Engineering Mathematics [For Semesters I and II] 9780071328777, 0071328777

Table of contents :
Title
Contents
Part 1 Mathematics-I
1 Matrices
2 Three Dimensional Analytical Geometry
3 Differential Calculus
4 Functions of Several Variables
5 Multiple Integrals
Part 2 Mathematics-II
1 Ordinary Differential Equations
2 Vector Calculus
3 Analytic Functions
4 Complex Integration
5 Laplace Transforms
Appendix A
Appendix B
Appendix C
Appendix D
Appendix E
Appendix F
Appendix G
Appendix H
Appendix I
Appendix J
Appendix k
Index

Citation preview

Engineering Mathematics [For Semesters I and II] Third Edition

About the Author

T Veerarajan is currently Dean, Department of Mathematics, Velammal College of Engineering and Technology, Viraganoor, Madurai, Tamil Nadu. A Gold Medalist from Madras University, he has had a brilliant academic career all through. He has 50 years of teaching experience at undergraduate and postgraduate levels in various established Engineering Colleges in Tamil Nadu including Anna University, Chennai.

Engineering Mathematics [For Semesters I and II] Third Edition

T Veerarajan Dean, Department of Mathematics, Velammal College of Engineering & Technology Viraganoor, Madurai, Tamil Nadu

Tata McGraw Hill Education Private Limited NEW DELHI McGraw-Hill Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

Published by Tata McGraw Hill Education Private Limited, 7 West Patel Nagar, New Delhi 110 008 Engineering Mathematics, 3e Copyright © 2012, by Tata McGraw Hill Education Private Limited No part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers, Tata McGraw Hill Education Private Limited. ISBN (13 digits): 978-0-07-132877-7 ISBN (10 digits): 0-07-132877-7 Vice President and Managing Director—MHE: Asia Pacific Region: Ajay Shukla Head—Higher Education Publishing and Marketing: Vibha Mahajan Deputy Manager—Acquisitions (Science, Engineering & Mathematics): H R Nagaraja Sr Editorial Executive—Acquisitions: Vamsi Deepak Sankar Executive—Editorial Services: Sohini Mukherjee Production Executive: Anuj K Shriwastava Sr Production Manager: Satinder S Baveja Marketing Manager—Higher Education: Vijay Sarathi J Sr Product Specialist: John Mathews General Manager—Production: Rajender P Ghansela Assistant General Manager—Production: B L Dogra Information contained in this work has been obtained by Tata McGraw-Hill, from sources believed to be reliable. However, neither Tata McGraw-Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither Tata McGrawHill nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that Tata McGrawHill and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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Contents Preface

xi

PART I: MATHEMATICS − I 1. Matrices

I-1.3

1.1 Introduction I-1.3 1.2 Vectors I-1.4 1.3 Linear Dependence and Linear Independence of Vectors I-1.5 1.4 Methods of Testing Linear Dependence or Independence of a Set of Vectors I-1.5 1.5 Consistency of a System of Linear Algebraic Equations I-1.5 Worked Example 1(a) I-1.7 Exercise 1(a) I-1.24 1.6 Eigenvalues and Eigenvectors I-1.27 Worked Example 1(b) I-1.32 Exercise 1(b) I-1.44 1.7 Cayley – Hamilton Theorem I-1.46 1.8 Property I-1.47 1.9 Calculation of Powers of a Matrix A I-1.49 1.10 Diagonalisation by Orthogonal Transformation or Orthogonal Reduction I-1.49 Worked Example 1(c) I-1.49 Exercise 1(c) I-1.62 1.11 Quadratic Forms I-1.64 Worked Example 1(d) I-1.67 Exercise 1(d) I-1.74 Answers I-1.75

2. Three Dimensional Analytical Geometry 2.1 Introduction I-2.1 2.2 Direction Ratios of the Line Joining Two Points I-2.4 Worked Example 2(a) I-2.6 Exercise 2(a) I-2.13 2.3 The Plane I-2.14 2.4 Plane through the Intersection of Two Given Planes I-2.19 Worked Example 2(b) I-2.19 Exercise 2(b) 1-2.26 2.5 The Straight Line I-2.28

I-2.1

Contents

vi

2.6 The Plane and the Straight Line I-2.29 2.7 Shortest Distance between Two Skew Lines I-2.30 2.8 Aliter I-2.31 Worked Example 2(c) I-2.32 Exercise 2(c) I-2.44 2.9 Sphere and Circle I-2.47 2.10 Standard Form of the Equation to a Sphere I-2.47 2.11 Equation of a Sphere Through a Circle I-2.50 Worked Example 2(d) I-2.50 Exercise 2(d) I-2.60 2.12 Cone and Right Circular Cone I-2.62 Worked Example 2(e) I-2.64 Exercise 2(e) I-2.72 2.13 Cylinder I-2.74 2.14 Right Circular Cylinder I-2.75 Worked Example 2(f) I-2.76 Exercise 2(f) I-2.82 Answers I-2.83

3. Differential Calculus

I-3.1

3.1 Curvature and Radius of Curvature I-3.1 3.2 Centre and Circle of Curvature I-3.6 Worked Example 3(a) I-3.7 Exercise 3(a) I-3.20 3.3 Evolutes and Envelopes I-3.21 Worked Example 3(b) I-3.23 3.4 Aliter I-3.33 Exercise 3(b) I-3.38 Answers I-3.40

4. Functions of Several Variables 4.1 Introduction I-4.1 4.2 Total Differentiation I-4.1 Worked Example 4(a) I-4.4 Exercise 4(a) I-4.20 4.3 Jacobians I-4.27 4.4 Differentiation Under the Integral Sign I-4.31 Worked Example 4(b) I-4.33 Exercise 4(b) I-4.48 4.5 Maxima and Minima of Functions of Two Variables Worked Example 4(c) I-4.52 Exercise 4(c) I-4.65 Answers I-4.67

I-4.1

I-4.50

Contents

vii

5. Multiple Integrals 5.1 Introduction I-5.1 5.2 Evaluation of Double and Triple Integrals I-5.1 5.3 Region of Integration I-5.2 Worked Example 5(a) I-5.3 Exercise 5(a) I-5.15 5.4 Change of Order of Integration in a Double Integral 5.5 Plane Area at Double Integral I-5.18 Worked Example 5(b) I-5.21 Exercise 5(b) I-5.38 5.6 Line Integral I-5.41 5.7 Surface Integral I-5.43 5.8 Volume Integral I-5.44 Worked Example 5(c) I-5.45 Exercise 5(c) I-5.57 5.9 Gamma and Beta Functions I-5.59 Worked Example 5(d) I-5.63 Exercise 5(d) I-5.77 Answers I-5.79

I-5.1

I-5.17

PART II: MATHEMATICS − II 1. Ordinary Differential Equations 1.1 Equations of the First Order and Higher Degree II-1.3 Worked Example 1(a) II-1.5 Exercise 1(a) II-1.14 1.2 Linear Differential Equations of Second and Higher Order with Constant Coefficients II-1.16 1.3 Complementary Function II-1.16 Worked Example 1(b) II-1.21 Exercise 1(b) II-1.34 1.4 Euler’s Homogeneous Linear Differential Equations II-1.35 1.5 Simultaneous Differential Equations with Constant Coefficients Worked Example 1(c) II-1.37 Exercise 1(c) II-1.49 1.6 Linear Equations of Second Order with Variable Coefficients II-1.50 Worked Example 1(d) II-1.55 Exercise 1(d) II-1.69 1.7 Method of Variation of Parameters II-1.73 Worked Example 1(e) II-1.75 Exercise 1(e) II-1.84 Answers II-1.84

II-1.3

II-1.37

Contents

viii

2. Vector Calculus 2.1 Introduction II-2.1 2.2 Vector Differential Operator ∇ II-2.1 Worked Example 2(a) II-2.4 Exercise 2(a) II-2.9 2.3 The Divergence of a Vector II-2.10 Worked Example 2(b) II-2.16 Exercise 2(b) II-2.24 2.4 Line Integral of Vector Point Functions Worked Example 2(c) II-2.28 Exercise 2(c) II-2.35 2.5 Integral Theorems II-2.37 Worked Example 2(d) II-2.40 Exercise 2(d) II-2.54 Answers II-2.56

II-2.1

II-2.25

3. Analytic Functions 3.1 Introduction II-3.1 3.2 The Complex Variable II-3.2 Worked Example 3(a) II-3.10 Exercise 3(a) II-3.22 3.3 Properties of Analytic Functions II-3.24 Worked Example 3(b) II-3.30 Exercise 3(b) II-3.41 3.4 Conformal Mapping II-3.43 3.5 Some Simple Transformation II-3.46 3.6 Some Standard Tranformations II-3.49 Worked Example 3(c) II-3.57 Exercise 3(c) II-3.74 3.7 Bilinear and Schwarz-Christoffel Transformations 3.8 Schwarz-Christoffel Transformations II-3.80 Worked Example 3(d) II-3.82 Exercise 3(d) II-3.91 Answers II-3.93

4. Complex Integration

II-3.1

II-3.77

II-4.1

4.1 Introduction II-4.1 4.2 Cauchy’s Integral Theorem or Cauchy’s Fundamental Theorem II-4.2 Worked Example 4(a) II-4.7 Exercise 4(a) II-4.17 4.3 Series Expansions of Functions of Complex Variable-Power Series II-4.20 4.4 Classification of Singularities II-4.25 Worked Example 4(b) II-4.31 Exercise 4(b) II-4.43 4.5 Contour Integration—Evaluation of Real Integrals II-4.47

Contents

ix

Worked Example 4(c) II-4.50 Exercise 4(c) II-4.67 Answers II-4.68

5. Laplace Transforms 5.1 Introduction II-5.1 5.2 Linearity Property of Laplace and Inverse Laplace Transforms 5.3 Laplace Transforms of Some Elementary Functions II-5.3 5.4 Laplace Transforms of Some Special Functions II-5.7 5.5 Properties of Laplace Transforms II-5.10 Worked Example 5(a) II-5.12 Exercise 5(a) II-5.33 5.6 Laplace Transform of Periodic Functions II-5.36 5.7 Derivatives and Integrals of Transforms II-5.37 Worked Example 5(b) II-5.40 Exercise 5(b) II-5.60 5.8 Laplace Transforms of Derivatives and Integrals II-5.63 5.9 Initial and Final Value Theorems II-5.66 5.10 The Convolution II-5.68 Worked Example 5(c) II-5.70 Exercise 5(c) II-5.92 5.11 Solutions of Differential and Integral Equations II-5.96 Worked Example 5(d) II-5.96 Exercise 5(d) II-5.108 Answers II-5.110 Appendix A: Partial Differentiation Appendix B: Solutions to the January 2011 Question Paper of Anna University, Chennai Appendix C: Solutions to the December2010/January 2011 Question Paper of Anna University, Coimbatore Appendix D: Solutions to the January 2011 Question Paper of Anna University, Madurai Appendix E: Solutions to the May/June 2011 Question Paper of Anna University, Tirunelveli Appendix F: Solutions to the December2010/January 2011 Question Paper of Anna University, Tiruchirapally Appendix G: Solutions to the June 2011 Question Paper of Anna University, Chennai Appendix H: Solutions to the April/May 2011 Question Paper of Anna University, Coimbatore Appendix I: Solutions to the June 2011 Question Paper of Anna University, Madurai Appendix J: Solutions to the May/June 2011 Question Paper of Anna University, Tirunelveli

II-5.1 II-5.2

II-A.1–21 II-B.1–14 II-C.1–10 II-D.1–11 II-E.1–13 II-F.1–7 II-G.1–8 II-H.1–10 II-I.1–11 II-J.1–12

x

Contents

Appendix K: Solutions to the May/June 2011 Question Paper of Anna University, Tiruchirapally

II-K.1–10

Preface

I am deeply gratified by the enthusiastic response shown to all the earlier editions of my book on Engineering Mathematics (for the first year) by the students and teachers throughout the country. This book has been revised to meet the requirements of the first-year undergraduate course on Engineering Mathematics offered to the students of engineering. The contents have been covered in adequate depth for both semester 1 and semester 2 Syllabi of various universities/ deemed universities across the country. The book offers a balanced coverage of both theory and problems. Lucid writing style supported by step-by-step solutions to all problems enhances understanding of the concepts. It has an excellent pedagogy with 1212 unsolved problems, 539 solved problems, 669 short answer questions and 914 descriptive questions. For the benefit of the students, the solutions to 2010-2011 question papers of the five Anna Universities are given in the Appendix. Additionally, the book is accompanied by a website which provides useful supplements including recent solved question papers. I hope that the book will be received by both the faculty and the students as enthusiastically as the earlier editions of the book and my other books. Critical evaluation and suggestions for the improvement of the book will be highly appreciated and acknowledged. T VEERARAJAN

Publisher’s Note: Tata McGraw Hill Education looks forward to receiving from teachers and students their valuable views, comments and suggestions for improvements, all of which may be sent to [email protected], mentioning the title and author’s name.

PART I Mathematics-I 1. Matrices 2. Three Dimensional Analytical Geometry 3. Differential Calculus 4. Functions of Several Variables 5. Multiple Integrals

Chapter

1

Matrices 1.1

INTRODUCTION

In the lower classes, the students have studied a few topics in Elementary Matrix matrix theory as well as the elementary operations on and properties of matrices. Though the concept of rank of a matrix has been introduced in the lower classes, matrix, as it will be of frequent use in testing the consistency of a system of linear algebraic equations, that will be discussed in the next section.

1.1.1

Rank of a Matrix

Determinant of any square submatrix of a given matrix A is called a minor of A. If the square submatrix is of order r, then the minor also is said to be of order r. Let A be an m × n matrix. The rank of A is said to be ‘r’, if (i) there is at least one minor of A of order r which does not vanish and (ii) every minor of A of order (r + 1) and higher order vanishes. In other words, the rank of a matrix is the largest of the orders of all the nonvanishing minors of that matrix. Rank of a matrix A is denoted by R(A) or (A). A, we may use the following procedure: A. Let their order be r. If any one of them does not vanish, then (A) = r. If all of them vanish, we next consider minors of A of next lower order (r – 1) and so on, until we get a non-zero minor. The order of that non-zero minor is (A). This method involves a lot of computational work and hence requires more time, a matrix A is given below: Reduce A to any one of the following forms, (called normal forms) by a series of elementary operations on A the normal form of A: Ir ; Ir | O ;

Ir Ir | O ; O O|O

Part I: Mathematics I

I – 1.4

Here Ir denotes the unit matrix of order r and O is zero matrix. By an elementary operation on a matrix (denoted as E-operation) we mean any one of the following operations or transformations: (i) Interchange of any two rows (or columns). (ii) Multiplication of every element of a row (or column) by any non-zero scalar. (iii) Addition to the elements of any row (or column), the same scalar multiples of corresponding elements of any other row (or column).

Note property that the rank of a matrix is unaltered by elementary operations. Finally we observe that we need not necessarily reduce a matrix A to the normal A to an equivalent matrix, whose rank can be easily found, by a sequence of elementary operations on A. The methods are illustrated in the worked examples that follow.

1.2 VECTORS A set of n numbers x1, x2, . . ., xn written in a particular order (or an ordered set of n numbers) is called an n-dimensional vector or a vector of order n. The n numbers are called the components or elements of the vector. A vector is denoted by a single letter X or Y etc. The components of a vector may be written in a row as X = (x1, x2, ..., xn)

or in a column as X

x1 x2

. These are called respectively row vector and

xn column vector. We note that a row vector of order n is a 1 × n matrix and a column vector of order n is an n × 1 matrix.

1.2.1 Addition of Vectors The sum of two vectors of the same dimension is obtained by adding the corresponding components. i.e., if X = (x1, x2, . . ., xn) and Y = (y1, y2, . . ., yn), then X + Y = (x1 + y1, x2 + y 2. . ., xn + yn).

1.2.2

Scalar Multiplication of a Vector

If k is a scalar and X = (x1 , x2, . . ., xn) is a vector, then the scalar multiple kX kX = (kx1, kx2, . . ., kxn).

1.2.3

Linear Combination of Vectors

If a vector X can be expressed as X = k1X1 + k2X2 + . . . + krXr then X is said to be a linear combination of the vectors X1, X2, . . ., Xr.

Chapter I: Matrices

1.3

I – 1.5

LINEAR DEPENDENCE AND LINEAR INDEPENDENCE OF VECTORS

The vectors X1, X2, . . ., Xr are said to be linearly dependent k1, k2, . . . kr, which are not all zero, such that k1Xl + k2X2 + . . . + krXr = 0. A set of vectors is said to be linearly independent if it is not linearly dependent, i.e. the vectors X1, X2, . . ., Xr are linearly independent, if the relation k1X1 + k2X2+ kl = k2 = . . . = kr = 0. . . . krXr

Note When the vectors X1, X2, . . ., Xr are linearly dependent, then k1X1 + k2X2 + . . . + krXr = 0, where at least one of the k’s is not zero. Let km Xm

Thus

k1 X1 km

k2 X2 km

kr Xr . km

Thus at least one of the given vectors can be expressed as a linear combination of the others.

1.4

METHODS OF TESTING LINEAR DEPENDENCE OR INDEPENDENCE OF A SET OF VECTORS

Method 1 Method 2 We write the given vectors as row vectors and form a matrix. Using elementary row operations on this matrix, we reduce it to echelon form, i.e. the one in which all the elements in the rth column below the rth element are zero each. If the number of non-zero row vectors in the echelon form equals the number of given vectors, then the vectors are linearly independent. Otherwise they are linearly dependent. Method 3 If there are n vectors, each of dimension n, then the matrix formed as in method (2) will be a square matrix of order n. If the rank of the matrix equals n, then the vectors are linearly independent. Otherwise they are linearly dependent.

1.5

CONSISTENCY OF A SYSTEM OF LINEAR ALGEBRAIC EQUATIONS

Consider the following system of m linear algebraic equations in n unknowns: a11x1 + a12x2 + . . . + a1nxn = b1 a21x1 + a22x2 + . . . + a2nxn = b2 am1x1 + am2x2 + . . . + amnxn = bm This system can be represented in the matrix form as AX = B, where

A

a11 a21 am1

a12 a22 am 2

a1n a2 n , X amn

x1 x2 xn

,B

b1 b2 bm

I – 1.6

Part I: Mathematics I

The matrix A is called of the system, X is the matrix of unknowns and B is the matrix of the constants. If B O, a zero matrix, the system is called a system of homogeneous linear equations; otherwise, the system is called a system of linear non-homogeneous equations. The m × (n + 1) matrix, obtained by appending the column vector B to the A as the additional last column, is called the augmented matrix of the system and is denoted by [A, B] or [A | B]. i.e.

A, B

a11 a21 am1

a12 a22 am 2

a1n b1 a2 n b2 amn bm

A set of values of x1, x2 . . ., xn. which satisfy all the given m equations simultaneously is called a solution of the system. When the system of equations has a solution, it is said to be consistent. Otherwise the system is said to be inconsistent. When the system has only one solution, it is called the unique solution. non-homogeneous equations is provided by a theorem, called Rouches’s theorem, which we state below without proof.

1.5.2

Rouche’s Theorem

The system of equations AX = B A and the augmented matrix [A, B] are of the same rank. Thus to discuss the consistency of the equations AX = B (m equations in n unknowns), the following procedure is adopted: R(A) and R(A, B). (i) If R(A) R(A, B), the equations are inconsistent (ii) If R(A) = R(A, B) = the number of unknowns n, the equations are consistent and have a unique solution. In particular, if A is a non-singular (square) matrix, the system AX = B has a unique solution. (iii) If R(A) = R(A, B) < the number of unknowns n, the equations are consistent

1.5.3

System of Homogeneous Linear Equations

Consider the system of homogeneous linear equations AX = O (m equations in n unknowns) a11x1 + a12x2 + . . . a1nxn = 0 i.e a21x1 + a22x2 + . . . + a2nxn = 0 –––––––––––––––– am1x1 + am2x2 + . . . amnxn = 0

Chapter I: Matrices

I – 1.7

This system is always consistent, as R(A) = R(A, O). A is non-singular, the system has a unique solution, namely, x1 = x2 = . . . = xn = 0. This unique solution is called the trivial solution, which is not of any importance. A is singular, i.e. if | A| number of non-zero or non-trivial solutions. equations is illustrated in the worked examples that follow. WORKED EXAMPLE 1(a) Example 1.1 Show that the vectors X1 = (1, 1, 2), X2 = (1, 2, 5) and X3 = (5, 3, 4) are linearly dependent. Also express each vector as a linear combination of the other two.

Method 1 k1X1 + k2X2 + k3X3 = 0 k1(1, 1, 2) + k2(1, 2, 5) + k3(5, 3, 4) = (0, 0, 0) k1 + k2 + 5k3 = 0 k1 + 2k2 + 3k3 = 0 2k1 + 5k2 + 4k3 = 0

Let i.e.

(2) – (1) gives k2 – 2k3 = 0

k2 =2k3

(4)

k1 = – 7k3

(5)

or

Using (4) in (3),

(1) (2) (3)

Taking k3 = 1, we get k1 = – 7 and k2 = 2. Thus –7X1 + 2X2 + X3 = 0 The vectors X1, X2, X3 are linearly dependent. From (6), we get

X1

2 X2 7

1 X3 , 7

X2

7 X1 2

1 X3 2

Method 2

(6)

and

X3 = 7X1 – 2X2

R2

R1 , R3

Writing X1, X2, X3 as row vectors, we get A

1 1 2 1 2 5

1 0

1 1

2 3 R2

5 3 4

0

2

6

1 1 2 0 1 3 R3 0 0 0

R3

R3

5 R1

2 R2

In the echelon form of the matrix, the number of non-zero vectors = 2 (< the number of given vectors). X1, X2, X3 are linearly dependent.

Part I: Mathematics I

I – 1.8 Now

0

R3 R3 2 R2 R3 5 R1 2 R2 7 R1 2 R2 R3

i.e.

–7X1 + 2X2 + X3 = 0

As before,

X1

2 X2 7

1 X3 , X2 7

7 X1 2

R1

1 X3 2

and

X3 = 7X1 – 2X2.

Method 3 |A| = 0 R (A The vectors X1, X2, X3 are linearly dependent.

R (A) = 2

Example 1.2 Show that the vectors X1 = (1, –1, –2, –4), X2 = (2, 3, –1, –1), X3 = (3, 1, 3, –2) and X4 = (6, 3, 0, –7) are linearly dependent. Find also the relationship among them.

A

X1 X2 X3 X4

1 2 3 6

1 3 1 3

2 1 3 0

4 1 2 7

1 0 0 0

1

1

2

4

0

1

0 0

4 9

3 5 9 12

1

1

2

4

0

1

0

0

0

0

3 5 33 5 33 5

7 5 22 5 22 5

1

1

2

4

0

1

0

0

0

0

3 5 33 5 0

1 5 4 9

7 5 R2 10 17

7 5 R4 22 5 0

2 3 9 12

4 7 R2 R2 2 R1 , R3 10 R3 3R1 , R4 R4 6 R1 17

1 R2 ; R3 5

R3

R3

R4

R3 ; R4

4 R2 ; R4

R4

R4

R3

Number of non-zero vectors in echelon form of the matrix A = 3. The vectors X1, X2, X3, X4 are linearly dependent.

9 R2

Chapter I: Matrices Now

0

R4

R4

I – 1.9

R3 R4

R4

9R2 R3

R4

4R2

R3

3R1

R2

6R1

R1

R3

R2

R3

The relation among Xl, X2, X3, X4 is – X1 – X2 – X3 + X4 = 0

R2

2R1

R4

or X1 + X2 + X3 – X4 = 0.

Example 1.3 Show that the vectors X1 = (2, –2, 1), X2 = (1, 4, –1) and X3 = (4, 6, –3) are linearly independent.

Method 1 Let k1 X1 + k2 X2 + k3 X3 = 0 i.e. k1 (2, –2, 1) + k2 (1, 4, –1) + k3 (4, 6, –3) = (0, 0, 0) 2k1 + k2 + 4k3 = 0 –2k1 + 4k2 + 6k3 = 0 k1 – k2 –3k3 = 0 k2 + 2k3 = 0 k2 = 0 k1 = 0 = k2 = k3.

From (1) and (2), From (2) and (3),

(1) (2) (3) (4) (5)

The vectors X1, X2, X3 are linearly independent.

Method 2 X1 X2 X3

A

2 1

2 4

1 1

1 2

4 2

4

6

3

4

6

1 0

4 10

1 3 R2

0

10

1

1

4

1

1 1 R1 3

R2

R2 ; R2

2 R1 ; R3

R1

R3

4 R1

0 10 3 R3 R3 R2 0 0 2 Number of non-zero vectors in the echelon form of A = number of given vectors, X1, X2, X3 are linearly independent. Example 1.4 Show that the vectors X1 X3 A

X1 X2 X3

1 2 7

1 1 2

1 2 7

3 1 4

X2 1 0 0

1 3 9

1 0 0

3 R2 7 17 R3

R2

2 R1 ;

R3

7 R1

Part I: Mathematics I

I – 1.10 1

1

1

3

0 0

3 0

0 0

7 R3 4

R3

3R2

Number of non-zero vectors in the echelon form of A = number of given vectors. X1, X2, X3 are linearly independent. Example 1.5 Test for the consistency of the following system of equations: x1 + 2x2 + 3x3 + 4x4 = 5 6x1 + 7x2 + 8x3 + 9x4 = 10 11x1 + 12x2 + 13x3 + 14x4 =15 16x1 + 17x2+ 18x3 + 19x4 = 20 21x1 + 22x2 + 23x3 + 24x4 = 25 The given equations can be put as 5 10 15 20

1 2 3 4 x1 6 7 8 9 x2 11 12 13 14 x3 16 17 18 19 x4 21 22 23 24

25

AX = B (say) A, B] by reducing it to the normal form

i.e. 1 6 A, B

2 7

3 8

4 9

5 10

11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

1 5

2 5

3 5

4 5

5 ( R2 5 R3 10 10 10 10 10 R4 15 15 15 15 15 R5 20 20 20 20 20

R2 R3

R1 R1

R4 R5

R1 R1 )

Note

If two matrices A and B are equivalent, i.e. are of the same rank, it is denoted as A ~ B. 1 2 3 4 5 1 1 1 1 1 R2 1 1 1 1 1 1 1 1 1 1 R5 1 1 1 1 1 1 0 0 0 0

2 1 1 1 1

3 2 2 2 2

4 3 3 3 3

1 R3 , R4 10

1 R2 , R3 5

1 R4 , 15

1 R5 20

5 4 R 4 2 R 4 4 4

R2 R4

R1 , R3 R1 , R5

R3 R5

R1 , R1

Chapter I: Matrices 1 0 0 0 0

0

0

0

0

1 1 1 1

2 2 2 2

3 3 3 3

4 (C2 4 C4 4 4

1 0 0 0 0

0 1 1 1 1

0 1 1 1 1

0 1 1 1 1

0 1 C2 1 C 1 4 1

1 0 0 0

0 1 0 0

0 1 0 0

0 1 0 0 0 0 0 0

0 1 R 0 3 R 0 5 0

1 0 0 0 0 0 1 0 0 0

C 0 0 0 0 0 3 C 0 0 0 0 0 5

C2 C4

C2 , C3 C4

I – 1.11

2C1 , C3 4C1 , C5

C3

C3 C5

3C1 , 5C1 )

2 ,

3 , C5

C5

4

R3 R5

R2 , R4 R2

R4

R2 ,

C3 C5

C2 , C4 C2

C4

C2 ,

0 0 0 0 0 I2 Now [A, B] has been reduced to the normal form 0

0 0

The order of the unit matrix present in the normal form = 2. Hence the rank of [A, B] = 2. A can be found as 2, in a similar manner. Thus R (A) = R [A, B] = 2 The given system of equations is consistent and possesses many solutions. Example 1.6 Test for the consistency of the following system of equations: x1 2x2 3x3 = 2; 3x1 2x2 = 1; 2x2 3x3 = 2; x2 + 2x3 = 1. The system can be put as 1 3 0 0

2 2 2 1

3 x1 0 x2 3 x3 2

2 1 2 1

Part I: Mathematics I

I – 1.12 i.e. AX = B (say)

A, B

1 3 0 0

2 2 2 1

3 0 3 2

2 1 2 1

1 0 0 0

0 4 2 1

0 9 3 2

0 7 C2 2 1

1 0 0

0 1 2

0 2 3

0 1 R2 2

0

4

9

7

1 0 0 0 1 2 0 0 0 0

1 1

1 0 0 0 1 0 0 0 1

0 1 4 11

11

1 0 0 0 1 0

0 0

0 0 1 0 0 0

4 15

1 0 0 1 0

0 0

0 0 1 0 0 0

0 15

1 0 0 0

0 1 0 0

0 0 1 0

0 0 R4 0 1

2 4 2 1

3 9 3 2

C2

2 7 R2 2 1

2C1 , C3

R4 , R4

R2

C3

R2

2 R2 , R4

R4

4 R2

C3

2C2 , C4

C4

C1

R4

R4

R3

C4

C4

4C3

1 R4 15

3R1

3C1 , C4

R3

0 0 C3 4

0 0 1

0

R3

1 0 0 0

C4

2C1

Chapter I: Matrices

I – 1.13

R[A, B] = 4 But R (A) 4, as A is a (4 × 3) matrix. In fact R (A) = 3, as the value of the minor 3 0 0

2 2 1

0 3 2

0

Thus R (A) [A, B] The given system is inconsistent. Example 1.7 Test for the consistency of the following system of equations and solve them, if consistent, by matrix inversion. 3y + 4

+ 1 = 0;

4y + 7

7

A

1 1

1 3

1 4

4 7

3 4

2 7

A, B

1

1

1

1

1

1

1 4 7

3 3 4

4 2 7

6 3 16

0 0 0

2 7 3

3 6 0

0

0

0

0 0 0

2 7 3

3 6 0

5 C2 1 9

1 0 0 0

0 1 5 9

0 6 3 0

0 7 R2 2 3

0 1 0 0

0 0 R3 0 0 27 33 C3 54 66

C2

C1 , C3

1 R2 5 R3 1 R4 9

C3

R3 and then C2

R3 C3

2 + 3 = 0;

16 = 0.

1

1

1 0 0 0

6 = 0; 4x + 3y

5 R2 , R4 and then 6C2 , C4

R2

R1 ,

R3 R4

4 R1 , 7 R1

C1 , C4

C4

C4

R4

9 R2

C4

7C2

C1

Part I: Mathematics I

I – 1.14 1 0 0 0

1 0 0 C3 0 1 1 0 2 2

0 0 0

1 0 0 0

0 1 0 0

0 0 1 0

1 C3 , C4 27

1 C4 33

0 0 R4 R4 2 R3 and theen 0 C4 C4 C3 0

R [A, B] = 3. Also R (A) = 3 The given system is consistent and has a unique solution.

i.e.

1 1

1 3

4

3

1 x 4 y 2

AX = B, say X=A B

i.e.

Let

1 6 3

A

1 1

1 3

1 4

4

3

2

a11 a21 a31

(1) a12 a22 a32

a13 a23 a33

Now A11 = co-factor of a11 in |A| = 6 A12 = 18; A13 = 15; A21 = 1; A22 = 6; A23 = 7; A31 = 1; A32 = 3; A33 = 2. Adj A

6 18

1 6

1 3

15

7

2

|A| = a11 A11 + a12 A12 + a13 A13 = 9

A

1

6 1 18 9 15

1 adj A A

1 6 7

Using (2) in (1), x y

6 1 18 9 15

1 6 7

1 3 2

1 6 3

1 3 2

(2)

Chapter I: Matrices 1 3 3 11 3

3 1 27 9 33 1 ,y 3

Solution of the system is x

I – 1.15

11 3

3,

Example 1.8 Test for the consistency of the following system of equations and solve them, if consistent: 3x + y + x+y 2 = 6; – 2x + 2y 3

Note

As the solution can be found out by any method, when the system is consistent, we may prefer the triangularisation method (also known as Gaussian elimination method) to reduce the augmented matrix [A, B] to an equivalent matrix. Using the equivalent matrix, we can test the consistency of the system and also only elementary row operations and convert the elements below the principal diagonal of A as zeros. 3 1 1 1

1 2

8 5

1 1 1 1

1 2

6 5

1 1 2 2

1 3

6 7

3 1 2 2

1 3

8 7

1 0 0

1 2 2

1 1 2

6 1 R2 10

0

4

1

A, B

1 1

R1 , R3

R3

R3

3R1 , R4

R4

2 R1

5

0 2 0 0

1 1 3

6 1 R3 9

0 0

1

3

1 0 0 0

1 1 3 0

6 1 R4 9 0

1 2 0 0

R2

R1

Now, Determinant of [A, B]

R3

R2 , R4

R4

1 R3 3

R4

2 R2

Determinant of the equivalent matrix = 0. ( R A, B

3

(1)

Two

Part I: Mathematics I

I – 1.16 1 1 Now 0 2 0 0

1 1 3

6

0

R [A, B] = R (A) = 3 = the number of unknowns. The system is consistent and has a unique solution. A system of equations equivalent to the given system is also obtained from the equivalent matrix in (1). The equivalent equations are 6,

2

1

and

3 =

9

Solving them backwards, we get x = 1, y = 2

3.

Example 1.9 Examine if the following system of equations is consistent and x

y

A, B

1, 2

2

1; x

3

y

2

1 2

1 1 1 2 3 1

1 0

1 4

1 1

1 3

1 2 5 1 1 2

0

2

1

0

2

2

1 0

1 4

0

0

0

0

1 0

1 4

0

0

0

1 1

1 2 0 0

1 1

1 1

1 2 5 2

9 R 3 2 1 2

R3

5; 3 1 R2 1 R3 4 R4 1

1 R2 , R4 2

2.

R2 R3 R4

R4

2 R1 , R1 , 3R1

1 R2 2

1 1 9 R4 2 22

R4

5 R3

It is obvious that det [A, B] = 4 and det [A] = 3 R [A, B R [A]. The system is inconsistent.

Note

The last row of the equivalent matrix corresponds to the equation 0 x 0 y 0 22, which is absurd. From this also, we can conclude that the system is inconsistent.

Chapter I: Matrices

I – 1.17

Example 1.10 Solve the following system of equations, if consistent: 3 1; 3x + 3 5 1. A, B

1 1 1 1 3 3

1 3 1 1 5 1

1 1 0 0 0 0

1 2 0

1 1 0 0 0 0 3 2 R3 0

1 2 8

1 0

R2

R1 , R3

R3

3R1

4 R2

R3

All the third order determinants vanish R [A, B] Consider

3 2 R2 8

3

1 , which is a minor of both A and [A, B]. 2

R (A) = R [A, B] < the number of unknowns. The system is consistent with many solutions. 3 and 2 i.e.

=1

x + y = 2.

and

The system has a one parameter family of solutions, namely x = k, y = 2 1, where k is the parameter. Giving various values for k, Example 1.11 Solve the following system of equations, if consistent: x1

2 x2 A, B

x3

5 x4

3 x2

2 x3

1 1

2 3

1 2

5 4 7 5

2

1

3

0 3

1 0

2 1

1 1

5 2

0

5

5

10 0

1 2 0 1 0 0 R [A, B] Similarly R (A)

4; x1

1 1 0

4 1 R2

7 x4

R2

5; 2 x1

R1 , R3

x2

3 x3

R3

2 R1

5

5 4 2 1 R3 0 0

R3

5 R2

the last row contains only zeros) 3.

1 2 0 , R (A) = R [A, B] = 2 < the number of unknowns. 0 1 The given system is consistent with many solutions. Since

3.

k,

Part I: Mathematics I

I – 1.18

x1

2 x2

x3

5 x4

4

x2

x3

2 x4

1

and

(1) (2)

As there are only 2 equations, we can solve for only 2 unknowns. Hence the other 2 unknowns are to be treated as parameters. Taking x3 = k and x4 = we get [from (2)] x2 = 1 + k + 2k' k + 2k') + k + 5k' [from (1)] and x1 i.e. x1 = 2 k + k' The given system possesses a two parameter family of solutions. Note From the Examples (10) and (11), we note that the number of parameters in the solution equals the difference between the number of unknowns and the common rank of A and [A, B]. Example 1.12 Find the values of k, for which the equations and x + 5y + 9 = k2 have a solution. For these values of k, x + 2y + 3 solutions also.

A, B

1 1 1 1 2 3

1 k

1 1 1 0 1 2

1 k

1

1 5 9 k2

0 4 8 k2

1

1 1 1 0 1 2

1 k

0 0 0 k

2

R3

1 4k

R3

R2 R3

R2 R3

1,

R1 , R1

4 R2

(1)

3

1 1 1 A

R A

0 1 2 0 0 0

2

If the system possesses a solution, R [A, B] must also be 2. The last row of the matrix in (1) must contain only zeros. k2 4k + 3 = 0 i.e. k = 1 or 3. For these values of k, R (A) = R [A, B] = 2 < the number of unknowns. The given system has many solutions. Case (i) k=1 and

y+2

1 0

(2) (3)

Puting x Case (ii) k=3 The equivalent equations are

and

1

(2)

Chapter I: Matrices and Putting

I – 1.19

y+2 2 the one-parameter family of solutions of the given system is 1, y = 2

Example 1.13 equations may have a solution: x+2 3

A, B

A

(4)

2 a, b, c, so that the following system of

3

5y + 8 = c.

2

3 a 2 b 8 c

1 3 1

2 1 5

1 0 0

2 7 7

a 3 11 b 3a R2 11 c a

1 0 0

2 7 0

3 a 11 R3 b 3a 0 2a b c

1 0

2 7

3 11

0

0

0

R A

R2

3R1 , R3

R3

R3

R2

R1

(1)

2

If the given system possesses a solution, R [A, B] = 2. The last row of (1) should contain only zeros. 2 have a solution.

a, b, c, the system will

Example 1.14 Find the value of k such that the following system of equations has (i) a unique solution, (ii) many solutions and (iii) no solution. 1

1

A

k 1 1 1 k 1 1 1 k

A

k k2

1

k

1 k2

k

1

2

k

1

k

k

2

2

|A| = 0, when k = 1 or k = –2 When k k A R(A) = 3 Then the system will have a unique solution.

1.

1

k

Part I: Mathematics I

I – 1.20

When k = 1, the system reduces to the single equation In this case, R(A) = R[A, B] = 1. The system will have many solutions. (i.e. a two parameter family of solutions) When 2,

A, B

2 1 1

1 2 1

1 1 1 1 2 1

1 2 1

2 1 1

1 1 1 1 R1 2 1

1 1 3 3 R2 3 0

R2

2 R1 , R3

1 0 0

2 3 3

1 0 0

2 1 1 3 3 3 R3 0 0 3

Now

1 0

2 1 3 3

0

0 0

1 0

2 3

R[A, B] = 3. Thus R(A The system has no solution.

A

1 2

0 1 1 2

3

2

1 2

and A

6 4 1

6 4 3

3

R1

0

R[A, B].

1 1 0 1

1 2

R A

a minor of A, B

1 1 1 6 1 2 3 10

1 1 0 1 0 0

R3

R2

R A

0

R2

, the equations x + y + = 6, have (i) no solution, (ii) a unique solution,

Example 1.15 x + 2y + 3 = 10 and x + 2y +

1 1 0 1 0 0

R3

0

2 1 1 3 3 3 0 0 3

A, B

= 1.

6

R3

R3

10 3

R2 R3

R2

R2 R3

R1 , R1

Chapter I: Matrices

I – 1.21

A|

R(A) = 3 takes any value, the system has a unique solution. 1 1 = 3, |A| = 0 and a second order minor of A, i.e. 0 0 1 R (A) = 2. 1 1 1 6 A, B 0 1 2 4 0 0 0 10

(1)

10, the last row of (1) contains only zeros. R[A, B] = 2. = 10, R(A) = R[A, B] = 2.

R[A, B

A, B], i.e. 1 1 1 2

6 4

0 0

10

0

10

R [A, B] = 3 R(A R[A, B]. The given system has no solution. Example 1.16 Test whether the following system of equations possess a non-trivial solution. x1 + x2 + 2x3 + 3x4 = 0; 3x1 + 4x2 + 7x3 + 10x4 = 0; 5x1 + 7x2 + 11x3 + 17x4 = 0;

6x1 + 8x2 + 13x3 + 16x4 = 0.

The given system is a homogeneous linear system of the form AX = 0.

A

1 1 3 4

2 7

3 10

1 1 2 0 1 1

5 7 11 17 6 8 13 16

0 2 0 2

1 1 0 1

2

3

0 0 0 0

1 1 1

1 R3 0 4

1 1

2

3

0 1 0 0 0 0

1 1 0

1 R4 0 4

|A| = 4 i.e. A is non-singular

1 1

3 R2 1 R3 2 R4 2

R2 R3 R4

R3

2 R2 , R4

R4

R3

3R1 , 5 R1 , 6 R1

R4

2 R2

Part I: Mathematics I

I – 1.22

R(A) = R[A, 0] = 4 The system has a unique solution, namely, the trivial solution. Example 1.17 Find the non-trivial solution of the equations x + 2y + 3 = 0, 3x + 4y + 4 = 0, 7x + 10y + 11 = 0, if it exists. 1 2 3 3 4 4 7 10 11

A

|A| = 0

and

1 0 0

2 2 4

3 R 5 2 R 10 3 2 R2

1 0 0

2 2 0

3 5 R3 0

R3

1 0

2 2

0

R(A) = 2

R2 R3

3R1 , 7 R1

(1)

given equations are equivalent to x + 2y + 3 = 0 y

(2)

=0

(3)

5 k from (3) and x = 2k. 2 2k. Thus the non-trivial solution is x = 4k, y = –5k and Example 1.18 Find the non-trivial solution of the equations 2 3w = 0, 3x + 0, 5x – 3y + 2 6w = 0, x – 9y + 14 2w = 0, if it exists. 2y – 4 we get y

Putting

A

1 3

1 2

2 4

3 1

1 0

1 5 2 8

2 10 8 12

5 1

3 9

2 14

6 2

0 0

1 0 0 0

1 1 2 8

2 2 8 12

3 2 R2 21 1

1 R2 5

1 0 0 0

1 1 0 0

2 2 4 4

3 2 R3 17 17

R3

3 R2 10 R3 21 R4 1

2 R2 , R4

R2 R3

3R1 , 5 R1 ,

R4

R1

R4

8 R2

Chapter I: Matrices

~

1

1

0 0

1 0

2 2 4

0

0

0

3 2 R4 17

I – 1.23

R4

R3

0

|A| = 0 i.e. R(A) < 4 The system has a non-trivial solution. The system is equivalent to x

y+2

w=0

(1)

+2w = 0

(2)

+ 17w = 0

(3)

y

Putting w = 4k, we get = 17k from (3), y = 26k from (2) and x = 4k. Thus the non-trivial solution is x = 4k, y = 26k, = 17k and w = 4k. Example 1.19 y 0, x

A

1 1 2 2

= 0, 2x

y

4 4 2 3

2 4

3

4

3

1 ~ 0 0

4 2

4

8

5

x y = = 0 have a non-trivial solution.

2 R 2 2 R3

R2 R1 , R3 2 R1

(1)

For non-trivial solution, |A| = 0 2

+ 16 = 0

x + 6y + 10 = 0 y Putting = 3k, we get y i.e. the solution is x = 0, y

= 0,

from (1)

k and x = 0 k and = 3k. x + 2y

=0

y + 4 = 0, Putting = k, we get y = k and x = 4k. i.e. the solution is x = 4k, y = k and = k.

from (1)

Part I: Mathematics I

I – 1.24

EXERCISE 1(a) Part A (Short Answer Questions)

3. If a set of vectors is linearly dependent, show that at least one member of the set can be expressed as a linear combination of the other members. 4. Show that the vectors X1 = (1, 2), X2 = (2, 3) and X3 = (4, 5) are linearly dependent. 5. Show that the vectors X1 = (0, 1, 2), X2 = (0, 3, 5) and X3 = (0, 2, 5) are linearly dependent. 6. Express X1 = (1, 2) as a linear combination of X2 = (2, 3) and X3 = (4, 5). 7. Show that the vectors (1, 1, 1), (1, 2, 3) and (2, 3, 8) are linearly independent. a 8. Find the value of a dependent. 9. What do you mean by consistent and inconsistent systems of equations. Give examples. 10. State Rouche’s theorem. 11. State the condition for a system of equations in n unknowns to have (i) one solution, (ii) many solutions and (iii) no solution. 12. Give an example of 2 equations in 2 unknowns that are (i) consistent with only one solution and (ii) inconsistent. 13. Give an example of 2 equations in 2 unknowns that are consistent with many solutions. 14. Find the values of a and b, if the equations 2x y = 5 and ax + by have many solutions. 15. Test if the equations x + y + = a, 2x + y + 3 = b, 5x + 2y + = c have a unique solution, where a, b, c are not all zero. x+y = 10, x y + 2 x y + 4 = 30 have a unique solution. 17. If the augmented matrix of a system of equations is equivalent to 1 0

2 5

1 3

0

0

0

2 2

solution. 18. If the augmented matrix of a system of equations is equivalent to 1 0

2 2

0

0

1

3

2

2 1

only one solution.

for which the system has 3

Chapter I: Matrices

I – 1.25

19. If the augmented matrix of a system of equations is equivalent to 1 0

2 5

0

0

3

4

4

2 2

for which the system 3

has many solutions. 20. If the augmented matrix of a system of equations is equivalent to 1

1

0 0

3 0

2 1 8

3 2 11

for which the system

has no solution. 21. Do the equations x y = 0, 3x + y = 0 and 2x + 5y + 6 = 0 have a nontrivial solution? Why? = 0 have a non22. If the equations x + 2y + = 0, 5x + y – = 0 and x + 5y 23. Given that the equations x + 2y

= 0, 3x + y

= 0 and 2x – y = 0 have

Part B Show that the following sets of vectors are linearly dependent. Find their relationship in each case: 24. 25. 26. 27. 28. 29.

30. 31. 32. 33.

X1 = (1, 2, 1), X2 = (4, 1, 2), X3 = (6, 5, 4), X4 X1 X2 X3 X4 X1 X2 X3 X1 = (1, 0, 4, 3), X2 X3 X1 X2 X3 X4 Determine whether the vector x5 = (4, 2, 1, 0) is a linear combination of the X2 X3 = (3, 1, 0, 0) and X4 = set of vectors X1 Show that each of the following sets of vectors is linearly independent. X1 = (1, 1, 1); X2 = (1, 2, 3); X3 X1 X2 X3 X1 X2 = (1, 3, 1, 2); X3 = (4, 2, 1, 0); X4 = (6, 1, 0, 1). X1 X2 X3 X4 = (0,

34. Test for the consistency of the following system of equations: 3 4 5 6 x1 4 5 6 7 x2 5 6 7 8 x3 10 11 12 13 x4 15 16 17 18

7 8 9 14 19

Part I: Mathematics I

I – 1.26

Test for the consistency of the following systems of equations and solve, if consistent: 35. 2 5y + 2 3 3y + 3 0; 1. 36. 3x + 5y 1; 4 x y+5 9; 7x = 4. 37. 2x + 2y + 4 = 6; 3x + 3y + 7 = 10; 5x + 7y + 11 17; 6x + 8y + 13z = 16. Test for the consistency of the following systems of equations and solve, if consistent: 38. x + 2 3; 2x + 3y + 2 = 5; 3x y+5 2; 3x + 9 4. 39. 2x + 6 3 18; 3x y+7 31; 5x + 3y + 3 48; 8x y+2 21. 40. x + 2y + 3 = 6; 5x y + 2 = 4; 2x + 4y = 5; 3x + 2y + 4 = 9. 41. x + 2y = 4; 10y + 3 3 = 5; 3x + 3y + 2 1. 42. 2x1 + x2 + 2x3 + x4 = 6; x1 x2 + x3 + 2x4 = 6; 4x1 + 3x2 + 3x3 3x4 x1 + 2x2 + x = 10 3 4 43. 2x + y + 5 5; x + y + 3 4 1; 3x + 6 2 + w = 8; 2x + 2y + 2 3w = 2. Test for the consistency of the following systems of equations and solve, if consistent: 44. 3 8 x + y = 4 ; 2x + 5y + 6 13. 45. 5x + 3y + 7 4; 3x + 26y + 2 9; 7x + 2y + 10 = 5. 46. 4 3 16 = 0; 2x + 7y + 12 = 48; 4 6 = 16; 5 5y + 3 0. 47. 2y + 3w = 1; 2 3y + 2 + 5w = 3; 3 7 2 10w = 2. 48. x1 + 2x2 + 2x3 x4 = 3; x1 + 2x2 + 3x3 + x4 = 1; 3x1 + 6x2 + 8x3 + x4 = 5. 49. Find the values of k, for which the equations 1, x + 2y + 4 and 2 x + 4y + 10 have a solution. For these values of k, also. x+2 4, 2 2 4 7 5 51. Find the condition on a, b, c, so that the equations x + y + = a, x + 2y + 3 = b, 3x + 5y + 7 = c may have a one-parameter family of solutions. 52. Find the value of k for which the equations kx y + = 1, 2 and x y + = 1 have (i) no solution, (ii) one solution and (iii) many solutions. the equations x + y + 2 2, 2 3 2 and 5 have (i) no solution, (ii) a unique solution, (iii) an 54. Find the values of a and b for which the equations x + y + 2 3, 2 3 = 4 and 5 = b have (i) no solution, (ii) a unique solution, (iii) many solutions. 55. Find the non-trivial solution of the equations x + 2y + = 0; 5x + y = 0 and x + 5y + 3 0, if it exists. 56. Find the non-trivial solution of the equations x + 2 2w = 0; x + 3y + 2 + 2w = 0; 2x + 4y + 3 6w = 0 and 3x + 7y + 4 6w = 0, if it exists. = 0, 4 2 3 = x + 4y ,

5

0, 10x

4 7 = 0, 7 0 possess a non-trivial solution. For

Chapter I: Matrices

1.6

I – 1.27

EIGENVALUES AND EIGENVECTORS

Let A = [aij] be a square matrix of order n. If there exists a X

column vector

AX = X A and X is called the eigenvector A, we proceed as follows: A and X be the corresponding eigenvector. Then, by AX = X = IX, where I is the unit matrix of order n. i.e.

i.e.

I) X = 0

(A

(1)

a11 a21

a12 a22

a1n a2 n

1 0 0 1

an1

an 2

ann

0 0

i.e.

a11

x1

a21 x1

a22

0 0 0

a12 x2

1 a1n xn

x2

a2 n xn

x1 x2

0 0

xn

0

0 0

(2)

..................................................... an1 x1

an 2 x2

ann

xn

0

Equations (2) are a system of homogeneous linear equations in the unknowns x1, x2, . . . , xn. x1 x2 Since X is to be a non-zero vector, xn x1, x2, . . . , xn should not be all zeros. In other words, the solution of the system (2) should be a non-trivial solution. The condition for the system (2) to have a non-trivial solution is a11 a21 an1 i.e.

a12 a22 an 2

a1n a2 n

0

(3)

I| = 0

(4)

ann |A

The determinant |A I| is a polynomial of degree n the characteristic polynomial of A. The equation |A I| = 0 or the equation (3) is called the characteristic equation of A.

Part I: Mathematics I

I – 1.28

When we solve the characteristic equation, we get n n roots of the characteristic equation are called the characteristic roots or latent roots or eigenvalues of A. , the equations (2) possess a non-zero (nontrivial) solution X. X is called the invariant vector or latent vector or eigenvector of . A

Notes 1. Corresponding to an eigenvalue, the non-trivial solution of the system (2) will be a one-parameter family of solutions. Hence the eigenvector corresponding to an eigenvalue is not unique. , . . . , n of a matrix A are distinct, then the cor1 2 responding eigenvectors are linearly independent. 3. If two or more eigenvalues are equal, then the eigenvectors may be linearly independent or linearly dependent.

1.6.2

Properties of Eigenvalues

1. A square matrix A and its transpose AT have the same eigenvalues. Let A = (aij); i, j = 1, 2, . . . , n. The characteristic polynomial of A is

A

I

a11 a21 an1

a12 a22 an 2

a1n a2 n

(1)

ann

The characteristic polynomial of AT is

T

A

I

a11 a12 a1n

a21

an1 an 2

a22 a2 n

(2)

ann

Determinant (2) can be obtained by changing rows into columns of determinant (1). | I| = |AT I| The characteristic equations of A and AT are identical. The eigenvalues of A and AT are the same. 2. The sum of the eigenvalues of a matrix A is equal to the sum of the principal diagonal elements of A. (The sum of the principal diagonal elements is called the Trace of the matrix.) The characteristic equation of an nth order matrix A may be written as n

D1

n 1

D2

n 2

n

1 Dn

0,

(1)

where Dr is the sum of all the rth order minors of A whose principal diagonals lie along the principal diagonal of A.

Chapter I: Matrices

I – 1.29

Dn = |A|). We shall verify the above result for a third order matrix.

(Note

a11 a21 a31

A

Let

a13 a23 a33

a12 a22 a32

The characteristic equation of A is given by a11 a12 a13 a23 a21 a22 a31 a32 a33

0

(2)

Expanding (2), the characteristic equation is 2

a11 a12

a22

3

i.e.

a23 a33

a21 a31

a21 a11

a22 a32

a33

a22

a11

a12

a21

a22

a23 a33 a21 a31

a13 a31

a22 a32

0

2

a33 a11 a31

a13 a33

a22 a32

a23 a33

A

0

3

D1 2 + D2 D3 = 0, using the notation given above. This result holds good for a matrix of order n.

Note

This form of the characteristic equation provides an alternative method for getting the characteristic equation of a matrix. be the eigenvalues of A. 1 2 n They are the roots of equation (1). D1 1

n

2

1 a11

D1

a22

ann

Trace of the matrix A. 3. The product of the eigenvalues of a matrix A is equal to |A|. are the eigenvalues of A, they are the roots of 1 2 n n

D1

n 1

D2

1

2

n

,

2

n

= Dn = |A|.

are the roots of |A

n

1 Dn 1

1.6.3 Aliter 1

1 Dn

n

Product of the roots 1

n

n 2

I| = 0

0

.

Part I: Mathematics I

I – 1.30 n

I

|A

1

n

2

n

n

), since L.S. is a nth degree polynomial

. n

A 1

1.6.4

2

1

2

)

n

= |A|. n

Corollary

If |A| = 0, i.e. A is a singular matrix, at least one of the eigenvalues of A is zero and conversely. are the eigenvalues of a matrix A, then 1 2 n (i) k 1, k 2, . . . k n are the eigenvalues of the matrix kA, where k is a nonzero scalar. (ii) 1p , 2p , , np are the eigenvalues of the matrix Ap, where p is a positive integer. (iii) 1 , 1 , 1 are the eigenvalues of the inverse matrix A r 1

2

n

A is non-singular. be an eigenvalue of A and Xr the corresponding eigenvector. r AXr x (1) r r Multiplying both sides of (1) by k, (kA)Xr = (k r) Xr (2) From (2), we see that k r is an eigenvalue of kA and the corresponding , namely Xr. r (ii) Premultiplying both sides of (1) by A, A2 X r

A AX r A

r

AX r

r 2 r 3 Similarly A X r

In general, A p X r

3 r

Xr

Xr

X r and so on. p r

Xr p

From, (3), we see that r is an eigenvalue of Ap with the corresponding eigenvector equal to Xr . r (iii) Premultiplying both sides of (1) by A , A (AXr) = A i.e.

Xr A 1Xr

r

r

Xr)

(A Xr)

1 r

Xr

(4)

Chapter I: Matrices 1

From (4), we see that

I – 1.31

is an eigenvalue of A–1 with the corresponding

r

eigenvector equal to Xr, . r 5. The eigenvalues of a real symmetric matrix (i.e. a symmetric matrix with real elements) are real. X be the corresponding eigenvector. Then AX = Premultiplying both sides of (1) by X T (the transpose of the conjugate of X), we get X T AX XT X (2) Taking the complex conjugate on both sides of (2), XT A X

X T X (assuming that may be complex )

XT AX XT X ( A i.e. Taking transpose on both sides of (3), X T AT X i.e.

XT AX

XT X

( AB )T

XT X

( A)T

A, as A is real)

(3)

BT AT A, as A issymmetric

(4)

From (2) and (4), we get XT X i.e.

(

XT X

)XT X

0

X T X is an 1 × 1 matrix, i.e. a single element which is positive 0 Hence all the eigenvalues are real. 6. The eigenvectors corresponding to distinct eigenvalues of a real symmetric matrix are orthogonal. x1 y1 x2 y2 and Y Note Two column vectors X are said to be xn yn … orthogonal, if their inner product (x1y1 + x2y2 + xnyn) = 0 i.e. if XTY = 0. be any two distinct eigenvalues of the real symmetric matrix A 1 2 and X1, X2 be the corresponding eigenvectors respectively. Then AX1 X (1) 1 1 and

AX2 =

Premultiplying both sides of (1) by, X 2T AX 1

X2

(2)

2

X 2T 1

we get

X 2T X 1

Part I: Mathematics I

I – 1.32

Taking the transpose on both sides, X 1T AX 2

1

X 1T X 2

( AT

A)

(3)

Premultiplying both sides of (2) by X1T , we get X 1T AX 2

2

X 1T X 2

(4)

From (3) and (4), we have 1

i.e.

X 1T X 2 ( 1

2

1

2

Since

X 1T X 2 T 2 ) X1 X 2

0

, X 1T X 2

0

i.e. the eigenvectors X1 and X2 are orthogonal.

WORKED EXAMPLE 1(b) 5 4 Example 1.1 Given that A , verify that the eigenvalues of A2 are the 1 2 squares of those of A. Verify also that the respective eigenvectors are the same. The characteristic equation of A is

5

4 1

0

2

2

The eigenvalues of A is given by (A 5

i.e.

4 1

2

x1 x2

I) X = 0

0

= 1, the eigenvector is given by the equations 4x1 + 4x2 = 0 and x1 + x2 = 0, which are one and the same. Solving, x1

x2. Taking x1 = 1, x2 =

The eigenvector is

1 1

Chapter I: Matrices

I – 1.33

x1 + 4x2 = 0 x1 4x2 = 0

and Solving, x1 = 4x2 Taking x2 = 1, x1 = 4 1 The eigenvector is 4 Now

5 4 5 4 1 2 1 2

A2

The characteristic equation of A2 is

29 7 2

29 28 7 8 28

0

8 + 36 = 0

The eigenvalues of A2 are 1 and 36, that are the squares of the eigenvalues of A, A2 is given by 28 28 x1 7 x1 7 x2 0 0. i.e. 28x1 28 x2 0 and 7 7 x2 Solving, x1 = –x2. Taking x1 = 1, x2 = –1. A2 is given by 28 x1 28 x2

7 7

0. i.e. 7 x1

28 x2

0

and

7 x1

28 x2

0.

Solving, x1 = 4x2. Taking x2 = 1, x1 = 4. Thus the eigenvectors of A2 are 1 4 , which are the same as the respective eigenvectors of A. and 1 1 Example 1.2 Find the eigenvalues and eigenvectors of the matrix 1 1 3 A 1 5 1 3 1 1 The characteristic equation of A is 1

1 1

3 1

5

3

1

0

1

2 3

2

3 2

The eigenvalues of A Case (i) The eigenvector is given by

2

+ 36 = 0

(1) =

Part I: Mathematics I

I – 1.34

3 1 3 x1 1 7 1 x2 3 1 3 x3

(2)

0

x1 + 7x2 + x3 = 0 3x1 + x2 + 3x3 = 0 Solving these equations by the rule of cross-multiplication, we have i.e.

x1 21 1 x1 20

i.e.

x2 3 3 x2 0

x3 1 21 x3 20

(3)

Note

To solve for x1, x2, x3, we have taken the equations corresponding to the second and third rows of the matrix in step (2). The proportional values of x1, x2, x3 obtained

From step (3), x1 = k, x2 = 0 and x3 = –k. Usually the eigenvector is expressed in terms of the simplest possible numbers, corresponding to k = 1 or – 1. x1 = 1,

x2 = 0,

X1

x3 = – 1

1 0 1

Case (ii) The eigenvector is given by

2 1 1 2

3 x1 1 x2

3 1

2 x3

0.

Values of x1, x2, x3 are proportional to the co-factors of –2, 1, 3 (elements of the i.e.

x1 5

x2 5

x3 or 5

x1 1

1 1 1

X2 Case (iii) The eigenvector is given by

5 1 3

1 1 1

3 x1 1 x2 5 x3

0

x2 1

x3 1

Chapter I: Matrices x1 4 x1 1

or

x2 8 x2 2 1 2 1

X3

I – 1.35

x3 4 x3 1

Note

Since the eigenvalues of A are distinct, the eigenvectors X1, X2, X3 are linearly independent, as can be seen from the fact that the equation k1X1 + k2X2 + k3X3 k1 = k2 = k3 = 0. Example 1.3 Find the eigenvalues and eigenvectors of the matrix 0 1 1 1 0 1 1 1 0

A The characteristic equation is given by 3

– D1

2

+ D2

D3 = 0, where

D1

A that lie along the main diagonal of A =0+0+0 =0 D2 = the sum of the second order minors of A whose principal diagonals lie along the principal diagonal of A. 0 1 0 1 0 1 1 0 1 0 1 0 =–3 D3= |A| = 2 Thus the characteristic equation of A is 3

+ l)2

– 2) = 0

The eigenvalues of A Case (i) The eigenvector is given by 1 1 1

1

1 1

x1 x2 x3

0

All the three equations reduce to one and the same equation x1 + x2 + x3 = 0. There is one equation in three unknowns. Two of the unknowns, say, x1 and x2 are to be treated as free variables (parameters). Taking x1 = 1 and x2 = 0, we get x3 = –1 and taking x1 = 0 and x2= 1, we get x3= – l

Part I: Mathematics I

I – 1.36

1 0 1

X1

and

0 1 1

X2

Case (ii) The eigenvector is given by 2 1 1

1 2 1

1 x1 1 x2 2 x3

0

Values of x1, x2, x3 x1 3 x1 1

i.e. or

x3 3 x3 1

x2 3 x2 1 1 1

x3

1

Note

Though two of the eigenvalues are equal, the eigenvectors X1, X2, X3 are found to be linearly independent. Example 1.4 Find the eigenvalues and eigenvectors of the matrix

A

2 1

2 1

2 1

1

3

1

The characteristic equation of A is 2 2 1 1 1

3

2 1

0

1

2

i.e.

(2 –

The eigenvalues of A Case (i) The eigenvector is given by 4 1 1 x1 8

x2 2

2 2 x1 3 1 x2 3 1 x3

0

x3 (by taking the co-factors of elements of the third row) 14

Chapter I: Matrices i.e.

x1 4

x2 1

X1

4 1 7

Case (ii) = 2. The eigenvector is given by 2 1 3

2 x1 1 x2 3 x3

x2 4

x3 4

or

X2

Note

x3 7

0 1 1 x1 0

I – 1.37

0

x1 0

x3 1

x2 1

0 1 1

X3

Two eigenvalues are equal and the eigenvectors are linearly dependent.

Example 1.5 Find the eigenvalues and eigenvectors of the matrix A

11 7

4 2

7 5

10

4

6

Can you guess the nature of A from the eigenvalues? Verify your answer. The characteristic equation of A is 11

4 7

7 5

2

10

4

6

2 3

2

The eigenvalues of A Case (i) 11 The eigenvector is given by 7 10 x1 8

x2 8

7 x1 5 x2 6 x3

4 2 4 x3 8

0

0

Part I: Mathematics I

I – 1.38 or

x1 1

x2 1

X1

1 1 1

x3 1

= 1.

Case (ii)

10 The eigenvector is given by 7 10 x1 1

0

x3 2

x2 1 1 1 2

X2 Case (iii)

7 x1 5 x2 7 x3

4 3 4

= 2. 9

4

7

4

10

4

The eigenvector is given by

or

7 x1 5 x2 8 x3

x1 12

x2 6

x3 12

x1 2

x2 1

x3 2

X3

2 1

0

2 Since one of the eigenvalues of A is zero, product of the eigenvalues = |A| = 0, i.e. A 11 7

4 2

7 5

10

4

6

11 12

20

4

42

50

7

28

20

0.

Example 1.6 Verify that the sum of the eigenvalues of A equals the trace of A and that their product equals |A|, for the matrix

A

1 0 0

0 3 1

0 1 3

Chapter I: Matrices

I – 1.39

The characteristic equation of A is 1

0 0 0

0 1

3 1

0

3

2

The eigenvalues of A Sum of the eigenvalues = 7. Trace of the matrix = 1 + 3 + 3 = 7 Product of the eigenvalues = 8. |A| = 1 × (9 – l) = 8.

Example 1.7 Verify that the eigenvalues of A2 and A–1 are respectively the squares and reciprocals of the eigenvalues of A, given that 3 1 4 0 2 6

A

0 0 5 The characteristic equation of A is 3

1 0

4 6

2

0

0

0

5

The eigenvalues of A Now

A2

3 1 4 3 1 4 0 2 6 0 2 6

9 5 38 0 4 42

0 0 5 0 0 5

0 0 25

The characteristic equation of A2 is 9

5 0 0

4 0

38 42

0

25

The eigenvalues of A2 are 9, 4, 25, which are the squares of the eigenvalues of A. Let

A

3 1 4 0 2 6 0 0 5

A11 = Co-factor of a11 = 10; A12 = 0; A13 = 0;

a11 a21 a31

a12 a22 a32

a13 a23 a33

Part I: Mathematics I

I – 1.40

A21 = – 5; A22 = 15; A23 = 0; A31 = – 2; A32 = – 18; A33 = 6 |A| = 30. A

10 1 0 30 0

1

1 3

1 6 1 2

0 0

2 18 6

1 15 3 5 1 5

0

The characteristic equation of A–1 is 1 3

5 15 0

1 6

1 15 3 5

1 2

0 0

0

0

1 5

1 1 1 0 3 2 5 1 1 1 The eigenvalues of A–1 are , , , which are the reciprocals of the eigenvalues of A. 3 2 5

i.e.

Example 1.8 Find the eigenvalues and eigenvectors of (adj A), given that the matrix 2 0 1 A 0 2 0 1 0

2

The characteristic equation of A is 2

0 0

1 0

2

1

0

0

2

3 2

The eigenvalues of A Case (i) The eigenvector is given by

1 0 0 1 1 0

1 x1 0 x2 1 x3

0

Chapter I: Matrices x1 1 X1

I – 1.41

x3 1

x2 0 1 0 1

Case (ii) 1 x1 0 x2 0 x3

0 0 0 0 1 0

The eigenvector is given by

0

i.e. –x3 = 0 and –x1 = 0 x1 = 0, x3 = 0 and x2 is arbitrary. Let x2 = 1 0 1 0

X2 Case (iii)

= 3.

The eigenvector is given by

x1 1

1 0

0 1

1

0

0.

x3 1

x2 0

X3

1 x1 0 x2 1 x3

1 0 1

The eigenvalues of A–1 are 1, Now i.e. adj A = |A| · A–1 = 6A–1 (

1 1 , with the eigenvectors X1, X2, X3. 2 3

adj A A A

A

1

6 for the given matrix A)

The eigenvalues of (adj A) are equal to 6 times those of A–1, namely, 6, 3, 2. The corresponding eigenvectors are X1, X2, X3 respectively. Example 1.9 Verify that the eigenvectors of the real symmetric matrix A are orthogonal in pairs. The characteristic equation of A is

3 1 1

1 5 1

1 1 3

Part I: Mathematics I

I – 1.42

3

1 1 1

1 1

5 1

3

0

3

2

The eigenvalues of A Case (i) The eigenvector is given by

x1 2

x2 0

1 1

1 3

1

1

x3 2

or

0 1

1 2

1

1

1 x1 1 x2 1 x3 x1 1

0

x3 1

x2 0

1 0 1

X1 Case (ii)

The eigenvector is given by

x1 1 X2

1 x1 1 x2 0 x3

0

1 x1 1 x2 3 x3

0

x3 1

x2 1 1 1 1

Case (iii)

= 6.

The eigenvector is given by

3 1

1 1

1

1 x1 2 X3

x2 4 1 2 1

x3 2

Chapter I: Matrices Now X1T X 2

I – 1.43

1 1 1 1

1 0

0

1 X 2T

X 3T

X3

1 1 1

X1

1

2 1

2 1

0 1 0 1

0

Hence the eigenvectors are orthogonal in pairs. Example 1.10 Verify that the matrix A

1 3

2 2

2 1 1 2

1

2 2

1

is an orthogonal matrix. Also verify that

is an eigenvalue of A

and that the eigenvalues of A are of unit modulus.

Note

A square matrix A is said to be orthogonal if AAT = ATA = I.

Now T

AA

1 3

2 2

2 1 1 2

1

2 2

9 0 0 1 0 9 0 9 0 0 9

2 1 2 3 1 1 0 0 0 1 0 0 0 1

2 2 1

1

2

3

1 2

2

2

The eigenvalues of 3A are given by

2

2

1

1 2

2 2

I

Similarly we can prove that ATA = I. Hence A is an orthogonal matrix. The characteristic equation of 3A is 2

2

0

Part I: Mathematics I

I – 1.44

2

4 2

36

1

i2 2

The eigenvalues of A are

1

Now

1

1,

1

1

1 2

i2 2 , 3

3

1 i2 2 3

1

1 2

and similarly

1

1 2

3 1 i2 2 i2 2 i2 2

3 i2 2

1 i2 2 3

3

.

3

1

is also an eigenvalue.

| = |1| =1.

1

2

1 3

i2 2 3

1 9

8 9

1

| = 1. 3 Thus the eigenvalues of an orthogonal matrix are of unit modulus.

EXERCISE 1(b) Part A (Short Answer Questions)

2. Prove that A and AT have the same eigenvalues. 3. Find the eigenvalues of 2A2, if A

4 1 . 3 2

4. Prove that the eigenvalues of (–3A–1) are the same as those of A

5. Find the sum and product of the eigenvalues of the matrix A

6. Find the sum of the squares of the eigenvalues of A

1 1 2

3 1 4 0 2 6. 0 0 5

1 2 . 2 1 2 0 1

2 3. 3

Chapter I: Matrices

7. Find the sum of the eigenvalues of 2A, if A

I – 1.45 8

6

6 2

7 4

2 4 3

.

2 2 1 1 3 1 are equal to 1 each. Find the 1 2 2

8. Two eigenvalues of the matrix A third eigenvalue.

9. If the sum of two eigenvalues and trace of a 3 × 3 matrix A value of |A|. 1 2 . 10. Find the eigenvectors of A 0 3 3 0 0 8 4 0. 6 2 5

11. Find the sum of the eigenvalues of the inverse of A

12. The product of two eigenvalues of the matrix A the third eigenvalue.

6 3

2 3

2 1 is 16. Find

2

1

3

Part B 13. Verify that the eigenvalues of A–1 are the reciprocals of those of A and that the respective eigenvectors are the same with respect to the matrix 1 5

A

2 . 4

14. Show that the eigenvectors of the matrix A

a

b

b a

are

1 i

1

and

i

.

Find the eigenvalues and eigenvectors of the following matrices: 2 2 15.

18.

21.

0 1

2 1 7 2

16.

3

1 1

2 2

2 1

1

1

0

2 2 1

2 1 2

3 6 0

2 2 1 19. 1 3 1 1 2 2

3 2 3

10 3 5

5 4 7

2 22. 1 1

2 1 3

2 2 17. 2 0

20.

2 1 1

7

1

2

1

3

6 2

2 3

2

1

2 1 0 23. 0 2 1 0 0 2

2 1 3

Part I: Mathematics I

I – 1.46 5 2 24. 0 0

2 2 0 0

0 0 5 2

0 0 2 2

8 6 2 25. Find the eigenvalues and eigenvectors of A 6 7 4 . 2 4 3 What can you infer about the matrix A from the eigenvalues. Verify your answer. 15 4 3 10 12 6 , verify that the sum and product of the eigen26. Given that A 20 4 2 values of A are equal to the trace of A and |A| respectively. 27. Verify that the eigenvalues of A2 and A–1 are respectively the squares and 3 0 0 reciprocals of the eigenvalues of A, given that A 8 4 0 . 6 2 5

28. Find the eigenvalues and eigenvectors of (adj A), when A

29. Verify that the eigenvectors of the real symmetric matrix A

2 1

1 2

1 1.

1

1

2

2 1

1 1

1

2

1 2 . 1

are orthogonal in pairs.

30. Verify that the matrix A

1 3

1 2 2 1

2 2

2 2

1

is orthogonal and that its

eigenvalues are of unit modulus.

1.7

CAYLEY-HAMILTON THEOREM

inverse of a matrix A. Also any positive integral power of A can be expressed, using this theorem, as a linear combination of those of lower degree. We give below the statement of the theorem without proof:

1.7.1

Statement of the Theorem

Chapter I: Matrices

I – 1.47

This means that, if c0 n + c1 n –1 + … + cn –1 cn = 0 is the characteristic equation of a square matrix A of order n, then (1) c0 An + c1 n –1 + … + cn –1 A + cn I = 0

Note: cn should be replaced by cn I to get the result of Cayley-Hamilton theorem, where I is the unit matrix of order n. Also 0 in the R.S. of (1) is a null matrix of order n.

1.7.2

Corollary

(1) If A is non-singular, we can get A–1, using the theorem, as follows: Multiplying both sides of (1) by A–1 we have c An –1 +c An –2 + … + c I + c A–1 = 0 0

A

n –1

1

1 c0 An cn

1

1

c1 An

2

n

cn 1I .

(2) If we multiply both sides of (1) by A, c0 An + 1 + c1 An + … + cn –1 A2 + cn A = 0 1 c1 An c2 An 1 cn 1 A2 cn A c0 Thus higher positive integral powers of A can be computed, if we know powers of A of lower degree. An

1.7.3

1

Similar Matrices

Two matrices A and B are said to be similar, if there exists a non-singular matrix P such that B= P–1 AP. When A and B are connected by the relation B = P–1 AP, B is said to be obtained from A by a similarity transformation. When B is obtained from A by a similarity transformation, A is also obtained from B by a similarity transformation as explained below: B= P–1 AP Premultiplying both sides by P and postmultiplying by P-1, we get PBP–1 = PP–1 APP–1 =A Thus A = PBP–1 Now taking P–1 = Q, we get A = Q–1 BQ.

1.8

PROPERTY

Two similar matrices have the same eigenvalues. Let A and B be two similar matrices. B = P–1 AP B

= P–1AP I –1 = P AP – P–1 IP

Part I: Mathematics I

I – 1.48

= P–1 (A I) P |B – I| = |P–1| |A – I| |P| = |A – I| |P–1P| = |A – I| |I| = |A – I| Thus A and B have the same characteristic polynomials and hence the same characteristic equations. A and B have the same eigenvalues.

1.8.1

Diagonalisation of a Matrix

M such that M–1 AM = D, where D is a diagonal matrix, is called diagonalisation of the matrix A. As M–1 AM = D is a similarity transformation, the matrices A and D are similar and hence A and D have the same eigenvalues. The eigenvalues of D M such that M–1 AM = D, D is not any arbitrary diagonal matrix, but it is a diagonal matrix whose diagonal elements are the eigenvalues of A. M for a given square matrix whose eigenvectors are distinct and hence whose eigenvectors are linearly independent.

1.8.2 Theorem If A is a square matrix with distinct eigenvalues and M is the matrix whose columns are the eigenvectors of A, then A can be diagonalised by the similarity transformation M–1 AM = D, where D is the diagonal matrix whose diagonal elements are the eigenvalues of A. be the distinct eigenvalues of A and X1, X2, . . . , Xn be the 1 2 n corresponding eigenvectors. Let M = [X1, X2, …, Xn], which is an n × n matrix, called the Modal matrix. AM = [AX1, AX2,…, AXn] [Note Each AXr is a (n × 1) column vector] , Since Xr is the eigenvector of A r AM

X 1 1

AXr= Xr (r=l,2,…n) X X] 2 2 n n

X1 , X 2 ,

, Xn

0

0 0

0 0

0

0

n

0 0

= MD (1) As X1, X2, . . . , Xn are linearly independent column vectors, M is a non-singular matrix Premultiplying both sides of (1) by M–1, we get M–1 AM = M–1 MD = D.

Note For this diagonalisation process, A need not necessarily have distinct eigenvalues. Even if two or more eigenvalues of A are equal, the process holds good, provided the eigenvectors of A are linearly independent.

Chapter I: Matrices

1.9

I – 1.49

CALCULATION OF POWERS OF A MATRIX A

Assuming A D = M AM A=MDM A2 = (M D M ) (M D M )

Similarly, Extending,

= MD(M M)DM = MD2M 3 A = MD3 M Ak = M Dk M

M

1.10

k 1

0

0

k 2

0 0

0 0

0

0

0

k n

M

1

DIAGONALISATION BY ORTHOGONAL TRANSFORMATION OR ORTHOGONAL REDUCTION

If A is a real symmetric matrix, then the eigenvectors of A will be not only linearly independent but also pairwise orthogonal. If we normalise each eigenvector Xr, i.e. divide each element of Xr by the square-root of the sum of the squares of all the elements of Xr and use the normalised eigenvectors of A to form the normalised modal matrix N, then it can be proved that N is an orthogonal matrix. By a property of orthogonal matrix, N = NT. The similarity transformation M A M = D takes the form NT AN = D. Transforming A into D by means of the transformation NT AN = D is known as orthogonal transformation or orthogonal reduction. Note: Diagonalisation by orthogonal transformation is possible only for a real symmetric matrix.

WORKED EXAMPLE 1(c)

Example 1.1 Verify Cayley-Hamilton theorem for the matrix A A . The characteristic equation of A is 1

3 4 1

7 3

2 2

1

0

1 3 7 4 2 3 1 2 1

Part I: Mathematics I

I – 1.50 2 3

2

Cayley-Hamilton theorem states that A3 A2

A2

Now,

A3

A A2

A

I=0

1 3 7 1 3 7 4 2 3 4 2 3 1 2 1 1 2 1

(1) 20 23 23 15 22 37 10 9 14

1 3 7 20 23 23 4 2 3 15 22 37 1 2 1 10 9 14

135 152 232 140 163 208 60 76 111

Substituting these values in (1), we get, 135 152 232 L.S. = 140 163 208 60 76 111

80 92 92 60 88 148 40 36 56

20 60 140 80 40 60 20 40 20

35 0 0 0 35 0 0 0 35

0 0 0 0 0 0 0 0 0 R.S. A , 2

A – 4A – 20I – 35A = 0 A

1

1 2 A 4A 35 20 23 1 15 22 35 10 9 1 35

20 I 23 37 14

4 1

11 6

5 25

6

1

10

Example 1.2 Verify that the matrix A A4. The characteristic equation of A is 2 1 1

4 12 28 16 6 8 12 4

8

2 1 1

1 2 1

1

2 1

2 1

2

4

2 1 2

0

20 0 0 20

0 0

0

20

0

Chapter I: Matrices

I – 1.51

2 3

2

According to Cayley-Hamilton theorem, A

A

Now

A3

2 1 1 2 1 1

2

A A2

1 2 1 1 2 1

A3

A2 + 8A

2 1 2 2 1 2

2 1 1 7 5 5

1 2 1 6 6 5

42 30 30

36 36 30

I=0

2 1 2 9 6 7

(2)

7

6

9

5 6 6 5 5 7 29 28 38 22 23 28 22 22 29

Substituting these values in (2), 29 22 22

L.S.

28 23 22

0 0 0 0 0 0

38 28 29

54 36 42

16 8 8

8 16 8

16 8 16

3 0 0 3 0 0

0 0 3

R.S.

0 0 0 Thus A Multiplying both sides of (2) by A, we have, A4 A3 + 8A2 A=0 4 3 2 A = 6A A + 3A 2

= 6(6A A + 3 I) = 28A2 A+18I A4 can be computed by using either (3) or (4). From (4),

A

4

Example 1.3 (A8 A7 + 7A6 A

2 1 1 0 1 0 . 1 1 2

196 140

168 168

252 168

140

140

196

124

123

95 95

96 95

A5 + 8A4

90 45 45

(3) 2

8A + 3A, using (2) (4)

45 90 0 45

90 45 90

18 0 0 0 18 0 0 0 18

16 62 123 124

A3 + 8A2

A + I), if the matrix

Part I: Mathematics I

I – 1.52

The characteristic equation of A is 2

1 0 1

1 0

1 1

0

2

2 3

2

3

2

A A + 7A I = 0, by Cayley-Hamilton theorem Now the given polynomial in A A2 + 7A 3I) + A(A3 A2 + 8A I) + I = A5(A3 3 2 2 = 0 + A(A A + 7A 3I) + A + A +I, by (1) = A2 + A +I, again using (1) 2 1 1 2 1 1 0 1 0 0 1 0 1 1 2 1 1 2

A2

Now

(1)

(2)

5 4 4 0 1 0 4 4 5

Substituting in (2), the given polynomial 5 4 4 0 1 0

2 1 1 0 1 0

1 0 0 0 1 0

4 4 5

1 1 2

0 0 1

8 5 5 0 3 0 5 5 8 An (n is a positive integer),

Example 1.4 Find the eigenvalues of A 1 2 . 4 3 The characteristic equation of A is 1 given that A

2 4

3

0

2

The eigenvalues of A n

2

Q

and the remainder be

(a + b). n

2

Q

+ (a + b)

(1)

n

(2)

a+b n

5a + b = 5

(3)

Solving (2) and (3), we get a

5n

( 1) n 6

and

b

5n

5( 1)n 6

Chapter I: Matrices

I – 1.53

A in (1), we have An

( A2 4 A 0 Q( A) 5n

5 I ) Q( A) aA bI aA bI (by Cayley-Hamilton theorrem)

( 1) n 6

5n

1 2 4 3

5( 1) n 6

1 0 0 1

For example, when n = 3, A3

125 1 1 2 6 4 3 21 84 41 84

42 63 42 83

125 5 1 0 6 0 1

20 0 0 20

2 2 2 1

Example 1.5 Diagonalise the matrix A

7 2 by similarity transforma-

0 1

3

7 2

0

A4. The characteristic equation of A is 2

2 2

1

0

1

3

2 3

Eigenvalues of A Case (i) = 1.

7 x1 2 x2

1 2

The eigenvector is given by 2 0 0 1 x1 2

X1

Case (ii)

+ 12 = 0

4 x3 x2 8 1 4 1

x3 2

0

I – 1.54

Part I: Mathematics I

The eigenvector is given by

1 2 0

x1 10

x2 12

X2

5 6 1

=

Case (iii)

7 x1 2 x2 6 x3

2 2 1

0

x3 2

. 7 x1 2 x2 1 x3

6 2 The eigenvector is given by 2 5 0 1 x1 3

x2 2

0

x3 2 3 2

X3

2 Hence the modal matrix is M a11

a12

M

a21 a22 a31 a32 Then the co-factors are given by Let

A11 = 14, A12 = 10, A31

1 5 4 6

3 2

1 1

2

a13 a23 a33

A13 = 2,

A32

A21

A22 = 5,

A33 = 26.

|M| = a11A11 + a12A12 + a13A13 = 70.

and

M

1

14 1 10 70 2

7 5

28 10

6

26

The required similarity transformation is M–1 A M = D

AM

2 2 2 1 0 1

7 2 3

1 5 4 6 1 1

3 2 2

A23

Chapter I: Matrices 1 15 4 18 1 3 M

1

AM

12 8 8

14 1 10 70 2

1 0 0 3 0 0 A4 is given by

28

1 15

12

5 6

10 26

4 18 1 3

8 8

0 0 280

0 0 4

A4 = M D4 M–1 4

1

D M

1 0 0 81 0

4

MD M

1

0 0

0 256

14 1 810 70 512 1 5 4 6 1 1 5600 1 3780 70 1820

i.e.

7

70 0 0 210 0 0

1 70

I – 1.55

A

4

80 54 26

37 79 38

(2)

14 1 10 70 2 7 405

7

28

5 6

10 26

28 810

1536 6656 14 3 1 810 2 70 512 2

7

28

405 15336

810

2590

15890

5530 266 60

18060 12530

227 258 179

6656

2 2 1 1 3 1 by Example 1.6 Find the matrix M that diagonalises the matrix A 1 2 2 means of a similarity transformation. Verify your answer. The characteristic equation of A is

Part I: Mathematics I

I – 1.56 2

2 1 1

ie

2

2

5

3 2

4

3

7

2

ie

0

2

21

ie

1

0

5

0

11 1

The eigenvalues of A Case (i)

1 1

2

5

0

= 5, 1, 1. 3 1 1

The eigenvector is given by

1 x1 1 x2 3 x3

2 2 2

x1 4

x2 4

X1

1 1

0

x3 4

1

Case (ii)

1 2 1 x1 The eigenvector is given by 1 2 1 x2 1 2 1 x3

0

All the three equations are one and the same, namely, x1 + 2x2 + x3 = 0 Two independent solutions are obtained as follows: Putting x2 = –1 and x3 = 0, we get x1 = 2 Putting x2 = 0 and x3 = -1, we get x1 = 1 2 1

X2

and

X3

0

1 0 1

Hence the modal matrix is M

1 1 1

2 1 0

Then the co-factors are given by A11 = 1, A12 = 1, A13 = 1, A21 = 2, A31 = 1, A32 = 1, A33 = –3 and

1 0 1

a11 a21 a31

A22 = –2,

a12 a22 a32

a13 a23 a33

A23 = 2

|M| = a11A11 + a12A12 + a13A13 = 4

Chapter I: Matrices

M

1 1 1 4 1

1

I – 1.57

2

1

2 2

1 3

The required similarity transformation is M

1

AM

D 5, 1, 1

(1)

We shall now verify (1).

AM

2 2 1 1 1 3 1 1 1 2 2 1 5 5 5

M

1

AM

2 2 2

1 1 1 4 1

2 2 2

1 1 3

1 5 0 5 1 5

2 2 2

1 1 3

2 1

1 1

1 2

1

2

1

1 1 3 2 1 0

20 0 0 1 0 4 0 4 0 0 4 5 0 0 0 1 0 0 0 1 D 5, 1, 1 .

Example 1.7 Diagonalise the matrix

A

by means of an

orthogonal transformation. The characteristic equation of A is 2

1 1 1

ie

2

2

2

1 2

3

1

ie

3

ie

1

The eigenvalues of A are 1 = –1, 1, 4.

4

2

1

1 2

0

1

0

4

0

1

4

0

I – 1.58

Part I: Mathematics I

Case (i) The eigenvector is given by

3 1

1 2

1

2

x1 0

x2 5

X1

0 1 1

1 x1 2 x2 2 x3

0

x3 5

Case (ii) 1 1

1 0

1

2

1 x1 2 x2 0 x3

x1 4

x2 2

x3 2

The eigenvector is given by

0

2 1

X2

1 Case (iii) The eigenvector is given by

2 1

1 3

1

2

x1 5

0

x3 5

x2 5

X3

1 x1 2 x2 3 x3

1 1 1

Hence the modal matrix M

0 1 1

2 1 1

1 1 1

Normalising each column vector of M by 2 , that of the second column by get the normalised modal matrix N.

6 and that of the third column by

3 , we

Chapter I: Matrices 2

1

1

6 1

3 1

2 1

6 1

3 1

2

6

3

0 N

Thus

I – 1.59

The required orthogonal transformation that diagonalises A is NT A N = D (–1, 1, 4)

2

1

1

6 1

3 1

2 1

6 1

3 1

2

6

3

0 2 1 1

AN

1 1 2

1 2 1

2

4

1

6 1

3 4

2 1

6 1

3 4

2

6

3

0

NT A N

2

4

1

6 1

3 4

6 1

2 1

6 1

3 4

3

2

6

3

1

1

2

2 1

2 1

6 1

6 1

3

3

0

(1)

0

1 0 0 0 1 0 0 0 4 D

1, 1, 4 .

Example 1.8 Diagonalise the matrix A transformation.

2 0 4 0 6 0 by means of an orthogonal 4 0 2

Part I: Mathematics I

I – 1.60

The characteristic equation of A is 2

0 0 4

ie

2

6

6 0

2

ie

6

ie

6

4 0

0

2 16 6

2

4

0 12

0 0

The eigenvalues of A Case (i) 2.

2, 6, 6.

4 0 4 x1 The eigenvector is given by 0 8 0 x2 4 0 4 x3 x1 32

0

x3 32

x2 0 1 0

X1

1 Case (ii) The eigenvector is given by

4 0 0 0 4 0

4 x1 0 x2 4 x3

0

We get only one equation, (1) i.e. x1 x3 = 0 From this we get, x1 = x3 and x2 is arbitrary. x2 must be so chosen that X2 and X3 are orthogonal among themselves and also each is orthogonal with X1. 1 Let us choose X2 arbitrarily as 0 1

Note

This assumption of X2 a Let X3 b c X3 is orthogonal to X1

x2 is taken as 0.

a

c=0

(2)

a+c=0

(3)

X3 is orthogonal to X2

Chapter I: Matrices

I – 1.61

Solving (2) and (3), we get a = c = 0 and b is arbitrary. Taking b

Note

1, X 3

0 1 0

Had we assumed X2 in a different form, we should have got a different

X3. For example, if X 2

1 2 , then X 3 1

1 1 1

1 1 0 0 0 1 1 1 0

The modal matrix is M

The normalised model matrix is

N

1

1

2 0 1

2 0 1

2

2

0 1 0

The required orthogonal transformation that diagonalises A is NT AN = D

AN

2 0 4 0 6 0 4 0 2

1

1

2 0 1

2 0 1

2 2

6

2 0 2

2 0 6

2

2

0 6 0

2

0 1 0

Part I: Mathematics I

I – 1.62

1 N T AN

2 1 2 0

0 0 1

1

2

6

2 0 2

2 0 6

2

2

2 1 2 0

0 6 0

2 0 0 0 6 0 0 0 6 D

2, 6, 6

Note

From the above problem, it is clear that diagonalisation of a real symmetric matrix is possible by orthogonal transformation, even if two or more eigenvalues are equal.

EXERCISE 1(c) Part A (Short Answer Questions) 1. State Cayley-Hamilton theorem. 2. Give two uses of Cayley-Hamilton theorem. 3. When are two matrices said to be similar? Give a property of similar matrices. 4. What do you mean by diagonalising a matrix? Ak, using the similarity transformation M 1 AM = D. 6. What is the difference between diagonalisation of a matrix by similarity and orthogonal transformations? 7. What type of matrices can be diagonalised using (i) similarity transformation and (ii) orthogonal transformation? 8. Verify Cayley-Hamilton theorem for the matrix A

5 3 . 1 3

A

A3, given that A

7 3 . 2 6 1 3 . 2 4

Chapter I: Matrices

I – 1.63

A4 – 4A3 – 5A2 + A + 2I), when A

1 2 . 4 3 5 3 1 3

12. Find the modal matrix that will diagonalise the matrix A Part B a b c d

13. Show that the matrix A A 1.

7 6 6

14. Verify Cayley-Hamilton theorem for the matrix A A

2 1 2

2 2 and 1

1

15. Verify Cayley-Hamilton theorem for the matrix A A 1. 16. Verify that the matrix A

1 2

2 1

3 4

3

1

1

1 1 2

1 2 1

1 3 and hence 3

A4. 17. Verify that the matrix A

1 2

0 1

3 1

1

1

1

A4. 18. Find An, using Cayley-Hamilton theorem, when A

5 3 1 3

7 3 19. Find An, using Cayley-Hamilton theorem, when A 2 6 1 0 3 1 1 , compute the value of (A6 20. Given that A 2 A5 + 8A4 1 1 1 2 9A + 31A I), using Caylay-Hamilton theorem. Diagonalise the following matrices by similarity transformation: 2 2 0 21. 2 1 1 7 2 3 22.

1 1 1 0 2 1 4 4 3

A4.

A3. A3

Part I: Mathematics I

I – 1.64 3 1 1

23.

1 5 1

1 1 3

1 24. 3 6

3 3 5 3 6 4

25.

6 2 2

2 3 1

2 1 3

2 1

1 2

1

1

1

0 1 1 26. 1 0 1 1 1 0 Diagonalise the following matrices by orthogonal transformation: 27.

10 2 5

2 2 3

5 3 5

2 2 1

2 1 2

3 6 0

30.

1.11

28.

3 1

1 2

0

1

0 1 3

29.

1 2

QUADRATIC FORMS

A homogeneous polynomial of the second degree in any number of variables is called a quadratic form. For example, x12 2 x22 3 x32 5 x1 x2 6 x1 x3 4 x2 x3 is a quadratic form in three variables. The general form of a quadratic form, denoted by Q in n variables is c11 x12

Q

c1n x1 xn

2 22 2

c21 x2 x1

c x

c311 x3 x1

c32 x3 x2

c3n x3 xn

cn1 xn x1

cn 2 xn x2

cnn xn2

n

i.e.

c12 x1 x2

c2 n x2 xn

n

Q

cij xi x j j 1 i 1

In general, cij

cji

xi xj = cij + cji. aij

1 cij 2

c ji , for all i and j, then aii = cii, aij = aji and

aij + aji = 2aij = cij + cji. n

n

Q

aij xi x j , where aij = aji and hence the matrix A = [aij] is a symmetric j 1 i 1

matrix. In matrix notation, the quadratic form Q can be represented as Q = XT AX, where

Chapter I: Matrices

A

aij , X

x1 x2

I – 1.65

and

XT

x1 , x2 ,

a11 a21

a12 a22

a1n a2 n

an1

an 2

ann

xn

The symmetric matrix A

, xn .

aij

is called the matrix of

the quadratic form Q.

Note

A

1 xi2 is placed in the aii position and 2 and aji positions. For example, (i) if Q

2 x12

3 x1 x2

A

(ii) if Q

then A

x12

3 x22

6 x32

1

1

3

1

3

5 2

3

5 2

6

2 x1 x2

coefficient xi x j is placed in each of the aij

4 x22 , then 2

3 2

3 2

4

6 x1 x3

5 x2 x3 ,

Conversely, the quadratic form whose matrix is 3 1 2 0

1 2

0

0

6 is Q

6

7

3 x12

7 x32

x1 x2

12 x2 x3

If A is the matrix of a quadratic form Q, |A| is called the determinant or modulus of Q. The rank r of the matrix A is called the rank of the quadratic form. If r < n (the order of A) or |A| = 0 or A is singular, the quadratic form is called singular. Otherwise it is non-singular.

Part I: Mathematics I

I – 1.66

1.11.2

Linear Transformation of a Quadratic Form T

Let Q = X AX be a quadratic form in the n variables x1, x2, . . . , xn. Consider the transformation X = PY, that transforms the variable set X = [x1, x2, . . . , xn]T to a new variable set Y = [y1, y2, . . . , yn]T, where P is a non-singular matrix. We can easily verify that the transformation X = PY expresses each of the variables x1, x2, . . . , xn as homogeneous linear expressions in y1, y2, . . . , yn. Hence X = PY is called a non-singular linear trans formation. By this transformation , Q X T AX is transformed to T Q = PY A PY Y T PT AP Y Y T BY , where B PT AP T

BT

PT AP PT AT P A is sysmmetric PT AP B B is also a symmetric matrix. Hence B is the matrix of the quadratic form YT BY in the variables y1, y2, . . . , yn. Thus YT B Y is the linear transform of the quadratic form XT AX under the linear transformation X = PY, where B = PT AP.

Now

1.11.3

Canonical Form of a Quadratic Form

In the linear transformation X = PY, if P is chosen such that B = PT A P is a diagonal 1

0

0

2

0 0

0

n

matrix of the form 0 Q

, then the quadratic form Q gets reduced as

Y T BY

, yn

y1 , y2 ,

2 1 1

y

2

y22

1

0

0

2

0

0

0 y1 0 y2 n

n

yn

yn2

This form of Q is called the sum of the squares form of Q or the canonical form of Q.

1.11.4 Orthogonal Reduction of a Quadratic Form to the Canonical Form If, in the transformation X = PY, P is an orthogonal matrix and if X = PY transforms the quadratic form Q to the canonical form then Q is said to be reduced to the canonical form by an orthogonal transformation.

Chapter I: Matrices

I – 1.67

We recall that if A is a real symmetric matrix and N is the normalised modal matrix of A, then N is an orthogonal matrix such that NT AN = D, where D is a diagonal matrix with the eigenvalues of A as diagonal elements. Hence, to reduce a quadratic form Q = XT AX to the canonical form by an orthogonal transformation, we may use the linear transformation X = NY, where N is the normalised modal matrix of A. By this orthogonal transformation, Q gets transformed into YT DY, where D is the diagonal matrix with the eigenvalues of A as diagonal elements.

1.11.5

Nature of Quadratic Forms

When the quadratic form XT AX is reduced to the canonical form, it will contain only r terms, if the rank of A is r. The terms in the canonical form may be positive, zero or negative. The number of positive terms in the canonical form is called the index (p) of the quadratic form. The excess of the number of positive terms over the number of negative terms in the canonical form i.e. p r -p) = 2p r is called the signature(s) of the quadratic form i.e. s = 2p r. The quadratic form Q = XT A X in n variables is said to be (i) positive definite, if r = n and p = n or if all the eigenvalues of A are positive. ( r = n and p = 0 or if all the eigenvalues of A are negative. r < n and p = r or if all the eigenvalues of A at least one eigenvalue is zero. r < n and p = 0 or if all the eigenvalues of A at least one eigenvalue is zero. A has positive as well as negative eigenvalues.

WORKED EXAMPLE 1(d)

Example 1.1 Reduce the quadratic form 2 x12

x22

x32

2 x1 x2

2 x1 x3

4 x2 x3 to

and nature of the quadratic form. Matrix of the Q.F. is A

2 1 1

1 1 2

1 2 1

Refer to the worked example (7) in section 1(c). The eigenvalues of A The corresponding eigenvectors are [0, 1, 1]T

T

T

respectively.

Part I: Mathematics I

I – 1.68

The modal matrix M

0 1 1

2 1 1

1 1 1 2

1

1

6 1

3 1

2 1

6 1

3 1

2

6

3

0 The normalised modal matrix N

T

D Hence N AN = D principal diagonal elements. The orthogonal transformation X = NY will reduce the Q.F. to the canonical form y12 y22 4 y32 Rank of the Q.F. = 3. Index = 2 Signature = 1 terms. Example 1.2 Reduce the quadratic form 2 x12

6 x22

2 x32

8 x1 x3 to canonical

form by orthogonal reduction. Find also the nature of the quadratic form. Matrix of the Q.F. is A

2 0 4 0 6 0 4 0 2

Refer to worked example (8) in section 1(c). The eigenvalues of A T

, [1, 0, l]T and [0, 1, 0]T respectively.

Note

Though two of the eigenvalues are equal, the eigenvectors have been so chosen that all the three eigenvectors are pairwise orthogonal. The modal matrix M

1 1 0 0 0 1

1 1 0 The normalised modal matrix is given by

N

1

1

2 0 1

2 0 1

2

2

0 1 0

Chapter I: Matrices

I – 1.69

Hence NT AN The orthogonal transformation X = NY 1 1 i.e. x1 y1 y2 2 2 x2 y2 1

x3

2

1

y1

2

y2 2 y12

will reduce the given Q.F. to the canonical form

6 y22

6 y32 .

negative terms. Example 1.3 Reduce the quadratic form x12 2 x22 x32 2 x1 x2 2 x2 x3 to the canonical form through an orthogonal transformation and hence show that it is x1, x2, x3) which makes this quadratic form zero. Matrix of the Q.F. is A

1 1

1 0 2 1

0

1 1 1

The characteristic equation of A is

i.e.

1

2

1 1 0

2 1

1

The eigenvalues of A

0

1

1

1 2

1

i.e.

0 1

0 3

0

= 0, 1, 3. x1

x2

x1 +2x2 +

x3 = 0 and x2 + x3 = 0. Solving these equations, x1 = 1, x2 = 1, x3 T

x2

x1 + x2 + x3 =

0 and x2 = 0. Solving these equations, x1 = 1, x2 = 0, x3 = 1. [1, 0, 1]T x1 x3 = 0 and x2 x3 = 0 Solving these equation, x1

Now the modal matrix is M

x2 = 2, x3 = 1. T . 1 1 1 0 1 1

1 2 1

x2

x1

x2 +

Part I: Mathematics I

I – 1.70

The normalised modal matrix is 1

1

3 1

N

1

2

6 2

0

3 1

6 1

1

3

2

6

Hence NT AN = Diag (0, 1, 3) The orthogonal transformation X = NY. i.e.

1

x1

3 1

x2

3

3

6 y1

1

y2

2 2

y1

1

x3

1

y1

6

y3

y3 1 2

1

y2

6

y3

will reduce the given Q.F. to the canonical form 0 y12 y22 3 y32 y22 3 y32 . As the canonical form contains only two terms, both of which are positive, the The canonical form of the Q.F. is zero, when y2 = 0, y3 = 0 and y1 is arbitrary. 3 , y2 = 0 and y3 = 0, we get x1 = 1, x2 = 1 and x3 Taking y1 These values of x1, x2, x3 make the Q.F. zero. Example 1.4 Determine the nature of the following quadratic forms without reducing them to canonical forms: i x12

3 x22

6 x32

2 1

5x

2 1

2 2

ii 5 x

iii 2 x

2 2

x

2 x1 x2 2 3

14 x 2 3

3x

2 x2 x3

2 x1 x2 12 x1 x2

4 x3 x1

16 x2 x3 8 x2 x3

8 x3 x1 4 x3 x1 .

Note alternative method uses the principal sub-determinants of the matrix of the Q.F., as explained below: Let A = (aij)n×n be the matrix of the Q.F. Let

D1 = |a11| = a11,

D3

a11 a21 a31

a12 a22 a32

D2 a13 a23 etc. a33

a11 a21

a12 , a22

and Dn = |A|

Chapter I: Matrices

I – 1.71

D1, D2, D3, … Dn are called the principal sub-determinants or principal minors of A. D1, D2, …, Dn are all positive i.e. Dn > 0 for all n. n

(i) Q

x12

3 x22

6 x32

D1, D3, D5, … are all negative and D2, D4, D6, Dn > 0 for all n. Dn 0 and least one Di = 0. n Dn 0 and at least one Di = 0.

2 x1 x2

4 x3 x1

1 1 2 1 3 1 2 1 6

Matrix of the Q.F. is A

1 1 1 3

D1 = |1| = 1; D2

Now

2 x2 x3

2 ;

D3 D1, D2, D3 are all positive. (ii) Q

5 x12

5 x22

14 x32

2 x1 x2 5 1

1 5

4 8

4

8

14

A

5 1 1 5

24;

A

5 70

64

30

18

Now D1 = 5; D2 D3

48

16 x2 x3

1 14

32

8 x3 x1 .

4

0

D1 and D2 are > 0, but D3 = 0 (iii) Q

2 x12

x22

3 x32

12 x1 x2

A

Now D1 = |2| = 2; D2

2 6 6 1

2 6 2

8 x2 x3 6 1 4 34;

2 4 3

4 x3 x1

8

20

Part I: Mathematics I

I – 1.72 D3

A

2

38

3 156

16

6

44

162

18

2 2

18 x3 x1 and 2 x 5x 4 x1 x2 real non-singular transformation.

2

6 x12

Example 1.5 Reduce the quadratic forms 2 1

8

3 x22

24

14 x32

2

4 x1 x2

4 x2 x3

2 x3 x1 simultaneously to canonical forms by a

Note

We can reduce two quadratic forms XT AX and XT BX to canonical forms simultaneously by the same linear transformation using the following theorem, (stated without proof): If A and B are two symmetric matrices such that the roots of |A B| = 0 are all distinct, then there exists a matrix P such that PT AP and PT BP are both diagonal matrices. The procedure to reduce two quadratic forms simultaneously to canonical forms is given below: (1) Let A and B be the matrices of the two given quadratic forms. (2) Form the characteristic equation |A B| = 0 and solve it. Let the eigenvalues . 1 2 n , (3) Find the eigenvectors Xi (i = 1, 2, …, n i using the equation (A B) X = 0. i i (4) Construct the matrix P whose column vectors are X1, X2, …, Xn. Then X = PY is the required linear transformation. (5) Find PT AP and PT BP, which will be diagonal matrices. (6) The quadratic forms corresponding to these diagonal matrices are the required canonical forms. 6 2 2 3

A

9 2

9 2 14 The matrix of the second quadratic form is 2 2 1 2 5 0

B The characteristic equation is |A

1 0 0 B| = 0 6 2 2 2 9

i.e.

3

1 1, , 1 5

2 2 3 5 2 2

9 2 14

+1=0

0

Chapter I: Matrices

I – 1.73

A B) X = 0 gives the equations. 8x1 + 4x2 + 10x3 = 0; 4x1 + 8x2 + 2x3 = 0; 10x1 + 2x2 + 14x3 = 0. Solving these equations,

x1 72

x3 48

x2 24

T

X1 1 , A 5

When

0 gives the equations.

B X

28x1 + 8x2 + 44x3 = 0; Solving these equations,

8x1 + 10x2 + 10x3 = 0; x1 360

x3 216

x2 72

T

X2 A

44x1 + 10x2 + 70x3 = 0.

B) X = 0 gives the equations 4x1 + 8x3 x2 + 2x3 = 0; 8x1 + 2x2+ 14x3 = 0 T

X3 Now P

3 1

5 1

2 1

2

3

1

[ X1 , X 2 , X 3 ]

PT AP Now

.

3 5 2

1 1 1

2 6 2 9 3 2 3 2 1 9 2 14

3 1 2

5 1 3

2 1 1

1 1 1

3 1 2

2 1 1

1 0 0 0 1 0 0 0 1

3 1 2

5 1 3

2 1 1

Hence the Q.F. XT AX is reduced to the canonical form y12

T Now P B P

3 5 2

1 1 1

2 2 2 1 3 2 5 0 1 1 0 0

2 5 1

1 5 1

3 5 2

3 1 2

5 1 3

3 1 2 2 1 1

5 1 3

y 22

y32 .

2 1 1 1 0 0 0 5 0 0 0 1

Hence the Q.F. XT B X is reduced to the canonical form y12 5 y22 y32 . Thus the transformation X = PY reduces both the Q.F.’s to canonical forms.

Note

X = PY is not an orthogonal transformation, but only a linear non-singular transformation.

I – 1.74

Part I: Mathematics I EXERCISE 1(d)

Part A (Short answer questions) 2

5 x22

2. Write down the matrix of the quadratic form 3 x1 6 x3 x1 .

5 x32

3. Write down the quadratic form corresponding to the matrix

2 x1 x2

2 x2 x3

2 1

1 2

2

2

2 2. 3

4. When is a Q.F. said to be singular? What is its rank then? 5. If the Q.F. XT AX gets transformed to YT BY under the transformation X = PY, prove that B is a symmetric matrix. 6. What do you mean by canonical form of a quadratic form? State the condition for X = PY to reduce the Q.F. XT AX into the canonical form. XT AX to the canonical form? 9. Find the index and signature of the Q.F. x12

Part B 11. Reduce the quadratic form 2 x12 12. Reduce the Q.F. 3 x12

3 x22

5 x32

5 x22 2 x1 x2

2 x22

3 x32

3 x32 .

4 x1 x2 to canonical form by an

6 x2 x3

6 x3 x1 to canonical form by

13. Reduce the Q.F. 6 x12 3 x22 3 x32 4 x1 x2 2 x2 x3 4 x3 x1 to canonical form by an orthogonal transformation. Also state its nature. 14. Obtain an orthogonal transformation which will transform the quadratic form 2 x12 2 x22 2 x32 2 x1 x2 2 x2 x3 2 x3 x1 also the reduced form. 15. Find an orthogonal transformation which will reduce the quadratic form 2 x1 x2 2 x2 x3 2 x3 x1 16. Reduce the quadratic form 8 x12 7 x22 3 x32 12 x1 x2 8 x2 x3 4 x3 x1 to the canonical form through an orthogonal transformation and hence show that it x1, x2, x3 that will make the Q.F. zero. 17. Reduce the quadratic form 10 x12 2 x22 5 x32 6 x2 x3 10 x3 x1 4 x1 x2 to a canonical form by orthogonal reduction. Find also a set of non-zero values of x1, x2, x3, which will make the Q.F. zero. 18. Reduce the quadratic form 5 x12 26 x22 10 x32 6 x1 x2 4 x2 x3 14 x3 x1 to a canonical form by orthogonal reduction. Find also a set of non-zero values of x1, x2, x3, which will make the Q.F. zero.

Chapter I: Matrices

I – 1.75

19. Determine the nature of the following quadratic forms without reducing them to canonical forms: (i) 6 x12 2 1

3 x22

14 x32 2 2

(ii) x

2 x1 x2

x

(iii) x12

2 x22

3 x32

4 x2 x3

18 x1 x3

4 x1 x2

2 3

x

2 x1 x2

2 x2 x3

2 x3 x1 x12

x22

x32

x1 x2 2 x2 x3 2 x3 x1 21. Find real non-singular transformations that reduce the following pairs of quadratic forms simultaneously to the canonical forms. (i) 6 x12

2 x22

3 x32

4 x1 x2

8 x3 x1 and 5 x12

(ii) 3 x12

3 x22

3 x32

2 1

2 2

x22

5 x32

2 x1 x2

8 x3 x1 .

2 x1 x2

2 x2 x3

2 x3 x1 and 4 x1 x2

2 x2 x3

2 x3 x1 .

2 3

4 x3 x1 and 2 x2 x3

2 x1 x2

x22 .

(iii) 2 x

2x

3x

2 x1 x2

4 x2 x3

(iv) 3 x12

6 x22

2 x32

8 x1 x2

4 x2 x3 and 5 x12

5 x22

x32

ANSWERS

Exercise 1(a) Part A 1 3 (6) X 1 X2 X3 2 2 (8) a = 8. (12) x + 2y = 3 and 2x – y = 1; x + 2y = 3 and 2x + 4y = 5. (13) x + 2y = 3 and 2x + 4y = 6. (14) a b = 6.

= 3. A x = k, y = 2k, = 5k. Part B X1 + X2 + X3 + X4 = 0; (25) 2X1 X2 X3 + X4 = 0; (26) 2X1 + X2 X3 = 0; (27) X1 X2 + X3 = 0; (28) X1 X2 + X3 X4 = 0; (29) Yes. X5 = 2X1 + X2 X3 + 0.X4. (34) R(A) = R[A, B] = 2; Consistent with many solutions.

8 x1 x2

2 x2 x3 .

Part I: Mathematics I

I – 1.76 (35) (36) (37) (38) (40) (42) (43) (44) (45) (46) (47) (48) (49)

R(A) = 3 and R[A, B] = 4; Inconsistent. R(A) = 3 and R[A, B] = 4; Inconsistent. R(A) = 3 and R[A, B] = 4; Inconsistent. Consistent; x = –1, y = 1, = 2. (39) Consistent; x = 3, y = 5, = 6. (41) Consistent; x = 2, y = 1, Consistent; x = 1, y = 1, = 1. Consistent; x1 = 2, x2 = 1, x3 x4 = 3. 1 4 Consistent; x 2 , y 0, , . 5 5 Consistent; x = 2k y k, = k. k 3 7 16k Consistent; x ,y , . 11 11 Consistent; x 16 9 k , y 16 6 k , . 3 5 3 5 k k y k + k = k, w = k Consistent; x Consistent; x1 k + 5k x2 = k, x k x4 = k k = 1, 2: When k = 1, x y When k = 2, x = 2 , y , = . x = k + 2, y k, = 5k. 1 1 x k 52 , y 3k 16 , . When 5 5

= 0. (51) a + 2 (52) No solution, when k = 1; one solution, when k when 2.

(54) If a = 8, b a a = 8 and b = 11, many solutions. (55) 2 3k. (56) x = k, y = 9k, 2k.

2; Many solutions,

= any = 6. b = any value, unique solution; If 4k, y = 2k, = 2k, w = k. , = 2k 9, x = 3k, y =

k, k, k k, k, 2k).

2k

k,

Exercise 1(b) (3) 2, 50. (6) 38. (8) 5. 1 and (10) 0 (12) 2.

(5) 2, 1. (7) 36. (9) 0. 1 . 1

(11)

47 . 60

(15) 1, 3 4; ( 2, 1, 4)T, (2, 1, 2)T, (1, 3, 13)T T

T

(16) 1, 5 , 5 ; 1, 0, 1 , 5 1, 1, 1 , 5 (17) 1, 3, 4; ( 1, 4, 1)T, (5, 6, 1)T, (3, 2, 2)T

1,

T

1, 1 .

,

Chapter I: Matrices (18) (19) (20) (21) (23) (24) (25) (26) (28) (29)

I – 1.77

5, 3, 3; (1, 2, 1)T, (2, 1, 0)T , (3, 0, 1)T 5, 1, 1, ; (1, 1, 1)T, (2, 1, 0)T, (1, 0, 1)T 8, 2, 2; (2, 1, 1)T, (1, 2, 0)T, (1, 0, 2)T 3,.2, 2; (1, 1, 2)T, (5, 2, 5)T 7)T, (0, 1, 1)T 2, 2, 2; (1, 0, 0)T. 1, 1, 6, 6; (0, 0, 1, 2)T, ( 1, 2, 0, 0)T, (0, 0, 2, 1)T and (2, 1, 0, 0)T 0, 3, 15; (1, 2, 2)T, (2, 1, 2)T, (2, 2, 1)T; A is singular Eigenvalues are 5, 10, 20; Trace = 25; |A| = 1000 1, 4, 4; (1, 1, 1)T, (2, 1, 0)T, (1, 0, 1)T 1, 1, 4; (0, 1, 1)T, (2, 1, 1)T, (1, 1, 1)T

Exercise 1(c) 1 6 36 2 3 2 (11) 4 5

3 7

(9)

(13)

(15)

(17)

1 11

bc

d c

b a 4 1

5 4

5

3

1

30 42 13 46

6

14 17

n (19) A

1 (14) 3

3 9

7 18

n (18) A

19 57 38 76 1 3 1 1

(12) M

1 ad

(10)

6n

2n

2 5

2 2

6

2

5

248 101 218 (16) 272 109 50 104 98 204

5 3 1 3

3.2n

6n 1 0 976 960 ; 0 1 320 336 2

4n 7 3 2 6 5

9.4n

4.9n 1 0 463 399 ; 0 1 266 330 5

4 9n

3 6

0 0 0 (20) 0 0 0 0 0 0 (22) D 1, 2, 3 ; M

(21) D 1, 3, 4 ; M 1 1 1 2 1 1; 2 0 1

A

4

99 115 65 100 116 65 160 160 81

2 1 4

2 1 2

1 3 13

Part I: Mathematics I

I – 1.78

1 1 0 1 1 1

(23) D 2, 3, 6 ; M

(24) D 4,

2,

1 2 1

1 1 1 1 2 0

2 ; M

2 1 1

(25) D 8, 2, 2 ; M

1 1 0 2 2 0 1 1 1

(26) D 2, 1, 1 ; M

(27) D 0, 3, 14 ; N

1 1 0 1

3

42 5

3 1

14 1

42 4

3 1

14 2

42

3

14

1

6 2

3,

3; N

3 1

0

6 1

3 1

1 2

3

1

1

1

3 1

2 1

6 1

3 1

2

6 2

0

3

(30) D 5,

1

2

6

(29) D 4, 1, 1 ; N

1 0 1

1

1

(28) D 1, 3, 4 ; N

1 0 1

6

1

2

1

6 2

5 1

30 2

6 1

5

30 5

6

0

30

Chapter I: Matrices

I – 1.79

Exercise 1(d) (2)

3

1 3

1

5 1 1 5

3 (3) (4) (6) (9)

2 x12 2 x22 3 x32 2 x1 x2 4 x2 x3 4 x3 x1. Singular, when |A| = 0; Rank r < n. PT AP must be a diagonal matrix. index = 2 and signature = 1. 1

(11) N

2

0

5 2

5 1

0

5 0

5 0

1 3

1

1

10

35 5

14 2

1

35 3

14 3

10

35

14

(12) N

0

2

(13) N

(14) N

0 1

3 1

6 1

2 1

3 1

6

2

3

1

1

1

3 1

2 1

6 1

3 1

2

6 2

0 1

1

3 1

2 1

5 1

3 1

2

5 2

0

;Q

6 y32 ; r

4 y12

y22

3; p

8 y32 ; r

3; s

3

3; p

;Q

8 y12

2 y22

1; s

1

2 y32 ; Q is positive definite

;Q

4 y12

y22

y32

;Q

2 y12

y22

y32 ; Q is indefinite

6

1

3

3 y22

1

6 1

3

(15) N

y12

;Q

5

Part I: Mathematics I

I – 1.80

(16) N

1 3 2 3 2 3

(17) N

(18) N

(21) (i) P

2 3 1 3 2 3

2 3 2 ;Q 3 1 3

15 y32 ; x1 = 1, x2 = 2, x3 = 2

3 y22

1

1

3

42 5

3 1

14 1

42 4

3 1

14 2

42

3

14

;Q

16

2

1

378 1

14 1

27 5

378 11

14 3

27 1

378

14

27

1 1 0 1 1 1 ; Q1

3 y22

14 y32 ; x1 = 1, x2 =

;Q

14 y22

y12

4 y22

5, x3 = 4.

27 y32 ; x1 = 16, x2 = 1, x3 = 11.

2 y32 ; Q2

y12

4 y22

y32

1 0 0

(ii) P

1 1 2

1 1 1 1 ; Q1 0 0

(iii) P

1 0 1 0 1 1 ; Q1 1 1 1

(iv) P

0 0 0 1 1 1

1 1 ; Q1 1

16 y12

y12

y22

2 y12

4 y22

y32 ; Q2

4 y22

8 y32 ; Q2

y22

y32 ; Q2

4 y12

4 y22

y32 .

y12

4 y22

y32 .

4 y32

Chapter

2

Three Dimensional Analytical Geometry 2.1

INTRODUCTION

Let X' OX, Y' OY and Z' OZ be three mutually perpendicular lines in space, that are concurrent at O (origin). [Refer to Fig. 2.1] These three lines, called respectively as x–axis, y-axis and z-axis (and collectively as co-ordinate axes), form the frame of reference, using which the co-ordinates of a

Note The positive parts of the co-ordinate axes, namely OX, OY, OZ should form a right-handed system. The plane XOY determined by the x-axis and the y-axis is called xoy-plane or xy-plane. Similarly the yz-plane and zx co-ordinate planes, divide the entire space into 8 parts, called the octants. The octant bounded by OX, OY, OZ

Z

x

L

y

M

P (x, y, z) z O

Y

N X Fig. 2.1

Part I: Mathematics I

I – 2.2

Let P be any point in space. Let PL (= x), PM (= y) and PN (= z) be the lengths of the perpendiculars drawn from P on the co-ordinate planes yoz, zox and xoy respectively. Then the triplet (x, y, z) are called the co-ordinates of P. Alternatively, the co-ordinates of P Draw PA, PB, PC perpendiculars to the x, y and z-axes intersecting them at A, B, C respectively. Then (OA, OB, OC) are called the co-ordinates of P.

Note

It can be proved that OA = PL= x, OB = PM = y and OC = PN = z. Using L, M, N are respectively (o, y, z), (x, o, z) and (x, y o). In fact, the x co-ordinate of any point in the yz-plane will be zero, the y co-ordinate of any point in the zx-plane will be zero and the z co-ordinate of any point in the xy plane will be zero. In other words, the equations of the yz, zx and xy-planes are x = 0, y = 0 and z = 0 respectively. The point A lies on the x-axis and hence in the zx and xy-planes. Hence the co-ordinates of A will be (x, o, o). Similarly the co-ordinates of B and C will be respectively (o, y, o) and (o, o, z). We give below two important basic formulas, the derivations of which are left as exercises to the students. 1. The distance between the two points P (x1, y1, z1) and Q (x2, y2, z2) is given by PQ

x1

x2

2

y1

y2

2

z1

z2

2

.

2. (i) If P (x1, y1, z1) and Q (x2, y2, z2) are two given points, then the co-ordinates of the point R which divides PQ 1 are x2

x1 y2 y1 z2 z1 , , 1 1 1

(ii) The co-ordinates of the point R', which divides PQ externally in the ratio 1 are

2.1.1

x2

x1 1

y2

y1 z2 z1 , . 1 1

Direction Cosines and Direction Ratios

made by it with the positive x-axis or equivalently by tan But in analytical solid geometry (of three dimensions), the direction of a , , made by it with the positive x, y and z-axis respectively.

Note

If a line L does not intersect OX, we draw a line L' parallel to L that intersects OX. The angle made by L' with OX L with OX. The angles , , are called the direction angles of the given line. The triplet (cos , cos , cos ,) are called the direction cosines (abbreviated as D.C.’s) of the line and usually denoted as (l, m, n). A set of paralled lines will make the same angles with the co-ordinate axes and hence will have the same D.C.’s.

Chapter 2: Three Dimensional Analytical Geometry

Note

2.1.2

1. If the D.C.’s of a line PQ are (l, m, n), then the D.C.’s of QP are (–l, –m, –n), as the angles made by QP with the co-ordinate axes are 180° – , 180° – , 180° – when the angles made by PQ with the axes are , , 2. The D.C.’s of OX, OY, OZ are respectively (1, 0, 0), (0, 1, 0) and (0, 0, 1).

Direction Cosines of a Line Through the Origin

Let (x, y, z) be the co-ordinates of P and let OP = r. [Refer to Fig. 2.2] If B is the foot of the perpendicular from P on the y-axis, then OB = y (1) Let OP make angles , , with the positive x, y and z-axes. Then D.C.’s of OP are (cos , cos , cos ). Now, from the triangle OBP, OB OP y , from (1) r x Similarly cos and cos r The D.C.’s of OP are x , y , z r r r

Z

O

Y

B

Fig. 2.2

z r

Relation Among the D.C.’s of a Line

Let (l, m, n) be the D.C.’s of a line AB. [Refer to Fig. 2.3] Consider the line OP parallel to AB. Then the D.C.’s of OP are also (l, m, n). Let P be (x, y, z) and OP = r.

Z

l

x , m r

y , n r

z r

Now OP2 = r2 i.e. x2 + y2 + z2 = r2, by distance formula. 2

2

2

B A

Then the D.C.’s of OP are x , y , z r r r Hence

P (x, y, z)

r

X

cos

2.1.3

I – 2.3

P (x, y, z)

O

Y

(1) X Fig. 2.3

x y z 1 r r r i.e. l2 + m2 + n2 = 1, using (1) Thus the sum of the squares of the D.C.’s of a line is equal to 1.

Part I: Mathematics I

I – 2.4

Direction Ratios of a line are any three numbers a, b, c that are proportional to its Direction cosines and are abbreviated as D.R.’s. a, b, c) be the D.R.’s and (l, m, n) be the D.C.’s of a line. a b c k, say. l m n l l2 1 2

i.e.

k k

a b , m and n k k 2 2 1 m n (a 2

b2

c2 )

a2

b2

Thus the D.C.’s are

c k

1 c2

a 2

2

2

2

2

,

b

a a

b

c

2

2

2

2

,

c

a b c a b c a b2 c2 In order to maintain the correspondence between the signs of a, b, c and the signs of l, m, n, the +ve sign is chosen. The D.C.’s of the line are 2

2

b

, a

2

b

c

2

c

, a

2

b2

c2

Note: 1. l2 + m2 + n2 = 1, whereas a2 + b2 + c2 2. To specify the direction of a line in space, its direction angles, direction cosines or direction ratios must be known. 3. The D.R.’s of two parallel lines are proportional.

2.2

DIRECTION RATIOS OF THE LINE JOINING TWO POINTS Q Z

Let P(x1, y1, z1) and Q(x2, y2, z2) be two given points. [Refer to Fig. 2.4] Let PQ make angles , , with the co-ordinate axes and let (l, m, n) be the D.C.’s of PQ. Then l = cos , m = cos , n = cos If PL and QM are drawn perpendicular to OY, then OL = y1, and OM = y2, so that PR = LM = y2 – y1. Now from triangle PQR, X y2 y1 PR cos PQ PQ x2 x1 z2 z1 and cos Similarly, cos PQ PQ The D.C.’s PQ are

x2 x1 y2 y1 z2 z1 , , PQ PQ PQ

P

O

L

Fig. 2.4

R

M

Y

Chapter 2: Three Dimensional Analytical Geometry

I – 2.5

Hence one set of D.R.’s of PQ are (x2 – x1, y2 – y1, z2 – z1). A (l1, m1, n1)

Z

2.2.1 Angle between Two Lines Let OA and OB be two lines through the origin parallel to two given lines whose D.C.’s are (l1, m1, n1) and (l2, m2, n2) [Refer to Fig. 2.5] Then the D.C.’s of OA are (l1, m1, n1) and the D.C.’s of OB are (l2, m2, n2) Let OA = 1 and OB = 1. If A is (x1, y1, z1), then D.C.’s of OA are x1 y1 z1 , , 1 1 1

B (l2, m2, n2) O

Y

X Fig. 2.5

Co-ordinates of A are (l1, m1, n1). Similarly, co-ordinates of B are (l2, m2, n2). OAB, AB2 = OA2 + OB2 – 2 . OA . OB cos i.e.

(l1 – l2)2 + (m1 – m2)2 + (n1 – n2)2 = 1 + 1– 2 cos

i.e.

(l12 + m12 + n12) + (l22 + m22 + n22) – 2(l1l2 + m1m2 + n1n2) = 2 – 2 cos 1 + 1 – 2 (l1l2 + m1m2 + n1n2) = 2 – 2 cos [ (l1, m1, n1) and (l2, m2, n2) are D.C.’s]

i.e.

l1l2

cos

n1n2 or

m1m2

= cos–1 (l1l2 + m1m2 + n1n2). Cor. 1: If the two lines are perpendicular, then = 90° or cos = 0. i.e.

l1l2

m1m2

n1n2

0.

We recall that, if the two lines are parallel, then l1 = l2, m1 = m2 and n1 = n2. Cor. 2: If the D.R.’s of the two lines are (a1, b1, c1) and (a2, b2, c2), then their D.C.’s are

a1 a12

,

b1 a12

,

c1 a12

a2

and

a22

,

If is the angle between the two lines, then a1a2 b1b2 c1c2 cos 2 a1 b12 c12 a22 b22

b2 a22

,

c2 a22

c22

Cor. 3: If the two lines whose D.R.’s are (a1, b1, c1) and (a2, b2, c2) are perpendicular, then = 90° or cos = 0, i.e. a1a2 b1b2 c1c2 0 We recall that if the two lines are parallel, then

a1 a2

b1 b2

c1 . c2

Part I: Mathematics I

I – 2.6

2.2.2

Projection of a Line Segment on a Given Line Q

Let P be (x1, y1, z1) and Q be (x2, y2, z2). [Refer to Fig. 2.6] P Let (l, m, n) be the D.C.’s of the given line R AB. segment PQ on the line AB. If PL, QM are drawn perpendicular from P A B and Q on the given line AB, L M of PQ on AB = LM = PR (where PR to QM) Fig. 2.6 = PQ cos , where is the angle between PQ and AB. (1) Now the D.R.’s of PQ are (x2 – x1, y2 – y1, z2 – z1) and the D.C.’s of PQ are x2 x1 y2 y1 z2 z1 , , PQ PQ PQ x2

x1

2

y2

y1

2

z2

, Also the D.C. s of AB are (l, m, n). Since is the angle between PQ and AB, 1 cos l x2 x1 m y2 PQ or

PQ cos

l x2

x1

m y2

y1 y1

2

z1

PQ

n z2 n z2

z1 (2)

z1

Using (2) in (1), PQ on AB = l (x2 – x1) + m (y2 – y1) + n (z2 – z1) WORKED EXAMPLE 2(a) Example 2.1 (i) Find the angle between any two diagonals of a cube. (ii) If a line makes angles , , , with the diagonals of a cube, prove that Z cos 2 a + cos 2 b + cos 2 c + cos 2 d = 4 . 3 C

B'

A'

O'

a a A

O

B a

C'

X

Fig. 2.7

Y

Chapter 2: Three Dimensional Analytical Geometry

I – 2.7

(i) Choose one of the vertices of the cube as the origin O and the three edges of the cube passing through O as the co-ordinate axes. Let the length of the edge of the cube be a. The vertices are named in such a way that the pairs (O, O'), (A, A'), (B, B') and (C, C') are opposite corners of the cube. [Refer to Fig. 2.7] Hence the diagonals of the cube are OO', AA', BB' and CC'. O (o, o, o); A a, o, o); B (o, a, o); C (o, o, a); O' a, a, a); A' o, a, a); B' a, o, a); C' (a, a, o). D.R.’s of OO' are (a, a, a); D.R.’s of AA' are (–a, a, a); D.C.’s of BB' are (a, –a, a); D.R.’s of CC' are (a, a, –a). D.C.’s of OO', AA', BB' and CC' are respectively 1 3

,

1 3

,

1 3

1

,

3

1

,

3

1

,

1

,

3

3

1

,

3

,

1

and

3

1 3

,

1 3

,

1 3

Let be the angle between two diagonals (one of which is OO'), say OO' and AA'. Then

1 3

cos

or

cos

1

1 3 1 3

1 3

1 3

If is the angle between two diagonals (none of which is OO'), say AA' and BB', then 1 3

cos or

cos

1 3

1 3

1 3

1 3

1

Thus, if is the angle between any two diagonals of a cube, then cos 1 cos 1 or 3

1 3

However, if the acute angle between any two diagonals is required, it is given by 1 cos 1 3 (ii) Let the D.C.’s of the line, say, L be (l, m, n). L makes an angle 1

diagonal OO', whose D.C.’s are

3 cos

l

1 3

m

1 3

n

1 3

,

1 3 1 3

,

with the

1 3 l

m

n

(1)

Part I: Mathematics I

I – 2.8

cos b =

Similarly,

cos c

1

l

3 1

l

3 1

m m

n

(2) (3)

n

(4)

l m n 3 Squaring both sides of (1), (2), (3) and (4) and adding, we get, cos2 + cos2 cos2 cos2 cos d

and

1 l m n 3 1 4 l 2 m2 3 4 3

2

l

m

2

n

n2

l

m

n

2

l

m

n

2

all the product terms of the form lm vanish [

(l, m, n) are the D.C.,s]

Example 2.2 If the vertices of a triangle ABC are (1, 8, 4), (4, 5, – 2) and (2, –3, 1) A B , D.R. s of AB are (3, – 3, – 6) and , D.R. s of AC are (l,–11,–3) A is the angle between the lines AB and AC. Cos A =

3 1

3 9

9

11

6

3

36 1 121 9

54 131 A

cos

1

54 131

50.06

, D.R. s of BC are (–2,–8, 3) B is the angle between the lines BC and BA. 2

Cos B

3 4 64

B

8 3 9 9

3 6 9

36

0 90

C is the angle between the lines CA and CB. 1 2

Cos C

11 8

1 121 9 4 C

cos

1

77 131

3

3

64

9

39.94 .

77 131

Chapter 2: Three Dimensional Analytical Geometry A (1, 8, 4)

Example 2.3 Find the co-ordinates of the foot of the perpendicular drawn from A (1, 8, 4) on the B (0, – 11, 4) and C (2, – 3, 1). Let D be the foot of the perpendicular. [Refer to Fig. 2.8] Since D lies on BC, it will divide BC in some ratio. : 1.

Note

> 0, D divides BC hence lies outsides BC. 2

Co-ordinates of D are

1 2

The D.C.’s of AD are i.e.

1 , 1

11

19 1

1,

1 ,

3

,

I – 2.9

B (0, –11, 4)

D

C (2, –3, 1)

Fig. 2.8

< 0, D divides BC externally and 11 , 1 3

4 1

11 1

(1) 4 1

8,

4

3 1

D.C.’s of BC are (2, 8, –3) AD is perpendicular to BC. 1 2 1 8 1

11

19

3

3

0

D are (4, 5, – 2) Example 2.4 Find the angle between the lines whose D.C.’s are given by the equations l + 3m + 5n = 0 and 2 l The given equations are

6 m

5 n

0.

l + 3m + 5n = 0 (1) and 2mn – 6nl – 5lm = 0 (2) If we solve (1) and (2), we will get two sets of values proportional to the D.C.’s of the two given lines. From(l), l = – (3m + 5n) (3) Substituting in (2), we get, 2mn + 6n (3m + 5n) + 5m (3m + 5n) = 0 i.e. m2 + 3mn + 2n2 = 0 i.e. (m + n) (m + 2n) = 0 m = – n or m = –2n m 1

i.e. Case (i)

m 1

n 1

k say

n m or 1 2

n . 1

Part I: Mathematics I

I – 2.10 m

k

and n

k

Using these values in (3), l = 2k. l 2 m 2

Case (ii)

n 1

m 1

k'

n 1

k

say m = 2 k' and n = –k'

Using these values in (3), we get l = – k' l

m 1 2 , Thus the D.R. s of the two lines are

n k' 1

(2, 1, –1)

and

(–1, 2, –1)

If is the angle between the two lines, then 2 2 cos = 4 1 1 1 or

= cos

1

1 4

1 6

1

1 6

, Example 2.5 Show that the straight lines whose D.C. s are given by the equations 2 2 2 al + bm + cn = 0 and pl + qm + rn = 0 are perpendicular if a2 (q + r) + b2(r +p) 2 b2 c2 + c2 (p + q) = 0 and are paralled if a p q r al + bm + cn = 0 pl2 + qm2 + rn2 = 0

0. (1) (2)

If we solve (1) and (2), we will get two sets of values proportional to the D.C.’s of the two lines. 1 l bm cn From (1), (3) a Using (3) in (2), we get, 1

i.e. i.e

2

qm 2 rn 2 a (pb2 + qa2) m2 + 2pbc mn + (ra2 + pc2) n2 = 0 p

pb 2

qa 2

m n

2

bm

cn

2 pbc

m n

2

ra 2

pc 2

0

0

(4)

Now, (4) is a quadratic equation in m . n m1 m and 2 If (l1, m1, n1) and (l2, m2, n2) are the D.C.’s of the two given lines, then n1 n2 are the roots of Equation (4).

Chapter 2: Three Dimensional Analytical Geometry

Product of the roots of (4) = m1m2 n1n2

i.e.

ra 2

pc 2

pb 2

qa 2

ra 2

pc 2

pb 2

qa 2

m1m2

or

ra 2

pc 2

n1n2 pb 2

I – 2.11

(5)

qa 2

Equations (1) and (2) are cyclically symmetric in a, b, c; p, q, r; l, m, n. Each of the ratios in (5) =

l1l2 qc

2

rb 2

k , say.

l1l2 = k (qc2 + rb2); m1m2 = k (ra2 + pc2); n1n2 = k (pb2 + qa2) The two given lines are perpendicular, if l1l2 + m1m2 + n1n2 = 0 i.e. k [(qc2 + rb2) + (ra2 + pc2) + (pb2 + qa2)] = 0 i.e. a2 (q + r) + b2 (r + p) + c2 ( p + q) = 0 The two given lines are parallel, if

l1 l2

i.e.

if

m1 m2

n1 n2

i.e.

m1 n1 m2 n2 m m2 if 1 n1 n2 if the roots of Equation (4) are equal

i.e.

if 4p2b2c2 – 4 (pb2 + qa2) (ra2 + pc2) = 0

i.e.

if a2 qr + b2rp + c2 pq = 0 a2 b2 c2 0. if p q r

i.e.

i.e.

Example 2.6 Prove that the four points A (2, 5, 3), B (7, 9, 1), C (3, – 6, 2) and D (13, 2, – 2) are coplanar. If the four points A, B, C, D are to be coplanar, the three lines AB, AC, AD must lie in a plane. If the lines AB, AC, AD lie in a plane, there must exist a common normal to the lines. D.C.’s of AB are (5,4, – 2) D.C.’s of AC are (1, – 11, – 1) D.C.’s of AD are (11, – 3, – 5) Let us assume that there exists a line with D.R.’s (a, b, c) which is perpendicular to each of the three lines. 5a + 4b – 2c = 0 (1) a – 11b – c = 0 (2) 11a – 3b – 5c = 0 (3)

Part I: Mathematics I

I – 2.12

If Equations (1), (2) and (3) are consistent and possess a non-trivial solution, it means that there exists a line with D.R.’s (a, b, c) which is to AB, AC and AD.

5 1 11

4 11 3

2 1 5

5 55

3

4

5

260 0

24

236

11

2

3

121

Equations (1), (2) and (3) are consistent and possess a non-trivial solution [Refer to Chapter 1 on Matrices] Hence the four points are coplanar. Example 2.7 If the distance between two points P and Q is d and the lengths of the pro 2 2 d 22 d32 . PQ on the co-ordinate planes are d1 d2, d3, show that 2d = d1 Let P be (x1, y1, z1), and Q be (x2, y2, z2). Then d2 = (x1 – x2)2 + (y1– y2)2 + (z1 – z2)2. P and Q on the xoy-plane are the feet of the rs drawn from P and Q on this plane. Let them be P' and Q'. Then P' is (x1, y1, 0) and Q' is (x2, y2, 0) Now d1 = PQ on the xoy-plane = P'Q' x1 or

d12

( x1

Similarly,

d 22

y1

y2

and

d32

z1

z2

d32

2 x1

d12

d 22

2

x2 x2 ) 2

y1

y2 ) 2

( y1 2

2

x2

z1

z2

x1

x2

2

2

y2

y1

2 2

y2

2

z1

z2

2

2d 2 . Example 2.8 (1, 1, 2) and D D.R.’s of CD are (–1, 4, 4)

AB on CD, if A

(1, 2, 3), B

B (–1, 0, 2) (1, 2, 3) A

C (1, 1, 2)

A'

B' Fig. 2.9

D (0, 5, 6)

C

Chapter 2: Three Dimensional Analytical Geometry

D.C.’s of CD are

1 33

,

4 33

,

4

I – 2.13

[Refer to Fig. 2.9]

33

x1, y1, z1) and (x2, y2, z2) on a line whose D.C.’s are (l, m, n) = l (x2 – x1) + m (y2 – y1) + n(z2 – z1) Here (x1, y1, z1) (1, 2, 3) and (x2, y2, z2) (–1, 0, 2) 1 33 10 33

4

2

33 10 33

2

4 33

1

units.

EXERCISE 2(a) Part A (Short Answer Questions) 1. Using direction cosines, prove that the points A (3, 1, 3), B (1, –2, –1) and C (–1, –5, –5) are collinear. 2. Find the angle made by a line with any of the co-ordinate axes, if it makes equal angles with the co-ordinate axes. 3. A line makes angles 30° and 60° with the x and y axes. Find the angle made by the line with the z-axis. 4. If a line makes complementary angles with any two co-ordinate axes, prove that it is perpendicular to the third axes. 5. If a line makes angles , , with the co-ordinate axes, prove that sin2 sin2 sin2 = 2. 6. A, B, C, D are the points (–3, 2, k), (4, 1, 6), (–1, –2, –3) and (13, –4, –1) respectively. Find the value of k, if AB is paralleled to CD. 7. A, B, C, D are the points (k, 3, –1), (3, 5, –3), (1, 2, 3) and (3, 5,7) respectively. Find the value of k, if AB is perpendicular to CD. 8. A, B, C, D are the points (1, 2, 4), (2, 4, 3), (2, 3, 4) and (0, 4, 5) respectively. Find the acute angle between AB and CD. 9. Find the direction cosines of a line perpendicular to the two lines whose D.R.’s are (1, 2, 3) and (–2, 1, 4). 3, 6. Part B 11. Find an expression for sin , where is the angle between the two lines whose D.C.’s are (l1, m1 n1) and (l2, m2, n2) and hence deduce the condition that the two lines are parallel. 12. If the edges of a rectangular parallelopiped are of lengths a, b, c, prove that the angles between the four diagonals are given by cos

1

a2 a2

b2 b2

c2 . c2

Part I: Mathematics I

I – 2.14

13. Find the angles of the triangle whose vertices are (1, 0, – 1), (2, 1, 3) and (3, 2, 1). (7, –2, 5). 15. Find the co-ordinates of the foot of the perpendicular drawn from the point 16. Prove that three concurrent lines with D.C.’s (l1, m1, n1), (l2, m2, n2), (l3, m3, n3) l1 m1 n1 are coplanar, if l2 m2 n2 l3 m3 n3

0

17. Prove that the four points (1, 2, 3), (–2, 4, –3) (4, 5, 7) and (4, 10, 5) are coplanar. Find also the D.C.’s of the normal to the plane in which they lie. 18. If A, B, C, D are the points (1, 0, 5), (– 1, 2, 4), (3, 4, 5) and (4, 6, 3) respecAB on CD. 19. Find the angle between the lines whose D.C.’s are given by the equations l + m + n = 0, 2lm – mn + 2ln = 0. 20. Prove that the lines whose D.C.’s are given by 3l + m + 2n = 0 and 4lm – 3mn – nl = 0 are perpendicular. 21. Show that the lines whose D.C.’s are given by the equations l + m + n = 0 and al2 + bm2 + cn2 = 0 are (i) perpendicular if a + b + c = 0 and (ii) parallel if 1 a

1 b

1 c

0

22. Show that the lines whose D.C.’s are given by the equations a2l + b2m + c2n = 1 1 1 0 and (ii) paral0 and lm + mn + nl = 0 are (i) perpendicular if 2 2 a b c2 lel if a ± b ± c = 0. 23. If two pairs of opposite edges of a tetrahedron are perpendicular, show that the third pair is also perpendicular. 24. If, in a tetrahedron OABC (where O is the origin), OA2 + BC2 = OB2 + CA2 = OC2 + AB2, prove that its pairs of opposite edges are at right angles.

2.3 THE PLANE lies on it, the surface is a plane surface or simply a plane. This is a characteristic property of a plane, that is not true for any other surface.

2.3.1

General Form of the Equation of a Plane

x, y, z, namely ax + by + cz + d = 0 (1) always represents a plane. Let A (x1, y1, z1) and B (x2, y2, z2) be any two points on the surface (1).

Note

An equation in (x, y, z) is assumed to represent a surface.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.15

Then ax1 + by1 + cz1 + d = 0 and ax2 + by2 + cz2 + d = 0 Let P be any point on the line AB and let it divide AB x2 x1 y2 y1 z2 z1 P is , , 1 1 1 Substituting these co-ordinates in the L.S. of (1), L S of 1

x2

a 1

1 1 1

x1 1

b

ax1

by1

0

y2

y1 1

cz1

c

d

(2) (3)

z2

z1 d 1

ax2

by2

cz2

d

0 , from 2 and 3

0 R S of 1 Thus the point P lies on the surface (1) Equation (1) represents a plane. Cor: If the plane ax + by + cz + d = 0 (1) passes through a given point (x1, y1, z1), (2) then ax1+ by1 + cz1 + d = 0 Using (2) in (1), Equation (1) can be written as a (x – x1) + b (y – y1) + c (z – z1) = 0.

Note

The general Equation (1) of a plane can be re-written as Ax + By + Cz + 1 = a 0, where A etc. This equation contains three independent constants. Thus, to d be known. Intercept form of the equation of a plane which cuts off intercepts a, b, c on the co-

Z R

Let the required equation of the plane be Ax + By + Cz + D = 0 (1) It makes intercepts a ( = OP), b (= OQ) and c (= OR) on the co-ordinate axes. [Refer to Fig. 2.10] i.e. It passes through the points P (a, o, o), Q (o, b, o) and R (o, o, c) Since P (a, o, o) lies on plane (1), we have Aa + D = 0 (2) Similarly Bb + D = 0 (3) and Cc + D = 0 (4) From (2), (3) and (4), we get A

D ,B a

D b

c b

O

Q

a P

X Fig. 2.10

and C

D c

Y

Part I: Mathematics I

I – 2.16

D x a

Using the values in (1), the required equation of the plane is D=0 x y z i.e. 1 a b c

D y b

D z c

Three points form of the equation of a plane which passes through the points (x1, y1, z1), (x2, y2, z2) and (x3,, y3, z3 Since the required plane passes through the point (x1, y1, z1), its equation can be assumed as a(x – x1) + b(y – y1) + c(z – z1) = 0 (1) The plane also passes through (x2, y2, z2) and (x3, y3, z3). and

a(x2 – x1) + b(y2 – y1) + c(z2 – z1) = 0

(2)

a (x3 – x1) + b (y3 – y1) + c(z3 – z1) = 0

(3)

Eliminating a, b, c from (1), (2) and (3), the equation of the plane is given by x x2 x3

x1 y x1 y2 x1 y3

y1 z y1 z2 y1 z3

z1 z1 z1

0

Note

Had we assumed the equation of the plane as ax + by + cz + d = 0, used the given conditions and eliminated a, b, c, d we would have got the equation of the plane as x y z 1 x1 y1 z1 1 x2 y2 z2 1 0 x3 y3 z3 1 N

Normal form of the equation of a plane in terms P (x, y, z) of p, the length of the perpendicular from the p origin on it and in terms of the direction cosines (or angles) of the perpendicular line. O Let the direction angles of the perpendicular ON drawn from O to the plane be , , Then Fig. 2.11 its D.C.’s are (cos , cos , cos ). [Refer to Fig. 2.11] Co-ordinates of N are (p cos , p cos , p cos ). Let P (x, y, z) be any point on the plane. Now if a line is to a plane, it is to every straight line in the plane. Thus ON to NP (1) Now D.R.’s of NP are (x – p cos , y – p cos , z – p cos ) From (1), we have Cos

(x – p cos ) + cos (y – p cos ) + cos (z – p cos ) = 0

i.e.

x cos

cos

( cos2

cos =p

+ cos2

cos2 )

Chapter 2: Three Dimensional Analytical Geometry Thus the equation of the plane is x cos + y cos

I – 2.17

cos

Length of the perpendicular from the origin on the plane ax + by + cz + d = Let the length of the perpendicular from the origin on the plane be p and let , x cos + y cos D.C.’s of the cos = p The equation of the given plane can be rewritten as ax + by + cz = – d Comparing (1) and (2), we have cos a p =-

Note

cos b

=

=

cos c

=

cos 2 + cos 2 + cos 2 a2 + b2 + c2

p = -d

the +z (1) (2)

(3)

d 2 a + b2 + c2

1. From (3), we see that a, b, c are proportional to cos , cos , Hence (a, b, c) are D.R.’s of the perpendicular line from the origin on the given plane, or (a, b, c) are D.R.’s of any normal of the plane ax + by + cz + d = 0. 2. The angle between any two planes is equal to the angle between the respective normals. Hence, if is the angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0, a1a2

then cos 2 1

2 1

a

b

b1b2 2 1

c

c1c2

a

2 2

b22

c22

3. If the two planes are perpendicular, their normals are also perpendicular. Hence the two planes are perpendicular if a1a2 + b1b2 + c1c2 = 0. 4. If the two planes are parallel, their normals are parallel. Thus the condition for parallelism of the two planes is a1 a2

b1 b2

c1 c2

1 k

Thus a2 = ka1, b2 = kb1, c2 = kc1. Using these values in the equation of the second plane, it becomes a1 x

b1 y

c1 z

d2 k

0 or a1 x

b1 y

c1 z

d1

0.

Thus the equations of two parallel planes i.e. a1x + b1y + c1z + d1 = 0 and a1x + b1y + c1z + d1 = 0 differ only in the constant term. Length of the Perpendicular from the point (x1, y1, z1) on the plane ax + by + cz + d = 0. Let Q (x2, y2, z2) be the foot of the perpendicular drawn from P (x1, y1, z1) on the given plane. [Refer to Fig. 2.12] D.R.’s of PQ are (x2 – x1, y2 – y1, z2 – z1) D.R.’s of any normal of the given plane are a, b, c.

Part I: Mathematics I

I – 2.18

Since PQ is a normal of the plane, we have x2

x1

y2

y1

z2

P (x1, y1, z1)

z1

k , say. a b c x2 = x1 + ka, y2 = y1 + kb, z2 = z1 + kc (1) Since Q (x2, y2, z2) lies on the given plane, ax2 + by2 + cz2 + d = 0

Q (x2, y2, z2)

i.e. a (x1 + ka) + b (y1 + kb) + c (z1 + kc) + d = 0 i.e. k (a2 + b2 + c2) = – (ax1 + by1 + cz1 + d) ax1

k

by1 a

2

cz1

b

2

d

(2)

2

PQ2 = (x2 – x1)2 + (y2 – y1)2 + (z2 – z1)2 = k2(a2 + b2 + c2), from (1)

Now

k a2

PQ

ax1

b2

c2

by1 a

2.3.2

c

Fig. 2.12

2

cz1 b

2

d

c

, using (2)

2

Positions of Points w.r.t. a Given Plane

ax + by + cz + d = 0 (1) P (x1, y1, z1) and Q (x2, y2, z2). Let the plane meet the line PQ at R. Let PR RQ = Co-ordinates of R are x2

x1 1

y2

P (x1, y1, z1)

R

y1 z2 z1 , 1 1

Q (x2, y2, z2) Fig. 2.13

Since R lies on the plane (1), we have a

x2

x1 1

b

y2

y1 1

c

z2

z1 1

d

0

ax2 + by2 + cz2 + d) = – (ax1 + by1 + cz1 + d) ax1

by1

cz1

d

ax2

by2

cz2

d

.

Now the points P (x1, y1 z1) and Q (x2, y2, z2) lie on the same (or opposite) side of the plane ax + by + cz + d = 0 if R lies outside (or within) PQ; (or positive); i.e. if (ax1 + by1 + cz1 + d) and (ax2 + by2 + cz2 + d) are of the same (or opposite) sign.

Chapter 2: Three Dimensional Analytical Geometry

2.4

I – 2.19

PLANE THROUGH THE INTERSECTION OF TWO GIVEN PLANES

Let the two given planes be P ax + by + cz + d = 0 and P ax+by+cz+d =0 The planes (1) and (2) intersect along a straight line, say, l. l or containing l. Consider the equation P + P (ax + by + cz + d) (a x + b y + c z + d = 0 x, y, z. Hence it represents a plane. Let A (x1, y1, z1) be any point on l. A lies on both the planes (1) and (2) ax1 + by1 + cz1 + d = 0 and a x1 + b y1 + c z1 + d = 0 Hence (ax1 + by1 + cz1 + d) (a x1 + b y1 + c z1 + d = 0 This means that A lies on plane (3) also. Thus all points on the line l lie on plane (3). i.e. Equation (3) represents a plane passing through the line of intersection of planes (1) and (2).

(1) (2)

(3)

(4) (5)

the

WORKED EXAMPLE 2(b) Example 2.1 Prove that the four points A (2, – 2, – 1), B (2, 3, 4), C (– 3, 5, 1), D (4, – 1, 2) are coplanar. Find also the distance of the point (1, –4, 5) from the plane ABCD. A, B and C and then verify that D lies on this plane. Since the plane passes through A (2, –2, –1), its equation may be assumed as a(x – 2) + b(y + 2)+c(z + 1) = 0

(1)

Plane (1) passes through B (2, 3, 4). 5b + 5c = 0 Plane (1) passes through C (–3, 5, 1)

(2)

–5a + 7b + 2c = 0 Eliminating a, b, c from (1), (2), (3), we get the equation of the plane ABC as

(3)

x

2 0 5

i.e. i.e.

y 1 z 1 5 5 7 2

0

– 25 (x – 2) – 25 (y + 2) + 25 (z + 1) = 0 x+y–z–1=0

(4)

I – 2.20

Part I: Mathematics I

Substituting the co-ordinates of D in (4), we have L.S. = 4 – 1 – 2 – 1 =0 = R.S. D also lies on the plane ABC or the four points are coplanar. Now distance of (1, – 4, 5) from plane (4) 1 4 5 1 1 1 1 units. 3 3 Example 2.2 Find the equation of the plane which passes through the points (6, 2, Since the required plane passes through (6, 2, – 4), its equation may be assumed as a (x – 6) + b (y – 2) + c (z + 4) = 0 Plane (1) passes through (3, –4, 1) –3a – 6b + 5c = 0

(1) (2)

Plane (1) is parallel to this line (–2, 2, 1) Any normal of plane (1) is to the line (–2, 2, 1). –2a + 2b + c = 0 (3) [ (a, b, c) are the D.C.’s of normal of plane (1)] We may eliminate a, b, c from (1), (2) and (3) to get the required equation as in From (2) and (3), by the rule of cross-multiplication, we have a b c 6 10 10 3 6 12 i.e.

a 16

b 7

c 18

k (say)

Using these values of a, b, c in (1), the equation of the plane is –16k (x – 6) –7k (y – 2) –18k (z + 4) = 0. i.e. 16x + 7y+ 18z – 38 = 0. Example 2.3 Find the equation of the plane passing through the point (1, 2, – 1) and perpendicular to the planes x + y – 2z = 5 and 3x – y + 4z= 12. Since the required plane passes through (1, 2, – 1), its equation may be assumed as a (x – 1) + b (y – 2) + c(z + 1) = 0 (1) Plane (1 ) is to the plane x + y – 2z = 5. Any normal of (1) [(a, b, c)] is to any normal of the given plane [(1, 1, –2)] a + b – 2c = 0 (2) Similarly, using the other condition, 3a – b + 4c = 0 (3)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.21

From (2) and (3), we have a 4 2

b 6 4

c 1 3

a b c k , say. 1 5 2 Using these values in (1), the required equation is (x – 1) – 5 (y – 2) – 2 (z + 1) = 0 i.e. x – 5y – 2z + 7 = 0

i.e.

Example 2.4 (1, 3, –2) and (3, 1, 6). A (1, 3, – 2) and B (3, 1, 6), it passes through the mid-point of AB, namely, (2, 2, 2). Since the plane cuts AB at right angles, AB is a normal of the plane. D.R.’s of AB are (2, – 2, 8) or (1, –1, 4) The equation of the required plane is (x – 2) – (y – 2) + 4 (z – 2) = 0. i.e. x – y + 4z – 8 = 0. Example 2.5 A variable plane is at a constant distance p from the origin and meets the co-ordinate axes in A, B and C. Through A, B and C, planes are drawn parallel to the co-ordinate planes. Prove that the locus of their point of intersection is given by 1

1

1

1

2

2

2

p2

x

y

z

x Let a typical position of the variable plane have the equation a Distance of plane (1) from origin = p 1 p i.e. 1 1 1 a2 b2 c2 i.e.

1

1

1

1

a2

b2

c2

p2

y b

z c

1

(1)

(2)

Since plane (1) meets the co-ordinate axes at A, B, C, we have OA = a, OB = b, OC = c. Equations of the planes through A, B, C, parallel respectively to yoz, zox and xoy–planes are x = a, y = b, z = c. Let the point of intersection of these three planes be (x1, y1, z1). Then x1 = a, y1 = b, z1 = c (3) Using (3) in (2), we have i.e.

1

1

1

1

x12

y12

z12

p2

Locus of the point (x1, y1, z1) is the surface

1

1

1

1

2

2

2

p2

x

y

z

.

Part I: Mathematics I

I – 2.22

Example 2.6 A plane meets the co-ordinate axes at A, B, C, such that the centroid of the triangle ABC is the point (a, b, c). Find the equation of the plane. Let the equation of the plane be

x p

y q

z r

1.

(1)

Then it intersects the co-ordinate axes at A (p, o, o), B (o, q, o) and C (o, o, r). The centroid of the triangle ABC is p

o 3

o o ,

q 3

o o ,

o 3

r

p q r , , 3 3 3 But the centroid is given to be (a, b, c)

i.e.

p q r a, b, c 3 3 3 or p = 3a, q = 3b, r = 3c Using (2) in (1), the required equation of the plane is x a

y b

z c

(2) 3.

Example 2.7 Find the foot of the perpendicular drawn from P (– 2, 7, – 1) to the plane 2x – y + z = 0 and also the image of P in the plane. Let Q(a, b, c) be the image of P in the plane 2x – y + z = 0. [Refer to Fig. 2.14] Then the mid-point M of PQ lies on the plane, P

Now

M

a 2 b 7 c 1 , , 2 2 2

Note

M is also the foot of the perpendicular drawn from P to the plane. M lies on 2x – y + z = 0.

M

Q

b 7

c 1 Fig. 2.14 0 2 2 i.e. 2a –b + c = 12 PQ is to the plane. i.e. PQ, whose D.R.’s are (a + 2, b –7, c + 1), is a normal to the plane. a 2

b 7 c 1 ( D.R.’s of any normal of the plane are 2, –1, 1) 1 1 = k, say a = 2k – 2, b = 7 – k, c = k – 1 Using (2) in (1), we get, 2 (2k – 2) – (7 – k) + (k – 1) = 12 i.e. 6k = 24 or k = 4.

(1)

a 2 2

(2)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.23

The co-ordinates of Q, from (2), are (6, 3, 3) Also M is (2, 5, 1). Example 2.8 Find the distance between the parallel planes 2x – 3y + 6z + 12 = 0 and 6x – 9y + 18z – 6 = 0. Find also the equation of the parallel plane that lies mid-way between the given planes. Let (x1, y1, z1) be any point on the plane 2x–3y + 6z +12 = 0 (1) Distance between the parallel planes = Length of the

from (x1, y1, z1) on the plane (6x– 9y + 18z – 6 = 0)

6 x1 9 y1 18 z1 6 62

9

2

182

3 2 x1

3 y1

6 z1

6

21 3

12

6

[

21

the point (x1, y1, z1) lies on plane (1)]

42 2. 21 Let the equation of the parallel plane that lies mid-way between the two given planes be 2x – 3y + 6z + k = 0 (3) Now distance of any point on plane (1) from plane (3) = 1. Let us take the point (– 6, 0, 0) on plane (1). Distance of (– 6, 0, 0) from plane (3) 12 4

9

k 36

k 12 7

1

12 – k = 7 k – 12 = 7 or k = 19 or k=5 Let us assume that k = 5 and verify that the distance of an obvious point on the plane 6x – 9y + 18z – 6 = 0 (2) is also 1. An obvious point on plane (2) is (1, 0, 0). Distance of (1, 0, 0) from plane (3)

i.e.

2 5 4 9 36 7 7

1

Thus the value of k = 19 is not admissible.

Part I: Mathematics I

I – 2.24

The equation of the required plane is 2x – 3y + 6z + 5 = 0. Example 2.9 Find the equation of the plane which passes through the line of intersection of the planes x + 5y – 2z = 6 and 5x – 4y + 5z = 2 and is parallel to the line The equation of any plane passing through the line of intersection of the two given planes is of the form. x + 5y – 2z – 6 + x – 4y + 5z – 2) = 0 Note The equations of the given planes must be rewritten in the form P = 0 and = 0 and then the equation of the required plane is P + = 0. x + (5 – y+ – 2) z – A (0, 3, 4) and B (1, 2, 3) are (1, –1, –1) Plane (1) is parallel to the line AB. 5– – 2) is to the line AB. – (5 –

–

– 2) = 0 –2=0

1 2 in (1), the equation of the required plane is 5 x 2 7x + 6y + z = 14. 1

i.e.

5

4 y 2

5 2

2 z

2 2

6

0

Example 2.10 The plane 4x – y + 5z – 7 = 0 is rotated through a right angle about its line of intersection with the plane 2x – 3y + 3z – 5 = 0. Find the equation of this plane in its new position. 4x – y + 5z – 7 = 0 New Position

90°

2x – 3y + 3z – 5 = 0

Fig. 2.15

Obviously the new position of the plane passes through the line of intersection of 4x – y + 5z – 7 = 0 (1) and 2x – 3y + 3z – 5 = 0 (2) [Refer to Fig. 2.15] Hence its equation is of the form 4x – y + 5z – (2x – 3y + 3z – 5) = 0 + 4)x – + 1)y + + 5)z – + 7) = 0 (3) Now plane (3) is to plane (1)

Chapter 2: Three Dimensional Analytical Geometry any normal of (3) is

I – 2.25

to any normal of (1). + 5) = 0 –

21 13

21 in (3), the equation of the required plane is l0x + 50y + 2z + 14 = 0 13 5x + 25y + z + 7 = 0. or Example 2.11 Find the equations of the planes through the line of intersection of the planes x + 3y + 6 = 0 and 3x – y – 4z = 0, whose distance from the origin is unity. The equation of any plane through the line of intersection of the two given planes is x + 3y + x – y – 4z) = 0 x + (3 – y – z+6=0 (1) Distance of (0, 0, 0) from (1) = 1. =–

6

i.e. 3

1

2

1

2

3

+ l)2 + (3 –

16 2

2

2

= 36

2

4x + 2y – 4z + 6 = 0 and – 2x + 4y + 4z + 6 = 0 i.e. the equations of the required planes are 2x + y – 2z + 3 = 0 and x – 2y – 2z –3 = 0. Example 2.12 Show that the planes 2x + 35y – 39z + 12 = 0, 6x + 6y – 7z – 8 = 0 and 12x – 15y + 16z – 28 = 0 have a common line. If the three planes

2x + 35y – 39z + 12 = 0

(1)

and

6x + 6y – 7z – 8 = 0 12x – 15y + 16z – 28 = 0

(2) (3)

have a common line, then plane (3) must be one of the planes through the intersection of planes (1) and (2). Now any plane through the intersection of planes (1) and (2) is of the form (2x + 35y – 39z x + 35) y –

i.e.

6

2 12

6

x + 6y – 7z – 8) = 0 z + (12 –

7

35 15

39 16

–

25 . 9

12

8 28

I – 2.26

Part I: Mathematics I

Hence the three planes have a common line.

11 . 9

EXERCISE 2(b) Part A (Short Answer Questions) 1. Find the intercepts made by the plane ax + by + cz + d = 0 on the co-ordinate axes. 2. Find the general equation of a plane parallel to the x-axis. 3. Write down the equation of the plane parallel to xoy-plane and 5 units above it. 4. Find the equation of the plane through the origin and parallel to the plane x + 2y = 3z + 4. 5. Find the equation of the plane through the point (–1, 2, –3) and perpendicular 6. Find the equation of the plane passing through the points (0, 0, 2), (0, – 1, 0) and (–3, 0, 0). 7. Find the angle between the planes 2x – y + z + 7 = 0 and x + y + 2z – 11 = 0. 8. Find the equation to a plane parallel to x – 2y + 2z = 0 and at a distance of 1 unit from the origin. 9. Find the distance between the parallel planes 3x + 6y + 2z = 22 and 3x + 6y + 2z = 27 10. Find the equation of the plane through the point P (2, 3, –1) and perpendicular to OP. 11. Find whether the points (2, 3, –5) and (3, 4, 7) lie on the same side of the plane x + 2y – 2z = 9. 12. Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and passing through the origin. 13. The D.R.’s of the perpendicular to a plane from the origin are (2, 1, –2) and the length of the perpendicular is 2. Find the equation of the plane. 14. The foot of the perpendicular from the origin to a plane is (2, 3, 6). Find the equation of the plane. 15. Find the equation of the plane through the line of intersection of x + y + z = 1 and 2x + 3y – z + 4 = 0 and parallel to z-axis. Part B 16. Prove that the four points (0, –1, –1), (4, 5, 1), (3, 9, 4) and (–4, 4, 4) are coplanar. Find also the intercepts made by the plane, in which the points lie, on the co-ordinate axes. 17. Prove that the four points (0, –1, 0), (2, 1, –1) (1, 1, 1) and (3, 3, 0) are coplanar. Find also the angles which the normal to the plane, in which the points lie, makes with the co-ordinate axes.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.27

18. Find the equation of the plane passing through the points (2, 2, 1) and (1, –2, 3) and parallel to (i) the x-axis, (ii) the y-axis and (iii) the z-axis. 19. Find the equation of the plane through the point (1, 0, –2) and perpendicular to the planes 2x + y – z = 2 and x – y – z = 3. 20. Find the equation of the plane through the point (–1, 1,–4) and perpendicular to the planes 2x – y – z = 2 and x + y – 3z + 1 = 0. 21. Find the equation of the plane through the points (–1, 1, 1) and (1, –1, 1) and perpendicular to the plane x + 2y + 2z = 5. 22. Find the equation of the plane through the points (1, –2, 2) and (–3, 1, –2) and perpendicular to the plane 2x + y – z + 6 = 0. 23. Find the equation of the plane which bisects perpendicularly the line segment 24. Two systems of rectangular axes have the same origin. If a plane cuts them at distances a1, b1, c1 and a2, b2, c2 from the origin, show that 1

1

1

a 12

b 12

c 12

1 a

1

2 2

b

2 2

1 c 22

.

x y z 1 meets the co-ordinate axes at A, B, C respectively. a b c If the angle BAC is equal to 60°, show that a2b2 + b2c2 + c2a2 = 3a4. 26. A variable plane which is always at a constant distance 3p from the origin cuts the co-ordinate axes at A, B, C. Show that the locus of the centroid of the 25. The plane

triangle ABC is

1

1

1

1

2

2

2

p2

x

y

z

.

a, b, c) and meets the coordinate axes at A, B, C. Show that the locus of the point of intersection of the planes through A, B, C respectively parallel to the co-ordinate planes is a b c 1. x y z 28. Find the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. Find also 29. Find the foot of the perpendicular drawn from the point (1, 0, 1) on the plane x + 2y + 3z = 6 and also its image in the plane. 30. Find the distance between the parallel planes x – 2y + 2z – 8 = 0 and x – 2y + 2z + 19 = 0. Find also the equation of the parallel plane which lies mid-way between the given planes. 31. Find the equation of the plane through the intersection of the planes x + y + z = 1 and 2x + 3y – z + 4 = 0, parallel to (i) y2, –3) and (2, 3, 4). 32. Find the equation of the plane passing through the line of intersection of the planes 2x – 5y + z = 3 and x + y + 4z = 5 and (i) parallel to the plane x + 3y + 6z = 1 and (ii) perpendicular to the plane 2x + y – z = 5. 33. The plane x – y – z = 2 is rotated through a right angle about its line of intersection with the plane x + 2y + z = 2. Find the equation to the plane in its new position.

Part I: Mathematics I

I – 2.28

34. The plane lx + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle of 45°. Prove that the equation of the plane in its new position, is lx

l 2 m2 z

my

0.

35. Prove that the planes x + 2y + 3z = 6, 3x + 4y + 5z = 2 and 5x + 4y + 3z + 18 = 0 have a common line of intersection.

2.5 THE STRAIGHT LINE In two dimensional analytical geometry, an equation in x and y represents a curve, but in three dimensional analytical geometry, an equation in x, y and z represents a surface. In three dimensions, a curve will be considered as the intersection of two surfaces and hence ax + by + cz + d = 0 and a'x + b'y + c'z + d , = straight line, i.e. the straight line of intersection of the two planes. For example, the x-axis is the line of intersection of the zox - plane whose equation is y = 0 and the xoy - plane whose equation is z = 0. Hence the equations of the x-axis are y = 0, z = 0. Similarly the equations of the y-axis are z = 0, x = 0 and those of the z-axis are x = 0, y = 0. Symmetric form of the equations of a straight line, passing through the point (x1, y1, z1) and having D.R.’s (l, m, n); Let P (x, y, z) be any point on the given line, that passes through A (x1, y1, z1). Z [Refer to Fig. 2.16] P (x, y, z) Now D.R.’s of AP are (x – x1, y – y1, z – z1). But D.R.’s of the same line are given as A (x1, y1, z1) (l, m, n). 0 Y Hence

x

x1 l

y

y1 m

z

z1 n

(1)

Equations (1) [Note 3 equal ratios give rise to 2 independent equations] are the equations of the given line. They are called the symmetric form of the equations of the line.

Note

X Fig. 2.16

1. Though the letters l, m, n have been used above, they represent only the D.R.’s of the line. 2. To write down the equations of a line in the symmetric form, we should know the co-ordinates of a point through which it passes and its D.R.’s. 3. If we equate each of the ratios in (1) to r, we get x = x1 +lr, y = y1 + mr, z = z1 + nr. Since (x, y, z) are the co-ordinates of any point on line (1), (x1 + lr, y1 + mr, z1 + nr) can be taken as the co-ordinates of any point on line (1), where r is a parameter.

Equations of the Line Joining the Points(x1, y1, z1) and (x2 , y2, z2) The D.R.’s of the line are (x2 – x1, y2 – y1, z2 – z1).

Chapter 2: Three Dimensional Analytical Geometry The line passes through the point (x1, y1, z1). Hence its equations are x x1 y y1 x2 x1 y2 y1

z z2

I – 2.29

z1 z1

2.6 THE PLANE AND THE STRAIGHT LINE Let us consider the plane x

and the straight line

ax + by + cz + d = 0 x1 y y1 z z1 l m n

(1) (2)

1. The straight line is normal to the plane, a l

if

b m

c n

the D.R.’s of any normal of the plane are a, b, c)

(

2. The straight line is parallel to the plane, if al + bm + cn = 0 ( the line is to the normal of the plane). 3. The straight line lies on the plane, if ax1 + by1 + cz1 + d = 0 (3) [ the plane also passes through the point (x1, y1, z1] and

al + bm + cn = 0

[

any line lying on a plane is to the normal of the plane] (1) – (3) gives the equation of the plane containing the line as a (x – x1) + b (y – y1) + c (z – z1) = 0

(4)

al + bm + cn = 0

provided the condition

(5)

4. If is the angle at which the line (2) is inclined to the plane (1), then the angle between the line (2) and the normal of the plane (1) is 90° – Thus

cos (90

)

al

sin (a

2.6.1

b

2

bm 2

c ) (l

x1

y

l1 x

m1 x2

l2

y1

y

z

z1 n1

y2 m2

z

z2 n2

are coplanar. The equation of a plane containing the line (1) is of the form a (x – x1) + b (y – y1) + c (z – z1) = 0 where

cn 2

m2

n2 )

Coplanar Straight Lines x

and

2

al1 + bm1 + cn1 = 0

(1) (2)

(3) (4)

Part I: Mathematics I

I – 2.30

Since the two lines are coplanar, plane (3) also contains line (2). Hence plane (3) passes through (x2, y2, z2) and any normal of (3) is Thus

to line (2).

a (x2 – x1) + b (y2 – y1) + c (z2 – z1) = 0

(5)

al2 + bm2 + cn2 = 0

(6)

and

Eliminating a, b, c from the Equations (5), (4) and (6) [the equations are assumed to be written in this order], we get x2

x1

y2

l1 l2

y1

z2

m1 m2

z1 n1 n2

(7)

0

(7) is the required condition for coplanarity of the two given lines. Eliminating a, b, c from the Equations (3), (4) and (6), we get x

x1

y

l1 l2

y1

z

m1 m2

z1 n1 n2

(8)

0

(8) is the equation of the plane in which the lines lie.

2.7

SHORTEST DISTANCE BETWEEN TWO SKEW LINES

Two straight lines which do not lie in the same plane are called non-planar or skew lines. Skew lines are neither parallel nor intersecting. Such lines have a common perpendicular. The length of the segment of this common perpendicular line intercepted between the skew lines is called the shortest distance (S.D.) between them. The common perpendicular line itself is called the S.D. line. lines

x

x1

y

l1 and

x

y1

z

m1 x2

l2

y

(1)

n1 y2

z

m2

z2

(2)

n2

Let L1, L2 be the skew lines passing through A (x1, y1, z1) and B (x2, y2, z2) respectively. [Refer to Fig. 2.17] Let MN be the S.D. between L1 and L2 and let (l, m, n) be the D.R.’s of the S.D. line. Since MN is to L1 and L2, we have

and

z1

ll1 + mm1 + nn1 = 0

(3)

ll2 + mm2 + nn2 = 0

(4)

A (x1, y1, z1)

M L1

B (x2, y2, z2)

N

Fig. 2.17

L2

Chapter 2: Three Dimensional Analytical Geometry

I – 2.31

Solving (3) and (4), we get l

m

m1 n2

m2 n1

n1 l2

n n2 l1

l1 m2

l2 m1

D.R.’s of MN are (m1n2 – m2n1, n1l2 – n2l1, l1m2 – l2m1) Hence D.C.’s of MN are m1 n2 (m1 n2

m2 n1 m2 n1 )

Now MN

2

n1 l2

,

n2 l1

(m1 n2

m2 n1 )

2

,

l1 m2

l2 m1 m2 n1 ) 2

(m1 n2

,

AB on MN 1 (m1 n2

{(m1 n2

m2 n1 ) 2

m2 n1 ) (x2

x1 ) + (n1l2

n2 l1 ) (y2

(l1 m2

y1 )

l2 m1 ) (z2

z1 )}

S.D. line is the intersection of the planes AMN and BMN. Now AMN is the plane containing the lines L1 and MN. Hence its equation is x

x1

y

y1

z

z1

l1

m1

n1

l

m

n

0

(5)

0

(6)

Similarly the equation of the plane BMN is x

x2

y

y2

z

z2

l2

m2

n2

l

m

n

2.8 ALITER length and equations of the S.D. The point M may be taken as (x1 + l1r1, y1 + m1r1, z1 + n1r1). and N may be taken as (x2 + l2r2, y2 + m2r2, z2 + n2r2) Then the D.R.’s of MN are found out. Then we use the fact that MN is to both L1 and L2 and obtain two equations in r1 and r2, solving which we obtain the values of r1 and r2. Using them, we know the co-ordinates of M and N. Then the length and equations of MN are found out.

Part I: Mathematics I

I – 2.32

WORKED EXAMPLE 2(c) Example 2.1 by the equations 3x – 2y + z – 1 = 0 and 5x + 4y – 6z – 2 = 0. The given line is the line of intersection of the planes 3x – 2y + z – 1 = 0 (1) and 5x + 4y – 6z – 2 = 0 (2) Let (l, m, n) be the D.R.’s of the line. Since it lies on plane (1), it is to the normal of the plane, whose D.R.’s are (3, – 2, 1). 3l – 2m + n = 0 (3) Similarly, 5l + 4m – 6n = 0 (4) Solving (3) and (4), we get l 12 4

m 5 18

n 12 10

l m n 8 23 22 To write down the symmetric form of the equations of the line, we require the co-ordinates of a point on it, apart from its D.R.’s. i.e.

z = 0 plane. Let it be (x1, y1, o) Since this point lies on the line, it lies on both the planes (1) and (2). 3x1 – 2y1 = 1

(5)

5x1 + 4y1 = 2

(6)

Solving (5) and (6), we get x1

4 and y1 11

1 22 4 11

x Hence the equations of the line in the symmetric form are x

8 y

l 25

8

m 12 5

n 4

30

2

1 22 23 z

z

0 22

4 and x + 2 1 1 5y – 2z = 6, 6x – 4y + 5z = 2. Find also the equations of the line passing through the origin and perpendicular to the given lines. Let (l, m, n) be the D.R.’s of the line given by the equations x + 5y – 2z = 6 and 6x – 4y + 5z = 2. Then l + 5m – 2n = 0 (1) and 6l – 4m + 5n = 0 (2) Solving (1) and (2), we get Example 2.2 Find the angle between the lines

6

y

Chapter 2: Three Dimensional Analytical Geometry

I – 2.33

l m n 1 1 2 Thus the D.R.’s of the second line given are (1, –1, 2).

i.e.

If is the angle between the two lines, then 2

cos 1

1

1

2

4 4

1

1

1 6 1 6 Let (a, b, c) be the D.R.’s of the required line. Since it is 2a + b – c = 0 or

cos

1

to each of the given lines, (3)

a – b + 2c = 0

and Solving (3) and (4), we get

a 2

(4)

b 1 4

1

a b 1 5 The required line passes through (0, 0, 0). Equation of the required line are

c 3

i.e.

x 1

c 2 1

y 5

z 3

Example 2.3 Find the foot, the length and the equations of the perpendicular drawn from the point (1, 0, – 3) to the line x

2 3

y

3 4

z

4 5

Since the foot of the perpendicular M lies on the given line, its co-ordinates may be assumed as (3r + 2, 4r + 3, 5r + 4). [Refer to Fig. 2.18] D.R.’s of PM are (3r + 1, 4r+ 3, 5r + 7) D.R.’s of the given line are (3, 4, 5) Since PM is to the given line,

P (1, 0, –3)

3(3r + 1) + 4(4r + 3) + 5(5r + 7) = 0 i.e. or

50r + 50 = 0 r = –1

M Fig. 2.18

I – 2.34

Part I: Mathematics I

Co-ordinates of the foot M are (– 1, –1, –1). Now the length of PM

1 1

2

0 1

2

3 1

2

3 units. Equations of PM are x 1 y 0 z 3 2 1 2 x 1 Example 2.4 Find the equations of the image of the line 3 in the plane 2x – y + z + 3 = 0. P

A

y

3 5

z

4 2

(1, 3, 4)

M

Q

(a, b, c)

Fig. 2.19

be advantageous if we choose one of the points as the point A of intersection of the line x 1 y 3 z 4 (1) 3 5 2 and the plane 2x – y + z + 3 = 0 (2) as the image of the point A in the plane is itself [Refer to Fig. 2.19]. Since A is a point on the line (1), its co-ordinates may be assumed as (3r + 1, 5r + 3, 2r + 4) (3) Since A lies on the plane (2), we have 2 (3r + 1) – (5r + 3) + (2r + 4) + 3 = 0 i.e. 3r + 6 = 0 or r = –2. Putting r = –2 in (3), we get the co-ordinates of A as (–5, –7, 0). Now let us consider another point on the line (1). We may choose the obvious point P (1, 3, 4) through which the line passes. Let the image of P be the point Q(a, b, c). M of PQ lies on the plane (2). a 1 b 3 c 4 Co-ordinates of M are , , 2 2 2 As M lies on plane (2), we have a 1 b 3 c 4 2 3 0 2 2 2 i.e. 2a – b + c + 9 = 0 (4)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.35

D.R.’s of PQ are (a – 1, b – 3, c – 4) PQ is normal to plane (2). a 1 2

b 3 1

c

4

k , say

1

(5)

From (5), a = 2k + 1, b = –k + 3, c = k + 4 Using these values in (4), we get 2(2k + 1) – (–k + 3) + (k + 4) + 9 = 0 i.e. 6k + 12 = 0 or k = –2 Putting k = –2 in (6), we get a = –3, b = 5, c = 2 Co-ordinates of Q are (–3, 5, 2) A (–5, –7, 0) and Q (–3, 5, 2) Hence the equations of the image line are x 5 y 7 z 0 3 5 5 7 2 0 x

z 1 6 1 Example 2.5 Find the equation of the plane that contains the parallel lines

i.e.

x 1 1

y

2 2

5

z

3 3

y

(6)

7

x 3 1

and

y

2 2

z

4 3

a (x – 1) + b (y – 2) + c ( z – 3) = 0 (1) [ the plane contains the obvious point of the line, namely, (1, 2, 3)] where a + 2b + 3c = 0 (2) [ the line is to any normal of the plane (1)] If plane (1) also contains the second line, it has to pass through the obvious point of the second line namely, (3, –2, –4)] 2a – 4b – 7c = 0 (3) Solving (2) and (3), we get a 14 12 i.e.

a 2

b 6 b 13

7 c 8

c 4 4 k , say

Using these values of a, b, c in (1), the equation of the required plane is –2k(x – 1) + 13k (y – 2) –8k (z – 3) = 0 i.e. 2x – 13y + 8z = 0 y 3 z 5 Example 2.6 Find the equation of the plane which contains the line x = 2 3 and which is (i) perpendicular to plane 2x + 7y – 3z = 1 and (ii) parallel to the line

I – 2.36

Part I: Mathematics I x – 2y + 3z – 4 = 0 = 2x – 3y + 4z – 5

(i) Equation of the plane containing the given line is of the form ax + b (y – 3) + c(z – 5) = 0 where a + 2b + 3c = 0 The required plane (1) is to the plane 2x + 7y – 3z=1 Any normal of (1) is to any normal of (3) 2a + 7b – 3c = 0 Solving (2) and (4), we have a 6 i.e.

b 21 a 9

6 b 3

(1) (2) (3) (4)

c 3 c 1

7

4 (5)

k

Using these values of a, b, c from (5) in (1), we have –9x + 3 (y – 3) + (z – 5) = 0 Thus the equation of the required plane is 9x – 3y – z + 14 = 0 (ii) Let the D.R.’s of the line given by x – 2y + 3z – 4 = 0 and 2x – 3y + 4z – 5 = 0 be (l, m, n) Then l – 2m + 3n = 0 (6) 2l – 3m + 4n = 0 (7) ( the line lies on each of the planes and hence plane)

to the normal of each

l m n 1 2 1 Thus the D.R.’s of the line are (1, 2, 1). Now required plane (1) is parallel to this line. Hence a normal of plane (1) is to this line. a + 2b + c = 0 (8) Solving (2) and (8), we have a b c (9) 4 2 0 Using (9) in (1), the equation of the required plane is –4 (x) + 2 (y – 3) = 0 i.e. 2x – y + 3 = 0.

Solving (6) and (7), we get

y z Example 2.7 Assuming that the line x 4 5 6 of the line of the greatest slope in the plane 3x – 4y + 5z = 5 passing through the point (3, –4, –4).

Note

Let be the given plane and AB be the line of intersection of horizontal plane

with a

Chapter 2: Three Dimensional Analytical Geometry

I – 2.37

Plane

P A M

Plane

B Fig. 2.20

Let P be any point on the plane and PM perpendicular to AB. Then PM is the line of the greatest slope in the plane through P. [Refer to Fig. 2.20] Thus the line of the greatest slope in a given plane is a line which lies in the plane and which is perpendicular to the line of intersection of the plane with a horizontal plane.) x 4

y 5

z is given to be a vertical line. 6

Hence any plane perpendicular to this line will be a horizontal plane. Let us consider the horizontal plane passing through the origin. Its equation is 4x – 5y + 6z = 0 (1) The given plane is 3x – 4y + 5z = 5 (2) Let (a, b, c) be the D.R.’s of the line of intersection of the planes (1) and (2). Then 4a – 5b + 6c = 0 and 3a – 4b + 5c = 0 Solving (3) and (4), we get a b 25 24 18 20

(3) (4) c 16 15

a b c 1 2 1 Thus the D.R.’s of the line of intersection of planes (1) and (2) are (–1, –2, –1). Let (l, m, n) be the D.R.’s of the required line of the greatest slope. Since the required line is to the line whose D.R.’s are (–1, –2, –1), we have –l – 2m – n = 0 or l + 2m + n = 0 (5) Also, the required line lies on the given plane and hence it is to any normal of the given plane. 3l – 4m + 5n = 0 (6) Solving (5) and (6), we get l m n 10 4 3 5 4 6 i.e.

Part I: Mathematics I

I – 2.38

l m n 7 1 5 Also the required line passes through the point (3, – 4, – 4) Hence the equations of the required line of the greatest slope are i.e.

x 3 7

y

Example 2.8 Show that the lines x 4 2 z 4 5

4 1

z

y 5 3

4. 5 z 6 and x 2 3 4

y 3 4

Let us verify whether the condition for coplanarity of the lines x xr lr z zr ; r 1, 2, nr i.e

x2

Now

x1 l1

y2 y1 m1

l2

m2

z2

y

yr mr

z1 n1

0

n2

2 4 3 5 4 6 2 3 4 3

4

2 15 16

2 10 12

2 8 9

x x1 l1

y1

5 2 4 2 0

are coplanar. The plane in which the two lines lie is given by

l2 i.e.

x 4 2 3

i.e. i.e. i.e.

y 5 z 6 3 4 4

y

z

z1

m1

n1

m2

n2

0

0

5

(15 – 16) (x – 4) – (10 –12) (y – 5) + (8 – 9)(z – 6) = 0. – (x – 4) + 2 (y – 5) – (z – 6) = 0 x – 2y + z = 0.

z 1 Example 2.9 Show that the lines x 4 y 3 z 2 and x 3 y 2 7 4 3 5 2 6 are coplanar. Find their point of intersection and the equation of the plane in which they lie.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.39

(5r1 + 4, –2r1 + 3, – 6r1 + 2) The co-ordinates of any point on the second line are

(1)

(4r2 + 3, – 3r2 + 2, –7r2 + 1)

(2)

For some values of r1 and r2, if the co-ordinates of the points given by (1) and (2) are identical, it means that the two given lines have a common point, i.e. they intersect and hence are coplanar. Hence we have to establish that there exist values of r1 and r2 which make the co-ordinates in (1) and (2) identical. Let us assume that the co-ordinates in (1) and (2) are identical. Then 5r1 + 4 = 4r2 + 3, i.e. 5r1 – 4r2 = – 1 (3) i.e. 2r1 – 3r2 = 1 (4) –2r1 + 3 = –3r2 + 2, i.e. 6r1 – 7r2 = 1 (5) and –6r1 + 2 = –7r2 +1, Solving (3) and (4), we get r1 = –1 and r2 = – 1. These values of r1 and r2 satisfy equation (5) also. Thus there exist values of r1 and r2 which make the co-ordinates in (1) and (2) identical. Hence the two given lines are coplanar. The point of intersection is obtained by putting r1 = –1 in (1) or r2 = –1 in (2) The point of intersection is (–1, 5, 8). x 4 y 3 z 2 The equation of the plane containing the two lines is 5 2 6 0 4

3

7

i.e. (14 – 18) (x – 4) – (–35 + 24) (y – 3) + (–15 + 8) (z – 2) = 0 4x – 11y + 7z + 13 = 0

i.e.

x 1 y 1 z 1 ; x + 2y + 3z = 14, 3x + 2 3 4 4y + 5z = 26 are coplanar. Find the co-ordinates of their point of intersection and also the equation of the plane in which they lie. The second line may be put in the symmetric form and then the method of the previous example may be adopted to solve this problem. We give an alternative Example 2.10 Prove that the lines

(2r – 1, 3r – 1, 4r – 1). If this point lies on the second line given by x + 2y + 3z = 14 and 3x + 4y + 5z = 26 it must lie on both the planes (1) and (2). Hence and i.e. and

(1) (2)

2r – 1 + 2 (3r – 1) + 3 (4r – 1) = 14 3(2r – l) + 4 (3r – l) + 5 (4r – l) = 26 20r = 20 38r = 38

(3) (4) r, i.e. r = 1.

Part I: Mathematics I

I – 2.40

Putting r = 1, the co-ordinates of the point become (1, 2, 3). This is the point of intersection of the two lines, as it lies on both the given lines. Hence the two lines are coplanar. The equation of any plane containing the second line is of the form x + 2y + 3z x + 4y + 5z – 26) = 0 (5) should lie on the plane. 10 19

x

10 3 x 4 y 5 z 26 0 19 11x + 2y – 7z + 6 = 0.

2 y 3 z 14

i.e.

Example 2.11 Show that the lines x + y + z – 3= 0 = 2x + 3y + 4z – 5 and 4x – y + 5z – 7 = 0 = 2x – 5y – z they lie. If the point of intersection of the two lines is required, we have to adopt the procedures of examples (9) and (10) after putting the equations of one or both the lines in the symmetric form. As the point of intersection of the two lines is not required, the following alternative x+y+z

x + 3y + 4z – 5) = 0 x y z Equation of any plane through the second line is 4x i.e.

5z 7

y

x

4 2

2x 5 y y

1 5

z 3

0

z

7 3

5

0

(2)

, then that plane contains both the given lines and hence the two given lines are exist. are proportional. i.e.

1 2 4 2

1 3 1 5

16

14

10

6

3 5

1 4 5

7 3

7 3

5 0 1 0

(4) × 5 – (5) × 8 gives 22

11

33 0

(3)

(4) (5)

Chapter 2: Three Dimensional Analytical Geometry 2

i.e. From (6), we have

I – 2.41 (6)

3 0

2

(7)

3

2 2

1 2

1,

When

= –1, from (7), we get 1 , from (7), we get 2

1 2

1 make all the ratios in (3) equal, each 1 1 equal to 2 are inadmissible. . The other set of values and 2 2 1 , Equations (1) and (2) represent the same plane. Hence the two given lines are coplanar. The equation of the plane in which they lie is 1 in (2). It is x + 2y + 3z = 2. Example 2.12 Find the equations of the line through (– 6, – 4, – 6) which cuts y 1 x y z z 1 and also the co-ordinates and x 2 each of the lines 2 2 1 3 of the points in which it meets them. Any point P r, r, 3r). Any point Q on the second line is ( r

2, 2r

1, r

1)

A(– 6, – 4, – 6) and P are (2r + 6, r + 4, 3r + 6). D.R.’s of AQ are ( r

4, 2r

5, r

5)

If PQ passes through A, the D.R.’s of AP and AQ must be proportional. 2r 6 r 4 3r 6 r 4 2r 5 r 5 5rr 6r 16r 14 0 From the second and third ratios in (1), we have 7 rr 10r 16r 10 0 From (2) and (3), eliminating rr term, we get r

4r

(2) (3) (4)

6

Using (4) in (2), we have 2r i.e.

2

7r

(1)

5 0

(r

1)(2r

r

1 or r

5) 0 5 2

Part I: Mathematics I

I – 2.42 When r

5 , r = –4 2

1 , r = 2; When r 5 2

The values r = – 4 and r Hence r = 2 and r

1.

For these values of r and r , we get P

4, 2, 6

Q

and

Also the equations of the required line AP are

1, 1, 0

x 6 4 6

y 4 2 4

z 6 6 6

x 6 y 4 z 6. 5 3 6 Example 2.13 A line with D.R.’s (2, 7, –5) is drawn to intersect the lines x 5 y 7 z 2 x 3 y 3 z 6 . Find the co-ordinates of the points ; 3 1 1 3 2 4 of intersection and the length intercepted on it. P and the second line at Q. r + 5, – r + 7, r – 2). Since P Similarly Q may be taken as ( 3r 3, 2r 3, 4r 6)

i.e.

D.R.’s of PQ are ( 3r

3r 8, 2r

r 4, 4r

r

8)

But D.R.’s of PQ are proportional to (2, 7, –5) 3r

3r 8

2r

r 4 7

2

4r

r 8 5

(1)

25r 23r 48 0 From the second and third ratios of (1), we get 38r

2r

36 0

Solving (2) and (3), we get r = –1 and r P

Hence and

PQ

(2)

1

2, 8, 3 2 0

(3)

and 2

8 1

Q 2

0,1, 2 . 3 2

2

78 .

Example 2.14 Find the length and equations of the shortest distance between the lines x 3 3

y 8 1

z 3 and x 3 1 3

y

7 z 6. 2 4 P and the second line at Q. P may be taken as (3r1 + 3, – r1 + 8, r1 + 3) Q may be taken as (–3r2 – 3, 2r2 – 7, 4r2 + 6) Hence D.R.’s of PQ are (–3r2 – 3r1 – 6, 2r2 + r1 – 15, 4r2 – r1 + 3) PQ is 3 (–3r2 – 3r1 – 6) – (2r2 + r1 – 15) + (4r2 – r1 + 3) = 0

Chapter 2: Three Dimensional Analytical Geometry i.e. PQ is also

I – 2.43

7r2 + 11r1 = 0 to the second line. –3(– 3r2 – 3r1 – 6) + 2 ( 2r2 + r1 – 15) + 4 (4r2 – r1 + 3) = 0

29r2 + 7r1 = 0 Solving (1) and (2), we get r1 = 0 and r2 = 0. Using these values of r1 and r2 in the co-ordinates of P and Q, we get i.e.

P Length of S.D.

3 3

3, 8, 3

2

8 7

Q

and 2

3 6

(1)

(2)

3, 7, 6

2

270 3 30 units Equations of the S.D. line are

i.e.

x 3 6

y 8 15

z 3 3

x 3 2

y 8 5

z 3 1

Example 2.15 Find the length of the shortest distance between the lines x 2 y 1 z ; 2 x 3 y 5 z 6 0 3x 2 y z 2 3 4

B P

3 A

line. Any plane containing the second line is of the form 2x + 3y – 5z x – 2y – z + 3) = 0. x y z+

D

Fig. 2.21

=0

–

Putting

Q

C

7 . 4 7 in (1), the equation of the required plane is 4 2

21 x 4

3

14 y 4

5

7 z 4

21 6 4

0

i.e. 13x – 26y + 13z + 45 = 0 (2) Now the required S.D. is the length of the perpendicular drawn from any point P on

Part I: Mathematics I

I – 2.44 Taking P S.D.=

26 26 2

13

26

45 2

97 2

13

units.

13 6

EXERCISE 2(c) Part A (Short Answer Questions) 1. Find the equations of the line passing through the origin and making equal angles with the co-ordinate axes. 2. Find the equations of the line passing through (1, 2, 3) and parallel to the line 3. Find the equations of the line passing through (1, –2, –1) and to the plane 2x – 3y + 5z + 4 = 0. 4. Prove that the points (3, 2, –4), (5, 4, –6) and (9, 8, –10) are collinear and write down the equations of the line of collinearity. 5. Find the angle between the lines

x 2

6. Find the value of k, if the lines x 1 1 z are perpendicular. k

y 2

z x 4 and 1 2

y 3 k

y 5 1

z 1 and x 1 4 5

z

6 2

y 1 3

7. Find the equations of the normal to the plane 2x – 3y + 6z = 0, passing through (2, –3, 4). x 1 y 1 z 6 8. Find the point where the line meets the xoy-plane. 1 2 3 9. Find the D.R.’s of the line given by the equations x + y – z = 1 and 2x – y + 7z + 1 = 0. x 2 y 4 z 5 10. Find the value of k, if the line is parallel to the plane k 1 2 x + y – z – 2 = 0. x 3 y 2 x 2 y 1 z 3 and 11. Find the values of k, if the lines k 3 3 2 k z 4 are coplanar. 5 12. If the lines

x 7 3

y 1

z

7 x 3 and 2 2

y 1 3

z

2 1

the equation of that plane. 13. Find the D.R.’s of the S.D. line between the lines

x 3 1

y 5 2

z 7 and 1

Chapter 2: Three Dimensional Analytical Geometry x 1 7

y 1 6

I – 2.45

z 1 1

x 4 y 2 z 3 lies in the plane 3x – 4y + z = 7. 2 3 6 x y z and 15. Find the equations of the line perpendicular to both the lines 1 2 3 x y z and passing through their point of intersection. 2 1 2

14. Prove that the line

Part B 16. Find, in the symmetric form, the equations of the line 3x + 2y – z = 4, 4x + y – 2z = –3. 17. Put the equations of the line given by x + y + z – 1 = 0, 2x – y – 3z + 1 = 0 in the symmetric form. 18. Show that the lines 2x + 3y – 4z = 0 = 3x – 4y + z and 5x – y – 3z + 12 = 0 = x – 7y + 5z – 6 are parallel. 19. Find the angle between the lines 3x + 2y + z – 5 = 0 = x + y – 2z – 7 and 2x – y – z – 16 = 0 = 7x + 10y – 8z – 15. 20. Find the distance of the point (1, – 2, 3) from the plane x – y + z = 5 measured x y z parallel to the line 2 3 6 21. Find the co-ordinates of the foot, the length and equations of the perpendicular x 13 y 8 from the point (– 1, 3, 9) to the line z 31 5 8 x 1 y 2 z 3 on the plane 2x + y + z = 6. 2 1 4 23. Find the equation of the plane which contains the two parallel lines x 3 y 2 z 1 x 2 y 3 z 1 and 1 4 5 1 4 5 24. Find the equation of the plane which passes through the point (1, 2, – 1) and which contains the line

x 1 2

y 1 3

z

2 1

25. Find the equation of the plane which contains the line

x 1 2

y 1 1

z 3 4

and is perpendicular to the plane x + 2y + z = 12. x 1 3 parallel to the line x – 2y + 3z – 4 = 0 = 2x – 3y + 4z – 5.

26. Find the equation of the plane containing the line

27. If the line

x 4

y 3

y 1 4

z 2 and 2

z 7

slope in the plane 2x + y – 5z = 12 and passing through the point (2, 3, – 1)

Part I: Mathematics I

I – 2.46

28. Show that the lines x 5 4

y 7 4

x 3 2

29. Show that the lines

z

y 5 3

3 and x 8 7 5

y 4 1

x 1 z 7 and 4 3

z 5 are 3

y 1 5

z 1 are 1

x 4 y 3 z 1 and x 1 y 1 z 10 are 1 4 7 2 3 8 coplanar. Find their common point and also the equation of the plane containing them.

30. Prove that the lines

x 3 1

31. Prove that the lines

y 1 3

x 2 z 1 and 2 2

y 3 1

z

4 3

are

coplanar. Find their point of intersection and the equation of the plane containing them. 32. Show that the lines

x 1 1

y 1 1

z 1 and 3x + 2y – 3z – 5 = 0 = 5x + 4y – 2

z – 7 are coplanar. Find their point of intersection and also the equation of the plane containing them. 33. Show that the lines

x

4 3

y

6 5

z 1 and 3x – 2y + z + 5 = 0 = 2x + 3y + 2

4z – 4 are coplanar. Find their point of intersection and the equation of the plane in which they lie. 34. Prove that the two lines x + 2y – z – 3 = 0 = 3x – y + 2z – 1 and 2x – 2y + 3z –2 = 0 = x – y + z ing them. 35. Show that the lines x + 2y – 5z + 9 = 0 = 3x – y + 2z – 5 and 2x + 3y – z – 3 = 0 = 4x – 5y + z them. 36. Find the equations of the straight line through the origin which intersects x 4 y 3 z 14 both the lines x 1 y 3 z 5 and 2 3 4 2 4 3 37. Find the equations of the line through the point (1, 0, 7) which intersects each of the lines x = y = z and x 4 y 5

z 1 2

38. A line with D.R.’s (2, 7, – 5) is drawn to intersect the lines

x 3 3

y 3 2

z 6 and x 8 y 6 z 1 Find the co-ordinates of the points of 3 1 1 4 intersection and the length intercepted.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.47

39. Find the equations of the straight line perpendicular to both the lines x 1 1 x 2 y 5 z 3 y 1 z 2 and passing through their and 2 1 2 2 3 point of intersection. 40.

x 10 1

y 9 3

41.

x 1 1

y 2 2

z

2 2

and

z 3 3

and

x 1 2 x 1 2

y 12 z 5 4 1 y z 1 1 3

x 1 y z and 42. x 1 y 1 z 1 3 4 5 2 3 4 x y z 3 5 7 x y 1 z 1 1 and 43. 1 2 1 7 6 1 x 1 y 2 z 44. and 3x + 2y – 5z – 6 = 0 = 2x – 3y + z – 3 3 2 4 45. x – y + z = 0 = 2x – 3y + 4z and x + y + 2z – 3 = 0 = 2x + 3y + 3z – 4.

2.9

SPHERE AND CIRCLE

2.9.1 point (called the centre of the sphere) is a constant (called the radius of the sphere). Equation of the sphere whose centre is (a, b, c) and radius is r. Let P(x1, y1, z1) be any point on the sphere. P (x1, y1, z1) [Refer to Fig. 2.22] CP = r or CP2 = r2 2 i.e. (x1 – a) + (y1 – b)2 + (z1 – c)2 = r2 C (a, b, c) The locus of (x1, y1, z1), i.e. the equation of the sphere is (x – a)2 + (y – b)2 + (z – c)2 = r2. Cor: The equation of the sphere whose centre is the Fig. 2.22 origin and radius r is x2 + y2 + z2 = r2.

2.10

STANDARD FORM OF THE EQUATION TO A SPHERE

The equation x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (called the standard form of the equation to a sphere) represents a sphere with centre at (– u, – v, – w) and radius u 2 v 2 w2 d The given equation is x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (1) can be re-written as (x2 + 2ux + u2) + (y2 + 2vy + v2) + (z2 + 2wz + w2) = u2 + v2 + w2 – d i.e. (x + u)2 + (y + v)2 + (z + w)2 = u2 + v2 + w2 – d 2 2 2 i.e. {x ( u )} { y ( v)} {z ( w)}

u 2 v 2 w2 d

(1)

2

(2)

Part I: Mathematics I

I – 2.48

Comparing equation (2) with the equation (x – a)2 + (y – b)2 + (z – c)2 = r2 (3) we see that equation (2) and hence equation (1) represent a sphere with centre at u2

(–u, –v, –w) and radius

v2

w2

d

Note of x2, y2, z2 are all equal and (ii) the product terms xy, yz and zx are absent. Hence the equation a (x2 + y2 + z2) + 2ux + 2vy + 2wz + d = 0 (4) also represents a sphere.

x2

y2

z2

2

u x a

2

v y a

Then centre of the sphere is u2

v2

w2

2

2

2

2

w z a

u , a

d a v , a

0

(5)

w and its radius a

d a a a a 3. The general equation of the second degree in x, y, z, namely ax2 + by2+ cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0 represents a sphere if(i) a = b = c (ii) f = g = h = 0. 4. Since the standard form of the equation to a sphere contains four independent constants, we should know four geometrical conditions Equation of a sphere

A (x1, y1, z1) and B (x2, y2, z2) as

Let P (x, y, z) be any point on the sphere. [Refer to Fig. P (x, y, z) 2.23] For the circle passing through the points A, B and P, AB is a diameter. A B APB 90 i.e. AP is to BP (1) D.R.’s of AP are (x – x1, y – y1, z – z1) D.R.’s of BP are (x – x2, y – y2, z – z2) Fig. 2.23 Hence, by (1), we have (x – x1) (x – x2) + (y – y1) (y – y2) + (z – z1) (z – z2) = 0 (2) (2) is the required equation of the sphere described on AB as diameter.

2.10.1 Tangent Line and Tangent Plane When a straight line intersects a sphere at two coincident points or when it touches a sphere at a point P, the line is called a tangent line of the sphere at P and is perpendicular to the radius of the sphere through P. There are many tangent lines through P which are perpendicular to the radius of the sphere through P. Also all these tangent lines lie on a plane through P, which is perpendicular to the radius through P. This plane is called the tangent plane of the sphere at P.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.49

Equation of the tangent plane Let P(x1, y1, z1) be a point on the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 The centre C of the sphere (1) is (–u, –v, –w). Since P lies on (1),

(1)

x12

y12

z12

2ux1

2vy1

2wz1

d

(2)

0

D.R.’s of CP are (x1 + u, y1 + v, z1 + w) to CP. The tangent plane at P passes through it and is Hence the equation of the required tangent plane is (x1 + u) (x – x1) + (y1 + v) (y – y1) + (z1 + w) (z – z1) = 0 i.e.

xx1

yy1

zz1

ux vy

wz

x12

y12

z12

ux1

vy1

(3)

wz1

Adding (ux1 + vx1 + wz1 + d) on both sides of (3), we have xx1 + yy1 + zz1 + u(x + x1) + v(y + y1) + w(z + z1) + d x12 y12 z12 2ux1 2vy1 2 wz1 d (4) Since the R.S. of (4) vanishes by (2), the equation of the tangent plane becomes xx1 + yy1 + zz1 + u(x + x1) + v(y + y1) + w(z + z1) + d = 0. Length of the tangent from an external point to a sphere. T Let P(x1, y1, z1) be a point lying outside the sphere x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0, whose centre is C (–u, –v, –w). [Refer to P (x1, y1, z1) C Fig. 2.24] Let PT be the tangent from P to the sphere. Then CT = radius of the sphere u2

v2

w2

Fig. 2.24

d

Now, from the right angled triangle CPT, PT2 = CP2 – CT2 = (x1 + u)2 + (y1 + v)2 + (z1 + w)2 – (u2 + v2 + w2 – d) x12 PT

2.10.2

x12

y12 y12

z12 z12

2ux1 2ux1

2vy1 2vy1

2 wz1 2 wz1

d d

Plane Section of a Sphere

The curve of intersection of a sphere and a plane is obviously a circle. The centre of the circle is the foot of the perpendicular from the centre of the sphere on the intersecting plane. The radius r C of the circle is given by r R 2 p 2 , where R is the radius p of the sphere and p is the length of the perpendicular from the centre of the sphere on the intersecting plane. [Refer to Fig. C' 2.25] The equation of a sphere and the equation of a plane the plane.

R

Fig. 2.25

P

Part I: Mathematics I

I – 2.50

Note

2.11

1. When the intersecting plane passes through the centre of the sphere, the circle of intersection is called a great circle of the sphere. 2. The intersection of two spheres is also a circle. The circle of intersection,

EQUATION OF A SPHERE THROUGH A CIRCLE

x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 and P px + qy + rz + s = 0 Consider the equation S + P x2 + y2 + z2 + 2ux + 2vy + 2wz + d + px + qy + rz + s) = 0 S

(1) (2) (3)

Equations (1) and (2) i.e. all points lying on the circle given by (1) and (2) also lie on the sphere (3). Thus Equation (3) represents any sphere passing through the circle given by (1) and (2).

WORKED EXAMPLE 2(d) Example 2.1 Prove that the two spheres x2 + y2 + z2 –2x + 4y – 4z = 0 and x2 + y2 + z2 +10x + 2z + x2 + y2 + z2 – 2x + 4y – 4z = 0

(1)

Centre of this sphere, C1 is (1, –2, 2) [Refer to Fig. 2.26] Radius r1

1 4 4

3 units.

x2 + y2 + z2 + 10x + 2z + 10 (2)

=0

P r1

Centre of this sphere, C2 is (–5, 0, –1) Radius r2 C1C2

25 0 1 10

( 5 1) 2

0 2

2

C1

4 units. 1 2

r2 C2

2

Fig. 2.26

= 7 units We see that C1C2 = r1 + r2. The two spheres touch each other externally. Also the point of contact P divides C1C2 internally in the ratio r1 r2 Hence P

15 4 0 8 3 8 , , 3 4 3 4 3 4

Chapter 2: Three Dimensional Analytical Geometry

I – 2.51

11 8 5 , , 7 7 7 Example 2.2 Find the equation of the sphere passing through the four points (1, 2, 3), (0, –2, 4), (4, –4, 2) and (3, 1, 4). Let the equation of the required sphere be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (1) (1) passes through (1, 2, 3) 2u + 4v + 6w + d = –14 (2) (1) passes through (0, –2, 4) –4v + 8w + d = –20 (3) (1) passes through (4, –4, 2) 8u – 8v + 4w + d = –36 (4) (1) passes through (3, 1, 4) 6u + 2v + 8w + d = –26 (5) (2) – (3) gives 2u + 8v – 2w = 6 (6) (4) – (3) gives 8u – 4v – 4w = –16 (7) (4) – (5) gives 2u – 10v – 4w = –10 (8) Eliminating w from (6), (7) and (8), we have i.e. P is the point

u – 5v = –7 (9) and u + 13v = 11 (10) Solving (9) and (10), we get u = –2 and v = 1. Using these values in (6), we get w = –1 Using the values in (2), we get d = –8. Substituting these values in (1), the equation of the required sphere is x2 + y2 + z2 – 4x + 2y – 2z – 8 = 0. Example 2.3 Find the equation of the sphere passing through the points ( 1, 1, –2) and (–1, 1, 2) and having its centre on the line x + y – z –1 = 0 = 2x – y + z – 2. Let the equation of the sphere be (1) x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (1) passes through the point (1, 1, –2) 2u + 2v – 4w + d = –6 (2) (1) passes through the point (–1, 1, 2) –2u + 2v + 4w + d = –6 (3) Centre of the sphere (1) is (–u, –v, –w). It lies on the line x + y –z – 1= 0 = 2x – y + z – 2. It lies on the two planes whose intersection is the given line. –u – v + w = 1 (4) and –2u + v – w = 2 (5) (2) – (3) gives 4u.–8w = 0 or u – 2w = 0 (6)

Part I: Mathematics I

I – 2.52 (4) + (5) gives From (6),

–3u = 3 or u = –1. 1 w . 2

From (4),

v = –u + w – 1 1 1 1 1 2 2 d = –6 – 2u – 2v + 4w

From (2),

= –6 + 2 + 1 – 2 = –5 Putting the values of u, v, w, d in (1), the equation of the required sphere is x2 + y2 + z2 – 2x – y – z – 5 = 0. Example 2.4 a, b, c) and cuts the axes in A, B, C. a b c 2. Show that the locus of the centre of the sphere OABC is x y z Let OA = p, OB = q and OC = r. Then the equation of the plane ABC is x p

y q

z r

1

(l)

Since plane (1) passes through (a, b, c), we have a p

b q

c 1 r

(2)

Let the equation of the sphere OABC be x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 (3) Since sphere (3) passes through O(o, o, o), A(p, o, o), B(o, q, o) and C(o, o, r), we have d=0 2up + d = –p2 2vq + d=–q2 2wr + d = –r2 From (4), (5), (6) and (7), we get, p q , v , w 2 2 The equation of the sphere OABC is u

(4) (5) (6) (7) r 2

and

d = 0.

x2 + y2 + z2 – px – qy – rz = 0 The centre of this sphere is p q r (8) , , 2 2 2 Since the locus of the centre of this sphere is required, let the co-ordinates of the centre be assumed as (9) (x1, y1, z1)

Chapter 2: Three Dimensional Analytical Geometry From (8) and (9), we have p = 2x1,

q = 2y1 and Using (10) in (2), we get the relation

The locus of (x1, y1, z1) is

a 2 x1 a x

b 2 y1 b y

c 2 z1 c z

I – 2.53

r = 2z1

(10)

1

2

Example 2.5 Show that the plane 2x – 2y + z + 12 = 0 touches the sphere x2 + y2 + z2 – 2x – 4y + 2z = x2+ y2 + z2 –2x – 4y + 2z – 3 = 0 (1) 2x – 2y + z +12 = 0 (2) 1 4 1 3 Centre of (1) is (1, 2, –1) and radius Length of the from (1, 2, –1) on plane (2)

3 C

2 4 1 12 4 4 1 9 3

3 = radius of the sphere.

P Fig. 2.27

The plane (2) touches the sphere (1). Let P be the point of contact. Then CP is normal to the plane (2). [Refer to Fig. 2.27] D.R.’s of CP are (2, –2, 1) Also CP passes through C(l, 2, –1) Hence the equations of CP are x 1 y 2 z 1 2 2 1 Any point on this line may be taken as (2r + l, –2r + 2, r – l) If this point lies on plane (2), it will be the point of contact. This point lies on plane (2), if 2(2r + 1) – 2(–2r + 2) + (r – 1) + 12 = 0

(3)

9r + 9 = 0 or r = –1 From (3), the point of contact is (–1, 4, –2) Example 2.6 Find the equations of the tangent planes to the sphere x2 + y2 + z2 – 4x – 2y – 6z + 5 = 0 which are parallel to the plane x + 4y + 8z = 0. Find also their points of contact. Centre of the sphere, C Radius of the sphere = 4 1 9 5 3. Any plane parallel to x + 4y + 8z = 0 is of the form x + 4y + 8z + k = 0 (1) i.e

Part I: Mathematics I

I – 2.54

If (1) is a tangent plane to the given sphere, length of the of the sphere 2 4 24 k i.e. 3 1 16 64

from C on (1) = radius

k 30 27 k = –3 or k = –57 i.e. There are two tangent planes parallel to (1) with equations x + 4y + 8z – 3 = 0 and x + 4y + 8z – 57 = 0 Let the point of contact of (2) with the sphere be P (a, b, c). P (a, b, c) lies on plane (2) a + 4b + 8c – 3 = 0 CP, whose D.R.’s are (a – 2, b – 1, c – 3) is normal to plane (2). a 2 b 1 c 3 k 1 4 8 From (5), a = k + 2, b = 4k + 1, c = 8k + 3 Using (6) in (4), we have k+ 2 + 4 (4k + 1) + 8 (8k + 3) – 3 = 0 1 i.e. 8lk + 27 = 0 or k 3 5 1 1 1 , . in (6), the required point of contact is , Substituting k 3 3 3 3

(2) (3)

(4) (5) (6)

Let the point of contact of (3) with the sphere be Q (p, q, r). Q (p, q, r) lies on plane (3). p + 4q + 8r – 57 = 0 CQ, whose D.R.’s are (p – 2, q – 1, r – 3) is normal to plane (3). p 2 q 1 r 3 1 4 8 From(8), p = + 2, q = 1, r Using (9) in (7), we have i.e. Substituting

(7) (8)

1 3 1 7 7 17 in (9), the required point of contact is , , 3 3 3 3 81 – 27 = 0

or

Example 2.7 Find the centre and radius of the circle given by x2 + y2 + z2 – 2x – 4y – 6z – 2 = 0 and x + 2y + 2z – 20 = 0. Centre of the given sphere x2 + y2 + z2 – 2x – 4y – 6z – 2 = 0

r

P

C' R C

(1)

is C Radius of the sphere is R

1 4 9 2

4

Fig. 2.28

Chapter 2: Three Dimensional Analytical Geometry

I – 2.55

C', the centre of the given circle is the foot of the r from C on the given plane x + 2y + 2z = 20 (2) Since CC' is a normal to plane (2), its D.R.’s are (1, 2, 2) Equations of CC' are x 1 y 2 z 3 (3) 1 2 2 Any point on (3) is (4) (r + l, 2r + 2, 2r + 3) If the co-ordinates in (4) represent C' , they satisfy Equation (2), since C' lies on plane (2) r + 1 + 2 (2r + 2) + 2 (2r + 3) = 20 i.e. 9r = 9 or r = 1 From (4), C' Now

CC

2 1

2

4 2

2

5 3

2

3

If r is the radius of the circle, r2 = R2 – (CC' )2 = 16 – 9 = 7 r 7 Example 2.8 (2, –1, 4) and (–2, 2, –2) as diameter. Find also the area of the circle in which the sphere is cut by the plane 2x + 2y – z = 3. The equation of the required sphere is (x – 2) (x + 2) + (y + 1) (y – 2) + (z – 4) (z + 2) = 0 i.e. x2 + y2 + z2 – y – 2z – 14 = 0 (1) 1 The centre of the sphere (1) is given by C 0 , ,1 2 [Refer to Fig. 2.29] r The radius of the sphere (1) is given by P C' 1 61 ' 1 14 . C , the centre of the circle is R 4 2 C the foot of the r from C' on the intersecting plane 2x + 2y – z = 3 (2) Since CC' is a normal to plane (2), its D.R.’s are (2, 2, –1) Equations of CC' are Fig. 2.29 1 y x 0 (3) 2 z 1 2 1 2 Any point on (3) is 1 2r ,2r , r 1 (4) 2 If the co-ordinates in (4) represent C', they satisfy (2), since C' lies on plane (2). 1 2 2r 2 2r r 1 3 2 R

0

Part I: Mathematics I

I – 2.56

i.e.

9r = 3

From (4),

Now

r

or

1 3

2 7 2 , , 3 6 3

C

2 3

CC

2

7 6

1 2

2

1

2 3

2

1. If r is the radius of the circle, then r2

R2 61 4

2

CC 1

57 4

57 4 Example 2.9 Find the circumcentre of the triangle ABC, where A, B, C are the points at which the plane 6x – 3y – 2z – 6 = 0 intersects the co-ordinate axes. Equation of the given plane 6x – 3y – 2z – 6 = 0 may be re-written as Area of the circle

x y z 1 (1) 1 2 3 From (1), we get OA = 1, OB = –2 and OC = –3 or A (1, 0, 0), B (0, 2, 0), C (0, 0, 3) Circumcircle of triangle ABC is the circle passing through A, B and C. [Refer to Fig. 2.30] This circle is the intersection of the plane ABC and any sphere through A, B, C. For convenience let us consider the sphere through O, A, r P B, C. C' Proceeding as in worked example (2.4) above, the equation R of the sphere OABC can be obtained as C x2 + y2 + z2 – x + 2y + 3z = 0 (2) 1 3 , 1, The centre of this sphere (2) is given by C 2 2 Fig. 2.30

1 9 7 1 and radius is given by R 4 4 2 ' Since CC is a normal of the plane (1), its D.R.’s are (6, –3, –2), [C' is the centre of the circle and the foot of the from C on plane (1)] 1 2

Equations of CC' are x

y

6 Any point on (3) is 6r

1 , 2

z

1 3

3r

1,

2r

3 2 2

(3)

3 2

(4)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.57

If the co-ordinates in (4) represent C', they satisfy (1), since C' lies on plane (1). 1 2

6 6r

3 ( 3r

1)

2r

2

3 2

6

3 49 13 , Using this value in (4), the co-ordinates of the centre of the circle are 98 135 . 98

i.e.

49r = –3

or

r

40 , 49

Example 2.10 Find the equation of the sphere passing through the circle given by x2 + y2 + z2 + 3x + y + 4z – 3 = 0 and x2 + y2 + z2 + 2x +3y + 6 = 0 and the point (1, –2, 3). Equation of any sphere passing through the circle of intersection of two given ' = 0. spheres S = 0 and S' = 0 is Hence the equation of any sphere through the circle given by the two given equations is x2 + y2 + z2 + 3x + y + 4z – (x2 + y2 + z2 + 2x + 3y + 6) = 0 (1) (1 + 4 + 9 + 2 – 6 + 6) = 0 3 2 3 Putting in (1), the equation of the required sphere is 2 2(x2 + y2 + z2 + 3x + y + 4z – 3) – 3(x2 + y2 + z2 + 2x + 3y + 6) = 0 i.e.

x2 + y2 + z2 + 7y – 8z + 24 = 0.

Example 2.11 Find the equation of the sphere having the circle x2 + y2 + z2 + 10y – 4z – 8 = 0, x + y + z = 3 as a great circle. The equation of any sphere passing through the given circle is of the form i.e.

(x + y + z – 3) = 0 x2 + y2 + z2 + l0y – 4z – 2 2 2 x + y + z + x + +10) y + – 4) z – + 8) = 0

(1)

If the given circle is a great circle of sphere (1), centre of (1) should lie on the plane x + y + z = 3 in which the given circle lies. Thus 2

1 2

10

1 2

= –4 in (1), the equation of the required sphere is x2 + y2 + z2 – 4x + 6y – 8z + 4 = 0 Example 2.12 Find the equation of the smallest sphere which contains the circle given by the equations x2 + y2 + z2 + 2x + 4y + 6z – 11 = 0 and 2x + y + 2z + 1 = 0.

Part I: Mathematics I

I – 2.58

Equation of any sphere containing the given circle is of the form x + y + 2z + 1) = 0 x2 + y2 + z2 + 2x + 4y + 6z 2 2 2 x y z i.e. x +y +z Radius of the sphere (1) is given by 2

R

1

2

2

2

3

2

11

9 2 9 25 4 Now, sphere (1 ) will be the smallest sphere, if its radius R takes the least value. R is least, when R 2 f ( ) 9 2 9 25 is least 4 9 9 Now f ( ) 9 and f ( ) 2 2 f' f '' f R2 x2 + y2 + z2 – 2x + 2y + 2z – 13 = 0 Example 2.13 Find the equation of the sphere which passes through the circle x2 + y2 + z2 = 4 and x = 0 and which is cut by the plane x + 2y + 2z = 0 in a circle of radius 4. The equation of any sphere passing through the given circle is 0 (1) x2 + y2 + z2 – 4 P

C'

Centre of the sphere (1) is given by C

2

[Refer to Fig. 2.31]

, 0, 0 R C

2

Radius of the sphere (1) is given by R Now CC' = length of the (x + 2y + 2z = 0)

4

4

from C' on the plane Fig. 2.31

2 6 1 4 4 The intersection of sphere (1) and the plane x + 2y + 2z = 0 is a circle of radius r = 4. From the right angled triangle CPC' , CC' 2 + C' P2 = CP2 2

i.e. 36 i.e.

2

2

16

54 or

4

4 3 6

Chapter 2: Three Dimensional Analytical Geometry Putting

I – 2.59

3 6 , the equations of the required spheres are

x 2 y 2 z 2 3 6 x 4 0. Example 2.14 Find the equations of the spheres which pass through the circle x2 + y2 + z2 = 5 and x + 2y + 3z = 3 and touch the plane 4x + 3y = 15. Equation of any sphere passing through the given circle is of the form x2 + y2 + z2 – x + 2y + 3z – 3) = 0 2 + 5) = 0 (1) i.e. x + y2 + z2 + 2 x + y + .z– Centre of the sphere (1), say C Radius of the sphere (1), say, R 14

2

4

2

6

2

9

2

6

5

5

Sphere (1) touches the plane 4x + 3y – 15 = 0 Length of the from C on (2) = R 4

i.e.

6 2

15

2

14

2

(2)

6

5

4 3 Squaring both sides, we have 10

15

2

14

25

2

6

5

2

2 5 Using these values in (1), the equations of the required spheres are x2 + y2 + z2 + 2x + 4y + 6z – 11 = 0 2 and 5(x + y2 + z2) – 4x – 8y – 12z – 13 = 0. Example 2.15 Find the equation of the sphere that passes through the circle x2 + y2 + z2 + 2x + 3y + z – 2 = 0 and 2x – y – 3z – l = 0 and cuts orthogonally the sphere x2 + y2 + z2 – 3x + y – 2 = 0. (Note on orthogonal spheres: Two spheres x2 + y2 + z2 + 2urx + 2vry + 2wrz + dr = 0; r =1, 2; with centres at C1 (u1, – v1, – w1) and C2 (u2, – v2, –w2) cut orthogonally at a point of intersection P, if the tangent planes at P to the two spheres are perpendicular, or equivalently, the radii C1P and C2P are r. From the triangle C1 PC2 , C1C22 i.e.,

C1 P 2

C2 P 2

(u1 – u2)2 + (v1 – v2)2 + (w1 – w2)2 u12

v12

w12

d1

u22

v22

w22

d2

i.e., 2u1u2 + 2v1v2 + 2w1w2 = (d1 + d2) Equation of any sphere passing through the given circle is of the form x2 + y2 + z2 + 2x + 3y + z x – y – 3z–1) = 0

Part I: Mathematics I

I – 2.60 i.e.,

x2 + y2 + z2 +

x

y+

Sphere (1) cuts

z

=0

(1)

x2 + y2 + z2 – 3x + y – 2 = 0

(2)

orthogonally. (2

2 )

3 2

(3

)

1 2

(1

3 )

0

2

2

x2 + y2 + z2 + 4x + 2y – 2z – 3 = 0

EXERCISE 2(d) Part A (Short Answer Questions) 1. Show that the sphere, whose centre is (1, 2, –2) and radius is 3, passes through the origin. 2. Find the equation of the sphere whose centre is (1, –2, 3) and which passes through the point (3, 1, –3). 3. Find the radius of the sphere whose centre is (4, 4, –2) and which passes through the origin. 4. Find the equation of the sphere having the points (2, –3, 4) and (–1, 5, 7) as the ends of a diameter. 5. Find the centre and radius of the sphere 2 (x2 + y2 + z2) + 6x – 6y + 8z + 9 = 0. 6. Find the equation of the sphere concentric with x2 + y2 + z2 – 4x + 6y – 8z + 4 = 0 and passing through the point (1, 2, 3). 7. Find the equation of the sphere with centre at (2, 3, 5) which touches the xoy-plane. 8. Find the equation of the sphere with centre at (1, 2, 3) which touches the z-axis. 9. Find the equation of the sphere whose centre is (1, 2, 3) and which touches the plane 2x + 2y – z = 2. 10. Find the equation of the sphere passing through the origin and the points (2, 0, 0), (0, 3, 0) and (0, 0, 4). 11. Find the equations of the circle which lies on the sphere x2 + y2 + z2 – 6x – 8y + 2z – 9 = 0 and whose centre is (1, 2, –1). (Hint: The circle is represented by the equations of the sphere and the intersecting plane. The intersecting plane passes through the centre of the circle and is 12. Find the equations of the circle lying on the sphere x2 + y2 + z2 – 2y + 4z = 11 and having its centre at (1, 3, 4)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.61

13. Find the equation of the tangent plane at the point (1, – 1, 2) to the sphere x2 + y2 + z2 – 2x + 4y + 6z – 12 = 0. 14. Find the length of the tangent drawn from the point (1, 2, 3) to the sphere x2 + y2 + z2 – 3x + 4y + 5z – 7 = 0. 15. A tangent plane of the sphere x2 + y2 + z2 = r2 makes intercepts a, b, c on the 1 1 1 1 a2 b2 c2 r 2 (Hint: length from the centre of the sphere on the tangent plane x a 1 is equal to the radius of the sphere.)

co-ordinate axes. Prove that

y b

z= c

Part B 16. Show that the two spheres x2 + y2 + z2 = 25 and x2 + y2 + z2 – 18x – 24y – 40z + 225 = 0 touch each other. Find their point of contact and also the equation of the common tangent plane. 17. Show that the two spheres x2 + y2 + z2 – 2x – 4y – 4z = 0 and x2 + y2 + z2 + 10x + 2z+ 10 = 0 touch each other. Find their point of contact and also the equation of the common tangent plane. 18. Find the equation of the sphere through the points (4, –1, 2), (0, –2, 3), (1, 5, 1) and (2, 0, 1). 19. Find the equation of the sphere passing through the points (3, 0, 1), (–1, 1, 1), (2, –5, 4) and having its centre on the plane 2x + 3y + 3z = 3. 20. Find the equation of the sphere which passes through the points (2, 1, 1) and (0, 3, 2) and has its centre on the line 2x + y + 3z = 0 = x + 2y + 2z. 21. Find the equation of the sphere which passes through the points (1, – 4, 3), 1, –5, 2) and has its centre on the line x

4

y

2

z

6

4 1 3 22. Find the equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0), (0, 0, 1) and has its radius as small as possible. 23. A sphere of constant radius k passes through the origin and meets the axes in A, B, C. Prove that the centroid of the triangle ABC lies on the sphere 9 (x2 + y2 + z2) = 4k2. 24. Show that the plane 2x – 2y + z = 9 touches the sphere x2 + y2 + z2 + 2x+ 2y – 7 = 0. Find the point of contact also. 25. Show that the plane 4x + 9y + 14z – 64 = 0 touches the sphere 3 (x2 + y2 + z2) – 2x – 3y – 4z – 26. Find the equations of the tangent planes to the sphere x2 + y2 + z2 – 4x + 2y – 6z + 5 = 0 which are parallel to the plane 2x + 2y – z = 0. 27. Find the centre and radius of the circle given by x2 + y2 + z2 + 2x – 2y – 4z – 19 = 0 and x + 2y + 2z + 7 = 0. 28. Find the area of the circle in which the sphere x2 + y2 + z2 + 12x – 2y – 6z + 30 = 0 is cut by the plane x – y + 2z + 5 = 0. x y z 29. The plane 1 meets the axes in A, B, C. Find the co-ordinates of a b c the circumcentre of the triangle ABC.

Part I: Mathematics I

I – 2.62

30. Find the equation of the sphere passing through the circle x2 + y2 + z2 – 2x + 4y + z – 2 = 0, x + 3y – z = 4 and through the point (1, 2, –1). Find also its centre and radius. 31. Find the equation of the sphere passing through the circle x2 + y2 + z2 – 2x – 3y + 4z + 8 = 0, 3x + 4y – 5z = 3 and having its centre on the plane 4x – 5y – z = 3. 32. Find the equation of the sphere for which the circle x2 + y2 + z2 + 7y – 2z + 2 = 0, 2x + 3y + 4z = 8 is a great circle. 33. Find the equation to the sphere which passes through the circle x2 + y2 + z2 = 4, z = 0 and is cut by the plane x + 2y + 2z = 0 in a circle of radius 3. 34. Find the equations of the spheres which pass through the circle x2 + y2 + z2 – 2x – 4y = 0, x + 2y + 3z = 8 and touch the plane 4x + 3y = 25. 35. Find the equation of the sphere which touches the sphere x2 + y2 + z2 + 2x – 6y + 1 = 0 at the point (1, 2, –2) and passes through the origin. [Hint: The required sphere is the one that passes through the point circle of intersection of the sphere x2 + y2 + z2 + 2x – 6y + 1 = 0 and the tangent plane to the sphere at (1, 2, –2)] 36. Find the equation of the sphere that passes through the circle x2 + y2 + z2 + x –3y + 2z – 1 = 0, 2x + 5y – z + 7 = 0 and cuts orthogonally the sphere x2 + y2 + z2 – 3x +5y – 7z – 6 = 0.

2.12 CONE AND RIGHT CIRCULAR CONE 2.12.1

cone. a right circular cone is called the vertex of the cone and the variable line is called a generator of the cone. In the case of the general cone, the given curve intersected by the generator is called the base curve or guiding curve. axis of the cone and the constant angle is called the semi vertical angle of the cone.

2.12.2 Equation of a Cone with Vertex at the Origin Let the equations of the base curve be ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy + 2ux + 2vy + 2wz + d = 0

(1)

and px + qy + rz = s

(2)

O

P (x, y, z) Fig. 2.32

Let P (x, y, z) be any point on the cone. Then OP is a generator of the cone. Equations of OP are x l

y m

z n

k

(3)

Chapter 2: Three Dimensional Analytical Geometry OP intersects the base curve. Using (3) in (1) and (2), we get k 2 al 2 2k 2

I – 2.63

OP, for some k, lies on it. fmn

2k ul

d

0

1 d 0 k k2 and k (pl + qm + rn) = s Eliminating k between (4) and (5), we have i.e.,

( al 2

2 fmn)

( al 2 s2

i.e.,

al 2

Replacing l, m, n by

(4)

2( ul )

pl s

2 fmn)

2 ul

2 fmn

2 s ( ul )( pl )

d

(5)

( pl ) 2 s2

0

( pl ) 2

0

(6)

x y z , , respectively in (6) and multiplying by k2, equation (6) k k k

becomes s 2 { ax 2 2 fyz} 2 s ( ux)( px) ( px) 2 (7) represents the required cone, as (x, y, z

0

(7)

Note 1. Equation (7) is a homogeneous equation of the second degree in x, y, z. 2. Equation (7) can be got by homogenising equation (1) using (2). 3. Conversely, any homogeneous equation of the second degree in x, y, z will represent a cone with vertex at the origin. 4. Equation of a cone with its vertex at ( , , ) is a homogeneous equation of the second degree in x – , y – , z x y z 5. From (6) and (7), we observe that if is a generator of the cone l m n represented by the homogeneous equation f (x, y, z) = 0, then f (l, m, n) = 0.

2.12.3

General Equation of a Right Circular Cone )

Let the vertex of the right circular cone be ( , , ), semi-vertical angle be and the equations of the axis be x

y l

V

z m

P(

. n

Let P (x, y, z) be any point on the right circular cone. Then the D.R.’s of the generator VP are (x – , y – , z VC are (l, m, n) Angle between VA and VC is given by l x m y n z cos 2 2 l 2 m2 n2 x y

A

,z x, y

C ( , , ) B Fig. 2.33

z

2

Part I: Mathematics I

I – 2.64

The locus of (x, y, z) or the equation of the cone is l2

m2

n2

2

x

lx

2

y

my

2

z 2

nz

cos 2

.

Corollary 1: If the vertex of the R.C. cone is the origin, its equation is (l 2

m2

n 2 ) (x 2 (lx

y2

z 2 ) cos 2 ,

nz ) 2 .

my

Corollary 2: If the vertex is the origin and the axis is the positive z-axis, then the equation of the R.C. cone is x2

y2

z 2 cos 2 x2

or

y2

z 2 since l

0, m

0, n

1

z 2 tan 2 .

Corollary 3: The curve of intersection of a R.C. cone and a plane perpendicular to its axis is a circle. For, if the cone is taken as (l 2 m 2 n 2 ) (x 2 y 2 z 2 ) cos 2 (lx my nz ) 2 and the perpendicular plane is taken as lx + my + nz = p, the curve p 2 sec 2 of intersection is given by x 2 y 2 z 2 and lx + my + nz = p, which 2 l m2 n2 is a circle.

WORKED EXAMPLE 2(e) Example 2.1 Find the equation of the cone whose vertex is the origin and guiding curve is x2 y 2 z 2 1, x y z 1 . 4 9 1 Since the vertex of the cone is the origin, the required equation of the cone is got by 2 y2 z2 homogenising x 4 9 1 The required equation is

1 , using x + y + z = 1

x2 4 i.e., i.e.,

y2 9

z2 1

x

y

z

2

9x2 + 4y2 + 36z2 = 36 (x + y + z)2 27x2 + 32y2 –72 (xy + yz + zx) =0

x y z 0 meets the a b c co-ordinate axes in A, B, C. Find the equation of the cone whose vertex is the origin and guiding curve is the circle ABC. Example 2.2 A variable plane parallel to the plane

Chapter 2: Three Dimensional Analytical Geometry Any plane parallel to x a

y b

z c

x a

y b

z c

I – 2.65

0 is

k ...(1), where k is a parameter.

Then (ka, 0, 0), B kb, 0), C kc) The sphere passing through A, B, C can be found out as that passing through O, A, B, C. Proceeding as in an example in the topic on ‘spheres’, the equation of a sphere through A, B, C can be found out as (2) x2 + y2 + z2 – k (ax + by + cz)= 0 The base curve is the circle of intersection of (1) and (2). Since the required cone has its vertex as the origin, its equation is got by homogenising equation (2) with the help of equation (1). The required equation is x2 i.e.,

y2

z2

x2

y2

k ax by z2

a b

i.e.,

cz

ax by

b xy a

b c 1 2 b bc

1 x k a

y b

z c

0

x a

y b

z c

0

a zx c

0

cz c yz b

c a

i.e.,

1 2 a ab

i.e.

c(a2 + b2) xy + a(b2 + c2) yz + b(c2 + a2) zx=0

b 2 xy

c 2 yz

1 2 c ac

a 2 zx

0

Example 2.3 Find the equation of a cone whose vertex is the origin and base curve is y2 + z2 = b2, x = a. Show that the section of the cone by a plane parallel to the xoy -plane is a hyperbola. The required equation of the cone is got by homogenising y2 + z2 = b2 by x = a. It is y 2

z2

b2

x a

2

i.e., b2x2 – a2y2 – a2z2 = 0 Any plane parallel to the xoy-plane is z = c The section of (1) by (2) is given by

(1) (2)

b2x2 – a2y2 – a2c2 = 0 and z= c i.e.,

x2 ac b

2

y2 c2

1 and z=c, which is a hyperbola lying on the plane z = 0.

Example 2.4 Find the general equation of the cone which passes through the coordinate axes. Find also the equation of such a cone that also passes through the lines x = – y = z and x = 2y = – 3z. Since the cone passes through the co-ordinate axes, the origin is the vertex of the cone. Hence its equation is a homogeneous equation of the second degree in x, y, z.

Part I: Mathematics I

I – 2.66 Let it be

ax2 + by2 + cz2 + 2fyz + 2gzx + 2hxy = 0 Since the x-axis

x 1

y 0

z lies on the cone, it is a generator. 0

a = 0. Similarly b = 0 and c = 0. General equation of the cone passing through the co-ordinate axes is fyz + gzx + hxy = 0 (Note

(1)

(2)

The cone with this property is called a quadric cone.)

Now the cone (2) passes through the line

x 1

y 1

z 1

i.e., – f + g – h = 0

(3)

Cone (2) passes through the line x = 2y = – 3z i.e.,

x 6

y 3

z . 2

f – 2g + 3h = 0 f g h Solving (3) and (4), we have 1 4 3 Using (5) in (2), the equation of the required cone is yz + 4zx + 3xy = 0.

(4) (5)

Example 2.5 Show that the cones ayz + bzx + cxy = 0 and fyz + gzx + hxy = 0 intersect in the co-ordinate axes and in the line (bh – cg)x = (cf – ah)y = (ag – bf)z pass through the co-ordinate axes, given by x 1

y 0

z x , 0 0

y 1

z 0

and

x 0

y 0

z 1

The two cones intersect along the three co-ordinate axes. To get the other line of intersection, we solve the equations of the cones simultaneously by the rule of cross multiplication. Solving we get yz zx xy bh cg cf ah ag bf i.e.,

xyz x bh cg

xyz y cf ah

xyz z ag bf

i.e., (bh – cg) x = (cf – ah) y = (ag – bf) z, which are the equations of the required line. Example 2.6 Find the equation of the cone whose vertex is (1, 2, 3) and whose guiding curve is the circle x2 + y2 + z2 = 4, x + y + z = 1.

Chapter 2: Three Dimensional Analytical Geometry

I – 2.67

Let a generator of the cone be x 1 y 2 z 3 (1) r l m n Any point on ( 1 ) is P(lr + 1, mr + 2, nr + 3) Since the generator intersects the guiding curve, P lies on it, for some value of r. and From (3),

(lr + 1)2 + (mr + 2)2 + (nr + 3)2 = 4

(2)

lr + 1 + mr + 2 + nr + 3 = 1 5 r l m n

(3)

Using this value in (2), we get 5l l i.e.,

2

2

5m l

1

5n l

2

2

3

4

(– 4l + m + n)2 + (2l – 3m + 2n)2 + (3l + 3m – 2n)2 = 4(l + m + n)2

i.e.,

5l2 + 3m2 + n2 – 2lm – 6mn – 4nl= 0

i.e.,

r2[5(x – l)2 + 3(y – 2)2 + (z – 3)2 – 2(x – l) (y – 2) –6 (y – 2) (z – 3) – 4 (z – 3) (x – l)] = 0, on using (l)

The equation of the required cone is 5x2 + 3y2 +z2 – 2xy – 6yz – 4zx + 6x + 8y + 10z – 26 = 0 Example 2.7 Find the equation of the cone whose vertex is the point (a, ß, 2 2 whose base curve is the ellipse x 2 y2 1, z 0 . a b Let a generator of the cone be x

y b

z m

Any point on (1) is P (lr + , mr + ß, P lies on it, for some value of r. 2

lr a

(1)

r

n

). Since (1) intersects the base curve, mr

(2)

1 b2 nr + = 0

2

and From (3),

r

(3) n

Using this value in (2), we get

i.e.,

1 a2

2

l n

1 a2 l

n

2

1 b2 1 b2

m n m

1

n

2

n2

(4)

Part I: Mathematics I

I – 2.68 Using (1) in (4), we get 1 a2

x

i.e., 1 a2

z

x

1 b2

2

1 b2

2

z z

y

y

2

2

z 2

z

2

z

, which is the equation of the required cone

Example 2.8 Prove that the equation x2 – 2y2 + 3z2 + 5yz – 6zx – 4xy + 8x – 19y – 2z – 20 = 0 represents a cone with vertex (1, – 2, 3) The given equation can be rewritten as (x – 1)2 – 2(y + 2)2 + 3(z – 3)2 + 5(y + 2)(z – 3) – 6(z –3)(x–1) – 4(x – l) (y + 2) = 0 (1) (1) is a homogeneous equation of the second degree in x – 1, y + 2, z – 3. (1) represents a cone with vertex at (1, – 2, 3) Example 2.9 Prove that the equation ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d = 0 2 v 2 w2 represents a cone, if u d . Find its vertex. a b c ax2 + by2 + cz2 + 2ux + 2vy + 2wz + d= 0 u 2 v 2 w2 d a b c Using the value of d from (2) in (1), it becomes

ax 2

by 2

cz 2

i.e.,

2ux

2vy u a

a x

u2 a

2 wz

2

v b

b y

v2 b

2

c z

w2 c w c

(1) (2)

0

2

(3) is a second-degree homogeneous equation in x

(3)

0 u ,y a

v ,z b

w . c u , a

Equation (3) and hence (1) represents a cone. Vertex of the cone

v , b

w c

Example 2.10 Find the equation of the right circular cone whose semi-vertical angle is 30°, vertex at the point (2, 1, – 3) and direction ratios of whose axis are (3, 4,–1) The equation of the right circular cone whose vertex is ( , , ), semi-vertical angle and D.R.’s of whose axis are (l, m, n) is l 2 [Refer to the book-work] Accordingly, the equation of the required cone is

i.e., i.e.,

2

26 x

2

39 x 2

y2

21x 2

7 y2

y 1 z2 37 z 2

2

4x

z 2y

x

3 4

3 x

6 z 14

2 3x

3

2

2

cos 2

2 4y

4 y 1 z 13

48 xy 16 yz 12 zx 130 y 182 z

2

l x

z 2

208

0

3

2

Chapter 2: Three Dimensional Analytical Geometry

I – 2.69

Example 2.11 Find the equation of the right circular cone whose vertical angle , which has its vertex at the origin and whose axis lies along the line x = – 2y = 2 z. Also show that the plane z = 0 cuts the cone in two straight lines inclined at an is

1

angle of cos

4 . 5

Equations of the axis are

x 2

y z 1 2 l = 2, m = – 1, n = 2

The equation of the required cone is 9 x 2

y2

z2

1 2

2x

2

2 z , using the

y

formula derived in the book-work. i.e., x2 + 7y2 + z2 +8xy + 8yz – 16zx = 0 The two lines of intersection of (1) with z = 0 are given by x2 + 7y2 + 8xy = 0 If (l, m, n) be the D.R.’s of a line of intersection, then l2 + 8lm + 7m2 = 0 i.e., (l + 7m) (l + m) = 0 l l 7 1 or i.e., m m 1 1 D.R.’s of the lines are (– 7, 1, 0) and (–1, 1, 0) If is the angle between the two lines, 4 1 4 4 cos = or cos 1 5 5 50 2

(1) (2)

Example 2.12 Find the equation of the right circular cone generated by revolving the line x = 0, y – z = 0 about the axis x = 0, z = 2. The revolving line is a generator whose equations are The equations of the axis are

x 0

y 1

z

x 0

y 1

z . 1

(1)

2

(2)

0

The point of intersection of (1) and (2), namely, (0, 2, 2) is the vertex of the cone. The semi-vertical angle of the cone is given by cos = The equation of the required right circular cone is l x i.e., i.e.,

1 2 l2

1 1

.

2 x

2

cos 2

2

x2 + (y –2)2 + (z – 2)2=2(y – 2)2 x2 – y2 + z2+ 4y – 4z = 0

Example 2.13 Lines are drawn through the origin with direction ratios (1, 2, 2), (2, 3, 6) and (3, 4, 12). Find the D.R.’s of the axis of the right circular cone passing through them and the semi-vertical angle. Find also the equation of the cone.

Part I: Mathematics I

I – 2.70

Since the cone passes through the three given lines, they are generators of the cone. The angles between them and the axis are equal. Let (l, m, n) be the D.R.’s of the axis of the cone. l

Then

2m

9

2m

2n

l l

i.e.,

2n

2

2l

3m l

2l

6n

2

3l

l2

49

3m 7

4m 12n

6n

3l

169

4m 12n 13

3 From these equal ratios, we get l + 5m – 4n = 0, 5l + 11m – 6n = 0, 4l + 14m – 10n = 0 l 1

m 1

n . These values satisfy the third equation also. 1

Semi-vertical angle is given by the angle between the axis and any generator. 1 2 2

cos

3

1

9

3

The equation of the cone is l2 3 x2

i.e.,

2

x

i.e.,

cos 2

2

l x

y2

z2

1 3

xy

yz

zx

x

y

z

2

0

Example 2.14 Find the semi-vertical angle, and the equation of the right circular cone having its vertex at the origin and passing through the circle y2 + z2 = 25, x = 4. Since the cone has its vertex at the origin, its equation is got by homogenising y2 + z2 = 25 with x = 4. y2

It is

z2

25

x 4

2

25x2 – 16y2 – 16z2 = 0

i.e.,

5

section, namely, (4, 0, 0) and the vertex O x y z . 4 0 0 The radius of the base circle = 5. The semi-vertical angle is given by is the axis.Its equation is

tan

5 or 4

tan

1

5 4

4

O

Fig. 2.34

Chapter 2: Three Dimensional Analytical Geometry

I – 2.71

Example 2.15 Find the equation of the right circular cone passing through the coordinate axis and passing through the origin. Obtain the semi-vertical angle and equations of the axis. By Example (4) given above, the equation of the general quadric cone is fyz + gzx + hxy = 0 (1) x l

Let the equations of the axis of this cone be

y m

z n

(2)

(2) makes equal angles with the co-ordinate axes whose D.R.’s are (1, 0, 0), (0, 1, 0) and (0, 0, 1) l

m l

2

l

n 2

l

1

cos

2

or

3

=cos

1

1

. Squaring, we

3

get l2 = m2 = n2. Taking l = 1, we get m = ± 1 and n Equations of the axis of the cone are x y z 1 1 1 Let P(x, y, z) be any point on the cone. Then OP is a generator with D.R.’s (x, y, z) Angle between OP and the axis = cos x

i.e.,

y

x

3 (x i.e., yz

zx

xy

z

y

1

1

3 1

2

3

z )2

x2

y2

z2

0 , which is the equation of the cone.

Example 2.16 Prove that the plane px + qy + rz = 0 cuts the cone ax2 + by2 + cz2 = 0 in perpendicular lines, if (b + c)p2 + (c + a)q2 + (a + b)r2 = 0. Let (l, m, n) be the D.R.’s of one of the lines of intersection. Since this line lies on the plane,

2

(1)

2

al + bm + cn = 0

Also

From (1), n

pl + qm + rn= 0 2

pl

(2)

qm

(3)

r

Using (3) in (2); al 2 i.e.,

ar 2

cp 2 l 2

bm 2

2cpq lm

c pl br 2

qm

2

r2 cq 2 m 2

0 0

Part I: Mathematics I

I – 2.72

i.e.,

ar

2

cp

(4) is a quadratic equation in

l m

2

2

2cpq

l m

br 2

cq 2

0

(4)

l l l , giving two values, say, 1 and 2 , where m2 m1 m

(l1, m1, n1) and (l2, m2, n2) are the D.R.’s of the two lines of intersection. From (4), i.e.,

l1 l2 m1 m2

br 2 ar 2

l1l2

m1m2 cp 2 ar 2

br 2

cq 2

Similarly each of the above ratios

cq 2 cp 2

n1n2 aq bp 2 2

Now the two lines are perpendicular, if l1l2 + m1m2 + n1n2 = 0 i.e.,

br2 + cq2 + cp2+ ar2 + aq2 + bp2 = 0

i.e.,

(b + c)p2 + (c + a)q2 + (a + b)r2 = 0

EXERCISE 2(e) Part A (Short Answer Questions) 1. What is the difference between a cone and a right circular cone?

4. What is the nature of the equation of a cone with its vertex at (i) the origin? (ii) the point ( , , )? 5. Write down the general equation of a right circular cone with its vertex at (i) the point ( , , ) (ii) the origin? 6. Derive the equation of a right circular cone whose vertex is the origin and axis is the positive z-axis. 7. Assuming the equation of a right circular cone in any form, prove that its section by a plane is a circle. 8. Find the equation of the cone whose vertex is the origin and which passes through the curve ax2 + by2 + cz2 = 1, lx + my + nz = p. 9. Find the equation of the right circular cone whose vertex is the origin, axis is the y-axis and semi-vertical angle is 30°. 10. Find the equations of the curve of intersection of the cone x2 = y2 + z2 and a plane parallel to the yoz-plane. 11. If 2l2 + 3m2 – 5n2 x y z . l m n

Chapter 2: Three Dimensional Analytical Geometry

I – 2.73

12. The direction ratios of a generator of a cone with vertex (1, 1, 1) satisfy the relation l2 + m2 = 3n2. Find the equation of the cone. Part B 13. Write down the general equation of a quadric cone. Find the equation of the x y z x y z . Is quadric cone passing through the lines and 1 2 3 1 1 1 x y z the line a generator of this cone? 5 4 1 , ) and base curve y2 – 4ax = 0, z = 0. 15. Find the equation of the cone whose vertex is (3, 1, 2) and base curve 2x2 + 3y2 = l, z = l. 16. Find the equation of the cone whose vertex is (1, 1, 3) and which passes through the ellipse 4x2 + z2 = 1, y = 4. 17. Find the equation of the cone whose vertex is (3, 4, 5) and base curve is the conic 3y2 + 4z2 = 16, z + 2x = 0. 18. Prove that the equation 2x2 + 2y2 + 7z2 – 10yz – 10zx + 2x + 2y + 26z – 17 = 0 represents a cone whose vertex is (2, 2, 1). 19. Prove that the equation 4x2 –y2 + 2z2 – 3yz + 2xy + 12x – 11y + 6z + 4 = 0 represents a cone whose vertex is (– 1, – 2, – 3). 20. Find the equation of the right circular cone, which passes through the point x y z (1,1,2) and has its vertex at the origin and the axis along the line . 2 4 3 21. Find the equation of the right circular cone whose vertex is the origin, whose axis makes equal angles with the co-ordinate axes and the direction ratios of a generator of which are (1, – 2, 2). 22. If the equation of a right circular cone with the vertex at the origin and the axis making equal angles with the co-ordinate axes in 4 (x2 +y2 + z2) + 9(yz + zx + xy) = 23. Find the equation of the right circular cone, if its vertex is (1, – 1, 1), axis is y parallel to x z and the direction ratios of one of its generators are 2 (2,2,1). 24. Find the equation of the right circular cone, if its vertex is (1, 2, – 3), axis is parallel to the y-axis and semi-vertical angle is 45°. 25. Find the equation of the right circular cone, if its vertex is the origin and three of its generators have direction ratios (1, 2, 2), (1, – 2, 2) and (2, – 1, – 2). 26. Find the semi-vertical angle, the axis and the equation of the right circular x y z x y z x y z . , and 3 6 2 2 2 1 1 2 2 27. Find the equation of the right circular cone generated by rotating the line cone passing through the lines

x 1

y 2

z x about the line 3 1

y 1

z . 2

Part I: Mathematics I

I – 2.74

28. Find the equation of the right circular cone formed by rotating the line 2y + 3z = 6, x = 0 about the z-axis. [Hint Vertex of the cone is (0, 0, 2)] 29. Find the equation of the right circular cone formed by rotating the line 2x + 3y = 6, z = 0 about the y-axis. 30. Find the equations of the lines in which the plane 3x + 4y + z = 0 cuts the cone 15x2 – 32y2– 7z2 = 0. 31. Find the equations of the lines in which the plane 2x +y – z = 0 cuts the cone 4x2– y2 + 3z2 = 0. 32. Find the angle between the lines of section of the plane 3x + y + 5z = 0 and the cone 6yz – 2zx + 5xy = 0. 33. Find the angle between the lines of section of the plane 6x – y – 2z = 0 and the cone 108x2 – 7y2 – 20z2 = 0. 34. Prove that the plane ax + by + cz = 0 meets the cone yz + zx + xy = 0 in 1 1 1 perpendicular lines, if 0. a b c 35. Prove that the angle between the lines of intersection of the plane x + y + z = 0 with the cone ayz + bzx + cxy = 0 is

2

, if a

b

c = 0 and

3

, if

1 a

1 b

1 c

0

2.13 CYLINDER 2.13.1 A cylinder is the surface generated by a straight line which is always parallel to a a given curve. The moving straight line is called the generator of the cylinder and the given curve is called the guiding curve of the cylinder. The equation of a cylinder whose generators are parallel to a given line and intersect a given curve. x y z l m n Let the generators of the cylinder intersect the curve whose equations are Let the generators of the cylinder be parallel to the line

ax2 + 2hxy + by2 + 2gx + 2fy + c = 0 and z=0 Let (x1, y1, z1) be a point on the required cylinder. Then the equations of the generator of the cylinder through (x1, y1, z1) are x

x1 l

y

y1 m

z

z1 n

r

(1)

(2) (3)

(4)

Any point on the line (4) is (x1 + lr, y1 + mr, z1 + nr) (5) If this point lies on the curve given by (2) and (3), it means that (4) intersects this curve. For the generator (4) to intersect the given curve,

Chapter 2: Three Dimensional Analytical Geometry z1 + nr = 0

I – 2.75

z1 n

r

For this value of r, the point (5) becomes lz1 mz1 , y1 ,0 n n For the generator (4) to intersect the given curve, the point (6) lies on (2) x1

a x1

lz1 n

2

2 h x1 lz1 n

2 g x1 a nx1 lz1

i.e.,

2

lz1 n

2 f y1

y1

mz1 n

mz1 n

mz1 n

2

c 0

2h nx1 lz1 ny1 mz1

2 g n nx1 lz1

b y1

(6)

b ny1 mz1 cn 2

2 fn ny1 mz1

2

0

The locus of (x1, y1, z1), viz, the required equation of the cylinder is a (nx – lz)2 + 2h (nx – lz) (ny – mz) + b (ny – mz)2 +2gn (nx – lz) + 2fn (ny – mz) + cn2 = 0

2.14

RIGHT CIRCULAR CYLINDER

2.14.1 Right Circular Cylinder is a surface generated by a straight line which is parallel to a the radius the axis. In other words, it is a surface whose guiding curve is a circle and whose generators are parallel to the normals of the plane of the circle. The equation of the right circular cylinder whose radius is r and whose axis is the line x y z l m n Let P (x, y, z) be any point on the cylinder and A ( , , ) is a point on the line x

y l

z m

n

(1)

P r A

M

Fig. 2.35

AP2 = (x –

2

–(

)2 – (z –

2

B

Part I: Mathematics I

I – 2.76 AM

AP on the line (1) m y m z

l x

l2 , i.e.,

m2 n2 2 2 = +

2

x

y

2

z

2 2

lx

m y l2

m2

n z n2

2

r2

which is the required equation of the right circular cylinder.

WORKED EXAMPLE 2(f) Example 2.1 Find the equation of the cylinder whose generators are parallel to the z-axis and which passes through the curve of intersection of x2 + y2 + z2 =1 and x + y + z =1. Let P(x1, y1, z1) be a point on the required cylinder. The generators are parallel to the z-axis whose D.C.’s are (0, 0, 1) The equations of the generator of the cylinder through P are x x1 y y1 z z1 (1) r 0 0 1 Any point Q on line (1) is (x1, y1, z1 + r) Since the generator intersects the guiding curve, Q will lie on the guiding curve for some value of r (2) i.e., x12 y12 (z1 r ) 2 1 and x1 + y1 + (z1 + r) = 1 (3) From (3), r = 1 – x1 – y1 – z1 Using this value of r in (2), we have x12 y12 (1 x1 y1 ) 2 1 2 x12 2 y12 2 x1 2 x1 y1 2 y1 0 The locus of (x1, y1, z1) or the equation of the cylinder is x2 + y2 + xy – x – y = 0.

i.e.,

Example 2.2 Find the equation of the cylinder whose generators are parallel to a line with direction ratios (1, –2, 3) and whose guiding curve is the ellipse x2 + 2y2 = 1 and z = 3. Let P (x1, y1, z1) be any point on the required cylinder. Then the equations of the generator through P are x x1 y y1 z z1 (1) r 1 2 3 Any point Q on (1) is (x1 + r, y1 – 2r, z1 + 3r). Since the generator (1) intersects the guiding curve, Q lies on it for some value of r. and From (3), r

z1

3 3

(x1 + r)2 + 2(y1 – 2r)2 = 1

(2)

(z1 + 3r) = 3

(3)

. Using this value of r in (2),

Chapter 2: Three Dimensional Analytical Geometry z1

3

2

2 2 y1 3 z1 We have x1 3 3 i.e., (3x1 – z1 + 3)2 + 2{3y1 + 2z1 – 6}2 = 9 i.e., 9 x12

18 y12

9 z12

24 y1 z1

6 z1 x1

18 x1

I – 2.77

2

1

72 y1

54 z1

72

0

The locus of (x1, y1, z1), viz., the equation of the required cylinder is 3x2 + 6y2 + 3z2 + 8yz – 2zx + 6x – 24y –18z + 24 = 0 Example 2.3 Find the equation of the cylinder whose guiding curve is given by x2 + y2 – 4x – 2y + 4 = 0 and z = 0 whose axis contains the point (0, 0, 3). The guiding curve is a circle in xy-plane. Its centre is (2, 1, 0). Since the axis of the cylinder is a line of symmetry, it will pass through the centre of the guiding circle. The D.R.’s of the axis are (–2, –1, 3) If P (x1, y1, z1) is any point on the required cylinder, the equations of its generator through P are x

x1 y y1 z z1 r 2 1 3 Any point Q on line (1) is (x1 – 2r, y1 – r, z1 + 3r) If Q is the point where the line (1) intersects the guiding curve, then (x1 – 2r)2 + (y1 – r)2 – 4 (x1 – 2r) – 2 (y1 – r) + 4 = 0 and z1 + 3r = 0 (3) Eliminating r between (2) and (3), we get 2 2 2 z1 z1 z1 2 z1 2 y1 4 0 x1 y1 4 x1 3 3 3 3 i.e., (3x1 + 2z1)2 + (3y1 + z1)2 – 12 (3x1 + 2z1) – 6 (3y1 + z1) + 36 = 0 i.e.,

9 x12

y12

5 z12

6 y1 z1

12 z1 x1

36 x1 18 y1

30 z1

36

(1)

(2)

0

The locus of P, viz., the required equation of the cylinder is 9 (x2 + y2) + 5z2 + 6yz + 12zx – 36x – 18y – 30z + 36 = 0 Example 2.4 Find the equation of the right circular cylinder whose axis is the line x

1 y 2 z 3 and whose radius is equal 2 1 2 to 5. Let P (x1, y1, z1) be any point on the required right circular cylinder. AP =

x1

1

2

y1

2

2

z1

3

2

P (x1, y1, z1)

M

(1, 2, 3) A

PM = radius of the cylinder = 5 AM AP on the axis whose D.R.’s are (2, 1, 2) 2 x1

y1

1 22

2 z1

2 12

22

3

Fig. 2.36

Part I: Mathematics I

I – 2.78

APM, AM2 + MP2 = AP2 1 2 2 2 2 x1 1 y1 2 z1 3 2 x1 y1 2 z1 10 25 9 4 x12 y12 4 z12 100 4 x1 y1 4 y1 z1 8 z1 x1 40 x1 20 y1 40 z1

i.e., i.e.,

225

9 x12 y12 z12 2 x1 4 y1 6 z1 14 Locus of (x1, y1, z1), viz. , the equation of the required cylinder is 5 x2 + 8y2 + 5z2 – 4xy – 4yz – 8zx + 22x – 16y – 14z – 199 = 0 Example 2.5 Find the equation of the right circular cylinder of radius a whose axis passes through the origin and makes equal angles with the co-ordinate axes. The D.C.’s of the axis are (cos , cos , cos ) 1 Since cos2 + cos2 + cos2 =1, we get cos 3 D.R.’s of the axis are (1, 1, 1). It passes through the origin. Its equations are x y z 1 1 1 Let P (x1, y1, z1) be any point on the required cylinder. Then OP 2 x12 y12 z12 OM OP on the axis. x1 y1 z1 and Now

3 PM = radius of the cylinder = a. OM2 + MP2 =OP2

i.e.,

x1

i.e.,

2 x12

y1

2

z1

3 y12

a2 z12

x1 y1

x12 y1 z1

y12

z12 z1 x1

3a 2

Locus of (x1, y1, z1), viz., the required equation of the right circular cylinder is 2(x2 + y2 + z2 – xy – yz – zx) = 3a2 Example 2.6 Find the equation of the right circular cylinder whose axis is x

y 1 z 0 and which passes through the point (0, 0, 3). 2 1 3 Any point on the axis of the cylinder is (2r + 2, r + 1, 3r). If this point is the foot of the perpendicular N drawn from Q(0, 0, 3) M then the D.R.’s of QN are (2r + 2, r + 1, 3r – 3) N Now QN is perpendicular to the axis (0, 0, 3) Q whose D.R.’s are (2, 1, 3) 2(2r + 2) + (r + 1) + 3(3r – 3) = 0 (2, 1, 0) A 2 r i.e., 14r = 4 7 2

Co-ordinates of N are

4 7

2,

2 7

1,

6 7

Fig. 2.37

P (x1, y1, z1)

Chapter 2: Three Dimensional Analytical Geometry

I – 2.79

18 9 6 , , 7 7 7

i.e.,

Now the radius of the cylinder = QN (= PM) 18 7

2

9 7

2

6 7

2

3

630 90 or 49 7 Let P (x1, y1, z1) be any point on the required cylinder Then AP =

x1

2

2

y1 1

AM

2

(1)

z12

AP on the axis whose D.R.’s are (2, 1, 3) 2 x1 2 x1

2 y1

y1 1 3 z1 5

22

3 z1

12

32

14

PM = radius of the cylinder

90 from (1) 7

Now AM2 + MP2 = AP2 i.e.,

2 x1

y1

3 z1 5

2

90 7

14

x1

2

2

y1 1

2

z12

Locus of (x1, y1, z1), viz., the required equation of the right circular cylinder is (2x + y + 3z – 5)2 + 180 = 14 (x2 + y2 + z2 – 4x – 2y + 5 ) i.e., 4x2 + y2 + 9z2 + 4xy + 6yz + 12zx – 20x – 10y – 30z + 25 + 180 = 14 (x2 + y2 + z2 – 4 x – 2 y + 5) i.e.,

10x2 + 13y2 + 5z2 – 4xy – 6yz – 12zx – 36x – 18y + 30z – 135 =0

Example 2.7 Find the equation of the right circular cylinder, one of whose sections is the circle given by x2 + y2 + z2 – x – y – z =0 and x + y + z =1. Since the section of the right circular cylinder is a circle, it should be a normal section, viz., it is the section of the cylinder by a plane perpendicular to the axis. We can consider the given circular section as the guiding circle of the required cylinder. Since the axis is normal to the guiding circle, it is perpendicular to the plane x+y+z=1 The D. R.’s of the axis are (1, 1, 1) The D. R.’s of any generator of the cylinder are (1, 1, 1) If P (x1, y1, z1) is any point on the required cylinder, the equations of the generator through P are x x1 1

y

y1 1

z

z1 1

r

(1)

Part I: Mathematics I

I – 2.80

Any point on this generator is (x1 + r, y1 + r1, z1 + r) Since the generator intersects the guiding circle, (x1 + r)2 + (y1 + r)2 + (z1 + r)2 – {x1 + r + y1 + r + z1 + r} = 0 (2) x1 + r + y1 + r + z1 + r = 1

and From (3),

x1

1

r

y1

(3)

z1

3 Using this value of r in (2), we have 1

2 x1

y1

z1

2

1

x1

3 i.e.,

6 x12

y12

z12

6 x1 y1

2 y1 3 6 y1 z1

z1

2

x1

1

6 z1 x1

3

y1 3

2 z1

2

1

9

The locus of (x1, y1, z1), viz., the required equation of the right circular cylinder is x2 + y2 + z2 – xy – yz – zx – 1 = 0 Example 2.8 Find the equation of the right circular cylinder described on the circle passing through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) as the guiding circle. The given three points are at a distance of 1 unit each from the origin. The sphere with centre at the origin and radius 1 can be considered as one sphere passing through the given points. (Note Any number of spheres can be constructed passing through 3 given points, but only one sphere passing through 4 given points exists.) The equation of the unique plane passing through the three given points is x + y + z = 1. (1) The guiding circle is the circle of intersection of the sphere x2 + y2 + z2 = 1 (1) and x+y+z=1 (2) Proceeding as in example 2.7 given above, the required equation of the right circular cylinder is x2 + y2 + z2 – xy– yz– zx = 1 Example 2.9 Find the equation of the enveloping cylinder of the sphere x2 + y2 + z2 – 2y – 4z – 11 = 0, having its generators parallel to the line x = – 2y = 2z. ( The cylinder whose generators touch a given surface is called the enveloping cylinder of that surface.) The generators are parallel to the line x = –2y = 2z x y z i.e., 2 1 1 If P (x1, y1, z1) is any point on the required cylinder, then the equations of the generator through P are x

x1

y

y1 z z1 (1) r 2 1 1 Any point on (1) is (x1 + 2r, y1 – r, z1 + r) If this point lies on the given sphere, then (x1 + 2r)2 + (y1 – r)2 + (z1 + r)2 – 2 (y1 – r) – 4 (z1 + r) – 11 = 0

Chapter 2: Three Dimensional Analytical Geometry i.e.,

6r 2

2 2 x1

y1

x12

z1 1 r

y12

z12

2 y1

4 z1 11

I – 2.81 (2)

0

(2) is a quadratic equation in r. If the roots of (2) are real and distinct, the two values of r will give two points of intersection of the generator (1) with the given sphere. If they are real and equal, the value of r will give the point of contact of (1) with the given sphere. In other words, if the roots of (2) are real and equal, the generator (1) will touch the given sphere. The generator (1) will touch the given sphere, if the discriminant of (2) = 0 i.e., i.e.,

4 2 x1 4 x12

y1 y12

z12

2 z1 i.e.,

2 x12

2

z1 1

4 x1 y1 2 1

5 z12

y12

2 y1 z1 2 1

6 x

1

5 y12

24 x12

z

4 x1 y1

2 y1

4 z1 x1

2 1

y

z12

2 y1

2 y1 z1

4 x1

4 z1 11

0

2 y1

4 z1 11

4 z1 x1

4 x1 14 y1

22 z1 67 0 The locus of (x1, y1, z1), viz., the required equation of the enveloping cylinder is 2x2 + 5y2 + 5z2 + 2 (2xy + yz – 2zx + 2x – 7y – 11z) – 67 = 0.

Example 2.10 Find the equation of the enveloping cylinder of the ellipsoid x y z x2 y 2 z 2 1 whose generators are parallel to the line . 2 2 2 l m n a b c If P(x1, y1, z1) is any point on the required cylinder, then the equations of the generator through P are x x1 y y1 z z1 (1) r l m n Any point on (1) is (x1 + lr, y1 + mr, z1 + nr) If this point lies on the given ellipsoid, then lr

x1

2

y1

a2

i.e.,

z1

b2

l2

m2

n2

2

2

2

a

2

mr

b

c

r2

x12

y12

z12

a2

b2

c2

nr

2

1

c2 2lx1 a

2my1

2

1

b

2nz1

2

c2

r (2)

0

Since the generator (1) touches the ellipsoid, the roots of (2) are equal. 4 i.e.,

lx1

my1

nz1

a2

b2

c2

x12

y12

z12

a2

b2

c2

4

1

l2

m2

n2

a2

b2

c2

Part I: Mathematics I

I – 2.82

The locus of (x1, y1, z1), viz., the equation of the enveloping cylinder is lx

my

nz

a2

b2

c2

2

l2

m2

n2

x2

y2

z2

a2

b2

c2

a2

b2

c2

1

EXERCISE 2(f) Part A (Short answer questions)

3. Find the equation of the right circular cylinder whose axis is the z-axis and whose guiding curve is a circle of radius ‘a’ in the xy-plane.

Part B 5. Find the equation of the cylinder whose generators are parallel to the x y z line and which passes through the circumference of the circle x2 + l m n z2 = a2 in the zox-plane. 6. Find the equation of the cylinder whose generators are parallel to the x-axis and intersect the curve given by ax2 + by2 + cz2 = 1 and lx + my + nz = p. 7. Find the equation of the cylinder whose generators are parallel to the y - axis and pass through the curve of intersection of x2 + y2 + 2z2 = 12 and x – y + z = 1. 8. Find the equation of the cylinder whose generators are parallel to the z-axis and intersect the curve ax2 + by2 = 2z and lx + my + nz = p. 9. Find the equation of the cylinder whose generators are parallel to the line x y z and pass through the circle x2 + y2 = 16, z = 0. 1 2 3 10. Find the equation of the cylinder which cuts the xoy-plane in the curve 4x2 +y2 = 1 and has its axis parallel to the line 3x = – 6y = 2z. 11. Find the equation of the cylinder whose generators are parallel to the line x = y = z and whose guiding curve is the circle given by x2 + y2 + z2 – 2x – 3 = 0 and 2x + y + 2z = 0. 12. The radius of a normal section of a right circular cylinder is 2 units and its x 1 y 3 z 2 . Find its equation. axis is given by 2 1 5 13. Find the equation of the right circular cylinder of radius 3 units whose axis is x 1 y 3 z 5 the line . 2 2 1

Chapter 2: Three Dimensional Analytical Geometry

I – 2.83

14. Find the equation of the right circular cylinder whose axis is the line x = 2y = –z and radius 4 units. 15. Find the equation of the right circular cylinder of radius 3 units whose axis passes through (2, 3, 4) and has D.C.’s proportional to (2, 1, –2). 16. Find the equation of the right circular cylinder which passes through the circle given by x2 + y2 + z2 = 9 and x – y + z = 3. 17. Find the equation of the right circular cylinder whose guiding circle is given by x2 + y2 + z2 = 4 and x + y + z = 2. 18. Find the equation of the right circular cylinder whose guiding circle is given by x2 + y2 + z2 + 2x – 2y – 4z – 19= 0 and x + 2y + 2z + 7 = 0. 19. Find the equation of the right circular cylinder of radius 1 whose axis passes through the origin and makes angles 30° and 60° with the x- and y-axis respectively. 20. Find the equation of the right circular cylinder which passes through the point (1, 0, – 3) and whose axis is given by x 2 y 3 z 4 . 3 4 5 21. Find the equation of the right circular cylinder which passes through the point (–1, 3, 9) and whose axis is given by x 13 y 8 z 31 . 5 8 1 22. Find the equation of the enveloping cylinder of the sphere x2 + y2 + z2 – 2x + 4y = 1 having its generators parallel to x = y = z. 23. Find the equation of the cylinder whose generators touch the sphere x2 + y2 + x y z z2 = a2 and are parallel to the line . l m n 24. Find the equation of the enveloping cylinder of the surface x2 + 2y2 + 3z2 = 6 whose generators are parallel to x = 2y = 3z.

ANSWERS

Exercise 2(a) (2) 54° 44'. (6) k=5. (8) cos

1

(3) 90°. (7) k = 2. 1 , (9) 6

1 . 6

(10) 7. (11) Sin 2 (13) 90 , cos

(l1m 2 1

l2 m1 ) 2 ; l1 = l2 , m1

2 , cos 3

1

1 . 3

m2 , n1

2 6

,

1 6

.

n2 .

(14) (–2, 4, 2). 26

(15) (8, –7, 8).

(17)

4 (18) 3

(19) 60°.

937

,

6 937

,

15 937

Part I: Mathematics I

I – 2.84

Exercise 2(b) (1)

d , a

d , b

d . c

(2) by + cz + d = 0.

(3) z = 5. (5) 8x + 2y – 3z – 5=0. (7) 60°.

(4) x + 2y – 3z = 0. (6) 2x + 6y– 3z + 6 = 0. (8) x – 2y + 2z ± 3 = 0

5 . 7 (11) No. (13) 2x + y – 2z = 6.

(12) x –3y = 0. (14) 2x + 3y + 6z – 49 = 0.

(15) 3x + 4y + 3 = 0.

(16)

(10) 2x + 3y – z – 14 = 0.

(9)

(17) cos (18) (19) (21) (23) (29)

4

1

29

, cos

1

(i) y + 2z = 4; 2x – y + 3z + 4 = 0. 2x + 2y – 3z + 3 = 0. x + y – z = 4.

3 29

, cos

1

2 29

(ii) 2x + z = 5;

8 2 10 9 4 13 . , , ; , , 7 7 7 7 7 7

(31) x + 4z = 7; 20x + 31y–13z + 46 = 0. (32) (i) x + 3y + 6z = 7; (33) 5x + 4y + z =10.

4 4 , 5 7

4 . 11

. (iii) 4x – y = 6. (20) 4x + 5y + 3z + 11 = 0. (22) x– 12y – 10z – 5 = 0. (28) (–3, 5, 2); (–1, 4, 3). (30) 9; 2x – 4y + 4z + 11 = 0.

(ii) y + z = l.

Exercise 2(c) (1) x = y = z.

(3)

x 1 2

(5) cos

(7)

1

x 2 2

y

2 3

z 1 5

4 . 9 y 3 3

(9) (2, –3, –1). (11) k = 1, 4.

(2)

x 1 1

y 2 5

z 3 . 3

(4)

x 3 2

y 2 2

z

(6) k = 2. z 4 . 6

(8) (–3, –3, 0). (10) 3. (12) x + y + z = 0.

4 . 2

Chapter 2: Three Dimensional Analytical Geometry

(13) (2, 3, 4).

(16)

x

y 5 2

2 3

z 0 . 5

(19) 90°.

x 7

y 4

z . 5

(17)

x 2

y 1 5

z . 3

(20) 1. x 1 4

(21) (3, 8, 29); 21; 3x 1 4

(22) (i)

3y

5 2

y 3 5

z 9 . 20

3z 8 ; 1

x

y

2 2

z 1 . 1

3 1

(ii)

6x

6 y 11 3

4 6

6 z 17 . 15

(24) x – y – z = 0. (26) y – 2z + 3 = 0.

(23) 11x – y – 3z = 28. (25) 9x – 2y – 5z + 4 = 0. (27)

(15)

I – 2.85

(28) 17x – 47y – 24z + 172 = 0.

(29) 6x – 5y – z = 0. (30) (5, – 7, 6); –11x + 6y + 5z + 67 = 0. (31) (4, 2, –1); –x + y + z + 3 = 0

18 , 11

(32)

4 , 11

3 , ; x + y + z = 1. 11

(33) (2, 4, –3); 45x – 17y + 25z + 53 = 0. (34) 7x – 7y + 8z + 3 = 0. (35) 83x – 2y – 7z – 21 = 0. (36)

x 1

y 3

z . 5

(37) 7x – 6y – z = 0 = 9x – 7y – z – 2.

(38) (0, 1, 2) and (2, 8, –3);

x 8 (40) 5 6; 11

(42)

1 6

(44)

(45)

5 3

x ;

1

97 13 6 13 66

y

3

78.

z

5

y

2 2

3 2

(39)

.

(41)

x 7

2 3

z 1

13 3 .

z 1 . 5

3 4

x ;

(43) 2 29 ;

; 2x + 11y – 7z – 20 = 0 = 13y – 13z – 3.

; 3x – y – z = 0 = x + 2y + z – 1.

y

2

25 9

y

3

y

14 9 1

1 x 2

5 3

75 9 . 1

z

z

7 4

.

Part I: Mathematics I

I – 2.86

Exercise 2(d) (2) (x– 1)2 + (y + 2)2 + (z – 3)2 =72.

(3) 6.

(4) x2 + y2 + z2 – x – 2y – 11z +11 = 0

(5)

3 3 , , 2 2

2 and 2.

(6) x2 + y2 + z2 – 4x + 6y – 8z + 2 = 0. (7) x2 + y2 + z2 – 4x – 6y – 10z + 13 = 0. (8) x2 + y2 + z2 – 2x – 4y – 6z + 9 = 0. (9)

x 1

2

y

2

2

z

3

1 . 9

2

(10) x2 + y2 + z2 – 2x – 3y – 4z = 0.

(11) x2 + y2 + z2 – 6x – 8y + 2z – 9 = 0, x + y – 3 = 0. (12) x2 + y2 + z2 – 2y + 4z = 11, x + 2y + 2z = 15. (14) 3 3.

(13) y + 5z – 9 = 0. (16)

9 12 , , 4 ; 9 x 12 y 5 5

(17)

11 8 5 , , and 6 x 7 7 7

(18) (19) (20) (21) (22) (24) (26)

2 y 3z

5 0.

x2 + y2 + z2 – 4x – 14y – 22z + 25 = 0. x2 + y2 + z2 + 4y – 6z – 1 = 0. 9 (x2 + y2 + z2) + 28x + 7y – 21z – 96 = 0. x2 + y2 + z2 – 4x + 7y – 3z + 15 = 0. 3 (x2 + y2 + z2) – 2 (x + y + z) – 1 = 0. (1, –3, 1). (25) (1, 2, 3). 2x + 2y – z – 8 = 0; 2x + 2y – z + 10 = 0.

(27)

7 , 3

(28)

40 . 3

5 , 3

2

c

2 a

2

a b (29)

20 z 125.

2 (30) 4 x

y2

2 ; 3. 3

2

2

a

2 a

2

b c ,

z2

2

2

b

2 a

2

c a ,

17 x 11 y 13 z

(31) x2 + y2 + z2 + 7x + 9y – 11z – 1 = 0. (32) x2 + y2 + z2 – 2x + 4y – 6z + 10 = 0.

28

2

0;

17 11 13 131 , , ; . 8 8 8 8

Chapter 2: Three Dimensional Analytical Geometry (33) (34) (35) (36)

I – 2.87

x2 + y2 + z2 ± 6z – 4 = 0. x2 + y2 + z2 + 6z = 16; 5 (x2 + y2 + z2) – 14x – 28y – 12z + 32 = 0. 4 (x2 + y2 + z2) + 10x – 25y – 2z = 0. 2 (x2 + y2 + z2) + 8x + 9y + z + 19 = 0.

Exercise 2(e) 6. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 20. 21.

x2 + y2 = z2 tan2 p2 (ax2 + by2 + cz2) = (lx + my + nz)2 3(x2 + y2) = z2 y2 + z2 = k2 , x = k 2x2 + 3y2 – 5z2 = 0 (x–1)2 + (y – l)2 = 3(z – l)2 5yz + 8zx – 3xy = 0 (ßz – y)2 = 4a( )( x) 2x2 + 3y2 + 20z2 – 6yz – 12zx + 12x + 6y – 38z +17 = 0 12x2 + 4y2 + 3z2 + 6yz + 8xy – 32x – 34y – 24z + 69 = 0 528x2 + 363y2 + 76z2 – 528xy – 264yz + 353zx + 784x + 1352z – 4436 = 0 4x2 + 40y2 + 19z2 – 48xy – 72yz + 36zx = 0 4(x2 + y2 + z2) + 9(yz + zx + xy) = 0

22. cos

1

1

3 3

23. 3y2 + 4yz – 2zx – 4xy – 2x + 6y + 6z + 1 = 0; 24. x2 – y2 + z2 – 2x + 4y + 6z + 6 = 0 25. 12x2 – 4y2 – 3z2 + 8zx = 0 26. cos

1

1

3

;x

y

z; xy

yz

zx

27. 5x2 + 5y2 – z2 + 4xy – 8yz + 8zx = 0 28. 4x2 + 4y2 – 9z2 + 36z – 36 = 0 29. 4x2 – 9y2 + 4z2 + 36y – 36 = 0 30.

x 3

y 2

z x and 1 2

y 1

31.

x 1

y 4

z x and 2 1

y 2

32. cos

1

33. cos

1

1 6 20 . 21

z 2 z 0

0

Part I: Mathematics I

I – 2.88

Exercise 2(f) (3) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)

x2 + y2 = a2 (mx –ly)2 + (mz – ny)2 = a2m2 a (my + nz – p)2 + l2 (by2 + cz2 – 1) = 0 2x2 + 3z2 + 2zx – 2x – 2z – 11 = 0 n (ax2 + by2) + 2 (lx + my – p) = 0 9x2 + 9y2 + 5z2 – 12yz – 6zx – 144 = 0 36x2 + 9y2 + 17z2 + 6yz – 48zx – 9 = 0 17x2 + 18y2 + 17z2 – 18xy – 18yz – 16zx – 30x + 10y + 20z – 75 = 0 26x2 + 29y2 + 5z2 + 4xy + 10 yz – 20zx + 150y + 30z + 75 = 0 5x2 + 5y2 + 8z2 – 8xy + 4yz + 4zx – 6x – 42y – 96z + 225 = 0 5x2 + 8y2 + 5z2 – 4xy + 4yz + 8zx –144 = 0 5x2 + 8y2 + 5z2 – 4xy + 4yz + 8zx – 40x – 56y – 68z + 179 = 0 x2 + y2 + z2 + xy + yz – zx = 9 x2 + y2 + z2 – xy – yz – zx = 4 8x2 + 5y2 + 5z2 – 4xy – 8yz – 4zx + 28x + 2y – 16z – 52 = 0

2 (19) x

(20) (21) (22) (23) (24)

3y2

z2

2 3 xy

4

0

41x2 + 34y2 + 25z2 – 24xy – 40yz – 30zx – 28x + 4y – 20z – 2444 = 0 (5x – 8y + z – 160)2 + 39690 = 90{(x – 13)2 + (y + 8)2 + (z –31)2} x2 + y2 + z2 – xy – yz – zx – 4x + 5y – z – 2 = 0 (lx + my +nz)2 – (l2 + m2 + n2) (x2 + y2 + z2 – a2) = 0 5x2 + 16y2 + 27z2 – 12xy – 12yz – 12zx – 66 = 0

Chapter

3

Differential Calculus 3.1 CURVATURE AND RADIUS OF CURVATURE Consider the two circles shown in the Fig. 3.1. It is obvious that the ways in which the two circles bend or ‘curve’ at the point P are not the same. The smaller circle ‘curves’ or changes its direction more rapidly than the bigger circle. In other words the smaller circle is said to have greater curvature than the other. This concept of curvature which

P

as follows:

Fig. 3.1 y Q P A x

O Fig. 3.2

Let P and Q be any two close points on a plane curve. Let the arcual distances of P and Q A on the given curve be s and s + s, so that PQ (the arcual length of PQ) s. [Refer to Fig. 3.2]

Part I: Mathematics I

I – 3.2

Let the tangents at P and Q to the curve make angles and line in the plane of the curve, say, the x Then the angle between the tangents at P and Q . s in the arcual length of the curve, the direction of the . Hence

is the average rate of bending of the curve (or average rate of change

of direction of the tangent to the curve in the arcual interval PQ or average curvature of the arc PQ. d is the rate of bending of the curve with respect to arcual distance lim = ds s 0 at P or the curvature of the curve at the point P. The curvature is denoted by k. a circle of radius at any point on it. [Refer to Fig. 3.3] Let the arcual distances of points on the C P circle be measured from A, the lowest point of the circle and let the tangent at A be chosen x as the x AP = s and let the tangent at A T P make an angle with the x Then s = a A C P Fig. 3.3 =a [ the angle between CA and CP equals the angle between the respective perpendiculars AT and PT.] 1 = as

or

d 1 ds = a Thus the curvature of a circle at any point on it equals the reciprocal of its radius. Equivalently, the radius of a circle equals the reciprocal of the curvature at any curvature. Radius of curvature 1 ds . = k d k or of a curve at any point on it, we should know the relation for that curve, which is not easily derivable in most cases.

curvature of the curve at that point and denoted by . Thus

Note between s and

=

equations. Hence formulas for in terms of cartesian, parametric or polar co-ordinates are necessary, which are derived below: Let P (x, y) and Q (x + x, y + y) be any two close points on a curve y =f(x). [Refer to Fig. 3.4.] Let AP s and AQ s s where A point on the curve. Let the chord PQ make an angle with the x

Chapter 3: Differential Calculus

I – 3.3

y Q

P

A

R x

O Fig. 3.4

PQR, sin

and

cos

RQ RQ s , where PQ PQ s PQ y s s PQ PR PQ

s

(1)

PR s s PQ

x s s PQ

(2)

When Q approaches P, chord PQ

tangent at P and hence

Thus in the limiting case when Q dx . cos = ds tan

Also

P, (1) and (2) become sin

=

1.

PQ

=

dy ds

and

dy .. dx

be the angle made by the tangent at any point (x, y) on the curve y = f (x). dy tan = (1) Then dx

Let

Differentiating both sides of (1) w.r.t. x, we get, sec 2 i.e. i.e.

sec 2 sec 2

d

=

d 2y

dx dx 2 d 2y d . ds = ds dx dx 2

. 1 . sec

=

d2y

dx 2 sec 3 = d2y dx 2

... cos

=

dx ds

Part I: Mathematics I

I – 3.4

=

) 3/2

(1 + tan 2 d2y dx 2

dy 1+ dx =

d2y dx 2

Note

2

3/2

, by (1).

As curvature (and hence radius of curvature) of a curve at any point is independent of the choice of x and y x and y can be interchanged in the formula for derived above. Thus is also given by dx dy

1

2 3/ 2

d2 x dy 2 This formula will be of use, when

dy dx

Let the parametric equations of the curve be x = f(t) and Then

x

dx dt

f (t ) dy dx d2 y dx

2

dy dt

y

and

y = g(t). g (t ).

y x d dy dx dx xy

yx x

2

2

dy dx

1 Now

d y dt x 1 x 3/ 2

d2 y dx 2 y x

1 xy

2

yx x3

3/ 2

dt dx xy

yx x

3

Chapter 3: Differential Calculus x2

y2 x

I – 3.5

x3

3

xy

yx

x2 y2 xy yx

Students are familiar with the following transformations from cartesian co-ordinates (x, y) to polar co-ordinates (r, ): and y = r sin (1) x = r cos We shall make use of (1) and the formula for the radius of curvature in cartesian coordinates, namely, 2 3/ 2 dy 1 dx (2) d2 y dx 2 and derive the corresponding formula for at the point (r, ) which lies on the curve r = f( ) dy dy/d r cos r sin Now dx dx/d r sin r cos dr where r (3) d r is a function of d2 y dx 2

d dx

dy dx

d d

r sin r cos r 2 sin r r sin cos

r 2 rr r sin Also

dy dx

dx d

r cos r sin

r sin 2r cos 2r sin r cos r sin r cos

r sin r cos

3rr sin cos rr sin 2 2r 2 cos 2 2 2 r cos 3rr sin cos rr cos 2 2r 2 sin 2 r r sin cos r sin r cos 3

2r 2 r cos 1

, where r 3 dy dx

2

1

d2r d 2 r cos r sin

r2 ( r sin

(4) r sin r cos r 2 r cos

2

(5) 2

Part I: Mathematics I

I – 3.6

Using (4) and (5) in (2), we get (r 2 r

3.2

2

2 3/ 2

r

)

rr

2r

2

CENTRE AND CIRCLE OF CURVATURE

Let P (x, y) be a point on the curve y = f (x). On the inward drawn normal to the curve at P, cut off a length PC = radius of curvature of the curve at P (namely ). The point C is called the centre of curvature at P for the curve. [Refer to Fig. 3.5] y

N C

Q

O

T

P (x, y)

C'

x

P' Fig. 3.5

The circle whose centre is C and radius for the curve. Let (–x, –y ) be the co-ordinates of C. Then

x

is called the circle of curvature at P

OC OP

QP (

= x - t sin }

angle between CP and CQ = angle between the respective perpendiculars PT and )

t = x - cosec } t 1 + cot 2 } tyl = x1 + yl 2 = x-

=x

y (1 + y 2 ) 3/2 . y" 1+y

i.e.

x

x

Now

y

CC PP

y (1 y QC

y 2)

cot

where y = 2

=

dx 1 = dy y

d 2y dy and y" = dx dx2

Chapter 3: Differential Calculus =y+ =y+

I – 3.7

cos sec

=y+

1 + tan 2 3

2 (1 + y' ) 2 . 1 =y+ y" 1 + y'

=y+

2

(1 + y' y"

2

)

Having found out the co-ordinates of the centre of curvature, the equation of the 2 circle of curvature is written as (x x ) 2 (y y ) 2 .

Example 3.1 Find the radius of curvature at the point 3

3

3a 3a , 2 2

on the curve

x + y = 3axy. Differentiating the equation of the curve with respect to x, 3 x2 (y 2

i.e.

y2

dy dx

ax)

dy dx dy dx

3a x

dy dx

ay

x2

ay

x2

y

(1)

2

y ax Again differentiating with respect to x, d2y

(y 2

ax) a

dy dx

dx 2

dy dx

3a 3a , 2 2

3a 2 2 9a 2 4

9a 2 4 3a 2 2

2x

(ay

(y 2

ax) 2

1

and

x2 ) 2 y

dy dx

a (2)

Part I: Mathematics I

I – 3.8

3 2 a 4

2

d y dx 2

3a

a

9a 4 16 dy as a function of x and y from (1) and dx 3a 3a dy and y , 1 , which may be used in (2). 2 2 dx

2 then evaluate d y . When x dx 2

d2 y dx 2

3a 2 4

3a

3a 3a , 2 2

Note

i.e.

a

6a 3 9a 4

3a 3a , 2 2

2

dy dx

1

32 3a

16 3/ 2

d2 y dx 2 3/ 2

1

1 32 3a

3a 3a , 2 2

3 2a . 16 Example 3.2 Find the radius of curvature at (a, 0) on the curve xy2 = a3 The equation of the curve is a3

y2

x3

3

.

(1)

x

Differentiating w.r.t. x, 3x 2

x

2 yy

y y

2 x3

1

dx dy d2 x dy 2

2

a3

x2 2 x3

a3

2x2 y

.

a, 0

The formula

x3

x2

i.e. Now

a3

3/ 2

should be used.

(2)

Chapter 3: Differential Calculus

I – 3.9

dx 2x2 y dy 2 x3 a3 Differentiating (3) w.r.t. y,

From (2),

2 x3

d2 x dy 2

a3

(3)

x2

2

2 x3

From (3), we get

dx dy

From(4), we get

d2 x dy 2

dx dy

y 2x

2

a3

2 3a 5 9a 6

a, 0

(4)

2 3a 3/ 2

0

3a 2

2 / 3a

y

dx dy

0

a, 0

1

Example 3.3 If

x2 y 6x2

is the radius of curvature at any point (x, y) on the curve 2/ 3

ax , show that 2 a a x

2

x y

2

y . x

ax

y

a

Differentiating w.r.t. x,

(1)

x a a

y

x

a

a2

x 2

x

a

x

2

(2)

Differentiating again w.r.t. x, 2a 2

y a

x

1

y

(3)

3 2 3/ 2

(only the numerical value of is considered)

y 3/ 2

a4

1 a

a

4

x

x

3

2a 2 a

2 a 2 a

2/ 3

x

4

a4

a3 a a

x

a2 a

x

4

a4 x

2

3/ 2

3

a

x a

2

2

a a

x

Part I: Mathematics I

I – 3.10 x y

2

y x

2

the point x , y lies on the curve

x Example 3.4 Show that the measure of curvature of the curve a ab at any point (x, y) on it is . 2 ax by The equation of the curve is

y x y b

a a

x

1

3 2

1

1

x

a

b

y

1

y'

0

(1)

Differentiating w.r.t. x, 1

1

2 a x

2 b y b y

y'

a x

Differentiating further w.r.t. x, x b

y"

1 2 y

y'

1 2 x

x

a b 1 a x b 2ax b

y

b

y

2 a

2 x

bx

3 2

, using 2

ay

ab , using 1

3 2

2ax b

2 ax 1 Now,

3 2

y'

by 1 ax b

3 2 2

y" 2 ax ab Curvature k

1

by

3 2

ab 2 ax by

3/ 2

3 2 3

2 a x2

Chapter 3: Differential Calculus

I – 3.11

Example 3.5 Find the co-ordinates of the real points on the curve y2 = 2x tangents at which are parallel to the x 1 of these point is . 3 x2) y2 = 2x Differentiating w.r.t. x, 2yy = 2 x2] 2 x) i.e. yy = The points at which the tangents are parallel to the x x2) = 0, from (2) i.e. x = ±1. 2 Putting x y = negative i.e. y is imaginary. The real points are given by x = 1. Putting x = 1 in (1), we get y2 = 4. y = ± 2. The points, the tangents at which are parallel to the x

(1)

(2)

y = 0.

3 1 x2

y

From(2),

x2), the

y 3 1 x2 2 3x

x3

Differentiating w.r.t. x, y

3x

3

x3

(1 x 2 )

2x 3x

2

y

1, 2

Example 3.6 Show that the curves y

3/ 2

x3

3

1 3

3

c cosh

3x

x3

3 ( 2 2) . 2 2 2 3/ 2 (1 y ) y 1 0

1, 2

d

x c

2

= 2c (y

curvature at the points where they cross the y x The point at which the curve y c cosh crosses the y c equation of the curve with x = 0.

c) have the same

Part I: Mathematics I

I – 3.12

Thus the point is (0, c). Similarly the point of intersection of the second curve with the y to be (0, c) y

c cosh

x . c

Differentiating w.r.t. x twice, we get x 1 . c c

y

c sinh

y

1 x cosh c c 2 3/2

y ) y

(1

sinh

x c

x c 1 x cosh c c

3/ 2

1 sinh 2

( )(0,c) = c. x2 2c

Equation of the second curve is y

c

Differentiating w.r.t. x twice, we get y

x and y c 1

0 ,c

x2 c2 1 c

1 c

3/ 2

c x 0

Thus ( )(0,c) is the same for both curves. 1 for both curves. (k)(0,c) is the same c Example 3.7 Find the radius of curvature at the point (a cos3 sin3 ) on the curve 2/3 2/3 2/3 x + y =a . The parametric equations of the given curve are x = a cos3 and y = a sin3 Differentiating twice w.r.t. x x y

dx dy sin ; y 3a cos 2 d d d2 x 2 cos sin 2 3a cos3 d 2 d2 y 3a 2 sin cos 2 sin 3 d 2

3a sin 2

cos

Chapter 3: Differential Calculus (x 2 y 2 )3/2 xy yx

cos 2 sin 2

27 a 3 sin 3 9a 2 sin 2

cos 2

9a 2 sin 4

cos 2 )3 / 2

si n

2 sin cos 2

sin 3

cos3

cos

cos3

cos 2

2cos 2

sin 2

3a sin cos cos 2

sin 2

(9a 2 cos 4 9a 2

I – 3.13

2 cos sin 2 3/ 2

sin 2 cos 2

2sin 2

3a sin cos

sin 2

3a sin cos . Example 3.8 Show that the radius of curvature at the point ‘ ’ on the curve x = 3a cos a cos 3 , y = 3a sin a sin 3 is 3a sin . a cos 3 y = 3a sin a sin 3 . x = 3a cos Differentiating w.r.t. , x dy dx

Now,

3a sin 3 y x

3a sin ; y

3a (cos 3a (sin 3

d d

tan 2 d dx

tan 2

2 sec 2 2

3a cos 3 ;

cos 3 ) sin )

2 sin 2 sin 2 cos 2 sin d2 y dx 2

3a cos

1 3a (sin 3

sin )

2

2 sec 2 6a cos 2 sin sec3 2 3a sin (1 tan 2 2 )3 / 2 3a sin sec3 2

y 2 )3/2 y 3a sin . (1

Example 3.9 Find the radius of curvature of the curve r = a (1 + cos ) at the point

. 2 r = a (1 + cos ) r = a sin and r (r r

2

2

2 3/2

r ) r 2r

2

a cos

Part I: Mathematics I

I – 3.14

2

a 2 1 cos 2

a 2 1 cos

a 2 cos

a 2 sin 2

3/ 2

1 cos

2a 2 sin 2

3/ 2

a 3 2 1 cos a 2 3 1 cos 2 2 a 1 cos 3 2

4a cos 3 4

1/ 2

4a cos 3 2

2 2 a. 3

Example 3.10 Show that the radius of curvature of the curve rn = an sin an the pole is . n 1 rn 1

at

rn = an sin n log r = n log a + log sin

(1)

Differentiating w.r.t. , n r n cot n r r = r cot r = r cot = r cot2 (r 2 r2

2) nr cosec2 nr cosec2

r 2 )3/2 rr 2r

2

r2 r2

r 2 cot 2 n

r 2 cosec 2 n

(3)

r 2 cot 2 n

3/ 2

nr 2 cosec 2 n

2r 2 cot 2 n

, using (2) and (3)

3/ 2

r 2 cosec 2 n n 1 r cosec n n 1 an n 1 rn 1

Example 3.11 Find the radius of curvature at the point (r, ) on the curve r2 cos 2 = a2 . r2 = a2 sec 2 (1) 2 log r = 2 log a + log sec 2 .

Chapter 3: Differential Calculus

I – 3.15

Differentiating w.r.t. , 2 r 2 tan 2 . r i.e. r = r tan 2 r = r tan 2 r sec2 2 2 = r tan 2 + 2r sec2 2 , using (2). (r 2 r2 r

2

r 2 )3/2 rr 2r

2

r 2 tan 2 2 )3 / 2 2r 2 sec 2 2 2r 2 tan 2 2

(r 2 r tan 2 2

)

2

, using (1) and (2),

r 3 sec3 2 r 2 sec 2 2

r3 a2

r sec 2

Example 3.12 Show that at the points of intersection of the curves r

and

a , their curvatures are in the ratio 3:1. r=

(1) a

r

and

(2)

a

Solving (1) and (2), a

i.e. = ± 1. The points of intersection of the two curves are given by = ± 1. For curve (1), r = a and r = 0. 1

1

For curve (2), r

a 2

a 2 )3/2 2a 2

2

2 2 a. 3

2a

r

2

2

(2a 2 )3/2 3a 2

1

and

(a 2 a2

3

a2 2

a2

a2

2

4

2a 2 4

3/ 2

2

a2 4

Part I: Mathematics I

I – 3.16

2

( 2a 2 )3 / 2 a2

=±1

1

:

2 2a

1:3

2

Ratio of their curvatures = 3:1 Example 3.13 If the centre of curvature 2

of the ellipse

x a2

y B (o, b)

2

y b2

1 at one end of O

x

prove that the eccentricity of the ellipse 1 is . B' (o, –b) 2 The centre of curvature of the ellipse at Fig. 3.6 b). [Refer to Fig. 3.6] B(0, b) lies at B We recall that if the centre of curvature of any curve at a point P is C, then PC equals the radius of curvature of the curve at P. Radius of curvature of the ellipse at b. (1) B = BB x2 a2

y2 b2

1

Differentiating w.r.t. x, x a2

y y' b2

0

b2 x a2 y

(2)

b 2 y xy a2 y2

(3)

y Differentiating again w.r.t. x, y From (2) and (3), we get y

o,b

0 1

Now,

y

and y

o,b

b a2

3 2 2

y 0 ,b

From (1)

a2 b

1 b a2 2b i.e. a 2

a2 b

2b 2

(4)

Chapter 3: Differential Calculus

I – 3.17

The eccentricity e of the ellipse is given by b2 = a2 (1 Using (4), we get, e 2

b2 2b 2

2

)

or

1 2

a2

e2

b2 a2

1

e

2

Example 3.14 Find the centre of curvature at

on the curve 2 x = 2 cos t + cos 2t, y = 2 sin t + sin 2t. x = 2 cos t + cos 2t; y = 2 sin t + sin 2t Differentiating w.r.t. t, x

2 sin t

2 sin 2t ; y

2 cos 2t

y

2 cos t

2 cos 2t

cos t

2 sin 2t

sin t

3t t cos 2 2 3t t 2 sin cos 2 2 3t cot 2 d d 3t dt y cot dx dt 2 dx 3 3t 1 cosec 2 2 2 2 sin 2t sin t 2 cos

y

3 t 2

8 sin 3 x

cos

1; y

t 2

2; y

2

cot

2

2

3 2

3

y 8 sin 3

2

x

Now

1 2

y

y

cos

y 1 y

x

x

4

1

y

4 2

1 3 2

1 1 y

1

y

2

4 3

1 3

3 4

1;

Part I: Mathematics I

I – 3.18

y

1

2

1 3 2

2

4 3

2

2 3

1 2 , . 3 3

Required centre of curvature is

Example 3.15 Find the equation of the circle of curvature of the parabola y2 = 12x at the point (3, 6). y2 = 12x Differentiating w.r.t. x, 2yy = 12

:. y

Differentiating again w.r.t. x, y

6 y y2

( ) (3, 6) = 1

and

y

1 6

3, 6

(1+ y 2 )3 / 2 y

x

x

x

3, 6

3

y

y 6

y 1 y

6 y

( )(3,6)

2 2 1 6

2

y

1 1 1 1 6 1 1 y2 y 1 1 1 1 6

15.

6

The equation of the circle of curvature is x

x

2

y

y

2

2

The equation of the circle of curvature at the point (3, 6) is (x 15) 2 + (y + 6) 2 = (12 2 ) 2 i.e. i.e.

x2

x + 225 + y2 + 12y + 36 = 288 x2 + y2 30x + 12y

12 2

Chapter 3: Differential Calculus

I – 3.19

Example 3.16 Find the equation of the circle of curvature of the curve a a , . x y a at 4 4 x Differentiating w.r.t. x,

1

y

1

2 x

y

y

2

a 0 y

y

x

Differentiating again w.r.t. x, 1

x

y

1

y

2 x

x

1

a a , 4 4

y

y

2

y

y

and 1 1 2 2 a 4

a a , 4 4

1

y

4 a

a a , 4 4

y

x

x

x

a a , 4 4

a 4

y

y

2 2 4 a

2 3/ 2

y 1 y

y

1 1 4 a 1 1 y

3a 4

2

y

; y

a 4

a a , 4 4

1 1 1 4 a

The equation of the circle of curvature is x

2

y

y

The equation of the circle of curvature at x

3a 4

2

2

1

x

a

2

y

2

2

a a , is 4 4

3a 4

2

a2 2

3a 4

Part I: Mathematics I

I – 3.20

Part A (Short Answer Questions)

2. Prove that the radius of curvature of a circle is its radius. 3. Find the curvature of the curve given by s = c tan at 0. Find the radius of curvature of each of the following curves at the points indicated: 4. y = ex at x = 0. 5. y e 3 x at x = 0. 6. y = log sec x at any point on it. 7. 8. 9. 10. 11. 12. 13. 14.

y = log sin x at x . 2 2 xy = c at (c, c). y2 = 4ax at y = 2a. x = t2, y = t at t = 1. at the pole. at any point on it. r = a cos at any point on it. r = e at any point on it.

Part B 15. x3 + xy2 16. 4ay 2

y2 = 0 at (3, 3). 2a

17. x3 + y3 = (y

3

a . 2 x) at (0, 0).

x at a , x) (y

18. xy2 = a2 (a 19. 20. 21. 22. 23.

x) at (a, 0) 7 a 4ay2 = 27 x 2a)3 at a, 3 2 y = x2 (x y = c cosh x at the point where it is minimum. c x2 = 4ay at the point where the slope of the tangent is tan x2 y2 1 at (a cos sin ) a2 b2

24. x = a (cos t + t sin t), y = a (sin t t cos t) at ‘t’. 25. x = a ( ), y = a (1 ) at ‘ ’. 26. x = et cos t, y = et sin t at (1, 0).

x

Chapter 3: Differential Calculus 27. x = a log

4

2

I – 3.21

, y = a sec at ‘ ’.

28. x = a log cot

cos , y = a sin at ‘ ’. 2 29. Find the radius of curvature at any point on the equiangular spiral r = . ae ) at any point on it. 30. Find the radius of curvature of the curve r = a 31. Find the radius of curvature of the curve rn = an cos at any point ( ). Hence prove that the radius of curvature of the lemniscate r 2 = a2 cos 2 is a2 . 3r 32. Find the radius of curvature at any point ( a.

) on the curve

r cos

2

33. Find the radius of curvature at any point ( ) on the curve r (1 + cos ) = a. 34. If 1 and 2 be the radii of curvature at the ends of any chord of the cardioid 16a 2 2 2 . r = a (1 + cos ), that passes through the pole, prove that 1 2 9 [Hint: The ends of any chord that passes through the pole are given by = 4a and = . Use the result cos .] 3 2 35. Find the centre of curvature of the curve y = x3 x2 + 3x + 1 at the point 36. Find the centre of curvature of the hyperbola

x2 a2

y2 b2

1 at the point

(a sec , b tan ). 37. Show that the line joining any point ‘t’ on the cycloid x = a(t + sin t). y = a(1 cos t) and its centre of curvature is bisected by the line y = 2a. 38. Find the equation of the circle of curvature of the parabola y2 = 4ax at the positive end of the latus rectum. 39. Find the equation of the circle of curvature of the rectangular hyperbola xy = 12 at the point (3, 4). 40. Find the equation of the circle of curvature of the curve x3 + y3 = 3axy at the point

3a 3a , . 2 2

Let Q be the centre of curvature of a given curve C at the point P on it. When P moves on the curve C and takes different positions, Q will also take different positions and move on another curve . This curve is called the evolute of the curve C. Thus When

is the evolute of the curve C, C is called the involute of the curve

.

Part I: Mathematics I

I – 3.22

below: Let the equation of the given curve be y = f (x) (1) If x , y is the centre of curvature corresponding to the point (x, y) on (1), then y x x 1 y2 (2) y y

y

1 1 y

y

2

(3)

By eliminating x and y from (1), (2), (3), we get a relation between x and y , which is the equation of the evolute.

Note

If the parametric co-ordinates of any point on the given curve are assumed, then we have to eliminate the parameter from Equations (2) and (3), which will simplify the procedure.

concept of envelope of a family of curves, which is discussed below: Consider the equation f (x, y, c) = 0, where c is a constant. If c takes a particular value, the equation represents a single curve. If c is an arbitrary constant or parameter which takes different values, then the equation f (x, y, c) = 0 represents a family of similar curves. If we assign two consecutive values for c, we get two close curves of the family. The locus of the limiting positions of the points of intersection of consecutive members of a family of curves is called the envelope of the family. It can be proved that the envelope of a family of curves touches every member of the family of curves.

Let f (x, y, c) = 0 be the equation of the given family of curves, where c is the parameter. Two consecutive members of the family (corresponding to two close values of c) are given by f (x, y, c) = 0 (1) and f (x, y, c + c) = 0 (2) The co-ordinates of the points of intersection of (1) and (2) will satisfy (1) and (2) and hence satisfy

f x, y , c

c c

f x, y , c

0

Hence the co-ordinates of the limiting positions of the points of intersection of (1) and (2) will satisfy the equation f x, y , c c f x, y , c lim 0 c 0 c f (3) i.e. x, y , c 0 c

Chapter 3: Differential Calculus

I – 3.23

These limiting points will continue to lie on (1) and satisfy f(x, y, c) = 0 If we eliminate c between (1) and (3), we get the equation of a curve, which is the locus of the limiting positions of the points of intersection of consecutive members of the given family, i.e. we get the equation of the envelope. Thus the equation of the envelope of the family of curves f(x, y, c) = 0 (c is the parameter) is obtained by eliminating c between the equations f f(x, y, c) = 0 and x, y , c 0. c 2

Very often the equation of the family of curves will be a quadratic equation in the parameter. In such cases, the equation of the envelope may be remembered as a formula. Let the equation of the family of curves be 2 0 (1) Differentiating partially w.r.t. 2 (2) B From (2), 2A Substituting this values of in (1), we get the eliminant of as A i.e. i.e. B2

B 2A

2

B

B 2A

C

0

B2 B2 C 0 4A 2A AC = 0, which is the equation of the envelope of the family (1).

The normals to a curve form a family of straight lines. The envelope of this family of normals is the locus of the limiting position of the point of intersection of consecutive normals. But the point of intersection of consecutive normals of a curve is the centre of curvature of the curve. Hence the locus of centre of curvature is the same as the envelope of normals. Thus the evolute of a curve is the envelope of the normals of that curve.

Example 3.1 Find the evolute of the parabola x2 = 4ay. The parametric co-ordinates of any point on the parabola x2 = 4ay are x = 2 at and y = a t2 dy y t x 2a ; y 2 at dx x

Part I: Mathematics I

I – 3.24

d2 y dx 2

dt dx

d t dt

1 2a

Let x , y be the centre of curvature at the point ‘t’ x

y 1 y y t 2at 1 1 2a

x

2

t2

2 at 3 y

(1)

1 1 y

y

y

1 1 1 2a

at 2

2

t2

3at 2

2a

(2)

To get the relation between x and y , we have to eliminate t from (1) and (2). From (1), we get t 3 From (2), we get t 2

y

x 2a

(3)

2a 3a

(4)

From (3) and (4), we get 2

x 2a

27 a x 2

i.e.

y

2a 3a

4 y

3

3

2a .

Locus of x , y , i.e. the equation of the evolute is 27a x2 = 4(y 2 y2 Example 3.2 Find the evolute of the hyperbola x 1. 2 a b2 The parametric co-ordinates of any point on the hyperbola are x = a sec and y = b tan x y y

a sec tan ; y b sec x a tan b a sin 2 b cos3 a 2 sin 3

cos

y

b sec 2

b a sin d dx

b cos a sin 2

cos 2 a sin

a)3.

Chapter 3: Differential Calculus x

y 1 y

x

y

2

y

a 2 sin 3 b cos3

b a sin

a sec a cos

1 a cos

a 2 cos 2

a 2 sin 2 a cos3

a 2 b2 a cos3 1 1 y y

3

a 2 sin 2 b2

(1)

y

2

1

b sin cos

sin b cos3

a 2 sin 2

sin b cos3

b 2 cos 2 b2

sec3

From (2),

tan 3

ax a

2

b2

a2

b2

b2

a 2 sin 2

b2 (2)

and

by a

2

b2

To eliminate , we use the identity sec2 2/ 3

b2 a sin 2 2

tan 3

b

ax

b2

a 2 sin 3 b cos3

From (1),

b2 a 2 sin 2

1

b tan

a2

I – 3.25

by 2 a b2

2

=1

2/ 3

1

The locus of x , y i.e. the evolute of the hyperbola is (ax)2/3

(by)2/3 = (a2 +b2)2/3

[

by)2/3 = (by)2/3]

Example 3.3 Find the evolute of the curve x2/3 + y2/3 = a2/3. The parametric co-ordinates of any point on the curve are x = a cos3 sin3

and y = a

Part I: Mathematics I

I – 3.26

3a cos 2

x y x

y

cos

tan d dx

sec 2

y

3a sin 2

y

sin ;

1 3a cos 2 sin

sec 2

1 3a cos x

4

sin

y 1 y

x

a cos

y

1 3a cos 4 sin

a sin y

sec 2

3

a cos

sin

y

a cos

x

y

2/3

x

a

y

3

(1)

sin

2

(2)

3 cos

2

sin

3 cos sin

2

3

sin 3

3 cos

a cos Now

3a cos 4 sin

3a cos 4

3a sin cos 3

a cos x

1 cos 2

sin cos

a sin 3

and

tan 2

3a cos sin 2 a cos3 1 1 y 2 y y

y

x

1

tan

3

a cos3

Now

2

2

(3) sin

3 cos sin

2

sin

3

3

sin

(4)

2/3

cos

a

2

sin

cos

sin

2

2 i.e.

x

y

2/3

x

y

2/3

2a 2 / 3

The equation of the evolute is x

y

2/ 3

x

y

2/ 3

2a 2 / 3 .

Example 3.4 Show that the evolute of the cycloid x = a( sin ), y = a(1 cos ) is another cycloid. Any point on the cycloid is given by x=a( sin and y = a(1 cos ) x

a 1 cos

;

y

a sin

Chapter 3: Differential Calculus 2 sin

a sin a 1 cos

y

2 sin

1 d cosec 2 2 2 dx

y

cos

2

2

cot

2

2 1

2 sin

I – 3.27

2 1

2

2

2 a sin

2

2

1 4 a sin 4 x

y

2

x

y 1 y

a

sin

cot

a

sin

4a sin

a a

sin sin

2a sin

y

1 1 y

2

y

a 1 cos a 1 cos a 1 cos

2

cosec2 2

cos

2

4a sin 4

2

2

(1) y

2

cosec 2

4a sin 4 2 2 2a 1 cos

(2) Elimination of from (1) and (2) is not easy. The locus of x , y is given by the parametric equations x = a ( + sin ) and y a ), which represent another cycloid. Example 3.5 Find the equation of the evolute of the curve x = a (cos t + t sin t), y = a (sin t t cos t) x = a(cos t + t sin t); y = a (sin t t cos t) x

a

y

at sin t at cos t

y

sec 2 t

x

x

sin t

dt dx

y 1 y

sin t

t cos t ; y

a cos t

t sin t

tan t sec 2 t y

1 at cos t

1 at cos3 t

2

a cos t

t sin t

tan t sec 2 t at cos3 t

a cos t

t siin t

a t sin t

a cos t

cos t

(1)

Part I: Mathematics I

I – 3.28 y

1 (1 y

y

a sin t

y 2) sec 2 t at cos3 t

t cos t

a sin t

(2)

Eliminating t between (1) and (2), we get x2

y2

a2

:. The evolute of the given curve is x2 + y2 = a2. Example 3.6 Prove that the evolute of the curve x x y= a sin is the catenary y a cosh . a x

a cos

x

a

log tan

1

sin

tan

a

2

a cos + log tan

2

,

; y = a sin .

sec 2

1 2 2

2 1

sin

2 sin cos 2 2 a and

y

a cos

y'

a cos

y"

sec 2

x

x

a cos 2 sin

1 sin

sin

sin a cos 2 d dx

sec 2

y' 1 y"

sin a cos 2

sin a cos 4

y' 2

a cos

log tan

a cos

log tan

a log tan

tan

2

2 2

tan

sec 2

a cos 4 sin

a cos

(1)

Chapter 3: Differential Calculus y

1 1 y"

y

y' 2

a sin

sec 2

a sin

a cos 2 sin

a cos 4 sin

a siin

(2)

a 1 tan 2

2

y

Now

2 tan

(3)

2

ex / a 2 From (3) and (4), we get, and

tan

y=

(4) [from (1)]

a 1 e2 x / a 2 ex / a a x/a e 2

The evolute is y

I – 3.29

a cosh

e

x/a

a cosh

x a

x a

Example 3.7 Find the envelope of the family of straight lines given by (i) y mx y sin sec , a 2 m 2 b 2 , where m is the parameter, (ii) x cos where is the parameter, (iii) the family of parabolas given by y = x tan gx 2 , where is the parameter. 2u 2 cos 2 (i) Rewriting the given equation, we have a2 m 2

=( )2 2 =y xym + m2x2 2 i.e. (x2 ) m2 2xym + (y2 + b2) = 0 This is a quadratic equation in ‘m’. The envelope is given by the equation ‘B2 4AC = 0’ 4x2y2

i.e. i.e.

x2y2

(x2y2 + b2x2

i.e. 2

i.e.

x a2

4(x2

2

a2) (y2 + b2) = 0 a2y2

a2b2) = 0

b2x2

a2y2 = a2b2

2

y b2

1, which is the standard hyperbola.

Part I: Mathematics I

I – 3.30

Note

The envelope touches every member of the given family of straight lines and vice versa. This is, in fact, obvious as the given family represents

2 the family of tangents to the hyperbola x a2

y2 b2

1.

y sin = a sec (ii) x cos Dividing throughout by cos we have sec2 x + y tan i.e. x + yt = a (1 + t2), where t = tan can be treated as the new parameter. i.e. at2 yt + (a x) = 0 This is a quadratic equation in ‘t’. :. The envelope is given by y2 a (a x) = 0 i.e. y2 a (x a) gx 2 2u 2 cos 2 Putting t = tan we get

(iii) y

x tan

gx 2 1 t2 2 2u i.e.

gx2t2

xt

y

0

2u2 xt + (gx2 + 2u2y) = 0

If we treat ‘t’ as the parameter, we see that this equation is a quadratic equation in the parameter The envelope is given by 4u4 x2 4gx2 (gx2 + 2u2 y) = 0 i.e. i.e.

g2x2 + 2u2 x2

u4 = 0

2u 2 y g

u2 . 2g

Example 3.8 Find the envelope of the family of straight lines (i) y cos cos 2 being the parameter, (ii) x cos y sin = c sin cos , the parameter, (iii) x sec2 + y cosec2 being the parameter. sin = a cos 2 (i) y cos Differentiating (1) partially w.r.t. sin (1)

(1)

cos

sin being (1)

2a sin 2

cos (2) sin gives y = a(cos 2 cos + 2 sin 2 sin ) = a(cos + sin 2 sin ) sin + (2) cos gives (cos 2 sin 2 sin 2 cos ) x= ( sin sin 2 cos ) = a(sin + sin 2 cos )

(3)

(4)

Chapter 3: Differential Calculus

I – 3.31

Adding (3) and (4), we get x + y = a{(sin + cos ) + sin 2 . (sin + cos )} = a(sin + cos ) (1 + sin 2 ) = a(sin + cos ) (sin2 + cos2 + 2 sin cos ) = a(sin + cos ) (sin + cos )2 = a(sin + cos )3 Subtracting (3) from (4), we get {(sin cos ) cos )3 = a(sin From (5) and (6), we get x

y

2/ 3

x

a

y

sin 2 (sin

(5)

cos )} (6)

2/ 3

= (sin + cos )2 + (sin

a

cos )2

=2 The envelope is (x + y)2/3 + (

Note

)2/3 =2a2/3

This is the evolute of the astroid x2/3 + y2/3 = a2/3

In this problem, we have found out the envelope of the normals of the astroid. (ii) x cos y sin Dividing by sin

sin cos

cos we get x sin

y cos

c

cos

y cos 2

sin

(1)

Differentiating (1) w.r.t. x sin 2 From (2), x cos sin 2

(2)

y sin cos 2 x sin 3

i.e.

y cos3 x k

sin 3

x k

2/ 3

k say.

and cos3

sin2

i.e.

0

y k

(3)

2

y k

2/ 3

k2/3 = x2/3 + y2/3

1 (4)

Part I: Mathematics I

I – 3.32 From (3), we have

x1/ 3

sin

y1/ 3

and cos

k 1/ 3 k 1/ 3 Using (5) in (1), the equation of the envelope is k 1/3 ( x 2/3 + y 2/3 ) = c x2 / 3

i.e.

y2/ 3

1/ 2

x2 / 3

x2 / 3

i.e.

y2/ 3 y2/ 3

x2 / 3

i.e.

c,

3/ 2

y2/ 3

from (4)

c c2 / 3

(iii) x sec2 + y cosec2 = c Differentiating (1) partially w.r.t. , y cosec2 cot = 0 2x sec2 tan x sin y cos i.e. 0 3 cos sin 3 From (2),

x

y

cos 4

x k

cos 4

x k

y k

and sin 4

Using the identity cos2 + sin2

1, we have y k

x

x

1

y

2

y x

(4) x

y

y y

c

c

c , which is the equation of the required envelope. x y Example 3.9 Find the envelope of the straight line 1 , where a and b are a b parameters that are connected by the relation a + b = c. x y (1) 1 a b a+b=c (2) From (2), b = c a. x y Using in (1), (3) 1 , where a is the only parameter. a c a i.e.

x

y

(2)

(3)

i.e. k ( x y )2 Using (3) and (4) in (1), we get

i.e.

(1)

k . say.

sin 4

x

(5)

Chapter 3: Differential Calculus

I – 3.33

Differentiating (3) w.r.t. a, we get x a From (3)

x a

c a

0

2

(4)

y

2

x a

y

2

2

c a y

x

y

c a

c

x

1 a

y

and

c x

x

1 c a

Using (5) in (3), the equation of the envelope is x

i.e.

y x

or

2

y

(5)

c y x c

x

y

y c

x

y

1

c

y

c.

3.4 ALITER Without eliminating one of the parameters, we may treat both a and b as functions of a third parameter t and proceed as follows: Differentiating (1) w.r.t. t, x da y db 0 a 2 dt b 2 dt x da a 2 dt

i.e.

y db b 2 dt

(3)

db dt

(4)

Differentiating a + b = c w.r.t. t da dt Dividing (3) by (4), we have x y a2

b2

x a

y

x

b

y

(5)

c

a b

Using (5) in (1), we get x c

x

y

y c

x

y

1

c

Part I: Mathematics I

I – 3.34 i.e. or

x

y x

2

c y

c.

Example 3.10 Find the envelope of the l m m are connected by the relation a b x l l and a Differentiating (1) and (2) w.r.t. t, x dl l 2 dt 1 dl and a dt

x y 1, , where l and l m 1 (l and m are the parameters). system of lines

y 1 m m 1 b

(1) (2)

y dm m 2 dt 1 dm b dt

0

(3)

0

(4)

by

(5)

From (3) and (4), we have

i.e.

or

x

y

l2 a

m2 b

x l l a

y m m b

x l l a

ax

by

2

m2

l

l

y m m b

1 1

1

ax

and m x a

Using (5) in (1), we get the envelope as

y b

Example 3.11 Find the envelope of the ellipse

1.

x2

y2

a 2 b2 2 2 2 are connected by the relation a + b = c , c being a constant.

and

x2

y2

a2

b2

1

a2 + b2 = c2

1, where a and b

(1) (2)

Chapter 3: Differential Calculus

I – 3.35

Eliminating b from (1) and (2), we get x2

y2

a2

c2

a2

1

2 i.e. (c2 a2)x2 + a2y2 = a2(c2 ) i.e. a4 a2(c2 + x2 y2) + c2x2 = 0 (3) is a quadratic equation in a2, which may be regarded as the parameter. The envelope is given by B2 4AC = 0 i.e. (c2 + x2 y2)2 4c2x2 = 0 i.e. [(c2 + x2 y2) + 2cx] [c2 + x2 y2 2cx] = 0 i.e. (x + c)2 y2 = 0; (x c)2 y2 = 0 x + c = ± y and x c = ± y i.e. x = c ± y and x = c ± y i.e. x ± y = ± c.

(3)

Example 3.12 The equation of the system of ellipses is x2

y2

1 a b2 a and b is =c Differentiating (1) and (2) w.r.t. ‘t’,

(1)

2

2 x 2 da a 3 dt b

(2)

2 y 2 db b3 dt

0

(3)

db dt

0

(4)

da dt

a

From (3) and (4), we have x2

y2

a3 b

ab3

x a

or From (5), a

x k xy

and

y b b

or

x2

y2

a2

b2

k , say y k

xy c k Using (5) and (6) in (1), the equation of the envelope is Using in (2),

2

(5)

c or k

(6) xy c

xy c

1

i.e. 2 Example 3.13 Find the evolute of the parabola y2 = 4ax, considering it as the envelope of its normals.

Part I: Mathematics I

I – 3.36

The normal at any point (at2, 2at) on the parabola y2 = 4ax is y + xt = 2at + at3

(1)

(1) represents the family of normals, where t is the parameter. Differentiating (1) w.r.t. ‘t’, x = 2a + 3at2 From (2),

x

t

(2)

1 2

2a 3a

(3)

Substituting (3) in (1), we get y

x

2a

x

2a

3a

1 3

1 2

3a

1 2

a

x

2a 3a

x

2a 3a

3 2

3 2

1 2

3 2

2a

4 x 27 a

y2

2a 3a

3 2

x

2 3 i.e.

x

1 2

2a

3

The evolute of the parabola is 27ay2 = 4 (x a)3 x2 Example 3.14 Find the envelope of the ellipse 2 a envelope of its normals.

y2 b2

The normal at any point (a cos , b sin ) on the ellipse ax cos

by sin

a2

1 , treating it as the

x2 a2

b2

y2 b2

1 is (1)

where is the parameter. Differentiating (1) w.r.t. ‘ ’, ax cos 2 From (2),

ax cos 3

by sin 3

cos

sin

by cos sin 2

0

(2)

k , say

ax k

1 3

and sin

by k

1 3

(3)

Chapter 3: Differential Calculus Using the identity cos2

sin2 = 1, we have ax k

2 3

by k

2

k3 k

i.e.

I – 3.37

2 3

ax

1 3

2 3

1 by

2 3

ax

2 3 1 2

2 3

by

(4)

Using (3) and (4) in (1), we have 2 3

ax

i.e.

ax

by 2 3

2 3

k3 3 2

2 3

by

ax

i.e.

1

2 3

by

a2

b2

a2

b2

2 3

a2

b2

Example 3.15

2 3

x

a cos

log tan

y = a sin , treating it as the envelope of its normals. x

a cos

log tan

2

, y

2

,

a sin .

Differentiating w.r.t. ,

x

a

1

sin tan

a y

and

dy dx

2

1 sin

sin

cos 2

1 2 2

a cos sin

2

a cos y x

tan .

Slope of the normal at ‘ Now the equation of the normal at ‘ ’ is y

a sin

cot

x

a cos

log tan

2 is the parameter.

(1)

Part I: Mathematics I

I – 3.38

Differentiating (1) w.r.t. ‘ ’, x cosec 2

a cos

x sin 2 x si n 2 x

a cot a cos sin 2

3

a cos

a log tan

cos 2 sin

a cosec 2

cos

a cos sin 2

a cosec 2

log tan

a sin 2

log tan

log tan

2

2

2 (2)

2

Rewriting (1), we have y

a sin

x cot

a

cos 2 sin

a cot

log tan

(3)

2

Using (2) in (3), we get y i.e.

y

a cos 2 sin

a sin

2

a cot log tan

2

a sin tan 2

a 1

2 tan

2

2

a tan 2 2 a e 2 i.e. y

a cot log tan

a cosh

x a

1 tan e

x a

2

, again using (2)

x a

Part A (Short Answer Questions)

y = f (x).

Chapter 3: Differential Calculus

f(x, y, ) = 0, being the parameter. 5. Obtain the equation of the envelope of the family f1(x, y) f3(x, y) = 0, where is the parameter.

I – 3.39

2

+ f2(x, y)

7. If the centre of curvature of a curve at a variable point ‘t’ on it is (2a + 3at2, at3 8. If the centre of curvature of a curve at a variable point ‘t’ on it is c c 3 cos3 t , sin t a b 9. If the centre of curvature of curve at a variable point ‘ ’ on it is a a log cot , 2 sin 10. Find the envelope of the family of lines y mx a 1 m 2 , m being the parameter. a , m being the parameter. 11. Find the envelope of the family of lines y mx m 12. Find the envelope of the family of lines y = mx + am2, m being the parameter. 13. Find the envelope of the family of lines y mx a 2 m 2 b 2 , m being the parameter. x 14. Find the envelope of the family of lines yt 2c , t being the parameter. t 15. Find the envelope of the lines x cos + y sin being the parameter. x y 16. Find the envelope of the lines cos sin 1, being the parameter. a b x y 17. Find the envelope of the lines sec tan 1, being the parameter. a b 18. Find the envelope of the lines x sec y tan = a, being the parameter. y cot = a, being the parameter. 19. Find the envelope of the lines x cosec 20. Show that the family of circles ( )2 + y2 = a2 (a is the parameter) has no envelope.

Part B 21. Find the evolute of the parabola y2 = 4ax. x2 y 2 22. Find the evolute of the ellipse 2 1. a b2 23. Find the evolute of the rectangular hyperbola xy = c2. 24. Show that the evolute of the cycloid x = a( sin ), y = a(1 cos ) is another cycloid, given by x = a( sin ), y a = a(1 + cos ). x y 25. Find the envelope of the family of lines 1 , where the parameters a a b and b are connected by the relation a2 + b2 = c2.

Part I: Mathematics I

I – 3.40

26. Find the envelope of the family of lines x y 1 , where the parameters a a b and b are connected by the relation ab = c2. x2 y 2 27. Find the envelope of the family of ellipses 2 1 , where the parameters a b2 a and b are connected by the relation a + b = c. x2 y 2 28. From a point on the ellipse 2 1 a b2 and the feet of these perpendiculars are joined. Find the envelope of the line thus formed. 29. Find the evolute of the parabola x2 = 4ay, treating it as the envelope of its normals. x2 y 2 30. Find the evolute of the hyperbola 2 1, treating it as the envelope of a b2 its normals.

ANSWERS

(5)

1 . c

8 . 3

(4) 2 2.

(5)

(6) sec x.

(7) 1.

(8) c 2 .

(9) 4a 2 .

(10)

5 5 . 2

(11)

(13)

a . 2

(14)

2 r.

(16)

125a . 24

(17)

5 5 . 18

(19)

97 97 a . 216

(20)

1 . 6

(12)

3 2 2

a1

.

4

(15) 5 5.

(18)

a . 2

(22) 2a sec3 .

(21) c. (23)

a 2 sin 2

(25) 4a sin

a . 2

2

b 2 cos 2 .

3 2

(24) at.

/ ab.

(26)

2.

(27) a sec 2 .

Chapter 3: Differential Calculus (29) r cosec .

(28) a cot .

I – 3.41

(30)

2 2ar . 3

3

(31)

an n 1 rn

(32)

2 (37) x

y2 43 6

(40)

x

21a 16

(5) f 22 ax

2 3

57 8

y 2

by

x2 a2

y2 b2

c3 .

1.

ax

(23)

x

2 3

by y

2 3

x

a2 2 3

y

(26) 4xy = c2. (28)

x a

(30)

ax

2 3

2 3

y b by

2 3

1. 2 3

a2

3

a2

,

b2

b

tan 3

.

2

2

9a 2 128

(9) y

.

4 x

3

2a .

x . a

a cosh

(11) y2 = 4ax.

(12) x2 + 4ay = 0.

(14) xy = c2.

(15) x2 + y2 = p2.

x2 a2

(21) 4 (x 2 3

.

125 . 24

(17)

(19) x2 y2 = a2. (22)

sec

2

2 3

3 2

0

(7) 27 ay 2

0.

(10) x2 + y2 = a2. x2 y 2 (13) 2 1. a b2 (16)

2

21a 16

y

4 f1 f 3

b2 a

3a 2

4ay

2

x

(33)

2r a

(36)

10ax

(39)

.

a a2

43 . 36, 6

(35)

(8)

.

1

2r 2

b2

2 3

a)3 = 27ay2.

. 2

2

2

(27) x 3

y3

c3

(29) 27ax2 = 4(y

b2

2 3

(18) x2 y2 = a2.

1.

. 2 3

4c

y2 b2

.

a)3.

2

2

(25) x 3

y3

2

c3 .

Chapter

4

Functions of Several Variables 4.1

INTRODUCTION

The students have studied in the lower classes the concept of partial differentiation of a function of more than one variable. They were also exposed to Homogeneous functions of several variables and Euler’s theorem associated with such functions. In this chapter, we discuss some of the applications of the concept of partial differentiation, which are frequently required in engineering problems.

4.2

TOTAL DIFFERENTIATION

In partial differentiation of a function of two or more variables, it is assumed that only one of the independent variables varies at a time. In total differentiation, all the independent variables concerned are assumed to vary and so to take increments simultaneously. Let z f (x, y), where x and y are continuous functions of another variable t. Let t be a small increment in t. Let the corresponding increments in x, y, z be x, y and z respectively. Then

z

z t

f (x { f (x

x, y x, y

f (x

x, y

f ( x, y

We note that x and y

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

0 as t

f ( x, y f ( x, y )

y )} { f ( x, y y)

y)

f ( x, y )}

x t

y t

0 and hence z

0 as t dz Taking limits on both sides of (1) as t 0, we have dt ( x, y and z are functions of t only and f is a function of x and y).

0 f dx x dt

f dy y dt

Part I: Mathematics I

I – 4.2 dz dt

i.e.,

dz dx dy and also and dt dt dt

z dx x dt

z dy [since f ( x, y ) y dt

(2)

z ( x, y )].

of z.

is called the

z. z Thus to differentiate z, which is directly a function of x and y, (where and y x x and y are functions of t) with respect to t, we need not express z as a function of t by substituting for x and y. We can differentiate z with respect to t via x and y using the result (2). Corollary 1: In the differential form, result (2) can be written as dz

z dx x

z dy y

(3)

dz is called the total differential of z. Corollary 2: If z is directly a function of two variables u and , which are in turn functions of two other variables x and y, clearly z is a function of x and y ultimately. z and z by Hence the total differentiation of z y x using the following results which can be derived as result (2) given above. z x

z u

u x

z

z y

z u

u y

z

(4) x (5) y

We note that the partial differentiation of z is performed via the intermediate z z are called variables u and , which are functions of x and y. Hence and x y partial derivatives of a function of two functions.

Note

Results (2), (3), (4) and (5) can be extended to a function z of several intermediate variables.

4.2.1

Small Errors and Approximations

Since lim x

0

y x

dy y , dx x

dy approximately or y dx

dy dx

x

(1)

If we assume that dx and dy are approximately equal to x and y respectively, result (1) can be derived from the differential relation.

Chapter 4: Functions of Several Variables dy dx dx

dy

I – 4.3 (2)

Though (2) is an exact relation, it can be made use of to get the approximate relation (1), by replacing dx and dy by x and y respectively. Let y f (x). If we assume that the value of x is obtained by measurement, it is likely that there is a small error x in the measured value of x. This error in the value of x will contribute a small error y in the calculated value of y, as x and y are functionally related. The small increments x and y can be assumed to represent the small errors x and y. Thus the relation between the errors x and y can be taken as y f ( x) x This concept can be extended to a function of several variables. If u u(x, y, z) or f (x, y, z) and if the value of u is calculated on the measured values of x, y, z, the likely errors x, y, z will result in an error u in the calculated value of u, given by u

u x x

u y y

u z, z

which can be assumed as the approximate version of the total differential relation du

u dx x

u dy y

u dz z

The error x in x is called the absolute error in x, while x is called the x relative or proportional error in x and 100 x is called the percentage error in x. x

Note

4.2.2

Differentiation of Implicit Functions

When x and y are connected by means of a relation of the form f (x, y) 0, x and y are said to be implicitly related or y is said to be an implicit function of x. When x and y are implicitly related, it may not be possible in many cases to express y as a dy can be found out in such cases as single valued function of x explicitly. However dx a mixed function of x and y using partial derivatives as explained below: f ( x, y )

Since i.e.,

f dx x

f dy y

0, df

0 x, we have

0,

f x

f dy y dx

0

Part I: Mathematics I

I – 4.4

f x f y

dy dx 2 2 If we denote f , f , f , f and x y x2 x y then dy dx

(1)

2

f by the letters p, q, r, s, t respectively, y2 p q

(2)

d2 y in terms of p, q, r, s, t as given dx 2 below. Noting that p and q are functions of x and y and differentiating both sides of (2) with respect to x totally, we have We can express the second order derivative

q

d2 y dx 2

dp dx

p

dq dx

q2 q x

p

q dy y dx

p dy y dx

p x

q q2

p q

p s t

q r s

p q

,

q2

p x

since

2

f 2

r;

p y

x p(qs pt ) q (qr q3

2

q x

f x y

s;

q y

2

f y2

t.

ps )

( p 2 t 2 pqs q 2 r ) q3

WORKED EXAMPLE 4(a)

Example 4.1 (i) If u

xy + yz + zx, where x

et, y

(ii) If u

sin 1 (x

3t and y

y), where x

1 , t

du dt d u 4t3, show that dt

e t and z

3 1

t2

Chapter 4: Functions of Several Variables (i) u

xy + yz + zx du dt

u dx x dt

u dy y dt

( y z )e t

e

1 t e t

t

2 sinh t t sin 1 (x

1 t2

1 t e t

t

1

e

1 t e t2

t

1 e t2

et

e

t

.

1 t2

t

2 cosh t. t2

y) du dt

dx y ) dt

1 1

(x

1

y)2

y)2

(x

dy dt

1

2

1

Now 1 ( x

u dz z dt

( z x) ( e t ) ( x y )

1 t 1 e e t t

1

(ii) u

I – 4.5

(x

1

y)

2

(1)

(3 12t 2 )

1 (3t 4t 3 ) 2 1 9t 2 24t 4 16t 6 (1 t 2 ) (1 8t 2 16t 4 ) (1 t 2 ) (1 4t 2 ) 2

(2)

Using (2) in (1), we get du dt

1 2

(1 4t ) 1 t 3

2

3(1

4t 2 )

1 t2

Example 4.2 If u

Let

x y z u , , , prove that x y z x x

f

r

x ,s y

y and t z

z x

y

u y

z

u z

0.

(1)

Part I: Mathematics I

I – 4.6 u

f (r, s, t), where r, s, t are functions of x, y, z as assumed in (1) u x

u r r x 1 u y r

u y

u r

r y

u s s y u 1 u r z s

u t

u r u s r z s z y u 1 u z2 s x t

u t

x y2 u z

u s u t s x t x u z u 0 2 s t x

(2)

t y (3) t z (4)

From (2), (3) and (4), we have x

u x

y

u y

z

u z

x u y r

z u x t x u y r

y u z s

y u z s

z u x t

0. eu + e and y

Example 4.3 If z be a function of x and y, where x that z u

z

z x

x

z u

z x eu

z

y x u

z e x

z x

z y z y

x

z y z e x

e

z

y z y

z (e x

(e u e ) x

z x

y

z y

e , prove

(1)

(2)

From (1) and (2), we have z u

u

y u z y

u

e

u

e )

z y

Chapter 4: Functions of Several Variables Example 4.4 If u

f (x, y), where x u x

r cos 2

and y

2

u r

1 r2

u

x r

u y

u x

u x

x

1 u r

. y r u y

(1)

u y y u r cos x

r sin i.e.,

2

u sin x

cos u

r sin , prove that

2

u y

u r

I – 4.7

u cos x

sin

u y u y

(2)

Squaring both sides of (1) and (2) and adding, we get u r

2

1 r2

2

u

u x

2

u

u y

2

2

u x2

Example 4.5 Find the equivalent of

2

u in polar co-ordinates. y2

u(x, y), where x

r cos

and y

r sin

u can also be considered as u(r, ), where r

x2

y2

tan

and

1

y x

u u via r and . and x y u x

u r

r x

u x

x x

2

cos

y

2

u r

u r

1 y2 1 2 x sin u r

y x2

u

(1)

Part I: Mathematics I

I – 4.8 From (1), we can infer that cos

x 2

Now

u

x

2

sin r

r

u x

x cos

sin r

r 2

cos

u

2

r

2

u sin r2 cos

u

u r

u r

sin

r

u y

Now

1 2u r r

cos 2

sin r

sin r2

u

sin

u u and are functions of r and r

are independent and cos 2

1 u r r

u r

cos

u

u sin r r

cos

sin cos

2

sin r

r and

(2)

r y

x

sin r2

2

sin

u

cos

2

u

(3)

y y

u r

1 y2 1 2 x u cos r

2

u r

siin

1 u r2

u

y 2

.

u

1 x

(4)

From (4) we infer that sin

y 2

u y2

y sin

cos r

cos r

(5)

u y r

cos r

sin

2

sin 2

r

u sin cos r2 2

sin

u r

u r

1 2u r r cos

u r

cos r

u

1 u r2 cos r2

2

cos

u 2

sin

u

(6)

Chapter 4: Functions of Several Variables

I – 4.9

Adding (3) and (6), we have 2

2

u x2

2

u y2

u r2

1 u r r

2

1 r2

u 2

Example 4.6 Given the transformations u ex cos y and function of u and and also of x and y, prove that 2

f

x

2

2

f

y

2

(u 2

f x

f u

2

)

u x

x

f

u

2

2

2

x f e x sin y u

f u

f

f

(1) (2)

u

u

2

f x2

f

f

e x cos y u

2

ex sin y and that f is a

u

f u

u

u 2

u u

f

u

2

2

f u

2

f

f

u

2

f

u

u

f u

f

(3)

2

f y

f u

u y

f y

e x sin y f u u y 2

f

y

2

u 2

u

(4) (5)

u

u

2

f f e x cos y u f

f 2

f u

u 2

u

u

f

u f

f 2

u

u 2

u2

f

f 2

f u (6)

Part I: Mathematics I

I – 4.10 Adding (3) and (6), we get 2

x Example 4.7 If z

2

f 2

f

y

(u 2

2

f (u, ), where u 2

x

z 2

2

y

2

)

2

f

u

2

2

2

cosh x cos y and 2

z

z u2

sinh 2 x sin 2 y

2

zx

f

zu u x

z

x

sinh x sin y, prove that 2

z 2

z etc. x cosh x sin y z

, where z x

sinh x cos y zu

Since z is a function of u and , zu and z are also functions of u and . Hence to differentiate zu and z with respect to x or y, we have to do it via u and . z xx

cos y cosh x zu sinh x zuu sinh x cos y zu cosh x sin y sin y sinh x z

i.e.,

cosh x z

u

sinh x cos y z

sinh 2 x cos 2 y zuu

z xx cosh x cos y zu sinh x sin y z

cosh 2 x sin 2 y z

2 sinh x cosh x sin y cos y zu zy z yy

cosh x sin y

(1)

zu cosh x sin y z sinh x cos y cosh x [cos y zu sin y {zuu

cosh x sin y

zu sinh x cos y } siinh x [ sinh y z cos y { z i.e.,

z yy

cosh x sin y z

u

sinh x cos y }]

cosh x cos y zu sinh x sin y z cosh 2 x sin 2 y zuu 2

2 sinh x cosh x sin y cos y zu

2

sinh x cos y z Adding (1) and (2), we get z xx z yy

(sinh 2 x cos 2 y cosh 2 x sin 2 y ) ( zuu z ) {sinh 2 x (1 sin 2 y ) (1 sinh 2 x)sin 2 y }( zuu z ) (sinh 2 x sin 2 y ) ( zuu z )

(2)

Chapter 4: Functions of Several Variables Example 4.8 Find dy , when (i) x3 dx x3

(i) f (x, y)

p

f x

3 x 2 6 axy

q

f y

3 y 2 3 ax 2

xy

3( x 2 2 axy ) 3( y 2 ax 2 )

p q yx

p

f x

yx y

q

f y

x y log x xy x

1

yx y 1 xy x 1

p q

ax2 + 2hxy + by2

r dy dx d2 y dx 2

x(2 ay x) y 2 ax 2

y x log y

Example 4.9 If ax + 2hxy + by2

p

c.

c

dy dx

f (x, y)

yx

y3 3ax2y

dy dx (ii) f (x, y)

3ax2y and (ii) xy

y3

I – 4.11

f x

y x log x . x y log x 1, show that

f

x

2

p q

d2 y dx 2

h 2 ab . 3 hx by

1 f y

2 ax hy ; q

2

1

2 hx by

2

f x y

2 a; s

2

f y2

2h; t

2b

ax hy hx by

p 2 t 2 pqs q 2 r q3 (Refer to differentiation of implicit functions) 8b ax hy

2

16 h ax hy hx by 8 hx by

8a hx by

2

3

1 [2 h {ahx 2 (ab h 2 ) xy bhy 2 } (hx by )3 {a 2 bx 2 2abhxy h 2 by 2 } {ah 2 x 2 2abhxy ab 2 y 2 }]

Part I: Mathematics I

I – 4.12

1 [a (h 2 ab)x 2 2h(h 2 ab ) xy b(h 2 ab) y 2 ] (hx by )3 (h 2 ab) (ax 2 (hx byy )3

(h 2 ab ) 1 (hx by )3

2 hxy by 2 )

h 2 ab . (hx by )3

du if (i) u sin (x2 + y2), where a2x2 + b2y2 c2 (i), u tan dx a2, by treating u as function of x and y only.

Example 4.10 Find where x2 y2 (i) u

y x

sin (x2 y2) du dx

u x

u dy y dx

2 x cos x 2 Now a2x2 + b2y2

y2

2 y cos x 2

y2

dy dx

(1)

c2 x, 2a 2 x 2b 2 y

or

dy dx dy dx

0 a2 x b2 y

(2)

Using (2) in (1), we get du dx

(ii)

2 x cos x 2

y2

2 y cos x 2

2 x cos x 2

y2

b2 a 2 / b2

u

tan

du dx

u x

1

y2 dy 2x 2 y dx

a2 x b2 y

y x u dy y dx

y 1 1 1 2 y2 x2 y x 1 2 1 2 x x y x dy 2 2 2 2 x y x y dx x2

y2

a2 0

dy dx (3)

Chapter 4: Functions of Several Variables dy dx

or

I – 4.13

x y

(4)

Using (4) in (3), we get du dx

y x

2

x y

2

x

2

y

x y

2

1. y Example 4.11 If u

x2

y2 and

x x y , , u u and

xy

y

x and y cannot be easily expressed as single valued functions or u and . x2

Given

y2

u

(1)

xy

and

(2)

Nothing that x and y are functions of u and and (2) partially with respect to u, we have x y 2y u u

2x

x u

y

1

(3)

0

(4)

y u

x

and differentiating both sides of (1)

Solving (3) and (4), we get x u

x 2 x

2

y

and

2

y u

y 2 x

2

y2 , we have

x

2x x

y

y

2y x

y

(5)

0

(6)

1

Solving (5) and (6), we get x

y x2

y2

and

y

x x2

y2

.

Part I: Mathematics I

I – 4.14

Example 4.12 If x2

y2

z2

2xyz x2

Let

y2

1 x

z 2 2 xyz 1

0

dz

0

2 x yz dx 2 y zx dy 2 z xy dz

0

x

dx

y

x yz

Now

2

dy

z

dy 2

1 y

dz 2

1 z2

0

d i.e., i.e.,

dx

1, show that

0. (1)

(2)

x 2 2 xyz y 2 z 2 1 y2

z2

y 2 z 2 , from (1)

1 y2 1 z2 x yz

1 y2 1 z2

Similarly,

y zx

1 z 2 1 x2

and

z xy

1 x2 1 y 2

Using these values in (2), we have 1 y 2 1 z 2 dx

1 z 2 1 x 2 dy

1 x 2 1 y 2 dz

0

1 x 2 1 y 2 1 z 2 , we get dx 1 x

dy 2

dz

1 y

2

1 z2

0.

W1 where W1 W1 W2 and W2 are the weights of the body in air and in water respectively. Show that if there is an error of 1% in each weighing, s is not affected. But if there is an error of 1% in Example 4.13

s of a body is given by s

W1 and 2% in W2, show that the percentage error in s is s log s

W1 W1 W2

log W1 log W1 W2

W2 . W1 W2

Chapter 4: Functions of Several Variables

I – 4.15

Taking differentials on both sides, 1 ds s

1 1 dW1 dW1 dW2 W1 W1 W2

The relation among the errors is nearly 1 s s

1 1 W1 W1 W1 W2

(i) Given that 100 W1 W1

W2

100 W1 1 100 W1 100 W2 W1 W1 W2

100 s s

or

W1

1 and

100 W2 W2

(1)

(2)

1

Using these values in (2), we have 100 s s

1

1 W1 W2 W1 W2

0

s is not affected, viz., there is no error in s. (ii) Given that

100 W1 W1

1 and

100 s s

100 W2 W2 1

1 W1 W2

2. Using these values in (2), we have

W1 2W2

W2 W1 W2 i.e., % error in s

W2 . W1 W2

Example 4.14 The work that must be done to propel a ship of displacement D for a distance s in time t is proportional to s2 D3/2 ÷ t2. Find approximately the percentage increase of work necessary when the distance is increased by 1%, the time is diminished by 1% and the displacement of the ship is diminished by 3%. Given that W ks2 D3/2/t2, where k is the constant of proportionality. 3 log W log k 2 log s log D 2 log t. 2 Taking differentials on both sides, dW W

2

ds s

3 dD dt 2 2 D t

Part I: Mathematics I

I – 4.16

The relation among the percentage errors is approximately,

Given that

100 W W

2

100 s s

1,

100 s s

3 100 D 100 t 2 2 D t

100 t t

1 and

100 D D

(1)

3.

Using these values in (1), we have 100 W W

3 2

2 1

3

2

1

0.5 0.5.

i.e., percentage decrease of work Example 4.15 given by T

The period T of a simple pendulum with small oscillations is l . If T is computed using l g

2

approximately the error in T, if the values are l also the percentage error. T

log T

2

6 cm and g 5.9 cm and g

980 cm/sec2 981 cm/sec2. Find

l g

log 2 log

1 1 log l log g 2 2

Taking differentials on both sides, 1 dT T dT

2

l g 1 lg

1 1 dl dg 2l 2g

1 1 dl dg 2l 2g dl 01

l g g

dg 5 9

5.9 981 981 981

i.e.,

Error in T

(1)

0.0044 sec.

1

Chapter 4: Functions of Several Variables % error in T

I – 4.17

100 dT T 50

dl l

dg , by (1) g

50

01 5 9

1 981

0.8984 Example 4.16 The base diameter and altitude of a right circular cone are measured as 4 cm and 6 cm respectively. The possible error in each measurement is 0.1 cm. Find approximately the maximum possible error in the value computed for the volume and lateral surface. Volume of the right circular cone is given by V

dV

12 12

i.e., Error in V

1 3

D 2

2

h

( D 2 dh 2 Dh dD) {16 0.1 2 4 6 0.1}

1.6755 cm3.

Lateral surface area of the right circular cone is given by D l 2

S

4

dS

4

1

D 2 D

2

4 4

16 144

1.6889 cm 2 .

4h 2

D D 2 4h 2

(2 D dD 8h dh

{4 0 1 24 0 1}

D 2 4h 2 dD

16 144 0 1

Part I: Mathematics I

I – 4.18

Example 4.17 The side c of a triangle ABC is calculated by using the measured values of its sides a, b and the angle C. Show that the error in the side c is given by c

cos B

a cos A b a sin B

C.

The side c is given by the formula c2

(1)

a 2 b 2 2ab cos C

Taking the differentials on both sides of (1), 2c c

2a a 2b b 2 b cos C a cos C

i.e.,

c

Now b cos C

a b cos C

c cos B

a b cos C c a cos C + c cos A b a cos C c Also

a

a b ab sin C

b a cos C c

C , nearly

b ab sin C

C

(2)

a (3)

cos B b

(4)

cos A

b sin B b sin C

c sin C c sin B

ab sin C c

a sin B

(5)

Using (3), (4) and (5) in (2), we get c

cos B

a cos A b a sin B

C

Example 4.18 The angles of a triangle ABC are calculated from the sides a, b, c. If small changes , , are made in the measurement of the sides, show that A

a 2

a

b cos C

c cos B

and and are given by similar expressions, where Verify that 0. In triangle ABC, cos A

b2 c2 a 2 2bc

is the area of the triangle.

(1)

Chapter 4: Functions of Several Variables

I – 4.19

Taking differentials on both sides of (1), 2 sin A

A

b2 c2 a 2 b c c b

bc 2b b 2c c 2a a b 2 c c3 a 2 c

b

bc 2 b3 a 2 b

b2 c2

c 2abc a

2 2

bc c a

2

b

2

c

2

b b c2 a 2 b2

c 2abc a

2 2

bc c 2 ab cos C

b b 2 ca cos B

c 2abc a

2 2

bc

,

by formulas similar to (1)

A

2a cos C b cos B c a bc a a cos C b cos B c bc sin A a 2

a cos C

b 2

b cos A c cos C

c 2

c cos B

1 bc sin A 2

b cos B c , since

(2)

Similarly, B

C

a

(3)

a cos A b

(4)

Adding (2), (3) and (4), we get 2

A

B

C

a b cos C c cos B

a

b a cos C c cos A b a a

a

b b

b

c c

c a cos B b cos A c c

b cos C c cos B a etc. 0 A

B

C

0.

Example 4.19 The area of a triangle ABC is calculated from the lengths of the sides a, b, c. If a is diminished and b is increased by the same small amount k, prove that the consequent change in the area is given by

Part I: Mathematics I

I – 4.20

2(a b)k c ( a b) 2 2

The area of triangle ABC is given by s(s

a )( s

1 {log s 2

log

b)( s

log( s

c) , where 2 s

a)

log( s

b)

a log( s

b

c c)}

Taking differentials on both sides, we get 1 2 Since

2s

s s

s a s a

a b c, 2 s

i.e.; 2 s

k

k

0

s b s b a

b

s c s c

(1)

c

0, by the given data.

s Using (2) in (1), we have 1 k 2 s a

k s b

k 2 2 2 b c a c a b

2s

a b c

(c a b) (b c a ) k 2 [c (a b)][c (a b)] 2 2k ( a b) c 2 ( a b) 2

EXERCISE 4(a) Part A (Short Answer Questions) 1. 2. 3. 4. 5.

What is meant by total differential? Why it is called so? If u sin(xy2), express the total differential of u in terms of those of x and y. If u xy.yx, express du in terms of dx and dy. If u xy log xy, express du in terms of dx and dy. If u axy, express du in terms of dx and dy.

Chapter 4: Functions of Several Variables 6. Find du , if u dt

x3 y2 + x2 y3, where x

at2, y

I – 4.21

2at.

7. Find

du , if u dt

e xy , where x

8. Find

du , if u dt

log (x + y + z), where x

9. Find

dy , using partial differentiation, if x3 + 3x2y + 6xy2 + y3 dx

10. If x sin (x

y)

y)

(x

dy , when u dx

12. Find

dy , if u dx

sin 3 t.

e t, y

sin t, z

cos t. 1.

0, use partial differentiation to prove that

dy dx 11. Find

a2 t 2 , y

y x 2 cos ( x y ) . x x 2 cos ( x y )

sin (x2 + y2), where x2 + 4y2

9.

x2y, where x2 + xy + y2 1. 15 . x4 + 7x3

8x2 + 3x + 1

when x 0.999. 16. What error in the common logarithm of a number will be produced by an error of 1% in the number? 17. The radius of a sphere is found to be 10 cm with a possible error of 0.02 cm. Find the relative errors in computing the volume and surface area. 18. Find the percentage error in the area of an ellipse, when an error of 1% is made in measuring the lengths of its axes. 19. Find the approximate error in the surface of a rectangular parallelopiped of sides a, b, c if an error of k is made in measuring each side. 20. If the measured volume of a right circular cylinder is 2% too large and the radius. Part B 21. If u

f (x

y, y

z, z

x), prove that

22. If f is a function of u, , w, where u that u

f u

x

f . x

u x

u y yz ,

u z

0. zx , and w

xy show

Part I: Mathematics I

I – 4.22 23. If f

y

f

x xy

,

z

x , show that 2 f x x zx

y2

u x

zx

x2

z z b x y

u y

yz

25. If f (cx – az, cy – bz)

z2

27. If z

f (x, y), where x

u2 +

z

z u

z

29. If x

x

r cos , y u r

to

2 ( x2

f (x, y), where x

28. If z

0.

2

,y

z . x

y2 )

u+ ,y

u , prove that

32. If z z x2

z y

u

4( x 2

z

y2

x2

y2 and

y2 )

x2

y2 )

0 is equivalent

z u

z u2

2

z 2

y, show that the equation 0.

2xy, prove that 2

z

y2 and 2

4( x 2

u y

0.

2

f (u, ), where u 2

u x

r sin , prove that the equation

1 tan r 4

2

z. y

z z 2y . x y

f (u, ), where u z x

z x

2u , prove that

f(u, ), where u x2 – 2xy y2 and z z z 0 is equivalent to x y x y x y

31. If z

z u

x – y, show that 2

30. If z

2

0.

c. x + y and

u

f z

0, where z is a function of x and y, prove that

f (u, ), where u

z u

u z

xy

26. If z

u

z2

f (x2 + 2yz, y2 + 2zx), prove that

24. If u

a

f y

y2

2

.

2xy, show that .

Chapter 4: Functions of Several Variables 33. If z

f (x, y) where x

2

2

z x2 34. If z 2

x

z y2

2

X cos a

2

2

y

z 2

and y

X sin

Y cos , show that

2

z X2

z . Y2

f (u, ), where u

z

Y sin

I – 4.23

lx

my and

2

2

z u2

(l 2 m 2 )

ly

mx, show that

z . 2

35. By changing the independent variables x and t to u and by means of the transformations u

x

at and

2 at, show that a

x

36. By using the transformations u

x

y and 2 z z 2 x y2

x

2

y

x

2

2

y

t

2

2

4a 2

y

u

.

y, change the independent

2

variables x and y in the equation 2

37. Transform the equation variables using u

x

x

z 2

2

y and

38. Transform the equation x 2

2

dent variables using u

x2

2

x + iy and z*

u x2

0 by changing the independent

y2

2

z

x

2

2

2 xy

2

z x y

y2

z

y2

0, by changing the

y2 . x

x and 2

5

2 z z 6 2 x y y

2x + y and

40. Transform the equation using z

z

z

x + y.

independent variables using u

39. Transform the equation

2

z x y

2

0 to u and .

3x

0, by changing the indepen-

y.

2

u y2

0, by changing the independent variables

x – iy, where i

1.

dy , when (i) xy dx (x

y)m + n; (iii) (cos x)

y

(sin y)x; (iv) (sec x)

y

d2 y , when x3 dx 2

yx; (ii) xmyn

(cot y)x; (v) xy y3 3axy

0.

ex - y.

Part I: Mathematics I

I – 4.24

d2 y , when x4 + y4 dx 2 2 44. Use partial differentiation to prove that d y

dx

when ax2 + 2hxy + by2 + 2gx + 2fy + c 45. Use partial differentiation to prove that x2. 46. If x2 – y2 + u2+ 2

2

4a2xy.

abc 2 fgh af 2 bg 2 ch 2

2

( hx by

f)

3

,

0. d2 y dx 2

1 and x2 + y2 – u2 –

b 2 ac , when ay2 + 2by + c (ay b)3 2

2, prove that

2x .

u x

3 x and u

x l and diameter d supported at its ends and loaded at the centre with a weight w is proportional to wl3/d4. What w, l and d are 3, 2 and 1 respectively. 48 The torsional rigidity of a length of wire is obtained from the formula 8 Il . If l is decreased by 2%, t is increased by 1.5% and r is increased t 2r 4 by 2%, show that the value of N is decreased by 13% approximately. N

49. The Current C measured by a tangent galvanometer is given by the relation C k tan , where C due to a given error in is minimum when 45°. 50. The range R of a projectile projected with velocity

at an elevation is given

2

by R

sin 2 . Find the percentage error in R due to errors of 1% in and g 1 in , when . % 6 2 g 2 T , where g and are con51. The velocity of a wave is given by 2 2 stants and and are variables. Prove that, if is increased by 1% and is 3 T . decreased by 2%, then the percentage decrease in is approximately 2 1 1 1 . If equal f u errors k are made in the determination of u and , show that the percentage

52. The focal length of a mirror is given by the formula

error in f is 100k 1 u

1

.

Chapter 4: Functions of Several Variables

I – 4.25

53. A closed rectangular box of dimensions a, b, c has the edges slightly altered in length by amounts a, b and c respectively, so that both its volume and a b c . surface area remain unaltered. Show that 2 2 2 a (b c) b (c a) c (a b) [Hint: Solve the equations dV 0 and dS 0 for a, b, c using the method of cross-multiplication] 54. If a triangle ABC is slightly disturbed so as to remain inscribed in the same circle, prove that a b cos A cos B

c cos C

0.

55. The area of a triangle ABC is calculated using the formula Show that the relative error in is given by b b

1 bc sin A. 2

c cot A A. c

If an error of 5 is made in the measurement of A the percentage error in ABC due to a small error in the measurement of c is given by 1 4 s

1 s a

1 s b

1 s c

c.

57. The area of a triangle ABC is determined from the side a and the two angles B and C. If there are small errors in the values of B and C, show that the resulting error in the calculated value of the area

Hint:

4.2.3

will be

1 2 (c B b 2 C ). 2

1 a 2 sin B sin C 2 sin( B C )

Taylor’s Series Expansion of a Function of Two Variables

Students are familiar with Taylor’s series of a function of one variable viz. f (x + h) = f ( x)

h h2 f ( x) f ( x) 1! 2!

,

can be extended to expand f (x + h, y + k

h. This idea h and k.

Part I: Mathematics I

I – 4.26

Statement If f (x, y

x, y), then

f ( x h, y k ) 1 h k x y 3!

1 h k f x y 1!

f ( x, y )

2

1 h k x y 2!

f

3

...

f

Proof: If we assume y to be a constant, f (x + h, y + k) can be treated as a function of x only. Then f ( x h, y k )

f ( x, y k )

h2 2!

h f ( x, y k ) 1! x

2

f ( x, y k ) ... x2

(1)

Now treating x as a constant,

f x, y k

2

k2 2!

k f x, y 1! y

f x, y

f x, y

...

y2

(2)

Using (2) in (1), we have f ( x h, y k )

h 1! x

f ( x, y )

h2 2 2! x 2 f ( x, y )

f f 1 h k x y 1!

f ( x, y )

1 h k 1! x y

k2 2!

k f ( x, y ) 1! y

f ( x, y )

f ( x, y ) ... y2

k f ( x, y ) y 1!

f ( x, y )

k2 2!

2

k2 2!

k f ( x, y ) 1! y

2 f 1 h2 2 2! x

f

2

f ( x, y ) ... y2 2

f ( x, y ) ... y2

2

2hk

1 h k 2! x y

2 f f k2 2 x y y

...

2

f

...

(3)

Interchanging x and h and also y and k in (3) and then putting h

f ( x, y )

f (0, 0)

1 f (0, 0) x 1! x

f (0, 0) y

y 2

2 xy

...

f (0, 0) x y

1 2 x 2! 2

y2

2

k

0, we have

f (0, 0) x2

f (0, 0) y2

...

(4)

Chapter 4: Functions of Several Variables

I – 4.27

Series in (4) is the Maclarin’s series of the function f (x, y) in powers of x and y. Another form of Taylor’s series of f (x, y) f ( x, y ) f ( a x a , b y b ) f (a h), (b k ), say f ( a, b) 1 f ( a, b) h k 1! x y

f ( a, b) 1 2 h 2! f ( a, b)

2

f ( a, b) 2kh x2

2

f ( a, b) 2 k x y

2

f ( a, b) y2

..., by (3)

f ( a, b) f ( a, b) 1 ( x a) ( y b) y x 1!

1 ( x a)2 2!

2

f ( a, b) 2( x a )( y b) x2

2

f ( a, b) ( y b) 2 x y

2

f ( a, b) y2

...

(5)

(5) is called the Taylor’s series of f (x, y) at or near the point (a, b). Thus the Taylor’s series of f (x, y) at or near the point (0, 0) is Maclaurins series of f (x, y).

4.3

JACOBIANS

If u and

are functions of two independent variables x and y, then the determinant u x

u y

x

y

is called the Jacobian or functional determinant of u, is written as (u , ) u, . or J ( x, y ) x, y Similarly the Jacobian of u, , w with respect to x, y, z

(u , , w) ( x, y , z )

u x

u y

u z

x w x

y w y

z w z

with respect to x and y and

Part I: Mathematics I

I – 4.28

Note n dependent variables, each of these must be a function of n independent variables. 2. The concept of Jacobians is used when we change the variables in multiple integrals. (See property 4 given below)

4.3.1

Properties of Jacobians

1. If u and Proof: Let u x Then

( x, y ) (u , )

1.

f(x, y) and g(x, y). When we solve for x and y, let (u, ) and y (u, ). u x u x

(u , ) ( x, y )

x u x

u y u y

x u x

x u x

Now

(u , ) ( x, y )

are functions of x and y, then

u x

( x, y ) (u , )

y u y

x

y

u x

u y

x

y

x u y u

0 0

u

1

x y

x y by interchanging the rows u u , and columns of the x y second determinant.

u x

x u

x

x u

u y

y u

y

y u

1 0 [by (1)] 0 1 1

1

(1) y u y

y u y

u y

u u u

u x

x

u y

x x

y y

y

Chapter 4: Functions of Several Variables 2. If u and

I – 4.29

functions of r and s, where r and s are functions of x and y, then u, x, y

u, r, s

r, s x, y

Proof: u, r, s

r, s x, y

r x s x

u u r s r

s

u r

u s

r

s

r s x x , by rewriting the second determinant. r s y y

u r . r x r

Note

u s . s x

r x

.

u x

u y

x

y

r y s y

s

.

s x

u r . r y r

.

r y

u s . s y s

.

s y

(u , ) . ( x, y )

The two properties given above hold good for more than two variables too.

3. If u, , w are functionally dependent functions of three independent variables x, y, z then

(u, , w) ( x, y , z )

0.

Note

The functions u, , w are said to be functionally dependent, if each can be expressed in terms of the others or equivalently f (u, , w) 0. Linear dependence of functions is a particular case of functional dependence. Proof: Since u, , w are functionally dependent, f (u, , w) 0 x, y and z, we have

(1)

fu . ux

f .

x

f w . wx

0

fu . u y

f .

y

f w . wy

0

(2) (3)

fu . uz

f .

z

f w . wz

0

(4)

Part I: Mathematics I

I – 4.30

Equations (2), (3) and (4) are homogeneous equations in the unknowns fu, f , fw. At least one of fu, f and fw is not zero, since if all of them are zero, then f (u, , w) constant, which is meaningless. Thus the homogeneous equations (2), (3) and (4) possess a non-trivial solution. fu, f , fw is singular. ux uy uz

i.e.,

i.e.,

wx wy wz

0

(u , , w) ( x, y , z )

0

x y z

Note

The converse of this property is also true. viz., if u, , w are functions of (u , , w) x, y, z such that 0 then u, , w are functionally dependent. i.e., there ( x, y , z ) exists a relationship among them. 4. If the transformations x integral

x(u, ) and y

f ( x, y ) dx dy, then f(x, y)

y(u, ) are made in the double

F(u, ) and dx dy

J du d , where

( x, y ) . (u, )

J

Proof: dx dy Elemental area of a rectangle with vertices (x, y), (x + dx, y), (x + dx, y + dy) and (x, y + dy) This elemental area can be regarded as equal to the area of the parallelogram with x y x x y y vertices (x, y), x and du , y du , x du d y du d u u u u x

x

y d v

d , y

, since dx and dy

Now the area of this parallelogram is equal to 2× area of the triangle with vertices x y x y (x, y), x du , y du and x d , y d u u x dx dy

2

1 x 2 x

y x du u x d

y y

1 y du 1 u y d 1

Chapter 4: Functions of Several Variables

x x du u x d

1 0

x du u x d

y du u y d

0

y u du d y

x u x

i.e.,

y y du u y d

I – 4.31

( x, y ) du d . (u, )

dx dy

Since both dx dy and du d are positive, dx dy

J du d , where J

( x, y ) (u , )

Similarly, if we make the transformations x

x(u , , w), y

y (u , , w) and z

z (u , , w)

f ( x, y, z ) dx dy dz , then dx dy dz

in the triple integral

J du d dw, where

J

( x, y , z ) . (u , , w)

4.4

DIFFERENTIATION UNDER THE INTEGRAL SIGN

When a function f(x, y) of two variables is integrated with respect to y partially, viz., b

treating x as a parameter, between the limits a and b, then

f ( x, y ) dy will be a a

function of x. Let it be denoted by F(x). F ' (x respect to x. F ' (x given below:

F(x) and then differentiate it with F(x), by using Leibnitz’s rules,

1. Leibnitz’s rule for constant limits of integration If f (x, y) and

f ( x, y ) are continuous functions of x and y, then x d dx

b

b

f ( x , y ) dy a

a and b are constants independent of x.

a

f ( x, y ) dy, where x

Part I: Mathematics I

I – 4.32 Proof: b

f ( x , y ) dy

Let

F ( x).

a

b

Then

F (x

x)

F ( x)

b

f (x

x , y ) dy

f ( x , y ) dy

a

a

b

x, y )

[ f (x

f ( x, y )]dy

a b

f (x

x

x, y ) x

a

by Mean Value theorem, viz., f ( x

F (x

x) x

F ( x)

b

f (x

a

b

x, y )

a

b

b

df ( x h) ,0 dx

1

(1)

dy

0,

f ( x, y ) dy x

f ( x, y )dy a

h

1,

f ( x, y ) dy x

F ( x)

d dx

f ( x)

x

Taking limits on both sides of (1) as x

i.e.,

h)

dy , 0

a

2. Leibnitz’s rule for variable limits of integration If f(x, y) and b( x)

f ( x, y ) are continuous functions of x and y, then d dx x

f ( x, y ) dy x

a( x)

f {x, b( x)}

db dx

f {x, a( x)}

b( x)

f ( x , y ) dy a( x)

da provided a(x) and b(x) possess , dx

Proof: Let

f ( x , y ) dy

F ( x, y ), so that

y

F ( x, y )

f ( x, y )

(1)

Chapter 4: Functions of Several Variables

I – 4.33

b( x)

f ( x , y ) dy

F{x, b( x)}

F{x, a( x)}

a( x)

b( x)

d dx

d F{x, b( x)} dx

f ( x, y )dy a( x)

d F ( x, y ) dx

x

d F{x, a( x)} dx

y

b( x)

F ( x, y )

x

d F ( x, y ) dx

y

F ( x, y )

F ( x, y )

y

dy dx

F ( x, y )

y

dy dx

y

a( x)

b( x)

y

a( x)

by differentiation of implicit functions y

x

b( x)

f ( x, y )

F ( x, y ) y

dy dx

y

b( x)

a( x)

f ( x, y )

y

x

f ( x, y ) dy x b( x)

a( x)

f ( x, y ) dy x

by (1) y

a( x)

b( x)

f {x, a( x)}a ( x)

f {x, b( x)}b ( x)

f ( x , y ) dy y

dy dx

a( x)

y

b( x)

y

a( x)

f {x, b( x)}b ( x) f {x, b( x)}b ( x)

f {x, a ( x)}a ( x)

f {x, a( x)}a ( x)

WORKED EXAMPLE 4(b) Example 4.1 degree.

Expand ex cos y in powers of x and y as far as the terms of the third

f ( x, y )

e x cos y;

f ( x, y ) x

f x ( x, y )

e x cos y;

Part I: Mathematics I

I – 4.34 f ( x, y ) y

f y ( x, y )

e x sin y.

f ( x, y ) x2

f xx ( x, y )

e x cos y;

f ( x, y ) y2

f yy ( x, y )

2

2

2

f xxx ( x, y )

Similarly

f ( x, y ) x y

e x sin y

f yx ( x, y )

e x cos y.

e x cos y; f xxy ( x, y )

e x sin y;

e x cos y; f yyy (xx, y )

f xyy ( x, y )

e x sin y

f (0, 0)

1; f x (0, 0)

1; f y (0, 0)

f xx (0, 0)

1; f xy (0, 0)

0; f yy (0, 0)

f xxx (0, 0)

1; f xxy (0, 0)

0; f xyy (0, 0)

0; 1; 1; f yyyy (0, 0)

0

Taylor’s series of f (x, y) in powers of x and y is

f ( x, y )

f (0, 0)

1 {xf x (0, 0) 1!

yf y (0, 0)}

1 2 {x f xx (0, 0) 2 xyf xy (0, 0) y 2 f yy (0, 0)} 2! 1 3 {x f xxx (0, 0) 3 x 2 yf xxy (0, 0) 3 xy 2 f xyy (0, 0) 3!

e x cos y

1

1 {x 1 1!

y 0}

1 2 {x 1 2!

e x cos y

Real part of e x

y 2 ( 1)}

2 xy 0

1 3 {x 1 3 x 2 y 0 3 xy 2 ( 1) 3! x 1 1 3 1 ( x2 y 2 ) (x 1! 2 ! 3!

y 3 f yyy (0, 0)}

y 3 0} 3 xy 2 )

iy

R.P. of 1 +

x + iy 1!

(x

iy ) 2 2!

(x

iy )3 3!

,

by exponential theorem

Chapter 4: Functions of Several Variables 1

x 1 2 (x 1! 2 !

I – 4.35

1 3 (x 3y2 ) 3!

y2 )

Expand ( x h)( y k ) in a series of powers of h and k upto the x h y k second degree terms. Example 4.2

Let

( x h)( y k ) x h y k

f ( x h, y k )

xy . x y

f ( x, y )

Taylor’s series of f (x + h, y + k) in powers of h and k is

f ( x h, y k )

f ( x, y )

f f 1 h k x y 1!

1 2 2f h 2! x2

Now

2

2hk

2 f f k2 2 x y y

fx

y

( x y) x ( x y)2

y2 ( x y)2

fy

x

( x y) y ( x y)2

x2 ( x y)2

2 y2 ; f xy ( x y )3

f xx

2{ y ( x y ) y 2 } (xx y )3

(1)

( x y ) 2 2 y y 2 2( x y ) ( x y)4 2 xy ( x y )3

2x2 . ( x y )3

f yy

Using these values in (1), we have ( x h)( y k ) x h y k

xy x y

hy 2 ( x y)2

kx 2 ( x y)2

h2 y 2 ( x y )3 2hkxy ( x y )3

k 2 x2 ( x y )3

Part I: Mathematics I

I – 4.36

y

Example 4.3 Find the Taylor’s series expansion of x near the point (1, 1) upto the second degree terms. 1 Taylor’s series of f (x, y) near the point (1, 1) is f x, y f 1,1 1! 1 2 x 1 f x 1,1 y 1 f y 1,1 x 1 f xx 1,1 2 x 1 y 1 f xy 1,1 2! y 1

2

f yy 1,1

(1) x y ; f x x, y

f x, y

yx y 1 ; f y x, y

f xx x, y

y y 1 x y 2 ; f xy x, y

f yy x, y

x y log x

xy

1; f x 1,1

f xx 1,1

0; f xy 1,1

yx y 1 log x

2

1; f y 1,1

f xx 1,1

1

x y log x;

0;

1; f yy 1,1

0

Using these values in (1), we get xy

1

x 1

x 1 y 1

Find the Taylor’s series expansion of ex sin y near the point

Example 4.4

1,

4

upto the third degree terms. Taylor’s series of f (x, y) near the point f x, y

f

1,

1 2!

1 1!

4

x 1

2

x 1 fx

f xx

1,

4

1,

1,

is

4 y

4

2 x 1 y

fy

4

f xy

4

1, 1,

4

4

2

y

f yy

4

1,

(1) 4

f x, y

e x sin y; f x

e x sin y; f y

f xx

e x sin y; f xy

e x cos y; f yy

f xxx

e x sin y; f xxy

f yyy

e x coss y.

f

1,

f xx

1,

1 4 4

e 2 1 e 2

e x cos y;

e x cos y; f xyy

; fx

1,

; f xy

1,

1 4 4

e 2 1

; fy

e 2

; f yy

e x sin y; e x sin y;

1,

1 4

1,

e 2

; 1

4

e 2

;

Chapter 4: Functions of Several Variables f xxx

1,

1 4

e 2 1 e 2

; f xxy

1,

; f yyy

1,

1 4

e 2 1

4

; f xyy

I – 4.37 1,

4

e 2

Using these values in (1), we get e x sin y

1 e 2

1 1!

1

x 1

y

4 2

1 2!

x 1

1 3!

x 1

2

2 x 1 y

y

4

4 2

3

3 x 1

2

y

3 x 1 y

4

3

y

4

4

Example 4.5 Find the Taylor’s series expansion of x2 y2 + 2x2 y 3xy2 in powers of (x + 2) and (y 1) upto the third powers. Taylor’s series of f (x, y) in powers of (x 2) and (y 1) or near ( 2, 1) is f x, y

f

1 x 1!

2,1 1 x 2! y 1

f x, y

2 2

2

f yy

2 fx

2,1

y 1 fy

2,1

2 x

2 y 1 f xy

f xx

2,1 2,1 (1)

2,1

x 2 y 2 2 x 2 y 3 xy 2

f

2,1

6

fx

2 xy 2 4 xy 3 y 2

fx

2,1

fy

2 x 2 y 2 x 2 6 xy

fy

2,1

4

f xx

2 y2 4 y

f xx

2,1

6

f xy

4 xy 4 x 6 y

f xy

2,1

10

f yy

2x2 6x

f yy

2,1

4

f xxx

0

f xxx

2,1

0

f xxy

4y 4

f xxy

2,1

8

f xyy

4x 6

f xyy

2,1

f yyy

0

f yyy

2,1

9

2 0

Part I: Mathematics I

I – 4.38

Using these values in (1), we have x 2 y 2 2 x 2 y 3 xy 2

6

1 1!

9 x 2

1 6 x 2 2!

2

1 24 x 2 3! Example 4.6

4 y 1 20 x 2 y 1

2

y 1

4 y 1

6 x 2 y 1

2

2

Using Taylor’s series, verify that

log 1 x y

1 x y 2

x y

1 x y 3

2

3

The series given in the R.H.S. is a series of powers of x and y. So let us expand f (x, y) log (1 + x + y) as a Taylor’s series near (0, 0) or Maclaurin’s series. fx

1 ; fy 1 x y

f xx

1 (1 x y ) 2

f xxx

1 1 x y f xy

2

f xxy

3

1 x y

f yy f xyy

f y 0, 0

f yyy

f 0, 0

0; f x 0, 0

f xx 0, 0

f xy 0, 0

f yy 0, 0

1;

f xxx 0, 0

f xxy 0, 0

f xyy 0, 0

f yyy 0, 0

1;

2.

Maclaurin’s series of f (x, y) is given by f x, y

f 0, 0

1 xf x 0, 0 1!

1 2 x f xx 0, 0 2!

yf y 0, 0

2 xyf xy 0, 0

Using the relevant values in (1), we have log 1 x y

x y

1 2

x 2 2 xy y 2

1 2 x 3 6 x 2 y 6 xy 2 2 y 3 6

y 2 f yy 0, 0

(1)

Chapter 4: Functions of Several Variables x y

Example 4.7

If x

r cos , y x

1 x y 2

r cos ; y

r sin

x x r y y r

cos sin

r cos 2

s in 2

r,

r2

2r

Similarly,

r x

2x

r x

x r

r y

y r

r, x, y

1 x y 3

3

x, y r,

r sin , verify that

x, y

Now

2

x2

r, x, y

1.

r sin r cos

r.

y 2 and

tan

1 x y2 1 2 x y x2 y 2

y

Similaarly

I – 4.39

1

y x

x

y

y r2 x r2

r r x y x y x r y r2

x, y r,

r, x, y

r

y r x r2 1 r

y2

x2 r3

r2 r3

1 r

1.

Example 4.8 If we transform from three dimensional cartesian co-ordinates (x, y, z) to spherical polar co-ordinates (r, , ), show that the Jacobian of x, y, z with respect to r, , is r2 sin .

Part I: Mathematics I

I – 4.40

The transformation equations are x x r x

r sin cos , y

Now

y r

sin cos , r cos cos ,

x

r sin sin , z

sin sin , y

r sin sin ,

z r

r cos sin , y

r sin cos ,

r cos .

cos z

r sin z

0..

x y z r r r x y z

x, y , z r, ,

x

y

z

sin cos r cos cos

sin sin r cos sin

r sin sin r 2 [sin cos

r sin cos 0 sin 2 cos

cos r sin 0 sin sin

(0 sin 2 sin ) cos (sin cos cos 2 r 2 [sin 3 cos 2 r 2 (sin 3

sin 3 sin 2

sin cos sin 2 )]

sin cos 2 ]

sin cos 2 )

r 2 sin . Example 4.9

If u

x2 – y2, x

2xy,

r cos

By the property of Jacobians, u, r,

u, x, y

x, y r,

ux u y

xr x

x

y

yr y

and y

r sin , compute

u, r,

.

Chapter 4: Functions of Several Variables 2y 2x 4( y 2

2x 2y

I – 4.41

r sin r cos

cos sin

x 2 ) r (cos 2

sin 2 )

4r 3 .

Example 4.10 Find the Jacobian of y1, y2, y3 with respect to x1, x2, x3, if

y1

x2 x3 , y2 x1

x3 x1 , y3 x2

x1 x2 x3

( y1 , y2 , y3 ) ( x1 , x2 , x3 )

y1 x1

y1 x2

y1 x3

y2 x1

y2 x2

y2 x3

y3 x1

y3 x2

y3 x3

x2 x3 x12

x3 x1

x2 x1

x3 x2

x3 x1 x22

x1 x2

x2 x3

x1 x3

x1 x2 x32

x2 x3 1 x 2 x3 x12 x22 x32 x2 x3

x3 x1 x3 x1

x1 x2 x1 x2

x3 x1

x1 x2

1 1

1 1

1 1

1

1

1

1 1 1 0 0 2 0 2 0

4

x12 x22 x32 x12 x22 x32

Example 4.11 Express xyz (1 x y z ) dx dy dz in terms of u, , w given that x + y + z u, y + z u and z u w.

Part I: Mathematics I

I – 4.42

The given transformations are x y z u y and

z=u

(2)

z

(3)

Using (3) in (2), we have y Using (2) in (1), we have x J

dx dy dz

(1)

u w u (1 – w) u(1 – )

u d dw, where

x u y u z u

( x, y , z ) (u , , w)

J

x y z

1 u (1 w) u (1 w) w wu u u

2

0 u u

){u 2 (1 w) u 2 w} u{u

(1 2

x w y w z w

(1

) u

2

2

(1 w) u

2

w}

2

dx dy dz = u 2 du d dw

(4)

Using (1), (2), (3) and (4) in the given triple integral I, we have u3

I

2

w(1

)(1 w)(1 u ) u 2 du d dw 1

u7 / 2 Example 4.12

2

1

w1 / 2 (1 u ) 2 (1

1

) 2 (1 w) 2 du d dw

Examine if the following functions are functionally dependent. If

(i) u

sin 1 x sin 1 y;

(ii) u

y z;

(i) u

sin 1 x sin 1 y;

x 2z2 ; w

x 1 y2

y 1 x2

x 4 yz 2 y 2 x 1 y2

y 1 x2

Chapter 4: Functions of Several Variables u x

1 1 x

u y

;

2

xy y

1 y

1 1 y

;

2

xy 1 x2

1 x2

2

1

1 x

(u, ) ( x, y )

1 y2

x

1 Now

I – 4.43

2

1 y2 xy

1 y2

1 x

xy 2

xy 2

1 x2

1 y2 1 1

2

(1 x )(1 y )

xy 2

(1 x )(1 y 2 )

0. u and Now

are functionally dependent by property (3). sin u

sin (sin 1 x sin 1 y ) sin (sin 1 x) cos (sin 1 y ) cos (sin 1 x) sin (sin 1 y ) x cos cos 1 ( 1 y 2 ) x 1 y2

cos cos 1 ( 1 x 2 )

y 1 x2

. The functional relationship between u and (ii) u

Now

y z;

is

x 2 z 2 ; w x 4 yz 2 y 2 u x u y u z

1;

(u , , w) ( x, y , z )

0 1 1

0;

1;

x y z

w 1 x w 0; 4 y 4z y w 4 z; 4y z 1;

1 0 4 y 4z

1 4z 4y

sin u.

y

;

Part I: Mathematics I

I – 4.44

4 y 4 y 4z u,

4 z 0.

and w are functionally dependent

Now

2 z 2 4 yz 2 y 2

w

2 y z

2

2u 2

The functional relationship among u, and w is 2u2 dx Example 4.13 Given that a b , a b cos x a 2 b2 0 dx 0

a b cos x

2

w.

cos x dx

and 0

a b cos x

dx a b cos x 0

a 2 b2

2

(1) a, we get

0

1 dx a a b cos x

a

a 2 b2

dx

i.e., 0

a b cos x

1 / 2 a 2 b2

2

0

3/ 2

2a

a

dx

i.e.,

, since the limits of integration are constants

a b cos x

2

a

2

b2

3/ 2

b, we get

0

1 dx b a b cos x 1

i.e., 0

a b cos x

2

0

a 2 b2

1 / 2 a 2 b2

cos x dx

cos x

i.e.,

b

a b cos x

2

b

dx a

2

b2

3/ 2

.

3/ 2

2b

Chapter 4: Functions of Several Variables

I – 4.45 x

Example 4.14 0 1

log 1 x

0

1 x2

1 y2

dy.

dx.

x

Let

log 1 xy

log 1 xy

f x

1 y2

0

(1)

dy

x, we have f

x

d dx

x

0

x

0

log 1 xy 1 y2

dx

log 1 xy x

1 y

dx

2

log 1 x 2 1 x

2

d x dx by Leibnitz's rule

x

0

y 1 xy 1 y 2

x

0

dy

x 1 x2

1 xy

1 log 1 xy 1 x2 1 1 log 1 x 2 2 1 x2

log 1 x 2 1 x2 1 1 x2

y x 1 y2

1 1 log 1 y 2 2 1 x2

dy

log 1 x 2 1 x2

x tan 1 y 1 x2

x

,

log 1 x 2

0

x tan 1 x 1 x2

(2)

Integrating both sides of (2) with respect to x, we have

f x

1 log 1 x 2 d tan 1 x 2 1 tan 1 x log 1 x 2 2

1 x2

x tan 1 x dx c 1 x2 tan 1 x

2x dx 1 x2

x tan 1 x dx c 1 x2

Part I: Mathematics I

I – 4.46

1 tan 1 x log 1 x 2 2 Now putting x

0 in (2), we get c

f(0) x

0, by (1)

log 1 xy

f x

1 tan 1 x log 1 x 2 2

dy

1 y2

0

Putting x

(2)

c

1 in (3), we get 1

log 1 y 1 y2

0

1 tan 2

dy

1

1

log 2

log 2

8

Since y is only a dummy variable, 1

log 1 x 1 x2

0

dx

d Example 4.15 Show that da d da

a

2

tan 0

1

x dx a

log 2

8 a2

tan

x dx a

1

0 a2

a

0

tan

1

2a tan

1

x dx tan a

1 log a 2 1 . 2

a

1

a2 a

d 2 a , da by Leibnitz's rule

a2

1 x2 0 1 2 a a2

0

x a

2

dx 2a tan 1 a

x dx 2a tan 1 a x2 a2

1 log x 2 a 2 2

a2 0

2a tan 1 a

(3)

Chapter 4: Functions of Several Variables a4 a2 1 log 2 a2

Example 4.16 If I

2

a x

e

dx, prove that

0

a x

x2

I

2a tan 1 a

1 log a 2 1 2

2a tan 1 a x2

I – 4.47

e

dI da

2I .

I.

2

dx

(1) .

0

a, we have dI da

a2

x2

a

0

x2

e

x2

e a2

2a dx x2

x2

0 a2

0

y2

e

dx

y2

2 dy, on putting x a y

y2

2 e

a or y y

a x

2

dy

0

i.e.,

Solving, we get log I

dI da

2I

dI I

2 da

log c – 2a I = ce

When

(2)

a

2a

(3)

0, I

e 0

Using (4) in (3), we get c

2

x2

dx

2

(4)

Part I: Mathematics I

I – 4.48

Hence

I

2

e

2a

.

EXERCISE 4(b)

Part A (Short Answer Question ) 1. Write down the Taylor’s series expansion of f (x h, y k) in a series of (i) powers of h and k (ii) power of x and y. 2. Write down the Maclaurin’s series expansion of (i) f (x, y), (ii) f (x + h, y k). 3. Write down the Taylor’s series expansion of f (x, y) near the point (a, b). 4. Write down the Maclaurin’s series for e x + y. 5. Write down the Maclaurin’s series for sin (x + y). 7. State any three properties of Jacobians. 8. State the condition for the functional dependence of three functions u(x, y, z), (x, y, z) and w(x, y, z). 9. Prove that

f x, y dx dy

f r cos , r sin

10. Show that

f x , y dx dy

f u (1

r dr d .

), u

u du d .

11. If x = u (1 ) and y = (1 u x, y with respect to u, . 12. State the Leibnitz’s rule for differentiation under integral sign, when both the limits of integration are variables. d 13. Write down the Leibnitz’s formula for dx stant.

b x

f ( x, y ) dy, where a is a cona

b

d 14. Write down the Leibnitz’s formula for f ( x, y ) dy, where b is a d xax constant. 1

15. Evaluate

d log( x 2 dy 0

y 2 ) dx, without integrating the given function.

Part B 16. Expand ex sin y in a series of powers of x and y as far as the terms of the third degree. 17. Find the Taylor’s series expansion of ex cos y in the neighbourhood of the point 1,

4

upto the second degree terms.

Chapter 4: Functions of Several Variables

I – 4.49

18. Find the Maclaurin’s series expansion of ex log (1 + y) upto the terms of the third degree. y 1 in powers of (x 1) and (y 1) 19. Find the Taylor’s series expansion of tan x upto the second degree terms. ) and (y 2) upto the third degree 20. Expand x2y 3y 2 in powers of (x terms. 21. Expand xy2 2x 3y in powers of (x 2) and (y 1) upto the third degree terms. 22. Find the Taylor’s series expansion of yx at (1, 1) upto the second degree terms. 23. Find the Taylor’s series expansion of e xy at (1, 1) upto the third degree terms. 24. Using Taylor’s series, verify that ( x y)2 2! 25. Using Taylor’s series, verify that cos ( x y )

tan 1 ( x y ) 26. If x

u (1

), y

u , verify that

27. (i) if x u2 – 2 and y and . (ii) if u x2 and y2 a cosh u cos

r cos , y

30. If F

xu +

r sin , z

( F , G, H ) . (u , , w) 31. If u xyz, xy

yz

1 x and y with respect to u

(u , ) ( x, y ) a sinh u . sin , show that

and y

– y, G

(u , ) ( x, y )

2u

a2 (cosh 2u cos 2 ). 2

( x, y ) (u , ) 29. If x

1 ( x y )3 3

( x y)

( x, y ) (u , )

28. If x

( x y)4 4!

1

z

( x, y , z ) (r , , z )

u2 +

y + w and H

zx and w

x

y

z

zu –

+

u, , w . x, y , z

32. Examine the functional dependence of the functions u xy . ( x y)2

w, compute

x y x y

and

Part I: Mathematics I

I – 4.50 33. Are the functions u

x y and 1 xy

34. Are the functions f1 x

y

tan 1 x tan 1 y functionally dependent?

x2

z, f2

y2

z2 and f3 xy + yz + zx functionf1, f2 and f3.

x

e

35. If

x

y

f ( y )dy

2

xe

x

e

prove that f (x)

x

. [Hint:

0

both sides with respect to x]. Use the concept of differentiation under integral sign to evaluate the following: x

x

dx ( x a 2 )2

Hint: Use

x m (log x) n dx

Hint: Use

36.

2

0

0

1

37.

1

0

x m dx 0

e

38.

dx x a2 2

x2

cos 2ax dx

0

e

39. 0

1

40. 0

4.5

ax

sin x dx and hence x

xm 1 dx, m log x

0

sin x dx x

0.

MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES

Students are familiar with the concept of maxima and minima of a function of one variable. Now we shall consider the maxima and minima of a function of two variables. A function f (x, y) is said to have a relative maximum (or simply maximum) at x a and y b, if f (a, b) f (a h, b k) for all small values of h and k. A function f(x, y) is said to have a relative maximum (or simply maximum) at x = a and y b, if f(a, b) f (a + h, b + k) for all small values of h and k. A maximum or a minimum value of a function is called its extreme value. We give below (x, y): (1) Find

f and x

f . y

Chapter 4: Functions of Several Variables (2) Solve the equations

f x

0 and

f y

I – 4.51

0 simultaneously. Let the solutions be

(a, b); (c, d);… f f 0 are called stationary 0 and y x points of the function f (x, y). The values of f (x, y) at the stationary points are called stationary values of f (x, y). 2 2 2 f f f A , B , C x dy x2 y2 AC B2. and (4) (i) If 0 and A (or C) 0 for the solution (a, b) then f (x, y) has a maximum value at (a, b). 0 and A (or C) > 0 for the solution (a, b) then f (x, y) has a minimum (ii) If value at (a, b). (iii) If < 0 for the solution (a, b), then f (x, y) has neither a maximum nor a minimum value at (a, b). In this case, the point (a, b) is called a saddle point of the function f (x, y). (iv) If 0 or A 0, the case is doubtful and further investigations are required to decide the nature of the extreme values of the function f (x, y).

Note

4.5.1

The points like (a, b) at which

Constrained Maxima and Minima

variables say f (x, y, z) which are not independent but are connected by some given relation (x, y, z) 0. The extreme values of f (x, y, z) in such a situation are called constrained extreme values. In such situations, we use (x, y, z) 0 to eliminate one of the variables, say z from the given function, thus converting the function as a function of only two [Refer to examples (4.8), (4.9), (4.10)]. When this procedure is not practicable, we use Lagrange’s method, which is comparatively simpler.

4.5.2 Let

Lagrange’s Method of Undetermined Multipliers u

f x, y , z

(1)

be the function whose extreme values are required to be found subject to the constraint x, y , z

(2)

0

The necessary conditions for the extreme values of u are f x

0,

f y

0 and

f z

0

Part I: Mathematics I

I – 4.52 f dx x

f dy y

f dz z

(3)

0

From (2), we have x

dx

y

dy

z

dz

(4)

0

is an unknown multiplier, called Langrange multiplier, gives f x

dx

x

f y

y

f x

x

f y

y

f z

z

f z

dy

z

0

(5)

Equation (5) holds good, if 0

(6)

0

(7)

0

(8)

Solving the Equations (2), (6), (7) and (8), we get the values of x, y, z, , which give the extreme values of u.

Note (1) The Equations (2), (6), (7) and (8) are simply the necessary conditions for ), where is also treated as the extremum of the auxiliary function (f a variable. (2) Lagrange’s method does not specify whether the extreme value found out is a maximum value or a minimum value. It is decided from the physical consideration of the problem.

WORKED EXAMPLE 4(c) Example 4.1 Examine f (x, y) = x3 f ( x, y )

3xy2

15x2

15y2

72x for extreme values.

x 3 3 xy 2 15 x 2 15 y 2 72 x

fx

3 x 2 3 y 2 30 x 72

fy

6 xy 30 y

f xx

6 x 30; f xy

6 y; f y

6 x 30

Chapter 4: Functions of Several Variables The stationary points are given by f x i.e.,

3( x 2

0 and f y

y 2 10 x 24)

0

6 y ( x 5)

0

and From (2), x 5 or y 0 When x = 5, from (1), we get y2 When y 0, from (1), we get x2

I – 4.53

0 (1) (2)

1 = 0; y = 1 10x 24 = 0 x

The stationary points are (5, 1), (5, 1), (4, 0) and (6, 0) At the point (5, 1), A fxx 0; B fxy 6; C fyy 0 Though AC B2 0, A 0 Nothing can be said about the maxima or minima of f (x, y) at (5, 1). At the point (4, 0), A = 6, B 0, C 6 B2 = 36

AC

0

and A

0

f (x, y) is maximum at (4, 0) and maximum value of f (x, y) At point (6, 0), A 6, B 0, C 6 B2 = 36

AC

0 and A

112.

0.

f (x, y) is minimum at (6, 0) and the minimum value of f (x, y) = 108. Example 4.2 Examine the function f (x, y)

x4 y 2

x3 y (24

x

f ( x, y )

12x 3 y 2

fx

36 x 2 y 2

fy

24 x 3 y

2 x 4 y 3x3 y 2

f xx

72 xy 2

12 x 2 y 2 6 xy 3

f xy

72 x 2 y

8 x3 y 9 x 2 y 2

f yy

24 x 3

2x

3y) = 0

y) for extreme values.

x3 y 3

4 x3 y 2 3x 2 y 3

2 x 4 6 x3 y

The stationary points are given by fx 0; fy i.e., x2 y2 (36 4x 3y) 0 and

x3y2(12

0 (1) (2)

Solving (1) and (2), the stationary points are (0, 0), (0, 8), (0, 12), (12, 0), (9, 0) and (6, 4). AC B2 0. Further investigation is required to investigate the extremum at these points. At the point (6, 4), A 2304, B 1728, C 2592 and AC B2 0. Since AC – B2 > 0 and A < 0, f (x, y) has a maximum at the point (6, 4). Maximum value of f (x, y) 6912.

Part I: Mathematics I

I – 4.54 Example 4.3 4xy – 2y2.

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

x4

x4

y4 2x2

y 4 2 x 2 4 xy 2 y 2 .

fx

4( x 3 x y )

fy

4( y 3

f xx

4(3 x 2 1); f xy

x y) 4(3 y 2 1)

4; f yy

The possible extreme points are given by fx

0 and f y

0

i.e.,

x3 x y

0

(1)

and

y3

0

(2)

Adding (1) and (2),

x3

Using (3) in (1):

y3

0

y

x3 2 x

0

x ( x 2 2)

i.e.,

x y

x

0

0,

2,

and the corresponding values of y are 0,

(3)

x

2, 2.

The possible extreme points of f (x, y) are (0, 0), ( At the point (0, 0), A

B

4, B

4 and C

4

AC

B2

2

2,

2 ) and (

0

The nature of f (x, y) is undecided at (0, 0). At the points ( 4, C 20 AC B2 0 f (x, y) is minimum at the points (

Example 4.4

2,

x2

2 ), A

2,

2 ), and minimum value of f (x, y)

Examine the extrema of f ( x, y ) f ( x, y )

2 , 2 ).

xy y 2

fx

2x y

1 x2

fy

x 2y

1 y2

x2

xy y 2

1 x

1 y

1 x

1 . y

20,

8.

Chapter 4: Functions of Several Variables f xx

2 ; f xy x3

2

1; f yy

0 and fy

The possible extreme points are given by fx

2

I – 4.55 2 y3

0.

i.e.,

2x y

1 x2

0

(1)

and

x 2y

1 y2

0

(2)

(1) – (2) gives

x y

1 y2

1 x2

0

i.e.,

x y

x2 y 2 x2 y 2

0

x y)

0

( x y )( x 2 y 2

i.e.,

x Using (3) in (1), 3x3 – 1

0 1 3

x

At the point

1 3

1 3

1 , 3

1 3

y

1 3

,A

8, B

1 3

1 3

1 , 3

1 and C B2

AC

f (x, y) is minimum at

(3)

y

8

0

1 3

4

and minimum value of f ( x, y )

Example 4.5 + x4 at the origin

f (x, y) f ( x, y ) fx fy f xx f xy

x 2 2 xy y 2

x3

2 x 2 y 3x 2 4 x3 2x 2 y 3y2 2 6 x 12 x 2 2;

f yy

2 6y

y3

x4 .

x2

2xy

33 .

y2

x3

y3

Part I: Mathematics I

I – 4.56

fx

0 and f y

0.

(0, 0) is a stationary point of f (x, y). At the origin, A 2, B –2 and C 2 AC

B2

f (x, y) at the origin. Let us consider the values of f (x, y) at three points close to (0, 0), namely at (h, 0), (0, k) and (h, h) which lie on the x-axis, the y-axis and the line y x respectively. f (h, 0)

h 2 h3 h 4 0.

f (0, k )

k2 k3

f (h, h)

h

4

k 2 (1 k ) 0, when 0 k

1

0

Thus f (x, y) > f (0, 0) in the neighbourhood of (0, 0). (0, 0) is a minimum point of f (x, y) and minimum value of f (x, y) Example 4.6

0.

Find the maximum and minimum values of f ( x, y ) f ( x, y )

sin x sin y sin ( x sin x sin y sin ( x

y ); 0

x, y

fx

cos x sin y sin ( x

y ) sin x sin y cos ( x

y)

fy

sin x cos y sin ( x

y ) sin x sin y cos ( x

y)

i.e.,

fx

sin y sin (2 x

and

fy

sin x sin ( x 2 y )

f xx

2 sin y cos (2 x

f xy

sin y cos (2 x

f yy

sin (2 x 2 y ) 2 sin x cos ( x 2 y )

y)

y)

y) y ) cos y sin (2 x

For maximum or minimum values of f (x, y), fx 0 and fy i.e., sin y sin (2x y) 0 and sin x · sin (x 2y) 0 i.e.,

1 [cos 2 x cos (2 x 2 y )] 2

0

and

.

y)

0

1 [cos 2 y cos (2 x 2 y )] 2

0

i.e.,

cos 2 x cos (2 x 2 y )

0

(1)

and

cos 2 y cos (2 x 2 y )

0

(2)

From (1) and (2), cos 2x cos 2y. Hence x y Using (3) in (1), cos 2x – cos 4x 0 i.e., 2 sin x sin 3x 0 sin x 0 or sin 3x 0 2 x 0, and 3x 0, , 2 i.e., x 0, , 3 3

(3)

Chapter 4: Functions of Several Variables 2 . 3 3 Thus the maxima and minima of f (x, y) are given by (0, 0)

The admissible values of x are 0,

At the point (0, 0), A AC B2

B

C

I – 4.57

,

and 2 , 2 , 3 3 3 3

0

Thus the extremum of f (x, y) at (0, 0) is undecided. At the point

,

3 3

,A

3 and C 2

3, B

As AC – B2 > 0 and A < 0, f (x, y) is maximum at Maximum value of f ( x, y )

At the point

3 3 3 2 2 2

2 2 , ,A 3 3

3, B

3 3 . 8

3 and C 2

3 and AC B 2

2

Example 4.7

3 . 0 4

3

. , 3 3

As AC – B > 0 and A > 0, f (x, y) is maximum at Minimum value of f ( x, y )

3 and AC B 2

2 2 , 3 3

3

3 4

0.

.

3 3 . 8

Identify the saddle point and the extremum points of f ( x, y )

x4

y 4 2x2 2 y 2 .

f ( x, y )

x4

y 4 2 x2 2 y 2

fx

4 x 3 4 x; f y

fxx

12 x 2 4; f x y

4 y 4 y3 f yy

0;

The stationary points of f (x, y) are given by fx

0 and fy

i.e.,

4( x 3 x)

0 and 4( y y 3 )

0

i.e.,

4 x( x 2 1)

0 and 4 y (1 y 2 )

0

x 0 or 1 and y 0 or 1. At the points (0, 0), ( 1, 1), AC The points (0, 0), (1, 1), (1, function f (x, y). At the point ( 1, 0), AC B2 f (x, y) attains its minimum at ( At the point (0, 1), AC B2

4 12 y 2 . 0

B2 0 1), ( 1, 1) and ( 1, 1) are saddle points of the 0 and A 0 1, 0) and the minimum value is 1. 0 and A 0

Part I: Mathematics I

I – 4.58

f (x, y) attains its maximum at (0, 1) and the maximum value is Example 4.8

Find the minimum value of x2 x y z

1.

y2 z2, when x y z 3a. x2 + y2 + z2, subject to the condition 3a

(1) x and y.

From (1), z 3a x y. Using this in the given function, we get f ( x, y )

x2

y 2 (3a x y ) 2

fx

2 x 2(3a x y )

fy

2 y 2(3a x

f xx

4; f xy

y)

2; f yy

4

The possible extreme points are given by fx 0 and fy

0.

i.e.,

2x + y

3a

(2)

and

x + 2y

3a

(3)

Solving (2) and (3), we get the only extreme point as (a, a) At the point (a, a), AC – B2 > 0 and A > 0 f (x, y) is minimum at (a, a) and the minimum value of f (x, y)

3a2.

Example 4.9 Show that, if the perimeter of a triangle is constant, its area is maximum when it is equilateral. Let the sides of the triangle be a, b, c. Given that

a+b+c

constant

2k, say Area of the triangle is given by A where

s

s ( s a ) ( s b) ( s c )

(1) (2)

a b c 2

Using (1) in (2), A

k ( k a ) ( k b) ( k c )

A is a function of three variables a, b , c Again using (1) in (3), we get A

k ( k a ) ( k b) ( a b k )

(3)

Chapter 4: Functions of Several Variables A is maximum or minimum, when f (a, b) maximum or minimum.

A2 k

k

a) k

fa

(k b){(k a ) 1 (a b k ) ( 1)}

fb

( k b) ( 2k 2a b) (k a ){(k b) 1 (a b k ) ( 1)}

I – 4.59 b)(a

b

k) is

(k a ) (2k a 2b) 2 (k b); f ab 3k 2a 2b;

f aa f ab

2 (k a)

The possible extreme points of f (a, b) are given by fa 0 and fb 0 i.e., (k b) (2k 2a b) 0 and (k a)(2k a 2b) 0 b k or 2a + b 2k and a k or a + 2b 2k Thus the possible extreme points are given by (i) a k, b k; (ii) b k, a + 2b 2k; (iii) a k, 2a + b a + 2b 2k.

2k and (iv) 2a + b

2k,

(i) gives a k, b k and hence c 0. (ii) gives a 0, b k and hence c k. (iii) gives a k, b 0 and hence c k. All these lead to meaningless results. Solving 2a + b 2k and a + 2b 2k, we get a At the point A

f aa

AC

B2

2k and b 3

2k 3

2k 2k , , 3 3 2k ;B 3

0 and A

f ab

k ;C 3

2k 3

f bb

0

f (a, b) is maximum at

2k 2k , 3 3

Hence the area of the triangle is maximum when a When a

2k ,b 3

2k ;c 3

2k ( a b)

2k and b 3

2k 3 .

2k 3

Thus the area of the triangle is maximum, when a triangle is equilateral.

b

c

2k , i.e., when the 3

Part I: Mathematics I

I – 4.60

Example 4.10 In a triangle ABC triangle ABC, A B C

A cos B cos C. In

Using this condition, we express the given function as a function of A and B Thus cos A cos B cos C cos A cos B cos { ( A B)} cos A cos B cos ( A B) Let

f ( A, B ) fA

cos A cos B cos ( A B ) cos B { sin A cos ( A B ) cos A sin( A B) }

fB

cos B sin (2 A B) cos A{ sin B cos ( A B) cos B sin( A B)}

f AA

cos A sin ( A 2 B) 2 cos B cos (2 A B)

f AB

cos B cos (2 A B) sin B sin (2 A B)

f BB

cos (2 A 2 B) 2 cos A cos ( A 2 B)

The possible extreme points are given by fA

0 and f B

0

cos B sin (2 A B)

i.e.,

(1)

0

and cos A sin (A 2B) 0 Thus the possible values of A and B are given by (i) cos B 0, cos A 0; (ii) cos B 0, sin (A + 2B) 0; (iii) sin (2A + B) 0, cos A 0 and (iv) sin (2A + B) 0, sin (A + 2B) 0 i.e., (i) A

2

,B

(iv) 2 A B

; (ii) B ,A 2 2 , A 2B or

0 or , (iii) A

2

,B

0 or

and

A

,B 3 3 A and B lead to meaningless results. Hence A At this point Also A

3

,B

give the extreme point.

3

, ,A 3 3

f AA

1; B

1 ; f BB 2

f AB

0

f (A, B) is maximum at A cos

B

3

3 cos

and the maximum value

3

cos

2 3

1 . 8

1 and AC

B 2 0.

Chapter 4: Functions of Several Variables

I – 4.61

Example 4.11 Find the maximum value of xm yn zp, when x + y + z a. x y z – a. Let f xm yn zp and Using the Lagrange multiple , the auxiliary function is g (f ). ) are given by gx 0, gy 0, gz 0 and g This stationary points of g ( f i.e.,

n

0

p

0

(1)

nx m y n 1 z p

0

(2)

px m y n z p

0

(3)

0

(4)

mx

m 1

y z

1

x y z a From (1), (2) and (3), we have mx m 1 y n z p m n x y

nx m y n 1 z p px m y n z p 1 . p m n p i.e., z x y z m n p , by (4) a Maximum value of f occurs, when x

am , y m n p

an ,z m n p

Thus maximum value of f

ap m n p

a m n p mm nn p p (m n p) m n p

Example 4.12 A rectangular box, open at the top, is to have a volume of 32 c.c. Find the dimensions of the box, that requires the least material for its construction. Let, x, y, z be the length, breadth and height of the respectively. The material for the construction of the box is least, when the area of surface of the box is least. Hence we have to minimise S

xy 2 yz 2 zx,

subject to the condition that the volume of the box, i.e., xyz 32. Here f xy + 2yz + 2zx; xyz 32. The auxiliary function is g f + , where is the Lagrange multiplier. The stationary points of g are given by gx 0, gy 0, gz 0 and g 0 i.e.,

y 2z

yz

0

(1)

x 2x

zx

0

(2)

2x 2 y

xy

0

(3)

xyz 32

0

(4)

Part I: Mathematics I

I – 4.62 From (1), (2) and (3), we have

1 z

2 y

(5)

1 2 z x 2 2 y x

(6) (7)

Solving (5), (6) and (7), we get 4

x

4

, y

2

and z

Using these values in (4), we get 32 3

32

i.e.,

0 1

x

4, y

4, z

2.

Thus the dimensions of the box and 4 cm; 4 cm and 2 cm.

Example 4.13 Find the volume of the greatest rectangular parallelopiped inscribed 2 y2 z2 in the ellipsoid whose equation is x 1. a 2 b2 c2 Let 2x, 2y, 2z be the dimensions of the required rectangular parallelopiped. By symmetry, the centre of the parallelopiped coincides with that of the ellipsoid, namely, the origin and its faces are parallel to the co-ordinate planes. Also one of the vertices of the parallelopiped has co-ordinates (x, y, z), which satisfy the equation of the ellipsoid. 2 8xyz, subject to the condition x a2

Thus, we have to maximise V Here f

x2 a2

8xyz and

y2 b2

z2 c2

1

z2 1 c2

The auxiliary function is g f + points of g are given by

i.e.,

y2 b2

, where is the Lagrange multiplier. The stationary

gx

0, g y

0, g z

0 and g

8 yz

2 x a2

0

(1)

8 zx

2 y b2

0

(2)

0

Chapter 4: Functions of Several Variables 8 xy x2 a2

2 z c2

0

(3)

z2 c2

1

(4)

y2 b2

2 Multiplying (1) by x, 2 x 2 a 2 Similarly 2 y b2

Thus

x2 a2

8 xyz

2 z2 c2

y2 b2

k say

1

k

a

x

8 xyz from (2) and (3)

z2 c2

Using in (4), 3k

3

I – 4.63

1 3

b

, y

3

c

and z

3

8abc

Maximum volume

.

3 3

Example 4.14 Find the shortest and the longest distances from the point (1, 2, 1) to the sphere x2 + y2 + z2 24. Let (x, y, z x, y, z) from (1, 2, 1) is given by d

x 1

2

y 2

2

2

z 1 . d or equivalently

d2

x 1

subject to the constant x2 +y2 +z2 – 24 Here

f

x 1 x2

2

y2

y 2

2

2

y 2

2

2

z 1 ,

0 2

z 1 and

z 2 24

The auxiliary function is g f + , where is the Lagrange multiplier. The stationary points of g are given by gx 0, gy 0, gz = 0 and g 0. i.e.,

2 x 1

2 x

0

(1)

2 y 2

2 y

0

(2)

2 z 1

2 z

0

(3)

24

(4)

x2

y2

z2

Part I: Mathematics I

I – 4.64 From (1), (2) and (3), we get 1

x

2

, y

1

1

1

,z

1

Using these values in (4), we get 6 1

2

1 or 2 When When

1 4

2

24 i.e., 1 3 2

1 , the point on the sphere is (2, 4, 2) 2 3 , the point on the sphere is ( 2, 4, 2) 2

When the point is (2, 4, 2), d

1

When the point is ( 2, 4, 2), d

2

2 3

Shortest and longest distances are

2

2

1 6

2

2

6 32

3 6

6 and 3 6 respectively.

Example 4.15 Find the point on the curve of intersection of the surfaces z and x + y + z 1 which is nearest to the origin. Let (x, y, z) be the required point. It lies on both the given surfaces. xy – z + 5 0 and x + y + z 1 x2

x, y, z) from the origin is given by d We have to minimize d or equivalently d 2 x2 y 2 z 2 , subject to the constraints xy – z + 5

0 and x + y + z – 1

y2

xy + 5

z2

0.

Note f (x, y, z) subject to the conditions 1 (x, y, z) 0 and 2 (x, y, z) 0, we form the auxiliary function g f + 1 1 + 2 2, where 1 and 2 are two Lagrange multipliers. The stationary points of g are given by gx 0, gy 0, gz 0, g 1 0 and g 2 0. In this problem, f x2 + y2 + z2, 1 xy – z + 5 and 2 x + y + z – 1. The auxiliary function is g f + 2 2, where 1, 2 are Lagrange multipliers. 1 1 The stationary points of g are given by 2x

1

y

2

0

(1)

2y

1

x

2

0

(2)

1

2

0

(3)

2z

Chapter 4: Functions of Several Variables

Eliminating

1

,

2

xy z 5

0

(4)

x y z 1

0

(5)

from (1), (2), (3), we have 2x 2y 2z

y 1 x 1 1 1

0

zx

0

y

0

y z 1

0

x y or x y z 1 y in (4) and (5), we have

0

x x 1

i.e.,

x

i.e.,

2

y y 2 y

2

x

i.e.,

Using x

I – 4.65

y

x y z x y x

z x2

(6)

5

z 1 2x

and

(7)

2

From (6) and (7), x 2x 4 0, which gives only imaginary values for x. Hence x y z 1 0

(8)

Solving (5) and (8), we get x

y

0

(9)

and

z

1

(10)

Using (10) in (4), we get Solving (9) and (11), we get x

xy

4 2 and y

(11) 2.

The required points are (2, 2, 1) and ( 2, 2 ,1) and the shortest distance is 3.

EXERCISE 4(c) Part A (Short Answer Questions) 2. State the conditions for the stationary point (a, b) of f (x, y) to be (i) a maximum point (ii) a minimum point and (iii) a saddle point. f (x, y). f (x, y, z) and (x, y, z), when we extremise f (x, y, z) subject to the condition (x, y, z) 0. 5. Find the minimum point of f (x, y) x2 + y2 + 6x + 12. 6. Find the stationary point of f (x, y) x2 – xy + y2 – 2x + y. 7. Find the stationary point of f (x, y) 4x2 + 6xy + 9y2 – 8x – 24y + 4.

Part I: Mathematics I

I – 4.66

8. Find the possible extreme point of f ( x, y )

x2

y2

2 x

2 . y

9. Find the nature of the stationary point (1, 1) of the function f (x, y), if fxx 6xy3, fxy 9x2 y2 and fyy 6x3 y. 10. Given fxx 6x, fxy 0, fyy 6y the function f (x, y). Part B Examine the following functions for extreme values: 11. x3 +y3 3axy 12. x3 + y3 12x 3y + 20 13. x4 + 2x2 y x2 + 3y2 14. x3 y 3x2 2y2 4y 3 15. x4 + x2 y + y2 at the origin 16. x3 y2 (a – x –y) 17. x3 y2 (12 – 3x 4y) 18. xy 27

1 x

1 y

19. sin x sin y sin ( x y ), 0 x, y 20. 21. 22. 23.

. 2 Identify the saddle points and extreme points of the function xy (3x + 2y + 1). Find the minimum value of x2 + y2 + z2, when (i) xyz a3 and (ii) xy + yz + zx 3a2. Find the minimum value of x2 y2 + z2, when ax by cz p. Show that the minimum value of (a3 x2 b3 y2 c3 z2), when 1 1 1 1 , is k 2 (a b c)3 . x y z k

the second and cube of the third may be minimum. 25. The temperature at any point (x, y, z) in space is given by T k x y z2, where k is a constant. Find the highest temperature on the surface of the sphere x2 + y2 + z2 a2. 26. Find the dimensions of a rectangular box, without top, of maximum capacity and surface area 432 square meters. 27. Show that, of all rectangular parallelopipeds of given volume, the cube has the least surface. 28. Show that, of all rectangular parallelopipeds with given surface area, the cube has the greatest volume. 29. Prove that the rectangular solid of maximum volume which can be inscribed in a sphere is a cube. 30. Find the points on the surface z2 xy 1 whose distance from the origin is minimum. 31. If the equation 5x2 6xy 5y2 8 represents an ellipse with centre at the

Chapter 4: Functions of Several Variables

32. 33. 34.

35.

I – 4.67

(Hint: The longest distance of a point on the ellipse from its centre gives the length of the semi-major axis. The shortest distance of a point on the ellipse from its centre gives the length of the semi-minor axis). Find the point on the surface z x2 y2, that is nearest to the point (3, 6, 4). Find the minimum distance from the point (3, 4, 15) to the cone x2 y2 4z2. Find the points on the ellipse obtained as the curve of intersection of the surfaces x + y 1 and x2 2y2 z2 1, which are nearest to and farthest from the origin. Find the greatest and least values of z, where (x, y, z) lies on the ellipse formed by the intersection of the plane x y z 1 and the ellipsoid 16x2 4y2 z2 16.

ANSWERS

Exercise 4(a) (2) (3) (4) (5) (6)

du cos (xy2) (y2dx 2xy dy) du xy 1 · yx (y x log y) dx + xy yx 1 (x du y (1 log xy) dx x(1 log xy) dy du (y log a) axy dx (x log a) axy dx 8 a5t6 (4t 7);

(7) e

a2 t 2

y log x) dy

sin 3 t 3 a 2 t 2 sin 2 t cos t t sin 3 t / a 2 t 2 e t – sin t)/(e

(8) (cos t

t

sin t

cos t)

x 2 xy 2 y ; x 2 4 xy y 2

(11)

3 x cos ( x 2 2

x(xy 4y2 – 2x2)/(x 2y); 4.984 0.006 cm3; 0.004 cm2 4(a b c)k 2 z 0; (36) u

(14) (16) (18) (20)

3.875 0.0043 2 1.5

2

(9)

2

(12) (15) (17) (19)

2

2 2

2

(38)

z u2 2

(40) (41)

z (i)

z

(37)

0 u z*

(39)

u

z

y2 )

0 0;

0

y ( y x log y ) x ( x y log x)

(iii) y tan x log sin y log cos x x cot y

(ii) y x (iv)

log cot y y tan x log sec x x sec y cosec y

Part I: Mathematics I

I – 4.68 x y x(1 log x)

(v)

(42) 2a3xy/(ax – y2)3 (47) 5%

(43) 2a2xy (3a4 (50) 2.3%

x2y2)/(a2x – y3)3

(55) 5 3 . 324

Exercise 4(b) (4) 1 ( x y ) (11) u

( x y ) 2 ... 2

(5) ( x y )

1; x2 y 2

(16) y xy e

(17)

(15) 2 tan

y2 2

(18) y xy

1 y

y 3 ... 6

1 ( x 1)

2

1

1 ( x y )3 ... 3!

y

4

( x 1) 2 2

1 2 1 2 x y xy 2 2

( x 1) y

4

2

1 y 2

4

1 3 y 3

1 1 1 1 ( x 1) ( y 1) ( x 1) 2 ( y 1) 2 4 2 2 4 4 (20) 10 – 4 (x 1) 4(y 2) – 2(x 1)2 + 2(x 1) (y

(19)

(21)

9

(22) 1

3 (x (y

2) – 7(y

1)

(x

1)

1) (y – 1)

(23) e 1 ( x 1) ( y 1)

2(x

2) (y

2

1 ( x 1) 2 2( x 1) ( y 1) ( y 1) 2 2

)

(ii) 4xy (28) r (30) x(y (31) (x

w) + z – 2u

1 y) (y

2

(32) u

(33) u tan

1

z) (z

x)

1)2

(x (x

1)2 (y 2) (y

…

3 3 (xx 1) 2 ( y 2) ( x 1) ( y 2) 2 2 2 (27) (i) 4(u2

1) – 2(y

2)

1 ( y 2) 3 6

1 ( x 1)3 6

2) 1)2

Chapter 4: Functions of Several Variables 2 (34) f 1

f 2 2 f3

(36)

1 x {tan 1 (ax) /( x 2 a 2 )} a 2a 3

(37)

( 1) n n ! (m 1) n 1

(38)

1 2

(39) tan

I – 4.69

a2

e 1

1 ; a 2

(40) log (1

m)

Exercise 4(c) (5) ( 3, 0).

(6) (1, 0).

(7)

(8) (1, 1).

(9) Saddle point.

0,

4 3

(10) Minimum point.

(11) Maximum at (a, a) if a < 0 and minimum at (a, a) if a > 0. (12) Minimum at (2, 1) and maximum at ( 2, 1). (13) Minimum at

3 , 2

1 4

(14) Maximum at (0, 1).

(15) Minimum at (0, 0).

(16) Maximum at a , a 2 3

(17) maximum at (2, 1).

(19) Maximum at

, 3 3

(20) Saddle point are (0, 0), (21) 3a2; 3a2.

and minimum at 3 1 , 0 and 0, 3 (22) a2

4 (25) ka . 8

,

(18) Minimum at (3, 3). . 3

1 ; maximum at 2

p2 b2 c2

(24) 4, 8, 12.

(26) 12, 12 and 6 metres.

(30) (0, 0, 1) and (0, 0, 1). (31) 4, 2. (33) 5 5 .

1 , 9

(34)

1 2 , , 0 ; (1, 0, 0). 3 3

(32) (1, 2, 5). (35) 8 ; 3

8 . 7

1 . 6

Chapter

5

Multiple Integrals 5.1

INTRODUCTION

When a function f(x) is integrated with respect to x between the limits a and b, we get b

f (x)d x . a

If the integrand is a funtion f (x,y) and if it is integrated with respect to x and y repeatedly between the limits x0 and x1 (for x) and between the limits y0 and y1 (for y), y1 x1

we get a double integral that is denoted by the symbol

f (x, y )dx d y . y0 x0

Extending the concept of double integral one step further, we get the triple integral z1

y1 x1

f (x, y, z ) dx dy dz z0 y0 x0

5.2

EVALUATION OF DOUBLE AND TRIPLE INTEGRALS y1 x1

To evaluate

f (x, y ) dx dy,

f (x, y) with respect to x partially,

y0 x0

i.e. treating y as a constant temporarily, between x0 and x1. The resulting function got after the inner integration and substitution of limits will be a function of y. Then we integrate this function of y with respect to y between the limits y0 and y1 as usual. The order in which the integrations are performed in the double integral is illustrated in Fig. 5.1.

Part I: Mathematics I

I – 5.2

y1

x1

y0

x0

f (x, y) dx

dy

Fig. 5.1

Note

Since the resulting function got after evaluating the inner integral is to be a function of y, the limits x0 and x1 may be either constants or functions of y. The order in which the integrations are performed in a triple integral is illustrated in Fig. 5.2.

z1

y1

x1 f (x, y, z) dx

z0

dy

dz

x0

y0

Fig. 5.2

x, we treat y and z as constants temporarily. The limits x0 and x1 may be constants or functions of y and z, so that the resulting function got after the innermost integration may be a function of y and z. Then we perform the middle integration with respect to y, treating z as a constant temporarily. The limits y0 and y1 may be constants or functions of z, so that the resulting function got after the middle integration may be a function of z only. Finally we perform the outermost integration with respect to z between the constant limits z0 and z1. y1

Note

y1

f ( x, y )dx dy is also denoted as y0

z1

x1

Sometimes x0

y0

y1 x1

z1

f (x, y, z )dx dy dz is also denoted as z0 y0 x0

y1

dz z0

f ( x, y )dx and x0

x1

dy y0

x1

dy

f (x, y, z ) dx . If these x0

notations are used to denote the double and triple integrals, the integrations are performed from right to left in order.

5.3

REGION OF INTEGRATION ( y)

d

2

c

1 ( y)

f (x, y ) dx dy. As stated above x varies from

Consider the double integral to 2(y) and y varies from c to d. i.e. 1(y x (y) and c y d. 2

(y)

1

Chapter 5: Multiple Integrals

I – 5.3

These inequalities determine a region in the xy-plane, whose boundaries are the curves x = 1(y), x = 2(y) and the lines y = c, y = d and which is shown in Fig. 5.3. This region ABCD is known as the region of integration of the above double integral. y

D

x=

C

y=d

x=

1 (y)

A

2 (y)

B

y=c

x

o Fig. 5.3 (x )

b

2

a

1 (x )

f (x, y )dy dx, the region of integration

Similarly, for the double integral ABCD, whose boundaries are the curves y = b, is shown in Fig. 5.4.

(x), y =

1

2

(x) and the lines x = a, x =

y y=

2 (x)

x=a

x=b

y=

o

1 (x)

x

Fig. 5.4 z2

2 (z)

For the triple integral

2 (y, z)

f (x, y, z) dx dy dz, the inequalities z1

1 (z)

(y, z

1

x

1 (y, z)

(y, z); 1(z y (z); z1 z z2 hold good. These inequalities determine a 2 2 domain in space whose boundaries are the surfaces x = 1(y, z), x = 2(y, z), y = (z), y = 2(z), z = z1 and z = z2. This domain is called the domain of integration of 1 the above triple integral. WORKED EXAMPLE 5(a) 2

1

Example 5.1 Verify that 1 2

(x 2 0

2

y 2 ) dx dy

0

1

L.S. = 1

1

(x 2

y 2 ) dx dy

(x 2 0

1

y 2 ) dy dx .

Part I: Mathematics I

I – 5.4 2

1

Note

x 1

x3 3

2

y x

y is treated a constant during inner integration with respect to x. 2

1

1 3

1

2

0

y 3

2

y dy

(x 2

R.S.

y 3

3

2

8 3

1

y2 ) d y d x

1

1

y

y3 3

2

x y 0

Note

dy x 0

2

dx y 1

x is treated a constant during inner integration with respect to y. 1

x

x3 3

7 dx 3

2

0

7 x 3

1

0

8 3

Thus the two double integrals are equal.

Note

From the above problem we note the following fact: If the limits of integration in a double integral are constants, then the order of integration is immaterial, provided the relevant limits are taken for the concerned variable and the integrand is continuous in the region of integration. This result holds good for a triple integral also. a

2

r 4 sin

Example 5.2 Evaluate 0

0

a

2

The given integral

dr d d .

0

d

r 4 sin d r

d

0

0

0

2

r5 5

d 0

0

a5 5

2

a5 5

2

d

a

sin d 0

sin d

0

0

( cos ) 0 d 0

2 5 a 5

2

4 5 a 5

d 0

Chapter 5: Multiple Integrals y2

1

1

Example 5.3 Evaluate 0

y2

1

1

The given integral 0

dx dy x2 y 2

1

0

1 dx dy y 2 ) + x2

(1

0

x

1

1

0

1

4

1

x

1

tan

y2

1

I – 5.5

y2

1

y2

dy x

0

dy y2

1

0

1

4

4

log y

y2

1

0

g (1 log 1

x

0

x

Example 5.4 Evaluate

2)

y ) dx dy.

xy (x

Since the limits for the inner integral are functions of x, the variable of inner integration should be y. Effecting this change, the given integral I becomes 1

x

0

x

1

y2 x 2

xy (x

I

1

0

x4 8 1 8

x3 2

y

y3 x 3

2

0

y ) dy dx

dx y

2 21

1 6

x

x4 2

1 5/ 2 x 3 2 7/ 2 x 21

x

x5 6 3 56

1

0

x4 3

dx

Part I: Mathematics I

I – 5.6

1 z

1

Example 5.5 Evaluate

1 y z

xyz dx dy dz . 0

0 z

1 1

0

dy dz 0

yz (1 z

z )2

yz {(1 0

y2 z) 2 2

z (1 0 1

1 2

z )y

2 (1

y 2 } dy dz

0

1

1 2

z ) 2 dy dz

y

0

0 1 1

1 2

z

z

1 1

1 2

y

1

x2 yz 2

The given integral 0

0

1 z (1 2

0

1 1 2 2

2 z (1 2 z (1 3

z )4

2 3

y3 z) 3 z )4

y4 z 4 1 z (1 4

y

1

y

0

z

dz z ) 4 dz

1

1 4

z (1

z ) 4 dz

z )} (1

z ) 4 dz

0

1

1 24

{1

(1

0

1 (1 z )5 24 5 1 1 24 5

6

1 6

log 2 x x

ex 0

0

1 720

y

Example 5.6 Evaluate 0

1

z )6

(1

y

z

dx dy dz .

0

Since the upper limit for the innermost integration is a function of x, y, the corresponding variable of integration should be z. Since the upper limit for the middle integration is a function of x, the corresponding variable of integration should be y. The variable of integration for the outermost integration is then x. Effecting these changes, the given triple integral I becomes, log 2 x x

y

ex

I 0

0

log 2

z

dz dy dx

dy e x

y

(e z ) zz

0 x

dx 0

x 0

y

0

log 2

x

(e 2 x

dx 0

y

0

2y

ex

y

) dy

Chapter 5: Multiple Integrals log 2

dx e

2x

0 log 2

0

1 4x e 2

1 4x e 8

e2 y 2

x

e e

3 2x e 2

3 2x e 4

I – 5.7

y

x

y

0

y

e x dx log 2

ex 0

5 8 Example 5.7 Evaluate

xy dx dy , where R is the region bounded by the line R

x + 2y We draw a rough sketch of the boundaries of R and identify R. The boundaries of R are the lines x = 0, y = 0 and the segment of the line x y 1 2 1 Now R is the region as shown in Fig. 5.5. y

B

C y x + =1 2 1 Q (x1, y)

(x0, y) P R O

A

x

Fig. 5.5

Since the limits of the variables of integration are not given in the problem and to Let us integrate with respect to x integral I becomes I=

y. Then the xy dx dy R

When we perform the inner integration with respect to x, we have to treat y as a x. Geometrically, treating y = constant is equivalent to drawing a line parallel to the x-axis arbitrarily lying within the region of integration R Finding the limits for x (while y of the x co-ordinate of any point on the line PQ. We assume that the y co-ordinates x0, y) of all points on PQ are y each (since y is constant on PQ) and P and Q x1, y).

Part I: Mathematics I

I – 5.8

Thus x varies from x0 to x1. Wherever the line PQ has been drawn, the left end P lies on the y-axis and hence x0 = 0 and the right end Q lies on the line x + 2y = 2, and hence x1 + 2y = 2 i.e. x1 y. y. When we Thus the limits for the variable x y. PQ, so that the region R is fully covered. To sweep the entire area of the region R, PQ has to start from the position OA where y = 0, move parallel to itself and go up to the position BC where y = 1. Thus the limits for y are 0 and 1. 2y

1 2

I

xy dx dy 0

0

1

x2 2

y 0 1

0

2

2y

dy 0

y (2 2

2 y ) 2 dy

1

y ) 2 dy

y (1

2 0

2

y2 2

2

y3 3

y4 4

1

0

1 6

5.3.1 Aliter Let us integrate with respect to y Then I

x.

xy dy dx R

y, we draw a line parallal to the y-axis (x = constant) in the region of integration and note the variation of y on this line y B Q (x, y1)

y x + =1 2 1

C

R o

P

(x, y0) Fig. 5.6

A

x

Chapter 5: Multiple Integrals P(x, y0) lies on the x-axis.

I – 5.9

y0 = 0

Q(x, y1) lies on the line x + 2y = 2.

1 (2 x) 2

y1

1 (2 x) . 2 To cover the region of integration OAB, the line PQ has to vary from OB (x = 0) to AC (x = 2) The limits for x are 0 and 2. i.e., the limits for y are 0 and

2

1 (2 x) 2

0

0

xy dy dx

I 2

0

1 8

y2 x 2

x (2 x) 2 dx 0

4

x3 3

2

x4 4

0

y

e R

dx 0

2

1 x2 4 8 2 1 6 Example 5.8 Evaluate

1 (2 x) 2

y

dx dy , by choosing the order of integration suitably,

given that R is the region bounded by the lines x = 0, x = y and y Q (x, )

y

y= (0, y) A

B (y, y)

x=0

R

y=x

P (x, x) x

o Fig. 5.7

Let

y

e

I

y

R

dx dy

Suppose we wish to integrate with respect to y Then

y

e

I 0

x

y

dy dx

Part I: Mathematics I

I – 5.10

We note that the choice of order of integration is wrong, as the inner integration cannot be performed. Hence we try to integrate with respect to x y

Then

0

y

0

(x)0y dy

y e

dx dy

y

e 0

y

e

I

y

dy

0 y 0

e

1

Note

From this example, we note that the choice of order of integration sometimes depends on the function to be integrated. Example 5.9 Evaluate

xy dx dy , where R is the region bounded by the parabola R

y2 = x and the lines y = 0 and x + y R is the region OABCDE shown in Fig. 5.8. y

y2 = x

D Q1

C x+y=2 Q2 F

E

P1

O

A

P2

B

x

Fig. 5.8

Suppose we wish to integrate with respect to y line parallel to y-axis (x = constant). We note that such a line does not intersect the region of integration in the same fashion throughout. If the line is drawn in the region OADE, the upper end of the line will lie on the parabola y2 = x; on the other hand, if it is drawn in the region ABCD, the upper end of the line will lie on the line x + y = 2. Hence in order to cover the entire region R, it should be divided into two, namely, OADE and ABCD and the line P1 Q1 should move from the y-axis to AD and the line P2 Q2 should move from AD to BF. Accordingly, the given integral I is given by 1

x

2

2 x

1

0

xy dy dx

I 0

0

xy dy dx

Chapter 5: Multiple Integrals

I – 5.11

the co-ordinates of D are (1, 1) and so the equation of AD is x = 1] 1 5 3 I 6 24 8 Note This approach results in splitting the double integral into two and evaluating two double integrals. On the other hand, had we integrated with respect to x problem would have been solved in a simpler way as indicated below. [Refer to Fig. 5.9] [

y

y2 = x D E

(y2,

C

y) P

Q (2 – y, y) x+y=2

O

x

B

Fig. 5.9 y

1 2

xy dx dy

I 0

1 2 1 2

y2 1

y {(2

y)2

y 4 }dy

0 1

(4 y 4 y 2

y3

y 5 ) dy

0

3 8 Note From this example, we note that the choice of order of integration is to be made by considering the region of integration so as to simplify the evaluation. Example 5.10 Evaluate

(x

y

z ) dx dy dz, where V is the volume of the

V

rectangular parallelopiped bounded by x = 0, x = a, y = 0, y = b, z = 0 and z = c. z C

A' Q

B'

O' O

B P

x

A

C'

Fig. 5.10

y

Part I: Mathematics I

I – 5.12

The region of integration is the volume of the parallelopiped shown in Fig. 5.10, in which OA = a, OB = b, OC = c. Since the limits of the variables of integration are not given, we can choose the order of integration arbitrarily. Let us take the given integral I as (x + y + z ) dz dy dx

I V

The innermost integration is to be done with respect to z, treating x and y as constants. Geometrically, x = constant and y = constant jointly represent a line parallel to the z-axis. Hence we draw an arbitrary line PQ in the region of integration and we note the variation of z on this line so as to cover the entire volume. In this problem, z varies from 0 to c. since P (x, y, 0) and Q (x, y, c) Having performed the innermost integration with respect to z between the limits 0 and c, we get a double integral. As P xy-plane, the line PQ covers the entire voulme of the parallelopiped. Hence, the double integral got after The limits for the double integral can be easily seen to be 0 and b (for y) and 0 and a (for x). a

b

c

0

0

0

a

b

0

0

a

b

(x + y + z ) dz dy dx

I

(x + y )z

c (x + y ) 0

0

a

c2 y cx + 2

0 a

dy dx z

0

c2 dy dx 2 y2 c 2

b

dx 0

bc 2 2

b2c dx 2

x2 2

bc (b 2

c)x

abc (a 2 Example 5.11 Evaluate

c

bcx 0

bc

z2 2

b

a

0

c)

dx dy dz , where V is the V

(terra- hedron) formed by the planes x = 0, y = 0, z = 0 and 2x + 3y + 4z = 12.

Chapter 5: Multiple Integrals

I – 5.13

z C

(x, y, z1)

Q O

y

B P (x, y, 0)

A x

Fig. 5.11

Let I = the given integral. dz dy dx

Let I = V

The limits for z, the variable of the innermost integral, are 0 and z1, where (x, y, z1) lies on the plane 2x + 3y + 4z = 12. [Refer to Fig. 5.11] 1 (12 2 x 3 y ) 4

z1

After performing the innermost integration, the resulting double integral is evaluated over the orthogonal projection of the plane ABC on the xy-plane, i.e. over the triangular region OAB in the xy-plane as shown in Fig.5. 12. In the double integral, the limits for y are 0 1 y and (12 2 x) and those for x are 0 and 6. 3 B E

1 (12 3

6

dx

I 0

1 4 1 4

dy

dx 0

3 y)

x y + =1 6 4

0 1 (12 3

6

2x

dz

0

O

(12

2x

dx (12

2 x)y

0 6

2 x) 2 dx

0

1 (6 x ) 3 6 3

A Fig. 5.12

3 y ) dy

0

(12

D

2x)

6

1 24

12

1 (12 4

2x)

6

0

3y2 2

y

1 (12 3

y 0

2x)

x

Part I: Mathematics I

I – 5.14

dz dy dx

Example 5.12 Evaluate

, where V is the region of space 1 x2 y 2 z 2 bounded by the co-ordinate planes and the sphere x2 + y2 + z2 = 1 and contained in the positive octant. V

z C (x, y, z1) Q O

y

B

P (x, y, 0) A x Fig. 5.13

Note In two dimensions, the x and y-axes divide the entire xy-plane into 4 quadrants. The quadrant containing the positive x and the positive y-axes is called the positive quadrant. Similarly in three dimensions the xy, yz and zx-planes divide the entire space into 8 parts, called octants. The octant containing the positive x, y and z-axes is called the positive octant. The region of space V given in this problem is shown in Fig. 5.13. dz dy dx Let I= 1 x2 y 2 z 2 V z, we draw a line PQ parallel to the z-axis cutting the voulme of integration. The limits for z and 0 and z1, where (x, y, z1) lies on the sphere x2 + y2 + z2 = 1 z1

1 x2

y2

the point Q lies in the positive octant)

(

After performing the innermost integration, the resulting double integral is evaluated over the orthogonal projection of the spherical surface on the xy-plane, i.e. over the circular region lying in the positive quadrant as shown in Fig. 5.14. y

B

E x2 + y2 = 1

O

D Fig. 5.14

A

x

Chapter 5: Multiple Integrals

1 x 2 and those for x are 0

In the double integral, the limits for y are 0 and and 1. 1 x2

1

1 x2

dx

I

0

dz (1 x

0

dx

dy sin

0

1 x

1 x2

2

dx 0

dy 0

1

2

1 x 2 dx 0

x 1 x2 2 2 2

4

1 sin 2

1 1

x 0

8 EXERCISE 5(a)

Part A (Short Answer Questions) 2

1

0

0

b

a

1

1

4 xy dx dy

1. Evaluate 2. Evaluate

dx dy xy

/2 /2

3. Evaluate

sin ( 0

0

1

x

0

0 sin

0

0

1

2

3

0

0

0

)d d

dx dy .

4. Evaluate

r dr d

5. Evaluate

xyz dx dy dz

6. Evaluate 1

z y

0

0

z

7. Evaluate

dz dy dx 0

y2 ) z2 1 x2

z

1

0 1

2

z

1 x2

1

2

y2

dy

0

I – 5.15

y2

z

0

y2

Part I: Mathematics I

I – 5.16

Sketch roughly the region of integration for the following double integrals: b

a

f ( x, y ) dx dy.

8. b

a

1

x

0

0

a

a2

f ( x, y ) dx dy.

9.

x2

10.

f ( x, y ) dx dy. 0

0

a (b b b

y)

f ( x, y ) dx dy.

11. 0

0

f ( x, y ) dx dy , where R is in the

Find the limits of integration in the double integral R

12. x = 0, y = 0, x + y = 1. x2 y 2 13. x 0, y 0, 2 a b2

1

14. x = 0, x = y, y = 1 15. x = 1, y = 0, y2 = 4x

Part B 4

16. Evaluate 0

y

y dx dy and also sketch the region of integration roughly. 2 2 x y 2 y /4

a

a2

x2

0

a

x

y dx dy and also sketch the region of integration roughly.

17. Evaluate 1

1

0

x

a

a2

18. Evaluate

y dx dy and also sketch the region of integration roughly. x2 y 2 x2

a2

19. Evaluate 0

x2

y 2 dx dy.

0

1 1 x1 x

y

xyz dx dy dz.

20. Evaluate 0

0

log 2 x x

0 log y x y

e

21. Evaluate 0

0 0

z

dz dy dx .

Chapter 5: Multiple Integrals

22. Evaluate

xe

x2 y

I – 5.17

dx dy , over the region bounded by x = 0, x

y = 0 and

y = x. 23. Evaluate xy dx dy , over the region in the positive quadrant bounded by the line 2x + 3y = 6. x dx dy , over the region in the positive quadrant bounded

24. Evaluate 2

2ax + y2 = 0.

by the circle x

(x

25. Evaluate

2

by the ellipse

x a2

y ) dx dy , over the region in the positive quadrant bounded y2 b2

( x2

26. Evaluate

1.

y 2 ) dx dy , over the area bounded by the parabola y2 = 4x

and its latus rectum. x 2 dx dy , where R is the region bounded by the hyerbola xy = 4,

27. Evaluate R

y = 0, x = 1 and x = 2. 28. Evaluate ( xy yz

zx) dx dy dz, where V is the region of space boun-

V

ded by x = 0, x = 1, y = 0, y = 2, z = 0 and z = 3. dx dy dz 29. Evaluate , where V is the region of space bounded by x y z 1)3 ( V x = 0, y = 0, z = 0 and x + y + z = 1. xyz dx dy dz, where V is the region of space bounded by the

30. Evaluate V

co-ordinate planes and the sphere x2 + y2 + z2 = 1 and contained in the positive octant.

5.4

CHANGE OF ORDER OF INTEGRATION IN A DOUBLE INTEGRAL

In worked example ( 1) of the previous section, we have observed that if the limits of integration in a double integral are constants, then the order of integration can be changed, provided the relevant limits are taken for the concerned variables. But when the limits for inner integration are functions of a variable, the change in the order of integration will result in changes in the limits of integration. d g2 ( y )

b h2 ( x )

f ( x, y ) dx dy will take the form

i.e. the double integral c g1 ( y )

f ( x, y ) dy dx , a h1 ( x )

when the order of integration is changed. This process of converting a given double integral into its equivalent double integral by changing the order of integration is often called change of order of integration. To effect the change of order of integration, the

Part I: Mathematics I

I – 5.18

5.5

PLANE AREA AS DOUBLE INTEGRAL

Plane area enclosed by one or more curves can be expressed as a double integral both in Cartesian coordinates and in polar coordinates. The formulas for plane areas in both the systems are derived below: (i) Cartesian System y

Let R be the plane region, the area of which is required. Let us divide the area into a large number of elemental areas like PQRS (shaded) by drawing lines parallel to the y-axis at intervals x and lines parallel to the x-axis at intervals y (Fig. 5.l5). Area of the elemental rectangle PQRS x. y. Required area A of the region R is the sum of elemental areas like PQRS. viz.,

A

x

lim ( x y

0 0

S

R

P

Q

A

o

x Fig. 5.15

y)

dx dy R

(ii) Polar System R

We divide the area A of the given region R into a large number of elemental curvilinear rectangular areas like PQRS (shaded) by drawing radial lines and concentric circular arcs, where P and R have polar coordinates (r, ) and (r r, ) (Fig. 5.16)

lim ( r

0 0

r r

Q P

O

Area of the element PQRS = r r ( PS = r and PQ r) Required area A

S

x Fig. 5.16

)

r dr d . R

5.5.1

Change of Variables

(i) From Cartesian Coordinates to Plane Polar Coordinates If the transformations x = x(u, v) and y = y (u, v) are made in the double integral ( x, y ) f ( x, y ) dx dy , then f ( x, y ) g (u , v) and dx dy = |J| du dv, where J . (u, v) [Refer to properties of Jacobians in the Chapter 4, “Functions of Several Variables” in Part I] .

Chapter 5: Multiple Integrals

I – 5.19

When we transform from cartesian system to plane polar system, and y = r sin

x = r cos

In this case,

x r y r

J

x y

r (cos 2

R

r cos

sin 2 ) r

f x, y dx dy

Hence

r sin

cos sin

g r,

r dr d

R

In particular, Area A of the plane region R is given by A

dx dy

r dr d

R

R

(ii) From Three Dimensional Cartesians to Cylindrical Coordinates y

of a point in space and derive the relations P (x, y, z) between cartesian and cylindrical coordinates (Fig. 5.17). Let P be the point (x, y, z) in Cartesian O y r coordinate system. Let PM be drawn z to the xoy-plane and MN parallel to Oy. Let r NOM and OM = r. The triplet (r, , z) are M N called the cylindrical coordinates of P. x Clearly, ON = x = r cos ; NM = y = r sin Fig. 5.17 and MP = z. Thus the transformations from three dimensional cartesians to cylindrical coordinates are x = r cos , y = r sin , z = z. In this case, ( x, y , z ) (r , , z )

J

xr yr zr

x y z

xz yz zz

cos sin 0

r Hence dx dy dz = r dr d dz f ( x, y, z ) dx dy dz

and V

g (r , , z ) r dr d dz V

r sin r cos 0

0 0 1

Part I: Mathematics I

I – 5.20

In particular, the volume of a region of space V is given by dx dy dz

r dr d dz

V

Note

V

f (x, y, z) dx dy dz is to be evaluated throughout the

Whenever

volume of a right circular cylinder, it will be advantageous to evaluate the corresponding triple integral in cylindrical coordinates. (iii) From Three Dimensional Cartesians to Spherical Polar Coordinates

r

Let relations between Cartesian and z spherical polar coordinates (Fig. 5.18). Let P be the point whose Cartesian P (x, y, z) coordinates are (x, y, z). Let PM be r to the xOy-plane. Let MN drawn be parallel to y-axis. Let OP = r, the y O angle made by OP with the positive z-axis = and the angle made by OM with x-axis = . M The triplet (r, , ) are called the N x spherical polar coordinates of P. Since OMP 90 , MP = z = r cos Fig. 5.18 , OM = r sin , ON = x = r sin cos and NM = y = r sin sin . Thus the transformations from three dimensional cartesians to spherical polar coordinates are x = r sin cos , y = r sin sin , z = r cos In this case, x, y , z J r 2 sin r, , [Refer to example (4.8) of Worked example set 4(b) in the Chapter 4 “Functions of Several Variables.” in Part I] Hence dx dy dz = r2 sin d

d

g (r , , ) r 2

f (x, y, z ) dx dy dz

and V

V

sin dr d d . In particular, the volume of a region of space V is given by r 2 sin dr d d .

dx dy dz V

Note

Whenever

V

f (x, y, z ) dx dy dz is to be evaluated throughout the

volume of a sphere, hemisphere or octant of a sphere, it will be advantageous to use spherical polar coordinates.)

Chapter 5: Multiple Integrals

I – 5.21

WORKED EXAMPLE 5(b) a a

x dx dy and then Example 5.1 Change the order of integration in 2 x y2 0 y evaluate it. The region of integration R . i.e. it is bounded by the lines x = y, x = a, y = 0 and y = a. The rough sketch of the boundaries and the region R is given in Fig. 5.19. y y=a After changing the order of integration, the given integral I becomes x x=y I= dy dx 2 2 x=a x y R Q (x, x) R The limits of inner integration are found by treating x as a constant, i.e. by drawing a line x parallel to the y-axis in the region of integration O (x, 0) P y = 0 as explained in the previous section. a x

Thus

x

I= 0 0

x

2

y

Fig. 5.19

dy dx

2

a

y2

x log y

y

x

y

0

x2

0

dx

a

x [log (x

x 2)

log (1 + 2))

x2 2

log x] dx

0 a

0

a2 log (1 + 2 ) 2 1 1

x dx dy and then evaluate it. 2 x y2 0 x Note Since the limits of inner integration are x and 1, the corresponding variable of integration should be y. So we rewrite the given integral I in the corrected form Example 5.2 Change the order of integration in

1 1

I 0 x

x dy dx x2 y 2

The region of integration R is bounded by the lines x = 0, x =1, y = x and y = 1 and is given in Fig. 5.20. The limits for the inner integration (after changing the order of integration) with respect to x x-axis (y = constant) 1 y

I 0 0

x dx dy x2 y 2

Part I: Mathematics I

I – 5.22 y

y=1

(0, y) P x=0

x=1 Q (y, y)

R y=x x

O Fig. 5.20 1

1 log (x 2 + y 2 ) 2

0

1 2

1

log 0

x

y

x

0

dy

2y 2 dy y2

1 log 2. 2 a (b b b

y)

xy dx dy

Example 5.3 Change the order of integration in 0

and then

0

evaluate it. The region of integration R is bounded by the lines x = 0, x x y 1 , y = 0 and y = b and is shown in Fig. 5.21. a b

a (b b

y ) or

y y=b

b (a – x) a

x,

Q y x + =1 a b

x=0 R O

P

x

y=0 (x, 0) Fig. 5.21

xy dy dx .

After changing the order of integration, the integral becomes I R b (a a a

x)

xy dy dx

I= 0

0

Chapter 5: Multiple Integrals a

0

y2 x 2

b2 2a 2

b (a a

I – 5.23

x)

dx 0

a

x (a

x ) 2 dx

0

2 b 2 x a 2 2a 2

x3 3

2

a 2b2 1 2 2

2a 2 3

x4 4

a

0

1 4

a 2b2 24 a

b 2 2 a x a

x 2 dy dx and then

Example 5.4 Change the order of integration in 0

0

integrate it. The region of integration R is bounded by the lines x = 0, x = a, y = 0 and the b 2 a a

2 x 2 i.e. the curve y 2 b and is shown in Fig. 5.22.

curve y

a2

2 x 2 , i.e. the ellipse x a2 a2

x2 + y2 = 1 a2 b2 (0, y) P x=0

a b

Q R

O

x=a x

y=0 Fig. 5.22

After changing the order of integration, the integral becomes x 2 dx dy

I R

b

a 2 b b

y2

x 2 dx dy

I 0 b

0

0

x3 3

a 2 b b

y2

dy 0

b2 – y2, y

y2 b2

1

Part I: Mathematics I

I – 5.24 a3 3b3

b

a3 3b3

/2

y 2 ) 2 dy

0

b 4 cos 4

(on putting y = b sin )

d

0

a 3b 3 16

3

(b 2

3 1 4 2

2

a 3b

a

a2

y2

0

a y

Example 5.5 Change the order of integration in

y dx dy and then

evaluate it. The region of integration R is bounded by the line x = a 2

2

y, the curve

x a y , the lines y = 0 and y = a. i.e. the line x + y = a, the circle x2 + y2 = a2 and the lines y = 0, y = a. R is shown in Fig. 5.23. y x, a2 – x2

Q

R x2 + y2 = a2 (x, a – x) P

x

O

y x + =1 a a Fig. 5.23

After changing the order of integration, the integral I becomes, y dy d x

I R a

a2 x2

0

a x

y dy dx a

y2 2

0

1 2

a2 x2

dx a x

a

(2ax 0

2 x 2 ) dx

Chapter 5: Multiple Integrals x2 a 2

I – 5.25

a

x3 3

0

a3 6

. 4

2 x

0

x2 4

Example 5.6 Change the order of integration in

dy dx and then evaluate it.

x2 i.e. the parabola 4 2 x i.e. the parabola y2 = 4x and the lines x = 0, x = 4. R is

The region of integration R is bounded by the curve y x2 = 4y, the curve y shown in Fig. 5.24. y y2 =

A

x2 = 4y

4x

y2 ,y P 4

Q (2

y, y)

O

Fig. 5.24

The points of intersection of the two parabolas are obtained by solving the equations x2 = 4y and y2 = 4x. 2

2 Solving them, we get x 4x 4 i.e x (x3 x = 0, x = 4 and y = 0, y = 4 i.e. the points of intersection are O(0, 0) and A(4, 4). After changing the order of integration, the given integral

I

dx d y R 4 2 y

dx d y 0

y2 4

4

2 y 0

y2 dy 4

Part I: Mathematics I

I – 5.26 4 32 y 3 32 3 16 3

4

y3 12

0

16 3

a a

a2

y2

0 a

a2

y2

Example 5.7 Change the order of integration in

xy dx dy and then

evaluate it. The region of integration R is bounded by the curve x circle (x 5.25.

a2

a

y 2 , i.e. the

a)2 + y2 = a2 and the lines y = 0 and y = a. The region R is shown in Fig.

y x, 2ax – x2 (x – a)2 + y2 = a2 Q

R

O (x, 0) P

C (a, 0) Fig. 5.25

After changing the order of integration, the integral I becomes xy dy dx

I= R 2a

2 ax

0

0

x2

xy d y d x 2a

0

1 2

y2 x 2

x2

2 ax

dx 0

2a

(2ax 2

x 3 )dx

0

1 x3 2a 2 3 2 4 a 3

x4 4

2a

0

x

Chapter 5: Multiple Integrals

I – 5.27 y

1 2

Example 5.8 Change the order of integration in

xy dx dy 0

and then

y

evaluate it. The region of integration R is bounded by the lines x = y, x + y = 2, y = 0 and y = 1. It is shown in Fig. 5.26. After changing the order of integration, y B (1, 1) the integral I becomes x+y=2

x=y

xy dy dx

I= R

y in the inner x integration, we have to draw a line parallel O C A to y-axis (since x = constant). The line drawn Fig. 5.26 parallel to the y-axis does not intersect the region R in the same fashion. If the line segment is drawn in the region OCB, its upper end lies on the line y = x; on the other hand, if it is drawn in the region BCA, its upper end lies on the line x + y = 2. In such situations, we divide the region into two xy dy dx

I=

xy dy dx

OCB

BCA x

2 2

x

1

xy dy dx 0

1

1

y2 2

x 0 1

0

x3 dx 2

x4 8 1 8

xy dy dx

0

1

0

x

0

2

dx

x 1

0 2

1

x (2 2

1 2x2 2

y2 2

x

2

dx 0

x ) 2 dx 4 3 x 3

x4 4

2

1

5 24

1 3

a 2a

0

evaluate it.

x2 a

x 2 , i.e. the parabola x2 = ay, a , i.e. x + y = 2a and the lines x = 0 and x = a. It is shown in Fig.

The region of integration R is bounded by the curve y the line y = 2 5.27.

x

xy dy dx and then

Example 5.9 Change the order of integration in

Part I: Mathematics I

I – 5.28

y D

y = 2a C Q2 (2a – y, y)

(0, y) P2 E

B (a, a) Q1

(0, y) P1

ay, y

A x

O Fig. 5.27

After changing the order of integration, the integral I becomes xy dx dy

I= R

When we draw a line parallel to x with respect to x, it does not intersect the region of integration in the same fashion. Hence the region R is divided into two sub-regions OABE and EBCD and then the xy dx dy

I

xy dx dy

OABE a

ay

0

0

EBCD 2a 2a

y

xy dx dy

xy dx dy a

0

Note

The co-ordinates of the point B are obtained by solving the equations x + y = 2a and x2 = ay. (a, a) and the equation of EB is y = a. a

I 0

1 2

x2 y 2

ay

2a

dy a

0

a

x2 y 2

2a

y

dy 0

2a

ay 2 dy

y ) 2 dy

a

0

1 y3 a 2 3

y (2a

a 2

2a y 0

2

4a 3 y 3

y4 4

2a

a

3 4 a . 8

Example 5.10 Change the order of integration in each of the double integrals 1

2

2 2 dx dy dx dy and hence express their sum as one double integral and 2 2 2 x y x y2 0 1 1 y and evaluate it. The region of integration R1 is bounded by the lines 1 x = 1, x = 2, y = 0 and y = 1.

Chapter 5: Multiple Integrals

I – 5.29

The region of integration R2 for the second double integral I2 is bounded by the lines x = y, x = 2, y = 1 and y = 2. R1 and R2 are shown in Fig. 5.28. y

E (2, 2)

(x, x) Q2

(x, 1) Q1

(1, 1) D

A P1 (x, 0)

O

C (2, 1)

P2 (x, 1)

B

x

Fig. 5.28

After changing the order of integration, 2

1

1

0

2

x

1

1

I1

and

I2

dy dx x2 y2 dy dx x2 y2

Adding the integrals I1 and I2, we get 2

1

1 x

dx 1 2

1 2

1

x2

0

2

0

y2

1

dy x2

y2

dy x

1 tan x dx 4 x

x

dy

dx

I

2

y2

1

y x

4

y

x

y

0

dx

log 2 .

Example 5.11 Find the area bounded by the parabolas y2 x and y2 = x by double integration. The region, the area of which is required is bounded by the parabolas (y 0)2 (x 4) and y2 = x and is shown in Fig. 5.29. dx dy

Required area OC AB

Part I: Mathematics I

I – 5.30

dx dy , by symmetry

2 o AB

y

y2

2 4

dx dy

2 y2

0

(y2,

2

2

y

(4

2

2 3 y 3

y ) dy

O

x

C

2

(2,– 2)

y2 = x

y2 = 4 – x

0

4 2 3

2 4 2

(4 – y2, y) A (4, 0)

2

0

2 4y

y)

(2, 2) B

Fig. 5.29

16 2 square units 3 Example 5.12 Find the area between the circle x2 + y2 = a2 and the line x + y = a lyThe plane region, the area of which is required, is shown in Fig. 5.30. y

Required area

dx dy

C

ABC

2

a

a

0

a

y

(a – y, y)

2

B ( a2– y2, y)

dx dy O

y

A

x

a

a2

y2

a

y dy

0

Fig. 5.30

y 2

2

a2

a2 2 2

y2

a2

a sin 2 a2 2

(

1

y a 2)

2 a

ay

y 2

0

a2 4

Example 5.13 Find the area enclosed by the lemniscate r2 = a2 cos 2 , by double integration. As the equation r2 = a2 cos 2 remains unaltered on changing , the curve is symmetrical about the initial line. The points of intersection of the curve with the initial line = 0 are given by r2 = a2 or r = ± a.

Chapter 5: Multiple Integrals

I – 5.31

Since r2 = a2 cos 2 = a2 cos 2 ( ), the curve is symmetrical about the line . 2 3 . Hence there is a loop On putting r = 0, we get cos 2 = 0. Hence , 4 4 3 and and another loop between of the curve between and 4 4 4 3 . 4 Based on the observations given above the lemniscate is drawn in Fig. 5.31.

D C

O

P

A

B

X

Fig. 5.31

Required area = 4 × area OABC (by symmetry) r dr d

4 OA B

When we perform the inner integration with respect to r, we have to treat as a r. Geometrically, treating = constant means drawing a line OP arbitrarily through Finding the limits for r (while of the r coordinate of any point on the line OP. Assuming that the coordinates of all points on OP are each (since is constant on OP), we take O (0, ) and P (r1, ); viz., r varies from 0 to r1. Now wherever OP be drawn, the point P(r1, ) lies on the lemniscate. Hence r12 = a2 cos 2 or r1 a cos 2 (since r coordinate of any point is + ve) Thus the limits for inner integration are 0 and a cos 2 . . OP so that it sweeps the area of the region, namely OABC. To cover this area, the line OP has to start from

Part I: Mathematics I

I – 5.32

the position OA ( = 0) and move in the anticlockwise direction and go up to OD

4

. Thus the limits for

are 0 and

4

.

4 a cos 2

Required area

r dr d

4 0

4

4 0

0

r2 2

a cos 2

d 0

4

2a 2

cos 2 d 0

a 2 (sin 2 )04

a2

Example 5.14 Find the area that lies inside the cardioid r = a (1 + cos ) and outside the circle r = a, by double integration. The cardioid r = a (1 + cos ) is symmetrical about the initial line. The point of intersection of the line = 0 with the cardioid is given by r = 2a, viz., the point (2a, 0). Putting r = 0 in the equation, we get cos . Hence the cardioid lies between the lines and = . The point of intersection of the line is a, . C 2 2 G Noting the above properties, the B cardioid is drawn as shown in Fig. 5.32. All the points on the curve r = a have A F O the same r coordinate a, viz., they are at the same distance a from the pole. D Hence the equation r = a represents a H circle with centre at the pole and radius E equal to a. y Noting the above points, the circle r = a is drawn as shown in Fig. 5.32. Fig. 5.32 The area that lies outside the circle r = a Both the curves are symmetric about the initial line. Hence the required area = 2×AFGCB

x

Chapter 5: Multiple Integrals

2

r2

0

r1

I – 5.33

r d r d , where (r1, ) lies on the circle r = a and (r2, )

2

lies on the cardioid r = a (1 + cos ) 2

a (1 cos )

0

a

r dr d

2 2

a (1 cos )

r2 2

2 0

d a

2

[(1 cos ) 2

a2

1] d

0 2

a2

1 cos 2 2

2cos 0

a 2 2 sin a2 2

2

1 sin 2 4

a2 ( 4

4 a

a

0

y

Example 5.15 Express

2 0

8)

x 2 dx dy x2

d

3/ 2

y2

in polar coordinates and then evaluate it.

The region of integration is bounded by the lines x = y, x = a, y = 0 and y = a, , r = a sec , 4 respectively. The region is shown in Fig. 5.33.

whose equations in polar system are

= 0 and r = a cosec

r = a cosec B 4

O

=0

r = a sec

A

x

Fig. 5.33

Putting x = r cos , y = r sin we get

and dx dy = r dr d in the given double integral I,

Part I: Mathematics I

I – 5.34

r 3 cos 2 dr d r3

I OAB

/ 4 a sec

cos 2 dr d 0

0 /4

cos 2

[r ]0a sec d

0 /4

a

a

a [sin ]0

cos d

2

0

a

a2 x2

dx dy

Example 5.16 Transform the double integral 0

ax x 2

a

2

x2

y2

in polar

coordinates and then evaluate it. The region of integration is bounded by the curves y and the lines x = 0 and x = a. y

2

a 2

x

i.e.,

x 2 is the curve x2 + y2

ax

(y

i.e. the circle with centre at

a2

y

a 2

0) 2

ax

x2 , y

a2

ax = 0

2

a a ,0 and radius 2 2

x 2 is the curve x2 + y2 = a2

i.e. the circle with centre at the origin and radius a. The polar equations of the boundaries of the region of integration are r2 cos

= 0 or r = a cos , r = a, r = a sec

shown in Fig. 5.34. Putting x = r cos , y = r sin we get /2

and

. The region of integration is

and dx dy = r dr d in the given double integral I,

r dr d /2

0

a cos

a2

ar

2

a

I 0

x2

r2

1 2 a2 2

a

r2

d , on putting a2 a cos

r2 = t

Chapter 5: Multiple Integrals

I – 5.35

r=a r = a sec 2 r = a cos a 2

O

a

x

Fig. 5.34 /2

a sin d

a cos

2 0

a

0

Example 5.17 By transforming into cylindrical coordinates, evaluate the integral (x 2

y2

z 2 ) dx dy dz

x2 + y2

1. z

The region of space is the region enclosed by the cylinder x2 + y2 = 1 whose base radius is 1 and axis is the z-axis and the planes z = 0 and z = 1. The equation of the cylinder in cylindrical coordinates is r = 1. The region of space is shown in Fig. 5.35. Putting x = r cos , y = r sin , z = z and dx dy dz = r dr d dz in the given triple integral I, we get (r 2

I

z 2 ) r dr d dz ,

z=1

z=0 y x

V

where V is the volume of the region of space. 1 2

1

(r 2 0

0

1 2

0

0

1 2

0

0 1

2 0

z 2 ) r dr d dz

0

r4 4

r2 z 2

1

2

d dz 0

1 4

1 2 z d dz 2

1 4

1 2 z dz 2

Fig. 5.35

Part I: Mathematics I

I – 5.36

2

1

z3 6

z 4

0

5 6

Note

The intersection of z = constant c and the cylinder x2 + y2 = 1 is a circle with centre at (0, 0, c) and radius 1. The limits for r and the area of this circle and then the variation of z has been used so as to cover the entire volume.] Example 5.18 Find the volume of the portion of the cylinder x2 + y2 = 1 intercepted z. between the plane z = 0 and the paraboloid x2 + y2 z

y

x Fig. 5.36

Using cylindrical coordinates, the required volume V is given by V

r dr d dz , taken throughout the region of space.

Since the variation of z to z and then with respect to r and . Changing to cylindrical coordinates, the boundaries of the region of space are r = 1, z = 0 and z r2. r2

1 4

2

V

dz r dr d 0

0

2

0

1

r (4 r 2 ) dr d 0

0

2

0 a

a

2

x

2

a

2

x

2

d 0 y

7 4

2

d 0

7 2

2

xyz dz dy dx , by transforming to spherical

Example 5.19 Evaluate 0

polar coordinates.

1

r4 4

2r 2

0

0

Chapter 5: Multiple Integrals

I – 5.37 a2

The boundaries of the region of integration are z = 0, z

x2

y 2 or x2 + y2 +

z2 = a2, y = 0, y a 2 x 2 or x2 + y2 = a2, x = 0 and x = a. From the boundaries, we note that the region of integration is the volume of the positive octant of the sphere x2 + y2 + z2 = a2. By putting x = r sin cos , y = r sin sin , z = r Z cos and dx dy dz = r2 sin dr d d , the given triple integral I becomes r

r 3 sin 2 cos sin cos . r 2 sin d r d d

I

P

V

O

where V is the volume of the positive octant of the sphere r = a, which is shown in Fig. 5.37. To cover the volume V, r has to vary from 0 to a, has to vary from 0 to vary from 0 to

2

x Fig. 5.37

.

2

Thus

2

M

has to

and

y

2

a

r 5 sin 3 cos sin cos d r d d

I 0

0

0

a

2

2

0

0

r5 dr

sin 3 cos d .

sin cos d .

0

[

sin 2 2

2 0

sin 4 4

2 0

r6 6

the limits are constants]

a

0

1 6 a 48 Example 5.20 Evaluate

1 x2

y2

z 2 dx dy dz , taken throughout the

volume of the sphere x2 + y2 + z2 = 1, by transforming to spherical polar coordinates. Changing to spherical polar coordinates, the given triple integral I becomes 1 r 2 r 2 sin d r d d

I V

Part I: Mathematics I

I – 5.38

z r

P (r, , )

O

y

M x Fig. 5.38

To cover the entire volume V of the sphere, r has to vary from 0 to 1, from 0 to and has to vary from 0 to 2 . 2

Thus

1

1 r 2 r 2 dr sin d

I 0

0

d

0

2

2

d 0

sin 2 t cos 2 t d t , by putting

sin d 0

0

r = sin t in the innermost integral 1 3 1 2 ( cos )0 2 2 4 2 2 1 1 2 4 4 4 4 EXERCISE 5(b) Part A (Short Answer Questions) a

x

0 1

0 1

0

y

a

a

0 1

x y

f (x, y ) dy dx.

1. Change the order of integration in

f (x, y ) dx dy .

2. Change the order of integration in

f (x, y ) dy dx.

3. Change the order of integration in 4. Change the order of integration in

f (x, y ) dx dy. 0

0

1 1 y

5. Change the order of integration in

f (x, y ) dx dy . 0

0

has to vary

Chapter 5: Multiple Integrals a a

I – 5.39

x

6. Change the order of integration in

f (x, y ) dy dx. 0

0

1

1 x2

0

0

a

a2

f ( x, y ) dy dx.

7. Change the order of integration in

y2

f ( x , y ) dx dy .

8. Change the order of integration in 0

0

1 2 x

f ( x, y ) dy dx.

9. Change the order of integration in 0

0 1/ y

10. Change the order of integration in

f ( x , y ) dx dy . 0

0

Part B Change the order of integration in the following integrals and then evaluate them: a

a

x dx dy x2 y 2

11. 0

y

x2 y

x

13.

xe 0

dy dx

2

15. 0

dy dx

(x

17. 0

3

y ) dy dx

4 x

3

y ) dy dx

y2

xy dx dy 0

1

y

0

y2

y dx dy x2 y 2

a2

y2

y 2 dx dy a

0

x 2 4/ x

dx dy

21. 0 5 2 x 9 1

4

dy dx

y

20.

1 3

2

a

(x

19. 5

x

18.

x2 4a

0

0

0

2a a

y

e

16.

0

y 2 ) dy dx

x

14.

0 y

( x2 0

1 1 x

e2 x

2

12.

xy dy dx.

22. 1

0

2 x2

dy dx

23. 0

x

1

x

0

0

xy dy dx

24. Change the order of integration in each of the double integrals 2 2 x

xy dy dx and hence express their sum as one double integral and

and 1

0

evaluate it.

Part I: Mathematics I

I – 5.40

0

1

( x2 + y 2 )

25. Change the order of integration in each of the double integrals 1

1

1 2

x

2

( x + y ) dy dx and hence express their sum as one double

dy dx and x

0

integral and evaluate it. 26. 27. 28. 29.

The area bounded by the parabola y = x2 and the straight line 2x y + 3 = 0. The area included between the parabolas y2 = 4a (x + a) and y2 = 4a(a x). The area bounded by the two parabolas y2 = 4ax and x2 = 4by. The area common to the parabola y2 = x and the circle x2 + y2 = 2.

30. The area bounded by the curve y 2 31. 32. 33. 34. 35.

x3

and its asymptote. 2 x The area of the cardioid r = a (1 + cos ). The area common to the two circles r = a and r = 2a cos . The area common to the cardioids r = a (1 + cos ) and r = a ). The area that lies inside the circle r = 3a cos and outside the cardioid r = a (1 + cos ). The area that lies outside the circle r = a cos and inside the circle r = 2a cos .

evaluate them: a2

a

x2

36.

e 0

x2

a

y2

dx dy

0

0

a

x

38. 0

0

x3 d x d y x

2

a

2

y

a

xdx d y

37.

x2

y

2a

2a x

0

0

x2

39.

2

y2 xdx d y x2

y2

dx d y

40.

x2

y2

3/2

coordinates: (x + y + z ) dx dy dz , where V is the region of space inside the cylinder

41. V

x2 + y2 = a2, that is bounded by the planes z = 0 and z = h. 42.

(x 2 + y 2 ) dx dy dz, taken throughout the volume of the cylinder x2 + y2 = 1 that is bounded by the planes z = 0 and z = 4.

43.

dx dy dz, taken throughout the volume of the cylinder x2 + y2 = 4 bounded by the planes z = 0 and y + z = 3.

Chapter 5: Multiple Integrals

I – 5.41

dx dy dz , taken throughout the volume of the cylinder x2 + y2 = 4

44.

bounded by the plane z = 0 and the surface z = x2 + y2 + 2. dx dy dz , taken throughout the volume bounded by the spherical

45.

surface x2 + y2 + z2 = 4a2 and the cylindrical surface x2 + y2 2ay = 0. Evaluate the following integrals (46-50) after transforming into spherical polar coordinates: d xd yd z 46. , taken throughout the volume of the sphere x2 + y2 + x2 y2 z 2 z2 = a2. d xd yd z 47. , taken throughout the volume contained in the 1 x2 y 2 z 2 positive octant of the sphere x2 + y2 + z2 = 1. z dx dy dz , where V is the region of space bounded by the

48. V

sphere x2 + y2 + z2 = a2 above the xOy-plane. a2

a

x2

a2

y2

49. 0

d xd yd z

50.

x dx dy dz 0

5.6

z2

0

0

0

0

a

2

x2

y2

z2

5/ 2

LINE INTEGRAL b

f (x) dx.

integral a

x-axis from a to b and the integrand a, b). In a line integral, we shall integrate along a C.

f(x) curve C

Let C be the segment of a continuous curve joining A(a, b) and B(c, d) (Fig. 5.39). y

B (c, d) Q

(xr, yr)

(xr –1, yr – 1) P

( r,

A (a, b)

r)

x

O Fig. 5.39

Part I: Mathematics I

I – 5.42

Let f (x, y), f1(x, y), f2(x, y) be single-valued and continuous functions of x and y, C. Divide C into n arcs at (xi, yi) [i = 1, 2, . . . (n Let x0 = a, xn= c, y0 = b, yn = d. Let xr xr = xr, yr yr yr and the arcual length of PQ (i.e. PQ ) sr , where P is (xr , yr ) and Q(xr, yr). Let ( r , r) be any point on C between P and Q. n

f(

lim

Then

n

r

,

r

) sr

r 1 n

f1 ( r , r )

lim

or

n

xr

f2 ( r ,

)

r

yr

r 1

C and denoted respectively as f (x, y ) ds

5.6.1

[f1 (x, y ) dx

or

C

f 2 (x, y ) dy ]

C

Evaluation of a Line Integral

Using the equation y =

c

dx

f 2 (x , y ) dy ] either in the form

d

g (x) dx or in the form a

If the line integral is in the form

C

h(y ) dy and evaluate b

f (x, y ) ds,

f (x, y )

C

f (x, y ) 1 C

5.6.2

dy dx

[f1 (x , y )

(x) or x = (y) of the curve C, we express

2

dx or as

f (x, y ) C

C

ds dy dy

f (x, y ) 1 C

dx dy

ds dx dx

2

dy and then

Evaluation when C is a Curve in Space C is a curve in [f1 (x, y, z ) dx

space. In this case, the line integral will take either the form f 2 (x, y, z ) dy

C

f 3 (x, y, z ) dz ] or the form

f (x, y, z ) ds. When C is a curve in C

space, very often the parametric equations of C will be known in the form x = 1(t), y = 2(t), z = 3(t). Using the parametric equations of C, the line integral can be f (x, y, z ) ds , it is rewritten as

C

ds f (x , y , z ) dt , where dt

ds dt

dx dt

2

dy dt

2

dz dt

C

2

.

Chapter 5: Multiple Integrals

5.7

I – 5.43

SURFACE INTEGRAL

The concept of a surface integral is a generalisation of the concept of a double integral. While a double integral is evaluated over the area of a plane surface, a surface integral is evaluated over the area of a curved surface in general. The formal Let S be a portion of a regular two-sided surface. Let f (z, y, z) be a S. Divide S into n s 1, n

S2

Sn. Let P( r, r,

Sr. Then lim

r

n Sr

f ( r,

r

, r ) Sr

0 r 1

is called the surface integral of f(x, y, z) over the surface S and denoted as f (x, y, z ) dS or S

5.7.1

f (x, y, z ) dS . S

Evaluation of a Surface Integral f (x, y, z ) dS , where S is the portion of the surface

Let the surface integral be S

whose equation is

(x, y, z) = c (Fig. 5.40). z

S

O

y R

x Fig. 5.40

Project the surface S orthogonally on xoy-plane (or any one of the co-ordinate planes) so that the projection is a plane region R. S A We can divide the area of the region R into elemental areas by drawing lines parallel to x and y y x respectively. Then = · . x · y = S cos , where is the angle between the surface S and the plane R (xoy-plane), i.e. is the angle between the normal to the surface S at the typical point (x, y, z) and the normal to the xoy-plane (z-axis). From Calculus, it is

Part I: Mathematics I

I – 5.44

known that the direction ratios of the normal at the point (x, y, z) to the surface (x, y, z) = c are

x

,

y

,

z

. The direction cosines of the z-axis are (0, 0, 1)

z

cos

2

2

x Thus

y

2 x

S

2

2 y

z

2 z

x y.

z

f (x, y, z ) dS

2 x

f ( x, y , z )

S

2 y

2 z

dx dy

z

R

Thus the surface integral is converted into a double integral by using the above as to cover the entire region R and the integrand is converted into a function of x and y, using the equation of S.

Note

Had we projected the curved surface S on the yoz-plane or zox-plane then the conversion formula would have been

f (x, y, z )dS S

or

f (x, y, z )

2 x

S

2 z

2 y

2 z

dy dz

x

R

f (x, y, z )dS

2 y

f (x, y, z ) R

2 x

dz dx ,

respectively.

y

5.8 VOLUME INTEGRAL Let V be a region of space, bounded by a closed surface. Let V. Divide V into n subVr by drawing planes parallel to the yoz, zox and xoy-planes at intervals of x, y z Vr is a rectangular parallelopiped with dimensions x y z. Let P( r , r r Vr . f(x, y, z

n

f ( r,

Then lim n Vr

r

, r ) Vr is called the volume integral of f(x, y, z) over the

0 r 1

region V (or throughout the volume V) and denoted as

Chapter 5: Multiple Integrals f (x, y, z ) dv

I – 5.45

f (x, y, z ) dx dy dz

or V

V

5.8.1 Triple Integral Versus Volume Integral A triple integral discussed earlier is a three times repeated integral in which the limits of integration are given, whereas a volume integral is a triple integral in which the limits of integration will not be explicitly given, but the region of space in which it V.

Note: of the line and surface integrals will be discussed in Part II, Chapter 2.

WORKED EXAMPLE 5(c)

Example 5.1

[(3xy 2

Evaluate

y 3 ) dx

(x 3

3 xy 2 ) dy ] where C is the

C

parabola y2 = 4ax from the point (0, 0) to the point (a, 2a) The given integral [(3xy 2

I y2

y 3 ) dx

(x 3

3 xy 2 ) dy ] y

4 ax

In order to use the fact that the line integral is evaluated along the parabola y2 = 4ax, we use this equation and the relation between dx and dy derived from it, namely, 2y dy = 4a dx and convert the body of the integral either to the form f(x) dx or to the form (y) dy. Then

A (a, 2a) C

(0, 0) O

x

between the concerned limits, got from the Fig. 5.41 end points of C. The choice of the form f(x) dx or (y) dy for the body of the integral depends on convenience. In this problem, x is expressed as 1 y 2 more easily than expressing y 4a as 2 ax .

Note

From y2 = 4ax, we get y

quadrant, y is positive and hence y 2a

2 ax . Since the arc C

2 ax .

y 1 2 2 1 1 2 2 dy y y y3 y6 3 y y dy 3 2 a 4 4 a a 64 a 0 (As the integration is done with respect to y, the limits for y are the y co-ordinates of the terminal points of the arc C). Thus I

3

Part I: Mathematics I

I – 5.46 2a

I 0

5 4 y 4a

1 5 y 4a 86 4 a 7 Example 5.2 Evaluate

[(2x

3 5 y 8a 2 1 y6 16a 2

y ) dx

1 y 6 dy 64a 3 1 y7 448a 3

2a

0

y ) dy ] , where C is the circle x2 + y2 = 9.

(x

C

In this problem the line integral is evaluated around a closed curve. In such a situation the line integral is denoted as [(2x

y ) dx

(x

y ) dy ] , where a small circle is put across the integral symbol.

C

When a line integral is evaluated around a closed curve, it is assumed to be described In the case of a line integral around a closed curve C, any point on C can be assumed to be the initial point, which will also be the terminal point. Further if we take x or y as the variable of integration, the limits of integration will be the same, resulting in the value ‘zero’ of the line integral, which is meaningless. Hence whenever a line integral is evaluated around a closed curve, the parametric equations of the curve are used and hence the body of integral is converted to the form f (t) dt or f ( ) d . In this problem, the parametric equations of the circle x2 + y2 = 9 are x = 3 cos and y = 3 sin . dx = d and dy = 3 cos d . y =

B

=

2

A'

=0 A =2

B' 3 = 2 Fig. 5.42 2

[(6 cos

The given integral = 0

3 sin ) ( 3 sin d )

x

Chapter 5: Multiple Integrals

I – 5.47

(3 cos + 3 sin ) (3 cos d )] 2

=9

(1 sin cos ) d 0 2

sin 2 2

9

0

18 xy ds , where C is the arc of the parabola y2 = 4x

Example 5.3 Evaluate C

between the vertex and the positive end of the latus rectum. Given integral

xy

I C

ds dx dx

Equation of the parabola is y2 = 4x dy dx

Differentiating with respect to x,

ds dx

dy dx

1

2

1

4 y2

y2 4 dx y

xy

I

2 y

C

1

x 4x

4 dx

0 2

2

(t 2

1) t 2t dt, on putting x + 1 = t2

1 2

(t 4

4

t 2 ) dt

1

4

t5 5

t3 3

8 1 15 (y 2 dx

Example 5.4 Evaluate

2

1

2

x 2 dy ), where C is the boundary of the triangle

C

C is made up of the lines BC, CA and AB.

Part I: Mathematics I

I – 5.48

Equations of BC, CA and AB are respectively y = 0, x + y = 2

x + y = 1.

2

(y dx x dy )

Given integral BC y 0 dy 0

CA x y 1 dy dx

AB x y 1 dy dx

y A (0, 1)

B (– 1, 0)

C (1, 0)

O

x

Fig. 5.43 0

1

[(1 x) 2

0

x 2 ]dx

[(1 x) 2

1

x 2 ]dx

0 1

0

(1 2 x

2 x 2 ) dx

(1 2 x) dx 0

1

x x

2 x3 3

2

0

(x

x 2 )0 1

1

2 3 [x 2 y dx (x z ) dy

Example 5.5 Evaluate

xyz dz ] , where C is the arc of the

C

parabola y = x2 in the plane z = 2 from (0, 0, 2) to ( 1, 1, 2). [x 2 y dx (x z ) dy

Given integral

xyz dz ]

2

y x z 2

[x 2 y dx (x 2) dy ] x2

y

[

dz = 0, when z = 2] 1

[x 4

(x 2) 2 x]dx

0

x5 5

2 x3 3

1

2x2 0

17 15 (x dx

Example 5.6 Evaluate

xy dy

xyz dz ) , where C is the arc of the

C

t 1. twisted curve x = t, y = t2, z = t3, The parametric equations of C are x = t, y = t2, z = t3 dx = dt, dy = 2t dt, dz = 3t2 dt on C. Using these values in the given integral I,

Chapter 5: Multiple Integrals

I – 5.49

1

t 3 2t

(t

I=

t 6 3t 2 ) dt

0

t2 2

2

t5 5

3

t9 9

1

0

17 30 (x 2

Example 5.7 Evaluate

y2

z 2 ) ds, where C is the arc of the circular helix

C

x = cos t, y = sin t, z = 3t from (1, 0, 0) to (1, 0, 6 ) The parametric of equations of C are x = cos t, y sin t, z = 3t. dx dy sin t, cos t , dt dx ds dt

dx dt

2

dy dt

sin 2 t

2

dz dt

cos 2 t

9

(cos 2 t

I=

sin 2 t

10 9t 2 )

0

Note

3 on C.

2

2

Given integral

dz dt

ds dt dt

The point (1, 0, 0) corresponds to t = 0 and (1, 0, 6 ) corresponds to t = 2 . I= t

3t 3

2

10

0

2 10 (1+12 Example 5.8 Evaluate

2

)

xyz dS , where S is the surface of the rectangular S

parallelopiped formed by x = 0, y = 0, z = 0, x = a, y = b and z = c (Fig. 5.44). z C

z=c

A' O'

B'

y=b y=0

O B A

x

z=0

y

C'

Fig. 5.44

Since S is made up of 6 plane faces, the given surface integral I is expressed as

Part I: Mathematics I

I – 5.50

(xyz dS )

I x 0

x

a

y

y b

0

z

z

0

c

Since all the faces are planes, the elemental curved surface area dS becomes the elemental plane surface area dA. On the planes x = 0 and x = a, dA = dy dz. On the planes y = 0 and y = b, dA = dz dx. On the planes z = 0 and z = c, dA = dx dy. I (xyz dy dz ) (xyz dz dx) x 0

x

a

z

0

y

0

y b

(xyz dx dy ) z

c

Simplifying the integrands using the equations of the planes over which the surface integrals are evaluated, we get c

b

I=a

a

yz dy dz 0

c

b

0

b

a

0

0

zx dz dx c 0

0

xy dx dy

Note On the plane face z = c), the limits for x and y are easily found to be 0, a and 0, b. Similarly the limits are found on the faces (x = a) and (y = b).] Now

b2 c2 c2 a2 b 2 2 2 2 abc (ab bc ca ) 4

I=a

Example 5.9 Evaluate

(y

c

a 2 b2 2 2

2 z 2) dS , where S is the part of the plane 2x +

S

3y + 6z = 12, that lies in the positive octant (Fig. 5.45). z C

B

O

y

A x Fig. 5.45

Rewriting the equation of the (plane) surface S in the intercept form, we get

Chapter 5: Multiple Integrals

I – 5.51

x y z 1 6 4 2 S is the plane that cuts off intercepts of lengths 6, 4 and 2 on the x, y and z-axes respectively and lies in the positive octant. We note that the projection of the given plane surface S on the xoy-plane is the triangular region OAB shown in the two-dimensional Fig. 5.46. y

B 2x + 3y = 12

4

O

x

A

6 Fig. 5.46

Converting the given surface integral I as a double integral, (y

I

2 z 2)

2 x

Here

2 z

dx dy ,

z

OAB

where

2 y

= c is the equation of the given surface S. = 2x + 3y + 6z

x

= 2;

y

= 3;

z

= 6.

(y

I

2 z 2)

OAB

7 6

(y

4 9 36 dx dy 6

2 z 2) dx dy

(1)

OAB

Now the integrand is expressed as a function of x and y, by using the value of z (as a function of x and y) got from the equation of S, i.e. from the equation 2x + 3y + 6z = 12 1 (2) z Thus (12 2 x 3 y ) 6 Using (2) in (1), we get I

7 6

1 (6 2x) dx dy 3 OAB

Part I: Mathematics I

I – 5.52 4

7 18

3 y 2

6

(6 2 x) dx dy 0

0

4

7 18

9 2 y dy 4

9y 0

28 3

z 3 d S , where S is the positive octant of the surface

Example 5.10 Evaluate S

of the sphere x2 + y2 + z2 = a2 (Fig. 5.47) z C

O

y

B

A x Fig. 5.47

The projection of the given surface of the sphere x2 + y2 + z2 = a2 (lying in the positive octant) in the xoy - plane is the quadrant of the circular region OAB, shown in the two-dimensional Fig. 5.48. y x2 + y2 = a2

B (0, y)

a2 – x2, y

O

x

A Fig. 5.48

Converting the given surface integral I as a double integral. z3

I

2 x

2 z

dx dy,

z

OAB

where = 2x; x

2 y

x2 + y2 + z2 = a2 is the equation of the given spherical surface. = 2y; z = 2z. y z3

I

4 (x 2

OAB

z 2 x2 OAB

y2 2z y2

z2 )

dx dy

z 2 dx dy

Chapter 5: Multiple Integrals (a 2

a

x2

y 2 ) dx dy

I – 5.53

(x, y, z) lies on x2 + y2 + z2 = a2]

[

OAB a2

a

y2

(a 2

a 0

y2

0

a

a

(a

2

x3 3

2

y )x

0

2 a 3

x 2 ) dx dy

a

a2

x

y2

dy x 0

3

(a 2

y 2 ) 2 dy

0

2 5 a 3

/2

cos 4 d , on putting x = a sin . 0

2 5 3 1 a 3 4 2 2 8

a5 y (z

Example 5.11 Evaluate

x) dS , where S is the curved surface of the

S

cylinder x2 + y2 = 16, that lies in the positive octant and that is included between the planes z = 0 and z = 5 (Fig. 5.49). z E

C

D O

B

y

A x z

Fig. 5.49

D

C

We note that the projection of S on the xoy-plane is not a plane (region) surface, but only the arc AB of the circle whose centre is O and radius equal to 4. For converting the given surface integral into a double integral, the projection of S must be a plane region. Hence we project S on the zox-plane (or yoz- O plane). The projection of S in this case is the rectangular region OCDA, which is shown in Fig. 5.50.

5

x 4

A

Fig. 5.50

Part I: Mathematics I

I – 5.54

Converting the given surface integral I as a double integral, 2 x

y( z + x )

I=

2 y

2 z

y

OADC

dz dx,

x2 + y2 = 16 is the equation of the given cylindrical surface. where 2y; z = 0. 4(x 2 + y 2 )

y (z + x)

I

2y

OADC

= 2x;

y

=

dz dx

(z + x) dz dx

4

x

[

(x, y, z) lies on x2 + y2 = 16]

OADC 5

4

0

0

(z + x) dx dz .

4 5

(4 z

4

8) dz

0

8( z 2 + 4z)50 360. Example 5.12 Evaluate

xyz dx dy dz , where V is the region of space inside the x y z tetrahedron bounded by the planes x = 0, y = 0, z = 0 and 1. a b c Vide worked Example 5.11 in the section on ‘Double and triple integrals’ for V

a

x a

b1

c1

x a

I=

y b

xyz dz dy dx 0

a

0

0 x a

b1

0

0

c2 2

a

c2 2

a

c2 2

a

z2 xy 2

bt

xy t 0

0

y2 xt 2 2

0

0

1 2

2 3

c1

x a

y b

dy dx 0

y b

2

dy dx , where t

2t y 3 b 3

1 y4 b2 4

1 2 4 b xt dx 4

bt

dx 0

1

x a

Chapter 5: Multiple Integrals b2 c2 24

a

b2 c2 24

a

x1

x a

a1

1

0

4

0

ab 2 c 2 24

x 1 a 5 a

I – 5.55

5

dx

x a

x a

1

x 1 a 6 a

6

4

dx

a

0

a 2 b2 c2 1 24 5

1 6

1 2 2 2 a b c 720 Example 5.13 Express the volume of the sphere x2 + y2 + z2 = a2 as a volume integral and hence evaluate it. [Refer to Fig. 5.51] z

y

O

x Fig. 5.51

Required volume = 2 × volume of the hemisphere above the xoy-plane. Vide worked example 5.12 in the section on ‘Double and Triple Integrals’. a

Required volume

a2

x2

2

2

x2

y2

dz dy dx

2 a

a

x a2

a

0 x2

(a 2

2 a

Taking a2

a2

a2

x2 )

y 2 dy dx

x2

x2 = b2, when integration with respect to y is performed, a

V

b

b2

2 a

b

y 2 dy dx

Part I: Mathematics I

I – 5.56 a

b

b2

4

y 2 dy dx

a

y 2 b 2

4 a

y

b2 sin 2

2

b2

[

a 0

1

y b

y 2 is an even function of y ]

b

dx 0

a

x 2 ) dx

( a2 a

x3 3

2

a x

2

a

0

4 3 a 3 (x

Example 5.14 Evaluate

y

z ) dx dy dz , where V is the region of space

V

inside the cylinder x2 + y2 = a2 that is bounded by the planes z = 0 and z = h [Refer to Fig. 5.52]. z (x, y, h)

Q

(x, y, 0)

P

y

x Fig. 5.52

Note

The equation x2 + y2 = a2 (in three dimensions) represents the right circular cylinder whose axis is the z-axis and base circle is the one with centre at the origin and radius equal to a. a

a2

x2

h

(x

I a a

a

2

a2

x

2

y

x2

(x + y) h a

a2

x2

a

a2

x2

2h

x a

0

z ) dz dy d x

0

h2 dy dx 2

h dy dx 2 [by using properties of odd and even functions]

Chapter 5: Multiple Integrals a

2h

x a

h 2

a2

I – 5.57

x 2 dx

a

2h 2

a2

x 2 dx

a2

x

[

x 2 is odd and

a2

x 2 is even ]

0

2h

2

2

x a2 2

x

2

a2 sin 2

1

x a

a

0

a 2 h 2.

EXERCISE 5(c) Part A (Short Answer Questions)

7. Write down the formula that converts a surface integral into a double integral. 8. Evaluate (x 2 dy y 2 dx) where C is the path y = x from (0, 0) to (1, 1). C

(x 2

9. Evaluate

y 2 ) ds, where C is the path y = x

C

(x dy

10. Evaluate

y dx) , where C is the circle x2 + y2 = 1 from (1, 0) to

C

(0, 1) in the counterclockwise sense. 11. Evaluate dS , where S is the surface of the parallelopiped formed by S

x = ± 1, y = ± 2, z = ± 3. dS gives the area of the surface S]

[Hint: S

dS, where S is the surface of the sphere x2 + y2 + z2 = a2.

12. Evaluate S

dS, where S is the curved surface of the right circular cylinder

13. Evaluate S 2

x2 + y2 = a , included between z = 0 and z = h. 14. Evaluate dV , where V is the region of space bounded by the planes V

x = 0, x = a, y = 0, y = 2b, z = 0 and z = 3c.

Part I: Mathematics I

I – 5.58

dV gives the volume of the region V]

[Hint: V

dV , where V is the region of space bounded by x2 + y2 +

15. Evaluate V

z2 = 1. 16. Evaluate

dV , where V is the region of space bounded by x2 + y2 = a2, V

z=

, z = h.

Part B (1, 3)

[x 2 y dx (x 2

17. Evaluate

y 2 ) dy ] along the (i) curve y = 3x2, (ii) line y = 3x.

(0, 0 )

[(x 2

18. Evaluate

y 2 + x) dx (2x y

y ) dy ] from (0, 0) to (1, 1), when C is

C

(i) y2 = x, (ii) y = x. (a , 0 )

( y 2 dx x 2 dy ) along the upper half of the circle x2 + y2 = a2.

19. Evaluate (- a , 0 )

(x dy

20. Evaluate

y dx) , where C is the ellipse

C

x2 a2

y2 b2

1 and described

in the anticlockwise sense. [(x 2

21. Evaluate

y 2 ) dx

2 xy dy ] , where C is the boundary of the rectangle

C

formed by the lines x = 0, x = 2, y = 0, y = 1 and described in the anticlockwise sense. [(3 x 2

22. Evaluate

8 y 2 ) dx ( 4 y 6 xy ) dy ], where C is the boundary of the

C

region enclosed by y2 = x and x2 = y and described in the anticlockwise sense. y 2 ) ds, where C is the arc of the circle x = a cos , y = a

(x

23. Evaluate C

sin ; 0

2

.

x ds, where C is the arc of the parabola x2 = 2y from (0, 0 ) to

24. Evaluate C

1 1, . 2 [ xy dx (x 2

25. Evaluate

z ) dy

(y 2

x) dz ] from (0, 0, 0) to (1, 1, 1) along

C

the curve C given by y = x2 and z = x3. [ (3x 2

26. Evaluate

6 y ) dx 14 yz dy

20 xz 2 dz ] , where C is the segment of

C

the straight line joining (0, 0, 0) and (1, 1, 1).

Chapter 5: Multiple Integrals [3 x 2 dx

27. Evaluate

(2 xy

y ) dy

I – 5.59

z dz ] from t = 0 to t = 1 along the curve

C

C given by x = 2t2, y = t, z = 4t3. xy ds along the arc of the curve given by the equations x = a tan

28. Evaluate

, y = a cot . 3 29. Evaluate

, z

2 a log tan from the point

4

to the point

z 2 )ds , where C is the arc of the helix x = cos t, y = sin t,

(xy C

). z=t x 30. Find the area of that part of the plane a

y b

z c

1 that lies in the

Hint: Area of the surface =

positive octant.

ds S

z dS , where S is the positive octant of the surface of the sphere

31. Evaluate S

x2 + y2 + z2 = a2. 32. Evaluate xy dS , where S is the curved surface of the cylinder x2 + y2 = a2 z k, included in the positive octant. 33. Find the volume of the tetrahedron bounded by the planes x = 0, y = 0, z = 0, x y z 1. a b c 34. Evaluate z dx dy dz , where V is the region of space bounded by the V

sphere x2 + y2 + z2 = a2 above the xoy-plane. ( x2

35. Evaluate

y 2 ) dx dy dz , where V is the region of space inside the

V

cylinder x2 + y2 = a2 that is bounded by the planes z = 0 and z = h.

5.9

GAMMA AND BETA FUNCTIONS e

x

xn

1

dx exists only when n > 0 and when it

0

exists, it is a function of n and called Gamma function “Gamma n”]. (n)

Thus

e

x

xn

1

n) [read as

dx

0 1

xm

1

(1

x) n

1

dx exists only when m > 0 and n > 0 and

0

when it exists, it is a function of m and n and called Beta function and denoted by (m, n) [read as “Beta m, n”]. 1

Thus

xm

(m, n) 0

1

(1

x) n 1dx:

Part I: Mathematics I

I – 5.60

Note

x

e

(1) =

dx

( e x )0

1.

0 1

dx = 1 .

(1, 1) = 0

5.9.1

Recurrence Formula for Gamma Function (n) =

e

x

xn

1

dx

0

(x n

1

e x )0

(n

1) e

x

xn

2

dx

[integrating by parts]

0

(n

1) (n

1), since lim

This recurrence formula (n) = (n exists only when n > 1.

n

xn 1 ex

(n

0 n > 1, as (n

Cor. n + 1) = n !, where n is a positive integer. n + l) = n n) (n = n (n = n (n n (n =..................... n (l) = n (n =n!( (l) = 1)

Note

1. (n n is 0 or a negative integer. 2. When n is a negative fraction, (n) formula. i.e. when n < 0, but not an integer, 1 (n) (n 1) n 1 For example, ( 3.5) ( 2.5) ( 3.5) 1 1 ( 1.5) ( 3.5) ( 2.5) 1 ( .5) (3.5) (2.5) ( 1.5) (0.5) (3.5) (2.5) (1.5) (0.5) The value of (0.5) can be obtained from the table of Gamma functions, though its value can be found out mathematically as given below. Value of

1 2

Chapter 5: Multiple Integrals 1 2

1 2

e tt

I – 5.61

dt

0

e

1 2 x dx x

x2

0 x2

e

2

(on putting t = x2)

dx

0

Now

1 2

2 x2

e

2

dx 2

e

0

y2

dy

[

0

is only a dummy variable] e

4 0

(x 2

y2 )

dx dy

(1)

0

[ the limits are constants]. x xy-plane. Let us change over to polar co-ordinates through the transformations x = r cos and y = r sin . Then dx dy = |J| d r d = r d r d r Then, from (1), we have 1 2

2

y

0

. 2

/2

e

4 0

r2

r dr d

0 /2

1 e 2

d

4 0

r2 0

/2

2

d 0

1 2

5.9.2

Symmetry of Beta Function (m, n) = (n, m) 1

xm

(m , n) =

1

(1

x) n

1

dx

0 a

a

f (x) dx

Using the property 0

f (a 0

x) dx in (1),

(1)

Part I: Mathematics I

I – 5.62

1

(m, n)

(1 x) m 1 {1

(1

xn

1

x) n

1

dx

0 1 1

(1 x) m

dx

0

(n, m).

5.9.3 Trigonometric Form of Beta Function 1

x m 1 (1 x) n

(m, n)

1

dx

0

Put x = sin2

dx = 2 sin cos d

The limits for are 0 and

2

.

/2

sin 2 m

(m, n)

cos 2 n

2

2

2 sin cos d

0 /2

sin 2 m

2

cos 2 n

1

1

d

0 /2

sin 2 m

Note

1

cos 2 n

1

1 (m, n) 2

d

0

sin and dividing the sum by 2. The second argument is obtained by adding 1 to the exponent of cos and dividing the sum by 2. /2

0

5.9.4

p

1 2

sin p cos q d

Thus

2

1 q ,

1 2

Relation Between Gamma and Beta Functions ( m) ( n) ( m n)

(m, n)

( m) ( n)

Consider

e

t

tm

1

dt

0

s

e

sn

0

t = x2 and in the second, put s = y2. ( m)

( n)

e

2 0

x2

x2m

1

dx 2

e 0

y2

y 2n

1

dy

1

ds

Chapter 5: Multiple Integrals ( x2

e

4 0

y2 )

x2m

y 2n

1

1

I – 5.63

dx dy

0

/2 r2

e

4 0

(r cos ) 2 m

(r sin ) 2 n

1

1

r dr d

0

[changing over to polar co-ordinates] /2

cos 2 m

4

sin 2 n

1

1

e

d

0

r2

r 2m

2n 2

r dr

0

(m, n)

e

r2

r 2( m

u

um

n 1)

2r dr

0

(m, n)

e

n 1

[putting r2 = u]

du

0

(m, n)

(m, n)

( m) ( n) ( m n)

(m, n)

2

Cor. Putting m

n

1 in the above relation, 2

1 1 , 2 2

2 (1)

2

1 2

1 1 , 2 2 /2

sin 0

2

cos 0 d

0

1 2 WORKED EXAMPLE 5(d)

Example 5.1 Prove that

e

ax

xn

1

(n) , where a and n are positive. an

dx

0 1

x 0

e

In 0

ax

xn

1

q 1

log

1 x

dx , put ax = t, so that dx

p 1

dx. dt a

Part I: Mathematics I

I – 5.64

e

ax

x

n 1

dx

e

0

0

1 an

x

In I

q 1

0

1 log x

dt a

e t t n 1 dt 0

1 an 1

n 1

t a

t

(1)

(n)

p 1

dx ,

1 ey x i.e. x=e y Then d Also the limits for y put

y

dy and 0. 0

I

e

(q 1)y

e

qy

yp

( e y ) dy

1

y p 1 dy

0

1 qp Example 5.2 Prove that

(p ) [by (1)] . xm 1 (1 x) m

(m , n) = 0 1

n

dx .

xm 1 xn 1 dx . (1 x) m n

Hence deduce that (m , n) = 0 1

t m 1 (1 t ) n 1 dt

(m , n) =

(1)

0

In (1), put t

x 1 x

1 dx (1 x) 2

. Then dt

When t = 0, x = 0; when t = 1, x Then (1) becomes, x

(m, n) = 0

= 0 1

= 0

x m 1

1 x xm 1 (1 x) m xm 1 (1 x) m

n 1

1 1 x

n

dx

n

dx

t 1 t

1 dx (1 x) 2 (2)

1

xm 1 (1 x) m

n

d

(3)

Chapter 5: Multiple Integrals

In 1

xm 1 (1 x) m

n

1 . Then dx y

dx , put x

I – 5.65

1 dy y2

When x = 1, y = 1; when x = , y = 0 1 x 1

0

m 1

x) m

(1

n

y

dx 1

( m 1)

1 y

1 1

ym y)m

(1

0 1

0

yn 1 (1 y ) m

1

n n

ym

1

dy

dy

n

xn 1 (1 x) m

0

1 dy y2

m n

n

dx

[changing the dummy variable]

(4)

Using (4) in (3), we have 1

(m, n) 0

xm (1

xn

1

x) m

1 n

dx .

1

x m (1 x n ) p dx

Example 5.3 Evaluate

in terms of Gamma functions and

0 1

0

dx

.

1 xn 1

x m (1 x n ) p dx,

I

In

0

xn = t; n nx dx = dt 1 dt dx n 1 1n t When x = 0, t = 0; when x = 1, t = 1. 1 m 1 1n 1 I t n (1 t ) p t dt n 0 put then

1 n 1 n

1

t

m 1 1 n

(1 t ) p dt

0

m 1 ,p 1 n

Part I: Mathematics I

I – 5.66

m 1 n m 1 n

1 n

1

1

dx 1 xn

0

1 2

x 0 (1 x n )

( p 1) (1) p 1

dx

0

1 . 2

Here m = 0, n = n, p Using (1); we have 1

0

dx 1 xn

1 n

1 n 1 n

1 2 1 2 1 n 1 n

n

Example 5.4 Prove that

1 2 ( n)

(n, n) 22 n

(or)

(n, n)

1 2

1

n,

2n 1

1 2

n

1 2

/2

(n, n)

sin 2 n

2

1

cos 2 n

1

d

[using trigonometric form]

0 /2

2

2n 1

sin cos

d

0 /2

2 0

sin 2 2

sin 2 n

2

1

2 d

0

1 2

d

/2

1 22 n

2n 1

sin 2 n

2n 2 0

1

d , putting 2 = 2

Chapter 5: Multiple Integrals /2

1 22 n

/2

sin 2 n

2

1

f (sin ) d

d

0

1 2

I – 5.67

1 2

2n 2

1 22 n

0

n, n,

1

22 n

0

1 2

1 2 1 2 1 2

( n)

1

f (sin ) d

2

1

n ( n)

2

2n 1

a2 x2

xne

Example 5.5 Show that

1 2

n

1 2a n

dx

0

Deduce that

e

a2 x2

2a cos ( x 2 ) dx

sin ( x 2 ) dx

0

xne

a2 x2

n 1 . 2

. Hence show that

dx

0

In I

1

0

dx , put ax

t ; then dx

0

1 2

2

dt 2a t

When x = 0, t = 0; when x = , t = . n

t a

I 0

1 2a n 1 2a n

e t

1

dt

t

n 1 2

2a t e t dt

0

n 1 2

1

(1)

In (1), put n = 0. e

Then

a2 x2

dx

0

In (2), put a

1 i 2

; then a 2

i

1 2 2a

2a

(2)

Part I: Mathematics I

I – 5.68 2

eix dx

2 (1 i )

0

(1 i )

2 2

Equating the real parts on both sides, 1 2

2

1 2

2

cos (x 2 ) dx 0

.

Equating the imaginary parts on both sides, sin (x 2 ) dx 0

.

Example 5.6 Evaluate b

(i)

(x

a)m

1

(a

x) m

1

x) n

(b

1

dx and

a a

(ii)

(a

x) n

a)m

1

1

dx in terms of Beta function.

a b

(i) In I1

(x

x) n

(b

1

dx ,

a

put = y; then dx = dy When x = a, y = 0; when x = b, y = b a

ym

I1

1

(b

a)

y

n 1

dy

0 b a

a)n

(b

ym

1

1

1

0

n 1

y b

dy

a

y

t ; then dy = (b a) dt b a When y = 0, t = 0; when y = b a, t = 1.

In (1), put

1

I1

a)m

(b

n 1

n 1

t m 1 (1 t ) 0

a)m

(b

n 1

(m, n)

a

(a

(ii) In I 2

x) m

1

(a

x) n

1

dx ,

a

put a + x = y; then dx = dy When x = , y = 0; when x = a, y = 2a. 2a

y m 1 (2a

I2 0

y)n

1

dy

dt

(1)

Chapter 5: Multiple Integrals 2a

(2a )

n 1

ym

1

n 1

y 2a

1

0

I – 5.69

(2)

dy

y t ; then dy = 2a dt. 2a When y = 0, t = 0; when y = 2a, t = 1. In (2), put

1

I2

(2a ) m

n 1

(2a ) m

n 1

t m 1 (1 t ) n

1

dt .

0

(m,n) x2

e

Example 5.7 Prove that

x

0

e

In I1 0

When x = 0, t = 0; when x t 1 e t dt I1 1/ 4 t 2 t 2 0

x2 e

In I 2

dt 2x

dx , put x2 = t; then dx

1 2 x4

t

e

2 t

dt

0

1 4 ds 4 x3

0

When x = 0, s = 0; when x se

s

0

1 4

4 2

dt

3 4

t

dx , put x4 = s; then dx

I2

dx

0

x2

x

x4

x2e

dx

ds 4s3 / 4

s= ds 4s3 / 4

1 4

s

1 4

e s ds

0

3 4 e

x2

x

0

x2e

dx

x4

1 8

dx

0

1 4

3 4

From Example 5.4; 1

(n, n) i.e.

(n)

(n) (n) (2n)

1 2 2n

1 2

22 n

n

2

2n 1

(n) 1

n

(2n) 1

1 2

n,

1 2

(1)

Part I: Mathematics I

I – 5.70 1 , we get 4

Putting n

1 4

1 2

3 4

2 2

1 2

(2)

Using (2) in (1); e

x2

x

0

Example 5.8 Evaluate 0

8

. 4 2

xm 1 dx and deduce that (1 x n ) p

. Hence show that 0

xm 1 dx , put t (1 x n ) p

0

2

dx

0

m n sin n In I

x4

x2e

dx

dx x4

1

0

xm 1 dx 1 xn

. 2 2

1 xn

1

1 1 t dt nx n 1 dx t2 t When x = 0, t = 1; when x = , t = 0 Then x n

1

I

m 1 n

t

(1 t ) t

0

1 n

1

t

m 1 n

p

m 1 n

dt

p

nt 2 t m

(n 1) n

(1 t )

n 1 n

1

(1 t ) n dt

0

1 n

m m , n n

p

m m p 1 n n (p ) n Putting p = 1 in (1), we get

0

x m 1 dx 1 xn

1 n

n

1

cosec

m n m n

(1)

m n H int : Use ( ) (1

)

(2) sin

Chapter 5: Multiple Integrals

I – 5.71

Taking m = 1 and n = 4 in (2), we get

0

dx 1 x4

4

cosec

4

2 2 xm

Example 5.9 Find the value of the ellipse

Put

x a

x2 a2

y2 b2

a

Then dx

y n 1 dx dy , over the positive quadrant of y

1 , in terms of Gamma functions. y b

X and

1

x

O

Y b

dX and dy

dY . Fig. 5.53 2 X 2 Y The region of double integration in the xy-plane is given by x y x2 a2

y2 b2

1 , shown in Fig. 5.53.

The region of integration in the XY-plane is given by X shown in Fig. 5.54. The given integral m 1

a X

I

b Y

O AB

ambn 4 m

a b 4

n

X

m 1 2

Y

n 1 2

B

ab

n 1

4 X Y

Y

X+Y

Y

dX dY

X+Y=1 O

dX dY

A

O AB 1 1 Y

X 0

m 1 2

Y

n 1 2

Fig. 5.54

dX dY

0

I

ambn 4

1

a mb n 2m

1

n

Y2

1

0 n

Y2

1

2 Xm2 m 1 Y

m2

0

a mb n 2m

n m , 2 2

a mb n 2m

n 2 m 2

1 m 1 2 n 1 2

1 Y 0

dY

dY

X

Part I: Mathematics I

I – 5.72

m m n a mb n 2 2 2 m n 2m 1 2 2 m n a mb n 2 2 m n 4 1 2 Example 5.10 Find the erea of the astroid x 2/3 + y tions. By symmetry of the astroid, required area A = 4 × area of OACB

= a 2/3, using Gamma func-

y

dx dy

4

2/3

B

OACB

x Put a

23

i.e. x = aX

y X and a 3/2

and y = aY

23

C

Y O

3/2

3 3 aX 1 2 and dy aY 1 2 dY 2 2 The region of integration in the xy-plane is

A

x

dx

23

23

Fig. 5.55

x y 1 , as shown in Fig. 5.55. a a The region of integration in the XY-plane is given by X Y as shown in Fig. 5.56. given by x

y

X+Y

Y

A

4

Q

9 2 a 4

X 1 2 Y 1 2 dX dY O PQ

O

1 1 Y

X 1 2 Y 1 2 d X dY

9a 2 0

0

1

Y12

9a 2 0

2 32 X 3

1 Y

Fig. 5.56

dY 0

1

Y 1 2 (1 Y ) 3 2 dY

6a 2 0

6a 2

6a 2

P

3 5 , 2 2 3 5 2 2 (4)

X

Chapter 5: Multiple Integrals 1 2

a2

1 2

3 2

1 2

I – 5.73

1 2

[

3 2 a 8

1 2

[xy (1 x y )]1 2 dx dy , over the area enclosed by the Example 5.11 Evaluate lines x = 0, y = 0 and x + y = 1 in the positive quadrant. 1 1 x

x 1 2 y 1 2 (1 x

Given Intergral I 0

y

0 a

1

x 1 2 dx

y 1 2 (a

0

where a

y )1 2 dy dx, y )1 2 dy

0

x.

(1)

a

y m 1 (a

Consider

y ) n 1 dy

x

O

0 a

an

ym

1

1

1

0

y a

Fig. 5.57

n 1

dy

1

an

a m 1 z m 1 (1 z ) n 1 a dz

1 0

am

n 1

putting

y a

z

(m, n)

(2)

Note

This result (2) will be of use in the following worked examples also. 3 Using (2) in (1) note that m n , 2 1

x1 2 (1 x) 2

I 0

3 3 , dx 2 2

3 3 3 , ,3 2 2 2 3 3 3 (3) 2 2 2 9 (3) 2 1 1 3 2 2 2 7 5 3 3 2 2 2 2 2 105

Part I: Mathematics I

I – 5.74

Example 5.12 Show that the volume of the region of space bounded by the x a

coordinate planes and the surface Required volume is given by

y b

z c

1 is

abc . 90

dz dy dx , where V is the region of space given.

Vol V

Put

x a

X,

y b

z c

Y,

Z

i.e. x = aX2, y = bY2, z = cZ2 dx = 2aX dX, dy = 2bY dY, dz = 2cZ dZ 8abc XYZ dZ dY dX , where V is the region of space in XYZ-space

Vol V'

X

Y

X+Y+Z Z 1

Vol

8 abc

X

1

dX 0

X

X dX 0

XYZ dZ

X dX 0

Z2 Y dY 2

1

X

Y

Y 0

X

1

1

Y

0 1

0

4 abc

X

dY 0

1

8 abc

1

Y (1

X

Y ) 2 dY

X

0

Fig. 5.58 1

4 abc

X (1

X )4

(2, 3) dX [by step (2) of Example (5.11)]

0

(2) (3) (2, 5) (5) 1 2 (2) (5) 4 abc 24 (7 ) abc 1 24 720 3 abc 90 4 abc

dx dy dz

Example 5.13 Evaluate 1

x2

y2

z2

, taken over the region of space

in the positive octant bounded by the sphere x2 + y2 + z2 = 1. Put x2 = X, y2 = Y, z2 = Z

Chapter 5: Multiple Integrals dX

dx

2 X

dY

, dy

2 Y

The region of integration in xyz z2 The region of integration V in the XYZ X+Y+Z 1 1 1 Given intergral I X 2Y 2 Z 8 V 1

1 8

X

dX

1 8

1 2

X

y

1 2

X

(1 1

X

x2 + y2 +

z

Y

Z

Y -

1

Z ) 2 dX dY dZ

Y

dY

0

1 2

(1

X

dX

1 2

Y

1

dY (1

X

Y )2

1 1 2

0

0

-

1

Z ) 2 dZ

Y

0 X

1

1 2

2 Z

X

Y

0 1

dZ

, dz x

X

1

1 2

I – 5.75

1 1 , 2 2

[by step (2) of Example (5.11)] 1

X

8

1 1

1 2

0

1 1 , 2 2

0

1

X

4

X

dX 2Y 2

1 2

X )1/2 dX .

(1

0

1 3 , 2 2

4

1 2

3 2 (2)

4 2

8 Example 5.14 Evaluate

a 2 b2 c2

b2 c2 x2

c2 a2 y 2

a 2 b 2 z 2 dx dy dz ,

V

where V Put i.e.

x a

x 2

X, x

dx

y b

2

Y,

a X, y a

z c

x2 a2

z

2

y2 b2

z2 c2

X

Y

1.

Z

b Y, z

dX , dy

y

c Z b

dY , dz

2 X 2 Y The region V of integration in the XYZ Y + Z 1.

c 2 Z

dZ Z

X+

Part I: Mathematics I

I – 5.76 abc

Integral

X

1

Y

abc dX dY dZ

Z

8 X

V ' 1

a 2 b2 c2 8 2 2 2

abc 8

1 2

X

X

1

dX

1 2

Y

0

Y X

1

dY

Z

X

X

1

1 2

dX

1 2

Y

0

1 2

1

X

(1

Y

Z ) 2 dZ

0

0

1

Z

Y

dY (1

X

1 3 , 2 2

Y)

0

[by step 2 of Example (5.11)] 2 2 2

abc 8

1

1 3 , 2 2

1 2

X

3

1 ,2 2

X )2

dX (1

0

[by step 2 of Example 5.11] 2 2 2

abc 8

1 3 , 2 2 1 2

a 2 b2 c2 8 2

1 ,2 2

1 5 , 2 2 1 2

3 2

1 2

(2 2) 5 2

(2)

5 2 (3)

a 2 b2 c2 32 xl

Example 5.15 Find the value of

1

ym

1

zn

1

(1 x

y

z ) p -1 dx dy dz , taken

over the interior of the tetrahedron bounded by x = 0, y = 0, z = 0 and x + y + z = 1, in terms of Gamma functions. x

1

1

Given integral

x

l

1

dx

y

0

0

1

1 x

x l 1 dx 0

1 x m 1

y

zn

dy

1

(1 x

y

z ) p - 1 dz ,

0

ym

1

(1 x

y)n

(n, p ) dy ,

p -1

0

[by step (2) of Example (5.11)] 1

x l 1 (1 x) m

(n, p )

n

p 1

(m, n

p ) dx ,

0

[by step (2) of Example (5.11)] (n, p ) (m, n p ) (l , m n p ) (n) (p ) (m) (n p ) (l ) (m n p ) (n p) (m n p ) (l m n p ) (l ) (m) (n) (p ) (l m n p )

Chapter 5: Multiple Integrals

I – 5.77

EXERCISE 5(d)

Part A (Short Answer Questions) x2

x4e

1. Prove that

3 8

dx

0 x3

xe

2. Evaluate

. 5 3

dx , given that

0

0.902 .

/2

sin 3 x cos5 / 2 x dx .

3. Find the value of 0 /2

sin 5 cos 7 d .

4. Find the value of 0 /2

5. Find the value of

tan d in terms of Gamma functions. 0

/2

6. Prove that

1 2

cot d 0

1 4

/2

7. Find the value of

/2

sin d 0

0

/2

/2

dx

cos x dx

8. Prove that

0

cos x

dx

(n) .

0 1

1 log x

9. Prove that 0

0

0

d sin .

n 1

xn dx ( n 1) . nx

10. Find the value of 11. Assuming that

3 . 4

x 1 dx 1 x

sin

, prove that

( )

(1

is neither zero nor an integer. [Hint: put x = tan2 ] 12 Find the value of

e

kx 2

dx .

13. Prove that

(m 1, n) (m, n 1)

m. n

14. Prove that

(m 1, n) (m, n 1) m n

(m, n) . m n

)=

sin

,

Part I: Mathematics I

I – 5.78

xm 1 dx in terms of a Beta function. ( a + bx ) m + n

15. Find the value of 0

a t b

H int: put x 16. Prove that (m + 1, n) + (m, n + 1) = (m, n). 2

17. Find the value of

1 3

x3 )

(8

dx in terms of Gamma functions.

0 a

x m (a

18. Prove that

x) n dx

am

n

1

(m 1, n 1) .

0

20. Derive the recurrence formula for the Gamma function. n + 1) = n! 21. When n 1 . 2

value of Part B 23. Prove that

e 0

(ax 2 + by 2 )

x2m

1

1 4a m b n

y 2 n 1dx dy

0

(m) (n) . 1

x m (log x) n dx

24. When n is a positive integer and m 0 n

( 1) n! . ( m + 1) n +1 1

25. Prove that 0

x m 1 (1 x) n 1 dx ( a + bx ) m +n

( m, n) . a n (a + b) m H int: Put

26. Express

n

1 ,n 2

xe

x2

e

dx

x8

0

29. Prove that 0

x2e

dx

2

x4

dx dx

0

x 2 dx 1 x

1

4 0

(2 n +1) . ( n +1)

2n

x2

x

0

xe 1

1 2

n

0

28. Prove that

a

z bx

a

b

1 in terms of Gamma functions in two different ways 2

and hence prove that 27. Prove that

x

. 2 2 . 16 2

dx 1 x4

4

.

Chapter 5: Multiple Integrals

30. Evaluate (i) 0

dx , (ii) 1 x4

0

x 2 dx and (iii) (1 + x 4 ) 2

I – 5.79

0

x 2 dx (1 + x 4 )3

[Hint: put x2 = tan2 ] y x+y x y dx dy , taken over the area x m

31. Find the value of

n

terms of Gamma functions, if m, n > 0. 32. Find the value, in terms of Gamma functions, of taken over the volume of the tetrahedron given by x y+z

x m y n z p dx dy dz y

0, z x a

x+

2/ 3

y b

2/ 3

1

and the co-ordinate axes. x m 1 y n 1 (1 x y ) p 1 dx dy , quadrant enclosed by the lines x = 0, y = 0, x + y = 1. x y z 1 meets the axes in A, B and C. Find the volume of the 35. The plane a b c tatrahedron OABC. x2 y 2 z 2 1. 36. Find the volume of the ellipsoid 2 a b2 c2 37. Find the volume of the region of the space bounded by the co-ordinate planes 34. Evaluate

x and the surface a 38. Evaluate

n

y b

n

xyz (1 x

z c y

n

1 z ) dx dy dz, taken over the tetrahedral x = 0, y = 0, z = 0 and x + y + z = 1.

x 2 yz dx dy dz ,

39. Evaluate

octant bounded by x = 0, y = 0, z = 0 and xyz dx dy dz,

40. Evaluate 2 and x a2

y2 b2

z2 c2

x a

y b

z c

1. x

y

1.

ANSWERS

Exercise 5(a) (1) 4 1 (4) 2

(2) log a log b (5)

4

(3) 2 9 (6) 2

Part I: Mathematics I

I – 5.80

y y=b

(7)

1 2

x=a x

(8) x = – a

O y=–b Fig. 5.59

y y y=x

x=1

(9) R y=0

O

(10) R

x=0

x

x

O

y=0

Fig. 5.60

Fig. 5.61

x y + =1 a b

(11) x = 0 O

y=0

x2 + y2 = a2

1 1 x

1 1 y

f ( x, y ) dy dx (or)

(12) 0

x

0

f ( x , y ) dx dy 0

0

Fig. 5.62 a

b 2 a a

0

0

x2

a 2 b b

0

0

y2

f ( x, y ) dy dx (or)

(13) y

1

f ( x, y ) dx dy

1

1

0

x

f ( x, y ) dx dy (or)

(14) 0

0

2

1

f ( x, y ) dy dx 1 2 x

f ( x, y ) dx dy (or)

(15) y2 4

0

(16) 2 log 2 a 6

(19)

(24)

2

f ( x, y ) dy dx 0

0

(17)

a3 6

(20)

1 720

3

8 19 (21) log 2 3 9 a3

(27) 6

(29)

b

1 (8 log 2 5) 16

(22) 1 (25)

1 ab (a b) 3

(28)

33 2

(30)

1 48

(18)

4

(23)

3 2

(26)

344 105

Chapter 5: Multiple Integrals

I – 5.81

Exercise 5(b) a

a

f ( x , y ) dx dy

(1) 0

y

a

y

f ( x , y ) dx d y

(3) 0

f ( x, y ) dy dx

1

0

x

f ( x, y ) dy dx y

f ( x , y ) dx dy

(6) 0

0

1

1 y2

a

a2

0

0

f ( x , y ) dx dy 2

f ( x, y ) dy dx 0

0

(14) 1 (17)

9 3 a 5

(20)

a4

(24)

32 3

(32) a

1 3 2 3

3 a2 4 a4 log 1 (38) 4

(44) 16

a 2 h2

(16) 2 (19)

3 8

(25)

2 3

16 2 (27) 3 a

16 (28) 3 ab

(30) 3

(31)

1 e

(36) 2

241 60

(22) 8 log 2

a2 (3 (33) 2

3 2

(35)

2

(13) 1

(21) 3

4

2

0

16 (12) 3 1 2 (15) 2 (e 1) 1 log 2 (18) 2

a 4

2

f ( x, y ) dy dx

(10)

y2 4

8

0 1/ x

f ( x , y ) dx dy

(11)

x2

(8)

1

0

(41)

1

f ( x, y ) dy dx

0

(9)

(29)

0

0

(7)

(26)

0

a a

1 1 x

(23)

x

(4)

0

(5)

1

(2)

8)

(34)

a2

(37)

(42) 2 16 3 a (3 9

a2

a 4 2 (40) a

a2

4 4 3 a (39) 3

(45)

3 2

(43) 12 4)

(46) 4

a

Part I: Mathematics I

I – 5.82 2

(47) (50)

(48)

8

4

a4

(49)

16

a4

6a 2

Exercise 5(c) (8)

2 3

(9) 1

(11) 88

(12) 4

(14) 6 abc

(15)

69 29 ; 10 4

(17) (20) 2

2 (13) 2

2

4 3

(16) 2

a 4

(26)

13 3

(29)

2 3

(31)

a3 4 4

1 2 2 1 . 3 13 (27) 6

(30)

a4

(35)

2

h

(19)

(24)

1 2 2 a b 2 ka 3 (32) 2

3

2

4 3 a 3 3 (22) 2

2 2 ; 3 3

(18) (21) 4

2 (23) a 1

(34)

(10)

b2 c2

(25) (28)

163 70 2a 3 3

c2 a2 (33)

abc 6

a4 h

Exercise 5(d) 8 (3) 77

(2) 0.456 (5)

(12) (22)

1 2

1 4

k

1 (4) 120 1 (log n) n

3 4

(15)

1 a nbm

(30)

5 2 ; ; 2 2 8 2 128

(m, n)

(17)

1 3

1 3

1

(n 1)

2 3

Chapter 5: Multiple Integrals (31)

(m 1) (n 1) / (m

(32)

(m 1)

3) .

(n 1) ( p 1) / (m

3 ab (33) 32 abc (35) 6 (37) abc

n

(34) (36) 1 n

(39) a3b2c2/2520.

3

3n 2

I – 5.83

n

p

4)

( m) ( n) ( p ) / ( m 4 3

n

p)

abc

3 n

(38) (40) a2b2c2/48.

2

/1920.

PART II Mathematics-II 1. Ordinary Differential Equations 2. Vector Calculus 3. Analytic Functions 4. Complex Integration 5. Laplace Transforms

Chapter

1

Ordinary Differential Equations 1.1

EQUATIONS OF THE FIRST ORDER AND HIGHER DEGREE

nth dy dx

n

fn If we denote

dy dx

dy dx

f1 x, y 1

n 1

x, y

f 2 x, y dy dx

f n x, y

dy dx

n 2

0.

p

pn + f1 (x, y) pn–1 + f2 (x, y) pn–2 + …+ fn–1 (x, y) p + fn(x, y) = 0

(1)

p. y. x.

1.1.1 Type 1—Equations solvable for p n linear factors.

II – 1.4

Part II: Mathematics II

(p – F1) (p – F2) . . . (p – Fn) F1, F2, … , Fn are functions of x and y. Each of these n

p = F1, p = F2 , … , p = Fn, where

n c) = 0,…,

1

(x, y, c) = 0,

2

(x, y,

(x, y, c) = n 1

(x, y, c)

2

(x, y, c) …

n

(x, y, c) = 0.

1.1.2 Type 2—Equations solvable for y y x and p. y = f(x, p)

(1)

x, p

x, p ,

dp dx

(2)

x and p. (x, p, c) = 0 ... (3), where c

p

If p p.

1.1.3 Type 3—Equations solvable for x x y and p. x = f (y, p)

(1)

y, 1 p

dp dy

(2)

(y, p, c) = 0

(3)

y , p,

y and p.

where c p If p p.

Chapter I: Ordinary Differential Equations

Note

II – 1.5 y = f(x, p) and x

= f(y, p).

y = px + f(p)

y = px+ f(p)

(1)

w. r. t. x, p dp dx

0

p

x

f

p f

(2) or

dp dx

p

x

(3)

0

p=c

(4) y = cx + f(c). p c

p

p the singular solution of the equation (1).

Note x

df dp

p

y = px + f(p) and c

0

y = cx + f(c) and x

df dc

0 y = cx + f(c)

WORKED EXAMPLE 1(a)

Example 1.1 Solve the equation

dy dx

2

p2 – 8p + The equation is (p – 3) (p – 5) = 0 dy 3 or dx

8

dy dx

15

0. p.

dy dx

5

II – 1.6

Part II: Mathematics II y = 3x + c and y = 5x + c.

y – 3x – c = 0 and y – 5x – c = (y – 3x – c) (y – 5x – c) = 0. Example 1.2 Solve the equation p (p + y) = x (x + y). p2 + yp – (x2 + xy) = 0. y

y2

p, p

4 x2

xy

2 y

y

2x

2

2 dy dx dy dx

i.e. or

x

(1) x

y

(2)

x2 c . 2 2 2y – x2 – c = 0 dy y x, (2), dx ex y

i.e.

i.e.

(3) y.

y ex = – x ex dx + c = – x ex + ex + c –x y+x–1–ce =0

(4)

(2y – x2 – c) (y + x – 1 – ce–x) = 0. Example 1.3 Solve the equation p2 – 2py tan x – y2. p2 – 2py tan x – y2 = 0. p, p

2 y tan x

4 y 2 tan 2 x 2

2 y tan x

4 y2 1

y tan x

i.e.

i.e.

4 y2

tan 2 x

(1)

2 sec x

dy dy = y (tan x + sec x) or = y (tan x – sec x). dx dx dy dy = (tan x + sec x) dx (1) and = (tan x – sec x) dx (2) y y y x x + tan x) + y = c sec x (sec x + tan x)

c

Chapter I: Ordinary Differential Equations

II – 1.7

c (1 sin x) cos 2 x c 1 sin x y (1 – sin x) – c = 0

i.e. y=

x

i.e.

(3)

x + tan x) +

c

c sec x sec x tan x c 1 sin x

y

y (l + sin x) – c = 0

i.e.

(4) y (1 – sin x) – c y (1 +

sin x) – c] = 0. Example 1.4 Solve the equation xp2 – 2py + x = 0. p, p

2y

4 y2 2x

y x

y x

dy dx

i.e.

4 x2 2

1

y = vx, dv v x v v2 dx dv dx 2 x 1 v

i.e.

v2

v

(1)

1 (2) x

1

c

Solutions are y

y2 x

i.e.

y

y2 y

x2

2

x2 y2

y

y2

0 and

y

c and cx 2 x2

cx 2

y

y2

x2

c

y2

x2

x2

c

c 0.

Example 1.5 Solve the equation p3 – (x2 + xy + y2) p2 + (x3y + xy3 + x2y2) p – x3y3 = 0.

0

II – 1.8

Part II: Mathematics II

p3 – ( + + ) p2 + ( where

2

,

and

)p–

+

0,

. )=0

( )( )( 2 (p – x ) (p – xy) (p – y ) = 0. 2

i.e.

dy dx y

i.e.

+

2

3y

x3 3

x2 ,

c ;l 3

x3

3y

x

c y

ce

y2 1 y

lo c

ce x

0; y

dy dx

xy and x2 2

y

c

3

dy dx

2

/2

x2 2

x

c.

0; x

1 y

c

1 y

c

0.

x

0.

Example 1.6 Solve the equation p2x – 2py – x – 0. p2 x x 2p

y

1 x px 2 p y.

(1)

x

p

i.e.

2p p3

i.e. i.e. i.e.

p

x

dp 2 p dx

p

dp x dx

1 p 2

dp dx

p

x

dp dx

x

dp dx

p2 p p p2

1

dp dx

p2

dp x dx xp 2

x

p

x

dp dx 1

0 (

p2

1

0)

Chapter I: Ordinary Differential Equations dp p

dx x

II – 1.9

lo c

p = cx

i.e.

(2) c2x2 = 2 c y + 1.

Example 1.7 Solve the equation 16x2 + 2p2y – p3x = 0. p, nor for x. equation is rewritten as p 3 x 16 x 2 y 2 p2 x, dp dp p 3 x 16 x 2 2 p p 2 p3 3 p 2 x 32 x dx dx 2p p4 i.e.

2 p5

p5

3 p4 x

i.e.

p5

32 p 2 x

i.e.

p 2 p3

dp dx

2 p4 x

dp dx

32 px 2

dp dx

px p 3

32 x

dp dx

dp dx

0

p4 x

32 x

32 p 2 x

p2

i.e. or (2) is differential equation in p p

px

dp dx

32 px 2

y (1)

dp dx

0 (2)

p3 + 32x=0

(3)

p = cx

(4)

p dp p

(2)

dx x

p 16x2 + 2c2x2y – c3x4 = 0 3 2

2

i.e. c x – 2c y equation, 16x2 + 2p2y + 32x2 = 0 i.e. p2y = – 24x2 i.e. p6y3 = – (24)3 x6 i.e. 1024x2y3 + (24)3 x6 = 0 i.e. 16y3 + 9x4

II – 1.10

Part II: Mathematics II

Example 1.8 Solve the equation y = (1 + p) x + p2. y dp dp 1 p x 2p dx dx dp x 2p 1 0 dx dx x 2p dp

p i.e. i.e.

x, (1)

(2)

This is a linear equation in x. xe p

2 pe p dp

c

= – 2 (pe p – e p) + c x =2 – 2p + ce–p p

i.e.

(3)

x = 2 – 2p + ce–p and y = (1 + p) (2 – 2p + ce – p) + p2 x = 2 – 2p + ce–p and y = 2–p2 + c (1 + p) e–p.

i.e.

Example 1.9 Solve the equation y = x + p2 – 2p. y and for x. We shall solve the y = x + p2 – 2p . . . x,

Method I:

i.e.

p

1

2 p

1

y.

2p dp dx

2 p

dp dx 1

(1) 0

p–1=0 or

2

dp dx

(2) (3)

1

p

y = x – 1. 2p = x + c

p

y

(4)

x

x

c

2

x

4

i.e. 4 (y + c) = (x + c)2, Method II:

c

x x = y + 2p – p2

(1)'

y, dx dy

1

2

dp dy

2p

dp dy

Chapter I: Ordinary Differential Equations

i.e.

1 p

1

i.e.

(1

p) 2 p

2 (1

p)

dp dy

II – 1.11

d d 0.

1

p=1 2p

or

dp dy

(5)

1

0

(6)

p p2 = y + c

(7)

x –y + y + c = 2p (x + c)2 = 4 ( y + c) Example 1.10 Solve the equation p2x + py – y4 = 0.

Note

p y x. y4

x

py p

(1)

2

y, p2 4 y3

dx dy

p

y

p 4 y3

i.e. i.e.

p2

i.e.

y 2 y3

p

y

dp dy

4 py 3 p

p2 dp dy

or

dp dy

2 y4

2 py

dp dy

2 y4 2 p 2 y3

py p

dp dy 0

dp 2p 0 dy 2y3 – p = 0

(2) (3)

2dy y

p i.e.

py 2 p

p3

y

p

y4

p4

1 p

(2) is dp

dp dy

y

c p = cy2

(4)

II – 1.12

Part II: Mathematics II p c2xy4 + cy3 – y4 = 0

2

i.e. c xy + c = y p 4xy6 + y4 = 0 2

i.e. 4xy

Example 1.11 Solve the equation p3 – 2 x yp + 4y2 = 0. p nor for y. p2 y

2x

y, p2 y2

2 p i.e. i.e. i.e.

2p y

4 y dp p 2 dy

2 p3 p2 y p3

2 y2

2 y2 2y

2 p dp y dy 2 p

dp dy

p

0

dp dy

p

2dp p

dy y

lo y

c

(1)

4 y dp p 2 dy

0

1 p3 py 2

or

p

4 p

dp dy

2y

i.e. 2 l

p2 y2

4y p

2 y2

0

0

(2)

p3 – 2y2 = 0

(3)

lo c

i.e.

p2 = cy

i.e.

p cpy –2xyp + 4y2 = 0 p (c –2x) = – 4y cy (c –2x)2 = 16y2

i.e. c (c – 2x)2 = 16y, y2 – 2 x yp + 4y2 = 0 i.e.

i.e.

2x3y2 = 27y3 2x3 = 27y,

xp = 3y

(4)

Chapter I: Ordinary Differential Equations

II – 1.13

Example 1.12 Solve the equaion p3x – p2y –1= 0. x. y y

1 p2

px

y

1 c2

cx

(1)

c, x

0

2 c3

(2)

2 x

(3)

c c3 c2y = c3x – 1 c6y3 = –27 4y3 = –27x2

i.e.

Example 1.13 Solve the equation y = 2px + yp2. x y2 = Y dy dY 2y = dx dx i.e.

2yp = P, say. y, y2 = 2ypx + y2p2

(1)

Y

Px

P2 , 4

(2)

Y

cx

c2 4

(3)

cx

c2 . 4 c 2

(4)

General solution of (2) is

2 General solution of (1) is y

c, 0 c

x

Y = – x2

II – 1.14

Part II: Mathematics II

i.e. x2 + y2 Example 1.14 Solve the equation (px – y) (py + x) = 2p. X = x2 and Y = y2 dX = 2 x dx and dY = 2y dy y dY y dy i.e. P x p (say) x dX dx x P y

p

or

x2 P i.e.

PX

i.e.

Y

Y PX

y2

2P P 1 2P P 1

x P 1 y

2

x P y

2c c 1 Example 1.15 Solve the equation e4x (p – 1) + e2y p2 = 0 Note I eax and eby ky Y = e where k a and b X = e2x and Y= e2y dY 2e 2 y dy dX 2e 2 x dx dY X i.e. p P, where P = . Y dX 2 General solution is y

cx 2

X2

X P 1 Y

Y

X2 2 P Y2

X = ekx and

0

i.e. XP – Y + P2 = 0 i.e. Y = PX + P2 e2y = ce2x + c2.

EXERCISE 1(a) Part A (Short Answer Questions) p – f1 (x, y)] p – f2 (x, y)]=0. y = f (x, p), x = f (y, p), solution.

y. x.

Chapter I: Ordinary Differential Equations

II – 1.15

y = px + f (p)? 7. 8. 9. 10. 11.

Solve the equation p2 – 5p + 6 = 0. Solve the equation p2 – (x + y) p + x y = 0. Solve the equation p2 – (ex + e–y) p + ex–y = 0. Solve the equation x yp2 – (x + y) p + 1 =0. Rewrite the equation (y – px) (p – 1) = p

12. Rewrite the equation p

px – y)

13. Rewrite the equation p = sin (y – px) y

px

1 . p

y = px – p2. Part B 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

yp2 + (x – y) p – x – 0 2p2 – (x + 2y2) p + xy2 = 0. xyp2 + (3x2 – 2y2) p – 6xy = 0 xyp2 –(x2 + y2) p + xy = 0. x y 1 p p y x p2 + 2py cot x = y2. x2p2 – 2xyp + (2y2 – x2) = 0. y = – px + x4p2. 4y = x2 + p2. y = 2px + pn. y = 2px – p2. y = x + 2 tan–1 p y = (1+ p) x + ep. y = 3x + p p3 – 4xyp + 8y2 = 0. x = y + p2. y = 2px + 4yp2 y = 2px + y2p3. y2 y = xyp + p2.

y and (ii) x.

y = (x – 1) p + tan–1 p. y p = px.

37. 38. 39. 40.

x2 (y – px) = yp2 X = x2, Y = y2] 2 y = 2px + p y X = 2x, Y = y2] 2 2 (y + px) = px Y = xy] (p – 1) e3x + p3e2y = 0 X = ex, Y = ey]

II – 1.16

Part II: Mathematics II

1.2 LINEAR DIFFERENTIAL EQUATIONS OF SECOND AND HIGHER ORDER WITH CONSTANT COEFFICIENTS 1.2.1

Introduction

nth order with constant

a0

dn y dx n

a1

dn 1 y dx n 1

an

1

dy dx

an y

X,

(1)

where a0 ( 0), a1, a2, . . . , an are constants and X is a function of x. D

d , D2 dx

d2 , dx 2

(a0Dn + a1 Dn–1 + ...+ an–l D + an) y = X or f (D) y = X, where f (D quantity. When X

Dn

dn , dx n (2)

D,

f (D) y = 0

(3)

General Solution of equation (2) is y = u + v, where y = u (3), that contains n y=v u is called the complementary function (C.F.) and v is called the particular integral

1.3

COMPLEMENTARY FUNCTION f (D) y = 0 or (a0D + a1Dn–1 + . . .+ an) y = 0 auxiliary equation (A.E.) n

f (m) = 0

or

nth

a0mn + a1mn–1 + . . . + an = 0 D m equation in m.

(3) (4)

Chapter I: Ordinary Differential Equations

Case (i) The roots of the A.E. are real and distinct. m1, m2, …, mn. mx Then the solution of Equation (3) is y c1e 1 c1, c2, …, cn c1e m1 x

u

II – 1.17

cn e mn x , where

c2 e m2 x

cn e mn x

c2 e m2 x

Case (ii) m1, m1, m3, m4, …, mn Then the solution of Equation (3) is y

c1 x

c2 e m1 x

c3 e m3 x

cn e mn x

If three roots of the A.E. are equal, i.e. if ml = m2 = m3 (say), then the solution is c1 x 2

y

c2 x

c3 e

m1 x

c4 e m4 x

cn e mn x

r y

(c1 x r

1

c2 x r

2

cr

1

x

cr ) e m1 x

cr 1e mr

1x

cn e mn x

Case (iii) m1 = + and m2 – . Then the solution of Equation (3) is y = e ax (c1 cos bx + c2 sin bx) + c3 e m3x + g + cn e mnx . Case (iv) i.e. ml = m3 = + and m2 = m4 = – . Then the solution of Equation (3) is y = e ax 7c1 x + c2h cos bx + ^c3 x + c4h sin bx A + c5 e m5x + g + cn e mn x .

1.3.1

Particular Integral f (D) y = X

(2)

is the function v, where y = v X and .I. i.e. f (D)

1 1 X , where f (D) f (D)

1 X f (D)

f (D).

X. X is equal

Rule I X= eax, where

is a constant. P.I. =

1 e ax = 1 e ax , f]ag ! 0 f]ag f ] Dg

II – 1.18

Part II: Mathematics II

When f( ) = 0, ( )r is a factor of f(D). f(D) = ( )r (D), where ( 1 Then e ax = r ^ D ah z ]Dg 1 1 e ax F = < z^ah ^ D ahr 1 $ x r e ax $ = z^ah r 1 e ax x = e ax and D a 1! 2 1 ax = x e ax . 2e 2! ^ D ah Rule 2 X = sin ax or cos ax, where are used.

When

(–

2

1 sin ax = ( D2 )

1 sin ax ( – a2 )

1 cos ax = ( D2 )

1 cos ax, if ( – a2 )

) = 0, (D2 +

(D2) = (D2

2

)

2

(D2)

) is a factor of

(D2), where

1 sin (D 2 )

x= =

( – a2 )

(–a2)

0.

1 (D 2 ) (D 2 + 1 (–

2

)

1 D2 +

2

)

x.

sin

2

sin

x

1 cos ax 1 ; 1 cos ax E = z ] D 2g }^ a2h D 2 + a2 1 D2 +

sin x = – 2

2

x

cos

x

x = 2 and D2

1 +

2

x.

cos x = x sin 2 =

x 2

x x.

Note

f(D) D2

(D2). – 2, D 3 (aD + b).

2

D, D4

4

f(D

Chapter I: Ordinary Differential Equations

II – 1.19

1 = ] g sin ax f D 1 sin ax =] aD + bg ]aD bg = 2 2- 2 sin ax, a D -b

Then

aD – b). Then

1

I. a

2

2

a

2

2

b2

1

,

1 cos x f D

b

1 a

2

2

b2

2

aD b sin x,

a cos x b sin x , since D

a sin x

d dx

b cos x .

Rule 3 X =xm, where m

. I.

1 xm f D

Rewrite f(D

(D)], f(D).

Thus

I.=

1 k

aD 1 1 aD

k

1

D

xm 1

D

xm

(D)]–1

D, 1

D

1

aD k xm. 1 Note f D , since Dm+1 (xm) = 0, Dm+2 (xm) = 0 and so on. Rule 4 X = e . V(x), where V

1

to Dm

I.= 1 f D

Note Rule 5

1 e xV x f D

e

x

Dm or xm.

, cos

1 f D

V x

V(x) Thie rule is referred to as Exponential shift rule. X = x. V(x), where V(x) or cos

.

II – 1.20

Part II: Mathematics II

I=

1 xV x f D

1 V x f D

x

d dD

1 f D

V x

or =x

f

1 V x f D

D 2

f D

V x X = xrV(x), where r

Note X = xr cos ax

when

1 x r cos x f D

or xr sin ax. 1 art of x r ei x ] f D

= R . of

1 x r ei f D

of ei

1 x r sin x I. f D Rule 6 X is any other function of x. 1 I X f D 1 X, D m1 D m2 D mn

ei

x

1

x

f D 1

x

f D

A1 D

m1

D

A2 m2

D

xr

i

f D into linear factors.

resolvi =

xr

i

An X, mn

(1) artial fractions..

1

Consider

i.e.

Its solution is u e

D

m

1

D

m

D

m u

mx

u , say

X

D

e

m X

mx

X

or

D du dx

mu

m u X.

X dx u

Chapter I: Ordinary Differential Equations u

I.

A1e m1 x

e

m1 x

1 D

X

m

A2 e m2 x

X dx

e mx

e

e

m2 x

mx

II – 1.21

X dx

X dx

(2)

An e m n x

e m n x X dx

WORKED EXAMPLE 1(b) Example 1.1 Solve the equation (D2 – 4D + 3) y = sin 3x + x2. A.E. is i.e.

m2 – 4m + 3 = 0. m = 1, 3 (m – 1) (m – 3) = 0; C.F. = c1ex + c2e3x. 1 I. = 2 (sin 3 x x 2 ) D 4D 3 1 1 = 2 (sin 3 x) + 2 D D 4D 3 4D .I.2 (say) I.1 I.1 = =

=

D

2

9

1 4D

3

1 4D

3

1 2 2D 1 2D 2 4D2 1 2D 90

sin 3 x

sin 3 x

3 3

9

sin 3 x

3 sin 3 x

1 6 cos 3 x 90

I.2

sin 3 x

3 sin 3 x

1 2 cos 3 x siin 3x 30 1 x2 2 D 4D 3 1 x2 D 4 D 31 3

3

x2

II – 1.22

Part II: Mathematics II 1 1 3

D 4

1 1 3

D 4

1 1 3

4 D 3

1

D

x2

3

D2 4

D 3

1 2 x 3

D

9

2

x2

13 2 2 D x 9

8 x 3

26 9 y

i.e.

y

c1 e x

1

1 2 cos 3 x sin 3 x 30

c2 e3 x

1 3

Example 1.2 Solve (D2 + 4) y = x4 + cos2 x. A.E. is m2 + 4 = 0. The roots are m = ± i 2. C.F. = A cos 2x + B sin 2x. I.1

1 4

1 1 4

D2 4

1 1 4

D2 4

1 4 x 4 I.2

x4

D2

4 1

D2

x4 D4 4 x 16 3 2

3x 2

1 D2

1

cos 2 x

1 4 2

1 os 2 x co 2

1 1 e0.x 2 2 D 4 1 1 2 4 1 1 8

x sin 2 x 2 2 x sin 2 x

1 D

2

4

cos 2 x

2

x2

8 x 3

26 9

Chapter I: Ordinary Differential Equations

y

A cos 2x

1 4 8

B sin 2 x

6x2

2x4

II – 1.23

x sin 2 x

Example 1.3 Solve (D3 + 8)y = x4 + 2x + 1 + cosh 2x. A.E. is m3 + 8 = 0. i.e. (m + 2) (m2 – 2m + 4) = 0. m

2

2, m C.F.

Ae

I.1

3

4

or 1 i 3 2 e x (B cos 3 x C sin 3 x)

2x

1 D

8

(x 4

2x

1)

(x 4

2x

1

1 1 8

D3 8

1 1 8

D3 4 (x 8

1 4 (x 8 1 4 (x 8 I.2

16

x

8

1)

3x

1) 1)

e2 x

1 D3

2x

2x

1)

e

2x

2

1 1 e2 x 3 2 D 8

1 2 D2

D

1 1 2x e 2 16

1 1 e 12 D 2

1 1 2x e 2 16

1 x e 12 1!

4

e

2x

2x

1

2x

1 (3 e 2 x 4 x e 2 x ) 96 General solution is y 1 2 Example 1.4 Solve (D4 – 2D3 + D2)y = x2 + ex. A.E. is m4 – 2m3 + m2 = 0. i.e. m2(m2 – 2m + 1) = 0 The roots are m = 0, 0, 1, 1.

2D

D

e

x

x e 1!

x

II – 1.24

Part II: Mathematics II C.F. = c1 x I.1 =

c2 1

2

D D 1

2

c3 x

c2 e x .

x2

1 1 D D2

2

x2

1 1 2 D 3D 2 4 D 3 5 D 4 x 2 D2 1 2 3 4 D 5D 2 x 2 D2 D x4 12 I.2

2 x3 3 1

3x 2

2

D D 1 1 12

2

ex

2

ex

1 D 1

8 x 10

1 2 x D

x 2 dx

8 x 10

x2 x e . 2

2

x x e 2! General solution is y

c1 x

c2

c3 x

Note

x4 12

c4 e x

x+

2 x3 3

3x 2

c1x +

1

c2) of the C.F.

y

c1 x

c2

x4 12

c4 e x

c3 x

2 x3 3

3x 2

x2 x e . 2

Example 1.5 Solve (D2 + 1 )2 y = x4 + 2 sin x cos 3x. A.E. is (m2 + l)2 = 0 The roots are m = i, i, –i, –i C.F. = (c1x + c2) cos x + (c3x+ c4) sin x. I.1

1 2 2

1

D

1

D2

2

1 2D2 x4 I.2

24 x 2 1

x4 3D 4 x 4

1 D2

x4

2

72. 2 sin x cos 3 x.

Chapter I: Ordinary Differential Equations 1 D

2

1

sin 4 x sin 2 x

2

1 16 1

II – 1.25

1

sin 4 x

2

4 1

1 n4x sin 225

1 sin 2 x. 9 y = C.F.

2

sin 2 x

.

1

2

Example 1.6 Solve (D2 + 6D + 9)y = e–2x x3 A.E. is m2 + 6m + 9 = 0 i.e. (m + 3)2 = 0 The roots are m = –3, –3 C.F. = (c1x+c2) e–3x 1 e 2 x x3 I. = 2 D 3 1

2x

e

D

2

= e

2x

1

e

2x

1 2D

D

3 2

2

x3

yE

l shift rule

x3 3D 2

4 D3 x3

e 2 x x 3 6 x 2 18 x 24 General solution is y = C.F. Example 1.7 Solve (D3 3D2 + 3D – 1) y = e–xx3. A.E. is m3 3m2 + 3m –1 = 0 i.e. (m l)3 = 0 The roots are m = 1, 1, 1 C.F. = (c1x2 + c2x + c3) ex 1 e x x3 I. 3 D 1 1 e x. x3 3 D 2 3

1 x D 1 e x3 8 2 1 x 1 D D2 e 12 2 3 3 4 8 12 2 4 1 x 5 e 2 3D 3D 2 D3 x3 16 2 1 x e 2 x 3 9 x 2 18 x 15 16 y = C.F.

4 5

D3 3 x 8

II – 1.26

Part II: Mathematics II

Example 1.8 Solve (D2 – 4) y = x2 cosh 2x. A.E. is m2 – 4 = 0 i.e. (m + 2) (m – 2) = 0 The roots are m = – 2, 2 C.F. = A e–2x + B e2x 1 I. x 2 cosh 2 x D2 4 1 2

D 1 2x e 2

x2 2 x e 4 2 1 D

2

2x

e

4

1 2x 1 e x2 2 2 D 4D

1 2x 1 e 1 8 D 1 1 D

D 4

1 2 x x3 e 8 3 1 e 8

2x

(the have

s

1

x2 D2 16

D 4 D2 16 x2 4

x3 3

x3 sinh 2x 12

1 e 2

D 4

1 2x 1 e 1 2 4D

1 e 2

x2

2

1

2x

D 1 e 2

2

D 2 2 2x

D3 2 x 64

x2 4

x2

4D

1 1 4D 1 e 8

D 4

1

x2

2x

D3 2 x 64 x 8

x2 4

1 32 x 8

x2 cosh 2x 16

1 2x 1 e and e 256 256 included in the C.F.)

1

2x

2x

1 32 x sinh 2x 32

are o itted, as they

considered to

Then the G.S. is y

Ae

2x

B e2 x

x 8 x 2 sinh 2 x 96

6 x cosh 2 x 3 sinh 2 x

Example 1.9 Solve (D4 – 2D2 + 1)y = (x + 1)e2x. A.E. is m4 – 2m2 + 1 = 0 i.e. (m2 – 1)2 = 0 The roots are m = ±l, ±l C.F. = (c1x + c2) ex + (c3x + c4) e–x.

Chapter I: Ordinary Differential Equations 1

I. D2

1

2

e2 x x 1

D 1

e2 x D

e

2

1 1 9

2x

1 2x 1 e 9

1

e2 x

4D

3

D D

2

2

x 1

1

x 1

2

2

4

x 1

3 2D D 3

2

II – 1.27

x 1

4

1 2x 8 e 1 D x 1 9 3 5 3

1 2x e x 9

y Example 1.10

Solve (D2 + 2D – 1)y = (x + ex)2.

A.E. is m2 + 2m –1 = 0. i.e. (m + 1)2 = 2 m

1

2 real roots C.F.

Ae

I.

D

2

D

2

I.1 I.1

1

2 x

Be

2 x

1

1 x ex 2D 1 1 x2 2D 1 I.2 1

D2

e2 x

x2

4 x 10 . 1 e2 x 2D 1

Ae

2x

Be

2x

2

1 2 D 5D 2 x 2

I.2

x

2 xe x

I.3 say

1 D D 2

x2

e

1 2x e 7

1 D( D 2) D 2 D 2

2

x2

II – 1.28

Part II: Mathematics II I3

D

1 2D

2

1

1

2e x D 2e x

1

e

x

2 D 2

1

1

x

D D

1 2e 1 2 ex 1

2

1 4D

D2

x

x

2 xe x

1

4

x

2

2D x

x

2

y = C.F.

1

.

2

2

3

–x

Example 1.11 Solve (D + 5D + 4) y = e sin 2x. A.E. is m2 + 5m + 4 = 0. i.e. (m + 1) (m + 4) = 0. The roots are m = 1, 4 C.F.=Ae–x + B e–4x. 1 e x sin 2x I. D 2 5D 4 e

1

x

D e

x

e

x

e

x

1

2

1 D

2

3D

5 D

sin 2x 1

4

sin 2x

1 sin 2x 3D 4 3D 9D

2

4 16

sin 2x

1 x e 6 cos 2x 52 1 x e 3 cos 2x 26 y = C.F.

4 sin 2x 2 sin 2x

Solve (D4 – l)y = cos 2x cosh x Example 1.12 A.E. is m4 – l = 0 i.e. (m – l) (m + l) (m2 + l) = 0 The roots are m = 1, 1, ± i C.F. = c1 ex + c2 e–x + c3 cos x + c4 sin x.

Chapter I: Ordinary Differential Equations

I. =

1 D

4

1

1 x e 2

cos 2x

D

1

2

1 1

4 D3

6D2

D

1 D4

4 D3

6D2

1 16 D

4D

24

1 x 3D e 8 9D2

2 4

6 sin 2x

4D

1 e 8

x

1 e 2

1

1

x

16

x

3D 9D

1 e 320

2

2

4 x

16 D

24

2 cos 2x

x

cos 2 x cosh x

The Example 1.13 A.E. is i.e.

2

y = C.F. 4D + 13) y= e2x cos 3x

Solve (D m2 4m +13 = 0 (m 2)2 = 9

The roots are m = 2 ± i3 C.F. = e2x (A cos 3x + B sin 3x) I.

D

2

1 4D

13

D e2 x

e 2 x cos 3x 1

e2 x 2

1 D

2

9

2

4 D

cos 3x

1 x e 2 x sin 3 x 6

4D

cos 2x 6 sin 2x

1 ex e cos 2x 2 80

1 3 sin 2 x sinh x 80

cos 2x

4

1 cos 2x D 3 2

x

2 cos 2x

3 ex e sin 2x 80 2

1

cos 2x

cos 2x

1 e 8

cos 2x

1

x

cos 2x

4D

1 x 1 e cos 2x D 8 3 2

1 x e 320

1 e 2

cos 2x

4

x

1 x e 2 16

x

e

1

1 x e 2 D4 1 e 2

ex

II – 1.29

cos 3x 2 e2 x

13 x sin 3x 2 3

coss 2x

II – 1.30

Part II: Mathematics II y = C.F.

The

Solve (D 2

Example 1.14 A.E. is

D

1) y

x

e

sin 2

m2 + m + 1 = 0 1 2

The roots are m

3 2

i

x 2

C.F.= e

I.

2

3 x 2

A cos

1 D

x 2

D 1

1 1 e 2 2 D D 1 1 e 2 1 e 2

1 e 2

x

1

x

D

2

D 1

e

x

1

x

(D 1)

1 e 2

x

3 x 2

1 cos x 2

x

e

B sin

2

(D 1) 1

1

x

D

2

D

1

cos x cos x

cos x

1 x 1 x 1 cos x e e D 2 2 1 x e (1 + sin x) 2 General solution is y = C.F. Solve (D2 + 2D + 5) y = ex cos 3 x. Example 1.15 A.E.is The roots are m

m2 + 2m + 5 = 0 1

i2 C.F.

I.

e

x

1 D2

2D 1

2

5

(A cos 2 x

B sin 2 x).

e x cos3 x ex

3 cos x 4

1 cos 3 x 4

2D 5 D 3 x 1 e cos x 2 4 2(D 1) + 5 (D 1) 1 x 1 e cos 3 x 2 4 2(D 1) + 5 (D 1) 1 1 x 1 3 x cos x e e 2 2 4 4 D D 4D 8 4D

8

cos 3 x

Chapter I: Ordinary Differential Equations

II – 1.31

3 x 1 e cos x 4 4D 7

1 x 1 e cos 3 x 4 4D 1 7) 1 x (4 D 1) cos x e cos 3 x 4 49 16 D 2 1

3 x (4 D e 4 16 D 2

3 x 1 x e (4 sin x 7 cos x) e (12 sin 3 x 260 580 y = C.F. .I. Solve (D2 + 4D + 8) y = 12e–2x sin x sin 2x

cos 3 x)

Example 1.16 A.E. is m2 + 4m + 8 = 0. The roots are m 2 i2

I.

D

2

= 6e

2x

= 6e

2x

= 6e

2x

1 4D

8

(D

2) 2

D

2 e 5

2x

6e

1 2

4

2x

e

C.F.

2x

(A cos 2 x

(cos x

1 4(D (cos x

2)

B sin 2 x)

cos 3 x)

8

(cos x

cos 3 x)

cos 3 x)

1 cos x 3

1 cos 3 x 5

(5 cos x

3 cos 3 x)

y = C.F. Solve (D3 – 1) y = x sin x.

Example 1.17

A.E. is m3 – 1 = 0 2 i.e. (m – 1) (m + m + 1) = 0 The roots are m

1,

C.F.

I.

1 D x

3

1

x

3

1

sin x

1 D

3 2

i

c1 e x

e

x/ 2

c2 cos

3 x 2

c3 sin

3 x 2

x sin x

1 D

1 2

1

sin x

3D 2 (D

3

1)

2

3 (D

1) 2

sin x sin x

1 xV f (D)

x

1 V f (D)

f (D) f (D)

2

V

II – 1.32

Part II: Mathematics II (D 1) sin x D2 1

x

D 3 cos x. 2

x (cos x sin x) 2 y = C.F. 1

I.

D

3

D

3

1

1

= I.

1

sin x

of x e ix )

(I

D

I. of e

3

1

x e ix 1

ix

I. of e ix

(D

i )3

D3

3i D 2

1 1

x

3D

I. of

1

I. of

e ix 1 1 i

3D x 1 i

I. of

e ix x 1 i i)

(11 2

D

1 3 (cos x 2 2 3 cos x 2

1

x

i

( 3

3iD

(cos x

i sin x) x

sin x)

sin x)

x (cos x 2

A.E. is m2 + 4 = 0 i2 The roots are m C.F. = A cos 2x + B sin 2x. 1 D

2

4

D )

x

i

1

x

of x 2 ei 2 x

3 (1 2

i (sin x 3 (sin x 2

sin x)

Solve the equation (D2 + 4) y = x2 cos 2x

I.

2

3

1 {(cos x 2

I. of

i

1

e ix 1 1 i

I. of

Example 1.18

1

x sin x

1

of

3 2D

2

i)

cos x)} x cos x)

3 (1 i ) 2

Chapter I: Ordinary Differential Equations of e i 2 x of e i 2 x

(D

1

iD 4

of

e i2x 1 4iD

iD 4

x2

i i 2 x x3 e 4 3 1 (sin 2 x 4

x3 3

x2

4iD

e i2x 1 4 iD

of

x2

4

1 D2

of

of

1 4

1 i 2) 2

II – 1.33

D2 16

iD 3 2 x 64

i 2 x 4 i cos 2 x)

x sin 2 x 8

1 2 x 4

x 8

i 32

x3 3

x 8

i 2 x 4

1 8

1 cos 2 x 8

General solution i s y = Solve (D2

4D + 4) y = 8x2 e 2x sin 2x m2 – 4m + 4 = 0 (m 2)2 = 0

Example 1.19 A.E. is i.e.

Roots are m = 2, 2. C.F. = (c1 x+ c2) e2x. I.

1 (D

2)

2

8 x 2 e 2 x sin 2 x cos 2 x 2

1 x2 D

8 e 2x

8 e 2x

1 D2

x 2 sin 2 x

sin 2 x 4

2x

2

yi

= e 2x

1 ( 4 x 2 cos 2 x) D

= e 2x

e

4 x2

sin 2 x 2

cos 2 x 4

4 x

cos 2 x 2

sin 2 x 4

2x

2 x 2 ) sin 2 x

4 x cos 2 x

(3

y=

Bernouilli's 1 (2 cos 2 x) D

1 (4 x sin 2xx) D 2x

cos 2 x 8

2

sin 2 x

sin 2 x 8

II – 1.34

Part II: Mathematics II

Solve (D2 + a2) y = sec a x Example 1.20 A.E. is m2 + a2 = 0 ia The roots are m C.F. = A cos a x + B sin a x. 1 I. = 2 sec a x D a2 1 D

ia D

1 2ia D ia

ia

sec a x

1 2ia sec a x D ia

1 eia x 2ia

e

ia x

1 e 2ia

sec a x d x 1 D

m

X

ia x

e ia x sec a x d x

e mx .

1 1 ia x e ia x 1 i tan a x d x e 2ia 2ia 1 1 ia x i lo ec a x e ia x x e x 2ia 2ia a x e iax e 2i a

ia x

x sin a x a

1

1

lo

ec a x

cos a x lo

ec a x

a2 a2

eia x

mx

Xe 1

i tan a x d x

i lo a e

dx

ec a x

ia x

2

General solution is y

EXERCISE 1(b)

Part A (Short Answer Questions) 1. Solve the equation (D2 – D + 1)2 y = 0. D – 1)3 y = 2 cosh x. D2 + a2) y = b cos ax + c sin ax. D2 + 4D + 4) y = x e–2x. D – 3)2 y = x e–2x. D + 1)2 y = e–x cos x. D2 – 2D + 5) y = e x sin 2x. D2 + 4D + 5) y = e–2x cos x.

Chapter I: Ordinary Differential Equations

II – 1.35

D2 – 2D + 6)y = e x (4 sin x + cos x). 1 f (x). D a Part B 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35.

(D3 + D2 + D + 1)y = x2 + 2e–x (D2 + 9)y = x2 + cosh x (D2 + 2D + 1)y = x3 + cos 2x (D2 _ 8D + 9)y = 8 sin 5x + x2 (D2 + 3D + 2)y = 2 sin2 x + 2x2 (D4 + D3 + D2)y = 12x2 + 2 cos 2x cos x (D2 _ l)y = 12ex (x + l)2 (D3 – 6D2 + 12D – 8)y = 16x3 e4x (D3 + 2D2 + D)y = x2e2x (D2 – 4) y = x sinh x (D2 + l)2y = 2x2e–x (D2 – 5D + 4)y = (2x + e–x)2 (D2 – 4D + 3)y = 8ex cos 2x (D4 – 1)y = cos x cosh x (D2 – 2D + 5)y = ex (sin x + cos x)2 x ( D 3 1) y e x cos 2 2 (D2 + 4)y = 4 e2x sin3 x (D2 – 4D + 3)y = sin 3x cos 2x (D2 – 2D + 1)y = x ex sin x (D2 + D)y = x cos x (D2 – 4D + 4)y = x sin x (D2 – l)y = x2 cos x (D2 + 4)2 y = cos 2x (D2 + l)y = x2 sin 2x (D2 + 4)y = 4 tan 2x.

EQUATIONS

a0x n

dn y dxn

a1 x n

1

d n -1 y dxn 1

a n 1x

where a0, a1, … , an are constants and X is a function of x linear differential equation.

dy dx

an y

X

(1)

II – 1.36

Part II: Mathematics II

Note

x and the order

x to t x = et or t

x

dy dx

dy dt dt d x

dy dx

dy dt

x 1 dy x dt (2)

x, 2

i.e.

x

d y dx 2

dy dx

x2

d2y d x2

x x2

i.e. d dx

D and

d dt

d 2y 1 dt 2 x

dy dx

d2y dt 2

d 2y d x2

d 2y dt 2

dy dt

2

(3)

,

and 2

x D

2

2

(

1) x3 D3 = ( x 4 D4 = (

) ( – 2) 1) ( – ) ( – 3) and so on. a0

a1

1

n

an y

2

1

n

1

0 , which is a linear differential equation with

s a0 a x

b an

1

n

d ny d xn

a1 a x

ax b

dy dx

b

an y

n 1

d n -1 y dxn 1

X

(2)

ax + b = et. Equation (2) is called

.

Chapter I: Ordinary Differential Equations

1.5

II – 1.37

SIMULTANEOUS DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

If x and y

t f1 (D) x + f2 (D) y = T1

(1) (2)

z1 ]Dgx + z2 ]Dgy = T2 where f1, f2,

1

,

2

d and T1 and T2 are functions dt .

D

,

of t

2

f2(D)

(D)

y. f1 D

2

D

f2 D

1

D x

2

D T1

f 2 D T2

f D x T

or

which is a linear equation in x and t The value of x y. x and y equal to the order of the resultant equation (3).

Note

x,

y and t

WORKED EXAMPLE 1(c)

2 Example 1.1 Solve the equation x

d 2y dy 4x 2y 2 dx dx

x 2 D 2 4 xD 2 y x 2 x = et or t =

x and denote 1

i.e.

d dt

x2

1 , where D x2

. 2 y

4 2

3

e2 t

2 y e 2t

e

2t

e

2t

1 . x2 d dx

(3)

II – 1.38

Part II: Mathematics II

A.E. is m2 + 3m + 2 = 0 i.e. (m + 1) (m + 2) = 0 The roots are m = 1, 2.

I.

(

t

Ae

C.F.

1 1) (

1 2t e 12 1 2t e 12

Be

2) 1 2

(e 2t

e

2

(

1)

1] y

sin 2t sin t

1) y

1 ( sin 3t 2

sin t )

x) + B

x).

1

1 (sin 3t sin t ) 12 t 1 1 sin 3t ( cos t ) 2 8 2 1 1 sin (3 lo x lo x os 16 4 2

General solution is y = C.F. Example 1.3 Solve (x2 D2 – 2 xD x = et or t =

x)

,

A.E. is m2 + 1 = 0. The roots are m = ± i C.F. = A cos t + B sin t = A cos I. =

d dt 1)

i.e. The roots are m = 4, 1.

x)2

4) y

x

i.e. A.E. is

x

x) ·

d dt

x

i.e.

)

1 l x2

1 2 x 12

y = C.F. Example 1.2 Solve (x2 D2 + xD + 1) y x = et or t

2t

B x2

2t

e

2t

te

A x

2t

(

32 t 2

4] y

2 2

,

3

4) y

32t 2

m2 3m 4 = 0 (m – 4)(m + l) = 0 C.F.= A e 4t

Be

t

x)

Chapter I: Ordinary Differential Equations B x

A x4 I.

1 3

4

8 1

(

2

32t 2 1

4

t2

3)

2

81

(

4

3)

3 4

81 8 t2

13 16

2

3 t 2

13 8

2

12

8 lo x

3) 2 t 2

(

16 t2

x) 13]

General solution is y Example 1.4 Solve(x 2 D 2

x = et or t =

d dt

x

(

2

2

, 2t

1)] y t 2 e

1) i.e.

2

lo x x.

xD 1) y

1) y t 2 e

2t

A.E. is m2 2m + 1 = 0. i.e. (m – l)2 = 0 The roots are m = 1, 1. B ) et

C.F. (At

(A l

x

B)

1 e 2t t 2 1) 2 1 t2 = e 2t ( 3) 2

I.=

(

=e

2t

1 = e 9 1 = e 9

1 1 9 3 2t

2t

1 t2

2 3 4 t 3

2

t2 2

3

9 2 3

t2

1 x

II – 1.39

II – 1.40

Part II: Mathematics II 1 3 lo x 27 x 2

2

x 2

4

General solution is y = C.F. Example 1.5 Solve (x 2D2 xD + 4) y = x 2 x

et or t

x).

d dt

x

4 y e 2 t sin t

1 i.e.

2

A.E. is The roots are m 1 i 3

m2

4 y e 2t sin t

2

2m + 4 = 0

C.F.= et A cos 3 t x A cos

..

4 2

e 2t

e 2t

B sin

2

1 2

2

1 2

3

sin t

2 4

2

4

sin t

sin t

2 3 sin t 4 2 9

1 2t e 2 cos t 3 sin t 13 1 2 x 2 cos l x 3 sin l 13 General solution is y = Example 1.6 Solve 2 x 3 x + 3 = et or

t

2

d2 y dy 2 2x 3 12 y 6 x. 2 dx dx x + 3)

Then i.e.

x

3l

e 2t sin t 1

e 2t

e 2t

x

3l

1 2

2

B sin 3 t

2x

3

dy dx

dy 2 dt 2 x 3

dy dx

2 y

x

Chapter I: Ordinary Differential Equations d2 y dx 2

4

12 y

3 et

3

8

12 y

t

3

2

3 y

2x 4

1

2

2

i.e. A.E. is The roots are m = 3,

2

2 2 4

i.e.

3

1 y

3 e

3 t e 4 3=0

m2 2m

3

1

C.F. Ae3t

Be

..

1 2

2

3 4

t

3

A 2x 3

B 2x 3

3 t e 34

1 t e 4

General solution is y = C.F. Example 1.7 Solve (x2 D2 + xD+1) y =

3

1

3 2x 3 16

x = et or t

II – 1.41

3 4

x

x)

d dt

x

1 y

1 2

i.e.

1 y t sin t 2

m +1=0

A.E.is The roots are m = ± i C.F. = A cos t + B sin t I.=

t sin t

1 2

1

of eit t

I

. . of eit

1 i

2

t

i it 1 i e 1 t 2 2

. . of

1 t2 sin t i cos t 2 2 1 2 t cos t 4

i 1 1 2i 2

. . of

i it t 2 e 2 2

1

. . of

t sin t 4

. . of eit

it 2

1

t it 2

II – 1.42

Part II: Mathematics II

1 lo x in lo x 4 General solution is y = C.F.

1 4

2

x cos l

x

Example 1.8 Solve (x2 D2 + 4xD + 2) y = sin x. x = et

1) + 4 + 2] y = sin (e–t )

( i.e. A.E. is The roots are

d dt

x=t

2

) y = sin (e–t ) m + 3m + 2 = 0 m = 1, 2. C.F. = A e–t + B e–2t. (

2

A x

B x2 1 1

I.= 1

1 1

e

sin e t

2

t

2

sin e t

sin e t e t dt e 1 X D m

e

t

sin u du e

e t cos u e t

e cos e

t

2t

sin e t e 2 t dt

em x 2t

2t

Xe

dx

u sin u du , u cos u

t

mx

e cos e

t

et

u

sin u e

2t

sin e t

1 sin x x2 General solution is y = C.F.

Example 1.9

d dt

dx 2 x 3 y 5t dt dy 3x 2 y 2 e2 t dt

D, (D + 2)x – 3y = 5t –3x + (D + 2) y = 2e2t (D + 2); (D + 2)2x – 3(D + 2)y = 5 + 10t –9x + 3(D + 2)y = 6e2t

(1) (2)

Chapter I: Ordinary Differential Equations

II – 1.43

and (2) (D2 + 4D – 5) x = 5(1 +2t) + 6e2 t

(3)

2

A.E. is m + 4m 5 = 0 The roots are m = 1, 5 C.F. = A e t + B e–5t. I.=

D2

1 5 1 2t 4D 5

1

1

1

1

D D

x

Aet

13 5

Be

1 2t

4D 1 2t 5 8 5

D2

1 e2 t 4D 5

1

4

5

1 2t 2t

6

6 2t e 7

6 2t e 7

6 2t e 7

6 2t e 7

5t

2t

13 5

6 2t e . 7

y, Method 1 x (D2 + 4D

y C et

5) y = 15t + 8e2 t

De

5t

3t

(4)

12 8 2t e 5 7

Note

In the solution of y used in the solution of x.

A and B x and y x and y should contain

C and D

A and B

x and y in Equation (1), Ae t 5 Be

5t

12 2t e 7

3Ce t 3De

i.e.

2 5t

2 Ae t

2 Be

24 2t 36 e 9t 7 5

3(A – C) et – 3(B + D) e–5t = 0 C = A and

5t

D = – B.

12 2t e 7 5t

4t

26 5

II – 1.44

Part II: Mathematics II

and

x

A et

Be

y

A et

Be

6 2t e 7

5t

2t

13 5

3t

12 5

8 2t e 7

5t

Method 2 Dy D2x + 2Dx

D;

3Dy = 5

9x + 6y + 3Dy = 6 e2t

6y y x

x

x

1 x 6

x

A et

x

Be

9 x 5 6e 2 t 9x

6e 2 t

6 2t e 7

5t

2t

5

(5)

13 5

(6)

x w.r.t. t; x

Aet

5e

5t

Aet

25 Be

12 2 t e 7

2

(7)

24 2 t e 7

(8)

t; x

i.e.

y

1 6

y

Aet

6 Aet

Be

Example 1.10 Solve Dx

D; D

2); (D2

i.e. 2

A.E. is m 2m + 2 = 0 The roots are m = 1 ± i

6 Be

8 2t e 7

5t

(D

5t

3t

5t

48 2t e 7

18t

72 5

12 5

2) y = cos 2t

2) x + Dy = sin 2t (D 2) y = cos 2t 2) x + Dy = sin 2t D(D 2) y = 2 sin 2t 2)2x + D(D 2)y = 2cos 2t 2sin 2t. D2 4D + 4) x = 2cos 2t 4sin 2t 2D + 2) x = cos 2t 2sin 2t (D Dx (D D2x (D

(1) (2)

(3)

Chapter I: Ordinary Differential Equations

II – 1.45

C.F. = et (A cos t + B sin t) I.

D2

1 2D

2

cos 2t 2 sin 2t

1 1 cos 2t 2 sin 2t . 2 D 1 1 10

1 D 1 . 2 5

2 sin 2t 4 cos 2t cos 2t 2 sin 2t

x et A cos t B sin t

cos 2t 2 sin 2t 1 cos 2t 2

1 cos 2t 2

(4)

2Dx – 2x + 2y = sin 2t + cos 2t 2y = 2x – 2x + sin 2t + cos 2t

(5)

t; x = et (A cos t + B sin t) + et (–A sin t + B cos t) + sin 2t

(6)

2y = 2A et sin t – 2B et cos t – sin 2t y = et A sin t – B cos t) –

1 sin 2t 2

(7)

Example 1.11 Solve D2x – Dy – 2x = 2t Dx + 4Dy – 3y = 0 (D2

2)x – Dy = 2t

Dx + (4D

3)y = 0

(1) (2)

AD – 3); (D2 – 2) (4D – 3) x –D(4D – 3) y = 8 – 6t

(1)

D2x + D(4D – 3) y = 0

(2)

D

and (2) (4D3 – 2D2 – 8D + 6)x = 8 – 6t i.e.

(2D3 – D2 – 4D + 3)x = 4 – 3t

(3)

II – 1.46

Part II: Mathematics II 2m3 – m2 – 4m + 3 = 0

A.E. is i.e.

(m – l) (2m2 + m – 3) = 0

i.e.

(m – l) (m – l) (2m + 3) = 0 3 2

The roots are m 1, 1,

1

I. D 1

2

B et

At

C.F.

1 1 3

4 3t 2D 3

1 2D 1 1 2 D 4 3t 3 3

3 t 2

Ce 2D 3

1 1 3

1

1 D

2

4 3t

4 D 4 3t 3

1 4 3t 4 3 t. x

B et

At

3 t 2

Ce

(4)

t

Dy 4D2x + Dx – 8x – 3y = 8t y

and

1 4x 3

x

8 x 8t 3 Ce 2

x

At

B et

Aet

x

At

B et

2 Aet

Example 1.12 Solve

3 t A B et

1 Ce 6

3 t 2

3 t 2

9 Ce 4

x, x' and x'' in (5), y

(5)

1 3

d2 x 3x 4 y 0 dt 2 d2 y x y 0 if x = y 1 dt 2

1 3 t 2

Chapter I: Ordinary Differential Equations dx dt

and

dy dt

0 , when t

II – 1.47

0

(D2 – 3)x – 4y = 0

(1)

x + (D2 + l)y = 0

(2)

D2 + 1); (D2 + 1) (D2 – 3)x – 4 (D2 + 1) y = 0

4x + 4(D2 + 1) y = 0 D4 – 2D2 + 1) x = 0 A.E. is (m2 – l)2 = 0 Roots are m = ±1, ±1 x = (At + B) et+ (Ct + D) e–t

(3)

x' = (At + B)et + A et – (Ct + D) e–t + C e–t

(4)

x'' = (At + B) et + 2 A et + (Ct + D) e–t – 2C e–t

(5)

Solution is x w.r.t. t;

y

1 x 4 1 4

3x 2 At

1 At 2

y

B et B et

1 At 2

2 Aet A t e 2

2 Ct 1 Ct 2

1 Ct 2

B et

D e D e

t

D e

t

t

2C e C e 2

t

t

(6)

(7)

x = 1 when t = 0 in (3), B+ D=1 y = 1 when t = 0 in (6), A B 2

C

D 2

1

(8)

II – 1.48

Part II: Mathematics II A–B–C–D=1

i.e

(9)

= 0 when t = 0 in (4), A+B+C–D=0

(10)

= 0 when t = 0 in (7), –B + D = 0 3 1 , B , C 2 2

A

and

x

1 3t t e 2

y

1 2

(11) 3 1 , D 2 2

1 3t e 2

3 t t e 4

1 2

t

3 t e 4

t

Example 1.13 Solve (D2 – 5)x + 3y = sin t –3x + (D2 + 5) y = t D4

y A.E. is m – 16 = 0 The roots are m = ±2, ±i 2.

16)x = 4 sin t

3t

4

C.F.= A e2t + B e–2t + C cos 2t + D sin 2t I.= 4.

1 D

4

16

3

sin t

D

4

16

4 3 D4 sin t 1 15 16 16 3 D4 4 1 sin t 16 16 15

t 1

t 1

t

4 3 sin t t 15 16

and

x

Ae 2t

x'

2 Ae 2t

2B e

x ''

4 Ae 2t

4B e

Be

2t

4 3 t sin t 15 16 4 cos t 2 C sin 2t 2 D cos 2t 15 4 sin t 4 C cos 2t 4 D sin 2t 15

C cos 2t 2t

2t

D sin 2t

(1)

Chapter I: Ordinary Differential Equations y

1 5 x x sin t 3 A 2t B 2t e e 3C cos 2t 3 3

1 sin t 5

3D sin 2t

II – 1.49

5 t 16

EXERCISE 1(c) Part A (Short Answer Questions) xy + y + 1 = 0 into a linear equation with constant 2. Solve the equation x2y 3. Convert the equation 4. Convert the equation x4 5. Solve the equation x2 6. Solve the equation x3 7. Solve for x

= 0. –1 y = x2 as a linear equation with constant

3 3

2 3x2

Part B D

2

1 as a linear equation with constant

(n + 1)y = 0. 0. and

= 1.

d dx

(x2D2 + 2xD – 20)y = (x2 + 1)2. (x4D3 – x3D2 + x2D)y = 1. x). (x3D3 – x2D2 + 2xD – 2)y = (x2D2 + xD – 9)y = sin 3 x). (x2D2 + 9xD + 25)y x)2. 4 4 3 3 2 2 x)2. (x D + 6x D + 9x D + 3xD + l)y 2 2 2 x) . (x D – 3xD + 4)y = x (x4D4 + 2x3D3 + x2D2 – xD + 1)y = x2 x. 1 2 2 16. x D xD 3 y cos 2 lo x x 17. (x2D2 + 3xD + 5)y = x cos x). x + 2)2 D2 + 3(3x + 2) D – 36] y = 3x2 + 4x + 1

8. 9. 10. 11. 12. 13. 14. 15.

x + 1)2D2 + (x + 1 ) D + 1 ] y =

x + 1) D

20. (D + 4) x + 3y = t. 2x + (D + 5) y = e2t.

d dt

II – 1.50

Part II: Mathematics II

21. (2D + 1) x + (3D + 1) y = et (D + 5)x + (D + 7) y = 2et. 22. Dx + y = sin t x + Dy = cos t 23. 2D2x Dy 4x = 2t. 2Dx + 4Dy – 3y = 0. 24. D2x + y = 3e2t Dx Dy = 3e2t. 25. (D2 + 4) x + y = 0 (D2 + 1) y – 2x= 1 + cos2 t. 26. D2x 2Dy x = et cos t. D2y + 2Dx y = et sin t. 27. (D2 + 4) x + 5y= t2. (D2 + 4) y + 5x = t + 1.

1.6

x = 2 and y = 0 at t = 0.

LINEAR EQUATIONS OF SECOND ORDER WITH VARIABLE COEFFICIENTS

x.

d2 y dx 2

p x

dy dx

q x y

r x ,

d 2 y is unity and p(x), q(x) and r(x) are functions of x. In dx 2 this section,

1.6.1

.

Method of Reduction of Order-Transformation of the Equation by Changing the Dependent Variable

d2 y dx 2

d2 y dx 2

p x

p x

dy dx

dy dx

q x y

r x

(1)

0

(2)

y = u (x)

(3)

q x y

Chapter I: Ordinary Differential Equations

II – 1.51

y = u (x) . v(x)

(4)

is a solution of equation (1). =

(5)

and

2

pu)

( i.e.

(

pu)

+ qu) v = r

+(

= r, v

i.e.

v

i.e. v

(6)

2

u u

p v

r u

2u u

r1 , where p1

p1v

r . u

p and r1

(7)

w in 7 , it dw dx

p1 w

r1

(8)

w y to w, we have reduced the order of the

Note

r(x) = 0, then equation 2u u

v i.e.

d v

p v 2u u

v

lo v

x,

0 p

2 lo Au

v

Au

2

u 2

pdx e

e

v and hence the solution of equation (1).

A second order differential equation in y canonical or normal form.

lo A

pdx

(9)

II – 1.52

1.6.3

Part II: Mathematics II

Reduction to canonical or normal form (x) (x)y = r(x) (x)v = g(x), where f (x) =

The second order linear differential equation 1 {p x }2 4

q x y

ve

1/ 2

1 p x and g x 2

p x dx

r x . e1/ 2

n

,

p x dx

Proof: y = uv is a solution of the equation y'' + p(x) y = uv

(x) y = r(x)(1) (2)

x, and The values of y, i.e. i.e.

= + + p( pu) + (

2 (

pu

i.e.

u u

i.e.

lo u c

0, we

tu=e

ny

Thus the s transfor

uv

u

and

u ''

u, u and 1 r it u

12

ve

) + quv = r + qu) v = r

+

(5)

0

p 2 pdx 2

c

pdx

pdx

12

equationn 1 into the canonical

When

v

+

and

2u

Assu

2

(3) (4)

s

e

12

pdx

p2 e 4

,u '

12

pdx

p e 2

12

pdx

p e 2

12

pdx

v

1 u u

pu

qu

Chapter I: Ordinary Differential Equations p2 4

v

p2 2

p 2 1 2 p 4

i.e.

v

q

i.e.

v

f x v

1/ 2

pdx

1/ 2

pdx

q v

re

1 p v 2

re

II – 1.53

g x , where 1 p x 4

f x

q x

g x

r x e

1/ 2

1 p x and 2

2

p x dx

1.6.4 Method of Reduction of Order—Special Types of Equations d2 y

Type 1. Equations of the form 2 dx absent. dy dx

p, w

et

d2 y dx 2

f x,

dy dx

, in which y is explicitly

dp dx

equation.

p

dp dx

f x, p ,

dy dx

x , c1

y=

(x, c1, c2).

Extension:

dn y dx n

f x,

dn 1 y can dx n 1

2 Type 2. Equation of the form d y2 dx absent.

dy dx

f y,

dn 1 y dx n 1

dy dx

p and treat p as a function of y. Then dp dy

f y, p ,

p and

in which x is explicitly

d2 y dx 2

dp dx

p

dp dy

II – 1.54

Part II: Mathematics II

dy dx

p

x=

y, c1

(y, c1, c2)

Extension: f{y, , y , ... y(n)}

dy = p dx

p as a function of y.

Type 3. Equations f (x, y, y', y'') = 0, which are homogeneous in y, y' and y'' (but not in x) y

e

solved. When y

zdx

e

, zdx

,y

ze

zdx

and y

z2

z e

zdx

. Thus, the order of the

z

Type 4. Equations f (x, y, , y ) = 0 which are exact, viz. which can be expressed as d 0 . x, y, y dx

d dx

x, y, y

x, y, y'

0 is

c1 , which is

a equation.

Note 0

p0(x)D2 + p1(x)D + p2(x)]y = r(x) 1

+ p2 = 0.

p0 D2 + p1 D + p2) y = D(q0 D + q1) y

Conversely, when

q0 D2 + (

p0 = q ; p1 + q1 and p2 0 p1 + q1, 0 = 0+ 1 1 = 0 + p2 – + p2 = 0 0 1 – 1 + p2 = 0, we have 0 (p0 D2 + p1D + p2)y = (p0 D2 + p1 = p0 + p1

0

+ q1)

]y

1

1

– 1 y – 1

)y y 0 0

Chapter I: Ordinary Differential Equations = (p0

) + (p1

0

1

= D(p0 ) + D(pl y) – D( = D(p0

l

0

0

II – 1.55

y) – (

0

0

)

y)

y)

2

Thus (p0 D + p1 D + p2)y =

0

1

+ p2 = 0.

WORKED EXAMPLE 1(d) 2(x + 1) + (x + 2)y = (x – 2)e2x,

Example 1.1 Solve the equation the order of the equation.

Important Notes p0(x)

(x) y + p2(x). y =

1

(i) If p0(x) + p1(x) + p2(x) = 0, y = ex is a solution of the equation. (ii) If p0(x) – p1(x) + p2(x) = 0, y e–x is a solution of the equation. (iii) If p1(x) + xp2(x) = 0, y = x is a solution of the equation. p0 = x, p1 = – 2(x + 1) and p2 = x + 2. The condition p0 + p1 + p2 = y = ex equation. y = vex = vex

Then y

x

v ex + 2v ex + vex

y, , x(ex

– 2 = (x – 2)ex

i.e.,

i.e.,

+ 2ex + exv) – 2(x + 1) (ex + exv) + (x + 2) exv = (x – 2)e2x

dp dx

2 p x

1

dv dx

2 x e , where p x

This is a I.F. = e

2 dx x

e

2 lo x

1 x2

Solution of this equation is p x2

1 x2

2 x e dx x3

3c1

II – 1.56

Part II: Mathematics II 1 x e x2

d

dv dx

i.e.

3c1

1 x e x2

3c1

3c1 x 2

ex

v =c1x3 + c2 +ex

y = ex (c1x3 + c2+ cx) Example 1.2 Solve the equation

d2 y dx 2

dy dx

1 cot x

y cot x

sin 2 x,

y cot x = sin2 x

(1)

+ (l – cot x) – y cot x = 0

(2)

+ (1 – cot x) p0=1; p1 = 1 – cot x; p2 = – cot x p0 – p1 + p2 = 0 y = e–x is a solution of the equation

y = ve–x Then

–x

– ve–x

and

–x

–2

y, (

–x

–2

–x

–x

+ve–x

and y

+ ve–x) + (1 – cot x) (

–x

– ve–x) – ve–x cot x = sin2 x x

(1+ cot x)

i.e.

dp dx

1 cot x p

I.F. = e e

sin2 x

(3)

e x sin 2 x

(4)

1 cot x dx

x lo

in x

e x sin x

Chapter I: Ordinary Differential Equations

II – 1.57

Solution of equation (4) is pe x sin x

cos x i.e.

e x dx sin x

e x sin 2 x

dv dx

p

c1

c1

e x c1 sin x sin x cos x

(5)

x, v

c1

c1 x e sin x 2

y

1 2

e x sin x dx cos x

A sin x

e x sin 2 x dx

c2

1 ex sin 2 x 2 5

2 cos 2 x

c2

1 sin 2 x 10

2 cos 2 x

Be x .

cos x

d2 y Example 1.3 Solve the equation 1 x 2 dx 2 reduction of order.

2x

dy dx

2y

2,

x2)

2

2y = 2

(1)

(1 – x2)

–2

+ 2y = 0

(2)

(1 p0 = 1 – x ; p1 = –2x; p2 = 2 2

p1 + p2x = 0. y = x is a solution of

y = vx Then y, , i.e. i.e.

i.e. where

=

+

2

satisfy equation (1) (1 – x2)(

2 ) – 2x(

x(1 – x2)v dp dx

2 x

) + 2vx = 2

x2) – 2x2]v = 2 2x p 1 x2

2 x 1 x2

(3)

II – 1.58

Part II: Mathematics II

Equation (3) is a 2 2x dx x 1 x2

I.F. e e

1 x2

2 lo x

x2 1 x2 Solution of equation (3) is 2

px 2 1 x 2

x 1 x2 x2

dv dx

i.e.

x 2 1 x 2 dx

c1

c1 c1

1 1 x2 1 1 x2

2

x 1 x2 1 1 c1 2 1 x2 x

(4)

x v

c1

1

c1

1

1 1 x lo 2 1 x

c1 x

c2

1 x 1 x

c1

c2 x

dy dx

y

Solution of equation (1) is y

2

x lo

d2y Example 1.4 Solve the equation x 2 dx a solution.

2

x

If y = u is a solution of the equation (x) + q(x)y = 0, then y = uv 0, where v

c1u 2 e

0

y

+ p(x)

1 is x

+ q(x)y =

p x dx

y

2 1 y x

1 y x

0

(1)

Chapter I: Ordinary Differential Equations p x

Since y

y

2 1 and q x x

II – 1.59

1 x

1 is a solution of equation (1), x

1 v is also a solution of (1), where x 2 1 dx x

2

v

c1 x e c1 x 2 e

2 lo x x

c1 x 2 x 2 e x v

1 c1e x x

y

c1e x

c2

c2

or

2 Example 1.5 Solve the equation d y dx 2

solution. y

c1e x

c1e x

xy

2 dy x dx

y

c2

1 sin x v x v

c1 c1

1 sin x is a x

y

0

, where x2

sin 2 x 2 x

2

sin x

e e

2 dx x ,

1 sin x and p x x

since u

2 x

c1

2 lo x

sin 2 x

v = c1 cot x + c2 or A cot x + B y

:. i.e. Example 1.6 Solve the equation

xy

A cos x

d2 y dx

2

2x

dy dx

1 sin x A cot x x

B sin x x2 y

0

B

II – 1.60

Part II: Mathematics II 2

y=0

+2

(1)

+ p(x)

(x) y = r(x)

(2)

We have p(x) = 2x; q(x) = x2; r(x) = 0. p x dx

y = uv, where u

2

e v

x2 2

e p2 4

q

,

p v 2

re

p 2d x

(x2 x2 1) v = 0

i.e.

=v

i.e.

(3)

= f(v, ) p as a function of v, we have v

dp dv

p

p

dp dv (4)

v

p2 = v2 + c 12 dv dx

p

or

1

sinh or

v

v2

v c1

x

c12

(5)

c2

c1 sinh x c2 ex

c1 Ae x

c2

e 2

Be

x

x c2

The required solution of equation (1) is y

Ae x

Example 1.7 Solve the equation 4 x 2

Be d2 y dx

2

x

e

4x

x2 / 2

dy dx

x2

1 y

0

Chapter I: Ordinary Differential Equations

1 y x

y 1 ,q x

p

1 1 4

1 x2

and r

1 1 4

1 x2

y

II – 1.61

(1)

0

0

y = uv u 1

y

x

1 2

e

pd x

dx x

1 2

e

1 x

v

v

1 1 4

v

i.e.

p2 4

q

1 x

p v 2

1

2

4x

1 2

v

v

A cos 1

Solution of equation (1) is y

x

x 2

A cos

d2 y Example 1.8 Solve the equation x dx 2 = 0.

dy dx

B sin x 2

v

0

1 v 4

0

2 x2

i.e.

re

p dx 2

(3)

x 2

B sin

x 2

x2e x

y(0) = – 1 and (0)

y dy dx

p

p as a function of x, we have

dx

x

i.e.

d2 y

dp dx

p

dp dx

1 p x

2

dp ; the equation dx

x2e x

xe x

(1)

II – 1.62

Part II: Mathematics II p.

Equation

p

Solution of (1) is

1 dx x

e

I.F. 1 x

1 x

1 xe x dx x

2c1

p = x (ex + 2c1)

i.e.

dy dx

or

xe x

xe x dx

y

2c1 x dx

c2

1) ex + c1x2 + c2.

y = (x

y(0) = 1, c2 = c1 = 0, the required solution is y = (x Example 1.9 Solve the equation

(2)

2c1 x

d2 y

dy dx

dx 2

(0) = 0 in (2), c1 is

1)ex.

2

1

y(0) = 0, (0) = 0

Method 1 +

2

=1

(1) y

dy dx

p

p as a function of x, we have

dp dx

1

p2

x

c1

x

c1

Solution of equation (2) is dp 1 i.e.

1 1 lo 2 1

p2 p p

Given that p = 0, when x = 0 c1 = 0

d2 y dx

2

dp dx

(2)

Chapter I: Ordinary Differential Equations Thus we have

1 p 1 p

II – 1.63

e2 x e2 x 1 e2 x 1

dy dx

p

ex ex y=

e e

x

(3)

tanh x

x

x + c2

y

c2 = 0

Solution of equation (1) is y

x.

Method 2 The equation 2

y

=1

(1)

x dy dx

p as a function of y, we have

p

p

dp 1 dy

p2

d2 y dx 2

p

dp dy

(2)

Solution of equation (2) is p dp 1 p2

y

c1

1 lo 1 p 2 2

i.e.

y

c1

Given that y = 0 and p = 0, when x = 0 or when y = 0, p = 0 c1 = 0 1 – p2 = e–2y

Thus i.e.

p

dy dx

1 e

dy 1 e

2y

2y

x c2

(3)

II – 1.64

Part II: Mathematics II e y dy

i.e.

x c2

e2 y 1

cosh–1 (ey)= x + c2

i.e.

x = 0, y = 0.

When

c2 = 0

The required solution of equation (1) is ey = cosh x or y = Example 1.10 Solve the equation x

y lo x

y

d2 y dx 2

dy lo dx

(1) y

p as a function of x, we have y

dp dx p = xev, we have

dp dx

p lo x ev

xev

ev 1 x

i.e.

1 dy x dx

y x

y =f (x, y =p

x.

x

dv dx

p x

dp . dx

(2)

dv dx

dv dx

ev lo ev

v 1

(3)

Solution of (3) is dv v 1

lo c

v = 1 + cx

i.e. i.e.

dx x

lo

p x

1 cx

p

dy dx

xe1

cx

(4)

Chapter I: Ordinary Differential Equations

y

cx

x e1 c

e1 cx c2

II – 1.65

c

The required solution of equation (1) is c2y = (cx – 1)e1+cx d2 y dx 2

Example 1.11 Solve the equation

dy 1 y dx 2

2 y 1 y

y

2

2

0

(1)

0

y = f(y, ), which does not contain x p as a function of y, we have y

p

i.e.

dp dy

1 y

p2

0

dp p

2 dy 1 y

0

2

p

dp . dy

(2)

solution of (2) is p

y dy dx

i.e.

p c1 1 y

dy 1 y

2

1

i.e.

1 y

c1 x c2

c1 x c2

The required solution of equation (1) is y

1

1 . c1 x c2

c1 2

(3)

II – 1.66

Part II: Mathematics II

Example 1.12 Solve the equation xy 2

xyy –

y (1), we have

e

zdx

d2 y dx 2

2

dy x dx

y

dy dx

0.

– 0 y, , y .

and hence y

(1)

zdx

ze

z2

and y

zdx

z e

in equation

2

e

zdx

x z2

xz 2

z

z

xz

i.e.

0 z=0

(2)

z = c1x y

e

c1 xdx

y = Ae Bx

or d2 y Example 1.13 Solve the equation y 2 dx yy

y

2

y

e

zdx

and hence y

c1

x2 2

c2

2

dy dx . 1 x2 y

2

1 x2

yy y,

dy dx

e

(1)

and y . ze

zdx

z2

and y

z1 e

zdx

in equation (1), we

have 2 zdx

z1

z2

i.e.

dz dx

2z2

i.e.

1 dz z 2 dx

i.e.

du dx

e

z2

z

0

1 x2 z 1 x2 1 z

1 1 x2 1 1 x2

u

0

(2)

2

2

(3)

Chapter I: Ordinary Differential Equations 1 z

u

where

II – 1.67

Equation 1

I.F. e e

lo

1 x2

dx

1 x2

x

1 x2

x

Solution of equation (3) is x

1 x2 u

x2

i.e.

1 x 2 dx c1

2 x

x 1 x2

x2

x 1 x2

1 d lo 2 dx

y e

y2

x 1 x2

c1

1 x2

x

c1

zdx

1 2 lo

x2

x2

x 1 x2

e

or

1 x2

x

lo

x2

c1

1 x2

x

z

1 x2

x

lo

A x2

x 1 x2

x

1 x2

lo

x

x 1 x2

Example 1.14 Show that the equation x

x

c2

1 x2

c1

1 x2

x

lo

d2 y dx 2

c1

2

dy dx

y

1 2

c3

B

0

hence solve it. xy + (x + 2) p0y + p1

y=0

2

y =0.

(1)

II – 1.68

Part II: Mathematics II p0 = x, p1 = x + 2, p2 = 1

We have

p0 – p1 + p2 = 0 – 1 + 1 = 0

(xy + y ) + (xy + y ) + y = 0 d xy dx

i.e.

d xy dx d xy dx

or

Solution of (2) is x

dy dx

x 1 y

0

y

0

(2)

c1 x

(3)

c1 dy dx

i.e.

xy

dy dx

1 y x

1

1

1

dx

x I.F. e x lo x e xe x

Solution of equation (3) and hence equation (1) is y x ex i.e. Example 1.15 Solve the equation sin x

d2 y dx 2

c1e x dx

xye x

c1e x

cos x

dy dx

c2 2 sin x y

p0 y + p1y + p2y = 0, we have p0 = sin x, p1 = – cos x and p2 = 2 sin x. p 0 – p 1 + p2 = – sin x – sin x + 2 sin x = 0 d p0 y dx

p1 y

p0 y

cos x

c2

cos x

Chapter I: Ordinary Differential Equations d dx

i.e.

sin x y

cos x y

cos x y

cos x

x) y y

i.e.

2 lo

x)y = 1 + c1 cosec x

cot x c1 2

cot x

y

i.e.

(1)

1 sin 2 x

in x

y sin 2 x

Solution of (1) is

x) y = sin x + c1

2 cot x dx

I.F. e e

II – 1.69

sin x cos x

c1

cosec3 x dx

cosec x cot x c1 sin 2 x l 2

c2

l

tan

tan x 2

x 2

c2 c2 sin 2 x

cos x

EXERCISE 1(d) Part A (Short Answer Questions) 1. If y = u(x) and y = u(x) v(x) are solutions of the equation y v(x).

q(x) y = r(x 4. When the equation y g(x

p(x) y y

p(x) y

q(x) y = r(x

v

q(x) y

y

p(x) y

v

f(x) v =

1 2 p x dx , what are the values of

f(x) and g(x)? y y

f(x, y

f(y, y f(x, y, y y

y, y y 7. Write down the condition for the equation p0(x) y 8. If the equation p0(x) y

p1(x) y

p2(x) y = r(x

p1(x) y

p2(x) y = r(x)

II – 1.70

Part II: Mathematics II y with y =ex as a solution (ii) y = e–x as a solution? y with y = x as a solution?

Part B

11.

x 1

12. x

13.

d2 y dx 2

d2 y dx 2

d2 y d x2

x 1

16. x

18.

d2 y d x2

d2 y d x2

x2

d2 y d x2 d2 y d x2

x2

20. x 2

d2 y d x2

3x

21.

2x

4x d y 2x 1 d x

19. x 2

d2 y d x2 solution.

2x dy dx

x2

2x x2

x 1y

ex .

x x

ex .

e x sin x .

0.

2 y

y 1

dy dx

2x 1

5y

dy dx

dy dx

x

x

1 cot x y

2x x 1

d2 y d x2

17. 1

dy dx

cot x

dy dx

3

dy dx

2x 1

2 14. x 2 d y d x2

15.

2 x

dy dx

x 1 y

2y

6(1

4 y 2x 1

0

dy dx

2 y

3y

x

0

x2 )

x3e x .

x 2 (2 x 1)

2 dy dx x

2x2 2x 2 y x3 x 2

0

y = x2 is a

Chapter I: Ordinary Differential Equations d2 y d x2

x

2 23. d y d x2

y

22.

24.

d2 y d x2

dy dx

4y

28. cos 2 x

2 29. sin x

y = cos x

y = sin 2x

4 tan 2 x

x2

26. x (1 3 x 2 )

1)

y = x3 is a solution.

0

sec x

2 25. x 2 d y d x2

2 27. (x

9y

x tan x

3 y 4

0

y

dy dx

6 xy

0

y

d2 y d x2

2

d2 y d x2

6y 1

d2 y d x2

2y

d2 y d x2

II – 1.71

1 x

cos x is a solution.

1 is a solution. x

y = x – x3

y = tan x is a solution.

2y

y = cot x is a solution.

30. Find the values of a and b if y = x x2

d2 y d x2

2x 1

x

dy dx

ax

dy dx

y cos x

x 3 . For

b y

these values of a and b 31. Solve the equation x sin x

cos x

d2 y d x2

that y = xm is a solution of the equation.

32. x 2

d2 y d x2

x

dy dx

x2

1 y 4

0.

33. x 2

d2 y d x2

x

dy dx

x2

1 y 4

0.

x cos x

0

II – 1.72

Part II: Mathematics II

2 34. x

d2 y d x2

35. 4 x 2

36.

2 x2

d2 y d x2

d2 y d x2

4x

dy dx x

x2 )

d2 y d x2

38. (1

x2 )

d2 y d x2

d2 y 39. d x 2

x

d2 y 40. y 2 dx 41. y

dy dx

2

43.

d2 y d x2

44. y

d2 y d x2

dy dx

45. y

d2 y d x2

dy dx

d2 y 46. y 2 dx

2)y

0.

4 x 2 sin x .

1)y

sin 2 x

y(0) = – 1 and y 2

dy dx

y(0) = 0 and y 1

1 2

1 0.

. 2

dy dx dy dx

dy dx

2x

3

d y d2 y d x d x2

d2 y 42. y d x 2

(16x 2

dy dx

dy dx

(x 2

3

dy dx

tan x

37. (1

dy dx

x

2

dy dx

3

0. d2 y d x2

2

.

dy . dx 2

a2

2

6 xy 2 . 2

dy 2 dx

3

2 y2 . 2

0.

y(0) = – 1 and y

4

.

Chapter I: Ordinary Differential Equations

47. x 2 .

d2 y d x2

x

dy dx

y

0.

48. 1 x 2

d2 y d x2

3x

dy dx

y

1.

49. x 2

d2 y d x2

3x

dy dx

y

1.

50. x

1.7

1

d2 y d x2

x

2

dy dx

II – 1.73

e x.

y

METHOD OF VARIATION OF PARAMETERS

f(D)y = x, f(D)y =

Solution of the equation

dy dx

Py

Q

(1)

where P and Q are functions of x.

dy dx i.e.

Py dy y

(2)

0

Pdx

lo y

P x

lo c

ce

Pdx

solution of Eq. (2) is y

ce

Pdx

c as a function of x required solution of (1). x, we have

(3)

II – 1.74

Part II: Mathematics II dy dx

cPe

Pdx

dc e dx

P x

c e

Pdx

cPe

Pdx

(4)

Pdx

Qe

c

Pdx

Q e

y

ye

P

Q

dc dx

i.e.

dc e dx

P

A

(5)

Pdx

Pdx

e

.dx

Qe

dx

A

Pdx

Pdx

Q e

.dx

A

(6)

Note

Solution of the equation

d2 y dx 2

P

dy dx

Qy

(1)

R

where P, Q and R are functions of x or constants.

d2 y dx 2

y

P

dy dx

Au

Qy

0

(2)

Bv

(3)

where A and B

y = u(x) and y = v(x) are A and B as functions of x x, we have dy dx

Au

Bv

Au

Bv

(4)

Chapter I: Ordinary Differential Equations

II – 1.75

We choose A and B such that Au+Bv=0

dy dx

(5)

Au

(6)

Bv

x, we have d2 y dx 2

Au

Bv

Au

(7)

Bv

Since (3) is a solution of Eq. (1), (3), (6) and (7) satisfy (1). (Au + i.e.

+

+

A(

) + P(

) + Q(Au + Bv) = R

)+B(

)

(8)

Since y = u is a solution of Equation (2) u + Pu + Qu = 0 v +

+ Qv = 0

(9) A values of A and B as functions of x.

Notes

the values of A and B

d2 y dx 2

differential equation is unity. A and B

x.

WORKED EXAMPLE 1(e)

Example 1.1 Solve the equation x 2

1

dy dx

4 xy

1 x

2

1

II – 1.76

Part II: Mathematics II

x2

1

dy y

i.e.

x2 +

y+2 c

y

i.e.

dy 4 xy xy 4x dx 2 1 x

x2

2

1

(1)

0 0

c is the solution of (1)

(2)

c as a function of x x2

dy dx

x2

x, we have 2

x2

2

1 c' x

2

4cx x 2 1

(x2 + l)2 c

i.e.

c 2 x2

1 c'

1

1 2x

(3)

4

1

4cx

3

x

2

1

1 2

x

2

1

cx (x2 + l) + 4cx (x2 + 1) = (x2 + 1)2

i.e.

1

c=x+k

(4)

x2 + 1)2 y = x + k, where k Example 1.2 Solve the equation

dy dx

2

sec 2 y

x sin 2y = x3 cos2 y

y

dy dx

2 x tan y

x3

(1)

x 3 , which is linear.

(2)

y=z dz dx

2x z

Chapter I: Ordinary Differential Equations dz dx

2 xz

dz z

or

0

2 xdx

II – 1.77 (3)

0

c – x2

z i.e. the solution of Eq. (3) is

z

ce

x2

(4)

c as a function of x

x, dz dx

– 2cxe–x2 +

–x2

x2

2cxe

ce

x2

(5)

+ 2cxe–x2 = x3 = x3 ex2

i.e.

c = x3 ex2 dx + k 1 2

t e t dt

1 t te 2 1 2 x 2

z

1 2 x 2

tan y

et 1 ex

ke

1

1 2 x 2

1

x2 = t

k,

k 2

k

(6)

x2

ke

x2

where k is Example 1.3 Solve the equation d2 y dx 2

d2 y dx 2 y

y

x cos x,

x cos x

(1)

II – 1.78

Part II: Mathematics II

d2 y dx 2

y

(2)

0

The solution of Eq. (2) is y = A cos x + B sin x,

(3)

where A and B A and B as functions of x, sin x +

– and

A

x+

and B

cos x = x cos x

(4)

sin x = 0

(5)

A

B

1 x sin 2 x and 2

x sin x cos x or x cos 2 x or

1 x 1 cos 2 x 2

(6)

(7)

x A

B

and

y

where c2

1 has 8

assu

x2 4

sin 2 x 4

(8)

c1

1 x sin 2 x 2 2

cos 2 x 4

c2

c1

x cos 2 x 4

1 sin 2 x 8

cos x

c2

x2 4

x sin 2 x 4

1 cos 2 x 8

y

c1 cos x

c2 sin x

1 sin x 8

y

c1 cos x

c3 sin x

x2 sin x 4

i.e. or

x cos 2 x 2

1 2

d as c3 .

(9)

sin x

x2 sin x 4 x cos x 4

x cos x 4

Chapter I: Ordinary Differential Equations

Example 1.4 Solve the equation

d2 y dx 2

a2 y

II – 1.79

tan ax

d2 y dx 2

a2 y

tan ax

(1)

d2 y dx 2

a2 y

0

(2)

The solution of Eq. (2) is y = A cos ax + B sin ax If we treat A and B as functions of x, –

and B

sin ax + A

and

cos ax = tan ax

ax +

B

(4)

sin ax =0

(5)

1 sin 2 ax a cos ax

A

and

(3)

(6)

1 sin ax a

(7)

x, A

1 sin ax a2

and

B

1 cos ax a2

i.e.

y

c1

c2 i.e.

y

sec ax

l

c1

(8)

c2

1 sin ax a2 1 cos ax a2

c1 cos ax

tan ax

(9)

l

sec ax

tan ax

cos ax

sin ax

c2 sin ax

1 cos ax l a2

sec ax

tan ax

II – 1.80

Part II: Mathematics II

Example 1.5 Solve the equation (2D2 – D – 3) y = 25e–x,

D2y D2

1 D 2

3 y 2

25 e 2

D2

1 D 2

3 y 2

0

3 2

0

m 1

0

1 m 2

m2 3 2

m

or

(1)

x

(2)

3 and 1. 2 Therefore the solution of Equation (2) is m

3

Ae 2

y A and B as functions of x,

and

3

A e2

A

y

2e

c1

Be

(3)

x

and

3 x 3 A e2 2

A

x

5e 5 x 2

2e

Be

x

5 x 2

Be

x

x

and B

3

e2

x

(4)

x

(5) 0

5

c1 and B

5 x 2

25 e 2

5x

c2

c2

5x e

x

Chapter I: Ordinary Differential Equations 3

c1 e 2

y

i.e.

x

c2 e

d2 y dx 2

Example 1.6 Solve the equation x 2

x

4x

x

2e dy dx

6y

5 xe

II – 1.81

x

x

sin l

. d2 y dx 2 d2 y dx 2

4 dy x dx

d2 y dx 2

6 y x2

4 dy x dx

d2 y dx 2

x2

4x

x=t

i.e.

1 sin l x2

6 y x2 6y

d dt

,

4

6 y

0

2

5

6 y

0

(1)

0 or

dy dx

1

x

0

(2)

(3)

m2 –5m + 6 = 0 m = 2, 3 Therefore the solution of Eq. (2) is

or

y

Ae 2t

Be3t

y

Ax 2

Bx 3

A and B as functions of x,

and

A

and

2A x

3B x 2

A x2

B x3

1 sin l x3

(4)

x and

1 sin (lo x x2

(5)

0

(6)

B

1 sin l x4

x.

II – 1.82

Part II: Mathematics II 1 sin l x3

A A

i.e.

e

2t

x dx

sin t dt

1 sin l x4

c1 and B =

c1 and B

e

3t

sin t dt

x dx

c2

c2 ,

x = t or x = et i.e.

A

i.e.

c1

e

2t

5

2 sin t

cos t and

3 sin t

cos t

B = c2

e 3t 10

A

c1

1 2 sin l 5x2

B

c2

1 3 sin l 10 x 3

x

cos l

x

cos l

x and x

A and B in (4), the required solution of Eq. (1) is y

1 2 sin lo x 5x2

c1

c2

i.e.

y

c1 x 2

c2 x 3

i.e.

y

c1 x 2

c2 x 3

2 5

os lo x

1 3 sin lo x 10 x 3 3 sin l 10

1 sin l 10

Example 1.7 Solve the equation x 2

os lo x

1 5

x

x

cos l

d2 y d x2

x

x2

1 cos l 10

x3

x

x .

dy dx

y

x log x

d2 y d x2 d2 y d x2

1 dy x dx

1 y x2

1 log x x

(1)

Chapter I: Ordinary Differential Equations d2 y d x2

1 dy 1 y x d x x2 d2 y dy x2 x y dx d x2

x = et or t

II – 1.83

0 or (2)

0

d dt

x

1y 0

1 i.e.

2

1

y

(3)

0

Therefore the solution of Eq. (3) is y

At

B et or y

A and B as functions of x, A 1 and

i.e.

lo x

(5)

x

(6)

1 x

1 lo x dx and B x

A

t dt and B

2

(4)

Bx

B x

1 lo x and B x

1 lo x 2

1 x

B

A

A

x

and

A x lo x

A

Ax l

2

x

1 x

x

2

t 2 dt on

x 1 3

c1 and B

dx

x

t

3

c2

A and B in (4), the required solution of Eq. (1) is

i.e.

y

1 lo x 2

y

c1 x lo x

2

c1 x

c2 x

1 x 6

1 lo x 3

x

x

3

3

c2 x

II – 1.84

Part II: Mathematics II EXERCISE 1(e)

1.

dy dx

e x sec x.

y tan x

dy dx

1

x y

3.

d2 y d2 x

y

x sin x

4.

d2 y d2 x

a2 y

4.

d2 y dx 2

2

dy dx

2y

e x tan x

6.

d2 y dx 2

6

dy dx

9y

1 3x e x2

2. x

x

e

sec ax

2 7. x 2 d y dx 2

4x

dy dx

2y

x2

d2 y dx 2

2x

dy dx

4y

32 l

8. x 2

1 x2 2

x .

ANSWERS Exercise 1(a) (7)

y

2x C y

3x C

0

( ) (y – ex – C) (ey – x – C) = 0 (11) y

px

p ;y p 1

Cx

C C 1

(8)

y

x2 2

(10)

y

lo x C

C lo y

x C

0

y2 2

x C

0

(12) y = px – ep; y = Cx – eC

(13) y = px + sin–1 p; y = Cx + sin–1 C

(14) y2 = 4x

(15) x2 = 4y

(16) (y – x – C) (x2 + y2 – C) = 0

(17)

y

x2 4

C x

1 y

C

0

(18) (y – Cx2) (3x2 – y2 – C) = 0

Chapter I: Ordinary Differential Equations (19) (y – Cx) (x2 – y2 – C) = 0 y (1 + cos x) – C

(23) y

y x

1

(22) sin

C x

(20) (xy – C) (x2 – y2 – C) = 0

y (1 – cos x) – C] = 0 1

x C sin

l

1 4x2

C2; y

y x

x C

l

0 y = x2 + p2 and lo

p

x

p

x x

n pn n 1 2p 3 C

x

p

2

p

x

C.

p2 3

2C . p

an 1 p 1

ep and y 2

p

x

x

C p2

1

p 1

lo

Ce

p

C and y p2

p

3x

0

p

p

(29) y

II – 1.85

C 1

p e

p

1 1 2

p ep

3 1 Ce3 x

lo

(30) 64y = C(C – 4x)2; 4x3 = 27y (31) of p y=C– p – 1) + 2p + p2] and x= C p – l) + 2p] (32) y2 = 2Cx + 4C2 (33) y2 = 2Cx + C3 y = Cx + C2 (35) y

x1 x

tan

1

x 1 x

(37) y2 =Cx2+ C2 (39) xy= Cx – C2

(36) y

x

(38) y2 = 2Cx + C2 (40) ey = Cex + C3

Exercise 1(b) x

(1) y

e 2 C1 x

(2)

x3 x e 6

(4)

x3 e 6

1 e 8 2x

(6) – e–x cos x

x

C2 cos

3 x 2

C3 x

C4 sin

(3)

3 x. 2

x b sin a x 2a

(5) (x – 2) e2x (7)

1 x xe cos 2 x 4

c cos a x .

II – 1.86

Part II: Mathematics II

1 x e 2 x sin x 2 (9) ex (4 sin x + cos x) (11) y = C1 e–x + C2 cos x + C3 sin x + x2 – 2x + xe–x (8)

(12) y (13) y

C1 cos 3 x Ax

e 4 x Ae

(15) y

C1e

(16) y

x

B e

(14) y

x

C1 x

x2 9

C2 sin 3 x x3

7x

Be

C2 e

2x

C2

e

6x2

2 81

18 x

1 cosh x 10 1 4 sin 2 x 3 cos 2 x 25 1 2 16 x x 2 sin 5 x 9 9

24

1 5 cos 5 x 29

7x

1 cos 2 x 3 sin 2 x 20 x 2

3 x 2

C3 cos

C4 sin

x2

3x

3 x 2

x4

110 81

4 4 x3

1 8 cos 3 x 3 sin 3 x sin x 657 (17) y = C1ex + C2e–x + xex + (3 + 3x + 2x2) (18) y = (C1 x2 + C2 x + C3) e2x + (2x3 – 9x2 + 18x – 15) e4x (19) y

C1

C2 x

C3 e

e2 x 2 x 18

x

7 x 3

11 6

2 x sinh x cosh x 3 9 (21) y = (C1x + C2) cos x + (C3x + C4) sin x + (x2 + 4x + 4) e–x (20) y

C1e 2 x

C2 e

2x

(22) y

C1e x

C2 e 4 x

1 e 18

(23) y

C1e x

C2 e 3 x

e x co 2 x

(24) y

C1e x

C2 e

(25) y

e x C1 cos 2 x

(26) y

C1e

x

x

7 10

x

C4 sin x

C2 sin 2 x

1 x e 4

e 2 C2 cos

3 x 2

5 x 2

x2

21 8

in 2 x

C3 cos x

x x

2 e 5

2x

C3 sin

1 sin x cosh x 5 1 x xe cos 2 x 4

3 x 2

1 xe 6

x

1 e 26

x

2 siin x

3 cos x (27) y

C1 cos 2x

C2 sin 2x

e2 x

3 7 sin x 4 cos x 65 1 12 cos 3 x sin 3 x 145

Chapter I: Ordinary Differential Equations (28) y

C1e x

(29) y

1 10 cos 5 x 11 sin 5 x 884

C2 e 3 x

Ax

B e

A

(31) y

Ax

B e2 x

(32) y

Ae x

Be

(33) y

C1 x

(34) y

A cos x

(35) y

A cos 2 x

Be

B

(6) y

A x

x

(7) x

A cos t

(8) y

Ax 4

(9) y

A

(12) y (13) y (14) y (15) y (16) y

B sin t

2

B x3

x

2

3 sin l 40

B

in 2 lo x

y

4

A lo x

x

x .

x

1 25

lo x

os lo x

C

x

D sin l

B

x

B x

3

B

x x

1 2 sin 2 l 20 x

2

2

4 lo x

C lo x x

e

7 os 2 lo x

1 sin 3 l 72

B sin 3 l

B

1 20

2

x

4

2

2

16 25

x

x

78 625

lo x

3

A lo x Ax 3

1 100

Cx 2

x

x A lo x 3

3 l 2

C sin

1 x 7 1 9x

C lo x

B x

A lo x

x

1 4 x lo x 9

A cos 3 l

lo x

2 x cos 2 x

(4)

3 l 2

x2 B

1 x4

x sin 2 x 64

(2) y

B cos

B x5

3

C4 sin 2 x

e 3t

Bx n 1 .

Ax

2 sin x

x sin x.

x

Ax n

(11) y

2 11 cos x 125

4 cos x

1 24 x cos 2 x 9 x 2 26 sin 2 x . 27 B sin 2 x cos 2 x l sec 2 x tan 2 x .

(5) y

A lo x

sin x 2

B sin x

1 y

(10) y

cos x

x 2 cos x C3 x

2 cos x .

2 cos x

cos x

1 1 2

C2 cos 2 x

4

x sin x

x 3 sin x 25

x

Exercise 1(c) (1) y A lo x 2

e

x

x sin x 2

x

(30) y

(3)

x

1 sin x 20

II – 1.87

6

D

x

cos 2 l

x

x2

x

4

2

2

t

II – 1.88

Part II: Mathematics II

(17) y

1 A cos 2 l x

(18) y

A 3x

2

2

(19) y = A (20) x

x B 3x

1 3x 108 x

2

2

x + 1) + B 2t

Ae

Be

y

2 Ae 3

(21) x

5 Ae 3

2t

2

2 lo 3 x

7 os lo x 2

x

1 x + 1)

5 31 1 2t t e 14 196 12 1 9 1 2t Be 7t t e 7 98 6

7t

4B t e 3

2t

x 4 in lo x 65

B sin 2 lo x

1 t e 3

y Ae 2t Bet (22) x = 2 cosh t; y = sin t – 2 sinh t (23) x

A Bt e

y (25) x y

A B cos t B cos t

t 2

C sin t

C sin t

A cos 2 t

C e 3

At

y

At

Aet Aet

y

1 3

9 2t e 10

6 2t e 10

B sin 2 t

C cos 3 t

B cos t

Ct

B sin t

Be

t

Be

1 t e 4 sin t 3 cos t 25 1 t D cos t e 3 sin t 4 cos t 25

Ct

C cos 3t t

5

A x 1 ex

4 2 5 37 t t . 9 9 81 5 2 4 44 D sin 3t t t . 9 9 81

D sin 3t

C cos 3t

(12) y = ex (c1

x x e 4

x + c2 + x).

1 1 cos 2t 4 D sin 3 t

D sin 3 t

D sin t

Exercise 1(d) (11) y

3t 2

1 2 A cos 2 t 2 B sin 2 t C cos 3 t

(26) x

(27) x

Ce

3t 2

2 A 6 B 2 Bt et

y (24) x

t

Be x .

Chapter I: Ordinary Differential Equations (13) y

Ae

x

1 x e cos x 2

Be x

2 sin x cos x

(14) y = (Ax3 + B)e–x (15) y = Axe–2x + Be–x (16) y = e–x (c1

x + c2)

c1 3 1 x x lo c1 2 4 1 x (18) y = c1e2x + c2x (19) y = c1 xex + c2x + x (x – 1)ex (20) y = c1 x3 + c2x + x3 x + x2 (17) y

3 2 x 2

c2 x

(21) y = c1 xex + c2x2 (22) y = c1 x3 + c2x–3 (23) y= c1 cos x + c2 sin x + x sin x + cos x (24) y= c1 cos 2x + c2 sin 2x – cos 2x (25) y

1 x

cos x A B x tan x lo

os x

1 x

2 (26) y c1 1 x

c2

(27) y c1 x x3

c2 4 3 x 6 x 2

3 x 3 lo

(28) y = c1 tan x – c2 (1 + x tan x) (29) y = c1 (1 – x cot x) + c2 cot x (30) y

Ae 2 x

x2 2

B x

(31) y = c1x – c2 cos x (32) y (33) y

1 x

A cosh x

B sinh x

1

A cos x B sin x x (34) y = xex (Ax + B) 1 1 (35) y sin x A cos 2 x B sin 2 x 3 x (36) y = 2 sin x – sin x cos x – x – 1 (37) y

1 sin 2

(38) y

1 x c1

1

x 1

x x + tan 2x)

2

4

sin

1

x

1 lo 1 c1 x c12

c2

1 x 1 x

1. 6

II – 1.89

II – 1.90

Part II: Mathematics II

(39) x= c1 sin (y – c2) (40) y + c1 y = x + c2 c1 x (41) y = c2 e + c1 1 (42) y = c2 ec1 x + c1 (43) y = a cos x – (a + 1) 3 (44) y = c1 e( x c2 x ) (45) y2 = c1 cosh (2x + c2) (46) y3 = Ax + B B (47) y Ax x (48) y

c1 sin

1

x/ 1

2 1 x2 (49) y 1 x (50) xyex = c1 ex + c2 – x.

x2

c2 / 1

x2

c1 sinh -1 x

c2

1

Exercise 1(e) x 2 2. xy ex = x + c

1. y cos x

5. 6. 7. 8.

c

1 1 2 x sin x x cos x 4 4 1 1 cos ax l cos ax y c1 cos ax c2 sin ax x sin ax. 2 a a y = ex (c1 cos x + c2 sin x) – ex cos x x + tan x) x. y = (c1x + c2) e3x – e3x c1 c2 1 2 1 lo x y x 2 12 x x x2 1 2 8 lo x 12 y c1 x 4 c2 x 13. x

3. y 4.

sin 2x 4

c1 cos x

c2 sin x

Chapter

2

Vector Calculus

2.1

INTRODUCTION

In Vector Algebra we mostly deal with constant vectors, viz. vectors which are constant

each value of scalar variable t a vector function of the scalar variable t and is denoted as F t.

F, then F is called F (t ) .

r t, i.e. r

r (t ) .

rectangular cartesian co-ordinates (x, y, z), then r is a function of the scalar variables x, y, z, i.e. r r (x, y, z ). A a scalar point function or a vector point function,

that region is called a

.

or a

2.2 VECTOR DIFFERENTIAL OPERATOR

, according

Part II: Mathematics II

II – 2.2

i

j

x

k

y

z

, vector) with i

x

x

,

y

,

i , j,k

z

, it should be noted that i and

. When writing

x

.

2.2.1

Gradient of a Scalar Point Function

Let (x, y, z) i

Note

1.

j

x

k i y z x and shortly denoted as

should not be written as

j

y .

k

z

.

2. When combines with , neither . and . 3. If is a constant, = 0. 4. (c1 1 ± c2 2 c1 1 ± c2 where c1 and c2 are constants and 1, 2 2

5. 6.

(

1

2

1

1

2

2

1

1 2

2

i.e. d

2

, if

2

0 .

(u)

u.

Directional Derivative of a Scalar Point Function (x, y, z)

Let P and Q origin O be r (= OP) and r = r. Let and + Then

1

2

7. If v = f (u), then

2.2.2

.

2

d dr

lim r 0

r

r(

OQ)

y, so that PQ

r and PQ P and Q

is called the directional derivative of in the direction OP.

gives the rate of change of dr direction of r .

Chapter 2: Vector Calculus ,

x

,

y

,

II – 2.3

are the directional derivatives of

z

at P(x, y, z) in the

.

2.2.3

Gradient as a Directional Derivative

Let (x, y, z (x, y, z) = c values of c, a family of surfaces, called the level surfaces of the function . , namely = c1 and = c2 r ) . [Refer to Fig. 2.1] P ( r ) and Q ( r R Let the normal at P to the level surface = c1 Q meet = c2 at R. = c2

surfaces = PR = n. If QPR , is almost a right angled triangle,

PQR, which

P

= c1

Fig. 2.1

n = r cos

(1)

If n is the unit vector along PR, i.e. in the direction of outward drawn normal at P to the surface = c1, then (1) can be written as where r PQ . n n r, or

n dr

d =

d dn dn

= Also

(2)

dn

d =

d n dr dn x

dx

x

[by using (2)]

y

i

y

dy j

z z

(3)

dz k (dx i

dr

dy j

dz k ) (4)

From (3) and (4), dr Since

PQ

d n.dr dn r (or d r ) is arbitrary, d n dn

From (1),

d dn

d dr (cos )

(5)

Part II: Mathematics II

II – 2.4

i.e.

d dr

cos

i.e.

d dr

d dn

d dn cos

i.e.

1 is

d , that is the directional derivative dn

in the direction of n. : is a vector whose magnitude is the greatest directional derivative of and whose direction is that of the outward drawn normal to the level surface = c. WORKED EXAMPLE 2(a) Example 2.1 Find the directional derivative of = x 2yz + 4xz2 PQ, where Q x 2 yz x

4 xz 2

i

y

)

(1, 2 , 1)

The magnitude of (

j

z

4z2 ) i

(2 xyz (

P (1, 2,

8i

j

k

x2 z j

(x 2 y

8 xz ) k

10k

)P is the greatest directional derivative of

at P.

at (1, 2 , 1)

64

1

100

165 units. PQ Directional derivative of

OQ

OP

2i

j

k

in the direction of PQ

along PQ . PQ PQ (8 i

j

10k ) (2 i 4

1

j

k)

1

27

units. 6 Example 2.2 Find the unit normal to the surface x3 (1, 1, 1).

Note

Unit normal to a surface

=c

+ z3

n in the

Chapter 2: Vector Calculus x3

xyz + z3

= x3

xyz + z3

II – 2.5 (x, y, z) = c.

is a vector acting in the direction of the outward drawn normal to the surface = c. (3 x 2

Now (

)(1,1,1)

2i

yz ) i j

n |n | 1 (2 i 3

n

2k

j

(3 z 2

xzj

xy ) k

n (= a vector in the direction of the normal)

2k )

Example 2.3 Find the directional derivative of the function = xy2 + yz2 x log z y2 y2

x log z

(x, y, z) = c. 2

(x, y, z) = x log z

y and c

The direction of the normal to this surface is the same as that of x Now (log z ) i 2 y j k z

(

(

)(

)( 2 ,

Directional derivative of

4j

1,2 ,1)

1,1)

k

b (say)

xy 2

yz 3

y2i

(2 xy

i

3j

.

z3 ) j

3 yz 2 k

3k

in the direction of b b |b | (i

3j

3k ) (

4j

k)

16 1 15 17

units.

Example 2.4 Find the angle between the normals to the surface xy = z2 , , 2) and (1, 9, . Angle between the two normal lines can be found out as the angle between the vectors acting along the normal lines. xy = z2 with (x, y, z) = c, we get = xy z2 and c = 0. yi

xj

2 zk

Part II: Mathematics II

II – 2.6 (

)( (

2i

2 , 2 ,2 )

)(1,9 ,

2j

9i

3)

j

4k

6k

n1 (say)

n2 (say)

n1 and n2 If n1 n2 n1 n2

cos

44 24

118

177

11

1

cos

11

177

Example 2.5 Find the angle between the surfaces x2

Identifying x2 we have

2

2

= x2

2

)(6 ,4 ,3) 12 i

Identifying we have

6 ,4 ,3

If

2

and c = 11. 2 zk

2y j 8j

6k

n1

18 with

= c´, and c´ = 18

(y z ) i

(z x) j (y x)k

i

9j

2k

is the angle between the surfaces at (6, 4, 3), then n1 n2 cos n1 n2 48

24

244 86

61 86

cos

24

1

5246 xz2

Example 2.6 Find the e 4x = Identifying 2xz2 we have

,

, 2). 4x = 7 with = c, = 2xz2 3 4x and

3

(2 z 2 ( (

18

= 11 with = c, 2 xi

(

y2 z2 = 11 and

)

)(1,

1,2 )

7i

3 y 4) i 3j

3x j

c = 7. 4 xzk

8k

is a vector in the direction of the normal to the surface

= c.

3

Chapter 2: Vector Calculus

II – 2.7

D.R.’s of the normal to the surface ( = c)

7(x

y + 1) + 8(z 7x y

i.e.

Example 2.7 Find the constants a and b, so that the surfaces 5x2 ax2y + bz3 =

Identifying 5x2

2yz

(10 x 9) i

1

(

1 )(1, 1,2 )

Identifying ax2y + bz3 = 4 with we have

2

i

4j

2z j

2 yk

n1 (say)

2k

= c´.

( (

2)

2axyi

2 )(1, 1,2 )

2ai

Since the surfaces cut orthogonally, n1

ax 2 j aj

3bz 2 k 12bk

n2 (say)

n2 .

n1 n2 0 a + 24b = 0 a + 4b = 0

i.e.

9x = 0 and

. = c, 1

9x = 0 with

we have

2yz

(1) ax2y + bz3 = 4

a + 8b = 4 Solving (1) and (2), we get a = 4 and b = 1.

(2)

Example 2.8 If r and

x, y, z), a is a constant vector

= x2 + y2 + z2,

(r a ) a and (ii) r grad r

xi

a

a1 i

a2 j

a3 k

r a

a1 x

a2 y

a3 z

Let

grad r a

yj

a1 i 2

grad r grad Example 2.9 If r (r n )

zk

a2 j 2

2 .

a3 k

a

2

x y +z 2 xi 2 yj 2 zk 2(x 2

y2

z2 ) 2

. x, y, z)

nr n 2 r .

-

Part II: Mathematics II

II – 2.8 r r

xi

2

|r | x

(r n )

x nr n

From (l),

2r

r x

Similarly,

zk 2

y2

(r n ) i

z2 (r n ) j

y

r i x

1

(1) z

r j y

(r n )k

r k z

(2)

2x r x r y

i.e.

yj

2

x r y r

and

r z

x i r

y j r

z r

(3)

Using (3) in (2), we have (r n )

nr n

1

nr n 2 ( xi nr

n 2

z k r zk )

yj

r.

Example 2.10 Find the function , if grad ( y2

2 xyz 3 ) i

(3

2 xy

x2 z3 ) j

(6 z 3

3 x 2 yz 2 )k .

( y2

2 xyz 3 ) i

(3

2 xy

x2 z3 ) j

(6 z 3

3x 2 yz 2 )k

x

x y z

y2 3 6z3

i

y

j

z

(2)

k

2 xyz 3 2 xy

(1)

(3)

x2 z3

(4)

3 x 2 yz 2

(5) x (i.e. treating y and z as

constants), = xy2

x2yz3 + a function not containing x

(6)

Chapter 2: Vector Calculus

II – 2.9

Note x we add an arbitrary function of the other variables y and z, i.e. an arbitrary function x. y. x2yz3 + a function not containing y z,

y + xy2 3 4 z 2

x 2 yz 3

(7)

a function not containing z

(8)

. The general form of

is obtained as

follows: be included in the value of . . The last terms indicate that there is a term of a constant. 3 4 3y z xy 2 x 2 yz 3 2

x, y, z, i.e. c.

EXERCISE 2(a) Part A (Short Answer Questions) grad and give its geometrical meaning. 2. If r

x, y, z

3. If r

x, y, z

4. If r

x, y, z

5. Find grad 7. Find the directional derivative of the x

1 r. r ( | r |2 ) 2r . 1 f (r ) f (r )r . r (r )

= 3x2y y3z2. = x3y2z = xy + yz + zx

= x2y2z4 T = xy + yz + zx

x, y, z

Part B 11. If = xy + yz + zx and F grad at the

x 2 yi

T(x, y, z

y 2 zj

z 2 xk

F · grad

2

+ y2 z.

and F

Part II: Mathematics II

II – 2.10

12. Find the directional derivative of = 2xy + z2 direction of i 2 j 2k . 13. Find the directional derivative of = xy2 + yz3 P direction of PQ where Q 14. Find a unit normal to the surface x2y + 2xz 15. Find the directional derivative of the scalar function = xyz in the direction of the outer normal to the surface z = xy 16. Find the angle between the normals to the surface xy3z2 17. Find the angle between the normals to the surface x2 = yz and (2, 4, 1). 18. Find the angle between the surfaces z = x2 + y2 x2 + y2 + z2 = 9 at the 19. Find the angle between the surfaces xy2z = 3x + z2 and 3x2

y2 + 2z = 1 at the

x log z y2 2

xy+z xz2 + x2y = z 22. Find the values of and , if the surfaces x2

= ( + 2)x and 4x2y + z3 = 4

23. Find the values of a and b, so that the surfaces ax3 by2z = (a + 3)x2 and 4x2y z3 (2 xy z 2 ) i (x 2 2 yz ) j 2 (y 2 zx) k . 25. If

2 xyz 3 i

x2 z3 j

3 x 2 yz 2 k

(x, y, z), given that

2.2 THE DIVERGENCE OF A VECTOR If F (x , y , z )

x, y, z) in F , denoted as div

F

F i

F , when F

F1 i

F

F x

F2 j

i

j

x

i Formula for

F

x

k

y F y

j

z k

F F z

F3 k

j

y

k

z

(F1 i

F2 j

F3 k )

Chapter 2: Vector Calculus F1 x

Note and hence

2.3.1

F3 z

F2 y

F

Since F

, F1, F2 .

Physical meaning of

(i) If V

Vx i

Vy j

II – 2.11

and F3

F

Vz k x, y, z), then .

(ii)

V

V

V

V

V

In general, if F

F F , it is called the divergence of F .

2.3.2

Solenoidal Vector

If F is a vector such that F 0 be a solenoidal vector in that region.

2.3.3

Curl of a Vector

If F (x, y, z) is a differentiable vector x, y, z) in the curl of F or the rotation of F , denoted as curl F or rot F F

F i

F x

i

Note

j

x

k

y j

F y

z k

F F z

F is also

Formula for functions):

F, when F

F

i

F1 i

x F1 (i x

F3 k (where F1, F2 and F3 are scalar

F2 j

(F1 i i)

F2 j F2 (i x

F3 k ) j)

F3 (i x

k)

Part II: Mathematics II

II – 2.12

F3 j x

F2 k x F3 y

2.3.4

F2 i z

i

j

k

x F1

y F2

z F3

F3 j x

F1 z

F2 x

F1 k y

Physical Meaning of Curl F

If F

x, y, z) of a rigid body that rotates , then curl F 2 .

2.3.5

Irrotational Vector

If F is a vector such that F 0 be an irrotational vector in that region.

2.3.6

Scalar Potential of an Irrotational Vector

If F function. Let

F

F1 i

Since F

F

0

i.e.

i.e.

F3 y

F2 i z F3 y

F1

F2 j

i

j

k

x F1

y F2

z F3

x

, F2

0

F3 j x

F1 z F2 ; z

F3 k

F2 x

F3 ; x

F1 z

y

and

F2 x

F1 k y F1 y

F3

0

(1)

z

Chapter 2: Vector Calculus F

i

j

x

If F is irrotational and F

2.3.7

k

y

II – 2.13

z

, then

F.

Expansion Formulae Involving Operations by ’s are given below: The

students. 1. If u and v

(u

2. If u and v

(u

3. If (

( F)

F) i

x

u

v.

( F) F x

x

F x

i

F i

F

x

F

F

F

4. If ( F)

(

F)

F

5. If u and v

u v (u v )

i

u v x u v u x

i

u v x

i i [ interchanged]

v)

v.

F

i

Proof:

u

F ( F)

Proof:

v)

u v x

v x v u x

i i

v u x

v curl u

u curl v.

Part II: Mathematics II

II – 2.14

u v x u u

i v

v u x

i v

6. If u and v (u v ) (

Note

v )u

(v

v and v

In this formula,

vx

v Thus v

)u

u

x u x

vx

(

u )v

)v

vy vy

vz

y u y

vz

z u z

, if v

u v

)u

v means div v , but

are not the same,

7. If u and v (v

(u

vx i

v curl u

. (

) = 2

2

x

2

y

2

2

2

2

x2

y2

z2

2

2

F

F x2

2

is called the Laplacian of .

2

9. If

)= i

i curl (grad )

x

y

x

y

y z

j

y j

2

(

) = 0.

k

z

k z z 2

z y

2

i

z x

2

x z

2

j

x y

2

y x

= 0.

Note

,

F resulting in F. z2

x

2

= 0 is called the

2

F y2

Proof grad

) =

is called the Laplacian operator and

z2

can also

2 2

u curl v

2

2

Note

vz k

)v .

(u

8. If where

vy j

This result means that (grad ) is always an irrotational vector.

Chapter 2: Vector Calculus 10. If F Proof:

II – 2.15

(curl F ) Let F

F1 i

F2 j

i

j

k

x F1

y F2

z F3

curl F

F3 y div curl F

x

(

F) 0 .

F3 k

F2 i z

F1 z

F2 z

y

F3 y

2

2

F3 x y

F3 j x F3 x

F1 z 2

2

F2 x z

F2 x

F3 y x

F1 y z

F1 k y F2 x

z 2

F1 y 2

F2 z x

F1 z y

0.

Note

This result means that (curl F ) is always a solenoidal vector.

11. If F curl (curl F ) Proof: Let Then

F

F1 i

F2 j

F3 y

curl F

(

F)

(

2

F)

F.

F3 k

F2 i z

F3 j x

F1 z

F1 k y

F2 x

curl (curl F ) i

j x

F3 y

y F2 z

2

F2 y x 2

F1 x2

F1 y

2

F3 z x F2 x y

F1 z

z

2

2

z F3 x

F1 z

F2 x

y

k

F1 y2

2

F3 x z

F2 x

F1 y

F3 i x 2

F1 i z2

2

F1 x2

2

F1 y2

2

F1 i z2

Part II: Mathematics II

II – 2.16 F1 x

x

i

2

F

x

F

x

j 2

F

2

F3 z

F2 y

x

2

2

y

2

2

F1 i

z2

F1 i F

y

k

2

F

z

F1 i

F2 j

F3 k

F.

12. If F grad (div F )

Note

(

F)

(

2

F)

F.

Rewriting the formula (11), this result is obtained. WORKED EXAMPLE 2(b)

Example 2.1 When (1, 2, 3).

= x3 + y3 + z3

xyz

.

,

and

= x3 + y3 + z3 3x yz i

x 3(x 2

yz ) i

3(y 2

3(x 2 yz )] [3 x 6(x y z )

y

i

yz ) 3(y 2

( 3x 3x) i

[3(z 2

xy )]

z zx) 3(z 2

( 3 y 3 y )j

xy ) ( 3z

3z )k

= 0, for any , ( ( . (x 2

(

F ) and

F,

(

)(1, 2 ,3)

15 i

)(1, 2, 3) = 36

Example 2.2 If F F ),

z

k

y

x

xy )k

[3(y 2 zx)]

j

3(x 2

Note

3(z 2

zx)j

y2

2 xz ) i (

and (xz F)

3j ( xy

×

21k )(1, 2, 3) = 0. yz )j

(z 2

x 2 )k ,

, 1, 1).

F,

Chapter 2: Vector Calculus

(

F

(x 2

F

(xz (x 2 y 2 2 xz ) x y (2 x 2 z ) ( x z ) 2 z x 5z

F)

x i

y2

2 xz ) i

(x 5 z ) i

(xz

y

2

y

2

2 xz

xz

y xy

yz

z

(y

z )k

(x

y) i

(2 x 2 x) j

(x

y) i

(y

(

F)

(

F ) 0 , for any F ,

x2 )

z

(0)

y

F)

z

x i

(

F )(1,1,1)

2i

(

F )](1,1,1)

k z y

z

F )](1,1,1)

i

2k ; [

(

a1 x

F )] 1,1,1

0;

yj

x, y, (a r ) a , (ii) div (a r ) = 0 and

zk

a3 k , where a1, a2, a3 are constants. a r

5k ;

k

Example 2.3 If is a constant vector and r z (iii) curl (a r ) 2a . xi

z)

k

6; [ (

i

(y

y 0

y

F )(1,1,1)

r

x2

j

x

(

2

z )k

[ (x y )] x 1 0 1 0

(

a2 j

(z 2

k

i

a1 i

z

(x 5 z )k

z

j x

x

Let a

yz )

xy

x 2 )k

5k

F

[

(z 2

yz ) j

(x 5 z ) j

i

Note

xy

II – 2.17

a2 y

a3 z

Part II: Mathematics II

II – 2.18 grad (a r )

x a1 i

(a1 x

a2 y

a2 j

a3 k

a3 z ) i

a a r

i a1 x

j a2 y

k a3 z

(a2 z a3 y ) i div (a r )

x

(a3 x a1 z ) j

(a2 z a3 y )

y

(a1 y a2 x)k

(a3 x a1 z )

z

(a1 y a2 x)

0 i curl (a r )

j

k

x y z a2 z a3 y a3 x a1 z a1 y a2 x (a1

a1 ) i

2(a1 i

(

a2 j

a2 ) j

a2

(a3

a3 )k

a3 k )

2a

(2x 2

Example 2.4 Show that u

x

(2 x 2

(3 x 3 y 3 xy ) j

(4 y 2 z 2

{ (4 y 2 z 2

2 x 3 z )}

2 x 3 z )k is

xyz 2 u is solenoidal.

not solenoidal, but v u

8 xy 2 z ) i

8 xy 2 z )

y

= (4x + 8y2z) + (3x3 = x3 + x

(3 x 3 y 3 xy ) x

z

y2z + 2x3)

x, y, z) u is not solenoidal. v xyz 2 u v

(2 x 3 yz 2

8x2 y3 z3 ) i

(6 x 2 yz 2

16 xy 3 z 3 ) (6 x 4 yz 2

=

(3 x 4 y 2 z 2

3x 2 y 2 z 2 ) j

(4 xy 3 z 4

6 x 2 yz 2 ) (16 xy 3 z 3

2 x 4 yz 3 )k

6 x 4 yz 2 )

x, y, z)

v is solenoidal. Example 2.5 Show that F (y 2 z 2 3 yz 2z )k is both solenoidal and irrotational.

2 x) i

(3 xz

2 xy ) j

(3 xy 2 xz

Chapter 2: Vector Calculus F

x

(y 2

=

z2 x

3 yz

2 x) +

y

(3 xz

II – 2.19

2 xy ) +

(3 xy

z

2 xz

2 z)

x+2

=

x, y, z)

F is a solenoidal vector. i F

j x

(y

2

z

(3 x

2

k

y

3 yz

3 x) i

2 x) (3 xz

(3 y

=

2z

z 2 xy ) (3 xy

2z

3 y) j

2 xz

(3 z

2z)

2y

2y

(2 x 2 z

y

3z ) k

x, y, z)

:. F is an irrotational vector. (y 2

Example 2.6 Show that F

2 xz 2 ) i

i F

(2 xy j

x (y

2

y

(4 xz

=

z 2

2 xz ) (2 xy 1) i

2 z ) k is

k

2

( 1

z) j

z ) (2 x z 4xxz ) j

(2 y

y

2z)

2 y) k

x, y, z)

F is irrotational. F be . F x x

i

y2

y

j

z

k

2 xz 2

x; = xy2 + x2z2 y

2 xy

x

(1)

z

y; = xy2 z

2 x2 z

yz

y y

2z

(2)

Part II: Mathematics II

II – 2.20

z; = x2z2 yz + z2 + From (1), (2), (3), we get = xy2 + x2z2 Example 2.7 (3 x 2

cz ) j

z

(3)

yz + z2 + c.

Find the values of the constants a, b, c, so that F (3 xz 2

(axy

b z3 ) i

y ) k may be irrotational. For these values of a, b, c, F.

F is irrotational. F i

0

j

k

i.e. x

y

(axy bz 3 ) (3 x 2 i.e.

(3 z 2

( 1 c) i c

,

3z2 a = 6,

0

z cz ) (3xz 2

3bz 2 ) j

y)

(6 x ax) k

b) = 0, x b= 1, c = 1.

0

a) = 0

Using these values of a, b, c, F

(6 xy

z3 ) i

Let

(3 x 2

z) j

(3 xz 2

y) k

F. F

x

x 6 xy

z3 ,

i

y y

j 3x 2

z

k

z,

3 xz 2

z

= 3x2y + xz3 = 3x2y = xz3

x

yz

y

yz +

z

y

(1) (2) (3)

From(l), (2) and (3), we get = 3x2y + xz3 Example 2.8 tional and (ii)

If

and is solenoidal. = ( )+ = =0 is irrotational

yz + c is irrota-

Chapter 2: Vector Calculus

II – 2.21

u v v curl u u curl v )= curl ( curl ( = = 0. ) is solenoidal.

(

If r = | r | , where r

Example 2.9 that

2

(r n ) nr n Now

2

x, y, z),

1 and hence deduce that r

(rn ) = n(n + 1) rn

2

r.

n

( r n)

(r )

(nr n

2

(r n

2

n n (n

r) rn

) r

2) r n

[since

4

r r

(rn)

= n [(n

4

r

= n (n+1) r

2

(

r)

3r n (xi

r x 3]

2

)

(x)

2

zk )

yj y

(y )

z

(z )

r2 + 3r 2]

2

n= 1 r

2

i.e.

( 1) (0) r

3

0

1 r

Example 2.10 If u and v (u v (u v

v u) = (u v =( u v+u =u

2

v

2

v

2

2

v u) = u

(v u) v v

u+v

2

2

u)

u.

F

v

F curl F uF F

v

u.

Example 2.11 If u and v such that uF Given

v

v 1 u

v

0.

Part II: Mathematics II

II – 2.22

1 u

F

v

1 ( u 1 u 1 u

F curl F

Now

1 u

v)

v

v , since 1 u

v

1 (0) , u

v = 0.

v

[

= 0. Example 2.12 If r = r , where r

x, y, z) with f (r)

2

(ii)

f (r )

f (r )

2 f (r ). r

f (r ) r and r

r2 = x2 + y2 + z2 2r r x

r x

2x

x . Similarly, r

r y

f (r )

Now

r z

y and r i

f (r )

x

r i x

f (r )

x i r

f (r )

2

f (r )

z . r

f (r ) r r f (r ) f (r ) r r f (r ) r r f (r ) r

r

f (r ) r

f (r ) 2

r

(r ) r

3

f (r ) r

r

3

Chapter 2: Vector Calculus r f (r ) r2

f (r ) 1 r r r

II – 2.23

3 f (r ) r r i x

(r ) r f (r ) 2

f (r ) 1 2 (r ) r

x i r

1 r r

3 f (r ) r

r 2 f (r ) f (r ) r

Example 2.13 Find f (r) if the vector f (r) r is both solenoidal and irrotational. f (r) r is solenoidal f (r ) r i.e. i.e. i.e. i.e.

f (r ) r

0 f (r )

r

0

f (r ) r r 3 f (r ) 0 r rf (r ) + 3f (r) = 0 f (r ) f (r )

3 r

0

Integrating both sides w.r.t. r, log f (r) + 3 log r = log c i.e. log r3 f (r) = log c c f (r ) r3 f (r) r is also irrotational f (r ) r i.e.

0 f (r ) r

f (r ) r r r

i.e.

(1)

f (r )

0 0

r

r

0 i

j

k

x x

y y

z z

0

f (r ) (0) 0 0 r

i.e. This is true for all values of f (r)

From (1) and (2), we get that f (r) r is both solenoidal and irrotational if f (r) Example 2.14

i.e,

If

= 0, only when 2 = 0. is solenoidal, only when 2 = 0

(2) c

r3 is both solenoidal and

(1)

Part II: Mathematics II

II – 2.24

i.e. is irrotational always From (1) and (2), is both solenoidal and irrotational, when

(2) 2

= 0, i.e. when

is a solution of

Example 2.15 If F is solenoidal, Since F is solenoidal, F 0 F

F

( 2

F

= curl curl ( = ( 2

[ { [ {

=

4

2

(

2

F)

4

F. (1) (2)

F

F [by (1)]

2

F)

2

F)

F}

2

F )}

( 4

2

F )] , by using (2)

F] ,

by interchanging the and

2

F {by using (1)}

EXERCISE 2(b)

Part A (Short Answer Questions) 2.

F F

7. If r x, y, z) r and curl r . 8. If F 3 xyz 2 i 2 xy 3 j x 2 yz k , F 9. If F (x 2 yz ) i (y 2 2 zx) j (z 2 3 xy ) k F 10. If F (x y 1) i j (x y ) k , show that F F. 11. If F zi x j yk F 0. 12. Show that F (x 2 y ) i (y 3 z ) j (x 2 z )k is solenoidal. 13. Show that F (sin y z ) i (x cos y z ) j (x y )k is irrotational. 14. Find the value of , so that F y 4 z 2 i 4 x3 z 2 j 5 x 2 y 2 k may be solenoidal. 15. Find the value of , if F 2x 5 y i x y j 3 x z k is solenoidal.

Chapter 2: Vector Calculus

II – 2.25

3 2 2 16. Find the value of a, if F (a x y z ) i (a 2)x j (1 a)xz k is irrotational. 17. Find the values of a, b, c, so that the vector F (x y a z ) i (bx 2 y z ) j ( x cy 2 z )k may be irrotational. 18. If and v (u v ) is solenoidal. 19. If 1 and 2 ( 1 2 ( 2 1). 20. If

Part B 21. If u = x2yz and v =

3z2,

( u

v) and

( u

v) at the

22. Find the directional drivative of ( of the normal to the surface xy2 z = 3x + z2 , where = x2 y2 z2 F , ( F ), F, ( F ) and 23. If F 3 x 2 i 5 xy 2 j x y z 3 k , ( F ) at the (1, 2, 3) 24. If is a constant vector and r is the vector of (x, y, z) w.r.t. the ori[(a r ) r ] a r . F 3 yzi tational. Find also 26. Show that F

(z 2

2 zx j

4 xy k is not irrotational, but (x2 y z3) F is irro-

2 x 3 y) i

(3 x

2y

z) j

(y

2 z x) k is irrotation-

27. Find the constants a, b, c, so that F (x 2 y az ) i (bx 3 y (4 x cy 2 z ) k may be irrotational. For these values of a, b, a, b, c, if F

axyz 3 i

z) j bx 2 z 3 j

cx 2 yz 2 k is irrotational. For these values of a, b, c, 29. If r

(x, y, z) w.r.t. the origin,

1 2 1 2 r and (ii) r r r3 r r n r is both solenoidal and irrotational, when 30. Find the value of n, if r . r xi y j zk (i)

2.4

LINE INTEGRAL OF VECTOR POINT FUNCTIONS C be a curve in that

region (Fig. 2.2). P and Q on C be r and r acts at P

Q

F

Let F (x, y, z)

C

P r+

r .Then PQ

r If F with PQ ,

r Fig. 2.2

Part II: Mathematics II

II – 2.26

then F

r

F ( r ) cos

In the limit, F dr

F dr cos

Note

F dr

. F through

dr . Now the integral

F dr

the line integral of F along the curve C.

C

Since

F cos dr , it is also called the line integral of the tangential

F dr C

C

F along C. B

Note

(1)

F dr

C but also on the terminal

A (C )

A and B. B

F dr

done by the force F in

A (C )

A to B along the curve C. B

F dr

(3) If the value of

curve C, but only

A

A and B, F is called a Conservative vector. F from A to B A and B, the force F is called a Conservative force. C is a closed curve, the line integral is denoted as

F dr . C

(5) When F

F1 i

F dr C

(F1 i

F2 j F2 j

F3 k , F3 k ) (dx i

dy j

dz k )

C

( (F1dx

F2 dy

r

xi

F3 dz )

yj

zk ) -

C

dr , where

(6) C

line integrals.

2.4.1

Condition for F to be Conservative

If F is an irrotational vector, it is conservative.

F C

dr are also

Chapter 2: Vector Calculus Since F

Proof:

II – 2.27 . i.e., F

B

B

F dr A

dr A B

x

A

i

j

y

k

z

(dx i

dy j

dz k )

B

x

A

dx

y

dy

z

dz

B

d A

[ ]BA (B)

(A)

F is conservative. Note If F F dr 0 . C

[

(B) , as A and B coincide]

(A)

2.4.2

Surface Integral of Vector Point Function

Let S be a two sided surface, one side of which is Let F

F

S. Let dS

S

x, y, z). Let n be the unit vector normal to the surface S at (x, y, z) direction) Let be the angle between F and n . F F n F cos

nˆ ds (x, y, z)

S

Fig. 2.3

dS over the surface S is called the surface integral of F over S and denoted as F cos dS or F n dS . S

S

If

is a vector whose magnitude is dS and whose direction is that of n , then n dS . F n dS can also be written as

S

F dS . S

Part II: Mathematics II

II – 2.28

Note

(1) If S is a closed surface, the outer surface is usually chosen as the F

dS and

(2)

dS , where

S

S

surface integrals. (3) When evaluating

F n dS S

To evaluate a surface integral in the scalar form, we convert it into a double integral and then evaluate. Hence the surface integral F dS is also denoted as S

F dS . S

WORKED EXAMPLE 2(c)

dr , where C is the curve x = t, y = t2, z =

Example 2.1 Evaluate

t) and

C

= x2 y (1 + z) from t = 0 to t = 1. r

xi

yj

dr

dx i

dy j

2

Hence the given line integral I

zk

x y (1

dz k

z ) (dx i

dy j + dz k )

C

x 2 y (1

i

z ) dx

C 1

i

z ) dy

1

t 4 ( 2 t ) dt

j

1

t6 6

7 i 30

8 j 21

If F

t6 j 4 6

0

xyi

z ) dz

C

t 4 (2

1

t ) 2t dt

k

0

t5 i 2 5

x 2 y (1

k

C

0

Example 2.2

x 2 y (1

j

t 4 (2

t ) ( dt )

0

t7 2 7

1

t5 2 5

k 0

t6 6

1

0

7 k. 30 zj

F

x 2 k , evaluate

dr , where C is the curve

C

x = t2, y = 2t, z = t3 from (0, 0, 0) to (1, 2, 1).

F

dr

i xy dx

j k z x2 dy dz

(z dz

x 2 dy ) i

(xy dz

x 2 dx) j

(xy dy

zd x ) k

Chapter 2: Vector Calculus

II – 2.29

The given line integral x 2 dy ) i

[ (zdz

x 2 dx) j

(xy dz

(xy dy

z dx ) k ]

C 1

[ (t 3 3t 2

t 4 2)) dt i

(2t 3 3t 2

t 4 2t ) dt j

(2t 3 2

t 3 2 t ) dt k ]

0

[

t 1

i (3t

5

1

4

2t ) dt

j

(6t

0

t = 1]

5

1

5

2t ) dt

k

(4t

0

i 3

t6 6

9 i 10

2

t5 5

1

j 4 0

2 j 3

3

4

2t ) dt

0

t6 6

1

k t4 0

2

t5 5

1

0

7 k 5

Example 2.3

F

(x 2

y2

x) i

xy y2 = x. W=

F

F dr C

[(x 2

y2

x) i

[(x 2

y2

x ) dx

(2 xy

y ) j ] (dx i

dy j

C

(2 xy

y ) dy ]

C

Case (i) C is the line y = x. [(x 2

W1 y (dy

y2

x) dx

(2 xy

y ) dy ]

x dx )

1

(

2 x 2 ) dx

0

2 3 Case (ii) C is the curve y2 = x. [(x 2

W2

y2

x) dx

(2 xy

y ) dy ]

y2 2 y dy )

x (dx 1

(2 y 5

2 y3

y ) dy

0

2 3 Comment force. In fact, F is a conservative force, as F is irrotational.

dz k )

(2xy y ) j y = x, (ii)

Part II: Mathematics II

II – 2.30

F 2 . 3

2

x =y F

Example 2.4 r

cos ti C,

zi

xj

sin t j

yk , when it moves

tk

from t = 0 to t = 2 .

curve as x = cos t, y = sin t, z = t. F

F dr C

(zi

xj

yk ) (dx i

dy j

dz k )

C

(z dx

x dy

y dz )

C 2

t ( sin t )

cos 2 t

sin t dt

0

t cos t (2 3

Example 2.5

sin t 1)

1 t 2

cos t 0

( 1)

F dr , where F

Evaluate

2

sin 2t 2

(sin y ) i

x (1

cos y ) j

zk and

C

C is the circle x2 + y2 = a2 in the xy Given integral

(sin y ) i

x(1

cos y ) j

zk

(dx i

dy j

dz k )

C

sin y dx x2

y2 z

cos y ) dy

z dz

0

sin y dx x2

x(1

a2

y2

x(1

cos y ) dy

a2

Since C y = a sin circle C

C, namely x = a cos , as the variable of integration. To move around the varies from 0 to 2 . (sin y dx

Now, given integral

x cos y dy )

x dy

d(x sin y ) x dy 2

d a cos 0

sin (a sin )

a 2 cos d

Chapter 2: Vector Calculus a2 2

a cos sin (a sin )

II – 2.31

sin 2 2

2

0

a2 Example 2.6

F

z3 )i

(2 xy

x2 j

3 xz 2 k ,

guess that the given force F is conservative. Let us verify whether F is conservative, i.e. irrotational. i F

j x

2 xy

k

y z

3

x

z

2

3 xz 2

(3 z 2

( 0 0) i 0

3z 2 ) j

(2 x

2 x) k

F is irrotational and hence conservative. F Since F is irrotational, let F . It is easily found that = x2 y + z3 x + c. (3 ,1,4 )

done by

F

F dr (1, 2 ,1) (3,1, 4)

dr (1, 2, 1) ( 3, 1, 4)

d (1, 2, 1) (3, 1, 4 ) 2, 1)

(x, y, z ) (1,

Example 2.7 j

2 x y zk

x2 y

z2 x

(201 202 .

c)

c

( 3, 1, 4) (1, 2, 1)

( 1 c) F

y (3 x 2 y

z2 )i

x(2 x 2 y

z2 )

C. C and the terminal

Part II: Mathematics II

II – 2.32 Since C x0, y0, z0

C is not given. Hence we guess that the given force F is i

F

j x

2

3y x

k

y

2

yz

( 2 xz 0

2

3

2x y

2 xz ) i

(

Since F is irrotational, let F F

z x

z 2 xyz

2 yz

2 yz ) j

2

(6 x 2 y

z2

6x2 y

z 2 )k

.

F dr C

dr C (x0 , y0 , z0 )

d (x0 , y0 , z0 )

(x0 , y0 , z0 )

(x0 , y0 , z0 )

0 12 x 2 yi

A dS , where A

Evaluate

Example 2.8

3 yz j

2 zk and S is

S

x+y+z= Given integral I =

A n dS, where n is the unit normal to the surface S given by z

= c, i.e. x + y + z = 1 x

y

i

j

k

(i

j

n

C

1 3

z B

O

k) A

Fig. 2.4

(12 x 2 yi

I

3 yzj

2 zk )

1

S

1 3

(12 x 2 y S

3 yz

2 z ) dS

3

(i

j

k ) dS

y x

Chapter 2: Vector Calculus

xoy -

S on the dA, where is the angle between the surface S and the

Then dS cos

xoy -

II – 2.33

, i.e. the angle between n and k . dA

dS

n k

dx dy 1 3

n k 1

I=

cos

(12 x 2 y

3

3 yz

2z)

OAB

[ 12 x 2 y

3 y (1

x

dx dy 1 3 S

y)

OAB]

2(1

x

y ) dx dy

OAB

Note

x and y only, z

function of x and y 1 1

y

(12 x 2 y

I 0

S. 3y2

3 xy

5y

2x

2) dx dy

0

1

4 y (1

3y (1 2

y )3

0

3 y 2 (1 y )

y)2

5 y (1

y)

(1

y)2

2(1

y ) dy

49 120 Example 2.9

F dS , where F

Evaluate

yzi

zxj

xyk and S

S

x2 + y2 + z2 Given integral I

F n dS , where n is the unit normal to the surface S given by S

= c i.e. x2 + y2 + z2 = 1. = x2 + y2 + z2 2 xi n

2 yj

2 xi

2 zk

2 yj

4(x

2

y

2(xi

2 zk 2

z2 ) zk )

yj

[

x, y, z) lies on S]

4 1 xi I= S

yj

zk .

(yzi

zxj

xyk ) (xi

yj

zk ) dS

Part II: Mathematics II

II – 2.34 3 xyz dS

z

S

3 xyz

dx dy , n k

R

where R is the region in the xy by the circle x2 + y2 =

B

O

y

A R

dx dy 3 xyz z

1

1

I=

x

y2

Fig. 2.5

3 xy dx dy 0

0

x2 3y 2 0

1

1

3 2

y2

dy 0

1

y 2 ) dy

y (1 0

3 y2 2 2

y4 4

1

0

3 8

Example 2.10

F dS if , F

Evaluate

yzi

2 y2 j

xz 2 k and S is the sur-

S

face of the cylinder x2 + y2 = 9 z = 0 and z = 2. F n dS , where n is the unit normal to the surface S given by Given integral I S

= c, i.e. x2 + y2 = 9 = x2 + y2 2 xi n

2 yj

2 xi 4(x 2(xi

2 yj 2

y2 ) yj )

4 9 1 (xi 3

yj )

[

x, y, z) lies on S]

Chapter 2: Vector Calculus I=

(yzi

2 y2 j

xz 2 k )

1 3

(xyz

2 y 3 ) dS

(xyz

2 y3 )

s

1 3

1 (xi 3

z

yj ) dS

C

B

s

R

II – 2.35

dx dz n j

O A

where R xoz S on the xoz

x Fig. 2.6

1 I= 3

(xyz R

dx dz 2 y3 ) y 3 2 y 2 ) dx dz

(xz R 2

3

[xz 0 2

0

x 2 )] dx dz

2(9

0

9 z 18 3 2 9 dz 2 2

9 z2 2 2

36 z

81 0

EXERCISE 2(c) Part A (Short Answer Questions) 2. 3. If F F dS is evaluated. S

6. Evaluate (2, 2).

r dr , where C is the line y = x in the xy C

F xi the curve 2y = x2 from (0, 0) to (2, 2). F

(2 x

yz ) i

(xz

2 yj 3)j

xyk is conservative.

y

Part II: Mathematics II

II – 2.36 9. Evaluate

(yzi

zxj

xyk ) dS , where S is the region bounded by x = 0,

S

x = a, y = 0, y = b and lying in the xoy a closed curve. Part B 11. Evaluate

dr , where

= 2xyz2 and C is the curve given by x = t2, y = 2t,

C

z = t3 from t = 0 to t = 1. 12. Evaluate F dr along the curve x = cos t, y = 2 sin t, z = cos t from t = 0 C

to t

2

, given that F

13. Evaluate

2 xi

yj

zk .

F dr along the curve x = cos , y = sin , z = 2 cos from = 0 C

to

2

, given that F

2 yi

zj

xk

F dr along the curve x = t2, y = 2t, z = t3 from t = 0 to t = 1,

14. Evaluate C

given that F

xyi

zj

x2 k . F

y2 j ,

3 xyi

along the curve y = 2x2 in the xy F

(y

3z ) i

(2 z

x)j

(3 x

x = a cos t, y = a sin t, z

2 y )k , 2at

a, 0, 0) and (0, a, a). 17.

F

(2 y

3) i

xzj

(yz

x)k when it

F dr , where C is the circle x2 + y2 = 4 in the xy

18. Evaluate C

F

(2 x

19.

y

z)i

(x

y

z 2 )j

(3 x F

2y

(x

2

4 z )k . 2

y )i

(x 2

z 2 )j

yk , when it

x2 + y2 =

(3 xz 2

2)k

F

(e x z

F

(y 2 cos x

2 xy ) i

(1 x 2 )j

z3 )i

(e x

(2 y sin x 2

z )k ,

4)j

, 1, 2 along any

Chapter 2: Vector Calculus y2 i C.

F F dS , where F

23. Evaluate

2(xy

x2 j

xyi

II – 2.37

(x

z) j

2 yk , when it

z )k and S

S

x + 2y + z F dS , where F

24. Evaluate

yi

xj

4k and S

zi

xj

3 y 2 zk and S is the surface of the

S 2

x2 + y2 + z = a2 25. Evaluate

F dS , where F S

cylinder x2 + y2 = z = 5.

2.5

z = 0 and

INTEGRAL THEOREMS

3. Gauss Divergence theorem double integral.

s theorem. Gauss Divergence theorem

2.5.1

Green’s Theorem in a Plane

If C (P dx

Q dy )

C

R

R in the xy R, then Q P d x dy x y

P(x, y), Q (x, y)

where C

2.5.2

Stoke’s Theorem C and if F is a S, then

If S F dr C

curl F dS

,

S

where C S.

Part II: Mathematics II

II – 2.38

2.5.3

Gauss Divergence Theorem

If S is a closed surface

V and if F is a vector V, then F dS

(div F ) dv .

S

2.5.4

V

Deduction of Green’s Theorem from Stoke’s Theorem F dr C

curl F dS

(1)

S

S curve C.

R in the xy F

F

Then curl

P(x, y ) i i

j

x P

y Q

Q x

Q(x, y ) j k z 0

P k y

[ P and Q are functions of x and y]

Inserting all these in (1), we get (Pi

Q j ) (dxi

dyj

Q x

dzk )

C

R

(P dx

i.e.

Q dy )

C

(Qx

P k dS y Py ) k n dS

(2)

R

Now n is the unit vector in the outward drawn normal direction to the surface R. R, that lies in the xy z z n , we get n k . Using this in (2), we get (P dx

Q dy )

C

(Qx

Py ) dx dy

R

[ dS =

xy

= dx dy]

Note

If the surface S

2.5.5 F

Scalar form of Stoke’s Theorem Pi

Qj

Rk

P, Q, R are functions of x, y, z.

Chapter 2: Vector Calculus Then

F dr F F dS

(Ry

P dx

Qz ) i

Q dy

(Pz

II – 2.39 (1)

R dz

Rx ) j

(Qx

Py ) k

curl F n dS (Ry

Qz )(n i )dS

(Ry

Qz ) dy dz

(Pz (Pz

Rx )(n j )dS Rx ) dz dx

(Qx

(Qx

Py )(n k )dS

Py ) dx dy

(2)

[ dS cos = dx dy i.e. dS (n k ) = dx dy] s theorem, it reduces to the scalar form (P dx

Q dy

R dz )

[(Ry

C

Qz ) dy dz

(Pz

Rx ) dz dx

(Qx

Py ) dx dy ]

S

(3)

Note

P and Q as functions of x and y only and R = 0 in (3), we get

2.5.6

Scalar form of Gauss Divergence Theorem

F x, y, z.

Pi

Qj

Rk in Divergence theorem, where P, Q, R are functions of

Then div

P x

F F dS

Q y

R z

(1)

F n dS P (n i ) dS Q (n j ) dS R (n k ) dS P dy dz Q dz dx R dx dyy

(2)

Inserting (1) and (2) in Divergence theorem, we get (Pdy dz

Qdz dx

Rdx dy )

(Px

S

Qy

Rz ) dx dy dz

(3)

V

which is the scalar form of Divergence theorem.

Note 2.5.7

(3) is also called Green’s theorem in space.

Green’s Identities

In Then div (

F = zd} , where =

(

and

) 2

Divergence theorem becomes dS S

( V

2

) dV

(1)

Part II: Mathematics II

II – 2.40 Interchanging

and ,

## }dz $ dS = ### (}d z + dz $ d})d V 2

(l)

S

V

(

) dS

(2)

(2) gives, S

Note

2

(

2

(3)

) dV

V

(1) or (2) is called Green’

. and (3) is called Green’

. WORKED EXAMPLE 2(d) (x 2

Example 2.1

y 2 ) dx +

C

2xy dy], where C is the boundary of the rectangle in the xoy lines x = 0, x = a, y = 0 and y = b. (OR) F rectangular region in the xoy F dr

(x y ) i 2 xy j in the x = 0, x = a, y = 0 and y = b. 2

2

curl F dS

C

S

i curl F

Now

x x

2

y

(2 y

2

j

k

y 2 xy

z 0

2 y) k

We have to verify that [(x 2

y2 )i

2 xy j ] (dxi

dy j

dzk )

4 yk n dS

C

S

[(x 2

i.e.

y 2 ) dx

2 xy dy ]

C

4 y dx dy [Fig. 2. 7]

(1)

R

n y

y=b

D

B

x=0

x=a

O

y=0 Fig. 2.7

A

x

k

Chapter 2: Vector Calculus

II – 2.41 [(x 2

L.S. of (1) OA y 0 dy 0

AB x a dx 0

b

x2d x

the boundary C consists of 4 lines] 0

(x 2

2ay d y

0

2 x y dy ]

DO x 0 dx 0

BD y b dy 0

[ a

y 2 ) dx

0

b2 ) d x

0

a

Note x or dy a

3

x 3 b

x 3

a (y 2 ) b0

2 ab

a

3

0 2

2

b x 0

a

R.S. of (1)

4 y dx d y 0 b

0

4 y (x)0a dy

2a (y 2 )

b 0

0

2 a b2 . [(3x 2

Example 2.2 Verify Green’s

8y 2 ) d x

(4y

6 xy )

C

dy ] , where C

x = 0, y = 0 and x + y = 1.

Green’s theorem is For the given integral, [3x 2

(P dx

Q dy ) =

C

8y 2 ] d x

(4y

(Qx

Py ) dx dy

R

6 x y ) dy ] =

C

10 y d x dy [Fig. 2.8] R

y B x+y=1

x=0

O

y=0

A

x

Fig. 2.8

[(3x 2

L.S. of (1) OA y 0 dy 0

AB x y 1 x 1 y dx dy

BO x 0 dx 0

8y 2 )d x (4y

6 x y ) dy ]

(1)

Part II: Mathematics II

II – 2.42 1

1

0

3x 2 d x

[{3(1 y )2

0 1

8y 2 }( dy ) + {4y 6 y (1 y )}dy ]+

0 1

3x 2 d x

1 1

(11y 2 + 4y

0

4 y dy

3) dy

0

4 y dy

1

0

11 3

2 3

2

5 3 1 1 y

R.S. of (1)

10 y dx dy 0 1

0

10 y (1 y) dy 0

5y

y3 10 3

2

1

0

5 3

Since L.S. of (1) = R.S. of (1),

. F dr , where F

Example 2.3

(sin x

y)

C

i

cos xj and C is the boundary of the triangle whose vertices are (0, 0),

and 2

2

,0

,1 .

Note

F dr

Evaluating C

integral. F dr

curl F dS , where S

C

S

surface bounded by C. [Fig. 2.9] S

R in the xoy

bounded by C. F dr C

curl F k dx dy [

For the xoy lane, n

R

i curl F

j

x (sin x y ) (sin x

1) k

y cos x

k z 0

k and dS

d x dy ]

Chapter 2: Vector Calculus he given line integral =

II – 2.43 y

(1 + sin x) d x d y

B

R

y=

1

2

0

y 2

2

x

(1 + sin x) d x dy

x= O

1

x

cos x

2

y 2

y dy cos 2

y2 4

2

0 1

0

2

2 y 2

y 2

4

,1

2

2 x

y=0

A

2

dy

,0

Fig. 2.9

y sin 2

1

0

. [(2x

Example 2.4

y ) dx

C

(x + y ) dy ] , where C is the boundary of the circle x2 + y2 = a2 in the xoy (P d x

Q dy ) =

(Qx

C

[(2x

y) d x

y

Py ) d x dy

R

(x + y ) dy ]

[1 ( 1)] d x dy

C

x O

R

2

d x dy[Fig. 2.10] R

2

Fig. 2.10

area of the region R

2 a2 Example 2.5

F

(2 x y

F dr

curl F n d S

C

S

i Now curl F

j x

2 xy (

2x 4 xk

x 2 ) i (x 2 y 2 ) j and C y2 = x and x2 = y.

k

y x2 2 x)k

x2

y2

z 0

Part II: Mathematics II

II – 2.44

[(2 x y x 2 ) d x (x 2

y 2 ) dy ]

4 x k k dx dy

C

R

4 x d x dy [Fig. 2.11]

(1)

R

B(1, I)

D

y = x2

A O x = y2

Fig. 2.11

L.S. of (1)

[(2 x y OAB y x2 dy 2 x d x

x 2 ) dx

(x 2

y 2 ) dy ]

BDO x y2 d x 2 y dy 0

1

(2 x

5

2

(3 y 4

x )dx

0

2 y5

y 2 ) dy

1

y = x 2 and

[ the coordinates of B x = y2] 3 5 1

y

0 y

2

R.S. of (1)

4x dx d y 1

(x 2 )

2

y y2

dy

0 1

(y

2

y 4 ) dy

0

3 5 S Example 2.6 1 (x dy y d x) 2 C

C is given by y2 = 4a x and its x2

y2

a2

b2

1.

Chapter 2: Vector Calculus (P dx

By Green’s theorem,

Q dy )

(Qx

C

P

Py )dx dy

R

y and Q 2

x , we get 2

1 2

y dx)

(x dy

II – 2.45

C

R

1 2

1 dx dy 2

dx dy R

=Area of the region R enclosed by C. 1 2

rectum

(x dy

y dx) [Fig. 2.12]

C

y (a, 2a)

L

1 2

(x dy LOL

y dx )

x=a

L SL

1 2

(x dy

2a

1 2

2a

2a

y dx )

L'

x a dx 0

(a

a)

Fig. 2.12 2a

y2 dy 4a

a dy 2a

2a

y2 dyy 4a

0

(x dy

y dx )

y2 x 4a y dy dx 2a

x

S

O

a dy 0

8 2 a . 3 y

1 2 1 2 1 2

=

(x dy

y dx)

=

2 =0 x =2

=

C

O

(x dy x a cos y b sin

y dx)

[Fig. 2.13] =

2

ab (cos 2 0

.

sin 2 ) d

Fig. 2.13

3 2

Part II: Mathematics II

II – 2.46 Example 2.7 (curl F )

) = 0 and (ii) div

0. curl F dS

F dr

S

F

grad

S2

S1

C

S3

, we have

curl (grad ) dS

grad

S

dr C

C

d

Fig. 2.14

C

=0 S C. [Fig. 2.14] curl (grad ) dS 0, [for any S and hence for any dS ] curl (grad ) =0 (ii) Gauss divergence theorem is F dS , where S is a closed surface enclosing a volume V.

(div F ) dv V

S

F by curl F , we have div (curl F ) dv

curl F dS

V

(1)

S

. However .S. of (1), as S is a closed surface. , we divide S S1 and S2 by , each of which is bounded by the same

. S1 and S2 closed curve C as given in Fig. 2.15. Now (1) becomes div (curl F ) dv

S1

curl F dS

V

S1

curl F dS

C

S2

F dr

F dr

C

S2

C

Fig. 2.15

Note

If C of the outer normal to S1 S2. div (curl F ) dv V

F dr C

= 0. Since V is arbitrary, div (curl F )

0.

F dr C

Chapter 2: Vector Calculus Example 2.8 Evaluate

(sin z dx

cos x dy

II – 2.47

sin y dz )

C

where C

x (P dx

Q dy

y R dz )

C

dy dz

(Pz Rx ) dz dx P = sin z, Q

(sin z dx

cos x dy

[(Ry

Qz )

S

(Qx Py ) dx dy ] x, R = sin y, we get sin y dz )

C

(cos y dy dz

cos z dz dx

sin x dx dy )

S

sin x dx dy

[

S

S is the rectangle in the z and hence dz = 0]

1

sin x dx dy 0

0

(cos x)0

=2 (x dy dz

Example 2.9 Evaluate

2 y dz dx

3 z dx dy ) where S is the closed

S

x2 + y2 + z2 = a2. Scalar from of divergence theorem is (P dy dz

Q dz dx

R dx dy )

(Px

S

Qy

Rz ) dV

V

P = x, Q = 2y, R = 3z, the given surface integral =

6 dv

6V

V

6

4 3 a 3

= 8 a3. F

Example 2.10

xyi

2 yz j

0, y = 2 and z = 3 above the xoy F dr C

Here

curl F

curl F dS S

i

j

k

x xy

y 2 yz

z xz

zxk where S is the x = 0, x = 1, y =

Part II: Mathematics II

II – 2.48 2 yi

zj

(xy dx

xk

2 yz dy

zx dz ) =

(2 yi

C

zj

(1)

xk ) dS

S

z C

A'

B'

O' ˆn = j

nˆ

j O

y

B

A C'

x Fig. 2.16

S x = 0, x = 1, y = 0, y = 2 and z = 3 and is bounded by the rectangle OAC B lying on the xoy L.S. of (1)

=

(xy dx

2 yz dy

zx dz )

OAC B

xy dx [Fig. 2.17]

( the boundary C

z = 0)

y

OAC B

(xy dx) OA y 0 dy 0

AC x 1 dx 0

C'

B

BO x 0 dx 0

CB y 2 dy 0

0

2 x dx

O

Fig. 2.17

=

. R.S. of (1) =

## + ## + ## + ## + ## (2yir + zjr - xkr ) $ nt ds

`ntx==-0ij

Note

x

A

1

1 ` xnt = = ij

y=0 ct m n = - rj

y=2 ct rm n=j

z=3 ct rm n=k

nt y nt

y Similarly nt for y Using the relevant value of nt R. S. of (1)

j.

j and so on. , we have

2 y dS x 0

y

2 y dS x 1

z dS y 0

z dS y 2

x dS z 3

Chapter 2: Vector Calculus 3

2

0

0

3

2

0

0

2 y dy dz

1

3

0

0

2 y dy dz

[ dz etc.]

II – 2.49 1

3

0

0

z dz dx

1

0

0

2

x2 dy 2 0

x dx dy

1

0

0

z dz dx

S on x = 0 and x 2

2

x dx dy

,

y

( the other integrals cancel themselves)

1

0

1. . F

Example 2.11 z

y 2 zi z 2 xj x 2 yk where S is the x = ±a, y = ±a and z = ±a, in which

a is cut. F dr

curl F dS

C

curl F

Here

i

j

k

x y2 z

y z2 x

z x2 y

(x 2 (y 2 z dx

z 2 x dy

x 2 y dz )

2 zx) i [(x 2

C

S

(y 2

(z 2

2 xy )j (y 2

2 zx) i

2 xy )j

2 yz )k (z 2

(1)

2 yz )k ] dS

S

F

E

O

H

z 0 D

x

y C

A

B Fig. 2.18

S

z

2.18) L.S. of (1)

(y 2 z dx (z

z 2 x dy

a (Fig.

x 2 y dz )

a)

( ABCD

ay 2 dx

a 2 x dy )

[ dz = 0, as z

a]

Part II: Mathematics II

II – 2.50

y=a dy = 0

C

B

x=a dx = 0

x a dx = 0

D

A

y a dy = 0 Fig. 2.19

(

L.S. of (1) AB

BC

CD

a

ay 2 dx

a 2 x dy ) Fig. 2.19

a

a

DA

a

a 3 dy

a 3 dx

a

a 3 dy

a

a 3 dx

a

a

4 a4 . R.S. of (1) x n

a i

x n

a i

y n

[(x 2 z n

2 zx) i

x 2 ) dS

a

x

(y 2 y

a j

(y 2

2 xy ) j

(x 2

2 zx) dS

a

(z 2 z

a 2 ) dy dz

2az (

2ax

(a 2

2 yz ) k ] ndS

a 2 ) dz dx

a

a a

(a 2

a

a

a

(a 2 a

2az ) dy dz

4ax dz dx a

2ay ) dx dy

a

= 0 + 0 + 4a4 = 4a4. .

a

2 yz ) dS

2ay ) dx dy

a

(2 xy

a

(a 2

4a z dy dz a

(z 2

y

2 xy ) dS

a

(

a

y n

a k

(2 zx x

a j

2ax) dz dx

y 2 ) dS

Chapter 2: Vector Calculus

II – 2.51

F yi 2 yz j y 2 k , where S is the x + y + z = a and C is the circular boundary on the xoy-

Example 2.12

2

2

2

2

j

k

y 2 yz

z y2

i F

x y k.

F dr

F dS

C

S

(

Here

ydx

2 yz dy

y 2 dz )

k dS

C

(1)

S

x2 + y2 = a2

C is the circle in the xoyx = a cos and y = a sin . L.S. of (1) =

( C lies on z = 0)

y dx x

2

y

2

a

2

2

a 2 sin 2

d

0

a2 2

sin 2 2

2

0

2

.

=

k n dS , where

R.S. of (1) = S

n

, where 2 (xi

yj

4 (x

2

xi

yj a

y

= x2

y2

zk ) 2

zk

z2 ) [

S

z dS a

R

z dx dy , where R a n k .

R.S. of (1)

z2

x, y, z) lies on

= a2]

S on the xoy-

dx dy , where R is the region enclosed by x2 + y2 = a2. =

R 2

.

Part II: Mathematics II

II – 2.52

x2 + y2 + z2 = 1, evaluate

Example 2.13 If S is (xi

2 yj

3 zk ) dS .

S

By Divergence theorem, F dS

(div F ) dv

S

V

F dS S

x

V

6

(x)

y

(2 y )

z

(3 z ) dV

dV V

= 6V, where V is the volume enclosed by S. 4 6 3 =8 . 2 y 2 j z 2 k , where Example 2.14 Verify Gauss divergence theorem for F x i S is x = 0, x=a, y=0, y=b, z = 0 and z = c. Divergence theorem is (1) F dS (div F ) dV S

V

S

. L.S. of (1) x n

0

x n

i

a i

y n

2

z n

0 k

2

y j

z k ) n dS x 2 dS

0

x

a

z 2 dS z

b j

c k

x 2 dS x

y n

j 2

(x i z n

0

0

c

z

0

a

0

dz dx 0

(2 x

0

2y

2 z ) dx dy dz

V c

b

a

0

0

0

(2 x

y

c

b2

= abc (a + b + c) R.S. of (1)

y 2 dS b

z 2 dS (on using the relevant values of n )

c

dy dz 0

y

b

a2

y 2 dS

2y

2 z ) dx dy dz

b

a

0

0

c2

dx dy

Chapter 2: Vector Calculus c

II – 2.53

b

(x 2 0

2 z x)0a dy d z

2y x

0

c

(a 2 y

ay 2

2 a z y )b0 d z

(a 2 b

ab 2

2ab z ) dz

0 c

0

a 2 bc abc (a

ab 2 c abc 2 b c) . x2 i

Example 2.15 Verify divergence theorem for F

zj

yz k over the cube

formed by x = ±1, Divergence theorem is

y = ±1, z = ±1.

F dS

(1)

(div F ) dV

S

V

L.S. of (1) = x n

1 i

x 1 n i

(x 2 i z n

y n

1 j

zj

y 1 n j

x

1

z

1

yz k ) n dS

x 2 dS x

1

z

1

yz dS

yz dS

R.S. of (1) =

(2 x

y ) d x dy dz

v 1

1

(2 x 1

1

y ) dx dy dz

1

1

( 2 y dy d z ) 1

1

0 .

z dS

z dS y

0

1

1 k

1 k

x 2 dS

1

z n

1

y

1

(using the relevant values of n)

Part II: Mathematics II

II – 2.54

EXERCISE 2(d) Part A (Short Answer Questions) (or) the connection between a line integral 1. State Green’s theorem in and a double integral. s theorem (or) the connection between a line integral and a surface integral. 3. State Gauss divergence theorem (or) the connection between a surface integral and a volume integral. s theorem. s theorem. 6. Give the scalar form of divergence theorem. 7. Derive Green’s identities from divergence theorem. s = 0. ( F) 0 . 10. Evaluate (yz dx zx dy xy dz ) where C is the circle given by x2 + y2 + C

z2 = 1 and z = 0. 11. Evaluate

(x dy d z

ydz dx

z d x dy )

S

x2+ y2 + z2 = a2. 12. If C

r

xi

yj

r dr

zk

0.

C

13. If C 0.

dr C

14. If S is a closed surface and r

xi

yj

zk , S

1 r

dS .

(r 2 ) dS .

15. If S is a closed surface enclosing a volume V, evaluate S

16. Evaluate

[(x

2 y )d x

(3 x

y )dy ] , where C is the boundary of a unit

C

(x dy

17. Evaluate

y d x) , where C is the circle x2 + y2 = a2.

C

18. If A

curl F

A dS

0 , where S is any closed surface.

S

19. If S is any closed surface enclosing a volume V and if A A dS

(a + b + c) V .

S

20. If r

xi r dS .

S

yj

zk and S

a xi

by j

c zk ,

Chapter 2: Vector Calculus

II – 2.55

Part B (x 2 dx

xy dy ) , where C

C

is the boundary of the

x = 0, y = 0, x = a, y = a. F

2

x i

xyj

xy-

x = 0, y = 0, x = 2, y = 2. x 2 (1

y ) dx

(x 3

y 3 ) dy

C

x = ± 1 and y = ± 1.

where C s

F dr , when F

(xy

x2 ) i

C

x 2 y j and C is the boundary of the triangle in the xoy y = 0, and y = x.

x = 1, (2 x 2

y 2 ) dx

C

(x 2 y 2 ) dy , where C is the boundary of the region in the xoyenclosed by the x x2 + y2 = 1. 2 2 s theorem for F (xy y ) i x j in the region in the xoyy = x and y = x2. y2 = 4ax and x2 = 4ay. xoy bounded by y3 = x2 and y = x. s theorem for F

(y

z

2) i (yz 4) j xzk , where S is x = 0, x = 2, y = 0, y = 2 and z = 2.

(x 2 y 2 ) i 2 xyj xyzk over the surface x = 0, x = a, y = 0, y = b and z = c.

30.

s theorem for F

31.

s theorem for F (2 x y ) i yz 2 j y 2 zk where S is the x2 + y2 + z2 = 1 and C is the circular boundary in the

xoy32. Verify Gauss divergence theorem for

(x 2

F

yz ) i

(y 2

zx) j

2

(z xy ) k by x = 0, x = 1, y = 0, y = 2, z = 0 and z = 3. F

33. Verify divergence theorem for (i) F

(2 x

z)i

x 2 yj

4 xzi

y2 j

yzk

and (ii)

xz 2 k , when S is the closed surface of the cube

formed by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1. (yz 2 i

34. Use divergence theorem to evaluate S

S is the closed surface bounded by the xoy x2 + y2 + z2 = a2

.

zx 2 j

2 z 2 k ) dS , where

Part II: Mathematics II

II – 2.56

2 y2 j

z 2 k ) dS , where S is

the closed surface bounded by the cylinder x2 + y2 = and z = 3.

z=0

35. Use divergence theorem to evaluate

(4 xi s

ANSWERS

Exercise 2(a) (5) (8)

12 i 1 19

(i

(11) 25; (14)

9j 3j

56 i

1 ( 3

-1 (17) cos

(6)

2j

47 k

1 3

1 (2 i 3 11 (13) 3

(10)

k)

-1 (16) cos

11 8

(18) cos -1

3 22

j

27

(15)

13

(i

14 3

(12)

2k )

(7) 5

14

(9)

3k ) 30 j

i

-1 (19) cos

3 21

1

-1 (20) cos

(23) a

16k

(21) 2x

30 7 ;b 3

64 9

(24) x2y

y

z+1=0

xz2 + y2z + c

2j

45 2299 3 7 6 5 ; 2

(22) x 2 yz 3

(25)

20

Exercise 2(b) (7) 3; 0.

(8) 4

(9) 2( i a=4

(17)

1, b= l, c

(23) 80; 80 i

37 j

3k 36k ; 27 i

(22)

54 j

20k ; 0; 74 i

124 21

27 j

(25) x3y2z4 (26) x2 + y2 + 3xy + yz + z2x (27) a = 4, b = 2,

x2 1; 2

3y2 2

z2

2 xy

yz

4 zx

(28) x2yz3

Exercise 2(c) (6) 3

(7) 6

(9)

a 2b2 2

j

k)

k)

1

Chapter 2: Vector Calculus

(10) 0

(11)

2

(13)

4

i

7 6

(15) (18) 8 (21) 4

5 2

(24)

8 i 11

1 j 2

4 j 5

k

II – 2.57 (12) (14)

i 51 70

(16) 2a2

(17) 8

(19) 2

(20)

(22) 0

(23) 27 4

19 2

(25) 90

Exercise 2(d) (10) 0 (16) 5

3

(11) 4 (17) 2

(23)

8 3

(24)

(28)

1 10

(34)

2

1 12 a4

(15) 6V (20) 4 (27) 16 a 2 3

1 j 2

3 k 2

Chapter

3

Analytic Functions 3.1

INTRODUCTION

Before we introduce the concept of complex variable and functions of a complex

If x and y are real numbers and i denotes 1, then z x iy is called a Complex x is called the real part of z z R z); y is called the imaginary part of z z Iz The Complex number x iy is called the conjugate of the Complex number z x iy z. z z x 2 y 2 r 2 , where r is z z

modulus of the Complex number z. Also z = z ; R z

z 2

, I z

z z 2i

Y P x, y) r, )

r y O

x

M

X

Fig. 3.1

The complex number z x iy P x, y) with reference to a pair of rectangular coordinate axes OX and OY, which are to each complex number, there is a unique point in the XOY corresponding to each point in the XOY-plane, there is a unique complex number. The XOY-plane, the points in which represent complex numbers, is called the Complex plane or Argand plane or Argand diagram.

Part II: Mathematics II

II – 3.2

If the polar coordinates of the point P modulus of z or z and Since x = r cos

MOP

tan

and y = r sin , z x iy r

1

r, y x

), then r amplitude of z

i sin )

or

OP

x2

y2

z

re i

r i sin ) is called the modulus-amplitude form of z and re i is called the polar form of z

3.2

THE COMPLEX VARIABLE z x iy is called a complex variable, when x and y are two independent

The Argand plane in which the variables z the z point z

3.2.1

z is referred to as the

Function of a Complex Variable

If z x iy and w u iv are two complex variables such that there exists one or more values of w, corresponding to each value of z in a certain region R of the z-plane, then w is called a function of z and is written as w f z) or w f x iy When w u iv f z) f x iy u and v are functions of the variables x and y w z2, then u iv x iy)2 x2 y2) i xy) Thus u x2 y2 and v 2xy w

f z)

u x, y)

iv x, y)

If z is expressed in the polar form, u and v are functions of r and value of z, there corresponds a unique value of w, then w is called a single-valued function of z 1 are single-valued functions of z z z, there correspond more than one value of w, then w is called a multiple valued function of z w z1/4 and w z) are 1/4 multiple valued functions of z w z is four valued and w z z w

3.2.2

z2 and w

Limit of a Function of a Complex Variable

The single valued function f z) is said to have the limit l i ) as z tends to z0, if f z z0 z0) such that the values of f z) are as close to l as desired for all values of z z0, but different from z0 lim f z )} = l z

z0

Chapter 3: Analytic Functions

II – 3.3

Note Neighbourhood of z0 is the region of the z-plane consisting of the set of points z for which z z0 , where z z0 as centre and lim { f z } = l ,

z

z0

, such that f z) l 0

z

, whenever

z0

Note

In real variables, x x0 implies that x approaches x0 along the x-axis or a line parallel to the x x complex variables, z z0 implies that z approaches z0 curved) joining the points z and z0 that lie in the z Y

z

z0 X O

Fig. 3.2

Thus, in order that lim { f z } z

z0

f z) should approach the same value l,

when z approaches z0 along all paths joining z and z0

3.2.3

Continuity of f z)

The single valued function f z) is said to be continuous at a point z0, if lim { f z } = f z0

z

z0

This means that if a function f z) is to be continuous at the point z0, the value of f z) at z0 and the limit of f z) as z z0 A function f z) is said to be continuous in a region R of the z-plane, if it is If f z) u x, y) iv x, y) is continuous at z0 x0 iy0, then u x, y) and v x, y) will x0, y0 x, y) is said to be continuous at x0, y0 ), if lim x y

3.2.4

x0 y0

x, y )] =

x0 y0) , in whatever manner x

x0 and y

Derivative of f z)

The single valued function f z) is said to be differentiable at a point z0, if

y0

Part II: Mathematics II

II – 3.4

lim z

f z0

z

f z0

z

0

This limit is called the derivative of f z) at z0 z0 z z or z z z0, f z0 On putting z0

f z

lim

z

f z0

z z0

z0

z f z

3.2.5

z0 On putting

f z

lim z

z

f z

z

0

Analytic Function

A single valued function f z) derivative at z0 A function f z at R

z0, if it possesses a z0 R of the z a regular function or a holomorphic

function A point, at which a function f z) is called a singular point or singularity of f z

3.2.6

Cauchy-Riemann Equations f z

Theorem If the function f z) u u v and , , x y x

Proof f z) u x, y)

u x, y)

R of the z

iv x, y)

R of the z-plane, then u x

v y

iv x, y) f' z

v and y

u y

v x

R z in R L

lim z

f z

0

L takes the same value, when z In particular, L takes the same value when z QSP

z z

f z

QRP and

Chapter 3: Analytic Functions

II – 3.5

Y Q z + z)

S

y P z)

R

x

X O Fig. 3.3

It is evident that when z takes the increment z, x and y take the increments x and y L

x

lim i

y

u x

x, y

y

iv x x

0

L,

L1

u x

lim x

x, y

iv x

u x

0

i

x, y u x, y x

L,

L2

u x, y

lim y

v x

x, y v x, y x

y

iv x, y

y

u x, y

1 u x, y i

0

u y

y u x, y y

v x, y

y y

v y

L L1 and L2 u u v and , , x y x L

iv x, y

i y

lim i

i

L2, corresponding to the path QSP, y

0

y

iv x y

v x

letting x Thus

u xy

x

lim u x

iv x, y

y

x, y

0

x

i

u x, y

L1, corresponding to the path QRP, x

letting y Thus

x, y + y

R v y

R

v x, y

Part II: Mathematics II

II – 3.6 L1

L2

u v i x x u x

v y

u y

and

v u i y y

v x

R

Note Cauchy-Riemann equations

Thus w

f z z) z) ux ivx or vy iuy , where ux , uy , vx, vy f z), then dw dz

Thus

L1 or L2.

z) is also denoted as d w dz u v i x x x w x

dw dz

Also

L,

i i

u iv

u y

y w i y

v y u iv

the above theorem, we have assumed that w f z)

derived conditions do not w f z). In other words, the two conditions w f z). w f z) R

are

Theorem The single valued continuous function w f z) u x, y) iv x, y) a region R of the z-plane, if the four partial derivatives ux, vx, uy and vy have the ux

vy and uy

vx

R.

Chapter 3: Analytic Functions Proof Consider

u

u x

x, y

[u x

y)

x, y

x ux x

u x, y)

y)

u x, y

x, y

1

II – 3.7

y)

u x, y

y . uy x, y

y)

y)

2

u x, y)]

y),

, 2 where 0 1 ux is a continuous function of x and y, ux x

x, y

1

0 as x

1

y)

ux x, y)

y

and

, where

0 or as z

uy is a continuous function of x and y, uy x, y

2

y)

uy x, y) 0 as z

2

u where

1

and

, where

2

0

x [ux x, y)

y[uy x, y)

]

],

2

0 as z

2

vx and vy, we have v

x [vx x, y)

where

and w

Now

u i v ux ivx) x

ux 1

1

w

z and

2

1

i

2

ux and uy

y

1

x z

x z

2

x

x ) 4

2

2

1 1

y

y z

z

2

y z

1

1

and

2

2

i

y, where z

vx

x

1

x z

1

ivy) y

ivx) z

w u x ivx z x

i

ux ivx) x vx iux) y ux x i y) ivx x i y) ux

Now

ivy) y

ivx) x uy i ) and vy

],

4

0 as z

4

uy

y [vy x, y)

]

y z

1

x x

2 2

y y

4

y,

Part II: Mathematics II

II – 3.8 x z

1

y z

2

z

0[

1

and

z

2

dw u x ivx dz w w

f z)

R.

f z)

3.2.7

R.

C.R. Equations in Polar Coordinates

When z is expressed in the polar form rei , where w u iv, are functions of r and u r, ) and v r, ), assuming that w

Theorem If the function w

f z)

u r, )

u and v, u r, )

iv r, ) is

iv r, )

z-plane,

u u v v , , , r r u 1 v r r

v r

and

1 r

u

Proof f z)

u r, ) z in R

z) L

L

lim z

f z

z z

0

lim i

re

u r

iv r, )

R

f z

r,

iv r

r, re

0

u r,

L take the same value, in whatever manner z In particular, L z

rei )

rei ) ei r, if

z z

iv r ,

i

r

L,

0, if z

L1,

Chapter 3: Analytic Functions

Thus L1

u r

lim r

e

r, e

0

u r,

i

i

r

e

i

v r, r

rei ) i

re i z

, if r

if r L

L2

r,

u v i r r

i

z

Thus

v r

II – 3.9

u r,

lim

u r, re

0

1 e r

i

u

i

Since L

L2,

i

z tending

i

i

v r,

v r, re

i

i

v

L1 and L2

R.

u u v v , , , r r

R.

Since L L1

e

i

L2

u v i r r

1 e r

i

u

i

v

u 1 v r r v r

and

1 u r

R

Note f z Thus, when f z) z) e

i

ur

ivr )

where ur , u , vr , v w f z), then

z) u r, ) or

L, iv r,

1 e r

i

v

iu ,

L1 or L2.

Part II: Mathematics II

II – 3.10 dw dz

e

dw dz

Also

i

u iv

r i e r i e r i e r

i

u

e

i

w r

iv

i

u iv w

i

f z r, ) in R ur , u , vr, v must exist

w

f z) u r, )

w iv

y

0

in the region R. WORKED EXAMPLE 3 a)

Example 3.1

li z

x2 y x y2

f z ) , when f z

0

2

f z) x Thus

li y x

0 0

f z)

lim x

0

This does not mean that li z

0

0 x2

lim 0 x

f z) li z

li z

f z

0

0

0

0

0

f z ) can be 0, as per the mathe-

f z)

0

z 0

, whenever 0

Now, using polar coordinates, f z

0

r cos 2

r cos

2

sin sin 2

r | cos 2 || sin | r, since cos f z

0

1 and sin

, when r
c

u

O

c

x

O

Fig. 3.14

The image of the region y v c u 2 v2 c u2

v2)

u2

1 2c

v

2

1 2c

v

u2

v or

u2

1 2c

c

2

v [ c

v2

1 2c

2

c < 0] 4)

2

, whose centre is 0,

y

1 2c

v

O

1 ) 2c

x y>c

u

O ) Fig. 3.15

y

1 4 v u

u2

and radius is 1 2c

v2

4v

2

v

1 4

2

or u2 v 2)2 22, u2 v 2)2 22

The image of y v 2

u2 v2 u2 v

2

1 2

u v v or u2 1)2

v

1)2

Chapter 3: Analytic Functions y

II – 3.65

v y = 1/2

u

y = 1/4 x

O

Fig. 3.16

Example 3.6 x2 w

1 is the lemniscate R2 z

1

1 under the transformation w

z2 is

cos w

x

1 under the transformation

cos 2 z

the cardioid R

y2

iy

1 1 or z z w 1 1 R R ei

i sin )

The transformation equations are x y

and

1 cos R 1 sin R x2

y2 w-plane in polar co-ordinates

1 as 2 cos 2 R

1 sin 2 R2

1

R2

cos 2 z 1 x 1 iy 1 x 1)2 y2 1 or x2 y2 2x 0 2 r 2r cos 0 or r 2 cos The transformation is w z2 R ei r2 ei2 R r2 and 2 Eliminating r and r2 R

4 cos2 cos2 ) cos

Part II: Mathematics II

II – 3.66

Example 3.7 w-plane of the region of the z the straight lines x 1, y 1 and x y 1 under the transformation w z2 w z2 u iv x iy)2 x2 y2 i2 xy The transformation equations are u x2 y2 and v 2xy w z2, we get the image of the line 2 x 1 as the parabola v u 1) and the image of the line y 1 as the parabola v2 u The image of the line x y x and y u v u v

and and Eliminating x v

u

1 1

u

x2 2x 2x 2x

x)2 x) x

1 2

1 u2 or u2 v 2 which represents a parabola in the w v

v2

1/2), x 1, y 1), v2

u

1 and x y 1 is the region u 1) and u2 v 1/2) v

v2 u–

y

B

y=1

B'

A' u

O

x =1

u + 1)

1)

2

= –2

y=

v2

u

x+

C

C'

1

1/

C"

A

v–

O

2)

Fig. 3.17

Note

The region corresponding region is that which contains the images of the points A and C Example 3.8 x y 1 and y

u 1, u

B

i) x 1, v 1 and v 2 under the transformation w z2

Chapter 3: Analytic Functions

II – 3.67 z2,

w images of the lines x 1, x y 1 and y 2 v2 u , v2 u , v2 u 1) and v2 The given rectangular region in the z 0 and /2 images of 0 and /2 the transformation equations of w

u

w-plane, since the 0 and , as one of z2 in the polar form is 2 v C'

y D y=2

C

x=1 A

D'

B'

x

A'

y=1 B

O

u

O

x

Fig. 3.18

z2 are u

w v 2xy. The images of u 1, u x2 y2 1, x2 y2

, v 1 and v xy 1/2 and xy

x2

y2 and

1 in the z y

D'

v D v=2 C u =1 u A v=1 B O

C'

A' B' B" u

A"

C" D"

)

Fig. 3.19

O

x

Part II: Mathematics II

II – 3.68

Both the region in the z-plane are valid images of the given rectangular region in the w This is because we are transforming from the w-plane to the z the transformation z w1/2, Example 3.9

ez:

w

y x, y half of the strip 0 y The transformation equations of the mapping w Rei R and log R

, y

ez

ex iy ex eiy ex y y x is or R e ,

y The image of the y 0 is The image of the region 0 y half of the w

x 0 is R 0 and that of the line y

w is ,

x semicircle w

y

0, between y

0 and y

is the

1, v v

y

u

O x O

|w|=1

Fig. 3.20 v

y

|w|=1

y= O x O

Fig. 3.21

u

Chapter 3: Analytic Functions y and R 1 or w w

circle w

y

II – 3.69 y

and x

is mapped onto the exterior of the w

unit circle w

v

y y=

u O x O

Fig. 3.22

Example 3.10 x

2

and y

c in the z-plane under the transformation w y y=c

C

v

B

x = – /2

sin z

x = /2 S2

S1 O

D

y=–c

C' D'

x

B' S2'

O

S1'

A'

A

) Fig. 3.23

The transformation equations of the mapping function w u sin x cosh y and v cos x sinh y The part AS1 of the side AB c y u cosh y and v Now u is a decreasing function of y y segment

1

AS1 of the u-axis in the w

c

y v

0, since

sin z are

du dy

0 and cosh c

x

/2 and

sinh y

0 , when

u S1 B

Part II: Mathematics II

II – 3.70

v 0 and 1 u cosh c, u The image of the line segment BC,

1

y

u sin x cosh c and v cos x sinh c, /2 BC is the elliptic arc u2 cosh 2 c [When

/2

x

1, v

c,

x

x

2

) of the

/2

/2

0

/2 and c 0, cos x 0 and sinh c BC is the upper half of the ellipse

u2 cosh 2 c line segments

v2 sinh 2 c

1

v2 sinh 2 c , 2

1 CS2, S2D and DA of the z-plane are the ) and the lower half of the 2

2

segments of the u w Example 3.11 Show that the transformation w cos z maps the segment of the x x /2 into the segment 0 u 1 of the u maps the strip y 0, 0 x /2 into the fourth quadrant of the w The transformation equation of w cos z u iv x iy) cos x cosh y i sin x sinh y u cos x cosh y v sin x sinh y The image of the x y u cos x and v 0 Since we consider the segment of the x x /2, 0 u 1 Thus the image of the segment of y 0, 0 x /2, is the segment of v 0 u x 0, x /2 and y The image of x u cosh y and v u 1 and v The image of x u 0 and v sinh y Since y 0 for the given strip, v 0 Thus the image of x /2, y 0 is u 0, v 0 The image of y 0, 0 x /2 is v 0, 0 u v

0, 0

u

l;

v

0, u

0 w

and

v

0, u u

0, v

1 and u 0

0, v

0

0,

Chapter 3: Analytic Functions

II – 3.71 v

y

x = /2

x=0

u

O

x

O y=0

v'

Fig. 3.24

Example 3.12 under the transformation w sinh z The transformation equations of w sinh z i sin i x iy) i sin i ix y) sinh x cos y i cosh x sin y u sinh x cos y and v cosh x sin y The image of x u 0 and v sin y u 0 and 1

v

2

1

x u

/4 iv

x

2

/4

y

/4

iy)

y

/4)

The image of x u u2 sinh 2

y and v

v2 os 2

1, u

sin y

0, since cos

The image of y u

1 2

sinh x

1

v

and

2

cosh x 1 2

v2 u 2 y

v

0)

4

v2 u2 the v

v-axis, the ellipse and the rectangular y D

y = /4

v C B' D'

x

x=0

x

O A

O A'

B y = – /4

Fig. 3.25

C' u B'

Part II: Mathematics II

II – 3.72 Example 3.13

a 2 b2 4z

Show that the transformation w z

transforms the

1 a b in the z-plane into an ellipse of semi-axes a, b in the w 2 k2 In the discussion of the transformation w z , we have proved that the z image of the circle z c r c is the ellipse whose equation is circle z

u2

v2 2

k2 c

c

c k2

Putting c

k2 c

c

k2 c

The image of w

z

a2

k2 c

b2

a2

b2 4

1 a b 2

1

2

and c

1 a b 2

we get

a and

1 1 a b a b b 2 2 1 1 z a b r a b 2 2

/ z is the ellipse

4

1 a b 2

u2 a2

v2 b2

under the transformation

1, whose semi-axes are a and b

Example 3.14 Prove that the region outside the circle z 1 whole of the w-plane under the transformation w z z

1 maps onto the k2 , we can prove z

Proceeding as in the discussion of the transformation w z that the image of the circle z r 2 u 2 and that the image of the circle z u2 a

1 a

v2 2

a

1 a

2

1 1 a2

r

a

1/a and g a

a2 1 0 , when a a2

1 and

a and hence when a f g a) are increasing functions of a, when a The region outside the circle r r a, where 1 a

a is the ellipse

1

The semi axes of this ellipse are f a) Now f a

a,

1 a

a a)

1

1/a2

0, for all

Chapter 3: Analytic Functions

II – 3.73

u2

v2

1 a

a

2

a

1 a

1, 1 a

2

, is

the entire w r

1 maps onto the entire w 1 ze 2

Example 3.15 Show that the transformation w maps the upper half of the interior of the circle z w

iv

and

1 rei e 2

1 e re

is real,

1 ze 2

1 ze

are

i

u

1 re 2

1 re

cos

v

1 re 2

1 re

sin

0,

and r

u

1 re 2

1 re

u

1 2

z z-plane consists of The image of

, where

e onto the lower half of the

The transformation equations of the mapping function w

u

e z

e or r e

e

and v 2

u

1 0 2

r e

2 and v

1

/2

/2

r e 0 or u

2 and v

0

1 and v u

1 and v

The image of r e u cos and v 0, 1 u 1 and v r

e

z-plane

is the entire u r When r

e , re

1 1 re

1

e and 0

Part II: Mathematics II

II – 3.74

1 re

re 1 re 2

1 re

0

0 , since sin is positive when 0

sin

v z|

e

z-plane

maps onto the lower half of the w v

y |z| = e

O

u

x

O

Fig. 3.26

EXERCISE 3 c) Part A

w z is w

w

f z)

e z

1 ze 2

w2 mation w w w

z

z z

) z ). 4 under the transfor-

sin z

2

i

i) z

z 2z

a under the following transformations w 4i) z i w 2z

w i

x transformation w

iz

i.

2 under the transformation w iz y-axis under the

Chapter 3: Analytic Functions y

w

II – 3.75

1 under the transformation w i)z xy 1 under the transformation w z x2 y2 a2 under the transformation

i)z y

w

1 z

w

1/z

c, when c

x

c, when c

0 under the transformation

0 under the transformation

i z in the z-plane under the transformation

y

mx under the transformation w

zn, where n

w

x

1 and y z

1 under the transformation w iz2 w z2

arg z under the transformation w z2 4 2 u and v under the transformation w iz2 x 1 under the transformation w x-axis under the transformation w cos z y-axis under the transformation w cos z x-axis under the transformation w sinh z y-axis under the transformation w sinh z w

a z 2

ez

1 z )

under the transformation w

z

1 z

Part B i) w i

z

4

w

az

2i

z i, z 1 i, z 1 i under w iz 2 i z 1 i i, 1 i and

b, where a

2 1 i and b

z|

i

a maps onto the interior of an ellipse ib y in the w-plane under the transformation w x , 0 b a. Is the a w w z-

16

16

1 maps the circle z z

Part II: Mathematics II

II – 3.76 i y

1, under the transformation w

1 2c

y

0

x, y z

Note

w

0 and

0, 1, 1

i

y

a

z-plane z is to

z maps the region of the z-plane a a/2 onto the upper half of the w x , y 0 under the transformation

a/2

x

sin

cos z /4

transformation w

cosh z

from 0 to i onto the segment 0 2 strip x 0, 0 y /2 /2 under the transformation w w z a of the u upper w 1 ze 2

e z

x

y

/4, 0

sin z w

w

1

The properties of the transformation w = log z are identical with ez, if we interchange z and w w cos z w sinh z w

y

x and y

z2

and i under the transformation w z2 a 0, show that w z/a) onto the upper half of the w w log z maps onto the horizontal strip v

those of w

0,

1 z y

under the transformation w

x

, where

u

i) the segment of the y-axis

1 of the u w

z sinh z a z 2

x 1 z

1 in the upper z

2

/2

y

where a is a positive constant a u z-plane onto the

is real, maps the interior of the circle z

1

onto the exterior of an ellipse whose major and minor axes are of lengths 2 cosh and 2 sinh

Chapter 3: Analytic Functions Hint: The image of the circle z

II – 3.77

1 is the ellipse

u2 cosh 2

z

v2 sinh 2

1

a, ), where a 1 is a

point outside the ellipse, since u a, )

u

) and v a, )

v , ) .

3.7 BILINEAR AND SCHWARZ CHRISTOFFEL TRANSFORMATIONS 3.7.1

Bilinear Transformation

az b , where a, b, c, d are complex constants such that cz d ad bc 0 is called a bilinear transformation. It is also called Mobius or linear fractional transformation. The transformation w

Now d w dz

ad bc cz d 2

ad

bc

z-plane becomes a critical

The transformation can be rewritten as w

a c

ad bc . c cz d

the transformation takes the form w

ad

bc

0,

a , which has no meaning as a mapping c ad bc ad bc)

is called the determinant The inverse of the transformation w

az b is z cz d

dw b , which is also a cw a

The images of all points in the z z

d c

for the point w

w-plane is a unique point, except a If we assume that the images of the points z c

d and w c

a c

w- and z

To discuss the transformation w

az b cz d

we express it as the combination of simple transformations discussed in the When c

Part II: Mathematics II

II – 3.78

a c

w

bc ad c

1 cz d

If we make the substitutions w1 cz d 1 w2 w1 then

w

Aw2 B

where

A

bc ad c

and

B

a c

z-plane onto the w1-plane, from the w1-plane onto the w2-plane and from the w2-plane onto the w

c

w

a z d

b d

d 0

w Az B.

If the image of a point z under a transformation w called a point or an invariant point w f z) w

f z) is itself, then the point is z f z). az b cz d

az b z cz d

As this is a quadratic equation in z,

Note w1, w2, w of z1, z2, z of the z Let the bilinear transformation required be az b cz d

w

w

a b z d d c z 1 d

or

Az B Cz 1

Chapter 3: Analytic Functions

II – 3.79

Since the images of z1, z2 and z are w1, w2 and w

and

w1

Az1 B C z1 1

w2

Az2 B C z2 1

w

Az Cz

B 1 A, B, C

Solving them we get the values of A, B, C

given,

If z1, z2, z , z4 are four points in the z-plane, then

z1 z2 z

z

z1 z4 z

z

is called the

cross- ratio of these points. Cross-ratio property of a bilinear transformation The cross-ratio of four points is w1, w2, w , w4 are the images of z1, z2, z , z4 transformation, then w1 w2 w

w

z1 z2 z

z

w1 w4 w

w

z1 z4 z

z

a bilinear

Proof az b Let the bilinear transformation be w cz d Then

wi w j

w1 w2 w

and

w

a zi b c zi d

azj b

ad bc zi

zj

czj

c zi d c z j

d

ad bc

d 2

z1 z2 z

c z1 d c z2 d c z ad bc

2

w1 w4 w

w

w1 w2 w

w

z1 z2 z

z

w1 w4 w

w

z1 z4 z

z

d c z4 d

z1 z4 z

c z1 d c z2 d c z

z

z

d c z4 d

Part II: Mathematics II

II – 3.80

Note To get the bilinear transformation that maps z2, z , z4 of the z-plane onto w2, w , w4, we assume that the image of z under this transformation is w and make use of the invariance of the cross-ratio of the four points z, z2, z , z4 w w

w

w4

z z

z

z4

w w

w

w2

z z

z

z2

z

in the form w

3.8

z 2, z , z 4 w w2, w , w4 Now, w, we get the required bilinear transformation

az b cz d

SCHWARZ-CHRISTOFFEL TRANSFORMATION

w-plane onto x-axis) z-plane is called Schwarz-

the x

Christoffel transformation. x1, x2 xn that are points on the x-axis such that x1 xn, are the images of the vertices w1, w2 wn and are n dw A z x1 dz

1

2

1

z x2

1

n

z xn

1

x2 x w-plane

,

where A Proof

v

4

y

A w)

P w) 2

A2 w2)

1

A1 w1)

A4 w4)

–

1

u

O

P'

A1'

A2'

A'

A4'

z

x1

x2

x

x4

0

Fig. 3.27

The given transformation is 1

dw

A z x1

1

2

z x2

1

n

z xn

1

dz

x

Chapter 3: Analytic Functions

am

n

d w)

1

amp A

2

z x1 )

1 am

II – 3.81

1 amp z x2

…+

1 amp z xn

z1z2 ...) z1) z2 ) .... zk) k z)] Let the image of P w) be z) When P moves towards A1, moves towards 1. As long as z) is on the left of x1), z x1, z x2 z xn 1 z x1) z x2) z xn) But once z) has crossed 1 x1), x1 z x2, is a positive real 1 z x1) z x2) z x) amp z xn) A z [ A is a constant and dz z) 0] Thus when z croses 1, z x1) to 0 or undergoes . [

Increase in amp

1

w

1

)

1

P w), moving along PA1, reaches A1, it changes its direction through an angle in the anticlockwise sense and then starts moving along A1A2 P w) reaches A2, it turns through an angle and then starts moving along A A 2 2 Pw z) moves along the x the w-plane in the anticlockwise sense, the interior of the z-plane should lie to the left of the person, when he or she walks along the corresponding path in the z x corresponding area is the upper half of the z w-plane is mapped onto the upper half of the z

Note 1

as w A z x1

1

2

z x1

1

n

z xn

1

dz B

where B z-plane onto the real axis of the w

z and w x1, x2,

xn can be chosen

Part II: Mathematics II

II – 3.82

xn, If we take A

n

xn

, where

is a constant, the transformation can be

1

written as n

dw dz

1

z x1

As xn dw dz

1

2

z x2

n 1

1

z xn

xn z xn

1

1

1

, the transformation reduces to 1

z x1

1

2

z x2

n 1

1

z xn

1

1 n

This means that, if xn

z xn

1

is absent in the

WORKED EXAMPLE 3 d) Example 3.1

w

2 z 4i Prove iz 1

z and its image w, form a set of four

2 z 4i iz 1

z iz2 z

z 4i i z i)

0 0

iz

4

i, the cross-ratio of the four points z1, z2, z and z4 z z1 , z2 , z , z

0

i. Taking z1 z , z2 w

The invariant points are 4i and z4

z2

or

2 z 4i iz 1

z i 4i i i z2 z i

4i i 2 z 4i iz 1

z 4i 2 z 8i

2 z 4i ,z iz 1

4i and

Chapter 3: Analytic Functions i i z2

z 4i

2

z 4i

i iz

II – 3.83

a constant, independent of z Example 3.2 the z-plane into the points 0, 1, i of the w Taking z1 1 i, z2 i, z i and w1 z, z1, z2, z ), we have w 0 1 i w i 1 0

0, w2

1, w

i

z 2 i 1 2i 1 i z 1 i 2 i z i 2 z 1 i

i w i w

i z i 2 z 2 2i i z i

1

2 z 2 2i w

w

Example 3.3 the z into the points w

i, w

w, w1, w2, w )

i

i z

i

2 z 2 2i z i i z 0, z 1 and z

of the w1 and w i z, z1, z2, z )

w w1 w2

w

z z1 z2

z

w w

w1

z z

z1

w

To avoid the substitution of w then put

2 z 2 2i i

z

w

i of

i and using the

z 1 i 2 z 2 i 1 2i

w i w 1

i,

1 w

Part II: Mathematics II

II – 3.84

w w1 w2 w ww Using the given values z1

1 w

1 w1

i, z2

z w 1 11

w i w i

z

z z

z

z1

1, w1

0, w2

1 and

z i 2 ,i z 1 1 i

w w1 w2

w

z z1 z2

z

w w

w1

z z

z1

Using the values z1 0, z2 w i 1 i w i 1 i

z z1 z2

z

w

1, z’

z

z z1 z z

1

zz

z1

0 and w1

1 z i, w2

w

0, we get 1 i z i z 1

, where z

1 and w

1 z

i, we get

1 1 1

1 i z 1 i

2w 2i

1 i z 1 i 1 i 1 i z

w

1 i z i 1 1 i 1 i z

Nr Dr Dr Nr w=

or

z i iz 1

Example 3.4 If a, b w a w b 1 w a

1 z a

k

z a , where k is a constant, if a z b

c , where c is a constant, if a a and b are a and b b

z a z

b

w b w

a

z b z

a

w a w b

z

b w

a

z

a w

b

b w, a, w , b)

w a w

z a z b

b;

z, a, z , b)

Chapter 3: Analytic Functions w a w b

k

z

a is w

a

cz

d)

c z a

z a , where k z b a, the bilinear transformation can assumed as

z a cz d

w a

z

II – 3.85

1]

0 d 1 c

z

0 a,

d 1 c

a

d

ca 1 z a cz ca 1

w a c z a 1 z a

1 w a 1

1

w a

z a

c, where c

Example 3.5 Show that the transformation w

z 1 maps the unit circle in the z 1

z

w The image of the unit circle w z 1 z 1 x

1) 2

z

1,

iy 2

x

x 1) y x 2x 2x or x 2x

2x or x

z

1)

iy

2

y2

1)

1

w 0, which is the right half of the z w

z

1

z

1

1 is the left half of the

z Example 3.6 Show that the transformation w circle z 1 onto the lower half of the w onto the interior of the circle w

z i 1 iz

i) the interior of the z-plane

Part II: Mathematics II

II – 3.86

z i 1 iz

w w z

iwz

z

i

iw)

w

i

w i 1 iw

z

z z z w i 1 iw

w

1 u

i v

2

u v

i < 1 v)

iu

2

2

1)
R, while it may or may not converge for |z – a| = R. This means that the power series converges at all points inside the circle |z – a| = R, diverges at all points outside the circle and may or may not converge on the circle. Due to this interpretation. R is called the radius of convergence of the above series and the circle |z – a| = R is called the circle of convergence. Power series play an important role in complex analysis, since they represent analytic functions and conversely every analytic function has a power series representation, called Taylor series that are similar to Taylor series in real calculus. Laurent’s series, which consist of positive and negative integral powers of the independent variable. They are useful for evaluating complex and real integrals, as will be seen later.

4.3.2 Taylor’s Series (Taylor’s Theorem) If f (z) is analytic inside a circle C0 with centre at ‘a’ and radius r0, then at each point z inside C0, f z

f a

f

a z a 1!

f

a z a 2!

2

f

a 3!

z a

3

Proof Let C1 be any circle with centre at a and radius r1 r0, containing the point z (Fig. 4.15). Let w be any point on C1. Then, by Cauchy’s integral formula,

Chapter 4: Complex Integration

f z

f w 1 dw 2 iC w z

(1)

f w n! dw 2 iC w a n 1 1

(2)

II – 4.21 C0 C1

1

f

and

n

a 1

Now

z a

1

w z

w a 1 w a 1

z a

Fig. 4.15

z a w a

1

w a

1

z a w a

1

z a w a

n 1

z a w a

n

n

i.e.

z a

1

1

w z

w a

w a

f w 1 dw 2 iC w z 1

2

2

w a

3

z a

n 1

w a

f w 1 dw 2 iC w a

z a

1

Rn

z a 1 2 iC w a

n

f z

f a

f a 1!

z a f

n 1

f

|z – a| = r and |w – a| = r1.

a 2!

a

n 1! Let

n

n 1

n

z a w a 2

1 w z

dw

f w 1 2 iC w a 1

n

dw Rn ,

f w dz w z

.

1

i.e.

1

f w 1 2 iC w a 1

z a

where

1

n 1

1

z a

1

and so

2

1

1 z a w a

n

1 1

n 1

2

1

2

z a w a

z a

z a

n 1

2

...

Rn ' by 1 and 2

(3)

Part II: Mathematics II

II – 4.22

|w – z| = | (w – a) – (z – a)| r1 – r Let M be the maximum value of |f (w)| on C1 rn 2

Rn

Then

2

0

r1M r r1 r r1

i.e. r r1

Since

1, lim n

M r1d r1 r r1n n

r r1

n

0

Rn

Taking limits on both sides of (3) as n f z

f a

r1ei

w

f

0 and so Rn

0 as n

.

, we get

a z a 1!

f

a z a 2!

2

(4)

Note

For any point z inside C0 C1. So the Taylor series representation of f (z) is valid for any z inside C0. The largest circle with centre at ‘a’ such that f (z) is analytic at every point inside it is the circle of convergence of the Taylor’s series and its radius is the radius of convergence of the Taylor’s series. Clearly the radius of convergence is the distance between ‘a’ and the nearest singularity of f (z). 2. Putting a = 0 in the Taylor’s series, we get f z

f 0

f 0 z 1!

f

0 2 z 2!

(5)

This series is called the Maclaurin’s series of f (z). 3. The Maclaurin’s series of some elementary functions, which can be derived by using (5), are given below: (i) e z

1

z 1!

z2 2!

z3 ..., when z 3!

.

z

z3 3!

z5 ..., when z 5!

.

(iii) cos z 1

z2 2!

z4 ..., when z 4!

.

(ii) sin z

z

z3 3!

z5 ... , when z 5!

(v) cosh z 1

z2 2!

z4 ..., when z 4!

(iv) sinh z

.

.

Chapter 4: Complex Integration

II – 4.23

(vi) (1 – z)–1 = 1+ z + z2 + z3 +…, when |z (vii) (1 – z)–2 = 1+ 2z + 3z2 + 4z3 +…, when |z

4.3.3

Laurent’s Series (Laurent’s Theorem)

If C1 and C2 are two concentric circles with centre at ‘a’ and radii r1 and r2 (r1>r2) and if f(z) is analytic on C1 and C2 and throughout the annular region R between them, then at each point z in R, f z

an z a

n

n

bn z a

n 0

,

n 1

where

an

f w 1 2 iC w a n 1

and

bn

f w 1 2 iC w a n 2

1

dw, n

1

0,1, 2,...

dw, n 1, 2,...

Proof By Extension of Cauchy’s integral formula to a doubly connected region, f w f w 1 1 dw dw Fig. 4.16 2 iC w z 2 iC w z

f z

1

2

C1

= I1+I2, say In I1 ,

1 w z

C2

1 w a 1

z

z a 1

w a

(1)

1

z a w a

1

z a

w a

w a

2

a

z a

2

w a

3

Fig. 4.16

z a

n 1

w a

since |z – a| < |w –a| when w is on C1 In I2,

1 w z

1 z a 1 z a

w a 1

w a z a

1

n

z a

n

w a

n

1 w z

2

Part II: Mathematics II

II – 4.24

n 1

w a

w a

1

z a z a2 z a since |w – a| < |z – a|, when w is on C2. Using the expansion (2) in I1, we have f w 1 dw 2 iC w a

I1

1

n

w a

n

z a

f w z a 2 i C w a 1

2

n

dw n 1

z a 2 i a0

z a 2 i

Rn

where

a1 z a

an

n

z a

1

(3)

,

z w

n 1

f w C1

w a

n

dw Rn

Rn ,

f w dw w a

C1

n

(4)

w z

Using the expansion (3) in I2, we have f w w a 1 1 dw f w dw 2 iC z a 2 iC z a 2 2 2

I2

n 1

w a f w 1 dw Sn . n 2 iC z a 2 b1 z a

2

b2 z a

bn z a

n

Sn

n

1

Sn

where

1

2 i z a

n C2

w a f w dw z w

(5)

If |z – a| = r, then r2 < r < r1. 0 as n , as in Taylor’s theorem. Rn Let M be the maximum value of |f(w)| on C2. Then i.e

| Sn |

2 r

= M r2 r2 r r2 r | Sn |

2

1 n

0

r2n n r2 d r r2

on C2 , w

r2 ei

n

0 and so Sn

0 as n

, since r2

Using (4) and (5) in (1) and taking limits as

r. , we have f z

an z a n 0

bn z a n 1

n

n

Chapter 4: Complex Integration

II – 4.25

Note 1. If f (z) is analytic inside C2, then the Laurent’s series reduces to the Taylor series of f (z powers in Laurent’s are zero. they are not usually found by the theorems given above, but by other simpler methods such as use of binomial series. n

an z a , consisting of positive integral powers of (z – a), is

3. The part n 0

called the analytic part of the Laurent’s series, while

bn z a

n

,

n 1

consisting of negative integral powers of (z – a) is called the principal part of the Laurent’s series.

4.4

CLASSIFICATION OF SINGULARITIES

We have stated, in Chapter 3, that a point at which f (z) is not analytic is called a singular point or singularity of f (z). We now consider various types of singularities.

1. Isolated Singularity The point z = a is called an isolated singularity of f (z), if there is no other singularity in its neighbourhood. In other words, z = a is called an isolated singularity of f (z), if such that the circle |z – a| = encloses no singularity other than a , then ‘a’ is called a non-isolated singularity. [If ‘a’ 0 such that encloses no singularity, then ‘a’ is called a regular point or ordinary point of f (z).]

Note singlarities are necessarily isolated. For example, z = 1 is an isolated singularity of f z

1 z 1

z 1

± i are isolated singularities of f z z

3

z2 1

2

and z = 0 and

.

2. If ‘a’ is an isolated singularity of f (z), then f (z) can be expanded in Laurent’s series valid throughout some neighbouhood of z = a (except at z = a itself), i.e., valid in 1. Here r2 is chosen arbitrarily small.

2. Pole If z = a is an isolated singularity of f (z) such that the principal part of the Laurent’s expansion of f (z) at z = a valid in 1 then z = a is called a pole.

Part II: Mathematics II

II – 4.26

f z

i.e. if in

an z a n 0

bm

n

n

bn z a

,

n 1

0, bm+1=0 = bm+2...., then z = a is called a pole of order m. simple pole. 1 , as the Laurent’s expansion. For example, z = 0 is a simple pole of f z 2 z z 1 1 1 z z

of f (z) valid in 0 < |z| < 1 is given by f z i.e.

2

1 1 2 z 3z 2 z

f z

1 z

2 3z 4 z 2

It has only one term i.e. 1/z in the principal part. Similarly, z =1 is a pole of order 2 of f (z), as the Laurent’s expansion of f (z), valid in 0 < |z – 1

f z

z 1

2

1 z 1

z 1

1

z 1

z 1

1 Here b2

2

1

z 1

1

z 1

2

z 1

3

1 z 1

2

1 2

z 1

0 and b3 = b4= ...= 0.

3. Essential Singularity If z = a is an isolated singularity of f (z) such that principal part of the Laurent’s expansion of f (z) at z = a, valid in 0 < |z – a| < r1 then z= a is called an essential singularity. For example, z = 1 is an essential singularity of f(z) = e expansion is given by f z

1

1 1 1 1 1! z 1 2 ! z 1 2

z – 1)

, as the Laurent’s

1 3! z 1

3

.

4. Removable Singularity If a single-valued function f (z

z = a, but lim f z z

a

exists, then z

= a is called a removable singularity. For example, z=0 is a removable singularity of

Chapter 4: Complex Integration f z

sin z , as f z

lim

sin z z

z3 3!

z5 5!

z

0

II – 4.27

1. The Laurent’s expansion of

f (z) is given by sin z z

1 z z 1

z2 3!

z4 5!

Note 1. If f (z) has a pole of order m at z = a, then its Laurent’s expansion is bm b1 b2 n f z . an z a , m 2 z a z a z a n 0 where bm

0. 1 z a

an z a

m

n m

bm

bm

1

z a

n 0

b1 z a

m 1

1

z ,say. m z a Clearly, (z) is analytic everywhere that includes z = a and (a) = bm 0. z Thus for a function of the form , z a is a pole of order m, provided that m z a (z) is analytic everywhere and (a) 0. f (z z-plane is called an entire function or integral function ez, sin z, cosh z are examples of entire functions. f (z meromorphic function. For example, f z

1

is a mermorphic function, as it has only two poles 2 z z 1 —a simple pole at z = 0 and a double pole at z =1.

4.4.1

Residues and Evaluation of Residues

If ‘a’ is an isolated singularity of any type for the function f (z 1 of (viz. b1) in the Laurent’s expansion of f (z) at z = a valid in 0 < |z – a| < r1 is z a called the residue of f (z) at z = a.

Part II: Mathematics II

II – 4.28

C1 C z r2

a

C2

Fig. 4.17

1 2 iC

We know, from Laurent’s theorem, that b1

f (w)dw, where C2 is any 2

circle |z – a| = r2< r1, described in the anticlockwise sense. [Fig. 4.17] Now if C is any closed curve around ‘a’ such that f (z) is analytic on and inside C except at ‘a’ itself, then by Extension of Cauchy’s integral theorem, f w dw C2

f w dw. C

1 f z dz , 2 iC where C is any closed curve around ‘a’ such that f (z) is analytic on and inside it except at z = a itself. Hence the residue of f (z) at z = a is also given by [Res. f (z)]z=a

4.4.2

Formulas for the Evaluation of Residues

1. If z = a is a simple pole of f (z), then Res. f z

lim

z a

z

a

z a f z .

Since z = a is a simple pole of f (z), then the Laurent’s expansion of f (z) is of the following form:

f z

an z a

n

z a

n 0

z a f z

b1

an z a

n 1

b1

n 0

lim

z

a

z a f z

b1

Res. f z

z a

.

Chapter 4: Complex Integration P z then [Ref.f (z)]z=a Q z

2. If z = a is a simple pole of f z P z Q z

lim

z

a

II – 4.29

..

By the previous formula,

[Res. f z ]z

z a P z

lim

a

z

Q z

a

z a P z Q z

lim

z

=

0 form 0

a

P a Q a

or

, by L’Hospital’s rule

P z Q z

lim

z

P z

a

3. If z = a is a pole of order m of f (z), then 1

Res. f z

z a

lim

m 1 !z

a

dm dz

1

m 1

z a

m

f z

Since z = a is a pole of order m of f (z), then the Laurent’s expansion of f (z) is of the following form: f z

an z a

b1

n

z a

n 0 m

z a

f z

an z a

bm

b2

n m

z a

2

z a

b1 z a

1 m 1

m 1

b2 z a

z a

m

bm z a

m

m 2

n 0

bm

z a

1

lim

z

b1 where

a

dm dz

bm . 1

m 1

Res. f z D

z a

z a

m

f z

m 1 !b1

1 , lim D m m 1! z a

1

f z

d . dz

Note The residue at an essential singularity of f (z) is found out using the Laurent’s expansion of f (z) directly.

Part II: Mathematics II

II – 4.30

4.4.3

Cauchy’s Residue Theorem

If f (z) is analytic on and inside a simple closed curve C singularities a1, a2,…, an lying inside C and if R1, R2,…, Rn are the residues of f (z) at these singularities respectively, then f z dz

2 i R1

R2

Rn .

C

Proof We enclose the singularities ai by simple closed curves Ci, which are not overlapping. f (z) is analytic in the multiply connected region bounded by the outer curve C and the inner curves C1, C2,… Cn. (Fig. 4.18) By Extension of Cauchy’s Integral theorem, we have f z dz C

f z dz C1

f z dz

f z dz

C2

2 i R1

R2

Cn

Rn , by the property of residue at a point.

a2

C’2

a1

Cn

C’1

an C Fig. 4.18

Note

Cauchy’s integral formulas for f (a) and f(n) (a) can be deduced from Cauchy’s residue theorem. Consider, C

f z dz , where f (z) is analytic on and inside C enclosing the point ‘a’. z a

f z Now has a simple pole at z = a with residue z a By Cauchy’s residue theorem, C

f z z a

n 1

f z dz z a

lim

z

a

2 if a .

has a ploe of order (n + 1) at z = a

z a

f z z a

f a .

Chapter 4: Complex Integration 1 lim D n n! z a

with residue

1 f n! f z C

z a

n

n

n 1

z a

a

dz 1

n

f

2 i

a

n!

f z n! 2 iC z a n

fn a

or

f z

n 1

z a

II – 4.31

1

dz.

WORKED EXAMPLE 4(b)

Example 4.1 Expand each of the following functions in Taylor’s series about the indicated point and also determine the region of convergence in each case: (i) e2z about z = 2i; (ii) cos z about 2; z 3 z 1 z 4

(iii)

at z = 2;

(iv) log

1 z at z = 0. 1 z

e2z = e2(z –2i) + 4i = e4i. e2(z – 2i) f (z) = e2(z – 2i). Then f (n) (z) = 2n e2(z – 2i) f (n) (2i) = 2n; n = 1, 2,… . f (2i) = 1. Taylor's series of f (z) at z = 2i is given by (i)

Let

f z

e

f

f 2i

2z

e

4i

2i z 2i 1!

2 z 2i 1 1!

f

2i z 2i 2!

22 z 2i 2!

2

2

23 z 2i 3!

3

The series converges in a circle whose centre is 2i and radius is the distance between 2i and the nearest singularity of f (z) which is . The region of convergence of the Taylor’s series in this case is |z – 2i| < (ii) Let f z cos z; f z sin z; f z cos z; f f f

z

sin z; f iv z

cos z , etc.

2

0 f

1 f

1 f

2

Taylor’s series of f (z) at z = f z

f

2

2 iv

2

2

0

0 etc.

2 is given by

f

2 1!

z

2

f

2 2!

z

2

2

Part II: Mathematics II

II – 4.32 z

cos z

z

2

2

1!

3

z

2!

The region of convergence is

2

5

z

2

5!

7

7!

2

z 3 z 1 z 4 The Taylor’s series of f (z) at z = 2 is a series of powers of (z – 2). Putting z – 2 = u or z = u + 2, we have (iii)

f z

u 5 u 1 u 2

f z

4 3 u 1

7 3 , u 2 on resolving into partial fractions

4 1 u 3

1

7 1 u 2 6

1

4 3n

n

1 un 0

u2

7 u 1 6 2

4 1 u u 2 u3 3 u

7 6n

0

22

n

2n

4 7 1 n 1 n 1 . n z 2 3 6 2 n 0 The Taylor’s series expansion was obtained using binomial series, which are convergent in |u| < 1 and |u/2| < 1, i.e., in |u| < 1. The region of convergence of the Taylor’s series of f (z) is |z – 2| < 1.

Note

The singularities of f (z) are at z = 1 and z = 4 which lie outside the region of convergence.

1 z = log (1 + z) – log (1 – z) 1 z Log (1 ± z) are many valued functions. We consider only those branches (values) which take the value zero when z = 0. Now consider f (z) = log (1 + z) (iv)

log

f

z

f 0

1 z 0; f 0

1

;f

z

1; f

1 z 1; f

0

2

;f 0

z

1

2 1 z

3

etc.

2 !, etc.

Taylor’s series of f (z) at z = 0 [or Maclaurin’s series of f (z)] is f z

log 1 z

f 0

z

z2 2

f 0 z 1! z3 3

z4 4

f

0 2 z 2! (1)

Chapter 4: Complex Integration

II – 4.33

The only singularity of log (1 + z) is – 1 and its distance from the origin, i.e. the radius of convergence of series (1) is 1. The region of convergence of series (1) is given by |z| < 1. Changing z into – z in (1), the Taylor series of log (1 – z) at z = 0 is given by log (1 – z) = – z – z2/2 – z3/3 – z4/4 – … The region of convergence being |z| < 1 Using (1) and (2), we get log 1 z 1 z

2 z

z3 3

z5 5

, which converges for |z| < 1.

Example 4.2 Expand each of the following functions in Laurent’s series about z = 0. Identify the type of singularity also. (i) ze

z2

(iii) z–1 e–2z

(ii) (1 – cosz)/z 1 (v) (z – 1) sin z (i)

ze

z2

z 1

z2 1!

z4 2!

(iv)

1 z

3

ez

2

z6 3!

z3 z5 z7 (1) 1! 2 ! 3! The Laurent’s series (1) does not contain negative powers of z and the circle of convergence does not include any singularity z = 0 is an ordinary point z

of ze

z2

(ii)

1 cos z z

1 1 z z 2!

1 z3 4!

z2 2!

z4 4!

z6 . 6!

z5 6!

(2)

1 cos z , the Laurent’s series of 1 cos z z z at z = 0 does not contain negative powers of z.

Though z = 0 appears to be a singularity of

z = 0 is a removable singularity of

Note

(iii)

The value of

z 1e

2z

1 cos z . z

1 cos z at z = z

1 2z 1 1! z 1 22 z 2 z 2!

22 z 2 2! 23 z 2 3!

lim z

0

1 cos z z

0.

23 z 3 3! (3)

Part II: Mathematics II

II – 4.34

The principal part in the Laurent’s series (3) contains the only term

1 . z

z = 0 is a simple pole of z–1 e–2z. 1

(iv)

z

3

ez

1

2

z

3

z2 1!

1

1

z 2!

1 z

z3

z4 2!

z6 3!

z3 3!

(4)

The last non-vanishing term with negative powers of z in the principal part of (4) is 1 . z3 z = 0 is pole of order 3 for the given function. 1 1 1 1 1 1 z 1 sin z 1 (v) z z 3! z 3 5! z 5 1 1 1 1 1 1 1 1 z 3! z 2 3! z 3 5! z 4

(5)

z = 0 is an essential singularity. Example 4.3 Find the Laurent’s series of f z |z +

< |z +

1 valid in the region (i) z1 z

z + 1| > 2.

1 ; Since Laurent’s series in powers of (z + 1) are required, z1 z put z + 1 = u or z = u – 1. f z

1 1 1 (1) u 1 2 u u 1 2 u (i) Since the region of convergence of required Laurent’s series is |u| < 1, the two terms in the R.H.S. of (1) should be rewritten as standard binomials whose u f z

f z

1 1 u

1

un n 0

1 n 0

1 u

u 2 1 2 1 2n 1 2

n 1

1

1 u 1 2 2

un 0

2n z 1

n

(2)

The Laurent’s expansion (2) is valid, if |u| < 1 and |u| < 2, i.e. |z + 1| < 1.

Chapter 4: Complex Integration

II – 4.35

(ii) Since the region of convergence of the required Laurent’s series is given by |u| >1 and |u a standard binomial in such a way that u occurs in the denominator of the second member and the second term as a standard binomial in such a way that u occurs in the numerator of the second member. f ( z)

1

1

1 u 1 u

u 2

2 1

1 1 1 u u

1 un 2 n 0 2n

1 1 u n 0 un 1 z 1

n 0

1

1 u 1 2 2

1

1 n 1

2

n 0

n 1

z 1

n

(3)

The Laurent’s expansion (3) is valid, if 1 u

1 and

u 2

1 i.e., u

1 and u

2,

z (iii) Since the region of convergence of the required Laurent’s series is |u| > 2, both the terms in the R.H.S. of (1) should be rewritten as standard binomials in which u f ( z)

1

1

1 u 1 u 1 1 1 u u 1 1 u n 0 un

1

1 2 1 u u

1

1 2n u n 0 un 1

(1 2n ) n 0

2 u

u 1

z 1

(4)

n 1

The Laurent’s expansion (4) is valid, if 1 u

1and

2 u

1, i.e., u

1 and u

2 i.e., z 1

2.

Part II: Mathematics II

II – 4.36

Example 4.4 Find all possible Laurent’s expansions of the function

f z

4 3z about z =0. Indicate the region of convergence in each case. Find z1 z 2 z also the residue of f (z) at z = 0, using the appropriate Laurent’s series. f z

4 3z z1 z 2 z

2 1 z 1 z

1 2 z

,

on resolving into partial fractions. Various Laurent’s series about z = 0 will be series 2 z unaltered and rewriting the other two terms as standard binomials in three different ways and expanding in powers of z. of ascending and descending powers of z. They can be obtained by keeping

Case (i) f z

2 z

1 z

2 z

zn

1 z 1 2 2

1

1 zn 2 n 0 2n

n 0

2 z

1

1

1

2

n 0

(1)

z n if z

n 1

1and z

2, i.e. z

1.

z = 0, which is a singularity of f (z). Laurent’s series expansion But |z is valid in an annular region which does not contain any singularity of the function expanded. z

Case (ii) f z

2 z

1 1 1 z z

2 z

1 zn

2 z if

1 z

1 and z

z 2

1, i.e. z

n 0

1 and z

1

1 1 z 2 2

1 n 0 z

1 zn 2 n 0 2n

1

zn n 1 02

zn 2

1 n

1

(2)

Chapter 4: Complex Integration

II – 4.37

Case (iii) 2 z

f z

2 z 2 z

1

1 1 1 z z 1 zn

n 0

1 n

1

1 2 1 z z 2n n 1 0 z 1

1 2n

z

n 0

(3)

n 1

1 2 1 and 1, i.e. z 1 and z 2 z z i.e. |z z Since z = 0 is an isolated singularity of f (z), the residue at z 1 of in the Laurent’s expansion of f (z z z (1) above. if

Thus

[Res. f (z)]z = 0 = 2. z2

Example 4.5 Find the residues of f z

z 1 z 2

at its isolated singularities

2

using Laurent’s series expansions. f z

z2 z 1 z 2

2

1 9 z 1

8 9 z 2

4 3 , on resolving into partial fractions. 2 z 2

Both z = 1 and z = – 2 are isolated singularities of f (z). f (z) at z = 1, we have to expand f (z) in series of powers of z

(z Thus

f z

1 in it. z 1

r 1 9 8 9 z 1 3 z 1

4 3 {3 z 1 }2

1 9 8 z 1 1 z 1 27 3

1

z 1 4 1 27 3 n

2

z 1 4 z 1 n 1 n 1 n 27 3 3n 0 n 0 The expansion (1) is valid in the region z 1 1, i.e. 0 z 1 3. 3 1 9 8 z 1 27 n

1

n

1

[Residue of f (z) at z 1 . 9

z 1

in (1)

n

(1)

Part II: Mathematics II

II – 4.38

f (z) at z = –2, we have to expand f (z) in series of powers z

of (z Thus

f ( z)

1 in it. z 2

r 1 9 z 2 3

8 9 z 2

z 2 3n

0

z 2

1

1 z 2 1 27 3 1 27 n

4 3 2

8 9 z 2

n

The expansion (2) is valid in the region z 2 1, i.e.0 3

4 3 z 2

8 9 z 2

z 2

z 2

3.

2

4 3

(2)

2

1 in (2) z 2

[Residue of f (z) at z 8 . 9 Example 4.6 Find the singularities of f z

z3

z2 4 and the corresponding 2z2 2z

residues. The singularities of f (z) are given by z3 z z

i.e

2z2 2

2z

2z 2

0, 0.i.e, z

0, 1 i

For these values of z, the numerator (z2 everywhere. z 0, 1 i are simple poles of f z . Res. f z

z 0

Res. f z

z

z f z

1 i

z2 z 0

z 1 i f z

z

2

z

4 2z 2

1 i

2. z2 4 z z 1 i

1 i

z

2 i 1 1 3i 1 i 2 Since z = –1– i is the conjugate of z = –1+ i, Res. f z

z

1 i

1 1 3i 2

Example 4.7 Identify the singularities of f z

z2 2

z 2 z2 9 residue at each singularity. The singularities of f (z) are given by

Chapter 4: Complex Integration z 2

2

z2

0, i.e. z

9

II – 4.39

2, i3.

Of these z = 2 is a double pole and z = ± i3 are simple poles. Res. f z

1 d z 2 1! dz

z 2

2

d z2 2 dz z 9 Res. f z

f z 18 z

z i3

z 2

z 2

2

36 169

2

z i3

z i3

3i 2 5 12i

9 i 6 i3 2

9

2

z

z i3 f z

z i3

z

z 2

2

2

3 12 5i 338 Res. f z

conjugate of

z i3

3 12 5i 338

3 12 5i 338 z2

Example 4.8 Find the singularities of f z

z 4 a4 each singularity. The singularities of f (z) are given by z4 + a4 = 0, i.e., z4 = ( 1) a4 = e i(2r + 1) . a4 z = ei(2r + 1)

i.e.

a, where r = 0, 1, 2, 3. a ei

The singularities are z simple poles. Res. f z

z a ei

lim

4 z

ei

1 4z Re s. f z

z aei 3

4

1 e 4a

Re s. f z

z aei 5

4

1 e 4a

Re s. f z

z aei 7

4

1 e 4a

4

4

, a e3

P z Q z

z aei i3

4

i 4

,a e 5

1 e 4a 1

4

4

1 4 2a

i7

4

i 4

, all of which are

, as f z is of the form

4 2a i5

and a e7

1 4 2a

i

4

1 4 2a

1 i 1 i

1 i

1 i

P z . Q z

Part II: Mathematics II

II – 4.40

Example 4.9 Evaluate the following integrals, using Cauchy’s residue theorem. (i)

cos z 2 sin z 2 dz , where C is | z | 3. z 1 ( z 2) C e z dz

(ii) (iii) C

(i)

, where C is z 2 2 z2 dz , where C is z 1 z sin z

cos z 2 sin z 2 dz z 1 z 2 C

4.

f z dz , say. C

The singularities of f (z) are z = –1 and z = 2, which are simple poles lying within the circle |z| = 3. Re s. f z

Re s. f z

z

1

z

2

cos z 2 sin z 2 z 2

z

1

cos z 2 sin z 2 z 1

z

2

1

1

By Cauchy’s residue theorem, 2 i sum of the residues

f z dz C

i.

= (ii)

e z dz

f z dz

Let C

C

z2

e z dz

2 2

The singularities of f (z) are z = i and of f (z) and lies within the circle |z| = 4. Res. f z

z i

i

z i

2

each of which is a double pole

1 d z i 1! dz

R1

2

f z z i

ez

d dz

z i

z i

2

2

ez z i

ei

2

z i

C

2i 8i

3 3

2

z i

2e z z i 4 z i

1 i 4i

1 3

1 4

3

i

Chapter 4: Complex Integration Res. f z

z

1

R2

i

i

3

4

II – 4.41

By Cauchy’s residue theorem, f z dz

2 i R1

R2

C

2 i 4 3 (iii) Let C

dz z sin z

i

2

f z dz C

The singularity of f (z) is given by z sin z z2 3!

z2 1

i.e.

z3 3!

0, i.e.z z

z4 5!

z5 5!

0

0

i.e. z = 0, which is a double pole of f (z) lying within the circle |z| = 1. Res. f z

1 d z2 1! dz z sin z

z 0

z 0

sin z z cos z sin 2 z =

z 2 cos z

z 0

0 form 0

, by L’Hospital’s rule z 0

=0 By Cauchy’s residue theorem. f z dz

2 i 0

0.

C

Example 4.10 Evaluate the following integrals, using Cauchy’s residue theorem. (i)

z 1 dz , where C is |z + 1 + i| = 2 z 2z 4 2

C

12 z 7 dz

(ii) C

(iii) C

z 1 dz z2

2

9

, where C is z i

2z 3 3

, where C is z i

3

3

Part II: Mathematics II

II – 4.42 f z dz

(i) Let C

z 1 dz z 2z 4 2

C

(– 1 + i 3 )

y

x

O (– 1 – i) (– 1 – i 3 )

Fig. 4.19

The singularities of f (z) are given by z2 + 2z + 4 = 0, i.e. (z + 1)2= 3, i.e. z = 1± i 3 , each of which is a simple pole. Of the two poles z = 1 i 3 alone lies inside the circle |z ( 1 i)| = 2, as shown in the Fig.4.19. 1 z 1 Res. f z z 1 i 3 z 1 i 3 z 1i 3 2 By Cauchy’s residue theorem, 1 f z dz 2 i i 2 C 12 z 7

(ii) Let C

z 1

2

dz

2z 3

f z dz . C

y

The singularities of f (z) are z = 1 and z = z = 1 is a double pole and z = the pole z = 1 lies inside the circle |z ( i)|= 3 , as shown in the Fig. 4.20.

2 z

O

2

2 z=1

3

3 z=–i

Res. f z

z 1

1 d 12 z 7 1! dz 2 z 3

50 2z 3

z 1

2.

2 z 1

By Cauchy’s residue theorem,

Fig. 4.20

x

Chapter 4: Complex Integration f ( z )dz

2 i 2

II – 4.43

4 i.

C

dz

f ( z )dz

(iii) Let C

C

(z

2

9) 3

The singularities of f (z) are z = ±3i, of which z = 3i lies inside the circle |z as shown in Fig. 4.21. y

z = 3i is a triple pole of f (z). [Res. f ( z )]z

3i

1 d2 1 2 ! dz 2 ( z 3i )3 1 2!

z = 3i z 3i

12 z 3i

6 5 5

6 i

i| = 3,

z=i 5 z 3i

x

O

1 1296i z = – 3i

By Cauchy's residue theorem, Fig. 4.21

f ( z )dz C

1 2 i 1296i

648

EXERCISE 4(b) Part A

2. State Taylor's theorem. 3. State Laurent's theorem. 4. What do you mean by analytic part and principal part of Laurent's series of a function of z? f (z). Give one example for each. f (z). Give one example for each.

12. Express the residue of a function at an isolated singularity as a contour integral.

Part II: Mathematics II

II – 4.44

pole. 15. State Cauchy's residue theorem. 16. Derive Cauchy's integral formula as a particular case of Cauchy's residue theorem. Find the Taylor series for each of the following functions about the indicated point: 17. sin z about z 18. cos z about z 19. ez about z = –i. 20. e–z about z = 1. 21.

z 1

about z = 1. z2 If each of the following functions is expanded as a Taylor's series about the

expanding. z 1 22. about z = 0 z 1

sin z about z = 0 z2 4 z 3 about z = 2 25. ( z 1)( z 4) 23.

z about z = 1 26.

27.

z

e about z = 4i. z ( z 1) (Hint: The radius of convergence is the distance between the centre of the Taylor's series and the nearest singularity of the concerned function.) Find the Laurent's series of each of the following functions valid in the indicated regions: 1

z

2

z ( z 2)

28.

1 z ( z 1)

29.

1 , valid in | z | > 1. z 3 (1 z )

z

30. z 1 , valid in | z – 1 | > 1. z2 Find the residue at the essential singularity of each of the following functions, using Laurent's expansions: z sinh z cos z 31. e z 34. 32 1 e 33. 35. 1 cosh z 4 2 z z z3 z

Chapter 4: Complex Integration

II – 4.45

Find the residues at the isolated singularities of each of the following functions: z z.2 36. 37. 2 ( z 1)( z 2) z a2 ze z 38. cot z (at z = 0) 39. ( z 1) 2

40.

z sin z . (z )3 Evaluate the following integrals using Cauchy's residue theorem.

41. C

z 1 dz, where C is |z| = 2 z ( z 1) z2

dz, where C is |z| = 2 z2 1 dz 43. dz, where C is |z| = 1 sin z C 42.

C

tan zdz , where C is |z| = 2

44. C

45.

e z dz , where C is |z| = 1 z2 C

Part B 46. Find the Taylor's series expansion of 1 f ( z) about z = –1. States also the region of convergence of the series. z (1 z ) 47. Find the Taylor's series expansion of z f ( z) about z = i. z ( z 1)( z 2) State also the region of convergence of the series. 48. Find the Laurent's series of f ( z ) z

z2 1 z 2 5z 6

valid in the region (i) |z

z| > 3.

49. Find the Laurent's series of f ( z ) z

z , valid in the region (i) |z + 2| ( z 1)( z 2) z + 2| >4.

7z 2 z ( z 2)( z 1) about z = –1. Indicate the region of convergence in each case. Find also the residue of f (z) at z = –1.

50. Find all possible Laurent's expansions of the function f ( z )

Part II: Mathematics II

II – 4.46

all possible Laurent's expansion of the function f ( z ) z 6z 1 about z = 3. Indicate the region of convergence ( z 1)( z 3)( z 2)

51. Find

3

in each case. Find also the residue of f (z) at z = 3. 52. Find the residues of f ( z )

z2 ( z 1) 2 ( z 2)

at its isolated singularities, using

Laurent's series expansions. 1

f ( z)

2

z ( z 2 1)

residue at the poles of f (z). 54. Indentify the singularities of f (z z

(i) f ( z )

z

2

(ii) f ( z )

2z 5

1 2

z (z

55. Find the singularities of the following: (i) z2

(ii) f ( z )

2

f ( z)

2 z 2) z ( z 1)( z 2 1)

2z

( z 1) 2 ( z 2 4) 56. Indentify the singularities of f (z) and the corresponding residues, when (i)

1

f ( z)

z

4

(ii) f ( z )

16 dz

ze z dz ; and (ii)

57. Evaluate (i) C

2

C

z sin z

z z

4

a4

,

using Cauchy's residue theorem, where C is the circle |z| = 1 in both cases. 58. Use Cauchy's residue theorem to evaluate the following dz , where C is the circle | z | = 4 sinh z C z dz , where C is the circle z (ii) z cos 2 C (i)

2

.

59. Use Cauchy's residue theorem to evaluate the following: (i)

z dz 2 C ( z 1)( z 2)

(3 z 2

(ii) C

( z 1)( z 2 (3 z 2

(iii) C

z )dz

(z

2

, where C is circle |z

, where C is the circle |z – 2| = 2.

9)

z 1)dz 1)( z 3)

, where C is the circle | z – i | = 2.

Chapter 4: Complex Integration

II – 4.47

60. Use Cauchy's residue theorem to evaluate the following: ( z 3)dz , where C is the circle | z + 1 – i | = 2 (i) 2 2z 5 C z dz

(ii) C

(iii)

4.5

(z

2

4)3

, where C is the circle |z – i| = 2

( z 1)dz 2 C ( z 1) ( z 2)

, where C is the circle |z + i| = 2.

CONTOUR INTEGRATION—EVALUATION OF REAL INTEGRALS

them in terms of integrals of complex functions over suitable closed paths or contours integrals is known as contour integration. in applications.

4.5.1 Type 1 P( x) dx, where P(x) and Q(x) are polynomials in x. This Q( x) integral converges (exists), if (i) The degree of Q(x) is at least 2 greater than the degree of P(x). (ii) Q(x) does not vanish for any real value of x. P( z ) dz , To evaluate this integral, we evaluate Q ( z) y C

Integrals of the form

|z|=R where C is the closed contour consisting of the S real axis from – R to + R and the semicircle S above the real axis having this line as diameter, R by using Cauchy's residue theorem and then letting . The contour C is shown in Fig. –R O R 4.22. Fig. 4.22 Now we consider a result, known as Cauchy's Lemma, which will be used in evaluation of f ( z )dz , where S is the semicircle |z| = R above the real axis.

y

S

Cauchy's Lemma If f (z) is a continuous function such that |zf (z

z

S, then f ( z )dz S

semicircle |z| = R above the real axis.

0, as R

S is the

Part II: Mathematics II

II – 4.48 Proof f ( z )dz

zf ( z )

S

S

zf ( z ) S

| zf ( z ) |

i. e. 0

1 dz z 1 dz |z|

1 Rd R When |z| = R, z = Re and dz = Riei d ]

f ( z )dz

lim

R

S

f ( z ) dz

lim | zf ( z ) |

R

Note

f ( z )dz

0 and hence

S

0, as R

.

S

f ( z )dz

Similarly, we can shown that

0 as r

0 , where S1 is the

S1

semicircle |z – a| = r above the real axis, provided that f (z) is a continuous function such that |(z – a) f (z r

4.5.2 Type 2 Integrals of the form

P ( x) sin m x dx or Q( x)

P( x) cos mx dx, where P(x) and Q (x) are Q( x)

polynomials in x. This integral converges (exists), if (i) m > 0 (ii) the degree of Q(x) is greater than the degree of P(x) (iii) Q(x) does not vanish for any real value of x. To evaluate this integral, we P( z ) imz e dz , where C is the same contour as in Type 1, by using evaluate Q( z ) C Cauchy's residue theorem and letting

the requirement.

P( x) imx e dx, Q( x)

Chapter 4: Complex Integration

II – 4.49 f ( z )eimz dz , S

where S is the semicircle |z| = R above the real axis, is given as follows.

Jordan’s Lemma If f (z) is a continuous function such that |f (z e

imz

f ( z )dz

0 as R

z

S, then

S is he semicircle |z| = R above the real axis and

S

m > 0. Proof |z| R, z = Re .

On the circle

eimz f ( z )dz S

eimR (cos

i sin )

. f ( z ).iRei d

eimR (cos

i sin )

| f ( z ) | Rd

S

0 mRsin

e

i.e.

| f ( z )| Rd

0 2

e

2

i.e.

mRsin

| f ( z ) | Rd

(1)

0

Now

2

sin

2mR

mR sin e

mR si

for 0

e

2 mR

2 , for 0

2

for 0

2

(2)

Using (2) in (1), we have 2

eimz f ( z )dz S

2 mR

e

2

| f ( z )| Rd

0 2

eimz f ( z )dz

lim

R

S

lim (2 Re

2 mR

R

0

) lim [ f ( z )]d R 2

i.e.

i.e.

lim [ f ( z )]

R

R

lim [ f ( z )] lim

R

2 Re

lim

R

2 mR

0

m

1 e

mR

d

Part II: Mathematics II

II – 4.50 i.e.

0, by the given condition.

m eimz f ( z )dz

0 and hence eimz f ( z )dz

S

0

S

as R

4.5.3 Type 3 2

Integrals of the form 0

P (sin , cos ) d , where P and Q are polynomials in sin Q (sin , cos )

and cos . To evaluate this integral, we take the unit circle |z| = 1 as the contour. z z 2i

When |z| = 1, z = e and so sin

1

and cos

dz . When iz circle |z| = 1.

id or d

2

Thus 0

z

z 2

1

z=e

z moves once around the unit

P(sin , cos ) d Q(sin , cos )

f ( z ) dz , C

where f (z) is a rational function of z and C is |z| = 1. Now applying Cauchy’s residue theorem, we can evaluate the integral on the R.H.S.

WORKED EXAMPLE 4(c)

Example 4.1

x 2 dx

Evaluate

x2

a2

x2 b2

, using contour integration, where

a > b > 0. z 2 dz

Consider C

z2

a2

, where C is the contour consisting of the segment

z 2 b2

of the real axis from R to + R and the semicircle S above the real axis having this segment as diameter. The singularities of the integrand are given by (z2 + a2) (z2 + b2) = 0 i.e. z = ± ia and z = ± ib, which are simple poles. Of these poles, only z = ia and z = ib lie inside C. (Fig. 4.23)

Chapter 4: Complex Integration

II – 4.51

y S

x R

O

R

Fig. 4.23

z2

R1 = (Res. of the integrand)z = ia =

Now

a

= 2ia b

2

2

z ia

a a

2

2i a

2

b2 b

R2 = Res. at (z = ib) =

Similarly

z ia z 2 b 2

2i a

2

b2

By Cauchy’s residue theorem, z2 z2

C

a 2 z 2 b2

dz

2 i ( R1 a b

= R

i.e., R

Now Similarly

b2

a b z 2 dz

(1) a b z 2 a 2 z 2 b2 ( on the real axis, z = x and so dz = dx) 2 2 2 z| a = R a2 on |z| = R. b2 on |z| = R.

a2

x2

|z2 + a2 |z2 + b2 R2 z.

a2

x 2 dx x2

R2 )

b2

S

z2 z2

a2

R3 z 2 b2

R2

a2

R 2 b2

Since the limit of the R.H.S. is zero as R z2

lim z.

R

z

2

z

2

a

2

a

2

z 2 b2

z 2 dz

By Cauchy’s lemma, S

z 2 b2

0, on | z | R. 0

Using (2) in (1) and letting R x 2 dx x2

a2

x2 b2

a b

(2)

Part II: Mathematics II

II – 4.52

Example 4.2 Evaluate

(x

x 2 dx , using contour integration. 1 )( x 2 2 x 2)

2

2

z 2 dz

Consider C

z 2 12 z 2

, where C is the same contour as in the previous 2z 2 z2

example. The singularities of f ( z ) z

2

1

2

z

are given by (z2 + 1)2

2

2z 2

(z2 + 2z + 2) = 0

i.e. z = ± i and z = 1 ± i, of which z = i and z = 1 + i lie inside C. z = i is double pole and z = 1 + i is a simple pole. R1

[Re s. f ( z ) z

1 d ( z i)2 z 2 1! dz ( z i ) 2 ( z i ) 2 ( z 2 2 z 2)

i

( z i)2 ( z 2

2 z 2).2 z z 2 {2( z i )( z 2

=

4

z i

z2

z i

2 z 2) ( z i ) 2 (2 z 2)} 2

2z 2

z i

9i 12 100 R2

[Re s. f ( z )]z

z2 1 i

z

2

2

1

z 1 i

z

3 4i 25

1 i

By Cauchy’s residue theorem, z 2 dz z2 1

C

2

z2

2 i ( R1

R2 )

2z 2 2 i

9i 12 100

3 4i 25

7 50 R

x 2 dx

i.e. R

x

2

x

2

2x 2

z. z 2

lim

R

1

2

z 2 dz

z2 1

2

z2

S

z

2

1

2

0 on | z | R. 2z 2

z

2

2z 2

7 50

(1)

Chapter 4: Complex Integration

II – 4.53

By Cauchy’s Lemma, when R z 2 dz S

z2 1

2

0

(2)

z 2z 2

Letting R x2d x x2 1

2

x2

2x 2 dx

Example 4.3 Evaluate 0

dz

Consider C

z

2

a2

3

7 50

x2

a2

, using contour integration, where a > 0.

3

, where C is the same contour as in example 4.1. 1

The singularities of f z

3

z 2 a2 the two poles, z = ia alone lies inside C. R [Re s. f ( z )]z

are z = ± ia,which are poles of order 3. Of

1 d2 2 ! dz 2

ia

6 z ia By Cauchy’s residue theorem,

R R

z2

a

dx x2

a

2 iR

2 3

dz 2 3

S

z2

z ia

a

2 3

z2

3 5

16a i

8a 5

(1)

8a 5 R

2 3

1

lim z.

3

3

a

z2

R

2 i.

1

z.

Now

3

16a 5i

z ia

dz

i.e.

z ia

3 5

C

1

a2

3

R2

a2

3

and so

0 on |z| = R

By Cauchy’s Lemma, when R dz S

z

2

a

2 3

0

(2)

Part II: Mathematics II

II – 4.54 Letting R

dx x

2

a

3 2 3

8a 5

.

1 Since the integrand dx 3 x2

0

a2

3

x2

is an even function of x, we have

3

a2

16a 5

Example 4.4 Use contour integration to prove that

0

x 2 dx x4 a4 z 2 dz

Consider C

z4

a4

2 2a

, where a > 0.

, where C is the same contour as in Example 4.1. z2

i.e.

are given by z4 = a4 = ei(2r + 1) . a4 z 4 a4 z = ei(2r + 1) . a, where r = 0, 1, 2, 3.

i.e.

z = aei , aei3 |4, aei5 , aei7 , all of which are simple poles.

The singularities of f ( z )

Of these poles, only z = aei R1

and z = aei3

[Res. f ( z )]z 1 e 4a

R2

i

aei

lie inside C.

4 2a

[Res. f z

z aei 3

aei 4

z

1

4

z2

lim

4

4

4z

lim

3

z

P( z )

a Q1 ( z )

(1 i ) 1 e 4a

i3

4

1 4 2a

1 i

By Cauchy’s residue theorem, z 2 dz C R

i.e. R

Now

x 2 dx x4

a4

4

z

a

4

2 i

1 4 2a

(1 i 1 i )

z 2 dz S

lim z.

r

z4

a4

z2 z4

a4

2a

2a (1)

0 on | z | R.

By Cauchy’s Lemma, when R z 2 dz s

z4

a4

0

(2)

Chapter 4: Complex Integration

II – 4.55

Letting R x 2 dx x4

a4

2a

Since the integrand is even,

0

x 2 dx x4 a4

2 2a

Example 4.5 Evaluate z4

Consider C

z

6

a6

x4 x6

dx, using contour integration where a > 0.

a6

dz, where C is the same contour as in example 4.1, with some

The singularities of f ( z )

z4 z6

a6

are given by r

i

z6 = a6 = ei2r a6, i.e. z = ae 3 , r = 0, 1, 2, 3, 4, 5. i.e. z = a, ae i , ae2 i , a, ae4 i , ae5 i . Of these singularities (simple poles), z = ae i and z = ae2 i lie inside C, but z = a and z = a lie on the real axis. But for the evaluation of the integrals of type 1, no singularity of f (z) should lie on the real axis. To avoid them, we modify C by introducing small indents, i.e., semi-circle of small radius at z = ± a, as shown in Fig. 4.24. y S S2

S1 –R

z=–a

O

z= a

R

x

Fig. 4.24

C contains only the simple poles z = ae i and ae2 i/3 R1

[Re s. f ( z )]z z4 6z

5

ae 3i

1 e 6a

1 1 3 i 6a 2 2

i 3

Part II: Mathematics II

II – 4.56

1 2 e 6a By Cauchy’s residue theorem, R2

Similarly,

z4 C

z

6

a

12 i

3 2

2 i 3 3 12 i 12 i 6a 2 2

dz

6

1 6a

i3

3a –a–r

i.e.

–R

a – r'

x4 z4 x4 z4 dx + dz+ dx+ dz 6 – a6 6 – a6 6 – a6 x6 – a6 z x z S –a+r S 1

2

R

x4 x6

a r

a6

z4

dx S

z6

a6

dz

3a

(1)

where r and r' are the radii of the semicircles S1 and S2 whose equations are |z + a| = r and |z – a| = r'. These two integrals taken along S1 and S2 vanish as r r' by the note under Cauchy’s Lemma. z4 S

Now, letting r

z

6

0 , as R

a6

r'

R

a

a

We get

a

a

x4

i.e. x

6

a6

x sin x

Example 4.6 Evaluate 0

Consider

x 4dx x6 a6

x2

a2

dx

3a

3a

dx , by contour integration.

zeiz dz , where C is the same contour as in Example 4.1. z 2 a2 C

zeiz . are z = ± ia, which are simple poles. Of z a2 these poles, only z = ia lies inside C. The singularities of f ( z )

[Res. f (z)]z = ia =

zeiz z ia

z ia

2

1 a e . 2

Chapter 4: Complex Integration

II – 4.57

By Cauchy’s residue theorem, zeiz z

C

2

a

2

dz

R

xeix dx x2 a2 R

i.e.

1 e 2

2 i.

S

a

zeiz dz z a2

ie

a

ie

a

(1)

z R z 2 a2 R2 a2 Since the limit of the R.H.S. is zero as R

Now

lim

R

z z2

0 on |z| = R.

a2

zeiz S

z2

a2

dz

R

Letting R xeix dx x2 a2 x

i.e. x

2

a

2

i e

a

(cos x i sin x)dx

i e a.

Equating the imaginary parts on both sides, x sin x dx x2 a2

we get

e a.

Since the integrand is an even function of x, x sin x dx x2 a2

0

2

e a.

cos xdx

Example 4.7 Evaluate x

2

a2

x2 b2

., using contour integration, where

a > b > 0. eiz

Consider C

z2

a2

z 2 b2

dz , where C is the same contour as in Example 4.1.

The singularities of the integrand f (z) are given by z = ± ia and z = ± ib, which are simple poles. Of these poles, z = ia and z = ib only lie inside C.

Part II: Mathematics II

II – 4.58

R1

[Res. f ( z )]z

eiz ia

z ia z 2 b 2 z ia

a

e

2ia a 2 b 2 R2

[Res. f ( z )]z

e ib

b

2ib(a 2 b 2 )

By Cauchy’s residue theorem, eiz z2

C

a2

dz

z 2 b2

2 i R1

R2 e b b

a 2 b2 R

eix

i.e.

x2

R

a2

x2 b2

S

z2

a2

z

a

2

dz

z 2 b2 e b b

1 2

a

eiz

dx

a 2 b2 Now

a

e

a

e

(1)

a

1 z

2

b

2

(R

2

2

a ) R 2 b2

R 1

lim

R

z

2

a

2

eiz S

z2

a2

0 on |z| = R.

z 2 b2

z 2 b2

dz

0 , as R

(2)

Letting R eix x2

a2

x2 b2

dx

a 2 b2

e b b

e a a

e

a 2 b2

e b b

Equating the real parts on both sides, we get cos x x2

a2

x2 b2

dx

a

a

.

Chapter 4: Complex Integration cos ax

Example 4.8 Evaluate 0

eiaz

Consider C

z

2

b

2 2

dx

x2 b2

4b3

ab

(1 ab)e

II – 4.59

, where a > 0 and b > 0.

dz , where C is the same contour as in Example 4.1. eiaz

The singularities of f ( z )

z 2 b2

are z = ± ib, which are double poles. Of these

2

poles, only z = ib lies inside C. Res. f ( z )

eiaz

1 d 1! dz

z ib

2

z ib

z ib

( z ib)ia eiaz z ib

2eiaz 3

1 ab 1 e 4ib3

z ib

ab

.

By Cauchy's residue theorem, R

R

eiax x

2

b

2

eiaz

dx 2

z

S

2

b

2 2

dz

2b3 1

Now z Since the R.H.S.

2

b

(ab 1)e

ab

ab

(1)

1 2 2

R

0 as R

2

b2

2

0 as R

, L.H.S. also

eiaz S

Letting R

1 (ab 1)e 4ib3

2 i

z 2 b2

2

dz

0 as R

on |z| = R.

, since a > 0

in (1) and using (2), we get eiax x

2

b

2 2

dx

2b3

ab 1 e

ab

.

(2)

Part II: Mathematics II

II – 4.60

cos ax Equating the real parts on both sides and noting that of x, we get x2 b2 cos ax x

0

2

b

2 2

dx

ab 1 e

4b3

ab

2

is an even function

.

Example 4.9 Use contour integration to prove that sin mx dx x 0 Consider

, when m > 0.

2

eimz dz , where C is the usual semicircular contour, but with an indent i.e. z C

a small semicircle at the origin, which is introduced to avoid the singularity z = 0, y S S' –R

–r O r

x

R

Fig. 4.25

C does not include any singularity of f ( z ) By Cauchy's residue theorem, –r –R

The Eqn. of

e imx dx e imz dz + + x z S'

is |z| = r.

R r

eimz . z

e imx dx e imz dz = 0 + x z S

(1)

z = re and dz = re id .

is described in the clockwise sense, varies from to 0. Thus

S

lim

r

0

S

1 z

eimz dz z eimz dz z 1 R

eimz dz z S

0 imrei

e rei id rei

0

lim eimr e

r

0 as R 0 as R

0

i

id

i

(2)

on |z| = R. , since m > 0

(3)

Chapter 4: Complex Integration Letting r

0 and R

II – 4.61

in (1) and using (2) and (3), 0

eimx dx i x

we get

eimx dx x 0

eimx dx x

i.e.

0

i

Equating the imaginary parts on both sides and noting that function of x, we get

sin mx dx x 0 sin x dx

Example 4.10 Evaluate 0

eiz

Consider C

z z2

a2

2

x x2

sin mx is an even x

.

, using contour integrations where a > 0.

a2

dz , where C eiz

4.9, which includes the only pole z = ia of f ( z )

[Res. f ( z )]z

eiz z z ia

ia

z z2

a2

e a 2a 2

z ia

By Cauchy's residue theorem, –r

–R

eix eiz dx + dz + 2 2 x x2 + a2 s' z z + a

R

eix eiz dx + dz 2 2 x x2 + a2 s z z +a

r

i a

S

0

a2

S

z z2

dz

eir e .rei id rei r 2 ei 2

a2

dz

lim

r

0

eir e id r 2 ei 2 a 2

i a2 1 z z

2

(2)

1 a

2

R R

a2 i

0

eiz

lim

r

z z2

(1)

i

0

eiz

a

e

2

2

a2

Part II: Mathematics II

II – 4.62 0 as R

Since R.H.S.

on |z| = R, L.H.S. also tends to 0 as R eiz z z2

S

0 and R

Letting r

dz

a2

(3)

0, as R

in (1) and using (2) and (3), we get eix x x

2

a

dx

2

a

i

i

2

2

a

e a.

Equating imaginary parts on both sides and noting that function of x, we get,

sin x x x2

0 2

a2

dx

d 1 2 x sin

Example 4.11 Evaluate 0

2a 2

x2

C

a2

is an even

1 z

z 2i

z2 1 . 2iz

, where C is |z| = 1

z2 1 1 2x 2iz

x

2

dz

I C

iz 1 xC

xz

2

x ix 2 z dz 1 z 1 x

z2 i x 1

The singularities of z ix Now |ix| = |x

x x2

x 1), using contour integration.

dz and sin iz

dz iz

The given integral I

sin x

(1 e a ).

(0

On the circle |z| = 1, z = e , dz = iei d or d

i.e.

.

z

i x

1 xC

dz z ix

(1)

i x

z

i , which are simple poles x

are z = ix and z

x

The pole z = ix lies inside C, but z

Res.

i lies outside C. x

1 z ix z i x

1 z ix

i x

1 x

ix 1 x2

Chapter 4: Complex Integration

II – 4.63

Using Cauchy's residue theorem, from (1) we get 1 x

I Note

ix 1 x2

2 i

2 1 x2

.

We have integrated w.r.t. and x is a parameter. 2

d (a a b cos

Example 4.12 Evaluate 0 2

d

Deduce the value of 0

z

cos

2

1

2z 2

I 0

2

a b cos

b

0), using contour integration.

. On the circle |z| = 1, z = e , d

dz and iz

. dz iz

d a b cos

C

a b

z

2

2i bC

1

z

2z

dz 2a z 1 b

2

(1)

where C is |z| = 1. The singularities of the integrand are given by z 2 i.e. z

a b

2a z 1 0, b

a2 1, which are simple poles. b2

a 1 and hence b outside |z| = 1.

Since a b,

z

a2 1 b2

a b

1 and so z

a b

a2 1 lies b2

a2 1 lies inside |z| = 1. b2

a b Res. of

z2

1 2a z 1 b

z

a b

a2 b2

1

1 2 z

a b

b 2 a

2

b2

z

a b

a2 b2

1

(2)

Part II: Mathematics II

II – 4.64

By Cauchy's residue theorem and using (2) in (1), we get 2i b

I 2

b

2 i 2 a

2

2

b

d a b cos

i.e. 0

2

a

2

b2

2 a

2

(3)

b2

Differentiating both sides of (3) partially with respect to ‘a’, we get 2

2

0

a

2

b2

32

2 a

d 2

a b cos

0 2

2

a b cos

0

Example 4.13 Evaluate

2 a

d

a2

b2

32

sin 2 d , using contour integration. 5 3 cos z 2 1 and cos 2iz

dz , sin iz

On the circle |z| = 1, z = e , d

z2 1 . 2z

2

2

I 0

z 2 1 dz . 4 z 2 iz , where C is |z| = 1 3 z2 1 C 5 2z

sin 2 d 5 3 cos

2

i z 2 1 dz 6 C z2 z 3 z 1 3

(1)

The singularities of the integrand which lie within C are z = 0 and z 1 z = 0 is a double pole and z is a simple pole. 3 R1 = [Res. of the integrand f (z)]z = 0 1 d 1! dz

R2

2

z2 1 10 z2 z 1 3

[Res. f ( z )]z

10 3 z 0

z2 1 13

2

z 2 ( z 3)

z 13

8 3

Chapter 4: Complex Integration

II – 4.65

By Cauchy's residue theorem, from (1), we get i 2 i R1 6

I

2

Example 4.14

Evaluate 0

a

10 3 3

R2

8 3

2 . 9

cos 2 d , using contour integration, where 1 2a cos a2

2

The given integral I

z2 1 and cos 2 2z

dz , cos iz

On the circle |z| = 1, z = e , d z4 1 . 2z2

ei 2

e 2

z 4 1 dz 2 z 2 iz , where C is |z| = 1 z2 1 2 C a 1 a z i 2a C

z 4 1 dz 1 z2 a z 1 a

z2

i z 4 1 dz 2a C z 2 z a z 1 a

(1)

The singularities of the integrand which lie inside C are z = 0 and z = a ( a2 z = 0 is a double pole and z = a is a simple pole. R1 = [Res. of the integrand]z = 0 1 d 1! dz z 2 a R2

(Res. of the integrand) z

R1

i2

R2

z4 1 a 1 a z 1

1 a a4 1

a

a

a2 a 1 a 1 a

a4 1 a2 a 1 a

2a 3 a2 1 By Cauchy's residue theorem and from (1), we get I

i 2a

2 i

2a 3

2 a2

a2 1

1 a2

z 0

Part II: Mathematics II

II – 4.66 2

Example 4.15 Evaluate 0

cos 3 d , using contour integration 5 4 cos

The given integral I

i.e.

z2 1 and cos3 2z

dz , cos iz

On the circle |z| = 1, z = e , d

ei 3

e 2

i3

z6 1 . 2z

z 6 1 dz . 2 z 3 iz , where C is |z| = 1. 2 1 C5 4 z 2z

I

i z 6 1 dz 4 C z3 z 2 z 1 2

The singularities of the integrand f (z) which lie inside C are the simple pole z = 0. z z6 1 z3 z 2

R1 = (Res. of the integrand)z

z

12

65 12 1 in the Laurent’s z

R2, the residue at z expansion of the integrand. f ( z)

z3

1 z3

z3

1 z3 1

z3

R2

z3

1

z 1 2z 2

1

z 2

1

5 z 2

1 in this expansion z

z2 4

1 2z 4z2

21 2 z 4

21 . 4

By Cauchy’s residue theorem. I

i

2 i

65 12

21 4

12

Chapter 4: Complex Integration

II – 4.67

EXERCISE 4(c)

Part A

2

f (sin ,cos )d

2. Explain how to convert

into a contour integral, where f is

0

a rational function. 3. Sketch the contour to be used for the evaluation of

sin mx dx. x 0

Part B Evaluate the following integrals by contour integration technique. x 2 dx

4. 0

x2 1 x2 x2

6.

x

4

4

0

x 2 dx 10 x 2 9

x 2 dx

8. 0

10. 0

x

2

x

2

4 x

2

x sin x

14.

1 x

2

sin x dx

16.

x

2

4x 5

sin x

18. 0

x x2 1

0

dx 4

a4

x6 1

x2 9

dx

dx

0; b

0)

x sin x dx x4 1 x sin x

15.

x2 1

2

dx

x cos x dx x 2x 5

17.

2

2

dx 2

x4

2

cos ax dx (a x2 b2

13. 9

4

x2

11.

cos x dx

12.

0

0

x 4dx x6 1

x2 dx

7.

9.

3

x2 1

x 2 dx

5.

19. 0

d 1 2 p cos

p2

,0

p 1

Part II: Mathematics II

II – 4.68 2

2

d (a a b sin

20. 0 2

ad

22.

a2

0 2

b

21. 0 2

(a

sin 2

0) 0)

23. 0 2

cos 3 d 5 4 cos

24. 0

25. 0

sin 2 d (a a b cos

b

0)

cos 2 d 5 4 cos cos 2 3 d 5 4 cos 2

ANSWERS

Exercise 4(a) 8.

10 (3 i ) 3

9.

i

3 – i2 4 3

11.

i i

16 i

13 . 15

i 16.

8 i 3

15.

10 i

8 3

0 in all cases.

(i) 511 49 i; (ii) 518 57i; (iii) 518 8i. 3 5 3 3 1 18. 96 5 a 5 80 3 a 3 30 a 15

17.

20.

8 i 3

i i; (ii) 0

24.

i.

2 3 8 i 3e 2

23.

sin t

25.

26.

27. 16

2

3 2i i

Exercise 4(b) 17.

sin z

1 2

1

z

4

z

4 2!

2

z

4 3!

3

i

i

i

Chapter 4: Complex Integration

18.

1 1 2

cos z

3 z

1 z 2!

3

3

2

II – 4.69

3 z 3!

3

3

1 z 4! 19.

i

1

z i 1!

z i 2!

2

e

e

1

1

z 1 1!

z 1 2!

2

20.

z 1 3!

4

3

3

21. (z – 1) –2(z–1)2 + 3(z–1)3 – 4(z–1)4 + ···· z

22. |z 24.

1 . 2

z 1

25. |z – 1

26. |z – 4i

2z 28.

1 z 1 1

29.

z4 1

30.

1

z 1

1

1 z

1

z 1 31. 1

2

z 1

z 1

2

1 2

2

z 1

z 1

2

32.

–1

34.

1 . 6

1 2

36.

Rz

37.

38.

1.

39.

ia 2 2e.

40.

–1

41.

2 i.

42.

0

43.

2 i.

44.

–4 i.

45. –2 i. 1 2 47.

1

n

n

z2 4

z2

35.

n 1

z 2

3

33. 1

46.

1

1 z 1 ;z 1 2 2 i

1

1 n 1

1 i

n

n 1

z i .

1

1 ;R 3 z

2

2 . 3

Part II: Mathematics II

II – 4.70 48.

(i)

1

1

(ii)

1 3

(iii)

1

1 2 z 1

(iii) 51.

1 3

2

3 z 1

1 2.3n

45 z 3

1

(iii)

1

45 z 3

1

2

1

1

n 1

n 2

5

2n

n

z 3 n

1 z 1 ( z 2) 2

i 2 z i

z n 1 ,| z 1 |

n

3

z 2 53. (i) 1 1 z2 2 i 2 (ii) z i

1 z 2 zn

2n

5n

z 1

3. Res. at z z 3

n

1 . z 3 5n 2

n 1

1

1

3.

z 3

n

1

n

z

, in |z – 3| > 5. Res. at z

( z 1) n ; ; Res. at (z

4

n

1 2

n 2

1 2

n 2

54. (i) Res. at the simple pole ( z and that at ( z

3

z z

1 i n z n , ; Res. at (z = 0) = 0.

(n 1) n 1

3.

z

n 1 (n 4) ( z 2) n ; Res. at (z

1

1

z 1 1

z 1 ;1 3n 1

2

n 1

z 1

1

3n

n

1

(ii)

(iii)

n

3

2.4n

(iii)

z 1 ;0

n 1

z 1

4

n 1

z 2

1

(ii)

1 zn 1

3n

n

45 z 3

52. (i)

1

n

1

(i)

3n

z 2 ;

n 1

1

zn

n

1

3.2n 8.3n

3

(ii)

8

zn 1

n

zn.

3

2n

z 2 4n 3

(i)

8 n 1

n

1 2.4n

(ii) 50.

2

1 1

49. (i)

3 n 1

n

1 2i )

i n (z–i)n; Res. at ( z 1

n

i)

n z i ; Res. at ( z

1 2i ) 1 (2 i) 4

1 (2 i) 4

i . 2 i)

i . 2

4 5

Chapter 4: Complex Integration (ii) z

II – 4.71

z = –1 + i is a simple pole with z = –1 – i

55. (i) z

z

14 (ii) z = –1 is a double pole; Res ; z 2i is a simple pole with 25 1 1 Res. 7 i ; z = –2i is a simple pole with Res. (7 i ) 25 25 56. (i) z = 2ei , 2ei3 , 2ei5 , 2ei7 are simple poles. Cor. residues are, 1 1 1 1 ( 1 i ), (1 i ), (1 i ), ( 1 i ). 32 2 32 2 32 2 32 2 (ii) z = aei , aei3 , aei5 , aei7 are simple poles. Cor. residues are i 4a i;

57. (i)

2

,

i 4a

2

,

i

4a (ii) i ; 3

2

,

i 4a 2

. 2

58. –2 i; (ii) 2

59. (i) –2 i; (ii) i; (iii)

i 4

i

60. (i) (i – 2); (ii) 3 ; 256

2 i. 9

(iii)

Exercise 4(c) 4.

5. 6 9.

8. 16 12. 14. 17. 21.

6. 200

3e 2 e

2 e3

13.

(e 1)

15.

2

18.

2 2 a b2 12

b

e

1

2

sin

b2

3 4e

ab

2

19. 22.

3 8

e

1

16.

2

25.

24.

2

2 2a 3

11. 3

2e

a2

7.

10. 6

3 30 e 2

5 12

e 2

sin 2

1 p 2

2

1 a2

2

20. a 23. 6

2

b2

Chapter

5

Laplace Transforms 5.1

INTRODUCTION

The Laplace transform is a powerful mathematical technique useful to the engineers and scientists, as it enables them to solve linear differential equations with given initial conditions by using algebraic methods. The Laplace transform technique can also be used to solve systems of differential equations, partial transform, we shall discuss below the properties of Laplace transforms and derive the transforms of some functions which usually occur in the solution of linear differential equations.

If f (t) is a function of t

t

e

st

f (t ) dt

Laplace

transform of f(t), provided the integral exists. Clearly the integral is a function of the parameters s. This function of s is denoted as f (s) or F(s) or (s). Unless we have to deal with the Laplace transforms of more than one function, we shall denote the Laplace transform of f (t) as s). Sometimes the letter ‘p’ is used in the place of s. The Laplace transform of f (t) is also denoted as L{f (t)}, where L is called the Laplace transform operator. Thus

L f (t )

(s)

e

st

f (t ) dt.

The operation of multiplying f (t) by e and integrating the product with respect to t Laplace transformation. The function f (t) is called the Laplace inverse transform of s) and is denoted by L { (s)}. Thus

f (t) = L {

s)}, when L {f (t)} =

s)

Part II: Mathematics II

II – 5.2

Note s

-

form. L (tan t) and L (e t ) for the existence of Laplace transform of a function f (t Conditions for the existence of Laplace transform If the function f (t for t t (ii) of the exponential order, then L{ f (t)} exists.

Note f (t a t b (i) f (t) is continuous at every point inside each of the sub-intervals and (ii) f (t t approaches the end points of each sub-interval from the interior of the sub-interval. f (t) is said to be of the exponential order, if | f (t M e , for all t some constants M and or equivalently, if lim{e t f (t )} a finite quantity. t

differential equations satisfy the conditions stated above and hence may be assumed to have Laplace transforms.

5.2

LINEARITY PROPERTY OF LAPLACE AND INVERSE LAPLACE TRANSFORMS

L{k f (t) ± k f (t)} = k L{f (t)} ± k L{ f (t)}, where k and k are constants. Proof: L{k f (t ) k f (t )}

{k f (t ) k f (t )}e k

f (t ) e

st

dt

k

st

dt f (t ) e

k L{f (t )} k L{ f (t )}. Thus L is a linear operator. L{k f (t)} = kL{f(t)}, where k is a constant.

st

dt

II – 5.3

Chapter 5: Laplace Transforms If we take L{f (t)} = (s) and L{ f (t)} = (s), the above property can be written in the following form. L{k f (t) ± k f (t)} = k L {k

(s) ± k

(s) ± k

(s).

(s)} = k · f (t) ± k · f (t) = k ·L { (s) ± k ·L { (s)}

Thus L is also a linear operator L {k

s)} = k L { (s)}, where k is a constant.

Note L{f (t) . f (t

L{f (t)} . L{f (t)} and

L { (s) .

L ( (s)} × L { (s)}

(s

n

n

we get

(i)

L

kr f r (t )

kr L{ f r (t )} r

r

n

n

(ii)

L

kr

r

(s)

kr L { r ( s )} r

r

a linear combination of elementary functions whose transforms are known. expressed as a linear combination of elementary functions whose inverse transforms are known.

5.3

LAPLACE TRANSFORMS OF SOME ELEMENTARY FUNCTIONS L{k}

k s

k is a constant,

L{k}

ke

k

e

k ( s k. s

st

dt

st

s )[ e

st

as t

, if s

]

Part II: Mathematics II

II – 5.4 In particular, L L L{e

at

=

L( ) =

s

.

s }

s

a L{e

, where a is a constant, at

}

e

st

e

at

e

(s

a )t

dt

(s

s Inverting, we get L L{e at }

s

Changing a L

s

a

a)

a e

a

(

e (s (s

a )t

a)

) , if (s + a

a.

, if s

at

dt

.

, where a is a constant, if s a

s > a.

a

s

L(t n ) =

e at

a (n sn

)

, if s

L(t n ) =

n

e

e

sn

st

x

t n dt

x s

e

n

x

dx , on putting st = x s

x n dx

II – 5.5

Chapter 5: Laplace Transforms (n sn

) , if s

n+

>

In particular, if n is a positive integer, (n

n!

)

n!

L(t n )

s

, if s

n

n! sn

Inverting, we get L L

sn

n is a positive integer.

n!

t n or

tn

L

Changing n to n If n

L

sn

In particular, L(t ) 5. L(sin at )

(n)

(n

)!

t n , if n is a positive integer.

tn L

or

s

sn

t.

s

a s

a

L(sin at )

e

st

e

st

s

sin at dt

s sin at

(

a

s s

e

a

st

a cos at )

sin at

a s

e

a

a s [ e

sin at and e

Inverting this result we get L

a cos at tend to zero at t a s

a

sin at .

s

st

cos at

Part II: Mathematics II

II – 5.6 s

6. L(cos at )

s

a

L(cos at )

st

e

cos at dt

st

e s

s cos at

(

a s s

a s

s

st

e

a sin at ) a

cos at

s

a

e

st

sin at

, as per the results stated above.

a

s

Inverting the above result we get L

s

cos at .

a

L(cos at + i sin at) = L(eiat) s ia s s s

i.e., L(cos at ) + iL (sin at )

s

ia a a

i

a

s

s

Equating the real parts, we get L(cos at )

s

Equating the imaginary parts, we get L(sin at ) 7. L(sinh at ) =

, by linearity property.

a a

. a

s

a

a s

L(sinh at )

a (e at

L

)

L (e at )

L (e

s

s

a

a s

at

e

a

at

a

, if s > |a|

) , by linearity property. , if s > a and

II – 5.7

Chapter 5: Laplace Transforms

L(sinh at

iL(sin i at) [

sin

i sinh ]

ia , by result (5) s i a

i

a s

a a

Inverting the above result we get L 8. L(cosh at )

s

sinh at.

a

s s

a

L(cosh at )

ea t

L

e

at

L ea t

L e

s

s

a

s s

a

at

, by linearity property,

, if s > |a|.

a

, if s > |a|.

L(cosh at) = L(cos iat) [ cos i = cosh s

]

s , by result (6) i a s

s

a

Inverting the above result we get L

5.4

s s

a

cosh at.

LAPLACE TRANSFORMS OF SOME SPECIAL FUNCTIONS

The function f (t )

, when t , when t

a a, where a

function and is denoted by ua (t) or u (t

a)

, is called Heavyside’s unit step

Part II: Mathematics II

II – 5.8

when t , when t

In particular, u (t ) Now

L{ua (t )}

st

e

ua (t ) dt

a

e s t ua (t ) dt

e s t ua (t ) dt a

st

e

dt ,

ua (t)

a

e

st

e

S

as

, assuming that s > .

S a

In particular, L{u (t )}

, which is the same as L

S

Inverting the above result, we get L

e

as

ua (t ) .

s

lim f (t ) , where f (t h

h

, when a

f (t )

t

h , otherwise

a

h is called Unit Impulse

Function or Dirace Delta Function and is denoted by Now L{ a (t )}

L lim f (t )

,where f(t

h

lim L{ f (t )} h

e

lim h

a

e

h

a

h

st

h

h

e h

f (t ) dt

h

lim

lim

st

st

a

s

a

dt

h

h

a

(t) or (t

a).

II – 5.9

Chapter 5: Laplace Transforms h

s a

lim

e

sh

h

as

e

lim

es h /

e sh

h

sinh as

e

lim

sh

s cosh

f (t )

u a

h

(t )

h

u

, when t

when

t

when

a

a h

L{f (t )}

h

h

a

(t )

h

h

h

h

a

sh

lim cosh h

(t ) ,since

h

a

h

a

h

a

(t )

(t ) = ,

(t )

u

h

a

L u

h

a

s

e

h

a

e s

s

sinh e

s

h

and u

as

sh

sh

sh , by L’ Hospital’s rule.

lim

(t )

h

L u

h

as

u

and hence u a

e

a

a

t

as

s h/

sh

h

e

h

s a

e

s

(t )

e

as

Part II: Mathematics II

II – 5.10

sinh L

a

(t )

lim L f (t )

as

e

h

lim h

sh

sh

e as Inverting the above result, we get L {e s}

5.5

a

(t). When a

(t).

PROPERTIES OF LAPLACE TRANSFORMS

If L{ f(t)} =

(s), then L{ f (at )}=

s t and L f a a

a

= a (as ) .

Proof (s )

L{ f (t )}=

and L{ f (at )}=

e

e

s

a

a

x a

st

st

f (t ) dt

f (at ) dt [ f(at) is a function of t]

f (x)

d x , putting x = at and making necessary changes. a

e

(s / a ) x

f (x) dx

e

(s / a ) t

f (t ) dt , changing the dummy variable x as t.

s

Thus

e

L{ f (at )}

a

(s a)

Changing a to a L{ f ( t a )}

a (as ) .

s in the integral a (s/a).

Chapter 5: Laplace Transforms

If L{ f (t)} =

(s), then

and

L{e f (t)} =

(s + a)

L{eat f (t)} =

(s

II – 5.11

a).

Proof L{ f (t )} and

L{ e

at

f (t )}

e

st

e

st

f (t ) dt at

[e

(s )

f (t )]dt [ e

e

(s

(s

a )t

f (t) is a function of t]

f (t ) dt

a)

Changing a a in the above result, at we get L{e f (t )}= (s a) .

Note L{e

at

f (t )}= (s

a)

=[ (s )]s

s +a

L{ f (t )}s ‘s

s +a

s + a’ means that s is replaced by (s + a).

which is e , we ignore e , a function of s and change s into (s + a) in it. Similarly, L{eat f (t)} = L{f (t)s s a

If L { (s )}= f (t ) , then L { (s

L { (s

a )}= e

a )}= e

at

at

f (t )

L { (s )}

s + a), inverse transform of the corresponding function of s and multiply it with e . Similarly, L { (s

a )}= e at L { (s )}

.

Part II: Mathematics II

II – 5.12

s-axis by a (or

a), i.e., replacing s by s + a (or s

).

The second shifting property, that follows, concerns shifting on the t-axis by a i.e., replacing t by

If

L{ f (t)} =

(s), then L{ f (t

a)ua(t)} = e

(s),

where a is a positive constant and ua(t) is the unit step function. Proof L{ f (t

a )ua (t )}=

st

e

f (t

a )ua (t )dt

a

e

st

f (t

a )ua (t ) dt

e

st

f (t

a )ua (t ) dt

a

e

st

e

s (x a )

e

as

e

e

as

(s )

f (t

a ) dt ,

ua(t).

a

st

f (x) dx , putting

and effecting consequent changes

f (t ) dt , changing the dummy variable x as t.

Note L{f (t a) ua(t)} = e as L{f (t)} . The above property can be stated in terms of the inverse Laplace operator as If

L { (s)} = f (t), then L {e

L {e

as

(s)} = L { (s)}t

as

t

one of which is e , we ignore e , factor as a function of t, replace t by (t

(s)} = f (t

a

a) ua(t).

· ua(t). a) in it and multiply by ua(t).

WORKED EXAMPLE 5(a) Example 5.1

II – 5.13

Chapter 5: Laplace Transforms

(i)

t

f (t )

(ii)

for t for

, ,

e k t, ,

f (t )

t

for t a for t a

sin t ,

t

for

f t

(iii)

for t

,

(iv)

t,

f (t )

(i)

t for for t

,

L{f (t )}

e

st

f (t ) dt

L{f (t )}

e

st

odt e

t

t st

s

e

e

s ,

t

u (t )

st

s

t

for t >

f (t) = (

Thus

e

s

fo

t

dt

e st s

t

s

s

st

e

u (t)

L{f (t)} = e · L(t2), by the second shifting property.

s

e

s

a

(ii)

L{f (t )}

e

st

e kt dt

e

st

dt

a

e

s k t

s k

a

s k

{

e

a s k

}

Part II: Mathematics II

II – 5.14

Consider e kt { Thus

e k t,

ua (t )}

,

kt

f (t) = e

kt

= e kt

ua (t)

k(

a) + ka

kt

ak

L{f (t)} = L(e )

s k

L{f (t )}

st

e

st

s sin t

te

st

e

st

t{ Thus

f (t)

u (t )} ( (

s

e

e

5e

st

e st s

s s

s

dt

s s

s

e

u (t )

cos t

s

e

t

}.

sin t dt

s

L{f (t )}

L{e kt } , by the second shifting property

as

a s k

e

s

(iv)

· ua(t)}

s k

{

e

a)

a s k

e

s k

· ua (t)

· L{e k (

e ak e

s k

(iii)

t a for t a

for

dt.

5

e

st

s

s

e

s

s

e

s

s

s

t, ,

5) u (t) u (t) + u (t)

for t for t

II – 5.15

Chapter 5: Laplace Transforms s

L (f (t )} L(t ) e s

L(t ) s

e

s

s

L{u (t )} e

s

.

Example 5.2 (i) (i)

t t

cos t . t

, (ii) sin t , (iii) L

t

L(t

t

L(t )

)

s

(ii)

sin t

s

s

s s

s

s

L sin t

!

t

!

!

t

5!

L(t )

!

n n

....

t

....

L (t )

5!

! s5

s

n

.

t t t

and

L(t ) .... ....

5! s

5 s s

s (iii)

!

t t

s

s

!

s

e

cos t t

s

!

!

t

t

t

!

!

!

t

.....

.....

5! ......

s

....

Part II: Mathematics II

II – 5.16

L

cos t

/

L(t

t

)

!

s

L t

! s

s

..... .....

s

! s

!

.....

s

!

s

e

s

.....

! s

s

s

L t

!

Example 5.3 (i) (t et (iii) e t cosh

t;

t (v) sinh sin (i)

t e t 5 sin t)e t}

L{(t = L(t

te t) ; (iv) cosh at cos at;

t)e

et

t)s !

s

s

5

s

s

s

s

6 s

s (ii)

L(l + te t) = L =L

te t L(t)s

s t

t e s

s

+ t e t) L(t )s

6 s iii

(s

(s

)

L(e t cosh t )

8 8 =

t

L{e L{e

8 s

(e

t

t

e e

s

s

6 )

L e

t

t

+ L(t )s

s

(s

)

t

e

e

t

t

e e

8t

}

s 8

e

t

6t

)}

s

II – 5.17

Chapter 5: Laplace Transforms

(iv)

L cosh at cos at

L cos at s s

ea t

L

s

cos at

L cos at

s a

a

s

a

a

s

at

e

s

a

a

(s

a) s

s as s

s s

(v)

L

as (s

a) s

a

a s

L sin

t

a

.

a

t L(sinh sin

a a

a

a

a

s as

s

a

a

s

s

t) et

e

L sin

t

t

sin

t s

s

s

s

s

s

s

s

s

s s

s

s

Example 5.4 (i) eat cos (bt + c); (ii) e t cos t; (iii) et sin t; (iv) e t t t; (v) e t t sin t. (i) L{eatcos (bt + c)}= L{cos (bt + c)}s = L{cos c cos bt c sin bt}s (s a ) cos c b sin c (s a ) b (s a ) b

s

a

s

a

as

Part II: Mathematics II

II – 5.18 (ii)

L{e t cos t}= L(cos t)s L

s +

cos 6t s

s

s s (iii)

s

6

L{et sin t} = L(sin t)s L

s

sin t

sin 6t s

s

6 s

s

s (iv)

L{e

t

t

s

s

t} = L L

s

s

s

t

7

t)s

sin 5t sin t

s

s

s

5 s

s 5 s

s (v)

L{e t

s

s

6

t sin t} = L L

s

t sin t)s

s s

cos t cos t s

s

s

s

s

s

s

s s

s

s s

s

s

8

Example 5.5 (i) (t · u (t); (ii) sin t · u (t); (iii) e · u (t) (i) L{(t u (t)} = e s · L{t }, by the second shifting property s (ii)

e

s

L{sint · u (t)} = L{sin (t = {sin (t

) · u (t)} ) · u (t)}

II – 5.19

Chapter 5: Laplace Transforms e

(iii)

s

L(sin t)

e s

s

L{e t u (t)} = L{e

t

e

6

s

e

u (t)} t

L {e

}=

s

e

.

s

Example 5.6 (i) t sin at; (ii) t cos at; (iii) te t e iat} (i) L (t sin at) = L{t L{t e iat} L(t)s I.

s

t

ia

of

I.

of

s ia

s

s

a

s

a

of

I.

s ia a

as

i s

a

I.

of

as s (ii)

a L{te iat}

L(t cos at

(iii) L(t e

t

s

a

s

a

t

L {t e

t

e i t} i t

L {t · e L(t)s

I.

}

(s +

of s

i s

I.

s

s

s s

i

of s

i

s

i

Part II: Mathematics II

II – 5.20 Example 5.7

L(sin t) and L(cos t), L

t).

L sin

s

L t sin t

and L t cost

s L t cos

s

s

L sin

t

L f

s t a

a L f (t )

s

as

, by the charge of sccale property

s s

L cos t

s s/

L co t

s/

L f at

a

L f t s

s

s a

, by the charge of scale property

.

s

s

L t sin t s L (t sin at )

a

L(at sin at ) s/ a

a a

s /a as

s

a

, by change of scale property

and

L(t sin at) and

s

t . a

(i) L sin t

t

II – 5.21

Chapter 5: Laplace Transforms s

L t cos t

s L t cos

t a

a L

t t cos a a as

a a

, by change of scale property

as a s

=

s

a

a

or

a s

s

.

a

Example 5.8 (i)

/

s (v)

s

L {e L

e

se

; (iii)

s

5

s

as

(s)} = L { (s)t

s

L

/

s

[ L { (s + e

a

; (iv)

e

a

L t

e

t

s

t

/

L { (s

t

/

L

t

t

/

t e

n

s /

t

t

e

t

u (t )

n n

/

/

tn

sn

/

s

s

ua(t)

/

e

s

u (t ) t

t

e

/

s

t

/

s

Now L

L

a

s

s

e

s

s

e

e s ; (ii)

n

e

bs

;

Part II: Mathematics II

II – 5.22

when t

, or

t

(ii) L

Now L

t

e

)

)

, when t

s

e ( s

L

)

( s

L

8

)

( s

u (t )

)

t

(s

t

)

t

8 8

s

Now L

!

e

se s (s )5

(iii) L

e

)

(s

)

5

(t

s

(iv) L

)5

e

s

( a s) e s a

u (t ) t

t

) )5 s5

s

t

!

e

) u (t )

s

e tL

(s

(t

)5

(s (s

L

e t

(s

e tL

L

)

t

t

s

L

e

s

t

s

e ( s

L

e L

s

t

!

t

e t( t

t )

(t

)

bs

)

{ (t

L

(t

a s

) } u (t )

s a

ub (t ) t

t b

II – 5.23

Chapter 5: Laplace Transforms a

Now L

s

s

a

L

a

s

a

at ( a s) e s a

L

(s s

(v) L

Now

bs

s

e

s s

L

b)

(s s

)

e

s

s

t

L

s

{cosh (t

b)]ub (t )

u (t t

t

L

a

cos a (t

s s

L

L

s

at

[ sin a (t )

s

L

s t

)} u (t )

sinh (t

)

Example 5.9 (i)

(v)

e s ) (s

(s (s s

(i) L

)

;

(ii)

e s (s

s

)

;

(iii)

e s (s

s

)

;

(iv)

e s ) (s

(s

A (s

L

)

(s

) (s

(s

)

) (s

)

, we resolve

;

u (t ) t

t

(s

B (s

By the usual procedure, we get A

s

s

.

) (s

fractions and then use the linearity property of L operator. A B Let (s ) (s ) s s

L

s

s

s

)e s

L

Then

e

s

L

s

5

s

, B

5

)

into partial

Part II: Mathematics II

II – 5.24

L

5

5

L

e s ) (s

(s

)

Now L

(e

t

e

t

e

(t

)

e

5 L

)

e s (s

s (s

)

s (s

)

Let Then

u (u

)

L

{(t

)

(t

)

u (t )

u (t )

)

u

u

L

s

t

t

s

s

t

) sin (t

} u (t )

s

s (s A(s +

L

)

s (s

A s

)

u (t )

)

t

Bs C s

and C

,B

s s (s

t

s (Bs + c) A

L

s

s

e s (s

(iii) L

L

)

s (s

=t

L

5

s

e s (s

(ii) L

s

)

L

s

(

s

cos t )

s

II – 5.25

Chapter 5: Laplace Transforms e s (s

L

s

Now

L

s

s

s

(s s

(v) L

L

e

s

L

t

L

e

t

e

t

(t

)

L

u (t )

s

(s

t

t

)

, by the shifting property

s

s

sin t

s

(s s

sin (t

s

)e s

e

t

=e

t

s

(s

t

t

) )

s

L

u (t )

s (s

L

s

) u (t )

s

L

s

Now L

s

s

e s

L

s

e

L

)} u (t )

s

e

L

(iv)

cos (t

{

)

, by the shifting property

s t

s

)e s

e

(t

)

) u (t ) .

cos (t

Example 5.10 (i)

(iv)

s s (s

s (s

s ) (s

) (s

)

; (ii)

)

s

; (v)

(s

s )

; (iii)

(s

s ) (s

s (s

) (s

) (s

)

.

)

;

Part II: Mathematics II

II – 5.26 s s (s

L

s ) (s

, we resolve the given function of s into partial

)

fractions and then use the linearity property of L operator. Let

s s (s

s ) (s

A s

)

B

C

s

s

, B

A

s s (s

L

s ) (s

and C

t

e

(ii) To resolve

s (s

s )

/ s

L

)

5

e

5

.

/ 5

/

s t

s

. x, so that

into partial fractions, we put s

s=x s (s

Then

s )

(x

(x

)

)

x x

x x

L

s (s

s )

x

x

x

(s

)

(s

(s

)

L

e e e

t

t

t

L t t

(s

) L

t

e

s !

t

t

!

t

t

)

(s

L

L

)

s

e

t

(s

L

s

)

II – 5.27

Chapter 5: Laplace Transforms e t( t

6

s

(iii) Let

(s

(s

;

B

t

u (u

e

t

e

t

C

D /

(s

L

u (u

s s

Thus

s L

s

s

sin t.

) (u

A u

)

B

s

C

u

u

, B / s

L t

(v) To resolve

/ )

A L

=s

into partial fractions and then replace u by s .

)

Now let

)

D

is a function of s , we put s = u, we resolve

s

) (u

Cs s

Cs + D) (s

L

) s

s s

(s

B(s

s

L

B

s

s

A

t )

A

) s

A(s

(iv) Since

t

8 /

/ s

s

sin t

8

and C

t.

s s

s

s

s

s

s

s

s

s into partial fractions as shown in (iv).

(u A u

) (u

B u

C u

) (u , say.

)

Part II: Mathematics II

II – 5.28 A

, B

and C

.

/ s

s

/ 5

s

s

s

s s

s

s

s

s

s

s

s

s

s

L

/

s

L

s

s

s

s L

s

s s L

cos t

cos t

cos t

Example 5.11 s

(i) (iii) (v)

(i)

s

s

s

a

s

s s

L

;

s

(ii)

;

s

s s

(iv)

5 7

s

s s

L

s

s 5 7

s 7

L s

7 s

7

7

7

L s

e

7

t

7

7

7 s

L s

7

7 7

s ;

s

7

s

s

;

s s

II – 5.29

Chapter 5: Laplace Transforms

7

t

cos t 7

s s s s (s s )

(ii) Let As(s

s

B(s

A

L

7

e

s s s

A s

s

B s

Cs s

D s

Cs + D)s

, B

s

sin t 7

s

,C

and D

L

L

s

s

s

s s

s

s

s

t

L

s

8

L

7

s

5 7 t

7

s

(iii)

Let

s

a

(s

s

a

s

et /

a) s A a

A(s + as + a ) + (s

A

s

cos

as

7

t

5 7

,B

7

sin

a

Bs C as a

a

and C

7

s

a)(Bs + C

a

7

s

et / L

t

s

a

.

t

Part II: Mathematics II

II – 5.30

L

s

a

L

a

s

e at

a

a

s

s

a

a

s

at

a

a

a

s

e at

a a

as

L

a

s

a

L

e

a

L a

s at

e at

a (iv)

s

s

s

s

s As

Let s

e

cos

at

s

s

s

B

sin

( s)

s Cs

D

s

s

s

s

(As + B) (s

s

Cs + D)(s

s

By the usual procedure, we get A

L

s

B

8

L

8 8

C s

s

L e

t

D

s

(s (s

) )

L

s s

8

L

s

s

s

L

(s (s

et L

s s

8

) )

at

II – 5.31

Chapter 5: Laplace Transforms t

e

8

sin t ) et (cos t

(cos t

et

et

t

e

sint

(sin t cosh t (v)

s s

s

e

t

cost

cos t sinh t )

s s

sin t )

s

s

s

s

s

s s Let

s s

s

s

As s

s B

Cs D s s

s

(As + B) (s s By the usual procedure, we get

L

s

s s

Cs + D) (s + s

B

A L

s

D

, C

s

s

L

s

s

.

s

L

L s

s

et / L

e

t/

s

s et / sin

L

e

nh t sin

t/

t

t

sin .

. Example 5.12 s

(i) If L s

t

sin t

s

L s

a

Part II: Mathematics II

II – 5.32

s

L

s

s L

(sin t

t os t

s

s a

a

L { (s/a)} = a L { (s)}t s

L

at

t

sin t

s

s/a

L

a

at

sin at

s a s

a L

i.e.,

s s

L s

a

t sin at

a

t sin at . a

a

(ii) By change of scale property, L{ f (t/a)} = a (as) L { (as )}= L ( (s )}t a s

L

t /a

t cosh t

s L

i.e.,

( s) [( s ) L

s

t ] t

cosh t

cosh t

s L

L

. s

(i) By change of scale property, L{ f (at )}=

s

L

t cosh t

s s

t cosh t .

Chapter 5: Laplace Transforms

L

(sin t

t os t

s

L

L s

s

i.e.

L

t

(sin t

t os t

(sin t

t os t

t

s

L s

EXERCISE 5(a) Part A

6. State the second shifting property in Laplace transformation.

L{ f (t)}, if f (t )

L{ f (t)}, if f (t )

e t , for t , for t , for

t

t , for , for t

t

L{ f (t)}, if f (t )

sin t , fo , for t

L{ f (t)}, if f (t )

cos t , fo t , for t

t

II – 5.33

Part II: Mathematics II

II – 5.34

,

t

for

L{ f (t)}, if f (t )= cos t

L{ f (t)}, if f (t )= t and L

L

sin t , fo t, for t

, for t t

t

L{(t u (t)} and L{u (t) sin L{t u2 (t)}

(t

at + b) t t cos t t t

t+ ) t t t t

et cosh t

in

t

s

e s

co

t

e e

)

s

se s

e s)/s s

s

e

s f (t) if L{ f (t s (s

s s

) s

s

s

s s

s (s + a )

s

s s

5

s

s

s

t

t sinh t

t cosh t e /s (a

t

e

e

6

s+

si

t

II – 5.35

Chapter 5: Laplace Transforms Part B e at (

et t e t sinh t

at )/ t

at cosh at co t et

et sin t) e kt sin ( t + et t t 56. t 58. t e t t

at sinh at

et t cos t 55. e t t t 57. te t cos t

t

s

L(t sin t )

t

L(t sin t)

s L(t cos t )

s s

L(t

t).

s

s

s

s

s

s

s

s 5

s

s

s

s

s

s

s s

65.

66. s

67.

s s

s

s

s

68.

s

s s

s

s

s s

ls s

as

s

s

s s

s

6 s s

s 75.

77.

s

s

76.

s s s

s

a

s

s

s

m bs c

s

Part II: Mathematics II

II – 5.36

78.

s

If L

t sinh t

s

L

s

s

L

(sin t

a

L

t cos t

. s

s s

L

t sin t 6

s

5.6

s

L s

LAPLACE TRANSFORM OF PERIODIC FUNCTIONS

f (t) is said to be a periodic function, if there exists a constant P that f (t + P) = f (t), for all values of t. Now f (t P) = f (t + P + P) = f (t + P) = f (t), for all t. In general, f (t + nP) = f (t), for all t, when n is an integer (positive or negative). P is called the period of the function. Unlike other functions whose Laplace transforms are expressed in terms of an t function f (t) with period P can be expressed in terms of the integral of e st f (t) over P), as established in the following theorem.

If f (t) is a piecewise continuous periodic function with period P, then P

L f (t )

e

st

e

Ps

f (t ) dt. .

Proof: L f (t )

e

st

f (t ) dt

e

st

f (t ) dt

P

e

st

f (t ) dt

P

t = x + P, e

st

f (t ) dt

e

s x

P

f (x

dt dx and the limits for x

.

P ) dx

P

e

sP

e

sx

f x dx

[ f(x + P) = f(x)]

II – 5.37

Chapter 5: Laplace Transforms

e

sP

e

sP

e

st

f (t ) dt , on changing the dummy variable x to t.

L f (t )

P

L f (t )

e

st

f (t ) dt

e

st

f (t ) dt

e

sP

L f (t )

P

e

(

Ps

) L f (t )

P

L f (t )

5.7

Ps

e

st

e

f (t ) dt

DERIVATIVES AND INTEGRALS OF TRANSFORMS

The following two theorems, in which we differentiate and integrate the transform function

s

L{f (t)} with respect to s

L{t f (t)} and L

t

f (t )

s L{tn f (t)} and L

If

tn

f t

, where n is a positive integer.

L{f (t)} =

s

then L{t f (t

' (s).

Proof : L{f (t)} = i.e.

s ...

e

st

f (t ) dt

(s ) s,

d ds

e

st

f (t ) dt

d (s ) ds

...

with respect to t and differentiation with respect to s d e ds

st

f (t ) dt

(s )

II – 5.38

Part II: Mathematics II

i.e.

te

st

f (t ) dt

(s )

i.e.

e st [t f (t )]dt

(s )

i.e.

L{t f (t )}

(s )

n times with respect to s, we get L{t n f (t )}

(

)n

dn ds n

(s )

or

(

)n

(n )

(s ) .

Note d (s ) ds d L{ f (t )} ds

L{t f (t )}

we ignore ‘t then we differentiate this function of s with respect to s Extending the above rule, L{t f (t )}

(

)

t’, s;

d L{ f (t )} and in general ds

dn L{ f (t )} . ds n . The above theorem can be stated in terms of the inverse Laplace operator as L{t n f (t )}

)n

(

If

L { (s)} = f (t),

then

L { (s)} = t f (t).

L { (s )}

t

L { (s )}

This rule is applied when the inverse transform of the derivative of the given function can be found out easily. In particular, the inverse transforms of functions of s that contain logarithmic functions and inverse tangent and cotangent functions can be found by the application of this rule.

II – 5.39

Chapter 5: Laplace Transforms

If L{ f (t )}

(s ) , then L

t

(s ) ds , provided lim t

f (t ) s

t

f (t ) exists.

Proof: L{ f (t)} =

i.e.

e

st

(s)

f (t ) dt

(s ) s between the limits s and

st

e

f (t ) dt ds

s

st

e

ds f (t ) dt

s

e

i.e.

t

(s ) ds s

st s

f (t ) dt

t i.e.

(s ) ds s

s s st

e

(

(s ) ds s

(s ) ds ,

) f (t ) dt s

assuming that s > i.e. i.e.

e

st

L

L

f (t ) dt t

s

f (t ) t

s

t

f (t )

(s ) ds (s ) ds

L

t t

f (t ) (s ) d s ds

s

s

(s )ds ds s

L

tn

s

(s ) (ds ) n

f (t ) s

s

s

, we have

Part II: Mathematics II

II – 5.40

Note

L

t

f (t )

L{ f (t )}d s s

t s and

we ignore

t integrate this function of s with respect to s between the limits s and Extending the above rule. We get; L

L

t tn

f (t )

,

.

L{ f (t )}ds ds and in general s

s

L{ f (t )}(ds ) n .

f (t ) s

s

s

. The above theorem can be stated in terms of the inverse Laplace operator as L { (s )}

If then

L

(s ) ds s

f (t ) ,

t

f (t ) .

L { (s )}

t L

(s ) d s s

This rule is applied when the inverse transform of the integral of the given function with respect to s between the limits s and can be found out easily. In particular, the inverse transforms of proper rational functions whose numerators s and denominators are squares of second degree expressions in s can be found by applying this rule WORKED EXAMPLE 5(b)

Example 5.1

f (t) f (t) = kt,

The graph of f (t P,

t< f (t) is

f (t )

k t P

t < P.

II – 5.41

Chapter 5: Laplace Transforms f (t) k

t Fig. 5.1

By the formula for the Laplace transform of a periodic function f (t) with period P, P

L{f (t )}

L{f (t )}

k e

dt e st s

s s

e s

e s s

s s

e

k e k s

st

st

e

t

s

k e

f (t ) dt

kt e

s

e

st

e

Ps

e

s

s s

e s

s s

ke

e s)

s(

Example 5.2

f (t) f (t) = k in

t

k in a and f (t

f (t

a t

a

a) = f (t) for all t.

a) = f (t) means that f (t

a. The graph of the function

f (t) with period P, P

L{f (t )} =

e

Ps

e

st

f (t ) dt

Part II: Mathematics II

II – 5.42 f (t) k

t O

a

a

a

a

–k

Fig. 5.2

a

L{f (t )}

as

e k e s((

=

k s

dt

k ) e st dt

( a

e as

a

st

e

s

k e k

s

a st

ke

as

s e

)

e as ) e as ) ( + e as

as

as

a

e

as

e

as

]

)

as /

e ) e as )

a

st

k (e s (e as /

e e

as / as

) )

k as tanh s f (t) whose

Example 5.3 y = [= f (t)] A

a

B t O

a

a

a

a

5a

Fig. 5.3

f (t) f (t) and AB.

a. t

a,

OA

II – 5.43

Chapter 5: Laplace Transforms OA Equation of OA is y = t

t

a

AB passes through the point B a, Equation of AB is y

=

t

y= a

or

a)

t in a

t

a.

f(t) [= y] can be taken as f(t) = t,

t

a

a

t, in a

t

a

f(t + a) = f(a).

and

a

Now L{f (t )} =

e

st

e

as

f (t ) dt

a as

e

dt

( a

st

t )e

dt

a

e

as

e

as

e s (

st

e

t

s a e s

as

as

e as s

as

as

e e

e ) e as )

tanh

a

e st s

( a

s

a s/

e ea s /

s

t)

as

e

e

a e s

s

e st s

st

s as

e as s

as

( s (

)

as

( s ( s

a st

te

e ) )( e

e

as

e e

a s/

as

)

a s/

as

Example 5.4 f (t) f (t)

a /

/

/ Fig. 5.4

/

t

a

a

Part II: Mathematics II

II – 5.44

graph, it is obvious that f(t

. The

graph of f (t)

, a and

f (t) is given by t = in t

f (t) = a sin and

f t

Now

L { f (t )}

t /

f (t ) .

e

s

e a e a e

e s

st

( s sin t

s

a (s

f (t ) dt

e st sin t dt

s

a )( e

(s

st

e

e

)

s

s

e

)

s

cos t

a s

(s

)

e

s

Example 5.5 f (t), We note that

f (t) = |sin ) = |sin

f (t +

t|, t (t + / )| t|

= |sin = f (t)

f (t) is periodic with period f (t) is always positive. The graph of f (t) is the sine curve as shown in

/

/

/ Fig. 5.5

t

II – 5.45

Chapter 5: Laplace Transforms

Now

L { f (t )}

s

e

st

e

| sin t | dt

s

e

st

e

sin t dt [ sin t

e

st

s

e (s

( s sin t

s

e e

s

s

e

)( s

Example 5.6 (i) t cosh t; (ii) t

s

,

s

s

coth

s

e e

s

e e

s

t cos t; (iii) t sin t; (iv) (t sin at) . et

(i) L{t cosh t} L t

8

e

L {t (e

et

d L (e 8 ds d 8 ds s 8 (s

t

t

)

t

e

et

t

e

s (s

e

t

t

e

)

(s

Note got the same result. (ii) L(t cos t cos t )

L

t

(cos t

cos t )

d L (cos t ds d s ds s

t

s )

cos t ) s s

]

cos t )

s

e

)

t

in

)

s )

(s

)

s s

Part II: Mathematics II

II – 5.46

s

s (s

s (s (iii)

L(t sin t )

L

s

s (s

) s (s

) sin t

)

)

sin t

d {L ( sin t sin t )} ds d ds s s s

s

(s

)

(s

)

(s

)

(s

)

cos at

L {(t sin at ) } L t

(iv)

(

)

d ds d ds

s

d {L( cos at )} ds s s s a s (s

s (s a )

s s s

a a ) s (s (s

a ) a )

s( a s ) (s a ) s( a s ) (s a )

Example 5.7 Use Laplace transforms to evaluate the following; (i)

(i)

t

te

e

sin t dt ; st

(t sin t ) dt

(ii)

L(t sin t )

te

t

cos t dt

(ss

a ) s

Chapter 5: Laplace Transforms d L (sin t ) ds d ds s

L (t sin t )

Now

6s (s

te

st

te

t

sin t t

)

6s (s

)

,s

s

(ii)

st

e

Now

.

sin t dt

(t cos t ) dt

L(t cos t ) d L(cos t ) ds

L(t cos t )

d s ds s s (s

te

st

cos t t

)

s (s

)

,s

s te

t

cos t dt

5

.

Example 5.8 (i) te t (i) L {te

t; (ii) t cosh t cos t; (iii) te t sin t} [ L(t sin t )] s s

t

t (iv) t e cos t.

II – 5.47

Part II: Mathematics II

II – 5.48 Now

d L(sin t ) ds

L(t sin t )

d ds s 6s (s

L {te

t

)

6s

sin t}

(s

)

s

(s (s

s

) s

5)

Note

The same problem has been solved by using an alternative method in Worked Example (6) in Section 5(a). t (ii) L {t cosh t cos t} L (et e t ) cos t [ L(t cos t ) s L (t cos t )

Now

)

(s (s

L (t cosh t cos t )

) s

s

Now

)

s

t}= L{t

)

(s

) t}s

6s (s

) s

s

d L (sinh t ) ds d ds s

L (t sinh t )

(s (s

s

(s t

]

s

d s ds s s (s

(iii) L{t e

L (t cos t ) s

s

s+

) s

s

)

II – 5.49

Chapter 5: Laplace Transforms L {te

t

L {te

t

sinh t}

(s {( s

)

t

(e

L { tet

(s

)

(s

s ) (s

L (t cos t ) (

)

d ds

e

5t

te

(iv) L{t e t cos t} = [L (t cos t)]s Now

(s

}

t

sinh t} L te

t

)

}

( s 5) )

(s ) ) (s )

(s

s+

d L (cos t ) ds s s s

d d s (s (s (s

) s) )

{( s ) (s (s s )

L {t e t cos t}

(s ) ) (s )

)

)}

Example 5.9 a ; s

(i) lo

(iii) log

s s( s

)

(i) L { ( s )}

L log

a s

(ii) log

s s

(iv) s log

;

t

L {

L log

a b s s

( s )} s a s

t

L

d s a log ds s

;

k (k is a constant)

Part II: Mathematics II

II – 5.50

t t t

L log

s s

a b

t t t

L log

s s (s

t t t

L

s log

s s

k

L

s a

(e at

L

)

s t

e at )

(

d [log(s ds

a ) log(s

s

L

s

a

b )]

s b

s

(cos bt cos at )

t

)

d log (s a ) log s ds

L

L

t t t

s

s

( cos t e

L L L L

) log s log(s

)]]

s

L

(

t

d [log (s ds

t

e

s t

cos t )

d [s log(s ds s

) s log (s

log(s

s s

)

s

s

s

t

s s

t

t

s s

)] L (k ) log(s

L [log(s L

s

)

k (t )

) log(s

s

k (t )

)] k (t )

II – 5.51

Chapter 5: Laplace Transforms

Note s

[L

s

and L

s

do not exist, as

s

s

s

s

s

s

and

s

as

s

s

are improper rational

s , which is a proper

s

rational function.

t t

cosh t

sinh t

(et

t t

e t)

cosh t

k (t )

k (t )

Example 5.10

(iii) cot (i)

L { (s )}

t

(as )}

L {

t t

t

L

tan

s

d cot (as ) ds

L

a a s

b

a s

L

a t

t

/a ( / a)

s

L

t s

d tan ds

a b

/b

L

s

a b

t

s

(s )}

L

a L t

a b

(iv) tan

;

s

L {cot

(ii)

s

(ii) tan

(i) cot (as);

L

(s

b a)

b

sin

t . a

;

Part II: Mathematics II

II – 5.52

t (iii)

L

cot

e

at

sin bt .

s

t

d cot ds

L

t

L

(s (s

t

.

s

L

(s

)

)

)

t

t (iv)

L

tan

e sin t.

t

s

t

d tan ds

L

s

.

L

s

s L

t

s s

s

Consider

s

s

s

s

(s

)

(s

s s

( s) s ) (s s

s

)

, by resolving into partial fractions

(s L

s

et sin t

s

(s

)

)

e t sin t

sin t sinh t

L

tan

s

t

sin t sinh t

II – 5.53

Chapter 5: Laplace Transforms Example 5.11 (i)

(v)

(i)

at

sinh t e ; (ii) t .

L

f (t ) t

s

sinh t t

s

; (iii)

L (sinh t ) ds ds

s log log

log

s s

s

s s

log

e

at

s

s s

log

s s

s

log

ebt t

log s

s

L

cosbt ; (iv) t

L {f (t )}ds

s

(ii)

e at

bt

t

sin t t

L

e

s s

log

L (e

at

e

bt

s

a

s b

s s

) ds

s

s

log

log

s a s b a s b s

ds log

s a s b

log

s b s a

s

s

sin t sin t ; t

Part II: Mathematics II

II – 5.54

lo

s b s a

= log

(iii)

L

e at cosbt t

L (e at

cos bt ) ds

s

s s

s b s a

log

s a

s

ds

b

log ( s a )

log ( s

b ) s

s a

log

log

b

s

b /s

sin t sin t t

s b s a

s

s b (s a )

log

(IV) L

log

s b (s a )

log

lo

b

s

s

a /s

log

s a

L( sin t sin t ) ds, by rul

)

s

L( cos t

cos t ) ds

s

s s

s

s log

log

s s s s s

ds

s

log s

s s

II – 5.55

Chapter 5: Laplace Transforms

s /s

log

log

s s

s

log s s

log

(v)

L

L

f (t ) t

s

s

sin t t

s

s

s s

log

L{ f (t )}ds ds

L (sin t ) ds ds cos t

L s

ds ds

s

s s

s

s

ds ds

s s

log

ds,

s

s

b

by putting a s

s log

s

s

s

s

s s

s

ds

by integrating by parts. s

s

L

L

log

s

s

log

s s

s

log

L

log

s

, say s

s

cot

s s/

Now

s

cot

s

s

log

s

s

8 s

s

s

ds

Part II: Mathematics II

II – 5.56

s /

log

s

s

s /

lim ( L)

log lim

s

s

s

log (e )

L

sin t t

s

log

s

log

s

s

cot

s

.

Example 5.12 e t sin t

(i)

dt ;

e t sin t

t

dt

L

t

t t

t

dt s

s

L(sin

t ) ds

s

s

s

(

)

ds s

cot

s

cot

s

e sin t

t

dt

cot

s

.

t

e t

t

si

t

sin

e

(iv) sin

st

e

L

Now

sin t dt ; t et

(ii)

cos at cos bt dt ; t

(iii)

(i)

t

t

dt .

II – 5.57

Chapter 5: Laplace Transforms

(ii)

sin t dt t et

e e

L

Now

sin t t

L

sin t dt t

t

sin t dt t

st

sin t t

s

s

cos t t

L

L(

cos t ) ds

s

s s

s

ds

s s

log s log

s

s

log

log

sin t dt t et

(iii)

cos at

cos bt t

/s log

s

s

s s

s

e L

s

log

s

log 5

dt

s

s

log

s

log

s

st

log 5

cos at

cos bt t

cos at

s

cos bt t

dt

s

Part II: Mathematics II

II – 5.58

Now

L

cos at cosbt t

L(cos at cos bt ) ds s

s s

s

a

log

log

cos at cos bt dt t e

(iv)

t

log

s s

log

b a

dt

e

L

L

e

t

e

e

log

t

dt

s

. e

t

e

t

dtt

t t

e

s

t

s

t

e

t

) ds

s

s

log

t

log

b a

st

L (e

t

e

a b

s

t

log

t

s s

t

s

e

a b

ds

b

s

s s

t

e t

Now

s

log

s s s s s s s

ds

s

log s

s s

s s

a b

II – 5.59

Chapter 5: Laplace Transforms Example 5.13 (i) (iii) (v)

s (s

;

a ) (s

(s

) s

)

s

)

s

(ii) ;

(s

)

;

s

(iv)

(s

s

)

;

s (s

(i)

L { (s )} = t L

(s ) ds s

L

s (s

s

=t L

a )

(s

s

t t

a

L

x

L

ds

dx , on putting s + a = x x

tL s

a )

s

s

a

a

t sin at. a (ii)

L

s (s

s

t L

)

(s

s

t

t

L

ds

)

, as in (i) above.

s

sinh t

Note

The inverse transform in this case can also found out by resolving the given function into partial fractions. (iii)

L

(s (s

) s

= et L

)

s s

(s

s

L

)

) +

Part II: Mathematics II

II – 5.60 et

t

sin t , as in problem (i)

= tet s

(iv) L

(s

s

L

L

(v)

e

t

=e

t

s

s

(s

)

sin t

s t

e

t

L t

t

) ( s s 5)

)

s (s

s

)

s {(s

) +

sin t , as in problem (i)

t) sin t. L

)

e

L

e

s (s

(s + s (s

L

)

L

t.

L

[(s

s )

]

s (s

, )

t sinh t , proceeding as in problem (ii) 6

t t e sinh t 6

EXERCISE 5(b) Part A

f(t) = t

t

f (t f(t) f(t) and t · f(t). (s) and

5. State the relation between the Laplace transforms of f(t) and

t 6. State the relation between the inverse Laplace transform of integral.

7.

a

t sin at

8. t cos at

kt

(s).

f (t ) .

kt cos kt

(s) and its

II – 5.61

Chapter 5: Laplace Transforms t a

kt + kt cos kt cos kt

log

s s

log

s s

cos at )

kt sin kt.

s

log a b

s s

log

a s et

t

e t

sin at t

s s

log

s

tan

s

a s

lo

t

sin t t

cos at t Part B f (t )

E, i , in

given that

f t

t

E

f (t ) = E

t

= , in T/ given that f (t + T) = f (t)

f (t ) = sin f (t) = |cos

n

f t

n

f (t) = et

t

E

t t

and

bt / a

kt a

m l

b a

a kt , sinh k a

k. m l

b t , if b a

ac = .

t.

cos t

sin

t.

II – 5.113

Chapter 5: Laplace Transforms t

e t

e

t

t).

et / cos

(sin at cosh at

a

t

t .

sin

cos at sinh at).

sin t sinh t t

75. 76. cos 78.

a

t

cosh

t t

.

t

t). t cosh

sin

77.

(sin t t cos t ) .

t sinh at

t sin t .

s

s e s

a

s

a

8.

s

.

s

7. s

k s

k

s

a

s s

a

ht

s s

e

t

bt

at

e

co

sin at t

at

e t

.

et

t t

.

t

k e t

. k

ks s

. a

.

t t

cos t

sin t . t (s/a).

.

t

.

Part II: Mathematics II

II – 5.114 s

log

s s

log

(s

e

s

e

s

).

e

s

s

s

(s

s

s

s

s

s s

e

s

(s

)

/

e

(s

)

) (e

s/

s

.

s

s

s

s

s s

s s 6s s

s 7 5)

(s s

s

)

s

(s

s

)

.

s (s

(s

s (s

.

s

s

s

s

e

)

s

(s

s

e

cosech

.

s )

s

/(s +

s

8 (s

s

s

e tanh

s

.

s

s e

s

s

s

e

]/(s

coth ( s)

s

s

E sT tanh . s

s/n

e

.

s

log

s/E

e

s

a s

E

s

log

s s

) .

) )

(s

.

5)

.

s

.

II – 5.115

Chapter 5: Laplace Transforms

t t

cos a t )

( (e

bt

cos a t ). a cosh at t

sinh a t

t t t

t

e

t a

(t )

t. t

sin

t

sinh

t

sin

t

e t ).

(cos t

bt

e

. sin at .

t

55. l

57.

b . a s log s

58. log

.

8 log log

65. 67.

t t

e

s

a

log

s s s

e t sin t .

68.

s

s.

cot

t sinh at . a a t

t) sin t.

t

e at (a sin b t

s t

b cos b t )

sin at t sin a t. e

t

sinh t..

s

t

s s b

.

s s

66.

8. No; * t

s

s log

sin t .

t

.

a

s t) e

t

a t

et

e

t

t

e t.

.

Part II: Mathematics II

II – 5.116 s.

cot

s

)

(

e

at

).

at

e

s

)

t

)

.

+t

s

a

s

a

a b

e

a

b

a

b

a

s

; s

s bt

at

e

t

);

t + e t.

b

sin b t

a s

a b

e t ; et

a

log

be

t

e

bt

).

e t.

sin a t ;

a

t

t ); (cos t

b

(cos b t

t t ); e

t

a t cosh a t ); (a t sinh a t (sin t cosh t

)

cos a t );

t

t.

cosh a t );

s

s s s

s s

6

t e a

at

sinnh at.

cos t sinh t ); cos t cosh t.

.

s

cot

at

(a e

b sin bt ).

(s

.

a

(a sin at t

cos a t ).

a

;

a

sin t sinh t;

s

s

as

(sinh a t

s s

. in t (

a

).

as

; a

;

(e

et

(cosh a t

a

t

as

.

s

t et

e t+t

a

s

;

e

s

s

.

t. (et

log

s

s

s (s

s (s

a

s

log

s

cot

s cot s

s

.

.

s

s

s s

s

.

II – 5.117

Chapter 5: Laplace Transforms e t.

t t

e

e

t

t t

e

57.

(sin t

.

t

(

t )].

t

e

t cos t .

56.

t cos t ).

58.

( tc

(cos t

e t ).

t

sin t ). t

t ).

sinh a t ) .

66.

t cos t ).

a

(sin a t

sinh a t ).

(sin a t

a t cos a t ).

t

e

t

t.

t

(t

t

t

)

(sin t cosh t

.

x = tet 5. y t)e t. t 8. x = e

t

y x=t 5

e e

e

t

cos t sinh t ).

y = et t 6. x t. y t. y

f (t) = cos t + sin t t

y x

c

(a t cosh a t

a

a

x=e t t + sin t y e t. 7. x = cos t.

x

t

t ).

cos 5 t )

(sin t

67.

(

c

a t sinh a t ).

t

sin t t

5

t

(

cosh a t

a

e

t

t

55.

65.

t

e

[

7

t

e

t

e e

t

cos t

et

t

y

t

cos t

sin t .

cos t sin t

t

( e

t. (sin t

cos t )

t

te

t

t

t . 6

e t)

Part II: Mathematics II

II – 5.118 y x

t cosh t t

x

t

t sinh t).

et

y

t

y

t

5

sin t

(t

sin t

t

cos t .

) sin t .

t

x

A cos a t

y x

B sin a t

sin a u f (t

u ) d u.

t + cosh t t

e

t et

x

(et + in t

y

( e

t

x

+e

t

sin t

x (t )

t

y (t) =

cos t

cosh a t ).

t sin t ).

sinh sin t. + e t; y = t y = t + cos t. t; y = sin t y = e t + e t.

t

t

e .

t. y

e

t

e

cos t}. 7

cos t

x t

(cos a t

cos t t cos t );

) e t.

t) e

{(

y e

t

y

et

sin t

x = cosh t; y x t t x = t + sin t; x = sin t x = t e t e t;

x

a

sin t

t sin t

t cos t .

y (t) = e t

8

t

t

t) .

t

(cos t

sin t ).

Appendix

A

Partial Differentiation A.1

INTRODUCTION

In many situations we come across a quantity whose value depends on the values of more than one variable. For example, (i) the volume of a right circular cylinder is a function of the base radius and height; (ii) the volume of a cuboid depends on its length, breadth and height. If the value of the variable quantity u depends on the values of several other variable quantities x, y, z,…, we say that u is a function of x, y, z,… and it is denoted as u = f(x, y, z,…); x, y, z,… are called independent variables and u is called the dependent variable.

A.2

PARTIAL DERIVATIVES

Let z = f(x, y z be the corresponding increment in z. given to x Then z + z = f(x + x, y)

Note

x be a small increment

(We do not make any change in y, viz., y is kept constant) z = f(x

Then lim x

0

z x

lim x

f (x

x, y

x, y ) x

0

f(x, y)

f ( x, y )

is called the partial derivative of z with respect to x and denoted as exists)

z (if the limit x

Smilarly, z y

lim y

0

z y

lim y

0

f ( x, y

y) y

f ( x, y )

z with respect to x is similar to that of x, but with the only difference that we treat the other independent variables as constants temporarily.

Part II: Mathematics II

II – A.2

Note

All elementary rules of ordinary differentiation hold good for partial differentiation too. For example, if u and v are both functions of x and y, then x

(uv)

u

v x

v

u x

Partial derivatives of higher order u u will be, in general, functions and x y of x and y. Hence each of them may be further differentiated partially with respect to x or y, giving partial derivatives of the second order.

When u = f(x, y), the partial derivatives

Thus

2

u x

x

u is a second order partial derivative of u x2

Similarly

and

2

y

u x

u y x

x

u y

u x y 2

y

u y

2

u y2 2

Note

u y x

The ways by which the two mixed second order partial derivatives 2

and

u x y

are different, but they are equal when they are continuous. In 2

almost all situations we may assume that

u y x

2

u . x y

Further partial differentiations will lead to third and higher order partial derivatives.

Alternative Notations 2 u u u and are denoted also as ux and uy· y x y x indicating that ux 2

u x y

x

u is denoted by uyx. y

y

u x

is denoted by uxy, y.

Appendix A 2

u and x2

A.3

II – A.3

2

u are denoted by uxx and uyy. y2

HOMOGENEOUS FUNCTIONS

When a function of x and y can be rewritten as x n f

y , then the function is called x

a homogeous function of the nth degree in x and y. The homogeneous polynomial in x, y of the nth degree, viz., a0xn + a1xn y + a2xn 2 y + … + an xyn + anyn, with which the students are familiar, may be considered a homogeneous function of the nth degree in x and y, as a0xn + a1xn y + … + an xyn + anyn x n a0 xn f

A.4

y x

a1

y x

a2

2

an

1

y x

n 1

an

y x

n

y x

EULER’S THEOREM ON HOMOGENEOUS FUNCTIONS

If u is a homogeneous function of degree n in x and y, then x

u x

u y

y

nu

Proof Since u is a homogeneous function of the nth degree in x, y, we may assume that y x Differentiating (1) partially with respect to x,

x

u

xn f

u x

xn f

u x

y x

(1)

y x2 y x

xn 1 y f

nx n 1 f nx n f

y x

y x (2)

Differentiating (1) partially with respect to y,

y

u y

xn f

u y

xn 1 y f

y x

1 x y x

(3)

Part II: Mathematics II

II – A.4 Adding (2) and (3), we have u x

x

Note

u y

y

y x nu , by (1) nx n f

If u is a function of several variables x, y, z,…, such that u = xn

y z , then u is said to be a homogeneous function of the nth degree in x, y, , , x x u u u z,…. In this case, Euler's theorem will be x y z nu. x y x f

A.5

EULER’S THEOREM FOR SECOND DERIVATIVES

If u is a homogeneous function of degree n in x and y, then 2

2

u x2

x2

2

u x y

2 xy

y2

u y2

n(n 1)u .

Proof By Euler’s theorem for x

u x

u y

y

nu

(1)

Differentiating (1) partially with respect to x, 2

x

u x2

2

u x

y

u x2

y

u x y

2

x

i.e.,

n

u x

2

u x y

(n 1)

u x

(2)

u y

(3)

Differentiating (1) partially with respect to y, 2

x

2

u y x

u y2

y

u y

2

i.e.,

x

n

u y

2

u y x

y

u y2

y

u y

(n 1)

(2) × x + (3) × y gives 2

x2

u x2

2

2 xy

u x y

2

(n 1) x = n(n

u y2

y2 u x

u, by Euler’s theorem

Appendix A

II – A.5

WORKED EXAMPLE u Example 1 If u = log (tan x + tan y + tan z), prove that sin 2 x x u sin 2 z 2. z

sin 2 y

u y

u = log (tan x + tan y + tan z)

sin 2 x

u x

sec 2 x tan x

u x

2 sin x cos x sec 2 x tan x 2 tan x tan x

(1)

Similarly,

sin 2 y

u y

2 tan y tan x

(2)

and

sin 2 z

u z

2 tan z tan x

(3)

Adding (1), (2) and (3), we get sin 2 x

u x

sin 2 y

u y

sin 2 z

2(tan x

u z

tan y tan z ) tan x

=2 Example 2 If u = log (x3 + y3 + z3 show that

(i)

u x

u y

u z

3xyz),

x

3 y

z

2

(ii)

x

y

u

z

(x

9 y z)2

u = log (x3 + y3 + z3 u x Similarly

u y

x3 x

3

xyz)

3( x 2 yz ) y 3 z 3 3 xyz 3( y 2 zx) y 3 z 3 3 xyz

Part II: Mathematics II

II – A.6 u z

and

x

3( z 2 xy ) y 3 z 3 3 xyz

3

Adding, we get u x

i.e.,

u y

x

3( x 2

u z

y

x3

u

z

y2

z2 y3

xy yz zx) z 3 3 xyz

(

the denominator

x

3 y

z

(x

y

z ){ x 2

x

3 y

z

xy}) (1)

2

x

y

z

u

x (x (x

Example 3 If x = r cos (i)

r x

(iv)

and

x

2

2

2

x2

y2

y

(x

3 y z )2

(x

3 y z )2

1 x r

2

(log r )

, by (1)

z

prove that

x

2 2

x

3 y z )2 9 y z )2

(ii) r

(log r )

z

y = r sin

x r 2

(iii)

y

3 y

cos 2 r2

x y

0. x = r cos y = r sin

(1)

r2 = x2 + y2

(3)

tan

and

(2)

1

y x

(4)

Differentiating (3) partially with respect to x, we have r 2r 2x x r x

x r

r cos r

cos

Appendix A

II – A.7

x cos r r x , x r which is the required result (i) Differentiating (4) partially w.r.t. x, y x2

1 x

y2 x2

1

r sin r2 i.e.,

r

x

y x2

y2

1 sin r (5)

sin

Differentiating (1) partially w.r.t. x 1 x r

From (5) and (6), r

x

r sin sin

(6)

1 x , which is required in (ii) r

From (3), 2 log r = log (x2 + y2) x

(log r )

1 1 2x 2 2 x y2

(log r )

( x2 y 2 ) 2x2 ( x 2 y 2 )2

2

x2

y 2 x2 ( x 2 y 2 )2

x x

2

y2

r 2 (sin 2

cos 2 ) r

4

1 cos 2 r2 Now

y

(log r )

1 1 2y 2 x2 y 2

(log r )

( x2 y 2 ) 2 y 2 ( x 2 y 2 )2

2

y2

(7) y x2 x2 ( x2

y2 y2 y 2 )2

Part II: Mathematics II

II – A.8

From (4), we get

1 cos 2 r2 1 1 2 y x 1 x2

y

2

(8) x x

2

y2

( x2 y 2 ) 2x2 ( x 2 y 2 )2

x y

1 cos 2 r2

(9)

From (7), (8), (9), the required result (iii) follows. y x2

1

From (4), we have

2

x

y x2

1

y x2

y2

2

1

y

y2

( x2

y 2 )2

2x

2 xy y 2 )2

(10)

2 xy (x y 2 )2

(11)

( x2

From (4) again, we have x y

x

2

y2

2

y

1

x

2

(x

2

2 2

y )

2y

2

From (10) and (11), the required result (iv) follows Example 4 If u = (x2 + y2 + z2) 2

2

u x2

, prove that

2

u y2

u z2

0

x2 + y2 + z2 = r2,

Putting we get

1 r

u u x 2

u x2

1 r2

(1)

r x

1 x , from (1) r2 r

r 3 1 x 3r 2 r6

r x

Appendix A r3

II – A.9

x r , from (1)

3 xr 2 r6

3x 2 r 2 r5 Since u is a symmetric function in x, y, z, 2

we get 2

2

u x2

u y2

u y2

3y2 r 2 r5

2

3( x 2

u z2

y2

Example 5 If u2 = (x u x2

2

2

u y2

u z2

u x u x

i.e.,

2

u x2

b)2 + (z

c)2, prove that

2 u

u2 = (x 2u

0 , by (1)

5

a)2 + (y

3z 2 r 2 r5

5

3r 2 r

u z2

z 2 ) 3r 2 r

3r 2

2

2

and

a)2 + (y

2( x x

b)2 + (z

(1)

a) a

(2)

u u

c)2

(x

u x

a)

u2 u

x

2

u u

u2

a 2

, by (2)

( x a)2 u3

(3)

Similarly, by symmetry, we have 2

and

u y2

u2

( y b) 2 u3

(4)

2

u2

( z c) 2 u3

(5)

u z2

Part II: Mathematics II

II – A.10

Adding (3), (4) and (5), we get 2

3u 2

u x2

{( x

a)2

b) 2

(y u3

(z

c) 2 }

3u 2 u 2 , by (1) u3 2 . u xz

Example 6 If V

2

x

2

2

V x2

y2

, prove that V

2

V y2

V z2 V

0 xz x2

V x

z

V x2

z

y2 ( x2 y 2 ) 2 x2 ( x 2 y 2 )2

2

( x2

z( y 2 x2 ) ( x 2 y 2 )2

y 2 ) 2 ( 2 x ) ( y 2 x 2 ) 2( x 2 ( x 2 y 2 )4

2 xz (3 y 2 x 2 ) ( x 2 y 2 )3

Note V y

(1)

V is not symmetric in x, y, z

xz

(x

2

2

V y2

2 xz

1 2y y 2 )2 ( x2

2 xyz (x y 2 )2 2

y 2 ) 2 1 y 2( x 2 ( x 2 y 2 )4

y2 ) 2 y

2 xz ( x 2 3 y 2 ) ( x 2 y 2 )3 Now

y2 ) 2x

(2) V z

x x

2

2

y

2

and

Adding (1), (2), (3), the required result follows.

V z2

0

(3)

Appendix A

II – A.11 2

z when x = y = z = 1. x y

Example 7 If xx · yy · zz xx · yy · zz = 1 Taking logarithms, we have x log x z log z

y log y

(1)

Differentiating (1) partially w.r.t. y, z

1 z z y

i.e.,

log z

z y

(1 log z )

z y

1 y

y

log y

(1 log y )

(2)

Differentiating (2) partially w.r.t. x, 2

z x y

(1 log z )

z 1 y z

z x

0

1 z z z x y (1 log z )

2

z x y

Differentiating (1) partially w.r.t. x, we can get (1 log z )

(3) z x

(1 log x)

(4)

Using (2) and (4) in (3), we have 2

z x y

(1 log x)(1 log y ) z (1 log z )3

2

z x y

x y z 1

1

xy + y2)

Example 8 If f (x, y x

11 1 13

f x

(1 x 2 )

y

y2

f y

, show that 0

xy + y2)

f (x, y f x

1 (1 2 xy 2

(1) y2 )

3 2

y (1 2 xy y 2 )3 2

x

(1 x 2 )

f x

(1 x 2 ) y x (1 2 xy y 2 )3 2

( 2 y)

Part II: Mathematics II

II – A.12

3 (1 2 xy 2 y 2 )3

y 2 )3 2 ( 2 x) (1 x 2 )

(1 2 xy y

(1 2 xy

y

2 x(1 2 xy y 2 ) 3(1 x 2 ) y (1 2 xy y 2 )5 2

y

2 x 3 y x 2 y 2 xy 2 (1 2 xy y 2 )5 2

y 2 )1 2 ( 2 y )

(2)

Differentiating (1) partially w.r.t. y, f y y2

f y

1 (1 2 xy 2 y2 (x

y )(1 2 xy

y

y

3 2

( 2x

y2 )

2 y)

3 2

y 2 )3 2 (2 xy

3y2 ) 3 ( y 2 x y 3 ) (1 2 xy 2 (1 2 xy y 2 )3

(1 2 xy 2

y2 )

f y

y 2 )(2 xy 3 y 2 ) 3( y 2 x (1 2 xy y 2 )5 2

(1 2 xy

y{ 2 x 3 y (1 2 xy

y 2 )1 2 (2 y

y 3 )( x

2 x)

y)

x 2 y 2 xy 2 } y 2 )5 2

(3)

Adding (2) and (3), the required result follows: 2

Example 9 Verify that

when

u , x y

y x

y 2 tan

1

y x

y 2 tan

1

u

x 2 tan

1

u

x 2 tan

1

u x

x2

1 2

1

2

u y x

y x2

y x2

x . y x y

2 x tan

(1)

1

y x

1

y2

2

1

x y2

1 y

Appendix A x2 y x y2

2 x tan

2

y 2

u y x

1 y2 x x2

2x2 x2 y 2

x2 x2

1 1

x

2

y2 (2)

1

1 2x

y3

y x

1

y x

1

2 x tan

II – A.13

y2 y2

(3)

Differentiating (1) partially w.r.t. y,

u y

1

x2

1 x

2

y x2

1 x3 x

2

x

2

x

2 y tan

1

y

2

u x y

y

2 y tan

1

1

x y

x y

1 y

x y2

2 y2 x2 y 2

2

x y2

2 y tan

x y

2

1 1

2

1

1 2y

2

1

xy 2

x y2

1

y2

x2 x2

y2 y2

(4)

From (3) and (4), we see that 2

u y x

2

u x y 2

Example 10 Verify that

V y x

V = xy · yx V = xy · yx Taking logarithms, log V = y log x + x log y

2

V , x y

when

(1)

Part II: Mathematics II

II – A.14

1 V V y

1 V y V x x Differentiating (2) partially w.r.t. x; and

1 2V V x y

V x

1 V2

(2)

log y

(3)

V y

1 x

1 y

V x y

V

1 V2

V x

1 x

1 y

1 V2

V x

2

i.e.,

x y

log x

V y

1 x

1 y

(4)

V y

1 x

1 y

(5)

Differentiating (3) partially w.r.t. y; 1 2V V y x

V y

1 V2

V x

2

V

y

V

x 2

V x y

From (4) and (5), we see that

2

V . y x

Example 11 Verify Euler’s theorem, when (i) u ax2 + 2hxy + by2 and (ii) u (i) u ax2 + 2hxy + by2 u x

2ax u x

x

2hy;

y

u y

u y (2ax 2

2hx

x2 a2

2hxy )

2h

ex

3

y x

y3

v = log u = x3 + y3 is a homogeneous functions of degree 3. v x

1 u u x

3x 2

y3

2by 2 )

(2hxy

b

u is a homogeneous function of degree 2. (ii) u

3

2by

= 2 (ax2 + 2hxy + by2) = 2u u

ex

(1) y x

2

x2 f

y x

Appendix A

v x

x

v y

1 u 2 3y u y

v y

3 x3

y

II – A.15

y3

= 3v

Note

(2)

(2) can be rewritten as x

1 u u x

1 u u y

y u x

i.e.,

x

Example 12 If u

sin

u y

y x2 x

1

3log u 3u log u

y2 u , prove that x y x

u y

y

tan u, (i), by

using Euler's theorem, (ii) without using Euler’s theorem (i) u

sin

1

x2 x

y2 y

x2 x

sin u

i.e.,

y2 y

y x

x2 1

y x

x 1

2

xf

y x

sin u is a homogeneous function of order (degree) 1. By Euler’s theorem, we have x

x

sin u

x

i.e.,

(ii) sin u

y

x2 x

y2 y

y u x

sin u

u y

y

or

x

1 sin u

tan u y sin u

x2

y2

(1)

Differentiating (1) partially w.r.t. x, u sin u x Differentiating (1) partially w.r.t. y, u x y cos u sin u y x

y cos u

2x

(2)

2y

(3)

Part II: Mathematics II

II – A.16 (2) × x + (3) × y gives, x

y cos u x

u x x

i.e.,

u x

u y

y

u y y

x

2 x2

tan u

x

x

u x

y

u y

y2

y2

y cos u

sin u cos u

2

2 x2

y sin u

or 2 tan u

tan u

Example 13 If u

x 2 tan

1

u

x 2 tan

1

2

y x

y 2 tan

1

y x

y 2 tan

1

y x

y x

1

x tan

x y

x

x y

2

tan

1

1

y x

y x

x2 f

u is a homogenous function of degree 2. By Euler’s theorem, x

Note

u x

u y

2u u x

From example (9),

u y

and

x

y

u x

y

x

u y

2 y tan

xy

1

2 x tan

1

y x

x y

1

2 x 2 tan

2 x 2 tan =2u

y

y x

1

y x

xy

y 2 tan

1

2 y 2 tan x y

1

y x

u x

y

u y

Appendix A Example 14 If u

tan

x2 x

1

y2 y

2

u

tan u

x2 x

2

u x2

x2

tan

2

u x y

2 xy x2 x

1

II – A.17

u . y2

y2

y2 y

y2 is a homogeneous function of degree 1. y

By Euler’s theorem, x

tan u

x

u x

x

i.e.,

y

u y

y

tan u

tan u

tan u sec 2 u

sin u cos u

y

1 sin 2u 2

(1)

Differentiating (1) partially w.r.t. x, 2

u x2

x

2

x2

u x2

2

u x

cos 2u

u x

2

u x

x

u x y

y

u x y

xy

x cos 2u

u x

(2)

u y

(3)

Differentiating (1) partially w.r.t. y, 2

2

u y x

x 2

xy

u y x

u y2

y 2

u y2

y2

y

u y

cos 2u

u y

y cos 2u

Adding (2) and (3), we get 2

x2

u x2

2

2 xy

u x y

2

u y2

y2

cos 2u x

u x

x y

u y

u x

y

u y

u y

Part II: Mathematics II

II – A.18 2

2

u x2

x2

2 xy

2

u x y

u y2

y2

1 sin 2u 2

1 sin 2u cos 2u, by (1) 2 x 2 u xx

i.e.,

1 sin 2u cos 2u 1 2

y 2 u yy

2 xyu xy

EXERCISE Part A (Short answer questions) 1. If u = (x

y)4 + (y

2. If u = ( x

y)(y

3

z)4 + (z

z)(z

u x

x)4, show that u x

x), prove that

3. If u = x y

xy , show that the value of

13 . 22 4. If z3 – 3yz

3x = 0, show that z

x y

u z

u z

1

u x

3

u y

u y

0. 1

at the point (1, 2)

is

z x

z . y 2

6. If u

log

x2

y2 xy

, verify that

7. Verify Euler’s theorem when u 8. Verify Euler’s theorem, when u

9. If u

f

10. If u

sin

functions.

y u , prove that x x x

1

x y

tan

1

2

u x y

u y x

2

2

5. If u = x cos y + y sin x, verify that

u x y

u . y x

x

y.

x 3 sin u y

y . x

0

y u , prove that x x x

y

u y

0

0

Appendix A

II – A.19

12. State Euler’s theorem on homogeneous functions for second order derivatives. y 13. If u xf , show that x 2

u x2

x2

14. If u

2

u x y

u y2

y2

2

xy x

2

2 xy

y

2

u x2

, prove that x 2

0

2

u x y

2 xy

2

r x

15. If x = r cos and y = r sin , prove that

u y2

y2

r y

0

2

1

Part B z x

16. If z(x + y) = x2 + y2, prove that 2

17. If u = log (x2 + y2), prove that

u x2

2

z y

z x

41

z y

2

u y2

0

18. If u = log (x2 + y2 + z2), prove that 2

u x2 19. If u 2

u x2 20. If V

2

2

u y2

u z2

1 , where r2 = (x r 2

u y2

x2

2 y2

z2

a)2 + (y

b)2 + (z

c)2, prove that

2

u z2

0. 2

1 , where r2 = x2 + y2, prove that r

2

V x2

V y2

1 . r3 2

21. If V = log r, where r2 = (x

22. If y = f (x

at) +

a)2 + (y

(x + at), show that

23. If z = tan (y + ax) + (y

2

y

t

2

ax) , prove that

2

a2 2

z

x

2

y

x2

2

V x2

b)2, prove that

V y2

0.

. 2

a2

z y2

0.

2

24. If xx yy zz = c, show that, when x = y = z,

z x y

{x (1 log x)}

1

Part II: Mathematics II

II – A.20

25. If x = r cos and y = r sin , prove that 2

2

r x2

r

r x

1 r

y2

2

r y

2

26. If V = ea cos (a log r), prove that 2

V r2

2

2

0.

u , when y x y2 + (y2

2x) .

xy , show that x2 y 2

1

tan

V

2

u x y u = 3xy

29. If u

2

u , when u = xy + yx. y x

2

28. Verify that

1 r2

2

u x y

27. Verify that

1 V r r

1 2

u x y

1 x2

(1

y 2 )3 2

30. Without using Euler’s theorem, prove that u x

x

31. If u

33. If u

1

sin

u x

x ex

3

y

x

x

32. If u

x

1

cos

y3

y

u x

y

x3 x

y

u y

y

1 , when u

log

, prove that

u y

1 cot u 2

0.

y3 , prove that y

u y

2 tan u.

, prove that x

u x

y

u y

3u log u

x3 x2

y3 y2

Appendix A

34. If u

sin

x

1

y

x

y

, prove that

2

x2

35. If u

tan

1

u x2

x3 x

2

2 xy

u x y

2

y2

u y2

sin u cos 2u 4 cos3 u

y3 , prove that y 2

x2

II – A.21

u x2

2

2 xy

u x y

2

y2

u y2

sin 4u

sin 2u

Appendix

B

Solutions to the January 2011 Question Paper of Anna University, Chennai Question Paper Code: 10003 B.E./B. TECH. DEGREE EXAMINATIONS : JANUARY 2011 First Semester Common to All Branches 181101 - MATHEMATICS I (Regulation 2010) Times: Three hours

Maximum; 100 marks Answer ALL Questions Part A - (10 ¥ 2 = 20 marks)

1.

Ê 1 2ˆ For what values of 'c' the eigen values of the matrix Á are real and Ë c 4˜¯ unequal, real and equal, complex conjugates?

2.

If A is an n ¥ n real symmetric matrix, D is an n ¥ n diagonal matrix whose diagonal elements are the eigen values of the matrix A and P is an n ¥ n orthogonal diagonalizing matrix whose columns are the normalized eigen vectors of the matrix A, satisfying the similarity transformation D = P–1 AP, then find the matrix Ak, where k is a positive integer?

3.

Find the equation of the sphere which has its centre at (6, –1, 2) and touches the plane 2x – y + 2z – 2 = 0.

4.

Find the equation of the right circular cone of revolution with vertex at the origin, axis the y-axis and semi-vertical angle 30°.

Part II: Mathematics II

II-B.2 5.

Find the envelope of the family of (x – a)2 + y2 = 2a, where a is a parameter.

6.

What is the radius of curvature of the curve x4 + y4 = 2 at the point (1, 1).

7.

Find

8.

If u =

9.

Evaluate a2.

10.

du Ê yˆ , if u = tan–1 Á ˜ where y = tan2 x. Ë x¯ dx y2 x2 ∂ ( x, y) and v = , find . x y ∂(u, v)

Ú Ú xy dx dy taken over the positive quadrant of the circle x

Find the area enclosed by the ellipse

x2 a2

+

y2 b2

2

+ y2 =

=1.

PART B-(5 ¥ 16 = 80 Marks) È 1 -1 1˘ 11. (a) (i) Show that the matrix Í0 1 0 ˙˙ satisfies the characteristics equation Í ÍÎ2 0 3˙˚ and hence find its inverse. (8) È 1 -1 4 ˘ (ii) Find all the eigenvalues and eigenvectors of the matrix ÍÍ 3 2 -1˙˙ ÍÎ2 1 -1˙˚ (8) Or (b)

Find a change of variables that reduces the quadratic form 3x12 + 5x22 + 3x32 – 2x1x2 + 2x1x3 – 2x2x3 to a sum of squares and express the quadratic form in terms of new variables. (16)

12 (a) (i) Fnd the equation of the sphere which passes through the circle x2 + y2 + z2 – 2x – 4y = 0, x + 2y + 3z = 0 and touch the plane 4x + 3y = 25. (8) (ii) Find the equation of the right circular cone generated when the straight line which is the intersection of the planes 2y + 3z = 6 and x = 0 revolves about the z-axis with constant angle. (8) Or (b)

(i) Find the equation of the tangent lines to the circle 3x2 + 3y2 + 3z2 – 2x – 3y – 4z – 22 – 0, 3x + 4y + 5z – 26 = 0 at the point (1, 2, 3). (8)

Appendix B

II-B.3

(ii) Find the equation of the right circular cylinder which passes through the circle x2 + y2 + z2 = 9, x – y + z = 3. (8) 13. (a) (i) Find the radius of curvature at the origin for the curve x3 + y3 + 2x2 – 4y + 3x = 0. (6) x 2 y2 = 1 at one end of the (ii) If the centre of curvature of an ellipse + a 2 b2 minor axis lie at the other end, prove that the eccentricity of the ellipse 1 (10) is 2 (b)

(i) Find the equation of the evolute of the parabola y2 = 4ax. (8) (ii) Find the envelope of the circles drawn on the radius vectors of the ellipse x2 a2

+

y2 b2

= 1 as diameter.

(8)

Ê x 2 + y2 ˆ 14. (a) (i) If u = sin–1 Á ˜ ; prove that Ë x+y ¯ (1) x

∂u ∂u +y = tan u ∂x ∂y

(2) x 2

∂2 u ∂x 2

+ 2 xy

∂2 u ∂2 u + y2 2 = tan 2 u ∂x ∂y ∂y

(ii) Test for an extreme of the function f (x, y) = x4 + y4 – x2 – y2 – 1.

(8) (8)

Or (b)

(i) If x = u cos a – v sin a, y = u sin a + v cos a and V = f(x, y), show that ∂2V ∂2V ∂2V ∂2V + 2 = + 2 . (8) ∂x 2 ∂y ∂u 2 ∂v (ii) A rectangular box open at the top is to hold a given volume. Find the relation between the dimensions so that the surface area of the tank is minimum. (8)

15. (a) (i) Evaluate Ú Ú ( x - y) dx dy over the region between the line y = x and the parabola y = x2. (4)

ÚÚÚx

2

yzdx dy dz taken over the tetrahedron bounded by the x y z planes x = 0, y = 0, z = 0 and + + = 1 . a b c

(ii) Evaluate

Or

Part II: Mathematics II

II-B.4

a

(b)

(i) Change the order of integration it.

a + a 2 - y2

Ú 0

Ú

2

a- a - y

2

xy dx dy and hence evaluate (8)

(ii) A circular hole of radius b is made centrally through a sphere of radius a. Find the volume of the remaining sphere. (8)

Solutions PART-A

1.

The C.E. is \ l=

1- l c

2 = 0 i.e., l2 – 5l + (4 – 2c) = 0 4-l

5 ± 25 - 4(4 - 2c) 2

9 \ The eigenvalues are real and unequal, if 9 + 8c > or c > - , real and equal 8 9 if 9 + 8c = or c = - ; 8 9 Complex conjugate if 9 + 8c < 0 or c < 8 2.

D = P–1AP \ A = PDP–1; A2 = (PDP–1)(PDP–1) = PD2P–1 A3 = PD3P–1 and so on. Then Ak = PDkP–1, where Dk = D(l1k, l2k, ...lnk )

3.

Radius of the required sphere = ^r length from C(6, –1, 2) on the touching plane 2x – y + 2z – 2 = 0 = |12 + 1 + 4 – 2| ∏ 4 + 1 + 4 = 15/3 = 5 \ Equation of the sphere is (x – 6)2 + (9 + 1)2 + (z – 2)2 = 52 i.e., x2 + y2 + z2 – 12x + 2y – 4z + 16 = 0

4.

The equation of right circular cone is Sl2 ◊ Sx2 ◊ cos2 q = (Slx)2 3 = y2 i.e., 1 ◊ (x2 + y2 + z2) 4 i.e., 3x2 – y2 + 3z2 = 0.

Appendix B

II-B.5

5.

(x – a)2 + y2 = 2a (1); Differentiating w.r.t. a, 2(x – a) (–1) = 0 (2) Eliminating a from (1) and (2), the envelope is y2 = 2a

6.

x4 + y4 + 2; 4x3 + 4y3 ◊ y¢ = 0 \ y¢ = -

8.

∂ (u, v) = ∂ ( x, y) vx

vy

=

y4

1

= -

uy

3 x 2 ( xy ¢ - y)

1 1 Ê yˆ ◊Á- ˜ + ◊ 2 tan x sec2 x y2 Ë x 2 ¯ y2 x 1+ 2 1+ 2 x x

du ∂u ∂u ∂y = + = = dx ∂x ∂y ∂x

ux

y3

; y ¢¢ =

(1 + y ¢ 2 )3/2 2 = | y ¢¢ | 3

(y¢)(1, 1) = –1; (y¢¢)(1,1) = –6; r =

7.

x3

y 2

x +y

y2 2

x 2x y

2

2y x -

x

2

+

2 x tan x sec 2 x x 2 + y2

= - 3. \

∂( x, y) 1 =∂(u, v) 3

y2

9. y

x

0

a

a2 - y2

0

0

Ú Ú xy dx dy = Ú

Ú

Ê x2 ˆ xy dx dy = Ú y Á ˜ 2¯ 0 Ë 0 a

a

a 1 1 Ê a2 y2 y4 ˆ 1 2 2 y ( a y ) dy = = a4 Á ˜ Ú 2 0 2Ë 2 4¯ 8 0

10. y

0

x

a2 - y2

dy

Part II: Mathematics II

II-B.6

a b2 - y 2 b b

Area = 4 Ú 0

Ú

dx dy = 4

0

a b

b

Ú

b2 - y2 dy

0

b

=

4a È y ◊Í b ÎÍ 2

=

4a p 2 ◊ b = p ab b 4

b2 - y 2 +

b2 y˘ sin -1 ˙ 2 b ˚˙ 0

PART B 11. (a) (i) C. E. of A is

-1 1- l 0 1- l 2 0

1 i.e., (1 - l ){(1 - l )(3 - l ) - 2} = 0 0 = 0; i.e., l 3 - 5l 2 + 5l - 1 = 0 3-l

To verify: A3 – 5A2 + 5A – I = 0 (1) Ê 3 -2 4 ˆ Ê 11 -5 15ˆ 3 Á ˜ A = 0 1 0 ; A = Á0 1 0 ˜ Á ˜ Á ˜ Ë 8 -2 11¯ Ë 30 -10 41¯ 2

Using these values in the L.H.S. of (1), Ê 11 -5 15ˆ Ê 15 -10 20ˆ Ê 5 -5 5 ˆ Ê 1 0 0ˆ L.H.S. = Á 0 1 0 ˜ - Á0 5 0 ˜ + Á 0 5 0 ˜ - Á 0 1 0˜ Á ˜ Á ˜ Á ˜ Á ˜ Ë 30 -10 41¯ Ë 40 -10 55¯ Ë 10 0 15¯ Ë 0 0 1¯ Ê 0 0 0ˆ = Á 0 0 0˜ = R.H.S. Á ˜ Ë 0 0 0¯ \ A satisfies the C.E. Ê 3 -2 4 ˆ Ê 5 -5 5 ˆ Ê 5 0 0ˆ 0 ˜ - Á 0 5 0 ˜ + Á 0 5 0˜ From (1), A = A – 5A + 5I = Á 0 1 Á ˜ Á ˜ Á ˜ Ë 8 -2 11¯ Ë 10 0 15¯ Ë 0 0 5¯ –1

2

3 -1ˆ Ê3 = Á0 1 0 ˜ Á ˜ Ë -2 -2 1 ¯

(ii)

1- l C.E. of the given matrix A is 3 2

-1 2-l 1

4 -1 = 0 -1 - l

Appendix B

II-B.7

i.e., l3 – 2l2 – 5l + 6 = 0 i.e., (l – 1) (l – 3) (l + 2) = 0 \ the eigenvalues of A are l = –2, l = 1 and l = 3, When l = –2, The eigenvector is given by 3x1 – x2 + 4x3 = 0 and 3x1 + 4x2 – x3 = 0 i.e., X1 = (–1, 1, 1)T When l = 1, the eigenvector is given by –x2 + 4x3 = 0 and 3x1 + x2 – x3 = 0 i.e., X2 = (–1, 4, 1)T When l = 3; the eigenvector is given by –2x1 – x2 + 4x3 = 0 and 3x1 – x2 – x3 = 0 i.e., X3 = (1, 2, 1)T Ê 3 -1 1 ˆ 11. (b) The matrix A of the quadratic form Q is A = Á -1 5 -1˜ Á ˜ Ë 1 -1 3 ¯ Refer to the worked example 1.9 given in page I-1.41 of the book, from which the eigenvalues are obtained as 2, 3 and 6 and the corresponding eigenvectors are obtained as X1 = (1, 0, –1)T, X2 = (1, 1, 1)T and X3 = (1, –2, 1)T Ê 1 Á 2 Á Á The normalised modal matrix M = Á 0 Á Á 1 ÁË 2

1 3 1 3 1

1 ˆ 6 ˜ ˜ 2 ˜ 6 ˜˜ 1 ˜ ˜ 6 ¯

3 The transformation of variables required to convert Q into the sum of square form is given by X = NY. i.e., x1 =

1 2 1

y1 + y1 +

1 3 1

y2 + y2 +

1 6

y3 ; x2 =

1 3

y2 -

2 6

y3 and

1

y3 2 3 6 Sum of squares form of Q is 2y12 + 3y22 + 6y32 x3 = –

12(a) (i) [Note: The equation of the plane x + 2y + 3z = 0 must have been given as x + 2y + 3z = 8. The problem is solved using the corrected equation.] Equation of any sphere passing through the given circle is x2 + y2 + z2 – 2x – 4y + 2l (x + 2y + 3z – 8) = 0

Part II: Mathematics II

II-B.8

i.e., Sx2 – 2(1 – l) x – 2(2 – 2l) + 6l z – 16l = 0 (1) C, centre of the sphere (1) is (1 – l, 2 – 2l, –3l) (l - 1)2 + (2 l - 2)2 + 9 l 2 + 16 l

R, radius of the sphere (1) =

(14 l 2 + 6 l + 5

=

Since the sphere (1) touches the plane 4x + 3y – 25 = 0 = (2), ^r length from C on (2) = R 4 - 4 l + 6 - 6 l - 25 4 2 + 32

=

14 l 2 + 6 l + 5

i.e., (2l + 3)2 = 14l2 + 6l + 5 i.e., 10l2 – 6l – 4 = 0 \l=

3 ± 9 + 40 10

or

5l2 – 3l – 2 = 0

i.e., l = 1 or -

2 5

Using these values of l in (1), the two required spheres are x2 + y2 + z2 – 6z = 16 and 5 (x2 + y2 + z2) – 14x – 28y – 12z + 32 = 0 12. (a) (ii) The generator, i.e,, the rotating line has the D.R.'s given by l = 0 and 2m + 3n = 0. i.e., (0, –3, 2) The vertex of the cone is the point of intersection of the axis (i.e., z-axis) and the generator. It is (0, 0, 2) The semi vertical angle is the angle q between the axis, whose D.R.'s are (0, 0, 1) and the generator whoe D.R.'s are (0, -3, 2) \ cos q =

2 1 ¥ 13

i.e., cos q =

2 3

\ The equation of the right circular cone is Sl2 ◊ S(x – a)2 . cos2q = {Sl(x – a)}2 i.e., 1{x2 + y2 + (z – 2)2} ¥

4 = (z - 2)2 13

i.e., 4(x2 + y2) – 9 (z – 2)2 = 0. 12.(b) (i) The tangent line to the circle lies on the tangent plane of the given intersecting sphere, whose equation is 3(x ◊ 1) + 3(y ◊ 2) + 3(z ◊ 3) – 3 (x + 1) – (y + 2) – 2(z + 3) – 22 = 0 2 i.e., 4 ◊ x + 9y + 14z – 64 = 0 (1)

Appendix B

II-B.9

Since the circle lies on the intersecting plane, the tangent line also lies on this plane whose equation is 3x + 4y + 5z – 26 = 0 (2) Equations (1) and (2) are the joint equations of the required tangent line. 12. (b) (ii) Axis of the cylinder is normal to the given circle, which can be considered as the guiding circle. \ D.R.'s of the axis are (1, –1, 1) \ D.R.'s of any generator of the cylinder are (1, –1, 1) If (x1, y1, z1) is any point on the required cylinder, the generator through it has equations. x - x1 y - y1 z - z1 = = =r 1 -1 1 \ Any point on the generator is (x1 + r, y1 – r, z1 + r) Since the generator intersects the guiding circle, (x1 + r)2 + (y1 – r)2 + (z1 + r)2 = 9 and (x1 + r) – (y1 – r) + (z1 + r) = 3,

(1) (2)

3 - x1 + y1 - z1 3 Using (2) in (1), we get

From (2), r =

(3)

6(x12 + y12 + z12 + x1y1 + y1z1 – z1x1) = 54 \ The equation of the cylinder is x2 + y2 + z2 + xy + yz – zx = 9 13. (a) (i) x3 + y3 + 2x2 – 4y + 3x = 0 (1); Diffg (1) w.r.t. x, we get 3x 2 + 4 x + 3 (2) 3x2 + 3y2y¢ + 4x – 4y¢ + 3 = 0 \ y¢ = 4 - 3 y2 Diffg. (2) w.r.t. x, y¢¢ = \ [y¢](0, 0) =

r =

(4 - 3 y2 )(6 x + 4) - (3 x 2 + 4 x + 3)(-6 yy ¢) (4 - 3 y2 )2 3 and [y ¢¢](0, 0) = 1 4 (1 + y ¢ 2 )3/2 \ [ r ](0,0) y ¢¢

9ˆ Ê ÁË 1 + 16 ˜¯ = 1

3/2

=

125 64

13. (a) (ii) [This problem is the worked example 3.13 given in page I-3.16 of the book]

Part II: Mathematics II

II-B.10

13. (b) (i) [Refer to W.E. 3.1 in page I-3.23 of the book. By inter changing x and y we get the required solution. Key steps are given below] –x = 3at2 + 2a (1) and y– = –2at3 (2) Eliminating t from (1) and (2), we get 27ay–2 = 4 (x– – 2a)3. \ Evolute of the parabola y2 = 4ax is 27 ay2 = 4(x – 2a)3. 13. (b) (ii) The equation of the typical circle drawn on the radius vector of the ellipse joining (0, 0) and (a cos q, b sin q) is x(x – a cos q) + y( y – b sin q) = 0 i.e., ax cos q + by sin q = x2 + y2 (1) Diffg. (1) w.r.t the parameter q, we get –ax sin q + by cos q = 0 (2) Squaring (1) and (2) and adding, a2x2 + b2y2 = (x2 + y2)2, which is the equation of the required envelope. 14. (a) (i) The first part of the problem is worked out in W.E. (12) given in page II-A.15 of the book. x

∂u ∂u +y = tan u ∂x ∂y

(1)

Diffg. (1) partially w.r.t. x; x uxx + ux + y uyx. = sec2u ◊ ux

(2)

Diffg. (1) partially w.r.t. y; x uxy + y uyy + uy = sec2 u.uy [(2) ¥ x + (3) ¥ y] gives x2uxx + 2xy uxy + y2uyy + (xux + yuy) = sec2 u (xux + yuy) i.e., x2uxx + 2xy uxy + y2uyy = (sec2u – 1) tan u, by (1) = tan3 u 14. (a) (ii) f(x, y) = x4 + y4 – x2 – y2 – 1 \ fx = 4x3 – 2x; fy = 4y3 – 2y fxx = 12x2 – 2; fxy = 0; fyy = 12y2 – 2 The stationary points are given by fx = 0 and fy = 0 i.e. 2x(2x2 – 1) = 0 and 2y(2y2 – 1) = 0

(3)

Appendix B

i.e., x = 0, ±

1 2

and y = 0, ±

II-B.11

1 2

Ê 1 ˆ Ê 1 ˆ i.e., the stationary points are (0, 0), Á 0, ˜¯ , ÁË 0, ˜ Ë 2 2¯ Ê 1 ˆ Ê 1 ˆ Ê 1 Ê 1 1 ˆ Ê 1 1 ˆ Ê 1 1 ˆ 1 ˆ , 0˜ , Á , ,, 0˜ + Á , ,ÁË ˜,Á ˜ , -Á˜ and ÁË ˜ 2 ¯ Ë 2 2¯ Ë 2 2¯ Ë 2 ¯ Ë 2 2¯ 2 2¯

B = fxy = 0 at all the points.

Ê 1 ˆ At (0, 0), AC = 4 > 0 and A < 0; At Á 0, ± ˜ , AC < 0; Ë 2¯ Ê 1 ˆ Ê 1 1 ˆ At Á , 0˜ , AC < 0; At Á ,± ˜ , AC = 16 > 0 and A (or C) > 0 Ë 2 Ë 2 ¯ 2¯ Ê 1 ˆ At Á , 0˜ , AC < 0; At Ë 2 ¯

Ê 1 1 ˆ ,± ÁË ˜ , AC = 16 > 0 and A (or C) > 0 2 2¯ Ê 1 1 ˆ \ f (x, y) is maximum at (0, 0) and is minimum at the points Á ,± ˜, Ë 2 2¯ Ê 1 1 ˆ and Á ,± ˜ Ë 2 2¯

Maximum of f(x, y) = –1 and Minimum of f(x, y) = –

3 . 2

14. (b) (i) x = u cos a – v sin a; y = u sin a + v cos a i.e., x = au – bv; y = bu + av, say Vu = Vx ◊ xu + Vy ◊ yu = aVx + bVy \

∂ ∂ ∂ ∫a +b ∂u ∂x ∂y

\ Vuu =

(1) (2)

Ê ∂ ∂ ∂ˆ (Vu ) = Á a + v ˜ (aVx + bVy ) ∂u ∂y ¯ Ë ∂x

= a2 Vxx + 2ab Vxy + b2Vyy

(3)

Vv = Vx ◊ xv + vy ◊ yv = –bVx + aVy

(4)

\

∂ ∂ ∂ = -b +a ∂ x ∂ y ∂v

(5)

Ê ∂ ∂ˆ + a ˜ (- b Vx + a Vy ) \ Vvv = Á - b ∂x ∂y ¯ Ë = b2 Vxx – 2ab Vxy + a2 Vyy

(6)

Part II: Mathematics II

II-B.12 (3) +(6) gives,

Vuu + Vvv = (a2 + b2) (Vxx + Vyy) = Vxx + Vyy (∵ a = cos a and b = sin a) 14. (b) (ii) [This problem is the same as the W.E. 4.12 given in page I-4.61 of the book. The given volume may be taken as k instead of 32.] The length, breadth and height of the box are respectively Ê kˆ 4Á ˜ Ë 32 ¯

1/3

Ê kˆ ; 4Á ˜ Ë 32 ¯

1/3

Ê kˆ ; 2Á ˜ Ë 32 ¯

1/3

.

i.e., they are in the ratio 4 : 4 : 2. 15. (a) (i) I =

y

Ú Ú ( x - y) dx dy

1

y

0

y

A 0

Ú Ú ( x - y) dx dy

=

y

=

1

1 Ê x2 ˆ Êy y2 ˆ Ú ÁË 2 - yx˜¯ dy = Ú ÁË 2 - y y + 2 ˜¯ dy 0 0 y

=

Ê y2 2 5/2 y3 ˆ 1 2 1 1 Á 4 - 5 y + 6 ˜ = 4 - 5 + 6 = 60 Ë ¯0

1

15. (a) (ii) y

z

x y + =1 a b

(x,y, z1) 0

y , ) (x,y0

x

I=

ÚÚÚx a

=

Ú 0

2

y z d zdy dx

xˆ Ê x yˆ Ê b Á 1 - ˜ cÁ 1 - - ˜ Ë a¯ Ë a b¯

Ú 0

B (1,1)

C

OABC

Ú 0

x 2 y zdzdy dx

0

x

x

Appendix B

=

xˆ Ê bÁ 1- ˜ Ë a¯

a

Ú 0

2

0

2 a

= c 2

Ú

0

2 a

= c 2

xˆ Ê bÁ 1- ˜ Ë a¯

Ú

0

xˆ Ê bÁ 1- ˜ Ë a¯

Ú

Ú

0

Ê x yˆ c 1- - ˜ a b¯

Ê z2 ˆ ÁË x yÁ ˜ Ë 2 ¯0

Ú

0

II-B.13

dy dx

2

x yˆ Ê x 2 y Á 1 - - ˜ dy dx Ë a b¯ 2 È 2Ê xˆ x ˆ y2 y3 ˘ Ê + x 2 ◊ 2 ˙ dy dx Íx Á 1 - ˜ y - 2x2 Á 1 - ˜ Ë a¯ a¯ b b ˙˚ ÍÎ Ë

3 È xˆ 3 Ê b 1 Í 2 2 a ÁË c2 Í 2 Ê x ˆ b2 Ê xˆ xˆ a ˜¯ 2 Ê x x 1 1 2 1 = Á ˜ Á ˜ Á ˜ Ë a¯ 2 Ë a¯ a¯ 2 Ú0 Í Ë 3b Í ÍÎ 4 xˆ ˘ Ê b4 Á 1 - ˜ ˙ Ë a¯ ˙ + x2 dx 2 ˙ 4b ˚ 4

a

=

=

=

=

b2 c 2 Ï 1 2 1 ¸ xˆ 2Ê Ì - + ˝ Ú 3 x Á 1 - ˜ dx Ë 2 Ó2 3 4˛ 0 a¯ b2 c 2 8a 4 b2 c 2 8a 4

a

Úx

2

(a - x)4 dx

0

a

Ú (a

4 2

x - 4a3 x 3 + 6a2 x 4 - 4ax 5 + x 6 ) dx

0

b2 c 2 a 7 È 1 1 1 1 1˘ 1 3 2 2 Í 3 - 4 ◊ 4 + 6 5 - 4 ◊ 6 + 7 ˙ = 840 a b c 4 8a Î ˚

15. (b) (i) (this problem is the worked example 5-7 given in page I-5.26 of the book) 15. (b) (ii) E

C A

F b{ 0

x2 + y2= a2 a2 – b2 D B

Part II: Mathematics II

II-B.14

The required volume is obtained by revolving the area EFD about the x-axis. V = 2 ¥ 2p

a 2 - b2

a2 - x 2

Ú

Ú

0

a 2 - b2

= 4p

Ú 0

Ê y2 ˆ Á 2˜ Ë ¯b

y dy dx

b

a2 - x 2

dx

a 2 - b2

= 2p

Ú

(a2 - x 2 - b2 ) dx

0

Ï x 3 Ô¸ = 2p ÔÌ(a2 - b2 ) x ˝ 3 Ô˛ ÓÔ 0

a 2 - b2

2p {3(a2 - b2 )3/2 - (a2 - b2 )3/2 } 3 4p 2 (a - b2 )3/2 = 3

=

Appendix

C

Solutions to the December 2010/January 2011 Question Paper of Anna University, Coimbatore B.E./B.TECH. DEGREE EXAMINATIONS: Dec 10/JAN 11 REGULATIONS 2008 FIRST SEMESTER 080030001 - MATHEMATICS I (Common to all Branches) Times: 3 Hours

Max. Marks:-100 PART - A ANSWER ALL QUESTIONS

1.

2.

3. 4.

È 3 -1 1 ˘ Two of the eigen values of ÍÍ-1 5 -1˙˙ are 3 and 6 Find the eigen values of A–1 ÍÎ 1 -1 3 ˙˚ È1 0˘ 3 If A = Í ˙ , express A interms of A and I using Cayley-Hamilton 4 5 ˚ theorem. Î Find the nature of the quadratic form 2x2 + 2xy + 3y2 sin q ˘ È cos q Show that Í is orthogonal. sin q cos q ˙˚ Î

5.

Find the equation of the sphere on the join of (2, –1, 4) and (–2, 2, –2) as diameter.

6.

Define a right circular cylinder.

7.

Find the equation to the tangent plane to the sphere x2 + y2 + z2 = 9 at (2, 2, 1).

8.

Find the equation of a sphere whose centre is (–6, 1, 3) and radius 4 units.

Part II: Mathematics II

II-C.2 9.

What is the radius of curvature at (3, 4) on the curve x2 + y2 = 25?

10.

Find the envelope of the family of straight lines x cos a + y sin a = r, where a is a parameter.

11.

Write down the equation of the circle of curvature of a given curve.

12.

Define evolute of a curve.

13.

Find the Taylor's series of xy near the point (1, 1) upto the first degree term.

14.

State Euler's theorem on homogenous function.

15.

If u = ex sin y, prove that

16.

If x = u (1 + v) and y = v(1 + u), find

17.

Evaluate

∂2 u ∂x

2

+

∂2 u ∂y 2

=0 ∂( x, y) ∂(u, v)

p cos q

Ú Ú 0

rdrd q

0

18.

Find the area of circle of radius 'a' by double integration in polar coordinates.

19.

Change the order of integration of

a a

3 4

20.

Evaluate

ÚÚ 1 3

4

Ú xyz dxdydz

Ú Ú f ( x, y) dxdy 0 y

1

PART - B (5 ¥ 12 = 60 Marks) ANSWER ANY FIVE QUESTIONS 21.

Reduce the quadratic form 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz to a canonical form by an orthogonal transformation

22. (a) Find the centre and radius of the circle which is the intersection of the sphere (6) x2 + y2 + z2 – 8x + 4y + 8z = 45 and the plane x – 2y + 2z = 3. (b) Find the equation to the right circular cone whose vertex is at (2, –3, 5) axis makes equal angles with the coordinate axes and semi-vertical angle is 30° (6) a a 23 (a) Find the radius of curvature at the point ÊÁ , ˆ˜ to the curve Ë 4 4¯

x + y = a. (6)

(b) Find the evolute of the parabola y2 = 4ax. 24. (a) Find the maximum and minimum values of x3 + 3xy2 – 3x2 – 3y2 + 4. (b) Find the maximum value of xmynzp when x + y + z = a.

(6) (6) (6)

Appendix C 25. (a) Evaluate

Ú Ú xy ( x + y)

II-C.3

dydx over the area between y = x2 and y = x.

(b) Change the order of integration and hence evaluate

a

2a - x

Ú0 Ú x

2

dydx

(6) (6)

a

È2 2 1˘ 26. (a) Find the eigen values and eigen vectors of A = ÍÍ 1 3 1˙˙ ÍÎ 1 2 2 ˙˚ È1 2˘ 3 (b) Verify Cayley-Hamilton for the matrix A = Í ˙ and hence evaluate A 4 3 –1 Î ˚ also find A . 27. (a) Find the equation to the sphere passing through the circle x2 + y2 + z2 = 9, x + y + z = 1 and cuts orthogonally the sphere x2 + y2 + z2 + – 4y – 16z + 17 = 0. (8) p (4) (b) Find the radius of curvature for r = a(1 + cos q) at q = 2 Ê u, v ˆ 28. (a) If u = x2 – y2, v = 2xy and x = r cos q, y = r sin q, compute J Á (6) Ë r, q ˜¯ (b) Evaluate

1

1 - x2

0

0

1 - x 2 - y2

Ú Ú

Ú

dx dy dz 1 - x 2 - y 2 - z2

1

Solutions PART-A 1.

l1 + l2 + l3 = Trace of the matrix A = 3 + 5 + 3 or 11 i.e., 3 + 6 + l3 = 11 \ l 3 = 2 1 1 1 , and 2 3 6 C.E. of A is (1 – l) (5 – l) = 0 or l2 – 6l + 5 = 0

Eigen values of A–1 are 2.

By Cayley-Hamilton theorem, A2 – 6A + 5I = 0 0 ˘ È1 0˘ È1 0˘ È 1 \ A3 = 6A2 – 5A = 6 Í ˙ - 5 Í4 5˙ = Í124 125˙ 24 25 Î ˚ Î ˚ Î ˚ 3.

È 2 1˘ Matrix of the Q.F. is A = Í ˙ ; The eigen values of A are given by (2 – l) Î 1 3˚ 5± 5 (3 – l) – 1 = 0 or l2 – 5l + 5 = 0. \ l = 2 Since both the eigen values are positive, the Q. F. is positive definite.

Part II: Mathematics II

II-C.4 4.

A ◊ AT = È cos q Í- sin q Î

- sin q ˘ È 1 = - cos q ˙˚ ÍÎ- 2 sin q cos q

sin q ˘ Ècos q - cos q ˙˚ ÍÎ sin q

- 2 sin q cos q ˘ ˙πI 1 ˚

\ The given matrix is not orthogonal È cos q Note: A must be Í Î- sin q

sin q ˘ to be orthogonal. cos q ˙˚

5.

The required equation of the sphere is (x – 2)(x + 2) + (y + 1) (y – 2) + (z – 4) (z + 2) = 0. i.e., x2 + y2 + z2 – y – 2z – 14 = 0

6.

Standard definition

7.

Equation of the tangent plane to the given sphere at (2, 2, 1) is x(2) + y(2) + z(1) = 9 i.e., 2x + 2y + z = 9.

8.

The required equation is (x + 6)2 + (y – 1)2 + (z – 3)2 = 16 i.e., x2 + y2 + z2 + 12x – 2y – 6z + 30 = 0

9.

x2 + y2 = 25 is a circle of radius 5 units. The radius of curvature at any point on a circle is its radius \ r = 5.

10.

x cos a + y sin a = p (1); Diffg. (1) partially w.r.t. a, –x sin a + y cos a = 0 (2); (1)2 + (2)2 gives x2 + y2 = p2, which is the envelope of (1)

11.

If ( x , y ) is the centre of curvature and r is the radius of curvature of the given curve at the given point, the equation of the circle of curvature is ( x - x )2 + ( y - y )2 = r 2

12.

Standard definition

13.

f(x, y) = xy ; fx(x, y) = yxy–1; fy(x, y) = xy log x f(1, 1) = 1 ; fx (1, 1) = 1 ; fy(1, 1) = 0 Taylor series is f(x, y) = f(1, 1) +

1 {(x – 1) fx(1, 1) + (y – 1) fy(1, 1)} + . . . 1!

\ x y = 1 + (x – 1) or x. 14.

Standard statement.

15.

u = ex sin y; ux = ex sin y ; uxx = ex sin y uy = ex cos y; uyy = –ex sin y \ uxx + uyy = 0

Appendix C

16.

∂x ∂ ( x, y) ∂u = ∂ (u, v) ∂y ∂u

17.

I=

= 18.

∂x u ∂v 1 + v = =1+u+ v ∂y v 1+u ∂v

cosq

p

Ê r2 ˆ Ú ÁË 2 ˜¯ 0 0 1 4

II-C.5

dq =

1 2

p

2 Ú cos q dq = 0

p

1 (1 + cos 2q ) dq 4 Ú0

p

1 p Ê ˆ ÁË q + 2 sin 2q ˜¯ = 4 0

Area of the circle whose polar equation is r = a is given by 2p a

Ú

A=

Ú r dr dq =

0

0

a2 ¥ 2p = p a2 2

19.

y

x

0 a a

a

0 y

0

I=Ú

20.

Ú f ( x, y) dx dy = Ú

Ú f ( x, y) dy dz 0

3 4

1 3 1

1

Ú Ú xyz dx dy dz = Ú 4

4

Ê x2 ˆ Ú ÁË 2 ˜¯ yz dy dz 3 1

3 4 4

I=Ú

=

x

3

3 15 Ê y2 ˆ 15 7 Ê z2 ˆ 15 7 zdz = ¥ ◊ = ¥ ¥ 4 = 105 2 Ú1 ÁË 2 ˜¯ 2 2 ÁË 2 ˜¯ 2 2 3 1

PART-B

21.

È 8 -6 2 ˘ Matrix of the given Q.F. is A = ÍÍ-6 7 -4 ˙˙ ÍÎ 2 -4 3 ˙˚ 8-l C.E. of A is -6 2

-6 7-l -4

i.e., l 3 - 18 l 2 + 45 l = 0 2 -4 = 0 i.e., l (l 2 - 18 l + 45) = 0 3-l i.e., l (l - 3)(l - 15) = 0

Part II: Mathematics II

II-C.6

\ Eigen values of A are 0, 3, 15 When l = 0, the eigen vector X1 = (1, 2, 2)T When l = 3, the eigen vector X2 = (2, 1, –2)T When l = 15, the eigen vector X3 = (2, –2, 1)T È 1 2 2˘ Model matrix M = ÍÍ2 1 -2 ˙˙ ÍÎ2 -2 1˙˚ 2 2 ˘ È1 3 3˙ Í 3 Í ˙ 2 1 2 Normalized model matrix N = 3 - 3˙ Í 3 Í1 1 ˙ -23 3 ˙˚ ÎÍ 3 The required orthogonal transformation is NTAN = D(0, 3, 15) The orthogonal transformation is X = NY 1 1 i.e., x1 = ( y1 + 2 y2 + 2 y3 ); x2 = (2 y1 + ye - 2 y3 ) and 3 3 1 (2y1 – 2y2 + y3) 3 The canonical form of the Q.F. is 0.y12 + 3y22 + 15y32 x3 =

22. (a) The centre C of the given sphere is (4, –2, –4); Radius of the sphere is R=

16 + 4 + 16 + 45 = 9

C¢C C

r R

Centre of the circle C¢ is the foot of the ^r from C on x – 2y + 2z – 3 = 0 DR¢s of CC ¢ are (1, –2, 2), since CC ¢ is normal to the plane. x-4 y+2 z+4 = = 1 -2 2 Any point on CC¢ is (r + 4, –2r – 2, 2r – 4)

\ Equations of CC¢ are

If these co-ordinates represent C¢, then

Appendix C

II-C.7

r + 4 + 2 (2r + 2) + 2 (2r – 4) = 3 i.e., 9r = 3 or r =

1 3

13 8 10 \ C¢ ∫ ÊÁ , - , - ˆ˜ Ë 3 3 3¯ 2

Also CC¢ =

2

2

13 ˆ 8ˆ 10 ˆ Ê Ê Ê ÁË 4 - 3 ˜¯ + ÁË - 2 + 3 ˜¯ + ÁË -4 + 3 ˜¯ = 1

If r is the radius of the circle, r2 = R2 – CC ¢2 = 81 – 1 = 80 \ r = 4 5 units. 22. (b) Since the axis of the cone makes equal angles with the co-ordinate axis, its D.C.'s are (cos a, cos a, cos a) 1 \ DR's of the axis are (1, 1, 1) Since 3cos2a = 1, cos a = 3 The equation of the right circular cone is S l 2 ◊ S( x - a )2 .cos2 q = S{l ( x - a )}2 i.e., 3{( x - 2)2 + ( y + 3)2 + (z - 5)2 } ◊

3 = ( x - 2 + y + 3 + z - 5)2 4

i.e., 9( x 2 + y2 + z2 - 4 x + 6 y - 10z + 38) = 4( x + y + z - 4)2 i.e., 5(x2 + y2 + z2) – 8(xy + yz + zx) – 4x + 86y – 58z + 278 = 0 23. (a) (Solution of this problem is available in the worked example 3.16 in page I-3.19 of the book.) (b) The parametric equations of the parabola are x = at2; y = 2at . x� = 2at ; y = 2a

and

dy 1 = dx t

d2 y

1 1 1 == - 2 ¥ 2at dx 2 t 2at 3 If (x–, y– ) is the centre of curvature at the point t, –x = x – y ¢ (1 + y ¢ 2 ) = at2 + 2 at2 y ¢¢

1ˆ Ê ÁË 1 + 2 ˜¯ t

= 3at2 + 2a 1 1ˆ Ê (1 + y ¢ 2 ) = 2at - 2at 3 Á 1 + 2 ˜ = - 2at 3 � y– = y + Ë y ¢¢ t ¯

(1) (2)

Part II: Mathematics II

II-C.8

2

y x - 2a ˆ Eliminating t, ÊÁ - ˆ˜ = ÊÁ Ë 2a ¯ Ë 3a ˜¯ From (2); t3 = -

3

y x - 2a ; from (3) , t 2 = 2a 3a

i.e., 27a –y 2 = 4(x–– 2a)3 \ The equation of the evolute is 27 ay2 = 4(x – 2a)3. 24. (a) f(x, y) = x3 + 3xy2 – 3x2 – 3y2 + 4 fx = 3x2 + 3y2 – 6x; fy = 6xy – 6y fxx = 6x – 6; fxy = 6y ; fyy = 6x – 6 The extreme points are given by fx = 0 and fy = 0 i.e., x2 + y2 – 2x = 0 and y (x – 1) = 0 \ x = 1 or y = 0 when x = 1, y = ±1 ; when y = 0; x = 0 or 2 \ The possible extreme points are (1, 1), (1, –1), (0, 0) and (2, 0) At the point (1, 1), A = 0, B = 6, C = 0; AC – B2 < 0 At the point (1, –1), A = 0, B = –6, C = 0; AC – B2 < 0 At the point (0, 0), A = –6, B = 0, C = –6; AC – B2 > 0 and A < 0 \ f(x, y) is maximum at (0, 0) At the point (2, 0), A = 6, B = 0, C = 6; AC – B2 > 0 and A > 0 \ f(x, y) is minimum at (2, 0) Maximum Value of f(x, y) = 4 Minimum value of f(x, y) = 8 – 12 + 4 = 0 24. (b) [This is the same as the worked example 4.11 in page I-4.61 of the book] 25. (a) y (1,1) x 0

I=

1

x

0

x2

Ú Ú xy( x + y) dy dx

Appendix C 1

=

Ú 0

1

=

Ú 0

=

II-C.9

x È Ê 2 ˆx Ê y3 ˆ ˘ Íx2 y ˙ + x Á 3 ˜ ˙ dx Í ÁË 2 ˜¯ 2 Ë ¯ 2 x x ˚ Î

Ê 5 4 1 6 1 7ˆ ÁË 6 x - 2 x - 3 x ˜¯ dx

5 1 1 1 1 1 1 1 1 3 ¥ - ¥ - ¥ = = 6 5 2 7 3 8 6 14 24 56

25. (b) (For changing the order of integration, refer to the worked example 5.9 given in page I-5.27 of the book) a

I=

Ú Ú 0

2a 2a - y

ay

dx dy +

0

a

Ú

ay dy +

0

=

=

dx dy

0

2a

a

=

Ú Ú

Ú (2a - y) dy a

a Ê 3ˆ 2 y

2 a◊ Á 3Ë

2a

Ê y2 ˆ ˜ + Á 2ay 2 ˜¯ ¯0 Ë a

2 2 3 7 a + 2a 2 - a 2 = a 2 3 2 6

26. (a) [Solution of this problem is available in worked example 1.6 given in page I-1.55 of the book] 26. (b) C.E. of A is (1 – l) (3 – l) – 8 = 0 or l2 – 4l – 5 = 0 As per C.H. theorem, A2 – 4A – 5I = 0(1) È 9 8˘ È 1 2˘ È 1 0 ˘ È0 0 ˘ L.S. of (1) = Í - 4Í - 5Í ˙ ˙ ˙=Í ˙ = R.S. of (1) Î16 17 ˚ Î 4 3˚ Î0 1˚ Î0 0 ˚ \ C.H. theorem is verified. È 9 8˘ È 1 2 ˘ È 41 42 ˘ From (1), A3 = 4A2 + 5A = 4 Í ˙ + 5 Í4 3˙ = Í84 83˙ 16 17 Î ˚ Î ˚ Î ˚ Again from (1), A–1 =

1 1 ÏÔ È 1 2 ˘ È4 0 ˘ ¸Ô {A - 4I } = ÌÍ ˙-Í ˙˝ 5 5 ÔÓ Î4 3˚ Î0 4 ˚ Ô˛ =

1 È-3 2 ˘ Í ˙ 5 Î 4 -1˚

Part II: Mathematics II

II-C.10

27. (a) Equation of any sphere through the circle Sx2 = 9 and Sx = 1 is x2 + y2 + z2 – 9 + l (x + y + z – 1) = 0 i.e., x2 + y2 + z2 + lx + ly + lz – (9 + l) = 0 (1) Sphere (1) cuts ^r by the sphere x2 + y2 + z2 + 2x – 4y – 16z + 17 = 0

(2)

\ 2uu¢ + 2 vv¢ + 2 ww¢ = d + d¢ l l l i.e., 2 ◊ - 4 ◊ - 16 ◊ = 17 - 9 - l 2 2 2 i.e., 8l = –8 or l = –1 \ The required sphere is x2 + y2 + z2 – x – y – z – 8 = 0. 27. (b) r = a(1 + cos q); r¢ = –a sin q; r¢¢ = –a cos q r=

(r 2 + r ¢ 2 )3/2 2

r - rr ¢¢ + 2r ¢

2

\ [ r]

p q= 2

=

(a2 + a2 )3/2 2

a + 2a

2

=

2 2 a 3

Ê u, v ˆ Ê u, v ˆ Ê x yˆ 28. (a) J Á = JÁ ¥ JÁ ˜ ˜ r , q x , y Ë r q ˜¯ Ë ¯ Ë ¯ =

=

ux

uy

vx

vy

¥

xr yr

xq yq

2 x -2 y cos q ¥ 2y 2x sin q

- r sin q r cos q

= 4(x2 + y2) ◊ r = 4r3 28. (b) (Solution of this problem is available in the worked example 5.12 given in Page I-5.14 of the book)

Appendix

D

Solutions to the January 2011 Question Paper of Anna University, Madurai B.E./B.Tech. DEGREE EXAMINATIONS, JANUARY 2011 (First Semester) 10AMA01 MATHEMATICS - I (Common to all Branches) Times: 3 Hours

Maximum:-100 marks INSTRUCTIONS

1.

Answer ALL Question in Part A and Part B.

2.

Part A and Part B questions should be answered separately in the same answer sheet.

3.

Answer the Questions EITHER/OR in Part B PART-A (10 ¥ 2 = 20)

1.

Prove that the eigen values of triangular matrix are just the diagonal elements of the matrix.

2.

Define similarity transformation.

3.

Find the center and radius of the sphere 7x2 + 7y2 + 7z2 – 15x – 25y – 11z = 0.

4.

Define Right circular cone.

5.

Find the radius of curvature for y = ex at the point where it crosses the y-axis.

Part II: Mathematics II

II-D.2

6.

Find the envelope of the family of lines y = mx + 1 + m 2 , m being parameter.

7.

Ê xˆ du Given u = sin Á ˜ , x = et and y = t2, find Ë y¯ dt

8.

Expand f(x, y) as a Taylor's series expansion about x = a and y = b.

9.

Solve (D3 + D2 + 4D + 4)y = 0

10.

Write down the Legendre's linear equation. PART-B (5 ¥ 16 = 80)

11. (a) Find the eigen values and eigen vectors of the matrix È2 -2 2 ˘ Í ˙ A = Í1 1 1 ˙ ÎÍ1 3 -1˚˙ (b)

Verify Cayley-Hamilton theorem for matrix È 2 0 -1˘ Í ˙ A= Í0 2 0˙ ÎÍ-1 0 2 ˙˚

12.

(8)

(OR)

Reduce the quadratic form 8x2 + 7y2 + 3z2 – 12xy + 4xz – 8yz to a canonical form by orthogonal reduction.

13. (a) Find the shortest distance between the lines x – 3 =

y-5 = z – 7 and -2

x +1 y +1 =z+1 = 7 -6 (b) Show that the plane 2x – 2y + z + 12 = 0 touches the sphere

x2 + y2 + z2 – 2x – 4y + 2z = 3 and also find the point of contact.

(8)

(8)

(OR) 14. (a) Show that the sphere x2 + y2 + z2 + 6y + 2z + 8 = 0 and x2 + y2 + z2 + 6x + 8y + 4z + 20 = 0 intersect at right angles. Also find their plane of intersection. (8)

Appendix D

II-D.3

(b) The radius of a normal section of a right circular cylinder is 2 units; the axis x -1 y+3 z-2 = = , find its equation lies along the straight line (8) 2 -1 5 Ê 3a 3a ˆ 15. (a) Find the radius of curvature at the point Á , ˜ on the curve x3 + y3 = 3axy Ë2 2¯ (8) (b) Find evolute of the rectangular hyperbola xy = c2.

(8)

(OR) 16. (a) Find the equation of the circle of curvature of the parabola y2 = 12x at the point (3, 6). x y + = 1, where a and b are a b parameters that are connected by the relation a + b = c. (8)

(b) Find the envelope of the straight line =

Ê x2 + y 2 ˆ ∂u ∂u 17. (a) If u = sin Á ˜ , prove x ∂x + y ∂y = tan u. x y + Ë ¯

(6)

(b) Determine the point where the function x2y + xy2 – axy has a minimum or a minimum. (10) (OR) 18. (a) If u = log(x3 + y3 + z3 – 3xyz), then prove

3 ∂u ∂u ∂u + + = x+ y+z ∂x ∂y ∂z

(8)

(b) Find the maximum and minimum of x2 + y2 + z2 subject to the condition ax + by + cz = p. (8) 19. (a) Solve (D2 + 6D + 9)y = e–2x x3 (b) Solve (2x + 3)2

d2y dx

2

(8)

- 2 (2 x + 3)

∂y - 12 y = 6x. ∂x

(8)

(OR) 20 (a) Solve (D2 – 9D + 20)y = 20x2 (b) Solve

d2y dx 2

+ 4 y = tan 2x by method of variation of parameters.

(6) (10)

Part II: Mathematics II

II-D.4

Solutions PART-A 1.

Ê a1 b1 Let us consider the upper triangular matrix A = ÁÁ 0 b2 ÁË 0 0 The C.E. is (a1 – l) (l2 – l) (c3 – l) = 0

c1 ˆ c2 ˜ ˜ c3 ˜¯

\ The eigenvalues are l = a1, b2, c3, that are the diagonal elements of A. 2.

When two matrices A and B are connected by the relation B = P –1AP, where P is a non-singular matrix, B is said to be obtained from A by a similarity transformation.

3.

The equation of the sphere is re-written

Âx

Ê 15 25 11 ˆ Center = Á , , ˜ and Radius = Ë 14 14 14 ¯

1 (225 + 625 + 121) = 196

2

-

15 25 11 xy- z =0 7 7 7

4.

Standard definition.

5.

The point of intersection of y = ex and y-axis is (0, 1)

971 . 14

3

(1 + y ¢ 2 ) 2 = 2 2 y¢ = e ; y¢¢ = e ; r = y ¢¢ x

6.

x

y = mx + 1 + m 2 i.e., (y – mx)2 = 1 + m2 i.e., m2(x2 – 1) – 2xym + (y2 – 1) = 0 This is a quadratic equation in m. \ The envelope is 4x2y2 = 4(x2 – 1)(y2 –1) i.e., x2 + y2 = 1

7.

∂u dx ∂u dy Ê xˆ Ê 1ˆ Ê xˆ Ê x ˆ du ◊ + ◊ = cos Á ˜ Á ˜ ◊ et + cos Á ˜ Á - 2 ˜ 2t = ∂ x dt ∂ y dt y y Ë ¯Ë ¯ Ë y¯ Ë y ¯ dt Ê e2 ˆ = cos Á 2 ˜ Ët ¯

8.

Ê et ˆ È et 2et ˘ et Í 2 - 3 ˙ = 3 cos Á 2 ˜ ◊ (t - 2) t ˙˚ t Ët ¯ ÍÎ t

f(x, y) = f (a, b) +

1 È( x - a ) f x ( a, b) + ( y - b) f y ( a, b) ˘ + ˚ 1! Î

1 È ( x - a ) 2 f xx (a, b) + 2 ( x - a ) ( y - b) f xy (a, b) + ( y - b) 2 f yy (a, b) ˘˚ + � 2! Î

Appendix D 9.

II-D.5

A.E. is m3 + m2 + 4m + 4 = 0 i.e., (m + 1) (m2 + 4) = 0 \ m = –1, ± 2i \ Solution is y = Ae–x + B cos 2x + C sin 2x, where A, B , C are arbitrary constants.

10.

Legendre's linear D.E. is {c0(ax + b)n Dn + c1 (ax + b)n – 1 Dn–1 + ... + cn–1 (ax + b) D + cn}y = R(x). where D =

d dx

PART - B È2 - l -2 11.(a) The C.E. of A is ÍÍ 1 1- l ÍÎ 1 3

2 ˘ 1 ˙˙ = 0 -1 - l ˙˚

i.e., (2 – l) (l2 – 4) –2 (l + 2) +2(l + 2) = 0 i.e., (l + 2) (l – 2)2 = 0 \ The eigenvalues of A are –2, 2, 2 When l = –2, the eigenvector is given by 4x1 – 2x2 + 2x3 = 0 and x1 + 3x2 + x3 = 0 Solving them, X1 = (4, 1, –7)T when l = 2, the eigenvector is given by 0. x1 – 2x2 + 2x3 = 0 x1 – x2 + x3 = 0 x1 – 3x2 – 3x3 = 0 T

Solving then, X2 = X3 = (0, 1, 1) È2 - l Í 11. (b) C.E. of A is Í 0 ÍÎ -1

0 2-l 0

-1 ˘ 0 ˙˙ = 0 2 - l ˙˚

i.e., (2 - l ){(2 - l ) 2 - 1} = 0 i.e., l 3 - 6l 2 + 11l - 6 = 0

To verify C.H. theorem, i.e., A3 – 6A2 + 11A – 6I = 0 (1) Ê 5 0 -4ˆ A = Á 0 4 0 ˜ and A3 = Á ˜ Ë -4 0 5 ¯ 2

Ê 14 0 -13ˆ Á 0 8 0 ˜ Á ˜ Ë -13 0 14 ¯

Using these values in the L.H.S of (1),

Part II: Mathematics II

II-D.6

Ê 14 0 -13ˆ Ê 30 0 -24ˆ Ê 22 0 -11ˆ Ê 6 0 0ˆ L.H.S. = Á 0 8 0 ˜ - Á 0 24 0 ˜ + Á 0 22 0 ˜ - Á 0 6 0˜ Á ˜ Á ˜ Á ˜ Á ˜ Ë -13 0 14 ¯ Ë -24 0 30 ¯ Ë -11 0 22 ¯ Ë 0 0 6¯ Ê 0 0 0ˆ = Á 0 0 0˜ = R.H.S. of (1) Á ˜ Ë 0 0 0¯ Hence C.H. theorem is verified. 12.

Ê 8 -6 2 ˆ Matrix of the given Q.F. is A = Á -6 7 -4˜ Á ˜ Ë 2 -4 3 ¯ 8-l -6 C.E. of A is -6 7 - l 2 -4

2 -4 = 0 3

i.e., l3 – 18l2 + 45l = 0 or l(l – 3) (l – 15) = 0 \ The eigenvalues of A are 0, 3, 15. when l = 0, the eigenvector X1 = (1, 2, 2)T when l = 3, the eigenvector X2 = (2, 1, –2)T when l = 15, the eigenvector X3 = (2, –2, 1)T È1 2 2 ˘ Í ˙ Modal matrix M = Í2 1 -2˙ ÍÎ2 -2 1 ˙˚ È1 Í3 Í Í2 The normalized modal matrix N = Í 3 Í Í2 ÎÍ 3

2 3 1 3 2 3

2 ˘ 3 ˙ ˙ 2 - ˙˙ 3 ˙ 1 ˙ 3 ˚˙

The required orthogonal transformation is X = NY i.e., x1 = x3 =

1 1 ( y1 + 2 y2 + y3 ) ; x2 = (2 y1 + y2 - 2 y3 ) and 3 3 1 (2 y1 - 2 y2 + y3 ) 3

2 2 2 The canonical form of the Q.F. is 0. y1 + 3 y2 + 15 y3 .

Appendix D

II-D.7

13. (a) Let the S.D. line cut the first line at P and the second line at Q. Then P ∫ (r1 + 1, – 2r1 + 5, r1 + 7) and Q ∫ (7r2 – 1, – 6r2 – 1, r2 – 1) \ D.R.'s of PQ are {7r2–n1 – r1 – 4, – 6r2 + 2r1 – 6, r2 – r1 – 8} PQ is ^r to the first line whose D.R.'s are (1, –2, 1) \ (7r2r1 – 4) – 2 (–6r2 + 2r1 – 6) + (r2 – r1 – 8) = 0 i.e., 20r2 – 6r1 = 0

(1)

PQ is also ^r to the second line whose D.R.'s are (7, –6, 1) \ 7(7r2 – r1 – 4) – 6 (–6r2 + 2r1 – 6) + (r2 – r1 – 8) = 0 i.e., 86r2 – 20r1 = 0

(2)

Solving (1) and (2), we get r1 = 0 and r2 = 0 \ P ∫ (3, 5, 7) and Q ∫ (–1, –1, –1) \ S.D. = PQ =

4 2 + 6 2 + 8 2 = 116 = 2 29 units.

Equations of the S.D. line are

x-3 y-5 z-7 = = 2 3 4

13. (b) (This problem is the same as the worked example 2.5, given in page I-2.53 of the book) 14. (a) The first sphere is Sx2 + 6y + 2z + 8 = 0 \ u = 0, v = 3, w = 1, d = 8 The second sphere is Sx2 + 6x + 8y + 4z + 20 = 0 \ u¢ = 3, v¢ = 4, w¢ = 2, d ¢ = 20 Now uu¢ + vv¢ + ww¢ = 2(0 + 12 + 2) = 28 i.e., the condition for orthogonal intersection of the two spheres is satisfied. \ The two spheres intersect at right angles. The plane of intersection of the two spheres S = 0 and S¢ = 0 is given by S – S¢ = 0. i.e., –6x – 2y – 2z – 12 = 0 or 3x + y + z + 6 = 0]

Part II: Mathematics II

II-D.8 14. (b)

P(x1, y1,z1)

M

A(1, –3,2)

Let P(x1, y1, z1) be any point on the cylinder AP =

( x1 - 1)2 + ( y1 + 3)2 + (z1 - 2)2

PM = radius of the cylinder = 2 AM = Projection of AP on the axis whose D.R.'s are (2, –1, 5) =

2 ( x1 - 1) - 1 ( y1 + 3) + 5(z1 - 2)

=

2 x1 - y1 + 5z1 - 15

2 2 + 12 + 52

30

From DAPM, AM2 + MP2 = AP2 1 i.e., (2x1 – y1 + 5z1 – 15)2 + 4 = (x1 – 1)2 + (y1 + 3)2 + (z1 – 2)2 30 i.e., (2x1 – y1 + 5z1 – 15)2 + 120 = 30 {x12 + y12 + z12 – 2x1 + 6y1 – 4z1 + 14} i.e., 26x12 + 29y12 + 5z12 + 4x1y1 + 10y1z1 – 20z1x1 + 150y1 + 30z1 + 75 = 0 \ The locus of (x1, y1, z1) or the equation of the cylinder is 26x2 + 29y2 + 5z2 + 4xy + 10yz – 20zx + 150y + 30z + 75 = 0 15. (a) This problem is the same as the worked example 3.1 given in page I-3.7 of the book] 15. (b) The parametric equations of the R.H. xy = c2 are x = ct, y = c/t . . \ x = c; y = –c/t2; y¢ = –1/t2 2 1 2 ◊ = 3 3 t c ct – Let ( x , y– ) be the centre of armature at ‘t’ y ¢¢ =

Appendix D

2

y ¢ (1 + y ¢ ) x– = x = ct + y ¢¢ = ct +

II-D.9

1 t2

1ˆ Ê ÁË 1 + 4 ˜¯ t 3 2/ct

ct 2

1ˆ 3 c Ê ÁË 1 + 4 ˜¯ = 2 ct + 3 t 2t

(1 + y ¢ 2 ) c ct 3 y– = y + = + y ¢¢ t 2

1 ˆ 3 c ct 3 Ê ÁË 1 + 4 ˜¯ = 2 t + 2 t

–x + y– = c ÏÌt 3 + 1 + 3 Ê t + ÁË 2Ó t3

1ˆ ¸ c Ê t+ ˝= t ˜¯ ˛ 2 ÁË

1ˆ t ˜¯

3

cÏ 1 1ˆ ¸ c Ê 1ˆ Ê y– – –x = Ìt 3 - 3 - 3 Á t - ˜ ˝ = Á t - ˜ Ë 2Ó t¯˛ 2 Ë t¯ t

3

\

Ê cˆ ( x– + y– )2/3 – (y– – –x )2/3 = Á ˜ Ë 2¯

2/3

Ê cˆ =4¥ Á ˜ Ë 2¯

ÏÔÊ ÌÁ t + Ë ÓÔ

2

1ˆ Ê - Át Ë t ˜¯

1ˆ t ˜¯

2¸

Ô ˝ ˛Ô

2/3

or (4c)2/3

\ The equation of the evolute is (x + y)2/3 – (x – y)2/3 = (4c)2/3 16. (a) [This problem is the same as the worked example 3.15 given in page I-3.18 of the book] 16. (b) This problem is the same as the worked example 3.9 given in page I-3.32 of the book] 17. (a) [This problem is the same as the worked example (12) given in page II-A.15 of the book] 17. (b) Let f(x, y) = x2y + xy2 – axy fx = 2xy + y2 – ay; fy = x2 + 2xy – ax fxx = 2y; fxy = 2x + 2y – a; fyy = 2x The possible extreme points are given by fx = 0 and fy = 0. i.e., 2xy + y2 – ay = 0 or y (2x + y – a) = 0 (1) and x2 + 2xy – ax = 0 or x(x + 2y – a) = 0 (2) Ê a aˆ Solving (1) and (2), we get the points (0, 0), (a, 0), (0, a) and Á , ˜ Ë 3 3¯ At the point (0, 0), A = fxx = 0; B = fxy = –a; C = fyy = 0 AC – B2 < 0, but A = 0 = C

Part II: Mathematics II

II-D.10

At the point (a, 0), A = 0, B = a; C = 2a AC – B2 < 0, but A = 0 and C > 0 At the point (0, a), A = 2a; B = a; C = 0 AC – B2 < 0, but A > 0 and C = 0 2a a 2a Ê a aˆ , B = and C = At the point Á , ˜ , A = Ë 3 3¯ 3 3 3 AC – B2 > 0, and A or C > 0. Ê a aˆ \ f(x, y) has a minimum at the point Á , ˜ Ë 3 3¯ 18. (a) [This problem is the same as the worked example (2) given in page II-A.5 of the book] 18. (b) Consider the auxiliary function f = f + lg, i.e., f(x, y, z) = x2 + y2 + z2 + l(ax + by + cz – p) The stationary points are given by fx = 0, fy = 0, fz = 0, fl = 0 i.e., 2x + la = 0

(1)

2y + lb = 0

(2)

2z + lc = 0

(3)

ax + by + cz – p = 0

(4)

\l= -

2x 2y 2z ==a b c

\ (a2 + b2 + c2)l = –2(ax + by + cz) = –2p \l= -

2p Sa

2

\ x=-

a ap bp cp l or ;y= 2 ;z= 2 2 2 Sa Sa Sa

\ Extreme value of f(x, y, z) =

1 ( Sa 2 ) 2

(a 2 + b 2 + c 2 ) p 2 =

p2 Sa 2

.

19. (a) [This problem is the same as the worked example 1.6 given in page II-1.25 of the book.] 19. (b) [This problem is the same as the worked example 1.6 given in page II-1.40 of the book] 20. (a) A.E. is m2 – 9m + 20 = 0 i.e., (m – 4)(m – 5) = 0 C.F. = Ae4x + Be5x

Appendix D

II-D.11

1 1 20 x 2 = x2 P.I. = D2 - 9 D + 20 D(D - 9) 1+ 20 D(D - 9) D2 (D - 9)2 ÔÏ Ô¸ + - �˝ x 2 = Ì1 20 400 ÓÔ ˛Ô 9 61 2 ˆ 2 9 61 Ê D+ D ˜ x = x2 + x+ = Á1 + Ë ¯ 20 400 10 200 \ G.S. is y = C.F. + P.I. 20. (b) {This problem is similar to the worked example 1.4 given in page II- 1.79 of the book. We have to replace 'a' by 2] The solution of the given problem is, therefore, y = C1, cos 2x + C2 sin 2x –

1 cos 2x. log (sec 2x + tan 2x). 4

Appendix

E

Solutions to the May/June 2011 Question Paper of Anna University, Tirunelveli M490 B.E./B.TECH. DEGREE EXAMINATIONS, MAY/JUNE-2011 REGULATIONS 2008 First Semester MA 12- MATHEMATICS CIVIL ENGINEERING (Common to all B.E/B.Tech) Times: Three Hours

Maximum:-100 marks ANSWER ALL QUESTIONS PART-A (10 ¥ 2 = 20 marks)

1.

Prove that x2 + 2y2 + 3z2 + 2xy + 2yz – 2zx is indefinite.

2.

È1 1 ˘ Find the eigen values of A = Í ˙ Î3 -1˚

3.

Show that the spheres x2 + y2 + z2 + 6y + 2z + 8 = 0 and x2 + y2 + z2 + 6x + 8y + 4z = 0 intersect at right angles.

4.

If

5.

Find the radius of curvature of x2 + y2 + 4x – 12y + 4 = 0

6.

Find the envelope of the straight line ty – x = at2 where t is the parameter.

7.

Find the stationary points of the function

x y z 2 2 2 = = a generator of the cone x + y – z = 0 find the value of K. 1 2 k

z = x3y2(12 – x – y).

Part II: Mathematics II

II-E.2

8.

If u = tan -1

9.

Evaluate

10.

x2 + y 2 ∂u find x+ y ∂x

p

a sin q

0

0

Ú Ú

Show that

2

3

ÚÚÚ 0

1

r dr dq

2

xy 2 z dy dx = 26

1

PART-B (5 ¥ 16 = 80 marks) 11. (a) Diagonalise the matrix and hence find A4

(16)

È 3 -1 1 ˘ A = ÍÍ-1 5 -1˙˙ ÎÍ 1 -1 3 ˙˚ Or (b)

È 2 -3 1 ˘ (i) Show that A = ÍÍ 3 1 3 ˙˙ satisfies the equation A(A – 5) (A + 2I) = 0 (8) ÍÎ-5 2 -4˙˚ È2 2 -7 ˘ (ii) Find the Eigen values and Eigen vectors of A = ÍÍ2 1 2 ˙˙ (8) ÍÎ0 1 -3˙˚

12.(a) (i) Find the equation of the circular cylinder having for its base the circle (8) 2 2 2 x + y + z = 9, x – y + z = 3 and passes through (–2, 1, 0). (ii) Find the equation of the tangent plane to the sphere

(8)

x2 + y2 + z2 + 2x – 4y + 6z – 7 = 0, which passes through the line 6x – 3y – 23 = 0 = 3z + 2 (b) (i) Find the center and radius of the circle x2 + y2 + z2 – 8x + 4y – 8z – 45 = 0 = x + 2y + 2z – 3

(8)

(ii) Find the equation of the sphere passing through the point (0, 0, 0), (1, 0 0), (0, 1, 0) and (0, 0, 1). (8) 13. (a) Find the equation of the circle of curvature of y2 = 12x at (3, 6) (b) (i) Find the evolute of the ellipse.

(8)

Appendix E

II-E.3

x y + = 1 where the parameters a, b are related by a b ab = c2 where c is known.

(ii) Find the envelope of

È x + 2 y + 3z 14. (a) (i) If u = sin - 1 Í Í x8 + y 8 + z 8 =0 Î

˘ ˙ show that x ∂u + y ∂u + z ∂u + 3 tan u ˙ ∂x ∂y ∂z (8) ˚

(ii) Expand x2y + 3y – 2 at(1, –2) using Taylor's theorem

(8)

Or (b) (i) If x = a cosh a. cosb, y = a sinh a. sinb show that

∂( x , y ) = ∂(a , b ) (8)

a2 [cos h 2a - cos 2 b ] 2 (ii) Find three positive numbers such that the sum is a constant and their product is maximum. (8) 15.(a) (i) By changing the order of integration evaluate

4a

2 ax

0

x2 4a

Ú Ú

xy dy dx

(8)

(ii) Using double integration find the area of the parallelogram whose vertices are A(1, 0), B(3, 1), C(2, 2), D(0, 1) (8) Or a

(b)

(i) Transform the integration

ÚÚ 0

y

dx dy to polar co-ordinates.

0

(ii) Show that 1

ÚÚ

0 0

1 - x2

(8)

Ú

0

1 - x2 - y 2

1

2 dz dy dx = p 8 1 - x2 - y 2 - z 2

Solutions PART-A 1.

Ê 1 1 -1ˆ Matrix of the given Q.F is A = Á 1 2 1 ˜ Á ˜ Ë -1 1 3 ¯ 1 1 = 1; D1 = |1| = 1; D2 = 1 2

Part II: Mathematics II

II-E.4

1 1 -1 D3 = 1 2 1 = 1 ¥ (6 – 1) –1 ¥ (3 + 1) – 1 ¥ (1 + 2) = 5 – 4 – 3 = –2 -1 1 3 Since D1 > 0, D2 > 0 and D3 < 0, the given Q.F. is indefinite

2.

C.E. of A is

1- l 3

1 = 0 i.e, l2 – 4 = 0 -1 - l

\ Eigen values of A are 2 and –2 3.

For the I sphere, u = 0, v = 3, w = 1, d = 8 For the II sphere, u1 = 3, v1 = 4, w1 = 2, d1 = 0 2uu1 + 2vv1 + 2ww1 = 28 π d + d1 Condition for orthogonality is not satisfied \ The two spheres do not interest at right angles. [The given problem is wrong]

4.

l = 1, m = 2, n = k should satisfy l2 + m2 – n2 = 0 \ 1 + 4 – k2 = 0

5.

\k= ± 5

x2 + y2 + 4x – 12y + 4 = 0 (1) represents a circle. 4 + 36 - 4 = 6

Radius of the circle =

Radius of curvature at any point of a circle is its radius and hence r = 6 6.

at2 – yt + x = 0. This is a quadratic equation is t, \ Equation of the envelope is B2 – 4AC = 0 i.e, y2 = 4ax

7.

8.

[This is problem is the same as the worked example 4.2, given in page I-4.53 of the book] Ê x2 + y 2 ˆ n = tan -1 Á ˜ Ë x+ y ¯ ∂u = ∂x

=

{( x + y )2 x - ( x 2 + y 2 )1}

1 Ê x2 + y 2 ˆ 1+ Á ˜ Ë x+ y ¯

2

x 2 + 2 xy - y 2 ( x 2 + y 2 )2 + ( x + y)2

( x + y)2

Appendix E p a sin q

9.

Ú Ú 0

p

r dr dq =

Ú 0

0

a2 a 2 sin 2 q dq = 4 2

p

Ú (1 - cos 2q ) dq 0

p

2

a Ê sin 2q ˆ p a2 = q 4 ÁË 2 ˜¯ 0 4

= 10.

II-E.5

The given problem contains an error. It must be taken I=

2

3

ÚÚÚ 0

=

3 2

1

Ú

2

0

2

2

2

xy z dz dy dx =

1

ÚÚ 0

3

1

2

Ê z2 ˆ xy 2 Á ˜ dy dx Ë 2 ¯1

3

Ê y3 ˆ 3 26 x Á ˜ dx = ¥ 3 2 3 Ë ¯1

Ú

2

2

x dx =

0

3 26 Ê x 2 ˆ = 26. ¥ ¥ 2 3 ÁË 2 ˜¯ 0

PART-B 11.(a) [Refer to the worked example 1.9 given in page I-1.41 of the book, in which the eigen values and eigen vectors of the given matrix A have been found] Ê 0 1 1ˆ Modal matrix M = Á 0 1 -2˜ ; M–1 = Á ˜ Ë -1 1 1 ¯

Ê 3 0 -3ˆ Á2 2 2 ˜ Á ˜ Ë 1 -2 1 ¯

Digitalization of A is given by M–1 AM = D(2, 3, 6) A4 is given by A4 = MD4 M–1 0 ˆ Ê16 0 Ê 3 0 -3ˆ 1Á Á ˜ 0 ¥ 2 2 2˜ Now D M = 0 81 Á ˜ 6Á ˜ Ë 0 0 1296¯ Ë 1 -2 1 ¯ 4

–1

0 -8 ˆ Ê 8 Á 27 27 ˜ = 27 Á ˜ Ë 216 -432 216¯ 0 -8 ˆ Ê1 1 1ˆÊ 8 Á ˜ Á MD M = 0 1 -2 27 27 27 ˜ Á ˜Á ˜ Ë -1 1 1 ¯ Ë 216 -432 216¯ 4

i.e,

–1

Ê 251 -405 235 ˆ A = Á -405 837 -405˜ Á ˜ Ë 235 -405 251 ¯ 4

Part II: Mathematics II

II-E.6

11. (b) (i) The given answer is wrong. It must have been A(A – I) (A + 2I) = 0, instead of A(A – 5) (A + 2I) = 0, as (A – 5) has no meaning. The solution is given for the corrected problem] 2-l -3 1 C.E. of A is 3 1- l 3 = 0, i.e, l3 + l2 – 2l = 0 -5 2 -4 - l or l (l + 2) (l – 1) = 0 By C.H. theorem, A satisfies its C.E. \ A(A + 2I) (A – I) = 0 11. (b) (ii) (The solution for this problem is available in the worked example 1.5 given in page I-1.53 of the book) 12 (a) (i) Assuming that the required cylinder is a right-circular cylinder, we find the equations of the axis of the cylinder. The center of the base circle is found as (1, –1, 1). It is the foot of the ^r from the center of the sphere Sx2 = 9, namely (0, 0, 0) on the intersecting plane. The axis of the cylinder is ^ r to the base circle or the intersecting plane x – y + z = 3 and passed through (1, –1, 1)

M

P(x1, y1,z1)

N

Q (–2,1,0)

C (1, –1,1)

\ The equations of the axis are

x -1 y +1 z -1 = = =r 1 -1 1

Any point on the axis is (r + 1, –r – 1, r + 1) If these co-ords represent N, then the D.R.'s of QN are (r + 3, –r – 2, r + 1) QN ^ r the axis \ r + 3 + r + 2 + r + 1 = 0; \ r = –2 \ N = (–1, 1, –1) \ Radius of the cylinder = QN = PM = 12 + 02 + 12 = 2 Let P(x1, y1, z1) be any point on the required cylinder Then CP =

( x1 - 1) 2 + ( y1 + 1) 2 + ( z1 - 1) 2

CM = Projection of CP on the axis

Appendix E

=

( x1 - 1) - ( y1 + 1) + ( z1 - 1) 2

2

2

1 +1 +1

II-E.7

=

x1 + y1 + z1 - 3 3

Now CM2 + MP2 = CP2 2 i.e., ( x1 - y1 + z1 - 3) + 2 = (x1 – 1)2 + (y1 + 1)2 + (z1 – 1)2 3 i.e., (x1 + y1 + z1)2 – 6(x1 + y1 + z1) + 9 + 6 = 3 {x12 + y21 + z21 – 2x1 + 2y1 – 2z1 + 3} i.e., 2(x12 + y21 + z21) + 2x1y1 + 2y1z1 – 2z1x1 – 6 = 0 \ The locus of (x1, y1, z1) or the equation of the required cylinder is Sx2 + xy + yz – zx – 3 = 0 12 (a) (ii) Center of the gives sphere, C is (–1, 2, –3) r, Radius of the sphere =

1+ 4 + 9 + 7 =

21

Equation of any plane passing through the line 6x – 3y – 23 = 0 = 3z + 2 is 6x – 3y + 3lz + (2l – 23) = 0 If this plane tonches the given sphere, the length of the ^ r CM from C on it = r C r

M

i.e.,

-6 - 6 - 9l + 2l - 23 36 + 9 + 9l 2

=

21

i.e., (7l + 35)2 = 21 ¥ 9 (l2 + 5) or 7(l + 5)2 = 27(l2 + 5) i.e., 20l2 – 70l – 40 = 0 or 2l2 – 7l – 4 = 0 i.e., (l – 4) (2l + 1) = 0 1 2 \ There are two tangent planes of the sphere passing through the given line. They are 6x – 3y + 12z – 15 = 0 or 2x – y + 4z – 5 = 0 and 4x – 2y – z – 16 = 0

\ l = 4, -

12 (b) (i) C, Center of the sphere = (4, –2, 4). R, Radius of the sphere = 16 + 4 + 16 + 45 = 9. C¢, the Center of the circle is the foot of the ^ r from C on x + 2y + 2z– 3 = 0

Part II: Mathematics II

II-E.8

C¢C

r R

C

Science CC¢ is normal to the plane, its D.R.'s are (1, 2, 2) \ Equations of CC¢ are

y+2 z-4 x-4 = = Any point of CC¢ is 2 2 1

(r + 4, 2r – 2, 2r + 4) If these co-ordinates represent C¢, (r + 4) + 2(2r – 2) + 2(2r + 4) – 3 = 0 5 i.e., 9r + 5 = 0 \ r = 9 28 26 ˆ Ê 31 \ C1 = Á , , Ë 9 9 9 ˜¯ 2

and CC1 =

2

Ê 5ˆ Ê 10 ˆ Ê 10 ˆ ÁË 9 ˜¯ + ÁË 9 ˜¯ + ÁË 9 ˜¯

2

=

5 units 3

12. (b) (ii) Let the equation of the required sphere be Sx2 + 2ux + 2vy + 2wz + d = 0 (1) passes through (0, 0, 0) \ d = 0 (1) passes through (1, 0, 0)

\ 2u + 1 = 0 or u = -

(1) 1 2

1 1 and w = - . 2 2 \ The equation of the required sphere is Similarly, v = -

x2 + y2 + z2 – x – y – z = 0 13.(a) [This problem is the same as the worked example 3.15 given in page I-3.18 of the book] 13. (a) (i) Parametric equation of the ellipse are x = a cos q, y = b sin q x˙ = –a sin q, y˙ = b cos q \ y¢ = -

b cot q a

Appendix E

and y¢¢ =

II-E.9

b 1 b cosec 2q ◊ = - 2 3 a a sin q a sin q

y¢ (1 + y ¢ 2 ) = a cos q – a cos q sin2 q x– = x y ¢¢

Ê b 2 cos 2 q ˆ Á1 + 2 2 ˜ a sin q ¯ Ë

1 cos q (a 2 sin 2 q + b 2 cos 2 q ) a

= a cos q –

Ê a 2 - b2 ˆ 3 = Á ˜ cos q a Ë ¯ a 2 sin 3 q Ê a 2 sin 2 q + b 2 cos 2 q ˆ 1 (1 + y ¢ 2 ) = b sin q y– = y + Á ˜ b a 2 sin 2 q y ¢¢ Ë ¯ Ê a 2 - b2 ˆ 3 = -Á ˜ sin q Ë b ¯ Ê ax ˆ \ Á ˜ Ë a 2 - b2 ¯

2/3

Ê - by ˆ +Á 2 ˜ Ë a - b2 ¯

2/3

= cos2q + sin2q = 1

\ Locus of (x–, y– ); i.e., the equation of the required evolute is (ax)2/3 + (by)2/3 = (a2 – b2)2/3 13. (b) (ii) The equation of the family of lines as Constraint is ab = c2

x y + =1 a b

(1) (2)

x ya + =1 a c2 i.e., ya2 – c2a + c2x = 0, (3) where 'a' is the parameter (3) is a quadratic equation in 'a'. \ The equation of the envelope is Using (2) in (1), we get

c4 – 4c2xy = 0 or xy = 14. (a) (i) sin2u =

( x + 2 y + 3z )2 S x8

c2 4 or Sx8. sin2u = (x + 2y + 3z)2

Diffg. (1) partially w.r.t. x; 2 Sx8 sin u cos u. ux + 8x7. sin2 u = 2(x + 2y + 3z) \ (Sx8 sin u cos u) xux + 4x8 sin2u = x(x + 2y + 3z) Similarly, (Sx8 sin u cos u) yuy + 4y8 sin2u = 2y(x + 2y + 3z)

(1)

(2) (3)

Part II: Mathematics II

II-E.10

and (Sx8 sin u cos u) z uz + 4z8 sin2u = 3z (x + 2y + 3z)

(4)

(2) + (3) + (4) gives Sx8 sin u cos u (xux + yuy + zuz) + 4(x + 2y + 3z)2 = (x + 2y + 3z)2 i.e., (Sx8)sin u cos u (xux + yuy + zuz) + 3 (x + 2y + 3z)2 = 0 \ xux + yuy + zuz +

3 ( Sx8 ) sin 2 u ( Sx8 ) sin u cos u

=0

i.e., xux + yuy + zuz + 3 tan u = 0 14. (a) (ii) f(x, y) = x2y + 3y – 2; fx = 2xy; fy = x3 + 3; fxx = 2y; fxy = 2x; fyy = 0; fxxx = 0; fxxy = 2, fxyy = 0 fyyy = 0

f(1, –2) = –2 – 6 – 2 = –10 fx(1, –2) = –4; fy(1, –2) = 4; fxx(1, –2) = –4; fxy(1, –2) = 2, fyy(1, –2) = 0 fxxx(1, –2) = 0, fxxy = 2; fxyy = 0 and fyyy = 0

Taylor series is f(x, y) = f(1, –2) +

1 {( x - 1) f x (1, - 2) + ( y + 2) f y (1, - 2)} + 11

1 {( x - 1) 2 f xx (1, - 2) + 2( x - 1) ( y + 2) f xy (1, - 2) + ( y + 2) 2 y yy (1, - 2)} + 2!

\ x2y + 3y – 2 = -10 + +

14, (b) (i)

xa ∂( x , y ) = ∂(a , b ) ya

1 {-4 ( x - 1) + 4( y + 2)} 1!

1 1 {-4 ( x - 1) 2 + 2 ( x - 1) ( y + 2)} + {2 ( x - 1) 2 ( y + 2)} 2! 3! xb yb

=

a sin h a cos b a cos h a sin b

- a cos h a sin b a sin h a cos b

= a2 sin h2 a cos2 b + a2 cos h2 a sin2 b =

a2 [(cos h 2a - 1) (1 + cos 2 b ) + (1 + cos h 2a ) (1 - cos 2 b )] 4

=

a2 (cos h 2a - cos 2 b + cos h 2a - cos 2 b ) 4

=

a2 (cos h 2 a - cos 2 b ) 2

14 (b) (ii) Let the 3 numbers be x, y, z To maximum P = xyz, s.t.S = x + y + z = 3a

Appendix E

II-E.11

Consider f = P + lS = xyz + l (x + y + z – 3a) The stationary points of f are given by yz + l = 0 (1); zx + l = 0 (2); xy + l = 0(3) and x + y + z = 3a

(4)

From (1), (2), (3); lx = ly = lz or x = y = z \ P is maximum, when x = y = z = a, from (4) Maximum of P = a3 15 (a) (i) The region of integration is bounded by x2 = 4 ay, y2 = 4 ax, x = 0 and x = 4a. y B (4a, 4a)

C A

x

0 4 a 2 ay

I=

Ú Ú 0

4a

=

Ú 0

=

1 2

xy dx dy

y2 4a 2 ay

Ê x2 ˆ yÁ ˜ Ë 2 ¯ y2

dy

4a 4a

Ú

y {4ay -

0

y4 16a 2

} dy 4a

1È y3 y6 ˘ 1 Ê 256 128 ˆ 4 a = Í 4a ˙ = Á 2 ÎÍ 3 96a 2 ˚˙ 2Ë 3 3 ˜¯ 0 =

64 4 a 3

15 (a) (ii) C(2,2)

2

x

– y= –2

x + y= 4

D(0,1) x + y= 1

B(3,1) x

0

A(1,0)

y= –2

1

x

Part II: Mathematics II

II-E.12

gm ABCD = Area of D ABD + Area of D DBC

Area of the

1 1+ 2y

=

Ú Ú

2

dx dy +

0 1- y

Ú

dx dy

2

Ú

[ x]11 +- 2y y dy + [ x]24 y- -y 2 dy

0

1

1

=

Ú Ú

1 2y - 2

1

=

4y

2

Ú 3 y dy + Ú (6 - 3 y) dy 0

1

1

2 Ê 3y2 ˆ 3y ˆ Ê 6 + y = Á Á ˜ 2 ˜¯ 1 Ë 2 ¯0 Ë

=

3 Ê + 62 ÁË

9ˆ = 3 units. 2 ˜¯

15 (b) (i) The region of integration is shown in the figure y

p 2 a cos eq

I=

Ú Ú p 4

r = a cosec q

r dr dq ,

0

q

since dx dy = r dr dq 1 b a

15 (b) (ii)

I=

dz dy dx

ÚÚÚ

a2 - z 2

0 0 0

1 - x2 - y 2

where a =

1 - x2

and b =

i.e., I =

=

=

p 2 p 2 p 2

1 b

Ú Ú dy dx 0 0

Ú

1

0

b

[ y ] dx 0

1

Ú 0

1 - x 2 dx

0 1 b

=

ÚÚ 0 0

a

Ê -1 z ˆ ÁË sin a ˜¯ dy dx 0

p q= 4

x

Appendix E

=

p 2

1

1

1 Èx 2 -1 ˘ Í 2 1 - x + 2 sin x ˙ dx Î ˚ 0

Ú

0

1

=

II-E.13

p p p2 ¥ ¥ dx = 2 4 8 0

Ú

Appendix

F

Solutions to the December 2010/January 2011 Question Paper of Anna University, Tiruchirapalli G 0751 B.E./B.Tech. DEGREE EXAMINATIONS, DECEMBER 2010/JANUARY 2011 First Semester MA 1101 Mathematics-I (Common to all Branches) REGULATION 2008 Times: Three Hours Maximum:-100 marks Answer ALL questions. PART-A (10 ¥ 2 = 20 marks)

1.

Ê 1 1 3ˆ If –2 and 6 are the two Eigen values of the matrix A = Á 1 5 1˜ , find the Á ˜ Eigen values of A–1. Ë 3 1 1¯

2.

Define a quadratic form.

3.

Find the equation of the sphere described on the line joining the points (2, –1, 4) and (–2, 2, –2) as diameter.

4.

Find the angle between two lines having direction ratios (1, 2, 1) and (2, –3, 6) respectively.

5.

Find the curvature of the curve y = x2 at (1, 1).

6.

Find the envelope of the family y = mx +

a . m

II-F.2

Part II: Mathematics II

7.

Define Euler's theorem.

8.

If x = r cos q and y = r sin q, then find

9.

Find the C.F of (D4 – 2D2 + 1)y = e–x.

10.

∂ ( x, y ) ∂ (r , q )

Find the P.I of (D4 – 4D + 3)y = x2. PART B - (5 ¥ 16 = 80 marks)

11.

Reduce the quadratic form 2x12 + 6x22 + 2x23 + 8x1x3 to canonical form by orthogonal reduction and hence find the nature of the quadratic form. Or

È 2 1 -1˘ 12. (a) Find the Eigen values and Eigen vectors of the matrix ÍÍ 1 1 -2˙˙ . ÍÎ-1 -2 1 ˙˚ (b) Verify Cayley-Hamilton theorem for the matrix A = find A4.

È 2 -1 2 ˘ Í-1 2 -1˙ and hence Í ˙ ÍÎ 1 -1 2 ˙˚

13. (a) Find the equation of the sphere having the circle x2 + y2 + z2 + 10y – 4z – 8 = 0, x + y + z = 3 as a great circle. (b) Find the equation of the cone whose vertex is at the point (1, 1, 3) and which passes through the ellipse 4x2 + z2 = 1, y = 4. Or 14. (a) Find the length and equations of the shortest distance between the lines x-3 x+3 y+7 z-6 y-8 z -3 and = = = . = 3 -3 2 4 -1 1 x = (b) Find the equation of the cylinder whose generators are parallel to 3 y z = and whose guiding curve is x2 + y2 = 25, z = 0. 1 6 15. (a) Find the equation of the circle of curvature of the parabola y2 = 12x at the point (3, 6) (b) Find the centre of curvature of the rectangular hyperbola xy = c2.

Appendix F

II-F.3

x y + = 1 where the parameters a a b and b are connected by the relation ab = c2.

16. (a) Find the envelope of the system of lines

(b) Find the radius of curvature of the curve xy2 = a3 – x3 at the point (a, 0) 17. (a) Discuss the maxima and minima of the function f(x, y) = x3 + 3xy2 – 15x2 – 15y2 + 72x. (b) Find the volume of the greatest rectangular parallelepiped that can inscribed 2 2 2 in the ellipsoid whose equation is x + y + z = 1. a 2 b2 c2 Or 18. (a) Find the Taylor's expansion of the function f(x, y) = ex sin y in powers of x and y upto third degree terms. (b) If u is a function of x and y and x and y are functions of r and q given by x = er cos q = er sin q, show that ∂2u ∂x 2

+

∂2u ∂y 2

Ê ∂2u ∂2u ˆ = e–2r Á 2 + 2 ˜ ∂q ¯ Ë ∂r

19. (a) Solve: (x2D2 + 4xD + 2)y = sinx. (b) Solve: (D3 – 1)y = x sinx. Or 20. (a) Use method of variation of parameters to solve (D2 + 1)y = tan x. (b) Solve (D2 – 4D + 13)y = e2x cos 3x.

Solutions

1.

PART-A Ê 1 1 3ˆ A = Á 1 5 1˜ ; l1 + l2 + l3 = 1 + 5 + 1 = 7; i.e., –2 +6 + l3 = 7 \ l3 = 3 Á ˜ Ë 3 1 1¯ 1 1 1 \ The eigenvalues of A–1 are - , and . 2 6 3

Part II: Mathematics II

II-F.4 2.

Definite given in page I-1.64 of the text book.

3.

I part of W.E. 2.8 given in page I-2.55 of the text book.

4.

If q is the angle between the given lines (1, 2, 1) and (2, –3, 6), Ê 6ˆ 2-6+6 2 -1 or q = cos Á = Then cos q = ˜ 6 ◊ 49 7 6 Ë 21 ¯

5.

y = x2; (y¢)(1, 1) = (2x)(1, 1) = 2; (y ¢¢)(1, 1) = 2 C(1, 1) =

6.

2 3 4) 2

=

2 5 5

.

(1 + Equation of the family is xm2 – ym + a = 0, that is a quadratic equation in m. \ Equation of the envelope is (–y)2 – 4ax = 0 or y2 – 4ax = 0

7. 8. 9.

Statement of Eulers theorem given in page II - A.3 xr ∂ ( x, y ) = yr ∂ (r , q )

xq cos q = yq sin q

- r sin q =r r cos q

(D4 – 2D2 + 1)y = e–x; A.E. is (m2 – 1)2 = 0 \ m = –1, –1, + 1, + 1 \ C.F = (Ax + B)e–x + (Cx + D)ex

10.

-1

1 Ï Ê D 4 - 4 D ˆ Ô¸ 2 x = ÔÌ1 + Á P.I. = 4 ˜˝ x D - 4D + 3 3 ÓÔ Ë 3 ¯ ˛Ô 1

2

=

˘ 1È D( D3 - 4) D 2 ( D3 - 4) 2 + - ...˙ x 2 Í1 3 ÎÍ 3 9 ˙˚

=

1 3

16 2 ˘ 2 1 È 2 8 32 ˘ È 4 1 (9 x 2 + 24 x + 32) Í1 + 3 D + 9 D ˙ x = 3 Í x + 3 x + 9 ˙ or 27 Î ˚ Î ˚ PART - B

11.

This is the worked example 1.2, given is page I - 1.68 of the text book.

12. (a) The solution of this problem is available in the worked example 1.7, given in page I - 1.57 of the text book. (b) This is the worked example 1.2, given in page I-1.50 of the text book.

Appendix F

II-F.5

13. (a) This is the worked example 2.11, given in the page I - 2.57 of the text book. (b) Let a generator of the cone be

x -1 y -1 z-3 = = =r l m n

(1)

Any point on (1) is P(lr + 1, mr + 1, nr + 3) Since (1) intersects the base curve, for some value of r, 4(lr + 1)2 + (nr + 3)2 = 1 and mr + 1 = 4

\r=

2 2 ÏÔ Ê 3l ˆ Ê 3n ˆ ¸Ô 3 + 3˜ ˝ = 1 and Ì4 Á + 1˜ + Á Ëm ¯ Ô m ÔÓ Ë m ¯ ˛

i.e., 4(3l + m)2 + 9(m + n)2 = m2 i.e., 36l2 + 12m2 + 9n2 + 24lm + 18mn = 0

(2)

Using (1) in (2), the equation of the cone is 12(x – 1)2 + 4(y – 1)2 + 3(z – 3)2 + 8(x – 1) (y – 1) + 6(y – 1) (z – 3) = 0 i.e., 12x2 + 4y2 + 3z2 + 8xy + 6yz – 32x – 34y – 24z + 69 = 0 14. (a) This is the worked example 2.14, given in the page I-2.42 of the text book. (b) Let P(x1, y1, z1) be any point on the cylinder Then the equations of the generator through P are z - z1 x - x1 y - y1 = = =r 6 3 1

(1)

Any point Q on (1) is (x1 + 3r, y1 + r, z1 + 6 r ) Since the generator intersects the guiding curve, (x1 + 3r)2 + (y1 + r)2 = 25 and z1 + 6 r = 0 Now r = -

2

2

Ê Ê 3z ˆ z ˆ and Á x1 - 1 ˜ + Á y1 - 1 ˜ = 25 6 Ë Ë 6¯ 6¯

z1

i.e., ( 6 x1 - 3 z1 ) 2 + ( 6 y1 - z1 ) 2 = 150 \ The equation of the cylinder is ( 6 x - 3 z ) 2 + ( 6 y - z ) 2 = 150 i.e., 6(x2 + y2) + 10z2 – 6 6 z x - 2 6 y z - 150 = 0

Part II: Mathematics II

II-F.6

15. (a) This is the worked example 3.15, given in the page I - 3.18 of the text book. (b) The parametric equation of the R.H. are x = ct and y = y¢ =

c t

2 1 y� = - 2 ; y¢¢ = ct 3 t x�

y¢ ct (1 + y ¢ 2 ) = ct + x– = x y ¢¢ 2

Ê 1ˆ 3e c ÁË1 + t 4 ˜¯ = 2 t + 2t 3

1 c c t3 (1 + y ¢ 2 ) = + y– = y + y ¢¢ t 2 – , The Centre of curvature is (x y–)

3e c 3 Ê 1ˆ + t ÁË1 + t 4 ˜¯ = 2 t 2

x y x ay c2 in + = 1, we get + 2 = 1 is ya2 – c2a + c2x = 0, which a b a c a is a quadratic equation in a

16. (a) Using b =

Equation of the envelope is c4 – 4c2 xy = 0 (b)

or 4xy = c2 This is the worked example 3.2, given in the page I - 3.8 of the text book.

17. (a) This is the worked example 4.1, given in the page I - 4.52 of the text book. (b) This is the worked example 4.13, given in the page I - 4.62 of the text book. 18. (a) f(x, y) = ex sin y; fx(x, y) = ex sin y; fy(x, y) = ex cos y; fxx(x, y) = ex sin y; fxy(x, y) = ex cos y; fyy(x, y) = –ex sin y; fxxx(x, y) = ex sin y; fxxy(x, y) = ex cos y; fxyy = –ex sin y; fyyy(x, y) = –ex cos y f(0, 0) = 0; fx(0, 0) = 0; fy(0, 0) = 1 fxx(0, 0) = 0; fxy(0, 0) = 1; fyy(0, 0) = 0 fxxx(0, 0) = 0; fxxy(0, 0) = 1; fxyy(0, 0) = 0; fyyy(0, 0) = –1 Taylor series of ex sin y =

1 1 1 y ◊1 + 2 xy ◊1 + {3 x 2 y ◊1 + y 3 (-1)} + ... 1 2 3

i.e., ex sin y = y + xy +

1 2 1 x y - y 3 + ... 2 6

Appendix F

(b) ur = uxer cos q + uyer sin q \

∂ ∂ ∂ +y = x ∂x ∂y ∂r

uq = ux (–er sin q) + uy (er cos q) \ urr =

II-F.7

∂ ∂ ∂ +x = -y ∂x ∂y ∂q

Ê ∂ ∂ ∂ˆ (ur ) = Á x + y ˜ ( xu x + yu y ) ∂y ¯ ∂r Ë ∂x = x2uxx + 2xy uxy + y2uyy + xux + yuy

uqq =

Ê ∂ ∂ ∂ˆ (uq ) = Á - y + x ˜ (- y u x + x u y ) ∂ ∂ ∂q x y¯ Ë = x2uyy – 2xy uxy + y2uxx – x ux – y uy

\ urr + uqq = (x2 + y2) (uxx + uyy) = e2r (uxx + uyy) \ uxx + uyy = e–2r (urr + uqq) 19 (a) This is the worked example 1.8, given in page II-1.42 of the text book (b) This is the worked example 1.17, given in page II-1.31 of the text book. 20. (a) This is the worked example 1.4, given in the page II-1.79 of the text book. We have to take a = 1. (b) This is the worked example 1.13, given in the page II-1.29 of the text book.

Appendix

G

Solutions to the June 2011 Question Paper of Anna University, Chennai B.E./B.Tech. DEGREE EXAMINATIONS, June 2011 (Common to all B.E./B.Tech) Second Semester 181202–Mathematics-II REGULATION 2008 Times: Three Hours Maximum: 100 marks Answer ALL questions. PART-A (10 ¥ 2 = 20 marks) 1.

Solve: (D3 + 1)y = 0.

2.

Reduce the equation x4 y¢¢¢ – x3y¢¢ + x2y¢ = 1 into linear equation with constant coefficients.

3.

Find the unit vector normal to the surface of the sphere x2 + y2 + z2 = 1. Æ Show that F = (y2 – z2 + 3yz – 2x)i + (3xz + 2xy)j +(3xy – 2xz + 2z)k is both solenoidal and irrotational.

4.

5.

Find the analytic function w = u + iv whose imaginary part is given by v = ex (x sin y + y cos y).

6.

Find the image of |z + 1| = 1 under the mapping w =

7. 8.

2z Find the residue of 1 - e at its pole. z4 1 at z =1 as a Taylor's series. Expand z+2

1 . z

Part II: Mathematics II

II-G.2

9. 10.

-2 s Find the inverse Laplace transform of e . s-3 1 Lt f (t ) . , find Lt f (t ) and t Æ• If L[f(t)] = 2 t Æ0 s(s + a 2 )

PART B-(5 ¥ 16 = 80 marks) 11. (a) (i) Solve : (D2 + 3D + 2)y = sin x + x2. (ii) Solve for x from the equation D2x + y =3e2t, Dx – Dy = 3e2t. Or 2 (b) Solve by the method of variation of parameters, the equation d y + a 2 y dx 2 = tan ax.

12. (a) Verify Green's theorem in the plane for

Ú (3x

2

- 8 y 2 ) dx + (4 x - 6 xy ) dy

C

Where C is the boundary of the region defined by x = 0, y = 0, x + y = 1.

Or (b) (i) Using Stock's theorem to prove that cure grad f = 0. �� Æ (ii) Evaluate ÚÚ F n ds where F = 2xyi + yz2j + xzk and S is the surface of s

the parallelopiped bounded by x = 0, y = 0, z = 0, x = 2, y = 1 and z = 3. 13. (a) (i) Prove that u = e –y cos x and v = e–x sin y satisfy Laplace equations, but that u + iv is not an analytic function of z (ii) Show that the families of curves rn = a sec nq and rn = b cose nq cut orthogonally. Or (b) Find the bilinear transformation which maps the points z = 1, i, – 1 into the points w = i, 0, – i. Hence find the image of |z| < 1. 14. (a) (i) Using Cauchy's integral formula evaluate circle |z + i| = 1. 2

z

Ú z 2 + 1 dz , where C is the

C

z -1 (ii) Expand, function f(z) = in Laurent's series for the ( z + 2) ( z + 3) region |z| > 3.

Appendix G

II-G.3

Or 2p

(b) Evaluate, by contour integration, the integral < 1.

dq

Ú 1 - 2a sin q + a 2 , 0 < a 0

15. (a) Using convolution theorem, find the inverse Laplace transform of s2 . (s 2 + a 2 ) (s 2 + b2 ) Or 2 (b) Solve, by Laplace transform method, the equation d y + 2 dy + 5 y = e–t dt dt 2 sin, t y (0) = 0, y¢(0) = 1.

Solution PART–A 1.

(D3 + 1)y = 0 ; A.E. is m3 + 1 = 0 i.e., (m + 1) (m2 – n + 1) = 0 \ m = –1, 1 ± i 3 2 2 x Ê 3 3 ˆ \ Solution is y = Ae–x + e 2 Á B cos x + C sin x 2 2 ˜¯ Ë

where A, B, C are arbitray constants. 2.

1 (1) x Putting x = et, so that x D = q, x2 D2 = q (q – 1) and x3 D3 = q¢(q – 1) (q – 2) d equation (1) gets reduced as Where q = dt [q (q – 1) (q – 2) – q (q – 1) + q]y = e–t The given D.E. is re-written as (x3 D3 – x2 D2 + xD)y =

i.e., (q3 – 4q2+ 4q)y = e–t. 3.

The given surface is f = 1, where f = x2 + y2 + z2. – – – – – – – N = —f = fx i + fy j + fz k = 2xi + 2yj + 2zk \ Unit normal n^ =

4.

2 (x i + y j + z k ) N – – – = = xi + yj + zk 2 2 2 |N| 4 (x + y + z )

[This problem is the same as the worked example 2.5, given in page II-2.18 of the book].

Part II: Mathematics II

II-G.4 5.

Let f (z) = u + iv be the analytic function of z required. f ¢ (z) = ux + ivx = vy + i vx, by C.R. equation = ex (x cos y + cos y – y sin y) + i{ex (x sin y + y cos y + sin y)} = ez (z + 1), by Milne. Thompson's rule \ f (z) = zez + C.

6.

7.

1 1 or z = z w 1 The image of |z + 1| = 1, is + 1 = 1 i.e., |w + 1| = |w| w 1 i.e., (u + 1)2 + v2 = u2 + v2 or the line u = 2 2 2z È 1 8 z 3 16 z 4 32 z 5 ÔÏ 2 z 4 z Ô¸˘ 1- e + + + + + ...˝˙ f(z) = = 4 Í1 - Ì1 + 1 2 3 4 5 z ÍÎ ÔÓ Ô˛˙˚ z4 w=

Ï 2 1 4 1 8 1 16 32 ¸ + = -Ì ◊ 3 + ◊ 2 + ◊ + z + ...˝ 2 z 3 z 4 5 Ó1 z ˛ {Res of f(z)}z = 0 = coefficient of

8.

-1 1 1 1 Ï Z - 1¸ = = Ì1 + ˝ z+2 z -1+ 3 3 Ó 3 ˛

\ Required Taylor's serves is or

9.

4 1 in (i) = 3 z

1 3n + 1

•

1 • Ê z - 1ˆ  (-1)n ÁË 3 ˜¯ 3n=0

n

 (-1)n ( z - 1)n

n=0

ÏÔ e-2 s ¸Ô -1 Ï 1 ¸ u2 (t ) L-1 Ì ˝ ˝= L Ì Ó s - 3 ˛t Æ t - 2 ÓÔ s - 3 ˛Ô = e3(t – 2). u2(t)

10.

L{f(t) =

1 2

s(s + a 2 )

= f(s)

f (t ) = lim [ s f ( s )] = lim Ê 1 ˆ By I.V.T., tlim Á ˜ Æ0 sÆ• s Æ • Ë s2 + a2 ¯ f (t ) = lim [ s f ( s )] = 1 By F.V.T., tlim Æ• sÆ0 a2

Appendix G

II-G.5

PART – B 11. (a) (i) (D2 + 3D + 2)y = sin x + x2 A.E., is m2 + 3m + 2 = 0 or (m + 1) (m + 2) = 0 \ m = –1, –2 + Be–2x

\ C.F, = Ae P.I1 =

1 2

D + 3D + 2

= -

sin x =

1 (3D - 1) sin x = sin x 3D + 1 9D2 - 1

1 (3cos x - sin x) 10

¸Ô 1 1 ÏÔ D D2 2 x 1 ( D + 3) + ( D + 3) 2 - ...˝ x 2 P.I2 = = Ì 2 ÓÔ 2 4 Ï D ( D + 3) ¸ ˛Ô 2 Ì1 + ˝ 2 Ó ˛ 1Ï 3 7 1 3 7 ¸ = Ì1 - D + D 2 ˝ x2 = x 2 - x + 2Ó 2 4 ˛ 2 2 4 \ Required solution is y = C.F. + P.I1 + P.I2 11. (a) (ii) D2x + y = 3 e2t fi D3x + Dy = 6 e2t 2t

2t

(1)

Dx – Dy = 3 e fi Dx – Dy = 3 e

(2)

(1) + (2) gives (D3 + D)x = 9 e2t

(3)

A.E. of (3) is m (m2 + 1) = 0 \ m = 0, ± i \ C.F, = A + B cos t + C sin t P.I. =

1 3

D +D

9e 2 t =

9 2t e 10

9 2t e . 10 11. (b) [This problem is the same as the worked example 1.4 given in page II–1.79 of the book] \ Solution is x = A + B cos t + C sin t +

12. (a) [This problem is the same as the worked example 2.2, given in page II– 2.41 of the book] 12. (b) (i) [This problem is the same as the worked example 2.7.(i), given in page II–2.46 of the book] 12. (b) (ii) [The given surface of the parallelopiped is a closed surfaces enclosing a volume V \ By Gauss's divergence theorem,

Part II: Mathematics II

II-G.6

ÚÚ F ◊ n ds = Ú Ú Ú (div F ) dv s V

312

=

Ú Ú Ú (2 y + z

2

+ x) dx dy dz

000

2

31

È x2 ˘ = Ú Ú Í(2 y + z 2 ) x + ˙ dy dz 2˚ 00Î 0 31

= 2 Ú Ú (2 y + z 2 + 1) dy dz 00 3

1

3

= 2 Ú ÈÎ y 2 + ( z 2 + 1) y ˘˚ dz = 2 Ú ( z 2 + 2) dz 0 0

0

3

Êz ˆ = 2Á + 2 z ˜ = 2(9 + 6) = 30. Ë 3 ¯0 3

13. (a) (i) u = e–y cos x; ux = – e–y sin x ; uxx = –e–y cos x uy = –e–y cos x ; uyy = e–y cos x \ uxx + uyy = 0 i.e.,u satisfies Laplace equation v = e–x sin y; vx = –e–x sin y; vxx = e–x sin y vy = e–x cos y; vyy = –e–x sin y \ vxx = vyy = 0. i.e., satisfies Laplace equation Now ux π vy and uy π –vx u and v do not satisfy C.R. equations. \ u + iv is not an analytic function of z. 13. (a) (ii) [This problem is the worked example 3.12, given in page II–3.39 of the book] 13. (b) By the invariance of cross-ratio property of the bilinear transformation,

( w - w2 ) ( w3 - w4 ) ( z - z2 ) ( z3 - z4 ) = ( w - w4 ) ( w3 - w2 ) ( z - z4 ) ( z3 - z2 )

(1)

(On putting z2 = 1, z3 = i, z4 = – 1 and w2 = i, = w3 = 0, w4 = –i in (1), we get

( w - i ) (0 + i ) ( z - 1) (i + 1) = ( w + i ) (0 - i ) ( z + 1) (i - 1)

Appendix G

II-G.7

w - i iz - i w-i z -1 = (-1) = (-i ) i.e., w + i z +1 w+i z +1 (i - 1) z + (i + 1) is the required transformation (1 - i ) z + (1 + i ) i (1 - w) The inverse transformation is given by z = 1+ w i (1 - w) \ The image of |z| < 1 is 0 14. (a) (i) y

z

are z = ± i, of z +1 which z = –i only lies inside C The singularities of

0

x C

Ê z ˆ ÁË z - i ˜¯ dz ∫ I= Ú z+i C

2

f ( z)

Ú z + i dz

C

= 2pi f(–i), by C.I.F

Ê -i ˆ = pi = 2p i Á Ë -2i ˜¯

14. (a) (ii) f(z) =

3 8 z2 - 1 = 1+ , on splitting into partial z + 2 z + 3 ( z + 2) ( z + 3)

fractions = 1+

3Ê 2ˆ ÁË1 + ˜¯ z z

-1

8Ê 3ˆ - Á1 + ˜ zË z¯

-1

3 • (-1) n ◊ 2n 8 • (-1) n ◊ 3n  zn - z  zn z n=0 n=0 2 < 1 or |z| > 2 The expansion is valid if |z| 3 < 1 or |z| > 3 and |z| = 1+

Part II: Mathematics II

II-G.8

i.e., if |z| > 3 \ The required Laurent's expansion in the region |z| >3 is f(z) = 1 +

•

1

 (-1)n {3 ¥ 2n - 8 ¥ 3n }◊ z n + 1

n=0

14. (b) [This problem is the worked example 4.11, given in page II–4.62 of the book In the W.E., we have to take x = a] 15. (a) [This problem is the worked example 5.11 (iii) given in page II–5.90 of the book.] 15. (b) y¢¢ (t) + 2y¢ + 5y(t) = e –t sin t Taking Laplace transforms on both sides of (1), we have

1 [s2 y–(s) – sy(0) – y¢(0)] + 2[sy–(s) – y(0)] + 5 y–(s) = 2 s + 2s + 2 1 2 – +1 i.e., (s + 2s + 5) y (s) = 2 s + 2s + 2 \ y–(s) =

s 2 + 2s + 3 2

2

( s + 2s + 2) ( s + 2 s + 5)

=

As + B 2

s + 2s + 2

+

Cs + D 2

s + 2s + 5

\ (As + B) (s2 + 2s + 5) + (Cs + D) (s2 + 2s + 2) = (s2 + 2 + 3) \ A + C = 0(1) ; 2A + B + 2C + D = 1 (2); 5A + 2B + 2C + 2D = 2 (3); 5B + 2D = 3(4) Solving the equations (1), (2),(3) and (4), we get

1 2 and D = 3 3 1 2 3 3 \ y–(s) = + ( s + 1) 2 + 12 ( s + 1) 2 + 22 A = 0, C = 0, B =

\ y(t) =

1 -t -1 Ê 1 ˆ 2 -t -1 Ê 1 ˆ e ◊L Á 2 2 ˜ + e ◊L Á 2 ˜ 3 Ë s +1 ¯ 3 Ë s + 22 ¯

2 1 Ê1 ˆ = e - t Á sin t + ◊ sin 2t ˜ Ë3 ¯ 3 2 =

1 -t e (sin t + sin 2t ) 3

Appendix

H

Solutions to the April/May 2011 Question Paper of Anna University, Coimbatore B.E./B.Tech. DEGREE EXAMINATIONS, May/June 2011 Second Semester 080030004 Mathematics-II Civil Engineering (Common to all Branches) REGULATION 2008 Times: Three Hours Maximum:-100 marks Answer ALL questions. PART-A (20 ¥ 2 = 40 marks) 1.

Solve (D3 – 3D2 + 4D – 2)y = 0.

2.

Find the particular integral of [D2 – 4D + 4]y = x3 e2x

3.

Write the general solution for the method of variation of parameters for second order differential equation.

4.

Eliminate x and y from the differential equation

5.

Prove that —(rn) = nrn–2. r

6.

Find the unit normal vector to the surface x2 + xy + z2 = 4 at (1, – 1, 2).

7.

State Stocke's Theorem.

8.

If r = xi + y j + zk then find curl r .

9.

Show that an analytic function with constant real part is constant.

10.

dy dx + 2 x = 0, - 2 y = 0. dt dt

Find the fixed point of the bilinear transformation w =

z-1 . z+1

Part II: Mathematics II

II-H.2 11.

Test the analyticity of the function 2xy + i(x2 – y2).

12.

15.

Find the image of the circle |z| = 3 under the transformation w = 5z. 2 dz where c is z = 2. Evaluate (z + 3) c p in a Taylor's series. Expand cos z at z = 4 Define essential singularity.

16.

Find the residue of f (z) =

17.

Find L [e– 2t sin 3t].

13. 14.

Ú

z 2

z +1

about each singularity.

19.

t If L[f(t)] = F(s) then find L ÈÍ f ˘˙ . Î 2˚ State Initial and Final value theorem in Laplace Transform.

20.

Find L

18.

- 1È

s ˘ Í ˙. + ( s 3) ˚ Î PART - B ANSWER ANY FIVE QUESTIONS

21. (a) Solve (D2 – 3D + 2)y = e–x sin 2x cos x. (b) Solve by the method of variation parameters

d2 y dx 2

+ y = x sin x

dx dy dx dy x + y = 10e t and + + x - y = 0 given x(0) = 2, dt dt dt dt y(0) = 3.

22. (a) Solve

(b) Find the scalar potential f if —f = (y2 + 2xz2) i + (2xz – z) j + (2x2z – y + 2z) k . 23.

Verify the Gauss's divergence theorem for F = 4 xzi - y2 j + yzk over the cube bounded by x = 0, y = 0, z = 0, x = 1, y = 1 and z = 1.

24. (a) Find the bilinear transformation which maps the point z = 1, i, –1 into the point w = i, 0, –1. (b) Show that the function u = x3 – 3xy2 + 3x2 + – 3y2 + 1 is harmonic and find it analytic function. 1 1 25. (a) Find the image of the infinite strip ≺ y ≺ under the transformation 1 4 2 w= . z

Appendix H

II-H.3

2p

(b) Using contour integration evaluate

dq

Ú 13 + 5 sin q . 0

26. (a) Evaluate using Cauchy's integral formula c is z = 3 .

sin p z2 + cos p z2 dz where c (z - 1)(z - 2)

Ú

z (b) Find Laurent's series expansion of f (z) = in the region (z - 1)(z - 3) 0 < |z – 1| < 2 È ˘ 1 27. (a) Find L - 1 Í ˙ using convolution theorem. + ( s 3)( s 1) Î ˚ ,0≺r≺a Ït (b) Find the Laplace transform of f (t ) = Ì with f(t + 2a) 2 a t , a ≺ t ≺ 2 a Ó = f(t) 28. (a) Using Laplace transform solve y¢¢ – 3y¢ + 2y = e– t given y(0) = 1, y¢(0) = 0. (b) Find L ÈÎte

-t

sin t.˘˚ SOLUTIONS PART-A

1.

(D3 – 3D2 + 4D – 2) y = 0 A.E. is m3 – 3m2 + 4m – 2 = 0 i.e., (m – 1)(m2 – 2m + 2) = 0 \m=1,

2. 3.

2± 4-8

or 1 ± i 2 \ Solution is y = A ex + ex(B cos x + C cos + sin x) 1 x5 e2 x 2x 3 2x 1 3 e x = e x = . P.I. = 20 ( D - 2) 2 D2

If the general solution of y¢¢ + Py¢ + Qy = 0 is y = Au + Bv and if u¢ and v¢ satisfy the equations A'u + B'v = 0 and A¢ u¢ + B¢v¢ = R, treating A and B as function of x, then the G.S. of the equation y¢¢ + Py¢ + Qy = R is y = Au + Bv , where A and B (treated as functions of x) contain two arbitrary constants C1 and C2.

4.

If we eliminate y from the D.E.’s 2x + Dy = 0 and Dx – 2y = 0, we get the D.E. (D2 + 4) x = 0. If we eliminate x from the given D.E.’s, we get D.E. (D2 + 4)

Part II: Mathematics II

II-H.4

5.

y = 0. [This problem is the same as the worked example 2.9 given in page II-2.7 of the book]

6.

f(x, y, z) = x2 + xy + z2;

\

—f = (2x + y) i + x j + 2z k

(—f)(1, –1,2) = i + j + 4k = n(say) \ n= 7.

8.

9. 10.

11.

12.

13.

n i + j + 4k 1 = ( i + j + 4k ) n 1 + 1 + 16 3 2

Stoke's theorem—Statement is available in § 2.5.2 given in page II-2.37 of the book. i Curl r = ∂ ∂x x

j ∂ ∂y y

[This problem is the same as the worked example 3.12(i) given in page II3.21 of the book] z-1 The fixed points of the bilinear transformation are given by z = z+1 i.e., z2 = –1 i.e., z = ± i Let u + iv = 2xy + i(x2 – y2); \ u = 2xy and v = x2 – y2 ux = 2y; uy = 2x; vx = 2x; vy = –2y Though ux, uy, vx, vy exist and are continuous every where, they do not satisfy C.R. equations at all points except at the origin. \ w = u + iv is not analytic everywhere; analytic only at the origin 1 The image of the circle |z| = 3 under the transformation w = 5z or z = w is 5 w the circle = 3 or w = 15. 5 2dz ; The only singularity of the integrand and z = –3 lies outside the z+3 c circle |z| = 2 i.e., the integrand is analytic on and inside C \ By Cauchy's

Ú

integral theorem, 14.

k ∂ =0 ∂z z

2dz

Ú z + 3 = 0.

f(z) = cos z; f ¢(z) = –sin z; f ¢¢(z) = –cos z; f ¢¢¢(z) = sin z etc. 1 1 1 1 Êpˆ Êpˆ Êpˆ Êpˆ fÁ ˜= ; f ¢Á ˜ = ; f ¢¢ Á ˜ = ; f ¢¢ Á ˜ = etc. Ë 4¯ Ë 4¯ Ë 4¯ Ë 4¯ 2 2 2 2

Appendix H

II-H.5

Taylor series of f(z) at z = p/4 is Êpˆ Êpˆ f¢Á ˜ f ¢¢ Á ˜ 2 Ë 4¯ Ê Ë 4¯ Ê pˆ pˆ pˆ Ê f(z) = f Á ˜ + z- ˜ + z- ˜ + Á Á Ë 4¯ 1! Ë 4¯ 2! Ë 4¯ i.e., cos z =

1 2

1 Ê pˆ 1 ÁË z - 4 ˜¯ 2 2! 2

-

•

2

pˆ 1 Ê ÁË z - 4 ˜¯ + 3! 2

pˆ Ê ÁË z - 4 ˜¯

2

or cos z =

1 Ï 1 Ê p ˆ¸ 1 Ê pˆ 1 Ê pˆ Ì1 - Á z - ˜ ˝ - Á z - ˜ + Á z - ˜ Ë ¯ Ë ¯ Ë 1! 4 ˛ 2! 4 3! 4¯ 2 Ó

15.

[Definition is available § 3 given in page II-4.26 of the book]

16.

The singularities of f(z)

z 2

z +1

3

•

3

•

are z = ±i, which are simple poles

Ê z ˆ 1 R1 = [Res. f ( z )] = lim Á = z Æ i Ë z + i ˜¯ 2 | z = i| Ê z ˆ 1 = R2 = [Res. f (z)] = lim Á zÆ -i Ë z - i ˜ ¯ 2 |z = - i| 17.

L{e–2t sin 3t} = {L(sin 3t)}s Æ s + 2 =

3 2

( s + 2) + 3

2

or

3 2

s + 4 s + 13

•

18.

F(s) =

Ú f (t ) e

- st

dt

0

Ï Ê t ˆ¸ \ L Ìf Á ˜˝ = Ó Ë 2¯ ˛

•

Ú 0

Ê tˆ f Á ˜ e - st dt = Ë 2¯

•

Ú f (u) e

- s ◊ 2u

0

du, on putting t = u 2

•

Ú

= 2 f (u) e -(2 s) u du = 2 F (2 s) 0

19.

20.

Statements of I.V.T. and F.V.T. are available in § 5.9.1 and § 5.9.2 given in page II-5.67 of the book ÏÔ ¸Ô s -1 L–1 Ì ( s + 3)2 ˝ = L ÓÔ ˛Ô

ÏÔ s + 3 - 3 ¸Ô = L-1 Ì 2 ˝ + ( s 3) ÓÔ ˛Ô

Ï1 3 ¸ = e -3t ◊ L-1 Ì - 2 ˝ Ós s ˛ –3t = e (1 – 3t).

ÏÔ 1 3 ¸Ô Ì 2˝ ÓÔ s + 3 ( s + 3) ˛Ô

Part II: Mathematics II

II-H.6

PART-B 21. (a) A.E. is m2 – 3m + 2 = 0 x

or

(m – 1)(m – 2) = 0

\ m = 1, 2

2x

C.F. = Ae + Be 1 1 P.I. = 2 e - x sin 2 x cos x = e - x 2 (D - 1) - 3(D - 1) + 2 D - 3D + 2 sin 2x cos x 1 Ï1 ¸ = e- x 2 Ì (sin 3 x + sin x)˝ 2 D - 5D + 6 Ó ˛ È ˘ 1 1 1 sin 3 x + sin x ˙ = e- x Í 2 5 - 5D Î -5D - 3 ˚ È (5D - 3) ˘ 1+ D sin 3 x + sin x ˙ Í 2 2 5(1 - D ) ÎÍ 9 - 25D ˚˙ 1 -x È 1 1 = e Í (15 cos 3 x - 3 sin 3 x ) + (sin x + cos x)˘˙ 2 10 Î 234 ˚ =

1 -x e 2

=

1 -x 1 -x e (5 cos 3 x - sin 3 x ) + e (sin x + cos x) 156 10

G.S. is y = C.F. + P.I. 21. (b) The solution of the corresponding homogeneous equation, namely, y¢¢ + y = 0 is y = A cos x + B sin x Treating A and B as functions of x, A¢ and B¢ are given by A¢ cos x + B¢ sin x = 0

(1) (2)

and –A¢ sin x + B¢ cos x = x sin x

(3) 2

Solving (2) and (3) we get A¢ = –x sin x and B¢ = x sin x cos x 1 1 \ A¢ = x (cos 2x – 1) and B¢ = x sin 2x 2 2 Integrating, we get 1 (2x sin 2x + cos 2x – 2x2) + c1 8 1 (sin 2x – 2x cos 2x) + c2 and B = 8 Using these values in (1), the required G.S. is 1 [2x sin 2x cos x + cos 2x cos x y = c1 cos x + c2 sin x + 8 – 2x2 cos x + sin 2x sin x – 2x cos 2x sin x] 1 x x2 cos x, where c¢1 = c1 + i.e., y = c¢1 cos x + c2 sin x + sin x – 8 4 4 A=

Appendix H

II-H.7

22. (a) The given equations are (D + 1)x + (D + 1)y = 10et . . . (1) and (D + 1)x – (D + 1)y = 0 . . . (2) (1) + (2) gives 2(D + 1)x = 10 et 5 I.F. = et and solution is x et = e 2 t + c1 2 5 or x = e 2 t + c1 e - t 2 Using x(0) = 2, c1 = –1/2 (1) – (2) gives (D + 1) y = 5 et 5 2t e + c2 e - t 2 1 Using y(0) = 3, we get c2 = 2 Solution is y =

\ Required solution is x = and y =

1 (5e t + e - t ) 2

1 (5e t - e - t ) 2

22. (b) —f = (y2 + 2x z2) i + (2xy – z) j + (2x2z – y + 2z) k = f x i + f y j + fz k \ f x = y2 + 2 xz2

f = xy2 + x 2 z2 + f1 ( y, z) f = xy2 - yz + f2 (z, x)

f y = 2 xy - z

fz = 2 x 2 z - y + 2z f = x 2 z2 - yz + z2 + f3 ( x, y) \ fx = xy2 + x2z2 – yz + z2 + c 23.

Gauss's Divergence theorem is

ÚÚ F . d s = Ú ÚÚ (div F ) dv s

(1)

v

L.S. of (1) =

ÚÚ

x=0 nˆ = - i

=

+ =

ÚÚ

ÚÚ

+

x =1 nˆ = i

ÚÚ

+

y=0 nˆ = - j

y =1 nˆ = j

+

ÚÚ

+

z =0 nˆ = - k

ÚÚ (4 xz i - y

z =1 nˆ = k

2

j

+ yzk ) ◊ nˆ ds

ÚÚ - 4 xzds + ÚÚ 4 x zds

x= 0

x =1

+

ÚÚ y

y= 0

2

ds -

ÚÚ y

y=1

2

ds -

ÚÚ yzds + ÚÚ yzds

z= 0

z=1

Part II: Mathematics II

II-H.8 1 1

=

ÚÚ

1 1

4z dy dz -

0 0

ÚÚ

1 1

dz dz +

0 0

Ú Ú y dx dy 0 0

1 3 =2–1+ = 2 2 R. S. of (1) =

ÚÚÚ (4z - 2 y + y) dx dy dz V 1 1 1

=

Ú Ú Ú (4z - y) dx dy dz 0 0 0 1 1

=

ÚÚ 0 0

1

1 (4z - y) dy dz = (4z - ) dz 2

Ú 0

1 3 =2– = 2 2

3 , G.D. theorem is verified . 2 24. (a) By the invarience of cross-ratio property, we have (w - w2 )(w3 - w4 ) (z - z2 )(z3 - z4 ) = (1) (w - w4 )(w3 - w2 ) (z - z4 )(z3 - z2 ) Since L.S. of (1) = R.S. of (1) =

Putting z2 = 1, z3 = i, z4 = –1 and w2 = i, w3 = 0, w4 = –1 in (1), we have

(w - i)(0 + 1) (z - 1)(i + 1) = (w + 1)(0 - i) (z + 1)(i - i)

Ê w - iˆ Ê z - 1ˆ i.e., Á i=Á (- i) ˜ Ë w + 1¯ Ë z + 1 ˜¯ i.e., w = (1 + i) - (1 - i) z is the required bilinear transformation. 2z 24. (b) [This problem is the worked example 3.1, given in page II-3.30 of the book] 25. (a) [This problem is the worked example 3.5 (iii), given in page II-3.63 of the book] 25. (b) On the circle |z| = 1, z = eiq, dq = 2p

\I=

dq

dz/iz = Ê z2 - 1 ˆ C 13 + 5 Á 2i z ˜ Ë ¯

Ú 13 + 5 sin q = Ú 0

dz z2 - 1 and sin q = iz 2i z

Ú 5z

C

2dz 2

+ 26 iz - 5

Appendix H 2 5

II-H.9

dz iˆ Ê C (z + 5i) Á z + ˜ Ë 5¯ The singularity of the integrand that lies inside =

Ú

C is z = –i/5 5 = i 24 i 5i 5 \ By Cauchy's residue theorem, 2 5 p = I = 2p i ¥ ¥ 5 24 i 6 (Res.)z = –i/5 =

26. (a)

1

1 1 1 = (z - 1)(z - 2) z - 2 z - 1 \I=

Ú

C

=

sin p z2 + cos p z2 dz z-2 f (z)

Ú

C

sin p z2 + cos p z2 dz (z - 1)

f (z)

Ú z - 2 dz - Ú z - 1 dz

C

C

f(z) is analytic on and inside C i.e., |z| = 3 The points z = 1 and z = 2 lie inside C \ By Cauchy's integral formula, I = 2pi ¥ [f(2) – f(1)] = 2pi {(sin 4p + cos 4p) – (sin p + cos p)} = 2pi (1 + 1) = 4pi 26. (b) f(z) =

3/2 1/2 3/2 1/2 3 = =z-3 z-1 u-3 u 4 =-

3 4

uˆ Ê ÁË 1 - 2 ˜¯

-1

-

2 3 ÏÔ u Ê uˆ Ê uˆ Ì1 + + Á ˜ + Á ˜ + Ë 2¯ 2 Ë 2¯ ÓÔ

The Laurent's expansion in (1) is valid in

1 2u ¸Ô 1 (1) •˝ 2u ˛Ô

u < 1 or 0 < |z – 1| < 2 2

\ Required Laurent's expansion is 1 3 f(z) = 2(z - 1) 4

•

 n=0

Ê z - 1ˆ ÁË 2 ˜¯

n

in 0 < |z – 1| < 2.

Ï 1 1 ¸ -3t t 27. (a) L– 1 Ì ◊ ˝ = e * e by Convolution theorem Ó s + 3 s - 1˛

Part II: Mathematics II

II-H.10 t

=

Úe

-3u

◊ e(t - u) du

0

t

t

= e

t

Úe

-4 u

0

Ê e -4u ˆ ◊ du = e Á ˜ Ë -4 ¯ 0 t

1 = e t (1 - e -4 t ) 4 27. (b) [This problem is the worked example 5.3, given in page II-5.42 of the book] 28. (a) y¢¢ – 3y¢ + 2y = e– t

(1)

Taking Laplace transforms of (1), we get [s2 –y (s) – sy (0) – y1 (0)] – 3[s –y (s) – y(0)] + 2y– (s) = i.e., (s2 – 3x + 2) –y (s) = i.e., –y (s) =

1 +s-3 s+1

1 s+1

s2 - 2s - 2 -2/3 3/2 1/6 = + + ( s - 2)( s - 1)( s + 1) s - 2 s - 1 s + 1

Inverting, we get y = –

2 2t 3 t 1 -t e + e + e 3 2 6

28. (b) L(te– t sin t) = [L(t sin t)]s Æ s + 1 L(t sin t) = -

(1)

d d L (sin t ) = ds ds

Using in (1); L(t e– t sin t) =

Ê 1 ˆ 2s ◊Á 2 ˜= 2 Ë s + 1 ¯ ( s + 1)2

2( s + 1) 2

( s + 2 s + 2)2

Appendix

I

Solutions to the June 2011 Questions Paper of Anna University, Madurai B.E./B.Tech. DEGREE EXAMINATIONS, June 2011 (Second Semester) 10177MA202 MATHEMATICS-II (Common to all Branches) Times: Three Hours Maximum:-100 marks INSTRUCTIONS 1. 2. 3.

Answer ALL Questions in Part A and Part B. Part A and Part B questions should be answered separately in the same answer sheet. Answer the Questions EITHER/OR in Part B (10 ¥ 2 = 20)

PART-A 1.

Find L(e–t sin 2t)

2.

Without integrating, evaluate

Ú

•

0

3. 4.

t cos t dt

Find the unit normal to the surface x2 + y2 + z2 = 1 at (1, 1, 1) � � � � Find the value of a, if the vector F = (x + 3y)i + (y – 2z)j + (x + az)k is solenoidal.

Part II: Mathematics II

II-I.2 5.

� Verify whether w = z is analytic or not.

6.

Define the critical point of the Bilinear transformation.

7.

Evaluate

1

0

8. 9. 10.

Evaluate Evaluate

x

Ú Ú

x

( x 2 + y2 ) dy dx

p /2

sin q

0

0

Ú Ú Úc

r dr dq

z2 + 1 ( z - 2)( z - 3)

Expand f(z) =

dz , where C:|Z| = 1

1 as a Taylor's series about z = 1. z PART-B

(5 ¥ 16 = 80)

11. (a) Find the Laplace transform of

Ï -E : 0 < t < p f(t) = Ì ÓE : p < t < 2p

and f(t) = f(t + 2p)

Ê ˆ s (b) Using Convolution theorem, find L-1 Á 2 2 2˜ Ë (s + a ) ¯

(8) (8)

(OR) 12. (a) Find (i) L(te–2t cos h 2t)

Ê 1 - cos t ˆ (ii) L Á ˜¯ Ë t

Ê ˆ 1 (iii) L-1 Á ˜ 2 Ë s + 2s + 5 ¯

(iv) L-1 (tan -1 ( 2 )) s

(b) Using Laplace transform, solve

d2 y dy –t -4 + 3 y = e , given that y(0) = 1 and y¢(0) = 1 2 dt dt 13. (a) Find the constants a, b c so that vector � � � � F = (x + 2y + az)i + (bx + 3y – z)j + (4x + cy + 2z)k is irrotational

(6)

Appendix I

II-I.3

� � � � (b) Verify Stoke's throrem for F = –yi + 2yzj + y2k , where is the upper half of the sphere x2 + y2 + z2 = 1 (10) (OR) � � � � 14. (a) Show that F = (6xy + z3)i + (3x2 – z)J + (3xz2 – y)k is irrotational and find its scalar potential (6) (b) Verify the Green's theorem in the plane for

Úx

2

(1 + y) dx + ( y3 + x 3 ) dy ,

c

where c is the square bounded by x = ± a, y = ± a 15. (a) If u =

(10)

sin 2 x , find the corresponding analytic function f(z) cos h 2 y + cos 2 x (8)

(b) Find the bilinear transformation which maps –1, 0, i of the z-plane into –1, –i, 1 of the w plane (8) (OR)

Ê ∂2 ∂2 ˆ 2 16. (a) If f(z) is analytic function of z, show that Á | f (z) |2 = 4|f ¢ (z)| + 2 2˜ ∂y ¯ Ë ∂x (8) (b) Find the image of the circle |z – 1| = 1 in the complex plane under of

1 z

mapping w =

(8) (OR)

17. (a) Evaluate

a

2a - x

0

x2 a

ÚÚ

xy dydx by changing the order of integration

(b) Find the volume of the tetrahedron bounded by the plane and the coordinate planes

(8)

x y y + + =1 a b c (8)

(OR) 18. (a) Find the area between the parabolas y2 = 9x and x2 = 9y 1

(b) Evaluate

ÚÚ 0

0

1 - x2

Ú

0

1 - x 2 - y2

dx dy dz 1 - x 2 - y 2 - z2

(8) (8)

Part II: Mathematics II

II-I.4

19. (a) Using Cauchy's integral formula, evaluate

Úz c

2

z+4 dz , where c is the circle (i) |z + 1 + i| = 2 (ii) |z| = 1 + 2z + 5

(8)

x 2 dx by method of residues. ( x 2 + 9)( x 2 + 4)

(8)

(b) Evaluate

Ú

•

0

(OR) 2p

dq using Contour integration (8) 0 17 - 8 cos q (7z - 2) (b) Find the Laurent's series expansion of f(z) = in 1 < |z + 1| z(z - 2)(z + 1) a PART - B

11. (a) (i) The solution of the corresponding homogeneous equation, namely, y¢¢ + 4y = 0 is y = A cos 2x + B sin 2x (1). Treating A and B as functions of x, they are given by A¢ cos 2x + B¢ sin 2x = 0 (2) and –2A¢ sin 2x + 2B¢ cos 2x = 4 tan 2x (3) Solving (2) and (3), we get sin 2 2 x and B¢ = 2 sin 2x cos 2 x \ A = sin 2x – log (sec 2x + tan 2x) + C1 and B = –cos 2x + C2 Using these values of A and B in (1), the required solution is y = C1 cos 2x + C2 sin 2x – cos 2x log (sec 2x + tan 2x)

A¢ = - 2

11. (a) (ii) Put 3x + 2 = et or t = log (3x + 2) dy dy 3 dy d = ¥ i.e., (3 x + 2) = 3q y, where q = dx dt 3 x + 2 dx dt and (3x + 2)2

d2y

= 9q (q – 1)y dx 2 \ The given equation becomes 1 t 4 t 2 [9q(q – 1) + 9q – 36] y = (e - 2) + (e - 2) + 1 3 3 i.e., (q2 – 4)y =

1 2t (e - 1) 27

C.F. = Ae2t + B e–2t; P.I. =

1 1 (e 2t - 1) = {te 2t + 1} 27 (q + 2) (q - 2) 108

\ G.S. is y = A(3x + 2)2 +

B (3x + 2)

(b) (i) (D2 + 5D + 6)y = 3x2

2

+

1 {(3 x + 2) 2 log (3 x + 2) + 1} 108

Part II: Mathematics II

II-J.6

A.E. is m2 + 5 m + 6 = 0 –2x

\ C.F. = Ae

\

m = –2, –3

–3x

+ Be

1 3 ÔÏ ( D + 5) D 2 ( D + 5) 2 Ô¸ 2 2 3 = + x Ì1 ˝x P.I. D ( D + 5) ¸ 6 ÔÓ 6 36 Ï Ô˛ 6 Ì1 + ˝ 6 Ó ˛ =

1Ï 5 19 2 ¸ 2 1 Ï 2 5 19 ¸ D ˝ x = Ìx - x + ˝ Ì1 - D + 2Ó 6 36 2Ó 3 18 ˛ ˛

1 (18 x 2 - 30 x + 19) 36 \ G.S. is y = C.F. + P.I. Ê (b) (ii) (x3D3 + 2x2D2 + 2)y = 10 Á x + Ë

1ˆ d , where D = ˜ ¯ x dx

Put x = et or t = log x Then x2D2 = q(q – 1); x3D3 = q(q – 1)(q – 2) where q =

d dt

\ The given D.E. becomes [q(q – 1)(q – 2) + 2q(q – 1) + 2]y = 10 (et – e–t) i.e., (q3 – 3q2 + 2)y = 10 (et – e–t) A.E. is (m – 1) (m2 – 2m – 2) = 0 \

2± 4+8

m = 1,

\ C.F. = Aet + Be(1 + = Ax + Bx (1 + P.I. =

or 1 ± 3

2 3)t 3)

10 3

2

q - 3q + 2

+ C e(1+ C x (1-

3)

(et + e - t ) =

10 10 - t {t et } + e 3 -2 10 x log x - 5/x = 3 \ G.S. is y = C.F. + P.I. = -

3)t

10 (q - 1) (q 2 - 2 q - 2)

(et + e - t )

Appendix J 12. (a) Gauss-Divergence theorem;

II-J.7

Ú Ú F . ds = Ú Ú Ú (div F ) dv S

V

ÚÚ F ◊ (-k ) dx dy + ÚÚ F ◊ k dx dy + ÚÚ F ◊ ds

L.S. =

z =0

=

z=3

ÚÚ - z

2

ÚÚ z

dx dy +

z =0

2

z =3

ÚÚ

= 9¥

= 36p + 2

dx dy + ÚÚ F ◊ ds S3

dx dy +

Ê ˆ 2 2 Á x + y = 4˜ z=0 ËÁ ¯˜

ÚÚ

S3

ÚÚ

F ◊ ds

x2 + y 2 = 4

F ◊ d s (1), where

2

2

x + y =4

ÚÚ

F ◊ d s is evaluated along

2

x + y =4

the cylindrical surface

2

ÚÚ

F ◊d s

2

2

x + y =4

ÚÚ

F ◊ nˆ ds

(2)

2

x + y =4

where nˆ is the unit normal to the surface x2 + y2 = 4 at the point (x, y) The surface is f = 4, where f = x2 + y2 fx = 2x; fy = 2y \ —f = 2 x i + 2 y j \ nˆ =

—f 2x i + 2 y j 2( x i + 2 yj ) = , 2 2 | —f | 4¥4 4 (x + y )

since (x, y) satisfies x2 + y2 = 4 1 (x i + y j ) 2 Using in (2) =

ÚÚ

(4 x i - 2 y 2 j + z 2 k ) ◊

ÚÚ

(2 x - y 3 ) dS = Ú

x2 + y 2 = 4

=

1 ( x i + y j ) dS 2

3 2p

x2 + y 2 = 4

Ú (2 ¥ 4 cos

2

q - 8 sin 3 q ) 2 d q dz

0 0

using cylindrical Co-ordinates 2p

=2¥3

Ú {4 (1 + cos 2q ) - 2 (3 sin q - sin 0

= 24 ¥ 2p = 48 p

3

q )} dq

Part II: Mathematics II

II-J.8

Using this value in (1), L.S. of G.D.T. = 36 p + 48 p = 84 p R.S. =

Ú Ú Ú (4 - 4 y + 2 z ) dv 2p 2 3

=

Ú Ú Ú (4 - 4r sin q + 2 z ) dz r dr dq , 0 00

using cylindrical Co-ordinates 2p 2

=

Ú Ú {(4 - 4r sin q ) 3 + 9} r dr dq , 0

0

2p

=

Ú 0

2

Ê r2 ˆ 3 Á 21 2 - 4 sin q r ˜ dq Ë ¯0

2p

=

Ú (42 - 32 sin q ) dq = 84 p 0

Since L.S. = R.S. = 84 p, G.D., Theorem is verified. 12. (b) As per Green's theorem, Ê ∂Q

=

Ú ( Pdx + Qdy) = Ú Ú ÁË ∂x

(1, 1)

C

2

=

R

-

∂P ˆ dx dy ∂y ˜¯

(1)

0

Ú

L.S. of (1) =

OAB y = x2 1

=

Ú 0

+

Ú

BDO y= x

È(3 x 2 - 8 y 2 ) dx + (4 y - 6 xy ) dy ˘ Î ˚

È(3 x 2 - 8 x 4 ) dx + (4 x 2 - 6 x3 ) 2 x dx ˘ Î ˚ 1 È ˘ 1 + Ú Í(3 x 2 - 8 x) dx + (4 x - 6 x3/2 ) dx ˙ 2 x ˙˚ Í 0 Î 1

2 3 4 2 = + Ú {(3 x + 8 x - 20 x ) - (2 - 11x + 3 x )} dx 0

8 20 11 3 -2+ = = 4 5 2 2

Appendix J

R.S of (1) =

ÚÚ

OABDO

II-J.9 1

(-6 y + 16 y ) dx dy = 10 Ú

y

Ú

0 y

y dx dy

2

1

= 10

Ú y(

y - y 2 ) dy

0

1

1 ˆ Ê2 Ê 2 1ˆ 3 = 10 Á y 5/2 - y 4 ˜ = 10 ¥ Á - ˜ = Ë5 Ë 5 4¯ 2 4 ¯0 Since L.S. of (1) = R.S. of (1), Green's theorem is verified. 13. (a) (i) f = 3x2y – y3 is the potential function. \ It is the real part of an analytic function w = f + iy. fx = 6xy; fy = 3x2 – 3y2 dw = fx + iyx = fx – ify = –3iz2, by M.T. Principle dz \ w = –iz3 + c Now

= –i(x + iy)3 + c = –i{x3 + 3x2 ◊ iy + 3x ◊ i2y2 + i3y3) + c i.e., f + iy

= (3x2y – y2) + i(3xy2 – x3) + ik

\ The stream function is y = 3xy2 – x3 + k, where k is an arbitrary constant. 13. (a) (ii) By the invariance of the cross ratio of the bilinear transformation. ( w - w1 ) ( w2 - w3 ) ( z - z1 ) ( z2 - z3 ) ( z - z1 ) ( z2 z13 - 1) , = = ( w - w3 ) ( w2 - w1 ) ( z - z3 ) ( z2 - z1 ) ( z z13 - 1) ( z2 - z1 ) 1 z Putting z1 = 0, z2 = 1, z31 = 0

where z13 =

and w1 = i, w2 = –1, w3 = –i in the above, we get ( w - i ) (-1 + i ) z (-1) = ( w + i ) (-1 - i ) (-1) (1) i.e.,

w-i (1 + i ) z == iz w+i -1 + i

\

2 w iz + 1 i-z = or w = is the required bilinear transformation 2i 1 - iz 1 - iz

Part II: Mathematics II

II-J.10

13. (b) (i) Let f(z) = u + iv be the analytic function of z. u = ex (x cos y – y sin y) ux = ex cos y + ex(x cos y – y sin y) uy = ex(–x sin y – y cos y – sin y) f ¢(z) = ux – iuy (∵ vx = –uy by C.R. equation) = (ez + zez) – i ¥ 0, by M.T. Principle \ f(z) = ez + zez – ez = zez + c, where c is a complex constant (b) (ii) [This problem is the same as the worked example 3.5 (iii), given in page II-3.63 of the book] 14. (a) (i) [This problem is the same as the worked example 4.10 (i) given in page II-4.41 of the book] (a) (ii) The singularities of f(z) =

z +1 2

z ( z - 2)

are z = 0, which is a double pole

and z = 2, which is a simple pole. {Res. of f(z)}z = 0 =

ÏÔ 1 z + 1 ¸Ô lim D Ì z 2 ¥ 2 ˝ 1! z Æ 0 ÓÔ z ( z - 2) ˛Ô

3 ÔÏ ( z - 2) ◊1 - ( z + 1) ◊1 Ô¸ = lim Ì ˝=2 zÆ0 Ô 4 ( z - 2) Ô˛ Ó ÔÏ ( z - 2) ( z + 1) Ô¸ 3 {Res. of f(z)}z = 2 = lim Ì 2 ˝= z Æ 2 Ô z ( z - 2) Ô Ó ˛ 4 14. (b) Consider f(z) = of order 2.

1 2

( z + 1) 2

whose singularities are z = ±i, which are poles

Of the two poles, z = i alone lies inside the semi-circular contour C, shown below: ¢

0

{Res. of f(z)}z = i =

1 ( z - i)2 ÔÏ Ô¸ lim D Ì 2 2˝ z Æ i 1! ÔÓ ( z - i ) ( z + i ) Ô˛

Appendix J

II-J.11

2 i ÔÏ -2 Ô¸ == lim Ì ˝=3 z Æ i Ô ( z + i )3 Ô 4 (2i ) Ó ˛ By Couchy's residue theorem, dz

Ú ( z 2 + 1)2

= 2p i ¥ -

C

R

i.e.,

dx

Ú

2 2 - R ( x + 1)

i p = 4 2

dz

+

Ú ( z 2 + 1)2

C¢

=

p 2

(1)

Taking limits on both sides of (1) as R Æ •, •

dx

Ú

-•

2

( x + 1)

2

+0=

p dz (∵ the Ú 2 = 0, 2 2 C ¢ ( z + 1)

by Cauchy's lemma) •

i.e., 2 Ú 0

dx ( x 2 + 1) 2

=

p or 2

•

dx

Ú ( x 2 + 1)2

=

0

p 4

15. (a) (i) [This problem is similar to the worked example 5.2 given in page II-5.41 of the book. In the W.E., we have to take k = 1 and to replace a by b. The required L { f (t ) =

(a) (ii) Let

5s 2 - 15s - 11 ( s + 1) ( s - 2)

3

=

1 Ê bs ˆ ˘ tanh Á ˜ ˙ Ë 2 ¯˚ s

A B C D + + + 2 s + 1 s - 2 ( s - 2) ( s - 2)3

\ A(s – 2)3 + B(s + 1)(s – 2)2 + C(s + 1)(s – 2) + D(s + 1) = 5s2 – 15s – 11 Putting s = 2 fi 3D = –21 \ D = –7 Putting s = –1 fi –27 A = 9 \ A = –

1 3

Equating coefficients of s3; A + B = 0 \ B = Equating constants; –8A + 4B – 2C + D = –11

1 3 \C=4

ÏÔ 5s 2 - 15 s - 11 ¸Ô Ï 1 ¸Ô 1 -1 Ê 1 ˆ -1 Ê -1/3 ˆ -1 Ô = + + L L 4 L \ L–1 Ì ˝ Ì Á ˜ Á ˜ 3 2˝ Ë s + 1¯ 3 Ë s - 2¯ ÓÔ ( s - 2) ˛Ô ÓÔ ( s + 1) ( s - 2) ˛Ô = -

1 - t 1 2t 7 e + e + 4t e 2t - t 2 e 2t 3 3 2

Part II: Mathematics II

II-J.12

15. (b) (i) [This problem is the same as W.E. 5.6 given in page II-5.99 of the book] (b) (ii) [This problem is the same as the worked example 5.11(ii) given in page II-5.90 of the book. We have to replace a in the W.E. by 1. Required answer is ÏÔ ¸Ô 1 s L-1 Ì 2 = t sin t ] 2˝ ÓÔ ( s + 1) ˛Ô 2

Appendix

K

Solutions to the May/June 2011 Question Paper of Anna University, Tiruchirapalli H 0456 B.E./B.Tech. DEGREE EXAMINATIONS, May/June 2011 Second Semester MA 1151 Mathematics-II (Common to all Branches) REGULATION 2008 Times: Three Hours Maximum:-100 marks Answer ALL questions. PART-A (10 ¥ 2 = 20 marks) at

1.

Lt

2.

Ê Ê s + a ˆˆ L-1 Á tan -1 Á Ë b ˜¯ ˜¯ Ë

3. f

4. 5.

fz

z ex

6. 7.

ze

Ú (z - a) c

8.

z

3

dz

y

Part II: Mathematics II

II-K.2

Ú Ú dx dy

9.

x

x

y

y

1 y

Ú Ú f (x, y) dx dy

10.

0 0

PART-B (5 ¥ 16 = 80 marks) 11. (a)

ft ft

a

Ïk 0 £ t £ a Ì Ó-k a £ t £ 2a f t "t

È ˘ s L Í 2 2 2˙ Î (s + a ) ˚

(b)

Or y¢¢

12. e

t

y

� F

a b c

13. (a)

y¢

� bz i

axy

x

(b)

y

x

� cz j

z

xy

� yk

xz

yz zx

Or � F

14.

V

15. (a) xy

� xy x i x y x x

w w

� y j y

C

z

u

i

(b) •

w Or

16.

y

y¢

w

z

ex

y

z

Appendix K

II-K.3

2 2 a a+ a + y

Ú

17. (a)

Ú

xy dx dy

0 a- a - y

ÚÚÚ

(b)

2

2

dzdy dx 1 - x 2 - y 2 - z2

V x

y

Or ••

ÚÚ

18.

e -( x

2

+ y2 )

•

Úe

dx dy

dx

0

0 0 •

Ú (x

19.

- x2

2

-•

cos x dx + a2 )( x 2 + b2 ) Or

Úz

20. (a)

c

fz

(b)

z+ + z+

dz

C

z z+ z+

z z z Solutions Part-A

1. 2.

Lt

at

-

d Ê a ˆ ds ÁË s + a ˜¯

as s +a

z

i

z

Part II: Mathematics II

II-K.4 3. 4.

5.

f

6.

ex fxx

ex

y fx

y fxx

ex

y fy

ex

y

fyy

z

z

a

a

C

ze z (z - a)3

C

C

Ú

\ 8.

ze z dz (z - a)3

§

9. y

R

x

ÚÚ

2 2

ÚÚ

dx dy

R

x

2

dx dy

0 0

x

10.

x

y y

y 1 1

x

x

I

y

Ú Ú f (x, y) dy dx 0 x

y

Ú 2 dy 0

x

y

y

y

\f

fyy

7.

y

ex

x

y

Appendix K

II-K.5

PART-B 11. (a)

11. (b)

12.

y¢¢ t

y¢ t

yt

e

t

s y s

sy

y¢

sy s s

y

s

s

y s

1 +s-3 s+1

y s

1 s+1

s2 - 2s - 2 s+1 \y s

s2 - 2s - 2 ( s - 2)( s - 1)( s + 1) 1 3 + 2 + 6 s-2 s-1 s+1 -2

3

\ yt 13. (a)

13. (b) 14.

15. (a)

15. (b)

2 3 1 - e 2t + et + e - t 3 2 6

A B C + + s-2 s-1 s+1

Part II: Mathematics II

II-K.6

w

16.

z

§

17. (a)

17. (b) z

q

P y

M

f x

dx dy dz I

x

r

q

f

y

r

q

f

z

r

q

qdf

r dr dq r

dzdy dx

ÚÚÚ

1 - x 2 - y 2 - z2

p p 2 2 1

ÚÚÚ

r 2 sin q dr dq df

0 0 0

1 - r2

p 2

1

p 2

Ú Ú

df sin q

0

0

p 2

p 2

Ú

1 - (1 - r 2 ) 1 - r2

0

dr

1

È ˘ 1 Ïr df sin q dq Ísin -1 r - Ì 1 - r 2 + sin -1 r ˙ 2 Ó2 Î ˚0 0 0

Ú Ú

p

p Ïp ¸ ◊ (- cos q )02 ◊ Ì ˝ 2 Ó4˛

p2 8

18. • •

I=

ÚÚe

- ( x 2 + y2 )

dy

0 0

£x