for the arc of Cthat joins the five plotted points and Figure 5 for the arc of C that is obtained by plotting r(t) for values oft between -311' and 311'. We call Ca helix. ..._ =
=
=
=
=
=
we study a particle moving through space, then it stands to reason that we will want to calculate its velocity and its acceleration. Therefore we will need to calculate derivatives. Before we learn to differentiate vector-valued functions, we must first define their limits. If
c
{{x,y,O) :x2 +I'= l} .&. Figure4
.&. Figure 3
Limits of Vector-Valued Functo i ns
.&. RgureS
If Lis
a vector and r is a vector-valued function, then we say that the limit of r(t) is Last tends to c, or r(t) tends to Last tends to c, if we can make r(t) be arbitrarily close to Lby takingt sufficiently close to c. The following definition states this idea precisely. We say that r(t) r1 (t)i + r2(t)j + 1'3(t)k converges to the vector L {Li, Li, L3) as t tends to c if, for any e > 0, there is a 5 > 0 such that =
=
0
llr(t) - Lii
converges to Last tends to c, then we write limr(t)
t-+c
=
L,
and we say that Lis the limit of r(t) as t tends to c.
784
Chapter 10
Vector-Valued Functions
Is there anything new in this definition? It looks just like the rigorous definition of limit in Section 2.2 in Chapter 2. The difference is that, because our function is vector-valued, we measure distance in the range by using the length II II instead of the absolute value I I· Figure 6 illustrates the geometry behind the definition. Z
Open ball of radiUSE
I
c-8 c
c+8
.A. Figure 6 The function r maps all points within 6 of c to an open ball of radius E centered at the terminal point of L.
Because the vector r(t) - L has a small magnitude if and only if its com ponents r1 (t) -Li, r2(t) -Lz, and r3(t) -L3 are all small in absolute value, we can establish the limit of a vector-valued function by showing that each of its scalar valued components has a limit. We state this useful strategy as the following theorem.
Let r(t) = r1 (t)i + r2(t)j + r3 (t)k be a vector-valued function. Then
THEOREM1
limr(t) =Lii + Lzj + 4k t--+e
if and only if
lim l'J.(t) =Li,
liml'2(t) = L,., t-oc:
t-oc:
and
lim'3(t) =4. t-oc:
We mention in passing that, if the limit of a vector-valued function exists, then it is unique. This assertion follows from reasoning that is similar to that used in Chapter 2 to see that the limit of a scalar-valued function is unique when it exists. Now we can legitimately calculate limits in the most convenient way, and the resulting answer will have the right physical significance. � EXAMPLE 3 Let
r(t)
=
(t2
-
4t)i +
t-
(12 39)j
+
sin('rrt) k. t- 3
What is linlt-.3r(t)? Solution We calculate
lim (t2-4t) = t->3
-3
and
29 = lim (t + 3) = 6. lim t t-.3 t 3 t->3 -
10.1
Vector-Valued Functions-Limits, Derivatives, and Continuity
notice that limt-+3sin(7rt )
To calculate the limit of the third component, sin(37r)
=
0
and limt-+3(t- 3)
=
0.
795
0/0,
To resolve the indeterminate form
=
we
apply l'Hopital's Rule as follows:
lim 3 t-+
sin(7rt)
t-3
=
lim t-+3
Thus according to Theorem
:t
sin(7rt) =
!!_ (t- 3) dt 1,
7rCos(7rt )
lim t-+3
1
we have limt-+3r(t)
=
=
7rCos(37r)
=
-7r.
-3i + 6j-7rk. <11111
The next theorem collects a number of results about calculating limits of vector-valued functions.
THEOREM 2
Let
function, let
c
f
and g be vector-valued functions, let ¢ be a scalar-valued
be a real number, and let .A be a constant. Assume that limt-+cf(t),
lim,--+eg(t), and lim,--+e¢(t) exist. Then, a.
lim,--+e(f
b.
lim,--+e(f-
c.
lim,--+e(f
d.
limt-+c .A f(t)
e.
lim,--+e(fXg )(t)
f.
limt-+c(cp(t)f(t))
(
INSIGHT
+ g)(t) ·
=
limt-+cf (t ) + lim,--+eg(t)
g)(t) limt-+cf (t)- lim,--+eg(t) g) (t) (lim,--+ef(t)) (limt-+cg(t)) =
=
)
=
·
.Alimt-+cf(t)
=
=
(limt-+cf(t)) X (lim,--+eg(t)) (lim,--+e¢(t)) (limt-+cf(t)) .
These formulas are similar to those in Theorem 2 of Section 2.4 in
Chapter 2. However, you should notice two things. First, the a dot product: It
must because f(t)
is a vector and
part e represents the cross product of f and
g(t)
"f g" in part c now signifies "f X g" in ·
is a vector. Second, the
g. Third, there is no result about the limit of a
quotient, simply because we may not take the quotient of two vectors.
�
EXAM PL E 4 Calculate lim,--+4 ((t2i-3tj)·
(4i-ytj + 5tk)).
Solution We calculate lim t--+4
((t2i-3tj +Ok)· (4i-Jlj + 5tk))
=
lim t-+4
(4t2 + 3t./i + 0)
=
64 + 24
=
88.
According to Theorem 2c, we obtain the same answer if we first compute the limits of the terms that appear in the dot product: lim t--+4
Continuity
(t2i-3tj) · lim (4i-./ij + 5tk) t--+4
=
(16i-12j +Ok)· (4i-2j + 20k)
=
64 + 24
=
88.
....
The definition of continuity for a vector-valued function is just the same as the definition given in Chapter 2 for a scalar-valued function. The essence of
796
Chapter 10
Vector-Valued Functions
c
the definition is that the actual value of the function at function approaches as
t � c.
agrees with the value the
To be precise, see the following definition.
We say that a vector-valued function r is
continuous at a point c of
its domain if
limr(t)
=
t-+C
If r is not continuous continuous at c.
at a point
c
r(c).
in its domain, then we say that
r
is
dis
Because of Theorem 1, we find that we can test for continuity componentwise, as the following theorem observes.
THEOREM 3
r.
Then,
r
is continuous
continuous at
c belongs to the domain of a vector-valued function at c if and only if each component function of r is
Suppose that
c.
� EX A M P L E 5 Discuss the continuity properties of the vector-valued func
r that is defined ln(lt - 2l)j + t3k.
tion
on the set V
Solution The functions
t f--t cos (t)
=
{t E
and
JR
:
t f--t t3
t f:- 2}
by the formula
and continuous at all real
=
are defined for every real
these functions are continuous everywhere. The function conclude that the function
r(t)
t f--t In ( It - 21)
cos ( t) i +
t.
Both of
is defined
t f:- 2. From these observations, we use Theorem 2 r is continuous at all points in its domain V.
to
INSIGHT
We do not say that r(t) is discontinuous at t 2. That is because r is undefined at 2; we discuss continuity and discontinuity only at points of a function's domain. =
The next theorem collects a number of results that are often convenient for establishing continuity.
THEOREM 4
Let >. be a constant. Suppose that ¢ is a scalar-valued function f and g are vector-valued functions. Suppose that these three functions continuous at a common value c of their domains. Then,
and that are a. b. c. d. e.
f + g and f - g are continuous at c f g is continuous at c >.f is continuous at c f X g is continuous at c ¢f is continuous at c. ·
10.1 Vector-Valued Functions-Limits, Derivatives, and Continuity INSIGHT
797
These facts about continuous functions are similar to those in Section 2.3
of Chapter 2 about scalar-valued continuous functions. Two things should be noted. First, there is no statement about quotients of continuous vector-valued functions because we do not have a way to form the quotient of two vectors. Second, there are three statements about products: one for the dot product, one for cross product, and one for scalar multiplication. Just as in Section 2.3, the proofs of these rules are immediate applications of the rules for limits.
�
3j
EXAM PL E 6 Discuss the continuity properties of the functions
f(t) = ltli -
t3k and g(t) = t l i i- sec ( t)k at the value 0. Next, discuss the continuity prop erties of f g and f x g at 0. +
·
and g are continuous at t = 0 because all of their component functions are. Therefore (f g)(t) and (f X g)(t) are continuous at 0 by Theorem 4b and 4d. <11111 Solution Both
f
·
INSIGHT
(f
·
g
Of course we can always calculate
) (t) _11_ 1 t+ =
The first of these,
-
f
3 t sec t
()
·
(f x g)(t)
and
=
()
3 sec t i +
(l l
()
t sec t +
L1)j t+
+
_l__l k. t+
g, is a scalar-valued function that can be shown to be continuous at
0 using the methods of Chapter 2. Theorem 3 can be used to show the continuity of the second function f X g at t 0 because each of its components is continuous at t 0.
t
=
=
Derivatives of Vector-
=
Now that we understand limits, it is a simple matter to define derivatives.
Valued Functions Suppose that r is a vector-valued function that is defined on an open interval that contains c. If the limit lim
�t--+O
l
ut
(r (c + �t) - r(c))
exists, then we call this limit the derivative of the function we denote this quantity by the symbols ' r c
( )
and
-dr dt
l
r
at the point
c,
and
. t=c
The notation r(c) is also used, especially when t represents time. The process of calculating r' is called differentiation of r. If the derivative r' (c) exists, then r is said to be differentiable at c.
798
Chapter 10
Vector-Valued Functions INSIGHT
Notice how similar the definition of
r
'
(c)
is to the definition of the deri
vative that is given in Section 3.2 of Chapter 3. The difference, of course, is that we are now considering vector-valued functions, so our limit process is a bit different. The quantity
(
r c
�t)
+
-
()
r c
is a vector, and the product
lt ( (
r c
+ �t)
-
( ))
r c
signifies
the operation of scalar multiplication. The result of this operation is a vector. Thus the derivative
r
'
(c)
is a vector.
By Theorem 1, the process of calculating a limit may be performed by calcu lating the limit in each component separately. It follows that a vector-valued function may be differentiated by differentiating each component separately,
provided that all of the c omponents are differentiable. THEOREM 5 entiable at
t
r(t)
ri(t)i + r (t)j + r3 (t)k is differ 2 c if and only if each component function of r is differentiable at c. A vector-valued function
=
=
In this case,
r'(c)
�
r(t)
EXAM PL E 7 Let
ferentiable? What is
r'(t)
Solution We calculate
We see that
r(t)
is
not
=
=
e2ti + ltlj
r'(t)
t
cos (t) k. For what values of
is
dif
r
by differentiating each component:
differentiable at
entiable fort> 0, and r'(t)
-
r3(c)k.
+
at these values?
not. However, for t> 0, we have
-1. It follows that
ri(c)i + r2(c)j
ltl
=
t
t
=
0 because the second component,
and
;t ltl
=
1.
It follows that
2e2ti + j + sin (t) k. Fort< 0, we have ltl
=
r(t) is differentiable
fort< 0 and
r'(t)
=
=
r(t)
ltl,
is
is differ-
;t ltl
-t and
2e2ti - j + sin (t) k.
=
.,..
The derivative of a vector-valued function is another vector-valued function. Therefore we can differentiate the derivative function to create a second deriva tive, and so on. Just as for real-valued functions, the second derivative of
r(t)
is
denoted by
r"(t) The notation �
r(t)
d2r dt2.
is also used, especially when
EXAM PL E
What is
or
8 Let
r(t)
=
t
represents time.
(3/t - t)i + cos ('rrt)j + e -tk.
Calculate
r'
and
r".
r"(l)?
Solution We
have
�cos(7rt)j + e tk . -
In
(-3/t2 - l)i- 7l"Sin(7rt)j - e-tk particular, r"(l) 6i + �j + (1/e)k . .,..
r'(t)
=
and
r"(t)
=
(6/t3)i
=
The next theorem gathers together several useful differentiation rules.
10.1 Vector-Valued Functions-Limits, Derivatives, and Continuity THEOREM 6
and
Let
taining the point a.
A
be a constant. Suppose that the vector-valued functions
and the scalar-valued function
g
(f + g)'(c)
c.
(f - g)'(c)
f
are defined on an open interval con
¢
Suppose further that
f'(c), g'(c),
and
¢'(c)
exist. Then
exists and
(f + g)'(c) b.
799
=
f'(c) + g'(c);
exists and
(f - g)'(c) = f'(c) - g'(c); c.
(f g)'(c) ·
exists and
(f g)'(c) ·
d.
=
f'(c) g(c) + f(c) g'(c); ·
exists and
(Af)'(c)
(Af)'(c) e.
(f X g)'(c)
(¢f)'(c)
=
'ljJ
A(f'(c))
f'(c) X g(c) + f(c) X g'(c).
exists and
(¢f)'(c) g. If
=
exists and
(f X g)'(c) f.
·
=
¢'(c)f(c) + ¢(c)f'(c).
is a scalar-valued differentiable function on an open interval I that
contains the point I, then
f o'ljJ
a,
if
'ljJ(a)
c, and if the at a, and
=
is differentiable
(f o'l/J)'(a)
INSIGHT
=
composition
f o'ljJ makes sense on
'l/J'(a)f'('l/J(a)).
These differentiation rules are similar to those of Chapter 3 for functions
with scalar values. However, two points should be noted. First, there is no quotient rule because we do not have a way to form the quotient of two vectors. Second, we have three product rules. Rule c is a product rule for the dot product. Rule e is a product rule for the cross product. Rule f is a product rule for scalar multiplication. Care should be taken to distinguish among these three rules.
Take particular note that it is important on the right-hand side of formula e that the order of
f
and
g
match the order of
f
and
g
on the left-hand side.
(Remember that the cross product is an anticommutative operation: Interchanging the operands of a cross product changes the sign of the result.) Also notice the form of the Chain Rule stated in part g. On the right side of the formula, and
f'('lj;(a))
'l/J'(a) is a scalar
is a vector. Thus the right side of the formula is the scalar multi
plication of a vector.
800
Chapter 10
Vector-Valued Functions
Let f(t)= cos(t)j - ln(t)k and g(t)= t2i- t-2j +tk. Calculate
� EXAMPLE 9
(f . g)'(t).
We apply formula Theorem 6c to obtain
Solution
(f . g)'(t)= f '(t) . g(t) +f(t) . g'(t) =
= =
-sin(t)j -
( (
�k)
·
(t2i- t-2j +tk) +(cos(t)j - ln(t)k) (2ti+2t-3j +k) ·
(o)(t2) +(-sin(t))(-t- 2) +
(-�) ) (
r2sin(t) -1+2r3cos(t) - ln(t). INSIGHT
can
first
·
g)'(t)
=
=
....
We are not obliged to use Theorem 6 in Example 9. As an alternative, we
calculate
(f
·
g)(t)
and then differentiate:
(f
)
(t) + (o)(2t) +(cos(t))(zt- 3) +(-ln(t))(l)
:
t
=
(O)(t2)
(-t-2cos(t) - tln(t))
=
(2t-3cos(t) + t-2sin(t)) +
: (
t
+ cos(t)(-t - 2 ) +(-ln(t)) (t)
(-t-2cos(t)) +
-ln(t) - t
( �))
=
-t-2cos(t) - tln(t)
:
t (-tln(t))
=
2t-3 cos(t) + t-2 sin(t) - ln(t) -1.
� EXAMPLE 1 0 Define f(t)= cos(t)i- sin(t)j, g(t)= t3j +t-1k, c/J(t)= t2, and.\= 5. Calculate (¢f +.\g)'(t). Solution
We use several of our differentiation rules: Theorem 6a
(¢f +ag)'(t) = (¢f)'(t) +(ag)'(t)
Theorem 6d, 6f
= c/J'(t)f(t) + c/J(t)f'(t) +a(g'(t))
= 2t(cos(t)i- sin(t)j) +t2( -sin(t)i- cos(t)j) +5(3t2j - t-2k) = (2t cos(t) - t2sin(t))i+(-2t sin(t) - t2cos(t) +15t2)j - st -2k. As an alternative, we can first calculate (¢f +.\g)(t) t2cos(t)i+(-t2sin(t) + 5t3)j +5t-1k. We can then directly differentiate this expression. As an exercise, verify that this second method results in the same formula for (¢f +.\g)'(t). ..,.. =
� EXAMPLE
(f
x
1 1
Let f(t)= tan(t)i- cos(t)k
and g(t)= sin(t)j. Calculate
g)'(t).
Solution
By Theorem 6e,
(f x g)'(t) = f'(t) x g(t) +f(t) x g'(t) = (sec2(t)i+sin(t)k) X sin(t)j +(tan(t)i- cos(t)k) X cos(t)j = (-sin2(t)i+sec2(t)sin(t)k) +(cos2(t)i+sin(t)k) =
(cos2(t) - sin2(t))i+(sin(t) + sec2(t)sin(t))k.
Verify this answer by first calculating the cross product f X g and then differentiating. ..,..
10.1 Vector-Valued Functions-Limits, Derivatives, and Continuity
801
We next turn to a statement relating differentiability and continuity of a vector valued function. The proof illustrates how we often analyze a vector-valued function by reducing the investigation to its scalar-valued components. THEOREM 7
If f is a vector-valued function that is differentiable at c, then f is
continuous at
c.
If f(t) fi (t)i + fz(t)j + f3(t)k is differentiable at c, then the scalar-valued expressions fi (t), fz(t), and f3(t) are all differentiable at c. Each of the functions Ji, fz, h is therefore continuous at c by Theorem 1 of Section 3.2 in Chapter 3. • Theorem 3 of this section now tells us that f is continuous at c. Proof.
Antidifferentiation
=
If f(t) f1 (t)i + f2(t)j + f3(t)k is continuous, then we may consider the anti derivative =
F(t)
=
=
J f(t)dt
(! f1(t)dt} (! fz(t)dt) j (! h(t)dt) k +
+
+ C.
Here it should be clearly understood that each of the antidifferentiations in the i, j, and k components will have a constant of integration. Therefore the constant of integration C is a vector of the form C C1i + Czj + C3k. =
� EXAM PL E 1 2 Find the antiderivative F(t) of f(t) satisfies the equation F(l) 2i - 3j + 2k.
=
3t2i - 4tj + 8k that
=
Solution
We have F(t)
=
=
J f(t)dt t3i - 2t2j + 8tk + C.
Now the additional condition that has been given tells us that 2i - 3j + 2k F(l) i - 2j + 8k + C. Solving for C gives C i - j - 6k. Therefore F(t) (t3 + l)i (2t2 + l)j + (8t - 6)k. .... =
=
Q UIC K
Q UIZ
=
=
1. What planar curve is described by r(t) cos(t)i + sin(t)j, 0 ::5 t < 27r? 2. If f(7r) (2,0,3), f'(7r) (2,3,0), g(7r) (0,1,4), and g'(7r) (-1,0,1), then what is (f g)'(7r)? 3. Referring to f and g of the preceeding question, what is (f X g)'(7r)? 4. What vector-valued function F satisfies F'(t) (3,sin(t),2 exp(t)) and F(O) (5,5,5)? =
=
=
=
=
·
=
=
Answers
1. The unit circle, traversed counterclockwise 4. F(t) (3t + 5, 6 - cos(t),3 + 2 exp(t)) =
2. 4
3. (12,-13 ,2)
802
Chapter 10
Vector-Valued Functions
EXERCISES Problems for Practice
In each of Exercises 1-6, describe and sketch the curve that is defined by the given vector-valued function. r(t) =ti + t2j r(t) =t2i+t2j r(t) =ti+t2j +2k 4. r(t)=t2i + t2j + 3k 5. r(t) =i + cos(t)j + sin(t)k 6. r(t) =ti+cos(t)j + sin(t)k 2. 3.
In each of Exercises 7-10, the position vector r(t) of a particle is given. Describe the particle's trajectory.
r(t) t2i + t2j + k 8. r(t)= t2i + t4j 9. r(t)=ti+ tj + sin(t)k 10. r(t) = cos(t2)i + sin(t2)j + tk =
-1sts2 -1sts2 Osts27r .,/iirsts2..fii
In each of Exercises 11-14, a vector-valued function r and a point c are given. Calculate the limit of r(t) as t c. --->
r(t) =cos(7rt)i + 2'j 12. r(t)=ln(l + t2)i + j ) 13. r(t)= (t2,f,e2In(t-1 ) t- 1 t+1 14. r(t)= i+ l. t2- 1 t2+1 11.
In 15. 16.
17. 18.
19. 20.
2L 22. 23. 24. 25. 26.
r(t) =
37. 38. 39.
40. 4L 42.
46.
each of Exercises 21-26, calculate r"(t).
r(t) = t2i- t312j + (t + 1)5/2k r(t) =(t2 +1r1i- sec(t)k r(t) =e-'i - ln(t2+t4)j+k r(t) =sec(t)j -cot(t)k r(t) =r113i - (1 + t2)-112j +t114k r(t)=arctan(t)i-tsin(t)j + arcsin(t)k
f(t2i - t-112j + cos(2t)k)dt f(e31i- e-21j +te1k)dt 29. f(ln(t)i-cos2(t)j + tan(t)k)dt 30. f(Vii- t-113j + sec2(t)k)dt 27.
r(t) =f( ..\t)
c=l
each of Exercises 27-30, calculate the indefinite integral 28.
35. 36.
45.
each of Exercises 15-20, calculate r'(t).
In
In each of Exercises 35-48, suppose that f(t)=e'icos(t)j + ln(l + t2)k, g(t) =t3i - tk,
c= -1 c=O c=2
r(t)=t112i- 3j + t2k r(t)=t312i- t2j-Sy'tk r(t)=tan(t)i-cot(t)j + sec(t)k r(t) =(1/t)i- tj+ln{t)k r(t) =ti+ (t+S)-1j + (t-7)-2k r(t) = ( v'l- t2, t;v'l - t2, arcsin(t)) In
f(t) =2ti - 3t2j + 4t3k F(2) =Si-2j + k F(2) =Si-2j f(t) =y'ti- t2j + k 33. f(t) =cos(t)i- sin(2t)j + sec(t)tan(t)k F(O) =3i+(3/2)j -2k F(e) = -2i + j-4k 34. f(t) = ln(t)j 3L
32.
1.
7.
In each of Exercises 31-34, find the antiderivative F(t) for f(t) that satisfies the additional condition that is given.
43. 44.
47.
48.
·
Further Theory and Practice In each of Exercises 49-52, suppose that f and g are the functions of Exercises 35-48. Calculate the derivative of the given function.
r(t) =(i f(t))(i x g(t)) 50. r(t) = (
SL 52.
(
·
)
·
1/;(t) =f(t) (f(t) x g(t)) r(t) = (¢(t)j X g(t)) X g(t)-¢(t)j X (g(t) X g(t)) ·
In
each of Exercises 53-56, calculate the given limit.
(
53.
limHo ltl'i + cos111(t)j+ (1- t)1k
54•
limi-..o .
55. lim,--.,,56.
)
tan(t) ) t (sin(t t) i+ _ + k cos(t)J t (-sin(t) ln(t2) , -- , lsec(t)I t-11" cos(t) J _.
(
)
lin1 -..o (1 + t)1/' i+ (1 + 2t)1/'j + (1- t)1/1k 1
Determine where each of the functions in Exercises 57-60 is continuous.
10.2
57. r(t) = (t)i - 7t2j, where A.(t) 'I' = 58. r(t)
{
1
if
t<5
-1
if
t2=5
68.
= tan(t)i - t3j + (t)k where
{
ifif:
69.
tf0 t=O
70.
= (t)i + ln(l + t2)j + ltlk where (t)=
{
sin(t)/t
if
cos(t)
if
t>0
7L
ts0.
{
t3 - 3t- 1
3t - 5
if if
for all s and t. Prove that f is continuous. t< 2
t2=2
6L Suppose that r(t)=p(t)i + q(t)j + s(t)k where p, q, and
Calculator/Computer Exercises s
are polynomials. Prove that there exists a positive integer N'1 derivative of r is identically the zero vector. What is the least N that will suffice? 62. Verify that the function r(t) = cos(t)i - sin(t)k satisfies the differential equation N such that the
d2 dt
r(t) + r(t)=0.
63. Verify that the curve described
lim1->
< llr(t) II = llLll· Prove that the converse is false. Suppose that t>--+ r(t) is continuous at t=c. Prove that t >--+ II r(t) 11 is continuous at t=c. Prove that the converse is false. Suppose r(c) f 0. Show that the differentiability of t >--+ r(t) at t=c implies the differentiability of t>--+ II r(t) II at t=c. Prove that the converse is false. Let f be a vector-valued function for which
llf(s) - f(t) II :5 Is-ti
60. r(t)=i + (t)j + (1/(1 + lt))lk where
(t)=
803
67. Suppose that r(t) is a vector-valued function. Prove that
(t)= sin(1/t) 1 59. r(t)
Velocity and Acceleration
by r(t)=cos2(t)i + cos(t)sin(t)j + sin(t)k lies on the sphere x2 + y2 + z2=1. 64. Let S be the surface that consists of all points (x, y, z) that satisfy the equation x2 + y2 = z2• What are the intersec tions of S with horizontal planes z=h? What is the intersection of S with the yz-plane x=O? Describe S. Describe the space curve C defined by r(t)=t cos(t)i + t sin(t)j + tk and verify that C lies on S. 65. Give an example of a vector-valued function that is continuous at a point c but is not differentiable at c 66. Formulate a definition of right-hand limit and a definition of left-hand limit for vector-valued functions. Prove that these limits may be calculated componentwise.
In each of Exercises 72-79, a vector-valued function r(t) = r1 (t)i + r2(t)j and a value to in the domain of r are given. Plot the planar curve associated with r. To your plot, add the straight line through r(to) with slope r2(to)/ri(to). Use the Chain Rule to explain the relationship of this quantity to dy/dx at r(t0). 72. r(t) = (5
+ 2 cos(t))i + (3 + sin(t))j, 0 :5 t :5 2ir; to= ir/3
73. r(t) = (t2 + t)i + (t2- t)j, -2 :5 t :5 2; to= -1 74. r(t) = tan(t)i + sec(t)j, - ir/4 < t < ir/4; to= ir/6 75. r(t)= 3t/(1 + t3)i + 3t2 /(1 + t3)j, -30
SL r(t) =(cos(t)/t)i + (sin(t)/t)j + (1/t)k
0sts2ir
4irsts12ir
1 0.2 Velocity and Acceleration Imagine a particle traveling through space with position vector r(t) r1(t)i + r2(t)j + r3(t)k. In Figure 1, we see that the vector r(t + �t) - r(t) is the displacement of the particle's position from r(t) to r(t + �t): It represents how far and in what direction the particle moves from time t to time t + �t. Therefore =
804
Chapter 10
Vector-Valued Functions
z
1
b.t
(r(t + b.t) - r(t))
[t, t + b.t] or [t, t + b.t]. Following the line of reasoning used in Chapter 3, we define the instantaneous velocity of the particle at time t to be represents the average change in position over the time interval
average velocity
of the particle over the time interval
v(t) r'(t) =
.& Figure 1
=
lim
�t-->O
! (r(t
LJ.t
+
b.t) - r(t))
provided that this limit exists.
INSIGHT
Our definition of instantaneous velocity is similar to the one we
adopted for scalar-valued functions in Section 3.1 of Chapter 3. Notice that the expression that defines average velocity and that appears in the formula for instanta neous velocity does
not contain the quotient of two vectors; it is actually the scalar r(t + �t) - r(t) by the number 1/ �t. Thus average velocity
multiplication of the vector
is a vector and instantaneous velocity is a vector. This is an important point to keep in mind. How could we have been so clumsy in Chapter 3 to have thought that velocity was a number, when now it turns out to be a vector? Notice that, in space, a vector has three components. In the plane, a vector has two components. But in one dimension, a vector has just one component; in other words, in one dimension a vector is just a number. Therefore in one dimension, there is no need to use vector language. Now that we are working in space, there is definitely a need.
The advantage of treating instantaneous velocity as a vector is that the direc tion of the vector is the instantaneous direction of motion of the moving body. If we wish to ignore direction and merely talk about magnitude, then we can proceed as follows: interval
llr(t + b.t) - r(t) II is the magnitude of the displacement vector over the time [t, t + b.t]. It represents distance traveled, without regard to direction tra
veled. Therefore
1 lt (r(t
+
b.t) - r(t))
I
represents the average rate of change of distance traveled over the time interval
[t, t + b.t]. This number is always non-negative. We define the instantaneous speed to be
v(t)
=
l��
1 lt (r(t
+
provided that this limit exists. Notice that summarize what we have learned.
b.t) - r(t)) v(t)
=
I
llv(t) 11
=
llr'(t) II·
Now we will
10.2
Veloclt:y and Acceleratlon
805
�.J@il@•@
Suppose that a particle moving through space has position r(t). Ifr is difrerentiabho at t, then lhe instanlaneous velocity of lhe particle at time I is the vectorv(t) =r' (t). We refer tov(t) = r'(t) as the velocity vector for short. The nonnegative number v (t)= llv(t)ll = llr'(t)ll is the instantaneous speed of the par ticle. If the instantaneous speed is pooitive, that is, if r'(t) � 0, then the direction vector
�
II r' t) II
r'(t) is said to be the instantaneous direction ofmotion of the particle.
Let the motion of a particle in space be given by r(t) = cos(t)i + sin(t)j + tk. Sketch the curve of motion on a set of axes. Calculate the velocity vector for any t. What value does the velocity have at time t = 7r/2? What is the speed at this time? Add the velociy t vectorv(7r/2) to your sketch, representing it by a directed line segment whose initial point is the terminal point of r(t). � EXAM PL E
1
Solution We have studied the curve of motion, a helix, in Example 2 of Section 10.1. Recall that the i andj components represent motion around a circle, and the t component represent.s a simple vertical motion. The velocity vector for any time tis
v(t) =r'(t) = -sin(t)i + cos(t)j + k. We therefore
have v(7r/2) = -i + k.
The speed
J
of the particle at time t = 1r/2 is v (t) = IIv(1r/2) II = ( -1)2 + fi + 12 = ../2. The curve of motion and this velocity vector are sketched in Figure 2. ....
.&. Figure 2
INSIGHT
Figure 2 suggests that the velocity vector r'(1r/2) is tangent to the curve of
motion at the point r(?r/2). Imagine that the curve is a roller-coaster track and that the particle is a roller-coaster car Suppose that the coaster loses its grip on the track and hurtles off into space. If we ignore air resistance and the effect of gravity. then the car will continue to move in the direction it was moving at the instant that it left the track; that direction is the direction of the instantaneous velocity vector. These ideas are developed in the remainder of this section and also in Section 10.S. .
The Tangent line to a Curve in Space
(a) .&. Figure 3
Let Cbe a space curve that is parameterized by t�r(t). Suppose that Po= r(to) is a point on C at which r'(to) exists. If tlJ is sufficiently small, then the secant line through the two point.s r(to) and r(to + �t) can be used as an approximation to the (as yet undefined) tangent line to Cat P0• Figure 3 illustrates this idea. As you look at panels (a), (b), and (c), notice that the approximation improves as �t decreases to 0. The secant line in panel (c) is very close to our intuitive conception of the tangent line, which is shown in panel (d).
(b)
{c)
806
Chapter 10
Vector-Valued Functions
r(to +D.t) - r(to) captures the direction of the secant line through the points r(to) and r(to +D.t) . We would like to let D.t tend to0 but, when we do so, the displacement vector r(to +D.t) - r(to) approaches 0 and Notice that the displacement vector
its direction becomes undefined . However, if D.t is positive, then the average velocity
1
(r(to +D.t)- r(to))
D.t
(a) the r'(to) as D.t
r(to +D.t)- r(to)
same direction as
limit
tends to 0. These ideas suggest the following definition for the
tangent line to C at the point
a typically nonzero
P0•
r is a para Po= r(t0). If r'(to) exists and is not the zero vector, then the tangent line to Cat the point Po= r(t0) is the line through Po that is parallel to vector r'(t0). The tangent line is parameterized by u f-+ r(t0) + ur'(t0). Let
Po
and
(b)
has both
be a point on a space curve C. Suppose that
meterization of C with
INSIGHT
Why do we introduce a new parameter
pass through the point
t f--> r(t)
u to
describe the tangent line?
u f--> r(to) + ur'(to) Po=r(to) at the values t=to and u=0, respectively. It is therefore
Notice that the curve described by
prudent to use different parameters for
and its tangent line
C and its tangent line.
� EXAM P L E 2 Consider the curve C defined by the vector-valued function
r(t)= (tcos(t), t sin(t), t). What are parametric equations for the tangent line to Cat the point Po = (0, 7f /2, 7f /2)? Solution We calculate
r'(t) = Therefore
(�t (t
cos (t)) ,
�t (
t sin (t)) ,
r'(7r/2) = (-7r/2,1,1).
�/) = (
cos (t) - t sin (t) , sin ( t)
The tangent line to C at
+t cos (t) ,1).
r(7r/2) = (0,7r/2,7r/2)
therefore parameterized by
()
()
)
)
)
Uf-+r � +ur' � =lo�� +ul-� 11 = 1-�u � +u � +u \ 2' ' \ 2 '2 2 2 '2 . \ '2'2
z
The parametric equations for the tangent line are z=
y
INSIGHT
A plot of
7f
Z
+u .
C may be found in Figure 4a. A static, two-dimensional image of
a space curve often does not impart a clear understanding of how the curve twists in
r(t)
=
(t cos(t), t sin(t), t)
.A Figure 4a
space. To help visualize the curve the equation x2 + y2
=
C of Example 2, we have plotted the solution set S of S comprises the two cones that are shown in
z2• The graph of
is
10.2 Velocity and Accelenition Figure 4b. (In Section 11.2 in surfaces.) Because the entries
s 07
Chapter 11, you will leam techniques for sketching such of r:t) {tcoslr),r sinl t),t) satisfy the equation =
we see that C lies on S. Superimposing the plots of C n nd S �n Figure 4c reveals that C winds around S. Finally, the tangent line that we have calculated in Example 2 has been added in Figure 4d.
...
y
I 'I 1(J) .&. Figure4b
(f cos(.!),' sin(I),')
.&. Figure 4c
Acceleration
AB we might expect, acceleration
.&. Figure .cd
obtained by differentiating velocity. Accel
is
eration is again represented by a vector.
l�l§3!mU.1ij
If i
velocity
� t)
=
a: t)
=
While the velocity vector points in the direction of motion of the moving body, the acceleration vector. usually does TWt. In fact, Newton's Second Law tells us that
force Fils equal to mass m times acceleration
a.
That is,
lli'=ma.
Now mass is a positive scalar, and force and acceleration are vectors. So Newton's S econd Law tells us that the acceleration
vector points in the same direction as the force
being applied to the body. The next example illustrates this point strikingly.
808
Chapter 10
Vector-Valued Functions
z
=cos(t)i + sin(t)j + tk as in Example 1. Calculate the
� EXAM PL E 3 Let r(t)
acceleration a(t). Evaluate the acceleration at t
Solution We have v(t) particular, a(O) a(27r)
=
=
=
-sin(t)i + cos(t)j +
- i, a(7r/2)
=
- j, a(37r/4)
=
k
=
0, 7r/2, 37r/4, 7r, and 27r.
and a(t)
(1/vtz)i
=
-cos(t)i - sin(t)j. In
+ (- 1/vtz)j,
a(7r)
=
i, and
-i. These acceleration vectors are exhibited in Figure 5. In each instance,
the acceleration vector points horizontally and into the helix. That is also the direction of the force that is required to hold the particle on the spiral path.
Because differentiation of a vector-valued function is performed component
.A Figure 5 r(t)
=
cos(t)i + sin(t)j+tk
.,..
wise, it follows that antidifferentiation is also performed componentwise. We saw this phenomenon at the end of Section 10.1. Now we use this idea in a physical example.
� EXAM PL E 4 Suppose that a particle moves through space with accelera tion vector that is identically 0. What does that tell us about the motion of the particle?
Solution Let r(t) be the position vector of the particle at time t. We are given that v'(t)
=
a(t)
=
Oi + Oj +Ok.
Antidifferentiating
componentwise yields v(t)
=
cii + c:ij + c3k,
each
where
side
ci, c2, c3
of
this
equation
are constants. Notice
here that we are using only the simple fact that the antiderivative of 0 is a constant. Although our computation is not finished, we can already observe that the speed of the particle is constant:
ll v(t) 11
=
Jcf+ c� + S for all t.
v(t) and v(t) cii + c:ij + c3k. We =cii + c:ij + c3k. Again we antidifferentiate componentwise,
Next, we work with the equations r'(t) equate to obtain r'(t)
=
=
this time obtaining
Here we have used the fact that the most general antiderivative of the constant
cit+ di,
and so on. Of course,
di, d2, d3
ci is
are new constants of integration. Our
calculation is complete. We have learned that r(t)
=(di, d2, d3) + t(ci, c2, c3), which
is the parameterization of a line. In summary, we have shown that if a particle moves through space with an acceleration vector that is identically 0, then the particle travels with constant speed along a line. INSIGHT
.,..
Example 4 is essentially Newton's First Law: A body moves at constant
speed along a straight line unless a force acts upon it. In particular, Example 4 tells us that, whenever a body travels along a path that is not a straight line, there must be some acceleration (and hence some force) present.
� EXAM PL E
5 Show that the acceleration vector of a particle moving
through space is always perpendicular to the velocity vector if and only if the particle travels at constant speed.
Solution Let r(t), v(t)
=
r'(t), and v'(t)
=
r"(t), respectively, denote the position
vector, the velocity vector, and the acceleration vector of the particle at time t. Let v(t) and
=
ll v(t) II
=
Jv(t)
2
·
v(t) denote the speed of the particle. Then, v(t)
=
v(t) v(t) ·
10.2 Velocity and Acceleration
(
d v(t)2 dt Of course
)
d (v(t) v(t)) dt
=
·
�t (v(t)2)
0
=
constant if and only if only if
v(t)
and
v'(t)
v(t)
=
if and only if
v(t) v'(t) ·
=
(
·
d d v(t) + v(t) dt dt
v(t)2
0 for
are perpendicular.
all
) ·
v(t)
=
2v(t) . v '(t).
is constant. We conclude that
t.
It follows that
v(t)
( Jzj + Jzk)
100
a=
i
y
15.5 200
_. Figure 6
speed of
7r/4
400
120 feet
(see Figure
is
is constant if and
� E X A M P L E 6 An arrow is shot into the air from a height of
-32k
v(t)2
..,..
z
12o
809
15
�
feet at a
per second. The initial angle of the arrow from the horizontal is
6).
Assuming that the acceleration due to gravity is
32
ft/sec2,
determine how far the arrow travels horizontally.
Solution The initial position vector is
Oi + Oj + (31/2)k.
The motion takes place
entirely in the yz-plane (there is no side-to-side motion). Writing the initial velocity as the scalar multiplication of its magnitude and direction vector, we have
120((1/v!z)j + (1/v!z)k) just that due to gravity:
Oi + 60v'2j + 60v'2k. a -32k. =
The acceleration in the problem is
=
We begin with the acceleration and work backward:
a(t) = -32k
=
Oi + Oj - 32k.
Antidifferentiating in each component separately (don't forget the constants of inte gration!) gives
v(t)
=
c1i + c2.i
Oi + 60v'2j This means that
+
(-32t + c3)k.
Now we have
60v'2k = v(O) = c1i + c2.i + (-32 0 + c3)k.
+
·
c1 0, c 60v'2, and c3 60J2. We conclude that v(t) 2 (-32t + 60v'2)k. Antidifferentiating once again gives =
=
=
=
60v'2j
+
r(t) = dii + (60v'2t + d )j + (-16t2 + 60v'2t + d3)k. 2 But then
Oi + We conclude that arrow is given by
Oj
+
(31/2)k
di = 0, d = 0, 2 r(t)
=
and
60v'2tj
+
=
r(O)
=
dii + d2.i
+ d3k.
d3 = 31/2. To summarize, the
trajectory of the
(-16t2 + 60v'2t + 31/2)k.
The motion of the arrow ceases when the arrow hits the ground, which is when the
k
component of
r is 0 .
We solve the quadratic equation
-16t2 + 60v'2t + 31/2 = 0
to obtain
t=
-60v'2±
J(-60v'2)2 - 4(-16)(31/2) -TI
The negative solution
=
-60v'2±J8192 -TI
=
-60v'2±64v'2 -TI
(-60v'2 + 64v'2)/(-32) -v'z/8 has no physical rele (-60v'2 - 64v'2)/(-32) = 31v!z/8, the j component of r is (60v'2) (31v!z/8) = 465. Thus the horizontal distance traveled by the arrow is 465 feet. ..,..
vance. Fort=
=
·
81 O
Chapter 10
Vector-Valued Functions
The Physics of Many experts consider the fastball to be the most important pitch in baseball. If the Baseball
pitcher imparts a certain backspin to the ball when it is released, then the ball can appear to abruptly sink shortly before it reaches the plate. The amount by which it sinks can be several inches. Because the sweet spot of a standard baseball bat has diameter 2.75 inches, and the batter must get his "fix" on the ball during the first 30 feet of the ball's 60-foot journey to the plate, the "sinking fastball" can certainly fool a batter. Let us use the ideas we have learned
to
understand why the fastball actually
sinks. Suppose that the pitcher stands at the origin of coordinates and throws in the direction of the positive x-axis with an initial velocity vector v0 = 130i - k. Here the units are distance measured in feet and time measured in seconds. The speed of 130 feet per second is about 88.6 miles per hour, a plausible speed for a fastball thrown by a good pitcher. Assume in addition that the pitcher releases the ball at a height of 5.5 feet, so x
Direction of forward motion
s =-SJ .A Figure 7
that the inii t al position is ro = 5.5k. Now the interesting part of this
analysis is the acceleration. Of course there is
a component of acceleration that is due to gravity; this component contributes
-32k. Because the ball is given an initial backward spin, there is also a component called "spin acceleration." Use the right-hand rule to see that the so-called spin vector S for the ball is pointing in the direction of the negative y-axis (Figure 7). We shall write S= -Sj. According to the theory of aerodynamics, the spin causes a difference in air pressure on the sides of the ball and results in a contribution of spin
acceleration
that is given by a"' = cS Xvo. Here
c
is a physical constant that is determined through experimentation. Let us
calculate this
cross
product:
j kl)
-S Vz
0 V3
= (-Sv3)i +
(Sv1)k.
Noting that v1=130 and v3= -1, we have thus determined that a•=c(Si +
130Sk)
and a= a• - 32k= c(Si + 130Sk) - 32k = cSi + (130cS
- 32)t.
Typically, the amount of spin placed on the ball is about S= 2.5 revolutions per second. Experimental evidence suggests that arrive at
c= 0.08
is plausible. Thus we
a= 0.2i-6k. As usual, we integrate this last equation to obtain v= 0.2ti-6tk + c. The value of C is determined by setting v(O)=v0• We find that C=130i - k. Hence
10.2 Velocity and Acceleration
811
(130 + 0.2t)i + ( -1 - 6t)k.
v=
Integrating a second time yields
r(t) = (130t + 0.1t2)i + (-t - 3r)k + D. The value of the constant vector Dis determined by setting
f(O)=r0. We find that
D= 5.5k. The final result is then
r(t)= (130t + 0.1t2)i + (5.5 - t - 3t2)k. What is the meaning of this formula for the motion of the ball? It is subtle.
60 feet. To find out 130t + O.lt2 = 60 to obtain t � 0.46.
The distance of the pitcher's mound from home plate is when the ball crosses the plate, we solve
Suppose that the batter gets his fix on the ball at the halfway point. Solving
130t + O.lt2=30, we obtain t � 0.23 seconds. At that time the height of the ball is 5.5 - 0.23 - 3 (0.23)2. Thus, the batter sees that, halfway through its course, the ball drops 0.23 + 3 (0.23)2 0.3887 feet. He would expect it to drop a similar amount on the second half of its journey. However, at time t = 0.46, the ball has height (5.5 - 0.46 - 3 (0.46)2). The ball has dropped a total of 0.46 + 3 (0.46)2= 1.0948 feet instead of the 0.3887 + 0.3887 = 0 .7774 feet that the batter expected. So, in effect, the ball has "sunk" 1.0948 - 0 .7774 = 0.3174 feet or 3.81 inches. Refer to Figure 8 to see the path of the pitched ball. It is worth noting that the vertical component of the trajectory of the ball is parabolic. Thus the ball does not travel a ·
·
=
·
·
linear path; it instead traces a parabola.
y
0.3887
s 4 3 2 1 x
10
20
30
40
so
60
_. Figure 8 The expected drop after the halfway point is 0.3887 ft.
Q UIC K
Q UIZ
1. 2. 3.
If
r(t)
=
(t, t2,t3),
what is
r'(2)? r(t) = (t, 2t2,2 + t2) at t = 1. r(t) = (t, v'4"+t, e 2t). What is its velocity when
Find symmetric equations for the tangent line to A particle's position is given by
-
t=O?
4.
A particle's position is given by
r(t)
=
(e
-
t
, cos (t), t2) .
What is its acceleration
when t=O?
Answers
1. (1,4,12)
2. x- 1=(y- 2)/4=(z- 3)/2
3. (1,1/4,-2)
4. (1,-1,2)
812
Chapter 10
Vector-Valued Functions
EXERCISES Problems for Practice
Further Theory and Practice
For each position function r in Exercises 1-8, calculate the corresponding velocity
L 2. 3. 4. 5. 6. 7. 8.
v,
speed
v,
and acceleration
izontal, and initial speed
r(t)=ti + t2j + t3k r(t)=(2 - t2)i - 5t4k r(t)=(1 - t2)i + 2tj + (1/(1 + t2))k r(t)= e'i+ e-'j - e -21k r(t)=ti - 5j + e'k r(t)=In(ltl)i+r1j - t-2k r(t)= -cos(t2)i + sin(t2)j + t3k r(t)=t112i + t113j+ t114k
how many seconds does the arrow hit the ground? How greatest height that the arrow achieves?
34. A rock is thrown from a 500-foot cliff, with a downward trajectory forming an initial angle of
vo= v(O)
-32k. The initial
r0=r(O)
are given in each problem. Using antidifferentiation,
and initial position
r(t).
at the given point
13. 14. 15. 16. 17. 18. 19. 20. 2L 22.
r
Po=(1,1,1) Po=(1,0,-5) Po=(1,1,-1) Po=(0,1,-1) Po=(2, -1,e) Po=(1,0,7r) Po=(5, -5,1) Po=(1, -4,1) Po=(1,4,0) Po=(1,0,0)
In each of Exercises 23-32, determine symmetric equa
23.
24. 25. 26. 27. 28. 29. 30. 3L 32.
at the given point
due to gravity is constantly
Derive a formula for
the position
r(t)
of the rock. When does the rock hit the
ground? How far does it travel horizontally?
35. A roller coaster is speeding along a track; its motion describes the path
t=2,
the roller coaster leaves the track. Ignoring
coaster will be at time
t=5.
36. An arrow is shot into the air from an initial height of and with trajectory making initial angle of
7r/3
6 ft,
radians
with the horizontal. If it were a calm day, the force
tions for the tangent line to the curve described by the given
r
radians with the
the effects of gravity and drag, determine where the roller
Po.
r(t)=ti+ t2j+ t3k r(t)=(2 - t2)i - (1 - t2)j - 5tk r(t)= e'i+e-tj- e-2.tk r(t)=ln(l +t)i +1/(1 +t)j- 1/(1 - t)2k r(t)=ti- j + e112k r(t)= -cos(t)i + sin(t)j + tk r(t)=ti- 5j +(5/t)k r(t)=(5 - t2)i + 2tj + (5/(1 + t2))k r(t)=(3 +t)/(1+t2)i+ t2j+ sin(7rt)k r(t)= e'i+te'j+t2e1k
function
7r/6
ft/s. The acceleration
At time
In Exercises 13-22, determine parametric equations for the tangent line to the curve described by the given function
50 -32k.
horizontal. The initial speed is
r(t)= -5t2i+ t3j+ (4t + 30)k.
ro=2i- 5k ro =2j + Sk r0 =3i- 2j- k ro=j
Vo=3i - 2j + k Vo=4i - 7j lL v0=i+j+k 12. Vo=k 9.
-32k, compute
far does the arrow travel horizontally? What is the
near Earth's surface has constant acceleration due to gravity
10.
radians with the hor
ft/s. Assuming that the
acceleration due to gravity is constantly
given by
compute a formula for the position
7r/6
100
the function r(t) describing the motion of the arrow. After
In each of Exercises 9-12, a body traveling through space velocity
33. An arrow is shot into the air with initial height 4 ft, initial trajectory forming an angle of
a.
Po.
Po=(-2, -4,-8) r(t)=ti- t2j+ t3k Po=(2,-5, e2) r(t)=ti- 5j+e1k Po=(5,-5, 1) r(t)=ti- 5j + (5/t)k Po=(6,2,-2) r(t)=(7 - t2)i +2j +(t+1/t)k Po=(1,4, 1) r(t)=(2/t)i+2tj +k r(t)=4arctan(t)i- cos(7rt)j+(1/t)k Po=(7r,1,1) Po=(1,0, ,/ff) r(t)= -cos(t2)i +sin(t2)j +tk P=(1,1,1) r(t)= e'i+e21j +e31k Po=(1,-4, 1) r(t)=(5 - t2)i + 2tj + (5/(1 + t2))k r(t)=ln(l +t2)i + arcsin(t)j + sec(t)k Po=(0,0, 1)
exerted by the bow would have launched the arrow with initial speed
120
ft/s. However, a sudden gust of wind
reduces the horizontal component of the arrow's initial velocity by
20
ft/s (the wind blows horizontally, and
dies out after its instantaneous gusting). The acceleration due to gravity is constantly
-32k.
What is the velocity
of the arrow when it strikes the ground? How far does the arrow travel horizontally before it strikes the ground?
37. A body moving through space has constant acceleration vector and zero initial velocity. What can you say about the path traveled by the body (give a geometric description)?
38. A body in motion through space has the property that its velocity vector always equals its acceleration vector. What can you say about the motion? Can you write a formula for it?
39. A body travels through space in such a way that its position vector is always perpendicular to its velocity vector. Show that this implies that the motion is on the surface of a sphere.
40. Redo Exercise
35,
but this time take into account the
effect of gravity.
4L Redo Exercise
36 if
the arrow is shot downward from a
height of 500 feet with an initial angle of
60 degrees from
the horizontal.
42. A particle moves along the curve that is parameterized by
=1+t,y=2t2, z=6t +t2.
x
Its vertical speed is
16
at all
10.3 Tangent Vectors and Ale Length times. What is the velocity vector of the particle when it passes through the point (2,2,7)? 43. Let Po= (0,0,2r) denote the north pole of the sphc:r:e r+r+(z - r)2=r. Let Pi be any other point on the sphere (see Figure 9). Suppose that a rig}.d straight tine joins Po to P1 and that a partic:le initially at rest is allowed to slide down the line from Po to P1 under a constant force equal to -mg t. Show that the elapsed time docs not depend on the choice of the point P1. z
Po .r:2 + y2 + (z -r)2=r2
813
always acute. Show that the speed of the particle is decreasing if the angle is always obtuse. 48. Let r(t) ={t3 -t}i +(r - 1)j. Plot the points r(-2), r(-1}, r(-1/2). r(O}, r(1/2}, r(1), r(2). Sketch the curve. Describe the tangent lines to the curve at the point (0,O}. 4'. Let Cbe the curve with parameteri7.ation r(t)= (t,t1,i'} fort>0. Find a point Po on Cforwbi.c:h the tangent tine to Cat Po passes through the point (0,-1/2,-1/./2). 50. ShOW that the C\IIVeS tH (t3+r- 3t+1,r,3t - 1} and tH (3t+1, 2t,P+t-1) intenect at two points. At each of these points, what is the angle between the tangent lines to the curves?
calculator/Computer Exercises In each of Exercises 51-!8, plot the planar curve as90ciated with the given vector-valued function r(t)=r1{t)i+ rz(t)J. Cakulate r'(to) at the given value of to and add the tangent tine to the curve at r(to) to your plot. SL Sl. 53. 54. SS. 56. 44.
Let ro and Yo be comtant vectors. Let constant. Show that
r(t) =cos(wt)ro + satisfies
sin(wt) --
w
w
be a positive
tJtZ r(t}+ w2r(t)=0,
.58.. 59.
Yo
the initial value problem
tJl
�.
r(O}= ro, r'(O) =To.
Let r{t} denote the position at time t of a body moving through space. Suppose that r(t) is parall el to a(t)=r"(t) for all t. Prove that r(t) X r'(t) is a comtant vector. 46. To each point P, =(cos(t},sin(t},t) with t�O, we anoci ate another point Q, as fullows: P1Q1 is tangent to the 45.
helix traced by P1, II P,Q, II = .,fb, and P,Q, t�0. Para meterize the curve traced by Q1• If'!. Show that the speed of a moving particle is decreasing if the angle between its velocity and a.cceleration vectors is ·
r(t) =(S+2cos(t))i+(3+sin(t))j, O:st:S211', to=1f/3 r(t) = (P + t)i + (P - t}j, -2 st s2, to =1 r(t) =tan{t)i+sec(t}j,-1f/3:St:Sw/3, to='lf/4 r(t)=3t/(1+P)i+3t2/(1+F}j,-o.6:st:s 15, to=1/3 r(t)= (e'+e- 1 )i+ (e-e- 1)j,-3:st:s3, to=2 r(t) = (t" -t)l+(P-t)j, -2s;1s;2, to=1/2 r(t)=2t/(1+P)I+(1-12)/(1 + P)j,-20:st:s20,to=1/2 r(t) = (t-sin(t))i +(1-cos(t))J, 0:St:S211', to=11'/6 Plot the planar curve r(t}=3t/(l+F)i+3t2/{l+i')j, t>-1/2. It is an arc of the Folium of Descartes. Fmd
all points at which the curve has a tangent tine parallel to the line y=x. Find all points at which the curve has a horizont&l tangent line. Find all pointa at wbic:h the curve has a vertical tangent line. 60. Plot r(t) =cos(3t)cos{t}i + sin(2t)sin(t)J forO < t «rr/2. Find the point Po where the curve crosses itself, and add the two tangents at Po to your plot. 61. Plot r(t)=(P-t,t4-1,(t- 1}1n(2+t)) for-6/S:sts; 6/5. To your plot, add the tangent lines at die point (0, 0,0). 62. The curve r(t) =(t cos(t),uin(t),t), Osts'fl'/6 has one tangent line that intersects the vertical line x = 1, y=1. What is the intersection point? Plot the curve, its tangent tine, and the vertical line.
1 0.3 Tangent Vectors and Arc Length In Section 102, we studied the derivative r' of a vector-valued function r. Our fOCWI also teamed
was on the velocity and accelerati on of the motion that r defines. We
how to use the derivative to obtain tangent lines to the graph of r. In this section,
814
Chapter 10
Vector-Valued Functions we will investigate the properties of tangents in greater depth. We will also learn an application to arc length.
Unit Tangent Vectors
To effectively study the geometry of a space curve C, we must use a parameterization r that provides us with information about the tangent lines of C. This forces us to eliminate certain parameterizations from our discussion. As an example of
r(t) = (cos(t3), sin(t3), 0), 1 }. Notice that the derivative 0. Because r'(O) 0, we obtain no
what we must avoid, consider the vector-valued function
-
which parameterizes the circle C =
r'(t) =
(
+ y2 =
{(x, y, 0) : x2
3t2sin(t3), 3t2cos(t3), 0) vanishes when t
=
=
information about the direction of the tangent line to C at the point r(O). Our next definition excludes such inconvenient parameterizations.
t:::; b.
r(t) = r1(t)i + r2(t)j + r3(t)k is continuous for a:::; smooth parameterization of the curve it defines if
Suppose that We say that r is a
a. the scalar-valued functions able on
b.
r'(t) =I= 0
(a, b),
and
for every
ri, r2,
r are all twice continuously differenti 3
t in (a, b).
[a, b] into finitely many subintervals [a, x1], [x1, x2], ... , [xN 1, b] such that the restriction of r to each subinterval is smooth, then we say that r is piecewise smooth. If we can divide the interval
Until Chapter
13, we will restrict our attention to smooth parameterizations. A
moving particle whose position vector r is a smooth parameterization of its path
r'(t), a positive speed a(t) r"(t). For such a para unit tangent vector to C at the point
has, at each instant of time, a differentiable velocity vector
v(t)
=
llr'(t) II,
and a continuous acceleration vector
meterization r of a curve C, we define the P
=
r(t)
to be the vector
T(t) x
.A Figure 1
=
As its name implies,
T(t)
�
C� ) 1r' 1) II
r'(t)
(10.3.1)
is a unit vector. Furthermore, if
-----+
T(t)
is represented by a
-----+
directed line segment PQ with initial point P, then PQ is tangent to C at P (see Figure
1).
Notice that a particle with position vector
indicated by
T(t).
r(t)
traces C in the direction
Finally observe that, by rewriting equation
express the velocity vector the speed:
(10.3.1), we may r'(t) as the scalar product of the unit tangent vector with r'(t) = llr'(t) llT(t) = v(t)T(t).
� EXAM PL E 1 Let
T(t). What
r(t)
=
eti + e-tj + v'zt k.
(10.3.2)
Calculate the unit tangent vector
are the unit tangent vectors at Po= r(O) and P1 = r(ln(2))?
r'(t) = eti - e -tj + VZk is a tangent vector to the curve defined by r at the point r(t). The length of r'(t)
Solution Differentiating each component of is given by
r(t),
we see that
10.3 Tangent Vectors and Arc Length
The unit tangent vector
T(t)
is therefore
z
T(t) =
2
1 e'i- e-t· + v'zk J e'+e-1
(
In particular, we have
T(ln (2)) -
T(O)
=
)
e-t · e' + .j2 k. ie1+e-t e1+e_,J e1+e-t
=
(1/2)i- (1/2)j + ( VZ/2) k -
r(t)
� EXAMPLE 2 Let
t= 1
corresponds to
and
1/2 . 2 . 4. 1 2./2k + .j2 k - S1 - 5. + 1 J 5 2 +1/2J 2 + 1/2 2 + 1/2
=
-
-
2.
These two unit tangent vectors are plotted in Figure
_. Figure 2
81 5
ti + t2j + (t3 -St) k.
·
.,..
The point
Po= (-1, 1, -4)
and lies on this curve. Using the unit tangent vector at this
point, write symmetric equations for the tangent line to the curve at
Solution The velocity vector at
t= 1
P0•
is
r' (l) = (- i+2tj + (3t2 -S) k) l,=1=-i+2j - 2k,
z
and the speed is
llr' (l)ll
J(
=
therefore
-
1) 2+ 22+ ( 2) 2 -
=
3
.
The unit tangent vector is
The tangent line we seek is
x ( 1) y-1 z-(-4) (-1/3) (2/3) (-2/3) ' -
-
_
_
2(x + 1) = 1 y=z+ 4. Figure 3 shows the plot of r(t) line at Po has also been plotted. .,..
or
_. Figure 3
Arc Length
-
If a curve C is defined by a smooth parameterization
ask, What is the length of C from
r(a)
r(b) ?
to
r(t)
for
for
0 ::5 t ::5 2. The tangent
a::::; t::::; b, then we may
An intuitive way to think about
length is this: Lay a piece of string along the curve from r( a) to r( b). Mark the string
where it touches
r(a)
and
r(b),
straighten the string out, and measure the distance
between the marks. That is the length. To obtain a method for actually calculating length from the function
r,
we follow the scheme that we used in Section 7.2 of
Chapter 7 to compute the lengths of planar curves. We divide the interval
[a, b]
into small subintervals by a uniform partition
a= to < t1 < Let Set
!1t= ( b a)/N be the common Pj r(tj) (xj, yj, Zj)· The basic -
=
=
·
·
·
<
tN
=
b.
width of the subintervals of this partition. idea is that the length of the curve
r
is
816
Chapter 10
Vector-Valued Functions
z
Po
approximated by the sum of the lengths of the directed line segments -------+
P2P3 , ..., PN-lpN (see
Figure
given by
SN= y
N
L j=l
llPj-lPj II=
4). Thus
-------+
-------+
P0P1 , P1P2 ,
the length of the curve is approximately
N
L J(xj -xj-1)2 + (yj -Yj-1)2 + (Zj -Zj-1)2. j=l
The Mean Value Theorem tells us that x
Xj - Xj-l = r1(tj) - r1(tj-1)=ri'(pj)f).t, Yj -Yj-l = r2(tj) -r2(tj-1)= r2'(af)f).t, Zj - Zj-l = r3(tj) - r3(tj-1)=r3'(Tj)f).t,
.& Figure 4
for some numbers formula for
SN
pj, Oj, Tj
in the subinterval
(tj-b tj)·
Then we can rewrite our
as
N SN= L j=l
2 2 2 r{(pj) + r2(uj) + ri,(Tj) !).t.
On the one hand, this sum approximates the length of the curve. On the other hand,
SN
approaches the Riemann integral
l: llr'(t) lldt as N tends to infinity. We
are led to the following definition.
1
•
The
Let C be a curve with smooth parameterization
arc length
of C is
l: llr'(t) lldt.
t f-+ r(t), a::::;; t::::;; b.
r, which has the r' is continuous, we ensure that the function t f-+ llr'(t) II is continuous. property in turn assures the existence of the Riemann integral l: llr'( t) lldt. It
By restricting our attention to a smooth parameterization property that This
should be noted that the formula for arc length can be used to find the arc length
any two llr'(t) lldt is the
between
1�2
P1=r(t1) P1 and P2.
points of a curve. If arc length between
and
P2=r(t2)
with
t1 < t2,
then
+ v'st k. Calculate the arc length (0,-1,0) and (1,0,v's7r/4). Solution Differentiating r(t), we obtain r'(t)=2cos (2t)i + 2sin (2t)j + v'sk. The point (0,-1,0) corresponds to t=O, and the point (1,0,v's7r/4) corresponds to t= 7r/4. The arc length of the portion of the curve between the point (0,-1,0) and (1,0, v's7r/4) is therefore
� EXAM PL E 3 Let
r(t)= sin (2t)i - cos (2t)j
of that portion of the curve between
r/4 lo
(2cos (2t)) 2 + (2sin(2t)) 2j + (v's)2dt v 1r/4 0 = r/4 y4 (cos2(2t) + sin2 ( 2t)) + Sdt
llr'(t) lldt =
lo = r/4 ../9dt lo = 37r/4. ....
10.3 Tangent Vectors and Arc Length
Reparameterization
81 7
It is important to bear in mind that many different vector-valued functions can describe the same curve in space. Imagine, for example, a railroad line between the city of New York and Boston. Each time a train travels this track from New York to Boston, it traces the same curve C. Yet, if the train travels at different speeds on different trips, then different parameterizations of Cresult. To make this idea concrete, let us consider a specific example. The vector valued functions
r(t) = cos(t)i +sin(tj) + tk, 0 :5 t :5 2rr and
p(s) = cos(2s)i + sin(2sj)
+
2sk,
0
<
s :5 7r s runs through [O,2rr]. Further
describe the same curve in space. To understand why notice that, as
[O,rr], the expression t = 2s every s in [O,rr], we have
the interval more, for
p(s) =p
runs through the interval
(�) =cos (2 (�) } +sin (2 (�) )j + 2 (�) k =cos(ti) + sin(tj) +tk = r(t).
This equation shows that
p
r
and
describe the same points in space (albeit at
different values of their domains). The functions parameterizations
of
the
same
curve.
Notice
r
and
that
p
are simply different
the
invertible
function
'I/;: [O,rr] ---+ [O,2rr] defined by 'l/;(s) = 2s = t affects the way we pass from one para
[O,rr], we have p(s) = r('l/;(s)) = r(t). Equivalently, given a value of t in [O,2rr], we have r(t) =p( 'l/;-1 (t)) =p(s). Finally, observe that 'I/; is increasing. As a result, r and p parameterize the curve in the same direction. That is, as t increases from 0 to 2rr, and as s increases from 0 to rr, the points r(t) and p(s) both trace the curve from the initial point r(O) =p(O) = (1,0,0) to the terminal point r(2rr) =p(rr) = (1,0,2rr). In general, passing from one vector-valued function r to another function p that meterization to the other: Given a value of sin
describes the same curve Cin the same direction is called reparameterization. If the domain of
r
is [a,b] and the domain of
the composition of
'I/;: [c,d]
---+
[a,b].
r
p
is
[c,d],
then the reparameterization pis
with a continuously differentiable increasing function
That is,
p(s) =(ra'lj;)(s) =r('l/;(s)),
c:5s:5d.
r is a smooth parameterization of a curve Cwith = ra'I/; is a reparameterization with 'l/;(s) = t. Show that point p(s) = r(t) = P will be the same whichever para
� EX A M P L E 4 Suppose that P
= r(t).
Suppose also that p
the tangent line to Cat the meterization we use.
Solution The parameterization reparameterization
p
yields
p'(s)
r
yields
r'(t)
as a tangent vector to C at P. The
as a tangent vector to Cat P. We relate the two
vectors by means of Theorem 6g of Section
10.1:
p'(s) =(ra 'l/;)'(s) = 'l/;'(s)r'('l/;(s)) ='!/;'(s)r'(t).
(10.3.3)
81 8
Chapter 10
Vector-Valued Functions Equation
(10.3.3) tells us that the vectors p'(s) and r'(t) are parallel.
the tangent line to C at the point parameterization we use.
p(s)= r(t) = P
It follows that
will be the same whichever
<11111
INSIGHT
Suppose that a differentiable, increasing function '!/; is used to obtain a reparameterization p r '!/;. From equation (10.3.3), we see that p'(s) 0 at anys for which 'l/;'(s) 0. Thus if pis a smooth parameterization, then 'l/;'(s) > 0 must hold for alls. =
o
=
=
The next theorem shows that the arc length of a curve does not depend on the chosen parameterization of the curve.
THEOREM 1
Suppose that C is a curve with smooth parameterization t r(t), 'ljJ : [c,d]-+[a, b] be a continuously differentiable increasing func tion, and let p = r 'ljJ be a reparameterization of C. Then the arc length of C when computed using the reparameterization p is equal to the arc length of C when computed using the parameterization r.
a :st :sb.
t-t
Let
o
Proof. We compute the arc length of C using the reparameterization equation
(10.3.3),
p.
Using
we have
1d llP'(s)lids= 1d 111/J'(s)r'('l/J(s)) lids= 1d 'l/J'(s)llr'('l/J(s)) lids. T = 'ljJ(s), dT = 'ljJ'(s)ds. The last integral then becomes l: 11r'(T) 11 dT, which is the arc length of C as determined by the parameterizationr. •
Now we make the substitution
Parameterizing a Curve by Arc Length
Suppose that the vector-valued function parameterization of C. Let
arc length function a
of
r defined on the interval [a, b]
L = l: llr'(T)lldT denote the
r is
is a smooth
(total) arc length of C. The
the increasing function from
[a, b]
to
[O, L],
which is
defined by
(10.3.4) For every tin the domain of the parameterization, C from the initial point
r(a) to
the point
r(t)
a(t) measures the distance along
(see Figure
Fundamental Theorem of Calculus to equation
(10.3.4),
5).
When we apply the
we obtain
a'(t) = llr'(t) II· If we think of a particle with position vector
(10.3.5)
r(t) tracing the curve,
then equation
(10.3.5) has a satisfying interpretation: The instantaneous rate of change of distance along the curve with respect to time (that is, particle (that is, .A Figure 5
a(t) [. ll•'(r)lldr is the r(a) to r(t). =
arc length of C from
llr'(t)ll ).
a'(t))
is equal to the speed of the
For some types of problems, it is convenient to reparameterize C by arc length. We will use the arc length function
a of r to accomplish this task. As a consequence
1 0.3 Tangent Vectors and Arc Length
81 9
r'(t) -# 0 for all t. From equation a'(t) > 0. This inequality implies that a is a strictly increasing function. Therefore a has an inverse function a-l, which is also increasing. We let p be the reparameterization of C defined by p(s) r (a-1(s) ) for 0
(10.3.5),
we deduce that
=
=
=
r () lo llP' u lldu
=
1a-l(s) a
llr'(r) lldr
=
a (a-1(s) )
s.
=
(10.3.6)
This means that the point p(s) is situateds units of arc length along C from its initial point-note the equality of the parameters of p(s) and the distance. Clearly, this property of the reparameterization p is very special. It is the basis for the following definition.
�·j3ijjji1Mil 0 ss s L,
Let L be the length of a curve C. A parameterization p(s),
of C is called the
arc length parameterization
of C if the arc length
(O) and p(s) is equal tos for everys in the interval [O, L]. We also say
between p
that p parameterizes C with respect to arc length. The next theorem tells us how to recognize an arc length parameterization of a curve without calculating arc length.
THEOREM 2
0
Let s� p(s),
curve C. Then,
llP'(s) II
=
the unit tangent vector T(s) for every s: T(s)
=
(10.3.7)
p'(s).
t� r(t), 0 < t::::; bis a smooth parameterization of a curve C such 1 for all t, then r is the arc length parameterization of C, and bis
Conversely, if that
llr'(t) II
=
the length of C.
Proof. Suppose thats� p(s), identity
J; II p'(u) IIdu
0
s then holds by definition. If we differentiate each side of
=
this equation with respect to s, then, according to the Fundamental Theorem of Calculus, we have
d llP'(s) II dds lor llP'(u) lldu d/ =
=
=
1,
which is the first assertion of Theorem 2. Applying formula parameterization p and substituting
which is formula
(10.3.7).
llP'(s) II
=
1, we obtain
(10.3.1)
with
820
Chapter 10
Vector-Valued Functions
Conversely, if t f-t r(t), 0:st:sb, such that llr'(t) II 1 for all t, then the arc length along C between r(O) and r(t) is J� llr'(r) lldr= f�ldr= t. This is precisely the requirement for r to be the arc length parametrization of C. Finally, we see that b is the arc length of C because b= Jg llr'(r) lldr, and the right side of this equation is the . ��� =
INSIGHT
Suppose that we observe the path C of a moving particle, beginning at time Theorem 2 tells us that the particle's position vector is the arc length para meterization of C if and only if the particle moves with constant speed 1.
s =
0.
� EXAMPLE 5 Are r(t)=cos(2t)i- sin(2t)k, O:st:'S?r and p(u)=cos(u)i sin(u)k, 0:su:s27r arc length parameterizations of the curves that they define?
The function r is not an arc length parameterization, for r'(t)= -2sin(2t)i- 2cos(2t)k and llr'(t) II = 2 -=J 1. On the other hand, p'(u)=-sin(u)icos(u)k so that llP'(u) II = 1 for all u. Therefore the function p(u)= cos(u)i- sin(u)k is an arc length parameterization. Notice that the functions r and p describe the same set of points in space. Indeed, we obtain p from r by composing with the increasing function 1/J(u)= u/2: p(u)=(ra1f;)(u)= r(1f;(u))= r(u/2). <1111 Solution
Reparameterize the curve r(t) so that it is parameterized with respect to arc length. � EXAMPLE 6
=
(cos(t), sin(t), t), 0:st:s27r
J
We calculate r'(t)=(-sin(t), cos(t),1) and llr'(t) II= sin2(t) +cos2(t) + 12= ../2. Therefore the arc length function of r is given by s=o-(t)= J� llr'(r) lldr= J� ../2 dr= ../2 t. Then t= s/../2=o--1 (s) and
Solution
\ Cri), sin (�) . �) ,
p(s)=r (u-1 (s)) = r(s/..fi) = cos
is the required parameterization by arc length. Unit Normal Vectors
r
0 :s s :s 2..fi?r
<1111
Suppose that t f-t r(t) is a smooth parameterization of a curve C. Fix a point P= r(t) on C. Figure 6 shows the plane that passes through point P is perpendicular to the tangent vector to C at P. It is called the normal plane of C at P. Although every vector in the normal plane is perpendicular to T(t), we will see that two of these normal vectors are of particular interest. Our starting point is the identity llT(t) 112=1, which we may write as T(t) T(t)=1. If we differentiate both sides of this last equation with respect to t, then we obtain T'(t) T(t) + T(t) T'(t)=0 or 2T(t) T'(t)=0. The resulting equation, ·
T(t)
·
·
·
(10.3.8)
T(t) T'(t)=0, ·
x
I
.& Figure 6
tells us that the vector T'(t) is perpendicular to the unit tangent vector T(t). Therefore T'(t) lies in the normal plane of C at P. Although the vector T(t) has unit length, its derivative T'(t) generally does not have this property. If T'(t) -=J 0, then we obtain a unit vector by scalar multiplying T'(t) by 1/ llT'(t) 11 ·
10.3 Tangent Vectors and Arc Length
If t then the vector
t-t r(t)
821
is a smooth parameterization of a curve Cwith T'(t) -=fa 0,
N(t)=
llT t) II
(10.3.9)
T'(t)
�
is called the principal unit normal to Cat r(t).
INSIGHT
equation
Ifs f-+
(10.3.7)
p(s) is the arc length parameterization of C, then we can use
to write the principal unit normal as
N(s)
z
=
llP" s) II
�
p"(s) .
(10.3.10)
The cross product
(10.3.11)
B(t)=T(t) X N(t) y x
_. Figure 7
B(t) T(t) X N(t)
of the unit tangent vector and the principal unit normal vector is called the binormal vector to C at the point r(t). Our work in Section 9.4 shows that the binormal vector is a unit vector that is perpendicular to both T(t) and N(t) (see Figure 7).
=
p(s) tracing a curve C with unit N(s) is the direction of the particle's acceleration vector, p"(s). Therefore N(s) points to the side to which C curves. (Refer again to
INSIGHT
Imagine a particle with position vector
speed. Formula Figure
7.)
(10.3.10)
tells us that
This observation will be important in the next section.
� EXAM PL E 7 Let C be the curve defined by r(t) =ti+ (t3/3)j + (1 -t2)/V'i,k. Calculate the principal unit normal and the binormal vector of C at the point P =(1, 1/3, 0). Solution
Observe that r'(t)= +t2j -V'i,tk and
i
2 llr'(t) 11 = 12+ (t2)2+ (-V2t) =v'1+2t2+t 4=1+t2.
J
It follows that
1 t T(t)= --i+ _f_.J - V'i, k 1+t2 1+t2 1+t2 and
T'(t)=
(-
2t (1+t2)2
)i+ ( (1+2tt2)2 ) - ..n( (1+-1t2)2 ) k. •
J
t2
The point P corresponds to t =1, and for this value of t we have
T(l) = _!i+ _! j - V'i, k 2 2 2
822
Chapter 10 Vector-Valued Functions
B(l)
and T'(l) = vector
N(l)
(-1/2)i + (1/2)j - Ok= (1/2)(-i + j). The required principal unit normal is the direction of T'(l), which is the direction of the vector -i + j. It is
quickly calculated to be p
N(l) Therefore
T(l)
_.
Figure 8 T(l), N(l), and B(l)
are mutually perpendicular.
In Figure 8, the vectors
T(l), N(l),
and
B(l)
have been added to a plot of
r(t)
for
o�t�2. ... INSIGHT
i, j, and k, the vectors T(t), N(t), and B(t) are orthogonal i, j, and k, the vectors T(t), N(t), and B(t) form a right-handed system. Unlike i, j, and k, the vectors T(t), N(t), and B(t) are not fixed directions; they change as r(t) traverses the curve C. The triple (T, N, B) is said to be a moving frame. It is Like the vectors
and have length
1.
Like
particularly useful for studying the geometry of space curves and the trajectories of moving bodies.
Q UIC K
Q UIZ
1. 2.
For
r(t)
=
t3i + t2j + t6k,
what is the unit tangent vector
T(l)?
What integral represents the arc length of the plane curve parametrized by
r(t) ti+ e1j + !e21k, 0�t�1? If r is an arc length parameterization of a curve of length 6, what is II r'(4) II? 4. If, at a point on a curve, (1/v'Z, 0, 1/vtz) is the unit tangent vector, and (-1/v'Z, 0, 1/vtz) is the principal unit normal vector, then what is the binormal vector? =
3.
Answers
1. (3/7, 2/7, 6/7)
2.
J� Vl + e21 + e41dt
4. (0, -1, 0)
3. 1
EXERCISES Problems for Practice In each of Exercises 1-8, a parameterization curve is given. Calculate the tangent vector unit tangent vector given point
Po.
T(t)
L r(t)=ti t2j + t3k 2. r(t)= -j + tk -
at
r(t),
r'(t)
at
r(t) of a r(t), the
and the tangent line at the
Po= (1, -1, 1) Po= (0, -1,5)
Po= (4, 0, -2) Po= (3, 9, 27) Po= (1, 3, 0) r(t)= (e1 + t2)i + 3j Po= (1, -e, e) r(t)=ln(t)i - t ln(t)j + tk Po= (0, 1, 7r/2) r(t)= cos(t)i + sin(t)j + tk r(t)=2 ln(cos(t))i + 2 ln(sin(t))j + 4tk, Po= ( ln(2) ln(2) 7r)
3. r(t)=ti - t112k 4. r(t)= e1i + e21j + e31k 5. 6. 7. 8.
-
, -
,
10.3 Tangent Vectors and Arc Length In each of Exercises 9-14, calculate the arc length of the
curve described by the given vector-valued function.
r(t)=sin(t2)i - cos(t2)j + t2k r(t)=e41i+ 8tj + 2e--41k 11. r(t)=cos3(t)i+ 3j - sin3(t)k 12. r(t)=sin(3t)i - cos(3t)j + t312k 13. r(t)=(1 + t) 312i + (1 - t)312j + t312k 14. r(t)=sin(t)i + cos(t)j + ln(cos(t))k 9.
10.
o::=;t::=;y'?r 0::=;t::=;ln(2) 7r::=; t::=;3n 7r/2 ::=; t ::=; 7r
0::=;t::=;1 O::=;t::=;n/4
by equation
18.
r(t)=cos2(t)i+ cos(t)sin(t)j + sin(t)k
sin(t)k
In each of Exercises 19-22, a parameterization
r(t)
t= o--1 (s )
(
In each of Exercises 23-30, a parameterization curve C and a point
Po
r
of a
are specified. Give symmetric equa
tions for the line through Po with direction vector equal to the principal unit normal vector to Cat
28. 29. 30.
Po. This line
is called the
P0•
r(t)=ti - t2j + t3k r(t)=t2i+ 2t2j + 3t3k r(t)=cos(t)i+ sin(t)j + tk r(t)=ti+ (2/t)j + t2k r(t)=e1i+ e-1j - t2k r(t)=e1i+ te1j + (t + e1)k r(t)= y'ti + (4/t)j + (4/v't)k r(t)=ln(t)i + arctan(t)j + (1/t)k
Po=(1, -1,1) Po=(1,2, -3) Po=(0,1,n/2) Po=(2, 1,4) Po=(1,1,0) Po=(1,0,1) Po=(2,1,2) Po=(O,n/4,1)
plane to the curve at the given point
33. 34. 35. 36.
given
formula
for
r(t)
41.
r(t)=e'i+ e-'j + v'Ztk r(t)=cos3(t)i + sin3(t)k
to
obtain
In each of Exercises 43-50, a parameterization
r(t)
of a
planar curve is given. Find Cartesian equations for the tan
43.
r(t)= (5 + 2cos(t))i + (3 + sin(t))j
to=�
44.
r(t)=(t2 + t)i + (t2 - t)j
45.
r(t)=tan(t)i + sec(t)j
to=1 to=� 4
3t . 3 t2 I+ 1 + t3 J 1 + t3 47. r(t)=(e1 + e-1)i+ (e1 - e--t )j
46
r(t)=ti - t2j + t3k r(t)=t2i+ 2t2j + 3t3k r(t)=ti+ (2/t)j + t2k r(t)=cos(t)i + sin(t)j + tk r(t)=e1i+ e-1j - t2k r(t)=e1i+ e21j + e31k
48.
r(t)
P0• Po=(l,-1,1) Po=(1,2, -3) Po=(2, 1,4) Po=(0,1,n/2) Po=(1,1,0) Po=(3, 9, 27)
(t)=
r
.
r(t)= (t3 - t)i + (t2 - l)j
2t . 1 - t2. + 1 + t2 I 1 + t2 J 50. r(t)=(t - sin(t))i + (1 - cos(t))j
49
r(t)=
3
to=1 to=O 1 to= 2 1 to= 2
to=� 2
r(t) of a (T,N,B) at
In each of Exercises 51-56, a parameterization space curve is given. Calculate the moving frame
r(to). 51. 52. 53. of a
curve is given. Find the Cartesian equation for the normal
32.
the
r(t)=ti+ tj + t312k r(t)=cosh(t)i - sinh(t)j + tk
•
In each of Exercises 31-38, a parameterization
31.
)
39. 40.
•
27.
into
defined in the instructions to Exercises 23-30.)
r(t)=t2i - ln(t)j + 2tk; (1,0,2) and (e2, -1,2e) 20. r(t)=(sin(t) - t cos(t))i+ t2j - (t sin(t) + cos(t))k; r(O) and r(2n) 21. r(t)=e1i+ e-1j + v'Ztk; (1,1,0) and (e,1/e, VZ) 22. r(t)=cos(t)i + sin(t)j - t312k; (1,0,0) and (0,1, -(n/2)3/2 )
26.
Calculate the inverse of o- by setting
gent and normal lines to the curve at r(t0). (The normal line is
19.
24. 25.
(10.3.4).
p(s)=r o--1 (s ) . The reparameterized curve will be para
of a
specified points:
23.
Find an
find the function 0--1 such that t= o--1 (s ). Now substitute
42.
curve is given. Calculate the arc length between the two
normal line to C at
[O,l].
meterized according to arc length.
r(t)=ti - (t2/2)j + (t3/3 )k 16. r(t)=cos(t/ VZ)i + sin(t/ VZ)j + (t/ VZ)k cos(t)j +
for t in the interval
s =o-(t) and solving for t as a function of s-in other words,
15.
�
r(t)
explicit formula for the arc length function o-(t) that is defined
defines.
�
Po=(2, 1,1) Po=(O,n/4,1)
In each of Exercises 39-42, a curve is parameterized by
the given function
function is an arc length parameterization of the curve it
r(t)= cos(t)i -
r(t)= y'ti + (4/t)j + (t2 - lS)k r(t)=ln(t)i + arctan(t)j + (1/t)k
Further Theory and Practice
In each of Exercises 15-18, determine whether the given
17.
37. 38.
823
54. 55. 56.
r(t)=cos(t)i + sin(2t)j + tk r(t)=(t2 + t)i + (t2 - t)j + t2k r(t)=tan(t)i + sec(t)j + tk r(t)=ti+ t2j + t3k r(t)=e1i+ e--tj + (e1 + e-1)k r(t)= (t3 - t)i + (t2 - l)j + t2k
to=n/4 to=1 to=n/4 to=1 to=ln(2) to=1
57. Given any positive number N, describe a way to define a piecewise smooth curve parameterized by
y(t)j
r(t)=x(t)i+
that has length greater than N and components that
satisfy
0::=;x(t),y(t)::=;1.
58. Show that the tangent lines to the helix parameterized by
r(t)= (cos(t), sin(t), t/2)
make a constant angle with
Show that the normal lines intersect the z-axis.
k.
824
Chapter 10
Vector-Valued Functions
Calculator/Computer Exercises
59. 60. 6L 62.
In each of Exercises 59-62, use Simpson's Rule to approximate to two decimal places-that is, with an error no greater than 0.005-the arc length of the curve defined by the given vector-valued function r over the given interval.
r(t) r(t) r(t) r(t)
=
=
=
=
t2i - t4j + t-1k ti e-tj + t5k cos(t)i + sin(2t)j + ytk ln(t)i + (2/t)j + t312k
1 0.4 Curvature In this section, we learn what the second derivative
r"
r" r.
of a vector-valued function
can tell us about the geometry of the curve C defined by that
1 :5t:54 0.4:::;; t:::;; 0.9 1 :5t:54 1 :5t:52
-
r
To be specific, we will see
can be used to measure the way C deviates from a straight line. The geo
metric tool for this study is the concept of curvature. What does it mean to say that a circle is curved while a line is not? One answer is that, as we move along a circle, the unit tangent vector changes direction, while if we move along a line, the unit tangent vector does not change direction. We define the curvature to be the magnitude of the instantaneous rate of change of direction of the unit tangent vector with respect to arc length.
l•l§ijWli[eHI
Suppose thats 1--+
a curve C. The quantity
p(s)
is a smooth arc length parameterization of
(s) "(s) II (s) � I �;s) I I !p' l l P r(s). =
=
(10.4.1)
=
is called the curvature of C at the point
INSIGHT
The definition of curvature involves differentiation with respect to arc length. That choice is influenced by geometric considerations, but there is a subtle motivation as well. As we saw in the preceding section, there are always many ways to parameterize a curve. The concept of curvature should depend only on the geometry of a curve and not on the choice of parameterization. By referring to the arc length para meterization, we avoid potential ambiguities that might arise from different choices of parameterization. Theorem 1 of Section 10.3 shows that any two parameterizations of C will give rise to the same arc length reparameterization.
� EXAMPLE 1
Let
r(s)
=
Solution The curve C described by
l r'(s) II
C is p, and the center of C is so that
=
s.
p cos(s/ p)i + p sin(s/ p)j for some positive constant
p. Calculate the curvature at each value of
r
(0, 0, 0).
r'(s)
is a circle in the xy-plane, z Notice that
Jsin2(s/p) + cos2(s/p)
=
1
=
=
0. The radius of
r(s)
(-sin(s/p))i + ( cos(s/ p))j,
for alls. We deduce that
length parameterization by Theorem 2 of the preceding section. Finally,
r"(s)
-
=
(s).
p1 cos p
1-
(s).
. p1 sm p
J
is an arc
10.4 Curvature
825
and therefore
( � / )) ( ?/ ) ) = �
�(s)= llr"(s)ll = Tz
-
Radius lOp T1 1 =lQp
2
s p
-sin
+
p
2
cos 2
(�)
+sin2
(�) = �·
.._
Straight line INSIGHT
K
cos
Example 1 shows that our notion of curvature is a sensible one because
it says that the curvature of a circle of radius pis 1 /p at all points of the circle. In particular, the curvature of a circle of large radius is small (the vector T changes
slowly with respect to arc length) , while the curvature of a circle of small radius is
K=O
_. Figure 1
large (the vector T changes rapidly with respect to arc length). See Figure 1, in which circles with radii pand 10pare plotted. The same pair of tangent vectors, Ti and T2, appears on each circle. The change of direction shown on the small circle is the same as that shown on the large circle, yet it is attained with a much smaller change of arc length. The rate of change of direction with respect to arc length is therefore greater on the smaller circle.
Calculating Curvature Without Reparameterizing
As we have discussed, there are good reasons for defining curvature by means of formula
(10.4.1). However, this approach does
have the drawback of requiring the
arc length parameterization of the curve, which is often tedious or difficult to calculate. It is therefore desirable to have an alternative expression for curvature that does not refer to arc length.
THEOREM 1
Suppose that
ris a smooth parameterization of a curve C. Then, �(tr )
=
llr(t' ) Xr"(t) II llr(t' ) 113
(10.4.2)
rt). ( The curvature atrt) ( may also be expressed as
is the curvature of C at the point
�(t)=
Proof. Let
(l rt(' ) II . llr"(t) II )z - (r(t' ).r"(t))2 IIrt(' ) 113
o-t( )= J� l rr (' )lldr denote the arc length function for the para r. Using equation (10.3.5) to express the speed l r(t' ) II asdo-(t)/dt, we
meterization
may rewrite equation
(10.3.2) as
(t) r(t' )= do---;ft T(t). When we differentiate each side of equation obtain
(10.4.3)
(10.4.4)
d 2o-(t) T do-(t) Tt (' ). () rt(" )= dt2 t+ dt
(10.4.4) with respect to time, we
(10.4.5)
826
Chapter 10
Vector-Valued Functions
Substituting the right sides of equations (10.4.4) and (10.4.5) into the expression 0, we obtain
r'(t) X r"(t) and noting that T(t) X T(t) r'(t) X r"(t)
=
=
(d�;t)) T(t) 2
X
T'(t).
(10.4.6)
Recall that T(t) and T'(t) are perpendicular by equation (10.3.8). Therefore llT(t) X T'(t)II llT(t)11 llT'(t)II llT'(t)II by Theorem 3b of Section 9.4. Comparing the magnitudes of each side of equation (10.4.6), we deduce that =
=
or
llT'(t)II
=
llr'(t) x r"(t)II llr'(t)112
(10.4.7)
Now let s a-(t) be the arc length of C between the points r(a) and r(t). Our work in Section 10.3 shows that the arc length reparameterization is p(s) r(a--1(s)) r(t). Observe that =
=
p'(s) T(t)
=
(10.4.8)
=
because each side is the unit tangent at the same point on the curve. As a result, we have
T'(t)
=
;tT(t) �� �T(t) (l0�.8) llr'(t)II � p'(s) =
=
llr'(t)llP"(s).
(10.4.9)
Finally, the curvature at the point in question is, by definition (10.4.1),
T (t)II (10�.7) llr'(t) x r"(t)II ii:(s) II p"(s)II (10�.9) ll ' llr'(t)ll llr'(t)f ' =
which is formula (10.4.2) for ii:r(t). Equation (10.4.3) can be derived from (10.4.2) by • elementary but tedious algebraic manipulations (Exercise 55). INSIGHT
(10.4.1)
Each formula for curvature has advantages and disadvantages. Equation
is simple, natural, and easy to remember. However, it is often useless for com
putation. Formula
(10.4.2) is
more complicated than
(10.4.1), but it is usually the easier
formula to use because it does not require reparameterization. The downside is that formula
(10.4.2) does not have any obvious connection with the concept of curvature and (10.4.3)
certainly seems to be dependent on the choice of parameterization. Formula
shares these drawbacks and has the further disadvantage of being less memorable. The upside is that formula
(10.4.3) does not require a cross
product calculation.
10.4 Curvature
827
� EXAM PL E 2 Let C be the helix parameterized by r(t) = cos(t)i + sin(t)j+ tk for 0 < t < 7r 4 . Calculate the arc length function u for r. Use u to find the arc length reparameterization p(s) of C. Using formula (10.4.1), show that the curvature is / 1 2 at each point of C for all s. Verify that formulas (10.4.2) and (0 1 .. 4 )3 both yield this same constant value for the curvature. Solution
Observe
that
J(-sin(t))2+(cos(t))2+12=
the curve is u(t)=
r'(t)= -sin(t)i+ cos(t)j+ k
and
ll r'(t) II =
-JZ.It follows that the arc length function u for
J� ll r'(T) ll dT= J� -J2 dT=
-J2t. Therefore if u(t)= s, then
-J2,t = s, or t = s/-J2.As we learned in Section 10.3, this implies that
is the arc length parameterization of C. We have
p'(s)
=
-
1 . sm -JZ
(s) -J2
I+ •
1 COS -JZ
(s) -J2
and
p"(s) = -
1 cos 2
(s) -J2
•
1-
1 . sm 2
J •
+
-J21 k
( s ). -J2
J.
Therefore
To use formula (0 1 .. 4 )2 , we calculate
r'(t) X r"(t) = (-sin(t)i + cos(t)j + k) X (-cos(t)i - sin(t)j) = det
which
([
t
-s n(t) -cos(t)
gives us r'(t) X r"(t) = sin(t)i - cos(t)j + k. Therefore
Jsin2 (t) + (-cos(t))2 +
!
co (t) -sin(t)
k 1 0
]),
ll r'(t) X r"(t) 11
=
12= -JZ.Formula (10.4.2) then tells us that curvature at r(t)
is equal to
-J2 ll r'(t) X r"(t) 11 ( -J2)3 II r'(t) 113 _
_
1 2,
which agrees with the result of our first calculation.Using formula (10.4.3), we find that the curvature at r(t) is
828
Chapter 10
Vector-Valued Functions
2 (llr'(t )II llr"(t )II ) - (r'(t ) r"( t ))2 _ llr'(t )f ·
·
J( J2
·
1
)2 - ((-sin( t )i+cos(t)j+k) (-cos( t )i - sin (t)j) ) 2 ( J2)3 ·
v2=D2 1 2v'2 - 2'
which agrees with the previously obtained value. <11111 In Example
2, the curvature calculation is feasible whether or not Theorem 1 is
used. The next example better illustrates the power of Theorem 1. �
EXAM PL E 3 Let C be the curve parameterized by
fi;r(t ) at point r(t ). calculate r'(t )= eti - e-tj+k.
Find the curvature Solution We function for
r is
r(t )= eti+e-tj+tk.
Notice that, because the arc length
it is not practical to find the arc length parameterization of C. We appeal to Theorem 1 instead. Because
r"(t )= eti+e-tj, it follows that
r'(t )Xr''(t ) (e'i - e-
�
det
( [ ::
Therefore the curvature is
2 ( t )Xr"(t )ll ll-e-ti+etj+2kll J2 = = e t+e- t+4 .,.. fi;r(t )= llr' 2 2 lleti - e-tj+kll3 (e t+e- t+1)3/2. llr'(t )113 The Osculating Circle
p(s)be a smooth arc length parameterization of a curve C. Recall that formula N(s)= llP"(s)r1p"(s). Using formula (10.4.1) for the curvature, we can rewrite this equation for N(s) in the form
Lets f-t
(10.3.10) expresses the principal unit normal vector as
p"(s)= fi;(s)N(s).
(10 4 . 10 ) .
4
p(s)traces C with unit p"(s) is the principal unit normal N(s), and the magnitude of its acceleration is the curvature fi;(s). If fi;(s) i- 0, then we let p(s)= 1/ fi;(s). By multiplying each side of equation (10.4.10) by p(s), we obtain
Equation (10. .10) states that, if a particle with position vector speed, then the direction of its acceleration vector
N(s) p(s)p"(s). =
The quantity
( 10 4 . 11 ) .
p(s) is known as the radius of convergence of C at p(s). The reason for
this terminology can be understood from the case in which C is a circle. We know
10.4 Curvature
1 that a circle of radius p has constant curvature ,.,;(s) 1/ p. Thus for radius p, we have p(s) 1/ ,.,;(s) p. In other words, the radius of cur
from Example a circle of
829
=
=
=
vature for a circle is the radius of the circle.
p(s) by a circle that is ,.,;(s), at p(s) as C. The radius of this approximating circle is p(s), and the center is a distance p(s) from p(s). If C were actually a circle, then the direction of acceleration, v'(s) p"(s), would point to the center of the circle (Example 5, Section 10.2). Therefore by equation (10.4.11), the vector from p(s) to the center of our approximating circle should have direction N(s). Finally, if C were a circle, then it would lie in a plane; In general, we can approximate C near the point
tangent to
T(s)
at
p(s)
and that has the same curvature,
=
that plane would contain its velocity and acceleration vectors. Therefore our approximating circle should lie in the plane determined by the vectors
N(s).
T(s)
and
We summarize all of these considerations with a precise definition of the
approximating circle.
Osculating plane
If
Let p be a smooth arc length parameterization of a space curve C. ,.,;(s) > 0, then the osculating circle (or the circle of curvature) of C at p(s) is the
unique circle satisfying the following conditions:
p(s) 1/ ,.,;(s). The circle has center p(s) + p(s)N(s). vature of C at p(s).
a. The circle has radius b.
=
This point is called the
c. The circle lies in the plane determined by the vectors plane is called the
osculating plane
of C at
N(s)
center of cur
and
T(s).
This
p(s).
_. Figure 2a
INSIGHT
By construction, the osculating circle has the same principal unit normal
vector and tangent vector as the curve Cat the point of contact
p(s).
It therefore has the
same binormal vector as well. The osculating circle and the curve Cjust touch at the point of contact
p(s) because they share the same tangent line. That is the reason for the name:
"Osculating" is a synonym for "kissing." An osculating plane for a helix is shown in Figure 2a. The unit tangent vector and the principal unit normal vector are also graphed. A different view of the same curve and osculating plane appears in Figure 2b. It shows how well the osculating plane "fits" the curve near the point of contact. The osculating circle is added in Figure 3.
_. Figure 2b
Although we have used the arc length parameterization to simplify the dis cussion, it should be noted that all of the ingredients needed for the osculating plane and osculating circle can be calculated without reparameterization. The next example will illustrate this point.
� EXAM PL E 4 Let C be the curve parameterized by the curvature
,.,;r(t)
at the point
r(t).
r(t)
=
Cartesian equation of the osculating plane at
Solution We calculate
r'(t)
=
Find
Also determine the radius of curvature, prin
Osculating plane cipal unit normal, and center of the osculating circle at the point
_. Figure 3
t2i + tj + tk.
2ti + j + k
and
r(t).
r(t)?
r"(t)
=
2i.
It follows that
What is the
830
Chapter 10
Vector-Valued Functions
r'(t) Xr"(t)
�
(2ti +j +k) X (2i)
Therefore the curvature at the point
�(t)
=
llr'(t) X r"(t) II llr'(t)3 II
=
r(t)
r(t)
T(t)
det
( [ � ! ! ])
�
2j-2k.
is
ll2j-2kll ll2ti +j +kf
In particular, the radius of curvature at at
�
r(t)
2v'z (2 +4t2)3/2
is
(1+2t2)3/2.
1 (1+2t2)3/2 .
The unit tangent vector
is
=
1 -- r'(t) r' ll (t) II
=
l (2ti +. +k) J v2 +4t2
=
2t l l .+ i+ k v2 +4t2 v2 +4t2J v2 +4t2 .
We calculate
T'(t)
=
4t 4t 4 k i(2 +4t2)3/2 (2 +4t2)3/2J (2 +4t2)3/2 •
=
4 ( . - -tk). (2 +4t2)3/2 I fJ •
The direction of T'(t) is the principal unit normal vector. This direction is the same as the direction of
i - tj - tk.
N(t) =
It follows that
t t l .ik J v1+2t2 v1+2t2 v1+2t2
is the principal unit normal vector at
r(t) +p(t)N(t)
r(t).
The center of curvature is therefore
(t2i +tj +tk) +(1+2t2)3/2
=
(
(t2i +tj +tk) + ((1+2t2)i-t(l +2t2)j-t(l +2t2)k) (1+3t2)i-2t3j-2t3k.
=
=
The binormal vector
B(t)
Osculating plane
Osculating circle
=
T(t) x N(t)
=
det
([
v�
+4t2 2t/ 1;v1+2t2
-
.& Figure 4
j 1;v2 +4t2 - t/vl +2t2
k 1;v2 +4t2 - t/vl +2t2
])
=
� (j-k)
is perpendicular to the osculating plane. Because the osculating plane passes through the point
r(t) = t2i +tj +tk, O(x - t2) +
Center of curvature ((1 + 3t2), -2t3, 213)
)
t t l ik jv1+2t2 v1+2t2 v1+2t2
its equation is
� (y - t) + (-¥) (z - t)
= 0,
which simplifies toy= z. The curve, its osculating plane, and osculating circle are shown in Figure
4. ..,.
10.4 Curvature
831
In Example 4, the entire curve r(t) t2i + tj + tk is contained in the plane y = z. It is therefore not surprising that the plane that contains the curve turns out to be the osculating plane at each point.
INSIGHT
=
Planar Curves
The theory that we have developed for space curves can also be applied to curves in the xy-plane. Such a curve C might be the graph y f(x) of a function f, or it might be given parametrically by functions x(t) and y(t): C {(x,y) : x x(t),y y(t), a< t < b }. In fact, the graph y f(x) of a function can be realized parametrically by means of the equations x x(t),y y(t) where x(t) t, and y(t) f(t). We can imagine the xy-plane to be the plane z 0 in xyz-space. Then the planar curve can be realized by the vector-valued function r(t) x(t)i + y(t)j +Ok, and our formula for the curvature of space curves applies. The next theorem will make this explicit. =
=
=
=
=
=
=
=
=
=
=
THEOREM 2
Suppose that x(t) and y(t) are twice differentiable functions that define a planar curve C by means of the parametric equations x x(t), y y(t). Then, at any point P ((x(t),y(t)) for which the velocity vector (x'(t),y'(t)) is not 0, the curvature of C is given by =
=
=
_(
�(t) - lx'(t)y"(t) - y'(t)x"(t)I 3/ . (x'(t))2 + (y'(t))2 2
)
In particular, the curvature of the graph of y �(x)
=
(10.4.12)
=
lf"(x)I
f(x) at the point (x,f(x)) is
. 32 1 + ( f'(x))2 I
(
(10.4.13)
)
Proof. With the xy-plane realized as the plane z 0 in xyz-space and with parameterized by the equation r(t) x(t)i + y(t)j +Ok, we have =
C
=
r'(t) X r"(t)
=
det
([ ;
�
x' t)
y' t)
x"(t)
y"(t)
�i)
=
(x'(t)y"(t) - y'(t)x"(t))k.
0
1 follows that llr'(t) II ((x'(t))2 + (y'(t))2) 12 and llr'(t) X r"(t) II lx'(t)y"(t) y'(t)x"(t)I. Formula (10.4.12) results when these values are substituted into equation (10.4.2). If x(t) t and y(t) f(t), then x'(t) 1, x"(t) 0, y'(t) f'(t), and formula (10.4.12) reduces to (10.4.13) •
It
=
=
=
=
=
� EXAM P L E 5
is called a cycloid. Repeat for t 37r.
=
=
The planar curve defined by r(t) (t - sin(t))i + (1 - cos(t))j Calculate its curvature at the point P that corresponds to t 7r/3. =
=
=
Solution Let x(t) t - sin(t) and y(t) 1 - cos(t). Then x'(t) 1 - cos(t), x"(t) sin(t), y'(t) sin(t), and y"(t) cos(t). Formula (10.4.12) becomes =
=
=
=
=
=
832
Chapter 10
Vector-Valued Functions
lx'(t)y"(t)- y'(t)x"(t)I I (1- cos(t)) cos(t) - sin(t)sin(t)I 3/2 3/ ' (1-cos (t))2 + (sin(t))2 2 (x'(t))2 + (y' (t))2
K,(t) _
y
_
)
(
)
(
which simplifies to
r(t)
=
(t - sin(t))i + (1 - cos(t))j K,(t)_
1 x
lcos(t)- 11 v'2 3/2 - 3/ 3/2 - 4y'1- cos(t) . 2 2 (1 - cos(t) ) 1- 2cos(t) + cos2t + sin2(t)2
)
_
_
7r/3, the curvature is 11,(7r/3) v'2/(4y'1- cos(7r/3)) 1/2. Similarly, we v'2/(4y'1- cos(37r)) v'2/(4y'1- (-1)) 1/4. Our calculations find that 11,(37r) tell us that the osculating circles at r(7r/3) and r(37r) have radii 2 and 4, respectively. The cycloid is plotted for 0 :5 t :5 47r in Figure 5, which also shows the osculating circles at the points corresponding to t 7r/3 and t 37r. ..,.. When
t
(
lcos(t)- cos2(t)-sin2(t)I
=
=
=
.& Figure 5
=
=
=
=
=
� EXAM P L E 6 Find the circle of curvature of the curve point
P
=
and
=
=
notice
=
-----+--------.-
x
that
4 (1 + 02)3/2
The radius of curvature is therefore
-
sin(2x) at the
f'(7r/4) 2cos(7r/2) 0 (10.4.13), we have =
=
and
according to formula
=
11,(7r/4)
=
=
(7r/4, 1).
Solution We set f(x) sin(2x) f'(7r/4) -4sin(7r/2) -4. Then,
y
y
1/4.
=
4·
Because
f '(7r/4)
=
0, the graph of
sin(2x) has a horizontal tangent. The principal unit normal therefore points
vertically. Because the graph of
y
=
sin(2x) is concave down at
P and the
principal
unit normal points in the direction to which the curve bends, we conclude that the principal unit normal points downward. The center of curvature is the point
(7r/4, 1-1/4), and the circle of curvature has Cartesian equation (x -7r/4)2 + (y- 3/4)2 (1/4)2. The graph off together with this osculating circle is shown in =
.& Figure 6
Q UIC K
Figure 6.
Q UIZ
1. 2.
..,..
The position
}z cos(7rs)k.
p(s) of a moving particle is such that p'(s)
=
sin(7rs)i +
}z cos(7rs)j? 2?
What is the curvature of the trajectory of the particle a s
As a particle passes through point
P,
its speed is
2,
=
the magnitude of its accel
eration is 6, and its acceleration is perpendicular to its velocity. What is the curvature of the trajectory of the particle at
3.
If
t f---t r(t) is
P?
unit vectors are perpendicular to the osculating plane of
4.
If the curvature at a point circle of Cat
Answers 1. 7r 2. 3/2
11,r(3) > 0, Cat r(3)?
a smooth parameterization of a curve Cand if
P on Cis 5 /4, then what is the radius of the osculating
P? 3. ±8(3)
then what
4. 4/5
1 0.4 Curvature
833
EXERCISES Problems for Practice
In each of Exercises 1-6, an arc length parameterization of a curve is given. Verify this assertion and then calculate a. T(s); b. N(s); and c. ,,;(s).
L r(s) =sin (s)i - cos(s)j + 3k
2. r(s) =(3 - cos(s))i - (6 + sin(s))j
3. r(s) =cos(s/v'z)i - sin(s/v'Z)j + (s/v'z)k 4. r(s) =sin(s/4)i + cos(s/4)j - (M/4)sk 5. r(s) =(3sin(s)/5)i + cos(s)j + (4sin(s)/5)k 6.
r(s) =
� - �sin(s)j + � cos(s)k
2
si
7. For the arc length parameterization given in Exercise 3, calculate the radius of curvature p(s) and the center of the osculating circle at each point of the curve. 8. For the arc length parameterization given in Exercise 4, calculate the radius of curvature p(s) and the center of the osculating circle at each point of the curve. 9. For the arc length parameterization given in Exercise 5, calculate the radius of curvature p(s) and the center of the osculating circle at each point of the curve.
In each of Exercises 10-15, a parameterization of a planar curve is given. Calculate the curvature at each point of the curve
10. r(t)=cos 2(t2)i + sin2(t2)j lL r(t) = e1i - e21j
In each of Exercises 28-32, a parameterization of a space curve is given. Find the Cartesian equation of its osculating plane at the given point Po. 28.
29. 30. 3L 32.
r(t) =(2 - t2)i+ 2t3j + t2k r(t) = e1i + tj + e21k r(t)=Vii+ (4/t)j + (t2/4)k r(t) =(t2 + t)i+ (t2 - t)j + t2k r(t)=sin(t)i+ cos(2t)j +6tk
Po=(l,2,1) Po=(1,0, 1) Po= (2,1,4) Po=(2,6,4) Po= (1/2, 1/2,.ir)
Further Theory and Practice
Given a curve C, the locus of its centers of curvature is called the evolute of C. In each of Exercises 33-40, find the evolute of the given plane curve.
33. 34. 35. 36. 37. 38. 39. 40.
r(t) =(2 - t2)i+ 2t3j r(t) =ti+ cos(t)j r(t)=ti+ e1j r(t) = 2cos(t)i+sin(t)j r(t) =(t - sin(t))i+ (1 - cos(t))j r(t) = tcos(t)i + tsin(t)j r(t) =ti+ ln(sec(t))j r(t) = e1cos(t)i + e1sin(t)j
In each of Exercises 41-44, find the point at which the graph y =f(x) of the given function f has the greatest curvature.
r(t)=t2i - t3j r(t) = ln(ltl )i - (1/t)j 14. r(t) = t112i+ t312j 15. r(t) =tan(t)i + sec(t)j
4L f(x)=x2 42. f(x) =� 43. f(x) = ln(2x2) 44. f(x)=x4/16+1/(2x2)
In each of Exercises 16-21, a function f is given. Calculate the curvature at every point on the graph (in the xy-plane) of the equation y =f(x).
In each of Exercises 45-48, find a parameterization for the curve that is the graph of the given equation. Find the cur vature at the specified point.
16. f(x) =x3 17. f(x)=x112 18. f(x) =In(lxl) 19. f(x) = ln(sec(lxl)) 20. f(x) =(� + e-x) /2 2L f(x)=x3 /6 + 1/(2x)
45. x2/a2+y2/b2=1 (a/v'2,b/v'2) 46. x3 + y3 =4xy; (2,2) Hint: Choose parameter t =y/x, divide the cubic equation by x3, and find x in terms of t. 47. x2/3 + y213 =1; (1/8,3V'3/8) Hint: Take advantage of the identity cos2(t) + sin2 �)=1. 48. x(x + y)2=9 + (x+ y) ; (2, 1) Hint: Try the parameter t =x+ y and solve for x in terms of t.
12.
13.
In each of Exercises 22-27, a parameterization of a space curve is given. At each point of the curve, calculate the radius of curvature and the center of curvature.
22. r(t) =ti+ tj + t2k 23. r(t) = t2i+ t2j + (1 + 2t)k 24. r(t)= Vii+(2 - t)j + 3tk 25. r(t) = e1i+ tj + e1k 26. r(t) =ln(t)i + tj + 2tk 27. r(t) =t4i+ tj + 2tk
x>0
49. Prove that a curve with curvature 0 at every point is a line. In Exercises
50-53, treat the concept of torsion.
50. Suppose that p(s) is an arc length parameterization of a curve. Prove that there is a scalar-valued function s r-t r(s) such that the unit tangent vector T(s), the prin cipal unit normal N(s), and the binormal B(s) satisfy the equations:
834
Chapter 10
Vector-Valued Functions
T'(s) N'(s)
/\:(s)N(s)
=
S6. Prove that
-/\:(s)T(s) + T(s)B(s)
=
_
/\:r(t) -
B'(s) = -T(s)N(s). The function
s
>-->
T(s)
torsion
is called the
of the curve.
The three equations are collectively known as the
Formulas.
Frenet
llr'(t) 114
.
Calculator/Computer Exercises In each of Exercises S7-62, a parameterization of a plane
T(s)
The resulting plane curve is called the
(p'(s) X p"(s)) p"'(s) llP'(s) X p"(s) II 2 ·
=
S2. Compute the torsion of the helix p(s) sin(s/V'Z)i + cos(s/V'Z)j + (s/V'Z)k. S3. The torsion of a curve measures the extent to which the =
curve is twisting out of the osculating plane. Prove that a
0 torsion lies in a plane. Conversely, if a curve is planar, it has 0 torsion. curve with
p(x)
·
curve C is given. Plot the locus of the centers of curvature of C.
Sl. Show that the torsion is given by the formula
S4. Let
I 11r'(t) ll2r"(t) - (r'(t) r"(t))r'(t) l
be a polynomial of degree
2
or greater. Prove
that there is a point on the plane curve y = p(x) where the curvature is greatest.
SS. Prove formula
(10.4.3).
S7. SS. S9. 60. 6L 62.
r(t) r(t) r(t) r(t) r(t) r(t)
=
=
=
=
=
=
evolute
of C.
(2 - t2)i + 2t3j
ti + cos(t)j ti + e'j 2cos(t)i + sin(t)j ( t - sin(t) )i + (1 - cos(t))j t cos (t)i + tsin(t)j
In each of Exercises 63-66, a parameterization of a space curve C is given. Plot the locus of the centers of curvature of C. The resulting space curve is called the
63. 64. 6S. 66.
r(t) r(t) r(t) r(t)
=
=
=
=
cos(t)i + sin(t)j +
evolute
of C.
tk
ti + t2j + t3k cos(t)i + sin(t)j + t2k (1 + t2) cos(t)i + (1 + t2) sin(t)j
+ t2k
1 0.5 Applications of Vector-Valued Functions to Motion r is a smooth parameterization of a space curve C. Then, the second r" has a physical interpretation: It is the acceleration of a moving particle has r as its position vector. There is also a geometric significance to r" that
Suppose that derivative that
shows up in the formula for curvature. We begin this section by establishing a connection between the two roles
r" plays. Let us try to anticipate the relationship.
When we go around a sharp curve in a car or a roller coaster, we feel a strong force in the outward direction. By Newton's second law, strong force means strong acceleration. So there is a correlation between large curvature and the acceleration vector having a large normal component.
T(t) and N(t) denote the unit r(t). The velocity and accel eration vectors are given by v(t) r'(t) and a(t) v'(t) r"(t), respectively. The magnitude of the velocity vector is the speed, v(t) llr'(t)ll· Let u(t) fo llr'(T)ll dT be the arc length function of r. Then, du/dt is another way to express the speed. To make these considerations precise, we let
tangent and principal unit normal vectors at the point =
=
=
Recall equation
=
=
(10.4.5),
r"(t)
=
d2;;t) T(t) d��t) T'(t), +
which decomposes the acceleration vector into orthogonal vectors in the osculating plane. We rewrite this equation as
10.5 Applications of Vector-Valued Functions to Motion
dv(t) (t) v(t) (t) a(t) ---cit T + T' .
(10.5.1)
=
Also recall formula
835
(10.3.9),
N(t)
�
llT' t) II
=
T'(t),
which we rewrite as
T'(t) llT'(t) llN(t).
(10.5.2)
=
Substituting equation
(10.5.2)
into
(10.5.1),
we obtain
dv(t) (t) v(t) (t) N(t) a(t) ---cit T + llT' ll .
(10.5.3)
=
Using formula
(10.4.7)
llT'(t) II in equation (10.5.3), we get dv(t) (t) v(t) llr'(t) X r"(t) II N(t) , dt T + llr'(t) 112
to substitute for
a(t)
=
or
a(t) Formula more,
(10.4.2)
llr'(t) II
=
dv(t) (t) v(t) r (t) 1 llr'(t) X r"(t) II N(t) . dt T + ll ' 1 llr'(t) f
tells us that the ratio preceding
(10.5.4)
N(t) in (10.5.4) is Kr(t). Further v(t). Substituting these
is an alternative way to express the speed
equivalent expressions into equation
a(t)
=
(10.5.4),
d��t) T(t)
+
we obtain
v(t)2Kr(t)N(t).
(10.5.5)
Let us summarize these computations with the following theorem .
Suppose that t r(t) is a smooth parameterization of a space v(t) llr'(t) II denote the speed of a particle moving along the curve with position vector r(t). Then, the acceleration vector a(t) r"(t) can be decomposed as the sum of two vectors, one with direction T(t), the unit tangent to Cat r(t), and the other with direction N(t), the principal unit normal to Cat r(t). The decomposition has the form
THEOREM 1
curve C. Let
t---t
=
=
a(t) aT(t)T(t) + aN(t)N(t) =
where
aT(t)
=
d��t)
and aN(t) v(t)2Kr(t). =
(10.5.6) (10.5.7)
836
Chapter 10
Vector-Valued Functions
The scalar ar(t) is called the tangential component of acceleration, and the scalar aN(t) is called the normal component of acceleration. Observe that the normal component of acceleration cannot be negative. Equation (10.5.6), which expresses acceleration as a linear combination of the unit tangent and principal unit normal directions, contains a great deal of information. The component of force in the direction of N(t) that is required to hold a moving body on its curved trajectory is called the centripetal force. By Newton's Second Law, force is mass times acceleration. Thus centripetal force is the mass of the body times aN(t)N(t), a vector that is called the centripetal or normal acceleration. Its magnitude aN(t) depends on the speed and the curvature (but not on the change of speed). The vector ar(t)T(t) is called the tangential acceleration. Its magnitude larl depends on the rate of change of speed with respect to time. If the path of the moving body becomes more curved, then the normal component of the acceleration increases in magnitude. For the body to follow such a path, the normal component of force must be increasing. Furthermore, if the speed of the body increases, then the normal component of acceleration also increases. Once again, we infer that such a trajectory requires a force with increasing normal component When calculating centripetal acceleration, the following theorem is often useful. THEOREM 2
a. b. c.
The tangential and normal components of acceleration satisfy
ar(t) a(t) T(t) aN(t) a(t) N(t) lla(t) ll2 (ar(t))2 + (aN(t)) 2 . =
·
=
·
=
Each of the three formulas is obtained by calculating a dot product and using the equations T(t) T(t) 1, N(t) N(t) 1, and T(t) N(t) 0. Thus part a is obtained by taking the dot product of each side of equation (10.5.6) with T(t). The result is a(t) T(t) ar(t)T(t) T(t) + aN(t)N(t) T(t) ar(t) + 0, which is part a. Part b is obtained in a similar way: We take the dot product of each side of equation (10.5.6) with N(t). Finally, Proot
·
·
lla(t) 112
=
=
=
=
=
=
·
=
·
·
·
=
=
a(t) a(t) (ar(t)T(t) + aN(t)N(t)) (ar(t)T(t) + aN(t)N(t)) ·
·
(ar(t))2T(t) T(t) + 2ar(t)aN(t)T(t) N(t) + (aN(t))2N(t) N(t) (ar(t))2 + (aN(t))2. ·
·
·
•
� EXAM PL E 1 Let r(t) sin(t)i - cos(t)j - (t2 /2)k. Calculate ar(t) and aN(t). Express a(t) as a linear combination of T(t) and N(t). =
We have r'(t) cos(t)j - k. Therefore
Solution
��
=
v(t)
=
=
cos(t)i + sin(t)j - tk
llr'(t) 11
=
Jcos2(t)
+
and
sin2(t) + t2
a(t) r"(t) =
=
J1 + t2 .
=
- sin(t)i +
10.5 Applications of Vector-Valued Functions to Motion
837
It follows that ar(t)= d2s =
dt2
!"! Vl +t2= t . dt J1 +t2
ll a(t) ll 2= ll -sin(t)i+cos(t)j -k ll 2= (-sin(t)) 2+ (cos(t)) 2+(-1)2=2.
Also,
According to Theorem 2,
aN(t) =
vll
a(t)
ll 2 (ar(t)) 2= -
v
2-
t2 1+t2
=
fi+tz Vf+tZ·
In conclusion, equation (10.5.6) becomes
a(t)=
INSIGHT
Equation
(10.5.6)
t
J1+t2 T(t)+ 1+t2
�
+t2 N(t). 1+t2
1, we
can use equation
The values for vectors yield
Central Force Fields
N(t).
a(t)
and
....
can be used in conjunction with Theorem 2b as an
alternative calculation of the principal unit normal vector Example
(10.5.8)
-
(10.5.8)
T(t)
to solve for
N(t):
N(t).
For instance, in
can then be substituted into this formula to
0 be a fixed point in space. We will use it as the origin of our coordinate axes. For each point P =I= 0, let rp= llDPll and up= dir(OP). We say that a force Fis a central force field if there is a scalar-valued function f of a real variable such that F(P) =f(rp)up for every P =I= 0 (see Figure 1). Iff(r) > 0, then the force is directed away from 0. If f(r) < 0, then the force is directed toward 0 and is said to be attractive. As an example of an attractive central force field, consider the sun as a point mass that is fixed at a point 0 in space. If a planet is at point P, then, Let
according to Newton's Universal Law of Gravitation, the gravitational force z
p
r/;
/)R Jf Unitvector � in direction of 0P
� rR y OP=
exerted by the sun on the planet is equal to
F(P) =- GMm Up rj
where G is a universal constant (i.e., the same for all planets), Mis the mass of the sun, m is the mass of the planet,
_. Figure 1 A central force field
F(P)
=
(10.5.9)
f(r)R acts in the direction -->
of the position vector 0P if f (r) > 0 and in the direction opposite to the position vector if f (r) < 0.
the direction of
rp is the distance of the planet to the sun, and Up is
OP. By settingf(r)=-GMm/r2, we see that the force of gravity is
an attractive central force field.
� E X A M P L E 2 Suppose that
a > b > 0. A force F acts on a particle of mass
m
in such a way that the particle moves in the xy-plane with position described by r(t) = acos(t)i+bsin(t)j. Show that Fis a central force field.
838
Chapter 10
Vector-Valued Functions Solution According to Newton's Second Law of Motion, F F=
mr"(t) = m(- a cos(t)i - bsin(t)j )
=
=
ma(t) and therefore,
-mr(t) = -m llr(t) 11
dir (r(t)).
This formula for F shows that it is a central force field directed toward the origin. <11111 A key fact about central force fields is that they always give rise to trajectories that lie in a plane.
THEOREM3
If a particle moving in space is subject only to a central force field,
then the particle's trajectory lies in a plane.
Proof. Let
r(t) be
m be its mass. The f(r)u where r llr(t) II and u (1/r)r(t). r(t) and u have the same direction. Newton's
the position vector of the particle, and let
force on the particle can be written as F Notice that
r(t) X F 0 =
because
=
=
=
Second Law of Motion, force equals mass times acceleration, allows us to conclude that
r(t) X (mr"(t)) 0, or r(t) X r"(t) 0. We can write this as =
=
r(t) X
d r'(t) = 0. dt
From this last equation, we deduce that
d (r(t) r'(t) X ) dt
=
r'(t) X r'(t) + r(t) X d r'(t) dt
=
0+0
=
0.
It follows that
(10.5.10)
r(t) X r'(t) = c for some constant vector c. This tells us that value oft may be. In other words,
r(t) is perpendicular to c whatever the
r(t) is in a plane that is normal to c.
•
When we study the trajectory of a body under a central force field, the motion is planar, so it is convenient to assume that the motion is in the xy-plane. As the body moves along its trajectory, its position vector sweeps out a region in the plane
2). Let A(t) denote r(T) for 0 :5 T :5 t.
of motion (see Figure .A Figure 2
the position vector
THEOREM 4
the area of the region swept out by
If a moving particle is subject only to a central force field, then
the particle's position vector sweeps out a region whose area A(t) has a constant rate of change with respect to
Proof. Let
b..t
t.
be an increment of time,
b..r(t)
the corresponding increment of
position, and Mthe increment of area swept out (Figure 3). We see that Mis
10.5 Applications of Vector-Valued Functions to Motion
0
839
x
_. Figure 3 approximately equal to half the area of the parallelogram determined by the
r(t) and r(t + �t) r(t) + �r(t). The area of this 11 r(t) X r(t + �t) 11 11 r(t) X (r(t) + �r(t)) 11 llr(t) X �r(t) 11 · Thus
vectors
=
=
�t
�t
A'(t)
-+
=
>:::J
! II r(t) X �r(t) II 2 �t
0 gives lim
ll.t--+O
� ut
We conclude that area
=
1 2
_
l r(t)
X lim ll.t--+O
=
� (t) � ut
l
! r(t) X �r(t) 2 �t
I = 2! llr(t)
X
·
II
r'(t) II (10�10) 1 2llcll · _
A(t) is swept out at a constant rate. This is Kepler's second •
law of planetary motion.
Ellipses
For the remainder of our work in this section, we will need an in-depth under standing of one particular planar curve, the ellipse. Fix two distinct points F1 and F2 in the xy-plane. Let
Y
_____,r--_
is
=
M
Letting
parallelogram
that
a
c denote half
of the distance between the two points. Suppose
is a fixed positive constant that is greater than
c.
The set of points P
(x, y)
with the property that the distance from P to F1 plus the distance from P to F2 equals
2a
is called an
ellipse
(see Figure
=
4).
Each of the points F1 and F2 is called a focus of the ellipse. Together they are
called the foci (plural of "focus") of the ellipse. The midpoint of line segment F1F2 _. Figure 4
llPF1ll
+
llPF2ll
=2a
center of the ellipse. The chord of the ellipse passing through the two major axis of the ellipse. The chord perpendicular to the major passing through the center is called the minor axis of the ellipse. These
is called the
foci is called the axis and
features are shown in Figure 5.
a>c>O
y
..--����-za����� Length of major axis
_. Figure 5
840
Chapter 10
Vector-Valued Functions
Let us derive the formula for an ellipse with major axis along the x-axis and center at the origin. Let c be a positive constant. We place the foci of the ellipse at points F1=(-c,0) and F2 =(c,0), symmetrically situated on the x-axis. Let a be a number that exceeds c. A point
(x, y) lies on the ellipse provided that
( (x- (- c )) 2 +(y-0)2 ) 1/2 + ((x- c) 2 + (y- 0)2 ) 1/2 distance to F1
or
=
2a,
distance to F2
( (x-(-c)) 2 +(y-0)2 ) 1/2 =2a- ((x- c) 2 +(y-0)2 ) 1/2 .
We square both sides and simplify to obtain
a2 - ex= a(x2 - 2cx +c2 +y2 )112. Squaring both sides again and simplifying gives
Because a > c > 0, it follows that each of the coefficients appearing in this equation is positive. Let us simplify matters by setting 1
( We then have b2x2
+a2y2 = a2b2
11 0 .5. )
or, equivalently,
x2 +y2 a2 b2 -
-
=
1
with a > b > 0. This last equation is the standard form for an ellipse with foci on the x-axis and center at the origin. By definition, the major axis will be a segment in the
y = 0 to find these), we see that the major axis is the segment connecting (-a, 0) to (a, 0). Similarly the minor axis is the segment connecting (0,-b) to (0,b) (see Figure 5).
x-axis. Because the x-intercepts of the ellipse are
(±a, 0)
(just set
If we were to repeat the preceding calculation with foci (0, -c) and (0, c) and
b = v'a2 - c 2 , then we would obtain the equation
In general, an ellipse with center
(h,k), major axis length 2a, minor axis length 2b,
and axes parallel to the coordinate axes has equation
(x-h)2 a2
(y-k)2 1 + b2 -
foci at (h±c,k)
-
or
(x- h)2 b2
(y-k)2 -1 , + a2 -
foci at (h,k±c)
(b2
+
c2
=
a2 ). (1 0. 5 .12 )
10.5 Applications of Vector-Valued Functions to Motion
841
Suppose that a> b>0. Show that the curve described by acos(t)i + b sin(t)j, 0 :st:s27f, is an ellipse. Where are the foci located? Solution If x acos(t) and y bsin(t), then x2/a2+ y2/b2 cos2(t) + sin2(t) 1. � EXAM P L E 3
r(t)
=
=
=
=
=
Therefore r(t) describes an ellipse. Although the half-distance c between the two foci does not explicitly appear in the equation x2/a2+ y2/b2 1, we can use the formula b (a2-c2)1/2to determine that c (a2-b2)1/2. Because a> b, the major axis of the ellipse lies along the x-axis. The foci are therefore located at (-v'a2-b2,0) and (v'a2-b2,0) . .,.. =
=
=
Let c denote the half-distance between the foci of an ellipse. Let a denote half the length of the major axis of the ellipse. The quantity e c/a is called the eccentir city of the ellipse. =
oo� Eccentricity Eccentricity Eccentricity 0.1
.A Figure 6
0.75
0.98
It is traditional to use the letter e for the eccentricity of an ellipse. In this context, the letter e has nothing to do with the base of the natural logarithm. Notice that the eccentricity e of an ellipse is a nonnegative number that is less than 1. The nearer that e is to0, the more the ellipse will look like a circle. The closer that e is to 1, the more the ellipse will look like a line segment (see Figure 6). � EXAM P L E 4 If the length of the major axis of an ellipse is double the length of its minor axis, then what is the eccentricity of the ellipse?
We suppose that 2a is the length of the major axis, and 2b is the length of the minor axis. We are given that b a/2. Therefore
Solution
=
Until now, we have described an ellipse by means of its two foci and the length 2a of its major axis. It is often convenient to have an alternative characterization of an ellipse. .,.. THEOREM 5
Let C be a closed curve in the plane. Then, C is an ellipse if and only if there is a real number e with0 < e < 1, a point F, and a line V such that C is the locus of all points P that satisfy ll PF II e IPVI. Here, IPFldenotes the distance between the points P and F, IPVI the distance of P to the line V. In other words, IPVI is the length of a perpendicular dropped from P to V. If C is the ellipse defined by the equation IPFI e IPVI, then =
=
·
a. The eccentricity of C is e. b. C lies on one side of V. c. F is a focus of C, the focus closest to V. d. The major axis of C is perpendicular to V.
·
842
Chapter 10
Vector-Valued Functions Proof. Rather than give a general proof, we will sketch the key algebraic procedure for establishing this theorem. Consider the ellipse x2/a2+y2/b2=1 with foci(-c,0) and F= (c, 0). The distance IPFI between F and a point P=(x,y) on the ellipse is given by
(
(x - c)2+(a2 - c2 ) 1 -
x2 a2
)
(a2 -xc)2 a2
Verify! =
_
la2 -xcl a
If we take the vertical line x=a2/c to be 'D, then we have IPFl=c
la2/c - xi a2 =ec-x =elP 'DI. a
I
I
The steps of this algebraic argument may be reversed to show that the locus of • points P such that I PF I=elPVI is an ellipse. If an ellipse is defined by the equation IPFI 'D is called a directrix for the ellipse.
=
elPVI then the line
� EXAM PL E 5 Suppose that F=(0, 0) and 'D is the line x=2. What is the Cartesian equation for the locus of points P=(x,y) for which IPFI=(1/2)IP'DI? What are the lengths of the major and minor axes? Where are the foci located? Solution If P=(x,y) and F=(0,0), then IPFI= Jx2+y2. The distance of the point (x,y) from the line x=2 is 12 -xl- The equation IPFI=(1/2)IPVI becomes
�
Jx2+y2= 12 -xl. Squaring each side results in the equation x2+y2=(2 -x)2/4. If we expand the right side of this last equation and bring the terms involving x to the left side, then we obtain 3x2/4+x+y2=1 or x2+ 4x/3+ 4y2/3=4/3. We now complete the square on the left side: (x2+ 4x/3+(2/3)2) + 4y2/3 4/3+(2/3)2 or (x+2/3)2+ 4y2/3=16/9 Finally, writing this last equation in standard form (10.5.12), we obtain =
(x+2/3)2 y2 =1 . + (4/3)2 (2/./3)2 This equation tells us that the center of the ellipse is at (-2/3,0), the length of the major axis is 2a 2(4/3) 8/3, and the length of the minor axis is 2b 2(2/./3) =
=
=
=
10.5 Applications of Vector-Valued Functions to Motion Directrix I I I x
=
21 Ix
843
4/../3. One focus is F (0, 0). The distance c between focus F (0, 0) and center (-2/3, 0) is 2/3. The second focus is also situated on the major axis a distance c from the center; it is therefore the point (-4/3, 0) (see Figure 7). <1111 =
=
I· I I I
DI
I
INSIGHT
natively, the equation IPFI = follows that
_. Figure 7
In Example 5, there are several other ways to calculate the value of
instance, we may use the formula c
V
= Ja2 - b2
to obtain
c
= Jl6/9 - 4/3= 2/3.
(1/2)IPVI tells us that the eccentricity
= e ·a= (1/2)(4/3)= 2/3.
� E X A M P L E 6 Let e and and
c
e
equals
c.
For
Alter
1/2. It
d be positive constants with e < 1. Suppose that U U and V by directed
are orthogonal direction vectors in space. Represent
line segments that have the origin 0 as their common initial point. These directed line segments determine a unique plane P. Figure 8a shows a curve C that lies in P. Suppose that, for every point P on C, the distance r of P to the origin 0 and the '----+
angle() that OP makes with
U are
related by the equation
r=
ed 1+ e cos(O) · ----
Show that C is an ellipse with one focus at 0 and eccentricity e.
p
( u, v) as the coordinates in P of the point whose position uU+ vV see Figure 8b). In these coordinates, elementary geometry shows that r = u2+ v2 and u = rcos(O). On cross-multiplication, the displayed equation becomes ../u2+ v2+ eu =ed. After isolating the radical and squaring, we Solution Let us write
r
_. Figure Sa
ed o �� ( O) =i�_+_e_c-s
vector is
obtain
u 2+ v 2 or
(ed -eu)2 e2d2- 2e2du+ e2u2
=
=
(1-e2)u2+ 2e2du+ v2 e2d2. =
Division by the nonzero coefficient of
u2 results
in the equation
e2d2 2e2d v2 u2+ u+ = . 1-e2 1-e2 1-e2 --
--
--
We complete the square by adding the square of one half of the coefficient of both sides of the equation:
(
_. Figure Sb
u 2+
(
))
(
)
2 2d 2 2d2 2d 2 2e2d u+ e + e + v = e 1-e2 1-e2 1-e2 1-e2 1-e2
--
This equation simplifies to
--
--
--
--
u to
844
Chapter 10
Vector-Valued Functions
which may be written in standard form as
e2d 2 1-e2 -.., + �-----d 2� 1 e2
(
u+
--
)
(� )
v2
(h)
2
- 1.
0 < 1- e2 < 1, we deduce that 1 - e2 < J1- e2. From that ed/(1-e2) > ed/J1-e2. Thus the major axis has
Because
this inequality, it
follows
direction U. The
distance between focus and center is
ed
(1- e2)
2
-
(J1ed- e2)
2 =
e2d e2
1-
·
Consequently the eccentricity of the ellipse is the quotient of this distance
e2d/(1- e2)
by half the length of the major axis, or
quotient is simply
Applications to Planetary Motion
e.
ed/(1- e2).
The value of this
<11111
In ancient times, it was believed that the planets traveled in circular orbits about Earth. The 16th century astronomer Copernicus (1473-1543) argued that the planets orbit the sun, but he believed that the paths of motion were circles. Early in the 17th century, Johannes Kepler (1571-1630), by studying many detailed pla netary observations of Tycho Brahe (1546 -1601), was able to devise three simple and elegant laws of planetary motion.
Kepler's Laws of Planetary Motion I. The orbit of each planet is an ellipse with the sun at one focus. II. The line segment from the center of the sun to the center of an orbiting planet sweeps out area at a constant rate. III. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its elliptical orbit, with the same constant of proportionality for any planet.
Kepler's First Law is that each planet travels in an ellipse. It turns out that the eccentricities of the ellipses that arise in the orbits of the planets are very small, so that the orbits are nearly circles, but they are definitely not circles. That is the importance of Kepler's First Law. Kepler's Second Law tells us that when the planet is at its aphelion (farthest from the sun), then it is traveling relatively slowly, whereas at its perihelion (nearest point to the sun), it is traveling relatively rapidly-Figure 9. Kepler's Third Law allows us to calculate the length of a year on any given planet from knowledge of the shape of its orbit. Let us see how to derive Kepler's Laws from Newton's inverse square law of gravitational attraction (10.5.9). To keep matters as simple as possible, we will assume that our solar system contains a fixed sun at the origin and just one planet. The problem of analyzing interactions of gravity among three or more bodies is incredibly complicated and is still not thoroughly understood.
10.5 Applications of Vector-Valued Functions to Motion
845
Slower Aphelion
Minimum Maximum value of r(t) value of r(t) ----------2a-------
• Figure 9
We denote the position of the planet at time t by r() t . We can write this position vector as
r() t = r(t)u() t where u() t is a unit vector pointing in the same direction as r(t) is a positive scalar representing the length of r() t . Because the gravitation force is a central force field, we know from Theorem 3 that the motion of the planet is contained in a plane. To be specific, equation (10.5.10) tells us that there is a vector c such that r() t Xr'() t = c. If Fis force, mis the mass of the planet, and a() t = r"(t) is the acceleration of the planet, then Newton's Second Law of Motion says that F= mr"(t). We can also use Newton's Law of Gravitation, formula (10.5.9), to express the force. By u( t ) ,
and
equating
the
two
different
mr"(t)=-(GMm/r(t)2)u() t
expressions
for
the
force,
we
conclude
that
or, equivalently,
GM r"(t)=- 2u(t). r(t) Therefore
(10.s.10) GM GM GM t ). t Xr'() t x (u() t )=-- u() t x (r(t) Xr'() t Xe = --2u() r"(t) Xc=--2u() r() t r(t) r(t) Now
r'(t) =
d t u'(t). t ) = r'() t u(t) + r() (r(t)u() dt
Therefore
r"(t) Xc = -
�� u(t) X (u()t X (r'(t)u()t
According to formula be written as
(9.5.10),
+
)
r(t)u'(t)) = -GMu(t) X (u(t) Xu'() t ).
the vector triple product in equation
(10.5.13)
(10.5.13)
may
846
Chapter 10
Vector-Valued Functions
()
( ) · ut ( ) · u' t ( ) - (ut ( ) ) u' t ( ) ) ut ( ). ( ) ) = (ut
( ()
(10.5.14)
ut X ut X u' t
Because ut ()
· u' t ( ) = 0 and ut ( ) = 1, equation (10.5.14) simplifies to ( ) · ut ()
( ()
( ))
ut X ut X u ' t
= -u' t ( ).
If we substitute this result into equation (10.5.13), then we obtain
( ) c =GM u' t ( ).
r" t X
By integrating each side of this last equation, we see that
( ) c= GMu(t) + h
r' t X
for some (constant) vector h. Taking the dot product of each side with results in
( ) ( ( ) c)
rt · r' t X
=GM rt ()
()
()
· ut + r t ·
h = GM r (t)
()
+ rt ·
h =GM r(t)
+
()
rt
r(t) ll h ll cos(O(t)),
where O(t) is the angle between r ( t) and h. After we set e = ll hll/(GM), our equation becomes
( ) ( ( ) c)
rt · r' t X
)
(
=GMr(t) 1 + ecos(O(t) ) .
(10.5.15)
But, according to formula (9.4.5), we may write the triple scalar product on the left side of equation (10.5.15) as ( ) · (r' t ( ) Xe rt
(9.4.s)
) =
( )) (rt ( ) Xr' t
(10.s.10) ·c = c·c= IIcII2 .
It follows that
(10.5.16) Observe that the left side of equation (10.5.16) is positive. We conclude that e < 1 and
r ( t) =
2 llcll /(GM) . 1 + e cos(O(t))
Because Example 6 shows that this equation describes an ellipse, the derivation of Kepler's First Law is complete. Kepler's Second Law about the area function A t ( ) is a special case of Theorem 4. The proof of Kepler's Third Law begins with an observation from Figure 9. The length 2a of the major axis of the elliptical orbit is equal to the maximum value of r( t) plus the minimum value of r(t) . From the equation for the ellipse, we see that these occur respectively when cos(O(t)) is -1 and when cos(O(t)) is 1. Thus
c 2 (GM) Za = ll ll / 1-e
+
llcll2/(GM) 1+e
=
2llcll2
GM(1 - e2 )
"
10.5 Applications of Vector-Valued Functions to Motion
84 7
We conclude that
llcll
=
JaGM(l - e2).
Now recall from the proof of Theorem 4 that the area function A satisfies A'(t) llcll/2. Then, by antidifferentiating, we find that A(t) llcllt/2. (The con stant of integration is 0 because A(O) 0.). It follows that =
=
=
A(t)
=
� VaGM(l - e2).
Let T be the time it takes to sweep out one orbit. In other words, T is the year for the particular planet. Because the area inside an ellipse with major axis 2a and eccentricity is e2, we have
e 7ra2v'l 7ra2v'l - e2
=
A(T)
=
� JaGM(l - e2).
Solving for T, we obtain
T
=
--2:!!_ _a3/2 V'GM
or
y2 -- 4� a3 GM. This is Kepler's Third Law.
Q UIC
K
Q UIZ
1. For two points P and Q on the trajectory of a particle, the curvature at Q is twice that at P. If the speed of the particle at Q is three times its speed at P, then by what factor is its normal component of acceleration at Q greater than it is at P? 2. At a point r(to) in its trajectory, the acceleration of a particle is r"(to) (6, 6, 7), and the normal component N (t0 ) of its acceleration is 4v'6. What is the rate of change of the particle's speed? 3. The major axis of an ellipse in the xy-plane is the line segment with endpoints ( 1 -2) and ( -1, 8). The eccentricity of the ellipse is 4/5. What is the right endpoint of the minor axis? 4. In a solar system, Aurie's year is 64 times that of Stavro. By what factor is Aurie's major axis greater than that of Stavro?
a
-
,
Answers
1. 18
2. 5
3. (2,3)
4. 16
=
848
Chapter 10
Vector-Valued Functions
EXERCISES Problems for Practice
Further Theory and Practice
is given. Calculate
r of a particle r'(t), r"(t), T(t), N(t), �(t), and the tan normal components, ar and aN, of acceleration
3L A particle moves at constant speed with position vector r. If r" fo 0, then what is the relationship of r" to r'?
gential and
32. A particle moves with position vector
In each of Exercises 1-8, the position vector
what can be said about the speed
for the planar motion.
1. 2. 3. 4. 5. 6. 7. 8.
r(t) r(t) r(t) r(t) r(t) r(t) r(t) r(t)
=
=
=
=
=
=
=
=
t2i - t3j (t - t2)i + (t + t2)j In(t)i + t112j (t - sin(t))i + (1 - cos(t))j 5 cos(3t + 7r/6)i+ 5sin(3t + 7r/6)j t312i + t112j cos3(t)i + sin3(t)j
r of a par ticle is given. Calculate r'(t), r"(t), T(t), N(t), �(t), and the tangential and normal components, ar and aN, of acceleration for the motion in space. 10. 11. 12. 13. 14. 15. 16.
r(t) r(t) r(t) r(t) r(t) r(t) r(t) r(t)
=
=
=
=
=
=
=
=
ti+ t2j + t3k (t - t2)i+ (t + t2)j + t2k (1 + t)3/2i + (1 - t)3/2j + tk t112i + t-1/ 2j + t-1k e1i+ e-1j + V'2tk e1i+ te'j +Uk cos(t)i + sin(t)j + t2k ln(ltl)i + r1j + tk
r of a a(t) r"(t). Calculate the tangen tial and normal components, ar and aN, of the decomposition a(t) arT(t) + aNN(t) without calculating T(t) and N(t). In each of Exercises 17-20, the position vector
particle is given. Calculate
=
=
r(t) r(t) r(t) 20. r(t) 17.
18. 19.
=
=
=
=
(t + 2)i + t2j + 2tk (t2 + l)i + (t2 - l)j + t2k e1i+ e-tj + tk cos(t)i+ sin(t)j + ln(cos(t))k
In each of Exercises 21-30, determine the Cartesian equation of the ellipse that is described.
(0, 0), major axis 6, minor axis 4, foci on y-axis foci (0, 3) and (0, -3), major axis 10 foci ( 4,0) and (-4, 0), minor axis 6 foci (5,0) and (-5,0), eccentricity 5/13 foci (1,14) and (1, -10), eccentricity 12/13 foci (2,4), center (2,1), minor axis 8 center (1,2), focus (1,10); eccentricity 0.8 center (1,2), focus (-5,2), minor axis 8 directrix x 6, focus nearest the directrix (2, 0), eccen tricity 1/2 directrix y 4, focus nearest the directrix (1, 1), eccen tricity 3/5
21. center 22. 23. 24. 25. 26. 27. 28. 29. 30.
=
=
·
=
0, then
33. Prove Huygens's law of centripetal force for uniform
ti - t2k
In each of Exercises 9-16, the position vector
9.
r. If r' r"
Ml ?
m moves with constant v along a circle of radius r, then the centripetal force F acting on the body has magnitude IFI mv2/r. 34. A car weighing 3,000 pounds races at a constant speed of 100 mph around a circular track that has radius 1 mi. circular motion: If a body of mass speed
=
Calculate the centripetal force, in pounds, needed to hold the car on the track.
35. A man-made satellite orbits Earth in an elliptical path with major axis 1850 km. The orbit of Earth's moon has semi major
axis 354, 400 km and period 27.322 days.
Calculate
the time for one orbit of the man-made satellite.
36. It is known that the eccentricity of Earth's orbit about
0.0167,
the sun is about
and the semi-minor axis of the
elliptical orbit is about 92,
943, 235 mi. Calculate the
semi
major axis. Of course, it is also known that the length of time for one orbit is one Earth year (by definition). Using Kepler's Third Law and this other information, calculate
GM where M is the sun's mass. 37. Earth's mass is known to be 5.976X1027 grams, and the gravitational constant G equals
6.637 X10-8 cm3/(g
·
sec2).
If a man-made satellite orbits Earth, with each orbit taking
15 h,
then what is the length of the major axis of the
elliptical orbit?
38. The sun's mass is
2 X 1033
grams, and the gravitational
constant is given in Exercise
37.
Use Kepler's Third Law
to calculate the length, in kilometers, of the semi-major axis of Earth's orbit around the sun.
39. A man-made satellite orbits Earth. It is known that, at a certain moment
in its orbit, 4,310 mi,
Earth's center is
the satellite's distance from and its speed is
900
mi/h.
Moreover, astronomical observations determine that, at that moment, the angle between the position vector and the velocity vector for this satellite is numerical value for
";.
7r/3.
Give a
2a 2b is 1!"ab. Show that the area can also be expressed as 7ra2v'l - e2, where e is the eccentricity of the
40. Show that the area inside an ellipse with major axis and minor axis
ellipse. Earth's semi-major axis is
365.256
149, 597, 887
km, and its year is
days. Use this information as needed in Exercises
41-44.
0.2056. Its year is 87.97 Earth days. Calculate Mercury's perihelion distance
4L The eccentricity of Mercury's orbit is to the sun.
10.5 Applications of Vector-Valued Functions to Motion 42. Pluto's orbit of the sun lasts 248.08 Earth years. Its perihelion distance to the sun is 4,436,824,613 km.
GM cos(8) + b j= c
Determine the eccentricity of the orbit.
43. The eccentricity of Jupiter's orbit is 0.04839. At its farthest, Jupiter is 816,081, 455 km from the sun. Deter mine the period of Jupiter's orbit.
44. Saturn's nearest and farthest distances to the sun are 1,349, 467, 375 km and 1,503, 983,449 km, respectively. Determine the duration of its orbit.
for some constants
47. Substitute x= rcos(9) and y rsin(9) into the equations of Exercise 46. Eliminate r from the resulting equations =
to obtain
r. iJ = GM - asin(O) + b cos (O). c
planets follow elliptic orbits. We assume that the motion is motion and that thex axis passes through the aphelion of the
Deduce that
planet, as in Figure 10. Let r= r(t) denote the distance of the planet to the sun, and let () be the angle between the
r=
position vector of the planet and the positive x axis. This
anomaly in ast�onomy. We assume C where C is a constant, and (} is the derivative of
angle is called the true that
,Z(}
=
the true anomaly with respect to time-as we will see in
M
and
m
gravitational constant.
.
ed ----=� r= -1 + ecos ( 8)
denote the masses of the sun and planet, respec
G denote the
c
GM/C- asin{O) + bcos ( O)
48. Use the Addition Formula for the cosine to show that
Chapter 12, this equation expresses Kepler's second law. Let tively, and let
a and b. Deduce that the hodograph of
planetary motion is a circle.
Exercises 45-50 outline an alternative demonstration that planar. Suppose that the sun is at the origin of the plane of
849
for some constants e and
d.
49. Show that y
di -= GMx r3 dt and
x
Aphelion (Planet is farthest from sun.) £
Figure
d"
10
Then, show that
!!_ (i:)2 = _GM !!_x2 r3 dt dt
45. After resolving gravitational acceleration into thex �d y directions, use the Chain Rule and the equation r28 = C to show that
and
di -
d()
=
GM
--cos(8)
c
and
50. Use the equations of the preceding exercise to show that
dy d()
= _
there is a constant E such that
GM sin(O). c
46. The hodograph of a motion tr-+(x(t),y(t)} is the para meterized velocity curve
tr+ (i(t),j(t)).
Integrate the
equations of the preceding exercise to show that
i= - G
�
and
GM
-1'.. =--y r3 . dt
sin(8) +
a
1 ·2+ y.2)- GMm +E. - m(x - -2 r (In Chapter 13, we will learn that this equation expresses the conservation of total energy.)
SL Suppose that the acceleration a(t) of a smooth trajectory is differentiable. Show that if l •(to)ll = aN(to) =A with tangent to each
j
tr-+ la(t)ll and tr-+aN(t) other at the point (to, A).
AfO, then the graphs of
are
850
t
r(t) (t)
Chapter 10 Vector-Value«1 Funns t
(t)
Calculator/Computer Exercises
In faeh tft-E�etrcii§es 8k-57, the ,ttajectqry 2>-> of a moving particle is given. Plot r> Ila 11, >->aT , and >--> aN in the specified viewing window W before answering the questions about the acceleration. 52. rq
t"
(
)t
+
,
W
[O ]X[O 12. ] At what
p�th.t on the curve defined by' are the normal and tan gential components of acc�leration equal? At what point does the normal component of acceleration have a local minimum? 53. r =t3i+ej 1 + 2k, W=[-1 l]X[-1 5] At what point ontthe curve cli:fin_f t Hi 6] 2 t + 3t Over what tfme intek'iils is the speed liecreasing? At what points do the graphs of normal acceleration and magni tude of acceleration touch? What condition must be s� fie� at su a port of nta$? 1 . ( ) (cos 55. r() cos( ) sin 5 )) ' w [ /4 l x [-30,35] On what interval is the particle slowing down? On what interval are both components of acceleration decreasing? t +cos t))i + t - sin t))j tk,W [O, 7r] x [-3/4,1] 56. r t On what interval is tangential acceleration greater than
t
=
r
57.
�
58.
59.
.
f
60.
=
Summary of Key Topics
in
nopi� accPglewtion? Fin,d � po!rts ;q �hich,)ariJ has� loca.\ extremurli. Show tllat d� i.as a 'lOcal extrem'um at eacll of these points. Explain why this behavior happens for r but cannot be expected to happen for other trajectories. ln cos )) w r() ew in )}i + exp cos )) [O 7]r x·[-7 /1 3] W�f!� do � thll abpolute maximum of normal acceleration i?ccur?t°Where does the absolute maximum value A of the magnitude of acceleration occur? (The quantities an aN, and A are required for Exercise 58.) Let aT ) , aN t, and A be as in Exerci 57. Plot the planar curve >--> '1T aN )j < < 271".�To your plot, acJ9 JP.e ,se�j'.1f.ct't; pf -!aqius t 4 that is cent�red at the _ an'd lies m fhe upper hal? p¥tne. wlam the rela 01\'gm ) j tionship be1f"1!en:f11e two plotted curves. J Calculate the 'tangential component aT ) and normal component aN(t of acceleration for the space curve r ) 1/ 1 + 2 + / 1 t2 j + 3 (the curve ofExercise54). Plot the planar curve t>->aT(t i aN(t )ft the viewing window [- 0 1 x [l.9,2.7]. Explain how' the loop that yoy s� · 11fins(fb the-iPnticw i moven:ient. Add vertical t plqt, ano explam how they ar1 es �) .the t�g�n lE J, OOtame ( Calculate the tange�Ja1 cokJonent a� ) and dormal component aN of acceleration for the space curve + /(1 + 2)j + ) . Plot the planar curve r() >--> aT )i+ aN 3/8 < < 9/8. Find the two values o and t1 for which aT to =aT t1 and aN to) =aN t1 .
Chapter
10
r A g vector-vatued fundlU6n (in space) of Pa real variallle has the form t r1 t i + r2 t j r3(t k. For each t, we think of r t) as t e pos t on vector ofa po nt n space. We represent r t) by a directed line segment w t t e origin as ts initial point. As runs through the domain of , the endpoints of the directed line se ments re resenting form a curve C in s ace. When t re resents time, we can hink of the curve C as the path of a partfcf� moving through space. We say that is a parameterization of C. If L is a vector, then we say that lim
L llr(t)
t-+C
if, for any
c:
>1'0, there is a 8 > ot such that 0 < I - cl < 8 im�6.es ( )
We say that
is continuous at
=
- L < r::
c if c is in the domain of r and limr t t-+C
=
rc.
r
at
We define the derivative of p r
=
1
t>11!1 �
r( t c to be
t)
Summary r( )
( c+D. - c)
of Key Topics
851
,
p
rovided that the limit exists A vector-valued function is continuous at c if and only if all component functions rj are continuous at c. It is differentiable at c if and only if all COJfl onent functions rj are differentiable at c. ������
Properties
of Limits
(Section 10.1)
�
����-1-���������������-
Limits of (vector-valied funttions satisfy(the following familiar properties: If f(t) and g(t) ate tector-'{alued �nctions and if lim1-+ef and lim1-+eg exist, the
a. lim1-+e f±g)(t)= lim1-+ef t) :±: limt-+cg t) b. limp..c f g) (t)y= lim1-+Cf t)) (lim1-+eg t)) c. li:rri;-+e Mt))= .A lim1-+ef t)) for any constant .A. n . It foll&ws that the set of continuous fun\!tions is closed under addition and under multi lication b scalars I f and g are vector-valued functions that are continuous at c, then the scalar valued functio f g, and the v��vilUed U.Unt:tiEfn f X�, will also be continuous at c. I f and g are differentiable at c, the f g and fXg will also be differentiable at c. We have the product rules ·
·
·
f ) gt)(
f't) ) . g(t)
f )(t� ( ( ) .
and (fXg'
d
=(f'( Xg )+(f
Xg',t)
There is also a sum-differen�e V\00 for '4,i}f�,Cntiati}>r(tind a rule for multiplication by scalars. If ¢ s a scalar-valued differentiable function then ¢f makes sense is ifferentiable, and ¢f ¢ t f + ¢(t f' (r )'(t) '(t) r (t) . Finally, if 't/Jis a differentiable scalar valued function and if r r 'ljJ is differentiable and
o
't/Jmakes sense, then
o
r
r (t) Velocity and Acceleration (Section 10_2)
0'1/J
=
't/J
'('t/J)
r (t)
If is a differentiable, vector-valued'f�ori6� a real variable, and if we think of as describing the motion of a body through space, then ' is velocity at time t, and is accelera tion at time t. Speed is a scalar-valued quantity and is given by "
llr'
r
r'(t) y 852
Chapter 10
"t( )
"t( )
't(
t
r
Vector-Valued Functions
If j and r exist, if r iis cpntinuous and if 'h ) is Never the 0 �ctor then we sa that isii_a smooth parameterizatioi of ;e 'i111Ye it oefines.
Tangent Vectors and Arc Length (Section 10.3)
If r s a smooth parameter zat on of a dlfrt,� C, t en t e vector r' ) can be inter prete as t e tangent vector to C at t e po nt t( . The unit tangent vector is then () ( 1 00 r ). T r(t ) - -t-1 The tangent line to C at the point r t is the line passing through the point rt) with direction T . 11 11 If f-+ , a :s :s b is a smooth parab�forization of a curve C then the arc length of C is f
s
(s
11s
' r
i T dT.
a
( ) ) The arc lengtli oft: does not depend on the choic& of parameterization. A vedl:�r) �unction f-+ p ) , ms :t L !;i. saidPto paran-iete�� a (chrvl C l'fitn respe& to arc length i pis a parameterization of C such that, for all s in the domain of parameterization, the arc length from p0 to p(s equals s. A smooth para meterization f-+ s( O:ss:sL o�� is the l rP'f �� th parameterization of C if and (u-1 s ) , only if II 's 1 for all s. Ot erw se, t e re' arameter zat 0 ps wher =
is the arc �gth parameterization.
------ ·'""�) IH-)i� l+� l --------=
curvature
(Section 10.4)
If c is a curve that is parameterized according to arc length by p then the curvature of C at p( is defided to b
s
T's (s) The vector p"(s) s always perpendicJtar normal to pat t to be =
1 (s) 'P'��) 1".s).
N
We have
ps . (s). t�I T(s). We define the principal unit
/1
(s)
( )t
() p"
e
(s)
=
,.,,
) i
(s)
N
f ,.,, =F 0. The The radius of curvature at p is defined to b p 1/ s( osculating plane at ps is the plane through ps hat contains the vectors Tand N.
Review Exercises
The
853
osculating circle (or circle of curvature) at the point p(s) is the unique circle p(s), center p(s) + p(s)N(s), and lying in the osculating plane. If r is any smooth parameterization of a curve, then the curvature at r(t) is
with radius given by
�(t)
=
llr'(t) X r"(t) II llr'(t) f
For a planar curve that is parameterized by r(t)
r(t)
=
x(t)i + y(t)j, then the curvature at
is given by
Kr(t) -
_
lx'(t)y"(t) - y'(t)x"(t)I (x'(t)z + y'(t)2)3/2
If the planar curve is the graph of the equation
(x,f(x))
Components of
If
r(t)
=
f(x),
then the curvature at
is
11:(x)
Tangential and Normal
y
=
(i
+
lf"(x) I
(f'(x))2
)
3
I
. 2
represents a motion through space, then the acceleration vector
Acceleration
r"(t)
(Section 10.5) where
can be
v(t)
=
llr'(t) II
=
(:t v(t)) T(t)
+
2 v(t) �(t)N(t)
is the speed of the motion.
Kepler's Three Laws of
A planet travels in an elliptical orbit with the sun at one focus. The rate at which
Planetary Motion
area is swept out by the ray from the sun to the moving planet is constant. The
(Section 10.5)
square of the period of revolution is proportional to the cube of the major axis of the elliptical orbit, with constant of proportionality independent of the particular planet.
Review Exercises for Chapter 1 0 In Exercises 1-3, sketch the space curve defined by the given vector-valued function. 1. 2.
r"(t)
decomposed as
r(t) =ti+ 3tj + sin(t)k r(t)=t2i+tj+2k 3. r(t)=ti+ sin(t)j +cos(t)k
0 ,st ,s27r 0'5t'52 0'5t'547r
In Exercises 4-6, a vector-valued function r and a point c are given. Calculate the limit of r(t) as t-+c.
4.
5. 6.
r(t) = sin(t) i - cos(t) j + tk r(t)=(et, t2, In(t - 1))
c = 7r c =2
r(t) =
c =2
In 7.
8.
i�! i - 1 �t2j + t2k
Exercises
7-9,
calculate r'(t).
r(t) = t113i - 3t2j + 2tk r(t)=(sin(t), tan(t), et)
+t
t+
854
9. r( )
Chapter ti 0
1
-1.
+ 3 +.t Z 3/ k --lJ
t t In Exercises 10-12, calculate the indicated indefinite integral. t + +t 10 J ( 2i +cps( � -,sin( )k)dt + 11. J (ft�_e¥l- e+lt}dt t + 12. J (vi - v'ij vk)dt t= t +t ' Jn£iceJCises 13-15, find the antiderivative G( ) of g( ) that satisfies th� additional given condition.
13. 14
g(tt) 2i-tt4j 6k g(� '='vi'- tlj tk j L3 tk g( � =y"1i =
••
15.
G(l) 2i -3 j k G(O) 4i 7j- k G(2)=2i - j 4k
In Exercises 16-18, let f( ) e3ti- cos( )j 2k r,o( � / , and >. 4. In each problem, calculate the deri= + t vative of r t t t+ + t t+ 16. r( ) -_f( r,o( ) ) t + t 11 rO r,o( )r( 18. r( ) >.f( ) + 3>.r,o( )f( ) a. In Exercises 19-21, calculate r" ( ) ·
/
19. r( j :e�i-e3fj :{ an( )k 20. r( I =1n( t �j ----rtj - vk 21. r( ) arcsin( )i- arccos( )j tan( )k In Exercises 22-24, calculate for each motion r the cor "
responding velocity v, speed v, and acceleration
22 r( � =3i- v'.13 -F1t4k4 23. r( t=e21i-e4tj e-31k+t 24. r( ) =ln( )1t e1;-.i -±5 5k+ In each of Exercises 25-27, a vector-valued function r and
a point Po are given. Determine parametric equations for the line that is tangent at Po to the curve described by r
25. r( ) 26. r( � 27. r(t �
k Po (3,-2,1) 3 2i- 2 � + t = 3k Po (-1,2/3,8) j i- s(?rt) i Ih(l_ t)j-sin(?r )k P0=(-1,ln(2),0) t= t + t In each of Exercises 28-30, a vector-valued function r and a point Po are given. Determine symmetric equations for the line that is tangent at Po to the curve described by r = at
28. r() 29. r() 30. r( )
:� lo
+
Vedl:or-Valued Functions
4/
Po
(2,-1/2,1)
Po=(16,-8,4) Po (1,1,-1)
In Exercises 31-33, a body is falling near the Earth's surface with constant acceleration due to gravity given by ( )=-32k. The initial height r0 r(O) and initial velocity
+
+
v0 v(O) are given in each problem. Using antidifferentation, calculate an exP,licit formula for r( ) T
31. ro i-2k 32. ro � 3i - 4t 2k 33. rot i- j 2Ii
vo -i 4j - k vo=i -1 vo j t= t t +t In each of Exercises 34-36, a vector-valued function r is given, as well as a point Po that lies on the curve that r parameterizes. At, Po calculate the tangent vector r', the unit tangent vector , and the tangent line.
34.
35.
36.
r( � ===(et te-t)i -f ( e+--e-t)j e21k rO =-t2i +t 4j - 3k t +t r( � sin(?r )i- cos(?r )j k
Po t(2, 0,1) Po t(-4,16,-8) Po t (O,-1,0)
In Exercises 37-39, calculate the arc length of that portion of the curve that is parametrized by the given function r over the specified interval.
0 � � ./2ir cos( 2)i- 2j sin(t 2)k 38. r( ) cos(S )i - sin(S )j ?r � � 2?r 512k 0 � � ln(2) 39. r( )=e2ti- 2v'2tj +e-2tk t= t t + t In yach of :qxercises 4P-.12t , a vector-valued function r is given, tas well as a �oj nt Po that lies on the curve that r parameterizes. Give parametric equations for the normal line through Po. This is the line that passes through Po that ha� the principal uriit normal vector to the curve at Po for its direction.
37. r( )
40.
r( )
3i- 2j
Po=(27, -9, 9)
3k
41. r( )= sin(3 )i - cos(3 )j 3 k Po=(0,1,3?r) Po=(1� -1, 0) 42. r( �=e2ti ...1e°"*j t k t= t + + t t= In �a� of :f xerc� es 4i3-45, a function rt aQ,d a value
o are given. At r( o), find parametric equations for the tangent and normal lines to the curve described by r. (Refer to the inst ructions to Exercises 40- 42 for the delniticn of the normal Unc.t). 0 0 43. r( '=ti - 2j 3 k t + t 44. r( §=cos(3 'i sin(3t'j +tah(3 )kt 0 ?r o 0 45. r( � eti e-tj � k In each of Exercises 46-48, a parametrized space curve is given. Calculate the moving frame (
N
B) at the point
0
46.
47. 48.
r( ' -=-sec(2 )f - tan�2 )j 4 k =to rO =coi(2 )i -sin(2 )j 2 k = o r( )=e21i-e-21j (e2t -e-21)k t o
2?r ?r 0
In each of Exercises 49-51, a function r is given, as well as a point Po on the curve parameterized by r. Find a Cartesian equation for the normal plane to the curve at Po
49. r( ) 50. r( )
3j - 3 4j - 4 sk ..; i- -1/3j tl/ k
Po Po
(1,-3,-4) (1,-1,1)
Review Exercises
51.
r(t)=cos(t)i - sin(t)j + tk Po=(-1,0,7r) In each of Exercises 52-54, verify that the given para
metrization p is an arc length parametrization. Then calculate T(s), N(s), and �(s).
p(s)=3i - sin(s)j - cos(s)k p(s)=3cos(s)/5i - 4 cos(s)/5j + sin(s)k 54. p(s)= (3 - sin(s))i + (6 + cos(s))j
52.
53.
In each of Exercises 55-57, a vector-valued function
r is
to in its domain. For the space curve r, calculate the radius of curvature and the center of curvature at r(to). given, as well as a value
parameterized by
56.
r(t)=e-1i - e'j + v'2tk r(t)=t2i - tj + tk
57.
r(t)=(1 - 2t)i + t2j - tk
55.
to=0 to=1 t0 =0
In each of Exercises 58-60, calculate the curvature at each
66.
r(t)=t113i - t2j + 3tk
Po=(1,0,27r) Po= (2, -64, 24)
r is to in its domain. For the planar motion that r describes, calculate r'(to), r"(to), T(to), N(to), Kr(to), and the tangential and normal components, ar and aN, of acceleration at r(to). In each of Exercises 67-69, a vector-valued function
given, as well as a point
67.
to=0
68.
to=1 to=0
r(t)=ti - t2j r(t)=ln(2t + l)i + y'tj 69. r(t)=e'i + e-'j
r is to in its domain. For the spatial motion that r describes, calculate r'(to), r"(to), T(to), N(to), Kr(to), and the tangential and normal components, ar and aN, of acceleration at r(to). In each of Exercises 70-72, a vector-valued function
r(t)=t2i - t3j + tk
to=0
r(t)=e1i - e---tj + v'2tk 72. r(t)=ti - ln(t)j + t-1k
to=0
70. 71.
f(x)=x4 59. f(x)=tr - e-x 60. f(x)=3.x2 - 4x 58.
to=1
r(t) of a a(t)=r"(t). Calculate the tangential and normal components, ar and aN, of the decomposition a(t) =arT(t) + aNN(t) without calculating T(t) and N(t). In each of Exercises 73-75, the parametrization
motion in space is given. Calculate
In each of Exercises 61-63, a vector-valued function
r is
given. Calculate the curvature at r(t) of the planar curve that r parameterizes.
62.
r(t)=cos(t2)i - sin(t2)j r(t)=e-1i + e1j
63.
r(t)=t312i - t112j
r(t)=(t2 - t)i + (t2 + t)j - tk 74. r(t)=e-21i + e2'j + 3 tk 75. r(t)=(t - l)i + (t + l)j + t2k 73.
r is Po that lies on the space curve
In each of Exercises 64-66, a vector-valued function given, as well as a point parameterized by
r(t)=cos(2t)i + sin(2t)j + 2tk
given, as well as a point
point of the graph of y=f(x).
61.
65.
855
r. Calculate a Cartesian equation of the Po.
osculating plane at
Po=(0,-1,1)
In each of Exercises 76-78, write the Cartesian equations of the ellipse that is described in words.
(4,0) and ( -4,0), major axis 6 (0,0), major axis 8, minor axis 6, foci on the x-axis 78. foci (2, 6), (2, 10), eccentricity 3/4
76. foci at 77. center
Tycho Brahe
awarded him a pension so that he could build and
On the night of November 11, 1572, Tycho Brahe, a young Danish nobleman with a hobbyist's interest in astronomy, cast his eyes
toward the constellation
Cassiopeia. To his astonishment, he sighted a new star, one that was much brighter than any other. Tycho was well aware that the appearance of a new star in the firmament, a nova as he called it, was extraordinary. He characterized his discovery as "the greatest wonder that has ever shown itself in the whole of nature since the beginning of the world." At the very least, Tycho understood that the spectacle he witnessed contra dicted the ancient astronomy of Aristotle. We now know that Tycho observed a cataclysmic explosion that is called a type I supernova. This celes tial event is the final evolutionary stage of a white dwarf star. It occurs when the core of the star collapses, releasing an enormous quantity of energy that blows
operate an observatory there. Tycho took possession of the island in 1576. For the next twenty-one years, Tycho watched and recorded the planets. Tycho's fortune took a tum for the worse when his royal
patron died. Frederick's
son
and successor,
Christian IV, was not inclined to continue the pension that his father had granted Tycho. Nor would he agree to honor his father's pledge to allow Tycho's children to inherit Hven. (Because Tycho had married a com moner, his children did not have automatic rights of inheritance.) Tycho abandoned Hven for Copenhagen, where he started over. When Christian had him dis mantle his new observatory because it obstructed the view from the royal palace, Tycho departed Denmark for good.
Johannes Kepler
the star to bits. In our galaxy, such supernovae occur
When Tycho began his exile in 1597, Johannes Kepler
very infrequently---only three have ever been observed.
was twenty-six years old and teaching in Graz. His
The first was recorded in 1054 A.D. by astronomers in
teaching duties were light enough that he had time to
China, Korea, and Japan. Tycho observed the second.
ponder the mechanics of the solar system. He published
Tycho Brahe was born in his family's ancestral seat,
his first book on the planets, the Mysterium Cosmo
Knutstorp Castle, in 1546. At the age of two he was
graphicum, in 1596. Though this work is more flight-of
abducted by a paternal uncle and childless aunt, who
fancy than science, it reveals Kepler's early interest in
brought him up. Tycho studied at the University of
the three mathematical questions that he would even
Copenhagen and at several renowned universities in
tually answer with his laws of planetary motion.
the Germanic territories. It was at Basel that he learned
By nature, Kepler was more a mathematician than a
the new theories of Copernicus, which he did not
stargazer. He did not have the instruments of Tycho, and
entirely accept. While he was a student in Rostok,
his eyesight was poor. Tycho, to whom Kepler sent a
Tycho engaged in a quarrel that escalated into a sword
copy of his book, commented that the observations of
fight. During the battle Tycho received a blow that left
Copernicus on which Kepler relied were not sufficiently
a diagonal slash across his forehead and hacked off a
accurate to be the basis of the conclusions Kepler
substantial chunk of his nose. (A nasal prosthetic can
reached. Moreover, Tycho did not think highly of
be seen in several subsequent portraits.)
Kepler's inclination to theorize. Tycho chided Kepler that
After his return to the Kingdom of Denmark,
the force behind the motion of the planets could only be
Tycho's social standing afforded him the leisure and
established "a posteriori, after the motions have been
wealth to pursue his interest in astronomy. In time, the
definitely established, and not a priori as you would do."
King of Denmark, Frederick II, offered Tycho a choice
Kepler was frustrated. "I did not wish to be discouraged,
of several fiefdoms. Tycho demurred. He wrote to a
but to be taught." He was aware that in order to make
friend "I am displeased with society here. Among
progress, he needed Tycho's planetary data, data that
people of my own class I waste much time." While
Tycho did not share with other scholars. The problem was
staying in the castle made famous by Shakespeare's
how to gain access. Referring to Tycho's wealth of data,
Hamlet,
Kepler wrote "Like most rich men Tycho does not know
Frederick
conceived
a
way
Denmark's leading scholar. From
of
retaining
his window, he
spotted the island of Hven. He granted it to Tycho and
856
how to make proper use of his riches. Therefore one must take pains to wring his treasures from him."
Genesis & Development
857
As it happened, fate brought the two men together.
no tools such as logarithms to ease the burden of
At the same time roy al hostility was driving T ycho from
numerical computation. After four years of intense
his native Denmark, religious persecution was driving
mathematical labor, Kepler discovered the first two of
Kepler from his post in Graz. Tycho made his way to
the planetary laws that now carry his name. He pub lished them in his Astronomia Nova of 1609.
Prague, the capital of the Holy Roman Empire, in 1599. Kepler arrived shortly afterwards. Although Tycho
For the most part, Kepler left remarkably candid
maintained a tight grip on his secrets at first, he even tually allowed Kepler to work on the orbit of Mars. Its
accounts of the steps that precipitated his discoveries.
eccentricity had caused great difficulties for all the circular motion theories that Ptolemy, Copernicus, and Tycho had put forth. In 1601, Kepler became Tycho's salaried assistant. Only a few days later, T ycho attended a banquet where, according to Kepler's account, he "drank a little over
In the case of his third law, however, Kepler was unu sually silent. He did record the date, March 8, 1618, when the idea first came to him. His initial efforts to confirm his theory were not successful, but on May 15, 1618, he was able to write "A fresh assault overcame the darkness of mY reason . . . I feel carried away and possessed by an unutterable rapture over the divine
generously and experienced pressure on his bladder.
spectacle of the heavenly harmony . . . I write a book
He felt less concern for the state of his health than for
for the present time, or for posterity. It is all the same
etiquette," which required guests to remain seated at the dinner table. On the basis of this document, Tycho's
to me. It may wait a hundred years for its readers, as
death has been traditionally attributed to a burst bladder. Other reports, originating soon after T ycho's
onlooker."
death, raised the suspicion of heavy metal poisoning. The modern verdict is that T ycho died of kidney failure brought on by enlargement of the prostate.
God
has
also
waited
six
thousand
years
for
an
Such moments of elation were brief. As one sci entist has remarked, Kepler's standard biography has passages that can bring its readers to tears. Poverty,
Tycho was laid to rest in the famous T:Yn Church of Prague. During the religious turmoil of the 1620s, many
religious persecution, war, smallpox, typhus, plague Kepler's life was an unrelenting struggle filled with hardship and sorrow. In his 58th Year, he had a pre
graves were desecrated. According to legend, T ycho's
monition of death and composed his own epitaph:
corpse was removed from its burial site during one of those desecrations. In 1901, the 300th anniversary of his interment,
Tycho's
crypt
was
refurbished.
That
restoration provided an opportunity to investigate the contents of the crypt. It was opened and the grave robbing story debunked: The male skeleton found there had a cranial wound that was consistent with the injury Tycho suffered during his sword fight. In the late 20th century, samples of T ycho's hair, stored since 1901,
I used to measure the heavens, now I measure the
shadows of the Earth. Although my mind was heaven-bound, the shadow of my body lies here. A few months after penning this distich, Kepler took ill and died. The peace that eluded Kepler in life, eluded his mortal remains as well. Scarcely a few years passed before war ravaged the churchyard of St.
were analyzed. The results indicate a very high level of
Peter's,
mercury present. There is no evidence, however, of criminal poisoning. It seems likely that T ycho ingested the mercury as a remedy for his urinary difficulties.
Kepler's burial site.
Kepler's Laws Kepler succeeded Tycho in the position of Imperial Mathematician. By compensating Tycho's heirs, the emperor was able to ensure that Tycho's observa tions were place d at Kepler's disposal. Those mea surements were all Kepler had to work with-there were no established physical theories such as gravita tion to guide him, no advanced mathematical tools such as analytic geometry and calculus to aid in calculation,
Regensburg,
obliterating
every
trace
of
Sir Isaac Newton The idea of a gravitational force originated in the early 16th century. By the last half of the 17th century, the foremost question of natural philosophy had become, How could Kepler's laws of planetary motion be derived from a theory of gravitation? The first step was to deduce the form of the gravity law. This was a problem with which Kepler wrestled unsuccessfully for thirty years. Ironically, his third law became one of the two keys that together were used to unlock the secrets of gravity. Christiaan Huygens (1629-1695) provided the
858
Chapter 10
Vector-Valued Functions
second when, in 1659, he published the law of cen trifug al force for uniform circular motion. If a body of mass m moves with constant speed v along a circle of radius r, then its centrifugal force Fis directed
away
from the center
of the circle
with
The result was Newton's
Principia,
the most important
scientific treatise ever written. Publication was always an anxious process for New ton. In 1672, he was forced to defend a paper on optics
magnitude
against several criticisms, most notably from Robert Hooke, who managed to question both Newton's con clusions and Newton's priority. Newton found the busi
mv2 IFI= r . To prevent a planet from flying out of its orbit, the sun must exert a gravitat ional force -F that exact ly coun terbalances the planet's centrifugal force. Now imagine planetary motion in an elliptic orbit with semi-major axis r an d nearly 0 eccentricity-an orbit that for all practical purposes is a circle of radius r. According to Kepler's third law, there is a constan t period T satisfies
k
such that the
On the other hand, a body traveling around a circle of r with constant speed v completes one orbit in
time
27rr v
) (Time = Distance Rate
4�r2 /v2 = kr3
·
v2 = 47r2 / (kr). If we now v into Huygens's formula for
or
= mv2 = 4�m. _!__ IFI k
phical matters finding [that they] tend to disputes and
Principia
interrupted Newton's quiet
life,
embroiling him in a new dispute with his old nemesis, Robert Hooke. As the Principia neared completion, Hooke began to stir things up. Halley, who was over seeing the publication of the Principia, communicated the problem to Newton: "Mr Hook has some preten
centrifugal force, we find
r
began for the sake of a quiet life to decline corre spondencies by Letters about Mathematical & Philoso
sions upon the invention of ye rule of the decrease of
By substituting this value of Tin Kepler's third law, we substitute this formula for
serene liberty." Many years later, when Newton looked back on the ensuing period of silence, he explained, "I
The
radius
T=
ness tiresome. Shortly thereafter, he declined to publish a more complete treatment of optics, remarking that with further use of the press "I shall not enjoy mY former
controversies."
T2 = kr3•
see that
out a general science of dynamics that included the Universal Law of Gravitation as well as Kepler's laws.
r2
The fall of an apple caused Newton to reflect on the nature of gravity, but it was by these considerations that he deduced, in 1666, the law of gravit y. In dependently,
Gravity . .. He sais you had the notion from him ... Mr Hook seems to expect you should make some mention of him in the preface." Newton was aggra vated: "Philosophy is such an impertinently litigious Lady that a man had as good be engaged in Law suits as have to do with her. I found it so formerly and now I no sooner come near her again but she gives me warning." Newton responded to the baseless charge of plagiarism by deleting some references to Hooke that were already in the manuscript. In private correspondence, he referred to his antagonist as an "ignoramus." After completing the Principia, Newton ra pidly lost interest in mathematical research. A number of docu
Sir Edmon d Halley, Sir Christopher Wren, and Robert
ments leave no doubt that he suffered a complete
Hooke came to the same conclusion several years later. The next step was to show that the inverse square
mental breakdown in 1693. The cause remains uncer tain, but mercury poisoning (resulting from carefree
law
implies Kepler's laws. During a visit to Cambridge in 1684, Halley asked Newton if he knew the curve that an inverse square law would entail. Without hesitation,
handling of substances in chemical experiments) is a plausible candidate. Samples of Newton's hair, taken from four preserved locks, were analyzed in 1979.
Newton answered that it would be an ellipse. Three months later, he sent Halley a short manuscript that derived the three laws of Kepler from the inverse
Although elevated levels of mercury, antimony, arsenic, gold, and lead were measured, a conclusive diagnosis is not possible 300 Years after the illness.
square law of gravitation. For the next fifteen months , Newton went into seclusion, devoting himself to setting
What is not in question is that in the remaining thirty four years of his life, Newton's scientific activity was
Genesis & Development
largely confined to polishing the exposition of earlier work. He retired from academic life in 1696, serving at the Mint until his death in 1727. In 1820, the apple tree that set Newton to think about the nature of gravity succumbed to disease and
859
remained hidden until 1936 when a large Portion of his estate was put up for auction. Of the volumes in Newton's personal library, 3% and 7% pertained to physics and mathematics, respectively, whereas 8% and 27.5% concerned alchemy and theology. Of Newton's
was felled. Scions were taken and the line lives on, both
personal papers, about 1 million words were devoted to
in England and the United States. The fruit is pear
scientific subjects compared to 2 million words on
shaped and, it is said, without flavor. Man y unexpected details concerning Newton's life came to light in the 20th century. Newton's interests in
alchemy, theology, and chronology. These revelations
alchem y, theology, and biblical chronology had long been known. However, the extent of those interests
led
the great economist John Maynard Keynes (1883-1946) to say of Newton, "Not the first of the age of Reason. He was the last of the magicians."
This page intentionally left blank
Functions of Several Variables P
R
E
V
E
W
Many of the quantities that we study in mathematics and other fields depend on two or more variables. For example, the output of a factory depends on the amount of capital that is allocated to labor and the amount that is allocated to equipment. The air pressure at a point in the atmosphere depends on the altitude of the point and also on the temperature at the point. The height above sea level of a point on the earth's surface depends on the longitude and latitude of the point. Functions of two or more variables are, indeed, commonplace. To analyze these functions, we need to develop appropriate tools of calculus. That is the purpose of this chapter. We will ask and answer the same types of questions that we posed for functions of one variable. We begin by learning how to plot a function
f of two variables: Its graph is a
surface in three-dimensional space. At a point at which the function is well behaved, its graph will have a tangent plane that is composed of tangent lines. The mathematical concept that is the key to obtaining the tangent planes to the graph of
f is the partial derivative. Iff is a function of variables x and y, then an increment
fu in x or an increment
D.y in y will generally cause a change in the value off The numerator of the quotient
f(xo + fu,yo) - f(xo,Yo) D.x represents the change off(x,y0) as
x varies from x0 to x0 + D.x while the variable y fu represents the change in x. The quotient therefore represents the average rate of change of f(x,y) as x varies from x0 to xo + D.x with y being held constant at yo. The limit is held constant at y0• The denominator
8f ax
- r
(Xo,Yo) - &�
f(xo + fu,yo) - f(xo,yo) b.x
represents the instantaneous rate of change of f with respect to x at(x0,y0). This quantity is called the
partial derivative off with respect to
x. Similarly, if we allow
861
y to
vary while holding
respect to
x
fixed at
xo,
then we obtain the partial derivative off with
y: 8 f -8 (xo,Yo) y
_ -
1.
un �y-+O
f(xo, Yo +
t::..y) /\ y L.l
-
f(xo,Yo) .
As in the one-variable case, the derivative is essential for understanding and answering many of the questions that arise in the analysis of functions of many variables. As an application, we will learn how to use the partial derivatives of f to identify and classify the local maxima and minima off
862
11.1 Functions of Several Variables
1 1 .1
863
Functions of Several Variables The first nine chapters of this text have been concerned with scalar-valued func tions that depend on one real variable. Such a functionf is evaluated at a real number x in its domain, and the result of the evaluation is a real numberf(x). Chapter 10 is devoted to vector-valued functions that depend on one real variable. Such a function
r
is evaluated at a real number t in its domain, and the result of the
evaluation is a vector r(t). Now we begin the study of functions that are evaluated at two or more real numbers. In this chapter, we will discuss the differential cal culus of scalar-valued functions of two or more real variables.
-·l§iHhi[.]�i
Suppose that 'D is a set of ordered pairs of real numbers and that
R is a set of real numbers. We say thatf is a
(scalar-valued) function of two variables with domain 'D and range R if, for every ordered pair (x,y) in 'D, there is associated a unique real number in R; we denote this number byf(x,y). The
numberf(x,y) is said to be the / is
image of the point (x,y) under f We also say that
at (x,y), or that/maps (x,y) to the value/(x,y). The notation (x,y) >-->f(x,y) is often used; the arrow-like symbol is read as "maps to." A
evaluated
function F of three variables is similarly defined: The only difference is that the domain of F is a set of ordered triples. The notation (x,y,z) >--> F(x,y,z) is used for a function F of three variables.
� EX A M PL E 1 Express the surface area A and volume V of a rectangular box as functions of the side lengths. Solution Suppose that the side lengths of the box are £, w, and h (see Figure 1). Then the top and bottom have area £w, the left and right sides have area wh, and the front
and
back
have
area
£h.
Thus
the
surface
Uw +2wh +2£h. The volume is, of course, V(£,w,h)
=
area
is A(£,w,h)
=
£wh. Each function has
for its domain the set of ordered triples (£,w,h) with all three entries positive.
<1111
e A Figure 1
It is often convenient to interpret an expression to be a function. When this is done, the domain of the function is taken to be the largest set of points for which the expression makes sense and evaluates to a real number. For example, the expression
J25 - (x2 +y2)
defines the function (x,y) >-->
domain consists of all ordered pairs (x,y) with
J25 - (x2 +y2)
x2 +y2 � 25.
J
whose
We employ the same
convention when we use an equation such asf(x,y) = 25 - (x2 +y2) to define a function without explicitly stating its domain. As in the one-variable case, it is convenient to think of a function as an input-output machine (Figure 2). The domain is thought of as the set of input values. Evaluation is the process of getting a unique output value from an input value that is fed into the machine. � EX A M PL E 2 Let
a
be any constant. Discuss the domains of the functions
f(x,y) =x2 +y2, g(x,y) =a/(x2 +y2) and h(x,y,z) =z/ (x2 +y2). Solution The expression
x2 +y2 is defined for
all values of x and y. Therefore the
domain off is the set of all pairs (x,y) of real numbers. For the function g, we observe that, whatever the value of a, the expression
a/ (x2 +y2) is defined if x and y
864
Chapter
11
Functions of Several Variables y
Domain of/
f
� � Rangeof/ ���--�------� - �
Input-output machine
x
f(x,y)
A Figure 2
0 and undefined if x= y=0. Therefore the domain of g is the set of all (x,y) other than (0, 0) For the same reason, the domain of h is the set ordered triples (x,y,z) for which x and y are not both 0. The set that is
are not both
.
ordered pairs of all
excluded from the domain of
h
is the z-axis in xyz-space. In other words, the
domain of his the set of all points not on the z-axis. .,.. Observe the distinction between functions g and h in Example 2. When we
define
g(x,y) a/(x2 + y2), =
we regard a as a parameter of
g, not
a variable. Although a
g(x,y). Indeed, the g(x,y) tells us to treat a as a constant. On the other hand, the notation h(x,y, z) z/(x2+y2) tells us that the numerator z of h(x, y, z) is a variable. Even though the values g(x,y) and h(x,y,a) are the same, the functions g and hare quite different.
may have any value, its value does not change when we consider notation
=
Combining Functions
Much of the elementary material that we have learned about functions in Chapter
1
still applies here. For instance, we can add, subtract, multiply, and divide functions of two variables to form new functions. If A is a constant and if f and g are functions with a common domain 1J, then we define the functions AJ,
(�)
on V by
c.
(Af)(x,y) = A f(x,y). (f+ g)(x,y)=f(x,y) + g(x,y). (/- g)(x,y) = f(x,y)- g(x,y).
d.
(fg)(x,y)=f(x,y) g(x,y).
e.
G) (x,y) ;�::��
a. b.
f + g, f- g, fg, and
=
provided
g(x,y) :f= O.
There are similar formulas for functions of three variables.
� EXAMPLE 3 Compute
Letf(x,y)=2x+3y2, g(x,y)=5+x3y,andh(x,y,z)=xyz 2•
(f + g)(l, 2), (.fg)(l,2),
Solution We have f(l, 2)=2(1)
(�) (1,2), (�h2) (1,2,3).
+ 3(2)2=14
and
and
g(l,2)=5 + (1)3 2=7. Therefore
(/ + 3 g)(1,2) =/(1,2) + 3 g(1,2) =14 + 3(7) =35, (fg)(l,2)=/(1,2) g(l, 2)=(14)(7)=98,
11.1 Functions of Several Variables
(l) (1 2) g
'
=
and
f(l, 2)
g(l, 2)
=
14 7
=
2
865
'
Now we want to discuss composition of functions. For this, a schematic dia gram is essential (see Figure
3). Consider a scalar-valued function (x,y)
f(x,y) of
t--t
two variables. If g is a function of one variable, and if the values of f lie in the domain of g, then we may consider the expression g(f(x,y)) for any point (x,y) in the domain off. We write (gof)(x,y) = g f(x,y) ,
(
(11.1.1)
and call the function gof defined by
)
(11.1.1)
the composition of g with f. The
domain of gof is the same as the domain off.
y
Domain off f
x
u
g(f(x,y))
f(x,y) Domain ofg _. Figure 3
Schematic diagram of (g f)(x,y) 0
Range ofg
=
g(f(x,y))
� EX A M P L E 4 Let g be defined on the set of all nonnegative numbers by the formula g( u) =
y'U.
Suppose that f is defined on the set 1J =
{ (x,y)
: x2 + y2 <
25}
by the formulaf(x,y) = 25 - (x2 + y2). Show that the composition gof is defined on
V, and calculate (go /)(2, 1J on which
hof
v's). If h(u)
=
1/../9 - u,
then what is the largest subset of
is defined?
(
)
Solution The composition (gof)(x,y) =g f(x,y)
is defined precisely when the
valuesf(x,y) are in the domain of g. In other words, (go f)(x,y) is defined provided f(x,y) > 0. This inequality holds if and only if x2 + y2 <
25. Because the coordinates
of every ordered pair (x,y) in the domain 1J of f satisfies this inequality, the composition gof can be formed. In particular, for (x,y) in V, we have
(
)
(gof)(x,y) = g f(x,y) =
Jf(x,y) = J25 - (x2 + y2).
v's), we first observe that, because 22 + ( v's)2 = 9 point (2, VS) is in the domain off, and
To calculate (gof)(2, not greater than
25,
the
(go /)(2,
v's)
=
J25 - (2
)
2 + ( ../5)2
=
../25 - 9
=
4.
is
888
Chapter 11
Functions of Several Variables For h
f)(x,y)
1
=
../9-f(x,y)
=
J
9-
1 25-
x2 + y2))
=
1 ../ x2 + y2)-16
to be defined, we must have x2 + y > 16. Therefore the largest subset of 1J on which f is defined is {(x,y): 16
h
Graphing Functions of When we graph a scalar-valued function of one variable, we use the two-dimensional Several Variables y
il-+Y.
2S
-3
2
is
for the domain of the function, and the other is for the
of the function and one dimension for the range. To be specific, the graph of the
3 2
plane: One dimension
range. Now we will learn to graph scalar-valued functions of two variables. This will be done in three-dimensional space because we need two dimensions for the domain function x,y) of , and z /:
1
=
-1
5
-2 -3
A Figure 4 The set 16 x +y2�25}
{(x,y):
z
element
x,y) consists of all points x,y,z) such that: x,y) is in the domain . In other words, we plot a point on the graph of f by locating an
x,y) of the domain of
f x,y) units.
in the .xy-plane and then, in space, going up or down
Some caution is in order here. In spite of all the sophisticated techniques that we know for graphing functions of one variable, we are often tempted to just plot
points and connect them with a curve. Sometimes we can get away with this because curves are often fairly simple. But now we are working in three dimen sions. Our graph will be a surface, and it is virtually impossible to plot some points by hand and "connect them with a surface." In fact, we need a new idea to graph functions of two variables. The idea is that a surface in three-dimensional space can be described by its two-dimensional slices.
For instance, suppose that every horizontal slice of a surface is a circle of radius 1 with center lying on the z-axis (Figure 5). The only thing that this surface could be is a cylinder (Figure 6). On the other hand, suppose that every horizontal slice of a surface is a circle with center lying on the z-axis, but that the radius of the circle increases with the height. Then, you may
have a mental picture of the object in
Figure 7a or perhaps the one in Figure 7b (there are other possibilities as well). We need to understand how to use horizontal slices effectively.
A Figure 6
A Figure 7a
A Figure 7b
11.1 Functions of Several Variables
867
Let (x,y) f-tf(x,y) be a function of two variables. If c is a con stant, then we call the set Le {(x,y):f(x,y) c} a level set off. =
=
Notice that, by definition, a level set of a function of x and y is a subset of the xy-plane. Level sets of functions of two variables are often curves; the terminology level curve is used in such cases. The terms contour, contour curve, contour line, and isoline are also used. If Le= {(x,y):f(x,y)=c} is a level set of a functionf, then the set {(x,y,c): (x,y) E Le} is the intersection of the graph of /with the horizontal plane z =c. This means that the level curves off are obtained by projecting the horizontal slices of the graph off into the xy-plane (see Figure 8). z
z -
z =
y
-
f(x,y)
_. Figure 8 Graph of z f(x,y) with six horizontal slices; the six horizontal slices; the level curves corresponding to the slices =
� EXAMPLE 5 Let f(x,y) x2+y2+4. Calculate and graph the level sets that correspond to horizontal slices at heights 20, 13, 5, 4, and 2. =
Solution The level set that corresponds to the slice at height 20 is the set
{(x,y) :f(x,y)
=
20)
{(x,y) :f(x,y)
=
13}
{(x,y) : x2+y2+4=20} = {(x,y):x2+y2=16 } . This is a circle of radius 4 centered at the origin of the xy-plane. The level set that corresponds to the slice at height 13 is {(x,y):x2+y2+4=13} = {(x,y):x2+y2=9 } . This set is the circle of radius 3 centered at the origin of the xy-plane. Similarly, we find that the level set that corresponds to the slice at height 5 is the circle of radius 1 centered at the origin of the xy-plane. The level set that corresponds to the slice at height 4 is {(x,y):x2+y2+4=4 } ={(x,y) :x2+y2=0}={(0,0)}. This level set is just a single point. The graphs of these level sets appear in Figure 9. The level set that corresponds to the slice at height 2 is {(x,y) : x2+y2+4=2} x
_ ____,__ ____,____-+-
{(x,y) :f(x,y)
=
{(x,y) :f(x,y)
=
_. Figure
9
5} 4}
=
{(x,y):x2+y2
=
-2}
=
0 (the empty set).
There are no points in this level set. In fact, all the level sets that correspond to slices at heights less than 4 are empty. .illl The level sets of a function tell us what the horizontal slices of its graph are. These in turn tell us a great deal about the graph of the function. � EXAMPLE 6 Letf(x,y)= x2+y2+4, as in Example 5. Use the level sets calculated in Example 5 to plot the horizontal slices at heights 20, 13, 5, 4, and 2. Then plot the graph off. Solution As we calculated in Example 5, the level set that corresponds to the slice z=
20 is {(x,y):x2+y2=16}. This means that the points {(x,y,20): x2+y2=16}
888
Chapter 11
Functions of Several Varlables
lie on the graph off. This set is a circle with center (0, 0, 20) and radius 4. It is located at height 20. Likewise, the slice of the graph off at height 13 is the circle with center (0, 0, 13) and radius 3. And the slice of the graph at height 5 is the circle with center (0, 0, 5) and radius 1. Finally, the slice of the graph at height 4 is just the point (0, 0, 4). It is important to note that, at heights below 4, the slice of the graph is the empty set. This means that the graph does not extend below the plane z = 4. All this information is amalgamated into a picture in Figure 10a. How do we go from a few horizontal slices to the plot of the entire surface? This is a critical step. We piece together the horizontal slices by taldng a slice in another direction: Imagine slicing the figure with the yz-plane. The shape of this last slice will tell us a lot about the shape of the figure. The yz-plane has equation x u. Now substituting x =O into z - flx,y) results in z =r + 4, which is the equation of a parabola in the yz-plane (Figure 10b). Because all the nonempty level sets off �re circles with centers at (0, 0), it follows that the graph of/has rotational symmetry about the z axis. Therefore the slice we have taken in the yz-plane suffices to confum that the graph should be as shown in Figure 10c. � -
c
z =20
z = 13
z=4 y
y
x
A Rgure 108 Four horizontal slices of
A Figure 1 Ob The slice with the
.z = x2 +y2 +4
yz-plane is added to the plot
A Figure 10c The plot of
.z =.t2 +y2 +4
Our new procedure for graphing has several steps. We first must draw several two-dimensional level sets. Then, we must merge these into a three-dimensional picture. One or more slices perpendicular to the xy-plane are used in this last step. Symmetries about a plane or rotational symmetry about an axis provide important shortcuts when present.
� EXAMPLE 7
Sketch the graph of/(x,y) =x2 +y.
Intersecting the graph of /with the plane z =c results in a horizontal slice of the form {(x>y,c): x2 +y =c). Several of these slices are exhibited in Figure lla. Notice that they are all parabolas that are symmetric in the y.z-plane. When we slice the graph off with the yz-plane by setting x = 0, we see that the vertices of the parabolas lie on the line x = O>z =f(O,y) =y. The plot appears in Figure 11b. Solution
11.1 Functions of Several Variables
889
y
x £. Figure 11a Horizontal slices of f(x,y)=xl-+y
£. Flgure11b/(.r,y)=xl-+y
INSIGHT
When you graph a function (x,y) 1-+f(x,y), you may use any slices that are convenient In Example 7, if we slice the graph of /with planes y = c that are parallel to the xz-plane, then we obtain a family of parabolas of the form y = c, z = xl-+ c (Figure 12a}. These vertical slices can be pieced together to obtain the plot off (Figure 12b). By com· paring Figures llb and 12b, we can see that different slices can provide different per spectives on a surface.
z y=c
z= x2+c y
....
£. Rgure 12a Vertical slices of f(x,y) =x2 +Y � EXAMPLE 8
£. Figure 12b/(x,y) =xl- +y
Sketch the graph of/(x,y) =y2-x2.
Several level sets are exhibited in Figure 13. Notice that they are all hyperbolas. These hyperbolas change orientation at height c =O. In fact, c =0 corresponds to the level set y2-x2 =0 or (y-x)(y + x) =O, which is a pair of crossed lines. The level sets are hyperbolas that open sideways (in the xy-plane) Solution
£. Figure 13 The family of y2 - xl- = k, where k is a constant
y
870
Chapter 11
Functions of Several Variables when c < 0 and are hyperbolas that open up and down (in the xy-plane) when c > 0. We use the slice corresponding to x = 0, which is a parabola, and the slice corresponding to y = 0, which is also a parabola, to complete our idea of the graph. The plot of /is shown in Figure 14. �
More
on
Level Sets
Level sets occur in many applications of mathematics. They are frequently used in geography, geology. and meteorology. For example, the contours that we see on a topographic map are level sets of the fJ eight function. Thus if h is the function that assigns to each point on the map the height above sea level at that point, then the curves hl.x,y) c that are drawn on the map give us an idea of where the moun tains, valleys, and ridges are located. An example is shown in Figure 15. A contour ..
that represents constant elevation is called an isohypse. A curve that represents a constant depth of a body of water is called an isobath.
A Figure 14 The graph of .z =
y2-x2
A Rgure 15 Topographic map of Stone Mountain, Georgia
In newspapers, we often see maps featuring level sets of the temperature function. That is, let T�x,y) be the function that assigns to each point on the map the temperature in degrees Fahrenheit at that point. The curves T\x,y) = c, called isothermal curves or isotherms for short, give us an idea of the changes in weather, the location of cold pockets, and so on (see Figure 16). Likewise, isoban are the level sets of the barometric pressure function, isohyets are level sets of the rainfall function, and so on. We have seen that the graph z =flx,y) of a function of two variables will generally be a surface. Although we cannot graph a function F of three variables, we Lan graph its level sets {(x,y,z): F(x,y,z) = c}. The level sets of a function of three variables are generally surfaces (and are called level surfaces�. A simple example illustrates this idea. .- EXAMPLE 9 Discuss the level sets of the function Fi;x,y,z) =x2 + y2 + z2• Solution Remember that this is a function of three variables: We may substitute values of z independently of the values of x and y. We therefore do not expect the level sets of F to be curves (as we would for a function of two variables). The level
11.1 Functions of Several Variables
871
_. Figure 16 Isotherms 13 August 2002 12AM EDT
sets have the form x2 + y2 + z2
0 and radius empty set.
ye; for c
=
=
c. For c > 0, the level set is the sphere with center
0, the level set is a single point; for c < 0, the level set is the
.,..
.... A LOOK BACK To understand the different types of functions that have been studied, keep in mind the geometry that each type of function describes. The graphs of scalar-valued functions of one variable are planar curves that meet the Vertical Line Test. Vector-valued functions of one variable describe curves in the plane or in space. The graphs of scalar-valued functions of two variables are sur faces in space. The level sets of scalar-valued functions of three variables describe surfaces in three-dimensional space.
Q UIC K
Q UIZ
1. 2.
Describe the domain of f(x,y) When the graph of f(x,y)
=
ln(3
x2) .
R
=
4-y2
x - y2 is sliced with planes that are parallel to the
yz-plane, what curves result?
3. Describe the level sets of f(x,y)
=
x2-y2
4. Describe the level sets of F(x, y, z)
=
•
x + 2y- 3z.
Answers
1.
The open rectangle
3. Hyperbolas
(-1, 3) X (-2, 2)
2.
4. Planes perpendicular to
Parabolas that open downward
(1, 2,-3)
872
Chapter 11
Functions of Several Variables
EXERCISES Problems for Practice
In each of Exercises 1-8, express the quantity that is described as a function of two or more variables. Describe the domain.
V(h,r) of a cylinder in terms of its height h and its radius r 2. The surface area S(h,r) of a cylinder in terms of its height h and its radius r 3. The volume V(h,r) of a cone in terms of its height hand its radius r 4. The area A(r,0) of a circular sector of radius rand interior angle(} 1. The volume
5. The area A(a,b) under the graph of y=x2 and over the interval [a,b] 6. The surface area S(x,y, V) of a rectangular box interms of two side lengths, x and y, and its volume V 7. The mass m(M, T, T) of a radioactive substance as a function of the initial mass M, the half-life T, and the amount of time T that has elapsed since the initial measurement 8. The magnitude aN(K-,v ) of centripetal acceleration of a moving body at a moment when its speed is v and the curvature of its trajectory is Kin each of Exercises 9-12, let f(x,y)=ex-y, g(x,y)=x2 -y3, h(x,y,z)=xy'+2, k(x,y,z)=y/(x+z), and
calculate the given quantity. 9. (2/ + g)(S,2)
(fg)(3,2) 11. (h -k/3)(1/2,3,0) 12. (h/k)(3,4,-l) 10.
In each of Exercises 13-18, let f(x,y)=3xy/(x2+y2), g(x,y)=x2 + 2y, (t)=.Ji, and calculate the indicated quantity. 13.
14. 15.
16. 17. 18.
( 0!)(3,4) ( 0g)(4,10) ( 0(2/+ l l g))( l,1) ((o/) g)( l,2) ((2) 0(fg))(3,4) (((2) 0!)((2) 0g))(3,4)
In each of Exercises 19-28, sketch the level sets off that correspond to horizontal slices at heights -6, -4, -2, 0, 2, 4, and 6. 19.
f(x,y)=3x
22.
f(x,y)=4x2 -y f(x,y)=1 +x -y2
20. f(x,y)=x+2y 21. f(x,y)=3x -2y
23.
f(x,y) =x2+y2/4 f(x,y) =1 -x2/9 -y2 26. f(x,y) =x2 -2x + y2 27. f(x,y)=x2-y2 28. f(x,y)=2 + y2 -x2
24. 25.
In each of Exercises 29-38, a function f of two variables is given. Sketch several level sets off. Then, sketch the graph of z=f(x,y).
29. f(x,y)=1 + x 30. f(x,y)= 6 -2x -3y 3L f(x,y) =x2+ y2-1
32. f(x,y) =2 -(x2 + y2)
33. f(x,y) = ,jx2 + y2
34. f(x,y)=3 - Jx2+yi 35. f(x,y)= ,/1 -x2 -y2
36. f(x,y)=1+ ,j4-x2 -y2 37. f(x,y)= y2 +2x + x2
38. f(x,y)=x2+y2/4 + 6
In each of Exercises 39-44, sketch level sets correspond ing to heights -8, -4, 0, 4, and 8 for the given function F. 39. F(x,y,z) =x+y+z 40.
F(x,y,z)= lO -x -y
4L F(x,y,z) =12 -z2 42. F(x,y,z)=x2+y2 + z2 - 8 43.
F(x,y,z)=x2+2x + y2 + 4y +z2+ 6z -25
44. F(x,y,z)=x2+y2 - l2
Further Theory and Practice 45. Let f(x,y) =xy/ ,jx2 + y2• Calculate f(f(x,y),f(x,y)). 46. Match the given functions to the correct set of level
curves in Figure 17.
f(x,y)=x2 -y f(x,y)=xy c. f(x,y) =Ix+YI d. f(x,y) =2x2 + y2 e. f(x,y) =x2 -y2 f. f(x,y) = (2x -y)2 a.
b.
47. How do the level sets of f1(x,y) =x -y and h(x,y)=
(x -y)2 differ?
T(h,vo) until ground impact of an object propelled downward from a height h with initial speed v0• (Assume gravity is the only force present.) 49. Calculate the x-intercept a(h,c) of the tangent line to the curve x >--> h+e' at x =c.
48. Calculate the time
11.1
Functions of Several Variables
y
y
x
x
(ii)
(i)
y
(iii)
y
�
x
x
x
(v)
(iv)
(vi)
_. Figure 17 50. Calculate the y-intercept b(k,c) of the tangent line to the 51.
curve y = k/ x at x = c. Let f(x,y)=x2+y 2 and
y
2 2 g(x,y)=4-2x -2y . Sketch
the set S ={(x,y):f(x,y) > g(x,y)}. x2+ y2. Sketch the set S {(x,y):f(x,y) � 2 f(x,y) }. 53. Let f(x,y) = x2 -y+4. Sketch the set S = {(x,y): 2 :-;:;
52. Let f(x,y)
873
=
=
x
f(x,y) <4}. In each of Exercises
54-59, describe the level sets of the
given function f.
54. f(x,y) =ln(xy). 55. f(x,y) =sin( y- x2). 56. f(x,y)=ysin(x) 57. f(x,y) =sin(x2+y2) 58. f(x,y) = y+sin(x) 59. f(x,y) = y2/(1+x2) 60. Figure 18 shows typical level curves of a function f.
tt t t I
2 1 o
-3/2
-1
r ;1 1 2
-1/2 3/2
_. Figure 18 Level curves off a.
Does f(O,1) /(1, O)? =
Answer the following questions about f, stating the rea
b. Does/(1/2,0) = f(O,1/2)?
sons for your answers.
c. Which is greater:/(-1,1) or/(1/2,1)?
874
Chapter 11
Functions of Several Variables
d. Ast increases, doesf(-1/2-t, -3/2+t) increase or
61. 62. 63.
64.
decrease? e. On what interval isf(x, -1) an increasing function of x ? f, On what interval is f(-1/2, y) an increasing function of y? Express the function f(x,y) = Ei�O (x/y)i without using sigma-notation. What is its domain? What is the domain of the function f(x,y) = E�1 nx+y? Sketch its level sets. Give an example of a function f(x,y) such that for each fixed yo, the graph of the function x>--+ f(x,yo) is a para bola, while for each fixed xo, the function y >--+ f(xo, y) is a cubic. Parameterize the level curves of f(x,y) = x2 + 2y2.
65. Parameterize the level curves of f(x,y) = x2 - 2xy + 2y 2• Calculator/Computer Exercises In each of Exercises 66-69, a function f and a viewing window R are given. Plot the level curves of the functionfin the given viewing window. Use the horizontal slices z =k/4 fork= -3,-2, ... ,3.
66. f(x,y) = x3 + x2y4 67. f(x,y)= x2 + xy + y3
R = [-1,1] x [-3,3] R = [-1.2,1]X[-1, 1]
68. f(x,y) = (1- x + y)/(1 + x2 + y2) R= [-6,4] x [-4,6] R= [-2,2]X[ 2,2] 69. f(x,y)= 2xy2 + x4y
11 .2 Cylinders and Quadric Surfaces You already know from your experience with functions of one variable that it requires some practice to feel comfortable with graphing. We have to do a lot of graphing and encounter a great many different functions and pictures before we can visualize ideas readily. The purpose of this section is to provide that vital experience for functions of two variables. Some of the graphs in this section will not be the graphs of functions. Part of what we will learn is to distinguish which graphs come from functions. We begin now with the simplest graphs that are encountered in the theory of several variables.
Cylinders
An equation in three-dimensional space has a graph that is said to be a cylinder if one (or more) of the variables x,y,z does not appear in the equation. The justifi cation for the term "cylinder" is apparent from the graph of the surface (x, y, z) : x2 + y2=1 In Example 2 of Section 12.1 of Chapter 12, we deduced
{
}.
that the plot of this equation in xyz-space is a cylinder (in the ordinary sense of the word). The next examples show that the mathematical meaning of "cylinder" is
z
quite a bit more general. � EXA M P L E 1
Sketch the set of points in three-dimensional space satisfying
the equation x2 +4y2= 16.
Solution We exploit the missing variable by taking slices that are perpendicular to its axis. In this case, z is missing, so we calculate the equations of some level sets z=c:
x
x2
+
4y2
=
_. Figure 1 Elliptic cylinder
16
z = -1
x2 +4y2=16
z=O
x2+4y2=16
z=2
x2+4y2=16.
11.2
Cytlnders and Quadric Surfaces
87&
Notice that all the level sets of the graph are the same ellipse. The reason for this is simple: The variable
z does not appear in the equation that we are graphing.
Substituting in different values of zhas no effect. Now we sketch the graph . We plot the ellipse in the xyp - lane and form the cylindrical surface with this ellipse (and its translates up and down) as cross-sections (see Figure
1). .,..
z
�EXAMPLE 2
Sketch the graph of x-3z2=3.
Solution The variable yis missing ,so we know that the graph will be a cylinder and that we should use slices obtained by setting y equal to a constant. We also x
.A. Figure
-
3.t2 3 =
know that no matter what value
c we take for y, the level
= x 3+ 3z• 2 The resulting graph is sketched in Figure
2
Quadric Surfaces
Quadratic
set will be the parabola
2.
equations provide a rich variety of examples on which to hone our
technique at sketching graphs. We will only discuss equations of the form
.Ax2 +By+ Cz2 +Dx+ Ey+ Fz =G.
(11.2.1)
The graphs of equations with terms such as xy,yz,xz in them are obtained by rotating the graphs of equations of the form
Ellipsoids
(11.2.1).
An ellipaoid is the set of points in space satisfying an equation of the form
x2
yz
+ zz=1 ff i1-
o(l +
(11.2.2)
with a, /J, "Y all positive constants. The distinguishing feature of such an equation is that all of its intersection s with planes parallel to the coordinate planes are ellipses.
More precisely , if we set = x c, then we obtain the set
c2 z2 y2 p2 + i1=1-o(l' This is an ellipse if lcl
if
lcl =a, and
empty otherwise. Similarly, if we set y= c, then we obtain the set
x2
zz
c2
o(l + ,2=1-ff'' This is an ellipse if lcl < {3, a point if lcl={3, and empty otherwise . The intersections with the planes
z= c are similar . It is worth noting that equation
(11.2.2)
is
unchanged when xis replaced by -x,or yby -y, or zby -z. Thus the graph of equation
(11.2.2) will be symmetric with respect to each coordinate
plane -that is ,
the graph looks the same on both sides of each coordinate plane. We put this information together to obtain the plot shown in Figure 3. Notice that. when
a, {3,
and "f are distinct, the three axes of the ellipsoid have different lengths. When they
are all equal, the ellipsoid is, in fact, a sphere.
876
Chapter
11
Functions of Several Variables
.A. Figure 3 Intersections of the ellipsoid coordinate planes, are ellipses.
z
.... EX A M PL E
y2 + 2z2 =4.
x2
4
12
.A. Figure
+
f3
3 Sketch the set of points satisfying the equation
Solution We divide through by
4x + ,• + .!z - 4
.� + � ;� = l, with planes parallel to the 4x2 +
to write the equation in standard form:
+
Y + 2
z2 .,2-1. v )
According to our discussion, this is the equation of an ellipsoid. Its graph is sketched in Figure
4
4. Observe that it is easy to read, from the standard form of the equation, :::!::: 1, , 0), 0, 2, 0), and , 0, ±J2). <411
that the intercepts of the surface are
Elliptic Cones
Elliptic cones arise from equations of the form
x2 .,
-+ with x
, /3, d
•
y2 = z2 (f -f
y2 +
or
&
µ
'Y
When z
or
positive. We will analyze the first of these equations. The remaining
two can be handled in a similar way.
=c,
the level set of
x2 /
= z2 'Y
has the form
y2_c -
x2 .A. Figure 5 Some level sets of x2 i-= c2 + n1ff- r
z2- x2 2- a2
2+(f
,Z"
The graph of this equation is an ellipse whose axes have ratio a :
{3. As 1-1 becomes
larger, the ellipse becomes larger, but the ratio of the axes remains the same. Some level curves are sketched in Figure 5.
It remains to see how the level sets stack up. To do this, we take some slices
parallel to the xz- and yz-planes. Setting x= , that is, slicing with the yz-coordinate plane, we obtain
y2· (f = z2Ii-
··r
y= ±(3 z, 'Y
which is a pair of lines. Likewise, setting y = 0, that is, slicing with the xz-coordinate
plane, we obtain the lines
x= ± .. 2 ,2 .L2 -+· -az p' .,z
.A. Figure 6 An elliptic cone
a
'Y
z.
Finally, we notice that the graph will be symmetric in all three coordinate
planes. Taking into account all this information, we exhibit the graph in Figure 6.
11.2
Cylinders and Quadric Surfaces
e. 77
..,. EX A M P L E 4 Sketch the set of points satisfying the equation x2 + 2z2 =2y2. Solution We divide through by 2 to write the equation in standard form: ---
According to our discussion, this surface is an elliptic cone. Its graph is sketched in .&.
Figure
Figure 7. <1111
7 The elliptic cone
xi +2z 2 =2y2
Elliptic Paraboloids
Elliptic paraboloids arise from equations of the form y2 z x2 - +� =
.,
/J
'Y
-
or
y2
ff
+
z2
r
=
x
-
a
or
z2
r
y x2 + - =- . ·,
.
As usual, we assume that a) (3) 'Y > 0. We will discuss only the first of these equa tions, the others yielding to a similar treatment. As in the case of the elliptic cone, the level set of x2/ � + y2/(32 = z/'Y obtained by setting z = c is the ellipse y2 c x2 - +- =-
a2 y
a ... Fi1gure 8'·
xi
2
y2 - z. +
Elliptic paraboloid
dl p
-
1,
ff
'Y
As c > 0 becomes larger, the ellipses get larger. For, c = 0, the level set consists only f the origin. The level sets for values c < 0 are empty.
Next we look at slices parallel to the xz- and yz-planes to see how to stack the
llipses. Setting x =0, that is, slicing with the yz-coordinate plane, we obtain
y2
ff =zh ··r
which is an upward-opening parabola. Likewise, setting y =0, that is, slicing with
the xz-coordinate plane, we obtain x2 d1- = z
'Y or
This, too, is an upward-opening parabola. Finally, we notice that the graph will be
symmetric in the xz-coordinate plane (replacing y by -y results in no change) and
in the yz-coordinate plane (replacing x by -x results in no change). However, it
will not be symmetric in the xy-coordinate plane. The resulting sketch appears in Figure 8.
The special case of the elliptic paraboloid when a =f3 � s called the·· ircular paraboloid. For th.is surface, the level sets are circles. The sketch is in Figure 9. Circular
.&.
Figure 9 xi
"2
+
i
ff
Circular paraboloid
=
paraboloids are of interest because they are the model for the shape of radio, television,
z;
n nd radar antennas. The reflecting properties of paraboloids are also crucial to the
1
construction of reflecting telescopes.
878
Chapter 11
Functions of Several Varlablu ketch the set of points satisfying the equation x2 +2z·
� EXAM PL E 5
Solution We divide through by
2
0,)2
Hyperboloids of One Sheet
According to our
y.
to put the equation in standard form:
x2
A. Figure 10 Elliptic paraboloid
=
+
z.2
�
y =
2·
discussion, the locus of this equation is an elliptic paraboloid. Its
graph is shown in Figure
10. �
The hyperboloid of one sheet arises from the equations z.2 We assume 1
.
1
-
· . .
:
or
are
f' + z2 - x2 ff- -y2 .,
positive.
Again,
-
or
1
z2 + x2 -
-y2
•,
yi.
-
1
.
.
we will discuss only the first of these
equations, the others two being similar. Following previous experience, we begin by looking at level sets, the ellipses
� These ellipses increase in size as
y + 2
I I
=
$
+1.
increases but the ratio of their axes remains
constant. To see how the ellipses stack up, we look at slices parallel to the xz- and
yz-planes. We slice with the yz-plane by setting x
-
·
lln.
1. This is a hyperbola Also, we slice by the xz-plane, !.-��eo,,
a-
JI -... p
-
0,
to
obtain
'Y
A. Figure 11 Hyperboloid of one sheet
This too is a hyperbola. The picture of our hyperboloid of one sheet is
in Figure 11.
The phrase "one sheet" means simply that the surface is all of one piece.
� EXAM PL E 6 Sketch the set of points satisfying the equation
3x2-
�
y +z 2
=
1
.
Solution We put the equation in standard form:
1. A. Figure 12 Hyperboloid of one sheet
According to our discussion, this surface is a shown in Figure 12. �
hyperboloid of one sheet. Its graph is
11.2
Hyperboloids of Two Sheets
Cyllnders and Quadric Surfaces s 79
Consider the equation
We begin by considering the level sets of the graph of this equation. Setting z=c, we obtain
y2 = c2 +13 2 2 -yi l.
x2
If lcl �/,then this level set is an ellipse. The ellipses obtained in this way increase in size with lcl, but the ratio of their axes remains constant. The level set corre sponding to
-
=1 contains only
-1 contains only the point
the point 0, 0, -y). The level set corresponding to
0, 0, -1). If
lcl < -y, then the
level set is empty.
We now slice by the coordinate planes. The slice with the yz-plane is a
hyperbola
z2 - y2 - 1 -y2 2- . imilarly, the slice with the xz-plane is the hyperbola -'
y
z2 - x2 =1 -y2 . 2
7? f� .& Figure 13 two sheets
yperboloid of
The picture of our hyperboloid of two sheets is in Figure soning, we see that the equations
x2 a2
y2_z2 =l -y2 rf
13.
By analogous rea
y2 _z _x.2 =1 2 -y2 a2
and
are also hyperboloids of two sheets.
� EX A M P L E 7
ketch the set of points satisfying
x2-2z2- y2=
4.
Solution We divide by 4 to put the equation in standard form: x2
-
2z2
-
4y2
=
4
z
J2)2
.& Figure 14 Hyperboloid of two sheets
-1
-
.
According to our discussion,this surface is a hyperboloid of two sheets. Its graph is shown in Figure 14. -4
The Hyperbolic Paraboloid
Finally, we consider the equation
880
Chapter 11
Functions of Several Variables As usual, a,/3,I are positive. As long as
_
+· ·
=f 0, the level set corresponding to z -
- is
the hyperbola
x2 a2 When
-
y2 c 132=,
c = 0, the level set is x2/ a2 -y2/f32
0 or x = ± y//3, which is a pair of lines. c < 0, c = 0, and - > . . The slices corresponding to x - 0 and y 0 are parabolas. The final picture is in Figure 16. The resulting surface is commonly known as a saddle surface. It is of particular
Figure 15 shows the family of level sets for
.A. Figure 15 Level curves of
interest when we solve maximum-minimum problems because the origin looks
a?-
y-direction.
like a local minimum in the x-direction and looks like a local maximum in the
x2_1C_z
2-'Y
� EXAM PL E 8 Sketch the set of points satisfying the equation z =2y2-4x2. Solution We divide through by 4 to put the equation in standard form:
_L_x2_z
12 - 4·
x? ±-� .. 2 -132 - ..,
·
According to our discussion, the surface is a hyperbolic paraboloid. The graph is
.A. Figure 16 Hyperbolic paraboloid (saddle surface)
shown in Figure
Recognizing the Graph of a Function
17.
..,..
Many of the quadric surfaces that we have discussed in this section are
not
the
graphs of functions, while a few of them are. How do we tell which are which? A function x,y) 1--t /:
y) has the crucial property that, for each (x0,y0) in its
domain, there is precise!�
ne value zo uch that zo = f x o,y ). This means that, if
x0,y0) is in the domain o . �and we cut the graph with the vertical line through the
z
point x0,y
,0), then the
line will intersect the graph just once. This is the analogue
of the Vertical Line Test for functions of one variable: A surface in space is the graph of a function z y z
2y2- 4x2
.A. Figure 17 Hyperbolic paraboloid
vertical line intersects the surface
at most
nee.
=f
x,y) if and only if every
� E XAM P L E 9 Is the set of points satisfying the equation x2 + 3y2 + z2 = 12
the graph of a function of x and y
What about the surface described by
x2 + 3y2 + z = 12? Solution The first surface is
not the graph of
line x
a function. For If we slice it with the
= O,y = 1, we see that the equation becomes z2 = 9 or z = ±3. Thus both the (0,1,3) and (0,1,-3) lie on the graph and have xy-coordinates (0,1). By contrast, the surface x2 + 3y2 + z = 12 is the graph of a function because when x = xo,y =Yo are specified, then z must be 2 z = 12 - : xo) 2 - . Yo) points
in order for :x0,y0,z) to lie on the surface. For the second surface, z is uniquely determined by x
:1
nd y. For the first, it
is not.
..,..
11.2 Cylinders and Quadric Surfllees
B-81
We have studied six
kinds of quadric surfaces. It is a good idea to go through similarities and differences between the equations of these surfaces. Note: Only one example of each type of quadric surface is listed. For each type, permuting the variables x, y, and z results in a quadric surface of the same type the following table, noting
.
ellipsoid elliptic cone �lliptic paraboloid hyperboloid of one sheet hyperboloid of
two sheets
hyperbolic paraboloid
Q UICK
Q UIZ
1. The graphs of which of the following equations are cylinders in space? a) x-
i1' + z2; b) y=f1' + z2; c) r+Zt=y; d) r+Zt- z2
2. The graphs of which of the following equations are cones in space? a) x2- y-2z2; b) x2=i1' + 2z2; c) x=i1' + z2; d) x=Y.- z2 3. The graphs of which of the following equations are hyperboloids of one sheet in space? a) x2-Y. + 2i:2 -1; b) x2-i1' + 2z2= O; c) x2-i1' + 2z2=1; d) x2- Y.-2z2 =1 4. The graphs of which of the following equations are hyperbolic paraboloids in space? a) x=f - 2z2; b) x =i1' +Zz2; c) z2 + i1' =x2; d) z + Y.- x2 -
Answers 1. b, d
2. b
3. c
4. a, d
EXERCISES
Problems for Pradice In each of Exercises 1-14, the equation of a cylinder in xyz-space is given. Sketch its plot. L
x+y=�
2. 4x2+z2=4
3. 4x2+z=4· 4. y2 +2.z2=4 5. x2+z=-4 6.y-x2=9
'· x" +y=4 z1 +.f=2y 9. x1-2.f=1 10. 4x2- 2y 2 = 0 ny1-4z2=1 12. z= sinly) 13. x-2.x}'=l 14. z' -y2 =1 8.
In each of Exercises 15-30, sketch some slices, and then give a complete plot of the quadric surface that is the graph of
882
Chapter 11
Functions of Several Variables
the given equation. Identify each quadric surface as to type
point. This is why parabolic mirrors and antennas are
(cylinder, paraboloid, etc.), and determine if the surface is the
important. (Hint: Reduce the problem to a two-dimensional
graph of a function of x and y .
problem with a slice, and use the tangent line to see how the particle bounces.)
15. x2- 4y2+ 9z2=36
33. Prove that the circular paraboloid is the only surface that
16.x2+y2=z2
has circular symmetry in the x and y variables and that
17. y-x2 z2 18. x2- 9z2- 4y2= 16
has the property described in Exercise 32 .
=
The projection in the xy-plane of a space curve C is the set
19. 4x2+y2+ 9y2=25
{(x,y) : (x,y,z) EC for some z}. In Exercises 34 and 35, find
20. 2x2 + 4z2 y 21. x2+y2+z=4
the projection in the xy-plane of the given space curve C.
22. y2+x2-z=4
34. C
=
4x2 + y2 + 16z2 25. y2-z2-x2=1
24.
=
35. C
16
=
z
intersection
of
the
elliptic
paraboloids
is
the
intersection
=
of
and
the
hyperbolic
the
paraboloid
elliptic
paraboloid
1-x2/2-5y2/4 .
x2
28. z2- 4x2- 9y2=36
Calculator/Computer Exercises
29. y2-x-z2=0
In each of Exercises
30.x2-y2- 2z2=6
ratic equation, then the surface is a quadric surface. Give
39. x2+ 2xy + 2y2-z2
bounces off a smooth surface, then the angle
40. Plot z
of the bounce is determined just as though it bounces off
-4
the tangent plane at that point: The incoming angle with
O
=
s: y s:
x2 +y2
-1
=
and
2x + 2y +z
=
2
for -4
s: x s:
3,
3. Parameterize the curve of intersection. Add
the graph of this curve to your plot.
the tangent plane is the same as the outgoing angle. Now
4L Plot
consider a circular paraboloid z =x2+y2 . Show that =
=
38. x2+ 2xy+ 2y2-z2= 1
reasons for your answer.
there is a point P
plot the given quadric
36. x2+ 2xy + 2y2 + z2= 1 37. x2+ 2xy + 2y2-z2
True or false: If every level curve of a surface is a quad
32. If a particle
36-39,
surface.
Further Theory and Practice 31.
the
z=x2!2 -y2
26. z2- 4x2=y
27. y2+ 8z2
is
z=x2+ 2y2- 8 and z = 12- 4x2- 3y2•
23. 4x2 + y2=z
z2
=
4(x2+y2)
for
0 s: z
s:
6.
Plot
the
plane
z=v'z(l-x) for -6/v'z+ls:xs:l , -3s:ys: 3. Dis
(0, 0,p) on the positive z - axis with the
play the two plots in the same viewing box. Parameterize
property that any particle falling vertically through the
the curve of intersection, graph it, and add it to your
paraboloid and striking the wall of the paraboloid will
previous plot.
bounce to the point P. Thus P plays the role of a focal
1 1 .3 Limits and Continuity Now that we have treated functions and graphs, we can discuss limits. Recall from Chapter 2 that the limit at c of a function of one variable is the value that we expect the function to take at
c.
The same philosophy will prevail for functions of several
variables, but there is an important difference. In one variable, the domain is linear, and we decide what we "expect" the function will do at c by approaching
c
from the
left and from the right. In two variables, however, the domain is planar, and we decide what we "anticipate" the function will do at c by approaching
c
along all
possible paths that tend to c (see Figure 1). In a sense, then, it is more difficult for a function of two variables to have a limit. If
P1
=
(xi,y1)
and
P2
=
(x2,y2) are points in the plane, then it is convenient to d(Pi, P2). This distance is the length of the
denote the distance between them by
11.3 Limits and Continuity
y
y
y
x
... , I
,.. ... --- ..... ,D(Pa, r) ' ' \
�
I I I 1 \ \ \
' ' ,
..... ___ .....
;
.,,,,
I 1 I I , ,
directed line segment -------+
P1P2
represents:
-------+
P1P2,
or, equivalently, the magnitude of the vector that
d(P1,P2)= x
x
_. Figure 1 Two possible paths that tend to a planar point
\I
Pa
883
If
Po= (x0,y0)
II Mii = V(x2 -x1)2
is a point in the plane and
r
+
(y2 -y1)2 .
is a positive number, then
D(P0,r)
denotes the open disk of positive radius r that is centered at P0• That is, D(P0,r)= { PE IR2: d(P, Po)
_. Figure 2
Limits
In many of the constructions of calculus, we must investigate the limit of a function at a point that is not in its domain. For example, we learned in Chapter the
y
...
...
, I I I I I I \ \ ' ' '
... --- ..... D.(Pa,r) ...
Pa
, \ \ \ I I I I I I , ,
formula
limx-+O
sin(x)
!
sin(x) =cos(x)
depends
=1. Yet the point 0 is
entirely
on
establishing
not in the domain of the expression
the
that limit
sin(x)
that x x is under consideration. To handle similar situations in the plane, it is convenient to
D·(Po,r) formed from the open disk D(P0,r) D·(P0,r)={PE IR2: O
refer to the "punctured" disk removing the center:
............ ____ ,,,,,,
3
by
Po=(xo,yo) be a D·(Po,r) intersects the limit of f(P) as P=(x,y)
Suppose that/is a function of two variables. Let fixed point in the plane such that every punctured disk x
domain off. We say that the real number f, is the approaches (or tends to)
Po,
and we write
_. Figure 3 lim f(P)=£ P-+Po
or, equivalently,
if, for any c > 0, there is a t5 > 0 such that
lim f(x,y)=£, (x,y)-+(xo,yo)
If(P) - £1
for all points
P
in the
domain off with 0
Po
be in the domain off in order for us
884
Chapter 11
Functions of Several Variables INSIGHT
y
Compare this definition of limit for a function of two variables with the
definition given in Section 2.2 of Chapter 2 for a function of one variable. The principal
difference is that the distance
d(P,Po) =
between P=(x,y)
Jex-xo)2 +(y-yo)2 rather than I
and
Po= (xo,y0) is measured using
I· What does this mean? Because the
triangle in Figure 4 is at least as long as each leg, we have Ix-.xol s d(P, Po) and IY -Yol s d(P,Po). These inequalities tell us that when Pis close to Po, the variable xis close to xo. and, simultaneously, the variable yis close to Yo· Also,
hypotenuse of the right
the Triangle Inequality tells us that
Hx-XoH
II "PoPll=ll(x-xo)i+(y-yo)jll
d{P,Po)=
A Figure4
From this
s
ll(x-xo)ill + ll(y-yo)jll =lx-xol + ly-yol·
inequality, we see that, when both Ix-.xol
and
IY- Yol
are
small, the distance
d(P,Po) must also be small. In summary, d(P,Po)= (x-xo)2 +(y-yo)2 is small pre
J
cisely when x is close to xo and, simultaneously, y is close to yo.
� EXAMPLE 1 Solution
z
Define f(x,y) =x2 + y2. Verify that lim(xJ1)-+(o,o)/(x,y) =0.
Let I.=0, P= (x,y), and Po= (0,0). Suppose that e > 0. Then, lf(P) -ii= lf(x,y) - 01 =lx2 + rl.
Y
Thus
lf(P) - I.I< e if lx2 +II< e. But this last inequality holds when
J
d(P,Po) = (x - 0)2 + (y - 0)2< .Je. Unwinding these observations, we see that we should set 6=.Je. It follows that if 0< d(P,Po)< 6 then If (P) - ii< e. � � EXAMPLE 2
Define
A Figure 5
z
Discuss the limiting behavior of f(x,y)
as
if
(x,y) =f: (0,0)
if
(x,y) = (0,0).
(x,y)-.(0,0).
Notice that limx-+0/(x,O) =limx-+00=0. So f(x,y) approaches 0 as (x,y) approaches the origin along the x-axis (see Figure 5). Also, limy...o/(O,y)= lim1...o0 =O. Thus f (x,y) approaches 0 as (x,y) approaches the origin along the y-axis. However, we find a different limiting behavior when we consider points (x,y)= (t,t) that approach the origin diagonally as t-.0. In this case, we have
Solution
limf(t ' t)= lim y
x
lim. f(t, t) r-+O
A Figure 6
=1 2
t-+O
t-+O
t2_ = lim ! = ! t-+O 2 2. t2 + t2
_
See Figure 6. In summary, if (x,y) is sufficiently close to the origin on either axis then f(x,y) is near 0, but if (x,y) is near the origin on the line y=x, then /(x,y) is near 1/2. For the limit of/(x,y) to exist at the origin, the values of/must approach the same number no matter how the point (x,y) tends to the origin. Hence lim(x,y)...(o,o).f(x,y) does not exist. �
Limits and Continuity
11.3 INSIGHT
(0, 0), we
Po=
885
Although the function fin Example 2 happens to have been defined at do not actually use the defined value f(Po) when we investigate the
existence of limp_p0f(P). That is because the definition of "limit" refers to the punctured disk D·(P0,
8)
from which the point Po has been removed. Our analysis would have been
exactly the same hadf(Po) been assigned a different value or, indeed, hadf(Po) not been defined at all.
Rules for Limits
Fortunately, there are a number of rules about limits of functions that often make it easy in practice to determine limits. These rules are similar to those that we learned in Chapter
2
for functions of one variable. We record the new rules here for
reference.
THEOREM 1
Suppose that f and
g
are functions of two variables such that
lim f(x,y) (x,y)-+(xo,yo)
and
lim
(x,y)-+(xo,yo)
g(x,y)
exist. Then,
b.
lim(x,y)-+(xo,yo)(f + g)(x,y) lim(x,y)-+(xo,yo) f(x,y)+lim(x,y)-+(xo,yo)g(x,y). lim(x,y)-+(xo,Yo)(f - g)(x,y) = lim(x,y)-+(xo,Yo) f(x,y) - lim(x,y) (xo,Yo)g(x y)
C.
lim(x,y)-+(xo,Yo)(fg)(x,y)
d.
hm(x,y)-+(xo,yo)
a.
e.
=
-+
.
(lim(x,y)-+(xo,Yo) f(x,y)) (lim(x,y)-+(xo,Yo)g(x,y)) .
(!) (x,y) =
_
g lim(x,y)-+(xo,Yo)g(x,y) =/=- 0. lim(x,y)-+(xo,yo) (.Af(x,y)) -
=
(x,y)
tends
.
lim(x,y)-+(x0 ,yo)f(x,y) provided that r ·
lm(x,y)-+(xo,yo)g( X, Y)
.A lim(x,y)-+(xo,yo)f(x,y) for any constant .A.
f(x,y) (x to ( 1,2)?
� EXAMPLE 3 Define
behavior off as
,
+
=
y + 1 )/(x2 -y2).
What is the limiting
Solution Because
x2 -y2)
( (x,y)-+(1,2) lim
=
x2 -
lim
(x,y)-+(1,2)
lim
(x,y)-+(1,2)
y2 = 1 -4 = -3,
which is nonzero, we have
x+y+1 (x,y)-+(1,2) x2 -y2 lim
=
lim(x,y)-+(1,2)(x + Y+1 ) lim(x,y)-+(1,2)(x2 -y2)
_
lim(x,y)-+(1,2)X
lim(x,y)-+(1,2)Y+1 -3
+
� EXAMPLE 4 Evaluate the limit
2x2 + 5xy + 3y2 2x+3y (x,y)-+(3,-2) lim
=
1 +2+1 = -3
4
3
·
886
Chapter 11
Functions of Several Variables Solution Notice that, as (x,y)-+(3, -2), both the numerator and the denominator tend to
0.
Using our experience from Chapter
2,
we factor the numerator to see if
anything cancels. Our limit problem then becomes
lim ) (3,-2) (x,y->
Continuity
2x2+5xy+3y2
_
2x+3y
lim (x,y-> ) (3,-2)
(2x+3y)(x+y) = 2x +3y
lim (x+ y)= 1. (x,y-> ) (3,-2)
The definition of continuity for a function of two variables is no different from the one in Section 2.3 in Chapter
2 for
functions of one variable, except that we now
use our new notion of limit. The philosophy is the same: We require that the value that we
anticipatef will
take at Po be the value that
factually
takes:
Suppose that f is a function of two variables that is defined at a point Po= (xo,yo). Iff(x,y) has a limit as (x,y) approaches (xo,yo), and if lim f(x,y)=f(xo,yo), (x,y)->(xo,yo) then we say that f is
continuous at P0• If f is not discontinuous there.
continuous at a point in its
domain, then we say thatf is
Notice that we discuss the continuity or discontinuity of a function only at a point of its domain. Continuity off at Po means a. f(Po) is defined, b. limp_.p0f(P) exists, and c. /(Po)= limp_.p0f(P). The function f(x,y) =x2+ y2 from Example 1 is continuous at
{
(0,0)
because
lim(x,y-> y)= 0=f(O,0). However, the function ) (O,of(x, )
f(x,y)=
from Example
2
is
not
xy x2
�
continuous at
y2
if (x,y)
=J (0, 0)
if (x,y) =
(0, 0)
exist, so it certainly cannot equal /(0,0).
(0,0)
because lim(x,y-> ) (O,O) f(x,y) does not
� EXAMPLE 5 Suppose that
if (x,y)
=J (0,0)
if (x,y) = Isf continuous at
(0, O)?
(0,0).
11.3 Limits and Continuity Solution Observe that, for
(x,y) =!= (0, 0),
0::51/(x,y)I=
887
we have
(x2-y2)2 (x2 + y2)2 ::5 2 + 2 =x2+y2. x2 + y2 x y
In Example 1, we proved that limcx,y )-+(O o) (x2 + , tends to 0 as (x,y)-+(0,0). Therefore
y2)= 0.
It follows that
f(x,y)
also
lim f(x,y)=0=/=1 =f(O, O). (x,y )-+(0,0) We conclude that/is not continuous at
Rules for Continuity
(0,0).
<11111
Our first rule for continuity tells us that a continuous function of one variable gives rise to a pair of continuous functions of two variables.
THEOREM 2
Suppose that
c inside the interval. Let f(x,y) = c.p(x) for a< x< b, -oo < y< oo and let g(x,y)=c.p(y) for-oo < x< oo, a< y< b. Then, for every yo, the function f is continuous at (c,y0), and, for every x0, the function g is continuous at (x0, c). point
Proof. We will prove that/is continuous at
Po = (c,y0).
Continuity of
g
is handled
in the same way. Let E be an arbitrary positive number. Because
c, there is a positive number 8 such that lc.p(x) - c.p(c)I< E P=(x,y) satisfies d(P, Po)< 8, then we have
if
Ix - cl< 8.
Now, if
Ix- cl= J(x-c)2::5 J(x-c)2 + (y-yo)2=d(P,Po)< 8. It follows that
lf(P)-f(Po)I=l f(x,y)-f(c,yo)I=lc.p(x)-c.p(c)I< E if
d(P, Po)< 8,
which establishes the continuity off at
Po.
•
We will use Theorem 2 in conjunction with the following rules for combining functions of two variables, which are similar to those that we learned in Chapter 2 for functions of one variable.
f and g are functions that are continuous at Po=(xo,yo). Then, f + g,f- g, and fg are also continuous at Po. If g(Po) =/= 0, then f / g is also continuous at P0. If is a function of one variable that is continuous at f(P0), then f is continuous at Po.
THEOREM 3
Suppose that
a
�
EX A M P L E 6 Discuss the continuity of 3 . U(x ' y)=y sm(x)-
cos(xy2) . (2x- y)2
888
Chapter 11
Functions of Several Variables
We first observe that V= {(x,y) : y =I= 2x} is the domain of U. Suppose that Po= (x0,y0) is a point in V. Because Xf--+Sin(x) and yf--ty3 are continuous functions of one variable, we know from Theorem 2 thatf0(x,y)= y3 andg0(x,y)= sin(x) are continuous at Po. Theorem 3 then tells us thatf(x,y)=fo(x,y) go(x,y)= y3sin(x) is also continuous at P0• By similar reasoning, we see that the continuous single-variable functions xf--tx, xf--t2x, and yf--ty2 can be combined to form the continuous functions /1(x,y)= xy2 and fz(x,y) = 2x-y. Because ¢1(u) = cos(u) and
Solution
°
°
g(x,y)=
cos(xy2) gl(x,y) = gz(x,y) (2x -y)2
is continuous at P0• Finally, by Theorem 3, U(x,y) f(x,y) -g(x,y) is also continuous at P0• In summary, U is continuous at every point in its domain. .,.. =
A function that is continuous at every point in its domain, such as the function U of Example 6, is said to be a continuous function.
Functions of Three Variables
We conclude by noting that the entire discussion of this section applies to functions of three variables x, y, z, with no change (except for notation). In particular, all of the rules and remarks are still true. � EXAMPLE 7
Show that V(x,y,z) =z3cos(xy2) is a continuous function.
The domain of V is all of xyz-space because the expression z3cos(xy2) makes sense for all values of x, y, and z. Therefore we must show that V is continuous at every point (x,y,z). In Example 6, we observed that the function (x,y) f--+Cos(xy2) of two variables is continuous at every point (x,y). The other factor of V(x,y,z), the function zf--tz3 of one variable, is also continuous. The extension of Theorem 2 to three variables tells us that the functions f(x,y,z)= z3 and g(x,y,z)= cos(xy2) are continuous at every point (x,y,z). The extension of Theorem 3 to three variables then tells us that the product V(x,y,z)= f(x,y,z)·g(x,y,z) is also continuous at every point (x,y,z). That is, V is a continuous function. .,.. Solution
Q UIC K
Q UIZ
1. True or false: If f is not defined at Po, then limP-+Pof(P) does not exist.
x2-y2 2. What is lim(x,y)-+( , ) 33 x y ? 3. True or false: If
Answers
1. False
2. 6
3.
True
4. True
a
11.3 Limits
EXERCISES
L 2. 3. 4. S. 6. 7. 8. 9. 10. lL
1-14, calculate
13. 14.
lim(x,y)--+( ,3jX(2 xy-3) 2 2 Y lim(x,y)--+(-l,6)sin 7r
(;)
23.
16. 17. 18. 19. 20.
lim(x,y)--+(4, ) Vx2y - X 2 x2-y2 . lim(x,y)--+(1,-1) x+ y
24.
lim(x,y)--+(-1,4) 1n(y2+ xy) lim(x,y)--+(4,-8)X-1/2yl/3 arccos(x - y) . lim(x,y)--+(1/4,1/4) ex-y
(
lim(x,y)--+(.-/ , ) arctan y'ycos 22
f(x,y)=
. (x-1)(y2-4) lim(x,y)--+(1, ) 2 (y 2)(x2-l)
ysin(x) . lim(x,y)--+(0,3) x( y2 + l)
y/ x
y(x2 + rrl/2
xy2/lxy21 (x2 + y2)j(x2-y2) (x + y)/(x -y) (x + y)/(x + 2y)
x4 + 3y4 x2 y2
:
(x,y) =I (0, 0) (x,y)=(0, 0) (x,y) =I (0, 0) (x,y)
=
(0, 0)
'if;(y) is continuous, as can be deduced from Theorems 2 and 3. Instead, give a direct, E:-8, proof. 26. Iff is continuous at (xo,yo), show that ¢(x) f(x,yo) is continuous at xo and that'if;(y)= f(xo,y) is continuous at yo. 27. Suppose that ¢(x) = f(x,yo) is continuous at Xo and 'if;(y)=f(xo,y) is continuous at Yo· Is it true that f is continuous at (xo,yo)? Give a reason for your answer. l/(xy) . 28. Calculate lim(x,yHl,OJ(l + xy) =
Calculate lim(x,yHO,o) x21n(x2 + y2). 2xy/(x2 + y2). Calculate limy--+0f(O,y). Also, 30. Let f(x,y) for any slope m with -a:i < m < a:i, calculate 1im x--+0f(x,mx). Show that f(x,y) has a limit as (x,y) approaches (0, 0) along any straight line approaching the origin, and that the limit can be any number in the interval [-1, 1]. 3L Letf(x,y) xy2/(x2 + y4 ). This function does not have a limit as (x, y) tends to (0, 0). However, you cannot detect this by looking only at straight-line paths approaching the origin. Calculate limy--+0/(0,y). Also, for any slope -a:i < m < a:i, calculate lim x--+0f(x,mx). Deduce that if (x,y) approaches the origin along any straight line, then the limit off(x,y) is 0. However, show thatf(x,y) has a different limit as (x,y) approaches (0, 0) along the para bolic path x y2• 32. Let f(x,y) xy3/(x2 + y6 ). Prove that along any straight line path approaching the origin, the limit of f(x,y) is 0. However, show that there is another path along whichf(x,y) has a different limit as (x,y) approaches (0,0). Therefore f(x,y) does not have a limit as (x, y) tends to (0, 0). =
=
=
Further Theory and Practice In each of Exercises 21-24, determine whether the given function is continuous at (0, 0).
=
;
(0,0)
=
_
f(x,y)
y2-2x2 x2 + 2
=
2 9.
. x2-xy-2y2 lim(x,y)--+(6,3) X 2Y 1 - exp(x2+y2) . lim(x,y)--+(0,0) X2 + y2
{
(x,y)
25. If
@))
lim(x,y)--+(0,.-/ )sin(ycos(x)) + cos(x + cos(y) ) 2 sin(x+y) . lim(x,y)--+(0,0) X+ Y
2L
{ {
f(x,y)=
In each of Exercises 15-20, the given expression does not have a limit at (0, 0). Let (x, y) approach the origin along two different paths, as in Example 2, to verify the nonexistence of the limit. 15.
(x,y) =I (0, 0)
the indicated limit.
_
12.
889
22.
Problems for Practice In each of Exercises
and Continuity
sin(x2 + y2) x2 y2
;
Calculator/Computer Exercises (x,y) =I (0,0) (x,y)= (0, 0)
33.
Plot f(x,y)=xy/(x2+y2) for (x,y) in a rectangle cen tered at (0, 0). Use your plot to verify visually thatf does not have a limit at (0, 0).
890
Chapter 11
FunctiOl'I• of Several V.riabln
Plot f(x,y) =(r -y2)/(r + y2) for (x,y) in a rectangle centered at (0, 0). Use your plot to verify visually that f does not have a limit at (0, 0). 3.S. Plot f(x,y) =(x + y)2/(r + y2) for (x,y) in a rectangle centered at (0, 0). Use your plot to verify visually that f does not have a limit at (0, 0). 34.
36. Plot f(x,y} =xy2 /(r + y') for
(x,y) in a rectangle cen tered at (0, 0). Use your plot to verify visually that f does not have a limit at (0, 0). Note, however, that the limit does CJcist and equals 0 along straight lines through the origin.
11 .4 Partial Derivatives determine the rate at which a function f of two variables changes. To do so, we will need an appropriate notion of differentiation. Because there are two independent variables, we shall develop a procedure for differentiating in each variable separately. The geometric idea behind these new differentiation techniques is simple: We take slices of the graph off. You are already familiar with the strategy of intersecting the graph of z=f(x,y) with planes. Up until now, we have relied primarily on horizontal planes as an aid in plotting. Such planes, however, are of no use for our cUITent purposes because the values of f(x,y) do not change on the level curves that they produce. Instead, we will now intersect the graph ofz= f(x,y) with planes that are parallel to the xz- and yz-coordinate planes. Let Po= (xo,Yo) be a point in the xy-plane. Suppose that f iir. a function whose domain contains a disk centered at Po. If we :fix Yo and only allow x to vary, then we obtain a function cp(x)=f(x,y0) of one variable. The graph of z =cp(x) can be reali7;ed by intersecting the graph of z =f(x,y) with the plane y =y0 as shown in Figure 1. Ordinarily, the graph ofz =ip(x) would be plotted in the xz-plane, which is the plane y =0 in xyz-space. Here it is realized in the plane y =y0, which is parallel to the xz-plane. Because cp is a function of one variable, we can differentiate it in the ordinary manner. Not only that, the number In this section, we will learn how to
x
A. Figure 1
'( ) = fun rp(Xo + �) - cp(Xo) = fun f(xo +�.Yo) - f(xo,Yo) Xo 'P 4J:-oO &1:...0 tu � has its usual interpretation-the instantaneous rate of change of cp at X(). Because ip is just the function/in which they variable has been fixed aty0, the value of <;/(Xo) is the instantaneous rate of change off at (Xo, Yo) as x changes and y =Yo remains fixed. In
an analogous manner, we fixx =.xo , define f/l(y) =f(x0,y), and use
1J/(yo) =
fun
.o.)HO
f/l(yo +a y) -1/J(yo) = ay
fun
�y...o
/(Xo,Yo + ay) -f(Xo,Jo)
to tell us the instantaneous rate of change of f at (.r0,y0) A. Figure 2
ay as
y changes and x =x0
remains :fixed. See Figure 2 for the geometry. These considerations give us a way to define differentiation for a function of two variables.
11.4 Partial Derivatives
�·Mll!!l[.Jll
891
Let Po (x0,y0) be a point in the xy-plane. Suppose that f is a function that is defined on a disk D (Po,r). We say that f is differentiable with �
respect to x at Po if
t(xo + � x,Yo) - t(xo,Yo) . 11m &--->O �x exists. We call this limit the partial
derivative
of t with respect to x at the point
Po and denote it by at ax
(Po).
differentiable
Similarly, we say that t is
with respect to y at Po if
t(xo,yo + �y) - t(xo,yo) 1.1 m �y
b.y--->O
���������
exists. We call this limit the partial
derivative
of t with respect to y at the point
Po and denote it by at ay
� EXAMPLE 1
(Po).
Define t(x,y) =xy2. Calculate the partial derivatives with
respect to x and with respect to y. What are values with the derivatives
d
I
cp dX x=l
and
8/x (1,
d 'lj; dY y=3
I
3) and
: (1,
where 'P and
3)? Compare these
'ljJ are the functions of
one variable that are defined by cp(x)= t(x,3) and 'l/;(y)= t(l,y). Solution When we calculate at /ax, we differentiate the expression t(x,y) with
respect to x, treating y, and any expression depending only on y, as if it were a
constant. In the case of xy2, that means we differentiate, with respect to x, the product of the constant y2 and the variable x. The result is y2 times
1.
Thus
at Iax =y2. Similarly' when we calculate at Iay' we think of x as a constant and differentiate the expression
t(x,y) with respect to y.
In other words, we
differentiate with respect to y, the product of the constant x and the expression y2• The result is the constant x times 2y. Thus at jay= 2 xy. By substituting x=
1
and
y= 3, we obtain
�� (1,
3)= llx=l,y = 3 = 9
and
: (1,
3)=
2xylx=l,y=3 = 6.
Next, consider the function cp(x) =t(x, 3) =(x)(3)2 =9 x. We calculate cp'(x) =9. Therefore cp'(l) =9, which is the same as at /ax evaluated at 'l/;(y)= t(l,y)= (l)(y2)= y2 and
(1, 3).
Similarly,
892
Chapter 11
Functions of Several Variables d
l
1
¢'(3) = d (y2) =6, which equals =2y y y=3 y=3
af ( 3) 1, . ya
....
All of our familiar rules for differentiation-the Product Rule, the Quotient Rule, the Chain Rule, and so on-are still valid for partial derivatives because partial differentiation is just the differentiation that we already know,applied one variable at a time. �
ExA M p L E
f(x,y) =Jx3 +y2. Solution
2
Calculate the partial derivatives
atI ax and
atI ya for
Applying the Chain Rule,we obtain
at (x,y) = .!!_ (x3 +y2)1/2 = ! (x3 +lrl/2 .!!_ (x3 +y2) = �x2(x3 +lrl/2 ax ax 2 ax 2 and
There are several common notations for the partial derivatives of a function (x,y) f(x,y) of two variables. Thus f-t
at , ax
Dxf,
fx, fi,
and
Dif
are all used for the partial derivative of /with respect to x. Similarly, af jay,Dy f,fy, h, and DLf all stand for the partial derivative of f with respect to y. �
EXA M P L E 3
Solution
Calculate fx and fy for the function f defined by x f(x,y) =ln(x)e cos(y).
We calculate fx using the Product Rule and Chain Rule:
fx(x,y) =
c� ln(x)) �cos(y) +ln(x) ! �cos(y) =� �cos(y) +ln(x) (�cos(y)cos(y)) .
When we calculate a partial derivative with respect to y,we treat x as a constant. Therefore ln(x) is treated as a constant when calculating fy: We do not need the Product Rule for this calculation: fy(x,y)
=
ln(x)
� (�cos(y))
=
ln(x)�cos(y)
� ( xcos(y))
=
-ln(x)�cos(y)xsin(y).
....
A string in the xy-plane vibrates up and down in the y-direction and has endpoints that are fixed at (0, 0) and (1, 0). Suppose that the displacement of the string at point x and time t is given by y(x,t) =sin(7rx)sin(2t). What is the instantaneous rate of change of y with respect to time at the point x =1/4? � EXA M P L E
4
11.4 Partial Derivatives
893
Solution We calculate
g
t
y(x, t) =
g( t
sin(rrx)sin(2t) ) =sin(rrx)
g
t
sin(2t) =2sin(rrx)cos(2t).
Therefore the instantaneous rate of change with respect to time of the string's displacement from equilibrium
at the point
x= 1/4 is 2sin(rr/4)cos(2t) or
v'2cos(2t). .,..
INSIGHT
Example 4 demonstrates that the partial derivative with respect to a
variable can be interpreted as the
rate of change
with respect to that variable. Many
physical quantities depend on several parameters, so that we need to have this new multivariate way to measure rates of change. For instance, if heat is being applied to the
T at a point x on the rod depends both {)TIOX and {)TIot tell us about the rates
ends of a rod, then the temperature the time
t.
The partial derivatives
on x and on of change
when one of the two independent variables is held constant.
Functions of Three Variables
The notion of partial derivative also applies to functions of three (or more) vari ables. If F(x, y,z) is an expression involving three independent variables x, y, and z, then we calculate aF /ax by holding y and z constant and differentiating with respect to x. The partial derivatives in y and z are calculated similarly. � EXAMPLE
5 Calculate the partial derivatives of F(x,y,z) =xzsin(y2z)
with respect to x, with respect to y, and with respect to z. Solution When we hold y and z constant, the expression zsin(y2z) is a constant and f(x,y,z) is this constant times x. Therefore aF/ax= zsin(y2z). Next, hold x and z constant to calculate the partial derivative with respect to y: aF ay
(x,y,z)
�( �(
xz sin(y2z))
xz
sin(y2z))
xz cos( y2z)
� �
(the constant slides outside the derivative)
( y2z)
xz2 cos( y2z)
( y2)
(using the Chain Rule) (the constant slides outside the derivative)
2xyz2 cos( y2z).
We calculate aF/az in a similar way, but for this partial derivative we require the Product Rule:
��
(x,y,z) =
!(
xzsin(y2z)) =x
!(
z sin(y2z)) =x sin (y2z) +xz
!(
sin(y2z)) .
894
Chapter 11
Functions of Several Variables
Continuing the calculation with an application of the Chain Rule, we have
�� (x,y ,z) =xsin(y2z) +xzcos(y2z) � (y2z) =xsin(y2z) +xy2zcos(y2z).
..,..
For a function (x,y ,z) f--tF(x,y ,z) of three variables, there are several alter native notations for the partial derivative with respect to x: aF ax' Similarly, aF jay, D y F, Fy, F , and D F can be used interchangeably for the partial 2 2 derivative with respect to y. Likewise, aF /az, DzF, Fn F3, and D3 F all refer to the partial derivative with respect to z.
Higher Partial Derivatives
Because the partial derivative of a given function f (x,y ) is a new function, we may attempt to differentiate that function again. The result, if it exists, is called a second partial derivative. Some examples are
( )
a2t a a which means ! ; ax ax axz .
. a2t a which means z ay ay
( )· a f ay
,
( )
a2t . a a which means f ; axay ax ay
( )
a2t a a which means ! ay ax . ay ax .
The last two of these higher order partial derivatives involve differentiation with respect to both x and y. They are therefore sometimes called mixed partial derivatives.
Of course we can also use the other partial derivative notations for second derivatives. For example, a2f/ax2 can be written as
and
:bx
Dxxf
or fxx
or !i,1
or
D1,i/,
or fxy
or !i,
or
D1,zf,
may be written as Dxyf
and so on.
2
11.4 Partial Derivatives
895
� EX A M P L E 6 Calculate all the second partial derivatives of
f(x,y) =xy- y3+x2y4. Solution We have
)
(
&t (x,y) = .E_ at (x,y) = .E...( y+2xy4) =2y4, ax ax
ax2
(
)
(
)
ax
&t = .E... at (x,y) = .E...(x-3y2+4x2y3) =-6y+ 12x2y2,
ay2
ay ay
ay
&t = .E... at (x,y) = .E...(x-3y2+4x2y3) =1+8xy3,
axay
ax ay
(
ax
)
&t = .E... at (x,y) = .E...( y+2xy4) = 1+8xy3.
ayax INSIGHT
ay ax
ay
and
.,..
A noteworthy occurrence may be observed in Example 6. The first order
partial derivatives for f(x,y)=xy
- y3+x2y4
:(x,y)=y+2xy4
and
turned out to be
:(x,y)=x-3y2+4x2y3.
At first glance, these expressions do not appear to have much in common. Yet, when we used
: and : to calculate
the mixed partial derivatives
::x :2;£y' and
we discovered
that
&t = 1+8xy3 = &t . axay 8y8x That is, differentiating first with respect toy and then with respect to x results in the same expression that is obtained by differentiating first with respect to
x and then with respect
toy. Is this a coincidence, or a particular case of something that is true all the time? The answer is that if we restrict our attention to a class of reasonably nice functions, then the order of differentiation is, indeed, irrelevant. The class of "nice" functions that we will study is described in the next definition.
A function
an open disk D (P0,
r)
if
f
(x,y) � f(x,y) fx
are continuous on D (P0,
entiable
on D (Po,
r)
is called
continuously differentiable
on
and its partial derivatives
r).
fy
and
The function
f
is called
twice continuously differ
if f and its partial derivatives
fx, are all continuous on D (P0,
fy, r).
fxx,
fyy,
fxy,
fyx
896
Chapter 11
Functions of Several Variables
THEOREM 1
disk
D(P0, r),
If
(x,y)
�
f(x,y) is twice continuously differentiable on an open
then its mixed partial derivatives are equal in the disk:
(11.4.1)
fxy(x,y) fyx(x,y) (x,y) ED(Po, r). =
INSIGHT
The proof of Theorem 1 is discussed in Exercise 78. Does Theorem 1
mean that we are going to have to sit down and check that every function we encounter is twice continuously differentiable? Fortunately, the answer is "No." The functions that we encounter in practice are built up using algebraic operations and composition involving functions that are obviously twice continuously differentiable. For the functions that we will be using in the rest of this book, the real meaning of Theorem 1 is that we
can
differentiate in any order we like.
Concluding Remarks about Partial Differentiation
The derivatives that we learned about in Chapter 3, namely derivatives of functions of one variable, are sometimes called
ordinary derivatives to distinguish them from
the partial derivatives that we discuss in this chapter. When we compute an ordinary derivative, we write write
a/ ax
or
a/ ay
d/dx,
but when we compute a partial derivative, we
(or one of the other notations that we have discussed).
Sometimes a partial derivative can turn out to be an ordinary derivative. For example, if
g is a function of one variable, then
a��y (g(x)y
+
4 sin(x))
=
! (� (g(x)y
+
) fx g(x).
4 sin(x))
=
In some applications, it is necessary to calculate derivatives of order higher than two. For example,
a ax axay
( &t)
is a third derivative obtained by applying
!
We write the resulting partial derivative as
where j + k
=
to the already familiar quantity
a��y·
In general, the expression
m, denotes a derivative of order m: Differentiation in
j times, and differentiation in
y
::ty"
x
is applied
is applied k times. To guarantee that the differ
entiations may be applied in any order with the same result, one usually assumes that the function is m times continuously differentiable (all partial derivatives up to and including order m exist and are continuous).
11 A Partial Derivatives � EXAMPLE
;
& 8Y
7 Calculate
xy - y3+x2y4 of Example 6.
&'f for the function f(x,y)= 8xZ8y
and
Solution Using the calculationfyy(x,y) = -6y+12x2y2 from Example
(
&t - 8 8y3 ay
&t (x,y) ayz
Using the calculation fxx(x,y)
=
)
6, we obtain
8 ( -6y+1 2y2) - -6+24x2 y - . 2x ay __
2y4 from Example 6, we obtain
a &t ( x,y) 8x28y By _
-
897
(
a2t (x,y) 8x2
)
_
a (2 4) Y -By3 By _
·
Alternatively, we might have used the calculation fxy(x,y) 1+8xy3 from Example 6 as our starting point for the last calculation. Because f is three times continuously differentiable, the result is the same: =
&t ( x,y) - a 8x28y 8x _
Q UIC K
Q UIZ
( ) &t 8x8y
<1111
a ( +Bxy3) -By3. 8x l
-
_
_
1. True or false: If r,o(x)=f(x,2) is differentiable, then r,o'(3) =
8/x
(3,2).
2. Calculate fx(3, -1) and fy(3, -1) for f(x,y)= x2y3+4x+10y2. 3. Calculate fyy(S, 3) and fxy(S, 3) for f(x,y)= x2ln(y). 4. Calculate
a��2
for the function f(x,y)
=
xy - y3+x2y4 of Example 6.
Answers
1. True fxy(S,3)
2. fx(3, -1) -2 and fy(3, -1) 10/3 4. 24 xy2 =
=
=
7
3. fyy(S, 3)
=
-25/9 and
EXERCISES Problems for Practice
9.
In each of Exercises 1-10, a function f and a point Po= (xo,yo) are given. Calculate ip(x) f(x,yo), 1/!(y) f(xo,y), (ji(xo) , and 1//(yo) . Then, calculate (8f/8x)(xo,yo) and (8f/8y)(xo,yo) . Verify that (ji(xo)= (8f/8x)(xo,yo) and 1f/(yo)= (8f/8y)(xo,yo) .
10. f(x,y) -
In each of Exercises 11-20, a function (x,y)>->f(x,y) s i given. Calculate f fyy, fxy, and fyx· Verify that the mixed partial derivatives are equal.
1.
lL
=
2. 3.
4.
5. 6.
7. 8.
f(x,y)=2x+y3 f(x,y)= 7r + sin(x) - cos(y) f(x,y)=7 - 3y4 f(x,y)= 1+3xy2 - y f(x,y)= cos(xy2) f(x,y)= xeY f(x,y)= x ln(xy) x2 f(x,y)= -x + 2y
Po=(S,1) Po=(7r/4,7r/3) Po=(2,-1) Po= (1,2) Po=(7r,1/2) Po=(1,0) Po=(1,e) Po=(4,2)
=
f(x,y) =xY
2x - y
1+3x+3y
xx,
x2y - xy + ys x 12. f(x,y) _ _ x+y 13. f(x,y) =x2 - x/y f(x,y)
=
=
14.
..jx2 + 2y sin(x) 15. f(x,y) - -cos( y) 16. f(x,y) sin(xy) 17. f(x,y) In(xy - y2 ) f(x,y)
=
_
=
=
Po=(2,2) Po=(3,-2)
898
Chapter 11
18. f (x,y) 19. 20.
Functions of Several Variables
y2ln(x+y) f(x,y) cos(x) sin(y) f (x,y) =ln(x+eY) =
=
46.
f(x,y) =eYcos(x)
47. f(x,y) =tan(x2y)
48. f(x,y) =xJx2+4y
For each of the functions in Exercises 21-30, calculate Dyf , Dyyf, Dyx/, fa, and fH. Verify that Dyx/ =fa.
49. f(x,y)
21. f(x,y) =xsec(y)
In each of Exercises 51-54, an expression f(x,y) and a point Po=(xo,yo,zo) are given. The plane x=xo (respec tively, y=Yo) intersects the graph of z=f(x,y) in a curve C1 (respectively, C ) with tangent line £1 (respectively, £ ) at Po. 2 2 Find the y-coordinate (respectively, x-coordinate) of the point of intersection of £1 (respectively, £ ) with the xy 2 plane.
x 22. f (x,y) =1 -xy
23. f (x,y) =sin(x3 +2y) 24. f (x,y)
=
tan(x2+y2)
25. f (x,y) =(sin(x) - cos(y))5 26. f (x,y)
=
e 2y -
27. f (x,y) =(x+ 3y) 4 28. f (x,y) =1/(4 + xy2) 3 29. f (x,y)
=
Jl + 3 x2+y4
30. f (x,y) =3y2 - sin(xy'.Y)
In each of Exercises 31-40, a function (x,y,z) � f (x,y,z) is given. Calculate fx, fy, fz, fxy, fxz, fyz, fxx, fyy, and fzz· 31. f(x,y,z) =2x3+y5 z8
32. f(x,y,z) =x3y-3zln(y) 33. f(x,y,z) =(x+2y + 3z) 5 34. f (x,y,z) =sin(x+yz) 35. f(x,y,z) =ey - yz
36. f(x,y,z) =x2/(y2+z2)
37. f(x,y,z) =yln(xz)
=
arctan(x+2y)
50. f (x,y) =xY
51. x3y-2xy3
(2,1,4)
52. 2x/(x+y)
(3, -2, 6)
53. x2vfl +2x+y2
(2, -2,12)
54. Sy+
(3,2,10)
� xln(x2-y3)
55. Express the volume V. of a cone as a function of height h.
and radius r. What is the rate of change of volume with respect to radius? With respect to height? 56. Express the surface area A of a box in terms of length x, width y, and height z. What is the rate of change of sur face area with respect to length? 57. The pressure P, volume V, and temperaure T of a sample of an ideal gas are related by the equation PV =kT for a constant k. Express each variable, P, V, and T, as a function of the other two variables. Then show that
38. f(x,y,z) =ln(x3y +2z) 39. f(x,y,z)
=
(x2+y3+z4) 3
40. f(x,y,z) =e zcos(x)sin( y)
In each of Exercises 41-44, a function (x,y,z) � f (x,y,z) is given. Calculate fxyz, fyxz, fxzy, fzxy, fyzx, and fzyx· (All six calculations should result in the same final expression but will involve different intermediate results.)
(Notice that the partial differentials cannot be formally canceled.) 58. Show that u(x,y) =x sin(x/ y) satisfies the equation xux(x,y) +yuy(x,y) =u(x,y). 59. Show that u(x,y) ln(e +eY) satisfies the equation ux(x,y) +uy(x,y) 1. 60. Show that u(x,y) arctan(y/ x), u(x,y) ln(y) -ln(x), and u(x,y) 2xy/(x2+y2) all satisfy the equation xux(x,y) +yuy(x,y) =0. =
=
41. f(x,y,z) =xy2z3
42. f(x,y,z) =xy+2xz
3+ycos(z) x 44. f(x,y,z) =(x+2y + 3zr2 43
•
/(
x,y,z)
=
=
=
=
The partial differential equation
Further Theory and Practice
In each of Exercises 45-50, a function (x,y) � f (x,y) is given. Calculate fxx, fyy, fxy, and fyx· Verify that the mixed partial derivatives are equal. 45. f(x,y) =
x-y x+y
is called Laplace's equation. It describes, for example, the steady state heat distribution in a thin metal plate. A function u(x,y) that satisfies Laplace's equation is said to be harmonic. Exercises 61-67 concern solutions of Laplace's equation.
11.4 Partial Derivatives
61. Which of the following functions is harmonic? a.
b. c.
d. e. f.
70. Let
u(x,t)
describe the height above or below the
x
time t of the point
u(x,y)=x2-y2 u(x,y)=xy u(x,y)=e"cos(y) u(x,y)=x2+y2 u(x,y)=x3 x u(x,y)=e- sin(y)
x=0
and
x=27r.
899
x axis at
on a vibrating string with ends fixed at
Using principles of elementary mechan
ics, it can be shown that u satisfies the wave equation
62. Verify that the function
where
c
is a constant that depends on the properties of the
string. Show that the functions
um(x,t)+Am sin(cmt) sin(mx), mEZ, is harmonic. It is called the unit disk
Poisson kernel
for the open
satisfy the wave equation.
D(O,1).
Suppose that fx(x,y)=2x
63. Verify that the function
the
half-plane
for the upper
64. Determine all nonzero solutions of Laplace's equation
u(x,y)=Ax2+Bxy+ Cy2• These are the spherical harmonics of degreee 2. 65. Suppose that A and E are given constants. Determine B and C so that u(x,y)=Ax3 +Bx2y+ C xy2+ Ey3 satisfies Laplace's equation. In this case, we call u a spherical harmonic of degree 3. 66. Let (xo,yo) be any fixed point in the plane. Show that u(x,y)=ln ((x-xo)2+ (y - Yo)2) is harmonic. 67. Let vn(x,y)=(x2 +y2r. Show that having the form
( a2 + a2 ) Vn(x,y)= --4n2 !l2 uy
with
andfy(x,y)=3y2• If we integrate to x, then we obtain
respect
If we differentiate this last equation with respect toy, holding
Poisson kernel
{(x,y):y>O}.
!l2 ux
equation
f(x,y) = x2+g(y). Notice that the constant of integration is a function ofy becausey is held constant when fx is calculated.
1 y P(x ,y)=- -7r x2+y2 is harmonic. It is called the
first
x fixed, we fy(x,y)=3y2. Here C is a expression for
Vn(x,y).
Deduce that vn(x,y) cannot be harmonic for any
fy(x,y)=g'(y). But we are given g'(y)=3y2 and g(y)=y3+ C.
constant of integration. If we substitute this
g(y)
into our formula for
given equations.
fx(x,y)=2xy fx(x,y)=2x+2 73. fx(x,y)=2x 74. fx(x,y)=3(x+y)2 -1
fy(x,y)=x2+1 fy(x,y)=2y fy(x,y)=1 - 2y fy(x,y)=3(x+y)2+2
71. 72.
P(x,y)=e-xy
an+m
0 xn 0ym
nomial of total degree
e"Y. Show that
m+ n in x
_
69. If
y2
uy
u(x,t)
denotes the temperature at position
t,
then
u
satisfies
8u u =ol& 8x2 8t The number
a
x in a the heat equation
and y. In fact,
where µ is the minimum of
m and n.
76. Define
long
if
(x,y)-/= (0,0)
if
(x,y)=(0,0).
Calculate fx and fy. (Note: To calculate these at (0, 0), you
·
is a constant determined by the heat con
ductivity properties of the rod. Show that the functions
un(x,t)=exp(-n2olt)sin(nx),
is a poly
z2
ux
thin rod at time
P(x,y)
n-/= 0.
2n(2n+1) (nz a2 a2 a2 ) + + nz nz Vn(x,y,z)- x + + vn(x,y,z).
2
then we see
method just outlined to find a functionf(x,y) that satisfies the
68. Show that vn(x,y,z)=(x2+y2+z2r satisfies
uz
f(x,y),
thatf(x,y)=x2+y3 + C. In each of Exercises 71-74, use the
75. Let
x2+y2
obtain
It follows that
nEZ,
have to resort to the
definition
of partial derivative, using
a limit, and you have to calculate the limit by hand.) Now calculate fxy(O,0) and fyx(O,0). They are unequal. Explain why Theorem
77. In Chapter
3,
1
is not contradicted.
we observed that if a function of one vari
able is differentiable at a point, then it is continuous at satisfy the heat equation.
that point. However, for a function of two variables,
900
Chapter 11
Functions of Several Variables
continuity does not follow from the existence of the
d. Subtract the two formulas for
partial derivatives. For the function
f(x,y) + prove that fx and
{
xy x2 + Y2 0
fy
if
(x,y)
if
(x,y)= 0
exist at (0,0), yet
f is
Calculator/Computer Exercises In Exercises 79 and 80, use numerical differentiation to cal
discontinuous
culate fx(Po) and fy(Po) for the given function f and the given point
1.
Fix a base point sion of order
(xo,yo).
1 in the x-variable about the point (xo,y).
first-order Taylor expansion the point c.
P1
f(x,y)= ( y'i + =
from part a in a
in the y-variable
Po.
eY-x)/-J3 + x2 + y4, Po+ (4, 3) 80. f(x,y) ..jx2cos2(y) + y4/ .Jl +y2cos2(x) +x4, Po= (7r/4,37r/4)
79.
Expand fin a Taylor expan
b. Expand the Taylor polynomial
you
(0, 0) to obtain the result.
=F (0, 0)
78. Complete the following outline to obtain a proof of a.
that
Divide through by (x -xv)· (y -yo), and let (x,y)-+
e.
at (0, 0). Theorem
f(x,y)
obtained in parts b and c, canceling like terms.
SL Use numerical differentiation to calculate
about
fyy(Po)
Yo·
for the function[ and the point
Po
82. Use numerical differentiation to calculate
Repeat steps a and b with the roles of x and y reversed.
fyy(Po)
for the function f and the point
Po
fxx(Po)
and
fxx(Po)
and
of Exercise 79. of Exercise 80.
1 1 .5 Differentiability and the Chain Rule Partial derivatives tell us how a multivariable function changes one variable at a time. In practice, it often happens that two or more of the independent variables are simultaneously changing. In this section, we will learn how to analyze the rate of change of a function when more than one of its variables is changing. Throughout this discussion, the notation scalars; there are
no
a
·
b will be used to signify the product of
dot products of vectors in this section.
Let us review what we know for a function ¢ of one variable that is differ entiable at c. By definition of ¢'(c), the quantity
c:
=
¢(x)-¢(c) x-c
_
¢'(c)
tends to 0 as x tends to c. After multiplying by x - c and isolating ¢(x), we obtain
¢(x)
where limx-+ec:
=
=
¢(c) + ¢'(c)
(x - c) +
·
c:
•
(x - c)
(11.5.1)
0. Equation (11.5.1) is the essence of the differential approxima
tion. The sum ¢'(c)
·
(x-c) + c:
•
(x-c)
measures how far ¢(x) is from ¢(c).
Because each of these two summands has the factor x-c, each summand is small when x is close to c. However, the term c: Consequently, as x -+c, the expression c:
·
•
(x-c) has the additional small factor c:. (x-c) becomes negligible compared to
the other terms on the right side of equation (11.5.1). Now we will develop similar ideas that are key to the concept of differentiability of functions of two or more variables. For simplicity, we will discuss the case of a function of two variables functions of three or more variables are treated analogously.
11.5 Differentiability and the Chain Rule
�·l@il!ii[•lll
Suppose that
P0
(x0,y0)
�
is the center of an open disk
901
D(Po, r)
on which a function f of two variables is defined. Suppose that fx (Po ) and f, (Po)
both exist. We say that/is
differentiable at the point Po
if we can express f(x,y)
by the formula
f(x,y) f(Po) + fx (Po ) (x - xo) + fy (Po ) (y - Yo) + c1 (x - xo) + c2 (y - Yo ) , =
·
·
·
(11.5.2)
·
where lim
(x,y)-+(xo,yo) We say that
c1
=
f is differentiable
0
and
lim c (x,y)-+(xo,yo) 2
=
0.
on a set if it is differentiable at each point of
the set.
It is clear that each of the last four summands on the right side of equation
(11.5.2) tends to 0 as (x, y) tends to (x0, y0). Therefore if/is differentiable at P0, we have lim(x,y)-+(xo,yo)/(x,y) f(Po). We state this observation as the following =
theorem.
THEOREM 1
If f is differentiable at
Po,
then
f is
continuous at
Po.
It is reasonable to wonder whether the existence of both partial derivatives at a point is enough to ensure the differentiability of the function at the point. As Example
1 will show, the answer is, Nol In fact, Example 1 shows that the existence not even ensure the continuity of the
of both partial derivatives at a point does function at the point. �
EX A M P L E 1
{
Show that the function
f(x,y)
=
xy x2
: y2
f defined
by
if
(x,y) =I- (0, 0)
if
(x,y)
=
(0, 0)
is not differentiable at the origin even though both partial derivatives
fy(O, 0)
fx(O, 0)
and
exist.
2 of Section 11.3. We lim(x,y)-+(O,o)/(x,y) does not exist. Therefore f is not continuous at of Theorem 1, this rules out differentiability at (0, 0). On the other
Solution We have considered this function in Example observed that
(0, 0).
In view
hand,
lim
ill-+D
/ (0 + Lix, 0) - f (O, 0) �x
Therefore fx(O, 0) exists and is and is
0
as well.
..,...
=
0.
lim __!__. ill-+D �x
(
(Lix). 0 (Lix)2 + 02
o)
=
lim _Q_ �x-+D Lix
=
0.
An analogous argument shows that fy(O, 0) exists
902
Chapter 11
Functions of Several Variables INSIGHT
In Example 1, we must use the definitions of the partial derivatives fx andfy to calculatefx(O, 0) andfy(O, 0). We cannot apply the Quotient Rule to xy/(x2 + y2)
at the point (0, 0) because the denominator would be 0.
Example
1
shows that we cannot take differentiability for granted; it must be
checked. Unfortunately, equation
(11.5.2)
is difficult to work with. Our next the
orem provides us with what we need: sufficient conditions for differentiability that are usually easy to verify.
THEOREM 2
Suppose that Po= (x0,y0) is the center of an open disk D(P0,r) on which a function/of two variables is defined. If bothfx(x,y) andfy(x,y) exist
and are continuous on D(P0,r) , then/is differentiable at P0• In other words, if f is continuously differentiable at Po (as defined in Section 11.4), then f is differentiable at P0•
We start with the identity
Proof.
f(x,y)=f(xo,Yo)+ (f(x,y) -f(xo,y)) + (f(xo,y) -f(xo,Yo))
(11.5.3)
and apply the Mean Value Theorem to the function of one variable x1-+f(x,y). Because we are holding y fixed, the derivative of this function with respect to x is the partial derivative fx· The Mean Value Theorem tells us that there is a point P1 on the line segment joining (xo,y) to (x,y) such that f(x,y) -f(xo,y)= fx(P1) ( x -xo). ·
See Figure
(xo.Y) Pz
1. A similar application of the Mean Value Theorem tells us that there is
a point P2 on the line segment joining (xo,yo) to (xo,y) such that f( xo,y) fy(P2) ( y -Yo). We can therefore rewrite equation (11.5.3) as
P1
f(xo,yo)
C
(x,y)
=
·
f(x,y)
Po= (xo.Yo)
f(xo,yo)+ fx(P1) (x -xo)+fy(P2) ( y -yo).
(11.5.4)
=
·
·
Let c1=fx(P1) -fx(Po) and c2=fy(P2) -fy(Po). Then, fx(P1)=fx(Po)+ c1 and fy(P2) fy(Po)+ cz. If we substitute these formulas into equation (11.5.4), then we =
obtain
f(x,y)
.A Figure 1
f(xo,yo)+fx(Po) (x -xo)+fy(Po) (y -Yo)+ c1
=
·
·
·
(x -xo)+ c2 (y -yo) . ·
Furthermore, P1 and P2 both tend to Po as (x,y) does (refer to Figure assumed continuity of fx and fy, we have
lim
(x,y)-+(xo,yo)
c1
=
lim
(x,y)-+(xo,yo)
(fx(P1) -fx(Po))
=
0
and
lim
(x,y)-+(xo,yo)
In conclusion, we have established equation
c2
=
lim
(x,y)-+(xo,yo)
(11.5.2)
By the
(fy(P2) -fy(Po))
=
0.
with c1 and c2 having the
required limit properties. This proves the differentiability of f at P0• 2
1).
•
� EXAM PL E 2 Show that f(x,y)= y / (1+ x ) is differentiable on the entire
xy-plane.
11.5
Solution We have
Differentiability and the Chain Rule
txx ( y , )= -2.xy/1 ( +x2)2
and
ty(x,y)=1/(1+x2).
903
Because
these partial derivatives exist and are continuous at each point of the plane, we may apply Theorem
2
to obtain the desired conclusion.
<1111
=
The Chain Rule for a Suppose thatz t(x,y)is a function of the two variablesxandy. Suppose also that ( ) and y= a (s) are functions of another variable s. The situation is repre Function of Two x=ps sented schematically in Figure 2. We see thatzdepends depends onxandy, each of Variables Each which in turn depends on s. So z is, indirectly, a function of s, and we may ask for Depending on Another the value of z d /s d . Naturally, the answer is given in terms of a Chain Rule. Variable THEOREM3
/ r� '�/'
z=tx ( y) , be a differentiable function ofxandy. Suppose that
Let
x=p(s)
are differentiable functions of s. Then tion of s and
y
z
=
f(x,y), x
_. Figure 2
=
p(s), y
=
y= a ( s)
and
z=tp ( (s), a (s))
is a differentiable func
z d = at dp+at od . s d ax ds ay ds
u{s)
When written entirely in terms of variables, equation
(11.5.5) (11.5.5)
takes the form
dz = az dx+az dy. ds ax ds ay ds Proof. Let
(11.5.2)
(11.5.6)
.
x=p(s+�s) , y= a ( s+�s) , xo=ps ( ) and Yo= a (s)
can be written as
t(p(s+�s) , a ( s+�s)) -t(p(s), a ( s)) =
(Z x( y, ) c1) +
+
·
Then equation
(p(s+�s) -p(s))
(%x ( y , )+c:2) (a (s+�s) -a (s)) . ·
Dividing each term by �s, we obtain
( +�s) , a (s+�s)) -t(p(s), a (s)) = at tp ( +�s) -p(s) . ps ( s x ( y + ) ) ) ( ( , c �s �s i ax +
(at(x,y)+c2) . ay
((
)
()
a s+ �s -a s �s
)
.
dz/ds. The quotients dp/dsand da/dswhile the coefficients c:1 and c:2 both tend
On letting �s tend to 0 the left side of this equation tends to on the right side tend to
to 0. The limiting form of the equation is therefore
dz= at dp+at da ds ax ds ay ds ' as asserted.
•
904
Chapter 11
Functions of Several Variables
z=f(x,y)=x2+y3, x=sin(s),
� EXAMPLE 3 Define culate
and
y=cos(s).
Cal
dz/ ds.
Solution The Chain Rule tells us that
dz ds
az ax
dx
- - -- + _
ds
az dy - - - (2.x) cos (s) ay ds _
+
(3y2) (-sm (s) ) . •
This is not a satisfactory form for our answer because it uses the x and
y
variables
to express a rate of change with respect to s. Therefore we substitute for
x
and
y.
The result is
��
INSIGHT
and
y
()
()
( ) (-sin (s) )
=2sin s cos s +3cos2 s
()
( ) (2-3cos (s) ) .
=sin s cos s
<11111
An alternative approach to Example 3 is to substitute the formulas for
in terms of
s
x
directly into the formula for z . By doing so, we express z
3 =f(x,y) =x2 +y3 = (sin( s)) 2 + (cos(s) )
explicitly as a function of
s so that we may differentiate it directly. (Verify that this differentiation yields the same answer obtained in Example 3.) In spite of the fact that the Chain Rule is not absolutely necessar y for calculations such as those of Example 3,
you will find that it is an excellent device for organizing complicated differentiation problems. In the next few sections, you will also see that the Chain Rule is essential for certain theoretical developments.
The Chain Rule for a Function of Two Variables Each Depending on Two Other Variables
Frequently we are faced with a function
y depends that x =p(s,t) and
(x,y) f-t f(x,y)
for which each variable
and
y =O"(s,t).
Let
z =f(x,y)
The Chain Rule for this setup is as follows in
Theorem 4.
THEOREM 4
be a differentiable function of
x
and
y.
Further
more, assume that
x=p(s,t) are differentiable functions of s and composition
z =f(p(s,t),O"(s,t))
is
and
t (as a
y= O" (s, t)
)
represented in Figure 3 . Then, the
differentiable
function
of s
and
t.
Furthermore,
z =
f(x,y), x
_. Figure 3
=
p(s, t), y
=
afap afaO" + ax as ay as
(11.5.7)
az at ap+at aO" = . at ax at ay at
(11.5.8)
az as
u(s, t)
x
on more than one independent variable. To begin, let us suppose
=
and
11.5 Differentiability and the Chain Rule When written entirely in terms of variables, equations
905
(11.5.7) and (11.5.8) take
the form
and
oz = oz ox oz oy + OS ox OS oy OS
(11.5.9)
oz = oz ox oz oy + . ot ox ot oy ot
(11.5.10)
(11.5.7) and (11.5.8) are essentially the same as (11.5.5). Notice that the form of the Chain Rule when
The derivations of equations the derivation of equation
there are two independent variables is similar to the form when there is just one variable: We simply treat each of the variables � EXAMPLE 4 Consider the functions
y = te-s.
Calculate
oz/os
and
Solution In this example, differentiate
z with
x
and
s.
t one
at a time.
z=f(x,y) =ln (x
y
+ y2) , x=set,
and
depend both on s and on
According to formula
oz oz ox oz oy + = = OS ox OS oy OS x
and
and
oz/ot.
respect to
Finally, we substitute for
s
y
(11.5.9),
t.
Let us first
we have
(-+1 y2) et+ (_q_ ) (t(-e-s)). x+ y2 x
to obtain
(
)
(
)
oz 1 2t e-s ( te-s) et+ · os - set+ t2e-2s set+ t2e-2s _
A similar calculation using formula
oz = ot
Function of Three
yields
s ( s et +1t2e-2s ) !...:;!( t) + ( 2te- 8 ) � ( - s) · set + t2e-2 ax at
fk ax
The Chain Rule for a
(11.5.10)
·
az ay
� at
Theorems 3 and 4 can be stated for functions of any number of variables. We consider the case of three variables next.
Variables Each Depending on Another Variable
THEOREM 5
Suppose that
w=f(x,y,z)
is a function of
x, y,
and
z,
and that
s. That is, suppose that there are x=p(s), y=u(s), and z=T(s). If the functions differentiable, then w=f(p(s),u(s),T(s)) is a differentiable
these variables are functions of the variable functions
f,p,u,
p, u,
and Tsuch that
and Tare
function of s and
dw = Of dp + Of du + Of dT . ds ax ds oy ds oz ds Schematically, we may write this as
dw aw dx + aw dy + ow dz = . ds ax ds ay ds oz ds
(11.5.11)
Chapter 11
Functions of Several Variables
/�
906
x
w
� .0 � / y
4 w=xy +
y = s ll
� EXAMPLE
5 Suppose that w=xy4+y2z, x=s2, y=s112, and z=s-1•
Calculate dw/ds. s
Solution Refer to Figure
z = s -'
4.
According to Theorem 5, we have
3 aw dz - 4 aw aw dy dw dx -+- -+- - y (2s)+ (4xy +2yz) � fu� �� &�
z
.& Figure 4
(
)
1 -s-1/2 +y2(-s-2). 2
Substituting for x,y,z and simplifying gives
You may verify this result by substituting the expressions for x, y, and z into w. It
3
turns out that w=s4+1 and the calculation dwj�=4s
The Chain Rule for a Function of Three Variables Each Depending on Two Other Variables
THEOREM 6
Suppose that w
=
follows immediately.
f(x,y,z) is a function of x, y, and z, and that
these variables are functions of the variables s and t. That is, suppose that there are functions p, er, and T such that x
u(s, t), and z T(s, t). If the functions f, p, er, and T are differentiable, then w f(p(s,t), u(s, t), T(s, t)) is a differentiable function of s and t. Furthermore, =
p(s,t), y
=
=
=
8w 8s
Bf 8p =
ax 8s
+
Bf Ber By 8s
+
Bf 8T 8z 8s'
and
Bf 8p
8w 8t
=
Bx 8t
+
Bf Ber By 8t
+
Bf 8T . 8z 8t
Schematically, we may write this as
aw as
=
Bf Bx ax as
+
Bf By By as
+
Bf 8z 8z 8s'
(11.5.12)
and
Bf ax
aw 8t
� EXAMPLE
=
Bx 8t
+
Bf By By 8t
+
Bf 8z 8z 8t
6 Suppose that w=xy+z2,
Calculate 8w/8s and 8w/8t.
.
(11.5.13)
x=2s -t, y=3t, and z=s.
Solution Theorem 6 tells us that
aw aw ax aw By aw 8z - = - - + - - + - - =( y)(2)+(x)(O)+(2z)(l)=(3t)(2)+(2s)(l) = 6t+2s as ax as By as 8z as and
aw
7ii
aw ax =
ax at
+
aw By By at
+
aw 8z az at
=
( y)(-1)+(x)(3)+(2z)(O)
=
(3t)(-1)+(2s - t)(3)
=
-6t+ 6s.
11.5 Differentiability and the Chain Rule
907
These formulas can also be obtained without the Chain Rule by substituting
x 2s - t , y 3t, and z= s + + w= (2s - t)(3t) s2= 6st - 3t2 s2, from
the formulas calculated.
=
=
w. The result is aw/at are easily
into the equation for which
aw/ as
and
INSIGHT
A schematic diagram can help you keep all terms straight, even in com
plicated examples. Figure 5 shows how this works in Example 6. There are three "paths" from w to t, which tells us that the formula for aw I 8t has three summands. Each path has two segments, which tells us that each summand is the product of two factors. Each factor is obtained by differentiating the variable that appears on the left of the segment with respect to the variable that appears on the right.
w=3xy+z2x
x = t·cos(s) y= t·sin(s) z=s
_. Figure 5
3-6 are special cases of a (11.5.6) and (11.5.9)-(11.5.13). We will
.... A LOOK BACK AND A LOOK FORWARD � Theorems general Chain Rule that includes equations
not state the general Chain Rule formula because it is best developed using the language of matrices. However, in Section
11.6,
we will introduce notation that
enables us to concisely unify several of our formulas. In the meantime, it is a good idea to review the formulas that we have learned and to notice that they follow the common pattern discussed in the preceding
Insight.
An understanding of that
pattern will help you to remember all forms of the Chain Rule without having to memorize several special cases. Furthermore, you will be able to generalize the Chain Rule to situations that have not been explicitly considered. For example, suppose that Vis a function of depends on
q,
r,
s, t, and
Several Variables
and
q,
r,
s, t,
av av aw = 8q aw 8q
+
av ax ax 8q
respect to any of the variables
Taylor's Formula in
w, x, y,
z,
and that each of these variables
u. Then, we may calculate the partial derivative of Vwith
and u. The first formula is +
av ay ay 8q
+
av az az 8q'
and the other four are obtained by replacing
q with
r,
s, t,
and u.
The basic ideas connected with Taylor's formula for functions of one variable have been developed in Chapter 8. Now we want to consider an analogous formula for functions of two variables. Whereas the Taylor formula in one variable was specified on an interval centered at the point about which the formula was expanded, we now formulate the theorem on a rectangle
I= {(x,y) : Ix - xol < r1, IY -Yol < r } that is 2 (x0,y0). We will state only the order 1
centered about the base point of expansion and order
2 cases
THEOREM 7
of the multivariable Taylor's Theorem.
Suppose that P (x, y) is a point in a rectangle /that is centered Po= (xo, Yo). Set h= x - xo and k= y - yo. If f is twice continuously differ entiable on I, then =
at
+
+
f(P)= f(Po) fx(Po)h fy(Po)k
+
Ri(P)
(11.5.14)
908
Chapter 11
Functions of Several Variables
where
Qi on the line segment between Po and P. If f is three times
for some point
continuously differentiable on I, then
f(P)
=
f(Po)+fx(Po)h+fy(Po)k +
ii ( fxx(Po)h2+2fxy(Po)hk (11.5.15)
where
Q on the line segment between Po and P. 2
for some point
Proof. We parameterize the line segment between Po and P by Pt= (x0+th,y0+tk) for 0::;; t::;; 1. We may apply the one-variable Taylor Theorem with base point 0 to F(t)= f(Pt)= f(x0 +th,y0 +tk). Using N= 1 in equation (8.8.4) yields a derivation of formula (11.5.14). Let us instead set N= 2 in equation (8.8.4) and prove (11.5.15). The result is
F(t)
=
F"(O) t2 pm(s) i3 F (O) t+ + F(O)+ ' 1! 2! 3!
5 16) (11 ..
0 and t. The next step is to express each derivative on the right side of equation (11.5.16) in terms of derivatives off. We use the Chain Rule (Theorem 3) to calculate for some number
s
between
F'(t)= fx(Pt)
�t (xo+th)+fy(Pt) �t (yo+tk),
or
(11.5.17) To calculate
F"(t), we apply the Chain Rule to equation (11.5.17), obtaining
or
(11.5.18) One further application of the Chain Rule (Theorem
3) yields (11.5.19)
On substituting
t
=
F'(O)= fx(Po)h+fy(Po)k
0 into equations (11.5.17) and (11.5.18), we obtain and
F"(O)= fxx(Po)h2+2fxy(Po)hk+fyy(Po)k2.
(11.5.20)
11.5 Differentiability and the Chain Rule On substituting t = s into equation
(11.5.19),
909
we obtain
(11.5.21) Substituting formulas denote
Ps,
(11.5.20) and (11.5.21) into equation (11.5.16), 1 yields equation (11.5.15).
letting
and setting t =
Q2 •
The terms Ri(P) and R2(P) on the right sides of equations (11.5.14) and (11.5.15) are called remainder terms, as in the one-variable theory. When we dis card R1(P) from the right side of (11.5.14), we obtain the order 1 Taylor polynomial T1(x,y) of f(x,y) with base point Po= (xo,yo). That is,
T1(x,y) = f(Po) + fx(Po)h + fy(Po)k,
(11.5.22)
h = x - xo and k = y - Yo· Similarly, if we discard Rz(P) from the right side of equation (11.5.15), then we are left with the order 2 Taylor polynomial of f(x,y) with base point Po= (xo,Yo):
where we continue to let
T2(x,y) = f(Po) + fx(Po)h + fy(Po)k +
i! (!xx(Po)h2
+
)
(11.5.23)
2fxy(Po)hk + fyy(Po)k2 .
T1 and T2, together with the R2, to approximate /near the base point.
As in Section 8.8, we use these Taylor polynomials, estimates of the remainder terms,
R1
and
� EXAMPLE 7 Find a quadratic polynomial function
f(x,y) = cos(x) cos(y)
Solution We choose (x0,y0)
= (0,0)
(11.5.23)
T2(x,y) = f(O,0) + (fx(O,O)x + fy(O,O)y)
+
that approximates the
for the base point in formula
this choice of base point, the increments y, respectively. Equation
T2(x,y)
near the origin.
(11.5.23).
Given
h = x - xo and k = y - Yo simplify to x and
becomes
i! ( fxx(O,O)x2
+
2fxy(O,O)xy + fyy(O,O)y2 . (11.5.24)
)
We calculate
f(x,y) = cos(x)cos(y) f(O,0)= cos(x)cos(y) l(o,o) = 1
I
fx(0,0) =
-sin(x)cos(y) (O,O)
fy(0,0) =
-cos(x)sin(y)
=0
I (o,o) = 0
fxx(0,0) = -cos(x)cos(y) I (O,O) fxy(0,0) = sin(x)sin(y) f (O,O)
=
=
-1
0
fyy(0, 0) = -cos(x)cos(y) f (o,o) = -1. Substituting these values into equation cos(x)cos( y) �
(11.5.24),
T2(x,y) = 1
we obtain the approximation
- � (x2
+
2 y ).
"'4
91 O
Chapter 11
Functions of Several Varlables
T (x,y) 2 (1,3).
� EXAM PL E 8 Use Taylor's formula to find a quadratic polynomial
that approximates the
/ function/(x,y) =xy + 3/x +9y
near the point
Solution We begin by calculating the first and second order partial derivatives of
/(x,y): fx(x,y) =y-3/r, f1(x,y) =x-9y / , fxy(x,y)= 1, fa(x,y) = 6!x3, and fy1(x,y�=18y / 3. Thus at the base point P0= (1,3), we have fx(1,3)= y (l,3)= 3-3/1 =0, /y(l,3) =1 -9/32=0, /xy(l,3)=1, /a(l,3)= 6/13= 6, and /y 18133 = 213. The required degree 2 Taylor polynomial is therefore /(1,3) + {fi(l,3)(x -1) +fy(1, 3)(y-3)) z =
+
T(x,y)
i, (r.a(l,3)(x- 1)2+2fxy(1,3)(x- l)(y-3) +fy,.(1,3)(y-3)2)
or
i, (6(x- 1)2 +2(x- l)(y-3)+ j Cy-3)2) 9 +3(x -1)2 + (x -1)(y-3) + � (y-3)2.
T (x,y)=9 + y
=
6 shows [ 8.985,9.05].
Figure
A Figure 6
the graphs of /and
Tin
the viewing box
[0.9,1.l]X[2.9,3. l]X
INSIGHT
It is important to have analytic techniques for determining the extrema of functions of several variables. When we graph a function of one variable in a small window containing a maximum or minimum value, we expect to have visual evidence of the extremum. However, the extrema of functions of two variables can be difficult to spot in a graph. It turns out that the Taylor polynomial T(x,y) of Example 8 has an absolute minimum at the base point (1,3), a fact that is not particularly evident from Figure 6. However, if we complete the square in the expression for T(x,y), then we obtain
T(xl y)
( 2 1 =9+3 (x-1+ 6(y-3)) +4(y-3)2, =
(
)
)
9 +3 (x -1)2+ !(x - l)(y-3) + .!.(y- 3)2 + ! - .!. (y-3)2 3 36 9 36 1
T(x,y) is the sum of T(1, 3)=9 and two nonnegative quantities. We T has an absolute minimum value of 9 at the point (1, 3). Because f(x,y) differs from T(x,y) by a remainder term that is small when the point (x,y) is close to the base point (1, 3), we expect/to also have a local minimum at the point (1,3), and it does. which shows us that
conclude that
We will return to these ideas in Section 11.8.
QUIC K
QUIZ
1.
True or false: If both fi(Po) and exist /y(Po), then f is continuous at Po.
3.
If/x(-1,2) =3./y(-1,2)=- 5, and
2. True or false: If f is differentiable at Po, then f is continuous at Po.
t=1?
z= f(7t- 8,2t),
then what is
dz/dt when
4. Give a quadratic polynomial that approximates the function x/(1 +y) near the
origin.
Answers
1.
False
2. True
3. 11
4.
x- xy
11.5
Differentiability and the Chain Rule
911
EXERCISES In each of Exercises 19-26, calculate
Problems for Practice In each of Exercises 1-10, calculate
dz/ds. Do this by two
methods: a.
19.
Use the Chain Rule.
b. Substitute the formulas for
x
and
y into
the formula for
z,
and calculate the derivative directly.
L 2. 3.
4. 5. 6. 7. 8. 9.
10.
z=f(x,y)=x2y-y3 y=u(s)=s-1 z=f(x,y)=x sin(y) y=u(s)= e-2s x z=f(x,y)= - -x2+y2 y=u(s)=sin(s) z = f(x,y) = xeY y=u(s)=s3+s z=f(x,y) = ln(x3 + y) y=u(s)=s312 z= f(x,y)= tan(x2y) y= u(s)=s2e' z=f(x,y)=e
20.
x=p(s)=s3-1 x=p(s)=e'
22. 23.
lL 12. 13. 14. 15.
16.
and
z=f(x,y)=x2-y1 y= u(s,t)= cos(3st) z= f(x,y)= ln(3x-y2) y= u(s,t)=s3t-3 z=f(x,y)=e3xy y=u(s, t)=tis z=f(x,y)=xly y= u(s,t)= cot(3st) z= f(x,y)= Jx2-y2 y= u(s,t)= (s2 +t2f1 1 z=f(x,y)= -x+y
y=u(s, t)=e8t-5s 17. z=f(x,y)=exp(2.x-y) y= u(s,t)=In(2s+3t) x+3y 18. z=f(x,y)= -y-5x y= u(s,t)=e--4s1
24.
x=p(s) = s112 x= p(s)= s e'
25.
·
x=p(s)=cos(s)
26.
x= p(s)= tan(s) x=p(s)=cos(s2)
·
az at.
x=p(s)=cos(s)
-
z=T(s)=tan(s) x= p(s)=3s3 z=T(s)=ln(s) x=p(s)=s5 z=T(s)=s9 +3 x= p(s)=e' z=T(s)=s1 x=p(s)=s2 z=T(s)=(s2 +8)1/3 x= p(s)=8' Z=T(s)= 4' x= p(s)=sin3(s) z=T(s)=s-1
In each of Exercises 27-32, for the given function f and
P0,
calculate the order
2
Taylor
polynomial.
f(x,y)=yln(x)-sin(xy) f(x,y)= cos(x2y) f(x,y)= 1/(1 +xy) 30. f(x,y)=�+2y 3L f(x,y)=tan(2.x- y) 32. f(x,y)= ../2+xy
Po=(1,7r) Po= (0,0) Po= (0,0) Po= (0,0) Po=(7r/4,7r/4) Po= (1,2)
27.
x=p(s)=s ln(s)
using the
x= p(s)= e' z=T(s)=(s-2)3
w= f(x,y,z)= xyz2 y=u(s)=s3 x2+y w=f(x,y,z)= z y=u(s)=sin(s) w=f(x,y,z)= ..jx2-y3+z y=u(s)=2s2 w=f(x,y,z)=yz sin(x) y=u(s)=s1 w=f(x,y,z)=sin(x3y2z) y=u(s)=s-2 w=f(x,y,z)=ln(x + y2 + z3) y=u(s)= (s2+3)1/2 x2 w=f(x,y,z)= - -Y2+z2 y=2' w=f(x,y,z)=x3 ln(z4 +y) y=u(s)=s-4
using the given base point
In each of Exercises 11-18, use the Chain Rule to calculate
az as
2L
x=p(s)=cos(s) x = p(s) = s2 -s
dw/ds
Chain Rule.
28. 29.
x= p(s,t)= sin(3st)
Further Theory and Practice
x= p(s,t)= s2-4t3
33. Suppose that
x=p(s,t)=sit
34. A rectangular box has sides that, due to temperature fluc
and
x=p(s,t)=tan(3st) x= p(s,t)= (s2-t2f1 x=p(s,t)=e'-t
z=h(x,y), x=p(s,t), y=u(s,t), s= a(u, v),
t={:J(u, v).
State a Chain Rule for
��
and
:
.
tuations, undergo changes in dimension. At a given moment, the length increases at a rate of width decreases at a rate of
0.02
0.01
cm/s, the
cm/s, and the height
increases at a rate of 0.005 emfs. At this particular moment,
the length is 15 cm, the width is 8 cm, and the height is 5 cm. Is the volume increasing or decreasing at this moment?
35. In
Exercise
34,
is
the
surface
area
increasing
or
z=f(x,y),x= p(s,t), and y=u(s,t), find (a2z)/(as2), (ffz)/(at2), and (a2z)/(asat).
for
decreasing?
x=p(s,t)=ln(s -t)
36. Given that mulas for
x=p(s,t)=e5'1
37. Let
be a continuously differentiable function of one
variable. Suppose that f is a differentiable function of two variables. Set
g(x,y)=(f(x,y)).
Prove that
912
Chapter 11
Functions of Several Variables for some twice continuously differentiable functions
g,(x,y) =¢l(f(x,y)) fx(x,y) and
g y(x,y)
=
x =rcosO ( ) and y =rsinO ( ). (Here r is the distance of P =(x,y) to the origin, and 0 is the angle that the position vector OP makes with the positive x-axis.) Given a twice
44. Let
Draw a schematic diagram for this version of the Chain Rule. Use the function
g(x,y)
=
(
sin .x2
In(y))
to illustrate
continuously differentiable function u of two variables, let
this Chain Rule. 38. Let
¢
and 'l/J be continuously differentiable functions of
f(x,t) for which f(x,t) v(x,t) ='l/J(f(x,t)). Show that unique value
=
x and t, there is a t+ x¢(t(x,t)). Let
u(x,y)
=¢(ax+ by) b
40. Let
¢
variable.
Show
that
denominator is nonzero.
46. Let c be a constant, and defines
satisfies the equation
u(x,y) =
satisfies
y(x,t) represents the displacement of a vibrating string at time t and position x, then y satisfies the wave equation
42. If
ax2
- 1 &y
(x,t) -
c2 at2
¢
Fx(x,y, z) Fz(x,y,z)
and
Fy(x,y,z) az (x y) = ay ' Fz(x,y,z)
be a continuously differentiable function of
f(t(x,y),f(x,f(x,y))). P =(x,y) withx >0 andy > Omay be written as (rcos(O),rsin(O)) where r = llOPll and() is the (acute) angle between
----->
OP
and
----->
OP',
where
P'
=
(x,O). If f(x,y) f(r cos() ( ),
has first quadrant for its domain, let u(r() , )
=
rsin() ( )). Prove the following formulas (in which each derivative of f is evaluated at
and 'l/J are any twice differ
entiable functions of a single variable, then
y(x,t) =
(r,()) and each partial (r cos() ( ), r sin(()))):
'l/J(x - ct)
d.
cos(())fx+ sin(())fy Urr = cos2(())fxx +2cos(()) sin(())fxy + sin2(())fyy uo = -r sin(())fx + r cos(())fy uoo =-ru,+r2 (sin2(())fxx-2sin(())cos(())fxy+cos2(())fyy)
e.
Urr+ -u, + 2uoo =fxx + fyy
a.
b. c.
u,
=
1 r
1 r
Formula e expresses the Laplacian
'l/J((x+c�t) -c(t+ �t)) ='l/J(x - ct). What physical interpretation of the term
f(x,y)
two variables. Calculate the derivative with respect to x of
partial derivative of u is evaluated at
(x, t)
where c is a positive constant that depends on the tension in the string. Show that if
_ =
48. Any point
av av (x,y) =x (x,y). ax ay
&y
F(x,y,z) =c y.
implicitly as a differentiable function of x and
at any point at which the denominator is nonzero . 47. Let
be continuously differentiable functions of one
y
z
az (x y) ax '
satisfies the equation
v(x,y) =¢(.x2+ y2)
a differentiable function of
Show that
au au (x,y) =y (x,y). ax ay
variable. Show that
F be
at any point at which the
three variables. Suppose that the equation
be a continuously differentiable function of one
x ¢
one
au au (x,y) =a (x,y). ax ay
variable. Show that
41. Let
of
y implicitly as a differentiable function of x. Show
y'(x) =-fx(x,y)/fy(x,y)
that
a and b be constants. Suppose that ¢ is a continuously function
45.
defines
av av =¢(f(x,t)) . ax at
differentiable
v(r, 0) u (r cosO ( ),rsinO ( )) . If u satisfies Laplace's equa tion u xx(x,y) + Uyy(x,y) =0, then what partial differential equation does v satisfy? Let c be a constant, and f be a differentiable function of two variables. Suppose that the equation f(x,y) =c =
one variable. Suppose that for each
39. Let
¢
and 'lj;.
of the results
polar coordinates r
and() of
fxx + fyy
P.
of
f in
terms
Polar coordinates
will be studied in greater detail in Chapter 12.
from this observation? What is the interpretation of
43. Let
e
Suppose that
y
is a twice con
tinuously differentiable function of two variables such that
Y xx(x,t)
=
(1/c2)yu(x, t).
Show that
v(e,'TJ) =y e
( ; 'TJ, e�'f})
satisfies
ve1J(e,'fJ) =0.
Solve this differential equation and
F(x,y), p(s,t), and f(s,t) =F(p(s,t),a(s,t)). For the given point (so,to), set xo =p(so,to), and Yo = a(so,to). Use central difference quotients to evaluate fs(so,to) and fr(so, to). Also, use central difference quotients to evaluate p.(so,to), pi(so,to), a5(so,to), a1(so,to), Fx(xo,Yo), and Fy(xo,Yo). Then, use these In each of Exercises 49-52, functions
a(s,t)
are given. Let
values together with the Chain Rule to obtain alternative
deduce that
y(x,t)
Calculator/Computer Exercises
=
evaluations of fs(so,to) and
fr(so,to).
11.6 Gradients and Directional Derivatives
913
SL F(x,y)= ln{l + x4+ y2), p(s, t) = (2s+ t)113, a(s, t) =tan(s+ t), (so, to)= (1, -1) 52. F(x,y) = sin(2x + 6y), p(s, t) = arcsin(s/t), a(s, t) =arcsin( (2s - 1)/t2), (s0, t0) =(1, ..,/2)
F(x,y) = (x+ y)/(x+ 1), p(s, t) =..,/st+ t, a(s, t) =2•-•, (so, to)= (3, 1) 50. F(x,y) =xe2Y-x, p(s, t) =s/(1 + t), a(s, t) =s', (so, to) = (2, 0) 49.
-
11 .6 Gradients and Directional Derivatives direction is another word for a unit vector.
Recall that a
In this section, we want to
discuss the notion of differentiating a function of several variables in a given direction. For motivation, look at the topographic map in Figure
1.
The level sets
are for the height functionf (measured in feet above sea level). They are plotted at
20 foot increments.
Imagine that you are standing at the point Pthat is the origin of
the superimposed coordinate axes. The level curve that passes through P tells us that Pis situated
1040 feet above sea level. If you look east, the height is increasing;
So the derivative off in that direction should be positive. If you look south or north (along the level curve), the height is not changing, so the derivatives off in those directions should be
0.
If you look west, the height is decreasing, so the derivative
in that direction should be negative. N
s _. Figure
1 Stone Mountain, Georgia
At what rate does your elevation change if you set out from P in a given direction? In which direction does elevation increase most rapidly? Decrease most rapidly? This section introduced methods for answering these questions.
The Directional Derivative
Let
u =
u1i + uij
be a direction vector and Po
=
(x0, y0) a point in the xy-plane. Let
Lu denote the line in the xy-plane that passes through Po and has direction
of Lu that passes through Po at
=
function
We
1 for every t, we note that t,.... r(t) is the arc length parameterization t 0 (see Figure 2).
=
vector-valued
u.
llr'(t) II
llnll
the
oPo +tu.
Lu
=
by
r(t)
parameterize
=
Because
914
Chapter 11
Functions of Several Varlables
z
;c:x �+tu +
Let x ( ,y) �f(x,y) be a differentiable function defined on a disk centered at Po. The :first step toward calculating the rate of change off at the point Po in the direction u u1i + uij is to restrictfto La by forming the compositionfor:
=
-4
u. = ut•
A Rgure2
U2J
-=--t
0r)(t) =f(OPo +tu) =f(xo +tu1,yo + tuz).
(f
This is a scalar-valued function of one variable. Its plot can be obtained by inter secting the surface z fx ( ,y) with the plane that contains the line La and that is
=
perpendicular to thexy-plane (refer to Figure 2). The rate of change that we seek is obtained from the derivative off"
r.
u
=u1i + u,j
The at the point
of the function f in the direction defined to be
directional derivative
DJ(Po)
Po is
=-ddtf(OPo +tu) I =-ddtf(xo +tu1,yo +tu2) I �
t=O
t=O
.
Now that we have defined the directional derivative, how can we calculate it? We use the Chain Rule to differentiate the function z with respect to
=for
t:
of d 'I) = -(Po)u1 8f 8f Of (dzl = (Po) (dxl) + -(Po)uz. dt lJx dt + -(Po) 8y dt 8x 8y t�O
Now we
can
t�O
t�O
summarize what we have found.
THEOREM 1
Let Po be a point in the plane, • u1i + u2j a unit vector, andfa differentiable function on a disk centered at Po. Then, the directional derivative off at Po in the direction u is given by the formula
=
Dv.f(Po)
=
:(Po)u1 + : (Po)ui.
INSIGHT
Notice that, if u = i, then u1=1 and uz = 0. Therefore Dif(Po) = h(Po) . Similarly, DJ f(Po) fy(Po). This tells us that the partial derivatives /;i(Po) and /,(Po) are
=
themselves directional derivatives: They correspond to the directions of the positive
x-axis and positive y-axis, respectively. From this point of view, the directional derivative may be regarded as a generalization of the concept of the partial derivative.
� EXAMPLE 1 Let fx ( ,y)
(1/2)i - (./3/2)j.
Calculate
=1- x2y.
D.,/(1, 3)
Let
v
=(J2/2)i + (J2/2)j
and Dv./(1, 3).
What is D-v.f(l, 3) ?
Solution To begin, notice that u and v are unit vectors, definition of the directional derivative. We calculate that of ox
1 ( '
3)
=-2xyl(1,3) =-6 and
of 8y
1 ( '
3)
as
and
u
=
required by the
=-rl(1,3) =-1.
11.6 Gradients and Directional Derivatives
915
Therefore
at Dvf(l,3) - ax (1,3)v1 _
Likewise,
D0/(1 ,3)
=
+
at v'2 v'2 v'2 ay (1,3)v2 - (-6)2 + (-1)2 - -72. _
: (1,3)u1 � (1,3)u2 +
Similarly,
=
(
(-6)
at at at D-u/(1,3) - ax (1,3)( - ui) + ay (1, 3)( - uz) - - ax (1, 3)u1 _
_
_
�
+
at ay (1,3)u2
+
(-¥) - (3 - ¥).
(-1)
=
) - -D0f(l,3) - 3 - 2. v'3 _
_
=
INSIGHT
The last computation in Example 1 shows that D-J(Po) -D0f(Po). There is a simple geometric explanation for this fact. Remember that the sign of a deri vative indicates whether the function increases (positive derivative) or decreases (negative derivative). Now think of a mountain path while looking at Figure 2. Follow the path downward. Relative to the ground, your path has a heading u. Because you are descending, your height function has a negative rate of change. Now follow the same path upward. You do this by reversing your ground direction from u to -u. The slope of the path is the same but, because you are now ascending, the rate of change of height reverses in sign.
�
f at P
=
(2,1)
in the direction
fy(P) =
fx(x,y)
=
=
=
2
fy(x,y)
3y2 .
Therefore fx(P) 2 and . The next step is to compute the required direction. The initial candidate, 3 12i + Sj, has length V122 + 52 , which simplifies to 13. Because PQ is not a
Solution We calculate
PQ
1 + 2x + y3. What is the directional derivative of from P to Q (14,6)?
EXAM P L E 2 Let f( x,y)
and
=
=
=
unit vector, we do not use it in the formula for the directional derivative. Instead,
unit vector in the (12/13)i + (5/13)j. The required we use the
The Gradient
direction of
P
to Q, namely
directional derivative is
---+
u
=
(1/13) PQ
=
Before proceeding, we introduce a notation that we will use to orgamze our calculations.
�.J@@h![.]11 function
of
Let
f is
f be
a differentiable function of two variables. The
the vector-valued function
\lf defined
by
gradient
\lf is a vector-valued function created from the scalar-valued \lf(P) is the function \lf evaluated at a point P. We say that \lf(P) is the gradient of f at the point P. The notation grad(/) is an alternative to \lf. In this notation,
function/and
_J
916
Chapter 11
Functions of Several Variables
We can use the gradient to rewrite the directional derivative as the dot product DJ(P)= \lf(P) · u .
(11.6.1)
0
0,
0
Notice that if \lf(P)=0, that is, if fx(P)= and fy(P)= then Duf(P)= for all unit vectors u. In other words, if the gradient at P is 0, then the rate of change off at P is in all directions .
0
3 Let f(x,y)= xsin(y). Calculate \lf(x,y). If 4 ( /S)j, then what is Duf(2, � EXAMPLE
Solution
rr/6)?
We have \lf(x,y)=
u=
(-3/S)i+
(� (x,y)} + (: (x,y))j =sin(y)i + x cos(y)j. 2
Therefore \1/(2,rr/ 6)=sin(rr/ 6)i+ cos(rr/6)j = (1/2)i+ v'3j and
Duf(2,�) =\1f(2,�) · u= (� i+ J3j) (- � i+ �j) = -:0 + 4� . ·
The Directions of Greatest Increase and Decrease
_...
Suppose now that the point Po and the function f are fixed. Which direction u will result in the greatest possible value for Duf(P0)? Which direction u will result in the least possible value for D0f(Po)? Because D0f(Po) measures the rate of change of f at Po in the direction u, we can pose our questions in this way: What is the direction of greatest increase for the function f at Po? What is the direction of greatest decrease for the function f at Po? Theorem answers these questions.
2
THEOREM 2
Suppose that f is a differentiable function for which \lf(Po) f:- 0. Then, Po) is maximal when the unit vector u is the direction of the gradient \lf(Po). For this choice of u, the directional derivative is D0f(Po)= ll\1/(Po)ll· Also, Duf(Po) is minimal when u is opposite in direction to \1f(P0). For this choice of u, the directional derivative is Duf(Po)= -ll\1/(Po)ll·
DJ(
To begin, fix an arbitrary direction u. Because \1f(P0) f:- 0, there is a well defined angle () between the nonzero vectors \1f(P0) and u. Expressing the dot product of these vectors in terms of(), we have Proot
Duf(Po)
=
\lf(Po) u ·
=
ll\1/(Po)ll llu llcos(O)
=
ll\1/(Po)ll cos(O).
Remember that f and Po are fixed. We are trying to select a direction u to make D0f(Po) as big or as small as possible. In other words, we want to determine() so that cos(O) is as big or as small as possible. Clearly,()= gives the greatest value for cos(O), namely 1. This value of () occurs when u is the direction of the gradient \lf(Po). For this choice of u, we have Duf(Po)= ll\1/(Po)ll(l)= ll\1/(Po)ll· Also observe that()=rr results in the least value of cos(O), namely - 1. This value of() occurs when u is the opposite direction of the gradient \1f(P0). For this • choice of u, we have Duf(Po)= ll\1/(Po)ll(-1)= -ll\1 /(Po)ll·
0
-2,
� EXAMPLE 4 At the point Po ( 1),what is the direction that results in the greatest increase for f(x,y)= x2 + y2, and what is the direction of greatest decrease? What are the greatest and least values of the directional derivative at Po? =
11.6
Solution The gradient of
-4i+ 2j.
equal to
u=
t at (x,y)
is
91 7
Gradients and Directional Derivatives
Vt(x,y)= 2xi+ 2yj.
At P0, this gradient is
The direction of greatest increase is therefore
�
llVt Po) II
Vt(Po)=
Jk (-4i+ 2j)= - Js i+ Js j.
The corresponding greatest value for the directional derivative at
Po
is
Dut(Po) = llVt(Po)ll = J20= 2VS. The direction of greatest decrease is just the negative
( 2/ v's)i+ ( -1 / v's)j
of the
Po
is
t is
a
vector u. The corresponding least value for the directional derivative at
-llVt(Po)ll = -2J5. The Gradient and Level Curves
<1111
We turn to another important application of the gradient. Suppose that
function that is differentiable at a point Po in the plane. Let c denote the value t(P0), and let r= (r1, r2 ) be an arc length parameterization of the level curve C= {(x,y) :t(x,y)=c} with r(so) =P0• Then, T=r'(s0) is the unit tangent vector to Cat P0• Because r(s) is a point on Cfor alls, we see that (t r)(s)=t(r(s))=c is ' constant. It follows that (f or) (s) = 0 for alls. In particular, for s =s0, we have 0
DTt(Po)=Vt(Po) T=Vt(Po) ·
·
r'
(so)=
y
Chain Rule .,,.I at at (Po)� (so)+ (Po)12 (so) = (f 0r) (so)= 0. x y a a 1
1
We gather together some conclusions that can be drawn from this equation:
THEOREM 3
Suppose that
t is
differentiable at
Po.
Let
T
be a unit tangent
vector to the level curve oft at P0• Then: a. b.
Dyt(Po) = 0. Vt(Po) is perpendicular
to
T.
x
_. Figure 3 In words, part a tells us that the instantaneous rate of change oft at
Po
is 0 in
the direction of the tangent to the level curve. Part b tells us that the gradient
Vt(Po)
is normal to the level curve oft at
INSIGHT
Po (see
In combination, Theorems 2 and
Figure
3).
3 tell us that a direction perpendicular to
a level curve offat Pis either the direction in whichfhas the greatest increase at Por the direction in whichfhas the greatest decrease at P. This principle is actually quite familiar to us. Imagine ascending the ramp shown in Figure 4. If we begin at point P, then our instinct is to walk toward Q, in the direction that is perpendicular to the level curves , rather than toward another point such as R.
918
Chapter 11
Functions of Several Variables y
x
2
x
.A Figure 4 The direction perpendicular to the level curves off is the direction of greatest increase or the direction of greatest decrease.
y
� EXAM P L E 5 Consider the curve C in the xy-plane that is the graph of the
x --+--1------1 1-+----+-1-_
equation x2+6y4=10. Find the line that is normal to the curve at the point
(2, 1).
Solution Let Po= (2, 1). We may f(x, y)= x2 +6y4• We calculate
where
regard C as the level curve
f(x,y)= 10
'\lf(Po)= (2xi +24y3j) (x,y)=(Z,l) = 4i +24j= 4(i +6j). .A Figure 5 A normal line to the graph of x2 + 6y4 10 =
Functions of Three or More Variables
According to Theorem 3, this gradient vector is normal to C at therefore
y = 6(x All
- 2) +1
of
(x,y,z)
write
the f-t
the
normal
line
as
in
this
(see Figure 5). -11111
ideas
F(x,y, z)
introduced
(x
- 2)/1= (y -1)/6
section
are
also
or,
valid
P0•
We may
equivalently,
for
functions
of three variables. The gradient of such a function is
'\1F= 8F i + 8F.J + 8F k.
ax
The directional derivative of
u=
(ui, u2, u3)
F
ay
at the point
az
Po= (x0,y0, zo)
in the direction
is
D0F(Po)= '\lF(Po) · u.
The direction that results in the greatest rate of increase of the function point
Po
f at
the
is =
u
1 llVF(Po)ll
'\?F(Po).
With this choice of direction u, we have results in the greatest rate of decrease -
u=
The level set
that
1
- llVF(Po)ll '\?F(Po).
With this choice of direction, we have
'\1F(P0) is
D0F(Po)= ll'\?F(Po)ll· The direction of the function F at the point Po is
D--uF(Po)= -llVF(Po)ll·
{(x,y, z) : F(x,y, z) = c} is a surface. At the point P0, the gradient F that passes through P0• We will discuss
normal to the level surface of
this point further in the next section.
11.6 Gradients and Directional Derivatives
91 9
� E X A M P L E 6 Find the directions of greatest rate of increase and greatest rate of decrease for the function
F(x,y,z) =xyz
at the point
(-1,2, 1).
Solution To do this, we calculate Y'F(x,y,z)= yzi+xzj+xyk and Y'F(-1,2,
2i-j-2k.
Then, the direction of greatest rate of increase is
1
1 Y'F(-1,2, 1)= llY'F(-1,2 , l) ll
y22+(-1)2+(-2)2
The direction of greatest rate of decrease at
(2i-j-2k)=
(-1,2, 1)
� i- � j- � k.
is then
1 F(-1,2, 1)= - i+ j+ k. llY'F(- 1,2, 1)11Y'
� � �
The Gradient and the Chain Rule
1)=
<11111
We conclude this section by observing that the gradient gives us another way to think about the Chain Rule. From this perspective, the multivariable Chain Rule will look a little bit more like the one-variable Chain Rule. Let us illustrate this point of view for functions of two variables. Namely, if and y
=a(t),
then we can think of
r(t) =p(t)i+o-(t)j as
z =f(x,y)
where x
= p(t)
the parameterization of a
curve. The Chain Rule tells us that
af dp af dodz = + dt ax dt ay dt
'
which we can write as
dz . = (Y'f) dt
(dr).
( 1 1.6.2)
dt
We see that the gradient allows us to write the Chain Rule in the more familiar form of a product, provided we interpret the"·" as the dot product. Notice that this last notation for the Chain Rule does not look any different when more indepen dent y
variables
are
= o-2(t),z= a3(t),
present.
For
if = F(x,y, z) where x = o-1(t), r(t)= o-1(t)i+o-2(t)j+o-3(t)k as the para
instance,
then we can think of
w
meterization of a curve inxyz-space. In this case, the Chain Rule is
dw = aF do-1 + aF do-2 + aF do-3 dt
ax
dt
ay
dt
az
dt
'
which can be written as
dw = (Y'F) . (dr). dt dt Q UIC K
Q UIZ
1. 2. 3. 4.
Suppose that u=
( 1 1.6.3)
(3/S)i+(4/S)j and f(x,y)= lOx +y5• What is D0/(2,-1)? For what vector u is D0f(Po)= fy(Po)? 3 What is Y'/(5,4)? Suppose that f(x,y)= 1+2x+y Suppose that fx(P0) 2VS and fy(P0) 4. What is the greatest directional derivative offat Po?
Answers 1. 10 2. j
•
=
3. (2,48)
=
4. 6
920
Functions of Several Variables
Chapter 11
EXERCISES
b. the directional derivative of f at Po in the direction of u
Problems for Practice
c.
Calculate the gradient of each of the functions in Exercises
1-10.
of
f(x,y) =xsin(y) f(x,y) =xln(x+3y) 3. f(x,y) =tan(xy2) 4. f(x,y,z) =xy2/z3 5. f(x, y, z) =cos(xy) sin(yz) 6. f(x, y) =x3/2y-5/2 - xy xy2 7. f(x,y) cos(y) 8. f(x,y) =y'Ycot(x+y) 9. f(x,y,z) =zsin(xy) 10. f(x,y,z) =xy2sin(yz2)
of greatest rate of decrease forfat the
f at Po
Po
in the given direction
Po=(2,- 3),
f(x,y) =x2y - y3x, u=( -6/10,8/10) 12. f(x,y)=xln(x2+y2), u= (4/5,3/5) 13. f(x,y) =cos(x2 - y2), u= (1/v'z, -1/v'z) x+y 14. f(x,y) - -, x- y u= (.../3/2,-1/2) 15. f(x,y) =e4x-7y, u= (1,0) 16. f(x,y) =tan(x/y), u= (-12/13,5/13) 1 17. f(x,y) = - -- , x2 +y2 u= (0,1) 18. f(x,y) =(x - y)ln(x +y),
11.
Po=(2,4),
Po=(4,7), Po=(0,ln(2)), Po=(7r/4,1), P=(2,2),
sin(xy) , x+y u= (1/v'z, -1/v'z) 20. f(x,y) =eXYe-Y, u= (0,-1) 2L f(x,y) =sin3(x+y), u=(-1,0) 22. f(x,y) =(x2+y3) 4, u= (-1/2,.../3/2)
Po=(0,1),
!(x,y) =
Po =(7r/2,7r/3) Po =(1,7r/3) Po =(1,7r) Po= (4,1) Po=(1, 1) Po =(7r/2,7r/3)
F at
the point
Po
Po=(1/2,3,7r) , Po =(6,3,2) , Po=(1,1,1), Po=(7r/6,7r/3,7r/2),
In each of Exercises 39-42, find the following: the direction function
F at
u
of greatest rate of increase for the
the point
Po
b. the directional derivative of c.
u
the direction
point
Po=(1,1),
Po
v
at
Po
in the direction
F
at
Po
F at the
in the direction
v
F(x,y,z)=x2(y3- zy) ln(x- y) 40. F(x,y, z)= z
39.
F
of greatest rate of decrease for
d. the directional derivative of of
Po
u.
Po=(4,2,2),
'
of
of greatest rate of increase for the
in the direction
Po =(2,1,-3),
F(x,y,z) =xyz3, u= (2/3, -2/3,1/3) x 34. F(x,y,z) = __, y+z u= (3/5,V7/5, -3/5) 35. F(x,y,z)=ysin(xz), u= (.../3/8, -6/8, 5/8) 36. F(x,y,z) = �. yz u= (5/7, -2v'2/7, -4/7) 37. F(x,y,z) =In(xy+z3), u= ( -1/2,3/4,.../3/4) sin(x- z) 38. F(x,y,z)cos( y+z) u= (1/2, -1/2,1/v'z)
a.
functionfat the point
p=(1,7r/4)
In each of Exercises 33-38, find the directional derivative
of the function
Po=(l,ln 2),
In each of Exercises 23-32, find the following:
u
Po =(1,1)
_
Po=(1, 1),
( y'3!2, -1/2)
the direction
Po=(2,1/2) Po=(7r,7r1/2/2)
f(x,y) =x3y2 - yx f(x,y) = cos(x - y2) ln(x+y) 25. f(x,y)= x+y 26. f(x,y) =x2 tan(y) sin(x+y) 27. f(x,y) = cos(x - y) 28. f(x,y) =tan(x2y) 29. f(x,y) =e-xcos(y) x 30. f(x,y) = _ _ x- y 3L f(x,y) =(x2+y) 5 32. f(x,y) =sec(x - y)
33.
_
u=
u.
in the direction
v
23.
In each of Exercises 11-22, calculate the directional
derivative off at
a.
v
24.
-
•
Po
d. the directional derivative of
1. 2.
19
the direction
point
Po=(2,-1,4) Po=(2,1,3)
11.7 Tangent Planes
4L
42.
P0 =(0,7r,27r) F(x,y,z) = sin(x2 - y + z) F(x,y,z) =sin(2x) +12 cos(yz) Po =(7r/6,1, 7r/6)
S6.
Laplace's equation for a function of two variables is given by
b.f(x,y) =
Further Theory and Practice 43.
Let f(x, y) be a continuously differentiable function, and suppose that 'Vf(P)=0. What does this say about the direction of greatest increase or greatest decrease of f at P? Give examples.
In each of Exercises 44-47, find the directional derivative of the given function fat Pin the direction from P to Q.
f(x,y) =x2+xy P = (1,2) 4S. f(x,y) =xyfl +xy P= (2,4) 4S. f(x,y) =exp(8x/y2+ y/x) P =(1, -2) p =(7r, 1/2) 47. f(x,y) =tan(xy2) Q =(-37r,1/2 + 37r) 44.
In each of Exercises 48-Sl, find the given information. 48. 49. SO. SL
921
(!{;2 + :2)t(x,y)=0.
Let u and v be any pair of orthogonal unit vectors in the plane. Prove that Laplace's equation is also given by
D.(DJ) +D-. (D..f)(x,y)=0. S7.
For a vector-valued function f(x,y) =f1(x,y)i + fz(x,y)j, we define the divergence of f to be
. f1 divf(x,y) - a (x,y)+ atz EJy (x,y). ox _
Q =(3,4) Q=(5,8) Q =(-5,6)
Show that if cp is a twice continuously differentiable function, then div (grad(cp) ) =/::,,.cp.
r::x (P) and r (P) using Y
= (0,1), v= (1/v'z ,1/v'z ), D0f(P)=2 v'z, D-.f(P)=3 v'z D0f(P) =1, u = (3/5,4/5), v = (1/v'z, 1/v'z ), D-.f(P)=v'z D0f(P) =19, u = (5/13, 12/13), v = (-12/13,5/13), Dvf(P) =22 D.f(P)=3, u= (1/2, -v'3/2), v= (1/v'z , 1/v'z ), D-.f(P) =0 u
Let f(x,y) =x2 - y3. At the point P=(3,7), which direction(s) u will have the property that D0f(P) =8? S3. Letf(x,y)=xy -y4 and P=(8, -2). Explain why there is no direction u such that D.f(P)=50. S4. You are standing at the bottom, (0,0,0), of a hole that is shaped like the graph of f(x,y)=x4+y2+x2y6. You need to climb out of the hole. What is the direction of steepest ascent? What is the direction of slowest ascent? SS. A heat-seeking missile can only sense heat in its immediate vicinity. In an effort to find the source of the heat, the missile always moves in the direction of greatest increase in temperature. Suppose that the temperature at any point in space is given by T(x,y,z) =1000 S2.
lO(x - 90)2 - 2(y - 12)4 - z2, and the missile is currently at ( 100, 15,8). In what direction will it move next?
Let f be a continuously differentiable function defined on a planar region. Suppose that Pis a point in the domain of f. Show that the set of all possible values for the direc tional derivative of f at P forms an interval. What are the endpoints of this interval? S9. Let f be a twice continuously differentiable function with Po in its domain. Let u=u1i + u j and v=vii+ vzi 2 be directions. Give formulas for D.(D.f)(P0), and D-. (D..f)(Po), and D0(D..f)(Po). What does D.(D0f)(Po) tell us about the concavity of the graph off!
SS.
Calculator/Computer Exercises the viewing window [-1,3.2] X [-2.2,2.2], plot the level curve of f(x,y) =4x+y - x2/8 - x4/8 +4y2 - y4 at height 1 . Calculate and sketch the directions of the gra dients at the six points of intersection with the coordinate axes. 6L In the viewing window [- 1.25,4.5] X [-5,0.75], plot the level curves Lc={(x,y):f(x,y)=c} of f(x,y)=l x2y/2 - x2 - y2 - 2y for c =-1/2,1/2,3/2,5/2, and 7/2. Calculate and sketch the directions of the gradients at the following points. a. (3, -3/2) on L1; 2 b. (2, -2 - v'2/2) and (2, -2+ v'2/2) on L1; 2 c. ( 1, -1) and (1, -3/2) on L3; 2 d. (3, -3) and (3, -7/2) on Ls;z e. (4, -5+ v'26/2) on L1; 60. In
2
11. 7 Tangent Planes If f is a differentiable function of one variable, then its derivative at a point
x
tells
us the slope of the tangent line at that point. Now we will learn that the gradient of a function of two variables tells us about the tangent lines to the graph off We will see that the gradient plays the role of a "total derivative."
922
Chapter
11
Functions of Several Varlables First we must address an important question: What geometric object should the tangent to the graph of a function of two variables be? Look at Figure 1, which exhibits the graph of f(x,y)= 1+x2+f. There are a great many tangent lines at the point (0, 0, 1), as the figure illustrates. The union of these tangent lines suggests that a tangent plane is the geometric object we seek (see Figure 2).
z
f(x. y)
=
+x2 + y2
1
f(x,y)= 1 + x2+ y2
.I
.I
Tangent plane at
(0,0,1)
y x
y
x
A Figure 1 Some tangent lines at the point (0, 0, 1)
A Figure 2 The union of all tangent
lines at a point is a tangent plane.
Now consider an arbitrary differentiable function f of two variables. Fix a point Po= (xo,Yo) in the domain off. Also fix two nonparallel directions u= u1i+uzj and v =vii+vzj in the xy-plane. Then, the curves C1 and C2 that are parameterized by
and
Ty(t) = (xo+tvi)i+(yo+tv2)j+f(xo+tv1,yo+tv2)k are subsets of the graph off. Each passes through the point geometry is shown in Figure 3. We know that the vectors A Figure 3
r
Po when t =0. The rJ(O) and r�(O) are
tangent to Ci and Ci. respectively. We use the Chain Rule to calculate these two tangent vectors:
r;(o)= u1i+uzj+
(!t(xo+tui,Yo+tuz) l,=J k=u1i+uzj+DJ(Po)k
r�(O) =vii+vzj+
(�f(xo+tv1,yo+tv2) l,�J k=vii+vzj+D.f(Po)k.
and
See Figure 4. Let us use the cross product to calculate a vector N that is normal x A Figure 4 Tangent vectors
A=u1i+uij+D.f(Po)k and B = V1i+vij+Dy/(Po)k
to both of these tangent vectors. We obtain N = {u1i+uz.i+D./(Po)k)
X
(vii+vz.i+D.f(Po)k) = det
([
i j k ui u2 Daf(Po) Vt Vz D.zf(Po)
])
11.7 Tangent Planes
923
or, equivalently, N
= (u2Dvf(Po) - v2Duf(Po))i- (u1Dvf(Po) - v1Duf(Po))j
+
(u1v2 - u2v1)k.
After substituting Duf(Po) = uifx(Po) + uz/y(Po) and Dvf(Po) = vifx(Po) + vz/y(Po) into the formula for N, we find that the resulting expression simplifies to N = c(/x(Po)i+
fy(Po)j -k)
where c is the scalar u2v1 -v2 u1. We have discovered a remarkable fact: No matter which two directions u and v we choose, any normal to the corresponding tangent vectors is a multiple of the vector fx(Po)i + fy(Po)j -k. Because the tangent lines are all perpendicular to the vector fx(Po)i+ fy(Po)j -k, they must form a plane that has this vector as its normal. These observations allow us to make the following definition.
�•HiImj[,jjI
If f is a differentiable function of two variables and (x0,y0) is in its domain, then the tangent plane to the graph of f at (xo,yo,f(xo,Yo)) is the plane that passes through the point (xo,yo,f(xo,yo)) and that is normal to the vector fx(xo,Yo)i + fy(xo,Yo)j -k. We say that the vector fx(xo,Yo)i+ fy(xo,Yo)j -k is normal to the graph off at the point (xo,yo,f(xo,yo)). Using equation ( 9. 5 .1), we see that fy(xo,yo)(y -Yo) - (z -f(xo,yo)) = 0,
(11.7.1)
z = f(xo,yo) + fx(xo,yo)(x -xo) + fy(Xo,Yo)(y -Yo)
(11.7.2)
fx(xo,yo)(x -xo)
+
or, equivalently,
is a Cartesian equation of the tangent plane to the graph of f at the point (xo,YoJ(xo,Yo)). INSIGHT
When we use equations
(11.7.1)
and
(11.7.2),
we calculate the derivatives
fx(x,y) andfy(x,y), and then we evaluate them at the point (xo,yo). Consequently, the coefficientsfx(x0,y0) andfy(x0,y0) are constants. When our calculation of equation
(11.7.1) or (11.7.2) is complete, 1, and is multiplied by a
power
z i variable
=
each of the variables x and y appears at most once, has constant. Equation
f(xo,yo) + fx(xo,yo) (x i �
......____.. constant
constant variable
(11.7.2),
for example, has the form
- xo) + fy(xo,yo) ( y '-..? � i constant
constant
variable
- Yo )
.._,....,,
constant
� EX A M P L E 1 Find a Cartesian equation of the tangent plane to the graph of f(x, y) = 2x -3 xy3 at the point (2, -1, 10). 2 Solution We calculate.fx(2, -1) = (2 -3y3)l(z,-i)=5 and_& (2, -1) = (-9xy ) l(z,-l = ) -18 . Also, /(2, -1) = 10. According to equation (11.7.1), S(x -2) + (-18)(y - (-1)) - (z - 10) = 0, or S x - 18y - z = 18 , is a Cartesian equation of the tangent plane to the graph of f at (2, -1, 10). <11111
924
Chapter 11
Functions of Several Variables
At this point in the theory, the variables x,y, andz do not have symmetric roles in the formula for the tangent plane to the graph of z =f(x,y). A partial expla nation is given by the fact that z is a dependent variable, whereas x and y are independent variables. However, a more geometric explanation is provided by the following discussion of level surfaces.
Level Surfaces
In Section
11.6,
we learned that the gradient '\lf(x0,y0) of a function of two
variables is perpendicular to the level curve of F that passes through the point (xo,y0). An analogous principle holds true for a function F of three variables. To understand why, fix a point Q0 (x0,y0,zo) in the domain of F. Let =
S = {(x,y,z) : F(x,y,z) = c} denote the level surface of F that passes through Q0.
We will show that
'\JF(Qo) is perpendicular to S at Q0 by demonstrating that it is
perpendicular to all curves that lie in S and that pass through Q0• Indeed, suppose that
r r1i + rL,j + r3k is a parameterization of such a curve C with r(to) =
�tF(r(t))lt=to by equation
(11.6.3).
=
=
'\JF(Qo) ·r'(to)
Q0 . Then
(11.7.3)
Because C is a subset of the level surface S, it follows that
F(r(t)) = c for all values oft. The left side of equation (11.7.3) is therefore 0. Thus '\JF(Qo) · r'(to) = 0, which implies that '\JF(Q0) is perpendicular to the velocity vector r'(t0). Because this velocity vector is tangent to C at Q0, we conclude that '\JF(Qo) is perpendicular to C. We state the result of this analysis as a theorem. THEOREM 1
If F is a differentiable function of three variables, then
'\JF(x0,y0,z0) is perpendicular to the level surface of F at (x0,y0,z0). We can now tie together the idea of tangent plane with our new insight into level surfaces. Suppose that (x,y) f-tf(x,y) is a function of two variables, and we
want to know a normal to the tangent plane to the graph at the point
Qo
=
(xo,yo,f(xo,Yo)). Define
(11.7.4)
F(x,y,z) =f(x,y) -z.
Then the graph offis just the set {(x,y,z) : F(x,y,z) = O}. That is, the graph offis a level surface of F. Therefore '\l F(Qo) is perpendicular to the graph off. But
'\JF(Qo)
=
ifx(xo,Yo),fy(xo,yo),
-1).
Thus we have rediscovered the fact that ifx(xo,yo),fy(xo,yo), graph of fat the point (xo,yo,f(xo,Yo)). �
-1) is a normal to the
EXAM PL E 2 Find the tangent plane to the surface x2 + 4y2 + 8z2 = 13 at
the point
(1, -1, 1).
Solution We do
not need to solve forz as a function of x andy . Instead, we define
the function of three variables F(x,y,z) =x2 + 4y2 + 8z2. Then, our surface is just the level set F(x,y,z) = 13. By Theorem 1, a normal to the surface F(x,y,z) = 13 at the point '\JF(l,
(1, -1, 1)
-1, 1) = (2.xi + 8yj
is given by
+
16zk) l (x, y, ) = (l,-l,l) z
= 2i- 8 j +
16k.
11.7 Tangent Planes
925
(xo,yo,zo) = (1,-1,1), we 2(x - 1) - 8(y - (-1)) +16(z - 1) = 0. This 2x - By+l6z= 26. ..,..
Because the tangent plane also passes through the point may write the equation of the plane as equation simplifies to
�
EX A M P L E
5 +x2 +2y4 . y2 +x4
z=
3 Let
S be
the
surface
defined
Find the tangent plane to Sat the point
by
the
equation
(1,-1,2).
z is given explicitly as a function of x and y, the calculation of 8z/8x and 8z/8y involves a lengthy application of the Quotient
Solution Although the derivatives
Rule. We may avoid this difficulty by observing that Sis a subset of a level surface
(x,y,z) of Ssatisfies the equation z2 = (5 +x2 +2y4)/(y2 +x4), or 5 +x2 +2y4 - (y2 +x4)z2 = 0. Therefore Sis a subset of a level set of F(x,y,z)= 5 +x2+2y4 - (y2+x4)z2. We calculate VF(x,y,z) = (2x - 4x3z2)i +(8y3 - 2yz2)j- 2(y2 +x4)zk. A normal to S at the point (1,-1,2) is therefore given by of a function that is easy to differentiate. Indeed, every point
VF(l,-1,2) = ((2x - 4x3z2)i+ (8y3 - 2yz2)j- 2(y2 + x4)zk)l(x,y,z) = (1,-1,2) = -14i- 8k . (x0,y0,zo) = (1,-1,2), we may write its equation as (-14)(x - 1) +(O)(y - (-1)) + (-8)(z - 2)= 0. This equation simplifies to 7x +4z= 15. ..,..
Because the tangent plane we seek passes through the point
Normal Lines
The normal line to a surface at a point
Q0= (x0,y0,f(x0,y0))
on the surface is the
line with direction vector given by a normal to the surface at through
Qo.
Q0
and passing
EX A M P L E 4 Find symmetric equations for the normal line to the graph of
�
f(x,y)= -y2 - x3 +xy2
at the point
(1,4,-1).
Solution We will solve this problem in two ways. A normal to the graph at
Q0
is
lfx(x,y),fy(x,y),-l)lx=l,y=4 = (-3x2 +y2, -2y +2xy,-l)lx=l,y= = (13,0,-1). 4 Because the desired line also passes through
Qo= (1,4,/(1,4))= (1,4,-1), we may
write the line parametrically as
x= 1+l3t, Solving for for
t
y= 4,
z= -1+(-l)t.
t in terms of x and also in terms of z, and then equating the expressions
leads to the symmetric equations
x - l z+l 13 ' ---=1
y= 4.
As an alternative, we can obtain the same result by defining
F(x,y,z)= f(x,y) - z= -y 2 - x3+xy 2 - z.
926
Chapter 11
Functions of Several Variables
{(x,y, z) : F(x,y, z)
Then the level set
=
O}
is precisely the graph off. Therefore a
normal to the graph off is
\i'F(l, 4, -3)
=
2 2 ((-3x + y )i+ (-2y+ 2xy)j - lk) lx=l,y=4,z=f(l,4)
=
The rest of the computation continues exactly as in the first solution.
Numerical Approximations Using the Tangent Plane
3.9,
As we learned in Section
1 3i -k. ..,..
the tangent line can be used as a means for
approximating a differentiable function of one variable. Now we apply the same idea to functions of two variables. Let T be the tangent plane to the graph of a
(x,y) f-t f(x,y)
differentiable function
(11.7.2)
(x0,y0,f(x0,y0)).
at the point
Equation
tells us that Tis the graph of the function
L(x,y)= f(Po)+ fx(Po)(x -xo)+ fy(Po)(y -Yo) where
Po= (xo,Yo).
Recall that the terms
f(Po), fx(Po),
and
(11.7.5)
fy(Po)
are just con
stants. After we have calculated these three fixed numbers, we need only perform two multiplications and a few additions to compute
L(x,y) for any x and y.
What is
significant is that this particularly simple function gives a good approximation to
f(x,y) for x and y close to x0 (11.5.2), which states that
and
y0,
respectively, as can be seen from equation
f(x,y)= f(Po)+ fx(Po)(x - xo)+ fy(Po)(y - Yo)+ c1 (x - xo)+ c2 (y - Yo). ·
·
Notice that the first three summands on the right side exactly constitute the formula for
L(x,y).
We conclude that
f(x,y)
=
L(x,y)+ c1 (x -xo)+ c2 (y -yo). ·
·
c1 and c2 both tend to 0 as x and y tend to xo and yo, respectively. This means that f(x,y) -L(x,y) is the sum of two terms, c1 (x -x0) and c2 (y -yo), each of which becomes small compared to the increments k x -xo and k y -y0 as x and y tend to x0 and y0• We summarize our analysis in the
Finally, recall that
·
=
·
=
following theorem. THEOREM 2
Let
f
be a differentiable function of two variables on a rectan
gular region centered at
Po= (x0,y0).
Let
L
be the linear function of two
variables defined by
L(x,y) Then, for every point
f(x,y)
�
L(x,y).
=
f(Po)+ fx(Po)(x -xo)+ fy(Po)(y -Yo).
(x,y) near the fixed
and
c2
P 0,
we have the approximation
L(x,y) is equal to c1 (x - x0)+ c2 (y - y0) where x and y tend to xo and yo, respectively.
and the approximating value
c1
point
The difference f(x,y) -L(x,y) between the exact value f(x,y)
both tend to 0 as
·
·
11.7 Tangent Planes
927
�•l@il@•lll Tue expression L(x,y) defined by equation (11.7.5) is called the linear approximation (or the tangent plane approximation) to f (x,y) at the point Po. Today with calculators and computers so common, we can easily compute the expression/(x,y) when it is given by an explicit formula that involves x andy. What is the point of an approximation? The answer is that f(x,y) may involve compli cated combinations and compositions of expressions that are difficult to analyze. Because the expression L(x,y) is either a constant or a degree
1
polynomial in x
and y, it is always easy to work with. It is natural to ask how good the approximation given by the theorem is. After all, we do not have exact expressions for the terms c1 and c2 that appear in the error f(x,y)-L(x,y). We will use the multivariable Taylor's Theorem to estimate this difference. If we set
h=x-xo
and
k=y-yo,
then formula
L(x,y)=f(Po) + fx(Po)h + fy(Po)k where
=
(11.7.5)
becomes
T1(h, k),
T1(h, k) is the order 1 Taylor polynomial of /with base point P0, as given by (11.5.22). Thus if f is twice continuously differentiable, then equation
equation
(11.5.14)
can be written as f(x,y)=L(x,y)+ Ri(x,y), or
for some point
Qi on the line segment between lf(x,y)-L(x,y)I
<
�(
Po and
(x,y). It follows that
x l -yol l -xol + y
for (x,y) in the rectangular region R = {(x,y): Ix-xol
<
)
2
(11.7.6)
r1, IY - Yol
<
r2 }
,
where M
is any number that is larger than the absolute values of all the second derivatives
lfxxl, lfyyl, lfxyl
in R.
=Jx2 +y4• Using Po= approximation L( 2 9 , 2.1) of /(2.9, 2 1)?
� EXAMPLE 5 Let/be defined by f (x,y)
is the tangent plane
.
Solution We use formula
(11.7.5)
.
with
(x,y)
=(2.9,2.1)
and
Because /(3,2)=5,
fx (3,2) =
I
x 2 + x y4 (3,2)
V
(3, 2), what
=� , 5
and /y (3, 2) _
V
2y3
(x0,y0)= (3,2).
I
2 x +y 4 (3,2)
16 - -5 , _
we have
/(2.9, 2.1) � L( 2.9 , 2.1)=/(3,2)+ fx(3, 2)(2.9 -3)+ fy(3, 2)(2.1 - 2) 16 =5+ (�)(-0.1)+ ( )(0.1)=5.26. 5 The true value of decimal place.
..,...
/(2.9, 2.1)
is
.
5.278. . ,
so our approximation is accurate to one
928
Chapter 11
Functions of Several Variables INSIGHT
If we assume that the rectangular region R on which Example
5 takes
{(x,y): Ix -31 :::::;O.l, ly-21 :::::;O.l}, then we may show that the second deri vatives off are not larger than M 10 in absolute value. Thus estimate (11.7.6) for the place is R
=
=
maximal size of the error tells us that it cannot be larger than
� (1-0.11+10.11)2
=
0.2.
In fact, the accuracy that we actually achieved is even better than this prediction. The error bound given by estimate
(11.7.6)
is a "worst case" value.
A Restatement of
The linear approximation is often stated using the notation of increments. Let us
Theorem 2 Using
write x = x0 +Lil, where Lil is a small increment. Similarly, we write y= y0 + �y. Continuing with this theme, we let f(x,y) = f(x0,y0) + �f. Then the approximation
Increments
L(x,y) = f(Po) + fx(Po)(x - xo) + fy(Po)(y - Yo), fx(Po)(x - xo) + fy(Po)(y - Yo), becomes f(x,y)
�
�f
�
or
f(x,y) - f(Po)
�
(11.7.7)
fx(Po)�x + fy(Po)�y.
This estimate is sometimes recorded as an equation among differentials:
df
=
(11.7.8)
fx(Po)dx + fy(Po)dy.
The quantity on the right side of equation (11.7.8) is called the
total differential of f at the point P0. In more advanced courses, a precise meaning is given to the differentials in equation (11.7.8) so that equality does hold. � E X A M P L E 6 A machinist is making a bearing that is in the shape of a small cone. It is being manufactured from a costly titanium-vanadium alloy. The
bearing is to have base of radius 3 millimeters and height 6 millimeters. The manufacturer can only allow errors of 1 % in the amount of material used in each bearing. How large an error is allowable in the dimensions of the bearing?
Solution The volume of the cone is V(r,h) = 1fr2h/3. The maximum allowable
ll.h
error is 1 % of volume or
0.06
Allowable error= (0.01) ·
�m·2hlr=3,h=6
= 0.187r.
According to Theorem 2, �V
�
h
Vr(3,6)�r + V (3,6)�h =
�
7rrh
lr=3 h=6 �7rrt=3 h=6 �r +
,
�h = 121f�r + 37r�h.
,
It is therefore required that �r and �h satisfy 127r�r + 37r�h:::; 0.187r or 4�r + �h:::; 0.06. ll.r 0.015
_. Figure
5
How do we interpret this inequality? See Figure 5 for a schematic of �r and �h. If a given bearing has
no error in
the height, then the formula reduces to �r:::; 0.015.
11.7
Tangent Planes
929
Thus a maximum error of 0.015 millimeter can be allowed in the radius of the bearing if there is no error in the height. Similarly, if there is no error in the radius, then the formula becomes �h ::5 0.06. We see that a maximum error of 0.06 mil limeters in the height can be allowed if there is no error in the radius. It is more realistic to suppose that the machinist can keep the height error down to 0.01 millimeters and ask how much tolerance can be allowed in the radius. To apply our error estimate, we set �h = 0.01 to obtain 4�r + 0.01 ::5 0.06 or �r ::5 0.0125. Thus the greatest variation in radius that we can allow is 0.0125 millimeters. <1111
Functions of Three or More Variables
The idea of linear approximation applies to functions of three or more variables. If (x,y,z) = (xo + �x,yo + �y,zo+ �z) is near the point Qo = (xo,yo,zo), then the linear approximation of F(x,y,z) is F(x,y,z)
�
F(Qo) + Fx(Qo)(x -xo)+ Fy(Qo)(y -Yo)+ Fz(Qo)(z -zo).
Stated another way, the difference �F between the values of Fat (x,y,z) and (xo,yo,zo) is approximately
Qo
=
Using the notation of differentials, this approximation is written as dF = Fx(Qo)dx + Fy(Qo)dy + Fz(Qo)dz,
(11.7.9)
and the right side is called the total differential of F at Q0• � EX A M P L E 7 The initial mass of a bacterial sample is measured to be m = 3 mg. The mass grows according to the exponential law mekt. One hour after the initial measurement, a second measurement = 3.3 mg is taken. The two measurements allow the researchers to determine the growth constant: k = ln(3.3/3) = ln(l.1). It is therefore predicted that 8 hours after the initial mea surement, the mass will be 3e8In(1.1) mg, or 6.43 mg. By about how much can the actual mass exceed this figure if the two mass measurements were as much as 0.05 mg off and the time measurement t = 1 hr was as much as 0.001 hr in error?
M
m
Solution If an exact initial mass measurement was made, and if an exact mass measurement was made exactly T hours after the first, then we would have
M=
M
mekT or k =
� In (�). The mass W at a fixed instant of time t hours after the
initial measurement would be
W(m,M, T) =mekt =mexp (� in (:)) =mexp (1n ((:)t/T)) = m(:)t/T. In particular, the mass would be
W(m,M,T)=m(mM)S/T
hours after the initial measurement. Using the values mo= 3, Mo= 3.3, and To = 1, we have 3.3, 1) = 6.43, as given in the statement of the problem. We calculate 8
W(3,
930
Chapter 11
Functions of Several Variables
( ( ) )I s;r
M a m Wm(3.0, 3.3, 1) - a m m _
s; (M/m) r M s;r s T m , =l m =3,M=3.3T
( )
_
(
M s;r a WM(3·0•3·3•l) m - aM m
( )
)I
I
_
m =3,M=33 . ,T=l
s M/m) ;r
- Sm ( =33T=l
m=3M '
.
TM
'
- -15.005,
I
l m=3M=33 . T= '
-
- l5.59'
'
and
Wr(3.0,3.3,1) _
(
M s;r a m m aT
( )
)I
_
m=3M=3.3T=l '
--
/ )8/T Sm (M m ln T2
'
(m ) I M
- -4.903. _
m=3M=33 . T=l '
'
The error �W is approximately �W�
aw am
(3,3.3,l)�m +
aw aw (3,3.3,l)�M + (3,3.3,l)�T aT aM
= (-15.005)�m + (15.590)�M + (-4.903)�T. Now the given bounds for the measurement errors are
-0.05::::;; �M
::::;; 0.05, and
-0.001::::;; �T::::;; 0.001.
-0.05::::;; �m::::;; 0.05,
In the worst case, the signs of �m,
�M, and �Twill cause the three summands of �W to have the same sign. When this happens, the individual measurement errors will accumulate and not cancel. Thus if � m
=
� w� Because
-0.05,
�M
=
+
0.05,
and �T
=
-0.001,
then
(-15.005)(-0.05) + (15.590)(0.05) + (-4.903)(-0.001)� 1.535.
1.535/6.43 � 0.23S, our error analysis tells 24 % of the predicted value. .,,..
us that, in the worst case, the
error can be about
<11111
A LOOK BACK In Section 3.9 of Chapter 3, we learned the linear approximation f � � f'(c)�x of a function/ of one variable. Equation (3.9.7), df f'(c)dx, expresses this linear approximation in the language of differentials. Equations (11.7.S) and (11.7.9) are the two-variable and three-variable analogues of equation (3.9.7). =
Q UIC K
Q UIZ
1. 2.
z x2 + 2y at (3, -2, 5)? Ax + By - z D of the plane that is tangent to the graph of z = x2 + 2y at the point (3,-2,5)? 3. What is the Cartesian equation x + By + Cz = D of the plane that is tangent to the graph of xz + 2y2 4 at the point (2, -1, l)? 4. If f(x,y) Jx + 3y, what is the tangent plane approximation to /(3.3, l.S) if the base point (3,2) is used? What is a normal vector to the graph of What is the Cartesian equation
=
=
=
=
Answers
1. (6,2, -1)
2. 6x + 2y - z
=
9
3. x - 4y + 2z
=
S
4. 2.95
11.7 Tangent Planes
931
EXERCISES 29.
Problems for Practice
In each of Exercises 1-12, a differentiable function given, as is a point Po in the domain off. Calculate a normal vector to the graph of f at the point (P0, f(P0)). Then, write the equation of the tangent plane to the graph off at the point (Po, f(Po)).
(x, y) ,_. f(x, y) is
1. 2.
f(x, y)= xy -y3 + x2 f(x, y)=sin(x-y) 3.f(x, y)=ln(2+x2 + y2) 4. f(x, y)=t?x-3Y x 5. f(x, y)= _ _ x+y 6. f(x, y)=sin(x)cos(y) 7. f(x, y)=xyeX+Y 8.
sin(x)
f(x, y)- (-) cosy x2 + 2 9. f(x, y)= Y x+y 10. f(x, y)=(x+y+1)1/3 lL f(x, y)=(x- yf113 12. f(x, y)=cot(x/y) _
__
Po=(1, 4) Po=(7r/2, 7r/3) Po=(1, 2) Po=(ln(2),ln(3)) Po=(2,1) Po=(7r/3, 7r/4) Po=(1, 2) Po=(7r/3, 7r/3) Po=(l, 1) Po=(6, 1) Po=(9, 1) Po=(7r/2, 2)
In each of Exercises 13-20, find a normal to the surface at the point Q0• Then, find the equation of the tangent plane to the surface at the point Qo. 13.
x2-4y2+z2=-14 14.xyz=8 15. _ Y_ =2 x+z 16. xy sin(z)=5 17. ln(l +x+y+z)=3 18. (x+y)(z -y2) =-14 19. (x+y + z2)6 =1 20. cos(xyz)=0
Q0=(1, 2, 1) Qo=(4, 1, 2) Qo=(1, 8, 3) Qo=(5, 2, 7r/6) Qo=(e2+2, e3 -e2, -3) Qo=(-2,4, 9) Qo=(3, -3, 1) Qo=(4, 7r, 1/8)
In Exercises 21-26, write symmetric equations for the nor mal line to the graph of the given functionfat the given point. 21. 22. 23.
f(x, y)=x3 -y5 f(x, y)=..jx + y-2 f(x, y)=xrY
t(x, y)= L x- y 25. f(x, y)=cos(x)-sin(y) 26. f(x, y)=xsin(y) 24.
(2, 1, 7) (2, 1, 1) (2, -2, 2)
28.
f(x, y)=..jx4+y2 f(x, y)=x/ ...j �l -� + x� 2 -+ y�2
(1, 2, -4) (0, 7r, 1) (7r/2, 7r/2, 7r/2)
p=(2.1, 2.9) p= (1.9, 2.1)
f(x, y)=xcos(2x- y) f(x, y)=ln(l+ft-y)
Po=(2, 3) Po=(2, 2)
Po= (1, 2) Po= (9, 3)
P=(0.9, 2.2)
p=(9.3, 3.2)
3L
A triangular plot of land is to have base 1000 feet and height 500 feet. Suppose that surveying errors are kept within 1%. Use the total differential to estimate the greatest error in area. 32. A rectangular beam with cross section of width w and thickness t is known to have stiffness a w3 t, where a is a physical constant. The beam is cut to have dimensions 10 cm (width) and 5 cm (thickness). Assuming neither cut is more than 1 % off, use the total differential to estimate how much the stiffness of the beam can vary. 33.A rectangular box is to be built with length 6 centimeters, width 4 centimeters, and height 3 centimeters. Suppose that measurement error can be kept accurate to half of 1 %. Use the total differential to estimate the greatest error in volume. 34.A machinist's jig is to be made in the shape of an isosceles triangle. The base is to measure 12 centimeters and the legs 8 centimeters. However, the man building the jig is a novice, and his measurements are off by 0.1 centimeter. Estimate the resulting error in the area of the jig. ·
·
Further Theory and Practice In each of Exercises 35-38, a function/and a point Po are given. Also given is an incorrect equation for the tangent plane to the graph of f at Po. Explain what error has been made, and state the correct equation for the tangent plane. 35.
f(x, y)=1+3x2+2y3 Po=(1, -1) z=6(x-1)+ 6(y+1) 36. f(x, y)=5+x2+y3 Po=(2, 1) z=l0+2x(x-2)+ 3y2(y-1) Po=(0, -3) 37. f(x, y)=3 + y + xy z=yx+(x+l)(y+3) Po=(l, 2) 38.f(x, y)=7x+4/y
z=9 + 7x-y
39. Let () be any number. Show that r6 (t) =tcos(O)i+ tsin(O)j +tk is a curve that passes through the origin and
on the surface z2 =x2 +y2• Show that the zero the only vector that is perpendicular to rb(O) for Use this observation to discuss the concept of vector and tangent plane to the surface z2=x2 +y2• Sketch the surface. What, if anything, does Theorem 1 tell us when we apply it to the differentiable function F(x,y, z)=x2 +y2 -z2 at the origin? 40. Define the angle between two surfaces at a point of intersection to be the angle between their tangent planes, which is the same thing as the angle between their nor mals. What is the angle between the surfaces x2 +y2 + that lies vector is every 0. normal
In each of Exercises 27-30, calculate the linear approx imation to f(P) using the tangent plane at (P0, f(P0)). 27.
30.
932
Chapter 11
z2= 16
and
Functions of Several Variables
x - y + 2z = 0
at their points of intersection?
40 for terminology. Calculate the angle x2 - 3xz + 4y = 0 and y2 + 3xyz Bx= 0 at the point (0,0,0). 42. Calculate the angle between the surface z = x2 - 4y and 41. Refer to Exercise
between the surfaces
the line
(3,1,5) (2,3,2) (-1,2,2) (2,-2,3) (2,1,2)
f(x,y)=(x + y +2xy2)/(x - y) f(x,y)= 4(x + y)/(x2 + 2y) 49. f(x,y)= (2x3 + x2y - xy3)1/3 50. f(x,y)= (1 + x6 + y4)1/4 47.
48.
x = 2t,y= -t + 1, z= 4t + 12 at their two points of intersection. (Hint: This is the same as the angle between the direction vector for the line and the tangent plane to the surface.)
43. Calculate the angle between the line x = 3t
z=St +5
and the sphere
+ 6, y= 4t + 8, x2 + y2 +z2=25 at their two
points of intersection.
(x,y) >-+ Po= (xo,yo) in the domain of f, and a u = u i + uzj. Consider the function (t) = 1
44. Fix a twice continuously differentiable function
f(x,y),
f near Po as the solution set of an F(x,y,z)= 0, find a Cartesian equation for the tangent plane to the graph off at Po. (One way to do this is to use equation (11.7.4). Instead, follow the method of Example 3 to avoid tedious differentiations.) expressing the graph of equation of the form
(Hint: There are many points of intersection.)
a point
direction
f(Po +tu). Write out the first order Taylor Expansion of¢> about t=0. What do the zero- and first-order terms that
(
)
f(x,y) = .J(x +2y)/(x - y) y2 - 3x 1/3 52. f(x,y) x2 y 3 2 + ln(y + 2) 5 f(x,y)= 1 +ln(x) 4 exp(x - 3y) 54 f(x,y)= ln(x - 2y)
SL
(1,2,-1)
_
(1,-1,2)
•
•
(3,1,4)
(;x-=_�)
(1,2,7!'/4)
55.
f(x,y)= arctan
56.
f(x,y)= xy2/(x - y) + 10x2y/(x + y)
(2,3,6)
you see in this expansion have to do with the equation for the tangent plane to the graph off at
Po?
45. A solid, round bead (with no hole drilled through it) of radius 4 mm is to be made of an alloy that is 603 gold. If the mass of the gold that is used is not to be greater than
1%
more than intended, then approxinlately what errors
in size and alloy percentage can be permitted?
46. Let S be any sphere and R, any line that intersects that sphere in two points. Show that the angle between the
line and the sphere is the same at the two points of intersection.
Po
f(x,y) and a point z= f(x,y) are given. By
In each of Exercises 47-56, a function on the graph of the equation
Calculator/Computer Exercises In each of Exercises 57-60, plot the given function f in a
suitable viewing box containing the given point
Po.
On the
same set of axes, plot the tangent plane at the point Po. Zoom in on the given point until it becomes apparent that the tan gent plane closely approxinlates the graph off.
f(x,y)= 3xy2/(1 + x2 +y4) f(x,y) = (3x - y)2/ .J6 + x2 + y4 59. f(x,y)=(x2 - y)e2x-y 60. f(x,y)=17 ln(l +x2)/(1 +x2y4) 57.
58.
Po=(2,1,1) Po = (3,1,16) Po= (2,3,e) Po=(1,2,ln(2))
1 1 .8 Maximum-Minimum Problems Throughout this section, we assume that f is defined on a planar set (/. We further assume that each point in contained in (/. Such a set
relative) minimum such that
g is the center of a disk of positive radius that is entirely g is said to be open. A point Po in g is called a local (or
for f if it is the center of a disk
f(Po) �t(x,y)
for all
D(P0, r)
with positive radius
r
(x,y)eD(P0,r).
Iff(Po) �t(x,y) for all (x,y) in(/, then Po is called an absolute (or global) minimum for f Similarly, a point Po in g is called a local (or relative) maximum for f if it is the center of a disk D(P0, r) with positive radius r such that f(Po)?. f(x, y)
for all
(x,y)eD(P0, r) .
11.8 Maximum-Minimum Problems
Hf Po)<::?:� x,y)
for all
:x,y)
in;;, then
Po
is
called an
absolute
(or
s- 33
global) max
mum for j �point that is either a local minimum or a local maximum is said to be ocal extremum.
The Analogue of Fermat's Theorem
Pictorially, we expect a local minimum
x0,y ,f Po))
where the graph of
Po= x ,yo) to
correspond to a point
looks like the plot in Figure la. We expect a
local maximum to correspond to a point
xo,yo,f Po))
where the graph of /looks
like the plot in Figure 2a. Figures lb and 2b suggest that if Po is a local extremum of f, then the tangent plane to the graph off at
xo,Yo.f Po)) will be parallel to the xy
plane. But this would happen only if the normal vector to the tangent plane,
· .....
f Po),/ Po),-1), were a multiple of f x Po)= and f (Po) 0
indicates that
,0, 1
. In other words, the geometry
must both hold if
Po is to be
a local
xtremum. Let us confinn these conjectures, supposing that both
Po= .xo,y0) is a local
Notice that if
x Po)
and
f Po)
!
xi.st.
minimum for/, then the one-variable function
x) =f x,yo)
has a local minimum at x=xo. It follows from the one variable xo)= 0. But vi x )= fx xo,Yo), as we have seen in Section 11.4. It follows thatfx xo,Yo) =O if xo,Yo) is a local minimwn for f By similar reasoning xo,yo) 'yo) . An analogous argument shows with 1/J y) = f x:o,y), we have that at a local maximum xo,Yo), it is also the case that both x xo,Yo) - 0 and f xo,yo)= 0. We summarize our findings in the following theorem. theory that
£. Figure 1a Local minimum at
Po=:.xo,yo)
r ( - )' "11)
A
£. Rgure 1 b Horizontal tangent plane at X{),yo,f Po))
£. Figure 2• Local maximum at
Po= (xo,y.)
£. Figure 2b H orizontal tangent plane at %0,yo/:po))
(x,y) ...... /[x,y) is a function of two variables defined g, that Po is a point in C, and that both partial derivatives fx Po)
�uppose that on an open set
nd
Po)
a. If
Po is a Po is a
b. If
xist. local minimwn for f, then fx(Po) = 0 and fy(Po)= 0. local maximum for/, then
In short, at a local extremwn
Po,
fx(Po) =0 and f1(Po)=O.
we have V/
Po) = .
934
Chapter 11
Functions of Seven1I Variables
i•l§i@O(.):@
Suppose that (x,y) 1-+ f(x,y) is a function of two variables defined on an open set g and that Po is a point in!}. We say that Po is a critical point forf if either a. both partial derivatives/:x:(Po) and /y(Po) exist and Vf(Po) =0, or b. f fails to have either a partial derivative with respect to x at Po or a partial derivative with respect to y at P0•
L
From Theorem 1, we see that the critical points off are the only points at which f might have a local extremum. /(x,y)
=
10 + (x -1)2 + (x
-
y)2
z
Let /(x,y)=10+(x-1) 2+(x-y) 2• Use Theorem 1 to locate all points that might be local extrema for/. Identify what type of critical points these are � EXAMPLE
1
.
Solution
.&. Figure 3a
/(x,y)
=
10 + (x - 1)2 + (x - y)2
<
[
-------
JbJ y
.&. Figure 3b
Saddle Points
Becausef is everywhere differentiable, a local extremum can occur only at a point for which V/(x,y) =O. We calculate Vf(x,y) ={2(x-1) + 2(x-y), -2(x-y)). We must therefore determine all points (x,y) where both 2(x-1)+ 2(x-y) =0 and -2(x-y)=0. The second of these equations tells us that y=x and, when we substitute this information into the first equation, we obtain x=1. Therefore (1,1) is the only critical point off. According to Theorem 1, (1,1) is the only possible point at whichf might have a local extremum. Theorem 1 does not tell us that (1,1) is a local extremum-it merely tells us that (1,1) is the only point that could be a local extremum. However, we may notice that/(x,y) is the sum of 10 and two nonnegative quantities. It follows thatf(x, y) ;2!:f (1, 1)=10 for all values of x and y. Therefore/ has an absolute minimum at the point (1, 1) . ..._ INSIGHT
Theorem 1 is analogous to Fermat's Theorem from Chapter 4. It gives us a
way to find that are candidates for local maxima and minima. After we isolate the
candidates, our next job is to identify them as local maxima, local minima, or neither. Figure 3a illustrates the absolute minimum of Example 1, viewed from a standard per spective. As Figure 3b demonstrates, it may be possible to rotate the plot to obtain a more convincing view. Nevertheless, it is desirable to have an analytic procedure for identifying a local extremum. Later in this section, we will learn the analogue of the Second Derivative Test for Local Extrema.
As in the one variable situation, a critical point need not be a local ext:remum. For example, the point Po=(0, 0) is a critical point for/(x,y)=1+xl+y2. However, if r is any positive number, then the disk D(Po,r) contains the points P1 =(-r/2,0) and P2 =(r/2, 0). The inequalities f(P1) =1-
r3
r3
1 /(Po) <1+ = /(P2) S< = S
rule out Po as either a local minimum or a local maximum (see Figure 4). The most important instance of what a critical point can be, besides a local maximum or a local minimum, is illustrated in the next example.
11.8 Maximum-Minimum Problems � EXAM PL E 2 Locate and analyze the critical points of fx l ,y)
. .
=
s- 35
x2 - r .
Solution Refer to Section 11.2 for details on plotting the graph off, which appears
'•
in Figure Sa. Because 'Vf(x,y)
.
=
{2x, -2y}, the only critical point is
(0,0). But this
point is neither a local maximum nor a local minimum. If we restrict attention to the x�ection by setting y 0, then we have f(x,0) =
x2. Thus the curve that is
=
formed by intersecting the graph of f and the plane y 0 is the parabola that =
consists of the points (x,O,x2 ). A minimum occurs whenx=O(see Figure restrict attention to theydirection by settingx 0, then we havef(O,y) =
Sb). Ifwe fl The
=
-
.
curve that is formed by intersecting the graph offand the plane x 0is a parabola =
that has maximum value when y 0. It follows that (00) , is neither a local =
� �·7)
£. Figure 4 Critical point Po that is not a local extremum.
maximum nor a local minimum for the functionf(x,y) of two variables in the sense introduced in the beginning of this section. Our surface, concave up in one direction and concave down in the perpendicular direction, looks like a saddle. ...._
£. Rgure 5a Saddle surface
DEFINITION
£. FlgureSb
Suppose that Pois a critical point for the function f. We call Po a
saddle point forfif /has neither a maximum value at Ponor a minimum value at Po. In particular , if there are two smooth curvesC andC in the domain of/that pass through Po, if f Po)< P) for all Pon C1 with P:f Po, and if Po)>f: P) for all P on C2 with P:f Po, then Po is a saddle point for f
The Second Derivative Test
The phenomenon of saddle points makes the analysis of critical points in the plane somewhat more complicated than in the one variable case. The following con struction is frequently useful for streamlining the analysis. DEFINITION
Let x,y) �fx,y) be a twice continuously differentiable func
tion. The scalar-valued function defined by
is called the discriminant off
936
Chapter 11
Functions of Several Variables THEOREM 2
(Second Derivative Test)
(x,y) f-tf(x,y)
Let
be a twice con
tinuously differentiable function, and suppose that Po is a critical point forf. Let
V (f, Po) a. If
be the discriminant off at
V(f, Po)> 0,
Po.
then fxx (Po) and fyy (Po) have the same sign.
i. If fxx (P0) and fyy (P0) are both positive, then ii. b. If c. If
Po is a local minimum for f. Iffxx(Po) and fv y (Po) are both negative, then Po is a local maximum for f. V (f, Po)< 0, then Po is a saddle point for f. V(f, Po) 0, then we can draw no conclusion from the discriminant. =
Proof. For the proof that we give, we will assume thatf is three times continuously
differentiable, which is a bit more than the stated hypothesis. Let A=fxx(Po), B=fxy (Po), C =fyy (Po), h=x-xo, and k=y-YO· We use Taylor's Theorem to expand f about the base point Po, noting that the first powers of h= (x-xo) and k= (y-yo) have coefficients fx(Po) and fy (Po), which are both zero. With this simplification, equation (11.5.15) can be written as
�
f (x,y)-f (Po)= (Ah2+ 2Bhk+Ck2)+ R (x,y), 2
(11.8.2)
where
for some point
�
Q
2 (Ah2+2Bhk+ ck2)
summand of
R 2
between
Po
and
(x,y).
Notice that each summand of
is a monomial of degree
is a monomial of degree 3 in
negligible compared to
� (Ah2+2Bhk+Ck2)
h
2
in
and
h k.
and
k,
whereas each
As a result,
when the point
(x,y)
R2 (x,y)
is
is restricted to a
D(P0,r) of sufficiently small radius. (Imagine, for example, that h k 10-3. Then, h2= hk=k2=10--6, whereas h3= h2k=hk2=k3=10--9.) From this observation and equation (11.8.2), it follows that the sign of f(x,y)-f(P0) is the disk
=
same as the sign of
� (Ah2+2Bhk+Ck2)
when
(x,y)
is in
=
D (P0,r).
This
observation is the key to both parts a and b.
A
The hypothesis of part a is that AC -B2> 0. Because AC> B2;::::: 0, we see that and
C
have the same sign. For now, let us assume that both
A> 0
and
C > 0.
VA" and .../C. By expanding the left side of the 2 inequality (VA°lhl-.../Clkl) ;:::: 0, we obtain Ah2-2.JA°.../Clhll kl+Ck2;::::o, or Ah2+Ck2;::::: 2.JA° .../Cihllkl. Using this inequality together with the inequality VAC - IBI > 0, which follows from the hypothesis AC-B2= V(f, Po)> 0, we Then we can form the expressions
obtain
� (Ah2+2Bhk+ ck2);::::: VA"VClhl lkl+Bhk;::::: VA"VClhl lkl-IBI lhl lkl= (v'AC-IBI) lhl lkl;:::o:: . Because f(x,y)-f (P0) and
D (Po,r), D (P0,r).
we conclude that
� (Ah2+2Bhk+Ck2) f(x,y)-f(Po);::::: 0
have the same sign for
hence
In other words, /has a local minimum at
P0•
f(x,y);::::f (Po)
for
(x,y)
in
(x,y)
in
This proves case i of part a.
11.8 Maximum-Minimum Problems If
f
satisfies the hypotheses of case ii of part a. then
g =-f
937
satisfies the
hypotheses of case i of part a. From what we have already established. this means that g bas a local minimum at P0• That is, g(x,y) 2!=g(P0) for (x,y) in some disk D(Po,r) . For these points (x,y), it follows that /(x,y) =-g(x,y) s -g(Po) =f(Po).
/(.x,y)= 2� +3xy+4y- Sx+2y+3 z
In other words, f has a local maximum at P0• Part b is proved by showing that, when 7J(/, Po) <0, there is a direction
(hi. k1} Ahf+ 2Bh1k1 +Clef <0 and a direction (hi,ki} for which Ah�+ 2Bh2k2+C�> 0. Only elementary algebra is needed to do this, but the formulas for which
appear unmotivated when they are not developed from a more advanced point of view. Students who study quadratic forms in a linear algebra class will encounter the missing details. • � EX A M P LE 3 Locate all local maxima, local minima, and saddle points for the function
f(x,y) = 2x2+ 3xy+4y2- 5x+ 2y+ 3. .&. Figure 6a
Solution We calculate that fx{x,y) =4x+ 3y-5,fy(x,y)= 3x+8y+ 2,/x:x(x,y) =4, fyy(x,y) = 8, and fry(x,y) =fyx(x,y) = 3. The critical points are determined by the
y 2
3 %
simultaneous solution of the equations
fx(x,y)=4x+3y-5=0
and
Solving simultaneously, we find that only critical point. Because
x =2
V(f
'
Po)=det
([ f;rx(Po)
Po) !:x:x(
f:i:y(Po) fyy(Po)
/1(x,y)= 3x+8y+2=0. and
])
=det
is, Po=(2, -1) is the
([ 3 83 ]) =23>0' 4
.&. Figure 6b Level curves of Figure 6a. Local minimum at
we see that we are in Theorem
Po= (2,-1}
shows a plot off. Several. level curves near the minimwn are plotted in Figure 6b. The pattern of curves enclosing the local extremum is typical. �
2a.
Because
y = -1. That
know that case i applies. Therefore the point
INSIGHT
fxx(2,-1)>0 and /yy(2,-1)>0, we (2, -1) is a local minimum. Figure 6a
The key to success with maximum-minimum problems is to be systematic.
The Second Derivative Test requires several pieces of information, and the only way to keep track of them all is to follow the recipe given in Example 3.
� EX A M PL E
4 Locate and identify the critical. points of the function
f(x,y) =�- 2y3- 4xy+5. Solution We calculate that
fx = 6x2 -4y,
fy =-6y2-4x,
fxx= 12x,
fyy = -12y,
and
fxy = fyx =-4.
The critical points are the simultaneous solutions of the equations
6x2-4y =0 The first equation yields
and
-6y2-4x =0.
y = (3/2)x2, which we may substitute into the second equation to obtain -6( (3/2)x?-)2-4x =0 or, equival.ently, 27x4 +8x=0. We factor this as x(27x3 +8)=0. It follows that x =0 and x =-2/3 are the only possible
938
Chapter 11
Functions of Several Variables
2 y = x 0, x = (0, 0) y= (x,y), ]) ([l2x -4 --412y ])--144xy - 16
solutions. We return to the rewritten first equation, value of
y
corresponding to
x
=
0 is
and the value of
is 2/3. Thus our critical points are Po
(3/ 2)
corresponding to
and P1
'D(f
- det )-
'p
([
fxx(P) fyx(P)
fxy(P) fyy(P)
1 =- 6<0
.A Figure 7 Saddle at Po= (0, 0), We see that 'D(f, Po)= Local maximum at P1=(-2/3,2/3) Next, we consider P1
= -2/3
(-2/3, 2/3). To identify
them, we need to examine the discriminant. Now for P= x
, to find that the
we have
- det -
and conclude that Po=
=(0, 0)
·
is a saddle point.
48 > 0, Theorem 2a applies. We calculate fxx(-2/3, 2/3)= -8 and fyy(-2/3, 2/3)= -8. These are both (-2/3, 2/3). Because 'D(f, P1)
negative and so we find ourselves in case ii. That is, (-2/3, 2/3) is a local maximum.
Level curves near the critical points are shown in Figure 7. The pattern of curves that enclose the local maximum is similar to what we saw near the local minimum of Example 3. The "crossing" that appears at the saddle point is typical for that type of critical point. ..,..
Applied Maximum Minimum Problems
Certainly part of the interest of learning to find local maxima and minima is their use in solving practical problems.
� E X A M P L E 5 A rectangular box, with a top, is to hold 20 cubic inches. The material used to make the top and bottom costs 2 cents per square inch, while the material used to make the front and back and the sides costs 3 cents per square inch. What dimensions will yield the most economical box?
Solution To solve this problem, we follow the general scheme for solving maximum-minimum problems that was introduced in Section 4.3 of Chapter
4. z y
We
introduce the function that we wish to minimize, namely cost. We use variable
to
denote the height of the box,
to
x
to represent the width of its front and back, and
denote the depth of the box. Then the following information becomes clear: a. Front and back each have area b. Sides each have area
yz
xz xy
square inches and cost
square inches and cost
c. Top and bottom each have area
3yz
cents.
3xz 2xy
cents.
square inches and cost
cents.
It follows that the total cost is given by
2(3xz) +2(3yz) +2(2xy)= 6xz +6yz +4xy. x= 0 y= 0, x 0 y 0. 4, xyz= z= 0 xy . C(x,y) 12y0 + 1x20 +4xy, (O
Notice that the factors of two appear because there are two of each item. If either
or
then the box degenerates and can hold no powder. So
we will assume in what follows that Section 4.3 of Chapter
>
and
>
Just as in the problems in
we now use information in the problem to eliminate one
of the variables. Namely, Volume=
information into the formula for C gives
20, or
=
2 /(
)
Substituting this
oo
as the function we are required to minimize. Now the methods developed in this section can be applied to analyze
We have
uy
11.8 Maximum-Minimum Problems
939
and
240 file (x,y) = ax2 x3 '
file 8y2
240 y3 '
file a2e (x,y) = 4 (x,y) = axay ayax ·
The critical points are the simultaneous solutions of
-120 +4y=O x2 We solve the first equation for
y
-120 +4x=O. y2
and
to obtain
y=30/x2.
Substituting this into the
second equation gives
-x4 +30x=0. x=0 or x= 30113. But x = 0 does not correspond to a point e(x,y). Thus the only critical point is Po=(30113,30113). We
As a result, we find that in the domain of
analyze this critical point by using the discriminant,
D(f,Po) - det _
Therefore
Po
([ 240/x 3 4
4 3 2401Y
)
JI
57600
- -33 x=y=301/3 X Y _
_
16
I
- 48.
_
x=y=301/3
is either a local maximum or a local minimum. But, because
exx (30113,30113) and eyy (30113,30113) are both positive, the point Po is a local minimum. Next, observe that if either x or y tends to 0, then e(x,y)=120/y +
120/x +4xy tends to infinity. This is not surprising. If, for example, y tends to 0, then xz= 20/y tends to infinity. So the areas of the front and back of the box become infinite, and the cost of material does as well. Because e(x,y ) ---+ oo as (x,y) tends to a point on the boundary of its domain, we deduce that the single local extremum in the domain of C must be an absolute minimum. Thus the optimal dimensions are
x=301/3'
y=301/3'
z= 20 . 30-2/
3
.
.....
We now consider a strategy that often eases the computational burden of locating extrema. Suppose that
cf>
is an increasing function of one variable and that
a differentiable function of two variables. Set g
f
is
= cf> af. Because increasing functions g(Po)= cf>(f(Po)) < cf>(f(P))=g(P) if and only if f(Po) < f(P). Thus Po is a local minimum for g if and only if it is a local minimum for preserve inequalities, we have
f The same reasoning is true for local maxima. We can benefit from this observation as follows: By composing f with a suitable increasing function ¢, we may create a
function that is easier to analyze. The next example illustrates this idea.
� EXAM P L E 6 Find the point on the plane
3x + 2y + z = 6
that is nearest to
the origin.
Solution The
function
to
be
minimized
is
the
distance
to
the
origin,
f(x,y,z)= Jx2 +y2 +z2. Because the partial derivatives off are ratios with the term Jx2 +y2 +z2 in the denominator, we see that it is simpler to minimize the square of the distance to the origin: f(x,y,z) 2=x2 +y2 +z2. (Here we have composed f with the increasing function cf>(u) = u2.) We make a substitution to eliminate one of the variables. Because we are only considering points in the plane
3x + 2y +z= 6, we
see that
z=6 - 2y - 3x.
Substituting this expression for
x2 + y2 + z2 yields the function we want to minimize:
z
into
940
Chapter 11
Functions of Several Variables
g(x,y)=x2 +y2+(6 -2y -3x) 2=10x2+12xy+5y2 -36x -24y+36. Now
gx(x,y)=20x+12y -36, gy(x,y)=12x+lOy -24 and
gxx(x,y)=20,
gyy(x,y) =10, gxy(x,y) =gyx(x,y) =12.
The only critical point is the simultaneous solution of the equations
20x+12y -36=0 which is
Po= (9/7, 6/7).
and
12x+lOy -24=0,
We calculate that
'D(g,Po) =det
( [ 1220 1210 ]) =(20)(10) -122=
56 >
0.
Thus our critical point is either a relative maximum or minimum. Because
gxx(9/7, 6/7) and gyy(9/7, 6/7) are both positive, the point Po is a relative minimum. The corresponding point on the plane is z =6
Thus
Least Squares Lines
(9/7, 6/7, 3/7)
3 -2y -3x= -;;/"
is the point in the plane that is nearest to the origin. <1111
The following table lists the weights and pulse rates of ten land mammals. The
x represents the natural logarithms of the weights, and the column y represents the natural logarithms of the pulse rates.
column labeled labeled
Mammal
Weight (g)
Pulse Rate (Vmin)
x
y
25
670
3.2189
6.5073
Rat
200
420
5.2983
6.0403
Guinea Pig
300
300
5.7038
5.7038
Rabbit
2000
205
7.6009
5.3230
Dog (small)
5000
120
8.5172
4. 7875
Dog (large)
30000
85
10.309
4.4427
Sheep
50000
70
10.820
4.2485
Human
70000
72
11.156
4.2767
450000
38
13.017
3.6376
3000000
48
14.914
3.8712
Mouse
Horse Elephant
Source: A. J. Clark, Comparative Physiology of the Heart, Macmillan, 1972.
11.8 Maximum-Minimum Problems Figure
8 shows a "scatter plot"
of the ten data points
941
(x,y). The plotted points
certainly suggest a straight line relationship, but what line should we use? The ten data points give rise to forty-five pairs of points, each of which determines a straight line. Whichever line we select, it will not pass through the other eight points. To choose between these possible lines, and other candidate lines, which may not pass through any of the points, we need a measure of how well a line £ "fits" the data. To each data point not on £, there is an error. If we quantify this error and sum over all data points (having made sure that all errors are taken to be positive so that there is no cancellation when we sum), then we obtain a measure of how well our line £ fits the data. The line for which the total error is minimized is called a
best-fitting line.
The first question, then, is how do we measure the amount by which a line
y mx + b misses a data point (xj,yj)? One possibility is to use the absolute value IYj- (mxj + b) I of the vertical distance that separates the point from the line. The =
absolute value is necessary so that errors on different sides of the line do not cancel. However, summing the absolute errors results in a function that is not differenti able. That prevents us from using calculus to identify the minimum. Instead, we use the
sum of the squares of the errors:
N SSE(m,b)
=
2
L (yj- (mxj + b) ) .
(11.8.3)
j=l
Although this function seems complicated, it is simply a quadratic in the variables
m and b. The best-fitting line that we obtain when we minimize SSE least-squares line. The term regression line is also used (see Figure 9).
6.5
•Mouse •Rat
6
•Guineapig
� 5.5
y
1-y
•Rab bit
� 5 � 4.5 ::§'
= mx
+bl
b
Dog (s mall)• Dog (large) • Sheep - Human Elephan t •
4
mx;+ b
• Horse
3.5
X; 4
6
8
10
12
14
l n(Weigh t)
THEOREM 3 observations
X
_. Figure 9 The least squares line minimizes
di + d� +
_. Figure 8
where
is called a
.
.
. +
d'J.t.
Suppose that N is an integer greater than or equal to 2. Given N
(x1,y1), (x2,y2), ... , (xN,YN),
the least squares line is
y
=
mx + b
942
Chapter 11
Functions of Several Variables
m=
N Ef:1 X;Y; - (Ef:1 x;) (Ef:1 Y;) N N N L;=1 xJ - (L;=1 x;) 2
(11.8.4)
and
(11.8.5)
INSIGHT
It is common to use vector notation to simplify the appearance of these
Ef=1 xm, x.xis used for the (Ef=1 xi)/N and (Ef=1 Yi)/N.
formulas. Thus x.y is used to denote the (dot product) sum sum
Ef=1 xJ, and x and y are used for the averages
Then, the formulas become m=
x·y-Nxy x·x-Nxx
b=y-mx.
and
(11.8.6)
Proot Because
a: (Y;-(mx;+ b))2 -2x;(Y;-mx;-b) =
and
! (Y;-(mx;+ b))2
=
-2(y;-mx;-b),
we have N N N N a SSE(m,b) = -2 L:x;(Y; = + + 2b 2m -m ; -2 -b) L L:x; L:xJ ; x; XY am j=l j=l j=l j=l and N N N a SSE(m,b) -2 L -m -b) (Y; 2m + 2Nb. -2 LY;+ x; L:x; ab j=l j=l j=l =
=
To find the critical point, we set
SSEm(m,b) =O and SSEb(m,b) =O.
Using the
vector notation discussed previously, we obtain
-2x y+ 2mx · x+ 2Nbx ·
and
Equation
=
0
-2Ny+ 2mNx+ 2Nb = 0.
(11.8.7) (11.8.8)
simplifies to b =y -m-x, which is one of the required equations. b with y -mx in equation (11.8.7) and solve form, then we obtain the
(11.8.8)
If we replace
other required equation after a little algebraic simplification. An application of the Second Derivative Test, outlined in Exercise
37,
shows that the critical point we
have found is a local minimum. From the nature of the function that the critical point corresponds to an absolute minimum.
SSE, we can reason •
11.8 Maximum-Minimum Problems
943
EX A M P L E 7 Use the ten observations in the table given previously to
�
formulate an explicit relationship between the body weight and the pulse rate of land mammals. Solution We calculate x·
y
=
ln(25) ln(670) +ln(200) In(420) + ln(300)ln(300) + x· x=
x=
ln(25)2 + ln(200)2 +ln(300)2 +
(
1 1n(25) + ln(200) +ln(300) + 10
·
·
+ln(3000000) In(48) 411.283, =
·
·
·
·
+ ln(3000000)2
·
·
·
+ln(3000000)
=
940.96,
)
=
=
4.88.
9.056,
and
y
=
� (I (670) +In(420) +ln(300) + n
1
The equation of the least squares line is y = mx
m
y-
lOxy x. x - xx
x. =
=
·
·
·
)
+ In(48)
+ b where
411.283 - 10(9.056)(4.88) 940.96 - 10(9.056)(9.056)
=
_0_254
and b
In Figure
=
10, we have
y - mx
=
4.88 - (-0.254)(9.056)
=
7.18.
superimposed this line on the scatter plot of the data. Plots
such as this are often called log-log plots because we plotted the log of the depen
dent variable p, the pulse rate (measured in beats per minute) , as a function of the
w, the body weight (measured in grams) . In terms of these variables, the equation that we have found is ln(p) = -0.254ln(w) + 7.18 or log of the independent variable
6.5 6 � 5.5 � � 5 "'
� :§'
4.5 4 3.5 4
6
8
10
ln(Weight) .& Figure 10
12
14
944
Chapter 11
Functions of Several Variables
P=
1312.91 w0.254
·
Roughly speaking, pulse rate is inversely proportional to the fourth root of weight. The first "mouse-to-elephant curve ," such as the line in Figure 10, was dis covered by the biologist Max Kleiber in the 1930s. In 1947, Kleiber found that th metabolic rate is proportional to the (3/4) power of weight. It has been a puzzling problem in physiology to convincingly explain the powers that appear in these laws. The mathematical theory of fractals is the basis of an explanation that has recently been advanced. """
<11111 A LOOK BACK
In Section 1.3 of Chapter 1, we used elementary algebra to find the least squares line that passes through a specified point. That task is a one variable problem in which only the slope
m must be determined. In practice, however , it rarely is the case that there is a preferred observation through which to
pass the best-fitting line. Now that we are able to minimize a function of two variables , we can determine the least squares line without making any assumption about the line.
Q UIC K
Q UIZ
Suppose that f(x, y) is twice continuously differentiable on an open disk centered at
Po.
What conclusion about the behavior of
f(x,y)
at
Po
can be drawn from the
given information? 1. 2. 3. 4.
fxx(Po) 4, fyy(Po) 9, fxy(Po) 1 fx(Po) = 0, fy(Po) = 0, fxx(Po) = 4, fyy(Po) fx(Po) = 0, fy(Po) = 0, fxx(Po) = 4, fyy(Po) fx(Po) = 0, fy(Po) = 0, fxx(Po) = 4, fyy(Po) =
=
=
= 9,
=6
= 9,
=5 =7
fxy(Po) fxy(Po) = 9, fxy(Po)
Answers 1. No local extremum
2. No conclusion
3. Local minimum
point
EXERCISES Problems for Practice In each of Exercises 1-20, find all the critical points of
the given function f. For each critical point P, determine
whether P is a local minimum, a local maximum, or a saddle point off.
f(x,y) -2x2 -2y2+4x+Sy+7 f(x,y) x2 -4xy+y2+6x+Sy+6 3. f(x,y ) =x3 - y3 -6xy+S 4. f(x,y) x cos(y) 5. f(x,y) 3x2+2y3 - 6xy 6. f(x,y) 3xy+3/x+2/y 7. f(x,y) (3 + x + y)xy L 2.
=
=
=
=
=
=
8. 9. 10.
f(x,y) f(x,y) f(x,y) lL f(x,y) 12. f(x,y) 13. f(x,y) 14. f(x,y) 15. f(x,y) 16. f(x,y) 17. f(x,y) 18. f(x,y) 19. f(x,y) 20. f(x,y)
=
=
=
=
= =
=
=
=
= =
=
=
x2+y2+4x - Sy+7
2x2 +4y2 -12x-24y+6
x2+2y2 -4xy +6x+12y+9 (x -2)(y+l)x x2 - 3y2 +6xy+3x+Sy+4 xy2 +yx2+Sxy+6 3y2+6x2 - 2xy+4x+3y+7 y3 +3x2y +3x2 -lSy +6 -2y3 + 3x2 +6y2 + 6xy x4 +y4 -Sxy sin(x) +cos(y) ln(l+x2 +y2) y2 -4xy +x2 +Sy -2x +7
4. Saddle
11.8 Maximum-Minimum Problems 21. Find three positive numbers the sum of which is 100 and whose product is as large as possible.
22. Which point on the plane x + 2y + 2z =6 is the nearest to the origin?
23. A box in the shape of a rectangular parallelepiped is made of three different types of material. The material for the top costs 2 cents per square inch, that for the sides costs 1/2 cent per square inch, and that for the bottom costs 1 cent per square inch. If the box is to hold 64 cubic inches, then what dimensions will result in the cheapest box?
24. Find the point(s) on the surface
z=x2 + 8y2
that are
nearest to the point (0, 0, 1)
25. A triangle is to enclose area 100. What dimensions for the triangle result in the least perimeter?
x
y
Australia
32.2
238.l
Belgium
17.0
118.l
Canada
33.5
211.6
West Germany
18.9
150.3
Ireland
27.7
187.3
18.14
124.7
Great Britain
27.9
194.1
United States
39
256.9
Netherlands
Plot the eight points
26. Which point on the sphere
(x,y). y =mx + b
Find the least squares line
(x - 4)2 + (y - 2)2 + (z - 3)2=1
945
that could be used
to model the relationship between cigarette consumption and mortality due to heart disease.
is nearest to the origin?
27. The following table records several paired values of automobile mileage
(x)
measured in thousands of miles
and hydrocarbon emissions per mile (y) measured in
Further Theory and Practice 30. A 20-inch piece of wire is to be cut into three pieces. From one piece is made a square and from another is
grams:
made a rectangle with length equal to twice its width.
x
5.013
10.124
15.060
24.899
44.862
y
0.270
0.277
0.282
0.310
0.345
From the third is made an equilateral triangle. How should the wire be cut so that the sum of the three areas is a maximum?
31. A radioactive isotope of gold is used to diagnose arthritis. Plot these points. Determine the regression line for this
Let y(t) denote the blood serum gold at time t (measured
data. If this pattern continues for higher mileage cars,
in days) as a fraction of the initial dose at time t=0.
about how many grams of hydrocarbons per mile would a
Measured values of y(t) for t=1,2, ...6 are as follows:
car with 100000 miles emit?
28. The following table displays total annual American income
(x)
and consumption
( y)
in billions of dollars for
the years 1985-1991.
t
Semm gold
0
1.0
1
0.91
Income (x)
Consumption (y)
2
0.77
1985
3325.3
2629.0
3
0.66
1986
3526.2
2797.4
4
0.56
1987
3776.6
3009.4
5
0.49
1988
4070.8
3296.1
6
0.43
1989
4384.3
3523.l
7
0.38
1990
4679.8
3748.4
8
0.34
1991
4834.4
3887.7 Plot the points (t,ln(y)). Find a least squares line of the
Plot these points. Determine the regression line for this data.
29. In the accompanying table,
of the initial dose that remains in the blood serum after
x
represents cigarette con
sumption per adult (in lOOs) for the year 1962 for eight countries. The variable y represents mortality per 100,000 due to heart disease in 1962.
form ln(y) =-mt. Use the model to predict the fraction 10 days. 32. If P1= (x1,y1,z1), P2 = (x2,y2,z2), ... , PN= (xN,JN,ZN) are points in space, then their "fit" to the plane
z =Ax + By + D
is measured by
Chapter 11
946
Functions of Several Variables
a.
Show that for any point P,
V(SSE,P) The plane of best fit is that plane that minimizes the expression
S.
best fit.
Find, in terms of P1,... ,PN, the plane of
=
Ax2 +
Cy2 + Dx + Ey + F.
c.
identify any local maxinia or minima.
c.
f(x,y)
d. f(x,y)
=
=
=
x2 +3y2 - 6x +Sy - 7 -3x2 + 6x- Sy- 5y2 + 6 4x2 - 7y2
35. If F(x,y,z)
=
SSE has
O is a surface in space, P is a point that does
Q
is the point on the surface
that is nearest to P, then prove that the line through P and
Q
is orthogonal to the surface at the point
36. Let f(x,y) a.
=
(x2 - y)(3x2- y)
=
Q.
3x4- 4x2y + y2.
Show that (0, 0) is a critical point. Show that the Sec
ond Derivative Test is inconclusive, however.
b. Plot y
=
x2 and y
=
3x2 in the xy-plane. These two
curves divide the plane into three regions. The function
f takes positive values on two of the regions, negative values on the other. Label these regions + and -
c.
accordingly.
By referring to your plot in part b, show that the
restriction off to any straight line through (0, 0) has a
minimum there.
d. Surprisingly, in view of part c, (0, 0), is a saddle point for
f. Show this by using your plot to find a parabolic curve C such thatf restricted to Chas a maxinlum value at (0,0).
37. Let
SSE(m, b)
)
xJ- 2 1,.�,. x;xi N
N
=
4
·
N
L L (xf + xJ). i=l j=i+l
=
L
l'5.i
2
(x; -Xj) > 0.
formulas (13.lS) and (13.19) define a point at which
+ 4x- 7y + 3
not lie on the surface, and
t
Then apply the Second Derivative Test to show that
5x2 + Sy2- 10x + 24y + 2
=
N
V(SSE, P)
quadratic functions: f(x,y)
(N-1)
Deduce that
34. Apply the method of Exercise 33 to analyze the following a.
(
(N-1) LxJ j=l
Explain how, by completing squares, one can immediately
b. f(x,y)
4
b. Show that
33. Consider a quadratic function of the form f(x,y)
=
be defined by equation (13.17).
a local minimum.
Calculator/Computer Exercises 38. Plot x4 + y2 - 10xy + y for 3:5 x :54 and
16:5y :5 1S.
Identify the coordinates of the local minimum that you see in the plot. Use the Second Derivative Test to verify
this local minimum.
39. Plot x4 - y5 +x2y + x for -1:5 x :50 and 0:5y :51. Iden
tify the coordinates of the saddle point that you see in the plot. Use the Second Derivative Test to verify that this critical point is not an extremum.
40. Plot y3 - xy- x5 for 0 :5 x
:5
1, -1 :5 y :5 0. Identify the
coordinates of the local maxinium that you see in the plot.
Use the Second Derivative Test to verify this local maxinium.
4L Plot
ln(l + x4)- sin(y)
for -1:5x :51
and
1:5y:52.
Identify the coordinates of a critical point that you see in the plot. Verify that the Second Derivative Test is
inconclusive at this critical point. What does your plot tell you about the behavior?
11 .9 Lagrange Multipliers Many times, when we extremize a function
f(x,y),
we are not interested in the
entire domain of the functionf In these applications, we constrain the point
(x,y)
to lie on a curve C in the domain off To allow for the greatest generality, we assume that the constraint curve Chas the form g(x,y)
= c
for some constant
c.
For
example, every point on a mountain has a height f(x, y) above its ground coordi nates
(x,y). But when we walk across the mountain and speak of the highest point
of our walk, we are only interested in the points along the path that we follow. This situation is illustrated in Figure la with a path whose ground coordinates satisfy the equation y
= c
for some constant
c.
Here we take
g(x,y) = y so that the constraint
11.9 Lagrange Mu!Upllers
19 47
curve has the form glx,y) =c. The problem is to maximize the expressionf(x,y) subject to the constraint that ( x,y) must satisfy the equation glx,y) ,._ In this section, we study an elegant mathematical procedure for determining the extrema of a function subject to a constraint. We begin by finding what might be called critical points for /b,y) subject to the constraint gb,y) =c. These points a.re analogous to the critical points that we studied in earlier parts of this book. That is, we locate the critical points by finding the points where derivatives involving f and g are 0. These are the potential extrema. We then examine the e ritical points to determine the extremum that we seek. -
Lagrange Multipliers A Geometric Approach
The example of a mountain path will provide you with a good geometric under standing of the technique. Look again at Figure ta and imagine yourself traversing the path from point A to point B. By following this path, you are constraining your x :1nd y coordinates to the curve g x,y) = c in the xy-plane. As you start out from point A toward point B, you begin to ascend the mountain. Your projection into the xy-plane moves along the curve g x,y) = c from point Ao toward point Bi�. r:rossing the level curves off (as seen in Figure lb}. That crossing behavior cor responds to passing from one level to another on the mountain. At the moment you pass through the high point P along your path, you break the pattern of passing to a higher level. That suggests that the curve g x, y) c will touch but not cross the level curve off at the corresponding point Po in the xy-plane. Indeed, Figure lb shows that the curve gl x,y) =c is tangent at Po to the level curve off through Po. The same holds true even when the constraint curve is not a straight line (Figure 2).
Leve' curves
.&. Figure 1a
.&. Figure 1b
- Level curves off
-g(x. y) = c
.&. Figure 2a
.&. Figure 2b
948
Chapter 11
Functions of Several Variables What we need, then, is an algebraic condition that tells us where g(x,y)
= c is
tangent to the level curve of/through (x,y). Now two curves are tangent at a point of intersection Po if and only if their normal vectors at the point are parallel. But we already know that \7g(P0) is normal to the curve g(x,y) =cat Po and that \7f(P0) is normal to the level curve of f at P0• So we want to find all points Po at which the vectors \7g(P0) and \7f(P0) are parallel. According to Theorem 2 of Section 9.1 in Chapter 9, the two vectors \7g(P0) and \7f(P0) are parallel if and only if either \lf(Po)
= >-.\7g(Po)
for some scalar >-. or \lg(Po)
= 0.
The critical points of f(x,y) subject to the constraint g(x,y) =care the places on the constraint curve at which one of these two equations holds. The factor >-. is known as
a Lagrange multiplier. The procedure for finding critical points by solving = >-.\lg(x,y) is called the method of Lagrange multipliers.
the equation \lf(x,y) INSIGHT
It is important to understand that a critical point off in the sense of
f(x,y) subject to the constraint g(x,y) = c. The surface z = f(x,y) that is plotted Figures 1and2 has a horizontal tangent plane only at the point (0, 0). That point is the only critical point of f(x,y), just as Po is the only critical point of f(x,y) subject to the constraint g(x,y) = c. There is no con nection between the set of critical points of f and the set of critical points of f subject to the constraint g(x,y) = c. Section 11.8 is not, in general, a critical point of
Let us see how the method of Lagrange multipliers is applied in practice. �
EXA M P L E 1
the point
Find the point on the hyperbola x2 - y2
= 4 that is nearest to
(0, 2).
Solution Our first job is to identify the function that we are maximizing or minimizing and to identify the constraint function. The constraint function is easy. = 4. So, let g(x,y) = x2 - y2.
We are only interested in points on the curve x2 - y2 Then, our constraint isg(x,y) only at the point critical point.
(0, 0),
= 4. We calculate \lg(x,y) = (2x, -2y). This vector is 0
which is not on the constraint curve. It is therefore not a
We turn to the Lagrange multiplier equation
\lf =>.\lg for a function/that we (0, 2), let us employ the strategy discussed in Section 11.8 and minimize the square of the distance to (0, 2),
will now specify. Rather than minimize the distance to
namely f(x,y) =x2 + (y- 2)2. The reason we want to do this is because the
expression x2 + (y - 2)2 is simpler than expressions x2 + (y - 2)2 and
Jx2
Jx2
+ (y - 2)2. We
can
do this because the
+ (y - 2)2 are both minimized at the same point.
According to the method of Lagrange multipliers, we find the critical points for this problem
by
solving
the
equation
\lf(x,y) = >-.\lg(x,y),
or
(2x, 2y- 4))
=
>-.(2x, -2y) ). Equating components, we convert this vector equation into the two scalar equations,
11.9
2x= lli
and
Lagrange Multipliers
949
2y- 4=>.(-2y),
which we need to solve forx, y and possibly>.. The first equation can be written as 2x(>.-1)=0, from which we see that eitherx=0 or>.=1. Butx =0 is not possible for points on the constraint curve x2- y2=4. Therefore >.=1. Substituting this into the second equation gives 2y- 4= (1)(-2y), or 4y- 4=0, or y=1. The points on the curve x2- y2=4 with y=1 are
(VS,1) and (-VS, 1).
These are the critical
points for our problem. We know from common sense that the problem of finding
=r-j������\�,.x._ \(x,y) =x2 -y2 = / 3
-3
Constraint curve
• Figure
(v's, 1)
the nearest point to (0,2) has a solution. Which of the critical points is it? Sym metry considerations (see Figure stant from (0,2). So both extremal problem.
(VS, 1)
3)
show that our two critical points are equidi-
and
(-VS,1)
are solutions to our constrained
<1111
3 Critical points at
and
(-v's, 1)
INSIGHT
in Example
Notice that, in the process of applying the method of Lagrange multipliers
1, we have solved for
>.. However, the actual value of >. is not a part of our
final solution of the problem. Although the parameter >. is part of the technique, and its value is often crucial to the solution process, the value of A. is usually of no interest. In some applications in economics and other constrained extremum problems, the actual value of A. does have significance. But, in general, A. is merely a proportionality factor between two parallel vectors and does not have any intrinsic importance.
Why the Method of Lagrange Multipliers Works
Let us say a few more words about why the Lagrange multiplier technique works. Look at Figure 4, which illustrates the level curve g(x,y)=c. Our job is to find the maxima and minima off among those points that lie on this curve. Say that we are standing at a point P1 on the curve, and looking for a maximum. If we are not already at a maximum, then we should move to a different point in our search. In which direction should we go? Ideally, we should go in the direction of '\lf(P1) because that is the direction of most rapid increase off. The trouble is that '\lf(P1) probably points out of the curve (as happens in Figure 4). The next best thing is that we can move in the direction of the projection of '\lf(P1) onto the tangent to the curve (as shown in Figure 4). Thus we move a little bit in that direction to a
Vf(P') = >..Vg(P') • Figure
4
point P , and repeat our program: If we are not already at a maximum, then we 2 move in the direction of the projection of '\lf(P ) onto the tangent to the curve at 2 P • When we find the maximum P' we seek, this procedure breaks down: There is 2 no direction to move that will increase the value off. Thus '\lf(P') must have zero projection into the tangent to the curve. In other words, at a maximum P', the vector '\lf(P') is normal to the curve. But '\lg(P') is normal to the curve g(x,y) =c at P', so '\lf(P') and '\lg(P') are parallel. Thus either '\lg(P')=0 or '\lf(P')= >.'\lg(P') for some constant >.. The same reasoning works at a minimum. This explains why we declare places at which '\lf(P')=>.'\lg(P') to be critical points. We state the results of our investigation as a theorem. A rigorous proof is outlined in Exercise 45.
THEOREM
1
(Method of Lagrange Multipliers).
Suppose that (x,y) f-Tf(x,y)
and (x,y) f-tg(x,y) are differentiable functions. Let c be a constant. If f has an extreme value at a point P' on the constraint curve g(x,y)=c, then either '\lg(P')=0 or there is a constant >. such that '\lf(P')= >.'\lg(P').
950
Chapter 11
Functions of Several Variables INSIGHT As noted, the points P' for which Vg(P') the solutions of the Lagrange multiplier equation
=
0 must be considered alongside
Vf(P') ).SJg(P') as potential extrema. Also, the method of Lagrange multipliers fails at points P' where either Vf(P') or Vg(P') =
is undefined. As with the theory of extrema of one variable, these situations must be handled with
ad hoc techniques. We will not treat them here.
� EXAM P L E 2 Find the maximum and minimum values of the function
f(x,y)=2.x2-y2
on the ellipse
x2 + 2(y-1)2=2.
Solution Let g(x,y) =x2 + 2(y-1)2. Our job is to maximize the function/subject to the constraint g(x,y)=2. Notice that \i'g(x,y)=(2.x,4(y-1)), which is 0 only at the point (0,1). Because this point is not on the constraint curve g(x,y)=2, it is not a critical point for our problem. The only critical points will be the solutions of the
\i'f(x,y)= AV'g(x,y) that lie on the constraint curve. Therefore we want to (4x,-2y)= A(2x,4(y-1)) or
equation solve
4x= A2x
(11.9.1)
-2y= >A(y - 1).
(11.9.2)
and
Equation
(11.9.1)
is equivalent to
2x(2 - A)= 0,
which holds when either
A= 2
or
x=O. x= 0, then the corresponding points on the curve g(x,y)= 2 are (0,0) and (0,2). We still must determine whether equation (11.9.2) is satisfied at either of these two points. In fact, by taking A=0, equation (11.9.2) holds for the point (0,0). Similarly, we find that the point (0,2) satisfies equation (11.9.2) for A=-1. Therefore (0,0) and (0,2) are both critical points. If A= 2, then equation (11.9.2) becomes -2y= 8(y - 1), or -lOy= -8, or y= 4/5. The corresponding points on the curve g(x,y)= 2 are (4v'3/5,4/5) and (-4v'3/5,4/5).
a. If
b.
Constraint curve g(x,y) = c
y
We have found four critical points:
(0,0), (0,2), (4v'3/5,4/5),
and
(-4v'3/5,4/5).
The maximum we seek will occur at one (or more) of these four points. The values
x
off(x,y) at these four points are 0,-4, 80/25, and 80/25, respectively. We conclude that the maximum value of f, when restricted to the curve g(x,y)=2, is 80/25. This maximum value is attained at the points (4vJ/5, 4/5), and (-4v'3/5, 4/5). The minimum value off when restricted to the curve g(x,y) =2 is -4. This minimum value is attained at the point (0,2). Figure 5 shows the location of the three extrema on the constraint curve. The critical point (0,0) yields neither a maximum nor a minimum. There is no contradiction in this: As in the extremization topics that
.A Figure 5
Level curves off were treated earlier, a critical point is a candidate for an extremum, but it may turn out not to actually be one. ..,.
11.9 Lagrange Multipliers INSIGHT
The Lagrange multiplier equation 'Vf(x,y)
=
951
>..'Vg(x,y) is a vector equa
tion that is equivalent to two scalar equations-equations (11.9.1) and (11.9.2) in Example 2. These two equations involve three unknowns: x, y, and>... When solving for three variables, we expect to work with three equations. It is the constraint equation g(x,y)
= c that provides
the third equation.
A Strategy for Solving
The algebra in Lagrange multiplier problems is often tricky. If we are not careful,
Lagrange Multiplier
then we may easily overlook or lose solutions. The following strategy can be helpful
Equations
'\lf(x,y) = >.'\1g(x,y) where '\1g(x,y) =/: 0. The '\lf(x,y) = >.'\lg(x,y) is equivalent to the two scalar equations
in looking for solutions of equation
vector
fx(x,y) = Agx(x,y)
(11.9.3)
fy(x,y) = Agy(x,y).
(11.9.4)
and
'\lg(x,y) =/: 0, we have gx(x,y) =/: 0 or gy(x,y) =/: 0. Suppose, for example, gx(x,y) =/: 0. On multiplying each side of (11.9.4) by gx(x,y), we obtain fy (x,y)gx(x,y) = Agx(x,y)gy(x,y), or, in view of equation (11.9.3), Because that
(11.9.5)
fy(x,y)gx(x,y) = fx(x,y)gy(x,y). Equation
(11.9.5)
together with the constraint equation,
equations in the two unknowns
x
and
y.
g(x,y) =c,
makes for two
It is a heuristic principle of algebra that,
when the number of unknowns equals the number of equations, the system will have a finite number solutions. �
EXAM PL E 3 Maximize
x + 2y
subject to the constraint
x2 + y2 = 5.
f(x,y) = x + 2y and g(x,y) = x2 + y2• Then, '\lf(x,y) = (1, 2) and '\1g(x,y) = (2x,2y ). Because '\1g(x,y) =0 only at the point (0, 0), which is not on the constraint curve, we may assume that '\lg(x,y) =/: 0. Then we may use equation (11.9.5), which gives us (2)(2x) =(1)(2y), or y =2x. Substituting this into the constraint, we obtain x2 + (2x)2 =5, or 5x2 =5, or x = ±1. There are therefore two critical points: (1,2) and (-1, -2). The maximum of f(x,y) =x + 2y clearly occurs for positive values of the variables and is /(1, 2), or 5 . ..,.. Solution Let
A Cautionary Example
Our next example emphasizes that it is important for us to know that an extremum exists before we leap to any conclusion about a candidate that is located by the method of Lagrange multipliers. �
EXAM PL E 4 What are the extreme values of f(x,y)
to the constraint
(x + y)2 =1?
= x2 + 2y + 16, subject
g(x,y) =(x + y)2 and '\lg(x,y) =2(x + y,x + Y{)· We observe that x + y =0. The constraint (x + y) = 1 rules out such points. We proceed to the vector equation '\lf =>.'\lg, which gives us the two scalar equations 2x =2>.(x + y) and 2 =2>.(x + y). Because the right sides of these two equations are the same, we see that 2x =2, or x =1. Substituting this value of x Solution Here,
'\lg(x,y) =0
if and only if
852
Chapter 11
Functions of Several Variables
)
into the constraint equation (x+y)2=1. we find that y=0 or y=-2. Thus (1, 0 and (1,-2) are the only critical points for f(x,y) subject to the constraint (x+y) = 1. Because 17=/(1,0) >/(1,-2) = 13, it is tempting to think that we have located a maximum at (1,0). That conclusion would be wrong, however. Notice that the constraint condition tells us that either x+y=1 or x+y=-1. Therefore on the constraint curve. either y=1- x or y=-1- x. Because f(x, 1- x)= x2+2(1-x)+16=(x-1)2+17 and f(x,-1- x)=x2+2(-1- x)+16=
(x- 1)2+13,
we see that, subject to the given constraint, f can assume arbitrarily
f does not have an extreme value at the point (1, 0). On our formulas for f(x, 1- x) and f(x,-1 - x) show that f(x,y) � 13=/(1,-2) at points on the constraint curve. Therefore subject to the large values. Therefore
the other
hand,
constraint,/has a minimum value of
• •
13, which occurs at (1,-2)
(see Figure
6).
�
(1,0, 17) (1, -2, 13)
.A. Rgure6
Lagrange Multipliers and Functions of Three Variables
The method of Lagrange multipliers works in much the same way for functions of
F(x,y,z) subject to the constraint P=(x,y,z) on the constraint curve for which the vector equation VF(P) =XV1G(P) holds. Any point P where VG(P) =0 or where VF(P) does not exist or where VG(P) does not exist is also included
three variables. If we want to extremize
G(x,y,z)=c,
then we must find all points
among the critical points. � EXAMPLE 5 The temperature at a point
F(x,y,z)=8x-4y+2z. sphere x2+r +z2= 21. Solution Let
(x,y,z)
in space is given by
Find the maximum and minimum temperatures on the
G(x,y,z) =x2 +y2 +:r:2. Thus we are finding extrema of f subject G(x,y,z) =21. First we calculate VG(x,y,z)={2x,2y,2z}). From that VG(x,y,z) =0 holds only at the origin, a point that may be
to the constraint
this, we see disregarded because it does not satisfy the constraint equation. We find the critical
VF(x,y,z) =.XVG(x,y,z). (8,-4, 2) = .X(2x, 2y,2z} or
points by setting explicitly gives
8= A2x
and
-4= >ay
Writing out the gradients more
and
The first and third equations yield
�=.A=! x z
or
x=4z.
2= A2z.
11.9 Lagrange MuHlpllers
953
The second and third equations yield
-2 y
-
>.
=
=
1 z
or
-
y
=
-2z.
G(x,y,z)=21. The result z= �1. If z=1, then x 4 and y -2, so we have found the critical point (4, -2, 1). Likewise, z=-1 yields the critical point (-4,2,-1). Because F(4,-2,1)=42 and F(-4,2,-1) =-42, we conclude the following: The minimum value of F on the surface G(x,y,z) 21 is -42, and this value is assumed at the point (-4,2,-1); the maximum value of Fon the surface G(x, y, z)=21 is 42, and this value is assumed at the point (4,-2,1). �
We may substitute these into the constraint equation
(4z;)2 +(-2z;)2 +z2=21.
of these substitutions is =
Therefore
=
=
Function Subject to Two Constraints
Extremizing
a
F(x,y,z) on the that is formed by the intersection of two surfaces G(x,y,z) =ci and H(x,y,z) c2• In this case, we use two Lagrange multipliers>. andµ. The Lagrange
It is sometimes necessary to find extreme values of an expression curve
=
multiplier equation is then
VF(P)
=
>.VG(P) +µVH(P).
(11.9.6)
By extracting components, we obtain from this vector equation three scalar equa
x, y, z, .A, and µ. The other two equations that we need are the constraint equations G(x,y, z) c1 and H(x,y,z) =cz. The reason that this
tions in the five unknowns
=
procedure works is discussed in Exercise 48. However, the algebra required to solve the resulting system of five equations in fi.ve unknowns is usually tedious and often complicated. 6 Let
� EXAM PL E
C be
the
curve
2x+y+ z =11/5 intersects the surface largest
z
that
x2+y+z=1.
results
when
the
plane
What point on Chas the
value?
Solution Because
z=1 - (x2 +y2) :s; 1 for each point (x,y,z)on C, we deduce that z on C. Figure 7. which shows the two surfaces, also
there is a maximum value of
makes it clear that our problem has a solution. It appears from the figure that there is also a minimum value of Y
should therefore expect two solutions of
F(x,y,z) =z. The G(x,y,z) =2x+y +z=11/5 and H(x,y,z) =x2+y+z=1. Equation (11.9.6) becomes {O, 0, 1) =.A(2,1,1) +µ(2x,2y,1) or
constraints
are
O .A. Rgure7
z. We
the Lagrange multiplier equations. The function to maximize is
=
2,\+ 2µx
The third equation gives us
and
O
=
>. +2µy
>. =1 - µ.
and
1
=
,\+µ.
Substituting this into 0=2>. + 2µx and
simplifying. we obtain µ(1 - x)=1. This equation tells us thatµ :/; 0and1 - x =1 /µ.
,\ with 1 -µ in 0 ,\+2µy, we obtain µ(1 - 2y) 1. Therefore 1 - x=1/µ=1 - 2y or x=2y. We can now eliminate x from the two constraints: G(2y,y,z) =2(2y)+y+z=11/5 or z=11/5-Sy and H(2y,y, z)=(2y)2 + y+z=1. Replacing z with 11/5-Sy in this last equation gives us Sy+(11/5 -Sy)=1or5y2 -Sy+ 6/5=0. Using the Quadratic Formula, we have Similarly, replacing
=
=
954
Chapter 11
Functions of Several Variables
- -(-5)± yBut we have found that z
=
.j(-5)2- 4(5)(6/5)_5±1_2 2(5)
11/5 - Sy. Therefore the largest value of z on the curve
of intersection is 11/5 - 5(2/5)
Q UIC K
Q UIZ
3
-10- 5'5'
=
1/5. The least value is 11/5 - 5(3/5)
=
-4/5.
1. True or false: If
f and g are differentiable functions of x and y, then an extreme f(x, y) subject to the constraint g(x, y) = c must occur at a point Po for "Vf(Po) = >.."Vg(Po).
value of which
2. True or false: If two critical points arise in the solution of a constrained extre mum problem, then a maximum occurs at one of the points and a minimum
occurs at the other.
3. Maximize, if possible,
4. Minimize, if possible,
x2 + y2 subject
to the constraint
x + 2y = 5.
x2 + y2 subject to the constraint x + 2y
=
5.
Answers
1. False
2. False
4. 5
3. No maximum
EXERCISES 16. Maximize the product
Problems for Practice In each of Exercises 1-12, find the extrema of f(x,y) subject to the constraint
g(x,y)=c.
1. f(x,y) =2x-
3y+6 -4y 3. f(x,y) =(x +1)2+y2 2. f(x,y) = 3x 4. f(x,y)
=x2 -y2
5. f(x,y) =3xy2 -24 6. f(x,y)
=xy f(x,y) =x + y2 7. 8. f(x,y) =y2 -x2 9. f(x,y) =4x2
+ 4y2
10. f(x,y) =sin2(x)+sin2 (y) 11. f(x,y,z) =xyz 12. f(x,y,z) =4x
- 7y + 6z
g(x,y) =x2+y2=16 g(x,y) =x2+9y2=18 g(x,y)=x2+y2= 9 g(x,y) =y2+2x2=4 g(x,y)=x4+y4=16 g(x,y)=x + y=7r g(x,y,z) =x2+y2+ z2=4 g(x,y,z)=x2+7y2 + 12y2=84
multipliers to solve the stated problem.
13. Minimize the functionf(x,y) =x2 +2y2+9 subject to the constraint 2x - 6y =5.
subject to the
subject
to
the
condition
that
require the total area of the window to be
+ 2y2=8.
6
square feet,
but the material used in the frame is very costly. What dimensions will minimize the perimeter of the window?
19. Minimize the surface area 2m2+27rrh of a cylindrical can
h and radius r, subject to the constraint 7rr2h of the cylinder is equal to a fixed
of height
that the
volume
positive
constant Vo.
x2 -y2 subject to the constraint x2 +2y2=8. x4 +y4+z4 on the sphere x2 + y2 + z2=12. 22. Find the point on the plane x -3y + 5z =6 that is nearest 20. Minimize
2L Maximize
to the origin.
23. Find the point on the ellipsoid x2+2y2+4z2 =4 that is
24.
nearest to
(1,0, 0).
The temperature of any point
(x,y,z) in
space is given by
T(x,y,z) =2x- 6y + 5z. Find the greatest and least temperatures on the surface
x2+6y2+4z2=24. 25. Maximize the product
constraint 2x+y2=4.
15. Minimizex2 +y2+4ysubject to the constraintx2
subject to the condition that
surmounted by an isosceles triangle. Building codes
g(x,y)=x2 +4y2=16 g(x,y) =x2+(y -2)2=9
f(x,y)=x -y2
xy
18. A window is to be constructed in the shape of a rectangle
g(x,y) =x2+2y2=4 g(x,y) =4x2+y2=7
In each of Exercises 13-25, use the method of Lagrange
14. Maximize the function
6x2+y2=8. 17. Maximize x3 +2y x2+y2=4/3.
x2+ 3y2
+ 2z2 =64.
xy2z
subject to the constraint
11.9 Lagrange Multipliers
Further Theory and Practice 26. By using the method of Lagrange multipliers, find two
level curves of f(x,y) = 5x+4y that are tangent to the hyperbola x2 - y2 = 1. Does an extreme value occur at either point of tangency? 27. What is the largest possible y-coordinate of a point on the curve 3x2 +2xy+3 y2 = 24? 28. The temperature on an ellipsoid 2x2+y2 +2z2 = 8 is given by the formula T(x,y,z) = 8z2+4xy- l2y+200.
29.
30. 31.
32.
What are the locations of the hottest and coldest points on the ellipsoid? A certain county consists of the region {(x,y) : 4x2+ 2y2�16}. The altitude at any point of this county is given by a(x,y) = 80x- 10y+150. What are the highest and lowest points in this county? If a, {3, "{are the angles of a triangle, then how large can sin(a)sin(f3)sin('Y) be? Some cylindrical cans, such as containers of juice con centrate, are manufactured with metal tops and bottoms but cardboard sides. Suppose that the cost per unit area of the metal is k times that of the cardboard. Minimize the cost of materials of such a can if its volume is to equal a fixed positive constant V0• A capsule has the shape of a cylinder that is "capped" at each end by a hemisphere. The cylinder has height h > 0 and radius r > 0. Is it possible to maximize or minimize the surface area 47rr2+27rrh subject to the constraint that the volume of the capsule is equal to a fixed positive constant Vo?
In each of Exercises 33-38, constraint equations G(x,y,z) = c1 and H(x,y,z) = c are given. As in Example 6, 2 use two Lagrange multipliers to find the requested extreme value of the given function F(x,y,z). 33. x+y+z=6, 34. 35. 36. 37. 38.
x+2y-3z =12;
minimize
F(x,y, z)=
x2+y2 +z2 x+y+z=6, x+3y-z=l2; minimize F(x,y,z)= x2+y2 maximize 5(x2 +y2 +z2) = 14; x+2y+3z = 6, F(x,y,z) = z 2x+y+z = 2, x2+y2/4+z2 = 1; maximize F(x,y,z) = x x+2y+z = 10, x2+y2- z = O; maximize F(x,y,z) = x2+y2 +z2 x2+2y2-z = 0, 2x2+y2 +z = 27; maximize F(x,y,z) = z
In economics, a utility function f(x,y) quantifies the satisfaction a consumer derives from x units of one item and y units of a second item. The level curves of f are called indifference curves. The consumer's budget line is the line
955
segment in the first quadrant of the xy plane that represents how many of the two items can be purchased for a total amount T. Exercises 39 and 40 concern these concepts. 39. A
Cobb-Douglas utility function has the form f(x,y) = CxPyq where C, p, and q are positive constants. If the first item costs A per unit and the second costs B per unit, and if the consumer has a total amount T that he can spend on the two items, then for what values of x and y is the consumer's utility maximized? 40. Using A=15, B=10, and T =120, plot the consumer's budget line in the viewing window [O, 12]) X [O, 12]). In this viewing window, add the plots (or sketches) of several indifference curves of the Cobb-Douglas utility function described in the preceding exercise. Use the values C = 1, p = 3/4, and q = 1/4. Include the indifference curve on which the consumer's maximum utility lies. What is the relationship of this indifference curve to the budget line? 41. A company allocates an amount T to produce an item. The
available capital can be divided into labor costs and plant costs. Let f(x,y) be the number of units of an item that can be produced with x units of labor and y units of plant expenditures. Suppose that the unit costs of these outlays are a and b, respectively. Show that when production is maximized. fx/fy = a/b. This equation can be stated as the following economic principle: When labor and plant costs are allocated optimally, then the ratio of their marginal productivities is equal to the ratio of their unit costs. 42. Suppose that f and g are differentiable functions of two variables. Suppose also that for every c there is a unique point (x(c),y(c)) that maximizes f subject to the con straint g(x,y) = c. Assume that 'Vg(x(c),y(c))# O' and let ..\(c) be the Lagrange multiplier defined by 'Vf(x(c), y(c)) = ..\(c)'Vg(x(c),y(c)). Let r(c) = (x(c),y(c)). Set M(c) = f(x(c),y(c)), the maximum value off subject to the constraint g(x,y)=c. a. Show that 1 = 'Vg(x(c),y(c)) r' (c). b. Show that ..\(c)='Vf(x(c),y(c)) r' (c). c. Deduce that ..\(c) =M'(c). ·
·
43. Let f(x,y)=xy314 be a production function (in the sense of
Exercise 41). Suppose that the equation 15x+ 12y = c describes the allocation of labor and plant costs. For an arbitrary positive constant c, find the point (x(c),y(c)) on the constraint curve at which f is maximized. Calculate the maximum value M(c) and the Lagrange multiplier ..\(c). According to Exercise 42, ..\(c) is approximately equal to the marginal productivity M(c+1) - M(c). Verify the approximation ..\(c) R:J M(c +1) -M(c) for c = 10000. 44. Suppose that A and B are positive constants, that f(x,y) =Ax+By, and that g(x,y) = x2+y2. Let (x(c), y(c)) be the point on the constraint curve g(x,y)=c at which f is maximized. Let M(c)=f(x(c),y(c)). Calculate the Lagrange multiplier ..\(c), and verify that ..\(c)=M'(c).
956
Chapter 11
Functions of Several Variables
45. The Milkmaid Problem: A river, in the shape of a smooth
d. We conclude that
curve, flows near a house Hand a barn B. Each morning,
V'f(Po)1-r'(to).
a milkmaid leaves the house, fills a bucket of water at a point
R on the river, and then goes to the barn. The
distance of this walk is IHRI)+ IRBI). If the point
Ro
minimizes this distance, then show that there is an ellipse with foci B and H and that is tangent to the river at Ro (see Figure 8).
e.
Because
r
was an arbitrary curve in the level set
we conclude that
g=c
at
Po.
f. Deduce that
V'f(Po)
g= c,
is normal to the level set
V'f(Po) = >-.V'g(Po) for some AEIR. G(x,y,z) =c1 and H(x,y,z)=c2
48. Suppose that surfaces
have the following two properties: they intersect in a smooth curve C containing a point
Po=(x0,y0,z0), P0•
and
V'F(Po)
is in
they do not have the same tangent plane at
Let
r(t)=x(t)i+y(t)j +z(t)k be a parametrization of C with r(to)=Po. Suppose further that Po is an extremum for F(x,y,z) subject to the constraint G(x,y,z)=c1 and H(x,y,z) c2. Show that tf--+F(r(t)) has an extremum at t =to. Deduce that =
_. Figure 8 46. Suppose that a, b, c, o:,(3,I are constants that satisfy the
a2 +b2+c2 > 0 and 0:(31 f= 0. f(x,y,z)=ax+by+cz be a linear function, and let
inequalities
Let
g(x,y,z)=o:2(x - e)2 + (32(y - /)2 +y(z - g)2. Prove that the extrema of
f on
and therefore
V'F(Po) 1-r'(to).
Conclude that
the plane that is normal to C at
Po.
Show that every vector
in this normal plane has the form
>-.V'G(Po)+µV'H(Po)
for some scalars ).. and µ.
49. Maximize x1+x + · · ·+XN subject to the constraint
xi+x�+
the region
·
·
·
2 +�=1.
Deduce the inequality
X1+X2 + · · ·+XN N
n={(x,y,z) :g(x,y,z):5l}
< -
xi+x�+ .. ·+x� N
50. Suppose x1, x , ..., XN are positive numbers. Minimize will always occur on the boundary of
n.
Prove that
there will always be exactly one maximum and one
2 x1+x + · · ·+XN subject to the constraint x1x · · ·XN=1. 2 2 Deduce the Arithmetic-Geometric Mean Inequality:
minimum, and they will occur at diametrically opposite points of the region
n.
47. Complete the following outline to obtain a more rigorous treatment of the method of Lagrange Multipliers. We will assume that
V'f and V'g exist at the critical points and that
V'g does not vanish at those points. We seek extrema of f(x,y) subject to the constraint g(x,y)=c. a. Let Po=(xo,Yo) satisfy g(Po)= c, and let r(t)=x(t)i+ y(t)j be a parametrization of a curve lying in g(x,y) =c that passes through Po. Say that r(to)=Po. b. If Po is an extremum for f subject to the constraint g=c, then the function of one variable /(r(t)) has an extremum at t=to. c. We have
Calculator/Computer Exercises 51. Find the largest value of xexp(x2 - xy) subject to the constraint
closest to the origin.
53. Find the minimum value of (1+x2+xy)/(l +x2 +y4) subject to the constraint
54. Minimize Jx2 +y2 + straint
d t (r(t)) d t
j
x2 +y2= 1.
52. Find the point on the curve xexp(x2 - xy)=1 that is
xy=2.
x2 +y2/4=1.
J(x - 2)2+y2 subject to the con
55. Find the extreme values of x+y+z subject to the con t=to
=o.
straints
z3 +z=e"
and
x2+y2 =1.
Summary of Key Topics
967
Summary of Key Topics in Chapter 11 Functions of Several Variables (Section 11.1)
A function of two variables has as its domain a set of ordered pairs of numbers; a function of three variables has as its domain a set of ordered triples. In this chapter, these functions will be scalar-valued. Functions of several variables may be added, subtracted, and multiplied in the usual way. They may be divided as long as we do not divide by zero. Iff is a scalar-valued function of several variables, ¢ is a scalar valued function of one variable, and the values of/lie in the domain of ti>, then we may form the composition ¢ f to obtain a new scalar-valued function of several variables. Functions of two variables are graphed in three-dimensional space. The graph off is a surface consisting of the points {:x,y,fti,y)}: fa',y)e uomain of/}. In general, a surface in.space is the graph of a function if and only if no verti.cal line intersects the surface more than once. We form the graph by sketching level curves
{(x,y) :l(x,y) =c} in the plane and then amalgamating them into the :final graph. Level curves arise in applications of mathematics as isotherms, isobars, isohyets, and so on. Cylinders (Section 11.�)
Quadric Surfaces
A cylinder is the set of points in space satisfying an equation that is missing one variable. Such surfaces have uniform sections, or level curves, and are easy to sketch.
The quadric surfaces are classified as follows:
(Section 11.�) ellipsoid
elliptic cone
,....-
elliptic paraboloid hyperboloid of one sheet hyperboloid of two sheets hyperbolic paraboloid
Limits and Continuity (Section 11.3)
If P ( P• , 112) is a point in the plane, let I>l Po,r) be the disk of radius r > 0 centered at Po. Let D.f Po,•'°) be the same disk but with the center removed. Let dr Po, P) Llenote the distance between the points P and Po. If the domain of /tx,y) contains D· lPo,r), then we say that =
958
Chapter 11
Functions of Several Variables
(x,y)
f (x,y)-+Po lim
if, for any
c > 0,
there is a
l5 > 0 such
O
f
that
lf(x,y)-£1
implies
Po= (xo,yo)
We say that f is continuous at
=
if the domain off contains
Po
and
lim f(x,y)= f(xo,Yo). (x,y)-+Po The standard limit rules apply to these new notions of limit; the standard results about continuity also hold.
Partial Derivatives (Section 11.4)
If the domain of f contains a disc D(Po,r) , then we define the partial derivatives of/ at
P
=
(xo,Yo)
to be
at (Po)= ax
lim
�-+O
f(xo + &,yo) - f(xo,Yo) D. x
and
(xo,Yo + D.y) - f(xo,yo) at (Po)= lim f ' Lly-+O ay D.y provided these limits exist. The partial differentiation process can be iterated to obtain higher partial derivatives. We will usually work with functions with con tinuous derivatives, called continuously differentiable functions. The familiar dif ferentiation rules also hold for partial differentiation. Other notations for the partial derivatives of
Dxf, D'lf,
f
are
fx, fi, fy, fz, fxy, fi2,
and so on.
All of these ideas can be adapted to functions of three variables.
The Chain Rule (Section 11.5)
If
z
=
f(x,y)
and in turn
x
=
p(s)
and
y
=
cr(s), all
functions continuously differ
entiable, then
dz ds
=
A similar Chain Rule applies when
az ax
dx ds
+
az dy ay ds .
x= p(z,t)
and
y= cr(s,t):
az az ay az ax = + as ax as ay as and
az az ax az ay = + at ax at ay at . There is an analogous Chain Rule when f is a function of three or more variables.
Summary of Key Topics
Gradients and
959
If f(x,y) is a continuously differentiable function, then the gradient off is
Directional Derivatives
grad/(x,y)= '\lf(x,y)= fx(x,y)i + fy(x,y)j.
(Section 11.6) If
u
is a unit vector, then the directional derivative of
point
(x,y)
f in
the direction
u
at the
is
Duf(x,y)= '\lf(x,y)
·
u.
If P o is fixed, then the greatest directional derivative of f at Po occurs when
'\f l (Po). In this case, D0f(Po)= 11 Vf(Po) II derivative of f at Po occurs when u is the direction Duf(P o)= -llV /(Po)ll· direction of
·
u
is the
Also the least directional of
-'\lf(P0).
In this case,
All of these ideas can be adapted to functions of three variables.
Normal Vectors and Tangent Planes
Iff(x,y) is
a continuously differentiable function, and Po=
(x0,y0) is
a point in its
domain, then
(Section 11. 7)
f x(Po)i + fy(Po)j - k is a normal vector to the graph at
(P0,f(P0))
(P0,f(P0)). The tangent plane to the graph at (P0,f(P0)) and normal to this vector.
will be the plane passing through
It will have equation
fx(Po)(x -xo) + fy(Po)(y -Yo) - (z -f(xo,Yo)) = 0. For any continuously differentiable function f of several variables, the gradient
of f is perpendicular to the level surfaces off.
Numerical Approximation
Because the tangent plane geometrically approximates a graph near the point of contact, we may approximate f(x,y) for
(Section 11. 7)
(x,y) near Po= (x0,y0) by
the function
L(x,y)= f(Po) + fx(Po)(x -xo) + fy(Po)(y -Yo). The error, or rate of approximation, can be estimated using the second derivatives off. This approximation is useful in numerical calculations and in error analysis.
Critical Points
Local maxima and minima are defined as for functions of one variable. A point at
(Section 11.8)
which a function has a local minimum when approached from one direction, and a local maximum when approached from another direction, is called a saddle point. If
f(x,y)
is continuously differentiable, then a critical point is a point where the
gradient vanishes. Let
Po= (x0,y0)
differentiable, then define
be a critical point. If
f is
twice continuously
960
Chapter 11
Functions of Several Variables
'D(f, Po)> 0, fxx(Po)> 0, and fyy(Po)> 0, then Po If 'D(f, Po)> 0, fyy(Po) < 0, and fyy(Po) < 0, then Po If 'D(f, Po) < 0, then Po is a saddle point for f. If 'D(f, Po) 0, then we can draw no conclusion.
f. a local maximum for f.
a. If
is a local minimum for
b.
is
c. d.
=
The ability to find and identify critical points is an aid in graphing and enables us to solve applied problems about extrema.
Lagrange Multipliers
To find the critical points for the problem of extremizing constraint
(Section 11.9)
g(x, y) = c,
f(x, y)
subject to the
we solve the equation
'\lf(x,y) = .A'\lg(x,y). This procedure will locate those critical points for the problem at which both and
'\lg exist
and
'\lg =I- 0.
'\lf
Review Exercises for Chapter 11
given. Sketch four level sets of
tion of the domain of this function.
space of
L The surface area A of a cylinder in terms of its height h and its radius
2. The time T required for a round trip of a total distance if the average speed of the first leg of the trip is average speed for the return is
s
2L
v.
£1 and £2 of its two legs. In
(
4. 5. 6.
Exercises ·
4-6,
y),h(x,y)=x2y3•
let
8. 9.
f(x,y)= (x -y)/(x2+1),g(x,y)=
11. 12.
f(x,y)=y2 + 6x2 f(x,y)=x2- 4y2 f(x,y)=x -Sy
of three variables that corresponds to
16. 17.
g(x,y,z)=x-y g(x,y,z)=x2+ 2y2 + 4z2 + 4 g(x,y,z)=6-x2-z2
f that
x2-y=4 y2-z2=1 2L x2+ 4z2= 2
19.
20.
In each of Exercises 22-24, sketch the graph of the given equation and identify the type of quadric surface you have drawn.
In Exercises 10-12, sketch (on a set of planar axes) the
10.
g(x, y, z)
level 5.
In each of Exercises 19-21, sketch the given cylinder on
Calculate the specified quantity.
level sets of the function
In each of Exercises 16-18, sketch the level set of the
axes in three-dimensional space.
{'Pog){ 4,2) ({'Po/)• {'Pog))(-4,2) ( 2'Pof)(O,-3)
0, 4, 8.
z=g(x, y).
g(x,y)=x2 +2y2 + 1 g(x,y) = 6 -y g(x,y) = 2 -(x2 + y2)
function
18.
In Exercises 7-9, let
7.
f(x,y)=xy/(x2 + y2),g(x,y)=
Calculate the specified quantity.
(!+ 3g)( 4, 6 ) (!. g)(l ,3) (f/( g+h+l0))(0,2)
x2 -y3, and 'P(t)=t312•
15.
and the
3. The perimeter P of a right triangle in terms of the lengths
sin x
13. 14.
r.
g of two variables is g, and then sketch the graph in
In each of Exercises 13-15, a function
In Exercises 1-3, express the quantity described as a function of two or more variables. Give an explicit descrip
correspond to values -8,- 4,
22.z-y2=x2 23. y2+ 4x2 = z 24. x2-y2+z2=9 In each of Exercises limit.
25-27, calculate the indicated
Review Exercises
25. 26. 27•
lim
(x,y )-+(1,1)
lim
(x,y)-+(2, 3)
lim
(x,y)-+(3,5)
x2-2x+l xy-2x-y+2 Jxy2 -y3
45.
46.
In each of Exercises 28-30, the provided function does not have a limit at the point (0, 0). Calculate the limit of the function along two different paths, and derive two different answers, to demonstrate that the limit does not exist. x2 -2y2 28. x2+4y2 x-y 29. x+y x 30. Jx2+y2 In each of Exercises 31-33, calculate 8f /8x and 8f /8y at the point P.
33. f (x,y) =xy2 e-X+ Y
P= (2,7r) P=(2,4) P= (0,1)
In each of Exercises 34-36, a function f of the two vari ables x and y is given. Calculate fxx,/yy,/xy, and fyx· In each case verify that fxy =fyx· 34. f (x,y)=tan(xy2)
sin(y) cos(x) 36. f(x,y)=e-XY sin(x2 -y) In each of Exercises 37-39, a function f(x,y, z) of three variables is given. Calculate fx,fy,fz,fxyJyzJxyz, and fzz· 35. f (x,y) -
z=f(x,y)= (x+y)(y-x ),x= p(s)= scos(s), y=a(s) =s sin(s)
In each of Exercises 46-48, use the Chain Rule to calcu late 8z/8s and 8z/8t.
sin(5x-3y) 5x-3y
31. f (x,y)= cos(yx2 ) 32. f (x,y)=ln(x2-xy)
961
47. 48.
z= f (x,y)= xy sin(x2-y2),x=p(s,t) =s cos(t), y=a(s,t) =t sin(s) z=f(x,y)=ln(x-y2),x=p(s,t)= set,y= a(s,t) = te' z=f(x,y)=x/y,x=p(s,t)=ln(s +t),y=a(s,t )=e>+t
In each of Exercises 49-51, use the Chain Rule to calcu late dwlds. 49. 50.
z x-y z=T(s )=s sin(s) w=f(x,y,z)=xyz tan(x),x=p(s)=s/(s+1), y a(s) s(s+1), z T(s) s2 w f (x,y,z) ln(x-y2+z3 ),x=p(s) ln(s)/s, y= a(s) s ln(s),z T(s) e?..
w=f(x,y,z)= __ ,x=p(s)=s ln(s),y=a(s)=se',
=
51.
=
=
=
=
=
=
=
=
=
In each of Exercises 52-54, give the Taylor polynomial of order 2 about the point P. P= (.fi,2 ) P= (2, 1) P= (0,0)
52. f(x,y)=sin(x2y) 53. 54.
f(x,y )= ln(x-y2) f(x,y )= e-X-zy
In each of Exercises given function.
55-57,
calculate the gradient of the
_
-
37. f(x,y,z)= x3yz-sin(yz)
+y2 x/Z 39. f(x,y,z)=xzeY2 x In each of Exercises 40-42, a function f(x,y, z) of three variables is given. Calculate fxyzJyxzJzyxJzxyJyzxJyxz· All six calculations should yield the same final answer. xyz 40. f( x,y,z)x2+y3 +z4 41. f(x,y,z)=z sin(xy) 42. f (x,y,z)=y2xln(x2-z2) In each of Exercises 43-45, calculate dz/ds by (i) using the Chain Rule and (ii) substituting the formulas for x and y into the formula for z and calculating the derivative directly. 38. f(x,y,z)=
55. f(x,y)=x2sin(xy) 56. f(x,y)=cos(xy)sin(y2x) 57.
g(x,y,z)=xyze-Xz-y2
In each of Exercises 58-60, calculate the directional deri vative of the functionf in the direction u at the point P. P=(l,4) u= (3/5 ,-4/5) P=(1,0) u= (1/v'z,-1/vtz) u= (6/10,-8/10) P= (0,0) In each of Exercises 61-63, calculate these four quantities:
58. f(x,y)=xy2sin(x-y) 59. f(x,y)=eY ln(x-y) 60. f (x,y) =xy2e-XY
•
_
43. 44.
z=f(x,y)=x2 -y2x,x= p(s) = s2 -s,y= a(s) = l/s x2-y z=f(x,y)= -,x=p(s)=s+e',y=a(s)=s2 x+y
• • •
the direction u of greatest increase of the function f at the point P; the directional derivative off at P in the direction u; the direction v of greatest decrease of the function f at the point P; the directional derivative off at P in the direction v.
61. f (x,y)= xy3
-x4
62. f (x,y)= ln(xy-y3) 63. f(x,y)=sin(x-y2)
(1,2) P= (3,1) p=(27r, .ft) p=
In each of Exercises 64-66, find the directional derivative of the function g(x,y,z) at the point P in the direction v.
962
Chapter 11
Functions of Several Variables
64.f(x,y,z)=x2 sin(yz),P=(1,0,1), v= ( 3/5, -V7/5,-3/5) 65.f(x,y,z)=y2z(l +x2),P=(1,1,5), v= (-2/3,1/3,2/3) 66.f(x,y,z)=xzeYz,P=(0,0,0), v= (-1/.../2,1/2,-1/2) In each •
• •
•
of Exercises 67-69, calculate these four quantities:
the direction u of greatest increase of the function F at the point P; the directional derivative of F at P in the direction u; the direction v of greatest decrease of the function F at the point P; the directional derivative of F at P in the direction v.
67. F(x,y,z)=xy(z3 -zx) 68. F(x,y,z)=sin(x2 +y2 +z4) 69. F(x,y, z)=ln(x2 +2yz +1)
P=(2,1,4)
p=(0, y'1i, 0)
P=(l,1,1)
In each of Exercises 70-72, calculate a normal vector to the graph of the function f(x,y) at the point P. Then write the equation of the tangent plane to the graph of f at the point P.
70.f(x,y)=y(x2-y2)
71. f(x,y)=sin(x2
-y2)
72.f(x,y)=ln(xy +1)
P=2,3)
p= (y'7i, 0)
P=(0,0)
In each of Exercises 73-75, find a normal vector to the given surface in three-dimensional space at the point Q. Then find the equation of the tangent plane to the surface at Q. 73. x2 +4y2 +8z2=13 74. (x2 +y2 +z)2=36 75. ln(l +x2 +z2)=y
Q=(1,1,1) Q=(1,2,1) Q=(1,ln(2),0)
In each of Exercises 76-78, find parametric equations for the normal line to the graph of the given function at the given point. x2 76.f(x,y)= - -x2-y2 77.f(x,y)=xytan(x +y) 78.f(x,y)= ln(xy +x2)
p=(2,1, 4/3) p=(7r/2,0,0)
P=(1,1,ln(2))
In each of Exercises 79-81, calculate the linear approx imation (given by the tangent plane at (P0,f(P0))) to f(P). P=(3.9,5.9) 79.f(x,y)= ..jy3 +x2 80.f(x,y)=y sin((2y- x)7r) P=(0.9,1.9) P=(l.1,0.1) 81.f(x,y)=ln(y'X+y)
Po=(4,6) P0=(1,2) P0=(1,0)
82. A cylindrical can is to be built with radius 3 inches and height 6 inches. Suppose that the error in measurement can be kept accurate to 0.3%. Use the total differential to estimate the greatest possible error in volume. In each of Exercises 83-85, find all the critical points of the function[. Determine whether each critical point is a local minimum, a local maximum, a saddle point, or none of these. 83.f(x,y)=4x2 +6y2- 6x +8y- 4 84.f(x,y)=3xy- 2/ x- 4/y 85.f(x,y)=(x-4)(y +l)y 86. Find the point(s) on the surface z=xy that are nearest to (1, 1,1). 87. Find three positive numbers whose product is 100 and whose sum is as small as possible. In each of Exercises 88-90, find the extrema of the function f(x,y) subject to the constraint given by the function g(x,y).
88.f(x,y)=cos2(x) +cos2(y),g(x,y)=x- y=7r 89.f(x,y)=(y +2)2 +x2,g(x,y)=y2 +6x2=36 90.f(x,y)=xy2z,g(x,y)=x2 +y2 +z2=9 In each of Exercises 91-93, use the method of Lagrange Multipliers to solve the problem.
91. Minimize the function f(x,y)=4x2 +8y2 +4 subject to the constraint x-4y=8. 92. Maximize the quantity xy2 subject to the constraint y2+2x2=4. 93. Minimize x2-3y2 subject to the constraint 5x2+y2=25.
The two founders of calculus, Newton and Leibniz,
Lindelof
both had occasion to employ partial derivatives. For
existing faulty proofs. Reinforcing the need for a valid
(1827-1908) published a critique of the
example, in a letter written to l'Hopital in 1694, Leibniz
proof, he also exhibited a nondifferentiable function
introduced the total differential of a function of two
for whc i hfxy(x,y) andfyx(x,y) exist but are unequal. At
variables. Although Leibniz wrote
6f and vf for the 8f /8y, respectively,
last, in 1873, Hermann Amandus Schwarz (1843-1921)
the practice of using notation to distinguish between
orem under appropriate assumptions on f. Thereafter,
ordinary and partial derivatives did not take hold
work on Clairaut's Theorem was directed toward
partial
derivatives
and
8f /8x
f
conclusively established the validity of Clairaut's The
among the mathematicians who followed. In fact, the
relaxing the hypotheses. Well-known mathematicians
notation,
such as Camille Jordan (1838-1922), Ulisse Dini
8f /8x
and
8f /8y,
introduced by Carl Gustav
Jacobi in 1841, did not receive widespread acceptance
(1845-1918), Axel Harnack (1851-1888), and Giu
until the end of the 19th century.
seppe Peano (1858-1932) all contributed to extending Clairaut's Theorem to a wider class of functions.
Clairaut's Theorem Today, we often refer to the identity
f y (x,y) = fyx(x,y)
as
Claude
Clairaut's
Theorem.
Alexis
The Vibrating String
x
Clairaut
A large part of the mathematical enterprise of the 18th
(1713-1765), the son of a mathematics teacher and the
century was devoted to applying calculus to natural
only one of twenty siblings to reach adulthood, was a
phenomena. One of the problems that first attracted
mathematical prodigy who published his first paper at
attention was the motion of a vibrating string (Figure 1).
the age of thirteen. When only sixteen, he initiated the
If
study of space curves, and, as a result, he became
string at point x (0
the youngest member ever appointed to the prestigious
shown that
y (x, t)
represents the vertical displacement of the
� x� L)
and time
t,
then it can be
Paris Academy of Sciences. In addition to geometry, Clairaut undertook important research in differential equations, the calculus of variations, mechanics, and astronomy. He participated in an expedition to Lap land to measure one degree of longitude, with the aim of verifying Newton's assertion that Earth is an oblate sphere. Clairaut's scientific activity continued unabated until his early death, but, not to the exclusion of all other interests. As one contemporary mathematician wrote, "Occupied by dinner parties and late-night dis cussions, burdened by unrestrained womanizing, and content to mix pleasure with business, Clairaut lost his repose, his health, and, finally, at the age of 52, his life." Before Clairaut, the equality of mixed partial derivatives had been observed in particular cases and
.._ Figure 1 A snapshot of a vibrating string at time t where
was tacitly assumed to be true in general. In 1740,
d'Alembert also argued that any solution of the one
Clairaut
published
fxy(x,y)
fyx (x y),
dimensional wave equation can be expressed in the form
=
,
a
derivation
of
the
equation
as did Euler. However, when the
standards of mathematical rigor tightened up half a century later, both proofs were judged inadequate. A succession
of
the Finnish
fi(x- ct )
+fz(x +ct)
(1)
including
where fi and fz are twice-differentiable functions of one variable. Letting Lh = c!:it, we observe that f1((x +!:ix) - c(t +!:it )) = fi(x- ct). This means that the displacement of the string at a point c!:it to the right of xis
mathematicians,
alternative demonstrations that were equally flawed. In 1867,
=
and Cauchy in 1823, advanced
eminent
Lagrange in 1797
y(x,t )
mathematician Leonard Lorenz
963
the same displacement that emted atx at a time tu units earlier.We conclude that/ represents a wave that travels to the right with speed c C see Figure 2). Similarly, f:l (x - .6.x) +c' t + �t))- f-111: +ct), so ft represents a wave that travels to the left with speed c.
+P:�--ct) --•z L y y= J
(z-c(t +At)) Xi +c�
A.
12+w
Figure 2
L
Suapshota at times wave on a vibrating atring
and r ,. t::.t of
a traveling
Six years later, Daniel Bernoulli (1700-1782), son
of Johann Bernoulli,
discovered an alternative method solving the wave equation. Hia idea was to represent the solution aa a superposition of fundamental waves (Figure 3). Following that approach, Bernoulli found
Other Equftlons of Matllematlcal Physics AB the 18th century progressed, mathematicians used
partial differential equations to study acoustics, hydrodynamics, elasticity, and celestial mechanics. One particular type of expression appeared repeatedly. If y depends on only one apace variable x, if z depends on only two space variableax and y, and if u depends on all three apace variables x, y, and z, then we write,
Using this notation, d'Alembert's one-dimensional wave equation takes the form Yrr 2�y. Similarly, the two-dimensional wave equation, which Euler intro duced in a 1764 study of the vibrating circular mem brane (Figure 4), has the form Zrt = c26.z.
of
y
that the general aolutioo of the wave equation has the form
ylx,t)= f: �,.cos:n�} +b,.sin C7) sin� 7) · u-o
2)
Now it often happens that different methods lead to different forms of an answer. Usually these differences can be easily reconciled. In the case of the vibrating string, however, there was no obvious way to relate the solutions of d'Alembert and Bernoulli. Calculus was left in something of a crisis for several decades.
(
I�
,_
A. Figure 4
A v.ibrating circular
membrane
Although the expression � :first appeared in a 17S2 paper of Euler that concerned :Duid flow, Pierre Simon Laplace (1749-1827) used the equation Au= 0 so extensively in his work on celestial mechanics that Au is now called the Laplacian f u. One partial dif ferential equation involving the Laplacian, the heat equation A u has had a particularly important impact on mathematical analysis. This equation, so-named becauae it describes the conduction of heat, was derived by Jean Baptiste Joseph Fourier (1768-1830) in 1807. Although Fourier's method of solution did not gain immediate acceptance, he undertook a more thorough exposition, titled heom Analytique de la Chaltur, which he published in 1822. When Fourier's boot came out, the controversy over the vibrating string was still very much alive even though the original participants in the dispute were long
,
I
dead. Fourier realized that his method coold finally
v
L
A. Figure a In Daniel Bernoulli's solution of the wave equation, every vtbration ii the superposition of waves that have fundamental frequencies.
"resolve all the difficulties that the analysis employed by Daniel Bernoulli presented." To that end, consider a wave y x, ) = f x - ct) such that y , t) = y L, ) = 0 for all . We have, on the one hand, y x, 0) = f x). On the other hand, Bernoulli's solution tells us that we can
Genesis & Development express
y(x, 0) by substituting t= 0 into the right side of
equation
(2). This leads to
the equation
(3)
.
=
Before
Fourier's
work
on
heat
conduction,
most
mathematicians did not believe that a general function could admit such an expansion as a trigonometric ser ies. Fourier was convinced that such skepticism would vanish once a concrete relationship between coefficients
an
There would be little point displaying an analogous
�
"Geometers only admit that which they cannot dispute.
discernible. Indeed, look at Figure
ies is slowest at the points
the coefficients
an
in equation
x = 1, x = 2, x = 4, and x = 7. f is not differentiable.
y
(3)
(3):
0.02.
0.015
complete is that which consists of actually resolving an
equation were exactly the right tools for determining
which shows the
Even so, the maximum error is only about
0.01
above] and assigning the values of the coefficients."
6,
y= f(x) - T100(x). The graph
These are the points at which
0.005
The methods Fourier brought to bear on the heat
Trno(x) =
suggests that the convergence of the trigonometric ser
Of all the derivations of Bernoulli's solution, the most arbitrary function into such a series [as equation
and
because no differences would be
plot of the error function
f and the
was made. As he phrased the matter,
f(x)
su erposition of the graphs of
En?:!i ansin(mrx/10)
00
f(x) = L: ansin (nlx) n l
965
8
6
10
x
-0.005 -0.01 -0.015 -0.02
. mrx {L a,, = L J f(x) sm (L) dx. 2
(4)
0
Fourier emphasized that his solution is "applicable
_. Figure 6 The
repercussions
of
Fourier's
work
touched
to the case where the initial figure of the string is that of
nearly every aspect of mathematical analysis. The
a triangle or a trapezoid, or is such that only one part of
modern definition of a function as a correspondence of
the string is set in motion while the other parts blend
real numbers (as opposed to an analytic expression)
with the axis." In other words, he asserted that his
ensued in large measure from Fourier's influence. The
method remains valid for a function that is defined by
introduction of Fourier series prompted Niels Henrik
different analytic expressions on different parts of its
Abel
domain. Figure interval
5 shows the graph of a function f on the
[O, 1OJ. This function is precisely of the type that
(1802-1829) and Peter Gustav Lejeune Dirichlet (1805-1859) to clarify the meaning of convergence of infinite series. By 1850, Fourier series were being used
partly
in number theory, far afield from mathematical physics.
"blended" with the axis. The trigonometric polynomial
Because many natural functions of number theory are
Fourier
mentioned:
partly
trapezoidal
and
Ts(x)= E�=l ansin(mrx/10), with coefficients specified
considerably more irregular than the garden-variety
Ts(x) already captures the general shape of the graph of f even though it is composed of a small number of terms.
Riemann
by equation (4), is also shown in Figure
5.
Notice that
functions
of
physics,
(1826-1866)
Georg
Friedrich
the meaning of equation (4) in cases where
f has
numerous discontinuities. The Riemann integral was the result of his investigations. In
(1845-1918)
3
Bernhard
thought it prudent to examine
1871,
Georg Cantor
studied the uniqueness of representation
by trigonometric series. The question he sought to answer was, for what sets X is it true that
2 1
(10,0) (1, 0)
_. Figure 5
4
6(7,0)
x
10
-T,(x) =n� 5 a.sin(nl�x) - f(x) -1
implies an= bn for all n? It was through these studies that Cantor came to create the subject of set theory.
This page intentionally left blank
Multiple Integrals P
R
E
V
E
W
In the preceding chapter, we learned how to differentiate functions of two or more variables. Now we will learn how to integrate such functions. To be specific, given a region R of the xy-plane and a continuous functionfdefined on 'R, we will define the integral offover 'R. This integral is denoted by II'Rf(x,y) dA and is called a double integral. Analogously, if Fis a continuous function of three variables that is defined on a solid region U in xyz-space, then we will define the triple integral III uF(x,y,z) dV. The main theorem of this chapter tells us that multiple integrals can be evaluated by iterating the familiar process of integration, one variable at a time. In Chapter 7, we studied certain physical applications of the integral: calcu lating centers of mass of planar regions, volumes of solids of revolution, and so on. Because we had only single integrals available to us at the time, the scope of our discussion was necessarily limited. For example, when we calculated volumes, we treated only solids with symmetries. We exploited the symmetry to express volume by means of a single integral. In this chapter, we will take a second look at some of the physical interpretations of the integral. As we will see, double and triple integrals allow us to handle these concepts in greater generality. Moreover, we will be able to treat some more substantial problems in Chapter 13 using multiple integrals.
967
968
Chapter 12
Multiple Integrals
1 2.1
Double Integrals Over Rectangular Regions Suppose that x ( ,y),_.fx ( ,y) is a positive, continuous function whose domain con-
z =
tains the rectangle n = {(x,y):
f(x,y)
a
:Sx :Sb, c :Sy :Sd}.
How should we calculate the
volume of the solid that lies under the graph off and above rectangle n? See Figure 1.
�dY� _,"7....-+... ---1�--=--+----J/
We begin by partitioning the domain of the function. A convenient way to do this is to partition the domain of the x variable: a
_. Figure 1
=xo
···
b
and also to partition the domain of the y variable:
y
c =Yo
YN d Yi Yj-1 Y2 Yi Yo
..,.
Ca
Q;,j
4y l_ T
b
···
d.
For simplicity, we use only uniform partitions, with the same number
N of intervals
in both thex-direction and they-direction. This partition breaks up the rectangle n
x
into smaller rectangles of equal sizes, as shown in Figure 2. Specifically, to each pair
X;-i,x; of consecutive points in thex-partition and each pairYi-1'Yi of consecutive points in they-partition, there corresponds a subrectangle The side lengths of
Qi,j· Q;,i are b.x =x; -x;-1 = (b - a)/N andb.y =Yi -Yi-1 = (d - c)/N. Over each subrectangle Q;,j, we erect a box with height f(�;,j), some point in Q;,i (see Figure 3). This box has volume
_. Figure 2
Vi,i = f(�;) b.xb.y = f(�;J) M
where
�i,j
is
.
Here the symbol M denotes the increment (b.x)(b.y) of (planar) area. The volume V of the solid under the graph of z f(x,y) and over rectangle n is then
=
approximated by the sum of the volumes of these boxes. That is, N
_. Figure 3 The volume V;,i f(e;j)t:ut::.y of the box of height f(e;j) and base area t::.A = t:ut::.y approximates the volume of the solid under the graph of z = f(x,y) and over the subrectangle Q;,i
N
N
N
v� L LVi.i= L Li(�;) M i=l j=l i=l j=l
( 12.1.1)
.
=
Approximation (12.1.1) is illustrated for Figure 4b. In Figure 4c, with
N=
2 in Figure 4a and for
N= 4
in
f(x,y). To
see how well the tops of the 162 boxes approximate the graph of [---compare Figure 4c with Figure 1.
/ _. Figure 4a
N=
16, we have omitted the graph of z =
_. Figure 4b
dY
_. Figure 4c
12..1
2
1 •
(1/4, 1/2) •
1
(3/4, 1/2)
y
•
(1/4,3/2) •
(3/4,3/2)
-'1y-ooj
� EXAM PL E
1 Use approximation
volwne of the solid
that
tangle
T 4x
Double Integrals Over Rectangular Regions
[O, 1) x [0,2)
(12.1.1)
lies under the surface z
=
6-x2-y2 with
N
=
2
969
to estimate the
and above the rec
in the xy-plane. Use the midpoint of each subrectangle for the
choice of points {{;;}.
_l_
that arise from the uniform [O, 1) and y-interval [O,2) with N= 2. The midpoints of these rectangles are also shown. The increment of area is M 6.xl:l.y (1/2)(1) = 1/2. We use /(x,y) = 6-x2 -y2 in approximation (12.1.1) to obtain Solution Figure 5 illustrates the four subrectangles partitions of the x-interval
x
=
• Figure 5 The rectangle [0, 1] x [O, 2] with two equal subdivisions of each side: The midpoints of the subrectangles constitute the choice of points {e,Jl·
v�
(6-(�)2-(�)2) (�) (6-(�)2-(�)2) (�) (6-(�f-(if) (�) (6-(�)2-(�f) (�)
=
+
+
+
=
Figure 6a illustrates the solid. Figure 6b shows the four approximating boxes. � We call the double swn in line (12.1.1) a Riemann sum. As we have seen, these swns can be used to estimate the volume under the graph off when f is a positive function. However, line Furthermore, it y
(1,2) • Figure 6a
can
be
(12.1.1)
makes perfect sense for any continuous function.
shown that the Riemann swns of any continuous function f
tend to a limit as N tends to infinity. We may therefore
make
the following
definition:
�·14!@1!.Jll � ! be �""!
linuous function that � defined on the rectangle . 'R.-{(x,y) .a-x-b,C-Y-dl· For each positive mteger N, let {xo, xi, ... ,xN} and {y0,yi,... ,yN} be uniform partitions of the intervals [a,b] and [c,d), respectively. Set t:u= (b-a)/N, ll.y = (d-c)/N, and M= (6.x)(ll.y). Let eii be any point in subrectangle Q11 = [x,-1,xt) X [Y1-1,Y1)· Then, the double integral of f over 'R. is defined to be the limit of the Riemann sums t
as Ntends to infinity. We denote the double integral of/ over the rectangle 'R. by the symbol ff'R. f(x,y) Thus
dA.
• Figure 6b
Iterated Integrals
f}
N
f(x,y)
dA= � k.
� N
N
f({,;)
N
to.A= N� � �f({;,;) �/;.y.
Now that we understand what quantity the double integral
(12.1.2)
ff'R.f(x,y) dA signifies,
we need to know how to calculate it The technique is to integrate the integrand
f(x,y)
one variable at a time.
The
idea is similar to partial differentiation: One
variable of f(x,y) is held constant while we apply an operation of calculus to the
970
Chapter 12
Multiple Integrals other variable. We continue to suppose that
'R.= {(x,y)
f is
continuous on the rectangle
: as xsb, csysd} =[a,b] X [c,4]. If we fix the value x=xo in the [a,b], then we can form the integral J:f(xo,y) dy. If y�f(xo,y) happens to be positive, then the integral fcf(xo,y) dy represents the area of the region in the x =x0 plane that lies under the slice z=f(x0,y) and above the xy pl an e (see Figure 7). Similarly, if we fix the value y=Yo in the interval [c,d ], then we can form the integral J:f(x,y0) dx. See Figure 8 for the area interpretation of this integral when x�f(x,yo) is a positive function. interval
-
A Figure 7 The area of the shaded planar region is
Cf(xo,y) dy.
Letf(x,y) =6-x2 -f' for Osxs 1 and Osy s2. For each [O, 1], calculate I(x)= J; /(x,y) dy. For each fixed y in the interval [0,2), calculate .J(y) = fJf(x,y) dx. � EXAMPLE 2
fixed
x
in the interval
Solution In the integral that Thus
the
antiderivative
(6-r)y-r/3. Therefore
A Rgure 8 The area of the shaded planar region is
J! f(:x.,yo)dx.
I(x)=
of
defines I(x), we treat the x variable as a constant. 6-r-r with respect to the y variable is
we have
12 (6-x2-y2)dy= ((6-r)y--y3 ) 11=2=2(6-x2)--8 =-28 -2x2. o
3
Similarly, in the integral that defines Thus the antiderivative of
x3/3. Therefore
y�o
3
3
.J(y), we treat they variable as a constant.
6-r-y2 with respect to the x variable
is
(6-y)x -
we have
1 ( :r=l=(6-y2)--1 =-17 -y2• 1 x1) 1 .J(y) = (6-x2-T) dx= (6-/)x-3 :r=O 3 3 O
Using the continuity off, it can be shown that the functionsI(x)=
�
fcf(x,y)dy
and .J(y) = J:f(x,y) dx are continuous for asxsb and csysd, respectively. In particular, each of these two functions can be integrated over its domain. Thus the quantities
are well-defined.
The expressions
called
iterated integrals.
.r: ctf(x,y) dy)dx and tu:f(x,y) dx)dy are
J: (fcf(x,y) dy)dx, we first calculate the inner integral fcf(x,y) dy. For this computation, we treat the variable x as if it were a constant. Because J:f(x,y) dy is a definite integral in the y variable, its value does not depend on y, but it does depend on x. We integrate this function of x to complete the calculation of the iterated integral. We handle the integral J: ( J:f(x,y) dx)dy In the iterated integral
in an analogous way.
12.1 Double Integrals Over Rectangular Regions �EXAMPLE in
Example
3 Let
2.
f(x,y)=6-x2-y2
Calculate
f� (J� f(x,y) dx) dy.
the
iterated
for
O:sy:s2, as JJ CJ: f(x,y) dy) dx and
Osxs1
integrals
J:
971
and
dx. 17/ J;yf(x,y) . 1 fl f1 f1 ) ( dx Jo ( Jo f(x,y)dy = Jo I(x)dx= lo ( 3 -2x2) dx= ( 3x-3x3) 1 o= 3
f(x,y) dy and .1(y)= Solution Let I(x)= showed that I(x)=28 3 - 2x2 and .J(y)= 3-
In Example 2, we
/
2
Therefore
28
28
26
2
and fl
lo(2 (lo
(17 )
) 12= T·
y3 f(x,y)dx dy= l{o 2 .J(y)dy= l{o 2 T-y2 dy= Ty-3
)
INSIGHT
In Example
(17
3. the integrands of the integrals
0
26
.,..
f� I(x) dx and J: .1(y) dy that
differ, as do the intervals of integration. Nevertheless, Example 3 demonstrates
the two integrals have the same value. In other words, the two iterated integrals
f� (f; f(x,y) dy) dx and J; (/� f(x,y) dx} dy are equal In the next subsection, we will see that these expressions are equal because they both represent the volume of the solid that lies under the graph off and over the rectangle IO, 1) X [O, 2].
to
will show
f is continuous and positive, the iterated integrals U: f(x,y) dx) dy both equal the volume of the solid U Calculate Double fa U: f(x,y) dy) dx and that lies under the graph off and above the rectangle 'R= [a,b] X [c,dj. Let Integrals I(x)= fc f(x,y) dy and .J(y)= faf(x,y) dx. Partition the interval [a, b] into N equal subintervals by means of the points a= x0, x1, ,XN-1, XN=b. As before, f(�.y) we set �= (b-a)/N. Figure 9 shows that, when � is small, the quantity f(x1,y) dy) �is a good approximation to the volume of the section of U that lies between the planes x=Xi-1 and x=Xi· It follows that the sum f(xh y) dy) .6x is a good approximation to the volwne of U when .6x is
Using Iterated Integrals
z
z
Next we
that, when
fc
• . •
=
(fc
Ef:1 {fc
A Figure 9 The area of the shaded solid is approximately
small. Therefore
ctt(x:;,y)dy)tu.
r
(ldf(x;,y) dy) �=Jlrf(x,y) N-+co t 1�1 lim
'It
c
On the other
hand,
Riemann integral
the sum
E!1 {fcf(x;,y) dy) .6x
dA.
is a Riemann sum for the
faI(x) dx. Therefore
t (ldf(x1,y) dy ) .6x= lb I(x) dx= 1" (ldf(x,y) dy) dx.
Nlim -+<:» i=l
c
(12.1.3)
a
a
c
(12.1.4)
972
Chapter
12
Multlple Integrals
From equations (12.1.3) and (12.1.4), we conclude that
t����*
The area of the shaded solid is approximately u: f(x,y,)dx)tl.y. A. Figure 1 O
=
Figure 10 reveals an analogous situation for the other iterated integral. If we partition the interval [c,d] into N equal subintervals by means of the points c and we set !:l.y y1, . - c)/N, then U: f(x,y;) dx)!:l.y is a good approximation to the volume of the section of U that lies between the planes and It follows that the sum u: f(x,y;) dx)6:y is a good
, YN-i. YN d, y =Y;-1 y =Yi· .
'4.y
ff f(x,y) dA
'R.
=
.
=
(d
=
Ef!.1
approximation to the volume of U, and it is also a Riemann sum for Letting N tend to infinity, we obtain
N�
t (J.'
)t.y =[Jf(x,y)
f(x,y;) dx
fc .J(y)
y0,
dy.
dA
and
d
d
N->«i t (lbf(x,y;) 11x) 1:1y =i .J(y) dy =i (lbf(x,y) 11x) dy. lim
j=l
a
c
c
=
a
dy.
It follows that ff'R. f(x,y)dA f (.r:f(x,y)dx) Although, our discussion has assumed that f is positive, the equalities we have obtained are valid for any continuous function as our next theorem asserts. ,
THEOREM 1
y
b, cs s d}
INSIGHT
=
=
Let f be continuous on the rectangle 'R. {(x,y): asxs [a,b) X [c,d]. Then, the double integral JJ'Jl. f(x,y) dA exists and
If you think back to the method of disks in Section 7.1 of Chapter 7, then
you will realize we were
using
a similar idea because we would calculate the areas of
slices in the y variable and then integrate out in the x variable, or vice versa.
Theorem 1 to calculate the integral of the function cos(x) sin(y) over the rectangle 'R. {(x,y): 1r/6sx S1r/2,1r/4 'Sf y S1r/3}.
� EXAM PL E 4 Use
f(x,y)
=
Solution
By Theorem 1,
[J
f(x,y) dA-
=
1.;: (J.;:'
)
cos(x)sin(y) dy tt..
12.1 Double Integrals Over Rectangular Regions
973
We evaluate an iterated integral by working from the inside out. Notice that the inner integral is an integral in the y variable. Just as in the theory of partial differentiation, we treat the x variable as a constant. Thus the antiderivative of cos(x) sin(y) with respect to they variable is cos(x)( -cos(y) ), or -cos(x)cos(y). We have /3 cos(x) sin(y) dy -cos(x) cos(y) 1y=7r/3 -cos(x) cos (� ) + cos(x) cos (4� ) 3 17r7r/4 y 7r/4 1 + 2cos(x) J2 ) = -2cos(x) = (../2-1 2 cos(x). =
=
To finish the calculation of the given double integral, we now do a straightforward integral in one variable. The result is
[Jf(x,y) dA
�
[:' (v'22-1)cos(x)dx
�
(v'22-1)sin(x{:: (v'22-l) 1 - (v'22-1) � ·
�
�
v'24-l
·
Integrate the function f(x,y) =x2y-y3+2x over the rec tangle R {(x,y): 1
We evaluate the inside integral in the x variable first. When calculating the anti derivative in the x variable, we treat y as a constant. Thus 3 (xy2 -y3+ 2x) dx= (y�3 -y3x+x2) 1x=3 = (9y-3y3+9 ) - ( -1y-y3+1) =-y 26 -2y3+8. 3 3 3 1 x=l Therefore 26y-2y3+ 8) dy= (13 y2 -1-y+ 25. 13 - 1- -8) =) 1y=O =0- (JO (4 8 = y d ) (x y A , f 3 3 2 3 2 6 ff y= 1
-1
-1
R
INSIGHT
Theorem 1 gives us a choice. We can calculate
grating first in they variable and then in the we can integrate first with respect to
x variable, as
Jf,Rf(x,y)
dA by inte
in Example 4. Alternatively,
x and then with respect toy,
as in Example 5.
97 4
Chapter 12
Multiple Integrals � EXAMPLE 6 Consider the integral ffn,x cos(xy) dA, where R= {(x,y): 0:sx:s1, 0:sy:s 7r}. Use Theorem 1 to convert this to an iterated integral. Eval uate the iterated integral by performing the y integration first. Solution
therefore
The antiderivative of x cos(xy) with respect toy is sin(xy). Our integral is
=
cos(7rx) 1x=l f1 (sin(7rx) - sin(O)) dx = - l 7r O
Jo
x=
=
_
.!_ (-1 - 1) 7r
=
�. 7r
INSIGHT
In Example 6, we could have integrated with respect to x first. However, then we would have had to use the method of integration by parts. Because we chose to do the y integration first, our calculation was much simpler.
Q UIC K
Q UIZ
1 . Estimate the volume under the graph of f(x,y) = 2x + 3y2 and above the rec tangle [O, 2] X [2, 6]. Use a Riemann sum with N 2. For the choice of points {t;,ij}, use the midpoint of each subrectangle. 2. The expression =
is a Riemann sum for the double integral ffn,(xy2) dA . The upper-right vertex of each subrectangle Q;j of R is used as the point t;,ij· What is R? 3. Calculate I(x) fz (2x + 3y2) dy and .J(y) = J; (2x + 3y2) dx. 4. Evaluate ffn,(2x + 3y2) dA, where R= {(x,y) : 0:sx:s 2, 2:sy:s 6}. =
Answers
1. 424
2. [3, 5]
x
[4, 5]
3 . I(x)
=
8x +
208; .J(y)
=
6y2
+
4
4. 432
EXERCISES Problems for Practice
In each of Exercises 1-4, approximate the volume of the solid region that lies under the graph of the given function f and above the given rectangle R in the xy-plane. Use a Rie mann sum with N =2 and, for the choice of points { �ij}, the midpoints of the four subrectangles of R. 1. f(x,y) =x + 2 y 2. f(x,y) =1 + 6xy2 3. f(x,y) =3x2 + 2y 4. f(x,y) = y/x
R =[0,2]X[1,5] R= [0,1] x [1,2] R =[-1,3] x [0,2] R = [3,7]X[2,4]
fc
In each of Exercises 5-10, calculate I(x) = t(x,y) dy and .:J(y) = f(x,y) dx for the given functionf and the given rectangle R =[a, b]X[c,d] in the xy-plane.
J:
5. f(x,y) =x + 2y 6. f(x,y) = 1+6xy2 7. f(x,y) = 3x2 + 2y 8. f(x,y) = y/x 9. f(x,y) =y exp(xy2) 10. f(x,y) =sin(x + 2y)
R = [0,2] x [1,5] R = [O,1] x [1,2] R = [-1,3]X[-2,2] R = [3,7] x [2,4] R=[O,l]X[0,1] R =[0,2rr]X[O,rr]
12.1 Double Integrals Over Rectangular Regions In each of Exercises 11-26, evaluate the double integral by converting it into an iterated integral.
11. 12.
13. 1 4. 15. 16.
17.
18. 1 9. 20. 21.
22. 23. 2 4. 25. 26.
IJ,R (x2- y) dA, R.={(x,y): -25x55,l5y54} IIn.(cos(x)- sin(y)) dA, R.={(x,y): 05x5?r/2- ?r/35y5?r} IIn.x cos(y) dA, R.={(x,y): 05x5?r,05y5?r} IIn.(x+2y) dA, R.={(x,y): 05x54,-25y50} IIn.e-y dA, R.={(x,y): 05x5l,-25y50} IIn,sin(x)tan(y )dA, R.={(x,y): 05x5?r/4,-?r/35y50} co (x) co (y) dA, R R.={(x,y): ?r/25x5?r,?r5y52?r} IIn.(x eY - y tr) dA, R.= {(x,y): 05x5l,l5y52} IIn.(y../X- xy'Y) dA, R.={(x,y): l5x54,l5y54}
JJ ( �
:
)
JL : z dA, R.= {(x,y): 05x52,l5y5v/3} 1 y fl ( ;2 + ;) dA,
In each of Exercises integral. 35. 36. 37. 38.
28.
39. 40. 41. 42. 43.
2 9. 30.
r
33. 34.
Suppose that ¢ and 'If; are continuous functions of one variable on the intervals [a,b] and [c,d] respectively. Show that if f(x,y) = ¢(x) 'lf;(y), then
Ik,(x+y)3 dA, R.={(x,y): l5x52,25y53}
31-34,
evaluate the given double
IIn.ln(xy) dA, R.={(x,y): l5x5e,l5y5e} IIn.ln(xY) dA, R.={(x,y): l5x5e,l5y52e} IIn.xyln(y) dA, R.={(x,y): l5x5e,l5y5e} IIn.y2eeY dA, R.={(x,y): -l5x5l,-25y50}
(lb ¢(x) ) (1d 'lf;(y) dy) . dx
and
IIn. x3y2 dA, R.={(x,y): 05x5l,05y5l}
jj
dx =
{1 x2(1- x2)3/2 dx = .!!.._ 32 Jo
Ik_cos3(x)sin(y) dA, R.={(x,y): 05x5?r/2,05y5?r}
In each of Exercises integral.
32.
f(x,y) =x2cos(xy), R.={(x,y): -l5x5l,-l5y5l} f(x,y) =y2 sin(xy) cos(xy), R.={(x,y): 05x5l,05y5?r/2} f(x,y) =xyln(xy) dA, R.={(x,y): l5x5e,l5y5e} (x,y) =y exp(x2+y2), R.={(x,y): O5x5l,-l5y5l}
Let R. denote the rectangle [O,1] X [O,2] in the xy-plane. Given that
JJ{R h dA, R.={(x,y): 05x5l/2,-l5y52} l- x2
Further Theory and Practice
31.
I� (J� xeXY dy) dx I�12 er; x sin(xy) dy) dx ti u;2x2yln(x2y2) dy) dx fo' (J� xy cos(xy2) dy) dx
lb (1d f(x,y) dy)
R.={(x,y): -35x5-1,-45y5-2} IIn.ecos(y) dA, R.={(x,y): 05x5l,O5y5?r/2} IIn.cos2(x) sin2(y) dA, {(x,y): ?r/25x5?r,05y5?r}
IIn.(x+y)4 dA, R.={(x,y): 05x5l,l5y52} 1 -- dA, R.={(x,y): 05x52,l5y5e} n,x+y IIn.sin(Zx+y) dA, R.={(x,y): -?r/45x5?r, ?r/65y5?r}
evaluate the given iterated
In each of Exercises 39-42, calculate IIn. f(x,y) dA for the given function f and the given rectangle R..
In each of Exercises 27-30, evaluate the given double integral twice by performing iterated integrations in both orders. Your answer should come out the same either way. 27.
35-38,
975
what is II R(xy)2 (l- x2)3(4- y2) dA? Suppose that f has continuous partial derivatives on an open set that contains the rectangle R.= [a,b] X [c,d] in the xy-plane. Evaluate IIR fxy(x,y) dA.
J
44.
In each of Exercises 45-48, functions f(x,y) and g(x,y) are given, along with a rectangle R. in the xy-plane. Calculate the volume of the solid U that is bounded above by z =f(x,y), below by z =g(x,y), and that is above R.. In other words, what is the volume of U ={(x,y,z):
45. 46. 47. 48.
(x,y)
E
R.,g(x,y)5z5f(x,y)}?
f(x,y) =1+x, g(x,y) =y f(x,y) =xy, g(x,y) =y f(x,y) =2x+y, g(x,y) =x+2y f(x,y) =e+y, g(x,y) =xy
R.= [1,2] x [0,1] R.= [1,3] x [1,2] R.= [1,2] x [0,1] R.= [0,1] x [0,1]
976
Chapter 12
Multiple Integrals
Calculator/Computer Exercises
[>
[>
By using a calculator or computer, we may use approx imation integral
(12.1) with a large value ffn.f(x,y) dA. Figure 11
(>
of N to estimate a double illustrates how this can be
done using the computer algebra system Maple. In the illus
a
:•
(x,y)
:• O:
Delta
-> sqrt(25-9*x�2-y�2):
b
:• 1:
=
o
:• 0:
: • ( (b-a) /H, (d-o) /111;
> sum(sum(evalf(
f(x,y) ,/25 -9x2 -y2, and the rectangle of integration is R [O,1] x [O,4]. Each side of the rectangle has been divided into N 100 equal sub
tration, the integrand is given by
f
6
d
:• 4:
M
=-
100:
tJ
=[ t�o·�J
f(a+(i-1/2)*Delta[l],c+(j-1/2)*Delta[2])), j�l.. H),i�l .. H)*Delta[l)*Delta[2);
=
=
16.05722143
intervals, and the midpoints of the subrectangles have been
{e;i}. In Exercises 49-52, ffn f(x,y) dA for the given function f and rectangle R. Use N=50, and take the mid points of the subrectangles for the choice of points {e;j}· (In used for the choice of points
_. Figure 11
approximate the double integral
R=[-1,l]X[0,1] R [1,3] x [O,1] R [-1,1] x [-1,1] R [0,2] x [O,1]
f(x,y) =cos(v'f+x + y) Jl + x + y3 50. f(x,y) SL f(x,y) exp(-x2 - y2) (1 + x + y)/(1 + x2 + y4) 52. f(x,y)
49.
=
=
each of these exercises, the suggested approximation is
=
=
accurate to at least three decimal places.)
=
=
1 2.2 Integration Over More General Regions Many integration problems are formulated over planar regions other than rec tangles. We need a technique for defining an integral over the interior of a triangle, a circle, an ellipse, or over even more complicated sets in the plane. We will restrict attention to regions bounded by a finite number of smooth curves. To begin, we need to know how to partition such a set. These are the considerations that we treat in the present section.
Planar Regions Bounded by Finitely Many Curves
Let R be a closed region in the plane that is contained in some bounded rectangle
Q
and that is bounded by finitely many continuously differentiable curves (see
Figure
1).
Suppose that
Q={(x,y) :a�x�b,c�y�d}.
Then, we can partition
the rectangle Qin the usual way by partitioning the intervals
a =xo
y
·
·
·
[a, b]
and
[c,d]:
and
d
c =Yo
·
·
·
[a, b] andyj-lo Yi in the partition of [c, d] there Qi,j relevant for
is a small subrectangle Qi,j as shown in Figure 2. The only rectangles
integration over the region R are the ones that are completely contained in R. Clearly the rectangles that lie completely outside R are irrelevant; the ones that
c a
_. Figure 1
x
b
overlap the boundary of R make a negligible error contribution when the sub rectangles of the partition are small. If f is a continuous function defined on R, then we consider Riemann sums
L:
t(ei,j)tuf).y =
Q;,;c'R. where
ei,j
is any point selected from
L:
f(ei,j)f).A,
Q;,;c'R.
Qi,j
for each i,j
12.2 Integration Over More General Regions
977
A theorem from mathematical analysis guarantees that, if we let N tend to
y
infinity, then these Riemann sums tend to a limit. We call this limit the integral off
d
over the region
R,
and we denote it by
JJ f(x,y) dA.
Yj Yj-1
'R,
c
x
The practical issue that we now need to address is how to evaluate integrals
R.
over general regions
To keep things from getting too complicated, we restrict
attention to two types of regions: the x-simple regions, which are regions of the
• Figure 2
form
y
for some continuously differentiable functions
a1(y):::::;, a2(y) on [c,d] (see Figure 3),
for some continuously differentiable functions
(31(x) :::::;, (32(x) on [a,b] (see Figure 4).
and they-simple regions, which are regions of the form
x
Notice that the first type of region is called x-simple because it is spanned from left
• Figure 3 The region R {(x,y): c�y�d,a1(y)�x� 0:2 (y)} is x-simple. =
to right by horizontal line segments; a similar reason motivates the name of the second type of region. The following two theorems give us a method for evaluating integrals over x-simple and y-simple regions. They tell us that double integrals over simple regions can be evaluated by iterated integrals
THEOREM 1
Suppose that
G ust as with the rectangles of Section 12.1).
a1(y) ::5a2(y)
for y E
[c,d].
Let
R={(x,y): c::5y::5d,a1(y)::5x::5a2(y)} be the corresponding x-simple region. If f is a continuous function on
R,
then
R,
then
x a
b
• Figure 4 The region R= {(x,y): a::5x::5b,(31 (x) ::5y::5 (32(x)} is y-simple. THEOREM 2
Suppose that
(31(x)::5(32(x)
for y E
[c,d].
Let
R= {(x,y): a::5x::5b,(31(x)::5y::5(32(x)} be the corresponding y-simple region. If f is a continuous function on
ff f(x,y) 'R,
dA
�
l (!,�,�) f(x,y)dy)ch
978
Chapter 12
Multiple Integrals
y
(3, 9)
You should think back to the method of disks that we used to calculate the
volumes of solids of revolution. Recall that in those integrations, we would cal
culate the areas of slices in the y variable and then integrate out in the x variable, or
vice versa. In Theorem 1, we are, in effect, taking slices in the x direction and then
integrating out in the y variable. In Theorem 2, we are taking slices in the y direction and then integrating out in the x variable.
� EXAM PL E .j...---1----1-�x .. i-::;..._ --1__;;:. -1 2 3 1 .& Figure 5 The region R is
1 Integrate the function f(x, y)=3xy over the region R
between the parabola y = x2 and the line y = 2x+ 3.
Solution To find the points of intersection of the parabola and the line, we solve
x2 = y = 2x+ 3, or x2 - 2x - 3 = 0. The left side of this equation factors to yield
y-simple.
(x+ l)(x - 3) = 0, from which we see that (-1,1) and (3, 9) are the points of intersection. Figure 5 reveals that R=
{(x,y): -1:5x:53,x2:5y:52x+ 3 }
y-simple region. According to Theorem 2, we have
ff
1_ (l{2x+3 1 � 1: ( 3
3xydA =
xz
n
=
)
3xydy dx =
1_1 ( ly-2x+3 ) 3
3 xy2 =x2 2 y
is a
dx
) dx.
x (2x + 3)2 - (x2)2
On expanding the integrand, we obtain
ff
n
) 1x=3
(
1
3 3 3 1 9 3xydA = (-x5+4x3+ 12x2+ 9x) dx = - - -x6+ x4 +4x3+ -x2 2 6 2 2 -1
(
3
x -l =
4
= - 108 - 2
3
)
= 160.
� EXAM PL E 2 Integrate the function f(x,y) = 2x -4y over the region R
y
between the curves x = y2 and x = y3 (see Figure 6).
(1, 1)
1
Solution We first notice that the curves intersect at the points (0,0) and (1,1). The region R= { (x,y) : 0:::; y:::; 1, y3:::; x:::; y2 } is x-simple. According to Theorem 1, the value of the integral is
ff
0.5
(2x -4y)dA =
Y
n
= (0, 0)
x 0.5
1
.& Figure 6 The region R is x-simple and y-simple.
11 (j:2 11 (
)
(2x -4y) dx dy =
11 (
(x2 -4xy)
i::::)
dy
)
(y4 -4y3) - (y6 -4y4) dy.
The last integrand simplifies to -y6+ 5y4-4y3 and therefore
ff
(2x -4y)dA=
n
INSIGHT
11 0
(
7
y= l
) 1 y=O
1 (-y6+sy4 -4y3)dy= --y7+ y5 - y4
1 =--.
Notice that the region R in Example 2 is also y-simple; indeed, R= {(x,y): O�x�l, x112�y�x113}.
From this point of view, the integral off over R is calculated as follows:
7
12.2
1
11'
x JrJR.{ (2x-4y) dA= 10 ( lx.1/2 1
+
1t'
)
{}01 (xy-y2) 1y=y=x.'x 12dx
x) dx=
(
(2x-4y)dy dx=2
10 (x4/3-x2/3-x3/2
=
979
Integration Over More General Regions
6 -x7f3
_
7
6 -x5f3 5
_
4 -x5f2 + x2
1
) 10
5
=
_
1 .
7
Changing the Order
Sometimes an iterated integral that is difficult to calculate becomes much easier if
of Integration
the order of integration is reversed. Theorems 1 and 2 taken together guarantee that if a region is both x-simple and y-simple, then we get the same answer no matter which order we choose to do the integration.
1
y
� EXAMPLE 3 Calculate the integral
(1, 1)
J�J: 6 y ex3 dxdy.
Solution The inner integral with respect to x is not one that we can do with the techniques that we have learned. Let us instead change the order of integration in an attempt to reformulate the problem. A glance at Figure 7 shows that the region
0.5
'R of integration is not only x-simple but also y-simple: 'R=
x
{(x,y )
}
:0:5x:5l,0:5y:5x .
Therefore the given iterated integral may be rewritten as
_. Figure 7 The region R is x-simple and y-simple. an integral that we easily evaluate as follows: 1
lof (lor
�dy
)
6y
INSIGHT
dx
=
1
lof (
�
3y2
1y=O=x) y
dx
=
lof
1 3x 2
� dx
=
�
1x=O=l x
=
e
-
1.
....
When you are changing the order of integration, there is no substitute for
drawing a good picture. It tells you immediately the new limits of integration and the order in which they should appear.
Part of our work in switching the order of integration in Example 3 was to re
y
express the boundary curves in terms of the x variable instead of the y variable.
y=9
8
Sometimes, the domain of a double integral must be broken up if it is to be evaluated in a certain order. However, it may be possible to avoid breaking up the
6
integral if we evaluate in the opposite order. Here is an example.
4
� EXAMPLE 4 Evaluate the integral shown in Figure 8.
2 1
.. Figure 8
2
3
4
s
x
ff'R.4xydA
where 'R is the region
Solution Notice that the region of integration is the union of two y-simple regions. If we choose to apply Theorem 2 and make y the inside variable, then we must note
(
)
that for points x, y in 'R with 0:::; x:::; 3, the ordinate y satisfies 0:::; y:::; x2, whereas
980
Chapter 12
Multiple Integrals when
3 ::s x ::s 5,
the ordinate y satisfies
y
over the interval
0 ::s y ::s 9. Because of the different formulas = x2 over the interval [O, 3] and
for y at the upper limits of integration, namely y
=9
two iterated integrals,
ff
4xy dA
�
1l
[3, 5],
we must write the given double integral as a sum of
l (t
4xy dy
}tx J,' ([ +
On the other hand, n is x-simple. In fact, n
Therefore by Theorem
ff
4xydA
1l
1,
� [ (£ =
19
4xy
4xy dy
zx'y
(50y - 2y2) dy= 25y2 -
(12.2.1)
= { (x,y) : 0 :5 y :5 9, VY :5 x :5 5}.
dx) � [ ( 1::� ) ( � ) 1: ay
)dx
y3
dy
�
=(25)(81)- 93=1539.
You may verify this answer by showing that the first integral on the right side of equation
The Area of a Planar Region y
(12.2.1)
evaluates to
243
and the second to
1296,
for a sum of
1539.
<11111
Suppose that n is a bounded region in the plane, as in Figure
9. Fix a rectangle Q Q into N2 congruent subrectangles {Qi,j}· Next, let/ be the function that is identically 1 on n. That is, define/by the formulaf(x,y) = 1 for every point (x,y) in n. Then, whatever choice of points {�i,j} is made, the that contains n and partition
Riemann sum for f over n is equal to
LlM.
(12.2.2)
Q;,ir.;R
This quantity is just the sum of the areas of the small rectangles x
partition is large, formula
_. Figure 9 A bounded region in the plane.
y
Qi,j that are 10). When the number N2 of subrectangles of the (12.2.2) gives a good approximation to the area of n. It
contained in n (see Figure
can be shown that, if the boundary of n is a piecewise smooth curve, then the Riemann sums in line
limit by
(12.2.2) tend to a limit as N tends to infinity. We
JJRl dA and make the following definition.
denote this
If n is a region in the plane bounded by finitely many con
tinuously differentiable curves, then the area of n is defined to be
ff 1
dA.
1l
x
_. Figure 10 The area of the shaded region is the Riemann sum �Qi,jf;;RlM.
� EX A M P L E 5 Suppose that
a
and
area A inside the ellipse x2 / a2 + y2 / b2
b
are positive constants. Calculate the
= 1.
Solution The given ellipse is a y-simple region:
12.2
Integration Over More General Regions
981
y
x
---t����+-���-+--
See Figure A=
ff
11. We have
ldA=
'R,
• Figure 11 The ellipse i2 Y2 = 1·is a y-s1mp . 1e region. . Z +
a
a( � l l-aa (l-bb� ) -a Y _1_- ( .Jl-xz/aZ b
ldy dx=
or
b2
A
(i2.
- -b
- )) dx , � (i2.
Y _1_
= 2b 1-aa J1- x� dx.
(12.2.3)
a
InSection6.3 ofChapter6,we learned that this type ofintegral can be handled by the change ofvariablesx =a sin(O),dx =a cos(O)dOwhich, using formula (6.2.4), leads to A= 2ab
1
7r/2
-ir/2
J1-
sin2(0) cos(O) dO =2ab
INSIGHT
(
7 r /2
2 . (20) 0 + sm cos2(0) dO =2ab 4 -ir/2
1
x = a u,
then the area computation goes as follows:
A= 2b =
= 7rab.
-ir/2
It sometimes happens that an integrand can be successfully treated by
different substitutions. If the change of variable
(12.2.3),
) 17r/2
la J - xa� dx
2ab
1
a
=
2ab
dx = a
du
£1 Vl - u2 du -1
(area of the upper half of the unit circle) =
2ab
is made in line
( �)
=
7rab.
� EX A M P L E 6 Suppose that 'R1 is the x-simple region given by 'R1= {(x,y) : c ::s y ::s d, a1(y) ::s x ::s a2(y )}. Suppose that 'R2 is the y-simple region given by 'R2= {(x,y): a::sx::sb,/31(x) ::sy::sf32(x)}. Using the double integral definition of planar area, express the areas of'R1 and 'R2 as "single" (as opposed to double) Riemann integrals. Solution
According to Theorem
1, the area of'R1 is given by
x az(y) ) dy d (a2(y) - a1(y))dy. ( y ) a 1 d d 2 ( x = 1 d 1 dx) dy ( l ff i x=a1(y) i a1(y) i A=
'R,
=
=
c
c
c
Similarly, using Theorem 2, the area of'R2 is given by
x) y z(x) ( 1 1 d = lb (1,B2 1 dy) dx = lb (y =,B ) dx = lb (/32(x)-f31(x))dx. ff a ,81(x) a =,81 x a A
'R,
y ()
982
Chapter 12
Multiple Integrals ..,... A LOOK BACK In Section 5.7 of Chapter 5, we learned to express the area between two curves using a single integral. Example
6 shows that, in such a situation, the
definition of area by means of a double integral leads to the very same formula that our earlier method prescribes. Of course it is essential that the two methods be consistent when they can be applied to the same region. The advantage of the double integral definition is that it has wider applicability than our earlier definition.
Q U IC K
Q U IZ
1.
R with vertices (0, 0), (2, 6), and (3,6) is x-simple. Specify c and d, and functions0:1(y) anda:2(y), for which R { (x,y) c :5y :5d,0:1(y) :5 x :5a:z(y)}. 2. Let R be the region bounded by the vertical line segment with endpoints 0 (0, 0) and P (0, 1), the horizontal line segment with endpoints P (0, 1) and Q (7r/2,1), and the arc of y sin(x) between 0 and Q. The area of R is J� J;(y) 1 dx dy for what functiona:(y)? 3. For what constant b and function (J(x) is J; Jj'-2f(x,y) dx dy= J� Jfl(�i);d(x,y) dydx? 4. Evaluate the integral JJR2x3y dA, where R= {(x,y): 0 :5 x :5 l O :s; y :s;x2 } The triangular region constants
=
=
=
:
=
=
=
,
.
Answers
1. c=O,d=6,0:1(y)=y/3,a:2(y)=y/2 2. a:(y)=arcsin(y) 3. b=4, (J(x)= yi 4. 1/8
EXERCISES Problems for Practice
In each of Exercises 1-6, sketch the region R that is described. Decide whether R is x-simple, y-simple, or both x-simple and y-simple. If R is x-simple (respectively, y simple), state the functionsy>->0:1(y) andy>->0: (y) (respec 2 tively,x >--> ,81 (x) andx >--> ,B (x)) whose graphs form part of its 2 boundary. L 2. 3. 4. 5.
The region between the curvesx =-y2 + 2 and -2y -1 The region between the curvesy=x andy=x2 The region between curvesx=y2 andx=- 2y2- 3y +18 The region between the curvesy=x2 andy=8x112 The region above the curve y=x32 1 , below the curve y= 6x- 1, and betweenx=1 andx=3 6. The region below y=2, above y=-1, and between the curvesy=x3 andy=x +8 In each of Exercises 7-14, evaluate the given iterated integral. 7. 8. 9.
rl J.8-(x+1)2 3xdydx J-l (x+l)' 4 �1/3 (y/x)dydx Jlr6 fx'I ' Jt I:/4 (Zx -1)./Ydydx
10. 1L 12.
f�
Jy"Y 2xeYdxdy
x 4x dydx J; Jo. () 12
fi J;+14x dxdy
f"/2 rsffi (y) 5 dxdy 13. J,,14 Jo r3 14. Jl
rx+1l Y1/2( X +1)d Yd X Jx-
In each of Exercises 15-22, calculate the integral of the given function f over the given region R.
15. f(x,y)= 6y -4x; R is the region between the curves
y=2x andy=x2. 16. f(x,y)= cos(x); R is the region between the curvesy =x
andy=2x, O�x�"lr. R is the region above y= 0, below y=2x- 6, and betweenx=4 andx= 7. 18. f(x,y)= 9.jY/4 ; R is the region belowy=4 , abovey= 1, and betweenx=y3 andx=y2 + 48. 19. f(x,y)=yv/X; R is the region between x=3 and x=4, abovey=x2- 3x and below y=x2 +8 . 20. f(x,y)=y2/x2; R is the region above y= -5, below y= -3, and betweenx= y2 andx= 5. 2L f(x,y)= �; R is the region between x =y and x = 6y, o�x�l. 17. f(x,y)=xeY;
12.2 Integration Over More General Regions
983
f(x,y) =y sin(x); R is the region bounded by y =cos(x) and y =sin(x), 0 :Sx :S7r/4.
In each of Exercises 45-54, calculate ffnf(x,y) dA for the given function f and region R.
In each of Exercises 23-26, switch the order of integration to make the integral manageable and then calculate the iterated integral.
f(x,y) =ax+by; R is the triangle with vertices (0,0), (1,3), (10,4). 46. f(x,y)=x2+y; R is the parallelogram with vertices (-1,0), (0,0), (1,1), (0,1). 47. f(x,y)=x+y; R is the region in the first quadrant that is below y=1+x2 and to the left of y=5(x- 1). 48. f(x,y)=x2y; R is the region bounded above by y=4+2x- x2 and below by y=x2. 49. f(x,y)=x+y; R is that part of the unit circle that lies in the first quadrant. 50. f(x,y) =x.jY; R is the region above y =1 - x and inside the unit circle. 51. f(x,y)=ysin(x); R is the region in the first quadrant that is bounded above by y=1 - cos(x), and that is bounded on the right by x 7r/2. 52. f(x,y)=1/x3; R= {(x,y) : 1:Sx:S2,x sin(l/x):Sy:S2}. 53. f(x,y)= IY- xi; R is the region in the first quadrant that is bounded above by y=2.x, and on the right by x =1. 54. f(x,y)=2y/(2- x); R is the region in the first quadrant that is bounded above by y=v'4 - x2 and on the right by x=l.
22.
23. 24. 25. 26.
10 7r/1x 7r/2 sin(y) dydx y J12 J1x2 y dydx 1o 2 1x24 vY dydx --
:!
ero y
Ji J exp(-x2) dxdy
:
Further Theory and Practice In
27. 28. 29. 30. 31. 32.
33. 34.
each of Exercises 27-34, calculate the given integral.
J/1' fx.,fi sin(y2) dydx
Ji J;'ccos(y) cos4(x) dxdy { l y(l- x2)2 dxdy lo x3 .,;y
J
Jg fov4' - x2 (4 - y2)3/2 dydx JJR(x- y) dA where R is the region bounded by y=cos(x) and y=sin(x), 0:Sx:S7r/4. JJR(y+sec2(x)) dA where R is the region bounded by y=tan(x) for O:Sx:S?r/4, y=-x for -l:Sx:SO, and y =1 for -1 :Sx :S7r/4 Ji x2sin(xy) dx dy
45.
=
55.
2 r exp(- y2) dy erf(x) = ..j7r l
o
satisfies
) (I: Jg ( J0.fi=Y2 v'9 - x2 dx) dy
1 erf(x) dx =zerf(z) - ..j7r (1 - exp(-z2)). lr o (Hint: Write the left side as an iterated integral and change the order of integration.) Deduce that
In each of Exercises 35-44, sketch the set R that is the union of the two regions of integration. Reverse the order of integrations and write the given sum as one iterated integral over R. Calculate this iterated integral. 35. 36. 37.
Given that limz->
�
JiJ; xy dydx+Ji Ji xy dyd x 39. JiJ xdxdy+J":J/!2xdxdy 1 - x+5 f f21dYdx+J21 Jx2 40 JoJ1 1 4 ldYdx 1 41. f� '2f x dxdy+J�1f,1 x dxdy y ._ - y - y2 2 y rlfv'Y r2J-.,IG ydxdy + 42. Jo .,;y Jl -y'Gydxdy rl/ ../2Jorarcsin(y) YdxdY+f l;../2Jorarccos(y) Y dxdY 43 Jo 1
�
•
�
•
44.
1 1- erf(x)) dx=z(l- erf(z))+ ..j7r (1 - exp(-z2)). lr o (
f-1f�xY dydx+JiJ;2Y dydx JiJt xydxdy+Jifg-y xy dxdy f_if�Yxy dxdy+JiJ xy dxdy
38.
Ji J�i lxl dxdy+ Ji J lxl dxdy
:
Show that the error function
Calculator/Computer Exercises In each of Exercises 56-59, a function/and a region R are given. Calculate JJRf(x,y) dA by writing the double integral as an iterated integral and finding the outer limits of integration.
f(x,y)=1+x+xy; R is the region that is bounded above by y=1 - x4 and below by y=x3. 57. f(x,y) =xy; R is the region that is bounded above by y =1+3x and below by y =eX.
56.
984
Chapter 12 MuHiple lnteerels
58. f x,y}=x;
R. is the region in the first quadrant that iii byy =4- x1/8 and below byy =1 + ,/i. 59. f(x,y)=x; R. is the region in the first quadrant that is bounded above byy =1 (1 +.%2) · nd below byy =.¥1°. In each of Exercises 60-63, a function f and a region R. are given. Approximate fk,f(x,y) dA to two decimal places by writing the double integral as an iterated integral for which the inner integration can be evaluated exactly. Use Simpson's Rule for the outer integration. bounded a bove
60. fx,y)=�; R. is
that part of the unit circle that lies in the first quadrant . 61. fx,y)=./xi+y; R. is that part of the unit circle that lies in the first quadrant. 62. fx y) = .,.1.%-FY; 'R. is the region in the first quadrant that is bounded above byy =x3 and on the right byx =1. 63. f x,y)= xp -y ); R. is the region in the first quadrant that is bounded on the left byy= an x) for Osxs /4, and on the right byy = .;- os {x} for 1f/4:5'x :5' 1r 2.
The iterated integral I= f (J".2<:>f x,y) dy)dx may be approx:imated by using an a ropriate Trapezoidal Rule . Let tu= b-a} N, x,=a+itu O:s :sN, .6.y},= ·p2 x;)-fJ1 Xi.)}/N, andyif={j (xi}+j(.6.y) 0:5'j:5'N. Then,
w
N
Ir::: where
cq=l
tu 2 L if
(.6.y) 2
N
HCIJ Xi,YtJ)
{i,J)C{O,N},
c11=4
if
{i, )c
.,N-1), and 11=2 otherwise (i.e., if exactly one of and is 0 or N}. In Exercises 64-67, approximate the given integral using N =4. Calculate the integral exactly. What is the approximation error'/ {1, 2,.
.
f f x-y) dydx 65. f_ f -rydydx ir) 66. f 1: r cot x) dydx 67. f j, ..r ydx 64.
12.3 Calculation of Volumes of Solids - r
'
In this section, we use double integrals to calculate volumes of solids in threeimensional space. To be specific, given a nonnegative continuous function efined on a region 'Tl f the xy·plane, we are interested in determining the volume f the solid that is above the region and below the graph off (see Figure 1). Recall that we motivated integration of functions of two variables by volume considerations. We now use the integral to precisely define what we mean by the
J.
volume of a solid.
A. Figu1 a planar
region R.
.: over
Suppose thatf is a nonnegative, continuous function defined on a f the xy-plane. Let U be the three�ensional solid that lies above and below the graph off The volume of U is defined to be region 'R,
jjI x,y)dA. 1t
y
� EXAM PL E 1 Figure below the planef
/ x
planar region
'Tl={ x,y) :2::!fx::!f5,-� ::!fy:!f-1}.
solid U that liee under the graph offand over the planar region R. A. Figure
2 illustrates the three�ensional solid U that lies
x,y) =2x- y and above the
1b The
Find the volume of
U.
12.3 Calculation of Volumes of Solids z
985
Because x > 0 and y < 0 on 'R, we have f(x, y) > 0 on 'R. The volume of U is therefore
Solution
=
1-1 (21-18y) dy = (21y-9y2)[� = 114. -3
x .A Figure 2 The solid that lies under the graph of f(x,y) =2t'-6y and over a planar rectangle
Sometimes a little geometric analysis volume as shown in the next example. �
is
required before we can calculate a
-
� EXAM PL E 2 Calculate the volwne below the plane 8x + 4y + 2z = 16, above the xy plane, and in the first octant. Solution
Solving for
z,
we
think
of the plane as the graph
of
the
function
z
=
f(x,y) =8-4x-2y. We notice that f(x,y) =0 when (x,y) lies on the line 4x + 2y = 8 in the .:ey-plane. Therefore if 'R is the region in the xy-plane that is bounded by the x-axis, the y-axis, and the line 4x + 2y = 8, then our problem is to find the volume of the solid that is below the graph of f and above 'R (see Figure 3). The region 'R is y simple:
-
'R= {(x,y):
Osxs2,0sys4-2x}. y 4
z
3 4x+ 2y =8, or
z=8-4x-2y
y=4-2x
2
1
y
x 1 .A Figure 3b
.A Rgure 3a The solid in the first octant that lies below the plane z = 8 - 4x - 2y
The required volume
JJJ(x,y) dA is therefore
12 (lo•-ix (8-4x-2y) dy)dx
2
= =
12 (csy-4xy-y2)1�:�-ix) dx 12 (16 - 16x 4x2) dx +
986
Chapter 12
Multiple Integrals
32
=
3
.
The Volume Between
So far, we have discussed the volume of a region lying below the graph of a
Two Surfaces
continuous function and above a region in the xy-plane. Now let us treat the volume below the graph of a continuous function f, above the graph of a continuous function g, and over a region
f(x. y) �-.
in thexy-plane. In this instance, the relevant
Riemann sums would be for the function f - g, and the integral representing the
=
desired volume is
jj
x,y)- gx,y)
)
dA.
12.3.1) :
'R,
The geometry is illustrated in Figure 4. The following is an example.
A Figure 4a
The graphs of and � over a planar region ··
.,._,
� E X A M P L E 3 Let U be the solid that is bounded below by the paraboloid 8 + x2 +y2 and above by the plane that is the graph of
that is the graph of g.x,y) fx : ,y)
=
2x +2y +15. Set up (but do not calculate) the integral for finding the
volume of U. surfaces intersect when 8 + x2 +y2=2x+2y+15 or, 2 ( - 2x) + (y -2y)= 7. After completing the squares, we obtain equivalently, x2 2 2 x ( -1) + (y -1) =32. Thus the solid we want to study "casts a shadow in thexy Solution The
two
plane"on the disk 'R that is bounded by the circle with center 1 ( ,1) Figure5 .'
------
and radius3.
shows the two boundary surfaces of U and the region 'R of integration .
The boundary of the region R of integration may be written asy -1
(x-1)2)112
or, equivalently, y
=
1±(8 +2x
-x2)1/2.
The solid U that lies under the graph off above the graph of 1 and over the planar region 'R., A Figure 4b
,
2
+� +
1
_j
y
A Figure 5
The paraboloid z
=
8 + x2
2
-� +
±
(9 -
We write R as they-simple
region given by
·,
=
�
I
2 x-x :sy
y2 and the plane z
=
2x + 2y + 15
12.3 calculation of Volumes of Solids
987
The desired volume is therefore
JJ (f(x,y)-g(x,y))
dA;
1: f::::�:
'R,
((2>:+2y
+15)-(8+x2+y2)) dyd<.
�
� EXAM P L E 4 Find the volume V of the solid U that is below the plane
11.
x+ 4y- 7, above the plane z = x+2y- 4, and bounded on the sides by the parabolic cylinder x=(y - 4)2 +3 and the plane x+2y =
z=
Solution The graph of U and its projection n in the xy-plane appear in Figure 6a. Observe that 'R, the region of integration, is the set of points in the xy-plane
11 -
+
11.
between the parabola x = (y- 4)2 +3 and the line x+ 2y = For the points of intersection, we solve 2y=x=(y - 4)2 +3, or Y2 -6y 8 =0. The roots y=2 and y = 4 are the ordinates of the points (7, 2) and (3, 4) of intersection. The plot of n in the xy-plane (Figure 6b) shows that 'R is x-simple: 'R = {( x, y) : 2< y <4, (y- 4)2+3
fl ( (x+4y- 7) - (x+ 2y- 4)) 'R,
dA =
11-2y (2y-3)dxdy (2y-3)x1x=ll-2y 12 4 12 41()'-4)2+3 x={y-4) =
2
+3
dy.
20 10
(a)
(b)
• Figure 6
On evaluating, expanding, and simplifying this integrand, we obtain (2y -3)x
2y x=lllx={y-4)
2
+3
=
(
(2y-3) (11- 2y)- ((y- 4)2+3)
)
=
-2y3+1Sy2 -34y +
24.
Therefore v=
k4 c-2y3+1sy2-34y+24) dy=(-!y4+sy3-11y2+24y)I: =16- 12=4.
�
988
Chapter 12
Muttiple Integrals ._ EXAM PL E 5 Calculate the volume cylinders :x:2+y = a2 and :x:2+42 = a2.
V
Solution Figure 7a shows the two cylinders. We
of the
solid
U enclosed by the
use symmetry to simplify the geometry of the problem. The left side of Figure 7b shows the part of the intersection of the two cylinders that lies in the first octant. We will compute the volume Vo of this smaller solid Uo and multiply by 8 to obtain V. Because Uo lies over the quarter-disk shown in the right side of Figure 7b, the region of integration is 'R = {(x,y) : 0 s x s a, 0 s y s ./a2-x2}, which is y-simple. The cylinder :x:2+z2 = a2is the upper surface of U0• By solving for z, we realize this surface as the graph of f(x,y) = Jaz -x2. The lower surface of U0 is the xy-plane, which we realize as g(x,y) =O. Therefore V = 8Vo = 8
can
J.'['"'--; (f(x, y)-g(x,y)) dydx = J.' (J.J"'-"' v'a2-x2 dy) dx. 8
z
y
a .&. Figure 7a
x
.&. Rgure 7b
Finally, we obtain V =8
QUIC K
QUIZ
(
x3) la = T16a3.
lr ((./a2-x2) y) 1y=.n=;J O dx = 8 lr (a2-x2) dx = 8 a2x - 3
y=
o
o
the volume of the solid that lies above the region 'R in the xy-plane, whichis bounded above by z = 6+:x:2-y2 and bounded below by z = 1+xl-+2y. Then, Vis equal to ffw. h(x,y) dA for what function h(x,y)? 2. Calculate the volume of the solid that lies below the graph off(x,y) = 2x+3y2 and over the rectangle [O, 1] x [O, 1] in the xy-plane. 3. Calculate the volume of the solid that lies below the graph off(x,y) = 1+x,above the graph of g(x,y) = 2x, and over the rectangle [O, 1] x [O,1] in the xy-plane. 4. Let 'R, be the region in the first quadrant of the xy-plane bounded by the curves y = :x:2 and y = x. Calculate the volume of the solid that lies above 'R, which is bounded above by z = 13x and below by z = x. 1. Let
V denote
0
Answers
1.
5-y2-2y
2. 2
3. 1/2
4.
1
12.3 Calculation of Volumes of Solids
989
EXERCISES Problems for Practice In each of Exercises 1-8, calculate the volume of the solid that lies below the graph off(x,y)=5+ x and over the given region in the xy-plane.
L The region bounded y=2x2+2x - 12
by
the
curves
y=x2+ x
and
2. The region bounded by the curves x=y2+y+2 and x=2y2+4y +4
22. f(x,y) =8- 2x+y; R is the region between the curves y=x3 and y =x3 + x2-2x- 3. 23. f(x,y) =sin(y); R is the region bounded by the y =-x+1, and x= 1.
curves
y=x+1,
In each of Exercises 24-30, calculate the volume of the solid U.
3. The region bounded by the curves y=x2- 2 and y=x 4. The region bounded by the curves x=y4 and x=16
24. U lies below the graph of x+2y+3z =6 and in the first
5. The region bounded by the curves y=Sx2- 7x+3 and y=x3
25. U lies below the graph of z=x+2y2+ 3, above the graph
6. The region bounded by the curves y=lxl and y=2 region
y=S- (x- 1)2
bounded
by
the
curves
y=x2
of z=x+y, and over the region U in the xy-plane that lies between x=y2 and x=-y2+2.
7. The region bounded by the curves x+y=2 and y=x2- 4 8. The
octant.
and
26. U lies below the graph of z=3x+y+7, above the graph of z= -x- 2y- 4, and over the region U in the xy-plane that is cut off by the line x+ 3y=12 in the first quadrant.
In each of Exercises 9-23, calculate the volume of the solidU that lies below the graph off(x,y) and over the region R in the xy-plane.
27. U is the solid that is bounded above by the cylinder y2+z2 =49, below by the xy-plane, and laterally by the cylinder x2+y2 =49.
28. U is the solid that is bounded by the two paraboloids z =x2+y2+1 and z = -x2 - y2+ 19. 29. U is the solid above the graph of z=x - y+1, below the
9. f(x,y) =x- 2y; R= {(x,y): 8:5x:510,2:5y:54}.
graph of z=3x+Sy+7, and bounded on the sides by the
10. f(x,y) =x2+y2;
vertical planes x=y,x=-y, and y=-4x+ 4. 30. U is the region bounded above by the plane 3x- 2y-
R={(x,y): 2:5x:54, 1:5y:56}.
1L f(x,y) =x2y2; R={(x,y):-4:5x:5-2,-5:5y:5-2}.
z = -8, below by the plane z= -2x - 3y - 6, and with vertical sides given by the parabolic cylinders x=y2 and
12. f(x,y) =sin(x);
x=-y2 - By +10.
R={(x,y):0:5x:57r,l:5y:53}.
13. f(x,y) =lOx- 3y +40; R={(x,y):3:5x:56,3x:5y :52x+8}. 14. f(x,y) =x-3f2ylf2;
Further Theory and Practice
R= { (x,y):1:5y:54,y2:5x :=;;y4 } .
3L Let R be the smaller region inside the circle x2+y2- 4y +
R={(x,y):3:5x:5 5, -6x:5y:5-2x}
the solid that lies above R and below the plane z =2x+8. 32. Let R be the bounded region between the line y=x+12
8x=6 that is cut off by the line y= x. Calculate the volume of
15. f(x,y) =x- y; 16. f(x,y) =x/y2;
R= { (x,y): -3:5y:5-2,y2:5x:5y4}
and the left branch of the hyperbola x2- 4y2=84. Cal
17. f(x,y) =x+2; R is the region between the curves y=x2+x and y=3x+8.
18. f(x,y) =12+ x+y; R is the region between the curves y= -x2+4 and y=x2+2x- 8.
19. f(x,y)=e"; R is the region bounded by the curves y=2x+ 3; y=x; and x=2. R is the region below y=1 and above y=exp(x2 - x).
2L f(x,y)=x+2y; is
the
region
the plane z = -x.
33. Find the volume of the solid that lies above the planar disk (x+ 4)2+y2 :5l and below the plane z=x+4.
34. Find the volume of the solid that lies below the graph of z =1/(x2+y2) and that is bounded laterally by the cylinder set y= lxl and the planes y=2 and y= 8.
35. Find the volume of the solid that is bounded by the two paraboloids z=x2+y2+3x+6y- 2 and z=-x2 - y2 5x+2y+6.
20. f(x,y) =x/y;
R
culate the volume of the solid that lies over R and under
bounded
x=O, y=x+2, and x=4.
by
the
curves
y=..ji,
36. Find the volume in the first octant that is below both the plane z=2x+y and the plane z=2- x- y.
37. Find the volume in the first octant that is inside the cylinder x2+y2 =1 and below the surface z =xy.
990
Chapter 12
Multiple Integrals Equate
the two expressions for I, and solve J: u(y)v'(y) dy. What formula have you found?
38. Find the volume of the solid that is bounded by the planes z = 0, y = 0, z = 1 - x + y and by the parabolic cylinder
for
y=l- x2.
39. Find the volume of the solid that is bounded above by the parabolic cylinder z = 1 y2 and below by the parabolic
Calculator/Computer Exercises
-
cylinder
z
= x2.
40. Find the volume of the infinite solid that lies below the surface
z = exp(-x2)
{(x,y,O): x2'
and
above
the
planar
region
IYI}.
4L Find the volume of the solid that is bounded above by the surface z = 1/(x + y), and that lies over the region in the xy-plane that is between y= 1/ x and y= 1 for 1 '5 x '5 2. 42. Find the volume of the solid that is bounded by the paraboloid z = 1 + x2 + y2 and by the planes x = 0, y = 0, z = 0, and x + 2y = 2. 43. Find the volume of the solid region that is bounded above by z = xy, below by z = 0, and laterally by the plane y= x and by the cylinder set y= x3•
a< b. Let R be the triangle with vertices (b, b ). Let u and v be continuously dif functions of one variable on [a, b]. Calculate
44. Suppose that
(a, a), (b, a),
ferentiable
and
I= JJ'l?.u'(x)v'(y) dA
45. Let Rbe the region in the first quadrant of the xy-plane that lies between the curves y lOx - 1 - 6x2 and y = x3 - 3x2 + 3x. Find the volume of the solid that lies above Rand below the plane z = x + y. 46. Let Rbe the region in the first quadrant of the xy-plane that lies between the curves y e" and y 1 + 3x. Find =
=
plane z = 10 - y. 47. Let Rbe the region in the xy-plane that lies between the curves y 2 - x3 and y 1 + x4• Find the volume of =
I
=
the solid that lies above Rand below the plane
z
=
y.
48. Let Rbe the region in the xy-plane that is bounded by the lines x= 0, y= 0, and y= 1 - x. Use the Trapezoidal Rule approximation with N= 4 to estimate the volume of the solid region that lies above R and below z = 1/(1 + xy). (The Trapezoidal Rule approximation for double inte
by using iterated integrals. Next,
reverse the order of integration. and calculate
=
the volume of the solid that lies above R and below the
grals is discussed in the headnote to Exercises 65-68 of Section 12.2.)
again.
1 2.4 Polar Coordinates You are familiar with the idea of locating a point in the plane with two real numbers: In the Cartesian setup, one of these numbers corresponds to the x displacement and the other to they-displacement. These represent signed distances to perpendicular axes. In certain situations, however, it is natural to express the location of a point P using the distance r
=
llOPll
to a fixed point 0, which is taken
to be the origin. For example, we found it convenient to refer to the coordinate r when we analyzed central force fields in Section
10.5
of Chapter
10.
Of course the
distance ralone does not determine the location of the point P. For that, we will need information that tells us the direction of
OP . The angle that OP
makes with
a fixed half-line emanating from 0 makes a convenient choice. This fixed half-line is called the polar axis. We will choose the positive x-axis of Cartesian coordinates to be the polar axis, as is conventional.
The Polar Coordinate
Suppose that r is a positive number and that(}is any real number. Then there is a
System
unique point P in the plane such that P is a distance r from the origin and such that the angle between
OP
and the polar axis measures (}radians. The angle is
measured counterclockwise when (}is positive (Figure la) and clockwise when (} is negative (Figure lb). We say that rand(}are polar coordinates for the point P. As with Cartesian coordinates, we write polar coordinates as an ordered pair, (r, 0). We must understand from the context which coordinate system is being used. Thus in Figure 2, point P has polar coordinates (2,
1),
whereas point Q has
12.4 Polar Coordinates
991
y
y
r
/';� OP
0
x
x
0
8 r
----t
OP
�
p
P has polar coordinates (r, 8) with 8 > 0
P has polar coordinates (r, 8) with 8 < 0
.& Figure 1a
.& Figure 1b
y 2
?-.p
1
I I I I I _____
',
Cartesian coordinates '
'
ordered pair '
I
2
'
\ :
Q
We must know the coordinate system to which the
(2, 1)
for its
In polar coordinates, once you have specified the radial coordinate r of a point P, then you have already restricted the point to lie on the circle of radius r
\ I
It('\
I
(2, 1).
refers before we can know which planar point has
coordinates.
...,__________ , :._ _ .
/
(2, 1)
\:1 1• �
1
/
I
centered at the origin (see Figure
3).
The angular coordinate () specifies which
point on the circle P is. Alternatively, once you have specified the angular coor x
dinate (), then you have restricted the point to lie on a certain ray. The radial
2
coordinate r then tells you how far out on that ray the point lies. These ideas are also shown in Figure
.& Figure 2
3.
y y
y
.& Figure 3a The circle r
=
r0
x
x
x
.& Figure 3b The ray ()
=
eo
.& Figure 3c The point whose polar coordinates are given by r r0 and () ()0 =
=
In polar coordinates, the number 0 is also a possible value for the radial coordinate r. For any (), the point in the plane with polar coordinates (0, ()) is the origin 0 because it is the only point a distance 0 from 0. Notice that the polar coordinates of the origin are ambiguous: The radial coordinate r must be 0, but the angular coordinate () may have any value.
992
Chapter 12
Multiple Integrals
y
� E X A M P L E 1 Plot the points whose polar coordinates are tabulated.
5
Point
p
R
Q
s
T
u
v
Polar coordinates (2,7r/4) (4,37r/2) (112, -7r) (5,37r/4) (l,47r/3) (3, -77r/6) (5/2, -77r/4) x
Solution The points are exhibited in Figure INSIGHT
In Example
1, compare
4. .,...
the first point P and the last point V. They lie
on the same ray emanating from the origin. That is because their angular coordinates differ by an integer multiple of
.A Figure 4
(2,11"/4) (1/2, TI) T= (1,411"/3) v (5/2, -711"/4)
p
=
R=
-
(4,311"/2) (5,311"/4) u = (3, -711"/6)
Q
=
s=
=
2rr: rr/4 = -7rr/4 + 2rr. As you can imagine from Figure 4,
if the radial coordinates of points P and V had been equal, then the points would have coincided despite their different angular coordinates. The polar coordinates of a point in the plane are
its rectangular coordinates) . Indeed, the polar
coordinates
the same point in the plane for all integer values
of
n.
not unique (unlike (r0, 00 + 2nrr) represent
Thus every point in the plane can be specified in infinitely many different ways
in the polar coordinate system.
Negative Values of the Radial Variable
It is often convenient to allow the radial coordinate
r<0
r to
be negative. Suppose that
(lrl, 0). lrl from the -----+ origin and that lies on the ray opposite to OP'. In other words, because lrl = -r, the ordered pairs (r, 0) and (-r, 0) are polar coordinates for points P and P' that are and that() is any number. Let P' be the point with polar coordinates
Then the polar coordinates
(r, 0)
refer to the point P that is a distance
reflections of each other through the origin. Another way to think of this rela tionship is that the ordered pairs same point. See Figure
5
(r, 0) and (-r,() + 7r)
are polar coordinates for the
in which P has polar coordinates
(r, 0)
with
r < 0.
y P' has polar coordinates (-r, 8) and (r, 8 + 'IT)
x
P has polar coordinates (r, 8) and (-r, 8 +
)
'IT
.A Figure 5 Polar coordinates with r < 0
� E X A M P L E 2 Give all possible polar coordinates of the point P that is a -----+ 4 from the origin and for which OP makes an angle of 57r/6 with the
distance
positive x-axis.
Solution In effect, we have been told that (4, 57r/6) is a pair of polar coordinates for P. However, we may add any multiple of 27r to the angle and obtain another valid set of polar coordinates for this point. Therefore
12.4
Polar Coordinates
993
y ...(4, -197!"/6), (4, -771"/6), (4, 57r/6), (4, 1771"/6), (4, 297!"/6)...
4
are all polar coordinates for P. We summarize this by saying that x
(4, 57r/6 + 2n7r),
n E Z
are polar coordinates for P. Is this a complete list of all possible polar coordinates for P? No, because we can allow r to be negative. Thus P may be thought of as the
-4
point (-4, -71"/6). See Figure 6. Again, we can add any multiple of 271" to the angle. Therefore the pairs
_. Figure 6
are also valid polar coordinates for P. Because P must have radial coordinate :±:4 and must have angular coordinate either 57!"/6 + 2n71" or -71"/6 + 2n71", we have accounted for all possible polar coordinate representations of P.
Relating Polar Coordinates to Rectangular Coordinates
Now we discuss the connection between polar coordinates and Cartesian (or rec tangular) coordinates. Look at Figure 7. The point Q has polar coordinates ( r, 0) and rectangular coordinates (x,y). By examining the right triangle in the figure, we see that x=rcos(O)
y
Q (x,y)
and
y =rsin(O).
(12.4.1)
As expected, the rectangular coordinates x and y of a point are uniquely deter mined by its polar coordinates r and 0. We also have equations that we can use to obtain polar coordinates r and 0 for the point with rectangular coordinates (x,y):
,.2 =x2 + y 2
and , if
x =I=
0,
tan(O) = :2::. x
(12.4.2)
As we expect from our earlier discussion, we see that these equations do not Y
determine r and 0 uniquely. The first formula in line (12.4.2) tells us the magnitude lrl = x2 + y 2 of the radial coordinate r but not its sign. The second formula in line
J
(12.4.2) is satisfied by infinitely many values of the angular coordinate 0.
� E X A M P L E 3 Compute the rectangular coordinates of the point P whose x
x _. Figure 7
polar coordinates are (4, 57!"/6).
Solution
Observe
that
r=4
and
0 =57r/6 so
x=4 cos(57r/6) =-2v'3
and
y =4 sin(57r/6) =2. Thus P has the ordered pair (-2v'3, 2) for its Cartesian coordinates.
Notice that Pis the same point in Examples
2 and 3.
In Example
2, we
determined infinitely many ways to express Pin polar coordinates. Whichever of these representations we choose to work with, the conversion to rectangular coordinates
( -2v'3, 2) that we obtained in Example 3. For instance, (-4, -7r/6) lead to x = (-4) cos ( -7r/6) = -2v'3 and y = (-4) sin ( -7r/ 6 ) = 2. will be the ordered pair polar coordinates
the
994
Chapter 12
Multiple Integrals � EXAM P L E 4 Calculate all possible polar coordinates for the point Q with rectangular coordinates
(-5, 5).
Solution We know that x take
r=
+ J50
= 5v'2,
y
�x -tH:a!����:ftl+J-ti
,.Z = x2 + y2 = 25 + 25 = 50. identities x = r cos(()) and y = r sin(()) yield
= -5, y = 5.
then the
cos(())
=
Therefore
- }i
and
sin(())
=
}i .
37r/4 + 2n7r, n E Z. So we have found the polar coordinates (5v'2, 37r/4 + 2n7r), n E Z. However, r could also have the value -JSO, or -5v'2. In this case, the equations x = r cos(()) and y = r sin(()) yield
It follows that ()=
}i
cos(())=
and
sin(())=
- }i·
77r/4 + 2n7r, n E Z. So we have found the (-5v'2, 77r/4 + 2n7r), n E Z. This accounts for all possible
It follows that () _. Figure 8
for Q .
Graphing in Polar Coordinates
If we
=
polar coordinates polar coordinates
..,..
Graphing in polar coordinates is different from graphing in rectangular coordi nates. We mention two new features in particular. The first is that the pictorial notions we have developed for graphing in rectangular coordinates-slope, x- and y-intercepts, and so on no longer apply. The second is that the ambiguity in the polar coordinates of a point means that we have to be careful not to forget part of the graph. Because of these twists, we begin with the most elementary concepts. � EXAM P L E 5 Sketch the graph of Solution The graph of y
=x
in
r = e.
rectangular
coordinates is a line, but do not expect
to see anything like that now. In Figure 8, we have plotted a few points corresponding to from () = 0, so
() = 0, 7r/4, 7r/2, 37r/4, 7r, and so on. We notice that, as() increases does r increase. Therefore it is logical to connect the points with a
smooth curve. Notice that the resulting curve is a spiral that grows with rotation in y
the counterclockwise direction. The graph so far is not complete because we have not considered negative values of e. We plot some points corresponding to
() = -7r/4, -7r/2, -37r/4, -7r,
and so on and connect them as we did for()> 0. Thus
the complete graph consists of two spirals, one expanding with clockwise rotation and the other expanding with counterclockwise rotation (Figure
-®�ll��lilr �x
Spiral of Archimedes (as polar coordinates is r = a() where a is a constant) . have graphed is called a
� EXAM P L E 6 Describe the graphs of
9).
The curve we
is any curve whose equation in ..,..
r cos(()) = 2
and
r sin(()) =
-1.
Solution The easiest thing to do is to convert these equations to rectangular coordinates: They become x _. Figure 9 Spiral of Archimedes r = (}
=2
and y
=
-1. The graphs are lines parallel to the
rectangular coordinate axes. ..,.. � EXAM P L E 7 Sketch the graph of
r = 2 sin(())
by two different methods.
12.4 Polar Coordinates
995
Solution For the first method, we create a table of values.
()
0
7r/6
7r/4
7r/3
7r/2
27r/3
37r/4
57r/6
7f
77r/6
r
0
1
v'2
J3
2
J3
v'2
1
0
-1
57r/4
47r/3
7r/2
57r/3
77f/4
-v'2
-J3
-2
-J3
-v'2
lh/6
27f
-1
0
The corresponding points are plotted in Figure lOa. These points are connected, in
y
increasing order of O, by a smooth curve. We see that a circular loop beginning and ending at the origin is traced when () increases from 0 to 7f. This loop is retraced when () increases from 7f to 27r.
An alternative method to sketch this graph is to multiply both sides of the x
-2
-1
1
equation by r to obtain
r2
=
2r sin( 0). Then, it is easy to convert to rectangular
coordinates: x2 +y2 2y. By completing the square, we obtain the equation 2 x2 +(y - 1) =1 of the circle with center (0, 1) and radius 1. .,.. =
2
.& Figure 10a
INSIGHT
The graph of
r
=
a is a circle of radius a. But
equation that has a circle of radius a
as
r
=
a is not the only polar
its graph. When generalized, the calculations of
Example 7 show that the graph of r = 2a sin ( 0) is a circle of radius a. In a similar manner,
we may show that the graph of r
=
2a cos ( O) is also a circle of radius a. Figure
lOb shows
these three circles in one viewing window.
x
(2a, 0)
� EXAMPLE 8 Sketch r=3(1 +cos(O)). Solution Notice that, as() increases from 0 to 7r/2, the values of cos(O) decrease from 1 to 0. Therefore r, being equal to 3(1 +cos(O)), decreases from 6 to 3.
r
.& Figure 10b
=
2acos(O)
Likewise, as () increases from 7f/2 to 7f, the values of r decrease from 3 to 0. This information is indicated in Figure 11. We now have the upper half of the curve. We could continue this type of analysis to get the lower half of the curve. Instead, we notice that, because cos(-0)= cos(O), the formula for r does not change when() is replaced by -(). So, for every point
(r, 0)
on the graph, there is a corresponding
point (r, -0), which means that the graph of r= 3(1 +cos(O)) is symmetric about the horizontal axis. The complete sketch appears in Figure 12. A curve whose polar equation has the form r = a(1 + cos( 0)) or r= a(l-cos( 0)) or r= a (1 +sin( 0)) or r= a(1-sin( 0)) is called a cardioid (because the locus is shaped like a heart).
.,..
y
y x
x
-6
2 4 -2 r = 3(1 + cos(O)), 0 ::s: 0 ::s:
6
-4
.& Figure 11
7T
r =
3(1 + cos(O))
.& Figure 12
Chapter 12
996
Multiple Integrals
r
� EXAMPLE 9 Sketch the graph of
y
4
=
4 sin(W).
Solution When 0 increases from 0 to 7r/4, we see that 0 then passes from 7r/4 to 7r/2, the value of
r
r increases from 0 to 4. When
decreases monotonically back to 0.
The corresponding sketch is in Figure 13. Figure 14 shows the added information obtained from considering 7r/2::; 0::; 7r: The value of
r decreases
from 0 to -4 and
then increases again back to 0. Notice that although the values of 0 correspond to
2
the second quadrant, this portion of the plot appears in the diagonally opposite quadrant, the fourth quadrant, because
r is
negative.
Now let us see what we can learn from symmetry. Notice that x
r
=
2 1 3 4sin(29), 0 s 9 s
(
sin 2(0 +
4
f
So to each point
_. Figure 13
(r, 0)
?r
))
=
sin(W).
on the curve, there is a corresponding point
(r, 0 + ?r) .
This
tells us that the graph is symmetric with respect to reflection through the origin. Glancing at Figure 14 and taking this symmetry into account, we complete our sketch as in Figure 15. This curve is called a
r2 =
� EXAMPLE 1 0 Sketch the curve Solution We notice that
four-leafed rose. .,.. -cos(W).
r is undefined when cos(W) is positive, so certain values of 0 r is undefined for 0 < 0 < 7r/4. However, when 0
must be ruled out. For instance,
increases from 7r/4 to 7r/2, the values of cos(W) decrease from 0 to -1. Therefore the values of
r2
will give rise to two branches of our curve: one r increasing from 0 to 1 and the other corresponding to r decreasing
increase from 0 to 1. This
corresponding to
from 0 to -1. These are illustrated in Figure 16. Once again, notice the phenomenon of plotting that part of a curve with
r < 0: It appears as the reflection through the origin r been positive.
of the arc that would have been plotted had the values of
y
y
y
x x
x
0.7 r2
=
-cos(29),
!!_<(J
r
=
4sin(29), 0 s 9 s
_. Figure 14
7f
r
=
4
2
r2 =
-cos(W)
4sin(28)
_. Figure 15
Next we observe that cos(2(7r - 0))
_. Figure 16
=
cos(W); hence the equation
is unchanged under the map 0 f-t ?r - 0. It follows that for each point the curve, there is a corresponding point
(r, 7r
-
(r, 0)
on
0) that is also on the curve. This
12.4
Polar Coordinates
997
property tells us that the graph is symmetric with respect to the y-axis (see Figure 17). The graph of
r2
INSIGHT
=
-cos(20) is called a lemniscate.
To be sure that we have not forgotten any part of the graph in the
previous example, we can perform a check. We have already considered 0 <(} < 7r/2.
7r/2 :5.(} :5. 37r/4, the quantity -cos (W) decreases from 1 to 0, which is indicated on 37r/4 <(} < 7r, -cos (W) is negative so that r is undefined. That is also indicated by our figure. If(} is between 7r and 27r, then we may write (} =(}0+ with 00 between 0 and 7r. Remembering that ( -r, 00) and (r, 00+ 7r) are polar coordinates for the 2 same point, and observing that r -cos(W) if and only if ( -r ) -cos(W) -cos(W0+ 27r) -cos(W) , we see that every point on the curve that has polar co ordinates ( r, 0) , or ( r, 00+ 7r), is a point on the curve with polar coordinates ( -r, 00) . In other words, the picture that is drawn as (} ranges from 0 to 7r is repeated as (} ranges from 7r to 27r.
For
Figure 17. For
7r
=
=
=
=
Symmetry Principles in Graphing
We have seen that symmetry can be a powerful tool in graphing polar equations. We summarize here the basic tests for symmetry.
y
Test
Symmetry with
respect to x-axis
Polar equation is unchanged when() is replaced with -().
y-axis
Polar equation is unchanged when() is replaced with 7r - 0.
origin
Polar equation is unchanged when r is replaced by -r, or polar equation is unchanged when() is replaced by()+ 7r.
� EXAM PL E 1 1 Test the equation r = 3 - sin(20) for symmetry. Solution Because
(
)
sin 2(7r - 0) = sin(27r - 20) = sin(-20) = -sin(20),
_. Figure 17
the test for symmetry in the y-axis fails. Because sin(2(-0)) f= sin(20), the test for
y
symmetry in the x-axis fails. Finally, sin(2(0+ 7r)) = sin(20+ 27r) = sin(20), so the equation is unaltered if() is replaced by()+ 7r. We conclude that there is symmetry in the origin. The curve is sketched in Figure
x
INSIGHT
18. ..,..
Notice that there are two tests for symmetry in the origin. The one that we
used in Example 11 is to replace(} by(}+ in Example 11
fails this second
7r. The other test is to replace r by
-r. The curve
test. Again, the fact that a single point has several
different polar representations is causing confusion. Always remember that our sym metry tests are
sufficient conditions for symmetry,
not necessary conditions. This means
that if a curve passes the test, then it is certainly symmetric; if it does not pass the test, it r =
3
_. Figure 18
-
sin(W)
may or may not be symmetric. In practice, you should just apply the tests that you know and use to advantage whatever information they give you.
998
Chapter 12
Q U IC K
Multiple Integrals Q
U
IZ
1. 2.
What are the polar coordinates of the point with rectangular coordinates
3.
Describe the curve that has polar equation
(1, v'3)?
What are the rectangular coordinates of the point with polar coordinates
(-2, 7r/2)? r
=
1 + 2()
for 0 < 0.
4. Iffis an even function, then what symmetry does the curve defined by
r =
f(O) have?
Answers 1.
(2, 7r/3)
2.
(0, -2)
3.
A spiral
4. Symmetry with respect to the x-axis
EXERCISES Problems for Practice
In each of Exercises 1-8, plot the four points whose polar coordinates are given. L 2. 3. 4. 5. 6. 7. 8.
(3, 7r/4), (-3, 7r/4) , (3, -7r/4), (-3, -7r/4) (2, 57r/6) , (-2, 57r/6), (2, -57r/6) , (-2, -57r/6) (0, 7r/2), (0, 7r), (0, -7r), (0, -7r/2) (6, 97r/4), (-6, 97r/4), (6, -97r/4) , (-6, -97r/4) (4, -77r/2) , (-4, -77r/2), (4, 77r/2) , (-4, 77r/2) (1, 7r/3), (-1, 7r/3) , (1, -7!"/3) , (-1, -7r/3) (2, 77r/3), (-2, 77r/3), (2, -77r/3) , (-2, -77r/3) (10, 0) , (-10, 0), (6, 0) , (0, 0)
In each of Exercises 9-16, the polar coordinates of a point are given. What are its rectangular coordinates? 9. 10. 11. 12. 13. 14. 15. 16.
{4,7r/4) (7, 97r/2) (0, 7r) (-8, -37r/4) (-4, -177r/6) (-3, 57r/6) (-6, 7r/3) (3, 0)
In each of Exercises 17-22, give all possible polar coordi nates for the point that is given in rectangular coordinates. 17. 18. 19. 20. 21. 22.
(4, -4) (-4v'2, -4v'2) (0, -9) (-5, -5) (-1, 0) (-8v'3, 8)
In each of Exercises 23-32, test the given equation for symmetry in the x-axis, the y-axis, and the origin. 23. 24. 25. 26. 27.
r2 =3 sin(ll) r=4 - 2 tan(W) r=3 0=7r/2 r=1 + cos2(ll)
28. 29. 30. 3L 32.
r sin(ll) =cos(ll) r=6 - cos(3ll) r=4 + sin(3ll) r=sin(ll) - cos(ll) r2=1 - 2 sin(ll)
In Exercises 33-44, sketch the curve that is described by the given equation in polar coordinates. {The name of each curve has been provided in parentheses.) 33. 34. 35. 36. 37. 38. 39. 40. 4L 42. 43. 44.
r=1 - cos(ll) r=3 + 3 sin(ll) r=3 cos(W) r2 =9 cos(W) r=4 - 2 sin(ll) r=2 sin(3ll) r=-1 +3 cos(ll) r2 =-3 sin(2ll) r=3 sin(ll) - 2 cos(ll) r= -3 sin(5ll) r=3 - 3 sin(ll) r=-20
(cardioid) (cardioid) (four-leafed rose) (lemniscate) (lima\:On) (three-leafed rose) (lima\:On) (lemniscate) (circle) (five-leafed rose) (cardioid) (spiral of Archimedes)
Further Theory and Practice 45.
46.
47.
48.
Give all possible polar coordinates for the points that have rectangular coordinates: a)(3, 7r/4); b)(-6, 77r/2) ; and c) (7r/4, -87r/3). Give rectangular coordinates for the points that have polar coordinates a) (2, -6); b) (-4, 7) ; c) (7r/3, 9) ; and d) (77r/3, -6). Give a formula in polar coordinates for the distance between two points in the plane with polar coordinates (r1, ll1) and (rz, llz). What is the polar equation of a circle of radius 1 with center given by (3, 7r/4) in polar coordinates? A regular N-gon(a convex polygon with N equal sides) is inscribed in a circle of radius r that is centered at the origin. If one of the vertices is on the positive x-axis, then what are the polar coordinates of the vertices?
In each of Exercises 49-54, find all points of intersection of the two polar curves that are given.
12.5 Integrating in Polar Coordinates 49. r=cos(IJ) and r=1 - cos(IJ) 50. SL 52. 53. 54.
65. r112=sin(IJ) 66. r3 =sin2(1J).
r=1 + sin(IJ) and r=3(1 - sin(IJ)) r
=
1 and r
=
999
tan(IJ)
r=3 sec(IJ) and r=4 cos(IJ) r=1 + cos(IJ) and
67. Sketch the graphs of r=fP and IJ r2. 68. Explain how and why the graphs of =
r2=(1/2)cos(IJ)
r=4 - 5 sin(IJ) and r=3 sin(IJ)
r
55. Find each ordered pair (a, b) that represents the same point in the plane in both polar and rectangular coordinates.
56. Suppose that
f
is a continuously differentiable scalar
valued function of one variable. If C is a curve given in
=
r2=sin2(1J)
sin(IJ) differ.
and
69. In a !Jr-plane, plot the Cartesian equation r=IJ + cos(IJ,
0�IJ �27r. Then, plot this equation by interpreting it as a polar equation in the xy-plane.
70. In a !Jr-plane, plot the Cartesian equation
polar coordinates by the equation r=f(IJ) for a�IJ�{J,
�
2 r= + , 1 + rr
then the arc length of C is
O�IJ�47r.
Then, plot this equation by interpreting it as a polar equation in the xy-plane.
Prove this formula. Hint: Describe the curve in terms of the position vector r(IJ) = (x(IJ),y(IJ)}, and use the formula for arc length that was developed in Chapter 10.
Refer to Exercise 56 for the arc length formula in polar coordinates.
In each
of
Exercises
57-60, calculate the
arclength of the given curve.
57. 58. 59. 60.
r= e8
O�IJ�27r
r
O�IJ�7r/4
r=sin3(1J/3) =
sec(IJ)
curves
r
=
sin(jlJ)
for j
=
1, 2,3, 4, 5.
Suppose that j is a positive integer. For what positive
=
=
sin(jlJ)?
j + sin(IJ) for j
=
0, 1, 2, 3, 5, 10,
20, 30. How does changing the additive constant affect the
0�IJ�7r/2
graph?
73. Plot the polar curves r
=
sin(IJ + j) for j
=
0, 1, 2, 3, 5, 10,
20, 30. How does changing the additive constant affect the
graph?
74. Plot Fay's Butterfly r=e00'(8l-2cos(41J)+sin5(1J/12) for 0�IJ�27r. Repeat, using the range 0�IJ�47r and again
r=exp(IJ/5), -37r�IJ�27r r=2 sin2(1J/2) =
polar
more petals than the graph of r
polar curve.
r2
the
Describe the graphs of r= sin(50 IJ) and r=sin(511J).
72. Plot the polar curves r
In each of Exercises 61-66, sketch the graph of the given
6L 62. 63. 64.
7L Plot
integer values of k does the graph of r=sin(k IJ) have
-7r/2�IJ �7r/2
r=a ( l -cos(IJ))
Calculator/Computer Exercises
for O�IJ�67r. Continue. For what range O�IJ�2N7r do
-3 sin(41J)
you obtain the complete plot without any repetition?
rl/3=sin(IJ)
1 2.5 Integrating in Polar Coordinates The theory of area in Cartesian coordinates rests on the basic notion that the area of any planar region can be thought of as a sum of areas of thin rectangles, up to a relatively small error. This is sensible because the Cartesian coordinate system is a rectangular one-the axes are perpendicular, and the notion of rectangular area
fits
the notion of rectangular coordinates. Now we are dealing with polar coordi
nates, and there is no longer a standard pair of perpendicular directions. In fact, we have a radial direction and an angular direction, which are qualitatively quite different from the information coming from rectangular coordinates. As a result,
sectors. sector is a portion of a circle cut off by two rays: look at Figure 1. The area A sector depends on the radius r of the circle and the angle (} between the two
the building blocks we use for area are not rectangles. Instead, they are A of the
1 000
Chapter 12
Multiple Integrals rays. In fact, by comparing the sector with the full circle and using proportionality,
A
we deduce that 2 'TfT
=
-
()
, or
27r
A=
(12.5.1)
�r20.
Our method for calculating area in polar coordinates will be determined by this formula.
Areas of More General Regions
We consider a region R in the xy-plane that is contained inside a polar curve
r
=
cp( O).
More precisely, if we set
S
=
{(r,
0)
for a given nonnegative function
:
0 :s r :s
cp(
0)
(12.5.2)
cp(O), a :s () :s {3}
defined for () between
a
and
{3,
then we
consider the region 'R, in the xy-plane defined by
'R, =
{(x, y)
:
x=
r cos ( O),y = r sin ( O) for
some
(12.5.3)
(r, ()) in S}.
(12.5.3) is in terms of the S that appears in equation
Notice that the description of R that appears in equation Cartesian coordinates (x,y ) of the points of R. The set
(12.5.2)
_. Figure 1 The area of the sector is A=
describes R in terms of the polar coordinates of the points of R. To avoid
ambiguities, we usually restrict shown in Figure
2.
a and {3 to differ by not more than 2rr. The region is
We estimate the area of the region R by forming a uniform partition,
�r2o.
y
r =
rp(O)
of the interval
[a, {3],
which is the domain of cp. The angular increment/),.() satisfies /),.() - ()j
x
_. Figure 2
for all
1 :s j :s N.
Figure
3 shows
{3-a - -- ()j-1 N
how the arc corresponding to
a :s () :s {3
is broken
into subarcs by the partition. Now each subarc determines (not a rectangle but) a sector. If
Sj
is the area of the J-th sector, then the sum of the areas of the sectors,
approximates the area of the region R.
Sj. The Sj forms at the origin is !:l.(). The radius is approximately given by rj = cp(�j), where �j is any number chosen from the interval [Oj-l, ()j]· See Figure 4. Thus according to formula (12.5.1), the area of the /h sector is approximately In order to calculate this sum, we need to know the area of the J-th sector
angle
_. Figure 3
12.5 Integrating in Polar Coordinates
1 001
The area of the region n is therefore approximately
y
However, this sum is a Riemann sum for the integral
(3 1 1 ()) 2 d(). a Z cp(
x
We summarize our findings in a theorem. _. Figure 4 THEOREM 1 where
Suppose that
{3-o: '5, 27r.
cp
is a nonnegative function on the interval
Let S = {(r, 0) : 0 '5, r '5,
cp( 0), o: '5, () '5, {3}
[o:, {3],
and let n be the
region in the xy-plane that is described in polar coordinates by S. Then, the area of n is given by the integral
!3 11 2 Z a cp( ()) d().
(12.5.4)
� EX A M P L E 1 Find the area inside the curve r = y !!._ (J ---i--- 2
Solution
=
(J
=
!!._ 3 (J
(J
= 7T
- !!._ 6 x
-1
(J
1
=
12.4, we learned that the curve r = 2 sin(O), shown in Figure 5, is a circle of radius 1 that is traced out once as() varies from 0 to 7r. Therefore we 2 expect to obtain 7r = 7r(1 ) as our answer. That is exactly what we obtain when we apply Theorem 1 using the correct interval of integration, [O, 7r]. Thus we have
0
In Section
Area
_. Figure 5 The curve described
by the equation r = 2 sin( O) in polar coordinates
INSIGHT y
2 sin(0).
=
111' ( . ( )2 2 sm ()) d () 2 0
=
1' . (6.2.8) 7r 21 sm2(()) d = 2 () 2 0 ·
=
7r.
Example 1 illustrates an important point. Because of the ambiguity built
into polar coordinates, the circle in Example 1 is generated twice as 0 ranges from 0 to 27!". If we attempt to calculate the area inside it by integrating from 0 to 27r, we count the (J
=
same area twice and obtain the incorrect value 27r as the answer. As a result of this
!!._ 4
observation, we must begin the solution to each area problem with a brief analysis to be sure we are avoiding duplication of area. A sketch will usual play a vital role in this x
analysis.
�
EX A M P L E 2 Find the area inside one leaf of the rose r
Solution
=
6 cos(W).
By the periodicity of the cosine function, each of the leaves has the same
area, so we choose for convenience the leaf on the right (look at Figure 6). We must _. Figure 6
integrate from()=
-7r/4
to()= 7r/4. By Theorem
1,
the area of the leaf is
1 002
Chapter
12
Multiple Integrals
/4 7r/4 A= -12 17r (6 cos(20))2 d(}=18 1 cos2(20) d(}. -7r/4 -7r/4 To handle this integrand, we use the method we learned in Section 6.2 of Chapter 6. Replacing(} with in formula (6.2.2), we obtain
20
cos2(20) It follows that the area
7r/4 A=lsj 7r (1+cos(40)) 2 - /4
d(}
=
7r/4 9 (o+ sin(40)) 1 4 -7r/4
� EXAM P L E 3
the polar curve
-
y �
'-= �4 +
=
� (1+cos(40)).
A of one leaf of the rose is =
9
((4
�
( �) ) + ( sin(7r) 4 4
-
-
_
sin(-7r)
4
))
=
97r. 2
<11111
Find the area A of the region Rin the xy-plane that is inside r cos(O).
4+cos(O) and outside the polar curve
=
2+
The boundary curves of Rare called lima(:ons (see Figure 7). To avoid duplicating our calculations, we will work with a general limac;on r cos( 0:::;(}:::; ; ; 27r where is a constant greater than The area inside this limac;on is
Solution cos(O)
c
1.
=c+
A(c)
0),
A(c) = 21 l{o 27r (c+cos(O)) 2 d(}= 21 l{o 27r (c2+2c cos(O) +cos2(0)) d(}
x
.& Figure 7
r
=
7r 7r 7r = 21 l{o 2 c2 d(}+c l{o 2 cos(O) d(}+ 21 l{o 2 cos2 (0) d(}. ?rc2,
The first summand on the right side evaluates to the middle summand evaluates to 0, and the last summand evaluates to 7r/2 by formula (6.2.9). Thus
A(c)
=
7rC
2+0+�
=
7rc2
+�.
(12.5.5)
The area A of R is obtained by subtracting the area inside the limac;on r cos(O) from the area inside the limac;on r cos(O). Taking and in formula we have
=2+ c 2 =
=4+ (12.5.5), A=A(4)-A(2)=(7r(4)2+�)-(7r(2)2+�)=127r.
INSIGHT
c=4
<11111
Example 3 illustrates the calculation of the area of a region bounded by two polar curves. In general, if 'Po ( 0) ::5 cp1 (0) for o: ::5 0 ::5 (3, then the formula 1
2
1
a
(3 'P ( 0) 1
2
dO
-
1
2
1
a
(3
'Po ( 0)
2
dO
12.5 Integrating in Polar Coordinates gives the area between the two curves
r =
\Oo (0) and
r =
r,o1 ( 0) for
o:
�
0 � f3.
1 003
This dif
ference of integrals is equivalent to
(12.5.6) Notice the error that we have avoided in Example 3: It would be a mistake to subtract first and then square-the integral
� 1(3 r,o1 (0)-r,o0 (0) )2 (
dO has nothing to do with the
area between the two curves.
Using Iterated Integrals
Suppose that R is a planar region given in polar coordinates by
S = {(r, 0) : a:::::;():::::; (3, 0:::::; r:::::; cp(O)}
to Calculate Area in Polar Coordinates
of R, it is instructive to derive an area formula that uses iterated integrals. Starting
where
(3-a:::::; 27!'. Although formula (12.5.4) suffices for the calculation of the area
from equation
(12.5.4),
Area of R =
2 l
we proceed as follows:
1/3
cp2(0)d() =
a
1/3 r= (IJ) 1r=Ocp a
r2 2 l
d() =
1/31 (IJ) cp a
0
rdrd().
It is just as easy to calculate the area of a region R that is bounded by two polar curves. Suppose that
S={(r,O) :a::;()::=;(J, cp0(0)::=;r::;cp1(0)} and that R is the S, as in Figure 8.
Then the area of R is clearly
region in the xy-plane that is described in polar coordinates by
1/31'Pi(IJ) a
0
rdrdO-
1/31'Po(IJ) a
0
r drdO=
1/3 (1'Pi(IJ) a
0
r dr -
1'Po(IJ) ) 0
r dr dO=
We state our observations as a theorem. THEOREM 2
1/31'Pi(IJ) ip0(1J)
r drdO.
a
Q::=;(J-a::=;27l' and Q::=;cp0(0)::=;cp1(0) for all 0. Let S= {(r,0) : a :::::; 0 :::::; (3, cp0(0) :::::; r :::::; cp1(O)}. Let R be the region in the xy-plane that is described in polar coordinates by S . That is, R = {(x,y) : x = r cos(O),y = r sin(O) for some (r, 0) in S}. Then, the area of R is given by Suppose that
fl 'R,
1 dxdy =
1/31'P (1J) 'Po(IJ1 )
rdrdO.
(12.5.7)
a
x
.A Figure 8
�
E X A M P L E 4 Calculate the area of the planar region lying between the
curves
r = cos(O) and r = 1
+ sin(O) for() ranging between 0 and
11'/4.
Chapter 12
1 004
Multiple Integrals Solution Notice that, for this range of (), we have cos(O):51 +sin(O). See Figure 9.
According to Theorem 2, this area is
A=
17r/41co1+ss(O)in(9) 0
y
r dr dO=
17r/4 (l 1r=l+sin(O)) r=cos(O) 2
0
2
,
dO=
l
2
17r/4 0
(1 +2sin(O) +sin2(0)-cos2(0)) dO.
We could proceed as in previous problems in this section by using the half-angle
1.2
formulas for sine and cosine. Here, however, it is simpler to exploit the
0.8
Formula for the cosine: cos(W)= cos2(0)-sin2(0). As a result,
combination of cos2(0) and sin2(0) that has arisen. We use the Double Angle
0.4
A= 0.8
0.4
1.2
1.6
=
_. Figure 9
Integrating Functions in Polar Coordinates
Because
1 r14 (� !) 1
210
8
-
(1+2sin(O)-cos(20)) d()=
v'2 2
4
(
e
2
-cos(O)-
-(0- - 0)= � + - v'2 4 8 2·
�
. sm(20) 4
) 17r/4 0
....
ff R1 dA and ffR1 dxdy represent the area of a planar region 'R when we
are working in rectangular coordinates, we sometimes use the expression
of area
element
to describe dA and dxdy. Theorem 2 tells us that, in polar coordinates, the
element of area is r drdO. Suppose that 'R is a planar region that is described in
terms of polar coordinates by a set S. Let (r,0) f-t g(r,0) be a continuous function defined on S. That is, g is a function of the polar coordinates of the points of 'R.
Then we can integrate g over S using the element of area r drdO. That is, we can consider the integral
r
=
ff8 g(r,
0) r dr d(). Observe that the element of area in polar
coordinates contributes a nontrivial factor, namely r, to the integrand.
cos(30)
y
� E X A M P L E 5 Let 'R be the region in the plane that is enclosed by the
Let S be the set of polar coordinates of the points of 'R. Integrate g(r,0)
over S using the element of area in polar coordinates.
��
---
=
rightmost petal of the three-leafed rose given in polar coordinates by r= cos(30).
� x
-----
r sin2(30)
Solution The plot of r= cos(30) appears in Figure 10. We need to find the range of () for the rightmost loop. Because the origin is the "initial" point and "endpoint"
of the loop, we must find the values of () in the first and fourth quadrants for which
cos(30)= 0. These are ()= 7r/6 and ()=-Jr/6, respectively. As () increases from
-Jr/6 to 0, the value of r = cos(30) increases from 0 to
1.
That is the bottom half of
the petal. As () increases from 0 to 7r/6, the value of r= cos(30) decreases from 1 to
0, generating the top half of the petal. The set of polar coordinates of the rightmost
_. Figure 10
petal is therefore S={(r,O): -7r/6:50:57r/6, 0:5r:5cos(30)}. This is the region
of integration. The requested integral is
11
s
g(r, 0) r dr d() =
1-7r/6/ 1cos(39) 7r117r6/6 -7r/6
= 3
0
r sin2(30) r drd()=
sin2(30)cos3(30) d().
1-7r7r/6/ ( 6
sin2(30)
1cos(39) ) 0
r2dr
d()
12.5 Integrating In Polar Coordinates
1005
We have learned the technique for evaluating such integrals in Chapter 6. We write os3( 36) = cos ( 36)cosl1'8) = ( 1 - sin2�3t.t) ) osl 38) 11.nd make the substitution sinf5tJ),du=3cos(36') dO. The integral becomes
2
c
r.1
-
i:
Sometimes an integration problem originally presented in rectangular coordinates i;; an be tackled more easily in polar coordinates. We cannot just write x = r cos1: fl') and y = r sinl.tJ) and use/( r cos! fJ ),,, Sin� 0)) instead of.f�x,y). We must take into account the i:i lements of area,which are different in the two coordinate systems. In other words, we must remember that the element of area dxdy in rectangular coordinates becomes r drdO in polar coordinates. It follows that, when we convert an integration problem from rectangular to polar coordinates, the iterated integral ff;;.. f(x,y),Jxdy becomes JJ8 fl. r � 9),r sinl 9)1 drd9. We will give a more formal tteatment of the change from rectangular coordinates to polar coordinates later in this section. For now, we state the resulting theorem and tum to some examples. THEOREM 3
Let tx,y) fb,y) be a continuous function that is defined on a planar region 'Fl. Suppose that 'R. .:an be described in polar coordinates by S={(r,O):asOsp, !f0[t>)s,,.s.;1tt1)} where Osp-asltr and Os ip0(6)scp�(9) for all 8. Then. the integral of/ over 'R, is given by L....+
flf(x,y) dA lt .. '1'1{� f(r coslrit),rsin[O)) =
n
'R.
� EXAM PL E 6 Calculate the volume boloid z = 2x2 + 2y2-8 and the xy-plane.
.&. Figure 11
r drdO.
V
of the
solid
2
121rf
2
_
J
bounded by the para
Look at Figure 11. The paraboloid intersects the xy-plane in the circle whose equation is 2x2 + 2y2-8 = 0, or x2- + y2 = 4. Thus we calculate the volume by integrating the height function f(x,y) =0-(2x2 + 2 y -8) =8-2(x2- + y2) over the disk S that is bounded by the circle with radius 2 that is centered at the origin of the xy-plane. We can accomplish this by integrating the function f(rcos(O), r sin (9)) =8-2(r2cos2(8) + r2 sin2(6) ) =8-2r2 over the region S= {(r,0): OsfJS27r,OSrs2} using the polar element of area. We obtain Solution
V
(12.S.8)
o(.,)
-2r2) rdrd(J
21rr2
fJ
Br-2�··) drd(•
/211"4 2
'2 __
4
,4
) 1,.=2a.· r=O
1 006
C: hapter 12
Multiple Integrals
INSIGHli
Rectangular coordinates can also be used to calculate the volume of the
solid in Example 6. The required integral is V=
/�2 J!j�:
8-2(x + y2)) dydx.
which. with some effort, does evaluate to 167r. The calculation is decidedly simpler in polar coordinates because of the symmetry about the z-axis. Problems involving circular symmetry are often best solved using polar coordinates.
� EXAM P L E 7 et up the integral for calculating the volume V f the solid U that consists of the points inside the lower branch of the cone (z-3)2 x2 + y2, above the xy-plane, and inside the circular cylinder x2 + y-1)2=1.
Look at Figure 12. Notice that the solidU lies entirely inside the cylinder, while its upper boundary is determined by the lower branch of the cone. and its lower boundary by the xy-plane. The equation for the lower half of the cone is
Solution
z=3-(x2 +
y2)112• Therefore we must integrate the function f(x,y)= 3-(x2+ y2)112 over the disk that is bounded by the circle x2+(y-1)2=1. This circle is the graph of the polar coordinate equation r=2 sin(0) for 0s0s"If, as we have seen in Example 7 of Section 12.4. Therefore we can calculate V by using polar coordinates, integrating the function f(r cos (0), r sin(0))=3-r over the region S= {(r, 0) : 0srs11', 0srs 2 sin(O)}. The volume of U is V=
J.'l
26in.(
3-r) r drd(}
-r2) drdO.
Therefore
The interested reader may verify that the value of Vis 37r-32 9. (Use equation 6.2.8) for the integration of sin2 ) and refer to Section 6.2 of Chapter 6 for the integration of sin3 0).) z
,Z
x
.A Figure 12a
Change of Variable and the Jacobian
.A Figure 12b Viewed from below
discussion of the element of area in polar coordinates suggests an important question: How does a change of variable affect a double integral? That is the problem that we now investigate. Suppose that S is a planar region on which two ur
12.5 Integrating in Polar Coordinates
1 007
differentiable functions ¢ and 'ljJ are defined. These are the functions that we use to effect the change of variables on S. Let R be the region defined by R = {(x,y) : x = ef>(u,v)
and
y = 'lj;(u,v) for some point (u,v) in S}.
Figure 13 shows that we can imagine ¢ and 'ljJ transforming the region S into the region n. In other words, we define a function T: S --+ IR2 by the equation T(u,v)= (ef>(u,v),'l/;(u,v)) and define R to be the image of Sunder T. That is, R= {(x, y) : (x,y)= T(u,v)= (ef>(u,v),'l/;(u,v)) for some point ( u, v) inS}. v
y
T •
(u, v)
u
x
_. Figure 13 The equation (x,y)
= (
We assume that T is one-to-one so that, given a point (x,y) inn, there is a unique point (u, v) in Sfor which T(u, v)= (x,y). In other words, T is an invertible function from s onto n. Because the components, ¢ and 'lj;, of T are differentiable, we can form the matrix
D(T)(u ' v)=
[
ef>u(u,v) 'l/Ju(u,v)
]
ef>v(u,v) 'l/Jv(u,v) '
which is called the Jacobian matrix of T. The determinant of this square matrix,
h(u,v) = det
( [ !:��'. �� ���'. ��])
= ef>u(u,v) 'l/Jv(u,v) - ef>v(u,v)'l/Ju(u,v) = (ef>u'l/Jv - ef>v'l/Ju)(u,v)
(12.5.9)
is called the Jacobian determinant (sometimes abbreviated to just the Jacobian) of T. The classical notation
��::��
for the Jacobian determinant is still encountered
frequently. The next example will reveal that the Jacobian determinant has been present all along in the integrations that we have been performing. �
EX A M P L E 8 Let n be a region in the plane. Let S be the set of polar
coordinates of points ofn. Let T: S--+R be defined by T(r,O) = (rcos(O),rsin(O)). Show that
jj R
1dxdy=
jj S
1 IJT(r,O)I drdO
.
(12.5.10)
1 008
Chapter 12
Multiple Integrals Solution We calculate the Jacobian matrix of
D(T)(r,0) =
[
a a (rcos(O)) ;
-(rsin(O))
ar
T as follows:
a (rcos(O)) ao a -(rsin(O)) ao
l
=
[
c?s(O) sm(O)
-rsin(O) rcos(O)
] ·
The Jacobian determinant is given by lr(r, 0) =det
( [ ��:f:? ::�:(�) ] ) =
cos(O)rcos(O) - (-rsin(O)) sin(O) =r(cos2(0) +sin2(0)) =r.
-
Using this result together with Theorem 2, we have
jj 1 dxdy =jj r drdO =jj 1 llr(r,O)I drdO. 'R,
s
s
The next example shows that the Jacobian determinant plays a role in area calculations even when polar coordinates are not involved. <1111
(u,v) in S={(u,v):0::5u::5l ,0::5v::5l}, let
9 For a point
jj 1 dx dy =jj lllr(u,v)I dudv. 'R,
(12.5.11)
s
R with coordinates (x,y ) =(2u+v,u - v) , then we can solve the equations x =2u+v,y =u - v simultaneously for u and v. We find u = (x+y) /3 and v = (x - 2y) /3. Because this solution is unique, we have shown that T: S--+R is invertible. If we write the position vector (x,y) of P as (2u+v,u - v), or u (2,1) +v (1, -1), then we see that, as u and v take on all values between 0 and 1, the point P runs through all points in the parallelogram R that has sides (2,1) and (1, -1). See Figure 14. If we write these two sides as spatial vectors that have 0 as the third coordinate, then we calculate Solution If Pis the point in
Area of R = II(2i +lj +Ok)
X (li +(-l)j +Ok)
11 = llO i+Oj +(-3)kll =3.
Also,
D(T)(u,v)
=
[
a
au
'lj;(u,v)
:v
(2u+v)
:v(2u+v)
a (u v) au -
a (u v) av -
l [! =
l
=
[
2 1
1 -1
l
·
12.5 Integrating in Polar Coordinates y
1 009
y
l i-----.
s x
x
1
_. Figure 14
Therefore JT(u, v) = (2)(-1) - (1)(1) = -3. Because the area of Sis 1, we have
JJ 1dxdy=3 = (3)(1) =3JJ 1dudv=JJ l lJT (u, v)Idudv, R
S
S
which is equation (12.5.11) . .,.. Theorem 4 shows that equations (12.5.10) and (12.5.11) are special cases of a general change of variable formula. As you read Theorem 4, you may find it helpful to refer to Figure 15 for a schematic diagram of the functions T, f, fa T, and JT that appear in the theorem.
v
y
T
u
x
f Image ofJT
Image of f
_. Figure 15
THEOREM 4
Suppose that T(u, v) = (
1 01 O
Chapter 12
Multiple Integrals
ff f(x,y) dxdy= ff f(¢(u,v),1j!(u,v))lh(u,v)I dudv,
n
(12.5.12)
s
where
is the Jacobian determinant of T.
Consider a partition of R. Let the lengths and widths of the rectangles in this partition be D.x and D.y, respectively. Let ei,j be a point in the i,J1.h rectangle in the partition of R. These choices determine a Riemann sum '''EJ(ei)&D.y for f over R. Let 'T/i,j be the point of S such that T('T/i) ei,j· Our job is to show that the Riemann sum L,f(T('TJi)) lh('T/ij)ID.uD.y over S corresponds to L,f(eij)&D.y. Suppose that D.u and D.v are small increments in u and v. Then, from formula (11.7.7), we have fu � ¢uD.u + ¢vD.v and D.y � 1f!uD.u + 1f!vD.v. Interpreted geo metrically, this means that a small horizontal displacement in the xy-coordinates, which we may represent by the vector(&,0), corresponds to a displacement in the uv-coordinates given by the vector (¢uD.u,¢vD.v). Similarly, a small vertical dis placement in the xy-coordinates, which we may represent by the vector (0, D.y), corresponds to a displacement in the uv-coordinates given by the vector (1f!uD.u,1f!vD.v). In Figure 16, we see that the rectangle with sides fu and D.y in the xy-plane corresponds to a parallelogram that is determined by the vectors(¢uD.u,¢vD.v) and (1f!uD.u,1f!vD.v) in the uv-plane. Therefore the element of area &D.y in the xy-plane corresponds to the area M of the parallelogram that is determined by the vectors (¢uD.u,¢vD.v) and (1f!uD.u,1f!vD.v) in the uv-plane. To calculate M, we think of the uv-plane as a coordinate plane in three-dimensional space. Then, as we learned in Section 9.4 of Chapter 9, M is equal to the length of the cross product of the two vectors (¢uD.u,¢vD.v,0) and (1f!uD.u,1f!vD.v,0). As a result, we have Proof.
=
y
v
u
!
B
A (
Figure 16 The parallelogram determined by vectors =
x
=
12.5 Integrating in Polar Coordinates
From this computation, we see that the Riemann sum
'L,f(�i,j)bi..xbi..y
1011
for f over R
corresponds to the Riemann sum in uv-coordinates given by
L.J(T(T/j)) I (c/JuV;v -V;uc/Jv)(T/i) lbi..ubi..v
·
(Remember that, in the algebra of functions, (c/>uV;v -V;uc/Jv)(ry) means c/Ju(T/)V;v(T/) -V;u(T/)c/Jv(T/).) Passing to the limit, we obtain equation (12.5.12). We have already used Theorem case of Theorem
4 several times. Indeed, Theorem 3 is the special
4 in which T transforms polar coordinates to Cartesian coordinates. 4 that is new to us.
We conclude this section with an application of Theorem �
EX A M P L E 1 0 Evaluate the integral
x+Y) 11(x/3)+2 (- dydx 2 x/3 0
using the transformation
u = x/3, v = (x+y)/2.
Solution In this problem, the region R is given by
2,0
{(x,y): x/3
We begin by rewriting the transformation in a form that fits the
x = 3u. The second equation is y = -x+2v, or y = -3u+2v. Our transformation T is therefore x = cp(u, v) = 3u, y = V;(u,v) = -3u+2v. To find the region S that corresponds to R, notice that 0
transformation is
JT(u,v) = (c/>uV;v -V;uc/Jv)(u, v) =(3)(2) - (-3)(0) = 6,
y
1
u
1
� Figure 17 The transformation x =
x
1
2
3
(u, v) = 3u, y = 'lj;(u, v) = -3u + 2v
1 01 2
Chapter 12
Multiple Integrals
the integral becomes
f31(x/3)+2 ( +2 y ) dydx = fl 12u+l 6v dvdu=6 fl ( ; 1v=2u+l) du lo x/3 lo Zu lo v=2u 1 = 3 1 (4u+l)du=3 ((2u2+u)I�) =9. x
<1111
Q UIC K
Q UIZ
1. 2. 3.
What is the area enclosed by the polar curve
4.
What is the Jacobian determinant
r= y'cos(9),
-
/2 '5, 9 '5, 7r/2?
7r
What is the element of area in polar coordinates? Use polar coordinates to calculate the integral of f(x, y)=x2 + y2 over the region that lies outside the unit circle and inside the circle
h(u, v) of the
x2+ y2 =4.
transformation
T(u, v)= (u3 + 4v, 3u+ 2v)? Answers
1. 1
2. r drde
3. l57r/2
4. 6u2- 12
EXERCISES Problems for Practice In each of Exercises 1-6, use formula (12.5.2 ) to evaluate the area enclosed by the given curve.
radius5 and outside the limac;onr=3 + cos(O).
16. R= {(r,O): 7r6 / s:Os:7r/4, sin(O) s:rs: cos(O)} 17. R is the region outside the cardioid r= 1 + cos(O) and
2. r= 7 -3 cos(O) 3. r=4sin(30)
inside the limac;on r=4 - sin(O).
4. r= 5 sin(W) 6.
18. R is the region inside the circle r=4cos(O) and outside
r2 = -sin(O) r2 = cos(0)
the right hand leaf of the rose r= cos(W).
In each of Exercises 7-12, use formula (12.5.4) to find the area of the region that is described.
r= 1 + cos(O).
8. The region in the first quadrant between the graphs of r=3 sin(W) and r=4 - cos(0). region
enclosed
by
the
In each of Exercises 19-26, integrate the given function f over the given planar region R, using a double integral in polar coordinates.
7. The region between the graphs of r=6 - sin(0) and
9. The
14. R is one leaf of the rose r=3 sin(4 0). 15. R is the region inside the circle with center the origin and
1. r= 1 - sin(O)
5.
13. R= {(r,O): -11/4s:Os:7r/4,2 s:rs:4cos(O)}
curves r= sin(O)
and
r= cos(O).
10. The "lune" outside the cardiod r= 1 + sin(O) and inside the circle r=3 sin(0).
19. f(r0) , = cos(O); R is the region inside the circler= cos(O). 20. f(r,0) = sin2(0); R is the region inside the limac;on r=3 - cos(O).
2L f(r0) , = cos(O); R is the region inside the leftmost petal of the rose r= cos(W).
, ) =0; R= {(rO , ): Qs:Os:7r/2,0 s:rs: cos112(0)} 22. f(rO 23. f(r0) , = sin(O); R is the region inside the circler= sin(O).
lL The region in the first quadrant that is inside the circle
, =r; R= {(r0) , 24. f(r0) : 0 s:0s:7r/2,0 s:rs: sin(O)} 25. f(r0) , =3r- 10; R is the region outside the circle r =3
12. The region that is inside the circle r= v'3 cos(0) and
26. f(r0) , = sin(O); R is the region in the first quadrant that
r= v'3cos(O) and outside the circler= sin(O). outside the cardioid r= 1 + sin(O).
In each of Exercises 13-18, calculate the area of the given region R using a double integral in polar coordinates.
and inside the limac;onr=4- cos(O).
is inside the circle r= 1 and outside the cardioid r = 1 -
cos(O).
Integrating in Polar Coordinates
12.5 In each of Exercises
27-38, use a double integral in polar
coordinates to determine the volume of the solid.
27. The solid bounded by the surface z =-x2-y2 + 4 and
50. The region between the two loops of r= J3 -2cos(O) 51. The region inside the smaller loop of r=1 -2 sin( 0) 52. The region in the first and second quadrants bounded by the x-axis, the spiral arc
xy-plane
the
arc
28. The solid bounded by cylinder (x-2)2 + y2=4, the paraboloid
z=x2 + y2 + 6,
and the
xy-plane
29. The solid bounded by the paraboloids z=3x2 + 3y2 -7 and z=-x2 -y2 + 9 30. The solid bounded by the paraboloids z=2x2 + 2 y2 + 4 and
z=-4x2 -4y2 -4
and by the cylinder
x2 + y2=16
r=0
Perform the integrations in Exercises
53. 54.
57. Integrate
34. The solid bounded by the upper nappe of the cone 35. 36. 37. 38.
z2= 4x2 + 4y2
and by the plane
z
12 The solid bounded by the cylinder x2 + y2=4 and by the hyperboloid z2 -x2 -y2=4 The solid bounded by the the sphere x2 + y2 + z2=20 and by the paraboloid z=x2 + y2 The solid outside the cone z2=x2 + y2 and inside the cylinder x2 + y2 =4 The solid bounded by the cylinder x2 + (y - 1)2 =1 and the sphere x2 + y2 + z2 =4 =
53-56 by converting
v'4-x2 J-2 2 J-v' 4-x2 (x2 + y2 + 1) -1/2 dydx 1/2 fo1 fo(1-x2) cos(x2 + y2) dydx
33. The solid bounded by the hyperboloid z2 -x2 -y2=1
z =4
for 0 � 0 � 7r, and the spiral
them to polar coordinates.
56.
and by the plane
r= 0
for 27r � 0 � 37r
31. The solid bounded by the upper nappe of the cone (z + 3)2=4x2 + 4y2 and by the paraboloid z=5 -x2-y2 32. The solid bounded by the paraboloids z =4x2 + 4y2 + 2
andz=7-x2-y2
1 013
1 � (x2 + y2) /3 dxdy
55. f-_3/1_
f�3f0� (1 + x2 + y2)3/2 dxdy R ={(r,
the
0):
f(r, 0)=0
function
58. Integrate the function f(r, 0) R= {(r,
0) :
over
the
region
0 � 0 � 7r/2, 0 � r � cos(O)}. =
cos(O) over the region
0 � 0 � 7r, 0 � r � O}.
59. Calculate the area in the first quadrant bounded by the
y=x, y=2x
lines
and by the curves
xy=2
and
xy=4.
After setting up the integral, simplify matters by using the change of variables
u=xy,v=y/x.
60. The convergent improper integral
I=
l: e(-x2) dx
39-44, perform the indicated change
arises in probability, statistics, error analysis, and many
of variable in the double integral and then evaluate the
other topics. It cannot be evaluated by the techniques of
integral.
Chapter
In each of Exercises
39. 40. 41. 42. 4 3. 44.
Jg J;/f4 (x-y)2 dxdy; u =(x-y)/2,v =y/2 0 16-x2 xy dydx; x=r cos(W), y=r sin(W) ta Jv' Ji Ji y2/x2 dxdy; u =l/x,v =y ta fy��2)+l (2x-y)/2 dxdy; u= (2x-y)/2,v=y/2 JgJ; J4-xy dydx; u=xy,v=y Jg6 fov'X ../YfXdydx; u2=x,v2=y
Convert to polar coordinates to evaluate this iterated integral. Then, solve for I.
61. Use polar coordinates to determine for which real values of
a
the improper integral
45-52, find the area of the region that
enclosed
by
both
r=1
+ sin( 0)
and
47. The region inside the cardioid r= 1- cos(O) and also inside the circle
r=cos(O)
lim
R-->oo
r=
1/2
11 R
R
-R
-R
1 dxdy. (l + X2 + y2)"
62. Prove that, for any real value of
1j
+ sin( 0)
49. The region inside both the rose r=cos(W) and the circle
oo
1 dxdy 1 + x ( 2 + y2)"
converges. Here, you should understand this integral to
mean
48. The region inside both the circle r = J3 cos( 0) and the
cardioid r=1
100100 oo
r=4- cos(O) region
oo
=
45. The region between the graphs of r=3 sin(W) and r=3 sin(O)
(l: e(-x2)dx) u� e<-rldy) l: (l: e<-x2Jdx) e<-Y2ldy l: l: e-(x2+y2) dxdy.
=
is described.
46. The
Instead, write
p=
Further Theory and Practice In each of Exercises
6.
converges.
oo
oo
-oo
-oo
exp (-(x2
a,
the improper integral
+ y2)) dxdy + 1 + x 2 y2)" (
1 014
Chapter 12
Multiple Integrals
Calculator/Computer Exercises Find the value 00 of 0 in [O,7!/2] for which 0 3 cos2(0). Calculate the area of the region in the first quadrant that is outside the curve r 0 and inside the curve r 3 cos2(0) for Os O s 00• 64. Find the value 00 of 0 in [O, 7l/2] for which 2 + 0 e9• Calculate the area of the region in the first quadrant that is between the curves r 2+0 and r e9 for 0 s0 s 00•
63.
=
=
=
for double integrals. We do so by using the product f(rcos(O) ,rsin(O)) r as the integrand in the approximation formula. In Exercises 65 and 66, use the Trapezoidal Rule with N = 4 to approximate the given integral. (Refer to the instructions for Exercise 64-67 from Section 12.2 for a dis cussion of the Trapezoidal Rule for double integrals.)
=
=
=
If an iterated integral J!Jt;(�/ f (rcos(O) ,rsin(O)) r drdO arises from integration in polar coordinates, then we may estimate its value by applying any approximation technique
The integral of f(r, 0) 1/(1+0 + r3) over the region bounded by the y-axis and the spiral r 0, 0 s0 s 7l/2 66. The integral of f(r,0) V7r + 0 + .ji over the half-disk Qsrsl, - 7r/2 sO s 7r/2
65.
=
=
=
1 2.6 Triple Integrals In this section we discuss the theory of the triple integral. In fact, integration can be done in any number of dimensions. We concentrate here on three dimensions because that is sufficient to study the basic properties of bodies in space.
The Concept of the Triple Integral
Let U be a solid in space and fa continuous function defined on U. We want to
f over U.
develop a concept of integrating
This will parallel the development in
Section 12.2, so we will be brief. Suppose that U is contained in a box
e � z � f} .
Q
=
{(x,y,z) : a� x � b,c� y � d,
We partition each of the intervals [a, b], [c, d], and [e,f] into N equal subintervals, giving rise to N3 small boxes Qj,k,t· We select a point �j,k,t in each Qj,k,t and form the Riemann sum
L f(�j,k,t) LUL'.lyL'.lz,
Q;,ktCU or
L f(�j,k,t) LlV,
Q;,ktCU where
LlV = (LU)(Lly)(Llz) is the increment of volume. As in Section 12.2, we sum
only those boxes that are contained in U. If the geometry of U is relatively simple (a
concept that is made precise in the upcoming Theorem
1), then it is known that the
Riemann sums will have a limiting value as N tends to infinity. The limiting value of these Riemann sums is called the
triple Riemann integral. This integral is denoted
by the symbol
jjj f(x,y,z)dV. u
The most important fact for us about triple integrals is that we have a means of evaluating them when the solid U has a regular form. The next theorem tells us how to calculate triple integrals in terms of iterated integrals.
12.6 Triple Integrals THEOREM 1
Suppose that
a1,a2,"Yt>
1015
and "Yz are continuously differentiable
{31
functions of one variable. Suppose that
and
{J2
are continuously differ
entiable functions of two variables. Assume that U1 and U2 are solids in space
given by
U1={(x,y,z): asxsb,a1(x) sy Saz(x),{31(x,y) szs{J2(x,y)}, and
If fit. a continuous function defined on U1, then
l�if/(x,y,z)
dV =
U1
1b1aoi(z)iP-i(z,y)f(x,y,z) tkdydx.
11
.81(x,y)
If fit. a continuous function defined on U2, then
l�if f(x,y,z)
dV=
Ui
r
11..,,(yia,Bi(z,y) f(x,y,z) dzdxdy. (y) c:
H1(%,y)
'Yt
Solids of the type described in Theorem 1 are called z-simple. Figure 1 shows a solid of the form U= {(x,y,z): csysd,-y1(y) sxs12(y),,81(x,y)
z-simple
s z s fi2(x,y)}. The
y
iterated
A Figure 1
cut-away image of U that is given in Figure
box of volume 6.V that arises in the definition of integral
J:.fri1g/ Jg:(�;{ f(x,y,z) dzdxdy,
the
2 reveals a typical f(x,y,z) fffu dV. In the inner
integral,
namely
J%:c �:/ f(x,y,z) dz, integrates up the vertical column that is shown in Figure 2. This result is then integrated, by means of J;:�{ . dx, along the strip that it. depicted in .
2. The result y=c to y=d.
Figure
.
of this calculation is then integrated over all such strips from
An analogue of Theorem 1 is true for x-simple and y-simple solids, which are
defined similarly. You will see all three of these types of solids in the examples that follow.
� EXAMPLE 1 Integrate the function f(x,y,z)=4x-12z over the solid U={(x,y,z): 1
Solution By Theorem 1, we have
fuff
[ lixl: (4x-12z) dzdydx. 2
f(x,y,z) dV =
x
We begin by evaluating the inside integral, and then we work our way outward. The last line equals
1018
Chapter 12
MuHiple Integrals
{:: 1212x = 12121-a ( = 12 12Jt; = 12 6xy) i::� = 12 -8x3) (4xz-6z2
_%
dydx
2)
(4xy-6y2) - 4x(y- x) -6(y- x )
dydx
(lox2-12xy) dydx
(1ox2y(
dx
dx
= -2x41� =-30. .... If we take the function f that is being integrated to be the constant function the Riemann swns
f(x,y,z) = 1. then
I:
1cej,k,t)1:1v=
Q1,1o1CR.
z
I:
1:1v
'21.k.tCR.
form a good approximation to the volwne of U. This observation motivates the following definition:
�.!@@i[.Jll
The volume of a solid U is defined to be
ffful dV
when the
integral exists.
Here is an example: � EXAM PL E 2 Compute the volume V of the three-dimensional solid U that is bounded laterally by the parabolic cylinders x y2- 2 and x -y2 + 6, above by
A Rgure3
y
=
the plane
+ 2y + 4, and below by the plane
Solution The solid U is shown in Figure
(2, -2)
3.
z =-x - 4.
Observe that the graphs of x
=y2- 2
and x = -f + 6 are vertical sides of U. The projection of U into the xy-plane is the region bounded by the parabolas x = y2- 2 and x =-y2 + 6. These parabolas, shown in Figure 4, intersect at the points (2, 2) and (2,-2). From Figure 4, we see
(2,2)
x
A Rgure4
z =x
=
that they-variable ranges between-2 and 2. So our outer integral will be
J� ..
. dy.
Now, for each given values of y, Figure 4 shows that x ranges from f-2 to -f +6. Thus the next integral is the integral in the x-variable, and we have
121-y -2 y2-22+6
...dxdy.
Finally, for each choice of x and y, the least value of z is -x-4 (because U is bounded below by the plane z -x - 4), and the greatest value of z is x + 2y + 4 =
12.6 Triple Integrals
1 01 7
(because U is bounded above by the plane z - x + 2y + 4). Our formula for the
volume finally takes the form
Working from the inside out, it is now straightforward to evaluate this integral: It quals
+6 ( j22j-y2 2-- 2 121 y2+6 2 2-2
V
x+ 2y+4)- -x- 4)) dxdy
2x+ 2y+ ) dxdy
=
After some simplification, we obtain v
=
1:
96+l6y- 24y2 - 4y3) dy
12 2
y2 -
96y+
=
x2+ 2. y+4)x)
sy3
-
=-y2+6 1xx=y22
y4) �
2
=
dy.
256.
.,...
-'.:!I
Take careful note of the analysis that we used in Example 2 to set up the triple integral. This is the key to success in triple integral problems: Work from the utside in to set up the integral, but work from the inside out to evaluate the integral.
..,_ EX A M P L E
.
=25
L
.& Figure 5
f z
3 Set up, but do not evaluate, the integral of
x,y,z) over the solid U bounded by the paraboloid z =x + 21+4y. Treat U
s a z-simple solid.
y2
x,y,z)
-+
and the plane
Solution The graph of U appears in Figure 5. When we consider U as a z-simple solid, we think of x and y as the outside variables of integration and z as the inside variable of integration. Let 'D be the projection of U to the xy-plane. If the point (x, y,
y20)
y2y2
belongs to 'D, then we integrate z from
x2 + =:: 21+4y and, on the boundary of x2+ (y- 2)2 25. Thus 'D is the disk in the =
with center
x2 +
'D, x2+
=
to 21 + 4y. In particular,
21+4y, or, equivalently,
xy-plane that is bounded by the circle
(0, 2) and radius 5 (see Figure 6). We see that y ranges from
-3 to 7.
We conclude that the integral will have the form
/7
•
•
•
3
dy.
For each choice of y in this interval of integration, x ranges from
-J25- . y -: )2,
r -J21+4y- y2, to J21+4y- y2• Therefore our integral will have the form
I=
.& Figure 6
...
-/
I
21+4y-y2...,. 111Jr-J21+4y-y2 -
3
After y
form
nd x are chosen, z ranges from
-·
·
.. �xdy.
x2 +
y2
to 21+4y. Our integral takes the
1 018
Chapter 12
Multiple Integrals
1
1 y'21+4y-y2
1-3
1
21+4y
� x2+y2
f(x,y, z) dzdxdy.
� EXAM P L E 4 Set up the integral from Example
3, treating Uas an x-simple
solid.
Solution Refer to Figure 5 again and imagine the projection 'P of U into the yz
z
plane. In this case, the projection Pis actually the intersection of U with the yz
x
0, we below by the parabola z
plane. Setting
see that P is the region of the yz-plane that is bounded
=
=
y2
x=O
z 21+4y (see Figure 7). Our y ranges from y -3 to y 7. We
and above by the line
analysis proceeds as follows: First, the variable
=
=
=
conclude that the integral will have the form
7
1-3 ... dy.
y
Next, for each choice of yin the interval of integration, the
z -z -z
=
y2
z
to
21+4y.
=
z
variable ranges from
Therefore our integral will have the form
=21 4y, x =0 =y2,x= 0 +
7
21+4y
1-3 1y2
.A Figure 7
...dzdy.
z, the variable x ranges from the x2 +y2 on which x - Jz - y2 to the other side of the Jz y2• Our integral therefore takes the form
Finally, for each choice of y and paraboloid on which
x
z
=
=
=
side of the paraboloid,
-
7
21+4 �
1-3 1y2 1-� f(x,y,z) dxdzdy. Q UIC K
Q UIZ
1. Let U be the solid in the first octant that lies underneath the plane
3x+4y+2z
=
12.
Set up (but do not evaluate) an iterated integral for the
dzdydx. (0,0,3) and radius 1. Express JJJuf(x, y, z)dV integral with volume element dzdydx (but do not
volume of U using the volume element
2.
Let Ube the solid ball with center by means of an integrated evaluate the integral).
3. 4.
Calculate
0:sz:s1 }.
Jffu12xy2z3dV,
where U
=
{(x,y, z): 1:sx:s2, -1:sy:s1,
Let U be the solid that lies below the paraboloid plane
z
=
5. Set up the triple integral
z
=
9
JJJuf(x, y, z)dV
- x2 - y2
and above the
as an iterated integral.
Answers r4 r3-3x/4 r6 1 . Jo Jo Jo
3. 3
-3x/2-2y 1 dzdydx
l Jv'1-x2 f3+y'1-x2-y2 f(x,y,z)dzdydx 2. f-1 -v'1-xz 3-y'l-xLyz
-x2-y2 f(x,y,z)dzdydx
4. t_ J!��::2J: 2
12.6 Triple Integrals
1 01 9
EXERCISES
JJJux dV
Problems for Practice
can also be expressed as an iterated integral
with y the inner variable of integration: Both
Evaluate the given iterated integral in each of Exercises 1-10.
L 2. 3. 4. 5.
6.
7. 8.
9. 10.
1216-3x13-3x/2-z/2 x dydzdx
fz J:+1J:+y (6x - 3z) dzdydx j!_2 fg+y J:-x 3x dzdxdy fi f:zg�: (x + 3) dxdydz Ji ft fo./zLx' 6y dydxdz ( J{j(z )fi (4/z) dzdxdy y Jg J:Y J;+y 8xyz dxdzdy _r;l2 J; fx z 2cos(y) dydzdx J�; Jgy J� l6y exp(-z2) dxdzdy .fo J!;fx 4ycos(x2) dzdxdy .fo/2 j!_5fo0s(y) 3J1 + sin(y) dxdzdy
0 0
and
16 12-z/313-3x/2-z/2 x dydxdz 0
Evaluate
the
volume
solid that is y-simple, then express the integral as an iterated integral with y the inner variable of integration. Draw a sketch of the domain of integration of each iter ated integral in Exercises 26-31.
by
means
of
iterated
integrals.
lL z = x2 + y2, z =8, z = 2
•
32.
integral. Describe the solid U over which the integration is performed.
24.
c
are positive constants. Use a
x2/a2 + y2/b2 + z2 /c2 = 1.
z =x2 + y2 and the z =ax + by+ c, a> 0, b> 0, c > 0.
the solid bounded by the paraboloid plane
c
be positive. Use a triple integral to find a
x/a + y/b + z/c = 1. f(x,y,z) =x2 + y2 + z2 over the solid that is bounded above by the cone z = mJx2 + y2 and below by the paraboloid z =x2 + y2. off by the plane
35. Integrate
36. Use a triple integral to find the volume of the solid
bounded by the cone z2 =x2 + y2 and the planes z =a and z =b, O
Further Theory and Practice
Jffux dV over a
and
formula for the volume of the solid in the first octant cut
X
integral
a, b,
33. Use a triple integral to give a formula for the volume of 34. Let a, b, and
Jg Jg-3x/2 to-3x-2y dzdydx 6 to to J;Y1 z dzdydx J� fov'l=X' J/l -x,_y, xz dzdydx J� f0.Jl=X2 J� -y' x dzdydx J� J; fov'l=X' z dzdydx J� Jo� J/x'+y' x(l + y) dzdxdy
25. The iterated integral of Exercise
Suppose that
triple integral to evaluate the volume inside an ellipsoid of the form
In each of Exercises 19-24, evaluate the given iterated
20. 2L 22. 23.
Jg fx2x+l J:f(x,y, z) dzdydx 27. t.1f;+3 J::;t(x,y,z) dzdxdy 4 28. f_1f,Jgx+ f(x,y,z) dydxdz 29. j!_3 J�;�;�5f(x,y,z) dzdxdy l J ./4-4x2 J2+x+z f(x,y,z) dydzdx 30 f-1 -./4-4x2 -4-x-z 3L fi h J: f(x,y,z) dzdxdy 26.
f�
z =x2 + y2, z =8 - x2 - y2 13. y=x2, z=l-y, z = -l + y 14. x + 2y + 4z =8, y = x/2, x = 0, z = 0 15. x2 + y2 = 4, z = x + y, z = -6 - x 16. y = x2' z = -y + 4, z = 0 17. z = -x2 - y2 + 9, z = 2x + 4y - 2 18. y = x2 + z2 - 3, y = -x2 - z2 + 5 12.
19.
0
do the job. Examine each iterated integral in Exercises
In each of Exercises 11-18, use a triple integral to express space.
0
20-24. If the integral represents a triple integral over a
the volume of the solid that is bounded by the given surfaces in
0
19
In Exercises 38 and
represents a triple
solid U that is z-simple, as we
can see from the inner variable of integration. However, the solid U is also y-simple, and, reflecting this property,
Jffu(xy + z) dV.
39, a solid U is described. Evaluate
38. U is the solid in the first octant that is bounded by the three coordinate planes, by the plane plane
3y + 2z = 6.
x =1,
and by the
1 020
Chapter 12
Multiple Integrals
39. U is the solid in the first octant that is bounded by the three coordinate planes, by the plane z = 2+7x, and by the parabolic cylinder y = 4 - x
2•
integration.
solid that is bounded below by the xy-plane, above by the
graph of
In Exercises 40 and 41, a solid U is described. Write
Jffuf(x,y,z) dV
45. Suppose that a, b, care positive constants. Let Ube the
in each of the six possible orders of
..jXfa+ JY7b+ JZ7C = 1,
and by the xz- and
yz-planes on the side. Show that the volume of U is
abc/90.
Also show that
40. U is the solid in the first octant that is bounded by the three
coordinate
planes
5x+3y+2z = 30.
and
by
the
plane
4L U is the solid in the first octant that is bounded by the planes x 0, x y, y 1, z 0, and z 1.
Calculator/Computer Exercises
42. Write the iterated integral
46. Let R be the region of the xy-plane that is bounded above
=
=
=
=
=
� (x2 -yz) 14125-x21 y'25 x2 y -4 9 -
-
by y =3-x2 and below by y dzdydx
y = 1+(x+1)3• Calculate
in all possible orders.
Integrate x3
that is bounded
z =by, and z =ay.
=
=
sin(x) and below by
bounded below by R, above by the plane z
yv'l -x dydzdx
Calculate
and by the planes x
12.7
=
x, and lat
erally by the cylinder sets y = sin(x) and y = 1/(1+x2).
c are positive constants with b >a. x2
above by y
y = 1/(1+x2). LetUbe the solid in the first octant that is
Jc2 - y2 over the solid that is bounded by the
parabolic cylinder y
Jffux2yz dV.
47. Let R be the region in the first quadrant of the xy-plane
43. Write the iterated integral
44. Suppose that a, b, and
1+(x+1)3• LetU be the
z = y, and laterally by the cylinder sets y = 3 - x2 and
-
in all possible orders.
101371v'X+f -1 2 x+l
=
solid that is bounded below by R, above by the plane
=
0, y
=
c,
Jffuz dV.
48. LetUbe the solid in the first octant that is bounded below by the plane z
=
x, above by the plane z
by the cylinder sets y = e' and y
Jffuz dV.
=
=
2, and laterally
6x-x2 -1. Calculate
Physical Applications Multiple integrals are useful for calculating a number of physical quantities. In this section, we learn how to calculate the mass of a body, the center of mass of a body, the first moment of a body about an axis, and the moment of inertia of a body about an axis.
Mass
Imagine a thin metal plate (sometimes called a
lamina) made of material that varies
in composition from point to point. We think of the plate as occupying a region n in the xy-plane and assume that it has mass density given by a continuous function
6(x,y) at the point (x,y) . In this context, "mass density" at a point (x,y) means mass
per unit area at the point (x,y). This physical concept is defined like a partial
derivative. To be specific, let M(x,y,h) be the mass of a square of side length h that is centered at the point (x,y). To obtain the density at (x,y), we divide this mass by
the area of the square and take the limit as
6(x,y)
=
1.un
+ h.....O
h tends
M(x,y,h)
h2
to 0:
(12.7.1)
12.7 Physical Applications
1 021
We now answer the following question: If the mass density of a plate is known at each point, then how can we calculate the mass of the entire plate? As usual, we think of the plate as lying inside a rectangle
c :s y :s d}.
Partitioning the intervals
we obtain rectangles
R . If
�i,;= (xi,Y;)
approximately
[a, b]
and
[c,d]
Q= {(x, y)
:
a :sx :s b,
into N subintervals each,
Qi,j· We consider only the rectangles that lie entirely inside Qi,;. then the mass Mi,; of the square Qi,; is
is a point in
Notice that we estimate the mass by multiplying
8(�i,;), which represents mass per
unit of area, by the area of the small rectangle. Limit formula (12. 71 . ) justifies this approximation. The total mass of the plate is about
Letting the number of subrectangles of the partition tend to infinity leads us to define the mass
M of the plate by M=
ff 8(x,y) dA.
(12.7.2 )
'R
� E X A M P L E 1 Suppose that a thin plate Ris in the shape shown in Figure1.
1/
y= x ( 2 -4 ) 2 and y = 2x-4 . Assume (x,y) is 8(x,y)=4x. Find the mass of R.
The bounding curves are density of Rat the point
also that the
Solution The curves intersect whenx= 2 andx=10/3. We may write Ras a simple region as follows:
Then the mass of Ris
2.2
.A Figure 1
2.4
2.6
2.8
3
3.2
M
=ff 8(x,y)dA 110/31 4xdydx 1/ {10/3 } ( ) - (2x - 1 ) dx 3 ) 1 0/3 � � ( 'R
(x2-4)1/2
2
2x-4
2
( 2 -4 4x x
2
x ( 2 -4)3/2
4)
x + 8x2
2
384
81 .
The mass of the plate is 384/81, or128/27.
-<1111
y
1 022
Chapter 12
Multiple Integrals
First Moments
Next we consider the first moments of a lamina about each axis. For a point mass, the first
moment about an axis is the product of the mass times the distance to the axis. For each i and j, the first moment of the small rectangle Qi,j about the x-axis is
y
approximately
Yi
-1
dx •ti,j Qi,j
Yj o(ei)M where ei,j is a point arbitrarily chosen from Qij· Look ·
at Figure 2. There are two approximations in estimating this first moment. The
Qi,j do not have constant distance to the x-axis, but, if Qij is small, then the distances of the constituent points do not vary greatly, and Yj can be used as an approximation. Also, we approximate the mass of Qij by o(ei)M, as usual. Summing over i and j, we obtain
�
different points that make up
Yi
L yjo(ei)M
x xi-1
xi
Q;,;r;R
.A Figure 2
as an approximation to the first moment of the plate about the x-axis. Letting the number of subrectangles tend to infinity leads us to define the the plate with respect to the x-axis, which we write as y My=O
=
JJ
=
first moment
of
0, to be (12.7.3)
y 8(x,y) dA.
'R,
Similar computations lead to the definition that the first momentof the plate with respect to the y-axis, which we write as x Mx O =
=
JJ
=
0, is (12.7.4)
x 8(x,y) dA.
'R,
� EX A M P L E 2 A metal plate 'R, sitting in the xy-plane is shown in Figure 3.
(4, 3)
(
Its density at a point (x,y) is 8 x,y) respect to the x-axis.
=
6x+9y. Calculate its first moment with
Solution First, write the region 'R, in x-simple form: 'R,
=
{(x,y)
: 1:::::; y:::::; 3,
y+1:5x:5 -y+7}. We have
(2, 1) .A Figure 3
(6, 1)
My=O
=
JJ
y8(x,y) dA -y 7
3
{ r + Ji }y+l { J1
3
3
y(6x+9y) dxdy
(3yx2+9y2x)
1 (�
1
x -y 7 = +
x=y l +
dy
(-18y3+6y2+144y) dy
-
y4 +2y3+72y2
) 1:
268. We conclude that the first moment of 'R, with respect to the x-axis is 268 units. For example, if 8 is measured in kilograms per square meter, and x and y are measured in meters, then our answer is measured in meter-kilograms.
-<1111
12.7 Physical Applications
Center of Mass
Suppose that
M, Mx=o, My=O x y Mx=O -M Mx=o, and
Define the numbers
and
1 023
denote the mass and first moments of a plate.
by
x-
(12.7.5)
Referring to the definition of
we see that x is the average value of the x
-
_
coordinate function on the plate, weighted according to the varying density func
tion 8(x,y). Similarly,
y
the mass density. The
defined to be the point
is the average y value on the plate, weighted according to
center of mass c=
(or
center of gravity)
of the plate is now
(x,y).
It is not difficult to see where the definition of "center of gravity" comes from.
The point c is determined by the property that the tendency of the plate to rotate about c in any given direction is balanced exactly by its tendency to rotate in the opposite direction. The physical significance of the center of gravity is that the plate would balance on the point of a pin placed at c. When the density function is not
mentioned explicitly, then we assume that mass is "uniformly distributed" (i.e.,
8(x,y)
=
80, a constant), and we speak of the "centroid" of a region. In this case,
notice that the constant density 80 can be brought outside each of the integrals that
M, M =o, My= · x O y My=olM.
define
and �
and
=
The constant 80 then cancels in the ratios
x=
Mx=o/M
That is why the value of a uniform density is often not specified.
EX A M P L E 3 Consider a plate R in the xy-plane bounded above by the
curve y = 12-3x2 and below by the x-axis. Assume that the plate has constant
density 8(x,y) = 1. Calculate the center of mass of the plate.
Solution By setting y = 0 in y = 12-3x2 and solving for x, we see that the curve
that forms the upper boundary of R crosses the x-axis at x = ±2. Thus R is the y simple region {(x,y): -2
M ff =
8(x,y) dA
'R,
=
1
_
2
f
0 2}
12-3x2
1 dydx
=
1
_
2
(12-3x2) dx
2
=
32.
Further, we know that the first moment with respect to the x-axis is
My=O ff 1(
y8(x,y)dA=
=
'R,
=
2
-2
12112-3x2 ) -2
2-x4 -36x2+12 2
0
dx =
21y=12-3x2 1 y=O 2
Y -2 2
y dydx=
768.
dx
5
Similarly, the first moment with respect to the y-axis is
Mx=O ff =
'R,
x8(x,y)dA=
{12-3x2 1 Jo 2
_
2
x dydx=
1
2
_ 2
(12-3x2)x dx = 0.
In physical terms, the first moment with respect to the y-axis is equal to 0 because
the plate is symmetric with respect to the y-axis. We conclude that
1 024
Chapter 12
Multiple Integrals
x
=
Mx=O M
= _Q_ =0 32
y=
and
My=O M
=
768/5 32
=
24 5.
Therefore the center of gravity is c= (x,y) = (0, 24/5). ..,..
Moment of Inertia
The
second moments, or moments of inertia, of a plate 'R about the coordinate axes
are defined as follows:
JJ JJ
Iy=O =
y28(x,y)
dA
1l
Ix=O =
x28(x,y)
dA.
1l
Here Iy=O is the moment of inertia about the x-axis (i.e., about the line y=0), and
Ix=O is the moment of inertia about the y-axis (the line x = 0). Notice that moments of inertia differ from first moments in that we use the square of the distance to the axis. Moments of inertia can also be defined about points. For example, the second
moment about the origin (0, 0), or the polar
JJ
Io= Ix=O+Iy=O=
moment of inertia,
(x2 +y2 ) 8(x,y)
I0, is defined by
dA.
1l
Second moments are used, for instance, to measure the kinetic energy of rotation of
a spinning body. These matters are studied in more detail in physics texts.
� EXAM PL E 4 Calculate the second moment Iy=O of the lamina with con
stant density 8=3 that occupies the region bounded by the curves y=-x2+4 and
y=O.
Solution The x limits of integration are -2 and 2. Thus according to the definition,
JJ
Iy=O =
y2
1l
= =
=
= = -
·
3 dA
-x2+4
2
11 1 ll 1 1: ( -x; �
3y2dydx - 0 2 -x2+4 2 dx 0 2 2 (-x2+4)3 dx 2 (-x6 +12x4 - 48x2 +64) dx +
4096
- 35 ·
1
5
- 16x3+64x
)[
2
12.7 Physical Applications
Mass, First Moment, Moment of Inertia, and Center of Mass in Three Dimensions
1 025
The reasoning that we used to derive double integral formulas for mass, first moment, moment of inertia, and center of mass in two dimensions can also be used to derive triple integral formulas for these quantities in three dimensions. If U is a solid in space and if the density of the solid at the point argument as in the first part of this section leads us to
M=
(x,y,z) is 8(x,y,z), then an define the mass of U to be
jjj 8(x,y,z)dV. u
Similarly, the first moments with respect to the xy-plane, yz-plane, and xz-plane are given by
Mz=o=
jjj z8(x,y,z)dV,Mx=o= jjj x8(x,y,z)dV, u
and
My=o=
u
jjj y8(x,y,z)dV. u
The coordinates of the center of mass of the body are then given by
x
-
_ -
Mx=O �,
- My=O y= �, Mz=O . z= � The center of mass is the point c= (x,y, z).
As in two dimensions, when the density function is not mentioned explicitly, we
centroid of a region. second moments, or moments of
assume that mass is uniformly distributed, and we speak of the Finally, if U is a solid with density 8, then its
inertia,
about the x-axis, the y-axis, and the z-axis are respectively given by
Ix=
j jj (y2
+ z2)8(x,y,z)dV,
u
ly =
j jj (x2
+ z2)8(x,y,z)dV,
u
and
lz=
j jj (x2
+ y2)8(x,y,z)dV.
u
Notice that the factors
y2 + z2, x2 + z2,
z-axis, respectively.
x2 + y2 of these moments of inertia (x,y,z) to the x-axis, the y-axis, and the
and
are the squares of the distances of the point
1 026
Chapter 12
Multiple Integrals �
EXAM P L E 5 Assuming that the body U
=
{ (x,y, z )
: -2::s y::s 0, y+3::sx::s y+5, 2x+y+ 4::sz::s2y+14
}
has a uniform mass distribution 8 (x,y, z = 3, calculate its first moment with respect
)
to the xz-plane.
Solution According to the definition, the first moment of U with respect to the xz plane is
My=O ff 1-02 1y+3y+512x+2y+y+4l4 1-20 1y+3y+5 1z==22yx+14y+4 z + y ° 1-2 1y++S3 ( 2 1: 3y dV
=
'R,
3y dzdxdy
3yz
dxdy
3y - 6xy+30y dxdy
)
-6y2+12y dy
-40 .
....
INSIGHT
A negative first moment simply means that the solid is positioned so that a coordinate of its center of mass is negative. By contrast, the second moments of a solid
cannot be negative, no matter where it is situated relative to the coordinate axes.
�
EXAM P L E 6 Assume that the solid body U=
{ (x,y,z )
has density function 8 (x,y, z
)
: -l::sz::sl, l::sy::s2,0::sx::sz+y =
2x+y. Compute the center of mass of U.
Solution First, we have M=
ff (
8 x,y,z dV
)
'R,
=
=
=
}
1-1l f1 21z+y (2.x ) 0 1 2 1-111 (x2 ) lz+y 0 2 1 f 2) 1-111 2
+y dxdy dz
+yx
dydz
(z +3zy+2y
dydz
( 1
=
1
-1
=
1
12.7 Physical Applications
3 2 z2y+ -zy2+ -y3 3 2
1
9
) 1y==Z yl
1 027
dz
14
z2+ -z+-dz 3 2 -1
=10. Similar calculations show that
Mx=O 1-ilorlJro +y My=o 1l1l1z+y Mz=O 1l1l1z+y 1
=
=
-1
0
0
-1
0
0
=
x(2x+y)dxdydz=
45 4,
y(2x+y)dxdydz
=
16,
z(2x+y)dxdydz
=
3.
Therefore
45/4 9 Mx=O ----x1 - 10 - 8' _
_
We conclude that the center of mass of U is the point
Q UIC
K
Q UIZ
1. A planar region 'R, has mass density 8(x,y)
=
(9/8, 8/5, 3/10).
3x2y3• What function do we
integrate over 'R, to calculate the first moment about the x-axis? 2. For a planar region R,
3. 4.
Mx=O
=
72,
=
24, and x
=
12.
What is
What is the polar moment of inertia of the unit disk {(x,y) if its mass density is 8(x,y)
=
1+x2+ y2?
What is the moment of inertia of the solid cylinder U
:
y? 0::::; x2+y2 :51}
= {(x,y,z): 1::::; z::::; 2,
O :5x2+y2 :51} about the z-axis if its mass density is 8(x,y, z)
Answers
1.
My=O
3x2y4
2.
4
3. 57r/6
4. 37r/2
=
2z?
1 028
Chapter 12
Multiple Integrals
EXERCISES Problems for Practice
19.
In Exercises 1-4, use a double integral to find the mass of the planar region
1. 2.
R
with density function
8(x,y).
20.
R={(x,y):Osxs2,x+lsys2x+3}, 8(x,y)=3x+4y R={(x,y):-2sxs2,0sys4-x2}, 8(x,y)=x+3
3. 4.
21. 22.
R={(x,y):Osxsl,1-x2sys2(1-x2)}, 8(x,y)=12xy R={(x,y):Osys2,ysxs4-y}, 8(x,y)=3y
23. 24.
In Exercises 5-12, find the first moment of the region with density function
5. 6. 7. 8. 9. 10. lL 12.
8(x,y)
R
about the indicated axis.
25. 26. 27. 28. 29. 30.
13. 14. 15. 16. 17. 18.
8(x,y).
R
with density function
8(x,y)
about
each coordinate axis.
3L 32. 33. and
In Exercises 19-24, calculate the mass of the solid U with
8(x,y,z):
about the three coordinate planes,
U.
In each of Exercises 31-34, calculate the second moments
R={(x,y):Osxsl,xsysy'X}, 8(x,y)=x+6 R is the region between the parabolas y=-x2 +4 y=x2-4, 8(x,y)=3-x R={(x,y): -lsysl,2sxsy+3}, 8(x,y)=x(y +1) R= {(x,y): 3sxs4,-2sysl}, 8(x,y)=x+y+2 R= {(x,y):1sys4,1sxs2/y'Y}, 8(x,y)=12/y'Y R= {(x,y):1sxs2,xsysS-x}, 8(x,y)=12x
density function
U
U={(x,y,z):-lsxsl,-v'l-x2sysv'1-x2, Qszs12}, 8(x,y,z) =z U={(x,y,z):Osxsl,Osys2, Oszsl +2x+3y}, 8(x,y,z)=3 U={(x,y,z):-lsxsO,xsysl, x+yszs2}, 8(x,y,z)=x+3 U= {(x,y,z):1szs2,z2sys4,2sxs2z}, 8(x,y,z) =y'Y U= {(x,y,z):1szs2,0sxsz,xsysz} 8(x,y,z)=y-x U={(x,y,z):x2+y2s4, Oszs4} 8(x,y,z)=x2 +yz+zz
of the planar region
In each of Exercises 13-18, find the center of mass of the with given density function
the first moments of
and the center of mass of
R= {(x,y):lsys4,y'Ysxsy2}, 8(x,y)=10, x-axis R= {(x,y):Osxsl,Osysx2}, 8(x,y)=y'X, y-axis R= {(x,y):-3sxs-1,2sys4}, 8(x,y)=-x/y, y-axis R= {(x,y):-lsysO,Osxsy+l}, 8(x,y)= VJYT, y-axis R
U={(x,y,z):Osxsl,Osysv'1-xz,x2+y2szs 2-x2-y2}, 8(x,y,z)= 60y In each of Exercises 25-30, calculate the mass of the solid
U,
R= {(x,y):Osxsl,1-x3sys2(1-x3)}, 8(x,y)=12xy, y-axis R={(x,y):Osys2,ysxs4-y}, 8(x,y)=3y, x -axis R={(x,y):lsxs2,xsys5-x}, 8(x,y)=12x, x-axis R={(x,y):1sys4,1sxs2/y'Y}, 8(x,y)=12/ y)I, x-axis
planar region
U={(x,y,z):Osxsl,Osys2,0szsxy}, 8(x,y,z)=x+l U={(x,y,z) :Osysl,ysxs4,yszsx}, 8(x,y,z) =y'X +y'Y U= {(x,y,z):Oszs2,zsys3,ysxsy+1}, 8(x,y,z) =x+z U={(x,y,z) :Osxsl,Osysl,Oszsx+y}, 8(x,y,z) =x+y+z U={(x,y,z):Osxsl,x2sysx,yszsl}, 8(x,y,z)=12x
34.
R={(x,y):Osxs2,0sys4-x2}, 8(x,y)=x+1 R={(x,y):1sxs2,xsys2x}, 8(x,y)=y R={(x,y):Qsxsl,1-xsys2(1-x)}, 8(x,y)=x +y R={(x,y):1sxs2,xsys2}, 8(x,y)=x In each of Exercises 35-38, calculate the polar moment of
inertia of the planar region
35. 36. 37. 38.
R with density function 8(x,y).
R={(x,y):Osxsl,Osys2}, 8(x,y)=6x R={(x,y):1sxs2,xsys2}, 8(x,y)=Sx R={(x,y):Osxsl,lsys2-x}, 8(x,y)=6y/(x2+y2) R={(x,y):lsxs2,xsys2}, 8(x,y)=xy In each of Exercises 39-42, calculate the moments of
inertia with respect to the three coordinate axes.
1 029
12.8 Other Coordinate Systems
U={(x,y,z) :3:5x:55,l:5y:54,l:5z:52} 8(x,y,z)=xy 40. U={(x,y,z): -l:5y:52,0:5x:52-y,0:5z:5x}, 8(x,y,z)=y+2 4L U={(x,y,z) :0:5x:53,1:5y:52,x +y:5z:5x + y +1}, 8(x,y,z)=4 42. U={(x,y,z): 1 :=;z :52,1:5x:5z,1:5y:5x} 8(x,y,z)=x 39.
mass of
U. The radius of gyration
of U about the x-axis is
the distance pfrom the x-axis of a point-mass that has mass Mu and kinetic energy KEu. Find a formula for pin terms of Ix and Mu. (Verify that your formula has the correct dimensions.)
SO. Pappus's Theorem says that if a region in the right half of the xy-plane (i.e., in the set
{(x,y,0): x > O},
is rotated
about the y-axis, then the volume of the resulting solid is the area of the region times the circumference of the circle traversed by the centroid of the region. Prove
Further Theory and Practice In each of Exercises 43-46, find the center of mass of the planar region R with given density function
43. 44. 45. 46.
8(x,y).
0:5x:57r,0:5y:5sin(x)}, 8(x,y)=Sy R ={(x,y) : 0:5x:5l,�:5y:5e}, 8(x,y) =1/y R= {(x,y):0:5x:5l,0:5y:5�}, 8(x,y)=8x R={(x,y): 1 :5x::; e, 0::; y:5ln(x)}, 8(x,y)=72x
R={(x,y):
47. Give an example of a planar region R such that the center of gravity R is not an element of R. 48. Suppose that a region R in the plane consists of each point that lies in one or more of the sets
{(x,y):x2 +y2 :5l}, {(x,y): (x-4)2+(y-3)2:5l}, {(x,y) :.x2+(y-12)2:54}. That is, R is the union of the three sets. Assume that the mass distribution is uniform in this region. Find the center of mass.
49. A solid
U
rotates about the x-axis with constant angular
speed w. Express the solid's kinetic energy KEu in terms of wand the moment of inertia Ix of U. Let Mu denote the
Pappus's Theorem.
SL Apply Pappus's Theorem (as stated in Exercise
50)
calculate the volume obtained when a disk of radius center
(R, 0) with r < R
r
to
and
is rotated about the y-axis.
Calculator/Computer Exercises S2. Let R be the region of the xy-plane that is bounded above by y = 3-x2 and below by
y=1+(x+1)3•
Calculate the
center of mass of R given that its mass density is
8(x,y)=x+3y. S3. Let R be the region in the first quadrant of the xy-plane that is bounded above by y =1+x+x2 and below by
y =exp(x). Calculate the center of mass of R given that its mass density is
8(x,y)=1 +x.
S4. Let R be the region in the first quadrant of the xy-plane that is bounded above by y =1 +x + 2x2 and below by y =1 +x3• The mass density of R is
8(x,y) =1+xy!y. x- and y
Calculate the second moments of R about the axes.
SS. Let R be the square in the xy-plane with vertices
(0,1), (1,1), and (1,0). The 8(x,y)=2 +sin(x2) + exp(-y4).
mass density
(0,0),
of R is
Approximate the center
of mass of R.
1 2.8 Other Coordinate Systems Earlier in this chapter, we learned that polar coordinates make the equations of planar curves with circular symmetry much easier to study. A similar situation occurs when we are doing calculus in space. In this section, we learn about two special coordinate systems for handling surfaces and solids with special symmetries.
Cylindrical Coordinates
The cylindrical coordinate system is created by using polar coordinates in the xy plane and the standard Euclidean z-coordinate in the vertical direction. The idea is illustrated in Figure 1. Notice that we locate the point Pin space by first looking at the projection P' of P into the xy-plane. We denote by
r
the distance of P'
from the origin. We denote by 9 the angle by which we rotate the positive x-axis
1030
P
Chapter 12
Multiple Integrals
(r, 8, z) cylindrical coordinates �
coordinates
(.r:, y, .t) r z
counterclockwise until it lies over the ray DP'. The third coordinate is simply z, the signed distance of Pfrom P'. We write the cylindrical coordinates for Pas (r, (), z). The formulas that we learned in Section 124 . for converting from rectangular to polar coordinates and back are valid in the context of cylindrical coordinates: x = rcos(O),
y
y = rsin(O),
x2 + y2 = r2,
and,
if
x � 0,
tan(O) = � x·
The point Phas rectangular coordinates (1,1,4). What are its cylindrical coordinates? � EXAMPLE 1
P' = (x,y, 0)
Solution The projection of P into the xy-plane is P' = (1, 1, 0), as shown in Figure 2. Then, we see that r = ...fi and 0 = 7r/4. Also , z = 4. Therefore the cylindrical coordinates for Pare (../2,11"/4,4 ). �
A Rgure 1
P = (1, 1, 4) rectangular coordinates z:
-----
-
4 (0, 1,0)
y
We can calculate many integrals easily by using cylindrical coordinates. To do so, we replace the rectangular element of area dxdy with the polar element of area r drdfJ. So our element of volume, which is dV = dxdydz. in rectangular coordinates , becomes dV = r
drdOdz
in cylindrical coordinates . A Rgure2
� EXAMPLE 2 Integrate the function/{x,y,z) x2 + y2 over the solid U that is inside the cylinder xl- + y2 = 4 , below the plane z= 3, and above the plane z=-2. =
Solution The solid U is drawn in Figure 3. It is convenient to convert the problem to cylindrical coordinates. Then/(rcos(O),rsin(O),z) = r2 and the solid over which
we wish to integrate becomes
z=3
U = {(r, 0, z) : 0 s rs 2, 0 s Os 2?r, -2s zs 3}. Y
Bearing in mind that the volume element is r drdOdz, we find that our integral is
z=-2 A Figure 3
=
2 y2
lr�2 dOdz= 16 -(27r)(3-(-2)) =401r . 4 1312w -2 o 4 r�o
� EXAM PLE 3 Calculate
z= Jx +
4
�
x2 - y2•
..._
the volwne of the solid U that lies above the cone and below the hemisphere z= 1 + J1 -
The solid U appears in Figure 4. We observe that, in cylindrical coordinates, the cone may be written as z= r, and the hemisphere may be written as z = 1 + v'l - r2. From Figure 4, we see that the projection of U in the xy-plane is a disk 'D. The boundary of 'Dis the projection of the circle that is the intersection of the cone and the hemisphere. Because the z-coordinate of each point on this circle
Solution
12.8 Other Coordinate Systems of intersection is given both by rand by 1+
2
v'l - r ,
we set r = 1 +
1 031
v'l - r•2
We
solve this equation by rewriting it as r-1= v'l - r2, or (r -1)2=1- r2, which
simplifies to 2r(r -1)=0. The solution is r =0or r =1, which means that 1J consists
of the points with cylindrical coordinates that satisfy z = 0 and the inequalities
0s ()<2?r and 0s rs1. These inequalities tell us the limits of integration for the outer and middle integrals. To determine the limits of integration for the inner integral, we observe that a point with cylindrical coordinates (r, (), z )lies in U if and
only if rs zs1 +
jjj • Figure
1 dV=
v'l - r2.
Thus the volume of U is
/2}/l+./hi
Hence we have
-/ ( ((
4
1hr ,
1dV-
.
w:
Spherical Coordinates
r2
-
1 + .J1=2° - r
2 ) 3 3 )) ) ( ; ;
1-r2)3
r=
-
) r drdO.
d()
r=O
1
= 1r
!2w:Jr
1)r dzdrd
-0 -
-
-
-0
= 1r.
..
For calculations involving objects in three-dimensional space that are symmetrical in the origin, spherical coordinates are often most convenient. Referring to
Figure 5, we see that, in this new coordinate system, we locate a point P according to its distance p from the origin, the angle
. .
OP
positive x-axis. We write the spherical coordinates for :: )
" y�
P
, , 0).
s
Every point in space can be located with spherical coordinates p, fying Osp, O<
<2?r.
an use the point
Po
, 0)
satis-
By convention, we limit the values of the
spherical coordinates to these ranges. Notice that
"'··
OP makes with the positive
makes in the xy-plane with the
and
have different ranges. We
with rectangular coordinates 0, -1,
) to understand why we
need not ever take
-�.. s
P
by the spherical coordinates p = 1,
escription is unnecessary because we can also describe inates
=1,
- a . . ,, .. .
= ?r.
2,
and()=:: ?r'
• Figure 5
(0,
• Figure 6
2
(see Figure 6).
Po
= 'Tr
2.
But such a
by the spherical coor
1 032
Chapter 12
Multiple Integrals � E X A M P L E 4 What are spherical coordinates for the point Po with rec tangular coordinates (1, 1, v'6)? Solution
Using the formula for the distance between two points, we calculate that
the distance of Po from the origin is
J(l - 0)2
+
(1 - 0)2 + ( v'6 - 0)2, or JS. Thus
p = J8=2./2. Because the projection of Po into the xy-plane is Pb=(1, 1, 0), we see that ()=7f/4. Next we use the equation z=p cos(¢), which can be deduced from Figure 5. This equation tells us that the spherical coordinate ¢ of Po satisfies v'6= 2VZ cos(¢), or cos(¢)= v'6/(2VZ)= ./3/2. It follows that ¢=7r/6. In summary, the spherical coordinates of Po are (2./2, 7f/6, 7f/4). <11111 Converting from spherical coordinates to rectangular coordinates and back again requires some calculation. The formula for the distance between the point with Cartesian coordinates (x, y, z) and the origin gives us P
=
Jx2 + y2 + z2.
(12.8.1)
Written in terms of cylindrical coordinates, equation (12.8.1) becomes p = Jr2 + z2. As can be seen from right triangle
OPP'
(12.8.2)
in Figure 5, we have
z =p cos(¢),
(12.8.3)
r=p sin(¢).
(12.8.4)
and
Equations (12.8.1)-(12.8.4) allow us to pass back and forth between spherical and cylindrical coordinates. If we substitute equation (12.8.4) into the equations x rcos(O) and y rsin(O), then we obtain =
=
x=p cos(O) sin(¢)
(12.8.5)
y=p sin(O) sin(¢).
(12.8.6)
and
Equations (12.8.3), (12.8.5), and (12.8.6) tell us the Cartesian coordinates of a point with spherical coordinates (p, ¢, 0). Observe that these formulas imply that x2 + y2 + z2=p2cos2(0)sin2(¢) + p2sin2(0)sin2(¢) + p2cos2(¢) p2sin2(¢) (cos2(0) + sin2(0)) + p2cos2(¢) =p2, =
which confirms that p is the radial coordinate in space, analogous to the coordinate r in polar coordinates in the plane.
12.8 Other Coordinate Systems
� EX A M P L E
(2,27r/3,57r/6).
5 A point
Po
1 033
in space has spherical coordinates given by
Calculate its rectangular and cylindrical coordinates.
Solution Denote the rectangular coordinates of
(12.8.3), (12.8.5), and (12.8.6), we calculate
Yo
=
Po by (xo,yo,zo).
( ) . (327f) 2 (1) (v'3\ Z T)
. 6 57r 2 sm Zo
=
sm
=
2 COS 7r 3
(2 ) 2 ( - �) =
=
Using equations
v'3 =
T
-1.
Po are (-3 /2, v'3/2, -1). It is now easy to ( r0, B0, z0). The cylindrical coordinate zo is the same as the rectangular coordinate z0, namely -1. The cylindrical coordinate Bo is
Thus the rectangular coordinates of convert to cylindrical coordinates
the same as the spherical coordinate B0, namely
z
So the cylindrical coordinates of
Po
are
57r/6. Finally,
(v'3,57r/6, -1).
Naturally we are interested in performing integrations in spherical coordinates. We need to determine the volume element in this new coordinate system. Figure 7 exhibits an increment of volume swept out by increments �p, �¢, and �B of the coordinates. This solid is approximately a box, with sides that are determined by arcs of circles and a radial height �p. One circular edge is an arc of a circle that is parallel to the xy-plane and that has radius psin(¢). Because this edge subtends an angle �B, it has arc length psin(¢) �B. Another circular edge of the box is an arc of a circle of radius p. This edge subtends an angle �¢ and therefore has arc length p�c/J. The height of the box in the radial direction is �p. Thus the product
_. Figure 7 A small "box" in
( psin(¢) �B) (p�c/J)(�p)
spherical coordinates
=
p2sin(¢) �B �cjJ�p
approximates the volume of the small box. We conclude, by using familiar limiting arguments, that the volume element for spherical coordinates is dV= p2sin(¢) dpd¢ dB.
INSIGHT
(12.8.7)
Together, Example 8 and Theorem 4 of Section 12.5 show how to apply a
T(r, (}) (r cos((}), r sin((})) r drd(}. Although we will not
change of variable formula in the plane to the transformation to obtain the polar element of area
dA
=
lh(r, (})ldrd(}
=
=
derive it, there is an analogous change of variable formula for space. The spherical coordinate transformation T(p, ¢, (})
Jacobian matrix
=
(pcos((J)sin(¢), psin((J) sin(¢), p cos(¢)) has
1 034
Chapter 12
Multiple Integrals
r :P D(T)(p,
=
(pcos(O)sin(
I :P :
l
(psin(O)sin(
:
(pcos(O)sin(
(pc<,,(¢)) p
o
: g0
o(psin(
(pcos(
(psin(
:
o(poo,(¢))
l
1,
j
or
D(T)(p,
=
r
cos(O)sin(
pcos(O)cos(
-psin(O)sin(
cos(¢)
-psin(¢)
0
sin(O)sin(
pcos(O)sin(
psin(O)cos(
1.
From this, we calculate the Jacobian determinant to be h(p,
The yz-plane, x
=
an
=
det(D(T)(p, ¢, 0))
p2sin(¢).
=
alternative approach to formula
(12.8.7).
� E X A M P L E 6 Use spherical coordinates to calculate the volume V of the
0
z
solid U that lies above the cone z
=
1+
y'l -x2-y2 .
z
=
Jx2 + y2
and below the hemisphere
Solution This solid is pictured in Figure 4, We have already calculated its volume to be 7r by using cylindrical coordinates in Example 3. The intersection of U with the yz-plane is shown in Figure 8. Because x 0 in 1+ -y2 or (z-1)2 1-y2 .
this plane, the equation of the hemisphere is z
y -1
1
y2 + z2 p2 and z pcos(c/J) =
.A Figure 8
=
=
y'l
=
=
After expanding the left side of this equation, we get y2 + z2
=
2z. But, as Figure 8
shows, in the yz-plane the relationships between y, z, p, and
=
=
p2 and
pcos(¢). Therefore the equation of the intersection of the hemisphere with the
yz-plane is p2
=
2pcos(
=
2cos(
spherical coordinates by the inequalities 0 ::5 0 ::5 27r, 0 ::5
V
=
f 27rr /4 { 2cos(¢)
lo lo lo
(1) p2sin(¢) dpd
We evaluate this integral as follows:
V
1
=
3
11 2
0
0
7r/4
8cos3(¢)sin(¢)d
8 =
3
1 (2r 7
0
cos4(¢)
4
1 ) ¢
/4 =7r
¢=0
dO
12.8 Other Coordinate Systems
1 035
� EX A M P L E 7 Calculate the center of gravity of the solid in the last example, assuming a uniform mass distribution 8(x,y, z) =1. (Refer to Section 12.7 for terminology and notation.)
It follows from symmetry considerations that x=y=0. So, we must calculate z. Now, substituting formula (12.8.3) for z, 1 for 8(x, y, z), and formula (12.8.7) for dV, we obtain
Solution
Mz=O =lufllff z8(x,y, z) dV=lo{2rrlor/4lo{2cos(rp) (p cos(¢)) (l)p2 sin(¢) dpdcpd(} ( 2cos(rp) p3dp) dcpdO. 2 /4 = r rrr cos(¢) sin(¢) f lo lo lo
Therefore
p= cos(rp) 1 4 /4 ( 2 1 4 2 rr ( r r Mz=O =lo lo cos(¢)sin(¢) 4p p=O ) dcp) d(}=4lor2rr (lor/ cos(¢)sin(¢)cos4(¢) dcp) d(}. Because the inner integral does not depend on (} , we have 1
Mz=O =27f . 4lor 4 cos5(¢)sin(¢) dcp.
We make the substitution u=cos(¢), du= -sin(¢) dcp, obtaining
From Examples 3 and 6, we know that the volume of the solid is 7r . The mass is therefore =III u8(x, y, z) dV =III ul dV =7r. We conclude that
M
77r/6 7 Mz=O Z- --,;t - -;- -6. __
_
_
Therefore the center of gravity is (0, 0, 7 I 6). ..,.. Q UIC K
Q UIZ
1. What are the cylindrical coordinates of the point with rectangular coordinates (1,2,3)? 2. What are the spherical coordinates of the point with rectangular coordinates (1,2,3)? 3. What are the rectangular coordinates of the point that has (2./6,7r/4,7r/3) for spherical coordinates? 4. What is the element of volume in spherical coordinates? Answers
1. (JS,arctan(2),3) 3. (J3,3,2v'3) 4.
p2
2. (v'I4,arccos(3/JI4),arctan(2)) sin(¢) dpdcpd(}
1 036
Chapter 12
Multiple Integrals
EXERCISES Problems for Practice
30.
In each of Exercises 1-8, a point P is given in rectangular coordinates. Give cylindrical coordinates for P. 1. p=(2v'3, 2,5) 2. p= (4,0,6) 3.
P=(-2,2, -1) (viz, -viz, 2) p= (0, -3, -2) P=(5,5,0) p= (1, -v'3, -2) P= (0,0, -3)
In each of Exercises 31-38, integrate the function f over the solid U using spherical coordinates. 3L
4. p= 5. 6. 7. 8.
In each of Exercises 9-16, rectangular coordinates are given for a point P. Give spherical coordinates for P. (1, 1,viz) p= (-1,v'3,2) P=(4,0,4) P=(0,0,3) p=(1, 1,v'6) p= (2,0,-2v'3) p= (-4,4,-4v'6) P=(v'3,-3, VU)
9. p 10. 11. 12. 13. 14. 15. 16.
17. 18. 19. 20. 21. 22. 23. 24.
P=(3,7r/3,27r/3) spherical P=(1,57r/6,77r/4) spherical P= (2,-2,-2vlz) rectangular P=(0,v'3,-1) rectangular P= (2,-37r/4,-2) cylindrical P=(6,-77r/4,-6) cylindrical P=(3,0,-3) rectangular P=(5,37r/4,27r/3) spherical
In each of Exercises 25-30, integrate the function f over the solid U using cylindrical coordinates. f(x,y,z)=x2 + y2; U is the region below the paraboloid z=-x2 -y2 and above the plane z =-4. 26. f(x,y,z)=x + y; U is the solid in the first octant bounded by z=x2+y2 and z= y'x2 +y2. 27. f(x,y,z) = y'x2 + y2 ; U is the solid bounded by the cylinder x2 +y2=4 and by z=x2+y2+1,z =-3. 28. f(x,y,z)= 1/(1 + x2 +y2); U is the solid bounded by the paraboloids z=x2 + y2 + 2 and z= 10 - x2 - y2• 29. f(x,y,z)= 1/(x2 +y2); U is the solid bounded by the cylinders x2+y2= 2 and x2+y2= 4 and by the planes z=8 and z= -2. 25.
32. 33. 34.
f(x,y,z)=5; U is the solid inside the sphere x2 +y2 + z2= 9. f(x,y,z)= x2 + y2 + z2; U is the solid outside the cone z2= x2 + y2 and inside the sphere x2 + y2 + z2=8. f(x,y,z)=(x2 +y2 +z2)1/2; Uis the solid inside the sphere x2+y2 + z2= 9 and outside the sphere x2+y2+z2= 4. f(x,y,z)= cos((x2 +y2 +z2)312); U is the solid inside the sphere x2+y2+z2 9 and above the xy-plane. f(x,y,z)= z; U is the solid below the surface z y'x2 +y2 and inside the sphere x2+y2+z2= 4. f(x,y,z)=x2 + y2; U is the region outside the cone z2=x2 + y2 and inside the sphere x2 + y2 + z2=4. f(x,y,z)=(x2 + y2 + z2)-1/2; U is the solid between the spheres x2+ y2 +z2= 2 and x2 +y2 +z2= 3. f(x,y,z)= (1 +x2 +y2 +z2)-2; U is the solid above the cone z= y'x2 + y2 and below the sphere x2 + y2 + z2=3. =
35.
=
In each of Exercises 17-24, rectangular, spherical, or cylindrical coordinates are given for the point P. Calculate coordinates in the other two coordinate systems.
f(x,y,z)=xy; U is the solid inside the cone z2=x2+y2, outside the cylinder x2 +y2= 9, and between z= -5 and z=5.
36. 37. 38.
=
Further Theory and Practice 39.
40.
4L
42. 43. 44. 45. 46. 47. 48.
Use cylindrical coordinates to calculate the volume out side the cone z2= x2 +y2 and inside the cylinder x2 + y2=4. A point P has rectangular coordinates (3,57r/6,27r/3). What are the cylindrical coordinates of P? What are the spherical coordinates of P? A point P has spherical coordinates (3,2,5). What are the rectangular coordinates of P? What are the cylindrical coordinates of P? Determine all points that have spherical coordinates equal to their rectangular coordinates. Determine all points that have cylindrical coordinates equal to their rectangular coordinates. Find formulas for converting directly from cylindrical to spherical coordinates and vice versa. Sketch the graph of the surface p=cos(
Calculator/Computer Exercises 49.
Let R be the planar region in the first quadrant that lies outside the curve r= sin(0)3 and inside the curve
Summary of Key Topics
r =
3
2cos( O) . Let Ube z = y. Calculate
-
below
!ff
the solid that lies above Rand
V(a)
be a constant between a
(x2+y2)1/3dV.
be the volume of that part of the unit ball that For what value of
a
is
V (a)
0 :5 0 :5 a and
0 and 1. The spherical
cap of
of the unit ball is the solid
u.
consists of points with spherical coordinates
0:5cfl:5a.
a
height
u
50. Let
SL Let
1 037
For what
=
a
{(x,y, z) : x2+y 2 + z2:5l,1 - a:5z}. is the volume of u. equal to one-fourth the
volume of the unit ball?
equal to one
fourth the volume of the unit ball?
Summary of Key Topics in Chapter 1 2 Double Integrals Over
The double integral of a continuous function
Rectangular Regions
f over a rectangular
region
R={(x,y) :asxsb,csysd}
(Section 12.1)
is defined to be the limit of a Riemann sum over the product of partitions of and
[c,d]. The
[a,b]
integral is written
jj f(x,y)
dA.
R
This double integral is evaluated by computing either of the iterated integrals
b l (idf(x,y) dy)
Integration Over More General Planar Regions (Section 12.2)
If
dx
or
b id (l f(x,y) ) dy. dx
R is a planar region bounded by finitely many smooth curves and contained in a
bounded rectangle Q={(x,y) :a sxsb,csy :5 d}, then for each positive integer N, we partition Q into N2 subrectangles by dividing each interval [a,b] and [c,d] into N equal subintervals. To integrate a continuous function
f over R,
we form
Riemann sums using only subrectangles of the partition that lie completely in
R.
The limit of the Riemann sums, as the number of subrectangles tends to infinity, gives the double integral; we denote the integral by
jj f(x,y)
dA.
R
Simple Regions
A region
R is x-simple if it has the
form
(Section 12.2) It is y-simple if it has the form
R={(x,y): asxsb, ,B1(x)sys,B2(x)}.
1 038
Chapter 12
Multiple Integrals The integral of a continuous f over an x-simple region R is calculated with the iterated integral
(y) id (lai a 1
c
(y)
)
f(x,y) dx
dy.
The integral of a continuous f over a y-simple region R is calculated with the iterated integral
lb ( 1/'lz(X)
)
f(x,y)dy dx.
P1(x)
a
If a region is both x-simple and y-simple, then the integral may be evaluated in either order; this observation is often useful in simplifying integration problems.
Area and Volume (Section 12.3)
The area of a bounded region R whose boundary consists of finitely many smooth curves is given by
ff
ldA.
'R,
The volume under the graph of a continuous function z
=
f(x,y) and above a region
R in the xy-plane is given by
ff
f(x,y) dA.
'R,
The volume under the graph of a function z = f(x,y) and above the graph of z = g(x,y),both defined on the region R in the xy-plane, is given by
ff (f
(x,y) - g(x,y) dA.
)
'R,
Polar Coordinates (Section 12.5)
A planar point Pis located by polar coordinates (r,0), where r is the distance of P ---+
from the origin, and 0 is the signed angle that the ray 0P makes with the positive x-axis. In the case r < 0, the point with polar coordinates (r,0) is the reflection in the origin of the point with polar coordinates (lrl,0). Polar coordinates
(r, 0) are related to Cartesian coordinates of a point Pby the
formulas x = rcos(O),
y = rsin(O),
and , if x =F 0,
tan(O) =
l. x
Graphing in Polar
To graph a curve r = cp(O), plotting points is a vital tool. Also we take note of
Coordinates
intervals of increase of cp(O) and intervals of decrease of cp(O). We test for symmetry
(Section 12.4)
as follows: symmetry in the x-axis if:
cp(0)
=
cp(-
0)
Summary of Key Topics
symmetry in the y-axis if:
cp(O) = cp(7r - 0)
symmetry in the origin if:
cp(0) = cp(0 + 7r)
If the curve is defined implicitly by an equation of the form curve is symmetric about the origin if
Area in Polar Coordinates
w(r,0)
=
0,
1 039
then the
'11(-r, 0)= w(r, 0).
cp( 0) � 0 for 0 E [a, ,B], if ,B - a :s 27r, and if R is plannar region described in polar {(r, 0): 0 :s r :s cp(O), a :s 0 :s ,B}, then the area of R is given by
If
coordinates by S =
1 fJ cp (O)dO. 1
(Section 12.5)
z
0 :s cpo(0) :s cp1 ( 0)
If
coordinates by S=
2
a
for 0 E [a, ,B] and if R is the plannar region described in polar {(r,0) : 'Po ( 0) :s r :s cp1 ( 0), a :s 0 :s ,B}, then the area of R is given
by
1 b ('Pi(O) - cp0(0)) dO. 1 2
2
Double Integrals in Polar Coordinates (Section 12.5)
If,
a
plannar
region
a
R
is
described
{(r,0) : a :s 0 :s ,B,'Po ( 0) :s r :s cp1 ( 0)}, with all 0 , then the area of R is given by
S
=
,B
in
polar
- a :s 27r
and
coordinates
0 :s 'Po ( 0) :s cp1 ( 0)
by for
1fJ1'Po('1'18()8) r drdO a
If f(r, 0) is a continuous function on R, then the integral off over R is given by
1fJ1\01( rp0(8)8) f(r,0) r drdO. a
Integrals in polar coordinates can also be used to calculate volumes.
Change of Variable in Double Integrals (Section 12.5)
Let S be a region in the uv-plane bounded by finitely many continuously differ entiable curves. Suppose that
T(u,v) = ( cp(u,v),'lj.J(u,v))
is a transformation of S to
a region R in the xy-plane that is also bounded by finitely many continuously differentiable curves. We assume that
T
is one-to-one and onto and that the
component functions
T
are as just described and if
f is
tinuous function on the region R, then
ff f(x,y) dxdy= ff f(
s
a con
1 040
Chapter 12
Multiple Integrals
The expression
lr(u, v)
= ( [ ��:: �� ��:: ��]) det
=
(¢u�v - �u
that appears in the change of variable formula is called the Jacobian the transformation T (or sometimes simply the
Triple Integrals
determinant of
Jacobian).
For a solid U in space contained in a cube
(Section 12.6)
Q
= {(
x,y,z)
:
a '5, x '5, b, c '5, y '5, d, e '5, z '5, f},
and a continuous function f(x,y,z) on U, the Riemann sums for f over U are defined using a product of partitions of the intervals
[a,b], [c,d], [e,f].
The triple
integral is a limit of the Riemann sums whenever the limit exists. The integral is denoted
jjj f(x,y,z)dV. u
When U is a solid of the form
then the triple integral may be evaluated as the iterated integral
y a ( x x 1la1(,x) l.B2( ,B1(x,y,) )
f(x,y,z)
dzdydx.
a
The roles of x,y, and z may be permuted in this last formula when they are similarly permuted in the definition of R.
Mass, Moment of Inertia, and Center of
If a planar plate (or lamina) R has density 8(x,y) at the point (x,y), then the mass of the plate is given by
M= !! 8(x,y)dA.
Mass (Section 12. 7)
'R,
The
=
first moment of 0) is
equation y
the plate with respect to the x-axis (the axis with Cartesian
My=o= jj y8(x,y)dA, 'R,
and the first moment with respect to the y-axis (the axis with Cartesian equation x
=
0)
is
Summary of Key Topics
Mx=O =
JJx8(x,y)
1041
dA.
'R
The respective second moments, or moments of inertia, are ly=O
=
JJ y28(x,y)dA 'R
and lx=O =
JJ x28(x,y)dA. 'R
The center of mass (or center of gravity) of the plate is the point c
=
(x, y) where
-- Mx=O x--M
and
Mass, moments, and centers of mass are defined similarly for three -dimensional solids using triple integrals.
Cylindrical Coordinates (Section 12.8)
For a point P in three -dimensional space, the cylindrical coordinates of P have the form (r,(}, z),where r andBare the polar coordinates of the projection P' of P in the xy-plane, and z is the vertical signed distance of the point from the xy-plane . Cylindrical coordinates are related to rectangular coordinates by the formulas x=rcos(B),
y=rsin(B),
x2+y2=r2.
The volume element dV in cylindrical coordinates is r dzdrdB.
Spherical Coordinates (Section 12.8)
For a point P in three-dimensional space, spherical coordinates have the form (p, ¢>, B), wherepis the distance of the point from the origin, ¢>is the angle that the ray OP makes with the positive z-axis, andBis angular polar coordinate of the projection P' of P in the xy-plane. Spherical coordinates are related to rectangular coordinates by the formulas y=psin(B)sin(¢),
x=pcos(B)sin(¢), and P
=
v'xz +yz +zZ.
z= pcos(ef>)
1 042
Chapter 12
Multiple Integrals
When symmetry about the z-axis is present, it is often easiest to integrate using cylindrical coordinates. The volume element in cylindrical coordinates is
r drd(}dz.
When symmetry about the origin is present, it is often easiest to integrate using spherical
p2sin( ¢)
coordinates.
dpd¢dfJ.
The
volume
element
in
spherical
coordinates
is
Review Exercises for Chapter 1 2 In Exercises 1-3, use a Riemann sum with N=2, and {�;j} the midpoints of the four resulting subrectangles of R, to approximate the volume of the solid that lies under the graph off and above the rectangle R in the xy-plane.
In each of Exercises integral.
L f(x,y)= 3x-y
17 .
2. f(x,y)=xy2-x
3. f(x,y)=x/y
R= [0,3] x [1,2] R= [1,2] X [O,1] R=[2,6] x [4,8]
In each of Exercises 4-6, calculate I(x)= fcf(x,y)dy and .7(y)= f:f(x,y) dx for the given function f and the rectangle R= [a,b] X [c,d] in the xy-plane. 4. f(x,y)=xe'Y'-
5. f(x,y)=
sin(2x-y)
6. f(x,y)=yx2-x
R= [0,2] x [0,2] R= [0,11"] x [0,211"] R= [0,3] x [0,2]
In each of Exercises 7-9, evaluate the double integral by converting it to an iterated integral.
ff1ly3-xdA ff 1lsin(x) cos(y)dA 9. ff'Tl�+YdA 7.
8.
R= [1,4] x [-2,1] R= [11",311"/2] x [11"/2,11"] R= [O,1] x [0,2]
In each of Exercises 10-12, evaluate the double integral by converting to an iterated integral in the order ff1ldxdy and also by converting to an iterated integral in the order ff1ldydx . The result of each calculation should of course give the same answer. 10. lL 12.
ff1ly3xdA ff'Tlcos(x-y)dA dA ff1l
eix-y
R= [0,1] x [1,3] R = [O,11"/2] x [11"/2,31!"/ 2] R = [0,2] x [O,1]
each of Exercises 13-15, draw a sketch of the region R. State whether the region is x-simple, y-simple, or both. If R is x-simple, then exhibit the functions y1-->a1( y) and y1->a ( y) 2 that describe the boundary of R. Perform the analogous task when R is y-simple. In
13. 14. 15.
The region between the curves y= x2-2 and y= x + 1 The region between the curves x=y2 and x =y The region between the curves x =y2 and x=2y112
16.
18.
16-18,
evaluate the iterated
f�1 f;:+6 xy2 -ydydx .fo J;/ yx113dydx 3 27 y dxdy x 1
y ! ly•!•'i'
In each of Exercises 19 -21, calculate the integral of the functionf over the region R. 19. f(x,y)=Sy - 3x,
R is the region bounded by x=y2 and x =2y. 20. f(x,y)= ../X.JY, R is the region bounded on the left and right by x=2 and x=4 and above and below by y=x2 +4 and y=x2 - 2. 2L f(x, y) =x, R is the region bounded by y=cos(x) and y=sin(x) for 0:5 x:511"/4
In each of Exercises 22-24, switch the order of integration to make the iterated integral calculable, and then perform the integration. 22
.fo/2 J;/2 cos(x)/x dxdy
23.
f� f;
•
24.
J Jy' ,):'_dxdy x e -x'
dxdy
4
1
1
In each of Exercises 25-27 , calculate the volume of the solid that lies below the graph of z=f(x, y) and over the region R in the xy-plane. 25. f(x,y)=x2
+2y2 +4, R= {(x,y): -1 :5x:5l, -2:5y:52} 26. f(x,y)=5-x +y,R= {(x,y):x2 +y2 :5l} 27. f(x,y) =x2 -y2 + 8, R = {(x, y): lxl :5y :54} In each of Exercises solid U.
28-30,
calculate the volume of the
is bounded by the paraboloids y= z2 + x2 + 4 and y=-z2 -x2 - 2.
28. U
Review Exercises
29. U is bounded on the sides by the cylinder x2+4y2 =4, above by
z=x+y+6, and
z=0. x - 3y+ 2z =6
53. 'R, is the region inside the circle r =V3 sin (0) and outside
below by
30. U is lies below the graph of
the rose and in the
r = sin(20).
54. 'R, = {(r,0) : 7r/6:50:57r/4,1:5r:52sin(O)}
fourth octant.
In each of Exercises
In each of Exercises
31-33, plot the given points on a
1 043
55-57, use a double (or iterated)
polar coordinate integral to integrate the function
f over
the
polar coordinate graph.
planar region 'R.
31. (2,7r/4), (-4,27r/3), (6, -7r), (l,37r/2) 32. (-5,97r/2), (-3,-27r/3), (4,117!"/3), (-2,77r/2) 33. (4,0), (-4,0), (0,7), (0,-7)
55. f(r, 0) = sin( 0), 'R, is the region inside the circle r = 2 sin(0) 56. f(r, 0) = 0, 'R, is the region inside the limai;on r = 4 - sin(0) 57. f(r, 0) = cos(O) , 'R, is the region in the second quadrant
In each of Exercises
and outside the cardioid
r = 1 -sin(O)
coordinates. Give the rectangular coordinates of the point.
In each of Exercises
34. (2,57r/4) 35. ( -3,-97!"/2) 36. (5,117!"/4)
r=3
that is inside the circle
34-36, you are given a point in polar
58-60, use a double integral in polar
coordinates to calculate the volume of the indicated solid.
58. The solid bounded by the surface z=-2x2 - 2y2+8 and
In each of Exercises
37-39, you are given rectangular
the plane
z=1
coordinates for a point. Give all possible polar coordinates for
59. The solid bounded by the upper nappe of the cone
that same point.
z2 2x2+ 2y2 and the plane z=100. 60. The solid bounded by the cylinder (x - 1)2+y2=1 and =
37. (7v12,-7v12) 38. (4,7r) 39. (7r, 4)
the sphere
x2+y2+z2=16
In each of Exercises
40-42, test the given polar equation for symmetry in the x-axis, in the y-axis, and in the origin. In each of Exercises
62. 43-45, sketch the polar curve.
variable
63.
the
61-63, perform the indicated change
double
integral.
In each of Exercises
to fo..JY ../X7Y dxdy Jg Ji J3+xy dxdy 4 3 2 \ dydx y
1 12 In
46-48, use formula (12.11) to calcu
late the area enclosed by the given curve.
64. 65. 66.
r=1+cos(O) 47. r2=cos(O) 48. r=3cos(20)
46.
u2=x
v2 =y
v=xy
u=x
v=1/y
u=x
each
of
Exercises
Ji J}r�4 J;+y 3y - xy dzdxdy fl Ji.; ft xy dydzdx ,;:;r .fa fz h 4zsin(y2) dxdydz In each of Exercises
67-69, use a triple integral to express
the volume of the solid that is bounded by the given surfaces
In each of Exercises
49-51, use formula (12.12) to calcu
in space. Then determine the volume by calculating the integral.
49. The region outside the curve r=1+cos( 0) and inside the
67. z=4x2+4y2,z=1,z=16 68. x=y4,z = -x+6,z = 0 69. z = 2x2 +2y2 - 4, z = x2+y2 +4
r = 3 cos(0)
50. The region in the second quadrant that is inside the circle and outside the circle
r = sin (0)
51. The region inside the circle r=3 sin(0) and outside the circle
r = sin(0)
In each of Exercises region
the
64-66, evaluate the iterated
late the area bounded by the two given curves.
r = v12 sin ( 0)
evaluate
integral.
44.
circle
Then
1
43. r=2 - 2 sin(0)
r=2cos(O) 45. r=30
in
integral.
61.
40. r2=3 cos( 0) 41. r=6+3 cot( 0) 42. r = 4 - cos(30) In each of Exercises
of
In each of Exercises
70-72, evaluate the iterated integral. U over which the integration is
Describe in words the solid calculated.
52-54, calculate the area of the
'R, by using a double (or iterated) integral in polar
coordinates.
52. 'R, is one leaf of the rose r = 5 cos(40)
70.
to to J6z/4 x dxdydz r2 rz rJ4-z2 y dydxdz J J J
71. o o o r1 rl-y/2 r4-2x-y y 72. Jo Jo dzdxdy Jo
1 044
Chapter 12
Multiple Integrals
In each of Exercises 73-75, use a double integral to find the mass of the planar region R with density function 8(x, y). 73. 74. 75.
R={(x,y):0:5y:53,y+2:5x:53y+5},8(x,y)=x+3y R={(x,y): -2:5x:53,0:5y:59 -.x2}, 8(x,y)=4+y R={(x,y): -1:5y:51, -y2 - 2:5x:5y2+1}, 8(x,y)=x2+y
In each of Exercises 76-78, find the first moment of the region R with density function 8(x, y) about the indicated axis. 76. 77. 78.
R={(x,y): -2:5y:5 -l,1:5x:54},8(x,y)= -y/x, x-axis R={(x,y):0:5y:52,0:5x:52y2},8(x,y)=.(JY, x-axis R={(x,y):2:5x:53,x113 :5y:5x112},8(x,y)=2, y-axis
In each of Exercises 79-81, find the center of mass of the planar region R with density function 8(x, y). 79. 80. 81.
R={(x,y):l:5y:52,y'y:5x:5y},8(x,y)=y+1 R={(x,y):1:5x:52,3:5y:55},8(x,y)=12 -x - y R={(x,y):-2:5x:5l,x+2:5y:52x+5},8(x,y)=y
In each of Exercises 82-84, calculate the mass of the solid body U having density function 8(x, y, z). 82. 83. 84.
U ={(x,y,z):1:5x:53,2:5y:5 4,x+y:5z:58}, 8(x,y,z)=y+x+4 U={(x,y,z):0:5z:53,1:5x:54,2:5y:55}, 8(x,y,z)=y'x U={(x,y,z):l:5x:54,2:5z:53, 0:5y:5x+2z},8(x,y,z)=2x+y+3z
In each of Exercises 85-87, calculate the three first moments and the center of mass of the given solid in space. 85. 86. 87.
U={(x,y,z):0:5x:5l,0:5y:52,0:5z:5y+1}, 8(x,y,z)=y+5 U ={(x,y,z):x2+4y2:5 4,0:5z:52}, 8(x,y,z)=x2 +y2 +2z2 U={(x,y,z):2:5z:53,1:5x:52,1:5y:5x}, 8(x,y,z)=z
In each of Exercises 88-90, a point P is given in rectan gular coordinates. Calculate cylindrical coordinates for P. 88. 89. 90.
P=(2, 0, 3) p=(-1,2,v'3) p=(-2,v'z, -v'z)
In each of Exercises 91-93, a point P is given in rectan gular coordinates. Calculate spherical coordinates for P. 91. p=(0,0,5) 92. p=(v'3,v'3,3v'2) 93. p=(-3,3v'3,6)
In each of Exercises 94-96, either rectangular or spherical or cylindrical coordinates are specified for the point P. Cal culate coordinates of the point in the other two coordinate systems. 94. 95. 96.
P=(5,7r/4,37r/4) p=(- 2v'2,2v'2, 4) p=(8, -7r/3, -8)
spherical rectangular cylindrical
In each of Exercises 97-99, integrate the function f over the solid U using cylindrical coordinates. f(x,y,z)=x2 +y2, U is the region below the paraboloid z=4 - 2x2 - 2y2 and above the plane z=2. 98. f(x,y,z) =1/(x2 +y2), U is the region bounded by the cylinders x2+y2=8 and x2+y2=16 and by the planes z=6 and z=1 99. f(x,y, z)=x2+y2, U is the solid inside the cylinder x2+y2=1 and bounded by z=x2+y2 and z=(x2+y2)114 97.
In each of Exercises 100-102, integrate the function f over the solid U using spherical coordinates. 100. 101.
102.
f(x,y,z)=8,U={(x,y,z):x2 +y2+z2:54} f(x,y,z)=x2 +y2+z2, U is the solid inside the sphere {(x,y,z):x2 +y2+z2=4} and above the xy-plane f(x,y,z)=(1+x2+y2+z2)-1, U is the solid above the cone z=4.jx2+y2 and below the sphere x2+ y2+z2=16
The double integral was introduced by Euler in
1769.
Indeed, at more or less the same time that Cellerier
Four years later, Lagrange used triple integrals to solve
was concocting his novel function, Bernhard Riemann
a problem of mathematical physics. Both of these
(1826-1866)
was thinking along the same lines. Some
mathematicians discovered the method for changing
time before
1861,
variables in an integral. In particular, Lagrange obtained
everywhere-continuous function
Riemann defined and studied the
the formula for integration in spherical coordinates and used it to study gravitational attraction. The general
R(x)
theory of changing variables in a multiple integral was developed by the Ukrainian mathematician, Mikhail Vasilevich Ostrogradskii
(1801-1862),
in
1836.
How
ever, Ostrogradskii's work did not receive wide dis semination in the West. Thus when Carl Gustav Jacobi covered much the same ground in
he received
1841,
credit for the "Jacobian," a determinant that already had appeared in the neglected papers of Ostrogradskii.
19th
f n��2 si
x)
.
n=l
Although Riemann's function is differentiable at some points, it is nondifferentiable at infinitely many points of any open interval. The graph of R over the interval is shown in Figure
1.
[O, 1]
The plotted curve does not seem
too irregular in some places, such as the small rectangle drawn in Figure
1, but when we zoom in, as in Figure 2,
we find that the apparent smoothness is an illusion.
Exotic Functions When
=
y
century mathematicians enlarged the con
0.8
cept of function to include any correspondence of real
0.6
numbers, they opened the door to some exceedingly
0.4
complicated functions. For some time, it was believed
0.2
that the assumption of continuity would be enough to
x
exclude the functions that are beyond the realm of differential calculus. Indeed, it is difficult to imagine a continuous function that has a graph without any smoothness. However, many efforts by notable math ematicians such as Andre Marie Ampere
0.2
0.4
0.6
0.8
� Figure 1 The graph of Riemann's function R(x)
=
I:�i sin��2x).
(1775-1836)
failed to demonstrate that a continuous function must
There is no record that any mathematician other
be differentiable on a "substantial" set. Eventually,
than Karl Weierstrass
some mathematicians began to suspect that continuity
mann's example. Weierstrass took Riemann's con
does not necessarily entail any differentiability.
struction a step further and, in
The
first
"everywhere-continuous
nowhere
differentiable" function was conceived by Bernhard Bolzano
(1781-1848)
in
1830.
(1815-1897)
W(x)
Because communication
=
was aware of Rie announced that
1872,
f �:7rx cos
)
n=O
beyond the established mathematical centers was then very spotty, Bolzano's work remained unknown until
1922.
Around
Cellerier
1860, the Swiss mathematician, Charles (1818-1890), also constructed an everywhere
continuous nowhere-differentiable function, C(x)
f in;:n s
=
x)
.
n=l
However, Cellerier's function was posthumously pub lished after a delay of thirty years, and by that time, his method of construction had become old hat.
0.764
0.772
0.78
� Figure 2 A detail of the graph of Riemann's function R, obtained by zooming in on the small viewing window shown in Figure 1
1045
1 046
Chapter 12
Multiple Integrals
is an everywhere-continuous nowhere-differentiable function. Weierstrass's function is plotted over the interval [O,1] in Figure 3. A detail appears in Figure 4. y 2
x
pose no difficulties for the Riemann integral. However, as mathematicians became familiar with ever more troublesome functions, they discovered some undesir able limitations of the Riemann integral. Above all, the Riemann integral does not "respect" some of the most fundamental processes of calculus. Let's consider one situation in which the Riemann integral comes up short. Set S ={O,1}, S2={O,1/2,1}, 1 S3 ={O,1/3,1/2,2/3,1}, and, in general,
-1
Sn = S n
-2
-1 {.!,�),. U
n
n
n
. .,
n
n
l}·
We define fn on the interval [O, 1] by the formula
_. Figure 3 The graph of Weierstrass's everywhere-continuous nowhere-differentiable function on the interval [O, 1]
fn(x)
y
=
{01
ifxES,, ifx�S,,
·
The graphs of /4 and f5 are shown in Figure 5. In gen eral, the graph of fn is obtained from the graph of fn-1
0.2 0.1 y
x
y lO---CX>-<:>-O-O--o-<:>-OD---0 y
-0.1 -0.2
1 1 43
_. Figure 4 A detail of the graph of Weierstrass's function, obtained by
In 1873, Gaston Darboux (1842-1917) defined yet another nondifferentiable continuous function,
t sin(x (� n=l
n.
f4(x) x
Y = fs(x) 1 s
2 3 34
2 s
3 s
4 s
x
1
_. Figure 5
zooming in on the small rectangle drawn in Figure 3.
G(x)=
1 2
=
+
by moving at most n - 1 points from the line y 1 to the line y Next, define the function f on [O, 1] by =
1)!) ,
and it was far from the last. By the end of the 19th century, such pathological functions had become so commonplace that Charles Hermite wrote, "I turn away with fright and horror from this lamentable plague of functions that do not have derivatives." Such reactions notwithstanding, the very existence of such intractable functions prompted Ulisse Dini (1845-1918) and others to undertake a deeper study of the theory of differentiation.
f(x) =
As badly behaved as they are for differential calculus, the continuous functions that we have been discussing
=
{�
if x is a rational number if x is an irrational number.
It is not hard to deduce that fn(x)-+f(x) for every x. Moreover, this convergence is as nice as can be: The values {fn(x)} decrease to f(x), and the graphs of all the functions are contained in the square [O, 1] x [O, 1]. Under these conditions, it is highly desirable that the limit function f have an integral and that we can interchange limit and integral: lim
n-+oo
The Lebesgue Integral
0.
f1 f1 Jo fn(x) dx = Jo ( lim fn(x)) n-+oo
dx =
Jof1 f(x) dx.
In the case of the sequence we have just constructed, however, the limit function f does not have a Riemann integral. No matter how many subintervals of [O, 1] we
Genesis & Development
1 04 7
take, some of the Riemann sums off will be 1, and
means of an iterated integral. For a simple planar
some will be 0.
region, the boundary is "infinitely thin" and therefore
Double integrals highlight another weakness of the
has zero area. Yet, in 1890, Peano constructed a con
Riemann integral. In a first course in calculus, we limit
tinuous curve that passes through every point of the
our attention to simple planar regions. These regions
square [O, 1] x [O, 1]. Such space-filling curves added to
are enclosed by a finite number of well-behaved curves.
the urgency of coming to a better understanding of
By the 1880s, serious attempts were underway to
multiple integrals.
understand the integral
ffRf(x,y) dA when Ris a very
irregular set of points in the plane. Simple, intuitive
Jordan's theory of measure did not prove to be the right instrument for resolving all the sore points that
concepts of area proved to be inadequate for the irre
beset the theory of integration. However, in the 1890s,
gular planar point sets being considered. Therefore
Emile Borel (1871-1956) published refinements to
Camille Jordan (1838-1922) and Giuseppe Peano
Jordan's theory, and those refinements guided the way
(1858-1932) initiated a program of measuring such
to a successful resolution. That step was accomplished
complicated point sets.
by Henri Lebesgue (1875-1941), who introduced a
Iterated Riemann integrals can also be proble
new integral in his doctoral thesis, which was published
matic. In the 1870s, mathematicians discovered that
in 1902. Within a few years of its introduction, the
changing the order of integration does not always result
Lebesgue integral became one of the primary tools of
in the same answers for some discontinuous integrands
mathematical analysis. It revitalized the study of tri
of two variables. Additional concerns arose. As we
gonometric series, paved the way for the study of
have learned, the boundary of a planar region plays an
function spaces, and played a fundamental role in the
important role when we calculate a double integral by
axiomatic treatment of probability theory.
This page intentionally left blank
Vector Calculus P
R
E
V
E
W
Imagine a fluid flowing through a planar region. Because velocity has both mag nitude and direction, the velocity of the flow at various points can be indicated by vectors. Suppose that we want to concentrate our attention on the flow in a certain region. In this chapter, we will learn to relate the flow through the boundary of the region to the flow in the interior of the region. In the end, we will see that these physical considerations are related to a remarkable multidimensional version of the Fundamental Theorem of Calculus. This chapter begins by developing the mathematical tools that are the basis of vector calculus. It is plain that a number of new ideas are needed. After we have learned the necessary tools in the first few sections, we will come to the theorems Green's Theorem, Stokes's Theorem, and the Divergence Theorem-that give precise formulations of the relationship between the behavior of a vector field on the boundary of a region in the plane or in space and its behavior in the interior. First we need functions that describe flows. Such a function would have to assign a direction and a magnitude to each point in the region; in other words, it would have to be a vector-valued function on a planar or spatial region. Next, we will need to develop certain integrals that involve vector-valued functions. They will enable us to integrate over curves and surfaces. Also, if we are going to use information on the boundary curve or surface to obtain information inside the curve or surface, then we need a mathematical way to tell the inside from the outside: This distinction will be closely connected to the idea of "orienting" a surface. Finally, we will need vector operations that help to measure the flow properties of a vector-valued function. Together with the gradient (which we have already seen in Chapter 11), we will use two new operations called divergence and curl.
1049
Chapter 13
1 050
Vector Calculus
13.1 Vector Fields Suppose that a fluid passes through a region Q in the plane or in space. At each point P, we may use a vector v(P) to represent the magnitude and direction of the - ,...
...
.... ____ _ __..
,.,
Ir'
__ ,...r,.,
,.,
��@���J1��i������ 1'( //..A' 1'( ,
,., ,
,,,
,...
velocity of the flow. Several of these vectors are sketched in Figure 1. By asso ciating a vector
v(P) to each point P in Q, we are in effect creating a vector-valued vector field on the region Q.
function with domain Q. Such a function is said to be a Let us make these concepts precise.
...
""
,
_. Figure 1
•·M@hi[.]:@
We say that a planar set Q is
open if each of its points is the center
of a disk with positive radius that is contained in Q. A set Q in three-dimensional
space is
open if each of its points is the center of a ball with positive radius that is path-connected if
contained in Q. A set Q in the plane or in space is said to be
every pair of points in Q can be connected by a continuous curve that lies entirely in Q. A set that is both open and path-connected is called a
region.
Figure 2 shows a set that is both open and path-connected: It is a region. Figure 3 shows an open set Q that comprises two components. This set is not a region because it is not path-connected: A point in one component cannot be connected to a point in the other component by means of a continuous curve that lies entirely within Q. In Figure 4, a new set S is formed by connecting the two components of Q with a line segment £. The set S is path-connected, but it is still
not a region because it is not an
open set: Each point of£ is in S but is not the center of any disk that is contained in S.
g
g
_. Figure 3 The set Q comprises two components. Q is open but not path-connected; it is not a region.
_. Figure 2 A set that is open
and path-connected: It is a region.
•·MiWMeU@
_. Figure 4 The set S Q u C is path-connected but not open; it is not a region. =
A vector field v on a region Q is a function that assigns to each point
v are planar vectors. That v1(x,y)i + v2(x,y)j for each point (x,y) in Q. If Q is a region in space, then the values of v are vectors in space, and v has the form v(x,y, z) v1 (x,y, z)i + v2(x,y, z)j + v3(x,y, z)k for each point (x,y, z) in Q. of Q a vector. If Q is a planar region, then the values of
is,
v has the form v(x,y) =
=
13.1
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \..
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \..
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. v(x,y)
_. Figure
INSIGHT
y \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. =
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. i-j
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \..
\.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \.. \..
\.. \.. \.. \.. \.. \,.x \.. \.. \.. \.. \.. \..
(1,0)
w(x,y)
1 051
10, we used vector-valued functions tr-+ r(t) of a scalar
In Chapter
variable t to parameterize curves. A vector field Pr-; v(P) is also a vector-valued function. It differs from a curve in that its domain consists not of scalars but of points P that
have two or three coordinates.
� EXAM PL E 1 Let g denote the entire plane (which is a two-dimensional region). Sketch the vector fields
v(x,y) =i- j
and
w(x,y) =xi- (y/4)j, which
are
defined on g.
Solution The vector field v assigns to each point (x,y) in the plane the vector i- j. It is a constant vector field. A sketch is indicated in Figure 5. Note that we make a sketch by drawing
some
of the vectors (if we drew them all, the picture would just
be solid ink). Now consider the vector field
w. This is a vector field that assigns to the point (x,y) in the plane the vector xi- (y/4)j. Some of the values of this vector field are indicated in Table 1.
5
(x,y)
Vector Fields
(0, 1)
(-1, 0)
(0,-1)
1. -J
-i
-J
4
1. 4
(1,1) .
1. 4
I - -J
(-1,1) .
-1-
1. 4
-J
(1,-1) .
1. 4
I+ -J
(-1,-1) 1. 4J
.
-· + -
(2,2) 1. 2.1- -J 2
(-2,2) -2.1-
1. 2
-J
Table 1 These values of the vector field
w, and
a few others, are shown in Figure 6.
In previous chapters, we have seen that it is convenient to deal with functions that are continuously differentiable. These functions have much nicer properties than arbitrary functions. The same is true for vector fields. As a result, we will deal primarily with continuously differentiable vector fields. We call a planar vector field v(x,y) =v1(x,y)i+v2(x,y)j or a spatial vector field w(x,y,z) =w1(x,y,z)i+ w2(x,y, z)j +w3(x,y, z)k continuously differentiable if each of the component functions v1, v2 or w1, w2, W3 is continuously differentiable.
Vector Fields in Physics
Vector fields arise naturally in many physical contexts. For example, gravitational, magnetic, and electrostatic forces generate vector fields that are called force fields. Another example from physics that will be important for us is the velocity vector field.
y � EXAM PL E 2 Let v denote the velocity of a wind that blows through a wind tunnel parallel to the sides of the tunnel. Suppose that, at each point the magnitude of
v(P) is
proportional to the height of
that the ground is at height z
x
-_+-4 -_-12--+---.i 2f-+4 •
_. Figure
6
P above the ground. Assume i points down the tunnel in 7). Describe the velocity as a
and that the vector
the direction that the wind is blowing (see Figure vector field.
magnitude of
//
=xi
=0
Solution We let c be the constant of proportionality. At P =(x,y, z), the
----'-1
w(x, y)
P =(x, y, z),
- (y/4)j
is therefore
v(P) is cz, and the v(x,y,z) =czi.
direction of
v(P)
is
i.
The vector field we seek
� EXAM PL E 3 Let p be a positive constant. Suppose that a solid ball 2 2 B = { (x,y, z) : x +y +z2 � p2 } is rotating about the z-axis with constant angular
1052
Chapter 13
Vector Calculus
speed w. Looking down at the xy-plane from the point (0,0, 2p) on the positive
z
z-axis, the rotation of 8 is counterclockwise. Show that there is a vector n such that the velocity vector field P �l'(P) is given by v(P) =fix Of> for each point
P=(x, y, z)
in 8. (The vector
fiis called
the angular velocity.)
P=(x, y, z) in B traces out a circleCp in a horizontal plane. A=(0,0,z) on the z-axis. The radius vector from A to Pis r=AP= ( x- 0,y- 0,z- z)={ x,y,O). Because Ptraces outCp, the velocity vector l'(P) at P=(x, y, z) is tangent to Cp. Because Cp is a circle, we deduce that v(P) is perpendicular to the radius vector r= ( x,y,O} of Cp (see Figure 8). It follows that the direction, dir(v(P)), of v(P)is equal to llrU-1(-y,x,O}. Letting v(P) denote the magnitude of v(P), we obtain Solution Each point
The center of the circle is the point
-
-
T(x,y,z)
=
c.d
A Figure 7
v(P) <9�·7> lll'(P)lldir(v(P)) = F(P)llrll-1(-y,x,O).
z
To determine v(P), notice that the arc length & that P traces during a time interval !l.t is equal to the product of the radius and the central angle subtended by the traced arc. That is, !l.s=llrll (w!l.t). This equation tells us that
v(P) = ds = dt
Substituting this value for A Figure 8 The velocity vector v(P) at P=(x,y,z) is perpendi cular to vector r= (x,y,O}.
fun & !l..t
at--..0
=
v(P) into the
fun
at--..0
llrll (w!l.t) = wllrll· !l.t
fonnula for
v(P), we
obtain
v(P)=(wllrll )llrlr1(-y,x,O) =w{-y,x,O) =-wy i+wxj+Ok. We
may now verify that fi =wkis a vector that satisfies the requirements: - =det Ox OP
([
i
� �j � ...
])
=-wyi+wxj+Ok=v(P).
�
� EXAMPLE 4 A mass is located at the point Po=(xo,yo,zo). It exerts a gravitational attraction F(P) at each point P f. Po in space. At P, the direction of the gravitational force is from P toward P0 and, according to Newton's law of gravitation, the magnitude of the force is proportional to the reciprocal of the square of the distance of as a vector field.
F(P)
Solution Let r=Of>
P to P0• Using a as the
proportionality constant, express
and ro=OPo. The distance of P=(x,y,z) to Pois
)
(
112 . Ur-roll = (x - xo) 2+(y -Yo)2+(z- zo)2 the magnitude of the gravitational force vector is allr-r01r2• The gravitational force vector points in the same direction as PP� =ro -r. Because the direction vector of PP� is (-1/llr-roll )(r-ro), we conclude that Newton's law tells us that
F(P)=llF(P)lldir(F(P)) =a:llr-rolr2
(IIr-roI ) (r-ro)=- r- ro -l
U
0
ll
3
(r-ro).
13.1 Vector Fields
1 053
In component form, the force is given by
F(x,y,z)- _
a(y - Yo) a(x-xo) 3/212 2 2 2 2 2 3/2l ((x-xo) +(y-yo) +(z-zo) ) ((x-xo) +(y-yo) +(z-zo) )
•
•
_
Integral Curves (Streamlines)
If
F
a(z-zo) k. 2 2 2 3/2 ((x-xo) +(y-Yo) +(z-zo) )
is a vector field on a region 9, and
t H r(t)
is a vector-valued function
whose values lie in 9, then we say that the curve C described by
curve
of
F
<111111
r
is an
integral
if
r'(t)
=
(13.1.1)
F(r(t)).
(13.1.1) tells us that for each point P= r(t) of the integral curve C, the vector F(P) is tangent to C at P. This behavior can be seen in Figure 9, which shows Equation
a vector field and three of its integral curves. Integral curves are often called by other names, such as field
lines.
(A curve that passes through a vector field is often
r as the position vector F may be interpreted as a velocity field. In this called flow lines or streamlines. As equation (13.1.1)
called a "line," whether or not it is straight.) If we think of of a point mass moving along C, then
_. Figure 9 A vector field and some of its integral curves
case, the integral curves are
suggests, determining the integral curves of a vector field usually amounts to sol ving a differential equation or a system of differential equations.
� EXAM PL E 5 Determine the field lines of the vector field
y
II/Ill/Ill I I I I I I I I I I
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
- x
that is shown in Figure
Solution We seek
10.
r(t)= x(t)i+y(t)j
F(x,y)= i+yj
such that
x'(t)i +y'(t)j = r'(t)= F(r(t)) = i+y(t)j= 1i+y(t)j. By equating the
i and j
components, we obtain two differential equations:
x'(t)= 1
y'(t)= y(t). The first of these differential equations can be integrated to give t+ C1 for some constant C1. The second differential equation, y'(t) y(t), is an equation of exponential growth: Each of its solutions is of the form y= C2et for some constant C2• (To derive this result, apply the technique for solving separable differential equations, as discussed in Section 7.6 of Chapter 7. Alternatively, set a= 0 and f3= -1 in differential equation (7.7.4) and its solution (7.7.5).) Com bining the two solutions x= t + C1 and y= C2et that we have found, we see that and
�
�
�
�
�
�
�
�
�
�
\ \ \ \ \ \ \ \ \ \
\\\\\\\\\\ F(x,y) _.
Figure 10
=
i+
yj
x
=
=
C c2e-c1• Thus the field lines of F(x,y) i+yj are the graphs of the equations y= cex where C is a constant. Several of these field lines are shown in Figure 11.
for the constant
=
� EXAM PL E 6 Let
=
w
be a positive constant. The differential equation
d2 u(t) +w2u(t) dt2
=
0
(13.1.2)
Chapter 13
1 054
Vector Calculus
is called the harmonic oscillator equation. It is known that u(t) is a solution of this equation if and only if u(t) =Acos(wt)+Bsin(wt) for some constants A and B. Use this fact to determine the integral curves of the vector field F(x,y) = -wyi+wxj.
y
Solution
We seek r(t) =x(t)i+y(t)j such that x'(t)i+y'(t)j = r'(t) =F(r(t)) = -wy(t)i+wx(t)j.
By equating i and j components, we obtain the following system of two differential equations: -w y(t)
(13.1.3)
y'(t) =wx(t).
(13.1.4)
x'(t) .A Figure 11
F(x,y)
=
i + yj
Field lines of
=
and
By differentiating equation (13.1.3) with respect to t and then using equation (13.1.4) to substitute for y'(t), we obtain x"(t) = -wy'(t) = -w (wx(t)) = -w2x(t), which may be rewritten as x"(t)+w2x(t) 0. In other words, x(t) is a solution of the harmonic oscillator equation. From the stated information, it follows that =
x(t) =Acos(wt)+ Bsin(wt)
(13.1.5)
for some constants A and B. By differentiating this formula for x(t), we find that x'(t) = -Aw sin(wt)+Bw cos(wt).
(13.1.6)
Now rewrite equation (13.1.3) as y(t) (-l/w)x '(t), and, into this equation for y(t), substitute formula (13.1.6) for x'(t). The result is =
+---+-----l f---1--+----+---+... x
y(t)
=
-Bcos(wt) +Asin(wt).
(13.1.7)
Putting together formulas (13.1.5) and (13.1.7), we conclude that the integral curves of the vector field F(x,y) = -wy i+wx j have the form .A Figure 12 The vector field F(x,y) = -w yi + w xj with three of its integral curves
r(t) = (x(t),y(t)) = (Acos(wt)+Bsin(wt), -Bcos(wt)+Asin(wt)). In the case that B is 0, it is clear that the integral curve r(t) =A(cos(wt),sin(wt)) describes the counterclockwise traversal of a circle of radius IAI. In general, the calculation
ll r(t)ll2 = (Acos(wt)+Bsin(wt)) 2+ ( -Bcos(wt) +Asin(wt)) 2
=A2 (cos2(wt)+sin2(wt)) +B2 ( cos2(wt)+sin2(wt)) +2ABcos(wt)sin(wt) - 2ABcos(wt)sin(wt) =A2+B2 shows that r parameterizes a circle of radius JA2+B2 (see Figure 12). ..,..
13.1 Vector Fields
Gradient Vector Fields and Potential Functions
In Section
11.6 of
11,
we introduced the gradient operator. When applied
to the scalar-valued function
in the vector-valued
function
Chapter
1 055
u(x,y), the gradient results ux(x,y)i + uy(x,y)j, and, when applied
V'u(x,y) to the scalar-valued u(x,y,z), the gradient results in the vector-valued function V'u(x,y,z) ux(x,y,z)i + uy(x,y,z)j + Uz(x,y,z)k. Notice that (x,y) V'u(x,y) is a planar vec tor field, and (x,y,z) V'u(x,y,z) is a spatial vector field. Thus whether u is a =
function
=
�
�
scalar-valued function of two or three variables, its gradient V'u is a vector field .
� EX A M P L E 7 Let the temperature at a point in the plane be given by
T(x,y)
=
2
x
+
2.
6y
Calculate the gradient of T, and give a physical interpretation.
Solution The gradient vector field of the temperature function is 2.xi +
12yj.
V'T(x,y)
=
Physically, this vector field tells us that if we are standing at a point Pin
the plane, then V'T(P) points in the direction of most rapid increase of temperature at P. The magnitude of V'T(P) indicates the rate of increase of the function T at the point Pin the direction of V'T(P).
<11111
g is a gradient vector field g for which F(P) V'u(P) at each point
We say that a vector field F on a region if there is a scalar-valued function P of
u
on
=
Q. A gradient vector field is also called a conservative vector field. If V is a g such that F(P) -V'V(P) for each point Pin Q, then
scalar-valued function on we say that V is a
=
potential function
for F.
A few remarks are in order concerning the term bit ambiguous. If
u
potential function, which -u, then
is a scalar-valued function, and we set V F(P)
is a
=
V'u(P)
(13.1.8)
-V'V(P).
(13.1.9)
=
if and only if F(P)
=
In other words, given a vector field F, equation
(13.1.8) has a solution u if and only (13.1.9) has a solution V. This means that a vector field F is conservative (equation (13.1.8) can be solved for u) if and only if F has a potential function (equation (13.1.9) can be solved for V). Of the two equivalent equations, mathe maticians tend to work with (13.1.8). In the mathematical literature, a scalar function u that solves equation (13.1.8) is often called a potential function for the if equation
vector field F. In physical applications, however, a function V that solves equation
(13.1.9) has energy. The
an important significance-it is the function used to define
potential
next example develops this background.
� EX A M P L E 8 Suppose that a force field F has a potential function V in a region
Q. Suppose that, due to the action of F, a particle of mass
with a continuously differentiable position vector
�mllr'(t)ll2 for some scalar constant C.
+ V
( (t) )
r
r t)
( ,
=
C
m
a :5 t :5 b.
moves through
g
Show that
(13.1.10)
1 056
Chapter 13
Vector Calculus Solution According to Newton's law, force equals mass times acceleration. We
therefore have F(r(t)) mr"(t). Because F has a potential function V, we may rewrite the force equation as -\7V(r(t)) mr"(t), or =
=
mr"(t) + \7V (r(t))
(13.1.11)
0.
=
If we form the dot product of each side of equation
(13.1.11)
obtain mr'(t) r"(t) + \7V (r(t)) r'(t) ·
Now llr'(t)ll2
=
·
=
(13.1.12)
0.
r'(t) r'(t). Therefore ·
:t llr'(t) 112 :t (r'(t) r'(t)) (:t r'(t)) =
·
with r'(t), then we
=
·
r'(t) + r'(t)
·
(: )
which allows us to write equation
t
r'(t)
=
r"(t) r'(t) + r'(t) r"(t) ·
·
+ \7V r(t)
(
)
·
r'(t)
=
V (r(t))
=
r"(t),
:t ( � m llr'(t)ll2
(13.1.13). The
\7V (r(t)) r'(t). ·
Taking this equation into account, we rewrite
or
·
(13.1.13)
0.
Next we look at the second summand on the left side of equation Chain Rule, in the form of equation (11.6.3), tells us that d
2r' (t)
(13.1.12) as
� m :t llr'(t)ll2
dt
=
(13.1.13) as
+ V r(t)
(
)
)
=
0.
It follows that
� m llr'(t)ll 2 which is equation
+ V r(t)
(
)
=
C,
(13.1.10). ..,..
Suppose that, as in Example 8, a conservative force field F acts on a particle that has mass m and position vector r(t). If Vo is a potential function for F, and if k is any scalar constant, then, because \7( V0 + k) \7V0 + \7k -F + 0 -F, we see =
that V
=
=
=
Vo + k is also a potential function for F. After a choice of potential
function V has been fixed, the quantity V(P) is said to be the potential energy of the mass at point P
=
r(t). The quantity
� m llr'(t)ll2, that is, half the product of the mass
and the square of the speed, is defined to be the kinetic energy of the mass at P r(t). Equation (13.1.10) states that the total mechanical energy of the particle kinetic energy plus potential energy-is constant. For this reason, equation =
13.1 Vector Fields
1057
(13.1.10) is called the Law of Conservation of Mechanical Energy for the con servative force F. � EXAMPLE 9 Suppose that F is the planar force field given by F(x,y)=-mgj. This is the force field for a particle of mass m that falls with constant acceleration a=-gj. Show that F is conservative. Find the potential function V that satisfies V(x,0)=0 for all x. Verify the Law of Conservation of Mechanical Energy for a particle of mass m that is dropped from the point (xo,Yo) .
If u(x,y)=-mgy + C and V(x,y)=mgy + C for any constant C, then we have F='\lu and F=-'\!V. Therefore F is a conservative force field. If we take C=0, then V(x,y)=mgy satisfies V(x,0)=0 for all x. Suppose that y0 > 0. If the particle is dropped from the point (x0,y0) and its position at time t is r(t), then we have Solution
a(t)=r"(t)=-gj, v(t)=r(t)=
l
t (- gj)dT= -gtj,
and r(t)=xoi+yoj+
l
t
r'(T )dT=xoi+yoj+
la
t
( � )
(-gT j)dT=xoi+ yo-
gt2 j.
At time t, the kinetic energy of the particle is
and the potential energy of the particle is
(
V(r(t))=v xo,yo-
� gt2)=mg (yo- � gt2)=mgyo- � mg2t2.
The total mechanical energy is
� mllr'(t) ll2 + V(r(t))=� mg2t2 + ( mgyo- � mg2t2)=mgyo, which does not depend on time, as asserted by the Law of Conservation of Mechanical Energy. <1111 Finding Potential Functions
In Section 13.3, we will learn how to recognize a conservative vector field. The next two examples show how to find a potential function for a vector field that is known to be conservative. The vector fieldF(x,y)=(y-3x2)i+(x+sin(y) )j is known to be conservative. Find a scalar-valued function u for which F='\lu. � EXAMPLE 10
Solution
We want u to satisfy the vector equation
(y-3x2)i+(x+sin(y))j=F(x,y)='\lu(x,y)=ux(x,y)i+uy(x,y)j.
1 058
Chapter 13
Vector Calculus
This vector equation is equivalent to the two simultaneous scalar equations Ux(x,y)=y- 3x2
(13.1.14)
uy(x,y)=x+sin(y).
(13.1.15)
and
In order to remove the partial derivative from u in equation (13.1.14), we anti differentiate each side of the equation ux(x,y)=y- 3x2 with respect to the x variable. Because we treaty as a constant when we calculate ux(x,y), we treaty as a constant when we integrate to undo the process. As a result, the constant of integration that we obtain can depend on y . Thus antidifferentiating each side of equation (13.1.14) yields u(x,y)=xy-x3+rj;(y),
(13.1.16)
where the constant of integration is some unknown function rj;(y). To verify our assertion concerning the constant of integration, observe that partial differentiating the right side of equation (13.1.16) with respect to x does result in the right side of equation (13.1.14). Our goal now is to determine the expression rjJ(y) that has arisen in the integration. To do so, we differentiate each side of equation (13.1.16) with respect to y , obtaining Uy(x,y)
=
(13.1.17)
X+r/J'(y).
Notice that the expression rjJ'(y) is an ordinary derivative because¢ depends only ony . Comparing equation (13.1.17) with equation (13.1.15), we see that ¢'(y) sin(y). After antidifferentiating with respect to y , we obtain rj;(y)=-cos(y)+C. Observe that, in this last antidifferentiation, which we performed to remove an ordinary deri vative, the constant of integration C is a numerical constant. By substituting our formula for rjJ(y) into equation (13.1.16), we obtain =
u(x,y) xy -x3 - cos(y)+C. =
...._
� EXAMPLE 11 The vector field F(x,y,z)=y2i+(2.xy+z)j+(y+3)k is known to be conservative. Find a scalar-valued function u for which F= V'u. Solution
We want to find a function u whose partial derivatives satisfy the
equations ( a) ux(x,y, z)=y2,
( b) uy(x,y,z)=2.xy+z,
and
( c) Uz(x,y,z)=y+3.
We antidifferentiate the first of these equations with respect to the x variable. The result is u(x,y,z)=xy2 + rjJ(y,z). Notice that the constant of integration rjJ(y,z) is constant only with respect to x . It can depend on y and z . Next, we determine the expression rjJ(y,z). To do so, we differentiate with respect toy each side of the equation u(x,y,z) xy2+rjJ(y,z). This differentiation yields the equation uy(x,y,z)=2.xy+¢y(Y,z). Comparing this last equation with =
13.1 Vector Fields
equation (b), we see that 2.xy + rPy(y, z)
=
2.xy +
z,
or rPy(y, z)
=
z.
1 059
We anti
differentiate this last equation with respect toy, noting that the constant of inte gration can depend on formula for
¢(y, z)
The result is
z.
into our equation for
u(x,y, z)
=
¢(y, z) yz + 'tf;(z). u, obtaining =
We substitute this
2 xy + yz + 'tf;(z).
(13.1.18)
Our next task is to identify the unknown expression partial derivative of each side of equation
Uz(x,y, z) =y + 'lj;'(z).
Comparing the right side of this last equation with the right
side of equation (c), we see that y +
'lj;(z)
=
3z + C
'lj;(z)
u(x,y, z)
Q U IZ
1.
Is
2. Is
3.
'tf;'(z)
y + 3, or
=
'lj;'(z)
=
3.
It follows that
where C is a numerical constant. We conclude our computation by
substituting our expression for
Q UIC K
'lj;(z). To do so, we take the (13.1.18) with respect to z. The result is
=
into equation
2 xy + yz +
(13.1.18),
3z + C.
obtaining
<1111
{(x,y, z) : 0
not?
True or false: If a particle is in motion due to a conservative force field, then the kinetic energy of the particle is conserved.
4. The vector field
valued function
F(x,y) yi + (x + 3)j is known to be conservative. Find a scalar u for which F =Vu. =
Answers
1.
Yes
2. Not path-connected
3.
False
4.
u(x,y)
=
xy + 3y + C
EXERCISES being the direction that the wind blows. Write the velocity
Problems for Practice In each of Exercises 1-8, sketch the given vector field in the plane by drawing the vectors that are assigned to several points.
1. 2. 3. 4. 5. 6. 7. 8.
vector field.
10. A cylindrical tank is 5 meters high with a radius of 4 meters. Water circulates about the tank's central axis, the line segment [O, 5] on the z-axis, in such a way that
F(x,y) = 3i+ j F(x,y) =xi+ yj F(x,y) = 2yi- 3j F(x,y) = 3xi + xj F(x,y) = yi- Sxj F(x,y) = x2i+ y2j F(x,y) =x2i- y2j F(x,y) i + sin(y)j
the flow is clockwise as you look down on it, and the speed at any point is proportional to the distance of the point from the center of the tank. Write the velocity vector field.
1L In the tank described in the preceding exercise, water circulates about the tank's central axis in such a way that the flow is clockwise as you look down on it, and the speed at any point is proportional to the distance of the point from the central axis of the tank. Write the
=
In each of Exercises 9-12, determine an explicit formula for the spatial vector field that is described.
takes place about the y-axis. Each particle rotates in a
9. A wind blows down a 10 foot high hall with vertical walls and width 8 feet. The speed of the wind at a point is proportional to the distance of the point from a line
velocity vector field.
12. Particles circulate in space in such a way that rotation
run
ning down the middle of the floor. The line down the middle of the floor is the y-axis, with the positive direction
plane perpendicular to the y-axis so that its oriented axis of rotation, determined by the right-hand rule, is the vector
j.
The speed of any particle is proportional to
the square of its distance from the y-axis. Write the velocity vector field.
1 060
Chapter 13
Vector Calculus 13-20, write the gradient vector field
In each of Exercises
associated with the given scalar-valued function.
u(x,y)=xy-y2+ 3x u(x,y) =sin(xy2) u(x,y)=ln(x3 -y2) u(x,y)=ezy-Sx u(x,y,z)=x2y3z sin(xy) 18. u(x,y,z) cos(z) ln(x -y) 19. u(x,y,z) ln(z) 20. u(x,y,z) =ecos(x)esin(y)ez
13. 14. 15. 16. 17.
Exercises
of
Exercises
43-46, the given vector field con-
(y2+zzrli -2xy(y2 +z2r2j -2 xz(y2+z2r2 k 45. xY-1yi+ xYln(x)j+ ln(z)k 46. ln(y y i+(xz/y)ln(y)z-lj + xln(yYln(ln(y))k 47. Show that F(x,y)=yi+2xj is not the gradient vector field of any continuously differentiable function.
21-26, the given vector field conservative. Find an expres
48. On separate sets of axes, sketch the following vector fields in a neighborhood of the origin in the plane: a. b.
21.i+j 22. xi+yj 23. yi+ xj 24. yi+(x+2)j 25. (y3+2 xy)i+ (3xy2+ x2)j 26. 2 eh-Yi -e2x-yj each
Exercises
43. z(y +zr1i -xz(y+zr2j + xy(y+zr2k
F(x,y)= M(x,y)i+ N(x,y)j is sion u(x,y) such that F =V'u.
In
of
44.
_
of
each
In
--
each
x x _) i+ _ _j (1n(x+y)+ _x+y x+y
F(x,y,z)= M(x,y,z)i+N(x,y,z)j+R(x,y,z)k is servative. Find an expression u(x,y,z) such that F=V'u.
_
In
42.
c.
d. e. f. g. h.
27-32, the given vector field
F(x,y)=xi+yj F(x,y)=-xi -yj F(x,y)=-xi+yj F(x,y)=xi -yj F(x,y)=yi+ xj F(x,y)=-yi+ xj F(x,y)=yi-xj F(x,y)=-yi-xj
Each of these exhibits qualitatively different behavior near
F
F(x,y,z)=M(x,y,z)i+N(x,y,z)j+R(x,y,z)k is conservative. Find an expression u(x,y,z) such that F=V'u.
the origin, which is the point where
27. 2xi+j+Ok 28. i+j+2zk
at the origin exhibit one of these eight modes of behavior.
29. 30. 31. 32.
In each of Exercises
33-38, find the integral curves t
of the given vector field
33. 34. 35. 36. 37. 38.
F(x,y)= M(x,y)i+ N(x,y)j.
satisfying a certain nondegeneracy condition and vanishing
49. Let r(x,y,z)= ll xi+yj+zkll
2 xyz3i+(x2z3+S)j+ 3x2yz2k yi+(x+2z)j +2yk (z3 -2 xy2)i+(z2 -2x2y)j+(3xz2+2yz)k (sin(y) -zsin(x))i+ xcos(y)j+cos(x)k
vanishes. It can be
proved, using advanced techniques, that typical vector fields
·
Calculate
Y'(l/r)
at all
points where the derivative exists.
50. Coulomb's Law says that the force F acting on a charge q at position
H
r
because of a charge
F=
r(t)
Q
at the origin is
cQq r 3 llrll
c is a positive constant that depends on the physical rII is a potential function for F. If Qq > 0, then F is said to be repu lsive, while if Qq < 0, then Fis said to be attractive. Explain why where
F(x,y)=i+j F(x,y)=i+2j F(x,y)=xi+2j F(x,y)=xi -yj F(x,y)=2yi+j F(x,y)=i - xj
units that are used. Show that V=cQq/II
this terminology is appropriate.
51. At the point (x,y), the elevation of the terrain is h(x,y) -x2+ 3xy+2y2+ 100 feet. Sketch the vector field =
Further Theory and Practice In
each
of Exercises
39-42, the given vector field
52. At the point (x,y) on a sheet of aluminum, the tem perature is T(x,y) 150 -xy+4y3. Sketch the vector
conservative. Find an expres
field that shows at each point the direction of greatest
F(x,y)= M(x,y)i+ N(x,y)j is sion u(x,y) such that F =V'u. 39. (e2x+2xe2x)i+j 40.
that shows the direction of quickest descent at each point.
(i+�2y2}+ (i+:2y2) j
41. y2cos(xy)i+ (sin(xy)+ xycos(xy))j
=
increase in temperature. If
ri and r 2
are distinct real numbers, then all solutions of
the differential equation
d2
dtZ
u(t) -(r1+ rz)
d u(t)+ rirzu(t)=0 dx
13.2 Line Integrals
u(t) =Aexp(r1t) + Bexp(r t) for some con 2 B. In each of Exercises 53-56, use this fact to determine the integral curves t 1--> r(t) of the given vector field F(x,y) M(x,y)i + N(x,y)j.
1061
have the form
"Combing the Hairy Sphere" Theorem, there is no such
stants A and
continuous vector field on the unit sphere in three dimensional space.
=
53. 54. 55. 56.
Calculator/Computer Exercises
F(x,y) (3x + y)i + 2yj F(x,y) = 2yi + (3y -x)j F(x,y) = (3x -4y)i -xj F(x,y) = (3x + 2y)i + (12x -7y)j =
58. Draw the given planar vector fields in the viewing window [-1,1) x [-1,1): a. (x + y)i + (1 + xy)j b. xyi + (1 + y -x)j c. (x + y)i + (x - y2)j d. (x2 - y2)i + (x2 + y2)j
a continuous vector field F(x,y) M(x,y)i + N(x,y)j such that, at each point (x,y) on the circle C { (x,y) : x2 + y2 = 1 }, the vector F(x,y) satisfies llF(x,y)ll 1, and F(x,y) is tangent to C. According to the
57. Find
=
=
=
1 3.2 Line Integrals In Chapters 5 through
11, the integrations we performed were over subintervals of 12, we learned to integrate over regions in the plane and
the real line. In Chapter
over solids in space. In this section, we will study how to integrate over curves in the plane and in space. By expanding the types of sets over which we can integrate, we will be able to learn new applications of the integral. For instance, if we apply a force at each point of a curve, then we need to integrate the force along the curve to calculate the work performed. Similarly, if a thin wire takes the shape of a curve, then we can integrate a mass density over the curve to obtain the total mass of the wire. In this section, we study the mathematical ideas that are the basis of these and other applications.
Work Along a Curved
In Section 7.5 of Chapter 7, we developed a method for computing the work
Path
performed in moving a mass along a line segment. Now we would like to calculate work along a curved path. To be specific, we suppose that
F is
a continuous force
field in a region of the plane or in space. If a particle moves through this force field
r(t) for a:::;; t:::;; b, then how much work is performed? See 1. The interesting feature here is that sometimes the force might be directly opposed to the path (point Pin Figure 1), whereas at other times it will be skew to the path (point Qin Figure 1). In other words, we do not assume that there is any with position vector
Figure
relationship between the direction of the force, curve,
T(t)= dir(r'(t)),
at a point
W
and the direction of the
by partitioning the domain
/
intervals
/ / I
I
then we have the approximation
/
dir(F(P)),
of the curve.
[a, b] of r into N sub t::..t= (b - a)/N. For each j, we let Wi denote the work performed as t passes from ti-l to ti. If �i is an arbitrary point in Ii> We estimate the work
I
'f{b )
P= r(t)
_. Figure 1
Wi
Rj
Ii= [tj-l, ti]
with equal lengths
(component of Force in direction of motion at r(�j)) · (distance between r(tj-1)r(tj)). Using the first equation of line tangent vector to the curve at
(9.3.8)
r(�j),
to evaluate the component of
we obtain
F
along the
1 062
Chapter 13
Vector Calculus
Because we can approximate the velocity vector
r'(ej) by the difference quotient
1 (r(tj) r(tj )), - -1 �t
llr(tj)- r(tj-1) II llr'(ej) ll�t. When we substitute this last expression into our approximation for Wj, we obtain Wj (F(r(ej)) r'(ej))�t. Therefore the work W performed in traversing the entire curve from r( a) to r(b) is approximately it follows that
�
�
·
N
l:F(r(ej)) ·r'(ej)�t, j=l which we notice is a Riemann sum. As N increases, the approximation becomes more accurate, and the sum tends to the integral of
[a, b]. This observation leads to the
�·liiiliiiMll a ::5 t ::5 b, IS
�
The work done by a force
·
F along a parameterized curve t
lb F(r(t)) ·r'(t) dt.
EXA M P L E
F(x,y)=(y, 1+x)
F(r(t)) r'(t) over the interval
definition of work.
1
What
is
the
work
done
>-+
r(t),
(13.2.1)
by
the
planar
force
field
r(t)=(t3,t2), 0:st:s1? If the r (t)= ((2- t)3' (2- t)2), 1::;; t::;; 2, then
moving a particle along the path
particle retraces its path with trajectory what is the work done by Solution We have
F during this "return trip"?
F(r(t))=F(t3,t2)=(t2,1+t3) and r'(t)=(3t2,2t). (13.2.1), the work performed is equal to
Therefore
according to formula
1 1F(r(t)) ·r'(t) dt
=
=
1 1(t2,1+t3) (3t2,2t) dt 1 1 1(3t4+2t(l+t3)) dt=1(5t4+2t) dt=(t5+t2) =2. 0 ·
11
The work performed along the return path is
12 F(r(t)). r'(t) dt
=
=
12 ((2- t)2,1+(2- t)3). (- 3(2- t)2,-2(2- t)) dt
-12 (5(2- t)4+2(2- t))dt
13.2 Line Integrals
1 063
and therefore
2
1 F(r(t)) . r'(t) dt INSIGHT
=
( (2 - t)5
+
(2 - t)2
)[
=
-2.
_,..
Work is a scalar quantity that can be positive, negative, or zero. In
Example 1, the values of work associated with the paths traced by
r and r are opposite in
sign. The physical meaning of "negative work" will be explored in Section 13.3. However,
F over r is negative. Observe r and r both trace the graph of y = x213, 0 :5 x :5 1 but in opposite directions (Figure 2). We can see from the plot that the angles between the vectors F{r(t)) and r'(t) are acute. The integrand F(r(t)) r'(t) is therefore nonnegative and F(r(t)) r'(t)dt>O. On the other hand, r' points in the opposite direction of r', the angles between F(r(t)) and r'(t) are obtuse, and the integrand F(r(t)) r'(t) is negative. It follows that F(r(t)) i'(t)dt < 0. we can understand right away why the work done by that the functions
·
Jg
·
·
Jg
·
y 1 ./ / / / /
/
;
/ / / / / /_,I 1 /---}_ I I - I I ! ! 1 I x I I
/p/ / I / / / I I I 0.6 I I I I I 0.4 I I I I I I I I I I 0.8
--- ------' - -""'--
r'f'
0 Po
0.2
0.4 y
_. Figure 2
Line Integrals
=
0.6
0.8
1
x2/3
smooth parametric curve is a continuously differentiable vector t�r(t),a
Let us recall that a valued function
term "curve" in its everyday sense, we really mean the geometric set of values, C
=
{ r(t) : a
domain . In fact, C has an
r imparts to C an initial point r(a), a terminal point r(b), and a direction for traversing Cfrom r(a) to r(b), we refer to Cas a directed or oriented curve. Whenever we speak of a parameterization of a directed curve C,
additional attribute. Because
we imply that the parameterization respects the direction of C. We now want to define the integral of a vector field
F over a directed curve C. F over C.
Our calculation of work motivates the definition of the integral of
�·jHiiillMll
Let C be a directed curve (in the plane or in space) with para
t H r(t), a:::;; t:::;; b. Suppose that F is a continuous vector field defined on C. Then, the line integral (or path integral) of F over C is denoted by the symbol fc F dr and defined to be
meterization
·
1 F dr lb F(r(t)) ·
=
·
r'(t) dt.
(13.2.2)
1 064
Chapter 13
Vector Calculus If C is a piecewise-smooth directed curve that is the union of smooth arcs C1,C2,...,Cn (as in Figure
{ F . dr
k
=
3),
fcF dr
then
·
{ F . dr + { F
k1
�
is defined by
. dr + ... +
{ F. dr,
�
where each summand on the right side of the equation is defined by formula (13.2.2).
(In each of the expressions fc;F dr, it is implicit that r is a parameterization of the particular curve Ci that appears in the expression. In other words, the symbol dr ·
signifies different parameterizations in the different summands.)
.& Figure 3
(13.2.1)
If Fis a force field, then equation
shows that the line integral
fcF dr ·
has a physical interpretation as the work performed in traversing C from initial point to terminal point. However, we may form the line integral
fcF dr ·
of
any
continuous vector field Fdefined on the directed curve C. The vector field Fdoes not have to be a force field, and the line integral
fc F dr does not have to represent ·
work. Notice that the value of a line integral is a scalar: The dot product notation F
·
dr
serves to remind us of this fact.
� EXAM P L E
r(t)
=
2
(t, -t2), 0 :5 t :51.
Let
C be the planar directed curve parameterized by
fcF dr
Calculate
·
for Fx ( , y)
Solution By definition, this line integral equals
F(r(t)) and
r'(t)
r lo
1
=
1i-2tj.
=
F(t, -t2)
=
=
e"i +xyj.
J� F(r(t))
e'i +t(-r)j
=
·
r'(t)dt.
Now
e1i - t3j.
Therefore the line integral equals
(e1i- t3j) . (li-2tj) dt
=
r lo
1
(e1 +2t4) dt
(e1+�t5) 1t=l t=O
=
5
=
e -�5
10, a directed curve C may be parameterized in several different ways. Formula (13.2.2) seems to imply that the line integral fc F dr of F over C depends on the parameterization r of C that is used in the As we know from Chapter
·
calculation. The next theorem demonstrates that the value of the line integral
fcF dr ·
does not actually depend on the particular parameterization of C that is
chosen (provided that the parameterization respects the direction of C).
THEOREM 1
Let
t H r(t), a ::::; t::::; b
be a continuously differentiable para
metrization for a directed curve C in the plane or in space. Suppose that Fis a
Lets H p(s), 0::::; s::::; L be the r(a) and p(L) r(b). Let at the point p(s). Then the line
continuous vector field whose domain contains C. arc
length
T(s)
=
p'(s)
parameterization
of C
with
p(O)
denote the unit tangent vector to C
=
=
integral of F over C is given by the equation
fc
F
·
dr
=
1L F(p(s))
·
T(s) ds.
(13.2.3)
13.2
Line Integrals
1 065
In particular, because the right side of equation parameterization r, the line integral of parameterization of C.
(13.2.3) does not involve the F over C does not depend on the chosen
Proot The derivation of equation
Theorem
1,
Section
10.3
in Chapter
p(s)=r(t),
T(s) =
Making the substitutions= side of equation
1L F(p(s)) ·
T(s) ds
=
(13.2.3),
(13.2.3) is very similar to 10. Ifs= J; llr'(T) II dT, then
�
llr' t)ll
r'(t),
the proof of
: = llr'(t)ll·
and
J; llr'(T)ll dT, ds = llr'(t)ll dt in the integral on the right
we obtain
lb F(r(t)) · ( ilr'�t)II r'(t)) llr'(t) II dt lb F(r(t)) · r'(t)dt 1 F · dr. =
=
•
� EX A M P L E 3 Let C be the quarter circle in the xy-plane that is para
meterized by
y
r(t)=ti+
F(x,y)=xi+xyj.
J1-t2j,
0 :5t :51.
Verify Theorem
1
for the vector field
Solution Because F(r(t))=F(t, J1 - t2)=ti+ tVl - t2j and r'(t)=i F(r(t)) ·r'(t)=t - t2• Therefore
we have
(t/Jl - t2)j,
1F· dr= 11 F(r(t)) ·r'(t)dt= 11 (t-t2)dt= ( --t2 t3 1 1 -. ) 0= 1
T
c
cos(s)
o
o
2
3
6
In this example, the arc length parameterization of C can be established from x
-'----+-���--���..-. 1 0.5
_. Figure 4
geometric principles: The length of an arc of a circle of radius a is a times the angle
at the center that is subtended by the arc. As Figure
4 shows, this implies p(s)=sin(s)i+ cos(s)j , 0 :5s :57f / 2 , is the arc length parameterization of C. Thus L=7r/2, F(p(s))=sin(s)i+ sin(s)cos(s)j, T(s)=p'(s)= cos(s)i-sin(s)j, and F(p(s)) · T(s)=sin(s)cos(s)-sin2(s)cos(s). We therefore have that
{lo L F(p(s)) ·T(s)ds= lo(
"/2
(sin(s)cos(s)-sin2(s)cos(s)) ds=
( - --- - ) 1 0 sin2(s)
sin3(s)
2
3
7r/2 1 = 6.
<11111
For every directed curve C from Po to P1 there is a directed curve, often
P1 as F over the
denoted by -C, that isopposite to C. It consists of the same points as C but has
its initial point and Po as its terminal point. The line integrals of a field two directed curves C and -C are related by the formula
{
f F ·dr= - F · di, 1-c le where rand r denote parameterizations of C and -C, respectively.
(13.2.4)
1 066
Chapter 13
Vector Calculus � EXAMPLE 4 Let t 1-+ r(t), a:$,t :$, b be a parameterization of a directed
curve C. Prove that t 1-+ r(t) = r(a+ b
-t), a:$,t :$, b
is a parameterization of the
opposite curve -C. Verify equation (13.2.4).
Solution Ast increases from a to b, the expression a+b It follows that the values r(t) = r(a+b
-t decreases from b to a.
-t), a:$,t:$, b, are the points of C traversed -t), we have
from r(b) to r(a). Because r'(t) = -r'(a+b
f-cF·di= 1bF(i(t)) ·i'(t)dt= - 1bF(r(a+b-t))·r'(a+b-t) dt.
By making the change of variable u= a+b obtain
-t, du= ( -1 )dt in this last integral, we
1-eF ·di= lbf" F(r(u)) ·r'(u) dud dt= lbf" F(r(u)) ·r'(u) du= -lb F(r(u)) ·r'(u) du= - lefF · dr. t
<11111
a
Equation (13.2.4) is a generalization of the formula J: f(t)dt J: f(t)dt, from one-variable calculus. Just as on the real axis, reversing the direction of integration in a line integral changes the sign of the integral's value.
INSIGHT
Dependence on Path
=
-
We have observed that, if C is a directed curve from P0 to P1, then the line integral
feF dr does not depend on the parameterization of C. It is reasonable to wonder if ·
the line integral depends on the path at all. That is, will a different path c. between P0 and P1 lead to a line integral
fe.F dr* ·
with the same value as
example shows that the answer to this question is generally No. � EXAMPLE 5 Let
F(x,y)= -yi +xj.
Set P
o = (1,0)
sider two paths from P0 to P1: C parameterized by
feF · dr? The next
and P1 =
(-1,0).
Con
r(t)= cos(t)i+ sin(t)j, 0 :$,t :$, 7r and c. parameterized by
r•(t)= cos(t)i - sin(t)j, 0 :$,t :$, 7r. Is the line integral of
F over C equal to
the line integral of
F over c.?
Solution We calculate
1F
·
dr=
1'/C F(r(t))
·
1'/C ( = 1'/C (
r'(t) dt =
-sin(t)i + cos(t)j sin2(t) + cos2(t)
) ( -sin(t)i+ cos(t)j ) dt ·
) dt= 7r.
However,
{ F · dr* = r F(r(t)) le. lo =
-1'/C
r'..(t)
dt= r (sin(t)i+ cos(t)j ) ( -sin(t)i - cos(t)j ) dt
lo (sin2(t)+ cos2(t)) dt= -7r. ·
The line integrals have different values.
·
1 067
13.2 Line Integrals INSIGHT
5 suggests why it is plausible that the two line integrals of Example 5 C is counterclockwise along the upper half of the unit circle: It is in the direction of the vector field F. However, the line integral over c. is Figure
should be different. The line integral over
clockwise along the lower half of the unit circle: It is against the direction of the vector field. Thus we would anticipate the values of the integrals to have opposite signs. It makes sense that the value of a line integral over a path from P0 to P1 ought to depend on the path between the points. But it turns out that line integrals of conservative vector fields are independent of path. Section
Other Notation for the Line Integral
13.3 treats these ideas in depth.
F= M(x,y,z)i + N(x,y,z)j + R(x,y,z)k is a vector field defined r(t) = (x(t),y(t),z(t)),a:5t:5b, integral f F· dr = J: F(r(t))· r(t)dt becomes If
on a directed
curve C that is parameterized by
then the line
c
1 F dr= lb (M(x(t),y(t),z(t)) : + N(x(t),y(t),z(t)) dr + R(x(t),y(t),z(t)) ��) dt ·
(13 .2 .5)
when expanded. Using the formalism of calculus,
dx -dt= dx' dt we may write equation
(13.2.5)
dy dt= dy' dt
and
dz -dt=dz' dt
more compactly as
1 F dr=1Mdx+ N dy+ Rdz.
x
(13.2.6)
·
� EXA M PL E
6 Let C be the directed curve that is parameterized by
r(t) = (1-t, 2 -t2,1), -1:::; t:::; 2. Calculate F(x,y)
=
-yi + xj
_. Figure 5
fcYZ dx +z dy+x dz.
(13.2.6) with x(t) = 1-t, y(t) = 2 -t2, and z(t) = 1. Then dx/dt= -1, dy/dt= -2t, and dz/dt= 0. As a result, we have
Solution We use formula
1xydx+zdy+xdz = 1: ( (1-t)( 2 -t2) : +l dr + (1-t) �� ) dt 2 = 1 ( (1-t)( 2 -t2)(-1)+ (1)(-2t)+ (1-t)(o)) dt. 1 On expanding the integrand, we obtain
j 2 (-2 +t2-t3) dt= (-2t+�3 -r_ )l 2 � 16 = ( + ) ( 2 !3 !) = 3
fxydx+zdy+xdz = le
4
Closed Curves
4
-1
_
_
4
_
_
_
4
-1
21 4·
.,..
If a directed curve C is parameterized by t f-t r(t) for a:::; t:::; band if r(a) =r(b), then we say that the curve C is We often write
closed. A closed curve begins and ends at the same point.
1 068
Chapter 13
Vector Calculus
t
or
for the line integral of a vector field
M dx+N dy+Rdz
F= (M, N, R) over a closed, directed curve C.
The small circle in the middle of the integral sign reminds us that the curve is closed.
fc(x+y)dx+zdy+x2dz, where C comprises the Po=(1,0,0) to P1 =(0,1,0), the line segment from P1 to P =(0,0,2), and the line segment from P to P0• 2 2 Solution Let C1 be the line segment from Po=(1,0,0) to P1=(0,1,0). We may parameterize it by r1(t)=OP0+t(P0P1)=(1 - t,t,0), 0::st::s1. Then dx/dt=-1, dy/dt=1, and dz/dt=0. We have � EXAMPLE 7 Calculate
line segment from
�
f (x+y)dx+zdy+x2dz= f lo lc1
1
((
(1 - t)+t)
-
)
1 dy dx dz dt= f (- 1+0+0) dt=-1. +0 +(1 - t)2 dt dt dt lo
Similarly, we parameterize the directed line segment C2 from
P2 =(0,0,2) by r2 (t)=OP1+t(P1P2 )=(0,l - t,2t), O::st::sl. dy/dt=-1, dz/dt=2, and �
l
(x+y)dx+zdy+x2dz= 2
1( 1
-
o+(l-t)
P1=(0,1,0) to dx/dt=O,
Then
: +2t dt +02 �� ) dt= 11 (0-2t+O)dt=-l.
P =(0,0,2) to 2 Po=(1,0,0) by r3 (t)= OPd +t(P P�)=(0,0,2)+t(l,0,-2)=(t,0,2 - 2t), 0::st::s1. 2 Then, dx/dt=1, dy/dt=0, dz/dt=-2, and
Finally, we parameterize the directed line segment C3 from
1 d d d (x+y)dx+zdy+x2dz= f (t+O) +(2-2t) +t2 c d d lo 3
(
1
It follows that
Other Applications of Line Integrals
�
ft
� ) dt= lof1 (t+0-2t2 )dt= -6.1
fc(x+y)dx+zdy+x2dz=-1+(-1)+(-1/6)=-13/6.
Suppose that C is a directed curve parameterized by arc lengths� Then equation
(13.2.3)
<11111
p(s), 0::ss::sL.
tells us that
1F
·
dr=
1L f(p(s))
ds,
(13.2.7)
f(p(s))=F(p(s)) T(s). Now the integral on the right side of equation (13.2.7) makes sense for any scalar-valued function/whose domain contains C. For example, if f(x, y,z) is identically 1, then integral (13.2.7) is the arc length of C, a topic that we studied in Chapter 10. We obtain a new application of the line integral by supposing that f(x, y,z) is the mass density of a thin wire that has the shape C. If we divide C into N arcs of where
·
13.2 Line Integrals
equal length
1 069
�s=L/N and choose a point p(sj) in the ;1h arc, then the mass of the L:j':iJ(p(sj))�s. The approximation tends to become more
wire is approximately accurate as
N tends
to infinity. At the same time,
{L
N
J�
�f(p(�))�s=lo
f(p(s)) ds.
Thus when/is a mass density, the right side of equation
(13.2.7) represents the total
mass of the curve. Of course similar interpretations can be made when type of density. The right side of equation Theorem
1 shows that, in
(13.2.7)
is often written as
fcf ds.
f is another
The proof of
general, if r is any parameterization of C, not necessarily
the arc length parameterization, then
it ds = 1b f (r(t)) llr'(t) II dt. � EXA M P L E
(13.2.8)
8 Suppose that a wire has the shape of the curve C that is
r(t)= (3t,4t,t2), 0:::;; t:::;; 6. The charge density of the wire is given q(x,y, z)=9x - y + vz. What is the total charge held by the wire?
parameterized by by
Solution We use formula
J32
+
(13.2.8).
42 + (2t)2 = ../25 + 4t2•
1qds = 16 q( (t)) II r
' r (t)
II dt =
Observe that
r'(t) = (3,4,2t)
16 (9(3t) - (4t) #) V25 +
+
4t2 dt =
16 24tV25
u=25 + 4t2, du=Stdt, and noting 25, and 25 + 4(6)2, or 169, for u, we have
25 + 4(0)2,
or
6 24t../25
4t2 dt = 3
f q ds = f lo le
Q UIC K
Q UIZ
1. 2.
+
f
+
4t2 dt.
the limits of
169 u112du= 2u312 1169=2(133 - 53)=4144.
12s
2s
What is the work done by the planar force field along the path
llr'(t)ll =
The total charge is therefore given by
Making the change of variable integration
and
r(t)= (t,t2), 1:5t:52?
F(x,y) = (y,x) moving a particle
True or false: The work performed by a force field
F over a directed curve C does
not depend on the choice of parameterization of C provided that the para meterization respects the direction of C.
fc2xdx + 2y dy + 5y'Xi.dz (t,t2,t3), 0:5t:51?
3.
What is
4.
If C is the curve that is parameterized by
f(x,y)=8x, then what is fc fds?
Answers
1.
7
2.
True
3. 5
4. 49
..,..
if C is the curve parameterized by
r(t)= (3t,2t2), 0:5t:5l
r(t)=
and if
1 070
Chapter 13
Vector Calculus
EXERCISES Po=(0,1), P1=(2,1), P =(3,4) 2
Problems for Practice In each of Exercises 1-8, calculate planar vector field
L 2. 3.
4.
5. 6.
7.
8.
F
fcF dr ·
In each of Exercises 9-16, calculate 9.
F
r.
fcF dr for ·
the given
and the given parametric curve
r.
F(x,y,z) =xzi- yxj +xyk, r(t) =ti+t2j + t3k, -1sts1 F(x,y,z)=yi - sin(4z)j+xk, r(t) =cos(t)i+sin(t)j+(t/4)k, -3rrsts-rr/2 F(x,y,z) = cos(y)i+ sin(z)j +xk, r(t) = cos(t)i+ tj +tk, rr/4stsrr/2 F(x,y,z)=(x/y)i- ( y/z)j+xzk, r(t) = t3i +t2j +(1/t)k, -2sts-1 F(x,y,z) =zi+k, r(t) =t i+j+ cos2(t)k, 0stsrr F(x,y,z) = yi+ (x - z)j +zk, r(t) = sin(t)i+ cos(t)j + sin(t)k, 0sts2rr
10. 11. 12. 13. 14.
F(x,y,z) = � k, r(t) = sin(t)i+ cos(t)j +tk, 0stsrr x +y 1 k, r(t)=ti+tan(t)j +tan(t)k, 16. F(x,y,z) =yzi+j+ 1 +y2 Ostsrr/4 15.
fcxdy fcxdx +dy 27. fc(2x +y)dx+ydy 28. fcydx - xdy 26.
In each of Exercises 29-32, calculate the line integral
fcfds
for the given function
In each of Exercises 17-20, calculate the work performed curve
r.
F
in moving a particle along the parametric
F(x,y)=xyi+x3j, r(t)=t112i+t114j F(x,y) = 3y2i - 2x3yj, r(t) = cos(t)i + sin(t)j 19. F(x,y,z) =x2i +y2j +z2k, r(t)=(1+t2)i+t2j +(2 + sin(rrt))k 20. F(x,y,z) = (x + y)i- (y - x)j +4zk, r(t) = sin(3t)i + cos(3t)j + 3tk 18.
1sts16 0stsrr 0sts1 0sts2rr
In each of Exercises 21-24, calculate the given line inte
t2i+t3j+t4k, 0sts1.
C
that is parameterized by
r(t)=
2L
fcxyzdx 22. fcdx+dy +dz 23. fc xzdx+yzdy+xydz 24. fc xeZ dx + yzdy In each of Exercises 25-28, calculate the given line inte fcMdx + Ndy where C is the triangle with vertices
gral
parametric curve
r.
0StS2rr 1StS2 1StS2 Ostsl
Further Theory and Practice In each of Exercises 33-36, a vector field Calculate
fc,F dri ·
for i=1,2,3 where
C1
F
is given.
is parameterized by
r1(t) = ( cos(2t),1 - 2t/rr),Ostsrr, C
2
is parameterized by
r (t) = (1 + sin(2rrt), cos(rrt)),0sts1, 2 and
C3
is parameterized by
r3(t) = (1,1 - 2t),Ostsl. Notice that all three paths begin at
(1,-1).
Verify that, for the given
F,
(1,1)
and terminate at
the values of the line
integrals are the same for all three paths. (The vector fields of
Exercises 33-36 are "path-independent." Section
13.3
is
devoted to this type of vector field.)
17.
gral over the directed curve
f and
r(t) = sin(t)i + cos(t)j r(t) =ti+t2j r(t) = t2i +t4j r(t) = ti +e'j
f(x,y) =x2+y2 f(x,y)=y/x 3L f(x,y) = x +VY 32. f(x,y) = yir 29.
30.
-
by a force
counterclockwise
2 5.
Ostsrr/3 0St'Srr Ostsft lStSe 1StS4 -1StS2 -4StS-1 -rr/2stsrr/2
F(x,y)=xyi+y2j, r(t)= cos(t)i+ sin(t)j F(x,y)=y2i- x2j, r(t)= sin(t)i+ cos(t)j F(x,y) = sin(x)i- cos( y)j, r(t) =t2i+t3j F(x,y) = ln(y)i- irj, r(t) = ln(t)i +t3j F(x,y) = xyi +x2y2j, r(t) = t112i + t-3/2j F(x,y) = y112i+x312j, r(t)=t2i+t4j F(x,y) = (x2 +lf1j - xyi, r(t) = ti+ t2j F(x,y) = yi- xj, r(t) = cos(t)i + sin(t)j
spatial vector field
for the given
and the given parametric curve
with
orientation.
33.
F(x,y) = 3i - 2j F(x,y) = i +yj F(x,y) =xi+yj 36. F(x,y) =yi+xj 34. 35.
37. Let
F(x,y)=2yi +xj. Calculate fc,F·dr; for i=l,2,3 C1, C , and C are the parametric curves described 3 2 in the instructions to Exercises 33 -36. The values will not
where
all be the same in this case.
38. Calculate
C
fcxeZ dx+yzdy +xeYdz over the directed curve r(t)= t2i+t3j +t4k, 0sts1.
that is parameterized by
In each of Exercises 39-42, a closed curve terclockwise orientation is specified. Calculate
J -ydx+xdy Jc xZ+yz .
C
with coun
13.3 Conservative Vector Fields and Path Independence
smooth closed curve that comprises the line segment from
39. C is the circle x2 + y2 = a2. 40. C is the square with vertices
(±1,±1). 4L C is the arc of the circle x2 + y2 2, from (1, -1) to (1,1) followed by the line segment from (1, 1) to (1,-1). 42. C is the arc of the circle x2 + y2 2, from (-1, -1) to (-1,1) followed by the line segment from (-1, 1) to (-1,-1). =
fi;Xyzdz
where C is the closed curve of inter
section of the cylinder
x2 + y2
=
1 and the plane z
=
1 + x,
oriented counterclockwise when viewed from above.
44. Calculate
fcxdx + x2 dy + ( y - z) dz
where C is the arc of
the curve of intersection of the cylinders
x2 + z2
=
the point
x2 + y2
1 that lies in the first octant and that (1,0, 0) and terminates at (0, 1,1).
=
1
and
begins at
(a, 0) to ( b, 0) and the arc of the graph of y (a, 0).
46.
y = x113 and y = 1 - x2 fcy3 dx + xdy where C
Calculate
fc dx + x2 dy
is the piecewise-smooth closed
y = x 113 from (0, 0) to P, the arc of the graph of y 1 - x2 from P to (0, 1), and the line segment from (0, 1) to (0, 0). =
47. Let
r(t)
C =
48. Let
r(t)
C =
be
the
planar
(t, t2), 0 5 t 51. be
the
curve
Calculate
spatial
(t, t2, t3), O 5 t 5 1.
whose
equation
is
equation
is
fcXeYds.
curve
Calculate
whose
fcvx + y + z ds.
49. Let C be the planar curve whose equation in polar coor dinates is
4 + 2x - 3x4
and b denote the real roots of
in the first quadrant. Calculate
curve that comprises the arc of the graph of
Calculator/Computer Exercises a a< b.
4 + 2x - 3x4
Let P denote the point of intersection of the graphs of
50. Calculate 45. Let
=
from (b, 0) to
=
43. Calculate
1071
r = 0, O 5 (} 5 27r. Calculate fcx2 ds. fc(x2 + y2) ds where C is the ellipse x2 + 2y2
=
1.
with
where C is the piecewise-
13.3 Conservative Vector Fields and Path Independence dr of a ti--+ r(t), a:=;. t:=;. b,
In the preceding section, we learned that to compute the line integral fcF vector field F over a directed curve C, we use a parameterization
·
of C and evaluate
lb F(r(t))
·
r'(t) dt.
r. does depend on the values F(r(t) ) that F assumes along C.
We observed that this integral does not depend on the choice of parameterization However, the integrand
Therefore if we were to use two different paths to connect a given pair of points, we would expect to obtain two line integrals with different evaluations. Indeed, Example 5 of the preceding section confirms this expectation: If we connect a pair of points by a path C, then the value of fcF
·
dr can depend on C. There is, however, independent of the
an important class of vector fields with line integrals that are
path. In this section, we will study and characterize such vector fields.
-·[email protected]
Suppose that Fis a vector field defined on a region(}. We say that
path-independent on(}if, for any two points Po and P1 in(}, the line integral fcF dr has the same value for all directed curves C in(}with initial point Po and
Fis
·
terminal point
P1.
Our first theorem characterizes path-independent vector fields by means of their line integrals over
closed curves.
1 072
Chapter
13
Vector Calculus THEOREM on
1
g if
and only if fcF
-C2
is a closed curve C (see Figure
Proot If
C1
dr
C
Let Fbe a vector field on a region Q. Then Fis path-independent
and
C2
·
=
0 for every closed curve
1).
P0
in Q.
P1,
C1
On the other hand, if C is a closed curve
are two directed curves from
to
then the union of
P0, then we may (arbitrarily) choose another point C and write C -C2 uC1 where C1 is the path along C from Po to P1 and -C2 the path along C from P1 to P0• Then
and
P1
with initial and terminal point is
on
=
i
F
·
dr
=
1
F
·
dr +
C1
C
1
F
·
dr
--C2
1 2
( 3. .4) =
1
F
·
C1
dr -
1
F
·
dr.
C2
Thus
JF Jc
_. Figure
·
dr
=
0
if and only if
{ F dr { F dr. lei le2 ·
=
•
·
1
At first glance, it might seem that few vector fields could have the remarkable property of path-independence. The next theorem produces an abundant supply of path-independent vector fields.
u is a continuously differentiable scalar-valued g in the plane or in space. Let Fbe the gradient vector field defined by F='Vu. Let C be a piecewise-smooth directed curve in g that has P0 as its initial point and P1 as its terminal point. Then
THEOREM
2
Suppose that
function on a region
{F. dr u(P1) - u(Po). le
(13.3.1)
=
fe F dr depends on the initial and P0 and P1 but not on the path C that joins them. In other words,
In particular, the value of the line integral terminal points
·
a gradient vector field is path-independent.
Proot Suppose that
C
parameterization of
Then, using the Chain Rule, we see that
C.
is
a
smooth
curve.
Let
t f-+ r(t),
a
:st :s b,
be
any
13.3 Conservative Vector Fields and Path Independence
1 073
lbF(r(t)) ·r'(t)dt lb\7u(r(t)) · r'(t)dt lbdtd (u(r(t))dt a
u(r(t))I� u(r(b)) - u(r(a)) u(P1) - u(Po). P0, P1,
The line integral depends on
and
u
alone, as asserted. This proves the
theorem in the smooth case. Now suppose that pieces,
C
is a continuous, directed curve comprising two smooth
C1 from P0 to P' and C2 from P' to P1. According to the smooth case that we
have just proved,
1
F
·
dr
C
=
1
F
C1
·
dr1 + f F dr2 (u(P') - u(Po)) + (u(P1) - u(P')) u(P1) - u(Po). lc2 ·
=
=
This proves the theorem in the case that the piecewise-smooth curve is made up of two smooth pieces. The case in which
C consists of an arbitrary number of smooth •
pieces is proved in the same way. INSIGHT
It is worth noting the similarity between equation (13.3.1) and the
Fundamental Theorem of Calculus. If F (x ,y, z)
=
\i'u (x ,y, z), then
u can be thought of as
an antiderivative of the first component of F with respect to x, the second component of F
with respect toy, and the third component of F with respect to z. Theorem 1 tells us that
the line integral of F over C can be calculated by evaluating the antiderivative u at the end points of C.
Recall from Section
conservative and that V
13.1 =
that we say that a gradient vector field F
-u is a potential function for F. Theorem 2
=
\7u
is
states that
a conservative vector field is path-independent. An equivalent formulation is that a vector field with a potential function is path-independent.
�
EX A M PL E 1
(x,y, z) yzi + xzj + xyk. Show Calculate the line integral of F from Po (0, - 1, 2) to
Consider the vector field F
that F is path independent.
P1 (2, 1, 4).
=
=
=
Solution Observe that F vector field. Equation
\7u where u(x,y,z) xyz. Therefore F is a conservative (13.3.1) tells us that, if C is any directed curve from P0 to Pi, =
=
then
1
F
· dr u(P1) - u(Po) (2)(1)(4) - (0)(-1)(2) =
=
Notice that we have evaluated
f F dr
=
8.
without actually having calculated a line
c integral. Let's verify our result by calculating fcF · dr as a line integral. For sim plicity, we take C to be a straight-line path joining P0 to P1. We can then para meterize C by ·
1 07 4
Chapter
13
Vector Calculus
r(t)=OPo +t PoP1 =(0, -1, 2)+t(2, 2, 2)=(2t, -1+2t, 2+2t), _____.
----+
0:st:s1.
With these choices, we have
F(r(t)) =(-1+2t)(2+2t)i+2t(2+2t)j +2t(-1+2t)k We
Therefore the line integral of F from
1 A Characterization of Path-Independent
Theorem
2
r'(t)=2i+2j +2k.
F(r(t)) · r'(t)=(-1+2t)(2+2t)(2)+2t(2+2t)2+2t(-1+2t)2=
calculate
-4+8t+24t2.
and
F
·
dr =
P0 to P1 is
11(-4+8t+24t2)dt=8.
<111i1
tells us that conservative vector fields are path-independent. The next
theorem tells us that there are no other path-independent vector fields.
Vector Fields THEOREM
3
Suppose that F is a continuous vector field on a region g in the
plane or in space. If Fis path-independent, then F='\Ju for some continuously differentiable function u.
Proof. Let F=(M, N, R) be a continuous, path-independent vector field in a spatial region Q. (The planar case is proved in the same way.) Fix a point P0 in Q. For each point P= (x,y,z) in Q, set
u(x,y,z)= {
z
where Cp is
any
curve that connects
F·dr,
lcp P0 to P and
lies entirely in Q. Such a curve
exists because g is a region (and therefore path-connected). Moreover, because Fis path independent, the value of
u(x,y,z) will
be the same no matter what
connecting path Cp we choose. This tells us that there is no ambiguity in the definition of u(x,y,z). For sufficiently small /).x, we can choose the path Cp• from
P0 P'=(x+&,y,z) to be the piecewise-smooth curve that comprises Cp and the directed line segment PP' (Figure 2). We can parameterize this segment by to
CP'
� Figure
=
2
--7 Cp UPP'
i(t)=(x+t&,y,z), 0:st:s1. F(r (t)) r '(t)=M(x+t/).x,y,z)&
Because
i'(t)=(&, 0, 0)
it
follows
and
f_F-di= f1M(x+t&,y,z){).xdt=& {1M(x+t[).x,y,z)dt .
}PP'
lo
lo
Thus
f_. ·d ) - fcpF·dr (lcpf F·dr+}fap• =& 11M(x+t[).x,y,z)dt
u(x+&,y,z) -u(x,y,z)=
F
r
that
13.3 Conservative Vector Fields and Path Independence
1 075
and
. u(x+�x y z) - u(x y z) ux(x y z)= hm ' ' ' ' = 0 ' ' Ax-+ �
.
hm �x 0 -+
1 1 M(x+t�x ' y z)dt. '
o
In courses on real analysis, it is shown that this limit can be taken inside the integral:
1 1 f M(x+t�x,y,z)dt= f Ax-+0 lo lo lim
lim
�x-+O
M(x+t�x,y,z)dt.
From this equality, it follows that
1 M(x+t�x,y,z)dt= M(x,y,z)dt=M(x,y,z). f ux(x,y,z)= f lo �x-+O lo 1
lim
Similar arguments show that Together these three equations
� EXAMPL E 2 Let
uy(x,y,z) = N(x,y,z) show that F = '\lu.
Uz(x,y,z) = R(x,y,z).
and
•
F be a path-independent force field on a region Q. If we
move a particle through the field from one point P0 in Q to another point P1 in Q, then how does the work that we do relate to the change in potential energy of the particle? Solution To move a particle from P0 to P1 along a path C, we must apply a force F. The work W that we do is therefore given by W= fc(-F) dr.
opposite to
2,
According to Theorem Theorem
1
=
u
for which
F= '\lu.
tells us that W=
where V
there is a scalar-valued function
·
-1 F dr= -(u(P1) - u(Po)) = V(P1) - V(Po)
(13.3.2)
·
-u. Because Vis a potential function for F, the potential energy at each
point P of Q may be defined to be V(P) , as discussed in Section
13.1. Observe that
the change in potential energy between any two points in Q does not depend on the choice of potential function that is used to define potential energy. Indeed, equation
(13.3.2) tells us that the work we do is equal to the change in the particle's
potential energy. � EXAMPLE
-'\!V(x,y)
<111111 3 Suppose that
V(x,y)
=
mgy
and
that
F(x,y)
is the conservative planar force field of Example 9, Section
move a particle of mass
m
from the point
Po= (xo,yo)
to the point
then how much work have we done? Solution The simplest solution is to use Example we have done is the change of potential energy,
=
-mgj
13.1.
=
If we
P1= (xi,y1),
2, which tells us that the work W
V(P1) - V(P0 ) ,
or
mgy1 - mgy0.
We can also obtain this result by calculating a line integral. The force we must apply is
F(x,y)=mgj ='\!V(x,y),
independent (Theorem
2),
which is conservative. Because F is path we may calculate W=fcF dr using any path j oining P0 ·
to P1. We will use the piecewise-smooth path C formed from the horizontal line
1 076
Chapter
13
Vector Calculus
segment ch parameterized by tf---t(xo+t(x1 -xo),yo), 0:::; t::5 l, followed by the vertical line segment Cv parameterized by tf---t(X1,yo+t(y1 -yo)), 0::5t::51. We have
1ch
1
d F(xo+t(x1 -xo),yo)·- (xo+t(x1 -xo),yo)dt dt o 1 1 = (0,mg)·(x1 -xo,0)dt= 0 dt = 0
F·dr=
1 1
1
and
1
.
F·dr = =
1
1 F(x1,yo+t(y1 -yo))·�t(x1,yo+t(y1 -yo))dt 1
1
1 (0,mg)·(0,Y1 -Yo)dt= 1 mg(y1 -Yo)dt=mg(y1 -Yo).
Therefore W=
Closed Vector Fields
fF Jch
·
1 F·dr=0+mg(y1 -Yo)=mgy1 -mgyo.
dr+
....
C,
Because it is not usually feasible to check path-independence directly, it is important to have a practical method for recognizing when a vector field is con servative. Our next theorem is a first step toward this goal: it provides us with an easily verifiable condition that will help us identify planar vector fields that are not conservative. THEOREM 4
Let F(x,y)=M(x,y)i+N(x,y)j be a continuously differentiable vector field in a region Q. If F is conservative, then BM_ 8N ay ax
-
(13.3.3)
-
at each point in Q.
Because F is conservative and continuously differentiable, there is a twice continuously differentiable function u such that
Proot
Bu. Bu. v U = -I+ M•I+N"J = F = "CT ax ay J -
.
In other words, M = 8u/8x and N = 8u/8y. It follows that •
13.3 Conservative Vector Fields and Path Independence A vector field
F(x,y) = M(x,y)i + N(x,y)j closed
each point of a region g is said to be
that satisfies equation
1 077
(13.3.3)
at
on g, Theorem 4 says that every
conservative vector field is closed. A logically equivalent reformulation is that any vector field that is not closed is not conservative. � EXAMPLE
4 Is the vector field
F(x,y)
=
(x2 - sin(y))i + (y3
+ cos(x))j
conservative?
Solution
Notice that
� (x2 -
sin(y)
) =-cos(y)
! (y3 +cos(x))=-sin(x).
and
Because these expressions are unequal, Theorem 4 guarantees that F is not a conservative vector field.
<11111
As we can see from Example 4, Theorem 4 gives us a handy way to tell when a planar vector field is
not conservative:
If the vector field does not satisfy equation
(13.3.3), then the vector field is not conservative. However, Theorem 4 does not say (13.3.3). In parti cular, Theorem 4 does not assert that a closed vector field is conservative. Indeed,
anything about a planar vector field that does satisfy equation
the next example shows that a closed vector field need not be conservative. � EXAMPLE 5 Show that the vector field
F(x 'Y) =
(x2-y+y2) i (x2+x y2) . +
is closed but not conservative on the region
J
y= {(x,y): (x,y) '# (0,0)}.
Solution Observe that F is defined and differentiable on y= {(x,y): (x,y) # (0,0)}. To verify equation (13.3.3) on g, we set M(x,y)=-y/(x2+y2), N(x,y)=x/(x2+y2), and calculate
Thus Fis closed on g, To show that Fis not conservative on g, it suffices to observe that F is not path-independent. Set paths from
P0
to
P1:
Po= (1,0),
and
C parameterized by
P1 = (-1,0),
and consider two
r(t)= cos(t)i +sin(t)j, 0
r(t)=cos(t)i - sin(t)j, 0
1
F
·
ln F(r(t)) r'(t)dt= ln (- �(t) i co�(t) j) = ln ( 2(t) +cos2(t)) dt=Jr. si
dr =
·
sin
+
·
(- sin (t)i + cos(t)j ) dt
1 078
Chapter
13
Vector Calculus However,
Jc
F dr ·
=
=
17r ( -17r (
F r(t) ) r' (t) dt ·
=
17r ( �
si (t)
i+
�
co (t)
j
)
.
( -sin(t)i - cos(t)j ) dt
2 2 sin (t) + cos (t) ) dt = -7r.
Our calculations show that F is not path-independent. We use Theorem 2 to conclude that F is not conservative. <11111 INSIGHT (J = :!!._ 2 y
If F is the vector field of Example
5 and if g is the region of Example 5,
then it is instructive to investigate why we cannot find a differentiable function u such that Vu(x,y)= F(x,y) on
Q. Using the method presented in Example 10 of Section 13.l to
solve the equations ux(x,y)= -y/(x2 + y2) and uy(x,y)=x/(x2 + y2), we integrate the equation ux(x,y) = -y/(x2 + y2) with respect to x and find that u(x,y) =arctan(y/x) + ¢(y). If we substitute this expression for u(x,y) into the equation uy(x,y) =x/(x2 + y2),
then we obtain ¢'(y) =0, or ¢(y) = Cfor some constant C. Thus if Vu(x,y) =F(x,y), then u(x,y) = arctan(y/x) + C. Now, on the region
Q ={(x,y)
: x > O}, the function u is
differentiable and F(x,y) =Vu(x,y). As a result, F is conservative on
Q.
Although we
cannot extend the expression arctan(y/x) + C to a larger region because it is not defined
for x =0, we can exploit an alternative formulation. Notice that arctan(y/x) is the polar coordinate () E
)
(
)
-?r, ?r
of each point (x,y) that does not lie on the negative x-axis
{(x,0 : x :5 O}. This observation allows us to define u(x,y)= () + Con the region
.A Figure 3
g. that
we obtain by removing the negative x-axis from the plane. But we cannot extend the definition of u from increases from Figure
3.
0 to
g. to all of Q. To understand why, notice that the polar angle () as we traverse the curve C from Po= (1, 0) to P1 = (-1, 0). See
7r
On the other hand, as we traverse the curve
the polar angle () decreases from
0 to
unambiguous way to define u(x,y) on
-?r.
C
from Po= (1, 0) to P1 = (-1, 0),
Consequently, there is no continuous and
g ={(x,y): (x,y) =I- (0, O)}.
A planar or spatial region g is called simply connected if any closed curve in it can be continuously deformed to a point in g while staying within the region.
Roughly speaking, a planar region is simply connected if it has no holes in it. Figure 4 illustrates some regions that are not simply connected: Any closed curve that encircles a hole in one of these regions cannot be continuously deformed to a point while staying within the region. By contrast, the regions in Figure 5 are simply connected.
.A Figure
4 The two shaded regions are not simply connected .
13.3 Conservative Vector Fields and Path Independence
_. Figure 5 The two shaded regions
are
1 079
simply connected.
If we remove one point from a simply connected planar region, then the resulting punctured region will not be simply connected. For instance, the region
g = {(x,y) : (x,y) =f (0, O)} of Example 5 is not simply connected. That example
shows that a closed vector field need not be conservative on a region that is punctured. It is a surprising fact that a closed vector field on a simply connected region
is
conservative. That is, we have the following theorem.
THEOREM 5
Let F=
M(x,y)i + N(x,y)j
be a continuously differentiable vec
aM /ay= aN /ax on Q, then there is a twice continuously differentiable function u on g such that Vu= F. In tor field on a simply connected planar region, Q. If
short, a closed vector field on a simply connected region is conservative.
= {(x,y)
Proof. We will outline the proof for a rectangle g
: a
For simplicity, we assume that the origin lies in Q. We define
u(x,y)=
lax M(s,y) ds la N(O,t) dt. y
+
Notice that, because our region g is a rectangle, we are integrating along paths that lie entirely in Q, and these integrals are guaranteed to make sense. We assert that
u
Vu= F. formula for u
satisfies
second expression in the
Ux(x,y) =
Let's see why. First we compute does not even depend on
a ax
x, so
lr o M(s,y)ds.
which is just what we want. Next we calculate
a ay
The
we have
The Fundamental Theorem of Calculus tells us that this last expression is
uy(x,y)=
Ux.
Uy.
We have
M(x,y),
a r a rx M(s,y) ds + = N(O, t) dt lr l l M(s,y) ds + N(O,y). o ay o ay o
It is a deep but plausible result that we may pass the differentiation under the integral sign to obtain
uy(x,y) =
taM
lo ay (s,y) ds + N(O,y).
1 080
Chapter 13
Vector Calculus Now we use the hypothesis that aM/ay= aN/ax to obtain
Uy(x,y)=
t lo
aN ax
(s,y) ds+N(O,y).
(
)
By theFundamentalTheorem ofCalculus,we obtainuy(x,y)= N(x,y)-N(O,y) + N(O,y)=N(x,y) as desired.
� EXAMPLE
6 Let M(x,y)= sin(y)+ysin(x)
•
and N(x,y)=xcos(y)
cos(x)+1. Is the vector field F(x,y)=M(x,y)i+N(x,y)j conservative on the rec tangle g= {(x,y) : -1
(x,y)
=
. cos(y)+ sm(x)
aN =
ax
(x,y).
Our vector field F passes the test: It is closed on Q. By Theorem 5, F is conservative. Let us find a functionu on g whose gradient is F. Proceeding as in Example 10 of Section 13.1, we first study the equation
��
=M(x,y) = sin(y)+ysin(x).
This equation tells us that u is an antiderivative with respect to x of the right-hand side. Thereforeu(x,y)=xsin(y)-ycos(x)+ ¢(y), where the constant of integration ¢(y) is independent ofx but possibly dependent ony. We differentiate this formula foru, with respect toy, obtaining au ay
=xcos(y)- cos(x)+ ¢ (y). I
Because Vu= F , we also know that au ay
=N(x,y)=xcos(y)- cos(x)+l.
On equating our two expressions for au/ay, we obtain ¢'(y)= 1. It follows that ¢(y)=y+ C, where C is a numerical constant. Putting this information together yields u(x,y)=xsin(y)-ycos(x)+y+ C.
Vector Fields in Space
<1111
Theorems 4 and 5 pertain to planar vector fields, but they have analogues that apply to vector fields in space. Our next theorem states a necessary condition for a spatial vector field to be conservative.
13.3 Conservative Vector Fields and Path Independence THEOREM 6
If
F(x,y,z)=M(x,y,z)i+N(x,y,z)j+R(x,y,z)k
uously differentiable conservative vector field on a region following equalities hold on
1 081
is a contin
Q, then all three of the
Q:
aM aN ay - ax'
aM aR az - ax'
_
(13.3.4)
To understand what Theorem 6 says, let's put it in words: If conservative, then the derivative of the first component of
F
F(x,y,z)
is
with respect to the
second variable equals the derivative of the second component of respect to the first variable; the derivative of the first component of to the third variable equals the derivative of the third component of
F
with
F with respect F with respect
F with respect to the F with respect to the second variable. A vector field F(x,y,z) =M(x,y,z)i+N(x,y,z)j+ R(x,y,z)k that satisfies the three equations of line (13.3.4) at every point of a region g is said to the first variable; the derivative of the second component of
third variable equals the derivative of the third component of
to be a closed vector field on
Q.
Theorem 6 states that every conservative vector
field on a spatial region is closed. Equivalently, a vector field that is not closed is not conservative.
�
EXAM PL E 7 Is the vector field in space given by
(y2/2)k
F(x,y,z) =xzi+yzj+
a conservative vector field?
F(x,y,z)=M(x,y,z)i+N(x,y,z)j+R(x,y,z)k, where M(x,y,z)=xz, N(x,y,z)=yz, and R(x,y,z)=y2/2. We notice that the partial derivative of the first component of F with respect to the third variable z does not equal the partial derivative of the third component of F with respect to the first variable x: Solution Let's
write
aR. aM = x#O= ax az Because one of the equations of line
F cannot
be conservative.
(13.3.4)
is not satisfied, the vector field
<11111
INSIGHT
Whereas a planar vector field must satisfy only one equation to be a conservative, spatial vector field must satisfy three. The vector field F(x,y, z) M(x,y, z)i + N(x,y, z)j + R(x,y, z)k of Example 6 does satisfy two of the three necessary equations to be conservative, namely My 0 Nx and Nz y Ry, but that does not suffice. =
=
=
=
=
Theorem 6 has a converse that will work on geometrically simple regions.
THEOREM 7
Suppose that
F(x,y,z) =M(x,y,z)i+N(x,y,z)j+R(x,y,z)k
continuously differentiable vector field on a simply connected region If
F
satisfies all three equations of line
tinuously differentiable function u on
g
(13.3.4),
is a
g in space.
then there is a twice con
such that '\lu
=F.
In short, a closed
vector field on a simply connected region in space is conservative.
1 082
Chapter 13
Vector Calculus
2i 2 F(x , ,)z y z y + (x z )z j + (2.xy z-y) k {(x,z , 0
� EXAMPLE 8 Is the vector field conservative on the box function u for which F
g
=
=
=
\lu.
Solution First we check whether Fis closed. Because
2) ..§__ (y z ay
=
2z
=
a2 a -( - (z 2.xy z-y) z)y - 2y a ax z
2z - z), ..§__ (x ax
_
_
,
and
2z ..§__ (x )z a z
2.x z-1
=
=
..§__ (2xy z- y), ay
we conclude that Fis closed. By Theorem 7, it is conservative. Thus F
=
some function u. Let us find u. First, u is the antiderivative in the
\lu for
variable x of
2• z y
It follows that x,y, u(
2z +
z ) =
(13.3.5)
Here the constant of integration does not depend on x, but it can depend ony and
. zNow we differentiate each side of equation (13.3.5) zas constants:
with respect toy, treating
and
au 2 a zx + y, z .
But the equality of the second components of the vector equation \lu
=
F also
gives us
�; =xz2 -z. We conclude that
2x + z
�
y, z)
=
2zx - z
or, equivalently,
a
By It follows that
z ), z+ '!/J(
and x y but may depend . ,)z into equation (13.2.5), then we obtain zIf we substitute our formula for
where now our constant of integration is independent of on
u(x,y ,)z
=
2 -zy+ '!jJ( xy z z ).
Finally, when we differentiate each side of equation obtain
(13.3.6)
(13.3.6) with respect to
, zwe
x
Conservative Vector Fields and Path Independence
13.3
1 083
8U = 2xyz-Y + 7/J (z). 8z I
But the equality of the third components of the vector equation
\Tu =Ftells us that
8u = 2xyz-y. 8z 8u/8z, we see that 7/J'(z) = 0. We conclude that 'ljJ(z) =C for some numerical constant C. Substituting this into equation (13.3.6), we arrive at our final formula, u(x,y,z) =xyz2-yz + C . ..,..
By equating the two expressions for
Summary of Principal Ideas
Let's summarize the key ideas that we have learned so far in this section. Let Fbe a continuously differentiable vector field either on a planar or spatial region
Q.
Then the following three properties are equivalent-if Fpossesses any one of them, then F possesses all three: a.
.fcF dr = 0 ·
for every closed curve C in
b. Fis conservative: F=
\Tu
for some
Q.
u.
fc1F dr = fc2F dr for any two directed curves in g that share the same initial point and the same terminal point.
c. Fis path-independent:
If F possesses any one
·
·
(and
hence all) of properties a, b, or c, then F also
possesses the following property: d. F is closed: F satisfies equation
(13.3.4)
(13.3.3)
if planar, the three equations of line
if spatial.
If g is simply connected, and Fpossesses property d, then Fpossesses any one
(and
hence all) of properties a, b, and c.
and equivalences in Figure
6
(The
schematic diagram of implications
may help you to organize these relationships. )
Fis conservative. (F =Vu for some u)
u
Fis path-independent.
PcF·dr
u0
=
for every closed curve in g _. Figure
Q UIC K
Q UIZ
====>
Fis closed.
·� =(=ifconnected g is simply
6
1.
True or false: If
2.
True or false: For a vector field on a region, the properties of being path
M(u) and N(u) are polynomials of one variable, .fcM(x) dx + N(y) dy =0 for every closed path in the plane.
then
independent and being closed are equivalent, even if the region is not simply connected.
1 084
Chapter 13
Vector Calculus 3. True or false: For a vector field on a region, the properties of being path independent and being conservative are equivalent even if the region is not simply connected. 4. True or false: If the first quadrant is removed from the plane, then the region that remains is not simply connected. Answers 1. True
2. False
3. True
4. False
EXERCISES Problems for Practice
In each of Exercises 1-12, a planar vec tor field F is given. De termine if F is closed on the square Q={(x,y): -1
5 x2 j, Po= (1, -1), i(5+xy)2 (5+xy)2
P1=(2,2)
3. F(x,y)=y3i+(3y2x - 7y6)j
x2 j y+2 5. F(x,y)=(y3 - xy2)i - (x2y2+x)j 6. F(x,y)= �i lj l +� x+ (x+ ) 7. F(x,y)=(x - 2f2i+(y - 1/2)2j 8. F(x,y)=xcos (xy)i+ysin (xy)j 9. F(x,y)=2xsin(x2 - y)i - sin(x2 - y)j � i+ l � j 10. F(x,y)= l +x +y2 +x +y2 11. F(x,y) =---z1'----zi+ � j x +y x +y
24. F(x,y) =(1+ln (xy))i+(1+x/y)j,
4. F(x,y)=2xln (y+2)i -
In each of
fcF·dr.
23. F(x,y) =
2. F(x,y)= (x2 - y2)i+(2xy - y3)j
12. F(x,y)=
In each of Exercises 23-28, a vec tor field Fis given , as are the initial point Po and the terminal point P1 of a direc ted curve C. Verify tha t F is closed in a sinlply connected region tha t con tains P0 and P1. Then , calcula te the line in tegral
(
( : - 2ln (3+y ) ) i+ 1n(3+x) 3 x
P1=( v'2, v'2)
Po=(1,1),
25. F(x,y) = (y2+ye'Y)i+(2xy+xe'Y)j,
Po= (0,0), Pi=(-1,1) 26. F(x,y,z) = (y2+z3)i+(2xy+2yz3)j+(3y2z2+3xz2)k, Po= (1,-1,-1), P1= (-1,1,0) 27. F(x,y,z)=eY 2Z(i+xj+2xk), Po=(0, 0,0), P1= (-1,2,1) 28. F(x,y,z) =(3x2z+1/x)i+(1/z - 1/y)j+(x3 - y/z2)k, Po=(1,1,1), P1=(4,2,2) +
3�J j
Exercises 13-22, a spa tial vec tor field Fis given. De termine if F is closed on the cube Q= {(x,y,z) : -1
In each of Exercises 29-32 a conserva tive vec tor field F and three poin ts P0, P1, and P are given. Verify tha t 2
f_. F dr+ f_. F ·dr+ f_. F dr= 0.
jPoP1
·
h1P2
jP2Po
·
29. F(x,y) =(y2+2y)i+(2xy +2x)j,
Po=(0,0), P1=(0,1), P2=(1,1) 30. F(x,y) = (2x - y)i -xj, Po= (1,1), P1 = (3,4), P = (2,5) 2 3L F(x,y,z) = 2xyi+x2j +2zk, Po= (0, 0,1), P1=(0,1, 1), P2=(1,1,1) 32. F(x,y,z) = yz2i+xz2j +2xyzk, Po= (0, 0,0), P1=(-1,2,1), P2 =(0,2,0) Further Theory and Practice
33. IfF(x,y,z)=yzi+xzj+xyk and if C is any directed curve
from Po=(0, -1,2) to P1 =(2,1,4), then fcF dr= 8, as shown in Example 1. Verify this equation for the para me terized curve ·
13.4 Divergence, Gradient, and Curl
U={(x,y,z): l
r(t)= (12t2 - lOt, st3 - st2 + 2t - 1, 4t2 - 2t + 2) defined for
38. Is the solid connected?
0 ,::; t ,::; 1.
The gravitational force that a point mass m at the origin 0 exerts
on
a
unit
point
mass
at
P=(x,y, z)=fO
is
F= - ( Gmr-3 ) 0P where r = llOPI. In Exercises 34 and 35 let Po= (x0,y0,z0) =f 0 be a fixed point and let V be a fixed choice of potential function for F. Let Wp be the work done in moving a unit mass from P0 to P=(x,y, z)=f 0. 34. Calculate the work Wp. For what points Q is WQ =Wp? 35. Show that W 00 =limP--+oo Wp exists in the following sense: For any c>0, there is an R such that IWP - W 00 I < c for all
P
with
IIOP II >R.
(Frequently in physics, a potential
function is chosen so that a point
P0
W 00
1 085
has a convenient value for
of interest.)
What
39. Sketch two simply connected spatial regions the inter section of which is not a simply connected region.
40. Verify that
F(x,y) =
� xp(-(x - y)2}(i e
is conservative. Calculate
(0, 0)
to
fcF dr ·
j)
for a curve that joins
(1,1).
Calculator/Computer Exercises F(x, y) = sin(7rxy2)(x-1i + 2y-1j) is con fcF · dr on a directed curve C from (1, 1/2) to (1/2,1). 42. The vector field F(x, y) =27r-1/2exp(-(x - y)2)(i -j) is conservative-see Exercise 40. Calculate fcF dr on a directed curve C from (0, 0) to (1, 2).
4L Verify
that
servative in the first quadrant. Calculate
Q ={(x,y): lxl<1, IYI <1}. Let Lc={(O,y):y>c} 9={(x, y)EQ :(x,y) ¢Lc}· For what values of c is the region Q simply connected? 37. Consider the square Q ={(x, y): lxl<1, IYI < l}. Let Lc,d={{0,y):c< y
·
connected.
1 3.4 Divergence, Gradient, and Curl In this section, we will learn about the differential operators that will be needed to formulate the fundamental theorems of vector analysis. Each of the constructs that we will see has an interesting physical interpretation.
Divergence ofa Vector Field
Suppose that
F(x,y)=M(x,y)i+N(x,y)j
defined on a region
Ny(x,y)
is a differentiable vector field that is
9 of the plane. The scalar-valued function
(x,y) 1--+Mx(x,y) +
is said to be the divergence of F. We denote this function by div (F) : div ( F)=
Similarly, if
F(x,y,z)
tor field on a region
=
oM oN + . ox oy
(13.4.1)
M(x,y,z)i+N(x,y,z)j +R(x,y,z)k is a differentiable vec F is defined to be
9 in space, then the divergence of div (F)
=
If div ( F) (P) = 0 at every point P of
oM + oN oy ox
+
oR oz .
(13.4.2)
9, then we say that F is solenoidal on(}. Gauss's
Law for magnetism states that a magnetic field B is solenoidal.
1 086
Chapter 13
Vector Calculus � EXAMPLE 1 Let F(x,y,z)=xyi+yzj-2xzk an d G(x,y,z)=2xyi+ (z-y2)j+xyk. Calculate div (F) an d div (G). Is either vector fiel d solenoi dal on JR3? Solution
We have div (F)(x,y,z) =
! (xy)+ � (yz)+ � (-2xz)=y+z - 2x
an d div (G)(x,y,z) =
! (2xy)+ � (z-y2)+ � (xy) = 2y-2y+0= 0.
Thus G is solenoi dal on JR3• Although div (F)(x,y,z) equals 0 at each point of the plane y +z - 2x=0, these are the only points at which div (F) vanishes. Conse quently, F is not solenoi dal on JR3. <11111 � EXAMPLE 2
Suppose that a point mass at the origin exerts a gravitational
fiel d
r, r3 .x
F(x,y,z) =
(x,y,z) =!= (0,0,0)
ll ll
where .Xis a proportionality constant an d ris the position vector xi+yj+zk ofthe point (x,y,z). Show that Fis solenoi dal on Q= {(x,y,z): (x,y,z) =!= (0,0,0)}. Solution
We calculate
x a ax (x2 +y2+z2 )3/2
1
(x2 +y2 +z2) 3 1
(x2 +y2 +z2) 3
((x2+y2+z2)3/2 ((x2+y2+z2)3/2
_
_
x.E_ ax
(x2+y2+z2)3/2)
3x2(x2+y2+z2)1;2
)
(x2+Y2 +z2)1/2 2+ 2+ 2 z ) -3x2) y ((x (x2+y2+z2)3 (y2+z2 2x2) (x2 +y2 +z2)s12 _
·
By symmetry, we de duce that . div (F)(x,y,z) = .X = .x
( (x(y22++yz22+z2x2)2s;)2 + (x(x22++yz22+zz2y)2s);2 + (x(x22++yy22+zz2z)2s;)2 ) ( 2+ 20+ 2 s12 ) =0 .... _
_
_
·
(x
y
z)
What is the significance of the divergence? Imagine that F(x,y,z) = M(x,y,z)i+N(x,y,z)j+R(x,y,z)k represents the velocity of a flui d flow.
Divergence, Gradient, and Curl
13.4
1 087
Recalling from Section 11.6 of Chapter 11 that a/ax is the directional derivative Di
in the i direction, aI ay is the directional derivative Dj in the j direction, and aI az is the directional derivative Dt in the k direction, we have
div(F) =
a M + aN + R = _Q_(F i)+ _Q_(F j)+ _Q_(F ax ay az ax ay az
a
·
·
·
k) =Di(F
·
i)+ D·(F j)+ Dt(F J ·
·
k).
Thus div(F) sums the rate of change of the i component of F in the i direction, the rate of change of the j component of F in the j direction, and the rate of change of the k component of F in the k direction. If, at a point P
=
(x,y,z ) , the divergence is
positive, then the fluid is tending to flow outward from P. If div(F) is negative at P, then the fluid is tending to flow inward at P. Thus the divergence measures the tendency of a vector field to expand or contract. These ideas are illustrated in
the next two examples.
� EXAMPLE 3 Define F(x,y) =xi+ yj, G(x,y) =-xi - y3j, and H(x,y) =
.x2i - y2j. Calculate the divergence of each vector field at the origin, and relate your answer to the physical properties of the flow that the vector field represents.
y
"' \ \
" "' \ "'-.. ' '
t II/
Solution We begin with F. Look at Figure 1. The flow at the origin seems to be
I
is the indication of an outward flow,
outward from
I //
\
/
/�
--x
--- /
/
I
I
"
\ "" ""'
/II
\\�
We calculate that div(F)(O,0) = 1+ 1 =2, which, being positive,
indication of an inward flow. Indeed, Figure 2 shows that the flow of G at the origin
(0, 1), Q (0, -1), and R (1,-1) are also plotted. Use Figure 4 to determine the sign of div(F)(P) and div(F)(Q).
I I/
"
,,
-
;
'
// I G(x,y)
-
...... ........_
--
\ \"
=
-xi - y3j
\ '
i
�
'
i
\ '
........_ ......
--
--x '
-- '
...... ........_
\
i
i
�
'
--
\
........_ ......
"\ \ i i \ \" H(x,y) =x2i - y2j
.A Figure 3
=
=
y
"\ \ i i \ \"
-
-
=
y
x
.A Figure 2
at the origin. The
sketch of the vector field (x,y) t--t (1/2)dir(F(x,y)), which shows us the directions of
F scaled to a convenient size. The points P
�\ \
-
0
..,...
y
-
we calculate that
-1, which, being negative, is the
� EXAMPLE 4 Define F(x,y) =2xyi+ (1+ 2xy- x2)j. Figure 4 represents a
.A Figure 1
-...
=
is inward. Finally, div(H)(x,y) = 2x- 2y =2(x- y), which is suggests (Figure 3) .
F{x,y) =xi+ yj
-
as expected. Next,
div(G)(x,y) =-1-3y2• Therefore div(G)(O,O)
(net) flow at the origin is neither outward nor inward, as the plot of vector field H
'-... -----...
//I
0.
////2 ///// I/// ///// fa. ///--//// 11/..-----\ I I/------x /""' " " ""'/ ---"' I I \..-----/11 ----.-/ ---///Q /�R/ I .///// ///I ////-2 ////
t dir(2xyi + (1 + 2xy - x2)j)
.A Figure 4
1 088
Chapter 13
Vector Calculus Confirm your deductions by calculating these values exactly. What can you say about div(F)(R)?
Solution Because the flow at P Similarly, the inward flow at Q
(0, 1) is outward, we deduce that div(F)(P) > 0. (0, -1) tells us that div(F)(Q) <0. However, the
=
=
figure is less revealing at R (and nearby points). Although the net flow does not appear to be strongly inward or strongly outward at such points, the visual evidence is not sufficiently accurate to determine whether the divergence is exactly
0
or
merely a small positive or small negative value. To be sure, we calculate
div(F)(x,y)
Thus div(F)(P)
=
!
=
(2xy)+
2 > 0 and div(F)(Q)
also discover that div(F)(R)
=
0.
�
=
(1+2xy
-
x2)
=
2y+ 2x.
-2 <0, confirming our deductions. We
The points just above R have a small positive
divergence, and the points just below have a small negative divergence. <1111 Many terms related to divergence have been coined to describe fluid flow. Suppose that the vector field v represents the velocity of a fluid in a region v (and the fluid) is called incompressible if div(v)
=
0 at
each point of
g,
Q. Then
After we
learn the Divergence Theorem in Section 13.8, we will understand that, if a fluid is incompressible in a region quantity exiting
Q.
Q,
then the quantity of fluid entering
g
equals the
There are circumstances in which the fluid can be compressed
so that the net flow into the region exceeds the net flow out. This situation cor responds to div(F) <0 on
g,
We then call the vector field (and the fluid) com
pressible. Finally, if div(F) > 0 on
Q,
then the vector field and the fluid are called
expanding.
The Curl of a Vector Field
If F(x,y)
=
M(x,y)i+N(x,y)j is a vector field in the plane, then we define its curl to
be the spatial vector field
curl(F)(x,y)
If F(x,y,z)
=
=
(
aN ax
-
aM ay
)
(13.4.3)
k.
M(x,y,z)i+N(x,y,z)j +R(x,y,z)k is a vector field in space, then we
define its curl to be the vector field
curl(F)(x,y,z)
=
(
aR
) (
aN . az • ay
aR
) (
aM . + az J ax
Formula (13.4.4) can be symbolically written as
a form which should be regarded as a memory aid.
aN
)
aM k. ax ay
(13.4.4)
13.4
Divergence, Gradient, and Curl
1 089
� EXAMPLE 5 Define F(x,y,z) =y2zi- x3j +xyk. Calculate curl(F). Solution We have aR
_
aN
ay
az
=x - o,
aR aM - - - - y - y2 , ax az _
and
aN
aM
ax
ay
=-3x2 -2yz.
- - -
Therefore curl(F)(x,y,z) =xi-(y -y2)j +(-3x2 -2yz)k. <11111 What is the geometric or physical meaning of the curl of a vector field? The vector field curl (F) measures the tendency of the vector field to "curl" or "rotate." More specifically, curl F points along the axis of rotation of F (determined by the right-hand rule), and the length of curl(F) measures the amount of curling. This is best seen by studying a simple example. � EXAMPLE 6 Sketch the vector field F(x, y,z) =-yi +xj +Ok and its curl. Solution Set M =-y, N =x, and R = 0. We calculate aR/ay = 0, aN/az = 0, aR/ax = 0, aMI az = 0, aNI ax =1, and aMI ay =-1. Substituting these values into formula (13.4.4), we get curl(F)(x,y,z) =2k. Both F and curl(F) are sketched in Figure 5. We see that the vector field F curls counterclockwise about the origin in the xy-plane. If you curl your right hand in the direction of the curl, then your thumb points in the direction k. Thus curl(F) coincides with our physical perception of the curl of a flow. If the vector field were rotating in the opposite direction-say that H =yi- xj-then we would expect the curl to have opposite sign. And, indeed, curl(H) = -2k. <11111 The notation rot(F) (abbreviation for rotation of F) is sometimes found in older texts instead of curl(F). In physics, a vector field for which curl(F) = 0 at all points is called irrotational. Comparing equation (13.3.3) with equation (13.4.3) in the planar case, and line (13.3.4) with equation (13.4.4) in the spatial case, we see that curl(F) = 0 on a region if and only if F is closed on the region. From our work in Section 13.3, it follows that, if F is conservative, or, equivalently, path-independent, then curl(F) 0. Also, on a simply connected region, curl(F) 0 if and only if F is conservative. This information is recorded in the diagram in Figure 6. It will be of use when we study Stokes's Theorem in Section 13.7. =
z
=
z Fis conservative. (F =Vu for some u) y
curl(F) = ii (Fis irrotational.)
u
Fis path-independent. x
F(x,y,z) = -yi +xj +Ok
.A Figure 5
curl(F)(x,y,z) = 2k
=====}
u
PcF·dr = 0 for every closed curve in
·� g
u
Fis closed.
=(_=if
g is simply connected
.A Figure 6 � EXAMPLE 7 Verify that the vector field F(x,y,z) = 2xi +(z2 -2y)j +2yzk is irrotational on all of three-dimensional space.
1 090
Chapter 13
Vector Calculus Solution We
must
show
curl(F)(x,y,z) =0
that
for
all
M(x,y,z) =2x, N(x,y,z) =z2- 2y, and R(x,y,z) =2yz, we have
oR oN - - - =2z-(2z) =O, oy oz
The Del Notation
oR - oM =0 - 0 =0' oz ox
x,y,z.
Setting
oN - oM =0 - 0 =0. ox oy
and
<11111
Let us introduce the formal notation '7 o. o. o v = -1+- +-k.
ox
oy
J
oz
This is not a vector, but it is a convenient notation in calculations. We are already accustomed to writing the gradient of a function u as
\i'u. The gradient of a function u may also be written as grad(u). Now we note that the divergence of a vector field F=Mi+Nj+Rk may be written as
(13.4.5)
div ( F) =V'·F.
This means that, if we treat V' algebraically as though it were a vector, then V'·F=
( fu
)
�i+ �j+ �k · (Mi+Nj+Rk) =
�
&
oM
fu
+
oN
�
+
oR
&
=div(F) .
Thus a field F is solenoidal on a region if and only if V'·F=0 on the region. Stated in this manner, Gauss's law for magnetism becomes V'·B =0, which is one of Maxwell's equations for classical electromagnetism. In a similar vein, we may write the curl of the vector field F=Mi+Nj+Rk as
(13.4.6)
cur( l F) =V' x F.
This means that, if we treat V' algebraically as though it were a vector, then
V'XF= det
([
� oy� oz�
ox M
N
R
l)
=
(
oR
oN
_
oy
oz
) ( i-
oR ox
_
oM oz
) ( J+ ·
oN ox
_
oM oy
)
k.
In physics books, it is quite common to see divergence and curl expressed using the notation V'. We call this the del notation. The symbol V' is also referred to as nabla. Probably the most important partial differential operator of mathematical physics is the Laplace operator, which is often denoted by the symbol a twice continuously differentiable scalar-valued function
6. Applied to
u of x and y, this operator
is given by the formula
l::,.
&u
u=
ox2
+
&u oy2.
For a twice continuously differentiable scalar-valued function Laplace operator (or laplacian) is given by
6v =
&v ox2
+
&v oy2
+
&v oz2
.
v
of x, y, and z, the
13.4 Divergence, Gradient, and Curl
1 091
We must be careful not to confuse the laplacian .6.v with its look-alike, .6.v, which represents a small increment in the value of v. Generally, we can infer which meaning is intended from the context. It is often convenient to define the laplacian of a vector-valued function F = (M, N, R) by applying .6. to the components of F: .6.F = (.6.M, .6.N, .6.R). Sometimes we write the laplacian as .6.v=
V7 V7 ·
because
a2v + a2v + a2v = _E__ av + _E__ av + _E__ av =V7. . (Vv) axz ay2 az2 ax ax ay ay az az
(13.4.7)
V7 V7
The notation is often contracted to V72. Many important differential equations in mathematical physics involve the laplacian. For example, if the edges of a metal plate are heated up, then the second law of thermodynamics implies that the steady state heat distribution h(x, y) in the plate satisfies the equation .6.h 0. A function u that satisfies .6.u 0 is called harmonic. Harmonic functions are important objects of study in mathematical analysis, physics, and engineering. ·
=
=
8 Show that the functions u(x,y)=x2 - y2 and v(x,y,z)= ( l e x+y cos ( J2z) are harmonic. � EXAMPLE
Proot
We have a2u + a2u = + - = 2 ( 2) 0 ax2 ay2
and
Identities Involving div, curl, grad, and 6.
The next theorem gathers together some basic identities that involve the operators we have been studying. Several other identities may be found in the exercises for this section. THEOREM 1
Let u be a twice continuously differentiable scalar-valued func tion on a region g in the plane or in space. Let F be a twice continuously differentiable vector field on Q. Then a. b. c. d. e.
div(grad(u)) = .6.u, div(curl(F)) =0, curl(grad(u)) =0, curl (curl (F) ) = grad(div(F) ) - .6. F, di v(uF) u div(F) + grad(u) F. =
·
1 092
Chapter 13
Vector Calculus Proot The first assertion is simply a restatement of equation (13.4.7). Part b
follows from the equality of the mixed partial derivatives of the components of
F=
(M,N,R):
)
(
div curl(F) =
( ay
)
( az
)
( ax
)
_E__ aR - aN + _E__ aM - aR + _E__ aN - aM ax
az
ay
ax
az
ay
or
(
div curl(F)
)
a2R - azR ) + ( a2N - a2N) + ( a2M - azM) (axay ayax azax axaz ayaz azay
=
Part c asserts that the conservative vector field
grad(u)
=
0
.
is closed, a fact which we
established in Section 13.3. Part d can be proved by a brute force calculation that is similar in spirit to part b. Finally, part e is obtained as follows:
a(uM) + -a(uN) +-a(uR) --
div (uF)
ax
ay
az
+Mau ) + (u aN+Nau ) + (u aR +R au ) (u aM ax ax ay ay az az aM+ aN + aR ) + ( au + au + au ) M u( N R ax ay az ax ay az udiv(F) +(\7u) F.
•
·
� EXAMPLE
(curl(F) )
=
9 For
grad (div(F) )
Solution We have curl(F)
=
=
(� (
-
+ez )
cos(x)
-
F(x ,y ,z)= (y,z3,cos(x)+ez ) ,
verify the identity curl
- 6F.
! z3}+ ( ! y - ! (
cos(x)+e
z
))j+
( gx z3- � y) k
3z2i+ sin(x)j+(-l)k
and therefore
(
curl curl(F)
)
=
=
(�
(-1)
-!
sin(x)
-6zj+cos(x)k.
On the other hand,
}+ ( ! (-3z2) - ! ) j+ ( ! (-1)
sin(x)
- � (-3z2)) k
13.4 Divergence, Gradient, and Curl
1 093
and 6F
=
(
Therefore grad (div(F)) - 6 F which equals curl (curl(F)) . <11111
Q UIC K
Q U IZ
1. 2. 3. 4.
)
(6y)i + (6z3)j + 6 (cos(x) + ez) k =
ezk-
=
6zj + (-cos(x) + ez)k.
( 6zj + (- cos(x) + ez)k)
=
-6zj + cos(x)k,
Calculate the divergence ofx2yzi + 7yj -xYk at the point (1, 2, -1). Calculate the curl ofx2i + (12/z)j -y3k at the point (-4, 3, 2). Calculate'V X ('Vu) for u (x,y, z) x2y3 +z4• True or false: F is conservative implies curl(F) 0. =
=
Answers 1.
3
2. -24i
3.
0
4. True
EXERCISES Problems for Practice
In each of Exercises 1-10, calculate the divergence of the given planar vector field F. L 2.
F(x,y) =3x2y3 i-xy4 j F(x,y) (Sx+2y)i+(Sx-2y)j
3.
F(x,y)
4.
F(x,y) =tan2(y3)i+ (cos2(x) +2y3)j F(x,y) =cos(x)i-sin2(xy)j F(x,y) In(x+y)i+f:'Y j 2y x 1. J F(x,y) x-y x+y F(x,y) xeYi-ye'j F(x,y)=tan (x/y)j+ cot(y/x)i F(x,y) =(x2 +y2) e+Y(i+j)
5. 6. 7
•
8. 9. 10.
=
=
x2sin(y)i+ (x2 +sin(y)) j
=
.
--
-
--
=
In each of Exercises 11-20, calculate the divergence of the given spatial vector field F. 1L 12. 13. 14. 15. 16. 17. 18. 19. 20.
F(x,y, z)=(2y -3z,4z-Sx, 6x+ 7z) F(x,y,z)=(xz-xy,xy-yz,yz-xz) F(x,y,z)=(x+2y +3z)(l,2,3) F(x,y,z) (zln(x),yz/x,z2 /x) F(x,y, z) x2i+y2j+ z2k F(x,y,z)=x3yi-y2zxj+xyzk F(x,y,z)=e-zi+ez-y j- eY-Xk F(x,y,z)=x2sin(y) (i-j+k) F(x,y, z)= cos(xy)i-sin(yz)j +cos(y)sin(x)k F(x,y,z)=xYi+y'j-x'k =
=
each of Exercises 21-26, calculate the curl of the given vector field F. In
2L 22.
F(x,y,z)=xzi- y3j+xyzk F(x,y,z)=cos2(z)i- sin(xy)j+x4zk
F(x,y,z) =In(x+z)i- eY'j+ xyk F(x,y,z)=z2i- xzj+x3k 25. F(x,y,z)=tan(xy)i+cos(xy)j-sin(xy)k 26. F(x,y,z)=xyzi+(xyz)2j+(xyz)3k
23. 24.
In each of Exercises 27-32, calculate the curl of the given vector field F. State whether F is closed or not.
F(x,y,z)=y2zi+2xyzj+xy2k F(x,y,z)=(y/x)i+(z/y)j+(x/z)k F(x,y,z)=cos(yz)i-xzsin(yz)j-xysin(yz)k Y Y 30. F(x,y,z)= __ i+ln(x+z)j+ -- k x+z x+z 3L F(x,y,z)=x cos(y)i-zsin(x)j+(xy)sin(z)k 32. F(x,y,z)=e'i-e'-'j+eY-xk
27. 28. 29.
In each of Exercises 33-38, a scalar-valued function u is given. Calculate div(grad(u)). 33. 34. 35. 36. 37. 38.
u(x,y) =x2- y2 u(x,y) e'cos(y) u(x,y,z) x/(z -y2) u(x,y,z)=ysin(x-z) u(x,y, z)=x(y +2z) u(x,y, z)=(2x- z)/(yz) =
=
In each of Exercises 39-42, a vector-valued function F is given. Calculate grad(div(F)). 39. 40. 4L 42.
F(x,y)=cos(x)i+sin(y)j F(x,y,z)=(x/y,-x/y2) F(x,y,z)=xyzi+ yz2j+xzk F(x,y,z) (x2,y3,z4) =
Further Theory and Practice In each of Exercises 43-46, calculate curl(curl(F)) for the given vector field F.
1 094 43. 44. 45. 46.
Chapter 13
Vector Calculus
F(x,y,z)= (xyz,xyz,xyz) F(x,y,z)= (xz + yz,yz + zz ,xyz ) F(x,y,z)=(x/y) i + (z/y)j + (x2 -y2) k F(x,y,z)=cos(y) i + sin(x + 2z)j + sin(x)cos(y)k Suppose that
g, h
56. Suppose that
G
57. 58.
are continuously
differentiable vector fields. Prove each of the identities in
potential function V. Prove that V is harmonic and
L.F=grad(div(F)). 1 tells us that the curl of any twice-continuously
47.
-
·
r
(x,y,z). Example 3. Verify
53. Let
60. Theorem
·
·
=
differentiable vector field on a region g is solenoidal on
Q. The converse is also true, provided g has certain
·
F(x,y,z) \1XF=0 at
yj + zk. Prove rX\lu is solenoidal. Prove that \1X(wXr)=2w when w is a constant vector and r(x,y,z)=xi+ yj + zk. Prove that F XG is solenoidal when both spatial vector fields F and G are irrotational.
59. Suppose that the curl of a spatial vector field F has a
Exercises 47-52.
\l(gh)=g\l(h) + h\l(g) 48. \1 (gF)=(\lg)· F + g(\1 F) 49. \1X(gF)=(\lg) XF + g(\1XF) F (\1XG) 50. \1 (FX G) =(\1XF)· G SL \1x(g\lh-h\lg)=2\lgx\lh 52. \1x(g\lh + h\lg)=0
properties. In this exercise, assume that g is an open ball containing the origin 0. Let F be a differentiable vector
Allrr3r
Define
=
as
that
all points of space
field such that div(F) let
G(P)=
Harmonic functions on the plane have a remarkable mean value property: The average value of a harmonic function on a circle is equal to the value of the function at the center. To
u
u(xo,Yo)
1
127r u(xo
27r 0
of
P (x,y,z) E Q, oP given by =
[ t F(rp(t))xrj,{t) dt.
Calculator/Computer Exercises
+ rcos(t),yo + rsin(t)) dt.
In Exercises 54 and 55, verify that the given function
0 on Q. For each
Show that F= curl(G).
is harmonic, then, for any r > 0,
=
=
rp be the parameterization rp(t)= (tx, ty, tz),0sts1. Define
in
except for the origin.
be precise, if
is a twice continuously differentiable
that
are twice continuously differentiable
scalar-valued functions and that F and
u
scalar-valued function. Set r(x,y,z) =xi +
u
In Exercises 61 and 62, verify the mean value property of
the given harmonic function
is
u at the given point (xo,yo). (See
harmonic, and then verify the mean value property at the
the instructions for Exercises 54 and 55). Use r=1.
given point
61.
54. 55.
(xo,yo).
u(x,y)=x2 - y2 u(x,y)=x3 - 3xy2 + 1
62.
(3,2) (2,1)
u(x,y)=trcos(y) u(x,y) ln(x2 + y2) =
(xo,Yo) (1,3) (xo,yo)=(3,4) =
1 3.5 Green's Theorem In this section, we learn our first theorem that relates the behavior of a flow at the boundary of a region to the behavior in the interior. There will be several inter esting applications. Everything in this section takes place in the two-dimensional plane. Throughout this section (and also in the remaining sections), we will write dA for the element of area in the plane.
We will begin our study with a simply connected bounded region R in the plane.
C 1).
Its boundary consists of a single closed curve _. Figure 1
valued function a. b. c. d.
tr---. r(t), as ts b
(see Figure
r is piecewise-smooth. r(a) r(b) (C is closed). r(t1) =f. r(t2) for as ti < tz s b. As r(t) traverses C with increasing t,
that we parameterize by a vectorWe assume four things about
r:
=
C is made up of finitely many smooth arcs. The r begins and ends at the same point. The third condition
The first condition states that second condition says that
the enclosed region R lies to the left.
13.5 Green's Theorem
_. Figure 3b Negative orienta tion of boundary
_. Figure 3a Positive orientation of boundary
_. Figure 2 A curve that inter sects itself is not simple.
1 095
C does not cross itself. It rules out curves like the one in Figure 2. A simple. Condition d says that C is oriented as in Figure 3a, not as in Figure 3b. This is said to be the positive orientation of C. If we regard the xy-plane as the z = 0 plane of xyz-space, then the
states that
closed curve that satisfies condition c is said to be
right- hand-rule for cross products tells us that k X
C is positively oriented. THEOREM 1
points into the region Rwhen
Suppose that Ris a simply connected region
(Green's Theorem).
with boundary
r'
C that is parameterized by a vector-valued function r with the four
properties previously listed. If F(x,y)=M(x,y)i + N(x,y)j is a continuously differentiable vector field on a region containing R and its boundary, then
(13.5.1) y
Proof. A number of subtleties arise in the proof of Green's Theorem in its full generality, and these subtleties are a proper topic for a course in advanced calculus. Therefore we will settle for a proof of Green's Theorem in a special case. We assume that our region is both x-simple and y-simple as in Figure 4, with
FGl L=sJ x
boundaries consisting of the graphs of two functions y = a(x) and y = J3(x). We can
C1 by r1 (t) = (t, a(t)),a :5 t :5 b, and the upper curve C2 r2(t) = (b +a t, J3(b +a t)), a :5 t :5 b. Now, we begin with the right side of the
parameterize the lower curve by
-
-
formula in Green's Theorem:
_. Figure 4 'R, is both x-simple and y-simple. We can calculate the second double integral on the right side by expressing it as an
11a(3(x(x)) a
iterated integral and using the Fundamental Theorem of Calculus: 8 M - dydx= 8Y
lb
M(x,y)
a
1yy=-(3a((xx)) _
dx=
lb
(
)
M x,j3(x) dx-
a
lb
(
)
M x,a(x) dx.
a
(13.5.2)
1 096
Chapter 13
Vector Calculus
Now,
i
M· i dr
{ Mi lei
·
dr1 +
{ Mi lc2
·
dr
2
l( �� d�;t)) la ( 1 l l la l l - Jl �� b
+
b b
M(t,a(t)) b
+0
dt
M(b+a-t,(3(b+a-t))
&
df3(b+a-t) +o &
)
dt
b
M(t,a(t))dt-
M(b+a-t,(3(b+a-t)) dt
M(t,a(t))dt+
M(x,(3(x))dx,(after substitution x=b+a-t, dx=-dt)
b
M(x,a(x))dxdA
d(b+a-t)
b
M(x,(3(x))dx
by (13.5.2).
This is half of the job. A similar calculation shows that
1
JJ ��
Nj dr = ·
dA.
'R
Adding the last two equations results in the identity that is Green's Theorem.
INSIGHT
•
Expressed in component form, Green's Theorem says that
j J'c
Mdx+Nd
y
=
{{ 11
(8N - aM)dA. 8x
8y
(13.5.3)
R
It is instructive to compare equation (13.5.3) with the Fundamental Theorem of Calculus, f(b) -f(a)
=
lb t dt, f' ( )
which relates the values of/ on the boundary of an interval [a, b], namely the points a and b, to the integral of the derivative off on the interval. Observe that the boundary orientation is important in the Fundamental Theorem of Calculus: The termf(b) appears with a"+" and the termf(a) appears with a" - ".It is plain that Green's Theorem is a two-dimensional version of the Fundamental Theorem of Calculus, in that it relates the integral of F on the oriented boundary of a region R to the integral of a certain derivative of F on the region.
� EXAMPLE 1
{(x,y) : x2+y2<1}.
Verify Green's Theorem for F(x,y) =-3yi + 6xj and
R
13.5 Green's Theorem
1 097
Solution The boundary of R is the unit circle C traversed counterclockwise when
f
F c
positively
oriented. We may use the parameterization
O
We have
·
dr =
But
r(t)=cos(t)i+ sin(t)j ,
2 2 l{o 1r ( -3 sin(t) ddt cos(t) + 6 cos(t) ddt sin(t)) dt = l{o 1r (3 sin2(t) + 6cos2(t)) dt.
l;'lr sin2(t)dt = l;'lr cos2 (t) dt = 7r by formula (6.2.9). It follows that fcF
·
dr = 97r.
Having calculated the left side of equation (13.5.1) for the data of this example, we turn to the right side. We have
(
y)
ff (°J: 8t:) =ff (8�:) ;: ) Jf -
-a
dA
R
dA
=
R
9dA = 9 x
(area of unit disk)= 9ir.
R
Thus we have verified Green's theorem for this particular example. <111111 At first Example 1 may seem pointless. Now we have two ways of calculating something instead of one. That's just twice as much work! But this example was just for mechanical practice, to help us to understand the
components
of Green's
Theorem. It is not the way that we usually use Green's Theorem. The next theorem shows one of the really striking applications. THEOREM 2
Suppose that R, C, and
r
satisfy the hypotheses of Green's
Theorem. Then we have
(Area of
Proof. For the vector field
(1/2) - (-1/2) = 1.
:
7?.) =
� i -ydx+ xdy.
(13.5.4)
F=Mi+ Nj = (-y/2)i+ (x/2)j,
we have
Nx -My=
Green's Theorem tells us that
h-y/2)dx+ (x/2)dy =
ff
(N,-M1)dA
R
This, of course, is just the area of R.
=ff
(l)dA.
R
•
� EXAMPLE 2 Use formula (13.5.4) to calculate the area A inside the ellipse
x2/a2 + y2/b2 = 1.
Solution We can parameterize the ellipse C by Then
A=
1
2
r(t)=acos(t)i+ b sin(t)j , 0
J (-y)dx + xdy= 2,1 l{271" ( (-b sin(t)) d acos(t) + (a cos(t)) d b sin(t))dt. o Jc dt dt
1 098
Chapter 13
Vector Calculus By simplifying the integrand, we obtain
A= Theorem
1 r2rr l ab ( cos2(t)+ sin2(t))dt= 1 lr2rr o ab · 1dt=7rab. Z o Z
5 of Section 13.3 asserts that a closed vector field on a simply connected
planar region is conservative. This is a deep result whose proof we have outlined for rectangular regions. Now we can use Green's Theorem to obtain a more general proof.
� EXAM PL E 3 Use Green's Theorem to show that a closed vector field
F= (M, N)
on a simply connected planar region
Solution Suppose that F = (M,N) is closed on in
Q.
conservative.
g and that C is a simple closed curve
Let 'R denote the region enclosed by C. Then,
i
F·dr=
ff(�� ��) -
dA=
'R
It follows from Theorems
A Vector Form of Green's Theorem
g is
1
and
3
of Section
ff
OdA=O.
'R
17.3
that F is conservative.
Green's Theorem has many variants, each of which reveals new information. In particular, there is a formulation of Green's Theorem for each of the differential
13.4. For example, for curl(F)= (Nx - My)k, and, therefore, Nx - My= curl(F)· k.
operators div, curl, and 6. that we studied in Section
F=Mi+ Nj,
we have
We can therefore state Green's Theorem as
i
ff
F·dr=
Notice that
curl(F)· kdA.
(13.5.5)
'R
k is a unit normal to the region n. The direction of this unit normal (as other unit normal to n, namely- k) is determined by the right-hand
opposed to the
rule: Curl your fingers of your right hand in the positive orientation of C, and your thumb points in the direction of
k.
Equation
(13.5.5)
tells us that we can compute
the line integral of F over a simple closed curve C by integrating the component of
curl(F) in the direction of the outward unit normal k to the
enclosed region n, the
integration being performed over n.
Next we will develop a divergence form of Green's Theorem. Suppose that n, C,
tf--+r(t)=x(t)i+ y(t)j,
and F(x,y)= M(x,y)i + N(x,y)j all satisfy the hypotheses of
Green's Theorem. Corresponding to each point r(t) of C, we define the unit vector
n( r(t)) =
y'(t)
(-x'(t))
i+
Jx'(t)2+ y'(t)2 Jx'(t)2+ y'(t)2
j=
--II ( 1 llr'(t)
Notice that
r'(t) n(r(t)) = ·
� (
llr' t)ll
)
y'(t)i-x'(t)j.
)
x'(t)y'(t)+ y'(t)(-x'(t)) = 0.
13.5 Green's Theorem
1 099
Because r'(t) is tangent to Cat the point r(t), we deduce that n(r( t)) is normal to Cat
r(t).
By observing the signs of x'(t) and y'(t) as r(t) traverses C in its positive orien
tation, we see that
n(r(t)) is
the
outward
unit normal to Cat r(t).
To obtain the divergence form of Green's Theorem, we apply formula (13.5.1) to the vector field
F(x,y) = -N(x,y)i + M(x,y)j.
f/
·
dr =
ff (�� 8;) +
We obtain
d xdy =
R
ff
div(F) dA.
(13.5.6)
R
Now let us look at the line integral on the left side of equation (13.5.6). We have
t
F dr
t (-N(x(t),y(t))x'(t) M(x(t),y(t))y'(t)) dt y'(t) (-x'(t)) ) +N(x(t),y(t)) llr'(t)ll dt M(x(t),y(t)) b llr'(t)ll llr'(t)ll l( lb (F·n)(r(t)) llr'(t)ll dt. +
·
a
=
But formula (13.2.8) allows us to identify this last expression as the line integral
.fcF n ds. ·
Thus we have
t
F nds = ·
ff
div(F) dA.
(13.5.7)
R
This form of Green's Theorem is sometimes called the "Divergence Theorem in two dimensions." The line integral on the left side of equation (13.5.7) is called the
flux of vector field F across C. If F is the velocity field
of a fluid, the flux is the sum
total of the scalar component of the fluid's velocity in the direction of the outward normal along the boundary curve. � EXAMPLE 4 Let
F(x,y)=xi+yj
and
R={(x,y) :x2+y2<1}.
Assume
that F represents the velocity of the flow of a fluid in the region. Explain what Green's Theorem tells us about this fluid flow. Solution Let C denote the positively oriented unit circle. The integral
measures the fluid flow across the boundary of
R.
If the integral is positive, then
more fluid is flowing out of the region than into it (indicating that there is a source
in the region); if the integral is negative, then more fluid is flowing into the region than out of it (indicating that there is a sink in the region). According to equation (13.5.6), we have
1F nds = ff ·
R
div(F) dA =
ff R
(1+ l)dA =Z(area
of
'R) = 2-ir.
11 00
Chapter 13
Vector Calculus
F
Because this number is positive, we see that the overall fluid flow is out of the region. A glance at the plot of
Green's Theorem for More General Regions y '"-"-\\�• H ++�ll;t'/
''"""'�••++lltfJ
\�I••#
11
tf tfJ<-'
in Figure
5
confirms our conclusion.
<1111
Now we want to discuss Green's Theorem on some regions different from those specified in Theorem 1. For example, the boundary of the region in Figure 6 consists of two smooth, simple closed curves. This new feature poses no problem. We orient
each curve so that the region Ris to its left as it is traversed (as in Figure 7). Then we calculate the line integral by integrating over each curve, one at a time, and adding up the results.
THEOREM 3
r1, r2,
Let R be a region in the plane bounded by one or more con
tinuously differentiable curves C1, ... , CN with parameterizations
=
+ ,
... , rN.
Assume that these curves are oriented so that the region Ris on the left as each curve is traversed. If tt;; ' "' ""'"' t tt;; """'"" "'
�J(J( JI ;;; /J( Jll
f;
F(x,y)
M(x,y)i
N(x y)j is a continuously
differentiable
vector field on a neighborhood of Rand its boundary, then
tJF·dr;= 11ff (
/Jl'lllfttt ;;;'"''"' F(x,y) =xi+ yj
j=1
_. Figure 5
C;
'R
)dA
aN aM By ax
.
(13.5.8)
Proot Rather than give a complete proof, we will simply indicate how to prove Green's Theorem for a region that is not simply connected. Look again at the region Rshown in Figure 6. Orient the boundary as in Figure curves by line segments as in Figure
8.
7.
Connect the two
Doing so creates two simply connected
subregions. By applying Green's Theorem to each subregion, we obtain
_. Figure 6
ff(�� ��)dA= ff(�� ��)dA+ ff(�� ��) = �C1f F·dr1 + �Cf2F dr2 = �CfF·dr. -
-
'R
-
�
�
·
�
------------
R
_. Figure 7
----
R1
C1
_. Figure 8
C2,
To obtain the last equality in this chain that we notice when we traverse the inserted line segments in C1 and
fc1F · dr1 fc2F · drz.
we travel in opposite directions, so these
portions of the line integrals cancel when we add
and
•
13.5 Green's Theorem
11 01
� EXAM PL E 5 Consider the region n (with piecewise-smooth boundary) that is shown in Figure 9. Verify equation (13.5.8) for
Nx -My= 2.x. )d = 11 jvl=xz 2.xdydx= 11 2.xy ly--vl=xzdx= 11 ( 2xV1-x2 - 2x(1-x))dx. 0 y=l-x 0 0 1-x Solution Here M=
�'i ( 'R,
N 8 - 8 X
M 8 8 Y
F= i+x2j.
1, N= x2,
We have
and
A
-
Making the substitution u
=
1 - x2, du= -2xdx,
we calculate
0 y'U u=--2 u312 lu--0= 1 ( 2.xV1-x2) dx= 11 d 3 u=l 1
y
-0-
o
Because
2 2 (--) = -.
3 3
x=l 1 2 1 -2x(1-x))dx= 1 ( 2x2 -2.x) dx= -x3 -x2 1 = --, ( 10 10 3 x =O 3
we conclude that x
ff 11 'R,
1
_. Figure 9
(
N 8
Bx
_
)
8M
By
dA
=
�+
3 (-!3 ) !.3 =
To calculate the left side of equation (13.5.8), we parameterize the boundary of n by using r1 (t)= cos(t)i+sin(t)j, 0 :5 t :57l'/2 for the circular arc and r (t)= ti+(1-t)j, 0 :5t :51 for the line segment C • Then
2
C1
2
3
3 An antiderivative of cos3(t), namely sin(t) - sin (t)/ , is calculated in Example Section
6.2
in Chapter
1
C1
6. It follows that
(
5
of
) 1t=7r/Z= (1- -1) -1= -1-. 3 3 3 t=O
3 sin (t)
. F·dr=cos(t)+ sm(t)---
Finally, we have
1
c2
F·dr=
( ) t=l 2 d d 1 (1-t+t2 -(1-t))dt= 1 (1-t 2)dt= t-- 1 = -. 1 1 dt dt 3 t=O 3 t3
o
o
Thus
{F·dr= { F·dr+ { F·dr=
jC
jC1
jC2
(-!3 ) + �3 = !,3
which is the same number we obtained when we evaluated the right side of equation (13.5.8). ..,..
11 02
Chapter 13
Vector Calculus � EXA M P L E
6 Let R be the region bounded by the circles x2 + y2 =1 and
x2 + y2=4. Verify Green's Theorem for the vector field
Solution Here, M=
-y3, N=2, (6.2.9), we have
formula
and
3
x
2
0
fo 1 1 � 7 2r
3 2 dA
'R,
- 3i +
=
y
Nx -My= 3y2. Using
ff y 1
y
F
=
2
polar coordinates and
3r2 sin2(0) r drd(}
7 r= 2 2r 4 45 sin 2(0) d(}= 4 r=l
Next we integrate over the oriented boundary
2j defined on R.
1
7 2r
0
45 sin2(0) d(}= 471".
C of R. If we let Ca
denote the circle
of radius a that is centered at the origin and that has counterclockwise orientation, then we observe that
C
=
C2 u ( -C1)
and
(13.5.9)
See Figure 10. We parameterize
_.. Figure 10
Ca
calculate
by r(t)=a cos(t)i +a sin(t)j, 0:::; t:::; 271" and
or r F le:.
We can calculate
lr F e.
·
dr=a4
lro
21f
·
dr=a4
lor
21f
2a lro
sin4(t) dt +
7 2r
cos(t)dt
� 0
J;'lf sin4(t)dt using formula (6.2.11) with n=4. We obtain
1 sin4(t)dt=a4 - sin3(t)cos(t) 4
(
1 t=2f7 r=o
+
r 7 3 r2 sin 2(t)dt 4
lo
(6 2 9) 3 �· 1l"a4• 4
)
(13.5.10)
We conclude by using formula (13.5.10) with a= 1 and a=
2 to evaluate the two
summands of the right side of (13.5.9). We obtain
1 C
F
·
dr =
1
C2
F
·
dr
2
-
1
C1
which agrees with our calculation of
Q UIC K
Q UIZ
F
·
2.
If
C is
and if
C
•
24 - -3 7l"
JJR(Nx -My)dA.
1. True or false: If a simple closed curve the positive orientation of
3 dr1 = - 7l" 4
•
4
14 =
45
-4
7l",
..,..
C is part of the boundary of a region, then
is always counterclockwise.
a positively oriented simple closed curve that encloses a region of area 3,
F(x,y) =5yi-2xj, then what is the value of fcF
·
dr?
13.5 Green's Theorem
11 03
3. If C is a positively oriented simple closed curve that encloses a region of area 3, if
F(x, y)
=
7.xi - Syj, and if n is the outward unit normal along C, then what is the
fcF nds? 4. Calculate fc - y dx + x dy where C is the triangle with vertices (4, 6), traversed counterclockwise. value of
·
(0, 0), (6, 6), and
Answers 1. False
2.
-21
3. 6
4.
12
EXERCISES Problems for Practice
In each of Exercises 1-6, calculate both sides of the for mula in Green's Theorem, and verify that they are equal. L 2. 3. 4. 5. 6.
F(x ,y)=y2i- 3xj, r(t)=sin(t)i- cos(t)j, 7r 5t537r. F(x,y)=xyi- x2j, r(t)=cos(t)i+3 sin(t)j, 05t527r. F(x,y)=(y - 2x)i+(3x - 4y)j, r(t) = cos(2t)i + sin(2t)j, 05t57r. F(x,y)=xy2i- xyj, r(t)=sin(2t)i- cos(2t)j, 7r/25t537r/2. F(x,y)=yi+2xj, r(t)=sin(t)i- cos(t)j, -7r5t57r. F(x,y)=yi+y2j, r(t)=cos(4t)i+sin(4t)j, -7r/25t50.
In each of Exercises 7-10, determine the region R bounded by the oriented curve C whose parameterization r is given. Then, use Green's Theorem to calculate fcF dr by integrating over R. ·
F(x,y)=xyi- x2j, r(t) = cos(t)i + sin(t)j, 05t527r. 8. F(x,y) =3yi- 4xj, r(t) =3 sin(t)i- 4 cos(t)j, 7r5t537r 9. F(x,y) =-y2i + x3j, r(t) = sin(t)i- cos(t)j, -7r5t57r 10. F(x,y)=(y - x)i+(x - 3y)j, r(t)=4 cos(t)i+3 sin(t)j, 05t527r 7.
each of Exercises 11-16, the given region has a piecewise-smooth boundary. Apply Green's Theorem to evaluate the line integral of F around the positively oriented boundary curve. In
lL 12. 13. 14. 15. 16.
{(x,y): -l5x5l, -25y52}, F(x,y) =xy2i- x3y2j the triangle with vertices (7r/4,0),(0,7r/4), and (0,0), F=cos(y)i+sin(x)j the square with vertices (1,0), (0,1), (-1,0), and (0, -1), F(x,y) =ln(3 + y)i- xyj the rectangle with vertices (-2,- 1), (2,-1), (2,1), and (-2,1), F(x,y)=sin3(y)i- cos3(x)j the region bounded by y=x2 and y = 4x + 5, F(x,y)=x2yi + xyj { (x,y): 05y5v'4 - x2, - 25x52}, F(x,y) =e'i- xyj
In each of Exercises 17-22, use formula (13.5.4) to cal culate the area of the given region.
17. The triangle with vertices (2,5), (3,-6), and (4,1) 18.The trapezoid with vertices (4,1), (-2,1),(3,3), and (-1,3) 19. The region bounded by the parabola y= -2x2 + 6 and the line y=4x 20. {(x,y): l/35y5l/x,1/25x53} 2L The area inside the circle x2 + y2 =1 and lying above the line y=(x +1)/\1'3 22. The region obtained by removing the disc with boundary x2 +y2=4 from the triangle with vertices (-14,16), (8, 12), and (1, - 20) In each of Exercises 23-25, compute the flux of F across the boundary of the given region.
F(x,y) =xy2i- xj, R={(x,y): 0< lxl
23.
_
•
Further Theory and Practice
Use formula (13.5.4) to calculate the area of the region bounded by r(t) = (cos(t) - 2 sin(t),cos(t)),05t527r. 27. Use formula (13.5.4) to calculate the area inside the cardioid that is described by the polar equation r=2+2 cos(l:I). 28. Use formula (13.5.4) to calculate the area inside the limac;on that is described by the polar equation r=5 - 3 sin(l:I).
26.
29.
Calculate
i (y
+ arcsin
( �)) dx + (2xy
+ In(l + y4 ))
dy
where C is the closed counterclockwise path consisting of the graphs y =x2, 0 5x5l and y = y'x, 05x5l. 30.
Calculate .fc(x - y sec2(x)) dx + arctan(y2) dy where C is the closed counterclockwise path consisting of the graphs y=tan(x)2, 05x57r/4 and y=tan(x)4, 05x57r/4.
11 04
Chapter 13
Vector Calculus
31. Sketch the curve C parameterized by
sin(2t)),0 ,st ,s'Ir.
r(t) = (sin(t),
area of the region enclosed by C.
&;ayz. a.
r(t) = (t-t2, t-t3),
32. Sketch the curve C parameterized by
Green's First Identity
0 ,st '51. Use formula (13 .5.4) to calculate the area of the region enclosed by C.
33. The Folium of Descartes defined by shown in Figure
11.
i3 + y3 = 3xy
ff
is
The loop in the folium is a simple
3t 3t2 J, + 1 + t3 • 1 + t3 •
0
•
<
-
formula (13.5 .4) does yield the correct area of the region enclosed by the loop. Calculate the area.
t
(u /'::,. v -v 6 u)dA =
fc
(uvxdy-uvydx).
fc
(uvx -vux)dy + (vuy -uvy)dx
Hint: Use part
a, by reversing the roles of
u
and
v.
35. Use Green's Second Identity (from the preceding exer cise) to show that
1
=
=
'/?.
Although the interval of the parameter is unbounded,
y
·
b. Green's Second Identity
ff
t< oo.
(u /'::,. v+\lu \lv)dA
'/?.
closed curve that can be parameterized by
r(t)-
/'::,.=ff/8x2 +
identities involving Laplace's operator,
Use formula (13 .5.4) to calculate the
if u
is harmonic
fc
uy dx-uxdy=O
(6u = 0
at all points).
Calculator/Computer Exercises
x
36. Plot the simple closed curve that is parameterized by
r(t) = sin(2t)i-sin2(3t)j, 7r/12
Calculate the
area that is enclosed.
In each of Exercises 37-40, verify Green's Theorem for the given vector field
r(t)
=
---1!._3 i + __lL3J· 1+t
1+t
_. Figure 11 34. Suppose that u and v are twice continuously differentiable functions
in
a neighborhood of a region
boundary curve C. Suppose that
R
R
and its
and C are as in the
statement of Green's Theorem. Prove the following
F
and region
R.
F(x,y) sin(x+y2)i, R {(x,y): x2
=
38.
=
=
{
R= (x,y): (x-1)2
}
1 3.6 Surface Integrals In this section, we develop the final concept that will be needed for the principal theorems of vector analysis. That is the concept of the surface integral, which enables us to integrate a function that is defined on a surface. We begin with the problem of calculating the area of a surface that is the graph of a function. Suppose that a surface Sis given as the graph of a function z
=
f(x,y) over a
region R, in the xy-plane as in Figure 1. How do we calculate the surface area of S? As you might suspect, we can develop a method for calculating surface area by using a double integral of a certain expression over the region R.
uppose that a surface xy-plane, as in Figure contains
-
11 05
is the graph of a function f over a bounded region
in the
Let
Q = {x
,y) :
. We partition thex-interval [a,b
a=xo
1.
1 3.6 Surface Integrals
.. ..J, ...
···
,
d}
be a rectangle that
s
wherex;
a+iili:, &=
b-a)/N,
nd we partition they-interval [c, ] as .s
c
Yo
·
·
·
d
whereY;
+ jfly
d- c) N.
fly
,
Corresponding to each pair Xi-1,X in the partition of thex-variable and each pair , in thexy Y;-1,Y; in the partition of they-variable, there is a small subrectangle Q ; plane. The only relevant rectangles
for our calculation are the ones that lie
A,; of the small Qi,j·Examine Figure 3. We estimate Ai,j by the area of a parallelogram that closely approximates the graph of f over Qi,j · To do this, we select a point (i; inside Q,; and use the parallelogram that is tangent to d at the point over ij and that has the vectors v tlx, O,f x ; n 0, fly,f (;;)fly for its sides. Refer to Figure 4. We know from � w entirely in 'R,
__ ....,,_, __ -----....
.A. Figure
Qi,j
see Figure 2). We want to approximate the area
piece of surface lying above
1
=
Chapter 9 that the area of this parallelogram is
ll vXw ll
=
l
=
J1+ x(i;)
det
([ L:
:
i
fx( )ili: fly f (i;)fly +f
l'
2 ij ) tlxfly.
If we sum over rectangles
L:Q
.i
n
J .
+
Q ,;
that lie in 'R., then we get an approximation the surface area of S. As the number
;.
w=
(0, y, ( . )Ay)
v=
(IU O,j, �, .)IU)
w
v
Ii ase
.A. Figure 2
.A. Figure 3
area
A�I
, .A. Figure
I I
�
·�
:r::.1 4
N
1 108
Chapter 13 Vector Calculus increases, our approximation improves. On the other hand, our approximation is a
J1+fz(x, y)2+fy(x,y)2 dA. We therefore
Riemann sum for the double integral ff'R. make the following definition.
Let S be the graph of a continuously differentiable function (x,y)- f (x y defined on a region 1l in the xy-plane. We refer to
,)
dS =
1+ (8xa t(x,y)) + ({Jya f(x ,y)) 2dA 2
as
the element of surface area on S. The surface area of the graph of f over 1l is defined by
Surface area of S =
ff 1 ff dS =
s
I
'R.
1+(ax8 f (x,y)) +({Jy8 f(x,y)) 2dA. (13.6.1) 2
3 + 2y+ =6 that
._ EXAM PL E 1 Find the area of the portion of the plane x
lies over the interior of the circle x2
+r 1 in the xy-plane. =
z
not itself a disk. It is the interior of an ellipse (see Figure 5). But we can do the area calculation without determining the exact nature of this ellipse. We need only notice that the plane is the graph of the function z = f x,y = We need to calculate the surface area of the graph over the region 1l = { (x, y) : x2 According to the definition, the required surface area is
Solution The surface whose area we are computing is
( ) 6 - 3x- 2y.
x2+y2=1
...:>c:::::::=:::: � ::=--
Y
x-
+ y2<1}.
ff Jt+(fx)2+(f1)2dA= ff V1+(-3)2+(-2)2dA=Vf4ffdA=v'f4·7r·12=v'f47r.
A Figure 5
'R.
'R.
�
'R.
._ EXAM PL E 2 Calculate the surface area of that portion of the graph of
v'x2+ y2 that lies inside the cylinder x2+r = 4. Solution See Figure 6. To determine the surface area of the graph of z = v' x2+ y2, we let f (x,y) = (x2+ y2)1/2 and calculate z=
x
R=
{(x,y) :r + y2 :5 4}
•f(x,y)
=
(r + j)112
•x2+f=4 A Figure 6
The region 1l that we integrate over is the disk in the xy-plane that is inside the cylinder 'Tl. The required surface area is therefore
1107
13.6 Surface Integrals
ff �
J1+ /)2+ JfdA=
ff
V2dA=Y2·(Area of
)=Y2·(7r·22)=4n-V2 .
....
�
Strictly speaking, the definition of surface area that we have given does not apply to the conical surface in Example
2. The
comer on the graph off at the
rigin means that the first derivatives off o not exist there. However, we may justify what we have done by calculating the area of the part of the graph over the ring
'Re ={(x,y)
:
}
nd then letting
e
tend to 0 (see Figure 7).
When a new formula is developed, it is always a good idea to verify that the results it
yields are consistent with those that are obtained by existing methods (when those methods can be applied). For the conical surface of Example
2, we can apply formula (72.7) for the
surface area of the frustum of a cone. In that formula, A
=
2w
� )s,
we set
r=
0
the radius at the vertex), R = 2 (the radius of the base), ands= 2vZl the slant height).
�
A Figure 7
We obtain
21T
:o ; ) 2-V , or 4T V'i, which agrees with the value obtained in Example 2. 1
..,.. EX A M P L E 3 Calculate the surface area of that part of the graph of
f: x,y)=
- x2 - y
that lies above the xy-plane.
Solution Look at Figure 8. We are required to calculate the surface area of the
graph of f over the region 'R= {(x,y)
=
ff J1+
:
x2
+y2:::; 4}:
2x)2 + 2y)2dA=
n
ff J1+4x2+y2}�A.
n
This integration is easiest if we use polar coordinates. The region of integration is { r, 0) : s; (J:::; 27r,0:::; r:::; 2} in polar coordinates. The integral iven by becomes
1
Integrating a Function Over a Surface
+4r2 drd
1
211"
1 1+4 2)3/2 r=2d() 12 r=O
1.
Once again, consider the surface S that is shown in Figure Suppose that, at each point x,y, z) of , we know the weight density x,y, z) of the surface. In order to calculate the total weight of the surface, we repeat the procedure that we used to find surface area. That is, we enclose the region in a rectangle we subdivide ach side of into N equal length subintervals (refer to Figure 2}, and we choose a point in each of the resulting small rectangles that are entirely contained within 'R. The weight of that part of S that lies over as shown earlier in Figure 3) is approximated by
i,;
Q,
Q
Q ,;
Qi,;
.: : i,j,f � i,;)) . .jt +.",: (;) +f i,;)2. Ax. dy.
.._..weight per unit area
approximate surface area of patch over Q ,;
11 08
Chapter 13
Vector Calculus
Summing over all the subrectangles Qi,j in
'R,
we obtain the Riemann sum
cp (ei,;J(ei) ) J1 + tx(ei,j )2+fy(ei,j )2 . Lll�y � i,j as an approximation of the weight ofS. This approximation tends to improve as the number N of subdivisions increases. Because our approximation is a Riemann sum for the double integral following definition.
JfRcp(x,y,f(x,y)) Vl +fx(x,y)2+fy(x,y)2dA, we make the
Let S be the graph of a continuously differentiable function f defined on a region 'R in the xy-plane. If cp is a continuous function onS, then the surface integral of cp over S, denoted by if8cpdS, is given by
ff cpdS= }} ff cp(x,y,f(x,y)) }}
1+
'R,
s
(aaxf(x,y)) 2+ (aya f(x,y)) 2dA.
( 13.6.2 )
Formula (13.6.2) is often used when cp is a density function (such as weight density, mass density, charge density, etc.). The surface integral then represents the total value of the quantity whose density is represented by cp. Of course formula (13.6.2) has meaning for any continuous function cp. If cp is the constant function that is identically 1 , then the surface integral reduces to the integral for surface area given by formula (13.6. 1). INSIGHT
It is worth taking a moment to understand the roles of the two functions
(x,y) f(x,y) is a function of two variables. It defines the surface S and gives rise to the factor Ji+fx(x,y)2+fy(x,y)2 that is part of the element of surface area. The function (x,y,z) rp(x,y,z) is a function of that appear in surface integral (13.6.2). The function
f-+
f-+
three variables that is defined on a region of space that includes the surface S. This is the
rp(x,y,z) at a general (x,y,z) does not appear in formula (13.6.2), as it does in the triple integrals Jffurp(x,y,z)dV of Section12.6 inChapter12 ( where we integrate rp over a solid U). Instead, when we integrate rp over the surface S that is the graph of z f(x,y), we evaluate cp only on S. Therefore only the values cp(x,y,f(x,y)) are used in formula (13.6.2).
function we integrate. Because the integration is over S, the value point
=
� EXAMPLE 4 Let the surfaceS be the graph of/(x,y)=x2+y2 for (x,y) in the planar region 'R { (x,y) : x2+y2 < 2}. Calculate the surface integral of cp(x,y, z) = z x2 over the surface S.
-
Solution
dS
=
=
The element of surface area is
J1 + fx(x,y)2+fy(x,y)2dA J1 + (2x)2+ (2y)2dA J1 +4(x2+y2)dA =
=
13.6 Surface Integrals
11 09
and
rp(x,y, f(x,y))
=
rp(x, y,z)lz=f(x,y) = ( z -x2)lz = f(x,y) = f(x,y) - y2 = (x2+ y2) -x2 = y2.
The surface integral therefore equals
ff
�ff
.,/
y2 1+4 (x2+ y2) dA
R
S
� [' 1,/2
r2 sin2 ( 8)(1 +4r2)112rdrd8,
where the last integral has been obtained by converting to polar coordinates. We
u= 1 +4r2, du= 8rdr. With r= 0 corresponds to u= 1+4 ( 0)2= 1 , and r= V'2 corresponds to u= 1 +4 ( ..;'2)2= 9. Because the equation u= 1+4r2 gives us r2 = ( u -1)/4, we compute this last integral by making the substitution
this substitution, obtain
ff
� ['[ ( � ) u 1
sin2 ( 8)u112
� dud8 � 3� ['
sin2 ( 8)
([ (
)
)
u312 - u112 du d8. (13.6.3)
s
The inner integral involving
u
is routinely determined:
(13.6.3)
Substituting this value into equation
and using formula
(6.2 .9) ,
1 . 1192 { 27r {{ 1 1192 149 }} rpdS= 32 Ll Jo sm2( O)dO= 32 . Ll 7r= 60 7r. •
we obtain
....
s
�
rp(x,y, z)= xy -3z over the surface S f(x,y)= xy over the region R= {(x, y): 0< x<1 ,0< y< 2}.
EX A M P L E 5 Integrate the function
given by the graph of
Solution The element of surface area is
and
rp(x,y,f(x,y))= rp(x, y,z)lz=f(x,y) = xy -3zlz=f(x,y) = xy -3f(x,y)= xy -3xy= -2 xy. Thus
ff s
�f, f, 2
To
1
-2.xy(l + y2+ x2)1i2 dxdy
calculate
du= 2xdx.
the
inner
� f, (J, -2
integral,
we
2
y
make
'
)
x(l + y2+ x2)112 dx dy. the
substitution
After finding the antiderivative with respect to
for x and use the original limits of integration. Thus
u,
(13.6.4)
u= 1 + y2+ x2,
we will resubstitute
111 O
Chapter 13
Vector Calculus
{ 1 x(l + y2+ x2)1/2 dx= ! 2 lo
1
x=l ul/2du=
x=O
! (� u3/2) 2
3
1
l x= =
x=O
! (1+y2+x2)3/2 3
1
l x= =
x=O
On substituting the result of this integration into
! 3
)
(
(2+ y2)3/2 - (1+ y2)3/2 .
(13.6.4),
we obtain
{{ cpdS= - � f2 y((2+ y2)3/2 - (1+ y2)312)dy= - � f2 y(2+ y2)3/2dy+ � f2 y(1+ y2)3/2dy. 3 lo 3 lo }} 3 lo s
The two integrals on the right side of equation
(13.6.5)
(13.6.5)
are quite similar and can be
handled by one integration, if performed in appropriate generality. We replace the constants
2
and
J; y(a+ y2)3/2dy.
note that
1 0
2
1
that are added to
y=0 corresponds to u=a
y(a+ y2)3/2dy=
Substituting
a=2
1
2
y2
a and consider u=a+ y2,du=2ydy, and y=2 corresponds to u=a+4. We have with a general constant
We then make the change of variable
and
1 2
_
1 a
a= 1
a+4
and
-·2-uS/2 u3/2du= 1 2 5
1
u=a+4
u=a
1 = -((a+4)s;2_as/2). 5
into this formula results in
y(2+ y2)3/2dy =
� ((2+4)s;2
_
) � (65;2
25;2 =
_
25;2
)
and
1
2
y(1+ y2)3/2dy=
� ((1+4)5/2 -15/2) = � (55/2 -1) .
Substituting the results of these two integrations into the right side of
ff
s
An Application
'PdS=
-
(13. 6.5) yields
� ( � (65/2 -25/2)) + � ( � (55/2 -1)) = 1� (25/2+55/2 -65/2 -1) .
We now give an application of the concept of surface integral to the determination of a center of mass. �
EXAM P L E 6 Assuming that it has uniform mass distribution 8, determine
the center of mass of the upper half of the sphere
x2+ y2+ z2=a2.
12.7 in Chapter 12 for the concept of center of mass. The surface to be studied is S= {(x,y,z): x2+ y2+ z2=a2,z >0}. By symmetry considerations, it is clear that x =0 and y=0. Thus we need only calculate z. We think of the surface as the graph of the function f(x, y)= Ja2 - x2 - y2 over the region n= {(x,y): x2+ y2 < a2}. We see that Solution Refer to Section
Mz=o= ff 'R
8·z·
J1+f (x,y)2+fy(x,y)2dA. x
13.6 Surface Integrals
Now
fx(x,y) It follows that
J1
+
fx(x,y)2 + fy(x,y)2
(
l+
=
=
-x 2 Va -x2 -y2
a2
_
x2 x2 yz _
fy(x,y)
and
) (2 +
a
_
y2 x2 yz _
=
)
1111
-y 2 Ja -x2 -y2
=
a
Ja2 xz y2 ·
(13.6.6)
_
_
Therefore
Mz=o=
rr; 8·z · J�-�-� h a
J z-xz-yz rr; h J�-�-� a
dA=a
n
rr; h
dA=a8
n
1dA=a8(area ofR)=7ra3 8.
n
We also need to calculate the mass of S. In Example 7 of Section 7 2 . in Chapter 7,
we learned that the surface area of a sphere of radius a is 47ra2• The surface area of the hemisphere Sis therefore 27ra2• The mass
Therefore the center of mass is INSIGHT
(0, 0, a/2).
M of Sis therefore 27ra28. It follows that
<11111
It is worth pausing to observe the difference between the calculation of
Example 6 and the calculations found in Section
12.7 of Chapter 12. In that section, the
geometric objects we treated were either planar regions or three-dimensional solids. For example, using the formulas of Section the
solid half-ball U defined
density
z
8, then its mass is � na38. Using cylindrical coordinates, we calculate z as follows:
� 1 3 r2" r (a2 r2" r r 8z dz rdrd() Jio 2na38/3 4na3 lo lo lo lo lo
M =
12.7, we are able to calculate the center of mass of x2 + y2 + z2 :5 a2, 0 :5 z. If U has uniform
by the inequalities
-
=
=
-
r2)rdrd()
=
3 Ba.
The calculation involves a triple integral, and the value of the z-coordinate of the center of mass is now different.
The Element of Area for
In many applications, it is most convenient to describe a surface parametrically.
a Surface Given
Because a surface is a two-dimensional geometric figure in three-dimensional
Parametrically
u and v as generic parameters and indicate the x, y, and z on u and v by writing x(u, v), y(u, v), z(u, v). Using these three functions of u and v, we obtain a vector-valued function r(u, v) (x(u, v),y(u, v),z(u, v)). Notice that we use the letter r as the position vector for a
space, a parameterization specifies the three space coordinates,
x, y, and z, in terms
of two parameters . We will use
dependence of
=
point on a parameterized surface, just as we do for a point on a parameterized
1112
c
hapter 13
Vector Calculus curve Of course, we know that r(t) refers to a point on a curve because there is one .
parameter, and
r:!..,v)
refers to a point on a surface because there are two
parameters. We now briefly descrie b how to calculate the element of area on a surlace S that is parameterized by a vector-valued function r u,v) = x u,v),y u,v), z: u, v)}, where u and v range over a parameter set 'R.. in the plane
(see Figure 9). Fix values
Uo and vo. Figure 10 shows a small rectangle Q with vertex at u ,vo), width l::iu, and height /::iv. The patch on that is the image, r Q), of Q is also shown. In analogy with our preceding investigation, we want to find the area of a parallelogram that approximates r Q). The functions
-
v
Vi->{x o,v),y o,v),z UQ,v))
nd ·J
.
z
r
r ---
.......--
-�------
---
42) �--..----
A Rgure9
·
nd hei!Jbt
A Figure 10
Po = r uo, vo). The ru o,vo) and rv uo, vo) are tangent to Sat P0, as are their scalar multiples 8 )ru uo, vo) and /::iv)rv o,vo). (Here the subscripts u nd v denote partial differentiation with respect to u and v, respectively). By reasoning similar to that used at the beginning of the section, we find that r Q) is closely approximated by the area of the parallelogram determined by the vectors l:iu)ru o, vo) and l::iv)rv UQ, vo) as shown in Figure 11. That area is just II l:iu)r o,v ) X !::iv)rv Uo, vo)ll, or II ru X r )(uo, vo)ll!::iul::iv. Adding these expressions, we obtain a Riemann sum for the integral ffR r rv)(u,v) dudv. We are therefore led to the describe curves on the surface S that pass through the point velocity vectors
y
..
A Figure 11
··.
The vectors
�u(ru uo, o)), .6.v lv o, vo)),
and the parallelogram that they determine
I
following definition of surlace area.
Let r( ,v) = (x u,v),y u, v),z u,v)} be a continuously differ in the plane. Let Sbe the surface entiable vector-valued function with domain that is the image of 'R.. under r. Then, the surface area of is given by the formula Surlace area of
S=
ff
ru Xrv)(u, v)
dudv.
13.6.7)
1l
We refer to the expression surface S.
ru X lv)( 0,v0)
udv as the element of area for the
13.6 Surface Integrals � EX A M P L E 7
(12.8.3), (12.8.5),
and
then we see that
111 3
If a is a positive constant, and if we set p = a in equations (12.8.6) for the spherical coordinates of the point (x, y,z),
r(O, ¢) = (a cos(O)sin(¢), a sin(O)sin(¢), acos(¢)), 0 :5 (} :5 27r, 0 :5 0. Solution
We have and
r9(0, ¢) = a(-sin(O)sin(¢), cos(O)sin(¢), 0) Thus
(ro x r
r
([
;
-sin(O sin(¢) cos(O)cos(¢)
!
cos(O sin(¢) sin(O)cos(¢)
� ])
-sin(¢)
.
'
or
(r9Xr
(r9Xr
=
-a2sin(¢)(cos(O)sin(¢)i + sin(O)sin(¢)j
+
cos(¢)k)
and so
It follows that the surface area of a sphere of radius a is 271"
11 0
7!"
a2sin(¢)d¢ d0=a2
0
1
271"
(-cos(¢))
0
1
d0=a2
1 0
271"
(-(-1) - (-l)) d0=47ra2,
which agrees with the value obtained in Example 7 of Section 7.2 in Chapter Surface Integrals Over Parameterized
7. .,..
Having already determined the element of surface area for a parametrically defined surface S, we may proceed directly to the definition of a surface integral on S.
Surfaces
Let r(u,v) (x(u, v),y(u,v),z(u,v)) be a continuously differ entiable vector-valued function with domain R in the plane. Let S be the surface that is the image of R under r. If r.p is a continuous function on S, then the surface integral of r.p over S is denoted by Jf8r.pdS and is given by the formula =
L
ff r.pdS= ff r.p(r(u,v)) ll (ruXrv)(u,v) ll dudv. s
n
111 4
Chapter 13
Vector Calculus � EXAMPLE 8 Integrate the function cp(x,y,z) =z-xy over the surface that is parameterized by
S
r(u,v) = (u + v, u- v,3u- 2v), o�u� 1,0� v�2.
Solution
We calculate that
II (r
u
ru(u,v) = (1, 1, 3) and rv(u,v) = (1, -1, -2). Then
X
rv)(u,v)II = Iii+ Sj-2kll = v'30.
The integral that we want to evaluate is
ff cpdS=
s
2 1 11 cp(u+ v,u- v,3u- 2v)v'30 dudv
2 = v'301 1 1
((3u- 2v)- (u+ v)(u- v)) dudv.
Performing the necessary algebra, we find that
ff cpdS
s
{2
=l u 1 ) ( 1 3 O J30 lo 2u2-2uv-3u3+uv2 u= dv ( � -2v+ v2 ) dv J30 ( + � v3) 1: J30 � v 1
2
-v2
J30. Q UIC K
Q UIZ
...
1. What is the element of area for a surface that is the graph of f(x,y)
2.
2x+ y2
(x,y) in a region of the xy-plane? Calculate Jf8cpdS where cp(x,y,z) =z ../2+ y2 and Sis the graph of f(x,y) x + y2 /2 for (x,y) in the unit square [O, 1] X [O, 1]. What is the element of surface area for a surface that is the graph of r(u,v) = (2u,-v,u+ v) for (u, v) in a region of the uv-plane? =
for
=
3.
Answers
1.
..js + 4y2dxdy 2. 8/5 3. 3dudv
EXERCISES Problems for Practice
1. 2. 3. 4. 5.
f(x,y) =xy, R ={(x,y): x2+y2<9} f(x,y) =x2/2, R={(x,y): 2:5x:56,0:5y:5x} 8. f(x,y) =2y2 -2x2, R= {(x,y): 16
In each of Exercises 1-12, calculate the surface area of the graph of the function f over the region R in the xy-plane. f(x,y) =3x - 4y+8, R={(x,y): 2�x�5, -3�y�-2} f(x,y) x2 + y2 + 6, R = {(x,y): x2+y2<16} f(x,y) = (x2 + y2)1/2, R {(x,y): 1
=
7.
In each of Exercises 13-20, calculate the area of the given surface.
13.6 Surface Integrals
13. The cone with base of radius 5 and height 7 (do not count the area of the base)
14. The portion of the sphere x2+y2+z2 = 9 that lies above the plane
z=2
15. The portion of the cone z2 = x2+ y2 that lies between z=4andz=l0
16. The portion of the surface cylinder
x2+y2=16
z=xy
that lies within the
z=(2/3)(x312+y312) that lies 1:Sx:S3,2:Sy:S5} 18. The portion of the surface z=(2/3)x312+y+2 that lies over the region R={(x,y): 1:Sx:S4,1:Sy:S4} 19. The portion of the surface 3x-3y+z=12 that lies over the interior of the ellipse x2+4y2 =4 20. The portion of the surface 2x+4y+z=11 that lies inside the paraboloid z=x2+y2 17. The portion of the surface
over the region R={(x,y):
In each of Exercises 21-28, integrate the function
e(f> x,y, z
over the surface given by the graph off over the region R.
21. 22. 23. 24. 25. 26. 27. 28.
ef>(x,y,z) =x- 2y+3z, f(x,y)=x2+y2+8, R={(x,y): x2+y2<6} ¢>(x,y,z)=3x2- 4y2+7z, f(x,y)=x- 3y, R={(x,y): lxl <2, IYI<3} ef>(x, y,z)=(3x2+3y2+z+1)112, f(x,y)=x2+y2, R={(x,y): 1(x,y,z)=x2-y2+z2, f(x,y)=2x-2y, R={(x,y): x2+y2<1} ef>(x,y,z)=y2+z, f(x,y)=8- y2, R={(x,y): 0(x,y,z)=sin(x)+sin(z), f(x,y)=9x-4y, R={(x,y) : 0(x,y,z)=x-xy+z, f(x,y) =3x- 7y, R={(x,y): 1<2y(x,y,z)=z, f(x,y)= Jx2+y2, R={(x,y): 0
36.
111 5
x=u2,y=v,z=u+3v, v'2:Su:SJ3,1:Sv:S3, ef>(x,y,z)=z-3y
Further Theory and Practice 37. Let f be a positive, continuously differentiable function of one variable. Let S be the surface of revolution that
[a, b], realized as {(x,f(x),O): a:Sx:Sb}, is rotated about
results when the graph of f over
the
space curve
the
x-axis. Parameterize S and obtain a formula for its surface area.
38. Suppose that
b > a > 0. Let S be the surface that results {(O,y,z): (y- b)2+z2=a2} is rotated
when the circle
about thez-axis-S is called a torus. Calculate the surface area of S.
f(x,y) over a planar n(x,y) be the unit upward normal to S at the point (x,y,f(x,y)). Let 1(x,y) be the angle between k and n. Give an argument to
39. Let S be the graph of a function region
R.
For each
(x,y) ER,
let
demonstrate that the surface area of S is given by
ff
sec(t)dA,
'R,
and the integral of a continuous function
ef>(x,y,z) over M
is given by
ff
ef>(x,y,f(x,y))sec(t)dA.
'R,
40. Define a concept of average temperature over a surface. If the temperature of any point in space is given by
T(x,y,z)=3x- 8y+z, then find the average temperature over the upper hemi sphere of the unit sphere centered at the origin.
41. Use an improper integral to integrate the function
the surface area of the parametrically defined surface. Do not
ef>(x,y,z)=
evaluate the integral.
1
( x 2+y2+1)3
x=u2- v,y=u+v2, z=v, 0:Su:S1,0:Sv:S4 x= cos(u),y=sin(v), z=u-v, 7r/6:Su:S7r/3,7r/4:Sv:S 37r/4 31. x= eu+v,y=eu-v, z=u, O:Su:S1,-1:Sv:S1 32. x=u/(u+v),y=v/(u+v), z=u+v, 1:Su:S2,1:Sv:S2
42. Use an improper integral to integrate the function
In each of Exercises 33-36, integrate the function¢> over
43. Assuming that it has uniform mass distribution, find the
29. 30.
the parametrically defined surface.
x=2u-v,y=v+2u, z=v-u, 0:Su:S2,0:Sv:S3, ef>(x,y,z)=x+y+z 34. x =u+3v,y=4u, z=-3v, -2:Su:S1,-1:Sv:S0, ef>(x,y, z)=sin (7r(x- y+z)/3) 35. x=v,y=u,z =u+v, 3:Su:S5,2:Sv:S3, ef>(x,y,z)=yf(z- x)y 33.
over the surface that is the graph of
f(x,y)=x2+y2/2.
ef>(x,y)=ze-(x2+y2) over the surface that is the graph of the function f (x, y)=x2-y2. center of gravity of the cone that is the graph of
z=2yfx2+y2
(0 :Sx2+y2:S4).
44. Assuming that it has uniform mass distribution, find the center
of
gravity
of
the
portion
x2+y2+z2=4 that lies above the plane
of
the
z=-1.
sphere
45. Assuming that it has uniform mass distribution, find the center of
gravity
z=4- x2- y2
of the
portion of the paraboloid
that lies above the xy-plane.
1118
Chaptw 13 Vec::tor Cak\tlus
ff, Let F be a continuously differentiable function of three variables. Let 'R. be a region in the xy-plane. Suppose that for each po.int x,y) E 'R., there i11 exactly one value of z such that F x,y,z} =O. urther assume that 8F 8.z is not 0 at each mch point. Prove that the area A of the portion of the graph of {(x,y,z}: F(x,y,z) =0} lying over 'R. is given by = tf7. Use
ff
v��"
dA.
the formula developed. in Exercise 46 to calculate the IW'face area of a hemilphere of radius a. 48. Use the formula developed. in Exercise 46 to calculate the surface area of a conewith base of radius r nd height . 49. A prolaJe spheroid is an ellipsoid with equal shorter axes. Calculate the surface area of the prolate spheroid r 2+y a2+.z2 c2 =1where0 a
�
yields · dxd8 for the element of surface area of the circular cylinder y2 z2 =a2.) In each
of Bxercilles 51-54, calculate the specified sur integral by using the parameterization de8Cribed in Exercise 5L face
.52. The surface integral
of (x,y, z} = x +2a ver that part of the cylinder x!- y2 = 4 that lies in the irst octant and under the plane z=3. x,y,z)=xyz � . ·er that part of 53. The !Ul'face integral of the cylinder x!- + z2 =16 that lies in the first octant between the planes y =2 and y =4. 54. The surface integral. of� x,y,z) =rz over that part of the cylinder y2 + z =9 that lies in the first octant between the planes x =0 and x =2. ··
calculator/Computer Exercises .55. Calculate the surface area of that portion of the graph of
f x y) = xp .x2) that lies over the interior of the square {(x y): -1 x 1 -1 y 1} .56. Calculate the surface area of that portion of the graph of f x,y) = os ./Y) that lies over the interior of the reotangle {(x,y}: x 1,0
1 3. 7 Stokes's Theorem When we studied Green's Theorem in Section 13.S, we used the notion of planar :flows to gain a physical understanding of the theorem. Now imagine a :flow through a surface (see Figure 1). We may ask how the fiow through the interior of the surface is related to the flow at the boundary of the surface. This is the subject of tokes's Theorem, which generalizes Green's Theorem from planar regions to surfaces in space. Orientable Surfaces and Their Boundaries
A crucial idea in formulating and using Stokes's Theorem is that of orientation. Intuitively, we want to say two things about the orientation of a surface and its boundary. The :first of our requirements is that we may continuously assign to each
point of S a "preferred" unit normal vector.
i•J9$WM•ili
A continuous vector field P- n P) defined on a smooth surface � i.s said to be an orientation of :; if, at each point P of S, the vector n P) is a unit normal to at P. We say that a surface S in space is orientable if it possesses an :.. rientation. An orientable surface together with a fixed choice of orientation is '·
called an oriented surface.
13.7 Stokes's Theorem INSIGH1i
111 7
If P t-t d P)is an orientation of !i , then so is the opposite vector field
Pt--t -nl P). We therefore say that an orientable surface is two-sided. If Sis oriented, then
we say that the side out of which the orientation n points is the positive side, and the side of which - n Points is the negative side.
°' ut
� EXAM PL E 1 Suppose that
sphere x1- + y2 + z2 = a2?
a> 0. What
are the two orientations of the
(x,y,z) is perpendicular to the sphere at the point (x,y,z). Because ll(x,y,z)ll = JxZ + y2 + zZ = a, it follows that the function (x,y,z)t-r(x/a,y/a,z/a} is a continuous unit normal on the sphere. It is the outward unit normal (see Figure 2). Solution By elementary geometry, we know that the position vector
A Rgure 2 'Ill& O'!Ml.'Wa.Jd orir.:11. tation of the sphere
The opposite orientation of the sphere is given by the inward unit normal
(x,y,z) � -(x/a,y/a,z/a}.
�
·Z
A surface with no boundary is said to be closed. The sphere is an example. With such a surface, we can speak of an inward normal or an outward normal vector field. When a surface is the graph of a function, we may speak of an upward normal vector field (as shown in Figure 3}
or
a downward nonnal vector field.
....
/ x
( --
'·
A Figure 3
-�
R)
- -------
Although familiar surfaces such as the sphere are orientable, nonorientable surfaces exist. For example, August Ferdinand Mobius (1790-1868) discovered a "one sided" surface, now called the Mob'ius strip, that is not orientable (see Figure 4). Notice that, when you move a unit normal continuously around a MLibius strip, returning to your original location (but on the opposite side), it winds up opposite to its initial position. You can make your own Mu bius strip by taking a strip of paper, Hiving one end a 180° twist, and then joining the ends. Suppose now that S is a smooth, bounded surface with orientation P'-+ nl.P) ;-ind with closed curve C as its boundary (see Figure Sa). We want to use the u rientation of S to induce a particular direction on C. Imagine yourself standing at a point Pon C, perpendicular to S with your foot-to-head direction given by nCP). There are two directions in which you may walk around C. Choose the direction for which S lies to your left tsee Figure Sb). We say that this direction around C is
,·Jo
A Figure4
D
D
A Figure Sa
A Figure 5b C is oriented so that S Hes to the left when C is
traversed in the direction of the orientation.
1118
Chapter 13
Vector Calculus
induced by the oriented surface . H C is a parameterized curve, we say that C is oriented consistently with S if the orientation of the parameterization is the same as the one induced by S. These notions can be formulated more precisely by observing that, if r is a parameterization of C, then the vector D P) r P) is perpendicular to C itnd tan gent to S'. It points either away from S r into
(see Figure 6). When C: is oriented
consistently with S, the vector nr P) X r P) points into S. Figure 7 shows two pos sible orientations of a disk in space with a consistently oriented boundary curve. Figure 8 shows another oriented surface with a consistently oriented boundary that
A Rgure 6a 'ij. Prx J: P points into S: S and C are consistently
comprises two disconnected pieces. We will discuss such surfaces at the end of this section. For now, just note that the definition of consistent orientation i.:an be applied to each component of a disconnected boundary.
oriented.
A Rgure 6b l')(Prx JP points away from S: S and C are not consistently oriented.
A Figure 7a Upward-pointing orientation of S with consistent
A Figure 7b Downward-pointing orientation of S with comistent boundary orientation
boundary orientation
The Component of
Recall that the curl operator, when applied to a vector field F, can be interpreted as
Curl in the Normal
a measure of the vorticity of the vector field about a given point. In our calcula tions, we saw that emit F) points in the direction of the axis of rotation. For example, if F(x,y) M'x,y)i + N x,y) is a planar vector field, then equation
Direction
=
C13.4.3) defines curl.F) to be a multiple ofk. In particular, the curl points out of the plane in which the circulation is taking place. These observations suggest that, if we want to measure the magnitude of the curl of a vector field as it relates to a surface S, then we should consider the component of the curl in the direction normal to the surface. We therefore use curl F P)) n P) to measure how much the ·
vector field is curling in the surface at
ff
P. The
ml
surface integral
l 13 7 1 )
F) ·ndS
.
.
s
represents the "total curl" of the vector field F on
•
Bear in mind that, in general,
the vector field F curls clockwise with respect ton P) at some points (the points P at which curl(F(P)) ·nlP) <0) and counterclockwise at other points (the points at which curll NP))· n(P) >0). So, surface integral (13.7.1) takes into account certain A Figure 8
cancellations.
13.7 Stokes's 'Theorem
1119
� EXAM PLE 2 Let S be the graph of z= 2x - 2y+10 over the circular region n. {(x,y):x2+y2<4}. Orient S with the upward pointing unit normal n (see Figure 9). Define F(x,y,z)=(y-z)i - (x + y)j+ 2xk. Calculate ff8curl(F)·ndS. =
Our surface is just that portion of the plane 2x - 2y - z= -10 that lies above the interior of the circle centered at the origin and with radius 2. We use the coefficients that appear in the equation of the plane, 2x-2y-z=-10, to obtain a vector 2i - 2j - lk that is normal to S. An upward normal at any point is then given by -2i+ 2j+1k (we know this is upward because the k component is positive).
Solution
Dividing by the length of this vector, which is
V(-2)2+22+12, or 3, we obtain
n{x, y, z)=(-2/3) i+(2/3)j+(1/3)k. Next we calculate
A Rgure9
)
curl(F)= (.£. (2x)-.£. (-(x+y)) i -(.£. (2x)-.£.(y-z))j oy
+ and
oz
ax
8z
( ! (-(x+y)) - �((y-z)))k= Oi-3j-2k.
mrl(F) ·n= (Oi - 3 j - 2k ) ·
�j+ !k)= -� (-�i+ 3 3 3 3·
Now the element of area dS on S is
dS=
J1+t�(x,y)2+fy(x,y)2dA=J1+'l?-+(-2)2dA=3dA.
Putting all this information together, and noting that'R is a disk of radius
ff curl(F)·ndS= ff (-
s
Stokes's Theorem
'R.
�·3 )dA=-
s ff
2, we have
ldA=-8·(area of'R)=-8·(11'·22)=-3211'.
.,..
'R.
Because the line integral §cF · dr sums the tangential component of F, namely F · r', over C, we can regard it as a measure of the tendency of F to flow about the boundary of S. It therefore makes good physical sense for this line integral over the boundary of S to be related to the surface integral ff8mrl(F) · ndS. Indeed, Green's Theorem, in the form of equation (13.5.5), states that
tF· dr= ff curl(F)-ndS
(13.7.2)
s
when S is a planar region with boundary C and upward pointing orientation n=k. The next example, which pertains to a surface that is not contained in the xy-plane, provides further evidence of this relationship. � EXAMPLE 3 As in Example 2, let F(x,y, z)=(y-z)i - (x+ y)j+ 2xk, and let S be the graph of z=2x - 2y+ 10 over the circular region n.={(x,y):x2+y2<4}. Let n be the upward pointing unit normal on S, and orient the boundary C of S consistently. Verify equation (13.7.2).
11 20
Chapter
13
Vector Calculus Solution In Example
2, we proved that the surface integral over S on the right side ( 13.7.2) is -327!". We now calculate the line integral over C on the left side. The orientation of C that is consistent with S is counterclockwise when viewed of equation
from above. The vector-valued function
r(t)
=
2 cos(t)i + 2 sin(t)j + (2(2 cos(t)) - 2(2 sin(t)) + 10)k, 0 :St :S271"
parameterizes C with the proper orientation. We calculate
F(r(t))
=
=
( 2sin(t) - (2(2 cos(t)) - 2(2 sin(t)) + 10) )i - ((2cos(t)) + ( 2sin(t)) )j + 2(2 cos(t))k (6 sin(t) - 4cos(t) - 10)i - ( 2cos(t) + 2sin(t))j + 4cos(t)k,
and
r'(t)
=
-2 sin( t)i + 2 cos( t)j + ( -4sin(t) - 4cos(t)) k.
After some simplification, we obtain F(r(t))
·
r'(t)
=
-12 sin(t)cos(t) + 20 sin(t) - 8 cos2(t) - 12
=
-6 sin(
2t) + 20 sin(t) - 8 cos2(t) - 12.
Therefore using formula (6.2.9) to evaluate the definite integral of cos2(t), we have
t
F dr ·
=
fo
2rr
( -6 sin(2t) + 20 sin(t) - 8 cos2(t) -12)dt 0 + 0 - 8 71" -12 . (27r) =
·
This equation together with the result of Example equation
(13.7.2)
for
F
and
-3271".
establishes the validity of
S. .,..
Our next theorem tells us that equation
THEOREM
2
=
(13.7.2)
holds quite generally.
1
(Stokes's Theorem). Suppose that (x,y,z) f---tF(x,y,z) M(x,y,z)i + N(x,y,z)j + R(x,y,z)k is a continuously differentiable vector field defined on an oriented surface S and its boundary C. Let Pf-tn(P) denote the orientation of S, and let r be a parameterization of C that orients C consistently =
with S. Then
t
F dr ·
=
ff
curl(F) ndS. ·
s
Proof. Our scheme is to reduce Stokes's Theorem to a calculation involving Green's Theorem. For ease of calculation, we will assume that our surface
S is the
graph of a continuously differentiable function f over a bounded planar region R. We suppose that the boundary of R is a continuously differentiable curve Co with a parametrization t f-tro(t) orientation
when
(ro(t),f(ro(t))),
=
�(t)i + 77(t)j,
viewed a :St :Sb
from is
a :St :Sb, that gives Co a counterclockwise
above.
then
a
The
vector-valued
parameterization of
consistently with the upward pointing orientation illustrates the geometry.
C
Pf-tn(P)
function that of
S.
r(t)
orients Figure
=
C
10
13.7 Stokes's Theorem
1121
....------
a
b
.A. Figure 10
Let us begin by rewriting the surface integral. The element of area on S is
Because Sis the graph of/, a normal to Sat each point P = x,y,
fx x,y)i
+/
- lk, as we learned in Section + lk
x,y)
upward normal is then -/ x,y)i-/ x,y)j
x,y)) is given by
11.7 of Chapter 11. Of course an
because the
k component is then
positive) . Thus
n
1
P)
� ------
J1
2
+ fx x,y) +
Using these expressions for dS nd :.•btain
ff
rr
curlF) ·ndS-
c
x,y) n
2
-
x x,y)i- f
x,y)j
+ 1t) .
P), and using formula (13 .4.4) for curl:F), we
;F). (-J;
Lk)dA
s
JJ
ly
-Nr.�,.
•
1
: Nx My))dA
13.7.3)
-
.
1t
In the last integration, each partial derivative of
M, N, and R is evaluated at the
points x,y,/ x,y)) of S. This is an explicit expression for the right-hand side of the formula in Stokes's Theorem. Now we look at the line integral portion of Stokes's Theorem. We have
fF· dr= j'Fr
t)) ·r' t)dt=
j' M r
)( ) +N r ))rf·
)
rt)) d / ( t),71 t)) t
)dt.
Using the Chain Rule, we obtain
fF · dr= j (M r )){'
)
+N r ))rf t) + R rt))! ro ) ) ;,
�(r 1)) (ro t) )rf
>)dt,
11 22
Chapter 13
Vector Calculus or
t
F· dr=
lb ( (
=
)
)
(
)
M(r(t))+R(r(t))fx(ro(t)) �'(t)+ N(r(t))+R(r(t))fy(ro(t)) rl(t) dt
JF !co
·
dro,
where
F(x,y)=(M(x,y,f(x,y))+R(x,y,f(x,y))fx(x,y))i+(N(x,y,f(x,y))+R(x,y,f(x,y))fy(x,y))j. fci· dr0
that has
- � (M(x,y,f(x,y))+R(x,y,f(x,y))fx(x,y)) ) dA.
(13.7.4)
We may apply Green's Theorem to the planar line integral arisen. Doing so gives us
t
F · dro =
ff ( ! (
'R
0
N(x,y,f(x,y)) +R(x,y,f(x,y))fy(x,y)
)
If these differentiations are written out, using the Product Rule and the Chain Rule, then
a (N+R · fy) = Nx+Nzfx+R xfy+Rzfxfy+Rfyx ax and
a (M+R · fx)=M y+Mz/y+R yfx +Rz/yfx+Rfxy· ay If we substitute these expressions into the right side of equation (13.7.4), then the integral reduces to the right side of equation (13.7.3), which establishes Stokes's Theorem.
•
Although the proof given here is limited, Stokes's theorem is not limited to surfaces that are the graphs of functions. In practice, it is often convenient to treat a general surface by breaking it into pieces, each of which can be realized as the graph of a function. In addition to its theoretical importance, Stokes's Theorem can be used to advantage whenever the computation of one side of equation (13.7.2) is substantially easier than the computation of the other side.
4 Let S be the graph of f(x,y) = 4x- By+5 over : (x - 1 ) 2+9(y-3)2 < 36}. Let n(P) be the upward unit normal at each point P of S. Consider the vector field F(x,y, z) = - 3zi+(x+y)j+yk. Let C be the boundary of S with consistent orientation. Evaluate fcF · dr. � EXAMPLE
R= {(x,y)
Solution To calculate the given line integral directly would involve an awkward parameterization of the boundary of S. With Stokes's Theorem, matters become much simpler. We calculate
13.7 Stokes's Theorem
dS=
1123
J1+b(x,y)2+f1(x,y)2 dA= J1+42+ (-8)2=9dA
and
([ !,
]) � �
i
curl(F)= det
k
j
-3z x+y
= i - 3j +k.
y
Finally, because our surface
is contained in the plane 4x - 8y - z = -5, the vector -4i+ 8j+ lk is an upward normal (the coefficient of k is positive) and n(x,y,z)= (-4/9)i+ (8/9)j+ (1/9)k. Thus
:l
F
·
dr
=
ff
curl(F)- ndS
ff
=
( � � �)
(i- 3j+ k)+ - i+ j+ k 9dA
'R.
s
=
ff
(-21)dA = -27 (area of 'R.). ·
'R.
But 'R. is an ellipse with semi-major area 6
·
2 1f, or ·
axis 6 and semi-minor a.xis 2. Therefore it fcF dr= -Zl 1211"= -3247r. �
121f. We conclude that
·
has
·
� EX A M P LE 5 Consider the surface S consisting of the part of the sphere of radius 3 centered at the origin that lies above the plane z= VS. Let P 1-+ n(P) denote the upward pointing normal on S. Define culate
. ff8cml(F) ndS.
F(x,y,z)= -yi+ xzj +y2t.
Cal
·
z= VS and the sphere r1-+y2+ z2= 9 {(x,y,z) :r+y2= 4,z =VS). See Figure 11. The surfaces is therefore the graph of the function f(x,y) = (9-r - y2)1/2 over the region 'R.= {(x,y): r1-+ y2 <4}. The counterclockwise-oriented boundary curve C can Solution The intersection of the level plane is the circle
be parameterized by
r(t)= 2cos(t)i+ 2sin(t)j + v'sk,
0 st s21f.
We have
F(r(t))= -2sin(t)i+2v'scos(t)j+ 4sin2(t)k Therefore
.&. Figure 11
ff
and
F(r(t)) r'(t)= 4 sin2(t)+4VS cos2(t) ·
curl(F)-ndS=
:l
F
·
dr=
fo
27r
r'(t)= -2sin(t)i+ 2cos(t)j+Ok. and, by Stokes's Theorem,
F(r(t)). r'(t) dt
s 27r
2w
f 4 sin2(t)dt+ lof 4VS cos2(t)dt (6�2 9) (4+ 4VS)1f. = lo
�
1124
Chapter 13
Vector Calculus
Stokes's Theorem on a Region with a Piecewise-Smooth Boundary
In many applications. the boundary of the surface S is not given by a single
curve
but rather by finitely many continuously differentiable curves. Stokes's Theorem is also valid in that setting.
The oriented surface S is the image of a continuously differentiable, vector valued function (u, v) - r( u, v) defined on a bounded region 'R, of the uv-plane. The boundary of S consists of finitely many continuously differentiable curves
r1, .. . , rN, each of which is consistent with the orientation P .-n(P) of S (see Figure 12). Now the statement of Stokes's Theorem is that if F is a continuously differentiable vector field on a region of space that
C1, ... , CN having parameterizations
contains S and its boundary, then
ff
s
cud(F) ·ndS= f;lF·dr1-
1 ( 3.7.5)
s
(Because the curve Ci might not be closed. we do not use the symbol fc . If Ci does happen to be closed, there is no harm in using the more general notatibn J'1.)
A Rgure 12
% f(x,y)
� EXAMPLE 6 Let S be the graph of f(x,y)=4x-8y +30 over the rec tangle 'R-= {(x,y): -2
=
4:r
-
8y + 30
�
Solution We could certainly calculate the required line integral directly, but to do so would involve parameterizing four separate line segments and evaluating four separate integrals. Using Stokes's Theorem, the
task becomes much simpler. We have
dS = V1 +fx(x,y)2 +fy(x, y)2dA = J1+4 2+ (-8)2 =9dA.
3
x
Also
A Figure 13
cml(F)(x,y , z) =det
i 8
ox
-x2
j
k
xz
yx
() () (Jy o z
=Oi-yj+ zk.
4x-8y - z = -30 , the vector -4i+8 j+ lk n(P)= (-4/9)i+ 8 ( /9)j+ 1 ( /9)k. Thus on S, we have
Finally, because our surface is upward normal and
curl(F)· n= (Oi-yj+ zk)·
�
is an
(-�i + �j+ !t) = !( z -8y) 9 9 9 9
= ((4x- By+30)-8y) = �(4x-16y+30).
We conclude
i
F· dr=
t,:
lF·dr;=
ff
s
that
curl(F)·DdS=
�ff 1t
(4x-16y+30)9dA=
l 1:
(4x-16y+30)dxdy.
13.7 Stokes's Theorem
11 25
For the inner integration, we have
�
)�3: 4x- 16y+30)dx = x· - 16xy+ 30x)1:: = -80y+ 100.
Finally,
J F· dr- f\-80y+160)dy h �
- oy2+160y)1y=2 -160+320
160 .....
rO
Example 6 involved a line integral over a piecewise-smooth boundary com prising several smooth curves. The next example concerns a surface that has a boundary consisting of disconnected curves. � EXAM PL E 7 Let S be the frustum of the cone y = 0-2Jx2+z.2 that lies between the planes y =8 and y = 14. Suppose that S and its boundary curves are : nsistently oriented as shown in Figure 14. Consider the vector field : x,y,z) zi+ 1j+z• k. Verify Stokes's Theorem for F nd .
Notice that S is not the graph of a function. We may use polar coordinates in the xz-plane as an aid in parameterizing S. If x =rcos(6) and z=rsin(O), then on S we have y =20-2Jxi+z2=20-2r, or r=10-y/2. Therefore the vector-valued function Solution
r
- - '.'! �xz �'
+
zZ
.&. Figure 14
parameterizes
S.
We calculate the partial derivatives
re 9,y) =- 10 -�)sin 9)i+Oj+ 10-�)cos O)k, A
normal vector at a point P =(x, y, z) on N=i:.�·,y)Xr1( ,y)=det
([
r1 6,y) =-�cos 9)i+1j- �sin O)k. is then given by
:
- 10- /2)sin ) - 1/2) OS )
= -10+�)cos )i+
t 1
7z
10-y )cos ) -: 1/2)sin 9)
-5+ �)j+
])
)
-10+ Y in O)k.
P is in the upper half of the yz-plane, then x =rcos ) = so 6 ='Tr 2, and the k-component of N is -10+y 2) sin 'Tr/2) y 2- 10 < 0. Therefore n = �correct orientation of S. We calculate ;.: -: 1/ llN If
;F)
11 26
Chapter 13
Vector Calculus
Therefore
Thus 1
rr url(F)· ndS r211" r 4 (s - �) dydO r271" lo ls lo ll c =
=
( Y2 Sy - 8
)1
1 4
8
dO
s
To compute the right side of equation (13.7.5), we must calculate two line integrals. Figure 14 shows the two components C1 and C of the boundary. Let 2 r1(0)
=
r(-0, 8)
=
6cos(O)i+ 8j- 6sin(O)k,
r (0) 2
=
r(O, 14)
=
3cos(O)i+ 14j+ 3sin(O)k
be parameterizations of C1 and C , respectively. (Verify that these parameteriza 2 tions orient C1 and C consistently with S.) We obtain 2 F(r1(0))· r1(0)=(-6sin(O)i+ lj+ 36sin2(0)k)· (-6sin(O)i+ Oj- 6cos(O)k)=36 sin2(0) -216 sin2(0) cos(O). Thus
Similarly, we have F(r (0))· rHO) 2
=
(3sin(O)i+ lj+ 9sin2(0)k) (-3sin(O)i+ Oj+ 3cos(O)k) ·
=
-9sin2(0)+27sin2(0)cos(O)
and 71"
{ F·dr2= f 2 lo lc2
(-9sin2(0)+27 sin2(0)cos(O))d0(6;;;9) -97r+0=-97f.
Thus
11
F dr1+ ·
1
2
F dr ·
2
=
367r - 97r
=
277r
which is the assertion of Stokes's Theorem. ..,..
=
ff
s
curl(F) . ndS,
13.7 Stokes's Theorem
11 27
llNll in Example 7. In general, when a surface Sis parameterized by (u, v) f-tr(u, v ) , (u, v ) ER,
INSIGHT
Notice that, because of cancellation, we did not need to calculate
we have
n= Therefore the scalar formula
1
±II ru Xrv II
II ru Xrv II
ff
ru Xrv
cancels when curl(F)
curl(F)
·
ndS
=±ff
·
curl(F)
ndS is
·
calculated, resulting in the
(ru Xrv) dudv.
(13.7.6)
'Tl
s
THEOREM 2
dS= llru Xrvlldudv.
and
Let S be an oriented surface with piecewise-smooth boundary
such as we have been discussing. Let the boundary curves be C1, ... , CN with consistent parameterizations r1, ... , rN. Suppose that F is a continuously dif ferentiable irrotational vector field on a neighborhood of S and its boundary. Then
(13.7.7)
13.4,
Proof. As discussed in Section
saying that F is irrotational means that
curl (F) = 0 at each point of S. We apply Stokes's Theorem to obtain
tL 1=1
F dri �
C1
·
ff curl (F) }}
·
ndS�
s
ff OdS� o. }}
•
s
If the irrotational vector field F represents a flow, then we see that the flow around the boundary integrates to 0. Though this is to be expected, there can be some subtleties involved. Remember that, if S is not simply connected, then the property "Fis irrotational" is weaker than the property "Fis conservative." If Fis
irrotational but not conservative, then a summand fc.F drj in equation ·
(13.7.7) can
be nonzero even if the path Cj is closed. The next exkmple will illustrate this point.
� EXAMPLE 8 Verify equation
F(x,y,z)= on the planar surface S pointing normal.
(13.7.7)
for the irrotational vector field
x 2) j+Ok (x2-+yy2) i+ (x2 +y
= { (x, y, z) : z = 0, 1
<
2 2 x +y
<
4}
oriented by an upward
11 28
Chapter 13
Vector Calculus Solution It is a routine matter to verify that F is irrotational. Let a be any positive constant.
The
vector-valued
function
r(t) = a cos(t)i
+ a sin(t)j +Ok, 0 :st :s27r,
parameterizes a counterclockwise circle C in the xy-plane of radius a and center
(0, 0, 0). We have F(r(t))= -(1/a) sin(t)i + (1/a) cos(t)j +Ok and r' (t)= +Ok. Therefore F(r(t)) r' (t)= sin2(t) + cos2(t) and
a cos(t)j
7r
7r
r · dr = lr 2 (sin2(t) + cos2(t))dt = lr 2 o o le F
z
-
a sin(t)i
+
·
ldt = 27r.
Notice that the value of this line integral does not depend on the radius a. If C1 and
x
C are the components of the boundary indicated in Figure 15, then C1 has 2 clockwise orientation, C has counterclockwise orientation, and 2
{(x,y,z) :z
Cz
.A Figure 15
=
0,1 :s
x2 + y2 :s 4)
An Application
The background material on curl that is presented in Section 13.4 provides ample
evidence that Stokes's Theorem is important in physics. Here we provide just one example.
� E X A M P L E 9 Imagine that B is a magnetic field in space. Let the surface S
together with its continuously differentiable boundary curve C be the graph of a
continuously differentiable function z = f(x,y). Suppose that Sis given the upward unit normal n and that C is parameterized by r with consistent orientation. What does Stokes's Theorem tell us about the flow of B around C?
Solution By Stokes's theorem,
t
B · dr
=
ff
curl(B) · ndS.
s
Let us normalize this equation so that we can introduce some standard physical terminology. Let c denote the speed of light (about 186,000 miles per second). The vector
J = -curl(B) c
47r
is called the current density for the magnetic field. It measures the current per unit area of the surface. Then the total current I for the field on the surface S is
I=
ff
J·ndS.
s
In summary, we see that
i c
47r B· dr= -I. c
(13.7.8)
13.7 Stokes's Theorem
11 29
Thus the flow of B around the boundary of Sis a physical constant times the total current I. Equation (13.7.8) is known as Ampere's Law, after its discoverer Andre Marie Ampere (1775-1836). It is geometrically obvious that a closed curve C bounds infinitely many different surfaces S. It is remarkable that Ampere's Law does not depend on which surface S we are considering.
Q UIC K
Q U IZ
1. True or false: For every surface Sin space, we can choose a unit normal n(P) to S at every point P on S such that P 1--+ n(P) is continuous on S. 2. Suppose that a surface Sis oriented by a unit normal vector field P 1--+ n(P) and
that a boundary curve C is oriented consistently. Suppose that you are standing at a point P on Cat the edge of S, that n(P) is the direction of the vector pointing from your feet to your head, and that you are facing in the direction of the tangent vector to C at P. On what side of you does S lie: right or left? 3. Stokes's theorem concerns an equation of the form
fcF dr ·
=
Jf8cpdS where cp is
a scalar-valued function that involves a unit normal vector field n on S that is used to orient S and C consistently. What is
cp?
4. True or false: Green's Theorem in the plane is a special case of Stokes's
Theorem. Answers
1. False
2. left
3.
curl(F)
·
n
4. True
EXERCISES Problems for Practice and a scalar-valued function fare given. Let S denote the graph offover the region R, and let n be the upward unit normal of S.
Calculate the surface integral of
curl(F) n ·
over
S
directly.
Then, verify the value you obtained by orienting the boundary of
S consistently and
using Stoke's Theorem.
F(x,y,z) =z3i- xyj + xzk, f(x,y)=2x-y, R={(x,y): lxl<2, IYI <1} 2. F(x,y,z) = (y- z)i-(x+z)j+(x+y)k, f(x,y)=-5x+2y, R = {(x,y):1
F(x,y,z)=xi+ (y- z2)j+k, f(x,y) =x+y+3, R= {(x,y): lxl
In each of Exercises 1-12, a planar region R, a vector field F,
In each of Exercises 13-18, use Stokes's Theorem to evaluate
JJ8curl(F) n dS ·
for the given vector field
oriented surface S.
F
and
F(x,y,z)=yi+z2j+ v'l+z4k; S={(x,y,z) :x2+y2+(z-4)2 =25,0--+ n(P) with n(O,0,9) =k 14. F(x,y,z)=xi +xj + exp(xyz)k; S={(x,y,z):z=Jx2+y2,z<1} oriented by the
13.
upward-pointing unit normal vector field
F(x,y,z)=zxj; S={(x,y,z) :x2+y2=9,-2--+ n (P) with n(3,0, 0)=-i 16. F(x,y,z)=2xi+(z+l)yj+k ; Sis the union of the cylin der {(x,y,z):x2+y2=4,0-+n(P ) with n (0,0,0)=-k 15.
Chapter 13
11 30
Vector Calculus
17.
F(x,y,z)=y2i+xj+sin(z2)k; S=l::iPQTul::iQRTu l::iRSTuMPT where P=(0,0,0), Q=(2,0,0), R=(2,2,0), S=(0,2,0), T=(1,1,2), oriented by the
18.
F(x,y,z)=(3x+y)i-zj+y2k; S is the union of the cylinder {(x,y,z):x2+y2=4,2
S={(x,y,z):x2+y2+(z-1)2=2,0
31. Let
upward-pointing unit normal vector field
required surface integral to a line integral, and then convert that to a surface integral over which integration is easy.
S={(x,y,z):z=1-x2/4 -y2,0
32. Let
.
vert the required surface integral to a line integral, and
Further Theory and Practice
then convert that to a surface integral over which inte gration is easy.
In each of Exercises 19-30, an oriented surface Sand a vector field
F
are given. Calculate
Jfscurl(F) · ndS
directly.
33. Consider the surface
Then, verify the value you obtained by orienting the bound ary of
S
normal on
consistently, and using Stokes's Theorem.
F(x,y,z)=y3i+z3j+x3k. Sis the hemisphere {(x,y,z): x2+y2+z2=1,y>O} oriented by Pf-+n(P) with n(O,1,0)=j. 20. F(x,y, z)=zi+xj+yk. S={(x,y,z):y2+z2=1,0
{
22.
23.
24. 25.
26.
}
oriented by Pf-+n(P) with n(P) · k<0. F(x,y, z)=-yi+xj+zk. S={(x,y,z):x2+y2+z2=4,00. F(x,y,z)=zi+yj+zk. S={(x,y,z) :z=x2+y2,z0. F(x,y,z)=yzi. S= {(x,y,z):x2+y2+z2=25,00. F(x,y,z)=x3i-zj+yk. S={(x,y,z):x2=y2+z2,l
S. Use Stokes's Theorem to show that if Fis a
continuously differentiable vector field on a neighbor
19.
S= (x,y,z):z= Jx2+y2,0
S, which is the unit sphere in space n denote the outward unit
centered at the origin. Let
hood of
S,
then
ff s
curl(F) · ndS=0.
In each of Exercises 34-37, Sis that part of the cylinder
x2+y2=9
between the planes
unit outward normal on
S.
z=0
the parameterization of Exercise calculate
Jfscurl(F) · ndS.
consistently
and
verify
and
z=2.
Let
n
be the
F, use 13.6 to boundary of S
For the given vector field
51
from Section
Then, orient the your
answer
by
using
Stokes's
Theorem.
34. 35. 36. 37.
F(x,y,z)=yz2 i F(x,y,z)=2yz i+xzj F(x,y,z)= (-7y,-xz,z) F(x,y,z)= ( x,2xz/(x2+y2),y)
Calculator/Computer Exercises In each of Exercises 38-41, an oriented surface vector field
F
are given. Orient the boundary of
sistently, and verify Stokes's Theorem.
S and a S con
·
27.
28.
29.
30.
F(x,y,z)=z3i+3yj. S={(x,y,z):z= x2+2y,(x-2)20.
38.
13.8 The Divergence Theorem
1131
1 3.8 The Divergence Theorem The Fundamental Theorem of Cal.culus relates the boundary data of a function at the endpoints of an interval with information about the derivative of the function :.• ·• � interval:
F b)-F a)=
lb F .t)dt.
s a one-dimensional theorem. Theorem relates the boundary data of a vector field on a planar region 'R with information about certain derivatives of the vector field Ml +Nj in the interior: (1 reen's
i.
Mdx+Ndy=
If (
aN -
��)dA.
:
II
D
This is a wo-dinumsional theorem. Stokes's Theorem is a similar two dimemional theorem for surfaces. Now we will learn a three-dimensional theorem in the same vein-the Divergence Theorem. As with Stokes's Theorem we must begin by considering the question of orientation. Suppose thatU is a bounded, three-dimensional solid in space whose boundary S consists of one or more continuously differentiable surfaces. To each point S iSsociate the unit norm.al vector that points out of the solid (see this vector field D. Suppose that F is a continuously differentiable lneighborhood ofU and its boundary. At a pointP :x,y,z} •nS, :x,y,z) ·n:x,y,z) is the normal component of F. F represents : P) ·n > 0, then the fl.ow at P is outward, while if F ·n < 0, then the fiow at P is inward The surface integral ,
� Figure 1
represents the overall normal component of fl.ow at the boundary. We call this : uantity the ftu.x. of the vector field F s cross the boundary. If the flux is positive, then, : ·verall, the flow is outward; if the flux is negative, then, overall. the flow is inward
.
The Divergence Theorem
It makes good physical. sense that the fiux of a vector field Fcan also be obtained by summing its divergence over the interior. This is in fact what the Divergence Theorem tells us.
ld'Y.l;l§.til
('ne DiTe11eate 1heorem) With
u, S,
a, and Fas descn1>ed
previously, we have
ff s
SF·nd
fff •'J
)dV. div(F
(13JU)
11 32
Chapter 13
Vector Calculus A proof of the Divergence Theorem in its full generality is far beyond the scope of this book. We therefore supply only a proof of a relatively simple case. Because even that simplified derivation is quite lengthy, we defer it to the end of the section. For now, we will concentrate on understanding the Divergence The orem in particular examples. � E X A M P L E 1 Let U be the unit ball in space. The boundary surface S is the
x2+y2+z2= 1. Verify the Divergence F(x,y,z)=3xi+3yj+3zk defined on U. unit sphere
Theorem for the vector field
Solution First, div(F)=3+3+3=9. Therefore
J JJ
div(F)dV
=
9
·
1)
(volume of ball of radius
=
9.
(� ) 7r .13
=
127r.
u
Turning our attention to the left side of equation
n(x,y,z)=xi+yj+zk
(13.8.1),
we observe that
is the unit outward pointing normal at the point
S. Therefore on S, where
x2+y2+z2=1,
(x,y,z)
of
we have
F n=(3xi+3yj+3zk) (xi+yj+zk)=3(x2+y2+z2)=3. ·
·
It follows that
JJ
F ndS= ·
s
JJ
3dS=3
·
(surface area of the unit sphere)=
3 .(47r -12)= 127r,
s
as required. <1111 � E X A M P L E 2 Let U be the cube centered at the origin with faces parallel to the coordinate planes and side length
-1
Verify
the
Divergence
2:
U= {(x,y,z)
Theorem
: -1
for
defined on U. Solution We have
!ff div(F)dV= !ff u
(4x-2y+ 1)dV=
u
1: 1: 1:
(4x-2y+1)dz dydx=2
1: 1:
Continuing,
2
11 1
1
1
_1
(4x-2y+ l)dydx =2
=
2
=2
=
1
1 1: 1:
_1
(4xy-y2+y)
dx
((4x- 1+1)-(-4x-1- 1))dx (8x+2)dx
2(4x2+2x
=8.
y 1y=l=-l
(4x-2y+ 1)dydx.
{��
1
13.8 The Divergence Theorem
1 1
33
Let S be the boundary of U. It consists of the six faces of the cube: top and bottom,
n(x,y, z)= k, the dxdy, and z = 1. Thus the surface integral of F nover the top of
front and back, left and right. On the top, the unit normal field is element of area is
·
the cube is
1-11 1-11 (2x2i- y2j + lk) k dxdy= 1-11 1-111dxdy=4. ·
- k, the element of area is dxdy, n over the bottom of the cube is
Similarly, we see that the normal to the bottom is and z=-1. Therefore the surface integral of F
·
111 1-11 (2x2i- y2j - lk). (-k) dxdy= 1-11 1-111dxdy=4. In the same fashion, it can be calculated that the integral over the front is 8, the integral over the back is -8, the integral over the left side is 4, and the integral over the right side is
-4 .
Summing these values, we see that
!!
F-ndS=4+4+8- 8+4- 4 =8,
s
which verifies the Divergence Theorem. From the positive value of the flux, we conclude that the net flow is out of the cube.
.,..
Of course there is no point in calculating the same quantity twice. Examples 1 and 2 are just for practice. The next example is a more typical application of the Divergence Theorem.
� EXAM PL E
3 Use the Divergence Theorem to calculate the flux of
F(x,y, z) = -x3i- y3j + 3z2k
across the boundary S of
U={(x,y,z) :x2+y2 <16 ,0
U
across S is defined to be
is a cylinder with radius
ffsF n dS. ·
4
and height
5.
The flux of
F
In order to evaluate this integral, we would
have to break up the calculation into an integral over the top, an integral over the bottom, and an integral over the cylindrical side. It is therefore considerably easier to use the Divergence Theorem to calculate the flux. Thus
Jf s
F-ndS=
/!! u
div(F) dV =
/!! (!
<-x3)+
� (-y')+ !3z2) dv= /ff (-3x2- 3y2+6z)dV.
u
u
Introducing cylindrical coordinates, we may rewrite this integral as
rs {27r {4
Jolo lo
(-3r2+6z)rdrd0dz,
11 34
Chapter 13
Vector Calculus
the evaluation of which is straightforward. We have
!Jr s
F
·
ndS=
5 21!" 3 - 4 r4 + 3zr2
11 0
0
(
) 1 =4d()dz r
r=O
= rs r27r (-192 + 48z)d()dz= 2n rs (-192 + 48z)dz= -720n.
lo
lo lo
The flux is negative, and we conclude that, overall, the flow is into the cylinder. Some Applications
The Divergence Theorem has many important uses in several branches of physics. We turn to a few representative applications. Let S be the boundary of a solid that contains the origin in its interior. Denote the outward unit normal of S by n. Suppose that a point charge at the origin exerts a gravitational field � EXAM P L E 4
k , (x,y,z)-=/= (0,0,0) F(x,y,z)= -3 r r ll ll where k is a proportionality constant, and r is the position vector xi + yj + zk of the point (x,y,z). Show that
ff F·ndS= 4nk.
(13.8.2)
s
Solution Let us first calculate J f8F n dS when S is the sphere Ea of radius a centered at the origin, oriented with outward pointing unit normal. (The symbol E is often used to denote a sphere, particularly when it cannot be confused with its familiar role as summation symbol.) At any point (x,y,z) of Ea, the outward unit normal is (1 /a)r. Therefore ·
F· n= k3r·a1r= k 11rll 2 = k a a2 a4 and
ff
F·ndS
=ff �dS= �
(surface area of sphere of radius
a) (13.8.3)
Equation (13.8.3) is a special case of equation (13.8.2). We will use this special case, together with the Divergence Theorem, to derive equation (13.8.2) in general. Of course, we cannot apply the Divergence Theorem to F on U because F is not differentiable at the origin. However, F is differentiable on the solid Ua which is obtained by removing a small ball centered at the origin and having radius a from U. Let Sa denote the boundary of Ua with outward normal n. Because div(F) = 0 at each point other than the origin (by Example 4 of Section 13.4), we have
13.8 The Divergence Theorem
!!
F·ndS=
s.
ff! div(F)dV= ff!
1 1 35
(13.8.4)
OdV=O.
u.
u.
But Sa comprises S and the sphere -Ea of radius a centered at the origin-the minus sign signifies that the sphere is oriented with the inward pointing normal.
Taking these observations into account in equation (13.8.4), we obtain 0=
ff
F·ndS=
S0
!!
F·ndS+
S
ff
F·ndS=
-E.
ff
F·ndS-
S
ff
F·ndS=
ff
Ea
F·ndS-4"k,
S
from which equation (13.8.2) immediately follows. <11111 INSIGHT
Equation (13.8.2) is known as Gauss's Law. Because the force between
two charged particles has the same mathematical form as Newton's Law of Gravitation, Gauss's Law is also applicable to the theory of electrostatics.
There are several variants of the Divergence Theorem. In the next example, we examine one that is often useful in applications. � EXAMPLE 5 Suppose that U, S, and n are as in the statement of the Divergence Theorem. Suppose that f is a continuously differentiable scalar-valued function on an open set containing U and its boundary. Prove that
ff
f dS= n
s
ff!
VfdV
(13.8.5)
u
Solution First of all, observe that the indicated equation is a vector equation. The
)
integrations are performed componentwise. Let c be a constant vector. Define the vector field F= f(x,y,z ) c. Theorem le, Section 13.4, tells us that div (F =
) ) ffJ(vt)
f(x,y,z div ( c + ('\lf) c. But div (c) = 0 because c is div (F) = ('\lf) c. The Divergence Theorem tells us that
constant.
·
Therefore
·
u
Therefore
Ji! div(F)dV= ff (ff! -ff )
·cdV=
F·ndS=
u
s
fndS
VfdV
u
for
every
constant vector
Ji
(jc) ·ndS.
s
c=
0
s
c.
Because
the zero
vector
1s
the
only
vector
perpendicular to every constant vector, we conclude that
!!!
VfdV-
u
from which the required identity follows.
ff
fndS=O,
s
<11111
Next we use the Divergence Theorem to derive a famous law of physics,
Archimedes's Law of Buoyancy.
1138
Chapter 13
Vector Calculus
� EXAM PL E 6 Let U be a solid body with piecewise-smooth boundary S. The body is immersed in water having constant mass density µ. For convenience,
z-axis points down into the water. z 0 (see Figure 2). The pressure
we use a coordinate system so that the positive The surface of the water corresponds to the plane
p=
=
exerted by the water at depth z is µgz, where g is the gravitational acceleration (we arranged for the positive z-axis to be the downward direction so that this formula for pressure would have no minus sign in it). Use the Divergence Theorem
to show that the buoyant force exerted on the body in the direction -k has
magnitude equal to the weight of the ft.uid displaced by the body. This statement is known as Archimedes's Law of Buoyancy.
n
k�
Solution Let denote the outward unit normal to S. Let By Newton's Third Law,
z
A Figure 2
B=-
// pndS=-fffVpdV,
�1JJ
pressure function is down, which is in the
{(.:i,y,z) :a(.:i,y) < z < JJ(.:i,y), x2 + y2 < r2}
A Rgure3
p=
the second equaliy t resulting from equation (13.8.5). But, because µgz, we have '\Ip= µgk. This makes sense because the direction of greatest change for the we have chosen. If we replace respectively, then we8o
U=
(13.8.6)
u
s
z = {J(:x.,y)
B be the buoyant force.
p
and
k direction in the coordinate system that Vp in line (13.8.6) with µgz and µgk,
µgkdV =
u
-(g ff! ) u
µdV t..
Because of the minus sign, we see that the vector
B
is directed upward.
(Remember, k points downward in our coordinate system.) By definition of den
sity, the value of the triple integral
is
just the total mass of water displaced. The
quantity in the parentheses is therefore the weight of the displaced water. Proof of the
Divergence Theorem
Here we will prove the Divergence Theorem in the case that U is the z-simple solid between
the
graphs
of
a(x,y) {3(x,y) ={(x,y,z): a(x,y)
two functions and over a disk is, U Look at FigtUe 3. The boundary S of U consists of three pieces. The upper
V= {(x,y): x2 +y2
is
the
surface
z =,8(x,y),x2 +y2
the lower boundary is the boundary on the side is the cylindrical surface
z =a(x,y),x2 +y2
Let us write our continuously differentiable vector field F in component form:
=Mi+Nj + Rk. Our job is to prove equation (13.8.1), which we may write as
ff (Mi+Nj+Rk)·ndS= !ff (M%+Ny+Ri)dV, s
or
u
We will show that the third summand on the left equals the third summand on the right. The equalities of the first summands and of the second summands obtained by similar calculations.
can
be
13.8 The Divergence Theorem
11 37
Let's begin with the surface integral ff8Rk· ndS. It consists of an integration over the top, an integration over the bottom, and an integration over the side. For the top, n points upward and n(x,y,z) =
(1 + ,Bx(x,y,z)/+ ,6y(x,y,z)2 )112
(-,6x(x,y,z)i- ,6y(x,y,z)j + lk).
2 Notice that k · n = (1 + ,Bx(x,y, z)2 + ,6y(x,y, z)2 r11 as a result. Also, we have 2 dS- ( 1 + ,Bx(x,y,z)
+
2 ) 1/2 dA, ,6y(x,y,z)
and so k · ndS= dA. The upshot of these observations is that
ff Rk·ndS= ff R(x,y,,B(x,y))dA. JJtopofS }}
(13.8.8)
'D
For the bottom, the analysis is the same except that the normal points downward (and of course we use z=a(x,y) instead of z=,B(x,y)). The result is Rk·ndS=- ff R(x,y,a(x,y))dA. ff JJbottomofS }}
(13.8.9)
'D
For the side, n is horizontal. Therefore k·n = 0. We conclude that
ff Rk·ndS=O. JJside ofS Summing equations (13.8.8), (13.8.9), and (13.8.10), we obtain
JI
Rk·ndS=
JI (
)
R(x,y, ,B(x,y))- R(x,y,a(x,y)) dA.
(13.8.10)
(13.8.11)
'D
s
We turn our attention now to the last summand on the right side of equation (13.8.7), applying the Fundamental Theorem of Calculus to the integrand in the z-variable to obtain
Ji (f!(�;] h� l
= =
)
Rz(x,y,z)dz dA
R(x,y,z)
'D
z=/9(x,y) dA= z a:(x,y)
h� (
=
'D
)
R(x,y,,B(x,y)) - R(x,y,a(x,y)) dA.
Because the double integral that terminates this chain of equalities is the right side of equation (13.8.11), we have proved the desired equality. Q UIC K
Q UIZ
1. Suppose that n is the outward unit normal on the boundary S of a solid. The flux 2.
ff
of a vector field F across S is 8cpdS for what scalar-valued function on S? The Divergence Theorem states that the flux of a vector field F across a surface S that bounds a solid U is equal to III uudV for some scalar-valued function u on U. What is u?
11 38
Chapter 13
Vector Calculus
3.
Let n be the unit outward normal on S, the boundary of
{(x, y, z) : 0:::;; x2 + y2:::;; 9, 1:::;; z:::;; 3). F(x,y,z) = (x,2y,z2)
U
=
Calculate
If8F
·
ndS for
Answers
1. F
·
n
2. div(F)
3. 1267r
EXERCISES Problems for Practice In each of Exercises 1-20, calculate the flux of the given vector field
U
F(x,y,z) across the boundary S of the given solid
F(x,y,z)=(x2-2xz)i-(3-y2)j +z2k, U={(x,y,z):x2+y2
in space by calculating the surface integral directly. Then,
use the Divergence Theorem to verify your answer.
F(x,y,z)=5k, U= {(x,y,z):x2 +y2<9, -1
Further Theory and Practice and set R=vfx2+y2+z2, K3(xi+yj+zk). For O
2LLet
across
the
boundary
of
a2 ,;;x2+y2+z2,;; ,ez}.
the
solid
F(x,y,z)= F U={(x,y,z): the flux of
21 to show that the flux of the F across any sphere E,={(x,y,z): x2+y2+ z2=r} is independent of r. (This statement is
22. Use the result of Exercise vector
field
less precise than Gauss's Law.)
23. Use the results of Exercises
21
and
22,
Divergence Theorem, to see that if U is
together with the
any
solid with no
holes that contains the origin, then the flux of
F
across the
boundary of U is the same as the flux across any �,. (This statement is again less precise than Gauss's Law.)
24. Suppose that
F(x,y,z)=(x-yz)i+ (y-xy)j+(xz -z)k.
Let
U={(x,y,z):x2+4y2<8, - 2(x,y)}, for some continuously differentiable function
(x, y).
If
the flux of F through the side of this solid is known to be 9 then what is the total flux through the top and bottom?
25. Suppose that
F
is a twice continuously differentiable
vector field and that G=
curl(F). Prove that the total flux
of G across the boundary of a solid such as the one in the statement of the Divergence Theorem is
26. Let
F
be a constant vector field and
U
0. a solid in space
that is bounded by finitely many continuously differenti able surfaces. What is the total flux of boundary of U?
F
across the
U be the solid in space between the two spheres {(x,y,z):x2+y2+z2=1} and {(x,y,z):x2+y2+z2=4}.
27. Let
Apply the Divergence Theorem to evaluate the flux of the vector field
F=xi-y2j+xzk
across the boundary of U.
28. Suppose that F is a continuously differentiable vector field defined on all of space and assume that div(F) =0 at all points. Suppose that S is a surface in the upper
Summary of Key Topics half-space
{(x,y,z): 0
and that the boundary of S is
xy
a simple closed curve C in the
-
ff
plane. Let Ube the solid
that is bounded above by S and below by the xy-plane.
11 39
B·ndS=O.
s
Suppose that S' is a surface in Uthat, like S, has C for its
Also, because div(B) = 0, there is a vector field
boundary. Let S and S' be oriented by their upward
F
such
that B = curl(F). Combining all this information yields
normal vector fields. Prove that
the equation
ff
F·ndS=
S
29. Let
ff
JJ
F·ndS.
S'
s
F(x,y,z) = (1/3)(xi + yj+ zk),
Now we apply Stokes's Theorem to see that
and let Ube a solid in
i
space bounded by finitely many continuously differenti able surfaces. Prove that the flux of
F across the boundary
of Uequals the volume of U.
F·dr=
continuously differentiable surfaces. Let
u
be a twice
for some function
continuously differentiable function whose domain con
u is harmonic, that is, Uxx +Uyy + Uzz = 0. Prove ff 8'\lu ndS= 0. 3L With U , S, n, and u as in Exercise 30, prove that that
JJJ
by Theorem
le,
F= grad(u)
But then
Section
13.4.
=
0
Of course this is absurd.
Where is the error?
In each of Exercises 34-37, calculate the vector
Jffu '\lf dV
for the given solid U and scalar-valued function f. Then,
IVul2dV.
verify the value you obtained by calculating the left side of equation
u
32. Let Ube a solid in space. Assume that the boundary S of U is piecewise continuously differentiable. Let
u.
B = curl(grad(u))
that
·
(vu· n)udS=
curl(F)·ndS=O.
Thus Fis path independent. We conclude that
tains a neighborhood of U and its boundary S. Assume
JJ
ff s
30. Let U be a solid in space bounded by finitely many
s
curl(F)·ndS= 0.
F
be a
continuously differentiable vector field with domain that
contains both Uand S. At any smooth point of S, let n be
(13.8.5).
f(x,y,z) = 2y+z2,U = {(x,y,z): -1
the outward unit normal. Prove that
Calculator/Computer Exercises
!!
nXFdS=
s
!!!
'\lXFdV.
u
the boundary S of the given solid
33. We now present a bogus proof that all magnetic fields are 0. (This nugget is taken from G. Arfken, Am. J. Physics 27
(1959), p. 526.) The mistake is purely a mathematical one. Your job is to fill in the details of the argument and then determine where the error lies. One of Maxwell's laws says that if B is a magnetic field, then div(B) =0. (Accept this fact-it is not the problem!) The Divergence Theo rem then says that
In each of Exercises 38-41, use the Divergence Theorem to calculate the flux of the given vector field
U.
F(x,y,z)
across
F(x,y,z)=x(l +z)sin(y2)i, U={(x,y,z): lxl<2,lyl<2,lzl
Summary of Key Topics in Chapter 13 A vector field in the plane is a function F(x,y)=M(x,y)i+N(x,y)j of two real variables with vector values in the plane. A vector field in space has the form F(x, y, z)=M(x,y, z)i+N(x,y, z)j + R(x, y, z)k. We usually assume that the
11 40
Chapter 13
Vector Calculus coefficient functions are continuously differentiable. Vector fields are useful for describing force fields and the velocity of a flow. They also arise when the gradient operator is applied to a scalar-valued function.
Line Integrals (Section 13.2)
If F is a vector field and if the interval
[a,b], then
t1-+ r(t),a ::5 t ::5 b
parameterizes a directed curve C over
the line integral of F over
1
F
·
dr
=
lb
F
r is
(r(t)) r'(t)dt. ·
If F is a force field, this integral represents work performed in traversing the curve. The line integral does not depend on the choice of the parameterization (as long as the parameterization traverses C in its direction).
conservative
Conservative and
A vector field is
Path Independent
conservative, then there is an integration procedure for finding a function u such
Vector Fields (Section 13.3)
if it is the gradient of a scalar-valued function. If F is
that Vu= F. We then say that V = -u is a potential function for F. If F is a conservative force field, then a potential function can be used to define potential energy. A vector field F is conservative in a region R if and only if
C in R that begin at the same point and end at path-independent. A curve C parameterized by t1-+ r(t), t E [a,b] is said to be closed if r(a) = r(b ) .
for every two directed curves C and
the same point. A vector field with this property is called
When we write the path integral over a closed curve, we often use the special integral sign
i F·dr. Closed Vector Fields
A planar vector field F = Mi
+ Nj is closed if
(Section 13.3)
Conservative vector fields are closed. If the domain of a closed vector field is simply connected, then the vector field is conservative. On regions with holes, this last assertion can be false. In space, all these statements are still correct, with the definition of a closed vector field F
=
Mi + Nj + Rk being aM
8Y
_
-
aN ax '
aM
8z
(all three equations must be satisfied).
-
aR ax '
aN aR az ay '
Summary of Key Topics
Divergence
If F
=
Mi + Nj
is a planar vector field, then its divergence is defined to be
(Section 13.4) div(F)
=
For a spatial vector field F
=
8M 8x
=
(Section 13.4)
If
F(x,y)
=
M(x,y)i + N(x,y)j
+
+
8N . 8y
the divergence is
8N 8y
8R 8z .
+
is a planar vector field, then the curl of Fis defined
to be
curl(F)(x,y,z) If
F
=
Mi + Nj + Rk is curl(F)
The Del Notation
8M 8x
Mi + Nj + Rk,
div(F)
Curl
1141
=
=
(! N(x,y)- � M(x,y)) k.
a spatial vector field, then the curl of F is defined to be
( 8R8y
_
) ( 8R8x
8N i 8z
_
) ( 8N 8x
8M . + 8z J
_
)
8M k. 8y
The del operator is
(Section 13.4)
The nabla symbol V has already been used in Chapter 11 to give an alternative notation,
Vu, for the
gradient
grad(u),
of a scalar-valued function
notation, we may write div(F) as V F and curl(F) as V X F.
u.
Using the del
·
Operations with Div and Curl (Section 13.4)
Green's Theorem (Section 13.5)
A spatial vector field F
=
Mi + Nj + Rk
is closed if and only if
curl(F)
=
0. In
physics, such a vector field is called irrotational. A closely related fact is that div(curl(F))
=
0 for any twice continuously differentiable vector field F.
This result says that if
F(x,y)
=
M(x,y)i + N(x,y)j
is a continuously differentiable
vector field defined on a neighborhood of a region R and its piecewise continuously
differentiable boundary curve C, then
This result relates the flux of a flow across the boundary to the flow across the interior . The parameterization
r
of C should have the property that the vector
obtained by rotating r' counterclockwise by 90
°
points into R. If
n
is the unit
11 42
Chapter 13
Vector Calculus
outward
pointing normal vector field on the boundary C, then Green's Theorem
may be rewritten as
L F nds =ff div(F)dA. ·
'R.
Surface Area and Surface Integrals
If R is a region in the plane and if f(x,y) is a continuously differentiable function on R, then the element of surface area on S {(x,y,f(x,y)): (x,y) ER} is =
(Section 13.6)
dS= J1 + fx(x,y)2 + fy(x,y)2 dA where dA is the element of area in the xy-plane . Thus the area of S is given by
A=
ff J1
2 2 + fx(x,y) + fy(x,y) dA .
'R.
If cp(x,y,z) is a continuous function defined on the surface S, then the surface integral of
cp over S is denoted by II 8cpdS and defined by
ff cp(x,y,f(x,y))Vl
2 2 + fx(x,y) + fy (x , y) dA .
'R.
If a surface S in space is given parametrically by the vector-valued function
(u,v)
f-7
r(u,v),(u,v) ER,
then the element of surface area is
llruXrv II dA where
dA is the element of area in the uv-plane . Thus the area of the surface is
If
cp is a continuous function defined on the surface,
then the integral of
cp over the
surface is
ff cpdS= ff cp(r(u,v))ll(ruXrv)(u,v) lldudv.
s
Stokes's Theorem (Section 13.7)
'R.
Let Sbe a surface in space with boundary curve C. We say that Sis oriented if it is possible to make a continuous assignment of a unit normal vector
n(P) to each P of S. We say that the boundary C of Sis oriented consistently with Sif it is parameterized by r so that, at each boundary point, nXr' points into the region S. If Sand C are consistently oriented, and if F is a continuously differentiable vector point
field defined on Sand its boundary, then Stokes's Theorem states that
i F dr =ff curl(F) ndS. ·
·
s
Review Exercises
The Divergence Theorem (Section 13.8)
11 43
If U is a solid in space bounded by finitely many continuously differentiable sur
faces, if the boundary of U is S, if n is the unit outward pointing normal vector field on S, and if F is a continuously differentiable vector field on
open set that
an
contains U and its boundary S, then the Divergence Theorem says that
!!
F ndS= ·
s
!ff
div ( F)dV.
u
Review Exercises for Chapter 1 3 In Exercises 1-3, sketch the given vector field in the plane.
13. F(x,y,z) =zi+lj +xk 14. F(x,y,z) cos(y)i-x sin(y)j+lk =
1. F(x,y) xi-yj 2. F(x,y) 4yi+3xj 3. F(x,y) =yi - xj
15. F(x,y,z)
=
=
y ln(z)i + x ln(z)j +
=
In each of Exercises 4-6, give an explicit formula for the spatial vector field that is described in words. 4. Particles move in space away from the origin with speed
equal to the cube of the distance to the origin. Write the velocity vector field. 5. Particles move in a wind tunnel parallel to a line running down the center of the tunnel. Think of that line as the x axis. The velocity of a particle is equal to the square of its distance to the line plus its height from the floor (which is at z = 5 ) The positive direction on the x-axis is opposite to the direction in which the wind blows. Write the velocity vector field. 6. Particles move outward inside the cone z2 =x2+y2 (inside both nappes!) at a speed equal to the square of the distance of the point from the origin. Write the velocity vector field. -
.
In each of Exercises 7-9, write the gradient vector field that is induced by the given scalar-valued function u. 7. u(x,y)=xy2-yx3+x 8. u(x,y) =In(x2-y3) 9. u(x,y)= sin (x3y)
In each of Exercises 10-12, the given vector field F(x,y)=M(x,y)i+N(x,y)j is conservative. Find a potential function u(x, y) so that 'Vu(x,y) = F(x,y). 10. F(x,y) =sin(y)i+x cos(y)j lL F(x,y) =xyi +
�2 j
12. F(x,y) =ln(y)i+�j
y
In each of Exercises 13-15, the vector field F(x,y,z) = M(x,y)i+N(x,y)j +R(x,y)k is conservative. Find a potential function u(x,y,z) such that 'Vu(x,y,z)=F(x,y,z).
xy z
k
In each of Exercises 16-18, find the integral curves tr-> r(t) of the vector field F(xy)=M(x,y)i+N(x,y)j. 16. F(t) =i - j 17. F(t) =2xi - 3j 18. F(t) =3zi - 3yj In each of Exercises 19-21, calculate .f F dr for the planar vector field F and the parametric curve r. ·
19. F(x,y)=y2i -xyj, r(t)= sin(t)i-cos(t)j,0:5t:57r/2 20. F(x,y) =x113i-y114j,r(t)=t3i+t2j,1:5t:52 2L F(x,y) =sin(x)i+cos(y)j,r(t)=ti+t2j,7r/2:5t:57r In each of Exercises 22-24, calculate .fcF dr for the spa tial vector field F and the parametric curve r. ·
22. F(x,y,z)=xzi-y2xj +x2yk, r(t)=t2i-tk+t3k, 0,;; t,;; 1 23. F(x,y,z) = sin(x)i +zj-cos(z)k,r(t) =ti-tj + t2k,
0:5t:57r
24. F(x,y,z) = (y/z)i-(x/y)j+yzk,r(t) = (1/t)i +t2j-tk,
1 :5t:52
In each of Exercises 25-27, calculate the work performed by a force F in moving a particle along the parametrized curve r. 25. F(x,y) =y2 i-x2j,r(t)= sin(t)i-cos(t)j,0,;; t,;; 7r 26. F(x,y,z)= (x-y + z)i+ (y + 2x-z)j+(z + x)k,r(t) =
cos(2t)i-sin(2t)j +2tk,7r/2 :5t:57r
27. F(x, y)=xy2i - yx2j, r(t) =t113i - t112j,1:5t:53 In each of Exercises 28-30, calculate the indicated line integral, where C is the triangle with vertices (1, 2), (3,S), and (4, 1) and with clockwise orientation. 28. J;:xy2dx 29. J;:ydx-xdy
Chapter 13
11 44
30.
Vector Calculus
fc(3y-x)dx-(x+y) dy
In each of Exercises 49-51, calculate the divergence of the
In each of Exercises 31-33, calculate the line integral
fc f ds
for the given function
parametrized by 31. 32. 33.
r (t).
f
and the parametric curve C
51.
35. 36.
In each of Exercises 52-54, calculate the curl of the vector
Q={(x, y): -l
If it is closed, then find a potential function.
In each of Exercises 55-57, calculate the curl of the vector
Q={(x, y, z): -1
If it is closed, then find a potential
55. 56. 57.
function. 37. 38. 39.
F(x, y) P1 for a
Po
and the terminal point
58. 59. is given.
F is closed in a simply connected region containing P0 and P1. Then calculate the line integral .fF dr for some (hence any) curve C that connects P0 to P1. ·
42.
F(x, y)=ln(xy)i+ �j, Po=(1, l), P1=(2, 2) y F(x, y, z)=y2zi+2xyzj+xy2k, Po=(2, 1, 3), P1=(4, 2, 5) F(x, y, z)=-y2zsin(xy2)i -2 xyz sin(xy2)j+cos(xy2)k, Po=(0, 0, 0), P1=(2, 1, 3) In each of Exercises 43-45, a conservative vector field
and three points
P0, P1, P2
60.
directed
curve C are specified. But the curve is not. Check that
41.
44. 45.
are given. Verify that
F
F(x, y)=y2i+2xyj, Po=(1, 1), P1=(2, 2), P2=(3, 3) F(x, y, z)=2xyz3i+x2z3j +3x2yz2k, Po=(0, 0, 1), P1=(0, 1, 0), P2=(1, 0, 0) F(x, y)=eYi+xeYj, Po=(2, 3), P1=(4, 6), P2=(1, 5) In each of Exercises 46-48, calculate the divergence of the
planar vector field 46. 47. 48.
F
is closed.
F(x, y, z)=xyi-yzj+xyzk F(x, y, z)=eYzi+xzeYzj+xyeYzk F(x, y, z)=sin(xyz)i+cos(yz)j+sin(xy)k
F.
F(x, y)=cos(xy)i-sin(x, y)j F(x, y)=ln(.x2-y)i+ln(y2-x)j F(x, y)=(x2-y3)i+(y2-x)j
u
is
u(x, y)=x3-xy2 u(x, y, z)=y2 sin(z2-x2) u(x, y)=xyln(y2-x2) In each of Exercises 61-63, calculate both sides of the
fomula in Green's Theorem and verify that they are equal. 61. 62. 63.
F(x, y)=x2i-3yj, r(t)=cos(t)i-sin(t)j, -7r:5t:57r F(x, y)=xyi-yx3j, r(t)=cos(2t)i-sin(2t)j, 7r:5 t:527r F(x, y)=.x2yi+y3j, r(t)= sin(8t)i-cos(8t)j, 0:5t:57r/4 In each of Exercises 64-66, determine the region that is
bounded by the oriented curve C. Then use Green's Theorem to calculate 64. 65. 66.
43.
State whether
In each of Exercises 58-60, a scalar-valued function
In each of Exercises 40-42, a vector field
40.
F.
given. Calculate div(grad(u)) .
F(x, y, z)=yi- zj+xk F(x, y, z)=(x2-y2)i+(y2-z2)j+(x2-z2)k F(x, y, z)=(yz-4)i+(xz+6)j+(xy+2)k
The initial point
F.
F(x, y, z)=z2yzi-x2zj+y2xk F(x, y, z)=ln(x-y)i+ey+zj-xyzk z x F(x, y, z)= _ _i+ _Y_j- _ _k x+y x+z y+z
field
In each of Exercises 37-39, determine whether the given
-1
52. 53. 54.
F(x, y)=(y- l)i+xj F(x, y)=y2i-.x2j F(x, y)=x cos(y)i-(x2/2) cos(y)j
vector field is closed on the cube
F.
F(x, y, z)=eX+2zi+e z-Zyj-eX+Yk F(x, y, z)=(.x2-y3)i+(y4 -xz)j+(z3-x)k F(x, y, z)=sin(xy)i-cos(yz)j +tan(xyz)k
field
In each of Exercises 34-36, determine whether the given vector field is closed on the region
34.
49. 50.
f(x, y)=x2-y3, r(t)=sin(t)i+ cos(t)j, 0:5t:57r/4 f(x, y)=xe2Y, r(t)=e-1i+tj,O:5t:5 l f(x, y)=3x/y, r(t)=t2i-tj, 2:5t:54
-2
spatial vector field
.fF dr. ·
F(x, y)=xyi-x2j, r(t)=sin(t)i-cos(t)j, 0:5t:527r F(x, y)=(x-3y)i+(y-4x)j, r(t)=cos(2t)i+sin(2t)j, 7r:5t:527r F(x, y)=y2i-x2j, r(t)=-sin(t)i-cos(t)j, -7r/2:5t:537r/2 In each of Exercises 67-69, the given region
R
has pie
cewise smooth boundary. Apply Green's Theorem to evalu ate the line integral of
F
around the positively oriented
boundary curve. 67. 68. 69.
R={(x, y): -1:5x:51, -2:5y:52}, F(x, y)=y2xi-x3y2j R is the triangle with vertices (0, 0), (1, 0), (0, 1), and F(x, y)=sin(y)i-cos(x)j R is the square with vertices (2, 0), (0, 2), (0, 0), (2, 2), and F(x, y)=eX-Yi+eX+Yj
Review Exercises
In each of Exercises 70-72, use formula
(15.36)
to calcu
late the area of the given region. 70. The triangle with vertices
(2,0), (0,2), (1,1) x2+y2=8 and
lying above
y=x2
and the line
71. The region inside the circle the line
y =x
89. 90.
72. The region bounded by the parabola
y=x+l
73. 74 . 75.
F
R, a vector F on R, and a scalar-valued function f on R are given. Let S denote the graph off. Calculate the surface integral of curl(F) n over S in two different ways:
across the boundary of the region
R.
F(x,y)=y2xi-x2yj,R={(x,y) : lxl<1, IYI <2} F(x,y)=yx312i-xy312j,R={(x,y): 0
field
·
•
Perform the integration directly as a surface integral.
•
Use Stokes's Theorem to reduce the problem to a line integral.
In each of Exercises 76-78, calculate the surface area of the graph of the function
f (x, y)
over the region
R
in the
xy
91.
plane. 76. 77. 78.
x=u+v,y=u-v,z=3v-u,O�u�3, l�v�2, rp(x,y,z)=x-y+z x=v,y=u,z=2u-v, l�u�4,2�v�3, rp(x,y,z)=y+2z x=3u-4v,y=u+v,z=u-v,-2�u� -1, -2�v�1, rp(x,y,z)=4x-3z In each of Exercises 91-93, a planar region
In each of Exercises 73-75, compute the flux of the vector field
88.
f(x,y)=x2+y2+4,R={(x,y):x2+y2< 9} f(x,y)=3x-6y,R={(x,y):-2
92. 93.
In each of Exercises 79-81, calculate the area of the given
F(x,y,z)=y2i-zyj +yzk,f(x,y)=x-3y, R={(x,y):lxl
If8curl(F) ndS S.
surface.
evaluate
79. The cone with base of radius 3 and height
oriented surface
the base itself) 80. The portion of the surface cylinder
x2+y2=1
81. The part of the surface paraboloid
z=x2+y2
z=xy
5 (not including
94 . that lies inside the
z=2x+2y+2
95. that lies inside the
82. 83. 84.
R.
rp(x,y,z) f over the
In each of Exercises 85-87, write the integral that gives
86.
96.
and the
F(x,y,z)=yxk,S={(x,y,z):y2+z2=1,l�x�2}, oriented by P1-+ n(P)withn(O,0,1)=k
vector field
In each of Exercises 88-90, integrate the function
•
calculating the surface integral directly using the Divergence Theorem 97. 98.
100.
over
across the boundary
•
101.
rp
F(x, y, z)
S of the given solid
U in space by the following two methods:
99.
x=u2-v2,y=u2+v2,z=uv,0�u�2,1�v�3 x=eu-v,y=ev-3u,z=v,0�U�1,2� V�4 x=sin(uv),y=cos(v),z=u+v,0�u�?r/2, 7f/3�v�27f/3
the parametrically defined surface.
F
F(x,y,z)=3zi-xyj+yk,S={(x,y,z) : x2+y2=4, 2�z�4}, oriented by P1-+ n(P)withn(2,0,0)=i F(x,y,z)=cos(y)i-zj+xk, S is the interior of the tri angle with vertices (1, 0,0), (0,1,0), (1,1, 0) oriented by
attempt to evaluate the integral.
87.
for the vector field
In each of Exercises 97-101, calculate the flux of the given
rp(x,y,z)=cos(y)+cos(z),f(x,y)=4x-3y, R={(x,y): 0
the surface area of the parametrically defined surface. Do not
85.
·
the upward-pointing unit normal vector field.
In each of Exercises 82-84, integrate the function over the surface which is the graph of the function region
11 45
F(x,y,z)=4zk,U={(x,y,z):x2+y2<4,2
Chapter
has been devoted to multivariable analo
13
gues-Green's Theorem, Stokes's Theorem, and the
solar system. Not knowing how to reconcile these observations with his theory of gravitation, Newton
Divergence Theorem-of the Fundamental Theorem
believed
of Calculus. In fact, each of these theorems is a special
necessary to keep the planets in their place. Laplace was
case of a single, very general formula that can be written in a remarkably concise form:
that
divine
intervention
was
occasionally
able to show that the planets could not have "secular "
(or constant direction) accelerations due to their mutual attractions. Laplace deduced that the inequality of
r df= r f. laM jM This
(1).
1752,
a2v
ay2
+
a2v
az2
(2)
0.
=
force
(2).
that a mass
(2) in
He noticed that the
M at
l
l lKJll
3
PoP
point Po exerts on a unit mass at point
Lagrange laid the foundation for a theory of the potential.
1782,
Laplace, then only twenty-four years old,
used the theory of potential functions to solve a problem about planetary motion that had baffled Newton, Euler, and Lagrange: the "great inequality of Jupiter and Saturn." According to all observations available at that the average angular velocity of Jupiter was
increasing, whereas that of Saturn was decreasing. The implication
was
that
Jupiter
would
eventually
be
pulled into the sun and that Saturn would fly off from the
1146
point of view; he sought subtleties everywhere, had only nitely small into administration." Regardless, Laplace's reputation as a scientist continued to grow with the publication of his
Traite de Mecanique Celeste,
appeared in five huge volumes from
1798
to
which
1825.
In
addition to the scientific accolades that Laplace received,
George Green
P can be expressed (using modem vector notation) as F(P) = -'\i'V(P) where V= -MG l l KJll -1 . Lagrange also observed that the potential function V satisfies equation (2). In a series of papers that followed,
time,
new cabinet (Minister of the Interior) but dismissed him
he was made a marquis.
F(P) =-MG
In
Revolution, Napoleon appointed Laplace to France's
doubtful ideas, and finally carried the spirit of the infi
provided him the occasion
for a deeper study of equation
Laplace ranks as one of the most important math
after only six weeks. Napoleon is alleged to have com
his own research on fluid dynamics. Lagrange's work on
1773
became so well
mented that "Laplace saw no question from its true
Nine years later, Lagrange encountered equation celestial mechanics in
(2)
Laplace's equation.
the young Napoleon Bonaparte. After the French
Euler introduced the partial differential equation
+
known that it was named
taire, Laplace had the opportunity to be the examiner of
while investigating a question in fluid flow,
D.V= a2v ax2
long period. Observation has born out the truth of his
ematicians of his era. As a professor at the Ecole Mili
The Beginning of Potential Theory In
Jupiter and Saturn was not secular but periodic of very deductions. His use of equation
will discuss the back
Genesis and Development
ground and history of equation
(1)
George Green
(1793-1841) was the self-taught son of a
baker and miller. He used the top floor of his father's mill as his study. Inspired by Laplace's mathematical success in celestial mechanics, Green attempted to establish a mathematical foundation for electricity and magnetism. The results of his research appeared in a
paper that was published by subscription in
1828.
Green announced his viewpoint with these words:
... by confining the attention solely on that peculiar function on whose differentials they all depend, I was induced to try whether it would be possible to discover any general relations, existing between this function and the quantities of electricity in the bodies producing it. Green coined the name
potential function
for the
"peculiar function." Because there were only fifty-two
Genesis & Development
11 4 7
subscribers to Green's publication, his work had no
first appeared in a postscript to an 1850 letter that
influence at first. Five years after its publication, in
Thomson sent to Stokes. Stokes liked Thomson's
1833, Green became an undergraduate at Cambridge at
equation enough to include it on several examinations
thirty-nine years of age. He obtained a B.A. degree and
at Cambridge, the first of which was given in 1854.
stayed on at Cambridge for a few years until poor
Thomson's theorem eventually came to be associated
health forced his return to Green's Mill in 1840. Green
with, and named after, Stokes.
died in obscurity one year later. It was not until the
Although Thomson published many mathematical
republication of his paper in 1846 that Green's research
papers, he is better known as a physicist, thanks to his
received
fundamental contributions to thermodynamics, elec
the
attention
it
deserved.
His
work
re
established the significance of the potential function
tricity, and magnetism. Thomson also made his mark as
when progress in the theory created by Lagrange and
a teacher. At the University of Glasgow, he established
Laplace was in danger of coming to a halt. Green also
the first teaching laboratory in Great Britain. His
discovered formulas that are equivalent to what we now
Treatise on Natural Philosophy, written with P. G. Tait,
call Green's Theorem. Additionally, he introduced a
first appeared in 1867 and remained the leading physics
construct, now called the Green's function, that is fun
textbook for decades.
damental to potential theory.
In the mid 1850s, Thomson served on the board of
Today, Green's Mill and Centre is a tourist attrac
directors of a consortium of British industrialists who
tion. What was once a windmill is now very much part
were planning to lay an underwater telegraph cable
of the 20th century. Included in the exhibits is a
from
weather satellite receiving system that displays data live
objections,
Ireland
to
Newfoundland.
the first two
Over
attempts
Thomson's
were based on
from orbiting and geostationary satellites. Among the
unsound
memorabilia that may be purchased there is a pencil
attempt, which followed Thomson's design, proved
with Green's Theorem inscribed on the side.
successful. Thomson's role in saving an enormous
electrical
practices
and failed. The third
capital investment made him a hero in British financial circles. A knighthood had long been the appropriate
William Thomson and George Stokes
reward for such a substantial contribution, and Thom
Only a few weeks before Green's death, William
son became Sir William. He later acquired the title
Thomson (1824-1907) entered Cambridge. Thomson's
Baron Kelvin of Largs, and you will usually find him
mathematical interests had already been formed by his
referred to as Lord Kelvin in physics texts.
reading of Fourier's Theorie Analytique de la Chaleur
As the twentieth century approached, the aging
had
Lord Kelvin rejected many of the scientific theories
acquired at the age of fifteen. As an undergraduate,
that were beginning to take hold. In biology, he was a
and
Laplace's
Mecanique
Celeste,
which
he
Thomson encountered a reference to Green's work on
staunch opponent of Darwin, and, in physics, he
potential theory but was not able to locate a copy until
rejected the atomic theory of matter. His scientific
after his graduation in 1845. Thomson recognized at
conservatism manifested itself in mathematics as well.
once the importance of this work and caused it to be
Although we now state Kelvin's best known mathe
republished the next year. He furthered the application
matical theorem in the language of vector analysis,
of potential theory to electromagnetic theory and
Kelvin's own opinion was that
developed his own integral formulas. Indeed, Thomson discovered Stokes's Theorem. George Gabriel Stokes (1819-1903) was Lucasian professor at Cambridge and a leading expert in fluid
"vector" is a useless survival, or offshoot, from quaternions, and has never been of the slightest use to any creature. The Divergence Theorem was first published by
dynamics. Stokes's Theorem, in the form
Mikhail Vasilievich Ostrogradsky (1801-1862) in 1831.
j(
adx + (3dy + 1dz)
+m
=
Ostrogradsky's ambition had been to become an officer
JJ { (?z �;)
±
£
-
(� �;) (�; �) } -
+n
-
in the military. Instead, because the life of an officer was too expensive for someone of modest means, he ds,
entered university in preparation for a career in civil service.
After
graduating
from
the
University
of
Kharkov, Ostrogradsky traveled to Paris, where he
Chapter 13
11 48
Vector Calculus
learned Fourier's theory of heat conduction. Research
Differental Form
Dimension Description ofM ofM
w
on the partial differential equation that describes heat conduction in three dimensions led Ostrogradsky to the Divergence Theorem. A few years later, in 1839, Gauss discovered the form of the Divergence Theorem that is named for him.
fidx + fzdy+ f3dz gi dx/\dy+ gz dy/\dz+ g3 dz/\dx hdx/\dy/\dz Here
The Fundamental Theorem of Calculus: An Advanced Point of View
f1,fz,f3,gi,g2,g3,
tions of
1
curve
2
surface
3
solid
and hare all real valued func
x, y, and z. The basic differentials dx, dy, and dz
are provided with an associative algebra structure in
Long after the separate discoveries of Green's Theo
which multiplication is denoted by the symbol"/\"and
rem, Stokes's Theorem, and the Divergence Theorem, mulas are all special cases of one general theorem, a
the following rules apply: dx /\dx = dy/\dy = dz/\dz= 0, dy/\dx = -dx /\dy, dz/\dy= -dy/\dz, and dx /\dz= -dz/\dx . The integration of F n on an
theorem that also contains the Fundamental Theorem
oriented surface S with normal n is written as
mathematicians came to understand that these for
·
of Calculus of one variable. To understand what is involved, we must first recast the Fundamental Theo rem of Calculus into an appropriate form. The equation
=
in this notation instead of
f(b) - f(a) ,
[ b].
f
and the
There are also two operations at
on the function. We have already used the notation
f'(x)dx . The other operation
df
with orientation induced from M. When
the interval is given its standard left-to-right orienta tion,
aM
"-"at
apply d to any function/to obtain the total differential df = (af;ax)dx + (af/8y)dy+ (af;az)dz. The differ ential operator d is applied to a differential form by applying d to each coefficient and then using the cal culus of differentials just described. For example
is a geometric one: To
the oriented interval M, we associate its boundary
8M = {a, b}
·
coefficients of the basic differentials are functions. We
work on them. One is the operation d of differentiation for
If5F ndS.
Notice that, in a general differential form, the
contains two basic objects, the function = a,
F1dy /\dz+ F dz/\dx + F3dx /\dy 2
s
lb f'(x)dx interval M
JJ
is oriented by assigning a "+ " at
b
and a
a.
d(xy2zdx /\dy+ sin(x)dy /\dz) = (y2zdx + 2xyzdy+ xy2dz) /\dx /\dy + cos(x)dx /\dy/\dz= (xy2 + cos(x))dx /\dy/\dz.
Recall that an integral over a region is the limit of Riemann sums associated with decompositions of the
Using these rules of differentiation, we can express
region. When that region contains only two points,
the Fundamental Theorem of Calculus, Green's Theo
there is no need to take limits. In view of the preceding
rem, Stokes's Theorem, and the Divergence Theorem
discussion, we may write
f(b) - f(a)
as
faMf
and
and many other related integral equations-in the form
reformulate the Fundamental Theorem of Calculus as
f df= f f. JM JaM A theory of integration has been worked out over very general oriented regions. The typical "differ entials"
w
that are integrated in three-dimensional
space vary in form according to the dimension of the geometric object M over which they are being inte grated. They may be written in the following forms:
f w= f dw JaM JM by specifying a differential form
w that is appropriate for
the theorem. A formal verification of the Divergence Theorem,
JJJ M
div(F)dxdydz =
JJ !JM
F
·
n dS,
Genesis & Development
runs as follows: Because F n dS F1 dy /\ dz+ F2dz /\ dx+ F3dx /\ dy, the calculus of differentials gives ·
d(F·ndS)
=
=
( fu 8F1 dx+ 8F1 + 8F1 ) dy azdz /\dy/\dz ay ( + 8F2 dx+ 8F2 dy+ 8F2 dz) /\dz/\ dx Bx By 8z 0F3 ) + ( 8F3 dx+ 8F3 dy+ dz /\dx/\dy ox oy oz ( 0F1 + 8F2+ 8F3 ) dxd dz oy ox oz
=
=
div(F).
Y
11 49
A complete understanding of the calculus of dif
ferential forms requires concepts from linear algebra, mathematical
analysis,
and
differential
geometry.
These subjects, like the theory of differential equations, make natural follow-ups to your study of calculus.
This page intentionally left blank
Table of Integrals
Integrals Involving L 2. 3. 4.
JI'
J�dx=ln(lxl)+C J dx= p� 1 + J_!_dx=-! x2 x +c J}xdx = 2y'x+ xf+l
xf
c,
(p =I -1)
c
Integrals Involving a +bx
5. j a Jbxdx= � ln(la+ bxl)+C (p =I - 1) j (a+bx'fdx= (pJl)b (a+ bxr1 + 7. j a:bxdx i2 ((a+bx) - aln(la+bxl))+C x 2dx 1 (m(la+bxl)+ - a ) +C j (a+bx) b2 (a+bx) j (a+xbx'f dx= _!_b1 ( (p -l)(aa+ bxr1 - (p -2)(a1+bxr2 ) + C' (p =I l, 2) 10. j a:bxdx= i3 G(a+bx)2 -2a(a+bx)+a21n(la+bxl)) +C j (a+bx) x2 dx= 1 ((a+ bx) -2aln(la+ bxl) -__±_ ) +C a+ bx 2 b3 a2 j (a+x2bx)3dx _!_b3 (m(la+bxl)+ � a+bx - 2(a+ bx)2 ) +C 2a 1 2 j (a+x2bx'f dx= _!_b3 (- (p - l)(aa+ bxr1 + (p - 2)(a+bxr2 (p - 3)(a+bxr3 ) +C' j a :bxdx i4 G(a+bx)3 - 3; (a+bx)2+3a2(a+bx) -a31n(la+bxl)) +C c,
6.
=
8.
=
9.
lL
12
=
•
13.
14.
(p=Jl,2,3)
=
T-1
T-2
1 5.
16.
17.
18.
19. 20.
21
•
Table of Integrals
) j (a+.x3bx)2dx= b\ (-21 (a+bx)2-3a(a+bx)+3a2ln(la+bxl) + ____t_ a+bx +C a3 )+C j (a+.x3bx)3dx= b\ ((a+bx)-3aln(la+bxl) - a+3a2bx + 2(a+bx) 2 3a2 + a3 )+c j (a+.x3bx)4dx= _!_b4 (in(la+bxl) + � a+bx 2(a+bx)2 3(a+bx)3 j (a+.x3bx)P dx= _!_b4 ( (p-l)(aa+3 bx)p-l - (p-2)(a3a2+bx)P-2
+ (p 3)(a3a+bx)P 3-(p 4)(a+1 bx)P 4)+c, (p#l,2,3,4) 1 dx=-!1n(la+bxl)+C x(a +bx) a x J ) j x(a+1bx)2dx= _!_a2 (in(la+xbxl)+ ____E!._ a+bx +C 1 + _1_- � 1n(la+bxl))+C j x2(a+1 bx)2dx=-b( a2(a+bx) x a2bx a3 _
Integrals Involving 22. 23 . 2 4. 2 5.
26 •
27.
28 •
29 •
30 •
x
+a and x + b, a =f=. b
Jxx+a +bdx=x+(a-b)ln(x+b)+C 1 dx= -1--in(lx+bl)+C j (x+a)(x+b) a b x+a j (x+a�x+b)dx= a� b (aln(lx+al) -bln(lx+bl))+C 1 lx+a 1 1 j (x+a)(x+b) 2dx= -(a-b)(x+b) + (a-b)2 ln( x+bl)+C b x a 1 lx+a1 J (x+a)(x+b)2dx= (a-b)(x+b) - (a-b)2 n( x+b )+C x2 dx= - b2 + a2 ln(lx+al)+ b2-2ab ln(lx+bl)+C (a-b)(x+b) (a-b)2 J (x+a)(x+b)2 (a-b)2 2 1 x+a 1 ( -1- _1_) 1 J (x+a)2(x+b)2dx=-(a-b)2 x+a + x+b + (a-b)3 n(lx+b1)+C x = 1 (-a +-b )- (a+b) 1n(lx+a dx J (x+a)2(x+b)2 (a-b)2 x+a x+b (a-b)3 x+b1)+C x2 =- 1 ( L + £_ ) + (2ab) 1n(lx+a1)+C dx (a-b)2 x+a x+b (a-b)3 x+b J (x+a)2(x+b)2 2
Integrals Involving a + x2 31. 32.
j aZ � xzdx= � ( �) +C j (a2+1 x2)2dx= 2a2(a:+x2) + 2a13 arctan
( �a ) +C
arctan
Integrals Involving a2 - x2
j(a2+1x2)3 dx= 4a2(a2x+x2)2 + 8a4(a2x3 +x2) + _l_ 8a5 arctan ( :!a ) +C x 1 Sx Sx 34. j (a2+x2)4 dx= 6a2(a2+x2)3 + 24a4(a2+x2)2 + 16a6(a2+x2) + -516a7 arctan ( :!a ) +C 35. j a2:x2 dx=x- a arctan ( �) +C x2 x 1 36 . j (a2+x2)2dx= 2(a2+x2) + 2a arctan ( :!a ) +C x2 x x 37. j ( :!a ) +C 8a3 (a2+x2)3 dx= 4(a2+x2)2 + 8a2(a2+x2) + _l_arctan x2 dx= - x + x x 1 -38. j (a2+x2)4 6(a2+x2)3 24a2(a2+x2)2 + 16a4(a2+x2) + 16a5 arctan ( :!a ) +C 1 39. j x(a2+x2) dx= 2_l_ a2 In(_£_ a2+x2 )+C 1 dx= 1 + _l_in(_£_ 4 0. j x(a2+x2)2 2a2(a2+x2) 2a4 a2+x2 )+c 41. Jx2(a2+1 x2)dx= - a-2x1- a1 arctan (xa )+C 1 x 3 (xa )+C 42. Jx2(a2+1 x2)2 dx= - a4x - 2a4(a2+x2) - 2-arctan as 33.
3
-
-
Integrals Involving a2 - x2
1 dx 1-1n(lx+al)+C ja2-x 2 2a x - a ja2 -=-x2 dx= � tanh ( �) +C 1 x a+x 45. j 4a3 in(la -xl)+C (a2 -x2)2 dx= 2a2(a2 -x2) + _l_ 1 a+x x x3 3 -46. j (a2 -x2)3 dx= 4a2(a2 -x2)2 + 8a4(a2 -x2) + 61 a5 ln(la -xl)+C 1 x Sx Sx S la+xl __ 47• j (a2-x2)4 dx= 6a2(a2-x2)3 + 24a4(a2-x2)2 + 16a6(a2-x2) + 23 a7 in( a-x )+C 48. j � dx= -x+ �ln(l:�:I)+C a2 x2 x2 dx= x -1-in(la+xl)+C 49 j (a2 -x2)2 2(a2 -x2) 4a a -x 2 5 0. J (a2-x x2)3 dx= 4(a2-x x2)2 - 8a2(ax2 -x2) __161_a3 1n( 1 a+x a -x 1 )+c x2 dx= x x x __l_ la+x 51. j (a2-x2)4 6(a2-x2)3 24a2(a2-x2)2 - 61 a4(a2 -x2) 23 a5 ln( a -xi)+C x2 ) 1 dx= -1 ln (I -52. 2) 2a2 a2 -x2 +c Jx(a2 -x 43.
44.
=
-l
•
I
T-3
T-4 53
•
54
·
55
•
Table of Integrals
x2 I) +C j x(a2 -1 x2)2dx=2a2(a21 - x2) + _!_1n (I 4 a2 - x2 2a a+x + _!_1n l 1 +C j x2(a2 - x2)dx= __!__ a2x 2a3 ( a - x ) x a+x 1) +C + j x2(a2 1- x2)2dx= __!__ + _l_1n ( l a4x 2a4(a2 - x2) 4a5 a-x 1
_
_
Integrals Involving 56
·
57. 58. 59. 60 . 61.
a4
± x4
x2+ax./2+a2 ax./2 1 1 ( n ( j a4+1 x4dx= + -) +C arctan ) l a2 - x2 4a3 ./2 x2 - ./2+a2 2a3 ./2 j a4 : x4dx=2�2 arctan ( �: ) +C + a2 ) +C ) + 2a./2 arctan (a2ax./2 j a4+x2 x4dx= - 4a./2 (x2x2+- ax./2 - x2 ax./2+a2 j a4 � x4dx= 4�31n(l :��I ) + �3arctan(�) +C 2 j a4 � x4dx= 4�2 ln (I:: ��I) +C x2 a+x i x dx= ) ( l n - arctan ( ) + l 4 4 a x 4a a x 2a J ax
1
_ _
1
ln
_ _
1
1
a
c
Integrals Involving .jX and a2 ± b2x 62. 6 3. 6 4. 65. 66. 67. 68
•
69
•
70. 71
•
b dx= � arctan( :) +C � J (a2+ 2x)y'x a + _!_ arctan(bv'x) +C j (a2+b2x)2 dx=a2(a2v'x +b2x) a3b a v'x j a2 �2x dx =2'f: - �� arctan(b:) +C v'x )2dx= - b2(a2v'x+b2x) + ab1 3arctan(bv'x j (a2+b2x +C a ) 1
arctan(bv'x) J v'x +C arctan(by'x j a ) 1 )y'xdx= ab_!_ ln(laa+- by'xl) +C j (a2 - b2x by'x j (a2 - b2x 2 dx= a2(a2v'x- b2x) + 2a3b ln (l aa+- by'xl) +C ) v'x by'x a+by'x ( +C dx= - 2y'x+ j a2 v'x l b2 b3 a - by'xl) - b2x by'xl) +C j (a2 -v'xb2x)2dx= b2(a2v'x- b2x) 2ab3 1n (l aa+- by'x 2x3/2 2a2 v'x 2a3 x3/2 - - +C - ----V- + b5 dx a a2+ b2x = 3 b2 a2 x3/2 2x3/2 a dx= + 3 - 3 2 b2(a2+ b2x) b4(a2+ b2x) bs (a2+b2x)
1
1
_ _
_!!_ ln
1
_ _
Integrals Involving
2a2..;x + a3s ln(la+b../XI) +C x3/2 dxJ a2-b2x --2x3b3/22 - � b a-b../X x312 dx=- 2b2x312 + 3a2..ji --ln 3a (la+b../XI) +C 73. J 2 b4(a2 -b2x) b4(a2 -b2x) 2 b5 a-b../X (a2-b2x)
72•
Integrals Involving v'a +bx
Va+IJX-y'a) +C j xv'a1+bx dx= __!__ ln( fa v'a+bx+fa 75. j �dx= ; (�(a+bx)312-av'a+bx) +C 2 a 76. j �dx= ; (�(a+bx)512-; (a+bx)312+a2v'a+bx) +C 3 2 +-1-m(Va+IJX-y'a) +C 77. j 1 dx= 2 3 x(a+bx) 1 av'a+bx a312 v'a+bx+fa x dx= 2 (va+bx+ �) +c 78. J a+bx (a+bx)3/2 b2 x2 dx= 2 (-1 (a+bx)312- 2av'a+bx- �) +C 79. j a+� + � � 31 2 b 3 3 80. J Va+bX dx= 2v'a+bx+vam(Va+bX .;a) +c x v'a+bx+fa 81. j xv'a+bxdx= ; (�(a+bx)512-�(a+bx)312) +C 2 82. j x2v'a+bxdx= ; (�(a+bx)712- 2; (a+bx)512+�(a+bx)312) +C 3 83. j v'f+x�dx= .�1n(v'a+bx+v'bv'l+x) +C +x a+bx vb (a 84. j i!iixdx= v'f+x� - b;�)In(v'a+bx+v'bv'l+x) +C (a b) 85. j v'l+xv'a+bxdx= � v'l+xv'a+bx+ �(l+x) 312v'a+bx- (��StIn(v'a+bx+v'bv'l+x) +c (a b) 86. JVJmdx=v'l+xv'a+bx+ � In(v'a+bx+v'bv'l+x) +C
74.
1
Integrals Involving v'a2 +x2
l dx= x +C J (a2+x2) 3/2 a2v'a2+x2 88. f kdx=m(x+v'a2+x2) +c a2+x2 89. j v'a2+x2dx= �xv'a2+x2+ �a2 (x+v'a2+x2) +C 90. j (a2+x2)3/2dx= �x(a2+x2)3/2+ �a2xv'a2+x2+ �a4 ln(x+v'a2+x2) +C 87•
ln
v'a+bx
T-5
Integrals Involving sin(x) 110. 111. 112. 113. 114. 115. 116. 117. 118. 119.
jh dx =ln(x+ v'x2-a2 ) +C x2-a2 j v'x2-a2dx =�xv'x2-a2-�a2ln(x+v'x2-a2) +C j x( 2-a2)3/2 dx=�x x( 2-a2)3/2- �a2xv'x2-a2+ �a4ln (x+ Vx2-a2) +C x2 j (x2-a2) dx=- � + ln(x+ v'x2-a2 ) +C 312 x2-a2 j h dx =�xv'x2-a2+ �a2ln(x+v'x2-a2 )+C j x2v'x2-a2dx =�x (x2-a2)3/2+ �a2xv'x2-a2-�a 4nI (x+v'x2-a2) +C j x x( 2-a2)3/2 dx= a_!_3arcsec (a! ) -a2\l'x2-a2 +C, O
1
_
1
a
c,
Integrals Involving sin(x) 120. Jsn i (x)dx= -cos(x)+C 121. 122. 123. 124.
jsin2 (x)dx =�x-�cos(x)sin(x)+C jsin3(x)dx =-�sin2 x( )cos(x)- jcos(x)+C jsin3(x)dx =�cos3(x)-cos(x)+C jsin4(x)dx =-�sin3(x)cos(x)- �cos(x)sin(x)+ �x+C
125. Jx sin(x)dx =sn i (x)-x cos(x)+C 126. J x2sn i (x)dx =2.xsin(x)- (x2- 2) cos(x)+C 127. J x3 sn i (x)dx =3(x2- 2)sn i x( )- x( 3- 6x)cos(x)+C 128' 129. 130. 131. 132.
J +sni (x) dx=- +tan(2 x/2) + 2 dx= +C . J sm( x) -tan(x/2) . xcos(x) x dx- ln( +sm(x) ) +C J +smx . . +smx( ) ( ) . xcos(x) x dx- ln( -sm(x) ) + J -smx +C ( ) -smx( ) atan(x/2)+b 2 dxarctan( ) +C, . Ja+bsmx ( ) � � a2-b2 a2-b2 1
1
1
c
1
1-
1
_
1
_
1
.
1
_
1
1
1
1
.
b
T-7
T-8
Table of Integrals
atan(x/2)+b-v'b2-a21) +C a
1
133 •
134.
135.
136.
137.
138.
139.
140.
141.
=
1
1
1
1
Integrals Involving cos(x) 142. 143. 144. 145. 146. 147. 148. 149. 150.
151.
152.
153.
154.
Jcos(x)dx= sin(x)+C j cos 2(x)dx= �x+ �cos(x)sin(x)+C j cos3(x)dx= � cos 2(x)sin(x)+ � sin(x)+C j cos3(x)dx= sin(x) -� sin3(x)+C j cos4(x)dx= �x+ � cos3(x)sin(x)+ �cos(x)sin(x)+C Jxcos(x)dx= cos(x)+xsin(x)+C Jx2 cos(x)dx=(x2-2)sin(x)+2xcos(x)+C Jx3cos(x)dx=(x3-6x)sin(x)+(3x2-6)cos(x)+C j +clos(x) dx=tan(�) +C j clos(x) dx=-cot(�) +C j +:os(x) dx=xtan(�) +2ln(l cos(�) I) +C j -:os(x) dx=-xcot(�) +2ln(lsin(�)I) +C l 2 arctan( (a-b)tan(x/2)) +C, b
1
_
1
1
Integrals Involving sin(x) and cos(x)
/2)+�1) +C a
1
1
=
Integrals Involving sin(x) and cos(x)
jsin2(x)cos2(x)dx= �x - � cos3(x)sin(x)+� cos(x)sin(x)+C 165. jsin2(ax)cos2(bx)dx= �( - �sin(2ax)+ �sin(2bx)2(a� b)sin(2(a - b)x)- 2(a� b)sin(2(a +b)x)) +C, a =I- b 166. j . smx( )lcos(x) dx= ln(ltan(x)I)+C 1 dx -csc(x)+lnJ (sec(x)+tan(x))J +C 167. j . sm2(x)cos(x) 168. j. sm(x)cos1 2(x) dx=sec(x)- ln(lcsc(x)+cot(x)I)+C 169. j . sm2(x)cos1 2(x) dx=sec(x)csc(x)-2cot(x)+C 1 dx= tan(x)-cot(x)+C 170. j . sm2(x)cos2(x) 1 dx=--21csc2(x)+ln(ltan(x)I)+C 171. j . 3 sm(x)cos(x) l 3(x) dx=-21sec2(x)+ln(ltan(x)l)+C 172. j. smx( )cos s (x) dx= ln(lsec(x)I)+C 173. j in cos(x) . +ln( I sec(x)+tan(x)I )+C sin2(x) dx -sm(x) 174. ( ) cos x J
164.
2x
=
--
=
T-9
T-1 O
175. 176. 177. 178. 179. 180. 181. 182. 183. 184.
Table of Integrals
sin3(x) 1 . ( ) - ln(I cos(x)I) + dx = -2sm2x cos(x)
J s(x) j csmx dx = ln(lsin(x)I) + ? ( )
C
C
cos2(x) dx = cos(x) - ln(lcsc(x)+ cot(x)I) + . smx ( )
J cos3(x) . J sin(x) dx = 2 cos2(x)+ ln(lsm(x)I) + sin2(x) J cos2(x) dx - tan(x) -x+ sin3(x) J cos2(x) dx = cos(x)+ sec(x)+ cos2(x) J sm2(x) dx= -cot(x) -x+ cos3(x) . dx= -sm(x) - csc(x)+ 2 J sm (x) _!_ arctan ( !?_ tan(x)) + ja2 cos2(x)+ b2sin2(x) dx = ab a x)+a l) + ja2 cos2(x)- b2sin2(x) dx= 2ab ln (l bb tan( tan(x) -a 1
_
C
C
C
C
.
C
.
1
C
1
1
_ _
Integrals Involving tan(x) 185.
J tan(x)dx = ln(lsec(x)I) + C
186.
J tan2(x)dx= tan(x)-x+
187. 188.
C
j tan3(x)dx = � tan2x( )+ ln(l cos(x)I) + j tan4(x)dx= � tan3(x)- tan(x)+x+
C
C
Integrals Involving cot(x) 189.
J cot(x)dx= - ln(lcsc(x)I) + C
190.
J cot2x ( )dx = - cot(x) -x+ C
191. 192.
C
j cot3(x)dx = -� cot2x( )+ ln(lcsc(x)I) + j cot4(x)dx= -� cot3(x)+ cot(x)+x+C
C
Integrals Involving sec(x) 193.
J sec(x)dx= ln(lsec(x)+ tan(x)I) + C
194.
J sec2x ( )dx = tan(x)+ C
C
Integrals Involving e"
195. 196.
j sec3(x)dx j sec4(x)dx
1 1 2sec(x)tan(x) + 21n(lsec(x) + tan(x)I)
+
C
1 3 tan3(x) + tan(x)+ C
Integrals Involving tan (x) and sec(x) 197. 198.
199. 200. 201. 202.
J sec(x)tan(x)dx
sec(x)+ C
=
j sec2(x)tan(x)dx j sec(x)tan2(x)dx j sec2(x)tan2(x)dx j sec3(x)tan(x)d.x j sec(x)tan3(x)d.x
=
=
� sec2(x)+ C � sec(x)tan(x) - � ln(lsec(x)+ tan(x)I) + C � tan3(x) C � sec3(x) C � sec3(x) - sec(x) C +
=
=
+
+
Integrals Involving csc(x) 203.
J csc(x)dx
204.
J csc2(x)dx
205. 206.
- ln(lcsc(x)+ cot(x)I) + C
=
=
j csc3(x)dx j csc4(x)dx
=
=
- cot(x)+ C
� � - � cot3(x) - cot(x)
- csc(x)cot(x) - ln(lcsc(x) + cot(x)I) +
+
C
C
Integrals Involving cot(x) and csc(x) 207.
J csc(x)cot(x)dx
208.
J csc2(x)cot(x)dx
209. 210. 211. 212.
214.
� - � cot(x)csc(x) � ln(lcsc(x) - � cot3(x) C - � csc3(x)+ C - � csc3(x)+ csc(x) C
- csc2(x)+ C
=
j csc(x)cot2(x)dx j csc2(x)cot2(x)dx j csc3(x)cot(x)dx j csc(x)cot3(x)dx
=
=
=
=
Integrals Involving 213.
- csc(x) + C
=
+
+
cot(x)I) + C
+
+
fiC
J ebx dx t ebx+ C j axdx ln�a)ax+c, =
a =f-1
(preceding formula with b
ln(a))
T-11
T-1 2
Table of Integrals
215. J xeaxdx= � (x - �) eax+ 216. J x2eaxdx= ;3 (a2x2 - 2ax+2)eax+ exp(x) 1 217. j a+bexp(x) dx=.!1n ( ) +c a a+bexp(x) 218. j exp(-ax2)dx= �erf(vfax)+ 219. f 00 exp(-ax2)dx= Va{i__ 220. j v'xexp(-ax)dx= - � v'xexp(-ax)+ � /ffserf(y'aX)+ 221. fo00 v'xexp(-ax)dx= �lffs 222. j }x exp(-ax)dx= faerl(../{iX) + 223.100 }x exp(-ax)dx= fa 224. 100 xnexp(-ax)dx= ann�1' (n = 2, ) 225. j x2exp(-ax2)dx= - 2� xexp(-ax2)+ � /ffserf(vfax)+ 226. j eaxsin(bx)dx= a2 ! b2 (asin(bx) - bcos(bx)) eax+c 227. J eaxcos(bx)dx= a2 ! b2 (acos(bx)+bsin(bx)) eax+ C
C
C
-oo
C
C
1,
3, ...
o
C
C
Integrals Involving In (x)
228. J In(x)dx=xln(x) - x+ 229. J In(ax)dx xln(ax) - x+ 230. j xln(ax)dx= � x2In(ax) - � x2+ 231. j x2In(ax)dx= � x3In(ax) - � x3+ 1- xb+1 In(ax) xb+l + 232. j xbln(ax)dx= b+ (b+1)2 233• j In(xbax) dx= b -1 In(ax) + xb-1 (b -1)2 xb-1 234. j In2(ax)dx= xln2(ax) - 2xln(ax)+2x+ 235. j In(a+bx)dx= � (a+bx)In(a+bx) - x+ a In(a+bx)+ x - 4x2+c 236. J xln(a+bx)dx= - (a2 -2bb2x2) 2b 2 C
C
=
C
C
1
C
1
_
1_
_
_
1
1
C
C
C
1
Integrals Involving Special Functions
ln(a2+x2)dx=xln(a2+x2) -2x +2a arctan
237. 238. 239.
( �) +C
j xln(a2+x2)dx=2(a2+x2)ln(a2+x2) -2(a2+x2)+C 2 2a2 2a3 x J x2ln(a2+x2)dx= 3x3 ( ln(a2+x2) - 3) + 3x - 3 arctan ( ) + 1
a
Integrals Involving Special Functions
244.
� (� 7r.x2) +C j FresnelS(x)dx=xFresnelS(x)+ � cos ( � ) +C j erf(x)dx=xerf(x)+ )rr exp(-x2)+C j xerf(x)dx= � x2erf(x)+ 2� xexp(-x2) - � erf(x)+C J erf(vx)dx=xerf(vx)+ )rr vxe-x- � erf(vx)+C
245.
f Si(x)dx=Si(x)+ cos(x)+C
246.
f Ci(x)dx =xCi(x) - sin(x)+C
240 241. 242. 243.
J
FresnelC(x)dx =xFresnelC(x) - sin
7r.x2
T-13
Formulas from Calculus: Single Variable
df(x)
=
cos2(0) =
j
cos2(x) dx
=
f'(x)d x.
(3.9.7)
1 + cos(W) 2
� � x +
(6.2.2)
sin(2x) + C
(6.2.4) (6.2.8)
lfo 2
� cos2(x)dx
2 = lf o
� sin2(x)dx
='If
(6.2.9) (6.2.11)
T7 l R
_l
Surface Area of frustum
� + (3y f(x)
=a
N
=
(n
n.
for some
T-14
27r
has general solution
f )() 1 C (x 2:-
n�o
=
s
er+
r
( �R )
y(t) =
(N+l)()
f
SI
(N + 1).
between c and x
s
� + ce-(31
N+l (x - c)
(7.2.7)
(7.7.4, 7.7.5)
(8.8.4)
Answers to Selected Exercises
17.
Section 9.1 L
(-1, -4)
7.
(6, -6)
3.
(-5,1)
s.
(-5,12) 13 (5,-12) (-5/13, 12/13) (5/13,-12/13) (4, -16) 4vTI (-4, 16) (1/vTI, -4/vTI) (-1/vTI,4/vTI) -7i+2j; -2i+9j;-4v=28i-8j; 3v -2w -17i-12j; 4v+ 7j -28i+ 15j v = 6i-2j; w = 9i-3j; -4v = -24i 8j; 3v -2w +Oj; 4v+7j 24i - j F= (3,4); magnitude: 5 F 3i-j; magnitude: v'IO (-1,0) (../3/2,1/2) 13 ( 12 - 5 ) 2 13' 13 v'I3(vbi- �j) (4,1) (7,9) 2v'IO 5/3 2 -6 10 3/2 -6 50\/2; 50(1+ ../3) 200 (-80../3 -200 -140\/2)i+ (80 -140\/2)j b.
a.
d.
c.
e.
(5, -11) 19.
b.
a.
d.
y
2L
c.
e.
v=
w=
=
23.
+
=
2S. 27.
x
29.
[ Figure for 9. lL 13. 15.
1, 3, 5, and
33
7
parallel VS; II RSll II PQ II = II RSll = v's3; not parallel
II PQ II
=
=
3VS;
3v'IO;
(9,1) See Figure for Exercise 15 (2,8) 8
y
4 2
-2
Figure for
I
'
13 and 15
37. 4L 49. S3. S7. S9.
Oi
=
=
3L
39.
43.
47. SL SS.
m
(each component in N) 63. )... (-b, a ) for all scalars)... 6S. � = (B2x0-ABy0-AD)!(A2 + B2) x
8
3S.
45.
. l l:
6
•
=
10
t t
- p1),y = p2 + t(q2 - p2)
67.
x
7L
3 (1.85325,1.44312) 1.89838
73. 7S.
= Pl
+ t(q1
A·1
A-2
Answers to Selected Exercises
77. Left x-intercept: -0.65421 at t=-0.98132; right
1 at t=O; y-intercept: 0.51447 t=-0.63598
x-intercept: at
y
37. (3,2,9)
35. (6/7,4/7, 12/7) 39.
a.
d.
(2, -3,6) b. (-2,3, -6) c. 7 (2/7, -3/7, 6/7) e. (24/7, -36/7,72/7)
(1, -1, -2) b. (-1, 1,2) c. J6 (1/v'6, -llv'6, -2/v'6) e. (2v'6, -2v'6, -4v'6) 43. -15i+20j - 5k 45. i - 6j +2k 41.
3
a.
d.
2
47. -4j +k
x
49. (4, -6,3), (2,2, -1)
-1
-2
51. (1,1, -2), (1, -3,0)
- tHg(r), 11(t)) - tHt,t3)
-3
53. (xo, Yo, 0), (0, y0, z0), and (x0, 0, z0), respectively 55. {(x,y,z) :y=O} 57. {(x,y,z): O
Section 9.2 1.
61. {(x,y,z): 3< Jx2 +y2 +z2 }
z
63. The set of all unit vectors
1
65. 404 /v'I05 ftIs 67. The sphere with center (-1,2, 3) and radius 6
(3,2,l)e
71. The ellipse x2+y2J9=1, z=0, the ellipse x2+z2/16=1,
y=0, and
3.
the ellipse
y2J9+z2/16=1, x=0
z
(- 4, 1, -2)•
y
7. 3J3
5. 3y'f4
9. Center: (-1, -2,-3); radius: ../22 11. Center: (-1/2, -1/2, -1/2); radius: ../5/2 13. Center: (1/2,0,O); radius: 1/2 15. The coefficients of x2, y2, and z2 are not all equal. 17. The equation has the term 4xy.
75. Do not span.
19. The equation has the term -2xy. 21. {(x,y,z): (x- 3)2 +(y+2)2 +(z- 6)2 <16} 23. {(x,y,z): (x -·n-)2+(y -11-)2+(z+?r)2=n2} 25. {(x,y,z) : 4<(x - 2)2 +(y- 1)2 +z2} 27. II PQ II = J4I; II RSll = 2v'4I; PQ and RS are parallel. 29. II PQ II = 3; II RSll =3../5; 31. (-12,3, -24)
PQ
and
73. Span.
RS are not parallel.
33. (0, -11,0)
77. -1.8136,-1, 0.4707, 2.3429 79. 0.6542 Section 9.3 1. 28 7. ?r/3
3. 4 9. 3?r/4
5. -31 11. ?r/2
13. Not perpendicular: Dot product is 1, not 0.
Section 9.5
15. Perpendicular: Dot product is 0.
39. 6 43. s=19, t=-6 47. 65
19. Pv(w)=(4,8,12); Pw(v) =(2,4,6); 4JI4; 2JI4 21. Pv(w)=(0,0,0); Pw(v)=(0,0,0); O;O
51.
17. Pv(w)=(-51/62,119/62,17/31); Pw(v)=(17/33, 68/33, -68/33); 171../62; 17/v'33
23. Direction: (v'3!2, -1/2,O); a=7r/6; f3=27r/3; 1=7r/2 25. Direction: (0,1,O); a=7r/2; f3=O; 1=7r/2
27. Direction: (-1/v'2,0,1/v'2); a=37r/4; f3=7r/2; 1=7r/4 29. Dot product: -16; lengths 6 and 3; l-16 I � 6 3 is true.
a.
(-3,3,3)
53. a. (-6,6,24) 59. -w
b.
v'3/2
b.
2v'2/3
4L 36 45. s=5, t=-1 49. 52 c.
3
d.
1/2
d. 1/3 9 61. 30 ft-lb c.
63. d(R,t) = 11 PR. x R(l II /II PO 11 69. 7x + 22y + 9z=78, the graph of which is a plane that passes through P0 and is perpendicular to vxw 71. -7.161655
·
31. Dot product: 9; lengths v's and 2v's; 191 � v's · 2v's is true.
33. P0(v)=(4,4, -2); v - P0(v) =(0, -3, -6); P0(v) (v - P0 (v))=0 ·
35. P0(v) =(-2/vf3, - 2/v'3,2/vf3); v - P0(v)=(8/vf3,2/vf3,10/vf3); P0(v) (v - P0(v)) =0 37. -2 ·
39. -4, 4
49.
a.
6/7
b.
-11/5
Section 9.5
1. Normals: .A(l,-3,4).AEJR; n=(-1,3,-4) 3. Normals: .A (3,0,-4).AEJR;n=(-12,0,16) 5. 7.
A. A.
(-5,7, -4) A.E JR (1,4, -2) A.E JR
9. -3x + 7y + 9z=79
11.
2x +
2y + 5z=18
13. -30x - y + 40z=119 15. 24x + 20y- 7z=-128
45. 100000V3 ft lb
c.
-1/5
d.
-9/25
51. & j =(V3/2) II &II 63. 0.158548,0.700449,2.282596; 3.141593 � 7r .
17. x=2 - 3t, y=1 + t, z=9 + 7t
19. x=2t, y=1 - t, z=t
21. (x - 2)/(-5)=(y-1)/8=(z -5)/2
65. 0.203888
23. x=y- 1=z/9
Section 9.4
27. x=-1 + t, y=-1 - t, z=-8 + t/3
1. (9,11,12) 5. (7,1,10)
9. (1/v'3,1/v'3, -1/v'3)
25. x=7 + t, y=-4 -3t, z=6 + 7t 3. (-12,4,10) 7. (32,0, -4)
15. v'212
17. 9/2
19. 2VI4
21. 7
29. x - 1=y= 1 - z
31. (x - 1)/2=2 - y=1 - z
33. x=2 - 39t, y=7t, z=l2t
35. x=6t, y=1/6 - St, z=-24t
11. (-11\1'6, -ll\1'6,21\1'6) y'2 13. 10 (4,3, -5)
23. uX(vXw)=(-1, -4,15) and (uXv)Xw=(40, -3,42) 25. uX(vXw)=(11, -33,11) and (uXv)Xw=(-2,3,6) 27. uxv=(1,11, -3); vxw=(-11, -1, -7); (uXv)·w=u·(vXw)=-40 29. uxv=(0, -2, -1); vxw=(7, -12,1); (uXv)·w=u·(vx w)=-7 33. -16 3L 5
A-3
37. (x - 4)/2=y- 2=-z
39. x=(y + 1)/2=(z + 3)/4 41. 43. 47. 51. 55. 59. 63.
(4, -2, -2) (2,2,7) (3,1,0) 2\1'6/5 9v'3/5
Not on line No intersection
45. (2,1,4) 49. 2/15 53. v'6 57. On line 61. (-2, -9, 2) 65. y'2
67. y'2
73. 3x + 4y + z=14
75. (x - 1) + 2(y - 2) + 3(z - 3)=0
A-4
Answers to Selected Exercises
71. x-intercept 3; y-intercept: 2; z-intereept: 6
99. x=(0.808827,0.617654,0.S73S19)
-1.373516, 3.560274)
and
(-0.186758,
6 REVIEW EXERCISES FOR CHAPTER 9 L 9
3. 9
S. Center: (2, -4, -1); radius: 6 7. Center: (-112, -1/2, 112); radius: 1 9. 9 (213. 2/3,1/3) 11. 2Vl (1/Vl, -1/V,l -1/V)l u. (6,3,8)
15. (-60, -66,48} 17. (2,2, 1) 19. -7 23. 3 27.
2L 12 25. 2,3
29.
arccos(-113)
31. (4,4,2)
arccos(5/6)
33. (63,36,36} 79. x-intercept: 5; y-intercept: 3; z-intercept: 1
z
w=62; llvll =7;llwll = 9 37. v w 220; llvll 15; llwll 15 4L (-2, -8,6) 39. (-3,-12,5}
35.
T·
·
43. 3¥'2
3x + Sy + 15.z =15
1
4� sv'2
49. 3M 55. 6
a.
83.
a.
paralle l
b. parallel
d. parall el
c.
not parallel
e. parallel no intersection b. the line lies in the plane c. (8,8, 3) is the point of intersection d. no intersection 85. 1/v'2 87. 1/v'2 9L 57./6/2 89. 10/\1'3
97.
r
.i + 2y + 2z = 17
53. -2
69. 213
65. 1/3 71. -sn
73.y-z=l 71. x= 1+t, y= 3, z=5 +3t 79. x 2t, y - 2t, z 1 - 3t =
=
47. 5./6
SL 9
59. -2,0,1
57. 16 63. -2
61. -7/5 67. -3n.
SL
=
=
=
75. 2x - 4y - z = 5
=
8L 2(x+2)=y-2=-(z+1)
3(y+3)12 - 3(z+ S)n. 85. (x - l)n. y- 1 (z+1)/(-2) 87. (x - 1)12 (y- 2)/2 z - 1 83. 3(x+2)
=
=
=
=
=
89. x
=
3t, y t, z 2 - 2t 9L x= t, y= 2 - 2t, z =2 - 3t 93. (-1,8,5) 97. (6, 3, -5) =
=
=
95. (2,1,4) 99. 2vs13
Section 10.2
A-5
25. 4/(9t713)i-( (2t2 -1)/(1+t2)5/2)j-3/(16t714)k
Section 10.1 1. The parabola y=x2
27. (t3 /3+ C )i -(2y'i+ C )j+ (sin(2t)/2+ C )k
1
y
2
29. (tln(t) -t + C )i- (cos(t)sin(t)/2
3
1 +t/2+ C )j +(ln(lsec(t)I)+ C )k 2 3 31. (t2+l)i+ (6 -t3)j+ (t4 -1 5)k 33. (3+sin(t) )i + (1+ (1/2)cos(2t))j+ (sec(t) -3)k 35. 3e31i+3 sin(3t)j+ (18t/ (1+9t2))k
37. 1 5(1+St) 2i -Sk
39. 3 (1+St) e1
41. -9t-10i+3t-4k
43.
6t5+2t
45. 5 (3t2e1+t3e1 -ln(l+t2) -2t2 /(1+ t2)) 49. (1+t)e'j 53. i+j+k
47. (9t2 +10t+l)j 51. 0 55. -i -2ln (rr)j+k 57. Continuous on {t
3. The parabola y=x2, z=2
59. Continuous on
z
r{t)
=
E JR
:
t =I- 5}
JR
65. r(t) = (t, ltl, 0), c= (0, 0, 0)
ti+t2j +2k
Section 10.2 1. v(t) =i +2tj +3t2k; v (t) = Jl+4t2+9t4; a(t) =2j+6tk y x
3. v(t) =-2t i+2j+ (-2t/(1+t2) 2)k;
J
v(t) =2 1+t2+t2 /(1 +t2) 4; a(t) =-2i + (2(3t2 -1)/(1+t2)3)k
5. The circle x=1, y2+z2=1
5. v(t) = i+e1k; v(t) = J1+e2t; a(t) =e1k
z
7. v (t) =2t sin(t2)i + 2t cos(t2)j + 3t2k;
v (t)=ltiJ4+9t2; a(t) = ( 4t2cos(t2) + 2sin(t2) )i+ (-4t2sin(t2) +2 cos(t2))j+6tk y x
9. (2+3t)i -2tj + (-5+t -16t2)k 11. (3+t)i+ (-2+t)j+ (-1 +t -16t2)k 13. x=1+s, y=1+2s, z=1+3s 15. x=l+s,y=1 -s, z=-1+2s
r(t) 7.
=
i+
cos(t)j
+
17. x=2+s,y=-1,z=e+(e/2)s
sin(t)k
straight line segment from P= (1, 1, 1) to Q= (0, 0, 1), followed by the line segment from Q to R= (4, 4, 1) A
9. One cycle of the graph of z= sin(x), located in the x=y
plane lL -i+ (1/2)j
13. (4, 4, 1)
15. (2t112r1i +2tk 17. sec2(t)i+csc2(t)j+sec(t) tan (t)k 19.
i -(t+sr2j -2(t -1r3k
2L 2i-3/(4t112)j+ (1 5/4)(t+1) 112k 23. e-1i -2 (2t4 +t2 +1) ; (t(l +t2))2 j
19. x= 5+s, y=-5, z=1- (1/S)s 21. x=1-(3/ 5)s, y=4+4s, z=1fs 23. x+2=(y+4)/4= (z+8)/12 25. x - 5= 5(1 -z), y=-5 27. 2 (1-x)=(y-4)/2, z=l 29. x=1, y= 27f -2...(irz 31. (x-1)=2(y+4)= 5(z-1) 33. Trajectory: r(t) =SOJ3ti+ (4+SOt -16t2)j; time until
the arrow hits the ground: (2 5 + v'689)/16 s; horizontal distance: 2SJ3(2 5+ v'689)/8 ft; greatest height: 689/16 ft 35. -80i +44j+SOk
Answers to Selected Exercises
A-6
41. Velocity when the arrow hits the ground:
40i -20v'107k ft/s; horizontal 25v'107-75../3 � 128.7 ft
59. Vertical tangent: at R=(2213,2113); horizontal tangents: at P=(0,0) and Q=(2113, 2213); tangents parallel to
distance:
at S=(1.2067,0.52540) and T=(0.52540, 1.2067)
y=x:
y
43.
2 -/ih
51.
r'(7r/3)=-../3i+(1/2)j y 4
x 5 53.
61.
z
r'(7r/4)=2i+ vlzj y y
Section 10.3 55.
1.
r'(2)= (e2 -e-2)i+ (e2+e-2)j
r'(t) i -2tj +3t2k; T(t) (1+4t2+ 9t4r 1t2 (i -2tj +3t2k); tangent line: (x-1) (y+1)/(-2) (z -1)/3 =
=
=
y
=
12
3.
r'(t)=i-1/(2t112)k; T(t)=(1+1/(4t)r112 (i -(t-1/2/2)k); tangent line: 1 -x/4 z+2,y =
x
5.
r'(t)= (e1+2t)i; T(t)=i;
tangent line:
r'(l/2)
=
0
y=3,z=0
r'(t)=-sin(t)i+cos(t)j +k; T(t)=(1/v'2) (-sin(t)i+cos(t)j +k); tangent line: x=7r/2-z,y =1 11. 6 9. v'27l" 15. Is not 13. 3 ../3 -2v'2 19. e2 17. Is 21. e-e - 1 7.
57.
=
(24/25)i -(32/25)j y
x
23.
(1 -x)/11=(y+1)/8=(z -1)/9
25.
x=0, z=7r/2
27.
x-l=y -1=-z/2
29.
(x-2)/5=(y-1)/4=z -2
31.
x-2y+3z=6
33.
x-y/2+4z=35/2
35.
x-y=O
37.
x-y+32z=33
Section 10.4 39.
p(s)
=
( (27s+l6;z)2/3 - 8 ) (i+j) (27s+ l6v'2)2/3 - 8 312 +( ) 9
k
2 j p(s) = s+ #+4 i+ 2 s+ vs2+4
(
) ( + +v'21n ( s v;+4)k
)
lcos3(t)l/(2 - cos2(t))312
19.
lcos(x) I
23.
Radius : v'2(2t2+1)3/2; center: (3t2+1,3t2+1,1 - 4t3)
25.
Radius: (2e21+1)312/(v'2e1); center: (2e' +e-t /2,t - 2e2t - 1,2e 1+e21/2)
27.
Radius: v'5(16t6+5)312 /(60t2);
29. 31.
17.
Tangent line: -(x - 6) v'3(2y- 6 - v'3); normal line: (x - 6) =v'3(2y- 6 - v'3)/12
=
=
=
(18t4+t2+ 2,8t3+2t/3) 35. (t - l - e21,2e 1+e-') 39.
(t+sin(t),- 1+cos(t)) (t - tan(t) ,ln(sec(t))+1)
41.
(0,0)
43.
(-v'2,ln(4)) and (v'2,ln(4))
45.
2v'2labl/(a2+b2)3/2
47.
4v'3/9
=
Tangent line: y =x/v'2+1/v'2; normal line: y =-v'2(x - 1)+v'2 47. Tangent line: x 2; normal line: y 0 49. Tangent line: y =- 4x/3+513; normal line: y =3x/4
4x - 2y- z =3 x+y- 2z 0
33.
37.
45.
51.
2/(4x+1)312 21. 8lxl3/(x4+1)2
15.
center: ((28t6 +5)/(12t2), 2t(5 - 8t6)/15, 4t(5 - 8t6)/15)
41.
43.
57. r
y
T(7r/4) (-1/vf3, 0, J2/3 ); N(7r/ 4) =(-v'6/21,-120/21,-vf3/21); B(7r/4) (4v'2/7,-1/7, 4/7) =
=
53.
T(7r/4) (2/./7,v'2/./7,1/./7); N(7r/ 4) =(0,1/vf3,-2/v'6); B(7r/4) (-3/ffi_,2v'2/ffi,2/ffi_) =
x
=
T(ln(2)) (4/v'f6,-1/v'f6,3/v'f6); N(ln(2)) (-2/-178,7/-178,5/-178); B(ln(2)) =(-1/vf3,-1/vf3,1/vf3) 59. Simpson's Rule approximation: 255.707
55.
=
=
6L
Simpson's Rule approximation: 4.69858
59.
Section 10.4 1. a. b.
3. a.
T(s) =(cos(s),sin(s),0) N(s) =(-sin(s),cos(s),0) T(s)
=
c.
K(s) =1
(1/v'2)(-sin(s/v'2),-cos(s/v'2),1)
(-cos(s/v'2),sin(s/v'2),0) c. K(s) 1/2 5. a. T(s) =((3/5)cos(s),-sin(s),(4/5)cos(s)) b. N(s) =(-(3/5)sin(s), -cos(s),-(4/5)sin(s)) b.
N(s)
C.
K(s) =1
=
=
p(s) =2; center: (-cos(s/v'2),sin(s/v'2),s/v'2) p(s) =1; center: (0,0,0) 13. t2/(t2 +1)3/2 lL 2/(1+4e2t)3/2 7.
9.
A-7
y
A-8
Answers to Selected Exercises
61.
y
N(t)=(e21+1r1(v'2e1, v12e1,1 - e21); llr(t)= v12/(e1+e-t)2; ar=e 1- e-1; llN =v'2
r"(t)=(-cos(t),-sin(t),2); T(t)=(1+4t2r112(-sin(t),cos(t),2t); N(t)=(1+4t2r112(s+4t2r112 (-cos(t)+4t sin(t)- 4t2cos(t),-sin(t)- 4t cos(t) - 4t2sin(t),2); llr(t)=VS+4t2/(1+4t2)312; ar=4t/vl +4t2; aN=vs+4t2/vl+4t2
15. r'(t)=(-sin(t),cos(t),2t);
17. ar=4t/VS+4t2; aN=2../5/vS+4t2 Section 10.5 1. r'(t)=(1,-2t); r" (t)=(0,-2);
T(t)=(1+4t2r112(1,-2t); N(t) =(1+4t2r1/2(-2t,-1); llr(t) = 2/(1+4t2)3/2; ar=4t/Vl+4t2; aN = 2/Vl+4t2 3. r'(t) = (1- 2t,1+2t); r" (t) = (-2,2); T(t) = (2+8t2r1/2(1- 2t,1+2t); N(t)=(2+8t2r1!2(-(l+2t),1- 2t); llr(t)=4/(2+8t2)3/2; ar=8t/V2+8t2; aN=4/V2+8t2 5. r'(t)=(1 - cos(t),sin(t)); r"(t)=(sin(t),cos(t)); T(t)=(2- 2 cos(t)r1!2(l- cos(t),sin(t)); N(t)=(2- 2 cos(t)r112(sin(t),-(1- cos(t))); llr(t)=1/(2.J2- 2 cos(t)); ar=sin(t)/ ..;= 2- -=-2- co-s("t); aN=(1- cos(t))/ J2- 2 cos(t) r"(t)=(3/(4./i),-1/(4 t-312)); N(t)=(9t2+1r112 (1,-3t); T(t)=(9t2+1r112 (3t,1); llr(t)=6./i/(9t2+1)3/2; ar = t-312(9t2 +1r1/2 (9t2- 1);
7. r'(t)=(3v'3/2,1/(2./i));
19. ar=(exp(2t)- exp(-2t))/ vl .,---e--- x-p,-(--=2--.,-t); � + _ _e_xp ---,- (2 =- t)-- +
aN= J4+exp(2t)+exp(-2t)/.Jl+exp(2t)+exp(-2t)
21. (x/2)2+(y/3)2=1 23. (x/S)2+(y/3)2=1 25. (x- 1)2/S2+ (y- 2)2/132=1 27. (x- 1)2/62+ (y- 2)2/102=1 29. (x- 2/3)2/(8/3)2+y2/(4/v'3)2=1 35. 3.22S x 10 - 3 days 37. 6.166 X 109 cm
39. 9697SOv'3 mi 2/h
41. 4.600 X 107 km 43. 11.87 Earth yr 53. The
graphs of tf--+aN(t) and tf--+ II a(t) II ar(t)=0.
touch when
y 5
�
llN= bt-lf2(9r+1rl/2 2 9. r'(t)=(1,2t,3t2); r"(t)=(0,2,6t); T(t)=(1+4t2+9t4r1/2(1,2t,3t2); N(t)=(1+4t2+9t4r112(1+9t2+9t4r112 (-t(2+9t2),1- 9t4,3t(2t2+1)); llr(t)=2v1+9t2+9t4/(1+4t2+9t4)312; ar =2t(2+9t2)/v1+4t2+9t4; aN=2V1+9t2+9t4/Vl+4t2+9t4 11. r'(t)=((3/2)v'f+t,-(3/2)v'f=t,1);
r"(t)=(3/4)((1+tr1l2,(1 - tr1l2,O); T(t)=2Z-112(3v'f+t,-3v'f=t,2); N(t)=( J(l- t)/2, J(l+t)/2,O); llr(t)=(3v'2)/(22V1- t2); ar=O; aN=3v'2/(4V1- t2)
13. r'(t)=(et,-e-t, v'z); r"(t)=(et,e-t,O);
T(t)=(e21+1r1(e21,-1, v'ze');
-1
1 - Y = lla( t)ll -y = ar(t) -y aN(t) =
55. The particle slows on the interval (0.608, 0.967). Both
components of acceleration decrease on (0.3S4, 0.6S49).
57. Absolute maximum of a N at 1.63782; absolute maximum of llall at 0 and 271'
Review Exercises for Chapter 10 REVIEW EXERCISES FOR CHAPTER 10
1.
z
x+3y+z=-4 x+4y+4 z=0
31. r(t) =(-t+l)i+( 4t)j +(-16t2-t- 2)k
10
x
29•
A-9
33. r(t) =i+(t- l)j +(-16t2+2)k 35. r'( 2)= - 4 i+3 2j - 12k, - 12 k T( 2) = ----=i_ i+ ___lL · + ./1184 J v'IT84 v'IT84
y
x=-4 + ---=±_t v'IT84 32 t y=16 + ./1184 -12 t z= -8+ v'IT84
3.
37. vts7r 39. 1 5/4 41.
{
x=O +Ot y =1- lt z=3 7r +Ot
43. tangent line:
5. lim t-+2 r(t) = (e2,4 ,0} 7. r'(t) = �t-213i- 6tj +2k
1 9. r'(t) = _!_i- - - · + �t112k t2 (t+1)2 J 2
normal line:
_
14 13 G(t)= ( � + �)i+ ( -t - )j + ( � + �)k . 3 3 5 5 7 7 15. G(t)= ( � (t+1)312+[2-�3 3/2])i
� - )j 5 + (- �t2+10)k
17. r'(t) = [ (3t3+3t2+3t+l)e31]i + [- (3t2+1) cos(t)+(t3 +t)sin(t)]j + [5t4+3t2]k 19.
r"(t) = 4e2ti- 9e31j +2 sec2(t)tan(t)k
21. r"(t) =-� ( 1-f) -3/2(-2t)i+
(-D
( 1-t2r3/2(- 2t)j +2 sec2(t) tan(t)k
23. v(t) =2e21i- 4e41j -3e-31k,v =v4e41+16e8t+9e-6t, a(t) = 4e21i- 16e41j +9e-31k 25.
27.
{
{
x=3 +6t y =-2- 6t z=1+4t x=-1+Ot y =ln( 2)+�t z=0+7rt
{ {�:� { { z=O
2t· + !.e3tk+c 11• e1t• - !.e2 J 3
+(
x=t y=O z=3t
45. tangent line:
normal line:
x=l+t y =l-t z=.../ii x=l+t y=l+t z=O
47. ((o, -1/v'z, 1/VZ}, (-1,0,0),(o, -1;vz,-1/v'2)) 49. 3x- 12y- 20z=119 51. y+z=7r 53. p'(s) =� (-sin(s))i+ �sin(s)j +cos(s)khence llr'(s)ll
=1. Then T(s) =-�sin(s)i+ �sin(s)j +cos(s)k,N(s) =-�cos(s)i+ �cos(s)j - sin(s)k,�(s) =1 55. radius of curvature=2v'2, center of curvature=(3, -3 , VZ) 57. radius of curvature=5 /2, center of curvature =( 1, 5 /2, 0) le-e-xl 59 • (3 +e2x+e- 2x)3/2
Section 11.1 27.
29.
3L
4
.r
33.
A.11
A-12
Answers to Selected Exercises 39.
35.
41. .. ,
..·'
r x
37.
y. �J
= -o
43
•
.•
z
The level sets are concentric spheres. Two are shown. 45. 47.
xy / J x2 + y2) The level sets offi are the lines y = x + c, c sets of are the pairs of lines y =x + c :2: a(h, c) = c - 1- h/exp(c) •
49.
E R The
nd
level y =x - · ,
Section 11.2 st.
69.
y 2
-2
-2 -2
53. \ \ \ \ \ \
' ' ' '
Section 11.2
I I I I I
8
\\
L
/ I
I
/y = x?-
" ,/
,_
x x
55. The level sets Le= {(x,y): f(x,y) = c}
are nonempty precisely when -1 ::$ c ::$ 1. For c in this interval, the level set Le is the union of all parabolas of the form y =x2 + arcsin(c) +2mr (n E .Z) and y =x2- arcsin(c) + (2n + 1)1r (n E Z). 57. The level sets Le= {(x,y) : f(x,y) =c} are nonempty precisely when -1 � c � 1. For c in this interval,the level set Le is the union of all circles of the form x2+ y arcsin(c) + 2mr (n E Z,n � -arcsin(c)/211") and x2+ y= -arc:sin(c) + (2n + 1)7r (n E Z,n ;a:arcsin(c)/211")-112). 59. The level sets Le= {(x,y) : f(x,y) = cJ are nonempty precisely when c � 0. For c positive, the level set Le is the hyperbola y2Jc-x2= 1. For c =0, the level set Le is the x-axis. =
y
x+y=S 3.
61. f(x,y) =y/(y - x); domain: {(x,y) : lxl < lyl} 63. f(x,y) =x2 + y3 65. The level sets Le= {(x, y): f(x, y) =c} are nonempty
precisely when c 2::: 0. For nonnegative c, we have Le= {(y'C(cos(t) + sin(t)),y'C sin(t)) : t E [O,211')}.
67.
y
z
5.
x
-f(x,y) = 314 -/(x,y)=O -f(x,y)= -314
y
Xl+z =-4
A·13
A-1 4
Answers to Selected Exercises
7.
13.
y
..1.2+ y- 4
15. Hyperboloid of one sheet, which is not the graph of a function
r x2-4y2 + 9:2
36
-
17. Circular paraboloid, which is not the graph of a function
x2-2y2-1
r
�x
11. !
.11-x
l =z
19. Ellipsoid, which is not the graph of a function
y2-4z2-1
4l +y2+9z2=25
Section 11.2 27. Elliptic cone. which is not the graph of a function
n. Circular paraboloid, which is the graph of
f(x,y) 4- (xl +y2)
..f.
=
I
L
.29. Hyperbolic paraboloid, which is not the graph of a function
•.
x
+,1.J -tz=4
23. Elliptic paraboloid, which is the graph of
4x2+y2
f(x, y)
=
31. False 35. The ellipse:
x2+yJ /4
=
1
37.
-----___.,.,..-- ._,.�� · · .
.
I
x
4} +]')
A-15
3:9.
=z
25. Hyperboloid of two sheets, which is not the graph of a function
:x?- + 2:xy + 2.Y.
-
r:- = -1
A-16
Answers to Selected Exercises
41. x =-1
v'ico s(O), y=sin ) , z=2v'2- 2cos ), 0S61S27r +
x)=10;'¢y)=7- y 1/1-1)=12=
y
cp
{) 2)=0= f (2 -1) x '
2,-l)
x)=COS X 4) '¢ y) = OS y2) 7r)=-v'/ =
I
:
) 1/ )
1/f 1/2)=- v'2 2= 8y 7r 1 2) ,••
x)=x ln ex) ?Py)=ln y); '1)=2= {)f 1/1 )=1/e= 8y 1 e) 9.
x)=x2 'lfiy)=2Y; 1/1 2)
L 162 5. ln(12)
•
2v'7
7. 7r/2 IL 2
2 13. 9 23. Not continuous •
o
2L Continuous 29. 0
(y2,y)=1/
33.
4ln 2)
lL fxx x,y)=2y
Sedion 11.3
3L limy
=
fxy x,y)=2x
=
� 2,2).
2)=4=
� 2,2)
y x,y)=20y3 1=f x x,y)
13. fxx x,y)=2 f y x,y)=-2x/y3
fxy f 15. xx fy fx y
x,y)=f x x,y)=1/y2 x,y)=-sin x)/cos y) x,y)=sin x)(l+sin2 y))/co s3' x,y)=cos x) sin(y)/ os2 y)=•1
'.
:
17. fxx x,y)=-1 x- y)2;
z
limf(x x)
x
0
=
1
l
x
f y x,y)=- x2- 2yx+ 2y )/ xy fxy x,y)=1 x- y)2=f x x,y) . ;in y) 19. fxx x,y)=-cos x ) sin y); f y x,y) cos y)= = x,y) x) x,y) sin f f 21. Dy x y)=x sec y) tan y) Dyy x,y)=x sec y)(l+ 2 tan2 y)) Dyx x,y)=sec(y) t an y)=fz x,y); f1 x,y)=0 ·
23. Dy x,y)=2cos x3+2y); Dyyf x,y) =-4sin x3 +2y) Dyx x,y)=f x,y)=- x2 sin x3 + 2y) 2 f 1 x,y)=-9x4 sin x3 + 2y)+6xco s x3 +2y)
35.
os y))4 Dyy x,y)=5 sin(x)- os y))3 4sin2 y)+sin x) os y)-cos2 y)); Dyx/ x,y) = Ocos x) sin y)(sin(x) - o s y))3= f 2 x,y); f 1 x,y)= 5 sin x)-cos y)) 3 4cos2 x)- sin2 x) +sin x)co s y))
25. Dy x,y)=5 sin y)(sin(x)
x m
0
,-
)
x,y) =12 (x + 3y)3 Dyyf x,y)= 10 8. Dyx x,y)= 36 x + y)2=f x,y) 2 fi x,y)=12 x+3y)2
27. Dy
Sedion 11.4 L
x)=2x+
y)=10+y3;
1/1 1)=3=-' , 1) ay
5)=2 =5' 1)·' ..
29. Dy x,y)=2y3
Jl+ x2+y4;
Dyy x,y)=2y2 y4+3+9x2) (1+3x2 +
')2;
Section 11.5
Dyxf(x,y) =-6y3x/(l+3.x2 +y4) 312=fo(x,y); fu(x,y) =3(1+y4)/(1+3x2+y4) 3/2 31.
33.
fx(x,y,z) =6x2; fy(x,y,z) =5y4z8; fz(x,y,z) =8y5z1; fxy(x,y,z) = O; fxz(x,y,z) = O; fyz(x,y,z) = 40y4z1; fxx(x,y,z) =12x; fyy(x,y,z) =20y3z8; fzz(x,y,z) =56y5z6 fx(x,y,z) =5(x+2y+3z) 4; fy(x,y, z) = lO(x+2y +3z) 4; .fz(x,y, z) = l5(x+2y +3z) 4; fxy(x,y, z) =40(x+2y +3z) 3; fxz(x,y,z) =60(x+2y +3z) 3; fyz(x,y, z) =l20(x+2y +3z) 3; fxx(x,y, z) =20(x+2y +3z) 3; fyy(x,y,z) =80(x+2y +3z) 3; fzz(x,y,z) =180(x +2y +3z) 3
fx(x,y, z) = y exp(xy-yz); fy(x,y,z) =(x - z)exp(xy - yz); .fz(x,y,z) =-y exp(xy-yz); fxy(x,y,z) =(1+xy-yz) exp(xy-yz); fxz(x,y,z) =-y2 exp(xy-yz); fyz(x,y,z) =-(1+xy-yz) exp(xy-yz); fxx(x,y,z) =y2 exp(xy-yz); fyy(x,y, z) = (x-z) 2 exp(xy-yz); fzz(x,y,z) = y2 exp(xy-yz) 37. fx(x,y,z) =y/x; fy(x,y,z) =ln(xz); fz(x,y, z) =y/z; fxy(x,y, z) = l/x; fxz(x,y, z) = O; fyz(x,y,z) = l/z; fxx(x,y,z) = -y/x2; fyy(x,y,z) = O; fzz(x,y,z) =-y/z2;
35.
fx(x,y,z) =6x(x2+y3+z4) 2; fy(x,y,z) =9y2(x2+y3+z4) 2; fz(x,y,z) =12z3(x2+y3+z4) 2; fxy(x,y, z) =36xy2(x2+y3+z4); fxz(x,y, z) =48xz3(.x2+y3+z4); fyz(x,y, z) = 72y2z3(x2+y3+z4); fxx(x,y,z) =6(x2+y3+z4)(5x2+y3+z4); fyy(x,y, z) =18y(x2+y3+z4)(x2+4y3+z4); fzz(x,y, z) =12z2(x2+y3+z4)(3x2+3y3+llz4) 41. 6yz2 43. sin(z)/x2 39.
45.
fxx(x,y) =-4y/(x+y) 3; fyy(x,y) =4x/(x+y) 3; fxy(x,y) =fyx(x,y) =2(x-y)/(x+y) 3
47.
fxx(x,y) =2y sec2(x2y) (1+4x2y tan(x2y)); fyy(x,y)=2x4 tan(x2y) sec2(x2y); fxy(x,y) = fyx(x,y) = 2xsec2(x2y)(l+2x2y tan(x2y))
49.
A-1 7
fxx(x,y) = -2(x+2y) (1+(x+2y) 2r2; fyy(x,y) =-8(x+2y) (1+(x+2y) 2r2; fxy(x,y) =fyx(x,y) =-4(x+2y)(l+(x+2y) 2r2
y-coordinate: 2; x-coordinate: 8/5 53. y-coordinate: 5/2; x-coordinate : 11/10
51.
Harmonic Not harmonic 65. B=- 3E; C=- 3A 61.
a.
b.
d.
e.
Harmonic Not harmonic
c.
f.
Harmonic Harmonic
f (x, y)=x2y+y+ C 73. f (x, y)=x2+y - y2+ C 79. fx(4,3) = -0.02126; fy(4,3) = -0.09108 81. fxx(4,3)=0.03337; /yy(4,3)=0.07633 71.
Section 11.5
5s4-4s- s-2+3s-4 5. 3/(2s) 3. -sin(s) 1.
7.
-(2sin(s)+3cos(s)) exp ( 2 cos(s)- 3sin(s))
4s/ (cos(s2)-sin(s2))2 11. Zs= 3tsin(3st) cos(3st)(2+7 cos5(3st)); Zt=3s sin(3st) cos(3st)(2+7 cos5(3st)) 13. Zs=O; Zt=0 15. Zs= -2t(3s4+t4)/(s4- t4)2; Zi=2s(s4+3t4)/(s4 - t4)2 9.
17. Zs= 2(s- t)(s+4t)/(2s+3t)2;
Zt = -(s- t)(7s+3t)/(2s+3t)2
19.
s2(s - 2)5t!(s2+7s - 6)
21.
� (6s5+l/s)/ Js6+ n(s) l
s
3s2 (s+l) e3s cos(s3e3 ) 1 25. 24s+ ln(2) 23.
27. 27r (X - 1) +(y - 7r) - 7r (X - 1) 2 /2+(x 29.
- l)(y - 7r)
l-xy
1+4(x-7r/4) - 2(y- 7r/4) +8(x- 7r/4) 2 -8(x- 7r/4)(y-7r/4) +2(y- 7r/4) 2 35. Decreasing (at the rate of 0.31 cm2/s) 31.
37. 47.
&(x,y)=2x ln(y) cos(x21n(y)); gy(x,y) = (x2/y) cos(x2 ln(y)) D1(f)(R) D1(f)(P)+D (f)(R) D1(f)(Q) 2 +D (f)(R) D (f)(Q) D1(f)(P) where P=(x,y), 2 2 Q=(x,f(P)), and R =(f(P),f(Q)) ·
·
·
·
A-18
Answers to Selected Exercises
Section 11.6 1.
Section 11.7
(sin(y),xcos(y))
(y2 sec2(xy2),2xy sec 2(xy2)) 5. (-y sin(xy) sin(yz),-x sin(xy ) sin(yz) +z cos(xy) cos(yz), y cos(xy) cos(yz))
1.
Normal vector: (6, - 47, - 1); tangent plane: z=6x - 47y+123
3.
Normal vector: (2/7,4/7, - 1); tangent plane: z=2x/7+4y/7+ln(7) - 1017
5.
Normal vector: (1/9, - 2/9, - 1); tangent plane: z=x/9 - 2y/9+2/3
7.
Normal vector: (4e3, 3e3, - 1); tangent plane: z=e3(4x+3y - S)
9.
Normal vector: (1/2,1/2, - 1); tangent plane: z=(x+y)/2
3.
(y2/cos(y),2xy/cos(y)+xy2sec(y)tan(y)) 9. (yzcos(xy),xzcos(xy),sin(xy)) 13. -(3/2)fo 11. -49 7.
15.
1/32
17.
19.
1;vz
21.
23.
a.
(5/13,12/13 ) (-5/13,-12/13 )
b.
(1/VZ,1/VZ) (-1/VZ,-1/VZ)
b.
(-1/v's,-2/v's) (1/v'5,2/v'5) 29. a. (1, 0) b. 1/e
b.
c.
25.
a. c.
27.
d.
d.
a.
d.
c.
(2/v's,1/v's) d. - sov's 33. 36 31.
a.
.
c
-1/16
-3v'3/S 13/2 - 13/2 VZ(l - ln(2))/4 - VZ(l - ln(2))/4 2v'5/ 3 - 2v'5/3
(-1 0)
sov's
b.
d.
,
35.
c.
- 1/e
(-2/v's,-1/v's)
-3/4
11.
Normal vector: (-1/4S,1/4S, - 1); tangent plane: z= ( -x+y+32)/4S
13.
Normal vector: (2, - 16,2); tangent plane: 2x - 16y+2z= - 2S
15.
Normal vector: (-1/2,1/4, - 1/2); tangent plane: 2x - y+2z=O
17.
Normal vector: (1,1,1); tangent plane: x +y+z= e3-1
19.
Normal 2z=2
21.
(x - 2)/12=(y - 1)/(-5)=(z - 7)/(-1)
23. 25.
x - 7r/4= y= z - 7r/4 x=O,y - 7r=1 - z
27.
5.26 0.9.
vector:
(6,6,12);
tangent plane:
x+y+
37.
(1+3v'3)/S
39.
a.
(3/v'IT,-1/v'IT,1/v'IT)
b.
4v'IT
29.
c.
(-3/v'IT,1/v'IT,-1/JiT)
d.
- 4v'IT
31.
3750 ft2
a.
(o,1;vz,-1/VZ)
V2
33.
1.0S cm3
-vz
35.
The summand f(l, - 1) has been omitted; the correct equation is z=2+6(x - 1)+6(y+1).
37.
The coefficients of x and y should be the evaluations fx(Po) and fy(P0), not fx(x, y) and fy(x, y); the correct equation is z= - 3x+(y+3).
41.
(o,-1;vz,1;V2) 45. 47/15 c.
b. c.
47.
-2/5+67r/5
49.
fx(P) = 3; fy(P) =-1
51.
fx(P) = 3y'3 - 3; fy(P) = -3y'3+3
41.
7r/2
55.
135s-1f2(-25,-27,-2)
43.
7r/4
45.
SOp +9r < 2112/25 where 100p is the percentage of gold and r is the radius
47.
x +9y+z= - 1
49.
6x - 13y+12z=- S
51.
3x - 6y+4z= S
53.
2x+y+z=3
55.
2x - y+z=7r/4
61.
y
Section 11.8 1.
Local maximum at (1,2)
3.
Saddle point at (0, O); local maximum at ( -2, 2)
5.
Saddle point at (0, O); local minimum at (1,1)
Review Exercises for Chapter 11 7. Saddle points at (0, 0), (0, -3), and ( -3, O); local max
imum at (-1, -1)
43. (x(lOOOO),y(lOOOO)) = (8000/21,2500/7);
M(lOOOO) Ri 31296.961; .\(10000) = (2500/7)3/4/15 Rl 5.477; (x(l0001),y(l0001)) = (40004/105,10001/28); M(lOOOl) Ri 31302.438; M(lOOOl)- M(lOOOO) Ri 5.477 (the same value as .\(10000))
9. Local minimum at (3, 3) 11. Saddle points at (0, - 1) and (2, - 1) 13. Saddle points at (0, 0), (0, - 8), and (-8, O); local max
imum at (-8/3,- 8/3) 15. Saddle points at (-2, -1) and (2, -1); local maximum at
(0, -VS); local minimum at (0, VS) 17. Saddle point at (0, O); local minima at (--/2, --/2) and
(-/2, -/2) 19.
51. 3.15694 53. 0.0342767 55. Minimum: -1.01006; maximum: 2.45745 REVIEW EXERCISES FOR CHAPTER 11
Local minimum at (0, 0)
1. A(h,r) = 27rrh, domain of A:{(h, r): 0 < h; 0 < r}
21. Each number is 100/3. 23. Base: a square with side length 4/3113 in.; height: 4. 3213 in.
3. P(£1,£2) =£1 +£2 +
25. Equilateral triangle with side length 20/3114
P:{(£1,£2):0
27. Regression line: y = 0.00l956x + 0.2577; y(lOO) Ri 0.4533
g/mi
5.
{0 sin (3)
Jti +£�,domain of
< £1; 0
<
£2}
7. 83/2
29. y = 6.12x + 21.2 31. Least squares line: ln(y) =- O.l4t + 0.0093; y(lO) Ri 0.25 39. Saddle point: (-0.50608337, 0.47573876)
9. 2 . 33/2 11.
41. Critical point: (0, 7r/2), which is a global minimum
4
y
x
Section 11.9 1. Maximum value: 6 + v'34; minimum value: 6- v'34 3. Maximum value: 25; minimum value: 11/3 5. Maximum
value: -128/v'3- 24
128/v'3- 24;
minimum
-4
value:
7. Maximum value: 37/4; minimum value: -3
13.
y
al/t.
9. Maximum value: 16-/2; minimum value: 16 lL Maximum value: 8/(3v'3); minimum value: -8/(3v'3) 13. 223/22
15. -4
17. 4v'3/3
19. 3
21. 144
23.
25. 256-/2/3
27. 3
�
J273
29. Highest: 150 + 180-/2; lowest: 150- 180-/2 31. 3
�27fkPo times the unit side cost
33. 15617
35. 1117
37. 420
39. x =pT/(A(p + q)), y = q T/(B(p + q))
A-19
x
A-20
Answers to Selected Exercises
15.
23.
y
a
x a
elliptic paraboloid 25. x
27. 29.
. x2-2.x +l lim(x,y)-+(1,1) xy 2x y + 2 _
_
=
. x-l lim(x,y)-+(1,1) y 2 _
=
0
. sin(t) . sin(Sx- 3y) - hm1)0 -- - 1 hm(x,y)-+(3,5) S t x 3y t -t . . -t 1 hm(1,t)-+(O,O) t + t = 0, lim(1,2t)-+(O,O) 3t = - 3 _
_
_
31. 0, 0
1 7.
33.
e,
35.
f, sin(x)sin(y) ti sin(y)[sin2(x) + 1] x = cos (x) ' xx = cos (x)
0
2
3
'
cos(y) -sin(y) ffyyy cos(x) , cos(x) _
_
Ji cos(y) sin(x) ti xy = yx =
x
cos2(x)
37. 19.
fx = 3x2yz,fy = x3 z - z cos(yz),fz = x3y - y cos(yz) fxy =3x2, cos(yz),fyz = x3- cos(yz) + yz sin(yz) fxyz 3x2Jzz y2 sin(yz) fx = zeY2x + xzyZey2x ,fy = 2x2yzeY2x ,fz = xelx fxy = 4xyzel x + 2x2y3zeY\fyz = 2.xzyerx fxyz = 4xyerx + 2x2y3eY\fzz = 0 fxyz fyxz fzyx fzxy fyzx fyxz cos(xy)- yx sin(xy) =
39.
y
21.
41.
=
=
=
=
43.
dz ds
45.
-2 (sin(s)- cos(s))2
47
•
49•
51•
dw ds
=
sZ
set - 2te'ls se1- t2ezY s2ln(s)cos(s) - s2e'cos(s) + s2 sin(s) - s sin(s) (sln(s)- se')
et-2t2e'ls {)z set- t2ezY ' 8t
{)z 8s
ds
=
4s3- 6s2 + 2s - _!_
=
- =
dw
=
=
=
1
...,,.--, ,. -= -
;
- ln(s) -2s3ln2(s) -2s3ln(s) + 6s2e6s sln(s)- s41n2(s) + s2e6s
Section 12.2
53. 55. 57. 59.
61.
1 (x-2)2 2 O+(x-2)-2(y-l)-2 2(x 2) (y 1) '\lf(x,y) = (2x sin(xy)+x2y cos(xy),x3cos(xy)) '\lf(x,y) = (yzexz-y2+xyz2ez-y2,xzez-y22xy2zez-y2,xyez-y2 +x2yzez-y2) D,.f(P) = VZ direction of greatest increase is u= Duf(P) = 4M, direction of greatest decrease is = Dvf(P) = -4M
-3(y-1) +
- -
\ Jm, Jin J,
\-Jm, Jin), 63. direction of greatest increase is -\ 1 2...fa ), Duf(P) v
-
u
v'T+47f 1+47r v'T+47f 1+47r direction of greatest decrease is 2...fa 1 f(P) v v'T+47f v'T+47f 1+47r Dv 1+47r 4 D f(P) = 3 -
-
,
-\
,
-
65. 67.
- r;-:-;o -v
-
)
,
1+47r,
-vr;-:-;o
1+47r
v
direction of greatest increase is I 48 112 92 \ D,.f (P) = v'23312 u= \ ' v'23312' v'23312' v'23312 I' direction of greatest decrease is I 48 - 112 - 92 \ = \v'23312' v'23312' v'23312 I' Dvf (P) = - v'23312 direction of greatest increase is v'3 /1 1 1\ u = \ v'3' v'3' v'3 / ' D,.f(P) = 2, direction of greatest decrease is v'3 I 1 - 1 - 1\ f( ) = = \ v'3' v'3' v'3 /' Dv P -2 v
71. 73.
75. 77. 79.
81. 83.
-
2...fax+z = 27r 2x Sy l6z = 26 x- y = 1- ln (2)
+ + + {:�:- ; +
z = 0 lt f (3.9, 5.9) 9.42 f(l.1, 0.1) 0.15 critical point is (3/4, -2/3), positive discriminant with equal, positive fxx, fyy• hence is a local minimum �
�
87. 89.
91. 93.
critical points (4,-1) and (4,0), at (4,-1) the discriminant is negative hence a saddle, at (4,0) the discriminant is -1 hence a saddle 100113, 100113, 100113 ( ffi6 125 )' critical points are ( ffi6 12 5 V ZS' ' V ZS' (0,6), (0,-6), the first two are minima and the third and fourth are maxima critical point is (8/9, -16/9), and it is the minimum critical points are (v's,O), (-v's,O), (0,5), (0,-5), the last two are minima
- ) -
_
Section 12.1
1. 5. 7.
v
69.
85.
A-21
9.
lL
15.
19.
23.
27. 29. 31. 37.
39. 41. 45. 49.
3. 72 56 I(x) 4x+24; :J(y) 2 4y I(x) = 12x2; :J(y) = By 28 I(x) = (e 1)/(2x) ; :J(y) = (ey2 -l)/y 161/2 13. 0 17. -ln(2) e3 - e2 - e+1 21. 5/3 0 25. 4/3 n2/8 1/12 (e+ 2)1n(e+2) - e - 3 ln(3) 2e- 2 33. (e4- 1)/8 35. e- 2 424 ln(2)-105 4(sin(l)- cos(l)) 43. n2!32 (e4 1)/4 2 47. 1 51. 2.231 0.2248 =
=
++
-
-
Section 12.2
1. R is x-simple only; boundary curves: a1 (y) = - 2y- l and az
(y) =-y2 2 for -1 :5y :53
+
y
A-22
Answers to Selected Exercises
3. n is x-simple only; boundary curves: 0:1 ( y)
=
y
y2 anda: (y) 2
=
-2y2-3y
Section 12.3
+ 18 for -3 ::5y::5 2
1. 1029/4
3. 99/4 7. 375/4
5. 48/5 9. 12
x
11. 728
13. 813/2
lS. 1960/3
17. 108 21. 1288/15
23. 2 sin(l) - sin(2)
25. 136/15
27. 2744/3
29. 4576/675
31. 64/3
33. 2/3
35. 817r
37. 1/8
39. Jr/2
19. 4e2 + e-3
41. ln(54/25) - 2 arctan(2) 43. 1/8
+
Jr/2 45. 4.7918
47. 3.4931 Section 12.4 5. n is y-simple only; boundary and,B2 (x) 6x -1for1::5x::53
curves: ,B1 (x)
=
x312
1.
s.
______
I I I I I I I I
=
y
y
_l_ .J2
--------
·p I I I I I I I IX
3
10
.J2
I I I I I
1 7. -8 11. 47r -4
3
2
S
=
13. 5/../2 17. e2(13e6 -7)/4 -33/2
19. 640 -145V3
21. 5/6
23. 1
25. e4 -1
27. 1
29. 1112
31. (../27r-5)/4 35. 11115
33. (1 - sin(l))/2 37. -1124
39. 9/4
41. 2/3
43. 114
45. (143a
47. 68/15
49. 2/3
51. 116
53. 113
y
3.
P Q =
+ 91b)/3
57. 6.6525
59. 0.27694
61. 0.62
63. 0.66 + 4../2 + 3vJ)/128;
P (o, f)
Q
R
S
=
65. Exact: 8/15; approximation: 17/32; absolute error: 0.00208 67. Exact: 4/35; approximation: (3
--------eR
R
9. 78../2/5
15. 112/15
lute error: 0.0060591
Q·----- - .J23 P=(3,*) Q=(-3,*) (3,-i) =(-3,-i)
abso
I I I I I
=
(0, -'IT)
=
=
=
R
=
(O, 'IT)
(o.-f)
S
x
Section 12.4 5.
4
y
y
33.
P=S
x x
-1
-4 Q=R p = (4 ,- 7 2) Q= (-4,-72) R = (4 , 72) s = (-4, 72)
r
=1
-
cos(B) y
35.
y
7. S•--- {3
----
•
I I I I I I I I
I I I I I I I I
-1
1
p=
(2, 7;) R = (2,7;)
x
x
I I I I I I I
I I I I I I I
{3 Q•-- -
P
____
r
= 3 cos(29)
_.R
( 7;) S = (-2,- 7; )
Q= -2,
y
37.
x
9. (2v'2 ,2v'2) 13. (2v'3,2)
11. (0,0)
-2
15. (-3,-3v'3)
17. (4v'2,-7r/4+2mr),n E Z; (-4v'2,37r/4+2n7r), n E Z
-4
19. (9, -7r/2+2n7r),n E Z; (-9, 7r/2+2n7r),n E Z
21. (1,7r+2n7r),n E Z; (-1,2n7r),n E Z
23. Symmetric with respect to origin, x-axis, y-axis 25. Symmetric with respect to origin, x-axis, y-axis 27. Symmetric with respect to origin, x-axis, y-axis 29. Symmetric with respect to x-axis 31. Not symmetric with respect to origin, x-axis, y-axis
r
4
= - 2sin(8)
A-23
A-24
Answers to Selected Exercises
39.
y
x
4
rectangular coordinates: (0,0), (1/4, v'3/4), and (1/4, -v'3/4)
49.
In
51.
In
rectangular coordinates: (±1/v'z, ±1/v'z)
53.
In
rectangular coordinates: (0,0) and (-1/4, ±v'3/4)
55. The points of the x-axis 57. v'2(exp (7r/2) - exp (-7r/2)) 59. 7r/4 - 3v'3/8 61.
r
=
-1 + 3
y
cos(O)
x 41.
y
r =exp 63.
(� ) y
x
r
3 sin(O)
=
43.
-
2
cos(O)
x
y x
-2
-4
r
45.
=
3
-
r2
3 sin(O)
( J9 + 7r2/4, arctan(7r/12) + 2n7r) and (- J9 + 7r2/4, arctan(7r/12) + (2n + 1)7r) for n E Z b. ( ./36 + 497r2/4, -arctan(77r/12) + (2n + l)7r) and (- ./36 + 497r2/4, -arctan(77r/12) + 2n7r) for n E Z c. (7rV1033/12, -arctan(32/3) + 2n7r) and (-7rV1033/12, -arctan(32/3) + (2n + l)7r) for n E Z
= -3 sin(48)
65.
a.
47. Distance:
circle:
V'1
+
� - 2r1r2 cos(01 - 02); polar equation of
Jr2 + 9 - 6r cos(O- 7r/4) =1
y
y
x
r
,- - -
= 92
9'lT < () < 9'lT g g
x
()
=
� 0 :5 () :5 2'lT
r
Section 12.7 67.
A-25
Section 12.6
y
1. 5. 9. 13.
27 3 cos(,,fl-)
16/15
3. 252 7. Jr- 4 11. 307r
-1
15.
247f
17. 1287r
1
19. 3; U is the region in the first octant cut off by the plane
2
3x+2y+z=6.
21. 1/15; U is that part of the unit ball x2+y2+z2 � 1 that lies in the first octant.
23. 1/8; U is that part of the cylinder x2+z2 � 1 that lies in the
x
1
first octant between the planes
1
-2
r112
=
2
sin(9)
41.
69.
y
7
6
x
4
43. 2
9 r
=
9+
cos(O)
y =0 and y =x.
35. 7rm6(3m2 +4)/60 39. 216
33. 7r (i2 +b2 +4c)2/32 37. (1 -cos (l))/24
Jg J: Jgf(x,y,z)dzdydx; Jg fci fgf(x,y,z)dzdxdy; Jg Jg J:f(x,y,z)dydzdx; Jg Jg J:t(x,y,z)dydxdz; Jg Jg fcif(x,y,z)dxdydz; Jg Jg fcif(x,y,z)dxdzdy r0 31 rv'X+I yv� 1 -xdydzdx; J -1 f2 Jx+l 31 r0 rv'X+I yvl � -xdydxdz; f2 J-1 Jx+l 3 1 v'X+I � r rO J-1 Jx+l f2 yvl -xdzdydx; 1 Jg J�-=_ 1 Ji1 yv'T=Xdzdxdy; rl 31 ,y-1 1 -xdxdzdy; Jy2-1yv� Jo f 1 1 Ji Jg J;2-=_ 1 yv'l -xdxdydz .
5 r
=
8 + cos(8)
2
Section 12.5 1. 37r/2
3. 47r
5. 2
7. 357r 11. 7r/6 +../3/4
9. 7r/8 -1/4 13. 7r +4
15. 317f/2
17. 157r
19. 2/3
21. -4v'2/15
23. 2/3
25.
29. 327r
27. 87r 31. 407r/3
33. 187r
35. 32(2v'2 -1)7r/3
37. 327r/3
39. 29/210
4L 248/27
43. 52/9 + (32/3) ln(2/3)
45. 127f
47. 77r/12 -../3
49. 7r/3 -../3/4
51. 7f -3../3/2
53. 2(vts-1)7r
55. 311137r/16
57. ,,fl- /32 -1/8
59. ln(2)
lh
63. 00
�
65. 0.24
0.967025;
area:
2.4393
47. 2.0333 Section 12.7 1. 108
3. 3
5. 2
7. 85 11. -(26/3)1n(2)
9. 1027/2 13.
(183/448,65/128)
15. (156/55, 29/55)
17. (k/2,2k/3) where k =(21n(2) -1)-1 19. 5/3
21. 13
23. 1/2
25. M =727r; Myz = O; Mxz = O; Mxy =5767r; center of mass
:
(0,0, 8)
27. M=67/8; Myz=-193/40; Mxz=25/24; Mxy = 389/60; center of mass:
(-193/335,25/201,778/1005)
29. M = 5/8; Myz = =31/120; Mxz =31/40; Mxy = 31/30; center of mass : (31/75, 31/25,124/75)
48 ly-o 1056 ; - = 5 35 33. fx=O =1/10; fy=O =13/15
31. /Fo =
A-26
Answers to Selected Exercises
35. 11
37. 4
47.
39. Ix = 650 I = 1160 lz = 1530 41. ix = 186 I = 194 I = 4
(1f 2,16 (9 ))
43.
45. ( e - 2, e2+1)/
)
47. 'R. ={(x,y):O
vi:;
-
7- 2
2 p= ./fx/M
51.'
55.
3.52217) J.48447)
Sec
is
53.
v,31f 4,-1)
1. 5.
·2)
9.
I)
7. (2, 5?r/3, - 2)
,, ,1f/ )
13.
11.
v,1f/ ,0)
15.
v,51f 6,31f/ )
17.
lar coordinates: -3v'3/4 / 3 2); cylindrical es: 3v'3 2,21f/3,3 2)
19. :··
:tl coordinates: 2./2,7/ 1f,-2./2); spherical es: 4, 7r/ ,7 /4)
21.
lar coordinates: -./2, -./2,-2); spherical es: 2VZ,37r/4,51f 4) :tl coordinates: (3, 0, - 3); spherical coordinates:
23.
!.6527
49.
REVIEW EXERCISES FOR CHAPTER 12 1. 9
3 ·
. 4 0) 25. 32?r/3
27. 51"'T /15
29. 101f ln(2)
31. H
�
33. 651f
35. -:
..
37. 21f
39. 327r/3
41. Rectangular coordinates:
3 cos(5) sin(2), 3 sin(5) sin(2), 3cos(2)); cylindrical coor dinates: (3 sin(2), 5, 3 cos(2)) 43. Points in the xz-plane 45.
z p
x
=cos(¢)
OS
=-i
- 33 32
2x - 7r) - COS 2x), OS 21f - y) i OS y) 9. e3 - e2
4 11. -2
13. y-simple, ,ih x) = x - ,/3 x) = x +1
2
+1
Section 13.1 15. either x-simple or y-simple,
0:1 (y)
=
y2,0:2(y)
=
2.JY,,81 (x)
=
(x/2)2,,82(x)
45. =
Vi
47. 1
49. 7r
51. 27r
53. �
8
�. 410/3
19•
4 15
21. 7r� -1
23. �(1 -e-1)
25. 56
27.
1 8 �
57.
V'2
59. 10007r 3
6L
63• ?__
65•
.
2 255 7r 67• 8
•
73
(2, 7T/4)
83. 42 (1, 3 7T/2)
•
87. (
(-4, 2 7T/3) (0,7)
33. •
(-4,0)
(0,-7)
35. (0, 3) 37. 77r/4 39. arctan(4/7r) 41. symmetric in the origin 43.
69. 327r
.
31.
•
-
441 ·2 1126 6 213/3 77• 75• 105 13 135 1723 -1632..;'2 63 1119 79• ( ) SL ( ) 576' 288 588 - 352..;'2' 2058 - 1232..;'2 7L 2
29. 127r
5./3 16 7r --- 29 6 4 -/ 32 2 15 19 5 - 1 3 - 7 6e 15 30 e +
3
55. 3 17.
A-27
�' �' i�)
89. (v's, arctan(-2),./3) (4,0)
85. (� ,
�;, �;)
9L (5,0 ,0 )
93. (6..;'2,7r/4,27r/3)
95. spherical: (4..;'2,?r/4,?r/4), cylindrical: (4, -?r/4,4) 97. 7r/6
99. 37r;10
647r lOL 5 Section 13.1
In the answers for this section, A, B, C, Ci, and C2 represent arbitrary constants.
A-28
Answers to Selected Exercises
y
7.
y
1.
x
F(x,y)
=
3i + j
y
3.
9.
av'x2+z2j for -4
a(yi - xj) for x2+y2< 16, 0
""'- ""'- ""'- ""'- ""'- ""'\
\
\
\
\
\
I
I
I
I
I
I
////// �� F(x,y)
=
2yi - 3j y
5.
I t t t \ \
" \ \ \ • ;
"' I i
(3x2 /(x3 -y2 ))i+( -2y /(x3 -y2 ))j
17.
2xy3zi + 3x2y2zj + x2y3k
19.
((x - y)ln(z)f1 i - ((x - y)ln(z)f1 j - (zln2(z )r11n(x -y)k
21.
x+ y+ C
23.
xy + C
25.
xy3 + x2y + C
27.
x2 + y + C
29.
x2yz3 + Sy + C
43.
xz(y + zt1 + C
31. xz3 -x2y2 + yz2 + C C1e1i+(2t+ C )j 35. C 33. (t+ C )i+(t+ )j 2 2 1 37. (t2 + 2C t+ C )i+(t+ C )j 1 2 1 41. ysin(xy) + C 39. x exp (2x)+ y+ C
' ' =
45.
xY + z ln(z) -z + C
49.
-r-3 (xi+yj+zk) 51. y
-/'//I
t f I I , I
\
F(x,y)
x
15.
..___
'
�
\\
+
t
f
'
yi - 5xj
t
I/ J.
I It ,,,_ ,., I t� t
---¥' /I
�\x
�
I
/
"'
"
" --...
t t /// /--- f /////�53. 55.
(Aexp(2t)+Bexp(3t), -Aexp(2t))
(Aexp( 4t)+Bexp(-t), (-A/4)exp(4t)+Bexp(-t)) 57. - yi+xj
Section 1 S.5
Section 13.2 1. 5. 9.
41.
0 67/896 -115
3.
1069/35
19. 813
21. 2111
25.
3
7.
y/ x+y)2 - x x -y)2
9.
- x/y2)s�(x/y) + y x:2) sc 2 y x) 13. 14
11. 7
15.
Potential function: V
-; - r/2+1rX+y)
3.
-:-x:y3+y7
5.
Potential function: V F is not closed on Q.
7.
Potential function; v·
x - r1 - y
1/2) 3I + c
r - y)+C
=C -z2/
Potential function: V
19.
F is not closed on Q. Potential function.: V x,y, z) = C -zta
--)
e'' 27. -
37.
f--,. F· dr= _:. F· r= 3'· f.. 'o 1P2 f.._ \ F · dr = ';f. �F · :-r= l;J-;:;t F · dr = -1 {(c,d): c;:!;; -1,
d
55.
is harm.a
61.
1 21r) J21r -2.69107S
�·
: ))cos(3+sin:: )) :• t= 3)
1. -37r-
3.
271"
s.
7.
0
9.
29. Jr r t F· dr= · �1 '
39.
39.-cos x) i 4L i+zj+ y
Section 13.5
exp( -1) - 2
1} {(c,d): -1
0
y+z x+ 45. -y-3 2Ki+
23. 3/4
_
25. -x os xy)i+yc os xy)j - (ysin xy)+xsec2(:xy))k 27. Oi+Oj+ k; closed 29. Oi+Oj+ k; closed 31. xsin z) +sin(x)) -ysin z)j+ x in y) - z: os x))k; closed 33. 0 3 z+1)/{z -y ) 35. 2K 3y2
43.
=C -(xyz+x+2y+3z)
17.
-e-y
x+ye>'Z)i+ 1 (x+ z) -y)j
23.
37.
:OS
x:-z
17.
-ysin(xy) -zcos(yz) 21. xzi + x yz)j
C represents an arbitrary
1.
2 (x+y+z)
19.
Section 13.3
31.
2Ksin(y) os(y) -sin{x) - 2Ksin(:xy)cos(xy)
-3
In the answers for this section, constant.
25.
3. 5.
27.
•
2L
2xy3
179/180
27f' 3t. 65v'65 - v) 6 33. 4 35. 37 .ft F·dr = -4;£ '" · r =f F· 3 = -2 1 ) 2 _ 39.27f' 41. 45. 3.33904 43. -'lr/4 47. 1.54122 49. 214.420
Potential function: V ll. F is not closed on Q. 13. Potential function: V 15. F is not closed on Q.
1.
23.
29.
9.
0.6118
Section 13.4
255/4+ln(1117) 11. 3/4 15. 2 7.
13. 7r/2 17.
2 -sin(• m)
A·29
13.
1r
3rr/4 121n(3) - 201n(2)
17,
c, max{c, 1}
d}
,.
·�:
21. .
ll.
-3113
15. i.4815 19. 1-413
I
3 -vJ/4
25. 811693 29. -1/30
23.
a
27. 1'11' 31. 413
• ot
A-30
Answers to Selected Exercises y
-22/3
11.
-24
13.
-97r
15.
637r
17.
-4
19.
-37r/4
21.
87r
23.
7r/2
25.
1757r/4
27.
-6Jr
29.
207r
31.
-Jr
35.
187r
37.
-47r
39.
-3.550999
41.
-0.4626517
9.
1
x
Section 13.8
-1
r(t) 5 sin(t)i + sin(2t)j 33.
3/2
37.
-0.070393
39.
5.3755
1.
0
3.
7r
5.
7r
7.
1
9.
-8
11.
4/15
13.
457/r 4
14.
65
17.
327r /3 - 4v'37r
19.
256/15
27.
287r/3
35.
37.
47rk
4L
1.68773
(7r/6)i 39. 2.81762
Section 13.6 REVIEW EXERCISES FOR CHAPTER 13
1.
3J26
5.
(17312 - 1)7r/6
9.
9v'38
11.
(145312 - 65312)7r/24
13.
57rJ74
15.
847rv'2
17.
4(275 - 6512 - 7512)/15 35747r/5
19.
27rv'I9
21.
23.
337r
25.
52/3
27.
135v'59/32
29.
l� l ;
31.
l� l�1 y'4exp(4u)+exp(2u+ 2v)+exp(2u - 2v)dvdu
33.
21J26
37.
27r l: f(x)
41.
27r/3
45.
(0,0,(289vTI- 41)/(170 vTI- 10))
47.
27ra2
49.
2Jra2+27r(ac/e)ar csin(e) where e = v'c2 - a2/c
11.
53.
192
u(x,y) =x2y/2
13.
55.
8.510466
u(x,y,z)=xz+y
15.
u(x,y,z)=xyln(z)
8v'27r 7. (373/2 - 5312)13
3.
1.
3.
J1+4u2+(4uv+1)2dvdu 35.
8v'3
43.
(0, 0, 8/3)
Vl+f'(x)2dx
57.
4.74121
v=
7.
\lu = (y2 - 3x2y+1,2xy - x3 )
9.
\lu = (3x2ycos(x3y),x3cos(x3y))
r(t) = e2'i - 3tj 19. 1 21. sin(7r2) - sin((7r/2)2)+1 17.
Section 13.7 1.
136
3.
2 - 2/e2+ln(2)
5.
-Jr
7.
-42
-(y2 + z2 + (z+5))i
5.
Review Exercises for Chapter 23. 2-i3 /3-sin('rr2)
53. ( - xz - e Y+z)i + (yz)j +
25. -413 27
•
55.
l._ . 32/3 + _!_
_
29. 11
10
10
57.
4 2 173/ 33. 4
12
653/2 4
___ _ _
35. 37. 39
8M
oy
8M
{)y
8M
8N
=2y, = 1,
ox
=
-2x,
2y #
-2x,
1 (-) k x-y
curlF =(xz + y)i - yzj - xk, not closed curlF =(xcos(xy) + y sin(yz))i - (ycos(xy) - xy cos(xyz))j
67.
=0, 1 # 0, not closed
8N
8M
8R
8N
8R
'
41. 74
lp;;p1 F·dr =4e6 -2e3, 1& F . dr =e5 2 45• l&o F·dr = -e5 + 2e3, the sum is 0 47. � + _q_
x2-y
y2-x 49. �2z -2e:-2Y
51. ycos(xy) + zsin(yz) + xysec2(xyz)
'
3i -
73. 8
75. 2 ln(3/4)
77.
24v'46
79. 3v'3'h
81.
127f
83. -4..ffi
1:
l� v'8u4 + l2v4 + 4u2v2 + 8uv3, dudv
87 .fo/2l;jf3 - 4e6,
-7f
69. e4 - 1
71. 47r
85.
lp;;p1 F·dr =7,1& F·dr=19, 2 43• l&o F·dr =-26, the sum is 0
+
65. the region is a disc, the integral equals
not closed
=z = = =y =x closed ox ' oz ox ' oz 8y u(x,y,z) =xyz - 4x + 6y + 2z
• 8y
=
ox
8N
A-31
(-xzcos(xyz))k, not closed 59. div(grad(u)) = [2 - 4x2 y2 - 4y2z2]sin(z2 - x2) 61. both sides equal 0 63. both sides equal -7f/ 4
31 � - ! - 5V2
• 8
13
•
89.
� v'6
93. -
Jsin2(v) + cos2(uv) ·[u2 + v2 -2uv + v2sin2(v)] dvdu 91.
0 95. -sin(l) + cos(l)
.� Vls[6 + e3 -e-3] 17 v
97. 647r
101. ln(2)
99.
0
This page intentionally left blank
A Abel,Niels Henrik,965
calculations, 1000-1003
cross product calculations,759-760
939, 1003-1004
basic principles,807-811,851-852
regions calculations,polar coordinate integration,1000-1003,1038
motion applications,834-847
scalar-valued functions,several
acoustics,mathematical physics,
variables,863-864
964-965
surface area,integrals for,1105-1107
addition,of vectors,724-725
Ampere,Andre Marie,1045, 1129 Ampere's Law, 1129 angles direction angles,dot products,749 between two planes,768-769 angular coordinate, 991-992 anticommutative cross product,756,787
parameterization,1112-1113 area function,planetary motion and,847 Aristotle,astronomy of,856 associative law of multiplication,788
Astronomia Nova (Kepler),857 attractive force,central force fields, 837-839 average rate of change,functions of two variables,861-862 average velocity,vector-valued functions, 804 axes,of ellipses,839-844 axis of rotation curl of vector fields, 1089-1090
antiderivative(s)
normal component of curl,1118-1119
iterated integrals,970-974 spatial vector fields,1083-1083 vector-valued functions,801 antidifferentiation acceleration vectors,808-811 potential function location,1058-1059 vector-valued functions,801 aphelion,54 approximations double integrals over rectangular regions,968-974 numerical,926-928,959 tangent plane,numerical approximation,926-928 Archimedes Law of Buoyancy,1135-1137 spiral of,994-997 arc length curvature parameterization,824
buoyance,Divergence Theorem and, 1135-1137
polar coordinate area calculations,
derivatives,793
of vectors,724-725
regions, 1124-1128
of planar region,980-982,1038,1047
vector-valued functions
exterior algebra,787
piecewise smoothing,bounded Brahe,Tycho,844,856-857
area
853
dot products,741-742
Stoke's theorem,1116-1118
vector field,1052-1053
tangential component of,836-837,
of cross product,755-756
1094-1102
polar coordinate integration,regions
acceleration
aerodynamics,baseball physics,810-811
negative orientation,1094-1102 positive boundary orientation,
osculating circle, 828-831
defined,932 least squares lines,942-944
algebra
Green's theorem
1068-1069
absolute minimum
normal component of,836-837,853
Divergence Theorem and, 1131-1137
directed curve,line integral,
absolute maximum,933
instantaneous,807
curve parameterization, 818-820 defined,814-815,852
c calculus. See also Fundamental Theorem of Calculus; vector calculus of variations,788 vector calculus,1049 Cantor,Georg,965 cardioid,995-997 Cartesian coordinates point distance,1032 polar coordinate system,990-997 transformation to,1011-1012 vectors in three-dimensional space, 732-733 Cartesian equations circle of curvature,832 lines in space,771-775 planes in space,766-769 tangent planes, 923-924 Cauchy,Augustin Louis,963 Cauchy-Schwarz inequality,dot products and,744 Cayley,Arthur,789
B
Cellerier,Charles,1045
baseball physics,acceleration and
center of ellipse,839-844
velocity, 810-811
center of mass,1023-1024,1040-1041
Bernoulli,Daniel,964-956
surface integral for,1111-1112
Bernoulli,Johann,964
three-dimensional space,1025-1027
best-fitting line,941-944 binormal vectors,821-822
central force fields,motion applications, 837-839
Bolzano,Bernhard,1045
centrifugal force,858-859
Bonaparte,Napoleon,1146
centripetal force,836-837
boundary curves
centroids, 1025-1027
Green's theorem,1094-1102 order of integration and,979-980 planar regions binding,976-979 Stoke's theorem,1116-1129 volume between two surfaces, 986-988 boundary orientation
Chain Rule conservative vector field,path independence, 1072-1083 differentiability and, 900-910 function of three variables, 905-907 function of two variables,903-905
1·1
Index
1-2
Chain Rule
(continued)
Taylor's formula, several variables, 907-910 directional derivative, 914 functions of several variables, 958 gradient and, 919 potential function, 1056-1057 partial derivatives, 892-893 functions of three variables, 893-896 Stoke's theorem and, 1122-1123 tangent planes, 922-924 Taylor's formula for several variables and, 908-910 vector-valued functions, differentiation, 799-801 Christian IV (King of Denmark), 856 circle(s) curvature of, 825, 832 cylindrical coordinates, 1030-1031 circular paraboloid, 877-878 Clairaut, Alexis Claude, 963
vector fields in space, 1080-1083 consistent orientation, Stoke's law and, 1117-1118 constant vector, baseball physics, 810-811 constraint curves, Lagrange multipliers, 946-954 functions of several variables, 952-954 geometric approach, 947-949 scalar equations, 949-951 solution strategies, 951-952 continuity defined, 886-887
independence, 1071-1083 Green's theorem, 1095-1102 line integrals, 1067-1068 closed surfaces, Stoke's theorem, 1117-1118 closed vector fields, 1076-1080, 1140 Green's theorem, 1098-1102 in space, 1081-1083 coefficients, tangent planes, 923-924 complex numbers, 787 components of vectors, orthogonal projection, 747-748 composition of functions, combined functions, 864-866 compressible fluid, vector field divergence, 1088
problems, 938-940 critical points. See also maximum minimum problems defined, 959-960 Lagrange multipliers and, 947-951 cross product area calculation, 759-760
vector-valued functions, 795-797 continuous functions
baseball physics, 810-811 determinants and, 754-755
continuity of, 886-888
geometric understanding, 756-759
scalar-valued functions as, 795-797
history of, 789
continuously differentiable function, partial derivatives, 895-896 contour, contour line, and contour curve, variables, 867-870
conservative vector field, path
to space curves, 806-807 cost function, maximum-minimum
algebraic properties, 755-756
graphing functions, of several
distance calculations, 735
region area calculations, 1001-1003 vector-valued functions, tangent line
rules for, 887-888
Clifford, William Kingdon, 789
closed curve
function integration, 1004-1006 graphing, 995-997
limits and, 957-958
Clairaut's Theorem, 963 closed ball with center set, vectors,
polar coordinates
convergence, radius of, 828-831 coordinate systems
physical applications, 760 planes in space, 770 two spatial vectors, 753-754, 784 vector-valued functions continuity of, 797 differentiation, 799-801 curl, 1049, 1141
angular coordinate, 991-992
Green's theorem, 1098-1102
Cartesian coordinates, 732-733
identity, 1091-1093
cylindrical coordinates, 1029-1031,
normal direction component,
1041 polar coordinates graphing in, 994-997 integration of, 999-1012 iterated integral area calculations, 1003-1006 radial variable, negative values, 992-993 rectangular coordinates and, 993-994
1118-1119 vector field, 1088-1090 current density, magnetic field, 1128-1129 curvature center of, 829-831 smooth parameterization, 825-828 vector-valued functions, 824-832, 852-853 osculating circle, 828-831
region area calculations, 1000-1003
planar curves, 831-832
symmetric graphing principles, 997
smooth parameterization, 825-828
variable and Jacobian changes, 1006-1012 rectangular coordinates, 1029-1031
curve(s) arc length parameterization, 818-820 area between, double integrals, 982
concavity, saddle points, 935
integration problem, 1005-1006
closed curve, 1067-1068
cones
polar coordinates and, 993-994,
conservative vector field, path
cylindrical coordinates, 1030-1031 elliptic, 876-877 polar coordinate integration, 1006 surface area, integrals for, 1107 conservative vector fields, 1055-1057, 1140 closed vector fields as, 1076-1080
1010-1013 spherical coordinates, 1031-1035, 1041 Copernicus, 844, 856-857 coplanar vectors, triple scalar product, 762-763 cosine(s)
independence, 1071-1083 constraint curves, 946-954 directed/oriented curve, 1063-1066, 1067 finite curves, planar regions binding, 976-979 Green's theorem, 1094-1102 integral (streamlines), 1053-1054
identities, 1091-1093
direction cosines, 749
path independence, 1071-1083
dot products, 743
isothermal, 870
potential function location in,
iterated integral calculations, 973-974
level curves
1057-1059
planar curves, 831-832
gradient and, 917-918
Index
Lagrange multipliers,949-951 line integrals,1061-1069 curved paths,1061-1063 path dependence,1066-1067 mouse-to-elephant curve,944 parametric curve line integrals,1063-1066 vector-valued functions,792 planar curves,831-832
identity,1091-1093
Taylor's formula, several variables,
theorem of,1131-1137
907-910 vector-valued functions,797-801
vector fields,1085-1088 del notation,1090-1091
reparameterization,817-818 differential equations
Divergence Theorem,1131-1137, 1143, 1147-1149
potential theory,1146
in physics,1134-1135
streamlines,1053-1054
proof of,1136-1137
differentiation acceleration vectors,808-811
domain,of functions,863-864
polar coordinate graphing,995-997
curvature definition,824
dot product and applications,
space-filling curves,1047
partial derivatives, 891-893, 896-897
motion applications,835-836 tangent line to,805-807 tangent vectors,814 vector-valued functions, tangent line to,805-807 velocity vector tangent,805 cycloid,planar curves,831-832 cylinders defined,957 graphing of,874-875 polar coordinate integration,1006 volume between two surfaces,988 cylindrical coordinates,1029-1031, 1041
centripetal force,836-837
vector-valued functions,797-801
direction cosines and angles,749
directed curve,1063-1066
geometric formula,742-745
closed curve,1067-1068
gradient,916
conservative vector field, path
line integrals,1064-1066
independence,1071-1083
planetary motion,846-847
line integrals,1067
projection,745-748
directed line segments,in vectors,
standard basis vectors,748-749 vector-valued functions
722-724, 782-783
continuity of,797
directional derivative 913-914 Chain Rule,914
spherical coordinate conversion,
functions of three or more variables,
1032-1034
algebraic definition,741-742
Dini,Ulisse,963, 1046
defined,959
surface integrals and,1112
741-750, 783
theories of,1148-1149
Divergence Theorem,1133-1137
limits of,795 Double Angle Formula,polar coordinate area calculations, 1004 double integrals
918-919
area between two curves,982
gradient,916
defined,967
greatest increase/decrease,916-917
Green's theorem,1095-1102
vector field divergence,1087-1088
history of,1045-1047
D
direction angles,dot products,749, 784
order of integration and,979-980
d'Alembert,Jean Le Rond,963-964
direction cosines,dot products,749, 784
over rectangular regions,
Darboux, Gaston,1046
direction vector. See also unit vector
decomposition,vector-valued functions, motion applications,835-836 degrees of freedom, planes in space, 768-769 del notation,vector fields,1090-1091, 1141 density function solids,1026-1027 surface area integration,1108-1111 derivatives. See also partial derivatives directional derivative,959
Dirichlet, Peter Gustav Lejeune,965
surface area,integrals for,1106-1107 variables of,982, 1039-1040
discriminant off, second derivative test,
volume of solids calculations,984-988
935-938 disks,method of planar regions binding,978-979
E
polar coordinate integration,1006
eccentricity,of ellipses,841-844
surface area,integrals for,1106-1107
element of area
displacement vector
derivative(s),vector-valued functions, 797-801 determinants, cross product of vectors and,754-755 differentiability Chain Rule and,900-910 function of three variables,905-907 function of two variables, 903-905
polar coordinates,999-1012
discontinuous functions,886-888
positive and negative,915
vector-valued functions,793
planar regions binding,977-979
in three-dimensional space,737-738 directrix of ellipse,842
Disney, Catherine,789
tangent planes,923-924
968-974, 1037 iterated integrals,968-974
defined,726-728, 783-784
ordinary derivatives,896-897 potential function location,1058-1059
1-3
tangent line to space curves,805-807 velocity as magnitude of,804-805 distance maximum-minimum problems, 939-940 vectors
polar coordinates,function integration,1004-1006 surface area parameterization, 1112-1113 ellipse(s) graphing of,875-876 Green's theorem,1097-1102 one sheet hyperboloids,878
lines and planes in space,775-778
planetary motion and,845-847
in three-dimensional space,
surface area,integrals for,1106-1107
733-735 divergence,1049, 1141 Green's theorem,1098-1102
two sheet hyperboloids,879 vector-valued functions and,839-844 y-simple region,980-981
1-4
Index region area calculations,1001-1003
ellipsoids,875-876, 881
dot products,742-745
elliptic cones, 876-877, 881
fractals theory,944
elliptic paraboloids,877-878,881
Frederick II (King of Denmark), 856
surface integrals and,1112
endpoints,polar coordinates,function
frustrum,surface area,integrals for,1107
vectors,in three-dimensional space,
integration,1004-1006
Lagrange multipliers,947-949
functions
735-737 vector-valued functions, limits of,
differentiation,963-964
error(s),sum of the squares of,941-944 Euclidean geometry,729,788
794-795
partial derivatives,891-893 exotic functions,1045-1046
global maximum,933
Euler,Leonhard,963-964,1045
Fourier's research on,964-965
global minimum, defined,932
"everywhere-continuous nowhere
gradient function,915-916
gradient, 915-916
cylindrical coordinates,1029-1031
differentiable" function,1045-1046 expanding fluid,vector field divergence, 1088
graphing,880-881
Chain Rule and,919
harmonic,1091
defined,959
over surface,integration of,
direction of greatest increase/decrease,
1107-1111
exterior algebra,787
916-917
polar coordinate integration,
extremum (extreme variables) Lagrange multipliers,946-951
functions of three or more variables, 918-919
1004-1006
functions of several variables, 952-954 solution strategies,951-952 local extremum,933
potential functions,1055-1059
identity,1091-1093
of several variables,863-871,957
level curves,917-918
combined functions,864-866
as total derivative,921-922
continuity of,886-888
vector fields, 1055-1057
gradients and directions,913-919
Taylor's formula for several variables,910
graphs and graphing
graphing functions,866-870
cylinders,874-875
Lagrange multipliers,946-954,
ellipsoids, 875-876
952-954
elliptic cones, 876-877
F
level sets,870-871
elliptic paraboloids,877-878
Fermat's theorem,maximum-minimum
limits and,883-886
of functions
problems and analogue of,
maximum-minimum problems,
933-934
recognition of,880-881
932-944
field lines,1053-1054 finite curves,planar regions binding, 976-979 first moments,1040-1041
several variables,866-870
partial derivatives,890-893
graph of,880-881
tangent planes,929-930
hyperbolic paraboloid,879-880
Taylor's formula,907-910
hyperboloid
of three variables,870-871,888
one sheet,878
lamina,1022
Chain Rule for,905-907
three-dimensional space, 1025-1027
gradient and,918-919
polar coordinates,994-997
partial derivatives,893-894
quadric surfaces,875
first octant,vectors in three-dimensional space,733
two sheets,879
surface area integration,1108-1111
"floating" vector notation,723
tangent planes,924-925
flow lines,1053-1054
of two variables,861-863
Divergence Theorem,1131-1137 fluid flow
gravity and gravitational force
Chain Rule for,903-905
buoyancy and, 1136-1137
tangent plane,922-930
center of, 1023-1024, 1041
Green's theorem,1099-1102
Fundamental Theorem of Calculus
Stoke's theorem,1116-1129
advanced theories,1148-1149
vector field divergence,1086-1088,
closed vector fields,1079-1080
1087-1088
surface area,integrals for,1106-1107 Grassmann,Hermann,787
central force fields,837 -839 planetary motion and,844-847, 857-859
conservative vector field,path
focus (foci),ellipses,839-844
vector field,1052-1053
independence,1073-1083
force
curve parameterization by arc length,
acceleration and,808
818-820
central force fields,837-839
Divergence Theorem and,1131-1137
cross product applications,760
Green's theorem and,1095-1102
divergence,1086-1088 Green,George,1146-1147 Green's function,1147 Green's theorem,1094-1102, 1141-1142,1147
dot product applications,750
divergence and,1131
planetary motion and,845-847 potential functions, 1055-1057 vectors and,728 Fourier,Jean Baptiste Joseph,964-965, 1147 four-leafed rose,polar coordinates, 996-997
Stoke's theorem and, 1120-1123
G
vector form of, 1098-1102
Gauss,Carl Friedrich,787 Gauss's Law of Magnetism,1085-1088 Divergence Theorem and,1135-1137 geometric constructions cross product,756-759
H Half-Angle Formula,polar coordinate area calculations, 1004
Index Halley, Edmond, 858 Hamilton, Archibald, 788 Hamilton, William Rowan (Sir),
787-789 harmonic function, del notation, 1091 harmonic oscillator equation, stream lines, 1054
instantaneous velocity, vector-valued functions, 804 integral curves (streamlines), 1053-1054 integrals Lebesgue integral, 1046-1047 line integrals, 1061-1069 applications, 967
height function, level sets, several
cylindrical coordinates, 1029-1031
variables, 870-871 acceleration vectors, 808
physical applications, 1020-1027
curvature parameterization, 827-828
polar coordinates, 990-997
osculating circle, 829-831
regional integration, 976-982
vector-valued functions, 793
solids volume calculations, 984-988
velocity vector, 805 hemisphere cylindrical coordinates, 1030-1031 spherical coordinates, 1034 Hermite, Charles, 1046
surface integrals, 1104-1114 integration multiple integrals, 976-982 order changes, 979-980
Hill, Thomas, 789
planar region area, 980-982
Hodge star operator, 787
planar regions, finitely bounded
horizontal slices, graphing functions, of several variables, 866-870 Huygens, Christiaan, 857-858 hyperbolas graphing functions, of several variables, 869-870
curves, 976-979, 1037 of polar coordinates, 999-1012
1003-1006
surface area, function integration,
hyperboloids
1107-1111 theories of, 1148-1149 in triple integrals, 1014-1018 inverse square law of gravitational attraction, 844-847, 858-859 inward oriented surface, Stoke's
divergence, 1088 double integrals over rectangular regions, 968-974 tangent planes, 928-929 inequalities Cauchy-Schwarz inequality, 744
theorem, 1117-1118 irrotational vector field, 1089-1090, 1141 Stoke's theorem, piecewise smoothing, bounded regions, 1127-1128 isobath, 870-871 isohypse, 870-871 isoline, graphing functions, of several variables, 867 -870
of ellipses, 843-844
isothermal curves/isotherms, 870
vectors, 883-885
iterated integral
inertia second moment of, 1024, 1040-1041 three dimensional
K Kelvin, Lord, 1147-1148 Kepler, Johannes planetary motion and, 844, 856-858 vector-valued functions and laws of,
791, 853 Kepler's First Law, 844-847, 853, 857 Kepler's Second Law, 844-847,
853, 857 Kepler's Third Law, 844-847, 853,
857-858 Keynes, John Maynard, 859 kinetic energy, 1056-1057 Kleiber, Max, 944
L Lagrange multipliers
1006-1012
hyperbolic paraboloid, 879-881
increments
Jordan, Camille, 963, 1047
variable and Jacobian changes,
spherical coordinates, 1033-1034
incompressible fluid flow, vector field
1006-1012 spherical coordinates, 1033-1034
Lagrange, Joseph Louis, 963, 1045, 1146
one sheet hyperboloids, 878
I
polar coordinate integration,
region area calculations, 1000-1003
potential function location, 1058-1059
one sheet, 878, 881
integration, 1007-1012 Jacobian matrix
iterated integral area calculations,
Lagrange multipliers and, 948-949
two sheets, 879, 881
Jacobian determinant, polar coordinate
spherical coordinates, 1031-1034 triple integrals, 1014-1018
Herschel, John, 789
Hooke, Robert, 858
Jacobi, Carl Gustav, 963, 1045
double integrals over rectangular regions, 968-974
helix
J
multiple integrals
Harnack, Axel, 963
double integral calculations, 971-974 double integrals over rectangular regions, 969-974
functions of several variables,
946-954, 960 two constraints, extremization of,
953-954 geometric properties, 947 -949 solution strategies, 951-952 lamina first moments, 1022, 1040-1041 mass density, multiple integrals,
1020-1021 Laplace, Pierre Simon, 788, 964, 1146 Laplace operator, del notation,
1090-1091, 1146 Laplacian of u, 964 Law of Buoyancy (Archimedes), Divergence Theorem and,
1135-1137 Law of Conservation of Mechanical Energy, 1057 Laws of planetary motion, 844-847 least squares lines, maximum-minimum problems, 940-944 Lebesgue, Henri, 1047
instantaneous acceleration, 807
order of integration and, 979-980
instantaneous rate of change
planar regions binding, 977-979
Lectures on Quaternions, 789
polar coordinate area calculations,
Leibniz, Wilhelm von, on partial
level curves, gradient and, 917-918 partial derivatives, 890-893 instantaneous speed, 804
1-5
1003-1004 triple integral, 1014-1018
Lebesgue integral, 1046-1047
derivatives, 963 lemniscate, polar coordinates, 997
1-6
Index
length (magnitude) of vector
log-log plots, least squares lines,
defined, 726
vector-valued function applications,
943-944
834-847
in three-dimensional space, 737
central force fields, 837-839 ellipses, 839-844
level curves gradient and, 917-918
M
Lagrange multipliers and, 949-951
magnetic field
level sets variables, 867-870
870-871 level surfaces, tangent planes, 924-925 l'Hopital's Rule, vector-valued functions, limits of, 795
1128-1129 vector field divergence, 1085-1088
applications, 967 cylindrical coordinates, 1029-1031
mass
double integrals over rectangular
center of, 1023-1024, 1040-1041
directed curve, line integral,
rules for, 885-886 vector-valued functions, 793-795, 851
regions, 968-974
central force fields, 837-839 on curved paths, 1061-1063
mass density of plate, 1021
multiple integrals
major axis, of ellipse, 839-844
lima�ons, 1002
of functions, 882-885
moving frame, unit normal vector, 822
magnitude, of vector, 726
limit(s) continuity and, 957-958
lines, 944
Stoke's theorem applications,
graphing functions, of several height function, several variables,
planetary motions, 844-847 mouse-to-elephant curve, least squares
iterated integrals, 968-974 integration over general regions,
976-982
1068-1069
integration order changes, 979-980 planar region area, 980-982
gradient vector field, potential function, 1055-1057
planar regions, finitely bounded
multiple integrals, 1020-1021,
curves, 976-979
1040-1041
physical applications, 1020-1027
LindelOf, Leonard Lorenz, 963
tangent plane measurements, 929-930
linear approximation
three-dimensional space, 1025-1027
first moments, 1022
vector field, 1052-1053
mass, 1020-1021
increments, 928-929 tangent planes, 927-928 two-variable and three-variable analogues, 930 linear equations, lines in space, 773-775 line integrals, 1061-1069, 1140 applications, 1068-1069
vector field divergence, 1086-1088
and center of mass, 1025-1027
maximum error, tangent plane
polar coordinates, 990-997
approximation, 928-929
graphing in, 994-997
maximum-minimum problems
conservative vector field, 1071-1083
Fermat's theorem analogue, 933-934
curved paths, 1061-1063
Lagrange multipliers and, 946-954
Green's theorem, 1098-1102
least squares lines, 940-944
notation, 1067
saddle points, 934-935
parametric smooth curve, 1063-1066
second derivative test, 935-938
path dependence, 1066-1067
variable functions, 932-944
regions, 1124-1128
three-dimensional mass, moments
mathematical physics, 964-965
applications, 938-940
piecewise smoothing, bounded
moment of inertia, 1024
mass density, of plate, 1020-1021
closed curves, 1067-1068
Stoke's theorem, 1119-1123
center of mass, 1023-1024
integration of, 999-1012 iterated integral area calculations,
1003-1006 radial variable, negative values,
992-993 rectangular coordinates and,
993-994 region area calculations, 1000-1003 symmetric graphing principles, 997
Mean Value Theorem arc length, 816
variable and Jacobian changes,
1006-1012
differentiability and Chain Rule and, 902
line(s) best-fitting line, 941-944 least squares lines, 940-944 normal lines cross product geometry, 757-759 tangent planes, 925-926 in space, 770-775, 784-785
Conservation of, 1057
948-954 minor axis, of ellipse, 839-844 mixed partial derivative, 894-896 Mobius, August Ferdinand, 1117 Mobius strip, 1117-1118
local extrema, 933 applied problems, 939-940 Fermat's theorem analogue, 934 Lagrange multipliers, 946-947 saddle points, 934-935 local maximum, defined, 932
spherical coordinates, 1031-1034 triple integrals, 1014-1018
Method of Lagrange multipliers,
Cartesian equations, 771-775
segment
volume between surfaces, 986-988
Mechanical Energy, Law of
parametric equations, 770-771 line segments. See Directed line
solids volume calculations, 984-988
multiplication, associative law of, 788 mutually perpendicular vector, 744-745 Mysterium Cosmographicum (Kepler), 856-857
moments first moments, 1022
N
negative moments, 1026
natural logarithms, least squares lines,
second moments (of inertia), 1024,
1040-1041 three-dimensional space, 1025-1027 motion instantaneous direction of, 805
940-944 navigation, dot product applications, 750 negative boundary orientation, Green's theorem, 1094-1102 negative derivative, 915
Index
negative moments, 1026
one-dimensional divergence theorem,
negative radial variables,992-993 negative work,on curved path,1063
1131-1137 one-dimensional wave equation,
Newton,Isaac
963-964
calculus and,791
one sheet hyperboloids,878,881
on partial derivatives,963
one-sided surfaces,1117-1118
planetary motion and laws of,857-859
open ball with center set,vectors,
Newton's First Law,acceleration and,808 Newton's Law of Gravitation
distance calculations,735 open sets,vector fields,1050 orbits,laws of planetary motion and,
central force fields,837-839 planetary motion and,844-847, 858-859
844-847,857 ordinary derivatives,896-897 orientation
potential functions,1056-1057
Divergence Theorem and,1131-1137
Newton's Second Law of Motion
Stoke's theorem,1116-1118
polar coordinate integration, 1010-1012 surface area,integrals for,1105-1107 surface area parameterization, 1112-1113 vectors in,724-725 operations in three-dimensional space,736 parallel planes, Cartesian equations, 767-769 parallel vectors direction,738 nonzero vectors as,727-728 parameterization curves,by arc length,818-820,852
acceleration,807
oriented curve,1063-1066
directed curves, 1063-1066
central force fields,838-839
oriented surface,Stoke's theorem,
Green's theorem verification,
planetary motion and,845-847
1116-1118
noncommutative algebra,789
piecewise smoothing,bounded
nonnegative numbers ellipses,841
regions,1127-1128 orthogonal vectors,744-745
velocity vectors,804
cross product of spatial vectors,
nonparallel vectors
753-754
cross product area calculations, 759-760
unit normal vectors,820-822
projection,745-748
direction,726-728,737-738 cosines and angles,749
oscilloscope,cross product applications in,760 osculating circle,828-831,853 osculating plane,829-831 Ostrogradskii,Mikhail Vasilevich,1045,
dot products and,741-743 planes in space,766-769
1147-1148 outward oriented surface,Stoke's
projection,745-748
theorem,1117-1118
normal component of curl,1118-1119 cross product geometry,757-759 tangent planes,925-926 normal plane,unit normal vectors, 820-822 normal vector cross product geometry,757-759
p graphing functions,of several variables,868-870 triple integrals,1018 volume between two surfaces, 986-988 paraboloids
tangent planes,923-924,959
hyperbolic,879-881
tangent plane,926-928
polar coordinate integration, 1005-1006 volume between two surfaces, 986-988 parallelepiped,triple scalar product,
0
761-763
octants vectors in three-dimensional space,733 volume between two surfaces,988
806-807 parametric curve,vector-valued functions,792 parametric equations lines in space,770-771 planes in space,769-770 partial derivatives
differentiability and Chain Rule,
parabolas
elliptic,877-878,881
defined,959
tangent line to space curves,
defined,890-893,958
planes in space,766-769 numerical approximations
722-723 vector-valued functions,792
Clairaut's theorem and,963
of acceleration,836-837,853 normal lines
surface integrals for,1113-1114 tangent vectors,814 vectors in plane,directed line segment,
work applications,750
nonzero vectors
planar curves,831-832 surface area,1112-1113
motion applications,834-847
distance calculations,777-778
778-779
1097-1102 osculating circle,829-831
planes in space,770
planes in space,769-770 triple vector product formulas,
1-7
parallelogram
900-910 differentiation applications,896-897 functions of three variables,893-894 functions of two variables,861-862 higher partial derivatives,894-896 potential function location,1058-1059 Stoke's theorem,piecewise smoothing, bounded regions,1125-1128 partial differential equations,potential theory,1146 partition functions double integrals over rectangular regions,968-974 mass density of plate,1021 planar regions binding,finite curves, 976-979
central force fields,839
path-connected sets,vector fields,1050
cross product area calculations,
path dependence,line integrals,
759-760
1066-1067
1-8
Index
path independence closed vector fields, 1076-1080 conservative vector fields, 1071-1083,
1140 path integral, 1063-1066
laws of, 844-847, 856-859
functions of three variables, Stoke's theorem and, 1122-1123
first moments of, 1022 mass density, 1020-1021 second moment of inertia, 1024
baseball physics, 810-811
893-896
center of mass, 1023-1024
Peano, Giuseppe, 963, 1047
810-811
partial derivatives, 892-893
plate
physics acceleration and velocity vectors and,
Product Rule
vector-valued functions and, 791
vector-valued functions, 799-801 projection
point(s), planes in space and, 768-769,
distance calculations and, 776-778
782
dot products, 745-748, 783
polar coordinates, 990-997, 1038
standard basis vectors and, 748-749
cross product applications in, 760
area calculations, 1003-1006, 1039
Divergence Theorem in, 1134-1137
double integral of, 999-1012, 1039
Divergence Theorem and, 1134-1137
mathematical physics, 964-965
graphing in, 994-997, 1038-1039
vector field, 1051-1053
multiple integrals, 1020-1027
integration of, 999-1012
center of mass, 1023-1024 first moments, 1022
iterated integral area calculations,
proportionality constant
vector field divergence, 1086-1088 proportionality factor, Lagrange
1003-1006
multipliers and, 949
mass, 1020-1021
region area calculations, 1000-1003
Ptolemy, 857
moment of inertia, 1024
variable and Jacobian changes,
punctured disk, limits of functions in,
three-dimensional mass, moments and center of mass, 1025-1027 Stoke's theorem in, 1128-1129 vector applications in, 728 vector fields in, 1051-1053 piecewise smoothing conservative vector field, path independence, 1073-1083
multiple integrals, 990-997 graphing in, 994-997
992-993
surface area, integrals for, 1107 variable and Jacobian changes,
planar curves, 831-832 line integrals, 1064-1066 planar regions area of, 980-982 closed vector fields, 1078-1080 finite curve binding, 976-979 integration over, 976-979, 1037 volume of solids calculation, 984-988 planar vectors closed vector fields, 1076-1080 conservative vector field, path independence, 1075-1083 continuous differentiability, 1051 plane(s) osculating plane, 829-831 in space, 766-769 Cartesian equations, 766-769 parametric equations, 769-770 vector fields, 1051 vectors in, 722-729, 784-785 length (magnitude), 726 special i and j vectors, 728-729
1006-1012 polar moment of inertia, 1024 in three-dimensional space, 737-738 polygonal paths, vectors in three dimensional space, 733 polynomials, Taylor's formula for several variables and, 909-910 tangent plane approximation, 927-928 position vectors
Q
quadrants, vectors in three-dimensional space, 733 quadratic equations acceleration and, 809 quadric surfaces, 875 quadratic polynomials, Taylor's formula for several variables and, 909-910 quadric surfaces, 875, 880-881 defined, 957 quaternions, 788-789 Quotient Rule differentiability and, 902
central force fields, 838-839
partial derivatives, 892-893
in geometric space, 735-736
functions of three variables,
vector-valued functions, 792-793
893-896
positive boundary orientation, Green's
tangent planes, 925
theorem, 1094-1102 positive derivative, 915 potential energy, 1055-1057 conservative vector field, path independence, 1075-1083 potential functions, 1146-1147
R radial coordinates, 991-992 radial variable, negative values, 992-993 radians, angle between two planes and,
768-769
gradient vector field, 1055-1057 location of, 1057-1059
radius
potential theory, 1146
of convergence, 828-831
principal unit normal vector, 821-822
curvature, 825
motion applications, 834-847
Principia Mathematica (Newton), 858
unit vectors and directions, 726-728
principle of superposition, vector
central force fields, 839
distance calculations, 733-734
polar vector, 727
triangle inequality, 729 planetary motion
vectors in three-dimensional space,
region area calculations, 1000-1003
line integrals, 1064-1066
1124-1128
distance calculations and, 776-778 lines in space, 774-775
rectangular coordinates and, 993-994
symmetric graphing principles, 997
tangent vectors, 814
Pythagorean Theorem
radial variable, negative values,
Green's theorem, 1100-1102 Stoke's theorem, bounded regions,
883-885
1006-1012
applications and, 728 probability theory, 1047
osculating circle, 829-831 planar curves, 832 vectors, distance calculations, 735 radius vector, 788-789 rate of change
Index differentiability and Chain Rule,
planar regions binding, finite curves,
900-910
976-979
partial derivatives, 893
polar coordinate integration
real numbers, scalars as, 723
region area calculations, 1001-1003
rectangles, surface area, integrals for,
variables and Jacobian matrix,
1010-1012
1105-1107 rectangular coordinates, 1029-1031
surface area, integrals for, 1106-1107
integration problem, 1005-1006
function integration over surface,
1108-1111
polar coordinates and, 993-994
triple integrals, 1014-1018
variables and Jacobian matrix,
1010-1012
volume between two surfaces,
986-988
spherical coordinate conversion,
1032-1034 vectors in three-dimensional space,
work on curved paths, 1062-1063 right-hand rule, standard unit normal
732-733 rectangular regions, double integrals over, 968-974, 1037
vector, 757-759
directional derivatives, 913-915 Divergence Theorem and, 1135-1137 instantaneous velocity, 804 Lagrange multipliers, 953-954 potential function location, 1058-1059 second derivative test, 935-938 several variables, 863-871 combined functions, 864-866 graphing of, 866-870 level sets, 870-871 tangent vectors, 814 scatter plots, least squares lines, 941-944 Schrodinger, Erwin, 788 Schwarz, Hermann Amandus, 963 secants, tangent line to space curves and,
Rig-Veda (Grassmann), 787 rotation, axis of, curl of vector fields,
805-807 Second Derivative Test, maximum
1089-1090
regions
minimum problems, 935-938 second moments (of inertia), 1024,
area calculations, polar coordinate integration, 1000-1003 bounded regions Green's theorem, 1094-1102 piecewise smoothing, Stoke's
1040-1041 s
three-dimensional space, 1025-1027
saddle points maximum-minimum problems,
integration over multiple regions,
Second Derivative Test, 935-938 saddle surface, hyperbolic paraboloid,
880
976-982 of multiple integrals, 967 planar region, area calculation,
980-982 planar region, boundary curves,
976-979 rectangular regions, double integrals over, 968-974, 1037 iterated integrals, 969-974 simple regions, 976-982, 1037-1038 closed vector fields, 1078-1080 regression line. See least squares lines relative maximum, 932 remainder term, Taylor's formula for several variables and, 909-910 reparameterization, vector-valued functions, 816 resultant force, vectors and, 728 Riemann, Georg Friedrich Bernhard,
965, 1045 Riemann integral, 965, 1046-1047 arc length, 816 iterated integral calculations,
971-974 planar region area, 981-982 triple Riemann integral, 1014-1018 Riemann sum double integrals over rectangular regions, 969, 1037 iterated integral calculations, 971-974 planar region area, 980-982
second partial derivative, 894-896 sector, polar coordinate integration,
934-935
theorem, 1124-1128 Green's theorem, 1099-1102
1-9
999-1000 region area calculations, 1000-1003 set theory, 965 simple regions, 976-982,
1037-1038, 1047
scalar multiplication acceleration and, 809 Lagrange multipliers and, 948-951 triangle inequality and, 729 vectors, 725 geometric construction, 736 length of vector and, 726 vector-valued functions
closed vector fields, 1078-1080 curl of vector field, 1089-1090 sine(s) planar curves, 831-832 polar coordinate integration, 1006 single vector, defined, 723-724 sink (in region), Green's theorem,
continuity of, 797 differentiation, 799-801
1099-1102 slicing calculations, iterated integrals,
972-974
scalar product dot product as, 742 history of, 789
smooth parameterization arc length, 817-818 curvature, 825-828
scalars cross product of two vectors, 755 dot products and, 742 projection, 747-748
tangent vectors, 814 unit normal vectors, 820-822 vector-valued functions, reparameteri-
real numbers as, 723, 726 triangle inequality and, 729 triple scalar product, 760-763, 785 distance calculations and, 776-778 formulas, 778-779 work on curved paths, 1063 scalar-valued functions conservative vector field, path independence, 1072-1083 continuity, 795-797 del notation, 1090-1091 differentiability and Chain Rule,
900-910
zation and, 817-818 solenoidal divergence, 1085-1088 del notation, 1090-1091 solids cylindrical coordinates, 1030-1031 Divergence Theorem and boundaries of, 1134-1137 triple integrals, 1015-1018 volume calculations, 984-988 solids of revolution, planar regions binding, 978-979 source (in region), Green's theorem,
1099-1102
Index
1-1 O
space
superposition principle,vector
surfaces in, Stoke's theorem, 1116-1129
applications and,728 surface area Divergence Theorem,1131-1137
vector fields in,1080-1083
function integration over,1107-1111
space-filling curves,1047
Thomson,William,1147-1148 three-dimensional solid surface integrals and,1112 volume calculation,984-988 three-dimensional space
motion applications,835-836
integrals for,1105-1107, 1142
curl of vector field,1089-1090
tangent line to,805-807
orientation,Stoke's theorem,
graphing functions of several
1116-1118
tangent vectors,814
parameterization,1112-1113
spatial vectors
surface integrals for,1113-1114
cross product of,753-754 triple vector products,763-764
surface graphs
vector fields,1080-1083 curl,1088-1090
i, j, and k vectors,738 i and j vectors,728-729
quadric surfaces,875
path-connected, 1050 vectors in, 732-739
surface integrals, 1104-1114, 1142 center of mass determination, 1111-1112
speed
mass in,1025-1027 open sets,1050
saddle surface,880 of several variables, 866-870
special unit vectors
variables,866-870 mass moments, inertia and center of
distance,733-734 length,737 operations,736 special i, j, and k vectors,738
instantaneous,804, 851-852
curl components,1118-1119
triangle inequality,738-739
kinetic energy and,1056-1057
Divergence Theorem,1131-1137
unit vectors and directions,
function integration over surface,
spheres
1107-1111
ellipsoids as,875-876
parameterized surfaces,1113-1114
vectors in three-dimensional space,
for surface area,1105-1107
distance calculations,734-735 spherical coordinates,1031-1035, 1041-1042
surfaces, volume between,986-988 symmetric equations
spin acceleration,baseball physics,
lines in space,772-775
810-811
polar coordinates, 996-997 vector field divergence,1086-1088
spiral of Archimedes,994-997 square matrix,determinant of,754-755
939-940 second moment of inertia,1024 standard basis vectors,projection and, 748-749, 784 standard unit normal vector, cross product geometry, 757-759 Stokes,George, 1147-1148 Stoke's theorem,1116-1129, 1142, 1147-1148 boundary orientation, 1116-1118 divergence,1131
T tangent line(s) component of acceleration, 836-837, 853 to curve in space, 805-807
functions of three or more variables, 929-930 increments,928-929
normal vectors and,959
subtraction, vectors,725 summands, functions of three variables, Chain Rule for,907 sum of the squares of the errors (SSE), 941-944 sums,of vectors,724-725 supernovas,856
work on curved paths,1062-1063
Treatise on Natural Philosophy (Thomson),1147 Triangle Inequality vectors
tangent planes,921-930
piecewise-smooth bounded regions,
iterated integral calculations,971-974
central force fields,838-839
limits of functions and,884
level surfaces,924-925
arc length,814-815
929-930 increments,928-929
Lagrange multipliers,948-949
normal lines,925-926
subintervals
approximation functions of three or more variables,
functions of two variables,861-862
physics applications,1128-1129
streamlines, vector fields,1053-1054
total differential, tangent plane
trajectories
Tait,P. G.,1147
line integrals,1119-1123
1124-1128
870-871 total curl,1118-1119
Traite de Mecanique Celeste (Laplace), 1146-1147
square of the distance maximum-minimum problems,
737-738 topographic mapping,height function,
numerical approximations,926-928 tangent vectors,813-814, 852 unit tangent vector,824 curvature parameterization,824 work on curved paths,1061-1063 Taylor's theorem maximum-minimum problems,Second Derivative Test,936-938 several variables,907-910 tangent plane approximation,927-928
Theorie Analytique de la Chaleur (Fourier),964, 1147
planar vectors, 729 in three-dimensional space, 738-739 triangles,cross product area calculations, 759-760 trigonometric functions dot products,743 mathematical physics and,965 triple integrals basic principles,1014-1018, 1040 defined,967 surface integrals and,1112 in three-dimensional space, 1025-1027 triple Riemann integral,1014-1018 triple products distance calculations and,776-778 formulas,778-779 scalar product,760-763, 785
Index vector products, 763-764 formulas, 778-779 planetary motion and, 845-847 triplet quaternions, 788 twice continuously differentiable func tion, partial derivatives, 895-896 two-dimensional theorems, Stoke's theorem, 1131 two sheet hyperboloids, 879, 881
u uniform partition arc length, 814-815 double integrals over rectangular regions, 968-974 polar coordinate integration, regions calculations, 1000-1003 unit normals Divergence Theorem, 1132-1137 oriented surfaces, 1117-1118 unit vector directional derivative, 914-915 directions, 726-728 cosines and angles, 749 normal vector cross product geometry, 757-759 parameterization, 820-822 Stoke's theorem, 1116-1118 projection and, 748-749 tangent vectors, 814 curvature parameterization, 824 motion applications, 834-847 osculating circle, 830-831 in three-dimensional space,
737-738 unknowns, lines in space and equations of, 774-775 upper limit of integration, multiple integrals, 979-980 upward-opening parabola, 877-878
surface area integration, 1108-1111 three variables, 870-871, 888,
893-894, 905-907, 924-925 two variables, 861-863, 903-905 gradients, 915-916 Chain Rule, 919 level curves, 917-918 radial variables, 992-993 for space, 1033-1034 variations, calculus of, 788 vector calculus, overview, 1049 vector fields, 1050-1059 closed vector fields, 1076-1080 conservative vector fields, 1071-1083 curl of, 1088-1090 del notation, 1090-1091 divergence, 1085-1088 gradient vector fields, 1055-1057 Green's theorem, 1097-1102 integral curves (streamlines),
1053-1054
directional derivatives, 913-915 greatest increase/decrease, 916-917 double integrals, 982, 1006-1012,
1039-1040 functions of Lagrange multipliers, 946-954,
952-954
length (magnitude), 726, 783 special i and j vectors, 728-729 triangle inequality, 729 unit vectors and directions, 726-728 planes in space, 766-769 Cartesian equations, 766-769 parametric equations, 769-770 polar form, 727 subtraction, 725 tangent planes, 922-924 tangent vectors, 813-814 in three-dimensional space, 732-739 distance, 733-734 length, 737 operations, 736 special i, j, and k vectors, 738 triangle inequality, 738-739 unit vectors and directions,
737-738 triple products
irrotational vector field, 1089-1090
distance calculations and, 776-778 formulas, 778-779
in physics, 1051-1053
scalar product, 760-763
potential functions, 1055-1059
vector product, 763-764
in space, 1080-1083 Stoke's theorem, 1116-1118 vectors
unit vector directions, 726-728, 749 normal vectors, 820-822
algebra, 724-725
projection and, 748-749
basic properties and applications, 721
in three-dimensional space,
calculus of, 787-788 components or entries of, 724
737-738 vector-valued functions
constant vector, 810-811
antidifferentiation, 801
cross product
applications, 791
algebraic properties, 755-756
arc length, 815-816
area calculation, 759-760
basic properties, 792-793, 850-851
determinants and, 754-755
continuity of, 795-797
geometric understanding, 756-759
curvature, 824-832
physical applications, 760 two spatial vectors, 753-754 displacement vector, 804-805 Divergence Theorem and, 1135-1137
variables
in plane, 721-729
line integrals, 1064-1066
distance calculations, 775-778
v
dot product and applications, 741-750 algebraic definition, 741-742
osculating circle, 828-831 planar curves, 831-832 smooth parameterization,
825-828 curve by arc length, 818-820 derivatives of, 797 -801 flow properties, 1049
direction cosines and angles, 749
limits of, 793-795
geometric formula, 742-745
motion applications, 834-847 central force fields, 837-839
projection, 745-748 standard basis vectors, 748-749 Green's theorem, 1098-1102
ellipses, 839-844 planetary motions, 844-847
history of, 787-789
reparameterization, 817-818
limits and, 883-886
instantaneous velocity as, 804-805
Stoke's theorem, 1120-1123
maximum-minimum problems,
least squares lines, 942-944
surface area parameterization,
932-944 partial derivatives, 890-893 several variables, 863-871,
907-910, 918-919, 929-930, 957
1-11
lines in space, 770-775
1112-1113
Cartesian equations, 771-775
tangent vectors, 813-815
parametric equations, 770-771
unit normal vectors, 820-822
planar vectors, 1051
vector fields as, 1051
1-12
Index
vector-valued functions
(continued)
velocity and acceleration applications,
scalar-valued functions, 871
Wessel, Caspar, 787
vertical slices, graphing functions, of
baseball physics, 810
870 vibrating string equation, 963 964
tangent line to curved space,
volume
852
regions, 968
instantaneous velocity, 804 of vector field, 1051
multiple integral calculations,
984 988
vector field divergence,
triple integrals, 1014
1086 1088 Green's theorem, 1099-1102 vector-valued functions baseball physics, 810-811
between two surfaces, 986-988 vorticity axis of rotation, 1089-1090
curve parameterization by arc
normal component of curl, 1118-1119
length, 819-820
805-809 graphs of functions, 880-881
y
z direction, 727-728
w Weierstrass, Karl, 1045-1046 weight density, function integration over surface, 1107-1111
980
y-simple regions, 977-981
zero vector, 725
derivatives, 793
Vertical Line Test
x x-simple regions, 977
1018
solids of revolution, 978-979
basic principles, 803-811
tangent line to curved space,
approximation, 928 Wren, Christopher, 858
974, 1038
of solids
1053
1063
dot product applications, 750
1031
double integrals over rectangular
average velocity, 804
on curved paths, 1061
worst case value, tangent plane
cylindrical coordinates, 1030 velocity, 851
work
several variables, 866
803 811
orthogonality, 744
