Elementary Vector Calculus and its Applications with MATLAB Programming 9788770223874, 9788770223867

Citation preview

Nita H. Shah Jitendra Panchal Sir Isaac Newton, one of the greatest scientists and mathematicians of all time, introduced the notion of a vector to define the existence of gravitational forces, the motion of the planets around the sun, and the motion of the moon around the earth. Vector calculus is a fundamental scientific tool that allows us to investigate the origins and evolution of space and time, as well as the origins of gravity, electromagnetism, and nuclear forces. Vector calculus is an essential language of mathematical physics, and plays a vital role in differential geometry and studies related to partial differential equations widely used in physics, engineering, fluid flow, electromagnetic fields, and other disciplines. ­Vector calculus represents physical quantities in two or three-­dimensional space, as well as the variations in these quantities. The machinery of differential geometry, of which vector calculus is a subset, is used to understand most of the analytic results in a more general form. Many topics in the physical sciences can be mathematically studied using vector calculus techniques. This book is designed under the assumption that the readers have no prior knowledge of vector calculus. It begins with an introduction to vectors and scalars, and also covers scalar and vector products, vector differentiation and integrals, Gauss’s theorem, Stokes’s theorem, and Green’s theorem. The MATLAB programming is given in the last chapter.

River Publishers

Elementary Vector Calculus and its Applications with MATLAB Programming

Nita H. Shah Jitendra Pancha

This book includes many illustrations, solved examples, practice examples, and multiple-choice questions.

Elementary Vector Calculus and its Applications with MATLAB Programming

Elementary Vector Calculus and its Applications with MATLAB Programming

River Publishers Series in Mathematical, Statistical and Computational Modelling for Engineering

River

Nita H. Shah Jitendra Panchal

Elementary Vector Calculus and its Applications with MATLAB Programming

RIVER PUBLISHERS SERIES IN MATHEMATICAL, STATISTICAL AND COMPUTATIONAL MODELLING FOR ENGINEERING Series Editors: MANGEY RAM Graphic Era University, India TADASHI DOHI Hiroshima University, Japan ALIAKBAR MONTAZER HAGHIGHI Prairie View Texas A&M University, USA Applied mathematical techniques along with statistical and computational data analysis has become vital skills across the physical sciences. The purpose of this book series is to present novel applications of numerical and computational modelling and data analysis across the applied sciences. We encourage applied mathematicians, statisticians, data scientists and computing engineers working in a comprehensive range of research fields to showcase different techniques and skills, such as differential equations, finite element method, algorithms, discrete mathematics, numerical simulation, machine learning, probability and statistics, fuzzy theory, etc Books published in the series include professional research monographs, edited volumes, conference proceedings, handbooks and textbooks, which provide new insights for researchers, specialists in industry, and graduate students. Topics included in this series are as follows:• • • • • • • • • • • • • • • •

Discrete mathematics and computation Fault diagnosis and fault tolerance Finite element method (FEM) modeling/simulation Fuzzy and possibility theory Fuzzy logic and neuro-fuzzy systems for relevant engineering applications Game Theory Mathematical concepts and applications Modelling in engineering applications Numerical simulations Optimization and algorithms Queueing systems Resilience Stochastic modelling and statistical inference Stochastic Processes Structural Mechanics Theoretical and applied mechanics

For a list of other books in this series, visit www.riverpublishers.com

Elementary Vector Calculus and its Applications with MATLAB Programming

Nita H. Shah Gujarat University, India

Jitendra Panchal Parul University, India

Published, sold and distributed by: River Publishers Alsbjergvej 10 9260 Gistrup Denmark www.riverpublishers.com

ISBN: 978-87-7022-387-4 (Hardback) 978-87-7022-386-7 (Ebook) c 2022 River Publishers 

All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, mechanical, photocopying, recording or otherwise, without prior written permission of the publishers.

Contents

Preface

ix

List of Figures

xi

1 Basic Concept of Vectors and Scalars 1.1 Introduction and Importance . . . . . . . . . . . . . . . . . 1.2 Representation of Vectors . . . . . . . . . . . . . . . . . . . 1.3 Position Vector and Vector Components . . . . . . . . . . . 1.4 Modulus or Absolute Value of a Vector . . . . . . . . . . . . 1.5 Zero Vector and Unit Vector . . . . . . . . . . . . . . . . . 1.6 Unit Vectors in the Direction of Axes . . . . . . . . . . . . 1.7 Representation of a Vector in terms of Unit Vectors . . . . . 1.8 Addition and Subtraction of Vectors . . . . . . . . . . . . . 1.9 Product of a Vector with a Scalar . . . . . . . . . . . . . . . 1.10 Direction of a Vector . . . . . . . . . . . . . . . . . . . . . 1.11 Collinear and Coplanar Vectors . . . . . . . . . . . . . . . . 1.11.1 Collinear Vectors . . . . . . . . . . . . . . . . . . . 1.11.2 Coplanar Vectors . . . . . . . . . . . . . . . . . . . 1.12 Geometric Representation of a Vector Sum . . . . . . . . . . 1.12.1 Law of Parallelogram of Vectors . . . . . . . . . . . 1.12.2 Law of Triangle of Vectors . . . . . . . . . . . . . . 1.12.3 Properties of Addition of Vectors . . . . . . . . . . . 1.12.4 Properties of Scalar Product . . . . . . . . . . . . . 1.12.5 Expression of Any Vector in Terms of the Vectors Associated with its Initial Point and Terminal Point . 1.12.6 Expression of Any Vector in Terms of Position Vectors . . . . . . . . . . . . . . . . . . . . . . . . 1.13 Direction Cosines of a Vector . . . . . . . . . . . . . . . . . 1.14 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 1 1 2 3 4 4 5 6 6 7 8 8 8 8 8 9 9 10

v

10 11 12 26

vi

Contents

2 Scalar and Vector Products 2.1 Scalar Product, or Dot Product, or Inner Product . . . . . 2.2 The Measure of Angle Between two Vectors and Projections . . . . . . . . . . . . . . . . . . . . . . . . 2.2.1 Properties of a Dot Product . . . . . . . . . . . . 2.3 Vector Product or Cross Product or Outer Product of Two Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Geometric Interpretation of a Vector Product . . . . . . . 2.4.1 Properties of a Vector Product . . . . . . . . . . 2.5 Application of Scalar and Vector Products . . . . . . . . 2.5.1 Work Done by a Force . . . . . . . . . . . . . . 2.5.2 Moment of a Force About a Point . . . . . . . . 2.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Vector Differential Calculus 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Vector and Scalar Functions and Fields . . . . . . . . . . 3.2.1 Scalar Function and Field . . . . . . . . . . . . 3.2.2 Vector Function and Field . . . . . . . . . . . . 3.2.3 Level Surfaces . . . . . . . . . . . . . . . . . . 3.3 Curve and Arc Length . . . . . . . . . . . . . . . . . . . 3.3.1 Parametric Representation of Curves . . . . . . . 3.3.2 Curves with Tangent Vector . . . . . . . . . . . 3.3.2.1 Tangent Vector . . . . . . . . . . . . . 3.3.2.2 Important Concepts . . . . . . . . . . 3.3.3 Arc Length . . . . . . . . . . . . . . . . . . . . 3.3.3.1 Unit Tangent Vector . . . . . . . . . . 3.4 Curvature and Torsion . . . . . . . . . . . . . . . . . . . 3.4.1 Formulas for Curvature and Torsion . . . . . . . 3.5 Vector Differentiation . . . . . . . . . . . . . . . . . . . 3.6 Gradient of a Scalar Field and Directional Derivative . . 3.6.1 Gradient of a Scalar Field . . . . . . . . . . . . 3.6.1.1 Properties of Gradient . . . . . . . . . 3.6.2 Directional Derivative . . . . . . . . . . . . . . 3.6.2.1 Properties of Gradient . . . . . . . . . 3.6.3 Equations of Tangent and Normal to the Level Curves . . . . . . . . . . . . . . . . . . . . . . 3.6.4 Equation of the Tangent Planes and Normal Lines to the Surfaces . . . . . . . . . . . . . . . . . .

. .

29 29

. . . .

30 30

. . . . . . .

. . . . . . .

37 38 39 45 46 46 52

. . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . .

55 55 55 56 56 56 57 57 58 59 60 61 61 64 67 70 73 73 73 74 75

. .

84

. .

85

Contents vii

3.7

3.8

Divergence and Curl of a Vector Field . . . . . . . . . . . . 86 3.7.1 Divergence of a Vector Field . . . . . . . . . . . . . 86 3.7.1.1 Physical Interpretation of Divergence . . . 86 3.7.2 Curl of a Vector Field . . . . . . . . . . . . . . . . . 89 3.7.2.1 Physical Interpretation of Curl . . . . . . . 89 3.7.3 Formulae for grad, div, curl Involving Operator ∇ . 96 3.7.3.1 Formulae for grad, div, curl Involving Operator ∇ Once . . . . . . . . . . . . . 96 3.7.3.2 Formulae for grad, div, curl Involving Operator ∇ Twice . . . . . . . . . . . . . 100 Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

4 Vector Integral Calculus 4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . 4.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Circulation . . . . . . . . . . . . . . . . . . . 4.2.2 Work Done by a Force . . . . . . . . . . . . . 4.3 Path Independence of Line Integrals . . . . . . . . . . 4.3.1 Theorem: Independent of Path . . . . . . . . . 4.4 Surface Integrals . . . . . . . . . . . . . . . . . . . . 4.4.1 Flux . . . . . . . . . . . . . . . . . . . . . . . 4.4.2 Evaluation of Surface Integral . . . . . . . . . 4.4.2.1 Component form of Surface Integral 4.5 Volume Integrals . . . . . . . . . . . . . . . . . . . . 4.5.1 Component Form of Volume Integral . . . . . 4.6 Exercise . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . .

111 111 111 112 112 113 113 122 123 123 124 129 129 131

. . . . .

. . . . .

135 135 137 146 154 163

6 MATLAB Programming 6.1 Basic of MATLAB Programming . . . . . . . . . . . . . . 6.1.1 Basic of MATLAB Programming . . . . . . . . . 6.1.1.1 Introductory MATLAB programmes . . 6.1.1.2 Representation of a Vector in MATLAB

. . . .

167 167 167 168 183

5 Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem 5.1 Green’s Theorem (in the Plane) . . . . . . . . . . . . . 5.1.1 Area of the Plane Region . . . . . . . . . . . . 5.2 Stokes’ Theorem . . . . . . . . . . . . . . . . . . . . 5.3 Gauss’ Divergence Theorem . . . . . . . . . . . . . . 5.4 Exercise . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . .

viii

Contents

6.2

6.1.1.3 Representation of a Matrix in MATLAB . 186 Some Miscellaneous Examples using MATLAB Programming . . . . . . . . . . . . . . . . . . . . . . . . . 188

Index

207

About the Authors

213

Preface

Vector calculus is an essential language of mathematical physics. Vector calculus plays a vital role in differential geometry, and the study related to partial differential equations is widely used in physics, engineering, fluid flow, electromagnetic fields, and other disciplines. Vector calculus represents physical quantities in two or three-dimensional space, as well as the variations in these quantities. The machinery of differential geometry, of which vector calculus is a subset, is used to understand most of the analytic results in a more general form. Many topics in the physical sciences can be mathematically studied using vector calculus techniques. Description of the book: This book is meant for readers who have a basic understanding of vector calculus. This book is designed to provide accurate information to readers. The language in the book is kept simple so that all readers can easily understand each concept. This book begins with the introduction of vectors and scalars in chapter 1. Chapter 1 contains essential basic definitions and concepts, vector in terms of unit vectors, geometric representation of vector sum, and direction cosines. The scalar and vector products, measurement of angle and projections, geometric interpretation of a vector product, and their applications are given in chapter 2. In chapter 3, vector and scalar functions and fields, curves, arc length, formulae for curvature and torsion, and its derivation, curl, divergence, and gradient with important properties and physical interpretation, and important results are given in vector differential calculus. Chapter 4 vector integral calculus includes line integrals, circulation, path independence, surface integrals, volume integrals, and its applications like flux and work done by a force are given. In chapter 5, derivation of Green’s theorem, Stokes’s theorem, and Gauss’ divergence theorem are given with various solve examples. MATLAB programming is given in the last chapter 6 includes basic information about MATLAB. Initially, basic examples are given with proper

ix

x

Preface

explanation wherever possible that helps readers to understand basic input and output, arithmetic operations, functions, plotting commands available in MATLAB. Variety of solved programs with MATLAB codes along with compiles and debug outputs. So, the reader can run the program using given codes and observe results. For MATLAB product information, please contact: The MathWorks, Inc. 3 Apple Hill Drive Natick, MA 01760-2098 USA Tel: 508-647-7000 Fax: 508-647-7001 E-mail: [email protected] Web: www.mathworks.com

List of Figures

Figure 1.1 Figure 1.2 Figure 1.3 Figure 1.4 Figure 1.5 Figure 1.6 Figure 1.7 Figure 1.8 Figure 1.9 Figure 1.10 Figure 1.11 Figure 1.12 Figure 1.13 Figure 1.14 Figure 1.15 Figure 1.16 Figure 2.1 Figure 2.2 Figure 2.3 Figure 2.4 Figure 2.5 Figure 3.1 Figure 3.2 Figure 3.3 Figure 3.4

Represents the geometrical representation of a vector . . . . . . . . . . . . . . . . . . . . . . Represents a position vector . . . . . . . . . . . . Represents equal and negative vectors . . . . . . . Represents unit vectors in xy-plane . . . . . . . . . Represents unit vectors in the direction of x-axis, y-axis, and z-axis . . . . . . . . . . . . . . . . . .  in terms of unit vectors . . Represents a vector OP Represents the product of a vector with a scalar . . Represents the law of parallelogram of vectors . . . Represents the law of the triangle of vectors . . . . Represents any vector in terms of various vectors associated with its endpoints. . . . . . . . . . . . . Represents any vector in terms of the position vector . . . . . . . . . . . . . . . . . . . . . . . . Represents direction cosines of a vector . . . . . . Represents a parallelogram . . . . . . . . . . . . . Represents a regular hexagon . . . . . . . . . . . . Represents a triangle ΔABC . . . . . . . . . . . . A space shuttle of 1000 tons weight hangs from two skyscrapers using steel cables . . . . . . . . . . . . Represents a scalar or dot product . . . . . . . . . Represents a vector or cross product . . . . . . . . Represents the geometric interpretation of a vector or cross product . . . . . . . . . . . . . . . . . . . Represents work done by a force F on a particle . . Represents the moment of a force about a point . . Represents a curve with a tangent vector . . . . . . Represents the plane of curvature of the curve . . . Represents the arc rate of rotation of binormal . . . ˆ Tˆ, and N ˆ orthogonal unit vectors . Represents B,

xi

1 2 3 4 5 6 7 9 9 11 11 12 14 15 16 25 29 38 39 46 47 59 60 65 66

xii

List of Figures

Figure 3.5 Figure 3.6 Figure 3.7 Figure 4.1 Figure 4.2 Figure 4.3 Figure 4.4 Figure 4.5 Figure 4.6 Figure 4.7 Figure 4.8 Figure 4.9 Figure 5.1 Figure 5.2 Figure 5.3 Figure 5.4 Figure 5.5 Figure 5.6 Figure 5.7 Figure 5.8 Figure 5.9 Figure 5.10 Figure 5.11 Figure 5.12 Figure 5.13 Figure 5.14

Represents the derivative of the vector v (t) . . . . Represents the directional derivative . . . . . . . . Represents the parallelopiped . . . . . . . . . . . . Representation of a vector function defined at every point of a curve C . . . . . . . . . . . . . . . . . . Representation of a closed curve C . . . . . . . . . Representation of parabola x = y 2 . . . . . . . . . Representation of the rectangle in xy-plane bounded by lines . . . . . . . . . . . . . . . . . . Representation of curved surface S and a plane region R . . . . . . . . . . . . . . . . . . . . . . . Representation of the projection of the plane in the first octant . . . . . . . . . . . . . . . . . . . . . . Representation of the positive octant of the sphere . Representation of the sphere . . . . . . . . . . . . Representation of the cylinder in positive octant . . Represents the region R bounded by the curve C . Represents the region R bounded two parabolas . . Represents the plane triangle enclosed by given lines . . . . . . . . . . . . . . . . . . . . . . . . . Represents the rectangle in the xy-plane bounded by lines . . . . . . . . . . . . . . . . . . . . . . . Represents the triangle in the xy-plane bounded by lines . . . . . . . . . . . . . . . . . . . . . . . . . Represents the region bounded by the parabola and lines . . . . . . . . . . . . . . . . . . . . . . . . . Represents an open surface bounded by a closed curve C . . . . . . . . . . . . . . . . . . . . . . . Represents the rectangle bounded by the lines . . . Represents the surface of a rectangular lamina bounded by the lines . . . . . . . . . . . . . . . . Represents the boundary of the triangle . . . . . . Represents the region bounded by a closed surface S . . . . . . . . . . . . . . . . . . . . . . Represents the cube . . . . . . . . . . . . . . . . . Represents the cube . . . . . . . . . . . . . . . . . Represents the circle . . . . . . . . . . . . . . . .

71 74 87 111 114 115 117 122 124 126 127 129 135 138 139 140 142 144 146 149 152 153 155 157 159 161

1 Basic Concept of Vectors and Scalars

1.1 Introduction and Importance The word “Vector” was first given by W. R. Hamilton. In the nineteenth century, Hamilton and Grassmann have formed vector analysis independently. Today, all physical quantities are classified into two different quantities. The physical quantities can be measured directly or indirectly. Some physical quantities are independent of each other or dependent. A quantity that has its value or a magnitude but no direction then it is called scalar quantity or scalar. For example, time, temperature, density, mass, length, power, distance, area, volume, speed, work, energy, electric charge, frequency, gravitational potential, etc., in this list of examples all quantities are having magnitudes but are independent of the direction. Whereas a quantity that has magnitude, as well as direction, is known as vector quantity or vector. Velocity, acceleration, magnetic field, force, momentum, lift, drag, thrust, displacement, fluid flow, the intensity of an electrical field, centrifugal force, etc., are examples of vector quantities. Vectors are generally denoted by capital bold letters or → − letters with an arrow-like A , A, or a.

1.2 Representation of Vectors A geometrical representation of a vector is given in Figure 1.1. Let O be any arbitrary point in the space and let M be any point in the space. A directed line segment joining −−→both the points is known as the −−→   vector OM . The length of the vector OM  is the magnitude of the vector

Figure 1.1

Represents the geometrical representation of a vector

1

2

Basic Concept of Vectors and Scalars

−−−→ −−→ OM and is denoted by |OM | or|OM |. The point O is called the initial point −−−→ and the point M is called the terminal point of the vector |OM |.

1.3 Position Vector and Vector Components Consider the cartesian coordinate system shown in Figure 1.2. Let P be any −−→ point in the three-dimensional system and let O be the origin then OP is −−→ → the position vector of the point P . If the vector OP is denoted by − a then → − the point is denoted by P ( a). All three axis are perpendicular to each other. The position vector can be obtained by taking perpendiculars on each axis. In the cartesian coordinate system, AN (= OB) represents x-coordinate of P , BN (= OA) represents y-coordinate of P , and P N represents Z-coordinate of P . Thus, the point P is denoted by P (x, y, z). Where x, y, and z are also known as components in the direction of the X-axis, Y -axis, and Z-axis respectively. Equal Vectors: Two vectors with the same direction and magnitude are called equal vectors irrespective of the position of their initial points. In Figure 1.3, → − → − a and b are equal vectors. Negative Vectors: Two vectors with the same magnitude but opposite in → direction are called negative vectors. In Figure 1.3, − c is a negative vector → − → − for both a and b .

Figure 1.2 Represents a position vector

1.4 Modulus or Absolute Value of a Vector

3

Figure 1.3 Represents equal and negative vectors

1.4 Modulus or Absolute Value of a Vector

−−→ In Figure 1.2, OP is the position vector of the point P (x, y, z). The modulus −−→ −−→ or absolute value of a vector OP is the length of the vector OP . i.e., OP 2 = BM 2 + M A2 + P M 2 = x2 + y 2 + z 2 −−→    ∴ OP  = OP = x2 + y 2 + z 2 Illustration 1.1: Find the modulus of the vector (−3, 4, −5). → Solution: Let − a = (−3, 4, −5) be a given vector. It is a three-dimensional → vector. Here, x = −3, y = 4, and z = −5 then the modulus of the vector − a is given by

  2 2 2 x + y + z = (−3)2 + (4)2 + (−5)2 √ √ = 9 + 16 + 25 = 50 √ =5 2 √ → Thus, the modulus of a vector − a is 5 2. − |→ a|=

Illustration 1.2: Find the modulus of the vector (6, 8). → Solution: Let − a = (6, 8) be a given vector. It is a two-dimensional vector. → Here, x = 6 and y = 8 then the modulus of the vector − a is given by   → |− a | = x2 + y 2 = (6)2 + (8)2 √ √ = 36 + 64 = 100 = 10

→ Thus, the modulus of a vector − a is 10.

4

Basic Concept of Vectors and Scalars

1.5 Zero Vector and Unit Vector Zero Vector: A vector with modulus zero is called a√zero vector. It is denoted by 0 or θ. Here, θ = 0 = (0, 0, 0). So, |θ| = |0| = 02 + 02 + 02 = 0. Note −→ −−→ that |θ| = 0. AA, BB etc. are zero vectors. Unit Vector: A vector with modulus unity (i.e., 1) is called a unit vector. Illustration 1.3: (−1, 0, 0) is a unit vector as its modulus is  (−1)2 + 02 + 02 = 1.    Illustration 1.4: √13 , 0, − 23 is a unit vector as its modulus is  2 

 1 2 2 1 1+2 3 2 √ 2 +0+ = = = 1. +0 + − = 3 3 3 3 3 3

1.6 Unit Vectors in the Direction of Axes Figure 1.4 represents a two-dimensional cartesian coordinate system (i.e., xyplane), in which (1, 0) and (0, 1) are called unit vectors in the direction of x and y axis respectively. The unit vector in the direction of x-coordinate is denoted by i = (1, 0) and in the direction of y-coordinate is denoted by j = (0, 1).

Figure 1.4 Represents unit vectors in xy-plane

1.7 Representation of a Vector in terms of Unit Vectors

Figure 1.5

5

Represents unit vectors in the direction of x-axis, y-axis, and z-axis

Figure 1.5 represents a three-dimensional cartesian coordinate system (i.e., xyz-axes), in which (1, 0, 0), (0, 1, 0), and (0, 0, 1) are called unit vectors in the direction of x, y, and z axes respectively. The unit vector in the direction of x-coordinate is denoted by i = (1, 0, 0), in the direction of y-coordinate is denoted by j = (1, 0, 0), and in the direction of z-coordinate is denoted by k = (1, 0, 0). Note that the modulus of each unit vector in the direction of each axis is unity.

1.7 Representation of a Vector in terms of Unit Vectors Let P (x, y, z) be a vector and i, j, and k be the unit vectors in the direction −−→ −−→ −→ of OX, OY and OZ respectively in the given Figure 1.6. Consider a perpendicular P M to the plane XOY . From Figure 1.6, we can observe that M L⊥OX and M N ⊥OY . Then OL = x, ON = y, and P M = z. −→ −−→ −−→ ∴ OL = xi, ON = yj, and OM = zk The vectors xi, yj, and zk are called the rectangular components of −−→ −−→ the vector OP and The magnitude of  we can write OP = xi + yj + zk. −−→ −−→ −−→ 2 2 2 OP = |OP | = x + y + z . Thus, P (x, y, z) and OP = xi + yj + zk represent the same vector.

6

Basic Concept of Vectors and Scalars

 in terms of unit vectors Figure 1.6 Represents a vector OP

1.8 Addition and Subtraction of Vectors → → − → If − x = (x1 , x2 , x3 ) and − y = (y1 , y2 , y3 ), then the sum vector → x +− y is given by → − → x +− y = (x1 + y1 , x2 + y2 , x3 + y3 ) . → → And the subtraction of vector − x −− y is given by → − → x −− y = (x1 − y1 , x2 − y2 , x3 − y3 ) . − → − Illustration 1.5: If → a = (4, −3, 2) and b = (−2, 5, 3).then → − − → a + b = (4 − 2, −3 + 5, 2 + 3) = (2, 2, 5) . − − Illustration 1.6: If → x = (4, 10, −2) and → y = (0, 1, −3).then − → → x −− y = (4 − 0, 10 − 1, −2 − 3) = (4 − 0, 10 − 1, −2 + 3) = (4, 9, 1) .

1.9 Product of a Vector with a Scalar → → Definition: Let k be a scalar and − a be a vector. Then k − a is defined as → a vector whose modulus is k times the modulus of the vector − a and whose → − → − direction is the same as that of a or opposite to that of a according to k is positive or negative (See Figure 1.7).

1.10 Direction of a Vector 7

Figure 1.7 Represents the product of a vector with a scalar

→ → − − → → → − → Note: − a  b ⇔− a =k b or b =k − a , k∈R → Illustration 1.7: If − a = (5, −3, 2), then → 3− a = 3 (5, −3, 2) = (15, −9, 6) . → − Here 3− a is a vector whose modulus is three times the modulus of → a and → whose direction is the same as that of − a.

1.10 Direction of a Vector In this section, we try to understand the concept of the direction of a vector. The direction of a vector in the direction from its initial point to its terminal point. Two vectors can be of the same direction, opposite direction, or different directions. → − → Definition: If − a and b are two non-null vectors and − → → − → → → (1) if there exists − a k > 0 such that − a = k b . then − a and b are of the same direction. → − → − → → → (2) if there exists − a k < 0 such that − a = k b . then − a and b are of opposite direction, → − → → (3) if there does not exist − a k ∈ R − {0} such that − a = k b , then the → − − directions of → a and b are different. → Illustration 1.8: Compare the directions of the vectors − a = (2, −5, 3) and → − b = (4, −10, 6). → Solution: Here − a = (2, −5, 3) = 12 (4, −10, 6) → − = 12 b and 12 > 0. → − → ∴− a and b are of the same direction.

8

Basic Concept of Vectors and Scalars

− Illustration 1.9: Compare the direction of the vectors → p = (3, −2, 1) and → − q = (−9, 6, −3) → Solution: Here − q = (−9, 6, −3) = −3 (3, −2, 1) → − = −3 p and −3 < 0. → → ∴− p and − q are of opposite direction. − → − Illustration 1.10: Compare the directions of → g = (2, 5, 7) and h = (3, 1, 6).

Solution: Here 23 = 51 = 76 i.e., there does not exist the same ratio between → − → the elements of − g and h . → − → − → → . . . We cannot express − g and h in the form − g = k h . (k = 0) → − → . . . The directions of − g and h are different.

1.11 Collinear and Coplanar Vectors 1.11.1 Collinear Vectors → − → − → − → → → If two vectors − a and b are such that − a = k b , or b = k − a , where k ∈ R− → − → − → − → → 1− {0}, then a and b are called collinear vectors. Thus a , → a , 3− a , −5− a are 2

2

collinear vectors. Collinear vectors can be represented by parallel lines or line segments of the same line. 1.11.2 Coplanar Vectors Any number of vectors, which are parallel to the same plane, are called coplanar vectors. → − → − → If two non-linear vectors − a and b are coplanar, then vector R given by → − → − → − → → R = x− a + y b is coplanar with the vectors − a and b for any x, y ∈ R.

1.12 Geometric Representation of a Vector Sum 1.12.1 Law of Parallelogram of Vectors → − −−→ −−→ → If − a is a vector represented by AB and b is a vector represented by AD → − − (the two vectors → a and b have a common initial point), then their sum → − −→ → − a + b is represented in magnitude and direction by AC, where ABCD is

1.12 Geometric Representation of a Vector Sum

9

Figure 1.8 Represents the law of parallelogram of vectors

a parallelogram (See Figure 1.8). This method of addition is called the law of the parallelogram of vectors. 1.12.2 Law of Triangle of Vectors → → If two vectors − u and − v are represented by the sides P Q and QR of ΔP QR, → − → − → → then the sum u + v is represented by P R. The direction of − u +− v is from P to R and its modulus is the length of the side P R of ΔP QR. This method of addition is called the law of the triangle of vectors. → Here the terminal point of the vector − u should be the initial point of the → − → − → vector v . The resultant (sum) vector u + − v can be obtained by joining the → − → initial point of the vector u and the terminal point of the vector − v as shown in Figure 1.9.

Figure 1.9 Represents the law of the triangle of vectors

1.12.3 Properties of Addition of Vectors → − → → Let − a , b , and − c be vectors then

10

Basic Concept of Vectors and Scalars

→ − − → → → (1) Commutative law: − a + b = b +− a   − → − → → → → → (2) Associative law: − a + b +− c =− a + b +− c → − (3) Identity vector for addition: θ = (0, 0, 0) is identity vector for addition. → → − − → → → − a + θ =− a = θ +− a (4) Opposite vector or negative of a vector: For a vector, there exists a → → − → → → vector (−− a ) such that − a + (−− a ) = θ . −− a is called the opposite → or negative of − a. → → → → a = (x, y, z), then −− a = (−x, −y, −z). And −− a = −1 · − a. Note: If − → − → − The moduli of a and (− a ) are equal but their directions are opposite to → − → − each other. We can define the difference of vectors a and b as the sum of  −    → → → − → − → − → a and − b i.e. − a − b =− a + −b . 1.12.4 Properties of Scalar Product → → − → Let − a , b ,− c be vectors and m, n ∈ R be scalars. → → (1) m− a =− am → → → (2) m (n− a ) = n (m− a ) = (mn− a) → → → (3) Distributive law: (m + n) − a = m− a + n− a   → − → − → → (4) m − a + b = m− a +mb → − − → (5) θ→ a =− aθ= θ 1.12.5 Expression of Any Vector in Terms of the Vectors Associated with its Initial Point and Terminal Point ←→ −−→ Let BC be a vector. Take a point A which is not on BC join AB and AC. −−→ −→ −−→ We now have three different vectors AB, AC, and BC (See Figure 1.10). By the law of the triangle of vectors, we have −−→ −−→ −→ AB+BC=AC −−→ −→ −−→ ∴ BC=AC−AB

1.12 Geometric Representation of a Vector Sum

Figure 1.10

11

Represents any vector in terms of various vectors associated with its endpoints.

−−→ In other words, BC= vector of point C− vector of point B. In general, any vector=vector of its terminal point-vector of its initial point. 1.12.6 Expression of Any Vector in Terms of Position Vectors The method of expression of a vector discussed above is true for any point A. We can take the origin O in place of the point A. As shown in Figure 1.11, −→ −−→ −−→ OA+AB=OB

Figure 1.11

Represents any vector in terms of the position vector

12

Basic Concept of Vectors and Scalars

−−→ −−→ −→ ∴ AB=OB−OA In other words, any vector=position vector of its terminal point –position vector of its terminal point –position vector of its initial point. Illustration 1.11: Let the position vectors of the points A and B are (2, 5, -3) and (3, -2, 5) respectively. Then AB= position vector of B− position vector of A = (3, −2, 5) − (2, 5, −3) = (1, −7, 8)

1.13 Direction Cosines of a Vector In two dimensions, a vector makes angles with only two axes, namely Xaxis and Y-axis. Hence it is easier to understand. But in three-dimensional space, a vector makes angles with three axes, and to understand the position of the vector, direction cosines of the vector are useful. In Figure 1.12, three important angles are shown considering OP . (i) Angle formed by OP with the X-axis is ∠P OX. It is denoted by α. (ii) Angle formed by OP with the Y-axis is ∠P OY . It is denoted by β. (iii) Angle formed by OP with the Z-axis is ∠P OZ. It is denoted by γ. The cosines of these angles are called direction cosines. Thus, we get three direction cosines.

Figure 1.12 Represents direction cosines of a vector

1.13 Direction Cosines of a Vector 13 x OP . Taking OP x2 + y 2 + z 2 .

(1) l = cos α = Now, r2 =

= r, we have cos a = xr .

∴ OP = r =

 x2 + y 2 + z 2

x ∴ l = cos α =  x2 + y 2 + z 2

(1.1)

(2) cos β is denoted by m. It is connected with the Y-axis. As explained above. y m = cos β =  (1.2) x2 + y 2 + z 2 (3) cos γ is denoted by m. It is connected with the Z-axis. z n = cos γ =  2 x + y2 + z2

(1.3)

Relation among l, m and n, Squaring and adding (1.1), (1.2), and (1.3), we get l 2 + m2 + n 2 =

x2 + y 2 + z 2 = 1. x2 + y 2 + z 2

Thus l2 + m2 + n2 = 1 i.e., cos2 α + cos2 β + cos2 γ = 1. Note that the direction cosines of a vector are the components of its unit vector. Illustration 1.12: If the position vectors of the vertices A, B, and Cof the → − → → parallelogram ABCD are − a , b and − c respectively, find the position vector of the vertex D. → − → → Solution: Here OA = − a , OB = b, and OC = − c . ABCD is a parallelogram (See Figure 1.13). → − → ∴ AD = BC = OC − OB = − c − b Now, ∴ AD = OD − OA  → → − → c − b +− a ∴ OD = AD + OA = −

14

Basic Concept of Vectors and Scalars

Figure 1.13 Represents a parallelogram

→ → − → =− a − b +− c Illustration 1.13: If the position vectors of the points A, B, C, D → → − → → − → − → are respectively − a , b , 3− a − b , −− a + 3 b , then express the vectors − → → AB, AC, BC, BD, and CD in terms of − a and b . → − → − → − → → → Solution: Here OA = − a , OB = b , OC = 3− a − b, and OD = −− a +3 b . → → − Now, AB = OB − OA = b − − a → → − → − → → AC = OC − OA = 3− a − b −− a = 2− a − b → → − → − → → AD = OD − OA = −− a +3b −− a = 3 b − 2− a → − − → → − → → BC = OC − OB = 3− a − b − b = 3− a −2b → − − → → → − → BD = OD − OB = −− a +3b − b =2b −− a   → − → − → → CD = OD − OC = −− a + 3 b − 3− a − b  → − → − → → − → → = −− a + 3 b − 3− a + b =4 b −− a

Illustration 1.14: If ABCDEF is a regular hexagon, prove that AB + AC + AD + AE + AF = 3AD. → − → → Solution: Suppose AB = − a , BC = b , and CD = − c. As ABCDEF is a regular hexagon, → CD = AF = − c, → − → AB = ED = − a and BC = F E = b .

1.13 Direction Cosines of a Vector 15

Figure 1.14 Represents a regular hexagon

→ − − − Now AB = → a , AC = AB + BC = → a + b , AD = AC + CD = → − → − → a + b +− c AE = AD + DC = AD − ED → → − − → → − → =− a + b +− c −→ a = b +− c → − AF = CD = c ∴ LHS = AB + AC + AD + AE + AF → → − − → → − → → − → → =− a +− a + b +− a + b +− c + b +− c +→ c   → − → → =3 − a + b +− c = 3AD = RHS Another method: From Figure 1.14, we have LHS = AB + AC + AD + AE + AF = ED + AC + AD + AE + CD (∵ AB = ED and CD = AF ) = (AC + CD) + (AE + ED) + AD = AD + AD + AD = 3AD = RHS Illustration 1.15: If the sides AB and AC of ΔABC (See Figure 1.15) represent two vectors and M is the mid-point of the side BC, then prove that AB + AC = 2AM

16

Basic Concept of Vectors and Scalars

Figure 1.15 Represents a triangle ΔABC

Solution: Here AB + BM = AM

(1.4)

AC + CM = AM

(1.5)

and Adding (1.4) and (1.5), we have AB + BM + AC + CM = 2AM

(1.6)

But BM and CM are opposite vectors. ∴ BM + CM = 0

(1.7)

Using (1.7) in (1.6), we get AB + AC = 2AM Illustration 1.16: Find position vectors, moduli, unit vectors, and direction cosines for vectors represented by the following points: (i) P (3, −4) (ii) Q (6, 2) (iii) R (−4, −6)

1.13 Direction Cosines of a Vector 17

Solution: → (i) OP = − r = (3, −4) = 3ˆi − 4ˆj  → Modulus = |OP | = |− r | = (3)2 + (−4)2 =

√ √ 9 + 16 = 25 = 5

→ − 3 4 3ˆi − 4ˆj r = ˆi − ˆj = Unit vector rˆ = → − 5 5 5 |r| Direction cosines: l = 35 , m = − 45 → (ii) OQ = − r = (6, 2) = 6ˆi + 2ˆj √ √ √ √ → Modulus = |− r | = 62 + 22 = 36 + 4 = 40 = 2 10 → − r 3 1 6ˆi − 2ˆj √ Unit vector rˆ = → = √ ˆi + √ ˆj = − |r| 2 10 10 10 Direction cosines: l =

√3 , m 10

=

√1 . 10

→ (iii) OR = − r = (−4, −6) = 4ˆi − 6ˆj  √ √ √ → − Modulus = | r | = (−4)2 + (6)2 = 16 + 36 = 52 = 2 13 → − r 2 3 −4ˆi − 6ˆj √ Unit vector rˆ = → = √ ˆi − √ ˆj = − |r| 2 13 13 13 Direction cosines: l = − √213 , m = − √313    →  1 1   → → Illustration 1.17: If − x = 1, 12 , − z = −2, − 32 then y = √2 , √2 and − → → (i) Find a unit vector in the direction of − x +− z, √ → → → y + 2− z. (ii) Find a unit vector in the direction of 2− x − 2− Solution:     → → (i) Here − x +− z = 1, 12 + −2, − 32

18

Basic Concept of Vectors and Scalars

=

1 − 2,

1 3 − 2 2



= (−1, −1) = −ˆi − ˆj  √ √ → → ∴ |− x +− z | = (−1)2 + (−1)2 = 1 + 1 = 2 → → . . . The unit vector in the direction of − x +− z  → − → x +− z 1  1 1 = → = √ −ˆi − ˆj = − √ ˆi − √ ˆj − → − |x + z | 2 2 2

 1 1 = −√ , −√ 2 2 √ → → − y + 2− z (ii) 2→ x − 2−

 



√ 1 1 1 3 = 2 1, − 2 √ ,√ + 2 −2, − 2 2 2 2 = (2, 1) − (1, 1) + (−4, −3) = (2 − 1 − 4, 1 − 1 − 3) = (−3, −3)    √ −   − 2 2 → → − → 2 y + 2 z 2 x −  = (−3) + (−3)  √ √ √ = 9 + 9 = 18 = 3 2 √ → → → y + 2− z 2− x − 2− . . . Required unit vector =  −  √ → − → − 2→ x − 2y +2z

 1 1 1 = √ (−3, −3) = − √ , − √ 3 2 2 2 → Illustration 1.18: Let − a = (2, −1, 2) be a given vector. → (i) Find the unit vector in the direction of − a. → (ii) Find a direction cosine of − a. → (iii) Find a vector of magnitude 6 in the direction of − a. → (iv) Find a vector of magnitude 4 in the opposite direction of − a.

1.13 Direction Cosines of a Vector 19

→ ˆ Solution: Here − a = (2, −1, 2) = 2ˆi − ˆj + 2k. (i) Unit vector in the direction of a  → − a 2ˆi − ˆj + 2kˆ 1 ˆ ˆ 2i − j + 2kˆ a ˆ= → = √ = − 3 |a| 4+1+4 2 1 2 = ˆi − ˆj + kˆ 3 3 3 (ii) Direction cosines: l = 23 , m = − 13 , n =

2 3

→ (iii) Vector of magnitude 6 in the direction of − a  6 ˆ ˆ = 6ˆ a= 2i − j + 2kˆ = 4ˆi − 2ˆj + 4kˆ 3 → (iv) Vector of magnitude 4 in the opposite direction of − a  8 4 4 = −4ˆ a = − 2ˆi − ˆj + 2kˆ = − ˆi + ˆj − 3 3 3

8ˆ k 3

Illustration 1.19: Answer the followings: → → → ˆ − ˆ then find (i) If − a = 3ˆi − ˆj − 4k, b = −2ˆi + 4ˆj − 3kˆ and − c = ˆi + 2ˆj − k, − → → − → − the direction cosines of the vector 3 a − 2 b + 4 c . → ˆ ˆ → → ˆ − ˆ then find the (ii) If − a = 2ˆi + ˆj − k, b = i − j + 2kˆ and − c = ˆi − 2ˆj + k, → − → − → − direction cosines of a + b − 2 c . → − → → (iii) If − a = (3, −1, −4) , b = (−2, 4, −3) and − c = (−1, 2, −1) then find  → −  −  → → − 3 a − 2 b + 4 c .   → − → −  →  → → → ˆ and − (iv) If − a = ˆi + ˆj, b = ˆj + k, c = kˆ + ˆi, then find 2− a − 3 b − 5− c . → → → ˆ − ˆ − ˆ ˆ ˆ ˆ (v) If − a = 5ˆi −3ˆj + 2k, b = 2ˆi +  3j − k and c = i + 2j + 3k, then find → −  →  → the value of 2− a − 3 b + 4− c . → → − → ˆ −  and − ˆ find ˆi − 2ˆj + k, (vi) If b = 2ˆi − 4ˆj − 3k, c = −ˆi + 2ˆj + 2k,  a = 3−  →  −  → a − 3 b − 5− c . 2→

20

Basic Concept of Vectors and Scalars

→ → − −c = ˆi − 2ˆj + kˆ then find ˆ − (vii) If 2ˆi + ˆj − k, b = ˆi − ˆj + 2kˆ and →  a = → − −  → a + b − 2− c . → → − → → (viii) If − a = (1, 2, 1), b = (2, 1, 1) and − c = (3, 4, 1) then find → − − −  → →  a + 2 b + c .  → − → −  → → a + 3 b . (ix) If − a = ˆj + kˆ − iand b = 2ˆi + ˆj − 3kˆ then find 2− → − → −c = (3, 2, −1) then find (x) If − a = (1, 2, 1), b = (1, −1, 2) and →  → −  −  → a + b − 2− c . 3→ Solution: → − → → (i) Here − a = (3, −1, −4), b = (−2, 4, −3) and − c = (1, 2, −1). → − → → → Let − x = 3− a − 2 b + 4− c. → ∴− x = 3 (3, −1, −4) − 2 (−2, 4, −3) + (1, 2, −1) = (9, −3, −12) − (−2, 4, −3) + (4, 8, −4) = (9 + 4 + 4, −3 − 8 + 8, −12 + 6 − 4) = (17, −3, −10)  → − ∴ | x | = 172 + (−3)2 + (−10)2 √ √ = 289 + 9 + 100 = 398 → If l, m, n are the direction cosines of − x , then l=

x1 17 −3 −10 x2 x3 =√ =√ =√ ,m = ,n = |x| |x| |x| 398 398 398

−8 −10 17 , cosβ = √ , cosγ = √ ∴ cosα = √ 398 398 398   → − → −c = 2ˆi + ˆj − kˆ + ˆi − ˆj + 2kˆ − 2 ˆi − 2ˆj + kˆ (ii) − a + b − 2→ = 3ˆi + kˆ − 2ˆi + 4ˆj − 2kˆ

1.13 Direction Cosines of a Vector 21

= ˆi + 4ˆj − kˆ = (1, 4, −1)    → → −  → → ∴ |− r | = − a + b − 2− c  = 12 + 42 + (−1)2 √ √ √ = 1 + 16 + 1 = 18 = 3 2 − → → → If l, m, n are the direction cosines of − a + b − 2− c , then x1 x2 1 4 l = cosα = → = √ , m = cosβ = → = √ , − − |r| |r| 3 2 3 2 x3 −1 n = cosγ = → = √ − |r| 3 2 → − − → (iii) 3→ a − 2 b + 4− c = 3 (3, −1, −4) − 2 (−2, 4, −3) + (−1, 2, −1) = (9, −3, −12) + (4, −8, 6) + (−4, 8, −4) = (9 + 4 − 4, −3 − 8 + 8, −12 + 6 − 4) = (9, −3, −10)    → −  −  → → − ∴ 3 a − 2 b + 4 c  = (9)2 + (−3)2 + (−10)2 √ √ = 81 + 9 + 100 = 190 → − → → ˆ − (iv) Here − a = ˆi + ˆj, b = ˆj + k, c = kˆ + ˆi = ˆi + kˆ → − → → ∴− a = (1, 1, 0) , b = (0, 1, 1) , − c = (1, 0, 1) → − − → ∴ 2→ a = (2, 2, 0) , 3 b = (0, 3, 3) , 5− c = (5, 0, 5)

→ − → → ∴ 2− a − 3 b − 5− c = (2, 2, 0) − (0, 3, 3) − (5, 0, 5) = (2 − 0 − 5, 2 − 3 − 0, 0 − 3 − 5) → − − → ∴ 2→ a − 3 b − 5− c = (−3, −1, −8)    → −  →  → ∴ 2− a − 3 b − 5− c  = (−3)2 + (−1)2 + (−8)2 (∵ Definition of magnitude) √ √ = 9 + 1 + 64 = 74

22

Basic Concept of Vectors and Scalars

→ − → → (v) Here − a = (5, −3, 2) , b = (2, 3, −1) , − c = (1, 2, 3). → − → → → Let− x = 2− a − 3 b + 4− c → − ∴ x = 2 (5, −3, 2) − 3 (2, 3, −1) + 4 (1, 2, 3) = (10, −6, 4) − (6, 9, −3) + (4, 8, 12) = (10 − 6 + 4, −6 − 9 + 8, 4 + 3 + 12) → ∴− x = (8, −7, 19)  → ∴ |− x | = 82 + (−72 ) + 192 (∵ Definition of magnitude) √ √ = 64 + 49 + 361 = 474 → − → → (vi) Here − a = (3, −2, 1) , b = (2, −4, −3) , − c = (−1, 2, 2) → − → → ∴ 2− a − 3 b − 5− c = 2 (3, −2, 1) − 3 (2, −4, −3) − 5 (−1, 2, 2) = (6, −4, 2) + (−6, 12, 9) + (5, −10, −10) = (6 − 6 + 5, −4, 12, 10, 2 + 9 − 10) → − → → ∴ 2− a − 3 b − 5− c = (5, −2, 1)    → −  →  → ∴ 2− a − 3 b − 5− c  = (5)2 + (−2)2 + (1)2 (∵ Definition of magnitude) √ √ = 25 + 4 + 1 = 30 → − → → (vii) Given that − a = (3, −1, −4) , b = (−2, 4, −3) , − c = (−1, 2, −5) → − → → ∴− a +2b −− c = 2 (3, −1, 4) + 2 (−2, 4, −3) − (−1, 2, −5) = (3, −1, −4) + (−4, 8, −6) + (1, −2, 5) = (3 − 4 + 1, −1 + 8 − 2, −4 − 6 + 5) → − → → ∴− a +2b −− c = (0, 5, −5)    → − →  → ∴ − a +2b −− c  = 0 + 52 + (−5)2 (∵ Definition of magnitude) √ √ = 25 + 25 = 50  √ → → − → ∴ − a +2b −− c=5 2

1.13 Direction Cosines of a Vector 23

→ ˆ (viii) Here − a = 2ˆi + ˆj − k, → ∴− a = (2, 1, −1) ,

− ˆ ˆ → b = i − j + 2kˆ → − → ˆ ∴ b = (1, −1, 2) and− c = ˆi − 2ˆj + k → ∴− c = (1, −2, 1)

→ − → → Now, − a + b − 2− c = (2, 1, −1) + (1, −1, 2) − 2 (1, −2, 1) = (2, 1, −1) + (1, −1, 2) + (−2, 4, −2) = (2 + 1 − 2, 1 − 1 + 4, −1 + 2 − 1) → − → → ∴− a + b − 2− c = (1, 4, −1)   → → −  → ∴ − a + b − 2− c  = |(1, 4, −1)|  = (1)2 + (4)2 + (−1)2 (∵ Definition of magnitude) √ = 1 + 16 + 1 √ = 18  = 9 (2) √ =3 2 → − → → (ix) Here − a = (1, 2, 1) , b = (2, 1, 1) , − c = (3, 4, 1) are given → → − → ∴− a +2b +− c = (1, 2, 1) + 2 (2, 1, 1) + (3, 4, 1) = (1, 2, 1) + (4, 2, 2) + (3, 4, 1) = (1 + 4 + 3, 2 + 2 + 4, 1 + 2 + 1) → → − → ∴− a +2b −− c = (8, 8, 4)    → − − −  → → ∴  a + 2 b + c  = (8)2 + (8)2 + (4)2 (∵ Definition of magnitude)  √ √ = 64 + 64 + 16 = 144 = (12)2

24

Basic Concept of Vectors and Scalars

 → → − → ∴ − a +2b +− c  = 12 → (x) Given that − a = ˆj + kˆ − ˆi = −ˆi + ˆj + kˆ → − → ˆ = (2, 1, −3) ∴− a = −ˆi + ˆj + kˆ = (−1, 1, 1) and b = 2ˆi + ˆj − 3k → − → ∴ 2− a +3b = 2 (−1, 1, 1) + 3 (2, 1, −3) = (−2 + 6, 2 + 3, 2 − 9) → − → ∴ 2− a + 3 b = (4, 5, −7)  → −  → ∴ 2− a + 3 b  = (4, 5, −7)  = (4)2 + (5)2 + (−7)2 (∵ Definition of magnitude) √ √ √ = 16 + 25 + 49 = 90 = 9 × 10  √ → −   → ∴ 2− a + 3 b  = 3 10 → − → → (xi) Given that − a = (1, 2, 1) , b = (1, −1, 2) , − c = (3, 2, −1)then find   → −  −  → a + b − 2− c 3→ → − → → ∴ 3− a + b − 2− c = 3 (1, 2, 1) + (1, −1, 2) − 2 (3, 2, −1) = (3, 6, 3) + (1, −1, 2) − (6, 4, −2) = (3 + 1 − 6, 6 − 1 − 4, 3 + 2 + 2) = (−2, 1, 7)   √ √ → −  −  → → − Hence, 3 a + b − 2 c  = 4 + 1 + 49 = 54 Illustration 1.20: If a (1, 0, 0) + b (0, 1, 0) + c (2, −3, −7) = (0, 0, 0), where a, b, c ∈ R, then find the values of a, b, and c. Solution: Here a (1, 0, 0) + b (0, 1, 0) + c (2, −3, −7) = (0, 0, 0)

1.13 Direction Cosines of a Vector 25

∴ a (a, 0, 0) + b (0, b, 0) + c (2c, −3c, −7c) = (0, 0, 0) ∴ (a + 0 + 2c, 0 + b − 3c, 0 + 0 − 7c) = (0, 0, 0) ∴ (a + 2c, b − 3c, −7c) = (0, 0, 0) ∴ a + 2c = 0, b − 3c = 0, −7c = 0 ∴ c = 0, a = 0, b = 0 or a = 0, b = 0, c = 0. Illustration 1.21: A space shuttle of 1000 tons hangs from two skyscrapers using steel cables as shown in Figure 1.16. Find the forces or tensions in both the cables attached with skyscrapers and also find their magnitude.

Figure 1.16 cables

A space shuttle of 1000 tons weight hangs from two skyscrapers using steel

Solution: − → − → Let F1 and F2 be two forces or tensions on the steel cables respectively. First, − → − → we represent F1 and F2 in terms of vertical and horizontal components.   − − − →  →  → F1 = − F1  cos50o ˆi + F1  sin50o ˆj  − − → − →  → F2 = F2  cos32o ˆi + F2  sin32o ˆj → − The gravity force acting on the space shuttle is F = −mgˆj = → − − → − (1000) (9.8) ˆj = −9800ˆj. Therefore, the counterbalance of F with F1 − → and F2 is given as → − → − → − → → − − → − ∴ F1 + F2 + F = 0 ⇒ F1 + F2 = − F = − (−9800) ˆj = 9800ˆj

26

Basic Concept of Vectors and Scalars

Thus,

  −  −   →  → − F1  cos50o ˆi + F1  sin50o ˆj + −   −   →  → F2  cos32o ˆi + F2  sin32o ˆj = 9800ˆj   −  −   →  → ∴ − F1  cos50o + F2  cos32o ˆi+ −   −   →  → F1  sin50o + F2  sin32o ˆj = 9800ˆj

Now, equation the components     − − − −  →  →  →  → − F1  cos50o + F2  cos32o = 0 ⇒ F1  cos50o = F2  cos32o − −    →  → o F1  sin50 + F2  sin32o = 9800 − → Solving for |F2 |, we get

 −  →  − F  1  cos50o  → o sin32o = 9800 F1  sin50 + cos32o  − 9800  → ≈ 8392 N ∴ F1  = sin50o + tan32o cos50o

−   → o  −  → F1  cos50 = ≈ 6361 N F  2 cos32o Thus, the force vectors are − → − → F1 ≈ −5394 ˆi + 6429 ˆj and F2 ≈ 5394ˆi + 3371ˆj. And

1.14 Exercise → → 1. If − x = (2, 1) and − y = (1, 3), then (i) find a unit vector in the direction → − → − → → of 3 x − 2 y , (ii) find direction cosines of 3− x − 2− y.   Answer : (i) 45 ˆi − 35 ˆj, (ii) 45 , − 35 2. If x (3, 2) + y (2, 3) = (17, 13), find the real values of x and y. (Answer : (x, y) = (5, 1))

1.14 Exercise

27

3. Find a, b ∈ R such that (i) (4, 7) + (a, b) = (17, 13) (ii) (a, −8) − 2 (3, b) = (−4, 6). (Answer : (i) (a, b) = (13, 20) , (ii) (a, b) = (2, −7)) 4. If x ˆ = (4, 7, 2) and yˆ = (−1, 3, 4), find the vectors 2ˆ x + 4ˆ y and 3ˆ x − yˆ. → − → − → − → − (Answer : (i) 2 x + 4 y = (4, 26, 20) , (ii) 3 x − y = (13, 18, 21)) 5. If for real values of x, y, z, x (1, 2, 0) + y (0, 3, 1) + z (−1, 0, 1) = (2, 1, 0), then find x, y, z. (Answer : (x, y, z) = (5, −3, 3)) 6. Find the vectors x, y ∈ R2 such that |x| = |y| = 1and |x − y| = 2. (Answer : (1, 0) and (−1, 0)) 7. If P (2, 4, −5) , and Q (1, 2, 3) are points in R2 , find the direction cosines of P Q.   Answer : − √169 , − √269 , √869 8. If a ˆ = (3, −1, −4) , ˆb = (−2, 4, −3) , cˆ = (−1, 2, −1), then find the direction cosines of the vector 3ˆ a − 2ˆb + 4ˆ c.   9 3 , − √190 , √−10 Answer : √190 190 9. If a ˆ = (1,2, 3) , ˆb = (2, −2, −5) , cˆ = (3, −2, −1), then find (i) a ˆ+   2ˆb − cˆ (ii) a ˆ + ˆb + cˆ. √ √   Answer : (i) 68 (ii) 41   10. Show that a ˆ = (2, −3, 2) , ˆb = 1, − 32 , 1 are parallel vectors

2 Scalar and Vector Products

2.1 Scalar Product, or Dot Product, or Inner Product → (1) Algebraic definition: If − x = x1ˆi + x2 ˆj + x3 kˆ = (x1 , x2 , x3 ) and → − ˆ ˆ ˆ y = y1 i + y2 j + y3 k = (y1 , y2 , y3 ) are two vectors in R3 , then their − → → → dot product → x ·− y (read − x dot − y ) is defined as − → − x ·→ y = x1 y 1 + x 2 y 2 + x 3 y 3

(2.1)

Note that the result of a dot product is a scalar i.e., it is a real number. → − → (2) Geometric definition: If the angle between the vectors − a and b is θ as shown in Figure 2.1 then −  → − → − → → − a . b = | a |  b  cos θ = ab cos θ,

Figure 2.1 Represents a scalar or dot product

29

(2.2)

30

Scalar and Vector Products

 − → → where, |− a | = a and  b  = b.

 → − → − − → Note: The angle θ is denoted by → a ˆ, b (read − a cab b ) also.

2.2 The Measure of Angle Between two Vectors and Projections

 → − → − → → Let OA = − a , OB = b , and − a ˆ, b = θ as shown in Figure 2.1. Now, 

→ − − → − OM → → − → → − → − =− a · OM a · b = a b cos θ = a (OB cosθ ) = a OB × OB → − → − → → → ∴− a · b = |− a | (Projection of b on the direction of − a)   → − → − → −   → − → − Similarly, a · b =  b  (Projection of a on the direction of b ) → →− → − − → Also, the projection − a on the direction of b = a . b and the projection → − b

→ − → of b on the direction of − a = | | → − → − Thus, geometrically, a · b denoted the product of the modulus of one vector and the projection of the second vector in the direction of the first vector. → − → − a·b An important result: From (2.2), we get cos θ = − → . − → | a || b | − → → − a.b . → − a

2.2.1 Properties of a Dot Product (1) The scalar product of two vectors is commutative. → − − → → → Thus, − a · b = b ·− a → − → We have − a · b = x 1 y 1 + x 2 y 2 + x 3 y3 = y1 x1 + y2 x2 + y3 x3 → → − = b ·− a (2) If two non-zero vectors are perpendicular to each other then their dot product is zero. → − → If − a ⊥ b , then θ = 90◦ and cos θ = 0. → − → ∴− a · b = ab cos θ = 0

2.2 The Measure of Angle Between two Vectors and Projections

31

→ − → Also, if − a · b = 0, then ab cos θ = 0. i.e., cosθ = 0 ⇒ θ = π2 . (3) Scalar product of two like vectors or two opposite vectors: → − → (i) If − a and b are of the same direction, then θ = 0 and cos θ = cos θ = 1. → − → − → → ∴− a · b = ab cos θ = − a b = Product of their moduli → − → (ii) If − a and b have opposite directions, then θ = p and cos θ = cos π = −1. → − → − → → ∴− a · b = ab cos θ = − a b (−1) → − → = −− a b = −(Product of their moduli) → − 2 2 → → → → → → → a ·− a =− a− a =− a = |− From (i), if − a = b , then − a| (4) From unit vectors i, j, k in the directions of the axes: ˆi · ˆi = ˆj · ˆj = kˆ · kˆ = 1, ˆi · ˆj = 0, ˆj · kˆ = 0, kˆ · ˆi = 0, Because ˆi, ˆj, kˆ are mutually perpendicular. (5) For p, q ∈ R,

 → − → − → →  − − → → → → p− a · p b = pq − a · b = (pq − a)· b =− a · pq b

(6) The scalar product of two vectors is distributive with respect to the vector addition. − → → − → → − − → − a · b +− c =→ a · b +− a ·→ c − → → − → → − − → Also, − a · b −− c =→ a · b −− a ·→ c −  − → → → → → → → Note: If − a = θ and − a · b −− c = 0, then − a⊥ b −− c or → − − b −→ c = θ. (7) The scalar product of two vectors is given by the sum of the products of their corresponding elements.

32

Scalar and Vector Products

→ − − If → a = (a1, a2 , a3 ) and b = (b1, b2 , b3 ), then → − → − a · b = (a1, a2 , a3 ) · (b1 , b2 , b3 ) = a1 b1 + a2 b2 + a3 b3 = Σai bi   → → Also − a = |− a | = a21 + a22 + a23 = Σa2i  −   − → → b =  b  = b21 + b22 + b23 = Σb2i → − → Now, − a · b = ab cos θ gives → − − → a.b  cos θ = → − → − | a | b  a 1 b1 + a 2 b2 + a 3 b3  = 2 a1 + a22 + a23 b21 + b22 + b23 Σai bi  = Σa2i Σb2i Σai bi = Σa2i × Σb2i ⎧ ⎫ ⎨ ⎬ Σai bi  ∴ θ = cos−1  ⎩ Σa2 Σb2 ⎭ i

i

Note: → (1) If (l1 , m1 , n1 ) and (l2 , m2 , n2 ) are the direction cosines of vectors − a → − and b respectively. Then cos θ = l1 l2 + m1 m2 + n1 n2 = Σl1 l2 (2)

sin2 θ = 1 − cos2 θ  2  2 a · b a2 b2 − a · b =1− = a 2 b2 a 2 b2

2.2 The Measure of Angle Between two Vectors and Projections



→2  →2 − →2 −  2 − − → − → − a · b |→ a| b − − a · b   − − = → → → → |− a | b  |− a | b  

a 2 b2 ∴ sin θ =

33

Remember the following two important results: 2 → → → → → → → (1) − x ·− x = |− x | (2)− x ·− y =0⇔− x ⊥− y → → → → Illustration 2.1: If − x = (1, 2, 3) and − y = (2, 3, 4),then find (i) − x ·− y and → − → − (ii) ( x ˆ, y ). Solution: − → (i) → x ·− y = (1, 2, 3) · (2, 3, 4) = (1) (2) + (2) (3) + (3) (4) = 2 + 6 + 12 = 20 2 − (ii) |→ x | = 1 + 4 + 9 = 14

√ → ∴ |− x | = 14

→ |− y | = 4 + 9 + 16 = 29 √ → ∴ |− y | = 29 2

Now,

→ − → x .− y 20 → → y)= − =√ √ cos θ = cos(− x ,∧ − → − → | x || y | 14 29

 20 → − → ∧− −1 √ √ ∴ θ = ( x , y ) = cos 14 29 → → ˆ then find (i) Illustration 2.2: If − x = ˆi + 3ˆj + 2kˆ and − y = 4ˆi − 2ˆj + k, → − → → → x ·− y (ii) the angle between − x and − y. Solution: (i) x · y = (1, 3, 2) · (4, −2, 1) = (1) (4) + (3) (−2) + (2) (1) =4−6+2=0 Hence, x and y are perpendicular.

34

Scalar and Vector Products

(ii) As x · y = 0, x⊥y. ∴The angle between x and y is 90◦ or π2 . Illustration 2.3: If x = (1, −2, 2) and y = (0, 0, −1), then verify that (i) |x.y| ≤ |x||y| (ii) |x + y| ≤ |x| + |y| (iii) |x + y| ≥ |x| + |y| Solution: Here, √2) and y = (0, 0,√−1). √ x = (1, −2, ∴ |x| = 1 + 4 + 4 = 9 = 3 and |y| = 0 + 0 + 1 = 1 (i) x · y = (1, −2, 2) · (0, 0, −1) = (1) (0) + (−2) (0) + (2)(−1) = −2 ∴ |x · y| = |−2| = 2

(2.3)

Also, |x| = 3 and |y| = 1 ∴ |x| · |y| = 3

(2.4)

Hence, from (2.3) and (2.4) we get 2 < 3 ⇒ |x · y| ≤ |x||y|. (ii) x + y = (1, −2, 2) + (0, 0, −1) = (1 + 0, −2 + 0, 2 + (−1)) = (1, −2, 1) √ √ ∴ |x + y| = 1 + 4 + 1 = 6

(2.5)

Also, |x| = 3 and |y| = 1 ∴ |x| + |y| = 3 + 1 = 4 √ Hence, from (2.5) and (2.6) we get 6 < 4 (∵ 6 < 16) ∴ |x + y| ≤ |x| + |y|.

(2.6)

2.2 The Measure of Angle Between two Vectors and Projections

35

(iii) x − y = (1, −2, 2) − (0, 0, −1) = (1 − 0, −2 − 0, 2 − (−1)) = (1, −2, 3) √ √ ∴ |x − y| = 1 + 4 + 9 = 14

(2.7)

Also, |x| = 3 and |y| = 1 (2.8) ∴ |x| − |y| = 3 − 1 = 2 √ Hence, from (2.7) and (2.8) we get 14 > 2 (∵ 14 > 4) Therefore, |x − y| ≥ |x| − |y|. → − → − → − → → → a | = | b |, then show that (− a + b) Illustration 2.4: If for − a , b ∈ R 3 , |− → − → and (− a − b ) are perpendicular to each other. → − → − → → Solution: To prove (− a + b ) and (− a − b )are perpendicular to each other,   → − → − → → we have to show that − a + b · − a − b = 0.  →  → − − → → − → − − → → − → − → − → ∴ → a + b · − a − b =− a ·→ a −− a · b + b ·− a + b · b − → − → → − →2 2 → → a · b + b ·− a +b = |− a| −− → − − → − →2 2 → → → (∵ − a ·− a = |− a | and b · b =  b  ) −  → − →2 2 → → (∵ |− a | = | b |) = |− a| −b =0  → → − − → → ∴ − a + b ⊥ − a − b → − → Illustration 2.5: If for − a = (2, 2, −1) and b = (6, −3, 2) , then find the → − → angle between − a and b . → − → Solution: We have − a = (2, 2, −1) and b = (6, −3, 2) , then → − − → a . b = (2, 2, −1) · (6, −3, 2) = 12 − 6 − 2 = 4 −  √ √ √ → √ → Also, |− a | = 4 + 4 + 1 = 9 = 3,  b  = 36 + 9 + 4 = 49 = 7,

36

Scalar and Vector Products

→ − → If θ is the angle between − a and b , then cos θ =

→ − − → 4 a · b 4 → = 3 × 7 = 21 = 0.1905 − → − | a || b |   ∴ sin 900 − θ = 0.1905

∴ 900 − θ = 100 58 (∵ Using trigonometric table) ∴ θ = 900 −100 58 = 790 2 → − → ˆ then find the Illustration 2.6: If − a = 3ˆi − 2ˆj + kˆ and b = ˆi − 2ˆj + k, → − → − projection of a on b . → − → Solution: The projection of − a on b is given as → − − → a · b  = − → b =

(3, −2, 1) · (1, −2, 1) √ 1+4+1

√ 8 4 6 3+4+1 √ = √ = = 3 6 6 Illustration 2.7: Show that (−1, 6, 6) , (−4, 9, 6) , and (0, 7, 10) are position vectors of the vertices of a right-angled triangle. Solution: Let A (−1, 6, 6) , B (−4, 9, 6) , and C(0, 7, 10) be the vertices of  ABC. Now, AB = OB − OA = (−4, 9, 6) − (−1, 6, 6) = (−3, 3, 0) BC = OC − OB = (0, 7, 10) − (−4, 9, 6) = (4, −2, 4) And CA = OA − OC = (−1, 6, 6) − (0, 7, 10) = (−1, −1, −4) AB · CA = (−3, 3, 0) · (−1, −1, −4) = 3 − 3 + 0 = 0 ∴ AB⊥CA ∴  ABC is a right-angled triangle. ˆ Illustration 2.7: Find the angle between the vectors 2ˆi + ˆj + 4kˆ and ˆi + ˆj + k.

2.3 Vector Product or Cross Product or Outer Product of Two Vectors

37

→ → Solution: Let − x = 2ˆi + ˆj + 4kˆ = (2, 1, 4) and − y = ˆi + ˆj + kˆ = (1, 1, 1) then we have → − → x ·− y = (2, 1, 4) · (1, 1, 1) = 2 + 1 + 4 = 7 and if (x ,∧ y) = θ, then √ → − → 7 7 x ·− y 7 ∴cos θ = − =√ √ =√ = → → − 3 |x|| y | 21 3 63 √ −1 7 ∴ θ = cos 3 Illustration 2.8: For what value of x, the vectors 2ˆi−3ˆj +5kˆ and xˆi−6ˆj −8kˆ are perpendicular to each other? → − → Solution: Let − a = 2ˆi − 3ˆj + 5kˆ = (2, −3, 5) and b = xˆi − 6ˆj − 8kˆ = (x, −6, −8). → − → − → − As − a⊥b ⇒→ a · b =0

∴ (2, −3, 5) · (x, −6, −8) = 0 ∴ 2x + 18 − 40 = 0 ∴ 2x = 22 ⇒ x = 11.

2.3 Vector Product or Cross Product or Outer Product of Two Vectors

→ − → The vector product of two vectors − a and b (See Figure 2.2) is denoted by → − → − a × b and it is defined as follow: → − → (1) Algebraic definition: If − a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ) then ⎡ ⎤ ˆi ˆj kˆ → − → − a × b = ⎣ a1 a2 a3 ⎦ (2.9) b 1 b 2 b3 → − − → a × b = (a2 b3 − a3 b2 ) ˆi − (a1 b3 − a3 b1 ) ˆj + (a1 b2 − a2 b1 ) kˆ = (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ) → − → Note: − a × b is also a vector.

38

Scalar and Vector Products

Figure 2.2 Represents a vector or cross product

(2) Geometric definition: → − → − → → If − a and b are given vectors, then their cross product − a × b is defined as

 − → − → − → → ˆ a × b = |− a |  b  sin θ n

(2.10)

→ − → where θ is an angle between − a and b , n ˆ is a unit vector perpendicular → − → − to both a and b . → − → ∴− a × b = (ab sin θ ) n ˆ (2.11)

2.4 Geometric Interpretation of a Vector Product

 → − → → − − The vector product of two vectors → a × b has modulus − a × b  which is → − → twice the area of the triangle whose two consecutive sides are − a and b or → equals the area of a parallelogram whose two consecutive sides are − a and → − b. → − → − → → Note that − a × b is a vector but |− a × b | is a scalar that gives the area of the parallelogram ABCD as shown in text is present for Figure 2.3.  → → − ˆ| Thus, − a × b  = |ab sinθ n = ab sinθ

(∵ |ˆ n| = 1)

2.4 Geometric Interpretation of a Vector Product

Figure 2.3

39

Represents the geometric interpretation of a vector or cross product

= Area of the parallelogram ABCD = 2 × Area of  ABC 2.4.1 Properties of a Vector Product 1. Vector product or cross product is non-commutative. → − − → → → i.e., − a × b = b × − a.

− → → Here if θ is the angle between directions of − a and b , then the angle → − → between the direction of b and − a is −θ. → − → ∴ − a × b = ab sin θ . n ˆ

→ → − and b × − a = ab sin (−θ) .ˆ n = −ab sin θ . n ˆ

(∵ sin (−θ) = −sin θ ) → − → = −(− a × b) 2. The vector product of two parallel vectors is a null vector. → − → If − a  b , then θ = 0. → − → ∴− a × b = ab sin θ · n ˆ = ab sin0 · n ˆ=θ

(∵ sin 0 = 0)

3. For the unit vectors ˆi, ˆj, kˆ : ˆi × ˆi = ˆj × ˆj = kˆ × kˆ = θ ˆ ˆj × kˆ = ˆi, kˆ × ˆi = ˆj, ˆj × ˆi = −k, ˆ kˆ × ˆj = −ˆi, and ˆi × ˆj = k, ˆi × kˆ = −ˆj.

40

Scalar and Vector Products

4. For any p, q ∈ R,

 → − − → → → − → − → → → p− a × q b = pq − a × b = (pq − a)× b =− a × (pq b )

→ − → Taking − a = 1 and b = −1, we get  − → → − → − → − → → a × − b = (−− a ) × b = −(− a × b) 5. Relation between cross product and dot product: →2   →2 − →2  → − 2 − → − → a · b = |− a |  b  − − a × b Or



→2 →2 − − 2 − →  2 − → − → + → a × b  = |− a| b a · b

6. Distributive law for vector product: → − → − → → − − → (i) − a × b +− c =→ a × b +− a ×→ c   → − → → − → → → → (ii) − a + b ×− c =− a ×− c + b ×− c ˆ 7. Expression of a cross product in terms of unit vectors ˆi, ˆj, k: → − → ˆ If − a = a1ˆi + a2 ˆj + a3 kˆ = (a1 , a2 , a3 ) and b = b1ˆi + b2 ˆj + b3 k, then    ˆi ˆj kˆ   →  − → − a × b =  a1 a2 a3   b 1 b 2 b3  = (a2 b3 − a3 b2 ) ˆi − (a1 b3 − a3 b1 ) ˆj + (a1 b2 − a2 b1 ) kˆ Remarks:  → → − 1. As − a × b  = absinθ, sinθ =

→ − → |− a×b| → − → a || b | |−

→ − → 2. If the direction cosines of − a and b are (l1 , m1 , n1 ) and (l2 , m2 , n2 ) respectively, then  sin θ = (m1 n2 − m2 n1 )2 + (l1 n2 − l2 n1 )2 + (l1 m2 − l2 m1 )2

2.4 Geometric Interpretation of a Vector Product

41

3. By definition of the cross product, → − → → − − → → − → a ⊥(− a × b ) and b ⊥(− a × b)  → − → → − − → → → ∴− a · − a × b = 0 and b · − a × b = 0.     Illustration 2.9: Find the modulus of 2ˆi − 3ˆj + kˆ × ˆi − ˆj + 2kˆ .    →  − → Solution: Consider − a = 2ˆi − 3ˆj + kˆ and b = ˆi − ˆj + 2kˆ   ˆ ˆ ˆ →  i j k − → − ∴ a × b =  2 −3 1  1 −1 2

     

= (((−3) × 2) − ((−1) × 1))ˆi − ((2 × 2) − (1 × 1))ˆj + ((2 × (−1)) − (1 × (−3)))kˆ = (−6 + 1) ˆi − 3ˆj + kˆ = −5ˆi − 3ˆj + kˆ   √ → √ → − ∴ − a × b  = 25 + 9 + 1 = 35 → − → Illustration 2.10: If − a = 2ˆi − ˆj, b = ˆi + 3ˆj − 2kˆ then compute  → → − − →  − a + b × − a − b .  → → − → Solution: Here, − a = 2ˆi − ˆj = (2, −1, 0) and b = ˆi + 3ˆj − 2kˆ = (1, 3, −2) . → − − ∴ → a + b = (2, −1, 0) + (1, 3, −2) = (3, 2, −2) and → − → − a − b = (2, −1, 0) − (1, 3, −2) = (1, −4, 2)    ˆi ˆj ˆ  k       → − → − → → ∴ − a + b × − a − b =  3 2 −2   1 −4 2  ˆ = ˆi (4 − 8) − ˆj (6 + 2) + k(−12 − 2) = −4ˆi − 8ˆj − 14kˆ = (−4, −8, −14)     √ √ → →  √ −  → − → ∴ − a + b × − a − b  = 16 + 64 + 196 = 276 = 2 69

42

Scalar and Vector Products

Illustration 2.11: Simplify:      ˆ . 10ˆi + 2ˆj + 3kˆ · ˆi − 2ˆj + 2kˆ × (3ˆi − 2ˆj − 2k) → − → Solution: Consider − a = 10ˆi + 2ˆj + 3kˆ = (10, 2, 3) , b = ˆi − 2ˆj + 2kˆ = → (1, −2, 2) and − c = 3ˆi − 2ˆj − 2kˆ = (3, −2, −2).    ˆi ˆj ˆ  k  → →  − ∴ b ×− c =  1 −2 2  = 8ˆi + 8ˆj + 4kˆ = 4(2, 2, 1)  3 −2 −2  The given expression − → − → − a · b ×→ c = (10, 2, 3) · 4 (2, 2, 1) = 4 (20 + 4 + 3) = 108. Illustration 2.12: Prove that −  → → − → → − → − → → → → a × b +− c + b × (− c +− a)+− c × − a + b = 0. −  → → − → → − → → → → → a × b +− c + b × (− c +− a)+− c × − a + b Solution: L.H.S. = − Using distributive law → → − − → → − − → → − → − → → → =− a × b +− a ×→ c + b ×− c + b ×− a +→ c ×− a +− c × b → → → − → → − − → → − → − → → → =− a × b −− c ×− a −− c × b −− a × b +− c ×→ a +− c × b = 0 = R.H.S. → → → → → → Illustration 2.13: Show that: (− x −− y ) × (− x +− y ) = 2(− x ×− y) → − → − → − → − Solution: L.H.S. = ( x − y ) × ( x + y ) → → → → → → =− x × (− x +− y)−− y × (− x +− y) → → → → → → → → =− x ×− x +− x ×− y −− y ×− x −− y ×− y → → → → → → → → x ×− x = θ, − y ×− y = θ and −− y ×− x =− x ×− y in the Now, we use − above expression → → → → → → =θ+− x ×− y +− x ×− y − θ = 2 (− x ×− y ) = R.H.S → → Illustration 2.14: If − x = (3, −1, 2) and − y = (2, 1, −1) are given vectors. → − → Find unit perpendicular vector to the x and − y both.

2.4 Geometric Interpretation of a Vector Product

43

→ → Solution: Given that − x = (3, −1, 2) and − y = (2, 1, −1)    ˆi ˆj kˆ   → − → −  ∴ x × y =  3 −1 2  = (1 − 2) ˆi − (−3 − 4) ˆj + (3 + 2) kˆ  2 1 −1  = −ˆi + 7ˆj + 5kˆ = (−1, 7, 5) √ √ √ → → |− x ×− y | = 1 + 49 + 25 = 75 = 5 3 → → ∴ Unit perpendicular vector to given vectors is − x and − y → − → − x × y 1 = − = √ (−1, 7, 5) . → |→ x ×− y| 5 3 Illustration 2.15: Find a unit vector of magnitude 10 which is perpendicular ˆ to vectors 2ˆi − ˆj − 2kˆ and 4ˆi − 3ˆj − 5k. → − → Solution: Consider the vectors − a = 2ˆi − ˆj − 2kˆ = (2, −1, −2) and b = 4ˆi − 3ˆj − 5kˆ = (4, −3,  −5)   ˆi ˆj ˆ  k  →  − → Now, − a × b =  2 −1 −2   4 −3 −5  = (5 − 6) ˆi − (−10 + 8) ˆj + (−6 + 4) kˆ = −ˆi + 2ˆj − 2kˆ = (−1, 2, −2)  √ → √ → − ∴ − a × b= 1+4+4= 9=3 → − → Now, the unit perpendicular vector to − a and b → − → − 1 a × b = → = 3 (−1, 2, −2) − → − |a × b| → − → A perpendicular vector with magnitude 10 to both − a and b =

10 (−1, 2, −2) . 3

→ → → ˆ then prove that − Illustration 2.16: If − x = ˆi + ˆj + kˆ and − y = 2ˆi − ˆj − k, x → − and y are perpendicular to each other. Also, find a unit vector perpendicular → → to the vectors − x and − y.

44

Scalar and Vector Products

Solution: Given that → − → x = ˆi + ˆj + kˆ = (1, 1, 1) and − y = 2ˆi − ˆj − kˆ = (2, −1, −1) → − → − → − − → Now, x . y = (1,1, 1) . (2, −1, −1)  =2−1−1=0⇒ x⊥y.  ˆi ˆj kˆ   → − → −  Also, x × y =  1 1 1   2 −1 −1  = (−1 + 1) ˆi − (−1 − 2) ˆj + (−1 − 2) kˆ = 3ˆj − 3kˆ = (0, 3, −3) √ √ √ → → |− x ×− y | = 0 + 9 + 9 = 18 = 3 2 → → Unit perpendicular vector to − x and − y    → − → x ×− y 1  ˆ ˆ = √1 ˆj − kˆ . √ 3 j − 3 k = → = → |− x ×− y| 3 2 2 Illustration 2.17: Find a unit vector that makes an angle of 600 with the ˆ vector ˆi − k. → ˆ Solution: Let − x = ˆi − k. → ∴− x = (1, 0, −1) → And suppose − y = (y1 , y2 , y3 ) be the required vector. 2 → ∴ |− y | = y12 + y22 + y32 = 1

→ → Now, (− x ˆ,− y ) = 600 − → → x · − y → → cos 600 = cos (− x ˆ, − y)= − → → − |x|| y | → (1, 0, −1). − y √ 2·1 y 1 − y3 1 = √ ∴ 2 2 1 ∴ y1 − y3 = √ 2 =

(2.12)

2.5 Application of Scalar and Vector Products

If we take y1 = 0, then y3 = − √12 . Then from (2.12), we get y22 = 1 − → 1 ⇒ y = √1 . Thus, the required vector − y = (0, − √1 , √1 ). 2

2

2

2

45 1 2

=

2

ˆ ˆ ˆ ˆ ˆ Illustration  2.18: Show that the angle between the vectors i+2j and i+j+3k is sin−1 46 55 . → − → Solution: Consider the vectors − a = ˆi + 2ˆj = (1, 2, 0) and b = ˆi + ˆj + 3kˆ = (1, 1, 3). √ − √ √ → √ → Also, |− a | = 1 + 4 + 0 = 5,  b  = 1 + 1 + 9 = 11    − → → − a × b =  

ˆi 1 1

ˆj 2 1

kˆ 0 3

    = (6 − 0) ˆi − (3 − 0) ˆj + (1 − 2) kˆ  

= 6ˆi − 3ˆj − kˆ = (6, −3, −1)  √ → √ → − ∴ − a × b  = 36 + 9 + 1 = 46

 → − − Now, if → a  , b = θ, then

√ → − → 46 46 |− a × b| sin θ = =√ √ = → − → 55 5 11 |− a || b |

46 −1 ∴ θ = sin 55

2.5 Application of Scalar and Vector Products The scalar and vector products are useful in many fields such as coordinate geometry, solid geometry, trigonometry, mechanics, physics, computer science, electric engineering, etc. In this section, we have discussed two basic applications of scalar and vector products (1) Work done by a force (2) Moment of a force about a point.

46

Scalar and Vector Products

 on a particle Figure 2.4 Represents work done by a force F

2.5.1 Work Done by a Force The scalar product of a force applied to a particle and the displacement of the particle in the direction of the force is known as work. Thus, work is a scalar quantity. → − → − If F denotes the force applied to a particle and d denotes the displace→ − ment of the particle in the direction of F , then work done (See Figure 2.4) → − by the force F is given by → → − − W =F · d → − Suppose a force F = (F1 , F2 , F3 ) acts on a particle at the point A(a1 , a2 , a3 ) and as a result, the particle moves to the point B (b1 , b2 , b3 ). → − ∴ The displacement d = AB = OB − OA = (b1 − a1 , b2 − a2 , b3 − a3 ) → − Now, work done by the force F is   → − W = Projection of F on AB × AB → −−→ − F · AB × AB = AM × AB = AB → −−→ − = F · AB → − −−→ = Dot product of F and AB = (F1 , F2 , F3 ) · (b1 − a1 , b2 − a2 , b3 − a3 ) = F1 (b1 − a1 ) + F2 (b2 − a2 ) + F3 (b3 − a3 ) → − → → − − → − → − → − → − Note: If forces F1 , F2 , F3 , . . . act on the point A, then F = F1 +F2 +F3 +. . . 2.5.2 Moment of a Force About a Point → − The moment of a force F about a point A is a vector and it is given by −→ − → → − AP × F , where P is a point on the line of the force F .

2.5 Application of Scalar and Vector Products

47

Figure 2.5 Represents the moment of a force about a point

→ − Thus, the moment of the force F about a point A (See Figure 2.5) is → − a vector which is perpendicular to the plane containing F and A having a modulus −→ − −  −  →  → → = AP × F  = AP  F  sinθ = AN  F  (∵ AN = AP sinθ ) , → − where AN is the line perpendicular to the line of the force F . → − Now let F = (F1 , F2 , F3 ) , A(a1 , a2 , a3 ) and P (p1 , p2 , p3 ). Then −→ −−→ −→ AP = OP − OA = (p1 − a1 , p2 − a2 , p3 − a3 ). → − Moment of F about the point A −→ − → = AP × F     ˆi ˆj kˆ    =  p1 − a1 p2 − a2 p3 − a3   F1  F2 F3 = {F3 (p2 − a2 ) − F2 (p3 − a3 )} ˆi − {F3 (p1 − a1 ) − F1 (p3 − a3 )} ˆj + {F2 (p1 − a1 ) − F1 (p2 − a2 )} kˆ Note: → − (1) The moment of the force F about a point on its line of action is always −→ zero, because AP is a null vector. → − → − → − (2) If concurrent forces F1 , F2 , F3 , . . . act at the point P , then their moment −→ − → → − − → − → − → about a point A = AP × F , where F = F1 + F2 + F3 + . . .

48

Scalar and Vector Products

Illustration 2.19: A particle moves from the point A (3, 2, −1) to the point → − ˆ Find the work done B (2, −1, 4) under the effect of force F = 4ˆi − 3ˆj + 2k. by the force. → − Solution: Here, the force vector is F = 4ˆi − 3ˆj + 2kˆ = (4, −3, 2) and the −−→ displacement is given by AB = (2, −1, 4) − (3, 2, −1) = (−1, −3, 5). So, the work done is obtained as follow: → − − → −−→ ∴ Work done by the force F = F · AB = (4, −3, 2) · (−1, −3, 5) = −4 + 9 + 10 = 15 units Note: In this illustration, the measurement unit for the force and displacement are not given so we have used a unit. Illustration 2.20: The constant forces ˆi + 2ˆj + 3kˆ and 3ˆi + ˆj + kˆ act on a particle. Under the action of these forces, a particle moves to 5ˆi + ˆj + 2kˆ from ˆj − 2k. Obtain total work done by these forces. − → − → Solution: Here, F1 = ˆi + 2ˆj + 3kˆ = (1, 2, 3) and F2 = 3ˆi + ˆj + kˆ = (3, 1, 1) are the forces acting on the particle. So, the total force or resultant force acting on the particle is given by → − − → − → ∴ Resultant force F = F1 + F2 = (1, 2, 3) + (3, 1, 1) = (4, 3, 4) Let A represent the initial position of the particle and B represents the final position of the particle after forces are applied. Therefore, A = ˆj − 2kˆ = (0, 1, −2) and B = 5ˆi + ˆj + 2kˆ = (5, 1, 2). Now, the displacement is given by −−→ ∴ Displacement AB = (5, 1, 2) − (0, 1, −2) = (5, 0, 4) As we have obtained the total forces acting on the particle and the displacement of the particle, we calculate work done as follow: → −−→ − ∴ Work done = F · AB = (4, 3, 4) · (5, 0, 4) =4×5+3×0+4×4 = 20 + 0 + 16 = 36 Thus, the total work done is by the forces is 36 units. Illustration 2.21: Forces 3ˆi − ˆj + 2kˆ and ˆi + 3ˆj − kˆ act on a particle and the particle moves from 2ˆi + 3ˆj + kˆ to 5ˆi + 2ˆj + kˆ under these forces. Find the work done by these forces.

2.5 Application of Scalar and Vector Products

49

− → Solution: The given forces acting on a particle are F1 = 3ˆi − ˆj + 2kˆ = − → ˆ (3, −1, 2) and F2 = i + 3ˆj − kˆ = (1, 3, −1). The total force acting on a particle is given by → → − − → − ∴ Resultant force F = F1 + F2 = (3, −1, 2) + (1, 3, −1) = (4, 2, 1) Let A represent the initial position of the particle then A = 2ˆi + 3ˆj + kˆ = (2,3,1) and let B denote the final position of the particle after forces are applied then B = 5ˆi + 2ˆj + kˆ = (5, 2, 1). So, the displacement of the particle from the point A to point B is given by −−→ Displacement = AB = (5, 2, 1) − (2, 3, 1) = (3, −1, 0) The work done by the forces is obtained as follow: → −−→ − Work done = F · AB = (4, 2, 1) · (3, −1, 0) = 4 × 3 + 2 × (−1) + 1 × 0 = 12 − 2 + 0 = 10 units. Illustration 2.22: A particle moves from point 3ˆi−2ˆj + kˆ to point ˆi+3ˆj −4kˆ ˆ ˆi + ˆj − 3k, ˆ 4ˆi + 5ˆj − 6k. ˆ Find under the effect of constant forces ˆi − ˆj + k, the work done by these forces on the particle. Solution: Let A be the initial point of the position of the particle then A = 3ˆi − 2ˆj + kˆ = (3, −2, 1) and let B be the final position of the particle then B = ˆi + 3ˆj − 4kˆ = (1, 3, −4). The displacement of the particle is obtained as −−→ Displacement = AB = (1, 3, −4) − (3, −2, 1) = (−2, 5, −5). − → The constant forces acting on the particle are F1 = ˆi− ˆj + kˆ = (1, −1, 1), − → ˆ ˆ − → F2 = i + j − 3kˆ = (1, 1, −3), and F3 = 4ˆi + 5ˆj − 6kˆ = (4, 5, −6). The resultant forces acting on the particle is given by → − → → − − → − ∴ Resultant forces F = F1 + F2 + F3 = (1, −1, 1) + (1, 1, −3) + (4, 5, −6) = (6, 5, −8) → −−→ − ∴ Work done = F · AB = (6, 5, −8) · (−2, 5, −5) = −12 + 25 + 40 = 53 units

50

Scalar and Vector Products

Illustration 2.23: Three forces of magnitudes 2, 1, and 5 units in the directions (1, 2, 3), (−1, 2, 3), and (−1, 2, −3) are applied on a particle. If a particle moves from the point (0, 1, −2) to the point (−1, 3, 2) under these forces, find the work done. → → → Solution: Let − x = (1, 2, 3), − y = (−1, 2, 3), and − z = (−1, 2, −3) are the given directions, and to find the forces in these directions, we have to find the unit vectors in these directions. √ √ √ √ → → ∴ |− x | = 1 + 4 + 9 = 14, |− y | = 1 + 4 + 9 = 14, √ → and similarly |− z | = 14. → − x 1 − ∴ The unit vector in the direction of → x = → = √ (1, 2, 3) − |x| 14 → − 1 y → = √ (−1, 2, 3) ∴ The unit vector in the direction of − y = − → |y| 14 → − 1 z → = √ (−1, 2, −3) ∴ The unit vector in the direction of − z = − |→ z| 14 − → ∴ F1 = Force of magnitude 2 in the direction of 2 → − x = √ (1, 2, 3) 14 − → ∴ F2 = Force of magnitude 1 in the direction of 1 → − y = √ (−1, 2, 3) 14 − → ∴ F3 = Force of magnitude 5 in the direction of 5 → − z = √ (−1, 2, −3) 14 → − → → − − → − ∴ The resultant force F = F1 + F2 + F3 1 = √ [2 (1, 2, 3) + (−1, 2, 3) + 5(−1, 2, −3)] 14 1 = √ [(2, 4, 6) + (−1, 2, 3) + (−5, 10, −15)] 14 1 = √ (−4, 16, −6) 14

2.5 Application of Scalar and Vector Products

51

Now, let A(0, 1, −2) be the initial position of the particle and B (−1, 3, 2) be the final position of the particle then the displacement of the particle is given by −−→ Displacement = AB = (−1, 3, 2) − (0, 1, −2) = (−1, 2, 4). Finally, the work done is given as → −−→ − 1 ∴ Work done = F · AB = √ (−4, 16, −6) · (−1, 2, 4) 14 1 12 = √ (4 + 32 − 24) = √ 14 14 Thus, the total work done is obtained as

√12 14

units.

Illustration 2.24: Find the moment about the point (2, 3, −1) of the force 3ˆi − kˆ acting through the point (1, −2, 1). Also, find the magnitude of the moment. Solution: Let A(2, 3, −1) and P (1, −2, 1) be the given points. → − Let F = 3ˆi − kˆ = (3, 0, −1) be the given force vector. −→ −−→ −→ ∴ AP = OP − OA = (1, −2, 1) − (2, 3, −1) = (−1, −5, 2) → − −→ − → Moment of the force F about the point A = AP × F    ˆi ˆj kˆ   =  −1 −5 2   3 0 −1  = (5 − 0) ˆi − (1 − 6) ˆj + (0 + 15) kˆ = 5ˆi + 5ˆj + 15kˆ = (5, 5, 15) √ √ √ ∴ Magnitude of the moment = 25 + 25 + 225 = 275 = 5 11. Illustration 2.25: Find the moment about the point (4, 0, 1) of the forces 2ˆi + ˆj − 3kˆ and 2ˆi − 2ˆj + kˆ acting through the point (−1, 3, −2). → − Solution: The resultant force F is given by → − F = 2ˆi + ˆj − 3kˆ + 2ˆi − 2ˆj + kˆ = (4, −1, −2). Let A(4, 0, 1) and P (−1, 3, −2) be the given points. −→ ∴ AP = (−1, 3, −2) − (4, 0, 1) = (−5, 3, −3)

52

Scalar and Vector Products

→ − The moment of the force F about the point A    ˆi ˆ  ˆj k   −→ − → = AP × F =  −5 3 −3   4 −1 −2  = (−6 − 3) ˆi − (10 + 12) ˆj + (5 − 12) kˆ = −9ˆi − 22ˆj − 7kˆ

2.6 Exercise

    1. Evaluate ˆi + 2ˆj + kˆ · 3kˆ − 2ˆj + 4ˆi . (Answer : 3)

− → → ˆ → ˆ and − ˆ then find 2. If − a= ˆi − ˆj + k, b = 2ˆi − ˆj + k, c = ˆi + ˆj − 2k, → − − → − → a · b + c . (Answer : 2) 3. For what value of p, the vectors 2ˆi + 3ˆj − kˆ and pˆi − ˆj + 3kˆ are perpendicular to each other? (Answer :p= 3) − → → 4. If the vectors − a = mˆi − 2ˆj + kˆ and b = 2mˆi + mˆj − 4kˆ are perpendicular to each other then find the value of m. (Answer : m = −1, 2) 5. If vectors (m, 2m, 4) and (m, −3, 2) are mutually perpendicular then find the value of m. (Answer : m = 2, 4) 6. Find the angle between vectors (1, 2, 3) and (−2, 3, 1).   Answer : π3 or 60o − − 7. For which value of p vectors → x = (1, −2, −3) and → y = (2, p, 4) are mutually perpendicular? (Answer : p = −5)

2.6 Exercise

53

− → − 8. Find the value of q if vectors → a = (q, 2, 1), and b = (2, q, −4) are perpendicular to each other. (Answer : q = 1) 



ˆ 9. Find the modulus of 2ˆi − 3ˆj + kˆ × (ˆi − ˆj + 2k).

√  Answer : 35



→ 10. Find a unit perpendicular vector to the given vector − x = (1, 2, 3) and → − y = (−2, 1, −2) .   Answer : 3√110 (−7, −4, 5) → 11. Find a unit perpendicular vector to the given vector − a = (1, −1, 1) and → − b = (2, 3, −1) both.   Answer : √138 (−2, 3, 5) → 12. Find a unit vector perpendicular to the plane containing the vectors − a = → − ˆ ˆ ˆ ˆ ˆ ˆ 3i + j + 2k and b = 2i + j − k.   Answer : √183 (−3, 7, 5)  + kˆ is 13. Show that the angle between the vectors ˆi + ˆj − kˆ and 2ˆi − 2j sin−1 26 27 . 14. Prove that the angle between two vectors 3ˆi + ˆj + 2kˆ and 2ˆi − 2ˆj + 4kˆ is sin−1 √27 . 2 2 2 2 → → → → → → x −− y | = 2(|− x | + |− y | ). 15. Prove that |− x +− y | + |−

→ → − → → → → → → 16. If − x +− y +→ z = θ, then prove that − x ×− y =− y ×− z =− z ×− x. 17. If the vectors ˆi − 2ˆj − 3kˆ and ˆi + 2kˆ represent two sides of the triangle then find the area of the triangle. √   Answer : 12 45 18. Find the area of the parallelogram whose adjacent sides are 3ˆi + ˆj − 2kˆ ˆ and ˆi − 3ˆj + 4k. √   Answer : 10 3

54

Scalar and Vector Products

− − 19. If |x − y| = 13 for the vectors → x = (7, p + 1, 1) and → y = (4, 1, 5), then find the value of p. (Answer : − 21) 20. Constant forces (3, −2, 1) and (−1, −1, 2) acts on the particle. Under the effects of these forces, a particle is displaced from (2, 2, −3) to the point (−1, 2, 4). Find the total work done by these forces. (Answer : 15 unit) 21. Under the effect of two forces (4, 1, −3) and (3, 1, −1) particle is displaced from origin to (1, 1, 1). Find the work done. (Answer : 5 unit) 22. A particle is displaced from point (0, 1, −2) to the point (5, 1, 2) under the effect of constant forces (1, 2, 3) and (3, 1, 1) then find the total work done. (Answer : 36 unit) 23. A particle moved from the point ˆi − ˆj to point 3ˆi + kˆ under the effect ˆ Then find the of the two constant forces −2kˆ + ˆi + ˆj and 2ˆj + 2ˆi − 4k. total work done. (Answer : 3 unit) 24. Force ˆj + ˆi + kˆ is action at −2ˆi + 3ˆj + 4kˆ find the moment of inertia ˆ and its magnitude above 2ˆi + 3ˆj + 5k. √   Answer : (1, 3, −4) and 26 25. Force (1, 1, 1) is acting on B(1, 2, 3) find the magnitude of the moment of inertia along A(−1, 2, 0). √   Answer : (−3, 1, 2) and 14

3 Vector Differential Calculus

3.1 Introduction In this chapter, we study vector functions in three dimensions and the applications of differential calculus to them. Vectors simplify many calculations considerably and help to visualize physical and geometrical quantities and relations between them. Consequently, vector methods are used extensively in applied mathematics and engineering. The impact of these methods on the study of physical phenomena such as fluid flow, elasticity, heat flow, electrostatics, electromagnetism, and waves in solids and fluids, which the engineer must understand as the foundation for the design and construction of systems such as aircraft, laser generators, robots, and thermo-dynamical systems, is critical to the engineer. Our goal is to acquaint readers with vector calculus, a branch of differential calculus that applies the basic notions of ordinary differential calculus to vector functions. The gradient, divergence, and curl are three physically and geometrically essential concepts connected to scalar and vector fields.

3.2 Vector and Scalar Functions and Fields A variable quantity whose value at any point in a region of space depends on the position of the point is called a point function. There are two types of point functions. A point function whose values are vector is called vector functions and is given as → − → v =− v (P ) = (v1 (P ) , v2 (P ) , v3 (P )) depending on the position of the points P in space. A point function whose values are scalars at any point P is called scalar functions and is given as f = f (P ) depending on the position of the point P in space.

55

56

Vector Differential Calculus

The domain of definition for such a function in applications is a region of space, a surface in space, or a curve in space. 3.2.1 Scalar Function and Field If to each point (x, y, z) of a region R in space corresponds a number or a scalar f = f (x, y, z) then f is called a scalar function of position or scalar (point) function and R is called a scalar field. For example, scalar functions are defined as quantities that take distinct values at various sites, such as the temperature field of a body, the pressure field of air in the earth’s atmosphere, density of a body, and potential. φ (x, y, z) = x2 − 2yz 3 defines a scalar field. A scalar field that is independent of time is called a stationary or steady-state scalar field. 3.2.2 Vector Function and Field If to each point (x, y, z) of a region R in space, there corresponds a vector → → − → v is v = − v (x, y, z) = [v1 (x, y, z) , v2 (x, y, z) , v3 (x, y, z)] , then − called a vector function of position or vector (point) function and R is called a vector field. For example, vector functions include the velocity of a flowing fluid at any given time, gravitational force, electric field intensity, magnetic field intensity, → and force. − v (x, y, z) = xy 2 ˆi−2yz 2 ˆj +x2 kˆ defines a vector field. A vector field that is independent of time is called a stationary or steady-state vector field. 3.2.3 Level Surfaces Let a scalar function f (x, y, z) be defined in a certain region R of space. Consider those points of the field for which scalar function f has a fixed value k. The totality of point satisfying the equation f (x, y, z) = k defines in general a surface and is called a level surface of the function since at every point of the surface f has a constant value k. For different values of k, we have different level surfaces and there will be no intersection of two-level surfaces. For example, if f (x, y, z) represents temperature in a medium, then f (x, y, z) = k represents a surface on which the temperature is a constant k. These surfaces are called isothermal surfaces. For example, the level surfaces of the scalar fields in space defined by the function f (x, y, z) = x2 + y 2 + z 2 is f (x, y, z) =√k i.e., x2 + y 2 + z 2 = k. Therefore, the level surfaces are spheres of radius k.

3.3 Curve and Arc Length 57

3.3 Curve and Arc Length First, we discuss the parametric representation of curves. 3.3.1 Parametric Representation of Curves If x and y are given as continuous functions x = f (t) , y = g (t) over an interval I of t−values, then the set of points (x, y) = (f (t) , g (t)) defined by these equations is a curve in the coordinate plane. The equations are parametric equations for the curve. The variable t is a parameter for the curve and its domain I is the parameter interval. If I is a closed interval, a ≤ t ≤ b, the point (f (a) , g (a)) is the initial point of the curve and (f (b) , g (b)) is the terminal point of the curve. If the equations of a curve are given in the plane in parametric equations and parametric interval then it is called a parameterized curve. The equations and interval constitute a parameterization of the curve. A curve C in the two-dimensional xy-plane can be parametrized by x = x (t) , y = y (t) , a ≤ t ≤ b. Then, the position vector of a point P on the C can be written as − → r (t) = x (t) ˆi + y (t) ˆj

(3.1)

Therefore, the position vector of a point on a curve defines a vector function. Similarly, a three-dimensional curve or a space curve C can be parametrized as − → ˆ a=t=b r (t) = x (t) ˆi + y (t) ˆj + z (t) k,

(3.2)

Illustration 3.1: Find a parametrization for the line segment with endpoints (−2, 1) and (3, 5) . Solution: Using (−2, 1), we crate the parametric equations x = −2 + at, y = 1 + bt. These represent a line, as we see by solving each equation for t and equating to obtain x+2 y−1 = a b

58

Vector Differential Calculus

This line goes through the point (−2, 1) when t = 0. We determine a and b so that the line goes through (3, 5) when t = 1. 3 = −2 + a ⇒ a = 5x = 3 when t = 1 5 = 1 + b ⇒ b = 4y = 5 when t = 1 Therefore, x = −2 + 5t, y = 1 + 4t, 0 = t = 1. or

− → r (t) = (−2 + 5t) ˆi + (1 + 4t) ˆj

is a parametrization of the line segment with initial point (−2, 1) and terminal point (3, 5) . Illustration 3.2: Find the parametric representation of the straight line through the point P (1, 2, 3) and has the direction b = i + 2j + 2k. Solution: In this case, a = (1 − 0) i + (2 − 0) j + (3 − 0) k = i + 2j + 3k ∴ a1 = 1, a2 = 2, a3 = 3 from b = i + 2j + 2k, we have b1 = 1, b2 = 2, b3 = 2 → ∴− r (t) = a + tb = (a1 + tb1 ) ˆi + (a2 + tb2 ) ˆj + (a3 + tb3 ) kˆ → gives − r (t) = (1 + t) ˆi + (2 + 2t) ˆj + (3 + 2t) kˆ → − or r (t) = (1 + t) ˆi + 2 (1 + t) ˆj + (3 + 2t) kˆ are required for parametric representation. 3.3.2 Curves with Tangent Vector → A curve is the locus of a point whose position vector − r relative to a fixed origin may be expressed as a function of a single variable parameter. Then its Cartesian coordinates x, y, z are also functions of the same parameter. Thus, the equation of the curve is usually expressed in parametric form x = x (t) , y = y (t) , z = z (t)

(3.3)

If P, Q are adjoining points (x, y, z) and (x + δx, y + δy, z + δz) , then their position vectors are given by (Figure 3.1)

3.3 Curve and Arc Length 59

Figure 3.1 Represents a curve with a tangent vector

→ − → → r = xˆi + yˆj + z kˆ and − r + δ− r = (r δx) ˆi + (y + δy) ˆj + (z + δz) kˆ → ∴ δ− r = (δx) ˆi + (δy) ˆj + (δz) kˆ 2 → ∴ |d− r | = δr2 = δx2 + δy 2 + δz 2

(3.4)

If δs is the length of the are P Q and in the limit as δt → 0, chord P Q (= δr) and are δs will be equal i.e., dr ds = 1

−  2  2  2 2 d→ r dx dy dz ∴ = + + =1 (3.5) ds ds ds ds and

→ dx ˆ dy ˆ dz ˆ d− r = i+ j+ k ds ds ds ds

 −    2  2  2  d→  dx dy dx r = ∴  + + =1  ds ds ds ds

(∵ from (3.4)) (3.6)

3.3.2.1 Tangent Vector → Limiting position of chord P Q (i.e., d− r ) as δt → 0 (as Q → P ) is along the tangent to the curve at P  −  →  d→ d− r r  ˆ  = T = Tˆ ( from (3.5)) ds ds  where unit vector Tˆ is along the tangent at P (t) . ∴

dr ds ds ˆ dr = · = ·T dt ds dt dt

where, Tˆ is a tangent vector.

(3.7)

60

Vector Differential Calculus

3.3.2.2 Important Concepts In Figure 3.2, the limiting position of the P QR passing through neighboring points P, Q, R on the curve and as Q, R approaches to P (i.e., the plane containing two consecutive tangents and therefore three consecutive points at P ) is called the osculating plane or the plane of curvature of the curve at point P. ˆ and is a line normal to the The binormal at the by B  point P is denoted  ˆ ⊥ Tˆ . And the principal normal at the point osculating plane at P i.e., B P denoted by n ˆ to the curve at point P is a line through the point P lying ˆ ⊥Tˆ in the osculating plane at P and it is perpendicular to the tangent line. N ˆ ˆ and N ⊥B. ˆ , it is parallel to Tˆ × N ˆ The binormal is perpendicular to both Tˆ and N ˆ, B ˆ form a right-handed system of mutually Hence trio unit vectors Tˆ, N perpendicular unit vectors, and therefore connected by the relations ˆ .B ˆ= B ˆ . Tˆ = 0 Tˆ . Tˆ = N and

ˆ × B ˆ = Tˆ , B ˆ × Tˆ = N ˆ, ˆ = B ˆ, N Tˆ × N

the cyclic order being preserved in the cross products. ˆ is an osculating plane of the (a) A plane through the point P , normal to B curve at P . (b) A plane through the point P normal to Tˆ is known as the normal plane of the curve at P . ˆ is known as rectifying plane (c) A plane through the point P normal to N of the curve.

Figure 3.2 Represents the plane of curvature of the curve

3.3 Curve and Arc Length 61

3.3.3 Arc Length The parameterization for a curve is a set of functions depending only on a parameter t along with the bounds for the parameter. When we parameterize a curve by taking values of t from some interval [a, b], the position vector → − r (t) of any point, t on the curve can be written as, → − r (t) = x (t) ˆi + y (t) ˆj + z (t) kˆ → The tangent vector − r  (t) is  → − r (t) = x (t) ˆi + y  (t) ˆj + z  (t) kˆ −    ∴ → r (t) = (x (t))2 + (y  (t))2 + (z  (t))2 The length of the curve is



b

  → − r (t) dt

l= a

The arc length function or arc length of the curve is obtained by considering a variable limit. i.e., a = t1 and b = t2 .  t2 −   → r (v) dv (3.8) l= t1

3.3.3.1 Unit Tangent Vector → The unit tangent vector of a curve − r (t) is → → d− r d− r /dt v Tˆ = = = ds ds/dt |v|

(3.9)

Illustration 3.3: Find the length of the arc between points (a, 0, 0) and (0, a, 12 πa tan α ) for the curve x = a cos θ , y = a sin θ , z = a θ tan α . Solution: Using result (3.8), we get

2 2 2 2 ds dx dy dz = + + dθ dθ dθ dθ = a2 sin2 θ + a2 cos2 θ + a2 tan2 α = a2 (1 + tan2 α) = a2 sec2 α ds ∴ = a sec α dθ

62

Vector Differential Calculus

For the given points the parameters are θ = 0[a, 0, 0] and θ = π2 [0, a, 12 πa tan α]

∴ The required length =

π/2 a 0

sec α · dθ =

π 2

a sec α .

Illustration3.4: Prove that the length of the curve  √ −1 2 x = 2a sin t + t 1 − t , y = 2 at2 , z = 4 at √ between the points, t = t1 and t = t2 are 4 2 (t2 − t1 ) a. Solution: Now, ! "

 1 dx 2t 2 = 2a √ + 1−t −1 − √ dt 1 − t2 2 1 − t2 ! " 2   1−t = 2a √ + 1 − t2 = 4a 1 − t2 1 − t2 dy dz = 4 at, = 4a dt dt

2 2 2 2 dx dy dz ds = + + ∴ dt dt dt dt   2 2 2 2 = 16a 1 − t + 16a t + 16a2 #  = 162 1 − t2 + t2 + 1] = 32a2 √ ds = 4 2a ∴ dt √  t2 √ ∴ s = 4 2a dt = 4 2 a (t2 − t1 ) t1

Illustration 3.5: Find the length of the arc of the curve y = log sec x from x = 0 to x = π/3 Solution: Here, dy 1 = · secx tanx = tanx dx secx !  2 " π/3 dy Length of the arc S = 0 1 + dx dx

3.3 Curve and Arc Length 63

 = 

π/3 

1 + tan2 x dx

0 π/3

π/3

sec x dx = [log(secx − tanx)]0 0 π    π  − tan − log (1 − 0) = log sec 3 3 √ = log(2− 3) − 0 √ = log(2 − 3 )

=

Illustration 3.6: Find the arc length of the Helix traced by → − ˆ a > 0, 0 ≤ t ≤ 2π . r (t) = a cost ˆi + a sinˆj + ct k,

→ Also, express the position vector − r (t) in terms of the arc length s. Solution: we have x (t) = a cost , y (t) = a sint , z (t) = ct and dx dy dz = −a sint , = a cost , =c dt dt dt

Therefore,



2π

[a2 sin2 t + a2 cos2 t + c2 ]1/2 s = Arc length = 0  2π   2 1/2 1/2 = dt = 2π a2 + c2 a + c2 0

Also, the arc length s is given as  t   2 1/2 1/2 s= dt = t a2 + c2 a + c2 0

or t= Therefore,



s

s (a2 + c2 )1/2 

 i + a sin (a2 + c2 )  s +c  k. (a2 + c2 )

r (s) = a cos



s (a2 + c2 )

 j

64

Vector Differential Calculus

   It can be verified that  dr ds = 1. Illustration 3.7: Find the unit tangent vector to the unit circle − → r (t) = (cos t) ˆi + (sin t) ˆj Solution: Here, r (t) = (cos t) ˆi + (sin t) ˆj dr → = (−sin t) ˆi + (cos t) ˆj ∴− v = dt √   −  v=1  → → − v (−sin t ) ˆi + (cos t) ˆj = (−sin t) ˆi + (cos t) ˆj ∴ Tˆ = → = − 1 |v|

3.4 Curvature and Torsion In this section, we discuss curvature, torsion, and Serret (1851) – Frenet (1852) Formulas. We may gather them together in the form. ˆ ˆ , (b) dBˆ = −τ N ˆ , (c) dNˆ = τ B ˆ − kN ˆ (a) ddST = k N dS dS These formulas give the derivatives of the direction cosines of the tangent, the binomial, and the principal normal. (I) Definition: Curvature k at point P on the curve is are the rate of rotation of tangent Tˆ at P ∴k=

dΨ dS

(i)

where δΨ is the angle through which Tˆ has turned when P moves to a dTˆ dTˆ dΨ dTˆ (ii) point Q . dS = dΨ . ds = k dΨ Now, Tˆ .Tˆ = 1 dTˆ dTˆ = 0 ⇒ Tˆ⊥ ∴ Tˆ · dS dS Since Tˆ and Tˆ + δ Tˆ lie in the osculating plane, δ Tˆ lies in the osculating ˆ plane i.e., ddST lies in an osculating plane and being perpendicular to, it is

3.4 Curvature and Torsion

Figure 3.3

65

Represents the arc rate of rotation of binormal

ˆ at P . along N

dTˆ ˆ ∴ = (1) N dψ

    dTˆ    ∵  =1  Dψ 

(iii)

From (ii) and (iii) we get, dTˆ ˆ = kN dS

(3.10)

(II) Definition: Torsion (τ ) at point P on the curve is arc-rate of rotation of binormal at P . If δ∅ is the angle of rotation when P moves to point Q (in Figure 3.3). ∴τ =

dφ dS

ˆ ˆ dφ ˆ dB dB dB = · = τ dS dφ ds d∅ ˆ ·B ˆ = 1, ∴ B ˆ· Now,B

ˆ dB dφ

(i) (ii)

=0 ˆ dB ∴ B⊥ dφ

ˆ and B ˆ +δ B ˆ (in limiting position as → 0 ) lie in a normal plane, Since B ˆ ˆ dB dB ˆ dφ lies in a normal plane and is perpendicular to B. Hence dφ is along ˆ at P . the normal N

66

Vector Differential Calculus

Figure 3.4

ˆ T,  and N ˆ orthogonal unit vectors Represents B,

If point P moves in the increasing direction of s and the right-handed screw is rotated in such a way as to advance in direction of s (i.e., Tˆ), ˆ is rotated in the clockwise direction and thus the direction the vector B ˆ dB ˆ. of ds is opposite to that of N     dB ˆ  ˆ dB  ˆ ∵ = −N (iii) ∴ =1  dφ  d∅ From (ii) and (iii), we get ˆ ˆ dB dB ˆ =τ = −τ N ds dφ Hence, ˆ dB ˆ = −τ N ds ˆ Tˆ, and N ˆ are orthogonal unit vectors (in Figure 3.4), (III) Since B, ˆ =B ˆ × Tˆ N ˆ ˆ ˆ dN ˆ × dT + dB × Tˆ =B ∴ ds ds ds     ˆ× kN ˆ + −τ N ˆ × Tˆ =B

(3.11)

3.4 Curvature and Torsion

67

Using results (3.10) and (3.11), we get     ˆ ×N ˆ −τ N ˆ × Tˆ =k B =k

    ˆ − − Tˆ − τ −B

Hence, ˆ dN ˆ − k Tˆ = τB ds

(3.12)

3.4.1 Formulas for Curvature and Torsion → The position vector − r (t) of the point [x (t) , y (t) , z (t)] on the curve is given by − → r (t) = x (t) ˆi + y (t) ˆj + z (t) kˆ

(i)

where x = x (t) , y = y (t) , z = z (t) represents parametric equations of → → the curve − r =− r (t) → → d− r ds ds ˆ d− r = · = ·T (ii) dt ds dt dt d2 s ˆ ds dTˆ d2 r → − . = r¨ = T+ dt dt dt2 dt2 

ds dTˆ ds d2 s = 2 Tˆ + . dt ds dt dt

2 ˆ ds dT d2 s ˆ = 2 T+ dt ds dt

2 2 ds d s ˆ ˆ = kN (iii) T+ 2 dt dt % ! " $ 2

2 ds ds s d → − → − ˆ ∴ r × ¨r = Tˆ × kN Tˆ + dt dt2 dt

3     ds ˆ k Tˆ × N ∵ Tˆ × Tˆ = 0 = dt

3 ds ˆ k B (iv) = dt → ∴− r =

68

Vector Differential Calculus

    →  → → ¨r  = k  ds 3 and − ∴ − r˙ × − r˙  = dt Equating the magnitudes, we get    3 −  −  → → − ˙ ¨ r × r = k r˙    →

ds dt

    − → → −  r˙ × ¨r  ∴ k =   3   −   r˙    →

(From (ii))

(v)

(3.13)

  Formula (3.13) determines the curvature k or radius of curvature = k1 at a point P on the curve. From equation (iii), we get $  %

2 ˆ → ... ds 2 ˆ ds dN d r d3 s ˆ d2 s dTˆ d3 − → − + k N +k = 3 T+ 2 r = 3 dt dt dt dt dt dt dt dt $  % % $ 

ˆ ds ds 2 ˆ ds 2 dN d3 s ˆ d2 s dTˆ ds d = 3 T+ 2 · + k · N +k dt dt ds dt dt dt dt ds dt Substituting we have

dTˆ ds

and

ˆ dN ds

from the results of the Serret-Frenet Formula,

$  %

3   3s 2 s ds ds 2 ˆ ds d d d → − ˆ − k Tˆ ˆ ˆ τ B k N + k N + k T + r = dt3 dt2 dt dt dt dt $ $  %' % & 

3 2 2 d2 s 2 ds ˆ + k ds d s + d k ds ˆ = − k T N dt3 dt dt dt2 dt dt $  % ds 3 ˆ + kτ B dt % $   ...  ... ds 3 ˆ → − → − → − → ˙ ¨ ∴ r · r × r = kB . − r (∵ From (iv)) dt

6   ds ˆ × Tˆ = B ˆ ×N ˆ =0 k2 τ ∵ B = dt ( ) → → ¨r 2 τ (∵ From (v)) = − r˙ × − ...

3.4 Curvature and Torsion

 ...  − → → → r · − r˙ × − r¨ ∴ τ= ( )2 → − → r˙ × − r¨

(3.14)

The formula (3.14) gives the torsion τ or radius of torsion σ = point P on the curve. Illustration 3.8: For the curve x = t, y = t2 , z = at point  t . Solution: − → r = t (i) ˆi + t2 ˆj +



69

2 3

1 τ

at a

, find (i) K and (ii) τ

 2 3 ˆ t k 3

− → r˙ = ˆi + 2tˆj + 2t2 kˆ → − ¨r = 2ˆj + 4t kˆ − → ¨r = 4kˆ

⎡

⎤ ˆi ˆj kˆ → − − ¨r = ⎢ 1 2t 2t2 ⎥ = 4t2  ˆi − (4t) ˆj + 2kˆ r˙ × → ⎢ ⎥ ⎢ 0 2 4t ⎥  2 2 → Hence, − r  = 1 + 4t2 + 4t4 = 1 + 2t2 ( ) → → ¨r 2 = 16t2 + 16t2 + 4 = 42t2 + 12 and − r˙ × −   → − → − → ˙ ¨ ¨r = ,4t2 i − 4t j + 2kj - . 4k = 8 and r × r · − (

Using formula k = we get 2(1+2t2 ) k = (1+2t2 )3 =

− → → ¨ r˙ × − r ( )3 → − ˙r

2 (1+2t2 )2

)



and τ =

and τ =

  ...  − → → → ¨ r˙ × − r · − r ( )2 → − → ˙r ×− ¨ r

8 4(1+2t2 )2

=

2 (1+2t2 )2

Illustration 3.9: Show that for the curve x = a cos θ , y = a sin θ , z = a θ cot β curvature and torsion are k = a1 sin2 β and τ = a1 sinβ cosβ . Solution: → Here, − r = (a cos θ ) ˆi + (a sin θ ) ˆj + (a θ cot β ) kˆ ⎡ ⎤ ˆi ˆj kˆ → → ¨r = ⎣ −a sin θ ∴ − r˙ × − a cos θ a cot β ⎦ −a cos θ −a sin θ 0

Vector Differential Calculus

70

and ...

→ − r = (a sin θ ) ˆi + (−a cos θ ) ˆj + 0 kˆ ( )2 → − = a2 sin θ + a2 cos2 θ + a2 cot2 β r˙ . # = a2 1 + cot2 β = a2 cosec2 β ( ) → − → ¨r 2 = a4 sin2 θ cot2 β + a2 cos2 θ cot2 β + a4 r˙ × − .



cot2 β + 1



= a4 cosec2 β and



→ − → ¨r r˙ × −



  ...    − r = a2 sin θ cot β ˆi + −a2 cos θ cot β ˆj + a2 kˆ · · →

[(a sin θ ) ˆi + (−a cos θ ) ˆj] = a3 sin2 θ cot β + a3 cos2 θ cot β = a3 cot β ( ) → − − ¨r r˙ × → 1 a2 cosec β = sin2 β = 3 ∴k= ( )3 3 a cosec β a → − r˙ and

  → − → → ¨r · − ¨r r˙ × − 1 a3 cot β = sin β cos β τ= ( )2 = 4 2 β a cosec a → − → ˙r × − ¨r

3.5 Vector Differentiation Next, we show the basic concepts of calculus, such as vector differentiation, differentiability defined for scalar, and vector functions in a simple and natural way. → The derivative of a vector function − v (t) with respect to the scalar, variable t is given as − → − → d→ v v (t + Δt) − − v (t)  → =− v (t) = lim Δt→0 dt Δt

(3.15)

provided the limit exists. → − The vector − v  (t) is called the derivative of → v  (t). (see Figure 3.5).

3.5 Vector Differentiation

71

Figure 3.5 Represents the derivative of the vector v (t)

In terms of components with respect to a given Cartesian coordinate → system, − v (t) is differentiable at a point t if and only if its three components → v1 (t), v2 (t), and v3 (t) are differentiable at t, and then the derivative − v  (t) is obtained by differentiating each component separately,    → − v (t) = v1 , v2 , v3 (3.16) Most of the familiar rules of differentiation yield corresponding rules for differentiating vector functions provided the order of factors in vector product is maintained. 1. The derivative of a constant vector is zero. → → 2. (c− v) =c− v  where c is constant 

   − → → → 3. (→ u ±− v) =− u ±− v   → − → − → → 4. (→ u ·− v) =→ u ·− v +− u ·− v   → → → → − → 5. (→ u ×− v) =− u ×− v +− u ×− v     → − → → → → → → → → → − 6. (− u ·− v ·− w) = − u ·− v ·→ w + (− u ·− v ·− w ) + (− u ·− v ·→ w )

→ − Illustration 3.10: If F (t) has a constant magnitude, then prove that → − perpendicular to F (t). Solution: → − F (t) hasa constant magnitude − →  ∴  F (t) = constant

→ − dF dt

is

72

Vector Differential Calculus

→ − → − ∴ F (t) · F (t) =

−  →  2  F (t) = constant → − → d − F ·F =0 dt → − → − → → dF − dF − + ·F =0 ⇒ F · dt dt → − → dF − =0 ⇒ 2F · dt → − → dF − =0 ⇒ F · dt ∴

→ − Since, F ·

→ − dF dt

= 0, dF dt is perpendicular to F (t). Hence proved. → − Illustration 3.10: If F (t) has a constant direction, then prove that → − → dF − F × = 0. dt Solution: −  → − → − →  Let  F (t) = φ (t). Let G (t) be a unit vector in the direction of F (t) so → − → − that F (t) = φ (t) G (t). → − → − → dF d G dφ − ∴ =φ + G (i) dt dt dt → − → − → − If F (t) has constant direction, so has G (t). Thus G (t) is a constant vector and

→ − dG dt

=0 From (i).

→ − → dF dφ − = G dt dt

 → − → − − − → dφ → dF = φG × G ∴F × dt dt − → − → dφ =φ G ×G =0 dt

Illustration 3.11: Find the angle between the tangents to the curve x = t2 , y = 2t, z = −t3 at the points t = 1 and t = −1.

3.6 Gradient of a Scalar Field and Directional Derivative

73

Solution: → Let − r be the position vector of any point (x, y, z) on the curve, then − → r (t) = (x (t) , y (t) , z (t)) = x (t) ˆi + y (t) ˆi + z (t) kˆ → ∴− r = t2ˆi + 2tˆj − t2 kˆ → → d− − r ∴ T = = 2tˆi + 2ˆj − 3t2 kˆ dt is a vector along the tangent at any point t. → − → − ∴ T 1 = 2ˆi + 2ˆj − 3kˆ and T 2 = −2ˆi + 2ˆj − 3kˆ are the vectors along the tangents at t = 1 and t = −1 respectively. → − → − If θ be the angle between T 1 and T 2 , Then → − − → T1· T2 2 (−2) + 2 (2) − 3 (−3) 9 cos θ = − →  − →  = √4 + 4 + 9 √4 + 4 + 9 = 17  T 1  T 2

 9 −1 ∴ θ = cos . 17

3.6 Gradient of a Scalar Field and Directional Derivative The vector differential operator is denoted by ∇ (read as del or nabla) and defined as ∂ ∂ ∂ ∇ = ˆi + ˆj + kˆ ∂x ∂y ∂z and this vector operator has properties analogous to those of ordinary vectors. It is very useful in defining three important quantities gradient, divergence, and curl. 3.6.1 Gradient of a Scalar Field For a given scalar function f (x, y, z) the gradient of f , is written as grad f or ∇f is the vector function defined by grad f = ∇f = ˆi

∂f ˆ ∂f ˆ ∂f +j +k ∂x ∂y ∂z

3.6.1.1 Properties of Gradient 1. The projection of ∇f in any direction is equal to the derivative of f in that direction.

74

Vector Differential Calculus

Figure 3.6 Represents the directional derivative

2. The gradient of f is a vector normal to the surface f (x, y, z) = c = constant. So, the angle between any two surfaces f (x, y, z) = c1 and g (x, y, z) = c2 is the angle between their corresponding normal given by ∇f and ∇g respectively. 3. The gradient at any point P gives the maximum rate of change of f in the direction of maximum increase of f at the point P . 3.6.2 Directional Derivative ∂f ∂f If f = f (x, y, z) then ∂f ∂x , ∂y , ∂z are the derivatives (rate of change) of f in the “direction” of the coordinate axes OX, OY , and OZ respectively. This concept can be extended to define a derivative of f in a given direction −−→ P Q (see Figure 3.6) → − Let P be a point in space and b be a unit vector from P in the given direction. Let s be the arc length measured from P to another point Q along → − the ray C in the direction b . Then the directional derivative of f at the point → − df → f or P in the direction of b is denoted D− ds and defined by (see Figure 3.6) b

f (Q) − f (P ) df = lim (s = distance between P and Q) s→0 ds s → − where Q is a variable point on the ray C in the direction of b and the ray C is given by → − r (s) = x (s) ˆi + y (s) ˆj + z (s) kˆ (s ≥ 0) → − Next, we use Cartesian coordinates, and for b a unit vector. Now consider the function →f = D− b

f (s) = f (x, y, z) = f (x (s) , y (s) , z(s))

3.6 Gradient of a Scalar Field and Directional Derivative

75

and it’s derivative with respect to the arc length s of C. Hence, assuming that f has continuous partial derivatives and applying the chain rule, we obtain ∂f ∂x ∂f ∂y ∂f ∂z df →f = D− = + + b ds ∂x ∂s ∂y ∂s ∂z ∂s 



dx ˆ dy ˆ dz ∂f ˆ ∂f ˆ ∂f ˆ ˆ +j +k · i +j +k = i ∂x ∂y ∂z ds ds ds !

 " → − ∂ ∂ ∂ = ˆi + ˆj + kˆ f · b ∂x ∂y ∂z

 → d− r dx ˆ dy ˆ dz ˆ ∵ =i +j +k ds ds ds ds → − − → df →f = ∴ D− = ∇f · b = b · grad f b ds Thus, the directional derivative of f at any point P is the dot product of → − unit vector b and grad f . → Note that the directional derivative of the vector − a of any length ( = 0) is

−  → a df 1 − → Daˆ f = = grad f · − = ∇f · − a = ∇f · a ˆ → → ds |a| |a| where a ˆ=

− → a → − a

| |

→ is the unit vector in the direction of the vector − a.

3.6.2.1 Properties of Gradient 1. The function f increases most rapidly when cosθ = 1 (i.e., when θ = → 0) or when − a is the direction of ∇f . The derivative in this direction is → D− a f = |∇f | cos(0) = |∇f | .

2. Similarly, f decreases most rapidly when in the direction of −∇f . The derivative in this direction is → D− a f = |∇f | cos(π) = − |∇f | .

→ 3. Any direction − a orthogonal to a gradient ∇f = 0 is a direction of zero change in f because θ then equals π/2 and π  → − D a f = |∇f | cos = |∇f | · 0 = 0 2

76

Vector Differential Calculus

These properties hold in both two as well as three dimensions. Illustration 3.12: (a) Gradient as a Surface normal  vector,  find a unit normal vector n of the cone of revolution z 2 = 4 x2 + y 2 at the point P : (1, 0, 2) . (b) If ∅ = 3x2 y − y 3 z 2 , find grad ∅ at the point (1, −2, −1). Solution: (a) The cone is a level surface   f = 0 for f (x, y, z) = 4 x2 + y 2 − z 2 . ˆ Thus, ∇f = grand f = 8x ˆi + 8y ˆj − 2z kˆ and at P , grad f = 8ˆi − 4k. Hence, a unit normal vector of the cone at P is n ˆ=

1 8ˆi − 4kˆ grad f =  |grad f | (8)2 + (−4)2

8ˆi − 4kˆ 2 1 ˆ √ = √ ˆi − √ k. 80 5 5   (b) Here, ∇φ = 6xyˆi + 3x2 − 3y 2 z 2 ˆj − 2y 3 z kˆ =

ˆ ∇at(1, −2, 1) = −12ˆi − 9ˆj − 16k. 3.13: Find the derivative of f (x, y) = x2 sin2y at the point Illustration  → 1, π2 in the direction of − v = 3ˆi − 4ˆj. Solution: Consider, and

  ∇f = (2xsin2y ) ˆi + 2x2 cos2y ˆj  π (∇f ) 1, = 2sinπ ˆi + (2cosπ ) ˆj 2 = 0 ˆi − 2ˆj = −2ˆj

→ − The direction of − v is the unit vector obtained by dividing → v by its length: → − v 4 3 vˆ = → = ˆi − ˆj − 5 5 |v|

3.6 Gradient of a Scalar Field and Directional Derivative

77

  The derivative of at 1, π2 in the direction of v is therefore    3 4ˆ ˆ ˆ ∇f |(1, π ) · vˆ = −2j · i− j 2 5 5 =0+

8 8 = 5 5

Illustration 3.14: (a) Find the derivative of f = (x, y, z) = x3 − xy 2 − z at → ˆ (b) In what directions v = 2ˆi − 3ˆj + 6k. P0 (1, 1, 0) in the direction of − does f change most rapidly at p0 , and what are the rates of change in these directions? Solution: (a) (∇f ) P0 (1, 1, 0) = 2ˆi − 2ˆj − kˆ → The direction of − v is → − 2 3 6 2ˆi − 3ˆj + 6kˆ v = ˆi − ˆj + kˆ = u ˆ= → − 7 7 7 7 |v| → v is The derivative of f and P0 in the direction of − 

  2 3 6 ˆi − ˆj + kˆ ˆ = 2ˆi − 2ˆj − kˆ · ∇f |(1,1,0) · u 7 7 7 4 6 6 4 = + − = 7 7 7 7 (b) The function increases most rapidly in the direction of ∇f = 2ˆi − 2ˆj − kˆ and decreases most rapidly in the direction of −∇f . The rates of change in the directions are, respectively  √ |∇f | = (2)2 + (−2)2 + (−1)2 = 9 = 3 and − |∇f | = −3. → ˆ Show that Illustration 3.15: If − r = xˆi + yˆj + z k, (i) ∇r = rˆ → (ii) ∇rn = nrn−2 − r

78

Vector Differential Calculus

Solution: ∂r + ˆj (i) grad r = ∇r = ˆi ∂x

= ˆi

∂r ∂y

+ kˆ

x

∂r ∂z

+ ˆj

y 

+ kˆ

z 

r r r → − ˆ ˆ ˆ xi + y j + z k r → = = =− r r r

 → − r ∵ − = rˆ unit vector |→ r| (ii) grad rn = ∇rn ∂ n ˆ ∂ n ˆ ∂ n r +j r +k r ∂x ∂y ∂z 





∂r ∂r ∂r n−1 n−1 n−1 ˆ + ˆj nr + k nr = ˆi nr ∂x ∂y ∂z      x y z = ˆi nrn−1 . + ˆj nrn−1 . + kˆ nrn−1 . r r  r  → n−2 n−2 − ˆ ˆ ˆ = nr xi + y j + z k = nr r = ˆi

Illustration 3.16: Find thegradient of f (x, y, z) = 2z 3 − 3 x2 + y 2 z + tan−1 (xz) at (1, 1, 1). Solution:

" ! 1 · z + ˆj [−6yz] ∇f = ˆi −6xz + 1 + x2 z 2 ! "   2 1 2 2 ˆ + k 6z − 3 x + y + ·x 1 + x2 y 2 " ! " ! 1 1 ∇f(1, 1, 1) = ˆi −6 + + ˆj [−6] + k 6 − 6 + 2 2 ! " −11 1 = ˆi − 6ˆj + kˆ 2 2   ∴ The gradient of f at (1, 1, 1) = 12 −11ˆi − 12ˆj + kˆ       ˆ determine φ. Illustration 3.17: If ∇φ = y 2 ˆi + 2xy + z 3 ˆj + 3yz 2 k,

3.6 Gradient of a Scalar Field and Directional Derivative

79

Solution: ∂φ ˆ ∂φ ˆ ∂φ +j +k ∇φ = ˆi ∂x ∂y ∂z  2     3 = ˆi y + ˆj 2xy + z + kˆ 3yz 2  ∂φ ∂φ  ∴ = y 2 (i) , = 2xy + z 3 ∂x ∂y

(ii),

∂φ = 3yz 2 ∂z

(iii)

Integrating (i), (ii) and (iii) partially with respect to x, y, z respectively, φ = xy 2 + f1 (y, z) φ = xy 2 + yz 3 + f2 (z, x) φ = yz 2 + f1 (x, y) φ = xy 2 + yz 3 + c (c is an arbitrary constant) Illustration 3.18: Find the magnitude and the direction of the greatest change of u = xyz 2 at (1, 0, 3). Solution:

∂u ˆ ∂u ∂u +j + kˆ ∂x ∂y ∂z  2  2 = ˆi yz + ˆj xz + kˆ (2xyz) ∇u = ˆi

∴ ∇u at (1, 0, 3) = 9ˆj ∴ The magnitude of the greatest change of u = |∇u| = 9 and its direction is along the y-axis. Illustration 3.19: The temperature at any point in space is given by T = xy + yz + zx. Determine the derivative of T in the direction of the vector 3ˆi − 4kˆ at the point (1, 1, 1). Solution: The derivative of T in the direction of the vector 3ˆi − 4kˆ at the point → → ˆ P (1, 1, 1) is given by (∇T )p · − a ; where − a = 3ˆi − 4k. Now, ∂ ∂ (xy + yz + zx) + ˆj (xy + yz + zx) ∂x ∂y ∂ (xy + yz + zx) + kˆ ∂z

∇T = ˆi

80

Vector Differential Calculus

= (y + z) ˆi + (z + x) ˆj + (x + y) kˆ ∴ (∇T )p(1, 1, 1) = 2ˆi + 2ˆj + 2kˆ

  ∴ Directional Derivative of T = (∇T )P · a ˆ = 2ˆi + 2ˆj + 2kˆ ·   3ˆi − 4kˆ √ 9 + 16 4 2 = (6 − 8) = − 5 5

Illustration 3.20: Find the directional derivative of g (x, y, z) − → ˆ 3ex cos (yz) at P0 (0, 0, 0) in the direction of A = 2ˆi + ˆj − 2k.

=

Solution: ∂g ∂g ∂g + ˆj + kˆ ∂x ∂y ∂z x ˆ ˆ = i [3 e cos (yz) ] + j [−3 ex sin (yz) · z] + kˆ [−3 ex sin (yz) · y]

∇g = ˆi

∴ ∇g at p0 (0, 0, 0) = ˆi (3) + 0 + 0 = 3ˆi 2ˆi + ˆj − 2kˆ 2ˆi + ˆj − 2kˆ Aˆ = √ = 3 4+1+4 ∴ Directional derivative = (∇g)p0 · Aˆ  ˆi + ˆj − 2kˆ 2 = 3ˆi · 3 =

6+0−0 =2 3

Illustration 3.21: Find the derivative of f (x, y) = x ey + cos (xy) at the point (2, 0) in the direction of A = 3ˆi − 4ˆj. Solution:

∂f ∂x

3ˆi − 4ˆj 3ˆi − 4ˆj Aˆ = √ = 5 9 + 16

= [ey − ysin (xy) ] and

∂f ∂y

= [x ey − xsin (xy) ]

3.6 Gradient of a Scalar Field and Directional Derivative

∴

∂f ∂x

at (2, 0) = e0 − 0 = 1 and

∂f ∂y

81

at (2, 0) = 2 e0 − 2.0 = 2

∇f(2,0) = ˆi (1) + ˆj (2) = ˆi + 2ˆj The directional derivative of f at (2, 0) in the direction Aˆ is therefore     3ˆi − 4ˆj 3−8 −5 = = = −1 (∇f )(2,0) · Aˆ = ˆi + 2ˆj · 5 5 5 Illustration 3.22: Find the direction in which f (x, y) =

x2 2

+

y2 2

(a) increases most rapidly and (b) decreases most rapidly at the point (1, 1). (c) What are the directions of zero change in f at (1, 1)? Solution: (a) The function increases most rapidly in direction of ∇f at (1, 1) The gradient is

  (∇f )(1, 1) = xˆi + yˆj (1, 1) = ˆi + ˆj

Its direction is → − ˆi + ˆj 1 1 u = √ ˆi + √ ˆj =  u ˆ= → − |u| 2 2 (1)2 + (1)2 (b) The function decreases most rapidly in the direction of −∇f at (1, 1) which is

1 1 → −− u = − √ ˆi − √ ˆj 2 2

(c) The directions of zero change at (1, 1) are the directions orthogonal to ∇f : 1 1 n ˆ = − √ ˆi + √ ˆj 2 2 and 1 1 −ˆ n = √ ˆi − √ ˆj 2 2 Since u ˆ·n ˆ = 0 and −ˆ u · (−ˆ n) = 0.

82

Vector Differential Calculus

Illustration 3.22: Find the directional derivative of φ = 4xz 3 − 3x2 y 2 z at the point (2, −1, 2) (i) in the direction 2ˆi + 3ˆj + 6kˆ (ii) towards the point (1, 1, −1) or vector ˆi + ˆj − kˆ (iii) along a line equally inclined with coordinate axes. (iv) along the tangent to the curve x = et cost, y = et sint, z = et at t = 0 (v) along the direction normal to the surface x2 + y 2 + z 2 = 9 at (1, 2, 3) (vi) along the z-axis. Solution: ∂φ ∂φ ∂φ = 4z 3 − 6xy 2 z , = −6x2 yz, = 12xz 2 − 3x2 y 2 ∂x ∂y ∂z     ∴ ∇φ = 4z 2 − 6xy 2 z ˆi − 6x2 yzˆj + 12xz 2 − 3x2 y 2 kˆ ∇φ at (2, −1, 2) = 8ˆi + 48ˆj + 84kˆ − (i) → a = 2ˆi + 3ˆj + 6kˆ 2ˆi + 3ˆj + 6kˆ 2ˆi + 3ˆj + 6kˆ ∴a ˆ= √ = 7 4 + 9 + 36 Directional derivative = ∇φ · a ˆ   2ˆi + 3ˆj + 6kˆ = 8ˆi + 48ˆj + 84kˆ · 7 =

664 16 + 144 + 504 = 7 7

(ii) an along the line joining say P (2, −1, 2) and Q (1, 1, −1) is − → a = (1 − 2) ˆi + (1 + 1) ˆj + (−1 − 2) kˆ = −ˆi + 2ˆj − 3kˆ

3.6 Gradient of a Scalar Field and Directional Derivative

∴a ˆ=

−ˆi + 2ˆj − 3kˆ √ 14

Directional derivative = ∇φ · a ˆ 





= 8ˆi + 48ˆj + 84kˆ ·

−ˆi + 2ˆj − 3kˆ √ 14

−164 8 (−1) + 48 (2) + 84 (−3) √ = √ 14 14

= − (iii) → a = ˆi + ˆj + kˆ

∴a ˆ=

ˆi + ˆj + kˆ √ 3

Directional derivative = ∇φ .ˆ a

  ˆi + ˆj + kˆ √ = 8ˆi + 48ˆj + 84kˆ · 3 



140 = √ 3 (iv)

dx dt



= et (cost − sint),

dy dt

= et (sint + cost ),

dz dt

= et

− → d→ − dx ˆ dy ˆ dz ˆ r = i+ j+ k ∴ T = dt dt dt dt ∴ Tat t=0 = ˆi + ˆj + kˆ ˆi + ˆj + kˆ √ ∴ Tˆ = 3 ∴ Directional derivative ∇φ · Tˆ

  ˆi + ˆj + kˆ √ = 8ˆi + 48ˆj + 84kˆ · 3 

140 = √ 3



83

84

Vector Differential Calculus

(v) Let f (x, y, z) = x2 + y 2 + z 2 = c ∴ ∇f = 2xˆi + 2yˆj + 2z kˆ → Let − a = ∇fat(1, 2, 2) = 2ˆi + 4ˆj + 4kˆ ˆi + 2ˆj + 2kˆ 2ˆi + 4ˆj + 4kˆ = ∴a ˆ= √ 3 4 + 16 + 16 ∴ Directional derivative = ∇φ · a ˆ   ˆi + 2ˆj + 2kˆ = 8ˆi + 48ˆj + 84kˆ . 3 =

272 8 + 96 + 168 = 3 3

(vi) Directional derivative ∇φ · a ˆ = ∇φ · kˆ   = 8ˆi + 48ˆj + 84kˆ · kˆ = 84 3.6.3 Equations of Tangent and Normal to the Level Curves At every point (x0 , y0 ) in the domain of f (x, y), the gradient of f is normal to the level curve through (x0 , y0 ). ∴ Equation of the tangent at (x0 , y0 ) to the level curve, f (x, y) is (x − x0 ) fx (x0 , y0 ) + (y − y0 ) fy (x0 , y0 ) = 0

(3.17)

and the equation of the normal is (x − x0 ) (y − y0 ) = fx (x0 , y0 ) fy (x0 , y0 )

(3.18)

Illustration 3.23: Find equations of the tangent and normal to the ellipse x2 2 4 + y = 2 at the point (−2, 1). Solution: The ellipse is a level curve of the function f (x, y) =

x2 + y2. 4

3.6 Gradient of a Scalar Field and Directional Derivative

85

The gradient f at (−2, 1) is ∇fat(−2, 1) =

x 2

 ˆi + 2y ˆj

at(−2, 1)

= −ˆi + 2ˆj

The tangent is the line (x + 2) (−1) + (y − 1) (2) = 0. or x − 2y = −4 Equation of the normal

x+2 y−1 = −1 2

or 2x + y = −3 3.6.4 Equation of the Tangent Planes and Normal Lines to the Surfaces The equation of the tangent plane at P (x0 , y0 , z0 ) to the surface z = f (x, y) or F (x, y, z) = f (x, y) − z is (x − x0 ) Fx (x0 , y0 , z0 ) + (y − y0 ) Fy (x0 , y0 , z0 ) + (z − z0 ) Fz (x0 , y0 , z0 ) = 0

(3.19)

and the equations of the normal lines to the surface through P (x0 , y0 , z0 )are x − x0 y − y0 z − z0 = = Fx (x0 , y0 , z0 ) Fy (x0 , y0 , z0 ) Fz (x0 , y0 , z0 ) Illustration 3.24: Find equations for the (a) tangent plane and (b) normal line at the point P0 (1, 1, 1) on the surface x2 + y 2 + z 2 = 3. Solution: F (x, y, z) = x2 + y 2 + z 2 − 3 ∴ ∇F = 2xi + 2yj + 2zk ∴ ∇FatP 0 (1, 1, 1) = 2i + 1j + 2k

86

Vector Differential Calculus

∴ The equation of the tangent plane is 2 (x − 1) + 2 (y − 1) + 2 (z − 1) = 0 or 2x + 2y + 2z = 6 or x + y + z = 3 Normal line y−1 z−1 x−1 = = =t 2 2 2 or x = 2t + 1, y = 2t + 1, z = 2t + 1

3.7 Divergence and Curl of a Vector Field 3.7.1 Divergence of a Vector Field → − Let V (x, y, z) = V1ˆi+V2 ˆj +V3 kˆ be a differentiable vector function, then the → − → − → − divergence of V is denoted by div V or ∇ · V is the scalar function defined by

   → − → − ∂ ∂ ∂ ˆ ˆ ˆ div V = ∇ · V = i +j +k · V1ˆi + V2 ˆj + V3 kˆ ∂x ∂y ∂z =

∂V1 ∂V2 ∂V3 + + ∂x ∂y ∂z

→ → − ˆ then div − V = 3z + 2x − 2yz. For example, if V = 3xzˆi + 2xyˆj − yz 2 k, 3.7.1.1 Physical Interpretation of Divergence → − Consider the motion of a fluid with velocity V = Vxˆi + Vy ˆj + Vz kˆ at a point A (x, y, z). Consider a small parallelepiped having one of its corners at A (x, y, z) and edges parallel to the coordinate axes and having magnitude δx, δy, δz respectively. (See Figure 3.7) The mass of the fluid entering through the face ABCD per unit time is Vy δx δz and  that flowing out through the opposite face EFGH is ∂V

Vy + δy δx δz = Vy + ∂yy δy δx δz by using Taylor’s series. Thus, the rate at which fluidflows out from the elementary volume along ∂V ∂V the y-direction is Vy + ∂yy δy δx δz − V, δx δz = ∂yy δx δy δz. Similarly, the rate of outward flow along x and z directions will be given ∂Vz x by ∂V ∂x δx δy δz and ∂z δx δz respectively.  which fluid flows out of the volume per unit time is  Thus, the rate at ∂Vy ∂Vx ∂Vz δx δy δz. ∂x + ∂y + ∂z

3.7 Divergence and Curl of a Vector Field 87

Figure 3.7 Represents the parallelopiped

Dividing this by the volume δx, δy, δz of the parallelepiped, we have the rate of outflow (or rate of loss of fluid) per unit volume per unit time is → − ∂Vx ∂Vy ∂Vz + + = div V . ∂x ∂y ∂z → − Hence, the divergence of V gives the rate of outward flow per unit volume at a point of the fluid. From this, we have the following observations. → − → − (1) If there is no gain of the fluid anywhere, then div V = 0 i.e., ∇ · V = 0. This is called the continuity equation for an incompressible fluid or condition of incompressibility. (2) Since the fluid is neither created nor destroyed at any point, it is said to have no sources or sinks. (3) If the flux entering any element of the space is the same as the leaving it, → − i.e., if div V = 0 everywhere then such a vector point function is called a solenoidal vector function or solenoidal → − (4) If V represents an electric flux, div V is the amount of flux that diverges per unit volume. → − → − (5) If V represents heat flux, div V is the rate at which heat flows from a point per unit volume. → ˆ show that div r = 3. Illustration 3.25: If − r = xˆi + yˆj + z k,

88

Vector Differential Calculus

Solution: We know that

 

 − → → − ∂ ∂ ∂ ˆ ˆ ˆ +j +k · V1ˆi + V2 ˆj + V3 kˆ div V = ∇ · V = i ∂x ∂y ∂z ∂V1 ∂V2 ∂V3 = + + ∂x ∂y ∂z

→ → ∴ div − r = ∇·− r =

∂ ∂x

(x) +

∂ ∂y

(y) +

∂ ∂z

(z)

=1+1+1=3 → − → − Illustration 3.26: If F = x2 z ˆi − 2y 3 z 2 ˆj + xy 2 z kˆ then find div F at the point (1, −1, 1) . Solution: We know that

   → − → − ∂ ∂ ∂ div V = ∇ · V = ˆi + ˆj + kˆ · V1ˆi + V2 ˆj + V3 kˆ ∂x ∂y ∂z =

→ − → − ∴ div F = ∇· F =

∂ ∂x

∂V1 ∂V2 ∂V3 + + ∂x ∂y ∂z  2    ∂ x z + ∂y −2y 3 z 2 +

∂ ∂z



xy 2 z



= 2xz − 6y 2 z 2 + xy 2 at point (1, −1, 1) → − ∴ ∇· F (1,−1,1) = 2 (1) (1) − 6(−1)2 (1)2 + (1) (−1)2 = −3. Illustration 3.27: Show that the vector → − V = (x + 3y) ˆi + (y − 2z) ˆj + (x − 2z) kˆ is solenoidal. Solution: A vector V is solenoidal if its divergence is zero. → − ∂ ∂ ∂ (x + 3y) + (y − 2z) + (x − 2z) ∇· V = ∂x ∂y ∂z = 1 + 1 − 2 = 0. − → → − Since ∇ · V = 0 therefore vector V is solenoidal.

3.7 Divergence and Curl of a Vector Field 89

3.7.2 Curl of a Vector Field → − Let V = (x, y, z) = V1ˆi + V2 ˆj + V3 kˆ be a differentiable vector function, → − → − → − then the curl or rotation of V , written curl V or ∇ × V is the vector function defined by   ˆj  ˆi kˆ  → − →  ∂ − ∂ ∂  curl V = ∇ × V =  ∂x ∂y ∂z   V V V  1

= ˆi

∂V3 ∂V2 − ∂y ∂z

2



3



 ∂V1 ∂V3 ∂V2 ∂V1 ˆ ˆ +j − +k − ∂z ∂x ∂x ∂y

3.7.2.1 Physical Interpretation of Curl To understand the curl of a vector physically, consider the motion of a rigid → body rotating about a fixed axis passing through O. Let − ω = ω1ˆi+ω2 ˆj +ω3 kˆ → − be the angular velocity of the rigid body, V the linear velocity of any point → → ˆ then P (− r ) on the body, where − r = xˆi + yˆj + z k, − → − → V =→ ω ×− r

   → − − → → − ∴ V = ω × r =  

 ˆi ˆj kˆ  ω1 ω2 ω3  x y z 

ˆ 1 y − ω2 x) = ˆi (ω2 z − ω3 y) + ˆj (ω3 x − ω1 z) + k(ω Now,  ˆi ˆj  kˆ → − →  − ∂ ∂ ∂ curl V = ∇ × V =  ∂x ∂y ∂z  ω z−ω y ω x−ω z ω y−ω x 2 3 3 1 1 2 = (ω1 + ω1 ) ˆi + (ω2 + ω2 ) ˆj + (ω3 + ω3 ) kˆ   = 2 ω1ˆi + ω2 ˆj + ω3 kˆ

     

→ − → − 1 → → ∴ curl V = 2 − ω ⇒− ω = curl V 2 Thus, the angular velocity of rotation at any point is equal to half the curl of the linear velocity at that point of the body. This indicates that the curl of a vector field has something to do with the → − rotational properties of the field. If the field V is that due to a moving fluid,

90

Vector Differential Calculus

for example, then a paddlewheel placed at various points in the field would → − → − tend to rotate in regions where curl V = 0, while if curl V =0 in the region → − there would be no rotation and the field V is then called irrotational. A field that is not irrotational is sometimes called a vortex field. → → ˆ show that curl − Illustration 3.28: If − r = xˆi + yˆi + z k. r = 0.  ˆj  ˆi kˆ  ∂ → − → − ∂ ∂ curl r = ∇ × r =  ∂x ∂y ∂z  x y z



∂z ∂ ∂y = ˆi − + ˆj ∂y ∂z ∂z

Solution:

      ∂ x− z ∂x



+ kˆ

∂y ∂x − ∂x ∂y



= ˆi (0) + ˆj (0) + kˆ (0) = 0 → − → ˆ find curl − Illustration 3.29: If F = xz 3ˆi − 2x2 yzˆj + 2yz 4 k, F at the point (1, 1, 1). Solution:

  ˆj  ˆi kˆ   → − ∂ ∂ ∂  ∇ × F =  ∂x ∂y ∂z   xz 3 −2x2 yz 2yz 4  " ! " !  3     → ˆ ∂  − ∂ ∂  ∂ 4 2 4 2yz − −2x yz + ˆj xz − 2yz ∴∇ × F = i ∂y ∂z ∂z ∂x ! "  ∂  ∂  3 2 ˆ +k −2x yz − xz ∂z ∂y   = 2x2 + 2x2 y ˆi + 3xz 2 ˆj − 4xyz kˆ → − ˆ ∴ curl F at (1, −1, 1) = 3ˆj + 4k.

Illustration 3.30: Find constants a, b, c so that → − V = (x + 2y + az) ˆi + (bx − 3y − z) ˆj + (4x + cy + 2z) kˆ is irrotational. Solution:

 → − → − − →  i j k  → − → − ∂ ∂ ∂ curl V = ∇ × V =  ∂x ∂y ∂z  x + 2y + az bx − 3y − z 4x + cy + 2z = (c + 1) ˆi + (a − 4) ˆj + (b − 2) kˆ

     

3.7 Divergence and Curl of a Vector Field 91

This equals zero when a = 4, b = 2, c = −1 and − → V = (x + 2y + 4z) ˆi + (2x − 3y − z) ˆj + (4x − y + 2z) kˆ → − → − Illustration 3.31: If a vector F is irrotational then show that F = ∇φ = → − → − grad φ, where φ is a scalar point function of F or scalar potential of F . → − → − Solution: Let curl F = 0 where F = F1ˆi + F2 ˆj + F3 kˆ   ˆj  ˆi kˆ   ∂ ∂ ∂  ∴  ∂x ∂y =0 ∂z   F F F  1 2 3 i.e., 





ˆi ∂F3 − ∂F2 + ˆj ∂F1 − ∂F3 + kˆ ∂F2 − ∂F1 = 0 ∂y ∂z ∂z ∂x ∂x ∂y ∴

∂F1 ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 − = 0, − = 0, − =0 ∂y ∂z ∂z ∂x ∂x ∂y

i.e., ∂F3 ∂F2 ∂F1 ∂F3 ∂F2 ∂F1 = , = , = ∂y ∂z ∂z ∂x ∂x ∂y These three conditions are satisfied when F1 =

∂φ ∂φ ∂φ , F2 = , F3 = , ∂x ∂y ∂z

where φ is a function of x, y, z → − ∂φ ∂φ ˆ ∂φ ∴ F = ˆi + ˆj +k = ∇φ ∂x ∂y ∂z → − φ is known as the scalar point function of F . A vector field F which can be → − derived from a scalar field φ so that F = ∇φ is called a conservative vector field and φ is called the scalar potential. → − → − Note that conversely if F = ∇φ, then ∇ × F = 0. → − Illustration 3.32: Prove that F = (y 2 cosx+z 3 ) ˆi+(2y sinx−4)ˆj +3xz 2 kˆ is irrotational and find its scalar potential.

92

Vector Differential Calculus

Solution:  ˆi  →  − ∂ ∇ ×F = ∂x  y 2 cosx + z 3 on simplification Hence,

∴ (i)

ˆj ∂ ∂y

2ysinx − 4

   ∂  ∂z  = 0 2 3xz  kˆ

− → ∂φ ∂φ ˆ ∂φ F = ∇φ = ˆi + ˆj +k ∂x ∂y ∂z

∂φ ∂φ ∂φ = y 2 cosx + z 3 , (ii) = 2ysinx − 4 , (iii) = 3xz 2 ∂x ∂y ∂z

∴ from (i), φ = y 2 sinx + xz 3 + f1 (y, z) . from (ii), φ = y 2 sinx − 4y + f2 (z, x) . from (iii), φ = xz 3 + f3 (x, y). ∴ φ = y 2 sinx + xz 3 − 4y + c , where c is an arbitrary constant. Illustration 3.33: A fluid motion is given by   − → ˆ V = (ysinz − sinx ) ˆi + (xsinz + 2yz) ˆj + xycosz + y 2 k. is motion irrotational? If so, find the velocity potential. Solution:   ˆ ˆi ˆj   k  →  − ∂ ∂ ∂  Curl V =   ∂x ∂y ∂z  ysinz − sinx xsinz + 2yz xycosz + y 2  = ˆi [x cosz + 2y − x cosz − 2y] + ˆj [ y cosz − y cosz] + kˆ [sinz − sinz] = ˆi (0) + ˆj (0) + kˆ (0) = 0 → − Since curl V = 0, the motion is irrotational. → − ˆ ∂φ ˆ ∂φ Hence V = ∇φ = ˆi ∂φ ∂x + j ∂y + k ∂z ∂φ ∂x ∂φ ∂y ∂φ ∂z

= y sinz − sinx (i) = y sinz + 2yz (ii)

= xy cosz + y 2 (iii) From (i), φ = xy sinz + cosx + f1 (y, z)

3.7 Divergence and Curl of a Vector Field 93

From (ii), φ = xy sinz + y 2 z + f2 (z, x) From (iii), φ = xy sinz + zy 2 + f3 (x, y) ∴ φ = xy sinz +y 2 z +cosx+c, where c an arbitrary constant is required velocity potential. → − → ˆ and ˆ − Illustration 3.34: If A = 2yzˆi − x2 yˆj + xz 2 k, B = x2ˆi + yzˆj − xy k, 2 3 φ = 2x yz , find −  → (i) A · ∇ φ → − (ii) A · ∇φ and compare them. −  → (iii) A × ∇ φ → − (iv) A × ∇φ and compare them. Solution: (i) "  ∂ ∂ ∂ ˆ + ˆj +k φ (A · ∇) φ = 2yzˆi − x yˆj + xz k · ˆi ∂x ∂y ∂z

  2 3 ∂ 2 ∂ 2 ∂ = 2yz − x y + xz 2x yz ∂x ∂y ∂z ∂  2 3 ∂  2 3 ∂  2 3 = 2yz 2x yz − x2 y 2x yz + xz 2 2x yz ∂x ∂y ∂z         = (2yz) 4xyz 3 − x2 y 2x2 z 3 + xz 2 6x2 yz 2 !

2

2ˆ

= 8xy 2 z 4 − 2x4 yz 3 + 6x3 yz 4 (ii)    ∂φ − → ∂φ ∂φ 2 2 ˆi + ˆj + A · ∇φ = 2yzˆi − x yˆj + xz kˆ · kˆ ∂x ∂y ∂z     = 2yzˆi − x2 yˆj + xz 2 kˆ · 4xyz 3ˆi + 2x2 z 3 ˆj + 6x2 yz 2 kˆ = 8xy 2 z 4 − 2x4 yz 3 + 6x3 yz 4 −  → Comparison with (i) illustrates the result A · ∇ φ = A · ∇φ.

94

Vector Differential Calculus

(iii) " !  ∂ − → ∂ ∂ 2 2 ˆ ˆ + ˆj +k φ ( A × ∇)φ = 2yzˆi − x yˆj + xz k × ˆi ∂x ∂y ∂z   ˆj  ˆi kˆ   =  2yz −x2 y xz 2  φ ∂ ∂   ∂ ∂x ∂y ∂z $



 ∂ ∂ ∂ ∂ 2 2 2 − xz + ˆj xz − 2yz = ˆi −x y ∂z ∂y ∂x ∂z

% ∂ ∂ + kˆ 2yz + x2 y φ ∂y ∂x



 ∂φ ˆ 2 ∂φ 2 ∂φ 2 ∂φ ˆ =− x y + xz i + xz − 2yz j ∂z ∂y ∂x ∂z

 ∂φ ∂φ ˆ + 2yz − x2 y k ∂y ∂x −      → A × ∇ φ = − 6x4 y 2 z 2 + 2x3 z 5 ˆi + 4x2 yz 5 ˆi     + 4x2 yz 5 − 12x2 y 2 z 3 ˆj + 4x2 yz 4 + 4x3 y 2 z 3 kˆ (iv)   ∂φ   − → ∂φ ˆ ∂φ + ˆj +k A × ∇ φ = 2yzˆi − x2 yˆj + xz 2 kˆ × ˆi ∂x ∂y ∂z   ˆj  ˆi kˆ   2  =  2yz −x y xz 2  ∂φ ∂φ   ∂φ ∂x ∂y ∂z 



∂φ ˆ 2 ∂φ 2 ∂φ 2 ∂φ ˆ − xz i + xz − 2yz j = −x y ∂z ∂y ∂x ∂z

 ∂φ ∂φ ˆ + 2yz + x2 y k ∂y ∂x     = − 6x4 y 2 z 2 + 2x3 z 5 ˆi + 4x2 yz 5 − 12x2 y 2 z 3 ˆj   + 4x2 y 3 z 4 + 4x3 y 2 z 3 kˆ −  → → − Comparison with (iii) illustrates the result A × ∇ φ = A × ∇φ.

3.7 Divergence and Curl of a Vector Field 95

Illustration 3.35: Find the scalar potential function f for → − A = y 2ˆi + 2xyˆj − z 2 kˆ .   ˆi → − →  ∂ − Curl A = ∇ × A =  ∂x  y2

Solution:

   ∂ ∂  ∂y ∂z  2xy −z 2  ˆj

kˆ

= ˆi (0) + ˆj (0) + kˆ (2y − 2y) = 0 → − → − → − Since and A = 0, A = ∇φ, where φ is a scalar potential function of A   → − ∂φ ˆ ∂φ ˆ ∂φ A = ∇φ =⇒ y 2ˆi + 2xyˆj − z 2 ˆj = ˆi +j +k ∂x ∂y ∂z ∂φ ∂x ∂φ ∂y ∂φ ∂z

= y 2 (i) = 2xy (ii)

= −z 2 (iii) from (i) φ = xy 2 + f1 (y, z) from (ii) φ = 2xy 2/2 + f2 (z, x) = xy 2 + f2 (y, z) from (iii) φ = −23 /3 + f3 (x, y) ∴ φ = xy 2 − z 3 /3 + C. where C is an arbitrary constant. → Illustration 3.36: Prove that rn − r is irrotational. Solution: → → → To prove rn − r irrotational, find ∇ × (rn − r ) i.e., curl (rn − r ). / → → ˆi × ∂ (rn , − r) r)= curl (rn − ∂x 0 1 / ˆi × nrn−1 rˆ x + rn ˆi =  r  / → n−2 = {nr × ˆi × − r + rn (ˆi × ˆi)}   / = n rn−2 x y kˆ − zˆj + 0 0      1 = n rn−2 x y kˆ − zˆj + y zˆi − xkˆ + z xˆj − yˆi =0 − ∴ rn → r is irrotational.

96

Vector Differential Calculus

  → − Illustration 3.37: Show that F = 2xy zˆi + x2 z + 2y ˆj + x2 y kˆ is → − irrotational and find a scalar function φ such that F = grad ∅ . Solution:

  ˆj  ˆi kˆ   → − ∂ ∂  ∂ ∇ × F =  ∂x ∂y ∂z   2xyz x2 z + 2y x2 y    = ˆi x2 − x2 + ˆj (2xy − 2xy) + kˆ (2xz − 2xz) = 0

→ − ∴ F is irrotational ∂φ ∂φ 2 2 F = grad φ =⇒ (i) ∂φ ∂x = 2xyz, (ii) ∂y = x z + 2y, (iii) ∂z = x y Integrating (i), (ii) and (iii) w.r.t. x, y, and z respectively, we get φ = x2 yz + f1 (y, z) φ = x2 yz + y 2 + f (z, x) φ = x2 yz + y 2 + f3 (x, y) ∴ φ = x2 yz + y 2 + c 3.7.3 Formulae for grad, div, curl Involving Operator ∇ 3.7.3.1 Formulae for grad, div, curl Involving Operator ∇ Once → → 1. If − u and − v are vector point functions and f , a scalar point function. → → → → (i) div (− u +− v ) = div − u + div − v → − → − → − → or ∇ · ( u + v ) = ∇ · u + ∇ · − v → − → − → (ii) div (f u ) = f (div u ) + (grad f )· − u → → or f (∇ · − u ) + (∇f ) · − u. 2.

− → → (i) ∇ × (f → u ) = (∇f ) × − u + f (∇ × − u) → − → − → u or curl (f u ) = (grad f ) × u + f curl − → − → − → − → − → − − − (ii) ∇ × ( u × v ) = (∇ × v ) u − (∇ · − u) → v + (→ v · ∇) → u − → − → − ( u · ∇) v

3. If u and v are scalar point functions (i) grad (u + v) = grad u+ grad v (ii) grad (uv) = u (grad v ) + (grad u ) v The proofs for (3) are very simple so left to the reader.

3.7 Divergence and Curl of a Vector Field 97

Proof: 1.

 → → ∂ + ˆj (i) ∇ · (− u +− v ) = ˆi ∂x

∂ ∂y

 → → ∂ + kˆ ∂z · (− u +− v)

/

→ → ˆi ∂ · (− u +− v) ∂x / ∂ / ∂ → → ˆi ˆi = · − u + · − v ∂x ∂x

 ∂ ∂ ∂ = ˆi + ˆj + kˆ ·u ∂x ∂y ∂z

 ∂ ∂ ∂ → ˆ + ˆi + ˆj + k ·− v ∂x ∂y ∂z → → =∇·− u +∇·− v →∂ − → → → → ∂ ∂ (ii) ∇ · (f − u ) = ˆi ∂x · (f − u ) + ˆj ∂y · (f − u ) + k ∂z · (f − u) =

∂ ∂ → → − ˆ ∂ (f → (f − u )+ˆj· (f − u )+ k· u) ∂x ∂y ∂z "

− / ! ∂f  ∂→ u → − ˆ = i u +f ∂x ∂x → − / ∂f / → ˆi ˆi · ∂ u = · − u +f ∂x ∂x → − → − = (∇f ) · u + f (∇ · u )   → − ∂ ∂ ∂ × f− u + ˆj ∂y + kˆ ∂z (i) ∇ × (f → u ) = ˆi ∂x = ˆi·

2.

/

→ ˆi ∂ × f − u ∂x / → ˆi × ∂ (f − = u) ∂x

 → − / ∂f ∂ u → − ˆi × = u +f ∂x ∂x → − / / → ˆi × ∂f − ˆi × f ∂ u = u + ∂x

/ 

/ ∂x−  ∂f ∂→ u → − ˆ ˆ = i × u +f i ∂x ∂x → − → − = (∇f ) × u + f (∇ × u )

=

98

Vector Differential Calculus

2 ∂ − → → → (ii) ∇ × (→ u ×− v ) = ˆi ∂x × (− u ×− v) /

→ → ˆi × ∂ (− u ×− v) ∂x

 → − → − / ∂ u ∂ v → − → − ˆi × = × v + u × ∂x ∂x

−  /

 → → / ∂u ∂− v → − → − ˆ ˆ = i × × v + i × u × ∂x ∂x

=

Now /

 / 3

 4

→  ∂− → → u ∂− u ∂− u − → − → − → ˆ ˆ × v = i· v − i· v ∂x ∂x ∂x −    → → → → − − → → − → → ∵− a × b ×− c = b (− a ·→ c)− − a · b − c   ∂−   ∂−   ∂−  ∂→ → → → − / u u u u → → → → ˆi · − = ˆi · − v + ˆj · − v + kˆ · − v v ∂x ∂x ∂y ∂z ∂u ∂u ∂u + v2 + v3 = v1 ∂x ∂y ∂z ˆi ×

→ where − v = v1ˆi + v2 ˆj + v3 kˆ

 ∂ ∂ ∂ → − = v1 + v2 + v3 u ∂x ∂y ∂u 3 4   ∂  ∂− → / u ∂ ∂ → − → − ˆ ˆ ˆi · v = v1ˆi + v2 ˆj + v3 k · ˆi + ˆj +k u ∂x ∂x ∂y ∂z → → = (− v · ∇) − u  

/ → − → / ∂− u − − → ˆi · ∂ u → ˆ v = i· v ∂x ∂x 

/ ∂ − → → ˆ · u − = i v ∂x  4 3

∂ ∂ ∂ → − → ˆ ˆ ˆ + j +k · u − = i v ∂x ∂y ∂z → → = (∇ · − u) − v 

→ / ∂− u → − → → → → ˆ × v = (− v · ∇) − u − (∇ · − u) − v ∴ i × ∂x

3.7 Divergence and Curl of a Vector Field 99

Also



− 

− / ∂→ v ∂→ v − → − ˆ ˆi × → =− × u i × u × ∂x ∂x − → → → = − (→ u · ∇) − v + (∇ · − v) − u

/

(Similar to the previous result) Hence → → → → → → → → → → ∇ × (− u ×− v ) = (− v · ∇) − u − (∇ · − u) − v − (− u · ∇) − v + (∇ · − v) − u → − → − → − → − → − → − → − → − = (∇ · v ) u − (∇ · u ) v + ( v · ∇) u − ( u · ∇) v → − → Illustration 3.38: Prove that V = rn − r is irrotational. Find n when it is also solenoidal. Solution:

→ − → ∇ × V = ∇ × (rn − r) → → r + rn (∇ × − r) = ∇ (rn ) × − −  → → → → using ∇ × (φ − u) = V φ ×− u + φ (∇ × − u) → − / / ∂ ˆi nrn−1 ∂ r ˆi ∇ (rn ) = rn = ∂x ∂x x / n−1 2 2 ˆi nr = since r = x + y 2 + z 2 r / ˆi rn−2 x =n   /   ˆi x = nrn−2 xˆi + yˆi + z kˆ ∴ ∇ r2 = nrn−2 → = nrn−2 − r → Since ∇ × − r =0 → − → → r ×− r + rn (0) Hence ∇ × V = nrn−2 − → → = 0, since − r ×− r =0

Hence the vector is irrotational → − → r) ∇ · V = ∇ · (rn − → → r ) + ∇ (rn ) ·− r = rn (∇ · −

100

Vector Differential Calculus

→ → → using ∇ · (φ − u ) = φ (∇ · − u ) + ∇φ · − u → − → − n n−2 since ∇ · r = 3 and ∇r = nr r Hence → − → → ∇ · V = 3rn + nrn−2 − r ·− r = 3rn + nrn−2 r2 = (n + 3) rn → − → − If the vector V is solenoidal ∇ · V = 0 ∴ (n + 3) rn = 0, i.e., n = −3. 3.7.3.2 Formulae for grad, div, curl Involving Operator ∇ Twice → − Given a scalar function f and a vector function V , the following combinations of ∇ twice are possible: (i) div (grad f) = ∇ · ∇f =

∂2f ∂x2

+

∂2f ∂y 2

+

∂2f ∂z 2

→ − → − (ii) (∇ · ∇) V =∇2 V  → − → − (iii) grad (div V ) = ∇ ∇ · V (iv) curl (grad f ) = ∇ × ∇f = 0  → − → − (v) div curl V = ∇ · ∇ × V = 0   → − → − − → → − (vi) curl curl V = ∇ × ∇ × V = ∇ ∇ · V − ∇2 V . Proof:  ∂ (i) ∇ · ∇φ = ˆi ∂x + ˆj

∂ ∂y

   ∂ ˆj ∂f + kˆ ∂f + kˆ ∂z + · ˆi ∂f ∂x ∂y ∂z

∴ ∇ · ∇φ = ∇2 φ = 2

2

2

∂2f ∂2f ∂2f + + ∂x2 ∂y 2 ∂z 2

∂ ∂ ∂ The operator ∇2 = ∂x 2 + ∂y2 + ∂z2 is called Laplace’s operator or Laplacian and ∇2 f = 0 is called Laplace’s equation. which frequently occurs in physical and engineering problems.

3.7 Divergence and Curl of a Vector Field 101 → → → → 2− − → − 2− 2− (ii) (∇ · ∇) V = ∇2 V = ∂∂xV2 + ∂∂yV2 + ∂∂zV2   ∂V2 ∂V3 1 (iii) ∇ (∇ · ∇) = ∇ ∂V + + = which is a vector ∂x ∂y ∂z

  ˆi   ∂ (iv) curl grad f = ∇ × ∇f =  ∂x  ∂f  ∂x

ˆj ∂ ∂y ∂f ∂y

 kˆ  ∂  ∂z  = 0 on simplification ∂f  ∂z

→ − (v) div curl V = ∇ · (∇ × V )   ˆ  ˆj  ˆi k  → − →  ∂ − ∂ ∂  , where V = V1ˆi + V2 ˆj + V3 kˆ ∇ × V =  ∂x ∂y ∂z   V V V  1 2 3 





→ ˆ ∂V3 ∂V2 − ∂V ∂V ∂V ∂V 1 3 2 1 ˆ − + ˆj − +k − ∴ ∇ ×V = i ∂y ∂z ∂z ∂x ∂x ∂y  → − → − ∴ div curl V = ∇ · ∇ × V





 ∂ ∂V3 ∂V2 ∂ ∂V1 ∂V3 ∂ ∂V2 ∂V1 = − + − + − ∂x ∂y ∂z ∂y ∂z ∂x ∂z ∂x ∂y

2  2  2  2 2 2 ∂ V3 ∂ V1 ∂ V2 ∂ V2 ∂ V3 ∂ V1 = − + − + − ∂x∂y ∂x∂y ∂y∂z ∂y∂x ∂z∂x ∂z∂y =0 → (vi) We have − a ×

−  −  → − → → − → → − → b ×→ c = b (− a ·→ c)− − a · b − c

Treating ∇ as a vector if we take → − → − − → → a = ∇ , b = ∇, and − c = V , we have  → − ∇ × ∇× V =∇ (∇· ∇) − (∇· ∇) V ∴

∇ × (∇ × V ) = ∇

→ − → − → − i.e., curl curl V = grad div V − ∇2 V



→ − − → ∇ · V − ∇2 V

102

Vector Differential Calculus

→ − → − Illustration 3.39: If ρ E = ∇φ, where ρ, φ are scalar fields and E is a vector → − → − field, prove that E · curl E = 0. Solution: → − 1 E = ∇φ ρ 

→ − → − 1 ∇φ curl E = ∇ × E = ∇ × ρ

 1 1 =∇ × ∇φ + ∇ × (∇φ) ρ ρ → − → → [∵ ∇ × (f u ) =∇f ×− u +f (∇×− u)

 1 × ∇φ + 0 [∵ ∇ × ∇f = 0 =∇ ρ  → − → − − →    ∴ E curl E = E × ∇ ρ1 × ∇φ

  → − 1 =∇ · ∇φ × E ρ − → → − → → − → [∵ − a · b ×− c = b · (− c ×→ a)

   → − − → 1 =∇ · ρE×E ρ → − − → = 0 since E × E = 0 Illustration 3.40: Prove that ∇2 f (r) = f  (r) +

2 r

f  (r).

Solution: ∇2 f (r) = ∇ · {∇ f (r)} =div {grad f (r)}

2  ∂ ∂2 ∂2 = f (r) + + ∂x2 ∂y 2 ∂z 2 Now

∂ ∂x

∂ ∂x ∂2

f (r) =

f (r) ·

∂r ∂x

= f  (r) ·

x r

since r2 = x2 + y 2 + z 2

. ∂ #  f (r) · x · r−1 ∂x x  −1 = f (r) r · 1 + f  (r) · · x r−1 − f  (r) x2 r−3 r x2 f  (r) x2 f  (r) f  (r) + = − r r2 r3

∴

∂x2

f (r) =

3.7 Divergence and Curl of a Vector Field 103

Similarly, ∂2 f  (r) y 2 f  (r) y 2 f  (r) + f (r) = − ∂y 2 r r2 r3 and

∂2 f  (r) z 2 f  (r) z 2 f  (r) + f (r) = − ∂z 2 r r2 r3

Adding, we have 

2 ∂ 3f  (r) ∂2 ∂2 f  (r)  + f f (r) = + + (r) − ∂x2 ∂y 2 ∂z 2 r r = ∴ ∇2 f (r) = div grad f (r) =

2f  (r) r

2f  (r) + f  (r) r

+ f  (r).

Illustration 3.41: Maxwell’s equations of the electromagnetic theory are − → → − − → → − → − → ∂E − ∂H ∇ · E = 0, ∇ · H = 0, ∇ × E = − , ∇×H = ∂t ∂t → − → − Show that E and H satisfy wave equations → → → 2− − → 2− − (i) ∇2 E = ∂∂tE and (ii) ∇2 H = ∂∂tH 2 2

Solution: 

− → ∇× ∇× E =∇×



→ − → − ∂  ∂H =− ∇ ×H − ∂t ∂t

∂ =− ∂t But



∂E ∂t

 =−

∂2E ∂t2

  → − → − → − → − ∇ × ∇ × E = ∇ ∇ · E − ∇2 E = −∇2 E

Then from (i) are (ii), we have → − ∂2 E ∇ E = ∂t2 → 2−

(i)

(ii)

104

Vector Differential Calculus

Now

but

→ − → − ∂E ∂  ∇× E = ∂t ∂t −  → − → ∂ ∂H ∂2 E = − =− 2 (iii) ∂t ∂t ∂t



− → ∇× ∇×H =∇×

  → − → − → − → − ∇ × ∇ × H = ∇ ∇ · H − ∇2 H = −∇2 H

(iv)

from (iii) and (iv), we have → − ∂2 H ∇ H= 2 ∂t → 2−

2

The equation ∇2 u = ∂∂t2u 2 2 2 i.e., ∂∂xu2 + ∂∂yu2 + ∂∂zu2 =

∂2u ∂t2

is called the wave equation.

3.8 Exercise 1. Determine k and τ for the following curve: (i) x = t cos t , y = t sin t , z =λt at t = 0 Answer: k = (ii) x = 3t, y = 3t2 ,z = 2t3 Answer: = σ =

3 2

2 , 1+λ2

τ=

3λ 2(1+λ2 )



 2 1 + 2t2 , = k1 , σ = τ1



  (iii) x = a 3t − t3 , y = 3 at2 , z = a(3t + t3 )  2   Answer: = σ = 3a 1 + t2 2. For the curve x = 3t, y = 3t2 , z = 2t2 at t = 1 Prove that = σ =   3 2 2. 2 1 + 2t 3. For the curve x = 2 log t, y = 4t, z = 2t2 + 1 at t = 1 prove that = σ = 9. → ˆ 4. Find curvature and torsion for the curve − r = cos t ˆi + sin t ˆj + tk. 2 2 Also, prove that 2 κ + τ = 1.   Answer: k = 12 , τ = 12

3.8 Exercise

105

5. Find curvature and torsion for the curve x = t cos t , y = t sin t , z = λt at t = 0.   2 3λ Answer: k = 1+λ 2 , τ = 2(1+λ2 ) 6. For the curve x = a cos θ , y = a sin θ , z = a θtan α , find ρ.   Answer:ρ = asec2 α 7. For the curve x = a cos θ , y = a sin θ , z = λθ, find k and τ .   a λ Answer: k = a2 +λ 2 , τ = a2 +λ2 8. Show that for the curve 2x2 z , = σ= . c c   log s2 + 1 , z = s − tan−1 s

x2 − y 2 = c2 , y = xtanh 9. For the curve x = tan−1 s , y = √

show that k = τ =

√1 2

2 . s2 +1

    10. For the curve x = a 3u − u3 , y = 3au2 , z = a 3u + u3 show that k = τ =

1 3a(1+u2 )2

.

11. For the curve x = 4a cos3 u , y = 4a sin3 u , z = 3c cos2u prove that k =

a 6(a2 +c2 )sin2u

.

12. Find the length of the curve x = et cost , y = et sint , z = et between t = t1 and t = t2 . √    Answer: 3 et2 − et1 13. Find the length of the arc of the curve x = 3t, y = 3t2 , z = 2t3 between t = 0 and t = 1. (Answer: 5) 3a 3 3 2 14. Find the lengthof the curve  x = a cos t, y = a sin t , z = 2 cos t 3a from the point a, 0, 2 to the point (0, a, 0).   3a Answer: √ 2

106

Vector Differential Calculus

→ 15. Find the length of the curves − r (t) = 2tˆi + 3sin2t ˆj + 3cos2t kˆ on the interval 0 ≤ t ≤ 2π. √   Answer: 2 10t 16. Find the magnitude of the velocity and acceleration of a particle which moves along the curve x = 2 sin3t , y = 2 cos3t , z = 8t at any time t > 0. Find the unit tangent vector to the curve.    1 Answer: 10, 18, 10 (6 cos3t ) ˆi − (6 sin3t ) ˆj + 8kˆ 17. A particle moves along a plane curve such that its linear velocity is perpendicular to the radius vector. Show that the path of the particle is a circle. 18. Find the magnitude of the tangential components of acceleration at any time t of a particle whose position at any time t is given by x = cost + tsint , y= sint − tcost . (Answer: 1) 19. Show that the length of the √ curve 2x = a (cos 3θ + cos θ ) , 2y = a (sin 3θ + sin θ ) z = 3 a cos θ measured from the point √   a, 0, 3 a in 2aθ.   d d2 V d3 V 20. Prove that dt × V. dV = V . dV 2 dt dt × dt3 . dt 21. Show that r = e−1 (a cos2t + b sin 2t) , where a and b are constant 2 vectors, is a solution to the differential equation ddt2r + 2 dr dt + 5r = 0. 22. What is the greatest rate of increase of u = x2 + yz 2 at the point (1, −1, 3)? √   Answer: 89 23. Find grad φ at the point (1, −2, 1) when φ is given by φ = 3x2 y − y 3 z 2 .   Answer: − 12ˆi − 9ˆj − 16kˆ → r · ∇φ = 3φ. 24. If φ = x3 + y 3 + z 3 − 3xyz, show that − 25. If u = x + y + z, v = x2 + y 2 + z 2 , w = yz + zx + xy then a. Prove that (∇ u) · [∇v × ∇w] = 0.

3.8 Exercise

107

b. Show that ∇u, ∇v, ∇w are coplanar.     ˆ determine φ. 26. If ∇φ = (2xyz) ˆi + x2 z ˆj + x2 y k,   Answer: φ = x2 yz + c  − → − → − → 27. If A is a constant vector then prove that ∇ r · A = A . 28.  Find a unit vector normal to the surface x2 + y 2 + z 2 = a2 at √a , √a , √a . 3 3 3   ˆ ˆ ˆ k Answer: i+√j+ 3 29. Find a unit vector normal to the surface x2 + y 2 − z = 1 at the point (1, 1, 1).   ˆ ˆ ˆ Answer: 2i−23j−k   30. Find a unit outward drawn normal to the surface x2 − 1 + y 2 + (z + 2)2 = 9 at the point (3, 1, −4).   ˆ ˆ ˆ k Answer: 2i+j−2 3 31. The temperature at a point (x, y, z) in space is given by T (x, y, z) = x2 + y 2 − z. A mosquito located at (1, 1, 2) desires to fly in such a direction that it will get warm as soon as possible. In what direction should it fly?    Answer: 13 2ˆi + 2ˆj − kˆ 32. What is the angle between the normals to the surface xy = z 2 at the points (1, 9, −3) and (−2, −2, 2)?    Answer: cos−1 √11 118 33. Show that ∇φ is a vector perpendicular to the surface φ (x, y, z) = 0. 34. Find the directional derivative of φ (x, y, z) = xy 2 + yz 3 at the point ˆ (2, −1, 1) in the direction of the vector ˆi + 2ˆj + 2k.   Answer: − 11 3

108

Vector Differential Calculus

35. Find the directional derivative of φ (x, y, z) = xy 2 + yz 2 at the point ˆ (2, −1, 1) in the direction of the vector ˆi + 2ˆj + 2k. (Answer: − 3) 36. If the directional derivative of φ = axy + byz + czx at (1, 1, 1) has maximum magnitude 4 in a direction parallel to the x-axis, find the values of a, b, c. (Answer: a = 2, b = −2, c = 2 ) 37. Prove that the angle between the surface x2 + y 2 + z 2 = 9 and   x2 + y 2 − z = 3 at the point (2, −1, 2) is cos−1 3√821 . 38. Find the angle between the tangent planes to the surfaces xlogz = y 2 − 1 and x2 y = 2 − z at the point (1, 1, 1).    Answer: cos−1 √130 39. In what direction form (3, 1, −2) is the directional derivative of φ = x2 y 2 z 4 maximum, and what is its magnitude?     1 Answer: 13 4ˆi + 3ˆj − 12kˆ , 1 40. Is there a direction u in which the rate of change of f (x, y) = x2 − 3xy + 4y 2 at P (1, 2) equals 14? Give reasons for your answer. √   Answer : No, the maximum rate of change is 185 < 14   → → ˆ − v = x2ˆi + 2yzˆj + 41. Given − u = xyz ˆi + 2x2 z − y 2 x ˆj + xz 3 k, → ˆ f = xy + yz + z 2 then find (i) ∇ · − v (ii) ∇ × v (iii) (1 + 2z) k, → → ∇ · (f − u ) (iv) ∇ × (f − v ) at (1, 0, −1)   Answer: (i) 3, (ii) 0, (iii) 5, (iv) − 2ˆj → → → → 42. If ω is a constant vector and − v =− ω ×− r prove that div − v = 0. → − 43. Prove that F =

xˆi+yˆ j x2 +y 2

is solenoidal.

 →  − ˆ 44. If F = y 2 − z 2 + 3yz − 2x ˆi+(3xz + 2xy) ˆj+(3xy − 2xz + 2z) k, → − Show that F is both solenoidal and irrotational.

3.8 Exercise

109

 →  − ˆ Show that 45. If F = z 2 + 2x + 3y ˆi + (3x + 2y + z) ˆj + (y + 2xz) k, F is irrotational but not solenoidal.      →  − 46. Show that F = 6xy + z 3 ˆi + 3x2 − z ˆj + 3xz 2 − y kˆ → − is irrotational. Find scalar φ such that F = ∇φ. → 47. In each case the velocity vector − v of a steady fluid motion is given, Find → − curl v . Is the motion incompressible? → → → (i) − v = z 2 ˆj , (ii) − v = yˆi − xˆj , (iii) − v = xˆi + yˆj ⎞ ⎛ → Answer: (i) curl − v = −2zˆj, incompressible → ⎠ ⎝ ˆ incompressible (ii) curl − v = −2k, → − → − (iii) curl v = 0, div v = 2, compressible → − → − 48. If the vector product of the vectors A and B be the curl of a third vector, → − → − → − prove that A · curl = B · curl A . → − → − 49. Prove that div (f curl F ) = (grad f ) · curl F . 50. If φ, ψ satisfy Laplace equation, prove that the vector (φ∇ψ − ψ∇φ) is solenoidal. 51. Show that #  . (i.) ∇2 ∇. rr2 = r24     (ii.) ∇ × a × ∇ 1r + ∇ a . ∇ 1r = 0. 52. If

→ d− u dt

→ → =− w ×− u and

→ d− v dt

→ → =− w ×− v , then prove that

d − → → → → (→ u ×− v)=− w × (− u ×− v ). dt  → 53. If − r = t3 ˆi + 2t3 −

1 5t2



ˆj, then show that rˆ ×

→ d− r dt

ˆ = k.

4 Vector Integral Calculus

4.1 Introduction Vector calculus deals with the differentiation and integration of vector functions. We have learned about the derivative of a vector function, gradient, divergence, and curl in vector differential calculus. In vector integral calculus, we learn about line integral, surface integral, and volume integral. It plays an important role in differential geometry and the study of partial differential equations. It is useful in the study of rigid dynamics, fluid dynamics, heat transfer, electromagnetism, theory of relativity, etc.

4.2 Line Integrals b

The line integral is a simple generalization of a definite integral a f (x)dx which is integrated from x = a (point A) to x = b (point B) along the x-axis. In a line integral, the integration is done along a curve C in space. →→ − Let F (− r ) be a vector function defined at every point of a curve C in → Figure 4.1. If − r is the position vector of a point P (x, y, z) on the curve C, →→ − then the line integral of F (− r ) over a curve C is defined by

Figure 4.1

Representation of a vector function defined at every point of a curve C

111

112

Vector Integral Calculus



− − → → F (→ r ) · d− r = C

 C

F1 dx + F2 dy + F3 dz

→ − → where, F = F1 ˆi + F2 ˆj + F3 kˆ and − r = xˆi + yˆj + kˆ If the curve C is represented by a parametric representation → − ˆ r (t) = xˆi + yˆj + z k, then the line integral along the curve C from t = a to t = b is  b  → → − − → d− − r → → − F (r)·dr = F · dt dt a C   b

dy dz dx = + F2 + F3 dt F1 dt dt dt a 9If C is a closed curve, then the symbol of the line integral C. → − → r and Two other types of line integrals are C F × d− is a scalar point function) which are both vectors.

C

is replaced by

C

→ φd− r , (where φ

Note: → → 1. The curve C is called the path of integration, the points − r (a) and − r (b) are called initial and terminal points respectively. 2. The direction from A and B along which t increases is called positive direction on C. 4.2.1 Circulation → − If F is the velocity of a fluid particle and C is a closed curve, then the line 9 − → → → − integral C F · d− r represents the circulation of F around the curve C. Note:

→ − 1. If the circulation of F around every closed curve C in the region R is 9B− → → → − → − r = 0, F is irrotational. zero, then F is irrotational, i.e., if A F · d−

4.2.2 Work Done by a Force → − If F is the force acting on a particle moving along the arc AB of the curve → → B− r represents the work done in displacing C, then the line integral A F · d− the particle from the point A to the point B.

4.3 Path Independence of Line Integrals

113

4.3 Path Independence of Line Integrals 4.3.1 Theorem: Independent of Path

→ → B− r be The necessary and sufficient condition that the line integral A F · d− → − independent of the path is that F is the gradient of some scalar function φ. → − Proof: Let F = F1ˆi + F2 ˆj + F3 kˆ be the gradient of a scalar function φ (i.e., → − → − F is conservative) i.e., F = ∇φ in components F1 =

∂φ ∂φ ∂φ , F2 = , F3 = ∂x ∂y ∂z

where φ is a scalar potential, then the line integral along the curve C from the point A to B is   B → − − → → F ·dr = ∇φ · d− r A C 

 B ∂φ ∂φ ∂φ = dx + dy + dz ∂x ∂y ∂z A  B = dφ A

= φ (B) − φ(A) where φ (A) and φ(B) are the values of φ at A and B respectively, That means the line integral depends only on the start and end values of the scalar potential, not on the path of the curve. Hence the condition is necessary. We show below that the condition is sufficient. → → B− Let A F · d− r depend on the end values A and B and not on the path of integration.  B → − − ∴ F · d→ r =φ (B) − φ (A) = [φ]B A

A

→ − − ∂φ ∂φ ∂φ → dx + dy + dz = ∇φ · d− r ∴ F · d→ r = dφ = ∂x ∂y ∂z → − Since this is true for all the curves between A and B, F = ∇φ. → − In Mechanics, F is called a conservative field and φ is the scalar potential, → − → − and in Potential theory, if φ is potential and F is a force then F = grad φ. → − Thus, the line integral is independent of the path in C iff F is the gradient of potential in C.

114

Vector Integral Calculus

→ − Corollary 4.3.1.1: If F = ∇φ, we have → − curl F = curl grad φ = ∇ × ∇φ = 0 → → − r be independent Hence, the necessary and sufficient condition that C F · d− → − → − of the path is the curl F vanishes identically and hence F is irrotational. → → − r is independent of the path of integration, Corollary 4.3.1.2: If C F · d− 9 − → − → F · d r along any closed path is zero. then C

→ − Corollary 4.3.1.3: Let F = ˆiP + ˆjQ, ∂Q its path, if ∂P ∂y = ∂x .

C

(P dx + Qdy) is independent of

Note: 1. As shown in Figure 4.2, → − if F is conservative and curve C is closed, then : → − − F · d→ r = φ (A) − φ (A) = 0. C

2. The work done in moving a particle from points A to B under a conservative force field is Work done = φ (B) − φ(A) .

Figure 4.2 Representation of a closed curve C

→ → − r along the parabola y 2 = x between the Illustration 4.1: Evaluate C F · d− → − points (0, 0) and (1, 1) where F = x2ˆi + xyˆj.

4.3 Path Independence of Line Integrals

Solution: → Let − r = xˆi + yˆj

115

→ ∴ d− r = ˆidx + ˆjdy

→ − Here, F = x2ˆi + xyˆj    → →  2ˆ − ∴ F · d− r = x i + xyˆj · ˆidx + ˆjdy = x2 dx + xydy

(4.1)

In the above expression, we can see that → − − F · d→ r contains both the variables x and y. To apply line integral in Equation (4.1), we require the expression either in terms of x or in terms of y variable. As we have discussed that the line integral is independent of the path, we can consider the expression in Equation (4.1) in terms of any one variable. In this illustration, we represent the Equation (4.1) in terms of y variable and as shown in Figure 4.3, the path of integration is the parabola x = y 2 ⇒ dx = 2ydy Substituting in the Equation (4.1) and integrating between the limits y = 0 to y = 1, we get   1 → − − → F ·dr = (y 4 · 2ydy + y 2 · ydy) 0 C  1  5  2y + y 3 dy = 0

  6 4 1  y y 7 1 1 = 2 +  = + = 6 4 0 3 4 12

Figure 4.3 Representation of parabola x = y 2

116

Vector Integral Calculus

→ → − → − r = 3π, where F = zˆi + xˆj + y kˆ and Illustration 4.2: Prove that C F · d− → C is the arc of the curve − r = cost ˆi + sint ˆj + tkˆ from t = 0 to t = 2π. → Solution: Here, we have − r = cost ˆi + sint ˆj + tkˆ ∴ x = cost, y = sint , z = t ∴ dx = −sint dt, dy = cost dt, dz = dt    → →  ˆ − ˆ ∴ F · d− r = z i + xˆj + y kˆ · ˆidx + ˆjdy + kdz = zdx + xdy + ydz = t (−sint ) dt + cost · cost dt + sint dt   = −tsint + cos2 t + sint dt → Now, the path of integration is the arc of the curve − r = cost ˆi + sint ˆj + tkˆ from t = 0 to t = 2π.  → − − F · d→ r ∴ C  2π   = −tsint + cos2 t + sint dt 0  2π (1 + cos2t ) 2π dt + |−cost |2π = −|t (−cost ) − (−sint )|0 + 0 2 0  2π t sin2t  = − (−2π) +  + − (cos2π − cos0 ) 2 4 0 2π = 3π = 2π + 2 → − ˆ Illustration 4.3 If F = (2x − y + 2z) ˆi+(x + y − z) ˆj +(3x − 2y − 5z) k, → − calculate the circulation of F along the circle in the xy-plane of 2 units radius and center at the origin. Solution: The circulation is given by

:

− − → F · d→ r

C

− Let → r = xˆi + yˆj + z kˆ then → ˆ d− r = ˆidx + ˆjdy + kdz

4.3 Path Independence of Line Integrals

117

→ → − Now, F · d− r   ˆ = (2x − y + 2z)ˆi + (x + y − z)ˆj + (3x − 2y − 5z)kˆ ·(ˆidx + ˆjdy + kdz) = (2x − y + 2z) dx + (x + y − z) dy + (3x − 2y − 5z) dz Next, the path of integration is the circle in xy-plane of radius 2 units and center at the origin, i.e., x2 + y 2 = 4 and in xy-plane z = 0. Therefore, the parametric equation of the circle is x = 2cosθ , y = 2 sinθ dx = −2sinθ dθ , dy = 2cosθ dθ For the complete circle, θ varies from 0 to 2π. → → − Substituting in F · d− r and integrating between the limits θ = 0 to θ = 2π,  2π ∴ Circulation = [(2 · 2 cos θ − 2 sin θ) (−2 sin θdθ) 0

+ (2 cos θ + 2 sin θ) (2 cos θ dθ)]  2π   =4 −2cosθ sinθ + sin2 θ + cos2 θ + cosθ sinθ dθ 0  2π   2π

  cos2θ sin2θ  = 8π dθ = 4θ + 1− =4 2 4 0 0  → → − →  − r where F = x2 + y 2 ˆi − 2xyˆj and C Illustration 4.4 Evaluate C F · d− is the rectangle in the xy-plane bounded by y = 0, x = a, y = b, x = 0. Solution:

Figure 4.4

Representation of the rectangle in xy-plane bounded by lines

118

Vector Integral Calculus

→ Consider, − r = xˆi + yˆj → ∴ d− r = ˆidx + ˆjdy     → → − Now, F · d− r = x2 + y 2 ˆi − 2xyˆj · (ˆidx + ˆjdy)   = x2 + y 2 dx − 2xydy Next, the path of integration is the rectangle OABD (See Figure 4.4) bounded by the four lines y = 0, x = a, y = b, x = 0.      → − − → − − → − − → − − → − − → → → → F ·dr = F ·dr + F ·dr + F ·dr + F · d→ r ∴ C

OA

AB

BD

DO

(4.2)

(i) Along OA : y = 0 ⇒ dy = 0 and x varies from 0 to a. 

− − → F · d→ r = OA



a 0

 3 a x  a3 x dx =   = 3 0 3 2

(ii) Along AB : x = a ⇒ dx = 0 and y varies from 0 to b.  b   b → − − → F ·dr = −2aydy = −ay 2 0 = −ab2 0

AB

(iii) Along BD : y = b ⇒ dy = 0 and x varies from a to 0.  3 0

3    0 x  → → − a − 2 2 2  2  F ·dr = +b a (x + y )dx =  + b x = − 3 3 a BD a (iv) Along DO : x = 0 ⇒ dx = 0 and y varies from b to 0.  0  → − − → F ·dr = 0dy = 0 DO

b

Substituting in the Equation (4.2), we get  → − − a3 a3 F · d→ r = − ab2 − − b2 a = −2ab2 3 3 C   → → − → − r where F = 3x2 + 6y ˆi − 14yzˆj + Illustration 4.5 Evaluate C F · d− 20xz 2 kˆ and C is the straight line joining the points (0, 0, 0) to (1, 1, 1).

4.3 Path Independence of Line Integrals

119

Solution: → → ˆ Consider, − r = xˆi + yˆj + z kˆ then d− r = ˆidx + ˆjdy + kdz  →  2 − 2 ˆ ˆ ˆ Here, we have F = 3x + 6y i − 14yz j + 20xz k   → →  2 − ˆ ∴ F · d− r = 3x + 6y ˆi − 14yzˆj + 20xz 2 kˆ · (ˆidx + ˆjdy + kdz) Next, the path of integration is the straight line joining the points A (0, 0, 0) to B (1, 1, 1). The equation of the line AB is x−0 y−0 z−0 = = 0−1 0−1 0−1 ∴x=y=z ∴ dx = dy = dz → − − Substituting in F ·d→ r and integrating between the limits x = 0 to x = 1, we get  1   . # 2 → → − − F ·dr = 3x + 6x dx − 14x2 dx + 20x3 dx 0 C  1 (20x3 − 11x2 + 6x)dx = 0

1   x4 x3 x2  13  = 20 − 11 + 6  = 4 3 2 0 3   → − ˆ then Illustration 4.6 If F = 2xyzˆi + x2 z + 2y ˆj + x2 y k, → − (i) If F is conservative, then find its scalar potential φ. (ii) Find the work done in moving a particle under this force field from (0, 1, 1) to (1, 2, 0). Solution: → − → − (i) Since F is conservative, we have F = ∇φ   ∂φ ˆ ∂φ ˆ ∂φ +j +k ∴ 2xyzˆi + x2 z + 2y ˆj + x2 y kˆ = ˆi ∂x ∂y ∂z

120

Vector Integral Calculus

Now, comparing the coefficients of ˆi, ˆj and kˆ on both sides, we have ∂φ ∂φ ∂φ = 2xyz, = x2 z + 2y, = x2 y ∂x ∂y ∂z Also, ∂φ ∂φ ∂φ dx + dy + dz ∂x ∂y ∂z   = 2xyzdx + x2 z + 2y dy + x2 ydz

dφ =

Integrating both sides,   dφ = y, z

 2xyzdx +

constant

 +

(x2 z + 2y)dy x, z constant (x2 y)dz x, y constant

Considering only those terms on the right-hand side of the integral which have not appeared in the previous integral, i.e., omitting the x2 yz terms in second and third integral, we get φ = x2 yz + y 2 + c where c is the integrating constant. → − (ii) F is conservation and hence the work done is independent of the path.  → − − F · d→ r ∴ Work done = 

C (1,2,0)

= (0,1,1)

(1,2,0)

dφ = |φ|(0,1,1)

 (1,2,0) = x2 yz + y 2 + c (0,1,1)

=3     → − ˆ then Illustration 4.7 If F = (x2 − yz)ˆi + y 2 − zx ˆj + z 2 − xy k, → − (i) If F is conservative, then find its scalar potential φ.

4.3 Path Independence of Line Integrals

121

(ii) Find the work done in moving a particle under this force field from (1, 1, 0) to (2, 0, 1). Solution: → − → − (i) Since F is conservative, we have F = ∇φ     ∂φ ˆ ∂φ ˆ ∂φ ∴ (x2 − yz)ˆi + y 2 − zx ˆj + z 2 − xy kˆ = ˆi +j +k ∂x ∂y ∂z Now, comparing the coefficients of ˆi, ˆj and kˆ on both sides, we have  ∂φ  2  ∂φ ∂φ  2 = (x2 − yz), = y − zx , = z − xy ∂x ∂y ∂z Also, ∂φ ∂φ ∂φ dx + dy + dz ∂x ∂y ∂z     = (x2 − yz)dx + y 2 − zx dy + z 2 − xy dz

dφ =

Integrating both sides,   dφ = y, z

 2xyzdx +

constant

 +

(x2 z + 2y)dy x, z constant (x2 y)dz x, y constant

Considering only those terms on the right-hand side of the integral which have not appeared in the previous integral, i.e., omitting the xyz terms in second and third integral, we get φ=

x3 y3 z3 − xyz + + +c 3 3 3

where c is the integrating constant. → − (ii) F is conservation and hence the work done is independent of the path.

122

Vector Integral Calculus



− − → F · d→ r

∴ Work done = 

C (2,0,1)

= (1,1,0)

(2,0,1)

dφ = |φ|(1,1,0)

(2,0,1)  3  x y3 z3  + + c =  − xyz + 3 3 3 (1,1,0) =

7 3

4.4 Surface Integrals The surface integral over a curved surface S is the generalization of a double integral over a plane region R given in Figure 4.5. → − Let F = F1ˆi + F2 ˆj + F3 kˆ be a continuous vector point function defined over a two-sided surface S. Divide S into a finite number of subsurfaces S1 , S2 , S3 , . . . , Sm with surface areas δS1 , δS2 , δS3 , . . . , δSm . Let δSr be the surface area of Sr and n ˆ r be the limit vector at some point Pr (in Sr ) in the direction of the outward normal to Sr . If we increase the number of subsurfaces, then the surface area δSr of each sub-surface will decrease. Thus, as m → ∞, δSr → 0 then,  m / → − → − F (Pr ) · n F ·n ˆ dS ˆ r δSr = lim m→∞

Figure 4.5

r=1

S

Representation of curved surface S and a plane region R

4.4 Surface Integrals

123

→ − This is called the surface integral of F over the surface S. The surface integral can also be written as  → − − → → − F · d S , Where d S = n ˆ dS S

If the equation of surface S is φ (x, y, z) = 0, then n ˆ=

∇φ |∇φ| .

4.4.1 Flux → − If F represents the velocity of the fluid at any point P on a closed surface S, → − → − ˆ dS represents the flux of F over S, i.e., the then surface integral S F · n volume of the fluid flowing out from S per unit time. Note: If

S

− → → − F ·n ˆ dS = 0, then F is called a solenoidal vector point function.

4.4.2 Evaluation of Surface Integral A surface integral is evaluated by expressing it as a double integral over the region R. The region R is the orthogonal projection of S on one of the coordinate planes (xy, yz or zx) . Let R be the orthogonal projection of S on the xy−plane and cosα, cosβ, cosγ are the direction cosines of n ˆ. Then n ˆ = cosα ˆi + cosβ ˆj + cosγ kˆ dxdy = Projection of dS on xy − plane = dScosγ dxdy dxdy  dS = =   cosγ ˆ · kˆ n Hence,



− → F ·n ˆ dS =



S

− → dxdy  F ·n ˆ   R ˆ · kˆ n

Similarly, taking projection on yz and zx-plane,   → − → − dydz  F ·n ˆ dS = F ·n ˆ   S R ˆ · ˆi n and



− → F ·n ˆ dS = S



− → dzdx  F ·n ˆ   R ˆ · ˆj  n

124

Vector Integral Calculus

4.4.2.1 Component form of Surface Integral     → − ˆ cosα ˆi + cosβ ˆj + cosγ kˆ dS F ·n ˆ dS = (F1ˆi + F2 ˆj + F3 k)· S  S = F1 dScosα + F2 dScosβ + F3 dScosγ  S = F1 dydz + F2 dzdx + F3 dxdy S

→ − − → ˆ dS, where F = 18zˆi − 12ˆj + 3y kˆ and Illustration 4.8: Evaluate S F · n S is the part of the plane 2x + 3y + 6z = 12 in the first octant. Solution: (i) The given surface is the plane 2x + 3y + 6z = 12 in the first octant (See Figure 4.6(a)). Let φ = 2x + 3y + 6z ∴n ˆ=

∇φ 2ˆi + 3ˆj + 6kˆ 2ˆi + 3ˆj + 6kˆ = = √ |∇φ| 7 4 + 9 + 36

(ii) Let R be the projection of the plane 2x + 3y + 4z = 12 (in the first octant) on the xy-plane, which is a triangle OAB bounded by the lines y = 0, x = 0 and 2x + 3y = 12 (See Figure 4.6(b)). (iii) Next, we find 7 dxdy  = dxdy dS =   6 ˆ · kˆ n

Figure 4.6

Representation of the projection of the plane in the first octant

4.4 Surface Integrals

125

(iv) Along the vertical strip P Q, y varies from 0 to 12−2x 3 , and in the region R, x varies from 0 to 6.  → − F ·n ˆ dS ∴ S     ˆi + 3ˆj + 6kˆ 7 2 = dxdy 18zˆi − 12ˆj + 3y kˆ · 7 6 R  1 = (36z − 36 + 18y) dxdy 6 R  "  !

12 − 2x − 3y =3 − 2 + y dxdy 2 6 R  6  12−2x−3y 6 (6 − 2x)dydx = 0 0  6 12−2x (3 − x) |y|0 3 dx =2 0  6 (12 − 2x) dx (3 − x) =2 3 0  4 6 2 (x − 9x + 18)dx = 3 0  6  4  x3 9x2 + 18x =  − 3 3 2 0

4 = (72 − 162 + 108) = 24 3 Illustration 4.9: Evaluate S (yzdydz + xzdzdx + xydxdy) over the surface of the sphere x2 + y 2 + z 2 = 1 in the positive octant. Solution: → − ˆ dS = yzdydz + xzdzdx + xydxdy (i) S F · n → − ∴ F = yzˆi + xzˆj + xy kˆ (ii) The given surface is the sphere x2 + y 2 + z 2 = 1. Let φ = x2 + y 2 + z 2 ∴n ˆ=

2xˆi + 2yˆj + 2z kˆ ∇φ = = xˆi + yˆj + z kˆ |∇φ| 4x2 + 4y 2 + 4z 2

126

Vector Integral Calculus

Figure 4.7

Representation of the positive octant of the sphere

(∵ x2 + y 2 + z 2 = 1) (iii) Let R be the projection of the sphere x2 + y 2 + z 2 = 1 (in the positive octant) on the xy−plane (z = 0), which is the part of the circle x2 +y 2 = 1 in the first quadrant (See Figure 4.7). (iv) dS = (v)

S

dxdy

|nˆ ·kˆ|

=

dxdy z

− → F ·n ˆ dS  = yzdydz + xzdzdx + xydxdy S      dxdy = yzˆi + xzˆj + xy kˆ · xˆi + yˆj + z kˆ z  R dxdy = (3xyz) z R =3 xydxdy R

Substituting x = rcosθ , y = rsinθ , the equation of the circle x2 +y 2 = 1 reduces to r = 1 and dxdy = rdrdθ. Along the radius vector OP , r varies from 0 to 1, and in the first quadrant of the circle, θ varies from 0 to π2 .  yzdydz + xzdzdx + xydxdy ∴ S

4.4 Surface Integrals



π 2

=3 0



1 0

127

rcosθ · rsinθ · r drdθ

 1 sin2θ dθ · r3 dr =3 2 0 0   π  1 3  −cos2θ  2  r4  3 3 =  ·  = (−cosπ + cos0 ) =  2 2 4 0 16 8 0 

π 2

→ − Illustration 4.10: Find the flux of F = ˆi − ˆj + xyz kˆ through the circular region S obtained by cutting the sphere x2 +y 2 +z 2 = a2 with a plane y = x. Solution: We know that



− → F ·n ˆ dS

Flux = S

(i) Here, the surface S (See Figure 4.8) is the intersection of the sphere x2 +y 2 +z 2 = a2 with a plane y = x, which is an ellipse 2x2 +z 2 = a2 . (ii) Normal to the ellipse 2x2 + z 2 = a2 is also normal to the plane y = x. Let φ = x − y n ˆ=

ˆi − ˆj ∇φ = √ |∇φ| 2

Figure 4.8 Representation of the sphere

128

Vector Integral Calculus

(iii) Let R be the projection of the surface S on the xz-plane, which is an ellipse 2x2 + z 2 = a2 √ (iv) dS = dxdz 2 dxdz = |nˆ ·ˆj |   ˆ ˆ  √ → → − √j (v) S F · − n dS = R ˆi − ˆj + xyz kˆ · i− 2 dxdz 2 

=

2dxdz R

Substituting x =

√a rcosθ 2

, z = arsinθ , the equation of the ellipse 2x2 + 2

a z 2 = a2 reduces to r = 1 and dxdz = √ rdr dθ . 2 Along the radius vector OP , r varies from 0 to 1, and for a complete ellipse, θ varies from 0 to 2π.   2π  1 2 → − − a √ rdr dθ F ·→ n dS = 2 2 0 0 S  2 1 2 √ 2 1   2a r 2 a · · 2π = √   |θ|2π 0 = 2 2 2 0 √ = 2 πa2

Aliter, 

− − → F ·→ n dS = 2

 dxdz ⎡

S

⎤

R

x2 y2 ⎢ ⎥ = 2 ⎣Area of the ellipse  2 + 2 = 1⎦ a √a a = 2 · π√ · a 2 √ 2 = 2 πa

Hence, flux =

√ 2 πa2 .

2

4.5 Volume Integrals

129

4.5 Volume Integrals If V be a region in space bounded by a closed surface S, then the volume → − → − integral of a vector point function F is V F dV . 4.5.1 Component Form of Volume Integral → − If F = F1ˆi + F2 ˆj + F3 kˆ then     → − F dV = F1ˆi + F2 ˆj + F3 kˆ dxdydz V V   ˆ ˆ ˆ =i F1 dxdydz + j F2 dxdydz + k F3 dxdydz Another type of volume integral is V φ dV , where φ is a scalar function. → − − → Illustration 4.11: Evaluate V F dV where F = xˆi + yˆj + 2z kˆ and V is the volume enclosed by the planes x = 0, y = a, z = b2 and the surface z = x2 . Solution: (i) V is the volume of the cylinder in positive octant with the base as OAB and bounded between the planes y = 0 and y = a. y varies from 0 to a (See Figure 4.9). (ii) Along the vertical strip P Q, z varies from x2 to b2 , and in the region OAB, x varies from 0 to b.   b  b2  a   → − F dV = xˆi + yˆj + 2z kˆ dxdydz V

Figure 4.9

x=0

z=x2

y=0

Representation of the cylinder in positive octant

130

Vector Integral Calculus

 2 a  y  a a ˆ   ˆ ˆ = xi|y|0 + j   + 2z k|y|0 dzdx 2 0 0 x2   b  b2

2 ˆ ˆixa + ˆj a + k2za dzdx = 2 0 x2   b

2  2 b 2 2 2 a b b ˆ z  2 dx ˆixa|z| 2 + ˆj |z| 2 + ka = x x 2 x 0   b

2 a 2 2 2 2 4 4 ˆ ˆixa(b − x ) + ˆj (b − x ) + ka(b − x ) dx = 2 0  2 2 



 5 b  b x x4 a2 2 x3 ˆ b4 x − x  − + ˆj b x− + ka = ˆia 2 4 2 3 5 0 





4 5 b b4 a2 3 b3 ˆ b5 − b − + ˆj b − + ka = ˆia 2 4 2 3 5  b

b2



3

b4 a2 b ˆ 4ab5 ˆ = a ˆi + j+ k. 4 3 5 

 − → →  − ∇ × F dV , where F = 2x2 − 3z ˆi −

Illustration 4.12: Evaluate V 2xyˆj − 4xkˆ and V is the closed region bounded by the planes x = 0, y = 0, z = 0 and 2x + 2y + z = 4. Solution: (i)

  ˆi ˆj kˆ  →  − ∂ ∂ ∂ ∇× F = ∂x ∂y ∂z   2x2 − 3z −2xy −4x

       

= (0 − 0) ˆi − (−4 + 3) ˆj + (−2y − 0) kˆ = ˆj − 2y kˆ (ii) Along the elementary volume P Q, z varies from 0 to 4−2x−2y. Along the vertical strip P  Q , y varies from 0 to 2−x, and in the region, x varies from 0 to 2.    2  2−x  4−2x−2y   → − ˆj − 2y kˆ dxdydz ∇ × F dV = V

0

0

0

4.6 Exercise

131

2  2−x 



 ˆj − 2y kˆ |z|4−2x−2y dydx 0

= 

0

0 2  2−x 



0

0 2  2−x 

 ˆj − 2y kˆ (4 − 2x − 2y) dydx

=

 (4 − 2x − 2y) ˆj − 2 (4 − 2x) y kˆ + 4y 2 kˆ dydx

= 0

0

2 $ 0 = 0

 2 2−x 1 y  ˆj − (4 − 2x) |y|2−x 0 0

 3 2−x ' % y   2 2−x kˆ dx − 2 (2 − x) y 0 − 4   3 0 "  2! 2 8 8 8 2ˆ 3ˆ ˆ = (2 − x) j − (2 − x) k dx = ˆj − kˆ = (ˆj − k) 3 3 3 3 0 &

4.6 Exercise

→ − → r along the curve x2 + y 2 = 1, z = 1 in the positive 1. Evaluate C F · d− → − direction from (0, 1, 1) to (1, 0, 1), where F = (yz + 2x) ˆi + xzˆj + ˆ (xy + 2z) k.

(Answer: 1) → → − → − r over the circular path x2 + y 2 = a2 where F = 2. Evaluate C F · d− siny ˆi + x (1 + cosy ) ˆj. (Answer : πa2 ) → − 3. Find the work done in moving a particle in the force field F = 3x2ˆi + (2xz − y) ˆj + z kˆ along the curve x2 = 4y and 3x3 = 8z from x = 0 to x = 2. (Answer: 16) 4. Find the work done in moving a particle from A(1, 0, 1) to B(2, 1, 2) → − along the straight line AB in the force field F = x2ˆi + (x − y) ˆj + ˆ (y + z) k.   Answer : 16 3

132

Vector Integral Calculus

5. Find the work done in moving a particle along the straight-line segments joining the points (0, 0, 0) to (1, 0, 0) , then to (1, 1, 0) , and finally to  →  − ˆ (1, 1, 1) under the force field F = 3x2 + 6y ˆi − 14yzˆj + 20xz 2 k.   Answer : 23 3 → − 6. Find the work done by the force F = xˆi − zˆj + 2y kˆ in displacing the particle along the triangle OAB, where OA : 0 ≤ x ≤ 1, AB : 0 ≤ z ≤ 1, BO : 0 ≤ x ≤ 1,

y = x, z = 0 x = 1, y = 1 y=z=x 

Answer :

3 2



  → − 7. Find the work done by the force F = 16yˆi + 3x2 + 2 ˆj in moving a particle once round the right half of the ellipse x2 + a2 y 2 = a2 from (0, 1) to (0, −1).   Answer : 8aπ − 4a2 − 4 → − → → − r , where F = 2xˆi + 4yˆj − 3z kˆ and C is the curve 8. Evaluate C F · d− → − r = cost ˆi + sint ˆj + t kˆ from t = 0 to t = π.   2 Answer : −3π 2 → − 9. Find the circulation of F = (x − 3y) ˆi + (y − 2x) ˆj around the ellipse in the xy− plane with the origin as a center and 2 and 3 as semi-major and semi-minor axes respectively. (Answer : 6π) → − 10. Find the circulation of F = yˆi + zˆj + xkˆ around the curve x2 + y 2 = 1, z = 0. (Answer : − π)   → − 11. If F = 2xy + z 3 ˆi + x2 ˆj + 3xz 2 kˆ is conservative then (i) find its scalar potential φ, (ii) find the work done in moving a particle under this force field from (1, −2, 1) to (3, 1, 4).   Answer : (i) φ = x2 y + xz 3 + c, (ii) 202

4.6 Exercise

133

→ − 12. If F = 3x2 yˆi + (x3 − 2yz 2 )ˆj + (3z 2 − 2y 2 z)kˆ is conservative then (i) find its scalar potential φ, (ii) find the work done in moving a particle under this force field from (2, 1, 1) to (2, 0, 1).   Answer : (i) φ = x3 y + z 3 − y 2 z 2 + c, (ii) − 7 → − 13. If F = 2xyezˆi + x2 ez ˆj + x2 yez kˆ is conservative then (i) find its scalar potential φ, (ii) find the work done in moving a particle under this force field from (0, 0, 0) to (1, 1, 1).   Answer : (i) φ = x2 yez + c, (ii) e → − → − 14. Evaluate S F · n ˆ dS where F = 3yˆi + 2zˆj + x2 yz kˆ and S is the surface y 2 = 5x in the positive octant bounded by the planes x = 3 and z = 4. (Answer : − 42) → − → − ˆ dS where F = (x + y 2 )ˆi − 2xˆj + 2yz kˆ and S is the 15. Evaluate S F · n surface 2x + y + 2z = 6 in the first octant. (Answer : 81) → − → − 16. Evaluate S ∇ × F · n ˆ dS where F = y 2ˆi + yˆj − xz kˆ and S is the upper half of the sphere x2 + y 2 + z 2 = a2 . (Answer : 0) → − 17. Find the flux of the vector field F through the portion of the sphere √ √ x2 + y 2 + z 2 = 36 lying between the planes z = 11 and z = 20 → − ˆ where F = xˆi + yˆj + z k. √  √  Answer : 72π 20 − 11  → − ˆj + x2 + y 2 − 1 kˆ through 18. Find the flux of the vector field F = xˆi + y  the outer side of the hyperboloid z = x2 + y 2 − 1 bounded by the √ planes z = 0 and z = 3. √   Answer : 2 3π → − 19. Find the flux of the vector field F = 2yˆi−zˆj +x2 kˆ across the surface of the parabolic cylinder y 2 = 8x in the first octant bounded by the planes y = 4 and z = 6. (Answer : 132)

134

Vector Integral Calculus

 20. Evaluate

V

→ − → − ∇ · F dV where F = 2x2 yˆi − y 2 ˆj + 4xz 2 kˆ and V is

the region in the first octant bounded by the cylinder y 2 + z 2 = 9 and the plane z = 2. (Answer : 180)  → − → − ∇ · F dV where F = 2xzˆi − xˆj + y 2 kˆ and V is the

 21. Evaluate

V

region bounded by the surfaces x = 0, y = 0, y = 6, z = x2 , z = 4.   Answer : 128ˆi − 24ˆj + 384kˆ 22. Evaluate V f dV where f = 45x2 y and V is the region bounded by the planes 4x + 2y + z = 8, x = 0, y = 0, z = 0. (Answer : 128)  → − → − 23. Evaluate V ∇ × F dV , where F = (x + 2y) ˆi−3zˆj+xkˆ and V is the closed region in the first octant bounded by the plane 2x+2y+z = 4.    Answer : 83 3ˆi − ˆj + 2kˆ

5 Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

5.1 Green’s Theorem (in the Plane) Double integrals over a plane region could also be transformed into line integrals over the boundary of the region and conversely. This is of practical interest because it’s going to help to form the evaluation of an integral easier. The transformation can be computed using a theorem known as Green’s theorem. ∂N Statement: If M (x, y), N (x, y) and their partial derivatives ∂M ∂y , ∂x are continuous in some region R of xy-plane bounded by a closed curve C, then  : 

∂N ∂M − dxdy (M dx + N dy) = ∂x ∂y C R

Proof: Let R be the bounded region by the curve C as shown in Figure 5.1. Let the curve C be divided into two parts, the curves EAB and BDE.

Figure 5.1

Represents the region R bounded by the curve C

135

136

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Let the equations of the curves EAB and BDE are x = f1 (y), x = f2 (y) respectively and are bounded between the lines y = c and y = d. Consider, %   d $ f2 (y) ∂N ∂N dxdy = dx dy c f1 (y) ∂x R ∂x  d f (y) |N (x, y)|f21 (y) dy = c  d = [N (f2 , y) − N (f1 , y)] dy c  d  d = N (f2 , y) dy + N (f1 , y) dy c c   = N (x, y) dy + N (x, y) dy BDE

EAB

: =

N (x, y)dy C



∴ R

∂N dxdy = ∂x

: N (x, y)dy

(5.1)

C

Similarly, let the curve C be divided into two parts, the curves ABD and DEA. Let the equation of the curves ABD and DEA are y = g1 (x) , y = g2 (x) respectively and are bounded between the lines x = a and x = b. Now, consider %   b $ g2 (x) ∂M ∂M dxdy = dy dx a g1 (x) ∂y R ∂y  b g (x) |M (x, y)|g21 (x) dx = a  b = [M (x, g 2 ) − M (x, g 1 )] dx a  b  b =− M (x, g 2 ) dx − M (x, g 1 ) dx a a ! "  =− M (x, y) dx + M (x, y) dx DEA

ABD

5.1 Green’s Theorem (in the Plane)

137

: =−

M (x, y)dx  C

∴ − R

∂M dxdy = ∂y

: M (x, y) dx

(5.2)

C

Adding Equations (5.1) and (5.2), we get  : 

∂N ∂M − dxdy (M (x, y) dx + N (x, y) dy) = ∂x ∂y C R 

:

or

(M dx + N dy) = C

R

∂N ∂M − ∂x ∂y

 dxdy.

→ − → → − → Note: Let F = M ˆi + N ˆj and − r = xˆi + yˆj, then F · d− r = M dx + N dy. Also,   ˆj  ˆi

 kˆ   ∂ → − → − ∂N ∂M ∂ ∂ curl F =  ∂x ∂y ∂z  = k − ∂x ∂y  M N O  → − The component of the curl F which is normal to a region R in xy-plane is  → ∂N → − − ∂M ∇× F · k = − ∂x ∂y Hence, the vector form of Green’s theorem is given as   : → → − − → − − → F ·dr = ∇ × F · k dxdy C

R

→ − → where, F = M ˆi + N ˆj, − r = xˆi + yˆj, kˆ is the unit vector along z-axis. 5.1.1 Area of the Plane Region Let A be the area of the plane region R bounded by a closed curve C. ∂N Let M = −y, N = x ⇒ ∂M ∂y = −1, ∂x = 1 Using Green’s theorem, :   (−ydx + xdy) = (1 + 1)dxdy = 2 dxdy = 2A C

R

∴A=

1 2

R

: (xdy − ydx) C

138

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Note: In polar coordinates, x = rcosθ , y = rsinθ dx = cosθ dr − rsinθ dθ, dy = sinθ dr + rcosθ dθ : 1 [rcosθ (sinθ dr + rcosθ dθ) − rsinθ (cosθ dr − rsinθ dθ)] ∴A= 2 C : 1 ∴A= r2 dθ 2 C Illustration 5.1: Using Green’s theorem, evaluate :  2  3x − 8y 2 dx + (4y − 6xy)dy C

where C is the boundary of the region bounded by x = y 2 and y = x2 . Solution: Here, C is the boundary of the region bounded by x = y 2 and y = x2 (See Figure 5.2). Therefore, solving both the parabolas, we get ∴ x4 − x = 0   ∴ x x3 − 1 = 0 ∴ x = 0 or x3 = 1 Therefore, for x = 0 ⇒ y = 0 and for x = 1 ⇒ y = 1. So, both the parabolas are intersecting at (0, 0) and (1,1).

Figure 5.2 Represents the region R bounded two parabolas

5.1 Green’s Theorem (in the Plane)

139

Next, N = 4y − 6xy M = 3x2 − 8y 2 , ∂M ∂N ∴ = −16y, = −6y ∂y ∂x  

: ∂N ∂M − dxdy M dx + N dy = ∴ ∂x ∂y C R  1  √x 10y dxdy = 0 x2  1   √x 5 y 2 x2 dx = 0  1   x − x4 dx =5 0

=5

x2 x5 − 2 5

1

=5

0

1 1 − 2 5

 =

3 2

Illustration 5.2: Applying Green’s theorem, evaluate : [(y − sinx ) dx + cosx dy] C

where C is the plane triangle enclosed by the lines y = 0, x = (See Figure 5.3).

π 2

and y = π2 x

Solution:

Figure 5.3

Represents the plane triangle enclosed by given lines

140

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Here, M = y − sinx , N = cosx ∂N ∴ ∂M ∂y = 1, ∂x = −sinx and we know that 

: M dx + N dy =



C

R

=  =

R π 2

x=0



π 2

∂N ∂M − ∂x ∂y

 dxdy

(−sinx − 1 ) dydx $ 2x % π (−sinx − 1)dy dx y=0 2x

[−ysinx − y]0π dx 0   π

2 2π 2π dx − sinx − = x x 0 ! 2 " π2 π x 2 π 2 2 =− − = − [−xcosx + sinx ]0 − π π 0 π 4 =

→ − Illustration 5.3: Verify Green’s theorem for the function F = (x + y) ˆi + 2xyˆj and C is the rectangle in the xy-plane bounded by x = 0, y = 0, x = a, y = b (See Figure 5.4). Solution:

Figure 5.4

Represents the rectangle in the xy-plane bounded by lines

5.1 Green’s Theorem (in the Plane)

We know that

:

:

− − → F · d→ r

M dx + N dy = :C

C

141

=

(x + y) dx + 2xy dy C

Along OA, y = 0 ⇒ dy = 0   M dx + N dy = ∴ OA

a

xdx = 0

a2 2

Along AB, x = a ⇒ dx = 0  b  2ab2 = ab2 M dx + N dy = 2ay dy = ∴ 2 0 AB Along BC, y = b ⇒ dy = 0   ∴ M dx + N dy =

0

(x + b)dx = −

a

AB

Along CO, x = 0 ⇒ dx = 0.   ∴ M dx + N dy =

a2 − ab 2

(i)

(ii)

(iii)

0

0 dy = 0

(iv)

b

CO

Adding (i), (ii), (iii), and (iv), we get : M dx + N dy = ab2 − ab

(v)

C

Now,



R

∂N ∂M − ∂x ∂y



a b

 dxdy = 0

0

(2y − 1) dxdy

b 2y 2 − y dx = 2 0 0  a  2  b − b dx = 0   a  2 dx = b −b  0 a  2 = b − b (x)0 = ab2 − ab 

a

(vi)

142

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

From (v) and (vi) we see that  : 

∂N ∂M − dxdy M dx + N dy = ∂x ∂y C R Hence, Green’s theorem is verified. Illustration 5.4: Use Green’s theorem to evaluate the integral :   2 y dx + x2 dy C

where, C: The triangle bounded by x = 0, x + y = 1, y = 0 (See Figure 5.5). Solution: Here, M = y 2 and N = x2 then ∂M ∂N = 2y, = 2x ∂y ∂x From Green’s theorem  : 

 1  1−y ∂N ∂M M dx + N dy = (x − y)dxdy − dxdy = 2 ∂x ∂y 0 0 C R % $ "1−y  1! 2  1 x (1 − y)2 =2 − xy − y(1 − y) dy dy = 2 2 2 0 0 0 ! " " ! 1 1 1 1 3 1 2 =2 y−y + y −1+ =0 =2 2 2 2 2 0

Figure 5.5

Represents the triangle in the xy-plane bounded by lines

5.1 Green’s Theorem (in the Plane)

143

→ − Illustration 5.5: Verify Green’s theorem for the function F = (x − y) ˆi+xˆj and the region R bounded by the unit circle → C :− r (t) = (cost ) ˆi + (sint ) ˆj, 0 ≤ t ≤ 2π → − → → Solution: Consider, − r = xˆi+yˆj ⇒ d− r = ˆidx+ ˆjdy and F = (x−y)ˆi+xˆj → → − ∴ F · d− r = (x − y) dx + xdy Now, M = x − y, N = x ∂M ∂N ∴ = −1, =1 ∂y ∂x "  ! : ∂N ∂M − dxdy M dx + N dy = ∴ ∂y C R ∂x  = [1 + 1] dxdy R  =2 dxdy = 2π : R : → − − F · d→ r = (x − y) dx + xdy C C : 2π . # (cost − sint ) (−sint ) + cos2 t dt = 0  2π # . −sint cost + sin2 t + cos2 t dt = 0  2π [−sint cost + 1] dt = 0 "  2π ! sin2t − + 1 dt = 2 0

 1 cos2t 2π + |t|2π = 0 2 2 0 1 = [cos4π − cos0 ] + 2π = 2π 4 ! "  : → − − ∂N ∂M → F ·dr = ∴ − dxdy = 2π ∂y C R ∂x

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

144

Hence, the theorem is verified.



 1 + dy where C is x C √ the boundary of the region bounded by the parabola y = x and the lines x = 1, x = 4, y = 1 (See Figure 5.6).

Illustration 5.6: Verify Green’s theorem for

1 y dx

Solution: (i) The point of intersection of the √ √ (a) parabola y = x and the line x = 1 is obtained as y = 1 = 1. Hence, A(1, 1) is the point of intersection. √ √ (b) parabola y = x and line x = 4 is obtained as y = 4 = 2. Hence, D(4, 2) is the point of intersection. (ii) M = y1 ,

N=

1 x

∂M 1 ∂N 1 = − 2, =− 2 ∂y y ∂x x (iii)

9 C

+

(M dx + N dy) = DQA (M dx

AB

(M dx + N dy) +

BD

(M dx + N dy)

+ N dy) (i)

(a) Along AB : y = 1 ⇒ dy = 0 and x varies from 1 to 4.   

1 1 dx + dy (M dx + N dy) = x AB AB y

Figure 5.6

Represents the region bounded by the parabola and lines

5.1 Green’s Theorem (in the Plane)



4

= 1

145

dx = |x|41 = 3

(b) Along BD : x = 4 ⇒ dx = 0 and y varies from 1 to 2.   

1 1 dx + dy (M dx + N dy) = x BD BD y  2 1 1 1 = dy = |y|21 = 4 4 1 4 √ (c) Along DQA : y = x ⇒ dy = 2√1 x dx and x vary from 4 to 1. 

  1 1 dx + dy (M dx + N dy) = x DQA DQA y   1

1 1 1 √ dx + · √ dx = x 2 x x 4 1   √ 1  = 2 x − √  x 4 1 5 =2−1−4+ =− 2 2 Substituting in equation (i), we get  3 1 5 (M dx + N dy) = 3 + − = 4 2 4 C

(ii)

√ (iv) Let R be the region bounded by the parabola y = x and the √ lines x = 1, x = 4, y = 1. Along the vertical strip, y varies from 1 to x and in the region R, x varies from 1 to 4.   

 4  √x

∂N 1 ∂M 1 − dxdy = − 2 + 2 dxdy ∂x ∂y x y 1 1 R √ x  4  1  − · y − 1  dx =  x2 y 1 1  4  3 1 −x− 2 − x− 2 + x−2 + 1 dx = 1 4    −1 1 1  2 2 = 2x − 2x − + x x 1

146

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

=1−4− =

1 +4−2+2+1−1 4

3 4

(iii)

From equations (ii) and (iii), we see that  : 

∂N ∂M M dx + N dy = − dxdy ∂x ∂y C R Hence, Green’s theorem is verified.

5.2 Stokes’ Theorem Statement: If S be an open surface bounded by a closed curve C (See → − Figure 5.7) and F be a continuous and differentiable vector function, then :  → − − → − → F ·dr = ∇× F ·n ˆ dS C

S

where n ˆ is the unit outward normal at any point of the surface S. Proof: → − → Let F = F1ˆi + F2 ˆj + F3 kˆ and − r = xˆi + yˆj + z kˆ 

→ − ∇× F ·n ˆ dS = S

Figure 5.7

 S

  ∇ × F1ˆi + F2 ˆj + F3 kˆ · n ˆ dS

Represents an open surface bounded by a closed curve C

5.2 Stokes’ Theorem

147

 

    ∇ × F1ˆi · n ∇ × F2 ˆj · n ˆ dS + ˆ dS

=

S  

+ S



S

∇ × F3 kˆ · n ˆ dS

(5.3)

Consider,  "    !

 ∂ ∂ ∂ ˆ ˆ ˆ ˆ ˆ ∇ × F1 i · n +j +k × F1 i · n i ˆ dS = ˆ dS ∂x ∂y ∂z S   S

∂F1 ˆ ∂F1 = +j ·n ˆ dS −kˆ ∂y ∂z S  

∂F1 ˆ ∂F1 ˆ = j·n ˆ− k·n ˆ dS (5.4) ∂y ∂z Let the equation of the surface S be z = f (x, y), Then, − → − r = xˆi + yˆj + z kˆ ⇒ → r = xˆi + yˆj + f (x, y)kˆ Differentiating partially with respect to y, we get → ∂f ˆ ∂− r = ˆj + k ∂y ∂y Taking dot product with n ˆ, → ∂− r ∂f ˆ ·n ˆ = ˆj · n ˆ+ k·n ˆ ∂y ∂y → ∂− r ∂y

is tangential and n ˆ is normal to the surface S. → ∂− r ·n ˆ=0 ∂y Substituting in the Equation (5.5), ∂f ˆ k·n ˆ 0 = ˆj · n ˆ+ ∂y ∂z ∂f ˆj · n ˆ = − kˆ · n ˆ ˆ = − kˆ · n ∂y ∂y

(∵ z = f (x, y))

(5.5)

148

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Substituting in the Equation (5.4),

 "  !    ∂F ∂z ∂F 1 1 − ˆj · n ˆ − kˆ · n ˆ dS ∇ × F1ˆi · n ˆ dS = ∂z ∂y ∂y S S  

∂F1 ∂z ∂F1 ˆ =− · + k·n ˆ dS (5.6) ∂z ∂y ∂y S The equation of the surface is z = f (x, y). F1 (x, y, z) = F1 [x, y, f (x, y)] = G(x, y) Differentiating partially with respect to y, ∂G ∂F1 ∂F1 ∂z = + · ∂y ∂y ∂z ∂y Substituting in the Equation (5.6),     ∂G ˆ k·n ˆ dS ∇ × F1ˆi · n ˆ dS = − S S ∂y Let R is the projection of S on the xy-plane and dxdy is the projection of dS on the xy-plane, then kˆ · n ˆ dS = dxdy Thus,   :   ∂G ˆ dxdy = Gdx ∇ × F1 i · n ˆ dS = − S R ∂y C1 (∵ Using Green s theorem) Since the value of G at each point (x, y) of C1 is the same as the value of F1 at each point (x, y, z) of C and dx is same for both the curves C1 and C, we get   :  F1 dx (5.7) ∇ × F1ˆi · n ˆ dS = S

C

Similarly, by projecting the surface S on to yz and zx planes, :    F2 dy ∇ × F2 ˆj · n ˆ dS = S

and

  S

(5.8)

C

:  F3 dz ∇ × F3 kˆ · n ˆ dS = C

(5.9)

5.2 Stokes’ Theorem

149

Substituting the Equations (5.7), (5.8), and (5.9) in (5.3),   :  :  → − − − → (F1 dx + F2 dy + F3 dz) = ∇× F ·n ˆ dS = F · d→ r . S

C

C

Note: If surfaces S1 and S2 have the same bounding curve C, then     :   → − → − → − − ∇× F ·n ˆ dS = ∇× F ·n ˆ dS = F · d→ r . S1

S2

C

  → − Illustration 5.7: Verify Stokes’s theorem for F = x2 + y 2 ˆi + 2xyˆj, taken round the rectangle bounded by the lines x = ±a, y = 0, y = b (See Figure 5.8). Solution: → → Consider, − r = xˆi + yˆj ⇒ d− r = ˆidx + ˆjdy and let ABCD be the given rectangle then      → − − → − − → − − → − − → − − → → → → F ·dr = F ·dr+ F ·dr+ F ·dr+ F · d→ r ABCD

and

AB

BC

CD

     − − → → 2 2 ˆ ˆ ˆ ˆ F · d r = x + y i + 2xy j · idx + jdy   = x2 + y 2 dx + 2xydy Along AB, x = a ⇒ dx = 0 and y varies from 0 to b   b → − − 1 → F · d r = −2a ydy = −2a · b2 = −ab2 2 0 AB

Figure 5.8 Represents the rectangle bounded by the lines

DA

150

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Similarly,  

→ − − F · d→ r =

BC



a



 2 x2 + b2 dx = − · a2 − 2ab2 3

−a

→ − − F · d→ r = 2a



0

ydy = −ab2

b

CD

and



− − → F · d→ r = DA



Thus,



a

x2 dx =

−a

2a3 3

− − → F · d→ r = −4ab2 . ABCD

− → ˆ Also, since curl · F = −4y k,   b → − ∴ curl · F dS = 0

S

= −4

−4y kˆ · kˆ dxdy

−a  b a

ydxdy 0

 = −4

a

0

−a b

|x|a−a ydy

 2 b y  = −8a  = −4ab2 2 0 Hence, the theorem is proved. → − Illustration 5.8: Verify Stokes’s theorem when F = (2x − y) ˆi − yz 2 ˆj − ˆ where S is the upper half surface of the sphere x2 + y 2 + z 2 = 1 and y 2 z k, C is the boundary. Solution: Stokes’s theorem is given as   → − − → − → F ·dr = curl F · n ˆ dS C

∴

S

  ˆi ˆj kˆ →  − curl F =  ∂/∂x ∂/∂y ∂/∂z  2x − y −yz 2 −y 2 z

    = kˆ  

5.2 Stokes’ Theorem

Let n ˆ = sinθ cosφ ˆi + sinθ sinφ ˆj + cosθ kˆ  → − ∴ curl F · n ˆ = cosθ   π  2π 2 → − ∴ curl F · n ˆ dS = cosθ sinθ dθ dφ = π S

θ=0

151

(i)

φ=0

Again C is the unit circle x2 + y 2 = 1, z = 0 ∴ x = cosφ , y = sinφ , z = 0 ∴ dx = −sinφ dφ, dy = cosφ dφ, dz = 0    → →  − ˆ ∴ F · d− r = (2x − y) ˆi − yz 2 ˆj − y 2 z kˆ · ˆidx + ˆjdy + kdz = (2x − y) dx − yz 2 dy − y 2 zdz = − (2cosφ − sinφ ) sinφ dφ   2π → − − → ∴ F ·dr =− (2cosφ − sinφ ) sinφ dφ 0 C  2π   2cosφ sinφ − sin2 φ dφ =− 0

=π

(ii)

From (i) and (ii), the theorem is verified. − → ˆ for Illustration 5.9: Verify Stokes’s theorem for F = xy 2ˆi + yˆj + z 2 xk, the surface of a rectangular lamina bounded by x = 0, y = 0, x = 1, y = 2, z = 0. → − Solution: Here, we have z = 0 ⇒ F = xy 2ˆi + yˆj : : # 2 . → − − → F ·dr = xy dx + ydy C

C

where C is the path OABCO as shown in Figure 5.9. Along OA, y = 0 ⇒ dy = 0 Along AB, x = 1 ⇒ dx = 0 Along BC, y = 2 ⇒ dy = 0 Along CO, x = 0 ⇒ dx = 0 :     → − − → 2 2 ∴ F ·dr = xy dx + ydy + xy dx + C

OA

AB

BC

ydy CO

152

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

Figure 5.9

Represents the surface of a rectangular lamina bounded by the lines

 =0+ 2

= 0





0

ydy + 0





2

0

4xdx + 1

x2 ydy + 4 2

0 1

 −

ydy 2

2 0

ydy = −2

Now,   ˆi  ∂ − → ∇ × F =  ∂x  xy 2

ˆj ∂ ∂y

y

 kˆ  ∂  ˆ ˆ ˆ ˆ ∂z  = i (0) + j (0) + k (−2xy) = −2xy k  0

ˆ dS = dxdy Normal to the surface, n ˆ = k.    → − ˆ (−2xy) kˆ · kdxdy ∇× F ·n ˆ dS = S S  1  2 xy dxdy = −2 x=0 1

 = −2

y=0 " y2

!

dx x 2

  4 = −2 dx x 2 0

2 1 x = −4 = −2 2 0 0 1

5.2 Stokes’ Theorem

Thus,

:

− − → F · d→ r =



C

153



→ − ∇× F ·n ˆ dS = −2

S

which verifies Stokes’s theorem. Illustration 5.10: Use Stokes’s theorem to evaluate → − − → → − ˆ ˆ ˆ C F · d r if F = (x + y) i + (2x − z) j + (y + z) k and C is the boundary of the triangle (2, 0, 0) , (0, 3, 0) and (0, 0, 6) (See Figure 5.10). Solution:  ˆj  ˆi kˆ →  ∂ − ∂ ∂ curl F =  ∂x ∂y ∂z  x + y 2x − z y + z

    = 2ˆi + kˆ  

Let φ = 3x + 2y + z = 6 ∴n ˆ=

3ˆi + 2ˆj + kˆ ∇φ √ = |∇φ| 14

  3ˆi + 2ˆj + kˆ → − 7 √ curl F · n ˆ = 2ˆi + kˆ · =√ 14 14 Projection on xy-plane: √ 14 dxdy   dxdy dS =  =  1 ˆ · kˆ n

Figure 5.10 Represents the boundary of the triangle

154

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

∴ By Stokes’s theorem we have, √  :  → → − → − 7 14 − √ · F ·dr = dxdy curl F · n ˆ dS = 1 14 C S S  = 7dxdy S

= 7 × Area of the triangle on xy-plane 1 = 7 × × 3 × 2 = 21 2

5.3 Gauss’ Divergence Theorem → − Statement: If F be a vector point function having continuous partial derivatives in the region bounded by a closed surface S, then   − → → − F ·n ˆ dS ∇ · F dV = V

S

where n ˆ is the unit outward normal at any point of the surface S. → − ˆ Proof: Let F = F1ˆi + F2 ˆj + F3 k,   

 → − ∂ ∂  ˆ ∂ ˆ ˆ ˆ +j +k F1 i + F2 ˆj + F3 kˆ dxdydz i ∇ · F dV = ∂x ∂y ∂z V   V

∂F1 ∂F2 ∂F3 = + + dxdydz (5.10) ∂x ∂y ∂z V Assume a closed surface S such that any line parallel to the coordinate axes intersects S at most at two points. Divide the surface S into two parts: S1 − the lower and S2 − the upper part. Let z = f1 (x, y) and z = f2 (x, y) be the equations and n ˆ 1 and n ˆ 2 be normal to the surfaces S1 and S2 respectively (See Figure 5.11). Let R be the projection of the surface S on the xy-plane. %   $ f2 (x,y) ∂F3 ∂F3 dxdydz = dz dxdy f1 (x,y) ∂z V ∂z R  F3 (x, y, z) |ff21 dxdy = R  = [F3 (x, y, f2 ) − F3 (x, y, f1 )] dxdy R

5.3 Gauss’ Divergence Theorem

155

Figure 5.11 Represents the region bounded by a closed surface S





= R

F3 (x, y, f2 ) dxdy −

R

F3 (x, y, f1 ) dxdy

(5.11)

Now, dxdy = Projection of dS on xy-plane = n ˆ · kˆ dS ˆ 2 · kˆ dS2 For surface S2 : z = f2 (x, y) ⇒ dxdy = n For surface S1 : z = f1 (x, y) ⇒ dxdy = −ˆ n1 · kˆ dS1 Substituting in the Equation (5.11),    ∂F3 ˆ ˆ dS1 dxdydz = F3 n ˆ 2 · k dS2 − F3 (− n ˆ 1 · k) V ∂z S2 S1   = F3 n ˆ 2 · kˆ dS2 + F3 n ˆ 1 · kˆ dS1 S2 S1  = F3 n ˆ · kˆ dS (5.12) S

Similarly, projecting the surface S on yz and zx− planes, we get   ∂F1 dxdydz = F1 n ˆ · ˆi dS (5.13) V ∂x S 

 ∂F2 F2 n ˆ · ˆj dS (5.14) dxdydz = V ∂y S Substituting Equations (5.12), (5.13), and (5.14) in the Equation (5.10),     → − ∇ · F dV = F1 n ˆ · ˆi dS + F2 n ˆ · ˆj dS + F3 n ˆ · kˆ dS

and

V

S

S

S

156

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

  = S

  = S

Hence,

V



F1ˆi · n ˆ + F2 ˆj · n ˆ + F3 kˆ · n ˆ

dS 

 F1ˆi + F2 ˆj + F3 kˆ · n ˆ dS =

− → ∇ · F dV =

− → F · n ˆ dS

S

− → ˆ dS S F · n

Note: The cartesian form of Gauss’ divergence theorem is  

 ∂F1 ∂F2 ∂F3 + + dxdydz = (F1 dydz + F2 dzdx + F3 dxdy) ∂x ∂y ∂z V S → − Illustration 5.11: Verify Gauss’ divergence theorem for F = 4xzˆi − y 2 ˆj + yz kˆ over the cube x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. Solution: By Gauss’s divergence theorem,   − → → − F · n ˆ dS ∇ · F dV = V

S

− → (i) F = 4xzˆi − y 2 ˆj + yz kˆ → − ∂  2 ∂ ∂ (4xz) + −y + (yz) ∇· F = ∂x ∂y ∂z = 4z − 2y + y = 4z − y (ii) For the cube (in Figure 5.12): x varies from 0 to 1 y varies from 0 to 1 z varies from 0 to 1 

− → ∇ · F dV = V

 

1 1 1

0

1 1

0



=

0

0

(4z − y)dxdydz

 2z 2 − yz 1 dxdy

0

0



1

1

dx

= 0

0

(2 − y)dy

1   y 2   = 2y − 2  0 1 3 =2− = 2 2 |x|10

(a)

5.3 Gauss’ Divergence Theorem

157

Figure 5.12 Represents the cube

(iii) Surface S of the cube consist of 6 surfaces, S1 , S2 , S3 , S4 , S4 , S5 , and S6 .    → − → − − → F · n ˆ dS = F · n ˆ dS + F · n ˆ dS+ ∴ 

S

− → F · n ˆ dS +



S3

S1

− → F · n ˆ dS + S4



S2

− → F · n ˆ dS S5



− → F · n ˆ dS

+

(b)

S6

ˆ dS = 1. On S1 (OABC) : z = 0, n ˆ = −k,

dxdy

|nˆ ·kˆ|

= dxdy

x and y both vary from 0 to 1.       → − F · n ˆ dS = 4xzˆi − y 2 ˆj + yz kˆ · −kˆ dxdy = 0 ∴ S1

S1

ˆ dS = 2. On S2 (DEF G) : z = 1, n ˆ = k,

dxdy

|nˆ ·kˆ|

= dxdy

x and y both vary from 0 to 1.       → − F · n ˆ dS = 4xzˆi − y 2 ˆj + yz kˆ · kˆ dxdy ∴ S2

S2

1



1

=

  1  2 1 y



ydxdy = 0

0

0

  = 1 |x|1 = 1 2 2 0 2 0

158

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

3. On S3 (OAF E) : y = 0, n ˆ = −ˆj, dS =

dzdx |ˆ n·j|

= dzdx

x and z both vary from 0 to 1.       → − F · n ˆ dS = 4xzˆi − y 2 ˆj + yz kˆ · −ˆj dzdx = 0 ∴ S3

S3

ˆ = ˆj, dS = 4. On S4 (BCDG) : y = 1, n x and z both vary from 0 to 1.   → − F · n ˆ dS = ∴ s4

dzdx |ˆ n·j|

= dzdx



   4xzˆi − y 2 ˆj + yz kˆ · ˆj dzdx

S4 1 1



−dzdx = −1

= 0

0

dydz |nˆ ·ˆi|

5. On S5 (OABC) : x = 0, n ˆ = −ˆi, dS =

= dydz

y and z both vary from 0 to 1.       → − F · n ˆ dS = 4xzˆi − y 2 ˆj + yz kˆ · −ˆi dydz = 0 ∴ S5

S5 dydz |nˆ ·ˆi|

ˆ = ˆi, dS = 6. On S6 (ABGF ) : x = 1, n y and z both vary from 0 to 1.   → − F · n ˆ dS = ∴ S6

= dydz



  4xzˆi − y 2 ˆj + yz kˆ · ˆi dydz

S6 1 1



4zdzdx = 2

= 0

0

Substituting in the equation (b),  → − 1 3 F · n ˆ dS = 0 + + 0 − 1 + 0 + 2 = ∴ 2 2 S From equations (a) and (c), we get   − → → − 3 F · n ˆ dS = ∇ · F dV = 2 V S

(c)

5.3 Gauss’ Divergence Theorem

159

Hence, the Gauss theorem is verified.

→ − Illustration 5.12: Verify Gauss’ divergence theorem for F = 2xzˆi + yzˆj + z 2 kˆ over the upper half of the sphere x2 + y 2 + z 2 = a2 . Solution: By Gauss’s divergence theorem,   → − ∇ · F dV = V

− → F · n ˆ dS S

→ − (i) F = 2xzˆi + yzˆj + z 2 kˆ → − ∂ ∂ ∂ 2 ∇· F = (2xz) + (yz) + (z ) ∂x ∂y ∂z = 2z + z + 2z = 5z (ii)



− → ∇ · F dV = V

 5z dxdydz V

Taking x = rsinθ cosφ , y = rsinθ sinφ , z = rcosθ , equation of the sphere x2 + y 2 + z 2 = a2 reduces to r = a and dxdydz = r2 sinθ dr dθ dφ. For the upper half of the sphere (hemisphere in Figure 5.13), r varies from 0 to a

Figure 5.13 Represents the cube

160

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

θ varies from 0 to

π 2

φ varies from 0 to 2π   → − ∇ · F dV = 5 V

2π φ=0 2π



π 2

θ=0



a 0

rcosθ · r2 sinθ drdθdφ

 4 a r  dφ cosθ sinθ dθ   =5 4 0 φ=0 θ=0  π 5 1  cos2θ  2 a4 − = πa4 = 5|φ|2π · · 0   2 2 0 4 4 



π 2

(a)

(iii) The given surface is open and closed by the circular surface S2 in xyplane. Thus, the surface S consists of two surfaces S1 and S2 .    → − → − → − F ·n ˆ dS = F ·n ˆ dS + F ·n ˆ dS (b) ∴ S

S1

S2

(1) Surface S1 (ABCEA) : This is the curved surface of the upper half of the sphere. Let φ = x2 + y 2 + z 2 ∴n ˆ=

2xˆi + 2yˆj + 2z kˆ ∇φ xˆi + yˆj + z kˆ = = |∇φ| a 4x2 + 4y 2 + 4z 2

Let R be the projection of S1 on the xy-plane, which is a circle x2 + y 2 = a2 . adxdy dxdy = dS =   z ˆ · kˆ n  → − F ·n ˆ dS S1

  = R



2xzˆi + yzˆj + z 2 kˆ ·



xˆi + yˆj + z kˆ a



adxdy z

5.3 Gauss’ Divergence Theorem



161



 2x2 + y 2 + z 2 dxdy

=  R =  R =



 2x2 + y 2 + a2 − x2 − y 2 dxdy



 x2 + a2 dxdy

R

Taking x = rcosθ , y = rsinθ ,the equation of the circle x2 +y 2 = a2 reduces to r = a and dxdy = r dr dθ. Along the radius vector OP , r varies from 0 to a and for the complete circle, θ varies from 0 to 2π (See Figure 5.14).  S1



2π



 r2 cos2 θ + a2 r dr dθ 0 0  2 a "  2π ! 4 a   r  2 2 r    =  4  cos θ + a  2  dθ 0 0  0 4"  2π ! 4

1 + cos2θ a a + dθ = 4 2 2 0    1 sin2θ 2π 5 4 5 = πa4 =a  θ+  8 8 2 0 4

− → ∇× F ·n ˆ dS =

a

Figure 5.14 Represents the circle

162

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

(2) Surface S2 (ABCDA) : This is the circle x2 + y 2 = a2 in xy−plane z = 0, n ˆ = −kˆ dxdy  = dxdy ∴ dS =   ˆ · kˆ n     → − F ·n ˆ dS = 2xzˆi + yzˆj + z 2 kˆ · −kˆ dxdy = 0

 ∴ S2

S2

Substituting in Equation (b),  ∴

− → 5 F ·n ˆ dS = πa4 4 S

From Equations (a) and (c), we have   → − ∇ · F dV = V

(c)

− → 5 F ·n ˆ dS = πa4 4 S

Hence, Gauss’ theorem is verified.   − → Illustration 5.13: Evaluate S yz ˆi + zx ˆj + xy kˆ · d S , where S is the surface of the sphere in the first octant. Solution: By Gauss’ divergence theorem,   → − − → F · dS = S

→ − ∇ · F dV

(i)

V

− → F = yz ˆi + zx ˆj + xy kˆ → − ∂ ∂ ∂ (yz) + (zx) + (xy) = 0 ∴∇ · F = ∂x ∂y ∂z From equation (i),   → − − → − → F · dS = ∇ · F dV = 0 S

V

Illustration 5.14: Evaluate S (lx + my + nz) dS, where l, m, n are the direction cosines of the outer normal to the surface whose radius is 2 units. Solution: By Gauss’ divergence  theorem,  → − → − F ·n ˆ dS = ∇ · F dV S

V

(i)

5.4 Exercise

163

→ − (i) F · n ˆ = lx + my + nz     = x ˆi + y ˆj + z kˆ · l ˆi + m ˆj + n kˆ → − ∴ F = x ˆi + y ˆj + z kˆ → − (ii) ∴∇ · F =

∂ ∂x

(x) +

∂ ∂y

(y) +

∂ ∂z

(z) = 3

(iii) 

− → ∇ · F dV = V



3 dV

V  Volume of the region bounded by =3 the sphere of 2 unit radius 4 = 3 · π(2)3 = 32 π 3

From Equation (i),

5.4 Exercise

 →  − 1. Verify Green’s theorem for the function F = y 2 − 7y ˆi+(2xy +2x)ˆj and curve C : x2 + y 2 = 1. (Answer: 9π) 9 

 2

2. Verify Green’s theorem for C 3x − 8y dx + (4y − 6xy)dy where C is the boundary of the triangle with vertices (0, 0) , (1, 0) and (0, 1).   Answer : 53 9 3. Using Green’s theorem, evaluate the linear integral C (siny dx +cosxdy) counterclockwise, where C is the boundary of the triangle with vertices (0, 0) , (π, 0) , (π, 1).   Answer : − 1 + π2 − π1 cos1  . 9 # 4. Evaluate C x2 − 2xy dx + (x2 y + 3)dy around the boundary C of the region y 2 = 8x, x = 2.   Answer : 128 5

164

Green’s Theorem, Stokes’ Theorem, and Gauss’ Theorem

 9  5. Verify Green’s theorem in the plane for C 2x − y 2 dx − xydy where C is the boundary of the region enclosed by the circles x2 + y 2 = 1 and x2 + y 2 = 9. (Answer : 60π)  6. Verify Green’s theorem for C x2 − 2xy dx + x2 y + 3 dy where C 2 is  the boundary  of the region bounded by y = x and the line y = x. 1 Answer : 4  →  − 7. Verify Stokes’s theorem for F = x2 − y 2 ˆi + 2xyˆj, in the rectangular region in the xy-plane bounded by the lines x = ±a, y = 0, y = b.   Answer : 4ab2 

9 



→ − 8. Verify Stokes’s theorem for F = (x + y) ˆi + (y + z) ˆj − xkˆ and S is the surface of the plane 2x + y + z = 2 which is in the first octant.

9. Evaluate by Stokes’s theorem curve x2 + y 2 = 4, z = 2.

(Answer : − 2)

9 C

(ex dx + 2ydy − dz), where C is the (Answer : 0)

→ − 10. Verify Stokes’s theorem for F = xzˆi+yˆj +y 2 xkˆ over the surface of the tetrahedron bounded by the planes y = 0, z = 0 and 4x + y + 2z = 4 above the yz-plane. (Answer : 0)  →  − 11. Verify Stokes’s theorem for F = x2 +y + 2 ˆi + 2xyˆj + 4zex kˆ over the surface S of the paraboloid z = 9 − x2 + y 2 above the xy-plane. (Answer : − 9π) − → → − ˆ dS where F = 3yˆi − 12. Evaluate using Stokes’s theorem S ∇ × F · n xzˆj + yz 2 kˆ and S is the surface of the paraboloid x2 + y 2 = 2z bounded by the plane z = 2 and C is its boundary traversed in the clockwise direction. (Answer : 20π) dS √ S a2 x2 +b2 y 2 +c2 z 2 c2 z 2 = 1.

13. Prove that b2 y 2 +

=

√4π , abc

where S is the ellipsoid a2 x2 +

5.4 Exercise

165

→ − − → → − 14. Evaluate S F · d S using divergence theorem where F = x3 ˆi+y 3 ˆj + z 3 kˆ and S is the surface of the sphere x2 + y 2 + z 2 = a2 .   5 Answer : 12 5 πa → − − → → − 15. Evaluate F · d S using Gauss’ divergence theorem where F 2xy ˆi + yz 2 ˆj +zx kˆ and S is the surface of the region bounded by x = 0, y = 0, z = 0, y = 3, x + 2z = 6.   Answer : 351 2 → − 16. Verify Gauss’ divergence theorem for F = (x2 − yz)ˆi + (y 2 − zx)ˆj + (z 2 − xy)kˆ over the region R bounded by the parallelepiped 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c. (Answer : abc(a + b + c)) → − 17. Verify Gauss’ divergence theorem for F = xˆi + yˆj + z kˆ over the region R bounded by the sphere x2 + y 2 + z 2 = 16. (Answer : 256π) → − 18. Verify Gauss’ divergence theorem for F = 2xyˆi + 6yzˆj + 3zxkˆ over the region bounded by the coordinate planes and the plane x+y +z = 2.   Answer : 22 3

6 MATLAB Programming

6.1 Basic of MATLAB Programming MATLAB is an acronym for MATrix LABoratory. MATLAB was created to make it simple to use the LINPACK (linear system package) and EISPACK (eigen system package) projects’ matrix tools. MATLAB, see The MathWorks Inc., 2005, and D. Houcque, 2005, is a high-performance technical computing language. It combines computing, visualization, and a programming environment into one package. MATLAB is also a contemporary programming language environment, with advanced data structures, built-in editing and debugging tools, and object-oriented programming capabilities. Because of these features, MATLAB is an outstanding teaching and research tool. In this section, we presume that the reader is familiar with basic terminology like different windows in MATLAB, basic input commands, creating MATLAB variables, error messages, functions, etc. 6.1.1 Basic of MATLAB Programming Many mathematical functions are predefined in MATLAB. By typing_help , and help users can access inbuilt commands for elementary and special mathematical functions. A large number of mathematical functions can be evaluated using MATLAB. There are two types of arithmetic operations in MATLAB: array operations and matrix operations. Users may execute mathematical computations using these arithmetic operations, such as adding two integers, raising the elements of an array to a specific power, or multiplying two matrices.

167

168

MATLAB Programming

6.1.1.1 Introductory MATLAB programmes In this section, we discuss some basic examples of MATLAB programming to recall basic operations, functions, inputs, and mathematical computations. And it will help us in further MATLAB programs. 1. Write a MATLAB program to find the addition, subtraction, and division of two numbers. Solution: Let a = 3 and b = 5 be two numbers for these operations. Assume c1, c2, and c3 variables for storing these results. So, c1, c2, and c3 are the outputs for these operations respectively. Therefore >> a=3; >> b=5; >> c1=a+b c1 = 8 >> c2=a-b c2 = -2 >> c3=a/b c3 = 0.6000 In this example, the semicolon at the end of the statements for a and b, suppress output to run the program smoothly. Whereas in the statement c1=a+b, c2=a-b, and c3=a/b are not followed by a semicolon so the stored values “dumped” as the output.

6.1 Basic of MATLAB Programming

169

2. Write a MATLAB program to evaluate the given expression for any two numbers. b 2a + b2 − ab + − 10 a Solution: In this MATLAB program, write a user-defined program. Users define the programs are those programs in which the input of the programs are taken from the user. The main advantage of the user define programs are, the user can give any input rather than a fixed input. In this program, we use the “input” command to take values from the user. The syntax is given as follow:

Input:

Output: Enter 1st number: -10 Enter 2nd number: 9 c= 140.1000 In this example, is given for the newline. i.e., once the program is run user can see the input statement, and the user can give input in the next line in the “command window”. 3. Write a MATLAB program to print “Welcome to MATLAB Programming” Solution:

170

MATLAB Programming

Input: >> fprintf (‘Welcome to MATLAB Programming \n’) Output: Welcome to MATLAB Programming In this program, “fprintf (‘ ’)” is used to print given input in a single quotation mark (‘ ’). 4. Write a user define MATLAB program to print square, cube, and square roots to preserve 4 digits after the decimal point for any number. Solution: Input:

Output: 1) If the number is 4 then Enter the number : 4 The square of 4 equal 16. The cube of 4 equal 64. The square root of 4 is 2.0000. 2) If the number is 5 then Enter the number : 5 The square of 5 equal 25. The cube of 5 equal 125. The square root of 5 is 2.2361.

6.1 Basic of MATLAB Programming

171

In this program, “%4u” represents 4 digit integer value of the input number, and to preserve 4 digits after the decimal point we use “%6.4f”. 5. Write a MATLAB program to plot sinx for 0 ≤ x ≤ 2π. Solution: Input:

Output:

In this program, to plot sinx for 0≤x≤2π, use the inbuilt command “plot(x,y)” then define x values using the command x=0:0.1:2*pi then store the values of sinx in variable y. 6. Write a MATLAB program to draw any curve. Solution: Input:

Output:

172

MATLAB Programming

7. Write a MATLAB program to plot user define line pattern, color, and thickness. Solution: Input:

Output:

6.1 Basic of MATLAB Programming

8. Write a MATLAB program to plot multiple curves. Solution: Input:

Output:

9. Write a MATLAB program to plot multiple panels. Solution: Input:

173

174

MATLAB Programming

Output:

10. Write a MATLAB program to plot a helix. Solution: Input:

6.1 Basic of MATLAB Programming

175

Output:

11. Write a MATLAB program to print any number consecutive 5 times. Solution: In this program, we take input from the user as we have to print any number and use “for loop” for printing any number consecutive 5 times. Input:

Output: Enter the number : 10 x= 10 x= 10 x= 10 x=

176

MATLAB Programming

10 x= 10 12. Write a MATLAB program to print the utility of dummy index for any number consecutive 6 times. Also, take index from user. Solution: Input:

Output: Enter the number : 2 ans = 2 ans = 4 ans = 8 ans = 16 ans = 32 ans = 64 Also, by user define index Input

6.1 Basic of MATLAB Programming

Output: Enter the number : 2 Enter the index : 3 ans = 2 ans = 4 ans = 8 13. Write a MATLAB program to find a sum upto 10. Solution: Input:

Output: sum1 = 55 By user define number

177

178

MATLAB Programming

Input:

Output: Enter a number upto which sum need to find: 5 a= 5 sum1 = 15 14. Write a MATLAB program to find the sum of the first four elements in an array of six elements. Solution: Input:

Output: sum1 = 5 By user define

6.1 Basic of MATLAB Programming

Output: Enter an array : [1 2 3 4 5 6 7 8 9 10] a= Columns 1 through 8 1 2 3 4 5 6 7 8 Columns 9 through 10 9 10 Enter a number to find sum upto the given number of elements: 7 b= 7 sum1 = 28 15. Write a MATLAB a program for a double loop. Solution: Input:

Output: n= n= n= n= n= n=

1m= 1m= 2m= 2m= 3m= 3m=

1 2 1 2 1 2

16. A MATLAB program for complicated use of loop and index. Solution: Input:

179

180

MATLAB Programming

Output: sum1 = 24 Summation performed by this program sum1 = a(b(1))+a(b(2))+a(b(3))+a(b(4)) = a(2)+a(3)+a(5)+a(7) = 5+7+9+3 = 24 17. Write a MATLAB program to find greater than or less than or equals to of any two numbers. Solution: Input:

Output 1: Enter 1st number : 10 a=

6.1 Basic of MATLAB Programming

181

10 Enter 2nd number : 12 b= 12 10 is less than or equal to 12 Output 2: Enter 1st number : 10

a= 10

Enter 2nd number : -1 b= -1

10 is greater than -1.000000e+00 18. Write a MATLAB program to represent the use of “if - elseif - else” loop. Solution: Input:

182

MATLAB Programming

Output: 4 is less than 5 but greater than 1 19. Write a MATLAB program to determine whether a given year is a leap year or not. Solution: Input:

Output: Enter a year in YYYY format : 2020 2020 is a leap year Enter a year in YYYY format : 2022 2022 is not a leap year

6.1 Basic of MATLAB Programming

183

6.1.1.2 Representation of a Vector in MATLAB In this section, we represent a vector as a special case of a matrix. An array of dimension 1×n is called a row vector and m×1 dimensional array is called a column vector. The elements of a vector are enclosed by square brackets. The elements of a row vector are separated using spaces or by commas, whereas the elements of a column vector are separated using a semicolon(;). For example, for a row vector v1 and v2 >> v1=[1 2 3 4 5] v1 = 1

2

3

4

5

>> v2=[1, 2, 3, 4, 5] v2 = 1

2

3

4

5

And for a column vector cv1 and cv2

>> cv1=[1;-4;3;9;-10] cv1 =

1 -4 3 9 -10

>> cv2=[1;2;-2;0.1]

MATLAB Programming

184

cv2 = 1.0000 2.0000 -2.0000 0.1000 The transpose of a row or a column vector is obtained using an apostrophe or a single quote (’). For example, transpose of a row vector v1 and a column vector cv1

>> w=v1’

w=

1 2 3 4 5 >> w1=cv1’ w1 = 1

-4

3

9 -10

To access elements, we use colon notation (:). For example, for the first three elements of the w1 vector, >> w1(1:3) ans = 1

-4

3

6.1 Basic of MATLAB Programming

185

Or, all elements after 3rd element >> w1(3:end)

ans = 3

9 -10

Similarly, we can find elements in the column vector also. 20 Write a MATLAB program to find the addition of two vectors. Solution: Input: a = [2 12 25]; b = [3 7 4]; c = a+b Output: c= 5 19

29

21 Write a MATLAB program to find the multiplication of a vector by a scalar. Solution: Input: a = [2 12 25]; b=4 c = a*b Output: c= 8

48 100

186

MATLAB Programming

6.1.1.3 Representation of a Matrix in MATLAB A matrix is an array of numbers. We use the following steps to enter a matrix in MATLAB. 1. We begin with a left square bracket. i.e., [ 2. Separate each element in a row with space or comma. i.e., (,) 3. Use a semicolon (;) to separate each row in the matrix. 4. After adding all rows in a matrix, end matrix with the right square bracket. i.e., ] For example, to enter a matrix

! A=

1 2 3 4

"

We use >> A=[1 2; 3 4] and the output matrix in MATLAB will be A= 1 3

2 4

For non-square matrix

⎤ 1 2 A=⎣ 3 4 ⎦ 5 6 ⎡

We have, >> A=[1, 2; 3, 4; 5, 6] A= 1 3 5

2 4 6

Once the matrix is stored in MATLAB, we can view a particular element of the matrix. For example, to view elements in row 1st and column 2nd, we use A(2,1)

6.1 Basic of MATLAB Programming

187

>> A(2,1) ans = 3 To find the dimension of a vector or a matrix, we use the size command. For example, to find the dimension of the matrix ⎤ ⎡ 1 2 A=⎣ 3 4 ⎦ 5 6 We have >> size(A) ans = 3

2

For a square matrix

! A=

1 2 3 4

"

We have >> size(A) ans = 2

2

i.e., size(A)=[m n] To find the transpose of a matrix, we can use an apostrophe or a single quote (’). For example, transpose of a matrix ! " 0 1 3 B= −2 0 7 We have

MATLAB Programming

188

B= 0

1

3

-2

0

7

>> B’ ans = 0 1 3

-2 0 7

6.2 Some Miscellaneous Examples using MATLAB Programming

Solution:

6.2 Some Miscellaneous Examples using MATLAB Programming

Output:

Solution:

Input: x-coordinate of a vector: 3 y-coordinate of a vector: 4 z-coordinate of a vector: -5 Output: The modulus of the vector (3,4,-5) = 7.071068

Solution: a = [3 -1 -4]; b = [-2 4 -3]; c = [1 2 -1];

189

190

MATLAB Programming

z=3*a-2*b+4*c Output: z= 17

-3

-10

Solution: a = [1 2 1]; b = [2 1 1]; c = [3 4 1];

z1 = a+2*b+c z2 = z1.*z1 z3 = sum(z2) z4 = sqrt(z3) Output: z1 = 8

8

4

64

64

16

z2 =

z3 = 144 z4 = 12

6.2 Some Miscellaneous Examples using MATLAB Programming

Solution:

Input-1: x-coordinate of a vector: 1 y-coordinate of a vector: 0 z-coordinate of a vector: 0 Output-1: The vector (1, 0, 0) is a unit vector. Input-2: x-coordinate of a vector: 1 y-coordinate of a vector: 1 z-coordinate of a vector: 0 Output-2: The vector (1, 1, 0) is not a unit vector. Input-3: x-coordinate of a vector: 0 y-coordinate of a vector: 0 z-coordinate of a vector: 0

191

MATLAB Programming

192

Output-3: The vector (0, 0, 0) is a zero vector. Input-4: x-coordinate of a vector: 1/sqrt(3) y-coordinate of a vector: 0 z-coordinate of a vector: sqrt(2/3) Output-4: The vector (5.773503e-01, 0, 8.164966e-01) is a unit vector.

Solution:

Input: The first vector in the form of [x y z]: [-1 6 8] The second vector in the form of [x y z]: [1 3 -4] Output: z1 = 0

9

4

3

12

z2 = -2

6.2 Some Miscellaneous Examples using MATLAB Programming

Solution:

Input: Enter x-coordinate of an arbitrary vector : 3 Enter y-coordinate of an arbitrary vector : 4 Enter z-coordinate of an arbitrary vector : 5 Enter an arbitrary scalar for a product with a given arbitrary vector : -1 Output: a= 3

4

5

-4

-5

da =

-3

193

194

MATLAB Programming

Solution:

Output: z1 = 20 z2 =

0.0000e+00 + 2.3220e+02i

x.y = 20 The angle between x and y is 0.000000.

Solution:

6.2 Some Miscellaneous Examples using MATLAB Programming

Input 1: Enter a first vector in the format [x y z] = [1 2 3] Enter a second vector in the format [x y z] = [-1 -2 -3] Output 1: z1 = -14 z2 = 1.8000e+02 - 1.9085e+02i x.y =-14 The angle between x and y is 180.000000 Input 2: Enter a first vector in the format [x y z] = [1 0 0] Enter a second vector in the format [x y z] = [0 1 0] Output 2: z1 = 0 z2 = 90 x.y =0

195

MATLAB Programming

196

The angle between x and y is 90.000000 Input 3: Enter a first vector in the format [x y z] = [1 -5 6] Enter a second vector in the format [x y z] = [6 3 -1] Output 3: z1 = -15 z2 = 1.8000e+02 - 1.8624e+02i x.y =-15 The angle between x and y is 180.000000

Solution:

6.2 Some Miscellaneous Examples using MATLAB Programming

Input 1: Enter a first vector in the format [x y z] = [1 -2 2] Enter a second vector in the format [x y z] = [0 0 -1] Output 1: (i) is true. 2.000000 < = 3.000000 (ii) is true. 2.449490 < = 4.000000 (iii) is true. 3.741657 > = 2.000000 Input 2: Enter a first vector in the format [x y z] = [1 -1 -1] Enter a second vector in the format [x y z] = [2 -4 7] Output 2: (i) is true. 8.306624 < = 14.387495 (ii) is true. 8.366600 < = 10.038675 (iii) is true. 8.602325 > = -6.574573

Solution:

197

198

MATLAB Programming

Input: Enter a first vector in the format [x y z] = [3 -2 1] Enter a second vector in the format [x y z] = [1 -2 1] Output: The projection of a on b is 3.265986. The projection of b on a is 2.138090.

Solution:

Input 1: Enter a first vector in the format [x y z] = [1 0 0] Enter a second vector in the format [x y z] =

6.2 Some Miscellaneous Examples using MATLAB Programming

[0 1 0] Output 1: c= 0 The vectors a and b are perpendicular to each other. Input 2: Enter a first vector in the format [x y z] = [1 0 0] Enter a second vector in the format [x y z] = [1 0 0] Output 2: c= 1 The vectors a and b are not perpendicular to each other. Input 3: Enter a first vector in the format [x y z] = [1 2 3] Enter a second vector in the format [x y z] = [-2 1 3] Output 3: c= 9 The vectors a and b are not perpendicular to each other.

Solution:

199

200

MATLAB Programming

Input: Enter a first vector in the format [x y z] = [2 -3 1] Enter a second vector in the format [x y z] = [1 -1 2] Output: c= -5 -3 1 The cross product is (-5, -3, 1).

Solution:

Output: The total work done is 10.000000 units.

Solution:

6.2 Some Miscellaneous Examples using MATLAB Programming

Output: The moment of the force F about the point A is (-3, 19, -4). The magnitude of the moment is 19.646883.

Solution: t = 0:0.1:3*pi; plot3(sin(2*t),cos(t),t) f = @(t) sqrt(4*cos(2*t).∧ 2 + sin(t).∧ 2 + 1); len = integral(f,0,3*pi) Output:

201

202

MATLAB Programming

len = 17.2220

xe−x Solution:

Output:

For the function: 3x2 y − y 3 z 2

2 −y 2

6.2 Some Miscellaneous Examples using MATLAB Programming

Solution: t = 0:0.1:2*pi; f = @(t)(-t.*sin(t)+cos(t).∧2+sin(t)); len = integral(f,0,2*pi) Output: len = 9.4248

Solution:

203

204

MATLAB Programming

Output:

Solution:

Output: q1 = 2.734244598320928e+03 q2 = 2.734244599944285e+03

Solution: fun = @(x,y) 1./(1 + x + y);

References

205

ymax = @(x) x; q = integral2(fun,0,1,0,ymax) Output: q= 0.261624071883185

References [1] The MathWorks Inc. MATLAB 7.0 (R14SP2). The MathWorks Inc., 2005. The MathWorks Inc. MATLAB 7.0 (R14SP2). The MathWorks Inc., 2005. [2] D. Houcque. Introduction to MATLAB for Engineering Students. Northwestern University, 2005.

Index

A

Absolute, 3 Acceleration, 1, 106 Access, 167, 184 Accurate, xi Acronym, 167 Addition, v, 5, 6, 9, 168 Adjacent, 38 Adjoining, 58 Advance, 66 Aircraft, 55 Algebraic, 29, 37 Analogous, 73 Analysis, 1 Analytic, ix Angle, vi, ix, 12, 29 Angled, 29 Angles, 12 Angular, 89 Applied, 46, 48, 49, 50 Arbitrary, 1, 92, 93, 95 Arc Length, vi, 57, 61 Arc rate, xi, 65 Area, vii, 1, 39, 122, 128 Arithmetic, x, 167 Arrow, 1 Atmosphere, 56 Axis, 2, 4, 5, 111, 137 B

Basic, v, 45, 55, 167, 168

Binormal, 60, 65 Bold, 1 Boundary, 135, 144, 153 Bounded, xii, 124, 129, 134 C

Cables, 29 Calculus, vi, 111 Capabilities, 167 Cartesian, 2, 4, 58 Centrifugal, 1 Charge, 1 Circle, 64, 117, 126, 143 Circular, 127, 133, 160 Circulation, vii, 112, 117 Clockwise, 66 Closed, 112, 114, 123, 129 Coefficients, 120 Collinear, v, 8 Command, 167, 168, 171, 187 Commands, x, 167 Common, 8 Commutative, 10, 30 Component, 71, 124, 128, 137 Components, v, 2, 5, 13, 25 Computations, 167 Concept, ix, 1, 7, 55 Cone, 76 Consecutive, 38, 60, 175, 176 Conservative, 91, 113, 114, 120 Constant, 48, 54, 56, 72, 74

207

208

Index

Construction, 55 Contemporary, 167 Coordinate, 2, 4, 58, 71 Coplanar, v, 8, 107 Cosine, 12 Cosines, vi, ix, 12, 16 Counterbalance, 25 Critical, 55 Cross, 37, 38, 39, 60 Cube, 156, 157, 159, 170 Curl, 55, 73, 86, 89 Curvature, 60, 64, 67, 105 Curve, 67, 73, 82, 84 Curves, 57, 58, 84, 106 Cylinder, 129, 134 D

Debugging, 134 Decimal, 170, 172 Decreases, 75, 81 Density, 1, 56 Dependent, 1 Derivation, ix Derivative, xii, 64, 71, 80 Derivatives, 64, 74, 75, 135 Difference, 10 Different, 1, 8, 10, 56, 167 Differentiable, 71, 86, 89, 146 Differential, vi, ix, 55, 73, 111 Differentiating, 147 Differentiation, vi, 111 Digit, 171 Digits, 170, 171 Dimensional, ix, 2, 3, 5, 57 Dimensions, 12, 55, 76 Directed, 1 Direction, ix, xi, 1, 2, Directional, xii, 73, 74, 75

Directions, 7, 8, 10, 31 Displaced, 55 Displacement, 1, 46, 48, 51 Displacing, 112 Distance, 1, 74 Distributive, 10, 31, 40, 42 Divergence, vii, 55, 73, 86 Dot, xi, 29, 30, 40 Drag, 1 Dumped, 168 E

Elasticity, 55 Electric, 1, 45, 56, 87 Electrical, 1 Electromagnetic, ix, 103 Electromagnetism, 55, 112 Electrostatics, 55 Elements, 8, 31, 167, 178 Ellipse, 84, 127, 128, 129 Endpoints, 11, 58 Energy, 1 Environment, 167 Equal, 2, 10, 38, 59 Error, 158 Essentials, ix Evaluation, vii, 123, 135 Execute, 167 Expression, 10, 40, 42, 115 F

Field, 1, 45, 55, 56, 73 Flowing, 56, 86, 123 Fluid flow, 3, 57, 88 Flux, 87, 123, 127, 128 Force, 1, 25, 26, 46 Forces, 25, 46, 48, 54 Frequency, 1

Index 209

Function, 55, 57, 61, 73 Functions, 55, 56, 167, 158

Intersection, 56, 127, 144 Irrotational, 90, 92, 95, 112 Isothermal, 56

G

Gauss, 154, 156, 159, 162 Generators, 55 Geometric, ix, xi, 1, 8 Geometrical, 1, 30 Geometrically, 30, 55 Geometry, ix, 45, 111 Gradient, 55, 7, 74, 81 Gravitational, 1, 56 Gravity, 25 Green, 135, 137, 138, 140 H

Hangs, 29 Heat flow, 55 Helix, 63 Hemisphere, 159 Hexagon, 14 High-performance, 167 Horizontal, 25 I

Incompressibility, 87 Incompressible, 87 Increasing, 66 Independence, ix, 113 Independent, vii, 113, 114, 115 Independently, 1 Initial, 1, 2, 7, 9 Input, 167, 159, 171, 175 Integral, 111, 122, 129, 135 Integrals, 111, 112, 113, 135 Integration, 111, 113, 114, 117 Intensity, 1, 56 Interpretation, vii, 38, 86, 69

L

Laser, 55 Law, 8, 9, 10, 11 Length, 1, 3, 57, 59 Level curves, 84 Lift, 1 Line, vii, 8, 46, 58 Linear, 8, 89 Lines, 8, 85, 124, 136 Locus, 42 M

Machinery, ix Magnetic, 1, 56 Magnitude, 1, 8, 18, 86 Magnitudes, 1, 50 Mass, 1, 86 Mathematical, ix, 167, 168 MATLAB, ix, 167, 170, 178 Matrices, 167 Maximum, 74, 108 Measure, vi, 30 Measured, 1, 74 Measurement, ix, 48 Messages, 167 Moduli, 10, 16, 31 Modulus, 3, 4, 9, 47 Moment, 1, 46, 47, 51 Momentum, 1 Mutually, 32, 52, 60 N

Negative, 2, 3, 6, 10 Neighboring, 60

210

Index

Newline, 169 Normal, 60, 64, 74, 84 Null, 7, 39, 47 O

Octant, 124, 125, 126, 129 Operations, x, 167 Operator, vii, 73, 100 Opposite, 2, 6, 7, 10 Order, 60 Ordinary, 55, 73 Origin, 2, 11, 58, 117 Orthogonal, 66, 75, 81, 123 Osculating, 60, 64 Output, 168, 179, 186, 202 Outputs, x, 158 Outward, 86, 87, 146, 154

Points, 1, 2, 12, 60 Position, 2, 3, 12, 14 Positive, 6, 112, 126, 129 Potential, 1, 91, 93, 133 Power, 1, 167 Preserve, 170, 171 Principal, 60, 64 Print, 169, 175 Product, x, xi, 7, 200 Products, 29, 31, 45, 60 Program, 167, 168, 169, 179 Projecting, 148, 155 Projection, 30, 46, 73, 198 Projections, vi, 30 Q

Quantities, ix, 1, 55, 73 Quantity, 1, 46, 55

P

Paddlewheel, 90 Parabola, 114, 115, 144, 145 Parallel, 8, 39, 60, 86 Parallelepiped, 86, 87 Parallelogram, 8, 9, 13, 38 Parallelopiped, 87 Parameter, 57, 58, 61, 62 Parameters, 62 Parametric, 57, 67, 112, 117 Particle, 46, 48, 49, 51 Path, 112, 113, 115, 117 Perpendicular, 2, 31, 35, 47 Perpendiculars, 2 Physical, 1, 55, 86, 89 Physics, 45 Plane, 4, 47, 60, 122 Planes, 123, 129, 134, 155 Plot, 171, 173, 174, 201 Point, 1, 7, 47, 56

R

Radius, 56, 117, 128, 162 Rectangle, 140 Rectangular, 5, 152 Region, 55, 90, 112, 154 Regular, 14 Relativity, 111 Representation, ix, 1, 8, 112 Resultant, 9, 48, 49, 50 Revolution, 76 Right, 60, 66, 186 Robots, 55 Rotation, 64, 65, 89, 90 Rotational, 89 S

Scalar, ix, 1, 29, 46 Scalars, 1, 10, 55 Section, 6, 45, 64, 167

Index 211

Segment, 1, 58 Segments, 8, 57 Semicolon, 168, 183, 186 Shuttle, 25 Sinks, 87 Skyscrapers, 25 Smoothly, 168 Solenoidal, 87, 98, 123 Sources, 87 Space, 1, 12, 25, 55 Special, 167, 183 Speed, 1 Sphere, 56, 125, 126, 159 Stationary, 56 Steady-state, 56 Steel, 25 Stokes, ix, vii, 135, 146 Straight, 58, 131, 132 Subset, ix Subsurfaces, 122 Subtraction, 6, 168 Sum, 6, 8, 10, 178 Surface, 56, 74, 76, 108 Surfaces, 56, 74, 79, 122 Syntax, 168 System, 2, 4, 55, 60 T

Tangent, 58, 61, 64, 84 Tangents, 60, 72, 73 Taylorâs series, 86 Temperature, 1, 56 Tensions, 25 Terminal, 1, 7, 9, 12 Theorem, 113, 135, 145, 146 Thermo-dynamical, 55 Three, 2, 4, 12, 50 Thrust, 1

Time, 1, 56, 86, 123 Times, 5, 7, 175, 176 Tons, 25 Torsion, 64, 65, 69, 105 Totality, 56 Traced, 63 Transformation, 135 Transformed, 135 Triangle, 9, 10, 16, 124 U

Understand, 7, 12, 55, 79 Unit, 4, 5, 13, 18 Unity, 4, 5 V

Value, 1, 3, 19, 37 Variable, 55, 58, 115, 167 Variables, 115, 168 Vector, 1, 9, 18, 154 Vectors, 1, 5, 14, 60 Velocity, 1, 86, 89, 123 Vertex, 13 Vertical, 25, 125, 129, 145 Vertices, 13 Visualization, 167 Visualize, 55 Volume, 1, 87, 129, 163 Vortex, 90 W

Work, 1, 46, 48, 112 “Work done”, 45, 48, 49, 122 Z

Zero, 4, 45, 82, 114

About the Authors

Nita H. Shah received her PhD in Statistics from Gujarat University in 1994. From February 1990 until now Professor Shah has been Head of the Department of Mathematics in Gujarat University, India. She is a postdoctoral visiting research fellow of University of New Brunswick, Canada. Professor Shah’s research interests include inventory modeling in supply chains, robotic modeling, mathematical modeling of infectious diseases, image processing, dynamical systems and their applications, etc. She has published 13 monograph, 5 textbooks, and 475+ peer-reviewed research papers. Four edited books have been prepared for IGI-global and Springer with coeditor Dr. Mandeep Mittal. Her papers are published in high-impact journals such as those published by Elsevier, Interscience, and Taylor and Francis. According to Google scholar, the total number of citations is over 3334 and the maximum number of citations for a single paper is over 177. She has guided 28 PhD Students and 15 MPhil students. She has given talks in USA, Singapore, Canada, South Africa, Malaysia, and Indonesia. She was Vice-President of the Operational Research Society of India. She is VicePresident of the Association of Inventory Academia and Practitioner and a council member of the Indian Mathematical Society. Jitendra Panchal is an Assistant Professor in the Department of Applied Sciences and Humanities, Parul University, India. He has 7+ years of teaching experience and 5+ years of research experience. His research interests are in the fields of mathematical control theory applied to various types of impulsive and/or fractional differential inclusions/systems with non-local conditions, and integer/fractional-order mathematical modeling of dynamical systems based on real-life phenomena. His 10 articles have been published in international journals indexed in SCIE, Scopus and Web of Science.

213

Nita H. Shah Jitendra Panchal Sir Isaac Newton, one of the greatest scientists and mathematicians of all time, introduced the notion of a vector to define the existence of gravitational forces, the motion of the planets around the sun, and the motion of the moon around the earth. Vector calculus is a fundamental scientific tool that allows us to investigate the origins and evolution of space and time, as well as the origins of gravity, electromagnetism, and nuclear forces. Vector calculus is an essential language of mathematical physics, and plays a vital role in differential geometry and studies related to partial differential equations widely used in physics, engineering, fluid flow, electromagnetic fields, and other disciplines. ­Vector calculus represents physical quantities in two or three-­dimensional space, as well as the variations in these quantities. The machinery of differential geometry, of which vector calculus is a subset, is used to understand most of the analytic results in a more general form. Many topics in the physical sciences can be mathematically studied using vector calculus techniques. This book is designed under the assumption that the readers have no prior knowledge of vector calculus. It begins with an introduction to vectors and scalars, and also covers scalar and vector products, vector differentiation and integrals, Gauss’s theorem, Stokes’s theorem, and Green’s theorem. The MATLAB programming is given in the last chapter.

River Publishers

Elementary Vector Calculus and its Applications with MATLAB Programming

Nita H. Shah Jitendra Pancha

This book includes many illustrations, solved examples, practice examples, and multiple-choice questions.

Elementary Vector Calculus and its Applications with MATLAB Programming

Elementary Vector Calculus and its Applications with MATLAB Programming

River Publishers Series in Mathematical, Statistical and Computational Modelling for Engineering

River

Nita H. Shah Jitendra Panchal