Vector Calculus and Differential Equations

Citation preview

ALBERT G. FADELL

VECTOR · CAL~CULUS and Differential Equations

VECTOR CALCULUS and Differential Equations by ALBERT

G.

FADELL

Associate Professor of Mathematics State University of Netv York at Buffalo

• ILLUSTRATIONS BY DON KURTIS

AMERICAN BOOK· VAN NOSTRAND· REINHOLD NEW YORK

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Copyright© 1968, BY AMERICAN BOOK COMPANY Published simultaneously in Canada by D. VAN NosTRAND CoMPANY (Canada), LTD.

No reproduction in any form of this book, in whole or in part (except for brief quotation in critical articles or reviews), may be made without written authorization from the publisher.

Library of Congress Catalog Card No. 65-1483

PRINTED IN THE UNITED STATES OF AMERICA

DEDICATED

To my father, Shickery Nassar Fadell and to my mother, Olga Sorour Fadell

Preface

This book is intended to fill the needs of a full year sophomore course to complete a two-year sequence of calculus with differential equations. It is properly the sequel to my Calculus with Analytic Geometry. However, it may be used effectively as a text generally for students who have studied a substantial first-year course in calculus, since with few exceptions the notation and terminology employed is standard. The spirit of the text is to supply the student with a sufficiently rigorous and comprehensive experience in elementary mathematics to enable him to continue into advanced calculus or introductory abstract analysis without the usual jump discontinuities. On the other hand, there is an abundance of routine illustrative examples and straightforward exercises to give the student a solid background in the manipulative applications of calculus and differential equations. The instructor will find that there is more than enough material in the first eight chapters to fill a third semester of calculus. Indeed, he will be expected to make some omissions, varying with the background, interests, and abilities of his class. For example, pressure of time may force him to pass over all or parts of Chapters 6, 7, and 8, which add topics not always covered in the standard three-semester calculus sequence, and are often left to advanced calculus. Chapters 9 to 13 constitute a one-semester course in elementary differential equations. Chapter 9 contains the traditional introduction to differential equations, culminating in a proof of the Picard existence and uniqueness theorem for first-order equations. Beginning with Chapter 10 the emphasis is on linear differential equations. Systems of linear differential equations are solved in §10.11 by matrix methods to introduce the student to eigenvalues. Chapter 11 on the Laplace transform is elementary in scope, but has abundant logical details. It provides an excellent opportunity to invite the student to rehearse his competence in applying the elementary theoretical tools of calculus. If for lack of time the instructor has passed over Chapter 8 on improper integrals and integrals with parameter, he is advised to interpose the results of Chapter 8 as they are needed for the Laplace transform theory. The Frobenius series method in Chapter 12 is not presented in exhaustive theoretical detail, but rather examples are allowed to be the vehicle of the vii

Vlll

PREFACE

salient ideas. The study of Fourier series in Chapter 13 has for its primary objective the solution of the vibrating string problem by the method of separation of variables. The author hopes that the instructor will present this chapter rather completely so that the student can gain at least a compelling appreciation of the power wielded by Fourier methods. It is abundantly clear that the problem of the vibrating string is a most exciting and vivid way of exposing the student initially to Fourier analysis. I am deeply grateful to Don Kurtis for his masterly execution of the figures, and to my sister Mary Rose Spendio for her expert typing of the manuscript. I am indebted to my brother Professor Edward R. Fadell for many valuable comments and criticisms. To the staff of D. Van Nostrand Company I am happy to express my gratitude for their constant patience and outstanding skill in handling the editing and publishing. ALBERT G. FADELL January 1968

Contents

Preface

Chapter 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

Chapter 2 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Chapter 3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14

v

Vectors Elementary vector geometry of 3-space Linear dependence and independence The dot product The vector (or cross) product Triple products of vectors The line and plane Vector spaces Vector spaces with inner products

Vector Functions Vector functions Limits of vector functions Derivatives of vector functions Curves, directed curves, and arclength Elementary kinematics Introductory differential geometry Curvature of arc The Frenet-Serret formulas

Differential Calculus of R" R"-Rm functions

Limits and continuity for R"-~ functions Surfaces Partial derivatives Higher-order partial derivatives M.V.T. for partial derivatives The gradient Differentiability The general chain rule for derivatives Gradient fields The direction derivative Implicit differentiation Taylor's theorem for R"-R functions Extrema for R"-R functions ix

1 7 11 17 24 27 31 36

41 41 43 46 52 61 66 70 74

77 77 78 89 98 105 110 114 116 127 134 139 146 154 157

X

CONTENTS

Chapter 4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Chapter 5 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Chapter 6 6.1 6.2 6.3 6.4

Chapter 7 7.1 7.2 7.3 7.4

7.5 7.6 7.7

7.8

Chapter 8 8.1 8.2 8.3 8.4

Chapter 9 9.1 9.2

Multiple Integrals Area and volume The double integral Iterated integration Double integrals in polar coordinates The triple integral Mass and moments Cylindrical and spherical coordinates The area of an explicit surface

Infinite Series Sequential limits Convergence and divergence of series Comparison tests for positive-termed series Alternating series Absolute and conditional convergence The ratio and root tests Power series Differentiation and integration of power series The binomial series

Line Integrals Line integrals with respect to arclength Line integrals of vector functions Green's theorem Exactness and path-independence

Advanced Concepts in Infinite Series Rearrangement of series Multiplication of series The substitution theorem for power series Inversion of power series Pointwise and uniform convergence of function sequences Termwise integration and differentiation Series of functions Complex series and elementary complex functions

Improper Integrals and Integrals with Parameter The definition of improper integral Convergence tests for improper integrals Proper integrals with parameter Improper integrals with parameter

Introductory Differential Equations Differential equations Families of curves and associated differential equations

168 168 171 176 183 189 195 204 210 217 217 226 234 240

243 245 248

259 267 271 271 276 283 288 294 294 298 303 306 308 318 324 329 341 341 350 355 359 367 367 370

CONTENTS

9.3 9.4 9.5 9.6 9.7 9.8 9.9 9.10 9.11

Chapter 1 0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 10.10 10.11

Chapter 11 ll.l 11.2 11.3 11.4 ll.5 11.6 11.7 11.8 11.9

Chapter 12

Initial-value problems and existence-uniqueness theorems Separable equations Homogeneous equations First-order linear equations Exact equations Orthogonal trajectories Applications to elementary problems in motion Picard's method of successive approximations Proof of the existence-uniqueness theorem

373 375 379 384 387 393 395 396 398

Linear Differential Equations

404

Existence and uniqueness Second-order equations Linear dependence and independence Differential operators with constant coefficients Non-homogeneous equations: undetermined coefficients Variation of parameters Reduction of order by known solutions The Euler equation Applications to vibrational systems Applications to electrical circuits Systems of differential equations

The Laplace Transform The definition of Laplace transform Table of transforms Algebraic properties Inverse transforms Transforms of derivatives Solving differential equations by transforms Systems of equations solved by Laplace transforms Derivatives of Laplace transforms Convolution theorem

Series Solutions

12.1 Power series solutions 12.2 Power series solutions: successive differentiations 12.3 Solutions about a regular singular point 12.4 Bessel's equation

Chapter 13 13.1 13.2

XI

Fourier Series Trigonometric Fourier series Convergence of Fourier series

404 408

415 424 431 437 441 444 448

455 458

466 466 469

473 476 479 481 483 484 486

493 493 499 501 504

510 510

518

:Xll

CONTENTS

13.3 Half-range Fourier series 13.4 Differentiation and integration of Fourier series 13.5 Fourier series with respect to orthogonal sets 13.6 Application to the vibrating string problem

523 527 531 533

Appendix A

541

Selected Answers

543

Index

553

VECTOR CALCULUS and Differential Equations

THE UNIVERSITY SERIES IN MATHEMATICS Editors: John L. Kelley, University of California, Berkeley Paul R. Halmos, University of Hawaii Frederick W. Gehring, University of Michigan PATRICK SuPPEs-lntroduction to Logic PAUL R. HALMOs-Finite-Dimensional Vector Spaces, 2nd Ed. EDWARD J. McSHANE & TRUMAN A. BoTTs-Real Analysis JOHN G. KEMENY & J. LAURIE SNELL-Finite Markov Chains PATRICK SuPPEs-Axiomatic Set Theory PAUL R. HALMOs-Naive Set Theory JoHN L. KELLEY-Introduction to Modern Algebra IVAN NIVEN-Calculus: An Introductory Approach A. SEIDENBERG-Lectures in Projective Geometry MAYNARD J. MANSFIELD-Introduction to Topology FRANK M. STEWART-Introduction to Linear Algebra LEON W. CoHEN &GERTRUDE EHRLICH-The Structure of the Real Number System ELLIOTT MENDELSON-Introduction to Mathematical Logic HERMAN MEYER-Precalculus Mathematics ALBERT G. F ADELL-Calculus with Analytical Geometry JOHN L. KELLEY-Algebra: A Modern Introduction ANNITA TULLER-A Modern Introduction to Geometries K. W. GRUENBERG & A. J. WEIR-Linear Geometry HowARD LEVI-Polynomials, Power Series, and Calculus ALBERT G. F ADELL-Vector Calculus and Differential Equations

VECTOR CALCULUS and Differential Equations

1 • Vectors

1.1 Elementary vector geometry of 3-space Scalar quantities are real numbers which measure physical phenomenapressure, speed, mass, and the like. The mathematical model of scalar quantity is then simply a real number. In contrast, vector quantities are representations of physical phenomena which have both magnitude and direction-displacement, velocity, acceleration, force, moments, and the like. To picture vector quantities, physicists (and mathematicians follow suit) use arrows: the length of the arrow indicates size or magnitude of the vector quantity, while the arrow head (together with the line along which the arrow lies) tells the direction. The arrow is especially suitable for describing pictorially a displacement of a particle, since the tail of the arrow would indicate initial position and the head terminal position. To handle the concept of arrow mathematically one properly begins with ordered pairs of points.* In Chapter 15 of Calculus with Analytic Geometry (Fadell), we treated vectors in 2-space from this point of view to show in small compass how one might proceed systematically (and without undue abstraction) to develop vector geometry with a reasonably high degree of rigor and clarity. We now make the natural generalization to 3-space, but in less detail and formality. Once the concept of vector is properly defined, we work informally in an intuitive geometric way in order to reach as quickly as possible the threshold of an elegant generalization called a vector space (§§ 1. 7 and 1.8), which is purely algebraic and has far-reaching consequences. Accordingly, many of the "proofs" which are supplied in the next few sections are in effect outlines of proofs which are designed more to convince the reader than to guarantee validity of the assertions. Indeed, quite often we appeal to the eloquence of a carefully selected picture. At other times, we presume all of the power of Euclidean geometry and trigonometry, without justifying these foundations rigorously. Now that the rules have been honestly spelled out, we can proceed. Given two points A and B in R 3 , we call the ordered pair AB a bound vector with initial point A and terminal point B. Then by a position vector we mean a bound vector with the origin as initial point. Thus every point P is the terminal point of the position vector OP. Two bound vectors are said to be vectorially equivalent if and only if their corresponding coordinate differences are equal. By a

* Anyone versed in elementary abstract set theory knows that an ordered pair of objects, usually written (A, B), may be defined to be the set {{A, B}, {A}}. (This definition of ordered pair is due to K. Kuratowski.) It is as if the set {A, B} denotes the two ends of an arrow (without distinction) and thesingleton{A} is the selection of A as the tail. Thus the notion of arrow (or displacement or bound vector) is perfectly and naturally carried by the concept of ordered pair of points. I

2

VECTORS

free vector U we mean the set of all bound vectors vectorially equivalent to some bound vector AB. Then we write U = AB, and say that AB is a representative of (or represents) U. Every free vector is represented by a position vector. In fact, two free vectors are alike if and only if their representative position vectors have the same terminal point. Thus there is a 1-1 correspondence between the set offree vectors in R3 and the set of points in R3 • If we picture a bound vector as an arrow with its tip at the terminal point and the tail at the initial point, we may view a free vector as a shower of arrows of equal length pointing in the same direction (Fig. l.IA). However, we usually ~

FIG. LIB. Vector addition.

FIG. l.IA. Free vector.

think of a free vector as a single arrow which may be moved to another position by a parallel displacement. This accounts for the adjective free. For simplicity let us agree to allow the single term vector to mean free vector in the sequel. Mter the concept of vector space is introduced in §1.7, the term vector will be generic for a host of other mathematical entities which are governed by a set of rules called the axioms of a vector space. ~ The length or magnitude lUI of a vector U = AB is defined by (I)

lUI = dist(A, B)

Vectors of length I are called unit vectors. The zero vector 0 is the vector oflength 0, corresponding to arrows which issue from and end at the same point. Thus ~ ~ 0 = AA = BB, etc. Vector addition is defined by the triangle law (Fig. l.IB)

(2) (3)

U + V

~

=

AB ---+

---+

+ BC

~

---+

=

AC ~

V + U = AD + DC = AC ---+

Notice (see Fig. I.IB) that the vector sumAC is also obtainable as the diagonal ~ ~ vector of a vectorial parallelogram with AD and AB as adjacent sides.

§ 1.1 Elementary vector geometry of 3-space

FIG. l.lD. Associative law.

FIG. l.lC. Commutative law.

From (2) and (3) it is clear that vector addition is commutative (Fig. 1.1C) :

W

U+V=V+U

Fig. 1.1D discloses that vector addition is also associative:

(5)

(U+V) +W=U+ (V+W)

Because of associativity we may write U + V + W without parentheses, and group at will. Since AA + AB = AB and AB + BB = AB we have

(6)

--

-- -- -- -- -- -

-

that is, the zero vector is an identity for vector addition. If AB = U, write BA = - U. Then from (2)

(7)

U + (- U) = 0 = (- U) + U

Thus, every vector u has an additive inverse - u which has the same length but points in the opposite direction (see Fig. 1.1E). The equality - U + U = 0 in (7) tells us that - (- U) = U. The difference U - V is defined by (8)

U-V=U+(-V)

that is, to subtract a vector, add its additive inverse (see Fig. 1.1F). By adding V to both sides of (8) we obtain

(9)

(U-V) +V=U

-

Thus U- Vis the vector which when added to V yields U (see Fig. l.IG). A vector given in the form AB may be expressed as a difference of position vectors (see Fig. l.IH) : (10)

-AB

=

~

~

OB- OA

For any real number c =I= 0, we denote by cU the vector which is lei times as long as U, and points in the same or the opposite direction as U according as c > 0 or c < 0. We agree that OU = 0. We call cU a scalar multiple of U. Scalar multiplication ofU by c may be thought of as a dilation or contraction

4

VECTORS

u-v

Fro. l.lE. Additive inverse.

Fm. l.lF. Subtraction by FIG. l.lG.Triangle law of subadditive inverses. traction.

ofU, followed by a reversal of direction if c < 0 (see Fig. 1.11). In particular, all vectors may be obtained from the unit vectors by scalar multiplication. The following rules hold for scalar multiplication:

(11)

!cUI = lei lUI -cU = c(-U)

(12)

(13)

1U = U

(14)

a(bU) = (ab)U

&

(ABSOLUTE HOMOGENEITY)

&

(-c)(- U) = cU

(-1)U= -U (AssOCIATIVITY)

B

FIG. l.lH. Vectors as differences of position vectors.

FIG. l.ll. Scalar multiplication.

The following two rules connect vector addition and scalar multiplication: (15) (16)

(a + b)U = aU + bU (Distributivity over scalar addition) a(U

+ V) = aU + aV (Distributivity over vector addition)

Formulas (11)-(16) follow from the definition of scalar multiplication. Fig. l.lj shows that (16) is the result of similar triangles. ~ We have already noted that every vector can be written as U = OP. The projections of P onto the X-, Y-, and Z-axes, say Ph P2 , and P3 , respectively, satisfy (17) OP1 + OP2 + OP3 = OP = U ~

~

~

~

§ 1.1 Elementary vector geometry of 3-space

5

y

FIG. l.lj. Distributive law.

-

FIG. l.lK. Vector components.

--~---since (see Fig. l.lK) OP = OP1 + P 1 P = OP1 + OP2 + OP3 • But OP1 = u 1i, O P2 = u:J, and OP3 = u3k, where i,j, and k are the unit vectors from the origin to the points (1, 0, 0), (0, 1, 0), and (0, 0, 1), respectively, and P = (u1 , u2 , u3 ). See Fig. l.lK. From (17) we therefore infer that U has the form (18)

U

=

-

OP = u1 i + u:J + u3 k

where

P = (u~o u2 , u3 )

Thus, every vector is determined by an ordered triple of real numbers, called the components of the vector. The unit vectors i, j, and k are called the coordinate base (or basis) vectors. Then in words, (18) reads: every vector is the sum of certain scalar multiples of the coordinate base vectors. This representation sets the stage for wholesale use of the algebraic rules of vector addition and scalar multiplication.

-

~ Example 1. Let P = (-1,4, -2) and Q= (2, -3,5). Then OP + OQ= ( -i + 4j- 2k) + (2i- 3j + 5k) = ( -1 + 2)i + (4- 3)j + (-2 + 5)k = i + j + 3k. ---'-+ ~ Similarly, PQ = OQ- OP = 3i - 7j + 7k.

Because the length ofU in (18) is the same as the distance between 0 and P, we can write lUI in terms of the components ofU: lUI = v'u~

(19)

+ u~ + u~

Example 2. Find the unit vector in the direction of U SOLUTION.

v

=

~ = 3i -

lUI

V3 2

IUI 2 =

u~

sj + 8k = + 52 + 82

1 (3i V98

+ u2 j + u3 k = 0 + u~ + u~ = 0 ~ u1 =

Example 3. Show that u1i SOLUTION.

= 3i- 5j + 8k.

The desired vector is

u2 , and u3 • Then the position vector corresponding to [uh u2, u3] has its tip at the point (ui> u2 , u3). See Fig. l.lL. In particular, (23)

i = [1, 0, 0],

j = [0, 1, 0],

k = [0, 0, 1]

I

(up u 2 , u 3 ]

I

u2 1 cu2 ~---t-~::>-1------p-

......

/

I

- - . . . . . . -:::J;' / - - - - _...........__..._J/

/

Fig. l.lL. Scalar multiplication by components.

Fm. l.lM. Vector addition by components.

Then (20) and (21) may be written as (see Figs. l.lL and M): (24) (25)

[uh u2 , u3]

+

[vh v2, v3]

=

[u1

+ vh u2 + v2 , u3 + v3]

c[u1 , u2 , u3 ] = [cui> cu2 , cu3 ]

Formulas (24) and (25) may be exploited (see §1.7) to lay a purely algebraic foundation for vector algebra based primarily on the field properties of the real numbers.

§ 1.2 Linear dependence and independence

7

Exercises §1.1 In Exercises l-4 form the vector 3U + 2V and find I3U + 2VI. 1. 3.

u = 2i - 3j + 4k, v = i - 2j + u = i + j, v = 3i - 2j + 4k

k

2. u = i - j + k, v = 2i - j + 3k 4. U=i-k, V=j+k

-- --

In Exercises 5-8, write P1 P 2 in the form [uu u2 , u3 ] and also in terms of the base vectors i, j, and k: 5. P 1 = (-1,2,3), P 2 = (0, -4,2) 7. P1 = (0, O,p), P2 = (I, -2, 3)

6. P 1 = (-3,0,2), P 2 = (2, I, I) 8. P1 = (-I, O, 4), P2 = (7, -II, I)

In Exercises 9-12 find P 1P 2 + P 2P 3 and P 1P 2 - P2P3 : 9. 10. 11. 12.

P1 = (0, 0, 0), P 2 = (I, I, I), P 3 = (-I, 4, 6) P 1 =(I, -I, I), P 2 = (0, -1,3), P 3 = (4,2, -I) P 1 = (-1,2,3), P 2 = (3,4, I), P 3 = (-4,2,3) P1 = ( -2, 2, 3), P 2 = (3, I, -2), P 3 = ( -2, -2, I) In Exercises 13-16, find the unit vector in the direction of the given vector:

13. 2i- 3j + 4k 14. i + j - k 15. i + j - k 16. 3i - k 17. Draw a figure to illustrate the following vector equations:

+V +W = 0 (b) U + V + W + X = 0 (c) U + U + U = 3U (d) (U + V) + (W +X)= (U + V + W) +X (e) U + V + W = W + V + U (f) 2U + 2V + 2W = 2(U + V + W) (a) U

18. Show that u1i + u2j + u3k = v1i + v2 j + v3k

(u1 = v1 & u2 = v2 & u3 = v3 ).

19. Using (24) and (25) of this section show that a[ul> u2 , u3 ] + b[vl> v2 , v3 ] = [au 1 + bv 1 , au 2 + bv2 , au3 + bv3 ]

20. Give a geometric meaning to the following inequalities: (a) IU + VI ~ lUI + lVI

(b) IU - VI ~ I lUI - lVII

1.2 Linear dependence and independence Given n vectors Uh ... , Un and n scalars all ... , an we say that n

~ a1U 1 = a 1U 1

+ · · · + anUn

I= 1

is a linear combination of the Ut's, and we call the at's the coefficients. We know for example that every vector in 3-space is a linear combination of i, j, and k, and that the coefficients are the components. The concept of linear combination plays a principal role in the following definition of linear independence of vectors:

(1)

{U1 ,

••• ,

Un} is linearly independent (a 1U 1 + · · · + anUn = 0

=>

a1 = · · · = an = 0)

8

VECTORS

In words, linear independence of a set of vectors means that the only way to make the zero vector out of a linear combination of the set is to take all coefficients to be zero. A good example of a set of independent vectors is {i, j, k}: see Example 3 in §1.1. We say that a set ofvectors is linearly dependent the set is not linearly independent. In other terms,

{U1 ,

Un} is linearly dependent there exist scalars ah ... , an not all zero such that a1U 1 + · · · + anUn = 0 Observe that any set of vectors containing the zero vector must be dependent. (Why?) Usage allows the expression "Uh ... , Un are linearly dependent [resp. independent]" to mean the "set {U1 , . . . , Un} is linearly dependent [resp. independent]". At least one of a set of linearly dependent vectors can be expressed as a linear combination of the remaining vectors of the set. In particular, if a1U 1 + · · · + anUn = 0 and a1 =/: 0, then U 1 = b2 U 2 + · · · + bnUn where b, = atfaH i = 2, ... , n. This leads to the phrase "U1 is linearly dependent on U 2 , . . . , Un" meaning that U 1 , U 2 , .. . , Un are linearly dependent. (2)

. .. ,

Example 1. Let U 1 = 6i + 9j + 15k and U 2 = 4i + 6j + lOk. Then U 1 and U 2 are linearly dependent, since 2U1 - 3U2 = 0. Notice that U 1 = (!)U 2 and u2 = (f)Ul. Example 2. Let U 1 = 2i- j + k and U 2 = i + 2j- 3k. To check on linear dependence or independence we look at the component form of an arbitrary linear combination ofU1 and U 2 : a 1U 1

+ a2U 2

=

(2a 1

+ a2 )i +

(2a 2

-

Then a 1U 1

+ a2U 2

=

0

=-

a1 )j

+

2a 1 { 2a 2

-

a1

-

+

a2 a1 3a2

(a 1 = =

=

-

3a2 )k

0 0 0

The sum of the last two equations gives us a2 = 0, whence the third equation requires a 1 = 0. Thus, U 1 and U 2 are linearly independent. This means that neither U 1 nor u2 is a scalar multiple of the other.

The concept of linear dependence and independence has vivid geometric ~

~

pictures. Suppose U 1 = AB and U 2 = AC are linearly dependent. This means that either U 1 or U 2 is a scalar multiple of the other, say AC = tAB. See Fig. 1.2A. Then A, B, and C are on the same line. Accordingly, two linearly dependent vectors are often also called collinear vectors, while two linearly independent vectors are called non-collinear. Suppose U 1 = AB, U 2 = AC, and U 3 = AD are linearly dependent. This means that one of the three vectors is a linear combination of the other two, say AD = sAB + tAC. See Fig. 1.2B. Then AD is the diagonal of a parallelogram with sAB and tAC as adjacent edges. Thus A, B, C, and D are on the same plane. Accordingly, three linearly dependent vectors are often called coplanar vectors, while three linearly independent vectors are called non-coplanar. ~

~

~

~

~

~

~

~

~

~

~

§ 1.2 Linear dependence and independence

9

D

A

A

FIG. 1.2A. Linear dependence of two vectors: collinearity.

of three vectors: coplanarity.

FIG. 1.2B. Linear dependence Fro. 1.2C. Linear dependence of four vectors.

All quadruples of vectors in 3-space are linearly dependent, since if three of them are non-coplanar, the fourth is the diagonal of a parallelepiped having three coterminal edges collinear with the three non-coplanar vectors (see Fig. 1.2C). As Example 2 illustrates, the question of linear dependence or independence rests on the problem of solving a system of linear equations. Consider in general the important question of linear dependence or independence of three vectors, say U = [uh u2, u3], V = [vh v2, v3], and W = [wb w2, w3]. We are inquiring into the nature of the real numbers x, y, and z (playing the roles of ah a2, and a3 in ( 1) and (2)) satisfying the vector equation (3)

x[ub u2, u2 ]

+ y[vb v2, v3 ] +

z[wb w 2, w 3 ]

= [0, 0, 0]

If we multiply (3) out and compare components we see that it is equivalent to the system of equations u1 x

(4)

u2x u3 x

+ v1 y + w1 z = 0 + v2y + w 2z = 0 + v3 y + w3 z = 0

We assert that (4) has a non-trivial solution (that is, one of the three numbers

x, y, or z is not zero) if and only if (5)

det

r:: :: ::] Ua

Va

=

0

Wa

That is, {U, V, W} is dependent [resp. independent] if and only if the determinant of the matrix of their components is zero [resp. is non-zero], where the matrix is formed by taking U, V, and Was rows or as columns.

10

VECTORS

To prove this assertion in one direction, suppose that (5) fails to hold, that is, the determinant of the coefficients in (4) is non-zero. Then at once we know by Cramer's rule that x = y = z = 0. (Why?) Now to complete the proof, suppose that (5) does in fact hold. There are two cases to consider. CASE 1. All minors in (5) are zero: Then it is easy to show that the second and third rows, for example, are multiples of the first, and so every solution (and there are infinitely many!) of the first equation in (4) also satisfies the other two. CAsE 2. At least one minor in (5) is not zero. Suppose for concreteness that u1 v2 - u2 v1 =F 0. Then by Cramer's rule the first two equations are satisfied by*

(6) for every real number t (note that z =F 0 for t =F 0). Substituting (6) into the third equation in (4) we get the Laplace expansion of (5), which by assumption is zero. Thus if (5) holds, not only does there exist a non-trivial solution of (4)-there are infinitely many. Hence our assertion is proved. In a similar way it is easy to see that every vector C = [cl> c2 , c3 ] is a linear combination of any three linearly independent vectors U, V, and W, since we are merely asking for a solution (x, y, z) of the system of equations representing the vector equation C =xU + yV + zW, namely, xu 1

(7)

xu 2 xu 3

+ yv1 + zw1 = c1 + yv2 + zw2 = c2 + yv3 + zw 3 = c3

Example 3. Show that the three vectors U = [1, -2, 0], V = [0, 1, 4], and = [0, - 1, - 3] are linearly independent, and express C = [ -1, 2, 3] as a linear combination ofU, V, and W. Why is {C, U, V, W} a dependent set? sOLUTION. U, V, and Ware linearly independent since

W

1 -2 [ det 0 1 0

-1

~]=l=FO

-3

To express Cas a linear combination ofU, V, and W we must solve the vector equation [ -1, 2, 3] = x[l, -2, 0]

+ y[O, 1, 4] + z[O,

-1, -3]

which may be written as the system of linear equations

=

X

-2x

+

y- z = 4y- 3z =

-1 2 3

By Cramer's rule (or by triangularization, or otherwise) we find that x = -1, y = 3, and z = 3. Thus, C = - U + 3V + 3W. This means that C is dependent on U, V, and W, or equivalently the set {C, U, V, W} is dependent. *For details see Calculus with Analytic Geometry (Fadell), Example 3, §16.12.

§ 1.3 The dot product

11

Exercises §1.2 In Exercises 1-4 determine whether the vectors are linearly dependent or independent: 1. u1 = 2i + 3j + k, u2 = 4i + 6j + 2k 2. u1 = i + j, u2 = i - j 3. U 1 = i + j + k, U 2 = i - j + 3k, U 3 4. U 1 = 2i + 4j, U 2 = i + 2j, U 3 = k

=i +k

In Exercises 5-8 solve the equations xU indicate that a solution fails to exist:

+ yV +

zW

= C for x, y, and z or

5. u = 2i + 3j - k, v = i - j + k, w = 2i + j - k, c 6. u = i + j, v = j + k, w = k + i, c = 3i - 2j + k. 7. u = 2i- j, v = i - 2j, = k, c = 2i- j - k. 8. u = i + j, v = j, w = 2i - k, c = i. 9. Show that if U 1 and U 2 are linearly independent, then

= 3i - 4j + 3k.

w

a1U 1

+ a2U 2 =

b1U 1

+ b2U 2 -=

(a1 = b1 & a2 = b2 ).

10. Show that ifU~> ... , U 11 are linearly independent, then II

II

1=1

1=1

L aiUI = L biUI -=

al

=

b~o

i

=

1, ... , n.

11. Show that any subset of a linearly independent set is a linearly independent set. 12. Show that if U 1 , ••• , U 11 are linearly independent, then so is any set of non-zero scalar multiples a1U1o ... , a11U 11 • 13. Show analytically that any four vectors in 3-space are linearly dependent. OuTUNE: Assume that four vectors are linearly independent, and then obtain a contradiction using the result of Exercise 11 and a suitable system of linear equations as in (7). 14. Indicate geometrically (using projections of points on suitable planes) how the edges rAB, sAC, and tAD are obtained to suit Fig. 1.20, given A, B, C, D, and E. 15. Using linear equations argue that two vectors U = [u 1 , u2 , 0] and V = [v1 , v2 , 0] are linearly independent if and only if u1 v2 - u2 v1 =F 0. 16. Show that U = [ul> u2 , u3] and V = [vl> v2 , v3 ] are linearly dependent if and only if

1.3 The dot product The dot or scalar product of two vectors U and V, denoted by U • V, is defined as the product of their magnitudes and the cosine of their included angle. Then if (U, V) denotes the angle between U and V with 0 :S (U, V) :S TT, we set

(1)

u. v =

{IUIIVI cos (U, V),

0

'

ifU=FO=FV if U = 0 or V

=0

Observe that special attention was given to the zero vector in (1), since it has no well-defined direction. However, in the sequel we will agree that the zero vector has all directions. Thus out of an ordered pair of vectors the dot product

12

VECTORS

produces a real number which is positive, negative, or zero according as the angle (U, V) is acute, obtuse, or right. Example 1. Show that i, j, and k are orthonormal in the sense that (2) SOLUTION:

i. i = lillil cos 0 =

and

i. j =

(1)(1)(1) = 1

lilljl cos (7r/2) =

(1)(1)(0) = 0,

and similarly for the other equalities.

From ( 1) we may derive several significant properties of the dot product. But each of these derivations is based on the notion of angle or some other intuitively acceptable geometric notion. In §1.8 we work on a purely algebraic basis and are able in an elegant way to get all the properties of the dot product on a high level of rigor and logical clarity. However, it is the geometric setting of the dot product which leads us to the threshold of an algebraic characterization. Basic to this geometric view is the idea of component (or proJection) of a vector (see Fig. 1.3A), which expresses the dot product as follows:

U · V = JUJ compu V

(3)

where compu V = JVJ cos (U, V). Verbalize compu Vas "the component of V in the direction of U ", or "the orthogonal projection of V onto U ". A particularly useful observation is that the operation of taking components in the direction of a given vector is additive. That is (see Fig. 1.3B), (4)

compu (V

+ W) = compu V + compu W

which gives us JUJ compu (V

+ W) = JUJ compu V + JUJ compu W

and in turn, by translating back into dot-product notation via (3), the distributive law of the dot product

(5)

U·(V+W) =U·V+U·W

FIG. 1.3A. The dot product of two vectors: U·V =lUI compu V.

FIG. 1.3B. Distributivity of dot product via additivity of components.

§ 1.3 The dot product

13

From ( 1) we have

(-U) ·V = U· (-V)

(6)

=

-U·V

since (see Fig. 1.3C) (- U, V) = (U, - V) = 1r - (U, V). real a and b

(7)

Indeed, for any

(aU) · (bV) = (ab)(U • V)

since for a > 0 and b > 0 we have (see Fig. 1.3D) (aU, bV) = (U, V) and laUIIbVI = labiiUIIVI, from which the case a < 0 andfor b < 0 is settled via (6). If a = 0 or b = 0, (7) holds since both sides are zero.

Fm. 1.3C. Angle relationships between veetors and additive inverses.

FIG. 1.3D. Angles between scalar multiples of vectors.

Since (U, V) = (V, U) (because we are using the absolute measure of the angle) we infer commutativity of the dot product from ( 1) :

U·V = V·U

(8)

We say that U and V are orthogonal (and write U j_ V) (U, V) = 1rj2. The dot product then gives us the important orthogonality criterion:

U_LV U·V=O

(9)

Observe that the zero vector is orthogonal to every vector.

(Why?)

Example 2. Let U = 3i + 4j + 2k and V = 2i - j - k. are orthogonal. SOLUTION. Applying (5), (7), (8), and (2), we have

Show that U and V

u .v

=

=

=

(3i + 4j + 2k) • (2i - j - k) (2i)(3i) + (2i)(4j) + (2i)(2k) + (-j). (3i) + (-j). (4j) + (-j). (2k) + (-k) . (3i) + (-k) . (4j) + (-k) . (2k) (2)(3) + (-1)(4) + ( -1)(2) = 0

Therefore by (9), U and V are orthogonal.

14

VECTORS

The computations in Example 2 above may be carried out for an arbitrary pair of vectors, say U = u1i + u2 j + u3 k and V = v1 i + v2 j + v3 k, using (5), (7), (8), and (2). Our result: U · V = u1 v1

(10)

+

u2 v2

+

UaV3

That is, the dot product is obtained by multiplying corresponding components and adding. From (10) and (1) we obtain a formula for the angle between two non-zero vectors in terms of their components:

(11) Example 3. Let P

=

(I, -2, 3), Q = (2, 0, 2), and R = (0, -I, I). Find the -+

-+

cosine of the angle between PQ and PR. What is the angle? SOLUTION. Let -+ U = PQ = i + 2j - k -+

+ j - 2k cos (U V) = U • V = (I) (- I) + (2)( I) + (- I) (- 2) = ! ' IUIIVI Vl + 4 + 1 Vl + 1 + 4 1 (U, V) = cos- (I/2) = 7Tf3. V = PR = - i

Therefore

The dot product gives us another way of expressing the magnitude of a vector via (10), namely,

(12)

U · U = IUI 2 = u~ + u~ + u~ lUI = VU · U = V u~ + u~ + u~

Example 4. Let U = 2i + 3j - k and V = 3i - 2j + k. Find the vectors of unit length which are orthogonal to both U and V. SOLUTION. If W = xi + yj + zk is a unit vector orthogonal to both U and V, then the following equations must hold: U • W = 2x + 3y - z = 0 V • W = 3x - 2y + z = 0 W·W = x2 + y 2 + z 2 = 1

Solving the first two equations for y and z in terms of x, we obtain y = - 5x and

z

= - l3x

whence the third equation gives x2 = l/195. Thus and z = + l3/Vl95, so that

X

w = ±.y~ 95 (i- sj-

= ± l/Vl95, y =

+s;VT95,

I3k)

Example 5. Show that every vector U may be expressed in the form (13)

U

=

(U • i)i

+

(U • j)j

+

(U • k)k

§ 1.3 The dot product

15

SOLUTION. Let U = u1i + u2j + u3k. Dot both sides by i and apply distributivity of the dot product (see (5)), and the orthonormality ofi, j, and k (see (2)). Then

U • i = u1 (i • i) + u2(j • i) + u3 (k • i) = u1 Similarly, U • j = u1 (i • j) + u2(j • j) + u3 (k • j) = u2 U • k = u1 (i • k) + u2(j • k) + u3 (k • k) = u8 Example 6. Show that every vector may be expressed in the form (see Fig. l.3E) (14)

U = IUir where r = (cos a:)i + (cos {J)j + (cos y)k where a: = (U, i), fJ = (U,j), y = (U, k) and cos 2 a: + cos 2 fJ + cos 2 y = I

FIG. l.3E. Radial unit vectors, direction angles, and direction cosines.

SOLUTION. By (I), U • i = lUI cos (U, i), U • j = lUI cos (U,j), and U • k = lUI cos (U, k). Substituting into (13) we have (14), if we observe that (see (12)) IUI 2 = (U • i) 2 + (U • j) 2 + (U • k) 2 so that ·

• • cos2 (U, 1) + cos 2 (U, J) + cos 2 (U, k)

(U. i)2

= 1Uf2 +

(U. j)2

(U. j)2

1Uf2 + 1Uf2 =

IUI2 IUI 2 = I

Exercises §1.3 1. Let U = 2i - 3j + k, V = i + j - 3k, W the following pairs of real numbers: (a) U·V + U·W and U· (V + W) (c) U • U and IUI 2

= i - j + 4k. Compute each of

(b) U • V and V · U (d) 3(U • V) and (3U) • V

2. Show that if A = aU + bV + cW, where U, V, and Ware pairwise orthogonal non-zero vectors, then

a

=

(A· U)/(U • U),

b = (A· V)/(V • V),

c = (A·W)f(W·W)

3. Show that if U = aT + bN + cB, where T • T = N • N = B • B = I and T • N = N • B = B • T = 0 (that is, T, N, and B are orthonormal), then

a = U • T, b = U • N, and c = U • B 4. Suppose aU+ bV + cW = 0 where U, V, Ware pairwise orthogonal non-zero vectors. Show that a = 0 = b = c. How does this show that triples of orthogonal non-zero vectors are always non-coplanar?

16

VECTORS

5. Show that the sum of the squares of the sides of a parallelogram is equal to the sum of the squares of the diagonals. OuTLINE: Consider (U - V) • (U - V) and (U + V) • (U + V). 6. Show that the diagonals of a rhombus (parallelogram with equal sides) are orthogonal. OUTLINE: Consider (U - V) • (U + V). 7. Let U and V be two non-collinear vectors orthogonal to a vector W. Show that if X is any vector coplanar with U and V, then W is orthogonal to X. What basic theorem in solid geometry is thus proved vectorially? OuTLINE: Use the dot product criterion for orthogonality along with the fact that all vectors coplanar with U and V are of the form aU + bV where a and b are scalars. 8. Prove vectorially that a parallelogram with equal diagonals is a rectangle. OUTLINE: Consider (U + V) • (U + V) = (U - V) • (U - V). 9. Prove vectorially that the diagonals of a square are equal and orthogonal. OuTLINE: Consider (U + V) • (U- V), (U + V) • (U + V), and (U- V) • (U- V) under the assumptions lUI = lVI and U • V = 0. 10. Let U, V, W be the vector sides of a triangle, where W = U - V. Deduce the law of cosines as a consequence of the identity W • W = (U - V) • (U - V). 11. Find cos (U, V) and (U, V), given U = 2i + j + k and V = i + 2j + 4k. 12. Let U = 3i + 4j - 6k. Find the angles (U, i), (U,j), and (U, k) as values of the inverse cosine function. Show these three angles on a diagram on which U is plotted as a position vector. 13. (a) Given U = 2i- j + 3k and V = i - 3j + k, find compu V and compv U. (b) Find (compu V)U/IUI and (compv U)V/IVI, and indicate their physical interpretation ifU or Vis a force vector. 14. (a) Find the three angles which the diagonal of a cube makes with the edges. (Express the answers as inverse cosine values.) (b) Find the three angles made by a diagonal of a rectangular box with coterminal edges a, b, and e, respectively. Show that the sum of the squares of the cosines ofthese angles is equal to l. 15. (a) Give a geometric meaning to the following inequalities, where U is any vector in 3-space: max CIU • il, IU • Jl, IU • kl) ~ lUI ~ IU • il + IU • Jl + IU • kl (b) Show that the above inequalities are a vectorial restatement of the following arithmetic inequalities: max (lal, lbl, lei) ~

v'a2

+ b2 + e2 ~ lal + lbl + lei

16. Show that (U • V)V - (V • V)U is orthogonal to V. Let U = 2i - 3j + k and V = i + j - 2k. Find W = aU + bV such that W is orthogonal to U. 17. (a) Let U =cos 8i +sin Bj, and V = cos (8 + TT/2)i +sin (8 + TT/2)j. Show that (U, V) = TT/2, using the dot product. (b) Show that if U = cos 8i +sin 8j and V = -sin 8i + cos Bj, then (U, V) = TT/2. 18. Using dot products, show that U .l V if and only if (aU) .l {bV) for every pair of non-zero real numbers a and b. 19. Show that IU +VI = lUI + lVI implies UIIV. OUTLINE: Square both sides; use IWI 2 = W·W. 20. Show that IU - VI = I lUI - lVI I implies UIIV. (See the preceding exercise.) 21. Show that IU + Vl 2 = IUI 2 + IVI 2 if and only if U • V = 0. What does this result have to do with right triangles?

§ 1.4 The vector (or cross) product

17

ll. Show that IU + Vl 2 23. l4. 25. 26. 27.

28. 29. 30.

::5: (lUI + IVI) 2 if and only if U · V ::5: IUI·IVI. OUTLINE: Use W·W = IWI 2 • If U • V = IUIIVI, what must be true about U and/or V? Find two vectors of unit length which are orthogonal to both U = i + j - 3k and V=2i+j-k. Let U be a unit vector, and let V be an arbitrary vector. Show that U is orthogonal to V- (V·U}U. Let U be a unit vector and letA be an arbitrary vector. Suppose that A= aU+ V and A= bU + W where both V and W are orthogonal to U. Show that a= band V = W. A triple of vectors {e1, e 2, e 3} is called orthonormal e 1 • e 1 = 0 if i =/: j but = I if i = j. Let U = a1e 1 + a2e 2 + a3e 3 and V = b1e 1 + b2e 2 + b3e 3, where {eh e 2, e 3} is an orthonormal triple. (a} Show that U • V = a1 b1 + a2 b2 + a3 b3 • (b} Show as a consequence of (a} that IUI 2 = a~ + a~ + a~. Criticize the "theorem": if U • V = U • W, then V = W. Show that in fact U•V = U•W implies U 1. (V- W}. Let U, V, and W be three linearly independent vectors. Find real numbers a, b, c, d, and e such that U, aU + bV, and cU + dV + eW are mutually orthogonal. (This is called the Gram-Schmidt orthogonalization process.} Prove: If X•V = X•W foreveryvectorX, .then V = W. (Comparethiswith Exercise 28.}

1.4 The vector (or cross) product In physics the effect of a given force F about a given point 0, called the moment of the force about the point, is measured as the product of the magnitude

of the force and the distance the line of action of the force is from the given point. Thus, in symbols ~ ~ Moment = IFIIOPIIsin (F, OP) I as illustrated by Fig. 1.4A, where P is the point of application of the force. The moment is directed along the line which is perpendicular to the plane determined by the line of action ofF and the point 0. The orientation of the moment is

Moment M

Fm. l.4A. Moment vector as a vector (or cross) product of a force vector and a position vector.

=F x OP

18

--

VECTORS

the direction a right-handed screw would move if it were turned from OP toward F. The mathematical model of moment offorce is the concept of vector product, which we now formulate. Given a pair of vectors U ~d V, we define the vector product (or cross product) of U and V to be a vector, denoted by U x V, which satisfies the following conditions: IU x VI = {IUIIVIIsin (U, V) 1, ~f U #: 0 #: V 0 , if U = 0 or V = 0

( 1)

MAGNITUDE:

(2)

ORTHOGONALITY:

(3)

ORIENTATION:

(UxV)_LU

(UxV)_LV

&

(U, V, U x V) and (i,j, k) have the same orientation

Fig. 1.4B illustrates the geometric meaning of vector product. Condition (1) tells us that the magnitude of U x V is the area of the parallelogram determined by U and V (see Fig. 1.4C). Condition (2) requires that U x V be normal to planes parallel to both U and V. Notice that (2) could be rephrased: (U

X

V) . u

=0=

(U

X

V) . v.

v ~

. .

.........•

I h =JY I sin (V + W).L}, which in turn dilates into the parallelogram {JUJV.L, JUJW.L, JUJ(V + W).L}, which finally rotates by 7T/2 into the parallelogram {JUJV~, JUJW~, JUJ (V + W)~}, Thus JUJ (V

+ W) ~ =

JUJV~

+

JUJW~

§ 1.4 The vector (or cross) product

23

u

lUI W1

I I I I I

Dilate

I

Project

.....,

I

~(V+WI1 " 1.!!_1 (V + WI 1 -....._

lU I (V +WJ~

Rotate

-----FIG. l.4G. Geometric proof of the distributivity of the cross product.

whence by Step 1 Ux(V+W)=UxV+UxW

Exercises §1.4 In Exercises 1-4 compute U x V, U x W, U x (V (3U) XV: 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

U. 12. 13. 14.

+ W),

u = 4i - j + 2k, v = 3i + 4j - k, w = 2i + 5j - 3k. u = 4i - j + 3k, v = 7i + 2j - k, w = 2i + j - k. u = i + 3j - k, v = i + 2j, w == 3i + k. u = i + 3j, v = 2i- 3j + k, w = -i.

3(U x V) and

Given U = u11 + u:J + uak, compute U x i, U x j, and U x k. Verify the formula IU X VF'1 + IU. Vl 2 = IUF.IIVI 2 • Verify that U• (V' x W) + U• (V" x W) = U• [(V' + V") x W]. Using cross products show that U is collinear with V aU is collinear with bV for every pair of scalars a and b. Show that UIIV IUIIVI = IU • VI. (Use Exercise 6, above.) Use cross products to prove the law of sines for triangles. OUTLINE: Let U, V, and W be the vector sides of the triangle with U + V + W = 0. Cross both sides successively by U, V, and W to get U x V = V x W = W x U. Now consider the magnitudes. Find the unit vectors orthogonal to both 2i - 3j + 4k and 3i - j + 2k. Find the area of the triangle with vertices at P( -1, 4, 3), Q(O, 1, 2), and R( -4, -1, 5). What inference can be drawn from the equality U x V = U x W? Verify formula (21) by matrix multiplication and some skillful algebraic manipulation.

24

VECTORS

15. Do Example 2 using the alternative method of solving two linear equations with an

additional condition. 16. Suppose formula (16), or equivalently (17) or (18), were accepted as the definition of cross product. Obtain (6)-(13) as consequences, using the properties of

determinants. 17. Infer (4) and (5) as consequences of (16).

1.5

Triple products of vectors

Given an ordered triple of vectors U, V, and W, there are only four meaningful triple products using the binary operations of cross and dot, namely U · (V x W), (U x V) · W, (U x V) x W, and U x (V x W). The other four apparent products (U · V) x W, U x (V · W), (U · V) · W, and U · (V · W) are meaningless since U · V and V · W are scalars. We first consider the scalar triple product U · (V x W), so-called because the product is scalar-valued. Suppose U = [u 1, u2, u3], V = [vl> v2, v3], and W = [W1o w 2 , w 3 ]. Then

(1)

Iw

U · (V x W) = u1 v2 2

I

v3 1 + u2 v3 w3 w3

I Iw

V1 + U3 V1

w1

1

V21

w2

so that recognizing the Laplace expansion of a determinant, we have the extremely versatile scalar triple product formula:

(2)

ul

u2

Ua

U · (V x W) = v1

v2

Va

w2 w2

Wa

From (2) we see from the antisymmetry of determinants that U · (V x W) invariant under a cyclic permutation :

(3)

U·(V

X

IS

W) =V·(W XU) =W·(U XV)

since a cyclic permutation of three rows is obtained by two transpositions ofrows. Furthermore, by commutativity of the dot product and by the extremes of (3) we have (U x V) · W = W · (U x V) = U · (V x W), so that equating extremes,

(4)

(U XV) ·W

=

U· (V

X

W)

What (4) tells us is that interchanging the cross and dot gives the same scalar triple product. This is a significant freedom, and leads to the notation [UVW] to mean (without ambiguity) either (U x V) · W or U · (V x W). Furthermore, we may write U x V · W and U · V x W without parentheses, since they have meaning only when the parentheses are properly placed. The scalar triple product [UVW] is the (signed) volume of a parallelepiped with vector edges U, V, and W (see Fig. 1.5A), since (5)

U·V

X

W = V

X

W·U = JV

X

WJ compvxwU

and JV x WJ is the area of the parallelogram base with adjacent edge vectors V and W, and compvxw U is the (signed) altitude of the parallelepiped on this

§ 1.5 Triple products of vectors

25

base. This volumetric view of [UVW] accounts for the alternative name box product. We now can argue geometrically that U, V, and W are coplanar if and only if [UVW] = 0. This proposition is used extensively. In particular, [UVW] = 0 if any two of the vectors aJ;,_e collinear.

FIG. l.5A. [UVW] is the signed volume of a parallelepiped.

compv+w U =!altitude

Example 1. Let U = [ -1, 2, 3], V = [0, I, 2], and W = [1, -3, -1]. Show that U, V, and W are non-coplanar and find the volume of the parallelepiped having U, V, and Was adjacent edges. Also find this parallelepiped's altitude on the parallelogram with edge vectors V and W, and the area of this parallelogram. SOLUTION. Since [UVW]

=

-1

2

3

0

I

2

-3

-1

-4

the vectors are non-coplanar and the parallelepiped has volume 4. Now since

IV

X

Wl2

=I

1

-3

2l2 + I 2 ol2 + lo -1 I 1

-1

2= 30

1 -3 1

we know from (5) and the computed value of [UVW] that compvxw U = -4/v30. Therefore the desired altitude and area are 4/v'30 and V30, respectively. Example 2. Prove the distributive law of cross products using basically the freedom to interchange dot and cross in the scalar triple product. SOLUTION. We prove that U X (V + W) = U XV+ U X W by showing that the components of both sides are equal. Treating the i-th components as a typical case we assert that i · U x (V + W) = i • (U x V + U x W). Indeed i . u X (V + W)

= i X u . (V + W) = (i X U) • v + (i X U) • w

= i. (U

X V) + i. (U X W)

= i. [(U

X V) + (U X W)]

Simple counterexamples show that the two vector triple products U x (V x W) and (U x V) x W are not in general equal. Thus in general associativity fails for the cross product. However, each of the triple products can be shown to be coplanar with the two parenthesized vectors. For example, U x (V x W) is

26

VECTORS

a linear combination ofV and W, as we now show (see (8) below). the components of i k j

(6)

u

X

(V

X

u2

ul

W) = IV2 w2

Val Wa

Iva Wa

We study

Ua

V1 I lv1 wl

wl

V21 w2

In particular the i-th component is given by

(7)

i · U x (V x W) = u2 lvl V21 w1 w2

= lv1

U2V21 w1 u2w2

= lvl

ulvl

w1 u1w1

-

ua Iva Wa

V1 1--lvl U2V21-1uava w1 w1 u2w2 u2wa

+ lv1

UaVal = lv1 U2V2 + uava I w1 UaWa w1 u2w2 + UaWa

+ u2v2 + uava I = + u2w2 + uawa

IV. i. U. VI W •1 U · W

= (U · W) (V · i) - (U · V)(W · i) [(U · W)V - (U · V)W] · i

=

Thus the i-th component of U x (V x W) and the i-th component of

(U·W)V- (U·V)W are equal. By showing in a similar way that their j-th and k-th components are equal, we arrive at the vector triple product formula

(8)

u

(V

X

X

W) = (U. W)V - (U. V)W

In view of (8), if we write (U x V) x W = - W x (U x V) we can easily deduce

(9)

(U XV)

X

w

= (U·W)V- (V·W)U

Combining the results on the scalar and vector triple products we obtain the famous Lagrange identity

(10)

(U

X

V) . (W

by studying U · V x (W x X). when U = W and V = X is

X

X) =

U·W U·X1 V·X

IV·W

A useful corollary of the Lagrange identity

(II) Exercises §1.5 1. Let V denote the volume of a tetrahedron with vertices at

P1(xH YH z1),

P 2(x2, y2, z 2), P3(x3, y 3, z3) and P,(x,, y,, z,)

§ 1.6 The line and plane

27

Show that V

=!

X2

Y2

Z2

6

Xa

Ys

Za

l

x4 y._ z._

l

2. Prove: (U x V) • (V x W) x (W x U) = (U • V x W) 2. 3. Assume that U, V, and Ware linearly independent, and define U 1 = a(V x W), V 1 = a(W xU), and W 1 = a(U x V) where a= (U•V x W)- 1 • Show that: (a) U 1 • U = V 1 • V = W 1 • W = l. (h) u. · v = u. · w = v2 · u = V2 • w = w. · v = w. · u = o. (c) (U1 ·V1 xW1 )(U·VxW)=l. 4. Prove that U, V, and W are linearly independent [UVW] :F 0.

1.6 The line and plane Let a non-zero vector V = [a, b, c] be prescribed along with a point P 0 In order that P = (x, y, z) be on a line containing P0 and parallel to V, we must have (see Fig. 1.6A)

= (x0 , y0 , z0 ).

---+

(1)

P0 P

=

tV for some t E R

FIG. l.6A. Collinearity.

---+

But P0 P = [x - x0 , y - y0 , z - Zo] and tV ponents we obtain

(2)

= [ta, tb, tc].

Equating com-

x - x0 = at, y - Yo = bt, z - z 0 = ct

We call (1) the vector equation, and {2) the parametric equations, of the line through P0 and parallel to V. Note that any non-zero vector which is collinear with V yields the same line. We therefore call any ordered triple of numbers which are

28

VECTORS

proportional to the components of V the direction numbers of the line. summary:

-

(3)

{P: P0 P

(4)

{(x,y, z): ( ; : ;: : :;)

=

tV &

z

=

te

In

R} is the line containing P0 and directed along V

&

tE

z0 + ct

R} is the line containing (x0 , y0 , z0) & having direction numbers [a, b, c]

We obtain the so-called symmetric equations of the line by eliminating tin (2), assuming of course that a, b, and c are non-zero:

x - Xo -a-

(5)

y - Yo -b-

=

z - Zo -c-

=

Example 1. Find the line L containing the points

-

and P 2 =(2,-1,5)

P 1 =(-1,4,3)

-

SOLUTION. If V = P 1 P 2 , then for any P = (x,y, z) on L we have P~ 2 P = tP 1 P 2 for some t. In component form this vector equation reads: [x- 2,y + 1, z- 5] = t[3, -5, 2]. Solving for x, y, and z we obtain the parametric equations of L: X

= 2

+ 3t,

y = - 1 - 5t,

Z

= 5

+ 2t

The symmetric form of the equations then is (just solve for t!) x-2

-3- =

y+l z-5 --=s= -2-

In 3-space lines which do not intersect and yet are not parallel are called skew Vector methods are particularly effective in finding the (shortest) distance between two skew lines.

lines.

Example 2. Find the distance between the following skew lines: ) x = t- I L 1 ={(x,y,z): ( y=-t+2 z

L2 = {(x,y, z):

= -2t-

teR}

&

teR}

I

(;:~::I) z

&

= -2t + 2

sOLUTION. Since the vectors U = [I, -I, -2] and V = [ -2, 2, -2] are the directions of L 1 and L 2 , respectively, we know (see Fig. 1.6B) that UxV=

i

j

I

-I

-2

2

k

-2 =6i+6j -2

§ 1.6 The line and plane

29

is normal to L1 and L 2 • Then UxV

i+j

1

N = IU x VI = v2 = v2 [1, 1, OJ

-

is the unit normal to L 1 and L2 • By setting t = 0 in the parametric equations we obtain the points P 1 = ( -1, 2, -1) and P 2 = (1, 2, 2) on L1 and L2 , respectively, which in turn determine the vector P1 P 2 = [2, 0, 3]. The required distance then is

PP ·N = [2 0 3] • [ 1•v2I, O] = ~ v2 = v· '2 1

2

,

,

The analytic definition of a plane is motivated by the geometric notion that a normal line is perpendicular to every line through its foot. Let N = [a, b, c] be a prescribed normal to the desired plane. Suppose P 0 = (x0 , y0 , Zo) is a given point of the plane and P = (x, y, z) any other point of the plane. The vector equation of the plane then is the orthogonality condition (see Fig. 1.6C) (6)

Fto. 1.6B. Distance between skew lines in Example 2.

FIG.

l.6C. Coplanarity: the definition of a ~

plane in vector form: P 0 P · N

= 0.

Translating (6) into component form we have

[x- x0 ,y- y0 , z- z 0 ] ·[a, b, c] = 0

(7) which reduces to (8}

a(x - x0 ) + b(y - y 0 ) + c(z - z 0 ) = Cl

Thus the plane containing P 0 and having N = [a, b, c] expressed in the following two forms:

(9) (10)

-

{P: P0 P • N {(x, y, z): a(x - x0 )

+ b(y -

=

0 &

y0)

=1=

0 as a normal may be

P E R3 }

+ c(z - Zo) = 0

&

(x, y, z)

E

R3 }

80

VECTORS

Notice that the equation of the plane in (8) may be put into the form ax+ by+ cz

(11)

+

d = 0

where d = -(ax0

+

by0

+ CZo)·

Conversely, for any prescribed quadruple of real numbers a, b, c, and d,

{(x,y, z): ax+ by+ cz + d

(12)

=

0

&

(x,y, z) eR3}

is a plane having [a, b, c] as a normal vector, assuming a 2 + b2 + c2 ::F 0. This is clear, since if (x0 , y0 , z 0 ) is in the plane, we know by substitution that d = - (ax0 + by0 + CZo) so that (12) is equivalent to (10). Example 3. Find the distance from the point P = ( -1, 4, 5) to the plane 3x- 2g + 4z = 12}. SOLUTION. N = [3, -2, 4] is normal to the plane. Take any point on the plane, say P0 = (0, 0, 3), found by setting x = g = 0 and solving for z. Then IPP0 • nl is the required distance if n = NfiNI (see Fig. l.6D). Therefore the distance is

{(x,g, z):

--

l 1[-1,4,2] •. ;-[3, -2,4]1 v29

3 v29

=.Inn

FIG. 1.6D. Distance from a point to a plane in Example 3.

{ Yt> z 1) and P 2 = (x2, y2, z2), the formula in ( 1) demands that

+ P2

be the point in R 3 obtained by adding corresponding coordinates ofP1 and P 2 • Similarly, if P = (x, y, z), formula (2) produces a point cP from P by multiplying each coordinate ofP by c. Thus, both sums and scalar multiples of

32

VECTORS

points in R3 are again points in R3 • It is a simple matter to verifY that addition and scalar multiplication of points in R3 conform to the basic rules of the elementary vectors discussed in §§1.1-1.2. In particular, they have the following properties, called the axioms of a real vector space, because they reflect in abstract form the behavior of elementary vectors. e

1.7A AXIOMS OF A REAL VECTOR SPACE*

D. We say that a set "'' is a real vector space with respect to two given binary operations (i) (ii)

+

and • the following axioms hold:

CLOSURE UNDER +: p E "f'" & Q. E "f'"

~

p + Q. E "f'"

1

AssociATIVITY : For every P, Q., and R E "'',

(P + Q.) + R = P + (Q. + R) (iii)

IDENTITY: There exists 0 E "'' such that

PE"F"

~

Vector addition

P+O=P

(iv)

INVERSES: For every P E "'' there exists - P E "'' such that P + ( - P) = 0

(v)

COMMUTATIVITY: For every P, Q.E "'',

P+Q.=Q.+P (vi)

CLOSURE UNDER • f:

P E "'' & c E R (vii)

(viii)

=>

cP E "''

AssOCIATIVITY : P E"'' & b, c E R => b(cP)

(bc)P

Scalar multiplication

IDENTITY: For every P E "'', IP = P

(ix)

(x)

DisTRIBUTIVITY OVER VECTOR ADDITION: For every P, Q.E"'' and c E R, c(P + Q.) = cP + cQ. DISTRIBUTIVITY OVER ScALAR ADDITION: For every P E "'' and cl> c2 E R,

Vector addition and Scalar multiplication

(c1 + c2 )P = c1 P + c2 P

* We can replace "real" by "complex" if we replace the scalar field R by the field of complex numbers. In abstract modern algebra one speaks of vector space over afield. Thus what we call a "real vector spac;e" is a "vector space over the real number field". t The symbol cP is understood to mean c • P, in agreement with the notation used already for scalar multiplication of elementary vectors.

§ 1.7 Vector spaces

33

Notice that (i)-(v) give the additive structure of?'", while (vi)-(viii) define the multiplicative structure of?'", and (ix)-(x) describe the connection between the two structures. In this context the elements of?'" are called vectors, the elements of R scalars, and R the scalar field. To verify that R3 satisfies (i)-(viii), where + and · are defined by (1) and (2), we merely have to invoke the field properties of the real numbers. For the following let P = (PI,P 2,Pa), Q. = (qh q2, q3), and R = (r1, r2, r3). (i')

CLOsURE UNDER + : P + Q. = (PI + q1, P 2 + q2, Pa + q3) is an ordered triple of real numbers: hence P + Q.e ?'".

(ii')

AssociATIVITY: (P + Q.) + R = ((PI,p2,Pa) + (qh q2, qa)) + (rl, r2, ra) = (PI + ql> P2 + q2, Pa + qa) + (rl> r2, ra) = ((PI + ql) + r1, (P2 + q2) + r2, (Pa + qa) + ra) = (PI + (ql + r1), P2 + (q2 + r2), Pa + (qa + ra)) = (ph P2, Pa) + (ql + r1, q2 + r2, qa + ra) = p + (Q. + R)

(iii')

IDENTITY: Let 0 P

+

=

(0, 0, 0).

Then

0 = (PI + O,P2 + O,Pa + 0) = (Phh,Pa) = P

(iv')

INVERSES: Let - P = ( -PI, - p2 , - Pa). Then P + ( -P) = (Pl + ( -pl),p2 + ( -P2),Pa + ( -Pa)) = (0, 0, 0) = 0 (see (iii') above).

(v')

CoMMUTATIVITY:

+ Q. = (PI + qh P2 + q2, Pa + qa) = (ql +PI, q2 + p2, q3 + Pa) = Q. + p (vi') ScALAR MuLTIPLICATION: cP = (cph cp 2 , cPa) is an ordered triple P

of real numbers, hence a member of?'". (vii')

AssoCIATIVITY : b(cP) = b(cph cp 2 , cp3) = (b(cPI), b(cp2 ), b(cPa)) = ((bc)PI, (bc)p 2 , (bc)Pa) = (bc)P.

(viii')-(x') See Exercise 1. Ordinary vectors and ordered triples of real numbers are but two of a host of objects which form a vector space. A few other important cases are: Example l. "Y = R", the set of all n-tuples of real numbers, forms a real vector space if we define (in analogy with (1) and (2)): (Pt> · · .,p,.) + (ql, · · ., q,.) = (PI + qu · · .,p,. + q,.) c(PI, ... , P,.) = ( cp 1 , . . • , cp,.) for any c e R In particular, the real number field and the complex number field form real vector spaces.

34

VECTORS

Example 2. .A,.: set of all square matrices of order n with real (or complex) entries. Here, matrix addition and scalar multiplication of matrices serve as the vector addition and scalar multiplication of the vector space. The following examples are real vector spaces if function addition is taken as vector addition, and function magnification (by real numbers) as scalar multiplication. Example 3. &': the set of all R-R polynomials. Example 4. &',.: the set of R-R polynomials of degree no greater than degree n. (For example, constant, linear, and quadratics, if n = 2.) Example 5. CC(S): the set of all R-R functions which are continuous at each point ofS. Example 6. CC"(S): The set of all R-R functions which have continuous derivatives of order n ~ 1 at each point of S. A real vector space "Y 1 is called a subspace of a real vector space "Y 2 if "Y 1 is a subset of "Y 2 and the vector operations of "Y 1 and "Y 2 are the same. For example&',. is a subspace of [IJ, and ~"(S) a subspace of~(S). The above list includes just a fraction of the possibilities. The merit of studying vector spaces is that, as the list shows, we have an abundance of customers for the results we produce. Whatever theorem we prove about vector spaces in general applies to each of the above examples of a vector space. The concept of linear combination for ordinary vectors, along with the associated notions oflinear dependence and independence (see §1.2), generalizes to arbitrary vector spaces. Let at and P1 be generic for scalars and vectors, respectively, in a given vector space "Y. Then for any n ~ 1

(3) is called a linear combination of the vectors P 1> ••• , P,. with scalar coefficients al> ... , a,.. We then say that {P1> ••• , P,.} is a linearly independent set [resp. linearly dependentset] a 1P 1 + · · · + a,.P,. = 0 holdsonlyif a 1 =···=a,.= 0 [resp. holds for at least one at =F 0]. Thus, every finite set of vectors is either linearly independent or linearly dependent, but not both. For any prescribed n ~ 1, the set of all vectors of the form (3) (i.e. all possible linear combinations of a prescribed set of P/s) is a subspace of "Y (see Exercise 38) and {P1> ••• , P,.} is said to span this subspace. Then {P1 , . . . , P,.} is called a basis of the subspace it spans if it is a linearly independent set. Example 7. The vector space R4 is spanned by the four vectors P 1 = (1, 0, 0, 0), P 2 = (0, 1, 0, 0), P 3 = (0, 0, 1, 0), and P 4 = (0, 0, 0, 1) since every (x1 , x2 , x3 , x4 ) in R 4 is a linear combination of these Pt's, namely (x1 , x2 , x3 , x4 ) = x1P 1

+ x2P 2 + x3P 3 + x4P 4 •

These Pt's in fact form a basis of R\ since they are linearly independent. This is a consequence of orthogonality, discussed in the next section.

§ 1.7 Vector spaces

35

Example 8. The monomials 1, x, x2 span the vector space &'2 of polynomials of degree at most 2, since every such polynomial is of the form a 1 • 1 + a2 x + a3 x2 • Indeed, {1, x, x2} is a basis for &' 2 since it is a linearly independent set. This may be verified in several ways. For example, by the fundamental theorem of algebra the equation (4)

can have at most two roots, and so (4) cannot hold for all x E R, unless a 1 = a2 = a3 = 0. To cite a pedestrian approach, substitute x = 0, I, 2 to obtain three equations in a 1 , a 2 , a 3 whose solution is a 1 = a 2 = a3 = 0. For a third method to be generalized later, differentiate (4) twice to give a2 + 2a3 x = 0 and 2a3 = 0; now look at (4) and the latter two equations for (say) x = 0.

Example 9. In the vector space rc(R) the functions sin x and cos x are linearly independent. For if a 1 sin x

+ a2 cos x =

0 on

R

we may set x = 0 to get a2 = 0, and then x = TT/2 to see that a 1 = 0. By a similar argument, {sin x, cos x, sin 2x, cos 2x} is a linearly independent set, and in general, {sin x, cos x, ... , sin nx cos nx} is a linearly independent set. On the other hand, {1, sin 2 x, cos2 x} is a linearly dependent set. (Why?)

Exercises §1. 7 Verify formally that the following sets in Exercises 1-14 form real vector spaces. Specify in particular what the zero element and the additive inverses are:

1. 3. 5. 7. 8. 9. 10.

11. 12.

13. 14.

R" in Example I. 2. .A,. in Example 2. &' in Example 3. 4. &',.in Example 4. rc(S) in Example 5. 6. P · Q. = 0. Example 7. In view of Example 2, two vectors P and Q in Rn are orthogonal + · ••+Pnqn = 0.

P1q1

Example 8. In view of Example 3, two functions! and gin~ ( [a, b]) are orthogonal = 0. In particular, for any non-negative integers n and m, the functions cos nx and sin mx are orthogonal in ~([ -TT, Tr]) since

J!Jg

J~,. cos nx sin mx dx

=

0

Similarly, for any two distinct non-negative integers m and n, the functions cos nx and cos mx are orthogonal in ~([ -TT, Tr]), and so are sin nx and sin mx. Exercises 8-11 list several important properties of orthogonality which generalize results already deduced geometrically for ordinary vectors in 3-space. One of the most important uses of orthogonal vectors concerns linear independence. It is easy to show that any set ofpairwise orthogonal vectors which are non-zero are linearly independent. Indeed, if P~o P 2 , •. • , Pn are non-zero and P 1 • P 1 = 0 for i =/: j, then the equality

(15) can hold only if a1

= ··· =

an

= 0, since for every

i we deduce from (15) that

P1 • (a 1P 1 + a2P 2 + · · · + anPn) = a1(P1 • P1) = 0 and so a1 = 0. Sets of unit vectors (i.e., norm = 1) which are pairwise orthogonal are called orthonormal. As with the base vectors i, j, k in 3-space, we often try to find an orthonormal basis for a given vector space. Example 9. In view of Example 8, the set {t, cos x, sin x, cos 2x, sin 2x, ... , cos nx, sin nx} is linearly independent in ~([ -TT, 7r]). By multiplying each of the functions in the braces by 1/\/Tr, we get an orthonormal set.

From any set of n linearly independent vectors {Pk}~=I we can construct an orthonormal set of vectors {Q.k}~=I by taking Q.k = Q.'t:/IQ.tl, where Q.t is defined to be a suitable linear combination of P ~o P 2 , ••• , Pk subject to the requirement that Q.t • Q.1 = 0 for j = 1, ... , k - 1. This is called the GramSchmidt orthogonalization process. See Exercise 15.

40

VECTORS

Exercises §1.8 1. Show that the following is equivalent to positive definiteness in (iv): P • P and P • P = 0 P = 0.

~

0,

2. Write the four axioms of an inner product using the notation (P, Q.) instead of p ·Q.. 3. Show that symmetry and linearity (see (ii) and (iii) of Definition 1.8A) imply that (aQ. + bR) • P = a(Q. • P) + b(R • P) for every P, Q.E 1' and every pair of real numbers a and b. 4. Show that linearity of an inner product may be expressed equivalently as the two conditions: (iii 1) HoMOGENEITY: P • (aQ.) = a(P · Q.) for every P, Q.E 1' and every real number a, and (iii 2 ) ADDITIVITY: P • (Q. + R) = P • Q. + P • R for every P, Q., R E 1'. 5. Show that -1 ~ P • Q./IPIIQ.I ~ I for every P, Q.E 1' with P =F 0 =F Q.. (This allows one to introduce the "angle" between the vectors P and Q.defined by cos- 1 (P • Q./IPIIQ.I). 6. Show that P • Q. = 0 if Q. = 0. 7. Show that laPI 2 = lai 2 IPI 2 • In Exercises 8-11 verifY the given property: 8. 9. 10. ll. 12. 13. 14. 15.

16.

17.

0 is orthogonal to every vector in 1'. Q.is orthogonal to P - aQ. if Q. =F 0 and a = P • Q./IQ.I 2 • If IPI = IQ.I, then P - Q. and P + Q. are orthogonal. P and Q.are orthogonal IP + Q.l 2 = IPI 2 + IQ.I 2 • Define the "distance" between P and Q. in 1' to be dist (P, Q.) = IP - Q.l. Show that (i) dist (P, Q.) = 0 P = Q., (ii) dist (P, Q.) = dist (Q., P), and (iii) dist (P, Q.) ~ dist (P, R) + dist (R, Q.). Show that IP • Q.l = IPIIQ.I aP + bQ. = 0 for some real numbers a and b, with a2 + b2 =F 0. HINT: Look at the proof of the Schwarz inequality. Show that the triangle inequality implies the Schwarz inequality. OuTUNE: Replace Pin (7) by -P. Let {P1 , ••• , P ,.} be any set of linearly independent vectors in a vector space 1' having an inner product. Show by construction that there exists an orthonormal set {Q. 1 , . . . , Q.,.} with the property that each Q.k is a linear combination of P 1 , . . . , Pk. OuTLINE: Let Q. 1 = P 1 /IP 1 1, and assume for k ~ I that an orthonormal set {Q. 1 , ••• ,Q.k} hasalreadybeenfoundsothateach Q. 1(i = l, ... ,k) is a linear combination of P 1 , .•• , P 1• Now show that there exist scalars a~o ... , ak such that Q.t+ 1 = Pk+ 1 - (a 1 Q. 1 + · · · + akQ.k) and Q.t+l • Q.1 = 0 (i = 1, ... , k). Finally take Q.k+l = Q.t+l/IQ.t+ll· By the Gram-Schmidt orthogonalization process in Exercise 15, construct an orthonormal triple out of a given triple of linearly independent vectors {P1o P 2 , P 3}. Speciali:t.e your result to the vectors P 1 = i + j, P 2 = i - j, and P 3 = j - 2k for 1' = R3 • Prove inequalities (9)-(14) as consequences of (4) and (5).

2 • Vector Functions

2.1

Vector junctions

This chapter is devoted to the study of R-R3 functions. Then by restricting the range to a coordinate plane in R3 , we obtain the corresponding theory for R-R2 functions. By definition an R-R3 function F is a set of ordered pairs (t, P) in which tis a real number and P a point in 3-space. In kinematics the domain PJF is an interval of time and for a prescribed instant t the image P = F(t) locates a moving object at that instant. Because R3 forms a vector space (see §1.7) we usually employ the notation and terminology ofvector theory to handle R-R3 functions, and we commonly call such functions vector functions. However, we must be careful to note that the name vector Junction applies to any function whose range is a set of vectors. In the sequel we will meet vector-valued functions whose domains are subsets of the plane and 3-space. For every tin PJF we can think of its image P = F(t) not only as the point P, but also the arrow (position vector) issuing from the origin and having Pas its tip (see Fig. 2.1A).

Fm. 2.IA. Vector function F, with components F 1 , F 2 , & F 3 , as a motion in 3-space.

I• I I I I I I I II I 1•••1• I I I• I •I t

The coordinates of the images ofF induce functions. For if we assemble all ordered pairs (t, x), (t, y), and (t, z) for which (x, y, z) is a point image ofF, we obtain three new R-R functions (say) Fh F 2 , and Fa, called the coordinate or component functions ofF. Conversely, any ordered triple of R-R functions F1 , F 2 , and Fa, with a common domain PJ induces a vector function F defined by (1)

F(t)

=

(F1 (t), F 2 (t), F 3 (t)) where 41

PJF

= PJ

42

VECTOR FUNCTIONS

and we speak of the function (or mapping)

z = Fa(t)

(2) or in base vector form

(3) Notice that the component functions and their associated vector function are related by the formulas

(4)

F 1 (t) = F(t) • i,

F 2 (t)

=

F(t) • j,

Fa(t) = F(t) • k

Example 1. The function F(t) =(cost, sint, 2t), teR, may be expressed as = sin t, z = 2t; or in base vector form F(t) = cos ti + sin tj + 2tk.

x = cost, y

Not only does F induce three R-R functions in a natural way, but also three R-R2 functions, defined on P.dF by the formulas

(5)

(F1 ,F2 }(t) = (F1 (t),F2 (t)), (F2 ,Fa)(t) = (F2 (t),Fa(t)}, (Fa,Fl)(t) = (Fa(t),F1 (t))

If we interpret F as motion in 3-space over an interval of time, then the coordinate functions Fl> F2 , and Fa describe the rectilinear motions exhibited by the projections of the space motion onto the X-, Y-, and Z-axes, respectively. Moreover, (Fh F 2 ), (F2 , Fa), and (Fa, F 1 } give the planar motions of the projections of the space motion onto the XY-, YZ-, and ZX-planes, respectively. Example 2. With reference to the vector function in Example I, the rectilinear motion on the X-axis is described by cos t, on the Y-axis by sin t, and on the Z-axis by 2t; on the XY-plane by (cost, sin t), on the YZ-plane by (sin t, 2t) and on the ZX-plane by (2t, cost). The space motion of this example is ascent with uniform upward velocity along a circular helix coaxial with the Z-axis. The motion of the projection onto the XY-plane is counterclockwise around the unit circle, while the projected motions on the YZ- and ZX-planes are both sinusoidal. Along the X-axis andY-axis we have simple harmonic motion, while along the Z-axis the motion is monotone increasing at constant speed (see Fig. 2.1B).

Fm. 2.1B. Helical motion described in Example 2.

§ 2.2 Limits of vector functions

43

Exercises §2.1 Describe the following motions in terms of the rectilinear motions on the coordinate axes, and the planar motions on the coordinate planes: 1. 3. 5. 7. 9.

F(t) = (cos 2t, sin 2t, t 2 ) 2. F(t) = (t cost, t sin t, 2t) 4. F(t) = (e-t sin t, e-t cos t, e- 1) F(t) = (2t, 3t, 4t) F(t) = 3 cos ti + 2 sin tj + 5tk 6. F(t) = (3t + 4)i + (5t - l)j F(t) = cos 2ti + sin 2tj + r 1k 8. F(t) = t2i + 3tj Suppose a particle is moving in accordance with the position function F(t) = ti

+ t 2j + t 3k.

How far from the origin is the particle at t = 5? At what velocity is it moving away from the origin at t = 5? 10. Two particles have position functions F(t) = (t, 2t, 3t 2 ) and G(t) = (3t, - t, 5t 2 ). Are they moving closer together or farther apart at t = 2? At what rate? 11. Show that the motion described by F(t) = (a sin t cos 2t, a sin t sin 2t, a cost) takes place on a sphere of radius a. 12. Show that the motion of an object described by F(t)

=

(v 0 cos Bt)i

+

(1-gt 2

+ v0 sin Bt)j

is along a parabola in the XY-plane, if v0 and 8 are constants. What is the maximum displacement from the X-axis in the first quadrant? Illustrate by a graph. What is the velocity and acceleration of the projection on the X-axis? on the Y-axis? When does the object cross the X-axis?

2.2 Limits of vector junctions The notion oflimit for vector functions is a rather direct generalization oflimit for real-valued functions and again builds upon the notions of distance and proximity. Thus, F will be said to have limit L at t0 if and only ifF(t) is within any prescribed distance of L for all t within some suitable distance of t 0 , except perhaps t 0 itself. Geometrically, in order that L qualify as the limit, F(t) should be within any prescribed spherical ball of radius £ > 0 centered at L for all t within some interval centered at t 0 , with t 0 deleted (see Fig. 2.2A). • 2.2A

DEFINITION OF LIMIT FOR VECTOR FUNCTIONS

D. Given an R-R 3 function F and a cluster point t 0 of ~F• we write

(1)

L = lim F(t) t-+to

In other terms, L = lim F

~

lim JF(t) - LJ = 0 t-+to

~

for every E > 0 there is a SF = SF(E, t0 ) > 0

t-+to

such that 0 < Jt - t0 J < SF ~ JF(t) - LJ < E. Geometrically, if F(t) and L are interpreted as position vectors, we are asking that the distances between the terminal points ofF(t) and L, namely JF(t) - LJ, become and remain as small as we please for all t (except perhaps t0 ) sufficiently close to t 0 • See Fig. 2.2A. As usual we use the notation F(t) -L as t-t0 to mean L = lim F(t). t-+to

44

VECTOR FUNCTIONS

FIG. 2.2A. Definition of limit.

Example 1. Let F(t) = cos ti L = cos t0 i

+ sin tj + 2tk. Show that lim

+ sin t0 j + 2t0 k

SOLUTION.

F(t) = L, where

t-+to

for any t0 in R. By the continuity of cosine, sine, the identity and root functions, we

have IF(t) - Ll = V (cost - cos t0 ) 2

+ (sin t - sin t0 ) 2 + 4(t -

t0 ) 2

-

0 as t - t0 •

The following theorem is a fundamental building block in that it allows us to apply all of our previous knowledge of limits, continuity, and derivatives of R-R functions to R-R3 functions. It tells us that we can check on the convergence of a vector function, and in fact compute the limit, by studying the coordinate functions separately. F3 (t)

La

;

FIG. 2.2B. Component theorem for limits.

L2

F2 to

+ (lim F 2)j + (lim F 3 )k

t->to

t->to

t->to

A basic implication of the component theorem for limits is that the algebra of limits for R-R functions generates an algebra of limits for vector functions. For example, ifF and G both converge at t 0 , then

lim (F ± G) = lim F

(4)

t->to

± lim G

t->to

t->to

lim cF = c lim F for any real number c

(5)

t->to

t->to

In fact, iff is an R-R function (that is, a scalar function) which converges at t0 , we have (6) lim fF = (lim f) (lim F) t->to

t->to

t->to

of which (5) is a special case. The validity of (4)-(6) is based on the sum, difference, magnification, and product rules in the algebra of limits for R-R functions applied to the following component representations, using (3) :

(7) (8)

(9)

F(t)

± G(t) = (F1(t) ± G1 (t)i + (F2 (t) ± G2 (t)j + (F3 (t) ± G3 (t)k cF(t) = cF1(t)i + cF2 (t)j + cF3 (t)k f(t)F(t) = f(t)F1(t)i + f(t)F2 (t)j + f(t)F3 (t)k

Furthermore, since the dot and cross products of two functions F and G are defined by (10)

F(t) · G(t) = F1(t)G 1(t)

(ll)

F(t)

G(t) IF2(t) G2 (t)

+ F2 (t)G 2 (t) + F3 (t)G3 (t)

Fs(t)l. G3 (t) 1

+ IF3 (t) F 1 (t)l. + IF1 (t)

G1(t) J G1(t) the component theorem for limits and the algebra of limits for R-R functions yield x

(12)

G3 (t)

lim F · G = lim F · lim G t->to

(13)

t->to

t->to

limF x G = limF x limG t->to

t->to

given the convergence ofF and G at t 0 •

t->to

46

VECTOR FUNCTIONS

We define continuity for vector functions in the usual way: F is continuous at t0 (i) t0 in ~F' (ii) F converges at t0 , and (iii) lim F(t) = F(t0 ). t->to

As an immediate corollary of the component theorem for limits we have a continuity counterpart. • 2.2C

COMPONENT THEOREM FOR CONTINUITY

H. As in theorem 2.2B above C. F is continuous at t 0 Fh F 2 , F 3 are continuous at t 0

A vector function is contmuous at a point if and only if its coordinate functions are continuous at the point.

The component theorem for continuity tells us, for example, that all vector functions having the circular functions or polynomials (or composites of them) as coordinate functions are continuous at each point in 1-space. In a more general way, the theorem asserts, in view of (4)-(6) and (12)-(13), that sums, differences, magnifications, continuous scalar function multiples, dot products, and cross products of continuous vector functions are continuous. Furthermore, the composite F o g of a continuous vector function F and a continuous scalar function g is continuous, since F[g(t)] = F 1 [g(t)]i

+ F 2 [g(t)]j + F 3 [g(t)]k

The above results on limits and continuity of vector functions find immediate application in developing the algebra of derivatives in the next section.

Exercises §2.2 Compute the following limits: 1. lim t sin (1/t)i 1->0

+ log (sin tft)j + e-l/t2k

tan t • . e' - 1 • + --J 3. 11m--1 1->0 t t

5. 6. 7. 8.

2.3

sinh +-t k t

2. lim

(e1' 1,

log (log t

1->1

+ e), it I)

. (cost,smt . ) 4. 1rm 1->71/2

Phrase definition 2.2A and theorems 2.2B and C for R-R 2 functions. Write the analogues of (7)-( 10) for R-R 2 functions. Is there an analogue for (11) ? Express continuity ofF at t0 in the epsilon-delta phraseology (see below (1)). VerifY the formulas at: (a) (4), (b) (5), (c) (6), (d) (12), and (e) (13).

Derivatives of vector junctions

Formally the development of the concept of derivative for vector functions is nearly identical to that for R-R functions. The geometric motivation is different in one significant detail: whereas the tangent line for R-R functions belongs properly to the graph of the function, the tangent line for vector functions is associated with the range of the function. Recall that the graph of a function is the set of all ordered pairs of pre-images and images, whereas the range is simply the set of all images of the function. The graph of an R-R 3 function is a subset of 4-space, whereas the range is a subset of 3-space. Obviously, for geometric purposes at least, we are compelled to focus attention on the range, since 4-space graphs are out of our reach except in purely theoretical constructs.

§_ 2.8 Derivatives of vector functions

47

As with R-R functions, we define derivability and derivative as convergence and limit, respectively, of difference-quotients. Then we proceed to develop an algebra of derivatives in terms of the various vector operations. Of course there will be no reciprocal or quotient rules; yet there are three product rules, corresponding to scalar multiplication, the dot product, and the cross product. Thanks to the component theorem for limits, we find all of our results are obtainable by appealing directly to our previous theorems stated for R-R functions. Many of these results, however, may be got directly, and for instructional purposes this direct approach will sometimes be used. Theorems on extreme values are not seen, since they are meaningful only for the magnitude of a vector function, inasmuch as no linear ordering relation is possible for vectors or points in 3-space. Likewise, questions concerning monotonicity of vector functions are out of place. Furthermore, concavity notions must be replaced by other characteristics, like curvature and torsion, except of course when the points in the range happen to be coplanar. In summary, the reader should be anticipating many straightforward generalizations (often looking formally the same) of the theory of derivatives already studied, but at the same time he should be acutely aware of the places in which generalizations are not expected to materialize. • 2.3A

DEFINITION OF DERIVATIVE FOR VECTOR FUNCTIONS

D. We say that an R-R3 function F is derivable (or differentiable) at t0 E !12F ~ there exists A E R3 such that

(I)

A

=

lim F(t) - F(t0 )

t - t0

t-+to

A vector function F is termed derivable (or differentiable) at any point t0 in its domain if and only if the difference quotient of F at t0 converges to some real number, which we call the derivative ofF at t0 , denoted by F' {t0 )

If A exists to satisfy (1), we write A = F'{t0 ). Briefly,

(2)

F'(to)

=

lim F(t) - F(to) t - to

t-+to

tangent

vector

FIG. 2.3A. Definition of derivative.

parameter

interval

48

VECTOR FUNCTIONS

As with derivatives of R-R functions we use the equivalent formulas - F(t) = lim F(t + Llt) - F(t) = lim .:lF h At-+0 ilt At->0 ilt and the notations dFfdt, (dFfdt)t=to' D 1F, (D1F)t=to' etc. What does the difference quotient (F(t) - F(t0 ))f(t - t0 ), t =F t0 , mean vectorially? First of all, F(t) - F(t0 ) is the vector which issues from the terminus of F(t0 ) and ends at the terminus of F(t). See Fig. 2.3A. Then (F(t) - F(t0 ))f(t- t0 ) is the result of magnifying this difference vector by the scalar (t - t0 ) - l . Thus the difference-quotient vector is collinear with the difference vector in its numerator. Ifwe think of the difference-quotient vector as a secant vector, then from a geometric standpoint the limit of the difference quotient, namely, the derivative, is reasonably accepted as the tangent vector to the range of the function F. Thus, ifF is interpreted as motion in 3-space, the derivative F' (t 0 ) (if it exists) is a vector pointing in the direction of the path of the moving object. If the forces which constrain the object to follow this path were suddenly removed, the object would follow a straight-line path in the direction ofF'(t0 ). The dynamics of a slingshot is a simple analogy. The magnitude of the difference IF(t) - F(t0 ) I measures the straight-line distance from the point F(t0 ) to the point F(t). Therefore, from a kinematic point of view, the magnitude of the difference quotient measures the average speed over the interval of time from t0 to t, or t to t0 • Accordingly, since*

F'(t) = lim F(t

(3)

+ h)

h-+0

lim IF(t) - F(to) I = I lim F(t) - F(to) I = IF' (to) I t->to t - to t->to t - to we are led to accept the magnitude of derivative IF' (t 0 ) I as the instantaneous speed of the object. In the next section we will see that IF' (t0 ) I is the time-rate of change of distance along the path of motion. The full development of this idea requires a study of arclength in 3-space. Accordingly, in view of the significance of its direction and magnitude, the derivative F' (t0 ) is a measure of the velocity of motion ifF is a position vector function. As a consequence of the component representation of the difference quotient F(t) - F(t0 ) F (t) - F1 (t 0 ) • F2 (t) - F2 (t0 ) • F3 (t) - F3 (t0 ) k --'--'::-------:-"--""'- = 1 I + J + -=-:...:....---=-'-=t - t0 t - t0 t - t0 t - t0 we have as an immediate corollary of the component theorem for limits (theorem 2.2B), the following counterpart for derivatives. e

2.3B

COMPONENT THEOREM FOR DERIVATIVES

H. An R-R3 function F = (Fh F 2 , F 3 ) t 0 E PJF C. (i) F is derivable at t 0 FhF2 ,F3 are derivable at t0

(ii) F'(to) = F~(to)i

+ F~(to)j + F~(to)k

A vector function is derivable at a point if and only if its coordinate functions are derivable there. Moreover, the components of the derivative are the derivatives of the components.

*To show that IG(t)I-+ILI if G(t)-+L as 1-+10 , observe that 0 :S IIG(t)I-ILII :S IG(t) - Ll, or else appeal to the component form ofG(t) and Land apply the component theorem for limits along with limit theorems for R-R functions.

§ 2.3 Derivatives of vector functions Example 1. Let F(t) = r F'(t)

21 i

+ cos 3tj

= -2r 21 i

- 2(t 2

+

I)'Y.k.

- 3 sin 3tj - 6t(t 2

+

49

Then

I)Y.k

As a corollary of the above theorem and the component theorem for continuity we at once observe that the proposition derivability implies continuity carries over from R-R functions to vector functions (see theorem 7.3A in Calculus with Analytic Geometry, and note that even its proof generalizes to vector functions). From the algebra of derivatives for R-R functions we know that the class of all R-R functions which are derivable at a given point is closed under the operations of addition and multiplication. Accordingly, the component theorem for derivatives tells us that the class of all vector functions which are derivable at a given point is closed under addition, scalar multiplication (by derivable scalar functions), dot products, and cross products. In fact, we can discover the rules for computing the derivatives of sums, scalar multiples, and dot and cross products by looking at what happens in the component representation. The rules in the vector algebra of derivatives are essentially the same as those in the R-R case, except for the absence of a reciprocal and quotient rule, and the presence of a variety of product rules. To illustrate we prove two product rules at once (without appeal to components), using an asterisk (*) to stand for either dot ( ·) or cross ( x ) . Assume that F and G are derivable at t 0 • Then:

F(t) • G(t) - F(t0 ) t - t0

•

F(t) • [G(t) - G(t0 )] + [F(t) - F(t0 )] t - t0

G(t0 )

_ F( )

-

t •

•

G(t0 )

G(t) - G(t0 ) F(t) - F(t ) G( ) + t - t0 0 • t 0 t - t0

Now passing to the limit on both sides by the vector algebra oflimits we have the product rule for •-products:

(4) [F • G]'(t0 ) = F(t0 ) • G'(t0 ) + F'(t0 ) • G(t0 ) The same procedure applies ifF is replaced by a scalar function, say j, and • is allowed to stand for scalar multiplication. Notice that whatever operation • stands for must have two properties: right and left distributivity, and homogeneity. Notice also that we were careful to avoid using commutativity because of the cross product. The procedure given above is exactly that given already in the proof of the product rule for R-R functions (see theorem 7.5A in Calculus with Analytic Geometry). For easy reference we now state the algebraic rules for vector derivatives : VECTOR ALGEBRA OF DERIVATIVES

(5)

(i) (aF ± bG)' = aF' ± bG' (ii) (F ·G)'

=

F · G'

+ F' · G

(a, be R)

(iii) (F

X

G)'

=

F

(iv) (JG)' = JG'

X

G'

+ F'

X

G

+ f'G

An immediate application of (ii) and (iii) is seen in the following example: Example l. Use the vector algebra of derivatives to verify the formula for the derivative of a scalar triple product: (F . G X H) = F . G X H' + F . G' X H + F' . G X H I

50

VECTOR FUNCTIONS

(This is a special case (n = 3) of the derivative of an nth-order determinant with derivable functions as entries. Write it out in determinant form to appreciate this view.) SOLUTION. (F. G X H)' = F. (G X H)' + F'. (G X H) = F • (G x H' + G' x H) + F' • (G x H) = F • G x H' + F • G' x H + F' • G x H The following example illustrating the vector algebra of derivatives turns out to be a basic tool in developing the Frenet-Serret formulas in §2.8. Example 3. Show that ifF is a vector function of constant magnitude and derivable on ~F• then F(t) • F'(t) = 0 on ~F Give a geometric interpretation. SOLUTION. Assume that IF(t) I = c on ~F• where c is some non-negative real number. Then F(t) • F(t) = c2 , so that by the dot-product rule for derivatives we have F(t) • F'(t) + F'(t) · F(t) = 0, whence by commutativity 2F{t) · F'(t) = 0 and finally F(t) • F'(t) = 0. In geometric terms we have shown that a vector function of constant magnitude is orthogonal to its derivative. Thus, ifF describes motion at a constant distance from the origin, the path is on a sphere, and the velocity vector F'(t) at any instant tis orthogonal to the position vector F(t), hence tangent to the sphere. See Fig. 2.3B.

z (O,O,cl

F10. 2.3B. Vector function of constant magnitude.

By the component theorem for derivatives (2.3B), the chain rule for R-R functions carries over to vector functions. Let F be a vector function and g a scalar function. Then if g is derivable at t0 and F derivable at g(t0 ), the composite vector function defined by [F o g](t) = F[g(t)] is derivable at t 0 , and (6)

[F o g]'(t0 ) = F'[g(t0 )]g'(t0 )

Formula (6) is more commonly seen in the form

dFdg dt=dudt

dF

where dFfdu is the derivative ofF at u = g(t). For the study of vector derivative equations the following result is basic, and is an immediate corollary of the component theorem for derivatives and the constant difference theorem (7.130 in Calculus with Analytic Geometry): If two

§ 2.3 Derivatives of vector Junctions

51

vector functions F and G are both derivable on (a, b) and continuous at the end points, then (7)

F'(t) = G'(t) on (a, b) ~ F(t) = G(t) + C on [a, b] where Cis a constant

Example 4. equation

Find all vector functions which are derivable on R and satisfy the Y' = sin ti + te1j + 3tv't 2 + 1 k

and find the particular solution satisfYing Y(O) = 0. soLUTION. Taking inverse derivatives by components, we see that one solution is -cos ti + el(t- l)j + (t 2 + l)%k so that by (7) above, ifF is any solution of the given equation, then F(t) = -cos ti + el(t- l)j + (t 2 + l)%k + C where C is some constant vector.

0

By setting F(O) = 0 we see that

=

-i- j

+k +c

so that the desired particular solution is F(t) = -cos ti + e1(t- l)j + (t 2 + l)%k + i + j - k = (1 -cos t)i + (tel- t + l)j + ((t 2 + 1)* - l)k

Exercises §2.3 In the following find the derivative F'(t) for the given F by limits of difference quotients. 1. F(t) = 3ti + t 2j + k 3. F(t) = cos ti + sin tj

2. F(t) = t 3 i - 3tj + (t 2 4. F(t) = 3i + 4j - k

-

l)k

In the following find F' (t), F"(t), IF'(t) I, and IF"(t) I by the algebra of derivatives: 6. F(t) = e2!i +log ltlk 5. F(t) =sin 3ti +cos 3tj + 4tk 7. F(t) = e-li + y't j - 2k 8. F(t) = sin 3 t(i + j + k) 9. Given F(t) = sin ti + cos tj + 3t2k and G(t) = ti - 3t 2j + e21k, D 1(F ·G), DtiFI 2 , and D 1(F x G) each in two ways. 10. Same as Exercise 9 above, with F(t) = log 2ti + e8101j + tk and

obtain

G(t) = t 2i - r 1j + log tk

11. Show that D 1IFI 2 = 2F • D 1F, assuming that F is derivable. 12. Show that F • dFfdt = IFI diFI/dt, assuming that F is derivable. 13. VerifY the formula

14. 15. 16. 17.

Derive a formula for D 1(F x (G x H)) assuming that F, G, and Hare derivable. Prove the dot-product rule for derivatives in two ways. Prove the cross-product rule for derivatives in two ways. Suppose F: [a, b]- R3 is derivable on (a, b) and continuous at the end points. Show that F' = 0 on (a, b) ~ F = constant on [a, b].

52

VECTOR FUNCTIONS

=G G' = F, find = 2ti + 3t 2j + cos tk = sin 2ti + cos 2tj + tk

Writing JF 18. F(t) 20. F(t)

-¢:>

JF for each of the following: 19. F(t) = sin 2 ti + cos 2 tj 21. F(t) = e-lj + t sin tj

2.4 Curves, directed curves, and arclength By definition a curve is any set in R 3 [ resp. R 2 ] which is the range of a continuous R-R3 [resp. R-R 2 ] function with an interval domain. Such a function is called a representation or parametrization of the curve, and its domain a parameter interval. Then an arc is defined to be any curve having a representation whose domain is a bounded closed interval. (Keeping in mind that vector functions on intervals describe motion, observe that a curve is like a road that can be traversed continuously by some traveler. Then if we can cite a beginning and ending of some such trip, the curve is an arc.) If some representation of an arc C has the form F: [a, b]-.- C such that F(a) = F(b), we say that Cis a closed curve. If, additionally, F is univalent on the semi-open interval (a, b], Cis termed a simple closed curve. Figs. 2.4A-D illustrate these four types of curves.

.,.n

--

FIG. 2.4A. A

\

I a

curve.

= F(b)

(~

I

I

t

F(a)

I

\

b

FIG. 2.4B. An arc.

a

I

I

I b

a

b

FIG. 2.4C. A closed

FIG. 2.4D. A simple

curve.

closed curve.

Example 1. (a) The parabola {(x, y): y = x2 & x e R} is a curve in R2 since it is the range of the continuous function F(t) = (t 2 , t), t e R. However, it is not an arc, for it is unbounded, whereas the range of a continuous function on a bounded closed interval is always bounded. (b) An ellipse {(x,y): x2 /a2 + y2 fb 2 = 1 & (x,y) e R} is an arc since it has the representation F(t) = (a cost, b sin t), t e [0, 21r]. It is a simple closed curve, since F(O) = F(21r), while F(t1 ) # F(t2 ) for lto t2 e (0, 21r]. (c) A segment joining (x0 , y 0 ) to (x1 , y 1 ) is an arc since every point (x, y) on it may be written (x,y) = (x 0 + t(x 1 - x0 ),y0 + t(y 1 - y0 )), t e [0, 1] The notion of "directed (or oriented) curve" is somewhat subtle. It seeks to combine the concept of curve as a set ofpoints with the idea ofvisiting points on the curve in a certain order-a kind of unscheduled itinerary which disregards the time interval of travel as well as the speed of traversal. It is designed to distinguish direction of travel (e.g., clockwise or counterclockwise motion in

§ 2.4 Curves, directed curves, and arclength

58

2-space) and to take into account repeated encounters with parts of the curve (e.g., a single lap of a closed curve should be a different directed curve from a double lap, etc.). The technical definition of directed curve depends on the concept of equivalent functions. Given two vector functions F and G having (possibly different) interval domains, we say that F is equivalent to G (and write F ,.., G) there is a monotone stricdy increasing continuous function p: £)6 --+--+ £)F such that F o p = G. Thus for F to be equivalent to G, not only must the range of both be the same but we must also be able to get every image G(t) via F by using a suitable change in parameter (see Fig. 2.4E).

.__.__.,- F(p(t)) = G(t)

·/~

---:-f=== 0. SOLUTION. Since f'(x) = sinh (xfa) is continuous on [0, a], we may apply (13) to give us fa fa e2-I t(C) = Jo VI + sinh 2 (xfa) dx = Jo cosh (xfa) dx = a~ If Cis the polar graph of r = F (8), 8 E [a, b], we obtain the integral formula for its length using the parametrization

R( 8) = (r cos 8, r sin 8) = (F( 8) cos 8, F( 8) sin 8)

60

VECTOR FUNCTIONS

on [a, b] with the polar angle 8 as parameter. ponents ofR(8), we have

Iff (8) and g ( 8) are the com-

f'(8) = F(8)( -sin 8) + F'(8) cos 8 g' (8) = F (8) cos 8 + F' ( 8) sin 8 so that IR'(8) 12

= [f'(8)]2 + [g'(8)] 2 = [F(8)] 2 + [F'(8)] 2

Now formula (6) gives us ~~----------------------------------,

t(C)

(15)

=

J:

v'[F(8)] 2

+

[F'(8)] 2 d8

if C is the polar graph ofF: [a, b] - R (i.e., (x,y)eC-