Multivariable and Vector Calculus: An Introduction [2 ed.] 1683929195, 9781683929192

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Multivariable and Vector Calculus

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license, disclaimer of liability, and limited warranty By purchasing or using this book (the “Work”), you agree that this license grants permission to use the contents contained herein, but does not give you the right of ownership to any of the textual content in the book or ownership to any of the information or products contained in it. This license does not permit uploading of the Work onto the Internet or on a network (of any kind) without the written consent of the Publisher. Duplication or dissemination of any text, code, simulations, images, etc. contained herein is limited to and subject to licensing terms for the respective products, and permission must be obtained from the Publisher or the owner of the content, etc., in order to reproduce or network any portion of the textual material (in any media) that is contained in the Work. Mercury Learning and Information (“MLI” or “the Publisher”) and anyone involved in the creation, writing, or production of the companion disc, accompanying algorithms, code, or computer programs (“the software”), and any accompanying Web site or software of the Work, cannot and do not warrant the performance or results that might be obtained by using the contents of the Work. The author, developers, and the Publisher have used their best efforts to insure the accuracy and functionality of the textual material and/or programs contained in this package; we, however, make no warranty of any kind, express or implied, regarding the performance of these contents or programs. The Work is sold “as is” without warranty (except for defective materials used in manufacturing the book or due to faulty workmanship). The author, developers, and the publisher of any accompanying content, and anyone involved in the composition, production, and manufacturing of this work will not be liable for damages of any kind arising out of the use of (or the inability to use) the algorithms, source code, computer programs, or textual material contained in this publication. This includes, but is not limited to, loss of revenue or profit, or other incidental, physical, or consequential damages arising out of the use of this Work. Images of ANSYS menus, dialog boxes and plots are copyright of ANSYS Incorporation, United States of America and have been used with prior consent. Commercial software name, company name, other product trademarks, registered trademark logos are the properties of the ANSYS Incorporation, U.S.A. The sole remedy in the event of a claim of any kind is expressly limited to replacement of the book, and only at the discretion of the Publisher. The use of “implied warranty” and certain “exclusions” vary from state to state, and might not apply to the purchaser of this product. Companion files for this title are available by contacting [email protected].

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Multivariable and Vector Calculus An Introduction Second Edition

Sarhan M. Musa, PhD

MERCURY LEARNING AND INFORMATION Dulles, Virginia Boston, Massachusetts New Delhi

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Copyright ©2023 by Mercury Learning and Information LLC. All rights reserved. This publication, portions of it, or any accompanying software may not be reproduced in any way, stored in a retrieval system of any type, or transmitted by any means, media, electronic display or mechanical display, including, but not limited to, photocopy, recording, Internet postings, or scanning, without prior permission in writing from the publisher. Publisher: David Pallai Mercury Learning and Information 22841 Quicksilver Drive Dulles, VA 20166 [email protected] www.merclearning.com (800) 232-0223 S. M. Musa. Multivariable and Vector Calculus: An Introduction, Second Edition. ISBN: 978-1-68392-919-2 The publisher recognizes and respects all marks used by companies, manufacturers, and developers as a means to distinguish their products. All brand names and product names mentioned in this book are trademarks or service marks of their respective companies. Any omission or misuse (of any kind) of service marks or trademarks, etc. is not an attempt to infringe on the property of others. Library of Congress Control Number: 2022947021 232425321  This book is printed on acid-free paper in the United States of America. Our titles are available for adoption, license, or bulk purchase by institutions, corporations, etc. For additional information, please contact the Customer Service Dept. at (800)232-0223(toll free). All of our titles are available in digital format at authorcloudware.com and other digital vendors. Companion disc files for this title are available by contacting [email protected]. The sole obligation of Mercury Learning and Information to the purchaser is to replace the disc, based on defective materials or faulty workmanship, but not based on the operation or functionality of the product.

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Dedicated to my late father Mahmoud, my mother Fatmeh, and my wife Lama

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Contents Prefacexiii Acknowledgmentsxv CHAPTER 1: VECTORS AND PARAMETRIC CURVES1 1.1 Points and Vectors on the Plane 1 Exercises 1.1 13 1.2 Scalar Product on the Plane 17 Exercises 1.2 25 1.3 Linear Independence 28 Exercises 1.3 31 1.4 Geometric Transformations in Two Dimensions 34 Exercises 1.4 42 1.5 Determinants in Two Dimensions 44 Exercises 1.5 51 1.6 Parametric Curves on the Plane 53 Exercises 1.6  67 1.7 Vectors in Space 70 Exercises 1.7 84 1.8 Cross Product 87 Exercises 1.8  96

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viii • Contents 1.9 Matrices in Three Dimensions Exercises 1.9 1.10 Determinants in Three Dimensions Exercises 1.10 1.11 Some Solid Geometry Exercises 1.11 1.12 Cavalieri and the Pappus–Guldin Rules Exercises 1.12 1.13 Dihedral Angles and Platonic Solids Exercises 1.13 1.14 Spherical Trigonometry Exercises 1.14 1.15 Canonical Surfaces Exercises 1.15 1.16 Parametric Curves in Space Exercises 1.16 1.17 Multidimensional Vectors Exercises 1.17 CHAPTER 2: DIFFERENTIATION 2.1 Some Topology Exercises 2.1 2.2 Multivariable Functions Exercises 2.2 2.3 Limits and Continuity Exercises 2.3 2.4 Definition of the Derivative Exercises 2.4 2.5 The Jacobi Matrix Exercises 2.5 2.6 Gradients and Directional Derivatives Exercises 2.6

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98 102 104 109 111 116 117 120 121 126 127 132 133 145 146 151 152 160 163 163 166 169 174 176 187 189 193 193 205 207 216

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Contents • ix

2.7 Levi-Civitta and Einste Exercises 2.7 2.8 Extrema Exercises 2.8 2.9 Lagrange Multipliers Exercises 2.9 CHAPTER 3: INTEGRATION 3.1 Differential Forms Exercises 3.1 3.2 Zero-Manifolds Exercises 3.2 3.3 One Manifold Exercises 3.3 3.4 Closed and Exact Forms Exercises 3.4 3.5 Two-Manifolds Exercises 3.5 3.6 Change of Variables in Double Integrals Exercises 3.6 3.7 Change to Polar Coordinates Exercises 3.7 3.8 Three-Manifolds Exercises 3.8 3.9 Change of Variables in Triple Integrals Exercises 3.9 3.10 Surface Integrals Exercises 3.10 3.11 Green’s, Stokes’, and Gauss’ Theorems Exercises 3.11

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218 221 222 227 229 233 235 235 240 241 242 243 248 250 256 257 267 273 280 283 287 290 295 297 301 303 306 307 313

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x • Contents APPENDIX A: MAPLE A.1 Getting Started and Windows of Maple A.2 Arithmetic A.3 Symbolic Computation A.4 Assignments A.5 Working with Output A.6 Solving Equations A.7 Plots with Maple A.8 Limits and Derivatives A.9 Integration A.10 Matrix

319 319 321 322 323 323 323 324 326 326 326

APPENDIX B: MATLAB B.1 Getting Started and Windows of MATLAB B.1.1 Using MATLAB in Calculations B.2 Plotting B.2.1 Two-dimensional Plotting B.2.2 Three-Dimensional Plotting B.3 Programming in MATLAB B.3.1 For Loops B.3.2 While Loops B.3.3 If, Else, and Elseif 3.3.4 Switch B.4 Symbolic Computation B.4.1 Simplifying Symbolic Expressions B.4.2 Differentiating Symbolic Expressions B.4.3 Integrating Symbolic Expressions B.4.4 Limits Symbolic Expressions B.4.5 Taylor Series Symbolic Expressions B.4.6 Sums Symbolic Expressions B.4.7 Solving Equations as Symbolic Expressions

327 327 329 336 336 347 353 355 356 357 359 361 361 364 365 366 366 367 367

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Contents • xi

APPENDIX C: ANSWERS TO ODD-NUMBERED EXERCISES Chapter 1 Chapter 2 Chapter 3

369 369 386 408

APPENDIX D: FORMULAS D.1 Trigonometric Identities D.2 Hyperbolic Functions D.3 Table of Derivatives D.4 Table of Integrals D.5 Summations (Series) D.5.1 Finite Element of Terms D.5.2 Infinite Element of Terms D.6 Logarithmic Identities D.7 Exponential Identities D.8 Approximations for Small Quantities D.9 Vectors D.9.1 Vector Derivatives D.9.2 Vector Identity D.9.3 Fundamental Theorems

417 417 419 420 422 424 424 424 425 425 425 426 426 427 428

BIBLIOGRAPHY431 INDEX435

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Preface Multivariable and vector calculus is an essential subject of the mathematical education for scientists and engineers. This book is aimed primarily at undergraduates in mathematics, engineering, and the physical sciences. Rather than concentrating on technical skills, it focuses on a deeper understanding of the subject by providing many unusual and challenging examples. Furthermore, this book can be used to impower the mathematical knowledge for Artificial Intelligence (AI) concepts. The topics of vector geometry, differentiation, and integration in several variables are explored. It also provides numerous computer illustrations and tutorials using Maple® and MATLAB®. The software applications allow the students to bridge the gap between analysis and computation. Mainly, this book compromises 3 chapters, 4 appendices and companion files. Chapter 1 provides vectors and parametric curves. It contains points and vectors on the plane, scalar products on the plane, linear independence, geometric transformations in two dimensions, determinants in two dimensions, parametric curves on the plane, vectors in space, cross products, matrices in three dimensions, determinants in three dimensions, some solid geometry, Cavalieri and the Pappus-Guldin rules, dihedral angles and platonic solids, spherical trigonometry, canonical surfaces, parametric curves in space, and multidimensional vectors. Chapter 2 provides differentiation of functions of several variables. This chapter mainly discusses some topology, multivariable functions, limits and continuity, definition of the derivative, the Jacobi matrix, gradients and directional derivatives, Levi-Civita and Einstein, extrema, and Lagrange multipliers.

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xiv • Multivariable and Vector Calculus 2/E Chapter 3 provides integrations of functions of several variables. It contains differentiation forms, zero-manifolds, one-manifolds, closed and exact forms, two-manifolds, change of variables in double integrals, change to polar ­coordinates, three-manifolds, change of variables in triple integrals, surface integrals, and Green’s, Stokes’, and Gauss’ Theorems. Finally, the book concludes with four appendices: Appendix A covers a basic tutorial on Maple software; Appendix B includes a basic tutorial on MATLAB; Appendix C provides the answers to odd-numbered exercises; Appendix D reviews the common, useful mathematical formulas. The companion files are available from the publisher. Contact [email protected].

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Sarhan M. Musa Houston, Texas January 2023

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Acknowledgments It is my pleasure to acknowledge the outstanding help and support of the team at Mercury Learning and Information in preparing this book, especially from David Pallai and Jennifer Blaney. Finally, the book is written based on the constant support, love and patience of my family.

..

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CHAPTER

1

Vectors and Parametric Curves In science and engineering, certain quantities such as force and acceleration possess both a magnitude and a direction. They are represented as vectors and geometrically drawn as an arrow. For example, in dealing with systems of linear equations, the solution can be pointed in the plane if the equations have two variables, points in three space if they are equations in three variables, points in four space if they have four variables, and so on. The solutions make up the subset of large spaces and the constructed spaces are called vector spaces, which are used in many areas of mathematics. We start this chapter with an introduction to some linear algebra necessary for the course. We mainly discuss points and vectors on the plane, scalar product on the plane, linear independence, geometric transformations in two dimensions, determinants in two dimensions, parametric curves on the plane, vectors in space, cross product, matrices in three dimensions, determinants in three dimensions, some solid geometry, Cavalieri, the Pappus–Guldin rules, dihedral angles and platonic solids, spherical trigonometry, canonical surfaces, parametric curves in space, and multidimensional vectors.

1.1 POINTS AND VECTORS ON THE PLANE Definition 1.1.1 A scalar a ∈  is simply a real number. Definition 1.1.2 A point r ∈  2 is an ordered pair of real numbers, r ∈ ( x, y ) with x ∈  and y∈  . Here the first coordinate x stipulates the

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2 • Multivariable and Vector Calculus 2/E location on the horizontal axis and the second coordinate y stipulates the location on the vertical axis. See Fig. 1.1.1. We will always denote the origin by O, that 0 O (= 0,0 )   .  2 is the set of all is, the point = 0 points = a

a1 , a2 ) (=

 a1   a  with real number  2

coordinates on the plane.

FIGURE 1.1.1  A point in  2 .

 a1   b1  Given = a   ∈  2 and = b   ∈  2 , we  a2   b2 

 a1   b1   a1 + b1  define their addition as a + = b  + =    . Similarly, we define  a2   b2   a2 + b2  the scalar multiplication of a point of  2 by the scalar a ∈  as  a1   a a1  a a a= =  a  a a  .  2  2 TIP!

The following sets have special symbols.  = {0,1,2,} denotes the set of natural numbers. =

{, −2, −1,0,1,2,} denotes the set of integers.

=

{

= 

{ x | −∞ < x < ∞} denotes the set of real numbers.

x| x =

}

a , a, b ∈ , b ≠ 0 denotes the set of rational numbers. b

{

}

= a + b : a, b ∈  and i = −1 denotes the set of complex numbers. f ={ TIP!

} denotes the empty set.

The symbol ⇒ is read “ implies”, and the symbol ⇔ is read “if and only if”. Definition 1.1.3  Given two points r and r′ in  2 the directed line segment with departure point r and the arrival point r′ is called the bi-point (fixed vector) r, r′, and is denoted by [r, r′]. See Fig. 1.1.2 for an example.

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Vectors and Parametric Curves • 3

The bi-point [r, r′] can be thus interpreted as an arrow starting at r and finishing, with the arrow tip, at r′. We say that r is the tail of the bi-point [r, r′] and that r′ is its head. TIP!

Let A be a set. If a belongs to the set A, then we write a ∈ A , read “ a is an element of A.” If a does not belong to the set A, then we write a ∉ A , read “ a is not an element of A.”  Definition 1.1.4  A Vector a ∈  2 is a codification of the movement of a bi-

FIGURE 1.1.2  A bi-point in  2 .

  x′ − x  point. Given the bi-point [r, r′], we associate to it the vector rr ′ =    y′ − y stipulating a movement of x′ − x units from ( x, y ) in the horizontal axis and of y′ − y units from the current position in the vertical axis. The zero vector  0  0 =   indicates no movement in either direction. Notice that infinitely 0  many different choices of departure and arrival points may give the same vector.

TIP!

0 Consider a ∈  2 , then [ a,a= ] O=   . 0 EXAMPLE 1.1.1

3  −1  The vector into which the bi-point with tail a =   and head at b =   2 4  3 − (−1)  4  falls = is ab =   .  4 − 2  2 

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4 • Multivariable and Vector Calculus 2/E EXAMPLE 1.1.2 Consider the following points: a1 = (1, 2), b1 = (3, −4), a2 = (3, 5), b2 = (5, −1), O = (0, 0), b = (2, −6). Though the bi-points [ a 1 ,b1 ], [ a 2 ,b2 ], [O,b ] are in different locations on the plane, they represent the same vector, as  3 −1   5 − 3   2 − 0   2  =  =  =   .  −4 − 2   −1 − 5   −6 − 0   −6  The instructions given by the vector are all the same: start at the point, go two units right and six units down. See Fig. 1.1.3. In more technical language, a vector is an equivalence class of bi-points, that is, all bipoints that have the same length, have the same direction. In this sense, points are equivalent and the name of this equivalence is a vector. As a simple example of an equivalence class, consider the set of integers  . According to their remainder upon division by 3, each integer belongs to one of the three sets:

FIGURE 1.1.3  Example 1.1.1.

{, −6, −3,0,3,6,}, 3 + 1= {, −5, −2,1,4,7,}, 3 + 2= {, −4, −1,2,5,8,}. 3=

The equivalence class 3 comprises the integers divisible by 3, and for example, −18 ∈ 3 . Analogously, in Example 1.1.2, the bi-point [ a 1 ,b1 ] belongs to 2 2 the equivalence class   , that is, [ a 1 ,b1 ] ∈   .  −6   −6   Definition 1.1.5  The Vector Oa that corresponds to the point a ∈  2 is called the position vector of the point a.  Definition 1.1.6  Let a ≠ b be the point on the plane and let ab be the line passing through a and b. The direction of the bi-point [a, b] is the direction of

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Vectors and Parametric Curves • 5

 the line L, that is, the angle q ∈ [ 0; p ] that the line ab makes with the positive x-axis (horizontal axis), when measured coun  terclockwise. The direction of a vector v ≠ 0 is the direction of any of its bi-point representatives. See Fig. 1.1.4. FIGURE 1.1.4  Direction of a bi-point.

FIGURE 1.1.5  Bi-points with the same sense.

Definition We say that [a, b] has the same direction as  1.1.7   [z, w] if ab = zw . Definition 1.1.8  We say that the bi-points [a, b] and [z, w] have the same sense if they have the same direction, if when translating one so its tail is over the other’s tail, and both their heads lie on the same half-plane made by the line perpendicular to their tails. See Fig. 1.1.5. Definition 1.1.9  We say that the bi-points [a, b] and [z, w] have the opposite sense if they have the same direction, if when translating one so its tail is over the other’s tail, and their heads lie on different half-planes made by the line perpendicular to their tails. See Fig. 1.1.6.

FIGURE 1.1.6 Bi-points with opposite sense.

TIP!

The sense of a vector is the sense of any of its bi-point representatives. Two bi-points are parallel if the lines containing them are parallel. Two vectors are parallel if bi-point representatives of them are parallel.

Bi-point [b, a] has the opposite sense of [a, b], so we write [ b, a ] = − [ a, b ] .   Similarly, we write ab = − ba . Definition 1.1.10  The Euclidean length (norm or magnitude) of bi-point [a, b] is simply the distance between a and b, and it is denoted by = [a,b]

( a1 -b1 )

2

+ ( a 2 -b2 ) . 2

A bi-point is said to have unit length if it has norm 1. The norm of a vector is the norm of any of its bi-point representatives. NOTE

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A vector is completely determined by three things: (i) its norm, (ii) its direction, and (iii) its sense. It is clear that the norm of a vector satisfies the following properties:

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6 • Multivariable and Vector Calculus 2/E  1. a ≥ 0    2. a = 0 ⇔ a = 0  Definition 1.1.11  A unit vector is a vector whose norm is 1. If v is a    v nonzero vector ( v ≠ 0 ), then the vector u =  is a unit vector in direction of v  v . The procedure of constructing a unit vector in the same direction as a given vector is called normalizing the vector. EXAMPLE 1.1.3

  1  Find the norm of the vector v =   and normalize this vector.  2  Solution:  v =

3, ( 1 )2 + ( 2 ) = 2

which is called the magnitude of the given vector.   1   1 The normalized vector is =    3  2    

1  3 . 2  3

We may use the software MATLAB in order to compute the norm of vectors. >> v = [1 sqrt(2)]; >> norm(v)   ans =  1.7321 >> u = v/norm(v) u=   0.5774 0.8165 Also, we may use the software MapleTM in order to compute the norm of vectors.

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Vectors and Parametric Curves • 7

  Definition 1.1.12  If u and v are two vectors in  2 , their vector sum   u + v is defined by the coordinate-wise addition:

 u1   v1   u1 + v1    u + v =  + =   . (1.1)  u2   v2   u2 + v2 

It is easy  to see that vector addition is commutative and associative, that the  vector 0 acts as an additive identity, and that the additive inverse of a is −a . To add two vectors geometrically, proceed as follows. Draw a bi-point repre  sentative of u . Find a bi-point representative of v having its tail at the tip of     u . The sum u + v is the vector whose tail is that of the bi-point for u and       whose tip is that of the bi-point for v . In particular, if u = AB and v = BC , then we have Chasles’ Rule:    AB + BC = AC . (1.2) See Figs. 1.1.7, 1.1.8, 1.1.9, and 1.1.10.

FIGURE 1.1.7 Addition of vectors.

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FIGURE 1.1.8 Commutative of vector addition.

FIGURE 1.1.9  Associative of vector addition.

FIGURE 1.1.10  Difference of vectors.

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8 • Multivariable and Vector Calculus 2/E  Definition 1.1.13  If a ∈  and a ∈  2 , we define scalar multiplication of a vector and a scalar by the coordinatewise multiplication:  a1  a a1   a a=a  =  . (1.3)  a2  a a2 



See Fig. 1.1.11. It is easy to see that vector addition and scalar multiplication satisfies the following properties:     1. a a + b = a a + a b    2. (a + b ) a = a a + b a   3. 1 a = a   4. (a b ) a = a ( b a )

(

FIGURE 1.1.11 Scalar multiplication of vectors.

)

We may use MATLAB in order to compute the sum of vectors and scalar multiplication of vectors. >> u = [2 5]; >> v = [3 4]; >> u + v ans =  5  9 >> 6∗u ans =  12  30 We may use the software MapleTM in order to compute the sum of vectors and scalar multiplication of vectors.

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Vectors and Parametric Curves • 9

  Definition 1.1.14  Let u ≠ 0. Put   2  u = {l u : l ∈  } and let a ∈  . The affine   u1  line with direction vector u =   and passing  u2   through a is the setoff points on the plane a +  u =  x   2 a1 + tu1 , y = a2 + tu2 , t ∈   . See   ∈  : x =  y   Fig. 1.1.12. FIGURE 1.1.12  Parametric equation of a line on the plane.

If u1 = 0 , the affine line previously defined is ver-

tical, as x is constant. If u1 ≠ 0 , then ( x − a1 ) x − a1 u u = t ⇒ y = a2 + u2 = 2 x + a2 − a1 ,2 u1 u1 u1 u1 u2 that is, the affine line is the Cartesian line with slope . Conversely, if u1

= y mx + k is the equation of a Cartesian line, then  x  1  0 =  y  m t +  k  ,       that is, every Cartesian line is also an affine line and one may take the vector 1    m  as its direction vector. It also follows that two vectors u and v are paral 

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10 • Multivariable and Vector Calculus 2/E

NOTE

    lel if and only if the affine lines  u and  v are parallel. Hence, u || v if   there exists a scalar l ∈  such that u = l v .     Because 0 = 0 v for any vector v , the 0 is parallel to every vector. EXAMPLE 1.1.4

   Let a ∈  , a > 0 , and let u ≠ 0 . Find a vector with norm a and parallel to u . Solution:

  u u = =   1 . Hence, the vector a u u  u  norm a and it is in the direction of u . One may also take − a  . u  u We know that  has norm 1 as u

 u  has u

EXAMPLE 1.1.5

  1  Find a vector of length 3, parallel to v =   , but in the opposite sense.  2  Solution:

 v 2 = 3 , the vector  has unit norm, and has the v  same direction and sense as v , so the vector sought is  2 Since v = (1 ) +

( )

2

 v 3  1  − 3  −3  = −  .  = v 3  2   − 6  EXAMPLE 1.1.6

1 Find the parametric equation of the line passing through   and in the  −1  2   direction of the vector   .  −3  Solution: The desired equation is plainly:  x  1   2   y  = −1  + t  −3  ⇒ x =1 + 2 t, y =−1 − 3 t, t ∈  .      

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Vectors and Parametric Curves • 11

Some plane geometry results can be easily proved by means of vectors. Here are some examples. EXAMPLE 1.17

   Given a pentagon ABCDE , determine the vector sum AB + BC + CD   + DE + EA . Solution: Utilizing Chasles’ Rule several times:        = 0 AA= AB + BC + CD + DE + EA . EXAMPLE 1.1.8 Consider a ∆ABC . Demonstrate that the line segment joining the midpoints of two sides is parallel to the third side and it is in fact, half its length. Solution: Let the midpoints of [A, B] and [C, A], be MC and MB, respectively. We will       demonstrate that BC = 2 MCMB . We have, 2 AMC = AB and 2 AMB = AC . Therefore,    BC = BA + AC   = − AB + AC   = − 2AMC + 2AMB   = 2MC A + 2AMB   = 2 MC A + AMB  = 2MCMB

(

)

as we were to show. EXAMPLE 1.1.9 In ∆ABC , let MC be the midpoint of [A, B]. Demonstrate that  1   = CMC CA + CB . 2

(

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12 • Multivariable and Vector Calculus 2/E Solution:   As AMC = MCB , we have,       CA + CB= CMC + MC A + CMC + MCB    = 2CMC − AMC + MCB  = − 2CMC from where the results follow. EXAMPLE 1.1.10 If the medians [ A, MA ] and [B, MB ] of the non-degenerate ∆ABC intersect at the point G, demonstrate that     AG = 2GMA ; BG = 2GMB . See Fig. 1.1.13. Solution:

FIGURE 1.1.13  Example 1.1.8.

 Since the triangle is non-degenerate, the lines AMA  and BMB are not parallel, and thus meet at a point G.   Therefore, AG and GMA are parallel and hence there   is a scalar a such that AG = aGMA . In the same fash  ion, there is a scalar b such that BG = bGMB . From

Example 1.1.6,   2MA MB = BA   = BG + GA   = bGMB − aGMA    =bGMA + bMA MB − aGMA , and thus 



( 2 − b) MAMB = ( b − a ) GMA .   Since ∆ABC is non-degenerate, MA MB and GMA are not parallel, where

( 2 − b) = 0, ( b − a ) = 0 ⇒ a = b = 2 .

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Vectors and Parametric Curves • 13

EXAMPLE 1.1.11 The medians of a non-degenerate triangle ∆ABC are concurrent. The point of concurrency G is called the barycenter or centroid of the triangle. See Fig. 1.1.14. Solution: Let G be as in Example 1.1.7. We must show  that the line CMC also passes through G. Let   the line CMC and BMB meet in G′ . By the aforementioned example,       AG = 2GMA ; BG = 2GMB ; BG′ = 2G′MB ;   CG′ = 2G′MC .

FIGURE 1.1.14  Example 1.1.9.

It follows that    GG′ = GB + BG′   = − 2GMB + 2G′MB   = 2 MBG + G′MB  = 2G′G .

(

)

Therefore,       GG′ =−2GG′ ⇒ 3GG′ =0 ⇒ GG′ =0 ⇒ G =G′ , demonstrating the result.

EXERCISES 1.1 1.1.1 Identify the following physical quantities as scalars or vectors: 1. time 2. pressure 3. acceleration 4. velocity 5. temperature

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14 • Multivariable and Vector Calculus 2/E 6. gravity 7. force 8. displacement 9. frequency 10. grade of a motor oil 11. sound 12. current in a river 13. speed 14. energy Is there is any truth to the statement “a vector is that which has magnitude and direction”?       −1   −2  1.1.3  = Let u = , and v   , be vectors in  2 . Find u + v , u − v ,  5  −4    2u , and normalization of vector u . 1.1.2

1.1.4

Name all the equal vectors in the parallelogram shown.

FIGURE 1.1.15  Exercise 1.1.4.

1.1.5

Copy the vectors in the figure and use them to draw the following vectors.

FIGURE 1.1.16  Exercise 1.1.5.

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Vectors and Parametric Curves • 15

  1. a + b   2. a − b   3. b + c    4. a + b + c  1  1 1.1.6 = Let u = and v   . Find:   −3  1   1. u + v   2. u + v  3. 3 u 1.1.7

1.1.8 1.1.9

1  Let a be a real number. Find the distance between   and  a 1 − a   1 .    1  1 Find all scales l for which l u = , where u =   . 2  −1   a+1  For which values of a will the vectors a =  2 ,  a − 1   2a + 5  b= 2  will be parallel? a − 4a + 3

1.1.10 In ∆ABC let the midpoints of [ A, B] and [ A, C] be MC and MB ,       respectively. Put MC B = x , MBC = y , and CA = z . Express:    1. AB + BC + MC MB ,    2. AMC + MC MB + MBC ,    3. AC + MC A − BMB    In terms of x , y , and z .

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16 • Multivariable and Vector Calculus 2/E 1.1.11 ABCD is a parallelogram. E is the of [B,C] and F is the  midpoint   midpoint of [D,C]. Prove that AC + BD = 2BC . 1.1.12 (Varignon’s Theorem) Use vector algebra in order to prove that in any quadrilateral ABCD, whose sides do not intersect, the quadrilateral formed by the midpoints of the sides is a parallelogram. 1.1.13 Let A,B be two points on the plane. Construct two points I and J    1  such that IA = −3IB , JA = − JB , and then demonstrate that for 3    any arbitrary point M on the plane MA + 3MB = 4MI and    3MA + MB = 4MJ . 1.1.14 Find the Cartesian equation corresponding to the line with parametric equation x =−1 + t , y= 2 − t . 1.1.15 Let x, y, z be points on the plane with x ≠ y and consider ∆xyz . Let Q be a point on side [x, z] such that [x,Q] : [Q,z] = 3 : 4 and let P be a point on [y, z] such that [y, P] : [P, Q] = 7 : 2. Let T be an arbitrary point on the plane.

   1. Find rational numbers α and β such that = TQ a Tx + b Tz .     2. Find rational numbers l, m, n such that TP =l Tx + m Ty + nTz .      1.1.16 Prove that if u and v are non-collinear then xu + y v = 0 implies x= y= 0 . 1.1.17 Prove that the diagonals of a parallelogram bisect each other as in Fig. 1.1.17.

FIGURE 1.1.17  Exercise 1.1.13.

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Vectors and Parametric Curves • 17

FIGURE 1.1.18  Exercise 1.1.14.

FIGURE 1.1.19  Exercise 1.1.14.

FIGURE 1.1.21  Exercise 1.1.14.

FIGURE 1.1.22  Exercise 1.1.14.

FIGURE 1.1.20  Exercise 1.1.14.

FIGURE 1.1.23  Exercise 1.1.14.

1.1.18 A circle is divided into three, four, or six equal parts (Figs. 1.1.18 through 1.1.23). Find the sum of the vectors. Assume that the divisions start or stop at the center of the circle, as suggested in the figures.

1.2 SCALAR PRODUCT ON THE PLANE We will now define an operation between two plane vectors, which provides a further tool to examine the geometry on the plane.   Definition 1.2.1  Let x ∈  2 and y ∈  2 . Their scalar product (dot product    x1   y1  or inner product) is defined and denoted by x=  y    =  x1 y1 + x2 y2 .  x2   y2  EXAMPLE 1.2.1   1  If a =   and b = 2 

  3   4  , then a  b = 1 × 3 + 2 × 4 = 11 .  

EXAMPLE 1.2.2  1   0     If x =   and y =   , then x  y = 0 and = x 0  1 

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 y= 1 .

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18 • Multivariable and Vector Calculus 2/E The following properties of the scalar product are easy to deduce from the definition. 1. Bilinearity            xz+y z , x ( y + z )= x  y + x  z (1.4) ( x + y )  z = 2. Scalar Homogeneity

      = x  (a y ) a ( x  y ) , a ∈  (a x )  y=

(1.5)

3. Commutativity

    xy=yx

  4. x  x ≥ 0      5. x  x = 0 ⇔ x = 0    6. x = x  x 

(1.6) (1.7) (1.8) (1.9)

The dot product of two vectors can be obtained using MATLAB. >> a = [1; 2]; >> b = [3: 4]; >> dot (a, b) ans =  11 The dot product of two vectors also can be obtained using MapleTM.

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Vectors and Parametric Curves • 19

  Definition 1.2.2  Given vectors a and b , we define the (convex) angle    between them, denoted by a , b ∈ [ 0;p ] , as the angle between the affine   lines  a and  b .

(

)

  Theorem 1.2.1  Let a and b be vectors in         2. Then, a b = a b cos a , b . See

(

)

Fig. 1.2.1.

FIGURE 1.2.1  Theorem 1.2.1.

Proof: From Fig. 1.2.1, using Al-Kashi’s Law of Cosines on the length of the vectors and (1.4) through (1.9), we have:  2 2 2      b − a = a b − 2 a b cos a , b      2 2     b−a  b− a a b − 2 a b cos a , b =       2 2     a a b − 2 a b cos a , b bb − 2a b + a= 2    2 2 2     b − 2a b + a = a b − 2 a b cos a , b        a b = a b cos a , b

(

(

)(

)

)

(

)

(

)

(

(

)

)

as what we wanted to show. EXAMPLE 1.2.3   If the vectors a and b have lengths 6 and 8, and the angle between them is      a , b = p / 3, find a  b .

(

)

Solution: Using Theorem 1.2.1, we have     a b = a b cos = (p / 3 )

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1 2

  6 )( 8 )   24 . (=

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20 • Multivariable and Vector Calculus 2/E The angle between two vectors can be obtained using MATLAB. >> a = [4; 3]; >> b = [2;5]; >> c= acos (dot(a,b) / (norm(a)∗norm(b))); >> (360∗c)/(2∗pi) ans =  31.3287 The angle between two vectors can be obtained using MapleTM.

   p Putting a , b = in Theorem 1.2.1, we obtain the following corollary. 2

(

)

Corollary 1.2.1  Two vectors in  2 are perpendicular if and only if their dot product is 0. NOTE



It follows that the vector 0 is simultaneously parallel and perpendicular to any vector!

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Vectors and Parametric Curves • 21

Definition 1.2.3  Two vectors be orthogonal if they are perpen are said to    dicular. If a is orthogonal to b , we write a ⊥ b. EXAMPLE 1.2.4    −2  Show that the vectors a =   and b = 3 Solution:   Since a  b =

3  2  are orthogonal.  





( −2 ) × ( 3 ) + ( 3 ) × ( 2 ) = 0, a and b are orthogonal.

   Definition 1.2.4 If a ⊥ b and = a

   b = 1 , we say that a and b are

orthonormal.   Definition 1.2.5  Let a ∈  2 be fixed. Then the orthogonal space to a is  defined and denoted by a ⊥ =∈ {x 2 : x ⊥ a} . Since cosq ≤ 1 , we also have the following corollary. Corollary 1.2.2 Cauchy–Bunyakovsky–Schwarz Inequality (CBS Inequality)     a b ≤ a b .   Equality occurs if and only if a || b .   b1    a1  If a =   and b =   , the CBS Inequality takes the form  a2   b2 

a1 b1 + a2 b2 ≤ ( a12 + a22 )

1/ 2

(b

2 1

+ b22 )

1/ 2

. (1.10)

Example 1.2.5 Let a, b be positive real numbers. Minimize a2 + b2 subject to the constraint a + b =. 1

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22 • Multivariable and Vector Calculus 2/E Solution: By the CBS Inequality, 1= a  1 + b  1 ≤ ( a2 + b2 )

1/ 2

(1

2

+ 12 )

1/ 2

⇒ a2 + b2 ≥

1 . 2

 a 1 Equality occurs if and only if   = l   . In this case, a= b= l , so equality  b 1 1 is achieved for a= b= . 2 Corollary 1.2.3  Triangle Inequality     a+b ≤ a + b . Proof:  2     a + b =+ a b  a+b      = a a + 2a b + bb   2   ≤ a +2 a b + b  2  = a + b ,

(

)(

(

)

2

)

from where the desired result follows. Corollary 1.2.4  Pythagorean Theorem    2 2 2 If a ⊥ b , then a + b = a + b . Proof:   Since a b = 0 , we have  2     a+b = a+b  a+b       =a a + 2 a b + b b     = a  a + 0 + b b 2 2 = a + b

(

)(

)

from where the desired result follows.

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Vectors and Parametric Curves • 23

EXAMPLE 1.2.6 2 ( x + y + z ) ≤ x2 + y2 +

Let a, b, z be positive real numbers. Prove that y2 + z2 + z2 + x2 . Solution:   x    y   z  Put a =   , b =   , c =   . Then,  y  z  x    a + b += c

 x + y + z  x + y + z=   

2 ( x + y + z).

Also,    a + b + c =

x2 + y2 + y2 + z2 + z2 + x2 ,

and the assertion follows by the triangle inequality       a+b+c ≤ a + b + c . We now use vectors to prove a classical theorem of Euclidean geometry.  Definition 1.2.6  Let A and B be points on the plane and let u be a unit   vector. If AB = l u , then l is the directed distance or algebraic measure of  the line segment [ AB] with respect to the vector u . We will denote this dis tance by ABu , or more routinely, if the vector u is patent, by AB . Observe that AB = − BA . Theorem 1.2.2  Thales’ Theorem   Let D y D′ be two distinct lines on the  plane. Let A, B, C be distinct points of D ,  and A ′ , B′ , C′ be distinct points of D′ , A ≠ A ′, B ≠ B′, C ≠ C′, A ≠ B, A ′ ≠ B′ . Let     AC A′C′ AA ′ || BB′ . Then, AA ′ || CC′ ⇔ =. AB A′B′ See Fig. 1.2.2.

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FIGURE 1.2.2  Thales’ theorem.

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24 • Multivariable and Vector Calculus 2/E Proof: Refer to Fig. 1.2.2. On the one hand, because they are unit vectors in the same direction,     AB AC A ′ B′ A ′C′ = = ; . AB AC A ′B′ A ′C′ On the other hand, by Chasles’ rule,        BB′ =BA + AA ′ + A ′B′ = A ′B′ − AB + AA ′.

(

)

   Since A ′= B′ AB + l AA ′. Assembling these results,     CC′ = CA + AA ′ + A ′C′  AC   A ′C′  = − ⋅ AB + AA ′ + AB + l AA ′ AB A ′B′  A ′C′ AC    A ′C′   =  −  AB +  1 + l  AA ′. A ′B′   A ′B′ AB     As the line AA ′ is not parallel to the line AB , the preceding equality reveals that

(

)

  AC A′C′ AA ′ || CC′ ⇔ − = 0, AB A′B′ proving the Theorem 1.2.2. From the preceding theorem, we immediately gather the following corollary (see Fig. 1.2.3).   Corollary 1.2.5 Let D and D′ be distinct lines, intersecting in the unique  point C. Let A, B, be points on line D and A′ , B′ be points on line  D′ . Then, FIGURE 1.2.3  Corollary to Thales’ theorem.

  CB CB′ AA ′ || BB′ ⇔ =. CA CA′

MVC.CH01_2PP-Part1.indd 24

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Vectors and Parametric Curves • 25

EXAMPLE 1.2.7 Prove that the altitudes of a triangle ∆ABC on the plane are concurrent. This point is called the orthocentre of the triangle. Solution:        Put a = OA , b = OB , c = OC . First observe that for any x , we have, upon expanding,             0 (1.10) ( x − a ) b − c + x − b ( c − a ) + ( x − c ) a − b =

(

) (

)

(

)

Let H be the point of intersection of the altitude from A and the altitude from B. Then           0 = AH  CB = OH − OA  OB − OC = OH − a  b − c , (1.11)

( )( ) (

)(

) (

)(

)

and

          0 = BH  AC = OH − OB  OC − OA = OH − b ( c − a ) . (1.12)

( )( ) (

)(

) (

)

  Putting x = OH in equation (1.10) and subtracting from it (1.11) and (1.12), we gather that       0= OH − c  a − b = CH  AB ,

(

)(

) ( )( )

Which gives the results.

EXERCISES 1.2 1.2.1

  Find a b , if:   4 1. a =   and b =  −1

3    5

 2  5 2. a =   and b =   0   −1

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26 • Multivariable and Vector Calculus 2/E

  2  3. a =   and b = 3 

1.2.2

1.2.3

5 6   

   −5  4. a =   and b =  −2    Find a b , if:

4   0 

1.

    a =5, b = 8 , and the angle between a and b is p / 3 .

2.

    a = 3, b = 4 , and the angle between a and b is p / 6 .

Find the angle between the vectors.   1   2  1. a =   , b =   0   3   0    −1 2. a =   , b =   1  1   4   1  3. a =   , b =    −2  2   0    1  4. a =   , b =   2   −1

1.2.4 1.2.5 1.2.6

 2 2    Prove that a a + b b = a 2 a + 2a b a b + b 2 b .  a  1 Find the value of a so that  be perpendicular to   .  1 − a   −1       Let a ≠ 0 and b ≠ 0 be vectors in  2 such that a b = 0 . Prove that    a a + b b =0 ⇒ a = b =0 .

1.2.7 1.2.8

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Prove that the diagonals of a rhombus (a parallelogram whose sides have equal length) are perpendicular.   What can we say about the relative magnitudes of vectors a and b ,     if a + b is perpendicular to a − b ?

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Vectors and Parametric Curves • 27

1.2.9

  27    −5  Show that the vectors a =   and b =   are perpendicular  3   15  (orthogonal).

 k   k 1.2.10 If a =   and b =   are orthogonal, find k. 3   −4        1.2.11 Prove that a + b = a − b if and only if a b = 0 .    2 2 2 1.2.12 Prove that if a and b are perpendicular, then a + b = a + b .      a b 1.2.13 Prove that c= a −   2  b  nonzero vector.

( )  b is orthogonal to b , where b is a  

   −3  1.2.14 Find all vectors a ∈  2 such that a ⊥   and a = 13 . 2   1.2.15 (Pythagorean Theorem) If a ⊥ b , prove that   2 2 a+b = a + b . 1.2.16 Let a, b be arbitrary real numbers. Prove that

(a

2

+ b2 ) ≤ 2 ( a4 + b4 ) .

  1.2.17 Let a , b , be fixed vectors in  2 . Prove that if        ∀ v ∈  2 , v a = v  b , a = b .   1.2.18 (Polarization Identity) Let u , v be vectors in  2 . Prove that

(

  1  2   u= v u +v − u−v 4

2

).

1.2.19 Consider two lines on the plane L1 and L2 with Cartesian equations L1= : y m1 x + b1 and L2= : y m2 x + b1 , where m1 ≠ 0 , m2 ≠ 0 . Using Corollary 1.2.1 from Section 1.2, prove that L1 ⊥ L2 ⇔ m1 m2 = −1 .

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28 • Multivariable and Vector Calculus 2/E

 −1  1.2.20 Find the Cartesian equation of all lines L′ passing through   , 2 p making an angle of radians with the Cartesian line L : x + y =. 1 6     1.2.21 Let v , w be vectors on the plane, with w ≠ 0 . Prove that the vector    v  w   a= v −  2 w is perpendicular to w . w

1.3 LINEAR INDEPENDENCE   Consider now two arbitrary vectors in  2 , x and y . Under which conditions   can we write an arbitrary vector v on the plane as a linear combination of x     and y , that is, when can we find scalars a, b such that v = a x + b y ? The answer can be promptly obtained algebraically. Operating formally,    v = a x + b y ⇔ v1 = ax1 + by1 , v2 = ax2 + by2 , v1 y2 − v2 y1 x v − x2 v1 = ⇔ a = ,b 1 2 . x1 y2 − x2 y1 x1 y2 − x2 y1 The previous expressions for a and b make sense only if x1 y2 ≠ x2 y1 . But, x1 x2 = l what does it mean for x1 y2 = x2 y1 ? If none of these are zero, then = y1 y2  x1   y1    and=  x  l  y  ⇔ x || y.  2  2   If x1 = 0 , then either x2 = 0 or y1 = 0 . In the first case, x = 0 , and a fortiori   x || y , since all vectors are parallel to the zero vector. In the second case, we        have x = x2 j, y = y 2 j , so both vectors are parallel to j and hence x || y . We have demonstrated the following theorem.

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Vectors and Parametric Curves • 29

   Theorem 1.3.1  Given two vectors in  2 , x , and y , an arbitrary vector v can be written as the linear combination    v = ax + by, a ∈ , b ∈     if and only if x is not parallel to y . In this last case, we say that x is linearly  independent from vector y . If two vectors are not linearly independent, then we say that they are linearly dependent. EXAMPLE 1.3.1 1  1 The vectors   and   are clearly linearly independent, since one is not a 0  1  a scalar multiple of the other. Given an arbitrary vector   , we can express it  b as a linear combination of these vectors as follows:  a 1  1 ( a − b)   + b   .  b =   0  1    Consider now two linearly independent vectors x and y . For a ∈ [ 0;1] , a x is   parallel to x and traverses the whole length of x : from its tip (when a = 1 ) to   its tail (when a = 0 ). In the same manner, for b∈ [ 0;1] , b y is parallel to y and    traverses the whole length of y . The linear combination a x + b y is also a vector on the plane. EXAMPLE 1.3.2 1  The vector 5   near combination of the vector 3 

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1  2 3  of  with a = 5.  

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30 • Multivariable and Vector Calculus 2/E EXAMPLE 1.3.3 0  1  The vector 2   + 3   is a linear combination of the vectors 1  2 

0  1  1  and 2  of    

 0   1   2 with= a 2,= b 3= , and x = and y   . By applying addition and sca 1  2  0  1  3  lar multiplication defined on  2 , we get 2   + 3   =   . Thus, we may 1  2  8   3   0   1  also say that the vector v =   is a linear combination = of x = and y    8  1  2  0  1  3  because there exists= a 2,= b 3 such that 2   + 3   =   . We can express 1  2  8   3  this relation by saying that the vector v =   is generated by the vectors 8   0   1  = x =  and y 2  . 1    Definition 1.3.1 (Fundamental parallelogram)  Given two linearly inde  pendent vectors x and y , consider bi-point representatives of them with the   tails at the origin. The fundamental parallelogram of the vectors x and y is   the set {a x + b y : a ∈ [ 0;1], b ∈ [ 0;1]} . Fig. 1.3.1 shows the fundamental parallelogram of  1  1     ,    , colored in brown, and the respective tiling  0  1  of the plane by various translations of it. Observe that the vertices of this parallelogram are  0   1   1   2     ,   ,   ,    . In essence then, linear independ 0   0   1   1  

FIGURE 1.3.1 Tiling and the fundamental parallelogram.

MVC.CH01_2PP-Part1.indd 30

ence of two vectors on the plane means that we may obtain every vector on the plane as a linear combination of these two vectors and hence cover the whole plane with all these linear combinations.

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Vectors and Parametric Curves • 31

EXERCISES 1.3 1.3.1

7  Show that   is a linear combination of the vectors 3 

1  3  0  and 1     

of  2 . 1.3.2

Determine whether the first vector of the set of vectors is a linear combination of the other vectors.  7   4   −1 1.   ;   ,    −1  2   1  15   4   −8  2.   ;   ,    −1  −1  2   0   2   4  2  3.   ;   ,   ,    4   −1  6  3   5   −1 1  4.   ;   ,   7   1   5 

1.3.3

 a Write an arbitrary vector   on the plane, as a linear combination  b 1  −1 of the vectors   and   . 1 1

1.3.4

Explain why the following are linearly dependent sets of vectors in  2 .

 1    −3  1. x =   and y =   2   −6   2   6    −5  = 2. x = , y =   and v    3  1  8 1.3.5

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 1   0  Show that the vectors x =   and y =   are linearly independent. 0  1 

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32 • Multivariable and Vector Calculus 2/E

1.3.6

Show that the following sets of vectors in  2 are linearly independent or linearly dependent.   −1 2   1   1.    ,   ,      2  3   4    1   3   2.    ,     3   −1   10   5   3.    ,     12  6   3   1   4.    ,      4   −1 

1.3.7

1.3.8

2  1  8  Consider the vectors   ,   , and   in  2 . Show that the set 1  2  7   2  1  8   are linearly dependent.   ,   ,     1  2  7     Are vectors x and y in Figs. 1.3.2 and 1.3.3 linearly independent or dependent? Explain the reasoning.

FIGURE 1.3.2 Exercise 1.3.

1.3.9

FIGURE 1.3.3 Exercise 1.3.

  Prove that for any two vectors x and y form a linearly dependent set if and only if they are parallel.

1.3.10 Prove the following sets of vectors are linearly dependent or linearly independent in  2 .   3   −9   1.    ,      −2   6  

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Vectors and Parametric Curves • 33

 3  5  2.    ,      −1 2     −4   2   3.    ,      2   −1   0  1   4.    ,     1  0    1  −1  5.    ,     1  1   1.3.11 Find the values of k for which the following sets of vectors are linearly dependent.   4 k   − k   1.   ,    2 k + 6   2     −4   2   2.    ,      k   −1  1.3.12 Prove that two non-zero perpendicular vectors in  2 must be linearly independent. 1.3.13 Let ABCD be a parallelogram.

 1   3  1. Let E and F be points such that AE = AC and AF = AC . 4 4 Show that the lines ( BE ) and ( DF ) are parallel. 2. Let I be the midpoint of [ A, D] and J be the midpoint of [ B, C] . Show that the lines (AB) and ( IJ ) are parallel.

1.3.14

ABCD is a parallelogram; point I is the midpoint of [ A,B] . Point  1  E is defined by the relation IE = ID . Prove that 3  1   = AE AB + AD , and prove that the points A, C, E are collinear. 3

(

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)

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34 • Multivariable and Vector Calculus 2/E

1.4 GEOMETRIC TRANSFORMATIONS IN TWO DIMENSIONS We are now interested in the following fundamental functions of sets (figures) on the plane: translation, scaling (stretching or shrinking) reflection about the axes, and rotation about the origin. A handy tool for investigating all of these (with the exception of translation) is a certain construct called matrices, which we will study in the next section. First, observe what is meant by a function F :  2 →  2 . This means that the input of the function is a point of the plane and the output is also a point on the plane. The following is a rather uninteresting example, but nevertheless an important one. EXAMPLE 1.4.1 The function I :  2 →  2 , I(x) = x is called the identity transformation. Observe that the identity transformation leaves a point untouched. We start with the simplest of these functions. Definition 1.4.1  A function Tv :  2 →  2 is said to be a translation if it is  of the form Tv (x)= x + v , where v is a fixed vector on the plane.

FIGURE 1.4.1 A translation.

A translation simply shifts an object on the plane rigidly by a given amount of units from where it was originally to form a copy of itself (that is, it does not distort its shape or re-orient it). See Fig. 1.4.1 for an example.

It is clear that the composition of any two translations commutes, that is, if Tv1 , Tv 2 :  2 →  2 are translations, then   Tv1  Tv 2 = Tv 2  Tv1 . Let Tv1 (a) = a + v 1 and Tv 2 (a) = a + v 2 . Then,

(T

    Tv 2 ) ( a ) =Tv1 ( Tv 2 ( a ) ) =Tv 2 ( a ) + v 1 =a + v 2 + v 1 ,

(T

    Tv1 ) ( a ) =Tv 2 ( Tv1 ( a ) ) =Tv1 ( a ) + v 2 =a + v 1 + v 2 ,

 v1

and  v2

from where the commutative claim is deduced.

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Definition 1.4.2  A function Sa, b :  2 →  2 is said to be a scaling if it is of  ax  the form Sa, b (r) =   , where a > 0, b > 0 are real numbers.  by  Fig. 1.4.2 shows the scaling   x    2x  S2,0,5     =  . y 0.5 y      

FIGURE 1.4.2 A scaling.

It is clear that the composition of any two scaling commute, that is, if Sa, b , Sa′, b′ :  2 →  2 are scaling, then Sa, b  Sa′, b′ = Sa′, b′  Sa, b . For

(S

  a′x    a ( a′x )   Sa′, b′ ) ( r ) Sa,= = S=  , and   b ( Sa′, b′ ( r ) ) a, b     b′y    b ( b′y ) 

a, b

  ax    a′ ( ax )  =  Sa, b ) ( r ) Sa= S= ′, b′ ( Sa , b ( r ) )    b′ by  , a′, b′   by    ( )  from where the commutative claim is deduced.

(S

a′, b′

Translation and scaling do not necessarily commute, however. Consider the   2 a1  translation Ti ( a )= a + i and the scaling S2,1 (a) =   . Then,  a2     −1      −2    −1   Ti  S=      T=    , i     0   0    0 

  −1  

(T  S = )  0    i

2,1



but

(S

2,1

   −1      −1   0 0  Ti = )   0   S2,1  Ti =      S= 0 . 2,1     0 0               

Definition 1.4.3  A function RH :  2 →  2 is said to be a reflection about  −x  the y-axis or horizontal reflection if it is of the form RH (r) =   .  y A function RV :  2 →  2 is said to be a reflection about the x -axis or vertical  x  reflection if it is of the form RV (r) =   . A function RO :  2 →  2 is said to  −y 

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36 • Multivariable and Vector Calculus 2/E be a reflection about origin if it is of the form  −x  RH (r) =   .  −y  Some reflections appear in Fig. 1.4.3.

FIGURE 1.4.3 Reflections. The original object (in the first quadrant) is yellow. Its reflection about the y-axis is magenta (on the second quadrant). Its reflection about the x-axis is cyan (on the fourth quadrant). Its reflection about the origin is blue (on the third quadrant).

A few short computations establish various commutative properties among reflection, translation, and scaling. See Exercise 1.4.4. We now define rotations. This definition will be somewhat harder than the others, so let us develop some ancillary results. Consider a point r with polar coordinates x = r cos a and y = r sin a as in Fig. 1.4.4.

FIGURE 1.4.4 Rotation by an angle θ in the levogyrate (counterclockwise) sense from the x-axis.

r Here=

x2 + y2 and a ∈ [ 0;2p ] . If we rotate it, in the levogyrate sense, by

an angle q , we land on the new point x′ with = x′ r cos (a + q ) and = y′ r sin (a + q ) . But r cos (a + q= a x cosq − y sin q , ) r cosq cosa − r sin q sin= and r sin (a + q= a y cosq + x sin q . ) r sin a cosq + r sin q cos=

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Vectors and Parametric Curves • 37

 x Hence, the point   is mapped to the point  y

 x cosq − y sin q   .  x sin q + y cosq 

We may now formulate the definition of a rotation. Definition 1.4.4  A function Rq :  2 →  2 is said to be a levogyrate rotation about the origin by the angle θ measured from the positive x-axis if  x cosq − y sin q  Rq (r) =  . Here= r x 2 + y2 .   x sin q + y cosq  Various properties of the composition of rotations with other plane transformations are explored in Exercises 1.4.5 and 1.4.6. We now codify some properties shared by scaling, reflection, and rotation. Definition 1.4.5  A function L :  2 →  2 is said to be a linear transformation from  2 to  2 if for all points a, b on the plane and every scalar λ , it is verified that L ( a + b ) = L ( a ) + L ( b ) , L(l a) = l L ( a ) . It is easy to prove that scaling, reflections, and rotations are linear transformations from  2 to  2 , but not so translation. Definition 1.4.6  A function L :  2 →  2 is said to be an affine transformation from  2 to  2 if there exists a linear transformation L :  2 →  2 and a  fixed vector v ∈  2 such that for all points x ∈  2 , it is verified that  A(x)= L(x) + v . It is easy to see that translations are then affine transformations. In this definition, where the linear transformation is L, we may take I :  2 →  2 , then the identity transformation I(x) = x . We have seen that scaling, reflection, and rotation are linear transformations. If L :  2 →  2 is a linear transformation, then     L(r ) = L( x i + y j) = xL( i) + yL( j) , and thus a linear transformation from R2 to R2 is solely determined by the  values L( i ) and L( j ) . We will now introduce a way to codify these values.

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38 • Multivariable and Vector Calculus 2/E

Definition 1.4.7 Let L :  2 →  2 be a linear transformation. The matrix AL associated with L is the 2 × 2 (2 rows, 2 columns) array whose columns 1 0 are (in this order) L     and L     . 0 1 EXAMPLE 1.4.2 (Scaling Matrices) Let a > 0 , b > 0 be real numbers. The matrix of the scaling transformation 0  a× 0  0 1  a×1   a a 0 Sa, b is  . For Sa, b = and Sa, b =     =    =   .      0 b   1    b×1   b 0  b× 0 0 EXAMPLE 1.4.3 (Reflection Matrices)  −1 0  It is easy to verify that the matrix for the transformation RH is   , that  0 1 1 0  the matrix for the transformation RV is   , and the matrix for the trans0 −1  −1 0  formation RO is  .  0 −1 EXAMPLE 1.4.4 (Rotating Matrices) cosq It is easy to verify that the matrix for a rotation Rq is   sin q

− sin q  . cosq 

EXAMPLE 1.4.5 (Identity Matrix) The

matrix

for

the

identity

linear

transformation

Id :  2 →  2 ,

transformation

N : 2 → 2 ,

1 0  Id (x) = x is I2 =  . 0 1  EXAMPLE 1.4.6 (Zero Matrix) The

matrix

for

the

null

linear

0 0  N (x) = O is 0 2 =  . 0 0 

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Vectors and Parametric Curves • 39

From Example 1.4.7, we know that the composition of two linear transformations is also linear. We are now interested in how to codify the matrix of a composition of linear transformations L1  L2 in terms of their individual matrices. Theorem

1.4.1 Let

L : 2 → 2

have the matrix representation

and let

L′ :  2 →  2

have the matrix representation

a b AL =   c d r AL′ =  t

s . Then the composition L  L′ has matrix representation u 

 ar + bt as + bu   cr + dt cs + du  .   Proof: 1 0 We need to find ( L  L′ )     and ( L  L′ )     . 0 1 We have 1



1

 r 





 a

 b   ar + bt 

0



0

 s 





 a

 b   as + bu 

( L  L′ )    = L  L′     = L    = rL(i) + tL(j ) = r   + t  =  ,  c   d   cr + dt  0  t   0 and

( L  L′ )    = L  L′     = L    = sL(i) + uL(j ) = s   + u  =  ,  c  d   cs + du  1  u  1  ar + bt as + bu  hence we conclude that the matrix of L  L′ is   , as we  cr + dt cs + du  wanted to show. The preceding motivates the following definition. a b r Definition 1.4.8 Let A =  and B =   c d t

s be two 2 × 2 matrices, u 

and l ∈  be a scalar. We define matrix addition as

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40 • Multivariable and Vector Calculus 2/E

a b r s  a + r A= +B  + =    c d   t u  c + t

b+ s . d + u 

We define matrix multiplication as  a b   r s   ar + bt as + bu  = AB =    .  c d   t u   cr + dt cs + du  We define scalar multiplication of a matrix as  a b  l a l b  = l A l=  c d  l c l d  .     NOTE

Since the composition of functions is not necessarily commutative, neither is matrix multiplication. Since the composition of functions is associative, so is matrix multiplication.

EXAMPLE 1.4.7 1 −1  1 2 Let M =  , N=  .  −2 1  0 1  Then, 1 −1  1 2   2 1  M= +N  + =   , 0 1   −2 1   −2 2  1 −1 2 −2  = 2 M 2= 0 1  0 2  ,     1 −1  1 2  1 × 1 + ( −1 ) × ( −2 ) 1 × 2 + ( −1 ) × 1  3 1 = MN = = .     0 × 2 + 1 × 1   −2 1 0 1   −2 1   0 × 1 + 1 × ( −2 ) EXAMPLE 1.4.8 2 1  2 1  2 1   4 5  2 1  10 5  2 5M 5= = =  , then M =  0 3   0 9  and 0 3   0 15  0 3 0 3           

Let M = 

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Vectors and Parametric Curves • 41

EXAMPLE 1.4.9 Find a 2 × 2 matrix that will transform the square in Fig. 1.4.5 into the parallelogram in Fig. 1.4.6. Assume in each case that the vertices of the figures are lattice points, that is, coordinate points with integer coordinates.

FIGURE 1.4.5  Example 1.4.8.

FIGURE 1.4.6  Example 1.4.8.

Solution: a b Let   be the desired matrix. Then, since c d  a b  0  0   c d  0  = 0  ,     

0 the point   is a fortiori, transformed to itself. We now assume, without loss 0 of generality, that each vertex of the square is transformed in the same order, counterclockwise, to each vertex of the rectangle. Then,  a b  1  2   a  2   c d  0 = 2  ⇒  c = 2  ⇒ a= c= 2 .          Using these values,  a b  1 1   a + b 1   c d  1= 3  ⇒  c + d = 3  ⇒ b= −1,d= 1 .          And so the desired matrix is 2 −1 2 1  .  

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42 • Multivariable and Vector Calculus 2/E

EXERCISES 1.4 1.4.1

a b  1 −1 2 If A =  , B= , and ( A + B) =A2 + 2 AB + B2 ,   1 −2  2 3  find a b.

1.4.2

1 2  1 2  Let M =  , N=   . Find M+N, 3M, and MN. 0 −1 3 −1

1.4.3

Find all matrices A ∈ M2×2 (  ) that A2 = 0 2 .

1.4.4

Find all linear transformations from  2 into  2 that 1. Carry the line x = 0 into the line x = 0 . 2. Carry the line y = 0 into the line y = 0 . 3. Carry the line x = y into the line x = y . 4. Carry the line x = 0 into the line x = 0 and carry the line y = 0 into the line y = 0 .

1.4.5

 x 1 Let L :  2 →  2 be defined by L   =   . Show if L is linear or  y  2  not linear.

1.4.6

 Let L be a linear transformation by reflecting each vector u in  2 with respect to the line y = − x . Determine a matrix for L.

1.4.7

Prove that the following transformation L :  2 →  2 defined by  x   x − y L  =   is linear.  y  3x 

1.4.8

Prove that the following transformation L :  2 →  2 defined by  x   2x  L  =   is linear.  y  x − y

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Vectors and Parametric Curves • 43

1.4.9

 −1   0 2 Consider ∆ABC = with A = , B = , C   , as in   2  −2  1 Fig. 1.4.7. Determine the effects of the following scaling transformations on the triangle: S2,1 , S1,2 , and S2,2 .

FIGURE 1.4.7 Exercises 1.4.2, 1.4.3, and 1.4.8.

1.4.10 Find the equation of the image of the line y = 3 x under scaling of factor 3 in the x direction and factor 5 in the y direction. 1.4.11 Find the effects of the reflections, Rp , Rp , R− p , and R− p on the 2

4

2

4

triangle in Fig. 1.4.7. 1.4.12 Find the effects of the reflections RH , RV , and RO on the triangle in Fig. 1.4.7. 1.4.13 Determine the matrix that defines a reflection in the y axis. Find 3  the image of   under this transformation. 2  x2 y2 1.4.14 Find the equation of the image of the ellipse + = 1 under a 16 25 rotation through an angle p / 2 . 1.4.15 Determine the matrix that can be used to define a rotation through p / 2 about the point (5,1). Find the image of the unit square under this rotation. 1.4.16 Let L :  2 →  2 be a counterclockwise rotation through p / 4 radians. Find the standard matrix representing L.

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44 • Multivariable and Vector Calculus 2/E     1.4.17 Let L1 x = A1 x and L2 x = A2 x be defined by the following

()

()

matrices A1 and A2 . Find L2  L1 , if: 1 3   1 4   2  = 1. A1 = , A2 =   , x 3  2 0   −1 0    1 2  2 −1     −2  2. A1  = , A2 = 0 3 ,x   =  0 1  1 2 −1  x   3x   x  0  1.4.18 Let L1   =   and L2   =   . Find L2  L1 .  y   −y   y  x + y

1.5 DETERMINANTS IN TWO DIMENSIONS We will now define a way of determining areas of plane figures on the plane. It seems reasonable to require that this area determination agrees with common formulae of areas of plane figures, in particular, the area of a parallelogram should be as we learn in elementary geometry and the area of a unit square is 1.  a From Figs. 1.5.1 and 1.5.2, the area of a parallelogram spanned by   and  b c  d  is    a  c   D    ,  =  ad − bc.   b  d  

FIGURE 1.5.1  Area of a parallelogram.

FIGURE 1.5.2  ( a + c )( b + d ) − 2

c (2b + d ) ab −2 = ad − bc. 2 2

This motivates the following definition.

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Vectors and Parametric Curves • 45

a b Definition 1.5.1  The determinant of the 2 × 2 matrix   is c d a b det  =  ad − bc. c d NOTE

a

The symbol for determinant of a matrix  c or

a b . c d

b a b can be written as det    d c d

EXAMPLE 1.5.1

3 1  Evaluate the determinants of M =  .  5 −2  Solution: det(M) = −11 . ( 3 )( −2 ) − (1 )( 5 ) = EXAMPLE 1.5.2  −2 4  Evaluate the det  .  5 7 Solution:  −2 4  det  (−2)(7) − (4 − 5) = −13 =  5 7 EXAMPLE 1.5.3

1  x − 2 Find the values of x for which det(M) = 0 when M =  .  −5 x + 4  Solution: det(M) =

x−2 −5

1 = ( x − 2 )( x + 4 ) − (1 )( −5 ) x+4

= x2 + 2 x − 3 =

( x − 1)( x + 3 )

Hence, det(M) = 0 ⇔ x = 1 or x = −3 .

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46 • Multivariable and Vector Calculus 2/E Theorem 1.5.1 If M and N are square matrices of same size, then det(MN) = det(M)det(N) . EXAMPLE 1.5.4 1 2  8 5  14 3  Let M =  and N =  , then MN =    , 3 −1 1 3  17 2  det(M) = 3 − 2 = 1 , det(N) =−8 − 15 =−23 , and det(MN) = 28 − 51 = −23 . Thus, det(MN) = det(M)det(N) as expected. Consider now a simple quadrilateral with vertices = r1

x1 , y1 ) , r2 ( = x2 , y2 ) , r3 ( = x3 , y3 ) , r4 ( x4 , y4 ) , listed in counterclockwise (=

order, as in Fig. 1.5.3. This quadrilateral is spanned by the vectors   x2 − x1    x4 − x1  r1 r2 =   , r1 r4 =  y − y  , 1  y2 − y1   4 and hence, its area is given by     x4 − x1  = D r2 − r1 ,r4 − r1 .  y4 − y1 

 x2 − x1 A = det   y2 − y1

(

)

Similarly, noticing that the quadrilateral is also spanned by FIGURE 1.5.3 Area of a quadrilateral.

  x4 − x3    x2 − x3  r3 r4 =   , r3 r2 =  y − y  , 3  y4 − y3   2

its area is also given by  x4 − x3 A = det   y4 − y3

    x2 − x3  = D r4 − r3 ,r2 − r3 .  y2 − y3 

(

)

Using the properties derived in previous theorems, we see that

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Vectors and Parametric Curves • 47

1 2 1 A= 2 1 A= 2 1 A= 2

A =

   

   

( D ( r − r ,r − r ) + D ( r − r ,r − r ))                 ( D ( r ,r ) − D ( r ,r ) − D ( r ,r ) + D ( r ,r )) + 12 ( D ( r ,r ) − D ( r ,r ) − D ( r ,r ) + D ( r ,r ))             ( D ( r ,r ) − D ( r ,r ) − D ( r ,r )) + 12 ( D ( r ,r ) − D ( r ,r ) − D ( r ,r ))         ( D ( r ,r ) + D ( r ,r ) + D ( r ,r ) + D ( r ,r )). 2

1

4

1

4

3

2

2

4

2

1

1

4

2

4

2

1

1

4

1

2

2

3

3

4

3

1

1

4

4

2

4

2

3

2

3

2

4

3

4

3

3

3

1

We conclude that the area of a quadrilateral with vertices (x1, y1), (x2, y2), (x3, y3), (x4, y4) listed in counterclockwise order is x3   x3 x4   x4 x1   + det  + det      . (1.11) y3   y3 y4   y4 y1      To find the area of a triangle of vertices, = r1 (= x1 , y1 ) ,r2 ( = x2 , y2 ) ,r3 ( x3 , y3 ) ,

 x1 1  det  2  y1

x2   x2 + det   y2   y2

listed in counterclockwise order as in Fig. 1.5.4, reflects about one of its sides, as in Fig. 1.5.5, creating a parallelogram.

FIGURE 1.5.4 Area of a triangle.

FIGURE 1.5.5 Area of a triangle.

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48 • Multivariable and Vector Calculus 2/E The area of the triangle is now half the area of the parallelogram, which, by virtue of equation (1.11), is         1 D r1 ,r2 + D r2 ,r + D r,r3 + D r3 ,r1 . 4 This is equivalent to

( ( ) ( ) ( ) ( ))

      1 D r1 ,r2 + D r2 ,r3 + D r3 ,r1 2           1 − D r1 ,r2 − D r2 ,r − D r,r3 + D r3 ,r1 + 2 D r2 ,r3 4

( ( ) ( ) ( )) ( ( ) ( ) ( ) ( )

(

)) .

We will prove that           D r1 ,r2 − D r2 ,r − D r,r3 + D r3 ,r1 + 2 D r2 ,r3 = 0.

(

)

( )

( )

(

)

(

)

To do this, we appeal once again to the bi-linearity properties derived in previ    ous theorems, and observe that we have a parallelogram, r − r3 = r2 − r1 , which     means, r = r3 + r2 − r1 . Thus,           D r1 ,r2 − D r2 ,r − D r,r3 + D r3 ,r1 + 2 D r2 ,r3 =               = D r1 ,r2 − D r2 ,r3 + r2 − r1 + 2 D r2 ,r3 − D r3 + r2 − r1 ,r3 + D r3 ,r1           = D r1 ,r2 − r3 + D r3 + r2 − r1 ,r2 − r3 + 2 D r2 ,r3       = D r3 + r2 ,r2 − r3 + 2 D r2 ,r3       = D r3 ,r2 − D r2 ,r3 + 2 D r2 ,r3     = D r3 ,r2 − D r2 ,r3

(

( ( ( ( (

)

( ) ( ) ( ) ( ) ) ( ) ( ) ( ) ( ) ( ) ) ( ) ) ( ) ( ) ) ( )

)

(

)

= 0, as claimed. We have proved then that the area of a triangle, whose vertices ( x1 , y1 ) , ( x2 , y2 ) , ( x3 , y3 ) are listed in counterclockwise order, is

 x1 1  det  2  y1

x2   x2 + det   y2   y2

x3   x3 + det   y3   y3

x1    . (1.12) y1  

In general, we have the following theorem.

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Vectors and Parametric Curves • 49

Theorem 1.5.1 (Surveyor’s Theorem)  Let ( x1 , y1 ) , ( x2 , y2 ) ,, ( xn , yn ) be the vertices of a simple (non-crossing) polygon, listed in counterclockwise order. Then its area is given by  x1 1  det  2  y1

x2   x2 + det   y2   y2

x3   x n −1 +  + det   y3   yn−1

xn   xn + det   yn   yn

x1    y1   .

Proof: The proof is by induction on n. We have already proved the cases n = 3 and n = 4 in (1.12) and (1.11), respectively. Consider now a simple polygon P with n vertices. If P is convex, then we may take any vertex and draw a line to the other vertices, triangulating the polygon, creating n − 2 triangles. If P is not convex, then there must be a vertex that has a reflex angle. A ray produced from this vertex must hit another vertex, creating a diagonal; otherwise, the polygon would have an infinite area. This diagonal divides the polygon into two sub-polygons. These two sub-polygons are either both convex or at least one is not convex. In the latter case, we repeat the argument, finding another diagonal and creating a new sub-polygon. Eventually, since the number of vertices is infinite, we end up triangulating the polygon. Moreover, the polygon can be triangulated in such a way that all triangles inherit the positive orientation of the original polygon but each neighboring pair of triangles have opposite orientations. Applying (1.12), we obtain that the area is  xi

∑ det  y

xj  , y j 

 i where the sum is over each oriented edge. Since each diagonal occurs twice, but having opposite orientations, the terms  xi det   yi

xj  x j + det   yj  y j

xi  0 = yi 

disappear from the sum and we are simply left with  x1 1  det  2  y1

MVC.CH01_2PP-Part2.indd 49

x2   x2 + det   y2   y2

x3   x n −1 +  + det   y3   yn−1

xn   xn + det   yn   yn

x1   . y1  

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50 • Multivariable and Vector Calculus 2/E We may use the software MapleTM in order to speed up computations with vectors. Most of the commands we will need are in the linalg package. For  1   2  example, let us define two vectors, a =   and b =   , and a matrix 2  1  1 2  A :=   . Let us compute their dot product, find a unit vector in the direc3 4   tion of a , and the angle between the vectors, the determinant of a matrixA. (There must be either a colon or a semicolon at the end of each statement. The result will not display if a colon is chosen.)

We may also use MATLAB in order to speed up computations with vectors. >> a = [1,2]; >> b = [2,1]; >> norm(a)

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Vectors and Parametric Curves • 51

ans =  2.2361 >> normalized = a/norm(a) normalized =  0.4472 0.8944 >> CosTheta = dot(a,b)/(norm(a)∗norm(b)); >> CosTheta = dot(a,b)/(norm(a)∗norm(b)); >> ThetaInDegrees = acos(CosTheta)∗180/pi ThetaInDegrees =  36.8699 >> A=[1 2;3 4] A=  1 2  3 4 >> det (A) ans =   −2

EXERCISES 1.5 1.5.1

Calculate 1 −2  1. det   6 3 

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52 • Multivariable and Vector Calculus 2/E

1 2  2. det   5 4  4 −2  3. det   5 3   −2 −7  4. det    6 −5   x − 2 −3  5. det  x − 3   5  3 6. det   6 1.5.2

4   2 

For what values of a, does 1 + a 1. det   1

1  =0 1 − a 

a + 4 −5  2. det  =0 a − 2   1 1.5.3

 cosq Compute the det   − sin q

1.5.4

cosq Compute the det   sin q

sin q  for any real number. cosq  sin q  when cosq 

1. q = p / 4 2. q = p / 3 1.5.5

Let k be a number, and let A be a 2 × 2 matrix. How does det(kA) differ from det( A)?

1.5.6

Let matrix B be formed from matrix A by interchanging two rows. Prove then that det ( A ) = − det ( B) .

1.5.7

Prove that if A and B are square matrices of the same size, then det( AB) = det(BA).

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Vectors and Parametric Curves • 53

1.5.8

1.5.9

Prove that if the sum of elements of each row (or column) of a square matrix A zero, then det( A) = 0.    a    c  Let the vectors r1 =   and r2 =   , and that the vector r3  b d     − b obtained by rotating r1 counterclockwise by p / 2 is r3 =   . a     Show that r3 r2 = D r1 ,r2 .

(

1.5.10

)

Prove that the area of the parallelogram spanned by the vectors     a    c  r1 =   and r2 =   is the D r 1 ,r 2 .  b d 

(

)

1.6 PARAMETRIC CURVES ON THE PLANE Definition 1.6.1 Let [ a;b ] ⊆  . A parametric curve representation r of a  x(t)  curve Γ is function r : a; b →  2 , with r ( t ) =   , and such that  y(t)  r ([ a;b ]) = Γ . r (a) is the initial point of the curve and r (b) its terminal point. A curve is closed if its initial point and its final point coincide. The trace of the curve r is the set of all images of r, that is, Γ . If there exists t1 ≠ t2 such that r(= t1 ) r(= t2 ) p , then p is a multiple point of the curve. The curve is simple if curve has no multiple points. A closed curve whose only multiple points are its endpoints is called a Jordan curve. Graphing parametric equations is a difficult art. A theory akin to the one studied for Cartesian equations in a first Calculus course has been developed. Our interest is not in graphing curves, but in obtaining suitable parameterizations of simple Cartesian curves. However, we mention in passing that MapleTM has excellent capabilities for graphing parametric equations. For example, the commands to graph the various curves in Figs. 1.6.1 through 1.6.4 follow.

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54 • Multivariable and Vector Calculus 2/E

FIGURE 1.6.1  = x sin2 = t , y cos6t.

t /10 FIGURE = 1.6.2  x 2= cos t , y 2t /10 sin t .

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Vectors and Parametric Curves • 55

x FIGURE 1.6.3  =

FIGURE = 1.6.4  x

t − t3 1− t 2 ,y = . 2 1+ t 1+ t 2

(1+ cos t )

= ,y 2

( sin t )(1+ cos t ) 2

.

Our main focus of attention is the following. Given a Cartesian curve with equation f ( x, y ) = 0 , we wish to find suitable parameterizations for them. That is, we want to find functions x : t → a(t), y : t → b(t) and an interval I such that the graphs of f ( x, y ) = 0 and f ( a(t), b(t= ) ) 0, t ∈ I coincide. These parameterizations may differ in features, according to the choice of functions and the choice of intervals.

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56 • Multivariable and Vector Calculus 2/E EXAMPLE 1.6.1 Suppose that m ∈  2 , b∈  2 . Write the Cartesian equation of the line = y mt + b in parametric form. Solution: We put x = t , then = y mt + b and so the desired parametric form is  x  1  y − b = t  m  .     EXAMPLE 1.6.2 Consider the parabola with the Cartesian equation y = x2 . We will give various parameterizations for portions of this curve. 2 1. If x = t and y = t 2 , then clearly y= t= x2 . This works for every t ∈  ,

and hence the parameterization = x t= , y t2 , t ∈  , works for the whole curve. Notice that as t increases, the curve is traversed from left to right.

x t,= y t2 , t ∈  FIGURE 1.6.5 =

2. If x = t and y = t , then again y= t=

( t) = 2

x2 . This works only for

t ≥ 0 , and hence the parameterization = x

t= , y t, t ∈ [ 0; +∞[ ,

gives the half of the curve for which x ≥ 0 . As t increases, the curve is traversed from left to right.

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Vectors and Parametric Curves • 57

x FIGURE 1.6.6 =

t, = y t, t ∈ [ 0; +∞[ .

3. Similarly, if x = − t and y = t , then again y= t=

( t) = 2

x2 . This works

only for t ≥ 0 , and hence the parameterization x = − t , y = t, t ∈ [ 0; +∞[ , gives the half of the curve for which x ≥ 0 . As t increases, x decreases, and so the curve is traversed from right to left.

FIGURE 1.6.7  x = − t , y = t, t ∈ [ 0; +∞[ . 2 4. If x = cos t and = y cos = t

2 t ) x2 . Both x and y are periodic with ( cos=

period 2p , and so this parameterization only agrees with the curve y = x2 when −1 ≤ x ≤ 1 . For t ∈ [ 0; p ] , the cosine decreases from 1 to −1 and so the curve is traversed from right to left in this interval.

FIGURE 1.6.8  x = cos t , y = cos2 t , t ∈ [ 0; p ] .

The identities 2 2 2 cos2 q + sin = 1, tan 2 q − sec= q 1, cosh 2 q − sinh= q 1 , are often use-

ful when parametrizing quadratic curves.

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58 • Multivariable and Vector Calculus 2/E EXAMPLE 1.6.3 Give two distinct parameterizations of the ellipse

( x − 1 )2 ( y + 2 ) 4

+

9

2

= 1.

1. The first parameterization must satisfy that as t traverses the values in the interval [ 0;2p ] , one starts at the point (3, −2), traverses the ellipse once counterclockwise, finishing at (3, −2). 2. The second parameterization must satisfy that as t traverses the interval [0; 1], one starts at the point (3, −2), traverses the ellipse twice clockwise, and returns to (3, − 2). Solution: What formula do we know where a sum of two squares equals 1? We use a trigonometric substitution, a sort of “polar coordinates.” Observe that for t ∈ [ 0;2p ] , the point ( cos t,sin t ) traverses the unit circle once, starting at (1, 0) and ending there. Put x −1 = cos t ⇒ x = 1 + 2 cos t, 2 and y+2 =sin t ⇒ y =−2 + 3 sin t . 3 Then x =1 + 2 cos t, y =−2 + 3 sin t, t ∈ [ 0;2p ] is the desired first parameterization. For the second parameterization, notice that as t traverses the interval [0; 1], ( sin 4p t,cos 4p t ) traverses the unit interval twice, clockwise, but begins and ends at the point (0, 1). To begin at the point (1, 0) we must make a shift: p p      sin  4p t + 2  ,cos  4p t + 2   will start at (1, 0) and travel clockwise twice,      as t traverses [0; 1]. Hence we may take

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Vectors and Parametric Curves • 59

p p   x =1 + 2 sin  4p t +  , y =−2 + 3 cos  4p t +  , t ∈ [ 0;1] 2 2    as our parameterization. Some classic curves can be described by mechanical means, as the curves drawn by a spirograph. We will consider one such curve. EXAMPLE 1.6.4 A hypocycloid is a curve traced out by a fixed point P on a circle C of radius ρ as C rolls on the inside of a circle with center at O and radius R. If the initial R position of P is   , and q is the angle, measured counterclockwise, with a 0 ray starting at O and passing through the center of C, which makes the x-axis, show that a parameterization of the hypocycloid is  (R − r )  x= ( R − r ) cosq + r cos  ,  r   (R − r )  y= ( R − r ) sin q + r sin  .  r  Solution: Suppose that starting from q = 0, the centre O′ of the small circle moves counterclockwise inside the larger circle by an angle q, and the point P = ( x, y ) moves clockwise an angle f . The arc length traveled by the center of the small circle is ( R − r )q radians. At the same time, the point P has rotated rf radians, and so ( R − r )q = rf . See Fig. 1.6.9, where O′B is parallel to the x-axis.

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60 • Multivariable and Vector Calculus 2/E

FIGURE 1.6.9  Construction of the hypocycloid.

p Let A be the projection of P on the x-axis. Then ∠OAP = ∠OPO′ = , 2 p ∠OO′P = p − f − q , ∠POA = − f , and OP = (R − r )sin (p − f − q ) . 2 p  x = ( OP ) cos ∠POA = (R − r )sin (p − f − q ) cos  − f  , 2  p  y = (R − r )sin (p − f − q ) sin  − f  . 2  Now, p  x = (R − r )sin (p − f − q ) cos  − f  2  (R − r )sin (f + q ) sin f = (R − r ) ( cosq − cos ( 2f + q ) ) 2 (R − r ) = (R − r )cosq − ( cosq + cos ( 2f + q ) ) 2 = (R − r )cosq − (R − r ) ( cos (q + f ) cos f ) .

=

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Vectors and Parametric Curves • 61

Also, cos (q + f ) =− cos (p − q − f ) =−

r r =− OO′ R−r

and  ( R − r )q  cos f = cos   , and so r    ( R − r )q  x= ( R − r ) cosq − ( R − r ) ( cos (q + f ) cosf ) = ( R − r ) cosq + r cos  , r   as required. The identity for y is proved similarly. Particular examples appear in Figs. 1.6.10 and 1.6.11.

FIGURE 1.6.10  Hypocycloid with= R 5,= r 1.

FIGURE 1.6.11  Hypocycloid with= R 3,= r 2.

Given a curve Γ how can we find its length? The idea, as seen in Fig. 1.6.12 is to consider the projections dx, dy at each point.

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62 • Multivariable and Vector Calculus 2/E

FIGURE 1.6.12  Length of a curve.

The length of the vector

 dx  dr =   dy

is

( dx )2 + ( dy ) . 2

dr = Hence the length of Γ is given by

∫



Γ

dr =

∫ ( dx ) + ( dy ) 2

Γ

2

. (1.13)

Similarly, suppose that Γ is a simple closed curve in  2 . How do we find the (oriented) area of the region it encloses? The idea, as seen in Fig. 1.6.13, borrowed from finding areas polygons, is to split the region into triangles, each area  x x + dx  1  x dx  1 1 = det  det  = ( xdy − ydx ) ,   2  y y + dy 2  y dy 2 and to sum over the closed curve, obtaining a total oriented area of  x dx  1 1 det =  y dy  ∫ ( xdy − ydx ) . (1.14) ∫ 2 Γ   2 Γ

Hence

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∫

Γ

denotes integration around the closed curve.

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Vectors and Parametric Curves • 63

FIGURE 1.6.13  Area enclosed by a simple closed curve.

EXAMPLE 1.6.5 Let ( A, B) ∈  2 , A > 0, B > 0 . Find a parameterization of the ellipse   x2 y2 Γ : ( x, y ) ∈  2 : 2 + 2 = 1 , A B   in Fig. 1.6.14. Furthermore, find an integral expression for the perimeter of this ellipse and find the area it encloses.

FIGURE 1.6.14  Example 1.6.4.

Solution: Consider the parameterization Γ : [ 0,2p ] →  2 , with  x   A cos t   y =  B sin t  .     This is a parameterization of the ellipse, for x2 y2 A2 cos2 t B2 sin 2 t + = + = cos2 t + sin 2 t= 1 . A2 B2 A2 B2

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64 • Multivariable and Vector Calculus 2/E Notice that this parameterization goes around once the ellipse is counterclockwise. The perimeter of the ellipse is given by

∫

Γ

 dr =

∫

2p

0

A2 sin 2 t + B2 cos2 t dt

The above integral is an elliptic integral, and we do not have a closed form for it (in terms of the elementary functions studied in Calculus I). We will have better luck with the area of the ellipse, which is given by 1 1 = ( xdy − ydx ) ∫ Γ ( A cos t d ( B sin t ) − B sin t d ( A cos t ) )  ∫ Γ 2 2 1 2p = ( AB cos2 t + AB sin2 t ) dt 2 ∫0 1 2p = ∫ AB dt 2 0 = p AB. EXAMPLE 1.6.6 Find a parametric representation for the astroid Γ : {( x, y ) ∈  2 : x2 / 3 + y2 / 3 = 1} , in Fig. 1.6.15.

FIGURE 1.6.15  Example 1.6.5.

Find the perimeter of the astroid and the area it encloses. Solution: Take  x  cos3 t   y =  3     sin t 

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Vectors and Parametric Curves • 65

with t ∈ [ 0;2p ] . Then x2 / 3 + y2 / 3 = cos2 t + sin 2 t = 1 The perimeter of the astroid is = ∫ dr Γ

∫

2p

9 cos4 t sin 2 t + 9 sin 4 t cos2 t dt

0 2p

= ∫ 3 sin t cos t dt 0

=

3 2p sin 2 t dt 2 ∫0 p /2

= 6∫

0

sin 2 t dt

= 6. The area of the astroid is given by

(

1 1 = ( xdy − ydx ) ∫ cos3 t d ( sin3 t ) − sin3 t d ( cos3 t )  ∫ Γ 2 2

1 2p ( 3 cos4 t sin2 t + 3 sin 4 t cos2 t ) dt 2 ∫0

=

=

2 3 2p sin t cos t ) dt ( ∫ 2 0

=

2 3 2p t sin 2 dt ( ) 8 ∫0

3 2p (1 − cos 4 t ) dt 16 ∫0

=

=

MVC.CH01_2PP-Part2.indd 65

)

3p . 8

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66 • Multivariable and Vector Calculus 2/E We can use MapleTM to calculate the above integrals. For example, if ( x, y) = ( cos3 t,sin3 t ) , to compute the arc length we use the path integral command and to compute the area, we use the line integral command with −y / 2 the vector field  .  x/2 

We include here for convenience, some MapleTM commands to compute various arc lengths and areas. EXAMPLE 1.6.7 To obtain the arc length of the path in Fig. 1.6.16, we type

FIGURE 1.6.16  Line path.

To obtain the arc length of the path in Fig. 1.6.17, we type

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Vectors and Parametric Curves • 67

FIGURE 1.6.17  Arc of the circle of radius 3, angle

p p ≤q ≤ . 6 5

To obtain the area inside the curve in Fig. 1.6.18, we type

FIGURE 1.6.18  Area inside the curve x =1+ (1+ cos t)(cos t), y =2 + (1+ cos t)(sin t) .

EXERCISES 1.6 1.6.1

A curve is represented parametrically by x(t= ) t 3 − 2 t , y(t= ) t3 + 2t . Find its Cartesian equation.

1.6.2

Give an implicit Cartesian equation for the parametric representation = x(t)

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t2 t3 = , y . 1 + t5 1 + t5

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68 • Multivariable and Vector Calculus 2/E 1.6.3

Let a, b, c, and d be strictly positive real constants. In each case give an implicit Cartesian equation for the parametric representation and describe the trace of the parametric curve. 1. x =at + b, y =ct + d 2. = x cos = t, y 0 = 3. x a= cosh t, y b sinh t  p p 4. = x a sec = t, y b tan t, t ∈  − ;   2 2

1.6.4

Parameterize the curve y = log cos x for 0 ≤ x ≤

p . Then find its arc 3

length. 1.6.5

Describe the trace of the parametric curve  x   sin t  =  y 2 sin t + 1 , t ∈ [ 0;4p ].    

1.6.6

Consider the plane curve defined implicitly by

x+ y= 1 . Give a

suitable parameterization of this curve, and find its length. The graph of the curve appears in Fig. 1.6.19.

FIGURE 1.6.19 Exercise 1.6.6.

1.6.7

Consider the graph given parametrically by x(t) = t 3 + 1, y(t) = 1 − t2 . Find the area under the graph, over the x-axis, and between the lines x = 1 and x = 2.

1.6.8

Find the arc length of the curve given parametrically by x(t) = 3 t 2 , y(t) = 2 t 3 for 0 ≤ x ≤ 1.

1.6.9

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Let C be the curve in  2 defined by

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Vectors and Parametric Curves • 69

= x(t)

t2 = , y(t) 2

( 2 t + 1 )3 / 2 3

 1 1 , t ∈ − ; +  .  2 2

Find the length of this curve.

1.6.10 Find the area enclosed by the curve 3 = x(t) sin= t, y(t) ( cos t ) (1 + sin 2 t ) . The curve appears in Fig. 1.6.20.

FIGURE 1.6.20  Exercise 1.6.10.

1.6.11 Let C be the curve in  2 defined by x(= t)

3t 3t2 = , y ( t ) , t ∈  \ {−1} , 1 + t3 1 + t3

which you may see in Fig. 1.6.21. Find the area enclosed by the loop of this curve.

FIGURE 1.6.21  Exercise 1.6.11.

1.6.12 Let P be a point at a distance d from the center of a circle of radius r . The curve traced out P as the circle rolls along a straight line, without slipping, is called a cycloid, as shown in Fig. 1.6.22. Find a parameterization of the cycloid.

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70 • Multivariable and Vector Calculus 2/E

FIGURE 1.6.22  Exercise 1.6.12, Cycloid.

1.6.13 Find the arc length of the arc of the cycloid x= r ( t − sin t ) , y = r (1 − cos t ) , t ∈ [ 0;2p ] . 1.6.14 Find the length of the parametric curve given by = x e t= cos t, y et sin t, t ∈ [ 0; p ] . 1.6.15 A shell strikes an airplane flying at a height h above the ground. It is known that the shell was fired from a gun on the ground with a muzzle velocity of magnitude V, but the position of the gun and its angle of elevation are both unknown. Deduce that the gun is situated within a circle whose center lies directly below the airplane and whose radius is V V 2 − 2 gh g

.

1.6.16 The parabola y2 = −4 px rolls without slipping around the parabola y2 = 4 px Find the equation of the locus of the vertex of the rolling parabola.

1.7 VECTORS IN SPACE Definition 1.7.1  The three-dimensional Cartesian Space is defined and denoted by  3 ={r =( x, y, z ) : x ∈ , y ∈ , z ∈ } In Fig. 1.7.1, we have pictured the point (2, 1, 3).

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Vectors and Parametric Curves • 71

FIGURE 1.7.1  A point in space.

   Definition 1.7.2  A system of unit vectors i , j , k is right-handed if the   shortest-route rotation which brings i to coincide with j is performed in a counter-clockwise manner. It is left-handed if the rotation is done in a clockwise manner. Having oriented the z-axis upwards, we have a choice for the orientation of the x and y-axis. Figs. 1.7.2 and 1.7.3 show the right-handed system and the right-hand, respectively. While Fig. 1.7.4 shows the left-handed system. Here, we adopt the convention right-handed coordinate system, as in Fig. 1.7.2.

FIGURE 1.7.2  Right-handed system.

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72 • Multivariable and Vector Calculus 2/E

FIGURE 1.7.3 Right-hand.

FIGURE 1.7.4  Left-handed system.

Let us explain. In analogy to  2 we put 1  0  0           i = 0  , j = 1  , k = 0  , 0  0  1  and observe that    r = ( x, y, z ) = xi + y j + zk . Most of what we did in  2 transfers to  3 without major complications. NOTE





The i is the unit vector in the x-axis direction, j is the unit vector in the y-axis direc-



tion, and k is the unit vector in the z-axis direction.

 a1      3 The rectangular coordinate form of a vector a in  is a =  a2  , then the  a3       unit vector form of vector a is a = a1 i + a2 j + a3 k . We also have

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Vectors and Parametric Curves • 73

  i= = j

       k= 1 and = i  j j=  k k  i = 0.

To calculate the magnitude of vectors in  3 , we need a distance formula for points in Euclidean space. Theorem 1.7.1  a1  b    1      The distance d a, b between vectors a =  a2  and b =  b2  in  3 is  a3   b3      2 2 2 d a, b = a − b = ( a1 − b1 ) + ( a2 − b2 ) + ( a3 − b3 )

( )

( )

  Definition 1.7.3  The dot (inner) product of two vectors a and b in  3 is   a  b = a1 b1 + a2 b2 + a3 b3 .  a1     The norm of a vector a =  a2  in  3 is  a3     2 2 2 a = a  a = ( a1 ) + ( a2 ) + ( a3 ) .     Just as in  2 , the dot product satisfies a b = a b cosq , where q ∈ [ 0; p ] is the convex angle between the two vectors. NOTE









Two vectors a and b in  3 are orthogonal (perpendicular) if a  b = 0 .

Theorem 1.7.1 (Pythagorean Theorem in  3 )    2 2 2 If a and b are orthogonal vectors in  3 , then a + b = a + b . Proof:     Since a and b are orthogonal, a  b = 0

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74 • Multivariable and Vector Calculus 2/E  2     a+b =a+b  a+b         = a  a + a  b + b a + b b 2 2 = a + b

(

)(

)

The Cauchy–Schwarz–Bunyakovsky Inequality takes the form     1/ 2 1/ 2 a b ≤ a b ⇒ a1 b1 + a2 b2 + a3 b3 ≤ ( a12 + a2 2 + a3 2 ) ( b12 + b2 2 + b3 2 ) , equality holding if and only if the vectors are parallel. EXAMPLE 1.7.1 8    Find the norm of the vector a =  5  in  3 .  −2  Solution:  2 2 2 a= ( 8 ) + ( 5 ) + ( −2 )=

64 + 25 + = 4

93 .

EXAMPLE 1.7.2  2   3      1  Normalize the vector a = and generate a unit vector.  2   1 −   4  Solution:  2   3    1   a  = =  2   1 −   4 

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2

2

2

2 1  1   +  +− =  3 2  4

4 1  1   + + =   9   4   16 

109 12

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Vectors and Parametric Curves • 75

    a  =  a   − 

8  109  6    The normalized vector of a is = . u 109   3  109    The normalized vector u is a unit vector since u = 1 . EXAMPLE 1.7.3 a     Let  have the Euclidean dot product. For which values of a are a =  5   3  3

a     and b = a  orthogonal.  2  Solution:   a  b=0 a  a      2  5   a  = a + 5a + 6 = 0  3   2  0 (a + 3 )(a + 2 ) = a = −3 or −2 EXAMPLE 1.7.4 Let x, y, z positive real numbers such that x2 + 4 y2 + 9 z2 = 27 . Maximize x+ y+ z. Solution: Since x, y, z are positive, x + y + z = x + y + z . By Cauchy’s Inequality,

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76 • Multivariable and Vector Calculus 2/E

1/ 2

1/ 2  1 1 1 1 x + y + z = x + 2 y   + 3 z   ≤ ( x 2 + 4 y2 + 9 z 2 )  1 + +  4 9 2 3 

=

7 7 3 27   = . 2 6

Equality occurs if and only if x  1  18 3 l l l2 l2     2 2 1 / 2 , , y l x l y z l . = ⇒ = = = ⇒ + + =27 ⇒ l =±     4 9 4 9 7  3 z  1 / 3  Therefore for a maximum, we take = x

18 3 9 3 2 3 = ,y = ,z . 7 14 7

  Definition 1.7.4  Let a be a point in  3 and let v ≠ 0 a vector in  3 . The  parametric line passing through a in the direction of v is the set  r ∈ 3 : r = a + tv .

{

}

EXAMPLE 1.7.5 1  −2      Find the parametric equation of the line passing through  2  and  −1  .  0 3     Solution: 1 − ( −2 )  3    The line follows the direction 2 − ( −1 )  =3  .  3 − 0  3   x   The desired equation is = y  z   NOTE

 1  3       2  + t 3  .  3  3     

Given two lines in space, one of the following three situations might arise: (i) the lines intersect at a point, (ii) the lines are parallel, and (iii) the lines are skew (non-parallel, one over the other, without intersecting, lying on different planes). See Fig. 1.7.5.

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Vectors and Parametric Curves • 77

FIGURE 1.7.5   1 ||  2 .  1 and  3 are skew.

    Consider now two non-zero vectors a and b in  3 . If a || b , then the set    sa + t b : s ∈ , t ∈ = l a : l ∈  ,

{

} {

}

  which is a line through the origin. Suppose now that a and b are not parallel. We saw in the preceding sections that if the vectors were on the plane, they would span the whole plane  2 . In the case at hand, the vectors are in space, they still span a plane, passing through the origin. Thus     sa + t b : s ∈ , t ∈ ,a || b

{

}

is a plane passing through the origin. We will say, abusing language that two vectors are coplanar if there exist bi-point representatives of the vector that lie on the same plane. We will say, again abusing language, that a vector is parallel to a specific plane or that it lies on a specific plane if there exists a bi-point representative of the vector that lies on the particular plane. All the above gives the following result.   Theorem 1.7.2: Let v , w in  3 be non-parallel vectors. Then every vec    tor u of the form = u a v + bw , where a, b are arbitrary scalars, is coplanar      with both v and w . Conversely, any vector t coplanar with both v and w    can be uniquely expressed in the form = t pv + qw . See Fig. 1.7.5.

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78 • Multivariable and Vector Calculus 2/E

FIGURE 1.7.6  Theorem 1.7.1.

 From the above theorem, if a vector a is not a linear combination of two   other vectors b , c , then linear combinations of these three vectors may lie   outside the plane containing b , c . This prompts the following theorem.    Theorem 1.7.3:  Three vectors a , b , c in  3 said to be linearly independent if     a a + b b + g c =0 ⇒ a = b =g =0 . Any vector in  3 can be written as a linear combination of three linearly independent vectors in  3 . A plane is determined by three non-collinear points. Suppose that a, b, and c  x   are non-collinear points on the same plane and that r =  y  is another arbi z     trary point on this plane. Since a, b, and c are non-collinear, ab and ac , which  are coplanar, are non-parallel. Since ax also lies on the plane, we have by theorem 1, that there exist real numbers p, q with    = ar pab + qac . By Chasles’s rule,       r =a + p b − a + q c − a ,

(

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) (

)

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Vectors and Parametric Curves • 79

is the equation of a plane containing the three non-collinear points a, b, and c,   where a , b , and c are the position vectors of these points. Thus we have the following theorem.   Theorem 1.7.4: Let u and v be linearly independent vectors. The parametric equation of a plane containing the point a, and parallel to the vectors   u and v is given by     r − a = pu + qv . Component wise this takes the for x − a1 = pu1 + qv1 , y − a2 = pu2 + qv2 , z − a3 = pu3 + qv3 . Multiplying the first equation by u2 v3 − u3 v2 , the second by u3 v1 − u1 v3 , and the third by u1 v2 − u2 v1 , we obtain,

( u2 v3 − u3 v2 ) ( x − a1 )= ( u2 v3 − u3 v2 ) ( pu1 + qv1 ) , ( u3 v1 − u1 v3 ) ( y − a2 )= ( u3 v1 − u1 v3 ) ( pu2 + qv2 ) , ( u1 v2 − u2 v1 ) ( z − a3 )= ( u1 v2 − u2 v1 ) ( pu3 + qv3 ) . Adding gives, 0. ( u2 v3 − u3 v2 ) ( x − a1 ) + ( u3 v1 − u1 v3 ) ( y − a2 ) + ( u1 v2 − u2 v1 ) ( z − a3 ) = Put a= u2 v3 − u3 v2 , b = u3 v1 − u1 v3 , c = u1 v2 − u2 v1 , and d = a1 ( u2 v3 − u3 v2 ) + a2 ( u3 v1 − u1 v3 ) + a3 ( u1 v2 − u2 v1 ) .   Since v is linearly independent u , not all of a,b,c are zero. This gives the following theorem. Theorem 1.7.5:  The equation of the plane in space can be written in the form ax + by + cz = d,

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80 • Multivariable and Vector Calculus 2/E

which is the Cartesian equation of the plane. Here a2 + b2 + c2 ≠ 0 , that is, at  a    least one of the coefficients is non-zero. Moreover, the vector n =  b is nor c  mal to the plane with the Cartesian equation ax + by + cz = d . See Fig. 1.7.7.

FIGURE 1.7.7  Theorem 1.7.4.

Proof: We have observe  already  proved the first statement. For the second  statement,  that if u and v are non-parallel vectors and r − a = pu + qv is the equation   of the plane containing the point a and parallel to the vectors u and v , then       if n is simultaneously perpendicular to u and v then r − a n = 0 for     un= 0= v n . Now, since at least one of a, b, c is non-zero, we may assume

(

)

a ≠ 0 . The argument is similar if one of the other letters is non-zero and a = 0 . In this case, we can see that d b c x = − y− z. a a a Put y = s and z = t . Then d   b  c  x− a − a  − a        y  s 1  + t  0  , =  z   0   1             

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Vectors and Parametric Curves • 81

is a parametric equation for the plane. We have c  b a  −  + b (1 ) + c ( 0= ) 0, a  −  + b ( 0 ) + c (1=) 0, a a      a   and so  b is simultaneously perpendicular to  c 

 b  c − a  − a       1  and  0  , proving the  0   1         

second statement. EXAMPLE 1.7.5 1   The equation of the plane passing through the point  −1  and normal to 2    −3    ­vector  2  is  4  −3 ( x − 1 ) + 2 ( y + 1 ) + 4 ( z − 2 ) = 0 ⇒ − 3 x + 2 y + 4 z = 3. EXAMPLE 1.7.6 Find both the parametric equation and the Cartesian equation of the plane 1 1  0       parallel to the vectors 1 and 1  , and passing through the point  −1  . 2 1 0    Solution: The desired parametric equation is  x  1 1         y + 1 = s 1 + t 1  .  z − 2 1 0   

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82 • Multivariable and Vector Calculus 2/E This gives s = z − 2,

t = y+1− s = y+1− z+ 2 = y− z+ 3

and x = s + t = z − 2 + y − z + 3 = y + 1. Hence the Cartesian equation is x − y = 1.   Definition 1.7.5  If n is perpendicular to plane Π 1 and n′ is perpendicular to plane Π 2 , the angle between the two planes is the angle between the   two vectors n and n′ . EXAMPLE 1.7.7 1. Draw the intersection of the plane z= 1 − x with the first octant. 2. Draw the intersection of the plane z= 1 − y with the first octant. 3. Find the angle between the planes z= 1 − x and z= 1 − y . 4. Draw the solid S which results from the intersection of the planes z= 1 − x and z= 1 − y with the first octant. 5. Find the volume of the solid S. Solution: 1. This appears in Fig. 1.7.8.

FIGURE 1.7.8  The plane z = 1− x .

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Vectors and Parametric Curves • 83

2. This appears in Fig. 1.7.9.

FIGURE 1.7.9  The plane z = 1− y .

1  0      3. The vector 0  is normal to the plane x + z =, 1 and the vector 1  is nor1  1  mal to the plane y + z =. 1 If q is the angle between these two ­vectors, then cosq =

1 ⋅ 0 + 0 ⋅1 + 1 ⋅1

1 +1 ⋅ 1 +1 4. This appears in Fig. 1.7.10. 2

2

2

2

⇒ cosq =

p 1 ⇒q = . 2 3

FIGURE 1.7.10  Solid bounded by the planes z = 1− x and z = 1− y in the first octant.

5. The resulting solid is a pyramid with square base of area A = 1 ⋅ 1 = 1 . Ah Recall that the volume of a pyramid is given by the formula V = , 3 where A is area of the base of the pyramid and h is its height. Now, the 1 height of this pyramid is clearly 1, and hence the volume required is . 3

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84 • Multivariable and Vector Calculus 2/E

EXERCISES 1.7 1.7.1 1.7.2

1.7.3

1.7.4

1.7.5

1.7.6

1.7.7

        Vectors a , b satisfy a= 13, b= 19, a + b= 24. Find a − b . 1   Find the equation of the line passing through  2  in the direction 3    −2    of  −1  .  0  6  1        Find the value of a when the vectors a = 1  and b =  4   5   k  orthogonal.  −5  2       Find the Euclidean distance between a =  4  and b =  −1 .  3   6  1   Find the equation of plane containing the point  1  and perpen1   dicular to the line x =+ 1 t, y = −2 t, z =− 1 t. 1   Find the equation of plane containing the point  −1  and contain −1    ing the line = x 2= y 3z . (Putnam Exam 1984) Let A be a solid a × b × c rectangular brick in three dimensions, where a > 0, b > 0, c > 0 . Let B be the set of all points which are at distance at most 1 from some point of A (in particular, A ⊆ B ). Express the volume of B as a polynomial in a, b, c .

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Vectors and Parametric Curves • 85

1.7.8

1.7.9

       It is known that= a 3,= b 4,= c 5 and that a + b + c = 0 . Find      a  b + b c + c a . Find the equation of the line perpendicular to the plane 0   2 3 ax + a y + a z = 0, a ≠ 0 and passing through the point  0  . 1  

1.7.10 Find the equation of the plane perpendicular to the line 1   ax = by = cz, abc ≠ 0 and passing through the point  1  in  3 . 1   1.7.11 Find the (shortest) distance from the point (1,2,3 ) to the plane x − y + z =. 1 1.7.12 Determine whether the lines  x   L1 :  = y    z



 1  2       1  + t 1  ,    1  1 

 x 0  2        L2 :  y =  0  + t  −1 ,  z 1  1        Intersect. Find the angle between them.

1.7.13 Let a, b, c be arbitrary real numbers. Prove that

(a

2

+ b2 + c2 ) ≤ 3 ( a4 + b4 + c 4 ) . 2

1.7.14 Let a > 0, b > 0, c > 0 be the lengths of the sides of ∆ABC. (Vertex A is opposite to the side measuring a, etc.) Recall that by Heron’s Formula, the area of this triangle is a+ b+ c S ( a, b, c ) = s ( s − a )( s − b )( s − c ) , where s = is the 2

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86 • Multivariable and Vector Calculus 2/E S ( a, b, c ) is a + b2 + c2 maximized when ∆ABC is equivalent, and find this maximum. semiperimeter of the triangle. Prove that f ( a, b, c ) =

2

1.7.15 Find the Cartesian equation of the plane passing through 1 0 0       0  ,  1  , and  0  . Draw this plane and its intersection with the 0 0 1      first octant. Find the volume of the tetrahedron with vertices at 0 1 0 0        0  ,  0  ,  1  , and  0  . 0 0 0 1       1.7.16 Prove that there do not exist three unit vectors in  3 such that the 2p angle between any two of them be > . 3    1.7.17 Let r − a n = 0 be a plane passing through the point a and per pendicular to vector n . If b is not a point on the plane, then the distance from b to the plane is    a − b n .  n

(

)

(

)

1.7.18 (Putnam Exam 1980) Let  S be solid in a three-dimensional space consisting of all points AC′ satisfying the following system of six conditions:



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x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 11, 2 x + 4 y + 3 z ≤ 36, 2 x + 3 z ≤ 24. Determine the number of vertices and the number of edges of S.

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Vectors and Parametric Curves • 87

1.8 CROSS PRODUCT We now define the standard cross product in  3 as a product satisfying the following properties.   Definition 1.8.1  Let x, y,z be vectors in  3 , and let a ∈  be a scalar. The cross product ( × ) is a closed binary operation satisfying 1. Anti-commutativity:

    x × y =− y × x

(

)

2. Bilinearity:               x + z × y = x × y + z × y , and x × z + y = x × z + x × y

(

)

(

) (

)

(

) (

) (

)

3. Scalar homogeneity: 







 

(a x ) × y = x × (a y ) = a ( x × y ) 4. Cross product of a vector with itself is always zero    x×x = 0 5. Right-hand Rule:          i ×=j k, j × = k i, k ×= i j It follows that the cross product is an operation that, given two non-parallel vectors on a plane, allows us to “get out” of that plane. NOTE

The cross product of vectors in  3 is not associative, since       i × i × j =i × k = − j

( )

but  

  



(i × i) × j = 0 × j = 0 . EXAMPLE 1.8.1  1  0      Find  0  × 1  .  −3  2 

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88 • Multivariable and Vector Calculus 2/E Solution: We have









 

 

 

 

( i − 3k ) × ( j + 2k ) = i × j + 2i × k − 3k × j − 6k × k     =k − 2j + 3i + 60    = 3i − 2j + k.

Hence  1  0   3         0  × 1  =  −2   −3  2   1  Operating as in Example 1.8.1, we obtain x  y    1   1 Theorem 1.8.1: Let x =  x2  and y =  y2  be vectors in  3 . Then,  x3   y3  ( x y − x3 y2 )        2 3  x × y = ( x2 y3 − x3 y2 ) i + ( x3 y1 − x1 y3 ) j + ( x1 y2 − x2 y1 ) k =  ( x3 y1 − x1 y3 )  or in  ( x1 y2 − x2 y1 )    determinant notation  x2   y2    x1 x × y = −  y1   x1  y1 

x3   y3  x3   y3   x2  y2 

Proof:

       Since i × i = j × j = k × k = 0 , we only worry about the mixed products, obtaining,

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Vectors and Parametric Curves • 89

  x × y=













( x i + x j + x k)×(y i + y j + y k) 1

2

3

1

2

3

            = x1 y2 i × j + x1 y3 i × k + x2 y1 j × i + x2 y3 j × k + x3 y1 k × i + x3 y2 k × j       = ( x1 y2 − y1 x2 ) i × j + ( x2 y3 − x3 y2 ) j × k + ( x3 y1 − x1 y3 ) k × i    = ( x1 y2 − y1 x2 ) k + ( x2 y3 − x3 y2 ) i + ( x3 y1 − x1 y3 ) j ,

proving the theorem. EXAMPLE 1.8.2 1 5         Let a =  2  and b = 0  , find a × b .  −3  1  Solution:

 We form the 2 × 3 matrix whose first row contains the components of a and  whose second row contains the component of b , that is,  a1 b  1

a2 b2

a3  1 2 −3  ⇒ b3   5 0 1 

−3   1   2  −3    = −16  .  1     −10  2  0  Using the cross product, we may obtain a third vector simultaneously perpendicular to two other vectors in space.  2   0  1   a × b = −  5   1  5 

      Theorem 1.8.2:  x ⊥ x × y and y ⊥ x × y , that is, the cross product of

(

)

(

)

two vectors is simultaneously perpendicular to both original vectors.

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90 • Multivariable and Vector Calculus 2/E Proof: We will only check the first assertion, the second verification is analogous.          = x  ( x × y ) x1 i + x2 j + x3 k  ( x2 y3 − x3 y2 ) i + ( x3 y1 − x1 y3 ) j + ( x1 y2 − x2 y1 ) k

(

) (

)

= x1 x2 y3 − x1 x3 y2 + x2 x3 y1 − x2 x1 y3 + x3 x1 y2 − x3 x2 y1 =0 completing the proof. Theorem 1.8.3:    a × b ×= c

(

)

 

  

( a c ) b − ( a  b ) c .

Proof:    a × b ×= c

   − b3 c2 ) i + ( b3 c1 − b1 c3 ) j + ( b1 c2 − b2 c1 ) k     = a1 ( b3 c1 − b1 c3 ) k − a1 ( b1 c2 − b2 c1 ) j − a2 ( b2 c3 − b3 c2 ) k + a2 ( b1 c2 − b2 c1 ) i   + a3 ( b2 c3 − b3 c2 ) j − a3 ( b3 c1 − b1 c3 ) i       = ( a1 c1 + a2 c2 + a3 c3 ) ( b1 i + b2 j + b3 i ) + ( − a1 b1 − a2 b2 − a3 b3 ) ( c1 i + c2 j + c3 i )      = ( a c ) b − a  b c ,

(







) ( a i + a j + a k ) × (( b c 1

2

3

)

2 3

( )

completing the proof. Theorem 1.8.4 (Jacobi’s Identity):            a × b × c + b × ( c × a )+c × a × b = 0

(

)

(

)

Proof: From Theorem 1.8.1, we have    a × b ×= c    b × ( c ×= a)    c × a ×= b

(

(

)

 

  





( a c ) b − ( a  b ) c ,

( ba ) c − ( bc ) a ,   ) ( cb ) a − ( ca ) b ,

and adding yields the result.

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Vectors and Parametric Curves • 91

   Theorem 1.8.5:  Let x, y ∈ [ 0;p ] be the convex angle between two vec  tors x and y . Then

( )

       x×y = x y sin x, y .

( )

See Fig. 1.8.1.

FIGURE 1.8.1  Theorem 1.8.4.

Proof: We have   2 2 2 x × y = ( x2 y3 − x3 y2 ) + ( x3 y1 − x1 y3 ) + ( x1 y2 − x2 y1 ) = y2 y32 − 2 x2 y3 x3 y2 + z2 y22 + z2 y12 − 2 x3 y1 x1 y3 + x2 y32 + x2 y22 − 2 x1 y2 x2 y1 + y2 y12 =

(x

 x  = x  = x =

+ y2 + z2 )( y12 + y22 + y32 ) − ( x1 y1 + x2 y2 + x3 y3 )   2 2  2 y − xy 2 2   2  2  y − x y cos2 x, y   2  2  y sin 2 x, y ,

2

2

( )

( )

( )

where the theorem follows.

  Theorem 4 has the following geometric significance: x × y is the area of the parallelogram formed when the tails of the vectors are joined. See Fig. 1.8.2.

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92 • Multivariable and Vector Calculus 2/E

FIGURE 1.8.2  Area of a parallelogram.

The following corollaries easily follow Theorem 1.8.5.      Corollary 1.8.1:  Two non-zero vectors x , y satisfy x × y = 0 if and only if they are parallel. Corollary 1.8.2 (Lagrange’s Identity):    x= ×y

 x

2

2   2 y − xy .

( )

The following result mixes the dot and the cross product.    Theorem 1.8.6: Let a,b,c , be linearly independent vectors in  3 . The    signed volume of the parallelepiped spanned by them is a × b c .

(

)

See Fig. 1.8.3.

FIGURE 1.8.3  Theorem 1.8.5.

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Vectors and Parametric Curves • 93

Proof: The area of the base of the parallelepiped is the area ofthe parallelogram  determined by the vectors a and b , which has area a × b . The altitude of     the parallelepiped is c cosq where q is the angle between c and a × b . The volume of the parallelepiped is thus       a × b c cosq = a × b c ,

(

)

proving the theorem. NOTE

Since we may have used any of the faces of the parallelepiped, it follows that          a × b c =b × c a = ( c × a ) b .

(

) (

)

In particular, it is possible to “exchange” the cross and dot products:       a  b × c = a × b c

(

) (

)

EXAMPLE 1.8.3

 2  2  2   Let x ∈  3 , and x = 1 . Find x × i + x × j + x × k . Solution: By Lagrange’s Identity,  2  x × i =x

2

2  2  2 i − ( x  i ) =− 1 ( x i ) ,

 2  x × j =x

2

2  2  2 j − ( x  j ) =− 1 ( x j ) ,

 2  x × k =x

2

2   k − x k

 2  2   and since ( x  i ) + ( x  j ) + x k

( )

MVC.CH01_2PP-Part3.indd 93

( )

2

2

  2 =− 1 x k ,

( )

2 =x = 1 , the desired sum equals 3 − 1 = 2.

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94 • Multivariable and Vector Calculus 2/E EXAMPLE 1.8.4 Consider the rectangular parallelepiped ABCDD′C′B′A′ , see Fig. 1.8.4 with vertices A ( 2,0,0 ) , B ( 2,3,0 ), C ( 0,3,0 ) , D ( 0,0,0 ) , D′ ( 0,0,1 ) , C′ ( 0,3,1 ) , B′ ( 2,3,1 ) , A′ ( 2,0,1 ) . Let M be the midpoint of the line segment joining the vertices B and C. 1. Find the Cartesian equation of the plane containing the points A, D′, and M. 2. Find the area of ∆AD′M .  3. Find the parametric equation of the line AC′ . 4. Suppose that a line through M is drawn cutting the line segment [ AC′] in   N and the line DD′ in P. Find the parametric equation of MP .

FIGURE 1.8.4  Example 1.8.2.

Solution: 1. Form the following vectors and find their cross product:  −2   −1  −3            ′ AD′ =  0  , AM =  3  ⇒ AD × AM =  −1  .  1   0   −6  The equation of plane is thus  x − 2   −3      y − 0    −1  =0 ⇒ 3 ( x − 2 ) + 1 ( y ) + 6 z =0 ⇒ 3 x + y + 6 z =6.  z − 0   −6 

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Vectors and Parametric Curves • 95

2. The area of the triangle is   AD′ × AM 1 2 2 2 3 + 1 + 6= = 2 2

46 . 2

 −2      3. We have AC′ =  3  , and hence the line AC′ has parametric equation  1   x   2   −2         y  =  0  + t  3  ⇒ x = 2 − 2 t , y = 3 t , z = t.  z 0  1        0   4. Since P in on the z-axis, P =  0  for some real number z′ > 0 . The para z′     metric equation of the line MP is thus  x   1   −1         y  =  3  + s  −3  ⇒ x = 1 − s, y = 3 − 3 s, z = sz′.  z   0   z′          Since N is on both MP and AC′ , we must have 2 − 2 t =1 − s, 3t =3 − 3 s, t =sz′. Solving the first two equations gives= s

1 2 = ,t . Putting this into the third 3 3

0   equation, we deduce z′ = 2 . Thus P =  0  and the desired equation is 2    x 1      y  =  3  + s [ 2 ] ⇒ x = 1 − s, y = 3 − 3 s, z = 2 s.  z 0    

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96 • Multivariable and Vector Calculus 2/E

EXERCISES 1.8 1.8.1

Prove that









 

( a − b ) × ( a + b ) = 2a × b . 1.8.2

1.8.3

1.8.4

1.8.5 1.8.6 1.8.7 1.8.8

   Prove that x × x = 0 follows from the anti-commutativity of the cross product. 0   Find the area of the triangle whose vertices are at A =  0  , 1   0 1     B =  1  , and C =  0  . 0 0     1 1      Find a vector simultaneously perpendicular to 1 and 1  , and 1 0  having norm 3. Prove or disprove! The cross product is associative.     Expand the product a − b × a + b .         If b − a and c − a are parallel and it is known that c × a = i − j and       a × b = j + k , find b × c .

(

) (

)

Redo Example 1.7.6 (Section 1.7), that is, find the Cartesian equa1 1      tion of the plane parallel to the vectors 1 and 1  , and passing 1 0  through the point ( 0, −1,2 ) , by finding a normal to the plane.

1.8.9

Find the equation of the plane passing through the points ( a,0, a ) ,

( a,0, a ) , ( − a,1,0 ) , and ( 0,1,2a ) in  3 . 1.8.10 Let a ∈  . Find a vector of unit length simultaneously perpendicu0 1        lar to v =  − a  and w =  a  .  a  0 

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Vectors and Parametric Curves • 97

   1.8.11 (Jacobi’s Identity) Let a , b , c be vectors in  3 . Prove that          a× b×c + b× c×a + c× a×b = 0.

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)

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  2  2   2 1.8.12 Let x ∈  3 , x = 1 . Find x × i + x × j + x × k .   1.8.13 The vectors a,b are constant vectors. Solve the equation       a × x × b =b × x × a .

(

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(

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  1.8.14 Let a and b be vectors in  3 and k be a scalar, prove that       k a × b = ka × b = a × k b .    1.8.15 Prove that the position vectors a,b, and c of  3 all lie in a plane if    and only if a  b × c = 0.    1.8.16 The vectors a,b,c are constant vectors. Solve the system of

(

)

(

equations

)

        2x + y = × a b, 3y + x = × a c.

    1.8.17 Let a,b,c,d , be vectors in  3 . Prove the following vector identity,            a × b = c×d a c b d − a d b c .

(

)(

) ( )( ) ( )( )

    1.8.18 Let a,b,c,d be vectors in  3 . Prove that  

 

 

 

 

 

0. ( b × c )( a × d ) + ( c × a )( b × d ) + ( a × b )( c × d ) = 1.8.19 Consider the plane Π passing through the points A ( 6,0,0 ) ,B ( 0,4,0 ) , and C ( 0,0,3 ) , as shown in Fig. 1.8.5. The plane Π intersects a 3 × 3 × 3 cube, one of whose vertices is at the origin and that has three of its edges on the coordinate axes, as in the figure. This intersection forms a pentagon CPQRS.   1. Find CA × CB .   2. Find CA × CB . 3. Find the parametric equation of the line LCA joining C and A, with a parameter t ∈  .

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98 • Multivariable and Vector Calculus 2/E

4. Find the parametric equation of the line LDE joining D and E, with a parameter s∈ . 5. Find the intersection point between the lines LCA and LDE . 6. Find the coordinates of the points P, Q, R, and S. 7. Find the area of the pentagon CPQRS.

FIGURE 1.8.5  Exercise 1.8.15.

1.9 MATRICES IN THREE DIMENSIONS We will briefly introduce 3 × 3 matrices. Most of the material will flow like that for 2 × 2 matrices. Definition 1.9.1 A linear transformation T :  3 →  3 is a function such that T ( a + b )= T ( a ) + T ( b ) ,

T ( l a )= l T ( a ) ,

for all points a,b in  3 and all scalars l . Such a linear transformation has a

3 × 3 matrix representation whose columns are the vectors T ( i ) , T ( j) , and

T (k) . EXAMPLE 1.9.1 Consider L :  3 →  3 , with   x1     L   x2   =   x3    

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 x1 − x2 − x3     x1 + x2 + x3  .   x3  

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Vectors and Parametric Curves • 99

1. Prove that L is a linear transformation. 2. Find the matrix corresponding to L under the standard basis. Solution: 1. Let a ∈  and let u, v be points in  3 . Then   u1 + v1    ( u1 + v1 ) − ( u2 + v2 ) − ( u3 + v3 )      L ( u + v= ) L   u2 + v2  =  ( u1 + v1 ) + ( u2 + v2 ) + ( u3 + v3 )     u3 + v3    u3 + v3      u1 − u2 − u3   v1 − v2 − v3      =  u1 + u2 + u3  +  v1 + v2 + v3      u3 v3       u1     v1       = L   u2   + L   v2     u3     v3       = L (u) + L ( v ), and also   a u1     L (a u ) = L   a u2     a u3      a ( u1 ) − a ( u2 ) − a ( u3 )    =  a ( u1 ) + a ( u2 ) + a ( u3 )    a u3   − − u u u  1 2 3    = a  u1 + u2 + u3    u3     u1     = a L   u2     u3     = a L (u), proving that L is a linear transformation.

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100 • Multivariable and Vector Calculus 2/E

1   2. We have = L 0 0  

1 0     1 , L  1  = 0 0    

 −1   0   −1        = L  0   1  , where the desired  1  , and 0 1  1       

matrix is 1 −1 −1   1 1 1  . 0 0 1  In addition, scalar multiplication, and matrix multiplication are defined for 3 × 3 matrices in a manner analogous to those operations for 2 × 2 matrices. Definition 1.9.2  Let A, B be 3 × 3 matrices. Then we define A + B=  aij + bij  ,

a A= a aij  ,

 3  AB=  ∑ aik bkj  .  k =1 

EXAMPLE 1.9.2 1 2 3  a b c     If A =  4 5 0  , and B =  a b 0  , then  6 0 0   a 0 0  1 + a  A + B = 4 + a  6 + a 3 6  3 A = 12 15 18 0

2 + a 3 + c  5+ b 0 , 0 0  9  0 , 0 

6 a 3 b c    AB = 9 a 9 b 4 c  , 6 a 6 b 6 c   a + 4 b + 6c 2a + 5b 3a   BA = 2a + 5b 3a .  a + 4b  a 2a 3 a 

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Vectors and Parametric Curves • 101

Definition 1.9.3 A scaling matrix is one of the form Sa, b, c

a 0 0   = 0 b 0  , 0 0 c 

where a > 0, b > 0, c > 0 . It is an easy exercise to prove that the product of two scaling matrices commutes. Definition 1.9.4  A rotation matrix about the z-axis by an angle q in the counterclockwise sense is cosq − sin q  Rz (q ) =  sin q cosq  0 0 A rotation matrix about the y-axis by an angle sense is cosq  Ry (q ) =  0  sin q

0  0 . 1  q in the counterclockwise

0 − sin q   1 0 . 0 cosq 

A rotation matrix about the x-axis by an angle q in the counterclockwise sense is 0 0  1   = Rx (q ) 0 cosq − sin q  . 0 sin q cosq  Easy to find counterexamples should convince the reader that the product of two rotations in space does not necessarily commute. Definition 1.9.5  A reflection matrix about x-axis is  −1 0 0    Rx =  0 1 0  .  0 0 1 

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102 • Multivariable and Vector Calculus 2/E A reflection matrix about y-axis is 1 0 0    = Ry 0 −1 0  . 0 0 1  A reflection matrix about z-axis is 1 0 0    Rz =  0 1 0  . 0 0 −1

EXERCISES 1.9

1.9.1

 −1 0 2  2 1 3     If A =  0 3 1  , and B =  −1 0 1  , find:  0 5 4   1 4 −3  1. A + B

1.9.2

2.  2A

3.  AB .

1 1 1   Let A ∈ M3×3 (  ) be given by A = 1 1 1 . Demonstrate, using 1 1 1 induction, that = A n 3 n−1 A for n ∈ , n ≥ 1.

1.9.3

 x   Consider L :  →  , L  y =  z  3

3

 x − y − z    x + y + z  . Prove that L is linear.  z 

1.9.4

Let L :  3 →  3 be a perpendicular projection of  3 onto the xy-plane. Show that L  L = L .

1.9.5

Consider the n × n matrix 1 1 1 0 1 1  A =0 0 1      0 0 0

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1  1 1 1  1 1  1  1 1 .      0  0 1 

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Vectors and Parametric Curves • 103

Describe A2 and A3 in terms of n. 1.9.6

Let x be a real number, and put

x  1 0  2  − x 1 − x  . m ( x) =  2 0 0 1    If a, b are real numbers, prove that 1. m(a)m= (b) m(a + b) . 2. m(a)m(− a) = I3 , the 3 × 3 identity matrix. 1.9.7

1.9.8

Determine the standard matrix representing the linear transforma  x1    x1 − x2      3 3 tion L :  →  which is defined as L   x2 =    x1 + x2  .   x3    x2 − x3        x1    x1      Find the matrix of the linear operator L   x2   =  3 x2  on  3   x3    5 x2      with respect to the standard basis of  3 , then use the matrix to find  −1    the image of the vector  4  . 2  

1.9.9

  x1    0      Find the matrix of the linear operator L   x2   =  x2  on  3 with   x3    0      respect to the standard basis of  3 , then use the matrix to find the

 −1    image of the vector  4  . 2   1.9.10 Find the matrix of each of the following linear transformations:

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104 • Multivariable and Vector Calculus 2/E

 x1    1. L  x2 =  x3 

[3 x1 − x2 + 5 x3 ]

 x1    2. L  x2  =  x3 

 x1 + 4 x2 − x3     x1 + 3 x2 + 2 x3  3 x1 + 2 x2 + 5 x3 

1.9.11 Determine whether the transformation L :  3 →  3 ;  x1   1      L  x2  =  x2  is linear. x  x   3  3 cosq  1.9.12 Let R(q ) =  sin q  0

− sin q cosq 0

0  0  . Describe geometrically the linear 1    transformation L :  3 →  3 given by Lx = R(q )x .

1.9.13 Find the standard matrix for the linear operator L :  3 →  3 , x    1 which maps a vector x =  x2  into its reflection through the x   3 xz-plane. 1.9.14 Find the standard matrix for the linear operator L :  3 →  3 , which rotates each vector p / 2 counterclockwise about y-axis (looking along the positive y-axis toward the origin).

1.10 DETERMINANTS IN THREE DIMENSIONS We now define the notion of determinant of a 3 × 3 matrix. Consider now the vectors

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Vectors and Parametric Curves • 105

a  b    1   1  = a = a2  , b = b2  , c  a3   b3  Since, in

 c1   a1    3  c2  , in  , and the 3 × 3 matrix A =  a2  c3   a3

b1 b2 b3

c1   c2  . c3 

In Theorem   1.8.5, the volume of the parallelepiped spanned by these vectors is a  b × c , we define the determinant of A ( det A ) to be

(

)

 a1     D a,b,c = det  a2  a3

(



c1  a1  = c2  a2 c3  a3

b1 b2 b3

)

b1 c1    b2 = c2 a  b × c . (1.15) b3 c3

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)

We now establish that the properties of the determinant of a 3 × 3 as previously defined are analogous to those of the determinant of 2 × 2 matrix defined in the preceding chapter. Theorem 1.10.1  The determinant of a 3 × 3 matrix A as defined by Equation (1.15) satisfies the following properties: 1. D is linear in each of its arguments.    2. If the parallelepiped is flat then the volume is 0, that is, if a,b,c , are lin   early dependent, then D a,b,c = 0 .   3. D i, j,k = 0 , and accords with the right-hand rule.

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Proof:       1. If D a,b,c = a  b × c , linearity of the first component follows by the dis-

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tributive law for the dot product:         D a + a ′,b,c =+ ( a a ′ ) b × c       = a  b × c + a ′ b × c       = D a,b,c + D a ′,b,c ,

(

)

( (

( ) ) ( ) ) ( )

and if l ∈  ,    D l a,b,c =

(

MVC.CH01_2PP-Part3.indd 105



 



 

c ) l ( ( a )( b ×= c )) ) ( l a )( b ×=

   l D a,b,c .

(

)

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106 • Multivariable and Vector Calculus 2/E The linearity on the second and third components can be established by using the distributive law of the cross product. For example, for the second component we have,      D a,b + b′,c = a   = a  = a

(

=







( ( b + b′ ) × c )

)

   

( b × c + b′ × c )      ( b × c ) + a( b′ × c )       D ( a,b,c ) + D ( a,b′,c ) ,

and if l ∈  ,             D a, l b,c = a  l b ×= c l a  b × c= l D a,b,c .

(

(( ) ) ( (

)

))

(

)

   2. If a,b,c are linearly dependent, then they lie on the same plane and the    parallelepiped spanned by them is flat, hence D a,b,c = 0 .             3. Since j × k = i , and i × i = 1 , D i, j,k = i  j × k = i × i = 1 .

(

) (

)

(

)

Observe that,



 a1 b1 c1       det  a2 b2 = c2  a  b × c (1.16)  a3 b3 c3      = a  ( b2 c3 − b3 c2 ) i + ( b3 c1 − b1 c3 ) j + ( b1 c2 − b2 c1 ) k (1.17)



= a1 ( b2 c3 − b3 c2 ) + a2 ( b3 c1 − b1 c3 ) + a3 ( b1 c2 − b2 c1 ) (1.18)



 b2 = a1 det   b3



(

)

(

)

c2   b1 + a2 det   c3   b3

c1   b1 + a3 det   c3   b2

c1  , (1.19) c2 

which reduces the computation of 3 × 3 determinants to 2 × 2 determinants. EXAMPLE 1.10.1 1 2 3    Find the det  4 5 6  . 7 8 9 

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Vectors and Parametric Curves • 107

Solution: Using Equation (1.19), we have 5 6 2 3  2 3  − 4 det  + 7 det  det A = 1det     8 9  8 9  5 6 = 1 ( 45 − 48 ) − 4 (18 − 24 ) + 7 (12 − 15 ) =−3 + 24 − 21 = 0. EXAMPLE 1.10.2 t  Find the det  t 2  t 3

1 a 1   1 b = t det  t  t 2 1 c 

1 a  1 b . 1 c 

EXAMPLE 1.10.3 1  3        Find the area of parallelogram with edges a = 2  and b = 2  . 3  1  Solution: 1 2 3  We have now   , then 3 2 1   2  det  2  1    a × b = − det   3   det 1 3    Area =

  a× b =

3   1    −4  3    = 8 1       −4  2  2  

( −4 )2 + ( 8 )2 + ( −4 )2 =

16 + 64 + 16 =

96 .

  Theorem 1.10.2:  The volume V of a parallelepiped with vectors a, b , and     c as adjacent edges = is V det a b × c .

( (

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))

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108 • Multivariable and Vector Calculus 2/E EXAMPLE 1.10.4

       Find the volume V of a parallelepiped having a = 2 i − 3j + k , = b 4 j − 3k ,     and c =i + 2 j + k as adjacent edges. Solution:    = V det a b × c

( (

))

2 −3 1   4 −3  0 −3  0 4    V det 0 = 4 −3  2 det  = + 3 det  + 1det     2 1  1 1  1 2  1 2 1  = V 2(10) + 3(3) + 1(−4) = 25 .    NOTE When a, b , and c have the same initial point, they lie in the same plane  a1 a2 a3       b1 b2 bc  0 . That is, the volume of the parallelepiped is = ⇔ a b × c det =  c1 c2 c3     0 ⇔ a, b , and c are coplanar.

(

)

Again, we may use the MapleTM packages linalg, LinearAlgebra, or Student [VectorCalculus] to perform many of the vector operations. An example follows with linalg.

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Vectors and Parametric Curves • 109

Also, we may use MATLAB package, using the following commands. >> a=[-2,0,1]; >> b=[-1,3,0]; >> CP=cross(a, b) CP =  -3 -1 -6 >> DP=dot(a, b) DP =  2 >> CosTheta = dot(a,b)/(norm(a)∗norm(b)); >> ThetaInDegrees = acos(CosTheta)∗180/pi ThetaInDegrees =  73.5701

EXERCISES 1.10 2 −1 3   1.10.1 Evaluate the det  −2 −4 1  .  5 8 0 

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110 • Multivariable and Vector Calculus 2/E

a 0  1.10.2 Evaluate the det  b e  c d

0  0. f 

a b c    1.10.3 Evaluate the det 0 e d  . 0 0 f  0 0 c    1.10.4 Show that det 0 b e  = − cba .  a d f  1.10.5 Solve for d in the equation Solve for d in the equation 1 − d 1 1    det  2 d −4  = 4.  3 1 0  0  z − 6 0   1.10.6 Determine the values of z for which the det  0 z −1  = 0.  0  4 z − 4 9   t −3   1.10.7 Evaluate the det 2 4 t + 1 . 1 t 2 3   a2 − b2  1.10.8 Show that det  a − b  a−b 

a + b a  1 1 = 0.  1 b

1.10.9 Solve the following equation for c in the determinant c − 1 1 0   det  2 1. −1 1  =  3 c 0 4  1  1.10.10 Prove det  a  a2

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1 b b2

1  c= ( b − c )( c − a )( a − b). 2 c 

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Vectors and Parametric Curves • 111

a b c   1.10.11 Prove det  b c a = 3 abc − a3 − b3 − c3 .  c a b 1 1   1   1.10.12 Evaluate the det  x y z .  x + 2 y + 3 z + 4  1 1 0    1.10.13 Find the values of l that satisfy det 0 l 1  = 0 . 0 1 l  0 −1  1 − l   1.10.14 Find the values of l that satisfy det  1 2−l 1  = 0. 3 −l   3            1.10.15 Find the a b × c of the vectors a = 2 i − j + 3k , b =i + 4 j − k , and    = c 2 j + 3k .     1.10.16 Find the constant a such that the vectors a = 2 i − j + k ,         b =i + 2 j − 3k , and c = 3 i + a j + 5k are coplanar.

(

)

1.11 SOME SOLID GEOMETRY In this section, we examine some examples and prove some theorems of three-dimensional geometry. EXAMPLE 1.11.1 Cube ABCDD′C′B′A ABCDD′C′B′A in Fig. 1.11.1 has side of length a. M is the midpoint of edge [BB′] and N is the midpoint of edge [B′C′] . Prove that   AD′  MN and find the area of the quadrilateral MND′A .

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112 • Multivariable and Vector Calculus 2/E

FIGURE 1.11.1  Example 1.11.1.

Solution:

 By the Pythagorean Theorem, AD′ = a 2 . Because they are diagonals that   belong to parallel faces of the cube, AD′  BC′ . Now, M and N are the mid  points of the sides [B′B] and [B′C′] of ∆B′C′B , and hence MN  BC′ by exam 1  a 2 ple 1.1.6. The aforementioned example also gives= MN AD′ . In = 2 2   consequence, AD′  MN . This means that the four points A, D′, M, N are all on the same plane. Hence MND′A is a trapezoid with bases of length a 2 and

a 2 , see Fig. 1.11.2. 2

FIGURE 1.11.2  Example 1.11.1.

From the figure   1   a 2 D′Q =AP = AD′ − MN = ⋅ 2 4 Also, by the Pythagorean theorem,

(

 D′N =

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 2  2 D′C′ + C′N =

)

a2 +

a2 a 5 = ⋅ 4 2

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Vectors and Parametric Curves • 113

The height of this trapezoid is thus  NQ =

5 a2 a2 3a − = ⋅ 4 8 2 2

The area of the trapezoid is finally,  a 2 a 2+ 3a  2 ⋅ 2 2 2   

  9 a2 = ⋅ 8   

Let us prove a three-dimensional version of Thales’ Theorem. Theorem 1.11.1 (Thales’ theorem):  If two lines are cut by three parallel planes, their corresponding segments are proportional. See Fig. 1.11.3.

FIGURE 1.11.3  Thales’ theorem in 3D.

Proof:

  Given the lines AB and CD , we must prove that AE CF = . EB FD  Draw line AD cutting plane P2 in G. The plane containing points A, B, and  D intersects plane P2 in the line EG . Similarly, the plane containing points A,

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114 • Multivariable and Vector Calculus 2/E  C, and D intersects plane P2 in the line GF . Since P2 and P3 are parallel   planes, EG  BD , and so by Thales’ Theorem on the plane (theorem 1.2.2) AE AG = . EB GD   Similarly, since P1 and P2 are parallel, AC  GF and CF AG = . FD GD It follows that AE CF = , EB FD as needed to be shown. EXAMPLE 1.11.2 In cube ABCDD′C′B′A ′ of edge of length a, as in Fig. 1.11.4, the points M  and N are located on diagonals [AB′ ] and [ BC′] such that MN is parallel to the face ABCD of the cube. If  5  MN = AB , find the ratios 3

  BN AM  and  . AB′ BC′

FIGURE 1.11.4  Example 1.11.2.

Solution: There  is a unique plane parallel P to face ABCD and containing M. Since MN is parallel to face ABCD, P also contains N. The intersection of P with the cube produces a lamina A′′B′′C′′D′′, as in Fig. 1.11.5.

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Vectors and Parametric Curves • 115

FIGURE 1.11.5  Example 1.11.2.

  ′ a 2 . Put First, notice that = AB′ BC =     AM MB′ AB′ − AM x ⇒  = = 1 − x.  = AB′ AB′ AB′ Now, as ∆B′AB  ∆B′MB′′ and ∆BC′B′  ∆BNB′′ .     MB′ B′′B′ MB′ MB′′   =  ,  =  ⇒ MB′′ = (1 − x ) a, AB′ BB′ AB′ AB   BB′′ AM  =  , BB′ AB′

  B′′N BB′′   =  ⇒ B′′N = xa. B′C′ BB′

 5 Since MN = a , by the Pythagorean theorem, 3

{ }

 2  2  2 5 1 2 2 MN = MB′′ + B′′N ⇒ a2 = (1 − x ) a2 + x2 a2 ⇒ x ∈ , . 9 3 3 There are two possible positions for the segment, giving the solutions     AM BN 1 AM BN 2 = = = = ⋅ ,     AB′ BC′ 3 AB′ BC′ 3 EXAMPLE 1.11.3

1   Find the direction ratios and direction cosines of a point   (4, 5, -2) in 3D     geometry

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116 • Multivariable and Vector Calculus 2/E

EXERCISES 1.11  1.11.1 In a regular tetrahedron with vertices A, B, C, D and with AB = a , points M and N are the midpoints of the edges [AB] and [CD] , respectively. 1. Find the length of the segment [MN]. 2. Find the angle between the lines [MN ] and [ BC]. 1.11.2 In cube ABCDD′C′B′A′ of edge of length a, find the distance between the lines that contain the diagonals [ A′B] and [ AC] . 1.11.3 Prove that the sum of the squares on the diagonals of a parallelepiped is equal to four times the sum of the squares on the three coterminous edges.

FIGURE 1.11.6  Exercise 1.11.2.

1.11.4 In a quadrilateral ABCD, prove that AB2 + BC2 + CD2 + DA2 = AC2 + BD2 + 4 PQ2 where, P and Q are the middle points of the diagonals AC and BD.

1.11.5 Let S be the solid in three-dimensional space consisting of all points (x, y, z) satisfying the following system of six conditions: x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 11, 2 x + 4 y + 3 z ≤ 36, 2 x + 3 z ≤ 24.

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Vectors and Parametric Curves • 117

Determine the number of vertices and the number of edges of S.

1.12 CAVALIERI AND THE PAPPUS–GULDIN RULES Theorem 1.12.1 (Cavalieri’s principle):  All planar regions with cross sections of proportional length at the same height have an area in the same proportion. All solids with cross sections of proportional areas at the same height have their volume in the same proportion. Proof: We only provide the proof for the second statement, as the proof for the first is similar. Cut any two such solids by horizontal planes that produce cross sections of area A( x) and cA( x), where c > 0 is the constant of proportionality, at an arbitrary height x above a fixed base. From elementary calculus, we x2 x2 know that ∫ A( x)dx and ∫ cA( x)dx give the volume of the portion of each x1

x1

solid cut by all horizontal planes as x runs over some interval [ x1 ; x2 ] . As

∫

x2

x1

x2

A( x)dx = ∫ cA( x)dx , x1

the

corresponding

volumes

must

also

be

proportional. EXAMPLE 1.12.1 Use Cavalieri’s principle in order to deduce that the area enclosed by the x 2 y2 ellipse with equation 2 + 2 = 1, a > 0, b > 0, is p ab . a b Solution: Consider the circle with equation x2 + y2 =, a2 as in Fig. 1.12.1. Then, for b y > 0 , y = a2 − x 2 , y = a2 − x 2 . a

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118 • Multivariable and Vector Calculus 2/E

FIGURE 1.12.1  Ellipse and circle.

The corresponding ordinate for the ellipse and the circle are proportional, and hence, the corresponding chords for the ellipse and the circle will be proportional. By Cavalieri’s first principle, Area of the ellipse =

b ( Area of the circle) a

b p a2 ) ( a = p ab . =

EXAMPLE 1.12.2 Use Cavalieri’s principle in order to deduce that the volume of a sphere with 4 radius a is p a3 . 3 Solution: The following method is due to Archimedes, who was so proud of it that he wanted a sphere inscribed in a cylinder on his tombstone. We need to recall that the volume of a right circular cone with base radius a and height h is p a2 h . 3 Consider a hemisphere of radius a, as in Fig. 1.12.2.

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Vectors and Parametric Curves • 119

FIGURE 1.12.2 Hemisphere.

Cut a horizontal slice at height x, producing a circle of radius r. By the Pythagorean theorem, x2 + r 2 =, a2 and so this circular slab has area = p r 2 p ( a2 − x2 ) . Now, consider a punctured cylinder of base radius a and height a, as in Fig. 1.12.3, with a cone of height a and base radius a cut from it. A horizontal slab at height x is an annular region of area p a2 − p x2 , which agrees with a horizontal slab for the sphere at the same height.

FIGURE 1.12.3  Punctured cylinder.

By Cavalieri’s principle, Volume of the hemisphere = Volume of the punctured cylinder p a3 = p a2 − 3 2p a3 = . 3  2p a3  4p a3 It follows that the volume of the sphere is 2  . = 3  3  Essentially, the same method of proof as Cavalieri’s principle gives the next result. Theorem 1.12.2 (Pappus–Guldin Rule):  The area of the lateral surface of a solid of revolution is equal to the product of the length of the generating

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120 • Multivariable and Vector Calculus 2/E curve on the side of the axis of revolution and the length of the path described by the center of gravity of the generating curve under a full revolution. The volume of a solid of revolution is equal to the product of the area of the generating plane on one side of the revolution axis and the length of the path described by the center of gravity of the area under a full revolution about the axis. EXAMPLE 1.12.3 Since the center of gravity of a circle is at its center, by the Pappus–Guldin rule, the surface area of the torus with the generating circle having radius r, and radius of gyration R (as in Fig. 1.12.4) is ( 2p r )( 2p R ) = 4p 2 rR . Also, the volume of the solid torus is (p r 2 ) ( 2p R ) = 2p 2 r 2 R .

FIGURE 1.12.4  A torus.

EXERCISES 1.12 1.12.1 Use the Pappus–Guldin Rule to find the lateral area and the volume of a right circular cone with base radius r and height h.

FIGURE 1.12.5  Generating a cone, Exercise 1.12.1.

1.12.2 Use the Pappus–Guldin rule to find the lateral area and the volume of a rectangular cylinder with base radius r and height h

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Vectors and Parametric Curves • 121

1.12.3 Use the Pappus–Guldin rule to find the lateral area and the volume of a semicircle sphere with base radius r and height h. 1.12.4 Find the volume for spherical cap with radius r and height h, with base area A. 1.12.5 A large plastic balloon with a thin metal coating used for satellite communications has a diameter of 60 m. Find its area and volume. 1.12.6 Tell whether the statement true or false. 1. A radius of a small circle of a sphere is a radius of the sphere. 2. A diameter of a great circle of a sphere is the diameter of the sphere.

1.13 DIHEDRAL ANGLES AND PLATONIC SOLIDS Definition 1.13.1  When two half-planes intersect in space they intersect on a line. The portion of space bounded by the half-planes and the line is called the dihedral angle. The intersecting line is called the edge of the dihedral angle and each of the two half-planes of the dihedral angle is called a face. See Fig. 1.13.1.

FIGURE 1.13.1  Dihedral angles.

Definition 1.13.2 The rectilinear angle of a dihedral angle is the angle whose sides are perpendicular to the edge of the dihedral angle at the same point, on each of the faces. See Fig. 1.13.2.

FIGURE 1.13.2  Rectilinear of a dihedral angle.

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122 • Multivariable and Vector Calculus 2/E All the rectilinear angles of a dihedral angle measure the same. Hence the measure of a dihedral angle is the measure of any one of its rectilinear angles. In analogy to dihedral angles, we now define polyhedral angles. Definition 1.13.3  The opening of three or more planes that meet at a common point is called a polyhedral angle or solid angle. In the particular case of three planes, we use the term trihedral angle. The common point is called the vertex of the polyhedral angle. Each of the intersecting lines of two consecutive planes is called an edge of the polyhedral angle. The portion of the planes lying between consecutive edges are called the faces of the polyhedral angle. The angles formed by adjacent edges are called face angles. A polyhedral angle is said to be convex if the section made by a plane cutting all its edges forms a convex polygon. In the trihedral angle of Fig. 1.13.3, V is the vertex, ∆VAB , ∆VBC , ∆VCA are faces. Also, notice that in any polyhedral angle, any two adjacent faces form a dihedral angle.

FIGURE 1.13.3  Trihedral angle.

Theorem 1.13.1:  The sum of any two face angles of a trihedral angle is greater than the third face angle. Proof:  Consider Fig. 1.13.3. If ∠ZVX is smaller or equal to in size than either ∠XVY or YVZ, then we are done, so assume that, say, ∠ZVX > XVY . We must demonstrate that ∠XVY + ∠YVZ > ∠ZVX.

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Vectors and Parametric Curves • 123

Since we are assuming that ∠ZVX > XVY , we may draw, in ∠XVY the line segment [ VW ] such that ∠XVW = ∠XVY . Through any point D of the segment [ VW ] , draw ∆ADC on the plane P containing the points V , X , Z . Take the point B ∈ [ VY ] so that VD = VB . Consider now the plane containing the line segment [ AC] and the point B. Observe that ∆AVD ≅ AVB . Hence AD = AB . Now, by the triangle inequality in ∆ABC, AB + BC > CA . This implies that ∠BVC > ∠DVC . Hence ∠AVB + ∠BVC = ∠AVD + ∠BVC

> ∠AVD + ∠DVC



= ∠AVC ,

which proves that ∠XVY + ∠YVZ > ∠ZVX , as wanted. Theorem 1.13.2  The sum of the face angles of any convex polyhedral angle is less than 2p radians. Proof:  Let the polyhedral angle have n faces and vertex V. Let the faces be cut by a plane, intersecting the edges at the points A1 , A2 , . . . An , say. An illustration can be seen in Fig. 1.13.4, where for convenience, we have depicted only five edges.

FIGURE 1.13.4  Polyhedral angle.

Observe that the polygon A1 A2 . . . An is convex and that the sum of its interior angles is p ( n − 2 ) .

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124 • Multivariable and Vector Calculus 2/E We would like to prove that ∠A1 VA2 + ∠A2 VA3 + ∠A3 VA4 +  + ∠An−1 VAn + ∠An VA1 < 2p . Now, let Ak −1 , Ak , Ak +1 , be three consecutive vertices of the polygon A1 A2 . . . An . This notation means that Ak −1 Ak Ak +1 , represents any of the n triplets A1 A2 A3 , A2 A3 A4 , A3 A4 A5 ,, An− 2 An−1 An , An−1 An A1 , An A1 A2 , that is, we= let A0 A= A= A2 , etc. Consider the trihedral angle with n , An +1 1 , An + 2 vertex Ak and whose face angles at Ak are ∠Ak −1 Ak Ak +1 , ∠VAk Ak −1 , and ∠VAk Ak +1 , as in Fig. 1.13.5.

FIGURE 1.13.5 

A, Ak , Ak +1 are three consecutive vertices.

Observe that as k ranges from 1 through n, the sum

∑ ∠A

k −1

1≤ k ≤ n

Ak Ak +1 = p ( n − 2) ,

be of the interior angles of the polygon A1 A2  An . By Theorem 1.13.1, ∠VAk Ak −1 + ∠VAk Ak +1 > ∠Ak −1 Ak Ak +1 . Thus

∑ VA A

1≤ k ≤ n

k

k −1

+ ∠VAk Ak +1 >

∑ ∠A

1≤ k ≤ n

k −1

Ak A= p ( n − 2 ). k +1

Also,

∑ VA A

1≤ k ≤ n

k

k +1

+ ∠VAk +1 Ak + ∠Ak VAk +1 =p n,

since this is summing the sum of the angles of the n triangles of the faces. But clearly

∑ VA A= ∑ ∠VA

1≤ k ≤ n

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k

k +1

1≤ k ≤ n

k +1

Ak ,

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Vectors and Parametric Curves • 125

since one sum adds the angles in one direction and the other in the opposite direction. For the same reason,

∑ VA A k

1≤ k ≤ n

k −1

=∠ ∑ VAk Ak+1 ⋅ 1≤ k ≤ n

Hence

∑ ∠A VA =

1≤ k ≤ n

k

k +1

pn−

∑ ( VA A

k +1

+ ∠VAk +1 Ak )

∑ ( VA A

k +1

+ ∠VAk Ak −1 )

k

1≤ k ≤ n

=p n −

1≤ k ≤ n

k

=p n − p ( n − 2 ) = 2p , as we needed to show. Definition 1.13.4  A Platonic solid is a polyhedron having congruent regular polygon as faces and having the same number of edges meeting at each corner. Suppose a regular polygon with n ≥ 3 sides is a face of a platonic solid with m ≥ 3 faces meeting at a corner. Since each interior angle of this polygon measures

p ( n − 2) , we must have in view of Theorem 1.13.2, n

 p ( n − 2)  m  < 2p ⇒ m ( n − 2 ) < 2 n ⇒ ( m − 2 )( n − 2 ) < 4. n   Since n ≥ 3 and m ≥ 3 , the preceding inequality only holds for five pairs ( n, m ) . Appealing to Euler’s formula for polyhedrons, which states that V + F = E + 2 , where V is the number of vertices, F is the number of faces, and E is the number of edges of a polyhedron, we obtain the values in the following table.

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m

n

S

E

F

Name of regular polyhedron

3

3

4

6

4

Tetrahedron or regular Pyramid.

4

3

8

12

6

Hexahedron or cube.

3

4

6

12

8

Octahedron.

5

3

20

30

12

Dodecahedron.

3

5

12

30

20

Icosahedron.

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126 • Multivariable and Vector Calculus 2/E Thus, there are at most five Platonic solids. That there are exactly five can be seen by explicit construction. Figs. 1.13.6 through 1.13.10 depict the Platonic solids.

FIGURE 1.13.6 Tetrahedron.

FIGURE 1.13.8 Octahedron.

FIGURE 1.13.7  Cube or hexahedron.

FIGURE 1.13.9 Dodecahedron.

FIGURE 1.13.10 Icosahedron.

EXERCISES 1.13 1.13.1 What are the five regular polyhedrons known as Platonic solids? 1.13.2 Each Platonic solid by pairs ( m, n ) , where m is the number of edges of each face (or the number of vertices of each face), and n is the number of faces meeting at each vertex (or the number of edges meeting at each vertex). Write the formula for the total number of vertices (V), edges (E), and faces (F), in terms of m and n. 1.13.3 Each Platonic solid by pairs ( m, n ) , where m is the number of edges of each face (or the number of vertices of each face), and n is the number of faces meeting at each vertex (or the number of edges meeting at each vertex). Write the formula for the dihedral angle, θ, of the solid ( m, n ) . 1.13.4 Each Platonic solid by pairs ( m, n ) , where m is the number of edges of each face (or the number of vertices of each face), and n is

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Vectors and Parametric Curves • 127

the number of faces meeting at each vertex (or the number of edges meeting at each vertex). Write the formula for the surface area A, and the volume V, of the solid ( m, n ) with edge length g and in radius r.

1.14 SPHERICAL TRIGONOMETRY Consider a point B ( x, y, z ) in Cartesian coordinates. From O ( 0, 0, 0 ) , we draw a straight line to B ( x, y, z ) , and let its distance be r . We measure its inclination from the positive z-axis. Let us say it is an angle of j , j ∈ [ 0; p ] radians, as in Fig. 1.14.1.

FIGURE 1.14.1  Spherical coordinates.

Observe that z = r cos j . We now project the line segment [OB] onto the xy- plane in order to find the polar coordinates of x and y. Let q be angle that this projection makes with the positive x-axis. Since OP = r sin j , we find x = r cos q sin j , y = r sin q sin j .

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128 • Multivariable and Vector Calculus 2/E

Definition 1.14.1  Given a point

( x, y, z )

in Cartesian coordinates, its

spherical coordinates are given by = x

r= cos q sin j , y

r= sin q sin j , z

r cos j .

Here j is the polar angle, measured from the positive z-axis, and q is the azimuthal angle, measured from the positive x-axis. By convention, 0 ≤ q ≤ 2p and 0 ≤ j ≤ p . Spherical coordinates are extremely useful when considering regions, which are symmetric about a point. Definition 1.14.2  If a plane intersects with a sphere, the intersection will be a circle. If this circle contains the center of the sphere, we call it a great circle. Otherwise, we talk of a small circle. The axis of any circle on a sphere is the diameter of the sphere, which is normal to the plane containing the circle. The endpoints of such a diameter are called the poles of the circle. NOTE

The radius of a great circle is the radius of the sphere. The poles of a great circle are equally distant from the plane of the circle, but this is not the case in a small circle. By the pole of a small circle, we mean the closest pole to the plane containing the circle. A pole of a circle is equidistant from every point of the circumference of the circle.

Definition 1.14.3  Given the center of the sphere, and any two points on the surface of the sphere, a plane can be drawn. This plane will be unique if and only if the points are not diametrically opposite. In the case where the two points are not diametrically opposite, the great circle formed is split into a larger and a smaller arc by the two points. We call the smaller arc the geodesic joining the two points. If the two points are diametrically opposite then every plane containing the line forms with the sphere a great circle, and the arcs formed are then of equal length. In this case, we take any such arc as a geodesic. Definition 1.14.4 A spherical triangle is a triangle on the surface of a sphere all whose vertices are connected by geodesics. The three arcs of great circles, which form a spherical triangle, are called the sides of the spherical triangle; the angles formed by the arcs at the points where they meet are called the angles of the spherical triangle.

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Vectors and Parametric Curves • 129

If A, B, C are the vertices of a spherical triangle, it is customary to label the opposite arcs with the same letter name, but in lowercase. NOTE

A spherical triangle has then six angles: three vertex angles ∠A, ∠B, ∠C, and three arc angles, ∠a, ∠b, ∠c. Observe that if O is the center of the sphere then

      ∠a =∠ OB,OC , ∠b =∠ OC,OA , ∠c =∠ OA,OB ,

(

)

(

)

(

)

and             ∠A =∠ OA × OB,OA × OC , ∠B =∠ OB × OC,OB × OA , ∠C =∠ OC × OA,OC × OB .

(

)

(

)

(

)

Theorem 1.14.1: Let ∆ABC be a spherical triangle. Then cosa cosb + sina sinb cosC = cosc. Proof: Consider a spherical triangle ABC with A ( x1 , y1 , z1 ) , B ( x2 , y2 , z2 ) , and let O be the center and r be the radius of the sphere. In spherical coordinates, this is, say, z1 = r cosq 1 ,

x1 = r sinq 1 cosj 1 ,

y1 = r sinq 1 sinj 1

z2 = r cosq 2 ,

x2 = r sinq 2 cosj 2 ,

y2 = r sinq 2 sinj 2 ;

By a rotation, we may assume that the z-axis passes through C. Then the following quantities give the square of the distance of the line segment [ AB] :

( x1 − x2 )

2

+ ( y1 − y2 ) + ( z1 − z2 ) , r 2 + r 2 − 2 r 2 cos∠ ( AOB) . 2

2

Since x12 + y12 + = z12 r 2 , x22 + y22 + = z22 r 2 , we gather that x1 x2 + y1 y2 + z1 z2 = r 2 cos∠ ( AOB) . Therefore we obtain cosq 2 cosq 1 + sinq 2 sinq 1 cos (j 1 − j 2 ) =cos∠ ( AOB) , that is, cosa cosb + sina sinb cosC = cos c.

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130 • Multivariable and Vector Calculus 2/E .

Theorem 1.14.2:  Let I be the dihedral angle of two adjacent faces of a regular polyhedron. Then p cos I n . sin = p 2 sin m Proof: Let AB be the edge common to the two adjacent faces, C and D the centers of the faces; bisect AB at E, and join CE and DE; CE and DE will be perpendicular to AB, and the angle CED is the angle of inclination of the two adjacent faces; we shall denote it by I. In the plane containing CE and DE draw CO and DO at right angles to CE and DE, respectively, and meeting at O; about O as center describe a sphere meeting OA, OC, OE at a, c, e respectively, so that cae forms a spherical triangle. Since AB is perpendicular to CE and DE, it is perpendicular to the plane CED, therefore the plane AOB which contains AB is perpendicular to the plane CED; hence the angle cea of the spherical triangle is a right angle. Let m be the number of ides in each face of the polyhedron, n the number of the plane angles which form each solid 2p p angle. Then the angle ace = ACE = = ; and the angle cae is half one of 2m m 2p p the n equal angles formed on the sphere round a, that is, cae = = . From 2n n the right-angled triangle cae cos cae = cos cOe sin ace, that is p p p I  = cos cos  −  sin ; 2 m  2 2 therefore, p cos I n. sin = 2 sin p m Theorem 1.14.3: Let r and R be, respectively, the radii of the inscribed and circumscribed spheres of a regular polyhedron. Then

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Vectors and Parametric Curves • 131

= r

p p a I a I cot = tan , R tan tan ⋅ m n 2 2 2 2

Here a is the length of any edge of the polyhedron, and I is the dihedral angle of any two faces. Proof: Let the edge AB = a , let OC = r and OA = R , so that r is the radius of the inscribed sphere, and R is the radius of the circumscribed sphere. Then = = cot ACE CE AE

p a cot , 2 m

I a p I = = = tan cot tan tan ; r CE tan CEO CE 2 2 m 2 also = r R= cos aOc R cot eca = cot eac R cot

p p cot ; m n

therefore p p a p I = tan tan tan tan ⋅ R r= m n 2 2 n From the previous formula, we now easily find that the volume of the pyramid, which has one face of the polyhedron for base and O for vertex is r ma2 p mFra2 p cot , and therefore the volume of the polyhedron is cot . ⋅ 3 4 m 12 m Furthermore, the area of one face of the polyhedron is fore the surface area of the polyhedron is

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ma2 p cot , and there4 m

mFa2 p cot . 4 m

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132 • Multivariable and Vector Calculus 2/E

EXERCISES 1.14 1.14.1 The four vertices of a regular tetrahedron are  0   −1 / 2   −1 / 2  1         V1 =  0 .  3 / 2  , V3 =  − 3 / 2  , V4 =  0  , V2 =   0  0   0         2

What is the cosine of the dihedral angle between any pair of faces of the tetrahedron?

1.14.2 Consider a tetrahedron whose edge measures a. Show that its a3 2 volume is , its surface area is a2 3 , and that the radius of the 12 a 6 inscribed sphere is . 12 1.14.3 Consider a cube whose edge measures a. Show that its volume is a3 , its surface area is 6a2 , and that the radius of the inscribed a sphere is . 2 1.14.4 Consider an octahedron whose edge measures a. Show that its a3 2 volume is , its surface area is 2 a2 3 , and that the radius of 3 a 6 the inscribed sphere is . 6 1.14.5 Consider a dodecahedron whose edge measures a. Show that its a3 volume is 15 + 7 5 , its surface area is 3 a2 25 + 10 5 , and 4

(

)

that the radius of the inscribed sphere is

a 1 . 10 + 22 4 5

1.14.6 Consider an icosahedron whose edge measures a. Show that its 5 a3 volume is 5 + 5 , its surface area is 5 a2 3 , and that the 12 a radius of the inscribed sphere is 5 3 + 15 . 12

(

)

(

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)

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Vectors and Parametric Curves • 133

1.15 CANONICAL SURFACES In this section, we consider various surfaces that we shall periodically ­encounter in subsequent sections. Just like in one-variable Calculus it is important to identify the equation and the shape of a line, a parabola, a circle, etc., it will become important for us to be able to identify certain families of often-occurring surfaces. We shall explore both their Cartesian and their parametric form. We remark that in order to parameterize curves (“one-dimensional entities”), we needed one parameter and that in order to parameterize surfaces we shall need two parameters. Let us start with the plane. Recall that if a, b, c are real numbers, not all zero,  a   then the Cartesian equation of a plane with normal vector  b and passing  c  through the point ( x0 , y0 , z0 ) is a ( x − x0 ) + b ( y − y0 ) + c ( z − z0 ) = 0.

  If we know that the vectors u and v are on the plane (parallel to the plane) then with the parameters p, q, the equation of the plane is x − x0 = pu1 + qv1 , y − y0 = pu2 + qv2 , z − z0 = pu3 + qv3 . Definition 1.15.1  A surface S consisting of all lines parallel to a given line ∆ and passing through a given curve Γ is called a cylinder. The line ∆ is called the directrix of the cylinder. NOTE

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To recognize whether a given surface is a cylinder we look at its Cartesian equation. If it is of the form f (A, B) = 0, where A, B are secant planes, then the curve is a cylinder. Under these conditions, the lines generating S will be parallel to the line of equation A = 0, B = 0.. In practice, if one of the variables x, y, or z is missing, then the surface is a cylinder, whose directrix will be the axis of the missing coordinate.

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134 • Multivariable and Vector Calculus 2/E EXAMPLE 1.15.1 Fig. 1.15.1 shows the cylinder with Cartesian equation x2 + y2 = 1 . One starts with the circle x2 + y2 = 1 on the xy-plane and moves it up and down the z-axis. A parameterization for this cylinder is the following: x= cos v, y= sin v, z= u,

u ∈ , v ∈ [ 0;2p ].

FIGURE 1.15.1  Circular cylinder x2 + y2 = 1.

The MapleTM commands to graph this surface are:

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Vectors and Parametric Curves • 135

The method of parameterization previously utilized for the cylinder is quite useful when doing parameterizations in space. We refer to it as the method of cylindrical coordinates. In general, we first find the polar coordinates of x, y in the xy-plane, and then lift ( x, y,0 ) parallel to the z-axis to ( x, y, z ) : = x r= cosq , y r= sin q , z z. See Fig. 1.15.2.

FIGURE 1.15.2  Cylindrical coordinates.

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136 • Multivariable and Vector Calculus 2/E EXAMPLE 1.15.2 Fig. 1.15.3 shows the parabolic cylinder with Cartesian equation z = y2 . One starts with the parabola z = y2 on the yz-plane and moves it up and down the x-axis. A parameterization for this parabolic cylinder is the following: x = u, y = v, z = v2 , u ∈ , v ∈ .

FIGURE 1.15.3  The parabolic cylinder z = y 2 .

The MapleTM commands to graph this surface are:

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Vectors and Parametric Curves • 137

EXAMPLE 1.15.3 Fig. 1.15.4 shows the hyperbolic cylinder with Cartesian equation x2 − y2 = 1. One starts with the hyperbola x2 − y2 = 1 on the xy-plane and moves it up and down the z-axis. A parameterization for this parabolic cylinder is the following: x= ± cosh v, y = sinh v, z = v, u ∈ , v ∈ .

FIGURE 1.15.4  The hyperbolic cylinder x 2 − y 2 = 1.

We need a choice of sign for each of the portions. We have used the fact that cosh2 v − sinh2 v = 1 . The MapleTM commands to graph this surface are:

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138 • Multivariable and Vector Calculus 2/E

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Vectors and Parametric Curves • 139

Definition 1.15.2  Given a point Ω ∈  3 (called the apex) and a curve (called the generating curve), the surface s obtained by drawing rays from Ω and passing through Γ is called a cone. NOTE

In practice, if the Cartesian equation of a surface can be put into the form

 A B f  ,  = 0, where A, B, C, are planes secant at exactly one point, then the surface C C is a cone, and its apex is given by A = 0, B = 0, C = 0.

EXAMPLE 1.15.4 2 The surface in  3 implicitly given by z= x2 + y2 is a cone, as its equation 2

2

 x  y can be put in the form   +   − 1 = 0 . Considering the planes  z  z = x 0,= y 0,= z 0 , the apex is located at ( 0,0,0 ) . The graph is shown in Fig. 1.15.5

FIGURE 1.15.5 Cone

x2 y2 z2 + =. a2 b2 c 2

Definition 1.15.3  A surface S obtained by making a curve Γ turn around a line ∆ is called a surface of revolution. We then say that ∆ is the axis of revolution. The intersection of S with a half-plane bounded by ∆ is called a meridian. NOTE

If the Cartesian equation of S can be put in the form f ( A, ∑ ) =0 , where A is a plane and ∑ is a sphere, then the surface is of revolution. The axis of S is the line passing through the center of ∑ and perpendicular to the plane A.

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140 • Multivariable and Vector Calculus 2/E EXAMPLE 1.15.5 Find the equation of the surface of revolution generated by revolving the hyperbola x 2 − 4 z2 = 1, about the z-axis. Solution: Let ( x, y, z ) be a point on S. If this point were on the xz plane, it would be on the hyperbola, and its distance to the axis of rotation would be = x

1 + 4 z2 .

Anywhere else, the distance of ( x, y, z ) to the axis of rotation is the same as the distance of ( x, y, z ) to ( 0,0, z ) , that is

x2 + y2 . We must have

x2 + y2 =+ 1 4 z2 , which is to say x 2 + y2 − 4 z 2 = 1. This surface is called a hyperboloid of one sheet. See Fig. 1.15.6.

FIGURE 1.15.6  One-sheet hyperboloid

z2 x2 y2 = + −1. c2 a2 b2

Observe that when z= 0, x2 + y2= 1 is a circle on the xy-plane. When x = 0, y2 − 4 z2 = 1 , is a hyperbola on the yz-plane. When y = 0, x2 − 4 z2 = 1 is a hyperbola o the xz-plane. A parameterization for this hyperboloid is x = 1 + 4 u2 cos v, y = 1 + 4 u2 sin v, z =u,

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u ∈ , v ∈ [ 0;2p ].

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Vectors and Parametric Curves • 141

EXAMPLE 1.15.6 The circle ( y − a ) + z2 =, r 2 on the yz-plane (a, r are positive real numbers) is 2

revolved around the z-axis, forming a torus T. Find the equation of this torus. Solution: Let ( x, y, z ) be a point on T. If this point were on the yz- plane, it would be on the circle, and of the distance to the axis of rotation would be y= a + sgn ( y − a ) r 2 − z2 , where sgn ( t ) (with sgn(t) = −1 if t < 0,sgn(t) = 1 if t > 0, and sgn(0) = 0 ) is the sign of t. Anywhere else, the distance from ( x, y, z ) to the z-axis is the distance of this point to the point ( x, y, z ) : x2 + y2 . We must have

(

x2 + y2 = a + sgn ( y − a ) r 2 − z2

)

2

= a2 + 2 a sgn ( y − a ) r 2 − z2 + r 2 − z2 .

Rearranging x 2 + y2 + z 2 − a 2 = − r 2 2 a sgn ( y − a ) r 2 − z2 , or

(x

2

)

+ y2 + z2 − ( a2 + r 2 ) = 4 a2 r 2 − 4 a2 z2 . 2

Since ( sgn ( y − a ) ) = 1 , (it could not be 0, why?). Rearranging again, 2

(x

2

+ y2 + z2 ) − 2 ( a2 + r 2 )( x2 + y2 ) + 2 ( a2 − r 2 ) z2 + ( a2 − r 2 ) = 0. 2

2

The equation of the torus thus, is of the fourth degree, and its graph appears in Fig. 1.15.7.

FIGURE 1.15.7  The torus.

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142 • Multivariable and Vector Calculus 2/E A parameterization for the torus generated by revolving the circle 2 r 2 around the z-axis is ( y − a ) + z2 = x= a cosq + r cosq cos a ,

y= a sin q + r sin q cos a , z = r sin a , with

(q ,a ) ∈ [ −p ;p ]2 . EXAMPLE 1.15.7 The surface = z x2 + y2 is called an elliptic paraboloid. The equation clearly requires that z ≥ 0 . For fixed z = c, c > 0, x2 + y2 = c is a circle. When = y 0,= z x2 is a parabola on the xz-plane. When= x 0,= z y2 is a parabola on the yz-plane. See Fig. 1.15.8. The following is a parameterization of this paraboloid: = x

u cos v= , y

u sin v= , z u,

FIGURE 1.15.8  Paraboloid = z

u ∈ [ 0; +∞[ , v ∈ [ 0;2p ].

x2 y2 + . a2 b2

EXAMPLE 1.15.8 The surface = z x2 − y2 is called a hyperbolic paraboloid or saddle. If z= 0, x2 − y2= 0 is a pair of lines in the xy- plane. When= y 0,= z x2 is a parabola on the xz- plane. When x = 0, z = − y2 is a parabola on the yz-plane. See Fig. 1.15.9. The following is a parameterization of this hyperbolic paraboloid: x =u, y =v,

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z =u2 − v2 ,

u ∈ , v ∈ .

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Vectors and Parametric Curves • 143

FIGURE 1.15.9  Hyperbolic paraboloid = z

x2 y2 − . a2 b2

EXAMPLE 1.15.9 The surface z2 = x2 + y2 + 1 is called a hyperboloid of two sheets. For x2 + y2 < 1, z2 < 0 , is impossible, and hence, there is no graph when x2 + y2 < 1 . When y= 0, z2 − x2 = 1 , is a hyperbola on the xz-plane. When x= 0, z2 − y2 = −1 is a hyperbola on the yz-plane. When z = c is a constant, then the x2 + y2 = c2 + 1 , are circles. See Fig. 1.15.10. The following is a parameterization for the top sheet of this hyperboloid of two sheets x =u cos v, y =u sin v,

z =u2 + 1,

u ∈ , v ∈ [ 0;2p ],

and the following parameterizes the bottom sheet, x= u cos v, y = u sin v,

z= − u2 − 1,

FIGURE 1.15.10  Two-sheet hyperboloid

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u ∈ , v ∈ [ 0;2p ].

z2 x2 y2 = + +1. c 2 a2 b2

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144 • Multivariable and Vector Calculus 2/E EXAMPLE 1.15.10 The surface z2 = x2 + y2 − 1 is called a hyperboloid of one sheet. For x2 + y2 < 1, z2 < 0 is impossible, and hence there is no graph when x2 + y2 < 1 . When y = 0, z2 − x2 = −1 is a hyperbola on the xz-plane. When x = 0, z 2 − y2 = −1 is a hyperbola on the yz-plane. When z = c is a constant, then the x2 + y2 = c2 + 1 are circles. See Fig. 1.15.11. The following is a parameterization for this hyperboloid of one sheet x=

u2 + 1 cos v, y =

u2 + 1 sin v,

z = u,

FIGURE 1.15.11  One-sheet hyperboloid

u ∈ , v ∈ [ 0;2p ].

z2 x2 y2 = + −1. c 2 a2 b2

EXAMPLE 1.15.11

x 2 y2 z 2 Let a, b, c be strictly positive real numbers. The surface 2 + 2 + 2 = 1 is a b c 2 2 y x called an ellipsoid. For = z 0, 2 + 2= 1 is an ellipse on the xy plane. When a b y2 z2 x 2 z2 y 0, 2 + = 1 is an ellipse on the xz-plane. When x 0, = = + = 1 is an a c2 b2 c2 ellipse on the yz-plane. See Fig. 1.15.12. We may parameterize the ellipsoid using spherical coordinates: x= a cosq sin f , y= b sin q sin f ,

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z= c cos f ,

q ∈ [ 0;2p ],f ∈ [ 0; p ].

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Vectors and Parametric Curves • 145

FIGURE 1.15.12  The ellipsoid

x2 y2 z2 + + = 1. a2 b2 c 2

EXERCISES 1.15 1.15.1 Find the equation of the surface of revolution S generated by revolving the ellipse 4 x2 + z2 = 1 about the z-axis. 1.15.2 Find the equation of the surface of revolution generated by revolving the line 3 x + 4 y = 1 about the y-axis. 1.15.3 Describe the surface parameterized by f ( u, v )  ( vcos u, vsin u, au ) , ( u, v ) ∈ ( 0,2p ) × ( 0, 1 ) , a > 0. 1.15.4 Describe the surface parameterized by f ( u, v ) = ( au cos v, bu sin v, u2 ) , ( u, v ) ∈ (1, +∞ ) × ( 0,2p ) , a, b > 0. 1.15.5 Use Maple to show the cylinder with Cartesian equation 3 x2 + 5 y2 = 1 . One starts with the circle 3 x2 + 5 y2 = 1 on the xy-plane and moves it up and down the z-axis, where −1 ≤ x ≤ 1 , −1 ≤ y ≤ 1 , and −10 ≤ z ≤ 10 . 1.15.6 Demonstrate that the surface in  3 S : e x 2 + y2 + z 2 − ( x + z ) e−2 xz = 0, implicitly defined is a cylinder. 1.15.7 Show that the surface in  3 implicitly defined by x 4 + y4 + z4 − 4 xyz ( x + y + z ) = 1 is a surface of revolution, and find its axis of revolution. 1.15.8 Show that the surface S in  3 given implicitly by the equation 1 1 1 + + = 1 is a cylinder and find the direction of its x−y y− z z− x directrix. 1.15.9 Show that the surface S in  3 implicitly defined as xy + yz + zx + x + y + z + 1 = 0 is of revolution and find its axis.

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146 • Multivariable and Vector Calculus 2/E

1.15.10 Demonstrate that the face in  3 given implicitly by z2 − xy = 2 z − 1 is a cone. 1.15.11 (Putnam Exam 1970): Determine, with proof, the radius of the x 2 y2 z 2 largest circle, which can lie on the ellipsoid 2 + 2 + 2 = 1, a b c a> b>c>0. 1.15.12 The hyperboloid of one sheet in Fig. 1.15.13 has the property that if it is cut by planes at z = ±2 , its projection on the xy plane produces y2 = 1 , and if it is cut by a at z = 0, its projection on 4 the xy-plane produces the ellipse 4 x2 + y2 = 1 . Find its equation. the ellipse x2 +

FIGURE 1.15.13  Exercise 1.15.12.

1.16 PARAMETRIC CURVES IN SPACE In analogy to curves on the plane, we now define curves in space. Definition 1.16.1 Let [ a; b] ⊆  . A parametric curve representation r of a  x(t)    curve Γ is a function r : [ a; b] →  , with r(t) =  y(t)  , and such that  z(t)    r ([ a; b]) = Γ. r ( a ) is the initial point of the curve and r(b) its terminal point. 3

A curve is closed if its initial point and its final point coincide. The trace of the  curve r is the set of all images of r, that is, Γ . The length of the curve is ∫ d r . Γ

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Vectors and Parametric Curves • 147

EXAMPLE 1.16.1

   The trace of r(t) = i cos t + j sin t + k t is known as a cylindrical helix. See Fig. 1.16.1.

FIGURE 1.16.1  Helix.

To find the length of the helix as t traverses the interval [0;2p ] , first observe that  = dx

1 dt ( sin t )2 + ( − cos t )2 +=

2 dt ,

and thus the length is

∫

2p

0

2 dt = 2p 2 .

The MapleTM commands to graph this curve and to find its length are:

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148 • Multivariable and Vector Calculus 2/E

EXAMPLE 1.16.2 Find a parametric representation for the curve resulting by the intersection of the plane 3 x + y + z = 1 and the cylinder x2 + 2 y2 = 1 in  3 . Solution: The projection of the intersection of the plane 3 x + y + z = 1 and the cylinder is the ellipse x2 + 2 y2 = 1 on the xy-plane. This ellipse can be parameterized as = x cos= t, y

2 sin t, 0 ≤ t ≤ 2p . 2

From the equation of the plane, z = 1 − 3 x − y = 1 − 3 cos t −

2 sin t. 2

Thus we may take the parameterization   cos t    x(t)   2    . r(t) = y(t) sin t =   2  z(t)    1 − 3 cos t − 2 sin t  2  

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Vectors and Parametric Curves • 149

EXAMPLE 1.16.3 Let a, b, c be strictly positive real numbers. Consider the region = ℜ {( x, y, z ) ∈  3 : x ≤ a, y= ≤ b, z c} . A point P moves along the ellipse x 2 y2 1, z =c + 1 , + = a2 b2 once around, and acts as a source light projecting a shadow of ℜ onto the xyplane. Find the area of this shadow. Solution: First consider the same problem as P moves around the circle x2 + y2 = 1, = ℜ′ {( x, y, z ) ∈  3 : x ≤ 1, y= ≤ 1, z c} . See z= c + 1, and the region Fig. 1.16.2.

FIGURE 1.16.2  Example 1.16.3.

For fixed P ( u, v, c + 1 ) on the circle, the image of ℜ′ ( a 2 × 2 square) on the xy- plane is a ( 2 c + 2 ) × ( 2 c + 2 ) square with center at the point Q ( − cu, − cv,0 ) (see Fig. 1.16.3).

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150 • Multivariable and Vector Calculus 2/E

FIGURE 1.16.3  Example 1.16.3.

As P moves along the circle, Q moves along the circle with equation x2 + y2 = c2 on the x2 + y2 = c2 on the xy-plane (Fig. 1.16.3), being the center of a ( 2 c + 2 ) × ( 2 c + 2 ) square. This creates a region as in Fig. 1.16.4, where each quarter circle has radius c, and the central square has side 2 c + 2 , of area p c2 + 4 ( c + 1 ) + 8 c ( c + 1 ) . 2

FIGURE 1.16.4  Example 1.16.3.

Resizing to a region = ℜ

( x, y, z ) ∈  3 :

x ≤ a, y= ≤ b, z c ,

and an ellipse x 2 y2 1, z =c + 1 , + = a2 b2 we use instead of c + 1, a ( c + 1 ) (parallel to the x-axis) and b ( c + 1 ) (parallel to the y-axis), so that the area shadowed is p ab ( c + 1 ) + 4 ab ( c + 1 ) + 4 abc ( c + = 1 ) c2 ab (p + 12 ) + 16 abc + 4 ab. 2

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2

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Vectors and Parametric Curves • 151

EXERCISES 1.16 1.16.1 Let C be the curve in  3 defined by x = t 2 , y = 4 t 3 / 2 , z = 9 t , t ∈ [ 0; +∞ ] .

Calculate the distance along C from (1, 4, 9) to (16, 32, 36).

1.16.2 Consider the surfaces in  3 implicitly defined by z − x2 − y2 − 1 =,  0 z + x2 + y2 − 3 =. 0

Describe as vividly as possible these surfaces and their intersection, if they at all intersect. Find a parametric equation for the curve on which they intersect, if they at all intersect.

 t4  1 + t2   3    t  1.16.3 Consider the space curve r : t   . Let t k , 1 ≤ k ≤ 4 non1 + t2   2   t  1 + t2   

zero real numbers.     Prove that r(t1 ), r(t2 ), r(t3 ) , and r(t 4 ) are coplanar if and only if 1 1 1 1 + + + = 0. t1 t2 t3 t 4

1.16.4 Give a parameterization for the part of the ellipsoid x2 +

y2 z2 + = 1, 9 4

which lies on top of the plane x + y + z =. 0 1.16.5 Find the parametric equations that represent the curve, which is the intersection of the surfaces y2 + z2 = 16 and x =8 − y2 − z .

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152 • Multivariable and Vector Calculus 2/E 1.16.6 Let a be a real number parameter, and consider the planes P1 : ax + y + z =− a, P2 : x − ay + az = −1. Let l be their intersection line. 1. Find a direction vector for l. 2. As a varies through  , l describes a surface S in  3 . Let ( x, y, z) be the point of intersection of this surface and the plane z = c . Find an equation relating x and y. 3. Find the volume bounded by the two planes, x = 0 , and x = 1 , and the surface S as c varies.

1.17 MULTIDIMENSIONAL VECTORS We briefly describe space in n-dimensions. The ideas expounded earlier about the plane and space carry almost without change. Definition 1.17.1   n is the n -dimensional space, the collection  x1       x2   n   : xk ∈   . =        xn         Definition 1.17.2 If a and b are two vector in  n their vector sum a + b is defined by the coordinatewise addition  a1 + b1  a + b    2 2 a+b=  .       an + bn 

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Vectors and Parametric Curves • 153

Definition 1.17.3  A real number a ∈  will be called a scalar. If a ∈   n and a ∈  , we define scalar multiplication of a vector and a scalar by the coordinatewise a a1  a a   2 a a=  .      a an  Definition 1.17.4 The standard ordered basis for  n is the collection of 0          vectors {e1 ,e2 , e n } with e k = 1  .     0  (a 1 in the k slot and 0’s everywhere else). Observe that a 1    n  a 2  a e = . ∑ k k   k =1   a n    Definition 1.17.5  Given vectors a , b of  n their dot product is  a1   b1  a  b    2 2 a  b =     = a1 b1 + a2 b2 +  + an bn =         an   bn  NOTE



The norm of a vector in  n is given by x =

n

∑a b . k =1

k k

  x x .

We now establish one of the most useful inequalities in analysis.

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154 • Multivariable and Vector Calculus 2/E  Theorem 1.17.1 (Cauchy–Bunyakovsky–Schwartz Inequality): Let x  and y be any two vectors in  n . Then we have     xy ≤ x y . Proof: Since the norm of any vector is non-negative, we have       x + ty ≥ 0 ⇔ ( x + ty )( x + ty ) ≥ 0      ⇔ x  x + 2 tx  y + t 2 y  y ≥ 0 2   2 ⇔ x + 2 tx  y + t 2 y ≥ 0 ⋅ This last expression is a quadratic polynomial in t, which is always non-­ negative. As such its discriminant must be non-positive, that is,  

( 2xy )

2

( )( y ) ≤ 0 ⇔ xy ≤ x

 −4 x

2



2

 

  y ,

giving the theorem. NOTE

The preceding proof works for any vector space (cf. below) that has an inner product. The form of the Cauchy-Bunyakovsky-Schwarz most useful to us will be

1/ 2

1/ 2

 n 2  n 2 ≤ x y ∑ k k  ∑ xk   ∑ yk  = k 1 =  k 1=  k 1  n

, (1.22)

or real numbers xk , yk .   Corollary 1.17.1 (Triangle Inequality): Let a and b be any two vectors in  n . Then we have     a+b ≤ a + b .

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Vectors and Parametric Curves • 155

Proof:  2     a + b =a + b  a + b      =a a + 2a b + bb 2    ≤ a +2 a  b + b   2 = a + b ,

(

)(

(

)

2

)

from where the desired result follows. Again, the preceding proof is valid in any vector space that has a norm. We  now consider a generalization of the Euclidean norm. Given p > 1 and x ∈  n , we put

 x

1

 n p p =  ∑ xk  (1.23)  k =1 

p

Clearly,  x p ≥ 0 (1.24)



 x

p

  = 0 ⇔ x = 0 (1.25)

 = ax p a

 x p , a ∈  (1.26)

We now prove analogues of the Cauchy–Bunyakovsky–Schwarz and the Triangle Inequality for ⋅ p . For this, we need the following lemma. 1 1 1.17.1 Lemma (Young’s Inequality): Let p > 1 and put + = 1 . Then p q p q 2 ( a ) ( b) for ( a, b ) ∈ ([ 0; +∞[ ) , we have ab ≤ . + p q Proof: Let 0 < k < 1 , and consider the function f:

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[0; +∞[ →  x → xk − k ( x − 1)

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156 • Multivariable and Vector Calculus 2/E Th= 0 f ′ ( x= x 1. ) kx k−1 − k ⇔=

f ′′( x) = k(k − 1) x k − 2 < 0

Since

for

0 < k < 1, x ≥ 0, x = 1 is a maximum point. Hence f ( x ) ≤ f (1 ) for x ≥ 0 , that

( a) 1 is x ≤ 1 + k( x − 1) . Letting k = and x = , we deduce q p b ( ) p  a 1  ( a) ≤ 1 + − 1  . Rearranging gives,  q q  p  ( b )  ( b) p p

k

p

ab ≤ ( b )

1+

p q

+

p

( a ) p ( b)1+ q − p ( b)1+ q −

p

p

From where we obtain the inequality.   Definition 1.17.6 Let x and y be two non-zero vectors in a vector space    over the real numbers. Then the angle x, y between the is given by the

( )

relation    cos = x, y

( )

  xy   ⋅ x y

This expression agrees with the geometry in the case of the dot product for  2 and  3 . EXAMPLE 1.17.1 Assume that ak , bk , ck , k = 1,, n , are positive real numbers. Show that 4

2

 n   n 4  n 4  n 2  a b c  ∑ k k k  ≤  ∑ ak  ∑ bk  ∑ ck  . = k 1 =  k 1=  k 1 =  k 1  Solution: Using CBS on

n

∑(a b ) c k =1

k k

k

, once we obtain 1/ 2

1/ 2

 n   n  ak bk ck ≤  ∑ ak2 bk2   ∑ ck2  ∑ = k 1=  k 1=  k 1  n

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.

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Vectors and Parametric Curves • 157

1/ 2

 n  Using CBS again on  ∑ ak2 bk2   k =1 

, we obtain 1/ 2

1/ 2

 n   n  ak bk ck ≤  ∑ ak2 bk2   ∑ ck2  ∑ k 1= =  k 1=  k 1  n

1/ 4

1/ 4

1/ 2

 n   n   n  ≤  ∑ ak4   ∑ bk4   ∑ ck2  =  k 1=   k 1=  k 1 

,

which gives the required inequality. We now use the CBS inequality to establish another important inequality. Lemma 1.17.2 Let ak > 0 , qk > 0 , with

n

∑q k =1

k

= 1 . Then

1/ x

n  n  lim log  ∑ qk akx  = ∑ qk log ak . x →0 = k 1  k 1= 

Proof: Recall that log (1 + x )  x as x → 0 . Thus 1/ x

 n  lim log  ∑ qk akx  x →0  k =1 

 n  log  ∑ qk akx   k =1  = lim x →0 x n

= lim

∑ q (a k =1

k

n

= lim ∑ qk x →0

− 1)

x k

− 1)

x

x →0



x k

(a

k =1

x

n

= ∑ qk log ak . k =1

Theorem 1.17.2 (Arithmetic Mean–Geometric Mean Inequality): Let ak ≥ 0 . Then n

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a1 a2  an ≤

a1 + a2 +  + an ⋅ n

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158 • Multivariable and Vector Calculus 2/E Proof: If bk ≥ 0 , then by CBS 2

1 n 1 n  bk ≥  ∑ bk  . (1.27) ∑ n k 1= = nk 1 



Successive applications of Equation (1.23) yield the monotone decreasing sequence 2

4

1 n 1 n  1 n  ak ≥  ∑ ak  ≥  ∑ 4 ak  ≥  , ∑ n k 1= = = nk 1  nk 1  which by Lemma 1.17.1 has a limit 1 n  exp  ∑ log ak  = n a1 a2  an , n  k =1  giving n

a1 a2  an ≤

a1 + a2 +  + an , n

as wanted. EXAMPLE 1.17.2 For any positive integer n > 1 , we have 1 · 3 ⋅ 5  ( 2 n − 1 ) < nn . For, by AMGM,  1 + 3 + 5 +  + ( 2 n − 1 )   n2 1 · 3 ⋅ 5  ( 2 n − 1) <  =  n    n n

n

 n = n . 

Notice that since the factors are unequal we have strict inequality. Definition 1.17.7 Let a1 > 0, a2 > 0, . . . , an > 0 . Their harmonic mean is given by n . 1 1 1 + + + a1 a2 an

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Vectors and Parametric Curves • 159

As a corollary to AMGM, we obtain Corollary 1.17.2 (Harmonic Mean–Geometric Mean Inequality) Let b1 > 0, b2 > 0, . . . , bn > 0 . Then n 1/ n ≤ ( b1 b2  bn ) . 1 1 1 + + + b1 b2 bn Proof: This follows by putting ak = 1/ n

1 1 1      b1 b2 bn 

1 in Theorem 1.17.2. Then bk 1 1 1 + + + b b2 bn . ≤ 1 n

Combining Theorem 1.17.2 and Corollary 1.17.2, we deduce Corollary 1.17.3 (Harmonic Mean–Arithmetic Mean Inequality) Let b1 > 0, b2 > 0, . . . , bn > 0 . Then n 1 1 1 + + + b1 b2 bn

≤

b1 + b2 +  + bn . n

EXAMPLE 1.17.3 Let ak > 0, and s = a1 + a2 +  + an . Prove that s n2 . ≥ ∑ n −1 k =1 s − ak n

and n

ak

∑ s−a k =1

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≥ k

n . n −1

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160 • Multivariable and Vector Calculus 2/E Solution: Put bk =

s . Then s − ak n s − ak 1 = = n −1 , ∑ ∑ bk k 1 n = k 1= n

and from Corollary 1.17.3, n

n ≤ n −1

s

∑ s−a k =1

k

n

,

from where the first inequality is proved. Since

a s − 1 = k , we have s − ak s − ak

n n  s  ak = − 1  ∑ ∑ s − ak k 1  s − ak = k 1=  n  s  = ∑ −n k =1  s − a k 

n2 −n n −1 n = ⋅ n −1 ≥

EXERCISES 1.17 1.17.1 The Arithmetic Mean Geometric Mean Inequality says that if ak ≥ 0 , then

( a1 a2  an )

1/ n



≤

a1 + a2 +  + an ⋅ n

Equality occurs if and only if a= a=  = an . In this exercise, you 1 2 will follow the steps of a proof by George Polya. 1. Prove that ∀x ∈ , x ≤ e x −1 .

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Vectors and Parametric Curves • 161

2. Put Ak =

nak , and Gn = a1 a2  an . Prove that a1 + a2 +  + an

A1 A2  An =

n n Gn

( a1 + a2 +  + an )

n

, and that

A1 + A2 +  + An = n. n

 a + a2 +  + an  3. Deduce that Gn ≤  1  . n   4. Prove the AMGM inequality by assembling the preceding ­results. 1.17.2 Demonstrate that if x1 , x2 ,, x n are strictly positive real numbers then 1 1 1  + +  +  ≥ n2 . xn   x1 x2

( x1 + x2 +  + xn ) 

1.17.3 (USAMO 1978): Let a, b, c, d, e be real numbers such that a + b + c + d + e =, 8 a2 + b2 + c2 + d 2 + e2 = 16. Maximize the value of e . 1.17.4 Find all the positive real numbers a1 ≤ a2 ≤  ≤ an such that n

n

n

= = = ∑ ak 96, ∑ ak2 144, ∑ ak3 1216.

k 1= k 1 = k 1 =

1.17.5 Demonstrate that for integer n > 1 , we have n

 n+1 n! <   .  2  n

1  1.17.6 Prove the sequence xn = 1,2, is strictly increasing. 1 +  , n = n  1.17.7 Let ak > 0 . Use the /Cauchy-Bunyakovsky-Schwarz Inequality to  n  n 1 show that  ∑ ak2   ∑ 2 = k =   k 1 ak

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 2 ≥n . 

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162 • Multivariable and Vector Calculus 2/E  1.17.8 Let a ∈  n be a fixed vector. Demonstrate that c= 0} is a subspace of  n . {x ∈  n : a x =  1.17.9 Let ai ∈  n ,1 ≤ i ≤ k, k ≤ n , be k non-zero vectors such that   ai  a j = 0 for i ≠ j . Prove that these k vectors are linearly independent. 1.17.10 Let ak ≥ 0, 1 ≤ k ≤ n be arbitrary. Prove that n ( n + 1 )( 2 n + 1 ) n ak2  n  . ∑  ∑ ak  ≤ 2 6 = k 1 k  k 1=  2

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CHAPTER

2

Differentiation

Based on the understanding of the concepts of vectors and parametric curves from the previous chapter, in this chapter, we focus on the differentiation of functions of several variables. We mainly discuss some topology, multivariable functions, limits and continuity, the definition of the derivative, the Jacobi matrix, gradients and directional derivatives, Levi-Civitta and Einstein, exterma, and Lagrange multipliers.

2.1 SOME TOPOLOGY Definition 2.1.1  Let a ∈  n and let e > 0 . An open ball centered at a of radius e is the set Be ( a ) = {x ∈  n : x − a < e }. An open box is a Cartesian product of open intervals

]a1 ; b1 [ × ]a2 ; b2 [ ×  × ]an−1 ; bn−1 [ × ]an ; bn [ , where ak , bk are real numbers. EXAMPLE 2.1.1 An open ball in  is an open interval, an open ball in  2 is an open disk (see Figure 2.1.1) and an open ball in  3 is an open sphere (see Figure 2.1.2). An

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164 • Multivariable and Vector Calculus 2/E

FIGURE 2.1.1  Open ball in 2 .

FIGURE 2.1.2  Open ball in  3 .

FIGURE 2.1.3  Open rectangle in 2 .

FIGURE 2.1.4  Open rectangle in  3 .

open box in  is an open interval, an open box in  2 is a rectangle without its boundary (see Figure 2.1.3) and an open box in  3 is a box without its boundary (see Figure 2.1.4). Definition 2.1.2  A set S ⊆  n is said open if for very point belonging to it we can surround the point by a sufficiently small open ball so that at this ball lay completely within the set. That is, ∀a ∈ S, ∃e > 0 such that Be ( a ) ⊆ S .

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Differentiation • 165

EXAMPLE 2.1.2 The region ]−1;1[ is open in  . The interval ]−1;1] is not open; however, as no interval centered at 1 is totally contained in ]−1;1] . EXAMPLE 2.1.3 The region ]−1;1[ × ]0; +∞[ is open in  2 . EXAMPLE 2.1.4

The ellipsoidal region {( x, y ) ∈  2 : x2 + 4 y2 < 4} open in  2 . You will recognize that open boxes, open ellipsoids and their unions, and finite intersections are open sets in  n . Definition 2.1.3  A set S ⊆  n is said closed in  , as its complement,  \ S is open. EXAMPLE 2.1.5

[ −1; 1] is closed in  , as its complement,  \ [ −1;1] = ]−∞; −1[ ∪ ]1; ∞[ is open in  . However, the interval ]−1;1] is nei-

The closed interval

ther open nor closed in  . EXAMPLE 2.1.6 The region [−1; 1] × [0; +1[ × [0; 2] is closed in  3 . Definition 2.1.3  A point P in a set S ⊂  n is called an interior point of S if and only if there is some open ball with center P, which contains only points of S. NOTE

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A set is called an open set if and only if all its points are interior points.

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166 • Multivariable and Vector Calculus 2/E

EXERCISES 2.1 2.1.1

Determine whether the following sub-sets of  2 are open, closed, or neither, in  2 . 1. = A 2. = B

{( x, y) ∈  {( x, y) ∈ 

2

: 2 ≤ x ≤ 4,1 ≤ y ≤ 5}

2

: 0 ≤ x2 + y2 ≤ 4}

  1  3. C= ( x, y ) ∈  2 : x > 0, y < sin     x   5. = E

{( x, y) ∈  {( x, y) ∈ 

6. = F

{( x, y) ∈ 

4. D=

( a, b) ∈  7. = G

2

: x > 0, y > 0}

2

: x2 + y2 < r 2 } ,for r > 0

2

: ( x − a ) + ( y − b ) < r 2 , for r > 0 and

}

2

2

2

{( x, y) ∈ 

: a < x < b, c < y < d} for a rectangle =

2

( a, b ) , ( c, d )  , where ( a, b ) ∈  2 and ( c,d ) ∈  2

{( x, y) ∈  : 2 ≥ y − 4 x > 1} 9. = I {( x, y ) ∈  : 2 y + x > −1} 10. = J {( x, y ) ∈  : x + y < 5, x − y > 1} 11. K= {( x, y ) ∈  : x + 4 y > 4} 12. L= {( x, y ) ∈  : x + 4 y ≥ 4, x + 4 y ≤ 16} 13. M = {( x, y ) ∈  : x ≥ 2 y, y ≥ 2 x} 14. N = {( x, y ) ∈  : x + y= 1} 8. H =

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

  x 2 y2 15. = L ( x, y ) ∈  2 : + < 1 2 3   16. T = 17. V =

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{( x, y) ∈  {( x, y) ∈ 

2

: x < 1, y < 1}

2

: x ≤ 1, y ≤ 1}

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Differentiation • 167

18. = V

2

: x2 + y ≥ 0}

19.

2

: x = y}

2

: xy > 0}

2

: y = x − 1 + 2 − x}

2

: x2 > y}

20. 21. 22. 2.1.2 HINT

{( x, y) ∈  V = {( x, y ) ∈  V ={( x, y ) ∈  V = {( x, y ) ∈  V ={( x, y ) ∈ 

{( x, y) ∈ 

Show that the set = A 2 in  .

2

: x2 + y2 < r 2 } , for r > 0 is open

If a > 0 and b > 0 , let d > 0 and e be the smaller of a and b, and consider Bd ( a, b ) .

2.1.3

Determine whether the following sub-sets of  3 are open, closed, or neither, in  3 . 1   1. = A ( x, y, z ) ∈  3 : 3 < ( x2 + y2 + z2 ) 2 < 6    3 2 2 2 2. = B {( x, y, z ) ∈  : x + 2 y + 3 z < 6}

3. = C

{( x, y, z ) ∈ 

4. = D

{( x, y, z) ∈ 

: x2 + y2 < 7, z ≡ 0}

3

}

: ( x − a ) + ( y − b ) + ( z − c ) < e 2 with center 2

2

3

2

( a, b, c ) and radius e 5. = E

{( x, y, z) ∈ 

}

: ( x − a ) + ( y − b) + ( z − c ) ≤ e 2

2

3

2

with center

( a, b, c ) and radius e 6. = F 7. = B

{( x, y, z) ∈ 

{( x, y, z ) ∈ 

}

3

: ( x − 1) + ( y − 2 ) + ( z + 3 ) ≤ 5

3

: x2 + y2 + z2 ≤ 5}

2

2

2

2   y2 z 2 3 x 8. = L ( x, y, z ) ∈  : 2 + 2 + 2 ≤ 1 a b c  

9. = V 10. = M

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z 0} {( x, y, z ) ∈  : x + y −= 4 z 0} {( x, y, z ) ∈  : x + y + z −= 3

3

2

2

2

2

2

2

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168 • Multivariable and Vector Calculus 2/E

11. = T

{( x, y, z ) ∈ 

3

: x ≥ 0, y ≥ 0, z ≥ 0}

xy   12. = G ( x, y, z ) ∈  3 :  z  3 13. C = {( x, y, z ) ∈  : xy > z} 14. = B 15. = T 2.1.4

{( x, y, z) ∈  {( x, y, z ) ∈ 

}

3

: ( x − y= ) z2

3

: 1 < x2 + y2 + z2 < 2}

2

Determine whether the following statements are true or false. 1. The set of {( x ) ∈  : 0 < x < 1} is open in  .

2. If a and b > 0, the open rectangle {( x, y ) ∈  2 : 0 < x < a,0 < y < b} is an open set in  2 . 3. The interaction of two open sets is an open set. 4. The union of two closed sets is a closed set. 5. The interaction of two closed sets is a closed set. 6. An open ball is a convex set. 7. A set that contains its boundary is called a closed set. 2.1.5 HINT

Prove that the union of two open sets is an open set.

A set is called an open set if and only if all its points are interior points.

2.1.6

Let p( x, y) be a polynomial with real coefficients in the real variables x and y, defined over the entire plane  2 . What are the possibilities for the image (range) of p( x, y) ?

2.1.7

A set of points of complex number ( z= x + iy ) is given. Determine whether the set is open, closed, or neither. 1. V is the set of all z satisfying z − 2 ≤ z + 5 i . 2. The set M consists of all z with Im ( z ) < 9 . 3. D is the set of all z such that 1 < Re ( z ) ≤ 6 . 4. The set C consists of all z such that Re ( z ) > ( Im ( z ) ) . 2

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Differentiation • 169

5. B is the set of all numbers x + iy with x and y any rational ­numbers 6. A is the set of all z satisfying z − i < 7 . 7. F is the set of all z satisfying z − 2 i < 9 . 8. The set T consists of all z= x + iy with x > 0 . 9. The set K consists of all z= x + iy with x ≥ 0 . 10. The set E consists of all z= x + iy with −1 ≤ x ≤ 1 and y > 0 .

2.2 MULTIVARIABLE FUNCTIONS Let A ⊆  n . For most of this course, our concern will be functions of the form f : A → m . If m = 1 , we say that f is a scalar field. If m ≥ 2 , we say that f is a vector field. We would like to develop a calculus analogous to the situation in  . In particular, we would like to examine limits, continuity, differentiability, and integrability of multivariable functions. Needless to say, the introduction of more variables greatly complicates the analysis. For example, recall that the graph of a function f : A →  m , A ⊆  n is the set

{( x, f (x)) : x ∈ A} ⊆ 

n+ m

.

If m + n > 3 , we have an object of more than three-dimensions. In this case, = n 2,= m 1, we have a tri-dimensional surface. We will now briefly examine this case. Definition 2.2.1  Let A ⊆  2 and let f : A →  be a function. Given c ∈ , the level curve (or contour) at z = c is the curve resulting from the intersection of the surface z = f ( x, y ) and the plane z = c , if there is such a curve. EXAMPLE 2.2.1 The level curves of the surface f ( x, y= ) x2 + y2 (an elliptic paraboloid) are 2 2 the concentric circles x + y = c, c > 0.

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170 • Multivariable and Vector Calculus 2/E EXAMPLE 2.2.2 Sketch the level curve for f ( x, y )=

1 12 − 6 x2 − 2 y2 . 3

Solution: The MapleTm commands to graph this function.

> >

The graph is shown in Figure 2.2.1(a).

FIGURE 2.2.1 (a) Level curve of f ( x , y )=

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1 12 − 6 x 2 − 2 y 2 in 2D with Maple. 3

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Differentiation • 171

>

The three-dimensional plot is presented in Figure 2.2.1(b).

FIGURE 2.2.1  (b) Level curve of f ( x , y )=

1 12 − 6 x 2 − 2 y 2 in 3D with Maple. 3

The MATLAB commands to graph this function. >> syms x y >> f =(1/3∗sqrt(12-(6∗x^2)-2∗y^2)) f= 1/3∗(12-6∗x^2-2∗y^2)^(1/2) >> ezcontour(f, [-2, 2, -2, 2])

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172 • Multivariable and Vector Calculus 2/E The graph is shown in Figure 2.2.2(a).

FIGURE 2.2.2 (a) Level curve of f ( x , y )=

1 12 − 6 x 2 − 2 y 2 in 2D with MATLAB. 3

>> syms x y >> f =(1/3∗sqrt(12-(6∗x^2)-2∗y^2)) f= 1/3∗(12-6∗x^2-2∗y^2)^(1/2) ezsurf(f, [-2, 2, -2, 2]) The surface plot is presented in Figure 2.2.2(b).

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Differentiation • 173

FIGURE 2.2.2  (b) Level curve of f ( x , y )=

1 12 − 6 x 2 − 2 y 2 in surface form with MATLAB. 3

Definition 2.2.2  Let A ⊆  3 and let f : A →  be a function. Given c ∈ , the level surface at w = c is the surface resulting from the intersection of the surface w = f ( x, y, z ) and the plane w = c , if there is such a surface. In other words, the level surface of a function w = f ( x, y, z ) is the surface in a rectangular three variables coordinate system ( xyz ) defined by f ( x, y, z ) = c , where c is any constant. EXAMPLE 2.2.3 Sketch the level surface for f ( x, y, z ) =x2 + 4 y2 + 16 z2 . Solution: The MapleTm commands to graph this function.

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174 • Multivariable and Vector Calculus 2/E

> > >

FIGURE 2.2.3  Level surface f ( x , y , z ) =x 2 + 4 y 2 + 16 z 2 .

EXERCISES 2.2 2.2.1

Sketch the level curves in 2D and 3D views for the following maps. 1. 2. 3. 4. 5. 6.

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( x, y)  x + y ( x, y)  xy ( x, y)  x3 − y ( x, y)  x2 + 4 y2 ( x, y)  y2 − x2 ( x, y)  min ( x , y )

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Differentiation • 175

12.

( x, y)  sin ( x2 + y2 ) ( x, y)  cos ( x2 − y2 ) ( x, y)  5 − x2 − y2 ( x, y)  ye x ( x, y)  sin x sin y ( x, y)  ln ( x2 + y2 − 1)

13.

( x, y)  tan −1 

7. 8. 9. 10. 11.

14. 15. 2.2.2

y    x +1

( x, y)  x2 / 3 + y2 / 3 ( x, y)  ( x + 1)2 + y2

Sketch the level surfaces for the following maps.

7.

( x, y, z )  x + y + z ( x, y, z )  xyz ( x, y, z )  min ( x , y , z ) ( x, y, z )  x2 + y2 ( x, y, z )  x2 + 4 y2 ( x, y, z )  sin ( z − x2 − y2 ) ( x, y, z )  x2 + y2 + z2

8.

( x, y, z )  cos−1

9.

( x, y, z )  x2 + y2 − z

1. 2. 3. 4. 5. 6.

y− z y+ z

 x+ z  1−y 11. ( x, y, z )  ln ( x − 2 y − 3 z + 4 )

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10.

( x, y, z )  sin 

12.

( x, y, z )  tan −1 

 x+ z   y 

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176 • Multivariable and Vector Calculus 2/E

2.2.3

Describe geometrically how a surface z = g ( x, y ) would have to be transformed in order to obtain each of the following surfaces z = f ( x, y ) , where is 1. f= ( x, y) g ( x, y) + 2 2. f ( x, y ) = 2 g ( x, y ) 3. f ( x, y ) = − g ( x, y ) 4. f ( x, y )= 2 − g ( x, y ) 5. f ( x, y= ) g ( − x, y) 6. f ( x, y ) = g ( 2 x, y ) 7. f ( x, y ) =− g ( − x, − y )

2.2.4

Let v(t) be a strictly increasing function of t , and let f ( x, y ) = v ( g ( x, y ) ) . How are the level curves of g( x, y) and f ( x, y) related?

2.3 LIMITS AND CONTINUITY We start this section with the notion of limit. Definition 2.3.1  A function f :  n →  m is said to have a limit L ∈  m at a ∈  n if ∀e > 0∃d > 0 such that 0 < x − a < d ⇒ f (x) − L < e . In such a case, we write lim f (x) = L. x →a

The notions of infinite limits, limits at infinity, and continuity at a point are analogously defined. Limits in more than one dimension are perhaps trickier to find, as one must approach the test point from infinitely many directions. EXAMPLE 2.3.1 x2 y Find lim . ( x, y) →( 0,0 ) x 2 + y2

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Differentiation • 177

Solution: We use the sandwich theorem. Observe that, 0 ≤ x2 ≤ x2 + y2 , and so x2 0≤ 2 ≤ 1 . Thus x + y2 lim 0 ≤

( x, y) →( 0,0 )

lim

( x, y) →( 0,0 )

x2 y ≤ lim y , x2 + y2 ( x, y)→( 0,0 )

And hence x2 y = 0. ( x, y) →( 0,0 ) x 2 + y2 lim

The MapleTM commands to graph this surface and find this limit appear in the following. Notice that Maple is unable to find the limit and so the limit unevaluated.

> >

FIGURE 2.3.1  Example 2.3.1 for 3D plot of ( x , y ) 

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x2 y . x2 + y2

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178 • Multivariable and Vector Calculus 2/E

>

EXAMPLE 2.3.2 x 5 y3 Find lim . ( x, y) →( 0,0 ) x 6 + y4 Solution: Either x ≤ y or x ≥ y . Observe that if x ≤ y , then If y ≤ x , then

x 5 y3 y8 ≤ = y4 . x 6 + y4 y4

x 5 y3 x8 ≤ = x2 . x 6 + y4 x 6

Thus x 5 y3 ≤ max ( y4 , x2 ) ≤ y4 + x2 → 0 , x 6 + y4 As ( x, y ) → ( 0,0 ) . Alternative: Let X = x3 , Y = y2 . x 5 y3 X 5/ 3 Y 3 / 2 ≤ . x 6 + y4 X2 + Y2 Passing to polar coordinate = X r= cosq , Y r sin q , we obtain x 5 y3 X 5/ 3 Y 3 / 2 = = r 5/ 3 + 3 / 2 − 2 cosq x 6 + y4 X 2Y2

5/ 3

sin q

3/2

≤ r 7 / 6 → 0,

as ( x, y ) → ( 0,0 ) .

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Differentiation • 179

The MapleTM commands to graph this surface appear as follows.

> >

FIGURE 2.3.2  Example 2.3.2 for 3D plot of ( x , y ) 

x5 y3 x6 + y4

EXAMPLE 2.3.3 Find

lim

( x, y) →( 0,0 )

1+ x+ y . x2 − y2

Solution: When y = 0 , 1+ x → +∞, x2

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180 • Multivariable and Vector Calculus 2/E As x → 0. When x = 0 , 1+ y → −∞, − y2 As y → 0. The limit does not exist. The MapleTM commands to graph this surface are as follows.

> >

FIGURE 2.3.3  Example 2.3.3 for 3D plot of ( x , y ) 

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1+ x + y . x2 − y2

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Differentiation • 181

EXAMPLE 2.3.4 xy6 Find lim . ( x, y) →( 0,0 ) x 6 + y8 Solution: 4 x t= , y t 3 , we find Putting=

xy6 1 = → +∞, 6 8 x +y 2t2 as t → 0 . But when y = 0 , the function is 0. Thus the limit does not exist. The MapleTM commands to graph this surface appear below.

> >

FIGURE 2.3.4  Example 2.3.4 for 3D plot of ( x , y ) 

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xy 6 . x + y8 6

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182 • Multivariable and Vector Calculus 2/E EXAMPLE 2.3.5 Find

lim

( ( x − 1)

( x, y) →( 0,0 )

2

)

+ y2 log e

( ( x − 1)

2

+ y2

x+y

).

Solution: When y = 0, have 2 ( x − 1 ) ln ( 1 − x ) 2

x

−

2x , x

And so the function does not have a limit at ( 0,0 ) . The MapleTM commands to graph this surface appear as follows.

> >

(( x − 1) FIGURE 2.3.5  Example 2.3.5 for 3D plot of ( x , y ) 

2

MVC.CH02_2PP.indd 182

)

+ y 2 log e

(( x − 1)

x+ y

2

+ y2

).

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Differentiation • 183

EXAMPLE 2.3.6 Find

lim

sin ( x 4 ) + sin ( y4 )

( x, y) →( 0,0 )

x 4 + y4

.

Solution: sin ( x 4 ) + sin ( y4 ) ≤ x 4 + y4 , and so sin( x 4 ) + sin(y4 ) x +y 4

4

≤ x 4 + y4 → 0,

as ( x, y ) → ( 0,0 ) . The MapleTM commands to graph this surface appear as follows.

> >

FIGURE 2.3.6  Example 2.3.6 for 3D plot of ( x , y ) 

MVC.CH02_2PP.indd 183

sin ( x 4 ) + sin ( y 4 ) x4 + y4

.

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184 • Multivariable and Vector Calculus 2/E EXAMPLE 2.3.7 Find

sin ( x ) − y . ( x, y) →( 0,0 ) x − sin(y) lim

Solution: When y = 0 , we obtain sin x → 1, x As x → 0. When y = x , the function is identically -1. Thus the limit does not exist. The MapleTM commands to graph this surface appear as follows.

> >

FIGURE 2.3.7  Example 2.3.7 for 3D plot of ( x , y ) 

MVC.CH02_2PP.indd 184

sin ( x ) − y . x − sin( y)

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Differentiation • 185

If f :  2 →  , it may be that the limits

(

)

(

)

lim lim f ( x, y ) , lim lim f ( x, y ) ,

y→ y0

x → x0

x → x0

y→ y0

Both exist. These are called the iterated limits of f as ( x, y ) → ( x0 , y0 ) . The following possibilities might occur. 1. If

lim

( x, y) →( x0 , y0 )

(

y→ y0

)

and lim lim f ( x, y ) exists. x → x0

(

f ( x, y ) exists, then each of the iterated limits lim lim f ( x, y )

y→ y0

(

)

(

x → x0

)

)

2. If the iterated limits exist and lim lim f ( x, y ) ≠ lim lim f ( x, y ) , then lim

( x, y) →( x0 , y0 )

f ( x, y ) does not exist.

lim

NOTE

y→ y0

)

x → x0

f ( x, y ) does not exist.

4. It may occur that does not exist.

lim

( x, y) →( x0 , y0 )

(

x → x0

y→ y0

lim

x → x0

)

y→ y0

f ( x, y ) exists, but one of the iterated limits

If you get two or more different values for along different paths, then

x → x0

lim lim f ( x, y ) = lim lim f ( x, y ) , but that

3. It may occur that ( x, y) →( x0 , y0 )

(

y→ y0

( x, y) →( x0 , y0 )

lim

( x, y) →( x0 , y0 )

f ( x, y ) as approach ( x0 , y0 )

f ( x, y ) does not exist.

EXAMPLE 2.3.8 Show that

lim

( x, y) →( 0,0 )

x−y does not exist. x+y

Solution:

(

)

(

)

We need to show that lim lim f ( x, y ) ≠ lim lim f ( x, y ) . y→ y0

x → x0

x → x0

y→ y0

  x−y x−y So, lim  lim = lim(1) lim(−1) = −1 . = 1 and lim  lim =  x → 0 y→ 0 x + y x →0 y→ 0 x → 0 x + y y→ 0     x−y Thus the iterated limits are not equal and therefore, lim , does not ( x, y) →( 0,0 ) x + y exist.

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186 • Multivariable and Vector Calculus 2/E

NOTE

A function f of two variables x and y is a rule that assigns to each ordered pair of real numbers ( x, y ) in a set of ordered pairs of real numbers (D) a unique real number f ( x, y ) . The set D is the domain of f , and the corresponding set of values for f ( x, y ) is the range of f , that is, { f ( x, y) : ( x, y) ∈ D} . EXAMPLE 2.3.9 Describe the domain of the function 1. f ( x, y= ) x 2 + y2 2. f ( x, y ) = ln xy 3. f ( x, y ) = 4. f ( x, y ) =

x2 + y2 − 16 x x 16 − x − y2 − z2 2

Solution: 1. The entire xy-plane. 2. The set of all points ( x, y ) in the plane for which xy > 0 . This consists of all points in the first and third quadrants. 3. The set of all points ( x, y ) laying on or outside the circle x2 + y2 = 16 except when x equal to zero, that= is D

{(x, y) : x

2

+ y2 − 16 ≥ 0, x ≠ 0} .

4. The set of all points ( x, y, z ) laying inside a sphere of radius 4 that is centered at the origin. Definition 2.3.2  A function f ( x, y ) is continuous at a point ( x0 , y0 ) in an open region R if all the following conditions hold: (a) f ( x0 , y0 ) exists (i.e., ( x0 , y0 ) is in the domain of f. (b) (c)

MVC.CH02_2PP.indd 186

lim

f ( x, y ) exists.

lim

f ( x, y ) = f ( x0 , y0 ) .

( x, y) →( x0 , y0 ) ( x, y) →( x0 , y0 )

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Differentiation • 187

If one or more of these three conditions fail to hold, then f is said to be discontinuous at ( x0 , y0 ) . The function f ( x, y ) is continuous in the open region R if it is continuous at every point ( x, y ) in R. Definition 2.3.3  A function f ( x, y ) defined in a domain D is continuous in D if it is continuous at each point of D. Theorem 2.3.1 If c is a real number and f ( x, y ) and g ( x, y ) are continuous at ( x0 , y0 ) , then the following functions are also continuous at ( x0 , y0 ) : 1. cf ( x0 , y0 ) 2. f ( x0 , y0 ) g ( x0 , y0 ) 3. f ( x0 , y0 ) ± g ( x0 , y0 ) 4. NOTE

f ( x0 , y0 )

g ( x0 , y0 )

, if ( x0 , y0 ) ≠ 0

The polynomial and rational functions are continuous at every point in their domain D.

EXAMPLE 2.3.10 The function f ( x, y ) =

x − 3y , is continuous at every point in its domain, x 2 + y2

which means that f ( x, y ) , is continuous at each point in the xy-plane except at the point ( 0,0 ) .

EXERCISES 2.3

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2.3.1

Sketch the domain of definition of ( x, y )  4 − x2 − y2 .

2.3.2

Sketch the domain of definition of ( x, y )  log( x + y) .

2.3.3

Sketch the domain of definition of ( x, y ) 

2.3.4

Describe the domain and the range of the function.

1 . x 2 + y2

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188 • Multivariable and Vector Calculus 2/E

1. f ( x, y) =

4 − x2 − y2

2. f ( x, y= ) ln(4 − x − y) 5

3. f ( x, y) =

x2 − y

 x 4. f ( x, y) = tan −1    y −1 5. = f ( x, y) cos ( x − y ) 6. f ( x, y, z) =

xy z

 1  + y2 ) sin   .  xy  sin xy 2.3.6 Find lim . ( x, y) →( 0,2 ) x 1 2.3.7 Find lim x2 + y2 sin 2 . ( x, y) →( 0,0 ) x + y2 2.3.5 Find

2.3.8 Find

lim

( x, y) →( 0,0 )

lim

(x

2

( xy − xy + 3 y ) . sin ( x + y ) . 3

2

( x, y) →( −2,1)

2

2.3.9

Find

lim

( x, y) →( 0,0 )

2

3 x 2 + 3 y2 2

2

ex +y − 1 . ( x, y) →( 0,0 ) x 2 + y2 x+ y+ z 2.3.11 Find lim . 2 ( x, y, z ) →( 0,0,0 ) x + y2 + z2 2.3.10 Find

lim

2.3.12 Demonstrate that x2 y2 z2 = 0. ( x, y, z ) →( 0,0,0 ) x 2 + y2 + z2 lim

MVC.CH02_2PP.indd 188

2.3.13 Prove that

x3 y = 0. ( x, y) →( 0,0 ) x 2 + y4

2.3.14 Prove that

2 xy2 = 0. ( x, y) →( 0,0 ) x 2 + y2

lim

lim

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Differentiation • 189

2.3.15 Show that

xy does not exist by considering the line ( x, y) →( 0,0 ) x + y2 lim

2

y = x as one path and the x-axis to the origin another. 2.3.16 Let f :  2 →  be such that f ( x, y ) =

x for y ≠ 0 . Show that y

f ( x, y ) is discontinuous at any point ( x0 ,0 ) ∈  2 . 2.3.17 Show that if g( x, y) is continuous or discontinuous where  x2 − y2  g( x, y) =  x2 + y2  0

( x, y) ≠ ( 0,0 )

.

( x, y) = ( 0,0 )

2.3.18 Describe the largest set L on which the f is continuous. 1. f ( x, y ) =

x−y+3

2. f ( x, y ) =

y + 5 xy + x2 x − y2 2

2.3.19 For what c will the function  1 − x2 − 4 y2 , if x2 + 4 y2 ≤ 1, f ( x, y )  c, if x2 + 4 y2 >1  be continuous everywhere on the xy-plane?

2.4 DEFINITION OF THE DERIVATIVE Before we begin, let us introduce some necessary notation. Let f :  →  be a function. We write f ( h ) = o ( h ) if f ( h ) goes faster to 0 than h , that is, f ( h) = 0 . For example, h3 + 2 h2 = o ( h ) , since h →0 h h3 + 2 h2 lim h 0. = lim h2 + 2= h →0 h →0 h We now define the derivative in the multidimensional space  n. Recall that in one variable, a function g :  →  is said to be differentiable at x = a if lim

the limit

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190 • Multivariable and Vector Calculus 2/E g ( x) − g ( a) = g′ ( a ) x→ a x−a exists. The limit condition above is equivalent to saying that lim

g ( x ) − g ( a ) − g′ ( a )( x − a ) = 0, x→ a x−a

lim or equivalently, lim h →0

g ( a + h ) − g ( a ) − g′ ( a )( h ) = 0. h

We may write this as g ( a + h) − = g ( a ) g′ ( a )( h ) + o ( h ) . The preceding analysis provides an analogue definition for the higher-dimensional case. Observe that since we may not divide by vectors, the corresponding definition in higher dimensions involves quotients of norms. Definition 2.4.1  Let A ⊆  n . A function f : A →  m is said to be differentiable at a ∈ A if there is a linear transformation, called the derivative of f at a , Da ( f ) :  n →  m such that lim x →a

f ( x ) − f ( a ) − Da ( f ) ( x − a ) x −a

= 0.

Equivalently, f is differentiable at a if there is a linear transformation Da ( f ) such that f ( a + h ) −= f ( a ) Da ( f ) ( h ) + o ( h ) , as h → 0 . NOTE

The condition for differentiability at a is equivalent to

f (x) −= f ( a ) Da ( f ) ( x − a ) + o ( x − a ) , as x → 0 .

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Differentiation • 191

Theorem 2.4.1  If A is an open set in definition 2.4.1, Da ( f ) is uniquely determined. Proof: Let L :  n →  m be another linear transformation satisfying the Definition 2.4.1. We must prove that ∀v ∈  n , L(v) = Da ( f ) (v) . Since A is open, a + h ∈ A for sufficiently small h . By definition, as h → 0 , we have f ( a + h ) −= f ( a ) Da ( f ) ( h ) + o ( h ) . and f ( a + h ) − f ( a= ) L ( h ) + o ( h ). Now, observe that, Da ( f ) ( v ) −= L ( v ) Da ( f ) ( h ) − f ( a + h ) + f ( a ) + f ( a + h ) − f ( a ) − L ( h ) . By the triangle inequality, Da ( f ) ( v ) − L ( v ) ≤ Da ( f ) ( h ) − f ( a + h ) + f ( a ) + f ( a + h ) − f ( a ) − L ( h ) = o( h ) + o( h

)

= o( h ), as h → 0 . This means L(v) − Da ( f ) (v) → 0, i.e., L(v) = Da ( f ) (v) , completing the proof. NOTE

If A = {a} , a singleton, then Da ( f ) is not uniquely determined. For x − a < d holds only for x = a, and so f ( x ) = f ( a ) . Any linear transformation T will satisfy the definition, as T (x − a= ) T ( 0= || 0 || = 0 , ) 0 , and || f ( x ) − f ( a ) − T(x − a) || = identically.

EXAMPLE 2.4.1 If L :  n →  m is a linear transformation, then Da ( L ) = L , for any a ∈  n .

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192 • Multivariable and Vector Calculus 2/E Solution: Since  n is an open set, we know that Da ( L ) uniquely determined. Thus if L satisfies Definition 2.4.1, then the claim is established. But by linearity L ( x ) − L ( a ) − L ( x − a ) =L ( x ) − L ( a ) − L ( x ) + L ( a ) =0 =, 0 hence the claim that follows. EXAMPLE 2.4.2 Let 3 × 3 →  f:   ( x, y )  x y be the usual dot product in  3 . Show that f is differentiable and that       D( x , y ) f h , = k x  y + hk .

(

)

Solution: We have             f x + h, y + k − f ( x, y ) = x + h  y + k − x  y           = x  y + x k + h y + hk − x  y       = x k + h y + hk .

(

)

(

)(

)

    As h , k → 0,0 , we have by the Cauchy–Buniakovskii–Schwarz inequality,      h,k ≤ h k = o h , which proves the assertion.

(

) ( )

( )

It is worth knowing that, just like in the one variable case, differentiability at a point implies continuity at that point. Theorem 2.4.2  Suppose A ⊆  n is open and f : A →  n is differentiable on A . Then f is continuous on A . Proof: Given a ∈ A , we must show that lim f ( x ) = f ( a ) . x →a

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Differentiation • 193

Since f is differentiable at a , we have f (x) −= f ( a ) Da ( f ) ( x − a ) + o ( x − a ) , and so

f ( x ) − f ( a ) → 0,

as x → a , proving the theorem.

EXERCISES 2.4 2.4.1

Let L :  3 →  3 be a linear transformation and 3 → 3 F:   ⋅ x  x × L(x)

Show that F is differentiable and that      Dx ( F ) h = x × L h + h × L ( x ) .   2.4.2 Let f :  n →  , n ≥ 1 , f ( x ) = x be the usual norm in  n , with   x = x  x . Prove that   xv  Dx ( f ) ( v ) =  , x    for x ≠ 0 , but that f is not differential at 0 .

()

()

2.5 THE JACOBI MATRIX We now establish a way that simplifies the process of finding the derivative of a function at a given point. Definition 2.5.1 Let A ⊆  n , f : A →  m , and put  f1 ( x1 , x2 ,, xn )    f2 ( x1 , x2 ,, xn )   . f (x) =       fm ( x1 , x2 ,, xn )  Here f i :  n →  . The partial derivative

MVC.CH02_2PP.indd 193

∂f i ( x ) is defined ∂x j

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194 • Multivariable and Vector Calculus 2/E

(

)

(

)

f i x1 , x2 ,, x j + h,, xn − f i x1 , x2 ,, x j ,, xn ∂f i , ( x ) = lim h →0 h ∂x j whenever this limit exists. To find partial derivatives with respect to the jth variable, we simply keep the other variables fixed and differentiate with respect to the jth variable. EXAMPLE 2.5.1 If f :  3 →  , and f ( x, y, z ) =x + y2 + z3 + 3 xy2 z3 then ∂f ( x, y, z )= 1 + 3 y2 z3 , ∂x ∂f ( x, y, z=) 2 y + 6 xyz3 , ∂y and ∂f z ) 3 z2 + 9 xy2 z2 . ( x, y,= ∂z The MapleTm commands to find these follow.

> > > > Since the derivative of a function f i :  n →  m is a linear transformation, it can be represented by aid of matrices. The following theorem will allow us to determine the matrix representation for Da ( f ) under the standard bases of  n and  m .

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Differentiation • 195

Theorem 2.5.1  Let  f1 ( x1 , x2 ,, xn )    f2 ( x1 , x2 ,, xn )   f (x) = .       fm ( x1 , x2 ,, xn )  Suppose A ⊆  n is an open set and f : A →  m is differentiable. Then each ∂f partial derivative i (x) exists, and the matrix representation for Dx ( f ) with ∂x j respect to the standard bases of  n and  m is the Jacobi matrix  ∂f1  ∂x ( x )  1  ∂f2 (x) f ′ ( x ) =  ∂x1     ∂fn ( x )  ∂x1

∂f1 ∂f ( x )  1 ( x )  ∂x2 ∂xn  ∂f2 ∂f2  (x)  ( x ) ∂x2 ∂xn .      ∂fn ∂fn  x x ( ) ( )  ∂x2 ∂xn

Proof:  Let e j , 1 ≤ j ≤ n be the standard basis for  n . To obtain the Jacobi matrix, we  must compute Dx ( f ) e j , which will give us the jth column of the Jacobi matrix. Let

( ) f ′(x) = ( J ) , and observe that ij

 J1 j  J   2j Dx ( f ) e j =   .       Jnj   and put y= x + e e j , e ∈ . Notice that,

( )

f ( y ) − f ( x ) − Dx ( f ) ( y − x ) y−x  f x1 , x2 ,, x j + h,, xn − f x1 , x2 ,, x j ,, xn − e Dx ( f ) e j

(

)

(

)

e

MVC.CH02_2PP.indd 195

( )

⋅

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196 • Multivariable and Vector Calculus 2/E Since the sinistral side → 0 as e → 0 , the so does the ith component of the numerator, and so,

(

)

(

)

f i x1 , x2 ,, x j + h,, xn − f i x1 , x2 ,, x j ,, xn − e J ij e

→ 0⋅

This entails that = J ij lim e →0

(

)

(

)

f i x1 , x2 ,, x j + e ,, xn − f i x1 , x2 ,, x j ,, xn ∂f i = (x) ⋅ e ∂x j

This finishes the proof. NOTE

Strictly speaking, the Jacobi matrix is not the derivative of a function at a point. It is a matrix representation of the derivative in the standard basis of  n . We will, however, refer to f ′ when we mean the Jacobi matrix of f .

NOTE

We will use the symbol J in the exercises to represent Jacobi determinant which is

J = det [ f ′].

EXAMPLE 2.5.2 Let f :  3 →  2 be given by f ( x,= y)

( xy + yz,log e xy) .

Compute the Jacobi matrix of f . Solution: The Jacobi matrix is the 2 × 3 matrix ∂f1 ∂f1  ∂f1   ∂x ( x, y ) ∂y ( x, y ) ∂z ( x, y )   f ′ ( x, y ) = = ∂f2 ∂f2  ∂f2   ∂x ( x, y ) ∂y ( x, y ) ∂z ( x, y )   

y 1   x

x + z y . 1 0 y 

EXAMPLE 2.5.3 Let f ( r ,q , z ) = ( r cosq , r sin q , z ) be the function, which changes from cylindrical coordinates to Cartesian coordinates. We have

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Differentiation • 197

cosq  f ′ ( r ,q , z ) =  sin q  0

− r sin q r cosq 0

0  0 . 1 

EXAMPLE 2.5.4 Let f ( r ,f ,q ) = ( r cosq sin f , r sin q sin f , r cos f ) be the function, which changes from spherical coordinates to Cartesian coordinates. We have cosq sin f r cosq cos f − r sin f sin q    f ′ ( r ,f ,q ) =  sin q sin f r sin q cos f r cosq sin f  .  cos f  − r sin f 0 The Jacobi matrix provides a convenient computational tool to compute the derivative of a function at a point. Thus differentiability at a point implies that the partial derivatives of the function exist at the point. The converse, however, is not true. EXAMPLE 2.5.5 Let f :  2 →  be given by  y if x = 0, f= x if y 0, ( x, y) = 1 if xy ≠ 0.  Observe that f is not continuous at ( 0,0 ) ( f ( 0,0 ) = 0 but f ( x, y ) = 1 for values arbitrarily close to ( 0,0 ) ), and hence, it is not differentiable there. We have, ∂f ∂f however, = ( 0,0 ) = ( 0,0 ) 1 . Thus even if both partial derivatives exist at ∂x ∂y

( 0,0 ) , there is no guarantee that the function will be differentiable at ( 0,0 ) . You should also notice that both partial derivatives are not continuous at ( 0,0 ) . We have, however, the following.

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198 • Multivariable and Vector Calculus 2/E

Theorem 2.5.2  Let A ⊆  n be an open set, and let f :  n →  m . Put  f1  f  2 f =   . If each of the partial derivatives D j f i exists and is continuous on A ,       fm  then f is differentiable on A . The concept of repeated partial derivatives is akin to the concept of repeated differentiation. Similarly with the concept of implicit partial differentiation, the following examples should be self-explanatory. EXAMPLE 2.5.6 Let f ( u, v, w ) = e u v cos w . Determine

∂2 p  f ( u, v, w ) at  1, −1,  . ∂u∂v 4 

Solution: We have ∂2 ∂ u = e u v cos w ) = e cos w ) e u cos w, ( ( ∂u∂v ∂u e 2 at the desired point. 2

which is

EXAMPLE 2.5.7 The equation z xy + ( xy ) + xy2 z3 = 3 defines z as an implicit function of x and z

y. Find

∂z ∂z and at (1,1,1 ) . ∂y ∂x

Solution: We have

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Differentiation • 199

∂ xy ∂ xy log z z = e ∂x ∂x xy ∂z  xy  =  y log z + z , z ∂x   ∂ ∂ z ( xy) = ezl og xy ∂x ∂x  ∂z =  log xy +  ∂x

z z  ( xy ) , x

∂ 2 3 ∂z xy= z y2 z3 + 3 xy2 z2 , ∂x ∂x Hence, at (1,1,1 ) , we have ∂z ∂z ∂z 1 + 1 + 1 + 3 =0 ⇒ =− ⋅ ∂x ∂x ∂x 2 Similarly, ∂ xy ∂ xy log z z = e ∂y ∂y  xy ∂z  xy =  x log z + z , z ∂y   ∂ ∂ z ( xy) = ezl og xy ∂y ∂y  ∂z =  log xy +  ∂y

z z  ( xy ) , y

∂ 2 3 ∂z = xy z 2 xyz3 + 3 xy2 z2 , ∂y ∂y

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200 • Multivariable and Vector Calculus 2/E Hence, at (1,1,1 ) , we have ∂z ∂z ∂z 3 + 1 + 2 + 3 =0 ⇒ =− ⋅ 4 ∂y ∂y ∂y Just like in the one-variable case, we have the following rules of differentiation. Let A ⊆  n , B ⊆  m be open sets f , g : A →  m ,a ∈ , be differentiable on A , and h : B →  l , be differentiable on B, and f ( A) ⊆ B . Then we have 1. Addition Rule: Dx ( ( f + a g ) ) = Dx ( f ) + a Dx ( g )

(

)

2. Chain Rule: Dx ( ( h  f ) ) = D f (x) ( h )  ( Dx ( f ) ) Since the composition of linear mappings expressed as matrices are matrix multiplication, the Chain Rule takes the alternative form when applied to the Jacobi matrix.

( h  f )′ = ( h′  f )( f ′) . (2.1)

EXAMPLE 2.5.8 Let  ue v    f ( u, v= )  u + v ,  uv     x2 + y h ( x, y ) =  .  y+ z  Find ( f  h )′ ( x, y ) . Solution: We have  ev  f ′ ( u, v ) =  1 v 

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ue v   1 , u 

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Differentiation • 201

And 2 x 1 0  h′ ( x, y ) =  .  0 1 1 Observe also that  ey+ z  f ′ ( h ( x, y ) ) =  1 y + z 

(x

2

+ y ) ey+ z   1 . 2 x + y  

Hence

( f  h )′ ( x, y) = f ′ ( h ( x, y) ) h′ ( x, y)  ey+ z ( x2 + y ) ey+ z    2 x 1 0  1 = 1  0 1 1   2 y + z  x +y    y+ z 2  2 xe 1 + x + y ) ey+ z ( x2 + y ) ey+ z  (   =  2x 2 1 . 2 2 2 xy + 2 xz x + 2y + z x + y    EXAMPLE 2.5.9 Let f :  2 → ,

f ( u, v ) = u2 + e v ,

u, v :  3 → , u ( x, y ) = xz, v ( x, y ) = y + z.  u ( x, y, z )  Put h ( x, y ) = f   . Find the partial derivatives of h .  v ( x, y, z )  Solution:  u ( x, y )   xz  Put g :  3 →  2 , g ( x, y ) =  =   . Observe that h = f  g . Now,  v ( x, y )   y + z 

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202 • Multivariable and Vector Calculus 2/E

 z 0 x g′ ( x, y ) =  , 0 1 1  f ′ ( u, v ) = 2 u e v  , f ′ ( h ( x, y ) ) = 2 xz ey+ z  . Thus  ∂h  ( x, y )  ∂x

∂h ( x, y) ∂y

 ∂h ( x, y) = h′ ( x, y) ∂z 

(

)

= f ′ ( g ( x, y ) ) ( g′ ( x, y ) )

 z 0 x = 2 xz ey+ z    0 1 1  = 2 xz2 ey+ z 2 x2 z + ey+ z  . Equating components, we obtain ∂h ( x, y) = 2 xz2 , ∂x ∂h ( x, y) = ey+ z , ∂y ∂h , y ) 2 x 2 z + e y+ z . ( x= ∂z Under certain conditions, we may differentiate under the integral sign. Theorem 2.5.3 (Differentiation under the integral sign):  Let f :[a, b] × Y →  be a function, with [a, b] being a closed interval, and Y being ∂ f ( x, y ) ∂x f ( x, y ) dy exists as a

a closed and bounded subset of  . Suppose that both f ( x, y) and are continuous in the variable x and y jointly. Then

∫ Y

continuously differentiable function of x on [ a, b] , with derivative ∂ d f ( x, y ) dy = ∫ f ( x, y ) dy. ∫ Y ∂x dx Y

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Differentiation • 203

EXAMPLE 2.5.10 Prove that 2 2 2 F ( x) = p log ∫ log ( sin q + x cos q ) dq = p /2

0

x +1 ⋅ 2

Solution: Differentiating under the integral ∂ x+1 log ( sin 2 q + x2 cos2 q ) dq = p log ⋅ 2 ∂x . p /2 cos2 q = 2 x∫ dq . 0 sin 2 q + x2 cos2 q The preceding implies that p /2

F′ ( x ) = ∫ 0

(x

2

− 1)

2x

( x − 1) cos q ⋅ F′ ( x ) = ∫ sin q + x cos q dq 2

p /2

2

0

=∫

p /2

0

2

2

2

x2 cos2 q + sin 2 q − 1 dq sin 2 q + x2 cos2 q

=

p /2 p dq −∫ 2 2 0 sin q + x2 cos2 q

=

2 p / 2 sec q dq p −∫ 2 0 tan 2 q + x2

=

p 1 tan q − arctan x 2 x

=

p /2

0

p p − , 2 2x

which in turn implies that for x > 0, x ≠ 1 . F ′= ( x)

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p 2x  p p  .  − =  x − 1  2 2x  x + 1 2

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204 • Multivariable and Vector Calculus 2/E

p /2

2 For x = 1 , one sees immediately= that F ′ ( x ) 2= ∫ cos q dq 0

p , agreeing 2

with the formula. Now F ′ ( x= ) p /2

Since = F (1 )

log1dq ∫= 0

p ⇒ F ( x= ) p log ( x + 1) + C. x+1 0, we gather that C = −p log 2 .

Finally thus x +1 . 2 Under certain conditions, the interval of integration in the preceding theorem need not be compact. F= = 2 p log ( x ) p log ( x + 1) − p log

EXAMPLE 2.5.11 Given that

∫

+∞

0

p sin x dx = , compute x 2

∫

+∞

0

sin 2 x dx. x2

Solution: sin 2 ax dx, with gather that a ≥ 0 . Differentiating both sides 0 x2 with respect to a, and making the substitution u = 2 ax . Put I ( a ) = ∫

+∞

+∞

I′ ( a ) = ∫

0

=∫

+∞

=∫

+∞

0

0

=

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2 x sin ax cos ax dx x2 sin 2 ax dx x sin u du u

p . 2

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Differentiation • 205

Integrating each side gives I (= a)

p a + C. 2

Since I(0) = 0 , we gather that C = 0 . The desired integral is I(1) =

p . 2

EXERCISES 2.5 ∂ ∂ f ( x, y ) , fy ( x, y ) = f ( x, y ) , ∂y ∂x

2.5.1 Find f x ( x, y ) = f x ( x, y, z ) =

∂ ∂ f ( x, y, z ) , fy ( x, y, z ) = f ( x, y, z ) , and ∂y ∂x

f z ( x, y, z ) =

∂ f ( x, y, z ) for the following functions: ∂z

1. f ( x, = y)

(x

3

− y2 )

3

1 2. = f ( x, y ) x3 sin   + 5 y2  x  3 xy if ( x, y ) ≠ ( 0,0 )  2 2 3. f ( x, y ) =  ( x + y )  if ( x, y ) = ( 0,0 )  0 4. f ( x, y, z ) = 2.5.2 2.5.3

x 3 − z2 1 + sin 3 y

∂f ∂f ∂f , , and . ∂x ∂y ∂z ∂f ∂f If f :  3 →  , and f ( x, y, z ) = ( x2 + z2 ) log ( x2 y2 + 1) . Find ∂x , ∂y ∂f and . ∂z If f :  3 →  , and f ( x, y, z ) = yx arctan ( zx ) . Find

2.5.4 Let u ( x, y ) and v ( x, y ) are defined by the equations x = u cos v and y = u sin v . Find

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∂u ∂v and . ∂x ∂x

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206 • Multivariable and Vector Calculus 2/E

f ( x, y ) = min ( x, y2 ) . Find

2.5.5 Let f :  2 → , 2.5.6

∂f ( x, y ) ∂x

and

∂f ( x, y ) ∂y

.

Prove that if an equation F ( x, y, z ) = 0 defines an implicit differentiable function f of two variables x and y such that z = f ( x, y ) for all

( x, y) in the domain of f , D, then Fy ( x, y, z ) Fx ( x, y, z ) ∂z ∂z . , = − = − Fz ( x, y, z ) ∂y Fz ( x, y, z ) ∂x 2.5.7 Let w =u3 + u2 v − 3 v, u =sin xy, v =y ln x . Find

∂w ∂w and . ∂y ∂x

2.5.8 Let f :  2 →  2 and g :  3 →  2 be given by  xy2   x − y + 2 z f ( x, y ) =  2  , g ( x, y )  . xy    x y Compute ( f  g )′ (1,0,1 ) , if at all defined. If undefined, explain. Compute ( g  f )′ (1,0 ) , if at all defined. If undefined, explain.  x − y  xy    2.5.9 Let f ( x, y ) =  and g ( x, y ) =  x2 y2  . Find ( g  f )′ ( 0,1 ) .   x + y  x + y 2.5.10 Let z be an implicitly defined function of x and y through the ∂z at (1,1,1 ) . ∂x 2.5.11 Let x = r cosq and y = r sin q . Find the Jacobi matrix f ′ ( r ,q ) and equation ( x + z ) + ( y + z ) = 8 . Find 2

2

the Jacobi determinant J ( r ,q ) . 2.5.12 Let x = e u sin v and y = e u cos v . Find the Jacobi matrix f ′ ( u, v ) and the Jacobi determinant J ( u, v ) . u v and y = 2 . Find the Jacobi matrix f ′ ( u, v ) and 2 u +v u + v2 the Jacobi determinant J ( u, v ) .

2.5.13 Let x =

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2

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Differentiation • 207

2.5.14 Let = y 2 v + u2 w , and z = uvw . Find the Jacobi matrix x u2 + vw , = f ′ ( u, v, w ) and the Jacobi determinant J ( u, v, w ) . 2.5.15 Let the transformation of coordinates x = f ( u, v ) and y = g ( u, v ) is one-to-one. Find the determinant

∂ ( x, y ) ∂ ( u, v )

.

∂ ( u, v ) ∂ ( x, y )

2.5.16 Let f :  3 →  3 be given by f ( r,q ,f ) = ( r sin f cosq , r sin f sin q , r cos f ) . Find the Jacobi matrix and Jacobi determinant. 2 d 1 2 x + t 3 ) dx . ( ∫ dt 0 2.5.18 Suppose g :  →  is continuous and a ∈  is a constant. Find the partial derivatives with respect to x and y of f :  2 → , f ( x, y ) =

2.5.17 Compute

∫

x2 y

a

g ( t ) dt .

2.5.19 Given that

∫

∫

∞

2.5.20 Evaluate

b

0

0

dx 1 b = arctan , evaluate x + a2 a a 2

∫

dx

b

0

(x

2

+ a2 )

2

.

e− ax sin x dx using differentiation under integral sign. x

2.6 GRADIENTS AND DIRECTIONAL DERIVATIVES A function f:

n → m x  f ( x)

is called a vector field. If m = 1 , it is called a scalar field. Definition 2.6.1  Let f:

n →  x  f ( x)

be a scalar field. The gradient of f is the vector defined and denoted by

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208 • Multivariable and Vector Calculus 2/E

 ∂f   ∂x ( x )   1   ∂f  ( x)  ∂ x ∇f ( x ) =  2 .       ∂f ( x )   ∂xn  The graduation operation is the operator  ∂   ∂x   1  ∂  ∇ = ∂x2  .       ∂     ∂xn  Theorem 2.6.1  Let A ⊆  n be open and let f : A →  be a scalar field, and assume that f is differentiable in A . Let K ∈  be a constant. Then ∇f (x) is orthogonal to the surface implicitly defined by f (x) = K . Proof: Let c:

 → n t  c(t)

be a curve lying on this surface. Choose t0 so that c(t0 ) = x . Then c ) ( t0 ) ( f =

f= ( c ( t ) ) K,

and using the chain rule f ′ ( c ( t0 ) ) c′ ( t0 ) = 0, which translates to 0. ( ∇f ( x ) )( c′ ( t0 ) ) =

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Differentiation • 209

Since c′(t0 ) is tangent to the surface and its dot product with ∇f (x) is 0, we conclude that ∇f (x) is normal to the surface. NOTE

Let

q

be

the

angle

∇f (x)

between

∇f ( x ) c′ ( t0 ) cosq , ( ∇f ( x ) )( c′ ( t0 ) ) =

and

c′(t0 ) .

Since

where ∇f (x) is the direction in

which f is changing the fastest. EXAMPLE 2.6.1 Find a unit vector normal to the surface x2 y + 3 xz2 = 8 at the point ( 2,0,1 ) . Solution: Given surface is x2 y + 3 xz2 = 8 , f ( x, y,= z ) x2 y + 3 xz2   ∂  ∂ ∂ 2 x y + 3 xz2 ) i + ( x2 y + 3 xz2 ) j + ( x2 y + 3 xz2 ) k ( ∂x ∂y ∂z    2 2 =( 2 xy + 3 z ) i + ( x ) j + ( 6 xz ) k    ∇f ( 2,0,1 ) = ( 3 ) i + ( 4 ) j + (12 ) k

∇f ( x, y, z ) =

Now, the unit vector normal to the surface x2 y + 3 xz2 = 8 at the point ( 2,0,1 ) is    ∇f ( 2,0,1 ) ( 3 ) i + ( 4 ) j + (12 ) k 3  4  12  i+ j+ k. = = ∇f ( 2,0,1 ) 9 + 16 + 1 26 26 26 EXAMPLE 2.6.2 Find a unit vector normal to the surface x3 + y3 + z = 4 at the point (1,1,2 ) . Solution: Here f ( x, y, z ) = x3 + y3 + z − 4 has gradient  3 x2   2 ∇f ( x, y, z ) = 3 y   1    3    which at (1,1,2 ) is 3  . Normalizing this vector, we obtain 1 

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210 • Multivariable and Vector Calculus 2/E

 3   19     3 .  19     1   19  For example, we can determine the gradient and the unit normal function to 3 xy graph of the function f ( x, y ) = 2 using MapleTm commands as 2 x +y

> >

> >

>

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Differentiation • 211

EXAMPLE 2.6.3 cos( x2 + y2 ) Determine the gradient for the function f ( x, y ) = using MapleTm 2 2 x +y and MATLAB commands. Solution: MapleTm commands:

>

>

MATLAB commands: >> syms x y >> f=(cos(x^2+y^2)/(x^2+y^2)) f= cos(x^2+y^2)/(x^2+y^2) >> gradf=jacobian(f,[x,y]) gradf = [-2∗sin(x^2+y^2)∗x/(x^2+y^2)-2∗cos(x^2+y^2)/(x^2+y^2)^2∗x, -2∗sin(x^2+y^2)∗y/(x^2+y^2)-2∗cos(x^2+y^2)/(x^2+y^2)^2∗y] EXAMPLE 2.6.4 Find the direction of the greatest rate of increase of f ( x, y, z ) = xye z at the point ( 2,1,2 ) .

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212 • Multivariable and Vector Calculus 2/E Solution: The direction is that of the gradient vector. Here  ye z   z  ∇f ( x, y, z ) =  xe   xye z     e2    which at ( 2,1,2 ) comes 2 e2  . Normalizing this vector, we obtain  2 e2    1  1   2 . 5  2  EXAMPLE 2.6.5 Let f :  3 →  be given by f ( x, y, z ) =x + y2 − z2 . Find the equation of the tangent plane to f at (1,2,3 ) . Solution: A vector normal to the plane is ∇f (1,2,3 ) . Now  1    ∇f ( x, y, z ) =  2y   −2 z  which is 1   4  −6  at (1,2,3 ) . The equation of the tangent plane is thus 1 ( x − 1) + 4 ( y − 2 ) − 6 ( z − 3 ) = 0, Or x + 4y − 6z = −9.

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Differentiation • 213

Definition 2.6.2  Let f:

n → n x  f ( x)

be a vector field with  f1 ( x )    f2 ( x )  . f ( x) =        fn ( x )  The divergence of f is defined and denoted by ∂f ∂f ∂f div f ( x ) = ∇ f ( x ) = 1 ( x ) + 2 ( x ) +  + n ( x ) . ∂x1 ∂x2 ∂xn EXAMPLE 2.6.6

(

If f ( x, y, z ) = x2 , y2 , ye z

2

) then div f ( x ) = 2 x + 2 y + 2 yze z . 2

Definition 2.6.3  Let gk :  n →  n ,1 ≤ k ≤ n − 2 be vector fields with gi = ( gi1 , gi 2 ,, gin ) . Then the curl of ( g1 , g2 ,, gn− 2 ) is e1 e2   ∂ ∂  ∂x2  ∂x1  g x g12 ( x ) curl ( g1 , g2 ,, gn− 2 ) ( x ) = det  11 ( )  g21 ( x ) g22 ( x )      g n−2 1 ( x ) g n−2 2 ( x ) ( )  ( ) EXAMPLE 2.6.7

(

)

     

en   ∂  ∂xn  g1 n ( x )  . g1 n ( x )     g( n− 2 ) n ( x ) 

If f ( x, y, z ) = x2 , y2 , ye z , then

MVC.CH02_2PP.indd 213

2

 2 curl f ( x, y, z ) = ∇ × f ( x, y, z ) = e z i.

( )

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214 • Multivariable and Vector Calculus 2/E EXAMPLE 2.6.8 = If f ( x, y, z, w )

e ,0,0, w ) ,g ( x, y, z, w ) (= xyz

2

( 0,0, z,0 ) , then

e2 e3 e4   e1  ∂ ∂ ∂ ∂     ∂x1 ∂x2 ∂x3 ∂x4  = curl ( f , g )( x, y, z, w ) det =  xyz  0 0 w2  e  0 0 0  z 

( xz e ) e . 2 xyz

4

EXAMPLE 2.6.9

    Find the curl of the vector v = 3 xy i+2 z j+x3 k . Solution:    i j  ∂ ∂  ∇× v = det   ∂x ∂y  3 xy 2 z

 k        ∂ = 0 − 2 ) i + ( 3 x2 − 0 ) j + ( 0 − 3 x ) k = −2 i + 3 x2 j − 3 xk. ( ∂z   x3 

Definition 2.6.4  Let A ⊆  n be open and let f : A →  be a scalar field,  and assume that f is differentiable in A . Let v ∈  n \ {0} be such that  x + t v ∈ A for sufficiently small t ∈ . Then the directional derivative of f in  the direction of v at the point x is defined and denoted by  f ( x + tv ) − f ( x ) = Dv f ( x ) lim ⋅ t →0 t Theorem 2.6.2  Let A ⊆  n be open and let f : A →  be a scalar field,   and assume that f is differentiable in A . Let v ∈  n \ 0 be such that   x + t v ∈ A for sufficiently small t ∈ . Then the directional derivative of f in   the direction of v at the point x is given by  ∇f ( x ) v.

{}

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Differentiation • 215

EXAMPLE 2.6.10 Find the directional derivative of f ( x, y, z ) = x3 + y3 − z2 in the direction 1    of 2  . 3  Solution: We have  3 x2   2 ∇f ( x, y, z ) =  3y   −2 z    and so  ∇f ( x, y, z ) v = 3 x2 + 6 y2 − 6 z. EXAMPLE 2.6.11 1   Find the directional derivative of f ( x, y,= z ) xy + yz at the point  −1  in    2    the direction of the vector a = 2 i+ j+ 3 k . 3

2

Solution: ∂f  ∂f  ∂f  i+ j+ k ∇f ( x, y, z ) = ∂x ∂y ∂z    ∇f ( x, y, z = ) y3 i+(3 xy2 + z2 ) j+(2 yz)k    ∇f (1, −1,2 ) = − i+ 7 j − 4 k     j 2i 3k a = + + 4 +1+ 9 14 14 14        2i j 3k  ∇f (1, −1,2 ) + + − i+ 7 j − 4 k  = 14 14   14 2 7 = − + 14 14

(

MVC.CH02_2PP.indd 215

    2i j 3k   + +  14 14   14 12 7 − = − . 14 14

)

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216 • Multivariable and Vector Calculus 2/E

EXERCISES 2.6

( )

2.6.1 Let g ( x, y ) = ( ln x ) ey . Find ∇g ( x, y ) . 2.6.2 Let f ( x, y, z ) = x 4 − xy + z2 . Find ∇f ( x, y, z ) . 2.6.3 Let f ( x, y ) = x3 − xy + y2 . Find ( ∇f ) (1,1 ) . 2.6.4 Let f ( x, y, z ) = xeyz . Find ( ∇f ) ( 2,1,1 ) . 2.6.5 Let f ( x, y, z ) = x2 y sin(yz). Find ∇f ( x, y, z ) .  xz    2.6.6 Let f ( x, y, z ) =  e xy  . Find ( ∇ × f ) ( 2,1,1 ) .  z   2.6.7 If f ( x, y, z ) = x2 ey and k ( x, y, z ) = y2 e xz . Find ∇f ( x, y, z ) , ∇k ( x, y, z ) , and ∇ ( fk ) . Verify that ∇ ( fk ) = f ∇k + k∇f . 2.6.8

Find the point on the surface x2 + y2 − 5 xy + xz − yz = −3 for which the tangent plane is x − 7 y = −6 .

2.6.9

Use a linear approximation of the function f ( x, y ) = e x cos 2 y at ( 0,0 ) to estimate f ( 0.1,0.2 ) .

2.6.10 Find the directional derivative of f ( x, y ) =x2 − 5 xy + 3 y2 at the point ( 2, −1 ) in the direction q = p / 4 . 2.6.11 Prove that the gradient ∇f points in the direction of most rapid increase for f . 2.6.12 Find the angles made by the gradient of f ( x,= y) x

3

+ y at the

point (1,1 ) with the coordinate axes. 2.6.13 Find the directional derivative of f ( x, y ) = 





( x − y) at the point ( x + y)

( 2, −1) in the direction v = 3i + 4j .

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Differentiation • 217

2.6.14 Find the directional derivative of= f ( x, y, z ) z2 arctan ( x + y ) at the 6     point ( 0,0,4 ) in the direction v = 0  . 1   x  y 2.6.15 Find the directional derivative of f ( x, y= , z)   −   at point  y  z P1 (0, −1,2) in the direction from P1 (0, −1,2) to P2 (3,1, −4) . Also determine the direction in which f ( x, y, z) increases most rapidly at P1 (0, −1,2) and find the maximum rate of increase. 2.6.16 Prove that ∇( u × v= ) v ( ∇ × u ) − u( ∇ × v ) .   2.6.17 Let f :  3 →  be a scalar field, and let U, V: 3 →  3 be vector fields. Prove that    1. ∇ f U =f ∇U + U∇f    2. ∇ × f U = f ∇ × U + ∇f × U       3. ∇ U × V = V ∇ × U − U∇ × V

( ) ( ) ( )

2.6.18 Find the tangent plane equation and the normal line equation to the graph of the equation 4 x2 − y2 + 3 z2 = 10 at the point ( 2, −3,1 ) . 2.6.19 Find the points on the hyperboloid x2 − 2 y2 − 4 z2 = 16 at which the tangent plane is parallel to the plane 4 x − 2 y + 4 z = 5. HINT

Parallel planes have proportional gradients.

2.6.20 Find the gradient vector of the function f ( x, y ) = cos p x sin p y + sin 2p y at point ( −1,1 / 2 ) . Then find the equation of the tangent plane.

 2.6.21 In what direction v does f ( x, y ) =1 − x2 − y2 decrease most rapidly at point ( −1,2 ) ?

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218 • Multivariable and Vector Calculus 2/E

2.6.22 Find the equation of the tangent plane to the surface z = xe−2 y at the point (1,0,1 ) .    2.6.23 Verify that ∇ × ∇ × f = ∇ ∇ f − ∇ 2 f for the vector field

(

) (

)

 3 xz2     f=  − yz  .  x + 2 z  

2.6.24 Find the directional derivative of f ( x, y ) =4 x + xy2 − 5 y at the point ( 2, −1 ) in the direction of a unit vector whose angle q with positive x-axis is p / 4 .

2.7 LEVI-CIVITTA AND EINSTE In this section, unless otherwise noted, we are dealing in the space  3 and so, subscripts are not in the set {1,2,3} . Definition 2.7.1 (Einstein’s Summation Convention):  In any expression containing subscripted variables appearing twice (and only twice) in any term, the subscripted variables are assumed to be summed over. NOTE

In order to emphasize that we are using Einstein’s convention, we will enclose any

terms under consideration with  .  . EXAMPLE 2.7.1

 Using Einstein’s Summation convention, the dot product of two vectors x ∈  n  and y ∈  n can be written as   = xy

n

xy ∑= i =1

i i

 xt yt  .

EXAMPLE 2.7.2

    Given that ai , b j , ck , dl are the components of vectors in  3 , a,b,c,d , respectively, what is the meaning of  ai bi ck dk  ?

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Differentiation • 219

Solution: We have = ai bi ck dk 

    3       b ∑ ck d k a b = c d a b = c d a = ∑ ii k k k k 3

=i 1 =i 1





( ab )( cd ).

EXAMPLE 2.7.3 Using Einstein’s Summation convention, the ij −th entry ( AB) ij of the product of two matrices A ∈ Mm×n (  ) and B ∈ Mn×r (  ) can be written as = ( AB) ij

n

Aik Bkj = Ait Bkj   Ait Btj . ∑= k =1

EXAMPLE 2.7.4 Using Einstein’s Summation convention, the trace tr ( A ) of a square matrix A ∈ Mn×n (  ) is tr= ( A)

n

A ∑= t =1

tt

 Att  .

EXAMPLE 2.7.5 Demonstrate, via Einstein’s Summation convention, that if A, B are two n × n matrices, then tr ( AB) = tr ( BA ) . Solution: We have

(

) (

)

(

) (

)

= tr ( AB) tr= Aik Bkj   Atk Bkt , ( AB) ij tr = and = tr ( BA ) tr= Bkj Aik   Btk Akt , ( BA ) ij tr = from where the assertion follows, since the indices are dummy variables and can be exchanged. Definition 2.7.2 (Kroenecker’s Delta):  The symbol d i, j is defined as follows:  0 if i ≠ j d ij =   1 if i = j.

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220 • Multivariable and Vector Calculus 2/E EXAMPLE 2.7.6 It is easy to see that  = d ikd kj 

3

d d ∑= ik kj

k =1

d ij .

EXAMPLE 2.7.7 We have that  d ij= ai b j 

3

3

d ab ∑ = ab ∑∑=

=i 1 =j 1

ij i

j

= k 1

k k

  a b.

Recall that a permutation of distinct objects is a reordering of them. The 3! = 6 permutations of the index set {1, 2, 3} can be classified into even or odd. We start with the identity permutation 123 and say it is even. Now, for any other permutation, we will say that it is even if it takes an even number of transpositions (switching only two elements in one move) to regain the identity permutation and odd if it takes an odd number of transpositions to regain the identity permutation. Since 231 → 132 → 123, 312 → 132 → 123, the permutations 123 (identity), 231, and 312 are even. Since 132 → 123, 321 → 123, 213 → 123, the permutations 132, 321, and 213 are odd. Definition 2.7.3 (Levi-Civitta’s Alternating Tensor):  The symbol e jkl is defined as follows:    0 if  e jkl = −1 if   +1 if  NOTE

{ j, k, l} ≠ {1,2,3} 1 j  1  j

2 3 is an odd permutation k l  2 3 is an even permutation k l 

In particular, if one subindex is repeated we have e= e= e srr . Also, rrs rsr

e= e= e= 1, e= e= e= −1. 123 231 312 132 321 213

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Differentiation • 221

EXAMPLE 2.7.8 Using the Levi-Civitta, alternating tensor and Einstein’s    conven   summation tion, the cross-product can also be expressed, = if i e= e2= , k e3 , 1, j Then    x×y =  e jkl ( ak bl ) e j  . EXAMPLE 2.7.9 If A =  aij  is a 3 × 3 matrix, then, using the Levi-Civitta alternating tensor, det A = e ijk a1 i a2 j a3 k  . EXAMPLE 2.7.10    Let x, y,z be vectors in  3 . Then    x ( y × z= )  xi ( y × z )i=  xie ikl ( yk zl )  .

EXERCISES 2.7 2.7.1

Use the Einstein’s summation convention and the Levi-Civitta’s          alternating tensor to show that x × ( y = × z) (x  z)y − (x  y)z.

2.7.2

Show that ∇ × ( ∇f ) = 0 using the Einstein’s summation convention

2.7.3

and the Levi-Civitta’s alternating tensor.  Show that ∇( ∇ × u ) = 0 using the Einstein’s summation convention

2.7.4

and the Levi-Civitta’s alternating tensor.    Show that ∇ × ( ∇ × u ) = ∇ ( ∇  u ) − ∇ 2 u using the Einstein’s sum-

2.7.5

mation convention and the Levi-Civitta’s alternating tensor.  v 2      Show that v ⋅ ∇v = ∇   + ( ∇ × v ) × v using the Einstein’s summa 2    tion convention and the Levi-Civitta’s alternating tensor.

2.7.6

Write True or false for the following statements: 1. e ijk = −e ikj 2. e ijk= eilm djl dkm − djm dkl

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222 • Multivariable and Vector Calculus 2/E   3. a b = ai bi   4. a × b ≠ e ijk a j bk

(

)

i

2.8 EXTREMA We now turn to the problem of finding maxima and minima for vector functions. As in the one-variable case, the derivative will provide us with information about the extrema, and the “second derivative” will provide us with information about the nature of these extreme points. To define an analog for the second derivative, let us consider the following. Let A ⊆  n and f : A →  m be differentiable on A . We know that for fixed x 0 ∈ A, Dx 0 ( f ) , Dx0 (f) is a linear transformation from  n to  m . This means that we have a function T:

A → l ( n , m ) x → Dx ( f )

,

Where l (  n ,  m ) denotes the space of linear transformation from  n to  m . Hence, if we differentiate T at x 0 again, we obtain a linear transforma-

tion = Dx 0 (T ) D= Dx20 ( f ) from  n to l (  n ,  m ) . Hence, given x 0 (Dx 0 ( f )) Dx20 ( f ) ( x 1 ) ∈ l (  n ,  m ) . Again, this means that given x 2 ∈  n , Dx20 ( f ) ( x 1 )( x 2 )

∈  m , thus the function T:

n × n → l ( n , m )

( x1 , x 2 ) → Dx2 ( f ) ( x1 , x 2 ) 0

is well defined, and linear in each variable x 1 and x 2 , that is, it is a bilinear function. Just as the Jacobi matrix was a handy tool for finding a matrix representation of Dx ( f ) in the natural bases, when f maps into  , we have the following analogue representation of the second derivative. Theorem 2.8.1 Let A ⊆  n be an open set, and f : A →  be twice differentiable on A . Then the matrix of Dx2 ( f ) :  n ×  n →  with respect to the standard basis is given by the Hessian matrix:

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Differentiation • 223

 ∂2 f (x)   ∂x1 ∂x1  ∂2 f (x)  H x f =  ∂x2 ∂x1    2  ∂ f (x)   ∂xn ∂x1

 ∂2 f ∂2 f (x)  ( x ) ∂x1 ∂x2 ∂x1 ∂xn  2  ∂ f ∂2 f (x)  ( x ) ∂x2 ∂x2 ∂x2 ∂xn .       ∂2 f ∂2 f (x)  ( x ) ∂xn ∂x2 ∂xn ∂xn 

EXAMPLE 2.8.1 Let f :  3 →  be given by f ( x, y, z ) = xy2 z3 . Then  0  H( x, y, z) f =  2 yz3  3 y2 z 2 

2 yz3 2 xz3 6 xyz2

3 y2 z 2   6 xyz2  . 6 xy2 z 

From the preceding example, we notice that the Hessian is symmetric, as the ∂2 ∂2 mixed partial derivatives f= f , etc., are equal. This is no coinci∂x∂y ∂y∂x dence, as guaranteed by the following theorem. Theorem 2.8.2  Let A ⊆  n be an open set and f : A →  be twice differentiable on A . If Dx20 ( f ) is continuous, then Dx20 ( f ) is symmetric, that is, ∀ ( x 1 , x 2 ) ∈  n ×  n have

Dx20 ( f ) ( x 1 , x 2 ) = Dx20 ( f ) ( x 2 , x 1 ) . We are now ready to study extrema in several variables. The basic theorems resemble those of one-variable calculus. First, we make some analogous definitions. Definition 2.8.1  Let A ⊆  n be an open set, and f : A →  . If there is some open ball Bx 0 ( r ) on which ∀x ∈ Bx 0 ( r ) , f ( x 0 ) ≥ f ( x ) , we say that f ( x 0 )

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224 • Multivariable and Vector Calculus 2/E is a local maximum of f . Similarly, if there is some open ball Bx1 ( r ) on which

∀x ∈ Bx 0 ( r ′ ) , f ( x 1 ) ≤ f ( x ) we say that f ( x 1 ) is a local minimum of f . A

point is called an extreme point if it is either a local minimum or local maximum. A point t is called a critical point if f is differentiable at t and Dt ( f ) = 0 . A critical point which is neither a maxima nor a minima is called a saddle point. Theorem 2.8.3  Let A ⊆  n be an open set, and f : A →  be differentia-

ble on A . If x 0 is an extreme point, then Dx 0 ( f ) = 0 , that is, x 0 is a critical point. Moreover, if f is twice differentiable with continuous second derivative and x 0 is a critical point such that Hx 0 f is negative definite, then f has a local maximum at x 0 . If Hx 0 f is positive definite, then f has a local minimum at x 0 . If Hx 0 f is indefinite, then f has a saddle point. If Hx 0 f is semi-definite (positive or negative), the test is inconclusive. EXAMPLE 2.8.2 Find the critical points of f:

2 →  ⋅ ( x, y)  x2 + xy + y2 + 2 x + 3 y

and investigate their nature. Solution: We have 2 x + y + 2 

( ∇f ) ( x, y) =  x + 2y + 3 , 



and so to find the critical points, we solve 2x + y + 2 = 0, x + 2y + 3 = 0,

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Differentiation • 225

2 1  1 4 which yields x = − , y = − . Now, H( x, y) f =   , which is positive defi3 3 1 2  2 1   1 4 nite, since ∆1 = 2 > 0 and ∆ 2 = det   = 3 > 0 . Thus x 0 =  − , −  is a 1 2  3 3   7  1 4 relative minimum and we have − = f  − , −  ≤ f ( x, y ) = x2 + xy + y2 3  3 3 +2 x + 3 y. EXAMPLE 2.8.3 Find the extrema of f:

→  ⋅ 2 2 ( x, y, z )  x + y + 3 z2 − xy + 2 xz + yz 3

Solution: We have 2 x − y + 2 z ( ∇f ) ( x, y, z=)  2 y − x + z  , 6 z + 2 x + y which vanishes when x= y= z= 0 . Now,  2 −1 2    Hr f =  −1 2 1  ,  2 1 6   2 −1 which is positive definite, since ∆1 = 2 > 0 and ∆ 2 = det   = 3 > 0 , and  −1 2   2 −1 2    ∆ 3 = det  −1 2 1  = 4 > 0 . Thus f has a relative minimum at ( 0,0,0 ) and  2 1 6  0 = f ( 0,0,0 ) ≤ f ( x, y, z ) = x2 + y2 + 3 z2 − xy + 2 xz + yz.

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226 • Multivariable and Vector Calculus 2/E EXAMPLE 2.8.4 Let f ( x, y ) = x3 − y3 + axy , with a ∈  a parameter. Determine the nature of the critical point of f . Solution: We have  3 x2 + ay  0  2 ax. − ay, 3 y2 = = 0  ⇒ 3 x = 2 − + 3 y ax    

( ∇f ) ( x, y) = 

If a = 0 , then x= y= 0 and so ( 0,0 ) is a critical point. If a ≠ 0 , then 2

 y2  33  = − ay ⇒ 27 y4 = − a2 y a   0 ⇒ y ( 27 y3 + a3 ) =

⇒ y ( 3 y + a ) ( 9 y2 − 3 ay + a2 ) = 0 a ⇒ y =0 or y =− ⋅ 3 2

a a 3  a If y = 0 x = 0 , so again ( 0,0 ) is a critical point. If y = − x = ×  −  = so 3 a  3 3 a a  , −  is a critical point. 3 3 Now, a  6 x 2 H f ( x, y) =   ⇒ ∆1 = 6 x, ∆ 2 = −36 xy − a . − a y 6   a a At ( 0,0 ) , ∆1 =0, ∆ 2 =− a2 . If a ≠ 0 , then there is a saddle point. At  , −  , 3 3 a a 2 ∆1= 2 a, ∆ 2= 3 a , hence  , −  will be a local minimum if a > 0 and a local 3 3 maximum if a < 0 . We can use the MapleTm commands to find the Hessian of functions. 5 xy For example, to find the Hessian H( x, y) f for f ( x, y ) = 2 , we obtain x + y2

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Differentiation • 227

> >

EXERCISES 2.8 2.8.1

Determine the critical points of f ( x, y ) = xy − x − y .

2.8.2

Determine the nature of the critical points of 2 f ( x, y ) = x 4 + y4 − 2 ( x − y ) .

2.8.3

Determine the nature of the critical points of f ( x, y, z )= 4 x2 z − 2 xy − 4 x2 − z2 + y.

2.8.4

Find the extreme of f ( x, y, z ) = x2 + y2 + z2 + xyz.

2.8.5

Find the extreme of f ( x, y, z ) = x2 y + y2 z + 2 x − z.

2.8.6

Determine the nature of the critical points of f ( x, y, z )= 4 xyz − x 4 − y4 − z2 .

2.8.7

Determine the nature of the critical points of f ( x, y= , z ) xyz ( 4 − x − y − z ) .

2.8.8

Determine the nature of the critical points of 2 2 2 g ( x, y, z ) = xyze− x − y − z . x2 + y

2.8.9 Let f ( x, y ) = ∫ 2

y −x

g ( t ) dt , where g is a continuously differentiable

function defined over all real numbers and = g ( 0 ) 0, g′ ( 0 ) ≠ 0 . Prove that ( 0,0 ) is a saddle point of f .

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228 • Multivariable and Vector Calculus 2/E

 144 − 16 x2 2.8.10 Find the minimum of F ( x, y ) =( x − y ) +  − 4 − y2  3  for −3 ≤ x ≤ 3, −2 ≤ y ≤ 2 . 2

2

  ,  

2.8.11 Find the extreme of f ( x, y ) = x2 − 3 xy − y2 + 2 y − 6 x . 2.8.12 Find the extreme of f ( x, y ) =4 x3 − 2 x2 y + y2 . 2.8.13 Find the extreme of f ( x, y ) =5 + 4 x − 2 x2 + 3 y − y2 . 2.8.14 Find the extreme of f ( x, y ) =

x . ( x + y)

2.8.15 Show that the critical points Pc for a function g ( x, y ) correspond to points P on the graph of g where the normal is vertical. 2.8.16 Find the critical points of f ( x, y) = x2 + 2 x + xy + 2 y + 1 . 2.8.17 Find the maximum and minimum of f ( x, y, z= ) xy + yz on the set of x points which satisfy y2 = 1 − x2 and y = . z 2.8.18 Find the lowest and highest points on the ellipse of intersection of the x2 + y2 = 1 (cylinder) and the plane x + y + z =. 1 2.8.19 Let a point P within a triangle in which the sum of the squares of the distances to the sides is a minimum. Determine this minimum in terms of the lengths of sides and area. 2.8.20 1 f x = 8 x3 − 2 x = 0 ⇒ x ( 8 x2 − 2 ) = 0 ⇒ x = 0 or x = ± , 2 fy = 6 y = 0 ⇒ y = 0;  f xx D =det   fyx

f xy  24 x2 − 2 0   −2 0  = ⇒ D ( 0,0 ) det  det     =−12 < 0 ⇒ fyy  6  0 6  0

So ( 0,0 ) is saddle point. 4 0  1   1  D  ± ,0  det  = 24 > 0 ⇒  ± ,0  is local minimum point.   2   2  0 6 

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Differentiation • 229

2.9 LAGRANGE MULTIPLIERS In some situations, we wish to optimize a function given a set of constraints. For such cases, we have the following. Theorem 2.9.1  Let A ⊆  n and let f : A →  , g : A →  be functions whose respective derivatives are continuous. Let g ( x 0 ) = c0 and let S = g −1 ( c0 )

be the level set for g with value c0 , and assume ∇g ( x 0 ) ≠ 0 . If the restriction of f to S has an extreme point at x 0 , then ∃l ∈  such that ∇f ( x 0 ) = l ∇g ( x 0 ) . NOTE

Theorem 2.9.1 only locates extrema, it does not say anything concerning the nature of the critical points found.

EXAMPLE 2.9.1 Optimize f :  2 → , f ( x, y ) = x2 − y2 given that x2 + y2 = 1. Solution: Let g( x, y) = x2 + y2 − 1 . We solve  x  x ∇f   = l ∇g   for x, y, l . This requires  y  y  2 x   2 xl   −2 y = 2 yl  .     From 2 x = 2 xl , we get either x = 0 or l = 1 . If x = 0 , then y = ±1 and l = −1 . If l = 1 , then y = 0 , x = ±1 . Thus the potential critical points are ( ±1,0 ) and 1 , then ( 0, ±1) . If x2 + y2 = f ( x, y = ) x2 − (1 − x2 =) 2 x2 − 1 ≥ −1, and f ( x, y ) =1 − y2 − y2 =1 − 2 y2 ≤ 1. Thus ( ±1,0 ) are maximum points and ( 0, ±1 ) are minimum points.

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230 • Multivariable and Vector Calculus 2/E EXAMPLE 2.9.2 Find the maximum and the minimum points of f ( x, y= ) 4 x + 3 y, subject to the constraint x2 + 4 y2 = 4 , using Lagrange multipliers. Solution: Putting g ( x, y ) =x2 + 4 y2 − 4 , we have 4 2 x  ∇f ( x, y ) = l ∇g ( x, y ) ⇒   = l  . 3  8 y Thus = 4 2= l x, 3 8l y , clearly then l ≠ 0 . Upon division, we find x 16 = . Hence y 3 x2 + 4 y2 = 4⇒

256 2 3 16 y + 4 y2 = 4⇒y= ± ,x = ± ⋅ 9 73 73

The maximum is clearly then  16   3  4 73 ,  + 3 =  73   73  and the minimum is − 73 . EXAMPLE 2.9.3 Let a > 0, b > 0, c > 0 . Determine the maximum and minimum values of f ( x, y ) =

x y z x 2 y2 z 2 1. + + and the ellipsoid 2 + 2 + 2 = a b c a b c

Solution: We use Lagrange multipliers. Put g ( x, y ) =

x2 y2 z2 + + − 1 . Then a2 b2 c2

 2 x / a2  1 / a      ∇f ( x, y, z ) = l ∇g ( x, y, z ) ⇔ 1 / b = l 2 y / b2  .  2 z / c2  1 / c   

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Differentiation • 231

a b c x 2 y2 z 2 It follows that l ≠ 0 . Hence = x = ,y = ,z . Since 2 + 2 + 2 = 1, 2l 2l 2l a b c 3 3 we deduce =1 l = ± . Since a, b, c are positive, f will have a maxi2 4l 2 mum when all x, y,z are positive and a minimum which all x, y,z are negative. Thus the maximum is when a = ,y 3

= x

b = ,z 3

c , 3

and 3 = 3 3

f ( x, y, z ) ≤ and the minimum is when

a b c x= − ,y = − ,z = − , 3 3 3 and f ( x, y, z ) ≥ −

3 =− 3 ⋅ 3

Alternative Method: Using the CBS Inequality 1/ 2

 x2 y2 z2  y x z ⋅1 + ⋅1 + ⋅1 ≤  2 + 2 + 2  a b c b c  a =

(1

x a

2

+ 12 + 12 )

1/ 2

y b

z c

(1 ) 3 ⇒ − 3 ≤ + + ≤ 3 ⋅

EXAMPLE 2.9.4 Let a > 0, b > 0, c > 0 . Determine the maximum volume of the parallelepiped with sides parallel to the axes that can be enclosed inside the ellipsoid x 2 y2 z 2 1. + + = a2 b2 c2

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232 • Multivariable and Vector Calculus 2/E Solution: Let 2 x,2 y,2 z , be the dimensions of the box. We must maximize f ( x, y, z ) = 8 xyz x2 y2 z2 subject to the constraint g ( x, y, z ) = 2 + 2 + 2 − 1 . Using Lagrange a b c multipliers,  8 yz    ∇f ( x, y, z ) = l ∇g ( x, y, z ) ⇔  8 xz  8 xy  2 x / a2  y x z   yz l 2 ,4= xz l 2 ,4= xy l 2 . = l 2 y / b2  ⇒ 4= a b c 2   2z / c  Multiplying the first inequality by x, the second by y, the third by z , and adding, 4 xyz = l

 x 2 y2 z 2 y2 x2 z2 ,4 xyz = l ,4 xyz = l ⇒ 12 xyz = l  2+ 2+ 2 a2 b2 c2 b c a

  = l. 

Hence y2 l x2 z2 = l= l = l ⋅ 3 a2 b2 c2 If l = 0 , 8 xyz = 0 , which minimizes the volume. If l ≠ 0 , then = x

a = , y 3

b = , z 3

c , 3

and the maximum value is 8 xyz ≤ 8

abc ⋅ 3 3

Alternative Method: Using the AM-GM Inequality

2

2 2 1/ 3

⇒ 8 xyz ≤

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1/ 3

 x 2 y z2  ⋅ 2⋅ 2 2 a b c 

yz ) (x =

( abc )2 / 3  8

3 3

2

x2 y2 z2 + 2+ 2 1 2 / 3 a2 b c ≤ ( abc ) ⋅ = 3 3

( abc ) ⋅

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Differentiation • 233

EXERCISES 2.9 2.9.1

A closed box (with six outer faces) has fixed surface area of S square units. Find its maximum volume using Lagrange multipliers. That is, subject to the constraint 2 ab + 2 bc + 2 ca = S , you must maximize abc .

2.9.2

Consider the problem of finding the closest point P′ on the plane Π : ax + by + cz = d, a, b, c non-zero constants with a + b + c ≠ d to the point P (1, 1, 1 ) . In this exercise, you will do this in three essentially different ways. 1. Do this by a geometric argument, arguing the point P′ closest to P on Π is on the perpendicular passing through P and P′ . 2. Do this by means of Lagrange multipliers, by minimizing a suitable function f ( x, y, z ) subject to the constraint g ( x, y, z ) = ax + by + cz − d . 3. Do this considering the unconstrained extrema of a suitable d − ax − by   function h  x, y, . c  

2.9.3

Given that x, y are positive real numbers such that x 4 + 81y4 = 36 , find the maximum of x + 3 y .

2.9.4

If x, y, z are positive real numbers such that x2 y3 z =

1 , what is the 62

minimum value of f ( x, y, z) = 2 x + 3 y + z ? 2.9.5

Find the maximum and the minimum values of f ( x, y= ) x 2 + y2 subject to the constraint 5 x2 + 6 xy + 5 y2 = 8.

2.9.6 Let a > 0, b > 0, p > 1 . Maximize f ( x, y= ) ax + by subject to the constraint x p + yp = 1. 2.9.7

Find the extrema of f ( x, y, z) = x2 + y2 + z2 , subject to the constraint ( x − 1)2 + (y − 2)2 + (z − 3)2 = 4.

2.9.8

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Find the axes of the ellipse 5 x2 + 8 xy + 5 y2 = 9.

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234 • Multivariable and Vector Calculus 2/E

2.9.9 Optimize f ( x, y, z) = x + y + z subject to x2 + y2 = 2 , and x + z =. 1 2.9.10 Let x, y be strictly positive real numbers with x + y = 1 . What is the maximum value of x + xy ? 2.9.11 Let a, b be positive real constants. Maximize f ( x, y) = x a e− x yb e− y on the triangle in  2 bounded by the lines x ≥ 0, y ≥ 0, x + y ≤ 1 . 2.9.12 Does there exist a polynomial in two variables with real coefficients p( x, y) such that p( x, y) > 0 for all x and y, and that for all real numbers c > 0 there exists (x0 , y0 ) ∈  2 such that p( x0 , y0 ) = c ? 2.9.13 Determine the maximum volume of the rectangular solid in the first octant with one vertex at the origin and the opposite vertex lying in x y z the plane + + = 1 where k1 , k2 , and k3 are positive conk1 k2 k2 stants, using Language multipliers. HINT

First octant ⇒ ( x ≥ 0, y ≥ 0, z ≥ 0 ) .

2.9.14 Find the local extreme of the function f ( x, y, z ) = x + y + z subject to constraint x2 + y2 + z2 =. 25

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CHAPTER

3

Integration

In this chapter, we focus on differentiation forms, zero-manifolds, one-­ manifolds, closed and exact forms, two-manifolds, change of variables in double integrals, change to polar coordinates, three-manifolds, change of variables in triple integrals, surface integrals, and Green’s, Stokes’, and Gauss’ Theorems.

3.1 DIFFERENTIAL FORMS We will now consider integration in several variables. In order to begin our discussion, we need to consider the concept of differential forms. Definition 3.1.1  Consider n variables x1 , x2 ,, xn in n-dimensional space (used as the names of the axes) and let  a1 j  a  2j a=j   ∈  n , 1 ≤ j ≤ k,       anj  be k ≤ n vectors in  n . Moreover, let { j1 , j2 ,, jk } ⊆ {1,2,, n} be a collection of k sub-indices.

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236 • Multivariable and Vector Calculus 2/E

An elementary k-differential form ( k > 1 ) acting on the vectors a j , 1 ≤ j ≤ k is defined and denoted by  ax j1 1  ax j 1 dx j1 ∧ dx j2 ∧  ∧ dx jk ( a 1 ,a 2 ,,a k ) = det  2     ax jk 1

ax j1 2 ax j2 2  ax jk 2

 ax j1 k   ax j2 k  .     ax jk k 

In other words, dx j1 ∧ dx j2 ∧  ∧ dx jk ( a1 ,a 2 ,,a k ) is the x j1 , x j2 ,, x jk component of the signed k-parallelotope in  n spanned by a 1 ,a 2 ,,a k . NOTE

By virtue of being a determinant, the wedge product ∧ of differential forms has the following properties:

1. Anti-commutative: da ∧ db = −db ∧ da . 2. Linearity: d ( a + b ) = da + db . 3. Scalar homogeneity: if l ∈  , dl a = l da . 4. Associative: ( da ∧ db ) ∧ dc =da ∧ ( db ∧ dc ) , notice that associative does not hold for the wedge product of vectors. NOTE

Anti-commutative yields, da ∧ da = 0.

EXAMPLE 3.1.1 Consider 1   = a  0  ∈ 3 .  −1 Then = dx ( a ) det = (1 ) 1, = = dy ( a ) det ( 0 ) 0, −1, dz ( a ) = det ( −1 ) =

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Integration • 237

Are the (signed) 1-volumes (that is, the length) of the projections of a onto the coordinate axes. EXAMPLE 3.1.2 In  3 , we have dx ∧ dy ∧ dx = 0 , since we have a repeated variable. EXAMPLE 3.1.3 In  3 , we have dx ∧ dz + 5dz ∧ dx + 4dx ∧ dy − dy ∧ dx + 12dx ∧ dx = −4dx ∧ dz + 5dx ∧ dy. NOTE

In order to avoid redundancy, we will make the convention that if a sum of two or more terms have the same differential form up to the permutation of the variables, we will simplify the summands and express the other differential forms in terms of the one differential form whose indices appear in increasing order.

Definition 3.1.2  A 0-differential form in  n is simply a differentiable function in  n . Definition 3.1.3  A k-differential form field in  n is an expression of the form = w

∑

1≤ j1 ≤ j2 ≤≤ jk ≤ n

a j1 j2  jk dx j1 ∧ dx j2 ∧  ∧ dx jk ,

where the a j1 , j2 ,, j k are differentiable function in  n . EXAMPLE 3.1.4 g ( x, y, z, w ) =x + y2 + z3 + w4 is a 0- form in  4 . EXAMPLE 3.1.5 An example of a 1-form field in  3 is w =xdx + y2 dy + xyz3 dz.

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238 • Multivariable and Vector Calculus 2/E EXAMPLE 3.1.6 An example of a 2-form field in  3 is w= x2 dx ∧ dy + y2 dy ∧ dz + dz ∧ dx. EXAMPLE 3.1.7 An example of a 3-form field in  3 is w=

( x + y + z ) dx ∧ dy ∧ dz.

We show now how to multiply differential forms. EXAMPLE 3.1.8 The product of the 1-form fields in  3 = w1 ydx + xdy, w2 = −2 xdx + 2 ydy, is w1 ∧ w2 =

(2x

2

+ 2 y2 ) dx ∧ dy.

Definition 3.1.4  Let f ( x1 , x2 ,, xn ) be a 0-form in  n . The exterior derivative d f of f = df

n

∂f

∑ ∂x i =1

dxi ⋅ i

Furthermore, if = w f ( x1 , x2 ,, xn ) dx j1 ∧ dx j2 ∧  ∧ dx jk is a k-form in  n , the exterior derivative dw of w is the ( k + 1 ) -form = dw df ( x1 , x2 ,, xn ) dx j1 ∧ dx j2 ∧  ∧ dx jk .

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Integration • 239

EXAMPLE 3.1.9 If in  2 , w = x3 y4 , then d ( x3 y4 ) = 3 x2 y4 dx + 4 x3 y3 dy ⋅ EXAMPLE 3.1.10 2 If in = , w x2 ydx + x3 y4 dy , then

= dw d ( x2 ydx + x3 y4 dy ) =

( 2 xydx + x dy) ∧ dx + ( 3 x y dx + 4 x y dy) ∧ dy 2

2

4

3

3

= x2 dy ∧ dx + 3 x2 y4 dx ∧ dy = ( 3 x2 y4 − x2 ) dx ∧ dy. EXAMPLE 3.1.11 Consider the change of variables x =u + v, y =uv . Then d= x du + dv, = dy vdu + udv, hence dx ∧ dy = ( u − v ) du ∧ dv. EXAMPLE 3.1.12 Consider the transformation of coordinates xyz into uvw coordinates given by u = x + y + z, v =

MVC.CH03_3PP.indd 239

y+ z z , w= ⋅ y+ z x+ y+ z

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240 • Multivariable and Vector Calculus 2/E Then du = dx + dy + dz, y z dv= − dy + dz, 2 2 ( y + z) ( y + z) y+ z x x dw = − dx + dy + dz ⋅ 2 2 2 ( x + y + z) ( x + y + z) ( x + y + z) Multiplication gives  y( y + z) zx  du ∧ dv ∧ dw =− −  ( y + z )2 ( x + y + z )2 ( y + z )2 ( x + y + z )2 

+

=

z ( y + z)

( y + z) ( x + y + z) 2

z2 − y2 − zx − xy

( y + z) ( x + y + z) 2

2

2

−

  dx ∧ dy ∧ dz 2 2 ( y + z ) ( x + y + z )  xy

dx ∧ dy ∧ dz.

EXERCISES 3.1 3.1.1

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Match the differential forms types with the elements forms to make the statement true. 1. 0-forms

A.  Surface elements

2. 1-forms

B.  Volume forms

3. 2-forms

C.  Functions forms

4. 3-forms

D.  Line elements

3.1.2

Let w1 = ydx ∧ dz + dx ∧ dt and w2 = ( x + 1) dy ∧ dt . Find w1 ∧ w2 .

3.1.3

Let w = xydx − xydy + xy2 z3 dz . Find the exterior derivative dw .

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Integration • 241

3.1.4

Let w= x2 ( y + z2 ) dx ∧ dy + z ( x3 + y ) dy ∧ dz. Find the exterior derivative dw.

3.1.5

Express the 2-form dx ∧ dy in polar coordinates.

3.1.6

u x2 − y2 and Consider f :  2 →  2 and ( x, y )  ( u, v ) , where = v = 2 xy . Find du ∧ dv in terms of dx ∧ dy .

3.1.7

Let the 1-form, w = fdx + gdy + hdz . Find the exterior derivative dw.

3.1.8

Let the 2-form, w = fdy ∧ dz − gdx ∧ dz + hdx ∧ dy . Find the exterior derivative dw.

3.1.9

Let w =+ ( x z2 ) dx ∧ dy . Find the exterior derivative dw .

3.1.10 Let the 2-form, w =( x + 2 z)dx ∧ dy + ydx ∧ dz and the vectors 1      = v = 5 , r  5  HINT

 = r) w ( v,

∑

1≤ i < j ≤ n

 −1     0  . Find w ( v, r ) .  3 

   Fij ( a ) dxi ∧ dx j ( v, = r)

   dxi ( v ) dxi ( r )   ∑ Fij ( a ) det dx j ( v ) dx j ( r ) . i, j > i  

3.2 ZERO-MANIFOLDS Definition 3.2.1 A 0-dimensional oriented manifold of  n is simply a point x ∈  n , with a choice of the + or − sign. A general oriented 0-manifold is a union of oriented points. Definition 3.2.2  Let M = + {b} ∪ − {a} be an oriented 0-manifold, and let w be a 0-form. Then = ∫ w w ( b ) − w ( a ). M

NOTE

− x has opposite orientation to +x and

∫

−x

MVC.CH03_3PP.indd 241

w = − ∫ w. +x

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242 • Multivariable and Vector Calculus 2/E EXAMPLE 3.2.1 Let M = − {(1,0,0 )} ∪ + {(1,2,3 )} ∪ − {( 0, −2,0 )} be an oriented 0-manifold, and let w =x + 2 y + z2 . Then

∫

M

NOTE

w =−w (1,0,0 ) + w (1,2,3 ) − w ( 0,0,3 ) =− (1 ) + (14 ) − ( −4 ) =17.

Do not confuse, say, − {(1,0,0 )} with − (1,0,0 ) = ( −1,0,0 ) . The first one means that the point (1,0,0 ) is given negative orientation, the second means that ( −1,0,0 ) is the additive inverse of (1,0,0 ) .

EXERCISES 3.2 3.2.1 Let M = − {(1,1,0 )} ∪ + {( 0,2,0 )} ∪ − {(1, −1,2 )} be an oriented

∫

0-manifold, and let w = 3 x − 2 y + z2 . Find

M

w.

3.2.2 Let M = + {(1,2,0 )} ∪ + {(1,0,3 )} ∪ − {( 7, −1,0 )} be an oriented

∫

0-manifold, and let w = −3x + y2 + z . Find

w.

M

3.2.3 Let M = − {(1,1,1 )} ∪ − {( 0, −2,5 )} ∪ − {(1,7,5 )} be an oriented

∫

0-manifold, and let w = 3 x − y + 4 z . Find

M

w.

3.2.4 Let M = + {( −1,2,1 )} ∪ − {( 0,2,1 )} ∪ − {( −3,1,6 )} be an oriented

∫

0-manifold, and let w =−2x2 + y − z . Find

w.

M

3.2.5 Let M = + {( 0,1,0 )} ∪ + {( −2,0,3 )} ∪ + {(1, −5,0 )} be an oriented 0-manifold, and let w = x + y − z2 . Find

∫

w.

M

3.2.6 Let M = − {(1,0,1 )} ∪ + {( −3,1,3 )} ∪ − {(1,2,0 )} be an oriented 0-manifold, and let w = −7 x − y2 + 3 z . Find

∫

M

w.

3.2.7 Let M = + {( 0,0,1 )} ∪ − {(1,3,2 )} ∪ + {(1, −2,1 )} be an oriented 0-manifold, and let w =x − 2 y + z3 . Find

∫

M

w.

3.2.8 Let M = − {(1,1,0 )} ∪ − {(1,3,1 )} ∪ + {( 6,2,0 )} be an oriented 0-manifold, and let w = 3x − y + z . Find

MVC.CH03_3PP.indd 242

∫

M

w.

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Integration • 243

3.3 ONE MANIFOLD Definition 3.3.1  A one-dimensional oriented manifold of  n is simply an oriented smooth curve Γ ∈  n , with a choice of a + orientation if the curve traverses in the direction of increasing t , or with a choice of a − sign if the curve traverses in the direction of decreasing t . A general oriented 1-manifold is a union of oriented curves. NOTE

The curve −Γ has opposite orientation to Γ and

∫

−Γ

w = − ∫ w. Γ

   dx  If f :  2 →  2 and if d r =   , the classical way of writing this is dy   f ∫  dr. Γ

We now turn to the problem of integrating 1-forms. EXAMPLE 3.3.1 Calculate

∫

Γ

xydx + ( x + y ) dy

where Γ is the parabola= y x2 , x ∈ [ −1;2 ] oriented in the positive direction. Solution: We parameterize the curve as= x t= , y t 2 . Then xydx + ( x + y ) dy= t 3 dt + ( t + t 2 ) dt 2=

( 3t

3

+ 2 t 2 ) dt,

hence = ∫w Γ

∫ ( 3t 2

−1

3

+ 2 t 2 ) dt 2

3  2 =  t3 + t4  3 4  −1  =

MVC.CH03_3PP.indd 243

69 ⋅ 4

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244 • Multivariable and Vector Calculus 2/E What would happen if we had given the curve above a different parameterization? First observe that the curve travels from (−1, 1) to (2, 4) on the parabola y = x2 . These conditions are met with parameterization. Then xydx + ( x + y ) dy=

=

=

(

) ( 3

t −1 d

(3 ( 1 2

)

) ((

t −1 +

3

t −1 + 2

(3 ( t

)

( 3

) (

t −1 +

t −1

) )d (

(

t −1

) ) dt,

(

t −1

t −1 + 2

2

t −1

t −1

) )d( 2

t −1

)

2

)

2

hence = ∫w

∫

Γ

1

9

0

2

(3 ( t

)

3

t −1 + 2

) ) dt 2

9

 3 t 2 7 t 3 / 2 5t  =  − + − t 3 2  4 0 =

69 , 4

as before. To solve this problem using MapleTm commands, you may use the following code.

> > NOTE

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It turns out that if two different parameterizations of the same curve have the same orientation, then their integrals are equal. Hence, we only need to worry about finding a suitable parameterization.

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Integration • 245

EXAMPLE 3.3.2 Calculate the line integral

∫

Γ

y sin xdx + x cos ydy,

where Γ is the line segment from ( 0,0 ) to (1,1 ) in the positive direction. Solution: This line has equation y = x , so we choose the parameterization x= y= t . The integral is thus

∫

Γ

1

y sin xdx + x cos ydy = ∫ ( t sin t + t cos t ) dt 0

1

=  t ( sin t − cos t )  0 − ∫ ( sin t − cos t ) dt 1

0

= 2 sin1 − 1, upon integrating by parts. To solve this problem using MapleTm you may use the following code.

> >

EXAMPLE 3.3.3 Calculate the path integral x+y x−y dy + 2 dx 2 2 Γ x +y x + y2

∫

around the closed square Γ = ABCD with A =(1,1 ) , B =(−1,1), C =(−1, −1) , and D= (1, −1) in the direction ABCDA .

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246 • Multivariable and Vector Calculus 2/E Solution: On AB = , y 1,d = y 0 , on BC, x = −1,dx = 0 , on CD, y = −1,dy = 0 , and on DA = , x 1,d = x 0 . The integral is thus

∫ w=∫ Γ

AB

=

∫

−1

1

= 4∫

w+∫ w+∫ w+∫ w BC

CD

DA

1 x +1 1 y+1 −1 y − 1 x −1 dx + ∫ 2 dy + ∫ 2 dx + ∫ 2 dy 2 1 1 1 y +1 − − x +1 y +1 x +1

1

−1

1 dx x +1 2

1

= 4 arctan x −1 = 2p . NOTE

When the integral is along a closed path, like in the preceding example, it is customary to use the symbol rather than . The positive direction of integration is that sense Γ Γ that when traversing the path, the area enclosed by the curve is to the left of the curve.

∫

∫

EXAMPLE 3.3.4 Calculate the path integral

∫

Γ

x2 dy + y2 dx,

where Γ is the ellipse 9 x2 + 4 y2 = 36 traversed once in the positive sense. Solution: Parameterize the ellipse = as x 2 cos = t, y 3 sin t, t ∈ [ 0;2p ] . Observe that when traversing this closed curve, the area of the ellipse is on the left-hand side of the path, so this parameterization traverses the curve in the positive sense. We have

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Integration • 247

= ∫ w Γ

=

∫ ( ( 4 cos t ) ( 3 cos t ) + ( 9 sin t )( −2 sin t ) ) dt 2p

2

0

∫ (12 cos 2p

0

3

t − 18 sin 3 t ) dt

= 0. To solve this problem, using MapleTm you may use the following code.

> > Definition 3.3.2  Let Γ be a smooth curve. The integral

∫ f ( x ) dx Γ

is called the path integral of f along Γ. EXAMPLE 3.3.5 Find ∫ x dx where Γ is the triangle starting at A : ( −1, −1 ) to B : ( 2, −2 ) , and Γ

ending in C : (1,2 ) , see Figure 3.3.1.

FIGURE 3.3.1  Example 3.3.5.

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248 • Multivariable and Vector Calculus 2/E Solution: The lines passing through the given points have equations LAB : y =

−x − 4 3

and LBC : y = −4 x + 6 . On LAB . 2

x 10 dx 1 2 x , ( dx )2 + ( dy )= x 1 +  −  d= 3  3

x dx= x and on LBC

x x 17 dx. ( dx )2 + ( dy )= x 1 + ( −4 )2 d= 2

x dx= x Hence

= ∫ x dx

∫

=

∫

Γ

LAB

2

−1

=

x dx + ∫

LBC

x dx

1 x 10 dx + ∫ x 17 dx 2 3

10 3 17 − ⋅ 2 2

EXERCISES 3.3 3.3.1 Consider

∫

xdx + ydy . Evaluate

C

∫

C

xdx + ydy where C is the

straight line path that starts at ( −1,0 ) goes to ( 0,1 ) and ends at

(1,0 ) , by parameterizing this path. 3.3.2 Consider

∫

xy dx . Calculate

C

∫

C

xy dx where C is the straight line

path that starts at ( −1,0 ) goes to ( 0,1 ) and ends at (1,0 ) , by parameterizing this path. 3.3.3 Evaluate

∫

C

xdx + ydy where C is the semicircle that starts at

( −1,0 ) goes to ( 0,1) and ends at (1,0 ) , by parameterizing this path. 3.3.4 Calculate

∫

C

xy dx where C is the semicircle that starts at ( −1,0 )

goes to ( 0,1 ) and ends at (1,0 ) , by parameterizing this path.

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Integration • 249

3.3.5 Find

∫

Γ

xdx + ydy where Γ is the path shown in Figure 3.3.2,

p p  starting at O ( 0,0 ) going on a straight line to A  4 cos ,4 sin  and 6 6  p p   continuing on an arc a circle to B  4 cos ,4 sin  . 5 5  

FIGURE 3.3.2  Exercises 3.3.5 and 3.3.7.

3.3.6 Solve Exercise 3.3.5 using Maple commands. 3.3.7 Find

∫

Γ

x dx where Γ is the path as shown in Figure 3.3.2.

FIGURE 3.3.2  Exercises 3.3.5 and 3.3.7.

3.3.8 Find

∫

Γ

zdx + xdy + ydz where Γ is the intersection of the sphere

x 2 + y2 + z 2 = 1 and the plane x + y = 1 , traversed in the positive direction. 3.3.9 Solve Exercise 3.3.7 using Maple commands. 3.3.10 Evaluate

∫ (x C

2

+ y2 ) dx + 2 x dy .

1. Where C consists of line segment from (1,2 ) to (1,8 ) and from

(1,8 ) to ( −2,8 ) . 2. Where C is the graph of y = 2 x2 form (1,2 ) to ( −2,8 ) .

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250 • Multivariable and Vector Calculus 2/E

3.3.11 Evaluate

∫ (x C

2

+ y2 ) dx + 2 xy dy along the curve

C : x= t, y= 2 t 2 ,0 ≤ t ≤ 1 , where t = sin u for 0 ≤ u ≤ p / 2 . 3.3.12 Compute the curve integral

∫ (x C

2

− 2 xy ) dx + ( y2 − 2 xy ) dy , along

the parabola y = x2 from ( −2,4 ) to (1,1 ) .

3.4 CLOSED AND EXACT FORMS Lemma 3.4.1 (Poincare Lemma):  If w is a p-differential form of continuously differentiable functions in  n then d ( dw ) = 0. Proof: We will prove this by induction on p. For p = 0 if w = f ( x1 , x2 ,, xn ) then ∂f dxk k =1 ∂xk n

dw = ∑ and dw =

n

k =1

=

=

 ∂f   ∧ dxk  k

∑ d  ∂x

 n  ∂2 f  ∑ ∑  j 1  ∂x j ∂xk k 1= =   n

   ∧ dx j  ∧ dxk    

 ∂2 f ∂2 f −  ∑  ∂xk ∂x j 1≤ j ≤ k ≤ n  ∂x j ∂xk n

  dx j ∧ dxk  

= 0,

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Integration • 251

since w is continuously differentiable and so the mixed partial derivatives are equal. Consider now an arbitrary p-form, p > 0 . Since such a form can be written as

∑

= w

1≤ j1 ≤ j2 ≤≤ jp ≤ n

a j1 j2  jp dx j1 ∧ dx j2 ∧  dx jp ,

where a j1 , j2  jp is the continuous differentiable functions in  n , we have = dw

∑

1≤ j1 ≤ j2 ≤≤ jp ≤ n

da j1 j2  jp dx j1 ∧ dx j2 ∧  dx jp

 n ∂a j1 j2  jp  dxi  dx j1 ∧ dx j2 ∧  dx jp ,  ∑  ∂xi 1≤ j1 ≤ j2 ≤≤ jp ≤ n  i = 1  it is enough to prove that for each summand

=

∑

(

)

d da ∧ dx j1 ∧ dx j2 ∧  dx jp = 0. But

(

)

(

d da ∧ dx j1 ∧ dx j2 ∧  dx jp = dda ∧ dx j1 ∧ dx j2 ∧  dx jp

)

(

+ da ∧ d dx j1 ∧ dx j2 ∧  dx jp

(

)

)

=da ∧ d dx j1 ∧ dx j2 ∧  dx jp , Since dda = 0 from the case p = 0 . But an independent induction argument proves that

(

)

d dx j1 ∧ dx j2 ∧  dx jp = 0, completing the proof. Definition 3.4.1  A differential form w is said to be exact if there is a continuously differentiable function F such that dF = w.

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252 • Multivariable and Vector Calculus 2/E EXAMPLE 3.4.1 The differential form xdx + ydy is exact, since 1  xdx + yd= y d  ( x 2 + y2 )  . 2  EXAMPLE 3.4.2 The differential form ydx + xdy is exact, since ydx + xdy = d ( xy.) EXAMPLE 3.4.3 The differential form y x dx + 2 dy 2 x +y x + y2 2

is exact, since y x 1  dx + 2 dy= d  log e ( x2 + y2 )  . 2 x +y x + y2 2   2

NOTE

Let w = dF be an exact form. By the Poincare Lemma 3.4.1, = dw dd = F 0 . A result of Poincare says that for certain domains (called star-shaped domains) the converse is also true, that is, if dw = 0 on a star-shaped domain then w is exact.

EXAMPLE 3.4.4 Determine whether the differential form = w

(

2 x 1 − ey

(1 + x )

2 2

) dx +

ey dy 1 + x2

is exact.

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Integration • 253

Solution: Assume there is a function F such that dF = w. By the Chain Rule dF =

∂F ∂F dx + dy ⋅ ∂x ∂y

This demands that

(

)

y ∂F 2 x 1 − e , = 2 ∂x (1 + x2 )

∂F ey = ⋅ ∂y 1 + x2 We have a choice here of integrating either the first, or the second expression. Since integrating the second expression (with respect to y) is easier, we find ey + f ( x), 1 + x2 where f ( x ) is a function depending only on x. To find it, we differentiate the F (= x, y )

obtained expression for F with respect to x and find ∂F 2 xey = − + f ′ ( x ). 2 2 ∂x + 1 x ( ) Comparing this with our first expression for f ′( x) =

∂F , we find ∂x

2x

(1 + x2 )

2

,

that is 1 − + c, f ( x) = 1 + x2

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254 • Multivariable and Vector Calculus 2/E where c is a constant. We then take F (= x, y )

ey − 1 + c. 1 + x2

EXAMPLE 3.4.5 Is there a continuously differentiable function such that dF = w= y2 z3 dx + 2 xyz3 dy + 3 xy2 z2 dz ? Solution: We have dw =( 2 yz3 dy + 3 y2 z2 dz ) ∧ dx + ( 2 yz3 dx + 2 xz3 dy + 6 xyz2 dz ) ∧ dy + ( 3 y2 z2 dx + 6 xyz2 dy + 6 xy2 zdz ) ∧ dz = 0, so this form is exact in a star-shaped domain. So put dF =

∂F ∂F ∂F dx + dy + dz = y2 z3 dx + 2 xyz3 dy + 3 xy2 z2 dz. ∂x ∂y ∂z

Then ∂F = y2 z3 ⇒ F= xy2 z3 + a ( y, z ) , ∂x ∂F = 2 xyz3 ⇒ = F xy2 z3 + b ( x, z ) , ∂y ∂F = 3 xy2 z2 ⇒ = F xy2 z3 + c ( x, y ) . ∂z

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Integration • 255

Comparing these three expressions for F , we obtain F ( x, y, z ) = xy2 z3 . We have the following equivalent of the Fundamental Theorem of Calculus. Theorem 3.4.1 Let U ⊆  n be an open set. Assume w = dF is an exact form, and Γ a path in U with starting point A and endpoint B. Then B

dF F ( B) − F ( A ) . ∫=

= ∫w Γ

A

In particular, if Γ is a simple closed path, then

∫ w = 0. Γ

EXAMPLE 3.4.6 Evaluate the integral

∫ x Γ

2

2y 2x dx + 2 dy 2 +y x + y2

where Γ is the closed polygon with vertices = at A = D

0,0 ) , B ( = 5,0 ) , C ( 7,2 ) , (=

3,2 ) , E (1,1 ) , traversed in the order ABCDEA. (=

Solution: Observe that  2x  2y 4 xy 4 xy − d 2 dx + 2 dy  = dy ∧ dx − dx ∧ dy = 0, 2 2 2 2 2 2 2 x +y x +y  (x + y ) ( x + y2 ) and so the form is exact in a start-shaped domain. By virtue of Theorem 3.4.1, the integral is 0. EXAMPLE 3.4.7 Calculate the path integral

∫ ( x Γ

2

− y ) dx + ( y2 − x ) dy,

where Γ is a loop of x2 + y3 − 2 xy = 0 traversed once in the positive sense.

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256 • Multivariable and Vector Calculus 2/E Solution: Since ∂ 2 ∂ x − y ) =−1 = ( y2 − x ) , ( ∂y ∂x the form is exact, and since this is a closed simple path, the integral is 0.

EXERCISES 3.4 3.4.1

Are the following statements true or false? 1. A form α is closed if dα = 0. 2. If w is exact and C is closed, then

3.4.2

∫

C

w = 0.

Are the following statements true or false? 1. Every exact form is closed, since d ( dw ) = 0 . On the other hand, there are closed but not exact forms. 2. A form α is exact if α = dβ for some form β .

3.4.3

Show that the differential form x2 dx + ydy is exact or not exact.

3.4.4

Show that the 1-form w = dq on  2 \ {0} , where q is the polar angle. In standard Cartesian coordinates: w=

xdy − ydx . Is the form exact or not exact. x 2 + y2

3.4.5

Prove that on a rectangular parallelepiped, Π , all closed forms are exact.

3.4.6

Consider the differential 1-form



y x − 2 dx + 2 dy. Defined on Ω =  2 − ( 0,0 ) , and be w= 2 x +y x + y2 closed curve at C:

[0,2p ] → q



2 ( cosq ,sin q ) .

Show that w is exact or not exact. Also, show that w is a closed 1-form.

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Integration • 257

3.5 TWO-MANIFOLDS Definition 3.5.1 A two-dimensional oriented manifold of  2 is simply an open set (region) D ∈  2 , where the + orientation is counter-clockwise and the − orientation is clockwise. A general oriented 2-manifold is a union of open sets. NOTE

The region − D has opposite orientation to D and

∫

−D

w = − ∫ w. D

We will often write

∫ f ( x, y) dA D

where dA denotes the area element. NOTE

In this section, unless otherwise noted, we will choose the positive orientation for the regions considered. This corresponds to using the area form dxdy .

Let U ⊆  2 . Given a function f : D →  , the integral

∫

D

f dA

Is the sum of all the values of f restricted to D. In particular,

∫

D

dA

is the area of D. In order to evaluate double integrals, we need the following. Theorem 3.5.1 (Fubini’s Theorem):  Let= D

[ a; b] × [ c; d ] ,

and let

f : A →  be continuous. Then = ∫ f dA D

( ∫ f ( x, y) dy) dx ∫ ( ∫ f ( x, y) dx) dy ∫= b

d

d

b

a

c

c

a

Fubini’s Theorem allows us to convert the double integral into iterated (­single) integrals.

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258 • Multivariable and Vector Calculus 2/E EXAMPLE 3.5.1

∫[

0;1] × [ 2;3 ]

xydA = ∫

1

0

( ∫ xydy) dx 3

2

  xy2  3  = ∫    dx 0 2 2     1

=

∫

1

0

 9x   − 2 x  dx 2   1

 5 x2  =   4 0 5 = . 4 Notice that if we had integrated first with respect to x we would have obtained the same result:

∫

3

2

(

  x2 y 1  ∫0 xydx dy = ∫2   2   dy 0  1

)

3

3 y  = ∫   dx 2 2  

3

 y2  =   4 2 5 = . 4

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Integration • 259

Also, this integral is “factorable into x and y pieces” meaning that

∫[

0;1] × [ 2;3 ]

xydA =

( ∫ xdx) ( ∫ ydy) 1

3

0

2

 1  5  =     2  2  5 = . 4 To solve this problem using MapleTm, you may use the following code.

> >

To solve this problem using MATLAB, you may use the following code. >> syms x y >> firstint = int(x∗y,x,0,1) firstint = 1/2∗y >> answer = int(firstint,y,2,3) answer = 5/4 EXAMPLE 3.5.2 We have xdy ∫ ∫ ( 2 x ∫ ∫ ( x + 2 y)( 2 x + y) d= 4

1

4

1

3

0

3

0

2

+ 5 xy + 2 y2 ) dxdy

2 5 2  + y + 2 y  dy 3 2   409 = . 12

=

MVC.CH03_3PP.indd 259

∫

4

3

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260 • Multivariable and Vector Calculus 2/E To solve this problem using MapleTm, you may use the following code.

> >

To solve this problem using MATLAB, you may use the following code. >> syms x y >> firstans = int(int((x+2∗y)∗(2∗x+y),x,0,1),y,3,4) firstans = 409/12 In the cases when the domain of integration is not a rectangle, we decompose so that, one variable is kept constant. EXAMPLE 3.5.3 Find ∫ xy dxdy in the triangle with vertices A : ( −1, −1 ) , B : ( 2, −2 ) , C : (1,2 ) . D

Solution: The lines passing through the given points have equations LAB : y = 3x + 1 −4 x + 6, LCA : y = . LBC : y = 2

−x − 4 , 3

Now, we draw the region carefully. If we integrate first with respect to y, we must divide the region as in Figure 3.5.1, because there are two upper lines, which the upper value of y might be. The lower point of the dashed line is (1, −5 / 3).

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Integration • 261

FIGURE 3.5.1  Example 3.5.3, integration order dydx .

The integral is thus

∫

1

−1

x

(∫

( 3 x +1 ) / 2

(− x−4) / 3

)

2

y dy dx + ∫ x 1

(∫

)

11 y dy dx = − . (− x−4) / 3 8 −4 x + 6

If we integrate first with respect to x, we must divide the region as in Figure 3.5.2, because there are two left-most lines, which the left value of x might be. The right point of the dashed line is (7 / 4, −1).

FIGURE 3.5.2  Example 3.5.3, integration order dxdy .

The integral is thus

∫

−1

−2

MVC.CH03_3PP.indd 261

y

(∫

( 6 − y) / 4

−4 − 3 y

)

2

x dx dy + ∫ y −1

(∫

( 6 − y) / 4

( 2 y −1 ) / 3

)

11 x dx dy = − . 8

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262 • Multivariable and Vector Calculus 2/E To solve this problem using MapleTm, you may use the following code.

> >

To solve this problem using MATLAB, you may use the following code. >> syms x y >> firstans = int(int(x∗y,x,-4-3∗y,(6-y)/4),y,-2,-1)+int(int(x∗y,x,(2∗y-1)/3,(6y)/4),y,-1,2) firstans = -11/8 EXAMPLE 3.5.4 Consider the region inside the parallelogram P with vertices at A : ( 6,3 ) , B : ( 8,4 ) , C : ( 9,6 ) , D : ( 7,5 ) , as in Figure 3.5.3.

FIGURE 3.5.3  Example 3.5.4.

Find

∫ xy dxdy. P

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Integration • 263

Solution: The lines joining the points have equations x LAB : y = , 2 LBC : = y 2 x − 12, LCD : y=

x 3 + , 2 2

LDA : = y 2 x − 9. The integral is thus 4

2y

3

y+ 9 ) / 2

∫ ∫(

xy dxdy + ∫

5

4

( y+12 ) / 2

∫(

y+ 9 ) / 2

xy dxdy + ∫

6

5

∫

( y+12 ) / 2

2 y− 3

409 xy dxdy = ⋅ 4

To solve this problem using MapleTm, you may use the following code.

> >

To solve this problem using MATLAB, you may use the following code. >> syms x y >> firstans = int(int(x∗y,x,(y+9)/2,2∗y),y,3,4)+int(int(x∗y,x,(y+9)/2,(y+12)/ 2),y,4,5)+ int(int(x∗y,x,2∗y-3, (y+12)/2),y,5,6) firstans = 409/4

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264 • Multivariable and Vector Calculus 2/E EXAMPLE 3.5.5 Find

∫x

2

D

y dxdy +1

where D =

{( x, y) ∈ 

2

| x ≥ 0, x2 + y2 ≤ 1}.

Solution: The integral is 0. Observe that if

( x, y) ∈ D ,

then

( x, − y) ∈ D .

Also,

f ( x, − y ) = − f ( x, y ) . EXAMPLE 3.5.6 Find 4



∫  ∫ 0

y

y/2

ey / x dx  dy. 

See Figure 3.5.5.

FIGURE 3.5.4  Example 3.5.6.

Solution: We have 0 ≤ y ≤ 4,

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y ≤ x ≤ y ⇒ 0 ≤ x ≤ 2, x2 ≤ y ≤ 2 x. 2

1/18/2023 12:05:19 PM

Integration • 265

We then have 4



∫  ∫ 0

y

y/2

2 ey / x dx  dy = ∫ 0 

2

(∫

2x

x2

(

)

ey / x dy dx

= ∫ xey / x 0

=

∫ ( xe 2

2

0

2x x2

) dx

− xe x ) dx

= 2 e2 − ( 2 e2 − e2 + 1 ) = e2 − 1. EXAMPLE 3.5.7 Find the area of the region = R

{( x, y) ∈ 

2

}

: x + y ≥ 1, 1 − x + 1 − y ≥ 1 .

See Figure 3.5.5.

FIGURE 3.5.5  Example 3.5.7.

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266 • Multivariable and Vector Calculus 2/E Solution: The area is given by 1  1 − (1 − 1 − x )  ∫D dA = ∫0  ∫(1− x )2 dy  dx 2

= 2∫

1

0

=

(

)

1 − x + x − 1 dx

2 ⋅ 3

EXAMPLE 3.5.8 + y2  dA, where R is the rectangle 0; 2  × 0; 2  . See Figure 3.5.6. Evaluate

∫  x

2

R

FIGURE 3.5.6  Example 3.5.8.

Solution: The function ( x, y )   x2 + y2  jumps every time x2 + y2 is an integer. For

( x, y) ∈ R , we have 0 ≤ x2 + y2 ≤ (

2

) +( 2) 2

2

= 4 . Thus we decompose R as

the union of the

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Integration • 267

Rk=

∫  x R

{( x, y) ∈  2

2

+ y2  d= A

: x ≥ 0, y ≥ 0, k ≤ x2 + y2 ≤ k + 1} , k ∈ {1,2,3}.

∑∫

Rk

1≤ k ≤ 3

 x2 + y2  dA  

∫∫

=

∫∫

1dA +

1≤ x2 + y2 < 2, x ≥ 0, y≥ 0

2dA +

2 ≤ x2 + y2 < 3, x ≥ 0, y≥ 0

∫∫

3dA.

3 ≤ x2 + y2 < 4, x ≥ 0, y≥ 0

Now the integrals can be computed by realizing that they are areas of quarter annuli, and so, 1 pk ⋅ kdA = k ⋅ ⋅ p ( k + 1 − k ) = 4 4 k ≤ x2 + y2 < k +1, x ≥ 0, y≥ 0

∫∫

Hence

∫  x

2

R

p 3p ⋅ (1 + 2 + 3 )= 4 2

+ y2  dA=

EXERCISES 3.5 3.5.1

Evaluate the iterated integral 3

x

1

0

∫∫

1 dydx. x

3.5.2 Let S be the interior and boundary of the triangle with vertices ( 0,0 ) , ( 2,1) , and ( 2,0 ) . Find ∫ ydA. 3.5.3 Let = S

{( x, y) ∈ 

S

2

: x ≥ 0, y ≥ 0,1 ≤ x2 + y2 ≤ 4}. Find ∫ x2 dA. S

3.5.4 Find ∫ xydxdy D

where D=

MVC.CH03_3PP.indd 267

{( x, y) ∈ 

2

| y ≥ x 2 , x ≥ y2 } .

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268 • Multivariable and Vector Calculus 2/E

3.5.5 Find

∫ ( x + y) ( sin x ) ( sin y) dA

D

where D = [ 0; p ] . 2

∫ ∫ min ( x 3.5.7 Find ∫ xydxdy 3.5.6 Find

1

1

0

0

2

+ y2 ) dxdy.

D

where D =

{( x, y) ∈  ∫

3.5.8 Evaluate

R

2

: x > 0, y > 0,9 < x2 + y2 < 16,1 < x2 − y2 < 16}.

xdA where R is the (un-oriented) circular segment in

Figure 3.5.7, which is created by the intersection of regions

{( x, y) ∈ 

2

: x2 + y2 ≤ 16}

and   3 2 x + 4 . ( x, y ) ∈  : y ≥ − 3  

FIGURE 3.5.7  Exercise 3.5.8.

3.5.9 Find

1

1

0

y

∫∫

3.5.10 Evaluate 3.5.11 Find

∫

R

2

2 e x dxdy.

∫[

0;1]

2

min ( x, y2 ) dA.

xydA , where R is the (un-oriented) ∆OAB in Figure 3.5.8

with O ( 0,0 ) , A ( 3,1 ) , and B ( 4,4 ) .

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Integration • 269

FIGURE 3.5.8  Exercise 3.5.11.

3.5.12 Solve Exercise 3.5.11 using Maple commands. 3.5.13 Find ∫ log e (1 + x + y ) dA D

where

{( x, y) ∈ 

= D

∫[

3.5.15 Evaluate

∫  x + y dA,

2

| x ≥ 0, y ≥ 0, x + y ≤ 1}.

 x + y2  dA.  

3.5.14 Evaluate

0;2 ]

2

R

where R is the rectangle [ 0;1] × [ 0;2 ] . 3.5.16 Evaluate ∫ xdA where R is the quarter annulus in Figure 3.5.9, R which is formed by the area between the circles x2 + y2 = 1 and x 2 + y2 = 4 in the first quadrant.

FIGURE 3.5.9  Exercise 3.5.16.

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270 • Multivariable and Vector Calculus 2/E

3.5.17 Evaluate

∫

R

xdA , where R is the E-shape figure in Figure 3.5.10.

FIGURE 3.5.10  Exercise 3.5.17.

3.5.18 Evaluate 3.5.19 Find

p /2

p /2

0

x

∫ ∫

cos y dydx. y

4 2  x px  px  sin d d + y x   ∫1  ∫ x 2 y  ∫2  ∫ x sin 2 y dy  dx. 2

3.5.20 Find ∫ 2 x ( x2 + y2 ) dA D

where D {( x, y ) ∈  2 : x 4 + y4 + x2 − y2 ≤ 1}. 3.5.21 Find the area bounded by the ellipses x2 +

y2 x2 = 1 and + y2 = 1. 4 4

as in Figure 3.5.11.

FIGURE 3.5.11  Exercise 3.5.21.

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Integration • 271

3.5.22 Find ∫ xydA D

where D {( x, y ) ∈  2 : x ≥ 0, y ≥ 0, xy + y + x ≤ 1}. 3.5.23 Find ∫ log e (1 + x2 + y ) dA D

where D {( x, y ) ∈  2 : x ≥ 0, y ≥ 0, x2 + y ≤ 1}. 3.5.24 Evaluate

∫

R

xdA , where R is the region between the circles

x 2 + y2 = 4 and x2 + y2 =, 2 y as shown in Figure 3.5.12.

FIGURE 3.5.12  Exercise 3.5.24.

3.5.25 Find

∫ x − y dA

D

where D {( x, y ) ∈  2 : x ≤ 1, y ≤ 1}. 3.5.26 Find

∫ ( 2 x + 3 y + 1) dA.

D

Where D is the triangle with vertices at A ( −1, −1 ) , B ( 2, −4 ) , and C (1,3 ) .

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272 • Multivariable and Vector Calculus 2/E 3.5.27 Let f : [ 0;1] → ]0; +∞ ] be a decreasing function. Prove that 1

1

∫ xf ( x ) dx ≤ ∫ f ( x ) dx ⋅ ∫ xf ( x ) dx ∫ f ( x ) dx 3.5.28 Find ∫ ( xy ( x + y ) ) dA . Where 2

0

D

2

0

1

1

0

0

D {( x, y ) ∈  2 : x ≥ 0, y ≥ 0, x + y ≤ 1}.

3.5.29 Let f , g : [ 0;1] → [ 0;1] be continuous, with f increasing. Prove that

∫ ( f  g ) ( x ) dx ≤ ∫ f ( x ) dx + ∫ g ( x ) dx. 3.5.30 Compute

1

1

1

0

0

0

∫ ( xy + y ) dA , where S = {( x, y) ∈  2

S

3.5.31 Evaluate 3.5.32 Find

∫

D

(

a

b max b2 x2 , a2 y2

0

0

∫∫

e

)

2

:x

1/ 2

+y

1/ 2

}

≤1 .

dydx, where a and b are positive.

xy dA.

Where

{

}

D ( x, y ) ∈  2 : y ≥ 0, ( x + y ) ≤ 2 x . 2

3.5.33 A rectangle R on the plane is the disjoint union R =  k =1 Rk of N

rectangles Rk . It is known that at least one side of each of the rectangles Rk is an integer. Show that at least one side of R is an integer. 1

1

1

0

0

1

1

1

0

0

0

 ∫ ( x1 x2  xn ) dx1 dx2  dxn .

∫∫ 3.5.35 Evaluate ∫ ∫  ∫ ( x 3.5.34 Evaluate

3.5.36 Find

0

1

∫ ( x + y)

4

1

+ x2 +  + xn ) dx1 dx2  dxn .

dA , where

D

D {( x, y ) ∈  2 | x ≥ 1, y ≥ 1, x + y ≤ 4}.

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Integration • 273

3.5.37 Find ∫ xdA. Where D

D {( x, y ) ∈  2 | y ≥ 0, x − y + 1 ≥ 0, x + 2 y − 4 ≤ 0}. 1

1

n →+∞ 0

0

3.5.38 Evaluate lim

p  cos2  ( x1 + x2 +  + xn )  dx1 dx2  dxn . 0  2n 

∫ ∫ ∫

1

3.6 CHANGE OF VARIABLES IN DOUBLE INTEGRALS We now perform a multidimensional analogue of the change of variables theorem in double integrals. Theorem 3.6.1  Let ( D, ∆ ) ∈ (  n ) open, bounded sets in  n with volume 2

and let g : ∆ → D be a continuously differentiable bijective mapping such that det g′ ( u ) ≠ 0 , both det g′( u) ,

1 bounded on ∆ . For f : D →  det g′( u)

bounded and integrable. f  g det g′( u) is integrable on ∆ and

∫ ∫

D

f = ∫ ∫

∆

( f  g ) det g′ ( u ) ,

that is

∫ ∫ f ( x , x ,, x ) dx 1

D

=

2

n

1

∧ dx2 ∧  ∧ dxn

∫ ∫ f ( g ( u , u ,, u ) ) det g′ ( u ) du ∆

1

2

n

1

∧ du2 ∧  ∧ dun .

One normally chooses changes of variables that map into rectangular regions, or that simplify the integrand. Let us start with a rather trivial example. EXAMPLE 3.6.1 Evaluate the integral

∫ ∫ ( x + 2 y)( 2 x + y) dxdy.

MVC.CH03_3PP.indd 273

4

1

3

0

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274 • Multivariable and Vector Calculus 2/E Solution: Observe that we have already computed this integral in Example 3.5.2. Put u =x + 2 y ⇒ du =dx + 2dy, v = 2 x + y ⇒ dv = 2dx + dy, giving du ∧ dv = −3dx ∧ dy. Now, 1 2   x    2 1   y

( u, v ) = 

is a linear transformation, and hence it maps quadrilaterals into quadrilaterals. The corners of the rectangle in the area of integration in the xy-plane are (0, 3), (1, 3), (1, 4), and (0, 4) (traversed counter-clockwise; see Figure 3.6.1). They map into (6, 3), (7, 5), (9, 6), and (8, 4), respectively, in the uv-plane (see Figure 3.6.2).

FIGURE 3.6.1  Example 3.6.1, xy -plane.

FIGURE 3.6.2 Example 3.6.1, uv-plane.

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Integration • 275

The form dx ∧ dy has opposite orientation to du ∧ dv so we use dv ∧ du = 3dx ∧ dy instead. The integral sought is 1 409 uv dvdu = , ∫ 3P 12 from Example 3.5.4. EXAMPLE 3.6.2 The integral 1 1  0. − y4 ) dA =∫  − y4  dy = 0;1 0 5   Evaluate it using the change of variables u = x2 − y2 , v = 2 xy .

∫[ ] ( x

4

2

Solution: First we find = du 2 xdx − 2 ydy, = dv 2 ydx + 2 xdy, and so du ∧ dv=

(4x

2

+ 4 y2 ) dx ∧ dy.

We now determine the region ∆ into which the square D = [ 0;1] is mapped. 2

We use the fact that the boundaries will be mapped into boundaries. Put

MVC.CH03_3PP.indd 275

AB =

{( x,0 ) : 0 ≤ x ≤ 1},

BC =

{(1, y) : 0 ≤ y ≤ 1},

CD =

{(1 − x, y) : 0 ≤ x ≤ 1},

DA =

{( 0,1 − y) : 0 ≤ y ≤ 1}.

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276 • Multivariable and Vector Calculus 2/E On AB, we have= u x= , v 0 . Since 0 ≤ x ≤ 1 , AB is thus mapped into the line segment 0 ≤ u ≤ 1 , v = 0 . On BC , we have u = 1 − y2 , v =. 2 y Thus u= 1 −

v2 . Hence,BC is mapped to 4

v2 , 0≤ v≤2 4 2 u =(1 − x ) − 1, v =2 (1 − x ) .

the portion of the parabola u= 1 − On

CD

we

have

This

means

that

v2 2 − 1,0 ≤ v ≤ 2 . Finally, on DA , we have u = − (1 − y ) , v =. 0 Since 4 0 ≤ y ≤ 1 , DA is mapped into the line segment −1 ≤ u ≤ 0, v =. 0 The region ∆ u=

is thus the area in the uv plane enclosed by the parabolas u ≤

v2 v2 − 1, u ≤ 1 − 4 4

with −1 ≤ u ≤ 1,0 ≤ v ≤ 2 . We deduce that

∫[ ] ( x 0;1

2

4

− y4 ) dA =∫ ( x 4 − y4 ) ∆

1 dudv 4 ( x + y4 ) 4

1 x2 − y2 ) dudv ( ∫ ∆ 4

=

=

1 u dudv 4 ∫∆

=

1 2  1− v2 / 4   ∫v2 / 4 −1 udu  dv ∫ 0 4  

= 0, as before.

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Integration • 277

EXAMPLE 3.6.3 Find ∫ e

( x3 + y3 ) / xy

dA ,

D

where = D

{( x, y) ∈ 

2

| y2 − 2 px ≤ 0, x2 − 2 py ≤ 0, p ∈ ]0; +∞[ fixed} ,

2 Using the change of variables = x u= v, y uv2 .

Solution: We have = dx 2 uvdu + u2 dv, = dy v2 du + 2 uvdv, dx ∧= dy 3 u2 v2 du ∧ dv. The region transforms into = ∆

{( u, v) ∈ 

2

| 0 ≤ u ≤ ( 2 p)

1/ 3

,0 ≤ v ≤ ( 2 p )

1/ 3

}.

The integral becomes

∫

D

 u 6 v 3 + u 3 v6  2 2 f ( x, y ) dxdy = ∫ exp   ( 3 u v ) dudv 3 3 u v   ∆ = 3 ∫ e u e v u2 v2 dudv 3

3

∆

1  ( 2 p)1/3 2 u3  = ∫ 3 u e du  3 0  =

2

2 1 2p e −1 . 3

(

)

2 As an exercise, you may try the (more natural) substitution = x3 u= v, y3 v2 u

and verify that the same result is obtained.

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278 • Multivariable and Vector Calculus 2/E EXAMPLE 3.6.4 In this problem, we will follow an argument of Calabi, Beukers, and Knock to prove that 1 p2 = ⋅ ∑ 2 6 n =1 n +∞

+∞ 1 3 1. Prove that if S = ∑ 2= , then S 4 n =1 n +∞ 1 2. Prove that = ∑ 2 n =1 ( 2 n − 1 )

1

1

0

0

∫∫

+∞

n =1

1

1

0

0

⋅

2

dxdy ⋅ 1 − x 2 y2 sin u sin v = ,y in order to evaluate cos v cos u

= 3. Use the change of variables x

∫∫

1

∑ ( 2 n − 1)

dxdy ⋅ 1 − x 2 y2

Solution: 1. Observe that the sum of even terms is +∞

1 1 +∞ 1 1 = = S, ∑ ∑ ( 2 n ) 2 4 n 1 n2 4 = n 1= a quarter of the sum, hence the sum of the odd terms must be three3 quarters of the sum, S. 4 2. Observe that 1 = 2n − 1

∫

1

0

x

2 n−2

2

 1  dx ⇒   =  2n − 1 

(∫ x 1

0

2 n−2

dx

)( ∫ y 1

0

2 n−2

)

dy =

∫ ∫ ( xy) 1

1

0

0

2 n−2

dxdy.

Thus +∞ +∞ 1 1 1 1 +∞ 1 2 n−2 2 n−2 = = xy d x d y = xy ) dxdy ( ) ( ∑ ∑ ∑ 2 ∫ ∫ ∫ ∫ 0 0 ( 2 n − 1) n 1 0 0 = n 1= = n 1

1

dxdy , 0 1 − x 2 y2

∫∫ 0

1

as claimed. sin u sin v = ,y then cos v cos u = dx ( cos u )( sec v ) du + ( sin u )( sec v )( tan v ) dv,

3. = If x

= dy

MVC.CH03_3PP.indd 278

( sec u )( tan v )( sin v ) du + ( sec u )( cos v ) dv,

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Integration • 279

from where

(

)

dx ∧ dy = du ∧ dv − ( tan 2 u )( tan 2 v ) du ∧ dv = 1 − ( tan 2 u )( tan 2 v ) du ∧ dv. Also, 1 − x2 y2 = 1−

sin 2 v sin 2 v ⋅ = 1 − ( tan 2 u )( tan 2 v ) . cos2 v cos2 u

This gives dxdy = dudv. 1 − x2 y2 We now have to determine the region that the transformation, sin u sin v = ,y forms in the uv-plane. Observe that x = cos v cos u 1 − y2 1 − x2 = u arctan = x , v arctan y ⋅ 1 − x2 1 − y2 This means that the square in the xy-plane in Figure 3.6.3 is transformed into the triangle in the uv-plane in Figure 3.6.4.

FIGURE 3.6.3  Example 3.6.4, xy-plane.

FIGURE 3.6.4 Example 3.6.4, uv-plane.

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280 • Multivariable and Vector Calculus 2/E We deduce, p /2

p / 2 p / 2− v p /2 dxdy p v2  = = p − = − d u d v / 2 v d v v ( )   ∫0 ∫0 1 − x2 y2 ∫0 ∫0 ∫0 2 0 2 1

1

=

p2 p2 p2 − = ⋅ 4 8 8

Finally, p2 p2 3 S= ⇒ S= ⋅ 4 8 6 EXAMPLE 3.6.5 Evaluate

1/ 5

 x  1 − 3 y  x + 3 y   

0

0

∫ ∫

e

dxdy using Maple.

To solve this problem using MapleTm, you may use the following code. > evalf(int(int(exp(x/(x+3*y)),x=0..1-3*y),y=0..1/5)); 0.2552649404

Maple automatically performs the change of variables.

EXERCISES 3.6 3.6.1 Let= D′

{( u, v) ∈  Φ:

: u ≤ 1, − u ≤ v ≤ u}. Consider,

2

→ 2

( u, v )

 u + v u − v .   ,  2   2



Find the image of Φ on D′ , that is, find D = Φ ( D′ ) .

3.6.2

Let= D′

{( u, v) ∈  Φ:

Find

MVC.CH03_3PP.indd 280

2

∫ ( x + y) e

2

2

→ 2

( u, v )

 u + v u − v .   ,  2   2

2 x2 − y2

D

: u ≤ 1, − u ≤ v ≤ u}. Consider,

dA.

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Integration • 281

3.6.3 Find

∫ f ( x, y) dA where

D

= D

{( x, y) ∈ 

2

| a ≤ xy ≤ b, y ≥ x ≥ 0, y2 − x2 ≤ 1, ( a, b ) ∈  2 ,0 < a < b}

and f ( x, y= ) y4 − x4 by using the change of variables = u xy, = v y2 − x 2 . 3.6.4

Use the following steps (due to Tom Apostol), in order to prove that 1 p2 = ⋅ ∑ 2 6 n =1 n ∞

1. Use the series expansion 1 =1 + t + t 2 + t 3 +  , 1−t

in order to prove (formally) that

t < 1, ∞

dxdy ∫0 ∫0 1= − xy 1

1

1

∑n n =1

2

⋅

2. Use the change of variables u =+ x y, v =− x y to show that 1 1 dxdy 1 u 2  2− u dv dv   = ∫0 ∫0 1 − xy 2 ∫0  ∫− u 4 − u2 + v2  du + 2 ∫1  ∫u−2 4 − u2 + v2  du. 3. Show that the preceding integral reduces to 1

2

0

4−u

2∫

2

arctan

u 4−u

2

2

2

1

4−u

du + 2 ∫

2

arctan

2−u 4 − u2

du.

p2 4. Finally, prove that the preceding integral is by using the sub6 u stitution q = arcsin . 2 3.6.5

Evaluate the integral

∫∫ sin ( x − y)( x + y) 2

2

dA using change of

D

variables, where D is the region bounded by square with vertices ( 0,1) , (1,2 ) , ( 2,1) , and (1,0 ) .

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282 • Multivariable and Vector Calculus 2/E

3.6.6

Evaluate the integral

∫∫D

cos ( ( x − y ) / 2 ) 3x + y

dA using change of vari-

ables, where D is the region bounded by the graphs y= −3 x + 3, y = −3 x + 6, y = x, y =− x p ; u =− x y , and = v 3x + y. 3.6.7

Evaluate the integral

∫∫D ( x

2

+ y2 ) sin xy dA using change of vari-

ables, where D is the region bounded by the graphs 2 2 y = , y =− , y2 =x2 − 1, y2 =x2 − 9; u =x2 − y2 , and v = xy . x x 3.6.8

Evaluate the integral

∫∫D y dA using change of variables, where D is

the triangle region with vertices ( 0,0 ) , ( 2,3 ) , and ( −4,1 ) ; = x 2 u − 4 v and = y 3u + v. 3.6.9

Evaluate the integral

∫∫D ( x + y)

3

dA using change of variables,

where D is the parallelogram region with sides x + y = k1 and x − 2y = k2 for appropriate choices of k1 and k2 ; x= u − y and x= v + 2 y .  −x + y   dA using change of variables, x+y  where D is the trapezoid region with vertices (1,1 ) , ( 2,2 ) , ( 4,0 ) , and

3.6.10 Evaluate the integral

∫∫D sin 

( 2,0 ) ; x= y − u and y= v − x . 3.6.11 Evaluate the integral

∫∫D ( x − y)

2

cos2 ( x + y ) dA using change of

variables, where D is the region by the square with vertices ( 0,1) , (1,2 ) , ( 2,1) , and (1,0 ) ; x= u + y and y= v − x . 3.6.12 Evaluate the integral

∫∫D

e

 x−y     x+ y 

D is the region defined= by D

MVC.CH03_3PP.indd 282

dA using change of variables, where

{( x, y) : x ≥ 0, y ≥ 0, x + y ≤ 1} .

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Integration • 283

3.7 CHANGE TO POLAR COORDINATES One of the most common changes of variable is the passage to polar coordinates where = x r cosq ⇒ = dx cosq dr − r sin q dq , = y r sin q ⇒ d= y sin q dr + r cosq dq , hence dx ∧ dy=

( r cos q + r sin q ) dr ∧ dq = 2

2

r dr ∧ dq .

EXAMPLE 3.7.1 Find ∫ xy x2 + y2 dA , where D

D=

{( x, y) ∈ 

2

| x ≥ 0, y ≥ 0, y ≤ x, x2 + y2 ≤ 1}.

See Figure 3.7.1.

FIGURE 3.7.1  Example 3.7.1.

Solution: We use polar coordinates. The region D transforms into the region = ∆

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p .  4 

[0;1] × 0;

1/18/2023 12:05:46 PM

284 • Multivariable and Vector Calculus 2/E Therefore the integral becomes

∫r

4

cosq sin q dr d= q

∆

(∫

p /4

0

cosq sin q dq

)( ∫ r dr=) 1

0

4

1 ⋅ 20

EXAMPLE 3.7.2 Evaluate ∫ xdA , where R is the region bounded by the circles x2 + y2 = 4 and R

x + y =. 2 y See Figure 3.7.2. 2

2

FIGURE 3.7.2  Example 3.7.2.

Solution: Since x2 + y2 = r 2 , the radius sweeps from r 2 = 2 r sin q to r 2 = 4 , that is, from 2 sinq to 2. The angle clearly sweeps from 0 to

∫

p /2

R

xdA = ∫ 0

∫

2 sin q

2

p . Thus the integral becomes 2

1 p /2 r 2 cosq drdq = 8 cosq − 8 cosq sin 3 q ) dq = 2. ( ∫ 0 3

EXAMPLE 3.7.3 Find

∫

D

e− x

2

− xy − y2

dA , where D =

{( x, y) ∈ 

2

: x2 + xy + y2 ≤ 1} .

See Figure 3.7.3.

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Integration • 285

FIGURE 3.7.3  Example 3.7.3.

Solution: Completing squares 2

2 y  3 y  x + xy + y =  x +  +   . 2   2   y 3y Put U = x + , V = ⋅ The integral becomes 2 2 2 − (U 2 + V 2 ) − x2 − xy − y2 e dxdy = dUdV. 2 2 ∫{x2 + xy+ y2 ≤1} e ∫ 3 {U + V ≤1} Passing to polar coordinates, the previous equals 2

2

2 2p 1 − r 2 2p r e d= r dq (1 − e−1 ). ∫ 0 ∫0 3 3 EXAMPLE 3.7.4 Evaluate

∫

R

(x

1

2

+ y2 )

3/2

dA , over the region

{( x, y) ∈ 

2

: x2 + y2 ≤ 4, y ≥ 1}

See Figure 3.7.4.

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286 • Multivariable and Vector Calculus 2/E

FIGURE 3.7.4  Example 3.7.4.

Solution: The radius sweeps from r =

∫

R

(x

1 to r = 2 The desired integral is sin q

1 2

+y

)

2 3/2

dA = ∫

5p / 6

p /6

=

∫

=

5p / 6

p /6

3−

∫

2

csc q

1 drdq r2

1   sin q −  dq 2  p ⋅ 3

EXAMPLE 3.7.5 Evaluate ∫ ( x3 + y3 ) dA , where R is the region bounded by ellipse R

x 2 y2 1, + = a2 b2

and the first quadrant, a > 0 and b > 0 . Solution: = Put x ar = cosq , y br sin q . Then = x ar cosq ⇒= dx a cosq dr − ar sin q dq , = y br sin q ⇒ = dy b sin q dr + br cosq dq ,

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Integration • 287

whence

( abr cos q + abr sin q ) dr ∧ dq=

dx ∧ d= y

2

2

abrdr ∧ dq .

Observe that on the ellipse x2 y2 a2 r 2 cos2 q b2 r 2 sin 2 q + = 1 ⇒ + = 1 ⇒ r = 1. a2 b2 a2 b2 Thus the required integral is + y ) dA ∫ ∫ ∫ ( x= 3

3

R

p /2

1

0

0

= ab

abr 4 ( cos3 q + sin 3 q ) drdq

( ∫ r dr ) ( ∫ 1

4

0

p /2

0

(a

3

cos3 q + b3 sin 3 q ) dq

)

3 3  1   2a + 2 b  = ab    3  5  

=

2 ab ( a3 + b3 ) 15

⋅

EXERCISES 3.7 3.7.1 Evaluate

∫

R

= R

MVC.CH03_3PP.indd 287

xydA where R the region is

{( x, y) ∈ 

2

: x2 + y2 ≤ 16, x ≥ 1, y ≥ 1} ,

as in the Figure 3.7.1. Set up the integral in the Cartesian and polar coordinates.

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288 • Multivariable and Vector Calculus 2/E

FIGURE 3.7.5  Exercise 3.7.1.

3.7.2 Find

∫(x

2

− y2 ) dA ,where

D

D = 3.7.3 Find

∫

D=

= D

∫

D

2

}

| ( x − 1 ) + y2 ≤ 1 2

xy dA ,where

D

3.7.4 Find

{( x, y) ∈ 

{( x, y) ∈  | ( x + y ) ≤ 2 xy} 2

2 2

2

f dA ,where

{( x, y) ∈ 

2

: b2 x2 += a2 y2 a2 b2 , ( a, b ) ∈ ]0; +∞[ fixed} and

f ( x, y= ) x 3 + y3 . 3.7.5 Find

∫

x2 + y2 dA ,where

D

{( x, y) ∈  | x ≥ 0, y ≥ 0, x + y ≤ 1, x + y − 2 y ≥ 0}. 3.7.6 Find ∫ f dA , where D = {( x, y ) ∈  | y ≥ 0, x + y − 2 x ≤ 0} and f ( x, y ) = x y. 3.7.7 Find ∫ f dA , where 1 D = {( x, y ) ∈  | x ≥ 1, x + y − 2 x ≤ 0} and = f ( x, y ) (x + y ) 3.7.8 Let D = {( x, y ) ∈  : x + y − y ≤ 0, x + y − x ≤ 0}. Find the integral ∫ ( x + y ) dA. D =

2

2

2

2

2

D

2

2

2

2

2

2

2

D

2

2

2

2

2

2 2

⋅

2

2

D

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Integration • 289

3.7.9 Let D =

{( x, y) ∈ 

3.7.10 Evaluate

2

| y ≤ x2 + y2 ≤ 1}. Compute 3

∫

{( x, y)∈2 , x≥ 0, y≥ 0, x4 + y4 ≤1}

∫

D

dA

(1 + x

2

+ y2 )

2

⋅

1 − x 4 − y4 dA , using

x3 y

= x2 r= cosq , y2 r sin q . 3.7.11 William Thompson (Lord Kelvin) is credited to have said: “A mathematician is someone to whom

∫

+∞

0



is as obvious as twice two is four to you. Liouville was a mathematician.” Prove that

∫

+∞

0



p 2

2

e− x dx =

p 2

2

e− x dx =

by following these steps. 1. Let a > 0 be a real number and put D= a

{( x, y) ∈ 

2

| x 2 + y2 ≤ a 2 } .

Find Ia = ∫ e

(

− x2 + y2

)

Da

dxdy.

2. Let a > 0 be arealnumberand put= ∆a Let Ja = ∫ e

(

− x2 + y2

)

Ia ≤ Ja ≤ Ia

3.7.12 Let D = Find

∫

+∞

0

{( x, y) ∈ 

2

2

| x ≤ a, y ≤ a}.

dxdy. Prove that

∆a

3. Deduce that

{( x, y) ∈ 

2

e− x = dx

2

⋅

p ⋅ 2

: 4 ≤ x2 + y2 ≤ 16} and = f ( x, y )

1 ⋅ x + xy + y2 2

∫ f ( x, y) dA.

D

3.7.13 Prove that every closed convex region in the plane of area ≥ p has two points which are two units apart.

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290 • Multivariable and Vector Calculus 2/E 3.7.14 In the xy-plane, if R is the set of points inside and on a convex polygon, let D( x, y) be the distance from ( x, y) to the nearest point R. Show that +∞

+∞

−∞

−∞

∫ ∫

e

− D( x , y)

dxdy = 2p + L + A ,

where L is the perimeter of R and A is the area of R.

3.8 THREE-MANIFOLDS Definition 3.8.1  A three-dimensional oriented manifold of  3 is simply an open set (body) V ∈  3 , where the + orientation is in the direction of the outward pointing normal to the body, and the − orientation is n the direction of the inward-pointing normal to the body. A general oriented three-manifold is a union of pen sets. NOTE

The region − M has opposite orientation to M and

∫

−M

w = − ∫ w. M

We will often write

∫

M

fdV

Where dV denotes the volume element. NOTE

In this section, unless otherwise noticed, we will choose the positive orientation for the regions considered. This corresponds to using the volume form dx ∧ dy ∧ dz .

Let V ⊆  3 . Given a function f : V →  , integral

∫ fdV V

is the sum of all the values of f restricted to V. In particular,

∫ dV V

is the oriented volume of V .

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Integration • 291

EXAMPLE 3.8.1 Find

∫

x2 ye xyz dV.

[ 0;1]3

Solution: The integral is

∫ (∫ (∫ 1

1

1

0

0

0

) )

∫ ( ∫ x ( e − 1) dy) dx ∫ ( e − x − 1) dx

x2 ye xyz d= z dy dx

1

1

0

0

1

=

xy

x

0

5 =e − ⋅ 2 EXAMPLE 3.8.2 Find ∫ z dV if R

R=

{( x, y) ∈ 

3

}

| x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1 .

Solution: The integral is 2 2 1  (1 − z )  (1 − z − x )   z d x d y d z = z dy  dx  dz ∫R ∫0  ∫0  ∫0   2 2 1  (1 − z )  = ∫ z ∫ 1 − z − x dx  dz 0  0  4 1 1 = z 1 − z dz 6 ∫0 1 = ⋅ 840

(

(

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)

)

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292 • Multivariable and Vector Calculus 2/E EXAMPLE 3.8.3 2 Prove that xdV = a bc , where V is the tetrahedron ∫V 24

x y z   = V ( x, y ) ∈  3 : x ≥ 0, y ≥ 0, z ≥ 0, + + ≤ 1 . a b c   Solution: We have c

b − bz / c

0

0

∫ xdxdydz = ∫ ∫ V

∫

a − ay / b − az / c

0

xdxdydz

2

ay az  1 c b− bz / c  = a− −  dydz  ∫ ∫ 0 0 2 b c   2 1 c a ( −z + c) b = ∫ dx c3 6 0 3

=

a2 bc ⋅ 24

∫ xdV

where S is the is the (unoriented) tetrahedron

EXAMPLE 3.8.4 Evaluate the integral,

S

with vertices ( 0,0,0 ) , ( 3,2,0 ) , ( 0,3,0 ) , and ( 0,0,2 ) . See Figure 3.8.1.

FIGURE 3.8.1  Example 3.8.4.

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Integration • 293

Solution: A short computation shows that the plane passing through ( 3,2,0 ) , ( 0,3,0 ) , 18 − 2 x − 6 y . We 9 must now figure out the xy limits of integration. In Figure 3.8.2, we draw the projection of the tetrahedron on the xy-plane. The line passing through AB and ( 0,0,2 ) has equation 2 x + 6 y + 9 z = 18. Hence, 0 ≤ z ≤

x 2 has equation y = − + 3 . The line passing through AC has equation y = x . 3 3

FIGURE 3.8.2 Example 3.8.4, xy -projection.

We find, finally, 3

3− x / 3

0

2x/3

=∫

3

3− x / 3

=∫

3

∫ xdV = ∫ ∫ S

0

0

MVC.CH03_3PP.indd 293

∫

2x/3

∫

(18 − 2 x − 6 y) / 9

0

xdzdydx

18 x − 2 x2 − 6 yx dydx 9 3− x / 3

18 xy − 2 x2 y − 3 y2 x dx 9 2x/3

=

 x3  2 ∫0  3 − 2 x + 3 x  dx

=

9 ⋅ 4

3

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294 • Multivariable and Vector Calculus 2/E To solve this problem using MapleTm, you may use the following code.

> >

To solve this problem using MATLAB, you may use the following code. >> syms x y z >> firstans = int(int(int(x,z,0,(18-2∗x-6∗y)/9),y,2∗x/3,3-x/3),x,0,3) firstans = 9/4 EXAMPLE 3.8.5 Evaluate

∫

R

xyzdV , where R is the solid formed by the intersection of the

parabolic cylinder z= 4 − x2 , the planes= z 0,= y x , y = 0 . Use the following orders of integration: 1. dzdxdy 2. dxdydz Solution: We must find the projections of the solid on the coordinate planes. 1. With the order dzdxdy, the limits of integration of z can only depend, if at all, on x and y. Given an arbitrary point in the solid, its lowest z coordinate is 0 and its highest one is on the cylinder, so the limits for z are from z = 0 to z= 4 − x2 . The projection of the solid on the xy-plane is the area bounded by the lines= y x= , x 2 , and the x and y axes.

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Integration • 295

2

y

4 − x2

0

0

0

∫∫∫

2 1 2 y 2 xy 4 − x dxdy ( ) 2 ∫0 ∫0 1 2 y = y (16 x − 8 x3 + x 5 ) dxdy 2 ∫0 ∫0 2 y7  = ∫  4 y3 − y5 +  dy 0 12  

xyz = dzdxdy

= 8. 2. With the order dxdydz, the limits of integration of x can only depend, if at all, on y and z . Given an arbitrary point in the solid, x sweeps from the plane to x = 2 , so the limits for x are from x = y to= x

4 − z . The projec-

tion of the solid on the yz-plane is the area bounded by z= 4 − y2 , and the z and y axes. 4

∫∫ 0

4− z

0

2

xyzdxdydz ∫= y

=

1 4 4− z ( 4 y − y3 ) zdydz 2 ∫0 ∫0  z3 − 2 z ∫0  8 4

  dz 

= 8.

EXERCISES 3.8 3.8.1 Compute

∫

E

zdV , where E is the region in the first octant bounded

by the planes x + z = 1 and y + z =. 1 3.8.2

Evaluate the integrals

∫

R

1dV and

∫

R

xdV , where R is the tetrahe-

dron with vertices at (0, 0, 0), (1, 1, 1), (1, 0, 0), and (0, 0, 1). 3.8.3 Compute

∫

E

xdV , where E is the region in the first octant bounded

by the plane y = 3 z and the cylinder x2 + y2 = 9.

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296 • Multivariable and Vector Calculus 2/E

3.8.4 Find

dV

∫ (1 + x z )(1 + y z ) where = D {( x, y, z ) ∈  : 0 ≤ x ≤ 1,0 ≤ y ≤ 1, z ≥ 0}. 2 2

2 2

D

3

3.8.5

Write an iterated integral for

∫∫∫ f ( x, y, z ) dV for the solid region S

  12 − 3 x − 2 y   = S ( x, y, z ) : 0 ≤ x ≤ 1,0 ≤ y ≤ 3,0 ≤ z ≤   . 6   

∫∫∫ ( 3 xy z ) dV for the solid region {( x, y, z ) : −1 ≤ x ≤ 3,1 ≤ y ≤ 4,0 ≤ z ≤ 2} . 3 2

3.8.6 Evaluate

S

= S 3.8.7

Find the volume of the solid bounded by the cylinders y = x2 and y = z2 , and the plane y = 1 .

3.8.8

Find the volume of the solid bounded by the graphs = z 2,= z 4 y2= , x 2 , and x = 0 .

3.8.9 Evaluate

∫∫∫ ( e

x + y+ z

) dV for the solid region S in 

3

bounded by

S

the planes z = 0, z = − x, x = 0, y = 1 , and y = − x . 3.8.10 Find the volume of the ellipsoidal solid = S {( x, y, z ) : 4 x2 + 4 y2 = + z2 − 16 0} . 3.8.11 Find the volume of the centroid of the tetrahedral defined by = S {( x, y, z ) : x + y + z ≤ 1, y ≥ 0, z ≥ 0} . 3.8.12 Find the volume of the region S bounded by the parabolic cylinder z= 4 − x2 and the planes x = 0, y − 0, y = 6 , and z = 0. 3.8.13 Find the volume V for the solid region S in  3 bounded by the graphs z = x 2 + y2 , z = 0, y = x, y =, 0 and x = 2 . 3.8.14 Find the volume V for the solid region S in  3 bounded by the graphs z =+ x y, y2 = − x2 + 4 , the coordinate plane, and first octant.

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Integration • 297

3.9 CHANGE OF VARIABLES IN TRIPLE INTEGRALS We demonstrate in this section change of variables in three integrals through examples. EXAMPLE 3.9.1 Find ∫ ( x + y + z )( x + y − z )( x − y − z ) dV , where R is the tetrahedron bounded R

by the planes x + y + z= 0, x + y − z= 0, x − y − z= 0 and 2 x − z =. 1 Solution: We make the charge of variables u = x + y + z ⇒ du = dx + dy + dz, v = x + y − z ⇒ dv = dx + dy − dz, w = x − y − z ⇒ dw = dx − dy − dz. This gives du ∧ dv ∧ dw = −4dx ∧ dy ∧ dz. These forms have opposite orientations, so we choose, say, du ∧ dw ∧ dv = 4dx ∧ dy ∧ dz which have the same orientation. Also, 2 x − z = 1 ⇒ u + v + 2 w = 2. The tetrahedron in the xyz-coordinate frame is mapped into a tetrahedron bounded by u= 0, v= 0, u + v + 2 w= 1 in the uvw -coordinate frame. The integral becomes 1 2 1− v / 2 2 − v− 2 w 1 uvwdudw= dv ⋅ ∫ ∫ ∫ 0 0 0 4 180

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298 • Multivariable and Vector Calculus 2/E Consider a transformation to cylindrical coordinates

( x, y, z ) = ( r cosq , r sin q ,z ) . From what we know about polar coordinates dx ∧ dy= r dr ∧ dq . Since the wedge product of forms is associative, dx ∧ dy ∧ dz= r dr ∧ dq ∧ dz. EXAMPLE 3.9.2 Find

∫

R

z2 dxdydz , if = R

{( x, y, z ) ∈ 

3

| x2 + y2 ≤ 1,0 ≤ z ≤ 1}.

Solution: The region of integration is mapped into = ∆

[0;2p ] × [0;1] × [0;1]

through a

cylindrical coordinate change. The integral is therefore

∫

R

( ∫ dq )( ∫ r dr )( ∫ z dz=)

f ( x, y, z ) dxdydz=

2p

1

1

0

0

0

2

p ⋅ 3

EXAMPLE 3.9.3 Evaluate

∫ (x

2

D

+ y2 ) dxdydz over the first octant region bounded by the cyl-

inders x2 + y2 = 1 and x2 + y2 = 4 and the planes= z 0,= z 1,= x 0,= x y. Solution: The integral is 1

p /2

2

0

p /4

1

∫∫ ∫

r 3 dr dq= dz

15p ⋅ 16

EXAMPLE 3.9.4 Three long cylinders of radius R intersect at right angles. Find the volume of their intersection.

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Integration • 299

Solution: Let V be the desired volume. By symmetry, V = 2 4 V ′ , where V ′ = ∫ dxdydz, D′

D′ =

{( x, y, z ) ∈ 

3

: 0 ≤ y ≤ x,0 ≤ z, x2 + y2 ≤ R 2 , y2 + z2 ≤ R 2 , z2 + x2 ≤ R 2 }.

In this case, it is easier to integrate with respect to z first. Using cylindrical coordinates    p = ∆′ (q , r , z ) ∈ 0;  × [ 0; R ] × [ 0; +∞ ],0 ≤ z ≤ R 2 − r 2 co2q  .  4   Now, V′ = ∫

p /4

0

=

∫

p /4

0

 R  R2 − r 2 cos2 q   dz  r dr  dq  ∫0  ∫0    

(∫ r R

0

)

R 2 − r 2 cos2 q dr dq

p /4 3/2 R 1 2 2 2 = ∫0 − 3 cos2 q ( R − r cos q )  0 dq

=

R 3 p / 4 1 − sin 3 q = dq , now let u cosq 3 ∫0 cos2 q

2  R3  1 − u2 p /4 2 = du   [ tan q ]0 + ∫1 2  u 3  

=

=

MVC.CH03_3PP.indd 299

2 R3  −1  1 −  u + u 12 3 

  

2 −1 3 R . 2

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300 • Multivariable and Vector Calculus 2/E Finally,

(

)

V 16= V ′ 8 2 − 2 R3 . = Consider now a change to spherical coordinates = x r= cosq sin f , y r= sin q sin f , z r cos f . We have dx =cosq sin f d r − r sin q sin f d q + r cosq cos f d f , dy = sin q sin f d r + r cosq sin f d q + r sin q cos f d f , = dz cos f d r − r sin f d f . This gives dx ∧ dy ∧ dz = − r 2 sin f dr ∧ dq ∧ df . From this derivation, the form dr ∧ dq ∧ df is negatively oriented, and so we choose dx ∧ d= y ∧ dz r 2 sin f dr ∧ df ∧ dq instead. EXAMPLE 3.9.5 Let ( a, b, c ) ∈ ]0; +∞[ be fixed. Find 3

∫

R

xyzdV if

  x 2 y2 z 2 = R ( x, y, z ) ∈  3 : 2 + 2 + 2 ≤ 1, x ≥ 0, y ≥ 0, z ≥ 0  . a b c   Solution: We use spherical coordinates, where

( x, y, z ) = ( ar cosq sin f , br sinq sin f , cr cosf ) . We have dx= ∧ dy ∧ dz abcr 2 sin f dr ∧ df ∧ dr .

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Integration • 301

The integration region is mapped into = ∆

p  p × 0; .  2   2 

[0;1] × 0;

The integral becomes

( abc )2

(∫

p /2

0

cosq sin q dq

)( ∫

1

0

r 5 dr

)( ∫

p /2

0

)

cos3 f sin= f df

( abc )2 48

⋅

EXAMPLE 3.9.6 Let = V

{( x, y, z ) ∈ 

3

: x2 + y2 + z2 ≤ 9,1 ≤ z ≤ 2}. Then 2p

p / 2 − arcsin1/ 3

0

p / 2 − arcsin 2 / 3

dz ∫ ∫ ∫ dxdy= V

∫

2 / cos f

1/ cos f

r 2 sin f dr df= dq

63p ⋅ 4

EXERCISES 3.9 3.9.1

Consider the region R below the cone= z

x2 + y2 and above the

paraboloid = z x2 + y2 for 0 ≤ z ≤ 1 . Set up integrals for the volume of this region in Cartesian, cylindrical, and spherical coordinates. Also, find this volume. 3.9.2

Consider the integral

∫

R

xdV , where R is the region above the

paraboloid = z x2 + y2 and under the sphere x2 + y2 + z2 = 4 . Set up integrals for the volume of this region in Cartesian, cylindrical, and spherical coordinates. Also, find this volume. 3.9.3

Consider the region R bounded by the sphere x2 + y2 + z2 = 4 and the plane z = 1 . Set up integrals for the volume of this region in Cartesian, cylindrical, and spherical coordinates. Also, find this volume.

3.9.4 Compute

∫

E

ydV where E is the region between the cylinders

x 2 + y2 = 1 and x2 + y2 = 4 , below the plane x − z =−2 and above the xy-plane.

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302 • Multivariable and Vector Calculus 2/E

3.9.5 Compute

∫

E

y2 z2 dV where E is bounded by the paraboloid

x =1 − y2 − z2 and the plane x = 0 . 3.9.6 Compute

∫

E

z x2 + y2 + z2 dV where E is the upper solid hemi-

sphere bounded by the xy-plane and the sphere of radius 1 about the origin. 3.9.7

Compute the four-dimentional integral x2 + y2 + u2 + v2 dxdydudv. ∫∫∫ e x2 + y2 + u2 + v2 ≤1

3.9.8 Find ∫ x1 y9 z8 (1 − x − y − z ) dxdydz , where 4

R

= R

{( x, y, z ) ∈ 

3

: x ≥ 0, y ≥ 0, z ≥ 0, x + y + z ≤ 1}.

3.9.9 Find the volume for the unit ball region S defined by = S {( x, y, z ) : x2 + y2 + z2 ≤ 1} . 3.9.10 Evaluate

∫∫∫ ( x

2

1

+ z2 ) 2 dV for the solid region S, which is bounded

S

by the paraboloid = y z2 + x2 and the plane y = 4 . 3.9.11 Find the volume V for the solid region S in  3 bounded by the graphs z = − x2 − y2 + 10 , and z = 1 . 3.9.12 Find the volume V for the solid region S in  3 bounded by the graphs z2 = − x2 + y and z2 = − x2 + 2 y − 4 . 3.9.13 Find the volume V for the solid region S in  3 bounded by the graphs z = 0, z2 = − x2 − y2 + 4, y = x, y =3 x , and first octant. 3.9.14 Find the volume V for the solid region S in  3 bounded by the graphs inside z2 = − x2 − y2 + 1 and outside z2 − x2 − y2 = 0.

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Integration • 303

3.10 SURFACE INTEGRALS Definition 3.10.1 A two-dimensional oriented manifold of  3 is simply a smooth surface D ∈  3 , where the + orientation is in the direction of the outward normal pointing away from the origin and the − orientation is in the direction of the inward normal pointing toward the origin. A general oriented 2-manifold in  3 is a union of surfaces. NOTE

The surface −Σ has opposite orientation on Σ and

∫

−Σ

NOTE

w = − ∫ w. Σ

In this section, unless otherwise noticed, we will choose the positive orientation for the regions considered. This corresponds to using the ordered basis

{dy ∧ dz,dz ∧ dx,dx ∧ dy}. Definition 3.10.2  Let f :  3 →  . The integral of f over the smooth surface Σ (oriented in the positive sense) is given by expression

∫f

d2 x .

Σ

Here, d2 x =

2 2 ( dx ∧ dy) + ( dz ∧ dx )2 + ( dy ∧ dz )

is the surface area element. EXAMPLE 3.10.1 Evaluate ∫ z d 2 x where Σ is the outer surface of the section of the paraboΣ

loid, z = x2 + y2 ,0 ≤ z ≤ 1 . Solution: We parameterize the paraboloid as follows. Let = x u,= y v,= z u 2 + v2 . Observe that the domain D of Σ is the unit disk u2 + v2 ≤ 1 . We see that

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304 • Multivariable and Vector Calculus 2/E

dx ∧ dy = du ∧ dv, dy ∧ dz = −2 udu ∧ dv, dz ∧ dx = −2 vdu ∧ dv, and so 2 2 ( dx ∧ dy) + ( dz ∧ dx )2 + ( dy ∧ dz )

d2 x =

= 1 + 4 u2 + 4 v2 du ∧ dv. Now,

∫ z d x = ∫(u 2

Σ

2

+ v2 ) 1 + 4 u2 + 4 v2 dudv.

D

To evaluate this last integral, we use polar coordinates and so

∫(u

2

dudv + v2 ) 1 + 4 u2 + 4 v2=

D

=

2p

1

0

0

∫ ∫

r 3 1 + 4 r 2 dr dq

p  1  5 5 + . 12  5

EXAMPLE 3.10.2 Find the area of that part of the cylinder x2 + y2 = 2 y lying inside the sphere x 2 + y2 + z 2 = 4. Solution: We have x2 + y2 = 2 y ⇔ x2 + ( y − 1 ) = 1. We parameterize the cylinder by 2

putting = x cos u, y= − 1 sin u , and z = v . Hence dx = − sin udu, dy = cos udu, dz = dv, whence dx ∧ dy= 0, dy ∧ dz= cos udu ∧ dv,dz ∧ dx= sin udu ∧ dv,

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Integration • 305

and so d2 x =

2 2 ( dx ∧ dy) + ( dz ∧ dx )2 + ( dy ∧ dz )

= cos2 u + sin 2 u du ∧ dv = du ∧ dv. The cylinder and the sphere intersect when x2 + y2 = 2 y and x2 + y2 + z2 = 4, that is, when z2= 4 − 2 y , i.e. v2 =− 4 2 (1 + sin u ) =− 2 2 sin u . Also, 0 ≤ u ≤ p . The integral is thus p

2 − 2 sin u

0

− 2 − 2 sin u

d x ∫ ∫ ∫= 2

Σ

= dvdu

= 2 2∫

p

0

∫

p

0

2 2 − 2 sin u du

1 − sin u du

(

)

= 2 2 4 2 −4 . EXAMPLE 3.10.3 Evaluate ∫ xdydz + ( z2 − zx )dzdx − xydxdy, where Σ is the top side of the triΣ

angle with vertices at ( 2,0,0 ) , ( 0,2,0 ) , ( 0,0,4 ) . Solution: Observe that the plane passing through the three given points has equation 2 x + 2y + z = 4. We project this plane onto the coordinate axes obtaining 4

2− z / 2

0

0

xdydz ∫ ∫ ∫= Σ

∫(z Σ

2

− zx ) dzdx=

( 2 − y − z / 2= ) dydz 2

4−2 x

0

0

∫∫

(z

2

8 , 3

− zx ) dzdx= 8,

2 2−y 2 − ∫ xydxdy = − ∫ ∫ xydxdy = − , 0 0 3 Σ

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306 • Multivariable and Vector Calculus 2/E and hence,

∫ xdydz + ( z

2

− zx ) dzdx − xydxdy = 10.

Σ

EXERCISES 3.10 3.10.1 Evaluate ∫ y d 2 x where Σ is the surface Σ

z = x + y ,0 ≤ x ≤ 1,0 ≤ y ≤ 2. 2

3.10.2 Consider the cone= z

x2 + y2 . Find the surface area of the part of

the cone which lies between the planes z = 1 and z = 2 . 3.10.3 Evaluate ∫ x2 d 2 x where Σ is the surface of the unit sphere Σ 2

x +y +z = 1. 2

2

3.10.4 Evaluate

∫

S

z d 2 x over the conical surface= z

x2 + y2 between

z = 0 and z = 1 . 3.10.5 You put a perfectly spherical egg through an egg slicer, resulting in n slices of identical height, but you forgot to peel it first! Show that the amount of egg shell in any of the slices is the same. Your argument must use surface integrals. 3.10.6 Evaluate ∫ xydydz − x2 dzdx + ( x + z ) dxdy, where Σ is the top of the Σ

triangular region of the plane 2 x + 2 y + z =, 6 bounded by the first octant. 2 3.10.7 Find the surface area of the part of conical z= x2 + y2 that is

directly over the triangle in the xy-plane with vertices ( 0,0 ) , ( 4,0 ) , and ( 0,4 ) . 3.10.8 Find the surface area of the part of paraboloid = z x2 + y2 that lies directly under the plane z = 9 .

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Integration • 307

3.10.9 Find the surface integral

∫∫ ( x + y + z ) dS , where S is the portion of S

the sphere x2 + y2 + z2 = 1 , that lies in the first octant using spherical polar coordinates. 3.10.10 Find the surface area of the part of the plane 2 x + 3 y + 4 z − 12 = 0 that lies directly above the region in the first octant bounded by the graph sin 2q = r . 3.10.11 Find the surface area of the part of the graph of z − x2 + y2 = 0 that lies directly in the first octant within the cylinder x2 + y2 − 4 =. 0 3.10.12 Find the surface area of the part of the graph of 4 z2 − x2 − y2 = 0 that lies directly within the cylinder ( x − 1 ) + y2 − 1 =. 0 2

3.11 GREEN’S, STOKES’, AND GAUSS’ THEOREMS We are now in a position to state the general Stokes’ Theorem in this section. Theorem 3.11.1 (General Stokes’ Theorem): Let M be a smooth oriented manifold, having boundary ∂M . If w is a differential form, then

∫

∂M

w = ∫ dw. M

In  , if w is a 1-form, this takes the name of Green’s Theorem. 2

EXAMPLE 3.11.1 Evaluate

∫ ( x − y ) dx + x dy where C is the circle x 3

3

C

2

+ y2 = 1.

Solution: We will first use Green’s Theorem and then evaluate the integral directly. We have dw = d ( x − y3 ) ∧ dx + d ( x3 ) ∧ dy

= ( dx − 3 y2 dy ) ∧ dx + ( 3 x2 dx ) ∧ dy = ( 3 y2 + 3 x2 ) dx ∧ dy.

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308 • Multivariable and Vector Calculus 2/E

The region M is the area enclosed by the circle x2 + y2 = 1 . Thus by Green’s Theorem, and using polar coordinates,

∫ ( x − y ) dx + x dy= ∫ ( 3 y 3

3

2

C

M

=∫

2p

0

∫

1

0

+ 3 x2 ) dxdy

3 r 2 dr dq

3p ⋅ 2

= Alternative Method:

We can evaluate this integral directly, again resorting to polar coordinates. dy ∫ ( cosq − sin q ) ( − sin q ) dq + ( cos q ) ( cosq ) dq ∫ ( x − y ) dx + x = 3

2p

3

3

3

0

C

=

∫ ( sin 2p

0

4

q + cos4 q − sin q cosq ) dq .

To evaluate the last integral, observe that 1 = ( sin2 q + cos2 q ) =sin 4 q + 2 sin2 2

q cos2 q + cos4 q , hence the integral equals

∫ ( sin 2p

0

4

q + cos4 q − sin q cosq ) dq =∫

2p

0

=

(1 − 2 sin

2

q cos2 q − sin q cosq ) dq

3p ⋅ 2

In general, let = w f ( x, y ) dx + g ( x, y ) dy

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Integration • 309

be a 1-form in  2 . Then = dw df ( x, y ) ∧ dx + dg ( x, y ) ∧ dy ∂  ∂  ∂ ∂ =  f ( x, y ) dx + f ( x, y ) dy  ∧ dx +  g ( x, y ) dx + g ( x, y ) dy  ∧ dy ∂y ∂y  ∂x   ∂x  ∂  ∂ f ( x, y )  dx ∧ dy =  g ( x, y ) − ∂y  ∂x  which gives the classical Green’s Theorem 



∂ ∂ ∫ f ( x, y) dx + g ( x, y) dy= ∫  ∂x g ( x, y) − ∂y f ( x, y)  dxdy.

∂M

M

In  3 , if w is a 2-form, the above theorem takes the name of Gauss’s Theorem or the divergence Theorem. EXAMPLE 3.11.2 Evaluate

∫ ( x − y) dydz + zdzdx − ydxdy where S is the surface of the sphere S

x +y +z = 9 and the positive direction is the outward normal. 2

2

2

Solution: The region M is the interior of sphere x2 + y2 + z2 = 9 . Now, dw =

( dx − dy) ∧ dy ∧ dz + dz ∧ dz ∧ dx − dy ∧ dx ∧ dy

= dx ∧ dy ∧ dz. The integral becomes xdydz ∫ d=

M

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4p = ( 27 ) 36p . 3

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310 • Multivariable and Vector Calculus 2/E Alternative Method: We could evaluate this integral directly, we have

∫ ( x − y) dydz = ∫ xdydz, Σ

Σ

since ( x, y, z )  − y is an odd function of y and the domain of integration is symmetric with respect to y. Now

∫

Σ

xdydz=

3

∫ ∫

2p

−3 0

r 9 − r 2 dr dq= 36p .

Also,

∫

Σ

zdzdx = 0,

since ( x, y, z )  z is an odd function of z and the domain of integration is symmetric with respect to z . Similarly, 0, ∫ −ydxdy = Σ

Since ( x, y, z )  − y is an odd function of y and the domain of integration is symmetric with respect to y. Now, In general, let = w f ( x, y, z ) dy ∧ dz + g ( x, y, z ) dz ∧ dx + h ( x, y, z ) dx ∧ dy be a 2-form in  3 . Then = dw df ( x, y, z ) dy ∧ dz + dg ( x, y, z ) dz ∧ dx + dh ( x, y, z ) dx ∧ dy ∂  ∂ ∂ f ( x, y, z ) dx + f ( x, y, z ) dy + f ( x, y, z ) dz  ∧ dy ∧ dz =  ∂y ∂z  ∂x  ∂  ∂ ∂ +  g ( x, y, z ) dx + g ( x, y, z ) dy + g ( x, y, z ) dz  ∧ dz ∧ dx ∂y ∂z  ∂x  ∂  ∂ ∂ +  h ( x, y, z ) dx + h ( x, y, z ) dy + h ( x, y, z ) dz  ∧ dx ∧ dy ∂y ∂z  ∂x  ∂  ∂ ∂ f ( x, y, z ) + g ( x, y, z ) + h ( x, y, z )  dx ∧ dy ∧ dz, =  ∂y ∂z  ∂x 

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Integration • 311

which gives the classical Gauss’s Theorem

∫ f ( x, y, z ) dydz + g ( x, y, z ) dzdx + h ( x, y, z ) dxdy

∂M

=

∂

∂



∂

∫  ∂x f ( x, y, z ) + ∂y g ( x, y, z ) + ∂z h ( x, y, z )  dxdydz.

M

Using classical notation, if  f ( x, y, z )      = a = g ( x, y, z )  , dS    h ( x, y, z ) 

dydz     dzdx  ,  dxdy 

then 





∫ ( ∇a ) dV = ∫ adS. ∂M

M

The classical Strokes’ Theorem occurs when w is a 1-form in  3 . EXAMPLE 3.11.3 Evaluate

∫

C

ydx + ( 2 x − z ) dy + ( z − x ) dz where C is the intersection of the

sphere x2 + y2 + z2 = 4 and the plane z = 1 . Solution: We have dw=

( dy) ∧ dx + ( 2dx − dz ) ∧ dy + ( dz − dx ) ∧ dz

= −dx ∧ dy + 2dx ∧ dy + dy ∧ dz + dz ∧ dx = dx ∧ dy + dy ∧ dz + dz ∧ dx. Since on C, z = 1 , the surface Σ on which we are integrating is the inside of the circle x2 + y2 + 1 =, 4 i.e., x2 + y2 = 3 . Also, z = 1 implies dz = 0 and so

∫ dw = ∫ dxdy. Σ

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Σ

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312 • Multivariable and Vector Calculus 2/E

Since this is just the area of the circular region x2 + y2 ≤ 3 , the integral evaluates to

∫ dxdy = 3p . Σ

In general, let w = f ( x, y, z ) dx + g ( x, y, z ) dy + h ( x, y, z ) dz Be a 1-form in  3 . Then = dw df ( x, y, z ) ∧ dx + dg ( x, y, z ) ∧ dy + dh ( x, y, z ) ∧ dz ∂  ∂ ∂ =  f ( x, y, z ) dx + f ( x, y, z ) dy + f ( x, y, z ) dz  ∧ dx ∂y ∂z  ∂x  ∂  ∂ ∂ +  g ( x, y, z ) dx + g ( x, y, z ) dy + g ( x, y, z ) dz  ∧ dy ∂y ∂z  ∂x  ∂  ∂ ∂ +  h ( x, y, z ) dx + h ( x, y, z ) dy + h ( x, y, z ) dz  ∧ dz ∂y ∂z  ∂x   ∂  ∂ =  h ( x, y, z ) − g ( x, y, z )  dy ∧ dz ∂z  ∂y  ∂ ∂  +  f ( x, y, z ) − h ( x, y, z )  dz ∧ dx ∂x  ∂z  ∂  ∂ f ( x, y, z )  dx ∧ dy +  g ( x, y, z ) − ∂y  ∂x 

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Integration • 313

which gives the classical Strokes’ Theorem.

∫ f ( x, y, z ) dx + g ( x, y, z ) dy + h ( x, y, z ) dz

∂M

 ∂



∂

∫  ∂y h ( x, y, z ) − ∂z g ( x, y, z )  dydz

=

M

∂ ∂  +  g ( x, y, z ) − f ( x, y, z )  dxdy ∂x  ∂z  ∂  ∂ +  h ( x, y, z ) − f ( x, y, z )  dxdy. ∂y  ∂x  Using classical notation, if  f ( x, y, z )      = a  g ( x, y= , z )  , dr  h ( x, y, z )   

 dx     dy , dS =  dz 

dydz     dzdx  ,  dxdy 

then 







∫ ( ∇ × a )  dS = ∫ a  dr.

M

∂M

EXERCISES 3.11 3.11.1 Evaluate

∫

C

x3 ydx + xydy where C is the square with vertices at

( 0,0 ) , ( 2,0 ) , ( 2,2 ) , and ( 0,2 ) . 3.11.2 Consider the triangle ∆ with vertices A : ( 0,0 ) , B : (1,1 ) , C : ( −2,2 ) . 1. If LPQ denotes the equation of the line joining P and Q , find LAB , LAC , and LBC . 2. Evaluate 3. Find

∫

∆

y2 dx + xdy.

∫ (1 − 2 y) dx ∧ dy where D is the interior of ∆ .

D

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314 • Multivariable and Vector Calculus 2/E   ∫ f ( a )da , where Γ is the boundary of the rectangle    D= {( x, y ) : x ∈ [ 0,2 ], y ∈ [ 0,1]} when f = ( a ) 5 x i − 6 xy j .       Prove that ∫ f  dr = 3p , where f ( x, y, z ) = z i + x j + yk and C is the

3.11.3 Find

Γ

3

3.11.4

C     curve r = cos t i + sin t j + tk when t ∈ [ 0,2p ] .

3.11.5 Use Green’s Theorem to prove that

∫ (x

2

Γ

+ 2 y3 ) dy = 16p , where Γ

is the circle ( x − 2 ) + y2 = 4 . Also, prove this directly by using a 2

path integral. 3.11.6 Let Γ denote the curve of intersection of the plane x + y = 2 and the sphere x2 − 2 x + y2 − 2 y + z2 = 0 , oriented clockwise when viewed from the origin. Use Stokes’ Theorem to prove that −2p 2 ⋅ Prove this directly by parametrizing the ∫ ydx + zdy + xdz = Γ

boundary of the surface and evaluating the path integral. 3.11.7 Let M ⊂  2 be the upper semi-disk of radius R. Find 2 ∫ ∂M ( x dx + 2 xydy) using Green’s Theorem.   3.11.8 Evaluate ∫ f  dr using Green’s Theorem, where    f ( x, y, z ) = ( 2 xy − x2 ) i + ( x + y2 ) j and C is the boundary of the region Γ defined by the curves y = x2 and y = x , 0 ≤ x ≤ 1 . 3.11.9 Evaluate

∫

C

4 x2 y dx + 2 y dy using Green’s Theorem, where C is

the boundary of the triangle with vertices in Figure 3.11.1.

Figure 3.11.1  Exercise 3.11.7.

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Integration • 315

3.11.10 Evaluate

∫ ( x + y ) dx + (1 + x ) dy using Green’s Theorem, 2

2

C

where C is the closed curve determined by y = x3 and y = x2 form

( 0,0 ) to (1,1 ) . 3.11.11 Evaluate

∫

C

e x sin y dx + e x cos y dy using Green’s Theorem, where

C is the ellipse 3 x2 + 8 y2 = 24 . 3.11.12 Evaluate

∫ ( y − sin x ) dx + cos x dy using Green’s Theorem in the C

plane, where C is the triangle of the Figure 3.11.2.

Figure 3.11.2  Exercise 3.11.10.

3.11.13 Evaluate

∫ ( 2 xy − x ) dx + ( x + y ) dy using Green’s Theorem in 2

2

C

the plane, where C is the closed curve of the region bounded by y = x2 and y2 = x intersect form ( 0,0 ) to (1,1 ) . 3.11.14 Evaluate

∫ ( ( 2 x C

2

)

+ 2 xy ) dx + ( x2 + xy + y2 ) dy using Green’s

Theorem in the plane, where C is the square with vertices ( 0,0 ) , (1,0 ) , (1,1 ) , and ( 0,1) , as in Figure 3.11.3.

Figure 3.11.3  Exercise 3.11.12.

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316 • Multivariable and Vector Calculus 2/E

3.11.15 Evaluate

∫ ( x + y ) dx + ( 2 x 2

C

2

− y ) dy using Green’s Theorem,

where C is the Boundary of the region determined by the graphs of y = x2 and y = 4 . 3.11.16 Evaluate

∫

C

∫

C

2

e x dx + 2 arctan x dy using Green’s Theorem, where C

is the Triangle with vertices ( 0,0 ) , ( 0,1 ) , and ( −1,1 ) . 3.11.17 Evaluate

xy2 dx + 3 cos y dy using Green’s Theorem, where C is

the boundary of the region in the first quadrant determined by the graphs of y = x2 and y = x3 . 3.11.18 Evaluate

∫

C

y3 dx + ( x3 + 3 xy2 ) dy using Green’s Theorem, where

C is the path from ( 0,0 ) to (1,1 ) along the graph y = x3 and from

(1,1 ) to ( 0,0 ) along the graph of y = x . 3.11.19 Use Gauss’s Theorem to find the outward flux of the vector field  x2     f = 2 yz  across the region M bounded by the parallelepiped  4 z3    0 ≤ x ≤ 1,

0 ≤ y ≤ 2,

0 ≤ z ≤ 3.

 x2 y     3.11.20 Use Gauss’s Theorem to find the flux of the vector field f = 2 xz   yz3    across the surface of the rectangle solid M as in Figure 3.11.4, where 0 ≤ x ≤ 1, 0 ≤ y ≤ 2, 0 ≤ z ≤ 3

Figure 3.11.4  Exercise 3.11.18.

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Integration • 317

3.11.21 Use Gauss’s Theorem to find the outward flux of the vector field 4 x    f =  y  across the region M bounded by the sphere  4 z  x 2 + y2 + z 2 = 4. 3.11.22 Use Gauss’s Theorem to find the outward flux of the vector field  xy2    2  f =  x y  across the region M bounded by the cone= z x 2 + y2 6 sin x    and the planes z = 2 and z = 4 . 3.11.23 Use Gauss’s Theorem to find the outward flux of the vector field  4xz    2 f =  − y  across the region M of the cube bounded by  yz  = x 0,= x 1,= y 0,= y 1,= z 0 , and z = 1 . 3.11.24 Use Gauss’s Theorem to find the outward flux of the vector field  y3 e z     = f  − xy  across the region M bounded by the coordinate  x tan −1 y   planes= x 0,= y 0,= z 0 and the plane x + y + z =. 1   3.11.25 Use Stokes’s Theorem to find ∫∫ ∇ × f  n dS for the vector field S

(

)

y    f =  − x  across the region S, which is paraboloid = z x2 + y2 with  yz  the circle x2 + y2 = 1 and z = 1 as its bounded.   3.11.26 Use Stokes’s Theorem to find ∫ f  dr for the vector field C

 x2    2 f =  y  across the region S, which is part of the cone  z2    = z

MVC.CH03_3PP.indd 317

x2 + y2 cutoff by the plane z = 1 .

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APPENDIX

A

Maple

Maple is interactive mathematical and analytical software designed to p ­ erform a wide variety of mathematical calculations as well as operations on symbolic, numeric entities, and modeling. In this appendix, we give a general overview of Maple. For more information on Maple, visit the maple website: www. maplesoft.com.

A.1 GETTING STARTED AND WINDOWS OF MAPLE When you double-click on the Maple icon, it opens as shown in Figure A.1. This figure shows Maple in the document mode. The worksheet mode is shown in Figure A.2, where the special [> prompt appears. This is the main area in which the user interacts with Maple. For general help, click on Help then Maple Help in menu bar as shown in Figure A.3. Also, Maple uses the question mark (?), followed by the command or topic name, to get help. For example, to get help on solve, you type ?solve. To terminate the Maple session, from the File menu, select Exit.

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320 • Multivariable and Vector Calculus 2/E

FIGURE A.1  Default environment (document mode).

FIGURE A.2  Worksheet mode.

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Maple • 321

FIGURE A.3  Maple help system.

A.2 ARITHMETIC Maple can do arithmetic operations like a calculator. Table A.1 provides Maple’s common arithmetic operations. To evaluate an arithmetic expression, type the expression and then press the Enter key. TABLE A.1  Maple common arithmetic operations.

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Operation

Descriptions

+

addition

-

subtraction

∗

multiplication

/

division

^

exponentiation

!

factorial

abs (n)

Absolute value of n

sqrt (n)

Square root of n

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322 • Multivariable and Vector Calculus 2/E Maple uses, pi, command to present p and uses, exp (1), command to present e . EXAMPLE A.2.1 Calculate 2 5 −

(8 + 6) 4

+ 3×9.

Solution:

>

EXAMPLE A.2.2 Simple numerical calculation 5 + −3 . Solution:

>

A.3 SYMBOLIC COMPUTATION Maple can do a variety of symbolic calculations. For example,

> Maple also makes simplifications to the expression when you use the command simplify. For example,

> The expand and factor commands are used to expand and factor the expression respectively. For example,

> >

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Maple • 323

A.4 ASSIGNMENTS To assign values to a variable, Maple uses colon equals (:=). For example,

> > To clear the value of the variable x, type

>

A.5 WORKING WITH OUTPUT One percent sign (%) refers to the output of the previous command. Two percent signs (%%) refer to the second-to-last output and three percent signs (%%%) to the third-to-last output. Maple remembers the output of the last three statements you entered. For example,

> > > >

A.6 SOLVING EQUATIONS Maple uses solve command to solve equations. For example,

>

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324 • Multivariable and Vector Calculus 2/E We can solve equations with more than one variable for a specific variable. For example,

> >

A.7 PLOTS WITH MAPLE Maple uses the basic plotting command, plot, to plot functions, expressions, list of points, and parametric functions. For example, to plot the graph of y= 3 x2 − x + 1 on the interval -1 to 1, type

>

Also, we can plot several functions or expressions on same graph. For example,

>

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Maple • 325

Maple allows you to annotate a plot by adding text and drawings by clicking on the plot. The Plot options tool bar will show up. Then click on Drawing button and the drawing tool bar will show up. For example,

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326 • Multivariable and Vector Calculus 2/E

A.8 LIMITS AND DERIVATIVES Maple can evaluate limits of lim f ( x) by using limit (f(x), x =a); command. x→ a

For example,

> Maple uses diff command to compute derivatives. For example,

>

A.9 INTEGRATION Maple uses int command to compute integrals. For example,

>

A.10 MATRIX Maple uses Matrix command to make a matrix. For example,

> > > >

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APPENDIX

B

Matlab

MATLAB has become a useful and dominant tool for technical professionals around the world. MATLAB is an abbreviation for Matrix Laboratory. It is a numerical computation and simulation tool that uses matrices and vectors. Also, MATLAB enables the user to solve wide analytical problems. A copy of MATLAB software can be obtained from: The Mathworks, Inc. 3 Apple Hill Drive Natick, MA 01760-2098 Phone: 508-647-7000 Website: http://www.mathworks.com This brief introduction of MATLAB (R2010b) is presented here to give a general idea about the software. MATLAB computational applications to science and engineering systems used to solve practical problems.

B.1 GETTING STARTED AND WINDOWS OF MATLAB When you double-click on the MATLAB icon, it opens as shown in Figure B.1. The command window, where the special >> prompt appears, is the main area in which the user interacts with MATLAB. To make the Command Window active, you need to click anywhere inside its border. To quit MATLAB, you can select EXIT MATLAB from the File menu, or by entering quit or exit at the Command Window prompt. Do not click on the X (close box) in the

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328 • Multivariable and Vector Calculus 2/E top right corner of the MATLAB window, because it may cause problems with the operating software. Figure B.1 contains four default windows, which are Command Window, Workplace Window, Command History Window, and Current Folder Window. Table B.1 shows a list of the various windows and their purpose of MATLAB.

Figure B.1  MATALB default environment. TABLE B.1  MATLAB windows Window

Description

Command Window

Main window, enter variables, runs programs

Workplace Window

Gives information about the variable used

Command History Window

Records command entered in the Command Window

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TABLE B.1  MATLAB windows (Continued) Window

Description

Current Folder Window

Shows the files in the current directory with details

Editor Window

Makes and debugs script and function files

Help Window

Gives help information

Figure Window

Contains output from the graphic commands

Launch Pad window

Provides access to tools, demos, and documentation

B.1.1 Using MATLAB in Calculations Table B.2 shows the MATLAB common arithmetic operators. The order of operations as first, parentheses ( ), the innermost are executed first for nested parentheses; second, exponentiation ^; third, multiplication ∗ and division / (they are equal precedence); fourth, addition + and subtraction -. TABLE B.2  MATLAB common arithmetic operators Operator symbols

Descriptions

+

Addition

-

Subtraction

∗

Multiplication

/ \

Right division (means Left division (means

a ) b

b ) a

^

Exponentiation (raising to a power)

‘

Converting to complex conjugate transpose

()

Specify evaluation order

For example, >> a = 11; b = -3; c = 5; >> x = 9∗a + c^2 - 2 x=  122

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330 • Multivariable and Vector Calculus 2/E >> y = sqrt(x)/6 y=  1.8409 Table B.3 provides a common sample of MATLAB functions. You can obtain more by typing help in the Command Window (>> help). TABLE B.3  Typical elementary math functions Function

Description

abs (x)

Absolute value or complex magnitude of x

acos (x), acosh (x)

Inverse cosine and inverse hyperbolic cosine of x (in radians)

angle (x)

Phase angle (in radians) of a complex number x

asin (x), asinh (x)

Inverse sine and inverse hyperbolic sine of x (in radians)

atan (x), atanh (x)

Inverse tangent and inverse hyperbolic tangent of x (in radians)

conj (x)

Complex conjugate of x (in radians)

cos (x), cosh (x)

Cosine and inverse hyperbolic cosine of x (in radians)

cot (x), coth (x)

Inverse cotangent and inverse hyperbolic cotangent of x (in radians)

exp (x)

Exponential of x

Fix

Round toward zero

imag (x)

Imaginary part of a complex number x

log (x)

Natural logarithm of x

log2 (x)

Natural logarithm of x to base 2

log10 (x)

Common logarithms (base 10) of x

real (x)

Real part of a complex number of x

sin (x), sinh (x)

Sine and inverse hyperbolic sine of x (in radians)

sqrt (x)

Square root of x

tan (x), tanh (x)

Tangent and inverse hyperbolic tangent of x (in radians)

For example, >> 9+3^(log2(4.25)) ans =  18.9077

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>> y=5∗cos(pi/4) y=  3.5355 >> z = exp(y+6) z=  1.3843e+004 In addition to operating on mathematical functions, MATLAB allows us to work easily with vectors and matrices. A vector (or one-dimensional array) is a special matrix (or two-dimensional array) with one row or one column. Arithmetic operations can apply to matrices and Table B.4 shows extra common operations that can be implemented to matrices. TABLE B.4  Matrix operations Operations

Descriptions

A'

Transpose of matrix A

det (A)

Determinant of matrix A

inv (A)

Inverse of matrix A

eig (A)

Eigenvalues of matrix A

diag (A)

Diagonal elements of matrix A

A vector can be created by typing the elements inside brackets [ ] from a known list of numbers. For example, >> A = [1 2 3 6 5 22] A=  1 2 3 6 5 22 >> B = [1 2 6; 8 9 11; 10 14 16]

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332 • Multivariable and Vector Calculus 2/E B=  1 2 6  8 9 11  10 14 16 Also, a vector can be created with constant spacing by using the command variable-name = [a: n: b], where a is the first term of the vector; n is spacing; b is the last term. For example, >> x = [1:0.6:5] x=  1.0000 1.6000 2.2000 2.8000 3.4000 4.0000 4.6000 Also, a vector can be created with constant spacing by using the command variable-name = linespace (a, b, m), where a is the first element of the vector; b is the last element; m is the number of elements. For example, >> x=linspace(0,4∗pi,6) x=   0 2.5133 5.0265 7.5398 10.0531 12.5664 Examples using Table B.4: >> B = [1 2 4; 7 8 9; 3 5 10] B=  1 2 4  7 8 9  3 5 10

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>> D=B^2 D=  27 38 62  90 123 190  68 96 157 >> C= A’ C=  1  2  3  6  5  22 >> E =[4 8;11 25]; >> inv(E) ans =  2.0833 -0.6667  -0.9167 0.3333 >> det(E) ans =  12

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334 • Multivariable and Vector Calculus 2/E Special constants can be used in MATLAB. Table B.5 provides special ­constants used in MATLAB. TABLE B.5  MATLAB named constants Name

Content

Pi

p = 3.14159...

i or j

Imaginary unit,

Eps

Floating-point relative precision, 2 −52

Realmin

Smallest floating-point number, 2 −1022

Realmax

Largest floating-point number,(2-eps). 21023

Bimax

Largest positive integer, 2 53 − 1

Inf or Inf

Infinity

nan or NaN

Not a number

Rand

Random element

Eye

Identity matrix

Ones

An array of 1’s

Zeros

An array of 0’s

−1

For example, >> eye(3) ans =  1 0 0  0 1 0  0 0 1 >> ones(3) ans =  1 1 1  1 1 1  1 1 1

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>> 1/0 ans =  Inf >> 0/0 ans =  NaN Arithmetic operations on arrays are done element by element. Table B.6 provides MATLAB common Arithmetic operations on arrays. TABLE B.6  MATLAB common arithmetic operations on arrays Operator symbols on arrays

Descriptions

+

Addition same as matrices

-

Subtraction same as matrices

.∗

Element-by-element multiplication

./

Element-by-element right division

.\

Element-by-element left division

.^

Element-by-element power

.’

Unconjugated array transpose

For example, >> A=[-1 2 0; 3 5 7; 8 9 9] A=  -1 2 0  3 5 7  8 9 9 >> A.∗A

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336 • Multivariable and Vector Calculus 2/E ans =  1 4 0  9 25 49  64 81 81 >> C=[0 1;2 4; 7 8]; >> D=[8 11;10 16;14 18]; >> C./D ans =   0 0.0909  0.2000 0.2500  0.5000 0.4444 >> C.^2 ans =  0 1  4 16  49 64

B.2 PLOTTING MATLAB has nice capability to plot in two-dimensional and three-dimensional plots. B.2.1 Two-dimensional Plotting First, we start with two-dimensional plots. The plot command is used to create two-dimensional plots. The simplest form of the command is plot (x,y). The arguments x and y are each a vector (one-dimensional array). The vectors x and y must have the same number of elements. When the plot command is executed a figure will be created in the Figure Window. The plot (x, y, ‘line

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specifiers’) command has additional optional arguments that can be used to detail the color and style of the lines. Tables B.7 through B.9 show various types of lines, points, and color types used in MATLAB. TABLE B.7  MATLAB various line styles Line types

MATLAB symbol

Solid (default)

-

Dashed

--

Dotted

:

Dash-dot

-. TABLE B.8  MATLAB various point styles

Point type

MATLAB symbol

Asterisk

∗

Plus sign

+

x-mark

x

Circle

o

Point

.

Square

s TABLE B.9  MATLAB various line color types

Color

MATLAB symbol

Black

K

Blue

B

Green

G

Red

R

Yellow

Y

Magenta

M

Cyan

C

White

W

For example, >> x=0:pi/50:2∗pi;%x is a vector, 0 y=3∗sin(2∗pi∗x);% y is a vector >> plot(x,y,’--b’)%creates the 2D plot with blue and dashed line

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The command fplot(‘function’, limits, line specifiers) is used to plot a function with the form y = f (x), where the function can be typed as a string inside the command. The limits are a vector with two elements that specify the domain x [xmin,xmax], or is a vector with four elements that specifies the domain of x and the limits of the y-axis [xmin, xmax, ymin, ymax]. The line specifiers are used the same as in the plot command. For example, >> fplot(‘x^3+2∗sin(3∗x)-2’,[-5,5],’xr’)

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Also, we can create a plot for a function y = f(x) using the command plot by creating a vector of values of x for the domain that the function will be plotted, then creating y with corresponding values of f(x). For example, >> x=[0:0.1:1]; >> y=cos(2∗pi∗x); >> plot(x,y,’ro:’)

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In MATLAB several graphs can be plotted at the same plot in two ways: first, using plot command by typing pairs of vectors inside the Plot command such as Plot(x,y, z, t, u, h), which will create three graphs: y vs. x, t vs. z, and h vs. u, all in the same plot. For example, the command to plot the function y = 2 x3 − 15 x + 5 , its first derivative = y′ 6 x2 − 15 , and its second derivative y′′ = 12 x , for domain −3 ≤ x ≤ 6 , all in the same plot, is as follows: >> x=[-3:0.01:6];% vector x with the domain of the function >> y=2∗x.^3-15∗x+5;% vector y with the function value at each x >> yd=6∗x.^2-15;% vector yd with the value of the first derivative

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>> ydd=12∗x; %vector ydd with the value of the second t derivative >> plot(x,y,’-r’,x,yd,’:b’,x,ydd,’--k’)

Second, using hold on, hold off commands. The hold on command will hold the first plotted graph and add to it extra figures for each time the plot command is typed. The hold off command stops the process of hold on command. For example, if we use the previous example, we get the same result using the following commands: >> x=[-3:0.01:6]; >> y=2∗x.^3-15∗x+5; >> yd=6∗x.^2-15; >> ydd=12∗x;

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342 • Multivariable and Vector Calculus 2/E >> plot(x,y,’-r’) >> hold on % the first graph is created >> plot(x,yd,’:b’) % second graph is added to the figure >> plot(x,ydd,’--k’) % third graph is added to the figure >> hold off

Plots in MATLAB can be formatted using commands that follow the plot commands, or by using the plot editor interactively in the Figure window. First, format the plot using commands as follows: „„

„„

„„

Labels can be placed next to the axes with the xlabel (‘text as string ‘) for the x-axis and ylabel (‘text as string ‘) for the y-axis. The command title(‘text as string ‘) is a title command which can be added to the plot to place the title at the top of the figure as a text. There are two ways to place a text label in the plot. First, using text (x,y,’text as string ‘) command which is used to place the text in the figure such that the first character is positioned at the point with the coordinates x, y according to the

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„„

axes of the figure. Second, using gtext (‘text as string ‘) command which is used to place the text at a position specified by the user’s mouse in the figure window. The command legend (‘string1’, ‘string2’,…,pos) is used to place a legend on the plot. The legend command shows a sample of the line type of each graph that is plotted and, places a label specified by the user, beside the line sample. The strings in the command are the labels that are placed next to the line sample and their order corresponds to the order that the graphs were created. The pos in the command is an optional number that specifies where in the figure the legend is placed. Table B.10 shows the options that can be used for pos.

The command axis is used to change the range and the appearance of the axes of the plot, based on the minimum and maximum values of the elements of x and y. Table B.11 shows some common possible forms of axis command. „„

The command grid on is used to add grid lines to the plot and the command grid off is used to remove grid lines from the plot. TABLE B.10  Options that can be used for pos

Pos value

Description

-1

Place the legend outside the axes boundaries on the right side

0

Place the legend inside the axes boundaries in a location that interferes the least with graph

1

Place the legend at the upper-right corner of the plot (this is the default)

2

Place the legend at the upper-left corner of the plot

3

Place the legend at the lower-left corner of the plot

4

Place the legend at the lower-right corner of the plot TABLE B.11  Some common axis commands

axis command

Description

axis ([xmin, xmax, ymin, ymax])

Sets the limits of both the x and y axes (xmin, xmax, ymin, ymax are numbers)

axis equal

Sets the same scale for both axes

axis tight

Sets the axis limits to the range of the data

axis square

Sets the axes region to be square

For example, >> x=0:pi/30:2∗pi;y1=exp(-2∗x);y2=sin(x∗3); >> plot(x,y1,’-b’,x,y2,’--r’) >> xlabel(‘x’)

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344 • Multivariable and Vector Calculus 2/E >> ylabel(‘y1 , y2’) >> title(‘y1=exp(-2∗x), y2=sin(x∗3)’) >> axis([0,11,-1, 1]) >> text(6,0.6,’Comparison between y1 and y2’) >> legend(‘y1’,’y2’,0)

In MATLAB, users can use Greek characters in the text by typing \name of the letter within the string as in Table B.12. TABLE B.12  Some common Greek characters Greek characters in the string

Greek letter

Greek characters in the string

Greek letter

\alpha

a

\Phi

Φ

\beta

b

\Delta

∆

\gamma

g

\Gamma

Γ

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TABLE B.12  Some common Greek characters (Continued) Greek characters in the string

Greek letter

Greek characters in the string

Greek letter

\theta

q

\Lambda

Λ

\pi

p

\Omega

Ω

\sigma

s

\Sigma

Σ

To get a lowercase Greek letter, the name of the letter must be typed in all lowercase. To get a capital Greek letter, the name of the letter must start with a capital letter. Second, format the plot using the plot editor interactively in the Figure window. This can be done by clicking on the plot and/or using the menus as illustrated in the following figure.

MATLAB can use logarithm scaling for a two-dimensional plot. Table B.13 shows MATLAB commands for logarithm scaling. TABLE B.13  Two-dimensional graphic for logarithm scaling

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Command

Description

Loglog

To plot log(y) versus log(x)

Semilogx

To plot y versus log(x)

Semilogy

To plot log(y) versus x

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346 • Multivariable and Vector Calculus 2/E Also, MATLAB can make plots with special graphics as in Table B.14. TABLE B.14  MATLAB plots with special graphics Command

Description

bar(x,y)

Vertical bar plot

barh(x,y)

Horizontal bar plot

stairs(x,y)

Stairs plot

stem(x,y)

Stem plot

pie(x)

Pie plot

hist(y)

Histogram plot

polar(x,y)

Polar plot

For example, >> t=[0:pi/60:2∗pi]; >> r=3+2∗cos(t); >> polar(t,r,’r.’)

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B.2.2 Three-Dimensional Plotting MATLAB has the capability to make a graph in three-dimensional plots using line, mesh, and surface plots. The command plot3(x,y,z) is used in a three-dimensional line plot which is a line that is obtained by connecting points in three-dimensional space. For example, >> x=linspace(0,10∗pi,100); >> y=cos(x);z=sin(x); >> plot3(x,y,z,’r’);grid on >> xlabel(‘x’);ylabel(‘cos(x)’);zlabel(‘sin(x)’)

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348 • Multivariable and Vector Calculus 2/E Another example, > t = -5:0.1:5; >> x = (3+t.^2).∗sin(50∗t); >> y = (3+t.^2).∗cos(50∗t); >> z = 5∗t; >> plot3(x,y,z,’g’) >> grid on >> xlabel(‘x(t)’),ylabel(‘y(t)’),zlabel(‘z(t)’) >> title(‘plot3 example’)

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Also, The command mesh(X,Y,Z) is used in a three-dimensional plot that is applied to plotting functions z = f(x,y). This can be done by creating a grid in the x-y plane that covers the domain of the function, then calculating the value of z at each point of the grid, and then creating the plot. For example, >> x=(-6:0.1:6);y=(-6:0.1:6);[X,Y]=meshgrid(x,y); >> Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2)); >> mesh(X,Y,Z) >> zlabel(‘Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2))’) >> xlabel(‘X’) >> ylabel(‘Y’)

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350 • Multivariable and Vector Calculus 2/E Table B.15 provides other common mesh plot types. TABLE B.15  Provides other common mesh plot types Mesh plot types

Description

meshz(X,Y,Z)

Mesh curtain plot which draws a curtain around the mesh

meshc(X,Y,Z)

Mesh and contour plot which draws a contour plot beneath the mesh

waterfall(X,Y,Z)

Draws a mesh in one direction only

For example, >> x=(-6:0.1:6);y=(-6:0.1:6);[X,Y]=meshgrid(x,y); >> Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2)); >> meshz(X,Y,Z) >> xlabel(‘X’) >> ylabel(‘Y’) >> zlabel(‘Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2))’)

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>> x=(-6:0.1:6);y=(-6:0.1:6);[X,Y]=meshgrid(x,y); >> Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2)); >> meshc(X,Y,Z) >> zlabel(‘Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2))’) >> ylabel(‘Y’) >> xlabel(‘X’)

Another command surf(X,Y,Z) is used in a three-dimensional plot that is applied to plotting functions z = f(x,y) as in Mesh. This can be done by the same step for the mesh. For example, >> x=(-6:0.1:6);y=(-6:0.1:6);[X,Y]=meshgrid(x,y); >> Z=sin(X.^2+Y.^2).∗exp(-0.6∗(X.^2+Y.^2));

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352 • Multivariable and Vector Calculus 2/E >> surf(X,Y,Z) >> xlabel(‘X’) >> ylabel(‘Y’) >> zlabel(‘Z=sin(X.^2+Y.^2).∗exp(-0.4∗(X.^2+Y.^2))’)

There are other common surface plot types as in Table B.16. TABLE B.16  Provides other common surface plot types Surface plot types

Description

surfl(X,Y,Z)

Surface plot with lighting

surfc(X,Y,Z)

Surface and contour plot which draws a contour plot beneath the mesh

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B.3 PROGRAMMING IN MATLAB So far we have used MATLAB commands and they were executed in the Command Window. This way is fine for a simple tasks, but for more complex ones, it becomes less convenient and difficult because the Command Window cannot be saved and executed again. Therefore, the commands and programs can be stored in a file. To begin, tell MATLAB to get its input from the file but this file must be created as an M-file. Do this by clicking on File/New/scripts to open a new file in the MATLAB Editor/Debugger or simple text editor, then type the program and save it by choosing save from the File menu. The file should be saved with an extension “.m”. For example, we created the program nano1.m using M-file as follows.

We typed nano1 in the Command Window, then hit enter to obtain the following figure.

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MATLAB uses flow control through its programs. To allow flow control in a program certain rational and logical operators are essential. These operators are shown in Tables B.17 and B.18. TABLE B.17  Rational operators Rational operators

Description




Greater than

=

Greater than or equal

==

Equal

~=

Not equal

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TABLE B.18  Logical operators Logical operator

Description

~

NOT

&

AND

|

OR

There are four kind of statements used in MATLAB to control the flow through the user code. They are for loops, while loops, if, else, and elseif constructions, and switch constructions. B.3.1 For Loops for loops allow a group of commands to be repeated a fixed number of times. The basic form of a for loop is: for index = start: increment: stop   statements end The increment can be omitted, but MATLAB will assume the increment is 1. Also, the increment can be positive or negative. For example, >> for n=1:7 x(n)=sin(n∗pi/10) end x = 0.3090 0.5878 0.8090 0.9511 1.0000 0.9511 0.8090 The general form of a for loop is: for x = array   commands… end

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356 • Multivariable and Vector Calculus 2/E For example, >> m =[1 5 8; 10 17 22] m=  1 5 8  10 17 22 >> for n = m x=n(1)-n(2) end x=  -9 x=  -12 x=  -14 B.3.2 While Loops while loop evaluates a group of statements an indefinite number of times in conjunction with a conditional statement. The general form of while loop is: while expression   commands… end For example, >> n=100; >> x=[];

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>> while (n>0) n=n/2-1; x=[x,n]; end >> x x=  49.0000 23.5000 10.7500 4.3750 1.1875 -0.4063 B.3.3 If, Else, and Elseif The form of an if statement is: if expression   statements end The expression can be either 1 (true) or 0 (false). The statements between the if and end statements are executed if the expression is true. If the expression is false the statements will be ignored and the execution will resume at the line after the end statement. An end keyword matches the if and terminates the last group of statements. For example, >> x= 20; >> if x>0 log(x) end ans =  2.9957 The optional elseif and else keywords provide for the execution of alternate groups of statements.

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358 • Multivariable and Vector Calculus 2/E The if and else can be presented as: if condition   statements else   statements end For example, using just if-else statement as: >> x= -24; >> if x>0 log(x) else ‘x is negative number’ end ans = x is negative number The if, else, and elseif can be presented as: if  condition 1    statements elseif condition 2    statements else  condition 3    statements end

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For example using if, else, and elseif statements as: >> x=’28’; >> if ~isnumeric(x) ‘x is not a number’ elseif isnumeric(x)&x> x= 9; >> switch x case 1 disp(‘x is 1’) case {7 , 8, 9} disp(‘x is 7, 8, and 9’) case 12 disp(‘x is 12’) otherwise disp(‘x is not, 1, 7, 8, 9 and 12’) end x is 7, 8, and 9 Considering the following tips can be helpful in working with MATLAB: 1. Variables and function names are case-sensitive. 2. Make comment in M-file by adding lines beginning with a % character. 3. Use a semicolon (;) at the end of each command to suppress output and the semicolon can be removed when debugging the file. 4. Retrieve previously executed commands by pressing the up ( ) and down ( ) arrow keys. 5. Use an ellipse (…) at the end of the line and continue on the next line, when an expression does not fit on one line.

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Matlab • 361

B.4 SYMBOLIC COMPUTATION In previous sections, you learned that MATLAB can be a powerful programmable calculator. However, basic MATLAB uses numbers as in a calculator. Most calculators and basic MATLAB lack the ability to manipulate math expressions without using numbers. In this section, you see that MATLAB can manipulate and solve symbolic expressions that make you compute with math symbols rather than numbers. This process is called symbolic math. Table B.19 shows some common Symbolic commands. You can practice some symbolic expressions in the following section. B.4.1 Simplifying Symbolic Expressions Symbolic simplification is not always straightforward; there is no universal simplification function because the meaning of a simplest representation of a symbolic expression cannot be defined clearly. MATLAB uses the sym or syms command to declare variables as symbolic variable. Then, the symbolic can be used in expressions and as arguments to many functions. For example, to rewrite a polynomial in a standard form, use the expand function: >> syms x y; % creating a symbolic variables x and >> x = sym(‘x’); y = sym(‘y’); % or equivalently >> expand (sin(x+y)) ans =      sin(x) ∗cos(y) + cos(x)∗ sin(y) You can use subs command to substitute a numeric value for a symbolic variable or replace one symbolic variable with another. For example, >> syms x; >> f=3∗x^2-7∗x+5; >> subs(f,2) ans =           3

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362 • Multivariable and Vector Calculus 2/E >> simplify (sin(x)^2 + cos(x)^2) % Symbolic simplification ans =           1 TABLE B.19  Common symbolic commands Command

Description

diff

Differentiate symbolic expression

int

Integrate symbolic expression

jacobian

Compute Jacobian matrix

limit

Compute limit of symbolic expression

symsum

Evaluate symbolic sum of series

taylor

Taylor series expansion

colspace

Return basis for column space of matrix

det

Compute determinant of symbolic matrix

diag

Create or extract diagonals of symbolic matrices

eig

Compute symbolic eigenvalues and eigenvectors

expm

Compute symbolic matrix exponential

inv

Compute symbolic matrix inverse

jordan

Compute Jordan canonical form of matrix

null

Form basis for null space of matrix

poly

Compute characteristic polynomial of matrix

rank

Compute rank of symbolic matrix

rref

Compute reduced row echelon form of matrix

svd

Compute singular value decomposition of symbolic matrix

tril

Return lower triangular part of symbolic matrix

triu

Return upper triangular part of symbolic matrix

coeffs

List coefficients of multivariate polynomial

collect

Collect coefficients

expand

Symbolic expansion of polynomials and elementary functions

factor

Factorization

horner

Horner nested polynomial representation

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Matlab • 363

TABLE B.19  Common symbolic commands (Continued) Command

Description

numden

Numerator and denominator

simple

Search for simplest form of symbolic expression

simplify

Symbolic simplification

subexpr

Rewrite symbolic expression in terms of common subexpressions

subs

Symbolic substitution in symbolic expression or matrix

compose

Functional composition

dsolve

Symbolic solution of ordinary differential equations

finverse

Functional inverse

solve

Symbolic solution of algebraic equations

cosint

Cosine integral

sinint

Sine integral

zeta

Compute Riemann zeta function

ceil

Round symbolic matrix toward positive infinity

conj

Symbolic complex conjugate

eq

Perform symbolic equality test

fix

Round toward zero

floor

Round symbolic matrix toward negative infinity

frac

Symbolic matrix elementwise fractional parts

imag

Imaginary part of complex number

log10

Logarithm base 10 of entries of symbolic matrix

log2

Logarithm base 2 of entries of symbolic matrix

mod

Symbolic matrix elementwise modulus

pretty

Pretty-print symbolic expressions

quorem

Symbolic matrix elementwise quotient and remainder

real

Real part of complex symbolic number

round

Symbolic matrix elementwise round

size

Symbolic matrix dimensions

sort

Sort symbolic vectors, matrices, or polynomials

sym

Define symbolic objects (Continued)

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364 • Multivariable and Vector Calculus 2/E

TABLE B.19  Common symbolic commands (Continued) Command

Description

syms

Shortcut for constructing symbolic objects

symvar

Find symbolic variables in symbolic expression or matrix

fourier

Fourier integral transform

ifourier

Inverse Fourier integral transform

ilaplace

Inverse Laplace transform

iztrans

Inverse z-transform

laplace

Laplace transform

ztrans

z-transform

B.4.2 Differentiating Symbolic Expressions Use diff ( ) command for differentiation. For example, >> syms x; >> f =-cos(5∗x)+2; >> diff(f) ans = 5∗sin(5∗x) >> y=8∗sin(x)∗exp(x); >> diff(y) ans = 8∗cos(x)∗exp(x)+8∗sin(x)∗exp(x) >> diff(diff(y))% second derivative of y ans = 16∗cos(x)∗exp(x)

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Matlab • 365

An example of a partial derivative is as follows: >> syms v u; >> f = cos(v∗u); >> diff(f,u)% create partial derivative

∂f ∂u

ans =           -sin(u v)∗ v >> diff(f,v) % create partial derivative

∂f ∂v

ans =           -sin(u v)∗ u >> diff(f,u,2) % create second partial derivative

∂2 f ∂u2

ans =           -cos(u v)^2 ∗v B.4.3 Integrating Symbolic Expressions The int(f) function is used to integrate a symbolic expression f. For example, >> syms x; >> f=cos(x)^2; >> int(f) ans =           1/2 ∗cos(x) ∗sin(x) + 1/2∗ x >> int(1/(1+x^2)) ans =           atan(x)

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366 • Multivariable and Vector Calculus 2/E B.4.4 Limits Symbolic Expressions The limit(f) command is used to calculate the limits of function f. For example, >> syms x y z; >> limit((sin(x)/x), x, 0) % lim x →0

sin x =1 x

ans =           1 >> limit(1/x, x, 0, ‘right’) % lim+ x →0

1 = ∞ x

ans =           infinity >> limit(1/x, x, 0, ‘left’) % lim− x →0

1 = −∞ x

ans =           -infinity B.4.5 Taylor Series Symbolic Expressions Use the taylor( ) function to find the Taylor series of a function with respect to the variable given. For example, >> syms x; N =4; 1 n f (0) n= 0 n ! N

>> taylor(exp(-x),N+1) % f ( x) ≅ ∑ ans =

          1 - x + 1/2 ∗x^2 - 1/6 ∗x^3 + 1/24 ∗x^4 >> f=exp(x); >> taylor(f,3) ans =           1 + x + ½∗ x^2

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Matlab • 367

B.4.6 Sums Symbolic Expressions Use the symsum( ) function to obtain the sum of a series. For example, >> syms k n; >> symsum(k,0,n-1) %

n−1

1

∑ k = 0 + 1 + 2 + ... + n − 1 = 2 n k=0

2

1 − n 2

ans =           1/2 n^2 - 1/2 n >> syms n N; >> symsum(1/n^2,1,inf) %

1 p2 = ∑ 2 6 n= 0 n N

ans =           1/6 ∗pi^2 B.4.7 Solving Equations as Symbolic Expressions Many MATLAB commands and functions are used to manipulate the vectors or matrices consisting of symbolic expressions. For example, >> syms a b c d; >> M=[a b;c d]; >> det(M) ans =           a ∗d - b ∗c >> syms x y; >> f=solve(‘5∗x+4∗y=3’,’x-6∗y=2’); % solve the system 5 x + 4 y = 3, x − 6 y = 2 >> x=f.x x=           13/17

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368 • Multivariable and Vector Calculus 2/E >> y=f.y y=           -7/34 >> syms x; >> solve (x^3-6∗x^2+11∗x-6) ans =           1           2           3 Use dsolve ( ) function to solve symbolic differential equations. For example, >> syms x y t; >> dsolve(‘Dy+3∗y=8’) % solve y / + 3 y = 8 ans =           8/3 -C1 +exp(-3 t) % C1 is undetermined constant >> dsolve(‘Dy=1+y^2’,’y(0)=1’) % solve y / = 1 + y2 with initial condition y(0) = 1 ans =           tan(t + 1/4 pi) >> dsolve(‘D2y+9∗y=0’,’y(0)=1’,’Dy(pi)=2’) % solve y / / + 9 y = 0 with initial conditions        % y(0) = 1 , y(p ) = 2 ans =           - 2/3 ∗sin(3 t) + cos(3 t)

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APPENDIX

C

Answers to Odd-Numbered Exercises CHAPTER 1 1.1

Points and Vectors on the Plane 1.1.1 1. Scalar. 3. Vector. 5. Scalar. 7. Vector. 9. Scalar. 11. Scalar. 13. Scalar. 1.1.3

   −3  u + v =  . 1   1  u − v =  . 9    −2  2u =   .  10 

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370 • Multivariable and Vector Calculus 2/E

 a

1.1.5

  ab

1.   b   bc

3. 

 b

 c

1.1.7

2 a2 − 2 a = 1.

1.1.9

a = ±1 or a = −8 .

  1.1.11 Since ABCD is parallelogram, AD = BC , Hence,      AC + BD = AD + BC = 2BC . 1.1.13 Since     3  IA + 3IA = −3AB ⇔ AI =AB . 4 Similarly,   1  1  JA = − JB ⇔ AJ =AB . 3 4  3   1  Thus, we take I such that AI = AB , and J such that AJ = AB . 4 4 Now,       MA + 3MB = MI + AI + 3IB = 4MI , and        3MA + MB = 3MJ + 3JA + MJ +JB= 4MJ . 4 3 1.1.15 1. a = , b = . 7 7          1.1.17 Since BD + u = v , BD= v − u . Then BE = x v −u .       Since AC= u + v , AE = y u+v .      But AB = AE + EB = AE − BE , i.e.        u = y u + v − x v − u = ( x + y) u + ( y − x ) v .

(

(

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) (

)

(

)

)

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Answers to Odd-Numbered Exercises • 371

  Since u and v are non-collinear we have by Exercises 1.1.11, 1 x+y= 1 and y − x = 0 , i.e., x= y= and E is the midpoint of both 2 diagonals. 1.2

Scalar Product on the Plane   1.2.1 1. a b =7 .   3. a b = 28 . 1.2.3

 1   1. cos a , b = . 2    Thus, a , b = 60 o .

(

)

(

)

   3. cos a , b = 0 .

(

)

   Thus, a , b = 90 o .

(

1.2.5

)

1 a= . 2

  1.2.7 Let a and b be the vectors of two adjacent sides of the rhombus.     Then a = b . Further, the diagonals are the vectors a + b and       a − b . Now compute a + b  a − b = a2 − b2 = 0 , since a = b.

(

1.2.9 1.2.11



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)(

)

  a b = − 675 + 675 = 0.      2  2 a+b = a−b ⇔ a+b = a−b .   ⇔ 4a b = 0.   ⇔ a b = 0. This proves what we want.

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372 • Multivariable and Vector Calculus 2/E    a b b     1.2.13 cb = a b − 2 b

( )

( )

2

    = a b − a b = 0.

( ) ( )

1.2.15 Since,   a b = 0 , we have        2 2  2 a+b = a + b  a + b = a a + 0 + bb = a + b .

( )

(

1.2.17







)(



)





( a − b )( a − b ) = a − b

2

= 0 , but the norm of a vector is 0 if and only  if the vector is the 0 vector.

1 1  1.2.19      = 0 ⇔ 1 + m1 m2 = 0 ⇔ m1 m2 = −1.  m1   m2   1.2.21  v −   1.3

  vw      = v  w 2 w  w w 

  vw  2  2 w= 0 . w

Linear Independence 1.3.1

We must show that there exist scalars a and b such that  3 b + a  7  1   3  7  a   + b   =   , we get the vector equation   =   . The 0  1  3   b  3  solution of this system is given by a = −2 and b = 3 . Thus, we can 7  1  3  write   = ( −2 )   + 3   . 3  0  1 

 a  b − a  −1 a + b 1 1.3.3 =    +  . 2 1 2 1  b 1.3.5

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1  0  0  a  + b   =   , = then a 0= and b 0 . Thus the vectors 0  1  0   1   0  x =   and y =   are linearly independent. 0  1 

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Answers to Odd-Numbered Exercises • 373

1.3.7

2a + b + 8c = 0 a + 2b + 7c = 0, By solving this homogenous system of linear equations with more unknowns than equations, we get a = −3 c, b = −2 c, thus for c = −1 , we= get a 3= and b 2 . Therefore, we have shown that the vectors 2 1     8  2 1  , 2  , and 7  in  linearly dependent because in addition to       solution a= b= c= 0 for the vector equation 2  1  8  0  a  + b   + c   =   , we also have the solution 1  2  7   0  a = 3, b = 2, and c = −1 .

  Suppose that vectors x and y are linearly dependent, that is,    a x + by = 0 , where a and b are not both zero. If a ≠ 0 , then   b a x = − y , and if b ≠ 0 , then y = − x ; in either case, the vectors are a b   parallel. Conversely, if x and y are parallel, then either        = x c= y or y c x . In the first case, 1 ⋅ x + ( − c ) y =0 ; in the second    case, c x + ( −1 ) y =0 ; in either case, the pair is linearly dependent.

1.3.9

1.3.11 1. −7 . 1.3.13 1. By Chasles’ Rule  1    1  AE= AC ⇔ AB + BE= AC 4 4 and  3    3  AF = AC ⇔ AD + DF = AC 4 4            AB + BE + AD + DF = AC ⇔ BE + DF = AD + DC − AB − AD .   ⇔ BE = − DF

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374 • Multivariable and Vector Calculus 2/E 1.4

Geometric Transformations in Two Dimensions 1.4.1

1 a= −3, b = − . 2

1.4.3

a b  2  c − a  , bc = − a .  

 x 1 1  1   2   1  L   = 1.4.5 From L   =   , it follows that =  , L     and also  y  2  1  2   2   2   1  2   3 1 3 1 L   =   . But, L    +    = L   =   and 3 2 3 2  1  2   1 2 1 1 2 L   + L   =   +   =   . Thus, L is not linear. 1 2 2 2 4 1.4.7

 x1   x2  We show that L preserves addition. Let   and   be elements  y1   y2  of  2 . Then

  x1   x2    x1 + x2  L  +   = L   y1 + y2    y1   y2    x1 + x2 − y1 − y2  =   3 x1 + 3 x2 



 x1 − y1   x2 − y2  = +   3 x1   3 x2   x1   x2  = L  + L .  y1   y2 

Thus L preserves vector addition.



Now, we show that L preserves scalar multiplication. Let k be a scalar.   x  kx  L k  = L   ky    y

 kx − ky  =    3 kx 

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Answers to Odd-Numbered Exercises • 375

 x − y = k    3x   x = kL   .  y Thus L preserves scalar multiplication. Thus, L is linear. 1.4.9

The desired transformations are in Figures C.3 through C.5.

1.4.11 The desired transformations are in Figures C.6 through C.9.

 −1 0   −3  1.4.13 RH =  , .  0 1  2  0 −1 0    1.4.15 The rotation through p / 2 about the point (5,1) is 1 0 0  . 0 0 1  The image of the unit square under this rotation is (6, -3), (5, -3), (5, -4), and (6, -4).  27  1.4.17 1. L2  L1 =  .  −11

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376 • Multivariable and Vector Calculus 2/E 1.5

Determinants in Two Dimensions 1.5.1

1. −9 . 3. 22 . 5. x2 − 5 x + 21 .

1.5.3

1.

1.5.5

det(kA) = k2 det( A) .

1.5.7 = det( AB) det( = A)det(B) det( = B)det( A) det(BA) . 1.5.9

1.6

  − b  c  a c  r3 = − bc + ad = det   a  d  = .    b d Since the dot product of two vectors is positive if the angle between them is less than p / 2 , the determinant is positive if the angle    between r2 and r3 is less than p / 2 . Thus r2 lies counterclockwise  from r1 .

Parametric Curves on the Plane 3

y− x  y− x  y− x and x =  − 2  . 4  4   4 

1.6.1

y − x = 4t ⇒ t =

1.6.3

1. ay − cx = ad − bc , this is a straight line with positive slope. 3.

1.6.5

x2 y2 − = 1, x > 0 . This is one branch of a hyperbola. a2 b2

Observe that = y 2 x + 1 , so the trace is part of this line. Since in the interval [ 0,4p ] , −1 ≤ sin t ≤ 1 , we want the portion of the line = y 2 x + 1 with −1 ≤ x ≤ 1 and thus −1 ≤ y ≤ 3 . The curve starts at the middle point ( 0,1 ) at t = 0 , reaches the high point (1,3 ) at p 7p , reaches its low point ( −1,1 ) at t = , and finishes in the 2 2 middle point ( 0,1 ) when t = 4p. t=

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Answers to Odd-Numbered Exercises • 377

1.6.7

2 . 5

1.6.9

1.

1.6.11

3 . 2

1.6.13 8r. 1.6.15 1.7

( gh − V )

2 2

≥ g 2 ( h2 + u2 ) ⇒ g 2 u2 ≤ V 2 ( V 2 − 2 gh ) .

Vectors in Space 1.7.1

22 .

1.7.3

−2 .

1.7.5

0. ( x − 1) − 2 ( y − 1) − ( z − 1) =

1.7.7

abc + 2 ( ab + bc + ca ) + p ( a + b + c ) +

1.7.9

 x  0   a       2  y = 0  + t  a  , t ∈  .  z  1   a2 

1.7.11

4p . 3

3 . 3

1.7.13 x1 y1 + x2 y2 + x3 y3 ≤ x12 + x2 2 + x3 2 ⇒ ( a2 + b2 + c2 ) ≤ ( a4 + b4 + c 4 ) ( 3 ) . 2

1.7.15 x + y + z = 1.

MVC.CH06_App-C_2PP.indd 377

1 . 6

1/10/2023 5:05:34 PM

378 • Multivariable and Vector Calculus 2/E

1.7.17 = proj

  r0 − b n

proj =

1.8





  r0 − b n

    r − b ) n r − b ) n  ( (= n 0

0

 n

    r0  n − b n =  n

 n

    a  n − b n =  n

( a

0

  − b n .  n

)

Cross Product 1.8.1









   

 

 



 

 



 

( a − b ) × ( a + b ) = a × a + a × b − b × a + b × b = 0 + a × b + a × b + 0 = 2a × b . 1.8.3 1.8.5

3.       It is not associative, since i × ( i × j ) =i × k = − j, but       (i × i )× j = 0× j = 0.

(

1.8.7

  −i − k.

1.8.9

2 ax + 3 a2 y − az = a2 .

   c 1.8.11 a × b ×=    c × a ×= b

( (

1.8.13







)





     a ) ( b a ) c − ( b c ) a , ) ( ac ) b − ( a b ) c , b × ( c ×=   ) ( c b ) a − ( ca ) b , and adding yields the result. 

{x : x ∈ a × b} .

 a1 a2 a3       1.8.15 = a  b × c det = b1 b2 b3  0 if and only if the row vectors  c1 c2 c3     a,b, and c are linearly independent, i.e., if and only if one vector lies in the plane of the other two vectors.

(

MVC.CH06_App-C_2PP.indd 378

)

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Answers to Odd-Numbered Exercises • 379

           1.8.17 Observe that, x ( y × z ) = ( x × y ) z , now putting x= a × b, y = c , and                   d ( c  a ) b − c  b a=  d ( c  a ) b d z = d , this gives, a × b × c=     − c  b a d .

( )( )

((

) )

(

( ))

( )

12    1.8.19 1. 18  . 24  3. x= 6 t, y= 0, z= 3 − 3 t . 3  5.  3,0,  . 2  7. 1.9

33 29 . 16

Matrices in Three Dimensions

1.9.1

 1 1 5   1.  −1 3 2  .  1 9 1   0 7 −9    3.  −2 4 0  .  −1 16 −7 

1.9.3

MVC.CH06_App-C_2PP.indd 379

 a  1  0  2 a − 3 b 2   −3         5 b=  a 0  + b  5  , so clearly the family,        a + 2b  1  2  a  1   0 

1/10/2023 5:05:36 PM

380 • Multivariable and Vector Calculus 2/E









 1   0         2   −3      0  ,  5    1   2         1   0   spans the subspace. To show that this is a linearly independent family, assume that 1   0  0  2   −3  0        0  . a 0  + b  5  =       1   2  0  1   0  0  Then it clearly follows that a= b= 0 , and so this is a linearly independent family.

1.9.5

1 0  A2 = 0   0



 1   0 A3 =  0    0

MVC.CH06_App-C_2PP.indd 380

2 3 4  n −1 n  1 2 3  n − 2 n − 1  0 1 2  n − 3 n − 2 .         0 0 0  0 1  3 6 10  1 3

6



0 1

3



  0 0

 0

 

( n − 1) 2

( n − 2 )( n − 1) 2

( n − 3 )( n − 2 ) 2  0

n ( n + 1)   2  ( n − 1) n   2 . ( n − 2 )( n − 1 )   2     1

1/10/2023 5:05:37 PM

Answers to Odd-Numbered Exercises • 381

1.9.7

1 −1 0    1 0 1  . 0 1 −1

1.9.9

0 0 0   0      0 1 0  ,  4  . 0 0 0   0 

1.9.11 Not linear, because   x1    a x1       L  a = x2   L = a x2  a x      3   x3  

 1   x1  1        a x2  ,but a L=  x2  a=  x2  a x  x  x   3  3  3

 a     a x2  . a x   3

1 0 0    1.9.13 0 −1 0  . 0 0 1  1.10

Determinants in Three Dimensions 1.10.1 −10 . 1.10.3 aef . 1.10.5 2. 1.10.7 − t 4 − t 3 + 18 t 2 + 9 t − 21 . 1.10.9 c = −5 . a b c  c a  b a b c   − b + c det  b c a  = a     1.10.11 .  a b  c b  c a  c a b = acb − a3 − b3 + abc + abc − c3 = 3 abc − a3 − b3 − c3 1.10.13 l = −1 and 1. 1.10.15 37 .

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382 • Multivariable and Vector Calculus 2/E 1.11

Some Solid Geometry 1.11.1

1.12

a p , . 2 4

Cavalieri and the Pappus-Guldin Rules 1.12.1 The lateral area is thus p r r 2 + h2 .

The volume of the cone is

1.12.3 Area =

1.13

1 2 pr . 2

Volume =

1.12.5

2 rh p p r × =r 2 h. 3 2 3

4 3 pr . 3

A = 11309.7 m 2 V = 113097.3 m 3

.

Dihedral Angles and Platonic Solids 1.13.1 1. Tetrahedron. 2. Cube or hexahedron. 3. Octahedron. 4. Dodecahedron. 5. Icosahedron. 1.13.3 sin

1.14

q cos (p / n ) . = 2 sin (p / m )

Spherical Trigonometry  1  1.14.1 cosq = , where q is the angle between a and b . 3 1 a 1.14.3 V =× ( 6 a2 ) ×   = a3 . 3 2

MVC.CH06_App-C_2PP.indd 382

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Answers to Odd-Numbered Exercises • 383

V= 1.14.5 = V 1.15

a3 4

(

 1  25 + 10 5 ×  10 + 22  5  .

)

a3 15 + 7 5 4

(

)

Canonical Surfaces 1.15.1 4 x2 + 4 y2 + z2 = 1. 1.15.3 A spiral staircase. 1.15.5 The MapleTM commands to graph this surface are:

> >

FIGURE C1.15.1 (a)  Exercise 1.15.5.

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384 • Multivariable and Vector Calculus 2/E

>

FIGURE C1.15.1 (b)  Exercise 1.15.5.

1.15.7 We may take A : x + y + z = 0, Σ : x2 + y2 + z2 = 0 , showing that the    surface is of revolution. Its axis is the line in the direction i + j + k . 1.15.9 We may take A : x + y + z = 0, Σ : x2 + y2 + z2 = 0 as our plane and    sphere. The axis of revolution is then in the direction of i + j + k . 1.15.11 Every plane through the origin, which makes a circular cross section, must intersect the yz -plane, and the diameter of any such y2 z2 + = 1. b2 c2 Therefore, the radius of the circle is at most b . Arguing similarly on the xy-plane shows that the radius of the circle is at least b . To cross section must be a diameter of the ellipse = x 0,

MVC.CH06_App-C_2PP.indd 384

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Answers to Odd-Numbered Exercises • 385

show that circular cross section of radius b actually exist, one may verify that the two planes given by a2 ( b2 − c2 ) z2 = c2 ( a2 − b2 ) x2 give circular cross sections of radius b . 1.16

Parametric Curves in Space 1.16.1 42 . t1 t2 t3 + t1 t2 t 4 + t1 t3 t 4 + t2 t3 t 4 0 = t1 t2 t3 t 4 1 1 1 1 ⇒ + + + = 0, t1 t2 t3 t 4

t1 t2 t3 + t1 t2 t 4 + t1 t3 t 4 + t2 t3 t 4 = 0⇒ 1.16.3



as required.

1.16.5 You can parameterize the cylinder y2 + z2 = 16 by y = 4 cos t , and z = 4 sin t . From the equation x =8 − y2 − z , you obtain that x= 8 − 16 cos2 t − 4 sin t . 1.17

Multidimensional Vectors 1.17.1 1. If f ′ ( x ) = 0 , then e x −1 = 1 implying that x = 1 . Thus f has a single minimum point at x = 1 . Thus for all real numbers x, 0 =f (1 ) ≥ f ( x ) =e x −1 − x, which gives the desired result. 3. Easy Algebra! 16 6 is reached when a= b= c= d= ⋅ 5 5 1.17.5 Applying the AM-GM inequality, for 1,2,, n : 1 + 2 + + n n + 1 1/ n = , with strict inequality n!1/ n = (1 × 2 ×  × n ) < n 2 for n > 1 . 1.17.3 The maximum value e =

MVC.CH06_App-C_2PP.indd 385

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386 • Multivariable and Vector Calculus 2/E

n

1.17.7 Observe that ∑ 1 = n. Then we have k =1

2



2

1   n  n   n  n 1 = n2 = 1   ∑ ( ak )  ≤  ∑ ak2   ∑ 2 ∑ ak = =  k 1 =  k 1=   k 1 ak k 1

 . 

1.17.9 The only linear combination giving the zero vector is the trivial linear combination, which that the vectors are linearly independent.

CHAPTER 2 2.1.1

1. Closed in  2 . 3. Open in  2 . 5. Open in  2 . 7. Open in  2 . 9. Open in  2 . 11. Open in  2 . 13. Closed in  2 . 15. Open in  2 . 17.  2 . 19. Closed in  2 . 21. Closed in  2 .

2.1.3

1. Open in  3 . 3. Neither open nor closed in  3 . 5. Closed ball in  3 . 7. Closed in  3 . 9. Closed in  3 . 11. Closed in  3 . 13. Open in  3 . 15. Open in  3 .

MVC.CH06_App-C_2PP.indd 386

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Answers to Odd-Numbered Exercises • 387

2.1.5 Suppose S1 and S2 are open sets and P is any point in S1 ∪ S2 . To prove that S1 ∪ S2 is an open set, we must show that P is an interior point of S1 ∪ S2 . From the defining property of P, the point is either in S1 or S2 . Suppose that P ∈ S1 as the proof is analogous when P ∈ S2 . Since P ∈ S1 , which is an open set, P is an interior point of S1 and there is some open ball with center P, which contains only points in S1 . Hence, P is also an interior of S1 ∪ S2 and thus this set is an open set. 2.1.7

1. V is closed set. 3. D is neither open nor closed set. 5. B is neither open nor closed set. 7. F is closed set. 9. K is closed set.

2.2

Multivariable Functions 2.2.1

MVC.CH06_App-C_2PP.indd 387

1.

> >

1/10/2023 5:05:49 PM

388 • Multivariable and Vector Calculus 2/E

>

3. >

>

>

MVC.CH06_App-C_2PP.indd 388

1/10/2023 5:05:49 PM

Answers to Odd-Numbered Exercises • 389

5. >

>

>

MVC.CH06_App-C_2PP.indd 389

1/10/2023 5:05:50 PM

390 • Multivariable and Vector Calculus 2/E

7.  >

>

>

MVC.CH06_App-C_2PP.indd 390

1/10/2023 5:05:50 PM

Answers to Odd-Numbered Exercises • 391

9.

> >

>

MVC.CH06_App-C_2PP.indd 391

1/10/2023 5:05:51 PM

392 • Multivariable and Vector Calculus 2/E

11. >

>

Or

MVC.CH06_App-C_2PP.indd 392

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Answers to Odd-Numbered Exercises • 393

>

In 3D

> >

MVC.CH06_App-C_2PP.indd 393

1/10/2023 5:05:52 PM

394 • Multivariable and Vector Calculus 2/E Or

>

Or

> > > > >

MVC.CH06_App-C_2PP.indd 394

1/10/2023 5:05:53 PM

Answers to Odd-Numbered Exercises • 395

13.

> >

In 3D

>

MVC.CH06_App-C_2PP.indd 395

1/10/2023 5:05:53 PM

396 • Multivariable and Vector Calculus 2/E

15.

> >

MVC.CH06_App-C_2PP.indd 396

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Answers to Odd-Numbered Exercises • 397

In 3D

>

MVC.CH06_App-C_2PP.indd 397

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398 • Multivariable and Vector Calculus 2/E

2.2.3

1. Shift g ( x, y ) upward 2 units. 3. Reflect g ( x, y ) about the xy-plane. 5. Reflect g ( x, y ) in the plane x = 0 . 7. Reflect g ( x, y ) in the origin.

2.3 Limits. 2.3.1

> >

FIGURE 2.3.8  Exercise 2.3.1 for ( x , y )  4 − x 2 − y 2 .

MVC.CH06_App-C_2PP.indd 398

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Answers to Odd-Numbered Exercises • 399

2.3.3

> >

FIGURE 2.3.10 Exercise 2.3.2 for ( x , y ) 

1 . x2 + y2

2.3.5 0. 2.3.7 0. 2.3.9

1 . 3

2.3.11 Does not exist. 2.3.13 Show that

MVC.CH06_App-C_2PP.indd 399

(x

x3 y 2

+ y2 )

≤ xy for ( x, y ) ≠ ( 0,0 ) . So the limit is 0.

1/10/2023 5:05:56 PM

400 • Multivariable and Vector Calculus 2/E

 x2  1  0  2.3.15 = = = ; lim f ( x,0) lim = lim f ( x, x) lim    2  0. x →0 x →0 x2 + x2 x → x → 0 0  x +0   2   x 2 − y2  x 2 − y2  2.3.17 Since lim  lim 2 1 , and lim lim =    = −1 . x → 0 y → 0 x + y2 y → 0 x → 0 x 2 + y2     Thus, the iterated limits are not equal and therefore, lim g( x, y) , ( x, y) →( 0,0 )

does not exist. Hence, g( x, y) is discontinuous at ( 0,0 ) . 2.3.19 c = 0 . 2.4

Definition of the Derivative 2.4.1

By the Cauchy-Bunyakovsky-Schwarz inequality, 1/ 2    n  2 L h ≤ h  ∑ L ( ek )   k =1 

()



and      h×L h ≤ h L h ≤ h

()

2.5

()

2

( L ( e ) )

2 1/ 2

k

 = o h ,

( )

as it was to be shown. The Jacobi Matrix

2.5.1

1. f x ( x= , y ) 3 ( x3 − y2 )( 3 x2 )

fy ( x, y ) = 3 ( x3 − y2 ) ( −2 y ) . f z ( x, y, z ) = 0

3. f x ( x, y ) =

3 y ( y2 − x2 )

(x + y ) 3x( x − y ) . ( x, y) = x + y ( ) 2 2

2

2

fy

2

2

2 2

f z ( x, y, z ) = 0

MVC.CH06_App-C_2PP.indd 400

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Answers to Odd-Numbered Exercises • 401

2 ( z2 + x2 ) xy2 ∂f 2.5.3 = 2 x log ( x2 y2 + 1 ) + x2 y2 + 1 ∂x 2 2 2 ∂f 2 ( z + x ) x y = ∂y x2 y2 + 1 ∂f = 2 z log ( x2 y2 + 1 ) ∂z

2.5.5

.

 1 ∂ f ( x, y ) =  ∂x 0

if x < y2 if x > y2

 0 ∂ f ( x, y ) =  ∂y 2 y 2.5.7

∂w = ∂x ∂w = ∂y

(3u

2

(3u

2

if x < y2 if x > y2

 y + 2 uv ) ( y cos xy ) + ( u2 − 3 )    x + 2 uv ) ( x cos xy ) + ( u − 3 ) ln x

2.5.9

2.5.11

cosq f ′ ( r ,q ) =   sin q



J ( r ,q ) = r .

2.5.13

 v2 − u 2  2 2 2 (v + u ) f ′ ( u, v ) =   −2 uv  ( v2 + u 2 ) 2 

− r sin q  . r cosq 

− ( u 2 + v2 ) J ( u, v ) = = 4 ( u 2 + v2 )

  (v + u )  . u 2 − v2  2 ( u2 − v2 )  −2 uv

2 2

2

2

MVC.CH06_App-C_2PP.indd 401

.

2

0 −1   0 0  . 2 1 



.

(u

−1 2

+ v2 )

2

.

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402 • Multivariable and Vector Calculus 2/E

2.5.15 1. 2.5.17 6 t 2 + 6 t 5 . 2.5.19 2.6

1 b b arctan + 2 2 . 3 2a a 2 a ( a + b2 )

Gradients and Directional Derivatives 2.6.1

 1 y   xe .   y  ln xe 

2.6.3

2  1  .  

2.6.5

2 xy sin(yz)    2   x ( y cos(yz)z + sin(yz) )  .   x2 y2 cos(yz)

2.6.7

2 xey   zy2 e xz   2 y   ∇f ( x, y, z ) = x e  , ∇k ( x, y, z ) = 2 ye xz  , fk =x2 y2 ey+ xz  0   xy2 e xz      y + xz 2 y + xz 2 2 2 xy e + x y ze   2 y+ xz  ∇= . ( fk )  2 x ye + x2 y2 ey+ xz    3 2 y + xz xye   2 xey   zy2 e xz      ∇k + k∇f x2 ey  2 ye xz  + y2 e xz  x2 ey  f=  0   xy2 e xz     

Therefore,

∇ ( fk ) = f ∇k + k∇f .

2.6.9

1.1.

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Answers to Odd-Numbered Exercises • 403

 2.6.11 Consider an arbitrary unit vector u . Then the directional derivative satisfies

∂f   ∇f  u = ∇f u cosq = ∇f cosq , = ∂u

 Where q is the angle between ∇f and u . Consequently,

∂f − ∇f ≤  ≤ ∇f . ∂u

∂f Thus  is the largest, when q = 0 (i.e, same direction of ∇f ), and ∂u the smallest when q = p (i.e, opposite direction of ∇f ). −2  −2  2.6.13 ∇f ( x, y ) = 2 y ( x + y ) i − 2 x ( x + y ) j;   ∇f ( 2, −1 ) = −2 i − 4 j;    u = (1 / 5 ) ( 3i + 4j ) ;

−22 / 5. Dv f ( −2,1 ) =

2.6.15

1     y    − x   1     1   1  , y, z)  2  −    ; ∇f ( 0, −1,2 ) =− i −   j −   k ; f ( x= 2 4  y   z     y   z2    3     P1 P2 . DP f ( 0,= P1 P2 =  2  , u = −1,2 ) 1 P2 7  −6 

( −6 − 2 + 3 ) −5

. The = 14 14

maximal direction is ∇f ( 0, −1,2 ) ; maximum rate is 21 ∇f ( 0, −1,2 ) = . 4

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404 • Multivariable and Vector Calculus 2/E     2.6.17 1. i  U   U1    U2    U3  x y z

 

U3 U U2    U3   1    . U1  U2  x y z x y z     iU  iU



3.

  ∇ U = ×V

(

)

∂ ∂ ∂ (U2 V3 − U3 V2 ) + (U3 V1 − U1 V3 ) + (U1 V2 − U2 V1 ) ∂x ∂y ∂z

 ∂U ∂U2   ∂U2 ∂U1   ∂U1 ∂U3  = V1  3 − − −  + V2    + V3  ∂z  ∂x  ∂y   ∂z  ∂y  ∂x  ∂V ∂V   ∂V ∂V   ∂V ∂V  − U1  3 − 2  + U2  1 − 3  + U3  2 − 1  ∂z  ∂x  ∂y   ∂z  ∂y  ∂x     = V ∇ × U − U∇ × V. 2.6.19 Parallel planes have proportional gradients. Therefore, if F ( x, y, z ) =x2 − 2 y2 − 4 z2 − 16 and G ( x, y, z ) = 4 x − 2 y + 4 z − 5 , then  2x    ∇F ( x, y, z ) =  −4 y and G ( x, y, z ) =  −8 z 

4    −2  must  4 



be proportional, i.e.,



k −k 2 x =4 k, −4 y =−2 k, −8 z =4 k, ⇒ x =2 k, y = , z = 2 2 2 2 k  k  4 2 ⇒ 4 k2 − 2   − 4   = ⋅ 16 ⇒ k =± 5  4   4 



 8 2 2 2 2 2  Thus, the points are ( x, y, z ) = ,± , ,. ± 5 5 5  

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Answers to Odd-Numbered Exercises • 405

 −2 x  2 2.6.21 ∇f ( x, y ) = −2 5  −2 y ; ∇f ( −1,2 ) =  −4  ; − ∇f ( −1,2 ) =      −2   −1   2 5   5   =  . Thus, the v direction is   −4   2   −2 5   5 



 y     2.6.23 ∇ × f 6 xz − 1 =  0   −6 x     ∇ × ∇ × f = 0  6 z − 1  ∇ f= 3 z2 − z + 2 .

(

)

 0     ∇ ∇ f = 0  6 z − 1

(

)

6 x     ∇ f= 0  0  2

Thus  −6 x       2 ∇ ∇ f − ∇ f =  0  = ∇ × ∇ × f . 6 z − 1

(



2.7

)

(

)

Levi-Civitta and Einstein

2.7.3

e ijk x j ( e klm yl zm ) = x j yi z j − x j y j zi = ( x j z j ) yi − ( x j y j ) zi .  ∇( ∇ × u ) = −e ijk ∂ i ∂ j hk = 0 .

2.7.5

va ∂ a vb =va ∂ b va − va ( ∂ b va ) + va ( ∂ a vb ) =va ∂ a vb .

2.7.1

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406 • Multivariable and Vector Calculus 2/E 2.8 Extrema 2.8.1

There is one critical point (1,1 ) .

2.8.3

The critical point is thus a saddle point.

2.8.5 2.8.7

1 1    1, −1,  , and  −1,1, −  are the critical points. 2 2   1 1    1, −1,  is a saddle point.  −1,1, −  is also a saddle point. 2 2   we have a local maximum at (1,1,1 ) . In some neighborhood of

( 0,0,0 ) is a saddle point. 2.8.9

By the Fundamental Theorem of Calculus, there exists a continuously differentiable function G such that f ( x, y= )



∫

x2 + y

y2 − x

g ( t ) dt= G( x2 + y) − G(y2 − x) .

∂f ∂f This gives, = ( 0,0 ) = ( 0,0 ) g(0), ∂x ∂y

0   − g′(0) So ( 0,0 ) is a critical point and H f ( 0,0 ) =  . g′(0)  0 Regardless of the sign of, g′(0) the determinant of this last matrix is − ( g′(0) ) < 0 , and so (0, 0) is a saddle point. 2

2.8.11 ∇f ( x, y ) ⇒ f (18 / 13, −14 / 13 ) is not extremum. 2.8.13 ∇f ( x, y ) ⇒ f (1,3 / 2 ) = 37 / 4 is a local maximum. 2.8.15 The graph of g is z − g ( x, y ) = 0 ; with G ( x, y, z )= z − g ( x, y ) , the

(

)

general normal is − g x , − gy ,1 so that the normal above a critical point for g is ( 0,0,1 ) , which is a vertical vector. Thus, the tangent plane is horizontal at any critical point.

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Answers to Odd-Numbered Exercises • 407

2.8.17 Let g ( x, y, z ) =x2 + y2 − 1,h ( x, y, z ) =x − yz. Hence y = −1 or y = 1 / 2 , y =−1 ⇒ x =0 and z = 0. Solution 3 2

0 . If y = 1 / 2 , x = ± ± 3. ,z = ( 0, −1,0 ) and f ( 0, −1,0 ) = 3 3 3 3 , minimum = − . 4 4 2.8.19 If x, y, and z are perpendicular distances to the sides of length Maximum =

L1 , L2 , and L3 , then it is necessary to maximize d, where d 2 = x2 + y2 + z2 , subject constraint area, A where A= 2.9

L1 x + L2 y + L3 z 4 A2 . Thus, minimum = . 2 L1 + L2 + L3

Lagrange Multipliers

( S) ( 6)

3

2.9.1

S We have a = , and the maximum volume is abc = 6

2.9.3

Using CBS, x + 3 y ≤ 2 5/ 4 3 .

2.9.5

The desired maximum is thus f − 2, 2 = f 2, − 2 , and the

(

)

(

3

.

)

minimum is

(

)

(

)



f 1 / 2,1 / 2 =− f 1 / 2, −1 / 2 = 1.

2.9.7

The first point gives an absolute maximum of 18 + second an absolute minimum of 18 −

12 14 and the 7

12 14 . 7

2.9.9

(0, 2,1) yields a maximum and that (0, − 2,1) yields a minimum.

2.9.11

a   b  −1 f= ( x, y) x a yb e−1 ≤     e .  a+ b  a+ b

a

2.9.13 The maximum =

MVC.CH06_App-C_2PP.indd 407

b

k1 k2 k3 . 27

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408 • Multivariable and Vector Calculus 2/E

CHAPTER 3 3.1

Differential Forms 3.1.1

1. 0-forms → C. Functions forms. 3. 2-forms → A. Surface elements.

3.1.3

dw = ( − x − y ) dx ∧ dy + y2 z3 dx ∧ dz + 2 xyz3 dy ∧ dz.

3.1.5

dx ∧ dy = rdr + dq .

3.1.7

 ∂h ∂g   ∂g ∂f   ∂f ∂h  dw =  −  dy ∧ dz −  −  dx ∧ dz +  −  dx ∧ dy .  ∂z ∂x   ∂y ∂z   ∂x ∂y 

3.1.9

dw= 2 zdx ∧ dy ∧ dz.

3.2 Zero-Manifolds 3.2.1

∫

w = −12.

3.2.3

∫

w = −44.

3.2.5

∫

w = −14.

3.2.7

∫

w = 4.

M

M

M

M

3.3 One-Manifold 3.3.1 Let L1 : y =x + 1, L2 : y =− x + 1 . Then, ∫ xdx + ydy = 0. C

3.3.3

 p p We put = x sin= t, y cos t, t ∈  − ,  .Then ∫ xdx + ydy = 0.  2 2 C

3.3.5 Let Γ1 denote the straight line segment path from O to

(

)

A = 2 3,2 and Γ 2 denote the arc of the circle centered at ( 0,0 ) and radius 4 going counterclockwise from q =

MVC.CH06_App-C_2PP.indd 408

p p to q = . Observe 6 5

1/10/2023 5:06:16 PM

Answers to Odd-Numbered Exercises • 409

 x that the Cartesian equation of the line OA is y = . Then on Γ1 , 3 4 xdx + ydy = xdx . Hence, ∫ xdx + ydy = 8 . On the arc of the circle, 3 Γ1 we may = put x 4= cosq , y 4 sin q and integrate from q =

p to 6

p . Observe that there xdx + ydy = 0 , and since the integrand is 5 0, the integral will be zero. Assembling these two pieces, ∫ xdx + ydy = ∫ xdx + ydy + ∫ xdx + ydy = 8 .

q=

Γ

Γ1

Γ2

3.3.7 Let Γ1 denote the straight line segment path from O to

(

)

A = 2 3,2 and Γ 2 denote the arc of the circle centered at ( 0,0 ) p p to q = . Observe 6 5  x that the Cartesian equation of the line OA is y = . 3 2 We find on Γ1 that x dx = xdx , hence ∫ x dx = 4 3 . 3 Γ1 and radius 4 going counterclockwise from q =



On Γ 2 that x dx = 16 cosq dq , hence ∫= x dx 4 sin Γ2



Assembling these we gather that ∫ x = dx 4 sin Γ

3.3.9



MVC.CH06_App-C_2PP.indd 409

p − 8. 5

p p − 8 + 16 sin . 5 5

> >

>

1/10/2023 5:06:20 PM

410 • Multivariable and Vector Calculus 2/E 3.4

Closed and Exact Forms 3.4.1

1. True.

3.4.3

1  1 Since x2 dx + ydy = d  x3 + y2  , the differential form is exact. 2  3

3.4.5 Let w= a 1 + Pdxdw + Qdydw + Rdzdw,



on Π , we can find the functions p, q, r so that = pw P= , qw Q= , rw R. Let b 1 = pdx + qdy + rdz and let w1= w + d b 1 . Now, let w1 = Adxdy + Bdxdz + Cdydz . Let = B b= cz and = b 2 bdx + cdy . Also, let fy = F , where f and F z ,C depend only on x and y. Then, let b 3 = fdx and notice that

w2 + d b 3 = 0 . Therefore, w = d ( − b 1 − b 2 − b 3 ) , completing the

proof. 3.5 Two-Manifolds 3.5.1

2.

3.5.3

15p . 16

3.5.5

4p .

3.5.7

21 . 8

3.5.9

e − 1.

3.5.11 18. 3.5.13

1 . 4

3.5.15 4.

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Answers to Odd-Numbered Exercises • 411

3.5.17 7. 3.5.19

4 (p + 2 ) . p3

3.5.21

 5 8 + 4 sin −1  . 5  5 

3.5.23

2 8 p log e 2 + − . 3 9 3

3.5.25

8 . 3 1 1

3.5.27

∫ ∫ f ( x ) f ( y)( y − x ) ( f ( x ) − f ( y) ) dxdy ≥ 0 . 0 0

3.5.29

1 1

1

1

0 0

0

0

∫ ∫ ( f ( x ) dx ) dy ≥ ∫ ( f  g )( y) dy − ∫ g ( y) dy. 2 2

3.5.31 3.5.33

ea b − 1 . ab

∫

R

3.5.35

n . 2

3.5.37

275 . 54

3.6

N

sin 2p x sin 2p y dxdy = ∑ ∫ sin 2p x sin 2p y dxdy. k =1 Rk

Change of Variables in Two Dimensions 1 du ∧ dv. 2

3.6.1

dx ∧ dy=

3.6.3

b−a . 4

3.6.5

13 ( 2 − sin 2 ) . 6

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412 • Multivariable and Vector Calculus 2/E

3.6.7

0.

255 . 4 1 1 1 3.6.11 + sin 6 − sin 2 . 3 12 12 3.6.9

3.7

Change to Polar Coordinates 3.7.1

49 . 2

3.7.3

p 2 . 12

3.7.5

p 16 − + 3. 18 9

3.7.7

p . 8

3.7.9

p 2 . 4

(

2

)

3.7.11 1. p 1 − e− a . 3. Since both Ia and Ia

2

tend to p as a → +∞ , we deduce that

Ja → p . This gives the result. 3.7.13 If no two points are farther than 2 units, their squares are no farther p /2 1 than 4 units, and so the area < ∫ 4dq = p , a contradiction. 2 0 3.8 Three-Manifolds 3.8.1

1 6

3.8.3

27 8

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Answers to Odd-Numbered Exercises • 413

1

3

0

0

3.8.5

∫∫∫

3.8.7

2

3.8.9

3−e

3.8.11

1 6

3.8.13

16 3

3.9

(12 − 3 x − 2 y)

f ( x, y, z ) dzdydx .

6

0

Change of Variables in Three Dimensions 3.9.1 Cartesian: 1 − y2

1



∫ ∫

x2 + y2

∫

dzdxdy

−1 − 1 − y2 x2 + y2

Cylindrical: 1 2p r



∫ ∫ ∫ rdzdq dr 0 0 r2

Spherical:  / 2 2  cos   /  sin  







 /4 0



2

r 2 sin drd d .

0

The volume is

p . 3

3.9.3 Cartesian: 3



3 − y2

∫ ∫

− 3 − 3 − y2

MVC.CH06_App-C_2PP.indd 413

4 − x2 − y2

∫

dzdxdy.

1

1/10/2023 5:06:33 PM

414 • Multivariable and Vector Calculus 2/E Cylindrical: 3 2p

4−r2

0 0

1

∫∫ ∫



rdzdq dr .

Spherical:  / 3 2

0

3.10

2

 



r 2 sin drd d .

0 1 / cos 



The volume is

3.9.5

p . 96

3.9.7

p2.

3.9.9

4p . 5

3.9.11

81p . 2

3.9.13

2p . 9

5p . 3

Surface Integrals 3.10.1

13 2 . 3

3.10.3

4p . 3

3.10.5

4p R 2 . n

3.10.7 8 2 .

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Answers to Odd-Numbered Exercises • 415

3.10.9

3p . 4

3.10.11

p 17 3 / 2 − 1 ) . ( 48

3.11

Green’s, Stokes’, and Gauss’ Theorems 3.11.1 −4 . 3.11.3 −6 . 3.11.5 16p . 3.11.7

4R3 . 3

2 3.11.9 − . 3 3.11.11 0. 3.11.13

1 . 30

3.11.15 −

96 . 5

3.11.17 −

1 . 20

3.11.19 240. 3.11.21 96p . 3.11.23

3 . 2

3.11.25 −2p .

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MVC.CH06_App-C_2PP.indd 416

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APPENDIX

D

Formulas

D.1 TRIGONOMETRIC IDENTITIES Right triangle definitions, where 0 < q
0 )

a x ln a

ln x ( x > 0)

1 x

c xa

− ca x a+1

log a x

log a e x

sin ax

a cos ax

cos ax

− a sin ax

tan ax

a sec 2 ax =

cot ax

−a − a csc 2 ax = sin 2 ax

sec ax

a sin ax cos2 ax

n

MVC.CH07_App-D_1PP.indd 420

a cos2 ax

1/6/2023 2:39:55 PM

Formulas • 421

y=

dy = dx

csc ax

− a cos ax sin 2 ax a

arcsin ax = sin −1 ax

1 − a2 x 2 −a

arccos ax = cos−1 ax

1 − a2 x 2

arctan ax = tan −1 ax

a 1 + a2 x 2

arc cot ax = cot −1 ax

−a 1 + a2 x 2

sinh ax

a cosh ax

cosh ax

a sinh ax

tanh ax

a cosh 2 ax a

sinh −1 ax

1 + a2 x 2 a

cosh −1 ax

MVC.CH07_App-D_1PP.indd 421

a2 x 2 − 1

tanh −1 ax

a 1 − a2 x 2

u( x) + v( x)

du dv + dx dx

u( x)v( x)

u

u( x) v( x)

1  du dv  v − u  v2  dx dx 

1 v( x)

−1 dv v2 dx

y(v( x))

dy dv dv dx

y(v( u( x)))

dy dv du dv du dx

dv du +v dx dx

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422 • Multivariable and Vector Calculus 2/E

D.4 TABLE OF INTEGRALS

∫ a dx= ax + c (c is an arbitrary constant) ∫  f ( x ) ± g ( x ) dx = ∫ f ( x ) dx ± ∫ g ( x ) dx ∫ x dy= xy − ∫ y dx n dx ∫ x=

1

x n +1 + c,(n ≠ −1) n+1

dx ∫ x= ax = ∫ e dx

ln x + c eax +c a

ax x = +c for (a > 0) a dx ∫ ln a = x ln x − x + c for ( x > 0) ∫ ln x dx − cos ax +c a sin ax dx +c ∫ cos ax= a − ln cos ax tan ax dx +c ∫= a ln sin ax = ax dx +c ∫ cot a  1 − sin ax  − ln    1 + sin ax  + c = ax dx sec ∫ 2a  1 − cos ax  ln    1 + cos ax  + c = csc ax dx ∫ 2a  x tan −1   1 a +c dx 2 ∫ x= + a2 a  x−a  x ln  tanh −1    1  x + a  + c or a +c ∫ x2 − a2 dx = 2a a ax dx ∫ sin=

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Formulas • 423

∫

 x+a ln   1  x−a +c = dx 2 2 ∫a −x 2a 1  x dx sin −1   + c ∫ a2 = 2 a −x  x sinh −1   1  a  + c or ln x + x2 + a2 + c = dx 2 2 a a +x 1 2 2 ∫ x2 − a2 dx = ln x + x − a + c  x sec −1   1 a +c dx ∫ x x2= 2 a −a

(

(

ax = ∫ xe dx

)

)

( ax − 1) eax

+c a2 cos ax + ax sin ax = +c ∫ x cos ax dx a2 sin ax − ax cos ax = +c ∫ x sin ax dx a2 x2 x2 x ln x dx = ln x − +c ∫ 2 4 eax (ax − 1) ax = xe dx +c ∫ a2 eax ( a cos bx + b sin bx ) ax e bx dx = cos +c ∫ a2 + b2 eax ( − b cos bx + a sin bx ) ax e sin bx dx = +c ∫ a2 + b2 x sin 2 x 2 − +c ∫ sin x dx = 2 4 x sin 2 x 2 + +c ∫ cos x dx = 2 4 2 ∫ tan x dx= tan x − x + c

∫ cot

MVC.CH07_App-D_1PP.indd 423

2

x dx =− cot x − x + c

1/6/2023 2:39:59 PM

424 • Multivariable and Vector Calculus 2/E

dx tan x + c ∫ sec x= − cot x + c ∫ csc x dx = 2

2

dx sec x + c ∫ sec x tan x= − csc x + c ∫ csc x cot x dx =

D.5 SUMMATIONS (SERIES) D.5.1 Finite Element of Terms N  1 − ( N + 1 ) a N + Na N +1  1 − a N +1 n = ; na a   ∑   1−a ( 1 − a )2 = n 0= n 0   N N + + N ( N + 1) N N 1 2 N 1 ( )( ) = n = ; n2 ∑ ∑ 2 6 n 0= n 0 = N N ( N + 1 )( N + 2 ) n ( n + 1) = ; ∑ 3 n=0 N NP N! NCn a N − n bn , where NCn =NCN − n = n = ( a + b)N = ∑ − n ! N ( n )! n! n=0 N

= ∑ an

D.5.2 Infinite Element of Terms ∞

∞ 1 1 = , ( x < 1) ; nx n = , ∑ 1− x ( 1 − x )2 0= n 0

∑x n

n

( x < 1)

∞ ( −1) 1 1 1 1 ∂k  x  , 1 ; n x =lim ( −1 ) x < =1 − + − + ... = p ( ) ∑ ∑   −a k a→0 3 5 7 4 ∂a  x − e  n 0= n 0 2n + 1 ∞ 1 1 1 1 1 =1 + 2 + 2 + 2 + ... = p 2 ∑ 2 n 2 3 4 6 n=0 ∞

n

k

k n

∞

xn 1 1 1 = 1 + x + x2 + x3 + ... 1! 2! 3! n= 0 n!

ex = ∑ ∞

( ln a ) n x n

n=0

n!

ax = ∑

2

∞

( ±1 )n x n

n =1

n

ln (1 ± x ) =−∑

MVC.CH07_App-D_1PP.indd 424

( ln a ) x ( ln a ) x2 ( ln a ) x3 = + + + ... 1+ 1! 2! 3! =± x −

x2 x3 ± − ..., 2 3

3

( x < 1)

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Formulas • 425

( −1)n x2 n+1 x 3 x 5 x7 sin x =∑ =x − + − + ... 3! 5! 7! n=0 ( 2 n + 1 )! n ∞ ( −1) x2 n x2 x 4 x6 cos x =∑ =1 − + − + ... 2! 4! 6! ( 2 n )! n=0 ∞

x3 2 x 5 + + ... , ( x < 1 ) 3 15 ∞ ( −1 )n x2 n+1 x 3 x 5 x7 tan −1 x =∑ =x − + − + ... , 2n + 1 3 5 7 n=0 tan x =x +

( x < 1)

D.6 LOGARITHMIC IDENTITIES log e a = ln a (natural logarithm) log10 a = log a (common logarithm) log= ab log a + log b a log = log a − log b b log a n = n log a

D.7 EXPONENTIAL IDENTITIES e x =1 + x +

x2 x3 x 4 + + + ... , where e  2.7182 2! 3! 4! e x e y = e x+ y

(e )

x n

= enx

ln e x = x

D.8 APPROXIMATIONS FOR SMALL QUANTITIES If a  1 , then ln(1 + a)  a

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426 • Multivariable and Vector Calculus 2/E

ea  1 + a sin a  a cos a  1 tan a  a

(1 ± a )n  1 ± na D.9 VECTORS D.9.1 Vector Derivatives 1. Cartesian Coordinates Coordinates

( x , y, z )

Vector

A = Ax a x + Ay a y + Az a z

Gradient

∇= A

Divergence

∇ ⋅= A

Curl

∂A ∂A ∂A ax + ay + az ∂x ∂y ∂z ∂Ax ∂Ay ∂Az + + ∂x ∂y ∂z

ax ay az ∂ ∂ ∂ = ∇× A = ∂x ∂y ∂z Ax Ay Az  ∂A ∂Ay   ∂Ay ∂Ax   ∂Ax ∂Az  = z − − −  ax +   az  ay +  y z z x ∂ ∂ ∂ ∂ ∂y       ∂x

Laplacian

∇2 A=

∂2 A ∂2 A ∂2 A + + ∂x2 ∂y2 ∂z2

2. Cylindrical Coordinates Coordinates

( r ,f , z )

Vector

A = Ar a r + Af af + Az a z

Gradient

∇= A

MVC.CH07_App-D_1PP.indd 426

1 ∂A ∂A ∂A ar + af + az ∂r ∂z r ∂f

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Formulas • 427

1 ∂ 1 ∂Af ∂Az Divergence = ∇⋅A ( r Ar ) + + r ∂r r ∂f ∂z

Curl

a r r af a z 1 ∂ ∂ ∂ = ∇× A = r ∂r ∂f ∂z Ar r Af Az  1 ∂Az ∂Af =  − ∂z  r ∂f

Laplacian

∂Ar   ∂Ar ∂Az  1 ∂ −  ar +   af +  (r Af ) − ∂r  ∂f r  ∂x  ∂z 

  az 

1 ∂  ∂A  1 ∂ 2 A ∂ 2 A + r + r ∂r  ∂r  r 2 ∂f 2 ∂z2

= ∇2 A

3. Spherical Coordinates Coordinates

( r ,q ,f )

Vector

A = Ar a r + Aq aq + Af af

Gradient

∇= A

1 ∂A 1 ∂A ∂A ar + aq + af ∂r r ∂q r sin q ∂f

Divergence = ∇⋅A

1 ∂ 2 1 1 ∂Af ∂ (r Ar ) + ( Aq sinq ) + r 2 ∂r r sin q ∂q r sin q ∂f

a r raq ( r sin q ) af 1 ∂ ∂ ∂ Curl = ∇× A = ∂f r 2 sin q ∂r ∂q Ar rAq ( r sin q ) Af  ∂  ∂Aq  1  1 ∂Ar ∂ − (rAf )  aq  ( Af sin q ) −  ar +  ∂f  r  sin q ∂f ∂r  ∂q  ∂Ar  1 ∂ +  (rAq ) −  af r  ∂r ∂q 

=

Laplacian

1 r sin q

= ∇2 A

1 ∂  2 ∂A  1 ∂  ∂A  1 ∂2 A r +  sin q + r 2 ∂r  ∂r  r 2 sin q ∂q  ∂q  r 2 sin 2 q ∂f 2

D.9.2 Vector Identity 1. Triple Products A ⋅ ( B × C ) =B ⋅ ( C × A ) =C ⋅ ( A × B ) A × ( B × C )= B ( A ⋅ C ) − C ( A ⋅ B )

MVC.CH07_App-D_1PP.indd 427

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428 • Multivariable and Vector Calculus 2/E 2. Product Rules ∇( fg) = f (∇g) + g(∇f ) ∇(A ⋅ B)= A × (∇ × B) + B × (∇ × A) + (A ⋅∇)B + (B ⋅∇)A ∇ ⋅ ( fA)= f (∇ ⋅ A) + A ⋅ (∇f ) ∇ ⋅ (A × B)= B ⋅ (∇ × A) − A ⋅ (∇ × B) ∇ × ( fA)= f (∇ × A) − A × (∇f )= ∇ × ( fA)= f (∇ × A) + (∇f ) × A ∇ × (A × B= ) (B ⋅∇)A − (A ⋅∇)B + A(∇ ⋅ B) − B(∇ ⋅ A) 3. Second Derivative ∇ ⋅ (∇ × A) = 0 ∇ × (∇f ) = 0 ∇ ⋅ (∇f ) = ∇ 2 f ∇ × (∇ × A) = ∇(∇ ⋅ A) − ∇ 2 A 4. Addition, Division, and Power Rules ∇( f + g) = ∇f + ∇g ∇ ⋅ ( A + B) = ∇ ⋅ A + ∇ ⋅ B ∇ × ( A × B) = ∇ × A + ∇ × B  f  g ( ∇f ) − f ( ∇g ) ∇  = 2 g g ∇f n= nf n−1∇f (n = integer) D.9.3 Fundamental Theorems 1. Gradient Theorem b

∫ (∇f ) ⋅ dl=

f (b) − f (a)

a

2. Divergence Theorem

∫

volume

MVC.CH07_App-D_1PP.indd 428

(∇ ⋅ = A)dv

∫

A ⋅ ds

surface

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Formulas • 429

3. Curl (Stokes) Theorem

∫

(∇ × A) ⋅= ds

surface

4.

∫

f dl =−

line

5.

f= ds

∫

A × ds = −

surface

MVC.CH07_App-D_1PP.indd 429

∇f × ds

surface

∫

surface

6.

∫

∫ A ⋅ dl

line

∫

∇f dv

volume

∫

∇ × Adv

volume

1/6/2023 2:40:04 PM

MVC.CH07_App-D_1PP.indd 430

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Bibliography 1. H. Anton and C. Rorres, Elementary Linear Algebra: Applications ­Version, 8th Edition. John Wiley& Sons, 2000. 2. B. O’Neill, Elementary Differential Geometry, Academic Press, 1966. 3. A. V. Pogorelov, Analytical Geometry, Mir Publishers, 1980. 4. H. Cartan, Differential Forms, Hermann Publishers, 1970. 5. D. Lovelock and H. Rund, Tensors, Differential Forms, and Variational Principles, Dover Publications, Inc., 1989. 6. R. Larson, R. P. Hostetler, B. H. Edwards, and D. E. Heyd, Calculus with Analytic Geometry, Houghton Mifflin Company, 2002. 7. R. T. Seeley, Calculus of Several Variables, Scott, Foresman and C ­ ompany, 1970. 8. W. Kaplan, Advanced Calculus, 2nd Edition, Addison-Wesley Publishing Company, 1973. 9. L. J. Goldstein, D. C. Lay, and D. I. Schneider, Calculus and Its Applications, 7th Edition, Prentice Hall, 1996. 10. G. James, Advanced Modern Engineering Mathematics, 3rd Edition, Pearson-Prentice Hall, 2004. 11. M. R. Spiegel, Schaum’s Outline Series Theory and Problems of Advanced Calculus, McGraw-Hill, 1963. 12. P. M. Fitzpatrick, Advanced Calculus: A Course in Mathematical ­Analysis, PWS Publishing Company, 1996.

MVC.CH08_Bib_1PP.indd 431

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432 • Bibliography 13. J. Stewart, Calculus, 5th Edition, Thomson Learning Inc., 2003. 14. E. J. Purcell and D. Varberg, Calculus with Analytic Geometry, 5th ­Edition, Prentice-Hall, Inc., 1987. 15. D. G. Zill and M. R. Cullen, Advanced Engineering Mathematics, 2nd Edition, Jones and Bartlett Publishers, 2000. 16. E. Kreyszig, Advanced Engineering Mathematics, 9th Edition, John Wiley & Sons, Inc., 2006. 17. H. E. Newell, Jr., Vector Analysis, McGraw-Hill Book Company, Inc., 1955. 18. P. V. O’Neil, Advanced Engineering Mathematics, 5th Edition, Brooks/ Cole, 2003. 19. D. G. Duffy, Advanced Engineering Mathematics with MATLAB, 2nd Edition, Chapman& Hall/CRC, 2003. 20. R. Osserman, Two-Dimensional Calculus, Harcourt, Brace & World, Inc., 1968. 21. S. Lang, Calculus of Several Variables, 3rd Edition, Springer-Verlag, 1987. 22. S. Dineen, Multivariate Calculus and Geometry, 2nd Edition, SpringerVerlag, 2001. 23. P. S. Clarke, Jr., Calculus with Analytic Geometry, D.C. Health and ­Company, 1974. 24. H. S. Wilf, Calculus and Linear Algebra, Harcourt, Brace & World, Inc., 1966. 25. J. H. Hubbard and B. B. Hubbard, Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach, Prentice Hall, 1999. 26. B. Willcox, Calculus of Several Variables, Houghton Mifflin Company, 1971. 27. M. R. Spiegel, Schaum’s Outline Series Theory and Problems of Vector Analysis and An Introduction to Tensor Analysis, McGraw-Hill, 1959. 28. K. A. Stroud and D. J. Booth, Engineering Mathematics, 6th Edition, Industrial Press, Inc., 2007. 29. S. Abbott, Understanding Analysis, Springer-Verlag, 2001.

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Bibliography • 433

30. M. N. O. Sadiku, Elements of Electromagnetics, 3rd Edition, Oxford ­University Press, 2001. 31. F. Ayres, Jr., Schaum’s Outline Series Theory and Problems of Modern Algebra, McGraw-Hill, 1959. 32. R. C. Buck and E. F. Buck, Advanced Calculus, McGraw-Hill, 1956. 33. E. W. Swokowski, Calculus with Analytic Geometry, 2nd Edition, ­Prindle, Weber & Schmidt, 1979. 34. S. I. Grossman, Multivariable Calculus, Linear algebra, and Differential Equations, 2nd Edition, Academic Press, Inc., 1986. 35. W. G. May, Linear Algebra, Scott, Foresman, and Company, 1970. 36. H. Schneider and G. P. Barker, Matrices and Linear Algebra, 2nd Edition, Holt, Rinehart and Winston, Inc., 1968. 37. M. O’nan, Linear Algebra, 2nd Edition, Harcourt Brace Jovanovich, Inc. 38. T. Blyth and E. Robertson, Basic Linear Algebra, 2nd Edition, Springer, 2002. 39. J. R. Munkres, Analysis on Manifolds, Westview Press, 1997. 40. J. E. Marsden and A. Tromba, Vector Calculus, 6th Edition, W. H. Freeman, 2001. 41. J. H. Hubbard and B. B. Hubbard, Vector Calculus, Linear Algebra, and Differential Forms: A Unified Approach, 2nd Edition, Prentice Hall, 2001. 42. M. Spivak, Calculus on Manifolds: A Modern Approach to Classical ­Theorems of Advanced Calculus, Westview Press, 1971. 43. H. M. Edwards, Advanced Calculus: A Differential Forms Approach, Birkhäuser, 1994. 44. B. Kolman, Elementary Linear Algebra, Macmillan Publishing Co., Inc., 1970. 45. L. Smith, Linear Algebra, Springer-Verlag, 1978. 46. G. Strang, Linear Algebra and Its Applications, Academic Press, 1976. 47. M. Rahman, Advanced Vector analysis for Scientist and Engineers, WIT Press, 2007.

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434 • Bibliography 48. M. Rahman and I. Mulolani, Applied Vector Analysis, 2nd Edition, CRC Press, 2008. 49. G. Williams, Linear Algebra with Applications, 5th Edition, Jones and Bartlett Publishers, 2005. 50. H. Anton, Elementary Linear Algebra, 10th Edition, John Wiley & Sons, 2010.

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Index A Approximations, 425–426 Arithmetic Mean-Geometric Mean Inequality, 157 Arithmetic operations in MapleTM, 321–322 in MATLAB, 329 C Canonical surfaces, 133–146 Cauchy-Bunyakovsky-Schwarz Inequality, 21, 154 Cavalieri’s principle, 117–119 Chasles’ rule, 7, 24 Cross product, 87–98 Cylindrical helix, 147 D Determinants in three dimensions, 104–111 in two dimensions, 44–53 Differentiation derivatives definition of, 189–193 of function, 190, 193 gradients and directional, 207–218

MVC.CH09_Index_1PP.indd 435

extrema, 222–228 Jacobi matrix, 193–207 Lagrange multipliers, 229–234 Levi-Civitta and Einste, 218–222 limits and continuity, 176–189 multivariable functions, 169–176 topology, 163–169 Dihedral angle, 121 vs. polyhedral angles, 122 rectilinear angle of, 121 vs. trihedral angles, 122 E Einstein’s Summation Convention, 218–219 Ellipsoid, 144–145 Exponential identities, 425 F Formulas approximations, 425–426 exponential identities, 425 hyperbolic functions, 419–420 logarithmic identities, 425 summations (series), 424–425

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436 • Index table of derivatives, 420–421 table of integrals, 422–424 trigonometric identities, 417–419 vectors, 426–429 Fubini’s Theorem, 257 Fundamental parallelogram, 30 G Gauss’ theorem, 307–313 Geometric transformations in two dimensions, 34–44 Green’s theorem, 307–313 H Harmonic Mean-Arithmetic Mean Inequality, 159 Harmonic Mean-Geometric Mean Inequality, 159 Hessian matrix, 222 Hyperbolic functions, 419–420 I Identity transformation, 34 Integration change of variables in double integrals, 273–280 in triple integrals, 297–301 closed and exact forms, 250–256 differential forms, concept of, 235–240 Green’s, Stokes’, and Gauss’ theorems, 307–313 one-manifold, 243–248 polar coordinates, 280–287 surface integrals, 303–306 three-manifolds, 290–295 two-manifolds, 257–267 zero-manifold, 241–242

MVC.CH09_Index_1PP.indd 436

J Jacobi’s Identity, 90 Jordan curve, 53 K Kroenecker’s Delta, 219–220 L Lagrange’s Identity, 92 Levi-Civitta’s Alternating Tensor, 220–221 Linear independence, 28–33 Logarithmic identities, 425 M MapleTM, 50 arithmetic operations, 321–322 assignments, 323 in document mode, 320 help system, 321 int command, 326 limit and diff command, 326 Matrix command, 326 percent sign, 323 plot command, 324–325 solve command, 323–324 symbolic calculations, 322 worksheet mode, 320 MATLAB arithmetic operators, 329 on arrays, 335 considerations, 360 default environment, 328 math functions, 330 Matrix operations, 331 named constants, 334 programming if, else, and elseif statements, 357–359

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Index • 437

for loops, 355 rational and logical operators, 354–355 switch statement, 359–360 while loops, 356–357 symbolic expressions, 361–368 three-dimensional plotting, 347–352 two-dimensional plotting axis command, 343 fplot command, 338 Greek characters, 344–345 hold on and hold off commands, 341 label command, 342 legend command, 343 lines, points, and color types, 337 logarithm scaling, 345 pos value, 343 with special graphics, 346 text and gtext command, 342–343 title command, 342 windows, 328–329 Matrices in three dimensions, 98–104 Matrix operations in MapleTM, 326 in MATLAB, 331 Multidimensional vectors, 152–162 O One-sheet hyperboloid, 140, 144 Orthocentre of the triangle, 25 P Pappus-Guldin Rule, 119–121 Paraboloid elliptic, 142 hyperbolic, 142–143 Parametric curves on the plane, 53–70

MVC.CH09_Index_1PP.indd 437

in space, 146–152 Platonic solids, 125–126 Plotting, MATLAB three-dimensional plotting, 347–352 two-dimensional plotting axis command, 343 fplot command, 338 Greek characters, 344–345 hold on and hold off commands, 341 label command, 342 legend command, 343 lines, points, and color types, 337 logarithm scaling, 345 pos value, 343 with special graphics, 346 text and gtext command, 342–343 title command, 342 Points and vectors on the plane, 1–17 Polyhedral angles, 122, 123 Programming in MATLAB if, else, and elseif statements, 357–359 for loops, 355 rational and logical operators, 354–355 switch statement, 359–360 while loops, 356–357 Pythagorean Theorem, 22 R Repeated partial derivatives, 198 S Scalar product on the plane, 17–28 Spherical trigonometry, 127–132 Stokes’ theorem, 307–313 Summations (series), 424–425 Surveyor’s Theorem, 49

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438 • Index T Table of derivatives, 420–421 Table of integrals, 422–424 Thales’ Theorem, 23, 113 Three-dimensional geometry, 111–117 Triangle Inequality, 22, 154 Trigonometric identities, 417–419 Trihedral angles, 122 Two-sheet hyperboloid, 143 V Vector normalization, 6 Vectors, 426–429 addition of, 7 difference of, 7 and parametric curves canonical surfaces, 133–146 Cavalieri’s principle, 117–119 cross product, 87–98 determinants, 44–53, 104–111

MVC.CH09_Index_1PP.indd 438

geometric transformations, 34–44 linear independence, 28–33 matrices in three dimensions, 98–104 multidimensional vectors, 152–162 Pappus-Guldin Rule, 119–121 on the plane, 53–70 points and vectors on the plane, 1–17 scalar product on the plane, 17–28 in space, 70–86, 146–152 spherical trigonometry, 127–132 three-dimensional geometry, 111–117 scalar multiplication of, 8 Vector spaces, 1, 70–86 Y Young’s Inequality, 155

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