Calculus III Text Book 2nd Edition by Blank & Krantz for Science & Engineering Majors. Vector Calculus with Solution Manuals Enjoy it!Descripción completa
Vector Calculus from University of Cambridge.
vector calculus 6th ed
Descripción: vector calculus 6th ed
vector calculus 6th ed
Descripción: Vector calculus: Jerrold Marsden, Anthony Tromba. 6th edition
Vector calculus: Jerrold Marsden, Anthony Tromba. 6th editionFull description
Vector calculus: Jerrold Marsden, Anthony Tromba. 6th editionFull description
Vector calculus: Jerrold Marsden, Anthony Tromba. 6th edition
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:
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
This page intentionally left blank
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 . Although this practice is common in the mathematical literature, it is infrequently found in calculus books. However,
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
•
Related media, examples, and sample practice items reinforce the learning
•
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
•
Algorithmically generated exercises in which students can click on a "help" button for hints, as well as a large cache of extra exercises with final answers
•
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
reinforce course goals inside and outside of the classroom as well as visibility into individual student progress. Precreated materials and activities help instructors optimize their time:
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
•
Classroom Response System (Clicker) Questions
•
Instructor's Solutions Manual
•
Gradable Reading Assignment Questions (embedded with online text)
•
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
� 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.
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
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
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?
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).
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 =
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
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
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
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.
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.
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.
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)
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.
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.
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 ·
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), ·
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
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.
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
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
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.
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).
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).
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)?
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
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
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 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.
(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
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.
£ 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.
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
(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
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),
(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
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
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.
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 _.
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
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
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 +
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
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
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
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
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
(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
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
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.
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
)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
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 =
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 =
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).
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
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.
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.
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.
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.
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
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
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
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
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'?
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 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
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.
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.
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.
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).
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
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) 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.
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,
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
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.
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
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
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 ·
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)&
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
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
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
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
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
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.
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
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
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.
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. ·
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
._ 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
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.
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
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.
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
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
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
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
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
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
- 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
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,
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