This comprehensive textbook explores the topics of vector functions and functions of several variables. With over 500 ex

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*Table of contents : PrefaceContents1 Vectors and Analytic Geometry 1.1 Vectors and Their Properties 1.2 Lines and Planes in Space 1.3 Cylinder and Surface of Revolution 1.4 Quadric Surfaces 1.5 Polar, Cylindrical and Spherical Coordinates 1.6 Solved Problems 1.7 Exercises2 Vector Functions and Parametric Equations 2.1 Definition and Calculus of Vector Functions 2.2 Velocity and Acceleration in Space 2.3 Unit Tangent Vector, Unit Normal Vector and Curvature 2.4 Radius of Curvature 2.5 Solved Problems 2.6 Exercises3 Functions of Several Variables, Limits and Continuity 3.1 Functions of Several Variables 3.2 Limits 3.3 Continuity 3.4 Solved Problems 3.5 Exercises4 Partial Derivatives, Directional Derivatives and Gradient Vectors 4.1 Partial Derivatives 4.2 Directional Derivatives and Gradient Vectors 4.3 Tangent Plane and Normal Line to a Surface 4.4 Solved Problems 4.5 Exercises5 Differentiability and Differential 5.1 Differentiability 5.2 The Chain Rule 5.3 Taylor Series 5.4 Solved Problems 5.5 Exercises6 Extreme of Functions 6.1 Derivative Tests for Local Extreme Values 6.2 Lagrange Multipliers 6.3 Solved Problems 6.4 Exercises7 Vector Fields 7.1 Vector Fields, Limits and Continuity 7.2 Linear Transformations 7.3 Total Derivatives 7.4 Conservative Vector Fields 7.5 Divergence and Curl 7.6 Solved Problems 7.7 Exercises8 Line Integrals 8.1 Work as a Line Integral 8.2 Line Integrals Independent of the Path 8.3 Solved Problems 8.4 Exercises9 Multiple Integrals 9.1 Double Integral over Rectangle Region 9.2 Double Integral over Bounded Non-rectangular Region 9.3 Triple Integrals 9.4 Masses and Moments in Three Dimensions 9.5 Solved Problems 9.6 Exercises10 Surface Integrals 10.1 Green's, Divergence and Stokes' Theorems 10.2 Surface Area 10.3 Solved Problems 10.4 ExercisesAppendix ReferencesIndex*

Bijan Davvaz

Vectors and Functions of Several Variables

Vectors and Functions of Several Variables

Bijan Davvaz

Vectors and Functions of Several Variables

Bijan Davvaz Department of Mathematical Sciences Yazd University Yazd, Iran

ISBN 978-981-99-2934-4 ISBN 978-981-99-2935-1 (eBook) https://doi.org/10.1007/978-981-99-2935-1 Mathematics Subject Classification: 26B12, 26Bxx © The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors, and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Singapore Pte Ltd. The registered company address is: 152 Beach Road, #21-01/04 Gateway East, Singapore 189721, Singapore

Preface

In this book, we study vector functions and functions of several variables. The major part of this book stems from the course in general mathematics given by the author for many years at Yazd University. The book is intended to serve as a textbook of Calculus (II and III) exercises with solutions for the calculus courses that are usually taken by first year students. Indeed, it can be used by all students in Mathematics, Statistics, Computer Science, Engineering and Basic Sciences. A total of 500 problems have been selected from many problems in different books. In choosing problems, three major criteria have been considered to be challenging, interesting and educational. So, most of the issues have technical and educational aspects. Over the years, in teaching general mathematics have encountered many challenging exercises which have been solved with much time. Most of these exercises have been collected over the last 30 years from different references. Although some of the references that have been used are listed at the end of the book, I do not know the sources and the main creators of most exercises and problems. So, my duty is to thank everyone who has designed these issues for the first time. The book is based on problems, although it is assumed that the students have learned the classical contents from other textbooks, at the same time, at the beginning of each chapter, some necessary definitions, concepts and theorems are listed. Moreover, as the reader will soon see, there are many exercises at the end of each chapter. They are divided into two categories: easier and harder. The purpose of these exercises is to allow students to test their assimilation of the material, to challenge their mathematical ingenuity and to be a means of developing mathematical insight, intuition and techniques. The book is organized into ten chapters. Chapter 1 discusses vectors and analytic geometry, like properties of vectors, lines and planes in space, cylinder and surface of revolution, quadric surfaces, polar, cylindrical and spherical coordinates. Vector functions and parametric equations, velocity and acceleration, unit tangent vector, unit normal vector and curvature are discussed in Chap. 2. Functions of several variables are presented in Chap. 3. In this chapter, we investigate the limits and continuity of these functions. In Chap. 4, we study partial derivatives, directional derivation, gradient vector, tangent plane and normal line to a surface. The next objective in Chap. 5 is differentiability and differential. In particular, applications v

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Preface

of chain rule and Taylor series are discussed. The main objective in Chap. 6 is the extreme of functions of several variables. We study derivative tests for local extreme and Lagrange multiplier. In Chap. 7, we investigate vector fields, calculus of vector fields, total derivative, conservative vector fields, divergence and curl. In Chap. 8, we study line integrals, work as a line integral and line integrals independent of path. The main subject of Chap. 9 is multiple integrals like double integral, triple integral, mass and moment of inertia. The aim of the last chapter (i.e., Chap. 10) is devoted to surface integrals. The study focuses on Green’s, Divergence and Stoke’s theorems. Finally, I hope the readers enjoy reading this book and carefully selected problems and hope they will find this issue informative and helpful. Yazd, Iran

Bijan Davvaz

Contents

1

Vectors and Analytic Geometry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Vectors and Their Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Lines and Planes in Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Cylinder and Surface of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Polar, Cylindrical and Spherical Coordinates . . . . . . . . . . . . . . . . . 1.6 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 3 5 6 7 12 59

2

Vector Functions and Parametric Equations . . . . . . . . . . . . . . . . . . . . . 2.1 Definition and Calculus of Vector Functions . . . . . . . . . . . . . . . . . . 2.2 Velocity and Acceleration in Space . . . . . . . . . . . . . . . . . . . . . . . . . 2.3 Unit Tangent Vector, Unit Normal Vector and Curvature . . . . . . . 2.4 Radius of Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65 65 67 69 70 71 98

3

Functions of Several Variables, Limits and Continuity . . . . . . . . . . . . 3.1 Functions of Several Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

103 103 105 108 109 137

4

Partial Derivatives, Directional Derivatives and Gradient Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Directional Derivatives and Gradient Vectors . . . . . . . . . . . . . . . . . 4.3 Tangent Plane and Normal Line to a Surface . . . . . . . . . . . . . . . . . 4.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

143 143 145 146 147 167

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5

Differentiability and Differential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.4 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

171 171 173 174 175 212

6

Extreme of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Derivative Tests for Local Extreme Values . . . . . . . . . . . . . . . . . . . 6.2 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

217 217 219 220 258

7

Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 Vector Fields, Limits and Continuity . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Linear Transformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 Total Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Conservative Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Divergence and Curl . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

263 263 265 266 268 269 270 293

8

Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Work as a Line Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Line Integrals Independent of the Path . . . . . . . . . . . . . . . . . . . . . . . 8.3 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

299 299 300 300 318

9

Multiple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 Double Integral over Rectangle Region . . . . . . . . . . . . . . . . . . . . . . 9.2 Double Integral over Bounded Non-rectangular Region . . . . . . . . 9.3 Triple Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.4 Masses and Moments in Three Dimensions . . . . . . . . . . . . . . . . . . 9.5 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

321 321 323 325 326 327 383

10 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.1 Green’s, Divergence and Stokes’ Theorems . . . . . . . . . . . . . . . . . . 10.2 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.3 Solved Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

391 391 393 393 414

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 419 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421

Chapter 1

Vectors and Analytic Geometry

1.1 Vectors and Their Properties We present the definitions and results related to vectors in space. A vector in space is an ordered triple of real numbers (x, y, z). The numbers x, y and z are called the component of the vector (x, y, z). Usually, we may use the notation (x, y, z) as a vector or as a point in the space, and must be careful to avoid confusing the notation. A vector can be represented by a directed line segment. If A = (a1 , a2 , a3 ), then the directed line segment having its initial points at origin and its terminal point at the point (a1 , a2 , a3 ) is called the position representation of A. The zero vector is the vector (0, 0, 0). The magnitude of A denoted byA is just the distance from the origin to terminal point (a1 , a2 , a3 ), i.e., A = a12 + a22 + a32 . Let A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ) be two vectors and c be a real number (scalar). Then A + B = (a1 + b1 , a2 + b2 , a3 + b3 ); A − B = (a1 − b1 , a2 − b2 , a3 − b3 ); c A = (ca1 , ca2 , ca3 ). If A, B and C are any vectors and c and d are any real numbers, then we have (1) (2) (3) (4) (5) (6) (7)

A + B = B + A (commutative law); (A + B) + C = A + (B + C) (associative law); A + 0 = 0 + A = 0 (existence of additive identity); A + (−A) = (−A) + A = 0 (existence of negative); (cd)A = c(d A); c(A + B) = c A + cB; (c+d)A=cA+dA.

By a basis, we mean any three fixed vectors e1 , e2 and e3 such that any arbitrary vector − → A has a unique representation of the form A = a1 e1 + a2 e2 + a3 e3 . If i = (1, 0, 0), − → − → j = (0, 1, 0) and k = (0, 0, 1), then we can write © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_1

1

2

1 Vectors and Analytic Geometry

− → − → − → A = a1 (1, 0, 0) + a2 (0, 1, 0) + a3 (0, 0, 1) = a1 i + a2 j + a3 k , − → − → − → and so i , j and k are basis vectors. The direction of a non-zero vector is given by three angles α, β and γ measured from the positive x, y and z axes, respectively, to the position representation of the vector. It can be shown that cos α =

a2 a3 a1 , cos β = and cos γ = , A A A

and cos2 α + cos2 β + cos2 γ = 1. The numbers cos α, cos β and cos γ are called the direction cosines of the vector A. − → − → − → If A = a1 i + a2 j + a3 k is a non-zero vector, then the unit vector U having the same direction as A is given by U=

a2 − a3 − a1 − → → → i + j + k. A A A

If A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ) are two vectors, then the dot product (or scalar product) of A and B is denoted by A · B and is given by A · B = a1 b1 + a2 b2 + a3 b3 . It is easy to see that A · A = A2 . If θ is the angle between the two non-zero vectors A and B, then A · B = A b cos θ. Two non-zero vectors are parallel if and only if one of the vectors is a scalar multiple of the other. If A and B are two vectors, then A and B are said to be orthogonal if and only if A · B = 0. If A = (a1 , a2 , a3 ) and B = (b1 , b2 , b3 ), then the cross product (or vector product) of A and B, denoted by A × B, is given by − → i A × B = a1 b 1

− → − → j k a2 a3 = a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 . b b 2

3

Let A, B and C be three vectors and c be a real number. Then, we have (1) (2) (3) (4) (5) (6) (7) (8)

A × A = 0; 0 × A = A × 0 = 0; A × B = −(B × A); A × (B + C) = A × B + A × C; (c A) × B = A × (cB) = c(A × B); A × B = A B sin θ, where θ is the angle between A and B; A · B × C = A × B · C; A × B is orthogonal to both A and B.

1.2 Lines and Planes in Space

3

Fig. 1.1 The projection of U onto V

U

V P rojV U

If S = {A1 , A2 , . . . , Ak } is a set of vectors, then the vector equation c1 A1 + c2 A2 + . . . + ck Ak = 0 has at least one solution, namely c1 = c2 = . . . = ck = 0. If this is the only solution, then S is called a linearly independent set. If there are other − → − → − → solutions, then S is called a linearly dependent set. For example, the set { i , j , k } is linearly independent. The vector projection of a vector U on a non-zero vector V is the orthogonal projection of U on a straight line parallel to V ; see Fig. 1.1. The vector projection of a vector U on a non-zero vector V is equal to Pr oj V U =

U ·V V (vector component of U along V ). V 2

1.2 Lines and Planes in Space A line in space is uniquely determined by specifying any fixed point P0 = (x0 , y0 , z 0 ) on and any non-zero vector A = (a, b, c) parallel to . Let P = (x, y, z) −−→ be a variable point in space. Then the point P belongs to if and only if the vector P0 P −−→ and A = (a, b, c) are parallel, or equivalently P0 P = t A, where t is any real number. −−→ Since P0 P = (x − x0 , y − y0 , z − z 0 ), it follows that x − x0 = ta, y − y0 = tb and z − z 0 = tc, or equivalent x = x0 + ta, y = y0 + tb and z = z 0 + tc.

(1.1)

Equations (1.1) represent the line , and these equations are called parametric equations of the line. The numbers a, b and c are called direction numbers of the line . If none of the numbers a, b or c is zero, then we can eliminate t from (1.1) and get y − y0 z − z0 x − x0 = = . a b c

(1.2)

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These equations are called symmetric equations of the line. If one of the numbers a, b or c is zero, we do not use the symmetric equations (1.2). For example, if b = 0, then the equations of the line are x − x0 z − z0 = and y = y0 . a c Distance between a point and a line in space: Given a line in space and a point P not on . Suppose that A is any vector parallel to and Q is any point on . Then, the distance d between P and is given by d=

−→ A × Q P . A

A plane in space is determined by knowing a point on the plane and a vector that is perpendicular or normal to the plane. Suppose that the plane S passes through a point − → − → − → P0 = (x0 , y0 , z 0 ) and is normal to the non-zero vector N = a i + b j + c k . Then −−→ S is the set of all points P = (x, y, z) for which P0 P is orthogonal to N ; see Fig. 1.2. −−→ So, we have N · P P0 = 0. This gives that a(x − x0 ) + b(y − y0 ) + c(z − z 0 ) = 0. − → Equations for a plane: The plane through P0 = (x0 , y0 , z 0 ) normal to N = a i + − → − → b j + c k has Vector equation: Component equation: Component equation simplified:

−−→ N · P P0 = 0; a(x − x0 ) + b(y − y0 ) + c(z − z 0 ) = 0; ax + by + cz + d = 0.

Two planes are parallel if and only if their normals are parallel. Two planes that are not parallel intersect in a line. Distance between a point and a plane: The distance d between the point P0 = (x0 , y0 , z 0 ) and the plane S with equation ax + by + cz + d = 0 is given by

Fig. 1.2 The standard equation for a plane in space is defined in terms of a vector normal to the plane

• N

• • P0 = (x0 , y0 , z0 )

P = (x, y, z)

•

1.3 Cylinder and Surface of Revolution

d=

5

|ax0 + by0 + cz 0 + d| . √ a 2 + b2 + c2

The angle between two intersecting planes is defined to be the (acute) angle determined by their normal vectors.

1.3 Cylinder and Surface of Revolution Let C be a plane curve, and let be a line which is not parallel to (or in) the plane of C. Then the surface S constructed from all lines through C parallel to is called a cylinder. The curve C is called the directrix of the cylinder S, and the infinitely many lines parallel to of which S is formed are called rulings (or generators) of S. In three-dimensional space, the graph of an equation in two of the three variables x, y and z is a cylinder whose rulings are parallel to the axis associated with the missing variable and whose directrix is a curve in the plane associated with the two variables appearing in the equation. Each surface in Fig. 1.3 is called an elliptic cylinder. The directrix for the left surface is the ellipse x 2 /a 2 + y 2 /b2 = 1 and the ruling is parallel to z-axis, while the directrix for the right surface is the ellipse y 2 /b2 + z 2 /c2 = 1 and the ruling is parallel to x-axis. The left surface in Fig. 1.4 shows a parabolic cylinder whose directrix is the parabola z = ax 2 in the x z-plane and the ruling is parallel to y-axis; and the right surface is a hyperbolic cylinder and its directrix is the hyperbola y 2 /b2 − z 2 /c2 = 1 in the yz-plane and the ruling is parallel to x-axis. The surface generated by revolving a plane curve about a line in its plane is called a surface of revolution. The fixed line is called the axis of the surface of revolution,

Fig. 1.3 Elliptic cylinders

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1 Vectors and Analytic Geometry

Fig. 1.4 From left to right, parabolic cylinder and hyperbolic cylinder

and the plane curve is called the generating curve. Below, we present the general equation for some surface of revolution: Generatingcurve

Axis of revolution

F(y, z) = 0 (x = 0)

y − axis

F(y, z) = 0 (x = 0)

z − axis

F(x, z) = 0 (y = 0)

x − axis

F(x, z) = 0 (y = 0)

z − axis

F(x, y) = 0 (z = 0)

x − axis

F(x, y) = 0 (z = 0)

y − axis

Surface of revolution √ F y, x 2 + z 2 = 0 F x 2 + y2, z = 0 F x, y 2 + z 2 = 0 F x 2 + y2, z = 0 F x, y 2 + z 2 = 0 √ F x 2 + z2, y = 0

1.4 Quadric Surfaces A sphere is the set of all points in space equidistant from a fixed point. The fixed point is called the center of the sphere and the measure of the constant distance is called the radius of the sphere. An equation of the sphere of radius r and center at (a, b, c) is (x − a)2 + (y − b)2 + (z − c)2 = r 2 .

1.5 Polar, Cylindrical and Spherical Coordinates

7

By a second degree equation in three variables x, y and z, we mean an equation of the form Ax 2 + By 2 + C z 2 + Dx y + E x z + F yz + Gx + H y + I z + J = 0, where the coefficients A, B, C, D, E and F are not all zero. The graph of any such equations is called a quadric surface. These surfaces correspond to the conics in the plane. The simplest types of quadric surfaces are the parabolic, elliptic and hyperbolic cylinders. There exist six other types of quadric surfaces, which are listed as follows: Qusric surface

Name

x2 y2 z2 + 2 + 2 =1 2 a b c

Ellipsoid

x2 y2 + =z a2 b2

Elliptic paraboloid

x2 y2 z2 + − =1 a2 b2 c2

Hyperboloid of one sheet

x2 y2 z2 + 2 − 2 = −1 2 a b c

Hyperboloid of two sheets

x2 y2 z2 + − =0 a2 b2 c2

Elliptic cone

x2 y2 − 2 =z 2 a b

Hyperbolic paraboloid

The graphs of the above quadric surfaces are shown in Fig. 1.5. If all three numbers a, b and c in the equation of an ellipsoid are equal, the ellipsoid is a sphere.

1.5 Polar, Cylindrical and Spherical Coordinates We usually use rectangular coordinates to represent a point in a plane. However, polar coordinates are more convenient for dealing with circles, arcs and spirals. In the polar coordinate, the frame of reference is a point O which we call the pole, and a ray that emanates from it we call the polar axis. Let P be a point in the plane, and r be the distance between O and P, and θ be the angle between the polar axis and the segment O P. Then, we describe the point P in polar coordinates by (r, θ); see Fig. 1.6. We also allow r to take negative values. Polar coordinates and rectangular coordinates are often used simultaneously, by choosing the pole and polar axis to

8

1 Vectors and Analytic Geometry

Fig. 1.5 Quadric surfaces from left to right, top to bottom: ellipsoid; elliptic paraboloid; hyperboloid of one sheet; hyperboloid of two sheets; elliptic cone; hyperbolic paraboloid

1.5 Polar, Cylindrical and Spherical Coordinates

9

Fig. 1.6 Polar coordinates

x r

P = (r, θ) • y

θ Polar axis

O

be the origin and the positive x-axis of the rectangular coordinates system. Then, it is apparent from Fig. 1.6 that the point P with polar coordinates (r, θ) has the rectangular coordinates x = r cos θ and y = r sin θ. If r > 0, it follows that r=

x 2 + y 2 and θ = tan

y (x = 0). x

The graph of a polar equation r = f (θ), or more generally F(r, θ) = 0, consists of all points that have at least one polar representation whose coordinates satisfy the equation. Symmetry tests for polar graphs: (Look at Fig. 1.7). (1) Symmetry about the pole axis (x-axis): If the point (r, θ) lies on the graph, the point (r, −θ) or (−r, π − θ) lies on the graph; (2) Symmetry about the y-axis: If the point (r, θ) lies on the graph, the point (r, π − θ) or (−r, −θ) lies on the graph; (3) Symmetry about the pole: If the point (r, θ) lies on the graph, the point (−r, θ) or (r, θ + π) lies on the graph. Let r = f (θ) be a polar equation of a curve, and let the tangent to the curve at the point P0 be the line inclination α (0 ≤ α < π), as in Fig. 1.8. The slope of T at the point P0 is dr +r tan θ dy dθ = . tan α = dr dx − r tan θ dθ

10

1 Vectors and Analytic Geometry

y

y

P (−r, −θ) •

x

θ

• Q(−r, θ)

P (r, π − θ) •

P (r, θ) •

P (r, θ) •

x

θ

• P (r, −θ)

• Q(r, π + θ)

• P (−r, π − θ)

Fig. 1.7 Symmetry tests for polar coordinates Fig. 1.8 Tangent line to a polar curve

y r = f (θ) •

θ

T P0

α x

O

In the polar coordinate system, the area of the region between the origin and the curves r = f (θ), α ≤ θ ≤ β is A=

β α

1 2 r dθ. 2

If r = f (θ) has a continuous first derivative for α ≤ θ ≤ β and if the point P = (r, θ) traces the curve r = f (θ) exactly once as θ runs from α to β, then the length of the curve is β

dr 2 L= r2 + dθ. dθ α In the cylindrical coordinate system, a point P in space (Fig. 1.9) is represented by ordered triple (r, θ, z), where (r, θ) are the polar coordinates of the point’s projection in the x y-plane and z is the usual z-coordinate, in the rectangular coordinate system.

1.5 Polar, Cylindrical and Spherical Coordinates

11

Fig. 1.9 Cylindrical coordinates

z P = (x, y, z) P = (r, θ, z) • z

y

O x

r θ y x Fig. 1.10 Spherical coordinates

z P = (x, y, z) P = (ρ, θ, ϕ) • ϕ ρ

z

y

O x θ y x

The rectangular coordinates (x, y, z) and the cylindrical coordinates (r, θ, z) of a point are related as follows: x = r cos θ, y = r sin θ, z = z, r 2 = x 2 + y 2 and tan θ =

y . x

Given a point P with rectangular coordinates x, y and z and suppose that ρ is the distance between origin O and P, and ϕ is the angle from the positive z-axis to O P. Also, suppose that θ is the same angle as in the case of cylindrical coordinates; see Fig. 1.10. Then the point P is said to have spherical coordinates P = (ρ, θ, ϕ). By placing the spherical coordinate system and the rectangular coordinate system together as shown in Fig. 1.10, we get the relation between these two coordinate systems as follows: x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ and z = ρ cos ϕ. (1.3) By squaring each of the equations in (1.3) and adding, we get x 2 + y 2 + z 2 = ρ2 .

12

1 Vectors and Analytic Geometry

Fig. 1.11 The segment joining the midpoints of two sides of a triangle is half as long as the third side

A

N M C

B

1.6 Solved Problems 1. Use vectors to prove that the segment joining the midpoints of two sides of a triangle is half as long as the third side and parallel to it (see Fig. 1.11). −→ −−→ −→ −−→ −−→ Solution. In the triangle M AN , we have M A + AN = M N . Since M A = 21 B A and −→ 1 −→ AN = 2 AC, it follows that −→ −→ −−→ B A + AC = 2 M N .

(1.4)

On the other hand, in the triangle ABC, we have −→ −→ −→ B A + AC = BC.

(1.5)

−→ −−→ By (1.4) and (1.5), we obtain BC = 2 M N . Moreover, by the definition of parallel −→ −−→ vectors we conclude that the vectors BC and M N are parallel. 2. The points A = (−2, 3) and C = (4, −5) are opposite vectors of a square, meaning that the line segment from A to C is a diagonal of the square. Find the other two vertices of the square. Solution. Consider Fig. 1.12. In this figure, we observe the square with known vertices A and C, and unknown vertices B and D. The two diagonals of the square have the −→ same length and bisect each other perpendicularly. Since AC = (6, −8), it is easy −→ −−→ to see that D B = (8, 6). Suppose that O M is the position vector of the midpoint M between A and C. Then, we can write

1.6 Solved Problems

13

Fig. 1.12 The square with known vertices A = (−2, 3) and C = (4, −5), and unknown vertices B and D

y A B x M D C

−→ −→ O A + OC −−→ −→ 1 −→ −→ 1 −→ −→ O M = O A + AC = O A + OC − O A = 2 2 2 1 1 = (−2, 3) + (4, −5) = (2, −2) = (1, −1). 2 2 So, we obtain −→ −−→ 1 −→ O B = O M + D B = (1, −1) + (4, 3) = (5, 4), 2 −−→ −−→ 1 −→ O D = O M − D B = (1, −1) − (4, 3) = (−3, −4). 2 Therefore, we deduce that the other vertices of the square are B = (5, 2) and D = (−3, −4). 3. Show that if the midpoints of the sides of a quadrilateral are connected, then the resulting quadrilateral is a parallelogram. Solution. We show that the quadrilateral M N P Q in Fig. 1.13 is a parallelogram. −→ −−→ −→ −−→ −−→ −−→ −−→ −−→ Since O M = O A + AM and O M = O D + D M, it follows that O M = 1/2 O A + −−→ −−→ −→ −→ O D . Similarly, we have O N = 1/2 O A + O B . Hence, we obtain −−→ −−→ −−→ 1 −→ −−→ 1 −→ M N = O N − O M = O B − O D = D B. 2 2

(1.6)

−→ −→ −−→ −→ −→ −−→ Again, since O P = 1/2 O B + OC and O Q = 1/2 O D + OC , it follows that −→ −→ −−→ 1 −→ −−→ 1 −→ Q P = O P − O Q = O B − O D = D B. 2 2

(1.7)

14

1 Vectors and Analytic Geometry

Fig. 1.13 The quadrilateral M N P Q is a parallelogram

A

Fig. 1.14 The altitudes of a triangle intersect in a common point

N •

M•

• O

D

• Q

B

•P

C

C P Q

A

S

R

B

−−→ −→ −−→ From (1.6) and (1.7), we conclude that M N = Q P. Similarly, we see that M Q = −→ N P. Since the opposite sides of the quadrilateral M N P Q are equal in length and are parallel, it follows that this quadrilateral is a parallelogram. 4. Use vectors to prove that the altitudes of a triangle intersect in a common point. Solution. Look at Fig. 1.14. In the triangle ABC, assume that the altitudes A P and B Q intersect in the point S. We show that the line C R passing through the point S is perpendicular to AB. We have − → −→ − → −→ AS · BC = 0 and B S · AC = 0. − → −→ Now, it is enough to show that C S · AB = 0. We can write

1.6 Solved Problems

15

Fig. 1.15 Position vectors

y B P

A x O

− → −→ − → −→ −→ C S · AB = C S · ( AC + C B) − → −→ − → −→ = C S · AC + C S · C B −→ − → −→ −→ − → −→ = (C B + B S) · AC + (C A + AS) · C B −→ −→ − → −→ −→ −→ − → −→ = C B · AC + B S · AC + C A · C B + AS · C B. −→ −→ Since C A = − AC, it follows that − → −→ −→ −→ −→ −→ C S · AB = C B · AC + (− AC) · C B = 0. This completes the proof. −→ −→ 5. Let O A and O B be the position vectors of two points A and B with respect to any −→ −→ origin O. Show that the final point P of the vector (1 − t) O A + t O B (0 ≤ t ≤ 1) divided the segment AB in the ratio t/(1 − t), and in particular that the midpoint of −→ −→ AB has position vector (1/2)( O A + O B). Solution. Look at Fig. 1.15. We observe that −→ −→ −→ −→ −→ −→ −→ −→ −→ O P = O B + B P, O P = O A + A P and O B = O A + AB. So, we can write −→ −→ −→ −→ −→ −→ −→ −→ (1 − t) O A + t O B = O B + B P and (1 − t) O A + t O B = O A + A P. This yields that −→ −→ −→ −→ −→ −→ (1 − t) O A + (t − 1) O B = B P and − t O A + t O B = A P,

16

1 Vectors and Analytic Geometry

A

• O

• O

B

Fig. 1.16 Two intersecting circles

−→ −→ −→ −→ −→ −→ or equivalently (1 − t) O A − (1 − t) O B = B P and t O A − t O B = P A. Hence, we −→ −→ −→ −→ −→ −→ −→ −→ −→ have (1 − t)( O A − O B) = B P and t ( O A − O B) = P A. Since O A − O B = B A, −→ −→ −→ −→ it follows that (1 − t) B A = B P and t B A = P A. This completes the proof. Clearly, if P is the midpoint of the segment AB, then t = 1/2. 6. Use vectors to show that the line joining the centers of two intersecting circles is perpendicular to the line joining the points of intersection. Solution. In Fig. 1.16, let the coordinates of A, B, O and O be as follows: A(x1 , y1 ), B(x2 , y2 ), O(0, 0), O (a, 0). −−→ −→ Then, we have AB = (x2 − x1 , y2 − y1 ) and O O = (a, 0). So, we obtain −→ −−→ AB · O O = (x2 − x1 )a.

(1.8)

On the other hand, we have −−→ O A = (x1 − a, y1 ), −−→ O B = (x2 − a, y2 ),

−→ O A = (x1 , y1 ), −→ O B = (x2 , y2 ).

−−→ −−→ Since | O A| = | O B|, it follows that (x1 − a)2 + y12 = (x2 − a)2 + y22 .

(1.9)

−→ −→ Since | O A| = | O B|, it follows that x12 + y12 = x22 + y22 .

(1.10)

1.6 Solved Problems

17

Subtracting both sides of (1.9) and (1.10) gives (x1 − a)2 − x12 = (x2 − a)2 − x22 . This implies that (1.11) − ax1 + ax2 = 0. −−→ −→ Now, from (1.8) and (1.11), we conclude that O O · AB = 0. Therefore, the vectors −−→ −→ O O and AB are perpendicular, as desired. 7. (1) Provethat the midpoint of the line segment from P = (x1 , y1 , z 1 ) to Q = x1 + x2 y1 + y2 z 1 + z 2 . (x2 , y2 , z 2 ) is , , 2 2 2 (2) Find the length of the medians of the triangle with vertices A = (1, 2, 3), B = (−2, 0, 5) and C = (4, 1, 5). −→ Solution. (1) We have P Q = (x2 − x1 , y2 − y1 , z 2 − z 1 ). Let M be the midpoint between P and Q, and O be the origin. We can write 1 −−→ −→ 1 −→ O M = O P + P Q = (x1 , y1 , z 1 ) + (x2 − x1 , y2 − y1 , z 2 − z 1 ) 2 2 x + x y + y z + z 1 2 1 2 1 2 = , , . 2 2 2 (2) We compute the midpoints using the part (1). Suppose that M AB , M BC and M AC are midpoints between A, B, between B, C and between A, C, respectively. Then, we have M AB = (−1/2, 1, 4), M BC = (1, 1/2, 5) and M AC = (5/2, 3/2, 4). Now, by using the distance formula, we obtain the distance between A and M BC is 5/2;√ the distance between B and M AC is 1/2√94; the distance between C and M AB is 1/2 85.

8. A wrench 30 cm long lies along the positive y -axis and grips a bolt at the origin. A force is applied in the direction (0, 3, −4) at the end of the wrench; see Fig. 1.17. Find the magnitude of the force needed to supply 100 N · m of torque to the bolt. Solution. We have to find the force F. Suppose that D = (0, 3, −4). The first task is to find the sine of the angle between the direction of the force and the y-axis. The force is applied to the wrench at position P, so the displacement vector is R = (0, 0.3, 0). Then, we can write − − → − → → j k i − → D × R = 0 3 −4 = 1.2 i . 0 0.3 0

18

1 Vectors and Analytic Geometry

Fig. 1.17 Wrench and force

z

0.3 m Wrench

P = (0, 0.3, 0) y

x

Direction of the force

Since D × R = D R sin θ, it follows that sin θ = 1.2/1.5 = 0.8. Finally, we obtain 100 R × F = = 416.66 N. F = R sin θ 0.3 × 0.8 − → − → − → − → 9. Find a unit vector that is orthogonal to both i + j and i + k . − → − → − → Solution. Let U = a i + b j + c k be the desired unit vector. Then, we have − → − → − → − → − → − → − → − → − → − → (a i + b j + c k ) · ( i + j ) = a + b and (a i + b j + c k ) · ( i + k ) = a + c.

− → − → − → We want to determine the unit vector U orthogonal to both i + j and i + − → k ; hence we put a + b = 0 and a + c = 0. This gives that a = −b = −c, and so − → − → − → − → − → − → − → the vector U is of the form a i − a j − a k = a( i − j − k ). Since i − √ √ − → − → j − k = 3, it follows that a = 1/ 3. Consequently, the desired vector is U = √ − → − → − → 1/ 3( i − j − k ). 10. Given three non-zero vectors A, B and C in R3 . Assume the angle between A and C is equal to the angle between B and C. Prove that C is orthogonal to the vector BA − AB. Solution. Let θ1 be the angle between A and C, and θ2 be the angle between B and C. We have A · C = A C cos θ1 and B · C = B C cos θ2 . Since θ1 = θ2 , it follows that

1.6 Solved Problems

19

B ·C A·C = , A B or equivalently B(A · C) = A(B · C). Therefore, we conclude that C · (BA − AB) = 0, as desired. 11. Show that the vector V = BA + AB bisects the angle between A and B. Solution. Let θ1 be the angle between A and V , and θ2 be the angle between B and V . Then, we have V · A = V A cos θ1 and V · B = V B cos θ2 .

(1.12)

On the other hand, we have V · A = (BA + AB) · A = B A2 + AB · A

(1.13)

V · B = (BA + AB) · B = BA · B + AB2 .

(1.14)

and

From (1.13) and (1.14), we conclude that V·B V·A = . A B

(1.15)

Finally, from (1.12) and (1.15) we get cos θ1 = cos θ2 , and so θ1 = θ2 . 12. Three vectors A, B and C in R3 satisfy all the following properties: A = C = 5, B = 1 and A − B + C = A + B + C. If the angle between A and B is π/8, find the angle between B and C. Solution. We can write (A + C) − B · (A + C) − B = (A + C) + B · (A + C) + B . This implies that −2(A + C) · B = 2(A + C) · B, and so we get −A · B = B · C. Now, let θ be the angle between B and C. Then, we obtain −A B cos(π/8) = B C cos θ, or equivalently cos θ =

π −A B cos(π/8) = − cos . B C 8

This yields that θ = 9π/8. 13. Let U and V be two unit vectors and the angle between them is π/3. If A = tU + V and B = U + t V , find the values of t for which the inner product A · B is maximum, minimum or zero if such values exist. Solution. We have U · V = cos(π/3) = 1/2. Hence, we obtain

20

1 Vectors and Analytic Geometry

A · B = (tU + V ) · (U + t V ) = tU · U + t 2 U · V + V · U + t V · V 1 1 3 = tU 2 + (t 2 + 1)U · V + tV 2 = t + (t 2 + 1) + t = (t + 2)2 − . 2 2 2 The function f (t) = (t + 2)2 /2 − 3/2 is a parabola and there is no maximum value. There exists a minimum value occurring √ at t = −2. The zero occurs when f (t) = 0. Hence, if f (t) = 0, then t = −2 ± 3. 14. Prove by using vectors that the points A = (2, 2, 2), B = (2, 0, 1), C = (4, 1, −1) and D = (4, 3, 0) are vertices of a rectangle. Solution. We obtain √ −→ AB = (2 − 2)2 + (0 − 2)2 + (1 − 2)2 = 5, √ −→ DC = (4 − 4)2 + (3 − 1)2 + (0 + 1)2 = 5, −→ AD = (4 − 2)2 + (3 − 2)2 + (0 − 2)2 = 3, −→ BC = (4 − 2)2 + (1 − 0)2 + (−1 − 1)2 = 3. −→ −→ −→ −→ −→ So, we see that AB = DC and AD = BC. Since AB = (0, −2, −1) −→ −→ −→ −→ and AD = (2, 1, −2), it follows that AB · AD = 0 − 2 + 2 = 0; and so AB is −→ −→ −→ −→ −→ orthogonal to AD. Similarly, we observe that B A · BC = 0, C B · C D = 0 and −→ −→ DC · D A = 0. This yields that ABC D is a rectangle. 15. Let A, B and C be three vectors in space. If A ·B = A · C and A × B = A × C, does it follows that B=C? Solution. Suppose that B − C = 0. By rearranging the equations, we have A · (B − C) = 0 and A × (B − C) = 0. Since A · (B − C) = 0, it follows that B − C is orthogonal to A. Since A × (B − C) = 0, it follows that B − C is parallel to A. This is a contradiction. Therefore, we conclude that B − C = 0, or equivalently B = C. 16. Let A, B and C be three vectors in space. (1) If A + B + C = 0, prove that A × B = B × C = C × A; (2) If A × B = C × D and A × C = B × D, prove that A − D and B − C are parallel. Solution. (1) We have A × B = (−B − C) × B = −B × B − C × B = −C × B = B × C, B × C = (−A − C) × C = −A × C − C × C = −A × C = C × A. (2) It is enough to show that (A − D) × (B − C) = 0. To do this, we have

1.6 Solved Problems

21

(A − D) × (B − C) = A × B − A × C − D × B + D × C = (A × B + D × C) − (A × C + D × B) = (A × B − C × D) − (A × C − B × D) = 0 − 0 = 0. 17. Given two vectors A and B. If A ·B = 0 and A · A = B · B = 1, show that (A × B) · (A × B) = 1. Solution. The fact that A · B = 0 tells us is that the vectors A and B are perpendicular. The fact that A · A = B · B = 1 tells us is that both the vectors A and B are unit vectors. The claim that (A × B) · (A × B) = 1 is equivalent to showing that the cross product is also a vector of magnitude 1. This is equivalent to saying the area of the parallelogram that is spanned by A and B has area 1. Put our information in the context of the parallelogram. The fact that A · B = 0 tells us is that the parallelogram is a rectangle. The fact that A · A = B · B = 1 tells us is that all sides of the parallelogram are of length 1. With all the information put together, we can conclude that the parallelogram that is created by A and B is in fact a square with side length 1, so it must have an area of 1, proving the claim. 18. Given two vectors A and B, do the equations V × A = B and V · A = A determine the vector V uniquely? If so, find an explicit formula of V in terms of A and B. Solution. We may assume that A and B are orthogonal, otherwise there is no solution. Suppose that θ is the angle between V and A. Since V × A = B, it follows that B = A V sin θ. Since V · A = A, it follows that A = A V cos θ. So, we conclude that B2 . V 2 cos2 θ + sin2 θ = 1 + A2 This implies that

V =

1+

B2 . A2

Moreover, we have θ = cos−1 1/V . Therefore, we deduce that V is the vector perpendicular to B with given length such that the angle between A and V is cos−1

A . A2 + B2

19. Use the vectors to show that any angle inscribed in a semicircle is a right angle (see Fig. 1.18). −→ −→ −→ −→ −→ −→ Solution. It is easy to see that A P = O P − O A and B P = O P − O B. Now, we obtain

22

1 Vectors and Analytic Geometry

P

A

O

B

Fig. 1.18 An angle inscribed in a semicircle

P A

B

θ

O

Q A+B

Fig. 1.19 Cosine rule for a triangle ABC

−→ −→ −→ −→ −→ −→ AP · B P = O P − O A · O P − O B −→ −→ −→ −→ −→ −→ −→ −→ = OP · OP − OP · OB − OA · OP + OA · OB −→ −→ −→ −→ −→ −→ −→ −→ = OP · OP + OP · OA − OA · OP − OA · OA −→ −→ = O P 2 − O A2 = 0. −→ Note that O P and O A are the radii of the semicircle. Therefore, the vectors A P and −→ B P are orthogonal, and the result follows. 20. Derive the cosine rule for a triangle ABC, that is, show that (BC)2 = (AB)2 + (AC)2 − 2(AB)(AC) cos θ.

Solution. Suppose that A and B represent two sides of a triangle and θ is the angle between them. Then, the third side is the vector A + B; see Fig. 1.19.

1.6 Solved Problems

23

Now, we have (A + B) · (A + B) = A · A + A · B + B · A + B · B, and so

A + B2 = A2 + B2 + 2 A · B = A2 + B2 + 2A B cos(π − θ).

Since cos(π − θ) = − cos θ, it follows that (BC)2 = (AB)2 + (AC)2 − 2(AB) (AC) cos θ, which is the cosine rule of trigonometry. 21. Let θ denote the angle between the following two vectors in Rn : A = (1, 1, . . . , 1) and B = (1, 2, . . . , n). Find the limiting value of θ as n → ∞. Solution. We have n

A2 = n, B2 =

k 2 and A · B =

k=1

To compute

n

n

k=

k=1

n(n + 1) . 2

k 2 , we can write

k=1

S=

n

k3 =

k=1

n−1 k=0

= S − n3 + 3

This implies that

n k=1

k2 =

(k + 1)3 =

n−1

(k 3 + 3k 2 + 3k + 1)

k=0

(n − 1)n + n. k2 + 3 2 k=1

n−1

(n + 1)(2n + 1)n . So, we have 6

1 n(n + 1) √ 1+ 3 A·B n 2 = , cos θ = =

A B 2 √ 1 1 (n + 1)(2n + 1)n 1+ 1+ n n 2n 6 which implies that

√ lim cos θ =

n→∞

π 3 or θ = . 2 3

24

1 Vectors and Analytic Geometry

22. Find all vectors that have a given length a and make an angle π/3 with the positive x-axis and the angle π/4 with the positive z-axis. Solution. We know that the direction of a vector in space is uniquely determined by three direction angles α, β and γ. These angles satisfy the equation cos2 α + cos2 β + cos2 γ = 1. Since α = π/3 and γ = π/4, it follows that 1 1 1 + cos2 β + = 1 or cos β = ± . 4 2 2 On the other hand, a unit vector with these three direction angles is of the form − → − → − → U = cos α i + cos β j + cos γ k . Since the√required vector must be of length a, we conclude that A = aU = a(1/2, ±1/2, 1/ 2). 23. A line makes angles α, β, γ and δ with the diagonals of a cube. Prove that cos2 α + cos2 β + cos2 γ + cos2 δ =

4 . 3

Solution. Let O ABC E F G D be the cube with each side of length a; see Fig. 1.20. The diagonals of the cube are O F, AE, C G and D B. The direction cosines of the diagonal O F which is the line joining two points O and F are d −0 d −0 d −0 , √ and √ , √ 2 2 2 2 2 2 2 d +d +d d +d +d d + d2 + d2 √ √ √ which are cosines of √ 1/ 3, √1/ 3, √1/ 3. √Similarly, √ the direction √ √ √ AE, C G √ and D B are −1/ 3, 1/ 3, 1/ 3; 1/ 3, −1/ 3, 1/ 3 and 1/ 3, 1/ 3, −1/ 3. Let a, b and c be the direction cosines of the given line which makes angles α, β, γ and δ Fig. 1.20 A cube with each side of length d

z D

E

G

F y O

A x

C

B

1.6 Solved Problems

25

with O F, AE, C G and D B, respectively. Then, we have cos α =

a+b+c , √ 3

cos β =

−a + b + c , √ 3

cos γ =

a−b+c , √ 3

cos δ =

a+b−c . √ 3

Now, since a 2 + b2 + c2 = 1, it follows that cos2 α + cos2 β + cos2 γ + cos2 δ =

4 1 2 4(a + b2 + c2 ) = . 3 3

24. Let A, B and C be three vectors in space. Verify by direct calculation that A × (B × C) = B(A · C) − C(A · B). Solution. Each of the vectors can be written in component form as follows: − → − → − → − → − → − → − → − → − → A = a1 i + a2 j + a3 k , B = b1 i + b2 j + b3 k and C = c1 i + c2 j + c3 k .

− → − → − → We have B × C = (b2 c3 − b3 c2 ) i − (b1 c3 − b3 c1 ) j + (b1 c2 − b2 c1 ) k , and so − → − → − → i j k A × (B × C) = a1 a2 a3 b c −b c b c −b c b c −b c 2 3 3 2 3 1 1 3 1 2 2 1 − → = (a2 b1 c2 − a2 b2 c1 − a3 b3 c1 + a3 b1 c3 ) i

(1.16)

− → −(a1 b1 c2 − a1 b2 c1 − a3 b2 c3 + a3 b3 c2 ) j − → +(a1 b3 c1 − a1 b1 c3 − a2 b2 c3 + a2 b3 c2 ) k . On the other hand, we have B(A · C) − C(A · B) − → − → − → = (b1 i + b2 j + b3 k )(a1 c1 + a2 c2 + a3 c3 ) − → − → − → −(c1 i + c2 j + c3 k )(a1 b1 + a2 b2 + a3 b3 ) − → = (b1 a1 c1 + b1 a2 c2 + b1 a3 c3 − c1 a1 b1 − c1 a2 b2 − c1 a3 b3 ) i − → +(b2 a1 c1 + b2 a2 c2 + b2 a3 c3 − c2 a1 b1 − c2 a2 b2 − c2 a3 b3 ) j − → +(b3 a1 c1 + b3 a2 c2 + b3 a3 c3 − c3 a1 b1 − c3 a1 b1 − c3 a2 b2 − c3 a3 b3 ) k , after canceling some terms, leaving us with the following expression:

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1 Vectors and Analytic Geometry

− → B(A · C) − C(A · B) = (b1 a2 c2 − c1 a2 b2 − c1 a3 b3 + b1 a3 c3 ) i − → +(−c2 a1 b1 + b2 a1 c1 + b2 a3 c3 − c2 a3 b3 ) j − → +(b3 a1 c1 − c3 a1 b1 − c3 a2 b2 + b3 a2 c2 ) k .

(1.17)

A term-by-term comparison of (1.16) and (1.17) shows they are the same, and we have proven the identity. 25. Let M be a parallelepiped of volume V. Find the volumes of all parallelepipeds whose adjacent edges are diagonals of the adjacent faces of M. Solution. Assume that A, B and C are the adjacent sides of M. Then, we have V = A · (B × C). The adjacent faces of M are parallelograms with adjacent sides being the pairs of these vectors. So, we deduce that the diagonals of the faces are D1 = A + B, D2 = A + C and D3 = B + C. Using the properties of product of vectors, the requested volume equals V = D1 · (D2 × D3 ) = D1 · (A × B + A × C + C × B + C × C) = (A + B) · (A × B + A × C + C × B + 0) = A · (A × B) + A · (A × C) + A · (C × B) +B · (A × B) + B · (A × C) + B · (C × B) = 0 + 0 + A · (C × B) + 0 + B · (A × C) + 0 = − A · (B × C) − A · (B × C) = 2A · (B × C) = 2V. 26. Heron’s formula for computing the area S of a triangle whose sides have length √ a, b and c states that S = s(s − a)(s − b)(s − c), where s = (a + b + c)/2. Give a vectorial proof of this formula. Solution. Suppose that the triangle has vertices O, A and B such that A = a, B = b and B − A = c. Then, we have A × B2 = A2 B2 − (A · B)2 and − 2 A · B = A − B2 − A2 − B2 . If we combine the above identities, then we obtain 4S 2 = 4

1 2

2 A × B = A × B2 = a 2 b2 − (A · B)2

1 2 1 = a 2 b2 − − (c2 − a 2 − b2 ) = a 2 b2 − (c2 − a 2 − b2 )2 2 4 1 = 2ab − c2 + a 2 + b2 2ab + c2 − a 2 − b2 . 4 Hence, we can write 16S 2 = (a + b)2 − c2 −(a − b)2 + c2 = (a + b + c)(a + b − c)(c + a − b)(c − a + b).

1.6 Solved Problems

27

This yields that

S=

1 (a + b + c)(a + b − c)(c + a − b)(c − a + b) = s(s − a)(s − b)(s − c), 16

where s = (a + b + c)/2, as desired. 27. Prove that if two vectors are linearly dependent, one of them is a scalar multiple of the other. Solution. To prove, suppose that the vectors U and V are linearly dependent. Since the vector equation c1 U + c2 V = 0 has a solution other than c1 = c2 = 0, this equation can be rewritten as −c −c 2 1 U or U = V. V = c1 c2 The above equalities yield that U is a scalar multiple of V or V is a scalar multiple of U . 28. Find three vectors in R3 which are linearly dependent, and are such that any two of them are linearly independent. Solution. Take A = (0, 0, 1), B = (0, 1, 0) and C = (0, 1, 1). We observe that A + B − C = 0. This means that A, B and C are linearly dependent. On the other hand, we have the following implications: c1 A + c2 B = 0 ⇒ c1 = c2 = 0, c1 A + c2 C = 0 ⇒ c1 = c2 = 0, c1 B + c2 C = 0 ⇒ c1 = c2 = 0. Therefore, any two vectors of the set {A, B, C} are linearly independent. 29. Let A, B and C be three linearly independent vectors. Prove that A+B, B+C and C+A are linearly independent. Solution. We consider the equation c1 (A + B) + c2 (B + C) + c3 (C + A) = 0. Then, we can rewrite the equation as (c1 + c3 )A + (c1 + c2 )B + (c2 + c3 )C = 0. Since A, B and C are linearly independent, it follows that c1 + c3 = 0, c1 + c2 = 0 and c2 + c3 = 0. This system of equations has only a trivial solution, i.e., c1 = c2 = c3 = 0. Therefore, we conclude that the vectors A + B, B + C and C + A are linearly independent. 30. Given two linearly independent vectors A and B in R3 . Suppose that C = (B × A) − B.

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(1) Prove that A is orthogonal to B + C; (2) Prove that the angle θ between B and C satisfies π/2 < θ < π; (3) If B = 1 and B × A = 2, compute the length of C. Solution. (1) We have A · (B + C) = A · B + A · C = A · B + A · (B × A) − B = A · B + A · (B × A) − A · B = A · (B × A) = 0. (2) We have A) − B = B · (B × A) − B · B = −B2 , B · C = B · (B × C2 = C · C = (B × A) − B · (B × A) − B = B × A2 + B2 . Now, we can write cos θ =

−B2 B ·C = 1/2 B C B B × A2 + B2

−B −1 , = 1/2 = 2 2 2 2 2 A sin2 α + 1 B A sin α + B where α is the angle between A and B. Hence, we conclude that −1 < cos θ < 0, which implies that π/2 < θ < π. (3) Since B = 1 and B × A = 2,√ it follows that C2 = B × A2 + B2 = 4 + 1 = 5, which implies that C = 5. 31. Find the equation in symmetric form of the line in which the planes 3x + 2y + z − 4 = 0 and x − 3y + 5z − 7 = 0 intersect. − → − → − → Solution. The normal vectors to the planes are N1 = 3 i + 2 j + k and N2 = − → − → − → i − 3 j + 5 k . So, a direction vector for the line of intersection is − − → − → → j k i − → − → − → N1 × N2 = 3 2 1 = 13 i − 14 j − 11 k . 1 −3 5 Now, we must find a point P0 belonging to both planes. If we take z = 0 in the system of equations 3x + 2y + z = 4 and x − 3y + 5z = 7, then the resulting equations are 3x + 2y = 4 and x − 3y = 7. The solution of the equations is x = 26/11 and y = −17/11. Hence, the point P0 = (26/11, −17/11, 0) lies in both planes, and consequently it lies in the line in which they intersect. Therefore, the equation in symmetric form for the required line is

1.6 Solved Problems

29

y + 17/11 z x − 26/11 = = . 13 −14 −11 32. Let U, V and W be three vectors in R3 such that U · V = 0 and U = V = 1. Let P denote the plane containing U , V and (0, 0, 0). (1) Show that every vector A in the plane P can be expressed as A = c1 U + c2 V , for some c1 , c2 ∈ R. Furthermore, show that c1 = A · U and c2 = A · V ; (2) Suppose that W is not in the plane P. Show that there exists a vector B in P such that W − B is perpendicular to both U and V . Solution. (1) Suppose that X is the projection of A on the line joining U and (0, 0, 0). We can write A = X + (A − X ). Then, we have (A · U ) U = (A · U )U, U 2 (A − X ) · U = A − (A · U )U · U = A · U − (A · U )U 2 = A · U − A · U = 0. X=

So, we take c1 = A · U . Since (A − X ) · U = 0 and V · U = 0, it follows that A − X = c2 V . (2) Note that (W · U )U + (W · V )V lies in the plane P and that W − (W · U )U + (W · V )V is perpendicular to both U and V . 33. Prove that the perpendicular distance between two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by |d1 − d2 | . √ a 2 + b2 + c2 Solution. At least one of the a, b or c is non-zero. Without loss of generality, suppose that a = 0. Then the point P0 = (−d1 /a, 0, 0) lies on the plane ax + by + cz + d1 = 0. So, it is enough to obtain the distance between the point P0 and the plane ax + by + cz + d2 = 0. By using the distance formula, this distance equals |a

−d 1

+ 0 + 0 + d2 | |d1 − d2 | a =√ , √ 2 2 2 a +b +c a 2 + b2 + c2

as desired. 34. Two lines in space are said to be skew if they are non-parallel and nonintersecting, indeed, they do not lie in one plane. Let 1 be the line through

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1 Vectors and Analytic Geometry

Fig. 1.21 Two parallel planes containing two skew lines

P1 A N

P2

U

B

C D

A = (1, 2, 7) and B = (−2, 3, −4), and let 2 be the line through C = (2, −1, 4) and D = (5, 7, −3). (1) Prove that 1 and 2 are skew lines; (2) Find the distance between the skew lines 1 and 2 . Solution. (1) It is enough to show that 1 and 2 do not intersect and are not parallel. Parametric equations of 1 are x = 1 − 3t, y = 2 + t and z = 7 − 11t,

(1.18)

and parametric equations of 2 are x = 2 + 3s, y = −1 + 8s and z = 4 − 7s.

(1.19)

Since the sets of direction numbers are not proportional, it follows that 1 and 2 are not parallel. To examine the lines intersect, we equate the right sides of (1.18) and (1.19) and we get 1 − 3t = 2 + 3s, 2 + t = −1 + 8s and 7 − 11t = 4 − 7s. Solving the first two equations simultaneously, we obtain t = −17/27 and s = 8/27. But these values do not satisfy the last equation. Thus, the two lines do not intersect. Consequently, 1 and 2 are skew lines. (2) Since 1 and 2 are skew lines, it follows that there exists parallel planes P1 and P2 containing the lines 1 and 2 , respectively; see Fig. 1.21. If d is the distance between P1 and P2 , then the distance between 1 and 2 is −→ −→ also d. The vector N = AB × C D is normal to the two planes. If U is a unit normal vector in the direction of N , then

1.6 Solved Problems

31

−→ −→ AB × C D U = −→ −→ . AB × C D −→ −→ Now, the scalar projection of C B on N is C B · U and −→ −→ d = |C B · U | = C B ·

−→ −→ AB × C D −→ −→ . AB × C D

Performing the computations required, we have −→ − → − → − → AB = −3 i + j − 11 k , − → → − −→ −→ i j N = AB × C D = −3 1 3 8

−→ − → − → − → CD = 3 i + 8 j − 7 k , − → k − → − → − → −11 = 27(3 i − 2 j − k ), −7

1 − → − → − → U = √ (3 i − 2 j − k ). 14 √ −→ − → − → − → −→ Thus, C B = −4 i + 4 j − 8 k , and so d = |C B · U | = 12/ 14. 35. Show that the lines 1 and 2 with symmetric equations x − x1 y − y1 z − z1 x − x2 y − y2 z − z2 = = and = = a1 b1 c1 a2 b2 c2 are skew lines if and only if the determinant of the matrix ⎡

x1 − x2 ⎣ a1 a2

⎤ y1 − y2 z 1 − z 2 b1 c1 ⎦ b2 c2

is non-zero. Solution. The lines 1 and 2 are parallel if and only if (a1 , b1 , c1 ) is a multiple scalar of (a2 , b2 , c2 ). This is equivalent to say that the determinant of the above matrix is zero. Now, we write the parametric equations of the lines 1 and 2 as follows: x = x1 + a1 t, y = y1 + b1 t and z = z 1 + c1 t; x = x2 + a2 s, y = y2 + b2 s and z = z 2 + c2 s. The lines 1 and 2 are intersecting if and only if x1 − x2 = a2 s − a1 t, y1 − y2 = b2 s − b1 t and z 1 − z 2 = c2 s − c1 t.

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1 Vectors and Analytic Geometry

In this case, we can write ⎡

⎤ a2 s − a1 t b2 s − b1 t c2 s − c1 t ⎦ b1 c1 a1 det A = ⎣ a2 b2 c2 ⎡ ⎤ ⎡ ⎤ a2 b2 c2 a1 b1 c1 = s ⎣ a1 b1 c1 ⎦ − t ⎣ a1 b1 c1 ⎦ = 0. a2 b2 c2 a2 b2 c2

According to the above argument, we conclude that the lines 1 and 2 are skew if and only if det A = 0. 36. Let A = (0, 2, −1), B = (4, 0, −1) and C = (7, −3, 0) be three points in space. Find a vector N which is orthogonal to the plane passing through A, B and C. Moreover, find the area of the triangle ABC. Solution. First, we determine two vectors on the plane and then compute the cross product of them. We obtain −→ −→ AB = (4, −2, 0) and AC = (7, −5, 1). These vectors lie on the plane. We know that a vector N is orthogonal to the plane if and only if it is orthogonal to any non-parallel vectors on the plane. So, if we consider −→ −→ AB × AC = (−2, −4, −6), then the vector N can be taken to be any scalar multiple of (−2, −4, −6), such as (1, 2, 3). −→ −→ −→ −→ Since AB × AC is equal to the area of the parallelogram formed by AB and AC, it follows that the area of the triangle ABC is half of the area of this parallelogram. That is √ 1 −→ −→ 1 AB × AC = (−2)2 + (−4)2 + (−6)2 = 14. 2 2 37. Let x + y + z = 1 and x − 2y + 3z = 1 be two planes in space. (1) Find the angle between these planes; (2) Find the symmetric equations for the line of intersection of these planes. Solution. (1) The normal vectors of these planes are N1 = (1, 1, 1) and N2 = (1, −2, 3). If θ is the angle between the planes, then cos θ =

2 2 N1 · N2 . =√ or θ = cos−1 √ N1 N2 42 42

(2) First, we determine a point on the line . It is easy to see that the point (1, 0, 0) lies in both planes, and hence it lies on . Since lies in both planes, we conclude that it is perpendicular to both of the normal vectors. Hence, a vector A parallel to is given by the cross product A = N1 × N2 = (−5, −2, −3). This implies that the symmetric equations of are

1.6 Solved Problems

33

y z x −1 = = . 5 −2 −3 38. Find the equation of the plane passing through the points (1, −2, 2) and (−3, 1, 2), and perpendicular to the plane 2x + y − z + 6 = 0. Solution. Since the required plane passes through the point (1, −2, 2), we let the equation of the plane be of the form a(x − 1) + b(y + 2) + c(z − 2) = 0.

(1.20)

Since the plane also passes through the point (−3, 1, −2), we get the equation a(−3 − 1) + b(1 + 2) + c(−2 − 2) = 0, or − 4a + 3b − 4c = 0.

(1.21)

On the other hand, the plane (1.20) is perpendicular to 2x + y − z + 6 = 0. It follows that 2a + b − c = 0. (1.22) From (1.21) and (1.22), we deduce b = −12a and c = −10a. Finally, using these values in (1.20) we obtain a(x − 1) − 12a(y + 2) − 10a(z − 2) = 0 or x − 12y − 10z = 0. 39. Find the equation of the plane perpendicular to the yz-plane, containing the point (2,1,1) and making an angle of radian measure cos−1 (2/3) with the plane 2x − y + 2z − 3 = 0. Solution. Suppose that a(x − 2) + b(y − 1) + c(z − 1) = 0 is the equation of the requested plane. Since the plane is perpendicular to the yz-plane, it follows that a = 0. If N1 = (0, b, c) and N2 = (2, −1, 2), then −b + 2c N1 · N2 −1 2 ⇒ cos cos =√ cos θ = 2 2 N1 N2 3 b + c 22 + (−1)2 + 22 −b + 2c 2 ⇒ = √ ⇒ −b + 2c = 2 b2 + c2 3 3 b2 + c2 −4c ⇒ b = 0 or b = . 3 If b = 0, then c(z − 1) = 0, which implies that z = 1. If b = −4c/3, then −4c(y − 1)/3 + c(z − 1) = 0. This gives that 4y − 3z − 1 = 0. − → − → − → −→ − → − → − → −→ −→ 40. Let O A = a1 i + a2 j + a3 k , O B = b1 i + b2 j + b3 k and OC = − → − → − → c1 i + c2 j + c3 k be the position vectors of points A = (a1 , a2 , a3 ), B = (b1 , b2 , b3 ) and C = (c1 , c2 , c3 ). Find an equation for the plane passing through A, B and C; see Fig. 1.22.

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1 Vectors and Analytic Geometry

Fig. 1.22 A plane passing through three points

z B

P C

A

y O

x

Solution. Suppose that A, B and C do not lie in the same straight line, and hence −→ − → − → − → they determine a plane. Let O P = x i + y j + z k denote the position vectors of any point P = (x, y, z) in the plane. We consider the vectors −→ −→ −→ −→ −→ −→ −→ −→ −→ AB = O B − O A, AC = OC − O A and A P = O P − O A, −→ −→ −→ which all lie in the plane. Then, we have A P · ( AB × AC) = 0, or equivalently we −→ −→ −→ −→ −→ −→ can write ( O P − O A) · ( O B − O A) × ( OC − O A) = 0. This yields that − → − → − → − → − → (x − a1 ) i + (y − a2 ) j + (z − a3 ) k · (b1 − a1 ) i + (b2 − a2 ) j − → − → − → − → +(b3 − a3 ) k × (c1 − a1 ) i + (c2 − a2 ) j + (c3 − a3 ) k = 0. Therefore, using the determinant notation we can write x − a1 b1 − a1 c −a 1 1

y − a2 b2 − a2 c2 − a2

z − a3 b3 − a3 = 0. c −a 3

3

41. Let A and B be two non-parallel vectors. Find the most general vector V that satisfies the conditions A · (V × B) = 0 and B · V = 0. Solution. We have A · (V × B) = V · (B × A) = 0. This yields that V is perpendicular to both A × B and B. The first condition implies that V lies in the plane spanned by A and B, and the other condition requires that V is orthogonal to B. By

1.6 Solved Problems

35

the first condition, we can write V = α A + β B for any real numbers α and β. By considering the second condition, we have V · B = (α A + β B) · B = α(A · B) + βB2 = 0, or β = −α(A · B)/B2 . Therefore, we conclude that V = αA −

α(A · B) B, B2

for any real number α which is the most general vector V satisfying the conditions. 42. Find an equation for the cylinder generated by a line through the curve x 2 + y 2 = − → − → − → 4x, z = 0 moving parallel to the vector i + j + k . Solution. The line passing through a point (x0 , y0 , 0) on the curve and parallel to − → − → − → the vector i + j + k lies on the cylinder. The equation of this line is x − x0 = y − y0 = z. This implies that x0 = x − z and y0 = y − z. Since (x0 , y0 , 0) satisfies the equation (x − 2)2 + y 2 = 4, an equation for the surface is (x − z − 2)2 + (y − z)2 = 4. 43. Find an equation for the surface generated by revolving the curve 4x 2 + 9y 2 = 36, z = 0 around the y-axis. Solution. Suppose that P = (x, y, z) is a point on the surface. Consider the point Q = (x0 , y, 0) on the curve. The distance from Q to the y-axis and the distance from P to the y-axis are the same. Consequently, we obtain x02 = x 2 + z 2 . An equation of the surface is 4(x 2 + z 2 ) + 9y 2 = 36. 44. Find an equation for the surface consisting of all points that are equidistant from the point (1,0,0) and the plane x = 1. Solution. Assume that P = (x0 , y0 , z 0 ) is an arbitrary point on the surface and Q = (1, 0, 0). Since the distance between P and Q is equal to the distance from P to the plane x = 1, it follows that (x0 + 1)2 + y02 + z 02 = (x0 − 1)2 . Thus, an equation of the surface is y 2 + z 2 = −4x. 45. Find the equation of the sphere has its center at (–1,2,3) and touch the plane 2x-y+2z-6=0. Solution. The equation of the sphere is (x + 1)2 + (y − 2)2 + (z − 3)2 = r 2 . We must determine r , the radius of the sphere. The radius of the sphere is equal to the length of the perpendicular line segment from the center (1, 2, 3) to the plane 2x − y + 2z = 6. So, it is |2 · (−1) + (−1) · 2 + 2 · 3| 4 |ax0 + by0 + cz 0 + d| = = . √ 2 2 2 2 2 2 3 a +b +c 2 + (−1) + 2 Therefore, we conclude that the equation of the sphere is (x + 1)2 + (y − 2)2 + (z − 3)2 = 16/9.

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1 Vectors and Analytic Geometry

46. Consider the points P such that the distance from P to the point P0 = (−3, 6, 9) is twice the distance from P to the origin. Show that the set of all such points is a sphere, and find its center and radius. Solution. Suppose that P = (x, y, z) is any point in space. The distance between P and P0 equals P P0 = (x + 3)2 + (y − 6)2 + (z − 9)2 , and the distance between P and the origin O is P O =

x 2 + y2 + z2.

Since the distance between P and P0 is twice the distance from P to the origin, it follows that (x + 3)2 + (y − 6)2 + (z − 9)2 = 2 x 2 + y 2 + z 2 . This yields that 3x 2 − 6x − 9 + 3y 2 + 12y − 36 + 3z 2 + 18z − 81 = 0, or equivalently 3(x 2 − 2x + 1) − 12 + 3(y 2 + 4y + 4) − 48 + 3(z 2 + 6z + 9) − 108 = 0. So, we obtain (x − 1)2 + (y +√2)2 + (z + 3)2 = 56. Therefore, it is a sphere with center (1, −2, −3) and radius 56. 47. Find the equation of the sphere which has the two planes x + y + z-3=0 and x + y + z-9=0 as tangent planes if the center of the sphere is on the planes 2x - y = 0 and 3x -z = 0. Solution. The planes x + y + z − 3 = 0 and x + y + z − 9 = 0 are parallel. Hence, we deduce that the center lies on the plane midway between x + y + z − 3 = 0 and x + y + z − 9 = 0, i.e., the plane x + y + z − 6 = 0. On the other hand, the center lies on the planes 2x − y = 0 and 3x − z = 0. Solving the system of equations x + y + z = 6, 2x − y = 0, 3x − z = 0 gives that x = 1, y = 2 and z = 3. Thus, the center is the point (1, 2, 3). The normal vector to x + y + z − 3 = 0 is N = (1, 1, 1). The points (1, 1, 1) on x + y + z − 3 = 0 and (3, 3, 3) on x + y + z − 9 = 0 differ by a vector, (2, 2, 2), which is a multiple of this normal vector. Therefore, the distance between the planes is

1.6 Solved Problems

37

Fig. 1.23 H is the foot of the perpendicular from C on the plane x + 2y + 2z + 15 = 0

C

H

P

√ √ (2, 2, 2) = 2 3 and the radius of the sphere is 3. Consequently, the equation of the desired sphere is (x − 1)2 + (y − 2)2 + (z − 3)2 = 3. 48. Find the center and radius of the circle in which the sphere x 2 + y 2 + z 2 + 2y + 4z − 11 = 0 is cut by the plane x + 2y + 2z + 15 = 0. Solution. We can write the equation of the sphere as x 2 + (y + 1)2 + (z + 2)2 = 16. So, the center of sphere is C = (0, −1, −2) and its radius is C P = 4. Suppose that H is the foot of the perpendicular from C on the plane x + 2y + 2z + 15 = 0; see Fig. 1.23. Then, we have C H =

0 + 2(−1) + 2(−2) + 15 = 3. √ 12 + 22 + 22

√ √ The radius of the circle is C P2 − C H 2 = 42 − 32 = 7, and the center of −→ the circle is H . Now, C H is perpendicular to the plane x + 2y + 2z + 15 = 0, and so its direction numbers are (1, 2, 2). Moreover, C H passes through C = (0, −1, −2). Therefore, the equation of the line segment C H is x −0 y+1 z+2 = = . 1 2 2 Any point on this line segment is (t, 2t − 1, 2t − 2). If this point is H , then it satisfies the plane x + 2y + 2z + 15 = 0. So, we must have t + 2(2t − 1) + 2(2t − 2) + 15 = 0. This gives that t = −1. Consequently, the coordinates of H are (−1, −3,√−4). Therefore, the center of the circle is the point (−1, −3, −4) and its radius is 7. 49. A plane is tangent to a sphere at a point P0 if it intersects the sphere in this point only. Find the equation of the tangent plane to the sphere

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1 Vectors and Analytic Geometry

x 2 + y 2 + z 2 − 2ax − 2by − 2cz + d = 0 at the point P0 = (x0 , y0 , z 0 ). Solution. It is easy to see that the line joining the point of tangency to the center of the sphere is perpendicular to the tangent plane. We apply this fact to obtain the equation of a tangent plane. Since P0 lies on the sphere, it follows that x02 + y02 + z 02 − 2ax0 − 2by0 − 2cz 0 + d = 0.

(1.23)

On the other hand, the center of the sphere is the point C = (a, b, c). So, the direction numbers of C P0 are (x0 − a, y0 − b, z 0 − c). Since the tangent plane passes through the point P0 = (x0 , y0 , z 0 ), its equation is α(x − x0 ) + β(y − y0 ) + γ(z − z 0 ) = 0,

(1.24)

for some α, β, γ ∈ R. Now, the line segment C P0 is perpendicular to (1.24), and so parallel to the normal to (1.24). Moreover, α, β and γ are direction numbers of the normal to the plane. This yields that (x0 − a, y0 − b, z 0 − c) and (α, β, γ) are proportional. Therefore, (1.24) implies that (x0 − a)(x − x0 ) + (y0 − b)(y − y0 ) + (z 0 − c)(z − z 0 ) = 0. Hence, we have x0 x + y0 y + z 0 z − ax − by − cz = x02 + y02 + z 02 − ax0 − by0 − cz 0 .

(1.25)

Using (1.23) and (1.25), we obtain x0 x + y0 y + z 0 z − ax − by − cz = ax0 + by0 + cz 0 − d. Therefore, the equation of the tangent plane at the point (x0 , y0 , z 0 ) is x0 x + y0 y + z 0 z − a(x + x0 ) − b(y + y0 ) − c(z + z 0 ) + d = 0. 50. Show that the plane 2x-y-2z-16=0 touches the sphere x 2 + y 2 + z 2 − 4x + 2y + 2z − 3 = 0, and determine the point of contact. Solution. We can write the equation of the sphere as (x − 2)2 + (y + 1)2 + (z + 1)2 = 9, i.e., the center of the sphere is (2, −1, −1) and its radius is 3. The length of the perpendicular line segment from the center to the plane 2x − y − 2z − 16 = 0 equals 9 |2 · 2 + 1 + 2 − 16| = = 3. √ 2 2 2 3 2 +1 +2

1.6 Solved Problems

39

Since this length is the same as the radius of the sphere, we conclude that the plane touches the sphere. Now, suppose that P0 = (x0 , y0 , z 0 ) is the point of tangency. Then, according to Problem 1.6, the equation of the tangent plane is x0 x + y0 y + z 0 z − 2(x + x0 ) + (y + y0 ) + (z + z 0 ) − 3 = 0. This yields that (x0 − 2)x + (y0 + 1)y + (z 0 + 1)z − 2x0 + y0 + z 0 − 3 = 0.

(1.26)

Equation (1.26) must be the same as the given plane 2x − y − 2z − 16 = 0. Thus, we must have x0 − 2 y0 + 1 z0 + 1 2x0 − y0 − z 0 + 3 = = = . 2 −1 −2 16 This implies that x0 = −2y0 and z 0 = 1 + 2y0 , and so we obtain y0 + 1 2x0 − y0 − z 0 + 3 −7y0 + 2 = = . −1 16 16 This gives that y0 = −2, and so x0 = 4 and z 0 = −3. Therefore, the point of contact is (4, −2, −3). 51. Cavalieri’s principle. Suppose that two regions in space (solids) are included between two parallel planes. If every plane parallel to these two planes intersects both regions in cross sections of equal area, then the two regions have equal volumes. Use Cavalieri’s principle to find the volume of the solid bounded by the ellipsoid x2 y2 z2 + + = 1. a2 b2 c2 Solution. Suppose that the plane z = k intersects the ellipsoid. To obtain the cross section of the surface with the plane z = k, replace z by k in the equation of the ellipsoid and get x2 y2 k2 + = 1 − . a2 b2 c2 If |k| < c, the the cross section is an ellipse and we have y2 x2 a √c2 − k 2 2 + b√c2 − k 2 2 = 1. c c This implies that the area of this cross section is S=

ab(c2 − k 2 ) π. c2

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1 Vectors and Analytic Geometry

B • O •

• E

A

B• O•

A

D• C

C

Fig. 1.24 Cavalieri’s principle

On the other hand, if we consider an elliptic cylinder where its directrix is the ellipse x 2 /a 2 + y 2 /b2 = 1 in the x y-plane and the ruling parallel to the z-axis bounded −c ≤ z ≤ c; see Fig. 1.24. The cross section of this cylinder with the plane z = k is again an ellipse. If we consider the region between the two ellipses in cross section (see Fig. 1.24) such that E B = ak/c and O B = bk/c, then the area of this region is S = abπ −

abk 2 ab(c2 − k 2 ) π = π. c2 c2

Since S = S , by Cavalieri’s principle the volume of the ellipsoid equals the difference of the volume of the cylinder and the volumes of the two cones in the figure. Hence, we obtain 4 1 V = abπ(2c) − 2 abcπ = abcπ. 3 3 52. Consider two skew lines in space and rotate one of them about the other. Determine the equation of the resulting surface of revolution and show that the surface is a hyperboloid of one sheet. Solution. Without loss of generality, suppose that one of the lines coincides with z-axis and the axis of rotation is z-axis. Let the other line be x = z/a and y = b. If we rotate the points (t, b, at) through the angle θ around the z-axis, then the surface of revolution has the following equations: x = t cos θ − b sin θ, y = t sin θ + b cos θ and z = t. Hence, we obtain x 2 + y 2 = (t cos θ − b sin θ)2 + (t sin θ + b cos θ)2 = t 2 + b2 = This yields that

z2 + b2 . a2

1.6 Solved Problems

41

Fig. 1.25 A barrel

x2 y2 z2 + − = 1, b2 b2 (ab)2 which is the equation of a hyperboloid of one sheet. 53. A barrel shown in Fig. 1.25. It is shaped like an ellipsoid with equal pieces cut from the ends by planes perpendicular to the z-axis. The cross-sections perpendicular to the z-axis are circular. The barrel is 2h units high, its midsection radius is R and its end radii are both r. Find a formula for the barrel’s volume. Then check two things. First, suppose the sides of the barrel are straightened to turn the barrel into a cylinder of radius R and height 2h. Does your formula give the cylinder’s volume? Second, suppose r=0 and h=R so the barrel is a sphere. Does your formula give the sphere’s volume? Solution. The equation of the ellipsoid is x 2 /R 2 + y 2 /R 2 + z 2 /c2 = 1. First, we determine c2 . We observe that the point (0, r, h) belongs to the surface of the barrel. So, we obtain r 2 /R 2 + h 2 /c2 = 1. This implies that c2 =

h2 R2 . R2 − r 2

Now, we evaluate the volume of the barrel by the disk methods. Hence, we can write

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1 Vectors and Analytic Geometry

h z2 z 2 (R 2 − r 2 ) dz V =π y dz = π R 1 − 2 dz = π R2 1 − c h2 R2 −h −h −h h (R 2 − r 2 ) 2 1 R 2 − r 2 3 h 2 =π R2 − dz = π R z z z − −h h2 3 h2 −h 2R 2 h r 2 h 1 . = 2π R 2 h − R 2 − r 2 )h = 2π + 3 3 3

h

h

2

2

If r = R, then V = 2π R 2 h, which is the volume of a cylinder of radius R and height 2h. If r = 0 and h = R, then V = 4π R 3 /3, which is the volume of a sphere. 54. (1) Find the volume of the solid bounded by the hyperboloid x2 y2 z2 + − =1 a2 b2 c2 and the planes z = 0 and z = h, for h > 0. (2) Express your answer in part (1) in terms of h and the areas A0 and Ah of the regions cut by the hyperboloid from the planes z = 0 and z = h. (3) Show that the volume in part (1) is also given by the formula V =

h (A0 + 4 Am + Ah ), 6

where Am is the area of the region cut by the hyperboloid from the plane z = h/2. Solution. (1) Assume that z is fixed. Then, we obtain x2 y2 a 2 (c2 + z 2 ) + b2 (c2 + z 2 ) = 1, c2 c2 a cross-sectional ellipse. The area of this cross-sectional ellipse is A(z) = π

a c

c2 + z 2

b πab c2 + z 2 = 2 (c2 + z 2 ). c c

So, the volume of the solid by the method of slices equals V = 0

h

A(z)dz = 0

h

πab 2 πab 2 1 3 h 2 z c (c + z )dz = z + 0 c2 c2 3

πab 1 πabh = 2 c2 h + h 3 = (3c2 + h). c 3 3c2 (2) We have

1.6 Solved Problems

43

A0 = A(0) = πab and Ah = A(h) =

πab 2 (c + h 2 ). c2

(1.27)

Hence, by part (1) we can write πabh 3+ 3 h 2πab + = 3

V =

h 2 πabh c2 + h 2 2+ = 2 c 3 c2 h πab 2 (c + h 2 ) = 2 A0 + Ah ). 2 c 3

(3) We have Am = A(h/2) =

πab 2 h 2 πab 2 = (4c + h 2 ). c + c2 4 4cr

(1.28)

Now, using (1.27), (1.28), part (1) and a simple calculation, the result follows. 55. (1) Identify the quadric surface given by the equation x 2 − z 2 − 2x + y + 6z = 10. (2) What is the trace of this surface in the yz-plane? (Write down the equation and identify the type of curve). Solution. We rearrange and complete the square. So, we can write (x 2 − 2x + 1) − (z 2 − 6z + 9) + y = 2. This yields that (x − 1)2 + (z − 3)2 + y = 2, or equivalently y − 2 = (z − 3)2 − (x − 1)2 . Consequently, the quadric surface is a hyperbolic paraboloid. (2) The yz-plane is given by x = 0. So, the equations of the trace are x = 0 and x 2 − z 2 − 2x + y + 6z = 10. If we substitute x = 0 into the second equation, we obtain −z 2 + y + 6z = 10, or y = z 2 − 6z + 10. This, we conclude that the curve is a parabola. 56. Identify the surface for each of the following cylindrical equations. (1) r = 3; (2) r = z;

√ (3) r = 5 z; (4) r 2 = z 2 + 1.

Solution. (1) Since r = 3, it follows that x 2 + y 2 = 9 with z arbitrary. Hence, r = 3 is a cylinder with axis of symmetry the z-axis and with radius 3. (2) Since z = r , we get z 2 = x 2 + y 2 . This is a circular cone with vertex at (0, 0, 0) and axis symmetry the z-axis. √ (3) The equation r = 5 z is equivalent to x 2 + y 2 = 25z. This is a paraboloid with axis of symmetry the z-axis. (4) The equation r 2 = z 2 + 1 is equivalent to x 2 + y 2 − z 2 = 1. This is a hyperboloid of one sheet with axis of symmetry the z-axis. 57. Identify the surface for each of the following spherical equations:

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1 Vectors and Analytic Geometry

(1) ρ = 4; (2) ϕ = π/3;

(3) θ = 2π/3; (4) ρ cos ϕ = 2.

Solution. (1) From ρ = 4, we obtain x 2 + y 2 + z 2 = 16. So, the surface is a sphere centered at (0, 0, 0) with radius 4. √ (2) Since tan(π/3) = tan ϕ = x 2 + y 2 /z, it follows that x 2 + y 2 /z = 3. This yields that x 2 + y 2 = 3z 2 . The surface is the top part of the cone with axis of symmetry the z-axis. √ √ (3) Since tan(2π/3) = tan θ = y/x, it follows that y/x = − 3. This implies that 3x + y = 0, which is equivalent to a vertical plane that forms an angle of 2π/3 with the positive x-axis. (4) The equation ρ cos ϕ = 2 is equivalent to z = 2. This is the equation of a plane parallel to x y-plane. 58. Describe the region x 2 + y 2 + z 2 ≤ a and x 2 + y 2 ≥ z 2 , in spherical coordinates. Solution. The region x 2 + y 2 + z 2 ≤ a 2 is the region inside the sphere of radius ρ ≤ a. The region x 2 + y 2 ≥ z 2 is the region outside a cone. The surface equation of the cone is x 2 + y 2 = z 2 , where it is a right angle cone and the region is given by π/4 ≤ ϕ ≤ 3π/4. Therefore, the region is given by ρ ≤ a and

3π π ≤ϕ≤ . 4 4

59. A rotation of axes in two dimensions is a mapping from an xy-Cartesian coordinate system to an x y -Cartesian coordinate system in which the origin is kept fixed and the x - and y -axes are obtained by rotating the x- and y-axes counterclockwise through an angle θ; see Fig. 1.26. Find the relation between these two coordinate systems. Solution. The equations defined in plane, which rotate x y-plane counterclockwise through an angle θ into x y -plane are derived as follows. If P = (x, y) has the polar coordinates (r, α), then x = r cos θ and y = r sin θ.

(1.29)

By considering the x y -plane, the point P = (x , y ) has the polar coordinate (r, α − θ). Using the standard trigonometric formula for difference angles, we can write x = x cos(α − θ) = r cos α cos θ + r sin α sin θ, y = r sin(α − θ) = r sin α cos θ − r cos α sin θ. Substituting (1.29) in (1.30), we obtain

(1.30)

1.6 Solved Problems

45

Fig. 1.26 An x y-Cartesian coordinate system rotated through an angle θ to an x y -Cartesian coordinate system

y

y

P = (x, y) P = (x , y ) • x

θ

x = x cos θ + y sin θ and y = −x sin θ + y cos θ.

x

(1.31)

Equations (1.31) can be represented in matrix form as follows:

or equivalently as

60.

x y

cos θ sin θ = − sin θ cos θ

x , y

x cos θ − sin θ x = . y y sin θ cos θ

Determine the quadric surface 3x 2 + 3y 2 − 2x y = 4.

Solution. The equation does not contain one variable (the z coordinate). The surface is a cylinder. To determine the type of the cylinder, we consider a rotation of coordinate system in the x y-plane and choose the rotation angle so that the coefficient at the mixed term vanishes. According to Problem 1.6, we can write 3x 2 + 3y 2 − 2x y = 3(x cos θ − y sin θ)2 + 3(x sin θ + y cos θ)2 −2(x cos θ − y sin θ)(x sin θ + y cos θ) = (3 − 2 sin θ cos θ)x 2 + (3 + 2 sin θ cos θ)y 2 −2(cos2 θ − sin2 θ)x y . Hence, it is enough taking θ = π/4. Then, we obtain 2x 2 + 4y 2 = 4, or equivalently x 2 2 + y = 1. 2 This yields that the surface is an elliptic cylinder. 61. By a translation and a rotation of the axes, reduce the equation 5x 2 + 6x y + 5y 2 − 4x + 4y − 4 = 0 to the simplest form and identify it.

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1 Vectors and Analytic Geometry

Solution. First, if we use x = x + a and y = y + b, then we can write 5(x + a)2 + 6(x + a)(y + b) + 5(y + b)2 − 4(x + a) + 4(y + b) − 4 = 0, or equivalently 5x 2 + 6x y + 5y 2 + (10a + 6b − 4)x + (10b + 6a + 4)y +5a 2 + 6ab + 5b2 − 4a + 4b − 4 = 0. Now, we put 10a + 6b − 4 = 0 and 10b + 6a + 4 = 0. By solving this system of equations, we get a = 1 and b = −1. So, the equation reduces to 5x + 6x y + 5y = 8. 2

2

(1.32)

Finally, to eliminate the x y term, similar to Problem 1.6, we use a rotation θ = π/4, then we obtain 1 1 1 1 x = √ x − √ y and x = √ x + √ y . 2 2 2 2

(1.33)

Substituting (1.33) in (1.32), we obtain x − y 2 x + y 2 x − y x + y 5 +5 +6 = 8. √ √ √ √ 2 2 2 2 Simplifying this equation gives x + 2

y 2 = 1. 4

This is an elliptic cylinder. 62. Determine the quadric surface 3y 2 + 3z 2 − 2yz = 1. Solution. We use rotation in the approximate coordinate plane to reduce the equation to one of the standard forms. Assume that y = y cos θ − z sin θ and z = z cos θ + y sin θ. Then, we obtain 3y 2 + 3z 2 − 2yz = 3(y cos θ − z sin θ)2 + 3(y sin θ + z cos θ)2 −2(y cos θ − z sin θ)(y sin θ + z cos θ) = (3 − 2 sin θ cos θ)y 2 + (3 + 2 sin θ cos θ)z 2 −2(cos2 θ − sin2 θ)y z .

1.6 Solved Problems

47

Similar to Problem 1.6, we are interested in finding an angle θ such that the mixed term is annihilated. So, we obtain θ = π/4. This yields that 2y 2 + 4z 2 = 1. This is an elliptic cylinder centered at the origin with axis parallel to the x-axis. 63. Determine the quadric surface x y − z 2 = 0. Solution. We use rotation in x y-plane. Suppose that x = x cos θ − y sin θ and y = y cos θ + x sin θ. Then, we can write x y = (x cos θ − y sin θ)(y cos θ + x sin θ) = x y cos2 θ + x 2 sin θ cos θ − y 2 sin θ cos θ − x y sin2 θ = x y (cos2 θ − sin2 θ) + (x 2 − y 2 ) sin θ cos θ. Now, if θ = π/4, then x y = x 2 /2 − y 2 /2. Substituting this into the given equation gives x 2 /2 − y 2 /2 − z 2 = 0, or equivalently x 2 = y 2 + 2z 2 . This is an elliptic cone parallel to x-axis. 64. Prove that the distance between two points P1 = (r1 , θ1 ) and P2 = (r2 , θ2 ) in the polar coordinates is

r12 + r22 − 2r1r2 cos(θ2 − θ1 ).

Solution Suppose that the Cartesian coordinates of the points are P1 = (x1 , y1 ) and P2 = (x2 , y2 ). Then, we have x1 = r1 cos θ1 , y1 = r1 sin θ1 , x2 = r2 cos θ2 and y2 = r2 sin θ2 . Therefore, the distance between them is P2 P1 = =

(x2 − x1 )2 + (y2 − y1 )2 (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2

r12 + r22 − 2r1r2 (cos θ1 cos θ2 + sin θ1 sin θ2 ) = r12 + r22 − 2r1r2 cos(θ2 − θ1 ).

=

65. Show that the polar equation of the circle of radius a centered at (r1 , θ1 ) is r 2 − 2r1r cos(θ − θ1 ) + r12 = a 2 . Solution. Let the Cartesian coordinates of the center of the circle be (x1 , y1 ). Then, the Cartesian equation of the circle is (x − x1 )2 + (y − y1 )2 = a 2 . Substituting x = r cos θ, y = r sin θ, x1 = r1 cos θ1 and y1 = r1 sin θ1 , we obtain a 2 = (r cos θ − r1 cos θ1 )2 + (r sin θ − r1 sin θ1 )2 = r 2 + r12 − 2r1r (cos θ1 cos θ + sin θ sin θ1 ) = r 2 + r12 − 2r1r cos(θ − θ1 ).

48

1 Vectors and Analytic Geometry

Fig. 1.27 Polar equation of L is r cos(θ − α) = d

L • P = (r, θ) A d θ α

Polar axis

O

66. Let L be a straight line which does not go through the pole O. Show that the polar equation of L is r cos(θ − α) = d, (1.34) where d is the length and α the inclination of the perpendicular drawn from O to L (see Fig. 1.27).What are the Cartesian coordinates of Eq. (1.34). Show that L has slope m = − cot α and y -intercept d/ sin α. Solution. By looking at Fig. 1.27, since O A P is a right triangle, we can write cos(θ − α) = d/r . This yields that r cos(θ − α) = d. Now, using the identity cos(θ − α) = cos θ cos α + sin θ sin α, we obtain r cos θ cos α + sin θ sin α = d. Substituting x = r cos θ and y = r sin θ, we get x cos α + y sin α = d. Dividing by cos α we have x + y tan α = d/ cos α. Simplifying this equation gives d . y = (−cotα)x + sin α This equation of the line implies that L has slope m = − cot α and y-intercept d/ sin α. 67. Find an equation in polar coordinates of the line L with the given description. (1) The point on L closest to the origin has the Cartesian coordinates (−2, 2); (2) L has slope 3 and is tangent to the unit circle in the fourth quadrant. Solution. (1) First, we convert the Cartesian coordinates (−2, 2) to polar coordinates (d, α). Since this point lies in the second quadrant, it follows that π/2 < α < π. Thus, we obtain √ √ (−2)2 + 22 = 8 = 2 2, 2 π 3π α = tan−1 = tan−1 (−1) = π − = , −2 4 4 d=

1.6 Solved Problems

49

Fig. 1.28 Tangent line to the unit circle in the fourth quadrant

φ O

A P0

√ √ and so we have (d, α) = (2 2, 3π/4). Substituting d = 2 2 and α = 3π/4 in the equation r = d sec(θ − α) yields √ 3π . r = 2 2 sec θ − 4 (2) In polar coordinates, we denote the point of tangency by P0 = (1, α). Since L is the tangent line to the circle at P0 , it follows that P0 is the point on L closest to the center of the circle at the origin. Thus, the polar equation of L is r = sec(θ − α). We need to find α. Suppose that φ is the angle shown in Fig. 1.28. By assumption, we know that tan φ = 3. Since the point of tangency lies in the fourth quadrant, it follows that φ is an acute angle. So, we have tan φ = 3 and 0 < φ < π/2. This implies that φ = 1.25 radian. Since 3π/2 < α < 2π, for the triangle O A P0 we can write (2π − α) + π/2 + 1.25 = π. This implies that α = 5.96 radian. Substituting it into r = sec(θ − α), we get r = sec(θ − 5.96) as the polar equation of the tangent line. 68. Prove that the conic with eccentricity e and focus to direct distance d has equation r = ed/(1 − e cos θ) in polar coordinates if the pole O is at the focus and the polar axis is perpendicular to the directrix L in the direction pointing away from L. Solution. In the case of an ellipse or a hyperbola, there exist two directrices and L is the one nearer the focus. The geometry of the situation is shown in Fig. 1.29, where the directrix lies to the left of the focus. Now, we have r O P = = e. P Q P Q

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1 Vectors and Analytic Geometry

Fig. 1.29 The geometry of polar equation of a conic

L

• P = (r, θ)

Q r

d

θ • O

A

Fig. 1.30 The graph of r = 2 + 2 sin θ

π 2 π 4

3π 4

π

0

1

2

5π 4

3

40

7π 4 3π 2

Since P Q = AO + r cos θ = d + r cos θ, it follows that r = e(d + r cos θ). This gives that r (1 − cosθ) = ed, as desired. 69. The graph of an equation of the form r = a ± b cos θ or r = a ± b sin θ is a limaçon. If b > a, the limaçon has a loop. If a ≥ b, the limaçon is a cardioid which is heart shaped. Draw a sketch of the graph of r = 2 + 2 sin θ. Solution. If (r, θ) is replaced by (r, π − θ), then an equivalent equation is obtained. So, we conclude that the graph is symmetric with respect to the π/2 axis. The following table gives the coordinates of some of the points on the graph: θ 0 r 2

π/6 3

π/3 √ 2+ 3

π/2 4

π 2

7π/6 1

4π/3 √ 2− 3

3π/2 0

Using the symmetry and the above table (note that r ≥ 0, for all θ), a sketch of the graph is shown in Fig. 1.30.

1.6 Solved Problems

51

Fig. 1.31 The graph of r = 1 − 3 cos θ

π 2 π 4

3π 4

π

0

1

2

3

40

7π 4

5π 4 3π 2

70. Draw a sketch of the graph of limaçon r = 1 − 3 cos θ. Solution. Replacing (r, θ) by (r, −θ), we obtain an equivalent equation. This means that the graph is symmetric with respect to the polar axis. The following table gives the coordinates of some points on the graph. θ 0 π/6 √ r −2 1 − 3 3/2

π/3 −1/2

π/2 1

2π/3 5/2

5π/6 √ 1 + 3 3/2

π 4

From these points, we draw half of the graph and the remainder is drawn from its symmetry with respect to the polar axis. A sketch of the graph is shown in Fig. 1.31. 71. The graph of an equation of the form r = a cos nθ or r = a sin nθ is a rose,having n leaves if n is odd and 2n leaves if n is even. Draw a sketch of the four-leafed rose r = 4 sin 2θ. Solution. We test for the symmetry of the graph with respect to the polar axis, the π/2 axis and the pole. We replace (r, θ) by (−r, π − θ), and obtain −r = 4 sin 2(π − θ), which is equivalent to the equation r = 4 sin 2θ. So, the graph is symmetric with respect to the polar axis. Now, if we replace (r, θ) by (−r, −θ), then we obtain −r = 4 sin(−2θ), which is equivalent to the equation r = 4 sin 2θ. Thus, the graph is symmetric with respect to the π/2 axis. Also, we see that the graph is symmetric with respect to the pole. Substituting 0 for r in the given equation, we have sin 2θ = 0 from which we get, for 0 ≤ θ < 2π, θ = 0, θ = π/2, θ = π and θ = 3π/2. The following table gives values of r for some values of θ from 0 to π/2: θ 0 r 0

π/12 2

π/6 √ 2 3

π/4 4

π/3 √ 2 3

5π/12 2

π/2 0

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1 Vectors and Analytic Geometry

Fig. 1.32 The graph of r = 4 sin 2θ

π 2 π 4

3π 4

π

0

1

2

5π 4

3

40

7π 4 3π 2

Fig. 1.33 The graph of r =θ

From these values and symmetry properties, we draw a sketch of the graph, as shown in Fig. 1.32. 72. Draw a sketch of the graph r = θ. Solution. If θ = nπ, where n is any integer, then the graph intersects the polar axis. If θ = nπ/2, where n is any odd integer, then the graph intersects the π/2 axis. If r = 0, then θ = 0, and so the tangent line to the graph at the pole is the polar axis. A sketch of the graph is shown in Fig. 1.33. The graph is called a spiral of Archimedes. 73. Draw a sketch of the graph of polar equation r θ = a, where a > 0. Solution. We observe that

1.6 Solved Problems

53

θ 0. 40. Find an equation in spherical coordinates of the graph of each of the equations: (a) x 2 + y 2 + 4z 2 = 4,

(b) 4x 2 + 25y 2 = 100.

41. Find a polar equation of the circle having its center at (r0 , θ0 ) and a radius a units. 42. Find a set of cylindrical coordinates for the point having spherical coordinates (5, π, π/4).

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1 Vectors and Analytic Geometry

Harder Exercises 43. Prove that the median of a triangle intersects in a common point which trisects each median. 44. Prove analytically that the four diagonals of a cube have the same length. 45. Let B = {V1 , V2 , V3 } be a set of three vectors in R3 . Prove that B is linearly independent if and only if B spans R3 . −→ −−→ −→ 46. Let P, Q and R be three non-collinear points in R3 and let O P, O Q and O R be the position representations of vectors A, B and C, respectively. Prove that the representations of the vector A × B + B × C + C × A are perpendicular to the plane containing the points P, Q and R. 47. Find an equation of the largest sphere that passes through the point (−1, 1, 4) and is such that each of the points (x, y, z) inside the sphere satisfies the condition x 2 + y 2 + z 2 ≤ 136 + 2(x + 2y + 3z). 48. Prove that an ellipse and a hyperbola with the same foci always intersect at right angles. 49. Find an equation of the plane containing the endpoints of the position represen− → − → − → − → − → − → − → − → tations of the vectors −2 i + j + 5 k , i − j + 2 k and −3 i + 5 j + − → 3k. 50. Find an equation for the surface consisting of all points P for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface. 51. Sketch the region bounded by the surfaces z = x 2 + y 2 and x 2 + y 2 = 1 for 1 ≤ z ≤ 2. 52. Let V and W be two non-parallel, non-zero vectors in R3 . If Z is any vector in R3 , verify that Z = aV + bW + c(V × W ) for some real numbers a, b and C. 53. Let V and W be two non-parallel, non-zero vectors in R3 . If Z is a non-zero vector in R3 that is perpendicular to V and W , show that Z is parallel to V × W . 54. Show that if A, B and C are vectors in R3 , then 2 (A × B) · (B × C) × (C × A) = A · (B × C) . 55. Identify and sketch the quadratic surface x 2 + 2z 2 − 6x − y + 10 = 0. 56. Find an equation for the surface obtained by rotating the parabola y = x 2 about the y-axis. 57. Sketch and describe the cylindrical surface of the given equation. (a) z = ln y,

(b) z = cos(π/2 + x).

58. Identify the surface z = x y by making a suitable rotation of axes in the x y-plane. 59. Suppose a rotation converts ax 2 + bx y + cy 2 into a x 2 + b x y + c y 2 . Prove that

1.7 Exercises

63

(a) a + c = a + c ,

(b) 4ac − b2 = 4a c − b 2 .

60. Show that using a rotation and then a translation, we can always reduce any quadratic equation to standard form. 61. Show that the spirals r = θ and r = 1/θ are perpendicular when they meet at θ = 1. 62. The tractrix t t x = t − a tanh and y = a sech a a from x = −a to x = 2a is revolved about the x-axis. Draw a sketch of revolution. 63. Prove that at the points of intersection of the two curves r = a sec2 (θ/2) and r = b csc2 (θ/2), their tangent lines are perpendicular. 64. Find the length of the polar curve θ=

1 1 r+ (1 ≤ r ≤ 3). 2 r

65. Show that the hyperbolic spiral r θ = 1 (1 ≤ θ < ∞) is not of finite length. 66. Use polar coordinates to show that the area enclosed by the loop of the Folium of Descartes x=

3at 3at 2 and y = (−∞ < t < ∞) t3 + 1 t3 + 1

is equal to (3/2)a 2 . 67. Draw three circles of radius 1 that touch each other and find the area of the curved triangle between them. 68. Consider the torus of equation (x 2 + y 2 + z 2 + R 2 − r 2 ) = 4R 2 (x 2 + y 2 ), where R > r > 0. (a) Write the equation of the torus in spherical coordinates. (b) If R = r , the surface is called a horn torus. Show that the equation of a horn torus in spherical coordinates is ρ = 2R sin ϕ. 69. A surface consists of all points P such that the distance from P to the plane y = 1 is twice the distance from P to the point (0, −1, 0). Find an equation for this surface and identify it. 70. Graph the surfaces z = x 2 + y 2 and z = 1 − y 2 on a common screen using the domain |x| ≤ 1.2, |y| ≤ 1.2 and observe the curve of intersection of these surfaces. Show that the projection of this curve onto the x y-plane is an ellipse. 71. Find the points in which the line y−3 z+2 x −4 = = −6 3 4

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1 Vectors and Analytic Geometry

1 2 1 2 1 2 x + y + z = 1. intersects the ellipsoid 36 81 9 72. Show that the elliptic cone y2 z2 x2 + 2 + 2 =0 2 a b c is asymptotic to both hyperboloids x2 y2 z2 x2 y2 z2 + − = 1 and + − = −1 a2 b2 c2 a2 b2 c2 in the sense that both hyperboloids get arbitrarily close to the cone as z → ±∞.

Chapter 2

Vector Functions and Parametric Equations

2.1 Definition and Calculus of Vector Functions When a particle moves through space during a time interval I , we think of the particle’s coordinates as functions defined on I: x = f (t), y = g(t) and z = h(t), for t ∈ I.

(2.1)

The points (x, y, z) = f (t), g(t), h(t) make up the curve in space that we call the particle’s path. The equations and interval in Eq. (2.1) parameterize the curve. A curve in space can also be represented in vector form. The vector −→ − → − → − → R(t) = O P = f (t) i + g(t) j + h(t) k

(2.2)

from the origin to the particle’s position P = f (t), g(t), h(t) at time t is the particle’s position vector; see Fig. 2.1. The functions f , g and h are the component functions (components) of the position vector. Equation (2.2) defines R as a vector function of the real variable t on the interval I . More generally, a vector function or vector-valued function on a domain set D is a rule that assigns a vector in space to each element in D. The domain of R is the set of values of t for which f (t), g(t) and h(t) are defined. Note that we can reduce the definition to x y-plane (R : R → R2 ) or generalize the definition to n-spaces (R : R → Rn ). − → Let R be a vector function whose function values are given by R(t) = f (t) i + − → − → g(t) j + h(t) k . Then the limit of R(t) as t approaches t0 is defined by lim R(t) =

t→t0

lim f (t)

t→t0

− → − → − → i + lim g(t) j + lim h(t) k , t→t0

t→t0

if lim f (t), lim g(t) and lim h(t) exist. t→t0

t→t0

t→t0

A vector function R is continuous at a point t = t0 in its domain, if

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_2

65

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2 Vector Functions and Parametric Equations

Fig. 2.1 The position vector R(t) of a particle moving through space is a function of time

z

• R(t)

P = f (t), g(t), h(t)

y

O

x

(1) R(t0 ) exists; (2) lim R(t) exists; t→t0

(3) lim R(t) = R(t0 ). t→t0

From the definitions, we deduce that the vector function R is continuous at t0 if and only if f , g and h are continuous there. The function R is continuous if it is continuous at every point in its domain. If R is a vector function, then the derivative of R is a vector function denoted by R and defined by R(t + Δt) − R(t) , R (t) = lim Δt→0 Δt if this limit exists. If R (t) exists, then R is differentiable at t. The notations Dt R(t) and d R/dt are sometimes used in place of R (t). A vector function R is differentiable if it is differentiable at every point of its domain. − → − → − → If R is a vector function, then R (t) = f (t) i + g (t) j + h (t) k , if f (t), g (t) and h (t) exist. Higher order derivatives of vector functions are defined as for higher order derivatives of real-valued functions. So, for the second derivative we have R (t) = Dt R (t) . The notation Dt2 R(t) can be used instead of R (t), and we can write R (t) = − → − → − → f (t) i + g (t) j + h (t) k , if f (t), g (t) and h (t) exist. Properties of the derivative: Let R and Q be differentiable vector functions of t, C a constant vector, c any real number and f any differentiable real function. Then we have (1) Dt (C) = 0; (2) Dt c R(t) = c R (t);

2.2 Velocity and Acceleration in Space

67

(3) Dt R(t) ± Q(t) = R (t) ± Q (t); (4) Dt f (t)R(t) = f (t)R (t) + f (t)R(t); (5) Dt R(t) · Q(t) = R(t) · Q (t) + R (t) · Q(t); (6) Dt R(t) × Q(t) = R(t) × Q (t) + R (t) × Q(t); (7) Dt R f (t) = f (t)R f (t) (Chain rule). Now, we define an indefinite integral of a vector function. If R is a vector function − → − → − → given by R(t) = f (t) i + g(t) j + h(t) k , then the indefinite integral of R(t) is defined by

R(t)dt =

f (t)dt

− → i +

g(t)dt

− → j +

h(t)dt

− → k.

By an antiderivative of a vector function R, defined on an interval I , we mean any other vector function Q(t) defined on I such that R (t) = Q(t). If F(t), G(t) and H (t) are antiderivatives of f (t), g(t) and h(t), respectively, then clearly Q(t) = − → − → − → F(t) i + G(t) j + H (t) k is an antiderivative of R(t). The analogue for vector functions of the fundamental theorem of calculus is of course b R(t)dt = Q(b) − Q(a). a

2.2 Velocity and Acceleration in Space The position of a particle is specified by the position vector − → − → − → R(t) = x(t) i + y(t) j + z(t) k . If during the time Δt the position vector of the particle changes from R1 = (x1 , y1 , z 1 ) to R2 = (x2 , y2 , z 2 ), then the displacement ΔR for that time is − → − → − → ΔR = R2 − R1 = (x2 − x1 ) i + (y2 − y1 ) j + (z 2 − z 1 ) k . If the particle moves through a displacement ΔR in a time Δt, then its average velocity for this time is V (t) =

Δx − ΔR(t) → Δy − → Δz − → = i + j + k. Δt Δt Δt Δt

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2 Vector Functions and Parametric Equations

A more useful quantity is the (instantaneous) velocity V (t), which is the limit of the average velocity when Δt → 0. So, it is the derivative of the position vector R, i.e., V (t) =

d R(t) d − → − → − → − → − → − → x(t) i + y(t) j + z(t) k = x (t) i + y (t) j + z (t) k . = dt dt

If the particle’s velocity changes by ΔV in a time Δt, the average acceleration is A(t) =

ΔV (t) , Δt

but a more interesting quantity is the result of Δt → 0, which gives us (instantaneous) acceleration A(t). It is the derivative of the velocity vector V (t), i.e., A(t) =

d V (t) − → − → − → = x (t) i + y (t) j + z (t) k . dt

The speed of the particle at time t is the magnitude of the velocity vector, that is, V (t). If V (t) is constant, then A(t) is orthogonal to V (t); see Fig. 2.2.

Fig. 2.2 At each point, A(t) and V (t) are perpendicular when V (t) is constant

2.3 Unit Tangent Vector, Unit Normal Vector and Curvature

69

2.3 Unit Tangent Vector, Unit Normal Vector and Curvature A unit vector is a vector with magnitude of 1, examples of which are the three unit − →− → − → vectors i j and k . At each point on a curve in space (or plane), we can associate the unit tangent vector and unit normal vector. If R(t) is the position vector of curve C at a point P on C, then the unit tangent vector of C at P denoted by T (t) is the unit vector in the direction of Dt R(t). Thus, we have T (t) =

Dt R(t) . Dt R(t)

Since T (t) = 1, it follows that T (t) · Dt T (t) = 0. This yields that Dt T (t) is orthogonal to T (t). If T (t) is the unit tangent vector of curve C at a point P on C, then the unit normal vector, denoted by N (t), is the unit vector in the direction of Dt T (t), that is, N (t) =

Dt T (t) . Dt T (t)

Since T (t) and N (t) are orthogonal, it follows that the angle between these two vectors is π/2, and so T (t) × N (t) = 1. This means that the cross product of T (t) and N (t) is a unit vector. Hence, the vector B(t) = T (t) × N (t) is a unit vector orthogonal to T (t) and N (t) and is called the unit binormal vector to the curve C at P. The three mutually orthogonal unit vectors T (t), N (t) and B(t) are called moving trihedral of C. − → − → − → If a curve C is defined by R(t) = f (t) i + g(t) j + h(t) k , where f , g and h are continuous and are not simultaneously zero and C traversed exactly once as t increases from t0 to t, then the length of C is the definite integral s(t) =

t

f (u)

2

2 2 + g (u) + h (u) du.

t0

It is clear that ds/dt = R (t). If the arc length s is chosen as the parameter, the position vector R = R(s), its final point P = P(s), the unit tangent vector T = T (s) = d R/ds and the unit normal vector N = N (s) =

dT /ds dT /ds

(dT /ds = 0)

(2.3)

are all functions of s, as indicated by the notation. It follows from (2.3) that dT /ds = dT /dsN .

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2 Vector Functions and Parametric Equations

The curvature vector of C at P, denoted by K (t), is defined by K (t) = Ds T (t) =

T (t) . R (t)

If C is the graph of R = R(t), then dT /ds is called the curvature of C at P, denoted by κ, and so, we can write dT /ds = κN . By the chain rule, we have

dT dT dt dT /dt

=

κ=

ds dt ds = ds/dt . Let B = T × N . The torsion function of a smooth curve is τ =−

dB · N. ds

If A is acceleration of a particle, then we can write A = aT T + a N N , where aT =

ds 2 d 2s d = = κ = κV . V and a N dt 2 dt dt

The aT and a N are called tangential and normal components of acceleration.

2.4 Radius of Curvature Suppose that κ(t) is the curvature of a curve C at a point P. Then by the radius of curvature of C at P, we mean the number ρ(t) =

1 , κ(t)

that is, the reciprocal of the curvature. If κ(t) is small, then ρ(t) is big. This yields that a very straight curve has a very big radius of curvature, and a straight line (κ(t) = 0) can be regarded as having an infinite radius of curvature. Let P be a point on C (in the plane). By the osculating circle of C at P, we mean the circle of radius ρ(t) = 1/κ(t) passing through P whose center Q c lies on the concave side of C along the normal to C at P; see Fig. 2.3. The center of curvature at P is the center of the osculating circle at P. The center of curvature at P, Q c (t) defined by Q(t) = R(t) + ρ(t)N ((t).

2.5 Solved Problems

71

C

ρ(t) P

Fig. 2.3 Osculating circle

2.5 Solved Problems 81. Find the domain of the given vector functions. √ 1 ; t + 4, t 2 + 1, (1) R(t) = t +5 √ (2) R(t) = t + 1, ln(4 − t 2 ), et . Solution. (1) The domain of R is {t ∈ R | t ≥ −4 and t = −5} = (−4, ∞). (2) The domain of R is {t ∈ R | t ≥ −1 and 4 − t 2 > 0} = {t ∈ R | t ≥ −1 and − 2 < t < 2} = [−1, 2). 82.

For the following vector functions, find (a) the domain of R; (b) lim R(t) and t→1

(c) Dt R(t).

√ 1 − t − 1− → → i + j; t +1 t −1 − → − → (2) R(t) = |t − 1| i + ln t j . (1) R(t) =

√ Solution. (1) Suppose that f (t) = 1/(t + 1) and g(t) = ( t − 1)/(t − 1). Then, we have D f = {t ∈ R | t = −1}, Dg = {t ∈ R+ | t = 1} and so D R = D f ∩ Dg = R+ − {1}. Since √ t −1 1 1 lim f (t) = and lim g(t) = lim √ = , √ t→1 t→1 t→1 2 2 ( t − 1)( t + 1) − → − → it follows that lim R(t) = (1/2) i + (1/2) j . Now, we compute the derivative of t→1

R(t). We obtain

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2 Vector Functions and Parametric Equations

t −1 √ √ − t +1 −1 − → − → − → − → 2 t i + j Dt R(t) = R (t) = f (t) i + g (t) j = (t + 1)2 (t − 1)2 √ −1 − → −t − 1 + 2 t − → i + j. = √ (t + 1)2 2 t(t − 1)2 (2) Assume that f (t) = |t − 1| and g(t) = ln t. Then, we see that D f = R, Dg = (0, ∞) and so D R = D f ∩ Dg = (0, ∞). Since lim f (t) = 0 and lim g(t) = 0, it t→1 t→1 − → follows that lim R(t) = (0, 0) = 0 . Finally, we evaluate the derivative of R(t). We t→1 get t −1 − → 1− → − → − → i + j. Dt R(t) = R (t) = f (t) i + g (t) j = |t − 1| t

83.

Prove that lim R(t) = A if and only if lim R(t) − A = 0. t→t0

t→t0

Solution. Suppose that R(t) = f 1 (t), f 2 (t), f 3 (t) and lim R(t) =

t→t0

lim f 1 (t), lim f 2 (t), lim f 3 (t) = (a1 , a2 , a3 ) = A.

t→t0

t→t0

t→t0

For each i = 1, 2, 3, we have lim f i (t) = a. So, for each > 0, there exists δi > 0 t→t0

such that | f i (t) − ai | < whenever |t − t0 | < δi . Assume that δ = min{δ1 , δ2 , δ3 }. Then, we have √ √ R(t) − A = | f 1 (t) − a1 |2 + | f 2 (t) − a2 |2 + | f 3 (t) − a3 |2 < 32 = 3, whenever |t − t0 | < δ. This shows that lim R(t) − A = 0. t→t0

Conversely, suppose that lim R(t) − A = 0. Then, for every > 0, there exists δ > 0 such that R(t) − A =

t→t0

| f 1 (t) − a1 |2 + | f 2 (t) − a2 |2 + | f 3 (t) − a3 |2 < ,

whenever |t − t0 | < δ. Since | f i (t) − ai | ≤ R(t) − A, for i = 1, 2, 3, it follows that | f i (t) − ai | < whenever |t − t0 | < δ. This completes the proof. 84. Find a vector function that represents the curve of intersection of the cylinder x 2 + y 2 = 4 and the surface z = x y. Solution. We parameterize the cylinder x 2 + y 2 by (3 cos t, 3 sin t, z). Since x = 3 cos t and y = 3 sin t, it follows that z = x y = 9 sin t cos t. Hence, the intersection

2.5 Solved Problems

73

of the cylinder x 2 + y 2 = 9 and the surface z = x y can be represented by R(t) = (3 cos t, 3 sin t, 9 cos t sin t). 85. Find a vector function that represents the curve of intersection of the cone z = x 2 + y 2 and the plane z = x + 2. Solution. First, we find the solution of equations z = x 2 + y 2 and z = x + 2. If x 2 + y 2 = x + 2, then x 2 + y 2 = x 2 + 4x + 4. This implies that x = y 2 /4 − 1. Now, we can determine the parametric equation for the curve C of intersection by choosing y = t. Then, we obtain x=

t2 t2 − 1 and z = x + 2 = + 1. 4 4

Therefore, a vector function representation is R(t) =

t2 4

− 1, t,

t2 +1 . 4

86. Find the angle between the curves R1 = (t, 1 − t, 3 + t 2 ) and R2 (t) = (3 − t, t − 2, t 2 ) where they meet. Solution. First, we determine the point of intersection. We need to find the solution of the simultaneous equations t = 3 − u, 1 − t = u − 2 and 3 + t 2 = u 2 . It is easy to see that t = 1 and u = 2 satisfy all three equalities. This implies that two curves meet at (1, 0, 4), the first when t = 1 and the second when t = 2. The derivatives are R1 (t) = (1, −1, 2t) and R2 (t) = (−1, 1, 2t), and so at the intersection point we have R1 (1) = (1, −1, 2) and R2 (2) = (−1, 1, 4). The angle between the two curves at the intersection point is the angle θ between their tangent vectors. Hence, we have cos θ =

87.

1 1 −1 − 1 + 8 = √ or θ = cos−1 √ . √ √ 6 18 3 3

Let R be a vector function given by R(t) =

2t − → 1 − t2 − → − → i + j + k. 1 + t2 1 + t2

Prove that the angle between R(t) and R (t) is constant, that is, independent of t. Solution. First, we evaluate R (t). We get −4t − 2 − 2t 2 − → → − → R (t) = 2 i + 2 j + 0 k . 2 2 1+t 1+t

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2 Vector Functions and Parametric Equations

Then, we have R(t) · R (t) =

2 − 2t 2 −4t 2t 1 − t2 4t − 4t 3 − 4t + 4t 3 + = = 0. 3 1 + t2 1 + t2 2 1 + t2 1 + t2 2 1 + t2

Since R(t) and R (t) are not zero vectors, for all t ∈ R, it follows that cos θ = 0 or θ = π/2, where θ is the angle between R(t) and R (t). 88. Prove that the vector function R(t) = et cos 4t, et sin 4t has the property that the angle θ between position vector and the tangent vector is constant. Solution. We have R (t) = (−4et sin 4t + et cos 4t, 4et cos 4t + et sin 4t). Then, we obtain R(t) · R (t) = −4e2t sin 4t cos 4t + e2t cos2 4t + 4e2t sin 4t cos 4t + e2t sin2 4t = e2t (cos2 4t + sin2 4t) = e2t , R(t) = et and (−4et sin 4t + et cos 4t)2 + (4et cos 4t + et sin 4t)2 = 16e2t (sin2 4t + cos2 4t) + e2t (sin2 4t + cos2 4t) √ √ = 16e2t + e2t = 17et .

R (t) =

Now, using the formula R(t) · R (t) = R(t) R (t) cos θ, we get e2t = cos θ. This yields that

√

17e2t

1 1 . cos θ = √ or θ = cos−1 √ 17 17 89. Show that the vector function R(t) = (1 + t 2 , 1 + t 2 , 1 + t) does not intersect the plane −2x + 3y + z = 1. Then, find the closest point of curve of the vector function to the plane. Solution. In order to see that they do not intersect, we put the formula of R(t) into the formula for the plane componentwise. We obtain 1 2 7 + > 1. −2(1 + t 2 ) + 3(1 + t 2 ) + (1 + t) = t 2 + t + 2 = t + 2 4 Hence, we conclude that the curve cannot intersect the plane. Now, by the distance formula between a point and a plane, we have D(t) =

t2 + t + 1 | − 2(1 + t 2 ) + 3(1 + t 2 ) + (1 + t) − 1| = . √ 14 (−2)2 + 32 + 12

2.5 Solved Problems

75

√ Taking the derivative of D(t), we obtain D (t) = (2t + 1)/ 14. If D (t) = 0, then we get t = −1/2. Therefore, the closest point is (5/4, 5/4, 1/2). 90. Find two parametric representations R1 (t) and R2 (t) for the line y = x in R2 such that R1 (0) = R2 (0) = (0, 0) and R1 (0) = (0, 0) but R2 (0) = (0, 0). Solution. It is enough if we consider R1 (t) = (t, t) and R2 (t) = (t 3 , t 3 ), where t ∈ R.

91. Show that the derivative of the norm is not equal to the norm of the derivative, in general. Solution. We consider R(t) = (t, 1, 1). First, we compute R(t) . We obtain R(t) =

d2 t . t +2= √ dt t2 + 2

On the other hand, we have R (t) = (1, 0, 0) and then R (t) = in this example that R(t) = R (t).

92.

√

1 = 1. It is clear

Use the dot product rule to prove the following: R(t) · R (t) d R(t) = . dt R(t)

Solution. Differentiating both sides of R(t)2 = R(t) · R(t), we obtain d R(t)2 = 2R(t) · R (t). dt On the left side, note that R(t)2 is a scalar function, and using the chain rule; we get d d R(t)2 = 2R(t) R(t). dt dt Hence, we have 2R(t) which implies that

d R(t) = 2R(t) · R (t), dt

d R(t) · R (t) R(t) = . dt R(t)

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2 Vector Functions and Parametric Equations

93. Let R be a non-zero differentiable vector function on an interval. Show that R(t) is constant for all t in the interval if and only if R(t) and Dt R(t) are orthogonal. Solution. Suppose that R(t) = c, where c is a constant number. Then, we have R(t) · R(t) = c2 . Differentiating on both sides with respect to t gives Dt R(t) · R(t) + R(t) · Dt R(t) = 0. This implies that R(t) · Dt R(t) = 0. Since the dot product of R(t) and Dt R(t) is zero, it follows that R(t) and Dt R(t) are orthogonal. Conversely, suppose that R(t) · Dt R(t) = 0. Since R(t)2 = R(t) · R(t), it follows that 2R(t) Dt R(t) = 2R(t) · Dt R(t). So, we conclude that R(t) Dt R(t) = 0. This implies that R(t) = 0 or Dt R(t) = 0. Since R is a non-zero function, it follows that Dt R(t) = 0, and hence R(t) is constant magnitude. 94.

Suppose that R(t) is differentiable three times. Show that

R(t) · R (t) × R (t) = R(t) · R (t) × R (t) .

Solution. Using the differentiation rule for dot products, we have

R(t) · R (t) × R (t) = R (t) · R (t) × R (t) + R(t) · R (t) × R (t) .

Since R (t) · R (t) × R (t) = 0, it follows that

R(t) · R (t) × R (t) = R(t) · R (t) × R (t) .

(2.4)

Next, using the differentiation rule for cross products, we can write

R (t) × R (t) = R (t) × R (t) + R (t) × R (t) = R (t) × R (t).

(2.5)

Putting (2.22) into (2.4), the result follows. 95. Consider the unit circle x 2 + y 2 = 1. By considering the parameter t = y/(x − 1) which is the slope of the line joining (1, 0) and the point (x, y) on the curve, show that t2 − 1 2t , 2 R(t) = 2 t +1 t +1

2.5 Solved Problems

77

is a parametric representation of the unit circle. This parametrization of the circle is called rational parametrization. Solution. We substitute y = t (x − 1) into the equation x 2 + y 2 = 1. Then, we get x 2 + t 2 (x − 1)2 = x 2 + t 2 x 2 + t 2 − 2xt 2 = 1. This implies that (t 2 + 1)x 2 − 2t 2 x + t 2 − 1 = 0. So, we obtain x=

2t 2 ±

√ 2t 2 ± 4t 4 − 4t 4 + 4 t2 ± 1 4t 4 − 4(t 2 + 1)(t 2 − 1) = = . 2(t 2 + 1) 2(t 2 + 1) t2 + 1

By ignoring the trivial solution x = 1, we have x=

t2 − 1 2t and y = 2 . t2 + 1 t +1

96. Parameterize the curve given by x 3 + y 3 = 3x y by considering the parameter t = y/x which is the slope of the line through the origin and the point (x, y) on the curve. Solution. We substitute y = t x into the equation x 3 + y 3 = 3x y. Then, we obtain x 3 + t 3 x 3 = 3xt x, which implies that x 3 (1 + t 3 ) = 3x 2 t. By ignoring the trivial solution x = 0, we obtain 3t x= . (2.6) 1 + t3 Substituting (2.6) in y = t x, we get y=

3t 2 . 1 + t3

(2.7)

Therefore, a parametrization for the curve is R(t) =

3t 3t 2 . , 3 1 + t 1 + t3

− → − → 97. For the curve R(t) = (cos t + t sin t) i + (sin t − t cos t) j with t > 0, find the unit tangent vector T (t) and the unit normal vector N (t) at t = π/3. − → − → Solution We obtain R (t) = t cos t i + t sin t j . Then, we have R (t) = t 2 cos2 t + t 2 sin2 t = t. Since T (t) = R (t)/R (t), it follows that − → − → T (t) = cos t i + sin t j . − → − → Differentiating T (t) with respect to t, we obtain T (t) = − sin t i + cos t j . This implies that T (t) = 1. Since N (t) = T (t)/T (t), it follows that

78

2 Vector Functions and Parametric Equations

− → − → N (t) = − sin t i + cos t j . Now, we determine T (t) and N (t) when t = π/3. We have √ π− π− 3− → → 1− → → = cos i + sin j = i + j, T 3 3 3 2 2 √ π π− π− 3− → → → 1− → = − sin i + cos j = − i + j. N 3 3 3 2 2 π

98. Let R(t) = (cos 2t, sin 2t, t), where t ∈ [0, 2π] and let C be the curve that is parameterized by R. Determine the unit binormal to C at R(π/4). Solution. We have R (t) = (−2 sin 2t, 2 cos 2t, 1), √ R (t) = 5, −2 R (t) 2 1 = √ sin 2t, √ cos 2t, √ , T (t) = R (t) 5 5 5 −4 −4 T (t) = √ cos 2t, √ sin 2t, 0 , 5 5 4 T (t) = √ , 5 N (t) =

T (t) = (− cos 2t, − sin t, 0). T (t)

Therefore, we obtain − → i −2 B(t) = T (t) × N (t) = √ sin 2t 5 − cos 2t

− → k 2 1 √ cos 2t √ 5 5 − sin t 0 − → j

1 1 2 − − → − → → = √ sin 2t i − √ cos 2t j + √ k . 5 5 5 √ − √ − → → Finally, we have B(π/4) = 1/ 5 i + 2/ 5 k .

2.5 Solved Problems

99.

79

Reparameterize the following curves in terms of arc length.

(1) R(t) = (2 + t, 3 − t, 5t), √ t ≥ 0; (2) R(t) = (2 cos t, 2 sin t, 5t), t ≥ 0. Solution. (1) By the definition s(t) =

t

x (u)

2

2 2 + y (u) + z (u) du =

0

t√ √ 27du = 27t. 0

This implies that

1 1 5 R(s) = 2 + √ s, 3 − √ s, √ s . 27 27 27

(2) Again, by the definition we obtain s(t) = 3t. This gives that √ s 5 s s . R(s) = 2 cos , 2 sin , 3 3 3 100. Suppose that a particle moves along the curve R(t) = (et , e2t , sin t) from t = 0 to t = 1, and then it moves on the tangent line to the curve at R(1) in the direction of the tangent vector. Find the position of the particle at t = 5. Solution. The tangent line at R(1) is defined by X (t) = (e, e2 , sin 1) + t (e, 2e2 , cos 1). Note that X (0) = R(1). So, the position vector of the particle at t = 5 is X (4). 101. Is there a point on the curve R(t) = (t 2 − t, t 3 /3, 2t) at which the tangent line is parallel to the vector B = (−5/2, 2, 1)? Solution. If there exists a point on the curve R(t) such that the tangent line is parallel to B, then there exist c and t0 such that B = c R (t0 ). This implies that

5 − , 2, 1 = c 2t0 − 1, t02 , 2 = 2ct0 − c, ct02 , 2c . 2

Hence, we conclude that 5 2ct0 − c = − , ct02 = 2 and 2c = 1. 2

(2.8)

The third equality in (2.8) gives that c = 1/2. From the second equality in (2.8), we obtain t0 = ±2. It is easy to check that only t0 = −2 satisfies the first equality in (2.8). Therefore, there is a point when t = −2, and this point is R(−2) = (2, −8/3, −4).

80

102.

2 Vector Functions and Parametric Equations

Find all solutions to R (t) = 2R(t), where R(t) is a vector function in R3 .

Solution. Suppose that R(t) = x(t), y(t), z(t) . Then, we haveR (t) = x (t), y (t), z (t) . Substituting in R (t) = 2R(t), we get x (t), y (t), z (t) =2 x(t), y(t), z(t) . Equating the corresponding components gives that x (t) = 2x(t), y (t) = 2y(t) and z (t) = 2z(t). It follows that x(t) = c1 e2t , y(t) = c2 e2t and z(t) = c3 e2t . Hence, we have R(t) = e2t (c1 , c2 , c3 ), where (c1 , c2 , c3 ) is a constant vector. 103. Let R(t) = f (t), g(t), h(t) be a continuous vector function on [a, b], and let C be any constant vector. Show that

b

b

C · R(t)dt = C ·

R(t)dt.

a

a

Solution. Suppose that C = (c1 , c2 , c3 ), where c1 , c2 and c3 are constant scalars. Then, we can write C·

b

R(t)dt = (c1 , c2 , c3 ) ·

a

b

f (t)dt,

a

b

= c1

f (t)dt + c2

a

g(t)dt,

a b

g(t)dt + c3

a

b

h(t)dt a

b

h(t)dt a

c1 f (t) + c2 g(t) + c3 h(t) dt

b

=

b

a

b

=

C · R(t)dt.

a

104.

− → − → − → Compute A · B, where A = 2 i − 4 j + k and

1

B=

− → − → − → te2t i + t cosh 2t j + 2te−2t k dt.

0

Solution. Using Problem 103, we obtain

1

A·B =

0 1

= 0

105.

2t 2te − 4t cosh 2t + 2te−2t dt e2t + e−2t + 2te−2t dt = 0. 2te2t − 4t 2

If R and R are integrable on [a, b], prove that

a

b

R(t)dt

≤

a

b

R(t)dt.

(2.9)

2.5 Solved Problems

81

Solution. Suppose that C =

b

R(t)dt. If C is the zero vector, then (2.9) obviously a

holds. So, let C be non-zero. Then, by Problem 103, we can write C2 = C · C = C ·

b

b

R(t)dt =

a

C · R(t)dt.

(2.10)

a

Since C · R(t) is real-valued, we have the following inequalities: a

b

C · R(t)dt ≤

b

|C · R(t)|dt ≤

a

b

C R(t)dt,

(2.11)

a

where in the last step we used the Cauchy-Schwartz inequality |R · C| ≤ C R(t). Combining (2.10) and (2.32), we get C2 ≤ C

b

R(t)dt.

a

Since C > 0, we can divide it by C to obtain (2.9). 106. A particle moves along the parabola x 2 + a(y − x) = 0 in such a way that the horizontal and vertical components of the acceleration vector are equal. If it takes T units of time to go from the point (a, 0) to the point (0, 0), how much time will it require to go from (a, 0) to the halfway point (a/2, a/4)? 2 Solution. Since x 2 + a(y − x) = 0, it follows that y = x − x /a. Suppose that 2 R(t) = x(t), x(t) − x (t)/a . Then, the first and second derivatives of R are

2x(t)x (t) R (t) = x (t), x (t) − , a 2 2 R (x) = x (t), x (t) − x (t) + x(t)x (t) . a Since the horizontal and vertical components of the acceleration vector are equal, it follows that 2 2 x (t) = x (t) − x (t) + x(t)x (t) . a Thus, we have x(t)x (t) + x 2 (t) = 0. This yields that x(t)x (t) = 0, and so x(t)x (t) = c is constant. Now, we can write x(t)d x = cdt, which implies that x 2 (t)/2 = ct. Now, using the conditions, we get (0 − a 2 )/2 = cT . This gives that c = −a 2 /(2T ). Next, we have 1 −3a 2 −a 2 1 a 2 2 = −a = (T f − 0). 2 2 2 4 2T

82

2 Vector Functions and Parametric Equations

Therefore, we conclude that T f = 3T /4. 107.

Suppose the motion of a particle is given by x = 4cost and y = sint.

(1) Describe the motion of the particle, and sketch the curve along which the particle travels. (2) Find the velocity and acceleration vectors of the particle. (3) Find the times t and the points on the curve where the speed of the particle is greatest. (4) Find the times t and the points on the curve where the magnitude of the acceleration is greatest. Solution. (1) We observe that (x/4)2 + y 2 = 1. This means that the path of particle is this ellipse. The motion is counterclockwise around the ellipse. − → − → (2) The position vector is R(t) = 4 cos t i + sin t j . Hence, the velocity is − → − → V (t) = R (t) = −4 sin t i + cos t j √ . (3) The speed is v(t) = V (t) = 16 sint + cost 2t. Then, we obtain dv 15 sin t cos t =√ . dt 16 sint + cos2 t If dv/dt = 0, then sin t cos t = 0. This implies that sin t = 0 or cos t = 0, and hence t = nπ or t = π/2 + nπ, where n is any integer. Next, we have v(nπ) = 1 and v(π/2 + nπ) = 4. So, the maximum speed occurs when t = π/2 + nπ. If n is even, then the maximum occurs at the point (0, 1). If n is odd, then the maximum occurs at the point (0, −1). (4) The magnitude of the acceleration is a(t) = A(t) = 16 cos2 t + sin2 t. Now, we have −15 sin t cos t da = . dt 16 cost + sin2 t If da/dt = 0, then sin t = 0 or cos t = 0. It follows that t = nπ or t = π/2 + nπ, where n is any integer. We see that a(nπ) = 4 and a(π/2 + nπ) = 1. Therefore, maximum magnitude of the acceleration occurs when t = nπ. If n is even, then the maximum occurs at the point (0, 4), and if n is odd, then the maximum occurs at the point (0, −4). 108. Show that if the velocity and position vectors of a particle remain orthogonal during the motion, then the trajectory lies on a sphere. Solution. Suppose that R(t) = x(t), y(t), z(t) is the position vector, and V (t) = R (t). Since R(t) and V (t) are orthogonal, it follows that V (t) · R(t) = 0, for all t. On the other hand, we have

R(t) · R(t) = R (t) · R(t) + R(t) · R (t) = 2R (t) · R(t) = 2V (t) · R(t) = 0.

2.5 Solved Problems

83

This yields that R(t) · R(t) is constant. Hence, let R(t) · R(t) = r 2 , for all t. Then, we obtain x 2 (t) + y 2 (t) + z 2 (t) = r 2 . This means that the particle remains at a fixed distance r from the origin all the time. 109.

Prove that the substitution rule:

b

R f (t) f (t)dt =

f −1 (b)

R(u)du, f −1 (a)

a

where f is a differentiable scalar function. Solution. Let R(t) = x(t), y(t), z(t) . Using the componentwise integration, we have b b b b R(u)du = x(u)du, y(u)du, z(u)du . a

a

a

a

Assume that u = f (t). Then, du = f (t)dt. It follows that a

b

=

f −1 (b) f −1 (a)

= =

R f (t) f (t)dt

f −1 (b) f −1 (a) f −1 (b) f −1 (a)

x f (t) f (t)dt,

f −1 (b) f −1 (a)

y f (t) f (t)dt,

f −1 (b) f −1 (a)

z f (t) f (t)dt

x f (t) f (t), y f (t) f (t), z f (t) f (t) dt x f (t) , y f (t) , z f (t) f (t)dt =

f −1 (b) f −1 (a)

R f (t) f (t)dt.

110. A differential equation of the form Y (x) + p(x)Y (x) = Q(x), where p is a given real function, Q a given vector function and Y an unknown vector function, is called a first- order linear vector differential equation. Prove that if p and Q are continuous on an interval I , then for each a ∈ I and each vector B there is one and only one solution Y which satisfies the initial condition Y (a) = B, and that this solution is given by the formula Y (x) = Be

−q(x)

+e

−q(x)

Q(t)eq(t) dt, a

where q(t) =

t

p(u)du. a

Solution. Suppose that

x

84

2 Vector Functions and Parametric Equations

Y (x) = y1 (x), y2 (x), y3 (x) , Q(x) = q1 (x), q2 (x), q3 (x) and B = (b1 , b2 , b3 ). (2.12) Then, for each i = 1, 2, 3, we have yi (x) + p(x)yi (x) = qi (x).

(2.13)

To solve (2.13), we would like to choose a function μi so that if (2.13) is multiplied by μi (x), then the left-hand side of (2.13) can be written as the derivative of the function μi (x)yi (x). That is, we want to choose μi , if possible, so that μi (x) yi (x) + p(x)yi (x) = μi (x)yi (x) = μi (x)yi (x) + μi (x)yi (x). Thus, μi (x) must satisfy μi (x) p(x)yi (x) = μi (x)yi (x). Assuming for the moment that μi (x) > 0, we obtain μi (x)/μi (x) = p(x). This implies that ln μi (x) = p(x) d x, or equivalently μi (x) = e p(x)d x . Returning to (2.13) and multiplying by μi (x), we obtain μi (x)yi (x) = μi (x)qi (x). So, we conclude that μi (x)yi (x) = μi (x)qi (x)d x + ci , or equivalently yi (x) = e

−

p(x)d x

qi (x)e

p(x)d x

d x + ci .

Since yi (a) = bi , it follows that yi (x) = bi e−q(x) + e−q(x)

x

qi (t)eq(t) dt,

(2.14)

a

for i = 1, 2, 3. Now, from (2.12) and (2.14), the result follows. 111.

Find a vector function R, continuous on the interval (0, ∞) such that R(x) = xe x A +

1 x

x

R(t)dt, 1

for all x > 0, where A is a fixed non-zero vector. Solution. Differentiating both sides of (2.15) with respect to x, we obtain

(2.15)

2.5 Solved Problems

85

R (x) = e x A + xe x A −

1 x2 1

x

R(t)dt +

1

1 R(x) x

1 xe x A − R(x) + R(x) x 1 1 = e x A + xe x A + e x A − R(x) + R(x) x x

= e x A + xe x A +

x

= 2e x A + xe x A = (2 + x)e x A. It follows that R(x) = 2e x A + xe x − e x A + C = xe x A + e x A + C,

(2.16)

where C is a constant vector. Taking the integral from both sides of this equation, we have x x x t te A + et A + C dt = A(tet − et ) + et A + Ct R(t)dt = 1

1

1

= A(xe − e ) + e A + C(x − 1) − e A = Axe + C(x − 1) − e A. x

x

x

x

Hence, we must have R(x) − xe x A = Ae x +

C(x − 1) e A − . x x

(2.17)

From (2.16) and (2.17), we get C = −e A. Therefore, we obtain R(x) = xe x A + e x A − e A = xe x + e x − e A. 112. A vector function R, which is never zero and has a continuous derivative R (t) for all t, is always parallel to its derivative. Prove that there is a constant vector A and a positive real function f such that R(t) = f (t)A, for all t. Solution. Since R(t) is parallel to R (t), we can write R (t) = g(t)R(t), where R(t) = f 1 (t), f 2 (t), f 3 (t) and g is a real function. So, we have f i (t) = g(t) f i (t), for i = 1, 2, 3, or equivalently f i (t)/ f i (t) = g(t). Taking the integral from both sides, we obtain t t f (t) t f i (x) i dx = = g(x)d x or ln g(x)d x. f i (a) a f i (x) a a This yields that t t f i (t) = e a g(x)d x or f i (t) = f i (a)e a g(x)d x , f i (a)

86

2 Vector Functions and Parametric Equations

for i = 1, 2, 3. Therefore, we have R(t) = e

t a

g(x)d x

f 1 (a), f 2 (x), f 3 (x) = f (x)A,

where A = f 1 (a), f 2 (x), f 3 (x) and f (t) =

t

g(x)d x. This completes the proof.

a

113. A plane curve C in the first quadrant has a negative slope at each of its points and passes through the point (3/2, 1). The position vector R from the origin to any − → point (x, y) on C makes an angle θ with i , and the velocity vector makes an angle − → φ with i , where 0 < θ < π/2, and 0 < φ < π/2. If 3 tan θ = 4cotφ at each point of C, find the Cartesian equation for C. Solution. Assume that R(t) = x(t), y(t) and R (t) = x (t), y (t) . Since dy/d x < 0 in the first quadrant and 3 tan θ = 4cotφ, it follows that y −4 x . = x 3 y Now, we can write

y dy 4 x dy/dt = =− . = x d x/dt dx 3 y

This implies that ydy = (−4/3)xd x, and so y 2 = (−4/3)x 2 + C, where C is constant. Since the curve passes through the point (3/2, 1), it follows that C = 4. Therefore, the Cartesian equation for C is x 2 /3 + y 2 /4 = 1. 114. Show that a moving particle will move in a straight line if the normal component of its acceleration is zero. Solution. If a N = 0, then we have κV = 0. Since the particle is moving, it cannot have zero speed. Hence, we have κ = 0, i.e., the curvature is zero. Therefore, the particle is moving along a straight line. 115. Prove that if R(t) = x(t), y(t), z(t) is a vector equation of curve C, and κ(t) is the curvature of C, then κ(t) =

R (t) × R (t) . R (t)3

(2.18)

Solution. We express R (t) and R (t) in terms of T (t) and T (t), then we compute their cross product. Since T (t) = R (t)/R (t) and ds/dt = R (t), it follows that

2.5 Solved Problems

87

R (t) =

ds T (t). dt

(2.19)

Taking the derivative with respect to t of (2.19) gives R (t) =

d 2s ds T (t). T (t) + dt 2 dt

(2.20)

From (2.19), (2.20) and using the properties of cross products, we obtain ds 2 ds d 2 s T (t) × T (t) + T (t) × T (t) dt dt 2 dt ds 2 = T (t) × T (t) . dt

R (t) × R (t) =

Hence, we can write R (t) × R (t) =

ds 2 dt

T (t) T t) sin θ,

where θ is the angle between T (t) and T (t). Since T (t) = 1, and T (t) and T (t) are orthogonal, we conclude that R (t) × R (t) =

ds 2 dt

This implies that T (t) =

T (t) = R (t)2 T (t).

R (t) × R (t) . R (t)2

Finally, since κ(t) = T (t)/R (t), the formula (2.18) follows. 116.

Show that κ and τ are both zero for the line − → − → − → R(t) = (x0 + at) i + (y0 + bt) j + (z 0 + ct) k .

− → − → − → Solution. We have V (t) = a i + b j + c k , and so A(t) = 0. This gives that V × A = 0. Now, using Problem 115, we deduce that the curvature is zero. Since the curve is a plane curve, it follows that τ = 0. 117. A particle moves along a path given by the curve R(t)= 2t 2 + 1, 2et + 2, t 3 . What are the normal and tangential components of acceleration when the particle is moving in the direction of the y-axis? Solution. The particle moves in the direction of the y-axis when its velocity vector is parallel to the vector (0, 1, 0). We observe that V (t) = R (t) = (4t, 2et , 3t 2 ).

88

2 Vector Functions and Parametric Equations

In order for this vector to be parallel to (0, 1, 0), it is necessary that the first and third components be zero. This happens at t = 0. So, we have R (0) = (0, 2, 0). Also, since R (t) = (4, 2et , 6t), it follows that R (0) = (4, 2, 0). Now, the tangential component of acceleration at time t = 0 is aT =

R (0) · R (0) (0, 2, 0) · (4, 2, 0) 4 = = = 2. R (0) (0, 2, 0) 2

The normal component of acceleration at t = 0 is aN = 118.

(0, 2, 0) × (4, 2, 0) (0, 0, −8) 8 R (0) × R (0) = = = = 4. R (0) (0, 2, 0) 2 2

Find the curvature κ of a plane curve C.

Solution. Without loss of generality, we assume that C lies in the x y-plane, and let R = (x, y). Then, we have R = (x , y ) and R = (x , y ). So, we can write − − → → j i R × R = x y x y

− → k → − 0 = (x y − y x ) k . 0

Now, in this case using Problem 115, we obtain κ= 119.

R × R |x y − y x | = 3/2 . R 3 x 2 + y2

Verify the curvature κ = κ(θ) of the polar curve r = r (θ) is given by κ=

|r 2 + 2r 2 − rr | 3/2 , r 2 + r 2

(2.21)

where r = dr/dθ and r = d 2 r/dθ2 . Solution. The curve can be expressed in the parametric form x(θ) = r (θ) cos θ and y(θ) = r (θ) sin θ. Calculating the first and second derivatives of x and y, we obtain x (θ) = r (θ) cos θ − r (θ) sin θ, y (θ) = r (θ) sin θ + r (θ) cos θ, x (θ) = r (θ) cos θ − 2r (θ) sin θ − r (θ) cos θ, y (θ) = r (θ) sin θ + 2r (θ) cos θ − r (θ) sin θ. Now, with the help of Problem 118 we obtain

2.5 Solved Problems

89

|x y − y x | |r 2 + 2r 2 − rr | κ= = 3/2 . 3/2 x 2 + y2 r 2 + r 2 120. Show that at the point of intersection of the curves r = aθ and r θ = a, the curvatures are in the ratio 3 : 1, where 0 < θ < 2π. Solution. Suppose that κ1 and κ2 are the curvatures of the curves r = aθ and r θ = a, respectively. The points of intersection of the given curves are aθ2 = a or θ = ±1. To evaluate the curvatures, we use the formula (2.21) in Problem 119. For the curve r = aθ, we have r = a and r = 0. So, at θ = ±1, we get κ1

at θ=±1

|r 2 + 2r 2 − rr | |a 2 θ2 + 2a 2 − 0| = 3/2 3/2 at θ=±1 at θ=±1 r 2 + r 2 a 2 θ2 + a 2 2 3/2 √ 3 √ 2a 2 2a 2 2 a. = = = 2 2 3a 3a 3 =

Similarly, for the curve r θ = a, we obtain r = −a/θ and r = 2a/θ3 . Thus, at θ = ±1, we have κ1

at θ=±1

a2 2a 2 2a 2 + − |r + 2r − rr | θ2 θ4 θ4 = = a2 3/2 at θ=±1 a 2 3/2 at θ=±1 r 2 + r 2 + θ2 θ4

3/2 √ 1 + θ2 = a = 2 2a. at θ=±1 θ4 2

2

Therefore, we observe that κ2 /κ1 = 3, as desired. 121.

For the curve r n = a n cos nθ, prove that (n−1)/n ds = a sec nθ . dθ

Solution. Taking logarithm on both sides of r n = a n cos nθ gives n ln r = n ln a + ln(sec nθ). By differentiating (2.22), we obtain sin nθ n dr = −n , r dθ cos nθ which implies that dr/dθ = −r tan nθ. Now, we have

(2.22)

90

2 Vector Functions and Parametric Equations

ds = dθ

r2 +

dr 2

=

√ r 2 + r 2 tan2 nθ = r sec nθ

dθ 1/n (n−1)/n = a cos nθ sec nθ = a sec nθ .

122. If C is a curve with equation y = f (x), where f is twice differentiable, prove that | f (x)| κ(x) = 2 3/2 . 1 + f (x) Solution. The proof follows easily from Problem 115. We parameterize the curve by x = x, y = f (x) and z = 0. Using x as the name of the parameter, then the position vector of C is R(x) = (x, f (x), 0). Hence, we have R (x) = (1, f (x), 0) and R (x) = (0, f (x), 0). Now, we get R (x) × R (x) = (0, 0, f (x)), which implies 2 that R (x) × R (x) = | f (x)| and R (x) = 1 + f (x) . Therefore, we conclude that R (x) × R (x) | f (x)| κ(x) = = 2 3/2 . R (x)3 1 + f (x) 123. Find the value of a such that the curvature of f (x) = eax at x = 0 is as large as possible. Solution. We have f (x) = aeax and f (x) = a 2 eax . Now, using the formula in Problem 122, we obtain a 2 eax κ(x) = 3/2 . 1 + a 2 e2ax So, the curvature at the origin is a2 κ(0) = 3/2 . 1 + a2 Since κ(0) and κ2 (0) have their maximum values at the same values of a, we maximize the function a4 h(a) = κ2 (0) = 3 . 1 + a2 We have h (a) =

4a 3 (1 + a 2 )3 − a 4 (3)(1 + a 2 )2 (2a) 2a 3 (1 + a 2 )2 (2 − a 2 ) = . 2 6 (1 + a ) (1 + a 2 )6

2.5 Solved Problems

91

√ √ If h (a) = 0, then the stationary points are a = 0, a = 2 and a = − 2. Since h(a) ≥ 0√and h(0) = 0, it follows √ that a = 0 is a minimum point. √ Since h (a) > 0 for a < 2 and h (a) < 0 for 2 < a, we conclude that a = 2 is a maximum √ point. On the other hand, since h is an even function, it follows that a = − 2 is a maximum √ point as well. Consequently, κ(x) takes its maximum value at x = 0 when a = ± 2.

124.

Find the radius of curvature for

x − a

y =1 b

(2.23)

at the point where it touches the coordinate axes. Solution. Taking the implicit differentiation from (2.23), we obtain 1 1 dy = 0. √ − √ 2 ax 2 by d x This implies that dy = dx

by . ax

(2.24)

The curve touches the x-axis if dy/d x = 0 or y = 0. If y = 0, then we have x = a. Hence, the given Eq. (2.23) touches x-axis at (a, 0). Similarly, the curve touches y-axis if d x/dy = 0 or x = 0. If x = 0, then we have y = b. So, the given Eq. (2.23) touches y-axis at (0, b). Now, from (2.24), we obtain d2 y = dx2 and so

b b 1 1 y − , a a 2x 2 x b d 2 y = 2. d x 2 at (a,0) 2a

Now, using Problem 122, we have ρ(t)

at (a,0)

Similarly, we have

3/2 1 + y2 2a 2 . = = at (a,0) |y | b d 2 x a = 2, dy 2 at (0,b) 2b

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2 Vector Functions and Parametric Equations

and so

ρ(t)

at (0,b)

3/2 1 + x 2 2b2 . = = at (0,b) |x | a

125. Show that the curvature of the parametrization R(t) = (a cos t, b sin t) of the ellipse x 2 /a 2 + y 2 /b2 = 1 is ab κ(t) = 3/2 . 2 2 b cos t + a 2 sin2 t Solution. We apply Problem 115. We compute the first and the second derivatives of R(t) as follows: R (t) = (−a sin t, b cos t) and R (t) = (−a cos t, −b sin t). Then, we obtain − → − → R (t) × R (t) = ab sin2 t + ab cos2 t k = ab k , which implies that R (t) × R (t) = ab. Moreover, we have R (t) = a 2 sin2 t + b2 cos2 t. Therefore, by using Problem 115, we get ab ab κ(t) = 3/2 . 3 = 2 2 2 2 2 2 a sin t + b2 cos2 t a sin t + b cos t 126.

In the notation of Problem 125, suppose that b ≤ a. Show that b a ≤ κ(t) ≤ 2 . a2 b

3/2 . Since Solution. In Problem 125, we proved that κ(t) = ab/ a 2 sin2 t + b2 cos2 t b ≤ a, it follows that ab ab 3/2 ≤ κ(t) ≤ 3/2 . 2 2 2 2 2 2 a sin t + a cos t b sin t + b2 cos2 t This implies that ab/a 3 ≤ κ(t) ≤ ab/b3 , and so b/a 2 ≤ κ(t) ≤ a/b2 , as desired. 127. Prove that if R(t) = x(t), y(t), z(t) is a vector equation of curve C, K (t) is the curvature vector of C at a point P, and s units is the arc length measured from an arbitrary chosen point on C to P, then 2 Ds R(t) · Ds3 R(t) = − K (t) .

2.5 Solved Problems

93

Solution. We know that K (t) = T (t)/R (t) and also K (t) = Ds T (t). Since T (t) and T (t) are orthogonal, it follows that T (t) and K (t) are orthogonal, which implies that T (t) · K (t) = 0. It follows that 0 = Ds T (t) · K (t) = Ds T (t) · K (t) + T (t) · Ds K (t). Since T (t) = Ds R(t), it follows that 2 0 = K (t) · K (t) + Ds R(t) · Ds2 T (t) = K (t) + Ds R(t) · Ds3 R(t). This gives the desired equality. 128.

Let s(t) =

t −∞

R (u)du for the Bernoulii spiral R(t) = (et cos 4t,

et sin 4t). Find the radius of curvature. Solution. First, we compute the curvature using the formula in Problem 118. We have x(t) = et cos 4t and y(t) = et sin 4t. Calculating the first and second derivatives gives x (t) = et (cos 4t − 4 sin 4t), y (t) = et (sin 4t + 4 cos 4t), x (t) = −et (15 cos 4t + 8 sin 4t), y (t) = et (8 cos 4t − 15 sin 4t). By Problem 118, we get κ(t) =

68e2t 4 |x (t)y (t) − y (t)x (t)| = = √ e−t . 2 3/2 3/2 3t 2 17 e 17 x (t) + y (t)

Since the radius of curvature is the reciprocal of the curvature, i.e., ρ(t) = 1/κ(t), it follows that √ 17 t e. (2.25) ρ(t) = 4 On the other hand, by the fundamental theorem of calculus we obtain s (t) = R (t) = This implies that s(t) =

√

x (t)2 + y (t)2 =

17et dt =

Thus, we have lim s(t) = lim

t→−∞

t→−∞

√

√

√ t 17e .

17et + C.

17et + C = C.

(2.26)

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2 Vector Functions and Parametric Equations

Moreover, since s(t) =

t

−∞

R (u)du, it follows that lim s(t) = 0.

t→−∞

(2.27)

From (2.26) and (2.27), we get C = 0, and so we conclude that s(t) =

√

17et .

(2.28)

Now, from (2.25) and (2.28) we obtain ρ(t) = s(t)/4, and we finished the solution. − → − → 129. Find an equation for the circle of curvature of the curve R(t) = t i + sin t j at the point (π/2, 1). (The curve parameterizes the graph y = sin x in the x y-plane.) √ − → − → Solution. We obtain R(t) = i + cos t j and R (t) = 1 + cos2 t. Also, we have 1 cos t R (t) − → − → =√ i +√ j, 2 2 R (t) 1 + cos t 1 + cos t sin t cos t − sin t → − → T (t) = i − j, (1 + cos2 t)3/2 (1 + cos2 t)3/2 T (t) =

T (t) =

| sin t| . 1 + cos2 t

At t = π/2, we get R (π/2) = 1 and T (π/2) = 1. Since κ(π/2) = T (π/2)/R (π/2), it follows that κ(π/2) = 1. This implies that ρ(π/2) = 1. Moreover, the center is (π/2, 0). Therefore, an equation for the circle of curvature of the curve is (x − π/2)2 + y 2 = 1, as desired. − → 130. Find an equation for the circle of curvature of the curve R(t) = 2 ln t i − 1 − → t+ j with e−2 ≤ t ≤ e2 , at the point (0, −2), where t = 1. t Solution. First, we compute the curvature. We have 1 − 2− → → R (t) = i − 1 − 2 j , t t 2 2 4 +1 1 t R (t) = + 1 − = , 2 2 t t t2 2t − R (t) → t2 − 1 − → = i − 2 j, T (t) = R (t) t2 + 1 t +1 4t −2(t 2 − 1) − → − → T (t) = i − 2 j, 2 2 2 (t + 1) (t + 1)

2.5 Solved Problems

95

T (t) = κ(t) =

4(t 2 − 1)2 + 16t 2 2 , = 2 (t 2 + 1)4 t +1

T (t) 2 t2 2t 2 . = = R (t) t2 + 1 t2 + 1 (t 2 + 1)2

Since κ(1) = 1/2, it follows that ρ(1) = 1/κ(1) = 2. Since the circle of curvature is tangent to the curve at the point (0, −2), it follows that the circle has the same − → tangent as the curve. The vector R (1) = 2 i is tangent to the circle, and so the center lies on the y-axis. If t = 1, then (t − 1)2 > 0, which implies that t 2 + 1 > 2t, or equivalently (t 2 + 1)/t > 2. Hence, t + 1/t > 2 or −(t + 1/t) < −2. This yields that y < −2 on both sides of the point (0, −2). Thus, we conclude that the curve is concave down, and so the center of the circle of curvature is (0, −4). Therefore, we deduce that an equation of the circle of curvature is x 2 + (y + 4)2 = 4. 131. We find the total curvature of the portion of a smooth curve runs from s = s0 to s = s1 > s0 by integrating κ from s0 to s1 . If the curve has some other parameter, say t, then t1 s1 t1 ds κds = κ dt = κR (t)dt, κT = dt s0 t0 t0 where t0 and t1 correspond to s0 and s1 . Find the total curvature of − → − → − → (1) The portion of the helix R(t) = 3 cos t i + 3 sin t j + t k , 0 ≤ t ≤ 4π; 2 (2) The parabola y = x , −∞ < x < ∞. − → Solution. (1) First, we compute the curvature. We have R (t) = −3 sin t i + 3 cos t √ − → − → j + k and R (t) = 10. Then, we get R (t) 1 − → − → − → =√ − 3 sin t i + 3 cos t j + k , R (t) 10

1 1

T (t) − → − →

= 3 . = κ(t) = − 3 cos t i − 3 sin t j √ √

10 R (t) 10 10 T (t) =

Now, the total curvature is equal to κT = 0

4π

3√ 12π 10dt = √ . 10 10

√ − → − → (2) Let y = x 2 and t = x. We have R (t) = i + 2t j and R (t) = 1 + 4t 2 . Using Problem 122, we obtain |y | 2 κ(t) = 3/2 = 3/2 . 2 1 + y 1 + 4t 2

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2 Vector Functions and Parametric Equations

Thus, the total curvature is equal to κT =

∞

−∞

2

1 + 4t 2 0

2 3/2 1 + 4t dt =

∞ −∞

2 dt 1 + 4t 2

b 2 2 dt + lim dt 2 a→−∞ a 1 + 4t b→∞ 0 1 + 4t 2 0 b π π = lim tan−1 2t + lim tan−1 2t = + = π. a→−∞ b→∞ a 0 2 2

= lim

132. Prove that the radius of curvature of the curve with parametric equations x = x(s), y = y(s) and z = z(s) is given by ρ=

2 d x 2 ds 2

+

d 2 y 2 ds 2

+

d 2 z 2 −1/2 ds 2

.

− → − → Solution. The position vector of any point on the curve is R(s) = x(s) i + y(s) j + − → z(s) k . Then, we have T (s) =

dR dx − → dy − → dz − → = i + j + k, ds ds ds ds

dT d2x − → d2 y − → d2z − → = 2 i + 2 j + 2 k. ds ds ds ds Since dT /ds = κ, it follows that κ=

d 2 x 2 ds 2

+

d 2 y 2 ds 2

+

d 2 z 2 ds 2

.

Since ρ = 1/κ, the result follows. 133. Let R be three times differentiable vector function that transverses a smooth curve whose curvature does not vanish. Then the torsion of the curve is

R (t) × R (t) · R (t) τ (t) = . R (t) × R (t)2 Solution. Suppose that v(t) = R (t). With this notation, we have R = vT and R = v T + κv 2 N . Using the definition of binomial vector, we can write

(2.29)

2.5 Solved Problems

97

R × R = vT × R = κv 3 B.

(2.30)

Differentiating on both sides of (2.29) gives R = v T + v T + κ v 2 + 2κvv N + κv 2 N . Since T = κv N and N = −κvT + τ v B, it follows that R = v − κ2 v 3 T + 3κvv + κ v 2 N + κτ v B.

(2.31)

From (2.30) and (2.31), we obtain (R × R ) · R = κv 3 B · R = κ2 v 6 τ B · B = κ2 v 6 τ . Consequently, we have τ=

(R × R ) · R . κ2 v 6

Now, using (2.18) in Problem 115, the result follows. 134. Show that

d3 R τ d R d2 R · × = 2. 2 3 ds ds ds ρ

Solution. We have dR d2 R d3 R dκ dT dN = T, = κN and + N. = =κ 2 3 ds ds ds ds ds ds

(2.32)

On the other hand, we have dN T dB =B× + × T = B × κN − τ N × T = −κT + τ B. ds ds ds Combining the third equality in (2.32) and (2.33), we get d3 R dκ dκ N = κτ B − κ2 T + N. = κ(τ B − κT ) + 3 ds ds ds Therefore, we obtain

(2.33)

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2 Vector Functions and Parametric Equations

d R d2 R d3 R dκ 2 · N × = T · κN × κτ B − κ T + ds ds 2 ds 3 ds dκ = T · κ 2 τ N × B − κ3 N × T + κ N × N ds 2 τ = T · κ τ T + κ3 B = κ2 τ = 2 . ρ

2.6 Exercises Easier Exercises 1. Let c be a constant. Show the angle between the position and the velocity vectors − → − → along the curve R(t) = ect cos t i + ect sin t j , t ∈ R, is constant. 2. An object moves with velocity vector (cos t, sin t, t), starting at (0, 0, 0) when t = 0. Find the function R giving its location. − → − → 3. Find the position vector R(t) if the acceleration vector A(t) = t 2 i − (1/t 2 ) j − → − → − → and V (1) = j , and R(1) = (−1/4) i + (1/2) j . 4. Suppose that an object moves such that its acceleration is given by A(t) = (−3 cos t, −2 sin t, 0). At time t = 0, the object is at (3, 0, 0) and its velocity vector is (0, 2, 1). Find V (t) and R(t) for the object. 5. At what point on the curve R(t) = (t 3 , 3t, t 4 ) is the plane perpendicular to the curve also parallel to the plane 6x + 6y − 8z = 1? 6. Find the equation of the line tangent to (cos t, sin t, cos(6t)) when t = π/4. 7. Suppose that the vector functions V (t) × U (t) and V (t) = 0 are continuous. Does this imply that the vector function U (t) is continuous? Support your arguments with examples. Find the limit if it exists: 10. lim t 2 , e2t , t 2 + 2t ; 8. lim cos t, t 2 + 3, tan t ; t→π/2 t→1 2 1 t − 2t − 3 t 2 − 5t + 6 1 et 1 , ln t, ; , , . 9. lim 2 − 2 11. lim t→0 t t→3 t sec t t t −3 3−t t − → − → − → − → − → − → 12. If R1 (t) = t 2 i − t j + (3t 5 + 1) k and R2 (t) = (3t 2 − 1) i + j − 2t k , find d(R1 ± R2 ) ; dt d(R1 · R2 ) ; (b) dt (a)

d(R1 × R2 ) ; dt d d R2 R1 × . (d) dt dt (c)

13. Find the magnitude of the velocity and acceleration of a particle which moves along the curve x = −2 sin 3t, y = −2 cos 3t and z = −5t at any time t > 0. Find unit tangent vector to the curve.

2.6 Exercises

99

14. Find an equation for the tangent plane to the surface z = x 2 + y 2 at the point (1, −1, 2). 15. Find the point of intersection of the plane y + z = 3 and the curve R(t) = (ln t, t 2 , 2t). Find the angle between the normal of the plane and the tangent line to the curve at the point of intersection. 16. At which points is the tangent to the curve x = t, y = t 2 and z = t 3 parallel to the plane x + 2y + z − 1 = 0? 17. Show that tangent lines to a circular helix have a constant angle with the axis of the helix. √ 18. Show that the arc length of the helix cos t, sin t, t), for 0 ≤ t ≤ 2π), is 2π 2, equal to the length of the diagonal of a square of side 2π. Show this graphically. − → − → − → 19. For the curve given by R(t) = (1/3)t 3 i + (1/2)t 2 j + t k , find (a) the unit tangent vector; (b) the unit normal vector; (c) the curvature. − → 20. Find the cosine of the angle between the vector j and the unit tangent vector − → − → − → to the curve R(t) = cos 2t i − 3t j + 2 sin 2t k at the point where √ t= √π. 3 21. Find the curvature of the curve x = cos t and y = sin3 t at point ( 2/4, 2/4). 22. Find the radius and center of the curvature of the curve y = x 4 at point (1, 1). 23. Show that the curvature of the curve y = ln x at any point (x, y) is equal√to x/(x 2 + 1)3/2 . Also, show that √ the absolute maximum of the curvature is 2/3 3, which occurs at the point ( 2/2, − ln 2/2). 24. What is the torsion of the curve x = cosh x, y = sinh x and z = t at the point which corresponds to t = ln 2? 25. Find the torsion τ of the general circular helix x = a cos ωt, y = a sin ωt and z = bt, where a > 0. 26. Show that the curvature κ and torsion τ of the curve x = 3t − t 3 , y = 3t 2 and z = 3t + t 3 are equal at every point. 27. Find the tangential and normal components of the acceleration vector of a particle − → − → − → with position function R(t) = t i + 2t j + t 2 k . Harder Exercises 28. Show that the curve with vector equation R(t) = a1 t 2 + b1 t + c1 , a2 t 2 + b2 t + c2 , a3 t 2 + b3 t + c3 lies in a plane and find an equation of the plane. 29. Show that dR dR d d R3 1 2 R1 · (R2 × R3 ) = · (R2 × R3 ) + R1 · × R3 + R1 · R2 × , dt dt dt dt

where R1 = R1 (t), R2 = R2 (t) and R3 = R3 (t) are differentiable vector functions.

100

2 Vector Functions and Parametric Equations

Fig. 2.4 A flyscreen

30. Prove the following rule for differentiating a 3 × 3 determinant each of whose rows consists of differentiable scalar functions: x1 y1 z 1 x y z x1 y1 z 1 x1 y1 z 1 1 1 1 d x2 y2 z 2 = x2 y2 z 2 + x2 y2 z 2 + x2 y2 z 2 . dt x y z x y z x y z x y z 3

3

3

3

3

3

3

3

3

3

3

3

(The prime denotes differentiation with respect to t.) 31. Prove that parametric equations of the catenary y = a cosh(x/a), where the parameter s is the number of units in the length of the arc from the point (0, a) to the point (x, y) and √ s ≥ 0 when x ≥ 0 and s < 0 when x < 0, are x = a sinh−1 (s/a) and y = a 2 + s 2 . 32. Define the ellipse C with parametric equations x = a cos t and y = b sin t, a and the points P = for positive constants b. For a fixed value of t, define a cos t, b sin t , Q = a cos(t + π/2), b sin(t + π/2) and Q = a cos(t − π/2), b sin(t − π/2) . Show that the vector Q Q (called the conjugate diameter) is parallel to the tangent vector to C at the point P. Sketch a graph and show the relationship among P, Q and Q . 33. In the x yz-space where a flyscreen lies along the plane with the equation 2x = y + 2z = 1, the trajectory of a bee as a function of time t is given by R(t) = − → − → − → t i + t 2 j + t 3 k with t ∈ R; see Fig. 2.4. (a) Find all times t when the bee is flying parallel to the screen; (b) Find all times t when the bee is flying perpendicular to the screen; (c) There are holes in the screen through which the bee passes. Find the coordinates of all of these holes. 34. Given the cycloid x = 2(t − sin t), y = 2(1 − cos t), express the arc length s as a function of t, where s is measured from the point where t = 0.

2.6 Exercises

101

35. Find the distance traveled by a thumbtack in the tread of a bicycle tire if the radius of the tire is 40 cm and the bicycle goes a distance of 50π m. Hint: The path of the thumbtack is a cycloid. 36. A projectile is shot from the top of a building 96 ft high from a gun at an angle of 30◦ with the horizontal. If the muzzle speed is 1600 ft/sec, find the time of flight and the distance from the base of the building to the point where the projectile lands. 37. Find the curvature of the curve with parametric equations

t

x= 0

t 1 sin πθ dθ and y = cos πθ2 dθ. 2 2 0 1

2

38. Let C be the twisted cubic x = t, y = (1/2)t 2 and z = (1/3)t 3 . (a) Find the maximum curvature of C; (b) Find the torsion τ and the vectors T , N and B of the moving trihedral of C at the origin; (c) Find the torsion τ and the vectors T , N and B at the point (1, 1/2, 1/3); (d) Find the maximum torsion of C. 39. Describe a situation in which the normal component of acceleration is 0 and the tangential component of acceleration is non-zero. Is it possible for the tangential component of acceleration to be 0 while the normal component of acceleration is non-zero? Explain. Finally, is it possible for an object to move (not be stationary) so that both the tangential and normal components of acceleration are 0? Explain. 40. Does the curve R(t) = (2t 2 , 2t, 2 − t 2 ) intersect the plane x + y + z = −3? If not, find a point on the curve that is closest to the plane. What is the distance between the curve and the plane. Hint: Express the distance between a point on the curve and the plane as a function of t, then solve the extreme value problem. 41. Consider a line through the origin. Any such line sweeps a circular cone when rotated about the z-axis and, for this reason, is called a generating line of a cone. Prove that the curve R(t) = (et cost, et sint, et ) intersects all generating lines of the cone x 2 + y 2 = z 2 at the same angle. Hint: Show that parametric equations of a line in the cone are x = s cos θ, y = s sin θ and z = s. Define the points of intersection of the line and the curve and find the angle at which they intersect. 42. The position R(t) of a particle moving in the plane is governed by the differential equation dR d2 R − 3R = 0. +2 dt 2 dt The initial conditions are R(0) = (5, 0) and V (0) = (1, 4), where V = d R/dt is the velocity of the particle. Solve for the motion of the particle and roughly plot its trajectory. In particular, describe the behavior of the trajectory for large positive t and for large negative t.

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2 Vector Functions and Parametric Equations

43. An epicycloid is the curve traced by a point P on the circumference of a circle of radius b which is rolling externally on a fixed circle of radius a. If the origin is at the center of the fixed circle, A = (a, 0) is one of the points at which the given point P comes in contact with the fixed circle, B is the moving point of tangency of the two circles and the parameter t is the radian measure of the angle AO B, prove that parametric equations of the epicycloid are x = (a + b) cos t − b cos

a+b a+b t and y = (a + b) sin t − b sin t. b b

44. A curve C in R3 has the following parametric equations in spherical coordinates: ρ = g1 (t), θ = g2 (t) and ϕ = g3 (t). Prove that if L is the length of arc C from the point where t = a to the point where t = b, then L=

b

(Dt ρ)2 + ρ2 sin2 ϕ(Dt θ)2 + ρ2 (Dt ϕ)2 dt.

(2.34)

a

45. A conical helix winds around a cone in a way similar to that in which a circular helix winds around a cylinder. Use the formula (2.34) to find the length of arc from t = 0 to t = 2π of the conical helix having parametric equations ρ = t, θ = t and ϕ = π/4.

Chapter 3

Functions of Several Variables, Limits and Continuity

3.1 Functions of Several Variables A real valued function of n variable is a function f : Rn → R. So, the domain D f of f is a subset of Rn and for each (x1 , . . . , xn ) ∈ D f , the value of f is a real number f (x1 , . . . , xn ). The set of all values taken by f at the points of D f is called the range of f . If f is defined by a formula, we usually take the domain D f to be as large as possible. When n = 1, we have a function of one variable; thus the domain is a subset of R. If n = 2, we have a function of two variables, and the domain is a set of points in R2 , for illustration see Fig. 3.1. If n = 3, we have a function of three variables, and the domain is a set of points in R3 . For example, if f is a function defined by f (x, y) = cos(x 2 + y 2 ) − 5 sin x + 3, we have a function of two variables defined for all (x, y) ∈ R2 . Hence, D f = R2 . However, if f is defined by f (x, y, z) =

1 x2

+ y2 + z2

,

then f is a function of three variables. The domain of this function is all (x, y, z) ∈ R3 except for (x, y, z) = (0, 0, 0). Likewise, a multi-variable function of n variables is a function f : Rn → Rm , where the domain D f of f is a subset of Rn . Hence, for each (x1 , . . . , xn ) ∈ D f , the value of f is a vector f (x1 , . . . , xn ) ∈ Rm . If f is a function of two variables, then the graph of f is the set {(x, y, z) ∈ R3 | z = f (x, y)}. So, the graph of a function f of two variables is a surface. Three-dimensional surfaces can be depicted in two dimensions by means of level curves or contour maps. By a level curve of a function f of two variables x and y, we mean the projection onto the x y-plane of the curve in which the graph of f intersects the horizontal plane z = c, © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_3

103

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3 Functions of Several Variables, Limits and Continuity

Fig. 3.1 Graph of a function of two variables

z

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

y (x0 , y0 ) x

Fig. 3.2 Contour map

where c is any constant in the range of f . Thus, the level curve corresponding to c has the equation f (x, y) = c, regarded as a curve in the x y-plane. By considering different values for the constant c, we obtain a set of level curves called a contour map, see Fig. 3.2. Level curves corresponding to different values cannot intersect. By a level surface of a function f of three variables x, y and z, we mean the graph of the equation f (x, y, z) = c, where c is any constant in the range of f . The element of Rn are called points, or vectors. If X = (x1 , . . . , xn ) and Y = (y1 , . . . , yn ) and if c is a real number, we define X + Y = (x1 + y1 , . . . , xn + yn ) and cX = (cx1 , . . . , cxn ). Hence, X + Y ∈ Rn and cX ∈ Rn . These two operators satisfy the commutative, associative and distributive laws. The point (0, . . . , 0) is the zero element. We also define the so-called inner product of X and Y by

3.2 Limits

105

X ·Y =

n

xi yi ,

i=1

and the norm of X by X = x12 + . . . + xn2 . Triangle inequality: The inequality X + Y ≤ X + Y holds for all X, Y ∈ Rn .

3.2 Limits If X = (x1 , . . . , xn ) and X 0 = (a1 , . . . , an ) are two points in Rn , then the distance between X and X 0 , denoted by X − X 0 , is given by X − X 0 =

(x1 − a1 )2 + . . . + (xn − an )2 .

(3.1)

If n = 1, X = x and X 0 = a, then (3.1) becomes x − a = (x − a)2 = |x − a|. For n = 2, 3, the formula (3.1) reduces to the formulas we have for distance on the line, in the plane and in the space. By a neighborhood N with the center X 0 and radius δ, we mean the set of all points X satisfying X − X 0 < δ. By a deleted neighborhood of a point X 0 , we mean any neighborhood of X 0 with the point X 0 itself cut. Let S be any set of points in Rn . Then a point X is called an interior point of S if there is some neighborhood of X containing only points in S. A point X is a limit point of S if every deleted neighborhood of X contains a point of S. A point P is called boundary point of S if every neighborhood of X contains both points in S and points not in S. These definitions are illustrated in Fig. 3.3 for the case where S is a set of pairs in the plane. A set S is said open if all its points are interior points, and closed if it contains all the boundary points. The set of all boundary points of S is called the boundary of S. Thus, a closed set contains its boundary, but an open set does not. Definition of limit: Let f be a function of several variables defined in a deleted neighborhood of a point X 0 . Then f (X ) is said to approach a limit L as X approaches X 0 if, for any > 0, we can find δ > 0 such that | f (X ) − L| < whenever 0 < X − X 0 < δ. This is expressed by writing lim f (X ) = L .

X →X 0

106

3 Functions of Several Variables, Limits and Continuity

Fig. 3.3 Interior and boundary points

A boundary point • S

An interior point •

Now, we state the definition of the limit of a function of two variables. It is the special case of the above definition where X 0 is the point (x0 , y0 ) and X is the point (x, y). Let f be a function of two variables that are defined in a deleted neighborhood of a pair (x0 , y0 ). Then the limit of f (x, y) as (x, y) approaches (x0 , y0 ) is L, written as f (x, y) = L , lim (x,y)→(x0 ,y0 )

if for any > 0, there exists δ > 0 such that | f (x, y) − L| < whenever 0 < (x − x0 )2 + (y − y0 )2 < δ. The limit of a function is unique if it exists. Properties of limits of functions of several variables: In the following, we list these properties for functions of two variables. Analogous properties hold for functions of more variables. Let L, M and c are real numbers and f, g are two variable functions, such that lim

(x,y)→(x0 ,y0 )

f (x, y) = L and

lim

(x,y)→(x0 ,y0 )

g(x, y) = M.

Then, the following statements hold: (1) We have (2) (3)

lim

(x,y)→(x0 ,y0 )

lim

lim

(x,y)→(x0 ,y0 )

(x,y)→(x0 ,y0 )

x = x0 ,

lim

(x,y)→(x0 ,y0 )

y = y0 and

f (x, y) ± g(x, y) = L ± M; f (x, y)g(x, y) = L M;

lim

(x,y)→(x0 ,y0 )

f (x, y) L = (if M = 0); (x,y)→(x0 ,y0 ) g(x, y) M (5) If h is a function of a single variable continuous at L, then (4)

lim

lim

(g ◦ f )(x, y) = g(L),

(x,y)→(x0 ,y0 )

c = c;

3.2 Limits

107

or equivalently lim

(x,y)→(x0 ,y0 )

g f (x, y) = g

lim

(x,y)→(x0 ,y0 )

f (x, y) .

Sandwich theorem: Let f, g and h be functions of two variables x and y such that f (x, y) ≤ g(x, y) ≤ h(x, y), for all (x, y) in some deleted neighborhood of (x0 , y0 ). f (x, y) = lim h(x, y)) = L, then If lim (x,y)→(x0 ,y0 )

(x,y)→(x0 ,y0 )

lim

(x,y)→(x0 ,y0 )

g(x, y) = L .

A point X 0 is said to be an accumulation point of a set S of points in Rn if every neighborhood of X 0 contains infinitely many points of S. Let f be a function defined on a set S of points in R2 , and let (x0 , y0 ) be an accumulation point of S. Then the limit of f (x, y) as (x, y) approaches (x0 , y0 ) in S is L, written as (3.2) lim f (x, y) = L , (x,y)→(x0 ,y0 ) (x,y)∈S

if for any > 0, there exists δ > 0 such that | f (x, y) − L| < whenever 0 < (x − x0 )2 + (y − y0 )2 < δ and (x, y) ∈ S. A special case of (3.2) occurs when S is a set of points on a curve containing (x0 , y0 ). In such cases, the limit in (3.2) becomes the limit of a function of one variable. Let the function f be defined for all points on a deleted neighborhood having its lim f (x, y) = L. Then if S is any set of points in R2 center at (x0 , y0 ) and (x,y)→(x0 ,y0 )

having (x0 , y0 ) as an accumulation point, then lim

(x,y)→(x0 ,y0 ) (x,y)∈S

f (x, y)

exists and always has the value L. The following is an immediate consequence of the above fact. If the function f has different limits as (x, y) approaches (x0 , y0 ) through two distinct sets of points having (x0 , y0 ) as an accumulation point, then f (x, y) does not exist. This is a very important consequence, one which lim

(x,y)→(x0 ,y0 )

makes computing limits for functions of several variables more difficult. Indeed, it does not even have to approach (x0 , y0 ) along a straight path as shown in Fig. 3.4. To show that a limit does not exist, it is enough to find two paths along which the limit are not equal. Infinite limits: Let f be a function of several variables. We say that f (X ) approaches ∞ as X approaches X 0 and we write lim f (X ) = ∞,

X →X 0

108

3 Functions of Several Variables, Limits and Continuity

Fig. 3.4 Possible paths through (x0 , y0 )

•

if X 0 is a limit point of D f and for any real number M there is δ > 0 such that f (X ) > M whenever 0 < X − X 0 < δ and X ∈ D f . We say that lim f (X ) = X →X 0

−∞ if lim (− f )(X ) = ∞. X →X 0

Limits as X → ∞: Let f be a function of several variables such that D f is bounded. We say that lim f (X ) = L , X →∞

if for any > 0, there exists a number N such that | f (X ) − L| < whenever X > N and X ∈ D f . We leave to reader to define lim f (X ) = ∞ and lim f (X ) = −∞. X →∞

X →∞

Let f : Rn → R be a function and ⊆ Rn . We say that f is bounded on if there exists a positive number M such that | f (X )| < M, for all X ∈ .

3.3 Continuity Continuity for functions of several variables is defined in the same way as for functions of a single variable. Thus, if f is a function of n variables and X 0 is a point in Rn , then we say f is continuous at X 0 if and only if the following conditions are satisfied: (1) f (X 0 ) exists; (2) lim f (X ) exists; X →X 0

(3) lim f (X ) = f (X 0 ). X →X 0

3.4 Solved Problems

109

If one or more of the above conditions fails to hold at point X 0 , then f is said to be discontinuous. So, a function f of two variables x and y is continuous at the point (x0 , y0 ) if and only if the following conditions satisfy: (1) f (x0 , y0 ) exists; f (x, y) exists; (2) lim (x,y)→(x0 ,y0 )

(3)

lim

(x,y)→(x0 ,y0 )

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

If a function f of two variables is discontinuous at the point (x0 , y0 ) but

lim

(x,y)→(x0 ,y0 )

f (x, y) exists, then f is said to have a removable discontinuity at (x0 , y0 ). If the discontinuity is not removable, it is called an essential discontinuity. If f and g are two functions that are continuous at the point (x0 , y0 ), then (1) f ± g is continuous at (x0 , y0 ); (2) f g is continuous at (x0 , y0 ); (3) f /g is continuous at (x0 , y0 ), provided that g(x0 , y0 ) = 0.

Since every polynomial p of two variables x and y is the sum of products of the continuous functions defined by f (x, y)x, g(x, y) = y and h(x, y) = c, where c is a real number, it follows that p is continuous at every point in R2 . Moreover, a rational function of two variables is continuous at every point in its domain. A function f of several variables is said to be continuous on an open set S in Rn if f is continuous in every point X in S. Uniform continuity: A function f : Rn → R is uniformly continuous on a subset S of its domain in Rn if for any > 0 there exists δ > 0 such that | f (X ) − f (Y )| < whenever X − Y < δ and X, Y ∈ S. We emphasis that δ must depend only on and S, and not the particular points X and Y . Every uniformly continuous function is continuous. Intermediate Value Theorem: Let f : Rn → R be a continuous function on a region S in Rn . Suppose that A and B are two points in S and f (A) < k < f (B). Then there exists C ∈ S such that f (C) = k.

3.4 Solved Problems 135. Let f be a function of two variables x and y. 2 (1) Find f (x, y) if f (x + y, √x − y) = x y + y . √ (2) Let f (x, y) = y + g( x − 1). Determine the functions f and g if f (x, y) = x when y = 1.

Solution. (1) Let x + y = u and x − y = v. Then, we have x = (u + v)/2 and y = (u − v)/2. Hence, we obtain f (u, v) =

u 2 − uv u + v u − v u − v 2 = · + . 2 2 2 2

110

3 Functions of Several Variables, Limits and Continuity

Now, it remains to name the variables u and v, x and y. Therefore, we obtain f (x, y) = (x 2 − x y)/2.√ √ √ (2) Since x = 1 + g( x − 1), it follows that g( x − 1) = x − 1. Now, if x − √ 1 = t, then g(t) = t 2 + 2t. This gives that f (x, y) = y + x − 1. 136. Let f (x, y) =

1−

y2 x2 − . 4 25

Find the domain and range of f . Solution. Because of the square root, the domain of f is all the pairs (x, y) such that x2 y2 x2 y2 1− − ≥ 0, or + ≤ 1. This describes an ellipse and its interior. We 4 25 4 25 can represent the domain of f in set notation as follows:

y2 x2 D f = (x, y) ∈ R2 | + ≤1 . 4 25 The range is the set of all positive output values. The square root ensures that all output is positive. Since the x and y terms are squared, then subtracted, inside the square root, the largest output value comes at x = 0 and y = 0, i.e., f (0, 0) = 1. Therefore, the range of f is the interval [0, 1]. 137.

Find the domain and range of the function f (x, y) =

1 4x 2

− y2

.

Solution. A point (x, y) lies in the domain of f if and only if 4x 2 − y 2 > 0. This happens if and only if y 2 < 4x 2 , or equivalently −2|x| < y < 2|x|. Figure 3.5 is a sketch showing a shaded region in R2 the set points in the domain of f . It consists of all points of the x y-plane that lie between the graph of y = −2|x| and y = 2|x|. On this set, 4x 2 − y 2 takes on all positive values, and hence does its f (x, y). Therefore, the range of f is (0, ∞). 138.

Find the domain and range of the function f (x, y) = sin−1

x + y √ + x y. 2

Solution. The first term of the function f is defined for |(x + y)/2| ≤ 1, or −2 ≤ x + y ≤ 2. The second term has real values if x y ≥ 0. So, we have two cases: (x ≥ 0

3.4 Solved Problems

111

Fig. 3.5 The domain of f (x, y) = 1/ 4x 2 − y 2

y = 2|x| y

x

y = −2|x|

Fig. 3.6 The domain of f (x, y) = √ sin−1 x+y + xy 2

y

x+y =2 x x + y = −2

and y ≥ 0) or (x ≤ 0 and y ≤ 0). In Fig. 3.6, we have a sketch showing a shaded region in R2 the set of points in the domain of f . The domain includes the boundary of the region too. The range of f is (−1, ∞). 139. Find the domain of the function f (x, y) = ln(12 − x 2 − y 2 ) + 3y − x 2 and sketch it in the coordinate plane. Solution. The function is defined if the following two conditions hold:

112

3 Functions of Several Variables, Limits and Continuity

Fig. 3.7 The domain of f (x, y)= ln(12 − x 2 − y 2 ) + 3y − x 2

y=

1 2 x 3

y

√ 2 3

x √ 2 3

12 − x 2 − y 2 > 0 and 3y − x 2 ≥ 0. The first condition yields that x 2 + y 2 < 12 and the second y ≥ (1/3)x 2 . The first condition holds√for the points in x y-plane which belongs to the disk centered at 90, 0) with a radius 2 3. Since we have no equality, it follows that the circle surrounding the disk does not belong to the set and we sketch the circle with a dashed line. The second condition holds for the points in x y-plane, which are above the parabola y = (1/3)x 2 . This condition contains equality, and so the parabola belongs to the set of domain and we sketch it with a continuous line. Finally, we conclude that the domain of f is the blue part of Fig. 3.7. 140. Let f (x, y) = |x|. (1) Draw a sketch of the graph of f ; (2) Draw a sketch of a contour map of f showing the level curves of f at 0, 1, 2 and 3. Solution. (1) The surface f (x, y) = |x| intersects the x z-plane along z = |x|. Since the function f does not depend on y, it follows that the intersection of the graph of f with any plane parallel to the x z-plane is a copy of z = |x|. Therefore, the graph of f is obtained by moving the graph of z = |x| along the y-axis, see Fig. 3.8. (2) Since f (x, y) = |x| = c, it follows that there are no level curves corresponding to the negative value of c. If c = 0, then x = 0. So, the level curve of value c = 0 is the y-axis. Let c be positive. From |x| = c, we conclude that x = ±c. This means that the level curve of value c consists of a pair of vertical lines x = ±c, see Fig. 3.9.

3.4 Solved Problems

113

Fig. 3.8 The graph of f (x, y) = |x|

c=3 c=2 c=1 c=0 c=1 c=2 c=3 y

x −3

−2

−1

0

1

2

3

Fig. 3.9 A contour map of f (x, y) = |x|

141.

Find

lim

(x,y)→(0,0)

5x 2 y if it exists. x 2 + y2

Solution. We observe that along the line x = 0, the function has value 0 when y = 0. Likewise, along the line y = 0, the function has value 0 provided x = 0. Hence, if the limit does exist as (x, y) approaches (0, 0), the value of limit must be 0. To prove the limit as 0, we apply the definition. Let > 0 be given, but arbitrary. We want to find δ > 0 such that | f (x, y) − 0| < whenever (x − 0)2 + (y − 0)2 < δ or 5x 2 |y| < whenever 0 < x 2 + y 2 < δ. 2 2 x +y Since x 2 ≤ x 2 + y 2 , it follows that 5x 2 |y| ≤ 5|y| = 5 y 2 ≤ 5 x 2 + y 2 . 2 2 x +y

114

3 Functions of Several Variables, Limits and Continuity

Hence, if we take δ = /5 and let 0

0, there exists δ > 0 such that |2x y − 3y 2 − 1| < whenever 0

0, we take δ = min{1, /9}, then from (3.3) and (3.4), |2x y − 3y 2 − 1| < 9δ ≤ whenever 0

0, there exists δ > 0 such that |x 2 + y 2 − 2x + 3y − 3| < whenever 0

0, it is enough to take δ = min{1, /7}. Then by (3.5) and (3.6), |x 2 + y 2 − 2x + 3y − 3| < 7δ ≤ whenever 0

1 + c2 1 · . 1−c

√ 1 + c2 1 Now, it is enough to take N = · . 1−c

(3.8)

116

145.

3 Functions of Several Variables, Limits and Continuity

Evaluate the following limit x 3 + x y + y2 . (x,y)→(0,0) 2 − x 3 + x y + y2 + 4 lim

Solution. It is worthwhile to multiply the numerator and denominator by the conjugate 2+

x 3 + x y + y 2 + 4.

So, the given limit can be written as follows: 2 + x 3 + x y + y2 + 4 x 3 + x y + y2 · lim (x,y)→(0,0) 2 − x 3 + x y + y 2 + 4 2 + x 3 + x y + y 2 + 4 (x 3 + x y + y 2 )(2 + x 3 + x y + y 2 + 4) = lim 3 2 (x,y)→(0,0) −(x + x y + y ) 3 2 = lim −(2 + x + x y + y + 4) = −4. (x,y)→(0,0)

146. Find

lim

(x,y)→(0,0)

x 3 + y3 if it exists. x 2 + y2

Solution. To evaluate this limit, we use polar coordinates. Suppose that x = r cos θ and y = r sin θ with r ≥ 0. Then, we have x 2 + y 2 = r 2 and x 3 + y 3 = r 3 (cos3 θ + sin3 θ ). Now, (x, y) → (0, 0) is equivalent to r → 0+ . Therefore, we obtain lim

(x,y)→(0,0)

147. Find

x 3 + y3 r 3 (cos3 θ + sin3 θ ) = lim = lim+ r (cos3 θ + sin3 θ ) = 0. r →0 x 2 + y 2 r →0+ r2 lim

(x,y,z)→(0,0,0)

x 3 + y3 + z3 if it exists. x 2 + y2 + z2

Solution. To compute this limit, we use spherical coordinates. Let x = ρ sin φ cos θ , y = ρ sin φ sin θ and z = ρ cos φ. Then, we can write lim

(x,y,z)→(0,0,0)

x 3 + y3 + z3 ρ 3 sin3 φ cos3 θ + ρ 3 sin3 φ sin3 θ + ρ 3 cos3 φ = lim 2 2 2 ρ→0 x +y +z ρ2 = lim ρ sin3 φ cos3 θ + sin3 φ sin3 θ + cos3 φ = 0. ρ→0

148. Let f (x, y) =

(x − 1)2 ln x . Find lim f (x, y) if it exists. (x,y)→(1,0) (x − 1)2 + y 2

Solution. It is easy to check that the limit along the paths x = 1, y = 0 and y = x − 1 is 0. Hence, we guess the limit we want, which might be 0. Now, we use the Sandwich theorem to prove it. In other words, we need a function g such that | f (x, y)| ≤ g(x, y)

3.4 Solved Problems

and

lim

(x,y)→(1,0)

117

g(x, y) = 0. In order to find g, we can write

(x − 1)2 ln x (x − 1)2 | ln x| ≤ | ln x|. | f (x, y)| = = (x − 1)2 + y 2 (x − 1)2 + y 2 Since | ln x| → 0 as (x, y) → (1, 0), it follows by Sandwich theorem that lim f (x, y) = 0.

(x,y)→(1,0)

149. Show that

lim

(x,y)→(0,0)

f (x, y) = 0 where f (x, y) =

x 3 y − 3x 2 y 2 . x 2 + y2 + x 4

Solution. We have x 3 y − 3x 2 y 2 |x 3 ||y| + 3x 2 y 2 4(x 2 + y 2 )2 ≤ ≤ 2 x + y2 + x 4 x 2 + y2 + x 4 (x 2 + y 2 ) + x 4 4(x 2 + y 2 ) = ≤ 4(x 2 + y 2 ). x4 1+ 2 x + y2 This implies that −4(x 2 + y 2 ) ≤ f (x, y) ≤ 4(x 2 + y 2 ). Since lim

(x,y)→(0,0)

by Sandwich theorem, it follows that 150.

Find

lim

(x,y)→(0,0)

±4(x 2 + y 2 ) = 0, lim

(x,y)→(0,0)

f (x, y) = 0.

cos(x y) − 1 . x 2 y2

Solution. The function in problem is g(t) = (cos t − 1)/t 2 for t = 0, where the argument t is replaced by the function h(x, y) = x y. Since the function h is a polynomial, it is continuous, and we have lim h(x, y) = h(0, 0) = 0. The function g is con(x,y)→(0,0)

tinuous for all t = 0, and its value at t = 0 is not defined. By using L’Hospital’s rule, we have 1 cos t − 1 − sin t =− . = lim lim 2 t→0 t→0 t 2t 2 Hence, the discontinuity of g is removable. By taking g(0) = −1/2, the function g becomes continuous at t = 0 and the conditions of the composition rule are satisfied. Consequently, the desired limit in the problem exists and equals −1/2.

118

151.

3 Functions of Several Variables, Limits and Continuity

Compute

lim

(x,y)→(0,0)

f (x, y) if

f (x, y) = (x 2 + y 2 ) sin

1 . xy

Solution. We change the variables to polar coordinates, i.e., x = r cos θ and y = r sin θ . Then, we have f (x, y) = f (r, θ ) = r 2 sin

r2

1 . cos θ sin θ

Since lim r 2 = 0 and | sin 1/(r 2 cos θ sin θ | ≤ 1, it follows that r →0

lim

(x,y)→(0,0)

152.

Show that

f (x, y) = lim r 2 sin r →0

1 = 0. r 2 cos θ sin θ

sin(x y) does not exist. (x,y)→(0,0) x + y lim

Solution. Evaluating this limit along the lines y = mx means replace all y’s with mx and evaluating the result limit. So, we have 1 sin(x · mx) sin(mx 2 ) sin(mx 2 ) = lim = lim · . x→0 x(m + 1) x→0 (x,mx)→(0,0) x + mx x m+1 lim

By applying L’Hospital’s rule, we observe that this limit is 0 except when m = −1, i.e., along the line y = −1. Note that this line is not in the domain of f . Now, we evaluate the limit along the curve y = − sin x. We have lim

(x,− sin x)→(0,0)

sin(−x sin x) sin(−x sin x) = lim . x→0 x − sin x x − sin x

By applying L’Hospital’s rule two times, we obtain sin(−x sin x) cos(−x sin x)(− sin x) − x cos x = lim x→0 x − sin x 1 − cos x − sin(−x sin x)(− sin x − x cos x)2 + cos(−x sin x)(−2 cos x + x sin x) = lim sin x x→0 . = −2 0 lim

x→0

Therefore, along any line y = mx in the domain of f , the limit is 0, while along the curve y = − sin x, which lies in the domain of f , for all x = 0, the limit does not exist. Since the limit is not the same along every curve passing through (0, 0), we conclude that the limit does not exist.

3.4 Solved Problems

119

153. Determine all values of the constant c > 0 for which the limit lim

(x,y)→(0,0)

|x| (x 2 + y 2 )c

exists. Solution. We consider three cases: c < 1/2, c = 1/2 and c > 1/2. Let c < 1/2. Since 0 ≤ x 2 ≤ x 2 + y 2 , it follows that 0≤

x 2 c |x| 1−2c ≤ |x| ≤ |x|1−2c , (x 2 + y 2 )c x 2 + y2

for all (x, y) = (0, 0). Since 1 − 2c > 0, it follows that

lim

(x,y)→(0,0)

|x|1−2c = 0. So,

|x| = 0 for c > 1/2. + y 2 )c For the case c = 1/2, suppose that S1 is the set of all points on the x-axes and S2 is the set of all points on the y-axes. Then, we have by Sandwich theorem, we conclude that

lim

(x,y)→(0,0) (x,y)∈S1

lim

(x,y)→(0,0) (x,y)∈S2

lim

(x,y)→(0,0) (x 2

|x| |x| |x| = lim 2 = lim = 1, 2 c 2 1/2 x→0 (x + 0 ) x→0 |x| +y ) |x| 0 = lim 2 = 0. 2 2 c y→0 (0 + y 2 )1/2 (x + y ) (x 2

Since these limits are different, it follows that the requested limit does not exist when c = 1/2. Now, let c > 1/2. Then, we have lim

(x,y)→(0,0) y=0

|x| |x| = lim 2 = lim |x|1−2c = ∞. x→0 (x + 02 )c x→0 (x 2 + y 2 )c

This yields that the requested limit does not exist for c > 1/2. 154.

Determine all values of the constant c > 0 for which the limit lim

x 2 y3 + |y|c

(x,y)→(0,0) |x|3

exists. Solution. Let c ≥ 0 and S be the set of all points on the path x = y 3 . Then we have lim

(x,y)→(0,0) (x,y)∈S

x 2 y3 y9 = lim 3 3 . c y→0 |y | + |y|c + |y|

|x|3

120

3 Functions of Several Variables, Limits and Continuity

To compute the second limit, we consider the left and right limits. So, we obtain y9 y9 1 = lim = lim+ lim+ 3 3 c 9 c + y→0 y→0 y→0 y +y 1 + y c−9

|y | + |y| 1 if c > 9 = 1/2 if c = 9 and lim−

y→0

y9 y9 1 = lim = lim− 3 3 c 9 c c − y→0 −y + (−1) y y→0 −1 + (−1)c y c−9 |y | + |y|

−1 if c > 9 = −1/2 if c = 9.

x 2 y3 does not exists for c ≥ 9. + |y|c Now, we suppose that c < 9. We can write

Therefore,

lim

(x,y)→(0,0) |x|3

0≤

a2 x 2 y 3 = |y|3−c/3 , |x|3 + |y|c a3 + 1

a2 |x| < 1. If . If 0 < a < 1, then a 2 < 1, which implies that 3 c/3 |y| a +1 a2 < 1. Consequently, we get a ≥ 1, then a 2 ≤ a 3 + 1, and hence 3 a +1 where a =

0≤ Finally, since lim

(x,y)→(0,0)

155. If

lim

(x,y)→(0,0) 2 3

x 2 y 3 ≤ |y|3−c/3 . |x|3 + |y|c

|y|3−c/3 = 0, by Sandwich theorem, we conclude that

x y = 0 for c < 9. |x|3 + |y|c lim

(x,y)→(x0 ,y0 )

f (x, y) = L and if the one dimensional limits lim f (x, y) and lim f (x, y)

x→x0

y→y0

both exist, prove that lim

x→x0

lim f (x, y) = lim lim f (x, y) = L .

y→y0

y→y0

x→x0

(3.9)

3.4 Solved Problems

121

The two limits in (3.9) are called iterated limits. This problem shows that the existence of the two-dimensional limit and of the two one- dimensional limits implies the existence and equality of the two iterated limits. Solution. Since

lim

(x,y)→(x0 ,y0 )

f (x, y) = L, it follows that for any > 0 there exists

δ1 > 0 such that (x − x0 )2 + (y − y0 )2 < δ1 ⇒ | f (x, y) − L| < /2. Let lim f (x, y) = gx0 (y). Then for any > 0, there exists δ2 > 0 such that x→x0

|x − x0 | < δ2 ⇒ | f (x, y) − gx0 (y)| < /2. Now, if we take δ = min{δ1 , δ2 }, then when |x − x0 | < δ and |y − y0 | < δ, we have | f (x, y) − L| < /2 and | f (x, y) − gx0 (y)| < /2. Applying the triangle inequality, we can write |gx0 (y) − L| = |gx0 (y) − f (x, y) + f (x, y) − L| ≤ |gx0 (y) − f (x, y)| + | f (x, y) − L| < /2 + /2 = . Therefore, we proved that for any > 0 there exists δ > 0 such that |gx0 (y) − L| < whenever |y − y0 | < δ. This means that lim gx0 (y) = L, or equivalently y→y0 lim lim f (x, y) = L. y→y0

x→x0

Now, by repeating the above argument switching x and y, we obtain lim

x→x0

lim f (x, y) = L .

y→y0

156. (1) Give an example of a function f of two variables such that lim

x→x0

lim f (x, y) = lim lim f (x, y) .

y→y0

y→y0

x→x0

(2) Use the result of part (1) and Problem 155 to deduce that f (x, y) does not tend to a limit as (x, y) → (x0 , y0 ). x−y , then Solution. (1) If we define f (x, y) = x+y

x lim f (x, y) = lim = 1, x→0 y→0 x→0 x −y lim lim f (x, y) = lim = −1. y→0 x→0 y→0 y lim

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3 Functions of Several Variables, Limits and Continuity

(2) It is straightforward. 157.

Show that the converse of Problem 155 is not always true.

Solution. We define f (x, y) =

x 2 y2

x 2 y2 . + (x − y)2

Then it is easy to see that lim

x→0

lim f (x, y) = lim lim f (x, y) = 0,

y→0

y→0

x→0

but that does not need to a limit as (x, y) → (0, 0); it is enough to examine f on the line y = x. 158. Give an example of a function f of two variables such that f (x, y) does tend to a limit as (x, y) → (x0 , y0 ) but the iterated limit does not exist. Solution. We consider the function f : R2 → R defined by 1 1 cos . f (x, y) = x sin x x Then we have | f (x, y)| ≤ |x| and we have sin

1 x

lim

(x,y)→(0,0)

f (x, y) = 0. Now, for a given x = 0,

= 0. It is clear that the limit 1 1 cos lim x sin y→0 x x

does not exist. Consequently, we have f (x, y) exists, but 1 1 cos does not exist. (2) the iterated limit lim lim x sin x→0 y→0 x x (1)

lim

(x,y)→(0,0)

Note that this example does not contradict Problem 155. 159. Let a, b, c ∈ R with a < c < b. If f is a monotone function from (a, b) to R, prove that lim+ f (x) and lim− f (x) exist in R ∪ {∞, −∞}. In particular, lim+ f (x) x→a

and lim− f (x) exist in R.

x→b

x→c

x→c

Solution. Without loss of generality, assume that f is increasing. Let S = { f (x) | x ∈ (a, b)}. We consider the following two cases: Case 1: S is bounded below. Case 2: S is not bounded below. In the case (1), S is bounded below and by the completeness of R, the set S has the greatest lower bound in R. We claim that

3.4 Solved Problems

123

lim f (x) = .

x→a +

Let > 0 be given. Since + > , the number + is not a lower bound for S. Hence, there exists a + δ ∈ (a, b) such that f (a + δ) < + . Since f is increasing, it follows that ≤ f (x) ≤ f (a + δ) < + , for all x ∈ (a, a + δ). This yields that f (x) → as x → a + . In the case (2), if S is not bounded below, then for each real number α there exists xα ∈ (a, b) such that f (xα ) < α. Since f is increasing, it follows that f (x) ≤ f (xα ) < α, for all x ∈ (a, xα ). This means that f (x) → −∞ as x → a + . Similarly, if S is bounded above and u is the least of upper bound of S, then lim f (x) = u,

x→b−

and if S is not bounded above, then lim f (x) = ∞.

x→b−

160.

A function f : R2 → R has the property that (x1 ≤ x2 and y1 ≤ y2 ) ⇒ f (x1 , y1 ) ≤ f (x2 , y2 ).

Prove that lim+ x→x0

lim+ f (x, y) = lim+ lim+ f (x, y) .

y→y0

y→y0

x→x0

Solution. Since f is increasing in each variable, by Problem 159, we conclude that lim f (x, y) and

x→x0+

lim f (x, y)

y→y0+

exist for all x, y ∈ R. Suppose that g(x) = lim+ f (x, y) and h(y) = lim+ f (x, y), y→y0

x→x0

for all x, y ∈ R. Furthermore, it is straightforward to see that the functions f and g are also increasing. So, again by Problem 159, the limits lim g(x) and

x→x0+

lim h(y)

y→y0+

exists. Let A := lim+ g(x) and B := lim+ h(y). We must prove that A = B. x→x0

y→y0

If y0 ≤ b ≤ y are arbitrary, then by hypothesis f (x, b) ≤ f (x, y), for all real number x. So, we obtain the following implications:

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3 Functions of Several Variables, Limits and Continuity

f (x, u) ≤ f (x, y) ⇒ g(x) = lim+ f (x, u) ≤ f (x, y) u→y0

⇒ A = lim+ g(x) ≤ lim+ f (x, y) = h(y) x→x0

x→x0

⇒ A ≤ lim+ h(y) = B. y→y0

Analogously, if x0 ≤ v ≤ x are arbitrary, then by hypothesis f (v, y) ≤ f (x, y), for all real number y. Next, we have the following implications: f (v, y) ≤ f (x, y) ⇒ h(y) = lim+ f (v, y) ≤ f (x, y) v→x0

⇒ B = lim+ h(y) ≤ lim+ f (x, y) = g(x) y→y0

y→y0

⇒ B ≤ lim+ g(x) = A. x→x0

Therefore, A = B, as desired. 161.

Let f (x, y) =

x 2 − y2 . x 2 + y2

(1) If 0 < < 1, prove that | f (x, y)| < if and only if y 2 1− 1+ < < . 1+ x 1− (2) Prove that f (x, y) → 0 provided that x → 0 and y → 0 in such a manner that 1 + |y| 1 − |x| y 2 < . < 1 + |y| x 1 − |x|

(3.10)

Solution. We can write x 2 − y2 x. Therefore, for any > 0, | f (x, y) − f (x0 , y0 )| = 1 − 1 = 0 < whenever This yields that

lim

(x,y)→(0,0)

(x − x0 )2 + (y − y0 )2 < δ.

f (x, y) = f (x0 , y0 ) = 1.

Case 2: If y0 < x0 , then a similar way can be applied to see the continuity of f at (x0 , y0 ). Hence, the function f is continuous at every point (x0 , y0 ) if y0 = x0 . Now, we consider the points on the line y = x. Let (x0 , x0 ) be a point on the line y = x. We have f (x0 , x0 ) = 1. An arbitrary neighborhood with the center (x0 , y0 ) divides into two parts by the line y = x. In the first part (y ≥ x), we have f (x, y) = 1, and in the second part (y < x), we have f (x, y) = 0. Consequently, for 0 < < 1, there is no neighborhood of radius δ > 0 in which | f (x, y) − f (x0 , x0 )| = | f (x, y) − 1| < , because | f (x0 , y0 ) − 1| = 1 for y < x in any such neighborhood. Therefore, the function is discontinuous along the line y = x in its domain. 167.

Suppose that the function f is defined by

f (x, y) =

x 2 + y2 i f x 2 + y2 ≤ 1 0 i f x 2 + y 2 > 1.

Discuss the continuity of f . Solution. Since f is defined at all points in R2 , it follows that f (x0 , y0 ) exists for every point (x0 , y0 ) ∈ R2 . Let (x0 , y0 ) be a point such that x02 + y02 = 1. If x02 + y02 < 1, then lim

(x,y)→(x0 ,y0 )

f (x, y) =

lim

(x 2 + y 2 ) = x02 + y02 = f (x0 , y0 ).

(x,y)→(x0 ,y0 )

If x02 + y02 > 1, then lim

(x,y)→(x0 ,y0 )

f (x, y) =

lim

(x,y)→(x0 ,y0 )

0 = 0 = f (x0 , y0 ).

128

3 Functions of Several Variables, Limits and Continuity

Therefore, we conclude that f is continuous at all points (x0 , y0 ) for which x02 + = 1. Now, we determine the continuity of f at points (x0 , y0 ) for which x02 + y02 = 1. Let S1 = {(x, y) | x 2 + y 2 ≤ 1} and S2 = {(x, y) | x 2 + y 2 > 1}. Then we have y02

lim

f (x, y) =

lim

f (x, y) =

(x,y)→(x0 ,y0 ) (x,y)∈S1 (x,y)→(x0 ,y0 ) (x,y)∈S2

This yields that

lim

(x,y)→(x0 ,y0 )

lim

(x 2 + y 2 ) = x02 + y02 = 1,

lim

0 = 0.

(x,y)→(x0 ,y0 ) (x,y)∈S1 (x,y)→(x0 ,y0 ) (x,y)∈S2

f (x, y) does not exist. Therefore, f is discontinuous at

all points (x0 , y0 ) for which x02 + y02 = 1. ⎧ n m ⎨ x y if (x, y) = (0, 0) f (x, y) = x 4 + y 6 ⎩0 if (x, y) = (0, 0),

168. Let

where m, n are non-negative integers. In each of the following items, determine whether there exist values of m and n for which f satisfies the given condition: (1) f (x, y) continuous at (0, 0); (2) f (x, y) → 1 as (x, y) approaches (0, 0) along the line y = x, and f (x, y) → −1 as (x, y) approaches (0, 0) along the line y = −x; (3) f (x, y) → 0 as (x, y) approaches (0, 0) along any line through the origin, and lim (x, y) → (0, 0 f (x, y) does not exist; (4) f (x, y) → 0 as (x, y) approaches (0, 0) along any line through the origin except the y-axis, and f (x, y) → 1 as (x, y) approaches (0, 0) along the y-axis. Solution. (1) If m = 4 and n = 1, then f is continuous at (0, 0). This follows by Sandwich theorem as x4 y x4 ≤ |y| ≤ |y|, 0 ≤ | f (x, y)| = 4 x + y6 x 4 + y6 for (x, y) = (0, 0). (2) If m = 3 and n = 1, then lim

(x,y)→(0,0) y=x

lim

(x,y)→(0,0) y=−x

x4 1 = lim = 1, x→0 x 4 + x 6 x→0 1 + x 2 −x 4 −1 f (x, y) = lim 4 = lim = −1. x→0 x + x 6 x→0 1 + x 2 f (x, y) = lim

3.4 Solved Problems

129

(3) If m = 2 and n = 3, then lim

(x,y)→(0,0) y=ax

lim

(x,y)→(0,0) x=0

a3 x 5 a3 x = lim = 0, x→0 x 4 + a 6 x 6 x→0 1 + a 6 x 2 f (x, y) = lim f (0, y) = 0, f (x, y) = lim

y→0

while lim

(x,y)→(0,0) y=x 2/3

and so

lim

(x,y)→(0,0)

f (x, y) = lim

x→0

x4 1 = = 0, 4 4 x +x 2

f (x, y) does not exist.

(4) If m = 0 and n = 6, then we have lim

(x,y)→(0,0) y=ax

lim

(x,y)→(0,0) x=0

a6 x 6 a6 x 2 = lim = 0, x→0 x 4 + a 6 x 6 x→0 1 + a 6 x 2 y6 f (x, y) = lim 6 = 1. y→0 y f (x, y) = lim

169. Suppose that the function f : Rn → R is defined by f (X ) = X , for all X ∈ Rn . Prove that f is a continuous function. Solution. By the triangle inequality, we obtain X − A ≤ X − A. Thus, given any > 0, we need only to choose δ = to make X − A < whenever 0 < X − A < δ. 170. Let f be a function defined in Problem 169. If X, Y belong to Rn , prove that f (X + Y ) = f (X ) + f (Y ) if and only if X = cY or Y = cX with c ≥ 0. Solution. Let X = cY with c ≥ 0. Then we have f (X + Y ) = X + Y = cY + Y = (c + 1)Y = cY + Y = X + Y = f (X ) + f (Y ). Conversely, suppose that f (X + Y ) = f (X ) + f (Y ). Then we have X + Y 2 = X 2 + Y 2 + 2X Y .

(3.12)

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3 Functions of Several Variables, Limits and Continuity

On the other hand, we can write X + Y 2 = (X + Y ) · (X + Y ) = X · X + X · Y + Y · X + Y · Y = X 2 + Y 2 + 2X · Y.

(3.13)

By (3.12) and (3.13), we conclude that X · Y = X Y . If Y = 0, we are done. So, assume that Y = 0. Then we obtain

Y X − X Y · Y X − X Y = Y 2 X 2 − 2X Y (X · Y ) + Y 2 X 2 = 0.

This implies that Y X − X Y = 0, or equivalently X=

X Y. Y

Now, if we take c = X /Y , then X = cY with c ≥ 0. 171.

If X, Y ∈ Rn , prove that X + Y 2 + X − Y 2 = 2X 2 + 2Y 2 .

Interpret this geometrically, as a statement about parallelograms. Solution. We have X + Y 2 + X − Y 2 = X 2 + 2X · Y + Y 2 + X 2 − 2X · Y + Y 2 = 2X 2 + 2Y 2 . The sum of the squares on the diagonals of a parallelogram equals the sum of the squares on the sides, for illustration, see Fig. 3.10.

Fig. 3.10 Geometrically interpretation of Problem 171

X − Y X

X + Y

Y

3.4 Solved Problems

172.

131

Suppose that A, B ∈ Rn . Find C ∈ Rn and δ > 0 such that X − A = 2X − B ⇔ X − C = δ.

Solution. We have X − A = 2X − B ⇔ X − A2 = 4X − B2 ⇔ X 2 − 2X · A + A2 = 4X 2 − 8X · B + 4B2 ⇔ A2 − 4B2 = 3X 2 + 2X · (A − 4B) A2 − 4B2 2 1 1 ⇔ + A − 4B2 = X 2 + X · (A − 4B) + A − 4B2 3 3 9 2 9 1 4 ⇔ B − A2 = X − (4B − A) 9 3 1 2 ⇔ B − A = X − (4B − A). 3 3 Now, it is enough to take C = 173.

1 2 (4B − A) and δ = B − A. 3 3

Let n ≥ 3, X, Y ∈ Rn , X − Y = δ > 0, and k > 0. Prove that

(1) If 2k > δ, then there exist infinitely many Z ∈ Rn such that Z − X = Z − Y = k; (2) If 2k = δ, then there exists exactly one such Z ; (3) If 2k < δ, then there is no such Z . Solution. Suppose that B = X − (X + Y )/2 and let A ∈ Rn such that A − B = A + B = k. So, we have A − B2 = A + B2 . This implies that A · B = 0. Hence, we obtain k 2 = A − B2 = A2 + B2 or A2 = k 2 − B2 . Now, we can write X +Y 1 1 B = X − = X − Y = δ, 2 2 2 which implies that 2B > δ > 0, and so B = 0. If B = (b1 , . . . , bn ), then we define the vectors U and V in the following two cases. Case 1: If there exists 1 ≤ i ≤ n such that bi = 0, then we take U = ei . Also, if there exists j = i and 1 ≤ j ≤ n such that b j = 0, then we take V = e j , and if b j = 0 for all j = i, we consider two distinct numbers 0 ≤ j, r ≥ n such that j, r = i. Then we take V = br e j − b j er . Case 2: If bi = 0 for all 1 ≤ i ≤ n, then we take U = b2 e1 − b1 e2 and V = b3 e1 − b1 e3 . Next, if we define An = U + nV = 0, then An · B = 0, for all n ∈ N, An = c Am , for all m, n ∈ N, m = n and c ∈ R.

132

3 Functions of Several Variables, Limits and Continuity

Now, we prove the statements (1), (2) and (3). (1) If 2k > δ, then k 2 > δ 2 /4 or k 2 − δ 2 /4 > 0. This means that k 2 − B2 > 0. For each positive integer n, we define Cn =

An 2 k − B2 . An

since Cn · B = 0 and Cn 2 = k 2 − B2 , it follows that Cn − B = Cn + B = k. Finally, for each positive integer n, we set Z n = Cn + (X + Y )/2. This gives that Z n − X = Z n − Y = k. (2) By Problem 170, we know that the equality can be held in the triangle inequality if one of the two vectors is a scalar multiple of the other. In addition, the scalar must be non-negative. Now, by the condition 2k = δ, we have X − Y = δ = X − Z + Z − Y . So, there exists c > 0 such that X − Z = c(Z − Y ). Then, by the hypothesis, we deduce that c = 1. Therefore, Z is uniquely determined by Z = (X + Y )/2. (3) Let 2k < δ. If Z satisfies the condition, then by the triangle inequality, we have X − Y = δ > 2k = X − Z + Z − Y , a contradiction. 174. Let f : Rn → R be a function. We say that f satisfies Lipschitz condition if there exists a positive number M such that | f (X ) − f (Y )| < MX − Y , for all X, Y ∈ Rn . If F satisfied the Lipschitz condition, prove that f is uniformly continuous. Solution. We can write −MX − Y < f (X ) − f (Y ) < MX − Y . By Problem 169, we obtain lim MX − Y = 0. Thus, we conclude that X →y lim f (X ) − f (Y ) = 0, or equivalently X →Y

lim f (X ) = f (Y ).

X →Y

Since X and Y are arbitrary, it follows that f is uniformly continuous.

3.4 Solved Problems

175.

133

Let f : Rn → R be a function. We say that f is a linear map if f (cX + Y ) = c f (X ) + f (Y ),

for all X, Y ∈ Rn and c ∈ R. If f is a linear map, prove that the following statements are equivalent: (1) f is continuous at the point X 0 ∈ Rn ; (2) f is bounded on the set = {X ∈ Rn | X = 1}; (3) f is uniformly continuous on Rn . Solution. (1⇒2): Suppose that f is continuous at X 0 ∈ Rn . So, for any > 0 there exists δ > 0 such that | f (X ) − f (X 0 )| < whenever X − X 0 < δ. Since f is linear, we have f (X ) − f (X 0 ) = f (X − X 0 ). Hence, we can say that for any > 0, there exists δ > 0 such that Z < δ ⇒ | f (Z )| < . Now, let Z ∈ Rn such that Z = 1. Consider X = δ Z . Then we have X = δZ = δ. This implies that | f (X )| = | f (δ Z )| = |δ f (Z )| = δ| f (Z )| < , and so | f (Z )| < /δ. Finally, if take = 1, we are done. (2⇒3): Suppose that X ∈ Rn is arbitrary. If X = 0, then X > 0. Let Z = X/X . Then Z = 1, and so | f (Z )| < M.

(3.14)

X 1 | f (X )|. | f (Z )| = f = X X

(3.15)

On the other hand, we have

By (3.14) and (3.15), we get | f (X )| < MX .

(3.16)

Note that (11S3) holds for X = 0 too. Now, by Problem 174, we deduce that f is uniformly continuous on Rn . (3⇒1): It is clear. 176.

If f : Rn → R is a linear map, prove that f is a continuous function.

Solution. Let X = (x1 , . . . , xn ) ∈ Rn such that X = 1. So, |xi | ≤ 1 for all i = 1, . . . , n. Now, we can write X = x1 (1, 0, . . . , 0) + . . . + xn (0, . . . , 0, 1). Then we obtain

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3 Functions of Several Variables, Limits and Continuity

f (X ) = f x1 (1, 0, . . . , 0) + . . . + xn (0, . . . , 0, 1) = x1 f (1, 0, . . . , 0) + . . . + xn f (0, . . . , 0, 1) ≤ x1 f (1, 0, . . . , 0) + . . . + xn f (0, . . . , 0, 1) = |x1 | f (1, 0, . . . , 0) + . . . + |xn | f (0, . . . , 0, 1) ≤ n max{ f (1, 0, . . . , 0), . . . , f (0, . . . , 0, 1)}. This yields that f is bounded on the set {X ∈ Rn | X = 1}. Now, by Problem 175, we conclude that f is continuous. 177.

n A sequence of points {X k }∞ k=1 in R converges to the limit A if

lim X k − A = 0.

k→∞

In this case, we write lim X k = A. k→∞

Let A ∈ D f ⊆ Rn and A is a limit point of D f . Show that f is continuous at A if and only if (3.17) lim f (X k ) = f (A) k→∞

whenever {X k }∞ k=1 is a sequence of points in D f such that lim X k = A.

k→∞

(3.18)

Solution. Assume that f is continuous at A and {X k }∞ k=1 is a sequence of points in D f satisfying (3.18). If > 0, there exists δ > 0 such that | f (X ) − f (A)| < whenever X − A < δ.

(3.19)

By (3.18), there is a positive integer N such that X k − A < δ whenever k ≥ N .

(3.20)

By (3.19) and (3.20), we conclude that | f (X ) − f (A) < whenever k ≥ N . This implies (3.17), as desired. For the converse, suppose that f is discontinuous at A. Then there is 0 > 0 such that for each positive integer k, there is a point X k that satisfy the inequality X k − A

0, there exist positive real numbers δ, δ1 and δ2 such that |x − x0 | < δ ⇒ | f (x, y0 ) − f (x0 , y0 )| < /2,

(3.23)

|y − y0 | < δ1 ⇒ | f (x0 + δ1 , y) − f (x0 + δ1 , y0 )| < /2,

(3.24)

|y − y0 | < δ2 ⇒ | f (x0 − δ2 , y) − f (x0 − δ2 , y0 )| < /2,

(3.25)

and the obvious products of intervals are contained in N . Let δ = min{δ1 , δ2 } and let (x, y) be any point in the rectangle [x0 − δ, x0 + δ] × [y0 − δ , y0 + δ ]. Note that the direction of monotonically of f in x may depend on the value of y. Assume that f is non-decreasing in x for given y. If f is non-increasing, the reverse inequalities will hold. Thus, we have f (x0 − δ, y) − f (x0 − δ, y) + f (x0 − δ, y0 ) − f (x0 , y0 ) ≤ f (x, y) − f (x0 , y0 ) ≤ f (x0 + δ, y) − f (x0 + δ, y) + f (x0 + δ, y0 ) − f (x0 , y0 ) ,

136

3 Functions of Several Variables, Limits and Continuity

and so (3.23), (3.24) and (3.25) we obtain −/2 − /2 < f (x, y) − f (x0 , y0 ) < /2 + /2. Consequently, f is continuous at (x0 , y0 ). n 180. Let {X k }∞ k=1 be a convergent sequence in R with limit X . If there exists n an element A ∈ R and a number δ > 0 such that X k − A ≤ δ for k sufficiently large, show that X − A ≤ δ.

Solution. The set U = {Z ∈ Rn | Z − A > δ} is an open subset of Rn . If X ∈ U , then U is a neighborhood of X and so X k ∈ U for a sufficiently large value of k, contrary to the hypothesis. This yields that X ∈ / U , and so X − A ≤ δ. n 181. A sequence {X k }∞ k=1 in R is said to be a Cauchy sequence in case for any > 0, there exists a natural number N such that if r, s ≥ N , then X r − X s < . ∞ n If {X k }∞ k=1 is a convergence sequence in R , prove that {X k }k=1 is a Cauchy sequence.

Solution. Suppose that lim X k = X . Then, for any > 0 there is a natural number k→∞

N such that if r ≥ N , then X r − X < /2. Thus, for r, s ≥ N , we can write X r − X s ≤ X r − X + X − X s

0 such that X k ≤ M, for all

3.5 Exercises

137

k ∈ N. For each k ∈ N, we write X k = (Ak , bk ), where Ak ∈ Rm and bk ∈ R. Now, we can write |bk | and Ak ≤ X k = Ak 2 + |bk |2 ≤ M. This yields that {Ak } is a bounded sequence in Rm and {bk } is a bounded sequence in R. By Bolzano-Weirstrass Theorem in Rm , it follows that {Ak } has a subsequence {Ak j } which is convergent to a point A in Rm . Since {bk } is bounded in R, it follows that {bk j } is also bounded in R. By Bolzano-Weirstrass Theorem in R, {bk j } has a subsequence {bk ji } which converges to b ∈ R. Since {Ak j } converges to A in Rm , it follows that its subsequence {Ak ji } also converges to A. Now, the subsequence {X k ji } of {X k } is converges to (A, b). 184. Let f : R3 → Rbe a continuous function and A be a point in R3 . If f (A) > 0, show that there exists a neighborhood B(A, δ) = {X ∈ R3 | X − A < δ} of A such that f (X ) > 0 for all X ∈ B(A, δ). Solution. Assume that there exists no such neighborhood. Then for each k ∈ N, there exists

1 1 X k ∈ B(A, ) = X ∈ R3 | X − A < k k such that f (X k ) ≤ 0. Since lim X k = A and f is continuous, by Problem 177, we k→∞

conclude that lim f (X k ) = f (A). Therefore, we get f (A) ≤ 0 which is a contrak→∞

diction.

3.5 Exercises Easier Exercises 1. Sketch sets of points satisfying the following inequalities. (a) |x 2 + y 2 − 2| ≤ 1, (b) (x − 3)2 + 4y 2 ≤ 4,

√ (c) y > x, (d) 2x < x 2 + y 2 < 4x.

Find the domain and range of each of the following functions f of two variables and draw a sketch showing as a shaded region in R2 the set of points in the domain of f . √ 2+ x−y 5. f (x, y) = 4 − x y 2 + 9, , 2. f (x, y) = x+y 6. sin−1 1 − x 2 − y 2 , 3. f (x, y) = ln(x y − 1), 1 1 4. f (x, y) = + , 7. f (x, y) = [x] + [ 1 − y 2 ]. x y

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3 Functions of Several Variables, Limits and Continuity

Find the domain and range of each of the following functions f of three variables. x y + x z + yz , x yz x 9. f (x, y, z) = , |y| − |z| 8. f (x, y, z) =

12. Let f (x, y) =

10. f (x, y, z) =

ln(1 −

x2

1 , − y2 − z2)

11. f (x, y, z) = e x y/z .

4x 2 + y 2 if x < 0 |y| if x ≥ 0.

Sketch the level curves f (x, y) = c for c = 0, 2 and 4. Determine whether each limit exists. If it does, find the limit and prove that it is the limit; if it does not, explain how you know. √ √ 3 x− 3y x 3 − 4y 2 x + 5y 3 ; 23. lim ; 13. lim (x,y)→(0,0) (x,y)→(0,0) x2 + √ y2 x−y √ x y e −e x −y+2 x −2 y 14. lim ; 24. lim ; √ √ (x,y)→(0,0) e−x − e−y (x,y)→(0,0) x− y √ xy 2x − y − 2 ; 15. lim ; 25. lim (x,y)→(0,0) x 2 + y 2 (x,y)→(2,0) 2x − y − 4 −1 y ; 16. lim tan |x| + |y| (x,y)→(2,2) x 26. lim tan−1 2 ; 2 2 (x,y)→(0,0) x + y2 e−x −y − 1 sin x 17. lim ; ; 27. lim (x,y)→(0,0) x 2 + y2 (x,y)→(0,π) sin y 3 3 x +y x 3 + sin(2x y) ; 18. lim 28. lim ; (x,y)→(0,0) x 2 + y 2 (x,y)→(0,0) x 2 2 x + sin y ex ez − e y ez ; 19. lim 29. lim ; (x,y)→(0,0) 2x 2 + y 2 (x,y,z)→(0,0,0) e2x − e2y x y 2 + ex x 2 + y2 − z2 ; 20. lim √ ; 30. lim (x,y)→(0,π/2) cos x + sin y (x,y,z)→(0,0,0) x 2 + y 2 + z 2 9 3 2 x y y + xz 21. lim ; 31. lim ; 2 (x,y)→(0,0) (x 6 + y 2 )2 (x,y,z)→(0,0,0) x + y 2 + z 2 2x 3 + 4x 2 y x 4 + yx 3 + x 2 z 2 32. lim ; . 22. lim 2 2 (x,y)→(0,0) (x,y,z)→(0,0,0) x +y x 4 + y4 + z4 Check the continuity of the following functions. If a function is discontinuous, suggest a way to change the function so that it becomes continuous. xy if (x, y) = (0, 0) 33. f (x, y) = x 2 + x y + y 2 if (x, y) = (0, 0); ⎧0 ⎨ sin(x y) if x y = 0 34. f (x, y) = xy ⎩1 if x y = 0;

3.5 Exercises

139

xy if (x, y) = (0, 0) x 4 − y4 0 if (x, y) = (0, 0);

1 − x 2 + y 2 if x 2 + y 2 < 1 f (x, y) = x 2 + y 2 − 1 if x 2 + y 2 ≥ 1; ⎧ −1/x 2 y ⎨ e f (x, y) = e2/x 2 + y 2 if x = 0 ⎩ 0 if x = 0. Let f and g be continuous functions on Rn to R and let h, k defined for X ∈ Rn by h(x) = max{ f (X ), g(X )} and k(x) = min{ f (X ), g(X )}. Show that h and k are continuous on Rn . A subset C of Rn is said to be convex if whenever X, Y belong to C and λ is a real number such that 0 ≤ λ ≤ 1, then the point λX + (1 − λ)Y also belong to C. Interpret this condition geometrically, and show that the subsets

35. f (x, y) = 36. 37. 38.

39.

= {(x, y) ∈ R2 | (x, y) ≤ 1}, C1 = {(x, y) ∈ R2 | 0 < x < y}, C2 = {(x, y) ∈ R2 | 0 ≤ y ≤ x ≤ 1}, are convex. Show that the subset S = {(x, y) ∈ R2 | (x, y) = 1} is not convex. 40. Show that (a) The intersection of any collection of convex subsets of Rn is convex; (b) The union of two convex subsets of Rn may not be convex. Harder Exercises Calculate the following limits using various techniques (decide for yourself which technique to use). y x 1 − cos(x 2 + y 2 ) 1+ 43. lim ; 41. lim ; (x,y)→(∞,0) x (x,y)→(0,0) (x 2 + y 2 )2 x2 1 1 x+y 42. lim (1 + x 2 + y 2 ) x 2 +y2 ; 1+ . 44. lim (x,y)→(0,0) (x,y)→(∞,1) x Evaluate lim

X →∞

f (X ) for the given function f , if it exists.

sin(x 2 + y 2 ) ; 45. f (x, y) = x 2 + y2 ln(x 2 + 2y 2 + 4z 2 ; 46. f (x, y, x) = x 2 + y2 + z2

140

3 Functions of Several Variables, Limits and Continuity

⎧ ⎨ sin(x 2 − y 2 ) if y = ±1 2 2 47. f (x, y) = ⎩ x −y 1 if y = ±1. 48. Find the iterated limits s lim lim logx (x + y) and lim lim logx (x + y) . x→1

y→0

y→0

x→1

What can be said about the corresponding two-variable limit? 49. Suppose that f and g are functions of two variables satisfying the following conditions: (a) f (t x, t y) = t n f (x, y) and g(t x, t y) = t n g(x, y), for some n and for all t; (b) g(1, 1) = 0 and g(1, 0) = 0; (c) g(1, 1) · f (1, 0) = g(1, 0) · f (1, 1). Show that

lim

(x,y)→(0,0)

f (x, y) does not exist. g(x, y)

50. Let f (x, y) =

(x 2 + y 4 )3 . 1 + x 6 y4

Show that lim f (x, t x) = ∞ for any real number t. Does |x|→∞

lim

(x,y)→∞

51. Does knowing that 1−

tell us anything about 52. Let

f (x, y) = ∞?

tan−1 (x y) x 2 y2 < 0 at which the function is continuous at the origin. 55. Suppose that f : R2 → R is a continuous function. Define the functions g1 , g2 : R → R by g1 (t) = f (t, 0) and g2 (t) = f (0, t), for all t ∈ R. Show that g1 and g2 are continuous. 56. Let f, g1 and g2 be related by the formulas in the preceding exercise. Show that from the continuity of g1 and g2 at t = 0, one cannot prove the continuity of f at (0, 0). 57. Let f : Rn → R be a function defined by f (x1 , . . . , xn ) =

|x1 |a1 . . . |xn |an . (x1 , . . . , xn )b

For what non-negative values of a1 , . . . , an , b does

lim

(x1 ,...,xn )→(0,...,0)

f (x1 , . . . , xn )

exist in the extended reals (i.e., it is a finite number or ∞)? 58. If A is a constant vector in Rn , show that the function f given by f (X ) = A · X , for all X ∈ Rn , is continuous. 59. Let = {X ∈ Rn | X < 1}. Is it true that a continuous function f : → R can be extended to a continuous function on = {X ∈ Rn | X ≤ 1} if and only if f is uniformly continuous on ? 60. The distance from a point X 0 to a non-empty set S is defined by d(X 0 , S) = inf{X − X 0 | X ∈ S}. (a) If S is closed and X 0 ∈ Rn , prove that there exists a point Y ∈ S such that Y − X 0 = d(X 0 , S). / S, then d(X 0 , S) > 0. (b) Show that if S is closed and X 0 ∈ (c) Show that the conclusions of (a) and (b) may fail to hold if S is not closed. 61. If S is any subset of Rn , prove that there exists a countable subset C of S such that if X ∈ S and > 0, then there is an element Y ∈ C such that X − Y < . 62. Let S be a non-empty closed subset of Rn and let X be a point outside of S. Prove that there exists at least one point Y belonging to S such that Y − X ≤ A − X , for all A ∈ S. n 63. If {X k }∞ k=1 is a sequence in R , prove that the following statements are equivalent:

(a) {X k }∞ k=1 does not converge to X ;

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3 Functions of Several Variables, Limits and Continuity

(b) There exists a neighborhood N of X such that if n is any natural number, then there is a natural number m ≥ n such that X m does not belong to N ; ∞ (c) There exists a neighborhood N of X and a subsequence {X rk }∞ k=1 of {X k }k=1 ∞ such that none of the elements of {X rk }k=1 belong to N . ∞ n 64. If {X k }∞ k=1 is a sequence in R , does {X k }k=1 converge to X if and only if the ∞ real sequence {X k }k=1 converges to X ? 65. Cauchy Convergence Criterion: Prove that a sequence in Rn is convergent if and only if it is a Cauchy sequence.

Chapter 4

Partial Derivatives, Directional Derivatives and Gradient Vectors

4.1 Partial Derivatives For a function of a single variable, y = f (x), changing the independent variable x leads to a corresponding change in the dependent variable y. The rate of change of df . A similar situation occurs y with respect to x is given by the derivative, written dx with functions of more than one variable. Let f be a function of n variables defined in a neighborhood of a point (x1 , . . . , xn ). By the partial derivative of f with respect to xi denoted by the expression ∂ f (x1 , . . . , xn ) ∂xi we mean limit lim

xi →0

f (x1 , . . . , xi−1 , xi + xi , xi+1 , . . . , xn ) − f (x1 , . . . , xi , . . . , xn ) , xi

where xi is given an increment xi , but all the other variables are held fixed, provided ∂f this limit exists and is finite. A prevalent notation is to write as f xi . There are ∂xi other notations for partial derivatives. If a dependent variable u = f (x1 , . . . , xn ) is ∂u ∂f introduced, we can write instead of . A function f of two variables has two ∂xi ∂xi partial derivatives as follows: ∂ f (x, y) f (x + x, y) − f (x, y) = lim x→0 ∂x x and

∂ f (x, y) f (x, y + y) − f (x, y) = lim , y→0 ∂y y

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_4

143

144

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

provided these limits exist and are finite. Geometric interpretation of the partial derivatives of a function f of two variables are similar to those of a function of one variable. Suppose that f is defined in a neighborhood of the point (x0 , y0 ) of the x y-plane and P0 = (x0 , y0 , f (x0 , y0 )) be the corresponding point of the graph of f , which is a surface S in R3 . The plane y = y0 intersects S in a curve going through P0 (see Fig. 4.1), and the plane x = x0 intersects S in another curve which also goes through P0 (see Fig. 4.2). Then,

Fig. 4.1 Slope in x direction

Fig. 4.2 Slope in y direction

4.2 Directional Derivatives and Gradient Vectors

145

∂ f (x, y) x=x0 ,y=y0 ∂x is the slope of the tangent line to the curve having equations y = y0 and z = f (x, y) at the point P0 = (x0 , y0 , f (x0 , y0 )) in the plane y = y0 . In an analogous fashion ∂ f (x, y) x=x0 ,y=y0 ∂y represents the slope of the tangent line to the curve having equations x = x0 and z = f (x, y) at the point P0 = (x0 , y0 , f (x0 , y0 )) in the plane x = x0 . Figures 4.1 and 4.2 show the portion of the curves and the tangent lines. The partial derivatives just introduced are first derivatives. Partial derivatives of higher order are defined in a natural way. For example, if f is a function of two ∂f ∂f and at every point of an open variables which has the first partial derivatives ∂x ∂y set. Then there are four second partial derivatives, namely ∂2 f ∂ ∂ f , = ∂x 2 ∂x ∂x 2 ∂ ∂f ∂ f = , = ∂x∂ y ∂x ∂ y

∂2 f ∂ ∂ f , = ∂ y2 ∂y ∂y 2 ∂ ∂f ∂ f = . = ∂ y∂x ∂ y ∂x

fx x =

f yy =

fx y

f yx

All this is based on the assumption that the limits defining these second partial derivatives exist and are finite. Partial derivatives of order greater than two, and higher order partial derivatives of functions of more than two variables are defined in similar ways. Clairaut’s theorem: Let f be a function of two variables. If f and its derivatives f x , f y , f x y and f yx are defined throughout a neighborhood of the (x0 , y0 ) and are all continuous at (x0 , y0 ), then f x y (x0 , y0 ) = f yx (x0 , y0 ).

4.2 Directional Derivatives and Gradient Vectors The partial derivatives f x (x0 , y0 ) and f y (x0 , y0 ) are the rates of change of z = f (x, y) at (x0 , y0 ) in the positive x- and y-directions. Rates of change in other directions are − → − → given by directional derivatives. If u = cos θ i + sin θ j is the unit vector, then the directional derivative of f in the direction of u, denoted by Du f , is given by Du f (x0 , y0 ) = lim

h→0

f (x0 + h cos θ, y0 + h sin θ) − f (x0 , y0 ) , h

if this limit exists. We can extend the definition of a directional derivative to a function of three variables. The direction of a vector in R3 is determined by the direction of − → − → cosines. Let f be a function of three variables x, y and z. If u = cos α i + cos β j +

146

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

− → cos γ k is the unit vector, then the directional derivative of f in the direction of u, denoted by Du f , is given by Du f (x0 , y0 , z 0 )= lim

h→0

f (x0 + h cos α, y0 + h cos β, z 0 + h cos γ) − f (x0 , y0 , z 0 ) , h

if this limit exists. The gradient of a function f of two variables at a point (x0 , y0 ) is the vector ∇ f (x, y) =

∂f− → ∂f− → i + j, ∂x ∂y

obtained by evaluating the partial derivatives of f at (x0 , y0 ). The definition of the gradient may be extended to functions of three variables as follows: ∇ f (x, y, z) =

∂f− → ∂f− → ∂f− → i + j + k. ∂x ∂y ∂z

The directional derivative can be written in the form Du f = ∇ f · u.

4.3 Tangent Plane and Normal Line to a Surface If an equation of a surface S is F(x, y, z) = 0, and Fx , Fy and Fz are continuous and not zero at the point P = (x0 , y0 , z 0 ) on S, then a normal vector to S at P is ∇ F(x0 , y0 , z 0 ). The concept of a normal vector is used to define the tangent plane to a surface at a point. If an equation of a surface S is F(x, y, z) = 0, then the tangent plane of S at a point P = (x0 , y0 , z 0 ) is the plane through P having as a normal vector ∇ F(x0 , y0 , z 0 ). An equation of the tangent plane of the above definition is Fx (x0 , y0 , z 0 )(x − x0 ) + Fy (x0 , y0 , z 0 )(y − y0 ) + Fz (x0 , y0 , z 0 )(z − z 0 ) = 0. (4.1) The normal line to a surface S at a point P on S is the line through P parallel to the gradient vector at P. Thus, the normal line to the surface S at P is the line with symmetric equations y − y0 z − z0 x − x0 = = . Fx (x0 , y0 , z 0 ) Fy (x0 , y0 , z 0 ) Fz (x0 , y0 , z 0 )

(4.2)

The denominators in (4.2) are components of ∇ F(x0 , y0 , z 0 ). Indeed, the normal line at a point on a surface is perpendicular to the tangent plane there. In particular, suppose that f is a function of two variables with continuous partial derivatives. In this case, (4.1) and (4.2) for the tangent plane and normal line at the point P = (x0 , y0 , z 0 ) reduced to

4.4 Solved Problems

147

Fig. 4.3 Tangent plane and normal line at P

z − z 0 = f x (x0 , y0 )(x − x0 ) + f y (x0 , y0 )(y − y0 ) and

x − x0 y − y0 z − z0 = = , f x (x0 , y0 ) f y (x0 , y0 ) −1

respectively, where z 0 = f (x0 , y0 ). Refer to Fig. 4.3, which shows the tangent plane and normal line to the graph of f at P.

4.4 Solved Problems 185. Let z = y k e−x following equation:

2

/(4y)

. Find a value of the constant k such that f satisfies the 1 ∂ ∂z = 2 ∂y x ∂x

x2

∂z ∂x

.

Solution. First, we compute the first partial derivatives of z as follows: −x −1 ∂z 2 2 = yk e−x /(4y) = x y k−1 e−x /(4y) ∂x 2y 2 and

x2 ∂z 2 −x 2 /(4y) = ky k−1 e−x /(4y) + y k e ∂y 4y 2 1 2 2 2 = ky k−1 e−x /(4y) + x y k−2 e−x /(4y) . 4

(4.3)

148

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

Then, we have ∂ 1 3 k−1 −x 2 /(4y) ∂ 2 ∂z x = − x y e ∂x ∂x ∂x 2 3 1 2 2 = − x 2 y k−1 e−x /(4y) + x 4 y k−2 e−x /(4y) . 2 4 This yields that 1 ∂ x 2 ∂x

∂z x ∂x

2

1 3 2 2 = − y k−1 e−x /(4y) + x 2 y k−2 e−x /(4y) . 2 4

(4.4)

If the left side of Eqs. (4.3) and (4.4) are equal, then we must have ky k−1 e−x

2

/(4y)

1 3 1 2 2 2 + x 2 y k−2 e−x /(4y) = − y k−1 e−x /(4y) + x 2 y k−2 e−x /(4y) . 4 2 4

This implies that 1 3 1 k(y − 1) + x 2 = − (y − 1) + x 2 , 4 2 4 and consequently, we obtain k = −3/2. 186. Given z = f (x, y)eax+by and ∂ 2 f /(∂x∂ y) = 0. Find values of the constants a and b such that ∂z ∂z ∂2 z − − + z = 0. (4.5) ∂x∂ y ∂x ∂y Solution. We obtain ∂z = ∂x ∂z = ∂y ∂2 z ∂x∂ y

∂ f ∂f (x, y)eax+by + a f (x, y)eax+by = (x, y) + a f (x, y) eax+by , ∂x ∂∂xf ∂f (x, y)eax+by + b f (x, y)eax+by = (x, y) + b f (x, y) eax+by , ∂y ∂y ∂f ∂f = a (x, y) + b (x, y) + ab f (x, y) eax+by . ∂y ∂x

If z satisfies (4.5), then (a − 1)

∂f ∂f (x, y) + (b − 1) (x, y) + (ab − a − b + 1) f (x, y) = 0. ∂y ∂x

This gives a = 1 and b = 1. 187. Show that the existence of ∂ 2 f /(∂x∂ y) does not imply the existence of ∂ f /∂x. Solution. It is enough to give an example. Suppose that f (x, y) = |x|,

4.4 Solved Problems

149

for all x, y ∈ R. Since ∂ f /∂ y is identically zero, it follows that ∂ 2 f /(∂x∂ y) is also identically zero. But it is clear that ∂ f /∂x does nor exist at (0, 0). 188. Let f and g be functions of two variables such that ∂f ∂g ∂f ∂g = and = , ∂x ∂y ∂y ∂x and suppose that ∂ f /∂x = 0, f (1, 2) = g(1, 2) = 5 and f (0, 0) = 4. Find f (x, y) and g(x, y). Solution. Since ∂ f /∂x = 0, it follows that f is a function of y only. Since ∂ f /∂x = ∂g/∂ y, it follows that ∂g/∂ y = 0. This yields that g is a function of x only. Assume that f (x, y) = h(y) and g(x, y) = k(x), for all x and y. Since ∂ f /∂ y = ∂g/∂x, it follows that h (y) = k (x), for all x and y. Consequently, h and k must be constant. If h (y) = k (x) = a, then h(y) = ay + b1 and h(x) = ax + b2 , where a, b1 and b2 are constant. Thus, we have f (x, y) = ay + b1 and g(x, y) = ax + b2 , for all x and y. Since f (1, 2) = g(1, 2) = 5, it follows that 5 = 2a + b1 = a + b2 .

(4.6)

Since f (0, 0) = 4, it follows that b1 = 4. Substituting this in (4.6), we get a = 1/2 and b2 = 9/2. Therefore, we have f (x, y) = (1/2)y + 4 and g(x, y) = (1/2)x + (9/2), for all x and y. 189. If

f (x, y) =

(x 2 + y 2 ) ln(x 2 + y 2 ) if x 2 + y 2 = 0 0 if x 2 + y 2 = 0.

Show that f x y (0, 0) = f yx (0, 0). Solution. By the definition of partial derivative, we can write f (0 + h, 0) − f (0, 0) 2 ln h = lim h ln h 2 = lim h→0 h→0 1/ h h 2/ h = lim −2h = 0 (by L’Hospital’s rule). = lim h→0 −1/ h 2 h→0

f x (0, 0) = lim

h→0

If y = 0, then f (0 + h, y) − f (0, y) (h 2 + y 2 ) ln(h 2 + y 2 ) − y 2 ln y 2 = lim h→0 h→0 h h 2h ln(h 2 + y 2 ) + 2h = 0 (by L’Hospital’s rule). = lim h→0 1

f x (0, y) = lim

Now, we evaluate f x y (0, 0). By using the definition, we have

150

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

f x y (0, 0) = lim

k→0

f x (0, 0 + k) − f x (0, 0) = 0. k

Since f is symmetry respect to x and y, it follows that f yx (0, 0) = 0 too. 190. If z = f (x, y) =

x3 y ∂z ∂z , show that x +y = 3z. x−y ∂x ∂y

Solution. Using the rules of derivative, we obtain ∂ ∂ 3 3 y) (x − y) − (x y) (x (x − y) ∂z ∂x = ∂x ∂x (x − y)2 (3x 2 y)(x − y) − (x 3 y)(1) 2x 3 y − 3x 2 y 2 = = (x − y)2 (x − y)2 and

∂ ∂ (x 3 y) (x − y) − (x 3 y) (x − y) ∂z ∂y ∂y = 2 ∂y (x − y) (x 3 )(x − y) − (x 3 y)(−1) x4 = = (x − y)2 (x − y)2

Now, we see that x

∂z 3x 4 y − 3x 3 y 2 3x 3 y ∂z 2x 4 y − 3x 3 y 2 + x 4 y = = +y = = 3z. ∂x ∂y (x − y)2 (x − y)2 x−y

191. Show that the function u=

1 x 2 + y2 + z2

, (where x 2 + y 2 + z 2 = 0),

is a solution of the three dimensional Laplace’s equation: ∂2u ∂2u ∂2u + + = 0. ∂x 2 ∂ y2 ∂z 2

(4.7)

Solution. First, we compute ∂u/∂x, ∂u/∂ y and ∂u/∂z. We find that −x −y −z ∂u ∂u ∂u = 2 = 2 = 2 , , . 2 2 3/2 2 2 3/2 ∂x (x + y + z ) ∂y (x + y + z ) ∂x (x + y 2 + z 2 )3/2

4.4 Solved Problems

151

Then, we obtain 2x 2 − y 2 − z 2 ∂2u 2y 2 − x 2 − z 2 ∂2u 2z 2 − x 2 − y 2 ∂2u = , = , = . ∂x 2 (x 2 + y 2 + z 2 )5/2 ∂ y 2 (x 2 + y 2 + z 2 )5/2 ∂z 2 (x 2 + y 2 + z 2 )5/2 Now, it is easy to see that ∂ 2 u/∂x 2 , ∂ 2 u/∂ y 2 and ∂ 2 u/∂z 2 satisfy (4.7). f (x + y) + g(x − y) , where f and g are arbitrary functions of single x variables with second derivatives f and g . Show that 192. Let u =

∂ 2 ∂u ∂2u x = x2 2 . ∂x ∂x ∂y

(4.8)

Solution. First, we evaluate ∂u/∂x and ∂u/∂ y. We obtain

x f (x + y) + g (x − y) − f (x + y) − g(x − y) ∂u = , ∂x x2

1 ∂u = f (x + y) − g (x − y) . ∂y x In continue, we find that

∂ 2 ∂u ∂ x = x f (x + y) + g (x − y) − f (x + y) − g(x − y) ∂x ∂x ∂x = f (x + y) + g (x − y) + x f (x + y) + xg (x − y) − f (x + y) − g (x − y) = x f (x + y) + xg (x − y), (4.9) and

1 ∂2u f (x + y) + g (x − y) . = (4.10) 2 ∂y x Now, from (4.9) and (4.10), we obtain (4.8), as desired. 193. The second order differential equation ∂2u 1 ∂2u = , ∂x 2 k 2 ∂t 2

(4.11)

where k is a positive constant, is called the wave equation. Let f and g be arbitrary functions of a single variable with second derivatives f and g . Show that the function u = f (x + kt) + g(x − kt) is a solution of the wave equation. Solution. Take r = x + kt and s = x − kt. First, we compute

∂2u . We can write ∂x 2

∂r ∂s ∂u = f (r ) + g (s) = f (r ) + g (s), ∂x ∂x ∂x

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4 Partial Derivatives, Directional Derivatives and Gradient Vectors

and so

∂ ∂2u f (r ) + g (s) = 2 ∂x ∂x ∂r ∂s = f (r ) + g (s) ∂x ∂x = f (r ) + g (s).

Thus, we have

Now, we evaluate

∂2u = f (r ) + g (s). ∂x 2

(4.12)

∂2u . We have ∂t 2 ∂r ∂s ∂u = f (r ) + g (s) = k f (r ) − kg (s), ∂t ∂t ∂t

and hence

∂2u ∂ k f (r ) − kg (s) = ∂t 2 ∂t ∂r ∂s − kg (s) = k f (r ) ∂t ∂t = k 2 f (r ) + k 2 g (s).

This yields that

∂2u = k 2 f (r ) + g (s) . 2 ∂t

(4.13)

Comparing (4.12) and (4.13) shows that the function u = f (x + kt) + g(x − kt) satisfies (4.11), as desired. 194. Let f (x, y) = e x sin y. Find ∂ n+m f (0, 0), ∂x n ∂ y m where m and n are natural numbers. Solution. It is clear that f satisfies Clairaut’s theorem. We may write f (x, y) = g(x)h(y), where g(x) = e x and h(y) = sin y. Then, we obtain ∂n f ∂m f (x, y) = g (n) (x)h(y) and (x, y) = g(x)h (m) (y). n ∂x ∂ ym Therefore, we deduce that ∂ n+m f (x, y) = g (n) (x)h (m) (y) = e x h (m) (y). ∂x n ∂ y m

4.4 Solved Problems

153

Now, we consider three cases: Case 1: If the number m is one greater than a multiple of 4, then h (m) (y) = cos y and h (m) (0) = cos 0 = 1. Case 2: If the number m is three greater than a multiple of 4, then h (m) (y) = − cos y and h (m) (0) = − cos 0 = −1. Case 3: If the number m is even, then h (m) (y) = ± sin y and h (m) (0) = ± sin 0 = 0. Consequently, we have ⎧ ⎨ 1 if m − 1 is a multiple of 4 ∂ n+m f −1 if m − 3 is a multiple of 4 (0, 0) = ⎩ ∂x n ∂ y m 0 if m is even. 195. Let f (X ) = X 2−n , where X = (x1 , . . . , xn ) ∈ Rn . If X = 0, prove that ∂ f2 ∂ f2 (X ) + · · · + (X ) = 0. ∂x1 ∂x1 ∂xn ∂xn Solution. Since X =

√

X · X , we can write f (X ) = (X · X )1−(n/2) . Then, we have

∂f (X ) = (2 − n)xi (X · X )−n/2 , for i = 1, . . . , n. ∂xi Next, we compute ∂ f 2 /∂xi ∂xi as follows: n ∂ f2 (X ) = (2 − n) (X · X )−n/2 + xi − (X · X )−(n/2)−1 (2xi ) ∂xi ∂xi 2 (2 − n)(X · X − nxi2 ) = , (X · X )(n/2)+1 for i = 1, . . . , n. Therefore, we obtain ∂ f2 ∂ f2 (X ) + · · · + (X ) ∂x1 ∂x1 ∂xn ∂xn (2 − n)(X · X − nxn2 ) (2 − n)(X · X − nx12 ) + ··· + = (n/2)+1 (X · X ) (X · X )(n/2)+1 n 2−n = (X · X − nxi2 ) (X · X )(n/2)+1 i=1 n n 2−n 2 1−n xi = X·X (X · X )(n/2)+1 i=1 i=1 2−n = (n X · X − n X · X ) = 0. (X · X )(n/2)+1

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4 Partial Derivatives, Directional Derivatives and Gradient Vectors

196. Two functions F and G of one variable and a function f of two variables are related by the equation

2 F(x) + G(y) e f (x,y) = 2F (x)G (y)

(4.14)

whenever F(x) + G(y) = 0. Show that the mixed partial derivative f x y (x, y) is never zero. (You may assume the existence and continuity of all derivatives encountered.) Solution. Since F(x) + G(y) = 0, by (4.14), it follows that F (x)G (y) = 0 and 2 F (x)G (y) f (x, y) = ln

2 + G(y)

F(x) = ln 2F (x)G (y) − 2 ln F(x) + G(y) . Now, we evaluate f x (x, y) as follows: F (x) 2 F (x)G (y) − 2 2F (x)G (y) F(x) + G(y) F (x) F (x) −2 . = F (x) F(x) + G(y)

f x (x, y) =

Next, we compute f x y (x, y) as follows: f x y (x, y) = −2

F (x)G (y)

2 .

F(x) + G(y)

Finally, since F (x)G (y) = 0, it follows that f x y (x, y) is never zero. 197. Suppose that f : R2 → R is defined by f (x, y) =

⎧ ⎨ ⎩

xy 0

x 2 − y2 x 2 + y2

if (x, y) = (0, 0) if (x, y) = (0, 0).

Show that (1) f x and f y exist and are continuous on R2 ; (2) f x y (0, 0) and f yx (0, 0) exist and are not equal. Solution. (1) The existence of f x (x, y) and f y (x, y) for (x, y) = (0, 0) follows from the fact that agrees with the function h(x, y) = x y

x 2 − y2 x 2 + y2

4.4 Solved Problems

155

near such (x, y). By the quotient rule, we obtain f x (x, y) = x y

(x 2 + y 2 )(2x) − (x 2 − y 2 )(2x)

(x 2 + y 2 )2 x 2 − y2 4x y , + y = xy 2 (x + y 2 )2 x 2 + y2

+y

x 2 − y2 x 2 + y2

2

(4.15)

for (x, y) = (0, 0). To check the existence of f x (0, 0), we observe that f (h, 0) − f (0, 0) 0 = lim = 0. h→0 h h

lim

h→0

This means that f x (0, 0) = 0. So, we have f (x, y) =

⎧ ⎨ ⎩

xy 0

x 2 − y2 4x y 2 if (x, y) = (0, 0) +y 2 2 2 +y ) x + y2 if (x, y) = (0, 0).

(x 2

Now, f x is continuous at (x, y) = (0, 0). In order to check the continuity at (0, 0), we must show that lim f x (x, y) = 0 = f x (0, 0). h→0

Since 2|x y| ≤ x 2 + y 2 , it follows that 4|x y|2 ≤ (x 2 + y 2 )2 . Then, for (x, y) = (0, 0), we have | f x (x, y)| ≤ x y

x 2 − y 2 4x y 2 +y 2 2 2 +y ) x + y2 4x y 2 x 2 − y 2 ≤ x y 2 + y 2 (x + y 2 )2 x + y2 x 2 − y2 4|x y|2 |y| = 2 + |y| 2 (x + y 2 )2 x + y2 x 2 − y2 (x 2 + y 2 )2 |y| ≤ + |y| 2 2 2 2 2 (x + y ) x +y ≤ |y| + |y| = 2|y|. (x 2

Since 2|y| → 0 as (x, y) → (0, 0), it follows that f x (x, y) → (0, 0). This yields that f x is continuous on R2 . Similarly, we can see that f y exists and is continuous on R2 . (2) By the argument in (1), we obtain f x (0, y) = −y, for all y ∈ R. On the other hand, for x = 0, we have f y (x, 0) = lim

y→0

f (x, y) − f (0, 0) x(x 2 − y 2 ) = lim = x. y→0 x 2 + y 2 y−0

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4 Partial Derivatives, Directional Derivatives and Gradient Vectors

If x = 0, then f y (0, 0) = lim

y→0

f (0, y) − f (0, 0) 0−0 = lim = 0. y→0 y−0 y

Since f y (x, 0) = x for x = 0 and f y (0, 0) = 0, it follows that f y (x, 0) = x, for all x ∈ R. Now, we compute f x y (0, 0) and f yx (0, 0). Indeed, we can write f x y (0, 0) = lim

h→0

f x (0, h) − f x (0, 0) −h − 0 = lim = −1. h→0 h h

Similarly, we get f yx (0, 0) = lim

h→0

f y (h, 0) − f y (0, 0) h−0 = lim = 1. h→0 h h

Therefore, f x y (0, 0) = f yx (0, 0), as desired. 198. Let f : R2 → R be defined by ⎧ ⎨

x2 y if (x, y) = (0, 0) f (x, y) = x 4 + y 2 ⎩ 0 if (x, y) = (0, 0). Show that the directional derivative in all directions at (0, 0) exists. Solution. Let u = (cos θ, sin θ) be a unit vector. Then, by using the definition of directional derivative, we obtain f (0 + h cos θ, 0 + h sin θ) − f (0, 0) h→0 h f (h cos θ, h sin θ) = lim h→0 h (h 2 cos2 θ)(h sin θ)

= lim 4 h→0 h h cos4 θ + h 2 sin2 θ cos2 θ sin θ = lim 2 h→0 h cos4 θ + sin2 θ cos θ cot θ if sin θ = 0 = 0 if sin θ = 0.

Du f (0, 0) = lim

√ 199. Suppose that f (x, y) = |x y| for all (x, y) ∈ R2 , and let u = (a, b) be a unit vector. Show that the directional derivative of f at the origin in the direction u exists if and only if (a, b) = (1, 0) or (a, b) = (0, 1). Solution. By the definition of directional derivative, we have

4.4 Solved Problems

157

f (0 + ah, 0 + bh) − f (0, 0) f (ah, bh) = lim h→0 h→0 h h √ h 2 |ab| |h| |ab| = lim = lim . h→0 h→0 h h

Du f (0, 0) = lim

It is clear that the above limit exists if and only if a = 0 or b = 0. 200. For y ≥ 0 put ⎧ √ if 0 ≤ x ≤ y ⎨x √ √ √ f (x, y) = −x + 2 y if y ≤ x ≤ 2 y ⎩ 0 otherwise and put f (x, y) = − f (x, |y|) if y < 0. (1) Show that f is continuous on R2 , and f y (x, 0) = 0, for all x ∈ R. (2) Define 1

g(y) =

−1

f (x, y)d x.

Show that g(y) = y if |y| < 1/4. Hence, g (0) =

1

−1

f y (x, 0)d x.

Solution. (1) According to the definition of f , we need only verify the continuity of √ √ f on the boundary curves x = y and x = 2 y in the first quadrant that separate the three different regions of definition. We observe that f (x, y) = 0, if x ≤ 0 and y ∈ R, f (x, y) = 0, if x > 0 and 0 ≤ y ≤

1 2 x . 4

Hence, by a simple calculation, we conclude that f y (x, 0) = 0. (2) Suppose that 0 < y < 1/4. Then, we have

2√ y √ xd x + √ (−x + 2 y)d x y 0 √ 1 √ y 1 √ 2 y 2 2 x + − x + 2 yx √ = 0 2 2 y 1 1 = y + (−2y + 4y) − − y + 2y = y. 2 2

g(y) =

√

y

Moreover, we have g(0) = 0 and g(y) = −g(−y) if y < 0. This yields that g (0) = 1, meanwhile 1

−1

f y (x, 0)d x = 0.

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4 Partial Derivatives, Directional Derivatives and Gradient Vectors

201. A skier is on a mountain with equation z = f (x, y) = 100 − 0.4x 2 − 0.3y 2 , where z denotes height. (1) The skier is located at the point with x y-coordinates (1, 1), and wants to ski downhill along the steepest possible path. In which direction (indicated by a vector (a, b) in the x y-plane) should the skier begin skiing? (2) The skier begins skiing in the direction given by the x y-vector (a, b) that we found in part (1), so the skier heads in a direction in space given by the vector (a, b, c). Find the value of c. (3) A hiker located at the same point on the mountain decides to begin hiking downhill in a direction given by a vector in the x y-plane that makes an angle θ with the vector (a, b) that we found in part (1). How big should θ be if the hiker wants to head downhill along a path whose slope is at most 0.5 (in absolute value)? Solution. (1) The direction of the greatest rate of decrease is opposite of the direc− → − → tion of the gradient. We obtain ∇ f (x, y) = −0.8x i − 0.6y j , and so ∇ f (1, 1) = → − → − −0.8 i − 0.6 j . Gradient vector is already a unit vector, so the unit vector in the − → − → opposite direction is u = −∇ f (1, 1) = 0.8 i + 0.6 j . (2) The directional derivative in the direction u (or (a, b)), Du f (1, 1) = ∇ f (1, 1) · u = −u · u = −1 gives the slope, which is the ratio of vertical √ change to horizontal change. In the direction of the vector (a, b, c), this ratio is c/ a 2 + b2 . Hence, we have c Du f (1, 1) = √ = c. 2 a + b2 This yields that c = −1. (3) If the hiker’s direction is given by a unit vector v, we want −0.5 ≤ Dv f (1, 1) ≤ 0. The angle v makes with u (that is, with (a, b)) is the same as the angle −v makes with −u (that is, with ∇ f (1, 1)). We have

Du f (1, 1) = ∇ f (1, 1) · v = − ∇ f (1, 1) · (−v) = − cos θ. Since we want −0.5 ≤ Dv f (1, 1) ≤ 0, it follows that 0 ≤ cos θ ≤ 0.5. This gives that π/3 ≤ θ ≤ π/2. 202. Find directions in which f (x, y) = x 2 + sin(x y) has directional derivative at (1, 0) with value 1. Solution. Indeed, we must find the unit vector u such that Du f (1, 0) = 1.

(4.16)

4.4 Solved Problems

159

− → − → Suppose that u = a i + b j such that a 2 + b2 = 1. We have ∂ ∂ 2 (x 2 + sin(x y)), (x + sin(x y)) ∂y

∂x = 2x + y cos(x y), x cos(x y) ,

∇ f (x, y) =

and hence ∇ f (1, 0) = (2, 1). Next, we obtain ∇ f (1, 0) · u = (2, 1) · (a, b) = 2a + b.

(4.17)

By (4.16) and (4.17), we conclude that 2a + b = 1. Therefore, we must determine the solutions to the following system of equations: 2a + b = 1 and a 2 + b2 = 1. We find the solutions as follows: 1 = a 2 + b2 = a 2 + (1 − 2a)2 = 5a 2 − 4a + 1, and so we get a = 0 or a = 4/5. Therefore, the unit vector u = (a, b) should be (0, 1) or (4/5, 3/5). 203. Let f (x, y) = ln(x 2 + x y + 1). Find a unit vector in which the value of f is not changing at the point (1, 1). Solution. Using technical terms, if a unit vector u is such that the directional derivative of f in the direction of u is zero, we have Du f = 0. This yields that ∇ f · u = 0, i.e., the directional derivative is zero along the directions perpendicular to the gradient vector. Intuitively, the gradient vector tells us the direction in which the change occurs and also tells us that no change occurs in its orthogonal direction. Hence, the directions in which f are not changing are those orthogonal to the ∇ f . Since ∇ f (x, y) =

x2

x 2x + y , 2 , + xy + 1 x + xy + 1

it follows that ∇ f (1, 1) = (1, 1/3). Hence, the vector (a, b) is orthogonal to ∇ f precisely when (1, 1/3) · (a, b) = 0. This yields that a + b/3 = 0, and hence b = −3a. This implies that the desired vectors are of√the form (a, −3a)√= a(1, −3). Consequently, there exist two such unit vectors 1/ 10(1, −3) and 1/ 10(−1, 3). 204. Find the direction from the point (1, 3) for which the value of f does not change y . if f (x, y) = e2y tan−1 3x Solution. We must find a direction such that the directional derivative is zero, i.e., Du (x, y) = 0. Let u = (cos θ, sin θ). Because

160

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

f x (x, y) = we obtain

y 3xe2y −3ye2y 2y −1 + and f (x, y) = 2e tan y 9x 2 + y 2 3x 9x 2 + y 2 1 − 1 − → → 1 ∇ f (1, 3) = − e6 i + πe6 + e6 j . 2 2 6

If Du f (x, y) = 0, then ∇ f (1, 3) · u = 0. this gives that 1 1

1 − e6 cos θ + πe6 + e6 sin θ = 0, 2 2 6 and consequently we get θ = tan−1

3 . 3π + 1

205. Find all points (x, y) at which the tangent plane to the surface 5 2 y − 2x + 3y 2

z = x 2 − 3x y +

is parallel to the plane given by the equation 2x − 4y + z = 6. Solution. The normal vector of the plane is (2, −4, 1). Now, we take z = f (x, y). So, z − f (x, y) = 0 We can find the normal vector to the tangent plane by using the gradient. We see that the gradient is

−

∂f ∂f ,− , 1 = − (2x − 3y − 2), −(−3x + 5y + 3), 1 . ∂x ∂y

Since the planes are parallel, it follows that their normal vectors are parallel, i.e., they are multiples of each other. Hence, we can write

− (2x − 3y − 2), −(−3x + 5y + 3), 1 = k(2, −4, 1).

From this equation, we conclude that k = 1. So, we must have 2 = −(2x − 3y − 2) and − 4 = −(−3x + 5y) + 3. Solving the equation, we get x = 3 and y = 2. 206. Show that √ the sum of the intercepts of the tangent plane to the surface √ √ y + z = a at any of its points is equal to a. Solution. A normal vector at (x0 , y0 , z 0 ) is

1 1 1 √ , √ , √ 2 x0 2 y0 2 z 0

or

1 1 1 √ ,√ ,√ x0 y0 z0

.

√

x+

4.4 Solved Problems

161

So, an equation of the tangent plane at (x0 , y0 , z 0 ) is 1 1 1 √ (x − x0 ) + √ (y − y0 ) + √ (z − z 0 ) = 0, x0 y0 z0 or equivalently √ √ √ 1 1 1 √ √ + √ + √ = x0 + y0 + z 0 = a x0 y0 z0 √ √ √ √ Hence, the x-intercept is a x0 , the y-intercept is a y0 and the z-intercept is √ √ a z 0 . Consequently, the sum of intercepts is √ √ √ √ √ √ a x0 + y0 + z 0 = a a = a. 207. Show that every tangent plane to the cone z 2 = x 2 + y 2 (see Fig. 4.4) passes through the origin. Solution. Suppose that f (x, y, z) = x 2 + y 2 − z 2 . The given surface is f (x, y, z) = 0. Hence, we have − → − → − → ∇ f (a, b, c) = 2a i + 2b j − 2c k is normal to the given surface at the point (a, b, c). Thus, an equation of the plane which is tangent to the given surface at the point (a, b, c) is 2a(x − a) + 2b(y − b) − 2c(z − c) = 0 or ax + by + cz − a 2 − b2 + c2 = 0.

(4.18)

Now, since (a, b, c) is the point of tangency, it must also lies on the surface. This implies that a 2 + b2 = c2 . Using this fact, the Eq. (4.18) of tangent plane can be written as ax + by + cz = 0, and (0, 0, 0) satisfies this equation. Since (a, b, c) is Fig. 4.4 Every tangent plane to the cone z 2 = x 2 + y 2 passes through the origin

162

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

an arbitrary point on the surface, and its tangent plane passes through the origin, we have all tangent planes to the surface passing through the origin. 208. Two surfaces are called orthogonal at a point of intersection if their normal lines are perpendicular at that point. Show that the sphere x 2 + y 2 + z 2 = 1 and the cone z 2 = x 2 + y 2 are orthogonal at all points of intersection. Solution. Assume that S1 : x 2 + y 2 + z 2 = 1 and S2 : z 2 = x 2 + y 2 intersect at (a, b, c). We find a normal vector to each surface at the point (a, b, c). In order to do this, suppose that f (x, y, z) = x 2 + y 2 + z 2 and g(x, y, z) = x 2 + y 2 − z 2 . Then, the vectors − → − → − → ∇ f (a, b, c) = 2a i + 2b j + 2c k − → − → − → ∇g(a, b, c) = 2a i + 2b j − 2c k are normal to S1 and S2 at the point (a, b, c), respectively. Consequently, these vectors are parallel to the normal lines to S1 and S2 at the point (a, b, c). Showing that the surfaces are orthogonal is equivalent to showing that ∇ f (a, b, c) · ∇g(a, b, c) = 0. To prove this equality, we have ∇ f (a, b, c) · ∇g(a, b, c) = 4(a 2 + b2 − c2 ). Since (a, b, c) is a point of intersection of S1 and S2 , it must satisfy both equations. In particular, since it satisfies S2 , it follows that a 2 + b2 = c2 . Using this fact, we see that ∇ f (a, b, c) · ∇g(a, b, c) = 4(a 2 + b2 − c2 ) = 4(a 2 − a 2 ) = 0. This yields that the surfaces are orthogonal to one another at the point of intersection. 209. A parametrization surface is a vector values function R(u, v) = x(u, v), y(u, v), z(u, v) , where x, y and z are three functions of two variables. The simplest type of parametric surfaces is given by the graph of a function of two variables z = f (x, y),

R(u, v) = x(u, v), y(u, v), f x(u, v), y(u, v) . (1) Let R : U → S be a parametrization of a surface S in some neighborhood of point P = (x0 , y0 , z 0 ). Show that the tangent plane to S at P has an equation as follows ∂R ∂u

(P) ×

∂R (P) · x − x0 , y − y0 , z − z 0 = 0. ∂v

4.4 Solved Problems

163

(2) Let f : R → R be a continuous differentiable function. Define a surface S as y − z = 0, x = 0 . Show that all tangent planes of S S = (x, y, z) | x f x pass through the origin. Solution. (1) Let R has partial derivative vectors ∂x ∂ y ∂z ∂x ∂ y ∂z ∂R ∂R = , , and = , , ∂u ∂u ∂u ∂u ∂v ∂v ∂v ∂v ∂R ∂R (P) × (P) is perpen∂u ∂v dicular to the tangent plane. Now, the point P = (x0 , y0 , z 0 ) itself clearly satisfies the equation. y . Hence, it has parametrization R(x, y) = (2) The surface is the graph z = x f x y ∂R ∂R x, y, x f . Now, we compute and as follows: x ∂x ∂y in the tangent plane at any point P. Thus, the vector

y y y y ∂R ∂R − f and . (x, y) = 1, 0, f (x, y) = 0, 1, f ∂x x x x ∂y x So, we obtain y y y y ∂R ∂R × (x, y) = − f + f ,− f ,1 . ∂x ∂y x x x x Consequently, an equation of the tangent plane at (x0 , y0 , z 0 ) ∈ S is

− f

y 0

x0

+

y

y0 y0 0 ,− f , 1 · x − x0 , y − y0 , z − z 0 = 0. f x0 x0 x0

This plane passes through the origin if and only if

− f

y 0

x0

+

y

y0 y0 0 ,− f , 1 · x0 , y0 , z 0 = 0. f x0 x0 x0

Indeed, taking into account that f

y 0

x0

=

z0 , we have x0

y y y y

0 0 0 0 + ,− f , 1 · x0 , y0 , z 0 − f f x x0 x0 yx0 z0 0 y0 0 = − x0 + y0 f − y0 f + z 0 = 0. x0 x0 x0

210. Find the point on the surface x 3 − 2y 2 + z 2 = 27 where the tangent plane is perpendicular √ to the line given parametrically as x(t) = 3t − 5, y(t) = 2t + 7 and z(t) = 1 − 2t.

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4 Partial Derivatives, Directional Derivatives and Gradient Vectors

Solution. We consider f (x, y, z) = x 3 − 2y 2 + z 2 − 27. Then, we have ∇ f (x, y, z) = (3x 2 , −4y, 2z). √ Now, the points in the line are given by √ (−5, 7, 1) + t (3, 2, − 2). Moreover, the direction vector of the line is (3, 2, − 2). We know that the line is perpendicular to the plane if the direction vector of the line is parallel to the normal vector of the plane. Therefore, we want to find all points (x, y, z) on the surface such that √ ∇ f (x, y, z) = (3x 2 , −4y, 2z) = k(3, 2, − 2), √ for some non-zero real number k. This yields that 3x 2 = 3k, −4y = 2k, 2z = − 2k, and so we get √ k k x = ± k, y = − and z = − √ . 2 2 Since (x, y, z) lies on the surface, it follows that 27 = x 3 − 2y 2 + z 2 . If x =

√

(4.19)

k, then by (4.19) we get 27 = k 3/2 −

2k 2 k2 + = k 3/2 , 4 2

which √ implies that k = 9. Hence, the corresponding point on the surface is (3, −9/2, −9/ 2). Indeed, in this case, we have one point on the surface in which the tangent plane is perpendicular to the line. √ If x = − k, then by (4.19), we conclude that 27 = −k 3/2 , which has no solution as k 3/2 ≥ 0. 211. At what point on the ellipsoid x 2 /4 + y 2 + z 2 = 9 is the tangent plane parallel to the plane x + y + z = 1? Solution. The ellipsoid is the level surface of the function f (x, y, z) =

x2 + y2 + z2. 4

Hence, we have f x (x, y, z) = x/2, f y (x, y, z) = 2y and f z (x, y, z) = 2z. Therefore, the tangent plane of the ellipsoid at (x, y, z) has normal vector (x/2, 2y, 2z). In order to make it parallel to the plane x + y + z = 1, which has normal vector (1, 1, 1), we must have x/2 = 2y = 2z. Putting this into the equation of the ellip2 2 2 2 y 2 + y 2 = 9, which implies that soid √ (2y) √ +√ √ √ √ y= √ x /4 + y + z = 9, we obtain ± 6/2. Therefore, (x, y, z) = (2 6, 6/2, 6/2) or (−2 6, − 6/2, − 6/2).

4.4 Solved Problems

165

At these two points on the ellipsoid, the tangent plane is parallel to the plane x + y + z = 1. 212. Let U denote an open and convex subset of Rn . Suppose that f : U → R is differentiable at every

u ∈ U . Fix u and v in U , and define g : [0, 1] → R by g(t) = f u + t (v − u) , for all 0 ≤ t ≤ 1. (1) Show that g is differentiable on (0, 1) and that

g (t) = ∇ f u + t (v − u) · (v − u), for all 0 < t < 1. (2) Use the mean value theorem to show that there exists a point z in the line segment connecting u to v such that f (v) − f (u) = Dw f (z)v − u, where w is the unit vector in the direction of the vector v − u. Solution. (1) We can write g = f ◦ γ, where γ : [0, 1] → Rn is the path given by γ(t) = u + t (v − u), for all t ∈ [0, 1]. By the chain rule, we conclude that g is differentiable in (0, 1) and

g (t) = ∇ f γ(t) · γ (t) = ∇ f u + t (v − u) · (v − u), for all t ∈ (0, 1). (2) The mean value theorem implies that there exists ξ ∈ (0, 1) such that g(1) − g(0) = g (ξ)(1 − 0). This yields that

f (v) − f (u) = ∇ f u + ξ(v − u) · (v − u). If we put z = u + ξ(v − u) and w = (v − u)/v − u, then we can write f (v) − f (u) = ∇ f (z) ·

1 (v − u) v − u = Dw f (z)v − u, v − u

as desired. 213. Let f (x, y, z) = e−(x+y) + z 2 (x + y). Suppose that a piece of fruit is sitting on a table in a room, and at each point (x, y, z) in the space within the room, f (x, y, z) gives the strength of the odor of the fruit. Furthermore, suppose that a certain bug always flies in the direction in which the fruit odor increases fastest. Suppose also that the bug always flies with a speed of 2 feet/second. What is the velocity vector of the bug when it is at the position (2, −2, 1)? 2

Solution. Since the bug flies in the direction in which the fruit odor increases fastest, it flies in the direction of ∇ f . The gradient of f is

166

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

Fig. 4.5 The angle between the tangent vector of the curve and the tangent plane of the surface at the point P

∇f (2, 2, −3)

R (2) R(t)

Φ P•

− −

− → → → 2 2 − 2(x + y)e−(x+y) + z 2 i + − 2(x + y)e−(x+y) + z 2 j + 2z(x + y) k ,

− → − → and so ∇ f (2, −2, 1) = i + j . The bug always has a speed of 2, so the velocity vector must have a magnitude of 2. A vector with magnitude 2 and in the same direction as the gradient is 2

∇ f (2, −2, 1) 2 − → − → = √ ( i + j ). ∇ f (2, −2, 1) 2

− → − → − → 214. The curve R(t) = t 2 /2 i + 4/t j + (t/2 − t 2 ) k intersects the hyperbolic paraboloid x 2 − 4y 2 − 4z = 0 at the point P = (2, 2, −3). What is the angle of intersection? Solution. Indeed, we must find the angle between the tangent vector of the curve and the tangent plane of the surface at the point of intersection (Fig. 4.5). It is clear that the curve passes through the point (2, 2, −3) at t = 2. Since − → − → − → R (t) = t i − 4/t 2 j + (1/2 − 2t) k , it follows that R (2) = (2, −1, −7/2). Tak− → − → − → ing f (x, y, z) = x 2 − 4y 2 − 4z, we get ∇ f (x, y, z) = 2x i − 8y j − 4 k , and so ∇ f (2, 2, −3) = (4, −16, −4). If θ is the angle between R (2) and ∇ f (2, 2, −3), it follows that cos θ =

∇ f (2, 2, −3) 19 √ R (2) · = 138 ≈ 0.539, R (2) ∇ f (2, 2, −3) 414

so that θ is approximately 1 radian. Since the gradient is normal to the tangent plane, we obtain = π/2 − θ ≈ 1.57 − 1.00 = 0.57 radian.

4.5 Exercises

167

4.5 Exercises Easier Exercises 1. Evaluate

∂3 f (x)g(y)h(z) ∂x∂ y∂z

where f , g and h are differentiable functions of a single variable. 2. Show that the function 1 2 u = √ e−x /4t (t > 0) t is a solution of the partial differential equation ∂ 2 u/∂x 2 = ∂u/∂t. Find a similar solution involving an exponential. 3. Suppose that ⎧ 3 ⎨ x + y3 if (x, y) = (0, 0) f (x, y) = x 2 + y 2 ⎩ 0 if (x, y) = (0, 0). Find f x (0, 0) and f y (0, 0). 4. Suppose that ⎧ 2 ⎨ x − xy if (x, y) = (0, 0) f (x, y) = ⎩ x+y 0 if (x, y) = (0, 0). Find

5. 6. 7. 8.

9.

(a) f x (0, y) if y = 0, and f x (0, 0); (b) f y (x, 0) if x = 0, and f y (0, 0). √ If f (x, y, z) = x y 2 z 3 + sin−1 (x z), find f x zy . Hint: Which order is easiest? √ of differentiation √ If g(x, y, z) = 1 + x z + 1 − x y, find gx yz . Hint: Use a different order of differentiation for each term. ∂ m+n u if u = x m y n . Find ∂x m ∂ y n Find the slope of the tangent line to the curve of intersection of the √ surface 36x 2 − 9y 2 + 4z 2 + 36 = 0 with the plane x = 1 at the point (1, 12, −3). Interpret this slope as a partial derivative. Verify that u(x, y, z) = e3x+4y sin 5z satisfies Laplace’s equation in R3 : ∂2u ∂2u ∂2u + 2 + 2 = 0. 2 ∂x ∂y ∂z

168

4 Partial Derivatives, Directional Derivatives and Gradient Vectors

10. Find the set of all points (a, b, c) in R3 for which the two sphere (x − a)2 + (y − b)2 + (z − c)2 = 1 and x 2 + y 2 + z 2 = 1 intersect orthogonally. (Their tangent planes should be perpendicular at each point of intersection.) 11. Let f (x, y) be differentiable on the interior of a circle or a rectangle D, and suppose that ∇ f (x, y) = 0 at every point of D. Show that f (x, y) is constant on D. 12. Show that the spheres x 2 + y 2 + z 2 = a 2 and (x − b)2 + y 2 + z 2 = (b − a)2 are tangent at the point (a, 0, 0). 13. Prove that every normal line to the sphere x 2 + y 2 + z 2 = a 2 passes through the center of the sphere. 14. Find the directions in which the directional derivative of f (x, y) = x 2 + sin(x y) at the point (1, 0) has the value 1. 15. The atmospheric pressure in a region of space near the origin is given by the formula P = 30 + (x + 1)(y + 2)e z . Approximately where is the point closest to the origin at which the pressure is 31.1? 16. What is the maximum rate of ascent of the surface z = f (x, y) = x y at the point (e, 1), and in which direction does it occur. 17. What is the maximum rate of increase of the surface f (x, y, z) = x y 2 z 3 at the point (2, 1, −1), and in which direction does it occur. 18. Give the equation of the tangent plane to each of these surfaces at the point indicated. (a) z = x y 2 at (1, 1, 1);

(b) z = y 2 /x at (1, 2, 4).

Harder Exercises 19. If u = ea1 x1 +a2 x2 +···+an xn , where a12 + a22 + · · · + an2 = 1, show that ∂2u ∂2u ∂2u + + · · · + = u. ∂xn2 ∂x12 ∂x22 20. Find the equation of the tangent plane to the cone z = x 2 + y 2 at the point P0 = (x0 , y0 , z 0 ) on the cone. Write parametric equations for the ray from the origin passing through P0 , and using them, show the ray lies on both the cone and the tangent plane at P0 . 21. Prove that if f is a function of two variables and all the partial derivatives of f up to the fourth order are continuous on some open disk, then ∂4 f ∂ 2 x∂ 2 y

=

∂4 f . ∂ y∂x∂ y∂x

4.5 Exercises

169

√ 22. Find the directional derivative of f (x, y) = x y at P = (2, 8) in the direction of Q = (5, 4). 23. Find the directional derivative of f (x, y, z) = x y + yz + zx at P = (1, −1, 3) in the direction of Q = (2, 4, 5). 24. A cylinder whose equation is y = f (x) is tangent to the surface z 2 + 2x y + y = 0 at all points common to the two surfaces. Find f (x). 25. If f (x, y) satisfies the partial differential equation ∂2 f (x, y) = 0, ∂x∂ y prove that f (x, y) = g(x) + h(y), where g(x) is a function of x alone and h(y) is a function of y alone. 26. Find the points at which the gradient of the function f (x, y) = ln(x −1 + y) is − → − → equal to (−16/9) i + j . 27. Let f be a function of two variables that has continuous partial derivatives, and consider the points A = (1, 3), B = (3, 3), C = (1, 7) and D = (6, 15). The −→ directional derivative of f at A in the direction of the vector AB is equal to 3, −→ and the directional derivative at A in the direction of the vector AC is equal to −→ 26. Find the directional derivative of f at A in the direction of the vector AD.

Chapter 5

Differentiability and Differential

5.1 Differentiability We know that if f is a differentiable function of the single variable x and y = f (x), then the increment Δy of the dependent variable can be expressed as Δy = f (x)Δx + Δx, where is a function of Δx and → 0 as Δx → 0. This means that if the function f is differentiable at x0 , the increment of f at x0 , denoted by Δf (x0 ), is given by Δf (x0 ) = f (x)Δx + Δx,

(5.1)

where lim = 0. Now, we will define differentiability for a function f (x, y) of Δx→0

two variables by generalizing formula (5.1). We begin by defining the increment of such a function. Let f be a function of two variables x and y. The increment of f at the point (x0 , y0 ) denoted by Δf (x0 , y0 ) is given by Δf (x0 , y0 ) = f (x0 + Δx, y0 + Δy) − f (x0 , y0 ). If the increment of f at (x0 , y0 ) can be written as Δf (x0 , y0 ) = f x (x0 , y0 )Δx + f y (x0 , y0 )Δy + 1 Δx + 2 Δy, where 1 and 2 are functions of Δx and Δy such that 1 → 0 and 2 → 0 as (Δx, Δy) → (0, 0), then f is said to be differentiable at (x0 , y0 ). The concept of differentiability is now extended to a function of n variables. If f x1 , . . . , xn ), then the is a function of the n variables x1 , . . . , xn and P0 is the point ( increment of f at P0 is given by © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_5

171

172

5 Differentiability and Differential

Δf (P0 ) = f ( x1 + Δx1 , . . . , xn + Δxn ) − f (P0 ). If the increment of f at the point P0 can be written as f x1 (P0 )Δx1 + · · · + f xn (P0 )Δxn + 1 Δx1 + · · · + n Δxn , where 1 → 0, . . . , n → 0 as (Δx1 , . . . , Δxn ) → (0, . . . , 0), then f is said to be differentiable at P0 . We call f differentiable if it is differentiable at every point in its domain. Standard linear approximation: The linearization of a function f of two variables at a point (x0 , y0 ), where f is differentiable, is the function L(x, y) = f (x0 , y0 ) + f x (x0 , y0 )(x − x0 ) + f y (x0 , y0 )(y − y0 ). The approximation f (x, y) ≈ L(x, y) is the standard linear approximation of f at (x0 , y0 ). The error in the standard linear approximation: If f has continuous first and second partial derivatives throughout an open set containing a rectangle R centered at (x0 , y0 ) and if M is any upper bound for the values of | f x x |, | f yy | and | f x y | on R, then the error E(x, y) incurred in replacing f (x, y) on R by L(x, y) satisfies the inequality 2 1 |E(x, y)| ≤ M |x − x0 | + |y − y0 | . 2 Analogous results hold for differentiable functions of more than two variables. Continuity of partial derivatives implies differentiability: If the partial derivatives f x1 , . . . , f xn of a function f are continuous throughout an open region S, then f is differentiable at every point of S. Differentiability implies continuity: If a function f of n variables is differentiable at P0 , then f is continuous at P0 . A function f of two variables is differentiable at (a, b) if (1) f is continuous at (a, b); and (2) there is a linear function L(x, y) such that lim

(x,y)→(0,0)

f (x, y) − L(x, y) (x − a)2 + (y − b)2

= 0.

If f is continuous at (a, b), and the partial derivatives f x and f y are both defined and continuous at (a, b), then f is differentiable at (a, b). The following theorem is the mean value theorem for a function of a single variable applied to a function of two variables. Let f be a function of two variables defined for all x ∈ [a, b] and y ∈ [c, d]. (1) If f x (x, y0 ) exists for some y0 ∈ [c, d] and for all x ∈ [a, b], then there exists ξ1 ∈ (a, b) such that f (b, y0 ) − f (a, y0 ) = (b − a) f x (ξ1 , y0 ).

5.2 The Chain Rule

173

(2) If f y (x0 , y) exists for some x0 ∈ [a, b] and for all y ∈ [c, d], then there exists ξ2 ∈ (c, d) such that f (x0 , d) − f (x0 , c) = (d − c) f y (x0 , ξ2 ). If f is a function of two variables x and y, and f is differentiable at (x, y), then the total differential of f is the function d f having function values given by d f (x, y, Δx, Δy) = f x (x, y)Δx + f y (x, y)Δy. Since the increments and differentials of the independent variables are equal, we can write ∂f ∂f df = dx + dy. (5.2) ∂x ∂y The generalization of formula (5.2) to the case of a function f of n variables is df =

∂f ∂f d x1 + · · · + d xn . ∂ x1 ∂ xn

5.2 The Chain Rule Chain rule for functions of two independent variables: If z = f (x, y) has continuous partial derivatives f x and f y , and if x = x(t) and y = y(t) are differentiable functions of t, then the composite z = f (x(t), y(t)) is a differentiable function of t and ∂ f dx ∂ f dy df = + . dt ∂ x dt ∂ y dt Chain rule for functions of three independent variables: If w = f (x, y, z) is differentiable and x, y and z are differentiable functions of t, then f is a differentiable function of t and df ∂ f dx ∂ f dy ∂ f dz = + + . dt ∂ x dt ∂ y dt ∂z dt Chain rule for two independent variables and three intermediate variables: Suppose that w = f (x, y, z), x = g(r, s), y = h(r, s) and z = k(r, s). If all four functions are differentiable, then f has partial derivatives with respect to r and s, given by the formulas ∂ f ∂x ∂ f ∂y ∂ f ∂z ∂f = + + , ∂r ∂ x ∂r ∂ y ∂r ∂z ∂r (5.3) ∂f ∂ f ∂x ∂ f ∂y ∂ f ∂z = + + . ∂s ∂ x ∂s ∂ y ∂s ∂z ∂s If f is a function of two variables instead of three, each equation in (5.3) becomes correspondingly one term shorter.

174

5 Differentiability and Differential

A formula for implicit differentiation: Let f be a function of two variables. If f is differentiable and that equation f (x, y) = 0 defined by y as a differentiable function of x, then at any point where f y = 0, dy fx =− . dx fy The technique of implicit differentiation can also be used to calculate partial derivatives. Given a function F(x, y, z) of three variables and suppose that the equation F(x, y, z) = 0 defines z as an implicit function of x and y, in the sense that there exists a function z = f (x, y) such that F x, y, f (x, y) = 0 on some open set. Then, we have Fy (x, y, z) Fx (x, y, z) ∂z ∂z =− and =− , ∂x Fz (x, y, z) ∂y Fz (x, y, z) provided that Fz (x, y, z) = 0.

5.3 Taylor Series Let f be a function of two variables all of whose nth partial derivatives exist at the point (a, b), for all n ∈ N. Then, the Taylor series for f at (a, b) is T (x, y) =

∞ n ∂n f (x − a)k (y − b)n−k (a, b) k!(n − k)! (∂ x)k (∂ y)n−k n=0 k=0

The Taylor series has the same partial derivatives as the original function at (a, b). Taylor’s formula for f (x, y) at the point (a, b): Suppose that f and its partial derivatives through order n + 1 are continuous throughout an open rectangular region R centered at a point (a, b). Then, throughout R, f (x, y) = f (a, b) + (x − a) f x + (y − b) f y (a,b) 1 2 (x − a) f x x + 2(x − a)(y − b) f x y + (y − b)2 f yy + (a,b) 2! 1 (x − a)3 f x x x + 3(x − a)2 (y − b) f x x y + 3(x − a)(y − b)2 f x yy + 3! ∂ ∂ n 1 (x − a) + (y − b) +(y − b)3 f yyy + ··· + f (a,b) (a,b) n! ∂ x ∂y n+1 ∂ ∂ 1 (x − a) + (y − b) + f (ξ1 ,ξ2 ) (n + 1)! ∂x ∂y The first n derivative terms are evaluated at (a, b). The last term is evaluated at some point (ξ1 , ξ2 ) on the line segment joining (a, b) and (x, y).

5.4 Solved Problems

175

We define the degree m Taylor polynomial of f to be the terms in the Taylor series up to degree m in x and y. Taylor remainder theorem: If f has continuous partial derivatives up to order n + 1 near (a, b), then and Tn (x, y) is the degree n Taylor polynomial for f at (a, b), then for any point (x, y), we have | f (x, y) − Tn (x, y)| ≤ M

(|x − a| + |y − b|)n+1 , (n + 1)!

where M is a constant such that | f | ≤ M for every (n + 1)th partial derivatives f at the segment joining (a, b) to (x, y).

5.4 Solved Problems 215. Let f (x, y) = |x y| for all x, y ∈ R. Show that (1) f is differentiable at (0, 0); (2) f x (0, b) and f y (a, 0) do not exist if a = 0 and b = 0. Solution. (1) Clearly, we have f x (0, 0) = 0 and f y (0, 0) = 0. Now, we can write √ f (h, k) − f (0, 0) − f x (0, 0)h − f y (0, 0)k h 2 + k 2 |k| |hk| =√ ≤ √ = |k|. √ h2 + k2 h2 + k2 h2 + k2 Since |k| → 0 as (h, k) → (0, 0), it follows that f is differentiable at (0, 0). (2) For b = 0, we observe that lim

h→0

f (h, b) − f (0, b) |h| |b| |h| = lim = |b| lim . h→0 h→0 h h h

Since the above limit does not exist, it follows that f x (0, b) does not exist. Analogously, f y (a, 0) does not exist if a = 0. 216. Let g be a continuous function on [a, b] and h be a continuous function on [c, d]. Show that x y f (x, y) = g(u)du h(v)dv a

c

is differentiable at (x, y) ∈ [a, b] × [c, d]. Solution. Since g and h are continuous, it follows that d dx

a

x

d y g(u)du = g(x) and h(v)dv = h(y). dy c

176

5 Differentiability and Differential

Now, we obtain df = g(x) dx

y

h(v)dv c

and

df = dy

x

g(u)du h(y).

a

Each of ∂ f /∂ x and ∂ f /∂ y is a product of two continuous functions and so they are continuous. Since the continuity of partial derivatives implies differentiability, it follows that f is differentiable. 217. Suppose that ⎧ ⎨ (x 2 + y 2 ) sin 1 if (x, y) = (0, 0) f (x, y) = x 2 + y2 ⎩ 0 if (x, y) = (0, 0) Show that the partial derivatives of f are not continuous at (0, 0) but f is differentiable at this point. Solution. By standard differentiation techniques, we obtain x cos (x 2 + y 2 )−1/2 f x (x, y) = − + 2x sin (x 2 + y 2 )−1/2 if (x, y) = (0, 0), x 2 + y2 f (x, 0) − f (0, 0) 1 = lim x sin √ = 0. f x (0, 0) = lim x→0 x→0 x −0 x2 Now, we claim that

lim

(x,y)→(0,0)

lim

f x (x, y) does not exist. It is easy to see that

(x,y)→(0,0)

2x sin (x 2 + y 2 )−1/2 = 0.

So, the second part of the limit is well-behaved. For the limit of the first part, we consider the path y = x. Then we have

lim

(x,y)→(0,0)

x cos (x 2 + y 2 )−1/2 x cos (x 2 + x 2 )−1/2 = lim √ 2 2 x→0 x 2 + y2 x + x 2 −1/2 x cos (2x ) = lim , √ x→0 |x| 2

which does not exist. In a similar way, we observe that

lim

(x,y)→(0,0)

exist. Thus, the partial derivatives of f fail of continuity at (0, 0).

f y (x, y) does not

5.4 Solved Problems

177

Now, we show that f is differentiable at (0, 0). We can write 1 f (0 + h, 0 + k) − f (0, 0) = f (h, k) = (h 2 + k 2 ) sin √ + k2 h2 1 1 = f x (0, 0)h + f y (0, 0)k + h sin √ h + k sin √ k h2 + k2 h2 + k2 = f x (0, 0)h + f y (0, 0)k + 1 (h, k)h + 2 (h, k)k, where

lim

(h,k)→(0,0)

1 (h, k) = 0 and

lim

(h,k)→(0,0)

2 (h, k) = 0. Hence, we have written

f (0 + h, 0 + k) is the required form, and so we conclude that f is differentiable at (0, 0). 218. Suppose that ⎧ ⎨ xy if (x, y) = (0, 0) f (x, y) = x 2 + y2 ⎩0 if (x, y) = (0, 0) Show that f is continuous at (0, 0) and has the partial derivatives f x (0, 0) and f y (0, 0) but it is not differentiable at (0, 0). Solution. The continuity is verified by using polar coordinates. If x = r cos θ and y = r sin θ , then lim

(x,y)→(0,0)

r 2 cos θ sin θ = lim r cos θ sin θ = 0. r →0 r →0 r

f (x, y) = lim

Since f (0, 0) = 0, it follows that f is continuous at (0, 0). Since f (x, 0) = 0 and f (0, y) = 0, it is clear that f x (0, 0) = 0 and f y (0, 0) = 0. Now, suppose that f is differentiable at (0, 0). Then the linear approximation would have to be L(x, y) = f (0, 0) + 0(x − 0) + 0(y − 0) = 0. Moreover, the following limit lim

(x,y)→(0,0)

f (x, y) − L(x, y) x 2 + y2

must be equal to 0. In our case, our limit is lim

(x,y)→(0,0)

xy . x 2 + y2

If this limit exists, it must be equal to the limit as (x, y) approaches the origin along the line y = x, which would be

178

5 Differentiability and Differential

lim

x→0

x2 1 = . x2 + x2 2

In particular, this value is not 0, so the original limit could not be equal to 0. Therefore, we conclude that f is not differentiable at (0, 0). 219. Let f : R2 → R be the function defined in Problem 178. (1) Prove that f is differentiable at (0, 0); (2) Prove that f is not differentiable at any point (x, y) = (0, 0); (3) At what points (x, y) ∈ R2 does ∂ f /∂ x exist, and at what points does ∂ f /∂ y exist? What are the values of the partial derivatives at points where they do exist? Solution. (1) For each non-zero real number h, we have f (h, 0) |h|2 = |h| → 0 as h → 0. ≤ h |h| So, the limit defining partial derivative of f with respect to x at (0, 0) is zero, i.e., f x (0, 0) = 0. Similarly, we obtain f y (0, 0) = 0. Now, we can write f (0 + h, 0 + k) − f (0, 0) = f (h, k) 1 f (h, k) 1 f (h, k) h+ k. = f x (0, 0)h + f y (0, 0)k + 2 h 2 k Since

it follows that

f (h, k) h 2 + k 2 ) ≤ → 0 as (h, k) → (0, 0), h h lim

(h,k)→(0,0)

f (h, k) = 0 and in a similar way we see that h

f (h, k) = 0. Therefore, f is differentiable at (0, 0). k (2) According to Problem 178, the function f is discontinuous at every point (x, y) = (0, 0), so it is not differentiable away from the origin. (3) In part (1) we observe that f x (0, 0) = 0 and f y (0, 0) = 0. Now, we consider the following cases. If y is non-zero and rational, then lim

(h,k)→(0,0)

f (x, y) =

x 2 + y 2 if x ∈ Q 0 if x ∈ / Q.

This function is discontinuous at every x ∈ R, and so ∂ f /∂ x does not exist. If y is irrational, then f (x, y) = 0 for all x ∈ R, and so ∂ f /∂ x = 0.

5.4 Solved Problems

179

The argument for the partial derivative with respect to y is the same. Therefore, we conclude that the partial derivative with respect to x exists if (x, y) = (0, 0) or y∈ / Q, and the partial derivative with respect to y exists if (x, y) = (0, 0) or x ∈ / Q. The partial derivatives are zero whenever they exist.

0.1 2 e−t dt using a linear approximation. 220. Approximate −0.1

Solution. We define

y

f (x, y) =

e−t dt. 2

x

Using the fundamental theorem of calculus, we obtain ∂ ∂x

y

e

−t 2

dt = −e

−x 2

x

∂ and ∂y

y

e−t dt = e−y , 2

2

x

and near the point (x0 , y0 ), we can write f (x, y) ≈ f (x0 , y0 ) +

or f (x, y) ≈

y0

∂f ∂f (x0 , y0 )(x − x0 ) + (x0 , y0 )(y − y0 ) ∂x ∂y

e−t dt − e−x0 (x − x0 ) + e y0 (y − y0 ). 2

2

2

x0

If (x0 , y0 ) = (0, 0), then f (x, y) ≈ −x + y. Hence, we get f (−0.1, 0.1) = 0.2. 221. Let f (x, y) = 0 defines y as an implicit function of x. Suppose that that the functions f (x, y) and y(x) are both differentiable. Show that d2 y b2 c − 2abd + a 2 e =− , 2 dx b3 where a =

∂f ∂2 f ∂2 f ∂2 f ∂f , b= , c= and e = , d= . 2 ∂x ∂y ∂x ∂ x∂ y ∂ y2

Solution. We use the chain rule to differentiate f (x, y) = 0 with respect to x: dx dy dy df = fx + fy = fx + f y = 0. dx dx dx dx Solving this equation for dy/d x, we get dy/d x = − f x / f y . Hence, we can write

180

5 Differentiability and Differential

Fig. 5.1 A labelled triangle

B c

a C

=− =− =−

A

d fy d fx − fx fy dy d fx d x dx =− = − dx dx fy f y2 dy dy − f x f yx + f yy f y f x x + f yx dx dx f y2 f x f x f y fx x + fx y − − f x f x y + f yy − fy fy 2 fy f y ( f y f x x − f x f x y ) − f x ( f y f x y − f x f yy ) f y3 2 f y f x x − 2 f x f y f x y + f x2 f yy b2 c − 2abd + a 2 e = − . f y3 b3

d2 y d = 2 dx dx =−

b

222. Consider the triangle shown in Fig. 5.1. ∂A ∂A and ; ∂a ∂b ∂a ∂a (2) Express a implicitly as a function of A, b and B, and calculate and . ∂A ∂B

(1) Express A implicitly as a function of a, b and c, and calculate

Solution. (1) By the cosine rule in the triangle, we have a 2 = b2 + c2 − 2bc cos A. ∂A , and so This implies that 2a = (2bc sin A) ∂a ∂A a = . ∂a bc sin A On the other hand, we obtain 0 = 2b − 2c cos A + (2bc sin A) ∂A c cos A − b = . ∂b bc sin A

∂A . This yields that ∂b

5.4 Solved Problems

(2) Since

181

b a = , it follows that sin A sin B (sin A)

This implies that (sin A)

∂a − a cos A ∂A = 0. 2 sin A

∂a − a cos A = 0 or ∂A a cos A ∂a = . ∂A sin A

Also, we obtain

1 ∂a = b(− csc B cot B), and so sin A ∂ B ∂a = −b csc B cot B sin A. ∂B z

223. Find the total differential for the function u = x y . Solution. First, we find the partial derivatives with respect to all three independent variables. Finding the partial derivative with respect to x, we have the power function with constant exponent y z . So, we obtain ∂u z = y z x y −1 . ∂x To find the partial derivative with respect to y, we use the chain rule. The outside function is the exponential function with constant base x and the variable exponent y z , which is the power function with respect to y. Hence, by the chain rule, we obtain ∂u z = x y (ln x)zy z−1 . ∂y To find the partial derivative with respect to z, we use the chain rule again. The outside function is the exponential function with a constant base x. The inside function is another exponential function y z with the constant base y. Therefore, by the chain rule, we get ∂u z = x y (ln x)y z (ln y). ∂z Now, using the above partial derivatives of u, we can write du = y z x y

z

−1

z

z

d x + x y (ln x)zy z−1 dy + x y (ln x)y z (ln y)dz z ln x 1 z = yz x y dx + dy + ln x ln ydz . x y

182

5 Differentiability and Differential

224. Let f be a twice differentiable function of two variables x and y. If x = au + bv and y = −bu + av where the constants a and b are connected by a 2 + b2 = 1. Prove that ∂2 f ∂2 f ∂2 f ∂2 f + 2 = + 2. 2 2 ∂x ∂y ∂u ∂v Solution. By using the chain rule, we obtain ∂ f ∂x ∂ f ∂y ∂f ∂f ∂f = + =a −b , ∂u ∂ x ∂u ∂ y ∂u ∂x ∂y and so ∂ 2 2 ∂2 f ∂ ∂ f ∂f ∂2 f 2∂ f 2∂ f − b a − b = a + b = a − 2ab . ∂u 2 ∂x ∂y ∂x ∂y ∂x2 ∂ x∂ y ∂ y2 Similarly, we get

(5.4)

∂f ∂ f ∂x ∂ f ∂y ∂f ∂f = + =b +a , ∂v ∂ x ∂v ∂ y ∂v ∂x ∂y

and hence ∂ ∂2 f ∂ ∂ f ∂f ∂2 f ∂2 f ∂2 f +a b +a = b2 2 + 2ab + a2 2 . = b 2 ∂v ∂x ∂y ∂x ∂y ∂x ∂ x∂ y ∂y

(5.5)

Since a 2 + b2 = 1, by (5.4) and (5.5), the result follows. 225. Suppose that f : R3 → R is a differentiable function, u is a unit vector in R3 , and R : R → R3 is a differentiable function representing the position of a moving object as a function of time. Let g(t) be the value of f at the object’s position at time t. Show that if at time t0 the object is at position (x0 , y0 , z 0 ) moving in the direction u with a speed of 1, then g (t0 ) = Du f (x0 , y0 , z 0 ). Solution. Assume that w = f (x, y, z) and write R in terms of its component R(t) = x(t), y(t), z(t) . Since x, y and z are functions of t, it follows that w is a function of t, i.e., w = f (t). Now, by the chain rule, we can write ∂w d x ∂w dy ∂w dz dw = + + dt ∂ x dt ∂ y dt ∂w ∂w ∂w d x∂zdydt dz , , · , , . = ∂ x ∂ y ∂z dt dt dt

(5.6)

Next, we rewrite (5.6) in terms of the functions and noting where these functions are to be evaluated g (t) = ∇ f x(t), y(t), z(t) · R (t) = ∇ f R(t) · R (t).

5.4 Solved Problems

183

When t = t0 we have R(t0 ) = (x0 , y0 , z 0 ). Also, R (t0 ) is the velocity of the moving object at time t0 . The velocity is a vector in the direction of motion (in this case, in the direction of u) whose magnitude is the speed (in this case, 1). Since u is a unit vector, a vector of length 1 in the direction of u is just u, and so R (t0 ) = u. Substituting back into our expression for g (t), we get g (t) = ∇ f R(t0 ) · R (t0 ) = ∇ f (x0 , y0 , z 0 ) · u = Du f (x0 , y0 , z 0 ), as desired. 226. Let f : R2 → R be a differentiable function and M be a real number such that | f x (x, y)| ≤ M and | f y (x, y)| ≤ M, for all (x, y) ∈ R2 . Show that | f (x, y) − f (a, b)| ≤ 2M (x − a)2 + (y − b)2 , for all (x, y), (a, b) ∈ R2 . Solution. Let (x, y) and (a, b) be arbitrary points in R2 . Since f is differentiable, by the mean value theorem, there exist ξ1 between x and a, and ξ2 between y and b such that f (x, y) − f (a, y) = (a − x) f x (ξ1 , y), f (a, y) − f (a, b) = (b − y) f y (a, ξ2 ). Now, by the triangle inequality and our assumptions, we obtain | f (x, y) − f (a, b)| = | f (x, y) − f (a, y) + f (a, y) − f (a, b)| ≤ | f (x, y) − f (a, y)| + | f (a, y) − f (a, b)| = |a − x| | f x (ξ1 , y)| + |b − y| | f y (a, ξ2 )|. ≤ |a − x|M + |b − y|M ≤ (x− a)2 + (y − b)2 M + (x − a)2 + (y − b)2 M = 2M (x − a)2 + (y − b)2 , as desired. 227. Suppose that f : R2 → R has continuous second partial derivatives. For (x0 , y0 ), (h, k) ∈ R2 , define F(h, k) = f (x0 + h, y0 + k) − f (x0 + h, y0 ) − f (x0 , y0 + k) − f (x0 , y0 ) . Show that (1) there exists ξ1 between x0 and x0 + h such that F(h, k) = f x (ξ1 , y0 + k) − f x (ξ1 , y0 ) h; (2) there exists ξ2 between y0 and y0 + k such that F(h, k) = f x y (ξ1 , ξ2 )hk;

184

5 Differentiability and Differential

1 F(h, k); (h,k)→(0,0) hk (4) f x y (x0 , y0 ) = f yx (x0 , y0 ).

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

lim

Solution. (1) We define g(x) := f (x, y0 + k) − f (x, y0 ). Then F(h, k) = g(x0 + h) − g(x0 ). By the mean value theorem there exists ξ1 between x0 and x0 + h such that (5.7) g(x0 + h) − g(x0 ) = g (ξ1 )h. Since g (ξ1 ) = f x (ξ1 , y0 + k) − f x (ξ1 , y0 ), the result follows. (2) The derivative of f x (ξ1 , y) relative to y exists. We apply the mean value theorem on [y0 , y0 + k]. Then, there exists ξ2 between y0 and y0 + k such that f x (ξ1 , y0 + k) − f x (ξ1 , y0 ) = f x y (ξ1 , ξ2 )k. This yields that g (ξ1 ) = f x y (ξ1 , ξ2 )k, and hence hg (ξ1 ) = f x y (ξ1 , ξ2 )hk. Now, by (5.7) we can write g(x0 + h) − g(x0 ) = f x y (ξ1 , ξ2 )hk. This means that F(x, y) = f x y (ξ1 , ξ2 )hk.

(5.8)

(3) Since f x y is continuous, it follows that f x y (x0 , y0 ) =

lim

(h,k)→(0,0)

f x y (x0 + h, y0 + k) =

lim

(h,k)→(0,0)

f x y (ξ1 , ξ2 ).

Now, by (5.7), we conclude that f x y (x0 , y0 ) =

lim

(h,k)→(0,0)

F(h, k) . hk

(5.9)

(4) By exchanging the rolls of x and y, we observe that f yx (x0 , y0 ) =

lim

(h,k)→(0,0)

F(h, k) . hk

(5.10)

Next, the result follows from (5.9) and (5.10). 228. Let f : [a, b] × [c, d] → R be a function of two variables, and ∂ f /∂ y exists on [a, b] × [c, d] and extends to a continuous function on [a, b] × [c, d]. If F(y) =

b f (x, y)d x, prove that a

d F(y) = dy

a

b

∂f (x, y)d x. ∂y

Solution. By the mean value theorem, there exists 0 < ξ < 1 such that f (x, y + h) − f (x, y) ∂f = (x, y + ξ h), h ∂y

5.4 Solved Problems

185

and (x, y + ξ h) − (x, y) = ξ |h| ≤ h. Since ∂ f /∂ y is continuous on [a, b] × [c, d], it follows that ∂ f /∂ y is uniformly continuous. Take δ > 0 such that (x, y) − (x , y ) ≤ δ, then ∂ f (x , y ) − ∂ f (x, y) ≤ . ∂y b−a ∂y Now, by using the above argument, we can write

b F(y + h) − F(y) ∂f − (x, y)d x a ∂y b h

b f (x, y + h) − f (x, y) ∂f = dx − (x, y)d x h ∂ y a

ab f (x, y + h) − f (x, y) ∂ f ≤ − (x, y) d x h ∂ y

a b ∂ f ∂ f (x, y + ξ h) − ≤ (x, y) d x ∂y ∂y

a b ≤ d x = (b − a) = . b − a b − a a This completes the proof.

x

x 1 1 1 −1 x 229. Using tan , find the value of d x = d x. 2 2 2 2 2 a a 0 x +a 0 (x + a ) Solution. Taking derivative from both sides of the given integral with respect to a we have

x ∂ 1 ∂ 1 −1 x tan . d x = ∂a 0 x 2 + a 2 ∂a a a Now, using Problem (228), we can write

x 0

and so we obtain

x 0

∂ 1 ∂ dx = 2 2 ∂a x + a ∂a

1 x tan−1 a a

,

∂ 1 −1 x 2a 1 ∂ −1 x tan + tan . d x = (x 2 + a 2 )2 a ∂a a ∂a a a

This implies that

x

2a 0

1 1 x x − dx = − tan−1 . (x 2 + a 2 )2 a(a 2 + x 2 ) a 2 a

If we divide the both sides of (5.11) by 2a, then

(5.11)

186

5 Differentiability and Differential

x 0

1 1 x x − 3 tan−1 . dx = − 2 2 2 2 2 +a ) 2a (a + x ) 2a a

(x 2 1

230. Evaluate 0

x −1 d x. ln x

1

Solution. Suppose that F(t) = 0

xt − 1 d x, where t ≥ 0. Then, we have ln x

1 t xt − 1 x ln x dx = dx ln x ln x 0

0 1 1 xtdx = . = t + 1 0

F (t) =

1

This implies that

∂ ∂t

F(t) =

1 dt = ln(t + 1) + C. t +1

Since F(0) = 0, it follows that C = 0. This yields that

1

0

∞

231. Evaluate F(t) =

x −1 d x = F(1) = ln 2. ln x

e−x

0

sin(t x) d x. x

Solution. We obtain F (t) =

0

∞

∂ ∂t

∞ sin(t x) e−x dx = e−x cos(t x)d x. x 0

By a simple computation, we find that

∞

e−x cos(t x)d x =

0

1 . 1 + t2

Hence, we have F (t) = 1/(1 + t 2 ), and so F(t) = tan−1 t + C. Since F(0) = 0, it follows that C = 0. Therefore, we conclude that F(t) = tan−1 t. 232. Let f : R2 → R be differentiable. (1) For (x0 , y0 ), (h, k) ∈ R2 , define g : [0, 1] → R by g(t) = f (x0 + th, y0 + tk), for all 0 ≤ t ≤ 1. Show that g is differentiable and g = h f x + k f y ; (2) If f (1, 2) = f (3, 4) = 0, show that there exists a point (a, b) lying in the segment joining (1, 1) to (3, 4) such that f x (a, b) = − f y (a, b).

5.4 Solved Problems

187

Solution. (1) Suppose that x(t) = x0 + th and y(t) = y0 + tk. By the chain rule, we conclude that g is differentiable and we get ∂ f dx ∂ f dy ∂f ∂f dg = + =h +k . dt ∂ x dt ∂ y dt ∂x ∂y (2) We consider (x0 , y0 ) = (1, 2) and (h, k) = (2, 2) in part (1). Then, we have g(t) = f (1 + 2t, 2 + 2t), for all t ∈ [0, 1]. According to our assumptions, we have g(0) = f (1, 2) = 0 and g(1) = f (3, 4) = 0. So, by Roll’s theorem there exists ξ ∈ (0, 1) such that (5.12) g (ξ ) = 0. On the other hand, since g (t) = 2 f x (1 + 2t, 2 + 2t) + 2 f y (1 + 2t, 2 + 2t), it follows that (5.13) g (ξ ) = 2 f x (1 + 2ξ, 2 + 2ξ ) + 2 f y (1 + 2ξ, 2 + 2ξ ). By (5.12) and (5.13) we conclude that 2 f x (1 + 2ξ, 2 + 2ξ ) + 2 f y (1 + 2ξ, 2 + 2ξ ) = 0. Finally, if we take a = 1 + 2ξ and b = 2 + 2ξ , then f x (a, b) = − f y (a, b). 233. Let f : [a, b] × [c, d] → R be a function of two variables, and ∂ f /∂ x exists on [a, b] × [c, d] and extends to a continuous function on [a, b] × [c, d]. If a ≤ u(x) ≤ v(x) ≤ b, show that d dx

v(x)

f (x, y)dy =

u(x)

v(x)

u(x)

dv du ∂f (x, y)dy + f x, v(x) − f x, u(x) . ∂x dx dx

Solution. Suppose that

F(x, u, v) =

v

f (x, y)dy,

u

where u = u(x) and v = v(x). Now, we use the chain rule. We get ∂ F dx ∂ F du ∂ F dv d F(x, u, v) = + + dx ∂ x d vx ∂u d x ∂v d x dv du ∂ + f x, v(x) . f (x, y)dy − f x, u(x) = ∂x u dx dx On the other hand, similar to Problem 228, we obtain

v

v d ∂f f (x, y)dy = (x, y)dy. dx u u ∂x Finally, by (5.14) and (5.15), the result follows.

(5.14)

(5.15)

188

5 Differentiability and Differential

x2

234. Evaluate x

sin(x y) dy. y

Solution. Let f (x, y) =

sin(x y) , u(x) = x and v(x) = x 2 . By Problem 233, we can y

write

du ∂ sin(x y) dv dy + f x, x 2 ) − f (x, x) ∂ x y d x d x x

x x 2 sin x 3 sin x 2 = cos(x y)dy + (2x) − x2 x x 2 1 sin x 2 2 sin x 3 x = sin(x y) + − x x 3 x x 2 sin x 2 sin x 3 sin x 2 sin x − + − = x x x x 1 = 3 sin x 3 − 2 sin x 2 . x 235. Let f : R → R and F : R2 → R be differentiable functions satisfy F x, f (x) = 0 and ∂ F/∂ y = 0. Prove that d dx

x2

sin(x y) dy = y

x2

f (x) = −

∂ F/∂ x , where y = f (x). ∂ F/∂ y

Solution. Since F x, f (x) = 0 (constant), it follows that ∂ F x, f (x) /∂ x = 0. Therefore, we obtain ∂x ∂ F x, f (x) ∂F ∂F 0= = x, f (x) + x, f (x) f (x) ∂x ∂x ∂x ∂y ∂F ∂F x, f (x) + x, f (x) f (x), = ∂x ∂y and the result follows. 236. Let f : R2 → R be a function such that f (x, y) depends only on the distance r of (x, y) from the origin, say f (x, y) = g(r ), where r = x 2 + y 2 . (1) Prove that for (x, y) = (0, 0) we have ∂2 f ∂2 f 1 + = g (r ) + g (r ). ∂x2 ∂ y2 r (2) Now assume further that f satisfies Laplace’s equation, ∂2 f ∂2 f + 2 = 0, 2 ∂x ∂y

5.4 Solved Problems

189

for all (x, y) = (0, 0). Use part (1) to prove that f (x, y) = a ln(x 2 + y 2 ) + b for (x, y) = (0, 0), where a and b are constant. Solution. (1) Take the partial derivative of r 2 = x 2 + y 2 on both sides with respect to x to obtain ∂(r 2 )/∂ x = 2x. Applying the chain rule on the left side we obtain 2r ∂r/∂ x = 2x, which leads to ∂r x = . ∂x r Hence, we have

∂f x ∂ f ∂r g (r ) = = g (r ) = x, ∂x ∂r ∂ x r r

or equivalently

1 ∂f g (r ) = . x ∂x r

(5.16)

Now, we take the partial derivative from (5.16) on both sides with respect to x. We get ∂ ∂x

1 ∂f x ∂x

∂ = ∂x

g (r ) r

1 ⇒− 2 x 1 ⇒− 2 x ∂2 f ⇒ ∂x2

1 ∂2 f ∂f ∂r g (r ) + = 2 ∂x x ∂x ∂x r 1 ∂2 f ∂f g (r )r − g (r ) ∂r + = ∂ x x ∂ x 2 r2 ∂x 1 ∂f g (r )r − g (r ) x 2 + . = r2 r x ∂x

Therefore, we have ∂2 f = ∂x2

g (r )r − g (r ) g (r ) 2 . + x r3 r

(5.17)

Similarly, by symmetry we obtain ∂2 f = ∂ y2

g (r )r − g (r ) r3

y2 +

g (r ) . r

(5.18)

The sum of (5.17) and (5.18) gives us ∂2 f ∂2 f g (r )r − g (r ) 2 g (r ) g (r )r − g (r ) 2 g (r ) + + = x + y + ∂x2 ∂ y2 r3 r r3 r g (r )r − g (r ) 2 2g (r ) 2 = x +y + r3 r g (r )r − g (r ) 2g (r ) + = r r (r ) g . = g (r ) + r

190

5 Differentiability and Differential

by part (1) we have rg (r ) + g (r ) = 0 f satisfies Laplace’s equation, (2) Since or rg (r ) = 0. This implies that rg (r ) = a, where a is constant. Since r = 0, it follows that g (r ) = a/r , and so

g (r )dr =

a dr. r

Thus, we get g(r ) = a ln r + b, where b is constant. Therefore, we conclude that f (x, y) = a ln(x 2 + y 2 ) + b, where a and b are constant. 237. Let f be a differentiable function of x, y and z = f (x, y), x = r cos θ and ∂z ∂z y = r sin θ . Express the partial derivatives and in terms of partial derivatives ∂r ∂θ ∂z ∂z and . ∂x ∂y Solution. The equations x = r cos θ and y = r sin θ gives us ∂x ∂y ∂x ∂y = cos θ, = sin θ, = −r sin θ and = r cos θ. ∂r ∂r ∂θ ∂θ Now, if we use the chain rule, then we obtain ∂z ∂z ∂z = cos θ + sin θ, ∂r ∂x ∂y ∂z ∂z ∂z = −r sin θ + r cos θ. ∂θ ∂x ∂y

(5.19)

These are the required formula. 238. Refer to Problem 237 and express the second order partial derivative terms of partial derivatives of f .

∂2z in ∂θ 2

∂z Solution. We begin the formula for in (5.19) and differentiate with respect to θ , ∂θ treating r as a constant. There exist two terms on the right, each of which must be differentiable as a product. Hence, we get ∂2z ∂ ∂z ∂z ∂(cos θ ) ∂ ∂z ∂z ∂(sin θ ) − r sin θ + r + r cos θ = −r ∂θ 2 ∂x ∂θ ∂θ ∂ x ∂y ∂θ ∂θ ∂ y ∂ ∂z ∂z ∂ ∂z ∂z − r sin θ − r sin θ + r cos θ . = −r cos θ ∂x ∂θ ∂ x ∂y ∂θ ∂ y (5.20) ∂z ∂z To compute the derivatives of and with respect to θ we must keep in mind ∂x ∂y ∂z ∂z that, as functions of r and θ , and are composite functions. Therefore, their ∂x ∂y

5.4 Solved Problems

191

derivatives with respect to θ must be determined by the use of the chain rule. We ∂z , to obtain again use (5.19), with z replaced by ∂x ∂z ∂z ∂ ∂ ∂ ∂z ∂x ∂x + ∂x ∂y = ∂θ ∂ x ∂x ∂θ ∂ y ∂θ ∂2z ∂2z (r cos θ ). = 2 (−r sin θ ) + ∂x ∂ y∂ x Similarly, using (5.19) with z replacing by

∂z , we find ∂y

∂z ∂z ∂ ∂ ∂ ∂z ∂y ∂x ∂y ∂y = + ∂θ ∂ y ∂x ∂θ ∂ y ∂θ ∂2z ∂2z (−r sin θ ) + 2 (r cos θ ). = ∂ x∂ y ∂y When these formulas are used in (5.20), we obtain ∂2z ∂z ∂z = −r cos θ + sin θ ∂θ 2 ∂x ∂y 2 z ∂ ∂2z ∂2z +r 2 sin2 θ 2 − 2 sin θ cos θ + cos2 θ 2 . ∂x ∂ x∂ y ∂y This is the required formula for

(5.21)

∂2z . ∂θ 2

239. Refer to Problem 237 verify the following formulas: 2 ∂z 1 ∂z 2 (1) ∇ f (r cos θ, r sin θ )2 = + 2 ; ∂r r ∂θ ∂2z ∂2z ∂2z 1 ∂2z 1 ∂z . (2) + = + + 2 2 2 2 2 ∂x ∂y ∂r r ∂θ r ∂r Solution. (1) By (5.19), we can write ∂z 2 ∂r

+

2 2 ∂z ∂z ∂z 1 ∂z 2 ∂z 1 cos θ + sin θ sin θ + r cos θ − r = + r 2 ∂θ r2 ∂x ∂y ∂∂zx 2 ∂z∂y2 = + ∂x ∂y = ∇z2 = ∇ f (r cos θ, r sin θ )2 .

(2) Already in Problem 238 we calculated ∂ 2 z/∂θ 2 . Now, we compute ∂ 2 z/∂r 2 . We have

192

5 Differentiability and Differential

∂z ∂z ∂ ∂ ∂ ∂z ∂x ∂x + ∂x ∂y = ∂r ∂ x ∂x ∂r ∂y ∂r ∂2z ∂2z . = cos θ 2 + sin θ ∂x ∂ y∂ x Similarly, we obtain ∂z ∂z ∂ ∂ ∂ ∂z ∂y ∂x ∂y ∂y = + ∂r ∂ y ∂x ∂r ∂y ∂r ∂2z ∂2z + sin θ 2 . = cos θ ∂ x∂ y ∂y Since

z ∂z ∂z = cos θ + sin θ , it follows that ∂r ∂x ∂y

∂ z ∂z ∂2z ∂ ∂z = cos θ + sin θ = ∂r 2 ∂r ∂r ∂r x ∂y ∂∂z ∂ ∂z + sin θ = cos θ ∂r ∂ x ∂y ∂z ∂2z ∂ 2 z ∂2z = cos θ cos θ + sin θ cos θ . + sin θ + sin θ ∂x2 ∂ y∂ x ∂ x∂ y ∂ y2 So, we conclude that ∂2z ∂2z ∂2z ∂2z + sin2 θ 2 . = cos2 θ 2 + 2 cos θ sin θ 2 ∂r ∂x ∂ x∂ y ∂y

(5.22)

Finally by (5.19), (5.21), (5.22) and using the obvious relation cos2 θ + sin2 θ = 1, the result follows. y z 240. Show that the function g(x, y, z) = x n f a , b , where f is a differentiable x x function, satisfies the equation x

∂g ∂g ∂g + ay + bz = ng. ∂x ∂y ∂z

(5.23)

Solution. Suppose that u = y/x a and v = z/x b . Then, we can write g(x, y, z) = x n f (u, v). Now, using the chain rule we obtain

5.4 Solved Problems

193

∂ f ∂u ∂ f ∂v ∂g = nx n−1 f (u, v) + x n + ∂x ∂x ∂v ∂ x ∂u −ay ∂ f −bz ∂ f , = nx n−1 f (u, v) + x n a+1 + b+1 x ∂u x ∂v n ∂ f ∂v x ∂f ∂g ∂ f ∂u = xn + = a , ∂y ∂u ∂ y ∂v ∂ y x ∂v n ∂ f ∂v x ∂f ∂ f ∂u ∂g = xn + = b . ∂z ∂u ∂z ∂v ∂z x ∂v Substituting the above quantities in (5.23), we are done. 241. If x x y y z z = c, show that −1 ∂2z = − x ln(ex) (at x = y = z). ∂ x∂ y Solution. We can write z z = c/(x x y y ). It follows that z ln z = ln c − x ln x − y ln y.

(5.24)

Taking partial derivative from both sides of (5.24) with respect to x, we get ∂z ∂x or equivalently

ln z + z

1 ∂z = −(1 + ln x), z ∂x

∂z (1 + ln z) = −(1 + ln x). ∂x

(5.25)

Now, taking partial derivative from both sides of (5.25) with respect to y, gives us ∂2z ∂z 1 ∂z (1 + ln z) + = 0. ∂ x∂ y ∂x z ∂x This implies that

1 ∂z ∂z ∂ z z ∂x ∂y =− . ∂ x∂ y 1 + ln z 2

Consequently, if x = y = z, then 1 −1 ∂2z −1 −1 =− x = = = − x ln(ex) . ∂ x∂ y 1 + ln x x(1 + ln x) x ln(ex) 242. A function f of two variables with domain D is said to be homogenous of degree k if

194

5 Differentiability and Differential

f (t x, t y) = t k f (x, y)

(5.26)

for all (x, y) in D and t > 0; here k can be any real number, and it is assumed that if (x, y) belongs to D, then so does every point (t x, t y)with t > 0. For example, the functions f (x, y) = x 2 ln(x/y) − x y and f (x, y) = x 2 + y 2 are homogenous of degree 2 and 1, respectively. Let f be homogenous of degree k and differentiable at every point of its domain. Show that (1) x f x (x, y) + y f y (x, y) = k f (x, y), a result known as Euler’s theorem on homogeneous functions; (2) If f has continuous second partial derivatives, then x 2 f x x (x, y) + 2x y f x x (x, y) + y 2 f yy (x, y) = k(k − 1) f (x, y). Solution. (1) First, we differentiate f (t x, t y) = t k f (x, y) with respect to t. To do this we let x(t) = t x and y(t) = t y and take derivative from the left side of (5.26) with respect to t. We have ∂ f dx ∂ f dy df = + . dt ∂ x dt ∂ y dt This is equal to x f x (t x, t y) + y f y (t x, t y). On the other hand, taking derivative from the right side of (5.26) gives us nt n−1 f (x, y). Consequently, we have x f x (t x, t y) + y f y (t x, t y) = kt k−1 f (x, y). Now, it is enough to consider t = 1 in (5.27). (2) We can write d fx d x d fx = + dt d x dt d fy dx d fy = + dt d x dt

d fx dy d fy dy

(5.27)

dy dt dy . dt

Therefore, if we take derivative from the left side of (5.27) with respect to t, then by using the above equalities, we obtain x 2 f x x (t x, t y) + x y f x y (t x, t y) + x y f yx (t x, t y) + y 2 f yy (t x, t y). On the other hand, taking derivative from the right side of (5.27) gives us k(k − 1)t k−2 f (t x, t y). Therefore, we conclude that x 2 f x x (t x, t y) + 2x y f x y (t x, t y) + y 2 f yy (t x, t y) = k(k − 1)t k−2 f (t x, t y). (5.28) Again, it is enough to put t = 1 in (5.28). x+y 243. If z = sin−1 √ √ , show that x+ y

5.4 Solved Problems

x2

195 2 ∂2z ∂2z cos 2z · sin z 2∂ z + y . + 2x y =− ∂x2 ∂ x∂ y ∂ y2 4 cos3 z

Solution. Note that z is not homogeneous. If we set u = sin z = √ f (x, y), then

x+y √ = x+ y

tx + ty 1/2 x + y 1/2 f (t x, t y) = √ √ = t f (x, y). √ =t √ x+ y tx + ty This means that f is homogeneous of degree 1/2. Therefore, by Problem 242, we have ∂u 1 1 ∂u +y = u = sin z. x ∂x ∂y 2 2 Since

∂u ∂z ∂u ∂z = cos z and = cos z , it follows that ∂x ∂x ∂y ∂y x cos z

This yields that x

∂z 1 ∂z + y cos z = sin z. ∂x ∂y 2 ∂z ∂z 1 +y = tan z. ∂x ∂y 2

(5.29)

Taking derivative from (5.29) with respect to x we obtain x or equivalently x

∂2z ∂2z 1 ∂z ∂z + y = sec2 z + 2 ∂x ∂x ∂ x∂ y 2 ∂x

1 ∂z ∂2z ∂2z 2 = sec . + y z − 1 ∂x2 ∂ x∂ y 2 ∂x

(5.30)

Again by taking derivative from (5.29) with respect to y we get y or equivalently x

∂2z ∂z 1 ∂2z ∂z + = sec2 z + x ∂ y2 ∂ x∂ y ∂y 2 ∂y

1 ∂z ∂2z ∂2z +y 2 = sec2 z − 1 . ∂ x∂ y ∂y 2 ∂y

Multiplying (5.30) by x, (5.31) by y and adding, we obtain

(5.31)

196

5 Differentiability and Differential

x2

1 ∂z 2 ∂z ∂2z ∂2z 2∂ z 2 + y sec + y + 2x y = z − 1 x ∂x2 ∂ x∂ y ∂ y2 21 1 ∂ x ∂ y 2 = sec z − 1 tan z 2 2 cos 2z · sin z . = 4 cos3 z

244. If z = tan−1

x 3 + y3 , show that x−y

∂z ∂z +y = sin 2x, ∂x ∂y ∂2z ∂2z ∂2z (2) x 2 2 + y 2 2 + 2x y = sin 4z − sin 2z. ∂x ∂y ∂ x∂ y (1) x

Solution. (1) We set u = f (x, y) = tan z. Then, we have f (t x, t y) =

t 3 x 3 + t 3 y3 x 3 + y3 = t2 = t 2 f (x, y). tx − ty x−y

So, f is a homogeneous function of degree 2. Hence, by Problem 242, we have x

∂ ∂ (tan z) + y (tan z) = 2 tan z. ∂x ∂y

It follows that x sec2 z

∂z ∂z + y sec2 z = 2 tan z. ∂x ∂y

Next, we conclude that x

∂z ∂z +y = 2 tan z cos2 z = 2 sin z cos z = sin 2z. ∂x ∂y

(2) We take partial derivative from both sides of (1) with respect to x. We get x

∂2z ∂z ∂2z ∂z +y = 2 cos 2z , + 2 ∂x ∂x ∂ x∂ y ∂x

x

∂2z ∂z ∂2z = (2 cos 2z − 1) . + y 2 ∂x ∂ x∂ y ∂x

(5.32)

y

∂2z ∂z ∂2z = (2 cos 2z − 1) . + x ∂ y2 ∂ x∂ y ∂y

(5.33)

which implies that

Similarly, we obtain

5.4 Solved Problems

197

Multiplying (5.32) by x, and (5.33) by y, and adding them, we obtain x2

∂z 2 ∂2z ∂2z ∂z 2∂ z + 2x y = (2 cos 2z − 1) x + y + y ∂x2 ∂ x∂ y ∂ y2 ∂x ∂y = (2 cos 2z − 1)(sin 2z) by part (1) = 2 cos 2z sin 2z − sin 2z = sin 4z − sin 2z.

245. A function f : Rn → R is said to be homogeneous of degree k if f (t X ) = t k f (X ), for any real number t and X = (x1 , . . . , xn ). Let f be continuously differentiable. Prove that f is homogeneous of degree k if and only if k f (X ) =

n ∂ f (X ) i=1

∂ xi

xi .

(5.34)

Solution. For any X ∈ Rn , let g X (t) = f (t X ) − t k f (X ). Note that g X is zero function for all X exactly when f is homogeneous of degree k. Assume that (5.34) holds. By the chain rule, we can write dg X (t) ∂ f (t X ) = xi − kt k−1 f (X ) dt ∂ x i i=1 n 1 ∂ f (t X ) = t xi − kt k f (X ) t i=1 ∂ xi 1 = k f (t X ) − kt k f (X ) t k = g X (t). t n

k g X (t), which means that y(t) = g X (t) satisfies the first order dift ky dy = . This equation has general solution y = ct k , and so ferential equation dt t g X (t) = ct k . Since g X (1) = f (X ) − f (X ) = 0, it follows that c = 0, which implies that g X is identically zero. Consequently, f is homogeneous of degree k. Conversely, let f be a homogeneous function of order k. Then g X (t) is identically zero, and so g X (t) = 0. On the other hand, we have Hence, g X (t) =

g X (t) =

n ∂ f (t X ) i=1

∂ xi

xi − kt k−1 f (X ).

Hence, we conclude that n ∂ f (t X ) i=1

∂ xi

xi = kt k−1 f (X ).

198

5 Differentiability and Differential

Finally, it is enough to set t = 1 in order to get (5.34). 246. If f is a homogeneous function of degree k, prove that each partial derivative ∂ f /∂ xi is homogeneous of degree k − 1. Solution. If f is a homogeneous function of order k, it follows that (5.34) holds, and we can differentiate both sides with respect to xi to obtain ∂2 f ∂f ∂f = xj + , k ∂x j ∂ x ∂ x ∂ xj j i i=1 n

(5.35)

where we used the fact that ⎧ 2 ∂ f ⎪ ⎪ x if i = j ∂ ∂ f ⎨ ∂ x j ∂ xi i xi = 2 ∂ f ∂f ⎪ ∂ x j ∂ xi ⎪ ⎩ xj + if i = j. ∂ x j ∂ xi ∂ xi Now, (5.35) is equivalent to n ∂2 f ∂f xi = (k − 1) , ∂ x ∂ x ∂ xj i j i=1

which by Problem 245 implies that ∂ f /∂ x j is homogeneous of degree k − 1. 247. Let f be a function of three variables x, y and z. If f is homogeneous of degree k = 1 having continuous second order partial derivatives, prove that fx x f yx f zx

fx y f yy f zy

fxz f yz f zz

(k − 1)2 = z2

fx x f yx fx

fx y f yy fy

fx fy kf k−1

.

Solution. By Problem 245, we have x f x + y f y + z f z = k f.

(5.36)

Now, we differentiate (5.36) with respect to x, y and z, then x f x x + y f x y + z f x z = (k − 1) f x , x f yx + y f yy + z f yz = (k − 1) f y , x f zx + y f zy + z f zz = (k − 1) f z . Hence, using (5.36) and (5.37) and determinant properties we obtain

(5.37)

5.4 Solved Problems

fx x f yx f zx =

=

=

=

199

fx x f x z fx y f x z 1 f yz = f yx f yy f yz z f zz z f zx z f zy z f zz fx x fx y 1 f yx f yy z xf + yf + zf x f x y + y f yy + z f zy xx yx zx f fx x fx y fxz fx y k − 1 xx k − 1 f yx f yy f yx f yy f yz = z z 2 fx fy fz fx fy fx x fx y x fx x + y fx y + z fxz k − 1 f yx f yy x f yx + y f yy + z f yz z 2 f fy x fx + y f y + z fz x f x x f x y (k − 1) f x 2 (k − 1) k−1 f yx f yy (k − 1) f y = 2 z 2 z fx fy kf fx y f yy f zy

fxz f yz x f x z + y f yz + z f zz z fxz z f yz zf

z

fx x f yx

fx y f yy

fx

fy

fx fy kf k−1

.

248. An equation of the form y = f (x, y) is said to be homogeneous whenever f does not depend on x and y separately, but only on their ratio y/x or x/y. Prove that if a homogeneous first order differential equation is written in polar coordinates, y−x . then it reduces to a separated equation. Use this method to solve y = y+x Solution. We have f (x, y) = f (r cos θ, r sin θ ) = f (1, tan θ ). Now, we compute

(5.38)

dy d = (r sin θ ). We obtain dx dx

d x = cos θ dr − r sin θ dθ and dy = sin θ dr + r cos θ dθ, and so we have 1 dr tan θ +1 sin θ dr + r cos θ dθ dy dθ . = = r 1 dr dx cos θ dr − r sin θ dθ − tan θ r dθ From (5.38) and (5.39) we get tan θ or equivalently

dr 1 dr + 1 = f (1, tan θ ) − tan θ f (1, tan θ ), dθ r dθ

(5.39)

200

5 Differentiability and Differential

tan θ − f (1, tan θ )

This yields that

1 dr r dθ

= − 1 + tan θ f (1, tan θ ) .

1 1 + tan θ f (1, tan θ ) dr = − dθ. r tan θ − f (1, tan θ )

(5.40)

The Eq. (5.40) is a separated differential equation. y−x Now, suppose that f (x, y) = . Then, by a simple calculation, we obtain y+x 1 + tan θ f (1, tan θ ) = 1. tan θ − f (1, tan θ ) Hence, from (5.40) we have 1 dr = −dθ. r This yields that ln r = −θ + C or ln(x 2 + y 2 ) = − tan−1 (y/x) + C. 249. If x 2 + y 2 + z 2 − 2x yz = 1, show that 1 1 1 dx + dy + √ dz = 0. √ 2 2 1−x 1 − z2 1−y Solution. Since x 2 + y 2 + z 2 − 2x yz = 1, it follows that (x − yz)d x + (y − zx)dy + (z − x y)dz = 0.

(5.41)

Moreover, we can write (x − yz)2 = (1 − y 2 )(1 − z 2 ). If we set (1 − x 2 )(1 − y 2 )(1 − z 2 ) = k, then (1 − x 2 )(x − yz)2 = k. This gives that x − yz =

k . 1 − x2

Similarly, we have y − zx =

k and z − x y = 1 − y2

k . 1 − z2

Thus, by putting these expressions in (5.41) we obtain √

k

√

1 1 − x2

1

1

dx + dy + √ dz 1 − z2 1 − y2

= 0.

5.4 Solved Problems

201

Since k is non-zero, the result follows. 250. A set of three variables x, y, z is connected with the another set u, v, w by the equations x + y + z = u, x y + yz + zx = v and x yz = w. Prove that

2(2x − y − z) ∂2x =− . ∂w 2 (x − y)3 (x − z)3

(5.42)

Solution. We compute du, dv and dw as follows: du = d x + dy + dz, dv = (y + z)d x + (x + z)dy + (x + y)dz, dw = yzd x + x zdy + x ydz. From the above equations we get dx =

x 1 x2 du − dv + dw (x − y)(x − z) (x − y)(x − z) (x − y)(x − z)

(5.43)

On the other hand, we can write dx =

∂x ∂x ∂x du − dv + dw. ∂u ∂v ∂w

(5.44)

Comparing (5.43) and (5.44) we obtain ∂x 1 = . ∂w (x − y)(x − z) Also, by the symmetry of relations we have ∂y 1 ∂z 1 = and = . ∂w (y − x)(y − z) ∂w (z − y)(z − x) Now, we conclude that ∂ x =− ∂w 2 2

(x − z)

∂x ∂y ∂z + (x − y) − ∂w ∂w ∂w ∂w . (x − y)2 (x − z)2

∂x

−

(5.45)

Now, by putting the expressions for ∂ x/∂w, ∂ y/∂w and ∂z/∂w in (5.45) we reach to (5.42).

202

5 Differentiability and Differential

251. Suppose that z = f (x, y) be a differentiable function such that f x (3, 3) = −2, f x (2, 5) = 7, f y (3, 3) = 11 and f y (2, 5) = −3.

(5.46)

If w is a differentiable function of u and v satisfying the equation f (w, w) = f (uv, u 2 + v 2 ),

(5.47)

for all u and v, find ∂w/∂u at u = 1, v = 2 and w = 3. Solution. We take the partial derivative of (5.47) with respect to u. Using the chain rule we obtain f x (w, w)

∂w ∂w ∂(uv) ∂(u 2 + v 2 ) + f y (w, w) = f x (uv, u 2 + v 2 ) + f y (uv, u 2 + v 2 ) . ∂u ∂u ∂u ∂u

This implies that

∂w = u f x (uv, u 2 + v 2 ) + 2u f y (uv, u 2 + v 2 ). f x (w, w) + f y (w, w) ∂u

When u = 1, v = 2 and w = 3 we get

f x (3, 3) + f y (3, 3)

∂w ∂u

= 2 f x (2, 5) + 2u f y (2, 5).

Now, using the quantities in (5.46), we obtain ∂w/∂u = 8/9. 252. Let f be a function of two variables and assume that the partial derivatives ∂ f /∂ x and ∂ f /∂ y are never zero. Let u be another function of two variables such that the partial derivatives ∂u/∂ x and ∂u/∂ y are related by equation f (∂u/∂ x, ∂u/∂ y) = 0. Assume that all second partial derivatives of u are continuous. Prove that a constant n exists such that ∂ 2 u n ∂ 2u ∂ 2u = , 2 2 ∂x ∂y ∂ x∂ y and find n. Solution. Suppose that ∂u/∂ x = r and ∂u/∂ y = s. Since f (r, s) = 0, it follows that ∂f ∂f dr + ds = 0. ∂r ∂s This yields that ∂ f ∂ 2u ∂ f ∂ 2u ∂ 2u ∂ 2u dy + d x + d x + dy = 0, ∂r ∂ x 2 ∂ y∂ x ∂s ∂ y∂ x ∂ y2

5.4 Solved Problems

203

or equivalently, ∂ f ∂ 2u ∂ f ∂ 2u ∂ f ∂ 2u ∂ f ∂ 2u d x + dy = 0. + + ∂r ∂ x 2 ∂s ∂ y∂ x ∂r ∂ y∂ x ∂s ∂ y 2 Since x and y are independent variables, we conclude that ∂ f ∂ 2u ∂ f ∂ 2u ∂ f ∂ 2u ∂ f ∂ 2u = 0 and + + = 0. ∂r ∂ x 2 ∂s ∂ y∂ x ∂r ∂ y∂ x ∂s ∂ y 2 Therefore, we can write ∂ 2u ∂ 2u ∂f − ∂s = ∂ y∂ x = ∂ y 2 . ∂f ∂ 2u ∂ 2u − ∂r ∂ x∂ y ∂x2 Now, the result follows and we conclude that n = 2. 253. Suppose that x and y are independent variables, and let f (x, y, u, v) = 0 and g(x, y, u, v) = 0, where the equations define the dependent variables u and v as (implicit) functions of x and y. Find (1)

∂u , ∂x

(2)

∂u , ∂y

(3)

∂v , ∂x

(4)

∂v . ∂y

Solution. We can write d f = f x d x + f y dy + f u du + f v dv = 0,

(5.48)

dg = gx d x + g y dy + gu du + gv dv = 0.

(5.49)

Furthermore, since u and v are functions of x and y, it follows that du = u x d x + u y dy,

(5.50)

dv = vx d x + v y dy.

(5.51)

Substituting (5.50) and (5.51) into (5.48) and (5.49), we get d f = ( f x + f u u x + f v vx ) d x + f y + f u u y + f v u y dy = 0,

(5.52)

dg = (gx + gu u x + gv vx ) d x + g y + gu u y + gv v y dy = 0.

(5.53)

Since x and y are independent, it follows that the coefficient of d x and dy in (5.52) and (5.53) are zero. Therefore, we conclude that

204

5 Differentiability and Differential

f u u x + f v vx = − f x and gu u x + gv vx = −gx ,

(5.54)

f u u y + f v v y = − f y and gu u y + gv v y = −g y .

(5.55)

Now, we use Cramer’s rule to solve the systems of Eqs. (5.54) and (5.55). We obtain − f x fv − f y fv −gx gv −g y gv ∂u ∂u (3) u y = = = (1) u x = , , ∂x ∂y fu fv fu fv gu gv gu gv fu − f x fu − f y gu −gx gu −g y ∂v ∂v (2) vx = (4) v y = = = , . ∂x ∂y fu fv fu fv gu gv gu gv 254. Estimate (2.97)2 + (4.02)2 . Solution. Suppose that f (x, y) = x 2 + y 2 . Then, we can write (2.97)2 + (4.02)2 = f (3 + Δx, 4 + Δy), where Δx = −0.03 and Δy = 0.02. So, we have Δf = f (3 + Δx, 4 + Δy) − f (3, 4) ≈ d f = f x (3, 4)Δx + f y (3, 4)Δy. This yields that f (3 + Δx, 4 + Δy) ≈ f (3, 4) + f x (3, 4)Δx + f y (3, 4)Δy. Now, we compute f x and f y as follows: f x (x, y) =

x x2

+

y2

and f y (x, y) =

y x2

+ y2

,

and hence f x (3, 4) = 3/5 and f y (x, y) = 4/5. Therefore, we conclude that

4 3 (2.97)2 + (4.02)2 ≈ 5 + (−0.03) + (0.02) = 4.998. 5 5

255. Use the linearization to approximate 3

(1.03)2 . √ 0.98 1.05

5.4 Solved Problems

205

√ Solution. If f (x, y, z) = x 2 / 3 y z, then f (1.03, 0.98, 1.05) is the quantity we want to approximate. We calculate the linearization at (1, 1, 1). Then, the partial derivatives are ∂f d x 2 = 2x = 2, (1, 1, 1) = x=1 ∂x d x 11/3 12 1/6 x=1 d 1 ∂f 1 1 (1, 1, 1) = = 4/3 =− , 1/3 1/6 y=1 y=1 ∂y dy y 1 3y 3 ∂f d 12 1 1 (1, 1, 1) = = − 7/6 =− . 1/3 1/6 z=1 z=1 ∂x dy 1 z 6z 6 Then, the linearization is 1 1 L(x, y, z) = 1 + 2(x − 1) − (y − 1) − (z − 1). 3 6 This yields that L(1.03, 0.98, 1.05) is an approximation for f (1.03, 0.98, 1.05). Hence, we obtain 1 1 f (1.03, 0.98, 1.05) ≈ L(1.03, 0.98, 1.05) = 1 + 2(0.03) − (0.02) − (0.05) 3 6 6.35 = ≈ 1.05833333333. 5 256. A cylinder has a radius r = 5 and a height h = 2. Approximate the change in the volume of the cylinder if the radius is increased by 0.1 and the height is decreased by 0.05. Solution. The volume of a cylinder is a function of its radius r and height h given by V (r, h) = πr 2 h. The exact change in volume can be calculated as ΔV = V (5 + 0.1, 2 − 0.05) − V (5, 2). Now, we want to compute the approximate differential of this function with the formula ΔV ≈

∂V ∂V (r, h)Δr + (r, h)Δh, ∂r ∂h

where r = 5, h = 2 and Δr = 0.1 is the change in the radius and Δh = −0.05 is the change in height. Note that Δh is negative since the height is decreased. The partial derivatives of the volume function are ∂V ∂V = 2πr h and = πr 2 . ∂r ∂h Evaluating these for the given radius and height gives ∂ V /∂r = 20π and ∂ V /∂h = 25π . Then, the change in V is approximated as ΔV ≈ (20π )(0.1) + (25π )(−0.05) = 0.75π ≈ 2.3562. Since the answer is positive, it means that the overall volume was increased.

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5 Differentiability and Differential

257. If |x| is much greater than |y|, |z| and |t|, to which of x, y, z and t is the value of the determinant x y f (x, y, z, t) = z t most sensitive? Give reason for your answer. Solution. We have f (x, y, z, t) = xt − yz. Then the first partial derivatives of f are f x (x, y, z, t) = t, f y (x, y, z, t) = z,

f z (x, y, z, t) = −y, f t (x, y, z, t) = x.

So, we can write df =

∂f ∂f ∂f ∂f dx + dy + dz + dt = td x − zdy − ydz + xdt. ∂x ∂y ∂z ∂t

Since |x| is much greater than |y|, |z| and |t|, it follows that the function f is most sensitive to a change in t. 258. The Wilson lot size formula: The Wilson lot size formula in economics says that the most economical quantity Q of goods√(radios, shoes, brooms, whatever) for a store to order is given by the formula Q = 2K M/ h, where K is the cost of placing the order, M is the number of items sold per week, and h is the weekly holding cost for each item (cost of space, utilities, security, and so on). To which of the variables K , M, and h is Q most sensitive near the point (K 0 , M0 , h 0 ) = (2, 20, 0.05)? Give reasons for your answer. Solution. First, we calculate the first partial derivatives of Q as follows: 1 ∂Q = ∂K 2

2K M −1/2 2M ∂Q 1 2K M −1/2 2K , = , h ∂M 2 h h h −1/2 −2K M ∂Q 1 2K M . = ∂K 2 h h2

So, we obtain ∂Q ∂Q ∂Q dK + dM + dh ∂ K ∂ M ∂h −1/2 1 2K M 2M 2K 2K M = dK + dM − dh . 2 h h h h2

dQ =

Then, we find that d Q

(2)(2) (2)(2)(20) 1 (2)(2)(20) −1/2 (2)(20) dK + dM − dh (2,20,0.05) 2 0.05 0.05 0.05 (0.05)2 = (0.0125)(800d K + 80d M − 32000dh). =

5.4 Solved Problems

207

Therefore, we conclude that Q is most sensitive to changes in h. 259. Determine Taylor series expansion of the function f (x + 1, y + π/3) in ascending power of x and y when f (x, y) = sin(x y), neglecting terms of degree higher than two. Solution. By Taylor’s formula, we can write ∂ ∂ +y f (1, π/3) f (x + 1, y + π/3) = f (1, π/3) + x ∂x ∂y 1 ∂ ∂ 2 = x +y f (1, π/3) + · · · . 2! ∂ x ∂y Here, we have f (1, π/3) = sin(π/3) = as follows:

√

3/2. The partial derivatives required are

∂f ∂f ∂2 f (x, y) = y cos(x y), (x, y) = x cos(x y), (x, y) = −y 2 sin(x y), ∂x ∂y ∂x2 ∂2 f ∂2 f (x, y) = cos(x y) − x y sin(x y) and (x, y) = −x 2 sin(x y). ∂ x∂ y ∂ y2 So, we obtain √ ∂f ∂f ∂2 f 2 3/18, (1, π/3) = −π (1, π/3) = −π/6, (1, π/3) = 1/2, ∂x ∂y ∂x2 √ √ ∂2 f ∂2 f (1, π/3) = 1/2 − π 3/6 and (1, π/3) = − 3/2. ∂ x∂ y ∂ y2 Neglecting terms of degree higher than two, we have √ √ 2 √ √ π π 1 π 3 3 3π 2 1 3 2 sin (x + 1)(y + ) = + x+ y− x + − xy − y + ··· . 3 2 6 2 36 2 6 4

260. Let f : R2 → R be a function of two variables such that all partial derivatives of order 3 exist and are continuous. Write down (explicitly in terms of partial derivatives of f ) a quadratic polynomial L 2 (x, y) in x and y such that | f (x, y) − L 2 (x, y)| ≤ M(x 2 + y 2 )3/2 , for all (x, y) in some neighborhood of (0, 0), where M is a number that may depend on f but not on x and y. Next, prove the above estimate. Solution. By the Taylor’s formula, we expand f around the point (0, 0) as follows:

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5 Differentiability and Differential

L 2 (x, y) = f (0, 0) + x f x (0, 0) + y f y (0, 0) 1 2 x f x x (0, 0) + 2x y f x y (0, 0) + y 2 f yy (0, 0) + 2! So, we have | f (x, y) − L 2 (x, y)| 1 = x 3 f x x x (ξ1 , ξ2 ) + 3x 2 y f x x y (ξ1 , ξ2 ) + 3x y 2 f x yy (ξ1 , ξ2 ) + y 3 f yyy (ξ1 , ξ2 ) , 3! for some (ξ1 , ξ2 ) on the line segment joining (a, b) and (x, y). Clearly, we have |x 3 | ≤ (x 2 + y 2 )3/2 , |x 2 y| ≤ (x 2 + y 2 )3/2 , |x y 2 | ≤ (x 2 + y 2 )3/2 , |y 3 | ≤ (x 2 + y 2 )3/2 . Since the third order partial derivatives are continuous, it follows that there exists a real number c such that max{ f x x x , 3 f x x y , 3 f x yy , f yyy }

0. We define ⎧ ⎨ f (x, y) − f (x0 , y) if x = x0 g(x, y) = x−x ⎩ f (x , y) 0 if x = x0 x 0 ⎧ ⎨ f (x0 , y) − f (x0 , y0 ) if y = y0 h (x0 ) (y) = y−y ⎩ f (x , y ) 0 if y = y0 . y 0 0

and

(1) Show that g : Nδ (x0 , y0 ) → R and h (x0 ) : (y0 − δ, y0 + δ) → R are continuous functions at (x0 , y0 ) and y0 , respectively; (2) Show that f (x, y) − f (x0 , y0 ) = (x − x0 )g(x, y) + (y − y0 )h (x0 ) (y), for all (x, y) ∈ Nδ (x0 , y0 ); (3) Show that f is differentiable at (x0 , y0 ). Solution. (1) Since lim h (x0 ) (y) = f y (x0 , y0 ) = h (x0 ) (y0 ), it follows that h (x0 ) is cony→y0

tinuous at y0 . For any (x, y) ∈ Nδ (x0 , y0 ) with x = x0 , by the mean value theorem, there exists ξ between x and x0 such that f (x, y) − f (x0 , y) = (x − x0 ) f x (ξ, y). This yields that g(x, y) = f x (ξ, y). Since f x is continuous at (x0 , y0 ), it follows that g is continuous at (x0 , y0 ). (2) It is straightforward. (3) For (h, k) ∈ R2 , we consider (h, k) =

f (x0 + h, y0 + k) − f (x0 , y0 ) − f x (x0 , y0 )h − f y (x0 , y0 )k . √ h2 + k2

By part (2), we can write |h| |g(x0 + h, y0 + k) − f x (x0 , y0 )| + |k| |h (x0 +h) (y0 + k) − f y (x0 , y0 )| h2 + k2 ≤ |g(x0 + h, y0 + k) − f x (x0 , y0 )| + |h (x0 +h) (y0 + k) − f y (x0 , y0 )| → 0

|(h, k)| ≤

5.4 Solved Problems

211

as (h, k) → (0, 0). This proves that f is a differentiable function at (x0 , y0 ). 264. Mean value theorem for functions of two variables: Let f : R2 → R be a function with continuous partial derivatives f x and f y . Prove that for all distinct pairs (x, y) and (u, v) ∈ R2 , there exists a point (ξ1 , ξ2 ) on the line segment joining the point (x, y) and (u, v) such that f (u, v) − f (x, y) = (u − x) f x (ξ1 , ξ2 ) + (v − y) f y (ξ1 , ξ2 ). Solution. Suppose that (x, y) and (u, v) are any points in R2 ; and h = u − x and k = v − y. Let be the line segment obtained by joining the point (x, y) and (u, v). The coordinates of any point on this line are given by (x + ht, y + kt), for some t ∈ [0, 1]. Now, we define a real single valued function F : [0, 1] → R by F(t) = f (x + ht, y + kt), keeping x, y, u and v fixed. The derivative of F is F (t) = h f x (x + ht, y + kt) + k f y (x + ht, y + kt).

(5.56)

Applying the mean value theorem to F gives F(1) − F(0) = (1 − 0)F (ξ0 ), for some ξ0 ∈ (0, 1). By the definition of F, we obtain f (u, v) − f (x, y) = F (ξ0 ). Next, using (5.56), we can write f (u, v) − f (x, y) = h f x (x + hξ0 , y + kξ0 ) + k f y (x + hξ0 , y + kξ0 ). This yields that f (u, v) − f (x, y) = (u − x) f x (ξ1 , ξ2 ) + (v − y) f y (ξ1 , ξ2 ), where (ξ1 , ξ2 ) is a point on the line segment whose coordinates are given by (x + hξ0 , y + kξ0 ). 265. Prove that a function f : R2 → R whose partial derivatives f x and f y exist and have value 0 at every point of the plane is constant. Solution. Problem 264 can be used to deduce the above fact. In order to prove this, suppose that (x, y) and (u, v) are two arbitrary points in R2 and apply Problem 264 to f . Then, we have f (u, v) − f (x, y) = (u − x) f x (ξ1 , ξ2 ) + (v − y) f y (ξ1 , ξ2 ), for some (ξ1 , ξ2 ) on the line segment joining the points (x, y) and (u, v). Since the partial derivatives are zero at every point, it follows that f (x, y) = f (u, v), for all x, y, u, v ∈ R. Therefore, f is a constant function on R2 . 266. Cauchy’s mean value theorem for functions of two variables: Suppose that f, g : R2 → R are two functions with continuous partial derivatives f x , f y , gx and g y . Prove that for any distinct pairs (x, y) and (u, v) in R2 , there exists a point (ξ1 , ξ2 ) on the line segment joining the points (x, y) and (u, v) such that

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5 Differentiability and Differential

f (u, v) − f (x, y) (u − v)gx (ξ1 , ξ2 ) + (v − y)g y (ξ1 , ξ2 ) = g(u, v) − g(x, y) (u − v) f x (ξ1 , ξ2 ) + (v − y) f y (ξ1 , ξ2 ) .

Solution. We define an auxiliary function f (u, v) − f (s, t) g(u, v) − g(x, y) − f (u, v) − f (x, y) g(u, v) − g(s, t) .

F(s, t) =

Then, we see that F(u, v) = F(x, y) = 0. Since f and g are differentiable, it follows that F is differentiable too. By the mean value theorem for functions of two variables (Problem 264), there exists a point (ξ1 , ξ2 ) at the segment line joining (x, y) and (u, v) such that (u − x)Fs (ξ1 , ξ2 ) + (v − y)Ft (ξ1 , ξ2 ) = 0. Carrying out the partial derivatives on F gives that (u − x) gx (ξ1 , ξ2 ) f (u, v) − f (x, y) − f x (ξ1 , ξ2 ) g(u, v) − g(x, y) +(v − y) f y (ξ1 , ξ2 ) g(u, v) − g(x, y) − g y (ξ1 , ξ2 ) f (u, v) − f (x, y) = 0. This completes the proof.

5.5 Exercises Easier Exercises 1. Let

⎧ ⎨ 3x 2 y 2 if (x, y) = (0, 0) f (x, y) = x 4 + y 4 ⎩ 0 if (x, y) = (0, 0).

Prove that f x (0, 0) and f y (0, 0) exist but f is not differentiable at (0, 0). 2. Let ⎧ ⎨ 3x 2 y if (x, y) = (0, 0) f (x, y) = x 2 + y 2 ⎩ 0 if (x, y) = (0, 0). Prove that f x (0, 0) and f y (0, 0) exist but f x and f y are not continuous at (0, 0).

5.5 Exercises

3. Let

213

⎧ ⎨ x y(x 2 − y 2 ) if (x, y) = (0, 0) 2 2 f (x, y) = ⎩ x +y 0 if (x, y) = (0, 0).

Prove that f is differentiable at (0, 0). 4. Let ⎧ x y2 z ⎨ if (x, y, z) = (0, 0, 0) f (x, y, z) = x 4 + y 4 + z 4 ⎩ 0 if (x, y, z) = (0, 0, 0). (a) Show that f x (0, 0, 0), f y (0, 0, 0) and f z (0, 0, 0) exist; (b) Prove that f is not differentiable at (0, 0, 0). 5. Let f (x, y, z) =

⎧ ⎨ ⎩

(x 2

0

x 2 y2 z2 if (x, y, z) = (0, 0, 0) + y 2 + z 2 )2 if (x, y, z) = (0, 0, 0).

Prove that f is differentiable at (0, 0, 0). 6. If z = 5x 2 + y 2 and (x, y) changes from (1, 2) to (1.05, 2.1) compare the values of Δz and dz. 7. Find the total derivative du/dt by using the chain rule. y , x = ln t, y = et ; (a) u = tan−1 x ex + e y (b) u = , x = 3 sin t, y = ln t; y − ex y (c) u = x y + x z + yz, x = t cos t, y = t sin t, z = t; (d) u = ln(x 2 + y 2 + t 2 ), x = t sin t, y = cos t. 8. Find the linearization L of f at the point P: +y ) and P = (1/2, 1/3), (a) f (x, y) = e−(x (b) f (x, y, z) = x 2 + y 2 + z 2 and P = (0, 1, 0), (c) f (x, y, z) = e−x sin(y + z) and P = (0, 0, π ). 2

2

9. Show that if f : Rn → R and g : Rn → R are differentiable functions, then so their product f · g : Rn → R, and ∇( f · g) = f · ∇g + g · ∇ f . 10. Find the value of m such that z = f (y + mx) is a solution of the partial differential equation ∂2z ∂2z ∂2z + c 2 = 0. a 2 +b ∂x ∂ y∂ x ∂y 11. Given the equation e z sin(x + y) + e y sin(x + z) + e x sin(y + z) = 0.

214

5 Differentiability and Differential

Find the first and second derivatives of z with respect to x and y at (π, 0, π/2). 12. The equation x + z + (y + z)2 = 6 defines z implicity as a function of x and y, say z = f (x, y). Compute the partial derivatives ∂ f /∂ x, ∂ f /∂ y and ∂ 2 f /(∂ x∂ y) in terms of x, y and z. 13. The equation sin(x + y) + cos(y + z) = 1 defines z implicity as a function of x and y, say z = f (x, y). Compute the second derivative ∂ 2 f /(∂ x∂ y) in terms of x, y and z. 14. Use differential to estimate (c) cos31◦ sin 58◦ ; √ √ (d) ln 1.03 + 3 1.08 − 1 .

(a) (0.97)1.05 ; (b) (1.05)2 + (1.98)3 ;

15. Find the plane through the point (0, 0, 1) tangent to the surface x 2 − y 2 + 3z = 0 and parallel to the line x/2 = y = −z/2. 16. The following equations define w implicitly as a function of the other variables. Find dw in terms of all the variables by taking the differential of both sides and solving algebraically for dw. 1 1 1 1 = + + ; w t u v (b) u 2 + 2v 2 + 3w 2 = 5. (a)

17. Suppose that f (x, y, z) = F g(x, y, z) . Give a formula for d f . 18. Let z = f (x, y), and make the change of variables x = u 2 − v 2 , y = 2uv. Show that (z u )2 + (z v )2 . (z x )2 + (z y )2 = 4(u 2 + v 2 ) 19. If f is a differentiable function of the variable u, let u = x 2 + y 2 and prove that z = x y + f (x 2 + y 2 ) satisfies the equation y

∂z ∂z −x = y2 − x 2. ∂x ∂y

Harder Exercises 20. Let g(s, t) = f u(s, t), v(s, t) , where f , u and v are differentiable and u(1, 0) = 2, v(1, 0) = 3, u s (1, 0) = −2, vs (1, 0) = 5, vt (1, 0) = 4, u t (1, 0) = 6, f u (2, 3) = −1, f v (2, 3) = 10. Find gs (1, 0) and gt (1, 0).

5.5 Exercises

215

21. Suppose that f is a differentiable function of x and y, and g(u, v) = f eu + sin v, eu + cos v . Use the following table of values (0, 0) (1, 0)

f 3 6

g 6 3

fx 4 2

fy 8 5

to calculate gu (0, 0) and gv (0, 0). 22. Given the system of equations u 2 − v + x 2 + y 2 = 0, u + v 2 − 2x y = 0. Find the first and second derivatives of u and v with respect to x and y. 23. The three equations x 2 − y cos(uv) + z 2 = 0, x 2 + y 2 − sin(uv) + 2z 2 = 2, x y − sin u cos v + z = 0, define x, y and z as functions of u and v. Compute the partial derivative ∂ x/∂u and ∂ x/∂v at the point x = y = 1, u = π/2, v = 0, z = 0. 24. The equation f (y/x, z/x) = 0 defines z as a function of x and y, say z = g(x, y). Show that ∂g ∂g +y = g(x, y) x ∂x ∂y ∂ y g(x, y) f , is not zero. ∂y x x 2 2 2 25. The equation F(x + y + z, x + y + z ) = 0 defines z implicity as a function of x and y, say z = f (x, y). Determine the partial derivatives ∂ f /∂ x and ∂ f /∂ y in terms of x, y, z and the partial derivatives ∂ F/∂ x and ∂ F/∂ y. 26. Two sides of a triangle are a and b, and θ is the included angle. The third side is c. at those points at which

(a) Give the approximation for Δc in terms of a, b, c, θ , and Δa, Δb, Δc; (b) If a = 1, b = 2, θ = π/3, is c more sensitive to small changes in a or b? 27. Given that f is a differentiable function with f (2, 5) = 6, f x (2, 5) = 6 and f y (2, −5) = −1, use a linear approximation to estimate f (2.2, 4.9). 28. Find the linear approximation of the function f (x, y) = 1 − x y cos π y at (1, 1) and use it to approximate (1.02, 0.97). Illustrate by graphing f and the tangent plane. 29. Find the linear approximation of the function f (x, y, z) = x 2 + y 2 + z 2 at (3, 2, 6) and use it to approximate the number

(3.01)2 + (1.98)2 + (5.97)2 .

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5 Differentiability and Differential

30. The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle. 31. Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick. 32. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick. 33. The dimensions of a rectangular box are 5, 10 and 20 cm., with a possible measurement error on each side of 0.1 cm. Use differentials to find what possible error should be attached to its volume. 34. Suppose that the equation F(x, y, z) = 0 implicitly defines each of the three variables x, y and z as functions of the other two: z = f (x, y) y = g(x, z) and x = h(y, z). If F is differentiable and Fx , Fy and Fz are all non-zero, show that ∂z ∂ x ∂ y = −1. ∂ x ∂ y ∂z 35. Show that the function z = f (x, y) defined implicitly by 2 zφ (t) = y − φ(t) , (x + t)φ (t) = y − φ(t) ∂z ∂z . ∂x ∂y Hint: Use differentials and eliminate dt. 36. Find the most general differentiable function f (X ) with domain R3 such that f (U + V ) = f (U ) + f (V ) for all U and V . 37. A farmer will earn P = P(x, t) dollars profit if he sells x pounds of cattle t weeks after the cattle market opens for the season. The total weight of his herd is a function x = x(t) of the time. Express the rate of change of his profit with respect to time in terms of the derivatives of P(x, t) and x(t) with appropriate units. (eliminate t) satisfies z =

Chapter 6

Extreme of Functions

6.1 Derivative Tests for Local Extreme Values Let f be a function of two variables. Then (1) f (a, b) is a local maximum value of f if f (x, y) ≤ f (a, b) for all domain points (x, y) in some neighborhood of (a, b). (2) f (a, b) is a local minimum value of f if f (a, b) ≤ f (x, y) for all domain points (x, y) in some neighborhood of (a, b). Local maximum corresponds to mountain peaks on the surface and local minimum corresponds to valley bottoms (Fig. 6.1). At such points, the tangent planes, when they exist, are horizontal. If the inequalities in the above definition hold for all points in the domain of f , then f has an absolute maximum (or absolute minimum) at (a, b). The discussion of the extremum of functions of two variables can be extended to functions of three or more variables. Extreme value theorem: Let f be a function of two variables which is continuous on a bounded closed set D. Then f is bounded on D, and f has both maximum M and a minimum m on D, i.e., there exist points (a, b) and (c, d) in D such that f (a, b) ≤ f (x, y) ≤ f (c, d), for all (x, y) ∈ D. First derivative test for local extreme values: If f has a local maximum or minimum value at an interior point (a, b) of its domain and if the first partial derivatives exist there, then f x (a, b) = 0 and f y (a, b) = 0. An interior point of the domain of a function f where both f x and f y are zero or where one or both of f x and f y do not exist is a critical point of f . A differentiable function f has a saddle point at a critical point (a, b) if every neighborhood of (a, b) contains points (x, y) such that f (x, y) < f (a, b) and also other points such that f (x, y) > f (a, b). For illustration, see Fig. 6.2. Second derivative test for local extreme values: Let f be a function of two variables with continuous second partial derivatives in a neighborhood of a critical point (a, b), and suppose that

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_6

217

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6 Extreme of Functions

Fig. 6.1 A local maximum is a mountain peak and a local minimum is a valley low

Fig. 6.2 Saddle point

A = f x x (a, B = f x y (a, b) = f yx (a, b), C = f yy (a, b), b), A B H = det = AC − B 2 , (this matrix is called theHessian matrix). B C Then (1) (2) (3) (4)

f has a local maximum at (a, b) if H > 0 and A < 0; f has a local minimum at (a, b) if H > 0 and A > 0; f has a saddle point at (a, b) if H < 0; The test is inconclusive at (a, b) if H = 0. In this case, we must find some other way to determine the behavior of f at (a, b).

A function f of three variables with continuous second partial derivatives has a local minimum at a critical point (a, b, c) if all three quantities

6.2 Lagrange Multipliers

219

f A = f x x , D = x x f yx

f x y , f yy

fx x f yx f zx

fx y f yy f zy

f x z f yz , f zz

are positive at (a, b, c), and a local maximum at (a, b, c) if A < 0, D > 0 and E < 0.

6.2 Lagrange Multipliers In many applied problems, the main focus is on optimizing a function subject to constraint; for example, finding extreme values of a function of several variables where the domain is restricted to a level curve (or surface) of another function of several variables. Lagrange multipliers are a general method which can be used to solve such optimization problems. Lagrange multiplier rule for functions of two variables: Let f and g be functions of two variables with continuous first partial derivatives on an open set containing the curve C which is the graph of the equation g(x, y) = k. Let ∇g(x, y) = 0 on C, and suppose that f has a constrained local extremum at a point (a, b) of C. Then there exists a number λ such that ∇ f (a, b) = λ∇g(a, b), that is, the gradients of f and g are parallel at (a, b). Lagrange multiplier rule for functions of three variables: Let f and g be functions of three variables with continuous first partial derivatives on an open set containing the surface S which is the graph of the equation g(x, y, z) = k. Let ∇g(x, y, z) = 0 on S, and suppose that f has a constrained local extremum at a point (a, b, c) of S. Then there exists a number λ such that ∇ f (a, b, c) = λ∇g(a, b, c), that is, the gradients of f and g are parallel at (a, b, c). The method of Lagrange multipliers can also be applied in situations with more than one constraint equation. Optimizing a function subject to two constraints Let f , g and h be functions of three variables with continuous first partial derivatives on an open set containing the surfaces S1 and S2 which are the graph of the equation g(x, y, z) = k and h(x, y, z) = l, respectively. Let ∇g and ∇h are not parallel, and suppose that f has a constrained local extremum at a point (a, b, c). Then there exist number λ and μ such that ∇ f (a, b, c) = λ∇g(a, b, c) + μ∇h(a, b, c), for illustration see Fig. 6.3.

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6 Extreme of Functions

Fig. 6.3 Lagrange multiplier rule subject to two constraints

6.3 Solved Problems 267. Find and classify all critical points for the function f (x, y) = x 4 + y 4 − 4x y. Solution. First, we calculate the first partial derivatives of f . We obtain f x (x, y) = 4x 3 − 4y and f y (x, y) = 4y 3 − 4x. If f x (x, y) = 0 and f y (x, y) = 0, then y = x 3 and x = y 3 . This implies that x 9 = x. Since x 9 − x = x(x 8 − 1) = x(x 4 − 1)(x 4 + 1) = x(x 2 − 1)(x 2 + 1)(x 4 + 1), it follows that 0, 1 and −1 are three real roots of the equation x 9 − x = 0. Hence, the critical points of f are (0, 0), (1, 1) and (−1, −1). Now, we compute the second partial derivatives of f as follows: f x x (x, y) = 12x 2 , f x y (x, y) = −4 and f yy (x, y) = 12y 2 . At the critical point (0, 0), we have 0 −4 = −16 < 0. H (0, 0) = −4 0 Hence, (0, 0) is a saddle point of the graph of f . At the critical point (1, 1), we have 12 −4 = 144 − 16 = 128 > 0. H (1, 1) = −4 12

6.3 Solved Problems

221

Since H (1, 1) > 0 and f x x (1, 1) > 0, it follows that f (1, 1) = −2 is a local minimum of f . Similarly, we see that f (−1, −1) = −2 is also a local minimum of f . 268. Find all critical points of the function f (x, y) = −x ye−(x whether f has a local maximum, minimum or saddle at them.

2

+y 2 )/2

and determine

Solution. The function f has the following partial derivatives: ∂f 2 2 2 2 2 2 (x, y) = −ye−(x +y )/2 + x 2 ye−(x +y )/2 = y(x 2 − 1)e−(x +y )/2 , ∂x ∂f 2 2 2 2 2 2 (x, y) = −xe−(x +y )/2 + x y 2 e−(x +y )/2 = x(y 2 − 1)e−(x +y )/2 . ∂y Setting both partial derivatives to zero, we get y(x 2 − 1) = 0 and x(y 2 − 1) = 0. The simultaneous solutions of these equations are x = 0, y = 0 or x = ±1, y = ±1. Hence, the critical points are (0, 0), (1, 1), (1, −1), (−1, 1) and (−1, −1). The second partial derivatives of f are as follows: ∂ f2 2 2 (x, y) = x y(3 − x 2 )e−(x +y )/2 , ∂x 2 ∂ f2 2 2 (x, y) = (x 2 − 1)(1 − y 2 )e−(x +y )/2 , ∂x∂ y ∂ f2 2 2 (x, y) = x y(3 − y 2 )e−(x +y )/2 . 2 ∂y Application of the second partial derivative test is outlined below. Point (0, 0)

A 0

B −1

C 0

H −1

(1, 1)

2/e

0

2/e

4/e2

Local minimum

(1, −1)

−2/e

0

−2/e

4/e2

Local maximum

(−1, 1)

−2/e

0

−2/e

4/e2

Local maximum

2/e

0

2/e

4/e2

Local minimum

(−1, −1)

Result Saddle point

269. Find a point within a triangle such that the sum of squares of its distance from the three vertices is maximum.

222

6 Extreme of Functions

Solution. Suppose that (x1 , y1 ), (x2 , y2 ) and (x3 , y3 ) are the vertices of the triangle and (x, y) is a point inside the triangle. Let f (x, y) denote the sum of the squares of the distance of (x, y) from the three vertices. Then, we have f (x, y) = (x − x1 )2 + (y − y1 )2 + (x − x2 )2 + (y − y2 )2 + (x − x3 )2 + (y − y3 )2 .

Now, we compute the partial derivatives of f . We obtain f x (x, y) = 2(x − x1 ) + 2(x − x2 ) + 2(x − x3 ), f y (x, y) = 2(y − y1 ) + 2(y − y2 ) + 2(y − y3 ). If f x (x, y) = 0 and f y (x, y) = 0, then x=

x1 + x2 + x3 y1 + y2 + y3 and y = . 2 2

Calculating the second partial derivatives, we get f x x (x, y) = 6 > 0, f x y (x, y) = 0, f yy (x, y) = 6, and so H (x, y) = 36 > 0. Therefore, f has a minimum value at x + x + x y + y + y 1 2 3 1 2 3 . , 3 3 270. Find all critical points of the function f (x, y) = − sin x sin y and determine whether f has a local maximum, minimum or saddle at them. Solution. Critical points are the solutions of the following system of equations: f x (x, y) = cos x sin y = 0 and f y (x, y) = sin x cos y = 0. From the first equation, we obtain x = π/2 + kπ or y = mπ, where k and m are integers. Hence, the system of equations is equivalent to the following systems of equations: π (6.1) x = + kπ and sin x cos y = 0 2 or y = mπ and sin x cos y = 0.

(6.2)

Since sin(π/2 + kπ) = (−1)k , it follows that the second equation in (6.1) is equivalent to cos y = 0 or y = π/2 + mπ. The second equation in (6.2) is equivalent to sin x = 0 or x = kπ. Therefore, we conclude that the critical points of f are π 2

+ kπ,

π + mπ and (kπ, mπ), 2

6.3 Solved Problems

223

for all m, k ∈ Z. Now, we apply the second derivative test to each critical point. First, we calculate the second partial derivatives of f . We get f x x (x, y) = − sin x sin y, f x y (x, y) = cos x cos y and f yy (x, y) − sin x sin y. For the critical points (π/2 + kπ, π/2 + mπ), we have fx x fx y f yy

π 2 π 2 π 2

π + mπ = −(−1)k+m , 2 π + kπ, + mπ = 0, 2 π + kπ, + mπ = −(−1)k+m , 2 + kπ,

and H

π 2

+ kπ,

−(−1)k+m π 0 2k+2m + mπ = = 1 > 0. k+m = (−1) 0 −(−1) 2

If k + m is even, then f (π/2 + kπ, π/2 + mπ) is a local maximum value of f . If k + m is odd, then f (π/2 + kπ, π/2 + mπ) is a local minimum value of f . Regarding the critical points (kπ, mπ), we have f x x (kπ, mπ) = 0, f x y (kπ, mπ) = (−1)k+m / and f yy (kπ, mπ) = 0. Since H (kπ, mπ) =

0 (−1)k+m 2k+2m = −1 < 0, = −(−1) (−1)k+m 0

it follows that the points (kπ, mπ) are saddle points of f . 271. Let (x, y) = sin x + sin y + cos(x + y). Find the local maximum and local minimum values and saddle points of f inside the region R = {(x, y) | 0 ≤ x ≤ 2π and 0 ≤ y ≤ 2π}. Solution. We find that f x (x, y) = cos x − sin(x + y) and f y (x, y) = cos y − sin(x + y). If f x (x, y) = 0 and f y (x, y) = 0, then cos x = sin(x + y) and cos y = sin(x + y). Hence, we obtain cos x = cos y. This implies that x = y or x = 2π − y. If x = y, then we get sin(2x) = cos x or equivalently cos x(2 sin x − 1) = 0. This yields that cos x = 0 or sin x = 1/2, and so x = π/2, 3π/2, π/6 or 5π/6. Thus, in this case, there are four critical points: (π/2, π/2), (3π/2, 3π/2), (π/6, π/6) and

224

6 Extreme of Functions

Fig. 6.4 The fence

(5π/6, π/5). If x = 2π − y, then cos x = cos y = 0. This gives us the critical points (π/2, 3π/2) and (3π/2, π/2). Therefore, we have six critical points. Now, we compute the second partial derivatives as follows: f x x (x, y) = − sin x − cos(x + y), f x y (x, y) = − cos(x + y), f y (x, y) = − sin y − cos(x + y). Application of the second partial derivative test is outlined below. Point (π/2, π/2)

A 0

B 1

C 0

H −1

Result Saddle point

(3π/2, 3π/2)

2

1

2

3

(π/2, 3π/2)

−2

−1

0

−1

Saddle point

(3π/2, π/2)

0

−1

−2

−1

Saddle point

(π/6, π/6)

−1

0

−1/2

3/4

Local maximum

(5π/6, 5π/6)

−1

−1/2

−1

3/4

Local maximum

Local minimum

272. Equal angle bends are made at equal distances from the two ends of a 100 m long fence so the resulting three segment fence can be placed along an existing wall to make an enclosure of trapezoidal shape. What is the largest possible area for such an enclosure? Solution. Figure 6.4 shows the fence. The area that it enclosed is A(x, θ) = (100 − 2x)(x sin θ) + 2(1/2)(x sin θ)(x cos θ) = (100x − 2x 2 ) sin θ + x 2 sin θ cos θ. First, we compute the partial derivatives as follows: ∂A (x, θ) = (100 − 4x) sin θ + 2x sin θ cos θ, ∂x

(6.3)

6.3 Solved Problems

225

∂A (x, θ) = (100x − 2x 2 ) cos θ + x 2 cos(2θ). ∂θ

(6.4)

If we set the partial derivatives in (6.3) and (6.4) equal to zero, then we have (100 − 4x) sin θ + 2x sin θ cos θ = 0 and (100x − 2x 2 ) cos θ + x 2 cos(2θ) = 0. Note that the fence of the maximum area cannot have x = 0 or sin θ = 0. So, from the first equation, we obtain (100 − 4x) + 2x cos θ = 0. This implies that cos θ = −

100 − 4x , 2x

and so cos(2θ) = 2 cos2 θ − 1 =

(100 − 4x)2 − 1. 2x 2

Substituting this in the second equation, we get −(100 − 2x)

(100 − 4x)2 100 − 4x +x − 1 = 0. 2x 2x 2

This leads to 6x 2 − 200x = 0. Therefore, we obtain x = 100/3 and θ = 60◦ , and √ the maximum area is A = 2500/ 3. 273. Find the absolute maximum and minimum of the function f (x, y) = x 2 − x y + y 2 + 1 on the closed triangular plate in the first quadrant bound by the lines x = 0, y = 4 and y = x. Solution. Since f is differentiable, the only places where f can assume these values are points inside the triangle (Fig. 6.5) where f x = 0 and f y = 0, and the points on the boundary. (a) Interior points: For these we have f x (x, y) = 2x − y = 0 and f y (x, y) = −x + 2y = 0, yielding the single point (0, 0), which is not an interior point of the region. (b) Boundary points: We consider each edge of the triangle. (1) On the segment line O A, x = 0. The function f (0, y) = y 2 + 1 may be regarded as a function g1 of y defined on the closed interval [0, 4]. If g1 (y) = 2y = 0, it follows that y = 0, and so x = 0. Hence, the extreme value of g1 may occur at the endpoints (0, 0) and (0, 4), where f (0, 0) = 1 and f (0, 4) = 17. (2) On the segment line AB, y = 4. The function f (x, 4) = x 2 − 4x + 17 may be considered as a function g2 of x on the closed interval [0, 4]. If g2 (x) = 2x − 4 = 0, then x = 2, and so (2, 4) is an interior point of AB. Hence, the extreme values of g2 may occur at the points (0, 4), (4, 4) or (2, 4), and we have f (0, 4) = 17, f (4, 4) = 17 and f (2, 4) = 13.

226

6 Extreme of Functions

Fig. 6.5 Closed triangular plate O AB

y (2, 4) •

A(0, 4) •

•

• B(4, 4)

x O

(3) On the segment line O B, we have y = x. So, the function f (x, x) = x 2 + 1 may be regarded as a function g3 of x on the closed interval [0, 4]. If g3 (x) = 2x = 0, then x = 0, and so y = 0. Here, (0, 0) is not interior of g3 . Moreover, we have already accounted for the values of f at the endpoints of O B. Finally, we list all the candidates: f (0, 0) = 1, f (0, 4) = 17, f (4, 4) = 17 and f (2, 4) = 13. Therefore, we conclude that the absolute maximum value is 17 at (0, 4) and (4, 4), and the absolute minimum is 1 at (0, 0. 274. Find the absolute extreme values of the function f (x, y) = (1/2)x 2 + 3 (1/3)y − x y in the rectangle D = [0, 2] × [0, 2]. Solution. Since f is a polynomial function, it is continuous, and so f attains its maximum and minimum in the rectangle (Fig. 6.6). First, we compute the partial derivatives of f . We obtain f x (x, y) = x − y and f y (x, y) = y 2 − x. Hence, the critical points satisfy the equations x − y = 0 and y 2 − x = 0. The point (0, 0) lies on the boundary of rectangle, while the point (1, 1) lies in the interior of rectangle and f (1, 1) = 6. The boundary of [0, 2] × [0, 2] consists of four segment straight lines. (1) Along the segment AB, the function f can be regarded as a function g1 (x) = f (x, 2) = (1/2)x 2 − 2x + 8/3, where 0 ≤ x ≤ 2. Since g1 (x) = x − 2 ≤ 0, it follows that g1 is decreasing and its maximum value is g1 (0) = f (0, 2) = 8/3, and its minimum value is g1 (2) = f (2, 2) = 2/3.

6.3 Solved Problems

227

Fig. 6.6 Closed rectangle D = [0, 2] × [0, 2]

y A(0, 2) •

• B(2, 2) • (2,

√

2)

•

•

O

• C(2, 0)

x

(2) Along the segment line OC, the function f can be considered as a function g2 (x) = f (x, 0) = (1/2)x 2 , where 0 ≤ x ≤ 2. Hence, the maximum of g2 is g2 (2) = f (2, 0) = 2, and its minimum is g2 (0) = f (0, 0) = 0. f (2, y) = (3) Along the segment line BC, we consider a function g3 (y) = √ ≤ 2. If g3 (y)√= y 2 − 2 = 0, then y = 2. Since (1/3)y 3 − 2y + 2, where 0 ≤ y √ g3 (0) = 2, g3 (2) = 2/3 and g3 ( 2) = 2 − 4 2/3, it follows √ that the maximum √ is g3 (0) = f (2, 0) = 2, and the minimum of g3 is g3 ( 2) = f (2, 2) = of g3 √ 2 − 4 2/3. (4) Finally, along the segment line AO, we consider g4 (y) = f (0, y) = (1/3)y 3 , where 0 ≤ y ≤ 2. It is clear that g4 is monotonically increasing. Hence, the maximum and minimum of g4 are g4 (2) = f (0, 2) = 8/3 and g4 (0) = f (0, 0) = 0, respectively. Therefore, the maximum value of f on the rectangle D is equal to max{0, −1/6, 8/3} = 8/3 = f (0, 2), and its minimum value is equal to min{0, −1/6} = −1/6 = f (1, 1). 275. Find two numbers a and b with a ≤ b such that

b

(6 − x − x 2 )d x

a

has its largest value. Solution. We define

f (a, b) = a

b

(6 − x − x 2 )d x,

228

6 Extreme of Functions

where a ≤ b. The boundary of the domain of f is the line a = b in the ab-plane, and f (a, a) = 0. This means that f is identically 0 on the boundary of its domain. For interior critical points, we must have ∂f ∂f = −(6 − a − a 2 ) = 0 and = 6 − b − b2 = 0. ∂a ∂b Solving the above equations, we obtain a = −3 or 2 and b = −3 or 2. Since a ≤ b, it follows that there exists only one interior critical point (−3, 2), and f (−3, 2) gives us the area under the parabola y = 6 − x − x 2 that is above x-axis. Therefore, we get a = −3 and b = 2. 276. Find two numbers a and b with a ≤ b such that

b 3

24 − 2x − x 2 d x

a

has its largest value. Solution. We define

f (a, b) =

b 3

24 − 2x − x 2 d x,

a

where a ≤ b. The boundary of the domain of f is the line a = b in the ab-plane, and on this line f is identically 0 on the boundary of its domain. For interior critical points, we must have ∂f ∂f 3 3 = − 24 − 2a − a 2 = 0 and = 24 − 2b − b2 = 0. ∂a ∂b From the above equations, we get a = 4 or −6 and b = 4 or −6. Since a ≤ b, it follows that there exists only one√interior critical point (−6, 4), and f (−6, 4) gives us the area under the curve y = 3 24 − 2x − x 2 that is above the x-axis. Therefore, we obtain a = −6 and b = 4. 277. The quantity which is conventionally used to measure the closeness between a line y = f (x) = mx + b in the plane and n points (x1 , y1 ), (x2 , y2 ), . . ., (xn , yn ) is the sum n n

2 f (xi ) − yi = (mxi + b − yi )2 , S= i=1

i=1

which will be recognized as the sum of the squares of the deviations between the given values of yi and the ordinates of the line y = mx + b corresponding to the

f (xi ) − yi given values of xi . This measure of closeness is better than the sum of the deviations themselves, in which positive and negative terms can cancel each other out, leading to the erroneous conclusion that the points lie near the line even

6.3 Solved Problems

229

when they actually do not. The line y = mx + b minimizing S is called the line of best fit to the n points. Find the slope of the line, m, and the y-intercept, b. Solution. Take the partial derivatives of S with respect to m and b. We get ∂S = 2(mxi + b − yi )xi = m 2xi2 + b 2xi − 2xi yi , ∂m i=1 i=1 i=1 i=1 n

n

n

n

∂S = 2(mxi + b − yi ) = m 2xi + b 2 − 2yi . ∂b i=1 i=1 i=1 i=1 n

n

n

n

Now, we solve the simultaneous equations ∂ S/∂m = 0 and ∂ S/∂b = 0. We define the constants σx =

n i=1

xi , σ y =

n

yi , σx 2 =

i=1

n

xi2

and σx y =

i=1

n

xi yi .

i=1

Then, we have the following equations: σx 2 m + σx b = σx y ,

(6.5)

σx m + nb = σ y .

(6.6)

Here, there are two linear equations on the unknowns m and b. We use (6.6) to solve for b in terms of m. So, we get 1 (σ y − σx m), n

b=

(6.7)

and then substitute this into (6.5) to obtain the equation σx 2 m +

1 σx (σ y − σx m) = σx y . n

This leads to (nσx 2 − σx2 )m = nσx y − σx σ y . If we solve this equation for m, then m=

nσx y − σx σ y . nσx 2 − σx 2

Substituting this into (6.7) gives b=

σx σx y − σ y σx 2 . nσx 2 − σx2

230

6 Extreme of Functions

Fig. 6.7 A tent on a square base

T

h H

G

E

F

y D

C x

A

B

x

278. A tent on a square base of sides x has its sides vertical of height y and the top is a regular pyramid of height h (Fig. 6.7). Find x and y in terms of h, if the canvas required for its construction is to be minimized for the tent to have a given capacity. Solution. Let V (x, y, h) be the volume of tent. So, we have h 1 . V (x, y, h) = x 2 y + x 2 h = x 2 y + 3 3 Also, let S(x, y, h) be the area of tent, i.e., the area of four faces of the cuboid together with the area of four faces of the pyramid. Hence, we can write S(x, y, h) = 4x y + 4 · = 4x y + 4 ·

1

base · (length T M)

2

x 2 x h2 + = 4x y + x x 2 + 4h 2 . 2 2

Now, we want to minimize S(x, y, h) subject to V (x, y, h) = c, where c is a constant value. Calculating the gradients of S and V , we get ∇ S(x, y, h) = 4y +

√

x2

+

4h 2

+√

x2

x 2 + 4h 2 h 2 1 2 ∇V (x, y, h) = 2x y + ,x , x . 3 3

, 4x, √

4hx x 2 + 4h 2

,

6.3 Solved Problems

231

Applying the Lagrange multiplier rule ∇ S = λ∇V , we have 4y +

x 2 + 4h 2 + √

h , = 2λx y + 3 x 2 + 4h 2 x2

4x = λx 2 , √

4hx x2

+

4h 2

=

(6.8) (6.9)

1 2 λx . 3

(6.10)

By (6.9) we obtain λ = 4/x. Substituting in (6.10) gives √

4hx x 2 + 4h 2

=

This implies that x 2 = 5h 2 . So, we have x = in (6.8), we get

4 x. 3

√

√ 5h and λ = 4/( 5h). Putting these

√ 4 5h 2 h 2 2 4y + 5h + 4h + √ = 2 · √ · 5h y + . 3 5h 2 + 4h 2 5h This yields that y = h/2. 279. Three cities A, B and C are located at (5, 2), (−4, 4) and (−1, −3), respectively, on the x y -plane. There is a rail track whose equation is y = x 3 + 1, and a station is going to be built on the track so that the sum of squares of the distances from each city to the station is minimized. Find the coordinates of the station. Solution. We notice that

(1) The distance from station to A is (x − 5)2 + (y − 2)2 ; (2) The distance from station to B is (x + 4)2 + (y − 4)2 ; (3) The distance from station to C is (x + 1)2 + (y + 3)2 .

Hence, we must minimize the following function: f (x, y) = (x − 5)2 + (y − 2)2 + (x + 4)2 + (y − 4)2 + (x + 1)2 + (y + 3)2 , subject to the constraint g(x, y) = y − x 3 − 1 = 0. By the Lagrange multiplier rule ∇ f = λ∇g and g = 0, we obtain the system of equations 2(x − 5) + 2(x + 4) + 2(x + 1) = −3λx 2 , 2(y − 2) + 2(y − 4) + 2(y + 3) = λ, y − x 3 − 1 = 0.

232

6 Extreme of Functions

This leads to 2x = −λx 2 , 6y − 6 = λ and y = x 3 + 1. Solving the system of equations, we obtain x = 0 and y = 1. Therefore, we deduce that the station must be located at the point (0, 1) to minimize the sum of squares of the distances. 280. For positive real numbers x and y such that x + y = 1, prove that

1 1 x + + y + ≤ 2. 2 2

1 1 Solution. Let f (x, y) = x + + y + and g(x, y) = x + y − 1 = 0. We want 2 2 to find the maximum of f subject to g = 0. First, we compute the gradients of f and g as follows:

⎞

⎛ ⎜ 1 ∇ f (x, y) = ⎜ ⎝

2 x+

,

1

⎟ ⎟ and ∇g(x, y) = (1, 1). 1⎠

1 2 y+ 2 2

By the Lagrange multiplier rule ∇ f = λ∇g, we obtain

1

1 2 x+ 2

= λ and

1

1 2 y+ 2

= λ.

This leads to x = y, and so x = y = 1/2. It is clear that f (1/2, 1/2) = 2 is the maximum value of f , and this completes the proof. 281. The Steiner problem: The lengths of all sides of a quadrangle are fixed, but the sides are linked freely at the vertices, so the angles between them can vary. Which position of the sides corresponds to the largest area of the quadrangle? Solution. We need to maximize f (α, β) =

1 (ab sin α + cd sin β) 2

subject to a 2 + b2 − 2ab cos α = c2 + d 2 − 2cd cos β, where a, b, c, d are the sides of the quadrangle, α is the angle between a and b, and β is the angle between c and d. The equality constraint comes from applying the cosine rule to both triangles with side a diagonal. Setting g(α, β) = a 2 + b2 − 2ab cos α − c2 − d 2 + 2cd cos β = 0 then by the Lagrange multiplier rule ∇ f = λ∇g, we get

6.3 Solved Problems

233

Fig. 6.8 Rectangular box

1 ab cos α = 2λab sin α, 2

(6.11)

1 cd cos β = −2λcd sin β. 2

(6.12)

By (6.11) we get tan α = 1/(4λ) and by (6.12) we have tan β = −1/(4λ). This yields that tan α = − tan β, and so α = π − β. This means that the quadrangle abcd is inscribed in a circle. Therefore, among all the quadrangles with given sides, the inscribed quadrangle possesses the largest area. 282. Determine the relative dimensions of a rectangular box without a top and having a special volume, if the least amount of material is to be used in the manufacture. Solution. If x, y and z are the edges (see Fig. 6.8), then (1) the volume of the box= V = x yz =constant; (2) the surface area of the box= x y + 2x z + 2yz. Each of the variables x, y and z is in the interval (0, ∞). From the items (1) and (2), we can write 2V 2V + . (6.13) S = xy + y x Take the partial derivatives of (6.13) to obtain ∂S 2V ∂ S 2V =y− 2, =x− 2, ∂x x ∂y y ∂2 S 4V = 3, 2 ∂x x

∂2 S = 1, ∂x∂ y

∂2 S 4V = 3. 2 ∂y y

0 and x y 2 − V = 0. Solving these If ∂ S/∂x = 0 and ∂ S/∂ y = 0, then x 2 y − 2V =√ √ 3 system of equations gives us x = 2V and y = 3 2V . For these values of x and y, we obtain

234

6 Extreme of Functions

∂2 S 4V = √ 3 = 2 > 0 and 3 ∂x 2 2V 4V 4V ∂ 2 S ∂ 2 S ∂ 2 S 2 − = √ 3 · √ 3 − 1 = 3 > 0. 3 3 ∂x 2 ∂ y 2 ∂x∂ y 2V 2V √ √ Thus, we conclude that S has a local minimum value when x = 3 2V and y = 3 2V . Note that S is very large when x and y are either close to zero or very large. Hence, we conclude that the √ √ local minimum value of S is an√absolute value of S. Since x = 3 2V and y = 3 2V , it follows that z = V /x y = 3 2V /2. Therefore, the box has a square base and a depth that is one-half of the length of a side of the base. 283. Consider F(x, y, z) subject to the constraint condition G(x, y, z) = 0. Prove that a necessary condition that F(x, y, z) have an extreme value is that Fx G y − Fy G x = 0 (when Fz = 0 and G z = 0 ). Solution. Since G(x, y, z) = 0, we consider z as a function of x and y, say z =

f (x, y). A necessary condition that F x, y, f (x, y) have an extreme value is that the partial derivatives with respect to x and y be zero. This yields that Fx + Fz

∂z = 0, ∂x

(6.14)

Fy + Fz

∂z = 0. ∂y

(6.15)

Gx + Gz

∂z = 0, ∂x

(6.16)

G y + Gz

∂z = 0. ∂y

(6.17)

Since G(x, y, z) = 0, it follows that

By (6.14) and (6.16), we have Fx G z − Fz G x = 0,

(6.18)

and by (6.15) and (6.17), we have Fy G z − F2 G y = 0. Finally, by (6.18) and (6.19), we obtain Fx G y − Fy G x = 0, as desired.

(6.19)

6.3 Solved Problems

235

284. Suppose that f (x, y) = x ye−(x

2

+y 2 )

, for all (x, y) ∈ R2 .

(1) Identify the points of local maximum and local minimum, and the saddle points; (2) Show that f is bounded on R2 . Solution. (1) Calculating f x and f y we obtain f x (x, y) = y(1 − 2x 2 ) and f y (x, y) = x(1 − 2y 2 ). Solving f x = 0 and f y = 0, we obtain five critical points as follows: √ √ √ √ √ √ √ √ (0, 0), (−1/ 2, −1/ 2), (−1/ 2, 1/ 2), (1/ 2, −1/ 2) and (1/ 2, 1/ 2). Now,√ by using√the second √ derivative √ test, we observe that (0, 0) is a saddle point; (−1/√2, −1/ √ 2) and (1/ √ 2, 1/ √2) are the points that f has a local maximum; (−1/ 2, 1/ 2) and (1/ 2, −1/ 2) are the points that f has a local minimum. (2) We have 2 2 2 2 2 | f (x, y) = x ye−(x +y ) ≤ (x 2 + y 2 )e−(x +y ) = r 2 e−r → 0 as r → ∞. Note that r2 2r 1 = 0 (by L Hospital s rule). 2 = lim 2 = r →∞ e−r r →∞ 2r er er 2

lim r 2 e−r = lim 2

r →∞

Hence, there is M > 0 such that (x, y) ≥ M ⇒ −

1 1 < f (x, y) < . 2e 2e

On the other hand, since f is a continuous function, it follows that f takes its maximum and minimum on the closed and bounded disk {(x, y) ∈ R2 | (x, y) ≤ M}. Therefore, we deduce that f is bounded on R2 . 285. Find the absolute maximum and minimum values of the function f (x, y) = 2x 3 + 2x y 2 − x − y 2 on the unit disk S = {(x, y) | x 2 + y 2 ≤ 1}. Solution. First, we determine the critical points of f in the interior of S. At a critical point, we have f x (x, y) = 6x 2 + 2y 2 − 1 = 0 and f y (x, y) = 4x y − 2y = 0. The second equation gives that√y = 0 or x = 1/2. Substituting these into the first in the second equation implies that y = ±1/ 6 in the first √ √ case, and no solution case. Consequently, the critical points are (1/ 6, 0) and (−1/ 6, 0). Moreover, we observe that both of these points lie in the interior of S. Boundary of S is the circle

236

6 Extreme of Functions

Fig. 6.9 The surface area of a right cylindrical can is the sum of the areas of two circles and one rectangle

√ x 2 + y 2 = 1. This yields that y = ± 1 − x 2 , where −1 ≤ x ≤ 1. √ solutions

These correspond to the upper and lower semicircles. We obtain g(x) = x, ± 1 − x 2 = x ≤ 1. If g (x) = 2x √ + 1 = 0, then x = −1/2. This gives the x 2 + x − 1, for −1 ≤ √ critical points (−1/2, 3/2) and (−1/2, − 3/2) of restriction of f to the boundary of S. On the other hand, we must consider the end points x = −1 and x = 1, i.e., (1, 0) and (0, 1).√Therefore, the and minimum √ absolute maximum √ √ of f on S occur at the points (1/ 6, 0), (−1/ 6, 0), (−1/2, 3/2), (−1/2, − 3/2), (1, 0) or (−1, 0). We observe that √ √ √ √ f (1/ 6, √ 0) = −1/3 2/3, f (−1/ 6, √ 0) = 1/3 2/3, f (−1/2, 3/2) = −5/4, f (−1/2, − 3/2) = −5/4, f (1, 0) = 1, f (−1, 0) = −1. Consequently, the absolute maximum value is 1 and the absolute minimum value is −5/4. 286. A right cylindrical can is to have a volume of 0.25 cubic feet (approximately two gallons). Find the height h and radius r that will minimize surface area of the can. What is the relationship between the resulting r and h? Solution. The surface area S is the sum of the area of two cycles of radius r and one rectangle with height h; see Fig. 6.9. Hence, S = f (r, h) = 2πr 2 + 2πr h. This is constrained by a volume of V = g(r, h) =πr 2 h = 0.25. The Lagrange multiplier rule ∇ f = λ∇g becomes 4πr + 2πh, 2πr = λ 2πr h, πr 2 . This leads to 4πr + 2πh = λ(2πr h) and 2πr = λ(πr 2 ). Note that r cannot be zero. The ratio of the two equalities is λ(2πr h) 4πr + 2πh = , 2πr λ(πr 2 ) and so (2r + h)/r = 2h/r . Thus, we get h = 2r . That is, all cans with minimal surface area have h = 2r , which means a height equal to the diameter. To determine which such can satisfies the constraint, we plug h = 2r in our constraint equation to

6.3 Solved Problems

237

obtain πr (2r ) = 0.25 or r = 2

3

0.25 2π

which leads to r = 0.3414 feet, with h = 0.6818 feet. To show that a minimum must occur at (r, h) = (0.3414, 0.6818), we observe that the constraint implies h = 0.25/(πr 2 ). So, we can consider S as a function of r in the form S = 2πr 2 + 0.5/r 2 . Since the second derivatives of S is positive for all r > 0, it follows that any extremum is a minimum. 287.

A function f from R2 to R is convex if

(1) its domain is a convex set in R2 ; (2) for all X, Y in the domain of f and λ ∈ [0, 1], f (λX + (1 − λ)Y ) ≤ λ f (X ) + (1 − λ) f (Y ).

Geometrically, if we take two points X, f (X ) and Y, f (Y ) on the graph of f , then the graph of f lies below the line segment joining the two points chosen. Suppose that f : R2 → R is a differentiable function. Prove that f is convex if and only if f (Y ) ≥ f (X ) + ∇ f (X ) · (Y − X ), (6.20) for all X, Y ∈ R2 .

Solution. Assume that f is a convex function. Then, by the definition we have f λY + (1 − λ)X ≤ λ f (Y ) + (1 − λ) f (X ), for all X, Y in the domain of f and 0 ≤ λ ≤ 1. After rewriting, we have

f X + λ(Y − X ) ≤ f (X ) + λ f (Y ) − f (X ) . This implies that

f X + λ(Y − X ) − f (X ) ≤ f (Y ) − f (X ), for all λ ∈ (0, 1]. λ Taking λ → 0+ , in the right side we get the directional derivatives along the vector Y − X . So, ∇ f (X ) · (Y − X ) ≤ f (Y ) − f (X ), and hence (6.20) is satisfied. Conversely, suppose that (6.20) holds. Take any X, Y in the domain of f and let Z = λX + (1 − λ)Y . We have f (X ) ≥ f (Z ) + ∇ f (Z ) · (X − Z ),

(6.21)

f (Y ) ≥ f (Z ) + ∇ f (Z ) · (Y − Z ).

(6.22)

Multiplying (6.21) by λ and (6.22) by 1 − λ and adding, we obtain

238

6 Extreme of Functions

λ f (X ) + (1 − λ) f (Y ) ≥ f (Z ) + ∇ f (Z ) · λX + (1 − λ)Y − Z = f (Z ) = f λX + (1 − λ)Y . 288. Let f : R2 → R be convex and differentiable. Prove that X is an absolute minimizer of f if and only if ∇ f (X ) = 0. Solution. First, we recall that the condition ∇ f (X ) = 0 is a necessary condition for X to be an absolute (and indeed already local) minimizer. Thus, we only need to show that this condition actually implies that X is an absolute minimizer. Therefore, assume that ∇ f (X ) = 0 and let Y be in the domain of f . Then, Problem 287 implies that f (Y ) ≥ f (X ) + ∇ f (X ) · (Y − X ) = f (X ). Thus, X is an absolute minimizer. a b 289. A matrix is said to be positive semi-definite if the matrix multiplication c d a b x [x y] = ax 2 + (b + c)x y + dy 2 ≥ 0, for all x, y ∈ R. c d y Prove that a twice differentiable function f : R2 → R is convex if and only if the matrix ⎡ ⎤ f x x (X ) f x y (X ) ⎣ ⎦ (6.23) f yx (X ) f yy (X ) is positive semi-definite, for all X ∈ R2 . Solution. First, suppose that f is convex and let X ∈ R2 . We define the function g : R2 → R by g(Y ) := f (Y ) − ∇ f (X ) · (Y − X ). Since f is convex, it follows that g is convex too. Moreover, we have ∇g(Y ) = ∇ f (Y ) − ∇ f (X ) and

⎡ ⎣

gx x (X ) gx y (X ) g yx (X ) g yy (X )

⎤

⎡

⎦=⎣

f x x (X ) f x y (X ) f yx (X ) f yy (X )

⎤ ⎦

(6.24)

for all Y ∈ R2 . In particular, we have ∇g(X ) = 0. So, Problem 6.3 implies that X is an absolute minimizer of g. Now, the second condition for a minimizer implies that the matrix

6.3 Solved Problems

239

⎡ ⎣

gx x (X ) gx y (X )

⎤ ⎦

g yx (X ) g yy (X ) is positive semi-definite. Since (6.24) holds, and X is arbitrary, it follows that ⎡ ⎣

f x x (X ) f x y (X )

⎤ ⎦

f yx (X ) f yy (X ) is positive semi-definite. Conversely, suppose that the matrix in (6.23) is positive semi-definite, for all X ∈ R2 . If X, Y ∈ R2 , then by Taylor’s formula, we have f (Y ) = f (X ) + ∇ f ⎡ (X ) · (Y − X ) ⎤ f x x (X + k(Y − X )) f x y (X + k(Y − X )) 1 ⎦ (Y − X )t , + (Y − X ) ⎣ 2 f yx (X + k(Y − X )) f yy (X + k(Y − X )) (6.25) for some 0 ≤ k ≤ 1. Note that the quadratic term in (6.25) is always non-negative. Thus, we can estimate f (Y ) ≥ f (X ) + ∇ f (X ) · (Y − X ). Now, Problem 287 proves the convexity of f . 290. Three hemispheres with radiuses 1, x and y, where 0 ≤ y ≤ x ≤ 1, are stacked on top of each other as shown in Fig. 6.10. Find the largest possible value of the total height h. √ Solution. Let f (x, y) = 1 − x 2 + x 2 + y 2 + y. We want to maximize f on the region S = {(x, y) | 0 ≤ y ≤ x ≤ 1}. First, we determine the critical points of f in the interior of S. We obtain f x (x, y) = − √

x 1−

x2

+

x x2

−

y2

and f y (x, y) = −

y x2

− y2

+ 1.

√ So, if f y (x, y) = 0, then y 2 = x 2 − y 2 or √ 2y 2 = x 2 , which implies that x = 2y because y) = 0 √ we get √Substituting x = 2y in the equation f x (x, √ √ x > 0 and y > 0. y = 3/3, and so x = 6/3. Hence, the only critical point of f is ( 6/3, 3/3) and it lies in S. Next, we consider the restriction of f to the boundary of S. Figure 6.11 shows the boundary of S. On the edge O A we have y = 0 and 0 ≤ x ≤ 1. So, we consider the function g(x) = f (x, 0) =

1 − x 2 + x, for 0 ≤ x ≤ 1.

240

6 Extreme of Functions

Fig. 6.10 Three hemisphere with radiuses 1, x and y

y

x

1

Fig. 6.11 Boundary of the region S = {(x, y) | 0 ≤ y ≤ x ≤ 1}

y B(1, 1)

x O

We obtain

g (x) = − √

x 1 − x2

A(1, 0)

+ 1.

√ If g (x) = 0, then x = √ 2/2. Hence, on the edge O A, by considering the endpoints, we have three points ( 2/2, 0), (0, 0) and (1, 0). On the edge AB, we have x = 1 and 0 ≤ y ≤ 1. We consider the function

6.3 Solved Problems

241

h(x) = f (1, y) =

1 − y 2 + y, for 0 ≤ y ≤ 1.

√ Similar to the previous case, this leads to the points (1, 2/2), (1, 0) and (1, 1). Finally, on the edge O B, we have y = x and 0 ≤ x ≤ 1. In this case, we consider the function k(x) = f (x, x) = 1 − x 2 + x, for 0 ≤ x ≤ 1. √ √ Again, similar to the previous case, we obtain the points ( 2/2, 2/2), (0, 0) and (1, 1). Next, we calculate the values of f at these seven points. We obtain f (0, √ 0) = f (1, 0) =√f (1, 1) = 1,√ √ √ f (√2/2, √ 0) = f (1,√ 2/2) = f ( 2/2, 2/2) = 2, f ( 6/3, 3/3) = 3. Consequently, the maximum possible for the total height of the three hemispheres is √ 3. 291. Find the shortest distance from the origin to the plane ax + by + cz + d = 0, where a, b, c and d are constant. Solution. We wish to find the minimize value of f (x, y, z) = x 2 + y 2 + z 2 subject to the constraint g(x, y, z) = ax + by + cz = d. The gradients of f and g are ∇ f (x, y, z) = (2x, 2y, 2z) and ∇g(x, y, z) = (a, b, c). It is clear that ∇g = 0. Thus, the Lagrange multiplier rule ∇ f = λ∇g becomes 2x, 2y, 2z) = λ(a, b, c), which implies that x=

1 1 1 λa, y = λb and z = λc. 2 2 2

Substituting these values of x, y and z into the plane equation, we obtain (1/2)λ(a 2 + b2 + c2 ) + d = 0. This implies that −d 1 λ= 2 . 2 a + b2 + c2 Hence, we get x=

a2

−ad −bd −cd , y= 2 and 2 . 2 2 2 2 +b +c a +b +c a + b2 + c2

(6.26)

Note the greatest distance from (0, 0, 0) to any point in the plane → ∞. So, the point we found is the least distance from the origin. Therefore, the minimum distance from the origin to the plane is the distance from the origin to the point (x0 , y0 , z 0 ), where x0 , y0 and z 0 are the values of x, y and z in (6.26). Then, the minimum distance is

242

6 Extreme of Functions

x02 + y02 + z 02 =

a2d 2 b2 d 2 c2 d 2 + + (a 2 + b2 + c2 )2 (a 2 + b2 + c2 )2 (a 2 + b2 + c2 )2

|d| =√ . 2 a + b2 + c2 292. Find the extreme of 4x 2 + y 2 = 4.

f (x, y) = x y if (x, y) is restricted to the ellipse

Solution. The constraint is g(x, y) = 4x 2 + y 2 − 4 = 0. Setting ∇ f (x, y) = λ∇ g(x, y, z), we obtain (y, x) = λ(8x, 2y). Hence, we have the following equations: y = 8λx, x = 2λy and 4x 2 + y 2 − 4 = 0. We begin by writing x = 2yλ = 2(8xλ)λ = 16xλ2 . This implies that x(1 − 16λ2 ) = 0, and so either x = 0 or λ = ±1/4. If x = 0, then using 4x 2 = y 2 − 4 = 0 we get y = ±2. Hence, the points (0, 2) and (0, −2) are possibilities for extreme of f . If λ = ±1/4, then y = 8xλ = 8x(±1/4) or√y = ±2x. Again, using the fact that 4x 2 + y 2 − 4 = 0 we obtain 8x 2 = 4 or x = √ ± 2/2. The 2 − 4 = 0 or y = ± corresponding√ values√for y √ satisfy 4(1/2) + y √ √ √ √ 2. This √ gives us the points ( 2/2, 2), ( 2/2, − 2), (− 2/2, 2) and (− 2/2, − 2). The value of f at each of the points we have found is listed below: f (0, √ 2) =√f (0, −2) =√0, √ f (√2/2, √ 2) = f (− √ 2/2, −√2) = 1, f ( 2/2, − 2) = f (− 2/2, 2) = −1. √ √ 2) or Now, that f takes a maximum 1 at either √ √ value √ √ ( 2/2, √ √ we conclude (− 2/2, − 2), and a minimum value −1 at ( 2/2, − 2) or (− 2/2, 2). These facts are indicated on the ellipse in Fig. 6.12. 293. (1) Two points A and B are situated in different optical media separated by a straight line (Fig. 6.13). The velocity of light in the first medium is v1 and in the second medium is v2 . Applying the Fermat principle, according to which a light ray is propagated along a path AM B which requires the least time to cover, derive the law of refraction of light rays. (2) Using the Fermat principle, derive the law of reflection of a light ray from a plane in a homogeneous medium (Fig. 6.14). Solution. (1) It is clear that the point M, at which the ray passes from one medium into another, must lie between C and D. It follows that

6.3 Solved Problems

243

Fig. 6.12 Ellipse 4x 2 + y 2 = 4

y Min

Max √2 √ •f , 2 =1 2

−√2 √ , 2 = −1 • f 2

x

f

−√ 2 2

√ ,− 2 = 1•

•f

Max

√2 2

√ , − 2 = −1

Min

Fig. 6.13 The law of refraction of a light ray A

a

α M

D

C β

b

B c

244

6 Extreme of Functions

A

B

a

α

b

β

C

M

D

c

Fig. 6.14 The law of reflection of a light ray

a = |AM| cos α, |C M| = a tan α,

b = |B M| cos β, |D M| = b tan β.

Hence, the duration of motion of the ray is b a + . v1 cos α v2 cos β Therefore, the problem reduces to determine the minimum of the function f (α, β) =

a b + v1 cos α v2 cos β

provided g(α, β) = a tan α + b tan β = c. We have ∇ f (α, β) =

b sin β a sin α , 2 v1 cos α v2 cos2 β

,

∇g(α, β) = a(1 + tan2 α), b(1 + tan2 β) . If we apply the Lagrange multiplier rule, then we get the following equations: aλ a sin α = aλ((1 + tan2 α) = , v1 cos2 α cos2 α b sin β bλ = bλ((1 + tan2 β) = . 2 v2 cos β cos2 β From the above equations, we have sin α/v1 = λ and sin β/v2 = λ. This yields that

6.3 Solved Problems

245

Fig. 6.15 A right circular cylinder surmounted by a right circular cone

θ s

h

r

v1 sin α = . sin β v2 (2) In this case, we have v1 = v2 . Hence, sin α = sin β, which implies that α = β. 294. A solid is to consist of a right circular cylinder surmounted by a right circular cone (Fig. 6.15). For a fixed surface area S (including the base), what should be the dimensions to maximize the volume V ? Solution. Let r and h be the radius of the base and the height of the cylinder, respectively, and let 2θ be the vertex angle of the cone. Looking at Fig. 6.15, we observe that 1 1 V = f (r, h, θ) = πr 2 h + πr 2 (r cot θ) = πr 2 h + πr 3 cot θ 3 3 S = g(r, h, θ) = πr 2 + 2πr h + πr s = πr 2 + 2πr h + πr 2 csc θ. We want to maximize f (r, h, α) subject to the constraints g(r, h, θ) = S. First, we determine the gradients of f and g. We get

246

6 Extreme of Functions

1 ∇ f (r, h, θ) = 2πr h + πr 2 cot θ, πr 2 , − πr 3 csc2 θ , 3 ∇g(r, h, θ) = 2πr + 2πh + 2πr csc θ, 2πr, −πr 2 csc θ cot θ . Applying the Lagrange multiplier rule ∇ f = λ∇g and g = S, we get the following system of equations: (1) 2πr h + πr 2 cot θ = 2πλ(r + h + r cscθ), (2) πr 2 = 2πr λ, 1 (3) − πr 3 csc2 θ = −πλr 3 csc θ cot θ, 3 (4) πr 2 + 2πr h + πr 2 csc θ = S. By (2), we get r = 2λ. Substituting this in (1) gives 2r h + r 2 cot θ = r (r + h + r csc θ), which implies that r (cot θ − csc θ) = r − h. By (3) and r = 2λ, we obtain 1 1 csc2 θ = csc θ cot θ. 3 2 √ √ √ This yields that cos θ = 2/3, and so sin√θ = 5/3, csc θ = 3/ √ 5 and cot θ = 2/ 5. Hence, we have cot θ − csc θ = −1/ 5. This yields that −r/ 5 = r − h. Therefore, we conclude that √ 1 5+1 h = r 1 + √ = √ r. 5 5 Substituting in the constraint (4) gives us √5 + 1 3 S = πr + 2πr √ r + πr 2 √ . 5 5 2

This leads to obtain r=

√ 5+1 S √ and h = √ √ . 3π + π 5 5 3π + π 5 S

1 295. Find the point of the paraboloid of revolution z = (x 2 + y 2 ) − 2 which is 4 closest to the point (0.1, 0). Solution. The square of the distance from the point (0, 1, 0) to the variable point (x, y, z) of the paraboloid is x 2 + (y − 1)2 + z 2 . Note that if the square of distance

6.3 Solved Problems

247

is minimized, so is the distance itself. Thus, we want the absolute minimum of the function f (x, y, z) = x 2 + (y − 1)2 + z 2 subject to the constraint g(x, y, z) = x 2 + y 2 − 4z = 8. Calculating the gradients of f and g, we obtain

∇ f (x, y, z) = 2x, 2(y − 1), 2z and ∇g(x, y, z) = 2x, 2y, −4, where it should be noted that ∇g = 0 on the domain of g. Hence, the Lagrange multiplier rule ∇ f = λ∇g implies that 2x = 2λx, 2(y − 1) = 2λy and 2z = −4λ. From the first equation, we get x = 0 or λ = 1. If λ = 1, then the second equation becomes −2 = 0, which is a contradiction. So, we must have x = 0. Note that y = 0 for the same reason. Eliminating λ from the second and third equations, we obtain z=−

2(y − 1) . y

(6.27)

Substituting of x = 0 and (6.27) into the equation of constraint leads to y 2 − 4z = y 2 +

8(y − 1) = 8, y

which implies that y = 2, and so (6.27) gives z = −1. Consequently, (0, 2, −1) is the only point satisfying the condition ∇ f = λ∇g. It is geometrically apparent that f has a minimum, but not a maximum on the paraboloid. Therefore, the minimum must be at (0, 2, −1) and this is the which is the closest point √ √ point of the paraboloid to the point (0, 1, 0), a distance f (0, 2, −1) = 2 away; see Fig. 6.16. 296. The plane x + y + 2z = 2 intersects the paraboloid z = x 2 + y 2 in an ellipse. Find the points on the ellipse that are nearest to and farthest from the origin. Solution. We have two constraints g(x, y, z) = x + y + 2z − 2 and h(x, y, z) = x 2 + y 2 − z. Let (a, b, c) be any point that satisfies both of the constraints, and so lies on the ellipse. We find the extremum of the square of distance from the origin. Hence, the objective function is f (z, y, z) = x 2 + y 2 + z 2 . Next, we obtain − → − → − → ∇ f (a, b, c) = 2a i + 2b j + 2c k , − → − → − → λ∇g(a, b, c) = λ i + λ j + 2λ k , − → − → − → μ∇h(a, b, c) = 2μa i + 2μb j − μ k . So, the system of equations we need to solve is (1) 2a = λ + 2μa, (2) 2b = λ + 2μb,

248

6 Extreme of Functions

Fig. 6.16 Paraboloid 1 z = (x 2 + y 2 ) − 2 4

z

z=

1 2 (x + y 2 ) − 2 4

(0, 1, 0) •

y • (0, 2, −1)

x

(3) 2c = 2λ + μ, (4) a + b + 2c = 2, (5) a 2 + b2 − c = 0. From (1) and (2), we get 2(a − b) = 2μ(a − b), which implies that μ = 1 whenever a = b. Substituting μ = 1 in (1) gives that λ = 0, and substituting μ = 1 and λ = 0 in (3) implies that 2c = −1 or c = −1/2. Substituting c = −1/2 in (4) and (5) implies that a + b − 3 = 0 and a 2 + b2 + 1/2 = 0. But the second equation has no solution. Consequently, we must have a = b. Since a = b, it follows that 2a + 2c = 2 and 2a 2 − c = 0. Hence, we obtain c = 1 − a and c = 2a 2 , and so 2a 2 + a − 1 = 0. The solutions of this equation are a = −1 and a = 1/2. Further substitution gives the critical points (−1, −1, 2) and (1/2, 1/2, 1/2). Finally, by substituting these points in the objective function, we obtain f (−1, √ −1, 2) = 6 and f (1/2, 1/2, 1/2) = 3/4. Therefore, the √ maximum distance of 6 occurs at (−1, −1, 2) and the minimum distance of 3/2 occurs at (1/2, 1/2, 1/2). 297. Find the extreme values of f (x, y, z) = x y + z if x 2 + y 2 = 1 and y + z = 2; see Fig. 6.17. Solution. Let g(x, y, z) = x 2 + y 2 − 1 = 0 and h(x, y, z) = y + z − 2 = 0. Then, we have ∇ f (x, y, z) = (y, x, 1), ∇g(x, y, z) = (2x, 2y, 0) and ∇h(x, y, z) = (0, 1, 1). Applying the Lagrange multiplier rule ∇ f = μ∇g + μ∇h, we get

6.3 Solved Problems

249

Fig. 6.17 The surfaces x 2 + y 2 = 1 and y + z = 2 intersect in a curve

y

x

z

(y, x, 1) = λ(2x, 2y, 0) + μ(0, 1, 1). This implies that y = 2λx, x = 2λx + μ and 1 = μ. So, y = 2λx and x = 2λx + 1. It is clear that x = 0, otherwise we obtain y = 0, a contradiction with g(x, y, z) = 0. From y = 2xλ, we find λ = y/(2x). Substituting this into x = 2λx + 1, we get x = y 2 /x + 1 or x 2 = y 2 + 1. Since x 2 + y 2 = 1, it follows that x 2 = 1 − x 2 + x. This yields that x = −1/2 or 1. If x = 1, then y = 0. √ If x = −1/2, then y = ± 3/2. Since z = 2 − y, we obtain the following values: f (1, 0, 1)√= 1, √ √ f (−1/2, √ 3/2, 2 − √ 3/2) = 2 − √ 3/4, f (−1/2, − 3/2, 2 + 3/2) = 2 + 3/4. Since f is a continuous function conclude that the √ on a closed and bounded set, we √ maximum value of f is 2 + 3/4 and the minimum value is 2 − 3/4. 298 Find the extreme values of f (x, y, z) = x(y + z) on the curve of intersection of the right circular cylinder x 2 + y 2 = 1 and the hyperbolic cylinder x z = 1. Solution. In this problem, we have two constraints g(x, y, z) = x 2 + y 2 − 1 = 0 and h(x, y, z) = x z − 1. We calculate ∇ f (x, y, z) = (y + z, x, x), ∇g(x, y, z) = (2x, 2y, 0) and ∇h(x, y, z) = (z, 0, x) so that ∇ f = λ∇g + μ∇ f . This leads to the following system of equations: (1) y + z = 2λx + μz, (2) x = 2λy,

250

6 Extreme of Functions

(3) x = μx, (4) x 2 + y 2 = 1, (5) x z = 1. From (3), we conclude that x = 0 or μ = 0. But the case x = 0 is impossible, since it contradicts with (5). So, we must have μ = 1. This implies that y + z = 2λx + z or y = 2λx. So, by (2), we get y = 2λ(2λy) = 4λ2 y. This yields that y = 0 or λ = ±1/2. If y = 0, then by (4) we get x 2 = 1 or x = ±1. Then, by (5), it follows that z = ±1. Thus, in this case, we obtain two points (1, 0, 1) and (−1, 0, −1). √ 2 x =√ ±1/ 2. If λ = −1/2, then y = −x. √ 1/2 or √ √ Hence, by (4) we obtain x = Now,√(5) gives √ that √ z = ± 2, and we get the points (1/ 2, −1/ 2, 2) and (−1/ 2, 1/ 2, − 2). √ 2 = 1/2 If λ√= 1/2, then y = x. It follows that √ x √ √ or x = ±1/√ 2. By (5), √ we√get z = ± 2 and we obtain two points (1/ 2, 1/ 2, 2) and (−1/ 2, −1/ 2, − 2). Finally, we evaluate f (x, y, z) at the above six points. We obtain f (1, √ 0, 1) = f√(−1,√0, −1) = 1, √ √ √ f (1/√2, −1/ √ 2, √ 2) = f ((−1/ √ 2, 1/√ 2, −√ 2) = 1/2, f (1/ 2, 1/ 2, 2) = f (−1/ 2, −1/ 2, − 2) = 3/2. Therefore, the absolute minimum is 1/2 and the absolute maximum is 3/2. 299. Find the plane x/a + y/b + z/c = 1 that passes through the point (2, 1, 2) and cuts off the least volume from the first octant. Solution. The volume of the pyramid in the first octant by the plane (see Fig. 6.18) is V = f (a, b, c) =

11 1 1 (base area) · (height) = ab c = abc. 3 3 2 6

Since the point (2, 1, 2) lies on the plane, it follows that 2/a + 1/b + 2/c = 1. Hence, we want to minimize f subject to the constraint g(a, b, c) = 2bc + ac + 2ab − abc. We have 1 (bc, ac, ab), 6 ∇g(a, b, c) = (c + 2b − bc, 2c + 2a − ac, 2b + a − ab). ∇ f (a, b, c) =

To do so, we find the values of a, b, c and λ for which ∇ f = λ∇g and g(a, b, c) = 0. Hence, we can write 1 (bc, ac, ab) = λ(c + 2b − bc, 2c + 2a − ac, 2b + a − ab), 6 and so bc/6 = λ(c + 2b − bc), ac/6 = λ(2c + 2a − ac) and ab/6 = λ(2b + a − ab). It follows that

6.3 Solved Problems

251

Fig. 6.18 The pyramid in the first octant formed by the plane

z

C = (0, 0, c)

y B = (0, b, 0)

O A = (a, 0, 0)

x

abc = λ(ac + 2ab − abc) = λ(2bc + 2ab − abc) = λ(2bc + ac − abc). 6 Hence, we obtain λac = 2λbc and 2λab = 2λbc. Since abc = 0, it follows that a = 2b = c. Substituting into the constraint equation gives 2/a + 2/a + 2/a = 1, and then we get a = 6, b = 3 and c = 6. Therefore, the desired plane is x + 2y + z = 6. 300. Find the rectangular box with the greatest volume that fits inside the ellipsoid x 2 /a 2 + y 2 /b2 + z 2 /c2 = 1, given that its sides are parallel to the axes. It is clear that the box has the largest volume if each of its corners touches the ellipse. Suppose that one of the corners of the box is (x, y, z) in the positive octant. Then, the corners of box are (±x, ±y, ±z) and its volume is V = f (x, y, z) = 8x yz. We want to maximize V subject to the constraint g(x, y, z) = x 2 /a 2 + y 2 /b2 + z 2 /c2 − 1 = 0. Since the constraint is bounded, it follows that a maximum/minimum does exist. Calculating the gradients of f and g, we get ∇ f (x, y, z) = (8yz, 8x z, 8x y) and ∇g(x, y, z) =

2x 2y 2z . , , a 2 b2 c2

Obviously, ∇g = 0. Thus, the Lagrange multiplier rule ∇ f = λ∇g gives that 8yz = 2λ

y z x , 8x z = 2λ 2 and 8x y = 2λ 2 . a2 b c

Suppose that V = 0. Then, we can write

252

6 Extreme of Functions

λ = −4a 2

yz xz xy = −4b2 = −4c2 , x y z

which implies that y 2 a 2 = x 2 b2 and z 2 b2 = y 2 c2 . Hence, we get x 2 /a 2 = y 2 /b2 = z 2 /c2 . This leads to x2 y2 z2 x2 1 = 2 + 2 + 2 = 3 2, a b c a √ √ √ √ √ or √ x = a/ 3, which yields that y = b 3 and z = c/ 3. Therefore, (a/ 3, b/ 3, c/ 3) is the required maximum point and the maximum value is a b c 8abc 8√ √ √ = √ . 3 3 3 3 3 301. Let z = f (x, y) = Ax α b1−α , where A > 0 and 0 < α < 1. This is a CobbDouglas production function. The price of capital is q, the price of labor is p, and we have a budgetary constraint g(x, y) = q x + py = B, where B is a positive constant. Maximize the value of the product subject to the budgetary constraint. Solution. We have ∇ f (x, y) = α Ax α−1 y 1−α , (1 − α)x α y −α and ∇g(x, y) = (q, p). We apply the Lagrange multiplier rule ∇ f = λ∇g. Then, we have the following system of equations: (1) α Ax α−1 y 1−α = λq, (2) (1 − α)x α y −α = λ p, (3) q x + py = B. Eliminating λ from (1) and (2), we get (1 − α)Ax α y −α α Ax α−1 y 1−α = . q p This yields that pαy = (1 − α)xq. Substituting this in the constraint (3), we obtain qx +

1−α α q x = B or x = B, α q

which implies that y=

1−α B. p

Consequently, the candidate for (x, y) in order to maximize the value of output is

6.3 Solved Problems

253

α 1−α B, B . q p

The corresponding value of f is f

α 1−α B, B q p

=A

α B q

α

1−α B p

1−α =

ABαα (1 − α)1−α . q α p α−1

302. Find 10 positive real numbers whose sum is 1000 and whose product is maximum. Solution. We want to determine the maximum of the function f (x1 , . . . , x10 ) = x1 . . . x10 subject to the constraint g(x1 , . . . , x10 ) = x1 + · · · + x10 − 1000 = 0. Computing the gradients of f and g, we obtain ∇ f (x1 , . . . , x10 ) =

f (x , . . . , x ) f (x1 , . . . , x10 ) 1 10 , ,..., x1 x10

∇g(x1 , . . . , x10 ) = (1, . . . , 1). Hence, the Lagrange multiplier rule ∇ f = λ∇g gives f (x1 , . . . , x10 ) f (x1 , . . . , x10 ) = ... = = λ. x1 x10 This implies that x1 = . . . = x10 . Combining with the constraint g(x1 , . . . , x10 ) = 0 we get x1 + · · · + x1 − 1000 = 0, which implies that x1 = . . . = x10 = 100. Finally, the product is 10010 = 1020 . This turns out to be the maximum. Note that as any of the variables approach 0, the product approach 0, without reaching it. Hence, in the domain x1 , . . . x10 > 0, the minimum does not exist. 303. If the sum of five values (not necessarily positive) is 1, and the sum of square is 13, what is the smallest possible value of the sum of the cubes? Solution. Suppose that f (x1 , . . . , x5 ) =

5 i=1

xi3 , g(x1 , . . . , x5 ) =

5

xi2 − 13 and h(x1 , . . . , x5 ) =

i=1

5

xi .

i=1

We need to determine the minimum of f subject to constraints g = 0 and h = 0. We have

∇ f (x1 , . . . , x5 ) = 3x12 , . . . , 3x52 , ∇g(x1 , . . . , x5 ) = (2x1 , . . . , 2x5 ), ∇h(x1 , . . . , x5 ) = (1, . . . , 1). So, using the Lagrange multiplier rule ∇ f = λ∇g + μ∇h, we obtain

254

6 Extreme of Functions

3xi2 = 2λxi + μ, i = 1, . . . , 5.

(6.28)

We observe that x1 , . . . , x5 satisfy the same quadratic equation. Equation (6.28) has at most two real roots, say a and b. Consequently, x1 = a or xi = b, for all i = 1, . . . , 5. Now, we consider the following cases. Case 1: Each xi equals to a. Case 2: Four of the xi ’s are a and remaining one is b. Case 3: Three of these values are a and two ones are b. In the case (1), using the constraint g = 0 and h = 0, we get 5a 2 = 13 and 5a = 1, but this is a contradiction. In the case (2), using the constraints g = 0 and h = 0, we obtain 4a 2 + b2 = 13 and 4a + b = 1. This gives us the following solutions: (a, b) = (1, −3) or

3 17 . − , 5 5

In the case (3), again using g = 0 and h = 0, we have 3a 2 + 2b2 = 13 and 3a + 2b = 1, which implies that (a, b) =

√ √ √ √ 3+8 6 1−4 6 3−8 6 1+4 6 , , or . 15 5 15 5

Substituting the above solutions into f (x1 , . . . , x5 ) and comparing the values, we observe that f has a minimum at the point (1, −3). Therefore, we have x1 = x2 = x3 = x4 = 1 and x5 = −3, and the minimum value is −23. Note that four other points of minimum are obtained by rearrangement of the variables xi . 304. Let n ≥ 2. (1) Suppose that f : Rn → R be given by f (x1 , x2 , . . . , xn ) = x12 x22 . . . xn2 and c ∈ R. Find the maximum value of f subject to the constraint g(x1 , x2 , . . . , xn ) = x12 + x22 + · · · + xn2 − c2 = 0; (2) Using part (1), prove that the inequality √ a1 + a2 + · · · + an n , a1 a2 . . . an ≤ n for any positive real numbers a1 , a2 , . . . , an . It is clear that f is a continuous function and the set D = {(x1 , x2 , . . . , xn ) ∈ Rn | g(x1 , x2 , . . . , xn ) = 0} is bounded in Rn . So, f has its maximum and minimum on D. The equation ∇ f = λ∇g implies that

6.3 Solved Problems

255

2 2x1 x22 x32 . . . xn2 = 2λx1 , 2x12 x2 x32 . . . xn2 = 2λx2 , . . . , 2x12 x22 . . . xn−1 xn = 2λxn .

For λ = 0, we obtain 2λx12 = 2λx22 = . . . = 2λxn2 . Since g = 0, it follows that g(x1 , x2 , . . . , xn ) = 0 = x12 + x22 + · · · + xn2 − c2 = xi2 + xi2 + · · · + xi2 − c2 . √ This gives that xi = c/ n, for i = 1, . . . , n. (2) In part (1), take x12 = a1 , x22 = a2 , . . . , xn2 = an and x12 + x22 + · · · + xn2 = cn , for some c. Then, we obtain (a1 a2 . . . an )1/n ≤

305. n

x 2 + x22 + · · · + xn2 a1 + a2 + · · · + an c2 = 1 = . n n n

Find the maximum of

n

xi yi subject to the constraints

i=1

n

xi2 = 1 and

i=1

yi2 = 1.

i=1

Solution. Assume that all numbers are non-negative. We need to maximize f (x1 , . . . , xn , y1 , . . . yn ) = x1 y1 + · · · + xn yn subject to the constraints g(x1 , . . . , xn , y1 , . . . yn ) = x12 + · · · + xn2 − 1 = 0, h(x1 , . . . , xn , y1 , . . . yn ) = y12 + · · · + yn2 − 1 = 0. Now, we use the Lagrange multiplier rule to maximize f . First, we calculate the gradients of f , g and h. We have ∇ f (x1 , . . . , xn , y1 , . . . yn ) = (y1 , . . . , yn , x1 , . . . , xn ), ∇g(x1 , . . . , xn , y1 , . . . yn ) = (2x1 , . . . , 2xn , 0, . . . , 0), ∇h(x1 , . . . , xn , y1 , . . . yn ) = (0, . . . , 0, 2y1 , . . . , 2yn ). If ∇ f = λ∇g + μ∇h, then (y1 , . . . , yn , x1 , . . . , xn ) = λ(2x1 , . . . , 2xn , 0, . . . , 0) + μ(0, . . . , 0, 2y1 , . . . , 2yn ). It is clear that λ = 0 and μ = 0. Hence, we can write 1 = y12 + · · · + yn2 = 4λ2 x12 + · · · + 4λ2 xi2 = 4λ2 (x12 + · · · + xn2 ) = 4λ2 .

256

6 Extreme of Functions

This gives that λ = ±1/2. Since yi = 2λxi and xi ≥ 0 and yi ≥ 0, it follows that λ = 1/2. Similarly, we obtain μ = 1/2. So, we obtain xi = yi , for i = 1, . . . , n, which implies that x1 y1 + · · · + xn yn = x12 + · · · + xn2 = 1. Therefore, the maximum value is 1. 306. Prove that Cauchy’s inequality n

ai bi ≤

i=1

We set A =

n

ai2

1/2

n

ai2

n 1/2

i=1

and B =

i=1

n

bi2

1/2

.

i=1

bi2

1/2

. If A = 0 or B = 0, the desired result

i=1

is obvious. Suppose that xi = ai /A and yi = bi /B. Thus, we obtain n i=1 n i=1

yi2 = 1. Now, by Problem 6.3, we conclude that

n

n

xi = 1 and

i=1

xi yi ≤ 1. Thus, we have

i=1

ai bi ≤ 1, or equivalently ai bi ≤ AB. A B i=1 n

307. A spring of natural length L extended to length L + x contains energy (1/2)kx 2 , where k is a constant called the stiffness of the spring. Suppose that n springs of natural lengths L 1 , . . . , L n and stiffnesses k1 , . . . , kn are stringed together and the resulting contraption is extended to length L 1 + · · · + L n + . Find the extensions of the individual strings, assuming that the system minimizes the total energy. Justify why the solution you found is an absolute minimum. Solution. Suppose the total extension is produced by extensions x1 , . . . , xn in the individual springs. Then x1 + · · · + xn = and the total energy in the system is 1 (k1 x12 + · · · + kn xn2 ). 2 The Lagrange multiplier condition gives ki xi = λ. So, we have xi = λ/ki . Since x1 + · · · + xn = , it follows that λ=

n i=1

1 ki

,

6.3 Solved Problems

257

and so x1 =

, . . . , xn = . n n 1 1 k1 kn k k i=1 i i=1 i

To show that this is indeed a minimum, consider the Lagrangian L=

n n 1 2 ki x i − λ xi − . 2 i=1 i=1

Its Hessian with respect to x1 , . . . , xn is the matrix with (k1 , . . . , kn ) on the diagonal, which is positive definite. Therefore, the Lagrangian is a convex function of x1 , . . . , xn . Therefore, the point we found is an absolute minimum. Note that the extension is inversely proportional to the stiffness, which makes sense. 308. Let a straight line and three points be given on the plane. Find (or characterize) the point on the line for which the sum of distances from this point to the three given points is minimal. Solution. We denote the given points by P1 , P2 and P3 , and the straight line by . We want to minimize f (X ) = X − P1 + X − P2 + X − P3 , where X ∈ . The gradient of X − P1 at X is a unit vector U1 with the same direction as the vector X − P1 . Similarly, we can define the unit vectors U2 and U3 . We write the condition X ∈ as a constraint g(X ) = (X − P0 ) · N = c, where P0 is a point on , N is a vector orthogonal to and c is a constant. Now, by the Lagrange multiplier rule ∇ f = λ∇g, we get U1 + U2 + U3 = λN . This yields that the sum U1 + U2 + U3 is orthogonal to . In other words, we can say that the sum of projections of vectors U1 , U2 and U3 onto is zero. This property characterizes the desirable point X .

258

6 Extreme of Functions

6.4 Exercises Easier Exercises 1. Does the function f (x, y) have a local maximum, local minimum or saddle point at the origin? (Use the first and second derivative tests; if all else fails, sketch a few level curves near the origin.) (a) f (x, y) = x 2 + 3y 2 ; (b) f (x, y) = −x 2 + 2x y − y 2 ;

(c) f (x, y) = x 2 y 2 − 3x 4 y 4 ; (d) f (x, y) = x 6 + y 6 − 6x 2 y 2 .

2. For the given functions, determine the local extreme of f , if there are any. (a) f (x, y) = 4x y 3 − 3x 2 y 2 + 5y; (b) f (x, y) = sin x + sin y, 0 ≤ x, y ≤ π;

(c) f (x, y) = e y sin x; 2x + 2y + 1 . (d) f (x, y) = 2 x + y2 + 1

3. Find the absolute extremum of the given function f on the specified region. (a) f (x, y) = x 2 − y 2 , R = {(x, y) | x 2 + y 2 ≤ 4}; (b) f (x, y) = 2x y + x 2 + 8y − 4x, R = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 1}; 2 2 (c) f (x, y) = (2x 2 + 3y 2 )e−(x +y ) , R = {(x, y) | 1 ≤ x 2 + y 2 ≤ 4}. 4. Find the absolute extremum of the function f (x, y, z) = (ax + by + cz)e−(x

2

+y 2 +z 2 )

.

5. Consider the function f (x, y) = x 2 + y 2 + 2x y − x − y + 1 over the square [0, 1] × [0, 1]. (a) Show that f has an absolute minimum along the line segment 2x + 2y = 1 in this square. What is the absolute minimum value? (b) Find the absolute maximum value of f over the square. 6. Determine the minimum of the function f (x, y) = x 2 + (y + 1)2 subject to the constraint g(x, y) = x 2 − y 3 = 0. 7. Find the points on the curve of intersection of the ellipsoid x 2 + 4y 2 + 4z 2 = 4 and the plane x − 4y − z = 0 that are closest to the origin, and find the minimum distance. 8. Find the points on the curve of intersection of the two surfaces x 2 − x y + y 2 − z 2 = 1 and x 2 + y 2 = 1 which are nearest to the origin. 9. Find the maximum value of y on the ellipse 6x 2 + 3x y + 2y 2 = 1. 10. What are the largest and the smallest possible values of the sum of squares of n numbers, if the sum of the fourth powers equals 1? 11. Find (or characterize) the point on a plane, for which the sum of distances from this point to k given points in this plane is minimal.

6.4 Exercises

259

12. Find (or characterize) the point on a plane, for which the sum of distances from this point to three given points in three-dimensional space is minimal. 13. Find three numbers whose sum is 100 and the sum of whose squares is√least. 14. Find the maximum and minimum points of the function f (x, y) = x + 3y + 2 on the unit disk R = {(x, y) | x 2 + y 2 ≤ 1}. 15. The temperature is T degrees at any point (x, y) of the curve 4x 2 + 12y 2 = 1, and T = 4x 2 + 24y 2 − 2x. Find the points on the curve where the temperature is the greatest and where it is the least. Also, find the temperature at these points. 16. If f (x, y, z) = x yz, find the maximum and minimum values of f subject to the constraints y = z and x 2 + y 2 = 1. 17. Determine constants a and b such that the integral

1

2 ax + b − f (x) d x

0

will be as small as possible if (a) f (x) = x 2 ; (b) f (x) = (x 2 + 1)−1 . 18. Using Lagrange multipliers, tell which point P in the first octant and on the √ surface x 3 y 2 z = 6 3 is closest to the origin. 19. Let f (x, y) = x 3/2 + 2y 3/2 . Maximize ∇ f on x 2 + (y − 1)2 ≤ 1. 20. Let a, b, c and d be positive real numbers. Find the distance from the origin to the plane ax + by + cz = d using (a) the distance formula; (b) the method of Lagrange multipliers. Harder Exercises 21. Suppose that f (x) has a local maximum at x = a and a local minimum at x = b, and g(y) has a local maximum at y = c and a local minimum at y = d. What can you say about h(x, y) = f (x) + g(y) at (a, c), (a, d), (b, c) and (b, d)? 22. A piece of wire L ft long is cut into three pieces. One piece is bent into the shape of a circle, a second piece is bent into the shape of a square, and the third piece is bent into the shape of an equilateral triangle. How should the wire be cut so that the combined area of the three figures is as small as possible, and the combined area of the three figures is as large as possible? 23. A rectangular produce box is to be made of cardboard; the sides of single thickness, the front and back of double thickness and the bottom of triple thickness, with the top left open. Its volume is to be 1 cubic foot; what proportions for the sides will use the least cardboard? 24. A manufacturer produces two lines of a product, at a cost of $2 per unit for the regular model and $3 per unit for the special model. If he fixes the price at x dollars and y dollars, respectively, the demand for the regular model is y − x and the demand for the special model is 14 + x − 2y in thousand units per week. What prices maximize his profits? 25. Find the greatest and least distances from the origin to the curve of intersection of the surfaces x 2 = 2yz and x 2 + 3y 2 + 2z 2 = 30.

260

6 Extreme of Functions

26. Prove that f (x, y) = x 2 − 6x y + 10y 2 has a positive minimum value p on the circle x 2 + y 2 = l. 27. If f (x, y) = ax 2 + bx y + cy 2 has a positive minimum p on x 2 + y 2 = 1, prove that f (x, y) > 0 for all (x, y) except (0, 0). Hint: Find the minimum of f (x, y) on x 2 + y 2 = r 2 . 28. Suppose that f (x, y) = 3x 4 − 4x 2 y + y 2 . Show that on every line y = mx the function has a minimum at (0, 0), but that there is no local minimum in any two-dimensional neighborhood of the origin. Make a sketch indicating the set of points (x, y) at which f (x, y) > 0 and the set at which f (x, y) < 0. 29. Suppose that f (x, y) = ax 2 + 2bx y + cy 2 + 2d x + 2 py + q, where a > 0 and b2 < ac. (a) Prove that a point (x0 , y0 ) exists at which f has a minimum; (b) Prove that f (x0 , y0 ) = d x0 + py0 + q at this minimum; (c) Show that a b d 1 b c p. f (x0 , y0 ) = 2 ac − b d p q 30. If a, b and c are positive numbers, find the maximum value of f (x, y, z) = x a y b z c subject to the side condition x + y + z = 1. 31. Find the minimum volume bounded by the planes x = 0, y = 0, z = 0, and a plane which is tangent to the ellipsoid y2 z2 x2 + + =1 a2 b2 c2 at a point in the octant x > 0, y > 0, z > 0. 32. Find the extremal point of x 2 + 2x y + 4y 2 + 6 and show it is a minimum point by completing the square. 33. Find the maximum and minimum values of the function f (x, y, z) = x 2 + y 2 + z 2 − 4(x + y + z) on R = {(x, y, z) | x 2 + y 2 + z 2 ≤ 16, z ≥ 0}. 34. Suppose that A does not lie on a closed surface F(X ) = 0 and B is a point of the surface that maximizes or minimizes the distance F(x) − A. Show that A − B is normal to the surface at B. Hint: Parameterize. 35. Find the rectangle of the largest area (with sides parallel to the coordinates axes) that can be inscribed in the ellipse x 2 + 2y 2 = 1. 36. Let 0 < p < q and b > 0. Find the maximum and minimum of x p + y q on x q + y q = bq for x ≥ 0 and y ≥ 0. 37. Let 0 < p < q and x ≥ 0 and y ≥ 0. Prove that 1 21/ p−1/q

x q + yq 2

1/q

≤

xp + yp 2

1/ p

≤

x q + yq 2

1/q .

6.4 Exercises

261

Hint: Set x q + y q = bq . 38. Let 0 < p < q. Find the maximum and minimum of x p + y p + z p on the surface x q + y q + z q = bq where x ≥ 0, y ≥ 0 and z ≥ 0. 39. Let 0 < p < q and x ≥ 0, y ≥ 0, z ≥ 0. Show that

x p + yp + zp 3

1/ p

≤

x q + yq + zq 3

1/q .

40. Suppose that x, y and z be the angles of a triangle. Find the maximum of sin x sin y sin z x yz and where this maximum is taken.

Chapter 7

Vector Fields

7.1 Vector Fields, Limits and Continuity Consider the flow vector that determines the direction of water flow at a certain point in a river (see Fig. 7.1). We can assign such a vector to any point in the river at any moment of time. Thus we arrive at a vector field that can be graphically represented by field lines that are tangential to the direction of the vector at each point. The difference between a vector and a vector field is that the former is one single vector while the latter is a distribution of vectors in space and time. The vector field exists in all points of space and at any moment of time. A function F : Rn → Rm is called a vector field. When m = 1, the function is called a scalar field. If F : Rn → Rm is a vector field, then for any X ∈ Rn , F(X ) = f 1 (X ), f 2 (X ), . . . , f m (X ) , for some functions f k : Rn → R, k = 1, . . . , n, the component, orcoordinate, functions of F. The contrast to the vector field F, we call the functions f 1 , f 2 , . . . , f m scalar fields. Let ek denote the kth unit coordinate vector, i.e., all the components of ek are 0 except the kth, which is equal to 1. Hence, f k (X ) is given by the inner product f k (X ) = F(X ) · ek . Moreover, we can write F(X ) =

m

f k (X )ek .

k=1

We can plot a vector field by choosing various sample points (x, y), evaluation the vector function F(x, y) and drawing the vector from that point. With enough vectors plotted, we start to get a sense of the vector fields. Figure 7.2 shows some examples of vector fields F : R2 → R2 . © The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_7

263

264

7 Vector Fields

Fig. 7.1 Flow vectors in a river as a natural example of a vector field

Fig. 7.2 Vector fields from left to right, top to bottom: F(x, y) = (x 2 − y 2 − 4, x y), F(x, y) = (x y + 4, y 2 + x), F(x, y) = (ln(x y) + 2, x 2 + y 2 ) and F(x, y) = (x 2 + y 2 , x 3 y)

7.2 Linear Transformations

265

We consider a function F : Rn → Rm . If A ∈ Rn and L ∈ Rm we write lim f (X ) = L ,

X →A

to mean that lim

X −A→0

f (X ) − L = 0.

(7.1)

The limit symbol in (7.1) is the usual limit of ordinary calculus. In this definition, it is not required that F be defined at the point A itself. We say a vector field F : Rn → Rm is continuous at a point A if F is defined at A and if lim f (X ) = F(A).

X →A

If a vector field F : Rn → Rm has component functions f 1 , f 2 , . . . , f n , then F is continuous at A if and only if f k is continuous at A, for k = 1, . . . , n. We say F is continuous on a set S if F is continuous at each point of S.

7.2 Linear Transformations In this section, we define the notion of a linear transformation between Rn and Rm and present the most basic properties about them. Let T : Rn → Rm be a vector field. We say that T is a linear transformation if T (cX + Y ) = cT (X ) + T (Y ), for all X and Y in Rn and all real number c. Let B = {e1 , . . . , en } and suppose that X ∈ Rn is an arbitrary n vector. Then, there ci xi . We call ci the is a unique n-tuple (c1 , . . . , cn ) of scalars such that X = i=1 ith coordinate of X relative to B. There is a one-to-one correspondence between the set of all vectors in Rn and the set of all n-tuples in Rn . We may show the coordinate matrix of X by ⎡ ⎤ c1 ⎢ .. ⎥ [X ] B = ⎣ . ⎦ . cn If T : Rn → Rn is a linear transformation, then the following conditions are equivalent: (1) T is invertible; (2) T is one to one; (3) T is onto;

266

7 Vector Fields

Representation of linear transformations by matrices: Let a linear transformation T : Rn → Rm is given by T (x1 , . . . , xn ) = (c11 x1 + c12 x2 + · · · + c1n xn , . . . , cm1 x1 + cm2 x2 + · · · + cmn xn ), for all (x1 , . . . , xn ) ∈ Rn . If we want to specify this linear transformation, all we need to do is specify ci j ’s. Indeed, xi ’s are just placeholders. So, we can correspond the following matrix to T : ⎡

c11 ⎢ c21 ⎢ ⎢ .. ⎣ .

c12 c22 .. .

cm1

cm2

... ... ... ...

c1n c2n .. .

⎤ ⎥ ⎥ ⎥. ⎦

cmn

If T : Rn → Rm and S : Rm → R p are linear transformations, then S ◦ T : Rn → R p is also a linear transformation.

7.3 Total Derivatives We say that a function F : Rn → Rm is differentiable at a point A ∈ Rn if there exists a linear transformation T A : Rn → Rm such that lim

H →0

F(A + H ) − F(A) − T A (H ) = 0. H

(7.2)

If such a linear transformation T A exists, then it is unique and we call it the total derivative of F at A. We denote T A to be D F(A). Hence, under the above definition, the total derivative D F(A) of F at A is not a number but rather a linear transformation. Eq. (7.2) can be written in several slightly different but equivalently ways. For instance, it can be written as lim

H →0

F(A + H ) − F(A) − T A (H ) = 0. H

If we take X := A + H , then we can write (7.2) as lim

H →0

F(X ) − F(A) − T A (X − A) = 0. X − A

Another useful way to restate (7.2) is to say that F(A + H ) = F(A) + T A (H ) + e(H ),

7.3 Total Derivatives

267

where the error term e(H ) satisfies lim

H →0

e(H ) = 0. H

Let F : Rn → Rm be a vector field with component function f 1 , . . . , f m : Rn → R. Then, F is differentiable at A ∈ Rn if and only if each component function f k is differentiable at A. Jacobian matrix: Our definition of the total derivative generalizes the classes we already studied, but an important question remains. How do we compute the total derivative of a vector field F : Rn → Rm ? Suppose that F : Rn → Rm is a vector field. Write F = ( f 1 , . . . , f m ), where f i : Rn → R, for i = 1, . . . , m. If for all i and j, ∂ f i /∂x j is defined and continuous at A, then F is differentiable at A, and the matrix for D F(A) is given by ⎡

∂ f1 ⎢ ∂x1 (A) ⎢ ⎢ ∂ f2 ⎢ ⎢ ∂x (A) 1 [D F(A)] = ⎢ ⎢ .. ⎢ ⎢ . ⎢ ⎣ ∂ fm (A) ∂x1

∂ f1 (A) ∂x2 ∂ f2 (A) ∂x2 .. . ∂ fm (A) ∂x2

... ... ... ...

⎤ ∂ f1 (A) ⎥ ∂xn ⎥ ⎥ ∂ f2 (A) ⎥ ⎥ ∂xn ⎥. ⎥ .. ⎥ ⎥ . ⎥ ⎦ ∂ fm (A) ∂xn

The matrix [D F(A)] is called the total derivative matrix or Jacobian matrix of F at A. The Jacobian matrix [D F(A)] is defined at each point where the mn partial derivatives ∂ f i /∂x j (A) exist. The total derivative T A is also written as f (A). The derivative f (A) is a linear transformation; the Jacobian [D F(A)] is a matrix representation for this transformation. If n = m, then the determinant of the Jacobian matrix is denoted by ∂( f 1 , . . . , f n ) or JF (x1 , . . . , xn ), ∂(x1 , . . . , xn ) and is called the Jacobian of F at (x1 , . . . , xn ). Let G : Rn → Rm and F : Rm → R p , and let H = F ◦ G. If G is differentiable at A and F is differentiable at B = f (A), then the chain rule states that H (A) = F (B) ◦ G (A). We can express the chain rule in terms of the Jacobian matrices D H (A), D F(B) and DG(A) which represents the linear transformations H (A), F (B) and G (A), respectively. Since composition of linear transformations corresponds to multiplication of their matrices, we can write [D H (A)] = [D F(B)][DG(A)],

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7 Vector Fields

where B = G(A). This is called the matrix form of the chain rule. A necessary and sufficient condition that the equations F(u, v, x, y, z) = 0 and G(u, v, x, y, z) = 0 can be solved for u and v (for example) is that ∂(F, G)/∂(u, v) is not identically zero in a region R. Similar results are valid for m equations in n variables, where m < n. If F : Rn → Rm is differentiable at A, then F is continuous at A.

7.4 Conservative Vector Fields Let F : Rn → Rm be a vector field. The vector field F is said to be conservative if it is the gradient of a function. In other words, there is a differentiable function f : Rn → R satisfying F = ∇ f . Such a function f is called a potential function for F. For example, according to Newton’s law of gravitation, the force exerted on a particle of mass m located at the point (x, y, z) by a particle of mass M located at the origin is given by G Mm R, F(x, y, z) = − R3 − → − → − → where R = x i + y j + z k , and G is the universal gravitational constant. It is not difficult to see that F is the gradient of the function f (x, y, z) =

G Mm R. R

When is a vector field F conservative? − → − → Suppose that a vector field F(x, y) = P(x, y) i + Q(x, y) j is defined on an open disk B in R2 , and ∂ P/∂ y and ∂ Q/∂x are continuous on B. Then, the vector field F is a gradient on B if and only if ∂Q ∂P = , ∂y ∂x at all points in B. We can extend the above result to functions of three variables. − → − → − → Suppose that F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k is defined on an open ball B in R3 , and ∂ P/∂ y, ∂ P/∂z, ∂ Q/∂x, ∂ Q/∂z, ∂ R/∂x and ∂ R/∂ y are continuous on B. Then, the vector field F is a gradient on B if and only if ∂Q ∂P ∂R ∂Q ∂R ∂P = , = and = , ∂y ∂x ∂z ∂x ∂z ∂y at all points in B.

7.5 Divergence and Curl

269

7.5 Divergence and Curl Divergence and curl are two measurements of vector fields that are very useful in a variety of applications. The del operator is denoted by ∇ which is called nabla. The del operator is a vector written as ∂ − ∂ − ∂ − → → → i + j + k. ∇= ∂x ∂y ∂z Let F : R3 → R3 be a vector field defined by − → − → − → F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k . Then, its divergence is the function div F : R3 → R which is defined by the rule ∂ − ∂ − ∂ − → → → − → − → − → i + j + k · P i +Q j +Rk ∂x ∂y ∂z ∂P ∂Q ∂R = + + . ∂x ∂y ∂z

div F = ∇ · F =

As a physical example, if F represents velocity field of a gas (or fluid) then div F represents the rate of expansion per unit volume under the flow of the gas (or fluid). The curl of F is the vector field curl F : R3 → R3 which is defined by the rule

− − → − →

→ j k

i

∂

∂ ∂

curl F = ∇ × F =

∂x ∂ y ∂z

P Q R ∂R ∂ Q − ∂ P − ∂ R − → ∂R → ∂Q → = − i + − j + − k. ∂y ∂z ∂x ∂z ∂x ∂y The curl of a vector field is a vector field. The vector field F : R3 → R3 is called rotation free if its curl is zero, and it is call incompressible (also known as a solenoidal vector field) if its divergence is zero. Note that the del operation makes sense for any n, not just 3. So, we can define the divergent in all dimensions. However, curl only makes sense when n = 3. So far in our work with ∇ we have only considered first partial derivatives. When we considered second partial derivatives, the following possible combination turns out to be meaningful and useful. If f : R3 → R is a function of three variables x, y and z, then ∇ · (∇ f ) = div(∇ f ), formally we find that ∇ · (∇ f ) =

∂2 f ∂2 f ∂2 f + + . ∂x 2 ∂ y2 ∂z 2

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7 Vector Fields

The operation ∇ · (∇ f ) is called the Laplacian of f and is written ∇ 2 f . When ∇ 2 f = 0, the function f is called harmonic. The Laplacian of a vector field F = (P, Q, R) is defined component wise, i.e., ∇ 2 F = (∇ 2 P, ∇ 2 Q, ∇ 2 R).

7.6 Solved Problems − → − → 309. A vector field on R2 is defined by F(x, y) = −y i + x j . Describe F by sketching some of the vectors F(x, y). Solution. According to the definition, we know that a vector field is a function that assigns a vector to each point. So, we pick several different points and then we try to determine the vector associated with that point: (x, y)

F(x, y)

(x, y)

(0, −2) (0, −1) (0, 2) (0, 1) (−1, 0) (−2, 0)

− → 2i − → i − → −2 i − → −i − → − j − → −2 j

(1, 0) (2, 0) (1, 1) (−1, 1) (1, −1) (−1, −1)

F(x, y) − → j − → 2 j − → − → −i + j − → − → −i − j − → − → i + j − → − → i − j

Now, we sketch all these vectors, see Fig. 7.3. The starting point of each vector is always the (x, y) value of the point that we pick.

Fig. 7.3 Vector field F(x, y) = (−y, x)

y − → 2j

− → −2 i − → −i

→ − − → −i + j

− → − → −i − j

− → j

− → −j − → − → i − j → − −2 j

x − → − → i + j

− → i − → 2i

7.6 Solved Problems

271

310. Let F : R2 → R2 be defined by F(x, y) =

x −y , x 2 + y2 x 2 + y2

.

Prove that (1)

lim

(x,y)→∞

F(x, y) = (0, 0);

(2)

lim

(x,y)→0

F(x, y) = ∞.

Solution. (1) To prove this limit, we have to show that for every > 0 there exists N > 0 such that (x, y) > N ⇒ F(x, y) − (0, 0) < . We observe that F(x, y)2 =

x 2 x + y2

2

+

−y 2 x + y2

2 =

x2

1 . + y2

Hence, it is enough to choose N = 1/. (2) To prove this limit we have to show that for every M > 0 there is δ > 0 such that 0 < (x, y) − (0, 0) < δ ⇒ F(x, y) > M. Let M > 0 be given. We require δ > 0 such that 0 < x 2 + y 2 < δ2 ⇒

1 > M 2, x 2 + y2

and the choice δ = 1/M.

311. Let F : R2 → R2 be defined by F(x, y) = f 1 (x, y), f 2 (x, y) , where f 1 (x, y) =

1 + x if y ≥ x y if y < x

and f 2 (x, y) =

1 + y if y ≥ x x if y < x.

What can be said about the limit of F as (x, y) → (0, 0) through (1) the set S = {(x, y) ∈ R2 | x < 0 and y > 0}; (2) the set T = {(x, y) ∈ R2 | x > 0 and y < 0}. Solution. (1) Through the set S we have y > x. So, we obtain lim

(x,y)→(0,0)

f 1 (x, y) = lim (1 + x) = 1 and x→0

lim

(x,y)→(0,0)

f 2 (x, y) = lim (1 + y) = 1. y→0

This yields that F(x, y) → (1, 1) as (x, y) → (0, 0) through the set S. (2) Through the set T we have x > y. Hence, we get

272

7 Vector Fields

lim

(x,y)→(0,0)

f 1 (x, y) = lim y = 0 and x→0

lim

(x,y)→(0,0)

f 2 (x, y) = lim x = 0. y→0

This gives that F(x, y) → (0, 0) as (x, y) → (0, 0) through the set T . 312. Let T : R3 → R2 be a function defined by T (x, y, z) = (x + z, y + z), for all (x, y, z) ∈ R3 . (1) Show that T is a linear transformation. (2) Determine the matrix representation of T . Solution. (1) We must check the property that defines a linear transformation. For any (x1 , y1 , z 1 ) and (x2 , y2 , z 2 ) in R3 and c ∈ R, we have T c(x1 , y1 , z 1 ) + (x2 , y2 , z 2 ) = T (cx1 + x2 , cy1 + y2 , cz 1 + z 2 ) = (cx 1 + x2 + cz 1 + z 2 , cy1 + y2 + cz 1 + z 2 ) = c(x1 + z 1 ) + x2 + z 2 , c(y1 + z 1 ) + y2 + z 2 = c(x1 + z 1 , y1 + z 1 ) + (x2 + z 2 , y2 + z2) = cT (x1 , y1 , z 1 ) + T (x2 , y2 , z 2 ). Thus, T is a linear transformation. (2) To find the matrix of T , we see that T (1, 0, 0) = (1, 0), T (0, 1, 0) = (0, 1), T (0, 0, 1) = (1, 1), and now we form the matrix A by concatenating these vectors: A=

1 0 1 , 0 1 1

and this matrix is the matrix representation of T . 313. Let T : Rn → Rm be a linear transformation. If T (X ) = T (Y ) for some X = Y , prove that for any U ∈ Rn there is V = U in Rn such that T (U ) = T (V ). Solution. Suppose that T (X ) = T (Y ) for some X = Y . Put Z = X − Y , and note that since X = Y , it follows that Z is non-zero. Now, we can write T (Z ) = T (X − Y ) = T (X ) − T (Y ) = 0. Consequently, for any U ∈ Rn we have T (U + Z ) = T (U ) + T (Z ) = T (U ) + 0 = T (U ). 314. Let T : Rn → Rm be a linear transformation. Prove that T is one to one if and only if the equation T (X ) = 0 has only the trivial solution. Solution. Suppose that T is one to one. Since T is a linear transformation, it follows that T (0) = 0. Now, let X 0 be a solution of the equation T (X ) = 0. Then, we have T (X 0 ) = 0. Since T (X 0 ) = T (0) and T is one to one, we conclude that X 0 = 0. Hence, the only solution to T (X ) = 0 is the trivial solution. Conversely, let T (X ) = 0 has only trivial solution. To prove T is one to one, suppose that T (X ) = T (Y ). Since T is linear, it follows that T (X − Y ) = 0. Hence, we conclude that X − Y = 0, or equivalently X = Y . So, T is one to one.

7.6 Solved Problems

273

315. Let T : R2 → R2 be a linear transformation with the following matrix representation: a b [T ] = . c d Prove that T 2 − (a + d)T + (ad − bc)I = 0. Solution. It is enough to show that the above equality holds for the matrix representation of T . To verify this we observe that

a b a c d c a 2 + bc = ac +cd 0 0 = . 0 0

b a b 1 0 − (a + d) + (ad − bc) d c2 d 0 1 ab + bd a + ad ab + bd ad − bc 0 − + bc + d 2 ac + cd ad + b2 0 ad − bc

316. Let F : R2 → R2 be given by F(x, y) = (x + y 2 , x 3 + 5y), for all (x, y) ∈ R2 . Show that the derivative of F at A = (1, 1) is the map T A (h 1 , h 2 ) = (h 1 + 2h 2 , 3h 1 + 5h 2 ). Solution. In order to observe this, we must prove that lim

(h 1 ,h 2 )→(0,0)

F(1 + h 1 , 1 + h 2 ) − F(1, 1) − T A (h 1 , h 1 ) = 0. h 21 + h 22

To verify this, we have F(1 + h 1 , 1 + h 2 ) − F(1, 1) − T A (h 1 , h 1 ) = lim (h 1 ,h 2 )→(0,0) h 21 + h 22 (1 + h 1 )+(1 + h 2 )2 , (1 + h 1 )3 + 5(1 + h 2 ) − (2, 6) − (h 1 + 2h 2 , 3h 1 + 5h 2 ) h 21 + h 22 (h 22 , 3h 21 + h 31 ) = lim (h 1 ,h 2 )→(0,0) h 21 + h 22 h2 3h 2 + h 31 2 1 = (0, 0), = lim , lim (h 1 ,h 2 )→(0,0) (h ,h )→(0,0) h 21 + h 22 1 2 h 21 + h 22 lim

(h 1 ,h 2 )→(0,0)

as desired. 317. Let F : Rn → Rm be a vector field and A be a point in Rn . Suppose that T A and L A are both linear transformation such that

274

lim

H →0

7 Vector Fields

F(A + H ) − F(A) − T A (H ) F(A + H ) − F(A) − L A (H ) = 0 and lim = 0. H →0 H H

Prove that T A = L A . Solution. If T A and L A are different linear transformations, then there exists B ∈ Rn such that T A (B) = L A (B). Now, we take derivative along the path H = t B. We obtain F(A + t B) − F(A) − T A (t B) F(A + t B)t−B F(A) − L A (t B) T A (t B) − L A (t B) + = lim t→0 t B t B T A (B) − L A (B) t T A (B) − t L A (B) t = lim . = 0 + lim t→0 t→0 |t|B B |t|

0 = lim

t→0

But the limit on the right side does not exist. Therefore, our assumption that T A (B) = L A (B) must have been false. 318. If u = x + y + z, uv = y + z and uvw = z, show that Solution. We have

∂(x, y, z) = u 2 v. ∂(u, v, w)

x = u − (y + z) = u(1 − v), y = uv − z = uv(1 − w), z = uvw.

Now, we obtain

∂x ∂x ∂x

∂u ∂v ∂w 1 − v −u 0

∂y ∂y ∂y ∂(x, y, z)

= v(1 − w) u(1 − w) −uv =

∂(u, v, w)

∂u ∂v ∂w

vw uw uv ∂z

∂z ∂z

∂u ∂v ∂w = (1 − v) u(1 − w)uv + uvuv + u (1 − w)uv + uv + uvvw = (1 − v)u 2 v + u 2 v 2 = u 2 v. ∂(x, y, z) , where (ρ, ϕ, θ) are the spherical coordi∂(ρ, ϕ, θ) nates and (x, y, z) are the Cartesian coordinates. 319. Compute the Jacobian

Solution. We know that x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ and z = ρ cos ϕ. Therefore, we get

7.6 Solved Problems

275

xρ xϕ xθ

∂(x, y, z) =

yρ yϕ yθ

∂(ρ, ϕ, θ) z zϕ zθ ρ

sin ϕ cos θ ρ cos ϕ cos θ −ρ sin ϕ sin θ

=

sin ϕ sin θ ρ cos ϕ sin θ ρ sin ϕ cos θ

cos ϕ −ρ sin ϕ 0

sin ϕ cos θ ρ cos ϕ cos θ

sin ϕ sin θ = (−ρ sin ϕ sin θ)

− (ρ sin ϕ cos θ) cos ϕ −ρ sin ϕ cos ϕ −ρ sin ϕ = (−ρ sin ϕ sin θ)(−ρ sin θ)(sin2 ϕ + cos2 ϕ) −(ρ sin ϕ sin θ)(−ρ sin θ)(sin2 ϕ + cos2 ϕ) 2 2 = ρ sin ϕ(sin θ + cos2 θ) = ρ2 sin ϕ, where we used the cofactor expansion with respect to the third column. 320. Chain rule for Jacobian: If u and v are functions of x and y, and x and y are functions of r and s, prove that ∂(u, v) ∂(x, y) ∂(u, v) = . ∂(x, y) ∂(r, s) ∂(r, s) Solution. Using the definition of Jacobian, we can write

∂u ∂u

∂(u, v) ∂(x, y)

∂x ∂ y

= ∂v ∂v

∂(x, y) ∂(r, s)

∂x ∂ y

∂u ∂x ∂u ∂ y ∂u ∂x

+

∂x ∂r ∂ y ∂r ∂x ∂s = ∂v ∂x ∂v ∂ y ∂v ∂x

+

∂x ∂r ∂ y ∂r ∂x ∂s

∂x

∂r

∂y

∂r ∂u + ∂y ∂v + ∂y

∂x ∂u

∂s

=

∂x ∂ y ∂v

∂s ∂x ∂ y ∂u

∂s = ∂r ∂ y

∂v

∂s ∂r

∂u

∂x ∂ y

∂ y ∂r ∂r

∂v ∂x ∂ y

∂ y ∂s ∂s ∂u

∂s

= ∂(u, v) . ∂v ∂(r, s)

∂s

Remark 1 In general, ∂(x1 , . . . , xn ) ∂(x1 , . . . , xn ) ∂(u 1 , . . . , u n ) = . ∂(u 1 , . . . , u n ) ∂(v1 , . . . , vn ) ∂(v1 , . . . , vn ) 321. Prove that

∂(u, v) ∂(x, y) = 1. ∂(x, y) ∂(u, v)

Solution. If we replace r and s by u and v in Problem 320, then we get the required result. Remark 2 In general, ∂(x1 , . . . , xn ) ∂(u 1 , . . . , u n ) = 1. ∂(u 1 , . . . , u n ) ∂(x1 , . . . , xn )

276

7 Vector Fields

322. If u = x yz, v = x 2 + y 2 + z 2 and w = x + y + z, find

∂(x, y, z) . ∂(u, v, w)

Solution. We have

∂u ∂u

∂u

∂x ∂y ∂z

yz zx x y

∂v ∂v

∂(x, y, z)

∂v =

= 2x 2y 2z

∂y ∂z ∂(ρ, ϕ, θ) ∂x

∂w ∂w ∂w 1 1i 1

∂x ∂y ∂z = 2(x − y)(x − z)(y − z). Since

∂(x, y, z) ∂(u, v, w) = 1, it follows that ∂(u, v, w) ∂(x, y, z) 1 ∂(x, y, z) = . ∂(u, v, w) 2(x − y)(x − z)(y − z)

323. Let F : R3 → R3 be a vector field. − → − → − → (1) If F(x, y, z) = x i + y j + z k , prove that the Jacobian matrix D F(x, y, z) is the identity matrix of order 3. (2) Find all differentiable vector fields F for which the Jacobian matrix D F(x, y, z) is the identity matrix of order 3. (3) Find all differentiable vector fields F for which the Jacobian matrix is a diagonal matrix of the form diag p(x), q(x), r (x) , where p, q and r are given continuous functions. Solution. (1) By using the definition of Jacobian matrix, it is easy to see that ⎡ ∂x ⎢ ∂x ⎢ ∂y ⎢ ⎢ ⎢ ∂x ⎣ ∂z ∂x

∂x ∂y ∂y ∂y ∂z ∂y

∂x ∂z ∂y ∂z ∂z ∂z

⎤ ⎥ ⎡ 1 ⎥ ⎥ ⎣ 0 = ⎥ ⎥ 0 ⎦

0 1 0

⎤ 0 0⎦. 1

(2) Suppose that the Jacobian matrix of the vector field F = ( f 1 , f 2 , f 3 ) is the identity matrix. So, we have ⎡∂f

1

(x, y, z)

⎢ ∂x ⎢∂f ⎢ 2 (x, y, z) ⎢ ⎢ ∂x ⎣ ∂ f3 (x, y, z) ∂x

∂ f1 (x, y, z) ∂y ∂ f2 (x, y, z) ∂y ∂ f3 (x, y, z) ∂y

⎤ ∂ f1 (x, y, z) ⎥ ⎡ ∂z 1 ⎥ ∂ f2 ⎥ (x, y, z) ⎥ = ⎣ 0 ⎥ ∂z 0 ⎦ ∂ f3 (x, y, z) ∂z

0 1 0

⎤ 0 0⎦. 1

7.6 Solved Problems

277

Since ∂ f 1 /∂ y and ∂ f 1 /∂z are identically zero, it follows that f 1 depends only on x. On the other hand, since ∂ f 1 /∂x = 1, it follows that f 1 (x, y, z) = x + α, where α is a constant. Similarly, we find that f 2 (x, y, z) = y + β and f 3 (x, y, z) = z + γ, where − → − → − → β and γ are constant. Therefore, we conclude that F(x, y, z) = x i + y j + z k , plus any constant vector. (3) Let the Jacobian matrix of the vector field F = ( f 1 , f 2 , f 3 ) is a diagonal matrix of the form diag p(x), q(x), r (x) . Thus, we have ⎡∂f

1

(x, y, z)

⎢ ∂x ⎢∂f ⎢ 2 (x, y, z) ⎢ ⎢ ∂x ⎣ ∂ f3 (x, y, z) ∂x

∂ f1 (x, y, z) ∂y ∂ f2 (x, y, z) ∂y ∂ f3 (x, y, z) ∂y

⎤ ∂ f1 (x, y, z) ⎥ ⎡ ∂z p(x) ⎥ ∂ f2 ⎥ (x, y, z) ⎥ = ⎣ 0 ⎥ ∂z 0 ⎦ ∂ f3 (x, y, z) ∂z

0 g(x) 0

⎤ 0 0 ⎦. r (x)

We deduce that f 1 depends only on x; f 2 depends only on y and f 3 depends only on z. Moreover, we have ∂ f1 ∂ f2 ∂ f3 (x, y, z) = p(x), (x, y, z) = q(y) and (x, y, z) = r (x). ∂x ∂y ∂z − → − → − → Hence, we obtain F(x, y, z) = P(x) i + Q(y) j + R(z) k , where P (x) = p(x), Q (y) = q(y) and R (z) = r (z). 324. Let u = f (x, y) and v = g(x, y), where f and g are continuously differentiable in some region R. Prove that a necessary and sufficient condition that there exists a functional relation between u and v of the form R(u, v) = 0 is the vanishing ∂(u, v) = 0 identically. of the Jacobian, i.e., ∂(x, y) Solution. Suppose that there exists a functional relation R(u, v) = 0 between u and v. Then, we have ∂R ∂R du + dv ∂u ∂v ∂ R ∂u ∂u ∂ R ∂v ∂v = dx + dy + dx + dy ∂x ∂y ∂v ∂x ∂y ∂u ∂ R ∂u ∂ R ∂v ∂ R ∂u ∂ R ∂v = + dx + + dy = 0. ∂u ∂x ∂v ∂x ∂u ∂ y ∂v ∂ y

dR =

It follows that

∂ R ∂u ∂ R ∂v = 0, + ∂x ∂v ∂x ∂∂u R ∂u ∂ R ∂v + = 0. ∂u ∂ y ∂v ∂ y

Since ∂ R/∂u and ∂r/∂v are not identically zero, it follows that

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7 Vector Fields

∂u

∂x

∂u

∂y

∂v ∂x ∂v ∂y

= 0 identically.

∂(u, v) = 0 identically. If both ∂u/∂x and ∂u/∂ y are ∂(x, y) zero, then the Jacobian is zero and u = c is a constant. This is a trivial functional relation. Now, suppose that at least one of the ∂u/∂x and ∂u/∂ y is non-zero, say ∂u/∂x = 0. Then, we solve the equation u = f (x, y) to get x = F(u, y). This yields that u = f F(u, y), y and v = g F(u, y), y . Conversely, assume that

So, we obtain ∂u ∂u dx + dy = ∂x ∂y ∂u ∂ F ∂u = du + ∂x ∂u ∂x

∂u ∂ F ∂ F ∂u du + dy + dy ∂x ∂u ∂ y ∂y ∂F ∂u + dy, ∂y ∂y

(7.3)

∂v ∂v dx + dy = ∂x ∂y ∂v ∂ F ∂v = du + ∂x ∂u ∂x

∂v ∂ F ∂ F ∂v du + dy + dy ∂x ∂u ∂ y ∂y ∂F ∂v + dy. ∂y ∂y

(7.4)

du =

dv =

From (7.3), we get ∂u ∂ F ∂u ∂ F ∂u = 1 and + = 0. ∂x ∂u ∂x ∂ y ∂y So, we conclude that

∂u − ∂F ∂y = . ∂u ∂y ∂x

If we put (7.5) into (7.4), then we obtain ⎛ dv =

⎜ ∂v ∂v ∂ F du + ⎜ ⎝ ∂x ∂x ∂u ⎛

∂u ⎜ ∂x ∂v ∂ F = du + ⎜ ⎝ ∂x ∂u

⎞ ∂u ∂v ⎟ ∂y ⎟ dy + ∂u ∂y ⎠ ∂x ⎞ ∂v ∂u ∂v − ∂y ∂ y ∂x ⎟ ⎟ dy. ⎠ ∂u ∂x

−

(7.5)

7.6 Solved Problems

279

∂u ∂v ∂v ∂ F ∂u ∂v − = 0, it follows that dv= du. Since v=g F(u, y), y , ∂x ∂ y ∂ y ∂x ∂x ∂u it follows that ∂v/∂ y = 0. This means that v does not depend on y and only it depends on u, i.e., v is a function of u, in other words the functional relation R(u, v) = 0 exists.

Since

325. Show that the functions u = x + y + z, v = x 3 + y 3 + z 3 − 3x yz and w = x 2 + y 2 + z 2 − x y − yz − zx are functionally dependent and find the relation between them. Solution. First, we compute the Jacobian. We get

∂u

∂x ∂(u, v, w)

∂v =

∂x ∂(x, y, z)

∂w

∂x

∂u ∂y ∂v ∂y ∂w ∂y

∂u ∂z ∂v ∂z ∂w ∂z

1 1 1

2 2 2

= 3(x − yz) 3(y − x z) 3(z − x y)

= 0.

2x − y − z 2y − z − x 2z − x − y

∂(u, v, w) = 0, it follows that the functions u, v and w are functionally depen∂(x, y, z) dent. Now, we can write Since

v = x 3 + y 3 + z 3 − 3x yz = (x + y + z)(x 2 + y 2 + z 2 − x y − yz − x z) = uw. Consequently, the relation between u, v and w is v = uw. 326. Show that the system of equations 3x + y − z + u 2 = 0,

(7.6)

x − y + 2z + u = 0,

(7.7)

2x + 2y − 3z + 2u = 0,

(7.8)

can be solved for x, y, u in terms of z; for x, z, u in terms of y; for y, z, u in terms of x; but not for x, y, z in terms of u. Solution. Adding (7.7) and (7.8) and subtracting (7.6) we obtain 3u − u 2 = 0. This implies that u = 0 or u = 3, and so the system of equations cannot be solved for x, y, z in terms of u. If u = 0 or u = 3, then we can solve (7.7) and (7.8) for any two of variables x, y, z in terms of the third. Hence, we find that

7z 9+z 3 + 7z z , y= , u=3 . x = − , y = , u = 0 or x = − 4 4 4 4

Also, we can have

280

7 Vector Fields

4y 60 + 4y 4y − 3 y , u = 0 or x = − , z= , u=3 . x =− , z= 7 7 7 7

Finally, we can also have

y = −7x, z = −4x, u = 0

or

y=

7x − 60 , z = 9 − 4x, u = 3 . 4

Now, if we consider F(x, y, z, u) = (3x + y − z + u 2 , x − y + 2z + u, 2x + 2y − 3z + 2u), then the Jacobian matrix of F is ⎡

3 1 [D F(x, y, z, u)] = ⎣ 1 −1 2 2

⎤ −1 2u 2 1 ⎦, −3 2

and any 3 × 3 submatrix containing the last column is invertible when u = 0 or u = 3. But it is straightforward to see that the first three columns of this matrix are not an invertible matrix. 2 3 327. Fix two real numbers a and b with 0 < a < b. Define F : R → R by F(x, y) = f 1 (x, y), f 2 (x, y), f 3 (x, y) , where

f 1 (x, y) = (b + a cos x) cos y, f 2 (x, y) = (b + a cos x) sin y, f 3 (x, y) = a sin x. (1) Describe the range R F of F. (2) Show that there exist exactly four points P in the range of F such that ∇ f 1 F −1 (P) = 0.

(7.9)

Find these points. (3) Show that one of the points found in part (2) corresponds to a local maximum of f 1 , one corresponds to a local minimum and the other are saddle points. Solution. (1) The range R F of F is a torus obtaining by moving a circle of radius a with center on a circle of radius b, always keeping the planes of two circles perpendicular and each plane passing through the center of the other circle, see Fig. 7.4. (2) If ∇ f 1 (x, y) = 0, then ∂ f1 ∂ f1 (x, y) = −a sin x cos y = 0 and (x, y) = −(b + a cos x) sin y. (7.10) ∂x ∂x

7.6 Solved Problems

281

Fig. 7.4 Torus

The second equation in (7.10) required x = kπ, and since these functions are periodic with period 2π in both x and y, we assume that x = 0 or x = π. In this case, from the first equation in (7.10), we get y = 0 or y = π. Consequently, the only points satisfying (7.9) are the image of points (0, 0), (0, π), (π, 0) and (π, π). The result is the points (a + b, 0, 0), (−a + b, 0, 0), (a − b, 0, 0) and (−a − b, 0, 0). (3) The point (a + b, 0, 0) is the maximum possible value of f 1 (x, y), and occurs when cos x = 1 and cos y = 1. The point (−a − b, 0, 0) is the minimum possible value of f 1 (x, y), and occurs when cos x = 1 and cos y = −1. The other two points which occur when (x, y) = (0, π) or (x, y) = (π, 0) lie near points of both larger and smaller values of f 1 (x, y). Indeed, if we consider two real functions g and h defined by g(x) = f 1 (x, π) = −(b + a cos x) and h(y) = f 1 (0, y) = b cos y, then x = 0 gives a maximum value of g and y = π gives a minimum value of h. Therefore, the point (0, π) is neither a maximum point nor a minimum point of f 1 (x, y). 328. Consider the function F defined in Problem 327 (1) Determine the set of all Q in the range of F such that ∇ f 3 F −1 (Q) = 0. (2) Which of the points Q found in part (1) correspond to maxima or minima? Solution. (1) If ∇ f 3 (x, y) = 0, then ∂ f3 ∂ f3 (x, y) = a cos x = 0 and (x, y) = 0. ∂x ∂x It follows that cos x = 0. This yields that x = π/2 or x = 3π/2. The image of these two conditions consists of two circles of radius b about z-axis in the plane z = ±a. It is easy to see that the points with z = a give the absolute maximum of f 3 (x, y), while those points with z = −a give the absolute minimum of f 3 (x, y).

282

7 Vector Fields

329. Let F : R2 → R2 and G : R3 → R2 be two vector fields defined as follows: − → − → F(x, y, z) = e x+2y i + sin(y + 2x) j , − → − → G(u, v, w) = (u + 2v 2 + 3w 3 ) i + (2v − u 2 ) j . (1) Compute each of the Jacobian matrix D F(x, v, w). y) and DG(u, (2) Compute the composition H (u, v, w) = F G(u, v, w) . (3) Compute the Jacobian matrix D H (1, −1, 1). Solution. (1) By the definition of Jacobian matrix, we have ⎤ ∂ x+2y ∂ x+2y (e (e ) ) ⎥ ⎢ ∂y [D F(x, y)] = ⎣ ∂ ∂x ∂ ⎦ sin(y + 2x) sin(y + 2x) ∂y ∂x x+2y x+2y e 2e = , 2 cos(y + 2x) cos(y + 2x) ⎡

⎤ ∂ 2 + 3w 3 ) ∂ (u + 2v 2 + 3w 3 ) ∂ (u + 2v 2 + 3w 3 ) (u + 2v ⎥ ⎢ ∂v ∂w [DG(u, v, w)] = ⎣ ∂u ∂ ⎦ ∂ ∂ 2 2 2 (2v − u ) (2v − u ) (2v − u ) ∂u ∂v ∂w 1 4v 9w2 . = −2u 2 0 ⎡

(2) We obtain H (u, v, w) = F G(u, v, w) = F u + 2v 2 + 3w 3 , 2v − u 2 − → → 2 3 2− = eu+2v +3w +4v−2u i + sin 2v − u 2 + 2u + 4v 2 + 6w 3 j . − → − → (3) We find that G(1, −1, 1) = 6 i − 3 j . So, we can write [D H (1, −1, 1)] = [D F(6, −3)][DG(1, −1, 1)] 1 2 1 −4 9 = 2 cos 9 cos 9 −2 2 0 −3 0 9 = . 0 −6 cos 9 18 cos 9 330. Let F : R3 → R2 and G : R3 → R3 be two vector fields defined as follows: − → − → F(x, y, z) = (x 2 + y + z) i + (2x + y + z 2 ) j , − → − → − → G(u, v, w) = uv 2 w 2 i + w 2 sin v j + u 2 ev k . (1) Compute each of the Jacobian matrix D F(x, y) and DG(u, v, w).

7.6 Solved Problems

283

(2) Compute the composition H (u, v, w) = F G(u, v, w) . (3) Compute the Jacobian matrix D H (u, 0, w). Solution. (1) By the definition of Jacobian matrix, we obtain ⎡ ∂ (x 2 + y + z) ⎢ ∂x [D F(x, y)] = ⎣ ∂ (2x + y + z 2 ) ∂x 2x 1 1 = , 2 1 2z

∂ 2 (x + y + z) ∂y ∂ (2x + y + z 2 ) ∂y

⎡ ∂ ∂ (uv 2 w 2 ) (uv 2 w 2 ) ⎢ ∂u ∂v ⎢ ∂ ∂ 2 2 [DG(u, v, w)] = ⎢ ⎢ ∂u (w sin v) ∂v (w sin v) ⎣ ∂ 2 v ∂ 2 v (u e ) (u e ) ∂v ⎡ 2∂u2 2 u w 2uvw 2uv 2 w 2 ⎣ 0 w cos v 2w sin v = u 2 ev 0 2uev

⎤ ∂ 2 (x + y + z) ⎥ ∂z ⎦ ∂ 2 (2x + y + z ) ∂z

⎤ ∂ (uv 2 w 2 ) ⎥ ∂w ⎥ ∂ 2 (w sin v) ⎥ ⎥ ∂w ⎦ ∂ 2 v (u e ) ⎤∂w ⎦.

(2) We have H (u, v, w) = F G(u, v, w) = F uv 2 w 2 , w 2 sin v, u 2 ev − → − → = (u 2 v 4 w 4 + w 2 sin v + u 2 ev ) i + (2uv 2 w 2 + w 2 sin v + u 4 e2v ) j . − → (3) We get G(u, 0, w) = u 2 k . Hence, we can write [D H (u, 0, w)] = [D F(0, 0, u 2⎡ )][DG(u, 0,⎤w)] 0 0 0 0 1 1 ⎣ 0 w2 0 ⎦ = 2 1 2u 2 2u u 2 0 2 2 2u w + u 0 . = 4u 3 w 2 + u 2 0 331. Let F : Rn → Rm be a function with components f 1 , . . . , f m and U be an open subset of Rn , where U consists of the points A, B and the segment line S joining A and B. If all partial derivatives of F are continuous on S, prove that F(B) − F(A) =

1

D F A + t (B − A) (B − A)dt.

0

defined by (t) = F A + t (B − A) . Then, Solution. Let : (−,1 + ) → Rm be we have (t) = D F A + t (B − A) (B − A). Now, if (t) = ϕ1 (t), . . . , ϕm (t) with ϕ j (t) = f j A + t (B − A) , then

284

7 Vector Fields

ϕ j (1) − ϕ j (0) = 0

1

ϕj (t)dt,

for all 1 ≤ j ≤ m. This completes the proof. 332. Lett F and G be functions from Rn to Rm . We say that F and G are tangent at a point A ∈ Rn if F(X ) − G(X ) lim = 0. X →A X − A Prove that if F is tangent to a linear transformation T : Rn → Rm at a point A, then there can be only one such. Solution. Assume that T1 and T2 are linear transformation from Rn to Rm , tangent to F at A. We show that T1 = T2 . It is clear that T1 (A) = T2 (A) = F(A). On the other hand, we can write T1 (X ) − T2 (X ) = T1 (X ) − T1 (A) + T2 (A) − T2 (X ) = T1 (X − A) − T2 (X − A).

Also, for every > 0, we have T1 (X ) − T2 (X ) ≤ X − A,

(7.11)

for all X , sufficiently close to A. Let bi j and ci j be the matrix representation of T1 and T2 , respectively. If X = (x1 , . . . , xn ) and A = (a1 , . . . an ), then X − A = (x1 − a1 , . . . , xn − an ). Hence, we have n n T1 (X ) − T2 (X ) = b1 j (x j − a j ), . . . , bm j (x j − a j ) j=1

j=1

n n c1 j (x j − a j ), . . . , cm j (x j − a j ) − j=1

j=1

n n = (b1 j − c1 j )(x j − a j ), . . . , (bm j − cm j )(x j − a j ) . j=1

j=1

Now, with δ a real number, choose X−A=

δ Z, Z

where Z = (z 1 , . . . , z n ). Then, for sufficiently small |δ|, by (7.11) we obtain

7.6 Solved Problems

285

n n δz j δz j ,..., T1 (X ) − T2 (X ) = (b1 j − c1 j ) (bm j − cm j ) Z Z j=1 j=1 Z ≤ X − A = |δ| = |δ|. Z

Therefore, we get n n zj z j ,..., (b1 j − c1 j ) (bm j − cm j ) ≤ . Z Z j=1 j=1

(7.12)

Since (7.12) holds for all positive number , it follows that bi j = ci j , for all 1 ≤ i ≤ m and 1 ≤ j ≤ n. 333. Let F = ( f 1 , f 2 ) be a function from R2 into R2 given by f 1 (x, y) = e x cos y and f 2 (x, y) = e x sin y. (1) What is the range of F? (2) Show that the Jacobian of F is not zero at any point of R2 . So, each point of R2 has a neighborhood in which F is one to one. Nevertheless, F is not one to one on R2 . (3) Take A = (0, π/3) and B = F(A), let G be the continuous inverse of F, defined in a neighborhood of B such that G(B) = A. Find an explicit formula for G, and compute D F(A) and DG(A). Solution. (1) Suppose that (r, s) = (0, 0) is arbitrary and take y such that s r and sin y = √ , cos y = √ 2 2 2 r +s r + s2 √ and let x = ln r 2 + s 2 . Now, we see that r = e x cos y and s = e x sin y. This means that (r, s) lies in the range of F. Note that the point (0, 0) does not belong to the range of F, because r 2 + s 2 = e2x > 0. Therefore, the range of F is all of R2 except the point (0, 0). (2) The Jacobian matrix is ⎡∂f

1

⎢ [D F(x, y)] = ⎣ ∂∂xf 2 ∂x and so

(x, y) (x, y)

⎤ ∂ f1 x (x, y) e cos y ⎥ ∂y ⎦ = e x sin y ∂ f2 (x, y) ∂y

∂( f 1 , f 2 ) = e2x , ∂(x, y)

−e x sin y , e x cos y

286

7 Vector Fields

which is never zero. Since F(x, y + 2π) = F(x, y), it follows that F is not one to one. √ (3) By the definition we get B = (1/2, 3/2). Hence, if y = tan−1 (s/r ), for (r, s) near B, then −π/2 ≤ tan−1 (s/r ) ≤ π/2. Consequently, we have G(r, s) = ln r 2 + s 2 , tan−1 (s/r ) . Then, we have F (x, y) =

x

e cos y e x sin y

⎡

r 2 ⎢ + s2 and G (r, s) = ⎣ r −s r 2 + s2

−e sin y e x cos y

x

⎤ s r 2 + s2 ⎥ ⎦. r 2 2 r +s

Taking r = e x cos y and s = e x sin y, we obtain

e−x sin y . e−x cos y

e−x cos y G F(x, y) = −e−x sin y

Now, it is easy to see that 1 0 G F(x, y) F (x, y) = . 0 1 Also, since

r −s , F G(r, s) = s r

it follows that

1 0 . F G(r, s) G (r, s) = 0 1

2

→ − → ex − j . Determine if the vector is a gradient. If 334. Let F(x, y) = 2xe ln y i + y it is, then find a function having this gradient. x2

2

2

Solution. Taking P(x, y) = 2xe x ln y and Q(x, y) = e x /y, we get Py (x, y) =

2xe x y

2

2

and Q x (x, y) =

2xe x . y

Hence, Py (x, y) = Q x (x, y); so the given vector field is a gradient ∇ f (x, y). Furthermore, 2 (7.13) f x (x, y) = 2xe x ln y,

7.6 Solved Problems

287 2

f y (x, y) =

ex . y

(7.14)

Integrating both members of (7.13) with respect to x we obtain 2

f (x, y) = e x ln y + g(y),

(7.15)

where g(y) is independent of x. We now partially differentiate both members of (7.15) with respect to y and have 2

f y (x, y) =

ex + g (y). y

(7.16)

We equate the right members of (7.16) and (7.14) and get 2

2

ex ex + g (y) = . y y This implies that g (y) = 0, and so g(y) = C, where C is constant. If we substitute 2 this value for g(y) into (7.15), we obtain f (x, y) = e x ln y + C. −y 1 − 1 − 2 − → −y → → j + 335. Let F(x, y) = + 2 i + + 2 k. 2 2 (x + z) x x+z (x + z) z Determine if the vector is a gradient. If it is, then find a function having this gradient. Solution. Suppose that P(x, y, z) =

−y −y 1 1 2 and R(x, y, z) = + 2 , Q(x, y, z) = + 2. (x + z)2 x x+z (x + z)2 z

Then, we obtain −1 , (x + z)2 −1 Q x (x, y, z) = , (x + z)2 −2y Rx (x, y, z) = , (x + z)3 Py (x, y, z) =

−2y , (x + z)3 −1 Q z (x, y, z) = , (x + z)2 −1 R y (x, y, z) = . (x + z)2 Pz (x, y, z) =

Since Py (x, y, z) = Q x (x, y, z), Pz (x, y, z) = Rx (x, y, z) and Q z (x, y, z) = R y (x, y, z), it follows that the given vector is a gradient ∇ f (x, y, z). Furthermore, f x (x, y, z) =

−y 1 + 2, 2 (x + z) x

(7.17)

1 , x+z

(7.18)

f y (x, y, z) =

288

7 Vector Fields

f z (x, y, z) =

−y 2 + 2. 2 (x + z) z

(7.19)

Integrating both members of (7.17) with respect to x we have f (x, y, z) =

y 1 − + g(y, z), (x + z)2 x

(7.20)

where g(y, z) is independent of x. We partially differentiate both members of (7.20) with respect to y and obtain 1 + g(y, z), x+z

f y (x, y, z) =

(7.21)

and so g y (y, z) = 0. Now, we integrate both members of this equation with respect to y and get g(y, z) = h(z), (7.22) where h is independent of x and y. Substituting from (7.22) into (7.20), we obtain f (x, y, z) =

1 y − + h(z). (x + z)2 x

(7.23)

Now, we partially differentiate with respect to z both members of (7.23), we get f z (x, y, z) =

−y + h (z). (x + z)2

(7.24)

Equating the right members of (7.24) and (7.19), we have −y −y 2 + h (z) = + 2. 2 2 (x + z) (x + z) z Hence, we conclude that h (z) = 2/z 2 , which implies that h(z) =

−2 + C, z

(7.25)

where C is constant. We substitute from (7.25) into (7.23) and obtain f (x, y, z) =

y 1 2 − − + C. x+z x z

− → − → − → 336. Let R(x, y, z) = x i + y j + z k and let f : R3 → R be a scalar field such that f (R) = 0 when R = 1. If ∇ f = R/R5 , show that

7.6 Solved Problems

289

f (R) =

1 1 1− . 3 R3

− → − → − → Solution. If R(x, y, z) = x i + y j + z k , then R = (x 2 + y 2 + z 2 )1/2 . Using the definition of gradient we must have ∂f− → ∂f− → ∂f− → − → − → − → i + j + k = R−5 (x i + y j + z k ). ∂x ∂y ∂z So, we conclude that ∂f ∂f ∂f = R−5 x, = R−5 y and = R−5 z. ∂x ∂y ∂z Integrating from the above equalities with respect to x, y and z, respectively, we get

1 (x 2 + y 2 + z 2 )−5/2 xd x = − (x 2 + y 2 + z 2 )−3/2 + g(y, z), 3 (7.26) 1 2 2 2 2 −5/2 2 2 −3/2 f (x, y, z) = (x + y + z ) ydy = − (x + y + z ) + h(x, z), 3 (7.27) 1 2 2 2 2 −5/2 2 2 −3/2 f (x, y, z) = (x + y + z ) zdz = − (x + y + z ) + k(x, y). 3 (7.28) From (7.26), (7.27) and (7.28) we deduce that g(y, z) = h(x, z) = k(x, y) = C, a constant value. Therefore, we have f (x, y, z) = (−1/3)(x 2 + y 2 + z 2 )−3/2 + C. Since f (R) = 0 when R = 1, it follows that C = 1/3. Consequently, we obtain f (x, y, z) =

1 1 1 1 1 1 1− . f (x, y, z) = − (x 2 + y 2 + z 2 )−3/2 + = − R−3 + = 3 3 3 3 3 R3 − → − → − → 337. Let R(x, y, z) = x i + y j + z k , prove that (1) ∇R3 = 3RR;

(2) ∇eR = 2eR R. 2

Solution. (1) By the definition we have ∂ ∂ ∂ − → − → − → R3 i + R3 j + R3 k ∂x ∂y ∂z ∂R − → → → 2 ∂R − 2 ∂R − i + 3R j + 3R2 k = 3R ∂x ∂y ∂z x y z − → − → − → = 3R2 i + 3R2 j + 3R2 k R R R − → − → − → = 3R x i + 3R y j + 3R z k − → − → − → = 3R(x i + y j + z k ) = 3RR.

∇R3 =

290

7 Vector Fields

(2) Again by the definition we can write ∂ R2 − ∂ R2 − ∂ R2 − → → → e e e i + j + k ∂x ∂y ∂z → → → 2 ∂R − 2 ∂R − 2 ∂R − i + 2R eR j + 2R eR k = 2R eR ∂x ∂y ∂z x − y − z − → → → 2 2 2 = 2R eR i + 2R eR j + 2R eR k R R R − → − → − → 2 = 2eR (x i + y j + z k )) = 2eR R.

∇eR = 2

338. Prove that (1) Every conservative vector field is rotation free; (2) The curl of a vector field is incompressible. Solution. Let f : R3 → R and F : R3 → R3 , with the components P, Q and R, be continuously differentiable. (1) We show that curl(∇ f ) = 0. To do this we compute:

− − → − →

→ j k

i

∂ ∂ ∂

curl(∇ f ) = ∇ × ∇ f =

∂x ∂ y ∂z

∂f ∂f ∂f

∂x ∂ y ∂z ∂2 f ∂2 f − ∂2 f ∂ 2 f − ∂ 2 f − → → ∂2 f → = − i + − j + − k = 0. ∂ y∂z ∂z∂ y ∂x∂z ∂z∂x ∂x∂ y ∂ y∂x

(2) We prove that div(curl F) = 0. To do this, we calculate:

− − → − →

→ j k

i

∂ ∂ ∂

div(curl F) = ∇ · (∇ × F) = ∇ ·

∂x ∂ y ∂z

P Q R ∂2 R ∂2 Q ∂2 R ∂2 P ∂2 Q ∂2 P = − − + + − = 0. ∂x∂ y ∂x∂z ∂ y∂x ∂ y∂z ∂z∂x ∂z∂ y 339. Let F = ( f 1 , f 2 , f 3 ) and G = (g1 , g2 , g3 ) be two vector fields from R3 into R3 and h : R3 → R be a scalar field, and a, b be real numbers. Prove that (1) curl(a F + bG) = a curl F + b curlG; (2) curl(h F) = hcurl F + ∇h × F; (3) div(F × G) = G · curl F − F · curlG.

7.6 Solved Problems

291

Solution. (1) By the definition of curl and properties of determinant, we have

− → − → − →

i j k

∂ ∂ ∂

curl(a F + bG) =

∂x ∂y ∂z

∂ ∂ ∂

∂x (a f 1 + bg1 ) ∂ y (a f 2 + bg2 ) ∂z (a f 3 + bg3 )

−

− − → − →

− → − →

→

→ j k j k

i

i

∂

∂

∂ ∂ ∂ ∂

= a

∂x ∂ y ∂z

+ b

∂x ∂ y ∂z

∂g ∂g ∂g

∂ f1 ∂ f2 ∂ f3 2 3

1

∂x

∂ y ∂z ∂x ∂ y ∂z = a curl F + b curlG. (2) We have h F = (h f 1 , h f 2 , h f 3 ). So, we can write

− − → − →

→ i j k

∂ ∂ ∂

curl(h F) =

∂x ∂y ∂z

∂ ∂ ∂

∂x (h f 1 ) ∂ y (h f 2 ) ∂z (h f 3 )

− → − → − →

i j k

∂ ∂ ∂

=

∂x ∂y ∂z

∂ f1

∂ f ∂ f ∂h ∂h ∂h 2 3

h

∂x + f 1 ∂x h ∂ y + f 2 ∂ y h ∂z + f 3 ∂z

− − → − →

→ j k

i

∂ ∂ ∂

= h curl F +

∂x ∂y ∂z

∂h ∂h ∂h

f1

∂x f 2 ∂ y f 3 ∂z ∂h ∂h − ∂h ∂h − → → = h curl F + f 3 − f2 i + f1 − f3 j ∂x ∂z ∂z ∂x ∂h ∂h − → − f1 k. + f2 ∂x ∂y ∂h − → ∂h − → ∂h − → i + j + k and using the formula for vector prod∂x ∂y ∂z uct, we see that the above equality is precisely curl(h F) = hcurl F + ∇h × F. (3) We have

Recalling that ∇h =

292

7 Vector Fields

div(F × G) = ∇ · (F × G)

− → − → − → = ∇ · ( f 2 g3 − f 3 g2 ) i + ( f 3 g1 − f 1 g3 ) j + ( f 1 g2 − f 2 g1 ) k ∂ ∂ ∂ ( f 2 g3 − f 3 g2 ) + ( f 3 g1 − f 1 g3 ) + ( f 1 g2 − f 2 g1 ) ∂x ∂ y ∂z ∂ f2 ∂g3 ∂ f3 ∂g2 + f2 − g2 − f3 = g3 ∂x ∂x ∂x ∂x ∂ f3 ∂g1 ∂ f1 ∂g3 + g1 + f3 − g3 − f1 ∂y ∂y ∂y ∂y ∂ f1 ∂g2 ∂ f2 ∂g1 + f1 − g1 − f2 + g2 ∂z ∂z ∂z ∂z ∂ f3 ∂ f1 ∂ f2 ∂ f2 ∂ f3 ∂ f1 = g1 − + g2 − + g3 − ∂y ∂z ∂z ∂x ∂x ∂y ∂g3 ∂g1 ∂g2 ∂g2 ∂g3 ∂g1 − − f2 − − f3 − − f1 ∂y ∂z ∂z ∂x ∂x ∂y = G · (∇ × F) − F · (∇ × G) = G · curl F − F · curlG. =

− → − → − → − → − → − → 340. Let R(x, y, z) = x i + y j + z k and A = a1 i + a2 j + a3 k be a constant vector. If F = A × R, prove that A = (1/2)curl F. Solution. We have − → − → − → F = A × R = (a2 z − a3 y) i + (a3 x − a1 z) j + (a1 y − a2 z) k . Now, we obtain

− → − → − →

i j k

∂ ∂ ∂

curl F = ∇ × F =

∂x ∂ y ∂z

a2 z − a3 y a3 x − a1 z a1 y − a2 z ∂ ∂ ∂ ∂ − → − → = (a1 y − a2 z) − (a3 x − a1 z) i − (a1 y − a2 z) − (a2 z − a3 y) j ∂ y ∂z ∂x ∂z ∂ ∂ − → (a3 x − a1 z) − (a2 z − a3 y) k + ∂x ∂y − → − → − → − → − → − → = (a1 + a1 ) i + (a2 + a2 ) j + (a3 + a3 ) k = 2(a1 i + a2 j + a3 k ) = 2 A.

This completes the proof. 341. Assume continuous differentiability of all vector fields involved. Suppose that H = F + G, where F is incompressible and G is rotation free. Then there is a vector field K such that F = curl K and a scalar field f such that G = ∇ f . Show that K and f satisfy the following partial differential equations: (1) ∇ 2 f = div H ; (2) ∇(div K ) − ∇ 2 K = curl H .

7.7 Exercises

293

Hence, every continuously differentiable vector field H can be expressed in the form H = F + G, where F is incompressible and G is rotation free. Solution. (1) Using the definitions we have div F = 0 and curlG = 0. Now, we can write div H = div(F + G) = div F + divG = divG = div(∇ f ) = ∇ 2 f. (2) We know that the following equality holds for any vectors A, B and C: A × (B × C) = B(A · C) − (A · B)C. If we substitute ∇ for A and B, and K for C, we obtain ∇ × (∇ × K ) = ∇(∇ · K ) − (∇ · ∇)K = ∇(∇ · K ) − ∇ 2 K , which holds if all mixed partial derivatives are continuous. In other words, curl(curl K ) = ∇(div K ) − ∇ 2 K .

(7.29)

On the other hand, we have curl H = curl(F + G) = curl F + curlG = curl F = curl(curl K ).

(7.30)

Finally, by (7.29) and (7.30) we get ∇(div K ) − ∇ 2 K = curl H .

7.7 Exercises Easier Exercises 1. Sketch the vector field given by − → − → (a) F(x, y) = y i + (1/2) j ; − → − → (b) F(x, y) = −y i + j ;

− → − → (c) F(x, y) = i + (x + y) j ; − → − → (d) F(x, y) = 2x i + 2y j .

2. Sketch the vector field given by − → (a) F(x, y, z) = k ; − → − → (b) F(x, y, z) = x i + z k ;

− → (c) F(x, y, z) = x k ; − → − → (d) F(x, y, z) = j − i .

3. Find the gradient vector field ∇ f of f and sketch it.

294

7 Vector Fields

(a) f (x, y) = x 2 − y;

(b) f (x, y) =

x 2 − y2.

− → − → 4. Prove that ∇ f (x, y) = a i + b j if and only if f (x, y) = ax + by + c, where a, b and c are constants. 5. Determine if the vector is a gradient. If it is, find a function having the given gradient. − → x − → − → (a) (2x y + y 2 + 1) i + (x 2 + j; (c) 3x + ln y) i + y 2 + − → y 2x y + x) j ; 1 1 − → 1 − 2x − → − → − → i + j ; + (b) (d) (sin 2x − tan y) i − x sec2 y j . x2 y2 y3 6. Determine if the vector is a gradient. If it is, find a function having the given gradient. − → − → − → (a) (2y − 5z) i + (2x + 8z) j − (5x − 8y) k ; − → − → − → (b) z tan y i + x z sec2 y j + x tan y k ; − → − → − → (c) e x (e z − ln y) i + (e y ln z − e x y −1 ) j + (e x+z + e y z −1 ) k ; x−z − 1 − x+y − → → → i − (d) j + k. 2 2 y+z (y + z) (y + z) 7. Find a vector field F(x, y) in the x y-plane with the property that at each point (x, y) = (0, 0), F is a unit vector pointing toward the origin. (The field is undefined at (0, 0).) 8. Find a vector field F(x, y) in the x y-plane with the property that at each point (x, y) = (0, 0), F points toward the origin and F is (a) the distance from (x, y) to the origin, (b) inversely proportional to the distance from (x, y) to the origin. (The field is undefined at (0, 0).) 9. Let A and B be constant vectors. Find the gradient of the given function f (R) − → − → − → of the position vector R = x i + y j + z k . (a) f (R) = A · R; (b) f (R) = (A · R)(B · R);

(c) f (R) = A · (B × R); (d) f (R) = R · (A × R).

10. If and are twice differentiable functions, show that ∇ 2 ( f g) = f ∇ 2 g + g∇ 2 f + 2∇ f · ∇g. − → − → − → 11. Let R = x i + y j + z k . If A and B are constant vectors, show that 1 A·R =− (a) A · ∇ ; R R3 1 3(A · R)(B · R) A·B = (b) B · ∇ A · ∇ − . 5 R R R3

7.7 Exercises

12. If

295

F(x, y, z) = x 2 y, −2x z 3 , x z 2

and

G(x, y, z) = 2z, y, −x 2 ,

find

∂2 (F × G) at (1, 0, −2). ∂x∂ y 13. Compute the Jacobian matrices of the given functions at the given points: (a) The function F : R2 → R2 defined by F(x, y) = (x 2 + y 3 , x y) at (−1, 1); (b) The function F : R2 → R3 defined by F(x, y) = (xe y , ye2x , x 2 + y 2 ) at (1, 1); (c) The function F : R3 → R defined by F(x, y, z) = x 2 + y 2 + z 2 at (1, 2, 3). 14. Take the vector field F(x, y) = (x 2 y 2 , x 2 + y 2 ). Show that this vector field is neither the curl nor the gradient of any function. 15. Which of the following functions T are linear transformation? Justify your answer. (a) The function T : R2 → R2 defined by T (x, y) = (−3x − 5y, x + y); (b) The function T : R2 → R3 defined by T (x, y) = (x y + yz, xz + yz); (c) The function T : R3 → R2 defined by T (x, y, z) = sin(π/5)x+ cos(π/5)y, −4z ; (d) The function T : R2 → R defined by T (x, y) = 2|x| + |y|. 16. Verify the property ∂( f g, h) ∂( f, h) ∂(g, h) = g+ f . ∂(u, v) ∂(u, v) ∂(u, v) 17. Show that if f (u, v) and g(u, v) are differentiable and if c is constant, then ∂( f, h) ∂(c f, h) =c . ∂(u, v) ∂(u, v) 18. A particle moves in a velocity field V (x, y) = (x 2 , x + y 2 ). If it is at position (2, 1) at time t = 3, estimate its location at 3.01.

Harder Exercises 19. If R1 and R2 denote the distance from a point (x, y) on an ellipse to its foci, show that the equation R1 + R2 = constant (satisfied by these distances) implies the relation T · ∇(R1 + R2 ) = 0, where T is the unit tangent to the curve. Interpret this result geometrically, there by showing that the tangent makes equal angles with the lines joining (x, y) to the foci.

296

7 Vector Fields

− → − → − → 20. If ∇ f (x, y, z) is always parallel to x i + y j + z k , show that f must assume equal values at the point (0, 0, a) and (0, 0, −a). 21. If F = ( f 1 , f 2 , f 3 ) is a vector field, its divergence is div F = ∂ f i /∂xi . Suppose that we take another rectangular coordinate system (with the same origin). Then its coordinates are x = ai j x j , where ai j ’s are constant. Express F in the new coordinate system, and show that div F = div F, that is, that div F is a geometric quantity, independent of the coordinate system. 22. Let F(r, θ) = (r cos θ, r sin θ). (This map takes a vector in polar coordinates and rewrites it in rectangular coordinates.) Compute F(1, π/4) and [D F(1, π/4)]. Use your computation to estimate F(1.01, 0.02 + π/4). 23. Let T be the function from R3 into R3 defined by T (x, y, z) = (x − y + 2z, 2x + y, −x − 2y + 2z). (a) Verify that T is a linear transformation; (b) If (a, b, c) is a vector in R3 , what are the conditions on a, b and c that the vector be in the range of T ? 24. Let T be the function from R3 into R3 defined by T (x, y, z) = (3x, x − y, 2x + y + z). (a) Verify that T is a linear transformation; (b) Is T invertible? If so, find a rule for T −1 like the one which defines T . 25. Let T1 be a linear transformation from R3 into R2 , and T2 be a linear transformation from R2 into R3 . Prove that the transformation T2 ◦ T1 is not invertible. 26. For F(x, y) = (x + 2y, 3x + 4y) and any A ∈ R2 , show that D F(A)(h 1 , h 2 ) = (h 1 + 2h 2 , 3h 1 + 4h 2 ). 27. Suppose that F : R2 → R2 is a differentiable function, F(1, 1) = (3, 7) and [D F(1, 1) =

1 1 . 3 5

Estimate F(1.1, 1.2). 28. Let F(x, y) = (x 3 + 4y, 2x + y 2 ) and G(x, y) = (y 2 , 3x 2 ). Compute [D(G ◦ F)] at (1, 1). 29. Let F(x; y) = (x 2 y, x y 2 ) and G(x, y) = (x/y, y/x). Use the chain rule to compute [D(G ◦ F)(1, 2)]. Check your answer by composing G and F and then differentiating.

7.7 Exercises

297

30. Let F : R2 → R2 be defined by F(x, y) =

1 x 2 + x 2 sin , y if x = 0 x (0, y) if x = 0.

Show that ∂ F/∂x exists at every point and ∂ F/∂ y exists and is continuous on a neighborhood of (0, 0). Show that F is differentiable at (0, 0). 31. Let F : R2 → R2 given by F(x, y) = (x + y, 2x + ay). (a) Calculate D F(x, y) and show that D F(x, y) is invertible if and only if a = 2; (b) Examine the image of the unit square {(x, y) | x, y ∈ [0, 1]} when a = 1, 2, 3. 32. Let ⊆ Rn and F : → Rm be differentiable at a point C interior to , and let V ∈ Rm . If we define g : → R by g(X ) = F(X ) · V , for all X ∈ , show that g is differentiable at C and that Dg(C)(U ) = D F(C)(U ) · V, for U ∈ Rn . 33. Let F : R2 → R2 be a function which sends the point (x, y) into the point (u, v) given by u = x 2 − y 2 and v = 2x y. What curves in the x y-plane map under F into the lines u = constant and v = constant? Into what curves in the uv-plane do the lines x = constant and y = constant map? Show that each non-zero point (u, v) is the image under F of two points. Into what region does F map the square {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}? What region is mapped by F into the square {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1}?

Chapter 8

Line Integrals

8.1 Work as a Line Integral A line integral is an integral where the function to be integrated is evaluated along a curve. Let C be a curve on an open disk B in R2 such that a vector equation of C is − → − → R(t) = f (t) i + g(t) j , where f and g are continuous on [a, b]. Moreover, let − → − → F(x, y) = P(x, y) i + Q(x, y) j be a force field, where P and Q are continuous on B. Then b P f (t), g(t) f (t) + Q f (t), g(t) g (t) dt, (8.1) W = a

where W is the measure of the work done by F in moving an object along C from f (a), g(a) to f (b), g(b) . Also, we can write (8.1) as follows: W =

b

F f (t), g(t) · R (t) dt.

a

The integral in (8.1) is called a line integral. A common notation for line integral3 is C P(x, y)d x + Q(x, y)dy. The concept of a line integral can be extended to R . Let P, Q and R be functions of three variables x, y and z such that they are continuous on an open disk B in R3 . Let C be a curve lying in B with the parametric equations x = f (t), y = g(t) and z = h(t) with a ≤ t ≤ b, such that f and g are continuous on [a, b]. Then P(x, y, z)d x + Q(x, y, z)dy + R(x, y, z)dz C

=

b

P f (t), g(t), h(t) f (t) + Q f (t), g(t), h(t) g (t) + R f (t), g(t), h(t) h (t) dt.

a

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_8

299

300

8 Line Integrals

A function f : [a, b] → Rn which is continuous on [a, b] is called a continuous path. The path is called smooth if the derivative f exists and is continuous in (a, b). The path is called piecewise smooth if the interval [a, b] can be partitioned into a finite number of subintervals in each of which the path is smooth. If the curve C consists of the smooth paths C1 , C2 , . . . , Ck , then the line integral is defined by P(x, y, z)d x + Q(x, y, z)dy + R(x, y, z)dz C k

=

i=1

P(x, y, z)d x + Q(x, y, z)dy + R(x, y, z)dz.

Ci

Let f be a scalar field and C be a smooth curve parameterized by R(t). Then, we define the line integral of f over C with respect to arc length to be

f ds =

C

b

f R(t) R (t)dt.

a

8.2 Line Integrals Independent of the Path − → − → Let the vector field F(x, y) = P(x, y) i + Q(x, y) j be continuous on a domain D. Then, the line integral C P(x, y)d x + Q(x, y)dy is path independent on D if and only if there is a function U = U (x, y) defined on D such that P = ∂U/∂x and Q = ∂U/∂ y. Let C be a piecewise smooth curve from the point (x1 , y1 ) to the point (x2 , y2 ). Then

P(x, y)d x + Q(x, y)dy = U (x1 , y1 ) − U (x2 , y2 ). C

The above result can be extended to functions of n variables. Fundamental theorem of calculus for line integrals: Let U be a differentiable scalar field with a continuous gradient ∇U on an open connected set D. Then, for any two points A and B joined by a piecewise smooth path R in D we have

b

∇U · R dt = U (B) − U (A).

a

8.3 Solved Problems 342. Find the value of the integral C

−y x dx + 2 dy, 2 +y x + y2

x2

8.3 Solved Problems

301

√ √ √ √ if Cis the arc of the circle x 2 + y 2 = 4from ( 2, 2)to (− 2, 2). Solution. The arc defined by the parametric equations x(t) = 2 cos t and y = 2 sin t,

3π π ≤t ≤ . 2 2

Now, we compute the value of the line integral by applying the definition. It is equal to

3π/4 3π/4 (−2 sin t)(−2 sin t) (2 cos t)(2 sin t) π + dt = dt = . 2 t + 4 sin2 t 2 t + 4 sin2 t 2 4 cos 4 cos π/4 π/4 343. Let f (t) be a differentiable and positive real function with a ≤ t ≤ b. Suppose − → − → − → that C is the path R(t) = t i + f (t) j with a ≤ t ≤ b,and F = y j . Is there any relation between the value of the work integral and the area of the region bounded by the t-axis, the graph of f ,and the lines t = a and t = b? Solution. We have F.d R =

b

− → − → − → f (t) i · i + f (t) j dt =

a

C

b

f (t)dt.

a

Since f (t) > 0, for a ≤ t ≤ b, it follows that the work and the area under the curve have the same value.

ye−x ds, where C is the plane curve x = ln(t 2 + 1)and y = 2 tan−1 √ t − twith 0 ≤ t ≤ 3. 344.

Find

C

Solution. We have ye C

−x

√

ds =

3

y(t)e−x(t)

0

x (t)

2

2 + y (t) dt

2

2 2t 2 = − 1 dt + 2 t2 + 1 t +1 0 √3 √3 √3 2 tan−1 t − t t = dt = 2 dt tan−1 t d(tan t) − 2+1 2+1 t t 0 0 0 √ √

3 1

3 π2 ln(t 2 + 1) = − ln 2. = tan−1 t − 0 0 2 9

√

3

2 tan−1 t − t t2 + 1

302

8 Line Integrals

345.

x(1 + 4y)ds, where Cis the curve y = x 2 , starting from (0, 0)

Evaluate C

and ending at (1, 1). Solution. We can consider x as a parameter. Along y = x 2 we have ds =

1+

dy dx

2 dx =

1 + 4x 2 d x.

Hence, the integral is equal to

x(1 + 4y)ds =

1

x(1 + 4x 2 ) 1 + 4x 2 d x =

0

C

1

3/2 x 1 + 4x 2 d x.

0

This integral can be computed by using the substitution u = 1 + 4x 2 , which implies that du = 8xd x. Therefore, we obtain 0

1

3/2 1 x 1 + 4x 2 dx = 8

5 1

u 3/2 du =

1 5/2

5 1 √ 25 5 − 1 . u = 1 20 20

346. The roof of a building is a paraboloid modeled by 10 − (1/8)x 2 − (1/12)y 2 = 0, with the highest point above the origin. Suppose that x and y represent distances from the origin along the floor (the x y -plane) orthogonal to one another, and that all measurements are in meters. A wall is to be built extending from the origin along the line y = 3x. The wall will rise from the floor up to the roof. What is the area of this wall? − → Solution. The path C is the line y = 3x. A parametrization of C is R(t) = t i + − → 3t j . The lower bound for t is 0, and the upper bound of t is where the roof meets the floor along this line. Since z = 0 at this point, and since y = 3x, we make substitutions and solve for x. Then, we get 10 − (7/8)x 2 = 0, which implies that √ x = 80/7 On the other hand, we have √ − → − → R (t) = i + 3 j and R (t) = 10. Consequently, the integral is

√80/7 √ 1 2 1 2 1 2 1 2 10 − x − 10 − t − (3t) y ds = 10dt 8 12 8 12 0 C

√80/7 √ 7 10 − t 2 dt = 10 8 0

8.3 Solved Problems

303

√ √ 7

80/7 10 10t − t 3 0 24

√ 3 √ 7 . = 10 10 80/7 − 80/7 24 =

Now, using a calculator, the area of the wall is about 71.27 m2 .

347. A glass rod of consistent thickness is in the shape of a quarter-circle of radius − → − → 3, modeled by R(t) = 3 cos t i + 3 sin t j , where0 ≤ t ≤ π/2. However, it is not uniformly dense. Suppose that its density x centimeters left and right of the origin, and y centimeters above and below the origin, is given by d(x, y) = x + y + x y, where d is grams per centimeter. Find the mass of this glass rod. Solution. This is equivalent to evaluating the scalar line integral (x + y + x y)ds. C

We have

− → − → R (t) = −3 sin t i + 3 cos t j and R (t) = 3.

The integrand is now written in terms of t and evaluated as follows:

π/2

(x + y + x y)ds = C

(3 cos t) + (3 sin t) + (3 cos t)(3 sin t) (3dt)

0

π/2

=9

cos t + sin t + 3 cos t sin t dt

0

π/2 3 63

= 9 sin t − cos t + sin2 t = . 0 2 2

− → − → − → (x + y) i + (y − z) j + z 2 k · d R, where C is given by C − → − → − → the vector function R(t) = t 2 i + t 3 j + t 2 k with 0 ≤ t ≤ 1. 348.

Compute

Solution. We have R = (2t, 3t 2 , 2t). Hence, we obtain − → − → − → (x + y) i + (y − z) j + z 2 k · d R C

1

= 0

2 t + t 3 , t 3 − t 2 , (t 2 )2 · (2t, 3t 2 , 2t)dt

304

8 Line Integrals

1

=

5 3 4 4 5 2t + 2t + 3t − 3t + 2t dt =

0

1

5 5t − t 4 + 2t 3 dt

0

5

1 1

1 17 = t6 − t5 + t4 = . 0 6 5 2 15 349.

(x + y)d x + (y + z)dy + (x + z)dz,

Calculate the value of line integral C

where C is the line segment from the origin to the point (1, 2, 4). Solution. The line segment can be represented parametrically by x(t) = t, y(t) = 2t and z(t) = 4t with 0 ≤ t ≤ 1. Then, we obtain x (t) = 1, y (t) = 2 and z (t) = 4. Therefore, we can write the line integral as follows:

(t + 2t) + (2t + 4t)(2) + (t + 4t)(4) dt =

1

0

1 0

35tdt =

35 2

1 35 t = . 2 0 2

350.

yd x + zdy + xdz, where C is the curve of

Compute the value of the line C

intersection of the two surfaces x + y = 2 and x 2 + y 2 + z 2 = 2(x + y). The curve is to be traversed once in a direction that appears clockwise when viewed from the origin. Solution. From the equation x + y = 2 we get y = 2 − x. Putting this into the equation x 2 + y 2 + z 2 = 2(x + y) we get x 2 + (2 − x)2 + z 2 = 4 or equivalently (x − 1)2 +

z2 = 1. 2

(8.2)

√ Now, we parameterize (8.2) by x(t) = cos t + 1 and z(t) = 2 sin t. Also, we obtain y(t) = √ − cos t and 0 ≤ t ≤ 2π. Hence, we have x = − sin t, y (t) = sin t and z (t) = 2 cos t. Next, we calculate the line integral as follows: yd x + zdy + xdz C

=

0

√ √ (−cost)(− sin t) + ( 2 sin t)(sin t) + (cos t + 1)( 2 cos t) dt

2π

=

0

2π

cos t sin t +

√

2+

√ 2 cos t dt

0 √ √ √ 1

= − cos 2t + 2t + 2 sin t = −2 2π. 2π 4

8.3 Solved Problems

305

351.

yd x + zdy + xdz, where C is the curve

Compute the value of the line C

of intersection of the two surfaces z = x y and x 2 + y 2 = 1, traversed once in a direction that appears counterclockwise when viewed from high above the x y-plane. Solution. A parametric equation of the curve of the intersection of the given surfaces is x(t) = cos t, y(t) = sin t and z(t) = (1/2) sin 2t. Then, we obtain x (t) = − sin t, y (t) = cos t and z (t) = (1/4) cos 2t. Therefore, we have yd x + zdy + xdz C

1 1 (sint)(− sin t) + ( sin 2t)(cos t) + (cos t)( cos 2t) dt 2 4 0 2π cos 2t − 1 1 + sin t cos2 t + cos t (1 − 2 sin2 t) dt = 2 4 0

2π 1 1 1 1 1

= sin 2t − t + cos3 t + sin t − sin3 t = −π. 0 4 2 3 4 6

=

2π

− → 352. Calculate the work done by the force field F(x, y, z) = (y − z) i + (z − − → − → x) j + (x − y) k along the curve of intersection of the sphere x 2 + y 2 + z 2 = 4 and the plane z = y tan α, where 0 < θ < π/2. The path is transversed in a direction that appears counterclockwise when viewed from high above the x y-plane. Solution. If we put z = y tan α into the equation of sphere, then we have x 2 + y 2 + y 2 tan2 α = 4. This implies that y2 x2 + = 1. 4 4 cos2 α So, we can consider x = 2 cos θ, y = 2 cos α sin θ and z = 2 sin α sin θ. This means − → − → − → that R(θ) = 2 cos θ i + 2 cos α sin θ j + 2 sin α sin θ k . It follows that R (θ) = − → − → − → −2 sin θ i + 2 cos α cos θ j + 2 sin α cos θ k . Therefore, the work done is equal to 2π W = 2 cos α sin θ − 2 sin α sin θ, 2 sin α sin θ, 2 cos θ − 2 cos α sin θ · 0

− 2 sin θ, 2 cos α cos θ, 2 sin α cos θ dθ

After a simple calculation, we obtain

2π

W = 0

4(sin α − cos α)dθ = 8π(sin α − cos α).

306

8 Line Integrals

353. Find the line integral of the vector field F = g R (A × R), where A is a constant vector and g is a real continuous function, along a straight line segment parallel to A. Solution. For a straight line segment parallel to A, the tangent vector d R is parallel to A. Since the triple product of co-planer vectors vanishes, it follows that F.d R = g R (A × R) · d R = 0. Therefore, we conclude that the line integral of F vanishes for any such straight line segment. 354.

3x yd x + (4x 2 − 3y)dy over the curve C con-

Evaluate the line integral C

sisting of the line y = 2x + 3 from (0, 3) to (3, 9) and then the parabola y = x 2 from (3, 9)to (5, 25). Solution. Fig. 8.1 shows the curve C composed of arcs C1 and C2 . The arc C1 is the line segment. So, C1 can be represented parametrically by x(t) = t and y(t) = 2t + 3 with 0 ≤ t ≤ 3. The arc C2 which is a part of the parabola Fig. 8.1 The curve C composed of arcs C1 and C2

y 25−

C2

y = x2

9− C1 3− 3

y = 2x + 3

5

x

8.3 Solved Problems

307

from (3, 9) to (5, 25) can be represented parametrically by x(t) = t and y(t) = t 2 with 3 ≤ t ≤ 5. Applying the definition for each of arcs C1 and C2 , we have 3x yd x + (4x 2 − 3y)dy = C1

3t (2t + 3) + (4t 2 − 6t − 9)(2) dt

C1

3

=

2 14t − 3t − 18)dt

0

3 14 3 3 2 117

= t − t − 18t = 0 3 2 2 and

3t 3 + 4t 2 − 3t 2 (2t) dt

3x yd x + (4x 2 − 3y)dy = C2

C2

5

=

5t 3 dt =

3

5 4

5 1360 t = . 4 3 2

Therefore, we have 117 1360 1477 3x yd x + (4x 2 − 3y)dy = + = . 2 2 2 C 355.

Compute the value of the line integral C

d x + dy , |x| + |y|

where C is the square with vertices (1, 0), (0, 1), (−1, 0) and (0, −1), traversed once in a counterclockwise direction. Solution. First, we determine the equations of the segment lines AB, BC, C D and D A in Fig. 8.2. By easy computation we get AB : BP : PQ : QA :

y =1−x y = x +1 y = −x − 1 y = x − 1.

We may use x as a parameter in each path. Therefore, we can write C

d x + dy = |x| + |y|

AB

d x + dy + x+y

BP

d x + dy + −x + y

PQ

d x + dy + −x − y

QA

d x + dy x−y

308

8 Line Integrals

Fig. 8.2 Arc C is the square with vertices A, B, P and Q

y B = (0, 1)

x A = (1, 0)

P = (−1, 0)

Q = (0, −1)

−1 1−1 1+1 dx + dx −x + (x + 1) 1 x + (1 − x) 0 0 1 1−1 1+1 + dx + dx −1 −x − (−x − 1) 0 x − (x − 1) 1 −1 2d x + 2d x = −2 + 2 = 0. =

0

=

0

0

− → − → − → 356. Let F(x, y, z) = x z i + 2y j + yz k be a vector field. Let C be a simple closed path that bounds the plane 2x + 2y + z = 4 and coordinate planes. The orientation is counterclockwise as shown in Fig. 8.3. Compute the line integral

F.d R. C

Solution. First, we parameterize the line segment C1 from A to P. The plane x + y = 2 contains C1 . A parametrization of C1 is R1 (t) = (2 − t, t, 0), where 0 ≤ t ≤ 2. Using the definition, we obtain

F.d R1 = C1

2

(0, 2t, 0) · (−1, 1, 0)dy = 4.

0

Now, we parameterize the line segment C2 from P to Q. Since the plane 2y + z = 4 contains C2 , a parametrization of C2 is R2 (t) = (0, (4 − t)/2, t), where 0 ≤ t ≤ 4. In this case, we have

4

F.d R2 = C2

0

0, 4 − t,

4t − t 2 −1 1 · 0, , t dt = 2 2 2

0

4

4 − t 2 + 5t − 4 dt = . 3

8.3 Solved Problems

309

Fig. 8.3 A triangular path in the plane 2x + 2y + z = 4

z

Q = (0, 0, 4)

C2

C3

y P = (0, 2, 0) C1 A = (2, 0, 0) x

Finally, we consider the line segment C3 from Q to A. A parametrization of C3 is R3 (t) = (t, 0, 4 − 2t), where 0 ≤ t ≤ 2. Hence, we get

2

F.d R3 = C3

(4t − 2t 2 , 0, 0) · (1, 0, −2)dt =

0

Therefore, we conclude that F.d R = F.d R1 + C

C1

C2

F.d R2 +

F.d R3 = 4 + C3

8 . 3

4 8 + = 8, 3 3

as desired.

357. Show that the value of the following integral is independent of the path and evaluate it e x sin yd x + e x cos ydy, C

where Cis any sectionaly smooth curve from the point (0, 0) to the point (2, π/2). Solution. Since

∂ x ∂ x e sin y = e cos y = e x cos y, ∂y ∂x

− → − → it follows that e x sin y i + e x cos y j is a gradient, and hence the value of the integral − → − → is independent of the path. If ∇U (x, y) = e x sin y i + e x cos y j , then

310

8 Line Integrals

From (8.3) we get U (x, y) =

∂U (x, y) = e x sin y, ∂x

(8.3)

∂U (x, y) = e x cos y. ∂y

(8.4)

e x sin yd x = e x sin y + g(y). This implies that

∂U (x, y) = e x cos y + g (y). ∂y

(8.5)

From (8.4) and (362.) we conclude that g (y) = 0, which implies that g(y) = c0 , a constant. So, we have U (x, y) = e x sin y + c0 . Consequently, using the fundamental theorem of line integral we can write

π π − U (0, 0) = e2 sin − e0 sin 0 = e2 . e x sin yd x + e x cos ydy = U 2, 2 2 C

− → − → − → 358. Find the work done by F = (x 2 + y 2 ) i + (y 2 + x) j + ze z k over the following paths from A = (1, 0, 0)to B = (1, 0, 1). (1) The line segment x = 1, y = 0 , 0 ≤ z ≤ 1; − → − → − → (2) The helix R(t) = cos t i + sin t j + t/(2π) k , 0 ≤ t ≤ 2π; (3) The x-axis from (1, 0, 0) to (0, 0, 0) followed by the parabola z = x 2 , y = 0from (0, 0, 0) to (1, 0, 0). Solution. Since ∂ ∂ 2 ∂ 2 ∂ (ze z ) = 0 = (y + x), (x + y) = 0 = (ze z ), ∂y ∂z ∂z ∂x ∂ 2 ∂ 2 (y + x) = 1 = (x + y), ∂x ∂y it follows that F is conservative. Thus, there is U such that F = ∇U . Now, we determine U . Let ∂U ∂U ∂U = x 2 + y, = y 2 + x and = ze z . ∂x ∂y ∂z

(8.6)

Then, from the first equation in (8.6) we get U (x, y, z) = (1/3)x 3 + yx + g(x, y), which implies that ∂g ∂U =x+ . (8.7) ∂y ∂y From (8.7) and the second equation in (8.6) we conclude that ∂g/∂ y = y 2 , and so g(y, z) = (1/3)y 3 + h(z). This yields that U (x, y, z) = (1/3)x 3 + yx + (1/3)y 3 +

8.3 Solved Problems

311

h(z). Taking the partial derivative from both sides of this equality relative to z gives that ∂U = h (z). (8.8) ∂z From (8.8) and the third equation in (8.6) we obtain h (z) = ze z , and so h(z) = ze z − e z + c0 . Therefore, we have U (x, y, z) =

1 3 1 x + yx + y 3 + ze z − e z + c0 . 3 3

Since F is conservative, it follows that the line integral does not depend on the path taken from A to B. Therefore, the work done over each path in (1), (2) and (3) is W = U (1, 0, 1) − U (1, 0, 0) = 1. − → − → − → 359. Show that the work done by a constant force field F = a i + b j + c k in −→ moving a particle along any path from A to B is W = F · AB. Solution. Suppose that A = (x1 , y1 , z 1 ) and B = (x2 , y2 , z 2 ). Since all partial derivatives of a, b and c are zero, it follows that the force F is conservative. We find that U (x, y, z) = ax + by + cz + c0 , the potential function. Hence, the work done by the force in moving a particle along any path from A to B is equal to W = U (x2 , y2 , z 2 ) − U (x1 , y1 , z 1 ) = (ax2 + by2 + cz 2 + c0 ) − (ax1 + by1 + cz 1 + c0 ) −→ = a(x2 − x1 ) + b(y2 − y1 ) + c(z 2 − z 1 ) = F · AB. 360.

F(x, y) · d R,where

Find C

F(x, y) = ye x y + cos x, xe x y +

1 y2 + 1

and Cis the portion of the curve y = sin xfrom x = 0to x = π/2. Solution. We find that ∂ xy ∂ xy 1 ye + cos x = xe + 2 = e x y + x ye x y . ∂y ∂x y +1 This shows that F is conservative. Now, we can use the fundamental theorem of line integral. If F(x, y) = ∇U (x, y), by a simple calculation we obtain U (x, y) = e x y + sin x + tan−1 y + c0 . Therefore, we have F(x, y) · d R = U C

π , 1 − U (0, 0) = eπ/2 + . 2 4

π

312

8 Line Integrals

361.

Let Cbe a curve represented by two parametric representations such that C = {R1 (w) | w ∈ [a, b]} = {R2 (t) | t ∈ [c, d]},

where R1 : [a, b] → R3 and R2 : [c, d] → R3 are two distinct one to one differentiable functions. Show that (1) There is a function g : [c, d] → [a, b] such that R2(t) = R1 g(t); F · d R1 = F · d R2 . (2) If R1 and R2 trace out Cin opposite direction, then C C If R1 and R2 trace out Cin same direction, then F · d R1 = − F · d R2 . C

C

Solution. (1) It is enough to consider g(t) = R2 (t) . (2) By the chain rule we have R2 (t) = R1 g(t) g (t). This implies that −1 R1

d

F · d R2 = c

C

F R2 (t) · R2 (t)dt =

d c

F R1 g(t) · R1 g(t) g (t)dt.

Next, changing variable from t to u we get

F · d R2 = C

g(d)

g(c)

F R1 (u) · R1 (u)du

=±

F R1 (u) ·

R1 (u)du

=±

F · d R1 . C

Let f be function of two variables xand y. Prove that the line integral 362. f (x, y)ds along the curve C with polar equation r = r (θ) (α ≤ θ ≤ β) is equal C

to

β

α

dr 2 f r cos θ, r sin θ r 2 + dθ. dθ

Solution. We have x = r cos θ and y = r sin θ. Hence, we obtain dx dr dy dr = cos θ − r sin θ and = sin θ + r cos θ. dθ dθ dθ dθ Since R(θ) = (r cos θ, r sin θ), it follows that

R (θ) = =

dx 2 dθ dr

=

dθ

r2 +

+

dy 2 dθ

cos θ − r sin θ dr 2 dθ

.

2

+

dr dθ

sin θ + r cos θ

2

8.3 Solved Problems

313

Fig. 8.4 The circle has its center at the point (1, 0)

P = (r, θ) • θ O

• 1

2

This completes the proof. 363.

(x − y)ds, where C is the circle x 2 + y 2 =

Use Problem 362. to evaluate C

2x traversed once in the counterclockwise direction. Solution. The equation of the circle can be written by (x − 1)2 + y 2 = 1. So, the circle has its center on the point (1, 0), see Fig. 8.4. If P = (r, θ) is any point on the circle, the angle at P inscribed in the circle is a right angle. Therefore, cos θ = r/2, or equivalently r = 2 cos θ. Now, by using Problem 362. we calculate the integral as follows:

(x − y)ds = C

π/2 −π/2

=4

2 cos2 θ − 2 sin θ cos θ 4 cos2 θ + 4 sin2 θdθ

π/2

−π/2

2 cos θ − sin θ cos θ dθ

1 + cos 2θ − sin θ cos θ dθ 2 −π/2 1 π/2 1 1

= 4 θ + sin 2θ − sin2 θ = 2π. −π/2 2 4 2

=4

364.

π/2

Prove that if F = ( f 1 , f 2 , f 3 ) is conservative, then the potential is

x x0

f 1 (t, y0 , z 0 )dt +

y

f 2 (x0 , t, z 0 )dt +

y0

where (x0 , y0 , z 0 )is any point in the domain of F.

z z0

f 3 (x, y, t)dt,

314

8 Line Integrals

Fig. 8.5 The path C consists of three straight line segments

z

(x0 , y0 , z0 ) C1 C3

C2

(x, y, z)

y

x

Solution. Suppose that the path C consists of three straight line segments; (x0 , y0 , z 0 ) −→ (x, y0 , z 0 ) −→ (x, y, z 0 ) −→ (x, y, z), see Fig. 8.5. We consider parametric equations for C1 , C2 andC3 as follows: C1 : R1 (t) = (t, y0 , z 0 ), x0 ≤ t ≤ x, C2 : R2 (t) = (x, t, z 0 ), y0 ≤ t ≤ y, C3 : R3 (t) = (x, y, t), z 0 ≤ t ≤ z. Hence, we obtain R1 (t) = (1, 0, 0), R2 (t) = (0, 1, 0) and R3 (t) = (0, 0, 1). This implies that F R1 (t) · R1 (t) = f 1 (t, y0 , z 0 ), , F R2 (t) · R2 (t) = f 2 (x, t, z 0 ), F R3 (t) · R3 (t) = f 3 (x, y, t). Therefore, we have

F · d R1 = C1

F · d R2 =

C2

F · d R3 = C3

x

f 1 (t, y0 , z 0 )dt,

x0 y

f 2 (x, t, z 0 )dt,

y0 z

f 3 (x, y, t)dt.

z0

365. A particle moves along the smooth curve y = f (x) from (a, f (a)) to (b, f (b)). The force moving the particle has constant magnitude k and always points away from the origin. Show that the work done by the force is

8.3 Solved Problems

315

F · T ds = k

2 2 b2 + f (b) − a 2 + f (a) .

C

− → − → − → − → Solution. We can write R(x) = x i + f (x) j , and then R (x) = i + f (x) j . The force k − → − → (x i + y j ) F(x, y) = 2 2 x +y has constant magnitude k and points away from the origin. Then, we have k x + f (x) f (x) ky f (x) F(x, y) · R (x) = + = 2 x 2 + y2 x 2 + y2 x 2 + f (x) 2 d =k x 2 + f (x) . dx

kx

Therefore, we obtain

2 d x 2 + f (x) dx a C

2 b 2 2 = k x 2 + f (x) = k b2 + f (b) − a 2 + f (a) ,

F · T ds = C

F · Rd x =

b

k

a

as desired. 366. A force field is given in polar coordinates by the equation F(r, θ) = − → − → −4 sin θ i + 4 sin θ j . Compute the work done in moving a particle from the point (1, 0) to the origin along the spiral whose polar equation is r = e−θ . Solution. Since in polar coordinates we have x = r cos θ and y = r sin θ, it follows that − → − → R(θ) = e−θ cos θ i + e−θ sin θ j , which implies that − − → → R (θ) = −e−θ cos θ − e−θ sin θ i + −e−θ sin θ + e−θ cos θ j . Hence, the work done is ∞ −θ 4e sin θ cos θ + 4e−θ sin2 θ − 4e−θ sin2 θ + 4e−θ sin θ cos θ dθ W = 0

∞

=4

e 0

−θ

sin 2θdθ = 4 lim

u→∞ 0

u

e−θ sin 2θdθ

316

8 Line Integrals

= 4 lim

u→∞

e−θ

= 4 lim

u→∞

e

− sin 2θ − 2 cos 2θ

u

0 5

−u − sin 2u

− 2 cos 2u 2 + 5 5

=

8 . 5

− → 367. Two-dimensional force field F is given by the equation F(x, y) = cx y i + − → x 6 y 2 j , where c is a positive constant. This force acts on a particle which must move from (0, 0) to the line x = 1 along a curve of the form y = ax b , where a > 0 and b > 0. Find a value of a (in terms of c) such that the work done by this force is dependent of b. − → Solution. Since y = ax b , then x can be used as a parameter. Let R(x) = x i + − → ax b j . So, the work done is

1

W =

− → − → − → − → cax b+1 i + a 2 x 2b+6 j · i + abx b−1 j d x

0 1

=

cax b+1 + a 3 bx 3b+5 d x =

0

=

ca b+2 a 3 b 3b+6

1 + x x

0 b+2 3b + 6

ca a3b 3ca + a 3 b + = . b + 2 3(b + 2) 3(b + 2)

Now, if we consider a =

√ 3c/2, then the work done is equal to

W =

2ca + cab 3ca + 3cab/2 = = ca/2, 3(b + 2) 2(b + 2)

as we see, this value of the work done is independent of b. 368. Let F be a conservative force field such that F = −∇U . Suppose a particle of constant mass m to move in this field. If Pand Q are two arbitrary points in space, prove that 1 1 U (P) − mv 2P = U (Q) − mv 2Q , 2 2 where v A and v B are the magnitude of the velocities of the particle at A and B, respectively. Solution. Newton’s second law of motion describes the relationship between an object’s mass and the force needed to accelerate it. Newton’s second law is often stated as F = m A, which means the force F acting on an object is equal to the mass m of the object multiply its acceleration A. We can write F = m A = md 2 R/dt 2 . This implies that

8.3 Solved Problems

317

d R d2 R dR m d =m · 2 = F· dt dt dt 2 dt

dR dt

2 .

Hence, we obtain

Q P

m F · dR = 2

Q

d dt

P

dR dt

2 =

m 2

Q 1 1 v = mv 2Q − mv 2P . P 2 2 2

(8.9)

Since F = −∇U , it follows that

Q

F · dR = −

P

P

∇U · ·d R = − U (Q) − U (P) = U (P) − U (Q). (8.10)

Q

Therefore, by (8.9)and (8.10) we conclude that U (P) − U (Q) =

1 2 1 mv Q − mv 2P , 2 2

and this completes the proof.

369.

A curve C is called closed if its terminal point is the same as initial point. Prove that F · d R is independent of path if and only if F · d R = 0 for any C

C

closed path C.

F · d R is independent of path, and let C be a closed curve.

Solution. Suppose that C

We consider two different points P and Q on C, and split C into two curves C1 and C2 with C1 going from P to Q along one part of C, and C2 going from Q to P along the second part of C. Let R1 and R2 be parametrization of C1 and C2 , respectively. Then, we have F · dR = F · d R1 + F · d R2 = F · d R1 − F · d R2 = 0, C

C1

C2

−C2

C1

because C1 and −C2 have the same initial and terminal points. F · d R = 0, for any closed path C. For any two paths

Conversely, assume that C

C1 and C2 with the same initial and terminal points, the path C = C1 − C2 is a closed path. This gives that

F · dR =

0= C

Therefore, we deduce that

F · d R1 −

C1

F · d R2 . C2

318

8 Line Integrals

F · d R1 = C1

F · d R2 , C2

this means that the line integral is independent of path.

8.4 Exercises Easier Exercises 1. Evaluate the following line integrals: (a) x x 2 − y 2 ds, where C is the right loop of the lemniscate r 2 = 4 cos 2θ C

(−π/4 ≤ θ ≤ π/4); ds (b) , where C is the helical arc x = cos t, y = sin t, z = t (0 ≤ 2 2 2 C x + y +z t ≤ 2π). 2. What is the work done by moving in the force field F(x, y) = 2x 3 + 1, 4π sin(π y 4 )y 3 along the curve y = x 4 from (−1, 1) to (1, 1)? 3. An object moves from A to B in a force field F = k R/R3 , k a constant. What is the work done by F? − → − → − → 4. Let R = x i + y j + z k . The function F(R) = −

mG R, R3

where G is the gravitational constant gives the gravitational force exerted by a unit mass at the origin on a mass m located at R. What is the work done by F if m moves from R1 to R2 ? (10x 4 − 2x y 3 )d x − 3x 2 y 2 dy, where C is the part of the curve x 5 −

5. Evaluate C

5x 2 y 2 − 7x 2 = 0 from (3, −2) to (3, 2). 6. An object traverses the curve y = ax(1 − x) from (0, 0) to (1, 0) subject to the − → − → force F(x, y) = (y 2 + 1) i + (x + y) j . What value of a minimizes the work done by F? 7. Let F(x, y, z) == (e y , xe y + sin z, y cos z). Find the work done by this force field on an object that moves from (0, 0, 0) to (1, −1, 3). 8. Let C be an oriented path from point A = (3, 1, 1) to B = (1, 1, 2) where y = 0. If z xz x ,− 2, − 1 F(x, y, z) = y y y

8.4 Exercises

319

F · d R by the fundamental theorem for line integrals.

is a vector field, evaluate C

9. Find a closed curve C with position vector R(t) for which the vector field F(x; y) = (x y, x 2 ) satisfies F R(t) · R (t)dt = 0. C

10. Give an example of a non-trivial force field F and non-trivial path R(t) for which the total work done moving along the path is zero. − → − → 11. The position of an object with mass m at time t is R(t) = at 2 i + bt 3 j , where 0 ≤ t ≤ 1. (a) What is the force acting on the object at time t? (b) What is the work done by the force during the time interval 0 ≤ t ≤ 1? − → 12. An object with mass m moves with position function R(t) = a sin t i + − → − → b cos t j + ct k , where 0 ≤ t ≤ π/2. Find the work done on the object during this time period. 13. Find the total work done in moving an object along the given arc C if the motion is caused by the given force field. − → − → (a) F(x, y) = −x 2 y i + 2y j ; C: the line segment from (a, 0) to (a, a) and then the line segment from (a, a) to (0, a); − → − → − → (b) F(x, y, z) = (x yz + x) i + (x 2 z + y) j + (x 2 y + z) k ; C: the line segment from the origin to the point (1, 0, 0) and then the line segment from (1, 1, 0) to (1, 1, 1).

Harder Exercises 14. If is a smooth curve C given by a vector function R(t), where a ≤ t ≤ b, show that V · d R = V · R(b) − R(a) . C

15. If C is a smooth curve given by a vector function R(t), where a ≤ t ≤ b, show that 1 R(b)2 − R(a)2 . R · dR = 2 C 16. Prove that a vector field F is conservative on a connected region R if and only if F · d R = 0 for all closed curve C lying in R. C 17. Evaluate y 2 d x + 3x ydy, where C is the boundary of the region between the C

circles x 2 + y 2 = 1 and x 2 + y 2 = 4 in the upper half plane, traversed counterclockwise.

320

8 Line Integrals

18. Let f (x) be continuously differentiable on (−∞, ∞). Show that the vector field − → − → − → − → F(x, y) = f (x 2 + y 2 )(x i + y j ) and G(x, y) = f (x y)(y i + x j ) are both conservative. ∞ F· 19. Let F = R/Rn for any n > 2, and suppose that A = 0. Show that A

d R taken along any path from A not passing through the origin and going out indefinitely depends only on A.Evaluate the integral. ∞

20. What value should be assigned to

F · d R, whereF = R/R2 , taken along

A

a path from A not passing through the origin and going out indefinitely? 21. Suppose that F and G are continuous on a simply connected open region R. Show that F · dR = G · dR C

C

for any smooth closed curve C in R if there is a function f with continuous partial derivatives in R such that F − G = ∇ f . 22. Set y x P(x, y) = 2 and Q(x, y) = − 2 2 x +y x + y2 on the punctured unit disk R = {(x, y) | 0 < x 2 + y 2 < 1}. (a) Verify that P and Q are continuously differentiable on R and that ∂Q ∂P (x, y) = (x, y), for all (x, y) ∈ R. ∂y ∂x − → − → (b) Verify that, in spite of (a), the vector field F(x, y)=P(x, y) i + Q(x, y) j is not a gradient on R. Hint: Integrate F over a circle of radius less than 1 centered at the origin. 23. A 160-lb man carries a 25-lb can of paint up a helical staircase that encircles a silo with a radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions climbing to the top, how much work is done by the man against gravity?

Chapter 9

Multiple Integrals

9.1 Double Integral over Rectangle Region Let f (x, y) be defined on a rectangular region R = {(x, y) | a ≤ x ≤ b and c ≤ y ≤ d}. We subdivide R into small rectangles, using a network of lines parallel to x-axis and y-axis; see Fig. 9.1. These lines divided R into n rectangular pieces, where the number of such pieces n gets large as the width and height of each piece get small. These rectangles form a partition Δ of R. The norm, denoted by Δ, is determined by the length of the largest diagonal of a rectangular subregion of the partition. A small rectangular piece of width Δx and height Δy has area ΔA = ΔxΔy. If we number the small pieces partitioning R in some order, then their area is given by ΔA1 , ΔA2 , . . . , ΔAn , where ΔAi is the area of the ith small rectangle. In order to form a Riemann sum over R, we consider an arbitrary point (xi , yi ) in the ith small rectangle, multiply the value of f at that point by the area ΔAi and add together we have n f (xi , yi )ΔAi . (9.1) Sn = i=1

There are many sums of the form (9.1). We are interested in what happens to these Remanian sums converge as Δ → 0. The resulting limit is written as lim

Δ→0

n

f (xi , yi )ΔAi .

i=1

As Δ → 0 and the rectangle becomes narrow and short, their number n increases. When a limit of the sum Sn exists, giving the same limiting value, no matter what choices are made, then the function f is said to be integrable and the limit is called the double integral of f over R, written as

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_9

321

322

9 Multiple Integrals

Fig. 9.1 Rectangular partitioning the region R into small rectangles

Fig. 9.2 Approximate solid volume

lim

n

Δ→0

f (xi , yi )ΔAi =

i=1

f (x, y)d A. R

Other symbols for the double integral in (9.2) are

f (x, y)d xd y and R

f (x, y)d yd x. R

(9.2)

9.2 Double Integral over Bounded Non-rectangular Region

323

If a function of two variables is continuous on a closed rectangular region R, then it is integrable on R. Fubini’s theorem: If f is continuous throughout the rectangular region R = {(x, y) | a ≤ x ≤ b and c ≤ y ≤ d}, then

d

f (x, y)d A = c

R

b

f (x, y)d xd y =

a

b

a

d

f (x, y)d yd x.

c

9.2 Double Integral over Bounded Non-rectangular Region Now, we direct our attention to integration over regions R in the plane which are between two graphs. If f (x, y) is continuous on the vertically simple region R defined by the inequalities a ≤ x ≤ b and g1 (x) ≤ y ≤ g2 (x), then

b

f (x, y)d A =

g1 (x)

a

R

g2 (x)

f (x, y)d yd x.

If f (x, y) is continuous on the vertically simple region R defined by the inequalities h 1 (y) ≤ x ≤ h 2 (y) and c ≤ y ≤ d, then

d

f (x, y)d A = c

R

h 2 (y) h 1 (y)

f (x, y)d xd y.

Let R be a region which is defined both by the inequalities a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x), and by the inequalities c ≤ y ≤ d, h 1 (y) ≤ x ≤ h 2 (y). If f (x, y) is continuous on R, then

f (x, y)d A = R

a

b

g2 (x) g1 (x)

d

f (x, y)d yd x = c

h 2 (y) h 1 (y)

f (x, y)d xd y.

If A is the area of a closed, bounded plane region R, then A =

d A. R

The average value of an integrable function f over a region R is 1 area of R

f (x, y)d A. R

Let f (x, y) and g(x, y) be integrable functions over a region R. Then

324

9 Multiple Integrals

(1) Linearity property: For every real numbers a and b, we have

a f (x, y) + bg(x, y) d A = a

R

f (x, y)d A + b

R

g(x, y)d A. R

(2) Additive property: If R is divided into two non-overlapping regions R1 and R2 , then f (x, y)d A = f (x, y)d A + f (x, y)d A. R

R1

R2

(3) Comparison theorem: If f (x, y) ≤ g(x, y) for every (x, y) ∈ R, we have

f (x, y)d A ≤ R

g(x, y)d A. R

In particular, if f (x, y) ≥ 0 for every (x, y) ∈ R, then f (x, y)d A ≥ 0. R

Mean value theorem for double integrals: Let f and g be continuous functions on a closed bounded region R. If g is non-negative on R, then there exists a point (ξ1 , ξ2 ) ∈ R for which f (x, y)g(x, y)d xd y = f (ξ1 , ξ2 ) g(x, y)d xd y. R

R

Changes of variables in double integrals: If f (x, y) is continuous on a region R, and if R is the image of a region R in the uv-plane under the transformation x = x(u, v) and y = y(u, v); see Fig. 9.3. Then

f (x, y)d xd y =

R

R

∂(x, y) dudv. f x(u, v), y(u, v) ∂(u, v)

In particular, in polar coordinates, we have

f (x, y)d xd y =

R

R

f r cos θ, r sin θ r dr dθ.

9.3 Triple Integrals

325

Fig. 9.3 Transformation of coordinates

9.3 Triple Integrals We can define triple integrals with a Riemann sum in a way similar to double integrals. Let f (x, y, z) be defined on a closed box D = {(x, y, z) | a ≤ x ≤ b, c ≤ y ≤ d, p ≤ z ≤ q}. Then, we have f (x, y, z)d V = lim

n

Δ→0

S

f (xi , yi , z i )ΔVi

i=1

and the limit is called the triple integral of f over S. Furthermore, we have

f (x, y, z)d V = a

S

b

c

d

q

f (x, y, z)dzdyd x.

p

This integral is also equal to any of the other five possible orderings for the iterated triple integral. Triple Integrals can also be used to represent a volume, in the same way that a double integral can be used to represent an area. That is f (x, y, z)d V = volume of S. S

If f (x, y, z) is continuous on S = {(x, y, z) | a ≤ x ≤ b, g1 (x) ≤ y ≤ g2 (x), h 1 (x, y) ≤ z ≤ h 2 (x, y)},

326

9 Multiple Integrals

then

b

f (x, y, z)d V =

g1 (x)

a

S

g2 (x)

h 2 (x,y) h 1 (x,y)

f (x, y, z)dzdyd x.

The linearity property, additive property and comparison theorem are valid for triple integrals too. Changes of variables in triple integrals: If f (x, y, z) is continuous on a region S, and if S is the image of a region S in the uvw-space under the transformation x = x(u, v, w), y = y(u, v, w) and z = z(u, v, w). Then f (x, y, z)d xd ydz S

= S

∂(x, y, z) dudvdw. f x(u, v, w), y(u, v, w), z(u, v, w) ∂(u, v, w)

In particular, in cylinder coordinates, we have

f (x, y, z)d xd ydz =

f r cos θ, r sin θ, z r dr dθdz.

S

S

Also, in spherical coordinates, we have

f (x, y, z)d xd ydz =

f ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ, cos ϕ ρ2 sin ϕdρdθdϕ.

S

S

9.4 Masses and Moments in Three Dimensions Let δ(x, y, z) be the density of an object occupying a region S in space (mass per unit volume), then the integral over S gives the mass of the object, i.e., δ(x, y, z)d V.

M= S

First moments about the coordinate planes are M yz =

xδ(x, y, z)d V, Mx z =

S

Center of mass is (x, y, z), where

yδ(x, y, z)d V and Mx y =

S

zδ(x, y, z)d V. S

9.5 Solved Problems

327

x=

M yz Mx y Mx z , y= and z = . M M M

Moments of inertia about the coordinate axes (second moments) are Ix =

(y 2 + z 2 )δ(x, y, z)d V, I y =

S

(x 2 + z 2 )δ(x, y, z)d V S

Iz =

(x 2 + y 2 )δ(x, y, z)d V. S

Moment of inertia about a line is I =

d 2 (x, y, z)δ(x, y, z)d V, S

where d(x, y, z) is the distance from the point (x, y, z) to the line .

9.5 Solved Problems 370. Let R = [a, b] × [c, d] and f be defined by f (x, y) = g(x)h(x), where g : [a, b] → R and h : [a, b] → R are continuous. Show that

b

f (x, y)d xd y =

g(x)d x

a

R

d

h(y)dy .

c

Solution. By Fubini’s theorem, we have c

d

b

g(x)h(y)d x dy =

a

d

c

=

h(y) b

b

g(x)d x dy

a

g(x)d x

a

d

h(y)dy .

c

371. If f (x, y) = esin(x+y) and R = [−π, π] × [−π, π], show that 1 1 ≤ e 4π 2

esin(x+y) d A ≤ e. R

Solution. Since −1 ≤ sin(x + y) ≤ 1, it follows that e−1 ≤ esin(x+y) ≤ e. So, we have 1 sin(x+y) e dA ≤ ed A, dA ≤ e R

R

R

328

9 Multiple Integrals

or equivalently 1 e

π −π

This implies that

π −π

d xd y ≤

1 2 4π ≤ e

esin(x+y) d A ≤ e

R

π

π

d xd y. −π

−π

esin(x+y) d A ≤ e 4π 2 ,

R

as desired. (1 + x − y)d A, where R = {(x, y) | |x −

372. Compute the double integral y| < 2/3, 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}.

R

Solution. We can write (1 + x − y)d A = dA + (x − y)d A. R

R

(9.3)

R

The region R is symmetric with respect to the line y = x, whereas the function f (x, y) = x − y changes sign under the reflection with respect to this line, i.e., f (x, y) = − f (y, x). This yields that the second integral in (9.3) is zero. On the other hand, the second integral in (9.3) is the area of R, and so we obtain

(1 + x − y)d A = R

d A = 12 −

1 2 3

=

8 . 9

R

373. Let R be a region between two squares with the center in the origin and the e x+y d A.

length of sides 2 and 4, respectively. Evaluate R

Solution. We consider R = R1 ∪ R2 ∪ R3 ∪ R4 as shown in Fig. 9.4. Then, we have

9.5 Solved Problems

329

Fig. 9.4 The region of integration in Example 373

e R

=

x+y

−1

dA =

2

e

x+y

dA +

R1

1

e

x+y

dA +

e

R22

R3

1

x+y

dA +

e x+y d A R4

−1

e x+y d yd x + e x+y d yd x + e x+y d yd x −1 1 −1 −2 2 2 e x+y d yd x + 1 −2 2 1 2 −1 ex d x e y dy + ex d x e y dy = −2 −1 2−2 2 −1 1 1 ex d x e y dy + ex d x e y dy + −2

−2

1

−2

−1

−2

= (e−1 − e−2 )(e2 − e−2 ) + (e − e−1 )(e2 − e) + (e2 − e)(e2 − e−2 ) +(e − e−1 )(e−1 − e−2 ) = e4 + e−4 − e2 − e−2 . 374. Find by double integral the area enclosed by the ellipse x 2 /a 2 + y 2 /b2 = 1. Solution. Look at Fig. 9.5. By the symmetry of the area, we can write A = 4

d A, R

where R is the region bounded in the first quarter. So, we find that

b√1−x 2 /a 2

x2 4b a 2 b 1 − 2 dx = a − x 2d x a a 0 0 0 0 4b x 2 x a a2 4b a 2 π sin−1 = 4ab. = a − x2 + = a 2 2 a 0 a 2 2

A=

a

d yd x = 4

a

330

9 Multiple Integrals

Fig. 9.5 The area of an ellipse

(x + 2y)d A, where R is the region bounded

375. Find the value of the integral R

by the parabolas y = 1 + x 2 and y = 2x 2 . Solution. The parabolas intersect at (−1, 2) and (1, 2). The region is shown in Fig. 9.6. We can write R = {(x, y) | − 1 ≤ x ≤ 1, 2x 2 ≤ y ≤ +x 2 }. Therefore, we have

(x + 2y)d A = R

−1

=

1

1+x 2

2x 2

(x + 2y)d yd x =

1 −1

1+x 2 (x y + y 2 ) 2 d x 2x

32 . −3x 4 − x 3 + 2x 2 + x + 1 d x = 15 −1 1

376. Find the value of the integral

xd A, where R is the region outside of the R

diamond and inside the circle in Fig. 9.7. Solution. First, we determine the equations of line segments AB, BC, C D and D A. We obtain AB : x + y = 2 BC : x − y = −2 C D : x + y = −2 D A : x − y = 2.

9.5 Solved Problems

331

Fig. 9.6 The region bounded by the parabolas y = 1 + x 2 and y = 2x 2

Fig. 9.7 The region outside of the diamond and inside the circle x 2 + y 2 = 4

So, we can write

xd A =

R

0

2

√4−y 2

xd xdy + 0

2−y

+

0 −2

−y−2

√

−

4−y 2

2

y−2

√

−

xd xdy 4−y 2

xd xdy +

0

√4−y 2 xd xdy.

−2

y−2

Finally, after a simple computation of the above iterative integrals, the sum of the results is zero.

332

9 Multiple Integrals

Fig. 9.8 The region of integration in Example 377

377. By reversing the order of integration, write the sum

3

0

x

6

f (x, y)d yd x +

0

3

6−x

f (x, y)d yd x

0

as just one iterated integral. Solution. The region of integration divides into two regions as shown in Fig. 9.8. So, the sum of given integrals is equal to 0

3

3

f (x, y)d xd y +

y

0

3

6−y

f (x, y)d xd y =

3

0

3

6−y

f (x, y)d xd y.

y

378. Derive the formula 0

a

y

a

em(a−x) f (x)d xd y =

0

(a − x)em(a−x) f (x)d x,

0

where a and m are constant with a > 0. Solution. The region of integration is shown in Fig. 9.9. We reverse the order of integration. Then, we have a a a y em(a−x) f (x)d xd y = em(a−x) f (x)d yd x 0 0 0 x a a = yem(a−x) f (x) d x x 0 a m(a−x) = (a − x)e f (x)d x. 0

9.5 Solved Problems

333

Fig. 9.9 The region of integration in Example 378

379. Which of the double integrals

x + 6x y + y 4

2 2

4

d A and

R

4x 3 y + 4x y 3 d A

R

is large? Solution. Since (x − y)4 ≥ 0, it follows that x 4 − 4x 3 y + 6x 2 y 2 − 4x y 3 + y 4 ≥ 0. This yields that x 4 + 6x 2 y 2 + y 4 ≥ 4x 3 y + 4x y 3 . Therefore, we conclude that

x 4 + 6x 2 y 2 + y 4 d A ≥

R

3 4x y + 4x y 3 d A.

R

380. Evaluate the integral

sin x d A, where R is the triangle {(x, y) | 0 ≤ x ≤ x

R

π, 0 ≤ y ≤ x}.

Solution. First, we try to evaluate the integral with respect to x. This yields that R

sin x dA = x

π 0

π y

sin x d xd y, x

but we cannot determine an indefinite integral sin x/x. Hence, we try doing the integration with respect to y. Then, we obtain

334

9 Multiple Integrals

sin x dA = x

π

0

x 0

sin x d yd x = x

π

0

sin x x y dx 0 x

π = − cos x = 1 − (−1) = 2.

R

0

y 2 e−x d A, where R = {(x, y) | 0 ≤ y ≤ x}. 2

381. Compute the integral R

Solution. We have 2 −x 2 y e dA =

∞

x

2 −x 2

y e 0

R

d yd x =

0 ∞

∞

0 ∞

1 3

2 x y 3 e−x d x 0

1 1 2 x 3 e−x d x = ue−u du 3 0 6 0 b b

b 1 1 ue−u du = lim − ue−u + e−u du = lim 0 6 b→∞ 0 6 b→∞ 0

1 −b −b = lim − be − e + 1 = 1. 6 b→∞

=

382. Evaluate the integral I = 1

2

ln x

(x − 1) 1 − e2y d yd x.

0

Solution. We change the order of integration. Since R = {(x, y) | 0 ≤ y ≤ ln x, 1 ≤ x ≤ 2} = {(x, y) | e y ≤ x ≤ 2, 0 ≤ y ≤ ln 2}, we can write ln 2 2y ln 2 2

e 2y − e y 1 + e2y dy (x − 1) 1 − e d xd y = − I = 2 0 ey ln 2 0 1 ln 2 2y =− e e y 1 + e2y dy. 1 + e2y dy + 2 0 0 To solve the first integral, we take u = e2y , and for the second integral, we put u = e y . Thus, we have 2 √ 1 + u 2 du 1 + udu + 1 1

4 1 2 1 u 1 + u 2 + ln 1 + u 2 + u = − (1 + u)3/2 + 1 6

11 2√

1 √

√ √ 1 3/2 2 − 53/2 + 2 5 + ln( 5 + 2) − 2 + ln( 2 + 1) = 6 2 2 √

1 √ √ 5 + 2 1 3/2 2 − 53/2 + = . 2 5 − 2 + ln √ 6 2 2+1

1 I =− 4

4

9.5 Solved Problems

335

∞

x

xe−x

383. Evaluate the integral 0

tion. Solution. We can write ∞ 0

x

2

/y

dyd x by changing the order of integra-

0

xe−x

2

/y

∞

d yd x =

0

0

∞

xe−x

2

/y

d xd y.

y

Now, we obtain 0

∞

∞

xe−x

2

/y

d x dy a→∞ 0 a b→∞ y

b y 2 lim − e−x /y dy = lim a→∞ 0 b→∞ y 2 a y y −y

−b2 /y lim − e = lim dy + e a→∞ 0 b→∞ 2 2 a a 1 1 = lim ye−y dy = lim −ye−y − e−y 0 2 a→∞ 0 2 a→∞ 1 1 = lim −ae−a − e−a + 1 = . 2 a→∞ 2

d xd y = lim

y

1

a

b

xe−x

2

/y

1

384. Evaluate the integral

x max(x, y)d yd x. 0

0

Solution. We can write 1 1 x max(x, y)d yd x = 0

lim

0

1

0

x

1

x 2 d yd x +

0

1

x ydyd x 0

x

1 − x2 dx 2 0 3 1 1 1 1 − = . = + 4 2 2 4 8

1

=

385. Evaluate the double integral

x3 + x

|y − x 2 |d A, where R = [−1, 1] × [0, 2]. R

2

Solution. First, we compute the integral

|y − x 2 |dy. We can write

0

2 0

2 |y − x |dy = 0

x2

x2

− ydy +

2 x2

y − x 2 dy.

336

9 Multiple Integrals

For the first integral we use the change of variable u = x 2 − y and for the second integral we consider u = y − x 2 . This yields that

2

|y − x 2 |dy = −

0

√

x2

0

2−x 2

udu + 0

√ 2 2 udu = x 3 + (2 − x 2 )3/2 . 3 3

Next, we compute the double integral as follows:

|y − x 2 |d A =

R

2

−1

0

1

=

1

−1

4 = 3

1

|y − x 2 |d yd x

2 3 2 2 3/2 dx x + (2 − x ) 3 3 3/2 2 − x2 dx

0

=

x 1 3/2 1 x 2 − x2 + 3x 2 − x 2 + 6 sin−1 √ 0 3 2

=

4 π + . 3 2

386. Let S be the solid bounded by the cylinder x 2 + y 2 = 1 and the plane y + z = 1 and z = 0. Find the volume of S. Solution. Let R = {(x, y) | x 2 + y 2 ≤ 1}. Then, the solid S lies above the region R and below the graph z = 1 − y. The volume of S is equal to

(1 − y)d yd x =

V = R

=

1

−1

1

−1

√

1−x 2

(1 √ − 1−x 2

− y)d yd x

1 2 1 − x 2 d x = x 1 − x 2 + sin−1 x

= sin−1 (1) − sin−1 (−1) =

−1

π π + = π. 2 2

387. Find the volume of the solid which is common to the cylinder x 2 + y 2 = 1 and x 2 + z 2 = 1. 2 2 Solution. The solid is √ enclosed by the cylinder x + y ≤ 1 and the surfaces z = √ − 1 − x 2 and z = 1 − x 2 . Let R = {(x, y) | x 2 + y 2 ≤ 1}. Then, the required volume is

9.5 Solved Problems

V =

R

=

1 −1

337

1−

x2

− −

1−

x2

d yd x =

1

−1

√

1−x 2

√ − 1−x 2

2 1 − x 2 d yd x

1

16 . 2(1 − x ) + 2(1 − x ) d x = 4 (1 − x 2 )d x = 3 −1 2

2

388. Find the volume of the solid S enclosed by the paraboloid z = 2 + x 2 + (y − 2)2 and the planes z = 1, x = −1, x = 1, y = 0 and y = 4. Solution. Suppose that R = [−1, 1] × [0, 4]. Then, the volume of the solid enclosed by the z = 2 + x 2 + (y − 2)2 and x y-plane is V1 =

zd A. R

Let V2 be the volume of the solid box S enclosed by z = 0 and z = 1. Since R is a 2 × 4 rectangle and the height of the box is 1, it follows that the volume of the box is V2 = 8. Therefore, the requested volume is equal to V = V1 − V2 =

1

−1

4

2 + x 2 + (y − 2)2 d yd x − 8

0

88 64 40

dx − 8 = −8= . 4x 2 + = 3 3 3 −1 1

[x + y]d A, where

389. Evaluate the integral R

R = {(x, y) | 1 ≤ x ≤ 3, 2 ≤ y ≤ 5}. 5

Solution. Suppose that R =

Ri , where

i=1

Ri = {(x, y) | i + 2 ≤ x + y < i + 3, 1 ≤ x ≤ 3, 2 ≤ y ≤ 5}, see Fig. 9.10. Then, we have [x + y]d A =

5

[x + y]d A =

i=1 R i

R

=

5 i=1

dA = Ri

(i + 2)d A

i=1 R i

(i + 2)

5

5 (i + 2)A(Ri ) i=1

338

9 Multiple Integrals

Fig. 9.10 The region of integration in Example 389

= 3A(R1 ) + 4 A(R2 ) + 5A(R3 ) + 6A(R4 ) + 7A(R5 ) 3

3

1

1

+4 + 5(2) + 6 +7 = 30. =3 3 2 2 2 390. Find the average value of the function f (x, y) = 4x on the region R between the parabolas y = 3x − x 2 and y = x 2 − 2. Solution. The region R is shown in Fig. 9.11. The x-bounds are determined by the intersection of two parabolas y = 3x − x 2 and y = x 2 − 2. If we set x 2 − 2 = 3x − x 2 , then we get x = 2. This implies that −1/2 ≤ x ≤ 2. The parabola y = x 2 − 2 is lower and y = 3x − x 2 is upper, and hence, we deduce that the y-bounds are x 2 − 2 ≤ y ≤ 3x − x 2 . Now, we compute the area of the region. We have A(R) =

2

−1/2

3x−x 2 x 2 −2

d yd x =

2

−1/2

(3x − 2x 2 + 2)d x =

125 . 24

Finally, we determine the average value of the function f (x, y) = 4x as follows: f ave

24 = 125

2 −1/2

3x−x 2 x 2 −2

24 4xdyd x = 125

2

−1/2

4x(3x − 2x 2 + 2)d x = 3.

1

391. Find the average value of the function f (x) = [0, 1].

x

cos(u 2 )du on the interval

9.5 Solved Problems

339

Fig. 9.11 The region R between the parabolas y = 3x − x 2 and y = x2 − 2

Fig. 9.12 A rectangular pool with its sub-rectangles

Solution. By the definition of the average of a function of one variable, we have f ave

1 = b−a

b

f (x)d x.

a

Therefore, we obtain

1

f ave =

0

0

1 x

1

=

cos(u 2 )dud x = 0

1

u

cos(u 2 )d xdu

0

1 1 1 u cos(u 2 )du = sin(u 2 ) = sin 1. 0 2 2

392. A 15 ft by 30 ft rectangular pool is filled with water. The depth is measured at the center of each sub-rectangle shown in Fig. 9.12. Each measurement is also shown on Fig. 9.12. Estimate the average depth of the water in the pool. Solution. The area of the pool is 15 × 30 = 450 square ft. The average value of the depth function f (x, y) can be computed as

340

9 Multiple Integrals

f ave =

1 450

f (x, y)d A, R

and the integral is equal to the volume of the pool and can be approximated using sums. The volume is V =

15 30 7 + 6 + 9 + 10 + 5 + 7 + 6 + 8 + 3 + 4 + 2 + 4 = 2775. 3 4

393. What region R in the x y-plane maximizes the value of

9 − x 2 − 2y 2 d A?

R

Solution. In order to maximize the integral, we need to determine the domain that includes all points where the integrand is positive andexcludes all points where the integrand is negative. These criteria are met by the points (x, y) such that 9 − x 2 − 2y 2 ≥ 0 or equivalently x 2 + 2y 2 ≤ 9, which is the ellipse x 2 + 2y 2 = 9 together with its interior. 394. What region R in the x y-plane minimizes the value of

x 2 + y 2 − 25 d A?

R

Solution. In order to minimize the integral, we need to determine the domain that includes all points where the integrand is negative and excludes all points where the integrand is positive. These criteria are met by the points (x, y) such that x 2 + y 2 − 25 ≤ 0 or equivalently x 2 + y 2 ≤ 25, which is the closed disk of radius 5 centered at the origin. 395. If R = [0, 1] × [0, 1], express the double integral I = cosh(x y)d xd y as R

a convergent series. Solution. We have 1 I = 0

1 0

cosh(x y)d xd y = R

Now, using the Maclaurin series sinh y =

1 sinh(x y) 1 sinh y dy. dy = 0 y y 0 ∞ n=0

y 2n+1 we obtain (2n + 1)!

9.5 Solved Problems

I =

341

1

0

=

1 ∞ ∞ 1 y 2n+1 y 2n dy = dy y n=0 (2n + 1)! (2n + 1)! 0 n=0

1 y 2n+1 1 . = (2n + 1)!(2n + 1) 0 n=0 (2n + 1)!(2n + 1)

∞

∞

n=0

1

1

1

1

1 d xdy is an improper integral and could be 0 0 1 − xy defined as the limit of double integrals over the rectangle [0, b] × [0, b] as b → 1− . But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that 396. The double integral

0

∞

1 1 . d xd y = 1 − xy n2 n=1

0

Solution. Since |x y| < 1, by the formula for the sum of geometric series, we have ∞

1 = (x y)n . 1 − xy n=0 Hence, we can write 0

1

0

1

1 d xd y = 1 − xy =

1

(x y)n =

0

0

n=0

1

y dy

=

0

1 = (n + 1)2

∞

(x y)n d xd y

0

n

x dx

1

1

n

0

n=0

0

n=0

=

n=0

∞ 1

∞

∞

∞ 1

∞ n=0

1 . n2

n=1

397. Let f be a continuous function on [a, b]. Prove that

2

b

f (x)d x

≤ (b − a)

a

b

2 f (x) d x.

a

Solution. We consider the iterated integral a

b

a

b

2 f (x) − f (y) d yd x ≥ 0.

1 1 n+1 n+1

342

9 Multiple Integrals

Then, we obtain 0≤ =

b b a

a

b b a

a

b b

2 f (x) d yd x +

a

a

b b

2

f (x) d yd x +

a

a

2 f (y) d yd x − 2 2

f (y) d xd y − 2

b b a

b

f (x) f (y)d yd x

a

f (x)d x

a

a

b

f (y)dy

b b b b 2 b 2 b = y f (x) d x + x f (y) dy − 2 f (x)d x f (x)d x a

a

= (b − a)

= (b − a)

b a

b a

= 2(b − a)

a

a

2

f (x) d x + (b − a) 2

f (x) d x + (b − a)

b a

2

f (x) d x − 2

a

b

a

2

f (y) dy − 2

a

b

2

f (x) d x − 2

a

b

2 f (x)d x

a

b

2 f (x)d x

a

2

b

f (x)d x

a

,

and this completes the proof. 398. Suppose that f (x) is an increasing function on a ≤ x ≤ b. Prove that

b

(b + a)

f (x)d x ≤ 2

a

b

x f (x)d x.

a

Solution. Since f (x) is increasing, it follows that (x − y) f (x) − f (y) ≥ 0, and so (x − y) f (x) − f (y) d xd y ≥ 0, where the integrand is taken over a ≤ x ≤ b R

and a ≤ y ≤ b. Now, we have 0≤

b

b

a

=

b

(x − y) f (x) − f (y) d yd x

a

a

b

a

−

=

b

a b

b

x f (x)d yd x −

a b

a

− a

a b

b

x f (y)d yd x

a

y f (x)d yd x +

(b − a)x f (x)d x −

a

b

b

a b

b

y f (y)d yd x

a

x f (y)d xd y

a

1 2 (b − a 2 ) f (x)d x + 2

a

b

a

b

y f (y)d xd y

9.5 Solved Problems

343

b

b

1 2 (b − a 2 ) f (y)dy 2 a a b b 1 2 2 (b − a ) f (x)d x + − (b − a)y f (y)dy a 2 a b b = 2(b − a) x f (x)d x − (b2 − a 2 ) f (x)d x,

= (b − a)

x f (x)d x −

a

a

and this completes the proof. 399. Use a double integral to show that if f and g are continuous functions on [a, b], then b b b 2 f (x)g(x)d x ≤ f (x)d x g 2 (x)d x. a

a

a

This is Cauchy-Schwarts inequality for integrals. Solution. We consider the double integral

2 f (x)g(y) − f (y)g(x) d xd y ≥ 0,

R

where R is the square bounded by the lines x = a, x = b, y = a and y = b. So, we have b b

f 2 (x)g 2 (y) + f 2 (y)g 2 (x) − 2 f (x)g(x) f (y)g(y) d xd y 0≤ a

a

b

=

a

b

f 2 (x)g 2 (y)d xd y +

a

−2

=

b

b

a

a

b

b

−2

a

f (x)g(x)d x

a

=2

b

b

f 2 (y)dy

a b

b

g 2 (x)d x

a

f (y)g(y)dy

a

f (x)d x 2

a

f 2 (y)g 2 (x)d xd y

g 2 (y)dy +

a b

b

f (x)g(x) f (y)g(y)d xd y

a

f 2 (x)d x

a

b

b

g (x)d x − 2 2

a

This implies Cauchy-Schwarts inequality.

a

b

2 f (x)g(x)d x

.

344

9 Multiple Integrals

400. Show that if the function h is positive and continuous on [a, b], then

b

b

h(x)d x a

√

Solution. Let f (x) =

a

1 d x ≥ b − a. h(x)

h(x) and g(x) =

1 . By Cauchy-Schwarts inequality, h(x)

we obtain b b b 1 1 dx ≤ d x, b−a = h(x) h(x)d x a h(x) a a h(x) as desired.

401. Evaluate the integral I =

2

−1 tan πx − tan−1 x d x by using the theory of

0

double integrals.

Solution. We write the integrand as an integral. So, we have I =

2

0

πx x

1 d yd x 1 + y2

2π 2 1 1 = d xd y + d xd y 2 1 + y 1 + y2 0 y/π 2 y/π 2 2π 1 − 1/π 2 − y/π = dy + dy 2 1+y 1 + y2 0 2 π − 1

2π 2 1 ln(1 + y 2 ) + 2 tan−1 y + ln(1 + y 2 ) = 0 2 2π 2π ln 5 1 ln 1 + 4π 2 + . = 2 tan−1 2π − 2 tan−1 2 − 2π 2 ∞ ax e − e−bx 402. Evaluate the integral I = d x by using the theory of double x 0 integrals. 2

y

Solution. We can write the given integral as follows: 0

∞

eax − e−bx dx = x

∞ 0

b

e−x y d yd x.

a

The domain of integration is the infinite strip {(x, y) | 0 ≤ x ≤ ∞, a ≤ y ≤ b}. Changing the order of integration, we obtain

9.5 Solved Problems

∞

I =

b

e

0

=

b

−x y

a

b

b

d yd x =

−1 y

e

−x y

∞

e a

lim

=

a

t→∞

a

345 −x y

0

t dy = 0

b

d xd y =

lim

lim

−1

t→∞

t→∞ 0

a

a

b

y

e

−t y

t

e

−x y

d x dy

1

dy + y

1 b dy = ln b − ln a = ln . y a

403. Suppose that f (x) and g(x) are continuous for x ≥ 0. Their convolution is the function x

( f g)(x) =

f (t)g(x − t)dt.

0

The Laplace transformation of f (x) is the function L f (s) =

∞

e−sx f (x)d x,

0

where convergent. Assuming everything in sight converges and changing the order of integration is valid. Prove that L f g (s) = L f (s) · L g (s). Solution. We have ( f g)(x) =

∞

e 0

x

f (t)

0

f (t)e−st

0 ∞

=

f (t)g(x − t)dt d x

f (t)e−st

0

∞

e t

∞

=

0

∞

=

−sx

−sx

∞

=

e

−sx

g(x − t)d x dt

∞

e−s(x−t) g(x − t)d x dt

∞

e−sy g(y)dy dt

t

0

∞

f (x)d x

0

e

−sy

f (y)dy

= L f (s) · L g (s).

0

404. Let R = [0, 1] × [3, 4] and suppose that f : R → R is defined by f (x, y) =

1 i f x is rational 0 i f x is irrational.

Prove that the double integral of f over R does not exist. Solution. Suppose that P is any partition of R into sub-rectangles Ri , 1 ≤ i ≤ n. If we consider any rectangle Ri , then there is a point (x, y) in Ri with x rational and

346

9 Multiple Integrals

there is another point (x , y ) in Ri with x irrational. Hence, we have m i = infimum of f in Ri = 0, Mi = supremum of f in Ri = 1. This shows that the sum of values of outer rectangular parallelepipeds is 1, and the sum of volumes of inner rectangular parallelepipeds is 0. Therefore, the double integral of f does not exist. 405. Let R = [−1, 1] × [−1, 1] and suppose that f : R → R is defined by f (x, y) =

1

1

Show that −1

−1

y i f x is rational 0 i f x is irrational.

f (x, y)dyd x exists, and the other repeated integral is not defined.

Solution. We have

1

−1

f (x, y)dy =

⎧ ⎪ ⎨ ⎪ ⎩

1

ydy if x is rational −1

0

if x is irrational.

1 1 f (x, y)dy = 0, and so f (x, y)d yd x = 0. Now, we show −1 −1 −1 1 1 the repeated integral f (x, y)d xd y does not exist. For this, we fix y, say 1

This yields that

−1

−1

y = 1, and define the function g on [−1, 1] by g(x) =

1 if x is rational 0 if x is irrational,

1

and we know that such a function is not integrable. Thus,

g(x)d x does not exist,

−1

and consequently, we cannot even talk about the repeated integral 406. Suppose that A = integral

1

e−u du and B = 2

0

1/2

1 −1/2

0

x

e−y d yd x 2

1

1

−1 −1

f (x, y)d xd y .

e−u du. Evaluate the iterated

0

I =2

2

9.5 Solved Problems

347

in terms of A and B. There are positive integers m and n such that I = m A − n B + e−1 − e−1/4 . Solution. Note that 1 x −1/2

e−y d yd x = 2

0

0

−1/2

x

e−y d yd x + 2

0

0

1

x

e−y d yd x. 2

0

So, we can write 1 I = 2

0

−1/2 0

−1/2

e−y d xd y +

y

2

0

1

1

e−y d xd y 2

y 1

1 2 2 − y e−y dy + (1 − y)e−y dy 2 −1/2 0 1

1 1 1 0 −y 2 2 0 2 2 1 −y e e−y e dy + + e−y dy + =− −1/2 0 2 −1/2 2 2 0 1 1 −1/4 1 1 −1 1 =− B+ − e + A+ e − 2 2 2 2 2 1 −1/4 1 −1 1 + A+ e . =− B− e 2 2 2 =

−

This gives that I = 2 A − B + e−1 − e−1/4 . 407. Show that π π 1 (1) dx = , y > 1; 2 y − cos x y −1 0

√ π b + b2 − 1 b − cos x (2) d x = π ln ln , a > 1 and b > 1. √ a − cos x a + a2 − 1 0 Solution. (1) If u = tan(x/2), then d x = 2du/(1 + u 2 ) and cos x = (1 − u 2 )/(1 + u 2 ). So, we obtain 2 1 1 + u 2 du = 2 I = du. y + 1 + (y − 1)u 2 1 − u2 y+ 1 + u2 √ √ 2 2 Now, let √ (y − 1)u2 = (y +√1) tan θ. This implies that y + 1 tan θ = y − 1u, and so y + 1 sec θdθ = y − 1du. This gives that

1 dx = y − cos x

348

9 Multiple Integrals

1 2 y + 1 sec2 θdθ 2 y + 1 + (y + 1) tan θ y−1 √ sec2 θ 2 y+1 dθ √ 1 + sec2 θ y − 1(y + 1) 2 2 θ+C dθ = √ √ 2 y−1 y+1 y −1

2 y−1 −1 tan u +C y+1 y2 − 1

y − 1 x 2 tan−1 tan + C. 2 y+1 y2 − 1

I =√ = = =

=

π

π 1 . dx = 2 y − cos x y −1 0 (2) Taking the integral with respect to y from both sides of the equality in (1) from a to b, we get b π b 1 π d xd y = dy. (9.4) 2 y −1 a 0 y − cos x a Therefore, we conclude that

Next, we compute the both integrals in (9.4). We obtain

b

a

π 0

1 d xd y = y − cos x

π

b

1 d yd x = 0 a y − cos x π b − cos x dx = ln a − cos x 0

π

b ln(y − cos x) d x a

0

(9.5)

and a

b

π y2 − 1

dy = π ln(y +

y2

b − 1) = π ln a

b+ a+

√

b2 − 1

. √ a2 − 1

(9.6)

Finally, from (9.4)–(9.6), the result follows. 408. Show that π 2π 1 dx = , y > 1; (1) y2 − 1 −π y + sin x π 9

5 + 3 sin x d x = 2π ln . (2) ln 5 + 4 sin x 8 −π Solution. (1) Let −π < x < π. We consider the change of variable u = tan x/2. Then, d x = 2du/(1 + u 2 ) and sin x = 2u/(1 + u 2 ). This gives that

9.5 Solved Problems

349

2 2 1 + u 2 du = du 2u y(1 + u 2 ) + 2u y+ 1 + u2 2 2 1 1 = =

y 2 − 1 2 1 2 1

2 y y 1 u+ + 1− 2 u+ + y y y y ⎞ ⎛

1

y u+ ⎜ y tan(x/2) + 1 2 y 2 y ⎟ −1 ⎜ −1 ⎟ tan ⎝ tan = = . y y2 − 1 y2 − 1 ⎠ y2 − 1 y2 − 1

1 dx = y + sin x

Therefore, we obtain

π

−π

2 1 dx = 2 y + sin x y −1 =

0

1 d x + lim t1 →π y + sin x

lim

t0 →−π

π

2 y2

−1

2

+

t0

π

2

=

2π y2 − 1

t1 0

1 dx y + sin x

.

(2) Taking the integral from both sides of the equality in part (1) with respect to y, we get b π b 1 2π dy. d xd y = y2 − 1 a −π y + sin x a Then, we have

2π y2 − 1

2π ln y +

y2

b − 1 = 2π ln

a

b+ a+

√

b2 − 1

. √ a2 − 1

(9.7)

Now, using Fubini’s theorem, we can write a

b

π

−π

1 d xd y = y + sin x

π

−π

a

b

1 d yd x = y + sin x

π

b + sin x ln a + sin x −π

From (9.7) and (9.8), we conclude that

π

b + sin x ln a + sin x −π

d x = 2π ln

b+ a+

√

b2 − 1

. √ a2 − 1

Finally, if we take a = 5/4 and b = 5/3, then we have

d x. (9.8)

350

9 Multiple Integrals

⎞ 5 25 + sin x π ⎜ 3 + 9 − 1⎟ ⎟ ⎜ ⎟. d x = 2π ln ⎜ ln ⎝ 3 ⎠ ⎝ ⎠ 5 −π 5 25 + sin x + − 1 4 4 16

This yields that

⎛

⎞

⎛5

π

4 5 + 3 sin x ln 3 5 + 4 sin x −π

d x = 2π ln

3

2

,

and so

π

ln −π

5 + 3 sin x 5 + 4 sin x

4 9

3 d x = 2π ln − ln = 2π ln , 2 3 8

as claimed. 409. Show that π/2 cos−1 y 1 (1) dx = , (0 ≤ y < 1); 1 + y cos x 1 − y2 0 (2) If 0 ≤ a < 1 and 0 ≤ b < 1, then

π/2 0

(3) 0

π/2

1 + b cos x sec x ln 1 + a cos x

dx =

1 −1 2 −1 2

cos a − cos b ; 2

5π 2 1 . sec x ln 1 + cos x d x = 2 72

Solution. (1) Taking u = tan x/2, we have d x = 2du/(1 + u 2 ) and cos x = (1 − u 2 )/(1 + u 2 ). Hence, we can write 2 2 1 + u 2 du = du 2 2 1 + u + y − yu 2 1−u 1+y 1 + u2 2 2 1 = du = du 1+y (1 − y)u 2 + (1 − y) 1−y u2 + 1−y √ √ √ 2 x 1−y 1−y 1−y 2 −1 −1 . tan u = tan tan = √ √ √ 1−y 1+y 2 1+y 1+y 1 − y2

1 dx = 1 + y cos x

Consequently, we obtain 0

π/2

2 1 dx = tan−1 1 + y cos x 1 − y2

√ 1−y . √ 1+y

(9.9)

9.5 Solved Problems

351

Fig. 9.13 A right √ triangle 1−y with tan θ = √ 1+y

Look at Fig. 9.13. We have √ √ 1−y 1+y and cos θ = √ tan θ = √ . 1+y 2 Since cos 2θ = 2 cos2 θ − 1 = y, it follows that 2θ = cos−1 y. This yields that tan

−1

√ cos−1 y 1−y . = √ 2 1+y

(9.10)

Now, from (9.9) and (9.10), the result follows. (2) Taking the integral from both sides of the equality in part (1) with respect to y, we obtain b b π/2 1 cos−1 y d xd y = dy. 1 + y cos x 1 − y2 a 0 a Then, using Fubini’s theorem, we have π/2 b 1 1 d xd y = d yd x 1 + y cos x a 0 0 a 1 + y cos x π/2 π/2 b 1 + b cos x = d x. sec x ln(1 + y cos x) d x = sec x ln a 1 + a cos x 0 0

b

π/2

On the other hand, we have a

b

cos−1 y 1 −1 2 −1 2

cos a − cos b , dy = 2 1 − y2

and this gives the results. (3) It is enough to put a = 0 and b = 1/2 in part (2). 410. (Mean value theorem for double integrals). Let f be a continuous function on a closed bounded region R. Prove that there exists (ξ1 , ξ2 ) ∈ R such that

352

9 Multiple Integrals

f (x, y)d A = f (ξ1 , ξ2 )A(R), R

where A(R) is the area of R. Solution. Since f is continuous on R, it has a maximum value M = f ( p, q) and a minimum value m = f ( p , q ), for some ( p, q), ( p , q ) ∈ R. Hence, we have m ≤ f (x, y) ≤ M, for all (x, y) ∈ R. This yields that

md A ≤

m A(R) = R

f (x, y)d A ≤

R

and so m≤

1 A(R)

Md A = M A(R), R

f (x, y)d A ≤ M. R

Since f is a continuous function on R, it follows that there is a point (ξ1 , ξ2 ) ∈ R such that 1 f (x, y)d A = f (ξ1 , ξ2 ), A(R) R

and this completes the proof. 411. Let f be a continuous function R = [a, b] × [c, d]. For a ≤ x ≤ b and x on y f (u, v)dvdu. Show that c ≤ y ≤ d, we define F(x, y) = a

c

∂2 F ∂2 F (x, y) = (x, y) = f (x, y). ∂x∂ y ∂ y∂x Use this fact to discuss the relationship between Fubini’s theorem and the equality of mixed partial derivatives. Solution. By the fundamental theorem of calculus, we have ∂F (x, y) = ∂x

y

f (x, v)dv,

c

and applying it once again, we get ∂2 F (x, y) = f (x, y). ∂ y∂x In the reverse order, we first apply Fubini’s theorem and then the fundamental theorem of calculus twice. Then, we have

9.5 Solved Problems

353

y

F(x, y) = c

x

f (u, v)dudv,

a

which implies that ∂F (x, y) = ∂y

x

∂2 F (x, y) = f (x, y). ∂x∂ y

f (u, y)du and

a

Fubini’s theorem is, in a sense, the integral version of the theorem on the equality of mixed partial derivatives. If ∂ F/∂x, ∂ F/∂ y, ∂ 2 F/(∂x∂ y) and ∂ 2 F/(∂ y∂x) are all continuous, then Fubini’s theorem and the fundamental theorem of calculus imply that x y 2 y x 2 ∂ F ∂ F (u, v)dvdu = (u, v)dudv a c ∂x∂ y c a ∂x∂ y y ∂F ∂F (x, v) − (a, v) dv = F(x, y) − F(x, c) − F(a, y) + F(a, c). = ∂y ∂y c

x

y

∂2 F (u, v)dvdu gives the same ∂ y∂x

y

∂2 F (u, v)dvdu. ∂ y∂x

A similar calculation of the iterated integral a

result. Consequently, we have

x

a

This yields that

c

y

∂2 F (u, v)dvdu = ∂x∂ y R

∂2 F dA = ∂x∂ y

c

x

a

c

∂2 F dA ∂ y∂x

R

for all rectangles R. Since R is arbitrary, it follows that ∂ 2 F/(∂x∂ y) = ∂ 2 F/(∂ y∂x). This shows how one may deduce the equality of mixed partial derivatives assuming Fubini’s theorem has been proved. 412. Suppose that f has continuous second partial derivatives on the region R = ∂2 f [0, 1] × [0, 1]. Determine (x, y)d A if f (0, 0) = 6, f (0, 1) = 3, f (1, 0) = ∂x∂ y 2 and f (1, 1) = 4.

R

354

9 Multiple Integrals

Solution. We have 1 1 1 1

∂2 f ∂ ∂ f ∂ (x, y)d A = (x, y) d xd y = (x, y) dy 0 ∂x∂ y ∂x ∂ y ∂ y 0 0 0 R

1

= 0

1 d f (1, y) − f (0, y) dy = f (1, y) − f (0, y) 0 dy

= f (1, 1) − f (0, 1) − f (1, 0) + f (0, 0) = 5. 413. Evaluate I =

∞

e−x d x by using the theory of double integrals. 2

−∞

Solution. We can write ∞ −x 2 2 e dx I = =

−∞

∞

e−y

2

−∞

∞ −∞

∞

e

−x 2

dx

−∞

e−x d xd y = 2

=

∞

e

−∞

∞

−x 2

dx

e

−∞ ∞

∞

−y 2

dy

−∞

e−(x

2

+y 2 )

d xd y.

−∞

Now, we transform to polar coordinates x = r cos θ and y = r sin θ. The region of integration is 0 ≤ r ≤ ∞ and 0 ≤ θ ≤ 2π. Therefore, we obtain I2 = =

2π ∞ 0

2π

0

2 r e−r dr dθ =

2π

b lim

b→∞ 0

0

2 r e−r dr

dθ

2π 2 b 2 1 2π 1 1 1 dθ = − e−r dθ = lim − e−b + dθ = π. 0 2 2 2 2 0 b→∞ 0 b→∞ lim

0

So, we conclude that I =

√

π.

414. Evaluate the integral I = 0

Solution. Suppose that u = 413, we obtain

∞

I = 0

√

∞

e−x √ d x. x

x or u 2 = x, and then 2udu = d x. Now, by Problem

e−u (2udu) = 2 u

∞

√ 2 e−u du = 2 π.

0

415. Using Problem 413, compute the integral 0

∞

e−x

2

x 2 + 1/2

2 d x.

9.5 Solved Problems

355

Solution. We apply the integration by parts in the indefinite integral. Then, we have

e−x

2

e−x − x(2x 2 + 1) 2

x 2 + 1/2

2 d x = −

xe−x +2 = 2 x + 1/2 2

e−x dx x2 2

e−x d x. 2

Next, by Problem 413 we deduce that

∞ 0

e−x

2

x 2 + 1/2

∞

416. Suppose that the integral

2 d x = 2

∞

e−x d x = 2

√

π.

0

f (x)d x converges. Prove that

0 ∞

e−t x f (x)d x converges for any t > 0; (1) the integral 0 ∞ ∞ −t x (2) lim e f (x)d x = f (x)d x. t→∞ 0

0

Solution. (1) For any positive numbers a < b, by the second mean value theorem, there is c ∈ [a, b] such that

b

e

−t x

f (x)d x = e

−ax

a

c

f (x)d x.

a

b c f (x)d x is small, and so the integral e−t x f (x)d x If a is sufficiently large, then a a ∞ ∞ f (x)d x is convergent, by the above fact it follows that e−t x is small. Since 0

f (x)d x is convergent. (b) We represent the difference between the integrals in the form

∞

0

0

∞

f (x)d x −

e−t x f (x)d x

0 a

0

1 − e−t x f (x)d x +

a

∞

f (x)d x −

∞

e−t x f (x)d x.

a

If a is sufficiently large, then the last two integrals are small, for any t > 0. The first integral can be made small when t → 0 for a fixed a.

356

9 Multiple Integrals

∞

417. Compute the integral 0

sin x √ d x. x

Solution. By Problem 416, we can write 0

∞

sin x √ d x = lim+ t→0 x

∞

1 √ (sin x)e−t x d x. x

0

Using Problem 413, it is easy to see that 1 2 √ =√ π x

∞

e−x y dy. 2

0

So, we have

∞

0

1 2 √ (sin x)e−t x d x = √ π x 2 =√ π 2 =√ π

∞

e

0 ∞ 0

0

∞ 0

−t x

∞

sin x

e

−x y 2

dy d x

0

∞

d x dy (t+y 2 )x sin x

e

1 dy. 1 + (t + y 2 )2

This gives that

∞ 0

sin x √ d x = lim+ t→0 x

∞ 0

1 2 √ (sin x)e−t x d x = √ π x

∞

0

1 dy. 1 + y4

∞ 1 Suppose that I = dy. By the substitution y = 1/u, we can write 2I = 1 + y4 0 ∞ 1 dy. Then, we have 4 −∞ 1 + y 4I = =

∞

u2 −

√

2u + 1 du √ √ − 2u + 1)(u 2 + 2u + 1) ∞ 1 1 2π du = du = √ , √ √ 2 2 u − 2u + 1 2/2) + 1/2 2 −∞ (u +

−∞ (u 2 ∞ −∞

√ and so we conclude that I = π/(2 2). Consequently, we have 0

∞

sin x π 2 √ dx = √ √ = π 2 2 x

π . 2

9.5 Solved Problems

357

418. A rectangular plate of constant density δ(x, y) = 1 occupies the region bounded by the lines x = 4 and y = 2 in the first quadrant. The moment of inertia Ia of 4

the rectangle about the line y = a is given by the integral Ia =

2

0

Find the value of a that minimizes Ia .

(y − a)2 d yd x.

0

Solution. Suppose that f (a) = Ia . Then, we obtain

4

f (a) =

(2 − a)3 a3 + 3 3

0

4 (2 − a)3 + a 3 . 3

dx =

So, we obtain f (a) = −4(2 − a)2 + 4a 2 . Now, if f (a) = 0, then a = 1. Since f (1) = 16 > 0, it follows that the value a = 1 gives a minimum value of Ia . 419. Let R be the plane region bounded by the line x = a and x = b, the x-axis and the curve y = f (x) (a ≤ x ≤ b), where f is continuous and non-negative on [a, b]. Show that the centroid of R is the point with coordinates 1 x= A

b

where A =

b

a

1 x f (x)d x and y = 2A

b

2 f (x) d x,

a

f (x)d x is the area of R.

a

Solution. First, we try to write

xd A and R

write

a

R

a

f (x)

f (x)

ydyd x =

0

dA =

R

b

x f (x)d x,

a

Similarly, we have A =

xdyd x =

0 b

yd A = R

R

b

xd A =

yd A as iterated integrals. So, we can

a

b

f (x)

1 2

b

2 f (x) d x.

a

b

d yd x =

0

f (x)d x. Therefore, by

a

using the above results, we obtain

xd A

R x =

= dA

R

1 A

yd A b

R x f (x)d x and y =

a

= dA

R

1 2A

a

b

2 f (x) d x.

358

9 Multiple Integrals

420. Let be the line through the region with inclination θ, and let I be the moment of inertia about of a plane region R. Show that I = Ix cos2 θ − 2J sinθ cos θ + I y sin2 θ in terms of the moment of inertia Ix , I y and the quantity J =

x yd A, R

known as the product of inertia. You may suppose δ(x, y) = 1. Solution. The distance between and the point (x, y) is |x tan θ − y| . √ 1 + tan2 θ So, by the definition, we have

I =

d 2 (x, y)d A = R

=

R

x 2 tan2 θ − 2x y tan θ + y 2 dA 1 + tan2 θ

x 2 sin2 θ − 2x y sin θ cos θ + y 2 cos2 θ d A

R

= sin θ

x d A − 2 sin θ cos θ

R

x yd A + cos θ

2

2

2

R

y2d A R

= I y sin2 θ − 2 sin θ cos θ J + cos2 θ Ix . ln(x 2 + y 2 )d xd y,

421. Evaluate by polar coordinates the double integral I = R

where R is the region bounded by the circles x 2 + y 2 = 4 and x 2 + y 2 = e; see Fig. 9.14. Solution. We have 2π e r ln r 2 dr dθ = I = 0

2

2π

=2 0

2π

e

2r ln r dr dθ 0

2

2π 1 2 1 2 1 2 1 e r ln r − r 2 dθ = 2 e − e − 2 ln 2 + 1 dθ 2 2 4 2 4 0

= π(e2 − 8 ln 2 + 4). 422. Determine whether the improper double integral R = [0, ∞) × [0, ∞), is convergent or divergent.

x2 R

1 d xdy, where + y2 + 1

9.5 Solved Problems

359

Fig. 9.14 The region bounded by the circles x 2 + y 2 = 4 and x 2 + y2 = e

Solution. Suppose that Rn = R ∩ {(x, y) | x 2 + y 2 ≤ n}. Using polar coordinates, we get Rn

1 d xd y = x 2 + y2 + 1 =

n

0

π 2

π/2 0

r dθdr = r2 + 1

π/2

dθ 0

0

n

r dr r2 + 1

1 π n ln(r 2 + 1) = ln(n 2 + 1). 0 2 4

Next, we have 1 1 π d xd y = lim d xd y = lim ln(n 2 + 1) = ∞, n→∞ n→∞ 4 x 2 + y2 + 1 x 2 + y2 + 1 R

Rn

and so the given integral is divergent. 423. Let p any number and R = (−∞, ∞) × (−∞, ∞). Prove that the double integral 1 d xd y (9.11) 2 (1 + x + y 2 ) p R

converges if p > 1 and diverges if p ≤ 1. Solution. Let Rn = R ∩ {(x, y) | x 2 + y 2 ≤ n}. Suppose that p = 1. Using polar coordinates, we obtain

360

9 Multiple Integrals

In = Rn

=

1 d xd y = (1 + x 2 + y 2 ) p

2π

0

(1 + r 2 )1− p 2(1 − p)

n dθ = 0

2π

0

n

0

r dr dθ (1 + r 2 ) p

π (1 + n 2 )1− p − 1 . 1− p

If p < 1, then lim In = ∞, and so the integral (9.11) is divergent. n→∞

If p > 1, then lim In = π/( p − 1), and this yields that the integral (9.11) is n→∞ convergent. If p = 1, then In = Rn

1 d xd y = 1 + x 2 + y2

2π 0

n 0

r dr dθ = π ln(1 + n 2 ). 1 + r2

This implies that lim In = lim π ln(1 + n 2 ) = ∞. So, if p = 1, then the integral n→∞ n→∞ (9.11) is divergent. 424. Determine whether the improper integral R

sin2 (x − y) d xd y, 1 − x 2 − y2

(9.12)

where R = {(x, y) | x 2 + y 2 ≤ 1}, is convergent. Solution. Since 0 ≤ sin2 (x − y) ≤ 1, it follows that sin2 (x − y) 1 ≤ . 2 2 1−x −y 1 − x 2 − y2

By Problem 423, since p = 1/2, we know that R

1 1 − x 2 − y2

d xd y is conver-

gent. So, by the comparison test, we conclude that the integral (9.12) is convergent. 425. If p is any positive and R = {(x, y) | x 2 + y 2 ≤ 1}, evaluate the double integral R

1 d xd y. (1 − x 2 − y 2 ) p

(9.13)

Solution. The integrand is not defined on the circle x 2 + y 2 = 1. Hence, we compute the integral on R = {(x, y) | x 2 + y 2 ≤ (1 − )2n }. We apply polar coordinates.

9.5 Solved Problems

361

If p = 1, then we have 2π 1− r r I = dr dθ = dr dθ 2 p (1 − r ) (1 − r 2 ) p 0 0 R 2π (1 − r )1− p 1− π = 1 − (2 − 2 )1− p . dθ = −2(1 − p) 0 1− p 0

If p = 1, then

2π 1− r r I = dr dθ = dr dθ 2 1−r 1 − r2 0 0 R 2π 1 1− = − ln(1 − r 2 ) dθ = π ln(2 − 2 ). 0 2 0 Therefore, we conclude that the integral (9.13) is divergent if p ≥ 1, and it is convergent if p < 1. In this case, the value of the integral is equal to π/(1 − π). 426. Compute the double integral sin θd A, where R is the region outside the R

circle r = 1 and inside the cardioid r = 1 + cos θ; see Fig. 9.15. Solution. Since the region is symmetric, we can restrict the integration to the first quadrant and multiply the result by 2. So, we have I =2

π/2

0

π/2

= 0

1+cos θ 1

π/2

(sin θ)r dr dθ = 0

1+cos θ (sin θ)r 2 dθ 1

−(1 + cos θ)3 4 π/2 2 + cos θ = . sin θ(1 + cos θ) − sin θ dθ = 0 3 3

427. Find the area lying inside the circle r = a sin θ and outside the cardioid r = a(1 − cos θ) (see Fig. 9.16), using the double integration. Solution. We consider r between the equations of two curves. Letting sin θ = 1 − cos θ we obtain (sin θ + cos θ)2 = 1, or equivalently sin 2θ = 0. This implies that θ = 0 or θ = π/2. Therefore, the required area is equal to

362

9 Multiple Integrals

Fig. 9.15 The region outside the circle r = 1 and inside the cardioid r = 1 + cos θ

Fig. 9.16 The region inside the circle r = a sin θ and outside the cardioid r = a(1 − cos θ)

r 2 a sin θ dθ 2 a(1−cos θ) 0 a(1−cos θ) 0

1 π/2 2 2 = a sin θ − (1 − cos θ)2 dθ 2 0

a 2 π/2 2 sin θ − 1 − cos2 θ + 2 cos θ dθ = 2 0

π

a 2 π/2 . − 2 cos2 θ + 2 cos θ dθ = a 2 1 − = 2 0 4 π/2

a sin θ

r dr dθ =

π/2

9.5 Solved Problems

363

Fig. 9.17 The region of integration in Example 428

428. Compute the double integral I =

1

√ 2/2

x

√

1−x 2

1 x2

+ y2

d yd x.

Solution. The region of integration consists of all points in the first quadrant above the circle x 2 + y 2 = 1 and under y = x; see Fig. 9.17. Using polar transformation, we obtain π/4 sec θ π/4 sec θ 1 r dr dθ = dr dθ I = r 0 1 0 1

π/4 √ π = ln | sec θ + tan θ| − θ = ln( 2 + 1) − . 0 4 429. Compute the double integral I = 1 ≤ x 2 + y 2 ≤ 4 and −x ≤ y ≤ x.

tan−1

y

d A, where R is the region x

R

Solution. To solve this problem, we use polar coordinates. We rewrite the function in terms of r and θ. Since x = r cos θ and y = r sin θ, it follows that tan−1

y

x

= tan−1

r sin θ

r cos θ

= tan−1 (tan θ) = θ.

Now, using the line segments to determine the bounds of r and θ, we observe that 1 ≤ r ≤ 2 and −π/4 ≤ θ ≤ π/4. Hence, the requested integral is equal to

364

9 Multiple Integrals

I =

π/4

−π/4

2

θr dr dθ =

π/4

θdθ

r dr

−π/4

1

2 1

π/4 1 2 3 π 2 π2

= 0. = θ2 r 2 = − 2 −π/4 2 1 4 16 16 1

430. Use a suitable transformation to evaluate the double integral I =

(x − y)2 sin(x + y)d xd y, R

where R is the parallelogram with vertices A = (π, 0), B = (2π, π, C = (π, 2π) and D = (0, π). Solution. The equations of the line segments are AB : x − y = π, C D : x − y = −π, BC : x + y = 3π, D A : x + y = π. The expressions x − y and x + y appear in the equations of the lines bounded R, and this suggests that we introduce the new variables u = x + y and v = x − y.

(9.14)

It follows that the curves forming the bounded of R correspond to the values u = π, u = 3π, v = −π and v = π, as shown in Fig. 9.18. Solving the system of equations (9.14) for x and y in terms of u and v, we get x=

u−v u+v and y = , 2 2

∂x ∂(x, y) ∂u = ∂(u, v) ∂ y ∂u

so that

1 2 = 1 2

∂x ∂v ∂y ∂v

1 1 2 =− . 2 1 − 2

Therefore, we have I =

π −π

π

3π

1 v sin ududv = 2

π

v dv 2

−π

3π

sin udu π

=

1 3 π . 3

9.5 Solved Problems

365

Fig. 9.18 The region of R in Problem 430

Fig. 9.19 The region of R in Problem 431

431. Using suitable transformation, show that

1

1−x

e 0

y x+y

0

d xd y =

1 (e − 1). 2

Solution. We use the transformation x + y = u and y = uv. Then, we get x + uv = u, or equivalently x = u(1 − v). It is easy to check that ∂(x, y)/∂(u, v) = u. The region of integration is shown in Fig. 9.19. Now, we obtain

1

1−x

e 0

y x+y

1

d xd y =

0

0

=

1 0

uev dudv =

1

1

udu 0

1 1 2 1 1 ev = (e − 1). u 0 2 0 2

0

ev dv

366

9 Multiple Integrals

Fig. 9.20 The region bounded by parabolas x y = 2, x y = 4 and the parabolas y 2 = x, y 2 = 3x

x2 d A, where R is the region bounded y4

432. Compute the double integral I = R

by parabolas x y = 2, x y = 4 and the parabolas y 2 = x, y 2 = 3x; see Fig. 9.20. Solution. We introduce the new variables u = x y and v = y 2 /x, then the curves forming the boundary of R correspond to the value u = 2, u = 4, v = 1 and v = 3. Moreover, we have ∂u ∂u y x 2 ∂y ∂(u, v) ∂x 3y = = 3v. = y2 = 2y ∂(x, y) ∂v x ∂v − x x ∂x ∂y This implies that ∂(x, y)/∂(u, v) = 1/(3v). Therefore, we obtain 1 I = 3

4 2

3

1

1 1 dvdu = 3 v 3

4

du 2

1

3

1 8 . dv = 3 v 27

433. Establish the equation

2

f (x y)d A = ln 2 R

f (u)du,

1

by introducing a suitable change of variables, where R is the region in the first quadrant bounded by the curves x y = 1, x y = 2, y = x and y = 4x. Solution. Taking u = x y and v = y/x, we observe that the curves forming the bounded of R correspond to the values u = 1, u = 2, v = 1 and v = 4. Since

9.5 Solved Problems

367

∂u ∂(u, v) ∂x = ∂(x, y) ∂v ∂x

y = ∂v − y x2 ∂y

x 2y 1 = x = 2v, x

∂u ∂y

it follows that ∂(x, y)/∂(u, v) = 1/(2v). Therefore, we have

4

f (x y)d A = 1

R

=

1

2

f (u) dudv = 2v

4

1 dv 2v

1

2

f (u)du

1

2 4 2 1 f (u)du = ln 2 f (u)du. ln v 1 2 1 1

434. Evaluate the double integral I =

2

0

(y+4)/2

y 3 (2x − y)e(2x−y) d xd y. 2

y/2

Solution. We introduce the new variables x = u + v/2 and y = v. Then, the Jacobian is ∂x ∂x 1 1 ∂(x, y) ∂u ∂v 2 = 1. = = ∂(u, v) ∂ y ∂y 0 1 ∂u ∂v The boundaries of the region in uv-plane are u = 0, u = 2, v = 0 and v = 2. Hence, we get

2

I = 0

=

2

2

2

v 3 (2u)e4u dudv =

0

v 3 dv

0

1 4u 2 2 1 4 2 v e = e16 − 1. 0 4 0 4

2

2

2ue4u du 0

435. Evaluate the double integral I = tical region defined by x 2 + y 2 /3 ≤ 1.

cos(3x 2 + y 2 )d A, where R is the ellipR

Solution. We use the change of variables x = r cos θ and y = is equal to ∂x ∂(x, y) ∂r = ∂(r, θ) ∂y ∂r

∂x ∂θ ∂y ∂θ

cos θ = √ 3 sin θ

√

3r sin θ. The Jacobian

−r sin θ √ = 3r. √ 3r cos θ

368

9 Multiple Integrals

Fig. 9.21 The region of the integration in Problem 436

So, we have

2π

1

√ cos(3r )( 3r )dr dθ =

2π

√

π sin(3) 3 sin(3)dθ = √ . 6 3

2

0

0

0

436. Compute the double integral I =

cos

y − x

R

y+x

d A, where R is the trape-

zoid region with vertices (1, 0), (2, 0), (0, 1) and (0, 2). Solution. We introduce the new variables u = x + y and v = y − x. It is easy to check that ∂(x, y)/∂(u, v) = 1/2. The region of the integration is shown in Fig. 9.21. The integration region R is bounded by the lines x + y = 1, x + y = 2, x = 0 and y = 0. The boundaries of integration region in uv-plane are u = 1, u = 2, v = −u and v = u. Therefore, we obtain I = 1

2

u

−u

v

1 1 cos dvdu = 2 u 2 2

= sin(1) 1

udu =

2

u sin 1

v u du u −u

3 sin(1). 2

437. Find √ the area of the region in the first quadrant that is bounded by the curve √ 4 x/a + 4 y/b = 1, where a and b are positive numbers.

9.5 Solved Problems

369

Solution. We introduce the new variables x = ar cos8 θ and y = br sin8 θ. The Jacobian of this transformation is ∂x ∂x a cos8 θ b sin8 θ ∂(x, y) ∂r ∂θ = = 7 7 ∂(r, θ) ∂ y ∂ y −8ar sin θ cos θ 8br cos θ sin θ ∂r ∂θ

= 8abr cos9 θ sin7 θ + cos7 θ sin9 θ = 8abr cos7 θ sin7 θ. Now, the required area is equal to

dA =

π/2

0

R

1

|8abr cos7 θ sin7 θ|dr dθ

0

ab π/2 7 sin 2θdθ 32 0 0 0 ab 1 ab ab π/2 (1 − cos2 2θ)3 sin 2θdθ = (1 − u 2 )3 du = = . 32 0 64 −1 70 π/2

= 8ab

1

cos7 θ sin7 θ

r dr

=

438. Let R be the region bounded by x = 0, x = 1, y = x and y = x + 1. Show that 1 2 1 1 d xd y = √ dx . x xy − x2 0 R

Solution. We consider the new variables u = x and v = y − x, which implies that y = u + v. In this case, the Jacobian is equal to 1. Therefore, we can write R

1 xy − x2

d xd y =

1 1 0

1 1 1 1 1 √ dudv = √ √ dudv uv u v 0 0 0

1 1 1 1 1 1 1 1 = √ du √ dv = √ dx √ dx , x x u v 0 0 0 0

as desired.

439. Find the value of a such that I = 0

1

4−a−x 2

0

Solution. First, we compute the integral. We obtain

a

4−x 2 −y

dzdyd x =

4 . 15

370

9 Multiple Integrals

I =

1 4−a−x 2 0

0

(4 − x 2 − y − a)d yd x =

1

1

(4 − x 2 − a)y −

0

1 2 4−a−x 2 y dx 0 2

1 1 1 (4 − a − x 2 )2 − (4 − a − x 2 )2 d x = (4 − a − x 2 )2 d x 2 2 0 0

1 1 1 2 1 1 (4 − a)2 x − x 3 (4 − a) + x 5 = (4 − a)2 − 2x 2 (4 − a) + x 4 d x = 0 2 0 2 3 5

1 1 2 (4 − a)2 − (4 − a) + . = 2 3 5 =

Now, we set

1

4 1 2 (4 − a)2 − (4 − a) + = . 2 3 5 15

It follows that 3(4 − a)2 − 2(4 − a) − 1 = 0. This gives that a = 3 or a = 13/3. 1 1 y f (x, y, z)d xdydz, express an iterative inte440. For the triple integral √ 0

z

0

gral in which the integration is performed in the order dzdyd x. Solution. The domain of integration is √ S = {(x, y, z) | 0 ≤ z ≤ 1, z ≤ y ≤ 1, 0 ≤ x ≤ y} = {(x, y, z) | 0 ≤ z ≤ y 2 , 0 ≤ x ≤ y ≤ 1}. In this region, x takes all values between 0 and 1. For each fixed x between 0 and 1, (y, z) takes all values in Rx = {(y, z) | 0 ≤ z ≤ y 2 , x ≤ y ≤ 1}. Figure 9.22 is a sketch of Rx . Hence, we can write

1 0

1 √

z

y

1

f (x, y, z)d xd ydz =

0

0

1

1−z

1−z

441. For the triple integral 0

0

1 x

y2

f (x, y, z)dzdyd x.

0

f (x, y, z)d xdydz, express an iterative

0

integral in which the integration is performed in the order dzdyd x. Solution. The domain of integration is S = {(x, y, z) | 0 ≤ z ≤ 1, 0 ≤ y ≤ 1 − z, 0 ≤ x ≤ 1 − z} = {(x, y, z) | x, y, z ≥ 0, x + z ≤ 1, y + z ≤ 1}.

9.5 Solved Problems

371

Fig. 9.22 A sketch of R x in Problem 440

Fig. 9.23 A sketch of R x in Problem 441

In this region, x takes all values between 0 and 1. For each fixed x between 0 and 1, (y, z) takes all values in Rx = {(y, z) | y, z ≥ 0, z ≤ 1 − x, y + z ≤ 1}. Figure 9.23 is a sketch of Rx . Looking at this figure, for each fixed y between 0 and x, we observe that z runs from 0 to 1 − x, and for each fixed y between x and 1, we find z runs from 0 to 1 − y. Therefore, we deduce that the integral in the new order is 1 1 1−y 1 x 1−x f (x, y, z)dzdyd x + f (x, y, z)d xd ydz. 0

0

0

a

0

z

442. Express 0

0

0

y

x

0

f (x)d xdydz as a simple integral.

372

9 Multiple Integrals

z

y

Solution. First, we consider the innermost double integral 0

f (x)d xd y. This

0

is over a triangle with x from 0 to y, and y from 0 to z. If we change the order of integration, then we have z 0

y

f (x)d xd y =

0

z 0

z x

z

f (x)d yd x =

f (x)(z − x)d x.

0

Now, applying this again, we obtain a z y 0

0

0

f (x)d xd ydz = =

a z 0

0

f (x)(z − x)d xdz =

a a 0

x

f (x)(z − x)dzd x

a a 1 1 a f (x)(z − x)2 d x = f (x)(a − x)2 d x. x 2 0 0 2

443. What domain S in space minimizes the value of the integral (2x 2 + 3y 2 + z 2 − 4)d V ? S

Give reason for your answer. Solution. In order to minimize the integral, we need the domain to include all the points where the integrand is negative and to exclude all points where it is positive. These conditions imply that 2x 2 + 3y 2 + z 2 − 4 ≤ 0 or 2x 2 + 3y 2 + z 2 ≤ 4. This is a solid ellipsoid centered at the origin. 444. What domain S in space maximizes the value of the integral (3 − x 2 − y 2 − z 2 )d V ? S

Give reason for your answer. Solution. In order to maximize the integral, we need the domain to include all the points where the integrand is positive and to exclude all points where it is negative. 2 2 2 2 2 2 These conditions imply √ that 3 − x − y − z ≥ 0 or 3 ≥ x + y + z . This is a solid sphere of radius 3 with center (0, 0, 0). 445. Find the volume of the parallelepiped defined by 0 ≤ 3x + 2y + z ≤ 4, 2 ≤ x + 3y ≤ 5 and −2 ≤ x − z ≤ 3. Solution. We introduce the new variables u = 3x + 2y + z, v = x + 3y and w = x − z. Then, we get

9.5 Solved Problems

373

∂u ∂x ∂(u, v, w) ∂v = ∂(x, y, z) ∂x ∂w ∂x

∂u ∂y

∂u ∂z

∂v ∂y

∂v ∂z

3 2 1 = 1 3 0 = −10. 1 0 −1

∂w ∂w ∂ y ∂z

This implies that the absolute value of the Jacobian is 1/10. Now, it is easy to compute the volume of the solid. We obtain

V =

dV =

4

0

S

5 2

3

−2

1 dwdvdu = 6. 10

2 y2 z2

x 446. Evaluate the triple integral I = d V , where S is the solid + + a2 b2 c2 S

x2 y2 z2 bounded by the ellipsoid 2 + 2 + 2 = 1. a b c Solution. Let x = au, y = bv and z = cw. Then, it is easy to check that the Jacobian is equal to abc. So, we have I =

abc(u 2 + v 2 + w 2 )dudvdw, S

where S = {u, v, w) | u 2 + v 2 + w 2 ≤ 1}. Now, using spherical coordinates, we obtain 2π π 1 ρ2 ρ2 sin ϕdρdϕdθ I = abc 0

0

0

2π

= abc

dθ 0

π

sin ϕdϕ

0

1

4 ρ dρ = πabc. 5 4

0

−1 447. Compute the triple integral I = dV . 1 + (x 2 + y 2 + z 2 )3 R3

Solution. Using spherical coordinates, we obtain

b

2π

π

1 ρ2 sin ϕdϕdθdρ 6 1 + ρ 0 0 b 2π π ρ2 = lim dρ dθ sin ϕdϕ 6 b→∞ 0 1+ρ 0 0

I = lim

b→∞ 0

374

9 Multiple Integrals

4π lim = 3 b→∞

b 3 1 4π 2π 2 −1 lim . du = tan u = 0 1 + u2 3 b→∞ 3

b3

0

448. Compute the triple integral I =

√ 1−x 2

1

√ − 1−x 2

−1

√

1+

√

1−

1−x 2 −y 2

1−x 2 −y 2

x 2 + y2 + z2

5/2

dzdyd x.

Solution. Note that√x runs from −1 to 1. For each fixed x in that range, we have √ 2 2 for each fixed − 1 − x 2 ≤ y ≤ 1 − x 2 , which implies that x + y ≤ 1. Also, 2 2 2 2 (x, y) with x + y ≤ 1, we have 1 − 1 − x − y ≤ z ≤ 1 + 1 − x 2 − y 2 , which implies that x 2 + y 2 + (z − 1)2 ≤ 1. Consequently, the domain of integration is S = {(x, y, z) | x 2 + y 2 + (z − 1)2 ≤ 1}. In spherical coordinates, the condition x 2 + y 2 + (z − 1)2 ≤ 1 is equivalent to (ρ sin ϕ cos θ)2 + (ρ sin ϕ sin θ)2 + (ρ cos ϕ − 1)2 ≤ 1. After simplification, we get ρ ≤ 2 cos ϕ. Therefore, the integral is I =

2π

π/2

2 cos ϕ

2π

ρ sin ϕdρdϕdθ =

7

0

= 32

0 2π

0

0

0

0

π/2

sin ϕ

28 cos8 ϕ dϕdθ 8

cos ϕ π/2

64π − dθ = . 9 0 9 9

449. Find the volume of a parallelepiped whose base is a rectangle in the z = 0 plane given by 0 ≤ y ≤ 2 and 0 ≤ x ≤ 1, while the top side lies in the plane x + y + z = 3. Solution. The parallelepiped is shown in Fig. 9.24. The z limits for the iterated integral are from 0 to 3 − x − y (the value of z on the plane). The y limits are from 0 to 2 and the x limits are from 0 to 1. If V cubic units is the required volume, then V =

1

0

=

2 0

1

3−x−y

dzdyd x =

0

0

1

2

(3 − x − y)d yd x

0

(4 − 2x)d x = 3.

0

450. Let S be the “ice cream cone” (see Fig. 9.25) bounded below by z = 3x 2 + 3y 2 and above by x 2 + y 2 + z 2 = 4. Find the volume of S.

9.5 Solved Problems

375

Fig. 9.24 The parallelepiped whose base is a rectangle in the z = 0

Fig. 9.25 An ice cream cone

Solution. The solid S has a simple description in spherical coordinates. Hence, we use spherical coordinates to write the triple √ integral as an iterated integral. The cone 2 can be written as z = 3r . In terms √ of spherical coordinates, we z = 3x 2 + 3y√ have ρ cos ϕ = 3ρ sin ϕ. This yields that tan ϕ = 1/ 3 or ϕ = π/6. The sphere x 2 + y 2 + z 2 = 4 is ρ = 2. Therefore, the volume V is equal to

2π

V =

π/6

2

0

0

ρ sin ϕdρdϕdθ =

= 2π − cos

0

π + cos 0 6

dθ 0

23

3

=

2π

2

√ 16 − 3 3

2

0

+ 1 π.

π/6

sin ϕdϕ

2

ρ dρ 2

0

376

9 Multiple Integrals

451. Find the volume and the centroid of the region S that is bounded above by the sphere ρ = a and below the cone ϕ = m, where 0 < m < π/2. Solution. The volume V is

2π

V =

m

a

2π

ρ sin ϕdρdϕdθ =

m

2

0

=

0

3

a 3

0

0

0

a3 sin ϕdρdϕdθ 3

m a 3 2π 2πa 3 (− cos ϕ) dθ = (1 − cos m)dθ = (1 − cos m). 0 3 0 3

2π 0

By symmetry, the centroid is on the z-axis. If (x, y, z) are Cartesian coordinates of a point, then z = ρ cos ϕ, and hence Mx y =

2π

zd V = 0

S

2π

m

0

a

ρ cos ϕ(ρ2 sin ϕ)dρdϕdθ

0

m

a4 a4 sin ϕ cos ϕdϕdθ = 4 4 0 0 2π 4 4 πa a sin2 m sin2 m. dθ = = 8 4 0 =

2π 0

1 2 m sin ϕ dθ 0 2

The centroid, in Cartesian coordinates, is (0, 0, z), where z=

Mx y 3 = a(1 + cos m). V 8

2 2 2 2 452. Find the mass of the solid bounded by the spheres x + y + z = 4 and x + y 2 + z 2 = 9 if the volume density at any point is k x 2 + y 2 + z 2 kg/m3 .

Solution. The mass of the solid is M =

k x 2 + y 2 + z 2 d V , where S is the region

S

bounded by x 2 + y 2 + z 2 = 4 and x 2 + y 2 + z 2 = 9. To evaluate the integral, we use spherical coordinates. Then, we obtain M= 2

=k

3

2π

0

π

0

34 − 24

4

3

kρ3 sin ϕdρdθdϕ = k

ρ3 dρ

2

0

2π

dθ

π

sin ϕdϕ

0

(2π)(− cos π + cos 0) = 65kπ kg.

453. Let the density of an object be given by x z, and the object occupies the tetrahedron with corners (0, 0, 0), (0, 1, 0), (1, 1, 0) and (0, 1, 1). Find the mass and center of mass of the object.

9.5 Solved Problems

377

Solution. The mass is the integral of density over the region. So, we have M= 0

=

1 2

1

1

0

1 x

y−x

1

x zdzdyd x =

0

1 x(1 − x) dx = 3 6 3

0

1

1 x

x(y − x)2 d yd x 2

(x − 3x 2 + 3x 3 − x 4 )d x =

0

1 . 120

Now, we compute the moments as follows:

1

1

1

Mx y =

0

Mx z = 0

M yz =

0

1

1

1

x

x

x

y−x

x z 2 dzdyd x =

1 , 360

x yzdzdyd x =

1 , 144

x 2 zdzdyd x =

1 . 360

0 y−x 0 y−x 0

Finally, the coordinates of the center of mass are x = M yz /M = 1/3, y = Mx z /M = 5/6 and z = Mx y /M = 1/3. 454. A solid cube with 2 units on a side is bounded by the planes x = ±1, z = ±1, y = 3 and y = 5. Find the center of mass and the moment of inertia about the coordinate axes. Solution. It is easy to see that M = 8. Moreover, we have Mx y = M yz =

1

−1 1

−1

Mx z =

1

5 3 5

3

−1

5

3

1

−1 1 −1 1 −1

zdzdyd x = 0, xdzdyd x = 0, ydzdyd x = 32.

So, we have (x, y, z) = (0, 4, 0). Also, we obtain Ix =

1 −1

Iy = Iz =

5

−1

3

1

1

5

(y 2 + z 2 )dzdyd x = 1

−1 3 −1 1 5 1

−1

3

−1

400 , 3

(x 2 + z 2 )dzdyd x =

(x 2 + y 2 )dzdyd x =

16 , 3

400 . 3

378

9 Multiple Integrals

455. Using a triple integral, determine the volume of the portion of the sphere 2 + y 2 and z = x of radius 2 center of the origin lying between the cones z = 3x 2 + 3y 2 and the above x y-plane. x 2 + y 2 makes the angle ϕ = π/4 with the positive z-axis Solution. The cone z = 2 and the cone z = 3x + 3y 2 makes the angle ϕ = π/6 with the positive z-axis. So, if we use spherical coordinates, the volume is equal to

2π

π/4

2

0

π/6

2π

ρ sin ϕdρdϕdθ = 2

π/4

dθ

0

sin ϕdϕ

= 2π − cos

ρ dρ 2

π/6

0

2 0

π 8

π + cos 4 6

3

=

√ 8 √ π( 3 − 2). 3

456. Consider the solid region S bounded by the x y-plane and the paraboloid z = 16 − x 2 − y 2 . What is the average height of a point in S above the x y-plane? Solution. Note that the height above the x y-plane of a point (x, y, z) is the function f (x, y, z) = z. Since there is rotational symmetry in x and y but not z, we evaluate the integrals in cylindrical coordinates. We have V (S) =

2π

dV =

0

S

0

4

16−r 2

4

r dzdr dθ = 2π

r (16 − r 2 )dr

0

0

r 4 4 = 2π 8r 2 − = 128π 4 0 and

2π

zd V = 0

S

4 0

4

=π

16−r 2

zr dzdr dθ = 2π

0

r 0

0

(16 − r 2 )2 r dr =

4

z 2 16−r 2 dr 2 0

2048π . 3

Therefore, the average value of f over S is 1 V (S)

f (x, y, z)d V =

16 1 2048π = . 128π 3 3

S

457. Find the volume enclosed by the ellipsoid x2 y2 z2 + + = 1. a, b, c > 0. a2 b2 c2

9.5 Solved Problems

379

Solution. We must find V =

d xd ydz, where S

y2 z2 x2 S = (x, y, z) | 2 + 2 + 2 ≤ 1 . a b c We introduce the new variables x = au, ∂x ∂x ∂u ∂v ∂(x, y, z) ∂y ∂y = ∂(u, v, w) ∂u ∂v ∂z ∂z ∂u ∂v

y = bv and z = cw. Then, the Jacobian is ∂x ∂w a 0 0 ∂ y = 0 b 0 = abc. ∂w 0 0 c ∂z ∂w

Hence, we conclude that the requested volume V is equal to V =

abc dudvdw, S

where S = {(u, v, w) | u 2 + v 2 + w 2 ≤ 1}. Therefore, we obtain V = (4/3)πabc. 458. In each of the following cases, describe the solid S in terms of the cylindrical coordinates. (1) Let S be the solid that is bounded by the z = x 2 + y 2 and z = 36 − 3x 2 − 3y 2 ; (2) Let S be the solid that lies within the cylinder x 2 + (y − 1)2 = 1 below the paraboloid z = x 2 + y 2 and above the plane z = 0; (3) Let D denote the torus generated by revolving the circle {(x, z) | (x − 2)2 + z 2 = 1} about the z-axis. Let S be the solid that is bounded above by the surface D and below by z = 0. Solution. (1) Solving x 2 + y 2 = 36 − 3(x 2 + y 2 ) we obtain x 2 + y 2 = 9. The projection of the solid S on the x y-plane is the circular disk {(x, y) | x 2 + y 2 = 9}. The solid is bounded by z = r 2 and z = 36 − 3r 2 . So, we conclude that S = {(r, θ, z) | 0 ≤ θ ≤ 2π, 0 ≤ r ≤ 3, r 2 ≤ z ≤ 36 − 3r 2 }. (2) The projection of S on the x y-plane is given by {(x, y) | x 2 + (y − 1)2 = 1} which is described in cylindrical coordinates as {(r, θ) | 0 ≤ θ ≤ π, 0 ≤ r ≤ 2 sin θ}. Consequently, we can write S = (r, θ, z) | 0 ≤ θ ≤ π, 0 ≤ r ≤ 2 sin θ, 0 ≤ z ≤ r 2 }.

380

9 Multiple Integrals

Fig. 9.26 The cross section in Problem 458 (3)

(3) The projection of the solid S on the x y-plane is the region between the circles r = 1 and r = 3. The angle θ runs from 0 to 2π and assume that the cross section of the solid is perpendicular to the x y-plane, corresponding to a fixed θ. The cross section is a circle which is shown in Fig. 9.26. The equation of the circle can be considered as (r − 2)2 + z 2 = 1 for 1 ≤ r ≤ 3. Therefore, we have S = {(r, θ, z) | 0 ≤ θ ≤ 2π, 1 ≤ r ≤ 3, 0 ≤ z ≤ 1 − (r − 2)2 }. 459. Find I =

z d xdydz, where S is described by 0 ≤ z, x 2 + y 2 ≤ b2 , x 2 + S

y 2 + z 2 ≤ a 2 and 0 ≤ b ≤ a. Solution. The domain is like a cylindrical can with a spherical cap at one end. We calculate the integral in cylindrical coordinates. Since x 2 + y 2 = r 2 , it follows that the domain of S can be described by 0 ≤ θ ≤ 2π, 0 ≤ r 2 + z 2 = a 2 . Consequently, √ r ≤ b and 0 ≤ z ≤ a 2 − r 2 . Therefore, we obtain

2π

I = 0

b

√ a 2 −r 2

0

0

2πdθ b

= 0

0

460. Show that 1 1 1 (1) 0

1

0 1

0 1

(2) 0

0

0

2π

b

1 r (a 2 − r 2 )dr dθ 0 0 2 1 2 1 2 2 1 4 b 1 (a r − r 3 )dr = π a r − r = πb2 (2a 2 − b2 ). 0 2 2 4 4 zr dzdr dθ =

∞

1 1 d xd ydz = ; 1 − x yz n3 n=1 ∞ 1 (−1)n−1 d xd ydz = . 1 + x yz n3 n=1

9.5 Solved Problems

381

Solution. (1) We have |x yz| < 1 except at the point (1, 1, 1). So, according to the formula for the sum of a geometric series, we have ∞

1 = (x yz)n . 1 − x yz n=1 Hence, we can write

1

0

1

0

0

1

1 d xd ydz = 1 − x yz =

1

0

0

∞ 1

∞ n=0

n=0

1

1

x n y n z n d xd ydz

0

0

∞

0

1 n

1

x dx

1

n

n

y dy

0

n=0

=

∞ 1

(x yz)n d xd ydz

0

n=0

=

1

z dz

0

0 ∞

1 1 1 1 = . n+1 n+1 n+1 n3 n=1

(2) We have | − x yz| < 1 except at the point (1, 1, 1). So, according to the formula for the sum of a geometric series, we have ∞

1 = (−x yz)n . 1 + x yz n=1 Hence, we can write

1 0

0

1

0

1

1 d xd ydz = 1 + x yz =

1 0

1

∞

∞ 1

(−x yz)n d xd ydz

0

∞ 1 n=0

=

0

0

∞ n=0

1

0

1

(−1)

(−1)n x n y n z n d xd ydz

0

n

1

1

n

z dz 0

∞

(−1)n

n

y dy 0

1

n

x dx 0

n=0

=

n=0

(−1)n 1 1 1 . = n+1 n+1 n+1 n3 n=1

461. Let m, n, p and q be positive integers. Prove that

382

9 Multiple Integrals

(1)

1

0

x m (1 − x)n d x =

m!n! ; (m + n + 1)!

x m y n z p (1 − x − y − z)q d xd ydz =

(2) S

m!n! p!q! , where (m + n + p + q + 3)!

S = {(x, y, z) | x, y, z ≥ 0 x + y + z ≤ 1}.

1

Solution. (1) Suppose that I (m, n) =

x m (1 − x)n d x. Then, by integration by

0

parts, we obtain

1 (1 − x)n−1 −x m (1 − x)n−1 1 dx mx m−1 I (m, n) = + 0 n+1 n+1 0 1 m = x m−1 (1 − x)n+1 d x n+1 0 1 m(m − 1) x m−2 (1 − x)n+2 d x = (n + 1)(n + 2) 0

= ... =

m!n! m(m − 1) . . . 1 = . (n + 1)(n + 2) . . . (n + m + 1) (n + m + 1)!

(2) We introduce the new variables x + y + z = u, y + z = uv and z = uvw. This yields that x = u(1 − v), y = uv(1 − w) and z = uvw. Then, the Jacobian is ∂x ∂u ∂(x, y, z) ∂y = ∂(u, v, w) ∂u ∂z ∂u

∂x ∂v ∂y ∂v ∂z ∂v

1−v −u 0 = v(1 − w) u(1 − w) −uv = u 2 v. vw uw uv

∂x ∂w ∂y ∂w ∂z ∂w

Hence, the given integral is equal to

1 0

1 0

1

u m+n+ p+2 (1 − u)q v n+ p+1 (1 − v)m w p (1 − w)n dudvdw

0 1

=

u m+n+ p+2 (1 − u)q du

0

0

1

1

v n+ p+1 (1 − v)m dv

w p (1 − w)n dw

0

=

n!(q + m + 1)! p!(q + m + n + 2)! m!q! (by the first part) (q + m + 1)! (q + m + n + 2)! (q + m + n + p + 3)!

=

m!n! p!q! . (m + n + p + q + 3)!

9.6 Exercises

383

9.6 Exercises Easier Exercises Evaluate the given iterated integral.

1

1

1. 0

0 π/2

x d xd y; 2 (1 + x + y 2 )3/2 y 3 d yd x;

0

sin x

4

x−1

3. 2

0

4

4.

2

√

0

0

1 d yd x; (x + y)2

sin(π y )d yd x;

2

5.

cos −4

−y/2

πy

d xd y; 4x

1−y

1 0

√

x + y(y − 2x)2 d xd y;

0

7.

2

1

√ 1

ye x d xd y; x3

y

1

2

e x d xd y;

8.

3

x

1 0

1

2.

6.

0 1

y cos−1 y

esin x d xd y;

9.

0

0 π/2

10. 0

2 sin θ

r 2 cos θdr dθ.

0

Evaluate the given double integral. ln(1 + x 2 + y 2 )d A, where R = {(x, y) | x 2 + y 2 ≤ 4, x ≥ 0, y ≥ 0}; 11. R

sin(y 2 )d A, where R is the triangle with the vertices (0, 0), (1, 3) and (9, 3);

12. R

e x+y cos 2xd A, where R = {(x, y) | 0 ≤ x ≤ π, 1 ≤ y ≤ 2};

13. R

x ln(x y)d A, where R = {(x, y) | 2 ≤ x ≤ 3, 1 ≤ y ≤ 2};

14. R

15.

cos R

y − x

d A, where R is the trapezoidal region with vertices (1, 0), y+x

(2, 0), (0, 2) and (0, 1); 16. Use double integration to find the area of the region in the first quadrant bounded by the parabola y 2 = 4x, the circle x 2 + y 2 = 5 and the x-axis by two methods: (a) integrate first with respect to x; (b) integrate first with respect to y. Compare the two methods of solutions. 17. Find by two methods the volume of the solid below the plane 3x + 8y + 6z = 24 and above the region in the x y-plane bounded by the parabola y 2 = 2x, the line

384

9 Multiple Integrals

2x + 3y = 10 and the x-axis: (a integrate first with respect to x; (b) integrate first with respect to y. Compare the two methods of solutions. 18. Use polar coordinates to evaluate R

1 − x 2 − y2 d xd y, 1 + x 2 + y2

where R is the unit disk x 2 + y 2 ≤ 1. Evaluate the given iterated integral by transforming it to polar coordinates. 1 x 1 √1−x 2 x2 2 d yd x. 20. 19. x x 2 + y 2 d yd x; x 2 + y2 0 0 0 0 21. Use polar coordinates to find the area A of the region R enclosed by the curve (x 2 + y 2 )2 = 2x 3 . Also, find the centroid (x, y) of the region R. Evaluate each of the following double integrals by making a suitable change of variables. (1 + 3x 2 )d A, where R is bounded by the lines x + y = 1, x + y = 2 and by 22. R

the curves y − x 3 = 0, y − x 3 = 1; √ y−x d A, where R is the square with vertices (0, 2a), (a, a), (2a, 2a) and 23. x+y R

(a, 3a) with a > 0; 24. (x + y)d A, where R is bounded by x 2 + y 2 = x + y; R

1−

25.

x2 y2 − d A, where R is bounded by the ellipse x 2 /a 2 + y 2 /b2 = 1; a2 b2

R

(9x 2 + 4y 2 )d A, where R is the region in the first quadrant bounded by the

26. R

ellipse 9x 2 + 4y 2 = 1; e(x+y)/(x−y) d A, where R is the trapezoidal region with vertices (1, 0), (2, 0), 27. R

(0, −2) and (0, −1). Evaluate each of the following triple integrals over the specified solid region by converting it to an appropriate iterated integral.

9.6 Exercises

385

(x 2 ye z + ze y )d V , where S = [0, 1] × [1, 2] × [0, 2];

28. S

29.

(1 + x + y + z)−3 d V , where S is bounded by the plane x + y + z = 1 and

S

by the coordinate planes; 30. d V , where S is the region bounded by the cylinders x 2 + z = 1 and y 2 + S

z = 1 and the x y-plane; √ 2 2 2 31. e x +y +z d V , where S is enclosed by the sphere x 2 + y 2 + z 2 = 9 in the S

first octant; 32. y cos(x + z)d V , where S is the region bounded by the cylinder x = y 2 and S

the planes x + z = π/2, y = 0 and z = 0; 33. (x 2 + 2x y)d V , where S = {(x, y, z) | |2x y| ≤ z ≤ 1 − x 2 − y 2 }; S

34. Use cylindrical coordinates to evaluate the iterated integral

2 0

√

2x−x 2

3

z x 2 + y 2 dzdyd x.

0

0

35. Use cylindrical coordinates to find the volume of the solid region between the paraboloid z = x 2 + y 2 and the cone z = x 2 + y 2 . 36. Evaluate the iterated integral

1

√ 1−x 2

0

0

√1−x 2 −y 2

(x 2 + y 2 )dzdyd x

0

by transforming it to spherical coordinates. 37. Find the partial derivatives ∂ I /∂x, ∂ I /∂ y, ∂ I /∂z of I =

z

z

y

y

d xd y and 0

0

0

0

x

f (x, y, z)d xd ydz.

0

f (x, y)d xd y = 0 on each rect-

38. Suppose f (x, −y) = − f (x, y). Prove that angle of the form a ≤ x ≤ b, −c ≤ y ≤ c.

R

386

9 Multiple Integrals

39. Let f be continuous on [0, 1]. Show that

1

f (x + y)d A =

u f (u)du if R

0

R

is the triangle with vertices (0, 0), (0, 1) and (1, 0). 40. Find the volume under the surface of f (x, y) = xe y over the rectangle R = [2, 3] × [0, 1]. 41. Compute the volume of the solid bounded by the surfaces (a) y = 0, z = 0, x = 0, z = −x − y + 1; (b) z = 2, x 2 + y 2 + z = 1, x 2 + 2y 2 − 1 = 0. 42. Find the common area to the circles r = a and r = 2a cos θ. 43. An object on the unit square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1 has density ρ(x, y) = ax + by + c. Find the mass and center of mass. 44. A homogeneous triangle with vertices (0, 0), (1, 0) and (1, 3). Find the coordinates of its center of mass. 45. An object on the plane region −1 ≤ x ≤ 1, x 2 ≤ y ≤ 1 has density ρ(x, y) = √ 1 + x + y. Find the mass and the work needed to stand the object up on the flat side. 46. An object of constant density k covers the polar region 0 ≤ θ ≤ π/2, 0 ≤ r ≤ b sin 2θ. Find the center of mass and the moment of inertia about the origin. 47. Find the moment of inertia about a diameter of the solid between two concentric spheres having radii a ft and 2a ft. The volume density varies inversely as the square of the distance from the center, and it is measured in slugs per cubic foot. Also, find the mass of the solid.

Harder Exercises 48. Compute the domain D f of the function f (x, y) = compute the integral f (x, y)d xd y.

4 − x 2 + y 2 and then

Df

x n y m d A = 0, where R is

49. Let n and m be positive integers. Prove that if R

bounded by an ellipse x 2 /a 2 + y 2 /b2 = 1, then at least one of the numbers n and m is odd. 50. Suppose that f (a) = 0. Express

b

y

2 a

in terms of a

b

f (x) f (y) 1 − f (x) d xd y

a

f (x)d x and a

b

3 f (x) d x.

51. If 0 = f (a) ≤ f (x) and f (x) ≤ 1 for a ≤ x ≤ b, prove that

9.6 Exercises

387

b

3 f (x) d x ≤

a

b

2 f (x)d x

.

a

52. Compute the double integral of f (x, y) =

e−x−y if x, y ∈ (0, ∞) x otherwise

over the domain R = {(x, y) | x + y ≥ 1}. 53. Suppose that level curves of a function f (x, y) are simple, closed and smooth. Let a region R be bounded by two level curves f (x, y) = a and f (x, y) = b. Prove that b f (x, y)d A = u F (u)du, a

R

where F(u) is the area of the region between the curves f (x, y) = a and f (x, y) = u. Hint: Partition the region R by level curves of the function f . 54. Prove that if f (x, y, z) is continuous in R and for any subregion D of R, f (x, y, z)d V = 0, S

then f (x, y, z) = 0 in R. 55. Use spherical coordinates to find the volume of a solid bounded by the given surfaces. (a) (x 2 + y 2 + z 2 )3 = 3x yz; (b) (x 2 + y 2 + z 2 )2 = a 2 (x 2 + y 2 − z 2 ) with a > 0. 56. Describe the solid with 0 ≤ ρ ≤ 1 − cos θ and find its volume. 57. Use the following steps to show that if (x, y) is not at the origin or on the negative real axis, then

y −1 −1 2 2 T (x, y) = x + y , 2 tan x + x 2 + y2 is the inverse of the polar coordinate transformation. 58. Find the volume of a solid bounded by the surfaces x 2 + z 2 = a 2 , x 2 + z 2 = b2 and x 2 + y 2 = z 2 , where x > 0 and 0 < a < b. 59. Let y if x isrational f (x, y) = 1 − y if x isirrational

388

9 Multiple Integrals

1

1

1 ; 2 0 0 (b) Show that for each constant y0 = 1/2, the function g(x) = f (x, y0 ) is everywhere discontinuous, so that the iterated integral (a) Show that

f (x, y)d yd x =

1

0

1

f (x, y)d xd y

0

is undefined. 60. Let f be defined on the rectangle R = [1, 2] × [1, 4] as follows: f (x, y) =

(x + y)−2 y if x ≤ y ≤ 2x 0 otherwise.

Indicate, by means of a sketch, the portion of R in which f is non-zero and f (x, y)d A, given that the integral compute the value of the double integral R

exists. 61. Let f be defined on the rectangle R = [0, 1] × [0, 1] as follows: f (x, y) =

1 if x = y 0 if x = y.

f (x, y)d A exists and equals 0.

Prove that the double integral R

62. Find the center of mass of an object of constant density filling the region x 2 + y 2 ≤ b, 0 ≤ z ≤ x 2 + y 2 . 63. The density at a variable point P of the solid hemisphere x 2 + y 2 + z 2 ≤ 2ay, z ≥ 0 of radius a is equal to the distance from P to the origin. Find the total mass of the hemisphere. 64. Find the moment of inertia of a solid sphere of mass M and radius a about any of its tangent lines. 65. Let R denote the unit square 0 ≤ x, y ≤ a. Suppose that f (x, y) = 0 at its four vertices. Prove that f (x, y)d xd y R

=−

1 2

y(a − y) f yy (x, y)d xd y − R

1 1 x(a − x) f x x (x, a) + f x x (x, 0) d x. 4 0

Also, if | f x x | ≤ M and | f yy | ≤ N on R, prove that

9.6 Exercises

389

1 1 f (x, y)d xd y ≤ M+ N. 12 12 R

66. A function f is said to have spherical symmetry if it depends on the distance to the origin only, that is, it can be expressed in spherical coordinates as f (x, y, z) = g(ρ), where ρ = x 2 + y 2 + z 2 . Show that

f (x, y, z)d V = 2π

b

ρ2 g(ρ)dρ

a

S

where S is the region between the upper concentric hemispheres of radii a and b centered at the origin, with 0 < a < b and f a spherical function defined on S. 67. Let S be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where 0 < a < b. Consider f a function defined on S whose form in spherical coordinates is f (x, y, z) = b g(ρ) cos ϕ. Show that if h(a) = h(b) and k(ρ)dρ = 0, then a

f (x, y, z)d V =

π2 ak(a) − bk(b) 4

S

where h is an antiderivative of g and k is an antiderivative of h. 68. Find the mean value of the function x 2 + y 2 + z 2 over the ball x 2 + y 2 + z 2 ≤ a 2 of radius a. , y = v 2 and z = w 2 to find the volume of the 69. Use the transformation x = u 2√ √ √ region bounded by the surface x + y + z = 1 and the coordinate planes. 70. Suppose that f is continuous on a disk that contains the point (a, b). Let Rc be the closed disk with center (a, b) and radius c. Use the mean value theorem for double integrals to show that 1 c→0 πc2

f (x, y)d A = f (a, b).

lim

R

71. Evaluate R

(x 2

1 d A, where n is an integer and R is the region bounded + y 2 )n/2

by the circles with center the origin and radii a and b with 0 < a < b. For what values of n does the integral have a limit as a → 0+ ? 1 d A, where n is an integer and R is the region 72. Evaluate (x 2 + y 2 + z 2 )n/2 S

bounded by the spheres with center the origin and radii a and b with 0 < a < b. For what values of n does the integral have a limit as a → 0+ ?

390

9 Multiple Integrals

− → − → − → 73. If A, B and C are constant vectors, R is the position vector x i + y j + z k , and S is given by the inequalities 0 ≤ A · R ≤ a, 0 ≤ B · R ≤ b and 0 ≤ C · R ≤ c, show that (abc)2 . (A · R)(B · R)(C · R)d V = 8 A · (B × C)| S

74. The plane

x y z + + = 1, a > 0, b > 0, c > 0 a b c

cuts the solid ellipsoid x2 y2 z2 + 2 + 2 ≤1 2 a b c into two pieces. Find the volume of the smaller piece. 75. What relationship must hold between the constants a, b and c to make

∞

−∞

∞

−∞

e−(ax

2

+2bx y+cy 2 )

d xd y = 1?

Hint: Let s = αx + β y and t = γx + δ y, where (αδ − βγ)2 = ac − b2 . Then ax 2 + 2bx y + cy 2 = s 2 + t 2 .

Chapter 10

Surface Integrals

10.1 Green’s, Divergence and Stokes’ Theorems Green’s theorem: Let R be a region in the x y-plane consisting of a piecewise smooth simple closed curve C and its interior, and let the functions P = P(x, y) and Q = Q(x, y) be continuously differentiable on R. Then C

∂P ∂Q − d A, Pd x + Qdy = ∂x ∂y R

where C is traversed in the counterclockwise direction. It turns out that Green’s theorem can be extended to multiply connected regions, which have one or more regions cut out from the interior, as opposed to discrete points being cut out. For such regions, the outer boundary and the inner boundaries are traversed so that R is always on the left side, see Fig. 10.1. Divergence theorem: Let T be the solid region consisting of a piecewise smooth closed surface S and its interior, let N be the outer unit normal to S, and let F = ( f, g, h) be a continuously differentiable vector field on T . Then

F · N dσ = S

div Fd V, T

that is, the flux of F across S equals the integral of div F over the region T bounded by S. Stokes’ theorem: Let S be a piecewise smooth surface bounded by a piecewise smooth simple closed curve C, and choose a unit normal N to S. Let C be traversed in the positive direction (the direction induced by N ), and let F = ( f, g, h) be a continuously differentiable vector field on S. Then

© The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1_10

391

392

10 Surface Integrals

Fig. 10.1 Multiply connected regions Fig. 10.2 Oriented surfaces with hole

F · dR = C

(curl F) · N dσ, S

that is, the circulation of F around C equals the flux of curl F across S. Stokes’ theorem can be extended to an oriented surface S that has one or more holes (Fig. 10.2), in a way analogous to the extension of Green’s theorem: The surface integral over S of the normal component of ∇ × F equals the sum of the line integrals around all the boundary curves of the tangential component of F, where the curves are to be traced in the direction induced by the orientation of S. Surface integral: If R is the shadow region of a surface S defined by the equation F(x, y, z) = c and g is a continuous function defined at the points of S, then the integral of g over S is the integral g(x, y, z) R

∇ F d A, |∇ F · V |

where V is a unit vector normal to R and ∇ F · V = 0. The integral itself is called a surface integral. Surface integrals behave like other double integrals, the integral of the sum of two functions being the sum of their integrals and so on.

10.3 Solved Problems

393

10.2 Surface Area The area of the surface F(x, y, z) = c over a closed and bounded plane region R is AS = R

∇ F d A, |∇ F · V |

where V is a unit vector normal to R and ∇ F · V = 0. If S is the graph of a function z = f (x, y, z), where R is a bounded region in the x y-plane and the partial derivatives ∂ f /∂x and ∂ f /∂ y exists and are continuous. Then the formula for the area of S is 2 2 ∂z ∂z + + 1d A. (10.1) AS = ∂x ∂y R

There exist similar formulas for the case where the surface S is a graph of a continuously differentiable function defined on a region in other coordinate planes. In order to obtain these formulas, it is enough to replace the integrand (10.1) by

∂ y 2 ∂ y 2 + +1 ∂x ∂z

if R is a region in the x z-plane, and by

∂x 2 ∂y

+

∂x 2 ∂z

+1

if R is a region in the yz-plane. − → − → − → The area of the smooth surface R(u, v) = f (u, v) i + g(u, v) j + h(u, v) k , where a ≤ u ≤ b and c ≤ v ≤ d is A= c

d

b

Ru × Rv dudv.

a

10.3 Solved Problems 462. Use Green’s theorem to evaluate

C

1 + x 3 d x + 2x ydy, where C is the

triangle with vertices (0, 0), (1, 0) and (1, 3) oriented clockwise.

394

10 Surface Integrals

Fig. 10.3 The boundary of the semi-annular region R in the upper half-plane between the circles x 2 + y 2 = 4 and x 2 + y 2 = 16

√ Solution. Let P(x, y) = 1 + x 3 and Q(x, y) = 2x y. Then, we have ∂ P/∂ y = 0 and ∂ Q/∂x = 2y. Now, using Green’s theorem the line integral is equal to

Pd x + Qdy = C

∂Q ∂P − ∂x ∂y

R

dA =

2yd A =

1

0

R

3x

2ydyd x = 3.

0

y 3 d x + 3x ydy, where C is the boundary of the semi-annular

463. Compute C

region R in the upper half-plane between the circles x 2 + y 2 = 4 and x 2 + y 2 = 16, see Fig. 10.3. Solution. In polar coordinate we have R = {(r, θ) | 2 ≤ r ≤ 4 and 0 ≤ θ ≤ π}. Then, by Green’s theorem we obtain

y 3 d x + 3x ydy = C

R

π

= 0

∂ ∂ 2 (3x y) − (y ) d A = yd A ∂x ∂y

4

(r sin θ)r dr dθ =

2

0

R π

π 1 4 112

r 3 = . = − cos θ

0 3 2 3

sin θdθ

4 2

r dr 2

464. Let C be the boundary of a region R on which Green’s theorem holds. Use Green’s theorem to calculate: (1) f (x)d x + g(y)dy; C

ayd x + bxdy, where a and b are constant.

(2) C

10.3 Solved Problems

395

Solution. (1) Since P(x, y) = f (x) and Q(x, y) = g(y), it follows that ∂ P/∂ y = ∂ Q/∂x = 0. This gives that f (x)d x + g(y)dy = C

∂Q ∂P d xd y = 0. − ∂x ∂y R

(2) Since P(x, y) = ay and Q(x, y) = bx, it follows that ∂ P/∂ y = a and ∂ Q/∂x = b. Hence, we have

∂P ∂Q − = ayd x + bxdy = (b − a)d xd y = (b − a)R(A), ∂x ∂y C R

R

where A(R) is the area of the region R. 465. Let C be a simple closed smooth curve and a be a real number. If

x ae y + e x d x + e x + ye y dy = 0,

C

find a. Solution. If R is the region enclosed by C, then by Green’s theorem we have

x ae y + e x d x + e x + ye y dy =

C

(1 − a)e x d xd y = 0. R

This implies that a = 1. 466. Let R be a region for which Green’s theorem holds. Suppose that u is a harmonic function, that is, u x x (x, y) + u yy (x, y) = 0, for all x, y ∈ R. If C is the boundu y d x − u x dy = 0.

ary of R, prove that C

Solution. By the assumption of the problem, we can apply Green’s theorem. Taking P(x, y) = u y (x, y) and Q(x, y) = u x (x, y) we get u y d x − u x dy = C

R

=−

− u x x (x, y) − u yy (x, y) d xd y

u x x (x, y) + u yy (x, y) d xd y = 0.

R

467. Evaluate the area of the region enclosed by the simple closed curve x 2/3 + y 2/3 = 1.

396

10 Surface Integrals

Solution. We can parameterize the curve by x(t) = cos3 t and y(t) = sin3 t with 0 ≤ t ≤ 2π. Now, the required area is equal to 1 A= 2

3 −yd x + xdy = 2 C

2π

3 sin t cos tdt = 8 2

0

2π

2

sin2 2tdt =

0

3 π. 8

468. Let C be any piecewise smooth simple closed in the counterclockwise direction. Show that −yd x + xdy = 2π. x 2 + y2 C Solution. Let P(x, y) = −y/(x 2 + y 2 ) and Q(x, y) = x/(x 2 + y 2 ). Then P and Q are continuously differentiable everywhere except at the origin, and we have ∂P ∂Q y2 − x 2 , (x, y) = (x, y) = 2 ∂y ∂x (x + y 2 )2 if (x, y) = (0, 0). Now, let C be a circle centered at (0, 0) and transverse once in the clockwise direction, which is small enough to be enclosed by C. Then P and Q are continuously differentiable and satisfy the condition ∂ Q/∂x = ∂ P/∂ y on the region R between C and C . Consequently, we can write

Pd x + Qdy + C

C

Pd x + Qdy =

∂Q ∂P − ∂x ∂y

d A = 0,

R

by the version of Green’s theorem for two boundary curves. This implies that

Pd x + Qdy = −

C

C

Pd x + Qdy =

−C

Pd x + Qdy,

where −C is transverse in counterclockwise direction. If the circle −C is of radius a, it has the parametric representation x = a cos t and y = a sin t with 0 ≤ t ≤ 2π. Therefore, we obtain C

Pd x + Qdy =

2π 2π (−a sin t)(−a sin t) + (a cos t)(a cos t) dt = dt = 2π. (a cos t)2 + (a sin t)2 0 0

469. Suppose that all the necessary derivatives exist and are continuous. Show that if f (x, y) satisfies the Laplace equation ∂2 f ∂2 f + = 0, 2 ∂x ∂ y2

10.3 Solved Problems

then C

397

∂f ∂f dx − dy = 0 for all closed curve C to which Green’s theorem ∂y ∂x

applies.

Solution. Let P = ∂ f /∂ y and Q = −∂ f /∂x. Then, we obtain ∂P ∂Q ∂2 f ∂2 f and = =− 2. 2 ∂y ∂y ∂x ∂x Therefore, we can write C

∂f ∂f dx − dy = ∂y ∂x

2 ∂ f ∂2 f − 2 − d xd y = 0. ∂x ∂ y2 R

470. Let the function f = f (x, y) and g = g(x, y) be continuously differentiable on a region R consisting of a piecewise smooth simple closed curve C and its interior. Show that ∂g ∂f ∂f ∂g g − + f − d A. f gd x + f gdy = ∂x ∂y ∂x ∂y C R

Solution. Suppose that P = f g and Q = f g. Then, we obtain ∂f ∂g ∂Q ∂f ∂g ∂P =g + f and =g + f . ∂y ∂y ∂y ∂x ∂x ∂x This implies that ∂ f ∂g ∂Q ∂P ∂f ∂g − =g − + f − . ∂x ∂y ∂x ∂y ∂x ∂y Now, the result follows from Green’s theorem. 471. Use Green’s theorem to show that the area enclosed by the simple closed polygonal path P1 P2 P3 . . . Pn P1 with vertices Pi = (xi , yi ) (i = 1, 2, . . . n) is equal to one-half the absolute value of the sum (x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + · · · + (xn y1 − x1 yn ). Solution. Let P1 P2 be the segment line from P1 to P2 . Then the equation of this segment line is y1 − y2 x2 . y = mx + y2 − x1 − x2 Now, we compute the line integral xdy − yd x. We can write P1 P2

398

10 Surface Integrals

y1 − y2 mx − mx − y2 − x2 d x x1 − x2 x1 y1 − y2 = y2 − x2 (x1 − x2 ) = x1 y2 − y1 x2 . x1 − x2

xdy − yd x = P1 P2

x2

In general case, we have P1 P2 P3 ...Pn P1

xdy − yd x

xdy − yd x +

= P1 P2

xdy − yd x + · · · + P2 P3

xdy − yd x

(10.2)

Pn P1

= (x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + · · · + (xn y1 − x1 yn ). On the other hand, by Green’s theorem we can write P1 P2 P3 ...Pn P1

xdy − yd x =

∂(−y) ∂x − dA = 2d A = 2 A(R), (10.3) ∂x ∂y R

R

where A(R) is the area enclosed by the simple closed polygonal path P1 P2 P3 . . . Pn P1 . Now, the result follows from (10.2) and (10.3). (x 2 sin2 x − y 3 )d x + (y 2 cos2 y − y)dy, where

472. Evaluate the line integral C

C is the closed curve consisting of x + y = 0, x 2 + y 2 = 25 and y = x and lying in the first and fourth quadrant. Solution. Suppose that R is the region enclosed by C and let P(x, y) = x 2 sin2 x − y 3 and Q(x, y) = y 2 cos2 y − y. Then, we have ∂ P/∂ y = −3y 2 and ∂ Q/∂x = 0. Hence, we can write (x 2 sin2 x − y 3 )d x + (y 2 cos2 y − y)dy = 3y 2 d xd y C

=

π/4

−π/4

5 0

3r 3 sin2 θdr dθ =

R

π/4

−π/4

sin2 θdθ

5

3r 3 dr

0

π/4 3 5 3π 1 3

θ − sin 2θ

r4 = − (625). = −π/4 4 0 2 4 16 8 1

473. Use Problem 471 to find the area enclosed by the quadrilateral with vertices (1, −1), (3, −2), (5, 1) and (2, 6).

10.3 Solved Problems

399

Solution. Suppose that

(x1 , y1 ) = (1, −1), (x2 , y2 ) = (3, −2), (x3 , y3 ) = (5, 1), (x4 , y4 ) = (2, 6).

Now, by Problem 471 the required area is equal to 1 (x1 y2 − x2 y1 ) + (x2 y3 − x3 y2 ) + (x3 y4 − x4 y3 ) + (x4 y1 − x1 y4 ) 2 1 1 (−2 + 3) + (3 + 10) + (30 − 2) + (−2 − 6) = (1 + 13 + 28 − 8) = 17. = 2 2 474. Let A be the area and x the x-coordinate of the centroid of a region R that is bounded by a piecewise smooth simple closed curve C in the x y-plane. Show that 1 2

x dy = − 2

C

C

1 x yd x = 3

x 2 dy − x yd x = Ax. C

Solution. We consider δ(x, y) = 1. Then, we have

xδ(x, y)d A R x =

δ(x, y)d A

R

xd A

R =

xd A =

dA

R

A

,

R

and so we conclude that 2 x Ax = dy, xd A = C 2 R

Ax =

xd A = −

Ax =

x yd x, C

R

R

xd A = −

= C

R

2 1 1 2 1 x + x d xd y = x dy − x yd x. 3 3 3 C 3

Now, the requested equalities follow. 475. Let I y be the moment of inertia about the y-axis of the region in Problem 474. Show that 1 1 3 2 x dy = − x yd x = x 3 dy − x 2 yd x = I y . 3 C 4 C C

400

10 Surface Integrals

Solution. We consider δ(x, y) = 1. By the definition we have Iy =

x 2 δ(x, y)d A =

R

Iy = R

Iy =

1 3

x 3 dy, C

R

x 2d A = −

x 2d A =

x 2 ydy, C

1 2 3 2 1 x + x d yd x = x dA = x 3 dy − x 2 yd x. 4 4 4 C 2

R

R

476. Let S be a “light-bulb-shaped region" as follows. Imagine a light-bulb cut off at the base so that its boundary is the unit circle x 2 + y 2 = 1 in the x y-plane, oriented with the outward-pointing normal. You can use an old-fashioned light-bulb as shown in Fig. 10.4. Suppose that 2 2 F = e z −2z x, sin(x yz) + y + 1, e z sin(z 2 ) . curl F · d S using Stokes’ theorem.

Compute the flux integral S

Solution. We use the parametrization R(t) = (cos t, sin t + 1, 0). Then, we obtain F R(t) = (cos t, sin t, 0) and d R = (− sin t, cos t, 0)dt. Therefore, we can write

curl F · d S =

S

Fig. 10.4 A light-bulb-shaped region

F · dR =

C

0

2π

cos tdt = 0.

10.3 Solved Problems

401

477. Let F(x, y, z) = (y, −x, 2z 2 + x 2 ) and S be the part of the sphere x 2 + y 2 + z 2 = 25 that lies below the plane z = 4. Evaluate the integral

curl F · N dσ, S

where N is the unit outward normal of S. Solution. The boundary C of the surface S is defined by (3 sin θ, 3 cos θ, 4). Note that C is oriented clockwise when viewed from above By Stokes’ theorem we have

curl F · N dσ = S

F · dR C

2π

=

3 cos θ, −3 sin θ, 32 + 9 sin2 θ) · (3 cos θ, −3 sin θ, 0)dθ = 18π.

0

478. Let F be a differentiable vector field defined on a region containing a smooth closed oriented surface S and its interior. Suppose that N is the unit normal vector field on S, and S is the union of two surfaces S1 and S2 joined along a smooth simple closed curve C. Can anything be said about (∇ × F) · N dσ? S

Give reason for your answer. Solution. We have (∇ × F) · N dσ = (∇ × F) · N dσ + (∇ × F) · N dσ. S

S1

(10.4)

S2

Since S1 and S2 are joined by the simple closed curve C, each of the integral on the right side of (10.4) is equal to a circulation integral on C. Note that for one surface the circulation is counterclockwise, and for the other surface the circulation is clockwise. (∇ × F) · N dσ = 0.

Since the the integrands are same, it follows that S

479. Among all smooth simple closed curves in the plane, oriented counterclockwise, find the one along which the work done by F(x, y) = is greatest.

1 4

x2 y +

1 3 − → − → y i +x j 3

402

10 Surface Integrals

Solution. Taking P(x, y) = (1/4)x 2 y + (1/3)y 3 and Q(x, y) = x we obtain curl F =

1 ∂P ∂Q − =1− x 2 + y2 , ∂x ∂y 4

which is positive in the interior of the ellipse (1/4)x 2 + y 2 = 1. The work done is F · dR =

W = C

1 1 − x 2 − y 2 d xd y. 4 R

Hence, W will be maximized over the region enclosed by (1/4)x 2 + y 2 = 1. 480. Use Stokes’ theorem to evaluate curl F · d S, where F = (z 2 , −3x y, x 3 y 3 ) S

and S is the part of z = 5 − x 2 − y 2 above the plane z = 1, and oriented upward. Solution. First, we find the boundary of S. Indeed, we need to determine where the plane intersects z = 5 − x 2 − y 2 . If we put 1 = 5 − x 2 − y 2 , then x 2 + y 2 = 4. This means that the boundary of S is the circle of radius 2, centered at (0, 0) and lying in the plane z = 1. We denote the boundary of S by C. Let R(t) = (2 cos t, 2 sin t, 1) with 0 ≤ t ≤ 2π, be a parametrization of C. Now, by Stokes’ theorem we have

curl F · d S = S

2π

F R(t) · R (t)dt

0

C

=

2π

F · ds =

1, −12 cos t sin t, 64 cos3 t sin3 t · (−2 cos t, 2 cos t, 0) dt

0

=

0

2π

2π 24

cos3 t = 0. −2 sin t − 24 sin t cos2 t dt = 0 3

481. Or a surface S with boundary C and a fixed vector A, prove that

A · N dσ =

2 S

(A × R) · d R, C

where R(x, y, z) = (x, y, z). Solution. Assume that A = (a, b, c) and R = (x, y, z). Then, we get

− − →

→ j

i

A × R =

a b

x y

− →

k

c

= (bz − cy, cx − az, ay − bx),

z

10.3 Solved Problems

403

and

− − → − →

→ i j k

∂

∂ ∂

= (2a, 2b, 2c) = 2 A. ∇ × (A × R) =

∂y ∂z

∂x

bz − cy cx − az ay − bx

Therefore, using Stokes’ theorem we can write

(A × R) · d R =

C

∇ × (A × R) dσ = 2

S

A · N dσ. S

482. Let S be the surface defined by z = 4 − 4x 2 − y 2 with z ≥ 0 and oriented outward. If F = (x − y, x + y, ze x y ), compute

(∇ × F) · dσ. S

Solution. Suppose that C is the boundary of S. Then, C is given by z = 0, 4x 2 + y 2 = 4. Since S is oriented with upward orientation, it follows that C has counterclockwise orientation when viewed from above. A parametrization of C is R(t) = (cos t, 2 sin t, 0), where 0 ≤ t ≤ 2π. So, by Stokes’ theorem we have

(∇ × F) · dσ = S

= =

C

F · dR =

2π 0

2π 0

2π 0

F R(t) · R (t)dt

cos t − 2 sin t, cos t + 2 sin t, 0 · − sin t, 2 cos t, 0 dt

2π 3

2 + 3 sin t + cos t dt = 2t + sin2 t = 4π. 0 2

483. For constants a, b, c and d consider the vector field − → − → − → F(x, y, z) = (ax + by + 5z) i + (x + cz) j + (3y + d x) k . (1) Suppose that the flux of F through any closed surface is 0. What does this tell us about the values of the constants a, b, c and d? (2) Suppose that the line integral of F around any closed curve is 0. What does this tell us about the values of the constants a, b, c and d? Solution. (1) If the flux of F through any closed surface is 0, then by divergence theorem, the vector field must have zero divergence. It follows that ∇ · F = 0. This gives that a = 0, but it does not tell us anything about b, c or d.

404

10 Surface Integrals

(2) If the line integral of F around any closed curve is 0, then the vector field has curl equal to zero everywhere. We have − → − → − → ∇ × F = (3 − c) i + (5 − d) j + (1 − b) k . This yields that c = 3, d = 5 and b = 1. It does not tell us anything about a. − → − → − → 484. Let F(x, y, z)x i + y j + z k and suppose that the surface S and region T satisfy the hypothesis of the divergence theorem. Show that the volume of T is given by the formula 1 F · N dσ. volume of T = 3 S

Solution. Using the divergence theorem, we have

F · N dσ = S

∇ · Fd V = T

3d V. T

This implies that 1 3

F · N dσ = S

d V = volume of T. T

485. Let C be the curve obtained by intersection the plane z = x and the cylinder x 2 + y 2 = 1, oriented counterclockwise when viewed from the above. Suppose that S be the inside of this ellipse, oriented with the upward pointing normal. If F = (x, z, 2y), verify Stokes’ theorem. Solution. Using z = x the curve C can be parameterized as R(t) = (cos t, sin t, cos t), for 0 ≤ t ≤ 2π. Then, we get F R(t) = (cos t, cos t, 2 sin t) and R = (− sin t, cos t, − sin t). Then, we have

2π

F · dR = 0

C

=

−

(− sin t cos t + 1 − 3 sin2 t)dt 2π 3 1 2 3

sin t + t − t + sin 2t = −π. 0 2 2 4

On the other hand, since the elliptical disk is a graph over the unit disk in the x yplane, we can parameterize it as R(x, y) = (x, y, x), where x 2 + y 2 ≤ 1. It is easy to check that curl F = (1, 0, 0) and Rx × R y = (−1, 0, 1). So, we conclude that

curl F · d S =

S

(1, 0, 0) · (−1, 0, 1)d xd y = −

S

d xd y = −π. S

10.3 Solved Problems

405

F · d R and

As we see the results of the two integrals C

curl F · d S are same. S

− → − → F · d R, where F = x 2 y i − x y 2 j + 486. Use Stokes’ theorem to compute C − → z 3 k , and C is the curve of intersection of the plane 3x + 2y + z = 6 and the cylinder x 2 + y 2 = 4, oriented clockwise when viewed from above. Solution. Assume that S is the part of the plane 3x + 2y + z = 6 that lies inside the cylinder x 2 + y 2 = 1, oriented downward. This yields that C is the boundary of S. We have

− − → − →

→ j k

i

∂ − → ∂ ∂

= (−y 2 − x 2 ) k . curl F =

∂ y ∂z

∂x

x 2 y −x y 2 z 3

A parametrization of S is R(u, v) = (u, v, 6 − 3u − 2v), where u 2 + v 2 ≤ 4. This gives that Ru × Rv = (1, 0, −3) × (0, 1, −2) = (3, 2, 1), where it is upward. Now, by Stokes’ theorem we can write

F · dR = C

curl F · d S =

S

u 2 +v 2 ≤1

curl F R(u, v) · (−Ru × Rv )dudv

=

u 2 +v 2 ≤1

(0, 0, −u 2 − v 2 ) · (−3, −2, −1)dudv

=

u 2 +v 2 ≤1

2π

(u + v )dudv = 2

2

2

r dr dθ =

2π

3

0

0

4dθ = 8π.

0

487. Suppose that S is the parallelogram with parametric equationsx = u + v, y = u − v and z = 1 + 2u, where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 1. Evaluate (x + y + S

z)d S. − → − → − → Solution. Indeed, we have R(u, v) = (u + v) i + (u − v) j + (1 + 2u) k , where 0 ≤ u ≤ 2 and 0 ≤ v ≤ 1. Hence, we have N = Ru × Rv = (1, 1, 2) × (1, −1, 0) = (2, 2, −2). √ This gives that d S = N dudv = 2 3dudv. Therefore, we obtain

406

10 Surface Integrals

(x + y + z)d S = S

1 2

√ (u + v) + (u − v) + (1 + 2u) (2 3)dudv

0

0

√

1 2

=2 3

0

0

2 √ 1 √

(1 + 4u)dudv = 2 3 (u + 2u 2 ) dv = 20 3. 0

0

F · ds, where F = (z 2 , y 2 , x) and C is the

488. Use Stokes’ theorem to compute C

triangle with vertices (1, 0, 0), (0, 1, 0) and (0, 0, 1) with counterclockwise rotation.

Solution. We have

−

→

i

∂ ∇ × F =

∂x

z2

− → − →

j k

− → ∂ ∂

= (1 − 2z) j . ∂ y ∂z

y2 x

We consider two vectors on the plane. The vector from (1, 0, 0) to (0, 1, 0) is given by A = (−1, 1, 0). The vector from (1, 0, 0) to (0, 0, 1) is given by B = (−1, 0, 1). This gives that N = A × B = (1, 1, 1). Hence, we deduce that the equation of the plane is x + y + z = 1. Now, a parametrization of this plane is R(u, v) = (u, v, 1 − u − v), where 0 ≤ u ≤ 1 and 0 ≤ v ≤ −u + 1. In addition, we have Ru = (1, 0, −1) and Rv = (0, 1, −1).So, we get Ru × Rv = (1, 1, 1). Finally, using Stokes’ theorem we obtain F · ds = (∇ × F) · d S = (0, 2z − 1, 0) · d S C

=

1

1

0

=

0

0

S

0, 2(1 − u − v) − 1, 0 · (1, 1, 1)dvdu

0 −u+1

1

(1 − 2u − 2v)dvdu =

0

1

=

S −u+1

0

−u+1

(v − 2uv − v 2 )

du 0

1 1 2 (−u + 1) − 2u(−u + 1) − (−u + 1) du (u 2 − u)du = − . 6 0

489. Let C be a curve with length L and F be vector field such that F ≤ M. Prove that

F · ds ≤ M L .

C

10.3 Solved Problems

407

Fig. 10.5 A hemisphere

Solution. Let R be the position vector of C. Using Cauchy-Schwarz inequality, we can write

b

b

F · ds =

F R(t) · R (t)dt

≤ |F R(t) · R (t)|dt

C

≤

a

a

b

F R(t) R (t)dt ≤ M

a

b

R (t)dt = M L .

a

490. Find the area of the top half of the sphere x 2 + y 2 + z 2 = a 2 . Solution. The hemisphere is shown in Fig. 10.5. Solving the equation of the sphere for z and setting this equal to f (x, y) we obtain f (x, y) = a 2 − x 2 − y 2 . If σ is the measure of the area of the surface, then we have f x2 (x, y) + f y2 (x, y) + 1d xd y σ= R

= R

x2 y2 + + 1d xd y a2 − x 2 − y2 a2 − x 2 − y2

=

a2 R

a d xd y. − x 2 − y2

Note that the boundary of the region R is the circle x 2 + y 2 = a, where f x and f y are not defined on it. Now, the double integral can be evaluated by an iterated integral in polar coordinates as follows:

408

10 Surface Integrals

Fig. 10.6 The surface obtained by rotating the curve r = cos u, z = sin 2u and −π/2 ≤ u ≤ π/2, around the z-axis

b

2π

a r r dθdr = 2πa lim− √ dr √ 2 2 2 b→a b→a a −r a − r2 0 0 b

= 2πa lim− − a 2 − r 2 = 2πa lim− − a 2 − b2 + a = 2πa 2 .

σ = lim−

b→a

b→a

0

Therefore, the area of the hemisphere is 2πa 2 square units. 491. Suppose that F is tangent to the closed surface S of a region T . Prove that div Fd V = 0. T

Solution. Since F is tangent to S, it follows that F is orthogonal to N . This yields that F · N = 0 everywhere on S. Now, by the divergence theorem we have

div Fd V =

T

F · N dσ = 0, S

as claimed. 492. Let S be the surface obtained by rotating the curve r = cos u, z = sin 2u and −π/2 ≤ u ≤ π/2, around the z-axis, see Fig. 10.6. Use the divergence theorem to find the value of the region inside of S. Solution. We want to compute the integral

d V , where T is the region inside of T

S. By the divergence theorem, we have

10.3 Solved Problems

409

F · d S,

dV = T

S

− → where F is any vector field whose divergence is 1. If we consider F = z k , then

dV =

T

− → z k · d S.

S

In the next step, we determine the surface integral

− → z k · d S. A parametrization

S

of the surface is x = cos u cos θ, y = cos u sin θ and z = sin 2u, where 0 ≤ θ ≤ 2π and −π/2 ≤ u ≤ π/2. Therefore, we have

− → − →

→ − j k

i − → − → − →

i j k

∂x ∂ y ∂z

dθdu d S =

∂θ ∂θ ∂θ

=

− cos u sin θ cos u cos θ 0

∂x ∂ y ∂z − sin u cos θ − sin u sin θ 2 cos 2u

∂u ∂u ∂u = 2 cos u cos 2u cos θ, 2 cos u cos 2u sin θ, cos u sin u dtdθ. Therefore, we have

− → z k · dS =

2π

−π/2

0

S

= 2π

π/2

π/2

−π/2

(sin 2u) cos u sin ududθ

sin 2u cos u sin udu =

π2 . 2

493. Let T = [0, 1] × [0, 1] × [0, 1]. Verify the divergence theorem for the region T boundary S oriented outward and the vector field F = (2x, 3y, 2z). Solution. Computing the divergence we havediv F = 7. So, we have

div Fd V =

T

7d V = 7(the volume of the box T ) = 7. T

410

10 Surface Integrals

Now, suppose that S1 = {(x, y, 0) | x, y ∈ [0, 1]}, S3 = {(0, y, z) | y, z ∈ [0, 1]}, S5 = {(x, 0, z) | x, z ∈ [0, 1]},

S2 = {(x, y, 1) | x, y ∈ [0, 1]}, S4 = {(1, y, z) | y, z ∈ [0, 1]}, S6 = {(x, 1, z) | x, z ∈ [0, 1]},

− → − → and the unit normal vectors to each face is N2 = −N1 = k , N4 = −N3 = i and − → N6 = −N5 = j . Next, we can write F · N dσ S

=−

2zdσ +

S1

=0+

2zdσ −

S2

1 1 0

0

2xdσ +

S3

2d xd y + 0 +

S4

1 1 0

0

2xdσ −

2dydz + 0 +

3ydσ − S5

S6

1 1 0

3ydσ

0

3d xdz = 2 + 2 + 3 = 7.

Note that we have plugged in the equation of the plane into the integrand. 494. Among the rectangular solids defined by the inequalities 0 ≤ x ≤ a, 0 ≤ y ≤ b − → and 0 ≤ z ≤ 1, find one for which the total flux of F(x, y, z) = (−x 2 − 4x y) i − − → − → 6yz j + 12z k outward through the six sides is greatest. What is the greatest flux? Solution. Since ∇ · F = −2x − 4y − 6z + 12, it follows that the flux is equal to 0

a

b

0

1

− 2x − 4y − 6z + 12 dzdyd x = ab(−a − 2b + 9).

0

We put f (a, b) = ab(−a − 2b + 9). Then, we have ∂f ∂f (a, b) = −2ab − 2b2 + 9b and (a, b) = −a 2 − 4ab + 9a. ∂a ∂b If ∂ f /∂a = 0 and ∂ f /∂b = 0, then b(−2a − 2b + 9) = 0 and a(−a − 4b + 9) = 0. This gives that (a = 0 or −a − 4b + 9 = 0) and (b = 0 or −2a − 2b + 9 = 0). If a = 0 or b = 0, then the flux is zero. If −2a − 2b + 9 = 0 and −a − 4b + 9 = 0, then a = 3 and b = 3/2. This yields that f (3, 3/2) = 27/2 is the maximum flux. 495. Let S be the sphere x 2 + y 2 + (z − 1)2 = 9. Find the unit outward normal vector to the surface S and evaluate the surface integral I =

S

x 2 sin y + y cos2 x + (z − 1)(y 2 − z sin y) dσ.

10.3 Solved Problems

411

Solution. Suppose that f (x, y, z) = x 2 + y 2 + (z − 1)2 . Then, the given sphere is f (x, y, z) = 9. The unit normal vector of S is N = ∇ f /∇ f = 1/3(x, y, z − 1). It is easy to check that N is the unit outward normal vector. Hence, we can write F · N dσ,

I =3 S

where F(x, y, z) = x sin y, cos x, y 2 − z sin y . Moreover, we have div F =

∂ ∂ ∂ , , · x sin y, cos x, y 2 − z sin y = sin y + 0 − sin y = 0. ∂x ∂ y ∂z

Finally, by the divergence theorem we get I =

div Fd V = 0. T

F · N dσ, where

496. Use the divergence theorem to calculate S

− → − → y3 − → + tan z j + (x 2 z + y 2 ) k F(x, y, z) = z 2 x i + 3 and S is the top half of the sphere x 2 + y 2 + z 2 = 1 oriented upward. Solution. Not that S is not a closed surface. Assume that S = {(x, y, 0) | x 2 + y 2 ≤ 1} oriented downward and take S = S ∪ S . Then, the surface integral over S can be obtained from integrals over S and S . If T = {(x, y, z) | x 2 + y 2 + z 2 ≤ 1, z ≥ 0}, then S is the boundary of T . So, by the divergence theorem we have

F · N dσ =

div Fd V.

S

T

Since div F =

∂ 2 ∂ y3 ∂ 2 (z x) + + tan z + (x z + y 2 ) = x 2 + y 2 + z 2 , ∂x ∂y 3 ∂z

by using spherical coordinates we get

(x 2 + y 2 + z 2 )d V =

F · N dσ = S

T

2π π/2 1 0

0

2π . ρ2 ρ2 sin ϕ dρdϕdθ = 5 0

412

10 Surface Integrals

− → F · N dσ we have N = − k and z = 0

On the other hand, for the surface integral S

on S . This gives that F · N = −y 2 . Now, we conclude that

F · N dσ = S

x 2 +y 2 ≤1

2π

−y 2 d A = 0

0

1

π − r 2 sin2 θ r dr dθ = − . 4

Consequently, we obtain

F · N dσ =

F · N dσ −

S

S

F · N dσ = S

13π 2π π + = . 5 4 20

497. Let C be a simple closed smooth curve in the plane 2x + 2y + z = 2, oriented 2yd x + 3zdy − xdz depends only on the area

as shown in Fig. 10.7. Show that C

of the region enclosed by C and not on the position or shape of C. − → − → − → Solution. Taking F(x, y, z) = 2y i + 3z j − x k we obtain

−

→

i

∂ ∇ × F =

∂x

2y

− → − →

j k

− → − → − → ∂ ∂

= −3 i + j − 2 k , ∂ y ∂z

3z −x

− → − → − → and N = (2 i + 2 j + k )/3. So, we have

Fig. 10.7 The simple closed smooth curve in Problem 497

10.3 Solved Problems

413

2yd x + 3zdy − xdz = C

(∇ × F) · N dσ =

S

−2dσ = −2

S

dσ, S

dσ is the area of the region enclosed by C on the plane S, that is, 2x +

where S

2y + z = 2. 498. Green’s first formula: Suppose that f and g are scalar functions with continuous first and second-order partial derivatives through a region T that is bounded by a closed piecewise smooth surface S. Show that f ∇g · N dσ = S

f ∇ 2 g + ∇ f · ∇g d V.

(10.5)

T

Solution. We apply the divergence theorem to the field F = f ∇g. Then, we obtain

∇ · f ∇gd V

f ∇g · N dσ = S

=

∇· T

=

f

T

∂g − ∂g − ∂g − → → → i + f j + f k dV f ∂x ∂y ∂z

∂ 2 g ∂ f ∂g ∂2 g ∂2 g ∂ f ∂g ∂ f ∂g + f + f dV + + + ∂x 2 ∂x ∂x ∂ y2 ∂y ∂y ∂z 2 ∂z ∂z

T

2 ∂ g ∂ f ∂g ∂ f ∂g ∂2 g ∂ 2 g ∂ f ∂g f + + dV = + + 2 + ∂x 2 ∂ y2 ∂z ∂x ∂x ∂y ∂y ∂z ∂z T

=

f ∇ 2 g + ∇ f · ∇g d V.

T

499. Green’s second formula: With the assumption in Problem 498, show that

f ∇g − g∇ f · N dσ = f ∇ 2 g − g∇ 2 f d V.

S

T

Solution. Interchanging the role of f and g in (10.5), we obtain the following similar formula g∇ f · N dσ = (10.6) g∇ 2 f + ∇g · ∇ f d V. S

T

Subtracting (10.5) from (10.6), the requested equality follows.

414

10 Surface Integrals

500. Suppose that F is given in cylindrical coordinates as follows − → F(r, θ, z) = r (2 + sin2 θ)er + (r sin θ cos θ)eθ + 3z k , and S is the quarter-cylinder r = 2, 0 ≤ θ ≤ π and 0 ≤ z ≤ 5. Evaluate the surface F · N dσ.

flux S

Solution. Since S is a closed surface, we can use the divergence theorem. We have 1 ∂ ∂ 1 ∂ 2 r (2 + sin θ) + r sin θ cos θ + (3z) r ∂r r ∂θ ∂z 1 = (2r )(2 + sin2 θ) + (cos2 θ − sin2 θ) + 3 r

∇·F =

= 4 + 2 sin2 θ + cos2 θ − sin2 θ = 5. Therefore, by the divergence theorem we obtain

F · N dσ = S

∇ · Fd V = 5

T

dV = 5

π(2)2 (5) = 25π. 4

T

10.4 Exercises Apply Green’s theorem in evaluating the line integral of F along the indicated curve C oriented in the counterclockwise direction. F · d R, where F(x, y) = (x 2 , 20x y), and C is a closed curve formed by 1. C √ y = x and y = x 2 ; 2. F · d R, where F(x, y) = (3y − esin x , 7x + y 4 + 1), where C is the circle C

x 2 + y 2 = 9; 3. F · d R, where F(x, y) = (x 2 , e x−y ), and C is a a parallelogram with vertices C

(0, 0), (1, 0), (2, 1) and (1, 1). Evaluate the given surface integral. 4. (x + y + z)dσ, where S is the cube in the first octant bounded by the coorS

dinate planes and planes x = 1, y = 1 and z = 1;

10.4 Exercises

415

(x 2 y 2 + x 2 z 2 + y 2 z 2 )dσ, where S is the part of the cone z =

5.

x 2 + y 2 cut

S

off by the cylinder x 2 + y 2 = 2x. 6. Let R be the region enclosed by a simple closed piecewise smooth curve C. Let F, Fx and Fy be continuous on an open set containing R. Show that

Fx d xd y =

R

Fy d xd y = −

Fdy and C

Fd x. C

R

7. Let R be the region enclosed by a simple closed smooth curve C. Show that

xdy = −

Area of R = C

yd x. C

8. Let f : [a; b] → R be a non-negative function such that its first derivative is continuous. Suppose that C is the boundary of the region bounded above by the graph of f , below by the x-axis and on the sides by the lines x = a and x = b. Show that b f (x)d x = − yd x. a

C

− → 9. Verify that Stokes’ theorem is true for the vector field F(x, y, z) = x 2 i + − → − → y 2 j + z 2 k , where S is the part of the paraboloid z = 1 − x 2 − y 2 that lies above the x y-plane and S has upward orientation. 10. Let S be the upper hemisphere x 2 + y 2 + z 2 = 1, z ≥ 0. → → y− y− (a) Find F such that curl F = xe i − e j ; 2 y (b) Evaluate x e − ye y dσ. S

11. Show that there is no vector field F such that − → − → − → curl F = 2x i + 3yz j − x z 2 k . ( f ∇g + g∇ f ) · d R = 0, where R describes a closed curve C to

12. Show that C

which Stokes’ theorem applies. 13. Find the area of the surface z = (1/2)(x 2 + y 2 ) that lies below the plane z = 2. 14. Find the surface area of the parameterized torus − → − → − → R(u, v) = (a + b cos u) cos v i + (a + b cos u) sin v j + b sin u k , where 0 ≤ u ≤ 2π, 0 ≤ v ≤ 2π and 0 < b < a are constants.

416

10 Surface Integrals

15. Find the area of the portion S of the cylinder x 2 + y 2 = 2y that lies inside the sphere x 2 + y 2 + z 2 = 4. − → √ − → − → 16. Consider the parameterized surface S: R(u, v) = u 2 i + 2uv j + v 2 k . Find the area of the portion of the surface S that lies inside the unit ball x 2 + y 2 + z 2 ≤ 1. F · d S, where

17. Use the divergence theorem to calculate the surface integral S

− → − → − → F(x, y, z) = x 3 i + y 3 j + z 3 k and S is the surface of the solid bounded by the cylinder x 2 + y 2 = 1 and the planes z = 0 and z = 2. 18. Let f (x, y, z) = (x 2 + y 2 + z 2 )−1/2 . Show that the clockwise circulation of the field F = ∇ f around the circle x 2 + y 2 = a 2 in the x y-plane is zero − → − → (a) by taking R(t) = a cos t i + a sin t j , 0 ≤ t ≤ 2π and integrating F · d R over the circle; (b) by applying Stokes’ theorem. 19. Show that the curl of F(x, y, z) =

−y − x → − → − → i + 2 j +z k x 2 + y2 x + y2

is zero but that

F · dR C

is not zero if C is the circle x 2 + y 2 = 1 in the x y-plane. (the theorem does not apply here because the domain of F is not simply connected. The field F is not defined along the z-axis so there is no way to contract C to a point without leaving the domain of F.)

Harder Exercises − → − → − → 20. If A is a constant vector, R = x i + y j + z k , and S is an oriented, smooth surface with a simple, closed, smooth, positively oriented boundary curve C, show that 2 A · d S = (A × R) · d R. S

C

21. Let F be a vector field from R3 to R3 with continuous first partial derivatives. If div F = 0, prove F = curlG for some vector field G.

10.4 Exercises

417

Fig. 10.8 S is the outwardly oriented surface

22. Find

− → − → − → F · N dσ, where F(x, y, z) = x i + y j + z k and S is the outwardly

S

oriented surface shown in Fig. 10.8 (the boundary surface of a cube with a unit corner cube removed). 23. A particle moves along the smooth curve from (a, f (a)) to (b, f (b)). The force moving the particle has constant magnitude k and always points away from the origin. Show that the work done by the force is F · T dσ = k

2 1/2 2 2 1/2 . b2 + f (b) − a + f (a)

C

24. Find the outward flux of the field F(x, y) = 3x y −

− x − → x → i + e + tan−1 y j 1+y

across the cardioid r = a(1 + cos θ), a > 0. − → − → − → 25. Find the flux of the vector field F = x i + y j + z k out of the closed surface 2 2 S bounded by the cylinder x + y = 1 and the planes z = 0 and z = 3. x y 2 d x + (x 2 y + 2x)dy around any square depends

26. Show that the value of C

only on the area of the square and not on its location in the plane. 27. Compute the surface area of that portion of the sphere x 2 + y 2 + z 2 = a 2 lying within the cylinder x 2 + y 2 = ay, a > 0. 28. Compute the area of that portion of the paraboloid x 2 + z 2 = 2ay which is cut off by the plane y = a.

418

10 Surface Integrals

29. Let T be a convex region in R3 whose boundary is a closed surface S and let N be the unit normal vector to S. Let F and G be two continuously differentiable vector fields such that curl F = curlG and div F = divG everywhere in T , and such that F · N = G · N everywhere on S. Prove that F = G everywhere in T . Hint: Let H = F − G, find a scalar field f such that H = ∇ f , and use a suitable ∇ f 2 d xd ydz = 0. From this deduce that H = 0

identity to prove that T

in T .

References

1. Apostol, T.M.: Calculus. Vol. I: One-variable Calculus, with an Introduction To Linear Algebra. Second edition Blaisdell Publishing Co. Ginn and Co., Waltham, Mass.-Toronto, Ont.-London (1967) 2. Apostol, T.M.: Mathematical Analysis. Addison-Wesley Publishing Company (1975) 3. Bartle, R.G.: The Elements of Real Analysis, 2nd edn. Wiley (1976) 4. Buck, R.: Creighton. 3rd edn. Waveland Pr Inc, Advanced calculus (2003) 5. Duren, W.L.: Calculus and Analytic Geometry. Xerox College Pub. (1972) 6. Ellis, R., Gulick, D.: Calculus with Analytic Geometry, 5th edn. Holt Rinehart & Winston (1998) 7. Kuhfittig, P.: Technical Calculus with Analytic Geometry, Boston. Brooks/Cole, Cengage Learning, MA (2013) 8. Leithold, L.: The Calculus with Analytic Geometry. Harper & Row, Publishers (1981) 9. Morris, C.C., Stark, R.M.: Fundamentals of Calculus. Wiley (2015) 10. Rudin, W.: Principles of Mathematical Analysis. International Series in Pure and Applied Mathematics, 3rd edn. McGraw-Hill Book Co., New York-Auckland-Dsseldorf (1976) 11. Silverman, R.A.: Calculus with Analytic Geometry. Prentice-Hall, Inc. (1985) 12. Simmons, G.F.: Calculus with Analytic Geometry, 2nd edn. McGraw-Hill Education (1996) 13. Stewart, J.: Calculus, 7th edn. Cengage Learning (2012) 14. Thomas, G.B., Finney, R.L.: Calculus and Analytic Geometry. Addison Wesley Publishing Company (1995) 15. Trench, W.F.: Advanced Calculus. Harper & Row, New York (1978)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1

419

Index

A Absolute maximum, 217 Absolute minimum, 217 Acceleration, 68 Accumulation point, 107 Angle between two planes, 5 Antiderivative of a vector function, 67 Approach a limit, 105 Average value of an integrable function, 323 Axis of the surface of revolution, 5 B Basis vectors, 1 Bernoulii spiral, 93 Bolzano-Weirstrass theorem, 136 Boundary of a set, 105 Boundary point, 105 Bounded function, 108 C Cardioid, 50 Cauchy convergence criterion, 142 Cauchey’s inequality, 256 Cauchy-Schwarts inequality for integrals, 343 Cauchy sequence, 136 Cauchy’s mean value theorem for functions of two variables, 211 Cavalieri’s principle, 39 Center of curvature, 70 Center of mass, 326 Chain rule for Jacobian, 275 Changes of variables in double integrals, 324 Changes of variables in triple integrals, 326

Clairaut’s theorem, 145 Closed curve, 317 Closed set, 105 Cobb-Douglas production function, 252 Component function, 65, 263 Component of a vector, 1 Conservative vector field, 268 Continuous at a point, 108 Continuous path, 300 Continuous vector field, 265 Continuous vector function, 65 Contour map, 104 Convergent of a sequence, 134 Convex function, 237 Convex set, 139 Convolution function, 345 Coordinate of a vector, 265 Critical point, 217 Cross product, 2 Curl, 269 Curvature, 70 Curvature vector, 70 Cylinder, 5 Cylindrical coordinate system, 10

D Deleted neighborhood, 105 Del operator, 269 Derivative of a vector function, 66 Directional derivative, 145 Direction cosines, 2 Direction numbers of a line, 3 Direction of a vector, 2 Directrix, 5

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Singapore Pte Ltd. 2023 B. Davvaz, Vectors and Functions of Several Variables, https://doi.org/10.1007/978-981-99-2935-1

421

422 Discontinuous, 109 Distance between a point and a line in space, 4 Distance between a point and a plane, 4 Distance between two points in Rn , 105 Divergence, 269 Divergence theorem, 391 Domain, 103 Dot product, 2 Double integral, 321

E Elliptic cylinder, 5 Epicycloid, 102 Error in the standard linear approximation, The, 172 Essential discontinuity, 109 Extreme value theorem, 217

F First derivative test for local extreme values, 217 First moments about the coordinates planes, 326 First-order linear vector differential equation, 83 First partial derivatives, 145 Folium of Descartes, 63 Fubini’s theorem, 323 Fundamental theorem of calculus for line integrals, 300

G Generating curve, 6 Gradient, 146 Graph of a function, 103 Graph of a polar equation, 9 Green’s first formula, 413 Green’s second formula, 413 Green’s theorem, 391

H Harmonic function, 270 Heron’s formula, 26 Hessian matrix, 217 Homogeneous first order differential equation, 199 Homogenous function of degree k, 193 Horn torus, 63 Hyperbolic cylinder, 5

Index Hyperbolic spiral, 53

I Increment, 171 Indefinite integral of a vector function, 67 Infinite limits, 107 Inner product, 104 Integrable function, 321 Interior point, 105 Intermediate Value Theorem, 109 Iterated limits, 121

J Jacobian, 267 Jacobian matrix, 267

L Lagrange multiplier rule, 219 Laplace transformation, 345 Laplace’s equation, 188 Laplacian, 270 Level curve, 103 Level surface, 104 Limaçon, 50 Limit of a vector function, 65 Limit point, 105 Linearization of a function, 172 Linearly dependent set, 3 Linearly independent set, 3 Linear map, 133 Linear transformation, 265 Line integral, 299 Line is space, 3 Line of best fit, 229 Lipschitz condition, 132 Local maximum, 217 Local minimum, 217

M Magnitude of a vector, 1 Mass of an object, 326 Matrix form of the chain rule, 268 Mean value theorem for double integrals, 324, 351 Mean value theorem for functions of two variables, 211 Moment of inertia, 327 Moments of inertia about a line, 327 Multi-variable function of n variables, 103

Index N Nabla, 269 Neighborhood, 105 Newton’s second law of motion, 316 Norm, 105 Norm of a partition, 321 Normal line, 146 Normal vector to a surface, 146 O Open set, 105 Orthogonal surfaces, 162 Orthogonal vectors, 2 P Parabolic cylinder, 5 Parallel vectors, 2 Parametric equations of a line, 3 Parametrization surface, 162 Partial derivative, 143 Partial derivatives of higher order, 145 Particle’s path, 65 Path independent, 300 Piecewise smooth path, 300 Plane, 4 Points in Rn , 104 Polar axis, 7 Pole, 7 Position representation of a vector, 1 Positive semi-definite matrix, 238 Potential function, 268 Product of vectors, 2 Q Quadric surface, 7 R Radius of curvature, 70 Range, 103 Rational parametrization, 77 Real valued function of n variable, 103 Removable discontinuity, 109 Representation of linear transformations by matrices, 266 Rose, 51 Rotation free, 269 Ruling, 5 S Saddle point, 217

423 Sandwich theorem, 107 Scalar field, 263 Scalar product, 2 Second derivative test for local extreme values, 217 Second moments, 327 Second partial derivatives, 145 Skew lines, 29 Smooth path, 300 Solenoidal vector field, 269 Speed, 68 Sphere, 6 Spherical coordinates, 11 Spiral of Archimedes, 52 Standard linear approximation, 172 Steiner problem, 232 Stokes’ theorem, 391 Subsequence, 136 Surface integral, 392 Surface of revolution, 5 Symmetric equations of a line, 4 Symmetry tests for polar graphs, 9

T Tangential and normal components of acceleration, 70 Tangent plane, 146 Taylor polynomial, 175 Taylor remainder theorem, 175 Taylor series, 174 Three-dimensional Laplace’s equation, 150 Torsion function, 70 Total curvature, 95 Total derivative matrix, 267 Total differential, 173 Tractrix, 63 Triple integral, 325

U Unit binormal vector, 69 Unit normal vector, 69 Unit tangent vector, 69 Unit vector, 2

V Vector, 1 Vector field, 263 Vector function, 65 Vector projection, 3 Vectors in Rn , 104 Velocity, 68

424 W Wave equation, 151 Wilson lot size formula, 206 Work, 299

Index Z Zero vector, 1