Engineering Mathematics For GATE [1 ed.] 7397052994
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Copyright © 2016 by Kaushlendra Kumar This work is subject to copyright. All rights are reserved by the Author, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Author’s location, in its current version, and permission for use must always be obtained from Author. Violations are liable to prosecution under the respective Copyright Law. Trademarked names, logos, and images may appear in this book. Rather than use a trademark symbol with every occurrence of a trademarked name, logo, or image we use the names, logos, and images only in an editorial fashion and to the benefit of the trademark owner, until no intention of infringement of the trademark. The use in this publication of trade names, trademarks, service marks, and similar terms, even if they are not identified as such, is not to be taken as an expression of opinion as to whether or not they are subject to proprietary rights. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the Author can accept any legal responsibility for any errors or omissions that may be made. The Author makes no warranty, express or implied, with respect to the material contained herein.
Authors Kaushlendra Kumar B.Tech. (Civil Engineering), Indian Institute of Technology Kanpur Masters in Environment Science, Jawaharlal Nehru University Rajesh Ram B.Tech. (Mechanical Engineering), Indian Institute of Technology Kanpur
Editor Shailendra Arya B.Tech. (Mechanical Engineering), Indian Institute of Technology Kanpur Please e-mail: [email protected] or contact @ +91 7397052994 for any suggestion or any other information.
Engineering Mathematics & General Aptitude For GATE – 2017
Volume – I Engineering Mathematics CONTENT AT A GLANCE Introduction
(i)
About the Authors and Editor
(ii)
Analysis
(iii)
Table of Contents
(v)
Introduction The book presents the subjects of Engineering Mathematics (as per GATE Examination Syllabus) in a systematic, structured and precise manner from three standpoints: To develop their calibre, aptitude and attitude for the engineering field and profession. To strengthen their grasp and understanding of the concepts of the subjects of study and their applicability at the grassroots level. The book aims to solve problems faced by aspirants in terms of extensive syllabus coverage, existing incongruity with syllabus, unavailability of a standard book, etc. Each topic in this book approaches the subject in a very conceptual and coherent manner. While it’s illustrative and solved examples (which are taken from previous year GATE papers from all departments) will facilitate easy mastering of the concepts and their applications. The solved problems will expose the students to the variety and nature of questions that they can expect to face in the GATE examination. This book covers all the important topics that are asked Engineering Mathematics in GATE examination. So this book is helpful to the GATE aspirants who are from following Engineering Streams: For AE, AG, BT, CE, CH, CS & IT, EC, EE, IN, ME, MN, MT, PE, PI, TF, and XE.
Salient Features Brief description of important theorems Derivations of important functions, relationships and equations Many times similar questions from previous years’ one GATE stream were asked in another or same GATE stream. In such a case a particular question is solved as an example and a note, like ‘Similar question was also asked in AE-2012’ is mentioned. From 2015 onwards, all GATE examination from all streams, numerical answer questions apart from multiple choice questions are asked. So the exercise questions have a mix of multiple choice questions and fill in the blanks (in which a numerical answer is filled). The book presents the subjects of Engineering Mathematics in a systematic, structured and precise manner. It intends to offer GATE aspirants a self-study and do-it-yourself approach by providing comprehensive and step-by-step treatment of each and every aspect of the GATE examination. In addition, the book aims to solve problems faced by aspirants in terms of extensive syllabus coverage, existing incongruity with syllabus, unavailability of a standard book, etc. The emphasis on fundamental concepts helps in developing the aptitude required for success in GATE.
(i)
About the Authors and Editor Kaushlendra Kumar is a senior Mathematics Faculty with more than 8 years of teaching experience in various coaching institutes in Delhi, UP and Bihar. Kumar achieved his Bachelor’s Degree in Civil Engineering at Indian Institute of Technology Kanpur, and Master’s degree in Environmental Sciences at Jawaharlal Nehru University. He has made competitive material for various GATE and also for JEE (Mains & Advance) Coaching Institutes. Many of the GATE aspirants, after getting guidance by Kumar, were able to secure SINGLE and TWO DIGIT Ranks in GATE Examination. In the past two years, he has focused on developing this book in such a ways that he is going to appear in GATE 2017 examination.
Rajesh Ram is a senior Mathematics Faculty with more than 8 years of teaching experience in various coaching institutes in Delhi, UP, Bihar and MP. Ram achieved his Bachelor’s Degree in Mechanical Engineering at Indian Institute of Technology Kanpur. He believes that a good teacher and a good book can make all the difference in the world in determining whether you enjoy and understand a subject; many things aren’t hard as they may seem, they just need the light to click on aspirants brain for a concept to become understandable. After getting guidance by Ram, many GATE aspirant were able to secure SINGLE and TWO DIGIT Ranks in GATE Examination.
Shailendra Arya is a Mathematics Faculty with more than 5 years of teaching experience in various coaching institutes in Delhi and UP. Arya achieved his Bachelor’s Degree in Mechanical Engineering at Indian Institute of Technology Kanpur.
(ii)
Stream wise analysis of weightage (out of 100 marks) of Engineering Mathematics in GATE Examination Year/Sets 2010 2011 2012 2013 2014 2015 2016 Average Year/Sets 2010 2011 2012 2013 2014 2015 2016 Average Year/Sets 2010 2011 2012 2013 2014 2015 2016 Average Year/Sets 2010 2011 2012 2013 2014 2015 2016 Average
AE Set: 1 14 11 13 12 15 13 13 13.0
AG Set: 1 15 13 13 9 13 13 13 12.7
Set: 1 16 14 16 13 13 11 11
CSE Set: 2 13 11 11 12.9
BT Set: 1 5 6 6 13 13 13 10 9.4
CE Set: 1 Set: 2 15 13 13 8 13 13 13 13 14 15 13.0
CH Set: 1 16 13 13 12 14 13 13 13.4
ECE Set: 2 Set: 3 13 13 14 15 13 13 12.6
Set: 3 13 13 -
Set: 1 13 8 11 10 13 15 13
Set: 1 13 11 10 13 17 12 16
EE Set: 2 11 12 16 13.2
Set: 3 14 -
IN Set: 1 12 10 15 10 10 10 15 11.7
Set: 1 9 13 13 14 13 15 13
Set: 2 Set: 3 13 13 15 15 13 13 13.2
MN Set: 1 16 13 9 8 14 15 12 12.4
MT Set: 1 15 10 13 14 11 13 13 12.7
PE Set: 1 15 15.0
PI Set: 1 17 15 15 15 10 13 13 14.0
TF Set: 1 15 9 10 8 14 13 14 11.9
XE Set: 1 15 15 15 15 15 15 15 15.0
(iii)
Set: 4 13 -
ME Set: 4 13 -
GATE: Engineering Mathematics Table of Content Page Number From To Chapter - 0 [1] Chapter - 0 [1] [0.1] [0.57] [0.1] [0.15] [0.16] [0.20] [0.21] [0.31] [0.31] [0.43] [0.44] [0.57] Chapter - 1 [1] - Chapter - 1 [10] [1.1] [1.98] [1.1] [1.26] [1.2] [1.6] [1.6] [1.7] [1.7] [1.9] [1.9] [1.12] [1.13] [1.13] [1.13] [1.17] [1.17] [1.26] [1.27] [1.38]
List of Chapters and their sub-topics
GATE - 2016: Chapter - 0: Prerequisite Chapter 0: Prerequisite 0.1 Equations and In-equations 0.2 Sequence, Series and Progression 0.3 Trigonometry 0.4 Two Dimensional Geometry 0.5 Three Dimensional Geometry GATE - 2016: Chapter - 1: Linear Algebra Chapter 1: Linear Algebra 1.1 Matrices and Determinants 1.1.1 Algebra of Matrices 1.1.2 Properties of Matrices 1.1.3 Special Matrices 1.1.4 Determinant of a Square Matrix 1.1.5 Operations in Determinants 1.1.6 Properties of Determinants 1.1.7 Adjoint and Inverse of a Square Matrix 1.2 System of Linear Equations Solution of a Non-Homogeneous System of Linear 1.2.1 [1.27] Equations Solution of Homogeneous Linear Equations 1.2.2 [1.35] Ax = O 1.3 Vector Algebra [1.38] 1.3.1 Scalar or Inner or Dot Product of two Vectors [1.38] 1.3.2 Vector or Cross Product of two Vectors [1.43] 1.3.3 Scalar Triple Product [1.46] 1.3.4 Vector Triple Product [1.47] Vector Space, Linear Dependence and Rank of a 1.4 [1.50] Matrix 1.4.1 Vector Space and Sub-Space [1.50] 1.4.2 Linear Dependence, Spanning Sets and Bases [1.51] 1.4.3 Dot – Product, Norm and Angle for n – Space [1.54] Rank of a Matrix 1.4.4 [1.60] Eigenvalue (Characteristic root) and Eigenvector 1.5 [1.72] (Characteristic Vector) 1.5.1 Eigenbases of a Matrix [1.86] 1.5.2 Similar Matrices [1.86] Matrix Polynomial 1.5.3 [1.90] 1.5.4 Quadratic Forms [1.94] GATE - 2016: Chapter - 2: Calculus Chapter - 2 [1] Chapter 2: Calculus [2.1] 2.1 Set Theory [2.1] 2.1.1 Venn – Euler Diagram [2.2] 2.1.2 Relations [2.5] 2.2 Function [2.11]
(iv)
-
[1.35]
-
[1.38]
-
[1.49] [1.43] [1.45] [1.47] [1.49]
-
[1.71]
-
[1.51] [1.54] [1.60] [1.71]
-
[1.98]
-
[1.86] [1.90] [1.93] [1.98] Chapter - 2 [13] [2.159] [2.11] [2.4] [2.11] [2.21]
Types of Functions Classification of Function Composite Function Limit, Continuity and Differentiability of a 2.3 Function 2.3.1 Limit of a Function 2.3.2 Continuity of a Function 2.3.3 Differentiability of a Function 2.4 Application of Derivatives 2.4.1 Tangent and Normal 2.4.2 Monotonicity of a Function 2.4.3 Mean Value Theorem 2.4.4 Partial and Total Derivatives of a Function 2.5 Integration 2.5.1 Indefinite Integrals 2.5.2 Definite Integrals Curve Sketching, Arc Length, Area and Volume of 2.5.3 a Curve 2.6 Vector Differential Calculus 2.6.1 Scalar and Vector Fields 2.6.2 Derivative of a Scalar Field 2.6.3 Derivative of a Vector field 2.6.4 Vector/Matrix Derivatives 2.7 Vector Integral Calculus 2.7.1 Line Integrals 2.7.2 Double Integrals 2.7.3 Green’s Theorem in a Plane 2.7.4 Surface Integrals 2.7.5 Volume Integrals 2.7.6 Gauss’s Divergence Theorem 2.7.7 Stokes’ Theorem 2.8 Sequence and Series 2.8.1 Sequence 2.8.2 Series 2.8.3 Power Series 2.8.4 Taylor’s Series and Maclaurin Series 2.8.5 Improper Integral GATE - 2016: Chapter - 3: Differential Equation Chapter 3: Differential Equation 3.1 Solution of a Differential Equation Solution of first order and first degree differential 3.1.1 equation 3.1.2 Exact Differential equation Solution of Special type of second order differential 3.1.3 equation 3.2 Higher Order Differential Equation 3.2.1 Preliminary theory: Linear Equations 3.2.2 Reduction of Order 2.2.1 2.2.2 2.2.3
(v)
[2.11] [2.15] [2.19]
-
[2.15] [2.19] [2.21]
[2.22]
-
[2.46]
[2.22] [2.30] [2.34] [2.47] [2.47] [2.49] [2.61] [2.63] [2.72] [2.72] [2.79]
-
[2.29] [2.33] [2.46] [2.71] [2.48] [2.61] [2.63] [2.71] [2.99] [2.79] [2.90]
[2.90]
-
[2.99]
[2.100] [2.100] [2.102] [2.107] [2.112] [2.115] [2.115] [2.119] [2.124] [2.126] [2.130] [2.132] [2.135] [2.140] [2.140] [2.141] [2.146] [2.148] [2.151] Chapter - 3 [1] [3.1] [3.1]
-
[2.114] [2.102] [2.107] [2.111] [2.114] [2.139] [2.119] [2.123] [2.126] [2.130] [2.132] [2.135] [2.139] [2.159] [2.141] [2.146] [2.148] [2.151] [2.159] Chapter - 3 [8] [3.99] [3.18]
[3.4]
-
[3.12]
[3.12]
-
[3.15]
[3.16]
-
[3.18]
[3.19] [3.19] [3.21]
-
[3.44] [3.21] [3.22]
3.2.3
Homogeneous Linear Equations with Constant Coefficients Non – Homogeneous Linear Equation with Constant coefficients Variation of Parameters Cauchy – Euler Equation Laplace Transform
[3.23]
-
[3.28]
[3.28]
-
[3.34]
[3.35] [3.37] [3.45]
-
[3.37] [3.44] [3.67]
Inverse Transforms and Transforms of Derivatives
[3.47]
-
[3.51]
Important Singularity Functions Operational Properties Fourier Series Waveform Symmetry Convergence and Sum of a Fourier Series Half Range Expansion Complex Form of Fourier Series Fourier Transform Partial Differential Equation Modelling and Solution of Vibrating String: Wave 3.5.1 Equation 3.5.2 Modelling and Solution of Heat Equation Solution of Steady Two – Dimensional Heat 3.5.3 Problem: Laplace’s Equation GATE - 2016: Chapter - 4: Complex Variables Chapter - 4: Complex Variables 4.1 Basic Concept of Complex Numbers 4.1.1 Complex Plane 4.1.2 Polar Form of a Complex Number 4.1.3 Complex Function 4.1.4 Logarithm of a Complex number 4.2 Complex Integration 4.2.1 Line Integral in the Complex Plane 4.2.2 Cauchy’s Integral Theorems 4.2.3 Derivatives of analytic function 4.2.4 Sequence and Series 4.2.5 Singularities, Zero and Infinity 4.2.6 Zeros of Analytic function 4.2.7 Residue Integration Method GATE - 2016: Chapter - 5: Numerical Methods Chapter - 5: Numerical Methods 5.1 Algebraic and Transcendental Equation 5.1.1 Location of real roots of an equation Exact Solution of Algebraic and Transcendental 5.1.2 Equation Numerical Solution of Algebraic and 5.1.3 Transcendental Equation 5.1.4 Rate of Convergence of Approximate Methods 5.1.5 Interpolation
[3.52] [3.54] [3.68] [3.69] [3.72] [3.75] [3.75] [3.76] [3.83]
-
[3.54] [3.67] [3.82] [3.72] [3.75] [3.75] [3.76] [3.82] [3.99]
[3.88]
-
[3.91]
[3.91]
-
[3.94]
[3.94]
-
[3.99]
Chapter - 4 [1] [4.1] [4.1] [4.2] [4.4] [4.7] [4.13] [4.16] [4.16] [4.17] [4.20] [4.21] [4.24] [4.25] [4.25] Chapter - 5 [1] [5.1] [5.3] [5.3]
-
Chapter - 4 [5] [4.34] [4.15] [4.3] [4.7] [4.13] [4.15] [4.34] [4.17] [4.20] [4.20] [4.24] [4.25] [4.25] [4.34] Chapter - 5 [4] [5.52] [5.25] [5.3]
[5.3]
-
[5.4]
[5.4]
-
[5.15]
[5.15] [5.16]
-
[5.16] [5.25]
3.2.4 3.2.5 3.2.6 3.3 3.3.1 3.3.2 3.3.3 3.4 3.4.1 3.4.2 3.4.3 3.4.4 3.4.5 3.5
(vi)
5.2
Numerical Solution of Definite Integrals, First Order ODEs, and System of Linear Equations
Numerical Solution of Definite Integrals
[5.26]
-
[5.52]
5.2.1 5.2.2
Numerical Solution of First Order ODEs
[5.26] [5.37]
-
[5.37] [5.43]
5.2.3
Numerical Solution of System of Linear Equations
[5.43]
-
[5.52]
Chapter - 6 [1]
-
Chapter - 6 [7]
[6.1] [6.1] [6.1] [6.5] [6.15] [6.15]
-
[6.88] [6.14 [6.5] [6.14] [6.27] [6.16]
[6.16]
-
[6.20]
[6.20]
-
[6.27]
[6.28]
-
[6.52]
[6.28]
-
[6.29]
[6.29] [6.43] [6.47] [6.53]
-
[6.43] [6.47] [6.52] [6.88]
[6.54]
-
[6.65]
[6.65] [6.71] [6.81] PTP [1]
-
[6.71] [6.81] [6.88] PTP [12]
GATE - 2016: Chapter - 6: Probability and Statistics Chapter - 6: Probability and Statistics 6.1 Permutation and Combination 6.1.1 Permutation 6.1.2 Combination 6.2 Statistics 6.2.1 Graphs for Qualitative and Quantitative Data Describing Data with Numerical Measures: 6.2.2 Measures of Centre Describing Data with Numerical Measures: 6.2.3 Measures of Variability 6.3 Probability Theory Basic Terminology in concept of Probability in Set 6.3.1 Theoretic Language 6.3.2 Definition of Probability 6.3.3 Conditional Probability 6.3.4 Total Probability and Baye’s Theorem 6.4 Random Variable Properties of Random Variable and their 6.4.1 Probability Distribution 6.4.2 Types of Discrete Random Variable Distribution 6.4.3 Types of Continuous Random Distribution 6.4.4 Jointly Distributed Random Variables 5 Practice Test Papers
(vii)
Engineering Mathematics
Chapter – 0: Prerequisite
Chapter – 0 [1]
GATE – 2016: Chapter – 0: Prerequisite Note: The following questions came in GATE – 2016 were based on Pre-requisite Chapter – 0. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 1. A straight line of the form y mx c passes through the origin and the point ( x, y ) (2, 6) . The value of m is _____. [IN-2016 (1 mark)] Solution: As the given straight line passes through the origin, so c 0 , thus we have y mx . So y mx passes through (2, 6) , thus we have 6 m(2) m 3 . 2. Equation of two planes are z 4 and z 4 3 x . The included angle between the two planes in degrees, is _____. [MN-2016 (1 mark)] Solution: The given planes can be written as: 0 x 0 y z 4 …(i), and 3x 0 y z 4 …(ii) The normal vector of plane (i) can be written as: n1 (0, 0,1) ; The normal vector of plane (i) can be written as: n2 ( 3, 0,1) . So the angle between the given two planes is 0 3 0 0 1 1
cos 1
2
2
2
2
0 0 1
2
cos 1
2
1
o
cos 1 ( 0.316) 71.56 o or 108.43 .
10
( 3) 0 1
3. The vector parallel to the plane 3 x 2 y z 1 is [MT-2016 (1 mark)] (a) iˆ ˆj kˆ (b) 3iˆ 2 ˆj kˆ (c) iˆ ˆj kˆ (d) 3iˆ 2 ˆj 2kˆ Solution: The normal vector of the given plane is n 3iˆ 2 ˆj kˆ . Now the dot product of any vector parallel to the given plane with n is always zero. So checking all the options, we get option (a) is the vector whose dot product with n is zero; since (iˆ ˆj kˆ ) n (iˆ ˆj kˆ ) (3iˆ 2 ˆj kˆ ) 3 2 1 0 . Hence option (a) is correct. 3
4
5
6
3
4. The coefficient of x12 in ( x x x x ) is _____. [CS-2016 (2 marks)] Solution: ( x3 x 4 x 5 x 6 )3 { x3 (1 x x 2 x 3 )}3 x 9 (1 x x 2 x 3 )3 x9 {(1 x) 1}3
( x3 x 4 x5 x 6 ) 3 x 9 (1 x ) 3 3
4
5
6
3
9
( x x x x ) x 1 ( 3)( x ) ( 3)( 3 1)
12
So coefficient of x
3
4
5
6
( x)2 2!
( 3)( 3 1)( 3 2)
3
( x )3 3!
3
in ( x x x x ) is ( 3)(3 1)( 3 2)(1 3!)(1) 10 .
5. Let S n 0 n n , where 1 . The value of in the range 0 1 , such that S 2 is _____.
[EE-2016 (2 marks)]
2
3
4
Solution: S n 0 n 0 2 3 4 S 2 3 4 …(i) n
2
3
4
S 2 2 3 3 4 4 5 …(ii) (i) (ii) (1 ) S 2 3 4 5 (1 ) S (1 ) 2 Now, as S 2 , so 2
2
2
2(1 ) 1 2 4 1 0 (1 ) 2 1.707, 0.292 . As (0,1) , so 0.292 .
Copyright © 2016 by Kaushlendra Kumar
4 ( 4) 2 4(2)(1) 2(2)
e-mail: [email protected]
Engineering Mathematics
Chapter 0: Prerequisite
[0.1]
Chapter 0 : Prerequisite 0.1
Equations and In-equations
A set of all real numbers can be expressed as: x ( , ) or x R or x x ( , a) (b, ) x R [a, b] x ( , a] [b, ) x R (a, b) x (, a ) [b, ) x R [ a, b) x ( , a] (b, ) x R ( a, b]
Intervals: The set of numbers between any two real numbers is called interval. There are three four types of interval: Closed interval: x [a, b] or { x : a x b} , as shown in Fig. 0.1(a). Open interval: x ( a, b) or { x : a x b} , as shown in Fig. 0.1(b). Open-Closed interval: x ( a, b] or { x : a x b} , as shown in Fig. 0.1(c). Closed-Open interval: x [ a, b) or { x : a x b} , as shown in Fig. 0.1(d). Figure 0.1: Types of Intervals
Inequalities If a and b are real and (i) ( a b) is positive then a b (ii) ( a b) is negative then a b If c is real and a b then a c b c and a c b c If m is real and a b , then
am bm if m 0 m m a b if m 0 a a ax a a ax If 0 1 , then 1 If 1 , then 1 For any x 0 b b bx b b bx x (1 x ) 2 for all x 0 , equality holds at x 1 x (1 x) 2 for all x 0 , equality holds at x 1 .
( a12 a22 an2 )(b12 b22 bn2 ) ( a1b1 a2 b2 an bn ) 2
Squaring an inequality: If a b , then it is not necessary that a 2 b 2 always holds true. For e.g. 2 2 2 2 2 2 2 5 2 5 ; but 5 2 (5) 2 . Thus a b a b follows only when a b . Law of reciprocal: If both sides of inequality have same sign, then after taking their reciprocal the sign of inequality gets reversed, i.e. if 0 a b (1 b ) (1 a ) ; or if a b 0 (1 b ) (1 a ) . If both sides of inequality have opposite sign, then after taking their
am bm if m 0 m m a b if m 0
reciprocal the sign of inequality remains same, i.e. if a 0 b (1 a ) (1 b ) .
Wavy-Curve Method: Generalized method of intervals for solving inequalities: Let f ( x ) ( x a1 ) k1 ( x a2 ) k2 ( x a n 1 ) k n1 ( x an ) kn …(i), where k1 , k 2 , , k n N and a1 , a2 , , an
are fixed numbers satisfying the condition a1 a2 an 1 an . First we mark numbers a1 , a2 , , an on the real axis and the plus sign in the interval of the right of the largest of these
numbers, i.e. on the right of an . If kn is even then we put plus sign on the left of an and if kn is odd then we put minus sign on the left of an . In the next interval we put a sign according to the following rule:
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 0: Prerequisite
[0.2]
1. When passing through the point an 1 the polynomial f ( x ) changes sign if kn 1 is an odd number and the polynomial f ( x ) has same sign if kn 1 is an even number. Then, we consider the next interval and put a sign in it using the same rule. Thus, we consider all the intervals. 2. The solution of f ( x ) 0 is the union of all intervals in which we have put the plus sign and the solution of f ( x ) 0 is the union of all intervals in which we have put the minus sign. It is to be noted that, we put plus sign in the interval of the right of the largest of these numbers, i.e. on the right of an , only when all the ‘ x ’ in (i) are positive. If some of the ‘ x ’ in (i) are negative then
Put plus sign in the interval of the right of largest of these numbers, i.e. on the right of an , if total number of ve ‘ x ’ are even. Then follow the rules given in 1 and 2 for finding the answer. Put minus sign in the interval of the right of largest of these numbers, i.e. on the right of an , if total number of ve ‘ x ’ are odd. If kn is even then put minus sign on left of an and if kn is odd then put plus sign on the left of an . Then follow the rules given in 1 and 2 for finding the answer.
It is to be always remember that while finding the values of ‘ x ’, such that f ( x ) 0 or f ( x ) 0 , we should not include those ‘ x ’ for which f ( x ) becomes undefined. Let us consider some example: Example 0.1: Find all values of x such that the inequality
2x
1
holds true? 2x 5x 2 x 1 2x 1 2 x2 2 x 2 x2 5x 2 3x 2 0 0 0 Solution: 2 2 x 5x 2 x 1 (2 x 1)( x 2)( x 1) (2 x 1)( x 2)( x 1) 3( x 2 3) ( x 2 3) 0 0 . Equating each factor equal to 0, we get ( x 1)( x 2)(2 x 1) ( x 1)( x 2)(2 x 1) 2
x 2, 1, 2 3, 1 2 . Marking the numbers on the real axis and putting signs in the interval according to the given rules in wavy-curve method, as shown in figure, we get x (2, 1) [ 2 3, 1 2) . Here we cannot put closed interval on ‘–2’, ‘–1’ and ‘ 1 2 ’, as at these values the given expression becomes undefined. Example 0.2: Find ‘ x ’ such that f ( x ) 0 , where f ( x ) ( x 1)( x 2)(1 2 x ) . Solution: Equating each factor equal to 0, we get x 1 2 ,1, 2 . Marking the numbers on the real axis and putting signs in the interval according to the given rules in wavy-curve method, as shown in figure, we get x (,1 2) (1, 2) for which f ( x ) 0 . It is to be noted that we put minus sign to the right of ‘2’ as number of negative is 1. Example 0.3 [TF-2014 (1 mark)]: The range of values of x satisfying the inequality x 2 3 x 2 0 is (d) cannot be determined (a) x 0 (b) 0 x 1 (c) 1 x 2 2 Solution (c): x 3 x 2 0 ( x 2)( x 1) 0 , now equating each factor to ‘0’ we get by applying wavy curve method for the given inequality we have, x (1, 2) or 1 x 2 .
Polynomial Equation: Algebraic expression containing terms of the form cx n , where n being a non-negative integer, is called a polynomial. A polynomial of degree ‘ n ’ can be represented as: f ( x ) an x n a n 1 x n 1 an 1 x n 1 a2 x 2 a1 x a0 , where x is a variable, a0 , a1 , a2 , , an are
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 0: Prerequisite
[0.3]
constants and an 0 . A polynomial is called to be real if the constants and variable are real; and is called to be complex if constants or variable or both are complex numbers. A polynomial of second degree is generally called a quadratic polynomial. Polynomials of degree 3 and 4 are known as cubic and biquadratic polynomials respectively. If f ( x ) is a polynomial, real or complex, then f ( x ) 0 is called a polynomial equation. The values of variable ‘ x ’ satisfying a given polynomial equation are called its roots, i.e. if x is a root of equation f ( x ) 0 , then f ( ) 0 .
A polynomial equation of n degree has n roots (real or imaginary). If 1 , 2 , , n 1 , n are n roots of a polynomial equation f ( x ) an x n an 1 x n 1 a2 x 2 a1 x a0 , then Sum of roots, 1 2 n 1 n an 1 an Sum of product of roots taking two at a time, 1 2 1 n 2 3 2 4 an 2 an Sum of product of roots taking three at a time, 1 2 3 an 3 an Product of all roots, 1 2 3 n 1 n (1) n a0 an
If all the coefficients are real then the imaginary roots always occur in pairs, i.e. number of complex roots is always even. If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at least one of the roots will be real. If is repeated root (repeating r times) of a polynomial equation f ( x ) 0 of degree n , i.e. f ( x ) ( x ) r g ( x) , where g ( x ) is a polynomial of degree ( n r ) and g ( ) 0 , then
f ( ) f ( ) f ( r 1) ( ) 0 and f r ( ) 0 If we divide a polynomial p( x ) by ( x ) , the remainder obtained is p ( ) . If p ( ) 0 , then ( x ) is a factor of p( x ) .
If a polynomial equation of degree n has ( n 1) roots, say, 1 , 2 , n , n 1 ( i j for i j )
then the polynomial is identically zero, i.e. f ( x ) 0 x . If f (a ) and f (b) , where a b , are of opposite signs, then f ( x ) 0 has odd number of roots in ( a, b) , i.e. it have at least one root in ( a, b) .
Descarte’s rule of sign for finding number of roots of a polynomial equation: Let f ( x ) 0 be a polynomial equation with real coefficients, The number of ve roots of f ( x ) 0 is equal to the number of variation of signs of the successive coefficients of f ( x ) or less by an even number. The number of ve roots of f ( x ) 0 is equal to the number of variation of signs of the successive coefficients of f ( x ) or less by an even number. Let f ( x ) 0 have no ‘zero’ root, and let the number of variations of signs of the successive coefficients of f ( x ) and f ( x ) be r and k , respectively, then The maximum number of real roots of f ( x ) is ( r k ) If ( r k ) n , then the equation f ( x ) 0 has exactly ‘ r ’ number of ve real roots and ‘ k ’ number of ve real roots. Example 0.4: Find how many ve real roots lie in (0,1) and (1, 2) for the polynomial equation f ( x ) 27 x 4 48 x 2 12 x 13 0 .
Solution: f ( x ) 27 x 4 48 x 2 12 x 13 0 two variation of signs, so f ( x ) 0 has two ve real roots or no ve real roots. Now f (0) 13 0 , f (1) 20 0 and f (2) 229 0 ; thus f ( x ) 0 has at least one ve real root in (0,1) and at least one ve real root in (1, 2) . Thus, we can say that f ( x ) 0 has exactly one ve real root in (0,1) and exactly one ve real root in (1, 2) .
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Example 0.5: Find the possible number of roots of the polynomial equation f ( x ) 0 , where f ( x ) 3 x5 2 x 4 5 x 3 4 x 2 6 x 7 .
Solution: f ( x ) 3 x 5 2 x 4 5 x 3 4 x 2 6 x 7 two change of signs, so f ( x ) 0 has two ve real roots or ‘zero’ ve real roots. f ( x ) 3x 5 2 x 4 5 x 3 4 x 2 6 x 7 three change of sign, so f ( x ) 0 has three ve real roots or one ve real root and two complex roots. The possibility of roots is given in table, in which only one row is possible.
ve real 2 2 0 0
ve real 3 1 3 1
complex
total
0 2 2 4
5 5 5 5
Example 0.6 [EE-1994 (2 marks)]: Find the number of positive real roots of the equation 3 x 2x 2 0 . Solution: As f ( x) x 3 2 x 2 we have two variation of sign so f ( x ) 0 has two ve real roots or no ve real root. Also f ( x ) x 3 2 x 2 we have one variation of sign so f ( x ) 0 2
has one ve real root. Now f ( x ) 3 x 2 0 x 2 3 ; also as x , f ( x ) and as x , f ( x) . Since
f (0) 2 and f
2 3 0 , so from x 0 to x 2 3 , the
graph of f ( x ) 0 not crosses the x axis. Hence the possible graph is shown in figure, which shows that f ( x ) 0 has no ve real roots and thus f ( x ) 0 has two complex roots and one ve real root. Thus f ( x ) 0 has no ve real roots. Example 0.7 [CS-1997 (2 marks)]: A polynomial p ( x ) is such that p (0) 5 , p (1) 4 , p (2) 9 and p (3) 20 . The minimum degree it can have is (a) 1 (b) 2 (c) 3 (d) 4 Solution (b): As x approaching from x 0 to x 1 , the value of p( x ) decreases; on the other hand as x approaching from x 1 to x 2 to x 3 , the value of p( x ) keeps on increasing; so from the given conditions, we can say that between x 1 and x 2 , the graph of p( x ) either cut the x axis two times (i.e. it has two real roots) or it takes the shape of concave downward with cutting the x axis (i.e. it has two complex roots); in either case we have minimum two roots. Example 0.8 [CS-2000 (2 marks)]: A polynomial p ( x ) satisfies the following: p (1) p (3) p (5) 1 ; p (2) p(4) 1 . The minimum degree of such a polynomial is (a) 1 (b) 2 (c) 3 (d) 4 Solution (d): As p (1) 1 0 and p (2) 1 0 , so we have one root of p ( x ) 0 in (1, 2) ; also p (2) 1 0 and p (3) 1 0 , so we have one root of p ( x ) 0 in (2,3) ; also p (3) 1 0 and p (4) 1 0 , so we have one root of p ( x ) 0 in (3, 4) ; also p (4) 1 0 and p (5) 1 0 , so we have one root of p ( x ) 0 in (4,5) ; thus from given conditions we have at least four real roots of p ( x ) 0 in x [1,5] and so the minimum degree of p ( x ) 0 is four. Example 0.9 [EE-2006 (2 marks)]: The algebraic equation F ( s) s5 3s 4 7 s 2 4s 20 is given. F ( s) 0 has (a) a single complex root with the remaining roots being real (b) one positive real root and four complex root, all with positive real parts (c) one negative real root, two imaginary roots, and two roots with positive real parts (d) one positive real root, two imaginary roots, and two roots with negative real parts
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Solution (c): The given equation has a total of five roots (real or imaginary). Applying Descarte’s rule of sign, we have F ( s ) s 5 3s 4 7 s 2 4s 20 we have two variation in sign so F ( s) 0 has two ve real roots or ‘zero’ ve real roots. Also F ( s) s 5 3s 4 7 s 2 4s 20 we have one variation of sign so F ( s) 0 has one ve real root. Now option (a) is not correct as complex roots always occur in conjugate pairs; options (b) and (d) are not correct as F ( s) 0 has either two ve real roots or ‘zero’ ve real roots. So we are left with option (c) which satisfy our calculated results. Example 0.10 [CE-2007 (2 marks)]: Given that one root of the equation x 3 10 x 2 31x 30 0 is 5, the other two roots are (a) 2 and 3 (b) 2 and 4 (c) 3 and 4 (d) –2 and –3 Solution (a): If 5 is a root of the given equation then ( x 5) is its factor. On dividing the given equation by ( x 5) , we get p ( x ) x 2 5 x 6 ( x 2)( x 3) , so the roots of given equation is the roots of p ( x ) 0 , i.e. x 2, 3 are its roots. Alternative method: As sum of roots of the given equation is ( 10 1) 10 . So from the given options, option (a) satisfies the equation. Example 0.11 [IN-2007 (2 marks)]: The polynomial p ( x ) x5 x 2 has (a) all real roots (b) 3 real and 2 complex roots (c) 1 real and 4 complex roots (d) all complex roots Solution (c): Applying Descarte’s rule to p ( x ) x5 x 2 , we have zero variation in sign so
p ( x ) 0 have no ve real roots. Also p ( x ) x5 x 2 , so we have one variation in sign and thus p ( x ) 0 have one ve real root. Hence the given equation have one ve real root and four complex roots (as the given equation have total five complex roots). Example 0.12 [IN-2008 (2 marks)]: It is known that two roots of the non-linear equation 3 2 x 6 x 11x 6 0 are 1 and 3. The third root will be (c) 2 (d) 4 (a) j (b) j Solution (c): Let the third root be . As the sum of roots be (6 1) 6 1 3 6 2 . Example 0.13 [EE-2009 (2 marks)]: A cubic polynomial with real coefficients (a) can possibly have no extrema and no zero crossings (b) may have up to three extrema and upto 2 zero crossings (c) cannot have more than two extrema and more than three zero crossings (d) will always have an equal number of extrema and zero crossings Solution (c): Let f ( x ) ax 3 bx 2 cx d f ( x) 3ax 2 2bx c f ( x ) 6ax 2b , as f ( x) 0 is a second order polynomial or quadratic equation so f ( x) 0 have at most two real roots; and at those values of x , where f ( x) 0 , f ( x ) either or or zero. So f ( x ) 0 have at most two extrema and thus f ( x ) 0 have at most three real roots. Thus option (c) is correct. Example 0.14 [EC-2013 (1 mark)]: A polynomial f ( x ) a4 x 4 a3 x 3 a2 x 2 a1 x a0 with all coefficients positive has (a) no real roots (b) no negative real root (c) odd number of real roots (d) at least one positive and one negative real root Solution (d): By applying Descarte’s rule of sign, we have one change of sign in f ( x ) 0 , so we have one ve real root ; also we have one change of sign in f ( x ) 0 , so we have, one ve real root. As the given equation have total four roots. So other two roots are complex roots. Hence from given options, option (d) is correct.
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Engineering Mathematics
Chapter 0: Prerequisite
[0.6]
Example 0.15 [CS-2014 (1 mark)]: A non-zero polynomial f ( x ) of degree 3 has roots at x 1 , x 2 and x 3 . Which one of the following must be true? (a) f (0) f (4) 0 (b) f (0) f (4) 0 (c) f (0) f (4) 0 (d) f (0) f (4) 0 Solution: It is given that for a third degree polynomial, f ( x ) 0 , f (1) f (2) f (3) 0 , i.e. f ( x ) 0 have three real roots. Now we have two cases, as shown in figure:
Case I: f (1 ) 0 , then f (1 ) 0 , f (2 ) 0 , f (2 ) 0 , f (3 ) 0 and f (3 ) 0 ; so f (0) 0 and f (4) 0 . Thus we have f (0) f (4) 0 . Case II: f (1 ) 0 , then f (1 ) 0 , f (2 ) 0 , f (2 ) 0 , f (3 ) 0 and f (3 ) 0 ; so f (0) 0 and f (4) 0 . Thus we have f (0) f (4) 0 . Hence in both cases we have f (0) f (4) 0 , so option (a) is correct. Example 0.16 [XE-2014 (1 mark)]: If a cubic polynomial passes through the points (0,1) , (1, 0) , (2,1) and (3,10) , then it also passes through the point (a) ( 2, 11) (b) (1, 2) (c) (1, 4) (d) ( 2, 23) Solution (b): Let the cubic polynomial is f ( x ) ax 3 bx 2 cx d . Now satisfying the given point, we get: (0,1) d 1 f ( x ) ax 3 bx 2 cx 1 …(i); (1, 0) a b c 1 …(ii); (2,1) 8a 4b 2c 0 …(iii); (3,10) 27 a 9b 3c 9 …(iv). From (ii), (iii) and (iv), we get 3 2 a 1 , b 2 and c 0 . So f ( x ) x 2 x 1 , now checking all the option we will find that the
point ( 1, 2) satisfy f ( x ) x 3 2 x 2 1 , so option (b) is correct.
Quadratic Equation: A quadratic equation in ‘ x ’ is given as 2 ax bx c 0 , where a , b, c and are constants
The roots of the quadratic equation (Eq. 0.1) are x
b b 2 4ac
2a 2 D b 4ac is called the discriminant of the quadratic equation (Eq. 0.1). If the two roots of a quadratic equation (Eq. 0.1) are and , then
(0.1) . The expression
Sum of the roots, b a Product of the roots, c a If a, b, c are all positive or all negative, the sum of the roots is negative while their product is positive. Hence both the roots are negative. If a and b are of the same sign and c is of the opposite sign, both the sum and the product of the roots are negative. Hence one root is positive and the other root is negative. If a and c are of the same sign and b is of the opposite sign, both the sum and product of the roots are positive. Hence both the roots are positive. If b and c are of the same sign and a is of the opposite sign, the sum of the roots is positive, while their product is negative. Hence one root is positive and the other root is negative. Nature of roots If D 0 then the roots are real and distinct If D 0 then the roots are real and equal
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If D 0 then the roots are complex If D is a perfect square then the roots are real rational and distinct If one root of a quadratic equation is irrational, the other root is its irrational conjugate (given that a, b, c are rational) If one root of a quadratic equation is complex, the other root is its complex conjugate (given that a, b, c are real) If sum and product of the roots of quadratic equation in x is given then, the quadratic equation is x 2 x(sum of the roots) (product of the roots) 0 If a1 x 2 b1 x c1 0 and a2 x 2 b2 x c2 0 are two quadratic equations. If both equations have only one common root then (c1a2 c2 a1 ) 2 (b1c2 b2 c1 )( a1b2 a2b1 ) ; and the common root is
c1a2 c2 a1 a1b2 a2 b1
or
b1c2 b2 c1 c1a2 c2 a1
If both equations have both roots in common then a1 a2 b1 b2 c1 c2
Sign of quadratic expression When a 0 , ( ax 2 bx c ) 0 , for all real values of x , if D 0 When a 0 , ( ax 2 bx c ) 0 , for all real values of x , if D 0 When a 0 , ( ax 2 bx c ) 0 when x lies between the roots of the eqn ax 2 bx c 0 When a 0 , ( ax 2 bx c ) 0 when x lies between the roots of the eqn ax 2 bx c 0 All the above four cases can be combined in a single statement as: The value of the quadratic expression ax 2 bx c has always the same sign as a , except when ax 2 bx c 0 has real and distinct roots and x lies between them.
Maximum and Minimum values When a 0 , the minimum value of ax 2 bx c is D 4a and is attained when x b 2a When a 0 , the maximum value of ax 2 bx c is D 4a and is attained when x b 2a [This point was asked in MT-2009 (1 mark)] Geometrically, the curve 2 y ax bx c , a 0 represent a parabola which opens upwards if a is positive and opens downward if a is negative. The coordinates of the vertex of the D b parabola are V , . 2a 4a The position of vertex for different set of values of a and D is shown in Fig. 0.2 Figure 0.2: Position of vertex of the parabola
Example 0.17 [CS-1995 (2.5 marks)]: Find the minimum value of 3 4 x 2 x 2 . Solution: Comparing the given equation with Eq. 0.1, we have a 2, b 4, c 3 . So the minimum 2
2
value will be D 4a (b 4ac) 4a ( 4) 4(2)(3)
4(2) 1 .
Position of roots of a quadratic equation (Eq. 0.1) Let f ( x ) ax 2 bx c , where a , b, c be a quadratic expression and k , k1 , k 2 be real numbers such that k1 k2 . Let , be the roots of the equation (Eq. 0.1). Then
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[0.8]
(a)
(b)
(c)
(d)
(e)
(f)
Figure 0.3: Condition for position of roots of Quadratic Equation (Eq. 0.1)
Condition for a number k , if both the roots of Eq. 0.1 are less than k : (i) D 0 (roots may be equal) (ii) a f ( k ) 0 (iii) k b 2a , where ; as shown in Fig. 0.3(a). Condition for a number k , if both the roots of Eq. 0.1 are greater than k : (i) D 0 (roots may be equal) (ii) a f ( k ) 0 (iii) k b 2a , where ; as shown in Fig. 0.3(b). Condition for a number k , if k lies between the roots of Eq. 0.1: (i) D 0 (ii) a f (k ) 0 , where ; as shown in Fig. 0.3(c). Condition for numbers k1 and k 2 , if exactly one root of Eq. 0.1 lies in the interval ( k1 , k 2 ): (i) D 0 (ii) f ( k1 ) f (k 2 ) 0 , where ; as shown in Fig. 0.3(d).
Condition for numbers k1 and k 2 , if both roots of Eq. 0.1 are confined between k1 and k 2 : (i) D 0 (roots may be equal)
a f ( k1 ) 0
(ii)
(iii)
a f ( k2 ) 0
(iv)
k1 b 2a k 2 , where and k1 k2 ; as shown in Fig. 0.3(e).
Condition for numbers k1 and k 2 , if k1 and k 2 lie between the roots of Eq. 0.1: (i)
D 0 (ii) a f ( k1 ) 0 (iii) a f ( k2 ) 0 , where ; as shown in Fig. 0.3(f).
x, if x 0 Modulus or Absolute value function: The function defined by f ( x) x 0, if x 0 is called x, if x 0 the modulus function. Its Domain: ; Range: [0, ) ; Period: Non – periodic; Nature: Even; Interval in which the inverse can be obtained: ( , 0] [0, ) . Example 0.18 [CE-2013 (1 mark)]: If 4 x 7 5 then the value of 2 x x is (a) 2, 1/3 (b) 1/2, 3 (c) 2/3, 1/3 (d) 2/9, 3 (4 x 7); 4 x 7 0 4 x 7; x 7 4
Solution (b): 4 x 7
0;
(4 x 7);
4x 7 0
0;
x7 4
4x 7 0 (4 x 7); x 7 4
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4 x 7 5; x 7 4 x 1 2; x 7 4 x 1 2 So, 4 x 7 5 0 5; x 7 4 0 5; x 7 4 x (4 x 7) 5; x 7 4 x 3; x 7 4 x 3 At x 1 2 2 x x 2 1 2 1 2 1 1 2 1 2 At x 3 2 x x 2 3 3 6 3 3 Please note that the questions on Equations and In-equations also came in previous years ‘General Aptitude’ section in GATE examination. These questions are given in ‘General Aptitude’ section. Exercise 0.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. 2. 3. 4.
5.
Both the roots of given equation ( x a )( x b) ( x b)( x c ) ( x c )( x a ) 0 are always (a) Positive (b) Negative (c) Real (d) Imaginary 2 If the roots of (b c ) x (c a) x ( a b) 0 are equal then a c (a) 2b (c) 3b (d) b (b) b 2 If 3 is a root of x 2 kx 24 0 , it is also a root of (a) x 2 5 x k 0 (b) x 2 5 x k 0 (c) x 2 kx 6 0 (d) x 2 kx 24 0 If the roots of equation 1 ( x p) 1 ( x q ) 1 r are equal in magnitude but opposite in sign, then ( p q) (d) None of these (a) 2r (b) r (c) 2r 2 For what values of k will the equation x 2(1 3k ) x 7(3 2k ) 0 have equal roots (a) 1, 10 9
6.
7.
8.
9. 10.
(b) 2, 10 9
(d) 4, 10 9
If the difference between the corresponding roots of x 2 ax b 0 and x 2 bx a 0 is same and a b , then (a) a b 4 0 (b) a b 4 0 (c) a b 4 0 (d) a b 4 0 2 If the sum of the roots of the quadratic equation ax bx c 0 is equal to the sum of the squares of their reciprocals, then a c , b a , c b are in (a) AP (b) GP (c) HP (d) None of these 2 2 Let , be the roots of x x p 0 and , be root of x 4 x q 0 . If , , , are in G.P., then the integral value of p and q respectively are (a) –2, –32 (b) –2, 3 (c) –6, 3 (d) –6, –32 2 If 1 i is a root of the equation x ax b 0 , then the values of a and b are (a) 2, 1 (b) –2, 2 (c) 2, 2 (d) 2, –2 2 If the roots of the equation x 5 x 16 0 are , and the roots of equation x 2 px q 0 are 2 2 , 2 , then (a) p 1, q 56 (b) p 1, q 56
11.
(c) 3, 10 9
(c) p 1, q 56
(d) p 1, q 56
12.
If , but 2 5 3, 2 5 3 , then the equation whose roots are and is (a) x 2 5 x 3 0 (b) 3 x 2 19 x 3 0 (c) 3 x 2 12 x 3 0 (d) None of these Let , be the roots of the equation ( x a )( x b) c , c 0 , then the roots of the equation ( x )( x ) c 0 are (a) a, c (b) b, c (c) a, b (d) a, d
13.
If and are roots of the equation x 2 ax b 0 and Vn n n , then
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(a) Vn 1 aVn bVn 1 14.
15.
18.
19.
(a) p 3 q 2 q(3 p 1) 0
(b) p 3 q 2 q (1 3 p ) 0
(c) p 3 q 2 q(3 p 1) 0
(d) p 3 q 2 q(1 3 p) 0
22.
23.
25.
26. 27.
28. 29.
30.
(b) x 2 x 1 0
(c) x 2 x 1 0
(d) x 2 x 1 0
If one root of a quadratic equation is 1 (2 5) , then the equation is (d) None of these (a) x 4 x 1 0 (b) x 4 x 1 0 (c) x 2 4 x 1 0 2 2 If one of the roots of the equation x ax b 0 and x bx a 0 is coincident. Then the magnitude of the value of ( a b) is _____
2
If a, b, c are in G.P. then the equations ax 2 2bx c 0 and dx 2 2ex f 0 have a common root if d a , e b , f c are in (a) AP (b) GP (c) HP (d) None of these The value of ‘a’ for which one root of the quadratic equation ( a 2 5a 3) x 2 (3a 1) x 2 0 is twice as large as the other is (b) 2 3
(d) 1 3
(c) 1 3 2
If x be real, then the minimum value of x 8 x 17 is _____.
If x is real, then greatest and least values of ( x 2 x 1) ( x 2 x 1) are (a) 3, 1 2 (b) 3,1 3 (c) 3, 1 3 (d) None of these If f ( x ) is quadratic expression which is positive for all real value of x and g ( x) f ( x) f ( x ) f ( x ) . Then for any real value of x (a) g ( x ) 0 (b) g ( x ) 0 (c) g ( x ) 0 (d) g ( x ) 0 If , ( ) are roots of the equation x 2 bx c 0 where (c 0 b) then (a) 0
24.
(d) Vn 1 abVn
Let , be the roots of the equation x 2 x 1 0 , the equation whose roots are 19 , 7 is
(a) 2 3 20. 21.
(c) Vn 1 aVn bVn 1
If one root of the equation x px q 0 is the square of the other, then
2
17.
(b) Vn 1 bVn aVn 1
[0.10]
2
(a) x 2 x 1 0 16.
Chapter 0: Prerequisite
(b) 0
(c) 0
(d) 0
If the roots of the equation x 2 2ax a 2 a 3 0 are real and less than 3, then (a) a 2 (b) 2 a 3 (c) 3 a 4 (d) a 4 2 The value of a for which 2 x 2(2a 1) x a ( a 1) 0 may have one root less than a and another root greater than a are given by (a) 1 a 0 (b) 1 a 0 (c) a 0 (d) a 0 or a 1 5 2 The maximum possible number of real roots of equation x 6 x 4 x 5 0 is _____. In solving a problem, a person makes a mistake in the coefficient of the first degree term and obtains –9 and –1 for the roots. Another person makes a mistake in the constant term of the equation and obtains 8 and 2 for the roots. The correct quadratic equation was (a) x 2 10 x 9 0 (b) x 2 10 x 16 0 (c) x 2 10 x 9 0 (d) None of these The value of x 2 4 x 11 0 can never be less than _____. If the roots of the equation ( a 2 bc ) x 2 2(b 2 ac ) x (c 2 ab) 0 are equal, where b 0 , then which one of the following is correct? (a) a b c abc (b) a 2 b 2 c 2 0 (c) a3 b3 c3 0 (d) a3 b3 c 3 3abc If the roots of the equation ax 2 bx c 0 are 1 and 1, then which one of the following is correct? (a) a and c are both zero (b) a and b are both positive (c) a and c are both negative (d) a and c are of opposite sign
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Engineering Mathematics 31. 32. 33. 34.
35.
36.
37. 38.
39.
40.
42.
[0.11]
If x 2 6 6 6 6 , then what is positive value of x ? _____ The difference in the roots of the equation 2 x 2 11x 5 0 is _____. The quadratic equation whose roots are 3 and –1 is (a) x 2 4 x 3 0 (b) x 2 2 x 3 0 (c) x 2 2 x 3 0 (d) x 2 4 x 3 0 If one root of the equation (1 a) x 2 (1 b ) x (1 c ) 0 is reciprocal of the other, then which one of the following is correct? (a) a b (b) b c (c) ac 1 (d) a c 2 2 If the roots of the equation x 2ax a a 3 0 are real and less than 3, then which one of the following is correct? (a) a 2 (b) 2 a 3 (c) 3 a 4 (d) a 4 2 Two students A and B solve an equation of the form x px q 0 . A starts with a wrong value of p and obtains the roots as 2 and 6. B starts with a wrong value of q and gets the roots as 2 and –9. What is the product of roots of the correct equation? _____ If one of the roots of 7 x 2 50 x k 0 is 7, then the value of k is _____. If sin and cos are the roots of the equation ax 2 bx c 0 , then which one of the following is correct? (a) a 2 b 2 2ac 0 (b) a 2 b 2 2ac 0 (c) a 2 c 2 2ab 0 (d) a 2 b 2 2ac 0 What is the condition that the equation ax 2 bx c 0 , where a 0 has both the roots positive? (a) a , b and c are of same sign (b) b and c have the same sign opposite to that of a (c) a and b are of same sign (d) a and c have the same sign opposite to that of b The equation (1 n 2 ) x 2 2ncx (c 2 a 2 ) 0 will have equal roots if (a) c 2 1 a 2
41.
Chapter 0: Prerequisite
(b) c 2 1 a 2
(c) c 2 1 n 2 a 2
(d) c 2 (1 n 2 )a 2
The equation whose roots are twice the roots of the equation x 2 2 x 4 0 is (a) x 2 2 x 4 0 (b) x 2 2 x 16 0 (c) x 2 4 x 8 0 (d) x 2 4 x 16 0 If and are the roots of the equation x 2 px q 0 , then 1 , 1 are the roots of which one of the following equations? (a) qx 2 px 1 0 (b) q 2 px 1 0 (c) x 2 px q 0 (d) x 2 px q 0
43.
If the equation ( a 2 b 2 ) x 2 2(ac bd ) x (c 2 d 2 ) 0 has equal roots, then which one of the following is correct? (a) ab cd (b) ad bc (c) a 2 c 2 b 2 d 2 (d) ac bd
44.
The solution of the equation
45.
The roots of the equation 4 x 3 2 x 2 32 0 is (a) 1, 2 (b) 3, 4 (c) 2, 3 (d 1, 3) 2 If and are the roots of the equation x x 1 0 , then what is the value of ( 4 4 ) ? _____ What is the least integral value of k for which the equation x 2 2( k 1) x (2k 1) 0 has both the roots positive? _____ When the roots of the quadratic equation ax 2 bx c 0 are negative of reciprocals of each other, then which one of the following is correct? (a) b 0 (b) c 0 (c) a c (d) a c 2 For what value of k , will the roots of the equation kx 5 x 6 0 be in the ratio of 2 : 3 ? _____ Students of a class are made to stand in rows. If one student is extra in a row, there would be
46. 47. 48.
49. 50.
x ( x 3) ( x 3) x 3 2 is _____.
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics
51.
Chapter 0: Prerequisite
two rows less. If one students is less in a row, there would be three rows more. Then, what is the number of students in the class? _____ x( x 1) ( m 1) x If the roots of the equation are equal, then what is the value of m ? ( x 1)( m 1) m (a) 1
52. 53.
54. 55.
56. 57. 58.
59.
60.
61. 62. 63.
64.
65. 66.
(b) 0
(d) 1 2
(c) 1 2
What is one of the roots of the equation 2 x (3 x) (3 x ) 2 x 3 2 ? (a) 1 (b) 2 (c) 3 (d) 4 2 If and are the roots of the equation x 3 x 2 0 , then which equation has the roots ( 1) and ( 1) ? (a) x 2 5 x 6 0 (b) x 2 5 x 6 0 (c) x 2 5 x 6 0 (d) x 2 5 x 6 0 For what value of k , does the equation kx 2 (2k 6) x 16 0 have equal roots? (a) 1 and 9 (b) –9 and 1 (c) –1 and 9 (d) –1 and –9 2 2 If the product of the roots of x 3kx 2k 1 0 is 7 for a fixed k , then what is the nature of the roots? (a) integral and positive (b) integral and negative (c) irrational (d) rational but not integral 2 2 If x kx 21 0 and x 3kx 35 0 have one common root, then sum of all the values of k is _____. If x 2 3x 5 0 and ax 2 bx c 0 have common root(s) and a, b, c N , then the minimum value of a b c is _____. Find the values of m for which the roots of an equation x 2 (m 3) x m 0 , m R , are equal in magnitude but opposite in sign. (a) m 0 (b) m 0 (c) m 1 (d) m 1 Find the values of m for which exactly one of the roots of an equation x 2 (m 3) x m 0 , m R , lies in the interval (1, 2) . (d) None of these (a) m 10 (b) m 10 (c) 10 m 10 Find the values of m for which one root is greater than 2 and the other root is smaller than 1 for the equation x 2 (m 3) x m 0 , m R . (a) m 10 (b) m 10 (c) 10 m 10 (d) There is no such m The number of the real solutions of the equation x 2 x 1 sin 4 x is _____. If a b c d , then the roots of the equation ( x a )( x c ) 2( x b)( x d ) 0 are (a) Real and Distinct (b) Real and Equal (c) Imaginary (d) None of these 2 If a 0, b 0, c 0 , then both the roots of the equation ax bx c 0 (a) Are real and negative (b) Have negative real parts (c) Are rational numbers (d) None of these If b a , then the equation ( x a )( x b) 1 0 , has (a) Both roots in [ a, b] (b) Both roots in ( , a ) (c) Both roots in (b, ) (d) One root in ( , a ) and other in (b, ) The real roots of the equation x 2 5 x 4 0 are (a) {4, 1} (b) {1, 4} (c) {4, 4}
(d) {4, 1,1, 4}
2
If x x 6 x 2 , then the values of x are (a) 2, 2, 4
67.
[0.12]
(b) 2, 2, 4
(c) 3, 2, 2
(d) 4, 4, 3
(c) {1, 2}
(d) {1, 2}
2
If { x R : x 2 x } , then x is (a) {1, 2} (b) {1, 2}
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 0: Prerequisite
[0.13]
68.
The sum of all real roots of the equation x 2 x 2 2 0 is _____.
69.
The number of real solutions of the equation x 2 4 x 3 2 x 5 0 is _____.
70.
The number of the real values of x for which the equality 3 x 2 12 x 6 5 x 16 holds good
71.
is _____. The minimum value of x 3 x 2 x 5 is _____.
72.
The sum of the real roots of the equation x 2 x 6 0 is _____.
73.
The solution set of ( x 1) x x 1 ( x 1) 2 x is (a) { x | x 0} (b) { x | x 0} {1} (c) x {1,1}
2
74.
For the equation x x 6 0 , the roots are
75.
(a) One and only one real number (c) Real with sum zero If x 2 , then the range of values of x 2 is (a) x 2 (0, ) (b) x 2 [0, )
76. 77. 78. 79.
(b) Real with sum one (d) Real with product zero (c) x 2 [4, )
(d) x 2 (4, )
If x 2 , then the range of values of x 2 is (a) x 2 (0, ) (b) x 2 [0, )
(c) x 2 [4, )
(d) x 2 (4, )
If x 2 , then the range of values of x 2 is (a) x 2 (0, ) (b) x 2 [0, )
(c) x 2 [4, )
(d) x 2 (4, )
If x 3 then the values of 1 x lies in the range of (b) (0,3) (c) (0,3] (a) (0,1 3) (b) ( 1 2 , 0)
(c) ( 1 2 , 0] (b) ( , 1 3) (1, ) (d) ( , 1) (3, )
83.
84.
85.
If x R , the range of values of 1 ( x 2 x 1) is (a) ( , 0] (4 5, )
82.
(d) [ 1 2 , 0]
If x ( 1,3) {0} then the values of 1 x lies in the range of (a) ( , 1) (1 3 , ) (c) ( ,1) (3, )
81.
(d) (0,1 3]
If x 2 then the values of 1 x lies in the range of (a) [ 1 2 , 0)
80.
(d) { x | x 1 or x 1}
2
(b) ( , 4 5) (0, )
(c) ( , 0] [4 5 , ) (d) ( , 4 5] (0, ) If (2 x 1)( x 3)( x 7) 0 , then which one of the following is correct? (a) x (, 7) ( 1 2 ,3)
(b) x ( , 7] [ 1 2 , 3)
(c) x ( , 1 2] [3, 7)
(d) x ( , 1 2) (3, 7)
If ( 2 x) 3 , then which one of the following is correct? (a) x (2 3, )
(b) x ( , 0) (2 3, )
(c) x ( , 0] (2 3, )
(d) x (, 2 3)
2x 3
3 , then 3x 5 (a) x (5 3,12 7)
If
(b) x [5 3,12 7]
(c) x (5 3,12 7]
(d) x [5 3,12 7)
If x 1 x , then
5 1 ,1 2
(a) x
5 1 ,1 2
(b) x
Copyright © 2016 by Kaushlendra Kumar
(c) x 1,
5 1 2
(d) x 1,
5 1 2
e-mail: [email protected]
Engineering Mathematics 86.
87.
88.
If
2
1
2
90. 91. 92. 93.
95.
97. 98. 99.
100. 101. 102. 103. 104.
(c) x [2, ) {1}
(d) x (, 2] {1}
4
(a) x [0,1 2] [3, )
(b) x [0,1) [3 2 , )
(c) x [0,1 2] [3, )
(d) x [0,1] [3 2, )
Find the range of values of y if y ( x 2 1) ( x 2 2) (b) y 1
(c) y 1 2 or y 1
(d) y 1 2 or y 1
If x 1 3 x , then (a) 2 x 3 (b) 2 x 3
(c) 2 x 3
(d) 2 x 3
If x 2 x 2 x 3 , then (a) x 0 (b) x 0
(c) x 3
(d) x 3
If ( x 2 4) x 2 1 0 , then (a) x (2,1) (1, 2) (b) x (2, 1] [1, 2)
(c) x ( 2, 1]
(d) x [1, 2)
2
If 2 x 1 x 1 3 , then the sum of values of x is (a) 1 (b) 2 (c) –2 2
(d) –1
2
If x x 2 2 x x , then x lies in the range of (b) ( 1, 2)
If 1 x x 2 2 x 1 , then (a) x 1 (b) x 1 If 3 x 2 x , then (a) x (1 2 ,1)
96.
5
If x( x 2) ( x 1) (2 x 3)( x 3) 0 , then
(a) [ 1, 2] 94.
[0.14]
2x 1
, then x x x 1 x3 1 (a) x ( , 2] {1} (b) x [2, ) {1} 2
(a) y 1 2 89.
Chapter 0: Prerequisite
(b) x [1 2 ,1]
(c) ( 2,1)
(d) [ 2,1]
(c) x 1
(d) x 1
(c) x {1 2 ,1}
(d) None of these
For x 1 , the numerical value of x 2 2 x 1 x 2 2 x 1 is _____. For 1 x 5 , in what range the values of 2 x 7 lies? (a) (0,5) (b) (c) (1, 4) (c) [1, 4] (d) [0,5] For x R , the possible values of 4 2 x 3 lies in the range of (a) ( , 4) (b) ( , 4] (c) ( ,5] The values of (a) (0, 2)
4 x 2 lies in the range of (b) [0, 2]
(d) ( ,5)
(c) ( 2, 2)
(d) [ 2, 2]
If x 3 x 2 1 then x lies in the range of (a) x (2,3] (b) x ( 2,5] (c) x [2,5)
(d) x [2,3]
2
For what values of x , the inequality x 4 x 3 0 holds true? (a) x (3, 1) (1,3) (b) x ( 3, 1) (c) x (2,3)
(d) x ( 3, 2) (2, 3)
For what values of x , the inequality 1 x 2 3 holds true? (a) x [1,1] [3,5] (b) x ( 1,1) (c) x [ 2, 1] [2,5] For what values of x , the inequality 0 x 3 5 holds true? (a) [ 2,8] {3} (b) ( 2,8] {3} (c) ( 3, 6] {2}
(d) x (2,5) (d) [ 3, 6] {2}
For what values of x , the inequality x 3 1 holds true? (a) ( 2, 2)
(b) ( , 4)
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(c) (4, )
(d) (a) or (b) or (c)
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Engineering Mathematics 105.
Chapter 0: Prerequisite
[0.15]
For what values of x , the inequality x 1 x 2 4 holds true? (a) ( ,1 2) [ 7 2 , )
(b) ( ,3 2) [ 5 2 , )
(c) ( , 1 2) [7 2 , )
(d) ( , 3 2) [5 2 , )
106.
The sum of values of x for which the inequality x 1 2 x 3 4 holds true, is _____
107.
For what values of x , the inequality ( x 3) ( x 1) 1 holds true? (a) x 1 (b) x 1 (c) x 2
108.
2
(d) x 2
2
For what values of x , the inequality x 2 x x 4 x 3x 4 holds true? (a) x (2, 0) (0, )
(b) x (6, )
(c) x (0, 2) (4, )
109.
If 2 x (2 x 2 5 x 2) 1 ( x 1) , then
110.
(a) 2 x 1 (b) 2 x 1 (c) 2 x 1 2 2 If for real values of x , x 3 x 2 0 and x 3 x 4 0 , then (a) 1 x 1 (b) 1 x 4 (c) 1 x 1 or 2 x 4
(d) x (0, 3) (6, )
(d) 2 x 1 (d) 2 x 4
111.
The number of integral solution of ( x 1) ( x 2) 1 4 is _____.
112.
If 2 a 3b 6 c 0 then at least one root of the equation ax 2 bx c 0 lies in the interval (a) (0,1) (b) (1, 2) (c) (2,3) (d) (3, 4)
113.
The set of all real numbers x for which x 2 x 2 x 0 , is
(a) ( , 2) (2, ) 114. 115. 116. 117. 118.
119.
2
(b) ( , 2) ( 2, )
121.
(c) ( , 1) (1, )
(d) ( 2, )
2 2
Product of real roots of the equation t x x 9 0 ( t 0 ) (c) does not exist (d) none of these (a) is always ve (b) is always ve 5 2 The maximum possible number of real roots of equation x 6 x 4 x 5 0 is _____ One root of the following given equation 2 x 5 14 x 4 31x 3 64 x 2 19 x 130 0 is (a) 1 (b) 3 (c) 5 (d) 7 4 3 2 The roots of the equation x 4 x 6 x 4 x 1 0 are (a) 1, 1, 1, 1 (b) 2, 2, 2, 2 (c) 3, 1, 3, 1 (d) 1, 2, 1, 2 3 2 If f ( x ) 2 x mx 13 x n and 2, 3 are roots of the equation f ( x ) 0 , then the value of m and n are (d) None of these (a) 5, 30 (b) 5,30 (c) 5, 30 Let Pn ( x ) 1 2 x 3 x 2 ( n 1) x n be a polynomial such that n is even. Then the number of real roots of Pn ( x) 0 is (a) 0 (b) n
120.
2
(c) 1
(d) None of these
2
The polynomial ( ax bx c )(ax dx c ) ac 0 , has _____ real roots. (a) four (b) at least two (c) at most two (d) no 4 3 2 If , , , are the roots of the equation x 4 x 6 x 7 x 9 0 , then the value of (1 2 )(1 2 )(1 2 )(1 2 ) is _____
122.
If x 1 is a factor of x 4 ( p 3) x 3 (3 p 5) x 2 (2 p 7) x 6 , then p is equal to _____
123.
For equation x 3 6 x 2 9 x k 0 to have exactly one root in (1,3) , the set of values of k is (a) ( 4, 0) (b) (1,3) (c) (0, 4) (d) None of these
124. 125. 126. 127.
The number of real roots of x8 x 5 x 2 x 1 0 is equal to _____. The number of positive real roots of x 4 4 x 1 0 is equal to _____. The number of negative real roots of x 4 4 x 1 0 is equal to _____. The number of complex roots of the equation x 4 4 x 1 0 is equal to _____.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
0.2
Chapter 0: Prerequisite
[0.16]
Sequence, Series and Progression Sequence: A set of numbers arranged in some definite order and according to some definite rule is called a sequence. Series: Terms of a sequence when added form a series. Progression: When terms of a sequence are written under specific conditions, then the sequence is called a progression. A progression is represented as a1 , a2 , , an , where a1 is the first term, an is the nth term of the progression. The number of terms can be finite or infinite. S n a1 a2 a3 an denotes the sum of corresponding series up to n terms. Clearly an S n S n 1 .
Arithmetic Progression: If the terms of a series successively increase or decrease by a constant quantity, the series is called an arithmetic progression (AP). The constant quantity is called the common difference. The general form of an AP is a , a d , a 2 d , , where a is the first term and d is the common difference. The nth term of an AP is, tn a ( n 1) d
Sum of n terms of an AP is, S n (n 2) 2a ( n 1) d ( n 2) a l , where l is the last term.
Three numbers in an AP can be taken as, a d , a , a d . Four numbers in an AP can be taken as, a 3d , a d , a d , a 3d . Five numbers in an AP can be taken as, a 2 d , a d , a , a d , a 2 d If a constant is added to (or subtracted from) every term of an AP, the resulting sequence will also be an AP with same common difference. If every term of an AP (with common difference d ) is multiplied by (or divided by) a fixed constant ( k ), the resulting sequence will also be an AP with common difference dk (in case of multiplication) and d k in case of division. If respective terms of two AP are added or subtracted, the new progression also in AP. The sum of the term which are equidistance from the beginning to the end is constant, and is equal to the sum of first and last term.
Example 0.19 [TF-2014 (2 marks)]: The 51st common term of the two arithmetic sequence 15, 19, 23, … and 14, 19, 24, …, is _____ Solution: The 1st AP: 15, 19, 23, 27, 31, 35, 39, 43, ; The 2nd AP is: 14, 19, 24, 29, 34, 39, . So the arithmetic sequence of common terms between the given two APs is 19, 39, . Thus 51st common term is t51 19 (51 1)20 1019 .
Geometric Progression: If the terms of a series increases or decreases by a common ratio then the 2 series is called a geometric progression (GP). The general form of a GP is a, ar , ar , , where a is the first term and r is the common ratio. The nth term of a GP is, t n ar n 1
Sum of n terms of the GP is, S n
a ( r n 1)
for all r 1 . ( r 1) Sum of an infinite GP with first term as ‘ a ’ and common ratio as ‘ r ’, where r 1 , is, S a (1 r ) [This point was asked in MN-2014 (2 marks)]
If a , b , c , d , are in GP then ak , bk , ck , dk , and a k , b k , c k , d k , are also in GP.
If a1 , a2 , a3 , and b1 , b2 , b3 , are two GPs then the sequence a1b1 , a2 b2 , a3b3 , is also in GP.
Three terms in a GP are taken as a r , a, ar . Four terms in a GP are taken as a r 3 , a r , ar , ar 3 .
Five terms in a GP are taken as a r 2 , a r , a, ar , ar 2 . If every term of a GP is raised by the same index then the new progression is also in GP.
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Engineering Mathematics
Chapter 0: Prerequisite
[0.17]
The product of the terms which are equidistance from the beginning and from the end is constant is equal to the product of 1st and last term. If a , b , c , d are in GP then (b a ) (c b) ( d c ) b 2 ac, c 2 bd , ad bc If every term of a GP is multiplied or divided by some non-zero fixed quantity, the resulting progression is also in GP.
Example 0.20 [ME-1998 (1 mark)]: For x 6 , sum of the series is (a)
(b) 3
(d) 1
(c) 2
Solution:
cos 2 n x cos 2 x cos 4 x n 1
2n cos ( n 1
2
4
6) cos ( 6) cos ( 6)
cos ( 6) 2
1 cos ( 6)
2
5
cos ( 6) 2
sin ( 6) 8
2
cot ( 6) 3 .
11
is 23 (a) 4.50 (b) 6.00 (c) 6.75 (d) 10.0 5 8 11 8 11 Solution: S 2 2 3 …(i); 2S 4 5 2 2 2 2 2 2 8 5 11 8 14 11 3 3 3 (ii ) (i ) S (4 5 2) 2 7 2 3 2 2 23 2 2 2 12 1 1 1 S 7 3 2 3 7 3 10 2 2 2 1 1 2
Example 0.21 [CE-2004 (1 mark)]: The summation of series S 2
2
2
2
…(ii).
Harmonic Progression: A set of terms is said to be in harmonic progression when their reciprocals are in arithmetic progression. The general form of a harmonic progression is
1
,
1
,
1
a a d a 2d where a and d are the first term and common difference, respectively, of arithmetic progression. The nth term of a HP is, tn 1 {a ( n 1) d }
, ,
Three terms in a HP can be taken as 1 ( a d ) ,1 a ,1 ( a d ) Four terms in a HP can be taken as 1 ( a 3d ) ,1 ( a d ) ,1 ( a d ) ,1 ( a 3d ) Five terms in a HP can be taken as 1 ( a 2 d ) ,1 ( a d ) ,1 a ,1 ( a d ) ,1 (a 2d )
If a1 , a2 , a3 , ( ai 0i ) are in GP then log x a1 , log x a2 , log x a3 , are in AP and vice-versa; also log a1 x, log a2 x, log a3 x, are in HP and vice-versa
Means: Let A be the arithmetic mean, G be the geometric mean and H be the Harmonic mean between two positive quantities a and b , then
ab
2ab
A, G , H are in GP, i.e., G 2 AH 2 ab If we have n quantities a1 , a2 , , an , then A
, G ab , H
AG H
Arithmetic mean of n quantities is, AM ( a1 a2 an ) n Geometric mean of n quantities is, GM (a1a2 an )1 n
1
Harmonic mean of n quantities is, HM
1
1
n an AM GM HM ( a1 a2 an ) n ( a1a2 an )1 n and a1
a2
equality
holds
when
a1 a2 an .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 0: Prerequisite
[0.18]
Natural Numbers Sum of first n natural numbers is, 1 2 n n (n 1) 2 Sum of squares of first n natural numbers is, 12 22 n 2 n( n 1)(2n 1) 6 Sum of cube of first n natural numbers is, 13 23 n3 n 2 ( n 1) 2 4 Example 0.22 [CS-2008 (2 marks)]: Let P
i and Q
1 i 2 K i odd
Then (a) P Q K
(b) P Q K
Solution (a): P
i 1 3 5 (2 K 1)
1 i 2 K i odd
Q
i 2 4 6 2k
1 i 2 K i even
K 2
i , where K is a positive integer.
1 i 2 K i even
(c) P Q
K 2
(d) P Q 2 K
{2 ( K 1)2} K 2
{4 ( K 1)2} K 2 K . Thus Q P K P Q K .
Please note that the questions on ‘Sequence, Series and Progression’ also came in previous years ‘General Aptitude’ section in GATE examination. These questions are given in ‘General Aptitude’ section. Exercise: 0.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1.
Let Tr be r th term of an A.P. whose first term is a and common difference is d . If for some positive integers m , n , m n , Tm 1 n and Tn 1 m , then a d equals (a)
2. 3. 4. 5. 6. 7. 8. 9.
10. 11.
12.
13.
1
1
(b) 1
(c)
1
(d) 0 m n mn The 19th term from the end of the series 2 6 10 86 is _____. In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then its 13th term is _____. If 7th and 13th term of an A.P. be 34 and 64 respectively, then its 18th term is _____. The nth term of the series 3 10 17 and 63 65 67 are equal, then the value of n is _____. 7th term of an A.P. is 40, then the sum of first 13 terms is _____ The first term of an A.P. is 2 and common difference is 4. The sum of its 40 terms will be _____. The sum of the first and third term of an A.P. is 12 and the product of first and second term is 24, the first term is _____ S S r 1 If Sr denotes the sum of the first r terms of an A.P., then 3 r is equal to S 2 r S 2 r 1 (d) 2 r 3 (a) 2r 1 (b) 2r 1 (c) 4r 1 If the sum of the first 2n terms of 2, 5,8, is equal to the sum of the first n terms of 57,59, 61, , then n is equal to _____ If the sum of the 10 terms of an A.P. is 4 times to the sum of its 5 terms, then the ratio of first term and common difference is (a) 1 : 2 (b) 2 : 1 (c) 2 : 3 (d) 3 : 2 150 workers were engaged to finish a piece of work in a certain number of days. 4 workers dropped the second day, 4 more workers dropped the third day and so on. It takes eight more days to finish the work now. The number of days in which the work was completed is _____. If the sum of first p terms, first q terms and first r terms of an A.P. be x , y and z
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
14. 15. 16.
(q r )
y
1
The sum of n terms of the series
1 3
18. 19.
1
2n 1
(b)
2n 1
1
3 5
1
is
5 7 1
( 2n 1 1) 2 2 After inserting n A.M.’s between 2 and 38, the sum of the resulting progression is 200. The value of n is _____ If a, b, c, d , e, f are six A.M.’s between 2 and 12, then a b c d e f is equal to _____.
(a)
17.
[0.19]
z ( r p) ( p q ) is p q r (b) pqr (c) xyz (a) 1 (d) 0 The sum of all odd numbers of two digits is _____. If sum of n terms of an A.P. is 3n 2 5n and Tm 164 , then m _____ respectively, then
x
Chapter 0: Prerequisite
(c)
(d)
2n 1
If a1 , a2 , a3 , , a24 are in arithmetic progression and a1 a5 a10 a15 a20 a24 225 , then a1 a2 a3 a23 a24 _____.
20. 21.
If a, b, c are in A.P., then 1 (bc) ,1 (ca ) ,1 ( ab ) will be in (a) AP (b) GP (c) HP n n If log 2, log(2 1), log(2 3) are in A.P., then n _____ (a) 5 2
22. 23. 24. 25. 26.
(b) log 2 5
(d) None of these
(c) log 3 5
(d) 3 4
The numbers ( 2 1),1, ( 2 1) will be in (a) AP (b) GP (c) HP (e) None of these th th th qr If the p , q and r term of a G.P. are a, b, c respectively, then a b r p c p q is equal to (d) pqr (a) 1 (b) 2 (c) abc If the third term of a G.P. is 4 then the product of its first 5 terms is 3 4 5 (d) None of these (a) 4 (b) 4 (c) 4 If x, 2 x 2, 3 x 3 are in G.P., then the numerical value of fourth term is _____ 100
Let an be the nth term of the G.P. of positive numbers. Let
100
a2 n n 1
and
a2 n 1 , n 1
such that , then the common ratio is (a)
27.
28. 29.
(b)
(c)
(b) 232 9990
(c) 0.232 990
(d) The sum of first two terms of a G.P. is 1 and every term of this series is twice of its previous term, then the first term will be (a) 1 2 (b) 1 3 (c) 1 4 (d) 1 5 The first term of an infinite geometric progression is x and its sum is 5. Then (a) 0 x 10 (b) 0 x 10 (c) 10 x 0 (d) x 10 The value of 0.234 is (a) 232 990
(d) 232 9990
30.
If a, b, c are in A.P. and a , b , c 1 , and x 1 a a , y 1 b b 2 ,
31. 32.
2 z 1 c c . Then x, y , z shall be in (a) AP (b) GP (c) HP (d) None of these The sum of two geometric means between the number 1 and 64 is _____. The G.M. of the numbers 3, 32 , 33 , , 3n is
33.
(a) 32 n (b) 3( n1) 2 (c) 3n 2 If a, b, c are in A.P. b a, c b, a are in G.P., then a : b : c is
2
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(d) 3( n1) 2
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34.
[0.20]
(a) 1: 2 : 3 (b) 1: 3 : 5 (c) 2 : 3 : 4 th th The 4 term of a H.P. is 3 5 and 8 term is 1 3 then its 6th term is (a) 1 6
35.
Chapter 0: Prerequisite
(b) 3 7
(d) 1 : 2 : 4
(c) 1 7
(d) 3 5
If the first two terms of an H.P. be 2 5 and 12 23 then the largest positive term of the progression is the nth term. The value of n is _____.
36. 37. 38. 39.
40.
The harmonic mean of the roots of the equation (5 2) x 2 (4 3) x 8 2 3 0 is ___. 3
(a)
41. 42. 43.
5
7
is equal to _____. 2 2 23 Sum of the series 1 2.2 3.2 2 4.23 100.299 is (a) 100.2100 1 (b) 99.2100 1 (c) 99.2100 1 (d) 100.2100 1 The sum of the series 3 33 333 n terms is (10n 1 9n 28) (10 n 1 9n 10) (10 n 1 10n 9) (d) None of these (a) (b) (c) 27 27 27 The sum of n terms of the following series 1 (1 x ) (1 x x 2 ) will be 1
2
1 xn
x(1 x n )
(c)
n (1 x) x (1 x n )
(b) (1 x ) 2 1 x 1 x The sum to n terms of the series 1 3 7 15 31 is (a) 2n 1 n (b) 2n 1 n 2 (c) 2n n 2 The sum of 13 23 33 43 153 is (a) 22000 (b) 10000 (c) 14400 th A series whose n term is ( n x) y , the sum of r terms will be (a)
r ( r 1) 2x
ry
(b)
r ( r 1) 2x
(c)
r ( r 1) 2x
(d) None of these
(d) None of these (d) 15000
ry
(d)
r ( r 1) 2x
44.
The sum of the series 1 (3 7) 1 (7 11) 1 (11 15) is
45.
(a) 1 2 (b) 1 3 (c) 1 6 (d) 1 12 The sum of the series 1 2 3 2 3 4 3 4 5 to n terms is (a) n ( n 1)( n 2) (b) ( n 1)( n 2)(n 3)
46. 47. 48.
rx
(c) n ( n 1)(n 2)( n 3) 4 (d) ( n 1)( n 2)( n 3) 4 If the A.M., G.M. and H.M. between two positive numbers a and b are equal, then (a) a b (b) ab 1 (c) a b (d) a b If the ratio of H.M. and G.M. of two quantities is 12 :13 , then the ratio of the numbers is (d) None of these (a) 1 : 2 (b) 2 : 3 (c) 4 : 9 In the given square, a diagonal is drawn and parallel line segments joining points on the adjacent sides are drawn on both sides of the diagonal. The length of the diagonal is n 2 cm. If the distance between consecutive line segments be 1 2 cm then the sum of the lengths of all possible line segments and the diagonal (in cm) is (c) n ( n 2) (a) n ( n 1) 2 (b) n2 (d) n 2 2
49.
50.
Arun ate half of a pizza on Monday. He ate half of what was left on Tuesday and so on. He followed this pattern for one week. How much (in %) of the pizza would he have eaten during the week? _____ If a1 , a2 , , an an are in A.P. with common difference d, then the sum of the following series sin d (cosec a1 cosec a2 cosec a2 cosec a3 cosec an 1 cosec an ) is
(a) sec a1 sec an
(b) cot a1 cot an
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(c) tan a1 tan an
(d) cosec a1 cosec an
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0.3
Chapter 0: Prerequisite
[0.21]
Trigonometry
Angle in Degrees: When two lines intersect, four angles are formed; if all four angles are equal, then each angle is called a right angle. When a right angle is divided into 90 equal parts, each part is called a degree. A degree is divided into 60 minutes. A minute is divided into 60 seconds; i.e., 1 right angle 90 degrees 90 o ; 1 degree 60 minutes 60 ' ; 1 minute 60 seconds 60 ''
Angle in Radians: The angle subtended at the centre of any circle by an arc of that circle is the ratio of length of an arc to the radius of the circle, i.e. (in radians)
length of arc
. Hence, one radian is radius of circle the angle subtended at the centre of any circle by an arc equal in length to the radius of that circle. Conversion of degree into radians: 1 degree ( 180) radians Conversion of radians into degree: 1 radian (180 ) degrees Angle measured in anti-clockwise is taken as ve ; Angle measured in clockwise is taken as ve .
Trigonometric function of an angle: The six trigonometric ratios sine, cosine, tangent, cotangent, secant and cosecant of an angle , 0o 90 o are defined as the ratios of two sides of a rightangled triangle with as one of the angles. The opposite side of the angle , for which we find the trigonometric ratios, is taken as ‘perpendicular’; the side which makes right angle with the perpendicular is taken as ‘base’; and the remaining third side is called hypotenuse, as shown in Fig. Perpendicular BC 1 Base AB 1 0.4; then by definition, sin ; cos ; Hypotenuse AC cosec Hypotenuse AC sec Perpendicular BC 1 tan Base AB cot
(a)
(b)
Figure 0.4: (a) The Right angled triangle (b) Sign of Trigonometric Functions
Trigonometric/Circular Functions and their Graphs Function Graph Sine Function: y f ( x ) sin x Domain: (, ) ; Range: [ 1, 1] Period: 2 ; Nature: Odd Interval in which the inverse can be obtained is , 2 2 Cosine Function: y f ( x ) cos x Domain: (, ) ; Range: [ 1, 1] Period: 2 ; Nature: Even Interval in which the inverse can be obtained is 0,
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Chapter 0: Prerequisite
[0.22]
Tangent Function: y f ( x ) tan x
Domain: (2 n 1)
, nI 2
Range: ( , ) Period: Nature: Odd Interval in which the inverse can be obtained is , 2 2 Cotangent function: y f ( x ) cot x Domain: n , n I Range: ( , ) Period: Nature: Odd Interval in which the inverse can be obtained is (0, )
Secant Function: y f ( x ) sec x
, nI 2 Range: ( , 1] [1, ) Period: 2 Nature: Even Interval in which the inverse can be obtained is 0, , 2 2 Cosecant Function: y f ( x ) cosec x Domain: n , n I Range: ( , 1] [1, ) Period: 2 Nature: Odd Interval in which the inverse can be obtained is , 0 0, 2 2
Domain: (2 n 1)
The table for the formulae for Allied angles is given as: o o o o 180 180 90 90 sin sin sin sin cos cos cos cos cos cos sin sin tan tan tan tan cot cot cot cot cot cot tan tan sec sec sec sec cosec cosec cosec cosec cosec cosec sec sec
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o
270 cos sin cot tan cosec sec
o
270 cos sin cot tan cosec sec
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Chapter 0: Prerequisite
[0.23]
The results in the above table are not to be memorised. We have to follow the following procedure: Step 1: First find the quadrant in which the angle lies and get the sign of the trigonometric function in o that quadrant. For e.g., for sin(270 ) , assume is very small, 270o lie in the 4th quadrant and sin is negative in 4th quadrant. Step 2: Now divide the given angle by 90 o and by dividing we get and even or odd number. If we get even number then the trigonometric function remains unchanged. If we get an odd number then the trigonometric functions interchange with each other according as: sin cos , tan cot , o sec cos ec . For e.g., for sin(270 ) , if we divide 270o by 90 o , we get 3 which is an odd o
number, so sin changes into cos . So sin(270 ) cos .
Values of trigonometric functions for some angles (where U Undefined) o
sin
0 0
cos
1
tan
0
o
30 12
45
3 2 1
o
60
1
2
1
2
90 1
3 2
o
o
0
sec
1
3
U
cosec
U
Some important formulae sin( A B) sin A cos B cos A sin B cos( A B) cos A cos B sin A sin B
cot
0 U
12
1
3
o
30
o
45 1
3 2
3
o
60 1
90 0
3
2
2
2
o
2
2
o
U
1
3
sin( A B) sin A cos B cos A sin B
cos( A B ) cos A cos B sin A sin B tan A tan B tan A tan B 2 tan A tan( A B ) tan( A B ) tan 2 A 1 tan A tan B 1 tan A tan B 1 tan 2 A 2 tan A 3 tan A tan 3 A sin 2 A 2 sin A cos A tan 3 A 1 tan 2 A 1 3 tan 2 A 1 tan 2 A 2 2 2 2 cos 2 A cos A sin A 1 2 sin A 2 cos A 1 1 tan 2 A 3 cos 3 A 4 cos3 A 3 cos A sin 3 A 3sin A 4 sin A A B A B A B A B sin A sin B 2 sin cos sin A sin B 2 cos sin 2 2 2 2 A B A B A B A B cos A cos B 2 cos cos cos A cos B 2 sin sin 2 2 2 2 2 sin A sin B cos( A B ) cos( A B ) 2 cos A cos B cos( A B ) cos( A B ) 2 sin A cos B sin( A B) sin( A B )
Identities relating to the angles of a triangle: In any ABC , the following results hold true: cot B cot C cot C cot A cot A cot B 1 sin 2 A sin 2B sin 2C 4sin A sin B sin C tan A tan B tan C (tan A)(tan B)(tan C ) A
B
B
C
tan
cos A cos B cos C 1 4 sin
2
tan
2
tan
2
tan
2
tan
C 2 A 2
tan sin
A 2 B 2
1
sin
C 2
sin A sin B sin C 4 cos
cot
A 2
cot
B 2
cot
C 2
cot
A 2 A 2
cos cot
B 2 B 2
cos cot
C 2 C 2
Relations between the sides and angles of a triangle: In the following relations, A, B, C are the angles and a, b, c are the sides of the triangle opposite to the angles A, B, C , respectively. a b c 2 R (Sine formula), where R is the circumradius sin A sin B sin C
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Engineering Mathematics
Chapter 0: Prerequisite
[0.24]
2 2 2 2 2 2 2 2 2 c a b 2ab cos C ; b c a 2ac cos B ; a b c 2bc cos A (cosine formula) a c cos B b cos C ; b a cos C c cos A ; c b cos A a cos B (Projection formula)
Half angle formula: In the following formula A, B, C are the angles of a triangle and a, b, c are the sides of the triangle opposite to the angles A, B, C , respectively. Also, perimeter of the triangle 2s a b c .
sin
cos
tan
tan
A 2 A 2 A 2
B C 2
( s b)( s c ) bc s ( s a) bc
; cos
( s b)( s c ) s( s a )
bc bc
; sin
cot
A 2
B 2
B 2
; tan
ca
s ( s b)
; tan
( s c )( s a )
ca B 2
CA 2
; cos
C 2
( s c )( s a ) s( s b)
ca ca
cot
; sin
B 2
C 2
( s a )( s b)
ab
s (s c) ab ; tan ; tan
C 2
( s a )( s b)
s(s c)
AB 2
a b
ab
cot
C 2
(Napier’s formula)
Area of triangles, cyclic quadrilateral and polygon 1 1 1 abc Area of a triangle ab sin C bc sin A ca sin B s ( s a )( s b )( s c ) 2 2 2 4R where, a, b, c are the sides of the triangle opposite to the angles A, B, C , respectively, and perimeter of the triangle 2s a b c .
Area of cyclic quadrilateral of sides a , b , c , d is given by
s ( s a )( s b)( s c )( s d ) , where
s (a b c d ) 2 .
2 n 2 2 R R sin , where r n 2 n and R , respectively, the inradius and circumradius of the regular polygon. Area of a regular polygon of n sides, with side a is nr 2 tan
Inradius and Exradii of a triangle: Let a, b, c are the sides of the triangle opposite to the angles A, B, C , respectively. Let r be the inradius of the triangle and r1 , r2 , r3 are the exradius of the triangle. An escribed circle is a circle which touches one of the sides of a triangle internally and the other two sides externally (sides produced). So r1 , r2 , r3 are the exradius of the triangle ABC opposite to angles A, B, C , respectively. In all the following formulae, is the area of triangle ABC , R is the circumcentre of the triangle ABC . A B C A B C r ( s a ) tan ( s b) tan ( s c) tan 4 R sin sin sin s 2 2 2 2 2 2 A A B C B A B C r1 s tan 4 R sin cos cos r2 s tan 4 R cos sin cos sa 2 2 2 2 s b 2 2 2 2 C A B C r3 s tan 4 R cos cos sin r1 r2 r3 r 4R sc 2 2 2 2
General solution of trigonometric equations n If sin sin then n (1) , n I , 2 , 2
If cos cos then 2n , n I , 0,
If tan tan then n , n I , 2 , 2
If sin 2 sin 2 , cos 2 cos 2 , tan 2 tan 2 then n
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Engineering Mathematics
Chapter 0: Prerequisite
[0.25]
Equation containing combination of trigonometric and non-trigonometric equations
For finding the solutions we have to take the inequalities involving trigonometric functions. For 2 2 e.g., consider the equation sec x 1 y , as we have two unknowns and one equation, since 2
2 sec x 1 for all x (, 1] [1, ) , also 1 y 1 for all y ; hence the given equation can
2
hold only if sec 2 x 1 x n and 1 y 1 y 0 . For finding the number of solution and not the values of solution, we take the help of graphs. For e.g., consider the equation cos x x , if we sketch the graph of y cos x and y x , we see they intersect at exactly one point; so this equation has one solution.
Example 0.23 [CS-1995 (1 mark)]: In the interval [0, ] the equation x cos x has (a) No solution (b) Exactly one solution (c) Exactly two solutions (d) An infinite number of solutions Solution (b): By sketching the graph of y x and y cos x , we have only one point of intersection. So the equation x cos x has exactly one solution.
Inverse Circular Function: The inverse of a function f : A B exists if f is one-one onto i.e., 1
a bijection and is given by f ( x) y f ( y) x . Consider the sine function with domain and range [ 1,1] . Clearly this function is not a bijection and so it is not invertible. If we restrict the domain of it in such a way that it becomes one–one, then it would become invertible. If we consider sine as a function with domain [ 2 , 2] and co-domain [ 1,1] , then it is a bijection and therefore, invertible. The inverse of all trigonometric function along with domain, range, nature, and their graphs is given in the following table: Inverse Trigonometric/Circular Functions and their Graphs Function Graph Inverse Sine Function: 1 y f ( x) sin x Domain: [ 1, 1]
, 2 2
Range:
Nature: Odd Inverse Cosine Function: 1 y f ( x) cos x Domain: [ 1, 1] Range: [0, ] Nature: Neither even nor odd Inverse Tangent Function: 1 y f ( x) tan x Domain: ( , )
, 2 2
Range:
Nature: Odd
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Chapter 0: Prerequisite
[0.26]
Inverse Cotangent function: 1 y f ( x) cot x Domain: ( , ) Range: (0, ) Nature: Neither even nor odd Inverse Secant Function: 1 y f ( x) sec x Domain: ( , 1] [1, )
, 2 2
Range: 0,
Nature: Neither even nor odd
Inverse Cosecant Function: 1 y f ( x) cosec x Domain: ( , 1] [1, )
, 0 0, 2 2
Range:
Nature: Odd
Relation between inverse function
; tan 1 x cot 1 x ; cosec 1 x sec 1 x 2 2 2 1 1 1 For all x 0 , we have, cosec 1 x sin 1 ; sin 1 x cosec 1 ; sec 1 x cos 1 ; x x x 1 1 1 cos 1 x sec 1 ; cot 1 x tan 1 ; tan 1 x cot 1 . x x x When x and y are positive then sin 1 x cos 1 x
1
tan x tan
1
x y x y if xy 1 and tan 1 x tan 1 y tan 1 if 1 xy 1 xy
1
y tan
1
y tan
xy 1
1
tan x tan
x y 1 xy
sin 1 x sin 1 y sin 1 x 1 y 2 y 1 x 2 ; sin 1 x sin 1 y sin 1 x 1 y 2 y 1 x 2 ;
2 2 1 1 1 2 2 cos x cos y cos xy 1 x 1 y ; cos x cos y cos xy 1 x 1 y ; 2 1 1 2 x 1 1 x 1 2 x 2 tan x sin cos tan 2 2 2 1 x 1 x 1 x 1
1
1
1
sin(sin 1 x) x for all 1 x 1
0 x 2 x, sin (sin x) x, 2 x 3 2 x 2 , 3 2 x 2
cot(cot 1 x) x for all x x , x 0
cot 1 (cot x ) x,
cos(cos1 x) x for all 1 x 1
1
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0 x
x , x 2 sec(sec1 x) x for all x 1
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x,
1
cos (cos x )
Chapter 0: Prerequisite
0 x
[0.27]
x, x 0, x 2
1
sec (sec x)
2 x, x 2 x, 0 x , x 2 1 tan(tan 1 x) x for all x cosec(cosec x) x for all x 1 0 x 2 0 x 2 x, x, 1 1 tan (tan x) x , 2 x 3 2 cosec (cosec x ) x, x ( 2 , 3 2] x 2 , 3 2 x 2 x 2 , 3 2 x 2 1 1 1 1 sin (sin x) , cos (cos x) , sec (sec x) and cosec (cosec x) are periodic with period 2 . 1
1
While, tan (tan x) and cot (cot x) are periodic with period .
Hyperbolic Function: The parametric co-ordinates of any point on unit hyperbola x 2 y 2 1 is e e e e 2 , 2
(cosh , sinh ) . It means that the relation which exists amongst cos ,sin and unit circle, that relation also exist amongst cosh , sinh and unit hyperbola. So, these functions are called as Hyperbolic functions. For any (real or complex) variable quantity x , x
sin hx
e e
x
x
2 sinh x e x e x tanh x cosh x e x e x sinh 0 0
cos hx
e e
x
2 cos hx e x e x cot hx x x sin hx e e cosh 0 1
1
sec hx
cosec h x
tanh 0 0
cosh x
1 sinh x
2 e e x x
2 e ex x
Graph of Real Hyperbolic Functions y sinh x
y tanh x y cosh x
y coth x
y cos ech x y sec h x
Domain and Range of Hyperbolic Functions: Let x be any real number Function sinh x cosh x tanh x
Domain
Range [1, ) ( 1,1)
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Function coth x sec hx cosec h x
Domain {0} {0}
Range [ 1,1] (0,1] {0}
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Chapter 0: Prerequisite
Formula for Hyperbolic Functions 2 2sin h x cos h 2 x 1
[0.28]
cos h2 x sin h2 x 1
cosh 2 2 2 sinh x cosh y sinh ( x y ) sinh ( x y )
cot h2 x cosec h2 x 1 sinh( x y ) sinh x cosh y cosh x sinh y tanh x tanh y tanh ( x y ) 1 tanh x tanh y x y x y sinh x sinh y 2 cosh sinh 2 2 x y x y cosh x cosh y 2 sinh sinh 2 2 x cosh x sinh x e
2 cosh x cosh y cosh ( x y ) cosh ( x y )
cos h x sin h x e x
2 cos h x sin h y sin h ( x y ) sin h ( x y)
2cos h2 x cos h 2 x 1
2 sinh x sinh y cosh ( x y ) cosh ( x y )
sinh 2 x 2 sinh x cosh x
2
2
2
2
sec h x tanh x 1
cosh x sinh x cosh 2 x
cosh ( x y ) cosh x cosh y sinh x sinh y
sinh x sinh y 2 sinh
cosh x cosh y 2 cosh
x y
cosh
2 x y
x y 2 x y
n
(cos h x sin h x) cos h nx sin h nx
cosh 2 x cosh x sinh x 2 cosh x 1 1 2 sinh x
tanh 3 x
sinh 3x 3sinh x 4sinh 3 x
cos h x sin h x e x sinh( x) sinh x coth( x ) coth x
2
2
2
3 tanh x tanh 3 x 2
2
2 tanh x 1 tanh 2 x
1 tan h 2 x 2
1 tan h x 2 tan h x
tan h 2 x
cosh 3x 4 cosh3 x 3cosh x
1 3 tanh x
2
1 tan h x n
(cosh x sinh x) cosh nx sinh nx cosh( x ) cosh x tanh( x ) tanh x sec h( x ) sec hx cosec h( x ) cosec hx 2
2
Transformation of Hyperbolic Function: Since cosh x sinh x 1 sinh x cosh 2 x 1 2
sinh x
1 sech x
sinh x
tanh x
sinh x
1
. In a similar manner we can 2 2 sec h x 1 tanh x coth x 1 express cosh x , tanh x , coth x , etc., in terms of other hyperbolic functions.
Relation between Hyperbolic and Circular Function: We have from Euler formulae, (1) ix ix ix ix e cos x i sin x (2) e cos x i sin x . Adding (1) and (2) cos x (e e ) 2 . Subtracting (2) from (1)
sin x (eix e ix ) 2i . Replacing
cos(ix ) (e x e x ) 2 cosh x
x by ix
in these values, we get
cos(ix) cosh x , sin(ix ) (e x e x ) (2i ) i (e x e x ) 2 sin ix i sinh x sin(ix) i sinh x . Also tan ix i tanh x . Similarly replacing x by ix in the cos ix cosh x definitions of sinh x and cosh x , we get cos h (ix) (eix e ix ) 2 cos x . Thus, we obtain the following relations between hyperbolic and trigonometrical functions: sin(ix) i sinh x ; sinh(ix ) i sin x ; sinh x i sin(ix) ; sin x i sinh(ix) ; cos(ix) cosh x ; cosh(ix ) cos x ; cosh x cos(ix ) ; cos x cosh(ix ) ; tan(ix ) i tanh x ; tanh(ix ) i tan x ; tanh x i tan(ix) ; tan x i tanh(ix) ; cot(ix ) i coth x ; coth(ix) i cot x ; coth x i cot(ix ) ; cot x i coth(ix ) ; sec(ix ) sec h x ; sec h (ix) sec x ; sec h x sec(ix ) ; sec x sec h(ix ) ; cosec(ix ) i cosec h x ; cos ech (ix ) i cosec x ; cosec h x i cosec (ix) ; cosec x i cosec h(ix ) .
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Engineering Mathematics
Chapter 0: Prerequisite
[0.29]
Expansion of Hyperbolic Function
sinh x
cosh x
tanh x
e x e x 2 e x e x 2
e x e x x
x
x 1
x
x3 3! x2 2!
x3
x5
5! x4 4!
2
5
x7 7! x6 6!
x
, x , x
17
7
x , x
e e 3 15 315 3 1 x x 2 5 coth x x , 0 x x 3 45 945 x 2 5 4 61 6 sec hx 1 x x , x 2 24 720 2 3 1 x 7x cosec hx , 0 x x 6 360
2
Example 0.24 [EC-2007 (1 mark)]: For x 1 , coth( x) can be approximated as (b) x 2
(a) x
Solution (c): From the expansion of coth x
(c) 1 x 1
x
x3
(d) 1 x
2
2
5
x , except the first term all
x 3 45 945 other terms tends to zero when x 1 , so For x 1 , coth( x) can be approximated as 1 x .
Period of Hyperbolic Functions: If for any function f ( x) , f ( x T ) f ( x ) , then f ( x) is called the Periodic function and least positive value of T is called the Period of the function. sinh x sinh(2 i x) , cosh x cosh(2 i x ) and tanh x tanh( i x ) . Therefore the period of these functions are respectively 2 i , 2 i and i . Also period of cos ech x , sec h x and coth x are respectively 2 i , 2 i and i . Remember that if the period of f ( x ) is T , then period of f ( nx ) will be T n .
Hyperbolic function are neither periodic functions nor their curves are periodic but they show the algebraic properties of periodic functions and having imaginary period.
Exercise 0.3 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. 2. 3.
4. 5. 6.
The circular wire of radius 7 cm is cut and bend again into an arc of a circle of radius 12 cm. The angle (in degrees) subtended by an arc at the centre of the circle is _____. The equation sec 2 4 xy ( x y ) 2 is only possible when (a) x y (b) x y (c) x y (d) None of these , if x cos 2 n , y sin 2 n , z (cos 2 n )(sin 2 n ) , then 2 n 0 n 0 n 0 (a) xyz xz y (b) xyz xy z (c) xyz x y z (d) Both (b) and (c) Which of the following number is rational? (a) sin 15 o (b) cos15 o (c) sin15o cos15o (d) sin15 o cos 75o If tan m ( m 1) and tan 1 (2 m 1) then _____ degrees. The value of tan x tan 3 x whenever defined never lie between
For 0
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(a) 1 3 and 3 7. 8. 9. 10. 11. 12.
13.
2
15. 16.
17.
18.
(c) 1 5 and 5
(d) 5 and 6
(c) 3 4 A 13 16
(d) 3 4 A 1
2
The number of solutions of the equation 3sin x 7 sin x 2 0 in the interval [0, 5 ] is _____ Number of solutions of the equation tan x sec x 2 cos x , lying in the interval [0, 2 ] is _____ General value of satisfying the equation tan 2 sec 2 is (a) m , n ( 3) (b) m , n ( 3) (c) m , n ( 6) (d) None of these The number of integral values of k , for which the equation 7 cos x 5 sin x 2 k 1 has a solution is _____. If the angles of a triangle are in the ratio 4 :1 :1 , then the ratio of the longest side to the perimeter is (b) 1: 6 (d) 2 : 3 (a) 3 : (2 3) (c) 1: (2 3) In a triangle ABC , B 3 and B 4 and D divides BC internally in the ratio 1: 3 . Then sin BAD sin CAD is equal to (b) 1
3
(c) 1
6
(d)
2 3
In a ABC , 2ac sin ( A B C ) 2 is equal to (a) a 2 b 2 c 2 (b) c 2 a 2 b 2 (c) b 2 c 2 a 2 (d) c 2 a 2 b 2 cos A cos B cos C In a ABC , and a 2 then square of the area of a triangle is _____ a b c In a triangle ABC , let C 2 . If r is the in-radius and R is the circum-radius of the triangle, then 2( r R) is equal to (a) a b (b) b c (c) c a (d) a b c A regular polygon of nine sides, each of length 2 is inscribed in a circle. The radius of the circle is (a) cosec( 9) (b) cosec( 3) (c) cot( 9) (d) tan( 9) From the top of a light house 60 m high with its base at the sea level the angle of depression of a boat is 15o. The distance (in metres) of the boat from the foot of the light house is
3 1 60 3 1
(a) 19.
(b) 1 4 and 4
[0.30]
4
If A cos sin , then for all values of (b) 13 16 A 1 (a) 1 A 2
(a) 1 3 14.
Chapter 0: Prerequisite
3 1 60 3 1
(b)
3 1 3 1
(c)
(d) None of these
A person observes the angle of deviation of a building as 30 o . The person proceeds towards the building with a speed of 25( 3 1) m/h. After 2 hours, he observes the angle of elevation
20.
21.
22. 23.
as 45o . The height of the building (in metres) is The shadow of a tower standing on a level ground is found to be 60 m longer when the sun’s altitude is 30 o than when it is 45o . The height (in metres) of the tower is (a) 60 m (b) 30 m (d) 30( 3 1) (c) 60 3 A person is standing on a tower of height 15( 3 1) m and observing a car coming towards the tower. He observed that angle of depression changes from 30 o to 45o in 3 sec. What is the speed (in km/h) of the car? _____ The angle of elevation of the top of a pillar at any point A on the ground is 15o. On walking 40 m towards the pillar, the angle becomes 30o. The height (in metres) of the pillar is _____. A man from the top of a 100 metre high tower looks a car moving towards the tower at an angle of depression of 30o. After some time, the angle of depression becomes 60o. The distance (in metre) travelled by the car during this time is
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Engineering Mathematics
Chapter 0: Prerequisite
(b) 200
(a) 100 3
[0.31]
(c) 100
3
3
(d) 200 3
24.
For a man, the angle of elevation of the highest point of the temple situated east of him is 60o . On walking 240 metres to north, the angle of elevation is reduced to 30 o then the height of the temple is (b) 60 (a) 60 6 (c) 50 3 (d) 30 6
25.
2 The principal value of sin 1 sin is 3 (a) 2 3 (b) 2 3
26.
1
(d) None of these 1
Considering only the principal values, if tan(cos x ) sin cot (1 2) , then x is equal to (a) 1
27.
(c) 4 3
(b) 2
5 2
x
2
If sin 1 x
x
3
4
(c) 3
5
2
cos x 1
x
4
2
x
6
4
(d)
5
2
5 3
for 0 x 2 , then x equals
_____ 28.
The number of real solutions of tan 1 x ( x 1) sin 1 x 2 x 1 2 is _____
29.
tan 2 tan 1 (1 5) ( 4) (a) 17 7 (b) 17 7 If a, b, c be positive
30.
1
31. 32. 33. 34.
tan a ( a b c) bc tan The value of sec h(i ) is (b) i (a) 1 The period of cosh 3 is (a) 6 i (b) 2 i The period of sinh( 2) is (a) 2 i (b) 2 The period of coth( n 4) is (a) i n
0.4
1
real
(c) 7 17 numbers
b( a b c) ca tan
(b) 4 i n
1
and
(d) 7 17 the value
of
c ( a b c) ab , then tan is __
(c) 0
(d) 1
(c) i
(d) 9 i
(c) 4 i
(d) 4
(c) n i 4
(d) i
Two Dimensional Geometry
Points: The distance between two points A( x1 , y1 ) , B ( x2 , y2 ) is AB ( x2 x1 ) 2 ( y2 y1 ) 2
If P ( x, y ) divides the join of A( x1 , y1 ) and B( x2 , y2 ) in the ratio m1 : m2 ( m1 , m2 0) Internal division: If P ( x, y ) divides the segment AB internally in the ratio of m1 : m2 PA PB
m1 m2
m1 x2 m2 x1 m1 y2 m2 y1 , . m1 m2 m1 m2
, as shown in Fig. 0.5. The co-ordinates of P ( x, y )
External division: If P ( x, y ) divides the segment AB externally in the ratio of m1 : m2
m1 x2 m2 x1 m1 y2 m2 y1 , . PB m2 m1 m2 m1 m2 If P ( x, y ) divides the join of A( x1 , y1 ) and B( x2 , y2 ) in the ratio :1( 0) , then PA
x
m1
, as shown in Fig. 0.6. The co-ordinates of P ( x, y )
x2 x1 y y1 ; y 2 . Positive sign is taken for internal division and negative sign is 1 1
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Chapter 0: Prerequisite
[0.32]
taken for external division. For finding ratio, use ratio : 1 . If is positive, then divides internally and if is negative, then divides externally.
Figure 0.5: Internal division
Figure 0.6: External division
x x y y2 The mid-point of A( x1 , y1 ) , B ( x2 , y2 ) is 1 2 , 1 (as m1 : m2 ::1:1 ). 2 2 Straight line ax by c 0 divides the join of points A( x1 , y1 ) and B( x2 , y2 ) in the ratio
ax1 by1 c . If ratio is ve then divides externally and if ratio is ve then divides ax2 by2 c internally.
Centroid of a triangle: The centroid of a triangle is the point of intersection of its medians, as shown in Fig. 0.7. The centroid divides the medians in the ratio 2 : 1 (Vertex : base). If A( x1 , y1 ) , B( x2 , y2 ) and C ( x3 , y3 ) are the vertices of a triangle. If G be the centroid upon one of the median
x1 x2 x3 y1 y2 y3 , . 3 3
(say) AD, then AG : GD 2 :1 Co-ordinate of G are
Circumcentre: The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle, as shown in Fig. 0.8. It is the centre of the circle which passes through the vertices of the triangle and so its distance from the vertices of the triangle is the same and this distance is known as the circum-radius of the triangle. Let vertices A, B, C of the ABC be ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) and let circumcentre be O ( x, y ) and then ( x, y ) can be found by solving (OA) 2 (OB ) 2 (OC ) 2 . If a triangle is right angle, then its circumcentre is the mid point of hypotenuse. If A, B, C are angles and A( x1 , y1 ), B ( x2 , y2 ), C ( x3 , y3 ) are vertices of ABC then its
x1 sin 2 A x2 sin 2 B x3 sin 2C y1 sin 2 A y2 sin 2 B y3 sin 2C , . sin 2 A sin 2 B sin 2C sin 2 A sin 2 B sin 2C
circumcentre is
Figure 0.7: Centroid of a triangle
Figure 0.8: Circumcentre of a triangle
Incentre: The incentre of a triangle is the point of intersection of internal bisector of the angles, as shown in Fig. 0.9. Also it is a centre of a circle touching all the sides of a triangle. Co-ordinates ax bx2 cx3 ay1 by2 cy3 of incentre 1 , , where a, b, c are the sides of triangle ABC . abc abc
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Chapter 0: Prerequisite
[0.33]
Excircle: A circle touches one side outside the triangle and other two extended sides then circle is known as excircle, as shown in Fig. 0.10. Let ABC be a triangle then there are three excircles with three excentres. Let I1 , I 2 , I 3 opposite to vertices A( x1 , y1 ) , B( x2 , y2 ) and C ( x3 , y3 ) then
ax1 bx2 cx3 ay1 by2 cy3 , ; a b c a b c ax bx2 cx3 ay1 by2 cy3 ax1 bx2 cx3 ay1 by2 cy3 I2 1 , , ; I3 . abc abc a b c abc
coordinates of three excentres can be given as: I1
Angle bisector divides the opposite sides in the ratio of remaining sides, i.e.
BD
AB
DC AC Incentre divides the angle bisectors in the ratio (b c) : a, (c a) : b and (a b ) : c
Figure 0.9: Incentre of a triangle
Figure 0.10: Excircle of a triangle
c b
Figure 0.11: Orthocentre of a triangle
Orthocentre : It is the point of intersection of perpendiculars drawn from vertices on opposite sides (called altitudes) of a triangle and can be obtained by solving the equation of any two altitudes, as shown in Fig. 0.11. Here O is the orthocentre since AE BC , BF AC and CD AB then OE BC , OF AC , OD AB . Solving any two we can get coordinate of O. If a triangle is right angled triangle, then orthocentre is the point where right angle is formed. If the triangle is equilateral then centroid, incentre, orthocentre, circum-centre coincides. Orthocentre, centroid and circum-centre are always collinear and centroid divides the line joining orthocentre and circum-centre in the ratio 2 : 1 In an isosceles triangle centroid, orthocentre, incentre, circum-centre lie on the same line.
Area of a triangle: The area of a ABC with vertices A( x1 , y1 ) , B( x2 , y2 ) and C ( x3 , y3 ) is given as
1 2
x1
y1
1
x2
y2
1
x3
y3
1
1 2
( x1 ( y2 y3 ) x2 ( y3 y1 ) x3 ( y1 y2 ) .
Three points A( x1 , y1 ) , B( x2 , y2 ) , C ( x3 , y3 ) are collinear (i) if area of triangle formed by A , B and C is zero, i.e. 0 or (ii) sum of two sides is equal to third side, i.e. AB BC AC or AC BC AB or AC AB BC .
In equilateral triangle, having sides a , area is ( 3 4) a 2 ; having length of perpendicular as ‘ p ’ area is p 2
3 . If area is a rational number, then the triangle cannot be equilateral.
Example 0.25 [CE-2014 (1 mark)]: With reference to conventional Cartesian ( x , y ) coordinate system, the vertices of a triangle have the following coordinates: ( x1 , y1 ) (1, 0) ; ( x2 , y2 ) (2, 2) ; and ( x3 , y3 ) (4, 3) . The area of the triangle is equal to
(a) 3 2
(b) 3 4
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(c) 4 5
(d) 5 2
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Engineering Mathematics
Solution (a):
1 2
1
0 1
2
2 1
4
3 1
Chapter 0: Prerequisite
1 2
1(2 3) 1(6 8)
[0.34]
3 2
Example 0.26 [MN-2010 (2 marks)]: The value of k for which the points (5, 5) , ( k ,1) , (10, 7) lie on a straight line is (a) 5 (b) 5 (c) 2 (d) 2 Solution (a): The given point are collinear if the area of triangle formed by given points is zero, i.e. 5 5 1 1 k 1 1 0 5(1 7) 5( k 10) 1(7k 10) 0 30 5k 50 7 k 10 0 k 5 . 2 10 7 1
Area of a quadrilateral: If ( x1 , y1 ) , ( x2 , y2 ) , ( x3 , y3 ) and ( x4 , y4 ) are vertices of a quadrilateral, 1 then its area is [( x1 y2 x2 y1 ) ( x2 y3 x3 y2 ) ( x3 y4 x4 y3 ) ( x4 y1 x1 y4 )] . 2 If two opposite vertex of rectangle are ( x1 , y1 ) and ( x2 , y2 ) , then its area is ( y2 y1 )( x2 x1 ) .
It two opposite vertex of a square are
A( x1 , y1 )
and C ( x2 , y2 ) , then its area is
1
1 AC 2 [( x2 x1 ) 2 ( y2 y1 ) 2 ] 2 2
Area of polygon: The area of polygon whose vertices are ( x1 , y1 ), ( x2 , y2 ), , ( xn , yn ) is
x1
y1
x2
y2
x3 {( x1 y2 x2 y1 ) ( x2 y3 x3 y2 ) ( xn y1 x1 yn )} . 1 2 Area of polygon : Stair method for finding area of a polygon: Repeat first 2 : co-ordinates one time in last for down arrow use positive sign and for up arrow use negative sign. Area of polygon xn 1 {( x1 y2 x2 y3 .... xn y1 ) ( y1 x2 y2 x3 .... yn x1 )} x1 2
y3
1
: : yn y1
Shifting and Rotation of axes
Shifting of origin without rotation of axes: Let P ( x , y ) with respect to axes OX and OY. Let O ( , ) with respect to axes OX and OY and let P ( x, y ) with respect to axes OX and OY , where OX and OX are parallel and OY and OY are parallel. Then x x , y y or x x , y y . Thus if origin is shifted to point ( , ) without rotation of axes, as shown in Fig. 0.12, then new equation of Figure 0.12: Shifting of Origin without curve can be obtained by putting x in place of x and Rotation of Axis y in place of y . Rotation of axes without changing the origin: Let O be the origin. Let P ( x , y ) with respect to axes OX and OY and let P ( x, y ) with respect to axes OX and OY are formed by rotating the axis without shifting of origin as shown in Fig. 0.13(a), where X OX YOY then following relations holds true: x x cos y sin , y x sin y cos ; and x x cos y sin , y x sin y cos .
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Engineering Mathematics
Chapter 0: Prerequisite
The above relation between ( x, y ) and ( x, y ) can be easily obtained with the help of following table
[0.35] y sin cos
x cos sin
x y
Figure 0.13: (a) Rotation of Axis without Shifting of Origin. (b) Rotation of Axis and Shifting of Origin
Change of origin and rotation of axes: If origin is changed to O ( , ) and axes are rotated about the new origin O by an angle in the anticlock-wise sense, as shown in Fig. 0.13(b), such that the new co-ordinates of P ( x, y ) become ( x, y ) then the equations of transformation will be x x cos y sin and y x sin y cos .
Reflection (Image of a point): Let ( x, y ) be any point, then its image with respect to x axis is ( x, y ) y axis is ( x, y ) Origin is ( x, y ) Line y x is ( y, x )
Locus: When a point moves in accordance with a geometric law, its path is called a locus. The equation to the locus is the equation connecting the x and y coordinates of every point on the curve. Equation of Straight Line: The general form of equation of straight line is ax by c 0 Slope form: The line making an angle with the positive direction of x axis and an intercept c on the y axis has the equation y mx c , where m tan slope of the line. Hence the slope of the line ax by c 0 is m a b
Intercept form: The line making an intercept a and b on the x and y axis is
x
y
1 a b Normal form: If the perpendicular from the origin to the line is of length p and makes an angle with the positive direction of the x axis, the equation of the line is x cos y sin p Line parallel to the x axis is y constant; line parallel to the y axis is x constant Line passes through the origin having slope m is y mx
Line passes through the point ( x1 , y1 ) with slope m is y y1 m( x x1 )
Line passes through the points ( x1 , y1 ) and ( x2 , y2 ) is y y1
the slope of the line The line passing thought a fixed point ( x1 , y1 ) and having inclination with the positive
direction of x axis is
x x1 cos
y y1 sin
y 2 y1 x2 x1
( x x1 ) , where
y2 y1 x2 x1
is
r , where r is the distance of any point ( x , y ) , on the
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Chapter 0: Prerequisite
[0.36]
line, from the given point ( x1 , y1 ) . The coordinates of any point ( x , y ) on this line are given by x x1 r cos , y y1 r sin
Position of a point with respect to a line: Two points ( x1 , y1 ) and ( x2 , y2 ) lies on the ax1 by1 c
same side of the straight line ax by c 0 according as
opposite sides of the straight line ax by c 0 according as
ax2 by2 c
0
ax1 by1 c ax2 by2 c
0
Length of perpendicular: Length of perpendicular from the point ( x1 , y1 ) to the line ax by c 0 is
ax1 by1 c a 2 b2
Two or more lines: The coordinates of the point of intersection of two straight lines are obtained by solving the equations of the two lines If is the acute angle between the lines
y m1 x c1
and
y m2 x c2 , then
m tan ( m2 m1 ) (1 m1m2 ) .
If the two lines are parallel then the angle between them is 0 o m1 m2 If the two lines are perpendicular then the angle between them is 90 o 1 m1m2 0
The equation of a line parallel to ax by c 0 is ax by k 0 , where k is found by given condition. The equation of a line parallel to ax by c 0 is bx ay k 0 , where k is found by given condition. The two lines a1 x b1 y c1 0 and a2 x b2 y c2 0 are Coincident if a1 a2 b1 b2 c1 c2
Parallel if a1 a2 b1 b2 c1 c2
Intersecting if a1 a2 b1 b2
Perpendicular is a1a2 b1b2 0
The equation of any line passing through the point of intersection of lines a1 x b1 y c1 0 and a2 x b2 y c2 0 is given as a1 x b1 y c1 ( a2 x b2 y c2 ) 0 , where is a constant and found by applying the given condition. Bisector of the angle between the lines Acute Obtuse Conditions angle angle a1 x b1 y c1 0 and a2 x b2 y c2 0 , bisector bisector where c1 , c2 0 have equations a1a2 b1b2 0 a1 x b1 y c1 a2 x b2 y c2 . a1a2 b1b2 0 a12 a22 a12 a22
The three lines a1 x b1 y c1 0 , a2 x b2 y c2 0 and a3 x b3 y c3 0 are concurrent if a1
b1
c1
a2
b2
c2 0 . If the point of intersection of two lines satisfies the third line then the three
a3 b3 c3 lines are concurrent.
Example 0.27 [MN-2007 (1 mark)]: If the slope of a diagonal of a rectangle is m , the slope of the other diagonal is (a) 1 2m (b) 1 2m (c) 1 m (d) 1 m
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Chapter 0: Prerequisite
[0.37]
Solution: Let the coordinated of any rectangle ABCD are A( a, b) , B (c, b) , C (c, d ) , D( a, d ) , as shown in figure. The slope of diagonal AC is d b d b mAC m (say) and slope of diagonal BD is mBD m ; thus ca ac given options are wrong. Example 0.28 [AG-2012 (1 mark)]: The line y x 1 can be expressed in polar coordinates ( r , ) as (a) r cos (b) r sin (c) r (cos sin ) 1 (d) r (cos sin ) 1 Solution (d): In polar coordinates, we have x r cos and y r sin . So the line y x 1 in polar coordinates will be r sin r cos 1 r (cos sin ) 1 .
Pair of Straight Lines: Let the equation of two lines be ax by c 0 …(i) and ax by c 0 …(ii). Hence ( ax b y c)( ax by c) 0 is called the joint equation of lines (i) and (ii) and conversely, if joint equation of two lines be ( ax by c)( ax by c) 0 then their separate equation will be ax by c 0 and ax by c 0 .
Equation of a pair of straight lines passing through origin: The equation ax 2 2hxy by 2 0 represents a pair of straight line passing through the origin where a, h, b are constants. Let the lines
m1
represented
h h 2 ab b
by
ax 2 2hxy by 2 0
and m2
straight lines represented by
h h 2 ab b
be
y m1 x 0 and
then, m1 m2
2h b
y m2 x 0 ,
and m1m2
a b
where,
. Then, two
are ax hy y h 2 ab 0 = 0 and
ax 2 2hxy by 2 0
ax hy y h 2 ab 0 .
The lines are real and distinct if h2 ab 0 The lines are real and coincident if h2 ab 0 The lines are imaginary if h2 ab 0 If the pair of straight lines ax 2 2hxy by 2 0 and ax 2 2hxy by 2 0 are common, then
( ab ab) 2 4( ah ah)( hb hb) . The equation of the pair of straight lines passing through origin and perpendicular to the pair of straight lines represented by ax 2 2hxy by 2 0 is given by bx 2 2hxy ay 2 0 . General equation of a pair of straight lines : An equation of the form, ax 2 2hxy by 2 2 gx 2 fy c 0 , where a, b, c, f , g , h are constants, is said to be a general equation of second degree in x and y . The necessary and sufficient condition for ax 2 2hxy by 2 2 gx 2 fy c 0
to
represent
a
pair
of
straight
lines
is
that
a
h
g
abc 2 fgh af 2 bg 2 ch 2 0 or h
b
f 0 ; the condition is remembered by the first
g f c letter of the sentence: ‘All Hostel Girls Having Boy Friend Go For Cinema’, i.e. first row represents the first letter of each word of ‘All Hostel Girls’; second row represent first letter of each word of ‘Having Boy Friend’; third row represent the first letter of each word of ‘Go For Cinema’. Separate equations from joint equation: The general equation of second degree be ax 2 2hxy by 2 2 gx 2 fy c 0 . To find the lines represented by this equation we proceed as follows:
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Engineering Mathematics
Chapter 0: Prerequisite
[0.38]
Step I: Factorize the homogeneous part ax 2 2hxy by 2 into two linear factors. Let the linear factors be ax by and ax by . Step II: Add constants c and c in the factors obtained in step I to obtain ax by c and ax by c . Let the lines be ax by c 0 and ax by c 0 . Step III: Obtain the joint equation of the lines in step II and compare the coefficients of x, y and constant terms to obtain equations in c and c . Step IV: Solve the equations in c and c to obtain the values of c and c . Step V: Substitute the values of c and c in lines in step II to obtain the required lines.
Angle between the pair of lines: The angle between the lines represented by 2
ax 2 2hxy by 2 2 gx 2 fy c 0 is given by tan
2 h ab ab
2
tan 1
2 h ab ab
.
The pair of lines ax 2 2hxy by 2 2 gx 2 fy c 0 and ax 2 2hxy by 2 0 are parallel to each other thus the angle between the pair of lines represented by ax 2 2hxy by 2 0 is given by tan 2 h 2 ab ( a b) . The lines are parallel if the angle between them is zero, i.e. 0 tan 0 h 2 ab The lines are perpendicular if the angle between them is 2 , i.e.
2 tan 1 0 a b 0
Point of intersection of pair of lines: The point of intersection of pair of line represented by ax 2 2hxy by 2 2 gx 2 fy c 0 is found by the following steps: Step I: Let ax 2 2hxy by 2 2 gx 2 fy c 0 Step II: Find x 2 ax 2 hy 2 g 0 (by keeping y as constant) Step III: Find y 2hx 2by 2 f 0 (by keeping x as constant) Step IV: Find the point of intersection of x 0 and y 0 , i.e. on solving
bg fh af gh , 2 . 2 h ab h ab
ax hy g 0 and hx by f 0 , we get ( x, y )
The point of intersection of lines represented by ax 2 2hxy by 2 0 is (0, 0) Example 0.29 [CS-1995 (1 mark)]: The value of k for which 4 x 2 8 xy ky 2 0 does not represent a pair of straight lines (both passing through the origin) is (a) 0 (b) 2 (c) 9 (d) 3 Solution: On representing the given equation in the form of quadratic equation in x , we have
4 x 2 8 xy ky 2 0 x
8 y 64 y 2 4(4)(ky 2 )
8 y 64 y 2 16ky 2
8y 4y 4 k
. Now 2(4) 8 8 the given equation will represent the pair of straight lines if 4 k 0 k 4 . Thus from given options, if k 9 4 then the given equations will not represent pair of straight lines.
Circle: The equation to the circle having its centre at O ( h, k ) and radius r , as shown in Fig. 0.14, is ( x h) 2 ( y k ) 2 r 2 . Any point P on this circle is ( h r cos , k r sin ) , where is the angle made by OP with the x axis. If the centre is the origin and radius is r , the equation of the circle is x 2 y 2 r 2 . Any point on this circle can be taken as ( r cos , r sin ) , where is the parameter of the point.
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Engineering Mathematics
[0.39]
The general equation to the circle is x 2 y 2 2 gx 2 fy c 0 . Its centre is
( g , f ) and radius is
Chapter 0: Prerequisite
g2 f 2 c .
Equation to the circle described on the line joining the points ( x1 , y1 ) and ( x2 , y2 ) as a diameter is ( x x1 )( x x2 ) ( y y1 )( y y2 ) 0
The second degree curve in two variables Figure 0.14: A circle 2 2 ax 2hxy by 2 gx 2 fy c 0 represents a circle if a b and h 0 . with centre and To find the centre and radius of this circle, reduce it to the form radius x2 y 2
2g
x
2f
y
c
f g , and radius is a a
0 centre
g 2 f 2 ca
a a a a A circle can be made to satisfy any three geometrical conditions such as passing through three given points or touching three given lines passing through two given points and touching a given line passing through a given point and touching two given lines passing through two given points and having its centre on a given line The length of tangent from point P ( x1 , y1 ) to the circle x 2 y 2 2 gx 2 fy c 0 is
x12 y12 2 gx1 2 fy1 c The point ( x1 , y1 ) lies outside, on, inside the circle x 2 y 2 2 gx 2 fy c 0 according as
p 0 , p 0 , p 0 , respectively, where p x12 y12 2 gx1 2 fy1 c .
Tangent to the circle: Equation of tangent to the circle x 2 y 2 2 gx 2 fy c 0 at ( x1 , y1 ) is xx1 yy1 g ( x x1 ) f ( y y1 ) c 0 .
Condition for the line y mx c to touch the circle x 2 y 2 a 2 is c 2 a 2 (1 m 2 ) .
The equation of tangent to the circle x 2 y 2 a 2 is y mx a 1 m 2
Two or more Circles: Let d be the distance between the centres of two circles with radii r1 , r2
If d r1 r2 , one circle lies completely inside the other circle
If d r1 r2 , the two circles do not intersect
If r1 r2 d r1 r2 , the two circles intersect
If d r1 r2 , the two circles touch one another externally and the point of contact divides the line joining the centres internally in the ratio r1 : r2 .
If d r1 r2 , the two circles touch one another internally and the point of contact divides the line joining the centres externally in the ratio r1 : r2 .
Let S x 2 y 2 2 gx 2 fy c 0 and S x 2 y 2 2 g1 x 2 f1 y c1 0 be the equations of two circles. Then the two circles will cut each other orthogonally if 2 gg1 2 ff1 c c1 If S 0 and S 0 are the equations of two circles, then S S 0 is the equation of the radical axis. However, if the two circles S 0 , S 0 coincides with the common chord. Therefore S S 0 is the equation of common chord of the intersecting circles. Equation of a circle passing through the points of intersection of the circles S 0 and S 0 can be written as S kS 0 , where k is a real number. The radical axis of three circles taken in pairs are concurrent and the point of concurrence is the radical centre of the three circles.
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Chapter 0: Prerequisite
[0.40]
Parabola: Standard equation of parabola is y 2 4ax . The following points hold true for y 2 4ax : x axis is the axis of the parabola; y axis is the tangent at the vertex
Vertex is A(0, 0) ; Focus is S (a, 0) ; Directrix is x a 0
Parametric form of representation of a point P on y 2 4ax is P ( at 2 , 2at )
At point P ( at 2 , 2at ) on y 2 4ax , slope of tangent is 1 t
Equation
of
tangent
at
P ( at 2 , 2at )
on
y 2 4ax
is
2
Figure 0.15: A Parabola
x yt at 0 2
2
3
Equation of normal at P ( at , 2at ) on y 4ax is y tx 2at at 0 ; slope of normal is t
If P ( at12 , 2at1 ) and Q ( at 22 , 2 at2 ) are the two point on the parabola y 2 4ax , then slope of the chord PQ is 2 (t1 t2 ) . The equation of the chord PQ is 2 x y (t1 t2 ) 2at1t2 0
If the tangent at P ( at12 , 2at1 ) and at Q ( at 22 , 2 at2 ) are perpendicular, then t1t2 1 . In this case, the tangents intersect on the directrix If PQ is a chord passing through the focus, then t1t2 1
If the normal at t1 meets the parabola, again at t2 , then t2 t1 (2 t1 )
If the point P on y 2 4ax is taken in Cartesian form, then
equation of tangent at ( x1 , y1 ) is yy1 2a ( x x1 ) 0 equation of normal at ( x1 , y1 ) is xy1 2ay ( x1 y1 2ay1 ) 0
y mx c touches the parabola y 2 4ax , the condition is c a m . So y mx (a m ) is a tangent to the parabola for all values of m .
If the line
The chord of contact of tangents to the parabola y 2 4ax from the point ( x1 , y1 ) is yy1 2a ( x x1 ) 0 , i.e., T 0
The chord of the parabola yy1 2 a ( x x1 )
y12
y 2 4ax
( x1 , y1 )
is
T S1 , i.e.,
4 ax1
A line which bisects a system of parallel chords of a parabola is called diameter of the parabola. If m is the slope of the parallel chords, y 2a m is the equation of the diameter
Ellipse: Standard equation of Ellipse is
whose mid-point is
x2
y2
1 . The following points hold true for
a2 b2 x axis is the major axis of length 2a ; y axis is the minor axis of length 2b
x2 a2
y2 b2
1:
b 2 a 2 (1 e 2 ) ( e 1 is the eccentricity of the ellipse) There are two foci, one S ( ae, 0) and the other S ( ae, 0) , respectively, the corresponding directrices
are x a e and x a e If P be any point on the ellipse, then (i) SP S P 2a ; and (ii) SP S P CD 2 , where CD is the semi-diameter parallel to the tangent at P Parametric form of representation of a point P on the x2 y 2 ellipse 2 2 1 is P ( a cos , b sin ) . a b x y Equation of tangent at P is cos sin 1 0 a b
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Figure 0.16: An Ellipse
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Engineering Mathematics
Chapter 0: Prerequisite
ax
Equation of normal at P is
by
[0.41]
a 2 b2
cos sin If CD be a semi-diameter parallel to the tangent at P , then D is ( a sin , b cos ) and 2
2
2
2
2
CD a sin b cos
The locus of point of intersection of perpendicular tangents of the ellipse
x2 a2
y2 b2
1 is the circle
(called the director circle) x 2 y 2 a 2 b 2 x2
y2
The circle x 2 y 2 a 2 is called the auxiliary circle of the ellipse
b 2 a 2 (e 2 1) ( e 1 is the eccentricity of the hyperbola)
There are two foci, one S ( ae, 0) and the other S ( ae, 0) , respectively, the corresponding
1 , and it is the locus a2 b2 of the foot of perpendicular from the centre of the ellipse to its tangents x2 y2 Hyperbola: Standard equation of the Hyperbola is 2 2 1 a b x axis is the transverse axis of length 2a ; y axis is the conjugate axis of length 2b
directrices are x a e and x a e
If P be any point on the Hyperbola, then SP S P 2a
Parametric form of representation of a point P on the hyperbola is ( a sec , b tan )
At P ( a sec , b tan ) on the tangent at is
x
x2 a2
sec
y2 b2
1 , equation to
y
tan 1 a b The locus of point of intersection of perpendicular x2 y2 tangents of the hyperbola 2 2 1 is the circle a b 2 (called the director circle) x y 2 a 2 b 2
The combined equation of the two asymptotes is
x2
Figure 0.17: A Hyperbola
y2
a2 b2 The angle between the asymptotes is given by 2 sec 1 (e)
Equation of the conjugate hyperbola is
x2
0
y2
1 a2 b2 The hyperbola is said to be rectangular if a b , the equation to the rectangular hyperbola is x2 y2 a2
Rectangular hyperbola: Standard equation of rectangular hyperbola is xy c 2 (axes are asymptotes). Parametric form of representation of a point P on xy c 2 is ct , c t
At P ct , c t on xy c 2 Slope of tangent is 1 t 2
Equation of the tangent is x yt 2 2ct 0
Slope of normal is t 2
Equation to the normal is xt 3 yt c ct 4 0
Eccentricity e of the rectangular hyperbola is e
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2
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Engineering Mathematics
Chapter 0: Prerequisite
[0.42]
Exercise: 0.4 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. A triangle with vertices (4, 0) , ( 1, 1) and (3,5) is (a) isosceles and right angled (b) isosceles but not right angled (c) right angled but not isosceles (d) neither right angled nor isosceles 2. The line x y 4 divides the line joining the points ( 1,1) and (5, 7) in the ratio (a) 2 : 1 , internally (b) 2 : 1 , externally (c) 1 : 2 , internally (d) 1 : 2 , externally 3. If the vertices of a triangle be (2,1) , (5, 2) and (3, 4) then its circumcentre is (a) (13 2 ,9 2) 4.
5.
6. 7.
8.
9.
10.
(b) (13 4 ,9 4)
(b) (2 3,1 3) (d) (1,1 3) (a) (1, 3 2) (c) (2 3, 3 2) Four points are A(6,3) , B ( 3, 5) , C (4, 2) and P ( x, y ) . Then ratio of area of PBC and ABC is (a) ( x y 2) 7 (b) ( x y 2) 2 (c) ( x y 2) 7 (d) None of these The points (0,8 3) , (1,3) and (82, 30) are the vertices of a/an _____ triangle. (a) equilateral (b) isosceles (c) right angled (d) none of these The point (2,3) undergoes the following three transformation successively, (i) Reflection about the line y x ; (ii) Transformation through a distance 2 units along the positive direction of y-axis; (iii) Rotation through an angle of 45o about the origin in the anticlockwise direction. The final coordinates of points are (a) (1 2 , 7 2) (b) ( 1 2 , 7 2) (c) (1 2 , 7 2) (d) None of these The ends of a rod of length l move on two mutually perpendicular lines. The locus of the point on the rod which divides it in the ratio 1 : 2 is (a) 36 x 2 9 y 2 4l 2 (b) 36 x 2 9 y 2 l 2 (c) 9 x 2 36 y 2 4l 2 (d) None of these One of the equation of the straight line passing through the point (4,3) and making intercept on the co-ordinates axes whose sum is 1 , is (a) x 2 y 2 (b) x 2 y 2 (c) x 2 y 2 (d) x 2 y 2 A line passes through (2, 2) and is perpendicular to the line 3 x y 3 . Its y intercept is (d) 4 3
3 x y 1 is
(a) y 2 0, (c)
13.
(c) 1
(b) 2 3
The equation of the lines which passes through the point (3, 2) and are inclined at 60o to the line
12.
(d) None of these
The incentre of the triangle with vertices (1, 3) , (0, 0) , and (2, 0) is
(a) 1 3 11.
(c) (9 4 ,13 4)
(b) x 2 0,
3x y 2 3 3 0
3x y 2 3 3 0
(d) None of these
3x y 2 3 3 0
If x1 , x2 , x3 and y1 , y2 , y3 are both in G.P. with the same common ratio, then the point ( x1 , y1 ) , ( x2 , y2 ) and ( x3 , y3 ) lie on a/an (a) straight line (b) ellipse (c) circle (d) parabola The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the triangle with vertices (0, 0) , (0, 21) and (21, 0) is _____
14.
If the sum of the slopes of the lines given by x 2 2cxy 7 y 2 0 is four times their product. Then the value of c is _____.
15.
The 2
point
of
intersection
of
the
lines
represented
by
the
equation
2
2 x 3 y 7 xy 8 x 14 y 8 0 is (a) (0, 2) (b) (1, 2)
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(c) ( 2, 0)
(d) ( 2,1)
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17. 18. 19.
20.
[0.43]
The lines 2 x 3 y 5 and 3 x 4 y 7 are the diameters of a circle of area 154 square units. The equation of the circle is (a) x 2 y 2 2 x 2 y 62 (b) x 2 y 2 2 x 2 y 47 (c) x 2 y 2 2 x 2 y 47 (d) x 2 y 2 2 x 2 y 62 If 3 x 4 y 4 0 and 6 x 8 y 7 0 are tangents to a circle, then radius of the circle is ____ If the chord y mx 1 of the circle x 2 y 2 1 subtends an angle of measure 45o at the major segment of the circle then absolute value of m is _____. If two circles ( x 1) 2 ( y 3) 2 r 2 and x 2 y 2 8 x 2 y 8 0 intersect in two distinct points, then (a) 2 r 8 (b) r 2 (c) r 2 (d) r 2 Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r . If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals (b) ( PQ RS ) 2
PQ RS
(a) 21.
Chapter 0: Prerequisite
(c) 2( PQ RS ) ( PQ RS )
(d)
PQ 2 RS 2 2
The equation of the directrix of the parabola y 2 4 y 4 x 2 0 is (a) x 1
(b) x 1
(c) x 3 2
(d) x 3 2
22.
The line x 1 0 is the directrix of the parabola y kx 8 0 . Then ve value of k is ___.
23.
The equation of the common tangent touching the circle ( x 3) 2 y 2 9 and the parabola
2
y 2 4 x above the x axis is
(a)
(b)
3 y 3x 1
(c)
3 y ( x 3)
(d)
3y x 3
3 y (3 x 1)
24.
If x y k is a normal to the parabola y 12 x , then k is _____.
25.
If P ( x, y ) , F1 (3, 0) , F2 ( 3, 0) and 16 x 2 25 y 2 400 , then PF1 PF2 equals _____.
2
26.
x2
(a) 2abe
(b) abe
27. Let E be the ellipse
28.
y
(d) None of these
(c) abe 2
2
1 and C be the circle x 2 y 2 9 . Let P and Q be the
9 4 points (1, 2) and (2,1) respectively. Then (a) Q lies inside C but outside E (b) Q lies outside both C and E (c) P lies inside both C and E (d) P lies inside C but outside E The locus of the middle point of the intercept of the tangents drawn from an external point to the ellipse x 2 2 y 2 2 between the coordinate axes, is 1 1 1 1 1 1 1 1 2 1 2 1 2 1 (a) 2 2 1 (b) (c) (d) 2 2 2 x 2y 4x 2y 2x 4y 2x y
29. If the foci of an ellipse 30.
x
2
y2
1 with foci F1 and F2 . If A is the area of a2 b2 the triangle PF1 F2 , then the maximum value of A is Let P be a variable point on the ellipse
x2
y2 2
1 and a hyperbola
x2
y2
1
coincide, then b 2 is ___.
16 b 144 81 25 2 If the circle x y a intersects the hyperbola xy c 2 in four points P ( x1 , y1 ) , Q ( x2 , y2 ) 2
2
, R( x3 , y3 ) and S ( x4 , y4 ) then 4
(a)
xi 0 i 1
4
(b)
yi 0 i 1
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(c) x1 x2 x3 x4 c 4
(d) None of these
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0.5
Chapter 0: Prerequisite
[0.44]
Three Dimensional Geometry
Please note that ‘Vectors and its Properties’ are given in ‘Section 1.3 of Linear Algebra’ Point in three dimension ( x1 x2 ) 2 ( y1 y2 ) 2 ( z1 z2 ) 2
Distance between two points ( x1 , y1 , z1 ) and ( x2 , y2 , z2 ) is
Section formula: Point which divides the segment joining P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) in the ratio m : n
mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn
Internally, as shown in Fig. 0.18 (a) is R
mx2 nx1 my2 ny1 mz2 nz1 , , mn mn mn
Externally, as shown in Fig. 0.18 (b) is R
Figure 0.18: Section Formula (a) Internal (b) External division
Co-ordinates of the general point: The co-ordinates of any point lying on the line joining
kx2 x1 ky 2 y1 kz2 z1 , , , which k 1 k 1 k 1
points P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) may be taken as
divides PQ in the ratio k :1 . This is called general point on the line PQ . Co-ordinates of the midpoint: When division point is the mid-point of PQ then ratio will be 1 : 1 , hence co-ordinates of the midpoint of PQ are ( x1 x2 ) 2, ( y1 y2 ) 2, ( z1 z2 ) 2 .
Triangle: If ( x1 , y1 , z1 ), ( x2 , y2 , z2 ) and ( x3 , y3 , z3 ) are the vertices of a triangle, then co-ordinates of its centroid are ( x1 x2 x3 ) 3, ( y1 y2 y3 ) 3, ( z1 z 2 z3 ) 3 . Area of triangle: Let A( x1 , y1 , z1 ) , B( x2 , y2 , z2 ) and C ( x3 , y3 , z3 ) be the vertices of a triangle,
1
then x
2 x
2
2 y
y1
z1
1
y2
z2
1 , y
y3
z3
1
2 z
1 2
x1
z1
1
x2
z2
1 , z
x3
z3
1
1 2
x1
y1
1
x2
y2
1 ; now, area of ABC is
x3
y3
1
1 1 . Also, Area of ABC AB AC 2 2
i
j
k
x2 x1
y2 y1
z2 z1
x3 x1
y3 y1
z3 z1
.
Condition of collinearity: Points A( x1 , y1 , z1 ), B( x2 , y2 , z2 ) and C ( x3 , y3 , z3 ) are collinear if x1 x2 x2 x3
y1 y2 y2 y3
z1 z2 z 2 z3
Tetrahedron: If ( xr , yr , zr ) , r 1, 2,3, 4 are vertices of a tetrahedron, then co-ordinates of its
x1 x2 x3 x4 y1 y2 y3 y4 z1 z2 z3 z4 , , . 4 4 4
centroid are
Volume of tetrahedron with vertices ( xr , yr , zr ) ; r 1, 2,3, 4 is V
Copyright © 2016 by Kaushlendra Kumar
x1
y1
z1
1
1 x2
y2
z2
1
6 x3
y3
z3
1
x4
y4
z4
1
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Chapter 0: Prerequisite
[0.45]
Direction Cosines and Direction Ratios
Direction Cosines: The cosines of the angle made by a line in anticlockwise direction with positive direction of coordinate axes are called the direction cosines of that line. If , , be the angles which a given directed line makes with the positive direction of the x, y , z co-ordinate axes respectively, as shown in Fig. 0.19, then cos , cos , cos are called the direction cosines of the given line and are generally denoted by l , m, n respectively. Thus, l cos , m cos , n cos .
Figure 0.19: Direction Cosine
By definition, it follows that the direction cosine of the x axis are respectively cos 0o , o o cos 90 , cos 90 i.e. (1, 0, 0) . Similarly direction cosines of the y and z axes are respectively (0,1, 0) and (0, 0,1) . Relation between the direction cosines: As shown in Fig. 0.19, let OP be any line through the origin O which has direction cosines l , m, n . Let P ( x, y , z ) and OP r . Then
OP 2 x 2 y 2 z 2 r 2 …(i). From P draw PA , PB , PC perpendicular on the co-ordinate axes, so that OA x , OB y , OC z . Also, POA , POB and POC . From x triangle AOP , l cos x lr . Similarly y mr and z nr . Hence from (i), r 2 2 2 2 2 2 r (l m n ) x y z 2 r 2 l 2 m2 n2 1 or, cos2 cos2 cos2 1 . [This point was asked in MT-2010 (1 mark)] If OP r and P ( x, y , z ) be any point, then d.c.’s of line OP are x r , y r , z r . Direction cosines of r a i b j c k are a r , b r , c r . Since 1 cos 1 , x R , hence values of l , m, n are such real numbers which are not less than –1 and not greater than 1. Hence d.c’s [1,1] . The direction cosines of a line parallel to any co-ordinate axis are equal to the direction cosines of the co-ordinate axis. The number of lines which are equally inclined to the co-ordinate axes is 4. If l , m, n are d.c.’s of a line, then the maximum value of lmn 1 3 3 . The angles , , are called the direction angles of line AB If direction cosines of a line AB are cos , cos , cos ; then direction cosines of line BA are cos( ), cos( ), cos( ) , where the angles , , are not coplanar. If
P ( x, y , z )
be a point in space such that
r OP
has d.c.’s
l , m, n
then
xl r , ymr , zn r . Projection of a
vector
r on the co-ordinate axes
are
l r , m r , n r , where
r r (l i m j n k ) and rˆ l i m j n k . Direction Ratios: Three numbers which are proportional to the direction cosines of a line are called the direction ratio of that line. If a , b and c are three real numbers such that l a m b n c k (say) l ak , m bk , n ck , then a , b and c are called direction ratios of
OP .
Since
l 2 m 2 n 2 1 (a 2 b 2 c 2 )k 2 1 k 1
a2 b2 c2 ,
so,
l a a 2 b 2 c 2 , m b a 2 b 2 c 2 , n c a 2 b 2 c 2 , where the signs should be taken all positive or all negative. Direction ratios are not unique, whereas d.c.’s are unique. i.e., a 2 b 2 c 2 1 Let r a i b j c k be a vector. Then its d.r.’s are a, b, c
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Engineering Mathematics
Chapter 0: Prerequisite
If a vector r has d.r.’s a, b, c then r
[0.46]
r
(a i b j c k ) 2 2 a b c D.c.’s and d.r.’s of a line joining two points: The direction ratios of line PQ joining 2
P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) are x2 x1 a , y2 y1 b and z2 z1 c (say). Then direction
cosines
are,
( x2 x1 )
l
( x2 x1 )
2
,m
( y2 y1 ) ( x2 x1 )
2
,n
( z2 z1 ) ( x2 x1 ) 2
,
i.e.,
l ( x2 x1 ) PQ , m ( y 2 y1 ) PQ , n ( z2 z1 ) PQ .
If P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) are two points in space, then the direction ratios of PQ are
x2 x1 , y2 y1 , z2 z1 . The angle between two lines with direction cosines l1 , m1 , n1 and l2 , m2 , n2 is given by cos l1l2 m1m2 n1n2 or sin
(m1n2 m2 n1 )2
Two lines are perpendicular if l1l2 m1m2 n1n2 0 and parallel if l1 l2 , m1 m2 , n1 n2 or a1 a2 b1 b2 c1 c2
Projection
Projection of a point on a line: As shown in Fig. 0.20, the projection of a point P on a line AB is the foot N of the perpendicular PN from P on the line AB . N is also the same point where the line AB meets the plane through P and perpendicular to AB . Projection of a segment of a line on another line and its length: As shown in Fig. 0.21, the projection of the segment AB of a given line on another line CD is the segment AB of CD where A and B are the projections of the points A and B on the line CD . The length of the projection AB AN AB cos .
Figure 0.20: Projection of a point on a line
Figure 0.21: Projection of a segment of a line on another line
Figure 0.22: Projection of a line joining the two points on another line
Projection of a line joining the two points on another line: As shown in Fig. 0.22, let PQ be a line segment where P ( x1 , y1 , z1 ) and Q ( x2 , y2 , z2 ) and AB be a given line with d.c.’s as l , m, n . If the line segment PQ makes angle with the line AB , then projection of PQ is PQ PQ cos ( x2 x1 ) cos ( y2 y1 ) cos ( z2 z1 ) cos , thus
PQ ( x2 x1 )l ( y2 y1 )m ( z2 z1 )n .
Angle between two lines
Cartesian form: As shown in Fig. 0.23, let be the angle between two straight lines AB and AC whose direction cosines are l1 , m1 , n1 and
l2 , m2 , n2 respectively, is given by cos l1l2 m1m2 n1n2 . If direction ratios of two lines a1 , b1 , c1 and a2 , b2 , c2 are given, then angle between two lines is given by cos
a1a2 b1b2 c1c2 a12 b12 c12 a22 b22 c22
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.
Figure 0.23: Angle between two lines
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Chapter 0: Prerequisite
[0.47]
sin 2 1 cos 2 (l12 m12 n12 )(l22 m22 n22 ) (l1l2 m1m2 n1n2 ) 2 sin 2 (l1m2 l2 m1 ) 2 ( m1n2 m2 n1 ) 2 (n1l2 n2 l1 ) 2 .
When d.r.’s of the lines are given if a1 , b1 , c1 and a2 , b2 , c2 are d.r.’s of given two lines, then angle between them is given by sin
( a1b2 a2b1 ) 2 a12 b12 c12 a22 b22 c22
Condition of perpendicularity: If the given lines are perpendicular, then 900 i.e. 90 o cos 0 l1l2 m1m2 n1n2 0 or a1a2 b1b2 c1c2 0 Condition of parallelism: If the given lines are parallel, then o 2 2 2 0 sin 0 (l1m2 l2 m1 ) ( m1n2 m2 n1 ) ( n1l2 n2l1 ) 0 , which is true, only when
l1m2 l2 m1 0 ,
m1n2 m2 n1 0
n1l2 n2l1 0
and
l1 l2 m1 m2 n1 n2 .
Similarly, a1 a2 b1 b2 c1 c2 .
Vector form: Let the vector equations of two lines be r a1 b1 and r a2 b 2 . As the lines are parallel to the vectors b1 and b2 respectively, therefore angle between the lines is same as the angle between the vectors b1 and b2 . Thus if is the angle between the given lines, then cos (b1 b 2 )
b
1
b2 .
If the lines are perpendicular, then b1 b 2 0 . If the lines are parallel, then b1 and b2 are parallel, therefore b1 b 2 for some scalar .
Equation of a line passing through a given point As shown in Fig. 0.24, the equation of a line passing through a given point is given as: Cartesian form: Cartesian equation of a straight line passing through a fixed point ( x1 , y1 , z1 ) and having direction ratios a, b, c is
x x1
y y1
z z1
. a b c Vector form: Vector equation of a straight line passing through a fixed point with position vector a and parallel to a given vector b is r a b . x x1 y y1 z z1 The parametric equations of the line are a b c x x1 a , y y1 b , z z1 c , where is the parameter. x x1
y y1
z z1
The co-ordinates of any point on the line
( x1 a , y1 b , z1 c ) , where . Since the direction cosines of a line are also direction ratios, therefore equation of a line passing through ( x1 , y1 , z1 ) and having direction cosines l , m, n is
x x1
y y1
z z1
a
.
b
c
are
Figure 0.24: Line through a point
passing
l m n Since x , y and z -axes pass through the origin and have direction cosines (1, 0, 0) , (0,1, 0) , and x0 y 0 z0 (0, 0,1) respectively. Therefore, the equations are x axis: or y 0 and 1 0 0 x0 y 0 z0 x0 y 0 z0 or x 0 and z 0 ; z axis: or z 0 ; y axis: 0 1 0 0 0 1 x 0 and y 0 .
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Engineering Mathematics
Chapter 0: Prerequisite
[0.48]
Equation of a line passing through two given points As shown in Fig. 0.25, the equation of a line passing through two given point is given as: Cartesian form: If A( x1 , y1 , z1 ), B( x2 , y2 , z2 ) be two given points, x x1
the equations to the line AB are
y y1
z z1
. The x2 x1 y2 y1 z2 z1 co-ordinates of a variable point on AB can be expressed in terms of x x y y1 z z a parameter in the form x 2 1 , y 2 ,z 2 1 1 1 1 , where {1} . In fact, ( x, y , z ) are the co-ordinates of the point which divides the join of A and B in the ratio : 1 . Vector form: The vector equation of a line passing through two Figure 0.25: Line through two points points with position vectors a and b is r a (b a)
passing
Angle between two lines The angle between the two lines be cos ( a1a2 b1b2 c1c2 )
x x1 a1
y y1 b1
a12 b12 c12 a22 b22 c22
z z1
x x2
and
c1
a2
y y2 b2
z z2 c2
is
Condition of perpendicularity: If the lines are perpendicular, then a1a2 b1b2 c1c2 0
Condition of parallelism : If the lines are parallel, then a1 a2 b1 b2 c1 c2 .
Reduction of Cartesian form of the equation of a line to Vector form and vice-versa
Cartesian to Vector: Let the Cartesian equation of a line be
x x1
y y1
z z1
. This is the a b c equation of a line passing through the point A( x1 , y1 , z1 ) and having direction ratios a, b, c . In vector form this means that the line passes through point having position vector a x1 i y1 j z1 k and is parallel to the vector m a i b j c k . Thus, the vector form of (i) is
r a m or r ( x1 i y1 j z1 k ) (a i b j c k ) , where is a parameter.
Vector to Cartesian: Let the vector equation of a line be
a x1 i y1 j z1 k , m a i b j c k r x i y j z k , a x1 i y1 j z1 k
and and
is
a
m ai b j ck
r a m , where
parameter. in
Now
r am,
we
put get,
x i y j z k ( x1 i y1 j z1 k ) (a i b j c k ) . Equating coefficients of i, j, k , we get x x1 a , y y1 b , z z1 c or
x x1 a
y y1 b
Intersection of two lines: Let the two lines be x x2
y y2
z z1 c
.
x x1 a1
y y1 b1
z z1 c1
…(i) and
z z2
…(ii). Now for finding the point of intersection the following steps should a2 b2 c2 be followed: Step I: Write the co-ordinates of general points on (i) and (ii). The co-ordinates of general points x x1 y y1 z z1 x x2 y y2 z z2 on (i) and (ii) are given by and , a1 b1 c1 a2 b2 c2
respectively, i.e., (a1 x1 , b1 y1 , c1 z1 ) and (a2 x2 , b2 y2 , c2 z2 )
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Chapter 0: Prerequisite
[0.49]
Step II: If the lines (i) and (ii) intersect, then they have a common point. a1 x1 a2 x2 ,
b1 y1 b2 y2 and c1 z1 c2 z2 . Step III: Solve any two of the equations in and obtained in step II. If the values of and satisfy the third equation, then the lines (i) and (ii) intersect, otherwise they do not intersect. Step IV: To obtain the co-ordinates of the point of intersection, substitute the value of (or ) in the co-ordinates of general point (s) obtained in step I.
Foot of perpendicular from a point to the line Cartesian form: If P is the foot of perpendicular from a point A( , , ) to the line x x1
y y1
z z1
r (say), as l m n shown in Fig. 0.26(a). If P be the foot of perpendicular, then P is (lr x1 , mr y1 , nr z1 ) . Find the direction ratios of AP and apply the condition of perpendicularity of AP and the given line. This will give the value of r and hence the point P Figure 0.26: Foot of perpendicular and image of a point on the line which is foot of perpendicular. Length and equation of perpendicular: As shown in Fig. 0.26 (a), the length of the perpendicular is the distance AP and its equation is the line joining two known points A and P . The length of the perpendicular is the perpendicular distance of given point from that line.
Reflection or image of a point in a straight line: As shown in Fig. 0.26 (b), if the perpendicular PL from point P on the given line be produced to Q such that PL QL , then Q is known as the image or reflection of P in the given line, as shown in Fig. 0.26. Also, L is the foot of the perpendicular or the projection of P on the line.
Vector form Perpendicular distance of a point from a line: As shown in Fig. 0.27 (a), let L is the foot of perpendicular drawn from P ( ) on the line r a b . Since r denotes the position vector of any point on the line r a b . So, let the position vector of L be a b . Then (a )b PL a b (a ) b . The length PL , is the magnitude of PL , and required 2 |b| length of perpendicular.
Figure 0.27: (a) Perpendicular distance of a point from a line (b) Image of a point in a line
Image of a point in a straight line: As shown in Fig. 0.27 (b), let Q( ) is the image of P in 2(a ) b r a b . Then, 2a b . 2 |b|
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Engineering Mathematics
Chapter 0: Prerequisite
[0.50]
Shortest distance between two skew lines Two straight lines in space which are neither parallel nor intersecting are called skew lines. Thus, the skew lines are those lines which do not lie in the same plane. If l1 and l2 are two skew lines, then the straight line which is perpendicular to each of these two non-intersecting lines is called the line of shortest distance, as shown in Fig. 0.28. There is one and only one line perpendicular Figure 0.28: Shortest between two lines to each of lines l1 and l2 .
Cartesian form: Let two skew lines be Therefore, x2 x1 d
l1 distance
the shortest y2 y1 z2 z1
l1
m1
n1
l2
m2
n2
x x1
y y1
m1 between
z z1 n1 the
and
x x2 l2 lines
y y2 m2 is
distance
z z2
n2 given
by
( m1n2 m2 n1 ) 2 ( n1l2 l1n2 ) 2 (l1m2 l2 m1 ) 2
Vector form: Let l1 and l2 be two lines whose equations are l1 : r a1 b1 and
l2 : r a2 b 2 . Then, shortest distance PQ
(b1 b 2 ) (a 2 a1 )
[b1 b 2 (a 2 a1 )]
| b1 b 2 | | b1 b 2 | Shortest distance between two parallel lines: The shortest distance between the parallel | (a 2 a1 ) b | lines r a1 b and r a 2 b is given by d . |b|
Condition for two lines to be intersecting i.e. coplanar Vector form: If the lines r a1 b1 and r a2 b 2 intersect, then the shortest distance between
them
is
zero.
[b1b 2 (a2 a1 )] 0
Therefore,
[(a2 a1 )b1b 2 ] 0
(a2 a1 ) (b1 b 2 ) 0
Cartesian form: If the lines x2 x1
y2 y1
z2 z1
l1
m1
n1
l2
m2
n2
x x1 l1
y y1 m1
z z1 n1
and
x x2 l2
y y2 m2
z z2 n2
intersect, then
0.
General equation of a Plane Every equation of first degree of the form Ax By Cz D 0 represents the equation of a plane. The coefficients of x, y , z , i.e., A, B, C are the direction ratios of the normal to the plane. Equation of co-ordinate planes: As shown in Fig. 0.29 (a), the equation of XOY plane: z 0 ; YOZ plane: x 0 ; ZOX plane: y 0
Vector equation of a plane: As shown in Fig. 0.29 (b), vector equation of a plane through the point A(a) and perpendicular to the vector n is (r a) n 0 or r n a n , which can also be written as r n d , where d a n . This is known as the scalar product form of a plane.
Normal form: As shown in Fig. 0.29 (c), vector equation of a plane normal to unit vector nˆ and at a distance d from the origin is r nˆ d . If n is not a unit vector, then to reduce the equation n d d or r nˆ . r n d to normal form we divide both sides by n to obtain r |n| |n| |n|
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Chapter 0: Prerequisite
[0.51]
Figure 0.29: (a) Equation of Coordinate plane; (b) Vector equation of a plane; (c) Normal form of a plane
Equation of a plane passing through a given point and parallel to two given vectors: The equation of the plane passing through a point having position vector a and parallel to b and c is r a b c , where and are scalars. Equation of plane in various form: Intercept form: If the plane cuts the intercepts of length a, b, c on co-ordinate axes, then its x y z equation is 1 . a b c Normal form : Normal form of the equation of plane is lx my nz p , where l , m, n are the d.c.’s of the normal to the plane and p is the length of perpendicular from the origin. Equation of plane parallel to co-ordinate planes or perpendicular to co-ordinate axes Equation of plane parallel to YOZ plane (or perpendicular to x axis) and at a distance ‘ a ’ from it is x a Equation of plane parallel to ZOX plane (or perpendicular to y axis) and at a distance ‘ b ’ from it is y b Equation of plane parallel to XOY plane (or perpendicular to z axis) and at a distance ‘ c ’ from it is z c Equation of plane perpendicular to co-ordinate planes or parallel to co-ordinate axes Equation of plane perpendicular to YOZ plane or parallel to x axis is By Cz D 0 Equation of plane perpendicular to ZOX plane or parallel to y axis is Ax Cz D 0 . Equation of plane perpendicular to XOY plane or parallel to z axis is Ax By D 0 . Equation of plane passing through the intersection of two planes Cartesian form: Equation of plane through the intersection of two planes P a1 x b1 y c1z d1 0 and Q a2 x b2 y c2 z d 2 0 is P Q 0 , where is the parameter. Vector form: The equation of any plane through the intersection of planes r n1 d1 and
r n2 d2 is r (n1 n 2 ) d1 d2 , where is an arbitrary constant. Equation of plane parallel to a given plane Cartesian form: Plane parallel to a given plane ax by cz d 0 is ax by cz d 0 , i.e. only constant term is changed. Vector form: Since parallel planes have the common normal, therefore equation of plane parallel to plane r n d1 is r n d2 , where d2 is a constant determined by the given condition. Equation of plane through three points: The equation of plane passing through three nonx x1 y y1 z z1 collinear points ( x1 , y1 , z1 ) , ( x2 , y2 , z2 ) and ( x3 , y3 , z3 ) is x2 x1
y2 y1
z2 z1 0 .
x3 x1
y3 y1
z3 z1
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Chapter 0: Prerequisite
[0.52]
Equation of plane passing through a given point : Equation of plane passing through the point ( x1 , y1 , z1 ) is A( x x1 ) B( y y1 ) C ( z z1 ) 0 , where A, B, C are d.r.’s of normal to the plane.
Example 0.30 [AG-2011 (2 marks)]: A plane contains the following three points: P (2,1,5) , Q (1, 3, 4) and R(3, 0, 6) . The vector perpendicular to the above plane can be represented as (a) 2iˆ ˆj kˆ (b) iˆ 2 ˆj 2 kˆ (c) 2iˆ 3 ˆj kˆ (d) iˆ 2 ˆj kˆ Solution (d): The equation of a plane passing though the given points is given as x x1 y y1 z z1 x 2 y 1 z 5 x2 x1
y2 y1
z2 z1 0 1 2
3 1
45 0
x3 x1
y3 y1
z3 z1
0 1
65
32
x 2y z 9,
where
( x1 , y1 , z1 ) (2,1, 5) , ( x2 , y2 , z2 ) (1, 3, 4) and ( x3 , y3 , z3 ) (3, 0, 6) . As any point on the plane can be given as r x i y j z k , so the equation of plane can be written as r n d , where n i 2 j k and d 9 .
Foot of perpendicular from a point to a given plane: Let P be the foot of perpendicular drawn from the point A( , , ) to a given plane ax by cz d 0 , then AP is parallel to the x
ax by cz d 0 is
y
z
r (say), on which the point P lies. So a b c the point P lies on line and plane thus putting the point ( ar , br , cr ) on the plane we find the value of r and hence the point P . Perpendicular distance Cartesian form : The length of the perpendicular from the point P( x1 , y1 , z1 ) to the plane
normal to the plane, so its equation is
ax1 by1 cz1 d
a 2 b2 c 2 Vector form : The perpendicular distance of a point having position vector a from the plane | an d | r n d is given by p |n| The distance between two parallel planes is the algebraic difference of perpendicular distances on the planes from origin, i.e. distance between two parallel planes D2 D1 Ax By Cz D1 0 and Ax By Cz D2 0 is 2 2 2 A B C Example 0.31 [CE-2003 (1 mark)]: If P , Q and R are three points having coordinates (3, 2, 1) , (1, 3, 4) , (2,1, 2) in XYZ space, then the distance from point P to plane OQR (O being the origin of the coordinate system) is given by (a) 3 (b) 5 (c) 7 (d) 9 Solution (a): The equation of a plane passing though the points is given as x x1 y y1 z z1 x0 y0 z0 x2 x1
y2 y1
z2 z1 0 1 0
30
40 0
x3 x1
y3 y1
z3 z1
1 0
2 0
20
10 x 10 y 5 z 0 ,
where
( x1 , y1 , z1 ) (0, 0, 0) , ( x2 , y2 , z2 ) (1, 3, 4) and ( x3 , y3 , z3 ) (2,1, 2) . Now distance from point P (3, 2, 1) to the plane 10 x 10 y 5 z 0 is given by
10(3) 10( 2) 5( 1) 0
3.
( 10) 2 (10) 2 ( 5) 2
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Chapter 0: Prerequisite
[0.53]
Position of two points w.r.t. a plane: Two points P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) lie on the same or opposite sides
ax by cz d 0
of a plane
according to
ax1 by1 cz1 d
and
ax2 by2 cz2 d are of same or opposite signs. The plane divides the line joining the points P and Q externally or internally according to P and Q are lying on same or opposite sides of the plane.
Angle between two planes
Cartesian form : Angle between the planes is defined as angle between normals to the planes drawn from any point. Angle between the planes a1 x b1 y c1 z d1 0 and
( a 2 b 2 c 2 )( a 2 b 2 c 2 ) 1 1 1 2 2 2 a1a2 b1b2 c1c2
a2 x b2 y c2 z d2 0 is cos 1
If a1a2 b1b2 c1c2 0 , then the planes are perpendicular to each other If
a1 a2
b1 b2
c1 c2
, then the planes are parallel to each other
Vector form : An angle
between the planes r1 n1 d1 and r2 n 2 d2 is given by
cos (n1 n 2 ) | n1 || n 2 | .
Example 0.32 [AG-2010 (2 marks)]: The angle of intersection between the planes x 3 y 2 z 10 and 2 x 4 y 5 z 0 is (a) 30o (b) 60o (c) 75o (d) 90o Solution (d): If is the angle between the planes x 3 y 2 z 10 and 2 x 4 y 5 z 0 , then cos
(1)(2) ( 3)(4) (2)(5) 2
2
(1) ( 3) (2)
2
2
2
(2) (4) (5)
2
0 90
o
Image of a point in a plane: Let P and Q be two points and let be a plane such that (i) Line PQ is perpendicular to the plane , and (ii) Mid-point of PQ lies on the plane , then either of the point is the image of the other in the plane , as shown in Fig. 0.30. To find the image of a point in a given plane, we proceed as follows: Write the equations of the line passing through P and x x1 y y1 z z1 normal to the given plane as . Figure 0.30: Image of a point in a plane a b c Write the co-ordinates of image Q as ( x1 ar , y1 br , z1 cr )
Find the co-ordinates of the mid-point R of PQ Obtain the value of r by putting the co-ordinates of R in the equation of the plane Put the value of r in the co-ordinates of Q .
Coplanar lines: Lines are said to be coplanar if they lie in the same plane or a plane can be made to pass through them. Every pair of parallel lines is coplanar Two coplanar lines are either parallel or intersecting The three sides of a triangle are coplanar Vector form : If the lines r a1 b1 and r a2 b 2 are coplanar, then [a1b1b 2 ] [a2b1b 2 ] and the equation of the plane containing them is [r b1 b 2 ] [a1 b1 b 2 ] or [r b1 b 2 ] [a2 b1 b 2 ] .
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Engineering Mathematics
Chapter 0: Prerequisite
Cartesian form : If the lines
x x1 l1
x2 x1
y2 y1
z2 z1
l1
m1
n1
l2
m2
n2
x x1
y y1
z z1
l1
m1
n1
l2
m2
n2
Then
0 or
y y1 m1
z z1 n1
and
[0.54]
x x2 l2
y y2 m2
z z2 n2
are coplanar
0 . The equation of the plane containing them is x x2
y y2
z z2
l1
m1
n1
l2
m2
n2
0.
Division by plane : The ratio in which the line segment PQ , joining P( x1 , y1 , z1 ) and
ax by1 cz1 d Q( x2 , y2 , z2 ) , is divided by plane ax by cz d 0 is 1 . ax2 by2 cz2 d Division by co-ordinate planes : The ratio in which the line segment PQ, joining P( x1 , y1 , z1 ) and Q( x2 , y2 , z2 ) is divided by co-ordinate planes are as follows: By yz plane: x1 x2
By zx plane: y1 y2
By xy plane: z1 z2
Equation of a plane through a given line: If equation of the line is given in symmetrical form as x x1
y y1
z z1
, then equation of plane is a( x x1 ) b( y y1 ) c( z z1 ) 0 , where a, b, c l m n are given by al bm cn 0 . If equation of lines are given in general form as a1 x b1 y c1 z d1 0 , a2 x b2 y c2 z d2 0 then the equation of plane passing through these line is (a1 x b1 y c1 z d1 ) (a2 x b2 y c2 z d2 ) 0 ., where is a parameter.
Intersection point of a line and a plane: As shown in Fig. 0.31, to find the point of intersection of the line x x1 y y1 z z1 and plane ax by cz d 0 . The col m n x x1 y y1 z z1 ordinates of any point on the line r Figure 0.31: Intersection of line and l m n plane (say) are given by ( x1 lr , y1 mr , z1 nr ) …(i). If it
lies
on the plane
ax by cz d 0 ,
then a( x1 lr ) b( y1 mr ) c( z1 nr ) d 0
(ax1 by1 cz1 d ) r (al bm cn) 0 . Thus r
( ax1 by1 cz1 d )
al bm cn of r in (i), we obtain the co-ordinates of the required point of intersection.
. Substituting the value
Angle between line and plane: As shown in Fig. 0.32, the angle between the line and plane is given as: Cartesian form : The angle between the line x y z , and the plane l m n ax by cz d 0 , is given by sin
al bm cn 2
2
( a b c 2 ) (l 2 m 2 n 2 )
.
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Figure 0.32: Angle between line and plane
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Engineering Mathematics
Chapter 0: Prerequisite
The line is perpendicular to the plane if and only if
a
[0.55]
b
c
. l m n The line is parallel to the plane if and only if al bm cn 0 . The line lies in the plane if and only if al bm cn 0 and a b c d 0 . Vector form : If is the angle between a line r (a b) and the plane r n d , then sin
bn
. | b || n | Condition of perpendicularity : If the line is perpendicular to the plane, then it is parallel to the normal to the plane. Therefore b and n are parallel. So, b n 0 or b n for some scalar . Condition of parallelism : If the line is parallel to the plane, then it is perpendicular to the normal to the plane. Therefore b and n are perpendicular. So, b n 0 . If the line r a b lies in the plane r n d , then (i) b n 0 and (ii) a n 0 .
Projection of a line on a plane: If P be the point of intersection of given line and plane and Q be the foot of the perpendicular from any point on the line to the plane then PQ is called the projection of given line on the given plane. x x1 y y1 z z1 Image of line about a plane : Let line is , plane is a1 b1 c1
a2 x b2 y c2 z d 0 . Find point of intersection (say P ) of line and plane. Find image (say Q ) of point ( x1 , y1 , z1 ) about the plane. Line PQ is the reflected line. Exercise: 0.5 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1.
A line makes the same angle with each of the x and z axis. If the angle , which it
2.
makes with y axis, is such that sin 3sin , then cos 2 equals _____ The lines x ay b , z cy d and x ay b , z cy d are perpendicular to each
2
2
other, if 3. 4.
5. 6. 7.
8.
(a) aa cc 1 (b) aa cc 1 (c) ac ac 1 (d) ac ac 1 x 1 y 1 z 1 x3 yk z If the line and intersect, then k _____. 2 3 4 1 2 1 A line with direction cosines proportional to 2, 1, 2 meets each of the lines x y a z and x a 2 y 2 z . The co-ordinates of each of the points of intersection are given by (a) (2a, 3a, 3a) , (2a, a, a) (b) (3a, 2a, 3a) , ( a, a, a) (c) (3a, 2a, 3a) , ( a, a, 2a) (d) (3a, 3a, 3a ) , ( a, a, a) The length of the perpendicular from the origin to line r (4 i 2 j 4 k ) (3 i 4 j 5 k ) is _____. The image of point (1, 2,3) in the line r (6 i 7 j 7 k ) (3 i 2 j 2 k ) is (a) (5, 8,15) (b) (5,8, 15) (c) ( 5, 8, 15) (d) (5,8,15) The line
x2
y 3
z4
and
x 1
y4
z 5
are coplanar, if 1 1 k k 2 1 (a) k 0, 1 (b) k 0,1 (c) k 0, 3 (d) k 3, 3 The ratio in which the plane x 2 y 3 z 17 divides the line joining the point ( 2, 4, 7) and
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Engineering Mathematics
9.
10.
13.
[0.56]
(3, 5,8) is (a) 10 : 3 (b) 3 :1 (c) 3 :10 (d) 10 :1 The equation of plane passing through the points (2, 2,1) and (9, 3, 6) and perpendicular to the plane 2 x 6 y 6 z 1 is (a) 3 x 4 y 5 z 9 (b) 3 x 4 y 5 z 0 (c) 3 x 4 y 5 z 9 (d) None of these Angle between two planes x 2 y 2 z 3 and 5 x 3 y 4 z 9 is (a) cos 1
11. 12.
Chapter 0: Prerequisite
3 2
(b) cos 1
19 2
(c) cos 1
9 2
(d) cos 1
3 2
10 30 20 5 Distance between two parallel planes 2 x y 2 z 8 and 4 x 2 y 4 z 5 0 is _____ A tetrahedron has vertices at O (0, 0, 0) , A(1, 2,1) , B (2,1,3) and C ( 1,1, 2) . Then the angle between the faces OAB and ABC will be 19 17 (a) cos 1 (b) cos 1 (c) (d) 6 2 35 31 A unit vector perpendicular to plane determined by the points P (1, 1, 2) , Q (2, 0, 1) and R (0, 2,1) is
(a)
2i jk
(b)
2i jk
(c)
2 i j k
(d)
2i jk
14.
6 6 6 The reflection of the point (2, 1, 3) in the plane 3 x 2 y z 9 is
15.
26 17 15 , , 7 7 7 The d.r.’s of normal to the plane through (1, 0, 0) and (0,1, 0) which makes an angle 4 with 26 15 17 , , 7 7 7
(a)
26 15 17 , , 7 7 7
15 26 17 , , 7 7 7
6
(b)
(c)
(d)
(b) 1,1, 2
(c) 1,1, 2
(d)
plane x y 3 , are (a) 1, 2,1 16.
x 1
Value of k such that the line
2
y 1 3
zk k
2,1,1
is perpendicular to normal to the plane
r (2 i 3 j 4 k ) 0 is 17.
18.
19. 20.
(a) 13 4 (b) 17 4 (c) 4 (d) None of these The equation of line of intersection of the planes 4 x 4 y 5 z 12 , 8 x 12 y 13z 32 can be written as x y 1 z 2 x y z2 x 1 y 2 z x 1 y 2 z (a) (b) (c) (d) 2 3 4 2 3 4 2 3 4 2 3 4 x3 y 6 z4 The plane which passes through the point (3, 2, 0) and the line is 1 5 4 (a) x y z 1 (b) x y z 5 (c) x 2 y z 1 (d) 2 x y z 5 The value of k such that
x4
1 The distance between the r (2 i j 3 k ) 5 is
(a)
5 14
(b)
y2 1 line
6 14
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zk
lies in the plane 2 x 4 y z 7 is _____. 2 r (i j 2 k ) (2 i 5 j 3 k ) and the plane
(c)
7 14
(d)
8 14
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Engineering Mathematics
Chapter 0: Prerequisite
[0.57]
Answers Keys 1 c 16 b 31 3 46 7 61 0 76 d 91 b 106 2 121 13
2 a 17 1 32 4.5 47 2 62 a 77 c 92 c 107 b 122 4
3 c 18 a 33 b 48 c 63 b 78 a 93 a 108 c 123 a
4 a 19 a 34 d 49 1 64 d 79 b 94 d 109 c 124 0
1 d 16 d 31 20 46 a
2 14 17 8 32 b 47 c
3 0 18 42 33 a 48 d
4 89 19 900 34 b 49 99.22
1 210 16 a 31 a
2 a 17 a 32 a
3 d 18 b 33 c
4 c 19 50 34 b
1 a 16 b
2 c 17 0.75
3 b 18 1
4 d 19 a
1 0.6 16 a
2 b 17 c
3 4.5 18 a
4 b 19 7
5 b 20 1 35 a 50 60 65 d 80 a 95 c 110 c 125 1
Answer Keys: Exercise 0.1 6 7 8 9 10 a c a b b 21 22 23 24 25 b b b a d 36 37 38 39 40 12 7 b d d 51 52 53 54 55 d b d a c 66 67 68 69 70 b d 4 2 2 81 82 83 84 85 d a b c a 96 97 98 99 100 2 d b b d 111 112 113 114 115 5 a b c 3 126 127 1 2
11 b 26 3 41 d 56 0 71 3 86 d 101 a 116 c
12 c 27 c 42 a 57 9 72 0 87 d 102 a 117 a
13 a 28 7 43 b 58 a 73 b 88 d 103 a 118 b
14 d 29 d 44 1 59 b 74 c 89 b 104 d 119 a
15 d 30 d 45 c 60 d 75 a 90 a 105 c 120 b
Answer Keys: Exercise: 0.2 6 7 8 9 10 520 3200 4 b 11 21 22 23 24 25 b b a c 13.5 36 37 38 39 40 4 6 b b c
11 d 26 a 41 b
12 25 27 b 42 c
13 d 28 b 43 a
14 2475 29 a 44 d
15 27 30 c 45 c
5 45 20 d
Answer Keys: Exercise: 0.3 6 7 8 9 10 a d 6 2 b 21 22 23 24 25 36 20 b a d
11 8 26 d
12 a 27 1
13 c 28 2
14 b 29 d
15 3 30 0
5 a 20 a
Answer Keys: Exercise 0.4 6 7 8 9 10 d b c d d 21 22 23 24 25 c 4 c 9 10
11 a 26 b
12 a 27 d
13 190 28 c
14 2 29 7
15 c 30 d
Answer Keys: Exercise 0.5 6 7 8 9 10 d c c c a
11 3.5
12 a
13 b
14 b
15 b
5 13 20 a 35 5 50 b
5 6 20 d
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [1]
GATE – 2016: Chapter – 1: Linear Algebra Note: The following questions came in GATE – 2016 were based on Linear Algebra Chapter – 1. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 1. Consider an eigenvalue problem given by Ax i x . If i represent the eigenvalues of the nonsingular square matrix A , then what will be the eigenvalues of matrix A 2 ? 4 i
2 i
(a)
[AE-2016 (1 mark)] (d) i1 4
12 i
(b)
(c)
Solution (b): If x is the eigenvector of matrix A corresponding to eigenvalues i , then Ax i x …(i). On pre-multiplying both sides of (i) with A we get A 2 x i Ax A 2 x i2 x …(ii). Thus from (ii), the eigenvalues of matrix A 2 is i2 . 2. If A and B are both non-singular n n matrices, then which of the following statement is NOT TRUE. Note: det represent the determinant of a matrix. [AE-2016 (1 mark)] (a) det( AB ) det( A) det(B) (b) det( A B) det( A) det(B) 1
T
(c) det( AA ) 1 Solution (b): For any 1
two
n n
(d) det( A ) det( A) matrices A and B , det( AB ) det( A) det(B) ;
1
det( AT ) det( A) .
det( AA ) det( A) det( A ) det( A) det( A) 1 ; det( A B) det( A) det(B) .
But
5 3 3. Eigen values of the matrix [AG-2016 (1 mark)] are 1 4 (a) –6.3 and –2.7 (b) –2.3 and –6.7 (c) 6.3 and 2.7 (d) 2.3 and 6.7 5 3 Solution (c): Let A and be the eigenvalues of A , then we must have A I 0 , i.e. 1 4 5
3
1
0
0
5
3
0 (5 )(4 ) 3 0 2 9 17 0 6.3, 2.7 .
1 4 0 1 1 4 [Similar question was also asked in PI-2016, MT-2016 (1 mark)]
2 4. The positive Eigen value of the following matrix 5 2
1 2
is _____.
[BT-2016 (1 mark)]
1
and be the eigenvalues of A , then we must have A I 0 , i.e. 5 2
Solution: Let A 2
1
5
2
1
0
0 1
0
2
1
5
2
0 (2 )( 2 ) 5 0 (2 )(2 ) 5 0
(4 2 ) 5 0 2 9 3 . So positive Eigen value of the given matrix is 3 . 5. If the entries in each column of a square matrix M add up to 1, then an eigenvalue of M is [CE-2016 (1 mark)] (a) 4 (b) 3 (c) 2 (d) 1 a b Solution (d): Let we have 2 2 matrix M , such that a c 1 and b d 1 . c d
Copyright © 2016 by Kaushlendra
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Engineering Mathematics
For
eigenvalues,
,
Chapter – 1: Linear Algebra
of
M I 0
M,
Chapter – 1 [2]
a
b
c
d
0 ( a )( d ) bc 0
2 ( a d ) ad (1 d )(1 a ) 0 2 ( a d ) a d 1 0 …(i) As (i) is satisfied only by when 1 , so 1 is the required eigenvalue. Hence option (d) is correct. In general: If the sum of the elements in each row or column of a square matrix is equal to k , then k is the eigenvalue of that matrix. 6. Two eigenvalues of a 3 3 real matrix P are (2 1) and 3. The determinant of P is _____. [CS-2016 (1 mark)] Solution: As one of the eigenvalue of real matrix P is (2 i) , so another will be (2 i ) ; and the given third eigenvalue is 3. Thus the determinant of P will be (2 i )(2 i )(3) (4 1)(3) 15 . 7. Consider the system, each consisting of m linear equations in n variables. I. If m n , then all such system have a solution II. If m n , then none of these systems has a solution III. If m n , then there exist a system which has a solution. Which one of the following is CORRECT? [CS-2016 (1 mark)] (a) I, II and III are true (b) Only II and III are true (c) Only III is true (d) None of them is true Solution (c): For statement I: Let we have the system which has 2 equations and 3 unknowns as: x y z 1 and x y z 1 , which has no solution. So, if m n , then all such system does not necessary to have a solution. Thus statement I is not correct. For statement II: Let we have the system which has 3 equations and 2 unknowns as: x y 2 , x y 0 and 2 x y 3 , which has a solution ( x 1, y 1 ). So, if m n , then there exist a system which have a solution. Thus statement II is not correct. For statement III: Let we have the system which has 2 equations and 2 unknowns as: x y 2 and x y 0 which has a solution ( x 1, y 1 ). So, if m n , then there exist a system which have a solution. Thus statement III is correct. Hence option (c) is correct. 1 T
8. Suppose that the eigenvalues of matrix A are 1, 2, 4. The determinant of ( A ) is _____. [CS-2016 (1 mark)] Solution: The determinant of matrix A is the product of all its eigenvalues. So A (1)(2)(4) 8 . ( A1 )T A1 1 A 1 8 0.125 . 9. Let M 4 I , (where I denotes the identity matrix) and M I , M 2 I and M 3 I . Then, for any natural number k , M 1 equals: [EC-2016 (1 mark)] 4 k 1 4k 2 4 k 3 (a) M (b) M (c) M (d) M 4k 4 1 Solution (c): Multiplying both sides of given relation M I with M , we get M 4 M 1 M 1 M 3 ( M ) M 1 M 1 M 3 M 1 M 1 M 3 M 1 M 3 M 1 IM 3 I 4 k M 3 ( I k I ), where k is any natural number. So M
1
I k M 3 ( M 4 ) k M 3 M 4 k 3 , thus option (c) is correct.
x 10. Consider a 2 2 square matrix A , where x is unknown. If the eigenvalues of the matrix A are ( j ) and ( j ) , then x is equal to [EC-2016 (1 mark)] (a) j (b) j (c) (d) Solution: We know that the product of eigenvalues of a matrix gives the determinant of that matrix. x 2 2 2 So we have, ( j )( j ) x x . Thus, option (d) is correct.
Copyright © 2016 by Kaushlendra
Kumar e-mail: [email protected]
Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [3]
4 3 2 11. The value of x for which the matrix A 9 7 13 has zero as an eigenvalue is _____. 6 4 9 x [EC-2016 (1 mark)] Solution: We know that product of eigenvalues of a matrix gives the determinant of that matrix. As one of the eigenvalues of the given matrix is zero, so the determinant of given matrix must be zero. Thus we have A 0 3{7( 9 x ) 13( 4)} 2{9( 9 x) (13)( 6)} 4{36 (7)( 6)} 0 33 21x 6 18 x 24 0 x 1 . 12. Consider the time-varying vector I xˆ 15 cos( t ) yˆ 5 sin( t ) in Cartesian coordinates, where
0 is a constant. When the vector magnitude I is at its minimum value, the angle that I makes with the x axis (in degrees, such that 0 180 ) is _____.` [EC-2016 (1 mark)] Solution: I {15 cos(t )}2 {5sin( t )}2 225 cos 2 (t ) 25sin 2 (t ) I
200 cos 2 ( t ) 25 100{1 cos(2 t )} 25 125 100 cos(2 t ) , which is minimum o
o
when cos(2t ) 1 2t 180 t 90 . We know that t , so the required value of in o
degrees for which I is minimum is 90 . 13. Consider a 3 3 matrix with every element being equal to 1. Its only non-zero eigenvalue is _____. [EE-2016 (1 mark)] Solution: For eigenvalues ‘ ’ of given square matrix A , we must have A I 0 . As
1 1 1 1 1 1 A 1 1 1 . So A I 0 1 1 1 0 1 1 1 1 1 1 2 (1 ){(1 ) 1} 1{(1 ) 1} 1{1 (1 )} 0 3 3 2 0 2 ( 3) 0 0, 3 . Thus non-zero eigenvalue of matrix A is 3. 14. A 3 3 matrix P is such that, P 3 P . Then the eigenvalues of P are [EE-2016 (1 mark)] (a) 1,1, 1 (b) 1, 0.5 j 0.866, 0.5 j 0.866 (c) 1, 0.5 j 0.866, 0.5 j 0.866 (d) 0,1, 1 Solution (d): If be an eigenvalue and x be the eigenvector of P , then Px x …(i) 2 2 2 2 On pre-multiplying both sides of (i), we get P x Px ( x ) x P x x …(ii) Again pre-multiplying both sides of (ii), we get
P 3 x 2 Px 2 ( x) 3 x P3 x 3 x
Px 3 x x 3 x ( 2 1) x 0 ( 2 1) 0 , as x 0 . Thus 0, 1 . 15. The vector that is NOT perpendicular to the vectors (i j k ) and (i 2 j 3k ) is _____. [IN-2016 (1 mark)] (a) (i 2 j k ) (b) ( i 2 j k ) (c) (0i 0 j 0k ) (d) (4i 3 j 5k ) Solution: The vector which is perpendicular to the vectors (i j k ) and (i 2 j 3k ) is parallel to
i
j
(i j k ) (i 2 j 3k ) 1 1
k 1 i (3 2) j(3 1) k (2 1) i 2 j k . So the vectors which
1 2 3 are parallel to (i 2 j k ) are given by options (a), (b) and (c). The vector (0i 0 j 0k ) is parallel to the vector (i 2 j k ) , as the direction of (0i 0 j 0k ) is any arbitrary direction. So the vector which is not perpendicular to the vectors (i j k ) , (i 2 j 3k ) is given by option (d).
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Kumar e-mail: [email protected]
Engineering Mathematics
a b 16. Let M b d c e
Chapter – 1: Linear Algebra
Chapter – 1 [4]
c e be a real matrix with eigenvalues 1, 0 and 3. If the eigenvectors
f T
T
corresponding to 1 and 0 are (1,1,1) and (1, 1, 0) respectively, then the value of 3 f is equal to _____. [MA-2016 (2 mark)] Solution: As the value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a n n real symmetric matrix A is always zero. So let ( x1 , x2 , x3 )T be the eigenvector corresponding to the eigenvalue ‘3’; hence (1,1,1)T ( x1 , x2 , x3 )T 0 x1 x2 x3 0 …(i); also (1, 1, 0)T ( x1 , x2 , x3 )T 0 x1 x2 0 …(ii); from (i) and (ii), we have x2 x1 , x3 2 x1 . Thus
eigenvector corresponding to eigenvalue of ‘3’ will be ( x1 , x2 , x3 )T ( x1 , x1 , 2 x1 )T x1 (1,1, 2)T or
(1,1, 2)T . Now we know that if X be the eigenvector corresponding to eigenvalue ‘ ’ of matrix M , then we must have MX X . So a b c 1 1 a b 0 a b (vi) T for 0 , X (1, 1, 0) , b
c
d
e 1 0 1 b d 0 b d (vii)
e
f 0
0
c e 0 c e (viii)
1 a b c 1 2a c 1 (iii) e 1 1 1 b d e 1 2b c 1 (iv) f 1 1 c e f 1 2c f 1 (v)
a b for 1 , X (1,1,1) , b d c e
c 1
T
a b 2c 3 2a 2c 3 (ix) 1 e 1 3 1 b d 2e 3 2b 2e 3 (x) f 2 2 c e 2 f 6 2c 2 f 6 (xi)
a b for 3 , X (1,1, 2) , b d c e
c 1
T
We know that sum of eigenvalues of any n n matrix is equal to its trace, so a d f 1 0 3 a a f 4 2a f 4 …(xii). Now from (ix) and (xi), we have 2 f 2a 3 …(xiii). From (xii) and (xiii), we have 3 f 7 .
2 5 x 2 17. The solution to the system of equations is 4 3 y 30 (a) 6, 2 (b) –6, 2 (c) –6, –2 2 Solution (d): The given system is written as AX B , where A 4
[ME-2016 (1 mark)] (d) 6, –2 x 2 , , X B . 3 y 30 5
So using Cramer rule, we have x 1 , y 2 , where,
2
2 2 5 2 6 20 26 , 1 6 150 156 and 2 60 8 52 . 4 3 30 3 4 30
5
Thus x 1 156 26 6 and y 2 52 26 2 . Thus option (d) is correct.
2 1 are positive is 1 k
18. The condition for which the eigenvalues of the matrix A (a) k 1 2
(b) k 2
Solution (a): Let 2
1
1
k
1
0
0 1
0
be the eigenvalues of 2
1
1
k
Copyright © 2016 by Kaushlendra
(c) k 0
[ME-2016 (1 mark)] (d) k 1 2
A , then we must have
A I 0 , i.e.
0 (2 )( k ) 1 0 (2 )( k ) 1 0
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [5]
2 (k 2) (2k 1) 0 …(i); f ( ) 0 , where f ( ) 2 ( k 2) (2k 1) We have to find values of k for which the roots of (i) are positive; so we must have {(k 2)}2 4(1)(2k 1) 0 k 2 4k 8 0 ( k 2) 2 4 0 k …(i); {( k 2) 2(1)} 0 k 2 …(iii); and (1) f (0) 0 (2k 1) 0 k 1 2 …(iv).
So from (ii), (iii) and (iv), we have k 1 2 . Hence option (a) is correct. 19. If [ A][ B] [ I ] , then T
(a) [ B ] [ A]
[MN-2016 (1 mark)] (b) [ A] [ B ] 1
Solution (c): Pre-multiplying with [ A] 1
1
T
1
(c) [ B ] [ A]
(d) [ B ] [ A]
to the given relation, we get
1
[ A] [ A][ B ] [ A] [ I ] [ I ][ B] [ A] [ B ] [ A]1 . Hence option (c) is correct. 20. A real square matrix A is called skew-symmetric if [ME-2016 (1 mark)] T T 1 T (a) A A (b) A A (c) A A (d) AT A A1 Solution (c): A square matrix A [ aij ] is called a skew-symmetric matrix if aij a ji , i, j [ A]ij [ AT ]ij , i , j . So option (c) is correct.
21. The number of solutions of the simultaneous algebraic equations y 3 x 3 and y 3 x 5 is [PI-2016 (1 mark)] (a) Zero (b) 1 (c) 2 (d) Infinite Solution (a): The given equations can be written as: y 3 x 3 …(i) and y 3 x 5 …(ii). It is clear that any set of values of x, y cannot satisfy both equations. Hence the number of solutions is zero.
1 1 2 1 22. Let A . The determinant of A is equal to 1 2 1 (a) 1 2
(b) 4 3
Solution (b): A 1 (1 2)(1 2) 3 4 . So A
(c) 3 4 1
[TF-2016 (1 mark)] (d) 2
1 A 4 3.
23. Consider the following system of linear equations: 2 x y z 1 , 3 x 3 y 4 z 6 , x 2 y 3z 4 . The system of linear equations has [AE-2016 (2 marks)] (a) no solution (b) one solution (c) two solution (d) three solution Solution (a): Using Cramer rule, we have 2 1 1
3 1 1
3 4 2( 9 8) 1(9 4) 1( 6 3) 2 5 3 0 . Now 2
3
1 1
1 6
3
4 1( 9 8) 1(18 16) 1(12 12) 1 2 1 . As 0 and 1 0 , so no need
4
2
3
to check for 2 and 3 ; and thus the given system of equations have no solution.
5 16 81 24. The value of determinant A given below A 0 2 2 is _____. 0 0 16
[BT-2016 (2 marks)]
Solution: As we have two zeros in the first column so expanding the given determinant along the first column we get A 5(2 16 2 0) 160 , which is the required value of the given determinant.
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Kumar e-mail: [email protected]
Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [6]
25. The magnitudes of vector P , Q and R are 100 kN, 250 kN and 150 kN, respectively as shown in the figure. The respective values of the magnitudes (in kN) and the direction (with respect to the x axis) of the resultant vector are [CE-2016 (2 marks)] (a) 290.9 and 96.0o (b) 368.1 and 94.7o o (c) 330.4 and 118.9 (d) 400.1 and 113.5o Solution (c): As shown in figure, the given forces can be written as: P 100 cos 60o i 100 sin 60o j Q 250 sin15o i 250 cos15o j R 150 cos15o i 150 sin15o j So the resultant of the above forces will be N PQR N (100 cos 60o 250 sin15o 150 cos15o ) i (100 sin 60o 250 cos15o 150 sin15o ) j N 159.59 i 289.26 j N ( 159.59) 2 (289.26) 2 330.4 kN; and the direction of N 1
289.26 o 118.9 . Thus option (c) is correct. 159.59
with respect to x direction is tan
26. Consider the following linear system: x 2 y 3z a , 2 x 3 y 3z b , 5 x 9 y 6 z c . This system is consistent if a, b and c satisfy the equation [CE-2016 (2 marks)] (a) 7 a b c 0 (b) 3a b c 0 (c) 3a b c 0 (d) 7 a b c 0 Solution (b): The given system is consistent if the system has one or more solution. 1 2 3 Using Cramer’s rule, we have 2
3
3 1(18 27) 2(12 15) 3(18 15) 0
5
9
6
Now if the system is consistent then we must have 1 2 3 0 .
a
2
3
Since 1 0 b
3
3 a( 18 27) 2( 6b 3c ) 3(9b 3c ) 0 3a b c 0 …(ii). We
c
9
6
get same condition (ii) on solving 2 0 & 3 0 . So option (b) is correct. 27. A set of simultaneous linear algebraic equations is represented in a matrix form as shown below. The value (rounded off to the nearest integer) of x3 is _____. [CH-2016 (2 marks)]
0 2 0 0 2
0 0 5 5 0
2 5
0 0 3
4 13 x1 2 10 x2 4
2 1
46 161 3 x3 61 5 x4 30 5 x5 81
Solution: The given system of linear equations in matrix form can be written as: 4 x4 13 x5 46 …(i); 2 x1 5 x2 5 x3 2 x4 10 x5 161 …(ii); 2 x3 5 x4 3 x5 61 …(iii); 4 x4 5 x5 30 …(iv);
2 x1 3x2 2 x3 x4 5 x5 81 …(v) From (i) & (iv), x4 5 and x5 2 ; from (iii), we get x3 15 . 28. Let A1 , A2 , A3 and A4 be four matrices of dimensions 10 5, 5 20, 20 10, and 10 5, respectively. The minimum number of scalar multiplications required to find the product A1 A2 A3 A4 using the basic matrix multiplication method is _____. [CS-2016 (2 marks)]
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [7]
Solution: There are five possible cases for finding the product A1 A2 A3 A4 , which are A{B (CD )} , A{( BC ) D} , {( AB)C}D , { A( BC )}D , {( AB)(CD )} . For A{B (CD )} , total number of multiplication is 20 10 5 5 20 5 10 5 5 1750 For A{( BC ) D} , total number of multiplication is 5 20 10 5 10 5 10 5 5 1500 For {( AB)C}D , total number of multiplication is 10 5 20 10 20 10 10 10 5 3500 For { A( BC )}D , total number of multiplication is 5 20 10 10 5 10 10 10 5 2000 For {( AB)(CD )} , total number of multiplication is 10 5 20 20 10 5 10 20 5 3000 Hence minimum number of multiplication for find the product A1 A2 A3 A4 is 1500. 29. If the vectors e1 (1, 0, 2) , e 2 (0,1, 0) and e3 ( 2, 0,1) from an orthogonal basis of the three3
dimensional real space 3 , then the vector u (4,3, 3) can be expressed as [EC-2016 (2 marks)] (a) u (2 5)e1 3e 2 (11 5)e3 (b) u (2 5)e1 3e 2 (11 5)e3 (c) u (2 5)e1 3e 2 (11 5)e3
(d) u (2 5)e1 3e 2 (11 5)e3
Solution: The given vectors e1 , e 2 , e3 form an orthogonal basis of the three-dimensional real space 3
3 . So any vector u can be expressed as u x1e1 x2e 2 x3e3 . Thus we have
(4, 3, 3) x1 (1, 0, 2) x2 (0,1, 0) x3 ( 2, 0,1) (4,3, 3) ( x1 2 x3 ), ( x2 ), (2 x1 x3 ) ( x1 2 x3 ) 4 …(i), x2 3 …(ii), (2 x1 x3 ) 3 …(iii)
(i) 2 (iii) 5 x1 2 x1 2 5 ; from (iii), we have x3 3 2(2 5) 3 (4 5) 11 5 . So we have, u ( 2 5)e1 (3)e 2 ( 11 5)e 3 . Hence option (d) is correct.
a 2 30. The matrix A 0 0 Solution:
0
3
7
5
1
3
0
2
4
0
0
b
Expanding 5 1 3 0
has det( A) 100 and trace ( A) 14 . The value of a b is ___.
the 3 7
given
determinant
along
[EC-2016 (2 marks)] first column,
det( A) a 0
2 4 2 0 2 4 10ab , as det( A) 100 10 ab 100 ab 10 …(i). sum 0 0 b 0 0 b
of diagonal elements of any matrix gives trace of that matrix, so a 5 2 b 14 a b 7 …(ii). From (i) and (ii), we have ( a b) 2 ( a b) 2 4ab 7 2 4(10) 49 40 9 a b 3 . 31. Let the eigenvalues of a 2 2 matrix A be 1, –2 with eigenvectors x1 and x2 , respectively. Then the eigenvalues and eigenvectors of the matrix A 2 3 A 4 I would, respectively, be [EE-2016 (2 marks)] (a) 2, 14; x1 , x2 (b) 2, 14; x1 x2 , x1 x2 (c) 2, 0; x1 , x2 (d) 2, 0; x1 x2 , x1 x2 Solution (a): If x is the eigenvector of 2 2 matrix A corresponding to eigenvalues , then 2
2
2
Ax x …(i). On pre-multiplying both sides of (i) with A we get A x Ax A x x …(ii) Similarly, on pre-multiplying both sides of (i) by ( 3) , we get 3 Ax 3 x …(iii) We know that eigenvalue of any identity matrix is ‘1’ and for any vector ‘ x ’, Ix x So eigenvalue of ‘ 4I ’ is ‘4’ and for any vector ‘ x ’, 4 Ix 4 x …(iv). Now on adding (ii), (iii), (iv), 2 2 2 2 we get A x 3 Ax 4 Ix x 3 x 4 x ( A 3 A 4 I ) x ( 3 4) x …(v)
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [8]
2
Hence from (v), we can say that eigenvector of ( A 3 A 4 I ) is x corresponding to eigenvalue of
( 2 3 4) , where x is the eigenvector of A corresponding to eigenvalue . 2
Thus x1 is the eigenvector of ( A 3 A 4 I ) corresponding to corresponding to eigenvalue
(12 3(1) 4) 2 . 2
Similarly, x2 is the eigenvector of ( A 3 A 4 I ) corresponding to corresponding to eigenvalue
( 2) 2 3( 2) 4 14 . So option (a) is correct. 32. Let A be a 4 3 real matrix with rank 2. Which one of the following statement is TRUE? [EE-2016 (2 marks)] T T (a) Rank of A A is less than 2 (b) Rank of A A is 2 T (c) Rank of A A is greater than 2 (d) Rank of AT A can be any number between 1 and 3 Solution (b): rank [ A43 ] 2 [ A3T4 ] 2 . As rank( A B ) min{rank( A), rank( B )} . Order of AT A is 4 4 ; and order of AAT is 3 3 . T T T T So rank( A A) min{rank( A), rank( A )} rank( A A) min(2, 2) rank( A A) 2 T
T
T
T
Similarly, rank( AA ) min{rank( A), rank( A )} rank( AA ) min(2, 2) rank( AA ) 2 . Thus from given options, option (b) is correct.
3 1 x a x 33. Let P . Consider the set S of all vectors such that a 2 b 2 1 , where P 1 3 y b y . Then S is [EE-2016 (2 marks)] (a) a circle of radius
(b) a circle of radius 1
10 T
10 T
(c) an ellipse with major axis along 1 1
(d) an ellipse with minor axis along 1 1
a x a 3 1 x a 3 x y a 3 x y Solution (d): P b y b 1 3 y b x 3 y b x 3 y a 2 b 2 1 (3 x y ) 2 ( x 3 y ) 2 1 10 x 2 12 xy 10 y 2 1 0 …(i), which is 2nd degree curve. 2
2
On comparing (i) with ax 2hxy by 2 gx 2 fy c 0 , we have a 10 , h 6 , b 10 , g f 0,
c 1 .
As
a
h
g
10
6
h
b
f 6
10
g
f
c
0
0
0 0 1(100 36) 64 0 ;
also
1
h 2 ab 62 (10)(10) 64 h 2 ab 0 . So equation (i) is an ellipse. For finding the centre of the ellipse (i), we have to find the point of intersection of the equations which are found by partial 2 2 derivatives of (i) w.r.t. x and y . Let f ( x, y ) 10 x 12 xy 10 y 1 ; so f x 20 x 12 y 0 , and f y 12 x 20 y 0 ; on solving these equations we get x 0 , y 0 . Thus the major and minor axis of the ellipse (i) passes through the centre of ellipse (i), i.e. (0, 0) . Because of the term ‘ xy ’ in (i), the major and minor axis of the ellipse are not parallel to x or y axis. To remove the ‘ xy ’ term from (i), we need to rotate the x y axis by an angle
. Hence we can conclude that the major or minor axis of 2 a b 2 10 10 2 2 4 o o the ellipse (i) is at an angle of 45 with the ve x axis; or in other words if major axis is at 45
1
tan 1
2h
1
tan 1
2(6)
1
o
o
with the ve x axis then minor axis is at (90 45) 135 with the ve x axis or vice-versa. So o
equation of line passing through (0, 0) making an angle of 45 with the ve x axis is y x 0
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [9] o
…(ii); and equation of line passing through (0, 0) making an angle of 135 with the ve x axis is y x 0 …(iii). The point of intersection of (ii) with (i) is A and B , which is given by putting (ii) in (i)
10 x 2 12 x 2 10 x 2 1 0 x 1 4 2 y 1 4 2 . Thus A(1 4 2 ,1 4 2) and B (1 4 2 , 1 4 2) ; and so AB (2 4 2) 2 (2 4 2) 2 1 2 . Similarly, point of intersection of (iii) with (i) is C and D , which is given by putting (iii) in 2
(i) 10 x 12 x 10 x 1 0 x 1 2 2 y 1 2 2 . Thus C (1 2 2 , 1 2 2) and 2
2
D( 1 2 2 ,1 2 2) ; and so CD (2 2 2) 2 (2 2 2 ) 2 1 . As AB CD , so the line CD is the major axis whose equation is given (iii), i.e. y x 0 y x ; and the line AB is the minor axis whose equation is given (ii), i.e. y x 0 y x . Hence option (d) is correct.
2 1 1 34. Consider the matrix A 2 3 4 whose eigenvalues are 1, –1 and 3. Then Trace of 1 1 2 ( A 3 3A 2 ) is _____.
[IN-2016 (2 marks)] 3 1 ,
3 2,
3 3
Solution: If 1 , 2 , 3 are eigenvalues of matrix A , then are eigenvalues of matrix A3 ; and 312 , 322 , 332 are eigenvalues of matrix A2 . So (13 312 ), (23 322 ), (33 332 ) are the eigenvalues
of matrix
( A 3 3A 2 ) . Hence 3
{13 3(1) 2 },{( 1) 3 3( 1) 2 },{(3)3 3(3) 2 } 2, 4, 0
are the
2
eigenvalues of matrix ( A 3A ) . 3
2
As sum of eigenvalues of any matrix gives trace of that matrix, . trace( A 3A ) 2 4 0 6 .
2 1 0 35. The number of linearly independent eigenvectors of matrix A 0 2 0 is _____. 0 0 3 [ME-2016 (2 marks)] 2 1 0 Solution: For eigenvalues of the given matrix, we must have A I 0
0
2
0
0
0 0 3 (2 ){(3 )(2 ) 0} 0 2, 2, 3 . We know that if we have n distinct eigenvalues then we have n linearly independent eigenvectors. As we have only two eigenvalues, i.e. 2, 3 , so we have only two linearly eigenvectors. So, answer is two. 36. A force P 2iˆ 5 ˆj 6kˆ acts on a particle. The particle is moved from point A to point B, where the position vectors of the position vectors of A and B are 6iˆ ˆj 3kˆ and 4iˆ 3 ˆj 2kˆ respectively. The work done is _____. [MN-2016 (2 marks)] ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ Solution: AB (4i 3 j 2k ) (6i j 3k ) 2i 4 j k .
So required work done will be W F AB (2iˆ 5 ˆj 6kˆ) ( 2iˆ 4 ˆj kˆ) 4 20 6 22 37. The value of x in the simultaneous equations 3 x y 2 z 3 , 2 x 3 y z 3 , x 2 y z 4 , is _____. [MN-2016 (2 marks)] Solution: 3 x y 2 z 3 …(i), 2 x 3 y z 3 …(ii), x 2 y z 4 …(iii) (ii) (iii) 3 x y 1 …(iv); (i) 2 (ii) 7 x 5 y 3 …(v). 5 (iv) (v) 8 x 8 x 1 .
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Engineering Mathematics
Chapter – 1: Linear Algebra
Chapter – 1 [10]
5 3 . The normalised eigen-vector corresponding to the smallest 3 5
38. Consider the matrix, M
eigen-value of the matrix M is (a) 3 2 1 2
T
[PE-2016 (2 marks)]
(b) 3 2
1 2
T
(c) 1
2
1
Solution (c): For eigenvalues ( ) of M , we must have M I 0
2
T
(d) 1
5
3
3
5
2
1
2
T
0
(5 )(5 ) (3)(3) 0 2 10 16 0 ( 2)( 8) 0 2,8 So smallest eigenvalue of the matrix M is 2 . Hence eigenvector ( X ) corresponding to eigenvalue 2 is given by MX X 5 x1 3 x2 2 x1 x2 x1 ...(i) x1 5 x1 3 x2 2 x1 5 3 x1 x 3 x 5 x 2 x 3 x 5 x 2 x x x ...(ii) 3 5 x2 2 1 2 2 1 2 2 2 1 T
Thus X x1 x1 is the required eigenvector corresponding to eigenvalue 2 . Hence from given options, option (c) is the required normalised eigenvector.
3 39. The eigen values and eigen vectors of 4 T
4 3
T
(a) 5 and 1 2 , 2 1 respectively T
T
(c) 4 and 1 2 , 2 1 respectively Solution
(a):
For
eigenvalues
‘ ’
of
are
[TF-2016 (2 marks)] T
T
(b) 3 and 1 2 , 2 1 respectively T
T
(d) 5 and 1 1 , 2 1 respectively
3
4
, 4 3
A
A I 0
3
4
4
3
0
2
(3 )( 3 ) 16 0 25 5 . Eigenvector corresponding to 5 is given by Ax x : x1 3 x1 4 x2 5 x1 x1 2 x2 x1 2 x2 3 4 x1 2 5 x x2 4 3 x2 1 x2 4 x1 3x2 5 x2 x1 2 x2 x2 x2 Eigenvector corresponding to 5 is given by: x1 3 x1 4 x2 5 x1 x2 2 x1 x1 x1 3 4 x1 1 5 x x1 4 3 x2 2 x2 4 x1 3x2 5 x2 x2 2 x1 x2 2 x1
1 1 40. Let M [XE-2016 (2 marks)] . Which of the following is correct? 0 1 (a) Rank of M is 1 and M is not diagonalizable (b) Rank of M is 2 and M is diagonalizable (c) 1 is the only eigenvalues and M is not diagonalizable (d) 1 is the only eigenvalue and M is diagonalizable 1 1 Solution (c): For eigenvalues ‘ ’ of M , M I 0 0 (1 )(1 ) 0 0 0 1 1,1 . As eigenvalues of M are not distinct, so M is not diagonalizable. So option (c) is correct.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.1]
Chapter 1 : Linear Algebra 1.1
Matrices and Determinants
A system of m n numbers (real or complex) arranged in a rectangular form along m rows and n columns and bonded by the brackets [] is called m by n matrix (or m n matrix). Thus (1.1) is a matrix of order m n ; and in compact form, (1.1) can be represented as, a11 a12 a1n A aij , 1 i m , 1 j n , or simply A aij . The
m n
a
numbers aij , 1 i m , 1 j n , of the rectangular array, th
th
A
21
am1
th
a22
a2 n
(1.1)
belongs to the i row and j column and is called the (i, j ) am 2 amn mn element of the matrix. Equal Matrices: Two matrices are said to be equal if they have the same order and each element of one is equal to the corresponding element of the other. Example 1.1 [CS-2000 (1 mark)]: An n n array v is v i, j (i j )i , j ,1 i, j n . The sum of the elements of the array v is (b) n 1
(a) 0
n
Solution (a): v[i, j ]
n
n
(i j ) i i , j 1
i 1
j 1
defined
as
follows:
(d) n 2 ( n 1) 2 (c) n 2 3n 2 n( n 1) n(n 1) j 0. 2 2
Classification of Matrices Vectors: A matrix having a single column (or row) is called a column (or row) matrix or column (or row) vector. Its entries are called the components of the vector. It is denoted as 2 a a j ,1 j n . For e.g., is a column vector; and 1 3 513 is a row vector. 3 21 Square Matrix: A m n matrix A is said to be a square matrix if m n , i.e., number of rows is same as number of columns. The diagonal from left-hand side upper corner to right-hand side lower 1 4 corner is known as principle diagonal. For e.g., A is a square matrix of order 2 2 and 2 3 principal diagonal is the diagonal which contains the elements 1, 3. Diagonal Matrix: A square matrix all of whose elements, except those in the leading diagonal, are zero is called a diagonal matrix, aij 0 , whenever i j . A diagonal matrix of order n n having
3 0 is a 0 5
d1 , d 2 , , d n as diagonal elements is denoted by diag d1 , d 2 , , d n . For e.g., A
diagonal matrix of order 2 2 to be denoted by A diag 3 5 . Scalar Matrix: A diagonal matrix whose all the leading diagonal elements are equal is called a scalar 0, i j matrix. For a square matrix A aij to be a scalar matrix, aij , where m 0 . For n n m, i j
5 0 is a scalar matrix of order 2 2 . 0 5
e.g., A
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.2]
Unit matrix or Identity matrix: A diagonal matrix of order n which has unity for all diagonal elements, is called a unit matrix of order n and is denoted by I n . For a square matrix A aij
0, i j
0
1,
1
be a unit matrix, aij
1 . For e.g., I 2 i j 0
n n
to
is a unit matrix or order 2 2 .
Triangular matrix: A square matrix in which all the elements below/upper the diagonal are zero is called upper/lower triangular matrix. Thus, for any square matrix A aij
n n
, we have Upper
Triangular matrix if aij 0, i j ; and Lower Triangular matrix if aij 0, i j . [This point was
a d asked in AG-2011 (1 marks)]. For e.g., 0 b 0 0
e
a f and d c e
0
0
b
0 are Upper and Lower
f c Triangular matrices, respectively. Diagonal matrix is both Upper and Lower Triangular matrix.
Example 1.2 [CS-1994 (1 mark)]: In a compact single dimensional array representation for lower triangular matrices (i.e. all the elements above the diagonal are zero) of size n n , non-zero elements (i.e. elements of the lower triangle) of each row are stored one after another, starting from the first row, the index of the (i, j )th element of the lower triangular matrix in this new representation is: (a) i j (b) i j 1 (c) j {i (i 1) 2} (d) i { j ( j 1) 2} Solution (c): Let a n n lower triangular matrix We have to convert A a ij nn into A ak nn a 0 0 11 such that the elements a11 a1 , a21 a2 , a a 0 21 22 , a22 a3 , a31 a4 , a32 a5 , a33 a6 , and so on. is given as, A aij 0 From the given options, option (c) is correct, as for the option (c) all the criteria are met. an1 an 2 ann Null matrix: If all the elements of a matrix (square or rectangular) are zero, it is called a null or zero 0 0 matrix. For A aij to be null matrix, aij 0, i , j . For e.g., , 0 0 are null matrices. m n 0 0
1.1.1 Algebra of Matrices Addition and Subtraction of Matrices: Two matrices are added or subtracted if and only if they both have same order. If A aij and B bij are two matrix of same order, their addition m n mn or subtraction, A B ( for addition; for subtraction) is defined to be a matrix of order m n such that A B ij aij bij for i 1, 2, , m & j 1, 2, , n .
Addition of matrix is commutative [ A B B A ] as well as associative [( A B) C A ( B C )] . Existence of additive identity: If O be m n matric each of whose elements are zero, then A O A O A for every m n matrix A .
Existence of additive inverse: Let A aij
inverse of A (since, ( A) A O A ( A) ). Cancellation law holds good in case of addition of matrices, i.e., if A X B X A B . The equation A X 0 has a unique solution, A, X be m n matrix.
m n
, then A aij
mn
Matrix A is additive
Scalar Multiplication: The matrix obtained by multiplying every element of a matrix A by a scalar is called the scalar multiple of A by .
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Chapter 1: Linear Algebra
[1.3]
Scalar multiplication of matrices distributes over the addition k ( A B) kA kB , where k is any scalar. If A be any matrix then, ( p q ) A pA qA , where, p, q are any scalar. p (qA) pqA , where, p, q are any scalar. If A be any matrix then, ( p ) A ( pA) p ( A) , where, p is any scalar.
of
matrices,
i.e.,
Matrix Multiplication: Two matrices A and B are conformable for the product AB , if the number of columns in A (pre-multiplier) is same as the number of rows in B (post-multiplier) [This point was asked in CE-1997 (1 mark)]. Thus, if A aij
m n
and B bij
n p
are two matrices of
order m n and n p , respectively, then their product AB is of order m p and is defined as,
b1 j b 2j th th ain ( i row of A ) ( j bnj
r n
AB ij air brj ai1b1 j ai 2b2 j ain brn ai1
ai 2
r 1
column of B ); where, i 1, 2, , m and j 1, 2, , p Commutative law does not necessarily hold for matrices. If AB BA , then matrices A and B are called commutative matrices. If AB BA , then matrices A and B are called anticommutative matrices. Matrix multiplication is associative, i.e., A( BC ) ( AB)C . Matrix multiplication is distributive w.r.t. addition, i.e., A( B C ) AB AC . Cancellation law does not necessarily hold, i.e., if AB AC , then in general B C even if AO. 2 Matrix multiplication A A is represented as A . Thus, An AAA n times. If AB O , then it is not necessary that at least one of the matrix should be zero. [This point was 0 2 1 0 0 0 asked in AE-2007 (2 marks)]. For e.g. for A , B ; AB while 0 0 0 0 0 0 neither A nor B is null matrix. If A diag ( a1 , a2 , , an ) and B diag (b1 , b2 , , bn ) , then A B diag ( a1b1 , a2 b2 , , anbn ) . Thus, A n diag ( a1n , a2n , , a nn ) .
If A and B are diagonal matrices of same order, then AB BA or diagonal matrices (of same orders) are commutative. ( A B) 2 ( A B )( A B) A2 AB BA B 2 [This point was asked in AE-2007 (2 marks), ME-2014 (1 mark)] If A aij
m n
and B bij
n p
are two matrices of order m n and n p , then total number of
multiplication in AB is m n p .
cos
Example 1.3 [CS-1996 (2 marks)]: The matrices A
sin
sin
a 0 and B commute cos 0 b
under multiplication. (b) Always (c) Never (d) If a cos b sin (a) If a b or n , n I Solution (a): The two matrices will cos sin a 0 a cos b sin AB commute under multiplication if and only 0 b a sin b cos sin cos if AB BA b sin a sin either a b or sin 0 n , n I a 0 cos sin a cos a sin BA . Thus option (a) is correct. 0 b sin cos b sin b cos
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.4]
Example 1.4 [CS-1997 (2 marks)]: Let a aij be an n rowed square matrix and I12 be the matrix obtained by interchanging the first and second rows of the n rowed identity matrix. Then AI12 is such that its first (a) row is same as its second row (b) row is the same as the second row of A (c) column is same as the second column of A (d) row is all zero a11 a12 a1n 1 0 0 0 1 0
a 21 Solution (c): Let A an1 a11 a12 a a22 21 AI12 an1 an 2
a2 n
0 and I an 2 ann 0 a1n 0 1 0 a12 a2 n 1 0 0 a22 ann 0 0 1 an 2 a22
st
1 0
1 0 0 I12 1 0 0 1 a1n a2 n . ann
0
a11
a21
an1
nd
Hence 1 column of AI12 is same as 2 column of A . Example 1.5 [CE-2004 (1 mark)]: Real matrices [ A]31 , [ B ]33 , [C ]35 , [ D]53 , [ E ]55 and [ F ]51 are given. Matrices [ B ] and [ E ] are symmetric. Following statements are made with respect to these matrices. (I) Matrix product [ F ]T [C ]T [ B][C ][ F ] is a scalar; (II) Matrix product [ D]T [ F ][ D ] is always symmetric. With reference these two statements, which of the following applies? (a) Statement I is true but II is false (b) Statement I is false but II is true (c) Both the statements are true (d) Both the statements are false Solution (d): Statement 1 is false, since the matrix product , when it exists, is always another matrix and not a scalar. Matrix product [ D]T [ F ][ D ] is not possible as number of columns in [ F ] is not same as number of rows in [ D ] . Hence both statements are false. Example 1.6 [CE-2005 (1 mark)]: Consider the matrices [ X ]43 , [Y ]43 and [ P ]23 . The order of
P( X
T
1
Y) P
T T
will be (b) 3 3
(a) 2 2
(c) 4 3 1
T
Solution (a): The order of A P ( X Y ) P
T T
(d) 3 4 1
((2 3)((3 4)(4 3)) (3 2))T
equal
1
T
T
T T A ((2 3)(3 3) (3 2)) ((2 3)(3 3)(3 2)) ((2 3)(3 2)) (2 2) (2 2) . equal
equal
cos
sin
, then f ( ) f ( ) cos (c) f ( ) (d) 2 2 zero matrix
Example 1.7 [AE-2007 (1mark)]: If f ( )
sin
(b) f ( )
(a) f ( ) Solution (b):
cos
sin cos
sin
sin
cos sin
cos
f ( ) f ( )
cos( )
sin( )
f ( ) sin( ) cos( )
1 1 3 1 Example 1.8 [MN-2007 (2 marks)]: If A B and A B , the value of A B , is 3 0 1 4
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Engineering Mathematics
1 1 (a) 4 0 3
Chapter 1: Linear Algebra
1 1 (b) 2 0 3
[1.5]
1 1 (c) 0 3
1 1 1 (d) 2 0 3
1 1 3 1 4 0 2 0 Solution (b): ( A B ) ( A B) 2A A 3 0 1 4 4 4 2 2 1 1 3 1 2 2 1 1 Also, ( A B ) ( A B) 2B B 3 0 1 4 2 4 1 2 2 0 1 1 2 0 2 0 2 2 1 1 2 . 2 2 1 2 2 2 2 4 0 6 0 3
Thus AB
Example 1.9 [TF-2011 (1 mark)]: X and Y are two matrices such that XY and X Y are both defined. The CORRECT statement from amongst the following is (a) X and Y are square matrices of same order (b) X and Y are rectangular matrices (c) X and Y are diagonal matrices of different (d) X is a square matrix; Y is a rectangular order matrix Solution (a): Let matrix X is of order m1 n1 ; and Y is of order m2 n2 . So, for existence of XY we must have n1 m2 ; also for existence of X Y we must have m1 m2 and n1 n2 . Hence from the three relations: n1 m2 , m1 m2 and n1 n2 m1 n1 m2 n2 X and Y are square matrices of same order.
2 1 1 3 0 , B 2 1 2 ; then P is 3 0 2 4 2 4 7 (c) (d) 0 3 9 3 9
Example 1.10 [AG-2013 (1 mark)]: If P A B , where A
7 2 4 7 2 (a) (b) 3 9 0 9 0 Solution (b): As A is a 2 2 and B is a 2 3 matrix, so the product P A B exist whose order is 2 1 1 3 0 2 2 6 1 0 2 4 7 2 2 3 . Hence, P . 3 0 2 1 2 3 0 9 0 0 0 3 9 0 [Similar question was also asked in BT-2013 (1 mark)] Example 1.11 [CE-2013 (2 marks)]: There are three matrices P (4 2) , Q (2 4) and R(4 1) . The minimum multiplication required to compute the matrix PQR is ………… Solution: If we first multiply PQ and then ( PQ) R then, the total number of multiplication in PQ is 4 2 4 32 and in ( PQ) R is 4 4 1 16 . So total number of multiplication in PQR is 32 16 48 . If we first multiply QR and then P (QR) then, the total number of multiplication in QR is 2 4 1 8 and in P (QR) is 4 2 1 8 . So total number of multiplication in PQR is 8 8 16 . Hence answer is min(48,16) 16 . Example 1.12 [CS-2011 (2 marks)]: Four matrices M1 , M 2 , M 3 and M 4 are dimensions p q , q r , r s and s t respectively can be multiplied in several ways with different number of total scalar multiplications. For example when multiplied as ( M 1 M 2 ) ( M 3 M 4 ) the total number of scalar multiplication is pqr rst prt . When multiplied as
( M1 M 2 ) M 3 M 4 ,
the total
number of scalar multiplication is pqr prs pst . If p 10 , q 100 , r 20 , s 5 and t 80 then the minimum number of scalar multiplications needed is (a) 248000 (b) 44000 (c) 19000 (d) 25000 Solution (c): The possible ways of multiplying of four matrices M1 , M 2 , M 3 and M 4 is given as:
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Chapter 1: Linear Algebra
[1.6]
Multiplication (M1 M 2 ) M 3 M 4
Total number of Multiplications pqr prs pst 10 100 20 10 20 5 10 5 80 25000
M 1 (M 2 M 3 ) M 4
qrs pqs pst 100 20 5 10 100 5 10 5 80 19000
( M1 M 2 ) (M 3 M 4 )
pqr rst prt 10 100 20 20 5 80 10 20 80 44000
M1 (M 2 M 3 ) M 4
qrs qst pqt 100 20 5 100 5 80 10 100 80 130000
rst qrt pqt 20 5 80 100 20 80 10 100 80 248000 M1 M 2 (M 3 M 4 ) So minimum number of scalar multiplications needed is 19000, when given matrices are multiplied as M 1 (M 2 M 3 ) M 4 .
1.1.2 Properties of Matrices Trace of Matrix: The sum of the elements of a square matrix A lying along the principal diagonal i n
is called the trace of A , i.e., tr ( A) , i.e., if A aij
n n
, then, tr ( A) aii a11 a22 ann . For i 1
2 2 3 e.g., if A 3 2 3 then tr ( A) 2 2 2 6 . [This example was asked in MN-2008 (1 mark)]. 4 1 2
If A is an n n identity matrix then aii 1 for all 1 i n , hence tr ( A) 1 1 n - times n . [This point was asked in MT-2009 (1 mark)].
If A aij
n n
, B bij
tr ( A) tr ( A)
n n
be any two matrices and be any scalar. Then tr ( A B ) tr ( A) tr ( B)
tr ( AB ) tr ( BA)
Transpose of Matrix: The matrix obtained from any given matrix A, by interchanging rows and T columns, is called the transpose of A and is denoted by A or A . If A aij and AT bij , m n n m
1 2 1 4 7 then bij a ji , i , j . For example, if A 4 5 then, AT 2 5 8 23 7 8 32
( AT )T A
( A B )T AT BT
( A)T AT , where, being a scalar
AT A
( AB)T BT AT . [This point was asked in CS-1994 (1 mark)].
Proof: Let A aij
m n
and B bij
n p
be any two matrices. The AB is an m p matrix and T
T
T
T
therefore ( AB)T is a p m matrix. Since A and B are n m and p n , therefore, B A is a T
T
T p m matrix. Thus, the two matrices ( AB) and B A
( AB) T
ij
r n
r n
r n
A B
( AB) ji a jr bri bri a jr B r 1 T
r 1
r 1
T
are of the same order such that,
T
T
ir
rj
T
T
A
ij
. Hence, by the definition
T
of equal matrices, ( AB) BT AT = B A . In general, ( ABC )T C T B T AT . Example 1.13 [CS-1994 (1 mark)]: Let A and B be real symmetric matrices of size n n . Then which one of the following is true? T 1 (d) ( AB )T BA (c) AB BA (a) AA 1 (b) A A Solution (d): As, A and B ( AB )T BT AT ( AB )T BA
be real symmetric matrices
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T
A A
and
BT B .
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Chapter 1: Linear Algebra
Example 1.14 [PI-2011 (1 mark)]: If matrix 2 4 4 6 and B A , the transpose of 1 3 5 9
[1.7]
28 19 (a) 34 47
19 (b) 47
48 33 product of these two matrices, i.e., ( AB)T is (c) 28 19 equal to 2 4 4 6 28 48 28 19 T Solution (d): AB ( AB ) 48 33 . 1 3 5 9 19 33
34 28
28 19 (d) 48 33
3 2 1 1 Example 1.15 [CE-2014 (1 mark)]: Given the matrices J 2 4 2 and K 2 , the product 1 2 6 1 T K JK is _____.
3 2 1 1 1 Solution: K JK 1 2 1 2 4 2 2 8 12 11 2 43 . Hence answer is 43. 1 2 6 1 1 T
[Similar question was also asked in CE-2001 (2 marks)]
Conjugate of Matrix: The matrix obtained from any given matrix A containing complex number as its elements, on replacing its elements by the corresponding conjugate numbers is called conjugate 1 2i 2 3i 1 2i 2 3i of A and is denoted by A . For example, if A then, A 4 5i 5 6i . 4 5i 5 6i
( A) A
If A A A is purely imaginary matrix
( A B) A B
If A A A is purely real matrix
( A) A , being any number (real or complex)
AB AB ,
A and B being conformable for
multiplication
Transpose Conjugate of Matrix: The transpose conjugate of a matrix A is called transposed
conjugate of A and is denoted by A . The conjugate of the transpose of A is the same as the transpose of the conjugate of A , i.e., A ( A) A . If A aij
m n th
, then, A b ji , where, n m
b ji aij , i.e., ( j , i )th element of A is equal to the conjugate of (i, j ) element of A . For example,
1 2i
2 3i
1 2i 4 5i , then, A . 4 5i 5 6i 2 3i 5 6i
if, A
( A ) A
( A B ) A B
( kA) kA , k or
( AB) B A
1.1.3 Special Matrices Symmetric Matrix: A square matrix A [aij ] is called a symmetric matrix if aij a ji i , j , A ij
a A i, j . For e.g., A h ij g T
h
g
b
f is a symmetric matrix as A A .
f
T
c
Skew-Symmetric Matrix: A square matrix A [aij ] is called a skew-symmetric matrix if
aij a ji , i, j A ij AT
ij
i, j . [This point was asked in CE-2009 (1 mark)]. For e.g.,
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Chapter 1: Linear Algebra
[1.8]
0 2 3 T A 2 0 4 is a skew-symmetric matrix as A A . [This example was asked in AG-2012 3 4 0 (1 mark)] T n 1 T If A is a symmetric matrix, then, A, kA, A , A , A , B AB are also symmetric matrices, where, n N , k R and B is a square matrix of order that of A . If A is a skew-symmetric matrix, then 2n kA is a skew-symmetric matrix for k R A is a symmetric matrix for n N 2 n 1 T A is a skew-symmetric matrix for B AB is skew-symmetric matrix, where B is a square matrix of order that of A nN If A , B are two symmetric matrices, then A B , AB BA are also symmetric matrices AB BA is a skew-symmetric matrix AB is a symmetric matrix when AB BA If A , B are two skew-symmetric matrices, then AB BA is a symmetric matrices A B , AB BA are skew-symmetric matrices
If A is a skew-symmetric matrix and C is a column matrix, then C T AC is a zero matrix. The elements on the main diagonal of a skew-symmetric matrix are all zero. If A be any square matrix, then T T A A is always a symmetric matrix A A is always a skew-symmetric matrix T T AA & A A are symmetric matrices. [This point was asked in CE-1998 (1 mark)] Every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skewsymmetric matrix. [This point was asked in CE-2011 (1 mark)]. Proof: Let A be a square matrix. Then, A {( A AT ) 2} {( A AT ) 2} , let P ( A AT ) 2 and Q ( A AT ) 2 . ( A AT ) is a symmetric matrix P is a symmetric matrix. ( A AT ) is a skew-symmetric matrix Q is a skew-symmetric matrix.
Unitary Matrix: A square matrix A aij is said to be unitary if A A1 A A I n AA . The determinant of unitary matrix is of unit modulus.
Hermitian and Skew-Hermitian matrix: A square matrix A aij is said to be Hermitian matrix if aij aij , i, j A A . A square matrix A aij is said to be skew-Hermitian matrix a
if aij aij , i, j A A . For example, A
b ic
b ic
2 i 0 and A are d 0 2 i
Hermitian and skew-Hermitian matrix, respectively. If A is a Hermitian matrix, then aii aii aii is real i . Thus every diagonal element of a Hermitian matrix must be real. A Hermitian matrix over a set of real numbers is actually a real symmetric matrix. If A is a skew-Hermitian matrix, then, aii aii aii aii 0 , i.e., aii must be purely imaginary or zero. A skew-Hermitian matrix over the set of real numbers is actually a real skew-symmetric matrix.
Orthogonal Matrix: Any square matrix A of order n is said to be orthogonal, if AT A 1 AAT AT A I n [This point was asked in MN-2009, MN-2013 (1 mark)]. Determinant of any orthogonal matrix is always 1 . [This point was asked in ME-2006 (1 mark)]. 2
Proof: AAT I n AAT I n A AT 1 A 1 A 1 ( A AT ).
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.9]
l 0 sin Example 1.16 [XE-2008 (2 marks)]: The matrix A 0 1 m is orthogonal, if n 0 cos (a) l sin , m cos , n 0 (c) l cos , m sin , n 0
(b) l sin , m 0, n cos (d) l cos , m 0, n sin 0 n l 0 sin l
1 0 Solution (d): For orthogonal matrix, AA I 0 1 m 0 1 0 0 1 n 0 cos sin m cos 0 0 l 2 sin 2 m sin nl sin cos 1 0 0 m sin 1 m2 m cos l 2 sin 2 1 l cos 0 1 0 2 2 nl sin cos m cos n cos 0 0 1 nl sin cos 0 (ii) , 1 m 2 1 m 0 (iii) , n 2 cos 2 1 n sin (iv) . From we can conclude that l and n must be of opposite signs; so values in option (d) is correct. T
0 0
1 (i) ,
(ii)
Idempotent Matrix: Any square matrix A of order n is said to be idempotent, if A2 A . Every idempotent matrix is either null matrix or identity matrix. A2 A A( A I ) O A 0 or A I .
Involuntary Matrix: Any square matrix A of order n is said to be involuntary, if A2 I . Nilpotent Matrix: A square matrix A is called a nilpotent matrix if there exist a positive integer m m1 m 2 such that Am O and A O, A O , A O . If m is the least positive integer such that m A O , then m is called the index of the nilpotent matrix A . Permutation matrices: It is a square matrix whose entries are all 0’s and 1’s, with exactly single 0 1 0 0 1 ‘1’ in each row and exactly single ‘1’ in each column. For e.g. 0 0 1 are permutation 1 0 1 0 0 matrices. Equivalently, permutation matrix is the identity matrix I with its rows rearranged. A T permutation matrix is non-singular, and satisfies AA I , thus the determinant is always 1 .
1.1.4 Determinant of a Square Matrix To every square matrix A aij of order n , where, aij (i, j )
th
element of A , a number (real or
complex) is associated called determinant of square matrix A . It is denoted by det( A) or A or . For non-square matrix, the determinant is not defined. [This point was asked in ME-2006 (1 mark)]. a d a d If A , then determinant of A is written as A . b c b c
A square matrix A aij is said to be singular if A 0 . [This point was asked in EE-1997, ME-2006 (1 mark)].
A square matrix A aij is said to be non-singular if A 0 .
If A and B are two singular matrices then A B may or may not be singular. If A and B are two non-singular matrices then A B may or may not be singular. [The above two points were asked in CS-2001 (1 mark)]
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.10]
8 x 0 Example 1.17 [ME-2004 (2 marks)]: If matrix 4 0 2 is singular, find value of x . 12 6 0 (a) 4
(b) 6
(c) 8 x 0
8
(d) 12
Solution (a): If given matrix to be singular, 4
0
2 0 8(0 12) x(0 24) 0 x 4
12
6
0
Example 1.18 [MT-2013 (2 marks)]: Which one of the following attributes is NOT correct for the cos sin 0 matrix? sin
0
cos
0 , where 60o
1 (b) Singular
0
(a) Orthogonal
(c) skew-symmetric
1 2 3 2 0 12 T Solution (b, c): At 60o , A 3 2 12 0 A 3 2 0 0 0 1
(d) positive-definite 3 2 12 0
0
0 A A is 1
not skew-symmetric.
1 2 3 2 0 1 2 T AA 3 2 12 0 3 2 0 0 1 0 12 A
3 2
3 2
12 0
0
1 0 0 0 0 1 0 A is orthogonal matrix. 1 0 0 1
0
1 3 0 1 1 A is non-singular matrix and positive definite. 4 4 1
12
0
3 2
0
[Please note that, from answer keys of MT 2013 from GATE website, marks to all was given]
Minors and Cofactors: If A is a square matrix, then the minor of the entry in the ith row and j th column is the determinant of the sub-matrix formed by deleting the ith row and j th column. This number is often denoted by M ij . The corresponding cofactor ( Cij ) is obtained by multiplying the minor by ( 1)i j , i.e., Cij (1)
a For, e.g., if A b c
i j
M ij .
d
g
e
h , then M 11
f
i
e
h
f
i
; M 12
b
h
c
i
; M 13
b
e
c
f
; and so on.
11
Also, C11 ( 1) M 11 ; C12 ( 1)1 2 M 12 ; C13 (1)13 M 13 and so on.
The value of the determinant is the sum of product of the elements of any row or column with the a11 a12 a13 corresponding cofactors. For e.g., A a21
a22
a23 then, in terms of notation of the cofactors,
a31
a32
a33
A a11C11 a12 C12 a13C13 a21C21 a22C22 a23C13 a31C31 a32C32 a33C33 A a11C11 a21C 21 a31C31 a12 C12 a22C 22 a32C32 a13C13 a23C23 a33C33
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e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.11]
The sum of product of the elements of any row or column with the cofactors of the corresponding elements of any other row or column is zero, i.e., a11C21 a12C22 a13C23 0; a11C31 a12 C32 a13C33 0
If A1 , A2 , , An are square matrix of same order, then A1 A2 An A1 A2 An
If k is scalar, then kA k n A , where, n is the order of matrix A . [This point was asked in CE-1999 (1 mark), MT-2012 (1 mark)] If A and B are square matrices of same order then AB BA even if AB BA
The determinant of a triangular matrix is the product of all elements along principle diagonal. If A and B are two n n square matrices then det( AB ) (det A)(det B) . [This point was asked in EC-2014 (1 mark)] If A and B are two n n square matrices and AB O (i.e., AB is a null matrix) then det( AB ) 0 (det A)(det B) 0 and so we have the following cases: (i) A 0 and B 0 (ii) A 0 and B 0 (iii) A 0 and B 0 . [This point was asked in IN-2010 (2 marks)].
6 8 0 2 Example 1.19 [CS-1997 (1 mark)]: The determinant of the matrix 0 0 0 0
1
1
4
6
4
8
is
0 1 (a) 11 (b) –48 (c) 0 (d) –24 Solution (b): Since the given matrix is triangular (upper) matrix, hence its determinant is the product of all elements along principle diagonal, which is equal to 6 2 4 1 48 .
5 3 2 Example 1.20 [CE-2001 (2 marks)]: Determinant of the matrix 1 2 6 is 3 5 10 (a) –76 Solution (c): 5 3 2 1
2
6 5
(b) –28 2
6
5 10
3
1
(c) 28 6
3 10
2
1
2
3 5
(d) 72
5(20 30) 3(10 18) 2(5 6) 28
3 5 10 [Similar questions were also asked in PI-2009, MT-2008 (1 mark)
3 4 3 Example 1.21 [TF-2010 (1 mark)]: Cofactor of ‘ a ’ in the following determinant 2 7 a is 5 9 2 (a) 3
(b) 5
Solution (c): a C23 ( 1) 2 3 M 23 1
(c) 7 3
4
5
9
(d) 9
1( 27 20) 7 .
1
3
0
2
6
Example 1.22 [ME-2014 (1 mark)]: If A 2
6
4 12 is, then 4
12
0 8 is
1 0
2 2 0 4 (a) –96 (b) –24 (c) 24 (d) 96 3 Solution (d): B 2 A B 2 A 2 A (as A is of order 3). Hence B 2 3 12 96 .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.12]
Value of nth-order determinant:
a
21
an1
When expanding along first row a11 a12 a1n a2 n
a22
a
j n
a1 j C1 j j 1 ann
an 2
When expanding along first column a11 a12 a1n
a22
21
an1
an 2
a2 n
in
ai1Ci1 i 1 ann
For easier calculations, we shall expand the determinant along that row or column which contains maximum number of zeros. Sarrus rule for Expansion (Valid only for determinant of order 3): The three diagonals sloping down to the right give the three positive terms and the three diagonals sloping down to left the three negative terms. a1 b1 c1 a2
b2
c2 a1b2 c3 b1c2 a3 c1a2 b3 a3b2 c1 b3 c2 a1 c3 a2 b1
a3
b3
c3
th
For n order determinant we have n! number of terms in its expansion. [This point was asked in CE-1999 (1 mark)] 0 0 0 0 0 1
0 0 Example 1.23 [CS-2014 (2 marks)]: Consider the matrix J 6 0 0 1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
1
0
0
0
0
0
0
0
0
0
which is
obtained by reversing the order of the columns of the identity matrix I 6 . Let P I 6 J 6 , where is a non-negative real number. The value of for which det( P ) 0 is …………… Solution: 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0
0 0 P I6 J6 0 0 0
0 1 0 0 1 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 Now the det( P ) 0 if and only if any two row or column are same which results 1
0
0
0
0 0
0
0
0
0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
0
1 4 Example 1.24 [BT-2011 (2 marks)]: Value of the determinant 1 2 (a) 24 (b) –30 Solution (c): Expanding along first row we get, 7 0 2 4 7 2 1 1
1 1 0 1 1
1
0
1 0
7
0
1
1 1
0
2
(c) –24
0
0
0
0
1
0
0
0
1
0
0 0 1.
1
2
is
1 (d) –10
1 7( 1 2) 0 2(2 0) 2(7 2) 0 1(4 7) 24
0 2 1 2 0 1 [Similar questions were also asked in CS-2000, CE-2014 (1 mark)]
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.13]
1.1.5 Operations in Determinants First, second and third rows of a determinant are denoted by R1 , R2 , and R3 , respectively; and the first, second and third columns of a determinant are denoted by C1 , C 2 , and C3 , respectively.
The interchange of ith and j th row is denoted by Ri R j . The interchange of ith and j th column is denoted by Ci C j
The addition of m times the elements of j th row of the corresponding elements of ith row is denoted by Ri Ri mR j
The addition of m times the elements of j th column of the corresponding elements of ith column is denoted by Ci Ci mC j . The addition of m times the elements of j th row to ntimes the elements of ith row is denoted by Ri nRi mR j
1.1.6 Properties of Determinants th
All the properties given below are applicable for n order of determinant 1. The value of the determinant is not changed when rows are changed into corresponding columns. Naturally, when rows are changed into corresponding columns, then the columns will change into corresponding rows. 2. If any two rows or columns of a determinant are interchanged, the sign of the value of the determinant is changed. 3. The value of the determinant is zero if any two rows or columns are identical. 4. A common factor of all elements of any row (or of any column) may be taken outside the sign of the determinant. In other words, if all the elements of the same row (or the same column) are multiplied by a certain number, then the determinant will become multiplied by that number. 5. If every element of some column (or row) is the sum of two terms, then the determinant is equal to the sum of two determinants; one containing only the first term in place of each sum, the other only the second term. The remaining elements of both determinants are the same as in the given a1 1 b1 c1 a1 b1 c1 1 b1 c1 determinant, i.e., a2 2
c2 a2
b2
b2
c2 2
b2
c2
a3 3 b3 c3 a3 b3 c3 3 b3 c3 6. The value of the determinant does not changed when any row or column is multiplied by a number or an expression and is then added to or subtracted from any other row or column, i.e., a1 b1 c1 a1 mb1 b1 c1 a2
c2 a2 mb2
b2
b2
c2
a3 b3 c3 a3 mb3 b3 c3 It should be noted that if the row or column which is changed is multiplied by a number, then the determinant will have to be divided by that number. If more than one operation like Ri Ri kR j is done in one step, care should be taken to see
that a row that is affected in one operation should not be used in another operation. Many times we use this operation to get as many zeros as we can. r n
r n
7.
r r 1
rn
f1 ( r )
r 1
r n
f 2 (r )
r 1
f3 (r ) r 1
a
b
c
d
e
f
, if r
f1 (r )
f 2 (r )
f3 (r )
a
b
c
d
e
f
, where, f1 ( r ), f 2 ( r ), f3 ( r )
are function of r and a, b, c, d , e, f are constant.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
q
q
q
p f1 ( x) p f 2 ( x) p f2 ( x) Also
q
p ( x )
[1.14]
a
b
c
d
e
f
f1 ( x )
f 2 ( x)
f3 ( x)
a
b
c
d
e
f
( x )
, if
where,
f1 ( x ), f2 ( x ), f3 ( x ) are function of x and a, b, c, d , e, f are constant. 1
2 7 8 Example 1.25 [CE-1997 (1 mark)]: If 0 5 6 26 , and A 0 5 6 then find determinant 2 7 8 1 3 2 of matrix A is (a) –26
(b) 26 2 7 8
Solution (a): Let, 0
5
3
2
1
3
(c) 0 2
(d) 52
6 0
5
6 [By applying R1 R3 and by using Property
1 3 2 2 7 8 no. 2], so from the given value of the determinant, we have 26 . 1 a
Example 1.26 [CS-1998 (2 marks)]: Consider the following determinant 1 b 1 c
bc ca . Which of ab
the following is a factor of . (a) a b (b) a b (c) a b c (d) abc Solution (b): Multiplying R1 by a ; R2 by b ; R3 by c and then taking the common term ( abc ) out a
a2
1
of the determinant (by using Property no. 4) from C3 we get, b
b2
1 . By applying
c
2
c
1
(a b )
(a b)( a b)
0
R1 R1 R2 and R2 R2 R3 , we get (b c )
(b c )(b c )
0 . Now by taking common
c
c
2
1
term (from R1 and R2 ) by using Property no. 4, and further solving it we get, (a b)(b c )(c a ) . So from the given options the common factor of is ( a b) . 2
1
1
Example 1.27 [CE-1999 (2 marks)]: The equation 1
1
1 0 , represent a parabola passing
y
2
x
x
through the points (a) (0,1), (0, 2), (0, 1) (b) (0, 0), ( 1,1), (1, 2) (c) (1,1), (0, 0), (2, 2) Solution (b): The value of the given determinant is calculated as, 2 1 1 3 2 0 1
1
1 0 R 1 R1 R2 1
1
(d) (1, 2), (2,1), (0, 0)
1 0 3( x x 2 ) 2( x y ) 0 0 2 y 3x 2 x
y x2 x y x2 x Now we have to check all the given options and we find that option (b) is the correct answer as (0, 0), ( 1,1), (1, 2) are the points which satisfies the equation of parabola 2 y 3 x 2 x .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.15]
1 3 2 Example 1.28 [MT-2007 (1 mark)]: The determinant of the matrix 2 6 4 is 5 3 1 (a) –10 1
Solution (c): 2
(b) –5 3 2 6
(c) 0 1 3
(d) 5 2
3 2 0 , since R2 R1 .
4 R2 R2 2 R1 1
5 3 1 5 3 1 [Similar question was also asked in TF-2008 (1 mark) 1 b
Example 1.29 [PI-2007 (1 mark)]: The determinant
b
b
1
1 b 1 evaluates to
1
(a) 0
2b 1 (c) 2(1 b)(1 2b)
(b) 2b(b 1) 1 b
Solution (a):
b 1
1
1
1
0
1 b 1
b
1 b
1
b 2b
1 b
1
(d) 3b(1 b) R1 R1 R2
[Applying
and
b 1 0
R3 R3 R2 ]. Now expanding along C3 0 1(b 1 1 b) 0 0 . 1
x
x2
Example 1.30 [CS-2013 (1 mark)]: Which one of the following does not equal 1
y
y2 ?
1
z
z
2
1
x ( x 1)
x 1
1
x 1
x2 1
0
x y
x2 y 2
2
x y
x2 y 2
(a) 1
y ( y 1)
y 1
(b) 1
y 1
y2 1
(c) 0
yz
y2 z2
(d) 2
yz
y2 z2
1
z ( z 1)
z 1
1
z 1
z 1
z
z
z
z
2
1
2
1
0
x y
x2 y 2
Solution (c): Applying R1 R1 R2 and R2 R2 R3 , we get 0
yz
y2 z2 .
1
z
z
2
2
Example 1.31 [MT-2014 (1 mark)]: If all the elements of one row of a 3 3 matrix are multiplied by 3, the determinant of the matrix changes by a factor of ……………. a11 a12 a13 3a11 3a12 3a13 a11 a12 a13 Solution: Let A a21
a22
a23 and A a21
a22
a23 3 a21
a22
a23 3 A . Hence the
a31
a32
a33
a32
a33
a32
a33
a31
a31
answer is 3.
2 Example 1.32 [CS-2014 (1 mark)]: If A 4 1 9 5 then A 7 2 18 2 2 18 10 Solution: A 4 1 9 5 4 36 20 A 4 36 7 63 7 7 63 35
………… 10
2
2
2
20 9 5 4
4
4
35
7
7
7
(Take common 9 from C 2 and 5 from C3 ) all the three columns of A are same A 0
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e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.16]
a1
b1
c1
1
1
1
Product of two determinants: Let, 1 a2
b2
c2 and 2 2
2
2 . Then row by row
a3
b3
c3
3
3
3
multiplication
1
of
2
and
is
a11 b11 c1 1
a1 2 b1 2 c1 2
a1 3 b1 3 c1 3
1 2 a21 b2 1 c2 1
a2 2 b2 2 c2 2
a2 3 b2 3 c2 3
given
by
a31 b3 1 c3 1 a3 2 b3 2 c3 2 a3 3 b3 3 c3 3 Multiplication can also be performed ‘row by column’ or ‘column by row’ or ‘column by column’ If 1 , 1 , 1 are the cofactors of the elements a1 , b1 , c1 of the determinant a1
b1
c1
1
1
1
a2
b2
c2 , 0, then, 2
2
2 2
3
3
3
a3 b3 c3 Proof: a1 b1 c1 1
1
1
a11 b11 c1 1
a1 2 b1 2 c1 2
a1 3 b1 3 c1 3
a2
b2
c2 2
2
2 a21 b2 1 c2 1
a2 2 b2 2 c2 2
a2 3 b2 3 c2 3
a3
b3
c3 3
3
3
a3 2 b3 2 c3 2
a3 3 b3 3 c3 3
a31 b3 1 c3 1
ai i bi i ci i , i 1, 2,3 and ai j bi j ci j 0
1
1
1
1
1
1
2
2 2
3
3
3
2
2
2 2
3
3
3
3
n 1
For n order determinant c , where, c is the determinant formed by the cofactors of and n is order of determinant. If two determinants of order n are multiplied then the resultant determinant is also of order n [This point was asked in MN-2010 (1 mark)].
Differentiation of a Determinant:
Let ( x ) be a determinant of order two. If we write ( x ) C1 the first and second column then, ( x ) C1
C2 C1
C2 , where, C1 and C 2 denotes
C 2 , where, Ci denotes the column
th
which contains the derivative of all the functions in the i column. R1 R1 R1 In similar fashion, if we write, ( x ) , then ( x ) , where, Ri denotes R2 R2 R2
the row which contains the derivative of all the functions in the ith row. Let ( x ) be a determinant of order three. If we write ( x ) C1 C2 C3 , where, C1 , C 2 and C3
denotes
( x) C1
C2
the C3 C1
first, C2
second
C3 C1
C2
and
third
column
then,
C3 , where, Ci denotes the column which
contains the derivative of all the functions in the ith column. R1
R1 R1 R1 In similar fashion, if we write, ( x ) R2 , then ( x) R2 R2 R2 , where, Ri R3 R3 R3 R3 denotes the row which contains the derivative of all the functions in the ith row.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.17]
Example 1.33 [TF-2012 (2 marks)]: Consider the following Assertion [A] and Reason [B]. 0 0 1 A. M is an orthogonal matrix, but not a skew-symmetric matrix. M 0
cos
0 sin
1
T
sin
cos
T
B. Because M M and M M Determine the correctness or otherwise of the above Assertion [A] and Reason [B] (a) A is right B is wrong (b) A is right B is right (c) A is wrong B is right (d) A is wrong B is wrong T Solution (b): If M is an orthogonal matrix then MM I 0 0 1 0 0 1 0 0 1 MM T 0
cos
0 sin
0 1 M 0 cos 0 sin T
sin 0
cos sin 0 1 0 I MM T I M T M 1 . cos 0 sin cos 0 0 1 0 sin M M is not skew-symmetric matrix. So (A), (B) both are correct. cos
Example 1.34 [EC-2013 (2 marks)]: Let A be an m n matrix and B an n m matrix. It is given that det( I m AB) det( I n BA) , where I k is k k identity matrix. Using the above property, the
2 1 determinant of the matrix 1 1
1
1
1
2
1
1
1
2
1
is
1 1 2 (b) 5 2 1 1 1
(a) 2
1 Solution (b): Let I n BA 1 1 1 1 1 1 1 1 1 1 A 1 BA 1 1 1 1 1 1 1 1
2 1 1
1 1
(c) 8 1 1 1 1
(d) 16 1 0
0 0 1 1 1 1 1 1 0 1 0 0 BA if I n I 4 . Now, 2 1 1 1 1 1 0 0 1 0 1 2 1 1 1 1 0 0 0 1 1 1 11 4 and B AB 4 and I m I1 1 . 1 1 41
So, det( I m AB ) det( I n BA) det( I n BA) 1 4 5 5 .
1.1.7 Adjoint and Inverse of a Square matrix Adjoint of a square matrix: Let A aij be a square matrix of order n and let Cij be cofactor of aij in A . Then, the transpose of the matrix of cofactors of elements of A is called the adjoint of A T
and is denoted by ‘ adjA ’. Thus, adjA Cij adjA ij C ji .
Inverse of a square Matrix: A non-singular square matrix of order n is invertible if there exist a square matrix B of same order such that AB I n BA . In such a case we say that the inverse of A 1
is B and write A B . From the properties of adjoint and inverse of a matrix A1 adj ( A) A . For A
1
to be defined, A 0 A must be non-singular matrix.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.18]
1 0 1 Example 1.35 [CS-1994 (2 marks)]: Find the inverse of the matrix A 1 1 1 . 0 1 0 Solution: Since A 1(0 1) 0 1( 1 0) 2 0 , hence A is invertible. Let Cij be cofactor of aij in A . Then the cofactors of elements of A are given by: C11
1 1 1 0
C23
1
1 , 0
0 1
C12
1 , C31
1 1 0
0
0 1 1 1
0,
C13
1 , C32
1 1 0
1
1 1
1 1
1 ,
C21
2 , C33
1
0 1 1 0
1 1
0
1,
C22
1
1
0
0
0,
1.
T
1 0 1 1 1 1 1 2 1 2 1 2 adjA 1 adjA 1 0 1 0 0 2 A 0 0 1 A 1 1 1 1 2 1 1 2 1 2 1 2 [Similar questions were also asked in CE-1997 (1 mark), MN-2012 (2 marks)]
a11
a12
b11
b12
1 0 and C A 1 1 be a21 a22 b21 b22 matrices such that AB I and CD 1 , D is any matrix. Express the elements of D in terms of the elements of B . 1 0 1 1 Solution: AB I A1 AB A1 I B A 1 . Let P C AP A C A AP 1 1 Example 1.36 [CS-1996 (5 marks)]: Let A
, B
1 1 1 BC P BCD PD BI PD ( CD I ). Now PD B P PD P B D P B .
T
1 1 1 0 Now we have to find P as, P (1 0) 1 and adjA Cij 0 1 1 1 b12 adjA 1 0 1 0 b11 b12 b11 1 1 . P D P B 1 1 b A 1 1 21 b22 b11 b21 b12 b22 T
1
Example 1.37 [CE-1998 (1 mark)]: In matrix algebra AS AT ( A, S , T are matrices of appropriate order) implies S T only if A is (a) Symmetric (b) Singular (c) Non-singular (d) Skew-symmetric 1 1 Solution (c): We have AS AT , multiplying both sides by A we get, A AS A1 AT . Now, if A1 A I then we have S T and this happens when A1 exists and for A1 existence, A must be a non-singular matrix. Example 1.38 [ME-2006 (2 marks)]: Multiplication of matrices E and F is G , where, cos sin 0 1 0 0 E sin
0
cos
0 , G 0 1
1
0 What is the matrix F ?
0 . 0 0 1
cos sin (a) sin cos 0 0 cos sin (c) sin cos 0 0
0 0
1 0 0 1
sin cos (b) cos sin 0 0 sin cos (d) cos sin 0 0
0 0
1 0 0
1
Solution (c): EF G E 1 EF E 1G F E 1G . Also since G is an identity matrix hence 1 1 1 1 F E G E I E . Hence we have to find E .
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e-mail: [email protected]
Engineering Mathematics
cos For E sin 0
Chapter 1: Linear Algebra
[1.19]
C11 cos , C12 sin , C13 0
sin
0
cos
0 , the cofactors can be written as, C 21 sin , C22 cos , C23 0
1
0 2
C31 0, C32 0, C33 1
2
E 0 0 1(cos sin ) 1 . T cos T adjE Cij 1 F E Cij sin E E 0
sin cos 0
T
cos 0 sin 1 0
0
sin
0
cos
0
0
1
Properties of adjoint and inverse of a matrix Let A be square matrix of order n , then A(adjA) A I n (adjA) A Proof: Let A aij , and Cij be cofactor of aij in A . Then ( adjA)ij C ji , 1 i , j n r n
n
A, i j
( A( adjA)ij ) Air (adjA) rj air C jr r 1
A 0 A( adjA) 0 0
0,
r 1
i j
0
0
0
A
0
0
0
A
0 A I n ; Similarly, ( adjA) A A I n can be proved.
0
0
0
A
(Reversal Law) If A and B are invertible matrices of the same order, then AB is invertible and ( AB ) 1 B 1 A1 . In general, if are invertible matrices, then A, B , C , ( A, B, C , ) 1 C 1 B 1 A1 . [This point was asked in CE-2000 (1 mark)]
Proof: It is given that A and B are invertible matrices A 0 & B 0 A B 0 AB 0 AB A B , Hence, AB is an invertible matrix.
Now, ( AB )( B 1 A 1 ) A( BB 1 ) A 1 ( AI n ) A1 AA1 I n Also, ( B 1 A1 )( AB ) B ( AA1 ) B 1 ( BI n ) B 1 BB 1 I n ( AB )( B 1 A 1 ) I n ( B 1 A 1 )( AB ) ( AB ) 1 B 1 A1
T
If A is an invertible square matrix, then A is also invertible and ( AT ) 1 ( A1 )T T
Proof: A is an invertible matrix A 0 AT 0 A is also invertible matrix. Now, AA 1 I n A 1 A ( AA1 )T ( I n )T ( A1 A)T ( A1 )T AT I n AT ( A 1 )T ( AT ) 1 ( A1 )T
If A is a non-singular square matrix of order n , then adjA A
A 0 Proof: A( adjA) A I n 0 n
0
0
A
0
0
0
A
n
A( adjA) A A ( adjA) A ( adjA) A
n 1
n 1
Reversal Law for adjoint: If A and B are non-singular square matrices of the same order, then
adj ( AB ) ( adjB )(adjA) ( AB ) 1 B 1 A1
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.20]
If A is an invertible square matrix, then adj ( AT ) ( adjA)T ( AT ) 1 ( A1 )T
If A is a non-singular square matrix, then adj ( adjA) A
n 2
A
Proof: B ( adjB ) B I n for every square matrix of order n . Replacing B by adjA , we get, ( adjA)[ adj ( adjA)] adjA I n A ( AadjA)[ adj (adjA)] A A I n [adj ( adjA)] A A [ I n adj ( adjA)] A
n 1 n 1
n 1
n 1
I n A(adjA)[adj ( adjA)] A A
n 1
In
( AI n )
A AI n A & AadjA A I n A [ adj ( adjA)] A
If A is a non-singular matrix, then A
1
A 1
1
n2
A
A1 1 A 1
Proof: A 0 A1 exist such that AA I A A AA1 I 1 A A1 1 A1 1 A A 0
Inverse of k th power of A is the k th power of the inverse of A , i.e., ( A1 ) k ( Ak ) 1
( Ak ) 1 ( A A A) 1 A1 A1 A1 ( A1 ) k . Example 1.39 [CS-2004 (1 mark)]: Let A, B, C , D be n n matrices, with each non-zero 1
determinant. If ABCD I , then B is 1 1 1 (d) Does not necessarily exist (b) CDA (c) ADC (a) D C A 1 Solution (b): Pre-multiplying both sides by A , A1 ABCD A1 I A1 BCD A1 Taking inverse on both sides, ( BCD) 1 ( A1 ) 1 D 1C 1 B 1 A Pre-multiplying both sides by D DD 1C 1 B 1 DA C 1 B 1 DA Pre-multiplying both sides by C CC 1 B 1 CDA B 1 CDA
2 0.1 1 2 a 1 and A 0 b . Then ( a b) 3 0
Example 1.40 [EC-2005 (2 marks)]: Let A (a) 7/20 Solution (a):
(b) 3/20
2 1 AA I 0
0.1 1 / 2 3 0
(c) 19/60
(d) 11/20
1 0 1 2a 0.1b 1 0 2a 0.1b 0 (i) b 0 1 3b 0 1 3b 1 (ii) 0 a
(ii) b 1 3 and thus (i) a 0.1 6 a b 1 3 0.1 6 7 20 . Example 1.41 [EC-2005 (2 marks)]: Given an orthogonal matrix 1 1 1 1
1
A
1
1 1 0 0
is
1 1 0 1
, then AAT 1 0 1
1 4 0 0 14 (a) 0 0 0 0 1 0 0 0 1 0 (c) 0 0 1 0 0 0
0
0 0 12 0 0 1 2 0 0 0 1
Solution (c): As A is an orthogonal matrix AAT I and I
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0
1
0 0 1 2 0 0 12 0 0 (b) 0 0 12 0 0 0 1 2 0 0 0 1 4 0 0 14 0 0 (d) 0 0 14 0 0 0 1 4 0
I.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.21]
1 0 1 1 Example 1.42 [EE-2005 (2 marks)]: If R 2 1 1 , the top row of R is 2 3 2 (a) 5
4
6
(b) 5
3 1
1
0
1
Solution (b): R 2
1
1 1
2
3
2
1
1
3
2
C11 C12 adj ( A) Cij 1 A C21 C22 A A C31 C32
hence top row of R
1
1
1
3
2
0 1
C13
2 1 2
3
2
T
5 , C21 (1)
is given as, R1 : 5
21
0
1
(d)
M 21
C21 C22
C32 .
C 23
C33
0
1
3
2
3 , C31 ( 1)
sin
cos
cos (b) sin
31
M 31
0
1
1
1
1,
3 1 .
cos
sin
1 1 / 2
C31
Example 1.43 [TF-2008 (1 mark)]: The inverse of the matrix A
cos (a) sin
2
(2 3) (6 2) 5 4 1
C11 1 C23 C12 A C33 C13
T
As C11 ( 1)11 M 11
(c)
sin cos
sin (c) sin
cos cos
sin
is cos
cos (d) sin
sin cos
Solution (b): A cos 2 sin 2 1 . So A is invertible. C11 ( 1)11 M 11 (1)11 cos cos , C12 ( 1)1 2 M 12 (1)1 2 sin sin C21 ( 1) 2 1 M 21 ( 1) 2 1 ( sin ) sin , C22 ( 1) 2 2 M 22 ( 1) 2 2 (cos ) cos T cos So, adj ( A) Cij sin 1
A
adj ( A) A
sin
T
cos sin cos
cos
sin
sin
cos
adj ( A)
sin cos
. 1
Example 1.44 [CE-2008 (1 mark)]: The product of matrices PQ P is 1 (b) Q 1 (c) P 1Q 1 (a) P Solution (b): [ PQ]1 Q 1 P 1 [ PQ]1 P Q 1 P 1 P Q 1 I Q 1 .
(d) PQP 1
3 5 4 5 Example 1.45 [ME-2009 (1 mark)]: For a matrix M , the transpose of the matrix is x 3 5 T
1
equal to the inverse of the matrix M M . The value of x is given by (a) –4/5 (b) –3/5 (c) 3/5 (d) 4/5 3 5 4 5 3 5 x 1 0 Solution (a): M T M 1 MM T I x 3 5 4 5 3 5 0 1 1
3 x 5 12 25
3x 5 12 25
1
12 4 1 0 3 0 1 5 x 25 0 x 5 .
Example 1.46 [TF-2009 (2 marks)]: If the determinant and trace of a 2 2 matrix M are –1 and 0 1 respectively, then the determinant of M M is (a) –4 (b) –2 (c) –1 (d) 0
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Engineering Mathematics
Chapter 1: Linear Algebra
a Solution (a): Let the matrix M c a M c
b
. Now, M 1 a
b d
adj ( M ) M
[1.22]
. As tr ( A) 0 a d 0 d a
adj ( M )
as
det( M ) 1 .
C11 ( 1)11 M 11 a , C12 (1)1 2 M 11 c , C21 ( 1) 2 1 M 21 b , C 22 (1) 2 2 M 21 a T
a c a b a b 2a 2b adj ( M ) Cij M 1 M M 1 b a c a c a 2c 2a T
M M 1 2 M det( M M 1 ) det(2M ) 22 det( M ) 4 1 4 .
3 2i
Example 1.47 [CE-2010 (2 marks)]: The inverse of the matrix A
i
(a) (c)
1 3 2i 12 i
i
C11 ( 1)
(d)
3 2i
Solution (b): A 11
(b)
3 2i
1 3 2i 14 i
i
3 2i
i
i
3 2i
1 3 2i 12 i
is 3 2i
i 3 2i
1 3 2i 14 i
i
i 3 2i
2 (3 2i )(3 2i) i 9 4 1 12 0 A is invertible. Now,
M 11 3 2i , C12 ( 1)1 2 M 12 i , C 21 ( 1) 21 M 12 i , C22 (1) 2 2 M 12 3 2i T
i i i 3 2i 3 2i adj ( A) 1 3 2i adj ( A) Cij A 1 3 2i 3 2i A 12 i 3 2i i i T
Equivalent Matrices: If a matrix B is obtained from a matrix A by one or more elementary transformations, then A and B are equivalent matrices and we write A B . 1 2 3 1 2 3 1 3 3 By Applying By Applying Let, A 2 1 4 A 1 1 1 A 1 0 1 . R2 R2 ( 1) R1 C2 C2 C1 3 1 2 3 1 2 3 4 2 An elementary transformation is called a row (or column) transformation or accordingly as it is applied to rows or columns. Every elementary row (column) transformation of an m n matrix (not identity matrix) can be obtained by pre-multiplication (post-multiplication) with the corresponding elementary matrix obtained from the identity matrix I m ( I n ) by subjecting it to the same elementary row (column) transformation. Let C AB be a product of two matrices. Any elementary row (column) transformation of AB can be obtained by subjecting the pre-factor A (post factor B ) to the same elementary row (column) transformation.
Method of Finding the Inverse of a Matrix by Elementary Transformations: Let A be a non-singular matrix of order n . Then A can be reduced to the identity matrix I n by a finite sequence of elementary transformation only. As we have discussed every elementary row transformation of a matrix is equivalent to pre-multiplication by the corresponding elementary matrix. Therefore there exist elementary matrices E1 , E2 , Ek such that Ek Ek 1 E2 E1 A I n Ek E k 1 E 2 E1 AA1 I n A1 E k Ek 1 E2 E1 I n A1
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.23]
If A and B are invertible matrices and AB I ABB 1 IB 1 AI B 1 A B 1 . [This 1 point was asked in AE-2007 (2 marks)]. Similarly we can also prove that B A by pre1 multiplying of A in AB I . Algorithm for Finding the Inverse of a Non Singular Matrix by Elementary Row Transformations: Let A be non-singular matrix of order n . Write A I n A , then perform a sequence of elementary row operations successively on A on the LHS and the pre factor I n on 1
the RHS till we obtain the result I n BA , A B . The following steps will be helpful to find the inverse of a square matrix of order 3 by using elementary row transformations. Step I: Introduce unity at the intersection of first row and first column either by interchanging two rows or by adding a constant multiple of elements of some other row to first row. Step II: After introducing unity at (1,1) place introduce zeros at all other places in first column. Step III: Introduce unity at the intersection of 2nd row and 2nd column with the help of 2nd and 3rd row. Step IV: Introduce zeros at all other places in the second column except at the intersection of 2nd row and 2nd column. Step V: Introduce unity at the intersection of 3rd row and third column. Step VI: Finally introduce zeros at all other places in the third column except at the intersection of third row and third column. 5 0 2 Example 1.48 [EE-1998 (2 marks)]: If A 0
3 0 . The inverse of A is 2 0 1 0 2 0 2 0 1 2 1 5 1 5 0 1 2 15 (a) 0 1 3 0 (b) 0 1 3 0 (c) 0 13 0 (d) 0 13 0 5 0 1 1 1 2 0 1 2 0 2 1 2 0 5 0 2 1 0 0 1 0 2 5 1 5 0 0 By Applying Solution (a): A IA 0 3 0 0 1 0 A 0 3 0 0 1 0 A R1 R1 5 2 0 1 0 0 1 2 0 1 0 0 1 1 0 2 5 1 5 0 0 By Applying 1 0 2 5 1 5 0 0 By Applying 0 3 0 0 1 0 A R2 R2 3 0 1 0 0 1 3 0 A R R 2 R 3 3 1 0 0 1 5 2 5 0 1 R R (1 5) 0 0 1 2 0 5 3 3 0 2 0 2 1 0 0 1 1 By Applying 1 1 0 1 0 0 1 3 0 A. I A A A 0 1 3 0 . R R (2 5) R 1 1 3 0 0 1 2 0 5 5 2 0
Similar questions were also asked in EE-1995, AG-2009, CH-2010 (1 mark), MN-2008, CE-2007, PI-2008 (2 marks) Exercise 1.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill.
1. If A 1
1 0 2 and , then value of for which A B is B 1 5 1
2. If A 2
2
0
(a) 1 (c) 2
(b) 2 (d)
3
and A 125 then maximum value of is _____.
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Engineering Mathematics
2 x 1
3. If A
0
Chapter 1: Linear Algebra
[1.24]
x 3 y 2 2 and B , then for what values of x and y s.t. A B ? 2 y 5 y 6 0 3y
(a) x 2 , y 2
(b) x 2 , y 3 xp y
x
y
4. The determinant yp z
y
z
xp y
yp z
0
(c) x 3 , y 2
(d) x 3 , y 3 x, y , z are in AP x, y , z are in GP x, y , z are in HP xy, yz , zx are in AP
(a) (b) (c) (d)
0 if
1
a2
a
5. The parameter on which the value of the determinant cos( p d ) x
cos px
cos( p d ) x does
sin( p d ) x
sin px
sin( p d ) x
not depend upon, is (a) a 1 6. If f ( x )
(b) p
(c) d
x
x 1
2x
x ( x 1)
( x 1) x
3 x( x 1)
x ( x 1)( x 2)
( x 1) x ( x 1)
(d) x
, then f (100) _____.
sin x
cos x
cos x
7. The number of distinct real roots of cos x
sin x
cos x 0 in
cos x
cos x
sin x
x is _____. 4 4
8. Let M and N be two 3 3 non-singular skew-symmetric matrix such that MN NM . If P denotes the transpose of P , then M 2 N 2 ( M T N ) 1 ( MN 1 )T is equal to 2 2 (d) MN (b) N 2 (a) M (c) M 9. Let
k and
let
2k 1 2 k A 2 k 1 2 k 2k
2 k
2k 1
0 B 1 2k k
and
2k 1
k
0
2 k.
2 k
0
T
If
det( adjA) det(adjB ) 106 , then [ k ] _____, where [] is the greatest integer k .
0 1 1 1 1 0 10. Let M be a 3 3 matrix satisfying M 1 2 , M 1 1 , M 1 0 , then the sum 0 3 0 1 1 12 of the diagonal entries of M is _____. 1 a
11. The value of the determinant 1 b 1 c
a 2 bc b 2 ca is _____. 2
c ab
1
log x y
12. For ve number x, y , z , the numerical value of the determinant log y x p
b
c
13. If a p , b q , c r and a
q
c 0 , then value of
b
r
a
p pa
q q b
log y z is ___.
1
log z x
log x z
log z y
r r c
1
is _____.
T
14. If M is a 3 3 matrix, where M M I and det( M ) 1 , then det( M I ) _____.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.25]
cos( A P)
cos( A Q )
cos( A R)
15. The value of determinant cos( B P)
cos( B Q )
cos( B R ) for all values of A, B, C , P, Q, R
cos(C P )
cos(C Q )
cos(C R )
is _____
a b c T 16. If matrix A b c a , where a, b, c are real ve numbers, abc 1 and A A I , c a b 3
3
3
a b c _____.
1 x 3 17. Find non-negative value of x for which the matrix A 3 1 1 x 3
2 x 2 is singular.
2
1 2 0 2 1 5 18. Let A 2 B 6 3 3 and 2 A B 2 1 6 then tr ( A) tr ( B ) is _____. 5 3 1 0 1 2 a b p 0 19. Let A and B . If AB B and a d 2 , then ( ad bc ) _____. c d q 0 20. If matrix A has m rows and ( n 5) columns, matrix B has m rows and 11 n columns. If both AB and BA exists, then the order of both matrices A and B is k k , where k _____. 0 1 1 21. If matrices A 4
3
3 3
4 B C , where B is symmetric matrix and C is skew-symmetric
4 matrix, then sum of diagonal elements of B and C is _____. 3 a 3 a 1 d
22. If A 2
5
b 8
c is symmetric and B b a
2
4 3 6 (a) 31 54 26 28 9 50
2
4 3 6 (b) 31 54 26 28 9 50
e 6
2b c is skew-symmetric, then AB is f
4 3 6 (c) 31 54 26 28 9 50
4 3 6 (d) 31 54 26 28 9 50
23. Let A, B, C , D (not necessarily square real matrices) such that AT BCD , BT CDA , T T 6 C DAB and D ABC for the matrix S ABCD . Then S (a) S (b) S 2 (c) S 3 1 1 3 (a) Orthogonal 24. The matrix A 5 2 6 is (c) Involuntary 2 1 3
(d) S 4 (b) Idempotent (d) Nilpotent of order 3 (a) (b) (c) (d)
1 0 k 25. The values of k for which the matrix A 2 1 3 is invertible is k 0 1
1 2 –1 All (a), (b), (c)
1
26. If A be a non-singular matrix, and I A A2 A3 An O , then A is 2 n 1 n (a) A (b) A (c) A (d) A 27. If A, B, C are n n matrix and A 2 , B 3 and C 5 , then A2 BC 1 _____.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.26]
1 1 1 4 2 2 28. Let A 2 1 3 and 10 B 5 0 . If B is the inverse of A , then _____. 1 1 1 1 2 3 2
29. If there are three square matrices A, B, C of same order satisfying the equation A A n
1
and let
( n2)
B A2 and C A2 , where n N , then B C _____. 30. The inverse of the matrix 1 2 1 2 1 2 0 1 2 (a) 4 3 1 A 1 2 3 is 5 2 3 2 1 2 3 1 1 1 2 1 2 1 2
(c) 4 3 1 5 2 3 2 1 2
31. If A aij
33
, B bij
32
1 2 1 2 1 2 (b) 4 3 1 5 2 3 2 1 2 1 2 1 2 1 2 (d) 4 3 1 5 2 3 2 1 2
and C cij , then which one of the following is not defined? 31
(a) ( AB)T C (b) C T C ( AB )T (c) C T AB (d) AT ABBT C 32. A determinant of 2nd order is made with the elements 0 and 1. The number of determinants with non-negative values is _____. 33. The largest value of a third order determinant whose elements are 0 or 1 is _____. 1 cos( ) cos( ) 34. The value of cos( ) cos( )
cos( )
1
x
x2
35. If x
x2
1 3 , then the value of
2
1
x
x
cos( ) is _____.
1
1 x3 1
0
0
x x4
xx
4
x x4 x 3 1 is _____.
3
x 1
0
36. The inverse of a skew-symmetric matrix of odd order is (a) a symmetric matrix (b) a skew symmetric matrix (c) diagonal matrix (d) does not exist 3 37. The number of diagonal matrix A of order n for which A A is n (a) 1 (b) 0 (d) 3n (c) 2 38. If A [ aij ]44 , such that aij 2 if i j , and aij 0 if i j , then det{adj (adjA)} 2k , where k _____. 39. If an odd-ordered square matrix A is orthogonal, then det{adj ( adjA)} is _____. k
40. If A and B are two non-singular matrices of the same order such that B I , for some positive 1 k 1 1 1 integer k 1 . Then A B A A B A (c) O (a) I (b) 2I (d) I 41. If A is order 3 square matrix such that det( A) 2 , then det[ adj{adj ( adjA)}] is _____. 42. For nth order square matrix, ( A) 1 (a) ( 1) n A1
(b) ( 1) n 1 A1
1
1
(c) A (d) A 9 43. If A is a square matrix of order n such that det{adj ( adjA)} {det( A)} , then n _____. 44. If A , B are symmetric matrices of same order and X AB BA , Y AB BA , then ( XY )T (d) None of these (a) XY (b) YX (c) YX 45. If A is a 3 3 skew-symmetric matrix, then tr ( A) (c) det( A) (a) 1 (b) –1 (d) None of these
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Engineering Mathematics
1.2
Chapter 1: Linear Algebra
[1.27]
System of Linear Equations
System of linear equations if number of equations is not equal to number of unknowns: A general system of m linear equations with n unknowns ( m n ) can be written as a11 x1
a12 x2
a1n xn
b1
a21 x1
a22 x2
a1n xn
b2
am 2 x2
amn xn
am1 x1
(1.2)
bm
Here x1 , x2 , , xn are the unknowns, a11 , a12 , , amn are the coefficients of the system, and b1 , b2 , , bm are the constant terms. The coefficients, constants and unknowns are real or complex numbers. The m equations of (1.2) may be written as a single vector equation, Ax B (1.3)
where, A a jk
m n
O is the coefficient matrix; x xk 1 n1 and B bk1 m1 are the column
vectors; A is the augmented matrix, are given as: a11 a12 a1n x1 b1
a11 a a a22 a2 n x b 21 , x 1 , B 2 and A 21 A am1 am 2 amn xn bm am1
a12
a1n
|
b1
a22
a2 n
|
b2
|
am 2
(1.4)
amn
| bm Matrix A is called the augmented matrix of the system (1.2), the dashed vertical line is merely a remainder that the last column of A does not belong to A . The augmented matrix A determines the system (1.2) completely because it contains all the given number appearing in (1.2).
System of linear equations if number of equations is equal to number of unknowns: A general system of n linear equations with n unknowns can be written as a11 x1
a12 x2
a1n xn
b1
a21 x1
a22 x2
a1n xn
b2
an 2 x2
ann xn
an1 x1
(1.5)
bn
Here also, x1 , x2 , , xn are the unknowns, a11 , a12 , , ann are the coefficients of the system, and b1 , b2 , , bn are the constant terms. The coefficients, constants and unknowns are real or complex numbers. Again, the n equations of (1.5) may be written as a single vector equation ‘ Ax B ’ (given
by Eq. 1.3); where, A a jk
n n
is the coefficient matrix; x xk 1 n1 and B bk1 n1 are the
column vectors; A is the augmented matrix, are given as: x1 b1 a11 a11 a12 a1n
a 21 A am1
a22 am 2
x b a a2 n , x 2 , B 2 and A 21 ann xn bn an1
a12
a22
a2 n
| b2
|
an 2
ann
a1n
|
b1
(1.6)
| bn
If all the b j in 1.2 (or 1.5) are zero, then 1.2 (or 1.5) is called homogeneous system.
If at least one b j in 1.2 (or 1.5) is non-zero, then 1.2 (or 1.5) is called non-homogeneous system.
Solution of System of Equations: A solution of (1.2) is a set of numbers x1 , x2 , , xn that satisfies all the m equations. A solution vector of (1.2) is a vector x whose components form a solution of (1.2).
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.28]
Consistent system: If the system of the equation (1.2) has one or more solutions, then it is said to be a consistent system of equation. If it has unique solution then consistent system is determinate. If it has more than one solution then consistent system is indeterminate. Inconsistent system: If the system of the equation (1.2) has no solutions, then it is said to be an inconsistent system of equation. Over-determined system: If the system of equation (1.2) has more equations than unknowns, i.e., m n , then it is said to be an over-determined system. In general, an over-determined system may be inconsistent (i.e., having no solution) or consistent (having unique or infinite number of solutions). [This point was asked in CE-2005 (1 mark)] Determined system: If the system of equation (1.2) has equal number equations and unknowns, i.e. m n , then it is said to be an determined system. In general, a determined system may be inconsistent (i.e., having no solution) or consistent (having unique or infinite number of solutions). Under-determined system: If the system of equation (1.2) has less equations than unknowns, i.e., m n , then it is said to be an under-determined system. In general, an under-determined system may be inconsistent (i.e., having no solution) or consistent (having infinite number of solutions).
1.2.1 Solution of a Non-Homogeneous System of Linear Equations There are various methods of solving a non-homogeneous system of simultaneous linear equations. Some of the methods discussed below are:
Cramer’s Rule: Cramer’s rule is applicable when number of equations is same as the number of unknowns, i.e., for (1.5), we have m n . Cramer’s rule express the solution in terms of the determinants of the coefficient matrix and of matrices obtained from it by replacing one column by the vector of right hand sides of the equations. Consider the system of equation given by (1.5), then a11 a12 a1n b1 a12 a1n a11 a12 b1 Let,
a2 n
a21
a22
an1
an 2
ann
and 1
a2 n
b2
a22
bn
an 2
ann
; , n
a21
a22
an1
an 2
b2
, where i
bn
is obtained by replacing the elements of i th column by the elements of B . Then, the solution of system of non-homogeneous linear equations is given by, x1 1 ; x2 2 ; xn n (1.7) Nature of Solution of System of Linear Equations If 0 then from (1.7), we have unique solution. If 0 then When at least one of 1 , 2 , n is non-zero. Let i 0 then from (1.7), i xi (where, i 1 or 2 or any number less than n ) will not be satisfied for any value of xi . Hence no value of xi is possible in this case. Thus, if 0 and at least one of 1 , 2 , , n is non-zero, then no solution is possible and hence the system is inconsistent. When 0 and 1 2 n 0 In this case, from (1.7), i xi (where, i 1 or 2 or any number less than n ) will be satisfied for any value of xi . Hence infinitely many value of xi is possible in this case. Hence, the given system of equation is consistent and it has infinitely many solutions.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.29]
Example 1.49 [CS-1998 (1 mark)]: Consider the following set of equations x 2 y 5 ; 4 x 8 y 12 ; 3 x 6 y 3 z 15 ; This set has _____ solutions. (a) unique (b) no (c) finite number of (d) infinite number of Solution (b): 1 2 0 5 2 0 1 5 0 1 2 5 We have, 4
8
0 0; 1 12
8
0 48; 2 4 12
0 24; 3 4
8 12 0
3
6
3
6
3
3
6 15
1
48
2
15 24
3
3 15
3
0
, which states that we have infinite values of z but we 0 0 0 have no values of x and y which satisfies the relation. So the given system is inconsistent. [Similar questions were also asked in EC-2008, CH-2012, AG-2014 (1 mark), ME-2003, AE2008, AE-2009, AE-2011 (2 marks)] x
; y
; z
Example 1.50 [ME-2002 (2 marks)]: The following set of equations 3 x 2 y z 4 ; x y z 2 ; 2 x 2 z 5 has (a) no solution (b) a unique solution (c) multiple solution (d) an inconsistency Solution (b): As we have non – homogeneous equations with 3 2 1 1
1 1 2(2 1) 0 2(3 2) 16 0 . So the given system has a unique solution.
2 0 2 [Similar questions were also asked in AG-2010, PI-2014 (1 mark), CS-2005, CS-2004 (2 marks)]
Example 1.51 [CE-2007 (2 marks)]: For what values of and the following simultaneous equations have an infinite number of solutions? x y z 5 ; x 3 y 3 z 9 ; x 2 y z . (a) 2, 7 (b) 3, 8 (c) 8, 3 (d) 7, 2 Solution (a): For infinitely many solutions, 1 2 3 0 1 1
We have, 1 3
1
5
1
1
3 2 4 0 2; 1 9
3
3 6 12 0 2;
1 2
1
5
1
2 1
9
3 4 2 6 0 7; 3 1 3
9 2 14 0 7
1
1 1
2
1 2
5
2
Example 1.52 [IN-2007 (1 mark)]: Let A be n n real matrix such that A I and y be an n dimensional vector. Then the linear system of equations Ax y has (a) no solution (b) more than one but finitely many independent solutions (c) a unique solution (d) infinitely many independent solutions Solution (c): A2 I A2 I AA 1 A A 1 A 1 0 . Hence 0 . Also for any value of 1 , 2 and 3 , we can say that the system Ax y has a unique solution. Example 1.53 [CE-2008 (2 marks)]: The following simultaneous equations x y z 3 ; x 2 y 3 z 4 ; x 4 y kz 6 will not have a unique solution for k equal to (a) 0 (b) 30 (c) 6 (d) 7 Solution (d): The linear system of equation will have not a unique solution if 0 , i.e.,
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1 1 0 1 2 1 4
Chapter 1: Linear Algebra
[1.30]
1 3 0 1(2k 12) 1(k 3) 1(4 2) 0 k 7 . k
Example 1.54 [CS-2008 (1 mark)]: The following system of equations x1 x2 2 x3 1 ; x1 2 x2 3 x3 2 ; x1 4 x2 x3 4 has a unique solution. The only possible value(s) for is/are (a) 0 (b) either 0 or 1 (c) one of 0, 1, –1 (d) any real number Solution (c): For the given system having unique solution the determinant of its coefficient matrix 1 1 2 3 0 1(2 12) 1( 3) 2(4 2) 0 5 . So from the
must be non-zero, i.e., 1 2
1 4 given options we can say that option (c) is correct.
Example 1.55 [ME-2008 (2 marks)]: For what values of a , if any, will the following system of equations in x, y and z have a solution? 2 x 3 y 4 ; x y x 4 ; x 2 y z a (a) any real number (b) 0 (c) 1 (d) There is no such value Solution (a): As the given system of equations are: 2 x 3 y 4 ; 2 x y 4 ; x 2 y z a 2
3
0
For finding the nature of solution, first we have to find , i.e., 2
1
0 1(2 6) 4
1
2
1
0 . So, it has a unique solution for any value of a . Example 1.56 [EC-2011 (1 mark)]: The system of equations x y z 6 ; x 4 y 6 z 20 ; x 4 y z has NO solution for values of and given by (a) 6; 20 (b) 6; 20 (c) 6; 20 (d) 6; 20 Solution (b): For non-homogeneous system of linear equations having no solution we must have 0 and one of i 0 , where i x or y or z . So, 1 1 0 1 4 1 4
1
6
1
1
6 0 6 ; also x 0 20
4
6 0 20
4
6
Example 1.57 [BT-2013 (2 marks)]: The solution of the following set of equations: x 2 y 3 z 20 ; 7 x 3 y z 13 ; x 6 y 2 z 0 ; is (a) x 2, y 2, z 8 (b) x 2, y 3, z 8 (c) x 2, y 3, z 8 (d) x 8, y 2, z 3 Solution (b): We have, 1 2 3 20 2 3 1 20 3 1 2 20 7
3
1 91; 1 13
3
1 182; 2 7
13
1 273; 3 7
3 13 728
1
6
2
6
2
0
2
6
x
1
182
0 2; y
2
273
1 3; z
3
1
0
728
8. 91 91 91 [Similar questions were also asked in PI-2009 (2 marks), AG-2014 (1 mark)]
Example 1.58 [AG-2014 (2 marks)]: Consider the following set of linear equations x1 x2 x3 6 , 2 x1 2 x2 3 x3 14 , 3 x1 x2 2 x3 14 . The solution for this set exists only when the value of x2 is _____.
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Chapter 1: Linear Algebra
1
1
1
6
1
1
Solution: 2
2
3 2; 1 14
2
3 6; 2 2 14
3 2; 3 2
2 14 4
3
1
2
1
2
2
1 14
x1
1
6 2
3; x2
14 2
2 2
1; x3
3
1
6
[1.31]
3 14
4 2
1
1 3
1
6
2 . Hence answer is 1.
1 2 0 1 Example 1.59 [XE-2014 (2 marks)]: Let r , s . If A 2 0 3 and b 1 , then the r s 0 s 1 system of linear equations AX b has (a) no solutions for s 2r (c) a unique solution for s 2r 2
(b) infinitely many solutions for s 2r 2 (d) infinitely many solutions for s 2r 2 1 2 0
Solution (d): For a given linear system of equations: 2
0
3 3(2r s ) ,
r
s
0
1
1
2
0
1
1
0
1
2
1
1
0
3 3( s 2) , 2 2
1
3 3( r s 1) , 3 2
0
1
3s 2r 4
s 1 s 0 r s 1 0 r s s 1 Now For a unique solution: 0 2r s 0 2r s . For no solution: 0 2r s 0 2r s and any one of 1 or 2 or 3 is non-zero. For infinitely many
solutions 1 2 3 0 s 2r and s 2 .
2 1 3 a 5 Example 1.60 [EC-2014 (2 marks)]: The system of linear equations 3 0 1 b 4 has 1 2 5 c 14 (a) a unique solution (b) infinitely many solutions (c) no solution (d) exactly two solutions Solution (b): As we have non-homogeneous equation with 2 1 3 5 1 3 3 1 2 2 3 1
0 1 4 14 18 0 ; 1 4
0
1 10 34 24 0 ;
2
2
5
2
1
5
4 1 68 70 138 0 ; 3 3
0
4 16 46 30 0
14
2 14
5 5
14 3 5
1
1 2 3 0 the given system of equations have infinitely many solutions.
Matrix Inverse Method: Matrix Inverse method is applicable when number of equations is same as the number of unknowns, i.e., for (1.5), we have m n . Consider a system of n linear equation in n unknowns; the system of equations can be written in matrix form Ax B where, the coefficient matrix A ; the column vectors x and B are given by Eq. 1.6. Multiplying both sides of Ax B by 1 the inverse matrix A , we get A 1 Ax A1 B Ix A1 B x A1 B . So for finding the solution, we have to find the inverse of the coefficient matrix and the solution given as adj ( A) 1 1 (1.8) x A B; A A
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[1.32]
Nature of Solution of System of Linear Equations If A is a non-singular matrix, then the system of equation given by Ax B as a unique solution given by x A1 B . If A is a singular matrix, and ( adjA) B 0 , then the system of equation given by Ax B is consistent with infinitely many solutions. If A is a singular matrix, and ( adjA) B 0 , then the system of equation given by Ax B is inconsistent and has no solutions. Example 1.61 [ME-1997 (1 mark)]: For the following set of simultaneous linear equations: 1.5 x 0.5 y 2 ; 4 x 2 y 3 z 9 ; 7 x y 5 z 10 . (a) the solution is unique (b) infinitely many solutions exist (c) the equations are incompatible (d) finite number of multiple solutions exist 1.5 0.5 0 x 2 Solution (a): A 4
7
2
3 , X y and B 9 are the coefficient matrix and column
1
5
z
0.5
1.5
vectors of the given system. A 4
2
10
0 3 1.5(10 3) 0.5(20 21) 10 0 A is non-
7 1 5 singular matrix. Hence the system has unique solution.
Example 1.62 [ME-2001 (1 mark)]: Consider the system of equation given: x y 2 ; 2 x 2 y 5 . The system has (a) One solution (b) No solution (c) Infinite solution (d) Four solutions 1 1 x 2 Solution (b): A , X and B are the coefficient matrix and column vectors of 2 2 y 5 the
given
A
system.
1
1
2
2
0,
so
the
coefficient
matrix
is
singular.
1 O . Hence the system of equation given by Ax B is 2 1 5 1 inconsistent and has no solutions. T
2
( adjA) B Cij B
1 2
Example 1.63 [ME-2012, PI-2012 (2 marks)]: The system of algebraic equation given x 2 y z 4 ; 2 x y 2 z 5 ; x y z 1 ; has (a) a unique solution of x 1 , y 1 and (b) only the two solutions of ( x 1 , y 1 and z 1 ) and ( x 2 , y 1 and z 0 ) z 1 (c) infinite number of solutions (d) no feasible solution 1 2 1 x 4 Solution (c): A 2
1 2 , X y and B 5 are the coefficient matrix and column vectors 1 1 1 z 1
of
the
given
system.
1
2
1
A 2
1
2 0 A
1 C12
2
2
1
1
0 , C13
2
1
1
1
is
singular.
Also,
C11
1 1
3 , C21
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2
1
1 1
3 , C22
1 1 1 1
0 , C23
1
2
1 1 1
2
1 1
3,
3,
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C31
2
1
1
2
3,
Chapter 1: Linear Algebra
C32
1 2 1 2
0,
C33
1
2
2
1
[1.33]
3 0 3 Cij 3 0 3 3 0 3
3
3 3 3 3 3 3 4 0 adj ( A) Cij 0 0 0 adj ( A) B 0 0 0 5 0 O . 3 3 3 3 3 3 1 0 T
Hence
the
system of equation given by Ax B is consistent and has infinitely many solutions.
Gauss Elimination and back substitution: This is a standard elimination method for solving linear system given by (1.2), where number of equations may or may not be equal to number of unknowns. To perform Gaussian elimination starting with the system of equations given by Ax B . a11 a12 a1n | b1
a
Composing the augmented matrix equation given by Eq. 1.4 as, A
21
am1
a22
a2 n
|
b2
|
.
am 2 amn | bm Here, the column vector in the variables x is carried along for labelling the matrix rows. Now perform elementary row operations to put the augmented matrix into the upper triangular form, i.e., a11 a12 a1n | b1 x1 a11 a12 a1n x1 b1
0 0
a22
a2 n
0
amn
Solve the equation of the m
th
0 | | bm xn 0 |
b2 x2
a22 0
b 2 xn bm amn
a2 n x2
(1.9)
row for xn ; then substitute back into the equation of ( n 1)th row to
obtain a solution for xn 1 ; then substitute back into the equation of (n 2)th row to obtain a solution for xn 2 and so on. Thus, in Gauss elimination method for solving a system of linear algebraic eqn, triangularization leads to upper triangular matrix. [This point was asked in ME-1996 (1 mark)]. Nature of Solution of System of Linear Equations: At the end of Gauss elimination the form of the coefficient matrix, the augmented matrix, and the system itself are called the row echelon form; i.e., at the end of Gauss elimination (before the back substitution) the row echelon form of the augmented matrix is given by Eq. 1.10. Here, r m , a11 0 , c22 0 , , kr r 0 and all the entries in triangle as well as in rectangle are zero.
a11
a12
a1n
|
c22
c2 n
|
b1 b2
|
kr r
kr n
br (1.10) | br1 | |
bm
| Now w.r.t solutions of the system with augmented matrix (1.10), there are three possible cases: Exactly one solution: If r n and br1 , , bm if present, are zero. To get the solution, solve
the nth equation corresponding to (1.10) for xn , then ( n 1)th equation for xn 1 , and so on. up the line. See Example 1.64. Infinitely many solution: If r n and br1 , , bm if present, are zero. To obtain any of these solutions, choose values of xr 1 , , xn , arbitrarily. Then solve the r
th
equation for xr , then the
th
(r 1) equation for xr 1 , and so on up the line. See Example 1.65.
No solution: If r m and one of the entries of br1 , , bm is not zero. See Example 1.66.
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[1.34]
Example 1.64 [CE-1998 (5 marks)]: Solve the following set of simultaneous equations by Gauss elimination method. x 2 y z 3 ; x 3z 11 ; 2 y z 1 . Solution: The augmented matrix of the given system is given as: 1 2 1 | 3 1 2 1 | 3 1 2 1 | 3 1 0 3 | 11 By Applying 0 2 2 | 8 By Applying 0 2 2 | 8 R R 3 R3 R 2 2 R2 R1 0 2 1 | 1 0 2 1 | 1 0 0 3 | 9 Restoring the transformed matrix equation, 1 2 1 x 3 x 2 y z 3 (i) 0
0
2
2 y 8 2 y 2 z 8
3 z
9
(ii) . (iii) z 3 ; Putting value of z in (ii), we get,
0 3z 9 (iii) y 1 ; Putting values of z and y in (i), we get, x 2 . Hence we get a unique solution. [Similar question was also asked in EE-1995 (5 marks)]
Example 1.65: Solve the following system of 6 x 15 y 15 z 54w 27 ; 12 x 3 y 3 z 24 w 21
linear
equations
3 x 2 y 2 z 5w 8 ;
3 2 2 5 | 8 Solution: The augmented matrix for the system is 6 15 15 54 | 27 12 3 3 24 | 21 By Applying 3 2 2 5 | 8 3 2 2 5 | 8 By Applying R2 R2 2 R1 0 11 11 44 | 11 0 11 11 44 | 11 R3 R3 R2 0 0 0 0 11 11 44 | 11 0 | 0 R R 4 R 3 3 1 x 3 2 2 5 8 3x 2 y 2 z 5w 8 (i) After restoring the y 0 11 11 44 11 11 y 11z 44 w 11 (ii) transformed matrix z 0 0 0 00 (iii) equation 0 0 w (ii) y 1 z 4w and (ii) & (i) x 2 w . If we choose a value for z and w we can get values for x and y . So, the given system is consistent and has infinitely many solutions. Example 1.66 [CE-2006 (1 mark)]: Solution for the system defined by the set of equations 4 y 3 z 8 ; 2 x z 2 ; 3 x 2 y 5 is (a) x 0; y 1; z 4 3 (b) x 0; y 1 2; z 2 (c) x 1; y 1 2; z 2 (d) Non-existent Solution (d): The augmented matrix for the system of linear equations is: 0 4 3 | 8 2 0 1 | 2 2 0 1 | 2 2 0 1 | 2 By Applying 0 4 3 | 8 By Applying 0 4 3 | 8 R2 R1 3 2 R3 3 R1 3 2 0 | 5 R 3 2 0 | 5 0 4 3 | 6 2x z 2 (i) 2 0 1 | 2 2 0 1 x 2 By Applying 0 4 3 | 8 0 4 3 y 8 4 y 3 z 8 (ii) . R 3 R3 R 2 0 0 0 | 6 0 0 0 z 6 06 (iii) As (iii) is not valid for any values of x , y and z thus the system has no solution so it is inconsistent.
Example 1.67 [XE-2008 (1 mark)]: The system of equations ax by a 2 0 , bx ay b 2 0 , x y a b 0
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[1.35]
(a) admits unique solution if a b 0 (b) admits unique solution if a b 0 (c) admits unique solution if a b 0 (d) does not admit unique solution Solution (d): The augmented matrix for the system of linear equation is
a b | a 2 By Applying a b | a2 2 2 2 b a | b R2 R2 (b a) R1 0 ( a b ) a | b(b a) 1 1 | b a (a b) a | b 2a R 3 R3 R1 a 0 a b | a2 By Applying 0 (a 2 b2 ) a | b (b a ) R 3 R3 ( a b ) R2 2 2 0 0 | b 2a b(b a ) Restoring the transformed matrix equation b a 2 0 (a b2 ) 0 0
a2 ax by a 2 (i) x 2 2 a b (b a ) ( a b ) y a b(b a ) (ii) y 2 2 b 2a b(b 2 a 2 ) 0 b 2a b(b a ) (iii)
From (iii) we can conclude that the system has no solution. Example 1.68 [AE-2012 (1 mark)]: The value of k for which the system of equations x 2 y kz 1 ; 2 x ky 8 z 3 has no solutions is (a) 0 (b) 2 (c) 4 (d) 8 Solution (c): The augmented matrix for the system of linear equations is 2 k | 1 1 2 k | 1 By Applying 1 2 k 8 | 3 R R 2 R 0 k 4 2( k 4) | 1 2 2 1 Now the system of linear equation have no solution if k 4 0 k 4 . Example 1.69 [EE-2014 (1 mark)]: Given a system of equations x 2 y 2 z b1 , 5 x y 3z b2 . Which of the following is true regarding its solutions? (a) The system has a unique solution for any given b1 and b2 (b) The system will have infinitely many solutions for any given b1 and b2 (c) Whether or not a solution exists depends on the given b1 and b2 (d) The system would have no solution for any values of b1 and b2 Solution (b): The augmented matrix of the given system of equation is b1 1 2 2 | b1 By Applying 1 2 2 | 5 1 3 | b R R 5R 0 9 7 | b 5b 2 2 2 2 1 1
After restoring the x x 2 y 2 z b1 (i) 1 2 2 b1 y transformed matrix 0 1 7 9 b2 5b1 y (7 9) z b2 5b1 (ii) z equation From (ii) we can conclude that for any value of b1 and b2 , the system of equation has infinitely many solutions.
1.2.2 Solution of Homogeneous Linear Equations
=
There are various methods of solving a homogeneous system of simultaneous linear equations. We are going to discuss only Cramer’s rule which is as follows: Cramer’s Rule: Consider the system of linear equation given by (1.5), where all b j 0 , 1 j n , let be the determinants of coefficient matrix A given by (1.6); and since all b j 0 , so
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[1.36]
1 2 n 0 where, i is obtained by replacing elements of i th column by the elements of T
O , i.e., elements of 0 0 0 . Then, solution of system of homogeneous linear equations is x1
0
; x2
0
; ; xn
0
(1.11) The solution given by Eq. 1.11 is called trivial solution if all the variables have values ‘0’. The solution given by Eq. 1.11 is called non-trivial solution if all the variables have values other than ‘0’. Nature of Solution of System of Linear Equations If 0 then A is non-singular matrix and from (1.11), we have only trivial solution, i.e. x1 x2 xn 0 , hence number of solution in this case is one. If 0 then A is singular matrix and from (1.11), i xi (where, i 1 or 2 or any number less than n ) will be satisfied for any value of xi . Hence infinitely many value of xi is possible in this case. Hence, the given system of equation has trivial solutions as well as infinite number of non-trivial solution and hence the system has infinitely many solutions.
Example 1.70 [AE-2007 (2 marks)]: Let a system of linear equation be as follows: x y 2 z 0 ; 2 x 3 y z 0 ; 2 x 2 y 4 z 0 . This system of equation has (a) No non-trivial solution (b) Infinite number of non-trivial solutions (c) An unique non-trivial solution (d) Two non-trivial solutions 1 1 2 Solution (b): 2
3
1 0 (since by applying R3 R3 2 R1 , all the elements of R3
2
2
4
becomes zero; so by expanding along R3 we get value of determinant be 0). Also the given system is homogeneous so 1 2 3 0 . Since, 1 2 3 0 The given system has trivial and infinite number of non-trivial solutions. [Similar question was also asked in ME-2011 (2 marks)] Example 1.71 [CH-2007 (1 mark)]: The value of ‘ a ’ for which the following set of equations y 2 z 0 ; 2 x y z 0 ; ax 2 y 0 have non-trivial solutions, is (a) 0 (b) 8 (c) –2 (d) 3 0 1 2 Solution (b): 2
1
1 0 1(0 a ) 2(4 a ) 8 a . Also 1 2 3 0
a
2
0
For non-trivial solution 1 2 3 0 0 8 a 0 a 8 [Similar questions were also asked in PI-2010 (1 mark), MN-2013 (2 marks)] Example 1.72 [MN-2008 (2 marks)]: The solution of the following system of linear equations x 4 y 3 z 0 ; 3 x 5 y 2 z 0 ; 8 x 10 y 12 z 0 is (a) (0, 0, 0) (b) (1, 1,1) (c) (2, 1, 2) (d) ( 3, 0,1) Solution (a): As the given system of linear equation is homogenous with 1 4 3 3
5
2 1(60 20) 4(36 16) 3(30 40) 10 0 . Hence the given system is
8 10 12 consistent with only trivial solutions.
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[1.37]
Exercise: 1.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. If the system of equations x ky z 0 , kx y z 0 , x y z 0 has a non-zero solution, then possible values of k are (a) 1, 2 (b) 1, 2 (c) 0,1 (d) 1,1 2. If the system of equations x ay 0 , az y 0 and ax z 0 has infinite solution, then the value of a is (b) 1 (c) 0 (d) No real values (a) 1 3. The number of values of k for which the system of equations ( k 1) x 8 y 4k , kx ( k 3) y 3k 1 , has infinitely many solutions, is _____. 4. Given, 2 x y 2 z 2 , x 2 y z 4 , x y z 4 , then the value of such that the given system of equations has no solution, is _____. 5. The number of 3 3 matrices A whose entries are either 0 or 1 and for which the system Ax b , T
T
where x x y z , b 1 0 0 has exactly two distinct solutions, is _____. 6. Consider the system of equations: x 2 y 3 z 1 , x 3 y 4 z 1 and x y 2 z k . Statement I: The system of equation has no solution for k 3 ; 1 3 1 Statement II: The determinant 1 2
k 0 for k 0
1 4 1 (a) Statement I and II both are true and Statement II is the correct explanation of Statement I. (b) Statement I and II both are true and Statement II is not the correct explanation of Statement I. (c) Statement I is true; Statement II is false (d) Statement I is false; Statement II is true 7. Consider the system of linear equations: (sin 3 ) x y z 0 , (cos 2 ) x 4 y 3 z 0 and 2 x 7 y 7 z 0 . Find values of for which this system has non-trivial solution; where, n Z . (a) n ( 1) n (c) n , n ( 1) n (d) n , n (b) n 6 6 6 8. Let and be real. Find the set of all values of for which the system of linear equations x (sin ) y (cos ) z 0 , x (cos ) y (sin ) z 0 , x (sin ) y (cos ) z 0 , has a non-trivial solution. For 1 , find all values of . (b) n (c) n , n (d) n , n (1) n (a) n 4 4 4 9. a, b, c are distinct real numbers and 1 . If ax y z 0 , x by z 0 and x y cz 0 have
10. 11.
12. 13.
a non-trivial solution, then the value of {1 (1 a)} {1 (1 b)} {1 (1 c)} _____. If the system of linear equations x y z 6 , x 2 y 3 z 14 , 2 x 5 y z ( , ) has a unique solution then value of should not be equal to _____. If , , are the angles of a triangle and the system of equations x cos( ) y cos( ) z cos( ) 0 , x cos( ) y cos( ) z cos( ) 0 , x sin( ) y sin( ) z sin( ) 0 has non-trivial solutions, then triangle is necessarily (a) acute angled (b) right angled (c) isosceles (d) equilateral If x ( y z ) a , y ( z x) b and z ( x y ) c , where x, y , z are not all zero, then the value of 1 ab bc ca _____. If pqr 0 and the system of equations ( p a) x by cz 0 , ax ( q b) y cz 0 ,
ax by ( r c ) z 0 has a non-trivial solution, then absolute value of {(a p ) (b q) (c r )} is _____.
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[1.38]
14. The system of equations kx y z k 1 , x ky z k 1 , x y kz k 1 , has no solution if k is (a) 1 (b) –2 (c) 2 (d) –1 15. The set of equations x y (cos ) z 0 , 3 x y 2 z 0 , (cos ) x y 2 z 0 , where [0, 2 ) , has non-trivial solutions (a) for no value of and (b) for all values of and (c) for all values of and only one value of (d) for all values of and no value of 16. If c 1 and the system of equations x y 1 0 , 2 x y c 0 , bx 3by c 0 is consistent, then one of the possible real value of b is (a) 2 (b) 3 (c) 4 (d) 5 17. If a, b, c are in GP with common ratio r1 and p, q, r are in GO with common ratio r2 , and equations ax py z 0 , bx qy z 0 , cx ry z 0 have only zero solution, then which one of the following is correct? (a) r1 1 (b) r2 1 (c) r1 r2 (d) All (a), (b), (c) 18. If a, b, c are non-zero real numbers and if the equations ( a 1) x y z , (b 1) y z x and (c 1) z x y have a non-trivial solution, then ab bc ca abc _____.
1 0 0 T 19. Let A 2 1 0 , if x1 and x2 are column matrices such that Ax1 1 0 0 and 3 2 1 T
Ax 2 0 1 0 , then x1 x 2 T
T
T
T
(a) 1 1 0 (b) 1 1 1 (c) 1 1 0 (d) 1 1 1 20. The solution of set of equations given by x y z 6 , x y z 2 , 3 x 2 y 4 z 5 is (a) x 2, y 1, z 3 (b) x 3, y 2, z 1 (c) x 2, y 3, z 1 (d) x 1, y 2, z 3 21. The system of equations x1 2 x2 x3 3 , 2 x1 3 x2 x3 3 , 3 x1 5 x2 2 x3 1 has (a) infinite number of solutions (b) exactly 3 solutions (c) a unique solution (d) no solution 22. The system of equations 2 x 6 y 11 0 , 6 x 20 y 6 z 3 0 , 6 y 18 z 1 0 is (a) consistent (b) inconsistent (c) unique solution (d) infinitely many solutions 23. The system of equations x 2 y 3 z 0 , 3 x 4 y 4 z 0 , 7 x 10 y 12 z 0 has (a) No non-trivial solution (b) Infinite number of non-trivial solutions (c) Only trivial solutions (d) An unique non-trivial solution 24. The system of equations 2 x y z 1 , x 2 y z 1 , x y 2 z 2 has (a) unique solution (b) infinitely many solutions (c) no solutions (d) a set of only two solutions 25. The system of equations x 2 y 3z 1 , 2 x y 3 z 2 , 5 x 5 y 9 z 4 has (a) a unique solution (b) infinitely many solutions (c) no solution (d) exactly 3 solutions 26. The system of equations x y z 0 , 2 x y z 0 , 3 x 2 y 0 has (a) no solution (b) infinite number of non-trivial solution (c) only trivial solution (d) trivial and infinite number of non-trivial solutions
1.3
Vector Algebra
Physical quantities are divided into two categories – scalar quantities and vector quantities. Those quantities which have only magnitude and have no direction in space are called scalar quantities, like mass, volume, etc. The quantities which have both magnitude and direction are called vectors quantities, like displacement, velocity, etc.
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[1.39]
Geometrical Representation of Vectors: The geometrical representation of a vector is a directed line segment or arrow where the length of the arrow is scaled according to the magnitude of the vector. We adopt boldface type ( a ) to signify that the quantity is a vector; alternative notations include the use of an overhead arrow as
in a . As shown in Fig. 1.1, a AB , where A is the initial point (the starting point of any vector) and B is the terminal point or tip (the end point of any vector). It Figure 1.1: Vector should be noted that the location of a vector is not specified, only its magnitude and Representation direction. Every vector AB has the following three characteristics:
Length: The magnitude or modulus or norm or length of ‘ a ’ is expressed as a AB . The magnitude of a vector is always a non-negative real number. Support: The line of unlimited length of which AB is a segment is called the support of the vector AB . Sense: The sense of AB is from A to B and that of BA is from B to A . Thus, the sense of a directed line segment is from its initial point to the terminal point.
Some Basic Definitions
Zero or null vector: A vector whose magnitude is zero is called zero or null vector and it is
represented by O . The initial and terminal points of the directed line segment representing zero vector are coincident and its direction is arbitrary. Unit vector: A vector whose modulus is unity, is called a unit vector. The unit vector in the direction of a vector a is denoted by aˆ , read as ‘a cap’. Thus, aˆ 1 aˆ a a . Unit vectors parallel to x axis, y axis and z axis are denoted by i , j and k or iˆ , ˆj and kˆ respectively. Two unit vectors may not be equal unless they have the same direction. Negative of a vector: The vector which has the same magnitude as the vector a but opposite
direction, is called the negative of a and is denoted by a . Thus, if PQ a , then QP a .
Position vectors: The vector OA which represents the position of the point A with respect to a fixed point O (called origin) is called position vector of the point A . If ( x, y , z ) are co-ordinates of the point A , then OA x i y j z k or OA x iˆ y ˆj z kˆ .
Equality of vectors: Two vectors a and b are said to be equal, if (i) a b (ii) They have the same or parallel support and (iii) The same sense.
Rectangular resolution of a Vector: In rectangular Cartesian coordinates the x , y and z axes are all mutually orthogonal and the ve sense along the axes is taken to be in the direction of increasing x , y , and z . The orientation of the axes will always be such that the ve direction along the z axis is the one in which a right-handed screw aligned with the z axis will advance when rotated from ve x Figure 1.2: (a) Rectangular resolution of a Vector in 3D axis to ve y axis. Space; (b) Rectangular resolution of a Vector in 2D Plane A system of axes with this property is called a right-handed system. As shown in Fig. 1.2(a), if the
coordinates of P are ( x, y , z ) then the position vector of r can be written as OP r x i y j z k . The vectors x i , y j and z k are called the right angled components of r . The scalars x , y , z are called the components or resolved parts of r in the directions of x axis, y axis and z axis respectively and ordered triplet ( x, y , z ) is known as coordinates of P whose position vector is r .
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x2 y2 z2 . Direction cosines of r are the cosines of angles that the vector r makes with the positive direction of x, y , z axes, i.e., cos l x r , cos m y r and cos n z r . Clearly, The magnitude or modulus or norm of r r
2 2 2 l m n 1 . Here POX , POY and POZ and i, j, k are the unit vectors along OX , OY , OZ axis, respectively. For 2D – plane as shown in Fig. 1.2(b), i.e., xy , yz , zx the z, x , y respectively, coordinate becomes zero and we get magnitude and direction cosine. For e.g., as shown in Fig. 1.3, in xy
plane, we have
2 2 z 0 , hence, magnitude of r r x y
and
cos l x r ,
o
cos m y r and cos n 0 90 .
(a)
(c)
(b) Figure 1.3: Addition of two vectors
Addition of two vectors: We define a process of addition between any two vectors u and v . The first step is to move v (if necessary), parallel to itself, so that its tail coincides with the head of u . Then the sum or resultant, u v is defined as the arrow from the tail of u to the head of v , as shown in Fig. 1.3(a). Reversing the order, v u is as shown in Fig. 1.3(b). Equivalently, we may place u and v tail to tail, as shown in Fig. 1.3(c) and comparing it with both Fig. 1.3(a) and 1.3(b), we see that the diagonal of the parallelogram is both u v and v u . Thus, Commutative law: u v vu (1.12) From above explanation, we can deduce that: Triangle law of addition: As shown in Fig. 1.3(a) and (b), if two vectors are represented by two consecutive sides of a triangle then their sum is represented by the third side of the triangle, but in opposite direction. Parallelogram law of addition: As shown in Fig 1.3(c), if two vectors are represented by two adjacent sides of a parallelogram, then their sum is represented by the diagonal of the parallelogram whose initial point is the same as the initial point of the given vectors. This is known as parallelogram law of addition of vectors. Addition in component form: If the vectors are defined in terms of i , j and k , i.e.,
if
u u1 i u2 j u3 k
as
and
v v1 i v2 j v3 k ,
then
their
sum
is
defined
u v (u1 v1 ) i (u2 v2 ) j (u3 v3 ) k . Let we have three vectors u , v and w then we can also show that the vector addition is associative as well, i.e., Associative law: (u v ) w v (u w ) (1.13) From Eq. 1.12 it follows for each vector u that, u0 0u (1.14) Corresponding to u we define a negative inverse ‘ u ’ such that if u is any nonzero vector; then u is determined uniquely as shown in Fig. 1.4(a), i.e., it is of same length as u but is directed in the opposite direction (again, u and u have the same length, the length of u is not negative). For the zero vector we have 0 0 . We denote ‘ u ( v ) ’ as ‘ u v ’ but emphasise that it is (b) (a) really the addition of u and v , as shown in Fig 1.4(b). Figure 1.4: Subtraction of Vector
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Position Vector: If a point O is fixed as the origin in space (or
plane) and P is any point, then OP is called the position vector of P with respect to O . If we say that P is the point r , then we mean that the position vector of P is r w.r.t. to some origin O .
AB in terms of the position vectors of points A and B : If a and b are position vectors of points A and B respectively, as
shown in Fig. 1.5(a). Then, OA a, OB b . In OAB , we
OA AB OB AB OB OA b a AB (Position vector of B ) – (Position vector of A ) AB (Position vector of head) – (Position vector of tail). Position vector of a dividing point Internal division: Let A and B be two points with position vectors a and b respectively, and let C be a point dividing AB internally in the ratio m : n , as shown in Fig. 1.5(b). Then the position vector of C is given by m b n a OC . mn External division: Let A and B be two points with position vectors a and b respectively and let C be a point dividing AB externally in the ratio m : n , as shown in Fig. 1.5(c). Then the position vector of C is given by Figure 1.5: (a) Position Vector (b) Internal division (c) External division m b n a OC . mn have
Scalar Multiplication of a Vector: The scalar multiplication, between a vector u and any scalar (i.e., a real number) : If 0 and u 0 then u is a vector whose length is times the length of u and whose direction is same as that of u if 0 , and opposite if 0 ; if 0 and/or u 0 , then u 0 . It follows from this definition that scalar multiplication has the following properties: ( u ) u (1.15) ( )u u u (1.16) (u v) u v (1.17) 1u u (1.18) where, , are any scalars and u and v are any vectors. 1.3.1 Scalar or Inner or Dot Product of two Vectors A product of two vectors u and v can be formed in such a way that the result is a scalar. The result is written u v and called the dot product of u and v . The names scalar product and inner product are also used in place of the term dot product. The angle between two nonzero vectors u and v is the ordinary angle between the two vectors when they are arranged tail to tail as shown in Fig. 1.6. Since there are two such angles, an interior angle ( ) and exterior angle ( ) ; for definiteness, we choose to be the Figure 1.6: Geometrical interior angle, i.e., 0 . Hence, the dot product of u and v can be Interpretation of Scalar Product defined as, u v u v cos (1.19) In component form, u v (u1 i u2 j u3 k ) (v1 i v2 j v3 k ) u1v1 u2v2 u3v3 (1.20) Since, i i j j k k 1 and i j j i j k k j k i i k 1 . If u v 0 is acute angle u v 0 either u 0 or v 0 or 90o If If u v 0 is obtuse angle [This point was asked in CE-2008 (2 marks)]
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Geometrical interpretation of Scalar product: Let u and v be two vectors represented by OA and OB respectively, as shown in Fig 1.6. Let be the angle between OA and OB . Draw BL OA and AM OB . From OBL and OAM , we have OL OB cos and OM OA cos . Here OL and OM are known as projection of v on u and u on v respectively. Hence, u v u v cos u (OB cos ) u (OL) (Magnitude of u)(Projection of v on u)
(1.21)
Also, u v u v cos v (OA cos ) v (ON ) (Magnitude of v)(Projection of u on v ) (1.22) Thus geometrically interpreted, the scalar product of two vectors is the product of modulus of either vector and the projection of the other in its direction.
Angle between two vectors: If u , v be two vectors inclined at an angle , then, uv uv u v u v cos In component form, if cos cos cos 1 . u v u v u1v1 u2 v2 u3v3 1 . u u1 i u2 j u3 k and v v1 i v2 j v3 k then cos u 2 u 2 u 2 v2 v 2 v 2 2 3 1 2 3 1 Example 1.73 [PI-2007 (1 mark)]: The angle (in degrees) between two planar vectors
3ˆ 1ˆ 3ˆ 1ˆ i j and b i j is 2 2 2 2 (a) 30 (b) 60 a
(c) 90
(d) 120
Solution (d): If is the angle between the vector a
cos 1
3ˆ 1ˆ 3ˆ 1ˆ i j and b i j then 2 2 2 2
cos 1 ( 1 2) 120o 2 2 ( 3 2) (1 2)
( 3 2) ( 3 2) (1 2) (1 2)
( 3 2) 2 (1 2) 2
[Similar question was also asked in ME-2004 (1 mark)]
Properties of Scalar Product
The scalar product of two vector is commutative i.e., u v v u . The scalar product of vectors is distributive over vector addition i.e., (a) u ( v w ) u v u w (Left distributivity) (b) (u v ) w u w v w (Right distributivity) If m is a scalar and u , v be any two vectors, then ( mu) v m(u v) u ( mv) If m , n are scalars and u , v be two vectors, then mu nv mn(u v ) ( mnu) v u ( mnv ) For any vectors u and v , we have (i) u ( v ) (u v ) ( u) v ; (ii) ( u) ( v ) u v
Components of a vector along and perpendicular to another vector: If u and v be two vectors represented by OA and OB . Let be the angle between u and v , as shown
in
Fig.
1.7.
Draw
BM OA .
In
OBM ,
we
have
OB OM MB v OM MB . Thus, OM and MB are components of v along u and perpendicular to u respectively. Now,
OM (OM )uˆ (OB cos )uˆ v cos uˆ v
(u v) u u v u
uv u u 2
uv u u 2 . Thus,
Figure 1.7: Vector along and perpendicular to another vector
and v OM MB MB v OM v
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Component of v along u :
Component of v to u :
Chapter 1: Linear Algebra
[1.43]
u v u 2 u u v MB v 2 u u OM
(1.23)
(1.24)
1.3.2 Vector or Cross Product of two Vectors Let u , v be two non-zero, non-parallel vectors. Then the vector product u v , in that order, is defined as a vector whose magnitude is u v sin where is the angle between u and v whose direction is perpendicular to the plane of u and v in such a way that u , v and this (a) (b) direction constitute a right handed system, as Figure 1.8: (a) Cross Product of two Vectors (b) Area of the shown in Fig. 1.8a. parallelogram OABC In other words, u v u v sin nˆ where is the angle between u and v , nˆ is a unit vector perpendicular to the plane of u and v such that u , v and nˆ form a right handed system.
Geometrical interpretation of vector product: If u , v be two non-zero, non-parallel vectors represented by OA and OB respectively and let be the angle between them. Complete the parallelogram OACB , as shown in Fig. 1.8b. Draw BL OB sin v sin u v u v sin nˆ (OA)( BL) nˆ . Hence,
BL OA .
In
OBL ,
u v (OA)( BL) nˆ (Base Height)nˆ (area of parallelogram OACB) (1.25) Thus, u v is a vector whose magnitude is equal to the area of the parallelogram having u and v as its adjacent sides and whose direction nˆ is perpendicular to the plane of u and v such that u and v and nˆ form a right handed system. Thus, area of parallelogram OACB u v . Also, area of
OAB (1 2)area of parallelogram OACB (1 2) u v (1 2) OA OB
The area of the triangle formed by the point O (origin), A and B is OA OB
The area of the triangle formed by the tips of vectors a , b and c is (1 2) ( a b) ( a c ) [This
point was asked in ME-2007 (2 marks)]
Properties of vector product
Vector product is not commutative i.e., if a and b are any two vectors, then a b b a , however, a b (b a) If a , b are two vectors and m is a scalar, then m a b m(a b) a m b If a , b are two vectors and m , n are scalars, then m a n b mn(a b ) m(a n b ) n(m a b ) Distributivity of vector product over vector addition. Let a, b , c be any three vectors. Then (i) a (b c) a b a c (Left distributivity); (ii) (b c) a b a c a (Right distributivity) For any three vectors a, b , c we have a (b c) a b a c The vector product of two non-zero vectors is zero vector iff they are parallel (Collinear) i.e., a b 0 a || b , where a , b are non-zero vectors. It follows that a a 0 for every non-zero vector a , which in turn implies that i i j j k k 0 Vector product of orthonormal triad of unit vectors i, j, k using the definition of the vector product, we obtain i j k , j k i , k i j , j i k , k j i , i k j [This point was asked in MN-2009 (1 mark)].
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Vector product in terms of components: If u u1 i u2 j u3 k and v v1 i v2 j v3 k . Then, i
j
k
u v u1
u2
u3 (u2 v3 u3 v2 ) i (u1v3 u3 v1 ) j (u1v2 u2 v1 ) k
v1
v2
v3
Angle between two vectors: If is the angle between u and v , then sin
(1.26)
u v u v
. In terms of
component, if u u1 i u2 j u3 k and v v1 i v2 j v3 k then sin
(u2 v3 u3v2 ) 2 (u1v3 u3v1 ) 2 (u1v2 u2v1 ) 2
(1.27)
(u12 u22 u33 )(v12 v22 v33 )
Right and Left handed system of vectors: Three mutually perpendicular vectors a, b , c form a right handed system of vector iff a b c , b c a , c a b , as shown in Fig. 1.9b. For e.g., the unit vectors i, j, k form a righthanded system, i j k , j k i , k i j , as shown in Fig. 1.9a. The vectors a, b , c , mutually perpendicular to one another form a left handed system of vector iff c b a , a c b , b a c , as shown in Fig. 1.9c.
Figure 1.9: (a), (b) Right – handed (c) Left – handed system
Example 1.74 [CS-1995 (2 marks)]: A unit vector perpendicular to both the vectors a iˆ 2 ˆj kˆ and b iˆ ˆj 2kˆ is (a) (iˆ ˆj kˆ )
(d) (iˆ ˆj kˆ) 3 Solution (a): Since a vector which perpendicular to both the vectors a 2iˆ 2 ˆj kˆ and b iˆ ˆj 2kˆ is found by taking their cross product, i.e.,
iˆ
(b) (iˆ ˆj kˆ ) 3
3
kˆ
ˆj
1 iˆ(4 1) ˆj (2 1) kˆ (1 2) 3iˆ 3 ˆj 3kˆ .
c a b 1 2 1
1
(c) (iˆ ˆj kˆ ) 3
2
As we have to find the unit vector, so c c
3iˆ 3 ˆj 3kˆ 2
2
3 3 3
2
3iˆ 3 ˆj 3kˆ
3 3
iˆ ˆj kˆ
.
3
Example 1.75 [MN-2007 (2 marks)]: Two sides of a triangle are represented by vectors a iˆ ˆj kˆ and b iˆ ˆj kˆ . The area (magnitude) of the triangle is (a) 1
(b) 1
2
2
(c)
(d) 2 2
iˆ
ˆj
kˆ
Solution (c): Since, a b (iˆ ˆj kˆ) ( iˆ ˆj kˆ ) 1
1
1 2(iˆ ˆj ) So, the area of the
1 1 1
triangle formed by a and b is (1 2) a b (1 2) 2(iˆ ˆj ) 12 12 2
Example 1.76 [AG-2008 (1 mark)]: The cross product of x 2iˆ ˆj and y iˆ 2 ˆj kˆ is (d) 2iˆ 4 ˆj (a) iˆ ˆj 2 kˆ (b) iˆ 2 ˆj 5kˆ (c) iˆ 2 ˆj 5kˆ
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iˆ
ˆj
kˆ
Solution (c): x y 2
1
0 iˆ(1 0) ˆj (2 0) kˆ (4 1) iˆ 2 ˆj 5kˆ
1
2
1
[Similar question was also asked in AG-2013 (1 mark)]
Example 1.77 [MN-2008 (2 marks)]: A force vector F (2iˆ 3 ˆj kˆ) in N is acting on a point, whose position vector r (iˆ ˆj 6 kˆ ) in m. The magnitude of the torque ( ) about the origin in Nm is (a) 20.85
(b) 21.42 iˆ ˆj kˆ
Solution (c): r F 1 2
(c) 21.97
(d) 22.27
1
6 17iˆ 13 ˆj 5kˆ
3
1
( 17) 2 132 5 2 21.97
Example 1.78 [PI-2011 (1 mark)]: If A(0, 4,3) , B(0, 0, 0) and C (3, 0, 4) are three points defined in x, y , z coordinate system, then which one of the following vectors is perpendicular to both the line
vectors BA and BC ? (a) 16iˆ 9 ˆj 12 kˆ
(b) 16iˆ 9 ˆj 12 kˆ
(c) 16iˆ 9 ˆj 12 kˆ (d) 16iˆ 9 ˆj 12 kˆ Solution (a): A vector perpendicular to both BA 4 ˆj 3kˆ and BC 3iˆ 4kˆ is found by taking their iˆ ˆj kˆ
cross product, i.e., BA BC 0
4
3 16iˆ 9 ˆj 12kˆ
3
0
4
Example 1.79 [CE-2011 (2 marks)]: If a and b are two arbitrary vectors with magnitude a and b ,
2
respectively, a b will be equal to 2 2
(a) a b (a b)
2
(b) ab a b
2 2
(c) a b (a b)
2
(d) ab a b
Solution (a): a b ab sin , where is the angle between a and b .
2 a b a 2b 2 ( a b ) 2 Also, a b ab cos cos sin 1 cos 1 ab ab ab 2 a 2 b 2 (a b ) 2 a b ab sin ab a 2b 2 ( a b ) 2 a b a 2b 2 ( a b ) 2
a b
2
ab
Example 1.80 [CE-2012 (2 marks)]: For the parallelogram OPQR shown in sketch, OP aiˆ bjˆ and
(a) (b) (c) (d)
OR ciˆ djˆ . The area of the parallelogram is
ad bc ac bd ad bc ab cd
iˆ
ˆj
kˆ
Solution (a): Since, OP OR ( aiˆ bjˆ ) (ciˆ djˆ ) a
b
0 kˆ (ad bc) . So, the area of the
c
d
0
parallelogram is OP OR ( ad bc )
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1.3.3 Scalar Triple Product If a, b, c are three vectors, then their scalar triple product is defined as the dot product of two vectors a and b c . It is generally denoted by a (b c) or [a b c] . It is read as box product of a, b , c . Similarly other scalar triple products can be defined as b (c a) and c (a b) .
Geometrical interpretation of scalar triple product: To observe an important fact about a (b c) , let us sketch the a, b , c parallelepiped, as shown in Fig. 1.10, for the case where a, b , c do not lie in the same plane. Now b c is the magnitude norm times the unit vector nˆ . Thus a (b c) a b c nˆ b c (a nˆ ) . But b c is the area of b , c parallelogram, i.e., the base area of parallelepiped; and a nˆ is the altitude Figure 1.10: Parallelepiped (dashed line). Hence a (b c) is the volume of the a, b , c parallelepiped.
Volume
of
Properties of scalar triple product
If a, b, c are cyclically permuted, the value of scalar triple product remains the same. i.e., a (b c) b (c a) c (a b ) or [a b c] [b c a] [c a b ] The change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude i.e., [a b c] [b a c] [c b a] [a c b ] In scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of the vectors remains same i.e., a (b c) (a b) c
Scalar triple product in terms of components: If a a1 i a2 j a3 k , b b1 i b2 j b3 k and c c1 i c2 j c3 k
be i
j
a (b c) ( a1i a2 j a3 k ) a2 a3
three k
a1
b1
c1
b2
c2 a2
b2
c2
b3
c3
b3
c3
a3
vectors.
Then,
Tetrahedron: A tetrahedron is a three-dimensional figure formed by four triangle OABC is a tetrahedron with ABC as the base, as shown in Fig. 1.11. OA, OB, OC , AB, BC and CA are known as edges of the tetrahedron. OA, BC , OB, CA and OC , AB are known as the pairs of opposite edges. A tetrahedron in which all edges are equal, is called a regular tetrahedron. The volume of a tetrahedron is Figure 1.11: A Tetrahedron 1 (area of the base)(corresponding altitude) . 3 If a, b, c are position vectors of vertices A , B and C with respect to O , then volume of tetrahedron OABC is
1
[a b c] . 6 If a, b, c, d are position vectors of vertices A , B , C , D of a tetrahedron ABCD , then its
volume is
1 6
b a
ca
d a
Example 1.81 [MN-2010 (2 marks)]: The volume of tetrahedron with vertices at (0, 0, 0), (1, 0, 0), (0, 1, 0) and (0, 0, 1) is (a) 1/2 (b) 1/4 (c) 1/6 (d) 1/8
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Solution (c): Let a 0iˆ 0 ˆj 0kˆ , b iˆ 0 ˆj 0kˆ , c 0iˆ ˆj 0 kˆ and d 0iˆ 0 ˆj kˆ ; so volume of tetrahedron is given as, (1 6) b a c a d a (1 6) (b a) (c a ) (d a ) (1 6) iˆ ( ˆj kˆ ) (1 6)(iˆ iˆ) 1 6
1.3.4 Vector Triple Product Let a, b, c be any three vectors, then the vectors a (b c) and (a b) c are called vector triple product of a, b, c . Since b c is perpendicular to the b , c plane so it must be some multiple of nˆ , as shown in Fig. 1.12. Similarly a (b c) is perpendicular to both a and b c , hence to both a and nˆ . Being perpendicular to nˆ , it must lie in the b , c plane so it must be expressible in the form b c . To determine and , let us introduce a convenient reference coordinate system and set of base vectors.
Figure 1.12: Vector Triple Product
Specifically, let the point O , as shown in Fig. 1.12, be the origin of a Cartesian coordinate system with the usual base vectors ˆi , ˆj and kˆ . And let the system be oriented so that ˆi is aligned with b and kˆ coincides with nˆ , as shown in Fig. 1.12. Then ˆj lust lie in the b , c plane so we can express
b b1 ˆi ,
c c1 ˆi c2 ˆj a a1 ˆi a2 ˆj a3 kˆ . b c b1c2 kˆ and Hence a b c (a1 iˆ a2 ˆj a3 kˆ ) (b1c2 kˆ ) a b c a1b1c2 ˆj a2b1c2 ˆi (a1c1 a2 c2 )b1 iˆ a1b1 (c1 ˆi c2 ˆj) (a1c1 a2c2 ) v a1b1 w
and
But a1c1 a2 c2 a c and a1b1 a b . Hence a (b c) (a c) b (a b )c
Properties of vector triple product
Vector triple product is a vector quantity and a (b c) (a b) c . The vector triple product a (b c) is a linear combination of those two vectors which are within brackets. The formula a (b c) (a c) b (a b ) c is true only when the vector outside the bracket is on the left most side. If it is not, we first shift on left by using the properties of cross product and then apply the same formula. Thus, (b c) a {a (b c)} {(a c) b (a b )c} (a b ) c (a c) b
If
a a1 i a2 j a3 k ,
b b1 i b2 j b3 k
i
j
k
a1
a2
a3
b2 c3 b3 c2
b3 c1 b1c3
b1c2 b2 c1
a (b c)
c c1 i c2 j c3 k
and
then
Example 1.82 [AE-2008 (2 marks)]: Which of the following is true for all choices of vectors p , q ,
r? (a) p q q r r p 0
(b) ( p q ) r (q r ) p ( r p ) q 0
(c) p ( q r ) q ( r p ) r ( p q ) 0 (d) p (q r ) q (r p ) r ( p q ) 0 Solution (d): Using vector triple product formula a (b c) (a c) b (a b ) c ; Option (d) can be written as,
p ( q r ) q (r p ) r ( p q ) ( p r ) q ( p q ) r (q p ) r (q r ) p ( r q ) p ( r p ) q 0 .
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[1.48]
Exercise: 1.3 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The direction cosines of the vector 3 i 4 j 5 k are (a)
3 4 1 , , 5 5 5
(b)
3
,
4
5 2 5 2
,
1 2
3
(c)
,
2
4 2
,
1
(d)
2
,
4
5 2 5 2
2. If ABCDEF is a regular hexagon, then AD EB FC
3
,
1 2
(c) 3AB (a) AB (b) 2AB (d) 4AB 3. The unit vector parallel to the resultant vector of 2 i 4 j 5 k and i 2 j 3 k is (a) (3 i 6 j 2 k ) 7 (b) (i j k ) 3 (c) (i j 2 k ) 6 (d) ( i j 8 k ) 69 4. The length of longer diagonal of the parallelogram constructed on (5 a 2 b) and (a 3 b) , it is given that a 2 2 , b 3 and angle between a and b is 4 , is k , where k _____. 5. The sum of two forces is 18 N and resultant whose direction is at right angles to the smaller force is 12 N. The magnitude of the two forces are (a) 13, 5 (b) 12, 6 (c) 14, 4 (d) 11, 7 6. The vector c , directed along the internal bisector of the angle between the vectors
a 7 i 4 j 4 k and b 2 i j 2 k with c 5 6 , is (5 3) d , where d (a) i 7 j 2 k (b) 5 i 5 j 2 k (c) i 7 j 2 k (d) 5 i 5 j 2 k 7. Let , , be distinct real numbers. The points with position vectors i j k , i j k , i j k (a) are collinear (b) form an equilateral triangle (c) form a scalene triangle (d) form a right angled triangle 8. The points with position vectors 60 i 3 j , 40 i 8 j , a i 52 j are collinear, if a _____ 9. Let a , b and c be three non-zero vectors such that no two of these are collinear. If the vector a 2 b is collinear with c and b 3 c is collinear with a ( being some non-zero scalar) then 10. 11. 12. 13.
a 2 b 6 c equals _____. If the vectors 4 i 11 j m k , 7 i 2 j 6 k and i 5 j 4 k are coplanar, then m is _____. (a i ) i (a j) j (a k ) k (a) a (b) 2a (c) 3a (d) 0 If a 3 , b 4 then a ve value of for which a b is perpendicular to a b is _____. A unit vector in the plane of the vectors 2 i j k , i j k and orthogonal to 5 i 2 j 6 k is
(a)
6 i 5k
(b)
3jk
(c)
2i 5 j
(d)
2i j 2k
61 10 29 14. If be the angle between the vectors a 2 i 2 j k and b 6 i 3 j 2 k , then
3
(a) cos 4 21 (b) cos 3 19 (c) cos 2 19 (d) cos 5 21 15. Let a, b and c be vectors with magnitudes 3, 4 and 5 respectively and a b c 0 = 0, then the values of (a b b c c a) _____. 16. The projection of a 2 i 3 j 2 k on b i 2 j 3 k is (a) 1
14
(b) 2
14
(c)
(d) 2
14
14
17. Let u , v , w be such that u 1, v 2, w 3 . If the projection v along u is equal to that of w along u and v , w are perpendicular to each other then u v w equals
k , where k _____.
18. Let b 3 j 4 k , a i j and let b1 and b 2 be component vectors of b parallel and perpendicular to a . If b1 1.5 i 1.5 j , then b 2
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(a) 1.5 i 1.5 j 4 k (b) 1.5 i 1.5 j 4 k (c) 1.5 i 1.5 j (d) None of these 19. A particle is acted upon by constant forces 4 i j 3 k and 3 i j k which displace it from a point i 2 j 3 k to the point 5 i 4 j k . The work done in standard units by the force is given by _____. 2
2
2
2
20. If a is any vector, then (a i) (a j) (a k ) k a , where k _____ 21. The sine of the angle between the vectors a 3 i j k , b 2 i 2 j k is (d) 5 41 (a) 74 99 (b) 25 99 (c) 37 99 22. The vectors c , a x i y j z k and b j are such that a, c, b form a right handed system, then c is (a) z i x k (c) y j (d) z i x k (b) 0 23. If a i j k , b i 3 j 5 k and c 7 i 9 j 11 k , then the area of the parallelogram having diagonals a b and b c is (a) 4 6 (b) 4 3 (c) 3 6 (d) 3 3 24. The position vectors of the vertices of a quadrilateral ABCD are a, b, c, d , respectively. Area of the quadrilateral formed by joining the middle points of its sides is (a) (1 4) a b b d d a (b) (1 4) b c c d a d b a (c) (1 4) a b b c c d d a (d) (1 4) b c c d d b 25. Three forces i 2 j 3 k , 2 i 3 j 4 k and i j k are acting on a particle at the point (0,1, 2) . The magnitude of the moment of the forces about the point (1, 2, 0) is (d) None of these (a) 2 35 (b) 6 10 (c) 4 17 26. If u, v, w are three non-coplanar vectors, then (u v w ) [(u v ) ( v w )] equals (b) u ( v w ) (c) u ( w v ) (d) 3u ( v w ) (a) 0 27. The value of ‘ a ’ so that the volume of parallelopiped formed by i a j k ; j a k and a i k becomes minimum is (a) 3 (b) –3 (d) 1 3 (c) 3 28. Let a, b and c be non-zero vectors such that (a b ) c (1 3) b c a . If is the acute angle between the vectors b and c , then sin equals (c) 2 3 (d) 1 3 (a) 2 2 3 (b) 2 3 29. Let the vectors a, b, c, d be such that (a b) (c d) 0 . Let P1 and P2 be planes determined by pair of vectors a, b and c, d respectively. Then the angle (in degrees) between P1 and P2 is ____. 30. If a i j k , a b 1 and a b j k , then b (b) i j k (c) 2 j k (a) i (d) 2i 31. If be the angle between the unit vectors a and b, then cos( 2) (a)
1
a b
(b)
1
(c)
ab
a b
(d)
a b
a b a b 2 2 32. If a and b are unit vectors and (a b) is also a unit vector, then the angle (in degrees) between a and b is _____.
33. If F1 i j k , F2 i 2 j k , F3 j k , A 4 i 3 j 2 k and B 6 i j 3 k , then the
scalar product of F1 F2 F3 and AB will be _____. 34. The horizontal force and the force inclined at an angle 60o with the vertical, whose resultant is in vertical direction of P kg, are (a) P, 2 P (d) None of these (b) P, P 3 (c) 2 P, P 3
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Engineering Mathematics
1.4
Chapter 1: Linear Algebra
[1.50]
Vector Space, Linear Dependence and Rank of a Matrix
1.4.1 Vector Space and Sub-Space Vector Space: An n - tuple of real numbers is an ordered list ( x1 , x2 , , xn ) of n real number. For, e.g., 1 - tuple, like ( x1 ); 2 - tuples, like ( x1 , x2 ); 3 - tuple, like ( x1 , x2 , x3 ), etc. The n - dimensional n
real space R is the set of all n - tuples ( x1 , x2 , , xn ) of real number. Similarly, an n - tuple of
vectors is an ordered list ( v1 , v 2 , , v n ) of n vectors. For, e.g., 1 - tuple, like, ( v1 ) or ( i ), creates
one-dimensional space; 2 - tuple, like, ( v1 , v 2 ) or (i , j ) , creates two-dimensional space; 3 - tuple, like
n
( i , j , k ), creates three dimensional space; etc. A vector space R consisting of all vectors with n n
components ( n real numbers) has dimension n . The n - dimensional vector space R is the set of all n n - tuples ( v1 , v 2 , , v n ) of vectors. The elements of R are called point vectors. If n 1, 2, 3 then we can visualize x R n as an arrow and vector addition corresponds to ‘adding’ arrows. Similarly, scalar multiplication of a vector corresponds to ‘stretching,’ shrinking,’ or ‘reversing’ it. Given a vector space V over a field K , we refer to the elements of the field K as scalars (real or complex). We will restrict our self to real vector space, hence, we use field R instead of field K . Given an element v of the vector space V , we shall refer to elements of V that are of the form c v for some scalar c as scalar multiples of v . A vector space over the field R consists of a set V on which is defined an operation of addition (denoted by ), associating to elements u and v of V an element u v of V ; and an operation of multiplication by scalars, associating to each element c of R and to each element v of V an element c v of V , where the following properties holds true: u, v V u v V u v v u u , v V (i.e., vector addition is commutative); (u v ) w u ( v w ) u, v, w V (i.e., vector addition is associative); an element 0 of V (known as the zero element) with the property that u 0 u u V ; given any element u V , there exists an element u V with the property that u ( u) 0 c (u v ) c u c v u, v V and c R ; (c d )u (c u d u) u V and c, d R c ( d u) (cd ) u u V and c, d R ; 1u u u V , where, 1 is the multiplicative identity element of the field R .
For justifying the above axioms, let, we have 2 - tuple vector (i , j ) , here this 2 - tuple vector is making a ‘two dimensional vector space V ,’ which can be considered as a plane, which contains
infinite vectors (in the form of i j , , R ). Now, justify yourself, by putting u i & v j . A vector space V over a field R is said to be trivial if it consists of a single element (which must then be the zero element of V ). A vector space with more than one element is said to be non-trivial. Let u , v be elements of a vector space V , then a unique element x of V satisfying x v u Let V be a vector space over a field R . Then c 0 0 and 0 v 0 c R , v V . Let V be a vector space over a field R . Then ( c ) v (c v ) , c ( v ) (c v ) c R , v V . Let V be a vector space over a field R . Then u v u ( 1) v u, v V . Let V be a vector space over a field R . Let c v 0 , then either v 0 or c 0 c R , v V . A vector space is free of any dependence of a ‘coordinate system.’
Let we have two ‘2 - tuple’ vectors, a i j & b 3i 2 j . Solving these two vectors in terms of
i & j , i 2a b & j 3a b . Let there is another vector, v 4i 5 j , then in terms of a & b ,
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[1.51]
we get, v 7 a b . Thus, v (4, 5) w.r.t. ( i , j ); v (7, 1) w.r.t. ( a , b ) . Hence, the vector v is
same but the two ‘2 - tuple’ are not same (as shown in Figure). So, we can conclude that, v ci d j
has unique representation w.r.t. ( i , j ) coordinate system; v a b has unique representation
w.r.t. ( a , b ) coordinate system. Hence, in general, v exist independently of choice of (i , j ) or ( a , b ) .
Sub-Space: Let V be a vector space over a field R . A non-empty subset U of V is said to be a subspace of V if u v U and c u U u, v U , c R . (Thus the sum of two elements of a subspace of V is required to be an element of the subspace, as is any scalar multiple of an element of
that subspace.) Let V (i , j ) be a 2-dimensional vector space; and we have a set of only two
individual vectors (not a plane), i.e., A {i , j} , then A is a subset of two dimensional vector space.
But i j A & 2i A . So, we cannot say that A is a subspace of 2-dimensional vector space.
Let, V (i, j , k ) be a three dimensional vector space; and we have a two dimensional vector space
i.e., A (i , j ) , since i j A & 2i A , hence we can say that A is a subspace of V . Let V be a vector space over a field R . Then any subspace of V is itself a vector space over R . If A is a m n matrix then the solution set U of the homogeneous linear system Ax 0 is a n subspace of R .
1.4.2 Linear Dependence, Spanning Sets and Bases Linear Combination of Vectors: Let v1 , v 2 , , v k be elements of some vector space V over a given field R . An element specified by an expression of the form c1 v1 , c2 v 2 , , ck v k 0 , where c1 , c2 , , ck are scalars (i.e., elements of the field R ) is said to be a linear combination of the
elements v1 , v 2 , , v k . A vector w V is a linear combination of vectors v1 , v 2 , , v n R n iff
v1
v2
v n x w has a solution, where x c1
Example 1.83: Can w 0
T
2 1
T
v 2 3 5 2 and v 3 2 6
c2
T
be expressed as a linear combination of v1 1 2 1 ,
T
4 ?
Solution: w is a linear combination of v1 and v 2 if v1
1 3 2 c1 0 system is 2 5 6 c2 2 1 2 4 c3 1 By Applying 1 3 2 | R2 R2 2 R1 0 1 2 | R R R 3 3 1 0 1 2 | By Applying 1 0 8 | R1 R1 3R2 0 1 2 | R R R 3 3 2 0 0 0 |
T
c3 .
v2
v 3 x w has a solution. The given
1 3 2 | 0 and its augmented matrix is given as 2 5 6 | 2 1 2 4 | 1 0 1 3 2 | 0 By Applying 2 0 1 2 | 2 R2 R2 / ( 1) 0 1 2 | 1 1 6
c1 8c3 6 (i) 1 0 8 c1 6 2 0 1 2 c2 2 c2 2c3 2 (ii) . From (iii), 1 0 0 0 c3 1 0 1 (iii) we conclude that the system is inconsistent, hence w is not a linear combination of the given vectors.
Example 1.84: Express the vector w 1 0
v1 7
T
0
T
1
as a linear combination of the vectors
T
6 4 5 and v 2 3 3 2 3 .
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Chapter 1: Linear Algebra
[1.52]
Solution: w is a linear combination of v1 and v 2 if v1
v 2 x w has a solution, i.e., the given
7 3 1 6 3 c 1 0 system is and its augmented matrix is 4 2 c2 0 5 3 1 1 0 | 1 By Applying 1 0 R2 R2 6 R1 0 3 By Applying 6 3 | 0 R R 4 R R R R 4 2 | 0 2 3 3 1 0 1 1 2 3 5 3 | 1 R 4 R4 5 R 1 0
7 3 6 3 4 2 5 3 | 1 | 6 | 4 | 6
1 0 |
|
1
|
0
|
0
|
1
.
1 0 | 1 1 0 | 1 By Applying By Applying 0 1 | 2 0 1 | 2 0 1 | 2 . R R 2 R R R 2 R R 3 3 1 3 3 1 0 2 | 4 0 2 | 4 0 0 | 0 R2 2 R 3 R4 R4 3R1 4 R4 3 R1 0 3 | 6 0 3 | 6 0 0 | 0 7 3 3 6 c1 1 To get c1 , c2 we have w 1 ( 2) w 1 v1 2 v 2 4 2 c2 2 5 3 Span of Vectors: Let V be a vector space. The span of vectors v1 , v 2 , , v n V is the set span( {v1 , v 2 , , v n } ) { w V : w is a linear combination of v1 , v 2 , , v n }. Also v1 , v 2 , , v n span V By Applying
1
if V span( v1 , v 2 , , v n ). The span of a subset of vector space is a subspace.
Any set of vectors in 2 which contains two non-collinear vectors will span 2 . Any set of vectors in 3 which contains three non-coplanar vectors will span 3 . Two non-collinear vectors in 3 will span a plane 3 .
Linear Dependence and Independence of vectors
Elements v1 , v 2 , , v n of V are said to be linearly dependent if there exist scalars c1 , c2 , cn , not all zero, such that
v1
v2
c1 v1 c2 v 2 ck v n 0 , i.e., any one of the solution of
v n x 0 is non-zero, where x c1
T
cn .
c2
Elements v1 , v 2 , , v n of V are said to be linearly independent if and only if there exist scalars c1 , c2 , , cn 0 , such that c1 v1 c2 v 2 ck v n 0 , i.e., the only solution of the equation
v1
v2
v n x 0 , where x c1
c2
T
cn , is the trivial solution in which the
scalars c1 , c2 , , cn are all zero.
If two vectors u and v in a plane are linearly independent, then, they cannot be linear [This point was asked in EE-1994 (1 mark)]. n If S is a set of more than n vectors in R then S is linearly dependent.
Example 1.85 [ME-2014 (1 mark)]: Which one of the following describes the relationship among the three vectors, iˆ ˆj kˆ , 2iˆ 3 ˆj kˆ and 5iˆ 6 ˆj 4 kˆ ? (a) The vectors are mutually perpendicular (c) The vectors are linearly independent
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(b) The vectors are linearly dependent (d) The vectors are unit vectors
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Solution (b): As the dot product of any two given vectors is not zero so the vectors are not mutually perpendicular; so option (a) is not correct. Also, any of the given vectors is not an unit vector so option (d) is not correct. Now we have to decide between option (b) or (d). Since, the vectors are linearly independent if there is a trivial solution of v1 v 2 v3 x 0 , where x c1 The system of equation and its augmented matrix is given as: 1 2 5 c1 0 1 2 5 | 0 By Applying 1 2 5 | 0
1 3 6 c 0 and 1 3 6 | 0 R R R 0 1 1 | 2 2 1 2 2 2 | 1 1 4 c3 0 1 1 4 | 0 R 3 R3 R 1 0 By Applying 1 0 3 | 0 1 0 3 c1 0 c1 3c3 0 R1 R1 2 R2 0 1 1 | 0 0 1 1 c2 0 c2 c3 0 0 0 0 | 0 0 0 0 00 R R 2 R c3 0 3 3 2
c2
T
c3 .
0
0 (i) (ii) (after
restoring
(iii)
the transformed matrix equation). From (i) and (ii) c1 3c2 3c3 and so we can say that all the scalars c1 , c2 and c3 are not zero. Thus the given vectors are linearly dependent. Example 1.86: Are the given vectors are linearly dependent: T
T
v1 0 1 1 , v 2 3 5 2 and v 3 2
T
3 4 ?
Solution: The vectors are linearly independent is there is a trivial solution of v1 where x c1
c2
v2
v3 x 0 ,
T
c3 . The system of equation and its augmented matrix is given as:
0 3 2 c1 0 0 1 5 3 c 0 and 1 2 1 2 4 c3 0 1 1 2 4 | By Applying 0 3 7 | R 2 R2 R1 0 3 2 | By Applying 1 0 2 3 R1 R1 2 R2 0 1 7 3 9 R R 3 R 3 3 2 0 0
3
2
| 0
1 2 4 | By Applying 5 3 | 0 1 5 3 | R1 R3 0 3 2 | 2 4 | 0 0 1 2 4 | 0 By Applying 0 1 7 3 | 0 0 R2 R2 / 3 0 3 0 2 | 0 | 0 1 0 2 3 | 0 By Applying | 0 0 1 7 3 | 0 . R 3 R3 / 9 | 0 0 0 1 | 0
0 0
0
Now restoring the
1 0 2 3 c1 0 7 transformed matrix equation we get, 0 1 7 3 c2 0 c3 0 , c2 c3 0 c2 0 , 3 0 0 1 c3 0 c1 (2 3)c3 0 c1 0 . As all c1 , c2 and c3 are zero column vectors are linearly independent.
Bases for vector space : A finite set S of vector space V is called a basis for V provided that (1) The vectors in S are linearly independent (2) The vectors in S span V , i.e. span (S) V . For e.g., the set
1
0
T
T
0 , 0 1 0 , 0
T
0 1
is a basis for R . 3
Suppose v1 , v 2 , , v n R n are linearly independent vectors. Consider any other vector w R n . Then v1 , v 2 , , v n , w are linearly dependent. n
If v1 , v 2 , , v n R n are linearly independent vectors then v1 , v 2 , , v n is a basis of R . Any two bases for a vector space have the same number of vectors. The number of vectors a basis of a vector space is called the dimension of the space.
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Engineering Mathematics
Let
Chapter 1: Linear Algebra
[1.54]
V be an n -dimensional vector space, and S a subset of V . Then, If S is linearly independent and consists of n vectors then S is a basis of V ; If S spans V and consists of n vectors then S is a basis of V ; If S is independent then S is contained in a basis of V ; If S spans V then S contains is a basis of V . 1 3 10 5
Example 1.87: Find the basis for A 1
1
4
11
2 .
3
8
1
Solution: The linear system of equation Ax 0 (where x c1
c2
x1 1 3 10 1 3 10 5 0 x 2 matrix is given as: 1 4 11 11 2 0 and 1 4 x3 1 3 1 3 8 8 1 0 x4 By Applying 1 3 10 5 | 0 1 3 By Applying R2 R2 R1 0 7 21 7 | 0 0 1 R2 R2 / 7 0 6 6 18 6 | 0 R 3 R3 R 1 0
c3 5
T
c4 ) and its augmented | 0
2 | 0
1 | 0
10
5
| 0
3
1 | 0
18
6 | 0
c1 1 0 1 2 | 0 1 0 1 2 0 c1 c3 2c4 0 (i) c2 R1 R1 3R2 0 1 3 1 | 0 0 1 3 1 0 c2 3c3 c4 0 (ii) . c3 0 0 0 | 0 0 0 0 0 0 R 00 (iii) 3 R3 6 R 2 0 c4 By Applying
We have two equations and four variable, we choose c3 s , c4 t . So, all solutions have the form:
c1 s 2t 1 2 c 3s t 3 1 2 s t . So, c3 s 1 0 0 1 c4 t
1 2 3 1 , is a basis of the solution space. 1 0 0 1
Example 1.88 [CS-2014 (1 mark)]: If V1 and V2 are 4-dimensional subspaces of a 6-dimensional vector space V , then the smallest possible dimension of V1 V2 is ……………. Solution: Let the basis of 6-dimensional vector space be {e1 , e2 , e3 , e4 , e5 , e6 } . In order for V1 V2 to have smallest possible dimension V1 and V2 could be, say, {e1 , e2 , e3 , e4 } and {e3 , e4 , e5 , e6 } , respectively. The basis of V1 V2 would then be {e3 , e4 } ; so smallest possible dimension is 2. Example 1.89 [EC-2014 (1 mark)]: For matrices of same dimension M , N and scalar c , which one of these properties DOES NOT ALWAYS hold? (a) ( M T )T M (b) (cM )T c ( M )T (c) ( M N )T M T N T (d) MN NM Solution (d): Since matrix multiplication is not commutative in general; so for matrices of same dimension M and N , MN NM does not always holds.
1.4.3 Dot – Product, Norm and Angle for – Space We define the norm of an n tuple vector, and the dot product and angle between two n tuple vector in terms of the components of the n tuples since the graphical and geometric arguments used for arrow vectors will not be possible for n 3 . Thus if u (u1 , u2 , , un ) , the norm or length or modulus of u is defined as,
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Chapter 1: Linear Algebra
2
2
[1.55]
n
2
2 ui i 1
u u1 u2 un
(1.28)
Similarly, the dot or scalar product of u (u1 , u2 , , un ) and v (v1 , v2 , , vn ) is defined as, n
u v u1v1 u2 v2 un vn ui vi
(1.29)
i 1
Also, angle between u (u1 , u2 , , un ) and v (v1 , v2 , , vn ) is defined as,
u v n ui vi u v i 1
cos 1
n
n
ui2
i 1
vi2 , 0
i 1
(1.30)
Other dot products and norms are sometimes defined for n space, but we choose to use above equations which are known as Euclidean dot product and Euclidean norm, respectively. To signify that the Euclidean dot product and norm have been adopted, we henceforth refer to the space as n Euclidean n space, rather than just n space. We will still denote it by the symbol .
Orthogonality of two vectors: If u (u1 , u2 , , un ) and v (v1 , v2 , , vn ) are two non – zero vectors such that u v 0 then the angle between the two vectors u and v becomes 2 then we say that the two vectors u and v orthogonal. If we have a set of non – zero vectors {u1 , u 2 , , u k } then we have orthogonal set of vectors if and only if every vector in the set is orthogonal to every other one, i.e., u i u j 0 i j .
Two vectors v1 and v 2 are orthogonal iff each is non-zero and the dot product v1 v 2 0 , i.e., ( v1 )T ( v 2 ) 0 . For e.g., the vectors T
3 3
2 2 3 6 6 0 ; the vectors
2 T
2
and
2
3
are orthogonal because
and 1 3 are not orthogonal because
T
2 1 3 3 6 0 .
Three vectors v1 , v 2 and v 3 are orthogonal iff each is non-zero and they are pairwise orthogonal,
2 0
3 3
0 , 0
0
T
i.e., 2
2 0 0 0
( v1 )T ( v 2 ) 0; ( v 2 )T ( v 3 ) 0; ( v 3 )T ( v1 ) 0 .
0 , 0
0
2
are
orthogonal
because,
For
2
e.g., T
0 0 0
the
vectors
2 0 0;
T
2 0; 0 0
2 2 0 0 0 . n
If n vectors ( v1 , v 2 , , v n ) each of which is in R are orthogonal, then they are surely linearly n
n
independent and hence span R and therefore form a basis for R . For, e.g., The vectors 2 0 0 , 0 2 0 , 0 0 2 are orthogonal and hence are linearly independent and hence 3
3
span R . They form a basis for R . Orthogonal Projection: Let u and v be two vectors (not orthogonal) in a dot product space V . uv If v 0 , then orthogonal projection (OP) of u onto v is given by (OP) v u v vv
uv v vv
The vector ( w ) which is orthogonal to v is given as: w u (OP) v u u
The orthogonal projection (OP) of w onto the subspace spanned by the orthogonal vectors u w u w v and v is given by w (OP)u w (OP) v w u v . uu v v Normalization: A vector of unit length is called a unit vector. Any non – zero vector u can be scaled to have unit length by multiplying it by 1 u so we say that the vector uˆ as, uˆ u u .
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Chapter 1: Linear Algebra
[1.56]
Orthonormal: A set of vectors {u1 , u 2 , , u k } is said to be orthonormal if it is orthogonal, i.e., u i u j 0 i j and if each vector is normalized, i.e., u j u j 1 j . [This point was asked
in MN-2011 (1 mark)]. Thus the set of vectors {u1 , u 2 , , u k } is orthonormal if u j u j ij i, j , where, the symbol ij (also known as Kronecker delta) is defined as,
i j
1,
ij
0, i j
.
Example 1.90 [EE-1994 (2 marks)]: Find the number of linearly independent solutions of the system 1 0 2 x1 of equations 1
1 0 x2 0 2 2 0 x3
1 0 2 | 0 Solution: The augmented matrix of the given system is 1 1 0 | 0 2 2 0 | 0 By Applying 1
0 2 | 0 1 0 2 | 0 By Applying R2 R2 R1 0 1 2 | 0 0 1 2 | 0 R2 R2 0 2 4 | 0 0 2 4 | 0 R R 2 R 3 3 1
1 0 2 | 0 1 0 2 x1 0 x1 2 x3 0 (i) 0 1 2 | 0 0 1 2 x 0 x 2 x 0 (ii) (after restoring 2 3 2 R R 2 R 3 3 2 0 0 0 | 0 0 0 0 x 0 0 0 (iii) 3 By Applying
the
transformation
X 2k
matrix).
Now
T
k k 2
2k
let
x3 k x1 x2 2k .
So
the
solution
set
is
T
2 1 we have only one independent solution.
Example 1.91 [ME-1995 (2 mark)]: Among the following, the pair of vectors orthogonal to each other is (a) [3, 4, 7],[3, 4, 7] (b) [1, 0, 0],[1,1, 0] (c) [1, 0, 2],[0, 5, 0] (d) [1,1,1],[ 1, 1, 1] Solution (c): The given two vectors are orthogonal if their dot product is zero. We can verify that the dot product of the vectors is zero if we take the vectors in option (c). Statement for Linked Answer Question 1.92 and 1.93: T
T
T
p 10 1 3 , q 2 5 9 , r 2 7 12 are three vectors. Example 1.92 [EE-2006 (2 marks)]: An orthogonal set of vectors having a span that contains p , q , r is (a)
6 4
T
3 6 , 4
T
2 3
T
(b) 4
T
2
T
4 , 5 7
T
T
T
11 , 8 T
2
T
3
T
(c) 3 11 , 1 31 3 , 5 3 4 (d) 6 7 1 , 3 12 2 , 3 9 4 Solution (a): We are looking for orthogonal vectors having a span that contains p , q and r . Taking T
option (a), 6 3 6 , 4 product 24 6 18 0 .
k1 6
T
3 6 k2 4
T
2 3 these are orthogonal, as can be seen by taking their dot The space spanned by these two vectors is T
2 3 . The span of 6
and r . We can show this by successively setting k1 6
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T
3 6 and 4 T
3 6 k2 4
T
2 3
contains p , q
T
2 3 to p , q and r
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[1.57]
one by one and solving for k1 and k2 uniquely. It should be noted that the vectors in option (b), (c) and (d) are none of them orthogonal, which can be seen by taking pairwise dot products. Example 1.93 [EE-2006 (2 marks)]: The following vectors is linearly dependent upon the solution to the previous problem T
(a) 8 9 3
T
(b) 2
17 30
(c)
4
T
4 5
(d) 13 2
T
3
T
Solution (b): We have to find which of the given vectors is linearly dependent with 6 and 4
3 6
T
2 3 . For linearly dependent vectors the determinant of the matrix, whose first two T
T
column elements are 6 3 6 , 4 2 3 and third column element is taken one of the given vectors in options, must be zero. By doing so we find option (b) is correct as 6 4 2 3
2
17 6( 60 51) 4( 90 102) 2( 9 12) 0
6
3
30
Example 1.94 [CS-2007 (2 marks)]: Consider the set of (column) vectors defined by
T
T
X x R 3 x1 x2 x3 0 , where x x1 , x2 , x3 . Which of the following is TRUE?
1, 1, 0 (b) 1, 1, 0
T
(a)
T
is a basis for the subspace X . , 1, 0, 1 is linearly independent set, but it does not span T
, 1, 0, 1
T
X and therefore is not a
basis of X 3 (c) X is not a subspace of R (d) None of these Solution (a): To be basis for subspace X , two conditions are to be satisfied: (i) the vectors have to be linearly independent (ii) they must span T
xT x1 , x2 , x3 . Since,
1, 1, 0
T
T
, 1, 0, 1
X . Here
X x R 3 x1 x2 x3 0 , where
is a linearly independent set because one cannot be
obtained from another by scalar multiplication. The fact that it is independent can also be established 1 1 0 by seeing the rank of is 2. Next, we need to check if the set spans X . Here 1 0 1
k1 k 2 X x R 3 x1 x2 x3 0 , the general infinite solution of X k1 . Choosing k1 , k2 as k1 k1 0 k1 k T k k and k 0 , we get two linearly independent solutions for X as, X k 0 k 2 2 or X k
k
T
T
0 . Now since both of these can be generated by linear combination of 1, 1, 0
T
and 1, 0, 1 , the set spans X . Since we have shown that the set is not only linearly independent but also spans X , therefore by definition it is a basis for the subspace X . Example 1.95 [EC-2007 (2 marks)]: It is given that X 1 , X 2 , , X M are M non-zero, orthogonal vectors. The dimension of the vector space spanned by the 2M vectors X 1 , X 2 , , X M , X 1 , X 2 , , X M , is (a) 2M
(b) M 1
(c) M
(d) dependent on the choice of X 1 , X 2 , , X M
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[1.58]
Solution (c): Since X 1 , X 2 , , X M are orthogonal, they span a vector space of dimension M . Since X1 , X 2 , , X M
are
linearly
dependent
X1 , X 2 , , X M ,
on
so
the
set
X 1 , X 2 , , X M , X 1 , X 2 , , X M will also span a vector space of dimension M only.
Example 1.96 [EE-2007 (2 marks)]: Let x and y be two vectors in a 3 dimensional space and
x, x x, y y, x y , y
x, y denote their dot product. Then the determinant det
(a) (b) (c) (d)
is zero when x and y are linearly independent is positive when x and y are linearly independent is non-zero for all non-zero x and y is zero only when either x or y is zero
x, x Solution (b): Let x x1iˆ x2 ˆj x3kˆ , y y1iˆ y2 ˆj y3kˆ , det y, x x12 x22 x32
x1 y1 x2 y2 x3 y3 y12
y22
y32
x, y y , y
( x1 y2 x2 y1 ) 2 ( x1 y3 x3 y1 ) 2 ( x2 y3 x3 y2 ) 2
x1 y1 x2 y2 x3 y3 Now, 0 if x and y are linearly dependent so option (a) , (c) and (d) is not correct. If x and y are linearly independent then 0 . So option (b) is correct.
Example 1.97 [EE-2008 (2 marks)]: Let P be a 2 2 real orthogonal matrix and x is a real vector
x1
T
x2 with length || x || ( x12 x22 )1/2 . Then which one of the following statement is correct?
(a) || Px || || x || where at least one vector satisfies || Px || || x ||
(b) || P x || || x || for all vectors x
(c) || Px || || x || where at least one vector satisfies || P x || || x ||
(d) No relationship can be established between || x || and || Px ||
cos
sin
Solution (b): Let a 2 2 real orthogonal matrix be P
. cos
sin
cos
So P x
sin
x1 cos x2 sin cos x2 x1 sin x2 cos
sin x1
Px ( x1 cos x2 sin ) 2 ( x1 sin x2 cos ) 2
Example 1.98 [AE-2013 (2 marks)]: Values of a , b and c , which render the matrix
1 / 3 1 / 2 Q 1 / 3 0 1 / 3 1 / 2
(a)
1
a
b c
orthonormal
are,
x12 x22 x for any vector x
(c)
,
2 1 3
1
,0
(b)
1
2 ,
1 3
,
1
(d)
,
6 1
3
6
2
,
1
,
6 2
6 ,
1
6
6
3 1
3 ,
respectively Solution (d): From the given matrix we have three vectors u1 1
T
T
3 1
T
u 2 1 2 0 1 2 and u3 a b c . For these set of vectors to be orthonormal if it is orthogonal and if each vector is normalized. For the set of vectors to be orthogonal,
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[1.59]
u i u j 0 i j u1 u 2 0 , u 2 u 3 0 and u 3 u1 0 . As u1 u 2 0 is true; and
u 2 u3 0
a
2
c
0ac
…(i)
2
u 3 u1 0 u 2 u 3 0
a
b
c
0 a b c 0 …(ii) 3 3 3 For each vector to be normalised, the given vectors must have unit modulus, i.e., 2
2
2
u1 u1 1 ,
u 2 u 2 1 and u 3 u3 1 . As u1 u1 1 , u 2 u 2 1 ; and u 3 u 3 1 a b c 1
…(iii)
Putting (i) and (ii) in (iii) we get a 2 1 6 a 1
6
b 1
and (i) c 1
6
and (ii)
6 . So from the given options we can say that option (d) is correct.
Example 1.99 [MN-2014 (1 mark)]: The value of k for which the vectors a 2iˆ 3 ˆj and b kiˆ 4 ˆj are orthogonal to each other is _____.
Solution: The given two vectors are orthogonal if their dot product is zero, i.e. a b 0 (2iˆ 3 ˆj ) ( kiˆ 4 ˆj ) 0 2k 12 0 k 6 . So the answer is k 6 . Example 1.100: Find the orthonormal basis of vectors 1 2 1 , 2 1 4 , 3 Solution:
3
1
Since,
T
2 1 2 1 4 0 ;
2
2 1 . T
1 4 3 2 1 0 ;
T
2 1 1 2 1 0 , hence the three vectors are orthogonal and hence they are linearly 3
3
independent and hence span R . To convert this set to an orthonormal basis of R , we need to divide each vector by its length. Hence,
v 3 32 (2)2 12 14 .
So,
v1 12 2 2 12 6 ; an
orthonormal
v 2 22 12 ( 4) 2 21 ; basis
of
R3
is
1 4 3 2 1 1 2 1 2 , , , , , , , . , 6 6 6 21 21 21 14 14 14
Rotation Matrix
Computing the rotation matrix when axis and angle are given: Given the unit vector axis a and angle , the rotation matrix is obtained by applying the following relation,
R I 2 sin 2 ( 2) aaT I sin A , where I is the identity matrix, aaT denotes a matrix
corresponding to a 3 3 Cartesian matrix with components
a1a1 [aa ]ij a2 a1 a3 a1 T
a1a2 a2 a 2
a1a3
0 a2 a3 . The matrix [ A ] a3 a3 a3 a2
a3 a2 matrix, associated with the axis a .
a3 0 a1
(aa )ij ai a j , written as
a2 a1 is the skew symmetric
0
Computing axis and angle when the rotation matrix is given: Given R , below an algorithm for finding the principal axis a and principal angle of rotation: 1 1 Step 1: Compute c tr ( R ) 1 ( R11 R22 R33 1) 2 2 Step 2: Decide whether the angles is an integral multiple of 180 o by checking the cosine and follow one of the following: If c 1 , proceed to step 3
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[1.60]
c 1 , then R I . Hence, the angle of rotation is 0 and the axis of rotation is arbitrary, so we can set it to {1, 0, 0} or any other convenient unit vector. Go to step 6.
If c 1 , then the angle of rotation is 180 o . To find the axis of rotation, simply normalize any nonzero column of R I . Go to step 6. Step 3: Compute the angle of rotation cos 1 c , 0 180 o
Step 4: Compute sin 1 cos 2 Step 5: Compute the axis of rotation a1
R32 R23 2 sin
, a2
R13 R31 2 sin
, a3
R21 R12 2 sin
Step 6: Stop
0 0 1 Example 1.101 [IN-2009 (2 marks)]: The matrix P 1 0 0 rotates a vector about the axis 0 1 0
1
1 1 by an angle of (a) 30o (b) 60o (c) 90o (d) 120o 1 1 1 1 Solution (d): c tr ( P) 1 (0 0 0 1) cos 1 120o 2 2 2 2
1.4.4 Rank of a Matrix The rank of a matrix A aij
m n
is the maximum number of linearly independent row vectors of A
(called the row rank) or maximum number of linearly independent columns vectors of A (called the column rank). [This point was asked in CE-2000 (1 mark)]. For every matrix, the column rank is 1 2 2 equal to the row rank. It is denoted by r ( A) . For e.g., the matrix 0
0 3 has rank 2, the first 0 0 0
two row vectors are linearly independent, whereas all three row vectors are linearly dependent. Echelon Form of a Matrix: A matrix A is said to be in Echelon form if either A is null matrix or A satisfies the following conditions: (i) Every non-zero row in A precedes every zero row; (ii) The number of zeros before the first non-zero element in a row is less than the number of such zeros in the next row. It can be easily proved that the rank of a matrix in Echelon form is equal to the number of non-zero rows (or linearly independent rows) of the matrix. For e.g., consider the two matrices 0 3 2 1 0 2 5 A 0
0 2 5 and B 0 0 1 . The rank of the matrix A is r ( A) 2 because it is in 0 0 0 0 0 0 4
Echelon form and it has two non-zero rows; but the matrix B is not in Echelon form, because the number of zeros in the second row is not less than the number of zeros in the third row. By applying, 0 2 5 R3 R3 4 R2 , matrix B B 0
0 1 , which is in Echelon form and contains two non-zero 0 0 0
rows and hence r ( B ) 2 . It is to be noted that ‘Elementary transformations do not alter the rank of matrix.’
Algorithm for finding the rank of a matrix: Let A aij mn be an m n matrix. The following are the steps for finding the number of non-zero row (or number of independent rows) which is equal to the rank of matrix.
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[1.61]
Step 1: Using elementary row transformations make a11 1 . Step 2: Make a21 , a31 , , am1 all zeros by using elementary transformation: R2 R2 a21 R1 , R3 R3 a31 R1 , , Rm Rm am1 R1 .
Step 3: Make a22 1 by using elementary row transformations. Step 4: Make a32 , a42 , , am 2 all zeros by using elementary transformation: R3 R3 a32 R2 , R4 R4 a42 R2 , , Rm Rm am 2 R2 .
The process used in Step 3 and 4 is repeated utpo ( m 1) th row. Finally we obtain a matrix in Echelon form which is equivalent to matrix A . The rank of A will be equal to number of non-zero rows in it. 0 0 3 Example 1.102 [CS-1994 (1 mark)]: The rank of the matrix A 9
3
3 1
(a) 0 (b) 1 Solution: For finding the rank 3 By Applying vectors. A 9 R 3 R1 0
5 is
1 (c) 2 (d) 3 of matrix, we have to find the number of linearly independent row 1 1 1 1 3 1 3 By Applying 3 5 A 9 3 5 R1 R1 3 0 3 0 0 3
1 1 3 1 3 By Applying 1 1 3 1 3 A 0 0 2 A 0 0 2 3 R R 9 R R R R 2 2 3 3 2 1 0 2 0 0 3 0 0 By Applying
As we have two independent rows, so its rank is 2. [Similar questions were also asked in ME-1994, CS-2002 (1 mark), EC-2006 (2 marks)] Example 1.103 [EE-1994 (1 mark)]: A 5 7 matrix has all its entries equal to –1. The rank of the matrix is (a) 7 (b) 5 (c) 1 (d) 0 Solution (c): In the given matrix if we choose operation Ci Ci C1 , i 2, 3, 4, 5 ; then we get C1 containing all elements equal to –1 and all other columns C i ( i 2, 3, 4, 5 ) containing all elements equal to 0. Hence we have only one non-zero column or independent column, so its rank is 1. Example 1.104 [CS-1995 (1 mark)]: The rank of the following ( n 1) (n 1)
1 1 A 1
a
a2
an
a
a2
an , where a is a real number is
matrix
2 n a a a (a) 1 (b) 2 (c) n (d) Depends on the value of a Solution (a): If we carefully notice the given matrix, we found we have only one independent row as all the elements in each row are same. So its rank is 1. 4 2 1 3
Example 1.105 [CE-2003 (1 mark)]: The rank of the matrix 6
3 4 7 is 2 1 0 1
(a) 4
(b) 3
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(c) 2
(d) 1
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[1.62]
Solution (c): For finding the rank of matrix, we have to find the number of linearly independent row 34 By Applying 1 1 2 1 4 3 4 By Applying 1 1 2 1 4 vectors.
6 R1 4 2 R1
3
4
1
0
7 R2 R2 6 R1 0
1
R R 2 R 3 3 1 0
0
5 2
5 2
0
1 2
1 2
By Applying 1 1 2 1 4 3 4 R2 0 0 5 2 5 2 . We have two independent row vectors, so its rank is 2. R3 R3 0 0 5 0 0 [Similar question was also asked in CE-2014 (2 marks)].
Example 1.106 [EE-2007 (1 mark)]: x x1
n n matrix V xxT (a) has rank zero (b) has rank 1 Solution: After 2 x1 x1 x2 x1 x3 x1 xn
x2 x1 V x3 x1 x x n 1
2 x2
x2 x3
x3 x2
x3
xn x2
xn x3
2
x2
T
xn is an n tuple non-zero vector. The (d) has rank n We
(c) is orthogonal multiplication, x12
By Applying x2 xn 0 Ri R1 x3 xn Ri V 0 xi x1 for all 1 i n 2 0 xn
x1 x2
x1 x3
0
0
0
0
0
0
have,
x1 xn
0 0 .
As
0
xi (1 i n) 0 . So we have only one row ( R1 ) which is independent, so its rank is 1.
Example 1.107 [PI-2007 (2 marks)]: q1 , , qm are n dimensional vectors, with m n . This set of vectors is linearly dependent. Q is the matrix with q1 , , qm as the columns. The rank of Q is (a) less than m (b) m (c) between m and n (d) n Solution (a): As Q is a m n matrix so rank (Q ) min( m, n) rank (Q ) m , as m n . But the column vectors are linearly dependent so rank (Q) must not be m rank (Q ) m .
Alternative method for finding the rank of a matrix Sub-matrix: Let A be an m n matrix. A matrix obtained by leaving some rows or columns or both 2 1 0 1 2 1 0 of A is called a sub-matrix of A . For e.g., for matrix A 3
2
2
2 5 3
4 , 3
1
2 2 and 2 5 3
2 2 5 3 are sub-matrices of matrix A . Any matrix A itself is a sub-matrix of A because it is obtained by leaving no rows or columns of A . Rank of a Matrix: A number r is said to be the rank of an m n matrix A if (i) Every square submatrix of order ( r 1) or more is singular; (ii) There exists at least one square sub-matrix of order r which is non-singular. Thus, the rank of a matrix is the order of the highest order non-singular square sub-matrix. For e.g., if A is a given matrix of rank 2, then every square sub-matrix of order 3 or more is singular and there exists at least one square sub-matrix of order 2 which is non-singular. If A is a non-singular square matrix of order n , then its rank is n . If A is a m n matrix, then r ( A) min(m, n) . Because the largest order square sub-matrix of A will be of the order equal to min( m, n) .
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Chapter 1: Linear Algebra
1 4 8 0 0 3 Example 1.108 [CS-1998 (2 marks)]: The rank of the matrix A 4 2 3 3 12 24
[1.63]
7 0
is
1
2 (a) 3 (b) 1 (c) 2 (d) 4 Solution (d): We have 4 4 matrix. For finding its determinant we will expand it along second row 1 4 7
as it contains maximum number of zeros. A 3 4
2
3 12
1 3 266 798 0 A is a non2
singular square matrix of order 4, so its rank is 4.
3 2 12 Example 1.109 [ME-1999 (1 mark)]: Rank of the matrix is A 6 4 8 12 8 36 (a) 1 (b) 2 (c) 3 (d) 2 Solution (b): We first check the determinant of the given matrix have 3 3 matrix. 3 2 12 3 2 12 A 6
4
12
8
8 2 4 3 36
3
2
4 (by taking common –2 and 3 from R2 and R3 respectively).
2
9
As we have C1 and C 2 are same, so A 0 A is a singular matrix and so its rank is not 3. Now let us choose the square sub-matrix of order 2 2 by leaving first column and third row we get 2 12 16 48 64 0 which is non-singular. As we get a non-singular square sub-matrix of 4 8 order 2 2 , hence the given matrix has rank 2. [Similar question was also asked in BT-2012 (2 marks)]
Some important points regarding Rank of a Matrix
If A is a m n matrix, i.e., matrix A have m row vectors with each having n components and also have n column vectors with each having m components; then (i) ≤ (ii) ≤ . The common values among row rank of A and column rank of A is simply called the rank of A . Hence, × ≤ ( , ), where, min ( , ) denotes the smaller of the two numbers and . [This point was asked in EC-1994 (1 mark)] Since, rank r of a matrix A equals the maximum number of linearly independent row/column T vectors of A . Hence A and its transpose A have the same rank. [This point was asked in MN2007 (1 mark)]. If a matrix A aij
m n
has m row vectors with n( m) components.
If the matrix A has rank m , then the row vectors are linearly independent. If the matrix A has rank m , then the row vectors are linearly dependent. In any matrix p vectors with each having q p components are always linearly dependent. The rank of matrix A is zero if and only if A O , i.e., if A is a null matrix. If A is a n n square matrix and is non-singular then its rank is n . [This point was asked in CE-2000 (1 mark)]. If A is a singular matrix then its rank is less than n , not n . [This point was asked in MT-2010 (1 mark)] Row/Column-equivalent matrices have the same rank. If matrix B is obtained from matrix A by elementary row/column operation then matrix B is called to be row/column equivalent to matrix A .
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[1.64]
The maximum number of linearly independent row/column vector of a matrix does not change if we change the order of rows/column or multiply a row/column by a non-zero constant or take a linear combination by adding a multiple of a row/column to another row/column. Hence rank is invariant under elementary row/column operations. If a row reduced echelon matrix A has r non-zero rows, then rank ( A) r . Let A be an m n matrix. n The row space Row( A ) of A is the subspace of R spanned by the rows of A . The dimension of Row( A ) is called row rank of A . m The column space Col( A ) of A is the subspace of R spanned by column of A . The dimension of Col( A ) is called the column rank of A . If A is m n matrix; and B is n k matrix, then rank( AB ) rank( A) rank( AB ) rank( B ) rank( AB ) rank( A) rank( B ) n rank( AB ) min rank( A), rank( B ) If A and B are both n n matrices with rank n , then then rank( AB ) n
Null-space: Let A be a m n matrix, then null-space of matrix A , N ( A) , is the set of all n dimensional column vectors x such that Ax 0 . The null-space N ( A) is the solution set of a system of linear homogeneous equations (with A as the coefficient matrix). n The null-space N ( A) is a subspace of the vector space . The dimension of null-space N ( A) is called the nullity of the matrix A . The rank of matrix A equals the number of nonzero rows in the row echelon form, which equals the number of leading entries. The nullity of A equals the number of free variables in the corresponding system, which equals the number of columns without leading entries, i.e. if A is m n matrix, then rank( A) nullity( A) n Example 1.110 [ME-2001 (2 marks)]: The rank of a 3 3 matrix C ( AB) , found by multiplying a non-zero column matrix A of size 3 1 and a non-zero row matrix B of size 1 3 , is (a) 0 (b) 1 (c) 2 (d) 3 Solution (b): a11 a11 a11b11 a11b12 a11b13 Let A a21 and B b11
a31
By Applying
b12
b13 , then C a21 b11
a11b11 R2 R2 a21 R1 a11 C 0 0 R R a R a 3 3 31 1 11
a11b12 0 0
b12 b13 a21b11 a21b12 a21b13 a31 a31b11 a31b12 a31b13 a11b13 0 we have one independent row, so rank is 1. 0
Example 1.111 [ME-2005 (1 mark)]: A is 3 4 real matrix and AX B is an inconsistent system of equations. The highest possible rank of A is (a) 1 (b) 2 (c) 3 (d) 4 Solution (c): rank( Amn ) min( m, n) Highest possible rank of A is min(3, 4) 3
2 4 6 Example 1.112 [CH-2007 (2 marks)]: A and B are two 3 3 matrix such that A 1 2 1 , 0 4 4 B 0 and AB 0 . Then the rank of matrix B is (b) r 3 (c) r 3 (d) r 3 (a) r 2 Solution (b): B 0 B is a null matrix and so its rank is 0. Thus that option (b) is correct.
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[1.65]
Example 1.113 [IN-2009 (1 mark)]: Let P 0 be a 3 3 real matrix. There exist linearly independent vectors x and y such that Px 0 and Py 0 . The dimension of range space of P is (a) 0 (b) 1 (c) 2 (d) 3 Solution (c): As we have two linearly independent vectors x and y such that Px 0 and Py 0 , so the rank of matrix P is 2 and so its dimension is 2. 0 1 1 Example 1.114 [IN-2013 (1 mark)]: The dimension of the null space of the matrix 1
1
1
0 is
0
1
(a) 0 (b) 1 (c) 2 (d) 3 Solution (b): For finding the dimension of the null space of the matrix, first we have to find the dimension of the matrix; and for the dimension of any matrix we have to find its rank. 1 1 0 1 1 0 1 1 0 By Applying By Applying By Applying 0 1 1 0 1 1 0 1 1 R R 3 R3 R1 3 R3 R 2 1 R2 1 0 1 R 0 1 1 0 0 0 So we have two independent row vectors so its rank is 2 and thus the given matrix is of 2 dimensional. Hence the dimension of the null space of the given matrix is 3 2 1 . p q Example 1.115 [EE-2014 (1 mark)]: Two matrices A and B are given as: A ; r s
p2 q2 B pr qs
pr qs
. If the rank of matrix A is N , then the rank of matrix B is
r 2 s2
(b) N 1
(a) N 2
(c) N
(d) 2N 2
2
p q q p q By Applying A r s C 1 pC1 qC 2 pr qs s By Applying p 2 q 2 pr qs A B . As matrix B is obtained by 2 2 C2 ( r p )C1 ( ps rq) p C 2 pr qs r s
Solution (c): The given matrix A
elementary row transformation from matrix A , so if rank ( A) N rank ( B ) N .
Number of Solutions for non-homogeneous Linear System in terms of rank: Consider the system of linear equation given by Eq. 1.2 and in matrix notation Eq. 1.2 can be written as Ax B , where A a jk
m n
O is the coefficient matrix; x x1k 1n and B b1k 1m are the column
vectors; A is the augmented matrix, are given by Eq. 1.4. Now, The system = is consistent and have unique solution if rank ( ) = rank ( | ) n . [This point was asked in CS-1996, EE-2005 (1 mark)] The system = is consistent and have infinite solution if rank ( ) = rank ( | ) n . The system = is inconsistent and have no solution if rank ( ) ≠ rank ( | ). From the above points we can say that for at least one solution of the system = , the rank of augmented matrix [ A | B] must have the same rank as matrix A . [This point was asked in EE2005 (1 mark)]. Example 1.116 [EE-1998 (1 mark)]: A set of linear equation is represented by the matrix equation AX B . The necessary condition for the existence of a solution for this system is: (a) B must be linearly dependent on the columns of A (b) A must be invertible (c) B must be linearly independent on the columns of A (d) None of the above Solution (a): For having a solution for the matrix equation AX B , the rank ( A) = rank ( A | B) n . If rank ( A) n , then A is invertible. If rank ( A) n then A is not invertible. Hence option (b) is
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[1.66]
not correct. If rank ( A | B) n then the column of B may or may not be linearly dependent on the columns of A . If rank ( A | B) n then the column of B must be linearly dependent on the columns of A . Hence option (a) is correct and (c) is not correct. Example 1.117 [CS-2003 (2 marks)]: Consider the following system of linear equations: 2 1 4 x
4 3 12 y 5 . Notice that the second and the third columns of the coefficient matrix are 1 2 8 z 7 linearly dependent. For how many values of , does this system of equations have infinitely many solutions? (a) 0
(b) 1 2 1
(c) 2 2 1
(d) Infinitely many
4 4 | Solution (b): Given: A 4 3 12 , A | B 4 3 12 | 5 . We have infinitely many 1 2 8 1 2 8 | 7
solutions for the system given if rank ( A ) rank ( A | B ) 3 (as we have 3 unknowns). Since rank ( A ) is 2 as we have two linearly independent column. The rank ( A | B ) is evaluated as:
R 1 R1 2
4
1
2
| 2
By Applying 1 1 2 2 3 12 | 5 R2 R2 4 R1 0 1 4 3 2 6 2 8 | 7 R R R 3 3 1 0 1 2 2 | 2 For rank ( A | B ) 1 4 | 5 2 0 0 | 11( 2) 7 2
1 1 2
By Applying
1 3 0 R3 R3 R1 2 0 have 11( 2) (7 2) 0 7 11 . Hence we have one value of . By Applying
| |
2 5 2
| 7 2
rank ( A ) 2, we must
Example 1.118 [IN-2007 (2 marks)]: Let A aij ,1 i, j n , with n 3 and aij i j . Then the rank of A is (a) 0
(b) 1 1 2 3
2 4 6 Solution (b): Let A 3 6 9 n 2n 3n
n
(c) n 1 1 2 3 n
2n
0 3n A 0 2 0 n
0
(d) n
0 0 By Applying
0 0 0 Rk Rk 2 Rk 1 2 k n 0 0 0
As we have only one independent row hence its rank is 1. Example 1.119 [XE-2007 (2 marks)]: Let Ax b be a system of m linear equation in n unknowns with m n and b 0 . Then the system has (b) either zero or infinitely many solutions (a) n m solutions (c) exactly one solution (d) n solutions Solution (b): From the given data we can say that A is m n matrix. m n rank( A) m . If rank( A) rank( A | b) m n The system have infinitely many solutions. Example 1.120 [EE-2008 (1 mark)]: If the rank of a (5 6) matrix Q is 4, then which one of the following statements is correct?
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Chapter 1: Linear Algebra
[1.67]
(a) Q will have four linearly independent rows and four linearly independent columns (b) Q will have four linearly independent rows and five linearly independent columns (c) QQT will be invertible (d) QT Q will be invertible Solution (a): The rank of a (5 6) matrix Q is 4 Q has four linearly independent rows and four linearly independent columns. Hence option (a) is correct and (c) is wrong. Also, as Q has rank 4 min(5, 6) so Q 0 Q 1 does not exist as Q is singular matrix; so (QQT ) 1 (QT ) 1 (Q ) 1
and (QT Q) 1 Q 1 (QT ) 1 both does not exists. So options (b) and (d) both are wrong. Example 1.121 [EE-2010 (2 marks)]: For the set of equations x1 2 x2 x3 4 x4 2 and 3 x1 6 x2 3 x3 12 x4 6 . The following statement is true
(a) Only the trivial solution x1 x2 x3 x4 0 (b) There are no solutions (c) A unique non-trivial solution exists (d) Multiple non-trivial solutions exists Solution (d): We have 2 equations and 4 unknowns. The coefficient matrix and augmented matrix of 1 2 1 4 1 2 1 4 | 2 the given system of equation is given as: A and A | B . 3 6 3 12 3 6 3 12 | 6
1 2 1 4 [By applying R2 R2 3R1 ]. As we have only one independent 0 0 0 0 1 2 1 4 | 2 r ( A) 1 . Similarly, For r ( A | B) : A | B [By applying 0 0 0 0 | 0
For r ( A) : A row so
R2 R2 3R1 ]. As we have only one independent row so r ( A) 1 . Now, as r ( A) r ( A | B ) 2 4 . Hence the given system has infinite number of solutions.
Example 1.122 [CS-2014 (1 mark)]: Consider the following system of equations: 3 x 2 y 1 , 4 x 7 z 1 , x y z 3 , x 2 y 7 z 0 . The number of solutions for this system is ……… Solution: The given system has 4 equations and 3 unknowns. The coefficient and augmented matrix 3 2 0 3 2 0 | 1
4
4 0 7 | 1 and A | B , respectively. 1 1 1 1 1 1 | 3 1 2 7 1 2 7 | 0 1 1 1 1 By Applying 1 1 1 1 1 By 4 0 7 R R 4 R 0 4 3 By Applying 0 1 3 4 2 1 2 For r ( A) : Applying 3 2 0 R3 R3 3R1 0 1 3 R R 4 0 1 3 2 2 R R 1 3 1 2 7 3 6 R 0 3 6 4 R4 R1 0 1 1 1 1 1 1 By Applying By Applying 0 1 3 4 0 1 34 r ( A) 3 . R3 R3 R2 R R 0 0 15 4 R 0 0 15 4 4 4 3 R 4 R4 3 R 2 0 0 0 15 4 0 0 1 1 1 | 3 By Applying 1 1 1 | 3 R2 R2 4 R1 0 4 3 | 11 By Applying 4 0 7 | 1 For r ( A | B ) : R R R R 3 R 3 2 0 | 1 0 1 3 | 8 1 3 3 3 1 3 6 | 3 1 2 7 | 0 R 4 R4 R1 0 for the given system of equation is A
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0
7
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Engineering Mathematics
1 By Applying 0 R R 4 2 2 0 0 1 By Applying 0 R R R 4 4 3 0 0
1
Chapter 1: Linear Algebra
1
1 1 1 By Applying 1 3 4 | 11 4 0 1 34 R3 R3 R2 1 3 | 8 0 0 15 4 R 4 R4 3R 2 3 6 | 3 0 0 15 4 1
|
[1.68]
1
3
|
|
3
| 11 4 | 21 4 | 21 4
3
1 3 4 | 11 4 . As we have 3 non-zero rows so r ( A | B ) 3 . 0 15 4 | 21 4 0 0 | 0
So, for the given system, r ( A) r ( A | B) 3 number of unknowns. Hence the system has unique solution. So the answer is 1.
Number of Solutions for homogeneous Linear System in terms of rank: Consider the system of linear equations given by Eq. 1.5, where all b j 0 , 1 j m . In matrix form the system of equation can be written as Ax O , where A a jk
m n
O is the coefficient matrix; x xk 1 n1 is
the column vector. Now, The system Ax O always has the trivial solution, i.e., x1 x2 xn 0 . Hence the system is always consistent. [This point was asked in CS-1996 (1 mark)]. The system Ax O have non-trivial solution if and only if rank ( ) n . Hence, in this case, the given system of equation has non-trivial solutions as well as trivial solution and hence the system has infinitely many solutions. [This point was asked in CS-1996 (1 mark), CH-2009 (1 mark)]. Example 1.123 [EE-2008 (2 marks)]: A is a m n full rank matrix with m n and I is an identity matrix. Let matrix A ( AT A) 1 AT . Then, which one of the following statement is FALSE? (b) ( AA ) 2 AA (a) AA A A (c) A A I (d) AA A A T Solution (d): Given order of A is a m n order of A is a n m For the order of A A ( AT A) 1 AT (( n m )( m n )) 1 (n m ) ( n n ) 1 (n m ) (n n )(n m ) n m equal
equal
Option (a): order of AA A ( m n)(n m)( m n ) (m n ) order of A , so option (a) is correct. equal 2
equal
Option (b): order of ( AA ) AA AA ( m n )( n m )( m n )( n m ) m m ; and order of equal
2
equal
equal
AA (m n )( n m ) m m , so order of ( AA ) is same as order of AA , so option (b) is correct. equal
Option (c): order of A A (n m)( m n ) n n , which is same as order of I which is n n , so equal
option (c) is correct. Option (d): since order of AA A is m n which is not same as the order of A which is n m , so option (d) is not correct. 2 2 x1 0 Example 1.124 [EE-2013 (1 mark)]: The equation has 1 1 x2 0 (a) No solution (c) Non-zero solution
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(b) Only one solution x1 x2 0 (d) Multiple solutions
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.69]
Solution (d): As we have homogeneous system of equation, whose coefficient matrix is: 1 2 2 2 2 . So for r ( A) : A [By applying R2 R2 R1 ]. As we have only one A 2 1 1 0 0 independent row so r ( A) 1 2 , so the given system of equation has non-trivial solutions as well as trivial solution and hence the system has infinitely many solution. [It is advised to solve the questions 1, 2, 3, 4, 10, 14, 15, 16, 17, 21, 22, 23, 24, 25 and 26 of Exercise: 1.2 by using rank method] Exercise: 1.4 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. Express u 1 2
v 2 1 0
T
and
T
2 ?
(a) u v1 2 v 2
(b) u 2 v1 v 2
2. Express u 1 2
v 2 1 0
T
1 , as a linear combination of two vectors v1 1 2 3 (c) u v1 v 2
(d) Cannot be expressed
T
T
0 , as a linear combination of two vectors v1 1 2 3
and
T
2 ?
(a) u v1 2 v 2
(b) u 2 v1 v 2
T
(c) u v1 v 2
T
(d) Cannot be expressed
T
T
3. Let v1 1 2 , v 2 0 1 and v 3 1 1 and v 4 1 0 . Is v 4 can be expressed as a linear combination of v1 , v 2 and v3 ? (a) cannot be expressed (c) can be expressed in infinite number of ways T
(b) can be expressed in only one way (d) None of these
T
4. If v1 1 2 3 and v 2 1 0 2 then span {v1 , v 2 } is (a) x y 2 z 0 (b) x 4 y 2 z 0 (c) 4 x y 2 z 0 T
T
5.
i e1 1 0 and j e2 0 1 span , where k _____.
6.
v1 1 1 and v2 2 1 span k , where k _____.
T
(d) 4 x 2 y z 0
k
T
T
T
T
k
7. The vectors v1 1 1 , v2 2 1 and v3 3 2 span , where k _____. T
8. Which one of the following statement if correct for the given three vectors: v1 1 2 3 , T
v 2 1 0 2 . (a) The vectors are mutually perpendicular (c) The vectors are linearly independent
(b) The vectors are linearly dependent (d) The vectors are unit vectors T
9. Which one of the following statement if correct for the given three vectors: v1 1 1 0 , T
T
v 2 1 0 1 and v 3 3 1 2 . (a) The vectors are mutually perpendicular (c) The vectors are linearly independent
(b) The vectors are linearly dependent (d) The vectors are unit vectors T
10. Which one of the following statement if correct for the given three vectors: v1 1 1 3 , T
T
T
v 2 1 3 1 , v 3 3 1 1 and v 4 3 3 3 . (a) The vectors are mutually perpendicular (b) The vectors are linearly dependent (c) The vectors are linearly independent (d) The vectors are unit vectors 3 11. The dimension of the plane x 2 z 0 in is _____.
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Engineering Mathematics
Chapter 1: Linear Algebra
[1.70]
T
12. Which one of the following is the basis for vector space V spanned by vectors v1 1 1 0 , T
T
T
v 2 0 1 1 , v 3 2 3 1 and v 4 1 1 1 ? (a)
2
T
1 1 0
T
(b) 2 1 1 0
T
2
(c)
1 1 0
(d)
T
2
1 1 0
a b 2c 2a 2b 4c d 13. The basis and dimension of the subspace W , where a, b, c, d are real are, b c d 3a 3c d _____ and _____ respectively. 1 1 0 1
2 2 1 ,3 0 1 1 3 0 1
(a)
1
2
1 2 (c) 0 3
2 2 4 ,3 0 1 1 3 0 3
(b)
14. The pair of vectors: (i) T
T
1
3
and
3
T
2
1
1 2 (d) 0 3
,4 1 1 0 1
T
T
(b) (ii), (iii) and (iv)
0
1 ; (ii)
2 4 1 and 4 1 2 ; (iv) x y T 1 1 1 which are orthogonal to each other are: (a) (i), (ii) and (iii)
1
y
and
T
1
3 T
and
x ; (v)
(c) (iii), (iv) and (v)
2
2
4
1
1
0
3
,4 T
10 a
1
10 3 ; (iii) T
a b
b
and
(d) (ii), (iv) and (v)
T
15. The two vectors which are orthogonal to v 1 4 7 , and are not scalar multiples of each other, are (a) (c)
T
10 1 2 and 1 T 3 1 1 and 10
T
3 1
T
1 2
(d)
16. Angle (in degrees) between two vectors u 4 0 T
17. The orthogonal projection of y 1 4 3 T
T
T
T
T
(b) 1 3 1 and 2 1 4
3 2
1 1 and 2 1 4 T
2 and v 2 0
T
1 1 is ____.
onto the subspace spanned by vectors
T
u 1 1 0 and u 1 1 0 is (a)
1
T
4 0
T
(b) 1 4 0
(c)
4
T
1 0
(d)
4
T
1 0
18. Suppose {v1 , v 2 , , v n } is a basis of V and {w1 , w 2 , , w m } is a linearly independent subset of V , then which one of the following is correct? (a) m n (b) m n (c) m n (d) m n 19. If {v1 , v 2 , , v m } is linearly dependent, then there exists _____ such that v j is a linear combination of {v1 , v 2 , , v j1} . (a) j m
(b) j m
(c) j m
(d) j m
1 0 2 1 20. The rank of the matrix A 0 2 4 2 is _____. 0 2 2 1 3 6 1 1 7 21. The nullity of the matrix A 1 2 2 3 1 is _____. 2 4 5 8 4
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Chapter 1: Linear Algebra
[1.71]
1 2 1 0 22. The rank of matrix A 2 4 2 0 is _____. 4 8 4 0 23. Which of the following statement is not correct for a matrix A aij (a) any 5 columns are dependent (c) any 6 rows are dependent 1 5 1
7 6
having rank 4?
(b) some set of 4 columns are independent (d) any 4 rows are independent
24. The rank of matrix A 2
1 1 is _____. 3 6 2
25. The rank of a null matrix is always _____. 26. For a matrix A aij
6 7
having six rows are dependent, if r be the rank of matrix A , then
which one of the following is true? (a) 0 r 5 (b) 0 r 5 (c) 0 r 5 (d) 0 r 5 27. If a 5 7 matrix has rank 3, then which one of the following statement is not correct? (b) every 4 4 sub-determinant is zero (a) every 5 5 sub-determinants is 0 (c) every 3 3 sub-determinants is not 0 (d) at least one of all 3 3 sub-determinant is not 0 28. Which of the following statement is needed for determining the rank of matrix A aij
6 6
?
(a) One of its 3 3 sub-determinant is 6 (b) The 3 3 sub-determinant in north-east corner is 0 and in south-west corner is 8 (c) One of its 3 3 sub-determinants is 0, another of its 3 3 sub-determinants is 8 and all the 4 4 sub-determinants are 0 (d) One of its 3 3 sub-determinants is 0 b a 29. For the matrix A 2 , which one of the following statement is not correct? 0 a (b a ) (a) If a b , then rank ( A) 1 (b) If a b , then rank ( A) 1 (c) rank ( A) 2 for all values of a and b (d) If a b , then rank ( A) 2 30. What is the rank of matrix A aij 31. If A aij
3 4
n n
, where aij i j 1 and n 1 ? _____.
have rank r1 ; and AB abij
correct regarding rank ( r ) of matrix B ? (a) r r1 r2 4 (b) r r1 r2 4 32. If matrices A aij
3 4
, AB abij
33. What is the rank of matrix A aij
35
n n
35
have rank r2 , then which one of the following is (c) r r2 r1 4
(d) r r2 r1 4
have full rank, then rank of B is?
, where aij 1 ? _____.
1 1 1 34. What is the rank of the matrix A b c a c a b if any two of a, b, c are equal and is bc ac ab
different from the third one? _____. 1 2
3 4 5 6 7 8 is _____. 35. The rank of the matrix A 9 10 11 12 13 14 15 16
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1.5
Chapter 1: Linear Algebra
[1.72]
Eigenvalue (Characteristic root) and Eigenvector (Characteristic Vector)
Any non-zero vector, X , is said to be a eigenvector of a matrix A , if there exist a number such that AX X ; where, is said to be a eigenvalue of the matrix A corresponding to the eigenvector X and vice-versa. Eigenvalues (vectors) are also often called proper values, latent values or eigenvalues (vectors). An eigenvector vector of a matrix cannot corresponds two different eigenvalues. If this statement AX 1 X ; AX 2 X , 1 2 1 X 2 X (1 2 ) X 0 X 0 , is wrong, then 1 2 . But, X 0 , then the statement is true.
An eigenvalue of a matrix can correspond to different eigenvectors. If AX X , then also A(kX ) (kX ) , so that kX is also a eigenvector of A corresponding to same eigenvalue . Thus we have truth of above statement. The value of the dot product of the eigenvectors corresponding to any pair of different eigenvalues of a n n real symmetric matrix A is always zero. [This point was asked in CS2014, ME-2014 (1 mark)]. Proof: Let X 1 and X 2 are the eigenvectors of their corresponding eigenvalues ( 1 and 2 ) of a
n n real symmetric matrix A . Then AX1 1 X1 …(i) and AX 2 2 X 2 …(ii). Taking transpose T
of (ii) ( AX 2 )T (2 X 2 )T X 2T AT 2 X 2T X 2T A 2 X 2T ( A A ) X 2T AX 1 2 X 2T X 1 T X 2T 1 X 1 2 X 2T X 1 1 X 2T X 1 2 X 2T X 1 (1 2 ) X 2 X 1 O , since 1 2 X 2T X 1 O .
Let be eigenvalue and X be its corresponding eigenvector of matrix A , i.e. AX X IX ( A I ) X 0 . X 0 , we deduce that the matrix ( A I ) is singular, hence A I 0 . Every eigenvalue of a matrix A is a root of its characteristic equation A I 0 .
Conversely, if be any root of the characteristic equation A I 0 , then the matrix equation ( A I ) X 0 necessarily possesses a non-zero solution X so that there exist a vector X 0 such that AX IX X . Every root of the characteristic equation of a matrix is a eigenvalue of the matrix. If A be n rowed matrix, then the characteristic equation A I 0 is nth degree equation, hence, it possesses n eigenvalues which may or may not be distinct. [This point was asked in PI-2007 (2 marks)]. If all n eigenvalues are distinct then we have n linearly independent eigenvectors. If all n eigenvalues are not distinct then there may or may not be n linearly independent eigenvectors. The eigenvalues of a matrix and its transpose are same. [This point was asked in CH-2013 (1 mark), EC-2014 (2 marks)] Proof: Let a matrix A has eigenvalues as det( A I ) 0 . We also know that the determinant of a matrix and its transpose are same; so applying this concept in the characteristic equation for the matrix A as, det( A I ) det( A I )T 0 det( AT I T ) T
T
det( AT I ) 0 ( I I ). Hence det( AT I ) 0 The eigenvalues of A is . The eigenvalues of hermitian matrices are real. The eigenvalues of a real symmetric matrices are all real; as real symmetric matrices are Hermitian matrices. [This point was asked in ME-2006, ME-2007, AE-2010, ME-2011, PI-2013 and ME-2013, CH-2013, EE-2014 (1 mark), EC-2014 (2 marks). Proof: Let be an eigenvalue of a Hermitian matrix. Then there exists a non-zero vector X s.t.
AX X …(i).
Taking
transpose
conjugate
( AX ) ( X ) X A X
X A X …(ii) ( A A ). Pre-multiplying (i) by X and post-multiplying (ii) by X we get, X AX X X X AX X X …(iii) and X AX X X …(iv). (iii) – (iv) ( ) X X 0 . X O X O X X O ( ) 0 is real.
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Chapter 1: Linear Algebra
[1.73]
The eigenvalues of skew-Hermitian matrices are either purely imaginary or zero. The eigenvalues of a skew symmetric matrices are either purely imaginary or zero; as skew-symmetric matrices are skew-Hermitian matrices. [This point was asked in EC-2010]. Proof: Let A be a skew-Hermitian matrix, iA is Hermitian. Let be an eigenvalue of A AX X , X 0 or (iA) X (i ) X i is an eigenvalue of iA . As iA is Hermitian and its eigenvalues are real i is real, which is possible only if is zero or purely imaginary. The eigenvalues of a orthogonal and unitary matrix are of unit modulus, i.e., 1 . Proof: Let A be an orthogonal matrix AAT AT A I . Let be an eigenvalue of A corresponding to an eigenvector X of matrix A , AX X , X 0 . Post-multiplying with AT AT AX AT X IX AT X ( AT A I ) X AT X AT X X 1 is a T
T
eigenvalue of A . Since, the eigenvalues of matrix A and A 1 2 1 1 . Similar proof was also given for unitary matrices.
same
The eigenvalues of an idempotent matrix are either zero or unity. Proof: Since A is an idempotent matrix A2 A . Let X be the eigenvector of the matrix A corresponding to the eigenvalue AX X …(i), such that X O . On pre-multiplying (i) by
are
A
A( AX ) A( X ) A2 X ( AX ) AX ( X ) X 2 X
2
( A A and
2 2 AX X ) ( ) X O 0 ( X O ) 0 or 1 The eigenvalues of an involuntary matrix is 1 or 1 . [This point was asked in CS-2014 (1 mark)]. Proof: Since A is an involuntary matrix A2 I . Let X be the eigenvector of the matrix A corresponding to the eigenvalue AX X …(i), such that X O . On pre-multiplying (i) 2 2 2 by A A( AX ) A( X ) A X ( AX ) IX ( X ) X X ( A I and
2 2 AX X ) ( 1) X O 1 0 ( X O ) 1
If 1 , 2 ,, n are the eigenvalues of a square matrix A , then
k 1 , k 2 ,, k n are the eigenvalues of kA . 1
1 1 ,1 2 , ,1 n are the eigenvalues of A (provided A is non-singular). k
1k , 2k , , nk are the eigenvalues of A .
A 1 , A 2 , , A n are the eigenvalues of adjA (provided A is non-singular).
T 1 , 2 ,, n are the eigenvalues of A .
If X be an Eigenvector corresponding to Eigenvalue of the matrix A then X be an 2
3
k
2
k
2
3
k
Eigenvector of A , A , , A corresponding to Eigenvalue , , , of A , A , , A , respectively. [A question based on this concept was asked in IN-2011 (1 mark)]. Proof: If X be an Eigenvector corresponding to Eigenvalue of the matrix A , then AX X . Pre-multiplying both sides with A , it gives AAX A X A2 ( X ) ( AX ) 2
A2 ( X ) ( X ) 2 X (i); so Eigenvector of A is X corresponding to Eigenvalue 2 of
A2 . Again pre-multiplying both sides with
A
in (i), it gives
AA2 X A 2 ( X )
3
A3 X 2 ( AX ) 2 ( X ) 3 X ; so Eigenvector of A is X corresponding to Eigenvalue
3 k k 3 of A . In general, Eigenvector of A is X corresponding to Eigenvalue k of A . Sum of eigenvalues of any square matrix A trace of A sum of diagonal elements of A . Product of all eigenvalues of matrix A is equal to A . Hence if A is a singular matrix then at least one eigenvalues of A is zero. In a triangular and diagonal matrix, eigenvalues are diagonal elements themselves. [This point was asked in EC-2014 (2 marks)]. If A and B are two n n matrices then matrices AB and BA both have same eigenvalues.
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Chapter 1: Linear Algebra
[1.74]
0 0 Example 1.125 [CS-1993, EC-1993 (1 mark)]: The Eigenvector(s) of the matrix A 0 0 0 , 0 0 0 0 is (are) (a) (0, 0, )
(b) ( , 0, 0)
0
Solution (b, d): For eigenvalues, A I 0 0
0
0
For eigenvector of A corresponding to 0 0 x1 0 x1 0 x2 x3 0 (i) 0 0 0 0
(d) (0, , 0)
(c) (0, 0,1)
00 0 x2 0 0 x1 0 x2 0 x3 0 (ii) 0 0 0 x3 0 x1 0 x2 0 x3 0 (iii)
0 0 3 0 0
(i) x3 0 , as 0 ; on the other hand we cannot say anything about x1 and x2 . Thus options (b) and (d) both are correct. 2 2 3 1
Example 1.126 [EE-1998 (1 mark)]: The Eigenvector of A 2
1
1 2
6 is 2 . One of the
1
0
Eigenvalues of A is: (a) 1 (b) 2 (c) 5 (d) –1 Solution (c): If X is the Eigenvector corresponding to the Eigenvalue ( ) of the matrix A , 2 2 3 1 1 5 1 1 1
AX X 2
1
1 2
6 2 2 10 2 5 2 2 5
0 1 1 5 [Similar question was also asked in PI-2008 (2 marks)]
1
1
1
23 13 23 Example 1.127 [CE-1999 (5 marks)]: Show that the matrix A 2 3 2 3 1 3 is orthogonal 1 3 2 3 2 3 and determine its Eigenvalues. Solution: Any square matrix A of order n is said to be orthogonal, if AT A 1 AAT AT A I n .
2 3 1 3 2 3 2 3 2 3 1 3 1 0 0 AA 2 3 2 3 1 3 1 3 2 3 2 3 0 1 0 I 3 A is an orthogonal matrix. 1 3 2 3 2 3 2 3 1 3 2 3 0 0 1 T
Now for Eigenvalues, A I 0
(2 3)
13
23
2 3
(2 3)
13
13
23
(2 3)
5
9 3 6 2 6 9 0 (9 2 15 9)( 1) 0 1,
6
0
11 i . 6
2 1 0 0 3 0 Example 1.128 [EC-2000 (1 mark)]: The Eigenvalues of the matrix 0 0 2 0 0 1 (a) 2, –2, 1, –1
(b) 2, 3, –2, 4
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(c) 2, 3, 1, 4
0 0
are
0
4 (d) None of these
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[1.75]
Solution (b): For the given matrix if we apply the operation R4 R4 R3 2 , then we get upper triangular matrix with diagonal elements 2, 3, –2, 4. So the Eigen values of the given matrix are 2, 3, 2, 4 ; so option (b) is correct. 2 1 0 Example 1.129 [CE-2000 5 marks]: A is a square matrix is given by A 1
1 . One of
2
0
1 2 the Eigenvalues of A is given to be equal to 2. Determine other Eigenvalues. Express the Eigenvector
corresponding to the lowest Eigenvalue in the form 1 a
T
b and determine a and b .
Solution: Let the other two Eigenvalues are 2 and 3 . Since, sum of Eigenvalues of any nonsingular square matrix A trace of A sum of diagonal elements of A . Also, product of all Eigenvalues of matrix A is equal to A . Hence, 2 2 3 2 2 2 2 3 4 …(i); and
223 2(4 1) 1(2 0) 4 23 2 …(ii). (i) and (ii) 2 (4 2 ) 2 22 42 2 0 2 2 2 3 2 2 . Thus minimum eigenvalue is min(2, 2 2, 2 2) 2 2 . For Eigenvector corresponding to 2 2 , ( A I ) X 0
2 1 0
1 2 1
0 x 1
2 x1 x2 0
(i)
1 x2 0 x1 2 x2 x3 0 (ii) . (i) x2 2 x1 . Putting value of (i) in x 2 3 x2 2 x3 0 (iii)
(iii) x3 x1 . Hence Eigenvector corresponding to 2 2 is X x1
T
2 x1
x1 .
T
Comparing this Eigenvector with 1 a b we have x1 1 a 2 x1 2 and b x1 1 . [Similar questions were also asked in CE-2006, AE-2012 (2 marks)] 5 3 Example 1.130 [CE-2001 (2 marks)]: The eigenvalues of the matrix are 2 9 (a) (5.13, 9.42) (b) (3.85, 2.93) (c) (9.00, 5.00) (d) (10.16, 3.84) 5 3 5 3 , A I 0 0 2 9 2 9
Solution (d): For Eigenvalues ( ) of the matrix A
(5 )(9 ) 6 0 2 14 39 0 10.16, 3.84 . [Similar questions were also asked in CE-2002, EE-1994, ME-2003, AG-2008, AG-2013, BT2013, MT-2012, AG-2013, EC-1998 (1 mark); CE-2004, CE-2008, CE-2012, CS-2005, ME-1999, AE-2007, AG-2012, AG-2014, PI-2011 (2 marks)] 1 2 34 49
0 2 43 94 Example 1.131 [CS-2002 (5 marks)]: Obtain the Eigenvalues of the matrix A 0 0 2 104 0 0 0 1 Solution: Since we have a triangular matrix and we know that the eigenvalues of a triangular matrix are its diagonal elements. Hence the eigenvalues of the given matrix are 1, 2, –2, –1. [Similar questions were also asked in CS-2011, EE-2002 (2 marks)] 0 0 0 1
100 Example 1.132 [EE-2002 (1 mark)]: The determinant of the matrix 100
1
0
0
200
1
0
is
100 200 300 1
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[1.76]
(a) 100 (b) 200 (c) 1 (d) 300 Solution (c): Since the given matrix is lower triangular matrix, so its diagonal elements (1, 1, 1, 1) are its Eigenvalues. So the determinant of the given matrix is the product of Eigenvalues which is 1. 1 1 3 Example 1.133 [ME-2004 (1 mark)]: The sum of the Eigen values of the matrix 1
5 1 is 3 1 1
(a) 5 (b) 7 (c) 9 (d) 18 Solution (b): Since, sum of eigenvalues of any square matrix A trace of A sum of diagonal elements of A . Hence the sum of its eigenvalues is 1 5 1 7 . [Similar questions were also asked in EE-1998, MN-2009, TF-2010, ME-2008, CE-2014 (1 mark)] Example 1.134 [CE-2005 (2 marks)]: Consider the system of equations Ann X n1 Bn1 . Let i , xi be an Eigen-pair of an Eigenvalue and its corresponding Eigenvector for real matrix. Let I be a n n unit matrix. Which one of the following statement is NOT correct? (a) For a homogeneous n n system of linear equations ( A I ) X 0 having a non-trivial solution, the rank of ( A I ) is less than n . m
(b) For matrix A , m being a positive integer, (im , xim ) will be the Eigen-pair for all i 1
T
T
(c) If A A , then i 1 i (d) If A A , then i is real i Solution (b): For a homogeneous n n system of linear equations ( A I ) X 0 having a nontrivial solution we must have A I 0 the rank of ( A I ) is less than n . So option (a) is correct. If is an eigenvalue of matrix A AX X . Now pre-multiplying by A 2
2 2 2 A X AX X Eigenvalue and Eigenvector pair of A is ( , X ) . Again pre-multiplying
3
3 by A A3 X 2 AX 3 X Eigenvalue and Eigenvector pair of A is ( , X ) . When pre-
multiplying the previous equation by A for m th we can easily get Am X m X Eigenvalue and m m Eigenvector pair of A is ( , X ) . Hence option (b) is not correct. For orthogonal matrices ( T T T 1 A A AA I A A ) the Eigenvalues are of unit modulus; so option (c) is correct. For T symmetric matrices ( A A ) the Eigenvalues are real; so option (d) is correct. 4 2 Example 1.135 [EC-2005 (2 marks)]: Given the matrix , the Eigenvector is 4 3
T
(a) 3 2
T
(b) 4 3
Solution (c): For eigenvalues A I 0
(c)
2
4
2
4
3
T
1
(d)
1
T
2
0 (4 )(3 ) 8 0
2 20 0 ( 5)( 4) 0 5, 4 . For eigenvector of A corresponding to x1 2 x2 0 (i) 2 x1 4 ( 5) 1 2 x1 5 0 0 4 8 x 4 x1 8 x2 0 (ii) 4 3 ( 5) x2 2 T
(i) and (ii) both x1 2 x2 . From the given options, choosing x2 1 x1 2 X 2 1 . [Similar questions were also asked in ME-1994 (2.5 marks), CE-1998 (5 marks), EC-2007 (4 marks), EE-2010, ME-2010 (2 marks), EE-1995, ME-2014 (1 mark)] 3 2 2 Example 1.136 [EE-2005 (2 marks)]: For the matrix P 0
0
2
1 , one of the eigenvalues is
0
1
equal to –2. Which of the following is an eigenvector?
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Engineering Mathematics
Chapter 1: Linear Algebra
T
T
[1.77]
T
T
(a) 3 2 1 (b) 3 2 1 (c) 1 2 3 (d) 2 5 0 Solution (d): For eigenvector ( X ) of P corresponding to eigenvalue 2 , we have (P I ) X 0 . 2 2 x1 3 ( 2) 5 0 2 ( 2) 1 x 0 0 2 0 0 1 ( 2) x3 0 From (iii) we have x3 0 , from (1) we can choose
2
5 x1 2 x2 2 x3 0 (i)
2 x1
0 1 x3 0 0 3 x3 x1 2 x2 5
x3 0
(ii)
x3 0
(iii)
[Similar question was also asked in XE-2012 (1 mark)] Example 1.137 [ME-2005 (2 marks)]: Which one of the following is an eigenvector of the matrix 5 0 0 0
0 5 5 0 ? 0 0 2 1 0 0 3 1
A
T
T
(a) 1 2 0 0
(b) 0 0 1 0
(c) 1 0 0
T
2
T
(d) 1 1 2 1
Solution (a): The characteristic equation for eigenvalues of the matrix A is A I 0 . 5
0
0
0
0
5
5
0
0
0
2
1
A
2
0 (5 )(5 )( 3 1) 0 5, 5,
0 0 3 1 The eigenvector corresponding 5 is ( A I ) X 0
0
0
0
0 x1 0 x2
00
0
(i)
0 0 5 0 5 x3 0 (ii) 0 0 3 1 x3 0 3x3 x4 0 (iii) 0 0 3 4 x4 0 3 x3 4 x4 0 (iv)
3 13 2
(iii) and (iv) x4 0 x3 0 . We don’t have any relation which exactly give values of x1 and x2 . Hence x1 and x2 can be anything. Hence from the given options we can say that option (a) is correct.
Example 1.138 [EC-2006 (2 marks)]: The Eigenvalues and the corresponding Eigenvectors of a T
T
2 2 matrix are given by 1 8 , v1 1 1 and 2 4 , v2 1 1 . The matrix is
6 (a) 2
2 6
4 (b) 6
6
2 (c) 4
4
a
b
c
d
Solution (a): Let the 2 2 matrix is A
4 2
4 (d) 8
8 4
. Since the Eigenvalue and its Eigenvector for a
given matrix can be related as AX X . 1 a b 1 1 a b 8 a b 8 (i) For 1 8 , v1 , AX X 8 1 c d 1 1 c d 8 c d 8 (ii) a b 4 (iii) 1 a b 4 4 d 1 1 c d 4 c d 4 (iv) 6 2 (i) and (iii) a 6 , b 2 . (ii) and (iv) c 2 , d 6 . Hence the matrix is . 2 6
1 a For 2 4 , v2 , AX X 1 c
b 1
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4 Example 1.139 [EC-2006 (2 marks)]: For the matrix 2
2 4
[1.78]
the Eigenvalue corresponding to the
T
Eigenvector 101 101 is: (a) 2 (b) 4 (c) 6 (d) 8 Solution (c): The matrix and its Eigenvector is given and we have to find the corresponding Eigenvalue. Since the Eigenvalue and its Eigenvector for a given matrix can be related as AX X . 4 2 101 101 606 101 101 101 6 6. 2 4 101 101 606 101 101 101 [Similar question was also asked in AE-2013 (1 mark)] 3 2 Example 1.140 [ME-2006 (2 marks)]: Eigen value of a matrix A are 5 and 1. What are 2 3 2
the Eigen values of the matrix A AA ? (a) 1 and 25 (b) 6 and 4
(c) 5 and 1
(d) 2 and 10
3 2 k k k are 5 and 1. We know that 1 , 2 , , n are the 2 3
Solution (a): Eigenvalues of the matrix A k
2
2
eigenvalues of A . So, eigenvalues of A are 52 and 1 , i.e., 25 and 1. Hence option (a) is correct. [Similar question was also asked in CH-2008 (2 marks)] 2 1 1 Example 1.141 [AE-2007 (2 marks)]: The eigenvalues of the matrix A , where A are 0 3 (a) 1 and 1/2 (b) 1 and 1/3 (c) 2 and 3 (d) 1/2 and 1/3 2 1 Solution (d): Eigenvalue of the matrix A is determined as, A I 0 0 3
2
1
0
3
0 (2 )(3 ) 0 0 2, 3 . 1 1 ,1 2 , ,1 n are the eigenvalues of
A1 (provided A is non-singular). As A 6 1 5 0 The Eigenvalues of A1 are 1/2 and 1/3. Example 1.142 [CE-2007 (1 mark)]: The minimum and maximum Eigenvalues of the matrix 1 1 3
1 5 1 are –2 and 6, respectively. What is the other Eigenvalue? 3 1 1 (a) 5 (b) 3 (c) 1 (d) –1 Solution (b): Let be the other Eigenvalue 2 6 1 5 1 3 Example 1.143 [ME-2007 (2 marks)]: The number of linearly independent Eigenvectors of 2 1 A is 0 2 (a) 0 (b) 1 (c) 2 (d) Solution (b): The characteristic equation for Eigenvalues of the matrix A is A I 0 . A I 0
2
1
0 (2 )(2 ) 0
0 2 2, 2 . Eigenvector corresponding 2 is 0 1 x1 0 x2 0 (i) (A I)X 0 0 0 (ii) 0 0 x2 0
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(i) x2 0 and we don’t have any relation which exactly give values of x1 . Hence T
x1 k (say). Hence X k 0 . Hence number of linearly independent Eigenvector is 1.
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Chapter 1: Linear Algebra
[1.79]
3 2 Example 1.144 [TF-2007 (1 mark)]: The Eigenvalues of the matrix are 1 1 and 1 2 2 4 . An Eigenvector of the given matrix is T
T
T
T
(a) 2 1 (b) 1 1 (c) 1 1 (d) 2 1 Solution (c): Eigenvector corresponding to every Eigenvalue is evaluated as, ( A I ) X O
3 1
For 1 1
1
2 x1 0 2 1 2 1 x2 0
0 2 x1 2 x2 0 (i) x1 x2 0 (ii) 1 x2 0 2 x1
x1 x As (i) and (ii) both x1 x2 x (say). Hence . From the given options, option (c) is correct x2 x as its components are equal. As this question is of one option correct so we need not check the Eigenvector corresponding to 2 4 . 1 1 1 Example 1.145 [XE-2007 (1 mark)]: Let M 0 1 1 , then the maximum number of linearly 0 0 1 independent Eigenvectors of M is (a) 0 (b) 1 (c) 2 (d) 3 Solution (d): As the given matrix is already in a row reduced echelon form and it has 3 non-zero rows, so its rank is 3. 1 0 1 Example 1.146 [AE-2008 (1 mark)]: The product of Eigenvalues of the matrix 0
2 1 is 1 1 3
(a) 4 (b) 0 (c) –6 (d) –9 Solution (d): As the product of Eigenvalues of any square matrix is the determinant of that matrix. So 1 0 1 product of Eigenvalues of the given matrix 0
2
1 1(6 1) 1(0 2) 7 2 9 .
1 1 3 [Similar questions were also asked in MN-2011, AE-2009 (2 marks)]
Example 1.147 [CS-2008 (2 marks)]: How many of the following matrices have an Eigenvalue 1? 1 0 0 1 1 1 1 0 0 0 , 0 0 , 1 1 and 1 1 . (a) One (b) Two (c) Three (d) Four 1 0 1 0 Solution (a): Eigenvalues of 0 (1 )( ) 0 1, 0 ; , A I 0 0 0 0
0 1 Similarly, Eigenvalues of are 0, 0 ; Eigenvalues of 0 0 1 0 1 1 are 1 . Thus only the matrix
1 1 1 1 are 1 i ; Eigenvalues of
1 0 0 0 have Eigenvalue as 1. p11
p12
p21
p22
Example 1.148 [EC-2008 (1 mark)]: All the four entries of the 2 2 matrix p
are
non-zero, and one of its Eigenvalues is zero. Which of the following statements are true? (a) p11 p22 p12 p21 1 (b) p11 p22 p12 p21 1
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[1.80]
(c) p11 p22 p12 p21 0 (d) p11 p22 p12 p21 0 Solution (c): As, the product of eigenvalues of any square matrix gives determinant of that matrix. As p11 p12 one of the eigenvalue is zero, so p 0 0 p11 p22 p12 p21 0 p21 p22
1 Example 1.149 [ME-2008 (2 marks)]: The Eigenvectors of the matrix 0 T
2
are written in the
2
T
form 1 a and 1 b . What is a b ? (a) 0 (b) 1/2 (c) 1 (d) 2 Solution (b): As the given matrix is upper triangular so its Eigenvalues are 1, 2 . If X is the Eigenvector corresponding to the Eigenvalue ( ) of the matrix A , then AX X . 1 2 x1 x1 x1 2 x2 x1 x1 2 x2 x1 (i) For 1, AX X 2 x x (ii) 1 0 2 x2 x2 2 x2 x2 2 2 (i) x2 0 and from the given Eigenvector we can say x1 1 a 0
x1 x1 2 x2 2 x1 x1 2 x2 2 x1 (i) 2 x 2 x (ii) 2 x2 x2 2 x2 2 x2 2 2 (i) x1 2 x2 and from the given Eigenvector we can say that x1 1 x2 b 1 2 . So a b 1 2 . 1 For 2 , AX X 0
2 x1
2
3
4
Example 1.150 [PI-2008 (2 marks)]: The Eigenvector pair of the matrix A
is 4 3 2 (d) and 1
2 1 2 1 2 1 (a) and (b) and (c) and 1 2 1 2 1 2 Solution (a): For Eigenvalues of the matrix, we have 3 4 2 A I 0 0 (3 )(3 ) 16 0 25 0 5 4 3 For Eigenvector corresponding to 5 , we have x1 3 x1 4 x2 5 x1 3x1 4 x2 5 x1 x1 2 x2 3 4 x1 AX X 5 4 3 x2 x2 4 x1 3x2 5 x2 4 x1 3 x2 5 x2 x1 2 x2
1 2
T
If we choose x1 2 x2 1 Eigenvector corresponding to 5 is X 2 1 For Eigenvector corresponding to 5 , we have x1 3 x1 4 x2 5 x1 3x1 4 x2 5 x1 2 x1 x2 3 4 x1 AX X 5 x 4 x 3 x 5 x 4 x 3 x 5 x 2 x x 4 3 x2 2 1 2 2 1 2 2 1 2 T
If we choose x1 1 x2 2 Eigenvector corresponding to 5 is X 1 2 . [Similar question was also asked in IN-2013 (2 marks)] Example 1.151 [XE-2008 (1 mark)]: If the characteristic equation of a 3 3 matrix is 3 2 1 0 , then the matrix should be (a) Hermitian (b) Unitary (c) Skew symmetric (d) Identity 3 2 2 Solution (b): 1 0 ( 1)( 1) 0 1, i As the determinant of any matrix is equal to the product of its Eigenvalues, so the determinant of the 2 given matrix is 1 (i ) (i) i 1 unit modulus. Hence the given matrix is unitary matrix. Example 1.152 [AE-2009 (1 mark)]: A non-trivial solution to the n n system of equations [ A] x 0 , where 0 is the null vector
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[1.81]
(b) may be found only if [ A] is not singular (a) can never be found (c) may be found only if [ A] is an (d) may be found only if [ A] has at least one Eigenvalue orthogonal matrix equal to zero Solution (d): A non-trivial solution of a homogeneous equation is obtained if the coefficient matrix is singular the determinant of the coefficient matrix is zero and this happens if its one of the Eigenvalue is zero, as the product of Eigenvalues gives the determinant of any square matrix. 1 2 Example 1.153 [CH-2009 (2 marks)]: The Eigenvalues of the matrix A are 5 and –1. 4 3 Then the Eigenvalues of 2 A 3I ( I is a 2 2 identity matrix) are (c) 1 7 and 1 5 (d) 1 7 and 1 5 (a) –7 and 5 (b) 7 and –5
1 2 1 0 2 4 3 0 1 4 3 . For Eigenvalues 4 3 0 1 8 6 0 3 8 3
Solution (a): 2 A 3I 2
of 2 A 3I , 2 A 3I I 0
1
4
8
3
0 (1 )(3 ) 32 0 5, 7 .
1 3 5 Example 1.154 [EC-2009 (2 marks)]: The Eigenvalues of the matrix A 3 1 6 are 0 0 3 (a) 3, 3 5 j , 6 j (b) 6 5 j, 3 j ,3 j (c) 3 j ,3 j, 5 j Solution (d): For the Eigenvalues, , of the given matrix, 1 3 5 A I 0
3
1
6
(d) 3, 1 3 j, 1 3 j
2
0 ( 3)( 2 10) 0 3, 1 3i .
0 0 3 [Similar questions were also asked in ME-2000 (1 mark), ME-1996 (2 marks)]
Example 1.155 [EE-2009 (1 mark)]: The trace and determinant of a 2 2 matrix are known to be –2 and –35 respectively. The Eigenvalues are (a) –30 and –5 (b) –37 and –1 (c) –7 and 5 (d) 17.5 and –2 Solution (c): Let 1 and 2 are the two Eigenvalues of the given matrix, then 1 2 2 and
12 35 .
From
these
two
equations
1 ( 2 1 ) 35 12 21 35 0
1 5, 7 2 7,5 Example 1.156 [IN-2009 (2 mark)]: The Eigenvalues of a 2 2 matrix X are –2 and –3. The 1 Eigenvalues of matrix ( X I ) ( X 5I ) are (a) –3 and –4 (b) –1 and –2 (c) –1 and –3 (d) –2 and –4 2 0 1 0 Solution (c): Let X diag( 2, 3) and I diag(1,1) 0 3 0 1
3 0 ; and 0 2 1 0 X I diag( 2, 3) diag(1,1) diag(2 1, 3 1) diag( 1, 2) 0 2 X 5I diag( 2, 3) diag(5,5) diag( 2 5, 3 5) diag(3, 2)
For finding the inverse of ( X I ) , we have X I 2 and C11 2; C12 0; C21 0; C22 1
2 Cij 0
0 T adj ( X I ) 1 2 0 1 adj ( X I ) Cij (X I) 1 X I 0 1 0 1 2 0
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1
Chapter 1: Linear Algebra
0 3 0 A (say) 0 1 2 0 2 0 1 3 0 So, for eigenvalue ( ) of A , we have A I 0 1
( X I ) ( X 5I )
[1.82]
0 3
0
1
0 3, 1
2
3
x
y
Example 1.157 [CS-2010 (2 marks)]: Consider the following matrix: A
. If the
Eigenvalues of A are 4 and 8, then (a) x 4, y 10 (b) x 5, y 8 (c) x 3, y 9 (d) x 4, y 10 Solution (d): trace of matrix A sum of its eigenvalues, 2 y 4 8 y 10 ; and determinant of matrix A product of its eigenvalues, A 4 8 32 2 y 3 x 32 x 4 . Example 1.158 [IN-2010 (1 mark)]: A real n n matrix A aij is defined as follows if i j
i
aij
0 if otherwise
. The summation of n Eigenvalues of A is
(a) n (n 1) 2 (b) n (n 1) 2 (c) n ( n 1)(2 n 1) 6 (d) n 2 Solution (a): As A is a diagonal matrix, so the Eigenvalues of matrix A are itself its diagonal n
elements and their sum is
i 1 2 n n( n 1) i 1
2.
Example 1.159 [XE-2010 (1 mark)]: Which one of the following matrices has the same Eigenvalues 1 2 as that of ? 4 3
3 4 1 4 4 2 2 4 (a) (b) (c) (d) 1 2 2 3 1 3 1 3 Solution (b): As the matrix in option (b) is the transpose of the given matrix, hence the matrix in option (b) and the given matrix has same Eigenvalues. 2 a Example 1.160 [AE-2011 (2 marks)]: Consider the matrix A where a and b are real b 2 numbers. The two Eigenvalues of this matrix 1 and 2 are real and distinct 1 2 when (a) a 0 and b 0 (b) a 0 and b 0 (c) a 0 and b 0 (d) a 0 and b 0 2 a Solution (c): For Eigenvalues ( ) of the given matrix, we have A I 0 0 b 2
2 4 (4 ab) 0 . Now this equation has real and distinct values of if its discriminant is 2 0 (4) 4 1 (4 ab) 0 ab 0 a 0 & b 0 or a 0 & b 0 . So option (c) is correct.
4 0 . Which one 0 4
Example 1.161 [CH-2012 (1 mark)]: Consider the following 2 2 matrix A of the following vectors is NOT a valid Eigenvector of the above matrix? T
T
T
T
(a) 1 0 (b) 2 1 (c) 4 3 (d) 0 0 Solution (d): The given matrix is a diagonal matrix so its eigenvalues, 4, 4 . For eigenvector x1 4 x1 4 x1 4 x1 4 x1 (i) 4 0 x1 corresponding to 4 , AX X . 4 0 4 x2 x2 4 x2 4 x2 4 x2 4 x2 (ii) (i) and (ii) x1 , x2 ; but both x1 and x2 cannot be zero at the same time, since any non-zero
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[1.83]
vector, X , is said to be a eigenvector of matrix A , if there exist a number s.t. AX X ; where, is an eigenvalue of matrix A corresponding to eigenvector X . Example 1.162 [CS-2012 (1 mark)]: Let A be the 2 2 matrix with elements a11 a12 a21 1 19 and a22 1 . Then the Eigen values of the matrix A are
(a) 1024, –1024 Solution
(b) 1024 2 , 1024 2 1 1 A , for its 1 1
(d):
(c) 4 2 , 4 2
(d) 512 2 , 512 2 1 1 A I 0 0 1 1
Eigenvalues,
(1 )(1 ) 1 0 2 . Now, we know that if 1 , 2 ,, n are the eigenvalues of a square matrix 19
A , then 1k , 2k , , nk
19
k
are the eigenvalues of A . So the Eigenvalues of
18
A ( 2) 2( 2) 512 2
5 3 Example 1.163 [ME-2012, PI-2012 (2 marks)]: For the matrix A , one of the normalized 1 3 Eigenvectors is given as (a) 1 2
T
T
T
T
3 2 (b) 1 2 1 2 (c) 3 10 1 10 (d) 1 5 2 5 Solution (b): For Eigenvalue of the given matrix we have, 5 3 A I 0 0 2, 6 . For Eigenvector corresponding to 2 we have 1 3
x1 5 x1 3 x2 2 x1 5 x1 3x2 2 x1 (i) 5 3 x1 AX X 2 x 3 x 2 x (ii) . From (i) or (ii) 1 3 x2 x2 x1 3 x2 2 x2 1 2 2 x x x1 x2 X 1 1 . From the given options if we choose x1 1 x2 x1
1
2
1
2
2 X
.
3 5 2 Example 1.164 [EC-2013 (1 mark)]: The minimum Eigenvalue of the matrix A 5 12 7 is 2 7 5 (a) 0
(b) 1
(c) 2
(d) 3 3
5
2
Solution (a): As the determinant of the given matrix is 5 12
7
By Applying C 2 C2 C 3
3
3
2
5
5
7 0.
2 7 5 2 2 5 Hence one of the eigenvalues of the matrix A is zero, as the product of eigenvalue of any matrix is equal to the determinant of that matrix. Thus from the given options, the minimum eigenvalue is 0.
Example 1.165 [EE-2013 (2 marks)]: A matrix has Eigenvalues –1 and –2. The corresponding T
T
Eigenvectors are 1 1 and 1 2 respectively. The matrix is
1 1 1 2 1 0 (a) (b) (c) (d) 2 4 0 2 1 2 Solution (d): If X is the Eigenvector corresponding to the Eigenvalue ( ) of a11 a12 AX X . Let A . a21 a22
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0 1 2 3 the matrix A , then
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[1.84]
a11 a12 1 1 1 a11 a12 1 a11 a12 1 (i) For 1 , X , 1 a a 1 (ii) 1 1 a21 a22 1 a21 a22 1 21 22 a11 a12 1 1 1 a11 2a12 2 a11 2a12 2 (iii) For 2 , X , 2 a 2a 4 (iv) 2 2 a21 2a22 4 a21 a22 2 21 22 (i) and (iii) a11 0 and a12 1 ; similarly (ii) and (iv) a22 3 and a21 2 . [Similar question was also asked in XE-2009 (2 marks)] 1 4 Example 1.166 [AG-2014 (1 mark)]: The eigenvalues of A are 2 3 (b) 1, 2 (a) 2 i (c) 1 2i (d) non-existent Solution (c): For eigenvalues ( ) of the given matrix, we have 1 4 2 A I 0 0 (1 )( 3 ) 8 0 2 5 0 1 2i . 2 3 Example 1.167 [CS-2014 (2 Solution: The given matrix is marks)]: The product of the non1 0 0 0 1 By Applying zero eigenvalues of the matrix 0 1 1 1 0 R5 R5 R1 1 0 0 0 1 0 0 0 0 0 it is a triangular matrix, 0 1 1 1 0 R3 R3 R2 0 0 0 0 0 0 1 1 1 0 is _____. R R R 4 4 2 0 0 0 0 0 0 1 1 1 0 so its Eigenvalues are its diagonal elements, i.e., 1, 1, 0, 0, 0. So 1 0 0 0 1 the product of the non-zero Eigenvalues of the matrix is 1. Example 1.168 [CS-2014 (1 mark)]: Which one of the following statements is TRUE about every n n matrix with only real eigenvalues? (a) If the trace of the matrix is positive and the determinant of the matrix is negative, at least one of its eigenvalues is negative (b) If the trace of the matrix is positive, all its eigenvalues are positive (c) If the determinant of the matrix is positive, all its eigenvalues are positive (d) If the product of the trace and determinant of the matrix is positive, all its eigenvalues are positive Solution (a): Since, product of eigenvalues is equal to the determinant of any square matrix. If determinant of the matrix is negative then we have at least one of the eigenvalue to be negative; and in this case we may have trace of the matrix to be positive. Example 1.169 [IN-2014 1 2 3 1 2 3
(2
marks)]:
For
the
matrix
A
satisfying
the
equation
[ A] 7
8 9 4 5 6 , the eigenvalues are 4 5 6 7 8 9
(a) (1, j, j ) Solution (c): 1 2 3
(b) (1,1, 0)
(c) (1,1, 1)
(d) (1, 0, 0)
1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 [ A] 7 8 9 4 5 6 A 7 8 9 4 5 6 A 7 8 9 7 8 9 A 1 . 4 5 6 7 8 9 4 5 6 7 8 9 4 5 6 4 5 6
(By applying R2 R3 in the right side of the determinant). So product of eigenvalues of A must be negative. Hence, from the given options, we can say that option (c) is correct.
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[1.85]
1 1 0 Example 1.170 [EE-2014 (2 marks)]: A system matrix is given as follows: A 6 11 6 . 6 11 5
The absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is _____. 0 1 1 Solution: For Eigenvalues of the matrix A , we have A I 0 6 6
11
6
11
5
0
3 6 2 11 6 0 ( 1)( 2)( 3) 0 1, 2, 3 . So maximum Eigenvalue is 1 and minimum Eigenvalue is 3 . Hence, the absolute value of the ratio of the maximum eigenvalue to the minimum eigenvalue is 1 3 . Example 1.171 [MA-2014 (1 mark)]: Let A M 3 () be such that det( A I ) 0 , where I denotes the 3 3 identity matrix. If the tr ( A) 13 and det( A) 32 , then the sum of squares of the eigenvalues of A is _____. Solution: We have 3 3 , so let 1 , 2 , 3 are the eigenvalues of the matrix A which can be assumed
1 0 as: A 0 2 0 0
0 0 A I 0 (1 1)(2 1)(3 1) 0 .…(i) tr ( A) 13 1 2 3 13
3
…(ii); det( A) 32 12 3 32 …(iii). Now by putting values of (ii) and (iii) in (i), we get 12 3 12 2 3 3 1 1 2 3 1 0 12 2 3 31 44 …(iv). Now squaring
both sides of (ii) 12 22 32 2(12 2 3 3 1 ) 169 12 22 32 169 2 44 81 [By putting value of (iv)]. 1 2 Example 1.172 [TF-2014 (2 marks)]: Of the two eigenvalues of the matrix A 4 3 (a) One is positive, one is negative (b) Both are positive (c) Both are negative (d) Both form a complex conjugate 1 2 Solution (a): For eigenvalues of the given matrix, A I 0 0 1,5 . 4 3 Example 1.173 [XE-2014 (1 mark)]: If 1, 0 and –1 are the eigenvalues of a 3 3 matrix A , then the trace of A2 5 A is equal to _____. 2
2
2
Solution: 1, 0 and –1 are the eigenvalues of a matrix A 1 , 0 , (1) are the Eigenvalues of a 2
matrix A sum of diagonal element of A2 1 0 1 2 . Also sum of diagonal elements of 5A 5(1 0 1) 0 . So sum of diagonal elements of A2 5 A sum of diagonal element of A2 sum of diagonal elements of 5A 2 0 2 . Hence answer is 2.
Block matrix: A block matrix is a matrix that is defined using smaller matrices called blocks. For 0 2 3 9 0 2 2 0 1 3 3 9 5 0 then matrix e.g., if A 2 0 , B , C 7 4 , then D M 0 5 4 6 5 0 1 3 4 6 7 4 0 5 A B M , where A, B , C , D are themselves matrices, is a block matrix. C D Copyright © 2016 by Kaushlendra Kumar
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[1.86]
A B If we have a block matrix defined by M , where A, B , C , D are of same order, and C D also, if C and D commute (i.e., CD DC ); then det M det( AD BC )
Example 1.174 [CS-2007 (2 marks)]: Let A be a 4 4 matrix with Eigenvalues –5, –2, 1, 4. Which A I of the following is an Eigenvalue of M , where I is the 4 4 identity matrix? I A (a) –5 (b) –7 (c) 2 (d) 1 A I I Solution (c): Let be the Eigenvalue of the matrix M then M I 0 0. I A I As we have block matrix and it is given that A and I both are 4 4 matrices, so ( A I ) is also 4 4
matrix; also
I ( A I ) ( A I ) I ; hence
A I
I
I
A I
2 2 0 ( A I ) I 0
( A I I )( A I I ) 0 A I I A I I 0 A I I 0 or A I I 0 . As, A be a 4 4 matrix with Eigenvalues –5, –2, 1, 4 so matrix A can be written as 5 0 0 0
0
A
0 0
2
0
0
0
1
0
diag ( A) ( 5, 2,1, 4) ; since I diag (1,1,1,1) and I diag ( , , , ) .
0 0 4 So, A I I diag (5 1, 2 1,1 1, 4 1) A I I 0 ( 4 )(1 )(2 )(5 ) 0 4, 1, 2,5 Also, A I I diag ( 5 1, 2 1,1 1, 4 1) A I I 0 ( 6 )( 3 )( )(3 ) 0 6, 3, 0, 3
1.5.1 Eigenbases of a Matrix n If n n matrix A has n distinct eigenvalues, then A has a basis of eigenvectors x1 , x2 ,, x n R . Proof: All we have to show is that x1 , x2 ,, x n are linearly independent. Suppose they are not. Let r be the largest integer such that {x1 , x2 ,, xr } is a linearly independent set. Then r n and the set
{x1 , x2 ,, xr , x r1} is linearly dependent. Thus there exists scalars c1 , c2 ,, cr 1 , not all zero, s.t. c1x1 , c2 x2 ,, cr 1xr 1 0
(1.31)
Multiplying both sides of Eq. 1.14 by A and using A x j j x j , we obtain,
c11 x1 c22 x2 cr 1r 1 xr 1 0
(1.32)
To get rid of the last term, we subtract r 1 times Eq. 1.14 from 1.15, obtaining
c1 (1 r 1 )x1 c2 (2 r 1 )x2 cr (r r 1 )xr 0
(1.33)
Here c1 (1 r 1 ) c2 (2 r 1 ) cr (r r 1 ) 0 since {x1 , x2 ,, xr } is linearly independent.
c1 c2 cr 0 , since all eigenvalues are distinct. But with this 1.14 reduces to cr 1 xr 1 0 , hence cr 1 0 , since xr1 0 . This contradicts the fact that not all scalars in 1.14 are zero. Hence the conclusion of the theorem must hold.
1.5.2 Similar Matrices 1
Two matrices A and B are said to be similar, if there exists a non-singular matrix P s.t. B P AP .
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1
When A and B are connected by the relation B P AP , B is said to be obtained from A by a similarity transformation. 1 1 1 1 B P AP , after pre- and post- multiplying by P on both sides, we get, PBP PP APP 1 A PBP ; hence, A is said to be obtained from B by a similarity transformation. Two similar matrices have same eigenvalues and hence same determinants (as product of all eigenvalues gives value of determinant), same rank, trace and characteristic polynomial. 1 Proof: Let A and B are said to similar matrices, then B P AP B I P 1 AP I 1
1
1 1 B I P AP P IP B I P ( AP IP) B I P ( A I ) P
B I ( A I ) P 1 P B I ( A I ) P 1 P B I ( A I ) I ( A I )
Thus, both A and B have same characteristic equations and hence both have same eigenvalues. 1 If A and B are two similar matrices, i.e., B P AP , and if x is an Eigenvector of A corresponding to nonzero Eigenvalue of A then P 1 x is an Eigenvector of B corresponding to same Eigenvalue (i.e., ) of B [This point was asked in XE-2007 (2 marks)]. Proof: As x is an Eigenvector of A corresponding to nonzero eigenvalue Ax x . 1 1 1 1 1 1 B P AP PBP A PBP x Ax PBP x x B ( P x ) ( P x) if x is an Eigenvector of A , with Eigenvalue , then P 1 x is an Eigenvector of B with same Eigenvalue.
Example 1.175 [XE-2009 (1 mark)]: Let A and B be two similar square matrices of order two. If 1 and –2 are the Eigenvalues of A , then the trace of B is (a) –2 (b) –1 (c) 1 (d) 2 Solution (b): As A and B are two similar square matrices Eigenvalues of A and B are same. So tr ( B) 1 2 1 . Hence option (b) is correct.
Diagonalization of Matrices: A diagonal matrix is a n n matrix with non-zero entries only along the main diagonal. Diagonal matrices are particularly convenient for eigenvalue problems since the eigenvalues of a diagonal matrix coincide with the diagonal entries aii and the eigenvector corresponding to the eigenvalue aii is just the i th coordinate vector. If any square matrix is not diagonal matrix then diagonalisation is the process of finding a corresponding diagonal matrix. Once we have a diagonal matrix, we can easily find the eigenvalues or eigenvectors. An n n matrix A is diagonalizable if and only if it has n linearly independent eigenvectors. Proof: Suppose A has n linearly independent eigenvectors. Then the matrix C formed by using these eigenvectors as column vectors will be invertible (since the rank of C will be equal to n ). On the other hand, if A is diagonalizable then, by definition, there must be an invertible matrix 1 C such that D C AC is diagonal. But the columns vectors of C must coincide with the eigenvectors of A . Since C is invertible, these n column vectors must be linearly independent. Hence, A has n linearly independent eigenvectors. If a n n matrix A is diagonalizable then it does not mean that matrix A has distinct Eigenvalues; as we have proved earlier that an Eigenvalue of a matrix can correspond to different Eigenvectors. However, if n n matrix A has n distinct Eigenvalues then A is diagonalizable. T If A is a real symmetric matrix (i.e., A A ) then (i) it has only real Eigenvalues (ii) it is always diagonalizable (iii) it has orthogonal eigenvectors [This point was asked in AE-2014 (1 mark)]. Let A be a n n matrix, let 1 , 2 ,, n be a set of n scalars, and let v1 , v 2 ,, v n be a set of n vectors. Let C be n n matrix formed by using v j for j
th
column vector, and let D be the
n n matrix whose diagonal entries are 1 , 2 ,, n . Then AC CD if and only if 1 , 2 ,, n are eigenvalues of A and each v j is an eigenvector of A corresponding to the eigenvalue j .
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[1.88]
| | | | Proof: Under the hypothesis AC A v1 v n A v1 A v n | | | | A v1 | | | 1 0 | CD v1 v n 1 v1 n v n and so, AC CD | | 0 n | | A vn
1v1
n v n and vice-versa. Now suppose AC CD , and the matrix C is invertible. Then we can write 1 D C AC and so we can think of the matrix C as converting A into diagonal matrix. Hence, An n n matrix A is diagonalizable if there is an invertible n n matrix C such that C 1 AC is a diagonal matrix. The matrix C is said to diagonalize A . 1 b a Example 1.176 [XE-2007 (2 marks)]: Let M 0
2 c , where a , b and c are real numbers. 0 0 1
Then M is diagonalizable (a) for all values of a , b and c (b) only when bc a (c) only when b c a (d) only when bc a Solution (d): We have to find the condition for which the given matrix has 3 linearly independent T
eigenvectors. Let an eigenvector of M is given as X x1 x2 x3 . As the Eigenvalues of the given matrix are 1, 2,1 . For Eigenvector corresponding to 2 , we have MX X (iii) x3 0 1 b a x1 x1 x1 bx2 ax3 2 x1 (i) 0
2 c x2 2 x2 2 x2 cx3 2 x2 x3 2 x3 0 0 1 x3 x3
So X 1 bk
k
T
(ii) (ii) x2 k (say) . (iii)
(i) x1 bk
T
0 k b 1 0 (which is one independent Eigenvector).
1 b a x1 x1 For Eigenvector corresponding to 1, we have MX X 0 2 c x2 1 x2 0 0 1 x3 x3 x1 bx2 ax3 x1
(iv) ; 2 x2 cx3 x2
(v) ; x3 x3
(vi) . Let x3 q (say); then from (ii), we get
x2 cq ; and putting values of x3 and x2 in (i), we get ( a bc) q 0 (vii) ; so we can let x1 p T
T
(say); hence X 2 p cq q . Now from (vii) if ( a bc ) 0 q 0 X 2 p 0 0 we have only one independent Eigenvector vector and so we have total two independent Eigenvectors; hence matrix M is not diagonalizable. On the other hand, if from (vii) if T
q 0 ( a bc) 0 X 2 p cq q we can have two independent Eigenvector vector and so we have total three independent Eigenvectors; hence matrix M is diagonalizable. So for matrix M to be diagonalizable we have the condition ( a bc ) 0 . Example 1.177 [XE-2011 (1 mark)]: For any positive number
a and b , the matrix
T
P 1 a b 4 5 6 is (a) Orthogonal (b) diagonalizable 4 4 5 6
(c) non-singular 5 6 1
(d) of rank 2 1
1
Solution (b): P 4a 5a 6a P 4a 5a 6a 4 5 6 a a a (By taking common 4, 4b 5b 6b b b b 4b 5b 6b
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[1.89]
5, 6 from C1 , C2 , C3 , respectively); so P 0 , as the two columns are identical. So P is a singular By Applying
4 5 6 matrix. R2 R2 aR1 , P 0 0 0 P has only one independent row so rank of P is 1. 0 0 0 R R bR 3 3 1 4 5 6 4 4a 4b 77 77 a 77b Also, PP 4a 5a 6a 5 5a 5b 77 a 77 a 2 77 ab I ; P is not orthogonal 4b 5b 6b 6 6a 6b 77b 77 ab 77b 2 T
matrix. As options (a), (c) and (d) are incorrect; so we have only one option (b) which is correct. 1 0 1 Example 1.178 [XE-2013 (2 marks)]: For the matrix M 0
1 1 . Consider the following 1 1 2 1
statements: (A) The characteristic equation of M is 3 0 . (B) M does not exist. (C) The matrix M is diagonalisable. Which of the above statement are true? (a) A, B and C (b) A and C but not B (c) A and B but not C (d) B and C but not A 1 0 1 Solution (a): For characteristic equation, we have M I 0
0
1
1
1
1
2
0
3
(1 ){(1 )( 2 ) 1} 1{0 (1 )} 0 0 . So statement (A) is correct. As the
characteristic equation is 3 0 ( 2 1) 0 0, 1 one of the Eigenvalue is zero so determinant of matrix M is zero, hence M is a singular matrix so it is not invertible; so statement (B) is correct. Now for finding the Eigenvector, we have MX X , where X x1
x2
T
x3 .
x1 x3 0 (i) 1 0 1 x1 x1 Now, Eigenvector corresponding to 0 0 1 1 x2 0 x2 x2 x3 0 (ii) 1 1 2 x3 x3 x1 x2 2 x3 0 (iii)
(i) and (ii) x1 x2 x3 k (say) X 1 k
k
T
T
k k 1 1 1
x1 x3 x1 1 0 1 x1 x1 Eigenvector corresponding to 1 0 1 1 x2 1 x2 x2 x3 x2 1 1 2 x3 x3 x1 x2 2 x3 x3
(i) and (ii) x3 0 and (iii) x2 x1 p (say) X 2 p
p
(i) (ii) (iii)
T
T
0 p 1 1 0
x1 x3 x1 1 0 1 x1 x1 Eigenvector corresponding to 1, 0 1 1 x2 1 x2 x2 x3 x2 1 1 2 x3 x3 x1 x2 2 x3 x3
(i) x1 x3 2 ;
(ii) x2 x3 2 ; T
(i) (ii) . So (iii)
(iii) x3 x3 r (say)
T
X 3 r 2 r 2 r r 1 2 1 2 1 . So we have three linearly independent Eigenvectors of matrix M so it is a diagonalizable matrix. So statement (C) is also correct.
Calculation of powers of matrix A using Diagonalisation: Assuming A satisfies the condition, D C 1 AC A CDC 1 . A2 CDC 1CDC 1 CDIDC 1 CD 2C 1 . Similarly, 3 2 2 1 1 2 1 3 1 n n 1 A A A CD C CDC CD IDC CD C ; and so on; Similarly, we have A CD C .
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[1.90]
Matrix Exponential: To find the exponential of matrix A , i.e., e A , let A be a diagonalizable 1
1
matrix such that there exist a matrix P for which P AP D or A PDP , where D is a diagonal A A2 A3 PDP 1 PDP 1 PDP 1 PDP 1 PDP 1 PDP 1 A matrix; then e I I 1! 2! 3! 1! 2! 3! PDP 1 PD 2 P 1 PD 3 P 1 D D 2 D3 1 A 1 D 1 e PIP P I P Pe P 1! 2! 3! 1! 2! 3! Therefore if we can find a P so that e D can be easily computed, e A can be easily found. 0 1 Example 1.179 [EC-2008 (2 marks)]: Consider the matrix A . The value of e A is 2 3
2e 2 3e 1
e 1 e 2
(a)
e 1 e 2
(b)
2 1 5e 2 e 1 2e 2e 5e 2 e 1 3e 1 e 2 (c) 2 1 4e 2 e 1 2e 6e
2e 2 e 1
1 2 3e 1 2e 2 2e 4e 2e 1 e 2 e 1 e 2 (d) 1 2 e 1 2e 2 2e 2e
1
Solution (d): If A PDP , where D is a diagonal matrix then e A Pe D P 1 , where P is the matrix which contains the Eigenvectors of matrix A . So for Eigenvalues of matrix A , we have A I 0
0
1
2
3
2 0 3 2 0 ( 1)( 2) 0 1, 2 . So the diagonal matrix
e 1 0 1 0 D D can be chosen as D e 0 e 2 . For Eigenvector corresponding to 1 0 2 x2 x1 (iii) x1 0 1 x1 . As (iii) and (iv) are identical equation so we 2 3 x2 x2 2 x1 3x2 x2 (iv) x1 x1 1 can choose X 1 x1 . For Eigenvector corresponding to 2 1 x2 x1 x2 2 x1 (i) x1 0 1 x1 . As (i) and (ii) are identical equation so we 2 2 3 x2 x2 2 x1 3 x2 2 x2 (ii) x1 x1 1 can choose X 2 x1 . So matrix P can be chosen as the independent 2 x2 2 x1 1 1 1 Eigenvectors of matrix A , i.e., P . Now we have to find P , as P 2 1 1 and 1 2 T
T
Cij 1 2 1 2 1 C11 2; C12 1; C21 1; C22 1 . So P P P P 1 1 1 1 e 1 e 2 1 1 e 1 0 2 1 2e 1 e 2 Now, e A Pe D P 1 e A 1 2 2 e 1 2e 2 1 2 0 e 1 1 2e 2e adj ( P )
1
1.5.3 Matrix Polynomial If
matrix
A
satisfies
the
polynomial
f ( x ) a0 a1 x a2 x 2 an x n ,
then
f ( A) a0 a1 A a2 A an A , where, a0 , a1 ,, an are constants. 2
n
Cayley-Hamilton Theorem: In linear algebra Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation.
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In other words if A be any n n matrix and I n be n n identity matrix, then the characteristic polynomial of A is defined as p ( ) A I . Cayley-Hamilton Theorem states that substituting the matrix A for in this polynomial result in zero or null matrix. The powers of A , obtained by substitution from powers of , are defined by repeated matrix 0 multiplication; the constant term of p ( ) gives a multiple of the power A , which power is defined as the identity matrix. The theorem allows n A to be expressed as a linear combination of the lower matrix powers of A . any matrix polynomial in A of order n n as a polynomial of degree n 1 in A . inverse of any square matrix as a linear combination of the matrix A .
Example 1.180 [CS-1993, EC-1993 1 0 0 1
0 1 0 1 (1 mark)]: If A 0 0 i i 0 0 0 i
1
0
0
1
0
1
0
1
0
0
i
i
0
i
Solution: A I 0
0 0 (1 )( 1 )(i )( i ) 0
4
then calculate matrix A by the use 4 1 0 . Then by Cayley-Hamilton of Cayley-Hamilton theorem or substituting A for , A4 I O A4 I . otherwise.
0
Theorem,
Example 1.181 [CS-2006 (2 marks)]: F is a n n real matrix. B is a n 1 real vector. Suppose there are two n 1 vectors, u and v such that u v , and Fu B , Fv B . Which one of the following statements is false? (a) det F 0 (b) There are infinite number of solutions to Fx B (c) F must have two identical rows (d) There is an x 0 such that Fx O Solution (c): From the given data we have the equation FX B do not have unique solutions, as X u , v ( u v ); hence the equation FX B have infinite number of solutions; and this appears when det F 0 ; so option (a) and (b) are correct. Since, if det F 0 then it does not necessarily that F must have two identical rows; so option (c) is not correct. Also if we subtract the two equation Fu B , Fv B , we get F (u v ) O but as u v u v O ; so option (d) is also correct. Statement for Linked Answer Question 1.182 & 1.183: Cayley-Hamilton Theorem states that a 3 2 square matrix satisfies its own characteristic equation. Consider a matrix A . 1 0 Example 1.182 [EE-2007 (2 marks)]: A satisfies the relation (c) ( A I )( A 2 I ) 0 (d) exp( A) 0 (a) A 3I 2 A1 0 (b) A2 2 A 2 I 0 3 2 2 Solution (a): A I 0 0 3 2 0 . Then by Cayley-Hamilton 1 0 Theorem, substituting A for , A2 3 A 2 I 0 . Now multiplying both sides of the above 1 relation with A , A1 A2 3 A1 A 2 A1 I 0 A 3I 2 A 1 0 . 9
Example 1.183 [EE-2007 (2 marks)]: A equals (a) 511A 510 I (b) 309 A 104 I 2
(c) 154 A 155I
(d) exp(9 A)
2
Solution (a): A 3 A 2 I 0 A 2 I 3 A . A4 A2 A2 (2I 3 A)(2 I 3 A) 4I 12 A 9 A2 4I 12 A 9(2I 3 A) 15 A 14I
A8 A4 A4 (15 A 14I )(15 A 14I ) 225 A2 420 A 196I A8 225(2I 3 A) 420 A 196I 255 A 254I
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[1.92]
A9 A8 A 255 A2 254 A 255(2 I 3 A) 254 A 511A 510I A9 511A 510 I . [Similar questions were also asked in EC-2012, EE-2012, IN-2012 (2 marks)] Example 1.184 [EE-2007 (2 marks)]: The linear operation L( x) is defined by the cross product T
T
L( x ) b x , where b 0 1 0 and x x1
x2
matrix M of this operation satisfies L( x ) M x1 (a) 0, 1 , –1 (b) 1, –1, 1 ˆj kˆ iˆ
x2 x3 . Then the Eigenvalues of M are (c) i , i , 1 (d) i , i , 0
Solution (c):
x3 are three dimensional vectors. The 3 3 T
L ( x) b x 0
1
0 x3iˆ 0 ˆj x1kˆ . In matrix notation,
x1
x2
x3
x1 x3 m11 L ( x ) M x2 0 m21 x3 x1 m31
m12 m22 m32
x3 L( x) 0 . x1
m11 m12 0; m13 1 0 0 1 m23 x2 m21 m22 m23 0 M 0 0 0 . m33 x3 m31 1; m32 m33 0 1 0 0 m13 x1
0
0
1
0
0
0
1
0
0
For Eigenvalues of M , M I 0
0 3 0 1, i .
1
Example 1.185 [XE-2007 (2 marks)]: Let M be a 2 2 matrix with eigenvalue 1 & 2. Then M is (a) ( M 3I ) 2 (b) (3I M ) 2 (c) (3I M ) 2 (d) ( M 3 I ) 2 Solution (c): As Eigenvalues of 2 2 matrix M are 1 and 2, so its characteristic equation can be 2 given as ( 1)( 2) 0 3 2 0 . Replacing M by , we get characteristic polynomials, 1
as, M 2 3M 2 I 0 . Now multiplying both of the characteristic polynomial with M , 1 2 1 1 1 M M M 3M M 2 I 0 M 3 I 2 M 0 1 2 1 ( M M M MM IM M ) M 1 (3I M ) 2 . Example 1.186 [EE-2008 (1 mark)]: The characteristic equation of a 3 3 matrix P is defined as ( ) I P 3 2 2 1 0 . If I denotes identity matrix, then inverse of matrix P will be 2
(a) ( P P 2I )
2
(b) ( P P I )
2
2
(c) ( P P I )
(d) ( P P 2I )
Solution (d): By Cayley-Hamilton Theorem, substituting P for , P3 P 2 2 P I 0 . 1 Pre-multiplying both sides by P P 1 P 3 P 1 P 2 2 P 1 P P 1 I 0
P 2 P 2I P 1 0 P 1 ( P 2 P 2 I )
1 1 Example 1.187 [XE-2010 (1 mark)]: Find P8 2 P 7 2 P 6 4 P5 3P 4 6 P 3 2 P 2 , if P , 1 1 (b) 2 P (c) 3 P (d) 4 P (a) P Solution (d): The characteristic equation for the given matrix can be found by P I 0 1
1
1
1
2 2 0 (1 ) 1 0 2 0 . Using Cayley-Hamilton Theorem, substituting P
3
2
2
2 2 for Now P P P 2 P P 2P 4P , , P 2P 0 P 2P . 4 2 2 2 5 6 7 8 P P P 2 P 2 P 4 P 8P , and similarly, P 16 P , P 32 P , P 64 P , P 128 P . 8 7 6 5 4 3 2 So, P 2 P 2 P 4 P 3P 6 P 2 P 128P 128 P 64 P 64 P 24 P 24 P 4 P 4 P
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Example 1.188 [XE-2011 (1 mark)]: The Eigen values of a 3 3 matrix P are 2, 2, –1. The P equal to (a) (3P P 2 ) 4 (b) ( P 2 2 P ) 2 (c) ( P 2 3P) 2 (d) ( P 2 2 P ) 4 Solution (a): The characteristic equation in is given as: ( 2)( 2)( 1) 0
1
is
( 2 4 4)( 1) 0 3 3 2 4 0 . Hence by Cayley-Hamilton Theorem, substituting P 1
for , P 3 3P 2 4 I 0 . Pre-multiplying by P , we get P 1 P 3 3P 1 P 2 4 P 1 I 0 2
P 3P 4 P
1
0 P
1
(3P P 2 ) 4 .
3
Example 1.189 [AE-2014 (2 marks)]: If [ A]
3
3
. Then [ A]2 7[ A] 3[ I ] is 4
(a) 0 (b) –324 (c) 324 (d) 6 Solution (a): The characteristic equation of the given matrix can be found as: A I 0
3
3
3
4
2 0 (3 )(4 ) 9 0 7 3 0 . Using Cayley-Hamilton Theorem,
substituting A for , A2 7 A 3I O ( A2 7 A 3I ) O O 0 .
2 0 3 1 Example 1.190 [MA-2014 (2 marks)]: Let A 3 1 3 . A matrix P such that P AP is a 0 0 1 diagonal matrix, is 1 1 1
1 1 1 (b) 0 1 1 1 1 0
(a) 0 1 1 1 1 0
1 1 1 (c) 0 1 1 1 1 0
1 1 1 (d) 0 1 1 1 1 0
Solution (a): For Eigenvalues of the given matrix, we have A I 0
2
0
3
3
1
3
0 (2 )( 1 ) 2 1, 2 . Now let X x1
x2
T
x3
0 0 1 Eigenvector corresponding to Eigenvalue of matrix A , then AX X . 2 x1 3x3 x1 2 0 3 x1 x1
is a
(i)
Eigenvector corresponding to 1 3 1 3 x2 1 x2 3x1 x2 3x3 x2 0 0 1 x3 x3 x3 x3
(ii) (iii)
(i) x1 x3 p (say), where p and putting value of x1 and x3 in (ii) we get x2 q (say), where q . Now as we have two independent elements p, q , so we have two independent Eigenvectors X 1 , X 2 p
T
p
q
corresponding to 1 ; but in both Eigenvectors x1 x3 . T
T
From the given options we can choose p 1 for q 0,1 . So X 1 1 0 1 and X 2 1 1 1 . 2 x1 3 x3 2 x1 2 0 3 x1 x1 Eigenvector corresponding to 2 3 1 3 x2 2 x2 3 x1 x2 3 x3 2 x2 0 0 1 x3 x3 x3 2 x3
(i) (ii) (iii)
(iii) x3 0 and putting value of x3 in (ii) we get x1 x2 r (say), where r . So
X 3 r
r
T
T
T
0 r 1 1 0 . From the given options taking r 1 X 3 1 1 0 . So the T
T
T
three linearly independent Eigenvectors are X 1 1 0 1 , X 2 1 1 1 and X 3 1 1 0 .
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Chapter 1: Linear Algebra
[1.94]
Quadratic Forms
By definition, a quadratic form Q in the components x1 , x2 ,, xn of a vector x is the sum of n 2 a11 x12 a12 x1 x2 a1n x1 xn n
T
n
terms, namely Q x A x a jk x j xk
a21 x2 x1 a22 x22 a2 n x2 xn
j 1 k 1
. A a jk is called the
an1 xn x1 an 2 xn x2 ann xn2 coefficient matrix of the form. We may assume A is symmetric, because we can take off – diagonal terms together in pairs and write the result as a sum of two equal terms as shown in example as: 3 4 x1 T 2 2 2 2 Let, Here x A x x1 x2 3 x1 4 x1 x2 6 x2 x1 2 x2 3x1 10 x1 x2 2 x2 . 6 2 x2
4 6 10 5 5 . From the corresponding symmetric matrix C c jk , where c jk
a jk akj 2
, thus
c11 3, c12 c21 5, c22 2 we get the same result. Example 1.191 [CE-1998 (2 marks)]: The real symmetric matrix C corresponding to the quadratic form Q x12 4 x1 x2 5 x22 , is
1 (a) 2
2
2 (b) 0
5
0 5
Solution (a): Let Q xT A x x1
a
x2
c
1 1 (c) (d) 1 2 b x1 2 2 ax1 (b c ) x1x2 dx2 . d x2
0 2 1 5 Now comparing it
with the given equation, we get a 1 , c d 4 , d 5 . Now the coefficient of x1x2 is 4 which can
1 be written as 2 2 . Hence, b c 2 . Hence the given symmetric matrix is given as 2
2 5
.
1 Example 1.192 [CH-2011 (1 mark)]: Let 1 1 and 2 3 be the Eigenvalues and V1 and 0 1 V2 be the corresponding Eigenvectors of a real 2 2 matrix R . Given that P V1 V2 , 1
1
which ONE of the following matrices represents P RP ? 0 1 0 3 3 (a) (b) (c) 3 0 1 0 0
a11
Solution (d): Let R
0 1
1 0 (d) 0 3
a12
a21 a22 1 a11 For 1 1 and V1 , RV1 1V1 0 a21 1 a11 For 2 3 and V2 , RV2 2 V2 1 a21 3 1 4 So, R . If we apply R1 R1 R2 4 0 3
1 a11 1 a11 1 1 a22 0 a21 0 0 a21 0 a12 1 1 a11 a12 3 a12 4 3 a22 1 1 a21 a22 3 a22 3 1 0 on matrix R , we get R . 0 3 a12 1
1
As R and R are similar matrices because R P RP . So option (d) is correct.
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[1.95]
Exercise: 1.5 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill.
2 1 is _____. 1 2 2 1 2. One of the eigenvector of the matrix A is 1 2 1. Maximum eigenvalue of the matrix A
T
T
(a) 1 1
(b) 2 3
T
(c)
3
2
(c)
5
2
T
(d) 1 2
5 2 is 7 4
3. An eigenvector of the matrix A (a)
2
T
5
(b) 7
T
2
T
(d)
2
T
7
3 4 1 4. The eigenvector of the matrix A 1 2 1 corresponding to maximum eigenvalue is 3 9 0 (a)
1
T
1 2
T
T
(b) 1 1 3
(c) 1 1 2
T
(d) 1 1 3
1 2 2 25 5. If A 1 2 1 , then eigenvalues of A are 1 1 0 (a) 1, 1,1
(b) 1, 0, 2
(c) 1,1,1 3 0
(d) 2, 0,1
0 6. For what value(s) of x if any does the matrix A 0 x 2 , has at least one repeated 0 2 x
eigenvalue? (a) x 2 or x 4 (b) x 1 or x 2 4 2 1
1
(c) x 1 or x 5 (d) x 2 or x 5 0 1 1
7. For matrix A 2
0 1 , and vector v1 1 , v 2 1 , v 3 2 and v 4 1 , which of 2 2 3 0 2 1 4
the given vectors are eigenvectors corresponding to matrix A ? (a) v1 and v 2 (b) v1 and v3 (c) v 2 and v3
(d) v3 and v 4
1 4 2 1 8. 2 is an eigenvector of matrix A 2 0 1 . What is the corresponding eigenvalue? _____. 2 2 3 2 4 7 1 9. What is the sum of eigenvalues of the matrix A 0 3 8 ? 0 0 2 2 5 6 10. Given that 1 is an eigenvalue of the matrix A 1 0 0 . The other two eigenvalues are 0 1 0 (a) 2 and 3
(b) 2 and 3
Copyright © 2016 by Kaushlendra Kumar
(c) 0 and 3
(d) 2 and 0
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.96]
4 1 6 11. The characteristic equation of the matrix A 2 1 5 is 2 1 0 (a) 3 5 2 4 0 (c) 3 5 2 4 0
(b) 3 5 2 21 42 0 (d) 3 5 2 14 0
1 2 4 12. Given that 5 is an eigenvalue of 1 4 8 , which of the following system of equations 0 1 1 should be solved to find the corresponding eigenvectors? 4 3 9 x 0 4 (a) 6 1 5 4
2 4 x x (b) 1 1 8 y y 0 1 6 z z
3 y 0 6 z 0
4 2 4 x 0 (c) 1 1 8 y 0 0 1 6 z 0
1 2 4 x 5 (d) 1 4 8 y 5 0 1 1 z 5
1 0 2 13. A particular 3 3 matrix A has an eigenvalue of 1 . The matrix A I reduces to 0 0 0 . 0 0 0 Corresponding to eigenvalue 1 , all the eigenvectors of A are non-zero vectors of the form: (where, s, t ) T
T
T
T
(a) 2t 0 t (b) 2t s t (c) t 0 2t (d) t s 2t 14. The vector x is an eigenvector of each of the matrices A and B , with corresponding eigenvalues and , respectively. Which of the following statement is correct? (a) x is an eigenvector of AB with eigenvalue (b) x is not an eigenvector of AB (c) x is an eigenvector of AB with eigenvalue ( ) (d) x is an eigenvector of AB with eigenvalue ( ) 15. The vector x is an eigenvector of each of the matrices A and AB , with corresponding eigenvalues 0 and , respectively. Which of the following statement is correct? (a) x is an eigenvector of B with eigenvalue (b) x is not an eigenvector of AB (c) x is an eigenvector of B with eigenvalue (d) x is an eigenvector of B with eigenvalue
3 2
16. If P
1 2
1 (a) 0
2005 1
1 0 17. If A 0 1 0 2
12
1 1 and Q PAPT , then PT Q 2005 P , A 0 1 3 2 2005 0 1 1 (b) (c) 1 2005 2005 1 0 1 and 6A1 A2 cA dI , then ( d c ) _____ 4 2
1 0 (d) 0 1
n
18. Matrix A is such that A 2 A I , where I is the identity matrix, then for n 2 , A
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics (a) 2n 1 A ( n 1) I
Chapter 1: Linear Algebra
(b) 2
n 1
AI
[1.97]
(c) nA (n 1) I
(d) nA I
0 a c and ( A I )50 50 A , then ( a b c d ) _____. 0 0 b d i i 1 1 20. If A , where i 1 , and B , then A8 kB , where k _____. i i 1 1 19. Let A
21. If P is an orthogonal matrix and Q PAPT and x PT Q1000 P , then x 1 is, where A is involuntary matrix. 1000 (d) None of these (a) A (b) I (c) A 22. If A and B are square matrices of same order and A is non-singular, then for a positive integer 1 n n , ( A BA) n
n
n
n
(a) A B A
n
n
(b) A B A
1
n
(d) n( A1 BA)
(c) A B A 1
23. If A and B are two square matrices such that B A BA , then ( A B ) 2 2 2 2 2 (b) O (d) A B (a) A B (c) A 2 AB B 24. If one of the eigenvalues of a square matrix A is zero, then (a) det A must be non-zero (b) det A must be zero (c) adjA must be a zero matrix (d) tr ( A) must be zero
0 1 T 8 6 4 2 and ( A A A A I ) B 0 11 , where I is a 2 2 identity matrix, then 3 0
25. If A
the product of all elements of matrix B is _____. 26. If the matrix A has as an eigenvalue with corresponding eigenvector x . The non-singular matrix Q is of the same order as A . Then (a) Q x is an eigenvector of the matrix B , where B QAQ 1 , and 1 is eigenvalue (b) Q x is an eigenvector of the matrix B , where B QAQ 1 , and is eigenvalue (c) B x is an eigenvector of the matrix Q , where B QAQ 1 , and is eigenvalue (d) B x is an eigenvector of the matrix Q , where B QAQ 1 , and 1 is eigenvalue 27. If X is a diagonalizable n n matrix with only one eigenvalue , then (a) X I n (b) I n X (c) I n n X (d)
the corresponding the corresponding the corresponding the corresponding
X n In
2 4 6 2 0 0 28. If matrices A 0 2 2 and B 2 6 0 , then 0 0 4 3 2 1 (a) only A is diagonalizable (c) both A and B are diagonalizable 2 0 0
(b) only B is diagonalizable (b) neither A nor B are diagonalizable
29. If A 1
1 2 1 , find the diagonal matrix D such that A PDP . 1 0 1
(a) diag 1 2
2
(b) diag 2 1 2
(c) diag 2
2 1
(d) diag 2
2
2
30. The real symmetric matrix corresponding to the quadratic form Q 8 x12 4 x1 x2 5 x22 is
8 (a) 2
2 5
0 (b) 2
2 5
Copyright © 2016 by Kaushlendra Kumar
8 (c) 5
2 0
8 (d) 2
2 5
e-mail: [email protected]
Engineering Mathematics
Chapter 1: Linear Algebra
[1.98]
Answer Keys 1 d 16 4 31 b
2 3 17 0 32 13
3 a 18 2 33 2
4 b 19 1 34 0
5 b 20 8 35 9
Answer Keys: Exercise: 1.1 6 7 8 9 10 0 1 c 4 9 21 22 23 24 25 1 b b d d 36 37 38 39 40 d d 36 1 c
1 d 16 a
2 a 17 d
3 1 18 0
4 1 19 d
5 0 20 d
Answer Keys: Exercise: 1.2 6 7 8 9 10 a c c 1 8 21 22 23 24 25 d b c b a
11 c 26 d
12 0
13 1
14 b
15 a
1 b 16 b 31 b
2 d 17 14 32 60
3 a 18 b 33 9
4 593 19 40 34 c
5 a 20 2
Answer Keys: Exercise: 1.3 6 7 8 9 10 a b 40 0 10 21 22 23 24 25 a a a c b
11 a 26 b
12 0.75 27 d
13 b 28 a
14 a 29 0
15 25 30 a
Answer Keys: Exercise: 1.4 6 7 8 9 10 2 2 c b b 21 22 23 24 25 3 1 d 2 0
11 2 26 b
12 b 27 c
13 a 28 c
14 d 29 c
15 c 30 2
Answer Keys: Exercise: 1.5 6 7 8 9 10 c a 2 3 b 21 22 23 24 25 b c a b 0
11 d 26 b
12 c 27 a
13 b 28 b
14 a 29 a
15 d 30 d
1 a 16 180 31 d
1 3 16 a
2 d 17 a 32 3
2 a 17 1
3 c 18 c 33 1
3 d 18 c
4 c 19 d 34 2
4 b 19 2
5 2 20 3 35 2
5 c 20 128
Copyright © 2016 by Kaushlendra Kumar
11 0 26 d 41 256
12 0 27 2.4 42 d
13 2 28 5 43 4
14 0 29 0 44 c
15 0 30 c 45 c
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Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [1]
GATE – 2016: Chapter – 2: Calculus Note: The following questions came in GATE – 2016 were based on Calculus Chapter – 2. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 2
2
1. Let x be a positive real number. The function f ( x ) x (1 x ) has its minima at x _____. [AE-2016 (1 mark)] 3 4 4 Solution: f ( x ) 0 2 x ( 2 x ) 0 x 1 x 1 . Also f ( x) 2 (6 x ) f ( x ) x 1 2 6 8 0 x 1 is the point of minima; and f ( x ) x 1 2 6 8 0 x 1 is the point of minima. But x must be positive, so answer is x 1 . 2. The vector u is defined as u y eˆx x eˆ y , where eˆx and eˆ y are the unit vectors along x and y directions, respectively. If the vector is defined as u , then ( )u _____.
[AE-2016 (1 mark)] Solution: As u , so is perpendicular to both and u . Thus 0 , since is perpendicular to . Hence ( )u 0 . 2
3. The function f ( x) x x 6 is
[AG-2016 (1 mark)]
(a) minimum at x 1 2
(b) maximum at x 1 2
(c) minimum at x 1 2
(d) maximum at x 1 2 2
f ( x ) x 2 x 6 x 2 x (1 4) (1 4) 6 x (1 2) (25 4) , which attains
Solution (a):
minimum value of 25 4 at x 1 2 . So option (a) is correct. 4. With n being a positive integer, the series (a) convergent
n1 (1 n p ) for
(b) divergent
p 1 is
[AG-2016 (1 mark)]
(c) asymptotic
Solution (a): Using p series test, we know that if k 0 then
(d) oscillatory
n k (1 n
p
) converges if p 1 and
diverges if p 1 . So option (a) is correct. 5. The value of the integral Solution: I
0
I ln
dx
0.9
x2
dx
0.9
0
(1 x )(2 x)
(1 x )(2 x)
0.9
0
is _____.
dx ( x 1)( x 2)
0.9
0
[BT-2016 (1 mark)] 1 1 0.9 dx {ln( x 2) ln( x 1)}0 x 2 x 1
0.9
0.9 2 0 2 ln ln ln11 ln 2 ln 5.5 1.7 x 1 0 0.9 1 0 1
0, 6. The Fourier series of the function, f ( x ) x,
x 0 0 x
in the interval [ , ] is
2 cos x cos 3x sin x sin 2 x sin 3 x 2 . The convergence of the 2 4 1 3 2 3 1 above Fourier series at x 0 gives [CE-2016 (1 mark)] 2 n 1 2 2 1 (1) 1 (1) n 1 (a) 2 (b) (c) (d) (2n 1)2 (2n 1) 4 6 n2 12 8 n 1 n n 1 n 1 n 1 f ( x)
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Solution (c): At x 0 , f (0)
0
4
21
12
1 32
Chapter – 2: Calculus
Chapter – 2 [2]
2 cos 0 cos 0 sin 0 sin 0 sin 0 2 2 4 1 3 2 3 1 1
0
12
1 32
2 8
n 1
1 (2n 1) 2
2 8
.
2
7. The optimum value of the function f ( x ) x 4 x 2 is [CE-2016 (1 mark)] (a) 2 (maximum) (b) 2 (minimum) (c) 2 (maximum) (d) 2 (minimum) Solution: f ( x) 0 2 x 4 0 x 2 ; f ( x) 2 f ( x ) x 2 2 0 x 2 is the point of 2
minima. So optimum value of given function is f (2) 2 4(2) 2 2 . Thus option (d) is correct. 8. What is the value of
lim
xy
? [CE-2016 (1 mark)] x y2 (a) 1 (c) 0 (d) Limit does not exist (b) 1 Solution (d): Limit along the path x 0 : First we find what the function becomes along this path xy xy (0) y lim (0) 0 . 0 lim 2 2 2 2 2 x 0, y 0 x y 2 x 0, y 0 x y x 0 (0) y x 0, y 0
2
Limit along the path y x : First we find what the function becomes along this path xy 2
x y
2 y x
x2 2x
2
1 2
lim x 0, y 0
xy 2
x y
2
lim x 0, y 0
1 2
1 2
.
As limit along the path x 0 is to the limit along the path y x and so given limit does not exist. 9.
sin( x 4)
_____ x4 sin( x 4) sin( x 4) sin h Solution: lim lim lim 1 x 4 x 4 0 h 0 h x4 x4 lim
[CS-2016 (1 mark)]
x 4
10. Let f ( x) be a polynomial and g ( x ) f ( x ) be its derivative. If the degree of { f ( x ) f ( x )} is 10, then the degree of { g ( x ) g ( x )} is _____. [CS-2016 (1 mark)] Solution: Let f ( x ) be a polynomial of even degree 10, then { f ( x ) f ( x )} is a polynomial of degree 10. Now g ( x) f ( x) g ( x ) be a polynomial of odd degree, i.e. 9, and so g ( x ) be a polynomial of odd degree, i.e. 9; so { g ( x ) g ( x )} is a polynomial of degree 9, because the coefficient of highest exponent of x are of opposite signs and they do not cancel each other. Note that: if f ( x ) be a polynomial of odd degree ‘ 2n 1 ’, then { g ( x ) g ( x )} is a polynomial of degree ‘ 2n 1 ’. 11. Given the following statements about a function f : , select the right option: P: If f ( x) is continuous at x x0 , then it is also differentiable at x x0 . Q: If f ( x) is continuous at x x0 , then it may not be differentiable at x x0 . R: If f ( x) is differentiable at x x0 , then it is also continuous at x x0 . [EC-2016 (1 mark)] (a) P is true, Q is false, R is false (b) P is false, Q is true, R is true (c) P is false, Q is true, R is false (d) P is true, Q is false, R is true Solution: We know that if a function f ( x ) is differentiable at x x0 , then it is also continuous at
x x0 . But if a function f ( x ) is continuous at x x0 , then f ( x ) may or may not be differentiable at x x0 . Thus statements Q and R are true, and P is false; so option (b) is correct.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [3]
12. Consider the plot of f ( x) versus x as shown below: x
Suppose F ( x) f ( y ) dy . Which one of 5
the following is a graph of F ( x ) ? [EC-2016 (1 mark)] (a)
(b)
(c)
(d)
Solution (d): F ( x) f ( x) , so from the given graph we have to find the shape of graph such that it will be the integration of given graph of function f ( x ) , which is shown as: From the figure we can say that option (d) is correct.
13. How many distinct values of x satisfy the equation sin( x ) x 2 , where x is in radians? [EC-2016 (1 mark)] (a) 1 (b) 2 (c) 3 (d) 4 or more Solution (c): From the graph shown in figure, the number of points for which sin( x ) x 2 , where x is in radians, is three. 14. The integral
1
0
dx
is equal to _____. 1 x Solution: Let 1 x t dx dt ; also at x 0 , t 1 ; and at x 1 , t 0 . 1
dx
0
1 x
Now I
0
1
dt t1 2
1
dt
0 t1 2
[EC-2016 (1 mark)]
1
t (1 2) 2{11 2 01 2 } 2 (1 2) 0
15. As x varies from –1 to 3 , which one of the following describes the behaviour of the function f ( x ) x3 3 x 2 1 ? [EC-2016 (1 mark)] (a) f ( x) increases monotonically (b) f ( x) increases, then decreases and increases again (c) f ( x) increases and then decreases (d) f ( x) decreases, then increases and decreases again
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [4]
3 2 2 Solution: f ( x ) x 3 x 1 f ( x) 3 x 6 x ; so f ( x) 0 3 x( x 2) 0 x 0, 2 So from wavy-curve method, f ( x ) increases in x ( , 0) (2, ) and decreases in x (0, 2) . So in x ( 1, 0) , f ( x ) increases; then in x (0, 2) , f ( x ) decreases; and then in x (2, 3) , f ( x ) increases. Hence option (b) is correct.
16. The maximum value attained by the function f ( x ) x ( x 1)( x 2) in the interval [1, 2] is _____. [EE-2016 (1 mark)] 2 2 2 2 Solution: f ( x) x 3x 2 x 2 x x x 3 x 6 x 2 ; 2
so f ( x) 0 3 x 6 x 2 0 x 1.57, 0.42 So from wavy-curve method, f ( x) changes its sign from ve to ve at x 0.42 . So x 0.42 is the point of maxima; and x 1.57 is the point of minima. Hence maximum value of f ( x ) is max{ f (1), f (2)} . As f (1) f (2) 0 . Hence maximum value of the given function is zero. 17. The value of the line integral
2
C (2 xy dx 2 x
and the point (1,1,1) is (a) 0 (b) 2 Solution
(b):
The
given
function
is
2
ydy dz ) along a path joining the origin (0, 0, 0)
[EE-2016 (1 mark)] (d) 6 (c) 4 2 2 ˆ ˆ ˆ V (r ) 2 xy i 2 x y j 1 k V1 iˆ V2 ˆj V3 kˆ . As
V3 y V2 z , V1 z V3 x , V2 x V1 y the given function is exact and so the
line integral V d r is path independent. Now we have to find the potential function ( f ) of V such
that V grad f . Since, f x V1 f 2 xy 2 dx x 2 y 2 c , where c h( y , z ) is a function of y and z . Now f y V2 2 x 2 y (c y ) 2 x 2 y c y 0 c is a function of z only, i.e., 2 2
c h( z ) . Now f z V3 0 (c z ) 1 (c z ) 1 c z ; so f x y z ; if A(0, 0, 0) and B(1,1,1) are endpoints of given line integral
V d r f ( B ) f ( A) (12 12 1) (02 02 0) 2 C
18. lim
n
n 2 n n 2 1 is _____.
Solution: L lim
n
[IN-2016 (1 mark)]
n2 n n2 1 n n n 1 lim 2 2 n n n n 1 2
2
n2 n n2 1
( n 2 n) ( n 2 1) ( n 1) 1 (1 n) lim lim n 2 n n 2 1 n n 2 n n 2 1 n 1 (1 n) 1 (1 n 2 )
L lim n
1 1 (0) 1 (0) 2 1 (0)
L lim n
1
19. Let a1 1 and an an1 4 , n 2 . Then lim n
a1a2
Solution: The numbers a1 , a2 , a3 , , an
1 a2 a3
is equal to _____. an 1an 1
[MA-2016 (1 mark)] form an AP, whose first term is a1 1 and common
difference d an an1 4 . So an a1 ( n 1)d 1 ( n 1)4 1 4n 4 4n 3 .
1
L lim n
a1a2
1 a2 a3
1 4 4 4 lim an 1an 4 n a1a2 a2 a3 an 1an 1
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2 [5]
1 1 1 1 1 1 1 lim n 4 a2 a3 an 1an 4 an 1 an a1a2 a1 a2 a2 a3 1 1 1 a a 1 1 4n 3 1 n 1 L lim lim n 1 lim lim n n n n 4 4n 3 (4n 3)(1) a1 an 4 a1an 4 1 (1 n) 1 (0) 1 L lim . n 4 (3 n ) 4 (0) 4 L
a2 a1
Chapter – 2: Calculus
1
lim
n
a3 a2
an an 1
20. Let be the triangular path connecting the points (0, 0) , (2, 2) and (0, 2) in the counter3 clockwise direction in 2 . Then I {sin( x )dx 6 xydy} is equal to _____.
Solution: In counter clock wise direction the given line integral I
3
[MA-2016 (2 marks)] {sin( x 3 ) dx 6 xydy} can be
3
written as I {sin( x ) dx 6 xydy} {sin( x )dx 6 xydy} {sin( x 3 ) dx 6 xydy} . Now OB
BA
3
along OB : y x dy dx , so I1 {sin( x ) dx 6 xydy}
x 0
OB
I1
x2
x 0
sin( x 3 ) dx 6
x 2
x 0
x 2 dx
x 2
x 0
sin( x 3 ) dx 2( x 3 ) 20 x 0
along AO : I 3 {sin( x 3 ) dx 6 xydy} A0
x 0
y2
sin( x 3 ) dx 16
{sin( x 3 ) 6 x 2 }dx
sin( x 3 ) dx 16
{sin( x 3 ) 0}dx
x2 y 0
BA
x 2
x 2
x 0
along BA : I 2 {sin( x3 ) dx 6 xydy}
Hence I I1 I 2 I 3
AO x2
x 2
x 0
sin( x 3 ) dx
{sin(03 ) 0}dx 0
x 2
x 0
sin( x 3 ) dx 0 16
21. The values of x for which the function f ( x )
x 2 3x 4 x 2 3x 4
is NOT continuous are
[ME-2016 (1 mark)] (a) 4 and –1 (b) 4 and 1 (c) –4 and 1 (d) –4 and –1 Solution (c): The given function is not continuous for all x where denominator term is zero, i.e.
x 2 3 x 4 0 ( x 4)( x 1) 0 x 4,1 . 22. lim
log e (1 4 x )
x 0
e3 x 1
is equal to
(a) 0
[ME-2016 (1 mark)]
(b) 1 12
(c) 4 3
log(1 x )
(d) 1
x
e 1
1 ; thus x 0 x x e (1 4 x)} 4 x ] 4 x[{log e (1 4 x )} 4 x ] 4 lim[{log 4 x 0 L lim . 3 x 3 x x 0 3x{(e 1) 3 x} 3 lim[{(e 1) 3 x}] 3
Solution (c): We know that lim x 0
1 and lim
x0
23. X 4 C is the general integral of 3
[MN-2016 (1 mark)] 3
(a) 3 x dx
(b) (1 4) x dx 3
(c)
3
(d) 4 x 3 dx
x dx
4
Solution (d): x dx (1 4) x c 4 x 3 dx x 4 4c x 4 C , C 4c . So option (d) is correct. 24. sinh( x) is
[MN-2016 (1 mark)] x
x
x
(a) (1 4)(e e ) x
x
(b) (1 2)(e e )
x
x
(c) (1 2)(e e )
x
x
(d) (1 4)(e e )
x
Solution (b): (1 2)(e e )
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [6]
25. If V x 2 y iˆ y 2 x ˆj xyz kˆ , the divergence of V is 3
3
2
2
[MT-2016 (1 marks)]
2
(c) 5xy (d) 0 (b) x y y x xyz ˆ ˆ j k. Solution (c): Divergence of V is V , where iˆ x y z ˆ ˆ 2 ˆ 2 2 2 So V iˆ j k ( x y i y x ˆj xyz kˆ ) ( x y ) ( y x ) ( xyz ) z x y z x y V 2 xy 2 xy xy 5 xy . (a) x y y x xyz
26. The value of the integral Solution: So
2
0
2
0
x sin x dx _____
d
[MT-2016 (1 marks)]
x sin x dx x sin xdx dx x sin xdx x cos x cos xdx x cos x sin x c
x sin x dx ( x cos x sin x c )0 2 (0 1 c ) (0 0 c) 1
ex 1 is equal to _____. sin x
27. The value of lim x 0
[PE-2016 (1 mark)]
Solution: At x 0 the given expression takes 0 0 form, so using L’Hospital rule, i.e. differentiating
ex 1 ex e0 1 Nr. and Dr. separately w.r.t. ‘ x ’, we get, L lim lim 1. x 0 sin x x 0 cos x cos 0 1 28. The function f ( x) 1 (1 | x |) is (a) continuous and differentiable (c) not continuous but differentiable Solution
(b):
The
given
[PE-2016 (1 mark)] (b) continuous but not differentiable (d) not continuous and not differentiable 1 (1 x ) , x 0
function
can
be
written
as:
f ( x)
1,
x 0.
As
1 (1 x ) , x 0 lim f ( x ) lim {1 (1 x )} 1 ;
x 0
lim f ( x ) lim {1 (1 x )} 1 ;
x 0
x 0
f (0) 1 ;
so
x 0
lim f ( x) lim f ( x) f (0) , thus f ( x ) is continuous at x 0 . Now
x 0
x 0
1 f (0 h) f (0)
11 h
1
1 lim 1 h lim 1 . Similarly h 0 h 0 h0 1 h h h 1 1 h 1 f (0 h) f (0) 1 1 (0 h) RHD x 0 lim lim lim 1 h lim 1 h 0 h 0 h 0 h 0 h h h 1 h As LHD x 0 RHD x 0 , so the given function is not differentiable at x 0 . So option (b) is correct. LHD x 0 lim
1 (0 h) lim h 0 h 1
29. The value of the definite integral
e
1 (ln x)dx
is equal to _____.
[PE-2016 (1 mark)]
d (ln x ) 1dx x ln x 1dx x ln x x dx
Solution: 1(ln x ) dx ln x 1dx So
e
e
1 (ln x)dx ( x ln x x)1 (e ln e e) (1ln1 1) 0 (0 1) 1 .
30. At x 0 , the function f ( x) sin{2 x L} , ( x , L 0) is [PI-2016 (1 mark)] (a) continuous and differentiable (b) not continuous and not differentiable Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [7]
(c) not continuous but differentiable Solution (d): LHL x 0
(d) continuous but not differentiable 2 (0 h) 2 h lim f (0 h) lim sin lim sin 0 0 ; h 0 h0 h 0 L L
RHL x 0 lim f (0 h) lim sin h 0
2 (0 h) L
h0
lim sin
2 h
h0
L
0 0 ; f (0) sin
2 (0) L
0
As LHL x 0 RHL x 0 0 , so f ( x ) is continuous at x 0 . Now
LHD x 0 lim
f (0 h) f (0) h
h 0
LHD x 0
2 L
sin lim h 0
2 (0 h) L h
0
sin
lim
h0
2 h
sin
2 h
L L lim L 2 h h 0 L 2 h 2 L 2 L
. Similarly sin
2 (0 h)
0
sin
2 h
2 2 L LHD x 0 . h0 h 0 h 0 L 2 h h L L 2 L As LHD x 0 RHD x 0 , so the given function is not differentiable at x 0 . So option (d) is correct RHD x 0 lim
f (0 h) f (0)
lim
3
L h
lim
2
2
3
31. For the two functions f ( x, y ) x 3xy and g ( x, y ) 3x y y , which one of the following options is correct? [PI-2016 (1 mark)] f g f g f g f g (b) (c) (d) (a) x y y x y x x x f f ( x 3 3 xy 2 ) 6 xy Solution (c): ( x 3 3 xy 2 ) 3 x 2 3 y 2 , x x y y g f g g (3 x 2 y y 3 ) 3 x 2 3 y 2 . As (3x 2 y y 3 ) 6 xy , , so option (c) is correct. x x y y y x 2
32. The volume of the solid obtained by revolving the curve y x , 0 x 1 around y axis is [XE-2016 (1 mark)] (a) (b) 2 (c) 2 (d) 5 Solution (d): The volume of the solid generated by the revolution, about y axis, of the area b
bounded by the curve x f ( y ) , the abscissa y a and y b is given by V x 2 dy . Here we a
b
2
1
have 0 x 1 0 y 1 ; and x f ( y ) y . So V x dy y dy (1 5)( y 5 )10 5 . a
2
4
0
2 2 2 33. The value of the surface integral F n dS over the sphere given by x y z 1 , where F 4 x iˆ z kˆ , and n denotes the outward unit normal, is [XE-2016 (1 mark)] (a) (b) 2 (c) 3 (d) 4
Solution (d): Using divergence theorem, div F dV F dS F n dA , we have V S S ˆ ˆ F 4 x i z k , so div F F ( x )(4 x) ( y )(0) ( z )( z ) 4 1 3 . Now
S F n dA V div F dV V 3 dV 3V 3(4 3) r
Copyright © 2016 by Kaushlendra Kumar
3
4 (since r 1 ). So option (d) is correct.
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Engineering Mathematics
Chapter – 2: Calculus
34. The value of definite integral
0 ( x sin x) dx
Chapter – 2 [8]
is _____.
[AE-2016 (2 marks)]
0
0
0
Solution: I ( x sin x ) dx {( x ) sin( x )}dx {( x ) sin x}dx
0
0
I sin x dx x sin x dx ( cos x)0 I 2 I 2 I 3.14 35. The value of the integral, I
4
x2 1
dx is _____. [AG-2016 (2 marks)] x2 1 2 2 4 x 1 4 x 1 2 4 4 4 2 1 1 dx dx 1dx 2 dx ( x ) 42 dx Solution: I 2 2 2 x 1 2 2 2 2 x 1 x 1 x 1 x 1 2
4
I (4 2) log( x 1) log( x 1)2 2 log 3 log 5 log1 log 3 2 log(9 5) 2.25 36. Let I , J and K are unit vectors along the three mutually perpendicular x , y and z axes, respectively. If F f I g J h K is a continuously differentiable vector point function, then curl F is [AG-2016 (2 marks)]
g
f g f h J K z y z x y x h g h f g f (c) I J K y z x z x y (a) I
h
h
g f h f J K y z x z x y g h f h f g (d) I J K z y z x y x (b) I
g
I
J
K
Solution (b): curl F F I J K ( f I g J h K ) x y z y z x f
h
curl F I
y
g
h
g
g f h f J K z x z x y 2
37. Consider the equation V aS {b S ( S c)} . Given a 4 , b 1 and c 9 , the positive value of S at which V is maximum, will be _____. [BT-2016 (2 marks)] 2 2 Solution: We have V 4 S {1 S ( S 9)} 36 S (9 9 S S )
dV 1(9 9S S 2 ) S (9 2 S ) 9 S2 ; so 0 9 S 2 0 S 3 36 2 2 dS dS 9 9S S 9 9S S 2 2 2 2 d V d 9S 2 S (9 9 S S ) (9 S )(9 2S ) 36 36 ; 2 2 2 dS dS 9 9 S S 9 9S S
dV
36
Thus ( d 2V dS 2 )
S 3
6 0 and ( d 2V dS 2 )
S 3
6 0 . So V is maximum if S 3 .
2
2
38. The angle of intersection of the curves x 4 y and y 4 x at point (0, 0) is [CE-2016 (2 marks)] (a) 0o (b) 30o (c) 45o (d) 90o Solution (d): Slope of curve C1 : x 2 4 y is given by 2 x 4 y y1 x 2 ( y1 ) (0,0) 0 ; so the o
tangent to the curve C1 at (0, 0) is making an angle of 0 with the x axis. Slope of curve C1 : y 2 4 x is given by 2 yy 4 y2 2 y ( y2 )(0,0) 2 0 undefined ; so the o
tangent to the curve C2 at (0, 0) is making an angle of 90 with the x axis. Hence angle between o
o
o
the two curves is the angle between the tangent at (0, 0) , which is 90 0 90 .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [9]
2
39. The area of the region bounded by the parabola y x 1 and the straight line x y 3 is [CE-2016 (2 marks)] (a) 59 6 (b) 9 2 (c) 10 3 (d) 7 6 2
Solution (b): The point of intersection of y x 1 and 2
2
x y 3 is given as: x 1 3 x x x 2 0 ( x 2)( x 1) 0 x 2,1 . So the two curves meet at A(1, 2) and B ( 2, 5) . Thus required area of shaded region is given by 1
1
A {(3 x) ( x 2 1)}dx {2 x x 2 }dx 2
2
2
3
1
x
x
2
3
2
3
1 ( 2) ( 2) 7 20 27 9 2( 2) 2 3 2 2 3 2 3 6 6 6 2 [Similar question was also asked in CE-2016 (2 marks)] A 2x
2
1
3
40. The Lagrange mean-value theorem is satisfied for f ( x ) x 5 , in the interval (1, 4) at a value (rounded off to the second decimal place) of x equal to _____. [CH-2016 (2 marks)] Solution: As the given function is continuous in [1, 4] and differentiable in (1, 4) , then there is at least one value x ( a, b) , such that
f ( x )
f (b) f ( a) ba
2
3x 2
f (4) f (1) 4 1
3x2
(43 5) (13 5) 4 1
2
9 x 63 x 7 x 7 2.64 . So x 2.64 (1, 4) . 41. A binary relation R on is defined as follows: ( a, b) R (c, d ) if a c or b d . Consider the following propositions: P: R is reflexive; Q: R is transitive Which one of the following statements is TRUE? [CS-2016 (2 marks)] (a) Both P and Q are true (b) P is true and Q is false (c) P is false and Q is true (d) Both P and Q are false Solution (b): It is reflexive as every ordered pair is related to itself as ( a, b) R (a, b) , since a a or b b . It is not transitive because (2, 4) R(3, 2) & (3, 2) R(1, 3) satisfies the condition but (2, 4) R (1, 3) not satisfying the condition. 42. The region specified by {( , , z ) : 3 5, 8 4 ,3 z 4.5} in cylindrical coordinates has volume of _____. [EC-2016 (2 marks)] Solution: The differential volume ‘ dV ’ of cylinder is given by dV d d dz . So 5
V
5
4
z 4.5
3 8 z 3
d d dz
5
3
d
4
8
d
z 4.5
z 3
2 4 4.5 dz 8 z 3 2 3
V (1 2) 52 32 {( 4) ( 8)}(4.5 3) 8 ( 8) 1.5 4.71
43. A triangle in the xy plane is bounded by the straight lines 2 x 3 y , y 0 and x 3 . The volume above the triangle and under the plane x y z 6 is _____. [EC-2016 (2 marks)] Solution: The required volume will be V
x 3
x 0
y 2 x 3
y2 V 6 y xy x 0 2 y 0 x 3
y2x 3
y 0
z dx dy
x 3
x 0
y2 x 3 y 0
(6 x y ) dy dx
3
8x2 2 8x3 8(3)3 2 dx 4 x dx 2 x 2(3) 10 . x 0 9 27 0 27 x 3
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [10] 2
2
44. The integral (1 2 ) ( x y 10) dx dy , where D denotes the disc: x y 4 , evaluates to D
_____. [EC-2016 (2 marks)] 2 2 2 Solution: D : x y 2 , converting the region D in polar coordinates, we have x r cos , y r sin , where r varies from 0 to 2; and varies from 0 to 2 . Now
I (1 2 ) ( x y 10) dx dy (1 2 )
2
2
r 0 0
D
( r cos r sin 10) rdrd
2
2
2
2
( r 2 cos r 2 sin 10r )d dr (1 2 ) ( r 2 sin r 2 cos 10 r ) 0 r 0 0 r 0 2 2 I (1 2 ) 20 r dr (1 2 )20 rdr 10(1 2)( r 2 )02 20 r 0 r 0 I (1 2 )
2
dr
2
45. Suppose C is the closed curve defined as the circle x y 1 with C oriented anti-clockwise. The value
( xy dx x 2
2
ydy ) over the curve C equals _____.
[EC-2016 (2 marks)]
F
F
Solution: Using Green’s theorem in a plane, which is
C ( F1dx F2 dy) R x2 y1 dxdy , where
F1
we
F2
and
are
functions
of
( x, y ) .
2
C ( xy dx x 2
2
So
F1 xy 2
have
F2 x 2 y ,
and
thus
2
( x y ) ( xy ) dxdy R 2 xy 2 xy dxdy 0 . y x
ydy ) R
46. The line integral of the vector field F 5 xz iˆ (3x 2 2 y ) ˆj x 2 z kˆ along a path from (0, 0, 0) to 2
(1,1,1) parameterized by (t , t , t ) is _____.
[EE-2016 (2 marks)]
2
Solution: We have x t dx dt , y t dy 2t dt , z t dz dt . Now F d r (5 xz iˆ (3x 2 2 y ) ˆj x 2 z kˆ) ( dx iˆ dy ˆj dz kˆ ) {5 xz dx (3 x 2 2 y ) dy x 2 z dz}
C
C
C
1
4 t3 t 5 11 53 F d r {5t dt 10t dt t dt} {5t dt 11t dt} 5 11 4.416 C t 0 t 0 4 0 3 4 12 3 t 1
2
3
t 1
3
2
3
[Similar question was also asked in ME-2016 (2 mark)] 3
4
47. Let f : [ 1,1] , where f ( x ) 2 x x 10 . The minimum value of f ( x) is _____. [IN-2016 (2 marks)] 3 4 2 3 2 Solution: f ( x ) 2 x x 10 f ( x ) 6 x 4 x f ( x) 12 x 12 x
f ( x) 0 6 x 2 4 x 3 0 x 0, 3 2 . So f ( x) x 0 0 , thus at x 0 , we do not have neither maxima nor minima; and f ( x ) x 3 2 9 0 . Thus the minimum value of f ( x ) is the min{ f ( 1), f (1)} min{ f ( 1), f (1)} min{13, 9} 13 2
48. Let be the curve which passes through (0,1) and intersects each curve of the family y cx orthogonally. Then also passes through the point [MA-2016 (2 marks)] (c) (1,1) (d) ( 1,1) (a) ( 2, 0) (b) (0, 2) Solution (a): Let : y f ( x ) be a curve which passes through (0,1) , so f (0) 1 …(i). As and
y cx 2 intersect orthogonally, so product of their slopes at intersection point must be 1 . 2 Since y f ( x) y f ( x) ; and y cx y 2cx y 2 y x ; So at the intersecting point, we have ( y )(2 y x ) 1 ( dy dx )(2 y x ) 1 2 ydy xdx …(ii) 2
2
2
2
On integrating (ii), y ( x 2) k , which passes through (0,1) , i.e. (1) (0 2) k k 1 . 2
2
Hence the required curve will be y ( x 2) 1 , which also passes through ( 2, 0) .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
n
49. Let S n k 1
1
Chapter – 2: Calculus
and I n
x2 (b) ln10 1
Solution (a): I10
10
x [ x]
1
10 [ x]
1
10 [ x ]
1
x
2
dx 2
1
1 x
2
x
2 [ x]
1
dx
x
2
dx
10
1
dx
2
3 [ x]
2
3
2
2
2
dx 2
10 [ x ]
1
x
2
x [ x]
1
k
(a) ln10 1
Now let
n
x
x
2
dx . Then S10 I10 is equal to
1 x
dx
4
3
3 x
2
10 [ x ]
1
4 [ x]
dx
3
[MA-2016 (2 marks)]
(c) ln10 (1 10)
3
dx
Chapter – 2 [11]
x
2
dx (ln x)110
dx
5
4
4
2
x
10 [ x ]
1
5 [ x]
4
dx
4
x
2
x2
dx 10
9
9
x2
5
2
dx ln10
x 10 [ x]
9
dx
(d) ln10 (1 10)
x2
10 [ x ]
1
x2
dx …(i)
dx
dx
10
1 2 3 4 9 dx 2 x x 1 x 2 x 3 x 4 x 9
1 1 2 3 4 9 1 1 1 1 dx 1 1 1 1 1 x 10 2 3 4 5 10 2 3 4 5 10 [ x ] 10 1 dx k 1 1 S10 1 . Thus from (i), I10 ln10 ( S10 1) S10 I10 ln10 1 . 2 1 x k
10 [ x ]
1
2
2
50. The number of roots of the equation x cos( x ) 0 in the interval [ 2 , 2] is equal to ____. [MA-2016 (2 marks)] Solution: From the given figure, the number of 2 roots of the equation x cos x is 2.
3
2
51. Consider the function f ( x ) 2 x 3 x in the domain [ 1, 2] . The global minimum of f ( x) is _____. [ME-2016 (2 marks)] 3 2 2 Solution: f ( x ) 2 x 3 x f ( x) 6( x x ) f ( x ) 6(2 x 1) 2 So for critical points of f ( x ) 0 , we have f ( x ) 0 6( x x ) 0 x 0,1 Thus f ( x ) x 0 6 0 x 0 is the point of maxima; f ( x ) x 1 6 0 x 1 is the point of 3
2
minima. As our function f ( x ) 2 x 3 x is defined in the closed interval, so global minimum of f ( x ) min{ f ( 1), f (1), f (2)} min{5, 1, 4} 5 . So answer is ‘ 5 ’. 52. The value of the line integral F r ds , where C is a circle of radius 4 units is _____. C Here F ( x, y ) y iˆ 2 x ˆj and r is the UNIT tangent vector on the curve C at an arc length s from a reference point on the curve. iˆ and ˆj are the basis vectors in the x y Cartesian
reference. In evaluating the line integral, the curve has to be traversed in the counter-clockwise direction. [ME-2016 (2 marks)] 2 2 2 Solution: Any point on the curve C : x y a is x a cos and y a sin with 0 2 , where a 4 . So we have r ( ) a cos iˆ a sin ˆj dr d a sin iˆ a cos ˆj . So the line 2 integral F r ds F r ( ) r ( )d {(a sin ) iˆ (2a cos ) ˆj} {a sin iˆ a cos ˆj}d C
C
0
2 2 1 3 3 sin 2 2 C F r ds a 2 0 { sin 2 2 cos 2 }d a 2 0 cos 2 d a 2 2 2 2 2 16 C F r ds a 2 16 .
Copyright © 2016 by Kaushlendra Kumar
2
0
e-mail: [email protected]
Engineering Mathematics
Chapter – 2: Calculus
Chapter – 2 [12]
x 2 x 1 x is
53. lim
x
[ME-2016 (2 marks)]
(a) 0 (b) Solution (d): Using rationalisation, we have x2 x 1 x
L lim
2
x
(c) 1 2
(d)
x2 x 1 x 2
2
( x x 1 x ) lim
2
x
x 1
lim
2
x x 1 x x x x 1 x 1 (1 x ) 1 0 1 L lim lim 2 2 x x 1 0 0 1) 2 ( x x 1 x) x 1 (1 x ) (1 x ) 1) x x 1 x ( x 1) x
54. The divergence of the velocity field V ( x 2 y )iˆ ( z 2 xy ) ˆj ( xy ) kˆ at (1,1,1) is _____. [PE-2016 (2 marks)] ˆ ˆ 2 Solution: div V V iˆ j k ( x y )iˆ ( z 2 xy ) ˆj ( xy ) kˆ x y z
div V V
x
( x2 y)
y
( z 2 xy )
z
( xy ) 2 x 2 x 0 0 2
2
3
4
55. The range of values of k for which the function f ( x) ( k 4) x 6 x 8 x has a local maximum at point x 0 is [PI-2016 (2 marks)] (a) k 2 or k 2 (b) k 2 or k 2 (c) 2 k 2 (d) 2 k 2 2 2 3 4 Solution (c): f ( x ) ( k 4) x 6 x 8 x
f ( x) ( k 2 4)(2 x ) 18 x 2 32 x 3 f ( x) ( k 2 4)(2) 36 x 96 x 2 Now, at f ( x) 0 ; so is the point of x 0, x0 2 2 2 f ( x ) x 0 0 (k 4)(2) 36(0) 96(0) 0 k 4 0 k (2, 2) .
local
maxima
if
2
e5 x 1 56. lim is equal to _____. x 0 x
[PI-2016 (2 marks)]
2
2
ex 1 e5 x 1 e5 x 1 Solution: L lim ; lim 25 25 lim 1. x0 x 0 x x 0 5 x x x 2 y 2 z 2 . The value of
57. Let f ( x, y , z ) 1
2 f x 2
2 f
y 2
2 f z 2
is equal to _____. [TF-2016 (2 marks)]
x2 y 2 z 2
Solution: f ( x, y , z ) 1
2 f x 2
Similarly, So
2 f x 2
f x
x 2
( x y 2 z 2 )3 2
1( x 2 y 2 z 2 )3 2 x (3 2)( x 2 y 2 z 2 )1 2 (2 x ) ( x2 y 2 z2 ) 2 f y 2 2 f y 2
2 y2 x2 z2 ( x 2 y 2 z 2 )5 2 2 f z 2 3
and
2x2 x2 z2 ( x 2 y 2 z 2 )5 2
2 f
z 2
2x2 y2 z2 ( x 2 y 2 z 2 )5 2
2z2 x2 y 2 ( x 2 y 2 z 2 )5 2
2 y 2 x2 z2 ( x 2 y 2 z 2 )5 2
2z 2 x2 y2 ( x 2 y 2 z 2 )5 2
0
2
58. Let f ( x ) 2 x 3x 69 , 5 x 5 . Find the point at which f attains the global maximum. [XE-2016 (2 marks)] 3 2 2 Solution: f ( x ) 2 x 3 x 69 f ( x) 6( x 1) f ( x) 12 x .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
3
Chapter – 2: Calculus
2
3
Chapter – 2 [13]
2
Also f ( 5) 2( 5) 3( 5) 69 256 ; and f (5) 2(5) 3(5) 69 244 2 Now f ( x) 0 6( x 1) 0 x 1 f ( x) x1 12 0 x 1 is the point of local maxima; also f ( x) x1 12 0 x 1 is the 3
2
point of local minima. As f ( 1) 2( 1) 3(1) 69 64 ; thus max{ f ( 5), f ( 1), f (5)} max{256, 64, 244} 244 , which comes at x 5 . 59. Calculate
C
1
global
maxima
2 F dr F dr , where C1 : r (t ) (t , t ) and C2 : r (t ) (t , t ) , t varying from 0
to 1 and F xy ˆj .
C2
[XE-2016 (2 marks)] 2
Solution: For C1 , x t dx dt and y t dy 2tdt , t 1
t 1 t 1 t5 2 so F dr ( xy ˆj ) ( dx iˆ dy ˆj dz kˆ) xy dy t 3 (2tdt ) 2 t 4 dt 2 C1 C1 C1 t 0 t 0 5 t 0 5
For C1 , x t dx dt and y t
12
dy (1 2)t 1 2 dt , so t 1
2 t 1 3 2 1 t 1 1t 1 1 2 ˆ ˆ ˆ ˆ C1 F dr C1 ( xy j ) (dx i dy j dz k ) C1 xy dy t 0 t (1 2)t dt 2 t 0 t dt 2 2 4 t 0 2 1 Thus F dr F dr 0.15 . C1 C2 5 4
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.1]
Chapter 2 : Calculus 2.1
Set Theory
The symbols used in defining the sets are given as: Symbol Meaning Symbol Meaning Implies There exist Belongs to iff If and only if A is a subset of B & And A B Implies and is implied by a|b a is a divisor of b Does not belong to Set of Natural numbers N s.t. Such that I or Z Set of Integers For every Set of real numbers R A set is a well-defined collection (i.e. a rule which decides whether a given element belongs or does not belong to the given collection) of elements. A set is described in the following two ways: In Roster method, a set is described by listing elements, in any order, separated by commas, within braces {} . The set of vowels of English alphabet is represented as {a, e, i, o, u} . In Set-builder method, a set is described by a characterizing property P ( x ) of its elements x as {x : P( x ) holds} . The set E of all even natural numbers can be written as E { x : x 2n, x N } .
Types of Sets
Null set: A set which contains no element at all is called ‘null or empty or void set’. It is denoted by the symbol or {} . A set which has at least one element is called a non-empty set. Singleton set: A set consisting of a single element is called a singleton set. Finite set: A set is called a finite set if it is either void set or its elements can be counted by natural number, and the process of counting terminates at a certain natural number n (say), which is called as the cardinal number or order of a finite set A and is denoted by n( A) or O ( A) . Infinite set: A set whose number of elements is infinite or cannot be counted is called infinite set. Equivalent set: Two finite sets A and B are equivalent if their cardinal numbers are same i.e. n( A) n( B ) . For e.g., A {1,3,5} ; B {6,9,11} are equivalent sets [ O( A) O( B ) 4] . Equal set: Two sets A and B are said to be equal, i.e. A B if x A x B . Equal sets are always equivalent but equivalent sets may need not be equal set. Universal set: A universal set is a set, denoted by U , containing of all possible elements which occur in the discussion. The universal set is not unique; it may differ from problem to problem. Power set: If S is any set, then the family of all the subsets of S , i.e. P( S ) {T : T S } , is called the power set of S . The power set of S is denoted by P( S ) . Obviously and S are both elements of P( S ) . Let S {a, b, c} , then P ( S ) {},{a},{b},{c},{a, b},{a, c},{b, c},{a, b, c} . If set A , then P( A) has one element n[ P ( A)] 1 , i.e. P ( A) is always non-empty. n
If set A has n elements, then P( A) has 2 elements.
P P ( ) ,{ } P P P ( ) ,{ }, { } , ,{ } . Hence n P P P ( ) 4 .
Subsets: Let A and B be two sets. If every element of A is an element of B , then A is called a subset of B , and we write A B . Thus, A B a A a B . The total number of subset of a n
finite set containing n elements is 2 . If A is a subset of B and A B then A is a proper subset of B . We write this as A B . The null set is subset of every set and every set is subset of itself, i.e., A and A A for every set A , they are called improper subsets of A .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.2]
Example 2.1 [CS-2002 (1 mark)]: Which of the following is true? (a) The set of all rational negative numbers forms a group under multiplication (b) The set of all non-singular matrices forms a group under multiplication (c) The set of all matrices forms a group under multiplication (d) Both (b) and (c) are true Solution (b): If non-singular matrices of same order are multiplied then we get a non-singular matrix. So option (b) is correct.
2.1.1 Venn – Euler Diagram In Venn-diagrams the universal set U is represented by points within a rectangle and its subsets are represented by points in closed curves (usually circles) within the rectangle, as shown in Fig. 2.1. Union of Sets: The union of sets A and B , denoted by A B , is the set of all elements which are in set A or in B , as shown in shaded region in Fig. 2.1. Thus, A B {x : x A or x B} . If n
A1 , A2 , , An is a finite family of sets, then their union is denoted by Ai or A1 A2 A3 An . i 1
Intersection of sets: The intersection of sets A and B , denoted by A B , is the set of all those elements that belong to both A and B , as shown in shaded region in Fig. 2.2. Thus, A B {x : x A and x B} . If A1 , A2 , , An is a finite family of sets, then their intersection is n
denoted by
Ai
or A1 A2 A3 An .
i 1
Disjoint sets: Two sets A and B are said to be disjoint, if sets A and B have no element in common, i.e. if A B , then A and B are called disjoint sets, as shown in Fig. 2.3. Difference of sets: Let A and B be two sets. The difference A B , is the set of all those elements of A which do not belong to B . Thus, A B {x : x A and x B} , as shown in Fig. 2.4, the shaded part represents A B . Symmetric difference of two sets: Let A and B be two sets. The symmetric difference of sets A and B is the set ( A B ) ( B A) and is denoted by A B , as shown in Fig. 2.5. Thus, A B ( A B) ( B A) { x : x A B} . Complement of a set: Let U be the universal set and let A be a set such that A U . Then, the c complement of A with respect to U is denoted by A or A or U A and is defined the set of all those elements of U which are not in A , as shown in Fig. 2.6. Thus, Ac {x U : x A} .
Figure 2.1: Union of Sets A and B
Figure 2.4: Difference of Sets ( − )
Figure 2.2: Intersection of Sets
and
Figure 2.5: Symmetric Difference of sets and
Figure 2.3: Disjoint Sets
and
Figure 2.6: Complement of Set
Some Important Results on Number of Elements in Sets: If A , B and C are finite sets and U
be the finite universal set, then n( A B ) n( A) n( B ) n( A B ) If A and B are disjoint non-void sets then n( A B ) n( A) n( B ) . n( A B) n( A) n( A B ) n( A B ) n( A B ) n( A) n( A B) n ( A B ) ( B A) n( A B ) n( B A) [ ( A B) & ( B A) are disjoint]
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.3]
n( A B ) n( A) n( A B) n( B ) n( B A) n( A) n( B) 2n( A B ) n( A B C ) n( A) n( B ) n(C ) n( A B ) n( B C ) n(C A) n( A B C )
n( Ac B c ) n( A B )c n(U ) n( A B )
n( Ac B c ) n( A B )c n(U ) n( A B )
Example 2.2 [CS-1998 (2 marks)]: In a room containing 28 people, there are 18 people who speak English, 15 people who speak Hindi and 22 people who speak Kannada. 9 persons speak both English and Hindi, 11 persons speak both Hindi and Kannada whereas 13 persons speak both Kannada and English. How many people speak all three languages? (a) 9 (b) 8 (c) 7 (d) 6 Solution (d): n( E ) 18 , n( H ) 15 , n( K ) 22 , n( E H ) 9 , n( H K ) 11 , n( K E ) 13 , and n( E H K ) 28 . We have to find n( E H K ) . We know that n( E H K ) n( E ) n( H ) n( K ) n( E H ) n( H K ) n( K E ) n( E H K ) . So 28 18 15 22 9 11 13 n( E H K ) n( E H K ) 6 . Example 2.3 [CS-2001 (2 marks)]: Consider the following statements: A. There exists infinite sets A, B, C such that A ( B C ) is finite. B. There exist two irrational numbers x and y such that ( x y ) is rational Which of the following is true about A and B? (a) Only A is correct (b) Only B is correct (c) Both A and B are correct (d) None of A or B is correct Solution (d): As B and C are infinite sets so B C is infinite set; as A is infinite set so A ( B C ) is an infinite set; thus statement ‘A’ is not correct. As the sum of two irrational numbers is an irrational number, thus statement ‘B’ is not correct. Example 2.4 [ME-2006 (1 mark)]: Let x denote a real number. Find out the INCORRECT statement. (a) S x : x 3 represents the set of all real numbers greater than 3
(b) S x : x 2 0 represents the empty set (c) S x : x A and x B represents the union of set A and set B (d) S x : a x b represents the set of all real numbers between a and b , where a and b are real numbers Solution (c): A B {x : x A or x B} .
Laws of Algebra of Sets: For any sets A , B and C Idempotent laws: Identity laws: Commutative laws:
A A A A A A A A A U A A B B A AB B A AB B A A B B A AB B A Associative laws: ( A B) C A ( B C ) ( A B) C A ( B C ) ( AB) C A( BC ) ( A B) C A ( B C ) ( A B) C A ( B C ) Distributive law: A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) De-Morgan’s law: A B A A B ( A B) ( B A) ( A B ) ( A B ) c ( A B) B A B A ( B C ) ( A B) ( A C ) B A B A ( A B) B A ( B C ) ( A B) ( A C ) ( A B ) c Ac B c
A B B c Ac
( A B ) c Ac B c
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A ( B C ) ( A B) ( A C )
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.4] c
c
Example 2.5 [CS-1996 (1 mark)]: Let A and B be sets and let A and B denote the complements of the sets A and B . The set ( A B ) ( B A) ( A B ) is equal to c c c c (a) A B (c) A B (b) A B (d) A B Solution (a): As shown in figure the total shaded region shows ( A B ) ( B A) ( A B ) which is equal to A B . Example 2.6 [CS-2005 (1 mark)]: Let A , B and C be non-empty sets and let X ( A B ) C and Y ( A C ) ( B C ) . Which of the following is true? (d) None of these (a) X Y (b) X Y (c) Y X Solution (a): X ( A B ) C X ( A C ) ( B C ) Y Example 2.7 [CS-2006 (2 marks)]: Let E , F and G be finite sets. Let X ( E F ) ( F G ) and Y E ( E G ) E F . Which one of the following is true?
(d) X Y and Y X (a) X Y (b) X Y (c) X Y Solution (c): From the Venn diagrams shown below, we can say that X Y .
[Similar question was also asked in CS-2008 (1 mark)]
Cartesian Product of Sets: Let A and B be any two non-empty sets. The set of all ordered pairs ( a, b) such that a A and b B is called the cartesian product of the sets A and B and is denoted by A B . Thus, A B [(a, b) : a A and b B ] . If A {a, b, c} and B { p, q} , then A B {( a, p), ( a, q), (b, p ), (b, q), (c, p), (c, q )} , which can be shown in terms of mapping in Fig. 2.7; also B A {( p, a ), ( p, b), ( p, c ), ( q, a), ( q, b), ( q, c )} . If A or B , then we define A B . Some important theorems on Cartesian Figure 2.7: Cartesian Product of product of sets are: If A and B are two finite sets, then n( A B ) n( A) n( B) . If Sets either A or B is an infinite set, then A B is an infinite sets. If A , B and C are finite sets, then n( A B C ) n ( A) n( B ) n (C ) For any three sets A , B and C A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) A ( B C ) ( A B) ( A C ) If A and B are any two non-empty sets, then A B B A A B If A B , then A A ( A B ) ( B A) If A B , then A C B C for any set C If A B and C D , then A C B D For any sets A , B , C , D , ( A B ) (C D ) ( A C ) ( B D ) For any three sets A , B , C A ( B c C c )c ( A B ) ( A C ) A ( B c C c )c ( A B ) ( A C ) Let A and B two non-empty sets having n elements in common, then A B and B A have 2 n elements in common.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.5]
2.1.2 Relations Let a relation R from set A to set B is a subset of the Cartesian product of A and B , i.e., A B . Thus R is a relation from A to B R A B . If R is a relation from a non-empty set A to a non-empty set B and if ( a, b) R , then we write aRb which is read as ‘ a is related to b by the relation R ’. If ( a, b) R , then we write a R b which is read as ‘ a is not related to b by the relation R ’. For e.g., if A {1, 2,3} and B {a, b, c} , then R {(1, b), (2, c ), (3, a )} being a subset of A B , is a relation from A to B . Here (1, b) , (2, c) and (3, a) R , so we write 1Rb , 2Rc , 3Ra . But
(2, b) R so we write 2 R b . A relation from set A to set B can be represented in following ways: (a) In Roster form, a relation is represented by the set of all ordered pairs belonging to R . The relation R from set A {1, 2} to B {1, 2, 3, 4} by the rule aRb a 2 b is R {(1,1), (2, 4)} . (b) In Set builder form, the relation R from set A to B is represented as R {( a, b) : a A, b B and satisfy the rule which associates a and b} . If A {1, 2,3} and
B {1,1 2 ,1 3,1 4} and R is a relation from A to B given by R {(1,1), (2,1 2), (3,1 3)} , then in set builder form R can be described as R {( a, b) : a A, b B and b 1 a} . It should be noted that it is not possible to express every relation from set A to set B in set builder form. (c) In arrow diagram, a relation from set A to a set B is represented by drawing arrows from first components to the second components of all ordered pairs belonging to R . If A {a, b, c} and B { p, q} , then a Figure 2.8: Arrow diagram representation of a relation relation R {( a, p ), (a, q ), (b, q ), (c, p )} , shown in Fig. 2.8. Domain and Range of a relation: Let R be a relation from a set A to a set B . Then the set of all first components of the ordered pairs belonging to R is called the domain of R , while the set of all second components of the ordered pairs in R is called the range of R . Inverse of a relation: Let A and B be two sets and let R be a relation from a set A to a set B . 1 Then the inverse of R , denoted by R , is a relation from B to A and is defined by 1 R 1 {(b, a ) : (a, b) R} . Clearly, ( a, b) R (b, a) R 1 ; also domain of R range of R 1
and range of R domain of R . Total number of relations: Let A and B be two non-empty finite sets s.t. n( A) m , n( B) n . mn
Then, A B consists of mn ordered pairs. So total number of subsets of A B is 2 [This mn point was asked in CS-1993 (2 marks), CS-1999 (1 mark)]. Among these 2 relations, the void ( ) and universal ( A B ) are the trivial relations from A to B .
Types of Relations
Void Relation: A relation R on a set A is called void or empty relation, if no element of A is related to any element of A . For e.g., a relation R on the set A {1, 2,3} is defined by R {(a, b) : a b 10} ; as a b 10 for any two elements of A , so ( a, b) R for any a, b A and thus R does not contain any element of A A R is an empty set. Universal relation: A relation R on a set is called universal relation if each element of A is related to every element of A . For e.g., a relation R on the set A {1, 2,3} is defined by R {( a, b) R : ( a b) 0} ; as a b 0 for all a, b A , so ( a, b) R for all ( a, b) A A , i.e., each element of set A is related to every element of set A and thus R A A R is a universal relation on set A . Identity relation: Let A be a set. Then the relation I A {( a, a) : a A} on A is called the identity relation on A , i.e., every element of A is related to itself only. For e.g., if A {1, 2,3} the
R1 {(1,1), (2, 2), (3,3)}
is
an
identity
relation;
but
R2 {(1,1), (2, 2)}
and
R3 {(1,1), (2, 2), (3,3), (1, 2)} are not identity relation as (3,3) R2 and in R3 1 is related to 2.
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.6]
Reflexive relation: A relation R on a set A is said to be reflexive if every element of A is related to itself. Thus R is reflexive (a, a ) R for all a A . A relation R on a set A is not reflexive if there exist an element a A such that ( a, a ) R . For e.g., if A {1, 2,3} the R1 {(1,1), (2, 2), (3,3)}
and
R2 {(1,1), (2, 2), (3, 3), (1, 2)}
are
reflexive
relations;
but
R3 {(1,1), (2, 2)} and R4 {(1,1), (2, 2), (1, 2)} are not reflexive relation as (3, 3) R3 and R4 . The identity relation on a non-empty set A is always reflexive relation on A . However, a reflexive relation on A is not necessarily the identity relation on A . For e.g., the relation R {(1,1), (2, 2), (3, 3), (1, 2)} is a reflexive relation on set A (1, 2, 3) but it is not identity relation on A . The universal relation on a non-empty set A is reflexive. For relations R1 and R2 on the set S , if either R1 or R2 is reflexive, then R1 R2 and R1 R2 both are reflexive.
Irreflexive relation: A relation R on a set A is said to be irreflexive if every element of A is not related to itself. Thus R is reflexive (a, a ) R for all a A . A relation R on a set A is not irreflexive if there exist an element a A such that ( a, a ) R . For e.g., if A {1, 2,3} the relations
R1 {(1, 2), (2, 3), (3,1)}
and
R2 {(2,1), (3, 2)}
are irreflexive relations; but
R3 {(1,1), (1, 2), (2,3)} is not irreflexive relation as (1,1) R3 .
Symmetric relation: A relation R on a set A is said to be symmetric relation iff ( a, b) R (b, a ) R for all a, b A . A relation R on a set A is not a symmetric relation if there are at least two elements a, b A such that ( a, b) R but (b, a) R The identity and the universal relations on a non-empty set are symmetric relations. A reflexive relation on a set A is not necessarily symmetric. For e.g., the relation R {(1,1), (2, 2), (1, 2)} is a reflexive relation on set A {1, 2} but it is not symmetric. For relations R1 and R2 on the set S , if both R1 and R2 are symmetric, then R1 R2 and R1 R2 both are symmetric.
Transitive relation: Let A be any set. A relation R on a set A is said to be transitive relation iff ( a, b) R and (b, c ) R then ( a, c ) R for all a, b, c A . The identity and the universal relations on a non-empty set are transitive. For relations R1 and R2 on the set S , if R1 and R2 are transitive, then R1 R2 is not transitive, but R1 R2 is transitive reflexive.
Anti-symmetric relation: Let A be any set. A relation R on set A is said to be an antisymmetric relation iff ( a, b) R and (b, a) R a b for all a, b A . The identity relation on a set A is an anti-symmetric relation. The universal relation on a set A containing at least two elements is not anti-symmetric because if a b are in A , then a is related to b and b is related to a under the universal relation will imply that a b but a b Equivalence Relation: A relation R on a set A is said to be an equivalence relation on A iff it is reflexive, symmetric and transitive relation on A . For relations R1 and R2 on the set S , if R1 and R2 both are equivalence, then R1 R2 is not equivalence and R1 R2 is equivalence. [This point was asked in CS-1998 (1 mark), CS2005 (2 marks)]
Example 2.8: Three relations R1 , R2 and R3 are defined on set A {a, b, c} as follows: (i) R1 {( a, a ), ( a, b), ( a, c ), (b, b), (b, c ), (c, a), (c, b), (c, c)}
(ii)
R2 {( a, b), (b, a ), ( a, c ), (c, a )}
R3 {(a, b), (b, c), (c, a )} . Find whether R1 , R2 and R3 is reflexive, symmetric and transitive.
Copyright © 2016 by Kaushlendra Kumar
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(iii)
Engineering Mathematics
Solution:
For
(i):
Chapter 2: Calculus
[2.7]
R1 {( a, a ), ( a, b), ( a, c ), (b, b), (b, c ), (c, a), (c, b), (c, c)} .
Clearly,
( a, a ), (b, b), (c, c ) R1 R1 is reflexive on A . As, ( a, b) R1 but (b, a ) R1 R1 is not symmetric
on A . As, (b, c) R1 and (c, a ) R1 but (b, a ) R1 R1 is not transitive on A . For (ii): R2 {( a, b), (b, a ), ( a, c ), (c, a )} . Clearly, ( a, a), (b, b), (c, c ) R2 R2 is not reflexive on A . As, ( a, b), (b, a ) R2 and ( a, c ), (c, a ) R2 so R2 is symmetric on A . As, ( a, b) R2 and (b, a) R2 but ( a, a ) R2 R2 is not transitive on A .
For (iii): R3 {(a, b), (b, c), (c, a )} . Clearly, ( a, a ), (b, b), (c, c) R3 R3 is not reflexive on A . As, (b, c ) R3 but (c, b) R3 R3 is not symmetric on A . As, ( a, b) R3 and (b, c ) R3 but ( a, c ) R3 R3 is not transitive on A .
Example 2.9 [CS-1994 (1 mark)]: Amongst the properties {reflexivity, symmetry, anti-symmetry, transitivity} the relation R {( x, y ) N 2 x y} satisfies ………. Solution: As x y ( x, x ) R R is not reflexive relation. If ( x, y ) R then ( y , x) R for all
x, y N R is symmetric relation. If ( x1 , y1 ) R and ( y1 , x1 ) R for all x1 , y1 N then for R to be anti-symmetric we must have x1 y1 which contradicts x1 y1 and so R is not anti-symmetric. If ( x1 , y1 ) R and ( y1 , x1 ) R for all x1 , y1 N then for R to be transitive ( x1 , x1 ) R which
contradicts x1 x1 so R is not transitive. Example 2.10 [CS-1995 (1 mark)]: Let R be a symmetric and transitive relation on a set A . Then (a) R is reflexive and hence equivalence relation (b) R is reflexive and hence partial order (c) R is reflexive and hence not an equivalence (d) None of these relation Solution (d): If R be a symmetric and transitive relation on a set A , then it is not necessary that R is reflexive. Let us consider an example, let a relation R {( a, b), (b, a ), ( a, a), (b, b)} on the set A {a, b, c} , as R is symmetric and transitive both but as (c, c ) R R is not reflexive. Example 2.11 [CS-1996 (2 marks)]: Let R be a non-empty relation on a collection of sets defined by A R B if and only if A B . Then, (pick the true statement) (a) R is reflexive and transitive (b) R is symmetric and not transitive (c) R is an equivalence relation (d) R is not reflexive and not symmetric Solution (d): As there is no common elements in set A and B so any element of A is not related to same element and thus R is not reflexive. So option (a) and (c) are not correct. If ( a, b) R then for R to be symmetric (b, a) R set A and set B have a common element ‘ b ’ which contradicts the statement A B thus R is not symmetric. So option (b) is wrong and option (d) is correct. Example 2.12 [CS-1998 (2 marks)]: The binary relation R {(1,1), (2,1), (2, 2), (2, 3), (2, 4), (3,1),
(3, 2), (3,3), (3, 4)} on the set A 1, 2,3, 4 is (a) reflexive, symmetric and transitive (b) neither reflexive, nor irreflexive but transitive (c) irreflexive, symmetric and transitive (d) irreflexive and antisymmetric Solution (b): As (4, 4) R R is not reflexive relation; so option (a) is wrong. As (2,1) R but (1, 2) R R is not symmetric relation, so option (c) is wrong. As (1,1), (2, 2), (3,3) R R is not irreflexive relation, so option (d) is wrong. As for every ordered pair if ( a, b) R and (b, c ) R then ( a, c ) R and thus R is transitive relation. So the given relation is neither reflexive, nor irreflexive but transitive.
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.8]
Example 2.13 [CS-2002 (2 marks)]: A binary relation S (empty set) on set A {1, 2,3} is (a) neither reflexive not symmetric (b) symmetric and reflexive (c) transitive and reflexive (d) transitive and symmetric Solution (d): As S and A {1, 2,3} , so in S A , there is no ( a, b) such that a S and b A . So S A {} . As (b, b) S A for all b A so it is not reflexive. As ( a, b) S A so we don’t require (b, a ) S A , hence it is symmetric. As ( a, b) S A and (b, c ) S A so we don’t require (b, c ) S A , hence it is transitive. Example 2.14 [CS-2009 (1 mark)]: Consider the binary relation R {( x, y ), ( x, z ), ( z , x ), ( z , y )} on the set { x, y , z} . Which one of the following is TRUE? (a) R is symmetric but not anti-symmetric (b) R is not symmetric but anti-symmetric (c) R is both symmetric & anti-symmetric (d) R is neither symmetric nor anti-symmetric Solution (d): R is not a symmetric relation as ( x, y ) R but ( y , x) R . R is not anti-symmetric as ( x, z ) R and ( z , x ) R but x z . Exercise: 2.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. Which of the following is the empty set (a) { x : x is a real number and x 2 1 0}
(b) { x : x is a real number and x 2 1 0}
(c) { x : x is a real number and x 2 9 0}
(d) { x : x is a real number and x 2 x 2}
2. The set A {x : x R, x 2 16 and 2 x 6} equals (a) (b) {14,3, 4} (c) {3} (d) {4} 3. Two finite sets have m and n elements. The total number of subsets of the first set is 56 more than the total number of subsets of the second set. The values of m and n are (a) 7, 6 (b) 6, 3 (c) 5, 1 (d) 8, 7 4. The number of proper subsets of the set {1, 2,3} is _____ 5. If X {8n 7 n 1: n N } and Y {49(n 1) : n N }, then (a) X Y (b) Y X (c) X Y
(d) None of these
c
6. If A and B are two given sets, then A ( A B) is equal to c c (a) A (b) B (c) A B (d) B A 7. If aN {ax : x N } and bN cN dN , where b, c N are relatively prime, then (d) None of these (a) d bc (b) c bd (c) b cd
8. If the sets A and B are defined as A {( x , y ) : y 1 x , 0 x R} and
9. 10. 11. 12.
13.
30
n
i 1
j 1
Ai B j , then
(c) A B (d) None of these (a) A B A (b) A B B Let A [ x : x R,| x | 1] , B [ x : x R,| x 1| 1] and A B R D, then the set D is (a) [ x :1 x 2] (b) [ x :1 x 2] (c) [ x :1 x 2] (d) None of these Sets A and B have 3 and 6 elements respectively. What can be the minimum number of elements in A B ? _____ Let n(U ) 700, n( A) 200, n( B ) 300 and n( A B ) 100, then n( Ac B c ) _____ In a town of 10,000 families it was found that 40% family buy newspaper A , 20% buy newspaper B and 10% families buy newspaper C , 5% families buy A and B , 3% buy B and C and 4% buy A and C . If 2% families buy all the three newspapers, then number of families which buy A only is _____. In a city 20% of the population travels by car, 50% travels by bus and 10% travels by both car and bus. Then persons (in %) travelling by car or bus is _____.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.9]
14. Suppose A1 , A2 , A3 , , A30 are thirty sets each having 5 elements and B1 , B2 , Bn are n sets each with 3 elements. Let
30
n
i 1
j 1
Ai B j S and each elements of S belongs to exactly 10 of the Ai ’s
and exactly 9 of the B j ’s. Then n is equal to _____ 15. In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics and Physics, 9 in Mathematics and Chemistry, 7 in Physics and Chemistry and 4 in all the three subjects. The number of students who have taken exactly one subject is (a) 6 (b) 9 (c) 7 (d) All of these 16. If A [ x : x is a multiple of 3] and B [ x : x is a multiple of 5], then A B is ( A means complement of A ) (a) A B (b) A B (c) A B (d) A B 17. If A {a, b}, B {c, d }, C {d , e} , then {(a, c), ( a, d ), ( a, e), (b, c ), (b, d ), (b, e)} is equal to (a) A ( B C ) (b) A ( B C ) (c) A ( B C ) (d) A ( B C ) 18. If A {x : x 2 5 x 6 0}, B {2, 4}, C {4, 5} , then A ( B C ) is (a) {(2, 4), (3, 4)} (b) {(4, 2), (4, 3)} (c) {(3,3), (4, 4)}
(d) {(2, 2), (3,3)}
19. If Q x : x 1 y , where y N , then 20. 21. 22. 23.
24.
25. 26. 27. 28. 29.
30.
(a) 0 Q (b) 1 Q (c) 2 Q (d) 2 3 Q Let S {0,1, 5, 4, 7} . Then the total number of subsets of S is _____. The smallest set A such that A {1, 2} {1, 2, 3,5, 9} is (a) {2, 3,5} (b) {3,5, 9} (c) {1, 2, 5,9} (d) None of these If A , B , C be three sets such that A B A C and A B A C , then (b) B C (c) A C (d) A B C (a) A B Let A and B be two non-empty subsets of a set X such that A is not a subset of B , then c (b) B is always a subset of A (a) A is always a subset of the B c (d) A and B are always disjoint (c) A and B are always non-disjoint The shaded region in the given figure is (a) A ( B C ) (b) A ( B C ) (c) A ( B C ) (d) A ( B C ) If A [ x : f ( x ) 0] and B [ x : g ( x ) 0] , then A B will be (c) g ( x) f ( x) (d) None of these (a) [ f ( x )]2 [ g ( x )]2 0 (b) f ( x) g ( x) Let A and B be two sets such that n( A) 0.16, n( B ) 0.14, n( A B ) 0.25 . Then n( A B ) is equal to _____. If A and B are not disjoint sets, then n( A B ) is equal to (a) n( A) n( B ) (b) n( A) n( B ) n( A B) (c) n( A) n( B ) n( A B ) (d) n( A) n( B) In a battle 70% of the combatants lost one eye, 80% an ear, 75% an arm, 85% a leg, x % lost all the four limbs. The minimum value of x is _____. Out of 800 boys in a school, 224 played cricket, 240 played hockey and 336 played basketball. Of the total, 64 played both basketball and hockey; 80 played cricket and basketball and 40 played cricket and hockey; 24 played all the three games. The number of boys who did not play any game is _____. A survey shows that 63% of the Americans like cheese whereas 76% like apples. If x % of the Americans like both cheese and apples, then (d) None of these (a) x 39 (b) x 63 (c) 39 x 63
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.10]
31. Of the members of three athletic teams in a school 21 are in the cricket team, 26 are in the hockey team and 29 are in the football team. Among them, 14 play hockey and cricket, 15 play hockey and football, and 12 play football and cricket. Eight play all the three games. The total number of members in the three athletic teams is _____. 32. In a college of 300 students, every student reads 5 newspaper and every newspaper is read by 60 students. The number of newspaper is (a) at least 30 (b) at most 20 (c) exactly 25 (d) None of these 33. In the village called Sultankari, only three TV channels are available: Soon Plus, Mony and See TV. Out of 4000 TV viewers in the village, 1500 watch Soon TV, 2000 watch Mony and 2500 watch See TV. Amongst these, 500 viewers watch Soon Plus and Mony, 800 watch Soon Plus and See TV, and 1000 watch Mony and See TV. How many viewers watch all three channels? _____ 34. There is a shortage of tube-lights, bulbs and fans in a village Sultankari. All houses do not have either tube-light or bulb or fan. Exactly 19% of houses do not have just one of these. Atleast 67% of houses do not have tube-lights. Atleast 83% of houses do not have bulbs. Atleast 73% of houses do not have fans. What percentage (in%) of houses does not have tube-light, bulb and fan? _____ 35. If A is a non-empty subset of a set E , then what is E ( A ) ( A ) ? c (c) (a) A (d) E (b) A 36. In a class of 110 students, x students take both Mathematics and Statistics, 2 x 20 students take Mathematics and 2 x 30 students take Statistics. There are no students who take neither Mathematics nor Statistics. What is the value of x ? _____ 37. In a school there are 30 teachers who teach Mathematics or Physics. Of these teachers, 20 teach Mathematics and 15 teach Physics, 5 teach both Mathematics and Physics. The number of teachers teaching only Mathematics is _____. 38. Which one of the following is correct? (a) {} {{},{{ }}} (b) {} {{ },{{ }}} (c) {{},{{ }}} (d) {{},{{}}} 39. If A {(22 n 3n 1) | n N } and B {9(n 1) | n N } then which one of the following is correct? (d) Neither (a) nor (b) is correct (a) A B (b) B A (c) A B c 40. {( A B ) A} ( A B ) is c (a) (b) A (c) B (d) B 41. If A {x : x 2 6 x 8 0} and B {x : 2 x 2 3 x 2 0} . Then which one of the following is correct? (a) A B (b) B A (d) Neither (a) nor (b) (c) A B 42. The set of natural number is closed under: (i) Addition; (ii) Subtraction; (iii) Multiplication; (iv) Division. (a) Only (i) (b) Both (i) and (ii) (c) (i), (ii) and (iii) (d) Both (iii) and (iv) 43. Let x {2, 3, 4} and y {4, 6, 9,10} . If A be the set of all ordered pairs ( x, y ) such that x is a factor of y . Then, how many elements does the set A contains? _____ 44. Which one of the following is null set? (a) A {x : x R, x 1 and x 1} (b) B {x : x 3 3 (c) C {} (d) D { x : x R, x 1 and x 1} 45. Let P be the set of all integral multiples of 3; Q be the set of all integral multiples of 4; R be the set of all integral multiples of 6. Consider the following relation: (i) P Q R ; (ii) P R ; (iii) R ( P Q) . Which of the relations given above is/are correct? (a) Only (i) (b) Only (ii) (c) Only (iii) (d) (ii) and (iii) 46. If A {1, 2, 3, 4} , then what is the number of subsets of A with atleast three elements? _____ 47. Which of the following is correct statement? (a) (b) P ( ) (c) P( ) (d) P ( ) 48. Let R be the relation on the set R of all real numbers defined by aRb iff a b 1 . Then R is
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49.
50.
51. 52.
53. 54. 55. 56.
57.
58.
2.2
Chapter 2: Calculus
[2.11]
(a) Reflexive and Symmetric (b) Symmetric only (c) Transitive only (d) Anti-symmetric only Consider the following statements: (i) Set of points of a given line is a finite set; (ii) Intelligent students in a class is a set; (iii) Good books in a school library is a set. Which of the above statements is/are not correct? (a) Only (i) (b) Both (ii) and (iii) (c) Both (i) and (ii) (d) (i), (ii) and (iii) In an examination, 52% candidates failed in English and 42% failed in Mathematics. If 17% candidates failed in both English and Mathematics, what percentage (in %) of candidates passed in both the subjects? _____ Which one of the following is a correct statement? (a) {a} {{a},{b},{c}} (b) {a} {{a}, b, c} (c) {a, b} {{a}, b, c} (d) a {{a}, b, c} Let X {1, 2,3, 4, 5} and Y {1,3, 5, 7,9} . Which one of the following is not a relations from X to Y ? (a) R1 {( x, y ) | y 2 x, x X , y Y } (b) R2 {(1,1), (2,1), (3, 3), (4,3), (5, 5)} (c) R3 {(1,1), (1,3)(3,5), (3, 7), (5, 7)} (d) R4 {(1,3), (2, 5), (2, 4), (7, 9)} The relation ‘less than’ in the set of natural numbers is (a) Only symmetric (b) Only transitive (c) Only reflexive (d) Equivalence relation With reference to a universal set, the inclusion of a subset in another, is relation, which is (a) Symmetric only (b) Equivalence relation (c) Reflexive only (d) None of these Let A {2, 4, 6,8} . A relation R on A is defined by R {(2, 4), (4, 2), (4, 6), (6, 4)} . Then R is (a) Anti-symmetric (b) Reflexive (c) Symmetric (d) Transitive Let N denote the set of all natural numbers and R be the relation on N N defined by ( a, b) R (c, d ) if ad (b c ) bc ( a d ) , then R is (a) Symmetric only (b) Reflexive only (c) Transitive only (d) An equivalence relation In order that a relation R defined on a non-empty set A is an equivalence relation, it is sufficient, if R (a) is reflextive (b) is symmetric (c) is transitive (d) is equivalence Let R and S be two relations on a set A . Then which one of the following is not correct? (a) R and S are transitive, then R S is also transitive (b) R and S are transitive, then R S is also transitive (c) R and S are reflexive, then R S is also reflexive (d) R and S are symmetric then R S is also symmetric
Function
A function is a relation whose every input corresponds with a single output. Every function is a relation; but every relation is not a function. Let X and Y be any two non-empty sets. “A function from X to Y is a rule that assigns to each element of set X , one and only one element of set Y ”. Let the rule be ‘ f ’ then mathematically we write f : X Y where y f ( x ), x X and y Y . We say that ‘ y ’ is the image of ‘ x ’ under f (or x is the pre image of y ). Two things should always be kept in mind: A mapping f : X Y is said to be a function if each element in the set X has its image in set Y (as shown in Fig. 2.9). It is also possible that there are few elements in set Y which are not the images of any element in set X . Every element in set X should have one and only one image. Functions cannot be multi-valued (A mapping that is multiFigure 2.9: A mapping : → valued is called a relation from X and Y ).
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Chapter 2: Calculus
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Testing for a function by vertical line test: A relation f : A B is a function or not it can be checked by a graph of the relation. If a vertical line is drawn which cuts the given curve at more than one point, as shown in Fig. 2.10 (i) and (ii), then the given relation is not a function; and when it cuts the curve at only one point, as shown in Fig. 2.10 (iii) and (iv), then the given Figure 2.10: Vertical line test relation is a function Number of functions: Let X and Y be two finite sets having n and m elements respectively. Then each element of set X can be associated to any one of m elements of set Y . So, total number of functions from set X to set Y is m n . [This point was asked in CS-1998 (1 mark)]
Example 2.15 [CS-2006 (1 mark)]: Let X , Y , Z be sets of sizes x, y , z respectively. Let W X Y and E be the set of all subsets of W . The number of functions from Z to E is xy
xy
x y
xyz
2 2 (b) Z 2 (d) 2 (a) Z (c) Z xy Solution (d): As W X Y has xy number of ordered pairs; also W has 2 number of subsets, so
E has 2
xy
number of elements. Thus number of function from Z to E is (2 xy ) z 2 xyz .
Domain, Co-domain and Range of Function: As shown in Fig. 2.11, if a function f is defined from set A to set B then for f : A B set A is called the domain of function f and set B is called the co-domain of function f . The set of all f images of the elements of A is called the range of function f . In other words, Domain All possible values of x for which f ( x ) exists; Range For all values of x , all possible values of f ( x ) ; Range Co-domain. Methods for finding domain of a function: The expressions under even root must be (i.e., square root, fourth root etc.) 0 ; the denominator must be 0 . If domain of y f ( x) and y g ( x) are D1 Figure 2.11: Domain, Coand D2 respectively then the domain of f ( x ) g ( x ) or f ( x ) g ( x ) is domain and Range of a Function
D1 D2 ; while domain of f ( x) g ( x) is D1 D2 {g ( x) 0} . Methods for finding Range of a function: Range of y f ( x) is collection of all outputs f ( x ) corresponding to each real number in the domain. If domain finite number of points range set of corresponding f ( x ) values. If domain or [some finite points], then express x in terms of y ; from this we find y for x to be defined (i.e., find the values of y for which x exists). If domain a finite interval, find the least and greatest value for range using monotonicity. There is no specific method to find out the range of a function, as the type of method varies with different types of functions. Two function f and g are said to be equal functions, if and only if (i) domain of f domain of g ; (ii) Co-domain of f co-domain of g ; (iii) f ( x) g ( x ) x their common domain.
Example 2.16 [PI-2013, ME-2013 (1 mark)]: Choose the CORRECT set of functions, which are linearly dependent (b) cos x , sin x and tan x (a) sin x , sin 2 x and cos 2 x 2 2 (d) cos 2x , sin x and cos x (c) cos 2x , sin x and cos x 2 2 Solution (c): cos 2 x cos x sin x , so cos 2x , sin 2 x and cos 2 x are linearly dependent. Example 2.17: Find the range of (1 x 2 ) x 2 is
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Solution:
Let
Chapter 2: Calculus
[2.13]
y (1 x 2 ) x 2 x 2 y 1 x 2 x 2 ( y 1) 1 x 2 1 ( y 1) .
Since,
2
x 0
1 ( y 1) 0 ( y 1) 0 y 1 y (1, ) . Example 2.18: Find the range of f ( x) 1 (8 3 sin x) . Solution: 1 sin x 1 3 sin x 3 5 8 sin x 11 1 11 1 (8 3 sin x ) 1 5 Example 2.19 [EC-2007 (1 mark)]: Which one of the following function is strictly bounded? 2 (a) 1 x 2 (b) e x (c) x 2 (d) e x Solution (d): For
y 1 x 2 , as x 0, y is
y 1 x2
unbounded. For
y e x , as
x , y y e x is unbounded. For y x 2 , as x , y y x 2 is unbounded. For 2
2
y e x , as x 0, y 1 and x , y 0 , also any value of x ( , ) , y e x lies between 2
y (0,1] ; and thus y e x is bounded. Example 2.20 [ME-2007 (2 marks)]: If y x (a) 4 or 1 (b) 4 only Solution (a): We have y x
x
x x x then y (2) (c) 1 only (d) Undefined
x y x
x
x yx
y ( y x ) 2 y . So
at x 2 , we have ( y 2) 2 y y 2 5 y 4 0 ( y 4)( y 1) 0 y 4 or 1
2.2.1 Types of Functions Algebraic functions: Functions consisting of finite number of terms involving powers and roots of the independent variable and the four fundamental operations , , , are called algebraic functions. For e.g., x 3/ 2 5 x ,
x 1 ( x 1) , x 1 .
Polynomial function: It is a function of the form a0 x n a1 x n 1 .... an 1 x an , where a0 0 and a0 , a1 , , an are constants and n N 3
is called a polynomial function of degree n . e.g.
2
f ( x ) x 2 x x 3 is a polynomial function. The polynomial of first degree is called a linear function and polynomial of zero degree is called a constant function. Transcendental function: A function which is not algebraic is called a transcendental function; for e.g., trigonometric, inverse trigonometric (discussed in section 0.3), exponential and logarithmic functions are all transcendental functions, which are given as: Exponential Function: If a 0 , a 1 , then the function defined by y a x , x is called an
exponential function with base a , i.e. the exponential function is y f ( x ) a x , a 0 , a 1 , as shown in Fig. 2.12. Domain: (, ) Range: (0, ) Period: Non – periodic Nature: Neither even nor odd Interval in which the inverse can be obtained: (, )
Logarithmic Function: If a 0 , a 1 , then the function defined by y log a x , x is called a logarithmic function with base a . For log a x to be real, x must be greater than zero. The logarithmic function is y loga x , a 0 , a 1 , as shown in Fig. 2.13. Domain: (0, ) Range: (, ) Period: Non – periodic Nature: Neither even nor odd Interval in which the inverse can be obtained: (0, ) The following are the logarithmic formulas: Product formula: loga (m n) log a m log a n , m 0 , n 0 , a 0 , a 1
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Division formula: log a ( m n ) log a m log a n , m 0 , n 0 , a 0 , a 1 Power formula: log a ( m n ) n log a m , m 0 , a 0 , a 1 Change of base formula: log a m (log b m)(log a b) , m 0 , a , b 0 , a , b 1 . In general, log a m (logb m)(log c b)(log d c) (log a k )
Interchange formula: log a b 1 (log b a ) , a , b 0 , a , b 1 Base power formula: log a p q m ( q p ) log a m , m 0 , a 0 , a 1 If a e , the logarithmic is denoted by ln x . Logarithmic function is the inverse of the exponential function.
Figure 2.12: Exponential Function
Figure 2.13: Logarithmic Function
Example 2.21: Find the domain of the function f ( x ) log 3 x ( x 2 1) . Solution: We know that, log a b is defined when, a , b 0 & a 1 . So, f ( x ) is defined when x 2 1 0 x 2 1 x1 ( , 1) (1, ) ;
3 x 0 x2 (3, )
and
3 x 1 x 2
x3 R {2} . Hence the solution is D x1 x2 x3 (3, 2) (2, 1) (1, ) .
Example 2.22 [IN-2008 (2 marks)]: The expression e ln x for x 0 is equal to (a) x (b) x (c) x 1 (d) x 1 Solution (c): Let y e ln x ln y ln e ln x ( ln x )(ln e) ln x ln x 1 y x 1 Please note that the questions on ‘Logarithms’ also came in previous years ‘General Aptitude’ section in GATE examination. These questions and exercise on it are given in ‘Section 9.4 of General Aptitude’.
Explicit and implicit functions: A function is said to be explicit if it can be expressed directly in terms of the independent variable. If the function cannot be expressed directly in terms of the independent variable or variables, then the function is said to be implicit. e.g. y sin 1 x log x is explicit function, while x 2 y 2 xy and x 3 y 2 ( a x ) 2 (b y ) 2 are implicit functions.
Figure 2.14: Constant function
Figure 2.15: Identity function
Figure 2.16: Modulus function
Constant function: Let k be a fixed real number. Then a function f ( x) given by f ( x) k for all x R is called a constant function. The domain of the constant function f ( x) k is the complete set of real numbers and the range of f is the singleton set {k} . The graph of a constant function is a
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straight line parallel to x axis as shown in Fig. 2.14 and it is above or below the x axis according as k is positive or negative. If k 0 , then the straight line coincides with x axis.
Identity function: The function defined by f ( x ) x for all x R , is called the identity function on . Clearly, the domain and range of the identity function is . The graph of the identity function is a straight line passing through the origin and inclined at an angle of 45o with positive direction of x axis, as shown in Fig. 2.15. x, if x 0
Modulus or Absolute value function: The function defined by f ( x) x 0, if x 0 is x, if x 0 called the modulus function, as shown in Fig. 2.16. Its Domain: ; Range: [0, ) ; Period: Non – periodic; Nature: Even; Interval in which the inverse can be obtained: ( , 0] [0, ) .
Greatest integer function: The function y f ( x) [ x] is called the greatest integer function and
x, if x is an integer
and its graph is shown in Fig. 2.17. integer just less than x if x is not an integer
is defined as, [ x ]
For e.g., [1.1] 1 , [2.2] 2 , [ 0.9] 1 , [ 2.1] 3 etc. Its, Domain: ; Range: I ; Period: Non – Periodic; Nature: Neither even or odd.
Fractional part of : The difference between the number ‘ x ’ and its integral value [ x] is called the fractional part of x and is symbolically denoted as { x} . Thus { x} x [ x ] and its graph is shown in Fig. 2.18. Since, x 1 [ x] x x [ x ] 1 x 0 x [ x ] 1 0 {x} 1 . Hence the fractional part of any number is always non – negative and less than one; for e.g., {4.92} 0.92 , {4.92} 0.08 . Its Domain: ; Range: [0,1) ; Period: 1; Nature: Neither even nor odd.
Figure 2.17: Greatest integer function
Figure 2.18: Fractional part of
Signum function: The function defined by
Figure 2.19: Signum function
1, x 0 |x| , x0 or f ( x ) 0, x 0 is called the f ( x) x 0 , x 0 1, x 0
signum function, as shown in Fig. 2.19. The domain is and the range is the set {1, 0,1} .
2.2.2 Classification of Function One – One and Many – One function: If no two elements of the domain have the same image in the co-domain, the function is said to be One-One, as shown in Fig. 2.20a. One-One functions are also called injective functions; for e.g., f : R R given by f ( x ) 3x 5 is OneOne function. On the other hand, if there are at least two elements in the domain whose images are the same, the function is known as Many-One, as shown in Fig. 2.20b; for e.g., f : R R given by Figure 2.20: (a) One - One function (b) Many One Function f ( x ) x 2 1 is Many – One function. Graphically, if the lines are drawn parallel to the x axis from the each corresponding image point
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shall intersect the graph of y f ( x) at one (and only one) point then the function is One – One. If there is at least one line parallel to x axis that will cut the graph at least two different points then the function is Many – One. A Many – One function can be made One – One function by redefining the domain of the original function. If the function is entirely increasing or decreasing in the domain, then f is One – One function. Any continuous function f ( x ) , which has at least one local maxima or local minima is Many – One function. All even function are Many – One function. All polynomials of even degree in have at least one local maxima or maxima and hence are Many – One in the domain . Polynomial of odd degree can be One – One or Many – One Number of one-one functions (injections): If A and B are finite sets having m and n
n Pm , if n m
elements respectively, then number of one-one functions from A to B
[This
0 , if n m
point was asked in CS-1993 (2 marks)]. The following steps should be followed for checking a function is One – One or not. Step I: Take two arbitrary elements x, y (say) in the domain of f . Step II: Put f ( x ) f ( y ) Step III: Solve f ( x ) f ( y ) , if f ( x ) f ( y ) gives x y only, then f : A B is a one-one function (or an injection). Otherwise it is Many – One function.
Onto and Into functions: We have another classification called Onto or Into functions. Let f : X Y be a function. If each element in the co-domain ‘ Y ’ has at least one pre – image in the domain ‘ X ’ that is, for every y Y there exist at least one element x X such that f ( x) y , then f is Onto. In other words range of f Y for Onto functions. On the other hand, if there exists at least one element in the codomain ‘ Y ’ which is not an image of any element in the domain X , then f is Into; for e.g., f : R R , where
f ( x) ax3 b is Onto, where a 0 , b R . Fig. 2.21 shows Onto and Into Figure 2.21: (a) and (b) are Onto function; (c) and (d) are Into function functions with the help of mapping. The following are the steps for determining whether a function is Onto or Into. Step 1: If Range Co-domain, then f is Onto. If Range is a proper subset of Co-domain, then f is Into. Step 2: Solve f ( x) y for x , say x g ( y ) . Now if g ( y ) is defined for each y Co-domain, then f ( x ) is Onto function. If this requirement is not met by at least one value of y in Codomain, then f ( x ) is Into. Number of onto function (surjection): The number of onto functions from set A (having m n
elements) to set B (having n elements), where 1 n m , is
(1)n r nCr r m . r 1
Any polynomial function f : R R is, Onto function if its degree is odd; Into function if its degree is even.
Example 2.23 [CS-2012 (2 marks)]: How many onto (or surjective) functions are there from an n element ( n 2) set to a 2 element set?
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n n n (d) 2(2n 2) (a) 2 (b) 2 1 (c) 2 2 Solution (c): The number of onto functions from set A (containing n elements) to set B (containing 2
2 elements) is
(1)2r 2Cr r n (1)21 2C11n (1)22 2C2 2n 2 2n . r 1
One-One and Onto function (Bijective): A function f : A B is said to be a Bijective function if it is one-one as well as onto. Number of One – One and Onto function (Bijective): If A and B are finite sets and f : A B is a bijection, then A and B have the same number of elements. If A has n elements, then the number of bijection from A to B is the total number of arrangements of n items taken all at a time i.e., n! .
Inverse of a function: If f : X Y be a function defined by y f ( x) such that f is both One – One and Onto, then there exist a unique function g : Y X such that for each y Y , g ( y ) x if 1
and only if y f ( x) . The function g so defined is called the inverse of f and denoted by f . Also if g is the inverse of f , then f is the inverse of g and the two functions f and g are said to be inverses of each other. The condition for existence of inverse of a function is that the function must be One – One and Onto both. Whenever an inverse function is defined, the range of the original function becomes the domain of the inverse function and domain of the original function becomes the range of the inverse function. It should be also noted that fof 1 ( x ) f [ f 1 ( x )] x always and roots of the equation f ( x ) f 1 ( x ) would always lie on the line y x . Graphs of f and f 1 are symmetrical about the line y x . The following are the steps for finding the inverse of a function: Firstly, we have to verify that the function is One – One and Onto both. Secondly, write y f ( x) and then solve x in terms of y . Let x g ( y ) , then interchange x and y . That is, y g ( x ) g ( x ) f 1 ( x ) .
Example 2.24: Find the inverse of a function f ( x) 3x 1 . Solution: As the given function is bijective so its inverse exists. f ( x ) y 3x 1 x (1 y ) 3
x f ( y ) , so interchanging x and y y (1 x) 3 f 1 ( x ) (1 x ) 3 Example 2.25 [CS-1996 (2 marks)]: Let R denotes the set of real numbers. Let f : R R R R be a bijective function defined by f ( x, y ) ( x y , x y ) . The inverse function of f is given by 1 (a) f ( x, y) 1 ( x y ) ,1 ( x y)
(b) f 1 ( x, y ) ( x y ) 2 , ( x y ) 2
1 (c) f ( x, y ) x y, x y
1 (d) f ( x, y) 2( x y ), 2( x y)
Solution (b): As the given function is bijection so its inverse exists f 1 f ( x, y ) f 1 ( x y, x y ) f 1 ( x y , x y ) ( x, y ) . Now let x y X and x y Y , on solving these two equations we
get, 1
x (X Y) 2
y (X Y) 2 .
and
Thus,
1
f ( x y, x y ) ( x, y ) f ( X , Y ) ( X Y ) 2, ( X Y ) 2 . Now putting x in place of X and y in place of Y , we get f 1 ( x, y ) ( x y ) 2 , ( x y ) 2 .
Odd function: If we put ( x) in place of x in the given function and if f ( x ) f ( x), x x
x
domain then f ( x ) is called odd function. For e.g. f ( x) e e is an odd function. The graph of odd function is always symmetric with respect to origin. First derivative of an (i) even function is an odd function; (ii) odd function is an even function.
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Zero function f ( x ) 0 is the only function which is even and odd both. Every function can be expressed as a sum of an even and odd function. Let f ( x ) be any function then
it
can
be
written
as,
f ( x ) f ( x ) f ( x) 2 f ( x ) f ( x) 2
f ( x ) h ( x ) g ( x ) , where h ( x ) f ( x ) f ( x ) 2 and g ( x) f ( x ) f ( x ) 2 is an even and odd function, respectively.
Even function: If we put ( x) in place of x in the given function and if f ( x) f ( x ) , x x
domain then function f ( x ) is called even function. For e.g. f ( x) e e graph of even function is always symmetric with respect to y axis.
x
is an even function. The
Example 2.26 [CE-1999 (1 mark)]: The function f ( x) e x is (a) Even (b) Odd (c) Neither even or odd (d) None of the above x x Solution (c): f ( x) e f ( x ) or f ( x ) . So f ( x ) e is neither even nor odd function.
Periodic Function: A function is said to be periodic function if its each value is repeated after a definite interval. So a function f ( x ) will be periodic if a positive real number T exist such that, f ( x T ) f ( x ), x domain. Here the least positive value of T is called the period of the function. Clearly f ( x ) f ( x T ) f ( x 2T ) f ( x 3T ) ..... . sin n x , cos n x , sec n x and cos ec n x has n
n
period, if n is even; 2 if n is odd or fraction. tan x, cot x has period if n is odd or even. sin x , cos x , tan x , cot x , sec x and cosec x are periodic with period of . The following are the rules for finding the period of periodic function: If f ( x ) is periodic with period T , then af ( x ) b , a ( 0), b R , is also periodic with period T .
If f ( x ) has a period T , then the function f ( ax b ) , a ( 0), b R , will have a period T a .
If f ( x ) is periodic with period T then 1 f ( x) ,
Let f ( x ) has period p m n ( m, n N and co-prime) and g ( x ) has period q r s ( r , s N
f ( x ) is also periodic with same period T .
and co-prime) and let t be the LCM of p and q , then t LCM( m, r ) HCF( n, s) . LCM of rational and irrational number is not possible. If m and r are irrational then LCM( m, r ) does not exist unless they have same irrational surd. Let us suppose that f ( x ) is periodic with period p and g ( x ) is periodic with period q . Let r be the LCM of p and q , if it exists. Then if f ( x ) and g ( x ) cannot be interchanged by adding a least positive number k r , to x , then r is the period of f ( x ) g ( x) . If f ( x ) and g ( x ) can be interchanged by adding a least positive number k r , to x , then k is the period of f ( x ) g ( x) . A constant function is periodic but its period is not defined. [This point was asked in AE-2011].
Example 2.27 [EC-2009 (1 mark)]: A function given by f (t ) sin 2 t cos 2t . Which of the following is true? f has frequency components at (a) 0 and 1 2 Hz
(b) 0 and 1 Hz
(c) 1 2 and 1 Hz
(d) 0, 1 2 and 1 Hz
2
Solution (b): f (t ) sin t cos 2t (1 cos 2t ) 2 cos 2t (1 cos 2t ) 2 . As cos t is of period 2
cos 2t is periodic with period 2 2 ; so (1 cos 2t ) 2 is periodic with period . So the time period ( T ) of the given function is and so its frequency is 1 T 1 Example 2.28 [IN-2009 (1 mark)]: The fundamental period of x (t ) 2 sin 2 t 3sin 3 t , with t expressed in second, is (a) 1 (b) 0.67 (c) 2 (d) 3
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Engineering Mathematics
Chapter 2: Calculus
[2.19]
Solution (d): period of 2sin 2 t is 2 2 1 ; period of 3sin 3 t is 3 2 3 2 ; so period of
x (t ) is LCM(1 1, 3 2) LCM(1,3) HCF(1, 2) 3 1 3 Example 2.29 [EE-2010 (1 mark)]: The period of the signal x (t ) 8 sin 0.8 t 4 is (c) 1.25 s (d) 2.5 s (a) 0.4 s (b) 0.8 s Solution (d): The period of the signal x (t ) 8 sin 0.8 t 4 is same as the period of
x(t ) sin(0.8 t ) . As period of sin t is 2 period of sin(0.8 t ) is 2 0.8 2.5
2.2.3 Composite Function Let f and g be two functions with domain D1 and D2 respectively. If range( f ) domain( g ) , we define gof by the rule ( gof )( x ) g{ f ( x )} for all x D1 . Also, if range( g ) domain( f ) , we define fog by the rule ( fog )( x ) f { g ( x )} for all x D2 . Let us say h( x ) ( gof )( x ) . To obtain h( x) , we first take the f image of an element x D1 so that f ( x) D2 , which is the domain of g ( x ) . Then take g image of f ( x ) , i.e., g{ f ( x )} which would be an element of D . The Fig. 2.22 shown clearly shows the steps to be taken. The function h defined is called composition of f and g and is denoted by gof . Thus Figure 2.22: Composition of Function ( gof )( x ) g{ f ( x )} .
Properties of Composite Function
Generally, the composition of function is not commutative, i.e., fog gof The composition of function is associative, i.e., if h : A B , g : B C and f : C D be three functions, then ( fog )oh fo( goh) The composition of any function with the identity function is the function itself, i.e., f : A B then foI A I B of f , where I A and I B are the identity functions of A and B , respectively.
If f : A B and g : B C are one-one, then gof : A C is also one-one Proof: Suppose gof ( x1 ) gof ( x2 ) g{ f ( x1 )} g{ f ( x2 )} f ( x1 ) f ( x2 ) as g is one-one, and f ( x1 ) f ( x2 ) x1 x2 as f is one-one. So gof is one-one.
If f : A B and g : B C are onto, then gof : A C is also onto. Proof: Given an arbitrary element z C , there exists a pre-image y of z under g such that g ( y ) z , since g is onto. Further for y B , there exists an element x in A with f ( x) y , since f is onto. Therefore, gof ( x ) g{ f ( x )} g ( y ) z which shows that gof is onto. If gof ( x ) is one-one, then f ( x ) is necessarily one-one but g ( x ) may not be one-one. Proof: Let f ( x1 ) f ( x2 ) for some x1 , x2 domain of f ; then taking g on both sides g{ f ( x1 )} g{ f ( x2 )} gof ( x1 ) gof ( x2 )
as
gof
is one-one so
gof ( x1 ) gof ( x2 )
x1 x2 ; thus we have for all x1 x2 , f ( x1 ) f ( x2 ) f is one-one.
If gof ( x ) is onto, then g ( x ) is necessarily onto, but f ( x ) may not be one-one. [This point was asked in CS-2005 (2 marks)] Proof: Let f : A B and g : B C then gof : A C . Let there is y C , as gof is onto function, then there exist a A such that gof ( a ) y g{ f (a )} y ; now let b f ( a) B , such that g (b) y . SO we have shown that for every y C , there is a b B such that g (b) y , so g is onto.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.20]
Exercise: 2.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. If A contains 10 elements then total number of functions defined from A to A is 10 10 (a) 10 (c) 1010 (b) 2 (d) 2 1 2. If f ( x ) x x x then f ( 1) _____. 3. If f ( y ) log y , then f ( y ) f (1 y) _____. 4. If f ( x ) log (1 x) (1 x) then f 2 x (1 x 2 ) (c) 2 f ( x ) (a) [ f ( x )]2 (b) [ f ( x )]3
(d) 3 f ( x )
5. If f ( x ) cos[ 2 ] x cos[ 2 ] x , where [] denotes greatest integer function, then (b) f ( ) 2 (c) f ( ) 1 (a) f ( 4) 2 (d) f ( 2) 1 n
6. If f : R R satisfies f ( x y ) f ( x ) f ( y ) for all x, y R and f (1) 7, then
f (r ) is r 1
(a) 7 n 2
(c) 7 n ( n 1)
(b) (7 2)( n 1)
7. If f ( x ) 1 x 2 2 x 4 1 x 2 2 x 4 denotes greatest integer function. 8. The domain of the function f ( x ) 1 (a) R
(b) R
9. The domain of the function (a) ( , )
(d) (7 2) n( n 1)
for x 2 , then [ f (11)] _____, where []
x x is
(c) R {0}
(d) R
2
log( x 6 x 6) is
(b) ( , 3 3) (3 3, )
(c) ( ,1] [5, ) x
(d) [0, )
y
10. The domain of definition of the function y ( x ) given by 2 2 2 is (a) (0,1] (b) [0,1] (c) ( , 0]
(d) ( ,1)
1
11. The domain of the function f ( x ) sin [log 2 ( x 2)] is (a) [1, 4] (b) [ 4,1] (c) [ 1, 4]
(d) None of these 1
12. The domain of the derivative of the function f ( x ) (a) R {0}
(b) R {1}
1 2 tan x
2 x 1
, ,
(c) R {1}
x 1
is x 1 (d) R {1, 1}
13. Domain of definition of the function f ( x ) 3 (4 x 2 ) log10 ( x3 x) , is (a) (1, 2)
(b) ( 1, 0) (1, 2)
(c) (1, 2) (2, )
(d) ( 1, 0) (1, 2) (2, )
2
14. The domain of the function f ( x ) log 3 x ( x 1) is (a) ( 3, 1) (1, ) (b) [ 3, 1) [1, ) (c) ( 3, 2) ( 2, 1) (1, ) (d) [ 3, 2) ( 2, 1) [1, ] 15. Domain of definition of the function f ( x ) sin 1 (2 x ) ( 6) for real value x is (a) 1 4 ,1 2 (b) 1 2 ,1 2 (c) 1 2 ,1 9 16. If f ( x ) a cos ( bx c ) d then range of f ( x ) is (a) [ d a, d 2a ] (b) [ a d , a d ] (c) [ d a, a d ] 17. The range of the function f ( x ) ( x 2) x 2 is (a) {0,1} (b) {1,1} (c) R
2
(d)
1 4 ,1 4
(d) [ d a, d a ] (d) R {2}
18. The range of f ( x ) sec ( 4) cos x , x is
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics
(a) [1, 2]
Chapter 2: Calculus
(b) [1, )
[2.21]
(c) [ 2, 1] [1, 2]
(d) ( , 1] [1, )
19. Range of the function f ( x ) ( x 2 x 2) ( x 2 x 1) ; x R is (a) (1, ) (b) (1,11 7) (c) (1, 7 3] (d) (1, 7 5] 20. The function f : R R defined by f ( x ) ( x 1)( x 2)( x 3) is (a) One-one but not onto (b) Onto but not one-one (c) Both one-one and onto (d) Neither one-one nor onto 21. The total number of surjection from A to B where A {1, 2,3, 4}, B {a, b} is _____. 22. If A {a, b, c}, then total number of one-one onto functions which can be defined from A to A is _____. 23. If f : R R, then f ( x ) x is (a) One-one but not onto (b) Onto but not one-one (c) One-one and onto (d) None of these x 24. The function f : R R defined by f ( x) e is (a) Onto (b) Many-one (c) One-one and into (d) Many one and onto 25. A function f from the set of natural numbers to integers defined by ( n 1) 2 , when n is odd , is f ( n) n 2 , when n is even (a) One-one but not onto (b) Onto but not one-one (c) One-one and onto both (d) Neither one-one nor onto 26. Which one of the following is an even function? ax 1 ax 1 a x ax (d) f ( x ) sin x (b) f ( x ) x (a) f ( x ) x a x 1 (c) f ( x ) x x a 1 a a
27. The function f ( x ) sin log( x x 2 1 ) is 28. 29. 30. 31.
(a) Even function (b) Odd function (c) Neither even nor odd (d) Periodic function The function f ( x) sin( x 2) 2 cos( x 3) tan( x 4) is periodic with period _____. If f ( x ) is an odd periodic function with period 2, then f (4) _____. The period of f ( x ) x [ x ] is _____. If a, b be two fixed positive integers such that
f ( a x ) b [b3 1 3b 2 f ( x ) 3b{ f ( x )}2 { f ( x)}3 ]1 3 for all real x then f ( x ) is a periodic function with period (a) a (b) b (c) 2a (d) 2b 2 2 32. f ( x ) sin x sin x ( 3) cos x cos x ( 3) and g (5 4) 1, then ( gof ) ( x) _____
33. If g ( x ) x 2 x 2 and (1 2)( gof )( x ) 2 x 2 5 x 2, then f 1 ( x ) (a) 2 x 3 (b) 2 x 3 (c) 2 x 2 3 x 1 (d) 2 x 2 3 x 1 34. Let g ( x ) 1 x [ x ] , where [] denotes greatest integer function, and f ( x ) sgn( x ) , then for all x , f ( g ( x )) _____ 35. If the function f : R R be such that f ( x ) x [ x ] where [ y ] denotes the greatest integer less than or equal to y , then f 1 ( x ) is (b) [ x ] x (a) 1 ( x [ x]) 36. If f : [1, ) [1, ) is defined as f ( x ) 2
(c) Not defined x ( x 1)
then f
1
(d) None of these
( x ) is equal to
(b) (1 2)(1 1 4 log 2 x ) (c) (1 2)(1 1 4 log 2 x ) (d) Not defined (a) (1 2) x ( x1) 37. Which one of the following function is invertible? (d) None of these (a) f ( x) 2 x (b) f ( x ) x 3 x (c) f ( x) x 2
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics
2.3
Chapter 2: Calculus
[2.22]
Limit, Continuity and Differentiability of a Function
2.3.1 Limit of a Function Let y f ( x) be a function of x . If at x a , f ( x ) takes indeterminate forms ( 0 / 0, / , , , 0 , 1 , 00 , 0 ), then we consider the values of the function which are very near to ‘ a ’ but not equal to ‘ a ’. If these values tend to a definite unique number as x tends to ‘ a ’, then the unique number so obtained is called the limit of f ( x ) at x a and we write it as lim f ( x ) . x a
Left Hand Limit (LHL) and Right Hand Limit (RHL): Consider the values of the functions at the points which are very near to a on the left of a . If these values tend to a definite unique number as x tends to a , then the unique number so obtained is called left-hand limit (LHL) of f ( x ) at x a and symbolically we write it as f ( a 0) lim f ( x ) lim f ( a h) , here we write a h in h 0
x a
place of x a ; where h is very-very small positive number. Similarly we can define right-hand limit (RHL) of f ( x ) at x a which is expressed as f ( a 0) lim f ( x ) lim f ( a h) , here we write h 0
x a
a h in place of x a ; here h is very-very small positive number. If both LHL and RHL exists and both are equal, i.e., lim f (a h) lim f (a h) , then we say that lim f ( x ) exists. h 0
h 0
x a
Example 2.30 [CH-2009 (2 marks)]: The value of the limit lim cos x {x ( 2)}3 is x 2
(b) 0
(a)
(c) 1
(d) cos( 2 h ) sinh Solution (a): RHL lim f ( x) lim f h lim lim 3 3 h 0 2 x 2 h0 ( 2 h 2) h0 h
cos( 2 h) sinh h lim lim 3 . LHL RHL 3 2 h0 ( 2 h 2) h0 h
RHL lim f ( x ) lim f h 0 x 2
[Similar question was also asked in, BT-2013 (2 marks)] Example 2.31 [CE-2014 (1 mark)] lim{( x sin x) x} equals to x
(b) 0 (c) 1 (a) (d) Solution (c): The given limit can be written as lim 1 {(sin x ) x} 1 lim{(sin x ) x} x
x
LHL 1 lim {sin( h)} ( h) 1 {(any values between 0 and 1) } 1 0 1 ; and h0
RHL 1 lim
sin( h)
1 {(any values between 0 and 1) } 1 0 1 . So, RHL RHL 1 ( h ) [Similar questions were also asked in ME-1994, CH-2008 (1 mark)] h 0
Example 2.32 [MN-2014 (1 mark)]: The lim x x is x0
(a) –1 (b) 0 (c) 1 (d) non-existent Solution (d): LHL lim f ( x) lim f (0 h) lim 0 h (0 h) lim{h ( h)} lim( 1) 1 ; x 0
h 0
h 0
h 0
h0
and RHL lim f ( x) lim f (0 h) lim{ 0 h (0 h)} lim(h h) lim(1) 1 . x 0
h 0
h 0
h 0
h 0
Hence lim x x does not exists as LHL RHL x0
Algebra of limits: The following hold true for lim f ( x ) l and lim g ( x ) m ( l , m R ). x 0
x 0
lim( f ( x ) g ( x )) l m
lim( f ( x ) g ( x )) l m
lim{ f ( x ) g ( x )} l m
lim k f ( x) kl
x a x a
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x a
x a
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.23]
lim g ( x )
lim log{ f ( x)} log{lim f ( x )}
lim[ f ( x )]g ( x ) {lim f ( x )}xa
lim ln[ f ( x )] ln(l ), only if l 0
If f ( x ) g ( x ) for all x , then lim f ( x ) lim g ( x )
lim { f ( x ) g ( x)} l m , m 0
If lim f ( x) or , then lim{1 f ( x )} 0
lim f ( g ( x)) f (lim g ( x )) f (m) if x a x a
x a
x a
x a x a
x a
x a
x a
x a
xa
x a
If p and q are integers, then lim( f ( x))
p/q
x a
f is continuous at g ( x ) m
l p/q ,
p/ q
provided (l ) is a real number. Some important expansions: In finding limits, use of expansions of some functions are useful: n( n 1) 2 ( x log a) 2 (1 x ) n 1 nx x a x 1 x log a 2! 2!
x
e 1 x
x2
x3
[This
series
was
2! 3! asked in EC-1995, CE-2012 (1 mark)] x 2 x 4 x6 cos x 1 2! 4! 6! [This series was asked in CE-1997 (1 mark)]
x2
x3
x4
, x 1 2 3 4 x2 x4 x6 cosh x 1 2! 4! 6! log(1 x ) x
sin
1
2
x x 1
x3 3!
2
2
3 1
x5 5!
x3
x5
[This series was 3! 5! asked in CE-1998 (1 mark)] x3 2 x5 tan x x 3 15 x2 x3 x 4 log(1 x ) x , x 1 2 3 4 3 5 x x sin h x x 3! 5! x3 5 tanh x x 2 x 3 sin x x
tan
1
x x
x3 3
x5 5
x7 7
2
1x
(1 x)
e
(1 x ) log(1 x )
x x 1 2 3
e
e 1
x 2
11 24
x2
Example 2.33 [ME-2007 (2 marks)]: lim e 1 x ( x 2 2) x 0
(a) 0
(b) 1/6 e 1 x ( x 2 2) x
Solution (b): L lim
3
x
x3
(c) 1/3 (d) 1 2 3 1 x ( x 2!) ( x 3!) 1 x ( x 2 2)
lim
3
x0 x x 3 L lim{( x 3!) ( x 4!) } x lim (1 3!) ( x 4!) (1 3!) 1 6 . x 0
3
4
x0
x 0
[Similar questions were also asked in CS-1993, EC-1993 (1 mark)]
Example 2.34 [CE-2014 (2 marks)]: The expression lim ( x 1) is equal to 0
(b) 0
(a) log x
(c) x log x
(d)
Solution (a): In the given limit, is the variable; using expansion of x by treating x as constant and being variable, we put in place of x , x in place of a in expansion of a x , 2 x 1 (log x ) 2 ( log x ) lim lim 1 log x 1 lim log x log x 0 0 0 2! 2!
Method of Evaluation of Limits: Following are indeterminate form: 0 0 , , 0 , , 0
0 0 , , 1 . We shall divide the ways of evaluation of limits in following categories.
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.24]
Evaluation of Algebraic limits: The following are the methods for evaluating algebraic limits: Direct substitution method: If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression. Example 2.35 [MN-2009 (1 mark)]: The value of lim 2 4 x 2 5 is x2
(b) 0
(a) 2 8 5
(d) non-existent
(c) 2 8 5
Solution (b): At x 2 , 2 4 x 2 5 0 , which is a finite number so lim 2 4 x 2 5 0 x 2
Factorisation method: In this method, numerator and denominator are factorised. The common factors are cancelled and the rest outputs the results. Example 2.36 [ME-2006 (2 marks)]: If f ( x ) (2 x 2 7 x 3) (5 x 2 12 x 9) , then lim f ( x ) will be x 3
(a) –1/3 (b) 5/18 (c) 0 (d) 2/5 Solution (b): Since at x 3 , the given limit takes the form 0 0 and so ( x 3) is a factor of numerator and denominator. lim
2x2 7x 3 2
lim
(2 x 1)( x 3)
lim
(2 x 1)
5
5 x 12 x 9 x 3 (5 x 3)( x 3) x 3 (5 x 3) 18 [Similar questions were also asked in ME-2000, CE-2004, TF-2008, (1 marks), TF-2010 (2 marks)] 1/3 Example 2.37 [ME-2008 (1 mark)]: The value of lim ( x 2) ( x 8) is x 3
x 8
(a) 1/16
(b) 1/12 (c) 1/8 (d) ¼ 1/3 1/3 x 2 ( x 2) 1 1 lim 1/3 Solution (b): lim 1/3 3 lim 2/3 3 2/3 1/3 2 1/3 2 x 8 ( x ) 2 x 8 ( x 2)( x 2 x 2 ) x8 x 2 x 2 12 [Similar question was also asked in MN-2010 (1 mark)]
Rationalisation method: Rationalisation is followed when we have fractional powers (like 1 2 ,1 3 etc.) on expressions in numerator or denominator or in both. After rationalisation the terms are factorised which on cancellation gives the result.
Example 2.38 [MN-2011 (1 mark)]: The value of lim(1 x ) x 0
(a) 0
(b) 1
Solution (b): lim
1 x 1 x x
x 0
1 x 1 x
is
(c) 2
(d) 3
lim (1 x) (1 x) 2 1 1 x x 1 x 1 x 2
1 x 1 x 1 x
x 0
Based on the form when x : In this case, expression should be expressed as a function of 1 x and then after removing indeterminate form, (if it is there) replace 1 x by 0. Step I: Write down the expression in the form of rational function, i.e. f ( x) g ( x ) , if it is not so. Step II: If k is the highest power of x in numerator and denominator both, then divide each term of numerator n and denominator by x k . Step III : Use the result lim (1 x ) 0 , where n 0 . x
Example 2.39 [CE-1999 (1 mark)]: Limit of the function lim n n
(a) 1/2
(b) 0
Solution (d): lim
n
n 2
lim
n n
n
n n 2
lim
n n n
n
(d) 1
(c) 1
n
2
n
n 2 n is
n
2
lim
n
1
1
1 1 n
1
1 0
[Similar questions were also asked in CE-1997 (5 marks), CE-2000 (2 marks)] Example 2.40 [CS-2008 (1 mark)]: lim ( x sin x ) ( x cos x ) x
(a) 1
(b) –1
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(c)
(d)
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Engineering Mathematics
Solution (a): lim
x
x cos x
lim
x 1
(cos x ) x
m,
lim
a0 x m a1 x m1 am 1 x am
x
1 lim(cos x) x
n
x
are positive integers and
b0 x n b1 x n 1 bn 1 x bn
[2.25]
1 lim(sin x ) x
1 (sin x) x
If
x
n
x sin x
Chapter 2: Calculus
a0 , b0 0
1 0 1 0
1
are non-zero real numbers, then
a0 b0 , if m n 0, if m n , if m n
n
If n Q , then lim{( x a ) ( x a )} na
n 1
x a
Proof: Since, (1 x) n 1 nx
n( n 1)
x2
n( n 1)(n 2)
x 3 , where x 1 , n Q (from
2! 3! binomial expansion). When x is infinitely small (approaching to zero) such that we can ignore n
higher power of x , then we have (1 x) 1 nx . n
LHL lim x a
x a x a
n
lim
xa n
LHL lim
n
( a h) a n
( a h) a
n 1
an
lim a 1 (nh a) 1 na
n 1
n
h (1 ( h a )) n 1
h 0
n
lim a 1 (nh a ) 1 na
a n (1 ( h a )) n 1
h
h 0
n
lim h 0 h0 aha h n n x a n 1 Hence, if n Q , then lim na x a x a n Example 2.41 [TF-2011 (1 mark)]: The value of lim{(1 x) 1} x is x a
xa
lim
lim
aha
h 0
n
n
h0
h
x 0
(b) n
(a) 0
(c)
(d) 1 n
Solution (b): Using binomial expansion, the given limit becomes (1 x ) n 1 n( n 1) 2 n( n 1)(n 2) 3 lim 1 nx x x 1 x x 0 2! 3! x
x 0
lim n
n( n 1) 2!
x
n(n 1)(n 2) 3!
x2 n
Example 2.42: Find the values of constants a and b so that lim
x
Solution: lim
x
( x
2
1) ( x 1) ax b 0 lim
x
x
2
( x
2
1) ( x 1) ax b 0
(1 a) x ( a b) 1 b ( x 1) 0
Since the limit of the given expression is zero, therefore degree of the polynomial in numerator must be less than that of denominator. As the denominator is a first degree polynomial. So, numerator must be a constant i.e., a zero degree polynomial. 1 a 0 and a b 0 a 1 and b 1 . Evaluation of Trigonometric limits: The following results holds true for f ( x) 0 when x a .
lim
sin f ( x)
1 lim
sin x
1 [This point
f ( x) x was asked in IN-2008, ME-2011, MT-2009 (1 mark), CS-1995 (2.5 marks)] sin 1 f ( x ) sin 1 x lim 1 lim 1 x a x 0 f ( x) x x a
x 0
lim sin 1 x sin 1 a, a 1 x a
lim x a
tan
1
f ( x)
f ( x)
x 0
tan x x
1
Copyright © 2016 by Kaushlendra Kumar
x a
lim x a
lim
x
x 0
tan f ( x ) f ( x) sin x x
1 lim
tan x
x0
lim
x
cos x x
x
1
0
lim cos 1 x cos 1 a; a 1
lim tan 1 x tan 1 a; a
1
1 lim
lim cos f ( x ) 1 lim cos x 1
x a
x a
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Engineering Mathematics
Proof of lim
sinf ( x ) f ( x)
x a
lim
sin f ( x )
xa
f ( x)
Chapter 2: Calculus
[2.26]
1 : Using expansion of sin x , we have
f ( x)3 f ( x)5 f ( x)7
lim f ( x)
3!
xa
5!
f ( x)
7!
f ( x )2 f ( x)4 f ( x )6 sin f ( x ) lim lim 1 1 , since as x a , f ( x) 0 . xa x a f ( x) 3! 5! 7! The other results can also be proved in a similar manner. Example 2.43 [ME-1994 (1 mark)]: The value of lim 1 sin x 1 tan x is x 0
(a) 0
(b) 1 1
(c) 2
(d) 2
2 sin ( x 2) 1 1 cos x lim lim lim tan( x 2) 0 sin x tan x x 0 sin x x 0 2 sin( x 2) cos( x 2) x0
Solution (a): lim x 0
Example 2.44 [ME-1995 (2 mark)]: Evaluate lim ( x 3 cos x ) ( x 2 sin 2 x) x
(b) 0
(a)
(c) 2
3
Solution (a): lim
x
3
x cos x 2
2
x sin x
lim
x
( x cos x) x 2
2
(d) Does not exist
2
( x sin x) x
2
lim
x
x (cos x ) x
2
0
2
1 (sin x) x
1 0
Example 2.45 [CE-1997 (1 mark)]: lim sin( m ) , where m is an integer, is: 0
(d) 1 (a) m (b) m (c) m Solution (a): If 0 m 0 , so lim sin( m ) m lim sin( m ) ( m ) m 1 m m 0
m 0
[Similar questions were also asked in EC-2007, CE-2010 (1 marks)] x3 x5 x 7 Example 2.46 [CE-2001 (1 mark)]: If f ( x ) x , then lim f ( x ) is x 2 3! 5! 7! (a) 2 3 (b) 2 (c) 3 (d) 1 Solution (d): sin x x
x3 3!
x5 5!
x7 7!
lim f ( x ) lim sin x 1 x 2
x 2
2
Example 2.47 [ME-2003 (1 mark)]: lim (sin x ) x x 0
(a) 0
(c) 1 (b) Solution (a): lim (sin x ) x lim (sin x ) x lim(sin x ) 1 0 0 x 0
2
x 0
x 0
Example 2.48 [PI-2007 (1 mark)]: What is the value of lim
x 4
(a)
(b) 0
2
Solution (c): LHL lim RHL lim
h0
cos x sin x x (
4 h 4
cos ( 4) h sin ( 4) h
4 h 4
lim
lim
4) ?
(d) limit does not exist
(c) 2
cos ( 4) h sin ( 4) h
h 0
(d) –1
(cosh sinh cosh sinh)
2
h
h0
(cosh sinh cosh sinh) h
h0
2
2
2.
So LHL RHL 2 [Similar question was also asked in XE-2008 (1 mark)]
Example 2.49 [MT-2010 (2 marks)]: The lim sin 2 ax x 0
(a) a 2
(b) 0
Copyright © 2016 by Kaushlendra Kumar
sin x is
(c) 1
2
(d) Undefined
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
sin 2 ax
[2.27] 2
sin 2 ax a 2 x 2
sin ax x 2 2 Solution (a): lim lim a lim lim a 1 1 a 2 2 2 2 x 0 sin x x 0 sin x a x x 0 ax x 0 sin x 2
Example 2.50 [BT-2014 (1 mark)]: The limit of the function e 2t sin(t ) as t , is ………… Solution: lim e
2 t
sin(t ) lim e 2t lim sin(t ) 0 (any value between 1 to 1) 0
t
t
t
sin[ x] , [ x] 0 Example 2.51: If f ( x ) [ x ] , then find lim f ( x ) . x 0 0 , [ x] 0 Solution: In closed interval of x 0 at right hand side [ x ] 0 and at left hand side [ x ] 1. Also
sin[ x] , ( 1 x 0) [0] 0. Therefore function is defined as f ( x) [ x ] 0 , (0 x 1) Left hand limit lim f ( x ) lim
sin[ x ]
sin( 1)
[ x] 1 Right hand limit 0 , Hence, limit doesn’t exist. x 0
sin1c
x 0
Evaluation of Exponential and Logarithmic limits based on series expansion: The following standard forms are used in case f ( x) 0 , when x a
1 f ( x)
1x
lim 1 f ( x)
e lim 1 x
x a
x 0
x
e or lim 1 1 x e [This limit was asked in ECx
2014 (1 mark)] b f ( x) 1 ax 1 ex 1 lim log e b ( b 0 ) lim log e a or lim 1 xa x 0 x 0 f ( x) x x
lim
log 1 f ( x) f ( x)
x a
1 f ( x)
f ( x)
1 1 f (x)
e : Using binomial expansion we have, lim 1 f ( x) x a
x a
f ( x) 1!
x
x0
Proof of lim 1 f ( x )
1
log 1 x
1 lim
1 1 2 1 f ( x ) f ( x) f ( x )
1 1 1 3 1 2 f ( x ) f ( x ) f ( x ) f ( x)
2! 3! 1 1 1 1 f ( x ) 1 f ( x ) 1 2 f ( x ) 1 f (x) lim 1 f ( x) lim 1 1 1 e xa x a 2! 3! 2! 3! 1! The other forms can be proved in a similar manner. lim 1 x a
Evaluation of Exponential and Logarithmic limits based on Form 0 0 and Let L lim f ( x )
g ( x)
Le
g ( x) log e L log e lim f ( x) lim g ( x) log e lim f ( x )
xa
x a
and ∞ form
,
g ( x ) 1 loge lim f ( x ) lim x a xa
x a
x a
. 0 (1/ ) form
0
For 0 form we have, lim f ( x ) lim g ( x ) 0 , hence L e x a
x a
0 0 form e which can
be solved using previous discussed methods. For
0
0 (1/ ) form
form we have, lim f ( x ) and lim g ( x) 0 , hence L e x a
x a
0 0 form e
which can be solved using previous discussed methods.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.28]
Example 2.52 [CE-1999 (2 marks)]: Value of the function lim( x a )
( x a )
x a
is (d) a
(a) 1 (b) 0 (c) 0 Solution: As by putting the given limit we get 0 form; so let L ( x a) ( x a ) log L ( x a) log( x a) L e ( x a )log( x a ) lim ( x a ) log( x a )
lim L lim e ( x a ) log( x a ) e xa xa
xa
lim ( x a ) lim log( x a )
e xa
x a
e 0(limit tending to ) e 0 1
Form 1 1 Type 1: lim 1 f ( x)
f ( x)
e (which was discussed earlier), where lim f ( x ) 0 .
x a
x a
Type 2: L lim f ( x )
g ( x)
x a
L lim f ( x )
, where lim f ( x) 1 and lim g ( x ) x a
x a
1 f ( x ) 1 f ( x ) 1 g ( x )
g(x)
lim 1 f ( x) 1
xa
x a
1 f ( x ) 1
since lim 1 f ( x) 1
e as lim f ( x) 1
xa
x a
lim f ( x ) 1 g ( x )
1 f ( x ) 1 xa
L lim 1 f ( x) 1
xa
Hence L e
0 0 form
lim f ( x ) 1 g ( x )
e xa
lim f ( x ) 1 1/ g ( x )
e xa
which can be solved using previous discussed methods. 2n
Example 2.53 [CS-2010 (1 mark)]: What is the value of lim 1 1 n ? n
(b) e 2
(a) 0
(c) e 1/ 2
(d) 1
Solution (b): The given limit is of the form 1 so let L 1 1 n
Le
2 n ln 11 n
lim L e
lim 2 n ln 11 n
e
n
1 1 1 lim 2 n 2 3 n 2 n 3n
n
e
2n
ln L 2n ln 1 1 n
1 1 1 lim 2
n 1 2 n 3 n2
e 2
n
Example 2.54: Evaluate lim ( x 3) ( x 1)
x2
x
Solution:
x3 lim x x 1
x2
2 lim 1 x x 1
( x 1) ( x 2)2 2 ( x 1)
x 1 2 2 lim 1 x x 1
1 (2 x ) 2. 1 (1 x )
Example 2.55: If a, b, c, d are positive, then lim 1 1 ( a bx )
e
1 (2 x ) 2 lim x 1 (2 x )
e2
c dx
x
Solution: lim 1 1 ( a bx ) x
lim 1 1 (a bx) x
a bx
c dx
lim
a bx
x
e and lim x
( c dx ) ( a bx )
1 1 (a bx)
ed /b
c dx d a bx b
L’Hospital’s rule: If f ( x) and g ( x) be two functions of x such that (i) lim f ( x ) lim g ( x ) 0 ; x a
x a
(ii) Both are continuous at x a ; (iii) Both are differentiable at x a ; (iv) f ( x ) and g ( x ) are continuous at the point x a , then lim f ( x ) g ( x ) lim f ( x ) g ( x ) provided that g ( a ) 0 . x a
xa
The above rule is also applicable if lim f ( x ) and lim g ( x ) .
If lim f ( x ) g ( x ) assumes the indeterminate form 0 0 or and f ( x ) , g ( x ) satisfy all
x a
x a
x a
the condition embodied in L’Hospital rule, we can repeat the application of this rule on f ( x) g ( x) to get, lim f ( x) g ( x) lim f ( x) g ( x) . Sometimes it may be necessary to x a
x a
repeat this process a number of times till our goal of evaluating limit is achieved. Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.29]
Example 2.56 [AE-2007 (2 marks), PI-2008 (1 mark)]: lim (sin x ) (e x x ) x 0
(a) 10 (b) 0 (c) 1 (d) Solution: As by putting x 0 the given expression take the form 0 0 ; so differentiating Nr and Dr
separately w.r.t. x , we get lim (sin x ) (e x x) lim (cos x) (e x xe x ) 1 1 1 . x 0
x0
Example 2.57 [TF-2010 (2 marks)]: lim ( x 3 64) {log e ( x 3)} x 4
(a) 24 (b) 48 (c) 72 (d) 96 Solution (b): As by putting x 4 the given expression take the form 0 0 ; so differentiating Nr and Dr separately w.r.t. x , we get lim
x 4
x 3 64 log e ( x 3)
lim
x 4 1
3x2
2
2
lim 3( x 3) x 3(4 3)4 48
( x 3)
x4
Example 2.58 [ME-2012, PI-2012 (1 mark)]: lim (1 cos x ) x 2 is x 0
(a) 1/4 (b) 1/2 (c) 1 (d) 2 Solution (b): At x 0 the given expression take 0 0 form; so differentiating Nr and Dr separately
w.r.t. x , we get lim (1 cos x) x 2 lim (sin x ) 2 x . Again at x 0 the evaluated expression takes x 0
x 0
0 0 form; so again differentiating Nr and Dr separately w.r.t. x , we get lim (cos x ) 2 1 2 . x 0
Example 2.59 [ME-2014 (1 mark)]: lim ( x sin x ) (1 cos x ) is x 0
(a) 0 (b) 1 (c) 3 (d) not defined Solution (a): As by putting x 0 the given expression take the form 0 0 ; so differentiating Nr and Dr separately w.r.t. x , we get lim ( x sin x ) (1 cos x ) lim (1 cos x) (sin x ) . Now again by x 0
x0
putting x 0 the given expression take the form 0 0 ; so again differentiating Nr and Dr separately w.r.t. x , we get lim (sin x ) (cos x ) 0 1 0 x 0
Example 2.60 [ME-2014 (1 mark)]: lim (e 2 x 1) sin(4 x ) is equal to x 0
(a) 0 (b) 0.5 (c) 1 (d) 2 Solution (b): As by putting x 0 the given expression take the form 0 0 ; so differentiating Nr and
Dr separately w.r.t. x , we get lim (e 2 x 1) sin(4 x ) lim 2e 2 x 4 cos(4 x) (2 1) (4 1) 1 2 x 0
x 0
Sandwich theorem: Sandwich theorem helps in calculating the limits, when limits cannot be calculated using the above discussed methods. According to this theorem, if f ( x ) , g ( x ) and h( x) are any three functions such that, f ( x ) g ( x ) h( x ) x neighbourhood of x a and lim f ( x) lim h( x) l ( say ) , then lim g ( x ) l. This theorem is normally applied when the lim g ( x) x a
xa
x a
x a
can't be obtained by using conventional methods as function f ( x ) and h( x) can be easily found. Example 2.61: Evaluate lim (log x) [ x] x
Solution:
x 1 [ x] x
1
1
1
log x
log x
log x
Let
f ( x)
log x
and [ x] x 1 x [ x] x 1 x log x log x 1x log x 1 x h( x ) . Since, lim f ( x ) lim lim 0 and lim h( x ) lim lim 0 by x x x 1 x x x 1 x 1 x 1 x using L’Hospital rule. Hence by Sandwich theorem, lim (log x ) [ x ] 1 . x
.
x
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.30]
2.3.2 Continuity of a Function ‘Continuous’ means without any break or gap. If the graph of a function has no break, or gap or jump, then it is said to be continuous. A function which is not continuous is called a discontinuous function. Cauchy’s definition of continuity: A function f is said to be continuous at a point a of its domain D if for every 0 0 (dependent on ) s.t. x a f ( x ) f ( a) .
Formal definition of continuity: The function f ( x ) is said to be continuous at x a , in its domain if for any arbitrary chosen positive number 0 , we can find a corresponding number depending on such that f ( x ) f ( a) x for which 0 x a .
Condition for a function to be continuous at a point: A function f ( x ) is said to be continuous at a point x a of its domain iff lim f ( x) f ( a) lim f ( x ) lim f ( x) f ( a) , i.e. LHL = x a
x a
xa
RHL value of the function at ‘ a ’. [This point was asked in ME-2014 (1 mark)]
Continuity of a Function in Open and Closed Interval
Open interval: A function f ( x ) is said to be continuous in ( a, b) iff (i) it is continuous at every point in that interval; (ii) RHL at x a exists; (iii) LHL at x b exists. Closed interval: A function f ( x ) is said to be continuous in a closed interval [ a, b] iff, (i) f is continuous in ( a, b) ; (ii) f is continuous from the right at ‘ a ’ i.e. lim f ( x ) f ( a) ; (iii) f is x a
continuous from the left at ‘ b ’ i.e. lim f ( x ) f (b) x b
Example 2.62 [CE-2011 (2 marks)]: What should be the value of such that the function cos x ( 2) x if x 2 defined below is continuous at x 2 ? f ( x) 1 if x 2 (b) 2
(a) 0
(d) 2
(c) 1
Solution (c): The turning points for f ( x ) is x 2 and f ( 2) 1 (given). So, cos( 2 h) sinh LHL at x 2 lim f ( 2 h) lim lim h 0 h 0 ( 2) ( 2 h ) h 0 h RHL at x 2 lim f ( 2 h) lim h 0
h 0
cos( 2 h) ( 2) ( 2 h)
lim
sinh h
h 0
If f ( x ) is continuous at x 2 , then LHL RHL f ( 2) 1 Example 2.63 [CS-2013 (1 mark)]: Which one of the following functions is continuous at x 3 ? x3 2, x3 4, (b) f ( x ) (a) f ( x ) x 1, x3 8 x , x 3 ( x 3) 3, x 3
x 3,
(c) f ( x )
x3
(d) f ( x )
x 4, x 3
1 3
x 27
, x3
Solution (a): For option (a), LHL at ‘ x 3 ’ lim f (3 h) lim (3 h) 3 3 2 ; RHL at ‘ x 3 ’ h 0
h 0
lim f (3 h) lim(3 h 1) 2 and f (3) 2 , so at x 3 , LHL RHL f (3) ; hence f ( x ) is h 0
h 0
continuous at x 3 . For option (b), LHL at ‘ x 3 ’ lim f (3 h) lim 8 (3 h) 5 ; RHL at ‘ h 0
h 0
x 3 ’ lim f (3 h) lim 8 (3 h) 5 and f (3) 4 , so at x 3 , LHL RHL f (3) ; hence h 0
h 0
f ( x ) is not continuous at x 3 . For option (c), LHL at ‘ x 3 ’ lim f (3 h) lim (3 h) 3 6 h 0
Copyright © 2016 by Kaushlendra Kumar
h 0
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.31]
; RHL at ‘ x 3 ’ lim f (3 h) lim (3 h) 4 1 , so at x 3 , LHL RHL ; hence f ( x ) is not h 0
continuous
at
lim f (3 h) lim
h 0
x 3. 1
For
option
(d),
LHL
at
‘ x 3’
1
lim 3 ; RHL at ‘ x 3 ’ lim f (3 h) 3 2 h0 (3 h) 27 h0 h 9h 3h 1 1 lim lim 3 , so at x 3 , LHL RHL ; hence f ( x ) is not h 0 (3 h) 3 27 h 0 h 9 h 3h 2 continuous at x 3 . h 0
h0
Example 2.64 [MN-2014 (2 marks)]: The value of a , for which the function below is continuous at
2 x ax 2 , x 1
x 1 is f ( x )
4 x 3,
x 1 (a) –5 (b) 0 (c) 5 (d) 10 2 Solution (c): The turning points for f ( x ) is x 1 and f (1) 2 1 a 1 2 a . So,
LHL at x 1 lim f (1 h) lim 2(1 h) a(1 h) 2 lim(2 2 h a 2ha ah 2 ) 2 a h 0
h 0
h 0
RHL at x 1 lim f (1 h) lim 4(1 h) 3 lim(4 4h 3) 7 h 0
h 0
h 0
If f ( x ) is continuous at x 1 , then LHL RHL f (1) 2 a 7 a 5 [Similar question was also asked in CS-1996 (5 marks)]
Important points regarding continuity of a function Let f ( x ) , g ( x ) be two continuous functions at x a . Then cf ( x ) , f ( x ) g ( x) , f ( x ) g ( x) are continuous at x a , where c is any constant f ( x ) g ( x) is continuous at x a , provided g (a ) 0
If f ( x ) is continuous in a closed interval [ a, b] then it is bounded on this interval Intermediate value theorem: If f ( x ) is a continuous function defined on [ a, b] such that f (a ) and f (b ) are of opposite signs, then there is at least one value of x for which f ( x ) 0 , i.e., if f ( a ) 0, f (b) 0 c ( a, b) such that f ( c) 0 . If f ( x ) is continuous on [ a, b] and maps [ a, b] into [ a, b] then for some x [a, b] we have f ( x ) x . Discontinuous Function: A function f ( x ) is said to be discontinuous at a point x a of its domain, in the following cases: LHL and RHL exists finitely but unequal. For e.g., at the integral points of y [ x ] , both the LHL and RHL exists but both are not equal; hence y [ x ] is discontinuous at integral points. LHL as well as RHL both may exist, but either of the two or both may not be equal to f (a ) .
( x 2 4) ( x 2) , x 2 For e.g., for f ( x ) , lim f ( x ) lim f ( x ) 2 but f (2) 5 . x 2 5, x 2 x 2 f (a ) is not defined. For e.g., y 1 x is not defined at x 0 LHL or RHL or both may not exist. For e.g., both the LHL and RHL of y sin(1 x) does not exist at x 0
Types of Discontinuities: Basically there are two types of discontinuities:
Removable discontinuities: If lim f ( x ) exists but is not equal to f (a ) , then f ( x ) has a x a
removable discontinuity at x a and it can be removed by redefining f ( x ) for x a .
Non – removable discontinuities: If lim f ( x ) does not exist, then we cannot remove this x a
discontinuity. So this becomes a non – removable or essential discontinuity.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.32]
Example 2.65: Redefine f ( x) [ x ] [ x] in such a way that it becomes continuous for x (0, 2) . Solution: Here lim f ( x) 1 but f (1) 0 . Hence, f ( x ) has a removable discontinuity at x 1 . If x 1
[ x ] [ x ], we redefine f ( x ) as, f ( x ) 1,
x (0, 2) {1} x 1
. Now f ( x ) is continuous for x (0, 2) .
Continuity of Composite Function: If u f ( x ) is continuous at x a, and y g (u ) is continuous at u f ( a) , then the composite function y ( gof )( x ) g f ( x ) is continuous at x a . If f is continuous at x c and g is discontinuous at x c , then the following result holds: f g and f g are discontinuous f g may be continuous If f and g are discontinuous at x c , then f g and f g may still be continuous. Point functions (domain and range consists one value only) is not a continuous function. Example 2.66: Find the points of discontinuity of y 1 (u 2 u 2) where u 1 ( x 1) . Solution: The function u f ( x ) 1 ( x 1) is discontinuous at the point x 1. The function y g ( x ) 1 (u 2 u 2) 1 (u 2) (u 1) is discontinuous at u 2 and u 1 when u 2 1 ( x 1) 2 x 1 2 ,
when u 1 1 ( x 1) 1 x 2 . Hence, the composite y g{ f ( x )} is discontinuous at three points 1 2 ,1, 2 .
Limit and Continuity of a Function of two variables: When we extend the continuity of a function of one variable to functions of two variables, we will see that the domain of functions of two 2 2 variables is a subset of , in other words it is a set of pairs. A point in is of the form ( x , y ) . So, the equivalent of x a will be ( x, y ) ( a, b) . For functions of three variables, the equivalent of x a will be ( x, y, z ) (a, b, c) , and so on. x could only approach a from two directions, from the left or from the right, but ( x , y ) can approach ( a, b) from infinitely many directions. In fact, it does not even have to approach ( a, b) along a straight path as shown in Fig. 2.23. For functions of several variables, we would have to show that the limit along every possible path exists and are the same. The problem is that there are infinitely many such paths. To show a limit does not exist, it is still enough to find two paths along which the limits are not equal. In view of the number of possible paths, it is not always easy to know which paths to try. We can try some suggested following paths: (i) Horizontal line through ( a, b) , the equation of such a path is y b ; (ii) Vertical line through ( a, b) , the equation of such a path is x a ; (iii) Any straight line through ( a, b) , the equation of the line with slope m through ( a, b) is y mx b am ; (iv) Quadratic paths; for e.g., a typical quadratic paths through (0, 0) is y x 2 .
Figure 2.23: Limit of a function in two variables
Properties of Limits of Functions of two (or more) variables: Let us assume that L , M and k are real numbers and that lim f ( x, y ) L and lim g ( x, y ) M , then ( x , y ) ( a ,b )
( x , y ) ( a , b )
lim
xa
lim
c c , c is any constant
lim
k f ( x, y ) kL
( x , y ) ( a ,b ) ( x , y ) ( a ,b ) ( x , y ) ( a ,b )
lim { f ( x, y) g ( x, y )} L M , if M 0
( x , y ) ( a ,b )
Copyright © 2016 by Kaushlendra Kumar
lim
( x , y ) ( a , b )
y b
lim { f ( x, y ) g ( x, y)} L M
( x , y ) ( a , b )
lim { f ( x, y)}{ g ( x, y)} LM
( x , y ) ( a ,b )
lim
{ f ( x, y )}r s Lr s , s 0
( x , y ) ( a , b )
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.33]
Method for evaluating the limits of two variables
Direct substitution method: If by direct substitution of the point in the given expression we get a finite number, then the number obtained is the limit of the given expression. For e.g., lim ( x 2 y ) ( x 4 y 2 ) (12 1) (14 12 ) 1 2 ( x , y ) (1,1)
Factorization method: In this method, numerator and denominator are factorised. The common factors are cancelled and the rest outputs the results. For e.g., 3 3 2 2 x y ( x y )( x xy y ) lim lim lim ( x 2 xy y 2 ) 0 ( x , y ) (0,0) x y ( x , y ) (0,0) ( x , y ) (0,0) x y Rationalization method: Rationalisation is followed when we have fractional powers (like 1 2 ,1 3 etc.) on expressions in numerator or denominator or in both. After rationalisation the terms are factorised which on cancellation gives the result. For e.g.,
x 2 xy
x( x
y )( x
y)
lim x( x y ) 0 ( x , y ) (0,0) x y ( x, y ) (0,0) x y Limit along a path: We usually use this method to prove the limit of a function in two variable does not exists. lim
( x , y ) (0,0)
lim
Example 2.67 [XE-2008 (1 mark)]:
( x 4 xy ) ( x 3 y 3 ) is
lim
( x , y ) (0,0)
(a) 0 (b) 1 (c) –1 (d) does not exist Solution (d): Limit along the path x 0 : First we find what the function becomes along this path
( x4 xy ) ( x3 y 3 )
x 0
0 ( y 3 ) 0
lim
( x , y ) (0,0)
(x
4
xy ) ( x3 y 3 )
lim
(0, y ) (0,0)
00
Limit along the path y x : First we find what the function becomes along this path
x 4 xy 3
x y
3 y x
x4 x 2 2x
3
x 4 xy
lim
( x , y ) (0,0)
3
x y
3
2x
( x , x ) (0,0)
lim( x 2) 0 and lim 1 2 x does not exists. So
lim
x 0
x 0
x4 x2
lim
( x , x ) (0,0)
3
(x
x 1 . ( x , x ) (0,0) 2 2x
As
lim
2) (1 2 x ) does not exist. So limit
along one path exists, but along another path it does not exists. So the given limit does not exists.
Change of coordinate: Sometimes changing coordinates may be useful in evaluating the limits of two variables. For e.g., putting x r cos and y r sin in
( x, y ) (0, 0)
limit
lim r 3 (cos3 sin 3 ) r 0
changes
r
2
lim
( x , y ) (0,0)
r 0
to
(x
and
3
y 3 ) ( x 2 y 2 ) , the
thus
we
get
(cos 2 sin 2 ) lim r (cos3 sin 3 ) 0 r 0
Sandwich Theorem: This theorem is also useful in evaluating the limits of two variables. For e.g., in evaluating the lim ( x 2 y ) ( x 2 y 2 ) , we need to find a function g ( x, y ) such that ( x , y ) (0,0)
0 f ( x, y ) 0 g ( x , y ) ; thus
(as x 2 ( x 2 y 2 ) )
lim
( x , y ) (0,0)
lim
( x , y ) (0,0)
x2 y
f ( x, y ) 0
0
x2 y
lim
( x , y ) (0,0)
( x2 y) ( x 2 y 2 ) 0
lim
x2 y2
( x , y ) (0,0)
2
x y
2
0
x2 y x2 y2
lim
( x , y ) (0,0)
2
y 0
x y x2 y 2
2
lim
( x , y ) (0,0)
x y x2 y 2
x2 y 2
x y
2
y
0
( x2 y) ( x 2 y 2 ) 0
A function f ( x, y ) is said to be continuous at a point ( a, b) if the following is true: ( a, b) is in the domain of f
lim
( x , y ) ( a ,b )
Copyright © 2016 by Kaushlendra Kumar
f ( x, y) exists
lim
( x , y ) ( a ,b )
f ( x, y ) f ( a , b )
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.34]
2.3.3 Differentiability of a Function Let y f ( x) be a given function. If at some point, abscissa is x1 and at other point abscissa is x2 , and ordinate can be represented by y1 and y2 respectively at those points. Change in ‘ y ’ is given as, y y2 y1 ; change
in
‘ x’
is
given
x x2 x1 ;
as,
then
y f ( x x) f ( x1 ) , clearly y can be positive, negative or may even be zero. The differential coefficient of y f ( x) , with respect to x is defined as the limiting Figure 2.24: Derivative of ( ) point P value of y x as x tends to zero. It is usually denoted by dy dx or f ( x ) symbolically. Now, consider the function f ( x ) defined on an open interval (b, c ) , let P ( a , f ( a )) be a point on the curve y f ( x) and let Q ( a h, f ( a h)) and R (a h, f ( a h)) (where h is very small positive number) be two neighbouring points on the left hand side and right hand side respectively of the point P as shown in Fig. 2.24. Then slope of chord PQ f (a h) f ( a ) ( a h) a f ( a h) f ( a) ( h) , and slope of chord
PR f ( a h) f ( a ) ( a h a) f ( a h) f ( a) h . As h 0 , point Q and R both tends to P from left hand and right hand respectively. Consequently, chords PQ and PR becomes tangent at point P . Thus, slope of the tangent at point P , which is limiting position of the chords drawn on the left hand side of point P , which is denoted as ‘Left Hand Derivative (LHD),’ i.e., LHD lim f ( a h) f ( a ) ( h) lim (slope of chord PQ) lim (slope of chord PQ). Similarly, h 0
h0
Q P
slope of the tangent at point P , which is the limiting position of the chords drawn on the right hand side of point P , which is denoted as ‘Right Hand Derivative (RHD),’ i.e., RHD lim f ( a h) f (a ) h lim slope of chord PR lim slope of chord PR h 0
h 0
R P
A function f ( x ) is said to be differentiable (finitely) at x a if LHD = RHD = finite, i.e., lim f ( a h) f ( a ) h lim f (a h) f (a ) ( h) finite and the common limit is called the h 0
h 0
derivative of f ( x ) at x a , denoted by f ( a) . f (a ) lim f ( x ) f ( a) ( x a ) { x a from x a
the left as well as from the right}. If a function f ( x ) is differentiable at in ( a, b) then it is continuous for all x ( a, b) . Proof: Given that f (c ) exists, we must show that lim f ( x ) f (c ) , or lim f (c h) f (c ) . x c
h 0
If h 0 , then, f (c h ) f (c ) f (c h ) f (c ) f (c ) f (c h ) f (c ) h h lim f (c h ) lim f (c ) lim f (c h ) f (c) h lim h lim f ( c) f (c) 0 f (c ) h 0
h 0
h 0
h 0
h 0
Similar argument with one – sided limits show that if f has a derivative from one side (right or left) at x c , then f is continuous from that side at x c . If f ( x ) is differentiable at x a LHD and RHD both exists and LHD RHD There is a unique tangent (which happens when the curve is smooth) at point P . A function whose graph is not smooth will fail to have a derivative where the graph has a corner, where the one sided derivative differ, as shown in Fig. 2.25a a cusp, where the slope of PQ approaches from one side and from the other, as shown in Fig. 2.25b. a vertical tangent, where the slope of PQ approaches from both sides or approaches from both sides, as shown in Fig. 2.25c a discontinuity, as shown in Fig. 2.25d.
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Engineering Mathematics
Chapter 2: Calculus
(b)
(a)
[2.35]
(d)
(c)
Figure 2.25: Types of Discontinuities
A function f ( x ) defined in an open interval ( a, b) is said to be differentiable or derivable in open interval ( a, b) if it is differentiable at each point of ( a, b) . A function f : [ a, b] R is said to be differentiable in [ a, b] if f ( x ) exists for every x such that a x b i.e., f is differentiable in ( a, b) Right hand derivative of f at x a exists Left hand derivative of f at x b exists
Some standard results on differentiability
Every polynomial and constant function is differentiable at each x R .
The exponential function a , a 0 is differentiable at each x R . The logarithmic function is differentiable at each point in its domain. Trigonometric and inverse trigonometric functions are differentiable in their domains. The sum, difference, product and quotient of two differentiable functions is differentiable. The composition of differentiable function is a differentiable function. If f is derivable in the open interval ( a, b) and also at the end points ‘ a ’ and ‘ b ’, then f is said to be derivable in the closed interval [ a, b] . A function f is said to be a differentiable if it is differentiable at every point of its domain. If a function is differentiable at a point, then it is continuous also at that point, i.e. Differentiability Continuity, but the converse need not be true. If a function ‘ f ’ is not differentiable but is continuous at x a , it geometrically implies a sharp corner or kink at x a . If f ( x ) is differentiable at x a and g ( x ) is not differentiable at x a , then the product function f ( x ) g ( x) can still be differentiable at x a . If f ( x ) and g ( x ) both are not differentiable at x a then the product function f ( x ) g ( x) can still be differentiable at x a . If f ( x ) is differentiable at x a and g ( x ) is not differentiable at x a then the sum function f ( x ) g ( x) is also not differentiable at x a . If f ( x ) and g ( x ) both are not differentiable at x a , then the sum function may be a differentiable function.
x
Example 2.68 [ME-2002 (1 mark)]: Which of the following function is not differentiable in the domain [ 1,1] ? (b) f ( x ) x 1 (c) f ( x ) 2 (d) f ( x ) max( x, x ) (a) f ( x) x 2 Solution (d): As the graph for the function in option (d) has a sharp turn at x 0 , so it is not differentiable at x 0 . All other functions in options (a), (b) and (c) have no sharp turn for all x , so they are differentiable for all x . sin x x 0
Example 2.69 [AE-2008 (1 mark)]: The function defined by f ( x ) 0
3x2
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x0 x0
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Engineering Mathematics
Chapter 2: Calculus
[2.36]
(a) is neither continuous nor differentiable at x 0 (b) is continuous nor differentiable at x 0 (c) is differentiable but not continuous at x 0 (d) is continuous but not differentiable at x 0 Solution (d): For continuity and differentiability of f ( x ) , the turning point of f ( x ) is x 0 . As
f (0) 0 and LHL x 0 lim f (0 h) lim sin(0 h) lim( sinh) 0 ; h 0
h 0
h 0
2
2
RHL x 0 lim f (0 h) lim 3(0 h) lim 3h 0 h 0
h0
h 0
So, LHL x 0 RHL x 0 f (0) 0 f ( x ) is continuous at x 0 . Now LHD x 0 lim f (0 h) f (0) ( h) lim sin(0 h) 0 ( h) lim (sinh) h 1 h 0
h0
h0
2
RHD x 0 lim f (0 h) f (0) h lim 3(0 h) 0 h 0
h0
h lim (3h 2 ) h lim 3h 0 h 0
h 0
So, RHD x 0 RHD x 0 f ( x ) is not differentiable at x 0 Example 2.70 [ME-2010 (1 mark)]: The function y 2 3 x (a) is continuous x R and differentiable x R (b) is continuous x R and differentiable x R except at x 3 2 (c) is continuous x R and differentiable x R except at x 2 3 (d) is continuous x R except at x 3 and differentiable x R Solution (c): As the graph of y f ( x) , where f ( x ) is a linear polynomial in x , is continuous for all x ; but it has a sharp turn at the point where f ( x ) 0 , so y f ( x) is not differentiable the point where f ( x ) 0 . In the given question f ( x ) 2 3x , so 2 3x is continuous for all x but it is not differentiable at those x where 2 3 x 0 x 2 3 . [Similar questions were also asked in CS-2007, CS-1998, ME-1995, ME-1997, ME-2001, ME2012, AE-2013, PI-2012 (1 marks)] (sin x ) x , x 0 Example 2.71 [XE-2010 (2 marks)]: Let f ( x ) . Then x0 1, (a) f is not continuous at x 0 (b) f is continuous at x 0 but not differentiable at x 0 (c) f is differentiable at x 0 and f (0) 0 (d) f is differentiable at x 0 and f (0) 1 Solution (c): For continuity and differentiability of f ( x ) , the turning point of f ( x ) is x 0 . As
f (0) 1 and LHL x 0 lim f (0 h) lim sin(0 h) (0 h) lim sinh h 1 ; h 0
h 0
h 0
RHL x 0 lim f (0 h) lim sin(0 h) (0 h) lim (sinh) h 1 h 0
h 0
h0
So, LHL x 0 RHL x 0 f (0) 1 f ( x ) is continuous at x 0 . Now
LHD x 0 lim f (0 h) f (0) ( h) lim sin(0 h) (0 h) 1 ( h) lim sinh h ( h2 ) h0
h 0
LHD x 0 lim h 0
h (h
h
LHD x 0 lim
h 0
h3
5
h
h 0
3!) ( h 5!) h
2
5
( h ) lim ( h 3!) ( h 5!) ( h 2 ) h 0
3
0 . Similarly,
3! 5! f (0 h) f (0)
h 0
RHD x 0 lim
3
lim
sin(0 h)
h0
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(0 h) 1 h
lim h 0
sinh h h2
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LHD x 0 lim
h (h
Chapter 2: Calculus
3
3!) ( h5 5!) h
lim
[2.37]
( h3 3!) ( h5 5!)
h 0 h2 h2 h h3 LHD x 0 lim 0 . LHD x 0 RHD x 0 0 f ( x ) is differentiable at x 0 . h0 3! 5! sin(1 x ), x 0 Example 2.72 [AE-2014 (1 mark)]: The function given by f ( x ) is x0 0, (a) unbounded everywhere (b) bounded and continuous everywhere (d) continuous and differentiable everywhere (c) bounded but not continuous at x 0 Solution (c): As ‘ sin ’ function is always bounded whose values lies between [ 1,1] . For the given function the turning point is x 0 . So LHL x 0 lim f (0 h) lim sin 1 (0 h) limsin(1 h) ; as 1 h sin(1 h) lies between h 0
h 0
h 0
h 0
[ 1,1] ; so lim sin(1 h) has oscillating limit and thus it does not exists. Similarly, h 0
RHL x 0 lim f (0 h) lim sin 1 (0 h) lim sin(1 h) ; as 1 h sin(1 h) h 0
h 0
h 0
lies between
[ 1,1] ; so lim sin(1 h) has oscillating limit and thus it does not exists. Hence the given function is not h 0
continuous at x 0 . Example 2.73: Find the value of the derivative of x 1 x 3 at x 2 is
( x 1) ( x 3) , x 1 2 x 4 , x 1 Solution: Let f ( x ) x 1 x 3 ( x 1) ( x 3) , 1 x 3 2 , 1 x 3 ( x 1) ( x 3) , x 3 2 x 4 , x 3 Since, f ( x ) 2 for 1 x 3 . Therefore f ( x) 0 for all x (1,3) . Hence, f ( x) 0 at x 2 . Example 2.74: If 5 f ( x) 3 f (1 x) x 2 and y x f ( x ) then find dy dx x 1 Solution: 5 f ( x) 3 f (1 x) x 2 …(i). Replacing x by 1 x in (i), 5 f (1 x) 3 f ( x) (1 x) 2 …(ii). On solving equation (i) and (ii), we get, 16 f ( x ) 5 x (3 x ) 4 , 16 f ( x ) 5 (3 x 2 ) y x f ( x ) dy dx f ( x ) x f ( x) (1 16){5 x (3 x ) 4} x (1 16){5 (3 x 2 )}
at x 1, dy dx (1 16)(5 3 4) (1 16)(5 3) 7 8 .
Some Standard Differentiation Formulas: In the following formulas, we have D d dx . D ( x n ) nx n 1 , x R, n R, x 0 D( a x ) a x log a , for a 0
D (sin x ) cos x
D (e x ) e x
D (cos x ) sin x
D (log a x ) 1 ( x log a ) , for x , a 0 , a 1
D(tan x) sec 2 x
D (sin 1 x ) 1
D (sec x ) sec x tan x
D (cos 1 x ) 1
D(cosec x ) cosec x cot x
D (tan 1 x ) 1 (1 x 2 ) , for x R
D (cot x ) cosec 2 x
D (sec 1 x) 1
D(cot 1 x) 1 (1 x 2 ) , for x R D (cosec h x) (cosec h x )( cot h x )
D (cosec 1 x ) 1
D (sin h 1 x ) 1
x
x 2 1 , for x 1
x
x 2 1 , for x 1
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1 x 2 , for 1 x 1 1 x 2 , for 1 x 1
1 x2
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Chapter 2: Calculus
[2.38]
D (sin h x ) cos h x
D (cos h 1 x) 1
D (cos h x ) sin h x
D (tan h 1 x ) 1 ( x 2 1)
D (tan h x ) sec h 2 x
D (cot h 1 x ) 1 (1 x 2 )
D (cot h x) cosec h 2 x
D (sec h 1 x) 1 x 1 x 2
D (sec h x ) sec h x tan h x
D (cosec h 1 x ) 1 x 1 x 2
x2 1
3
Example 2.75 [IN-2007 (2 marks)]: Consider the function f ( x ) x , where x is real. Then the function f ( x ) at x 0 is (a) continuous but not differentiable (b) once differentiable but not twice (c) twice differentiable but not thrice (d) thrice differentiable
x 3 , x 0 x, x 0 3 Solution (c): x 0, x 0 f ( x ) x 0, x0. x, x 0 x3 , x 0 3
3
As, LHL x 0 lim f (0 h) lim{(0 h) } 0 ; RHL x 0 lim f (0 h) lim{(0 h) } 0 ; and h 0
f (0) 0 .
So,
h0
LHL x 0 RHL x 0 f (0) 0
h0
the
h 0
function
is
continuous
3x 2 , x 0 6 x, x 0 6, x 0 f ( x ) 0, x 0 f ( x ) 0, x 0 f ( x ) 0, x 0 . 3x 2 , x 0 6 x, x 0 6, x 0 f ( x)
LHD x 0 3(0) 2 0 RHD x 0 3(0) 2 0
for f ( x )
LHD x 0 6 RHD x 0 6
f ( x ) exists; for f ( x)
at
x 0;
Now,
LHD x 0 6(0) 2 0 RHD x 0 6(0) 2 0
for
f ( x) exists;
f ( x) does not exists. So, f ( x ) is twice differentiable but not thrice.
Example 2.76 [XE-2014 (1 mark)]: Let the function
f :[0,5] R
be defined by
0 x 1 2 x 5, 2 f ( x ) 2 x 5, 1 x 2 . The number of points where f is not differentiable in (0, 5) , is 3 2 x 3 23 3, 2 x 5
_____.
2, 0 x 1 Solution: For the given function, f ( x ) 4 x, 1 x 2 . Now, the turning points for f ( x ) are 2 x2 , 2 x 5 x 1, 2 . As LHD x1 2 and RHD x1 4 1 4 LHD x 1 RHD x 1 at x 1 f ( x ) does not 2 exists. Similarly, LHD x 2 4 2 8 and RHD x2 2 2 8 LHD x 2 RHD x 2 at x 2
f ( x ) exists. f ( x ) is not differentiable at only one point, i.e., at x 1 .
Some standard substitutions for evaluating differentiation of given expression: Function Substitutions Function Substitutions
a2 x 2
x a sin or a cos
(a x ) ( a x )
x a cos 2
x2 a2
x a tan or a cot
(a 2 x 2 ) (a 2 x2 )
x a cos 2
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2
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Engineering Mathematics
Chapter 2: Calculus
x 2 a2
x a sec or a cos ec
ax x 2
x a sin
x ( a x)
x a tan
[2.39]
2
2
2
2
( x a)( x b )
x a sec b tan
2
( x a)(b x )
x a cos b sin
2
x ( a x)
x a sin
2
Theorems on Differentiation: Let f ( x), g ( x) and u ( x) be differentiable functions
If at all points of a certain interval. f ( x ) 0, then the function f ( x ) has a constant value within this interval. Chain rule Case I: If y is a function of u and u is a function of x , then derivative of y with respect to dy dy du dy du or y f (u ) f (u ) x is dx du dx dx dx Case II: If y and x both are expressed in terms of t , y and x both are differentiable with dy dy dt respect to t then . dx dx dt d
Sum and difference rule: Using linear property
Product rule: d d d ( f ( x) g ( x)) f ( x) g ( x ) g ( x ) f ( x) dx dx dx d dw du dv (u.v.w.) u.v. v.w. u.w. dx dx dx dx d d Scalar multiple rule: (k f ( x)) k f ( x) dx dx d f ( x) d d Quotient rule: g ( x ) ( f ( x )) f ( x ) ( g ( x )) dx dx dx g ( x )
dx
( f ( x) g ( x))
d dx
( f ( x ))
d dx
( g ( x))
2 ( g ( x )) , provided g ( x ) 0
Example 2.77 [ME-2005 (2 marks)]: With a 1 unit change in b , what is the change in x in the solution of the system of equation x y 2 , 1.01x 0.99 y b ? (a) Zero (b) 2 units (c) 50 units (d) 100 units Solution (c): Solving the given two equations, we get 0.02 x b 0.02dx db dx db 0.02 50(db) 50 units (as db 1 unit is given). Example 2.78 [MN-2007 (2 marks)]: The function f ( x ) and g ( x ) satisfy f ( x 0) 3 , f ( x 0) 5 , g ( x 0) 2 , g ( x 0) 10 . The value of ( d dx ) f ( x ) g ( x) x 0 is (a) –35.0 (b) –5.0 (c) 0.5 (d) 5.0 Solution (d): d f ( x) g ( x) f ( x) f ( x ) g ( x ) g (0) f (0) f (0) g (0) 2 ( 5) 3 10 5 2 2 2 dx g ( x ) x 0 {g ( x)} {g (0)} 2 x 0 Example 2.79 [IN-2008 (1 mark)]: Given y x 2 2 x 10 , the value of ( dy dx) x 1 is equal to (a) 0 (b) 4 (c) 12 Solution (b): dy dx 2 x 2 (dy dx) x 1 2 1 2 4
(d) 13
Example 2.80 [PI-2010 (1 mark)]: If f ( x) sin x , then the value of df dx at x 4 is (a) 0
(b) 1
2
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(c) 1
2
(d) 1
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Chapter 2: Calculus
x 0,
Solution (c): As for
x x f ( x ) sin x sin( x ) sin x
df dx cos x for all x 0 ; hence df dx x
⁄
Relation between
[2.40]
4
cos( 4) 1
for all
x 0 . So
2.
⁄
and
: Let x and y be two variables connected by a relation of the form f ( x , y ) 0 . Let x and y be a small change in x and y , respectively. dy dx lim y x , dx dy lim ( x y ) . ( y x).(x y ) 1 lim (y x ).( x y ) 1 y 0
x 0
x 0
lim ( y x ). lim ( x y ) 1 [ x 0 y 0 ] ( dy dx) (dx dy) 1 dy dx 1 {dx dy}
x 0
y 0
Differentiation of implicit functions: If y is expressed entirely in terms of x , then we say that x
y is an explicit function of x . For e.g., y sin x , y e , etc. If y is related to x but cannot be expressed in the form of y f ( x) but can be expressed in the form f ( x , y ) 0 , then we say that y
is an implicit function of x . In case of implicit differentiation, dy dx may contain both x and y . Working rule 1: Differentiate each term of f ( x , y ) 0 with respect to x . Collect the terms containing dy dx on one side and the terms not involving dy dx on the other side. Express dy dx as a function of x or y or both.
Working rule 2: If f ( x, y ) constant, then dy dx (f x) (f y) , where f x and f x are partial differential coefficients of f ( x, y ) with respect to x and y respectively. Partial differential coefficient of f ( x, y ) with respect to x means the ordinary differential coefficient of f ( x, y ) with respect to x keeping y constant.
Example 2.81: If sin( x y ) log( x y ) , then find dy dx . d
Solution by Rule 1:
cos( x y )
dx
sin( x y )
d
1 dy log( x y ) cos( x y ) 1 dx dx x y
dy 1 dx
1 dy dy dy 1 0 1 . 1 0 dx dx x y dx
Solution by Rule 2: It is an implicit function, so
dy dx
f x
f y
cos( x y ) 1 ( x y ) cos( x y ) 1 ( x y )
1 .
Logarithmic differentiation: If differentiation of an expression or an equation is done after taking log on both sides, then it is called logarithmic differentiation. This method is useful for the function having the forms: (i) y f ( x )
g ( x)
(ii) y f1 ( x ). f2 ( x)... g1 ( x). g 2 ( x )... where gi ( x) 0 , where
i 1, 2, 3, , fi ( x) and gi ( x) both are differentiable
g ( x)
1 dy y dx
g ( x)
Case I: y [ f ( x] where f ( x ) and g ( x ) are functions of x . Let y [ f ( x)] . Taking logarithm of both the sides, we have log y g ( x ).log f ( x ) . Differentiating w.r.t. x , we get g ( x ).
Case
II:
1
df ( x )
f ( x ) dx
y
log{ f ( x)}.
f1 ( x). f 2 ( x) g1 ( x).g 2 ( x)
dg ( x ) dx
Taking
dy dx
[ f ( x)
logarithm
g(x)
of
dg ( x ) g ( x ) df ( x ) f ( x ) dx log[ f ( x)] dx .
both
the
sides,
we
have
log y log[ f1 ( x)] log[ f2 ( x)] log[ g1 ( x)] log[ g 2 ( x)] . Differentiating with respect to x , we get 1 dy y dx
f1( x) f1 ( x)
f2( x) f2 ( x)
g1 ( x) g 2 ( x)
g 2 ( x) g 2 ( x)
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dy dx
f1( x )
y
f1 ( x )
f 2( x ) f2 ( x)
g1 ( x ) g1 ( x )
g 2 ( x )
.
g 2 ( x)
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[2.41]
Example 2.82 [AG-2014 (2 marks)]: If y x x , then dy dx is (a) x x ( x 1) Solution
(c) x x (1 log x)
(b) x x 1 y x x log y x log x ,
(c):
differentiating
both
(d) e x (1 log x ) sides
x,
w.r.t.
we
get
x
(1 y )( dy dx ) log x ( x x) (1 log x) dy dx x (1 log x )
Differentiation of parametric functions: Sometimes x and y are functions of a single variable, for e.g., x (t ) , y (t ) are two functions and t is a variable, so x and y are called parametric functions and t is called the parameter. To find dy dx , we first obtain the relationship between x and y by eliminating the parameter t and then we differentiate it w.r.t. x . But every time it is not convenient to eliminate the parameter, so
dy dx
can be obtained by the following formula
dy dx
dy dt
.
dx dt
Example 2.83 [ME-2004 (1 mark)]: If x a ( sin ) and y a(1 cos ) then dy dx (a) sin( 2) (b) cos( 2) (c) tan( 2) (d) cot( 2) Solution (c): dx d a (1 cos ) 2a sin 2 ( 2) and dy d a sin 2 a sin( 2) cos( 2) dy dx ( dy d ) ( dx d ) {2a sin( 2) cos( 2)} {2a sin 2 ( 2)} tan( 2)
Differentiation of infinite series: If y is given in the form of infinite series of x and we have to find out dy dx then we remove one or more terms, it does not affect the series.
If y 2y
dy dx
1 dy y dx
f (x)
f ( x ) , then y
f ( x)
f ( x )
If y f ( x)
f ( x)
dy dx
f ( x ) f ( x )
y f ( x ) f ( x)
dy dx
f ( x ) 2 y 1 y
then y f ( x) log y y log f ( x)
log f ( x )
dy
f ( x)
1
dy
y 2 f ( x )
f ( x )[1 y log f ( x )] 1 dy yf ( x ) y f ( x) y dx 2 y f ( x )
dx
1
If y f ( x )
2 f ( x ) y y f ( x) y
dx
f ( x) ....
Example 2.84 [AG-2010 (2 marks)]: The derivative of y y 0 is (a) –1 (b) 0 (c) 1 Solution
(a):
y
x
x x ........to
x x x with respect to x at
(d) 2
y
x y
y2 x y
2 y ( dy dx ) 1 ( dy dx ) ( dy dx )(2 y 1) 1 dy dx 1 (2 y 1) ( dy dx ) y 0 1
Differentiation of composite function: Suppose function is given in form of fog ( x) or f [ g ( x )] . d
f [ g ( x)] f [ g ( x )] g ( x ) dx Example: If g is inverse of f and f ( x) 1 (1 x n ) , then find g ( x ) . d Solution: Since g is inverse of f fog ( x ) x x { fog ( x )} 1 x f ( g ( x )).g ( x ) 1 dx n n f {g ( x)} 1 g ( x) 1 1 [ g ( x )]n 1 g ( x ) f ( x) 1 (1 x ) g ( x) 1 [ g ( x)]
Then differentiate by applying chain rule
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Engineering Mathematics
Chapter 2: Calculus
[2.42]
Differentiation of a function w.r.t. another function: Let u f ( x ) , v g ( x) be two functions of x . Then, to find the derivative of f ( x ) w.r.t. g ( x ) i.e., to find du dv we use the following du
du / dx
. Thus, to find the derivative of f ( x ) w.r.t. g ( x ) we first differentiate both dv dv / dx w.r.t. x and then divide the derivative of f ( x ) w.r.t. x by the derivative of g ( x ) w.r.t. x . formula
Example 2.85: Find the differential coefficient of tan 1 2 x (1 x 2 ) w.r.t. sin 1 2 x (1 x 2 ) .
Solution: Let y1 tan 1 2 x (1 x 2 ) and y2 sin 1 2 x (1 x 2 ) . Putting x tan 1
1
2
y1 tan tan 2 2 2 tan x dy1 dx 2 (1 x ) , and y2 sin 1 sin 2 2 tan 1 x dy2 dx 2 (1 x 2 ) . Hence dy1 dy2 (dy1 dx) (dy2 dx ) 1
Successive Differentiation or Higher Order Derivatives: If y is a function of x and is differentiable w.r.t. x , then its derivative dy dx can be found which is known as derivative of first order. If the first derivative dy dx is also a differentiable function, then it can be further 2
2
differentiated with respect to x and this derivative is denoted by d y dx which is called the second 2
2
derivative of y with respect to x further if d y dx is also differentiable then its derivative is called 3
3
third derivative of y which is denoted by d y dx . Similarly nth derivative of y is denoted by d n y dx n . All these derivatives are called as successive derivative and this process is known as
successive
differentiation.
2
2
n
We
n
dy dx , d y dx , , d y dx , or
also
use
the
symbols:
y1 , y2 ,..., yn ,...
or
n
f ( x), f ( x),..., f ( x),... for the successive derivatives of
y f ( x) . If y f ( x) , then the value of the nth order derivative at x a is usually denoted by
d
n
y dx n
n
xa
n or ( yn ) x a or ( y ) x a or f (a) .
Example 2.86 [AE-2012 (2 marks)]: The nth derivative of the function y 1 ( x 3) is (a)
1n n ! n 1 x 3
Solution (a): y
(b)
1 ( x 3) 2
1n1 n ! n 1 x 3 y
1 2 ( x 3)3
(c)
y
1n (n 1)! n x 3 1 2 3 ( x 3) 4
y(n)
(d)
1n n ! n x 3
( 1) n n ! ( x 3) n 1
Example 2.87 [TF-2007 (2 marks)]: If yn d n y dx n and y ( x 2 1) n , then the expression ( x 2 1) yn 2 2 xyn 1 is equal to
(c) n( n 1) yn (d) n( n 1) yn (a) ( n 2 1) yn (b) ( n 2 1) yn Solution: The easiest way of solving these type of question is to choose n 1. Thus we have to find ( x 2 1) y3 2 xy2 in terms of y1 for y x 2 1 . As for y x 2 1 , y1 2 x , y2 2 and y3 0 . So ( x 2 1) y3 2 xy2 4 x . Now we have to see from the given option which one is equal to 4x for n 1
. So for option (a) we have (12 1) y1 2 y1 4 x ; for option (b) we have (12 1) y1 0 ; for option (c) we have 1(1 1) y1 0 ; for option (d) we have 1(1 1) y1 4 x . As choosing n 1 we came to know that option (b) and (c) are not correct; on the other hand we have option (a) and (d) are seem to be correct, but it is not possible that both are correct. So now we will choose n 2 , so we have to find ( x 2 1) y4 2 xy3 in terms of y 2 for y ( x 2 1) 2 x 4 2 x 2 1 . Thus for n 2 , y1 4 x 3 4 x ,
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Engineering Mathematics
Chapter 2: Calculus
[2.43]
y2 12 x 2 4 , y3 24 x , y4 24 . So ( x 2 1) y4 2 xy3 ( x 2 1)24 2 x (24 x) 72 x 2 24 . Now we
have to see from the options which one is equal to (72 x 2 24) for n 2 . For n 2 , option (a) gives (2 2 1) y2 3(12 x 2 4) 36 x 2 12 2
which
is
not
correct;
and
option
(d)
gives
2
2(2 1) y2 6(12 x 4) 72 x 24 which is correct. So option (d) is the correct answer.
Example 2.88 [CS-2014 (2 marks)]: The function f ( x) x sin x satisfies the following equation: f ( x ) f ( x) t cos x 0 . The value of t is _____ Solution: f ( x ) x sin x f ( x ) sin x x cos x f ( x ) 2 cos x x sin x . So f ( x ) f ( x) t cos x 0 2 cos x x sin x x sin x t cos x 0 (t 2) cos x 0 t 2 .
Derivative as a rate of change: If a variable quantity y is some function of time t i.e., y f (t ), then small change in time t have a corresponding change y in y . Thus, the average rate of change y t . When limit t 0 is applied, the rate of change becomes instantaneous and the rate of change w.r.t. t , i.e., lim y t dy dt . It is clear that the rate of change of any variable with t 0
respect to some other variable is derivative of first variable with respect to other variable. The differential coefficient of y w.r.t. x , i.e, dy dx is nothing but the rate of increase of y relative to x . Example 2.89: If the volume of a spherical balloon is increasing at the rate of 900 cm2/sec, then find the rate of change of radius of balloon at instant when radius is 15 cm [in cm/sec] Solution: V (4 3) r 3 , Differentiate with respect to t , we get dV 4 dr dr 1 dV dr 1 1 7 3r 2 900 . 2 dt 3 dt dt 4 r dt dt 4 15 15 22 Exercise: 2.3 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1.
lim x a
a 2 x 3x
(a) 2 a 3 3 2.
3a x 2 x equals
(b) 2 3 3
(c) 0
(d) None of these
lim 1 cos 2( x 1) ( x 1) x 1
(c) 0 (d) Does not exist (b) 2 sin[ x ] [ x], [ x] 0 3. If f ( x ) , where [] represents greatest integer function then lim f ( x ) x 0 0 , [ x] 0 (a) 1 (b) 0 (c) 2 (d) Does not exist 3 4. lim (tan x sin x ) x _____ (a)
2
x 0
5.
lim
x 2
1 tan( x 2) [1 sin x] 1 tan( x 2) [ 2 x]3
(a) 1 8 (b) 0 (c) 1 32 (d) [( a n) nx tan x]sin nx 6. If lim 0, where n is non-zero real number, then a is equal to x 0 x2 (a) 0 (b) ( n 1) n (c) n (d) n (1 n) 7.
lim log e (1 2h) 2 log e (1 h) h 2 h 0
(a) 1
(b) 1
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(c) 2
(d) 2 e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.44]
8. If lim{log(3 x ) log(3 x )} x k , then the value of k is x 0
(a) 0 9.
1x
lim (1 x ) x 0
e
(a) 2
(b) 1 3
(c) 2 3
(d) 2 3
(b) 0
(c) 2 e
(d) e 2
(c) e 1
(d) 1
x
m
10. lim cos( x m) _____. m
11. lim ( n 2 n 1) ( n 2 n 1) n
(a) e
n ( n 1)
(b) e2
x
12. The integer n for which lim (cos x 1) (cos x e ) x 0
(a) 1
(b) 2
n
x is a finite non-zero number is
(c) 3
13. lim (sin cos ) ( 4)
(d) 4
x , where x _____.
4
14. If G ( x) 25 x 2 , then lim G ( x ) G (1) ( x 1) _____. x 1
(d) None of these (b) 1 5 (a) 1 24 (c) 24 15. If f ( a ) 2, f ( a ) 1, g (a ) 1, g ( a ) 2, then lim g ( x ) f ( a) g ( a) f ( x ) ( x a) _____. x a
16. Let f ( a) g ( a) k and their n
th
derivatives f n ( a ) , g n ( a ) exist and are not equal for some n .
If lim f ( a ) g ( x ) f ( a) g ( a) f ( x) g ( a ) g ( x ) f ( x ) 4, then the value of k is _____. x a
f (h h
17. Given that f (2) 6 and f (1) 4 , then lim f (2h 2 h 2 ) f (2) h 0
2
1) f (1)
_____.
x a 2 2 sin x , 0 x 4 18. If the function f ( x ) x cot x b , 4 x 2 , is continuous in the interval [0, ] b sin 2 x a cos 2 x , 2 x then the values of ( a, b) are (a) ( 1, 1) (b) (0, 0) (c) ( 1,1) (d) (1, 1) 19. The function f ( x ) log(1 ax ) log(1 bx ) x is not defined at x 0 . The value which should be assigned to f at x 0 so that it is continuous at x 0 , is (c) log a log b (d) log a log b (a) a b (b) a b 20. Let f ( x ) be defined for all x 0 and be continuous. Let f ( x ) satisfy f ( x y) f ( x) f ( y ) for all x, y and f (e) 1, then (a) f ( x ) ln x (b) f ( x ) is bounded (d) x f ( x) 1 as x 0 (c) f (1 x) 0 as x 0 21. The function f ( x ) x x x is (a) Continuous at the origin (b) Discontinuous at the origin because x is discontinuous there (c) Discontinuous at the origin because x x is discontinuous there (d) Discontinuous at the origin because both x and x x are discontinuous there 22. If the function f is defined by f ( x ) x (1 x ) , then at what points f is differentiable (a) Everywhere (b) Except at x 1 (c) Except at x 0 23. The value of the derivative of x 1 x 3 at x 2 is _____.
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(d) Except at x 0 or 1
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Engineering Mathematics
Chapter 2: Calculus
[2.45]
(sin x 2 ) x , x 0
24. The function f defined by f ( x )
0
(a) Continuous and derivable at x 0 (c) Continuous but not derivable at x 0 25. If f ( x ) log x , then
, x0 (b) Neither continuous nor derivable at x 0 (d) None of these
(a) f ( x ) is continuous and differentiable for all x in its domain (b) f ( x ) is continuous for all x in its domain but not differentiable at x 1 (c) f ( x ) is neither continuous nor differentiable at x 1 (d) None of these 26. The function f ( x) x x 1 is (a) Continuous at x 1 , but not differentiable (b) Both continuous and differentiable at x 1 (d) None of these (c) Not continuous at x 1
ax 2 b , x 1 27. If the derivative of the function f ( x ) 2 is everywhere continuous and bx ax 4 , x 1 differentiable at x 1 then (a) a 2, b 3 (b) a 3, b 2 (c) a 2, b 3 (d) a 3, b 2 28. Let f be twice differentiable function such that f ( x ) f ( x ) and f ( x) g ( x ), h( x ) { f ( x )}2 {g ( x )}2 . If h(5) 11, then h(10) is equal to _____.
2 x 3 [ x] , x 1 29. The function f ( x ) sin( x 2) , x 1 (a) Is continuous at x 2 (c) Is continuous but not differentiable at x 1
(b) Is differentiable at x 1 (d) None of these x2
30. The set of points where the function f ( x) 1 e is differentiable (a) ( , ) (b) ( , 0) (0, ) (c) ( 1, )
(d) None of these
x
31. The function f ( x ) e is (a) Continuous everywhere but not differentiable at x 0 (c) Continuous and differentiable everywhere
(b) Not continuous at x 0 (d) None of these
2
32. If f ( x) 1 1 x , then f ( x ) is (a) Continuous on [ 1,1] and differentiable on ( 1,1) (b) Continuous on [ 1,1] and differentiable on ( 1, 0) (0,1) (c) Continuous and differentiable on [ 1,1] (d) None of these 33. The function f ( x ) ( x 2 1) x 2 3 x 2 cos x is not differentiable at x _____.
xe (1
34. If f ( x )
x ) (1 x )
,
x0
, then f ( x ) is , x0 Continuous as well as differentiable for all x Continuous for all x but not differentiable at x 0 Neither differentiable nor continuous at x 0 Discontinuous every where f : R R be a function. Define g : R R by g ( x ) f ( x ) for all x . Then g is Onto if f is onto (b) One-one if f is one-one Continuous if f is continuous (d) Differentiable if f is differentiable
0
(a) (b) (c) (d) 35. Let (a) (c)
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Engineering Mathematics
Chapter 2: Calculus
[2.46]
36. If f (2) 4, f (2) 1, then lim xf (2) 2 f ( x ) ( x 2) _____. x 2
37. If f ( x y ) f ( x ) f ( y ) for all x and y and f (5) 2, f (0) 3, then f (5) will be _____. 38. If 5 f ( x) 3 f (1 x) x 2 and y xf ( x) then ( dy dx ) x 1 _____, where [] denotes greatest integer function. d log e x ( x 2) ( x 2) 3/ 4 is 39. dx (a) 1 (b) ( x 2 1) ( x 2 4)
(c) ( x 2 1) ( x 2 4)
(d) e x ( x 2 1) ( x 2 4)
40. If y cos 1 (5cos x 12sin x) 13 , x 0, 2 , then dy dx _____. 41. If xe xy y sin 2 x , then at x 0 , dy dx _____. 42. If ln( x y ) 2 xy , then y (0) _____. 43. If x e y e
y
, then dy dx
(a) (1 x) x
(c) (1 x) x
(b) 1 x
(d) x (1 x)
44. If g is inverse of f and f ( x) 1 (1 x n ) , then g ( x ) (b) 1 [ f ( x )]n
(a) 1 x n
(c) 1 [ g ( x)]n
(d) None of these
45. The differential coefficient of tan 1 2 x (1 x 2 ) w.r.t. sin 1 2 x (1 x 2 ) is _____.
x
46. The first derivative of the function cos 1 sin
x
1 x 2
with respect to x at x 1 is
_____.
47. If y x 1 x
2
n
2
then (1 x )
(a) n 2 y
d2y dx
2
x
dy dx
(b) n 2 y
48. If f ( x) x n then the value of f (1)
(d) 2x 2 y
(c) y f (1) 1!
f (1) 2!
f (1) 3!
......
( 1) n f n (1) n!
n n1 (c) 0 (d) 1 (a) 2 (b) 2 49. If f r ( x), g r ( x ), hr ( x ), r 1, 2, 3 are polynomials in x such that f r (a ) g r ( a) hr (a ), r 1, 2,3
f1 ( x )
f 2 ( x)
f3 ( x)
and F ( x ) g1 ( x )
g 2 ( x)
g3 ( x ) , then F ( x ) at x a is _____.
h1 ( x )
h2 ( x)
h3 ( x )
x3
50. Let f ( x ) 6 1
(a) p
sin x
cos x
1
0
2
p
p
where p is a constant. Then
3
(b) p p 2
d3 dx 3
[ f ( x )] at x 0 is
(c) p p 3
(d) Independent of p
51. The values of x , at which the first derivative of the function x (a) 2
(b) 1 2
(c) 3 2
1
2
3
w.r.t. x is , are 4 x (d) 2
3
52. If y logsin x (tan x ) , then ( dy dx) / 4 (b) 4 log 2 (a) 4 log 2 (c) 4 log 2 (d) None of these 53. f ( x ) and g ( x ) are two differentiable function on [0, 2] such that f ( x ) g ( x) 0 , f (1) 2 , g (1) 4 , f (2) 3 , g (2) 9 , then { g ( x ) f ( x)} at x 3 2 is _____.
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Engineering Mathematics
2.4
Chapter 2: Calculus
[2.47]
Application of Derivatives
2.4.1 Tangent and Normal Let P( x1 , y1 ) be any point on the curve y f ( x) . If a tangent at P and normal (which is a line perpendicular to tangent and passing through P ) makes an angle and with x axis, as shown in Fig. 2.26, then Slope of tangent at P( x1 , y1 ) tan ( dy dx ) P ( x1 , y1 ) If tangent is parallel 0 ( dy dx ) P ( x1 , y1 ) 0 If tangent is perpendicular 2 ( dy dx ) P ( x1 , y1 ) 1 0
to to
x axis x axis Figure 2.26: Tangent and Normal on the curve = ( )
Tangent to a curve at a point P( x1 , y1 ) can be drawn, even if dy dx at P does not exist.
Slope of normal at P( x1 , y1 ) tan 1 tan 1 ( dy dx) P ( x1 , y1 ) . If normal is parallel to x-axis (dx dy) P ( x1 , y1 ) 0 or ( dx dy) ( x1 , y1 ) 0 If normal is perpendicular to x axis (for parallel to y axis) ( dy dx ) P ( x1 , y1 ) 0
Equation of tangent at P ( x1 , y1 ) is given as, y y1 ( dy dx ) P ( x1 , y1 ) ( x x1 ) If at any point P ( x1 , y1 ) on the curve y f ( x) , the tangent makes equal angle with the axes, then at the point P , 4 or 3 4 . Hence, at P , tan ( dy dx) 1 .
( x x )
Equation of normal at P ( x1 , y1 ) is given as, y y1 1 ( dy dx ) P ( x1 , y1 )
Angle of intersection of two curves: The angle of intersection of two curves is the angle between the tangents to the two curves at their point of intersection. Let slope of tangent at P ( x1 , y1 ) of
1
two curves y1 f1 ( x) and y2 f 2 ( x ) is m1 ( dy dx )1( x1 , y1 ) , m2 (dy dx) 2( x1 , y1 ) . As the angle 1 between two straight lines having slopes m1 and m2 is tan (m1 m2 ) (1 m1m2 ) ; so angle
between the tangents of the two curves is found by putting values of m1 and m2 . Orthogonal curves: If the angle of intersection of two curves is right angle, the two curves are said to intersect orthogonally. The curves are called orthogonal curve if m1m2 1 .
Length of Tangent, Normal, Sub-tangent and Sub-normal: As shown in Fig. 2.26. PA and PB are called length of tangent and normal, respectively, at point P . If PC be the perpendicular from P on x axis, then AC and BC are length of subtangent and subnormal, respectively, at P . Length of tangent PA y cosec y
1 dy dx
Length of normal PB y sec y 1 dy dx
2
dy
dx
2
Length of sub-tangent AC y cot y dy dx Length of sub-normal BC y tan y dy dx
Length of perpendicular from origin to the Tangent: Length of perpendicular from origin
(0, 0) to the tangent at P( x1 , y1 ) of y f ( x) is p y1 x1 (dy dx) ( x1 , y1 )
1 ( dy dx) 2 .
Example 2.90 [ME-2006 (2 marks)]: Equation of a line normal to f ( x ) ( x 8) 2 3 1 at P (0,5) is (a) y 3 x 5 (b) y 3 x 5 (c) 3 y x 15 (d) 3 y x 15
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Engineering Mathematics
Solution 1
(b):
Chapter 2: Calculus
f ( x ) 2 3( x 8)1 3 ,
f ( x) 3(0 8) 13
(0,5)
so
[2.48]
slope
of
normal
P (0,5)
at
2 3 . Thus, the required equation is y 5 3( x 0) y 3 x 5 .
Example 2.91 [TF-2009 (2 marks)]: Radius of the circle passing through P (2, 4) and the point of intersection with the x axis of the tangent and normal drawn at P to the curve y 2 8 x is (a) 1 (b) 2 (c) 3 (d) 4 Solution (d): Slope of tangent at P (2, 4) is ( dy dx ) (2,4) (4 y ) (2,4) 4 4 1 ; Slope of normal at
P (2, 4) is 1 ( dy dx ) (2,4) 1 . So eqn of tangent having slope 1 and passing through P (2, 4) is y 4 1( x 2) y x 2 ; it meets the x axis at A(2, 0) . Eqn of normal having slope of –1 and passing through P (2, 4) is y 4 1( x 2) y x 6 ; it meets the x axis at B (6, 0) . Let the general equation of a circle with centre ( g , f ) and radius g 2 f 2 c is x 2 y 2 2 gx 2 fy c 0 , at P (2, 4) , A(2, 0) and B (6, 0) , through which the circle passes, we get, at P (2, 4) 4 g 8 f c 20 ; at A(2, 0) 4 g c 4 ; at B (6, 0) 12 g c 36 ; on solving these, we get g 2, f 0, c 12 and so radius of the circle is
( 2) 2 (0) 2 ( 12) 4
Example 2.92 [AE-2010 (2 marks)]: If e is the base of the natural logarithms then the equation of the tangent from the origin to the curve y e x is (a) y x (b) y x (d) y ex (c) y x e Solution: Any point on the given curve is ( x, e x ) ; slope of a line joining origin to the point ( x, e x ) is (e x 0) ( x 0) e x x …(i); Slope of tangent at ( x, e x ) on the given curve is y ( x ,e x ) e x
( x ,e x )
ex
…(ii). We have e x x e x e x (1 x) 0 x 1 ( e x 0 ) slope of a line joining origin to the point ( x, e x ) is e1 1 e . So, the required equation of tangent is y 0 e( x 0) y ex . Example 2.93 [CH-2011 (1 mark)]: Which ONE of the following y ( x ) has the slope of its tangent equal to ax y ? (Note: a and b are real constants) (b) y ax b
(a) y ( x b) a
(c) y ( x 2 b) a
(d) y ax 2 b
Solution (d): Slope of lines for the equations given in options (a) and (b) are 1 a and a , respectively. Differentiating eqn in option (c) w.r.t. x , slope dy dx x Differentiating eqn in option (d) w.r.t. x , slope dy dx ax
a x2 b x
a y a x ( ay ) .
ax 2 b ax y .
Example 2.94 [AG-2012 (1 mark)]: The tangent line to y f ( x) at ( x0 , y0 ) , assuming f ( x) 0 , intersects the x axis at (a) x0 y0 f ( x0 ) , 0 (b) x0 y0 f ( x0 ) , 0 (c)
x0 f ( x0 )
y0 , 0
(d)
x0 f ( x0 )
y0 , 0
Solution (a): Slope of line joining the points ( x0 , y0 ) and ( x, 0) is m y0 ( x0 x ) f ( x0 ) x x0 y0 f ( x0 ) . So, tangent to
x0 y0
y f ( x)
at
( x0 , y0 )
intersects the
x
axis at
f ( x0 ) , 0 , where f ( x) 0 .
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.49]
2.4.2 Monotonicity of a Function Functions are said to be monotonic if they are either increasing or decreasing in their entire domain, x x for e.g., f ( x) e is an increasing function; and f ( x) e is a decreasing function. Functions which are increasing as well as decreasing in their domain are said to be non – monotonic. For e.g., f ( x ) sin x is non-monotonic in x R , but, it is increasing is (0, 2) and decreasing in ( 2 , ) .
Monotonicity of a Function at a point: For small positive h , a function f ( x) at x a
is said to be monotonically increasing or simply increasing if f ( x ) satisfies f (a h) f ( a ) and f (a h) f ( a) , as shown in Fig. 2.27a. is said to be monotonically decreasing or simply decreasing if f ( x ) satisfies f (a h) f (a ) and f (a h) f (a ) , as shown in Fig. 2.27b. We can talk of monotonicity of f ( x ) at x a only if x a lies in domain of f , without any consideration of continuity or differentiability of f ( x ) at x a , as shown in Fig. 2.27c and d.
(a)
(b)
(c)
(d)
Figure 2.27: Monotonicity of a Function at a point: (a) and (c) are monotonically increasing; (b) and (d) are monotonically decreasing at =
Monotonicity of a Function in an interval: Let I be an open interval contained in the domain of a real – valued function f . Then f is said to be:
increasing on I if f ( x1 ) f ( x2 ) for all x1 x2 , x1 , x2 I , as shown in Fig. 2.28a.
strictly increasing on I if f ( x1 ) f ( x2 ) for all x1 x2 , x1 , x2 I , as shown in Fig. 2.28b.
decreasing on I if f ( x1 ) f ( x2 ) for all x1 x2 , x1 , x2 I , as shown in Fig. 2.28c.
strictly decreasing on I if f ( x1 ) f ( x2 ) for all x1 x2 , x1 , x2 I , as shown in Fig. 2.28d.
(a)
(b)
(c)
(d)
Figure 2.28: Monotonicity of a Function in an interval
Necessary and Sufficient condition for Monotonic Function: Let f be a differentiable real function defined on an open interval ( a, b) . From Fig. 2.29a, if f ( x ) is strictly increasing function on ( a, b) , then tangent at every point on the curve y f ( x) makes an acute angle with the ve direction of x axis tan 0
dy dx 0 or f ( x ) 0 x ( a, b) . If there are some points in ( a, b) such that f ( x) 0 then, f ( x) 0 x ( a, b) f ( x ) is monotonically increasing function, as shown in Fig. 2.29b.
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Engineering Mathematics
Chapter 2: Calculus
(b)
(a)
[2.50]
(d)
(c)
Figure 2.29: Monotonicity of a Function in ( , )
From Fig. 2.29c, if f ( x ) is a decreasing function on ( a, b) , then tangent at every point on the curve y f ( x) makes an obtuse angle with the ve direction of x axis tan 0 dy dx 0 or f ( x ) 0 x ( a, b) . If there are some points in ( a, b) such that f ( x) 0 then, f ( x) 0 x ( a, b) f ( x ) is monotonically decreasing function, as shown in Fig. 2.29d.
Example 2.95 [EC-2006 (2 marks)]: As x increased from to , function f ( x ) e x (1 e x ) (a) monotonically increases (b) increases to a maximum value and then decreases (c) monotonically decreases (d) decreases to a minimum value and then increases x x x x (1 e x ) 2 e x (1 e x )2 0 for all Solution (a): For the given function, f ( x) e (1 e ) e e
x , as e x 0 for all x , so the given function is monotonically increasing function. Example 2.96 [EE-2006 (2 marks)]: A continuous-time system is described by y(t ) e y (t ) is the output and x (t ) is the input. y (t ) is bounded (a) only when x (t ) is bounded (b) only when x (t ) is non-negative (d) even when x (t ) is not bounded (c) only for t 0 if x (t ) is bounded for t 0 Solution (d): y (t ) e
x (t )
x (t )
, where,
as x (t ) or , x(t ) x(t ) y (t ) 0 .
Properties of Monotonic Function
1
If f ( x ) is strictly increasing function on an interval [ a, b] , then f exists and it is also a strictly increasing function. 1 If f ( x ) is strictly increasing function on an interval [ a, b] such that it is continuous, then f is continuous on [ f ( a ), f (b)] . If f ( x ) is continuous on [ a, b] such that f (c ) 0 ( f ( c) 0) for each c ( a, b), then f ( x ) is monotonically (strictly) increasing function on [ a, b] . If f ( x ) is continuous on [ a, b] such that f (c ) 0 ( f ( c) 0) for each c ( a, b), , then f ( x ) is monotonically (strictly) decreasing function on [ a, b] . If f ( x ) and g ( x ) are monotonically (or strictly) increasing (or decreasing) functions on [ a, b] , then gof ( x ) is a monotonically (or strictly) increasing function on [ a, b] . If one of the two functions f ( x ) and g ( x ) is strictly (or monotonically) increasing and other a strictly (monotonically) decreasing, then gof ( x ) is strictly (monotonically) decreasing on [ a, b] .
Example 2.97: Find the interval in which the function sin 4 x cos 4 x increase in 0, 2 . 4
4
2
2
2
2
2
Solution: f ( x) sin x cos x (sin x cos x) 2sin x cos x (3 4) (1 4) cos 4 x . For f ( x ) to be increasing, f ( x) 0 f ( x ) sin 4 x 0 sin 4 x 0 4 x 3 2 4 x 3 8
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Engineering Mathematics
Chapter 2: Calculus
[2.51]
Critical Points and Point of Inflection of a function
Critical Points of a function: Critical point of a function of a real variable is any value in the domain where either the function is not differentiable or its derivative is zero. Point of Inflection of a function: A point of inflection is a point at which a curve is changing concave upward to concave downward or vice-versa. For continuous function f ( x ) , if f ( x0 ) 0 or f ( x0 ) does not exist at points where f ( x0 ) exists and if f ( x) changes its sign when passing through x x0 then x0 is called the point of inflection [This point was asked in MT-2011 (1 mark)]. For e.g., for f ( x) x 3 1 , at x 0 , f ( x) 0 and f ( x ) 0 , so x 0 is the point of inflection [This example was asked in ME-2012, PI-2012 (1 mark)]. The later condition may be replaced by f (c ) 0 , when f (c ) exists. Thus, x c is a point of inflection if f (c) 0 and f (c ) 0 . If f ( x ) 0 , x ( a, b) , then the curve y f ( x) is concave downward in ( a, b) , as shown in Fig. 2.30a. If f ( x ) 0 , x ( a, b) , then the curve y f ( x) is Figure 2.30: Point of Inflection concave upward in ( a, b) , as shown in Fig. 2.30b.
Example 2.98 [CE-1999 (1 mark)]: Number of inflection points for the curve y ( x 2) x 4 is (a) 3
(b) 1
(c) n
(d) (n 1) 2
Solution (b): As y ( x 2) x 4 y 5 x 4 8 x 3 y 4 x 2 (5 x 6) y 60 x 2 48 x For point of inflection, y 0 4 x 2 (5 x 6) 0 x 0, 6 5 . Now y x 0 0 x 0 is not a point of inflection. Also, y x 6 5 144 5 0 x 6 5 is the point of inflection
Maximum and Minimum value of a Function: The maximum (or minimum) value of function f ( x ) is the local or regional maximum (or minimum) and not the greatest (or least) value attainable by the function. It is also possible in a function that local maximum at one point is smaller than local minimum at another point. We also use the word extreme for maxima and minima. A function f ( x ) is said to have a maximum at x a if f (a ) is greatest of all values in the suitably small neighbourhood of a where x a is an interior point in the domain of f ( x ) . Similarly, a function f ( x ) is said to have a minimum at x b , if f (b ) is smallest of all values in the suitably small neighbourhood of b where x b is an interior point in the domain of f ( x ) . Analytically (also shown in Fig. 2.31) x a is a maximum point of f ( x ) if f (a ) f (a h) and f (a ) f (a h) x b is a minimum point of f ( x ) if f (b) f (b h) and f (b) f (b h) x c is neither a maximum point nor a minimum point of f ( x ) if { f (c ) f (c h)} and { f (c) f (c h)} Figure 2.31: Maximum and Minimum both have opposite signs. point of a function Monotonic functions do not have extreme points. A function can have several maximum and minimum values, and a minimum value may even be greater than a maximum value. The maximum and minimum values of a continuous function occur alternatively, and between two consecutive maximum values there is a minimum value and vice versa. A maximum (minimum) value of a function may not be the greatest (least) value in a finite interval.
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Engineering Mathematics
Chapter 2: Calculus
[2.52]
Tests for Local Maximum/Minimum when ( ) is Differentiable at
=
First – Order Derivative Test in Ascertaining the Maxima or Minima: As shown in Fig. 2.32, for the interval ( a h, a ) , we find ( dy dx) 0 , i.e., f ( x ) is increasing; for the interval ( a, a h) , we find ( dy dx) 0 , i.e., f ( x ) is decreasing. Hence x a is the point of maxima and
(dy dx) x a 0 . Also, for the interval (b h, b) , we find ( dy dx) 0 , i.e., f ( x ) is decreasing; for the interval (b, b h) , we find ( dy dx) 0 , i.e., f ( x ) is increasing. Hence x b is the point of maxima and ( dy dx) x b 0 . Hence, if f (a h) 0 and f (a h) 0 then x a is a point of local maxima, where f ( a) 0 . It means f ( x ) should change its sign from positive to negative. Similarly, if f (b h) 0 and f (b h) 0 then x b is a point of local minima, where f (b) 0 . It means f ( x ) should change its sign from negative to positive. However, if f ( x ) does not change its sign in a certain neighbourhood of c , then f ( x ) is either increasing (if f (c h) 0 and f (c h) 0 ) or decreasing (if f (c h) 0 and f (c h) 0 ) throughout its neighbourhood implying that f (c ) is not an extreme value.
Figure 2.32: First Order Derivative Test for finding Maxima and Minima point
Figure 2.33: Second Order Derivative Test for finding Maxima and Minima point
Second – Order Derivative Test in Ascertaining the Maxima or Minima: As shown in Fig. 2.33, it is clear that as x increases from a h to a h , the function dy dx continuously decreases, i.e., the function f ( x ) or dy dx continuously decreases which means in the neighbourhood of x a we have d 2 y dx 2 0 . Similarly, as x increases from b h to b h , the function dy dx continuously increases, i.e., the function f ( x ) or dy dx continuously 2
2
increases which means in the neighbourhood of x b we have d y dx 0 . Hence we have: 2
2
At local maxima, i.e., at x a , dy dx 0 and d y dx 0 2
2
At local minima, i.e., at x b , dy dx 0 and d y dx 0 2
2
However, if d y dx 0 , then the test fails. In this case, f can still have a maxima or minima or point of inflection (neither maxima nor minima), we revert back to the first-order derivative check for ascertaining the maxima or minima. Derivative test: It is nothing but the general version of the second – order derivative test. It n n 1 says that if f (a) f (a) f (a) f (a) 0 and f (a) 0 , then f ( x ) would have a local maximum or minimum at x a if n is an odd natural number and that x a would be a n 1
point of local minima if f (a) 0 and would be a point of local minima if f However, if n is even, then f has neither a maximum nor a minimum at x a .
n 1
(a) 0 .
Tests for Local Maximum/Minimum when ( ) is not Differentiable at = When f ( x ) is continuous at x a and f (a h) and f (a h) exist, and are non – zero, then f ( x ) has a local maximum or minimum at x a if f (a h) and f (a h) are of opposite signs.
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[2.53]
If f (a h) 0 and f (a h) 0 , then x a will be the point of local maximum, as shown in Fig. 2.34a. If f (a h) 0 and f ( a h) 0 , then x a will be the point of local minimum, as shown in Fig. 2.34b.
(a)
(b)
(c)
Figure 2.34: Maxima and Minima of a Function when it is not differentiable
When f ( x ) is continuous and f (a h) and f (a h) exist but one of them is zero, then we can infer the information about the existence of local maxima/minima from the basic definition of local maxima/minima. If f ( x ) is not continuous at x a , then compare the values of f ( x ) at the neighbouring points of x a . It is advisable to draw the graph of the function in the vicinity of the point x a , because the graph would give us the clear picture about the existence of local maxima/minima at x a , as shown in Fig. 2.34c.
Absolute or Global Maxima/Minima of a function in a given interval: If a function f ( x) is defined in an interval then Global maxima/minima is basically the greatest/least value of the function in that interval. Let c1 , c2 ,, cn be different critical points of the function y f ( x) in ( a, b) . Then
Global maxima/minima in [ a, b] will be given as, Global maxima: M1 max{ f (a), f (c1 ), f (c2 ),, f (cn ), f (b)} Global minima: M 2 min{ f (a), f (c1 ), f (c2 ),, f (cn ), f (b)}
Global maxima/minima in ( a, b) will be given as, Global maxima: M1 max{ f (c1 ), f (c2 ),, f (cn )} , if and only if lim f ( x ) M 1 and x a
lim f ( x) M 1 , as shown in Fig. 2.35a.
x b
Global minima: M 2 min{ f (c1 ), f (c2 ),, f (cn )} , if and only if
lim f ( x) M 2 and
x a
lim f ( x ) M 2 , as shown in Fig. 2.35b.
x b
(a)
(b)
Figure 2.35: Global Maxima and Minima if a function in open interval
Example 2.99 [CE-1999 (5 marks)]: Find the maximum and minimum values of the function f ( x ) sin x cos 2 x over the range 0 x 2 . Solution: For the critical points, f ( x ) 0 cos x 2 sin 2 x 0 cos x (1 2 sin x ) 0 cos x 0 or sin x 1 2 . In 0 x 2 , cos x 0 x 2 , 3 2 ; and sin x 1 2 x 6 ,5 6 . So critical points are x 6 , 2 ,5 6 , 3 2 . As f ( x ) sin x 4 cos 2 x . Thus
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[2.54]
f ( x) x
6
5 2 0 x 6 is the point of maxima and thus f ( 6) 1
f ( x) x
2
3 0 x 2 is the point of minima and thus f ( 2) 0
f ( x ) x 5
6
5 2 0 x 5 6 is the point of maxima and thus f (5 6) 1
f ( x) x 3
2
5 0 x 3 2 is the point of minima and thus f (3 2) 2
Thus, local minima of f ( x ) in 0 x 2 occur at x 2 ,3 2 , whose values are 0, –2, respectively; and local maxima of f ( x ) in 0 x 2 occur at x 6 , 5 6 , whose values are 1, 1, respectively.
Now
lim f ( x ) lim f (0 h) lim{sin(0 h) cos 2(0 h)} 1
x 0
h0
h 0
and
lim f ( x ) lim f (2 h) lim{sin(2 h) cos 2(2 h)} 1 . So, we have no global maxima as
x 2
h0
h0
max(1,1) 1 and lim f ( x ) 1 , x 0
lim f ( x) 1 ; but we have global minima having value –2, as
x 2
min(0, 2) 2 and lim f ( x) 2 , lim f ( x ) 2 . x 2
x 0
Example 2.100 [CE-2000 (2 marks)]: The maxima and minima of the function f ( x ) 2 x3 15 x 2 36 x 10 occur, respectively, at (a) x 3 and x 2 (b) x 1 and x 3 (c) x 2 and x 3 (d) x 3 and x 4 2 Solution (c): For the critical point of f ( x ) , f ( x) 0 6 x 30 x 36 0 x 2, 3 are the critical points. As, f ( x ) 12 x 30 , f ( x) x 2 6 0 x 2 is the point of maxima; and
f ( x) x 3 6 0 x 3 is the point of minima. [Similar questions were also asked in CS-1998 (5 marks), CE-2004, ME-1995, TF-2008, EE-2011 (2 marks), CE-2002 (1 mark)] Example 2.101 [ME-2000 (2 marks)]: From an equilateral triangular plate of side ‘ a ’, a square plate of maximum size has to be cut. The side of such a square plate is: (a) 3a 2 (b) (2 3) a 4 (c) (1 3) a 8 (d) 3a (2 3) Solution (d): In the figure shown, let FIHGF is a square of side x ; and ABCA is an equilateral triangle of side a. As, AD
AEF
AB 2 BD 2
and
a 2 ( a 2 4) 3a 2 ; also as both the triangles
ADB
are
similar
so
AD AE BD EF
( 3a 2) ( 3a 2 x) ( a 2) ( x 2) x ( 3a ) (2 3) Example 2.102 [EC-2005 (2 marks)]: The derivative of the symmetric function drawn in figure will look like
(a)
(b)
(c)
(d)
Solution (c): The given symmetric function gives (i) dy dx 0 for x 0 ; and (ii) dy dx 0 for x 0 . It is to be also noted that, (iii) for x 0 , dy dx increases from zero to ve maximum value
and then decreases to zero; (iv) for x 0 , dy dx decreases from zero to ve minimum value and then increases to zero. The four conditions are met by the figure given in option (c). Example 2.103 [EE-2005 (2 marks)]: For the function f ( x ) x 2 e x , the maximum occurs at x (a) 2 (b) 1 (c) 0 (d) –1
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Solution (a): For critical point of f ( x ) , f ( x) 0 xe x (2 x ) 0 x 0, 2 are the critical points. Also f ( x ) (2 x )(e x xe x ) xe x f ( x) x 0 2 0 x 0 is the point of minima; and f ( x ) x 2 2e
2
0 x 2 is the point of maxima.
[Similar question was also asked in EE-2014 (1 marks)] Example 2.104 [ME-2005 (2 marks)]: The right circular cone of largest volume that can be enclosed by a sphere of 1 metre radius has a height (in metres) of (a) 1/3 (b) 2/3 (d) 4/3 (c) 2 2 3 Solution (d): Let ABCA be a sphere of radius OC OA R and ABDCA be a cone of height AD h and radius CD r . AD AO OD R OD and from OCD , we have
OD OC2 DC2
R 2 r 2 AD h R R 2 r 2 1 1 r 2 (as R 1 is given). Volume
of the cone is V (1 3) r 2 h (1 3) r 2 (1 1 r 2 ) . For maximum or minimum volume,
dV dr
0
r2 r 2 2 1 r 2 3 1 r2
2 2 0 2 2 1 r (r
1 r2 ) 0
(as r 0 is not possible) or 2 1 r 2 2(1 r 2 ) r 2 2 1 r 2 3r 2 2 4(1 r 2 ) 9r 4 12r 2 4 r 2 (9r 2 8) 0 9r 2 8 0 (as r 2 0 is not
possible) r 2 2 3 ( ve value of r is not possible). As d 2V dt 2
r2 2 3
0 r 2 2 3 is the
point of maxima, i.e., at r 2 2 3 and thus h 4 3 , the cone have maximum volume. Example 2.105 [MN-2007 (2 marks)]: The cost of diesel is Rs. 25 ( x 90) per km to drive a dump truck at a speed of x km/hour. The maintenance cost of the truck is Rs. 10 per hour. To minimize the cost per km, the truck speed in km/hour is (a) 5 (b) 20 (c) 25 (d) 30 Solution (d): Speed of dump truck is x km/hr, thus the truck covers one km in 1 x hours; so per km the maintenance cost is Rs. 10 x . So the total cost per km is C 25 x 90 10 x ; for maximum or
minimum,
dC dx 0 1 90 10 x 2 0 x 30 2
d 2 C dx 2 20 x 3 d C dx
2 x 30
which
is
the
critical
point.
As
20 303 0 x 30 is the point of minima. n
Example 2.106 [AG-2008 (2 marks)]: If
( x ai ) 2 i 1
mean of the series (a) a1 a2 a3 ( 1) n 1 a n
has a minima at A , then A is the arithmetic
(b) a1 a2 a3 an
(c) (1 a1 ) (1 a2 ) (1 a3 ) ( 1) n
n 1
(1 an )
(d) (1 a1 ) (1 a2 ) (1 a3 ) (1 an )
n
Solution (b): ( x ai ) 2 ( x 2 2ai x ai2 ) nx 2 2( a1 a2 an ) x ( a12 a22 an2 ) . i 1
i 1
As the given expression on the RHS is a quadratic equation in x . We know that y ax 2 bx c has a minimum value of D 4a at x b 2a . So if the given expression has a minima at x A then 2( a1 a2 an ) a1 a2 an A , thus A is the arithmetic mean of a1 a2 a3 an 2n n Example 2.107 [AG-2008 (1 mark)]: If log e ( y ) x log e ( x ) , then the maximum value of y is (a) e
(b) e x
2
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(c) ee
1
(d) e x
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[2.56]
Solution (c): The given function can be written as log e ( y ) log e ( x x ) y x x . For maximum value of y , y 0 dy dx y (log e x 1) 0 log e x 1 0 x e 1 is critical point for the
e (e )
given function. As, d 2 y dx 2 y (log e x 1) y x y (log e x 1) 2 1 x x x (log e x 1) 2 1 x
d 2 y dx 2 xe
y x x
1
(e 1 ) e
x e 1
1
(log
e
e 1 1) 2 1 e 1 (e 1 ) e
1
(1 1)
2
1 e
1
e0
is the point of maxima for log e ( y ) x log e ( x ) ; thus maximum value of y is
x e1
1 x x 1 (e 1 ) e
1
e
e 1
.
Example 2.108 [AE-2008 (2 marks)]: The function f ( x ) x 2 5 x 6 (a) has its maximum value at x 2.0 (b) has its maximum value at x 2.5 (c) is increasing on the interval (2.0, 2.5) (d) is increasing on the interval (2.5, 3.0) Solution (d): f ( x ) 2 x 5 f ( x) 0 for all x ( , 2.5) ; and f ( x) 0 for all x (2.5, ) . Thus f ( x ) is decreasing function in x ( , 2.5) ; and f ( x ) is increasing function in x (2.5, ) . Example 2.109 [CS-2008 (2 marks)]: A point on a curve is said to be an extremum if it is a local minimum or a local maximum. The number of distinct extrema for the curve f ( x ) 3 x 4 16 x 3 24 x 2 37 is (a) 0 (b) 1 (c) 2 (d) 3 Solution (b): For the critical point of f ( x ) , we have f ( x ) 0 12 x 3 48 x 2 48 x 0 x ( x 2) 2 0 x 0, 2, 2 .
Now
f ( x ) 36 x 2 96 x 48 ,
so
f ( x ) x 0 48 0
and
f ( x ) x 2 0 . So number of different extrema for the given curve is one which is at x 0 .
Example 2.110 [EC-2008 (1 mark)]: For real values of x , the minimum value of the function f ( x ) exp( x ) exp( x ) (a) 2 (b) 1 (c) 0.5 (d) 0 x x 2x Solution (d): For critical values of f ( x ) , f ( x ) 0 e (1 e ) 0 e 1 2 x 0 x 0 . x
Also, f ( x) e e
x
f ( x ) x 0 e 0 e 0 2 0 x 0 is the point of local minima; thus
minimum value at x 0 is f (0) e 0 e 0 1 1 0 . [Similar questions were also asked in AE-2007, ME-2007 (1 mark), EC-2014, EE-2014 (2 marks)] Example 2.111 [EE-2008 (2 marks)]: Consider function y ( x 2 4) 2 where x is a real number. Then the function has (a) only one minimum (b) only two minima (c) three minima (d) three maxima 2 Solution (a): For critical point of y , y 0 2 x( x 4) 0 x 0, 2 . As y 4(3x 2 4) , so y x 0 16 0 x 0 is the point of minima; y x 2 32 0 x 2 are points of maxima. Example 2.112 [PI-2008, IN-2007 (2 marks)]: For real x , the maximum value of e sin x e cos x is (b) e
(a) 1 Solution (c): y e
sin x
e cos x esin x cos x e
(c) e 2 sin( x 4)
2
(d)
. So the maximum value of y occurs when we
have maximum value of sin( x 4) , i.e., 1. Thus maximum value of y is e
2
.
Example 2.113 [MN-2008 (2 marks)]: The volume of a cone is given by V ( 3)l 3 sin 2 cos where, l is the slant height and is the semi-vertical angle. The angle ( ), for which the volume of cone becomes maximum is
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(a) sin 1 (1
Chapter 2: Calculus
(b) cos 1 (1
3)
[2.57]
(c) cos 1 ( 2)
3)
Solution (b): For maximum or minimum,
dV d 0
(d) sin 1 ( 2)
( 3)l 3 (2 sin cos 2 sin 3 ) 0
( 3)l 3 sin (3 cos2 1) 0 (3cos 2 1) 0 (as sin 0 is not possible) cos 1 2
1
3) . As d V d
cos (1
2
cos1 1 3
0 cos 1 (1
3
3) is the point of maxima.
Example 2.114 [AE-2009 (2 marks)]: In the interval 1 x 2 , the function f ( x ) e x sin x is (d) monotonically decreasing (a) max at x 1 (b) max at x 2 (c) max at x 1.5 x x Solution (b): f ( x ) e cos x (e cos x ) 0 for x [1, 2] , as for x [1, 2] , e x [e , e 2 ] and cos x [ 1,1] . So, in x [1, 2] , f ( x ) has no critical point and it is monotonically increasing function. So, the maximum value of f ( x ) is found at the endpoints, i.e.,
maximum value of f ( x ) is max{ f (1), f (2)} max{e , e 2 } e 2 , which occurs at x 2 . So the given function has a maximum value in x [1, 2] at x 2 . Example 2.115 [EC-2010 (2 marks)]: e y x1 x , then y has a (a) max at x e (b) min at x e (c) max at x e 1 (d) min at x e 1 Solution (b): e y x1 x y (ln x ) x , for critical point of y , y 0 (1 ln x ) x 2 0 ln x 1
x e . Also y {1 2 x 2 (1 ln x )} x 5 , so y x e 1 e5 0 x e is the point of minima. [Similar question was also asked in ME-2001 (2 marks)] Example 2.116 [EE-2010 (2 marks)]: At t 0 , the function f (t ) (sin t ) t has (a) a minimum (b) a discontinuity (c) a point of inflection (d) a maximum Solution (d): The given function is f (t ) (sin t ) t . So, LHL t 0 lim sin(0 h) (0 h) 1 ; h 0
RHL t 0 lim sin(0 h) (0 h) 1 . Using expansion of sin t , h0
t2
f (t ) t t 3 3! t 5 5! t
t4
f (0) 1 0 0 1 . Hence LHL t 0 RHL t 0 f ( x) f (t ) is 3! 5! continuous at t 0 . Now, for maxima or minima or point of inflection, f (t ) 0 f (t ) 1
(t cos t sin t ) t 2 0 t cos t sin t 0 (tan t ) t 1 and this happens only when t 0 which
is the critical point. Now, f (t ) f (t ) t 0
t 2 (cos t t sin t cos t ) 2t (t cos t sin t ) 4
sin t
t t sin t 2 cos t 2 sin t sin t sin t t cos t lim 2 3 lim 2 lim t 0 t0 t t t t0 t t3
t
2
2 sin t t3
1
3
1 5 1 1 3 1 1 2 1 1 t t 1 2 lim t t 0 2! 3! 4! 5! 4! 5! 2! 3!
f (t ) t 0 1 2 lim t 3 t 0
2 cos t
t
f (t ) t 0 1 2 lim (t t 3 3! t 5 5! ) t (1 t 2 2! t 4 4! ) t 0
f (t ) t 0 1 2 (1 2!) (1 3!) 1 (2 3) (1 3) 0 . Hence t 0 is the point of maxima.
Example 2.117 [MT-2010 (2 marks)]: Determine the radius (in m) of a cylinder of volume 200 m3 that has the least surface area (a) 2.302 (b) 3.142 (c) 3.169 (d) 7.233 2 Solution (c): For a cylinder, Volume V r h ; and whole surface area S 2 r ( r h ) , where r is the radius of the base and h is its height. As the volume V
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is given h V ( r 2 )
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S 2 r r (V r 2 ) 2 r 2 2V r . For maximum or minimum surface area, dS dr 0
4 r 2V r 2 0 r (V 2 )1 3 ; as d 2 S dr 2
13
13
r V (2 )
0 r V (2 )
is the point of
13
minima as. So r 200 (2 ) 3.169 [Similar question was also asked in TF-2010 (2 marks)] Example 2.118 [AE-2011 (1 mark)]: The function f ( x1 , x2 , x3 ) x12 x22 x32 2 x1 4 x2 6 x3 14 has its minimum value at (a) (1, 2, 3) (b) (0, 0, 0) (c) (3, 2, 1) (d) (1, 1, 3) 2 2 2 Solution (a): f ( x1 , x2 , x3 ) ( x1 1) ( x2 2) ( x3 3) . f ( x1 , x2 , x3 ) has minimum value of ‘0’ which occurs when every expression in bracket becomes zero, i.e., when x1 1 , x2 2 and x3 3 . Example 2.119 [MN-2011 (1 mark)]: The largest area of a rectangular shaft for a given perimeter is obtained when length is (a) 2.5 times of breadth (b) 1.5 times of breadth (c) 2 time of breadth (d) Equal to breadth Solution (d): Let x and y are the length and breadth of the rectangular shaft; so perimeter
P 2( x y ) and area A xy x ( P 2) x . For maximum or minimum area of the rectangular shaft
dA dx 0 P 2 x x 0 x P 4 .
Also
d 2 A dx 2 1 ( d 2 A dx 2 )
xP 4
1 0
x P 4 is the point of maximum area. So x P 4 2( x y ) 4 2 x x y x y
3 x 2 , x 1 Example 2.120 [XE-2011 (2 marks)]: The function f ( x ) 3 x, 1 x 2 has x 1, x2 (a) a local maxima at x 3 and a local minima at x 0 (b) a local maxima at x 0 and no local minima (c) a local maxima at x 0 and a local minima at x 2 (d) no local maxima and a local minima at x 1 Solution (c): From the graph of the given function we can see that x 0 is the point of local maxima and x 2 is the point of local minima Example 2.121 [CS-2012 (1 mark)]: Consider the function
f ( x ) sin x in the interval
x 4 , 7 4 . The number and location(s) of the local minima of this function are
(a) One, at 2 (b) One, at 3 2 Solution (b): From the graph
(c) Two, at 2 & 3 2 of y sin x in
(d) Two, at 4 & 3 2
x 4 , 7 4 , we can see that 2 is the point of local
maxima and 3 2 is the point of local minima. So in x 4 , 7 4 , we have only one point of minima.
Example 2.122 [XE-2012 (2 marks)]: If f ( x) x sin x and g ( x ) x sin x , then (b) g ( x ) is an even function (c) g ( x ) is differentiable at x 0 (a) g ( x ) f ( x ) (d) x coordinates corresponding to the various local maxima are identical for both f ( x ) and g ( x ) Solution (c): For option (a), f ( x ) x sin x x sin x , so option (a) is not correct. For option (b), g ( x ) x sin( x ) x sin x g ( x ) , so g ( x ) is an odd function, not an even function; thus
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[2.59]
option (b) is not correct. For option (c), f ( x) ( x ) sin( x ) x sin x f ( x ) so f ( x ) is an even function, thus the graph of f ( x ) is symmetrical about y axis; but the graph of g ( x ) is symmetrical in opposite quadrant (as g ( x ) is an odd function); so if x a is the point of maxima in the graph of f ( x ) and g ( x ) for x 0 , then x a is the point of maxima for f ( x ) , but it is a point of minima for g ( x) ; thus (c) is not correct. For option (d),
LHD x 0 lim h 0
RHD x 0 lim
g (0 h) g (0) h g (0 h) g (0)
lim
h0
0 h sin(0 h) 0 h 0 h sin(0 h) 0
lim h 0 h h LHD RHD and thus g ( x ) is differentiable at x 0 . h0
lim h 0
lim
h sinh h h sinh
h0
h
lim sinh 0 h 0
and
lim sinh 0 . Thus at x 0 , h 0
Example 2.123 [IN-2012, EE-2012, EC-2012 (2 marks)]: The maximum value of f ( x ) x 3 9 x 2 24 x 5 in the interval [1, 6] is (a) 21 (b) 25 (c) 41 (d) 46 2 Solution (c): For the critical point of f ( x ) , we have f ( x) 0 3 x 18 x 24 0 x 2, 4 are the critical point of f ( x ) . So in [1, 6] , maximum value (i.e., global maxima) of f ( x ) is max{ f (1), f (2), f (4), f (6)} max{21, 25, 21, 41} 41 . [Similar questions were also asked in CS-1997, EC-2007, IN-2008, EC-2014 (2 marks)] Example 2.124 [MT-2012 (2 marks)]: At x 0.5 , the polynomial x 2 (1 x ) 2 has (a) No extrema (b) A saddle point (c) A minima (d) A maxima 2 2 Solution (d): Let y x (1 x ) , so for critical points of y , y 0 2 x (1 x )(1 2 x ) 0 x 0,1, 0.5 are the critical points of y . Now y 4 x(1 x ) (1 2 x )(2 4 x ) y x 0.5 1 0
x 0.5 is the point of maxima. Example 2.125 [MT-2013 (2 marks)]: From a 2 m 1.2 m sheet, squares are cut out from each of the four corners as shown in figure and then the sides are bent to form an open box. The maximum possible volume (in m3) of the box is …… Solution: Let box be of length l 2 2 x , breadth b 1.2 2 x and height h x . So, the volume of the open box is V (2 2 x)(1.2 2 x ) x 4( x 3 1.6 x 2 0.6 x ) . For maximum and minimum, Now
dV dx 0 3 x 2 3.2 x 0.6 0 x 0.824, 0.243 .
d 2V dx 2 4(6 x 3.2) d 2V dx 2
x 0.824
6.976 0
2 2 x 0.824 , the box has minimum volume; also d V dx
x 0.243 2
at
6.968 0 at x 0.243 , the box
has maximum volume, which is V 4(0.2433 1.6 0.243 0.6 0.243) 0.262 m3. Example 2.126 [EC-2014 (1 mark)]: The maximum value of the function f ( x ) ln(1 x ) x (where x 1 ) occurs at x ……………. Solution: For critical points of f ( x ) , f ( x ) 0 1 (1 x ) 1 0 x (1 x ) 0 x 0 . Also f ( x ) 1 (1 x ) 2 , so f ( x) x 0 1 0 x 0 is the point of maxima.
Example 2.127 [EC-2014 (2 marks)]: The maximum value of the determinant among all 2 2 real symmetric matrices with trace 14 is ……………. Solution: Let x , y be the eigenvalues of a 2 2 real symmetric matrices, then sum of eigenvalues of any square matrix is equal to its trace x y 14 …(i). Also, product of eigenvalues of a square
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.60]
matrix gives determinant ( ) of that matrix xy x (14 x ) . So, for maximum or minimum value of , d dx 0 14 2 x 0 x 7 is the critical point. Also d 2 dx 2 at x 7 is 2 0 , so x 7 is the point of maxima for . Thus 7(14 7) 49 . Example 2.128 [EC-2014 (2 marks)]: For a right angles triangle, if the sum of the lengths of the hypotenuse and a side is kept constant, in order to have maximum area of the triangle, the angle between the hypotenuse and the side is (a) 12o (b) 36o (c) 60o (d) 45o Solution (c): From the given triangle y h sin and x h cos ; and as given in the question, h x k h h cos k h(1 cos ) k , where k constant. so area of the triangle ABC is given as A (1 2) x y (1 2) h 2 sin cos (1 4) k 2 sin 2 (1 cos ) 2 . For maximum or minimum area we have to find the critical points of A , i.e.,
dA d 0 (k 2 4) (1 cos ) 2 2 cos 2 (sin 2 )2(1 cos ) sin
(1 cos ) 4 0
2
(1 cos ) 2 cos 2 (sin 2 )2(1 cos ) sin 0 (1 cos ) cos 2 sin 2 sin 0 (as (1 cos ) 0 for the given triangle) (cos 2 cos 2 cos sin 2 sin ) 0 cos 2 cos(2 ) 0
cos 2 cos 0 2 cos(3 2)sin( 2) 0 cos(3 2) 0 or cos 2 0
cos(3 2) 0 3 2 90o 60o and cos( 2) 0 2 90o 180o 2
(which is not possible for the given triangle). Now if we check d A d
2
60o
, we
find it is 0 60o is the point of maxima for the required area A . Alternative method: Put the values of from the given options in A (1 4) k 2 sin 2 (1 cos ) 2 and decide from which option we get maximum area. Example 2.129 [EE-2014 (1 mark)]: Minimum of the real valued function f ( x ) ( x 1) 2 3 occurs at
x (b) 0 (c) 1 (a) (d) 13 Solution (c): f ( x) 0 2 {3( x 1) } at x 1 the derivative of the given function does not exists.
Also
RHL x 1 lim f (1 h) lim(1 h 1) h 0
LHL x 1 lim f (1 h) lim(1 h 1) 2 3 h 2 3 0 ,
f (1) (1 1) 2 3 0 , h 0
23
h 0
lim h h0
23
h 0
0 LHL x 1 RHL x 1 f (1) f ( x)
is
continuous at x 1 . So for maxima or minima of f ( x ) at x 1 , we have to check the sign of f (1 h) and f (1 h) . So, f (1 h) lim
f (1 h) f (1)
lim
(1 h 1) 2 3 0
lim
1
h 0 h 0 ( h )1 3 h h similarly f (1 h) . So f (1 h) 0 and f (1 h) 0 x 1 is the point of minima. h 0
;
Example 2.130 [XE-2014 (2 marks)]: The perimeter of a rectangle having the largest area that can x2 y 2 be inscribed in the ellipse 1 is ……………. 8 32 Solution: Let ABCD be a rectangle, of length 2x and width 2 y , inscribed in a ellipse whose equation is x2
y2
1 . As the centre of the given ellipse is at the 8 32 origin, thus any rectangle inscribe inside the ellipse is symmetrical about y axis; so coordinates of point A,
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics B,
C
and
D
are
Chapter 2: Calculus
shown
in
figure.
So
the
[2.61] area
of
the
required
rectangle
is
2
A 2 x 2 y 4 xy 4 x 32 4 x ; for maximum or minimum area, we have to find its critical point, i.e.,
dA dx 0 4 32 4 x 2
32 4 x 2 4 x 2 2 0 32 8 x 0 x 2 2 2 2 32 4 x 32 4 x x (8 x)
2
y 32 4 2 4 . So (2, 4) is the critical points and if we check d 2 A dx 2
(2,4)
, we will find it
is 0 . Thus, (2, 4) is the point of maxima for A and hence perimeter of rectangle ABCD is P 2(2 x 2 y ) 2(4 8) 24 .
2.4.3 Mean Value Theorem Rolle’s Theorem: Let f be a real valued function defined on the closed interval [ a, b] such that, (i) f ( x ) is continuous in the closed interval [ a, b] ; (ii) f ( x ) is differentiable in the open interval ( a, b) ; and (iii) f (a ) f (b) . Then there is at least one value c (a, b) for which f (c) 0 . Geometrical interpretation of Rolle’s theorem: If the graph of a function y f ( x) is continuous at each point from point A a , f ( a ) to point B(b, f (b)) and the tangent at each point between A and B is unique, i.e., y coordinates of points A and B are equal, then there will be at least one point P on the curve between A and B at which tangent will be parallel to the x axis. In Fig. 2.27a, there is only one such point P where tangent is parallel to x axis; in Fig. 2.27b, there are more than one such points where tangents are parallel to the x axis.
(b)
(a) Figure 2.36: Geometrical meaning of Rolle's Theorem
Let y f ( x) be a polynomial function of degree n . If f ( x ) 0 has real roots only, then
f ( x) f ( x) f n1 ( x) 0 would have only real roots, because if f ( x ) 0 has all real roots, then between two consecutive roots of f ( x ) 0 , exactly one root of f ( x) 0 would lie. Converse of Rolle’s theorem is not true, i.e., if a function f ( x ) is such that f (c) 0 for at least one c in the open interval ( a, b) , then it is not necessary that (i) f ( x ) is continuous in [ a, b] ; (ii) 3 2 f ( x ) is differentiable in ( a, b) ; and (ii) f (a ) f (b) . For e.g., if f ( x) x x and the interval is [1,3] , then f ( x ) is continuous in [1,3] and it is differentiable in ( 1,3) but f (1) f (3) .
Lagrange’s Mean Value Theorem: If a function f ( x) is (i) continuous in the closed interval [ a, b] , and (ii) differentiable in the open interval ( a, b) . Then there is atleast one value c ( a, b) , such that; f (c) f (b) f ( a ) (b a )
Geometrical Interpretation: Let f ( x ) be a function defined on [ a, b] and let APB be the curve represented
Figure 2.37: Geometrical Meaning of Lagrange's Mean value theorem
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.62]
by y f ( x) . Then co-ordinates of A and B are a, f ( a ) and b, f (b) respectively. Let the
AB makes an angle with the axis of x . Then from triangle ARB , tan BR AR tan f (b) f (a ) (b a ) . By Lagrange’s Mean value theorem, we have,
chord
f (c ) f (b) f (a ) (b a ) tan f (c ) slope of chord AB slope of tangent at (c, f (c )) . Thus geometrical meaning of the mean value theorem states: If y f ( x) is continuous and differentiable in ( a, b) , then at least one point P on the curve in ( a, b) , where the tangent will be parallel to chord AB . As shown in Fig. 2.28a, there is only one such point P where tangent is parallel to chord AB ; but in Fig. 2.28b, there are more than one such points where tangents are parallel to chord AB .
Example 2.131 [ME-1994 (1 mark)]: The value of
in the mean value theorem of
2
f (b) f ( a) (b a ) f ( ) for f ( x ) ax bx c in ( a, b) is (a) b a (b) b a (c) (b a ) 2 2
(d) (b a) 2
nd
Solution (c): The function, f ( x ) y ax bx c , is 2 degree polynomial which is continuous and differentiable for all x ; thus it is also continuous in x [a, b] and differentiable in x ( a, b) ; so we can apply Lagrange’s MVT there is ( a, b ) such that f ( ) f (b) f ( a) (b a ) f (b) f ( a ) (b a ) f ( ) ab 2 b 2 c a3 ab c (b a )(2a b ) (b a ) 2 .
Example 2.132 [CE-2005 (2 marks)]: A rail engine accelerates from its stationary position for 8 seconds and travels a distance of 280 m. According to the Mean Value Theorem, the speedometer at a certain time during acceleration must read exactly (a) 0 km/hr (b) 8 km/hr (c) 75 km/hr (d) 126 km/hr Solution (d): Let the position of rail engine x (t ) is continuous and differentiable function, so according to Lagranges mean value theorem there exist a ‘ t ’, 0 t 8 such that x(t ) v (t ) x (8) x (0) (8 0) 280 0 (8 0) 280 8 m/sec (280 8) (18 5) 126 km/hr. Example 2.133 [EE-2013 (2 marks)]: A function y 5 x 2 10 x is defined over an open interval
x (1, 2) . Atleast at one point in this interval, dy dx is exactly (a) 20 (b) 25 (c) 30 (d) 35 Solution (b): As the given function, f ( x ) y 5 x 2 10 x , is 2nd degree polynomial which is continuous and differentiable for all x . So the given function is continuous in x [1, 2] and differentiable in x (1, 2) ; thus there is at least one value c (1, 2) , such that f (c) (dy dx) x c f (2) f (1) (2 1) f (2) f (1) 40 15 25 . sin
cos
tan
Example 2.134 [CS-2014 (1 mark)]: Let the function f ( ) sin( 6)
cos( 6)
tan( 6) where
sin( 3)
cos( 3)
tan( 3)
6, 3 and f ( ) denote the derivative of f with respect to . Which of the following statements is/are TRUE? (A) 6 , 3 s.t. f ( ) 0 . (B) ( 6 , 3) s.t. f ( ) 0 . (a) A only (b) B only (c) Both A and B (d) Neither A nor B Solution (c): As the given function is continuous and differentiable in 6 , 3 , so by applying Lagrange’s MVT, we can say that both the given statements (A) and (B) are correct. Example 2.135 [CS-2014 (2 marks)]: A function f ( x ) is continuous in the interval [0, 2] . It is known that f (0) f (2) 1 and f (1) 1 . Which one of the following statements must be true?
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Engineering Mathematics
Chapter 2: Calculus
[2.63]
(a) There exist a y in the interval (0,1) such that f ( y ) f ( y 1) (b) For every y in the interval (0,1) , f ( y ) f (2 y ) (c) The maximum value of the function in the interval (0, 2) is 1 (d) There exist a y in the interval (0,1) such that f ( y ) f (2 y ) Solution (a): Let g ( x ) f ( x ) f ( x 1) in [0,1] , so g (0) f (0) f (1) 1 1 2 0 and g (1) f (1) f (2) 1 1 2 0 . By intermediate value theorem there is y (0,1) such that g ( y ) 0 , i.e., f ( y ) f ( y 1) . Example 2.136 [XE-2014 (2 marks)]: Let the function
f : [0, ) R
be such that
2
f ( x ) 8 ( x 3x 4) for x 0 and f (0) 1 . Then f (1) lies in the interval (a) [0,1] (b) [2, 3] (c) [4, 5] (d) [6, 7] Solution (b): We have to find the range of f (1) , so let us take the interval x [0,1] , as in x [0,1] the function is continuous and differentiable so we can apply Lagrange’s MVT, as c [0,1] s.t. 2 2 f (c) { f (1) f (0)} (1 0) f (1) f (0) f (c ) . f (c ) 8 (1 3 1 4) , 8 (0 3 0 4) (as
c [0,1] ). So f (c ) 1, 2 f (1) f (0) 1, f (0) 2 f (1) 2,3 . [Similar question was also asked in AG-2012 (2 marks)]
2.4.4 Partial and Total Derivatives of a Function Partial Derivative of First Order: Let z f ( x, y ) be a function of two independent variables x and y . Partial derivative at z , w.r.t. x is the ordinary derivative of z by keeping y as constant. Partial derivative of z w.r.t. y is the ordinary derivative of z by keeping x as constant. Similarly, if z f ( x1 , x2 ,, xn ) is a function of n independent variables x1 , x2 ,, xn , then the
partial derivative of z with respect to any one of the variables is the ordinary derivative of z w.r.t. that variable when other variable are regarded as constant. For z f ( x, y ) , then if lim f ( x x, y ) f ( x, y ) x exists, it is called the partial derivative x 0
x . It is denoted by z x f x lim f ( x x, y ) f ( x, y ) x .
of
f
w.r.t.
or
f x
or
fx
or
f x ( x, y ) , Hence
x 0
The partial derivative of f w.r.t. x at any point ( a, b) is denoted by (z x )( a ,b ) or (f x )( a ,b ) or f x (a, b) . Thus f x ( a, b) lim f (a h, b) f ( a, b) h . h0
For z f ( x, y ) , then if lim f ( x, y y ) f ( x, y ) y exists, it is called the partial derivative y 0
of
f
w.r.t.
y . It is denoted by z y
or
f y
or
fy
or
f y ( x, y) , Hence
f y lim f ( x, y y ) f ( x, y ) y . y 0
The partial derivative of f w.r.t. y at any point ( a, b) is denoted by (z y ) ( a ,b ) or (f y )( a ,b ) or f y (a, b) . Thus f y ( a, b) lim f (a, b h) f ( a, b) h . h0
Geometrical Interpretation of Partial Derivatives: We know that z P ( x , y ) represents a surface. Also z f ( x, b) represent a curve which is the intersection of the surface z f ( x, y ) and the plane y b . Now z f ( x, b) is a function of one variable and we know that ( d dx ) f ( x, b) at x a represents the slope of tangent to the curve z f ( x, b) at the point
a, f ( a, b) .
Now ( d dx ) f ( x, b) at x a is f x (a, b) . Hence partial derivative f x at ( a, b)
represents the slope of tangent to the curve z f ( x, y ) , y b . Similarly f y ( a, b) represents the slope of the tangent to the curve z f ( x, y ) , x a .
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Engineering Mathematics
Chapter 2: Calculus
Partial Derivatives of Second Order: If z f ( x, y ) and the first
[2.64]
f 2 f f x ( x x, y ) f x ( x , y ) 2 f xx lim x 0 x x x x 2 f y ( x, y y ) f y ( x, y ) f f f lim yy 2 y0 y y y y
partial derivatives f x and f y exists, then they are themselves the functions of x and y . Their partial derivatives w.r.t. x and y , if they exist are called second order partial derivatives w.r.t. x and y . Hence we have the following results:
f 2 f f ( x, y y ) f x ( x, y ) f yx lim x y 0 y x yx y 2 f y ( x x, y ) f y ( x, y ) f f f xy lim x 0 x y xy x
Example 2.137 [BT-2013 (1 mark)]: If u log(e x e y ) , then (u x) (u y) (d) 1 (a) e x e y (b) e x e y (c) 1 (e x e y ) Solution (d):
u
x
ex
y
log(e e )
x
y
u
and
x
x x e e y y [Similar question was also asked in AG-2014 (1 mark)]
log(e e )
2 f
x
Example 2.138 [ME-2008 (2 marks)]: Let f y . What is (a) 0
xy
x
e e
y
u x
u y
1
at x 2 , y 1 ?
(c) 1
(b) ln 2
ey
y
(d) 1 ln 2
Solution (c): 2 f f y x 2 f x 1 x 1 x 1 ( xy ) y xy log y e xy x y x y x xy
12 1 2 12 1 log e 1 1 (2,1)
Example 2.139 [EC-2014 (1 mark)]: If z xy ln( xy ) , then (a) x
z
y
z
0
(b) y
z
x
z
(c) x
z
y
z
(d) y
z
x
z
x y x y x y x y Solution (c): z z xy ln( xy ) ln( xy ) ( xy ) xy ln( xy ) ln( xy ) y y x xy ln( xy ) xy x x x x x z z xy ln( xy ) ln( xy ) ( xy ) xy ln( xy ) ln( xy ) x x y xy ln( xy ) xy y y y y y 2
2 1 2
2
Example 2.140 [CE-2000 (2 marks)]: If f ( x, y , z ) ( x y z ) equal to (a) Zero
(b) 1
Solution (a):
f x
2 f x
2
x
x
2
2
2 32
(x y z )
2 f x
2
f x
2
x 2
2 f y 2
and
2 f z 2
( x 2 y 2 z 2 ) 1 2 ( x 2 y 2 z 2 ) ( x2 y 2 z 2 )
x
f x 2 2 2 32 x x x ( x y z )
. Now using quotient rule of
( x ) ( x ) ( x 2 y 2 z 2 )3 2 ( x 2 y 2 z 2 )3 x x
( x 2 y 2 z 2 ) 3 2
( x 2 y 2 z 2 )3 2 3 x 2 ( x 2 y 2 z 2 )1 2 2
2
is
(d) 3( x 2 y 2 z 2 )5 2
(c) 2
( x 2 y 2 z 2 ) 1 2
x 2
differentiations,
f
then
2 f
0
2 3
(x y z )
Copyright © 2016 by Kaushlendra Kumar
( x 2 y 2 z 2 )1 2 ( x 2 y 2 z 2 ) 3x 2 2
2
2 3
(x y z )
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Engineering Mathematics
2 f x
2
Thus,
2 x2 y 2 z 2 2
2
Chapter 2: Calculus
. Similarly,
2 52
(x y z )
2 f x 2
2 f y 2
2 f z 2
2 f y 2
[2.65]
2 y 2 x2 z2
and
( x 2 y 2 z 2 )5 2
2x2 y2 z 2 ( x 2 y 2 z 2 )5 2
2 y 2 x2 z2 ( x 2 y 2 z 2 )5 2
2 f z 2
2z2 x2 y2 ( x 2 y 2 z 2 )5 2
2z 2 x2 y2 ( x 2 y 2 z 2 )5 2
0
Homogeneous Functions: A function z is said to be homogeneous of x and y if it can expressed in
the 3
xn f ( x y) .
form
3
3
Here
z ( x y ) ( x y ) x 1 ( y x)
3
n
is
the
degree
of
x 1 ( y x) x f x y . 2
the
function.
For
e.g.,
Clearly z is a homogeneous
function of degree (or order) 2. If z f ( x, y ) is a homogeneous function of degree n , then z x and z y are homogeneous functions of degree ( n 1) . Euler’s Theorem on Homogeneous: It states that if z be a homogeneous function of x and y of order n , then x
z
y
z
nz , for all ( x, y ) domain of the function [This point was asked x y in ME-1994 (1 mark)]. Proof: Since z is a homogeneous function of x , y of order n z x n f y x z x nx n 1 f ( y x ) x n f ( y x )( y x 2 ) x (z x) nz yx n 1 f ( y x )
Also, z y x n 1 f ( y x ) y (z y ) yx n 1 f ( y x) . Hence, x(z x) y( z y) nz Euler’s Theorem on Homogeneous Function for three variables: If u is a homogeneous u u u y z nu . The proof of the statement is function of degree n in x , y and z , then x x y z similar to the previous statement. Similarly, If u is a homogeneous function of degree n in u u u x1 , x2 , , xm , then x1 x2 xm nu . x1 x2 xm
Composite Functions: If z is a function of x , y and x , y are themselves functions of t , the z is said to be a composite function of t . Again if z is a function of x , y and x , y are themselves functions of u , v , respectively; then z is said to be a composite function of u and v .
Differentiation of Composite Functions: If a function z f ( x, y ) has continuous partial derivatives w.r.t. x and y and x , y have derivatives w.r.t. t , then
dz
z dx
z dy
. dt x dt y dt Proof: Let z f ( x, y ) , and t be increment in t and x , y and z be the corresponding increments in x, y and z , respectively. Then z z f ( x x, y y ) , z f ( x x, y y ) f ( x, y ) z f ( x x, y y ) f ( x, y y ) f ( x, y y ) f ( x, y ) z f ( x x, y y ) f ( x, y y ) x f ( x, y y ) f ( x, y ) y t x t y t Let t 0 x, y 0 . Also, lim x t dx dt and lim y t dy dt . Hence, t 0
dz
lim lim dt y 0 x 0
dz
t 0
f ( x x, y y ) f ( x, y y ) dx f ( x, y y ) f ( x , y ) dy lim x y dt y0 dt
f ( x, y y ) dx f ( x, y ) dy f ( x, y ) dx f ( x , y ) dy z dx z dy dt y dt x dt y dt x dt y dt x
lim dt y 0
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Engineering Mathematics
Chapter 2: Calculus
[2.66]
Corollary: If z f ( x, y ) has continuous partial derivatives w.r.t. x and y and x , y both have continuous partial derivatives w.r.t. u and v , then we have z z x z y z z x z y and . u x u y u v x v y v
Implicit Functions: Assuming that f ( x , y ) 0 be an implicit relation between x and y , satisfied the conditions under which y defined as a derivable function of x then
dy dx f x f y , f y 0
d 2 y dx 2 f xx ( f y )2 2 f yx f x f y f yy ( f x )2
( f y )3 , f y 0
Proof: Here f is a function of two variables x and y . Therefore differentiating f ( x , y ) 0 w.r.t. x , we get (f x)( dx dx) (f y )( dy dx) 0 dy dx (f x) (f y) f x f y , f y 0 .
Again differentiating above equation w.r.t. x d 2 y dx 2 f y
d dx
( fx ) fx
d dx
( fy )
( f y )2
2
dx dy dx dy 2 f y ( fx ) ( fx ) fx ( f y ) ( f y ) ( f y ) x dx y dx x dx y dx dx
d y 2
d2y
dy dy 2 f x f xy f yy ( fy ) dx dx dx 2 f f d y 2 f y f xx f yx x f x f xy f yy x ( f y ) 2 dx f y f y
2
d2y dx 2
f y f xx f yx
( f y ) 2 f xx f yx f x f y f x f y f xy f yy ( f x ) 2 ( f y )3
d 2 y dx 2 ( f y )2 f xx 2 f x f y f xy f yy ( f x )2
( f y ) 2 f xx 2 f x f y f xy f yy ( f x ) 2 ( f y )3
( f y )3 , f y 0 f xy f yx
Total Derivative
If z f ( x, y ) possesses continuous partial derivatives of first order; and x g1 (t ) and y g 2 (t ) , then the total differential of z is given by dz (z x) dx (z y) dy [This point was asked in ME-2000 (1 mark)]. Proof: Let z f ( x, y ) has continuous partial derivatives w.r.t. x and y ; and x , y have derivatives w.r.t. t , then we have dz dt (z x)(dx dt ) (z y )(dy dt ) . Multiplying both
sides with dt we get, dz (z x) dx (z y) dy . In the same way, if z f ( x , y , w) possesses continuous partial derivatives of first order; and x g1 (t ) ,
y g 2 (t )
and
w g3 (t ) , then the total differential of
dz (z x) dx (z y) dy (z w)dw . The above two results can be extended to any number of variables. If z f ( x, y ) and y g ( x ) then, ( dz dx) (z x) (z y )(dy dx)
If z f ( x, y ) and x g1 (t1 , t2 ) and y g 2 (t1 , t2 ) , then
is given by
z x z y z z x z y t1 x t1 y t1 t2 x t2 y t 2 Example 2.141 [TF-2008 (2 marks)]: The total derivative of a function u f ( x, y, z ) is expressed
as du
z
z
f x
dx
f y 2
dy
f z
dz . If u exp( x 2 y 2 ) sin z , then the expression for du is given by,
2
where k exp( x y )
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Engineering Mathematics
Chapter 2: Calculus
(a) du k[2 xdx 2 ydy ]cos z sin zdz (c) du k [(2 xdx 2 ydy ) cos z sin zdz ] Solution (d): u x ( x )(e x
u y ( y )(e
x2 y 2
u x ( x)(e
du e
x2 y 2
y2
(b) du k[2 xdx 2 ydy ]sin z cos zdz (d) du k [(2 xdx 2 ydy ) sin z cos zdz ]
sin z ) sin z ( x) e x
sin z ) sin z ( y )e
x2 y2
du (sin z )(2 x)e
2
sin z ) e
x2 y 2
x2 y2
[2.67]
x2 y2
2
(sin z )(2 y )e
( z ) sin z (cos z )e
dx (sin z )(2 y )e x
2
y2
y2
dy (cos z )e x
(2 x sin zdx 2 y sin zdy cos zdz ) e
x2 y2
(sin z )(2 x ) e x
x2 y2
2
y2
;
; and
x2 y2
2
y2
dz
(2 xdx 2 ydy) sin z cos zdz
Example 2.142 [PI-2009 (1 mark)]: The total derivative of the function ‘ xy ’ is (a) xdy ydx (b) xdx ydy (c) dx dy (d) dx dy Solution (a): Let z xy , then total derivation of the function z is given as dz
z
dx
x
z y
dy dz
( xy ) x
dx
( xy ) y
dy ydx xdy
Example 2.143 [XE-2010 (2 marks)]: Let u ( x, y ) tan xy ( x y ) . Then, (a) x (c) x
u x u
y y
u y u
1 3
xy ( x y ) sec 2 xy ( x y )
(b) x
3 xy ( x y ) sec 2 xy ( x y )
(d) x
u x u
y y
u y u
1 3
xy ( x y ) sec 2 xy ( x y )
3xy ( x y ) sec 2 xy ( x y )
x y x y Solution (d): u tan xy ( x y ) sec 2 xy ( x y ) xy ( x y ) sec 2 xy ( x y ) 2 xy y 2 ; x x x u tan xy ( x y ) sec 2 xy ( x y ) xy ( x y ) sec 2 xy( x y ) 2 xy x 2 y y y u u sec 2 xy ( x y ) 2 xy 2 x 2 y x sec 2 xy ( x y ) 2 x 2 y xy 2 and y x y u u x y sec 2 xy ( x y ) 2 x 2 y xy 2 2 xy 2 x 2 y 3 xy sec 2 xy ( x y ) ( x y ) x y
Example 2.144 [XE-2012 (1 mark)]: Let f (u , v) u ln(v) and F ( x, y ) f u ( x, y ), v ( x , y ) , where u x y and v x y . Then F y is
(a)
x y
2
Solution u y
y
x
ln( x y ) (b): ( x y)
(b)
x 2
ln(v )
y( x y ) y F f u f v , y u y v y x
y
2
,
v y
y
u v
so
( x y ) 1
(c)
F y
f u
x y u
2
ln(v )
u v
u ln(v ) ln(v ) ,
(d) f v
x y
2
ln(v ) v
x yv
u ln(v )
u v
,
x u x u ( 1) 2 ln(v ) 2 y v y v
ln(v )
Maxima and Minima of Functions in Two Variables: Let z f ( x, y ) be a function with two independent variables ( x and y ) with continuous second order partial derivatives 2 z x 2 or z xx , 2 z y 2 or z yy , 2 z (xy ) or z xy and 2 z (yx) or z yx . For a well behaved function 2 z (xy ) 2 z (yx ) or z xy z yx . The critical values, which occur at z x 0 and z y 0 ,
are of three types:
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Engineering Mathematics
Chapter 2: Calculus
[2.68]
A critical point is a maximum if the value of z at that point is greater than its value at all sufficiently close surrounding points. A critical point is a minimum if the value of z at that point is less than its value at all sufficiently close surrounding points. A critical point is a saddle point if the value of z at that point is greater than its value at some sufficiently close surrounding points and less than its value at other sufficiently close surrounding points.
Procedure for finding maxima and minima of a function = ( , ): Find the critical points ( x0 , y0 ) by solving z x 0 and z y 0
At each critical point evaluate the second order partial derivatives z xx , z yy and z xy .
Calculate the discriminant D
z xx
z xy
z yx
z yy
z xx
z xy
z xy
z yy
2 z xx z yy z xy
z xy z yx . Now if
D 0 and z xx
( x0 , y0 )
0 then ( x0 , y0 ) is the point of maxima.
D 0 and z xx
( x0 , y0 )
0 then ( x0 , y0 ) is the point of minima.
D 0 then z has a saddle point at ( x0 , y0 ) [This point was asked in CE-1998 (1 mark)] D 0 then no conclusion can be drawn.
Example 2.145 [EC-1993, CS-1993 (1 mark)]: The function f ( x, y ) x 2 y 3 xy 2 y x has (a) no local extremum (b) one local minimum but no local maximum (c) one local maximum but no local minimum (d) one local minimum and one local maximum 2 Solution (a): As, f x 2 xy 3 y 1 , f y x 3 x 2 , f xx 2 y , f yy 0 and f xy 2 x 3 ; so for critical
points
of
f ( x, y ) ,
we
f y 0 x 2 3x 2 0 x 1, 2
have
and
fx 0
2 xy 3 y 1 0 y x 1 1, y x 2 1 . So we have two critical points (1,1) and (2, 1) . Also, f xx
(1,1)
2 , f yy
(1,1)
point. Again, f xx
0 , f xy
(2, 1)
2
(1,1)
2
1 D (1,1) f xx f yy f xy 0 ( 1) 1 0 (1,1) is a saddle
2 , f yy
(1,1)
0 , f xy
2
(2,1)
2
1 D (2, 1) f xx f yy f xy 0 (1) 1 0
( 2,1) is also a saddle point. Example 2.146 [ME-2002 (2 marks)]: The function f ( x, y ) 2 x 2 2 xy y 3 has (a) only one stationary point at (0, 0) (b) two stationary points at (0, 0) & 1 6 , 1 3 (c) no stationary point (d) two stationary points at (0, 0) & (1, 1) 2
Solution (b): As, f x 4 x 2 y , f y 2 x 3 y , f xx 4 , f yy 6 y , f xy 2 . So for critical or 2
stationary points of f ( x, y ) , f x 0 4 x 2 y 0 and f y 0 2 x 3 y 0 ; on solving we get
x 0,1 6 y x 0 0 and y x 1 6 1 3 . Thus two critical points: (0, 0) and 1 6 , 1 3 . 2
Example 2.147 [PI-2007 (2 marks)]: The function f ( x, y ) x 2 y 2 defined on R , the point [0, 0] is (a) a local minimum (b) neither a local minimum nor a local maximum (c) a local maximum (d) both a local minimum and a local maximum Solution (b): As, f x 2 x , f y 2 y , f xx 2 , f yy 2 , f xy 0 . For the critical points of f ( x, y ) , we have f x 0 2 x 0 x 0 and f y 0 2 y 0 y 0 ; so (0, 0) is the critical point of
f ( x, y ) . Also D f xx f yy f xy2 4 02 4 0 (0, 0) is a saddle point.
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.69]
Example 2.148 [XE-2007 (2 marks)]: Let f : R R be twice differentiable real valued function such that f 1 n 1 for n 1, 2, 3, . Then (a) f (0) 0 (b) f (0) 1 (c) 0 f (0) 1 (d) f (0) 1 Solution (a): We have f 1 n 1 (d dn) f (1 n) (d dn)1 f (1 n) ( 1 n 2 ) 0 f (1 n) 0 . So, as n f (1 ) 0 f (0) 0 Example 2.149 [ME-2009 (2 marks)]: The distance between the origin and the point nearest to it on the surface z 2 1 xy is (a) 1
(b)
(c)
3/2
(d) 2
3 2
Solution (a): Let P ( x, y , z ) be any point on the surface z 1 xy , then distance between P and origin is PO ( x 0) 2 ( y 0) 2 ( z 0) 2
x2 y 2 z 2
x 2 y 2 1 xy . So the distance
between origin and point nearest to the surface z 2 1 xy is the minimum value of PO , i.e., minimum value of f ( x, y ) x 2 y 2 1 xy . Now f x 2 x y f xx 2 and f y 2 y x f yy 2 , also f xy f yx 1 . So for critical points we have f x 0 2 x y 0 and f y 0 2 y x 0 an solving these two equation we get x 0, y 0 , so ( x, y ) (0, 0) is the critical point. As the 2
2
discriminant D f xx f yy f xy 4 1 3 0 and f xx 2
(0,0)
2 0 , thus (0, 0) is the point of minima
2
for f ( x, y ) and also for PO . Hence PO (0,0) 0 0 1 0 1 Example 2.150 [AE-2010 (2 marks)]: The function f ( x, y ) x 2 y 2 xy 3 y has an extremum at the point (a) (1, 2) (b) (3, 0) (c) (2, 2) (d) (1, 1) Solution (a): As, f x 2 x y , f y 2 y x 3 , f xx 2 , f yy 2 , f xy 1 . For critical points we have f x 0 2 x y 0 and f y 0 2 y x 3 0 ; solving these two equations we have
x 1, y 2 ; so (1, 2) is the critical point. As the discriminant D f xx f yy f xy2 4 1 3 0 (1, 2) is not a saddle point. Example 2.151 [CE-2010 (2 marks)]: Given a function f ( x, y ) 4 x 2 6 y 2 8 x 4 y 8 . The optimal value of f ( x, y ) (a) is a minimum equal to 10/3 (b) is a maximum equal to 10/3 (c) is a minimum equal to 8/3 (d) is a maximum equal to 8/3 Solution (a): As, f x 8 x 8 , f y 12 y 4 , f xx 8 , f yy 12 , f xy 0 ; so for the critical points of
f ( x, y ) , f x 0 x 1 and f y 0 y 1 3 ; so (1,1 3) is the critical point. Also the discriminant D f xx f yy f xy2 96 . As D 0 and f xx
(1,1 3)
8 0 , thus (1,1 3) is the point of minima and its
value is f (1,1 3) 4 12 6 (1 3) 2 8 1 4 (1 3) 8 10 3 . Example 2.152 [AG-2011 (2 marks)]: The stationary points of f ( x, y ) (1 3) x 3 xy 2 2 y are (a) (1, 1) & ( 1,1) (b) (1,1) & ( 1, 1) (c) (2, 2) & ( 2, 2) (d) (2, 2) & ( 2, 2) Solution (a): As, f x 2 x y , f y 2 y x 3 , f xx 2 , f yy 2 , f xy 1 . For critical points we have f x 0 2 x y 0 and f y 0 2 y x 3 0 ; solving these two equations we have
x 1, y 2 ; so (1, 2) is the critical point. As the discriminant D f xx f yy f xy2 4 1 3 0 (1, 2) is not a saddle point.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.70]
Exercise: 2.4 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The distance travelled s (in metre) by a particle in t second is given by s t 3 2t 2 t . The speed (in cm/s) of the particle after 1 sec will be _____. 2. A man of height 1.8 m is moving away from a lamp post at the rate of 1.2 m/sec. If the height of the lamp post be 4.5 meter, then the rate (in m/sec) at which the shadow of the man is lengthening _____. 3. The line x y 2 is tangent to the curve x 2 3 2 y at its point (b) ( 1,1)
(a) (1,1)
(d) (3, 3)
(c) ( 3, 0)
4. The equation of the normal to the curve y sin( x 2) at (1,1) is (a) y 1 (c) y x (b) x 1
(d) y 1 (2 )( x 1)
5. If the normal to the curve y f ( x) at the point (3, 4) makes an angle 3 4 with the positive x axis then f (3) is equal to _____. 6. The point(s) on the curve y 3 3 x 2 12 y where the tangent is vertical (parallel to y axis), is are (a) ( 4
(c) (0, 0)
(b) ( 11 3,1)
3 , 2)
2
(d) ( 4
3 , 2)
2
7. The angle between the curves y x and x y at (1,1) is (a) tan 1 (4 3)
(b) tan 1 (3 4)
(c) 90o
(d) 45o
8. The length of sub-tangent to the curve x 2 y 2 a 4 at the point ( a, a ) is (a) 3a (b) 2a (c) a 9. The sum of intercepts on co-ordinate axes made by tangent to the curve (a) a
(b) 2a
(d) 4a
x
y a is
(d) None of these
(c) 2 a
10. The length of perpendicular from (0, 0) to the tangent drawn to the curve y 2 4( x 2) at point (2, 4) is (a) 1
(b) 3
2
(c) 6
5
(d) 1
5 5
4
3
11. What are the minimum and maximum values of the function x 5 x 5 x 10 ? (b) It has 2 minimum and 1 maximum values (a) 37, 9 (d) It has 2 maximum and 1 minimum values (c) 10, 0 12. If y a log x bx 2 x has its extremum value at x 1 and x 2 , then ( a, b) (a) (1,1 2)
(b) (1 2 , 2)
(c) (2, 1 2)
(d) ( 2 3, 1 6)
(c) (e) e
(d) (1 e)e
x
13. Maximum value of (1 x ) is (a) (e) e
(b) (e)1 e
14. Maximum slope of the curve y x3 3 x 2 9 x 27 is _____. x
15. The function f ( x ) t (et 1) (t 1) (t 2) 3 (t 3)5 dt has a local minimum at maximum x 1
_____. 16. If the function f ( x ) 2 x 3 9ax 2 12a 2 x 1 , where a 0 attains its maximum and minimum at p and q respectively such that p 2 q , then non-zero value of a equals _____.
17. The maximum and minimum values of x 3 18 x 2 96 in interval (0,9) are (a) 160, 0 (b) 60, 0 (c) 160, 128 (d) 120, 28 2 18. On [1, e] the greatest value of x log x is (a) e2
(b) (1 e) log(1
Copyright © 2016 by Kaushlendra Kumar
e)
(c) e 2 log e
(d) None of these
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Engineering Mathematics
Chapter 2: Calculus
[2.71]
19. x and y be two variables such that x 0 and xy 1 . Then the minimum value of x y is _____. 20. The real number which most exceeds its cube is (a) 1 2 (d) None of these (b) 1 3 (c) 1 2 21. The radius of the cylinder of maximum volume, which can be inscribed a sphere of radius R is (a) 2 R 3 (c) 3R 4 (b) 2 3R (d) 3 4R 22. The ratio of height of a cone having maximum volume which can be inscribed in a sphere with the diameter of sphere is (a) 2 3 (b) 1 3 (c) 3 4 (d) 1 4 23. On the interval (1,3) the function f ( x) 3 x (2 x) is (a) Strictly decreasing (b) Strictly increasing (c) Decreasing in (2,3) only (d) Neither increasing nor decreasing 24. For which value of x , the function f ( x ) x 2 2 x is decreasing (a) x 1 (b) x 2 (c) x 1 x 25. The function x is increasing, when (c) x 0 (a) x 1 e (b) x 1 e 26. The function f ( x ) cos x 2 px is monotonically decreasing for (c) p 2 (a) p 1 2 (b) p 1 2
(d) x 2 (d) For all real x (d) p 2
27. The value of a for which the function ( a 2) x 3 3ax 2 9ax 1 decrease monotonically throughout for all real x , are (a) a 2 (b) a 2 (c) 3 a 0 (d) a 3 28. Function f ( x) ( sin x 6 cos x) (2 sin x 3cos x) is monotonic increasing if (a) 1 (b) 1 (c) 4 (d) 4 29. The function f ( x ) ln( x ) ln(e x ) is (a) Increasing on [0, )
(b) Decreasing on [0, e) and increasing on [ e , )
(c) Decreasing on [0, )
(d) Increasing on [0, e) and decreasing on [ e , )
30. f ( x ) xe x (1 x ) then f ( x ) is (a) Increasing on [ 1 2 ,1]
(b) Decreasing on R
(c) Increasing on R
(d) Decreasing on [ 1 2 ,1]
31. If the function f ( x ) x 3 6 x 2 ax b satisfies Rolle’s theorem in the interval [1, 3] and
f (2 3 1)
3 0 then
(a) a 11 (b) a 6 (c) a 6 32. In the mean-value theorem f (b) f (a ) (b a) f (c ) , f ( x ) x ( x 1)( x 2), the value of c is (a) 1 ( 15 6)
(c) 1 ( 21 6)
(b) 1 15
if
(d) a 1 a 0 , b 1 2
and
(d) 1 21
3
33. The abscissae of the points of the curve y x in the interval [ 2, 2] , where the slope of the tangent can be obtained by mean value theorem for the interval [ 2, 2] are (a) 2
3 1
(b) 3 2
(c) 3
(d) 0
3
34. If u sin ( x y ) , x 3t , y 4t then du dt (b) 3 1 t 2 (d) 3 (c) 3 1 t 2 x u u u y z y z _____. 35. If u f ( r , s , t ) and r , s , t , then x y z x x y z (a) 3 1 t 2
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Engineering Mathematics
2.5
Chapter 2: Calculus
[2.72]
Integration
2.5.1 Indefinite Integrals Integration is the reverse process of differentiation. If g ( x ) is a differentiable function such that g ( x ) f ( x ) , then integral of f ( x ) w.r.t. x is g ( x) c . Symbolically,
c is any constant. Here
f ( x)dx g ( x) c , where
is the integral sign, f ( x ) is the integrand, x is the variable of integration,
g ( x ) is a primitive or an anti-derivative of a function f ( x ) , and dx is the element of integration. If a function f ( x ) possesses a primitive, then it possesses infinitely many primitives which are given by g ( x) c , where c is a constant. For e.g. x 2 , x 2 2 , x 2 1 , etc. are primitives of 2x .
Based upon the definition ( d dx){ g ( x )} f ( x ) f ( x ) dx g ( x ) c , we have:
k dx kx c , k is any constant n n 1 x dx x (n 1) c , n 1 2 cosec x dx cot x c sin x dx cos x c x x e dx e c
dx 1 x
1 sin 1 x c1 or cos x c2
2
ax
x
a dx
log e a
c
2
sec x dx tan x c cos x dx sin x c sec x tan x dx sec x c cosec x cot x dx cosec x c (1 x)dx log x c , x 0 dx
1 x 2 tan
dx x
1
x c1 or cot 1 x c2
1 1 sec x c1 or cosec x c2
2
x 1
Two integrals of the same function can differ only by a constant
The differentiation of an integral is the integrand itself, i.e. ( d dx ) f ( x ) dx f ( x ) .
If c is any constant, then
Integral of the sum or difference of two functions is equal to the sum or difference of their integrals, i.e.,
c f ( x) dx c f ( x)dx .
f1 ( x) f 2 ( x) dx f1 ( x) dx f 2 ( x) dx .
Method of Integration: The following are some methods of integration which are given as:
Direct substitution When integrand is a function, i.e.,
= ∫ { ( )} ′( )
: If I f (t ) dt g (t ) c , then
we put h( x) t h( x ) dx dt I f (t ) dt g (t ) c g{h( x )} c . Integral of a function of the form = ∫ f ( x ) t f ( x ) dx dt I t dt . Integral of a function of the form
=∫ (
( ) ( ) + )
: In this case we put
: Put ax b t dx (1 a )dt
I (1 a) f (t ) dt . Obviously if
f ( x)dx g ( x), then, f (ax b)dx (1 a ) g (ax b) c Integral of a function of the form I f ( x ) f ( x ) dx : Put f ( x) t f ( x)dx dt and convert it into standard integral as, I (1 t ) dt log t c log f ( x ) c . Integral of a function of the form I { f ( x )} f ( x )dx : Put f ( x) t f ( x)dx dt , so we have I { f ( x )}n f ( x ) dx { f ( x )}n 1 ( n 1) c , n 1
Copyright © 2016 by Kaushlendra Kumar
n
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Engineering Mathematics
Chapter 2: Calculus
Example 2.153 [CH-2013 (1 mark)]: Evaluate (a)
e
x
(e x 1) c
(b)
Solution (d): Let I
dx x
e 1
ln (e
x
1 (e
1) e x c
x
(c)
[2.73]
ln e
1) dx (Note: c is constant of integration) x
(e x 1) c
x
(d) ln(1 e ) c
ex ex 1
ex ex dx x 1 dx x dx dx . In first integral, x e 1 e 1 e 1 x
x
x
put e x 1 t e x dx dt , I (1 t )dt dx ln t x c ln(e 1) ln e c ln(1 e ) c Example 2.154 [MT-2008 (1 mark)]:
1 (a bx) dx
(b) ln( a bx) c
(a) (1 b) ln( a bx) c
(c) b ln( a bx ) c
Solution (a): Let a bx t bdx dt dx
is
dt
dx
(d) (1 a) ln( a bx) c
1 dt 1 1 ln t c ln(ax b) c t b b
a bx b
2 Standard Substitution: The following standard substitutions are used according the type of integrand as: Integrand Substitution : x a sin , x a cos 2 2 2 2 2 2 a x , 1 a x , (a x ) : x a tan x2 a2 , 1 x2 a2 , x2 a 2 : x a sec or x a cosh x 2 a 2 , 1 x 2 a 2 , x2 a2 : x a tan 2 x (a x ), ( a x) x , x (a x ),1 x ( a x )
x ( a x ), ( a x) x , x( a x ),1
x(a x)
:
x a sin
x ( x a ), ( x a ) x , x( x a ),1
x( x a)
:
x a sec
(a x) ( a x), (a x ) (a x)
:
x a cos 2
( x ) ( x ), ( x ) ( x), ( )
:
x cos2 sin 2
2 2
Indirect Substitution: If integrand
f ( x) f1 ( x) f2 ( x) , where f 2 ( x ) is a function of integral of f1 ( x) , then put integral of f1 ( x) t . u and v are two functions of x , then Integration by Parts: If
uI IIv dx u vdx (du
f ( x ) can be written as product of two functions
dx ) vdx} dx , i.e., the integral of the product of two functions (First
function) (Integral of second function) – Integral of {(Differentiation of first function) (Integral of second function)}. Integration with the help of this rule is called integration by parts. Before applying this rule proper choice of first and second function is decided as: In the product of two functions, one of the function is not directly integrable (i.e., log x ,sin 1 x, cos 1 x, tan 1 x ...etc), then we take it as the first function and the remaining function is taken as the second function. If there is no other function, then unity is taken as the second function, for e.g. In the integration of sin 1 x dx, log x dx,1 is taken as the second function. If both of the function are directly integrable then the first function is chosen in such a way that the derivative of the function thus obtained under integral sign is easily integrable. Usually, we use the following preference order for the first function: Inverse, Logarithmic, Algebraic, Trigonometric, exponential. This rule is simply called as I L A T E . Example 2.155: If I n (log x) n dx, then find the value of I n nI n1 .
Solution: I n (log x ) .1 dx (log x ) n x n (log x ) n 1 (1 x ) xdx (log x) n x n (log x) n 1 dx . n
I n (log x ) n dx I n 1 (log x) n 1dx I n x (log x ) n n I n 1 I n n I n 1 x (log x) n
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Engineering Mathematics
Chapter 2: Calculus
[2.74]
form ∫ { ( ) + ′( )} : If the integral is of the form e f ( x) f ( x) dx, then by breaking this integral into two integrals integrate one integral by
Integral is
of the
x
parts and keeping other integral as it is, by doing so, we get e x f ( x) f ( x ) dx e x f ( x ) c e mx mf ( x ) f ( x ) dx e mx f ( x) c
mx mx e f ( x) (1 m) f ( x) dx (1 m)e f ( x) c
( ) + ( )}
Integral is of the form ∫{
: If the integral is of the form [ x f ( x ) f ( x )] dx
then by breaking this integral into two integrals, integrate one integral by parts and keeping other integral as it is, by doing so, we get, [ x f ( x ) f ( x )] dx x f ( x) c
Integrals of the form ∫
e
ax
, ∫
: To evaluate
e
ax
sin bx dx or
cos bx dx , (i) Put the given integral equal to I . (ii) Integrate by parts, taking e ax as the first
function. (iii) Again, integrate by parts taking e ax as the first function. This will involve I . (iv) Transpose and collect terms involving I and then obtain the value of I . Let I e ax sin bx dx . ax
Then I e sin bx dx e ax (1 b) cos bx ae ax (1 b)( cos bx )dx I
II
ax
ax
I (1 b)e cos bx ( a b) (e cos bx ) dx I
II
ax
I (1 b)e cos bx ( a b) (1 b)e ax sin bx ae ax (1 b) sin bx dx I (1 b) eax cos bx (a b2 )eax sin bx (a 2 b2 ) I I (1 b) e ax cos bx ( a b 2 )e ax sin bx ( a 2 b 2 ) e ax sin bx dx
I I ( a 2 b 2 ) (e ax b 2 )( a sin bx b cos bx ) I e ax ( a 2 b 2 ) (a sin bx b cos bx ) c
ax ax 2 2 e sin bx e (a b ) (a sin bx b cos bx) c ax ax 2 2 e cos bx dx e (a b ) (a cos bx b sin bx) c ax ax 2 2 e sin(bx c) dx e (a b ) a sin(bx c) b cos(bx c) k ax ax 2 2 e cos(bx c) dx e (a b ) a cos(bx c) b sin(bx c) c
Integrals of the form
dx
ax2 bx c
where ax 2 bx c cannot be resolved into factors: To
evaluate integrals of the above form (i) Make the coefficient of x 2 unity by taking ‘ a ’ common from ax 2 bx c (ii) Express the terms containing x 2 and x in the form of a perfect square by adding and subtracting the square of half of the coefficient of x . (iii) Put the linear expression in x equal to t and express the integrals in terms of t . (iv) The resultant integrand will be either in standard form which are given as: dX 1 X dX 1 X A 2 tan 1 c 2 ln c 2 2 X A A A X A 2A X A
Integral of the form
px q
ax 2 bx c dx : Breaking
px q into two parts such that one part is the
differential coefficient of the denominator and the other part is a constant. If M and N are two 2
px q px q M (d dx)(ax bx c) N M (2ax b) N constants, then as px q (2aM ) x Mb N . Comparing the coefficients of x and constant terms on both sides, we have, p 2 aM M p 2 a and q Mb N N q Mb q ( p 2 a )b . Hence,
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Engineering Mathematics
the given integral is
Chapter 2: Calculus
px q
2ax b
Integral of the form x
2
unity dx
ax 2 bx c
by 1
a
2
bx c ;
dx
ax2 bx c
dx
p
dx
where,
is evaluated by the method discussed already.
: To evaluate the given integrals make the coefficient of
2
ax bx c from ax 2 bx c . Then, dx b c . Put x 2 x , by the method of completing a a x 2 (b a) x (c a )
a
taking
2
2ax b
p
ax2 bx c dx 2a ax2 bx c dx q 2a b ax 2 bx c ,
ax 2 bx c dx log ax
[2.75]
common
2
2
2
2
2
the square in the form, A X or X A or X A where, X is a linear function of x and A is a constant. After this, use any of the following standard formulae according to the case under consideration dx dx x log x x 2 a 2 c sin 1 c 2 2 2 2 a x a a x dx log x x 2 a 2 c 2 2 x a px q Integral of the form dx : To evaluate this type of integrals, proceed as follows: ax 2 bx c d Let px q M ( ax 2 bx c ) N px q M (2ax b) N , where M and N are dx constants. By equating the coefficients of x and constant terms on both sides, we get p 2 aM M p 2 a and q bM N N q (bp 2 a ) . So, integral breaks up into two parts:
px q
dx
p
2a
2ax b
dx q
bp
dx
I1 I 2 , (say) 2a ax 2 bx c
ax 2 bx c ax 2 bx c p 2ax b 2 Now, I1 dx . Putting ax bx c (2ax b)dx dt , we have, 2 2a ax bx c I1
p
2a section.
t
1 2
dt
Integrals of the form
p
t1 2
2a (1 2)
C1
p
2 ax bx c C1 . I 2 is calculated as in the previous
a
f ( x)
ax 2 bx c dx , where
f ( x ) is a polynomial of degree 2 or greater than
2:
To evaluate this type of integrals, divide the Nr by Dr, we get, f ( x) R( x) ax 2 bx c dx Q( x)dx ax 2 bx c dx , where, Q( x) is a polynomial and R( x) is a linear polynomial in x . The integrals on R.H.S. can be obtained by the methods discussed earlier. x2 1 x2 1 Integrals of the form 4 dx and x 4 kx 2 1dx : To evaluate the integral of the form x kx 2 1 x2 1 1 (1 x 2 ) 2 I 4 dx , divide the Nr and Dr by to get x I x 2 k (1 x 2 ) dx Put x kx 2 1
x (1 x) t 1 (1 x 2 ) dx dt and x 2 (1 x 2 ) 2 t 2 x 2 (1 x2 ) t 2 2 . Then, the given integral reduces to the form I
dt 2
t 2k
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, which can be integrand as usual. To evaluate
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Engineering Mathematics
I
x2 1 x 4 kx 2 1
Chapter 2: Calculus
[2.76]
dx, first divide the Nr and Dr by x 2 and we get I
2
2
2
2
1 (1 x 2 ) 2
2
x k (1 x )
2
2
dx . Now,
2
let x (1 x) t 1 (1 x ) dx dt and x (1 x ) 2 t x (1 x ) t 2 . Thus, we have t
dt 2
t 2k
, which can be evaluated as usual.
Integrals of the forms
2
ax bx cdx : To evaluate this form of integrals, express
2 2 2 ax bx c in the form a ( x ) by the method of completing the square and apply the standard result discussed in the above section according to the case as may be
ax 2 bx cdx
(2 ax b ) ax 2 bx c 4a
4ac b 2 8a
dx
ax 2 bx c
. The following are the
2
result based on the form
ax bx cdx :
2
2
2
2
2
2
2
2
2
2
2
2
x a dx ( x 2) x a ( a 2) log( x x a ) c 1
a x dx ( x 2) a x ( a 2) sin ( x a) c
Integrals of the form
2
( px q)
ax bx cdx : To evaluate this form of integral, first put,
( px q ) as px q M ( d dx )( ax 2 bx c ) N px q M (2 ax b ) N where, M and N are constant. Compare the coefficients of x and constant terms on both sides, will get p 2 a M M p 2 a and q Mb N N q Mb q ( p 2 a )b . Now, the given integral is,
I
p
(2ax b) 2a
ax 2 bx c dx q
pb
p pb 2 ax bx c dx I1 q I 2 . To 2a 2a 2a
evaluate I1 , put ax 2 bx c t and to evaluate I 2 , follows the method discussed previously.
dx
P
, (where P and Q are linear or quadratic expressions in x ): To Q evaluate such types of integrals, we have following substitutions of P and Q in x : Integrals of the form
When Q is linear and P is linear or quadratic, we put Q t
2
When P is linear and Q is quadratic, we put P 1 t When both P and Q are quadratic, we put x 1 t
Integral of the form
dx
dx
a b cos x a b sin x ,
2
: Put cos x
1 tan 2 ( x 2) 2
1 tan ( x 2)
, sin x
2 tan( x 2) 1 tan 2 ( x 2)
2
and then replace 1 tan ( x 2) in the numerator by sec ( x 2) and then put tan( x 2) t so that dt (1 2) sec2 ( x 2)dx dt . Now we get 2 which is solved by method discussed earlier. at bt c dx Integral of the form : To evaluate, put b r cos and c r sin . a b cos x c sin x dx dx 2 2 2 . r b c and tan 1 ( c b ) . I a r (cos cos x sin sin x) a r cos( x ) Let x t dx dt , I
dt a r cos t
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, which is evaluated by the method discussed earlier.
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Engineering Mathematics
Chapter 2: Calculus
Integral of the form I a r cos ,
I
b r sin
dx a sin x b cos x
and
: To evaluate this type of integrals we substitute
r a 2 b 2 , tan 1 (b a ) .
so
x
1
1
[2.77]
cosec( x ) dx log tan r r 2 2
1 2
a b
The given integral can also be evaluated by using cos x
Integral of the following forms: dx dx dx a b cos2 x , a b sin 2 x , a sin 2 x b cos2 x ,
I
1
dx
r sin( x )
x 1 1 b tan c. a 2 2
log tan
1 tan 2 ( x 2) 2
1 tan ( x 2)
and sin x
dx
2 tan( x 2) 2
1 tan ( x 2)
.
dx
(a sin x b cos x)2 , a b sin 2 x c cos 2 x . To
evaluate above forms of integrals divide both the Nr and Dr by cos 2 x . Replace sec 2 x in Dr, if 2 2 any by (1 tan x) . Put tan x t sec xdx dt . Now, evaluate the integral thus obtained, by the method discussed earlier. a sin x b cos x Integrals of the form dx : Such rational functions of sin x and cos x are c sin x d cos x integrated by expressing the Nr of the integrand as: Nr M (Differentiation of Dr) N (Dr), i.e., d a sin x b cos x M (c sin x d cos x ) N (c sin x d cos x ) . The arbitrary constants M , N dx are determined by comparing the coefficients of sin x and cos x . So, the given integral is a sin x b cos x M (c cos x d sin x ) N (c sin x d cos x) I dx dx c sin x d cos x c sin x d cos x c cos x d sin x I M dx N 1dx M log c sin x d cos x Nx c. c sin x d cos x a sin x b cos x q Integrals of the form dx : To evaluate this type of integrals, express the Nr. c sin x d cos x r as, (c sin x b cos x q) M (c sin x d cos x r ) N (c cos x d sin x ) P , where M , N , P are constants to be determined by comparing the coefficients of sin x, cos x and constant term.
2
So,
a sin x b cos x q Diff.of denominator dx dx M dx N dx c sin x d cos x r Denominator c sin x d cos x r a sin x b cos x q c sin x d cos x r
dx Mx N log Denominator P
Integral of the form
sin
m
dx c sin x d cos x r
n
x cos x dx : (i) To evaluate the integrals of the form
I sin m x cos n x dx, where m and n are rational numbers.
Substitute sin x t if n is odd Substitute cos x t if m is odd
Substitute tan x t if m n is a ve even integer Substitute cot x t if ( n 1) 2 is an integer.
If m and n are rational numbers and (m n 2) 2 is a negative integer, then substitution cos x t or tan x t is found suitable.
Integrals of the form
R (sin x, cos x) dx,
where R is a rational function of sin x and cos x, are
transformed into integrals of a rational function by the substitution tan( x 2) t , where x , or cot( x 2) t for 0 x 2 . The above substitution enables us to integrate any function of the form R (sin x, cos x). However, it sometimes leads to extremely complex rational function. In some cases, the integral can be simplified by:
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Engineering Mathematics
Chapter 2: Calculus
[2.78]
Substituting sin x t , if the integral is of the form
R (sin x) cos x dx. Substituting cos x t , if the integral is of the form R (cos x) sin x dx. Substituting tan x t , i.e., dx 1 (1 t 2 ) dt , if the integral is dependent only on
tan x.
Substituting cos x t , if R ( sin x, cos x) R (sin x, cos x) Substituting sin x t , if R (sin x, cos x ) R (sin x, cos x) Substituting tan x t , if R ( sin x, cos x ) R (sin x, cos x)
Integrals of the form
x
m
(a bx n ) p dx : If p N (Natural number). We expand the integral
with the help of binomial theorem and integrate. If p I , write x t k where k is the LCM of denominator of m and n . n
k
If {( m 1) n} I and p fraction put (a bx ) t , where k is the Dr of the fraction p . If
(m 1) n P I
n
k n
and P fraction. We put (a bx ) t x , where k is the
denominator of the fraction P .
Integration of Rational Function by using Partial Fraction: Functions of the form f ( x) g ( x) , where f ( x ) and g ( x ) are polynomial and g ( x ) 0, are called rational functions of x . If degree of f ( x ) is less than degree of g ( x ) , then f ( x) g ( x) , is called a proper rational function. If degree of f ( x ) is greater than or equal to degree of g ( x ) , then f ( x) g ( x) , is called an improper rational function and every improper rational function can be transformed to a proper rational 3 2 function by dividing the numerator by the denominator. For example, x ( x 5 x 6) is an improper
rational function and can be expressed as ( x 5) (19 x 30) ( x 2 5 x 6) which is the sum of a
polynomial ( x 5) and a proper function (19 x 30) ( x 2 5 x 6) . Partial fractions: Any proper rational function can be broken up into a group of different rational fractions, each having a simple factor of the denominator of the original rational function. Each such fraction is called a partial fraction. If by some process, we can break a given rational function f ( x) g ( x) into different fractions, whose denominators are the factors of g ( x ) , then the process of obtaining them is called the resolution or decomposition of f ( x) g ( x) into its partial fractions. Depending on the nature of the factors of the denominator, the following cases arise. When the denominator consists of non-repeated linear factors: To each linear factor ( x a ) occurring once in the denominator of a proper fraction, there corresponds a single partial fraction of the form A ( x a ) , where A is a constant to be determined by equating the coefficients of like powers of x in the resulting identity and solving the equations. When the denominator consists of linear factors, some repeated: Each linear factor ( x a ) occurring r times in the denominator of a proper rational function, corresponds a sum of r A1 A2 Ar partial fractions of the form, , where A ' s are constants to be 2 r ( x a ) ( x a) (x a) determined by equating the coefficients of like powers of x in the resulting identity and solving the equations. When the denominator consists of quadratic factors: To each irreducible non repeated Ax B 2 quadratic factor ax bx c, there corresponds a partial fraction of the form 2 , where ax bx c A and B are constants to be determined. To each irreducible quadratic factor ax 2 bx c occurring r times in the denominator of a proper rational fraction there corresponds a sum of r
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Engineering Mathematics
Chapter 2: Calculus
partial fractions of the form
A1 x B1
2
( ax bx c )
[2.79]
A2 x B2 2
( ax bx c )
2
Ar x Br
2
( ax bx c )
r
, Where,
A1 , A2 ,, Ar and B1 , B2 , Br are constants to be determined by equating the coefficients of like powers of x in the resulting identity and solving the equations. 3x 1 Example 2.156: Evaluate 2 ( x 2) ( x 2) Solution: 3x 1 A B C 3 x 1 A( x 2) ( x 2) B( x 2) C ( x 2) 2 …(i); 2 2 ( x 2) ( x 2) ( x 2) ( x 2) ( x 2) Putting x 2, 2 successively in equation (i), we get B 7 4 , C 5 16 . Now, we put x 0 and get 3x 1 5 dx 7 dx 5 dx dx A 5 16 . I 2 2 ( x 2) ( x 2) 16 x 2 4 ( x 2) 16 x 2 I
5 16
log( x 2)
Example 2.157: If
7
1
5
4 ( x 2) 16
log( x 2) c
2x2 3
5 16
log
x2 x2
x 1
( x2 1)( x 2 4) dx a log x 1 b tan
2 Solution: Put x y
2x2 3 2
2y 3
2
1
7 4( x 2)
x c then find values of a and b . 2 2y 3
c
A
B
( x 1)( x 4) ( y 1) ( y 4) ( y 1) ( y 4) ( y 1) ( y 4) 2 y 3 A( y 4) B ( y 1) . Comparing the coefficient of y and constant terms A B 2 ,
4 A B 3 A 1 , B 1 . I I
1 2
log
x 1 x 1
1 2
tan 1
x 2
1 y 1
dx
1 y4
dx
1 2
x 1
dx
1 2
x 4
dx
c a 1 2 and b 1 2 .
Some integrals which cannot be found: Any function continuous on interval (a, b) has an antiderivative in that interval. In other words, there exists a function F(x ) such that F' ( x ) f ( x ). However not every anti-derivative F ( x ), even when it exists is expressible in closed form in terms of elementary functions such as polynomials, trigonometric, logarithmic, exponential etc. function. Then we say that such anti-derivatives or integrals ‘cannot be found.’ Some typical examples are: dx cos x x2 3 2 sin( x 2 ) dx x tan x dx dx 1 x dx e dx log x x
sin x x dx
x2
1 x5 dx
e
x2
dx
cos( x
2
) dx
sin x dx
1 x3 dx
2.5.2 Definite Integrals Let g ( x ) be the primitive or anti-derivative of a function
f ( x ) defined on [ a, b] i.e.,
( d dx )[ g ( x )] f ( x ) . The definite integral of f ( x ) over [ a, b] is denoted by
defined as [ g (b) g ( a)] , i.e.
b
a
b
a
f ( x ) dx and is
f ( x )dx g (b) g (a ) . This is also called Newton Leibnitz formula.
The numbers a and b are called the limits of integration, ‘ a ’ is called the lower limit and ‘ b ’ the upper limit. The interval [ a, b] is called the interval or range of integration. To evaluate the definite integral there is no need to keep the constant of integration.
If f ( x ) is not defined at x a, b , and defined in ( a, b) , then
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b
a
f ( x ) dx can be evaluated.
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Chapter 2: Calculus
[2.80]
f ( x ) dx 0 , then the equation f ( x ) 0 has at least one root lying in ( a, b) , provided f is
If
a continuous function in ( a, b) . Evaluation of Definite Integral by Substitution Method: When the variable in a definite integral is changed, the substitutions in terms of new variable should be effected at three places: (i) In the integrand, (ii) In the differential say, dx, (iii) In the limits. For example, if we put
a
g ( x) t in the integral
b
a
f { g ( x )}g ( x ) dx , then
b
a
f { g ( x )} g ( x ) dx
g (b )
g ( a )
f (t ) dt .
1
0 x ln x dx .
Example 2.158 [CE-2002 (2 marks)]: Evaluate
(a) 1 4 (b) 0 (c) 1 4 (d) 1 st Solution (c): By using product rule of integration, let us take ‘ ln x ’ as 1 function and ‘ x ’ as 2nd
d ln x dx
function; thus I x ln x dx ln x xdx 2nd 1st
2
2
xdx dx (ln x) x2 x4 c
1
x2 x2 1 1 0 0 1 Thus, x ln xdx (ln x) c (ln 1) c (ln 0) c 0 2 4 2 4 2 4 4 0 1
[Similar question was also asked in EE-2010 (1 mark)]
0 sin
Example 2.159 [EC-2006 (2 marks)]: (a) 1 2
3
d is given by
(b) 2 3
(c) 4 3 3
(d) 8 3
3
Solution (c): As sin 3 3sin 4sin sin (3sin sin 3 ) 4 , thus
cos 3 1 1 1 4 0 sin d 4 0 (3sin sin 3 )d 4 3cos 3 0 4 3 3 3 3 3 [Similar question was also asked in CE-2001 (1 mark)]
1
3
1
Example 2.160 [EC-2007 (1 mark)]: The following plot shows a function y which varies linearly with x . The value
(a) (b) (c) (d)
2
of the integral I y dx is 1
1.0 2.5 4.0 5.0
Solution (b): As the straight line is passing through (0,1) and (1,0) . So equation of the line is y 1
0 1 1 0
2
( x 0) y x 1 . Thus, I
2
1
Example 2.161 [AG-2009 (1 mark)]: I
x2 4 1 5 ( x 1) dx x 2 1 2.5 2 2 2 1 2 cos x
2
dx is 2 (1 sin x ) (a) –0.5 (b) 0 (c) 0.5 (d) 1 Solution (c): Let sin x t cos xdx dt , also at x 0 t 0 and at x 2 t 1 , so limit 0
changes from t 0 to t 1 I
2
0
cos x (1 sin x )
1
2
dx
dt
1
0 (1 t )2 (1 t )
t 1
t 0
1 2
1
1
1 2
Example 2.162 [CH-2010 (2 marks)]: For a function g ( x ) , if g (0) 0 and g (0) 2 , then lim
g ( x)
x 0 0
(a)
(2t x ) dt is equal to
(b) 2
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(c) 0
(d)
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Engineering Mathematics
Solution (c): Let L lim
Chapter 2: Calculus g(x)
x0 0
(2t x ) dt lim (t 2 x ) x 0
g ( x) 0
[2.81]
lim { g ( x )}2 x . As x 0 , L takes the x0
0 0 form, so using L’Hospital rule, L lim 2 g ( x ) g ( x ) 1 (2 0 2) 1 0 x 0
1
1 x
0
1 x
Example 2.163 [MN-2011 (2 marks)]: Value of the integral I (a) ( 2) 1
(b) ( 2) 1
dx is
(c) 1
(d) 1
Solution (b): Let x cos dx sin d and at x 0 2 and x 1 0 1
1 x
0
1 x
I
I
2
0
dx
0
2
sin
2 sin( 2) cos( 2) 2 0
I sin
1 cos 1 cos
cos( 2) sin( 2)
d
2
0
d
2
0
1 cos
sin
1 cos
d
0
2 cos 2 ( 2)d
2
0
2
sin
2 cos ( 2) 2 sin 2 ( 2)
d 2
(1 cos ) d sin 0
( 2 1) (0 0) ( 2) 1 1
1
Example 2.164 [MN-2012 (1 mark)]: The value of
0 sin
(a) ( 1) 2
(b) ( 1) 2
(c) (2 1) 2
Solution
1
1
sin x cos x 2
(a):
2
(cos x )dx is
(d) (2 1) 2
1
cos(cos x) x
and
for
1
1
0
0
all
1 x 1 .
sin 1 (cos x) ( 2) cos 1 (cos x) ( 2) x . So I sin 1 (cos x ) dx ( 2) x dx
2
I ( x 2) ( x 2)
1
( 0
2) (1 2) 0 0 ( 2) (1 2) x2
5
Example 2.165 [AE-2013 (1 mark)]: The value of
4 x 2 4 x 21 dx is
(a) ln 24 / 11
(c) ln 2
(b) ln 12 / 11
(d) ln(12 11)
2
Solution: Let t x 4 x 21 dt (2 xdx 4dx) 2( x 2)dx ( x 2)dx dt 2 ; also, x 4 t 11 and at x 5 t 24 . So limit changes from t 11 to t 24 . Thus, 5 x2 1 24 1 1 1 1 t 24 I 2 dx dt ln t t 11 (ln 24 ln11) ln(24 11) ln 24 11 4 x 4 x 21 11 2 t 2 2 2 Example 2.166 [CE-2013 (2 marks)]: The value of (a) 0
6
0
cos 4 3 sin 3 6 d is
(c) 1
(b) 1 15
at
(d) 8 3
Solution (b): Let 3 t 3d dt ; at 0, t 0 and at 6 , t 2 , so limit is changing from t 0 to t 2 . So I (1 3)
2
0
I (8 3)
2
0
cos 4 t sin 3 2t dt (8 3)
2
0
0
cos 7 t (1 cos 2 t ) sin t dt ( 8 3) k 7 (1 k 2 ) dk 1
cos 4 t sin 3 t cos 3 t dt
(By
putting
cos t k
sin t dt dk and limit is changing from k 1 to k 0 ). 0 8 0 7 I ( k k 9 ) dk ( 8 3) (k 8 8) ( k 10 10) ( 8 3) (0 0) (1 8 1 10) 1 15 1 1 3 Example 2.167 [ME-2013, PI-2013 (2 marks)]: The value of the definite integral (a) (4 9) e 3 (2 9)
(b) (2 9) e3 (4 9)
(c) (2 9) e 3 (4 9)
e
1
Thus,
x ln x dx is
(d) (4 9) e3 (2 9)
Solution (c): By using product rule of integration, let us take ‘ ln x ’ as 1st function and ‘ x ’ as 2nd d ln x function; thus I x ln xdx dx x dx ln x xdx st dx 2nd 1
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Engineering Mathematics 2 x3 2
I (ln x )
3
2 x1 2
3
Chapter 2: Calculus
2 x3 2
3
dx (ln x )
2 2 x3 2 3
3
[2.82]
c . Thus,
e
e
1
32 2 x3 2 4 x3 2 2e 3 2 4 e 3 2 2 4 4 2e x ln x dx (ln x ) c (ln e) c (ln 1) c 3 9 3 9 3 9 9 9 1
2
0
Example 2.168 [AG-2014 (2 marks)]: The value of
1 sin x (c) 2
(b) ( 2) 1
(a) 0 Solution (b): Let I
2
0
I
2
0
2
cos x 1 sin x
dx
2 1 sin
0
2
(1 sin x ) dx x cos x 0
cos 2 x
2
x
dx
2
dx is
(1 sin x )(1 sin x )
0
1 sin x
(d) ( 2) 1 1 sin x
dx
( 2) 0 0 1 ( 2) 1
Geometrical Interpretation of Definite Integral: In general,
b
a
f ( x ) dx represents an algebraic
sum of areas of the region bounded by the curve y f ( x) , the x axis and the ordinates x a and x b , as shown in Fig. 2.38a. The area above the x axis are taken positive, while those below the
x axis are taken negative. For e.g., in Fig. 2.38b,
b
a
f ( x ) dx A1 A2 A3 .
(b)
(a)
(c)
(d)
Figure 2.38: Geometrical interpretation of Definite integral
Similarly,
b
a
f ( y ) dy represents an algebraic sum of areas of the region bounded by the curve
x f ( y ) , the y axis and abscissa y a and y b , as shown in Fig. 2.38c. The area left the y axis are taken positive, while those right the y axis are taken negative. For e.g., in Fig. 2.38d, b
a
f ( y ) dy A1 A2 A3
Definite Integral as the Limit of Sum: Let f be a continuous function (whose graph lies in 1st quadrant) defined on a closed interval [ a, b] . The definite integral b
a
f ( x ) dx is the area bounded by the curve y f ( x) , the
ordinates x a , x b and the x axis. To evaluate this area, consider the region FPDEF between the curve, x axis and the ordinates x a , x b , as shown in Fig. 2.39. Divide the interval [ a, b] into n equal sub-intervals denoted
by
[ x0 , x1 ],[ x1 , x2 ],, [ xr 1 , xr ],,[ xn1 , xn ] ,
Figure 2.39: Definite integral as Limit of Sum
where, x0 a, x1 a h, x2 a 2h,, xr a rh,, xn a nh b n (b a ) h , where, h is separation between the successive intervals. From the figure, area of rectangle BCIHB area of the region BCIJHB area of the rectangle BCJGHB. Since, Area of the region BCIJHB b
r n 1
A f ( x ) dx ; Area of the rectangle BCIHB sn h[ f ( x0 ) f ( x1 ) f ( xn 1 )] h a
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r0
f ( xr )
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Engineering Mathematics
Chapter 2: Calculus
[2.83]
r n
; Area of the rectangle BCJGHB S n h[ f ( x1 ) f ( x2 ) f ( xn )] h f ( xr ) . r 1
Since, as n , h 0 the strips becomes narrower and all the three areas have the common b
limiting value, i.e. lim S n lim sn area of the region BCIJHB f ( x ) dx n
a
n
r n 1
b
f ( x) dx lim h a
n
r 0
r n
r n 1
f ( xr ) h f ( xr ) lim h
r 0
n
r 1
r n
f ( a rh) h f ( a rh ) , where nh b a r 1
r n 1
1
1
r0
1 rn
f (r r 1
If a 0 , b 1 nh 1 h 1 n f ( x ) dx lim
The average value of a function f ( x ) over the interval [ a, b] is f avg {1 (b a )} f ( x ) dx .
0
n
n
f ( r n)
n
n) . b
a
[This point was asked in AG-2011 (1 mark)] Some useful results for evaluation of definite integrals as limit for sums 2 2 2 2 1 2 3 n n ( n 1) 2 1 2 3 n n(n 1)(2n 1) 6 2
13 23 33 n 3 n (n 1) 2
a ar ar n1 a(r n 1) (r 1) , r 1
cos (ei ei ) 2 , sin (e i e i ) 2
cosh (e e ) 2 , sinh (e e ) 2
1
1
1
1
2
1
52 62 6 2 1 2 2 3 5 8 1
22 1
32 1
42
1
22
42
1 62
2
24 1 1 1 1 1 2 1 2 2 2 2 2 2 3 4 5 6 12 n 1 n 1 nh h h sin sin a sin[ a (n 1)h] [sin(a nh)] sin a sin 2 2 r 0 2 n 1
1
n 1 h sin nh sin h 2 2 2
cos a cos[ a (n 1)h] [cos(a nh)] cos a r 0
2
Example 2.169: If S n 1 (1 n ) 1 (2 2n ) 1 ( n n ) then find lim Sn . n
Solution:
lim Sn lim
n
n
1 r rn
lim
n
1
n (r n) r n
1
1
0
x (1 x )
dx 2[log(1 x )]10 2 log 2
Properties of Definite Integral
Property 1: Changing Dummy Variables
b
a
b
f ( x ) dx f (t ) dt , i.e. the value of a definite a
integral remains unchanged if its variable is replaced by any other symbol provided the limits of integration remains the same.
Property 2: Interchanging Limits
b
a
a
f ( x )dx f ( x ) dx , i.e. by the interchange in the limits b
of definite integral, the sign of the integral is changed.
Property 3: Splitting Limits
b
a
outside the interval [ a, b] . Proof: Let c
a
c
b
a
c
f ( x) dx f ( x) dx f ( x ) dx , i.e. where c may lies inside or b
f ( x)dx F ( x) a
f ( x ) dx F (b ) F ( a )
and,
b
f ( x ) dx f ( x ) dx F (c ) F ( a ) F (c) F ( a ) F (b) F (a ) . Here we have not used any c
condition whether c ( a, b) or c ( a, b) . This property is useful when f ( x ) is not continuous in [ a, b] because we can break up the integral into several integrals at the points of discontinuity so that the function is continuous in the sub-intervals.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
Example 2.170 [AE-2010 (1 mark)]: The definite integral
[2.84]
1
dx
1 x 2
(a) does not exist (b) is equal to 2 (c) is equal to 0 (d) is equal to –2 2 Solution (d): 1 x is not continuous in ( 1,1) , but continuous in (1,0) and (0,1) . Thus,
1
dx
1 x 2
0
1
dx x
2
1
dx
0
2
x
1 x
0
1
x
1
1
1 0
x
0
Example 2.171 [CS-2014 (1 mark)]: If
2
1
1 1
1 1 1 1 2 . x 0 0 1 1 0
x sin x dx k , then the value of k is equal to ………
0
2
x [0, ]
x sin x,
Solution: x sin x
2
0 x sin x, x [ , 2 ] 0 dx rule, x sin x dx x sin x dx sin x dx dx x cos x cos x dx x cos x sin x c dx
2
0
x sin x x sin x dx
x sin x dx . Using product
x sin x ( x cos x sin x c )0 ( x cos x sin x c )2 4 k k 4 .
Example 2.172 [CS-2014 (2 marks)]: The value of the integral
0
x 2 cos x dx is
(b) (c) (a) 2 (d) 2 Solution (a): Using product rule of integration, d x 2 cos x dx x 2 cos x dx x 2 cos x dx dx x 2 sin x 2 x sin x dx dx
d
x sin x dx x sin x dx dx x sin x dx dx x cos x cos x dx x cos x sin x x 2 cos x dx x 2 sin x 2( x cos x sin x ) x 2 sin x 2 x cos x 2 sin x c
Thus,
0
x 2 cos x dx ( x 2 sin x 2 x cos x 2 sin x c )0 (0 2 0 c ) (0 c) 2
Property 4:
b
a
b
f ( x )dx f ( a b x ) dx a
Proof: In R.H.S., put a b x t dx dt ; x a t b; x b t a a
a
b
b
R.H.S. f (t )( dt ) f (t ) dt f (t ) dt f ( x ) dx L.H.S. b
b
a
a
b
b
0
0
If a 0 , then above property f ( x) dx f (b x )dx . It is generally used for those complicated integrals whose denominators are unchanged when x is replaced by ( a x ) .
f ( x) dx
b
a
f ( x) f (a b x)
1
2
(b a) 2
Example 2.173 [EE-2007 (2 marks)]: The integral (1 2 )
0
(a) sin t cos t
b
a
(c) (cos t ) 2
(b) 0
Solution (d): Let
2
0
2I
2
0
(d) (sin t ) 2
2
I (1 2 ) sin(t ) cos d , then using the definite integral property 0
b
2
a
0
f ( x) dx f (a b x ) dx , we get, I
2I
sin(t ) cos d equals
sin(t ) cos d
2
0
sin(t 2 ) cos(2 ) d
0
sin(t ) cos d
2
0
2 sin t cos cos d sin t
2
0
2
2
sin(t ) cos d
sin(t ) sin(t ) cos d
2 cos d sin t
2
0
(1 cos 2 ) d sin t
2
sin 2 0
2 I (sin t )(2 0 0 0) (1 2 ) I (sin t ) 2
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Engineering Mathematics
Chapter 2: Calculus
(a) 0
dx evaluates to (1 tan x ) (c) ln 2
(b) 1 4
(1 tan x )
0
(1 tan x)
Solution (d): Let I
the definite integral property
I
(1 tan x )
4
0
Example 2.174 [CS-2009 (2 marks)]:
b
1 tan x I 4 tan x dx x . Using 0 4 1 tan x 4
b
a
f ( x) dx f (a b x ) dx , we get a
4
tan
0
1 1 1 ln 2 ln(1) ln 2 2 2
ln
Example 2.175 [MN-2010 (2 marks)]: The value of the given integral (a) (1 10) sin( 8)
sin x
3 10
5
I
b
a
dx , then using
sin x cos x sin{(3 10) ( 5) x}
3 10
5
sin{(3 10) ( 5) x} cos{(3 10) ( 5) x}
I
sin( 2 x )
3 10
5
2I
dx
5
sin( 2 x ) cos( 2 x ) sin x
3 10
5
dx
cos x sin x
cos x
3 10
5
sin x cos x
sin x cos x (d) 3 10
cos x sin x
dx
3 10
5
dx is
b
f ( x) dx f (a b x ) dx , we get, a
dx
cos x
3 10
sin x
3 10
5
(c) (1 10) sin( 5)
(b) 20
Solution (d): Let I
(d) (ln 2) 2
dx ; tan
4 0 x dx tan x dx ln cos x 0 4 4
4
0
[2.85]
dx
sin x cos x cos x sin x
dx
3 10
5
dx
3 10
5 10
I 20 Example 2.176 [CE-2011 (2 marks)]: What is the value of the definite integral, (a) 0 Solution (c): Let I
x
a
0
I
(c) a
(b) a 2 dx , then using
x ax
a0 x
a
0
a 0 x a (a 0 x)
2I
x
a
0
x ax
dx
ax ax x
0
ax x
a
0
(d) 2a b
a
a
dx
dx x ax ax x
Example 2.177 [ME-2014 (1 mark)]: The value of the integral (a) 3 Solution (b): Let I
(b) 0 (c) –1 2 2 (2 0 x 1) sin(2 0 x 1)
0
(2 0 x 1) cos(2 0 x 1)
2
(1 x ) sin(1 x )
0
(1 x ) 2 cos(1 x )
I
2
2
Property 5:
a
2
( x 1) 2 sin( x 1)
0
( x 1) 2 cos( x 1)
dx
dx ?
x ax
f ( x) dx f (a b x ) dx , we get,
0
dx ax
a
b
a
x
a
0
a
a
dx dx x 0 a 0 a 0
2
( x 1) 2 sin( x 1)
0 ( x 1)2 cos( x 1) dx
is
(d) –2 dx . Since,
b
a
b
f ( x) dx f (a b x ) dx , a
dx I I 2 I 0 I 0
a
a f ( x)dx 0 { f ( x) f ( x)}dx , this property is generally used when integrand is
either even or odd function of x . Proof:
a
0
a
a f ( x)dx a f ( x)dx 0
f ( x ) dx . Put, x t in first term on R.H.S. dx dt ;
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
also, x a t a; x 0 t 0
a
0
a
a
[2.86] a
f ( x) dx f ( t )( dt ) f ( x ) dx a
0
a
a
a
a
0
0
0
0
f ( x )dx f ( t ) dt f ( x) dx f ( x ) dx f ( x ) dx
a
a
a
a
2 f ( x ) dx, if f ( x ) is an even function or f ( x ) f ( x) 0
f ( x ) dx
0 , if f ( x) is an odd is odd function or f ( x ) f ( x ) [This property was asked in ME-2011 (1 mark)]
If f ( t ) is an odd function, then g ( x ) x
Proof: g ( x ) f (t ) dt g ( x ) a
x
a
g ( x)
a
f ( t ) dt is an even function.
a
x
f (t ) dt
a
Let t y g ( x )
x
f ( y )( dy )
x
x
x
a
a
[ f ( y ) is an odd function]
f ( y )dy
a
f ( y )dy f ( y )dy 0 f ( y ) dy g ( x ) g ( x) is an even function.
a
If f ( t ) is an even function, then for a non-zero ‘ a ’, function. It will be odd function if
a
0
x
x
a
g ( x)
a
a
Now if
a
0
x
f (t ) dt
a
Let t y g ( x )
f ( y )( dy ) x
f ( y ) dy f ( y )dy a
0
[ f ( y ) is an even function]
f ( y ) dy a
x
0
a
f ( y ) dy f ( y )dy f ( y )dy
a
a
f (t ) dt 0
x
a
0
f ( t )dt is not necessarily an odd
f ( t )dt 0 .
Proof: g ( x ) f (t ) dt g ( x ) a
x
a
a
f ( y )dy 0, f ( y ) dy 0 0
x
g ( x) f ( y ) dy g ( x ) . Hence, g ( x ) is an odd function. a
If f ( t ) is an even function, then g ( x )
x
a
f ( t )dt is an odd function. Proof is similar to
the above property.
x cos x dx
Example 2.178 [CS-1998 (5 marks)]: Evaluate
Solution: Let f ( x ) x cos x f ( x) ( x ) cos( x ) x cos x f ( x) f ( x ) is an odd function and thus
2
2 f ( x)dx 0 .
Example 2.179 [CE-2002 (1 mark)]: Evaluate (b) 2
(a) 2 ln 2 Solution (c): Let function and thus
2
f ( x)
sin 2 x
2 1 cos x dx (c) 0
sin 2 x 1 cos x
f ( x)
(d) (ln 2)
sin( 2 x )
1 cos( x )
sin 2 x 1 cos x
2
f ( x ) f ( x ) is an odd
2
2 f ( x)dx 0 .
[Similar question was also asked in PI-2008 (1 mark)] Example 2.180 [ME-2005 (1 mark)]: a
(a) 2 sin 6 x dx 0
a
a (sin
a
(b) 2 sin 7 x dx 0
6
6
x sin 7 x ) dx a
(c) 2 (sin 6 x sin 7 x ) dx 0
(d) zero
7
Solution (a): Let f ( x) sin x sin x , then using property 5 of definite integral, we have
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics a
a (sin
Chapter 2: Calculus
[2.87]
a
6
x sin 7 x ) dx {sin 6 x sin 7 x sin 6 ( x ) sin 7 ( x)}dx 0
a
a
0
0
(sin 6 x sin 7 x sin 6 x sin 7 x) dx 2 sin 6 x dx 5 2
Example 2.181 [XE-2011 (1 mark)]: The integral
5 2 f ( x)dx , where
equals (a) 4
(c) 5 2
(b) 8 5 2
Let
I1
x
5 2 (e
Solution (a):
5 2
5 2
2
sin 3 x 4 cos x) dx
5 2
5 2
2
(e x sin 3 x ) dx
I2
and
(d) 5 2
2
(e x sin 3 x ) dx
5 2
5 2
5 2
5 2
2
f ( x ) e x sin 3 x 4 cos x ,
(4 cos x ) dx . As
(4 cos x )dx 2
2
e ( x ) sin 3 ( x ) e x sin 3 x
2
e x sin3 x is an odd function I1 0 ; also 4 cos( x) 4 cos x 4 cos x is an even function I2 2
5 2
0
(4 cos x )dx (4 sin x)50
Property 6:
2a
0
In particular,
2
4 sin(5 2) sin(0) 4(1 0) 4 I 4
a
a
0
0
f ( x ) dx f ( x) dx f (2a x )dx 2a
0
0
f ( x ) dx
f (2a x ) f ( x )
, if a
2 0 f ( x ) dx , if
f (2a x ) f ( x )
It is generally used to make half the upper limit. 2a
0
Proof:
a
2a
0
a
f ( x) dx f ( x ) dx
x 2a t dx dt in the second
f ( x) dx , putting
integral on R.H.S. Also, x a t a and x 2a t 0 .
2a
0
2a
0
a
0
a
a
0
a
0
0
f ( x ) dx f ( x) dx f (2a t )( dt ) f ( x) dx f (2a t ) dt a
a
0
0
f ( x ) dx f ( x) dx f (2a x) dx
Example 2.182 [MN-2013 (2 marks)]: The value of (a) ( 2) log 2
(b) ( 4) log 2
Solution (a): Let I
2
0
2I
2
0
0
2I
2
0
2I
1 2
log(sin 2 x ) dx
2
0
2
2
0
log(sin t ) dt
2
0
2
log(cos x ) dx is
(c) ( 2) log 2
log cos xdx I
(log sin x log cos x )dx
2
0
b f ( x )dx b f (a b x ) dx a a
log sin xdx
log(sin x cos x )dx
2
0
log 2dx
log 2 2
1
log (sin 2 x) 2 dx
(d) ( 4) log 2
log(sin t ) dt log 2 2 0 2
[Putting 2 x t dx dt 2 ]
2 a f ( x ) dx 2 a f ( x )dx, if f (2a x ) f ( x ) 0 0
b f ( x ) dx b f (t ) dt log 2 a a 2 1 2 I I log 2 I log 2 log (log1 log 2) log 2 2 2 2 2 2 2 2I
2
0
log(sin x) dx
Property 7:
2a
0
a
f ( x )dx { f ( a x ) f (a x )}dx 0
a
a
a
0
0
Proof: R.H.S. { f ( a x ) f ( a x )}dx f ( a x ) dx f ( a x ) dx , 0
a
aa
0
0 a
R.H.S. f a ( a x ) dx a
2a
0
a
R.H.S. f ( x ) dx
f ( x )dx
f ( x )dx
2a
0
Copyright © 2016 by Kaushlendra Kumar
In second integral replace x a by x
f ( x ) dx L.H.S.
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Engineering Mathematics
b
[2.88]
1
a
Property 8:
Chapter 2: Calculus
f ( x )dx ( b a ) f {( b a ) x a )}dx 0
1
Proof: R.H.S. (b a ) f (b a ) x a dx 0
Let, (b a) x a t dx dt (b a) . Also, x 0 t a and x 1 t b . 1
dt
0
(b a )
R.H.S. (b a) f (t )
1
1
0
0
f (t ) dt f ( x) dx L.H.S.
Property 9: If f ( x ) is a periodic function with period T , then
nT
0
T
f ( x )dx n f ( x )dx , 0
where n I . Proof:
nT
0
nT
0
nT
0
0
a nT
2T
f ( x ) dx
nT
f ( x ) dx
( n 1) T
T
T
T
T
0
0
0
T
T
0
0
f ( x )dx f ( x ) dx f ( x ) dx n (times) f ( x T ) f x ( n 1)T T
f ( x )dx n f ( x) dx 0
Property
a
f ( x ) dx
f ( x )dx f ( x ) dx f ( x T )dx f x (n 1)T dx
f ( x ) nT
T
0
f ( x ) dx
10:
f ( x)
If
is
T
a
f ( x )dx n f ( x )dx , a R
periodic
function
with
period
then
0
Proof: I
a nT
a
0
nT
a
0
f ( x )dx f ( x )dx
put, x y nT dx dy
a nT
nT
0
nT
a
0
I f ( x) dx
f ( x )dx
a nT
nT
f ( x ) dx . In the third integral on R.H.S.
a
a
a
0
0
f ( x )dx f ( y nT )dy f ( y ) dy f ( x) dx 0
a
nT
0
0
f ( x ) dx f ( x ) dx
T
f ( x) dx n f ( x ) dx 0
Hence, if f ( x ) is a periodic function with period T , then, we can notice that
T,
independent of a . Property 11: If nT
f ( x) T
mT f ( x )dx ( n m )0
is
a
periodic
function
with
a nT
a
period
f ( x) dx is
T,
then
f ( x )dx , n,m I
Proof: [Putting x y mT dx dy , x mT y 0; x nT y ( n m)T ] Hence, L.H.S.
nT
mT
f ( x ) dx
( n m )T
0
f ( y mT ) dy
( n m )T
0
T
T
0
0
f ( y ) dy
( n m) f ( y ) dy ( n m) f ( x ) dx R.H.S. f ( y mT ) f ( y )
Property 12: If f ( x ) is a periodic function with period T , then
b nT
a nT
b
f ( x )dx f ( x )dx , a
n I Proof: [Putting x y nT dx dy , x a nT y a; x b nT y b ] L.H.S.
b nT
a nT
b
b
b
a
a
a
f ( x )dx f ( y nT )dy f ( y )dy f ( x ) dx R.H.S. f ( y nT ) f ( y )
Gamma Function: If m, n non-negative integers, then
2
0
sin m x cos n xdx
(m 1) 2 ( n 1) 2
, where, (n) is 2 ( m n 2) 2 called gamma function which satisfied the following properties: (i) ( n 1) n( n) n ! ; (ii)
(1) 1 ; (iii) (1 / 2) .
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics
In 2
0
place
Chapter 2: Calculus
[2.89]
of
gamma function, we can also use the (m 1)( m 3) (2 or 1)(n 1)(n 3) (2 or 1) sin m x cos n xdx ( m n)( m n 2) (2 or 1)
following
formula
:
Note that, it is important to note that we multiply by ( 2) ; when both m and n are even.
Reduction formulae Definite Integration b a n! e ax sin bxdx 2 e ax cos bxdx 2 e ax x n dx n 2 2 0 0 0 a b a b a 1 Walli’s Formula n 1 n 3 n 5 2 when n is odd n . n 2 . n 4 ... 3 , 2 2 n n sin xdx cos xdx 0 0 n 1 . n 3 . n 5 ... 3 . 1 . , when n is even n n 2 n 4 4 2 2 ( m 1) ( m 3)...(n 1) ( n 3)... If both m, n are odd I 2 ( m n) ( m n 2) or one odd I sin m x cos n x 0 (m 1) (m 3)...(n 1) (n 3) . If both m, n are even I (m n) (m n 2) 2 Leibnitz’s Rule
If f ( x ) is continuous and u ( x ) , v ( x ) are differentiable functions in the interval [a, b] , then,
d
d
v( x)
d
f (t )dt f v ( x) v ( x) f u ( x ) u ( x) dx u ( x ) dx dx Proof: Let
d dx
F ( x) f ( x)
v( x)
u ( x)
f (t ) dt F {v ( x)} F {u ( x)}
d v( x ) d d d f (t ) dt F {v ( x )} F {u ( x)} F {v ( x )} v ( x) F {u ( x )} u ( x) u ( x ) dx dx dx dx v ( x ) d d d f (t )dt f {v ( x )} v ( x) f {u ( x )} u ( x ) u( x) dx dx dx If functions ( x) and ( x) are defined on [a, b] and differentiable at a point x ( a, b), and f ( x, t ) is continuous, then,
( x) d d ( x) d ( x) d ( x) f ( x, t ) dt f ( x, t ) dt f ( x, ( x)) f ( x, ( x)) ( x ) dx dx ( x ) dx dx
Let f ( x, t ) be a function s.t. both f ( x, t ) and its partial derivatives f x ( x, t ) are continuous in t and x in some region of ( x, t ) plane, including a( x) t b( x ) , x0 x x1 . Let the functions
a ( x ) and b ( x ) are both continuous and both have continuous derivatives for x0 x x1 . Then for x0 x x1 ,
d
dx
b( x )
a(x)
f ( x, t ) dt f x, b ( x ) b( x ) f x , a ( x ) a ( x )
b( x )
a(x)
f x ( x, t ) dt .
x2
Example 2.183 [XE-2007 (2 marks)]: Let f ( x ) sin tdt for x 0 . Then f ( / 2) is equal to 0
(a) 0
(b)
(c) 1
(d) 2
Solution (b): Using Leibnitz rule, we have df d 2 d f ( x) sin x 2 x sin 0 0 (sin x )2 x 0 2 x sin x f 2 1 dx dx dx 2 2 [Similar question was also asked in ME-1998 (1 mark)]
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e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.90]
Important results
Every continuous function defined on [a, b] is integrable over [a, b] . Every monotonic function defined on [a, b] is integrable over [a, b] .
If f ( x ) is a continuous function on [a, b] , then c ( a, b) such that b
a
If at every point x of an interval [a, b] , the inequalities g ( x ) f ( x ) h( x ) satisfies, then b
b
a g ( x)dx a
f ( x) dx f (c )(b a ) .
f (c ) 1 (b a) f ( x) dx is called the mean value of f ( x ) for the interval [a, b] .
b
a
b
f ( x) dx h ( x) dx , where, a b . a
If m and M are the smallest and greatest values of a function f ( x ) on an interval [a, b] , then b
m(b a ) f ( x )dx M (b a ) . a
b
b
b
b
a
a
a
a
Proof: m f ( x ) M mdx f ( x ) dx Mdx m (b a ) f ( x ) dx M (b a )
b
b
a
f ( x)dx f ( x) dx a
b
b
b
a
a
a
b
a
f ( x ) dx f ( x ) dx f ( x ) dx
b
a
b
b
b
a
a
a
f ( x) f ( x) f ( x) , x [a, b] f ( x ) dx f ( x )dx f ( x) dx
Proof:
f ( x ) g ( x ) dx
b
a
b
f ( x)dx f ( x) dx a
g ( x) dx , if f b
f 2 ( x ) dx
2
a
2
b
( x) and g 2 ( x) are integrable on [a, b]
b
2 f ( x) g ( x) dx 0 a
Proof: Let, F ( x ) f ( x ) g ( x) 0, R b
2
b
2 g 2 ( x ) dx 2 f ( x ) g ( x) dx f 2 ( x )dx 0 , hence, its discriminant is non-positive. a
4
a
b
a
a
a
2
4 f ( x) dx g ( x) dx f ( x ) g ( x ) dx f ( x) dx g ( x ) dx b
b
f ( x) g ( x ) dx
b
2
a
b
2
a
b
2
a
2
a
Example 2.184 [XE-2009 (2 marks)]: Let f ( x ) be continuous and satisfy m f ( x ) M in 10
1 x 10 . Then (a) 333m
10
2 1 f ( x) x dx 1 x dx 2
satisfies
(b) 333 M 2
(c) m M 2
10
2
2
(d) m M 333
Solution (c): As m f ( x ) M mx f ( x ) x Mx mx dx 1
10
10
10
1
1
10
m x 2 dx f ( x) x 2 dx M x 2 dx m f ( x ) x 2 dx 1
1
10
1
10
1
10
f ( x ) x 2 dx Mx 2 dx 1
x 2 dx M m M
2.5.3 Curve Sketching, Arc Length, Area and Volume of a Curve Curve Sketching: To find the approximate shape of a curve, following procedure is given as:
Step 1: Symmetry Symmetry about x-axis: If all powers of y in equation of the given curve are even, then it is 2
symmetric about x axis. For example, y 4ax is symmetric about x axis. Symmetry about y-axis: If all power of x in the equation of the given curve are even, then it 2
is symmetric about y axis. For example, x 4ay is symmetric about y axis. Symmetry in opposite quadrants or symmetry about origin: If by putting x for x and y for y , the equation of a curve remains same, then it is symmetric in opposite quadrants.
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Engineering Mathematics
Chapter 2: Calculus
[2.91]
Symmetry about the line = : If the equation of a given curve remains unaltered by interchanging x and y then it is symmetric about the line y x which passes through the
origin and makes an angle of 450 with the positive direction of x axis. Step 2: Curve passing through Origin and Points of intersection with the axes If the equation of curve contains no constant terms then it passes through the origin. For 2 2 example, x y 4ax 0 passes through origin. If we get real values of x on putting y 0 in the equation of the curve, then real values of x and y 0 give those points where the curve cuts the x axis. Similarly by putting x 0 we can get the points of intersection of the curve and y axis. Step 3: Special points The points at which dy dx 0 , the tangent to the curve is parallel to x axis. The points at which dx dy 0 , the tangent to the curve is parallel to y axis. Step 4: Region of the curve: Find minimum and maximum values of x which determine the 2
2
region of the curve y f ( x) . For example for the curve xy a ( a x) y ( a x) x . Now y is real, if 0 x a , So its region lies between the lines x 0 and x a . Step 5: Regions where the curve does not exist: Determine the regions in which the curve does not exists. For this, find the value of y in terms of x from the equation of the curve and find the value of x for which y is imaginary. Similarly find the value of x in terms of y and determine the values of y for which x is imaginary. The curve does not exist for these values of x and y . 2
For e.g. the values of y obtained from y 4ax are imaginary for ve value of x , so the curve 2 2
2
does not exist on the left side of y axis. Similarly the curve a y x (a x) does not exist for x a as the values of y are imaginary for x a. 2
2
Example 2.185 [CE-1997 (1 mark)]: The curve given by the equation x y 2axy , is (a) symmetric about x axis (b) symmetric about y axis (c) symmetric about the line y x (d) tangential to x y a 3 x Solution (c): As by interchanging and y , the given equation remains unchanged, so the curve is symmetrical about the line y x . Example 2.186 [AG-2014 (2 marks)]: The graph of the function F ( x ) x ( k1 x 2 k 2 x 1) for 0 x is
(a)
(b)
(c)
(d)
Solution (a): For the given function, there is no symmetry along any axis or any line. As x ,
F ( x) 2 0 thus option (c) and option (d) are not correct; also at x 0 , F ( x ) 0 , which is 2
satisfied by options (a) and (b). Now, F ( x) F ( x) 0 if k1 x 2 1 0 x 1 k1 1
1(k1 x k 2 x 1) x(2k1 x k2 ) 2
( k1 x k2 x 1)
2
2
k1 x 1 2
( k1 x k 2 x 1)
k1 . So only option (a) has zero slope at x 1
hence option (a) is correct and option (b) is wrong.
Copyright © 2016 by Kaushlendra Kumar
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2
k1 ,
Engineering Mathematics
Chapter 2: Calculus
[2.92]
Arc Length of a Curve b
1 (dy dx) 2 dx
The length of arc of the curve y f ( x) from x a to x b is given as s
The length of arc of the curve x f ( y ) from y a to y b is given as s
The length of arc of the curve in parametric form, i.e., x f (t ) and y f (t ) from t t1 to t t2
a
a
is given as s
t2
1 ( dx dy ) 2 dy
( dx dt ) 2 (dy dt ) 2 dt
t1
b
The length of arc of the curve in polar form, i.e., r f ( ) and from 1 to 2 is given as
s
1
2
r 2 ( dr d ) 2 d 32
Example 2.187 [ME-2008 (2 marks)]: Length of the curve y (2 3) x between x 0 and x 1 is (a) 0.27 (b) 0.67 (c) 1 (d) 1.22 b 1 1 dy d 2 3 2 Solution (d): x x1 2 s 1 ( dy dx ) 2 dx 1 ( x1 2 ) 2 dx 1 xdx a 0 0 dx dx 3 Now, let I
1
(1 x ) dx , putting 1 x t dx dt and at x 0 , t 1 and at x 1 , t 2 ; thus,
0
I
2
tdt (2 3) t 3 2
1
2 1
(2 3) (2)3 2 (1)3 2 1.219 . Hence s 1.219 1.22
Example 2.188 [CE-2010 (2 marks)]: A parabolic cable is held between two supports at the same level. The horizontal span between the support is L . The slag at the mid-span is h . The equation of 2
2
the parabola is y 4h( x L ) , where x is the horizontal coordinate and y is the vertical coordinate with the origin at the centre of the cable. The expression for the total length of the cable is (a) (c)
L
2 2
4
0 1 64h x (1 L )dx L2 2 2 4 0 1 64h x (1 L )dx
(b) 2
L
1 64 h3 x 2 (1 L4 ) dx
0
(d) 2
L2
1 64 h 2 x 2 (1 L4 ) dx
0
Solution (d): As the length of the curve y f ( x) between x a and x b is given by L
b
2
the cable so L
2
2
1 ( dy dx ) 2 dx . So y 4h( x L ) dy dx 8h( x L ) ; also as the origin is at the centre of
a
xL 2
x L 2
x varies from x L 2 to x L 2 . So total length of the cable is
1 64 h 2 x 2 (1 L4 ) dx ; since
a
a
a g ( x)dx 2 0 g ( x)dx
g ( x) 1 64h 2 x 2 (1 L4 ) is an even function so L 2
xL 2
if g ( x ) is an even function. As
1 64h 2 x 2 (1 L4 )dx .
x 0
Example 2.189 [AG-2009 (1 mark)]: A curve is having the equation r a(1 cos ) . The perimeter of the curve between 0 to 2 is (a) 2a (b) 4a (c) 6a (d) 8a Solution (d): dr d {d d }a (1 cos ) a sin s
2
1
s
2
0
a 2 (1 cos ) 2 (a sin ) 2 d 2 a
2
0
r 2 ( dr d ) 2 d 2
sin( 2) d (2 a)2 cos( 2) 0 4 a(1 1) 8a
Area of the Bounded Curve
The area bounded by a cartesian curve y f ( x) , x axis and ordinates x a and x b is given b
b
a
a
by Area y dx f ( x) dx , as shown in Fig. 2.40. If the curve y f ( x) lies below x axis, then the area bounded by the curve y f ( x) the
x axis and the ordinates x a and x b is negative. So, area is given by
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b
a y dx .
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Engineering Mathematics
Chapter 2: Calculus
[2.93]
The area bounded by a cartesian curve x f ( y ) , y axis and abscissa y c and y d , as d
d
c
c
shown in Fig. 2.41, is given by Area x dy f ( y ) dy .
If the
equation of a b
t
a
t1
curve is
in parametric form,
say
x f (t ), y g (t )
then,
2 Area y dx g (t ) f (t ) dt where t1 and t 2 are the values of t , respectively, corresponding
to the values of a and b of x . If the equation of the curve is given in polar coordinates, say Area
1
2
r 2
d
1
Figure 2.40: Area bounded by = ( ), x-axis, = and =
2
{ f ( )} d , where 2
r f ( )
then,
and are the lower and upper limits of .
Figure 2.41: Area bounded by = ( ), y-axis, = and =
Figure 2.42: Area under the curve = ( ) with the −axis when the curve changing its sign many times
Sign convention for finding the Areas using Integration: While applying the discussed sign convention, we will discuss the three cases. Case I: In the expression
b
f ( x ) dx if b a and f ( x ) 0 for all a x b , then this
a
integration will give the area enclosed between the curve f ( x ) , x axis and the line x a and x b which is ve . No need of any modification. Case II: In the expression
b
a
f ( x ) dx if b a and f ( x ) 0 for all a x b , then this
integration will calculate to be ve . But the numerical or the absolute value is to be taken to mean the area enclosed between the curve y f ( x) , x axis and the lines x a and x b . Case III. If in the expression
b
a
f ( x ) dx where b a but f ( x ) changes its sign a numbers
of times in the interval a x b , then we divide the region [ a, b] s.t. we get the points lying between [ a, b] where f ( x ) changes its sign. For the region where f ( x ) 0 we integrate to get the area in that region and then add the absolute value of the integration calculated in the region where f ( x ) 0 to get the desired area between the curve y f ( x) , x axis and the line x a and x b . Hence as shown in Fig. 2.42, the shaded area is given by, c
A f ( x) dx a
d
c
e
f ( x ) dx f ( x ) dx d
e
f
b
f ( x ) dx f ( x ) dx f
Symmetrical Area: If the curve is symmetrical about a coordinate axis (or a line or origin), then we find the area of one symmetrical portion and multiply it by the number of symmetrical portions to get the required area. 2 2 2 Example 2.190: Find the whole area of circle x y a . Solution: The required area is symmetric about both the axis as shown in figure ∴ Required area
a
0
a
a 2 x 2 dx 4 ( x 2) a 2 x 2 (a 2 2) sin 1 ( x a) 4( 2)(a 2 2) a 2 .
0
Area between Two Curves When both curves intersect at two points and their common area lies between these points: If the curves y1 f1 ( x) and y2 f 2 ( x), where f1 ( x) f 2 ( x) intersect in two points
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Engineering Mathematics
A( x a )
Chapter 2: Calculus
B ( x b)
and
b
b
a
a
then,
[2.94]
common
area
between
the
curves
is
( y1 y2 ) dx [ f1 ( x) f 2 ( x )] dx , as shown in Fig. 2.43.
When two curves intersect at a point and the area between them is bounded by x-axis:
b
a
Area bounded by the curves y1 f1 ( x), y2 f 2 ( x) and x axis is, f1 ( x ) dx f 2 ( x ) dx , where, P( , ) is the point of intersection of the two curves, as shown Fig. 2.44. Positive and negative area: Area is always taken as positive. If some part of the area lies above the x-axis and some part lies below x-axis, then the area of two parts should be calculated separately and then add their numerical values to get the desired area.
Figure 2.44: Area between the two curves
Figure 2.43: Area between the two curves
2
Example 2.191 [CE-1997 (1 mark)]: Area bounded by the curve y x and the lines x 4 and y 0 is given by (b) 64 3
(a) 64
(c) 128 3
(d) 128 4
2
Solution (b): Putting x 4 in y x we get y 16 ; so the line and curve meet at (4,16) . So, the area of the shaded region bounded by the curve y1 x 2 and the lines y2 0 and x 4 , given in figure, is given as 4
4
0 ( y1 y2 )dx 0 ( x
2
0) dx ( x3 3) 04 4 3 3 64 3
Example 2.192 [ME-2003 (2 marks), ME-2012, PI-2012 (1 mark)]: The area enclosed between 2 parabola y x and the straight line y x is (a) 1 8
(b) 1 6
(c) 1 3
(d) 1 2
2
Solution (b): Solving y x and y x we get x 0, y 0 and x 1, y 1 ; so the line and curve meet at O (0, 0) and A(1,1) . Thus area of the shaded region given in figure, is given as 1
1
0 ( y1 y2 )dx 0 ( x x
2
1
1
0
0
) dx ( x 2 2) ( x 3 3) (12 2) (13 3) 1 6
Example 2.193 [CE-2006 (2 marks)]: What is the area common to the circle r a and r 2a cos ? (a) 0.524 a 2 (b) 0.614 a 2 (c) 1.047 a 2 (d) 1.228 a 2 Solution (d): The point of intersection of the two curves is a 2a cos cos 1 2
3, 5 3 . So the point of intersection of the two curves is (a, 3) and ( a, 5 3) . So the area of common to the two curves r1 a and r2 2a cos is A (1 2)
5 3
3
A (1 2)
5 3
3
A ( a 2 2)
(4a 2 cos2 a 2 ) d (1 2)
5 3
3
5 3
3
2a (1 cos 2 ) a d 2
2
5 3
(1 2 cos 2 ) d ( a 2 2) ( sin 2 )
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( r22 r12 ) d
3
1.228 a 2
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Engineering Mathematics
Chapter 2: Calculus
[2.95] 2
2
Example 2.194 [ME-2009 (2 marks)]: Area enclosed between the curves y 4 x and x 4 y is (a) 16 3
(c) 32 3
(b) 8 2
(d) 16
2
Solution (a): Solving y 4 x and x 4 y we get x 0, y 0 and x 4, y 4 ; so the curves meet at O (0, 0) and A(4, 4) . Thus required area is
4
4
0 ( y1 y2 )dx 0 2
x ( x 2 4) dx 2 x 3 2 (3 2)
32
4 0
(1 4)( x 3 3)
4 0
3
4(4 3) (1 4)(4 3) (32 3) (16 3) 16 3 [Similar question was also asked in MN-2014 (2 marks)] Example 2.195 [TF-2011 (1 mark)]: The area of ellipse with ‘ a ’ and ‘ b ’ as the length of major and minor axis, respectively, is (b) ( a b ) 2 (c) ab 4 (d) ab 2 (a) ab Solution (c): As the area of ellipse having ‘ A ’ and ‘ B ’ as length of semi-major and semi-minor axis is AB ; so we have 2A a and 2B b , thus the required area of ellipse is AB ab 4 .
Surfaces of Solid of Revolution: The curved surface of the solid generated by the revolution, about the x axis, of the area bounded by the curve y f ( x ) , the ordinates at x a , x b and the x axis is equal to 2
x b
xa
y ds , as shown in Fig. 2.45.
If the arc of the curve y f ( x ) revolves about y axis, then the area of the surface of revolution (between proper limits) 2
2 x ds, where ds 1 (dy dx) dx .
Figure 2.45: Area of surface of
If the equation of the curve is given in the parametric form revolution x f1 (t ) and y f 2 (t ) , and the curve revolves about x axis, then we get the area of the surface of revolution
2
t t2
t t1
yds 2
t t2
t t1
t2
f 2 (t ) ds 2 f 2 (t ) t1
(dx dt )
2
( dy dt ) 2 dt ,
where t1 and t 2 are the values of the parameter t corresponding to x a and x b .
If the equation of the curve is given in polar form then the area of the surface of revolution about A 2 y ds 2 ( r sin )( dS d )d 2 r sin
x axis is given by,
r
A 2 r sin
2
( dr d )
2
d
r
2
(dr d )
2
d
between proper limits.
Example 2.196 [AG-2008 (2 marks)]: Cycloid is formed by x a ( sin ) and y a(1 cos ) . The surface area of the curved plane obtained from the rotation of the cycloid around x axis, from 0 to 2 , is 2
2
(a) 16 a 3
2
(b) 32 a 3
(c) 64 a 3
2
(d) 128 a 3
Solution (b): dx d a (1 cos ) and dy d a sin , so the surface area of the curved plane from the rotation of the cycloid x a ( sin ) and y a(1 cos ) around x axis is s 2
2
0
(dx dt )
y
s 2 a 2
2
s 2 a 2
2
0
0
s 2 a
2
2
( dy dt ) 2 d 2
2
0
a (1 cos ) {a (1 cos )}2 (a sin ) 2 d
2 sin 2 ( 2) 2(1 cos ) d 2 a 2
2
0
2 sin 2 ( 2) 4 sin 2 ( 2) d
2
4 sin 3 ( 2) d 2 a 2 {3sin( 2) sin(3 2)}d
6 cos(
0
2
2) (2 3) cos(3 2) 0 2 a 2 6 (2 3) 6 (2 3) (64 3) a 2
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Engineering Mathematics
Chapter 2: Calculus 2
[2.96]
2
Example 2.197: The part of circle x y 9 in between y 0 and y 2 is revolved about y axis. Find the volume of generating solid. 2 2 Solution: The part of circle x y 9 in between y 0 and y 2 is revolved about y axis. Then a frustum of sphere will be formed. The volume of this frustum 2
2
2
0
0
0
x 2 dy (9 y 2 )dy 9 y (1 3) y 3 (46 3) cubic unit.
Volume of Solid of Revolution: If a plane curve is revolved about some axis in the plane of the curve, then the body so generated is known as solid of revolution. The surface generated by the perimeter of the curve is known as surface of revolution and the volume generated by the area is called volume of revolution. For e.g. a right angled triangle when revolved about one of its sides (forming the right angle) generates a right circular cones. The volume of the solid generated by the revolution, about the x axis, of the area bounded by b
the curve y f ( x) , the ordinates at x a , x b and the x axis is equal to
in Fig. 2.46. The volume of the solid generated by the revolution, about y axis, of the area bounded by the curve x f ( y ) , the abscissa y a and y b is given by (interchanging x and y in the above formulae)
2
a y dx , as shown
b
2
a x dy .
If the equation of the generating curve be given by x f1 (t ) and y f 2 (t ) and it is revolved about x axis, then the formula corresponding to
b
2
a y dx
becomes
t2
t
1
{ f 2 ( t )}2 d { f1 ( t )} ,
where f1 and f 2 are the values of t corresponding to x a and x b .
If the curve is given by an equation in polar co-ordinates, say r f ( ) , and the curve revolves b
a
about the initial line, the volume generated y 2 dx y 2 ( dx d ) d , where and are the values of corresponding to x a and x b . Now x r cos , y r sin . Hence the
volume r 2 sin 2 ( d d )( r cos )d .
The volume of the solid generated by revolving the area bounded by the curve r f ( ) and
the radii vectors and about the initial line is (2 3) r 3 sin d .
The volume in the case when the above area is revolved about the line 2 is
(2 3) r 3 cos d .
Figure 2.46: Volume of the solid generated by the revolution of the curve about −axis
Figure 2.47: Volume of the solid generated by the revolution of the curve about the line
If the generating curve revolves about any line AB (which is different from either of the axes), then the volume of revolution is ( PN ) 2 d (ON ) , as shown in Fig. 2.47.
Volume of an ellipsoid object having three semi axis whose length are a, b, c is V (4 3) abc .
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Engineering Mathematics
Chapter 2: Calculus
[2.97]
Example 2.198 [AG-2007 (2 marks)]: An ellipsoid object has three axes measuring 40 cm, 20 cm and 20 cm respectively. The volume of the object is (a) 4.23 litres (b) 8.38 litres (c) 12.63 litres (d) 17.05 litres Solution (b): The length of semi axis is 20 cm, 10 cm and 10 cm. So V (4 3) (20)(10)(10) 8.377 103 cm3 8.377 litres. H
Example 2.199 [EE-2006 (2 marks)]: The expression V R 2 (1 h H ) 2 dh for the volume of a 0
cone is equal to (a) (c)
R
2
2
0 R (1 h H ) dr H 0 2 rH (1 r R )dh
(b) (d)
R
2
2
0 R (1 h H ) dh R 2 0 RH (1 r R ) dr
Solution (d): The given integral represents the volume of the cone, whose base is centred at the origin, which is the integral of an infinite number of infinitesimally thin circular discs of thickness dh and each of radius r , which is found by similar triangles r R ( H h ) H r R ( H h) H ; so the 2
surface area of the circular disc is A r 2 R 2 ( H h ) H . So, volume V
H
0
2
2
2
H h R H H h 2 R . Now, we consider dh 0 R 1 dh 3 H H 2
options (a) and (d) only, because these contains variable r , as the variable of integration. By integrating
option
2
(d),
we
3
V ( H R) R r (r 3) Rr
R
2 R
0
R
V RH (1 r R ) 2 dr ( H R )( R 2 r 2 2 Rr )dr
get,
0
0
3
3
( H R ) R ( r 3) R
3
(1 3) R H . 2
Example 2.200 [ME-2010 (1 mark)]: The parabolic arc y x , 1 x 2 is revolved around the x axis. The volume of the solid of revolution is (a) 4 (b) 2 (c) 3 4 (d) 3 2 Solution (d): The required volume of the solid generated by the revolution about x axis, of the area x , the ordinated at x 1 , x b and the x axis is equal to
bounded by the curve y 2
2
2
V y 2 dx xdx ( x 2 2) ( 2)(2 2 1) 3 2 1
1
1
Example 2.201 [AE-2012 (2 marks)]: The volume of a solid generated by rotating the region between semi-circle y 1 1 x 2 and straight line y 1 , about x axis, is 2
2
(a) (4 3)
2
(b) 4 ( 3)
2
(c) (3 4)
(d) (3 4)
Solution (a): The curve y 1 1 x 2 is a circle with centre (0,1) and radius 1, which is shown in figure. The required volume of a solid generated by rotating the shaded region between semi-circle y1 1 1 x 2 and straight line y2 1 , about x axis is the volume generated by the region AOBCDA minus the volume generated by
DOCD, i.e., V
x 1
x 1
V
x 1 1 x 1
2
V 2
x 1
x 0
y12 dx
x 1
x 1
y22 dx
x 1
x 1
(1 1 x 2 ) 2 dx
x 1
x 1
( y12 y22 ) dx
( 1 x 2 2 1 x 2 ) dx
( 1 x 2 2 1 x 2 ) dx 2 x ( x 3 3) 2 ( x 2) 1 x 2 (12 2) sin 1 ( x 1) 3
1
V 2 1 (1 3) 2 0 (1 2)sin 1 0 2 (2 3) ( 2) (
Copyright © 2016 by Kaushlendra Kumar
2
x 1 x 0
4) (4 3) .
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Engineering Mathematics
Chapter 2: Calculus
[2.98]
Exercise: 2.5 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1.
(x
tan 1 x 3 ) (1 x 6 ) dx is equal to
2
1
1
3
(a) tan ( x ) c 2.
sin
1
x cos
1
x
(a) (2 )[(2 x 1) sin
e
x
sin
1
(c) ( 2)[(2 x 1) sin 3. If 4.
( x
1
x cos
1
1
x dx is equal to
x (1 x) ] x c (b) cos x sin x
4
2
x2 x 1
(b) k
x2 x 1
(c) tan x c
(d) None of these
(c) k
x2 x 1
(d) k
x2 x 1
x2 x 1 x2 x 1
(c) a 4 , b arbitrary constant
(d) a 4 , b arbitrary constant
cos x (1 sin x) (2 sin x) dx (a) log[(1 sin x ) (2 sin x)] c
(b) log (2 sin x) (1 sin x) c
(c) log (1 sin x) (2 sin x) c
(d) None of these
1 (sin cos ) d 3
2
(b) log tan (1 2) tan c
2
11.
(d) None of these
(b) a 4 , b 3
(c) log tan (1 2) tan c
10.
x x (1 x)] x c
(a) a 4, b 3
2
9.
1
(a) log tan tan c
8.
(b) (2 )[(2 x 1) sin
1) ( x x 1) dx (1 2) log k c , where k
x2 x 1 x2 x 1 5. If 1 (1 sin x ) dx tan ( x 2) a b then
7.
3 2
(d) (1 2)(tan x ) c
sin x dx (1 2)e x a c, then a
(a) k
6.
1
3 2
(c) (1 2)(tan x ) c
x x (1 x )] x c
(a) sin x cos x 2
1
3 2
(b) (1 6)(tan x ) c
(d) None of these
2 1 x 1 x 1 tan tan 1 2 dx k , where k _____. x 1 x 3
0 sin
2
x dx k , where k _____.
e2
e (loge x) x dx _____. 2 0 1 (1 sin x) dx _____. 1
f
12. Suppose
is such that
f ( x ) f ( x) for every real
x and
1
0 f ( x) dx 5 ,
0
13. 14.
1 f (t ) dt k , where k _____. 2 2 2 1 x dx _____. 1.5 2 0 [ x ]dx , where [] denotes the greatest integer function. (a) 2 2
e
15. If f ( x )
(b) 2 2 cos x
2,
sin x , x 2 otherwise
, then
(c) 1 2
(d)
2 1
3
2 f ( x)dx _____.
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then
Engineering Mathematics
16. 17. 18. 19.
sin 2 x
Chapter 2: Calculus
3
0 e cos xdx _____. 12 1 2 [ x] ln (1 x) (1 x) dx k , where [] denotes greatest integer function, then k ____.
1
1 log x
(1 x
2
x 2 1 dx _____.
) sin x cos 2 xdx _____.
0 e
20. If n is any integer, then 21.
[2.99]
cos 2 x
cos3 (2n 1) x dx _____.
3 4
4 1 (1 cos x) dx _____.
0
22. If
x f (sin x ) dx k f (sin x) dx , then the value of k _____. 0
23. For n 0,
2
0
(a) 2
x sin 2 n x
dx sin 2 n x cos 2 n x (b) 2 2
(c) 3 2
(d) 4 2
24. If f ( x ) is a continuous periodic function with period T , then the integral I
a T
a
(a) 2a 1n 25. lim{(n !) n}
(b) 3a
f ( x ) dx
(d) None of these
(c) Independent of a
n
(b) e 1
(a) e 26. 27.
2
4
6
0 sin x cos xdx 3 2 7 0 sin xdx _____.
(c) 1
(d) None of these
k , where k _____.
x
2
2
28. Let f : (0, ) R and f ( x ) f (t ) dt . If f ( x ) x (1 x) then f (4) _____. 0
29. Let f ( x )
x
1
2
2 t 2 dt . Then the real roots of the equation x f ( x) 0 are
(a) 1 30.
2
0 [ x
2
(b) 1
(c) 1 2
2
(d) 0 and 1
x 1]dx , where [] denotes greatest integer function
(a) (7 5) 2
(b) (7 5) 2
(c) ( 5 3) 2
(d) None of these
2
31. The area (in square units) enclosed by the curve x y 36, the x axis and the lines x 6 and x 9 is _____. 2 32. The area of the region bounded by the curve y x x between x 0 and x 1 is _____. (a) 1 6
(b) 1 3
(c) 1 2
(d) 5 6
2
33. The area bounded by the parabola y 4 x and its latus rectum is approximately _____. 2
34. The area enclosed by the parabola y 8x and the line y 2 x is approximately _____. 2
2
35. The part of circle x y 9 in between y 0 and y 2 is revolved about y axis. The volume of generating solid will be (a) (46 3) (b) 12 (c) 16 (d) 28 36. The part of straight line y x 1 between x 2 and x 3 is revolved about x axis, then the curved surface of the solid thus generated is (c) 37 (a) (37 3) (b) (7 2 ) (d) 7 2 37. The part of straight line y x 1 between x 2 and x 3 is revolved about x axis, then find the curved surface of the solid.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
2.6
Chapter 2: Calculus
[2.100]
Vector Differential Calculus
2.6.1 Scalar and Vector Fields
Scalar Function: A scalar function that assigns a real number, i.e. a scalar, to a set of real variables, i.e. u u ( x1 , x2 , , xn ) , where x1 , x2 , , xn are real numbers.
Vector Function: A vector function is a function that assigns a vector to a set of real variables, i.e. F f1 ( x1 , x2 , , xn ) i f 2 ( x1 , x2 , , xn ) j f3 ( x1 , x2 , , xn ) k , where x1 , x2 , , xn are real numbers. For e.g. R (t ) x (t ) i y (t ) j z (t ) k be a radius vector to a point P ( x, y , z ) in space which moves as ‘ t ’ increases; it traces out a curve in space; the parametric representation of space curves is x x (t ) , y y (t ) , z z (t ) . For e.g. R (u, v ) x (u , v ) i y (u , v ) j z (u , v ) k represents a surface in space; the parametric representation of surface in space is x x (u , v) , y y (u, v ) , z z (u , v) . Point Function: A point function u f ( P) is a function that assigns some number or value u to each point P of some region R of space. Scalar Point Function: A scalar point function is a function that assigns a real number (i.e. a scalar) to each point of some region of space. If to each point ( x, y , z ) of a region R in space there is assigned a real number u ( x, y , z ) , then is called a scalar point function. For e.g. the temperature distribution within some body at a particular point in time; the density distribution within some fluid at a particular point in time. Scalar Field: A scalar point function defined over some region is called a scalar field. A scalar field which is independent of time is called a stationary or steady-state scalar field. A scalar field that varies with time would have the representation u ( x, y , z, t ) . Vector Point Function: A vector point function is a function that assigns a vector to each point of some region of space. If to each point ( x, y , z ) of a region R in space there is assigned a vector F F ( x, y , z ) , then F is called a vector point function. Such a function is represented as
F f1 ( x, y , z ) i f 2 ( x, y, z ) j f 3 ( x, y, z ) k .
Vector Field: A vector point function defined over some region is called a vector field, for e.g. velocity at different point within a moving fluid. A vector field which is independent of time is called a stationary or steady-state vector field. A vector field that varies with time would have the representation F f1 ( x, y, z , t ) i f 2 ( x, y , z , t ) j f3 ( x, y , z, t ) k .
Limits, Continuity and Differentiability of vector functions of a single real variable: F (t ) f1 (t )i f 2 (t ) j f3 (t )k (2.1) A vector function of a single variable is given by Eq. 2.1 is said to have limit L , where L L1 i L2 j L3 k , at t t0 , if lim f1 (t ) L1 , lim f 2 (t ) L2 and lim f3 (t ) L3 , i.e. lim F (t ) L . t t0
t t0
t t0
t t0
In addition, if the vector function is defined at t0 and lim F(t ) F(t0 ) then F (t ) is said to be t t0
continuous at t0 . A vector function F (t ) that is continuous for each t in the interval a t b is said to be continuous over the interval. A vector function of a single real variable that is not continuous at a point t0 is said to be discontinuous at t0 . From Eq. 2.1, if t is increased to t t , the change F produced in F is given as F F (t t ) F(t ) f1 (t t )i f 2 (t t ) j f 3 (t t )k f1 (t )i f 2 (t ) j f3 (t )k . So the differentiability of vector function of a single real variable is given as, F f (t t ) f1 (t ) f 2 (t t ) f 2 (t ) f3 (t t ) f3 (t ) lim lim 1 i lim j lim k t 0 t t 0 t 0 t 0 t t t
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.101]
If functions f1 (t ) , f 2 (t ) and f 3 (t ) are differentiable, by letting t 0 it follows that the derivative of F (t ) , denoted by dF dt , can be expressed in terms of the derivatives of the components of F (t ) as dF (t ) df1 (t ) df (t ) df (t ) i 2 j 3 k (2.2) dt dt dt dt If F (t ) is continuous over a t b , but dF dt is discontinuous at a point t0 in the interval, the derivative dF dt will only be defined in the one – sided sense to the left or right of t0 .When higher order derivative exists then we have
d nF dt
n
d d n 1F
dt dt
n 1
, for n 2 . Geometrically, if
F (t ) is taken to be a differentiable position vector r (t ) , then dr dt is a vector that is tangent to the point r (t ) on the curve traced out by Figure 2.48: As ∆ → , so the the tip of the vector as t increases from t a to t b . Fig. 2.48 shows vector ∆ tends to coincidence the relationship between r (t t ) , r (t ) and r before proceeding the with the tangent line to the space curve Γ at ( ) limit as t 0 . As t 0 , so r tend to coincidence with the tangent line T to the curve at the point r (t ) . Also, if r (t ) is a position vector in space and t is the time, dr dt is the velocity of the point with position vector r (t ) and d 2r dt 2 is its acceleration. Let u(t ) and v (t ) be differentiable functions of t over some interval a t b , with C an arbitrary constant vector and c be an arbitrary constant scalar. Then rules for differentiation of vector functions of a single real variable over the interval a t b are: ( f u ) f u f u , where f f ( ) (u v ) u v u v ( u v ) u v , where , are constants (u v) u v u v (u v w ) u v w u v w u v w d du df ( ) u f ( ) [u ( v w )] u ( v w ) u ( v w ) u ( v w ) d df d Differentiation of a constant vector: C 0 If u(t ) is a differentiable function of t and t t ( s) is a differentiable function of s , then du du dt du du1 dt du dt du dt ; or if u (t ) u1 (t )i u2 (t ) j u3 (t )k then, i 2 j 3 k ds dt ds ds dt ds dt ds dt ds Example 2.202 [MN-2012 (2 marks)]: The angle between the tangents to the curve R t 2 iˆ 2t ˆj at the point t 1 is (a) 2 (b) 3 (c) 4 (d) 6 2 Solution (a): R t iˆ 2t ˆj d R dt 2t iˆ 2t ˆj . Thus slope of curve at t 1 is
m1 d R dt
t 1
2 iˆ 2 ˆj and m2 d R dt
points t 1 is cos m1 m2
m1
t 1
2 iˆ 2 ˆj . So, the angle between the tangents at the
m2 ( 4 4)
m1
m2 0 2 .
Vector Differential: The vector differential dF for the function given in Eq. 2.1 is defined as, dF ( df1 dt )i (df 2 dt ) j (df 3 dt )k dt (2.3) If s is the arc length measured from some fixed point, along space curve , which is defined by the position vector r (t ) x1 (t ) i x2 (t ) j x3 (t ) k for t t0 , then dx1 ( dx1 dt ) dt , dx2 ( dx2 dt ) dt and dx3 ( dx3 dt ) dt . As shown in Fig. 2.49, the differential element of arc length ds along is 12
12
2 2 2 dx 2 dx 2 dx 2 ds dr dx1 dx2 dx3 ds 1 2 3 dt dt dt dt dt dt dt dt dt
Copyright © 2016 by Kaushlendra Kumar
(2.4)
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.102]
When t is the time and r (t ) is a position vector in space, ds dt dr dt is the speed with which the tip of position vector r (t ) traces out a space curve .
The unit tangent vector T along as a function of t is given by T ( dr dt ) dr dt . Also, ds dt dr dt dr dt ( ds dt )T .
Since T is the unit tangent vector along T T 1 , after differentiating it w.r.t. s we get (since dot ( dT ds ) T T ( dT ds ) 0 T ( dT ds ) 0 product is commutative) which shows that T and d T ds are orthonormal. The unit vector N in the direction of d T ds at a point r r ( s) on is called the principal normal to at r ( s ) , and so N (dT ds ) ( dT ds ) for dT ds 0 .
When the relation between dT ds and N at a point r r ( s) on Figure 2.49: is written in the form dT ds k ( s ) N , the k ( s ) is called the relationship curvature of the curve at r r ( s) , and ( s ) 1 k ( s ) is called
the radius of curvature of the curve at r r ( s) . As N is a unit vector, so k ( s) dT ds . In case of a smooth plane curve , the circle of curvature at a point P on is tangent to at P with radius ( s ) 1 k ( s ) , and such that its centre lies on the concave side of . If the curvature is required in terms of the parameter t , the relationship between k ( s ) and k (t ) is given from the chain rule as:
differentials
The geometrical between the , , , and
dT
dT ds
dT
k (t )
ds
k (t )
dT
dr
dt dt dt ds dt dt dt The unit binormal vector to the curve at r r ( s) is defined as, B T N The three unit vectors T , N and B at a point r r ( s) on the space curve form a triad of mutually orthogonal unit vectors whose orientation depends on the location of the point on the curve .
2.6.2 Derivative of a Scalar Field Consider a scalar point function w f ( x, y , z ) with continuous first order partial derivatives with respect to x , y and z that is defined in some region R of space, and let a curve in R have the parametric equations x x (t ) , y y (t ) and z z (t ) . Then from the chain rule,
(2.5)
( dw dt ) (f x)( dx dt ) (f y)( dy dt ) (f z )(dz dt ) As from Eq. 2.5, dw dt
can be interpreted as the scalar product of the two vectors
(f x) i (f y ) j (f z ) k and ( dx dt ) i ( dy dt ) j ( dz dt ) k . The first vector (called the gradient of the scalar function
f ) is denoted by
grad f f (f x ) i (f y ) j (f z ) k , where, ( x) i ( y ) j ( z ) k is the gradient operator expressed in terms of Cartesian coordinates. Clearly, gradient of a scalar function is always a vector [This point was asked in CH-2014 (1 mark)]. The second vector ( dx dt ) i ( dy dt ) j (dz dt ) k is seen to be a vector that is tangent to the space curve and is denoted by dr dt ( dx dt ) i ( dy dt ) j ( dz dt ) k . Hence dw dt is the scalar product of grad f and dr dt at the point x x (t ) , y y (t ) and z z (t ) for any given value of t .
Also,
2
x
i
y
j
k
i
z x y 2 operator is also called the Laplacian operator.
Copyright © 2016 by Kaushlendra Kumar
j
2
z
x
k
2
2 y
2
2 z
2
,
where,
the
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Engineering Mathematics
Chapter 2: Calculus
[2.103]
Gradient in Cylindrical and Spherical Coordinates
Cylindrical Polar : Coordinates
Spherical Coordinates
1 f ˆ f zˆ , where x r cos , y r sin and z z r r z f 1 f ˆ 1 f ˆ f rˆ , where x r sin cos , r r r sin y r sin sin , z r cos f
:
f
rˆ
Example 2.203 [TF-2007 (1 mark)]: If f ( x, y , z ) 4( x 2 y 2 ) z 2 , then f at a point (1, 0, 2) is given by (a) 8iˆ 4 ˆj (c) 8 ˆj 4 kˆ (b) 8iˆ 4kˆ (d) 8kˆ 4iˆ Solution (b): As the scalar point function is f ( x, y , z ) 4( x 2 y 2 ) z 2 So, grad f f (f x ) i (f y ) j (f z ) k 8 x i 8 y j 2 z k grad f
(1,0,2)
8i 4k
[Similar question was also asked in CE-2009 (1 mark)] Example 2.204 [PI-2011 (2 marks)]: If T ( x , y , z ) x 2 y 2 2 z 2 defines the temperature at any location ( x, y , z ) , then the magnitude of the temperature gradient at point P (1,1,1) is (b) 4
(a) 2 6
(c) 24
(d)
6
So, grad T T (T x ) i (T y ) j (T z ) k 2 x i 2 y j 4 z k grad T
(1,1,1)
2
2
Solution (a): As the scalar point function is T ( x , y , z ) x y 2 z
grad T
2
2 i 2 j 4k
2 i 2 j 4 k 22 22 42 24 2 6
(1,1,1)
[Similar questions were also asked in ME-1998, EC-2014 (1 mark), EE-2005, MT-2014 (2 marks)] Example 2.205 [MT-2012 (2 marks)]: The temperature field of a slab is given by T 400 50 z exp( t x 2 y 2 ) . The temperature gradient in y direction is (a) 100 yz exp( t x 2 y 2 )
(b) 100 yz exp( t x 2 y 2 )
(c) 100 xz exp( t x 2 y 2 )
(d) 100 xz exp( t x 2 y 2 )
Solution (a): grad T T (T x ) i (T y ) j (T z ) k , so temperature gradient in y
direction is T y ( y ) 400 50 ze (t x
2
y2 )
50 ze
(t x2 y2 )
( 2 y ) 100 yze (t x
2
y2 )
Example 2.206 [PI-2014 (2 marks)]: If 2x3 y 2 z 4 then 2 is (a) 12 xy 2 z 4 4 x 2 z 4 20 x 3 y 2 z 3
(b) 2 x 2 y 2 z 4 x 3 z 4 24 x 3 y 2 z 2
(c) 12 xy 2 z 4 4 x 3 z 4 24 x 3 y 2 z 2
(d) 4 xy 2 z 4 x 2 z 4 24 x 3 y 2 z 2
Solution (c): 2
x
x
2 x
2
2 y
2
(2 x 3 y 2 z 4 ) 6 x 2 y 2 z 4 3
2 4
3
4
(2 x y z ) (4 x yz )
2 z
2 x
2
2
2
. x x y y z z
2 2 4 2 4 (6 x y z ) 12 xy z ; x x x
3
4
3 4
(4 x yz ) 4 x z y y y y y y 2 3 2 3 3 2 2 (2 x 3 y 2 z 4 ) 8 x 3 y 2 z 3 2 (8 x y z ) 24 x y z z z z z z z 2
So, 2 12 xy 2 z 4 4 x3 z 4 24 x3 y 2 z 2 [Similar question was also asked in ME-1999 (1 mark)] Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.104]
Directional Derivatives and its properties: The rate of change of a function f ( x, y, z ) in the direction of the unit vector vˆ l i m j n k , is called the directional derivative and is denoted by
Dvˆ f . Thus
Dvˆ f ( x, y , z ) lim f ( x lh, y mh, z nh ) f ( x, y , z ) h .
The
h 0
directional derivative Dvˆ f can be interpreted in terms of the ordinary Figure 2.50: The directional operation of differentiation by considering Fig. 2.50. In the diagram, a derivative straight line T in space in the direction of a given vector vˆ passes through a fixed point P and Q is a general point on the line T at a distance s from P . The direction derivative Dv f is then given by Dvˆ f df dv lim f (Q ) f ( P ) s . Let v be a unit vector defined as vˆ l i m j n k , where l , s 0
m and n are the direction cosines of the tangent to the space curve , so we have dx l dt
dr dy , m dt dt
dr dz and n dt dt
2 2 2 dr dx dy dz dr with dt dt dt dt dt
12
. Then as the
scalar product of a vector F and the unit vector vˆ is the projection of F in the direction vˆ , it follows that, directional derivative of f in the direction of vˆ , is given as, Dvˆ f vˆ grad f l (f x ) m(f y ) n(f z )
(2.6)
The most rapid increase of a differentiable function f ( x, y, z ) at a point P in space occurs in the direction of the vector vˆ P grad f ( P ) . The directional derivative at P is then given by
2
2
2 12
Dvˆ f ( P ) grad f ( P ) (f x ) P (f y ) P ( z ) P
.
The most rapid decrease of a differentiable function f ( x, y, z ) at a point P in space occurs when the vector vˆ P
just defined in previous point and grad f are oppositely directed, so that
vˆ P grad f ( P ) . The directional derivative at P is then the negative of the result in the
2
2 12
2
previous point and hence, Dvˆ f ( P ) grad f ( P ) (f x) P (f y ) P ( z ) P
.
There is a zero local rate of change of a differentiable function f ( x, y, z ) at a point P in space in the direction of any vector vˆ P that is orthogonal to grad f at P , so that vˆ P grad f ( P ) 0 .
When a scalar function f defined over a region D of space is suitably differentiable, the vectorvalued function grad f defines a vector field over D in terms of the scalar field defined by f .
Example 2.207 [CE-2009 (2 mark)]: For a scalar function
f ( x, y , z ) x 2 3 y 2 2 z 2 , the
directional derivative at the point P (1, 2, 1) in the direction of a vector i j 2k is (a) 18
(b) 3 6
(d) 18
(c) 3 6 2
2
Solution (c): As the scalar point function is f ( x, y , z ) x 3 y 2 z
2
grad f f (f x) i (f y ) j (f z ) k 2 x i 6 y j 4 z k grad f
(1,2, 1)
2 i 12 j 4 k
f at (1, 2, 1) in the direction of v i j 2 k is v (i j 2 k ) 2 12 8 18 Dvˆ f vˆ grad f grad f (2 i 12 j 4 k ) 3 6 2 2 2 v i j 2k 6 1 ( 1) 2 [Similar questions were also asked in CE-2002, CE-2006, TF-2009, ME-2008 (2 marks), XE2008 (1 mark)] . So the directional derivatives of
Example 2.208 [PI-2007 (1 mark)]: f ( x) x is a function defined for real numbers x . The directional derivative of f at x 0 in the direction d 1 is (a) 1
(b) 0
Copyright © 2016 by Kaushlendra Kumar
(c) 1 2
(d) –1
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.105]
Solution (a): As the given function is single variable function so, direction derivative of f ( x) x at f (0 1 h ) f (0)
x 0 in the direction d 1 is Dd f ( x ) x 0 lim
h
h0
0h 0
lim
h
h 0
lim1 1 . h 0
Example 2.209 [CH-2011 (2 marks)]: Unit vectors in x and z directions are i and k respectively. Which ONE of the following is the directional derivative of the function F ( x, z ) ln( x 2 z 2 ) at the
point P : (4, 0) , in the direction of (i k ) ?
(a) (1 2 2 )i
(c) 1
(b) i
(d) 1 2 2 2
2
Solution (d): As the scalar point function is F ( x, z ) ln( x z )
grad F F (F x) i (F y ) j (F z ) k 2 x ( x 2 z 2 ) i 2 z ( x 2 z 2 ) k grad F
(1 2) i . So the directional derivatives of F at (4, 0) in the direction of v i k is
(4,0 )
(i k ) 1 12 1 i v ik 2 12 12 2 2 [Similar question was also asked in CH-2007 (2 marks)] Dvˆ f vˆ grad f
v
grad f
Example 2.210 [EC-2014 (1 mark)]: The directional derivative of f ( x, y ) xy ( x y ) 2 at (1,1) in the direction of the unit vector at an angle of 4 with y axis, is given by ……………. Solution: As the scalar point function is f ( x, y ) xy ( x y )
2
grad f f (f x )i (f y ) j (2 xy y ) grad f
(1
(1,1)
of
unit
Dvˆ f vˆ grad f
2
2
2 i ( x 2 xy )
2
2 j
2)(3 i 3 j) . Also, unit vector in at an angle of 4 from y axis is
v v sin( 4) i cos( 4) j ( i j) direction
2
2 ( x y xy )
v v
vector
grad f
2 . So the directional derivatives of f at
(i j)
an
1
angle
3 3
(3 i 3 j)
2
2
of
2 2
with
4 6 2
at (1,1) in the y axis
is
3
[Similar question was also asked in AE-2013 (1 mark)] Example 2.211 [XE-2012 (1 mark)]: For f x 4 5 xy 2 , the direction of maximum increase of f ( x, y ) at the point (2, 2) is along (a) 3iˆ 10 ˆj (b) 12iˆ 40 ˆj (c) 3iˆ 10 ˆj (d) 12iˆ 40 ˆj Solution (c): The direction of fastest variation in any scalar function f at point P is grad f the
direction
grad f
P
f
of P
maximum
increase
of
f x 4 5 xy 2
at
point
(f x )iˆ (f y ) ˆj P (4 x 5 y ) iˆ ( 10 xy ) ˆj 3
2
(2,2)
P
. So
(2, 2) is given by 4(3 iˆ 10 ˆj )
[Similar questions were also asked in ME-1997, AE-2011, CH-2009 (1 mark), AE-2010, PI-2014 (2 marks)] Example 2.212 [ME-2000 (2 marks)]: The maximum value of the directional derivative of the function 2 x 2 3 y 2 5 z 2 at a point (1,1, 1) is (a) 10 Solution (c): As
(b) –4 grad
P
(c) P
(d) 152
152
( x ) iˆ ( y ) ˆj ( z ) kˆ
P
is the direction of
maximum increase, grad (1,1, 1) (4 x iˆ 6 y ˆj 10 z kˆ )(1,1, 1) 4 iˆ 6 ˆj 10 kˆ 152 .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.106]
Example 2.213 [XE-2011 (1 mark)]: Consider the function f ( x, y , z ) x 3 e y sin z and the point P 1, 0, 2 . The value of f DOES NOT change due to small displacement of P along the direction of (a) 1, 0, 2 (b) 1, 1,1 (c) 1, 3, 0 (d) 2, 0, 1 Solution (c): Let we have a unit vector vˆ l i m j n k , then directional derivative of f ( x, y , z ) x 3 e y sin z at point P 1, 0, 2 along vˆ l i m j n k must be zero when the value of
f does not change due to small displacement of P along the direction of vˆ ; i.e., Dvˆ f
P
0 f
(3 x e
P
vˆ 0 (f x ) i (f y ) j (f z )k P (l i m j n k ) 0
( x )( x 3e y sin z )i ( y )( x 3e y sin z ) j ( z )( x 3 e y sin z )k 2 y
sin z )i ( x 3e y sin z ) j ( x3 e y cos z )k
1,0, 2
P
(l i m j n k ) 0
(l i m j n k ) 0
3 i 1 j 0 k (l i m j n k ) 0 3l m 0 . So from the given options we can say that option (c) is correct as option (c) satisfies the condition 3l m 0 .
Rules for the gradient operator: Let the gradients of f and g be defined over a region D , then grad (cf ) c grad f c f , where c is a scalar constant.
grad ( f g ) grad f grad g f g grad ( f g ) ( f g ) f grad g g grad f f g g f
grad f g g grad f f grad g g 2 g f f g g 2 , g 0
grad f ( g ) ( df dg )g The above results are obtained by the usual rules for partial differentiation to each component of the gradient function on the left, and then recombining the results to obtain the expression on the right. Let us prove the third result, grad ( f g ) ( fg ) x i ( fg ) y j ( fg ) z k grad ( f g ) f (g x ) g (f x ) i f (g y ) g (f y ) j f (g z ) g (f z ) k grad ( f g ) f (g x)i (g y ) j (g z )k g (f x )i (f y ) j (f z )k grad ( f g ) f grad g g grad f f g g f
Gradient as a Surface Normal Vector and Tangent plane to the surface : Let S be a surface represented by f ( x, y, z ) c , where c is any constant and f is differentiable function. Such a surface is called a level surface of f , and for different c we get different level surfaces. Now let C be a curve on S through a point P of S . As a curve in space, C has a representation r (t ) x (t ) i y (t ) j z (t ) k . For C to lie on the surface S , the components of r (t ) must satisfy f ( x, y, z ) c 0 . Now a tangent vector of C is dr (t ) dt r (t ) x(t ), y (t ), z(t ) and the tangent vector of all curves on S passing through P will Figure 2.51: Gradient as a Surface Normal generally form a plane, called the tangent plane of S at P Vector (exceptions occur at the edges of S ). The normal of this plane (the straight line through P perpendicular to the tangent plane) is called the surface normal to S at P . A vector in the direction of the surface normal is called a surface normal vector of S at P , which is obtained by differentiating f ( x, y, z ) c 0 w.r.t. with the help of chain rule as, t (f x)( dx dt ) (f y)(dy dt ) (f z )(dz dt ) (grad f ) r 0 . Hence grad f is orthogonal to all the vectors r in the tangent plane, so that it is a normal vector of S at P .
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Engineering Mathematics
Chapter 2: Calculus
[2.107]
The vector equation of a plane with normal n containing the point P0 with position vector r0 is (r r0 ) n 0 , where r is the position vector of an arbitrary point on the plane. If we set
r x i y j z k and r0 x0 i y0 j z0 k
and identity n with grad f
at P0 , where
grad f ( P0 ) (f x) P0 i (f y ) P0 j (f z ) P0 k .
The required tangent plane to the surface at
P0 ( x0 , y0 , z0 )
is to be given by
( x x0 )(f x ) P0 ( y y0 )(f y ) P0 ( z z0 )(f z ) P0 0 .
Lagrange Multipliers: Optimization with Constraints: Finding the extrema of a function f ( x, y ) subject to a constraint g ( x, y ) k are called constrained optimization problems. For e.g. suppose that the constraint g ( x, y ) k is a smooth closed curve parameterized by r (t ) x (t ) i y (t ) j on [ a, b] ; and let f ( x, y ) is differentiable at each point on the constraint. Then finding the extrema of f ( x, y ) subject to g ( x, y ) k is equivalent to finding the absolute extrema of the function z (t ) f {x (t ), y (t )} for t in [ a, b] . As the extrema of z(t ) over [ a, b] must exist and occur either at the critical points or the endpoints of [ a, b] . Since the curve is closed, we only need consider the critical points of z(t ) in [ a, b] , which are solutions to dz dt f v 0 , where v is the velocity of r (t ) , i.e. critical points of z(t ) occur when f v . Also g v , it follows that the extrema of f ( x, y ) subject to g ( x, y ) k occur when f is parallel to g . If f is parallel to g , then there is a number for which f g . Thus, the extrema of f ( x, y ) subject to g ( x, y ) k must occur at the points which are the solution to the system of equations f x i f y j ( g x i g y j) , g ( x, y ) k (2.7) Eq. 2.7 is a Lagrange multiplier problem and we call a Lagrange Multiplier. To solve a Lagrange multiplier problem, first eliminate the Lagrange multiplier using the two equations f x g x and f y g y . Then solve for x , y by combining the result with the constraint g ( x, y ) k ; thus gives
the critical points. Finally, as the constraint g ( x, y ) k is a closed curve, the extrema of f ( x, y ) over g ( x, y ) k are the largest and smallest values of f ( x, y ) evaluated at the critical points. Example 2.214 [XE-2007 (2 marks)]: The maximum value of the function 2 x 3 y 4 z on the ellipsoid 2 x 2 3 y 2 4 z 2 1 is (a) 2 (b) 3 (c) 6 (d) 9 2 2 Solution (b): Let f ( x, y , z ) 2 x 3 y 4 z and g ( x, y, z ) 2 x 3 y 4 z 2 1 , so we have to maximize f ( x, y, z ) subject to g ( x, y , z ) . Using Lagrange multiplier , we have f g
2 i 3 j 4 k (4 x i 6 y j 8 z k ) x y z 1 2 .
Putting
values
of
x,
y,
z
in
g ( x, y , z ) 0 , 2 (4 2 ) 3 (4 2 ) 4 (4 2 ) 1 0 9 (4 2 ) 1 3 2 . So critical points are x y z 1 3 . Thus f ( x, y , z ) ( 1 3, 1 3, 1 3) 3 , f ( x, y , z ) (1 3,1 3,1 3) 3 . So maximum value of the function 2 x 3 y 4 z on the ellipsoid 2 x 2 3 y 2 4 z 2 1 is 3 at x y z 1 3 .
2.6.3 Derivative of a Vector field When we came to consider the rate of change of a vector point function F (r ) , we see that there are two ways of combining the gradient operator with the vector F , i.e., F and F .
Divergence of a Vector: The operation, F , is called the divergence of a vector and it associates a scalar function with a differentiable vector field F . In terms of Cartesian coordinates if F F1 i F2 j F3 k is a differentiable vector field, the divergence of F , written as div F , given as
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Chapter 2: Calculus
[2.108]
F1
F2
F3
(2.8) x y z x y z A more general way of defining the divergence of a vector field F (r) at the point r is to enclose the point in an elementary volume V and find the flow or flux out of V per unit div F F
i
volume. Thus, div F F lim
j
k ( F1 i F2 j F3 k )
flow out of V
.
V A non-zero divergence at a point in a fluid measures the rate, per unit volume, at which the fluid is flowing away from or towards that point. That implies that either the density of the fluid is changing at the point or there is a source or sink of fluid there. When div F 0 , there is a net flow of F out of the volume; and when div F 0 , there is a net flow of F into the volume. In the case of a nonmaterial vector field, for example temperature gradient in heat transfer, a nonzero divergence indicates a point of generation or absorption. When the divergence is everywhere zero, the flow entering any element of the space is exactly balanced by the outflow. This implies that the lines of flow of the field F (r ) where div F 0 must either form closed curves or finish at boundaries or extend to infinity. Vectors satisfying this condition are sometimes termed solenoidal. V 0
Properties of the divergence operator: Let the vector fields F , G and the scalar fields , be a suitably differentiable; let a , b are constants. Then, divergence operator has the following properties: div ( aF ) a div F a F div (aF bG ) a div F b div G a F b G
div (grad ) 2 div ( ) 2 grad grad 2 div ( F) div F F F F
div ( ) div ( ) 2 2
Divergence of a vector in Cylindrical and Spherical Coordinates
Cylindrical Polar Coordinates: Spherical Coordinates:
:
u
1 (r u r )
u
1 ( r 2u r )
r
r
1 u r
1
u z z
, where x r cos , y r sin and z z
(u sin )
1
u
, where
r 2 r r sin r sin x r sin cos , y r sin sin , z r cos
:
Example 2.215 [ME-2001, EE-2010 (1 mark), MT-2014 (2 marks)]: Divergence of the three dimensional radial vector field r is (a) 3 (b) 1 r (c) iˆ ˆj kˆ (d) 3(iˆ ˆj kˆ) Solution (a): As any 3-D radial vector field r can be represented as r x iˆ y ˆj z kˆ . So
div r r ( x ) i ( y ) j ( z )k ( x iˆ y ˆj z kˆ) (x x ) (y y ) (z z ) 3
Example 2.216 [CE-2007 (2 marks)]: A velocity vector is given as V 5 xyi 2 y 2 j 3 yz 2 k . The divergence of this velocity vector at (1,1,1) is (a) 9 (b) 10 (c) 14 (d) 15
2
2
Solution (d): div V V ( x) i ( y ) j ( z )k (5 xyi 2 y j 3 yz k )
div V ( x )(5 xy ) ( y )(2 y 2 ) ( z )(3 yz 2 ) 5 y 4 y 6 yz div V
(1,1,1)
5 4 6 15
[Similar questions were also asked in EC-1998 (5 marks), MN-2009, MT-2007, TF-2010 (2 marks) (2 marks), ME-2008, ME-2009, EC-2013 (1 mark)]
Example 2.217 [MN-2014 (1 mark)]: The divergence of the vector b ( x y )( y iˆ x ˆj ) is
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Chapter 2: Calculus
(b) x y
[2.109]
(c) x 2 y 2
(d) y 2 x 2
Solution (b): div b b ( x ) i ( y ) j ( z )k ( yx y 2 ) i ( x 2 xy ) j 0 k
2
2
div b ( x )( yx y ) ( y )( x xy ) ( z )(0) y x [Similar question was also asked in ME-2014 (1 mark)]
Example 2.218 [IN-2012, EE-2012, EC-2012 (2 marks)]: The direction of vector A is radially n outward from the origin, with A kr where r 2 x 2 y 2 z 2 and k is constant. The value of n for which A 0 is (a) –2 (b) 2 (c) 1 (d) 0 2 2 2 2 Solution (a): As r x y z is the equation of a sphere of radius r and thus in spherical 1 2 1 2 n k n 2 coordinates, for u u 0 , A 2 ( r ur ) 2 (r kr ) 2 ( r ) k ( n 2)r n 1 . Thus r r r r r r for A 0 , we must have n 2 0 n 2 . Example
2.219
[EE-2007 (1 mark)]: Divergence 2 2 2 V ( x, y , z ) ( x cos xy y )iˆ ( y cos xy ) ˆj (sin z x y ) kˆ (b) sin xy 2 z cos z 2
(a) 2 z cos z 2 Solution (a):
of
the
(c) x sin xy cos z
vector
field
(d) None of these
2 2 2 div V V {( x )iˆ ( y ) ˆj ( z ) kˆ} {( x cos xy y )iˆ ( y cos xy ) ˆj (sin z x y ) kˆ} 2 2 2 div V V ( x )( x cos xy y )iˆ ( y )( y cos xy ) ˆj ( z )(sin z x y ) kˆ
2
div V V cos xy xy sin xy cos xy xy sin xy 2 z cos z 2 z cos z [Similar question was also asked in AG-2011 (2 marks)]
2
Example 2.220 [EC-2014 (2 marks)]: If r xaˆ x yaˆ y zaˆ z and r r , then div{r 2(ln r )} …
Solution:
In
spherical
coordinates
for
ˆ ˆ 0 ,
2
f (f r ) rˆ ;
so
for
f ln r ,
2
(ln r ) ( r )(ln r ) 1 r div{r (ln r )} div( r r ) div( r ) ( x ) x ( y ) y ( z ) z 3
Example 2.221 [EE-2014 (1 mark)]: Let ( f v ) x 2 y y 2 z z 2 x , where f and v are scalar and vector fields respectively. If v yiˆ zjˆ xkˆ , then v f is (a) x 2 y y 2 z z 2 x
(b) 2 xy 2 yz 2 zx
(c) x y z
(d) 0
Solution (a): ( f v ) div ( f v ) f div v v f f ( v ) v f .
2
2
2
v ( x) y ( y ) z ( z ) x 0 v f ( f v ) x y y z z x
Curl of a Vector: The operation, F , is called the curl of a vector and it associates a vector function with a differentiable vector field F . If F F1 i F2 j F3 k is a differentiable vector field, the curl of F , written as curl F , is the scalar function defined in terms of Cartesian coordinates as, i j k F F F F F F curl F F x y z 3 2 i 1 3 j 2 1 k (2.9) y z z x x y F1 F2 F3 More generally, the component of the curl of a vector field F (r ) in the direction of the unit vector nˆ at a point L is found by enclosing L by an elementary area S that is perpendicular to L , as shown in Fig. 2.52, and calculating the flow around S per unit area. Thus, (curl F) nˆ lim (flow around S ) S . S 0
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[2.110]
Properties of curl and other operators: Let F and G be vector functions and let be a scalar function, all of which are suitable differentiable. The operators grad, div, and curl can be combined in various ways that lead to identities, the results which are listed as follows: curl (grad ) 0 , where 0 is vector of zero magnitude [This point was asked in ME-1996, EE-2013, MT-2014 (1 mark)] Figure 2.52: div (curl F) ( ) 0 Circulation around the element ∆ grad ( F ) ( F) F F curl ( F ) (curl F ) F (grad ) ( F) F ( ) [This point was asked in ME-1995 (1 mark)] grad (F G ) F (curl G ) G (curl F) (F )G (G )F div (F G ) (F G ) G (curl F ) F (curl G ) G ( F ) F ( G ) curl (F G ) (F G ) F (div G ) G (div F ) (G )F (F )G F ( G ) G ( F ) (G )F (F )G
curl (curl F) ( F ) ( F) 2 F grad (div F ) 2 F [This point was asked in EC2006 (1 mark)] grad (div F ) ( F ) 2 F
Curl of a vector in Cylindrical and Spherical Coordinates
Cylindrical Polar Coordinates: 1 (u z ) (u ) ˆ u r u z ˆ 1 ( ru ) u r ˆ u where x r cos , r z, z r r r r z y r sin and z z . Spherical Coordinates: 1 (u sin ) u 1 1 u r 1 ( r u ) u r ˆ u rˆ ( r u ) ˆ , where r sin r sin r r r x r sin cos , y r sin sin , z r cos
Example 2.222 [ME-2010 (2 marks)]: Velocity vector of a flow field is given as V 2 xy iˆ x 2 z ˆj . The vorticity vector at (1,1,1) is (a) 4iˆ ˆj (c) iˆ 4 ˆj (b) 4iˆ kˆ (d) iˆ 4kˆ i j k i j k
Solution (d): Vorticity vector of V or curl V V x
y
v1
2
z x
v2
v3
2 xy
y
z
2
x z
0
2
curl v i 0 ( x ) j 0 0 k 2 xz 2 x x i 2 xz 2 x k
curl v
(1,1,1)
i 2 2 k i 4 k
[Similar questions were also asked in ME-1999, AE-2011, ME-2014 (1 mark)]
Example 2.223 [ME-2003 (2 marks)]: The vector field F x i y j (where i and j are unit vectors) is: (a) divergence free, but not irrotational (b) irrotational, but not divergence free (c) divergence free and irrotational (d) neither divergence free nor irrotational Solution (c): div F F i j k x i y j 0 k x ( y ) 1 1 0 . x y x y z
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[2.111]
i
j
So the vector field F is divergence free. As, curl F F x
y
z 0 i 0 j 0 k . So
y
x
the vector field F is irrotational.
k 0
Example 2.224 [AE-2008 (2 marks)]: The function f ( x, y , z ) (1 2) x 2 y 2 z 2 satisfies (a) grad f 0 (b) div (grad f ) 0 (c) curl (grad f ) 0 (d) grad div (grad f ) 0
Solution (c): As grad f f (1 2) ( x)( x 2 y 2 z 2 ) i ( y )( x 2 y 2 z 2 ) j ( z )( x 2 y 2 z 2 ) k i
i x
2 2
grad f xy z
2
yz
2
j x
2
2
j
y z k . So curl(grad f ) x
xy z
k
y
2 2
2
x yz
2
z x2 y 2 z
curl(grad f ) i 2 x 2 yz 2 x 2 yz j 2 xy 2 z 2 xy 2 z k 2 xyz 2 2 xyz 2 0 i 0 j 0 k
Example 2.225 [AG-2010 (2 marks)]: The curl of the vector A xy iˆ yz ˆj zx kˆ ( iˆ , ˆj and kˆ represent unit vectors along the three orthogonal axes) is (a) x iˆ y ˆj z kˆ (b) x iˆ y ˆj z kˆ (c) y iˆ z ˆj x kˆ (d) y iˆ z ˆj x kˆ
i
Solution (a): curl V V x
j
k
y
i
z x
j
k
y
v1 v2 v3 xy [Similar question was also asked in CH-2010 (2 marks)]
yz
z y i z j x k zx
Example 2.226 [TF-2007 (1 mark)]: Given a vector u ( x, y, z ) xy iˆ ( z x) ˆj y kˆ , the points where the u vanishes lie on the plane (a) y 2 (b) y 1 (d) x 1 (c) z 1
i
j
Solution (d): u x
y
k z (1 x ) k . So u 0 when (1 x ) 0 x 1 .
( z x)
xy
y
Example 2.227 [XE-2009 (2 marks)]: Let u y iˆ x ˆj and v z ˆj y kˆ be two given
vectors, where is a constant. Then div(u v) equals (b) 2 2 y
(a) 0
i
j
Solution (d): u v y x
0
z
(c) 4 2 y
(d) 4 2 y
k 0
2 ( xy i y 2 j yz k ) . So div(u v) (u v )
y
div(u v ) ( x )i ( y ) j ( z )k 2 ( xy i y 2 j yz k ) 2 y 2 y y 4 2 y
Example 2.228 [MT-2013 (1 mark)]: For scalar fields and , the value of ____ Solution: Let F and G , also, (F G ) div (F G ) and we know that
div ( F G ) G ( F ) F ( G ) G ( F ) F ( G ) .
G ( ) F ( ) G (0) F (0) 0 , since answer is 0 .
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So,
0 . Hence the
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2.6.4 Vector/Matrix Derivatives
Derivative of Vectors w.r.t. Scalars: The derivative of the vector y ( x ) y1
with respect to the scalar x is the vector y x y1 x y2 x yn x . The second or higher derivative of a vector with respect to a scalar is likewise a vector of the derivatives of the individual elements; that is, it is an array of higher rank. Derivatives of Matrices with Respect to Scalars: The derivative of the matrix Y ( x ) [ yij ] w.r.t
y2
yn
the scalar x is the matrix Y ( x ) x (yij x ) . The second or higher derivative of a matrix with
respect to a scalar is likewise a matrix of the derivatives of the individual elements. Derivatives of Scalars with Respect to Vectors (The Gradient): The derivative of a scalarvalued function w.r.t. a vector is a vector of the partial derivatives of the function with respect to the elements of the vector. If f (x) is a scalar function of the vector x x1 x2 xn , then f x f x1 f x2 f xn if those derivatives exist. This vector is called the gradient of the scalar-valued function, and is denoted by f ( x ) or simply f . Thus
f ( x) f x . The notation f ( x ) implies differentiation with respect to “all” arguments of f , hence, if f is a scalar-valued function of a vector argument, they represent a vector. Derivatives of Vectors with Respect to Vectors (The Jacobian): The derivative of an m vector-valued function of an n vector argument consists of nm scalar derivatives. These derivatives could be put into various structures. Two obvious structures are an n m matrix and an m n matrix. For a function f : S n m , f T x to be the n m matrix, and f xT to be its transpose, the m n matrix. If all the derivatives exists, then we have f1 x1 f 2 x1 f m x1 f
f T
f 1 x x x
f 2 x
f x 1 2 x f1 xn
f m
f2 x2
f m x2
. This derivative is xn
f 2 xn f m called the matrix gradient and is denoted by f for the vector-valued function f . The m n
matrix f xT (f )T is called the Jacobian of f . Derivatives of Matrices with Respect to Vectors: The derivative of a matrix with respect to a vector is a 3-D object that results from applying eqn f x f x1 f x2 f xn to each of the elements of the matrix. So, it is simpler to consider only the partial derivatives of the matrix Y with respect to the individual elements of the vector x , that Y xi . Using the rules for differentiation of powers from definitions, we can write the partial derivatives of the inverse of the matrix Y as ( x )Y 1 Y 1 ( x )Y Y 1 . A simple result of the linearity of the operation is the rule for differentiation of the trace: ( x )tr (Y ) tr ( x )Y . The following table lists formulas for the vector derivatives of some common expressions. The derivative f xT is the transpose of f x . In the table below x is an n vector, a is a constant scalar, b is a constant vector; and A is a constant matrix. f (x ) f (x ) f x f x T T ax aI x Ab b A T
b x T x b T
x x x xT T
b Ax
b
xT Ax
( A AT ) x
1 exp xT Ax 2
1 exp xT A x A x 2 if A is symmetric
T
b I x xI
2 xT T
A b
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Example 2.229 [IN-2014 (2 marks)]: A scalar valued function is defined as f ( x) xT Ax bT x c , where A is a symmetric positive definite matrix with dimension n n ; b and x are vectors of dimension n 1 . The minimum value of f ( x ) will occur when x equals (a) ( AT A) 1 b (b) ( AT A) 1 b (c) ( A1b ) 2 (d) ( A1b ) 2 Solution (c): For extremum value of the given function we have to differentiate the given function T w.r.t. the vector x , f (x ) ( x Ax) (bT x) (c) ( A AT ) x b 0 2 Ax b (as A x x x x T is symmetric matrix so A A ) and now equating it to 0 for getting the critical point, i.e., ( x) f ( x) 0 2 Ax b 0 2 A1 Ax A1b x (1 2) A1b . Now we have to check, whether
2 x 2
f ( x)
x (1 2) A1b
x
(2 Ax)
x
is
the
(b) 2 A 0 2 A
point
2 x 2
of
maxima
or
minima;
so
0 (as A is a positive matrix).
f ( x) x (1 2) A1b
Thus x (1 2) A1b is the point of minima. Exercise: 2.6 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. Consider the pressure field f ( x, y ) 9 x 2 4 y 2 . The pressure at point P (0.5, 3.25) is _____. 2. The vector field given by F x i y j is given by which one of the following?
(a)
(b)
None (d) of these
(c)
3. The region in which the pressure field f ( x, y ) 9 x 2 4 y 2 varies between 36 and 144 is between (a) two ellipses (b) two parabolas (c) two hyperbolas (d) two straight lines 4. The parametric representation of the straight line through a point A(3,1,5) in the direction of a vector b 4 i 7 j k is (a) (5 t ) i (3 4t ) j (1 7t ) k (b) (3 4t ) i (5 t ) j (1 7t ) k (c) (3 4t ) i (1 7t ) j (5 t ) k (d) 3 i j 5 k 5. The curve given by the parametric representation r (t ) t i (t 3 2) j 0 k is (a) y x 2 2 , z 0
(b) y x 3 2 , z 0
(c) y x 2 2 , z 2
(d) y x 3 6 , z 2
6. For a curve given by r (t ) t i t 3 j , the unit tangent vector u(t ) at a point P (1,1, 0) is (a) i 3 j
(b) (1
10) i (3
(c) i 3 k
10) j
7. The length of the curve given by r (t ) t i t
32
(d) (1
10) i (3
10) k
j from (0, 0, 0) to (4,8, 0) is _____.
8. The magnitude of tangential acceleration for the curve given by r (t ) 5t 2 k is _____. 9. If w ( x 2 y 2 z 2 ) 1 2 , x cos t , y sin t , z t then dw dt at t 1 is _____. 10. The grad f for the function f ln( x 2 y 2 ) at (2, 0) is (b) j (c) i j (a) i
(d) i j
11. The direction of maximum decrease of flow of heat in a temperature field T x y at P (8, 1) is (a) 8 i j (b) 8 i j (c) i 8 j (d) i 8 j
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12. The unit normal vector for the surface z ( x 2 y 2 ) 1 2 at the point P (6,8,10) is parallel to (a) 6 i 8 j 10 k (b) 8 i 6 j 10 k (c) 8 i 6 j 10 k (d) 6 i 8 j 10 k 13. For the vector field v 2 x i 4 y j 8 z k , the potential field f such that v grad f is (a) 2 x 2 y 2 4 z 2 (b) x 2 y 2 4 z 2 (c) x 2 2 y 2 4 z 2 (d) x 2 y 2 z 2 14. For the vector field v xy i 2 xy j , the potential field f such that v grad f is (a) xy
(b) x 2 y 2
(c) 2x 2 y 2 2
2
(d) f has no potential 2 1 2
15. The magnitude of directional derivative of f ( x y z ) direction of a i j k is _____.
at the point P(3, 0, 4) in the
16. The magnitude of directional derivative of f e x cos y at the point P (2, , 0) in the direction of a 2 i 3 j is _____. 17. The divergence of vector field given by v e x (cos y i sin y j) is (a) 2e x cos y
(b) 2e x sin y
(c) e x cos y
(d) e x sin y
18. The divergence of vector field given by v ( x 2 y 2 ) 1 ( y i x j) is _____. 19. Consider the flow with velocity vector v y i . This flow is (a) Compressible and Irrotational (b) Incompressible and Irrotational (c) Incompressible and not Irrotational (d) Compressible and not Irrrotational 20. Which of the following is correct? (a) div ( ) div ( ) (b) curl (curl F ) grad (div F ) F (c) div ( ) div ( ) 2 2
(d) curl (curl F) div (grad F ) 2 F
21. If f tan 1 ( y x ) , then 2 f is _____. 22. If the vector field is given by v xyz ( x i y j z k ) , then curl v at point P (1,1,1) is (a) i j k (b) i j (c) j k (d) 0 23. If u y i z j x k and v yz i zx j xy k , then u curl v (a) i j k (b) i j (c) j k (d) 0 24. Which one of the following depends on the choice of a particular coordinate system? (a) Gradient (b) Divergence (c) Gradient and Curl (d) None of these 25. The directional derivative of a scalar point function is a function of (a) position (b) direction (c) both (a) and (b) (d) either (a) or (b) 26. The necessary and sufficient condition that the force field F is conservative is (a) grad F 1 (b) curl F 0 (c) div F 0 (d) curl F 1 27. The magnitude of maximum directional derivative of f x 2 y xyz z 3 x in a direction from the point P (1, 0,1) is _____. 28. The maximum value of the directional derivative of f x 3 y 2 z 2 from P (1, 1,1) is _____. 29. If velocity vector of a flow is v ( x 2 y az ) i (bx 3 y 2 z ) j (3 x cy 2 z ) k , then the value of a, b, c such that the flow is irrotational is, respectively, _____. (d) None of these (a) 2, 3, 2 (b) 3, 2, 2 (c) 2, 2, 3 30. The angle ( ) between the two surfaces x 2 y 2 z 16 and x 2 2 y z 2 9 at P (1,1, 2) is (a) 0 90o (b) 90o 180 o (c) 0o (d) 90o 31. The relationship between a, b, c such that f ax 2 by 2 cz 2 satisfies Laplace equation is (a) a b c 0 (b) a b c 0 (c) a b c 0 (d) a b c 0 32. Which one of the following satisfies Laplace’s equation? (a) f 2 x 2 y 2 z 2 (b) f x 2 2 y 2 3 z 2 (c) f x 2 2 y 2 z 2
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(d) f x 2 3 y 2 2 z 2
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2.7
Chapter 2: Calculus
[2.115]
Vector Integral Calculus
The vector integral calculus extends integrals as known from calculus to integrals over curves (line integrals), surfaces (surface integrals) and solids. These different kinds of integrals can be transformed into one another by powerful formulas of Green, Gauss and Stokes.
2.7.1 Line Integrals In a line integral we integrate a given function (or integrand), along a curve C in space (or in the plane). Hence ‘curve integral’ or ‘line integral’ would be a better name. We represent the curve C by a parametric representation as r (t ) [ x (t ), y (t ), z (t )] x (t )i y (t ) j z (t )k , where a t b . The curve C is called the path of integration, A : r (a ) is its initial point and; B : r (b) is its terminal point. C is now oriented. The direction from A to B , in which t increases, is Figure 2.53: Oriented Curve called the positive direction on C and can be marked by an arrow, as shown in Fig. 2.53a. The points A and B may coincide, as shown in Fig. 2.53b, then C is called a closed path. C is called a smooth curve if it has at each point a unique tangent whose direction varies continuously as we move along C . Technically, r (t ) is differentiable and the derivative r(t ) dr dt is continuous and different from the zero vector at every point of C . A line integral of a vector function F (r ) over a curve C : r (t ) is defined by, b
C F (r ) dr a F r(t ) r(t )dt , where r dr
(2.10)
dt
In terms of components, with dr dx i dy j dz k and F (r ) F1 i F2 j F3 k , Eq. 2.10 becomes, b
C F (r) dr C ( F1 dx F2 dy F3 dz ) a ( F1 x F2 y F3 z)dt ,
where, x dx dt ,
y dy dt
and, y dz dt .
C
C
If the path of integration C in Eq. 2.10 is closed curve, then instead of
The integrand in Eq. 2.10 is a scalar, not a vector, because we take the dot product. F r r is the tangential component of F .
The integrand on the RHS of Eq. 2.10 is a definite integral of a function of t taken over a t b on the t axis in the ve direction (the direction of increasing t ). This definite integral exists for continuous F and piecewise smooth C , because this makes F r piecewise continuous. Simple general properties of line integral
we write
.
C k (F dr ) k C F dr , ( k as constant) (F G) dr F dr G dr F dr F dr F dr , where the path C is subdivided into two arcs C1 and C2 C
C
C1
C
C
C2
that
have the same orientation as C . If the sense of integration along C is reversed, the value of the integral is multiplied by –1. The line integral in Eq. 2.10 generally depends not only on F and on the endpoints A and B of the path, but also on the path itself along which the integral is taken. Example 2.230 [EC-2008 (2 marks)]: The value of the integral of the function g ( x, y ) 4 x 3 10 y 4 along the straight line segment from the point (0, 0) to the point (1, 2) in the x y plane is (a) 33 (b) 35 (c) 40 (d) 56 Solution (a): The equation of straight line segment from the point (0, 0) to the point (1, 2) in the x y
Thus
plane is x 1
3 4 3 4 y 0 {(2 0) (1 0)}( x 0) y 2 x g ( x, y ) 4 x 10(2 x) 4 x 160 x . x 1
x 0 g ( x, y )dx x 0 (4 x
3
160 x 4 ) dx {x 4 160(1 5) x 5 }xx 10 1 (160 5) 0 33 .
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Chapter 2: Calculus
[2.116]
Example 2.231 [TF-2008 (2 marks)]: A curve in space is represented by a vector 2 r (t ) x (t )iˆ y (t ) ˆj z (t ) kˆ . Given a vector function F ( r ) 5 z iˆ xy ˆj x z kˆ and r (t ) t iˆ t ˆj t kˆ
(a) 7 12
1
0 F {r (t )} (d r
, 0 t 1 , the value of the integral
dt ) dt is
(b) 17 12 (c) 27 12 2 3 Solution (d): F {r (t )} 5t iˆ t ˆj t kˆ and d r dt iˆ ˆj kˆ thus 1
(d) 37 12
1 1 dt ) dt (5t iˆ t 2 ˆj t 3 kˆ ) (iˆ ˆj kˆ ) dt (5t t 2 t 3 )dt
0 F {r (t )} (d r
0
2
0
3
I (5 2)t (1 3)t (1 4)t
4 t 1
37 12
t 0
Example 2.232 [ME-2009 (2 marks)]: A path AB in the form of one quarter of a circle of unit radius is shown in the figure. Integration of ( x y ) 2 on path AB traversed in a counter-clockwise sense is
(a) ( 2) 1 (b) ( 2) 1 (c) 2 (d) 1
Solution (b): The curve is C : x 2 y 2 1 in first quadrant, and in parametric form this curve is represented as x cos , y sin for 0 2 . So the integrand ( x y ) 2 x 2 y 2 2 xy ( x y ) 2 cos 2 sin 2 2 sin cos 1 sin 2 .
2
0
2
(1 sin 2 ) d (1 2) cos 2 0
Thus
the
required
integral
is
( 2) 1 .
Example 2.233 [EE-2009 (2 marks)]: F ( x, y ) ( x 2 xy )aˆ x ( y 2 xy )aˆ y . Its line integral over the straight line from ( x, y ) (0, 2) to ( x, y ) (2, 0) evaluates to (a) –8 (b) 4 (c) 8 (d) 0 Solution (d): The curve is a straight line whose equation is y 2 (0 2) (2 0) ( x 0)
y x 2 , on which any point can be taken as r (t ) t aˆ x ( t 2) aˆ y r (t ) aˆ x aˆ y with
0 t 2 . So, F r (t ) r (t )
t
2
t ( t 2) aˆ x (t 2) 2 t ( t 2) aˆ y {aˆ x aˆ y }
F r (t ) r(t ) t 2 t (t 2) (t 2) 2 t (t 2) 2t 2 4t . Thus, line integral of F ( x, y ) over a
C
straight line is
2
2
2
0
0
0
F (r ) dr F r2 (t ) r2 (t ) dt (4t 4)dt (2t 2 4t ) 8 8 0
[Similar question was also asked in PI-2009 (2 marks)] Example
[EC-2010 (2 2 marks)]: If A xyaˆ x x aˆ y then
2.234
C A dl
Solution
We
have
C A d l y 1 A d l x 2
x2
C A d l x 1
3
xy plane
(d) 2 3
3
Adl
y 3
Adl
3 3
x dx
y 3
4
y 1
3
dy
x 1 3
x 2
3
3 x dx
dl dx aˆ x dy aˆ y A dl xy dx x 2 dy .
so
x 1
3
y 1 1 y 3
3
Adl .
3 , A dl (4 3) dy ; y 3 , A dl 3 x dx ; x 1
(c) 1
(c):
x2
(b) 2
over the path shown in
the figure is
(a) 0
dy
y 1,
Along
So,
A dl x dx ;
3 , A d l (1 3) dy . Thus,
x
2 x2
2
3
x 1
3
4 3
y 3
y y 1
3x
2 x 1 3
2
x2
3
1 3
y 1
y y 3
C A d l (1 2) (4 3) (1 3) (4 3)(3 1) (3 2) (1 3) (4 3) (1 3)(1 3) 1
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Engineering Mathematics
Chapter 2: Calculus
[2.117] 2
2
Example 2.235 [EE-2013 (1 mark)]: Given a vector field F y xax yza y x az , the line integral evaluated along a segment on the x axis from x 1 to x 2 is
F dl
(a) –2.33 (b) 0 (c) 2.33 (d) 7 Solution (b): The curve is a straight line whose equation is y 0 , on which any point can be taken as 2
r (t ) 0 a x t a y with 1 t 2 . As ( d dt )r (t ) r (t ) a y F r (t ) r (t ) ( t az ) ( a y ) 0 . Thus, 2
2
C F (r) dr 1 F r2 (t ) r2 (t)dt 1 0 dt 0
line integral of F over a straight line y 0 is
Example 2.236 [EE-2014 (2 marks)]: The line integral of function F yz iˆ , in the counterclockwise direction, along the circle x 2 y 2 1 at z 1 is (b) (c) (a) 2 (d) 2 2 2 Solution (b): As any point on x y 1 can be taken as x cos , y sin with 0 2 ; at z 1, So the line integral r cos i sin j k ( dr dt ) r(t ) sin i cos j . I F r ( ) r ( )d C
2
0
I (1 2)
2
0
(sin i ) ( sin i cos j)d
2
0
( sin 2 )d
2
(cos 2 1) d (1 2) (1 2) sin 2 0 (1 2) 2 (0)
[Similar question was also asked in AE-2014 (2 marks)] Example
2.237 [XE-2013 (2 mark)]: The work done by the force 2 3 F ( x x )iˆ ( x y ) ˆj in moving a particle once along the triangle with vertices (0, 0) , (1, 0) and (0,1) in the anti-clockwise direction is
2
(b) 1 6
(a) 0 Solution
(c) 1 3
(d) 5 3
The work done by the force F along is r x iˆ y ˆj 2 ˆ 2 3 ˆ 2 2 3 ( x x )i ( x y ) j dx iˆ dy ˆj ( x x )dx ( x y )dy . Now along OA ,
(c):
C F d r C
C
2
y 0 dy 0 F d r ( x x ) dx OA
OA
x 1
x 0
( x x 2 ) dx (1 2) x 2 (1 3) x 3
F dr
x 0
x 1
AB
5 6
x 0
2
2
3
AB
( x 3 3x 2 4 x 1)dx (1 4) x 4 x 3
x 1
( x x )dx ( x ( x 1) )(dx) 2 x x 1 4 . Similarly, along
Now along AB , y x 1 dy dx F d r AB
y 0
x 0
2
x 1
BO ,
y 0
x 0 dx 0 F d r y 3 dy y 3dy (1 4) y 4 1 4 . So, the work done by the y 1 BO BO y 1 force F ( x x 2 )iˆ ( x 2 y 3 ) ˆj in moving a particle along the given path is
OABO F d r OA F d r AB F d r BO F d r (5 6) (1 4) (1 4) 1 3 [Similar question was also asked in CH-2007 (2 marks)] Example
2.238
[AG-2014
(2
marks)]:
The
2
value
of
the
integral
2
I ( x y ) dx ( x z ) dy ( xyz ) dz , where C is the arc of the parabola y x in the plane z 2 C
from P(0, 0, 2) to Q (1,1, 2) is (a) 43 15
(b) 43 15
Solution (c): The given curve y x
(c) 17 15 2
(d) 17 15
in the plane z 2 from P(0, 0, 2) to Q (1,1, 2) can be
represented in parametric form as C : r (t ) t i t 2 j 2 k from t 0 to t 1 , so r (t ) i 2t j . Thus 2 2 4 3 x t , y t , z 2 and the given integral F ( x y ) i ( x z ) j ( xyz ) k t i (t 2) j 2t k . 1
1
1
0
0
0
So F (r ) dr F r (t ) r (t ) dt {t 4 2t (t 2)}dt {(1 5)t 5 (2 3)t 3 2t 2 } 17 15 C
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Chapter 2: Calculus
[2.118]
Example 2.239 [XE-2014 (2 marks)]: If the work done in moving a particle once around a circle x 2 y 2 4 under the force field F ( x, y ) (2 x ay ) iˆ (2 y ax) ˆj is 16 , then a is equal to …… Solution: Any point on x 2 y 2 2 2 can be taken as x 2cos , y 2 sin with 0 2 , So the line integral r 2 cos i 2 sin j k ( dr dt ) r (t ) 2 sin i 2 cos j . I F r ( ) r ( )d C
2
0
I
2
0
(4 cos 2a sin ) i (4 sin 2a cos ) j ( 2 sin i 2 cos j)d
(8sin cos 4a sin
2
2
2
0
0
) (8sin cos 4 a cos 2 ) d 4 ad 4 a
8a
I 8a 16 a 2 .
Path Independence of Line Integrals: The line integral given by Eq. 2.10 is path independent in a domain D in space if every pair of endpoints A , B in D , the integral Eq. 2.10 has the same value for all paths in D that begin at A and end at B , as shown in Fig. 2.54, i.e. a line integral (Eq. 2.10) with continuous F1 , F2 and F3 in a domain D in space is path independent in D if and only if F F1 i F2 j F3 k
f
is the gradient of some function
F grad f F1 f x , F2 f y , F3 f z
and
in
in
D , i.e., Figure 2.54: Path this case Independence
C ( F1dx F2 dy F3 dz) f ( B) f ( A)
Vector Fields that are Gradients of Scalar Fields (Potentials): Some vector fields can be obtained from scalar fields. Such a vector field is given by a vector function F( P) , which is obtained as the gradient of a scalar function, F ( P) grad f ( P ) . The function f ( P ) is called a potential function or a potential of F( P) . Such a F( P) and the corresponding vector field are called conservative because in such a vector field, energy is conserved; that is no energy is lost (or gained) in displacing a body from a point P to another point in the field and back to P . Path Independence and Integration around Closed Curves: The integral of Eq. 2.10 is path independent in a domain D if and only if Figure 2.55: Closed Path its value around every closed path in D is zero. Path Independence and Exactness of Differential Forms: The path independence to the exactness of the differential form given by F dr F1 dx F2 dy F3 dz (2.11) under the integral sign (2.10) is called exact in a domain D in space if it is the differential df (f x) dx (f y )dy (f z )dz (grad f ) dr of a differentiable function f ( x, y, z ) everywhere in D . Let F1 , F2 , F3 in the line integral
C F(r ) dr ( F1 dx F2 dy F3 dz) C
be continuous first
partial derivatives in a domain D in space, then, if the differential form F dr F1 dx F2 dy F3 dz is exact in D , and Eq. 2.10 is path independent, then in D i curl F curl (grad f ) 0 f 0 x F1
j y F2
k z 0 F3
F3 y F2 z , F1 z F3 x , F2 x F1 y (2.12) Hence, if Eq. 2.12 holds in D and D is simply connected, then Eq. 2.11 is exact in D and thus Eq. 2.10 is path independent.
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Engineering Mathematics
Chapter 2: Calculus
[2.119]
Example 2.240 [ME-2005 (2 marks)]: The line integral
V d r
of the vector function
2 2 V ( r ) 2 xyz iˆ x z ˆj x y kˆ from the origin to the point P(1,1,1) is (a) 1 (b) 0 (c) –1 (d) cannot be determined without specifying the path 2 2 Solution (a): The given function is V ( r ) 2 xyz iˆ x z ˆj x y kˆ V iˆ V ˆj V kˆ .
1
2
3
As
V3 y V2 z , V1 z V3 x , V2 x V1 y the given function is exact and so the
line integral V d r is path independent. Now we have to find the potential function ( f ) of V such
that V grad f . Since, f x V1 f 2 xyz dx x 2 yz c , where c h( y , z ) is a function of y and z . Now f y V2 x 2 z (c y ) x 2 z c y 0 c is a function of z only, i.e.,
c h( z ) . Now f z V3 x 2 y (c z ) x 2 y (c z ) 0 c is also not a function of z , 2
thus c is just a constant and not a function of x or y or z ; so f x yz c ; if A(0, 0, 0) and
B(1,1,1) are endpoints of the given line integral V d r f ( B ) f ( A) (12 1 1 c) (0 c) 1 C
[Similar question was also asked in PI-2011 (2 marks)] Example 2.241 [EC-2008 (2 marks)]: Consider points P and Q in the x y plane, with P(1, 0) and Q
Q (0,1) . The line integral 2 ( xdx ydy ) along the semi-circle with the line segment PQ as its P
diameter is (a) –1 (c) 1
(b) 0 (d) depends on the direction (clockwise or anti-clockwise) of the semicircle Q
Q
P
P
Solution (b): The given integral can be written as I (2 xdx 2 ydy ) F (r ) dr , where
F (r) (2 x i 2 y j) F1 i F2 j and dr dx i dy j . As F2 x F1 y F (r ) is exact, so the line integral
Q
P F (r) dr
is path independent. Now we have to find the potential function ( f ) of
F (r ) such that F (r ) grad f . As, f x F1 f 2 x dx x 2 c , where c h( y ) is a function 2
2
of y . Now f y F2 y c y c 2 y dy y 2 , thus f x y . Now if P(1, 0) and
Q(0,1) are endpoints of the given line integral
C F (r ) dr f (Q ) f ( P ) (0
2
12 ) (12 0 2 ) 0 .
2.7.2 Double Integrals f ( x)
The definite integral of a function b
a
n
f ( x ) dx lim i 1 f ( xi )xi ,
where
all
n
of one variable is defined by the limit
xi 0
and
a x0 x1 x2 xn b
and
xi xi xi 1 . This integral is represented by the area between the curve y f ( x) , the x axis and between x a and x b . Now consider x f ( x , y ) and a region R of the xy plane as shown in Fig. 2.56. The integral of f ( x, y ) over region R is
R f ( x, y )dA
lim
n ,all Ai 0
n
i 1 f ( xi , yi )Ai ,
where Ai ( i 1, 2, , n ) is a partition of R into n elements of area Ai and ( xi , yi ) is a point in
Ai . Now z f ( x, y ) represents a surface, and so f ( xi , yi )Ai zi Ai is the volume between z 0 and z zi on the base Ai . The integral
R f ( x, y)dA
is the limit of the sum of all such volumes,
and so it is the volume under the surface z f ( x, y ) above the region R . The partition of R into elementary areas can be achieved using grid lines parallel to the x and y axes as shown in Fig. 2.60a. Then Ai xi yi and we can write
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Chapter 2: Calculus
[2.120] n
i1 f ( xi , yi ) xi yi R f ( x, y )dA R f ( x, y )dx dy nlim
(2.13)
Other partitions may be chosen, for example a polar grid as shown in Fig. 2.60b. Then the element of area is (ri i )ri Ai and we write,
R f ( x, y )dA R f (r cos , r sin )r dr d
Figure 2.56: Volume as an integral
(2.14)
Figure 2.57: Grid for the partition of R (a) Rectangular Cartesian (b) Polar
We can evaluate integrals of the type
R f ( x, y)dx dy
as a repeated single integrals in x and y .
Consequently they are usually called double integrals.
Figure 2.58: Evaluation of double integrals for a Region
Consider the region R shown in Fig. 2.58a with boundary ACBD. Let the curve ACB be given by
y g1 ( x) and the curve ADB by y g2 ( x) , then we can evaluate
R f ( x, y)dx dy
by summing for
y first over the yi , holding x constant, from y g1 ( xi ) to y g2 ( xi ) and then summing all such strips from A to B , i.e. from x a to x b . Thus
i1 j 1 f ( xi , y j )y j xi x a y g ( x ) R f ( x, y )dA n , alllim x , y 0 n2
i
n1
xb
y g2 ( x ) 1
j
f ( x, y ) dy dx
(2.15)
Here the integral inside the brackets is evaluated first, integrating w.r.t. y , keeping the value of x fixed, and then the result of this integration is integrated w.r.t. x . Alternatively, as shown in Fig. 2.61b, we can sum for x first and then y . If the curve CAD is represented by x h1 ( y ) and the curve CBD by x h2 ( y ) , thus
i1 j 1 f ( xi , y j )x j yi y c x h ( x ) R f ( x, y )dA n , alllim y ,x 0 n1
j
n2
i
y d
x h2 ( x ) 1
f ( x, y )dx dy
(2.16)
In both Eqs. 2.15 and 2.16, n min(n1 , n2 ) . If the double integral exists then these two results are equal, and in going from one to the other we have changed the order of integration. Notice that the limits of integration are also changed in the process. 1 1x x dx dy . Example 2.242 [CS-1993, EC-1993 (1 mark)]: Find value of the double integral 0 0 1 y2
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Solution: I
x
1 1x
0 0
1 y
Chapter 2: Calculus
1
x
1x
dx dy 0 0
2
2
1 y
1
0
[2.121]
dy dx x tan 1 y
1x
0
1
dx x tan 1 (1 x ) 0 dx 0
1
1
1 1 2 2 2 I x cot x dx (cot x )(1 2) x {1 (1 x )}(1 2) x dx 0 0 II
I
Let I1 {1 (1 x 2 )}(1 2) x 2 dx (1 2) 1 {1 (1 x 2 )} dx (1 2)( x tan 1 x ) 1
2
1
1
I (cot x )(1 2) x (1 2)( x tan x ) ( 4) (1 2) 1 ( 4) 0 (1 2) ( 8) 0 Example 2.243 [ME-2005 (1 mark)]: Changing the order of integration in the double integral
I
8
2
f ( x, y ) dy dx 0 x 4
leads to I
s
q
f ( x, y )dx dy . Then what is q ? r p
2 (c) x (d) 8 (b) 16y Solution (a): For the given integral, we integrate first w.r.t. y in which the vertical length of element
(a) 4y
dx , as shown in figure, vary from y 0 to y x 4 ; and then integrate w.r.t. x in which this element dx shifts from x 0 to x 8 . Now, if we integrate first w.r.t. x in which the horizontal length of element dy , as shown in figure, vary from x 0 to x 4 y ; and then integrate w.r.t. y in which this element dy shifts from y 0 to y 2 , thus I
y 2
x4 y
y 0 x 0
f ( x, y )dx dy
s
q
f ( x, y ) dx dy q 4 y r p
Example 2.244 [CE-2008 (2 marks)]: The value of (a) 13.5
(b) 27.0
3
x
0 0 (6 x y)dx dy (c) 40.5
is (d) 54.0
2 x 0 (6 x y )dx dy 0 0 (6 x y )dy dx 0 {6 y xy (1 2) y }0 dx (1 2) x dx 6 x (1 2)3 x dx 3 x (1 2) x 3 3 (1 2)3 0 13.5
Solution (a): I
3
x
3
x
3
0
I
3
0
6x x
2
3
2
2
3 3
2
2
3
0
0
[Similar questions were also asked in ME-2000, AG-2010 (2 marks)] Example 2.245 [ME-2008 (2 marks)]: Consider the shaded triangular region P
P
shown in figure. What is
xy dx dy ?
(a) 1 6
(b) 2 9
(c) 1 16
(d) 1
Solution (a): The equation of the line joining the points (0,1) and (2, 0) is y 1 {(0 1) (2 0)}( x 0) y ( x 2) 1 . Let us take the horizontal strip of width dy , which lies between x 0 and x 2 2 y ; also the strip moves from y 0 to y 1 . So I xy dx dy P
I
y 1 y 0
I 2
y 1 y0
(1 2)(2 2 y )
y 1 y 0
x 2 2 y
x 0
2
xy dx dy
y 0 dy
y 1 y 0
y 1 y0
(1 2)( x 2 y) xx 202 y dy
2(1 y ) y dy 2 2
( y y 3 2 y 2 ) dy 2 (1 2) y 2 (1 4) y 4 (2 3) y 3
y 1 y 0
(1 y 2 2 y ) y dy
y 1 y 0
(1 2) (1 4) (2 3) 0 1 6
[Similar question was also asked in ME-1997 (2 marks)] Example 2.246: Evaluate
R x
2
2 2 ydA , where R is the region x y 1 .
Solution: As the domain of integration is a circle so, x r cos and y r sin and dA r d dr
I x 2 ydA R
r 1
r 0
2
0
r 2 cos 2 r sin r d dr
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r 1
r 0
2
0
r 4 cos 2 sin d dr .
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Chapter 2: Calculus
[2.122]
Note that in this integral the integration is such that we can separate the variables r and and hence we can write, I
r 1
r 0
2
0
cos 2 sin d dr . As the limit of integration with respect to do not
involve r , we can write, I
r 1 4
r 0
r dr
2
cos2 sin d {(r 5 )10 } (1 3) cos3
0
2 0
0.
Example 2.247 [EE-2009 (2 marks)]: f ( x, y ) is a continuous function 2
defined over ( x, y ) 0,1 0,1 . Given the two constraints, x y and
y x2 , the volume under f ( x, y ) is y 1
x y
(a)
y 0 x y
(c)
y0 x0 f ( x, y)dx dy
y 1
f ( x, y ) dx dy
2
y 1
x 1
(b)
y x x y
(d)
y 0 x0
2
x 1
y x
f ( x, y )dx dy
2
x y
f ( x, y) dx dy 2
2
Solution (a): The shaded region in the figure satisfy the two constraints x y and y x for all ( x, y ) 0,1 0,1 , as shown in figure. So the volume under f ( x, y ) is V
y 1
x y
y 0 x y 2
Example 2.248 [XE-2013 (1 mark)]: The integral
1
x
0 x
2
f ( x, y ) dx dy
2
( x y )e x y dy dx equals
(a) (e 2) e (b) (e 1) 2 e (c) (e 1) 2 Solution (d): As the given integral is first integrated w.r.t. y with y
(d) (e 2) 2e
2
changing from y x to y x ; and then w.r.t. x with x changing from x 0 to x 1 . Now for changing the order of integration, i.e., if we integrate first w.r.t. x in which the horizontal length of element dy , as shown in figure, vary from x y to x y1 2 ; and then integrate w.r.t. y in which this element dy shifts from y 0 to y 1 , thus
I
1
x
0 x
2
2
( x y )e x y dy dx
I (1 2)
y 1 y 0
e
1
y 1 y 0
x y
2
( x y)e x y dx
x y
y 1 y 0
(1 2)e x2 y
x y
dy x y
e y dy ( 1 2)(e 1 y e y ) yy 10 ( 1 2){(2 e) 1} (e 2) 2e
Example 2.249 [MA-2014 (2 marks)]: The value of
R xy dx dy , where
R is the region in the first
2
quadrant bounded by the curves y x , y x 2 and x 0 is ……………. Solution: The bounding curves intersect where x 2 x 2 , which gives x 1 (with y 1 ) and x 2 (with y 4 ) in which A(1,1) lies in the first quadrant. So the region lies in the first quadrant is AOBA with B (0, 2) and O (0, 0) . In this question, we choose to take y first because the formula for the boundary is easier to deal with
I xy dx dy
x 1
x 0
R
I (1 2)
x 1
x 0
x
5
y 2 x y x
2
xy dy dx
x 1
x 0
y x2
rather that
2 y 2 x
(1 2) xy
y x
2
dx (1 2)
2
(a) (e 1) 2
x(2 x)
2
2
(b) (e 1) 2
0 0 e 2
x 0
2
x
x y
dy dx
2
0 (e
2
(c) (e e) 2 x y y x ) y 0
2
x
0 0 e
x y
xx 4 dx
x 3 4 x 2 4 x dx (1 2) (1 6) x 6 (1 4) x 4 (4 3) x 3 (4 2) x 2
Example 2.250 [ME-2014 (2 marks)]: The value of the integral
Solution (b): I
x 1
x y1 2 . Thus we get,
x 1 x0
0.375
dy dx is
(d) (1 2) e (1 e)
2
x2
dx 0 (e 2 x e x ) dx (1 2)e 2 x e x x 0
I {(1 2)e 4 e 2 } (1 2) 1 (1 2)e 4 e 2 (1 2) (1 2)(e 4 2e 2 1) (1 2)(e 2 1) 2
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Engineering Mathematics
Chapter 2: Calculus
[2.123]
Gaussian Integration: Gaussian integration is simply integration of the exponential of a quadratic. We cannot write a simple expression for an indefinite integral of this form but we can find the exact
2
answer when we integrate from to . The basic integral we need is G e x dx . The trick to
calculate this is to square this using integration variables x and y for the two integrals, i.e.,
2
2
G 2 e x dx e y dy
e
( x2 y 2 )
dxdy , which looks like an integral over the area dx dy .
Now, we can evaluate G 2 using the polar coordinates, whose differential area can be written as r dr d , with limits r 0 to r and 0 to 2 . Thus G2
2
0
0 e
r2
with
r dr d
2
0
limit
2
G e 0
k
2
2
2 d e r r dr 2 e r r dr . Now by letting r k 2rdr dk 0
0
changing
dk (1 ) (e
k
)
k k 0
k0
from
k ,
to
G . Thus
e
x2
dx
(2.17)
For a general quadratic exponent we simply complete the square and then integrate using a similar change of variables, i.e.,
e
ax 2 bx c
dx e
a x (b 2a )
2
a
e (b
2
4a ) c
dx ( a )e( b
2
4 a )c
a
x2
a f ( x)dx 2 0 f ( x)dx if f ( x) is an even function; as e is an even function thus from Eq. 2.14, we have e x dx 2 e x dx . Thus 0 x (2.18) 0 e dx (1 2) Example 2.251 [ME-1994 (1 mark)]: Evaluate ye y dy . 0 Now, using definite integral property
2
2
2
3
2 t
3 2 2 Solution: Let y t 3 y dy 2tdt dy
t . Thus, I
0
3
ye y dy t1 3e t 0
2
3y
2 t 3t
43
2
dt
dt
2 t 3t
2
43
t 2
e 3 0
dt ; also at y 0 , t 0 and at y ,
dt
21 32
(b)
Solution
(a):
From
(c) 2
Eq.
2.18
0
1 8,
with
2 ) exp x 2 8 dx is
Example 2.252 [EC-2005 (2 marks)]: The value of the integral I (1 (a) 1
. 3
0
(d) 2 1 exp x 2 8 dx (1 8) 2 2
x2 y2
I (1 2 ) 2 1 [Similar question was also asked in PI-2010 (1 mark)]
Example 2.253 [IN-2007 (2 marks)]: The value of the integral (a)
2
Solution (c): Using Eq. 2.17 with 1, I
0
0
e( x
2
2
y )
2
(b)
dx dy is
(d) 4 dx dy
Example 2.254 [XE-2013 (1 mark)]: The value of the integral (a)
e
(c)
(b)
0 0 e
e dx e dy G
x
2
0
1
0 1
y
2
0
( log e t ) dt is
(d) 2
(c)
2
Solution (b): Let log e t k (1 t ) dt 2kdk dt 2 k t dk 2ke 0
2
k
2
2
dk ( t e k ); also at
2
t 0 , k 2 k and at t 1 , k 2 0 k 0 . I 2 e k dk 2 e k dk 2
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2
0
1 2
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Engineering Mathematics
Chapter 2: Calculus
[2.124]
2.7.3 Green’s Theorem in a Plane This theorem shows the relationship between line integrals and double integrals, and will also provide a justification for the general change of variables in a double integral. Let R be a closed bounded region in the xy plane whose boundary C consists of finitely many smooth curves (like, LMN and LKN). Let F1 ( x, y) and F2 ( x, y) be functions that are continuous and have continuous partial derivatives F1 y and F2 x everywhere in some domain containing R , then Figure 2.59: Green's Theorem in a Plane F2 F1 (2.19) dx dy ( F dx F dy ) 1 2 R x y C Here we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration, as shown in Fig. 2.59 Proof: Consider the first on the left side of the integral, (F2 x) dx dy R
yd y c
x g1 ( y )
(F2 x ) dx dy R
x g2 ( y )
LMN
( F2 x) dx dy
F2 ( x, y )dy
LKN
yd
y c
F2 g 2 ( y ), y F2 g1 ( y ), y dy
F2 ( x, y ) dy
LMNKL
F2 ( x, y ) dy F2 ( x, y )dy C
Similarly, (F1 x ) dx dy F1 ( x, y )dy . R
C
It follows from the Green’s theorem (Eq. 2.19) that the area A enclosed by the closed curve is 1 given by A 1 dx dy C x dy C y dy 2 C ( y dx x dy) R Under the transformation of coordinates x x (u , v) and y y (u, v ) , the curve becomes C , enclosing an area A . Then, A du dv udv u (v x)dx (v y )dy R
C
C
A u ( v x) dx u (v y ) dy ( x ) u (v y ) ( y ) u (v x ) dx dy ; C
R
u v
Eq. 2.19, A R
x y
This
implies
that
the
u
from
2 v u v 2v u v u v u dx dy R dx dy . xv y x yx x y y x
element
of
area
( du dv )
is
equivalent
to
the
element
u v u v dx dy . The modulus sign is used to preserve the orientation of the curve. x y y x
x y x y . (u , v ) u v v u Hence, this enables us to make a general change of coordinates in a double integral as,
Similarly ‘ dx dy J ( x, y ) du dv ,’ where J ( x , y )
( x, y )
R f ( x, y)dx dy R f x(u, v), y (u, v) J ( x, y) du dv
(2.20)
The Green’s theorem (Eq. 2.19) in a plane may be generalised to three dimensions as
A curl ( P, Q, 0) k dx dy C ( P, Q, 0) dr . For a general surface S with bounding curve C , as shown in Fig. 2.60, this identity becomes curl F(r ) dS F(r ) dr , where dS nˆ dS is the S
C
2.60: Three-dimensional vector element of surface area and nˆ is a unit vector Figure generalization of Green’s theorem along the normal. This generalisation is called Stokes’ theorem, and will be discussed later, after introducing the concept of surface integral.
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Engineering Mathematics
Chapter 2: Calculus
Example 2.255 [CE-2005 (2 marks)]: Value of the integral
[2.125]
C ( xydy y dx) , where C 2
is the square
cut from the first quadrant by the line x 1 and y 1 will be (use Green’s theorem to change the line integral into double integral) (a) 1 2 (b) 1 (c) 3 2 (d) 5 3 Solution (c): Here F1 ( x, y ) y 2 and F2 ( x, y) xy , so that F1 y 2 y and F2 x y
F2 x F1 y y (2 y ) 3 y . Thus the line integral transforms into an easy double integral by using Green’s theorem in a plane, I ( xydy y dx) (F2 x F1 y ) dx dy 2
C
I
y 1
x 1
y 0 x 0
3 y dx dy
y 1 y0
R
(3 yx ) xx 10 dy
y 1
y 0
y 1
3 y dy (3 2) y 2
y 0
3 2
[Similar question was also asked in XE-2009 (1 mark)] Example 2.256 [XE-2007 (2 marks)]: The value of
C
( xy 2 2 x)dx ( x 2 y 4 x) dy along the circle
C : x 2 y 2 4 in the anticlockwise direction is (a) 16 (b) 4 (c) 4 (d) 16 2 2 Solution (d): The parametric form of C : x y 4 is x 2 cos dx 2 sin d and y 2 sin dy 2 cos d , thus substituting them in the given integral and integrating from 0
to
I ( xy 2 2 x ) dx ( x 2 y 4 x ) dy
2 ,
2
0
C
( 4 sin 2 4 sin 4 8 8 cos 2 ) d
I (2 cos 2 cos 4 8 4 sin 2 ) 20 16 [Similar question was also asked in XE-2008 (2 marks)]
Example 2.257 [AE-2008 (2 marks)]: The value of the integral (1 2 ) ( xdy ydx) taken anticlockwise along a circle of unit radius is (a) 0.5 (b) 1 Solution (b): Here F1 ( x, y ) y and
(c) 2 (d) F2 ( x, y ) x , so that F1 y 1 and F2 x 1
F2 x F1 y 1 (1) 2 . Thus the line integral transforms into an easy double integral by using Green’s theorem in a plane, I (1 2 ) ( xdy ydx) (1 2 ) (F2 x) (F1 y) dx dy , C
where
R
R
is the region around a circle of unit radius whose area is
12
I (1 2 ) 2 dx dy (1 ) dx dy 1 R
R
[Similar question was asked in ME-2014 (2 marks)] Example 2.258 [EE-2014 (2 marks)]: To evaluate the double integral
y 2 8
( y 2) 1
0
(1 2)(2 x y ) dx dy
, we make the substitution u (1 2)(2 x y ) and v y 2 . The integral will reduce to (a)
0 0 2udu dv 4
2
(b)
0 0 2udu dv 4
1
(c)
0 0 udu dv 4
1
(d)
0 0 udu dv 4
2
Solution (b): At x y 2 , u 0 ; at x ( y 2) 1 , u 1. At y 0 , v 0 ; at y 8 , v 4 . Now and
v y 2 y 2v
J
x u
x v
y u
y v
u (1 2)(2 x y ) x u v ;
1
1
0
2
2;
8
0
( y 2) 1
y 2
2x y 2
so
dx dy
4
0
the
Jacobian
transformation
u J du dv 2u du dv 1
4
1
0
0
0
Example 2.259 [ME-2005 (2 marks)]: By a change of variables x (u , v ) uv , y (u , v) v u in a double integral, the integrand f ( x, y ) changes to f uv, v u (u , v ) . Then (u , v ) is
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Chapter 2: Calculus
[2.126]
(d) 1 (c) v 2 Solution (a): As by the change of variables x (u , v ) uv , y (u , v) v u in a double integral, the (a) 2v u
(b) 2uv
integrand f ( x, y ) changes to f uv , v u J , where J
x u y u
x v y v
v v u
u 2
1u
v u
v u
J 2v u J 2v u (u , v ) 2v u .
Example 2.260 [CH-2011 (1 mark)]: R is a closed planar region as shown by the shaded area in the figure below. Its boundary C consists of circles C1 and C 2 . If F1 ( x, y ) , F2 ( x , y ) , F1 y and F2 x are all continuous everywhere in R, Green’s theorem states that
R (F2
x ) (F1 y ) dx dy F1dx F2dy .
Which
C
ONE
of
the
following alternatives CORRECTLY depicts the direction of integration along C ? (a)
(b)
(c)
(d)
Solution (c): As given in Eq. 2.19, we integrate along the entire boundary C of R in such a sense that R is on the left as we advance in the direction of integration; this criteria is shown by the figure given in option (c).
2.7.4 Surface Integrals In surface integrals, we integrate over surfaces in space, such as a sphere or a portion of a cylinder.
Representation of a Surface: A surface S in xyz space is represented as e (1) z f ( x, y ) or (2) g ( x, y , z ) 0 ; for e.g., z a 2 x 2 y 2 or x 2 y 2 z 2 a 2 0 , z 0 , represents a hemisphere of radius a with centre (0, 0, 0) . As, for curves C in line integrals, we use parametric representation r r (t ) , where a t b , on the t axis, onto the curve C in xyz space, as shown in Fig. 2.61(a). Similarly, for surfaces S in surface integrals, we use a parametric representation. Surfaces are 2-D; hence we need two parameters, (say) u and v . Thus a parametric representation of a surface S in Figure 2.61: Representation of a (a) Curve and (b) Surface space is of the form, r r (u , v ) x (u , v ), y (u , v ), z (u, v) x(u , v ) i y (u , v ) j z (u, v) k (2.21) where, (u, v) varies in some region R of uv plane, as shown in Fig. 2.64b. Parametric Coordinated of Sphere: When evaluating surface integral over the surface of a sphere of radius a , as shown in Fig 2.62, the parametric coordinates can be given as, x a sin cos , y a sin sin , z a cos and dS a 2 sin d d . Parametric Coordinates of a Cylinder: When evaluating surface integral over the surface of a cylinder of radius a , as shown in Fig. 2.63, the parametric coordinates can be given as, x a cos , y a sin , z z , dS a dz d . A surface S is called a smooth surface if its surface normal depends continuously on the points of S . S is called piecewise smooth if it consists of finitely many smooth portions.
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Engineering Mathematics
Chapter 2: Calculus
Figure 2.62: Surface element in spherical polar coordinates
[2.127]
Figure 2.63: Surface element in cylindrical polar coordinates
Tangent Plane and Surface Normal: As the tangent vectors of all curves on a surface S through a point P of S form a plane, called the tangent plane of S at P (except at points where S has a sharp corner). Also, a vector perpendicular to the tangent plane is called a normal vector of S at P. As S can be given by r r (u , v ) x(u, v) i y (u , v) j z (u , v) k ; we can also get a curve C on S by taking a pair of differentiable functions, u u (t ) , v v (t ) whose derivatives u du dt and Figure 2.64: Tangent plane and Normal v dv dt are continuous. Then C has the position vector Vector r (t ) r u (t ), v (t ) . By differentiation and the use of the chain rule, we obtain a tangent vector of C
on S , as r (t )
dr
dr
u
dr
v . Hence the partial derivatives ru
dr
and rv
dr
at P are dt du dv du dv tangential to S at P . We assume that they are linearly independent, which geometrically means that the curves u const and v const on S intersect at P at a non-zero angle. Then ru and rv span the tangent plane of S at P . Hence their cross product gives a normal vector N of S at P , as N ru rv 0 (2.22) The corresponding unit normal vector n (as shown in Fig. 2.67) of S at P is given as, n 1 N N 1 ru rv (ru rv ) (2.23) Also, if S is represented by g ( x, y , z ) 0 , then n can also be given as,
n 1 grad g (grad g )
(2.24)
Surface Integrals of a vector function: To define a surface integral, we take a surface S , given by a parametric representation as given by Eq. 2.21. We also assume S to be piecewise smooth so that S has a normal vector given by Eq. 2.22 and unit normal vector given by Eq. 2.23. For a given vector function F , we can now define the surface integral over S by
S F ndA R F r(u, v) N(u, v) du dv
(2.25)
Here N N n and N ru rv is the area of the parallelogram with sides ru and rv , by the definition of cross product; and ndA n N du dv N du dv dA N du dv is the element of area of
S . Also, F n is the normal component of F . In component form, F F1 i F2 j F3 k , N N1 i N 2 j N 3 k and n cos i cos j cos k , ( , , are the angles between n and the coordinate axes x , y , z axes, respectively), we can write Eq. 2.25 as,
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Chapter 2: Calculus
[2.128]
S F n dS A ( F1 cos F2 cos F3 cos )dA R ( F1 N1 F2 N2 F3 N3 )du dv .
We can write
cos dA dy dz , cos dA dz dx and cos dA dx dy , then above integral becomes,
S F n dS S ( F1 dy dz F2 dz dx F3 dx dy )
(2.26)
We can use Eq. 2.26 to evaluate surface integrals by converting them to double integrals over regions in the coordinate planes of the xyz coordinate system. But we must carefully take into account the orientation of S (the choice of n ). For e.g. if the surface S is given by z h( x, y ) with ( x , y ) varying in a region R in the xy plane and if S is oriented so that cos 0 then,
S F3 cos dA S F3 dx dy R F3 x, y, h( x, y) dx dy ; but if
cos 0 , the integral on the right
side gets a minus sign. The surface integral arises naturally in flow problems, where it gives the flux across S ( mass of fluid crossing S per unit time) when F v , where is the density of the fluid and v is the velocity vector of the flow. We may thus call the surface integral as the flux integral. Example 2.261: Evaluate the surface integral
S F n dA ,
where F x 2 i 3 y 2 k and S is the
portion of the plane x y z 1 in the first octant, as shown in Figure. Solution: Writing x u and y v , we have z 1 x y 1 u v . Hence we can represent the plane x y z 1 in the form
r (u , v) u i v j (1 u v ) k and F u 2 i 3v 2 k . We obtain the first octant portion S of this plane by restricting x u and y v to the projection R of S in the xy plane. R is the triangle bounded by the two coordinate axes and the straight line x y 1 , obtained from x y z 1 by setting z 0 . Thus 0 x 1 y , 0 y 1 . Now ru dr du i k , rv dr dv j k N ru rv (i k ) ( j k ) i j k F N (u 2 i 3v 2 k ) (i j k ) u 2 3v 2 .
S F n dA R (u
2
3v 2 )dudv
v 1 u 1 v
v 0 u 0
(u 2 3v 2 ) du
v 1
v 0
(1 3)(1 v)
3
So,
3v 2 (1 v) dv 1 3
Surface integral for a scalar function: If a surface S is given by z f ( x, y ) , then setting u x , v y and r x i y j f ( x , y ) k u i v j f (u, v) k gives, ru dr du i (f u )k i fu k
and
rv dr dv j (f v )k j f v k , where fu f u
and
fv f v N ru rv (i fu k ) ( j fv k )
N fu i fv j k 1 fu2 fv2 . Since fu f x and
fv f y
N 1 (f x ) 2 (f y ) 2
Figure 2.65: Surface integral for a Scalar function
dA N dx dy 1 (f x ) 2 (f y ) 2 dx dy .
Hence, the surface integral for a scalar function is given by, 2
2
f f dx dy x y
S G(r)dA R* G x, y, f ( x, y) N dx dy R* G x, y, f ( x, y ) 1
(2.27)
Here R * is the projection of S into the xy plane, as shown in Fig. 2.68 and the normal vector N on S points up. If it points down, the integral on the right is preceded by a minus sign. Eq. 2.27 with G (r ) 1 gives the area A( S ) of S : z f ( x, y ) and thus, A( S ) 1 dA S
R*
1 (f x ) 2 (f y ) 2 dx dy
(2.28)
where R * is the projection of S into the xy plane.
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Engineering Mathematics
Chapter 2: Calculus
Example 2.262: Evaluate the surface integral
[2.129]
S ( x y z )dS , where
S is the portion of the sphere
x 2 y 2 z 2 1 that lies in the first quadrant. Solution: The surface S is illustrated in the Figure (a). Taking z 1 x 2 y 2 , we have
2
1 x 2 y 2 and z y y
z x x
S ( x y z )dS A x y
2
2
1 x 2 y 2 1 ( zx ) ( z y ) 1
1 x2 y 2 1
2
(1 x y )
1 x 2 y 2 dx dy , where A is the quadrant of a
circle in the ( x , y ) plane as shown in Figure (b). 1
I ( x y z ) dS S 0 0
1
I x sin 1 y 0
1 x
2
1 x 2
x
1 x y y 0 2
2
1 x 2
1 x 2 y 2 1 dy dx
1 x2 y 2 y
1
dx ( 2) x 2 1 x 2 dx 0
1
I ( 4) x 2 x 1 x 2 sin 1 x 3 4
0
Second Method: An alternative approach to evaluating the surface integral in above example is to evaluate it directly over the surface of the sphere using spherical polar coordinates. We have a 1 , hence I ( x y z )dS S
I ( x y z ) dS
I
S 2
0
I
2
0
2
0 sin cos sin sin cos sin d d 1 (1 4) cos (1 4) sin 2 d 3 4
Evaluation of a Surface Integral by Second Method: The extensions of the idea of an integral to line and double integrals are not the only generalizations that can be made. We can also extend the idea to integration over a general surface S . Two types of such integrals occur: (a)
S f ( x, y, z)dS
(b)
S F(r) nˆ dS S F(r) dS
In case (a) we have a scalar field f (r ) and in case (b) a vector field F (r ) . Note that dS nˆ dS is the vector element of area, where nˆ is the unit outward-drawn normal vector to the element dS . In general, the surface S can be described in terms of two parameters, u and v say, so that on S , r r (u , v ) x (u , v ), y (u , v ), z (u, v) x(u , v ) i y (u , v ) j z (u, v) k . The surface S can be specified by a scalar point function C (r ) c , where c is a constant. Curves may be drawn on that surface, and in particular if we fix the value of one of the two parameters u and v then we obtain two families of curves. On one, Cu r (u , v0 ) , the value of u varies while v is fixed and on the other, Cv r (u0 , v) , the value of v varies while u is fixed, as shown in Fig. 2.66, then the vector element of area dS is r r r r x y z x y z dS du dv dudv , , , , dudv u v u v u u u v v v
dS ( J1 i J 2 j J 3 k )du dv where, J1
y z u v
y z v u
, J2
z x u v
z x v u
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and J 3
x y u v
(2.29) x y v u
. Hence,
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[2.130]
F(r ) dS ( PJ1 QJ 2 RJ 3 )du dv 2 2 2 S f ( x, y, z )dS A f (u, v) J1 J 2 J 3 du dv S
(2.30)
A
(2.31)
where, F(r ) ( P, Q, R) and A is the region of the (u, v) plane corresponding to S .
Figure 2.66: Parametric curves on a Surface
Figure 2.67: A surface described by = ( , )
In particular, u and v can be chosen as any two of x , y and z . For example, if z z ( x, y ) describes a surface as in Fig. 2.67, then r x, y, z ( x, y ) with x and y as independent variables. This gives, J1 z x , J 2 z y and J3 1. Hence, Eq. 2.30 Eq. 2.31
S F(r) dS A P(z x) Q(z y) R dx dy f ( x, y, z )dS f x, y, z ( x, y) 1 (z x) (z y ) 2
S
A
(2.32) 2
(2.33)
dx dy
Example 2.263 [ME-2012, PI-2012 (1 mark)]: For the spherical surface, x 2 y 2 z 2 1 , the unit outward normal vector at the point P (1 2 ,1 2 , 0) is given by (a) (1 2)iˆ (1 2) ˆj (b) (1 2)iˆ (1 2) ˆj (c) kˆ Solution (a):
For
f ( x, y , z ) x 2 y 2 z 2 1 ,
the surface
3)iˆ (1
(d) (1
n f (f x)i (f y ) j (f z )k 2 x i 2 y j 2 z k n P 1
the outward 2,1
2,0
3) ˆj (1
normal
3) kˆ
vector
2 i 2 j . So the
unit outward normal vector at P (1 2 ,1 2 , 0) is nˆ n n (1 2) i (1 2) j [Similar questions were also asked in CH-2008, IN-2009 (1 mark), TF-2008 (2 marks)] Example 2.264 [MN-2008 (2 marks)]: The direction of gradient vector at a point (1,1, 2) on a surface S ( x, y , z ) x 2 y 2 z is (a) (1 3)(2iˆ 2 ˆj kˆ) (b) (1 3)( 2iˆ 2 ˆj kˆ)
(c) (1 3)(2iˆ 2 ˆj kˆ ) (d) (1 3)(2iˆ 2 ˆj kˆ ) Solution (d): The direction of gradient vector is the outward unit normal vector of the surface at a point (1,1, 2) on a surface S ( x, y , z ) x 2 y 2 z . n S (S x)i (S y ) j (S z )k n S 2 x i 2 y j z k n P 1,1,2 2 i 2 j 1 k nˆ (1 3)(2 i 2 j 1k ) .
2.7.5 Volume Integrals The define the integral of a function of three variables through a region T of 3-D space by the limit n
lim i1 f ( xi , yi , zi )Vi T f ( x, y, z )dV n,all V 0
(2.34)
i
where Vi (i 1, 2, n) is a partition of T into n elements of volume, and ( xi , yi , zi ) is a point in Vi , as shown in Fig. 2.68a. In terms of rectangular Cartesian coordinates the triple integral, as shown in Fig. 2.68b, is written as, x b
y g2 ( x )
z h2 ( x , y )
T f ( x, y, z) xa dx y g ( x ) dy z h ( x , y ) 1
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f ( x, y , z ) dz
(2.35)
1
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Engineering Mathematics
Chapter 2: Calculus
[2.131]
Note that there are six different orders in which the integration in Eq. 2.35 can be carried out. As we saw for double integrals, the expression for the element of volume dV dx dy dz under the transformation x x (u , v, w) , y y (u, v, w) and z z (u , v, w) may obtained using the Jacobian as, J
( x, y , z ) (u , v, w)
x u
y u
z u
x v
y v
z v , so dV dx dy dz J du dv dw
x w
Figure 2.68: (a) Partition of region coordinates.
y w z w
into volume elements ∆
(b) The volume integral in terms of rectangular Cartesian
In case of cylindrical polar coordinates ( r , , z ) , as shown in Fig. 2.69a, x r cos , y r sin cos
sin
0
and z z J r sin
cos
0 r dV r dr d dz
0
(2.36)
0
1
In case of spherical polar coordinates ( r , , ) , as shown in Fig. 2.69b, x r sin cos , sin cos
sin sin
y r sin sin and z r cos J r cos cos
r cos sin
r sin sin
r sin cos
cos r sin r 2 sin 0
2
dV r sin dr d d
Figure 2.69: Volume element in (a) cylindrical (b) spherical polar coordinates
Example 2.265 [ME-2004 (2 marks)]: The volume of an object expressed in spherical co-ordinates is given by V
2
0
(a) 3
3 1 2
0 0 r
sin dr d d . The value of the integral is
(b) 6
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(c) 2 3
(d) 4
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Engineering Mathematics
Solution (a): V
2
0
2
V 0
Chapter 2: Calculus
3
1
0
0
2
r
cos
2
0
3 0
sin dr d d 3 r 1
(1 3)r
d
3
[2.132] 1
sin d r 2 dr
0
0
2 0 cos( 3) cos 0 (1 3) 0 3
r 0
Example 2.266 [EC-2014 (2 marks)]: The volume under the surface z ( x , y ) x y and above the triangle in the x y plane defined by { 0 y x and 0 x 12 } is ……………. Solution: Taking elemental strip of length dx whose height vary from z 0 to z x y ; as this strip lies in the triangle in xy plane which is defined by 0 y x and 0 x 12 ; so its width is changing from y 0 to y x ; and whose length is changing from x 0 to x 12 . So V
x 12
x 0
V
x 12
x 0
y x
z x y
y 0
z 0
xy (1 2) y
dz dy dx
2
0
y x y 0
x 12
x 0
dx
x 12
x 0
dx
y x
z zz 0x y dy y 0
x
2
x 12
yx
x 0
y0
(1 2) x 2 0 dx (3 2)
x 12
x 0
( x y ) dy dx . Thus
x 2 dx (1 2) x 3
x 12 x 0
864
[Similar question was also asked in XE-2007 (2 marks)] Example
2.267
[MA-2014
(2
marks)]:
g : 3 3
Let
be
defined
by
g ( x, y , z ) (3 y 4 z , 2 x 3 z, x 3 y ) and let S {( x, y , z ) 3 : 0 x 1, 0 y 1, 0 z 1} . If
g ( S ) (2 x y 2 z )dx dy dz S zdx dy dz , then
is …………….
Solution: Let g ( X , Y , Z ) (3 y 4 z , 2 x 3 z , x 3 y ) , so X 3 y 4 z , Y 2 x 3z and Z x 3 y . Thus the integrand of the left hand side of the given integral becomes 2 X Y 2Z 5z
g (S )
J
(2 X Y 2 Z ) dX dY dZ 5 z J dx dy dz S
( X , Y , Z ) ( x, y , z )
g(S)
X x
Y x
Z x
0
2
1
X y
Y y
Z y 3
0
3 0 2(0 12) 1(9 0) 15
X z
Y z
Z z
3 0
4
(2 X Y 2 Z ) dX dY dZ 5 z 15 dx dy dz 75 z dx dy dz 15 S
S
2.7.6 Gauss’s Divergence Theorem Triple integrals can be transformed into surface integrals over the boundary surface of a region in space and conversely. Such a transformation helps in establishing fundamental equations in fluid flow, heat conduction, etc. The transformation is done by the divergence theorem, which involves the divergence of a vector function F F1 i F2 j F3 k . Consider the closed volume V with surface area S , as shown in Fig. 2.68. The surface integral
F dS S
may be interpreted as the flow of a
liquid with velocity field F (r ) out of the volume V . The divergence of F could be expressed as flow out of V div F F lim . In terms of differentials, this may be written as, V 0 V div F dV flow out of dV . Now, consider a partition of volume V given by Vi (i 1, 2, , n) . Then the total flow out of
V
is the sum of the flows out of each
n
S
Vi , i.e.
n
F dS lim (flow out of Vi ) lim (div FVi ) . Hence n
i 1
n
i 1
V div F dV S F dS div F dV F n dA ( dS ndA) V S
Copyright © 2016 by Kaushlendra Kumar
(2.37) (2.38)
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Engineering Mathematics
Chapter 2: Calculus
[2.133]
This result is known as the Divergence theorem or Gauss’s divergence theorem, which was asked in ME-1994, EE-2002 (1 mark). It enables us to convert surface integrals into volume integrals [This point was asked in ME-2001 (1 mark)] and often simplifies their evaluation. In component form, F F1 i F2 j F3 k and of the outward unit normal vector n cos i cos j cos k of S , Eqs. 2.38 and 2.39 becomes, F F F3 (2.39) V x1 y2 z dx dy dz S ( F1 cos F2 cos F3 cos )dA
F
V x1
F2
F3
(2.40) dx dy dz S ( F1 dy dz F2 dz dx F3 dx dy ) y z If curl (grad ) 0 , then is an irrotational. [This point was asked in MT-2012 (1 mark)] If div (grad F) 0 then it gives Laplace equation. [This point was asked in MT-2012 (1 mark)] If grad ( ) F , then is called a potential function of F . [This point was asked in MT-2012 (1 mark)] If div F 0 then F is incompressible. [This point was asked in MT-2012 (1 mark)]
Example 2.268 [CS-1993, EC-1993 (1 mark)]: V x cos 2 y iˆ x 2 e z ˆj z sin 2 y kˆ and S the surface of a unit cube with one corner at the origin and edges parallel to the co-ordinate axes, find the value of
integral
S
V dS .
Solution: div V V ( x )iˆ ( y ) ˆj ( z ) kˆ ( x cos 2 y iˆ x 2 e z ˆj z sin 2 y kˆ )
2
2 z
2
2
2
div V ( x ) x cos y ( y ) x e ( z ) z sin y cos y 0 sin y 1 Using divergence theorem, we can change the given surface integral into volume integral as
1 1 1
S V dS V div V dV V 1 dV 0 0 0 dx dy dz 1 . Common
Data for Questions 2.269 & 2.270: Consider the vector field 3 3 3 3 2 2 2 A ( y z )iˆ ( x z ) ˆj ( x y ) kˆ defined over the unit sphere x y z 1 . Example 2.269 AE-2009 (4 marks): The surface integral (taken over the unit sphere) of the component A normal to the surface is (b) 1 (c) 0 (a) (d) 4
3
3
Solution (c): As div A A ( x)( y 3 z 3 ) ( y )( x 3 z 3 ) ( z )( x 3 y 3 ) 0 , so by using
divergence theorem
S
A ndA div A dV 0dV 0 . V
V
Example 2.270 AE-2009 (4 marks): The magnitude of the component of A normal to the spherical
surface at the point 1 (a) 1 3
3 ,1
3 ,1
3 is
(b) 2 3
(c) 3 3
(d) 4 3
Solution (b): As normal to the given spherical surface,
n f ( x)i ( y ) j ( z )k 2 x i 2 y j 2 z k n 1
the given point, A {2 (3 3 ) (i j k ) . As n A ( 2 n (2
3,1
f ( x, y , z ) x 2 y 2 z 2 1 , is 3,1
3
3)( i j k ) ; also at
(2
3 )(i j k ) {2 (3 3 )}(i j k ) 4 3 and
3) 12 12 12 2 . Now, the component of A normal to the spherical surface is same as
the component of A along n which is
n A
nA
nA
n
2
n , whose magnitude is
Copyright © 2016 by Kaushlendra Kumar
n
2
n
n
43 2
2 3
.
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Engineering Mathematics
Chapter 2: Calculus
[2.134]
Example 2.271 [IN-2014 (1 mark)]: A vector is defined as f y iˆ x ˆj z kˆ where iˆ , ˆj and kˆ are
f dS
unit vectors in Cartesian ( x, y , z ) coordinate system. The surface integral
over the closed
surface S of a cube with vertices having the following coordinates: (0, 0, 0) , (1, 0, 0) , (1,1, 0) , (0,1, 0) , (0, 0,1) , (1, 0,1) , (1,1,1) , (0,1,1) is …………….
Solution: div f f ( x )i ( y ) j ( z )k y i x j z k 1
S
So using divergence theorem,
f dS div f dV 1dV 1dV volume of the V
V
V
given cube which is 1 (as side of cube is of unit length). So answer is 1. Example 2.272 [XE-2011 (1 mark)]: A vector field is called solenoidal if its divergence is zero. Consider the vector fields P and Q given by P ( x, y , z ) k iˆ k ˆj k kˆ , where k (2 x 2 8 xy 2 z ) 3
2 2
1
3
2
2
3
1
; k2 (3 x y 3 xy ) ; k3 (4 y z 2 x z ) and Q ( x, y , z ) xyz P ( x, y , z ) . Then
(a) P and Q are both solenoidal
(b) Both P and Q are not solenoidal
(c) P is solenoidal but not Q
(d) Q is solenoidal but not P
Solution (d): div P ( x )(2 x 2 8 xy 2 z ) ( y )(3 x 3 y 3xy ) ( z )( 4 y 2 z 2 2 x 3 z ) div P 4 x 8 y 2 z 3 x 3 3 x 8 y 2 z 2 x 3 x x 3 P is not solenoidal.
divQ ( x) xyz 2 (2 x 2 8 xy 2 z ) ( y ) xyz 2 (3x 3 y 3 xy ) ( z ) xyz 2 (4 y 2 z 2 2 x 3 z )
3
2
2
3 3
4
2 2
2
2 2
3 4
4
3
div Q ( x )(2 x yz 8 x y z ) ( y )(3 x y z 3x y z ) ( z )( 4 xy z 2 x yz )
2 2 3 3 4 2 2 2 3 3 4 2 div Q 6 x yz 16 xy z 6 x yz 6 x yz 16 xy z 6 x yz 0 Q is solenoidal.
S e
Example 2.273 [XE-2012 (2 marks)]: Evaluation of
x
ˆ over a surface iˆ 3 y ˆj ze x kˆ ndA
S : x 2 y 2 z 2 1 using Gauss divergence theorem gives
(b) 4
(a) 0
(d) 12
(c) 4 3
x x Solution (b): The given spherical surface has radius r 1 . For F e iˆ 3 y ˆj ze kˆ div F F ( x)iˆ ( y ) ˆj ( z ) kˆ e x iˆ 3 y ˆj ze x kˆ 3 . Thus using divergence
theorem,
ˆ F dS div F dV 3dV 3V 3(4 3) r S F ndA S V V
3
4
[Similar questions were also asked in CH-2009, AE-2013 (2 marks)] Example 2.274 [ME-2013, PI-2013 (2 marks)]: The following surface integral is to be evaluated over a sphere for the given steady velocity vector field, F x iˆ y ˆj z kˆ defined with respect to a Cartesian coordinate system having iˆ , ˆj and kˆ as unit base vectors.
S (1 4) F n dA , where S
is
the sphere, x 2 y 2 z 2 1 and n is the outward unit normal vector to the sphere. The value of the surface integral is (a) (b) 2 (c) 3 4 (d) 4 Solution (a): div F F ( x )i ( y ) j ( z )k x i y j z k 3 . Using divergence theorem,
1
1
1
1
3
34
S 4 F ndA 4 S F dS 4 V div F dV 4 V 3dV 4 V 4 3 r
3
4 (Since r 1
for the given sphere) [Similar question was also asked in PI-2008 (1 mark)]
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Engineering Mathematics
Chapter 2: Calculus
[2.135]
Example 2.275 [EC-2011 (1 mark)]: Consider a closed surface S surrounding volume V . If r is the position vector of a point inside S , with nˆ the unit normal on S , the value of the integral
S
5r nˆ dS is
(a) 3V
(b) 5V
Solution
(d):
(c) 10V
5r nˆ dS 5r dS 5 r dS ; S
S
(d) 15V
S
so
using
divergence
theorem
and
div r r 3 , we get, 5 r dS 5 div r dV 5 3 dV 5(3) dV 5(3V ) 15V S
V
V
V
[Similar question was also asked in CH-2008 (1 mark)]
2.7.7 Stokes’ Theorem Stokes’ theorem is the generalisation of Green’s theorem, and relates line integrals in three dimensions with surface integrals [This point was asked in ME-2005 (1 mark)]. In section 2.7.3, we saw that the curl of the flow around S vector F could be expressed in the form (curl F ) nˆ lim . S 0 S In terms of differentials, this becomes curl F dS flow around dS . Consider the surface S , as shown in Fig. 2.70, bounded by the curve C . Then the line integral
C
F dr can be interpreted as the total flow of a fluid Figure 2.70: Surface S bounded by the curve C
with velocity field F around the curve C . Partitioning the surface S into elements Si (i 1, 2, , n) ,
n
n
i 1 (flow around Si ) nlim i 1 (curl F Si ) . Hence C F dr nlim (2.41) C F dr S (curl F) dS
This result is known as Stokes’ theorem, and was asked in EC-2006 (1 mark), EC-2009 (2 marks). Eq. 2.42 also says that the line integral of a vector ‘ F ’ around a closed path ‘ C ’ is equal to the integral of curl of vector ‘ F ’ over the open surface ‘ S ’ enclosed by the closed path ‘ C ’ [This point was asked in EC-2013 (1 marks)]. Eq. 2.42 can also be written as, (2.42) (curl F) n dA F r( s)ds S
C
Here n is a unit normal vector of S and, depending on n , the integration around C is taken in the sense shown in Fig. 2.71. Furthermore r dr ds is the unit tangent vector and s the arc length of C . Figure 2.71: Direction of and orientation of In component form, Eq. 2.43 can be written as, F3 F F F F F3 R y z2 N1 z1 x N 2 x2 y1 N3 du dv C ( F1dx F2 dy F3dz )
(2.43)
where, F F1 i F2 j F3 k , N N1 i N 2 j N 3 k , n dA N du dv and r ds dx i dy j dz k , and R is the region with boundary curve C in the uv plane corresponding to S represented by r (u , v) . The Stokes’ theorem provides a condition for a line integral to be independent of its path of integration. For e.g., if the integral the path of integration then
C F dr C F dr , where 1
different paths joining
C F dr C F dr , 1
2
A
B
A F dr
and
2
is independent of
C1 and C 2 are two
Figure 2.72: Two and , joining points A, B
B , as shown in Fig. 2.72. Since paths:
where C 2 is the path traversed in the opposite
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Engineering Mathematics
direction, we have
Chapter 2: Calculus
[2.136]
C F dr C F dr 0 C F dr 0 , where C 1
is combined, closed curve
2
formed from C1 and C 2 .
C F dr 0 then S (curl F) dS 0 for any surface
S bounded
Stokes’ theorem implies that if
by C . Since this is true for all surfaces bounded by C , we deduce that the integrand must be zero, i.e., curl F 0 . In component form, if F F1 i F2 j F3 k , we then have that F dr F1dx F2 dy F3 dz is an exact differential if curl F 0 , i.e.,
F1 z
F3 x
,
F1 y
F2
and
x
F2 z
F3 y
. Thus there is a
function f ( x, y, z ) f (r) s.t. F1 f x , F2 f y and F3 f z , i.e., F (r ) grad f .
When F (r ) represent a field of force, the field is said to be conservative (since it conserves rather than dissipates energy). When F (r ) represents a velocity field for a fluid, the field is said to be curl – free or irrotational.
F y i z j xk
Example 2.276: Verify Stokes’ theorem for 2
S
and
the paraboloid
2
z f ( x, y ) 1 ( x y ) , z 0 . Solution: The curve C , oriented as shown in Fig. 2.76, is the circle r ( s ) cos s i sin s j . Its unit tangent vector is r ( s) sin s i cos s j .
The function F y i z j x k on C is F r ( s) sin s i cos s k . 2
2
C F dr 0 F r (s ) r(s )ds 0
Hence,
sin 2 s ds . We now
consider the surface integral. We have F1 y , F2 z and F3 x so that, i
j
curl F x
y
y
z
k z i j k .
A
normal
vector
of
S
is
x
N grad z f ( x, y ) 2 x i 2 y j k (curl F) N 2 x 2 y 1 . Now n dA N dx dy . Using polar coordinates r , defined by x r cos , y r sin and denoting the projection of S into the xy plane by R , thus, I (curl F ) ndA (curl F) N dx dy ( 2 x 2 y 1) dx dy S
I
2
0
Hence,
R
R
2
r 1
r 0 2r (cos sin ) 1 r dr d 0
1 2 (cos sin ) d 2 3
S curl F dS C F dr and thus Stokes’ theorem is verified.
Exercise: 2.6 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. If C is a line segment from (3, 4, 0) to (1, 4, 2) , then
C ( z y
2
) ds _____.
2. If C is the part of the curve x cos y from (1, 2 ) to (1, 0) and F y i 2 x j , then
C F dr k , where k _____. 3. Find the work done by the force F ( x, y ) x 2 i xy j in moving a particle along the curve which runs from (1, 0) to (0,1) along the unit circle and then from (0,1) to (0, 0) along the y axis. (a) 1 3
(b) 2 3
Copyright © 2016 by Kaushlendra Kumar
(c) 1 3
(d) 2 3
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Engineering Mathematics
Chapter 2: Calculus
4. Evaluate the line integral
C y dx xdy , where 2
[2.137]
2 C is the arc of the parabola x 4 y from
( 5, 3) to (0, 2) . _____. 5. Evaluate
C (3 y e
sin x
(a) 4
y 4 1)dy , where C is the circle x 2 y 2 9 .
) dx (7 x (b) 8
(c) 16
C (3x 5 y )dx ( x 6 y)dy ,
6. Evaluate
where C
anticlockwise direction. (a) 4 (b) 8 7. Evaluate
2
S z ds , where S
(a) 3
(d) 36
is the ellipse
(c) 12
(1 4) x 2 y 2 1 in the
(d) 16
2
2
2
is the hemisphere given by x y z 1 with z 0 .
(b) 2 3
(c) 3 3
(d) 4 3
8. The area of the ellipse cut on the plane 2 x 3 y 6 z 60 by the circular cylinder x 2 y 2 2 x . (a) 7 6
(b) 5 6 4
2 2
(c) 3 6 2
I ( x y y z xz )dS , where
9. Evaluate
S
S
(d) 6
is the entire surface of the sphere
x2 y 2 z2 1 .
(a) 2 15
(b) 4 15
(c) 6 15
(d) 8 15
10. Find I F ndS , where F 2 x i 2 y j k , n is the outward pointing unit normal and S is S
the entire surface consisting of S1 the part of the paraboloid z 1 x 2 y 2 with z 0 together with S 2 disc {( x , y ) : x 2 y 2 1} . (a) (b) 2 (c) 3 (d) 4 11. Which of the following vector fields are (a) Only V conservative? (b) Only W V (2 x 3 y z ) i ( 3x y 4 z ) j (4 y z ) k and (c) Both V and W (d) Neither V nor W W (2 x 4 y 5 z ) i ( 4 x 2 y ) j ( 5 x 6 z ) k 12. Find the potential function ( ) for F (2 x 4 y 5 z ) i ( 4 x 2 y ) j ( 5 x 6 z ) k . (a) x 2 4 yz 5 zx y 2 x 3 z 2
(b) y 2 4 yz 5 zx y 2 x 3 z 2
(c) x 2 4 yz 5 zx y 2 x 3 y 2
(d) x 2 yz zx y 2 x 3 z 2
13. Evaluate
C F dr , where C
is the curve from A(1, 0, 0) to B (0, 0,1) in which the plane x z 1
cuts the hemisphere given by x 2 y 2 z 2 1 , y 0 . _____ 14. Evaluate
R e
( x y )
dxdy , where R is the region in the first quadrant in which x y 1 .
(a) 1 2e 15. Evaluate
(b) 2e 1
R e
( x2 y )
(a) 1 3 16. Evaluate
R ( x y 1)dxdy , where R (a) 7 5
18. Evaluate (a) 3 8
(c) 1 5
(d) 1 6
y 2 ) dxdy , where R is the region 0 x y L . (b) L4 2
4
(a) L 17.
2
(d) 2e 1 1
dxdy , where R is the region in the first quadrant in which x y . (b) 1 4
R ( x
(c) 1 2e 1
(c) L4 3
(d) L4 4
is the region inside the unit square in which x y 0.5 .
(b) 5 7
(c) 8 7
(d) 7 8
(c) 8 7
(d) 7 8
1 1
0 0 x max( x, y )dydx . (b) 8 3
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Engineering Mathematics
Chapter 2: Calculus
[2.138]
19. The volume of the solid bounded above by the plane z 4 x y and below by the rectangle R {( x, y ) : 0 x 1, 0 y 2} is _____. 20. Evaluate
D (3 x y )dA , where
D is the triangle in the ( x, y ) plane bounded by the x axis
and the lines y x and x 1 .
D (4 x 2)dA , where D is the region enclosed by the curves y x 3 22. Evaluate ( xy y ) dA , where D is the region consisting D 21. Evaluate
2
and y 2 x . of
the
square
{( x, y ) : 1 x 0, 0 y 1} together with the triangle {( x , y ) : x y 1, 0 x 1} . (a) 23 40 23. Evaluate
(b) 23 40
(d) 40 23
(c) 40 23
D (sin x) x dA , where D is the triangle {( x, y) : 0 y x, 0 x } .
24. Find the volume of the tetrahedron that lies in the first octant and is bounded by the three coordinate planes and the plane z 5 2 x y . _____
D ( x y )
25. Evaluate
2
dxdy , where D is the parallelogram bounded by the lines x y 0 ,
x y 1 , 2 x y 0 and 2 x y 3 . 26. Let D be the region in the first quadrant bounded by the hyperbolas xy 1 , xy 9 and the lines y x , y 4 x ; then evaluate
(a) 8 (52 3) ln 2
D
(b) 8 (52 3) ln 3
27. Change the order of the integral (a) (c)
1
ey
0 1
1 1
0 x
(b) (d)
2
e y dydx
(b) (e 1)
sin 1 y 2
1 sin 1
1
y
sin 1 y 2
0 sin
1
y
(d) (52 3) ln 2
e
log x
0 1 f ( x, y )dydx e 1 0 log x f ( x, y)dydx
[Hint: evaluate by changing order of integration]
29. Change the order of integration 1
(c) (52 3) ln 3
f ( x, y )dxdy .
log x
(a) (e 1) 2
(c)
1
0 1 f ( x, y)dydx e 1 1 log x f ( x, y)dydx
28. Evaluate
(a)
y x xy dxdy .
5 2
(c) (e 1) 3
(d) (e 1) 4
1
2 sin x f ( x, y )dydx
f ( x, y ) dxdy
(b)
f ( x, y ) dxdy
(d)
1
sin 1 y 2
1 sin 1
1
y
sin 1 y 2
0 sin
8xyzdV from (2,1, 0) to (3, 2,1) . _____ 31. Evaluate 2 xdV where E is the region under the plane E
1
y
f ( x, y ) dxdy f ( x, y ) dxdy
30. Evaluate
2 x 3 y z 6 that lies in the first
octant. _____ 32. Determine the volume of the region that lies behind the plane x y z 8 and in front of the region in the yz plane that is bounded by z (3 2) y and z (3 4) y . _____ 33. Evaluate
E
2 2 3 x 2 3 z 2 dV , where E is the solid bounded by y 2 x 2 z and the plane
y 8 . _____ 34. Evaluate
C ( xydx x
2
y 3dy ) where C is the triangle with vertices (0, 0) , (1, 0) , (1, 2) with
positive orientation. _____ 35. If C is the positively oriented circle of radius 2 centred at the origin then
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C ( y dx x dy ) 3
3
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(a) 24 36. Evaluate
Chapter 2: Calculus
(b) 24
C ( y dx x dy) , where 3
3
(d) 42
(c) 42
C are the two circles of radius 2 and radius 1 centred at the
origin with positive orientation. (a) 20 (b) 20 37. Evaluate
[2.139]
S curl F dS , where F z
(c) 22.5 2
(d) 22.5
3 3
i 3 xy j x y k and S is the part of z 5 x 2 y 2 above
the plane z 1 . Assume S is oriented upwards. _____. 38. Evaluate
S curl F dS ,
where F x 2 i (1 2) y 2 j z k and the surface consists of the three
surfaces z 4 3 x 2 3 y 2 , 1 z 4 on the top, x 2 y 2 1 , 0 z 1 on the sides and z 0 on the bottom. (a) (b) 1.5 (c) 2 (d) 2.5 39. Evaluate 40. 41.
42.
43.
S ( y
2
z i y 3 j xz k ) dA , where S is the boundary of the cube defined by 1 x 1 ,
1 y 1 and 0 z 2 . _____ In Gauss’s Divergence theorem, the vector point function should be (a) continuous (b) monotonic (c) differentiable (d) continuously differentiable Which of the following equations represent the tangent plane to the surface f xy yz zx at (1, 1, 2) ? (a) xy yz zx 0 (b) 2 xy 2 yz 2 zx x 3 y (c) 2 xy 2 yz 2 zx x y z (d) None of these Which of the following criteria, a curve must possess to satisfy Green’s theorem? (i) positive orientation; (ii) piecewise smoothness; (iii) simple; (iv) closed. (a) (ii), (iii) and (iv) (b) (i), (iii) and (iv) (c) (iii) and (iv) (d) (i), (ii), (iii) and (iv) If f ( x, y, z ) 0 represents a surface S , then the unit outward normal to S is given by nˆ (a) (grad f ) grad f
(b) grad f
(c) (grad f )
(d) None of these
grad f
44. If C is a closed curve in space and r is a position vectors of a point P ( x, y , z ) , then (a) ( x 2 y 2 z 2 )3 2
(b) 0
(c) Non-zero constant
C r dr
(d) None of these
S x( y z ) i y( z x) j z( x y) k dS kV , where k _____. is the closed surface of the sphere with radius a , then ( xdydz ydzdx zdxdy) S
45. For a closed surface S , 46. If S
(a) (4 3) a 3
(b) 4 a 3
(c) (4 5) a 3
47. If S is the closed surface and V is the volume enclosed by S , then
(d) None of these
V
(div n)dV
(a) S (b) 2S (c) 3S (d) 4S x 48. The function e over the interval [0, 1] is to be evaluated using the Taylor series 1 x (1 2!) x 2 (1 3!) x3 to an accuracy of 0 . The number of terms in the series that is considered for this accuracy is n . Then (a) for a given x [0,1] and a given , there is no finite n that is valid (b) for a given 0 , there is a valid n that is finite for a given x [0,1] , but there is no finite n that is valid for all x [0,1] (c) for a given 0 , there is a finite n that is valid for all x [0,1] (d) there is a finite n that is valid for all x [0,1] and all 0 49. A surfaces S ( x, y ) 2 x 5 y 3 is integrated once over a path consisting of the points that satisfy ( x 1)2 ( y 1)2
2 . The integrals evaluates to _____.
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2.8
Chapter 2: Calculus
[2.140]
Sequence and Series
2.8.1 Sequence A sequence is a list of numbers written in a specific order. The list may or may not have an infinite number of terms in them. General sequence terms are denoted as follows: a1 (first term), a2 (second term), , an ( nth term), an 1 ( ( n 1)th term), . As we are dealing with infinite sequences, each term in the sequence will be followed by another term as given above. There are many ways for denoting a sequence; the sequence {a1 , a2 , , an , an 1 , } can also be denoted as {an } or {an }n1 , ; while, the sequences denoted by {an } or {an }n1 , an is usually denoted by a formula. For e.g., the
sequence {an }n1 , an ( 1) n 1 2n is written as 1,1 2 , 1 4 ,1 8, 1 16 , .
If an an 1 for every n , then the sequence is increasing; on the other hand if an an 1 for every n , then the sequence is decreasing. If a number m s.t. m an n , then the sequence is bounded below; also if there exists a number M such that an M for every n , we say that the sequence is bounded above.
If the sequence is both bounded below and bounded above we call the sequence bounded.
Example 2.277: Determine if the following sequences are monotonic (i.e., increasing or decreasing) 2
and/or bounded. (a) n
n 0
2
(b) ( 1)
n 1
n 1
2
Solution: (a) As n ( n 1) n , i.e., an an 1 , so the sequence is a decreasing sequence and hence monotonic. Also, as the sequence terms will be either zero or negative, thus the sequence is bounded above but not bounded below and hence the sequence is not bounded. (b) For the sequence
given by ( 1)
n 1
n 1
, the sequence terms are alternate 1 or 1 so the sequence is neither increasing
nor decreasing so it is not a monotonic sequence. Also the given sequence has terms either 1 or 1 so the sequence is bounded below by 1 and bounded above by 1 .
Limit of a sequence
If lim an L , where L is a finite number, then the value of the an L as n .
If lim an , then it means that the value of the an gets larger and larger as n approaches to .
If lim an , then it means that the value of the an gets negative larger and larger as n .
n n n
Formal definition of Limit
lim an L if for every number 0 there is an integer N s.t. an L , whenever n N .
n
lim an if for every number M 0 there is an integer N s.t. an M whenever n M .
n
lim an if for every number M 0 there is an integer N s.t. an M whenever n N .
n
Convergence and Divergence of a sequence
If lim an exists and is finite we say that the sequence is convergent.
If lim an does not exist or is infinite then the sequence is diverges. If lim an we will say the
n
n
n
sequence diverges to . If lim an we will say that the sequence diverges to . n
Example 2.278 [CE-2002 (2 marks)]: The limit of the sequence xn n1 n as n is (a) 0 (b) 1 (c) (d)
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Chapter 2: Calculus
[2.141]
lim (1 n ) ln b
Solution (b): Let L lim n1 n ln L lim(1 n) ln n L e n n
n
. As lim (1 n) ln n , takes n
form, so from L’Hospital rule, lim (1 n) ln n lim (1 n) 0 . Thus L e 0 1 n
n
Example 2.279: Determine if the following sequences converge or diverge. If the sequence converges
2
Solution: (a) As
(3n
2
lim
2
1) (10n 5n )
3n 1
n 10 n
2n
n 2
n 1
2
2
5n 2
n
2
2
n 2 (10 n ) 5
n
(10
n) 5
5
so the sequence
converges and its limit is 3 5 . (b) As using L’Hospital rule we get
n 2
L lim e 2 n n lim 2e 2 n , so the sequence e n
(b) e n n 3 (1 n ) 3 (1 n ) 3 , lim lim 2
determine its limit. (a) (3n 1) (10n 5n )
n
2n
n
n 1
diverges (to ).
Theorems on Sequences
Given the sequence {an } , if we have a function f ( x ) s.t. f ( n) an and lim f ( x ) L then x
lim an L , i.e. we take the limits of sequences much like we take the limit of functions. Thus all
x
algebraic properties from the limit of functions will also hold for the limit of sequences, i.e., lim (an bn ) lim an lim bn lim can c lim an n
n
lim (a n bn ) lim an
n
p
n
n
lim( an ) lim an
n
n
n
lim b n
n
n
lim an bn lim an
n
p
n
lim b , n
n
provided lim bn 0
, provided an 0
n
Squeeze theorem for Sequences: If an cn bn for all n N , for some N , and lim an lim bn L then lim cn L
n
n
n
If lim an 0 then lim an 0 n
n
The sequence
{r n }n 0
converges if 1 r 1 and diverges for all other value of r . Also,
lim r n 0 for 1 r 1 ; and lim r n 1 for r 1 .
n
n
If both lim a2 n L and lim a2 n 1 L then the sequence {an } is convergent and lim an L
If the sequence {an } is bounded and monotonic then {an } is convergent
n
n
n
2.8.2 Series Series is the sum of all terms in a sequence. Let a sequence given by {an }i1 , then s1 a1 , n
s2 a1 a2 , , sn a1 a2 an i 1 ai , . The sn are called partial sums and they will
form a sequence {sn }n1 . The limit of the sequence of partial sums {sn }n1 is defined as
n
lim sn lim i 1 ai i 1 ai , where
n
n
i1 ai
is called an infinite series (note that the series starts at
i 1 as the original sequence {an }i1 also starts at i 1 ; if the original sequence starts at i 2 then the series will also starts at i 2 ). Example 2.280: Write
n1n2
(1 3n1 ) as a series that starts at n 3 .
Solution: As we want to increase the initial value by 2 and so all the n ’s in the series term must decrease by 2, i.e.,
n1n2
(1 3n1 ) n3 ( n 2) 2 (1 3( n 2)1 ) n3 ( n 2) 2 (1 3n1 ) .
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Chapter 2: Calculus
[2.142]
Convergence and Divergence of a Series
If the sequence of partial sums {sn }n1 is convergent and its limit exists and is finite then the infinite series
i1 ai
is convergent, thus if lim sn s i 1 ai s . If the sequence of partial n
sums is divergent (i.e., its limit does not exist or is ) then the infinite series is also divergent. If an and bn are both convergent then
can , where c is any number is also convergent and can c an n k an n k bn is also convergent and n k an n k bn n k ( an bn ) If an converges then lim an 0 ; but if lim an 0 then an converges or diverges. n n Divergence test: If lim an 0 then an will diverge; but if lim an 0 then an converges n n
or diverges. Integral test: Suppose that f ( x ) is continuous, positive and decreasing function on the interval
[ k , ) and that f ( n) an then If
n k an is also convergent is divergent then n k an is also divergent −series test: If k 0 then n k (1 n p ) converges if p 1 and diverges if
k f ( x)dx If f ( x) dx k
is convergent then
p 1
The
Comparison test: As it is not possible to integrate every function so integral test is not use-full all the times. We have another test for comparing the convergence or divergence of a series called comparison test or limit comparison test which states that: if we have two series an and bn with an , bn 0 for all n and an bn for all n , then
bn is convergent then an is convergent If an is divergent then bn is divergent
If
Limit Comparison test: Suppose that we have two series
an
and
bn
with an 0 , bn 0
for all n . Let us define c lim (an bn ) , if c is positive and finite then either both series converge n
or both series diverge. Alternating series test: Suppose that we have a series
an
and either an ( 1) n bn or
an ( 1) n 1 bn where bn 0 for all n . Then if lim bn 0 and {bn } is a decreasing sequence then n
the series
an
is convergent.
Ratio test: Suppose we have the series
an 1 an . Let us define L nlim
an . Then
If L 1 the series is convergent If L 1 the series is divergent If L 1 the series may be convergent or divergent
Root test: Suppose we have the series
an
and let us define L lim
n
n
an lim an n
1n
. Then
If L 1 the series is convergent If L 1 the series is divergent If L 1 the series may be convergent or divergent
Strategy for conducting the tests on Series
In a quick view, if it look like the series terms don’t converge to zero in the limit, i.e. does lim an 0 ? If so, use the Divergence Test. Note that we should only do the Divergence Test if a n
quick glance suggests that the series terms may not converge to zero in the limit.
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Engineering Mathematics
Chapter 2: Calculus
(1 n ) p
If the series is a p series
[2.143]
or a geometric series
n 0
ar
n
then use the fact that
p series will only converge if p 1 ; and a geometric series will only converge if r 1 . Sometimes algebraic manipulation is required to get a geometric series into the correct form. If the series similar to a p series or a geometric series then we can try the Comparison Test. If the series a rational expression involving only polynomials or polynomials under radicals (i.e. a fraction involving only polynomials or polynomials under radicals, then we can try the Comparison Test and/or the Limit Comparison Test. It is to be noted that in order to use the Comparison Test and the Limit Comparison Test the series terms all need to be positive. If the series contain factorials or constants raised to powers involving n , then we can try the Ratio Test. If the series term contains a factorial then the Ratio Test is the only test that will work. If the series terms is to be written in the form an ( 1) n bn or an ( 1) n 1 bn , then we can use the Alternating Series Test. If the series terms can be written in the form an (bn ) n , then we can try the Root Test.
If an f ( n) for some positive, decreasing function and
a
f ( x) dx is easy to evaluate we can use
the Integral Test. Example 2.281 [CE-1998, MN-2011 (1 mark)]: The infinite series 1 (1 2) (1 4) (1 8) is (a) Convergent (b) Divergent (c) oscillatory (d) semi-convergent Solution (a): The given series can be written as
an 1 n 2 . So we can convert this
n1 an , where
series as a function f ( x ) 1 x 2 which is continuous, positive and decreasing function on the interval
1
(1 x 2 ) dx converges so
n1 (1 n 2 ) 1
[1, ) , thus using integral test,
L lim
n
(b) Diverges
an
lim
(n 1)!2 (2n 1)!
n
Thus
(2n)! ( n !)
2
n 1(n !)2
(2n)!
(c) Is unstable
n1 an ,
Solution (b): Using the ratio test, the series an 1
1
n1 (1 n 2 ) also converges.
Example 2.282 [CE-1999 (1 mark)]: The infinite series (a) Converges
f ( x ) dx (1 x 2 ) dx 1 x 1 1 .
lim
n
( n 1) 2 (2n 1)
lim
n
(d) Oscillates where
an (n !)2 (2n)! . Now
(1 1 n)
lim( n 1) 1 . Thus the (2 1 n) n
given series diverges. Example 2.283 [TF-2009 (2 marks)]: If S is the largest possible set of real numbers x for which the series
n1 x n n is convergent, then S
(a) ( 1,1)
is
(b) [ 1,1)
(c) [ 1,1]
Solution (a): Using the ratio test, the series L lim
n
an 1 an
1 lim
n
x n 1
n
n 1 x
n
1 lim
n
x 1 1 n
n1 an ,
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where an x n n , converges if
1 x 1 x ( 1,1)
Example 2.284 [XE-2009 (2 marks)]: The infinite series (a) Divergent for all x (c) Convergent for all x
(d) (, )
m1(1)m x2
(1 x 2 ) m is
(b) Convergent only for x 1 (d) Divergent only for 1 x 1
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[2.144]
Solution (c): The given infinite series can be written as
m1 am m1 (1)m bm ,
where
bm x2 (1 x 2 )m . As lim bm lim x 2 (1 x 2 ) m x 2 (1 x 2 ) 0 ; also x 2 1 x 2 (1 x 2 ) m m
m
for all x and m [1, ) thus x 2 (1 x 2 ) m x 2 (1 x 2 ) m 1 for all x and m [1, ) thus
bm x 2 (1 x 2 )m is a decreasing sequence thus the given sequence converges for all x . Example 2.285 [IN-2011 (2 marks)]: The series (a) 2 x 2
(b) 1 x 3
m
am1 am
( x 1) 2( m1)
1 lim
4
m
m 1
converges for
(c) 3 x 1
Solution (b): Using the ratio test, the series L lim
m0 (1 4m )( x 1)2 m
4m ( x 1)
2m
n1 an ,
(d) x 3 m
where an (1 4 )( x 1) 2 m , converges if
1 lim
( x 1) 2
m
4
1
( x 1) 2
1
4
1 (1 4)( x 1) 2 1 . Now we have two inequalities (i) (1 4)( x 1) 2 1 x and
(ii) (1 4)( x 1) 2 1 ( x 1) 2 4 x 2 2 x 1 4 x 2 2 x 3 0 ( x 1)( x 3) 0 . using wavy curve method ( x 1)( x 3) 0 x ( 1, 3)
Now
Example 2.286 [MT-2013 (2 marks)]: Which one of the following series is divergent? (a)
n 11 3n1
(b)
n1 (1 n)
(c)
n 0 (1 2n )
(d)
n1 (1 n n )
Solution (b): For option (a): As infinite geometric series, which converges to a (1 r ) , where a is the first term and r is the common ratio. For the given series The given series
n 11 3n1 , a 1
and r 1 3 . Thus the given series converges to 1 (1 1 3) 3 2 . The option (b) diverges which is described in Example 2.289. For option (c): Again we have infinite geometric series whose first term is a 1 and common ratio r 1 2 . Thus the given series converges to 1 (1 1 2) 2 . For option L lim
n
(d): an 1 an
lim
n
We n
have
n
(n 1)
n 1
lim
n
an 1 n n ,
n
n
so
n
( n 1) ( n 1)
using
the
ratio
test,
n
1 lim 1 0 0 . So we (1 1 n) n n 1
lim n
1
have L 1 and hence the given series is convergent. Example 2.287 [AG-2014 (2 marks)]: Which of the following statements is true for the series given below? sn 1 (1
2) (1
3) (1
4) (1
n)
(a) sn converges to log( n )
(b) sn converges to
(c) sn converges to exp( n )
(d) sn diverges
Solution (d): Using the p series test we have
n
n k (1 n p ) , where k 1 and
p 1 2 1 , thus the
given series diverges.
Absolute and Conditionally Convergent Series: A series
an conditionally convergent. If an if
an
is convergent. If
is convergent and
an
an
is called absolutely convergent
is divergent then
an
is said to be
is absolutely convergent then it is also convergent.
Example 2.288 [MT-2014 (2 marks)]: The following power series P ( x ) will converge absolutely if
x is less than: P ( x) n 0 4n x n
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Engineering Mathematics (a) 0.25
Chapter 2: Calculus
(b) 0.5
(c) 2
n
Solution (a): The given series P ( x) n 0 4 x
n
(d) 4
will converge absolutely if P ( x) n 0 4 n x n
converges. So using ratio test, for the convergence of lim
n
an 1 an
1 lim
n
4n 1 x n 1 n
4 x
n
[2.145]
1 lim
n
Value of a Series: Let we have a series
4
n 1
x
n
4 x
where an 4n x n , we have
n 0 an ,
n 1
1 lim 4 x 1 4 x 1 x
n
n
n1 an
1 4
which converges to s , i.e., if we take the partial
n
sums then sn i 1 ai and this will form a convergent sequence and its limit will be s , i.e., lim sn s . Now limit converges means that we can make the partial sums, sn , as close to s as we
n
want simply by taking n large enough. In other words, if we take n large enough, then sn s . This is one method of estimating the value of a series. We can just take a partial sum and use that as an estimation of the value of the series. Now there are two questions: (i) How good is the estimation? (ii) Is there any way to make the estimate better? So, for determining how good the estimation is, consider the full series and strip out the first n terms, i.e.,
n
i1 ai i1 ai in 1 ai . The first series is the
partial sum sn ; and the second series (the one starting at i n 1 ) is called the remainder and denoted by Rn . Thus s sn Rn Rn s sn . So the remainder Rn tells us the error, between the exact value of the series and the value of the partial sums that we are using as the estimation of the value of the series. As we do not have the actual value of series so we cannot get actual value of the remainder. However, we can use some of the tests that we have got for convergence to get a good estimate of the remainder provided we make some assumptions about the series. Once we have got an estimate on the value of the remainder we will also have an idea on just how good a job the partial sum does of estimating the actual value of the series. There are several tests that will allow us to get estimates of the remainder, which are described as: Integral Test: In this case we will need to assume that the series terms are all positive and will eventually be decreasing. In definite integral, using rectangle method, we derived the integral b
a
f ( x) dx is the area under the
curve y f ( x) bounded by x axis, x a and x b . Similarly, Figure 2.73: Remainder for Integral Test the series could be thought of as an estimation of the area under the curve of f ( x ) where f ( n) an . We can do something similar
with the remainder, as Rn i n 1 ai an 1 an 2 . If we start at x n 1 , taking rectangles of width 1 and use the left endpoints as the height of the rectangle we can estimate the area under f ( x ) on the interval [ n 1, ) as shown in Fig. 2.73(a). We can see that the remainder, Rn , is the area estimation and it will overestimate the exact area. So, Rn
n 1
f ( x )dx . Next, we could also
estimate the area by starting at x n , taking rectangles of width 1 again and then using the right endpoint as the height of the rectangle. This will give an estimation of the area under f ( x ) on the interval [ n, ) , as shown in Fig. 2.73(b). Again, we can see that the remainder, Rn , is again this
estimation and in this case it will underestimate the area. Thus Rn f ( x ) dx . Combining the n
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Engineering Mathematics
two inequalities
Chapter 2: Calculus
n 1
Thus sn
n 1
[2.146]
n
n 1
f ( x ) dx Rn f ( x ) dx s n
f ( x) dx sn Rn s n f ( x ) dx . n
f ( x) dx s sn f ( x) dx . n
Comparison Test: In this case, unlike with the integral test, we may or may not be able to get an idea of how good a particular partial sum will be as an estimate of the exact value of the series. Much of this will depend on how the comparison test is used. As in comparison test, given a series an , we find a second series bn that converged and an bn for all n . Actually we
an , whose remainder is Rn i n1 ai , using the n partial sum sn i 1 ai . Let the remainder of the series bn is Tn i n 1 bi . As an bn so are estimating the actual value of the series
Rn Tn . When using the comparison test it is often the case that the bn are fairly nice terms and
that we might actually be able to get an idea on the size of Tn . For instance, if our second series is a
p series we can use the results from above to get an upper bound on Tn as:
Rn Tn g ( x) dx , where g ( n) bn . Also if the second series is a geometric series then we n
will be able to compute Tn exactly. If we are unable to get an idea of the size of Tn then comparison test is not helpful. Alternating Series Test: The integral test and comparison test requires all the terms must be positive. But in alternating series test there is no restriction for the terms to be positive. Let we n have a convergent series an ( 1) bn , which satisfies the condition of alternating series test, so bn 0 for all n . We want to know how good of an estimation of the actual series value will the partial sum, sn , be. If the series,
an , converges to s
will lies between sn and sn 1 for
any n and thus s sn sn 1 sn bn 1 Rn s sn bn 1
Ratio Test: Let the series
an
is convergent. To get an estimate of the remainder let us first
define the sequence rn an 1 an . We now have two possible cases: If {rn } is a decreasing sequence and rn 1 1 then Rn an 1 (1 rn 1 ) If {rn } is an increasing sequence then Rn an 1 (1 L) , where L lim an 1 an n
Example 2.289 [ME-1997 (1 mark)]: The sum of the infinite series, 1 (1 2) (1 3) (1 4) is (a)
(d) 2 4
(c) 4
(b)
Solution (b): The given series can be written as
n1 an , where
an 1 n . So we can convert this
series as a function f ( x ) 1 x which is continuous, positive and decreasing function on the interval
[1, ) , thus using integral test,
n1 (1 n) 1
f ( x ) dx (1 x) dx (ln x)1 ln ln1 . 1
Hence the given series is diverges. So the sum of the given series is .
2.8.3 Power Series A power series about a , or just power series, is any series that can be written in the form
n 0 cn ( x a )n , where
a and cn are numbers. The cn ’s are often called the coefficients of the series. Note that the power series is a function of x . The convergence of a power series will depend upon the values of x . Let there is a number R such that the power series will converge for x a R and will diverge for x a R . This number, R , is called the radius of convergence for the series. Note that the series may or may not converge if x a R . Also the interval of all x ’s, including the endpoints if need be, for which the power series converges is called the interval of convergence of
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[2.147]
the series. So if we know that the radius of convergence of a power series is R , then we have the following: Convergence of power series: a R x a R Divergence of power series: x a R and x a R The interval of convergence must then contain the interval a R x a R since we know that the power series will converge for these values. We also know that the interval of convergence cannot contain x ’s in the ranges x a R and x a R since we know the power series diverges for these value of x . Therefore, to completely identify the interval of convergence all that we have to do is determine if the power series will converge for x a R or x a R . If the power series converges for one or both of these values then we’ll need to include those in the interval of convergence. At x a , the power series becomes
n0 cn (a a )n n 0 cn (0)n c0 (0)0 n1 cn (0)n c0
and
so the power series converges. Thus it is important to note that no matter what else is happening in the power series we are guaranteed to get convergence for x a . The series may not converge for any other value of x , but it will always converge for x a . Example 2.290 [XE-2008 (2 marks)]: The radius of convergence of the real power series
m0 (m !)2
(2m 1)! x m is
(a) 4
(b) 3
(c) 2
(d) 1
Solution (a): By applying the ratio test, for the convergence of the given power series where am L lim
m
( m !) 2 (2m 1)!
am 1
m
x , we have L lim
m
( m 1) 2 x (2m 3)(2m 2)
lim
m
am
1 L lim
m
(1 1 m) 2 x (2 3 m)(2 2 m)
lim
m
m1 am ,
(m 1)!2 x m1 (2m 1)! 2(m 1) 1 ! ( m !) 2 x m (1 0) 2 x
(2 0)(2 0)
x 4
x 4
Now for the convergence of the given power series we have L 1 (1 4) x 1 x 4 . Thus the given power series will converge for
x 4 and diverge for
x 4 and thus the radius of
convergence is R 4 . Also the interval of convergence is x 4 x (4, 4) . Example 2.291 [CS-1993, EC-1993 (1 mark)]: Find The radius of convergence of the power series
m1(3m)! (m!)3 x3m . Solution: By applying the ratio test, for the convergence of the given power series am
(3m)! (m !)
3
L lim
m
x
3m
, we have L lim
m
(3m 3)! x 3m 3
(m 1)!
3
3
am 1
1 L lim
am
( m !)3 (3m)! x 3m
m
L lim
m
m1 am , where
3(m 1) ! x3( m1) (m !)3 3 (3m)! x3 m (m 1)!
(3m 3)(3m 2)(3m 1) x 3 m3 ( m 1)3
(3m 3)(3m 2)(3m 1)
3 L if L x lim (3m 3)(3m 2)(3m 1) 3 m (1 1 m) x 0 . But if x 0 then L 0 0 1 . So the given power series will converge only if x 0 . Thus the radius of convergence is R 0 and the interval of convergence is x 0 .
L x lim
m
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Engineering Mathematics
Chapter 2: Calculus
[2.148]
2 m m Example 2.292 [AE-2014 (1 mark)]: The series s m 1 m 3 ( x 2) converges for all x
with x 2 R given by (a) R 0 (b) R 3
(c) R
(d) R 1 3
Solution (b): By applying the ratio test, for the convergence of the given power series where am L lim
m2 3m
am1
m
( x 2) , we have L lim
am
m
( m 1) 2 ( x 2) 3m
m
2
lim
m 1
3
m
(1 1 m) 2 ( x 2) 3
m
(m 1) 2 ( x 2) m 1
1 L lim
(1 0) 2 ( x 2) 3
( x 2) 3
m1 am ,
3m 2
m ( x 2)
m
L 1
if
( x 2) 3 1 x 2 3 ; also L 1 if ( x 2) 3 1 x 2 3 . Thus for x 2 3 the given
series converges and for x 2 3 the given series diverges. Thus the radius of convergence R 3 ; also the interval of convergence is x 2 3 3 x 2 3 1 x 5 .
Power Series and Functions: Every power series can be represented as a function of x . For e.g., in the infinite geometric series,
n0 ar n a (1 r ) , provided
r 1 (the infinite geometric series
converges for r 1 and diverges for r 1 ) if we take a 1 and r x , the series becomes
n0 x n 1 (1 x) , provided
x 1 . So we can represent the infinite geometric series as a function
f ( x ) 1 (1 x ) with the power series
n 0 x n
provided x 1 . We can clearly plug any number
other than x 1 into the function, however, we will only get a convergent power series if x 1 . This means the equality, in
n0 x n 1 (1 x) , will holds if
x 1 and for any other value of x the
equality will not hold. Also the radius of convergence of the infinite geometric series is R 1 and the interval of convergence is x 1 . Thus we can represent many functions as power series and it will be important to recognize that the representations will often only be valid for a range of x ’s and that there may be values of x that we can plug into the function that we cannot plug into the power series representation.
Differentiation
and Integration
of a Power Series:
n
2
Let
3
f ( x ) n 0 cn ( x a) c0 c1 ( x a ) c2 ( x a ) c3 ( x a ) .
we
a
power
series
If we differentiate it we get
f ( x ) c1 2c2 ( x a ) 3c3 ( x a ) 2 n 1 ncn ( x a ) n 1 ,
also
f ( x) 2c2 3 2c3 ( x a ) n 2 n( n 1)cn ( x a ) n 2 ,
f ( n ) ( x ) n r 1 n( n 1)( n 2) ( n r )cn ( x a ) n ( r 1) .
n
n
and
Now
if
so we
( x a)
integrate
on; it,
thus we
get
n 1
f ( x)dx n0 cn ( x a) dx n0 cn ( x a) dx C n0 cn n 1 . If f ( x ) n 0 cn ( x a ) n dx has a radius of convergence of R0 n 1 ( n 1) both also f ( x ) n 1 ncn ( x a) n 1 and f ( x )dx C n 0 cn ( x a )
then, have a
radius of convergence of R .
2.8.4 Taylor’s Series and Maclaurin Series Let
f ( x)
be a function which is represented by a power series about
x a , i.e.,
f ( x ) n 0 cn ( x a) n c0 c1 ( x a ) c2 ( x a ) 2 c3 ( x a )3 . Let f ( x ) has derivatives of
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Engineering Mathematics
Chapter 2: Calculus
[2.149]
every order. Now we need to determine what are the coefficients cn . Let us first evaluate everything at x a , so using the power series and its derivatives we find that: f ( a ) c0 , f ( a ) c1 , f (a ) 2c2 c2 f (a ) 2 , f ( a ) 3(2)c3 c3 f ( a) {3(2)} , , cn f ( n ) ( a ) n ! . So, a power series representation for the function f ( x ) about x a is the Taylor series for f ( x ) about x a is f ( n ) ( a) f (a ) f ( a ) n 2 3 f ( x ) n 0 ( x a ) f ( a ) f ( a)( x a) ( x a) ( x a) (2.44) n! 2! 3! If we use a 0 , then the Taylor series about x 0 is called a Maclaurin series for f ( x ) , i.e.,
f ( x) n 0
f ( n ) (0)
n ( x ) f (0) f (0) x
f (0)
2
x
f (0)
3
x
(2.45) n! 2! 3! The Taylor series expansion of functions about x 0 is given in Section 2.3.1 under the heading of Some important expansion. To determine a condition that must be true in order for a Taylor series to exist for a function let us first define the nth degree Taylor polynomial of f ( x ) as n
f (i ) (a )
i
f ( n ) (a )
n
i0 n ! ( x a) , the nth i! degree Taylor polynomial is just the partial sum for the series. Next the remainder is defined to be Rn ( x) f ( x ) Tn ( x ) . So the remainder is really just the error between the function f ( x ) and the nth degree Taylor polynomial for a given n ; with this definition note that we can write the function as f ( x ) Tn ( x ) Rn ( x ) . Tn ( x ) i 0
Suppose
( x a ) . Notice that for the full Taylor series
f ( x ) Tn ( x ) Rn ( x ) ,
that
f ( x ) n 0
f ( n ) (a )
if
xa R
for
lim Rn ( x ) 0
n
then
n
( x a ) on x a R .
n! Example 2.293 [CE-2000 (2 marks)]: The Taylor expansion of sin x about x 6 is given by (a)
1 2
3
x
2
2
3
1 3 x x 6 4 6 12 6 3
5
(b) x
x3 3!
x5 5!
x7 7!
7
1 1 1 (d) 1 2 x x x 6 3! 6 5! 6 7! 6 Solution (a): The Taylor expansion of f ( x ) sin x about x 6 is given as:
(c) x
2
3
1 1 f ( x) f f x f x f x . 6 2! 6 6 3! 6 6 6 6 As, f ( 6) sin( 6) 1 2 , f ( 6) cos( 6) 3 2 , f ( 6) sin( 6) 1 2 , and so on. 2
3
Thus, f ( x ) (1 2) ( 3 2) x ( 6) (1 4) x ( 6) ( 3 12) x ( 6) . [Similar question was also asked in EC-2009 (2 marks)] Example 2.294 [EC-2007 (1 mark)]: For the function e x , the linear approximation around x 2 is (a) (3 x )e 2
(b) 1 x
(c) 3 2 2 1 2 x e 2
(d) e 2
Solution (a): f ( x) e x f ( x ) e x f ( x ) e x f ( x ) e x , so Taylor series expansion of f ( x ) about x a is f ( x ) e 2 e 2 ( x 2) (1 2!)e 2 ( x 2) 2 (1 3!)e 2 ( x 2) 3 . Now neglecting the higher power of ( x 2) , we get f ( x ) e 2 e 2 ( x 2) f ( x ) (3 x )e 2
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.150]
Example 2.295 [EC-2008 (2 marks)]: In the Taylor series expansion of exp( x ) sin( x) about the 2
point x , the coefficient of x is (a) exp( ) (b) 0.5 exp( )
(c) exp( ) 1
(d) exp( ) 1
x
Solution (b): The Taylor series expansion of f ( x) e sin x about the point x is given as: f ( x ) f ( ) f ( )( x ) (1 2!) f ( )( x ) 2 (1 3!) f ( )( x )3 . So the coefficient of ( x ) 2 is f ( ) 2! . As f ( x) e x cos x f ( x ) e x sin x f ( ) e 0 e . Thus the
coefficient of ( x ) 2 is e 2! e 2 0.5e [Similar question was also asked in ME-2008 (1 mark)] Example 2.296 [EC-2008 (1 mark)]: Which of the following functions would have only odd powers of x in its Taylor series expansion about the point x 0 ? (a) sin( x3 ) (b) sin( x 2 ) (c) cos( x3 ) (d) cos( x 2 ) Solution (a): The Taylor series expansion of f ( x ) sin x and g ( x ) cos x about x 0 is given as: f ( x ) sin x x 3
3
x3 3!
3
f ( x ) sin( x ) x 3
3
g ( x ) cos( x ) 1
x5
5! x9 3!
x6 2!
and
x15 5!
x12 4!
g ( x ) cos x 1 x6
2 2 2 , f ( x ) sin( x ) x
2 2 , g ( x ) cos( x ) 1
3!
x4 2!
x10
x8 4!
5!
x2 2!
x4 4!
.
So
,
Thus the Taylor series expansion of f ( x 3 ) sin( x 3 ) have only odd powers of x .
k 2 X , where k1 , k2 and k3 are k3 X
Example 2.297 [MT-2011 (2 marks)]: Y k1 1 exp
constants. If k 2 X k3 X , the value of Y up to first order approximation would be
(a) Y k1 1
k 2 X
(b) Y k1 1
k 2 X
(c) Y k1
k 2 X
(d) Y k1
k 2 X
k3 X k3 X k3 X k3 X Solution (d): e x 1 x (1 2!) x 2 (1 3!) x 3 ; now let x ( k 2 X ) ( k3 X ) and for first order
approximation, the higher powers of x should be neglected as k2 X k3 X and so e x 1 x , thus Y k1 (1 e x ) k1{1 (1 x))} k1 x k1 (k 2 X ) ( k3 X ) .
Example 2.298 [CH-2012 (1 mark)]: For the function f (t ) e t , the Taylor series approximation for t is (a) 1 (t ) (b) 1 (t ) (d) 1 t (c) 1 (t 2 2 2 ) Solution (b): Firstly we have to find the Taylor series expansion of f (t ) e t about t 0 . As f (t ) e t f (t ) (1 )e t f (t ) (1 2 )e t f (t ) (1 3 )e t
f (0) 1 ,
f (0) 1 ,
f (t ) f (0) f (0)(t 0)
1
2
f (0) 1 , 2
f (0)(t 0)
1
f (0) 1
3
3
f (0)(t 0) 1
2! 3! for t , the higher powers of t 0 . Thus f (t ) 1 t
and t
1 t2 2! 2
and
so
so
on.
Thus
Now
1 t3 3! 3
on.
So
Example 2.299 [XE-2012 (1 mark)]: Taylor series of f ( x ) 1 (1 x ) about x 0 is given as
TS
f
1 x x 2 x3 . This series can be used to evaluate f ( x ) for
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Chapter 2: Calculus
[2.151]
(a) x 1 (b) x 1 (c) x 1 (d) 1 x 1 Solution (c): As the given Taylor series is infinite geometric series whose common ratio r x , which converges when r 1 x 1 . Example 2.300 [MT-2013 (2 marks)]: Taylor series expansion of the function f ( x) x (1 x ) around x 0 will be (a) 1 x x 2 x3 (b) 1 x x 2 x 3 (c) 0 x (1 2) x 2 (1 3) x 3 (d) 0 x x 2 x 3 Solution (d): For f ( x) x (1 x ) , f ( x ) 1 (1 x) 2 f ( x ) 2 (1 x) 3 f ( x) 6 (1 x) 3 and so on. Thus f (0) 0 , f (0) 1 , f (0) 2 , f (0) 6 , and so on. So the Taylor series of the given function about x 0 is given as: f ( x ) f (0) f (0)( x 0) (1 2!) f (0)( x 0) 2 (1 3!) f (0)( x 0)3 0 x x 2 x3 Example 2.301 [EC-2014 (1 mark)]: The series (a) 2 ln 2
converges to (d) e
(c) 2
2
(b)
n 0 (1 n !)
Solution (d): The Taylor series expansion of e x at x 1 is e
1
1
1
1
0! 1! 2! 3!
n 0
1 n!
. Thus
the given series converges to e . Example 2.302 [CH-2014 (1 mark)]: If f ( x ) is real and continuous function of x , the Taylor series expansion of f ( x ) about its minima will NEVER have a term containing (a) first derivative (b) second derivative (c) third derivative (d) any higher derivative Solution (a): As for a real valued function y f ( x) , the Taylor series expansion about x a is f ( x ) f ( a) f ( a )( x a ) (1 2!) f ( a)( x a ) 2 (1 3!) f ( a )( x a)3 . Now for minima at
x a , f ( a) 0 . So f ( x ) f ( a) (1 2!) f ( a)( x a ) 2 (1 3!) f ( a )( x a)3 , which does not contain the first derivative term. Example 2.303 [EC-2014 (2 marks)]: The Taylor series expansion of 3sin x 2 cos x is (a) 2 3 x x 2 (1 2) x3 (b) 2 3 x x 2 (1 2) x3 (c) 2 3 x x 2 (1 2) x 3 (d) 2 3 x x 2 (1 2) x3 Solution (a): The Taylor series expansion of f ( x ) sin x and g ( x ) cos x about x 0 is given as: sin x x (1 3!) x 3 (1 5!) x 5 ; cos x 1 (1 2!) x 2 (1 4!) x 4 . So the Taylor series expansion
x3
3!
of h( x) 3sin x 2 cos x 3 x
x5 5!
x2
2!
2 1
x4 4!
2
2 3x x
x3 2
x4 12
2.8.5 Improper Integral A definite integral
b
a
f ( x ) dx is called an improper integral, if the range of integration is finite and the
integrand is unbounded and/or the range of integration is infinite and the integrand is bounded. For e.g., (1) The integral
1
0 (1 x
2
) dx is an improper integral, because the integrand is unbounded on [0,
1]. In fact, 1 x 2 as x 0 . (2) The integral
2 0 1 (1 x ) dx
is an improper integral, because
the range of integration is not finite. There are following two kinds of improper definite integrals:
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
Improper integral of first kind: A definite integral
b
a
[2.152]
f ( x ) dx is called an improper integral of
first kind if the range of integration is not finite (i.e., either a or b or a and b )
is bounded on [ a, b] . For e.g.,
f ( x)
and the integrand
2 1 (1 x ) dx, 1 3x
2 2 1 1 x dx, 0 1 (1 x ) dx,
(1 2 x )3 dx are improper integrals of first kind. In an improper integral of
first kind, the interval of integration is one of the following types [ a, ) , (, b] , (, ) .
If
t
a f ( x)dx
every t b then,
b
or lim
t t
t a
t
b
f ( x) dx exists for
t t
a
or
f ( x) dx
b
is said to be convergent, if lim f ( x)dx t
f ( x)dx
t a
f ( x) dx exists finitely and this limit is called the value of the improper integral.
t
t a
If
t
f ( x) dx lim f ( x ) dx . If
b
If lim f ( x ) dx or lim
a
f ( x) dx lim f ( x ) dx .
The improper integral b
exists for every t a then,
c
f ( x)dx
b
t t
c
and
f ( x) dx is either or , then the integral is said to be divergent.
f ( x) dx are both convergent then
where c is any number. The improper integral
f ( x)dx
c
f ( x)dx f ( x)dx c
f ( x ) dx ,
is said to be divergent if either of the
right hand side is divergent. a
Example 2.304 [CE-2000 (1 mark)]: The following integration lim x 4 dx a 1
(c) Converges to 1 a 3
(b) Converges to 1 3
(a) Diverges
a
Solution (b): lim x 4 dx lim a 1
a
x
3
( 3)
a 1
(d) Converges to 0
lim 1 (3a ) (1 3) 1 3 3
a
[Similar question was also asked in EE-2005 (1 mark)]
Example 2.305 [CH-2012 (2 marks)]: If a is constant, then value of the integral a 2 xe ax dx is 0
(b) a
(a) 1 a
(c) 1
(d) 0
k
Solution (c): Let I a 2 xe ax dx a 2 lim xe ax dx a 2 lim x e ax dx (d dx ) x e ax dx dx 0
2
e
I a lim x k
I a
2
ax
a
e
k 0
ax
a
k 2
e
dx a lim x k
0
k
ax
a
ax
k
ak
k 0
ak
ke e 1 2 2 0 2 a klim a 0 a a a
e
2
0 (1 a ) 1 2
Improper integral of second kind: A definite integral
b
a
f ( x ) dx is called an improper integral of
second kind if the range of integration [ a, b] is finite and the integrand is unbounded at one or more points of [ a, b] . If
b
a
f ( x ) dx is an improper integral of second kind, then a , b are finite real
numbers and there exists at least one point c [ a, b] such that f ( x) or f ( x) as x c i.e., f ( x ) has at least one point of finite discontinuity in [ a, b] . For example: 3
1. The integral
1 ( x 2) . 1 1 ( x 2) dx is an improper integral of second kind, because lim x 2
2. The integral
0 log xdx; is an improper integral of second kind, because log x as x 0 .
1
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Engineering Mathematics
3. The integral
Chapter 2: Calculus
2
0 1 (1 cos x) dx,
[2.153]
is an improper integral of second kind since integrand
1 (1 cos x) becomes infinite at x [0, 2 ] . 1
4.
(sin x) x 1 . 0 (sin x) x dx, is a proper integral since lim x 0
If
f ( x) b
a
is continuous on the interval
[a, b)
x b then,
and not continuous at
t
f ( x ) dx lim f ( x ) dx provided the limit exists and is finite. Note that we have to use a LHL a
t b
here as the interval of integration is entirely on the left side of the upper limit. If f ( x ) is continuous on the interval (a, b] and not continuous at b
a
xa
then,
b
f ( x ) dx lim f ( x ) dx provided the limit exists and is finite. Note that we have to use a RHL t
ta
here as the interval of integration is entirely on the right side of the lower limit.
If f ( x ) is not continuous at x c , where a c b and b
a
convergent then
c
b
a
c
c
a
f ( x) dx and
b
c
f ( x) dx are both
f ( x) dx f ( x) dx f ( x ) dx . Now the improper integral
b
a
f ( x) dx will
converge if both of the improper integral on the right hand side converges. If either of the two integrals on the right hand side is divergent then the improper integral
If f ( x ) is not continuous at x a and x b and if convergent then
b
a
c
b
a
c
c
a
b
a
f ( x) dx diverges.
f ( x) dx and
b
c
f ( x) dx are both
f ( x) dx f ( x) dx f ( x ) dx , where a c b . Again this requires both
integrals on RHS to be convergent in order for improper integral
b
a
f ( x) dx to also be convergent.
Some tests on convergence/divergence of improper integral − integral test:
convergent if
p 1 and divergent is p 1 Standard comparison test: If 0 g ( x ) f ( x ) on the interval [ a, ) then If
p a 1 x dx is convergent if p 1 and divergent is p 1 b If 0 b then 1 x p dx is convergent if p 1 and divergent is p 1 0 b b If a b then 1 ( x a ) p dx and 1 (b x ) p dx both are a a
If 0 a then
a f ( x)dx converges then a g ( x)dx also converges If g ( x )dx diverges then f ( x) dx also diverges a a
Limit comparison test: Let (i) f ( x ) 0 and g ( x ) 0 on [ a, ) ; (ii) f ( x ) and g ( x ) both are continuous on [ a, ) ; (iii) lim f ( x ) g ( x ) L , where L is some finite positive number. Then x
a
f ( x) dx and
a
g ( x )dx behave the same way, i.e., either both converges or diverges.
Similar results also hold for other three basic type of improper integrals; suppose f ( x ) and
g ( x ) are both positive and continuous, then (i) on (, b] and lim
x
f ( x)
g ( x ) L 0 ;
(ii) on [a, b) and lim f ( x ) g ( x ) L 0 ; (iii) on (a, b] and lim f ( x ) g ( x ) L 0 ; x b
x a
where L is some finite ve number. Then the improper integrals of f ( x ) and g ( x ) with same limits of integration behave the same way, i.e., either both converges or diverges.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 2: Calculus
[2.154]
Absolute and Conditional convergence: Let the function f ( x) be defined for all x a . If the integral
a
a
convergent; and in this case
a
a
f ( x ) dx converges, then the integral
f ( x) dx
f ( x) dx . If the integral
a
f ( x ) dx diverges, then the integral
a
f ( x) dx also converges and is called absolutely
a
f ( x ) dx converges and
f ( x) dx is called conditionally convergent.
Example 2.306 [CS-1993, EC-1993 (1 mark)]: Which of the following improper integrals is/are convergent? 4 1 sin x 1 2(1 cos x ) x 1 3sin (2 x ) (a) dx (c) dx (d) dx dx (b) 2 0 1 cos x 0 1 x 0 1 x5 2 x Solution (d): Option (a) I
sin x
1
0 1 cos
x
1
2 sin( x 2) cos( x 2)
0
2
dx
dx
1 cos( x
0
2 sin ( x 2)
2)
sin( x 2)
dx
1
1
I 2 ln sin( x 2) 0 2(ln sin(1 2) ln 0) sin x (1 cos x) dx diverges. 0
4
4
Option (b) 0 sin (2 x) 1 1 1 sin (2 x) 2 1
1 1 4 1 1 3sin (2 x)
[1, ) .
As,
Option (c)
x dx (2 x )1 ,
by
comparison
x 1 sin 4 (2 x)
x dx diverges as 1
x
so
1
2 0 1 x 2 dx 2 ln 1 x 0
1
x {1 sin 4 (2 x)}
ln 1 2
test
we
can
say
x in that
x.
2
x2
ln 1
x
dx diverges. 1 x2 Option (d) As for the given integral we have an improper point x 0 so we have the interval 2 2 1 2(1 cos x ) 1 4 sin ( x 2) 1 sin ( x 2) 1 x (0,1] . As dx dx 0 x5 2 0 ( x 2)2 x1 2 dx ; and for all x (0,1] , we 0 x5 2 0
4 sin 2 ( x 2) x 5 2 sin 2 ( x 2) have 0 sin ( x 2) 1 ; also lim lim 1 0 . So using the limit 2 x 0 1 x1 2 ( x 2) x0 2
lim f ( x ) g ( x ) 1 0
comparison test we have g ( x ) 1 x1 2 1
x 0
behave same (i.e., both converges or both diverges on 1
12 1 2 x 1 0 g ( x)dx 0 1 x dx (2 x ) x 0 2 g ( x) 1 1 f ( x ) 4 sin 2 ( x 2) x 5 2 dx 2(1 cos x ) 0 0
f ( x ) 4 sin 2 ( x 2)
so both
converges
x5 2
and
x (0,1] ). Since, and
so
x 5 2 dx also converges.
Example 2.307 [CS-2005 (2 marks)]: Let G ( x ) 1 (1 x) 2 i 0 g (i ) x i , where x 1 . What is
g (i ) ? (a) i
(b) i 1
(c) 2i
i
(d) 2 Solution: 1 (1 x ) 1 x x x x x ; differentiating both sides w.r.t. x , we get 2
3
4
5
1 (1 x ) 2 1 2 x 3 x 2 4 x 3 5 x 2 i 0 (i 1) x i i 0 g (i ) x i g (i) i 1
Example 2.308 [ME-2008 (2 marks)]: Which of the following integral is unbounded? 4 1 1 1 (a) tan x dx (c) xe x dx (b) 2 dx (d) dx 0 0 0 x 1 0 1 x Solution (d): Option (a)
4
0
tan x dx ln sec x
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4 0
ln 2 ln1 ln 2
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Engineering Mathematics
Option (b)
1
2
0
x 1
Chapter 2: Calculus
dx tan 1 x
[2.155]
tan 1 tan 1 0 2 0 2
0
Option (c) xe x dx xe x e x ( e e ) ( 0 e 0 e 0 ) 0 (0 1) 1 0
0
1
1
0
1 x
Option (d)
1
dx ln 1 x 0 ln(0) ln(1) . So option (d) is unbounded
Example 2.309 [ME-2010 (1 mark)]: The value of the integral
2 1 (1 x ) dx is
(a)
(b) 2 (c) 2 b dx dx 1 Solution (d): Let I lim lim tan x 2 1 x a a 1 x 2 a b b
(d) b
a
1
1
lim tan b tan a a b
I tan 1 ( ) tan 1 ( ) tan 1 () tan 1 () 2 tan 1 () 2( 2) [Similar question was also asked in CH-2011 (2 mark)] 1
4 3
Example 2.310 [XE-2010 (1 mark)]: The integral
1 x
(a) an improper integral converging to –6 (c) not an improper integral but has value –6
(b) an improper integral converging to 0 (d) a divergent improper integral
dx is
1
Solution (a): Let I x 4 3 dx , then x 0 is the only point of infinite discontinuity of x 4 3 in 1
0
[ 1,1] , then I x
4 3
1
1
dx x
4 3
0
dx
3
0
13
x
1
1
3
13
x
0
3 0
3 13
( 1)
3 13
(1)
3 (0)
6 . Thus the
given integral is converging to –6. Example 2.311: Find at which point the improper integral
0
e x dx is convergent or divergent.
k
Solution: I e x dx lim e x dx I lim[ e x ]k0 lim[ e k e0 ] 0
k 0
k
k
I lim(1 e ) 1 0 1 [ lim e k
k
k
e
k
0]
k
Thus, lim e x dx exists and is finite. Hence the given integral is convergent at 1. k 0
Exercise: 2.7 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. Which of the following sequences converges? (i) (iii)
n2 (3n2 1) (10n 5n2 ) n1(1)n n
(ii) (iv)
n1e 2n n n0 (1)n
(a) (i), (ii) and (iii) (b) (i) and (iii) (c) (ii), (iii) and (iv) 2. Which of the following sequences are monotonic? (i)
n 0 ( n 2 )
(a) (i) and (ii) 3. The series
1 n 2
( n 2) 2 n 1 n 3 (1 3 )
(ii)
(b) (ii) and (iii)
4. The series
(c)
(c) (i) and (iii)
n 5 (2
n2 )
(d) (i), (ii) and (iii)
(1 3n1 ) which starts at n 3 is
(a)
n 1 (1)n1
(d) (ii) and (iv)
( n 2) 2 n 1 n 3 (1 3 )
(b)
(n 2) 2 n 1 n 3 (1 3 )
(c)
( n 2) 2 n 1 ) n 3 (1 3
(d)
n2 1 (n2 1) converges to _____.
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Engineering Mathematics
[2.156]
n11 3n1 converges to _____. 2 3 Which of the following series diverges? (i) n0 (4n n )
5. The series 6.
Chapter 2: Calculus
(a) Only (i)
(b) Only (ii)
(c) Both (i) and (ii)
n0 (4)
7. Which of the following series converges? (i)
3n
8. If the series
2
2
n0 1 (n
(a) 1
; (ii) 9
) 5
n0 1 (n
n2 1 (n ln n)
(a) (i) and (ii)
(d) The series diverges
4n 3) converges, then its value is
(ii)
n0 ne n
(c) 0 2
(d) The series diverges
n 4 1 n7
(iii)
(iv)
(c) (iii) and (iv)
11. Which of the following series diverges? (i)
n1n
(b) Only (ii)
2
2
( n cos n) ; (ii)
n0 1 (3
n
n) ; (ii)
n1(n2 2)
(ii)
n1(1)n2
(a) Only (i) (b) Only (ii) 15. Which of the following series diverges?
n1(10)n
(i)
{42 n 1 ( n 1)}
(a) Only (i)
(ii)
n2 (4n2 n)
(iii)
(iii)
(c) Only (iii)
n1n
n
2 n 1
3
3
n 7 n3
cos(n ) n
n2
(d) (i), (ii) and (iii)
n0 n! 5n
( n 2 5)
(d) Neither (i) nor (ii)
n1(sin n) n3
(c) Both (1) and (iii)
(b) Only (ii)
16. Which of the following series diverges? (i)
n2
n
(d) Neither (i) nor (ii)
(a) (i) and (ii) (b) (iii) and (i) (c) (ii) and (iii) 14. Which of the following series is absolute convergent?
(a) Only (i) (b) Only (ii) (c) Both (i) and (ii) 13. Which of the following series converges? n 2 n 3 ( 1) n n ( 1) (iii) (i) n 1 2 (ii) n 0 n 4 n 5
n1(1)n n
n11
(d) (iv) and (i)
(c) Both (i) and (ii)
12. Which of the following series diverges? (i)
(i)
. The
(d) (ii), 50.2
(c) 2
(b) (ii) and (iii)
(a) Only (i)
4
(a) 12 5 (b) 5 12 10. Which of the following series converges? (i)
n 2 n 1
n 1
3n 2) converges, then its value is (b) 0
9. If the series
n 1 n
(d) Neither (i) not (ii)
n 1
series which converges find the value of the series. (a) (i), 259.2 (b) (ii), 259.2 (c) (i), 50.2
(10 2n3 ) ; (ii)
(d) Both (ii) and (iii)
n 2 n 2
(2n 1)!
(d) All (i), (ii) and (iii)
; (ii) (5n 3n ) n 0
3
3
(7 n 2)
n
(a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) Neither (i) nor (ii) 17. The radius of convergence and interval of convergence for the power series given by n ( 1) n n1 4n ( x 3)n is, respectively. (a) 4, 7 x 1 (b) 4, 1 x 7 (c) 3, 7 x 1 (d) 3, 7 x 0 18. The radius of convergence and interval of convergence for the power series given by
n1(2n
n)(4 x 8) n is, respectively.
(a) 1 , 1 x 2
(b) 1 8 , 15 8 x 17 8
(c) 1 4 , 1 x 17 8 (d) 1 8 , 15 8 x 17 8 19. The radius of convergence and interval of convergence for the power series given by
n0 n !(2 x 1)n is, respectively. (a) 0, x 1 2
(b) 1, x 1 2
Copyright © 2016 by Kaushlendra Kumar
(c) 0, x 1 2
(d) 1, x 1 2
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Engineering Mathematics
Chapter 2: Calculus
[2.157]
20. The radius of convergence and interval of convergence for the power series given by
n1( x 6)n
n n is, respectively.
(a) 0, x (b) 1, x (c) 2, x (d) , x 21. The radius of convergence and interval of convergence for the power series given by
n1 x 2 n
(3) n
is, respectively.
(a)
3, 3x0
(b)
2, 0 x 3
(c)
3, 3x 3
(d)
2, 3 x0
22. The power series representation of the function f ( x) x (5 x) and its interval of convergence is, respectively. (a) (c)
0 xn 0 xn
,
5n 1 , x 5
(b)
5n 1
(d)
x 4
0 x n1 0 x n1
,
5n1 , x 5 5n1
x 4
23. The power series representation of the function g ( x ) 1 (1 x ) 2 and its radius of convergence is, respectively. (a)
n1 nx n 1 , 2
n1 nx n 1 , 1
(b)
(c)
n1 nx n , 1
(d)
n1 nx n , 2
24. The Taylor series expansion for f ( x ) ln x about x 2 is
ln 2 ( 1)
n 1 ( n2 n ) ( x 2) n (a) ln 2 n1 ( 1)
(c)
n 1
n ( x 2) n
n 1
ln 2 ( 1)
( n2 ) ( x 2)
n 1 (2n ) ( x 2) n (b) ln 2 n1 ( 1)
(d)
n
n 1
n
n
25. The Taylor series expansion for f ( x ) x 3 10 x 2 6 about x 3 is (a) 87 33( x 3) ( x 3) 2 ( x 3)3
(b) 57 33( x 3) ( x 3) 2 ( x 3)3
(c) 57 33( x 3) ( x 3) 2 ( x 3)3
(d) 57 33( x 3) ( x 3) 2 ( x 3)3
26. The Taylor series expansion of f ( x )
dx about x 0 is
(b) C n 1
(2 n 1)(2n 1)! n
( 1) x
(c) C n 0
x
(1) n x 2 n
(a) C n 0
sin x
2 n 1
(d) C n 1
(2 n 1)(2n 1)!
( 1) n x 2 n 1 2n(2n 1)! ( 1) n x 2 n 1 (2n 2)!
x
0 (1 2 x)e dx is _____. 4 28. The value of the improper integral x ( x 2 9) dx is 0 27. The value of the improper integral
(a) 3 (b) 2 (c) 1 29. Determine which of the following integral diverges? (i)
1 1 ( x
3
3 x
(ii)
1) dx
2
(a) Only (i) (b) Only (ii) 30. Which of the following integral converges? (i)
1 1 x dx 5
0
(ii)
2
xe 3 x dx
(i)
1
0
{1 x 2 }dx
(a) (i) and (iii)
(ii)
2
x 1
2
(iii)
(c) Only (iii)
1
(b) (i), (ii) and (iii)
(iii)
(iv)
(1 x) dx
1 1 ( x 2
(c) (i) and (ii)
4 e
x
x dx
(d) All (i), (ii) and (iii)
(c) (ii), (iii) and (iv)
( x 3 8) dx
Copyright © 2016 by Kaushlendra Kumar
( x 3 1) dx
(iii)
(a) (i) and (ii) (b) (ii) and (iv) 31. Which of the following integral diverges?
(d) The given integral diverges
x 2 1 dx
10 {1 ( x ln x)}dx
(d) (i), (ii) and (iii) (iv)
1
0 1
3
x dx
(d) (iii) and (iv)
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Chapter 2: Calculus
[2.158]
Answers Keys 1 b 16 b 31 43 46 5
2 a 17 c 32 c 47 d
3 b 18 a 33 300 48 a
4 6 19 b 34 42 49 d
5 a 20 32 35 c 50 23
Answer Keys: Exercise: 2.1 6 7 8 9 10 c a c b 6 21 22 23 24 25 b b c d a 36 37 38 39 40 20 15 b a a 51 52 53 54 55 a d b d c
1 c 16 a 31 c
2 2 17 b 32 1
3 0 18 a 33 a
4 c 19 c 34 1
5 d 20 b 35 c
Answer Keys: Exercise 2.2 6 7 8 9 10 d 0 b c d 21 22 23 24 25 14 6 d c c 36 37 b a Answer Keys: Exercise: 2.3 6 7 8 9 10 d a c d 1 21 22 23 24 25 c a 0 a b 36 37 38 39 40 2 6 0 c 1 51 52 53 a c 5
1 b 16 4 31 a 46 0.75
2 d 17 3 32 b 47 a
3 d 18 b 33 2 48 c
4 0.5 19 b 34 b 49 0
5 c 20 a 35 c 50 d
1 8 16 2 31 a
2 0.8 17 c 32 c
3 a 18 a 33 a
4 b 19 2 34 d
5 1 20 b 35 0
1 b 16 0 31 2
2 a 17 0.5 32 a
3 a 18 0 33 2.67
4 c 19 0 34 1.34
5 d 20 0 35 a
Answer Keys: Exercise: 2.5 6 7 8 9 10 c c 2 0.5 2.5 21 22 23 24 25 2 0.5 a c b 36 37 d 31.1
5 b 20 c
Answer Keys: Exercise: 2.6 6 7 8 9 10 b 9.07 10 0.35 a 21 22 23 24 25 0 d d d c
1 44.5 16 4.01 31 a
2 a 17 a 32 d
3 a 18 0
4 c 19 b
Answer Keys: Exercise 2.4 6 7 8 9 10 d b c a c 21 22 23 24 25 b a b c a
Copyright © 2016 by Kaushlendra Kumar
11 300 26 c 41 d 56 d
12 3300 27 b 42 b 57 d
13 60 28 10 43 6 58 a
14 45 29 160 44 a
15 d 30 c 45 c
11 a 26 b
12 c 27 b
13 d 28 12
14 c 29 0
15 a 30 1
11 b 26 a 41 1
12 c 27 a 42 1
13 2 28 11 43 c
14 a 29 c 44 c
15 3 30 b 45 1
11 a 26 b
12 d 27 d
13 b 28 d
14 12 29 c
15 3 30 a
11 1 26 512
12 5 27 0.46
13 4 28 4
14 b 29 a
15 2 30 a
11 d 26 b
12 d 27 3.74
13 c 28 4.12
14 d 29 b
15 0.32 30 a
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Engineering Mathematics
1 48.08 16 c 31 9 46 b
1 b 16 a 31 c
2 2 17 d 32 9.8 47 a
3 4 d 40.83 18 19 a 5 33 34 92.87 0.67 48 49 b 0
2 c 17 a
3 c 18 d
4 0.75 19 c
Chapter 2: Calculus
5 d 20 1 35 a
5 1.5 20 d
[2.159]
Answer Keys: Exercise: 2.7 6 7 8 9 10 c b a b b 21 22 23 24 25 8 a 2 10.42 0.34 36 37 38 39 40 d 0 d 8 d
Answer Keys: Exercise: 2.8 6 7 8 9 10 c b a b b 21 22 23 24 25 c b b a d
Copyright © 2016 by Kaushlendra Kumar
11 b 26 a 41 b
11 a 26 c
12 a 27 c 42 d
12 b 27 3
13 2 28 a 43 a
13 c 28 d
14 c 29 b 44 0
14 d 29 b
15 d 30 15 45 3
15 b 30 a
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Engineering Mathematics
Chapter – 3: Differential Equation
Chapter – 3 [1]
GATE – 2016: Chapter – 3: Differential Equation Note: The following questions came in GATE – 2016 were based on Differential Equation Chapter – 3. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. u 2u 2 , where is a positive constant, is 1. The partial differential equation t x [AE-2016 (1 mark)] (a) Circular (b) elliptic (c) hyperbolic (d) parabolic 2u 2u 2u u u C 2 D E Fu G , we Solution (d): Comparing the given PDE with A 2 B x xt t x t 2 2 have B C D F G 0 and A , E 1 , so B 4 AC (0) 4( )(0) 0 , thus given PDE is parabolic in nature. 2. The general solution of the differential equation 2y
(1 3)(e3 x x 3 )
2y
(1 2)(e3 x x 2 )
(a) C (1 2)e (c) C (1 3)e Solution (a):
dy dx
dy
e3 x 2 y x 2 e 2 y is ___. [AG-2016 (1 mark)] dx 2y 3x 2 (b) C e (1 3)(e x )
(d) C e
2y
(1 3)(e3 x x3 )
e3 x 2 y x 2 e 2 y e 2 y (e 3 x x 2 ) e 2 y dy (e3 x x 2 ) dx …(i). On integrating both
sides of (i), we get
2y 3x 2 e dy (e x )dx
e2 y
2
e3 x 3
x3 3
C C (1 2)e 2 y (1 3)(e3 x x 3 ) .
3. The Laplace transform F ( s ) of the function f (t ) cos(at ) , where a is constant, is _____. [BT-2016 (1 mark)] 2 2 2 2 2 2 2 2 2 (a) s ( s a ) (b) a ( s a ) (c) s ( s a ) (d) s ( s a ) 2
2
Solution (c): {cos(at )} s ( s a ) 4. The type of partial differential equation
2P x
2
2P y
2
3
2P xy
2
P x
P y
0 is
[CE-2016 (1 mark)] (c) hyperbolic (d) none of these 2P 2P 2 P P P C 2 D E FP G , we Solution (c): Comparing the given PDE with A 2 B x xy y x y (a) Elliptic
(b) parabolic
2
2
have A 1 , B 3 and C 1 , so B 4 AC (3) 4(1)(1) 5 0 , thus given PDE is hyperbolic in nature. 5. The solution of the partial differential equation (a) C cos( kt ) C1e
( k )x
( (c) Ce kt C1e
k )x
C2 e
C2 e
( k ) x
( k ) x
Solution (c): The given PDE, u ( x, t ) ( A cos px B sin px )Ce
p 2 t
u t
2u x 2
is of the form
[CE-2016 (1 mark)]
(b) Ce kt C1 cos( k ) x C2 sin( k ) x (d) C sin( kt ) C1 cos( k ) x C2 sin( k ) x
u t
2u x 2
, is 1-D heat equation, whose solution is
…(i) [Refer to Chapter – 3, Section – 3.5.2]
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Engineering Mathematics
Now let p 2 k p
Chapter – 3: Differential Equation
k
Chapter – 3 [2]
k i …(ii), where i 1 .
So putting (ii) in (i), we get u ( x, t ) { A cos(i k x ) B sin(i k x )}Ce kt u ( x, t ) { A cosh( k x ) B sinh( k x )}Ce kt { A cosh( k x ) B sinh( k x )}Ce kt kx kx e k x e k x e e B 2 2 k x A B kt k x A B kt u ( x, t ) Ce e e Ce 2 2
u ( x, t ) Ce kt A
c e
kx
1
c2 e
kx
,
where c1 ( A B ) 2 and c2 ( A B ) 2 . at
6. The Laplace transform of e sin(bt ) is
b
(a)
2
(s a) b
(b)
2
[CH-2016 (1 mark)]
( s a) 2
(s a) b
(c)
2
(s a) 2
(s a) b
2
b
(d)
(s a)2 b2
at
Solution (a): From ‘First Shifting Theorem’ {e f (t )} F ( s a ) , where { f (t )} F ( s ) . Here 2
2
at
2
2
f (t ) sin(bt ) and {sin(bt )} b ( s b ) F ( s ) . So {e sin(bt )} b {( s a) b } . 2
7. Which one of the following is a property of the solutions to the Laplace equation: f 0 ? [EC-2016 (1 mark)] (a) The solutions have neither maxima nor minima anywhere except at the boundaries. (b) The solutions are not separable in the coordinates. (c) The solutions are not continuous. (d) The solutions are not dependent on the boundary conditions. 2 Solution (a): The solutions to the Laplace equation: f 0 are separable in the coordinates; continuous; dependent on the boundary conditions; and cannot have local maxima or minima, but extreme values must occur at the end points (i.e. at the boundaries). So option (a) is correct. 2t
8. The Laplace Transform of f (t ) e sin(5t )u (t ) is 2
2
(a) 5 ( s 4 s 29)
[EE-2016 (1 mark)] 2
(b) 5 ( s 5)
(c) ( s 2) ( s 4 s 29)
Solution (a): We know that {g (t )u (t a )} e
as
(d) 5 ( s 5)
{g (t a )} . We have a 0 , g (t ) e 2t sin(5t ) , so
{e2t sin(5t )u (t )} e (0) s {e 2t sin(5t )} {e 2t sin(5t )} 5 {(s 2) 2 52 } 5 {s 2 4 s 29} , at
2
since
2
{e sin(bt )} b {( s a) b } . 1
9. A function y (t ) , such that y (0) 1 and y (1) 3e , is a solution of the differential equation
d2y
2
dy
y 0 . Then y (2) is [EE-2016 (1 mark)] dt dt (a) 5e 1 (b) 5e 2 (c) 7e 1 (d) 7e 2 Solution (b): We have homogeneous linear equation with constant coefficients, so its solution will be mx mx 2 mx of the form y e y me y m e , putting all these in given DE, we get 2
emx ( m 2 2m 1) 0 ( m 2 2m 1) 0 , as emx 0 ; so ( m 2 2m 1) 0 ( m 1) 2 0 m 1 Thus we have real and equal roots and so the general solution of given DE is y (c1 c2 x )e mx (c1 c2 x)e x . Now y (0) 1 1 c1 ; and y (1) 3e 1 3e 1 {1 c2 (1)}e 1 c2 2 x
Hence y {1 2 x}e , so y (2) {1 2(2)}e
2
Copyright © 2016 by Kaushlendra Kumar
5e 2 .
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Engineering Mathematics
10. The value of
e
t
Chapter – 3: Differential Equation
Chapter – 3 [3]
(2t 2) dt , where (t ) is the Dirac delta function, is
(a) 1 (2e)
(b) 2 e
(c) 1 e
2
[EE-2016 (1 mark)] 2
(d) 1 (2e )
Solution (a): Let 2t 2 k 2dt dk dt (1 2) dk , also t , k and t , k ; so
e
t
(2t 2) dt
e
t
e 2
( k 2) 2
( k ) dk
1
f ( k ) ( k 0) dk , where 2
f ( k ) e ( k 2) 2 .
f ( x) ( x a)dx f (a) , so f (k ) (k 0)dk f (0) e
We know that Hence
1
(2t 2)dt
1
(0 2) 2
e 1
1
f ( k ) ( k 0) dk 2 2e
11. The solution of the differential equation, for t 0 , y (t ) 2 y (t ) y (t ) 0 with initial conditions y (0) 0 and y (0) 1 , is { u (t ) denotes the unit step function} [EE-2016 (1 mark)] t
t
t
t
t
t
(a) te u (t ) (b) (e te )u (t ) (c) ( e te )u (t ) (d) e u (t ) Solution (a): We have homogeneous linear equation with constant coefficients, so its solution will be mt mt 2 mt of the form y e y me y m e , putting all these in given DE, we get
emt ( m 2 2m 1) 0 (m 2 2m 1) 0 , as emt 0 ; so ( m 2 2m 1) 0 ( m 1) 2 0 m 1 Thus we have real and equal roots and so the general solution of given DE is y (c1 c2 t )e mt (c1 c2 t )e t y c2 e t (c1 c2t ) e t Now y (0) 0 0 c1 ; and y (0) 1 1 c2 e 0 (0 c2 (0))e 0 c2 1
0, t 0
t
Hence y te ; as u (t )
1, t 0
, but given DE is valid for t 0 , so required solution will be
y te t u (t ) . Thus option (a) is correct. 12. Let f ( x) be a real, periodic function satisfying f ( x ) f ( x) . The general form of its Fourier series representation would be [EE-2016 (1 mark)]
(b) f ( x) k 1 bk sin( kx)
(d) f ( x ) k 0 a2 k 1 sin(2k 1) x
(a) f ( x ) a0 k 1 ak cos( kx )
(c) f ( x ) a0 k 1 a2 k cos( kx )
Solution (b): If f ( x ) be a piecewise continuous function (i.e. having finite number of removable or jump discontinuities) on [ T , T ] . Then the Fourier series of f (t ) is the series given by
f (t ) a0 n 1 ( an cos n t bn sin nt ) , where 2 (2T ) . If f ( x ) is an odd function then the
Fourier series expansion of f ( x ) contains only the sine terms, the cosine and the constant term being zero [Refer to Chapter – 3, Section – 3.4.1]. So option (b) is correct. 13. Let u ( x, t ) be the d’Alembert’s solution of the initial value problem for the wave equation: utt c 2 u xx 0 , u ( x, 0) f ( x ) , ut ( x, 0) g ( x) , where c is a positive real number and f , g are smooth odd functions. Then u (0,1) is equal to _____. [MA-2016 (1 mark)] Solution: The given problem is Cauchy Problem for the infinite string, D’Alembert’s solution. On referring to Section 3.5.1, the solution of utt c 2 u xx 0 , u ( x, 0) f ( x ) , ut ( x, 0) g ( x) is given by
u ( x , t ) (1 2) f ( x ct ) f ( x ct ) {1 (2c )}
x ct
x ct
So
g ( s )ds . c
u (0,1) (1 2) f (0 c) f (0 c ) {1 (2c)} g ( s) ds (1 2) f (c) f (c ) {1 (2c)}(0) 0 ,
since g ( x ) is an odd function so
c
c
c g (s)ds 0 ; also
Copyright © 2016 by Kaushlendra Kumar
f ( x ) is an odd function so ‘ f ( x ) f ( x) ’.
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Engineering Mathematics
Chapter – 3: Differential Equation
Chapter – 3 [4]
14. If f (t ) is a function defined for all t 0 , its Laplace transform F ( s ) is defined as [ME-2016 (1 mark)] (a)
0
Solution
e st f (t )dt
(b):
(b)
If
e st f (t ) dt
0 be
f (t )
a
(c)
function
0
eist f (t )dt
defined
for
(d) t 0;
0
e ist f (t ) dt
then
the
integral
b
{ f (t )} e st f (t ) dt lim e st f (t ) dt is said to be the Laplace transform of f (t ) provided that 0
b 0
the integral converges. 15. Laplace transform of cos(t ) is 2
2
[ME-2016 (1 mark)] 2
(a) s ( s )
2
2
(b) ( s ) 2
2
2
(c) s ( s )
2
(d) ( s )
2
Solution (a): {cos t} s ( s ) 2
2
16. The differential of the equation, x y 1 , with respect to x is (a) x y
[MN-2016 (1 mark)]
(c) y x
(b) x y
(d) y x
Solution (a): On differentiating both sides w.r.t. ‘ x ’, we get 2 x 2 y ( dy dx) 0 ( dy dx) x y . 17. The solution of the DE x
(a) y e c
d2y
dy
is ___. [where, c, c1 , c2 are constants] dx dx x (c) y c1e x c2 (b) y e c 2
MT-2016 (1 marks)] (d) y c1e x c2
d dy dy dy dt dt t , so (i) t dx …(ii) …(i); now let dx dx dx dx dx dx dx t dy On integrating both sides, we get ln t x ln c1 t c1e x c1e x dy c1e x dx …(iii). Again dx x on integrating (iii) both sides, we get y c1e c2 . Solution (d):
2
dy
d2y
18. The Laplace transform of the function e 2t is (a) 1 (2s ) (b) 2 s
[PE-2016 (1 mark)] (d) e 2 s
(c) 1 ( s 2)
at
Solution (c): {e } 1 ( s a) . Here a 2 , so {e
2 t
} 1 ( s 2)
19. The following partial differential equation U xx U yy 0 is of the type (a) Elliptic
[TF-2016 (1 mark)]
(b) Parabolic
(c) Hyperbolic (d) Mixed type 2 2 U U U U U C 2 D E FU G , Solution (a): Comparing the given PDE with A 2 B x xy y x y 2
2
2
we have A 1 , B 0 and C 1 , so B 4 AC (0) 4(1)(1) 4 0 , thus given PDE is elliptic in nature. dy 20. Let y ( x) be the solution of the initial value problem 2 xy x ; y (0) 0 . Find the value of dx [XE-2016 (1 mark)] lim y ( x ) . x
Solution: Given DE is in Leibnitz’s Linear form, whose I.F. e e x . On multiplying both sides 2 2 2 2 2 d x2 x 2 dy of given DE with IF, we get e 2 xe x y xe x (e y ) xe x d (e x y ) xe x dx …(i). dx dx 2 xdx
On integrating both sides of (i), we get y (0) 0 e
(0 ) 2
(0) (1 2)e
( 0) 2
d (e
x2
x2
x2
2
x2
y ) xe dx e y (1 2)e c …(ii)
c c (1 2) .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics So
the
general
x2
Chapter – 3: Differential Equation solution
of
x2
given
e y (1 2)e (1 2) y (1 2) (1 2)e
x2
DE
Chapter – 3 [5]
with
given
. Thus lim y ( x ) lim{(1 2) (1 2)e x
x
21. Which of the following is a quasi-linear partial differential equation? 2
2
u
u u (b) 0 t x
2
condition x2
2
is
} 1 2 0.5 .
[XE-2016 (1 mark)]
2
u u (c) 0 t x
4
3
u u (d) 0 t x
u 0 t 2 Solution (a): A PDE is called a quasi-linear if all the terms with highest order derivatives of dependent variables occur linearly, i.e. the coefficients of such terms are functions of only lower order 2 2 2 derivatives of the dependent variables. So among given PDEs the PDE ( u t ) u 0 is quasilinear PDE and others are non-linear. So option (a) is correct. (a)
2
2
22. Consider a second order linear ordinary differential equation ( d y dx ) 4( dy dx) 4 y 0 , with the boundary conditions y (0) 1 ; ( dy dx) x 0 1 . The value of y at x 1 is _____ [AE-2016 (2 marks)] (c) e (a) 0 (b) 1 (d) e 2 Solution (a): We have homogeneous linear equation with constant coefficients, so its solution will be mx mx 2 mx of the form y e y me y m e , putting all these in given DE, we get
emx ( m 2 4m 4) 0 (m 2 4m 4) 0 , as emt 0 ; so ( m 2 4m 4) 0 (m 2) 2 0 m 2 . Thus we have real and equal roots and so the general solution of given DE is y (c1 c2 x)e mx (c1 c2 x )e 2 x y c2 e 2 x 2(c1 c2 x )e 2 x . Now y (0) 1 1 {c1 c2 (0)}e 2(0) 1 c1 ; and y (0) 1 1 c2 e 2(0) 2{1 c2 (0)}e 2(0) c2 1 2x
2(1)
Hence y ( x ) (1 x )e ; so y (1) (1 1)e 0 . Thus option (a) is correct. [Similar question was also asked in EC-2016, EE-2016 (2 marks)] 2
2
23. ( d y dx ) y 0 . The initial conditions for this second order homogeneous DE are y (0) 1 and dy dx 3 at x 0 . The value of y when x 2 is _____. [BT-2016 (2 marks)] Solution: We have homogeneous linear equation with constant coefficients, so its solution will be of mx mx 2 mx the form y e y me y m e , putting all these in given DE, we get
emx ( m 2 1) 0 ( m 2 1) 0 , as emt 0 ; so ( m 2 1) 0 m 1,1 . Thus we have real and distinct roots and so the general solution of given DE is y c1e x c2 e x y c1e x c2 e x . Now
y(0) 1 1 c1 c2 …(i); and y(0) 3 3 c1 c2 …(ii). From (i) and (ii), we have c1 1 and c2 2 . Hence y ( x ) e 24. The value of
2e x ; so y(2) e 2 2e 2 14.642 .
2 0 1 (1 x ) dx 0 (sin x) x dx
(a) 2 Solution (b):
x
(b)
sin t t
(c) 3 2
(d) 1
1 2
s 1
F (s)
sin t t
2 . The 2nd integral is found by
s
s
F ( s ) ds
1 2
s 1
1
ds (tan s ) s
st sin t s e dt s . Now taking limit s 0 both sides we 0 2 t 2 sin t sin x sin t 1 dt lim s dt . So dx dx . 2 0 0 0 s 0 2 t t 2 1 x x 2 2 1
1
tan tan s
get lim e s 0 0
[CE-2016 (2 marks)]
2 1 1 1 0 1 (1 x ) dx (tan x)0 tan () tan (0)
Integral of Transform as: {sin t}
is
st
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 3: Differential Equation
Chapter – 3 [6]
25. The respective expression for complimentary function and particular integral part of the solution d4y d2y 3 108 x 2 are of the differential equation [CE-2016 (2 marks)] dx 4 dx 2 (a) c1 c2 x c3 sin 3 x c4 cos 3x and 3 x 4 12 x 2 c (b) c2 x c3 sin 3x c4 cos 3 x and 5 x 4 12 x 2 c (c) c1 c3 sin 3x c4 cos 3 x and 3 x 4 12 x 2 c (d) c1 c2 x c3 sin 3 x c4 cos 3x and 5 x 4 12 x 2 c Solution (a): The homogeneous part of the given DE is
y e mx y me mx y m 2e mx y m3e mx
d4y
3
d2y
0 which is solved by putting
dx 4 dx 2 y m 4 e mx ,
emx ( m 4 3m 2 ) 0
so
( m 4 3m 2 ) 0 , as emx 0 , so m 0, 0, 3 i, 3i , thus we have two real and equal roots, i.e. m1 m2 0 ; and two complex conjugate roots, i.e. m3 3i , m2 3i . So complementary solution of the given DE is given (0) x (0) x yc (c1 c2 x )e e (c3 cos 3x c4 sin 3 x) yc c1 c2 x c3 cos 3 x c4 sin 3 x .
as:
2
Now, as the expression ‘ 108x ’ is linear 2nd degree polynomial but our differential equation is of 4th order, so for particular solution if we must choose y p Ax 4 Bx 3 Cx 2 Dx E , then yp 4 Ax3 3 Bx 2 2Cx D y p 12 Ax 2 6 Bx 2C y p 24 Ax 6 B y p 24 A ;
on
2
2
substituting these in given DE we get, 24 A 3(12 Ax 6 Bx 2C ) 108 x , and on comparing the 2
coefficient of x , x and constant terms, we get 36 A 108 A 3 , 18B 0 B 0 , 4 2 24 A 6C 0 C 12 . Thus particular solution is y p 3 x 12 x Dx E ; so the general 4
2
solution is y yc y p c1 c2 x c3 cos 3 x c4 sin 3 x 3x 12 x Dx E y c1 (c2 D ) x c3 cos 3 x c4 sin 3 x 3 x 4 12 x 2 E , which is given by option (a). 2
2
26. What is the solution for the second order differential equation ( d y dx ) y 0 , with the initial conditions y x 0 5 and ( dy dx ) x 0 10 ?
[CH-2016 (2 marks)]
(a) y 5 10 sin x (b) y 5 cos x 5sin x (c) y 5 cos x 10 x (d) y 5 cos x 10 sin x Solution (d): We have homogeneous linear equation with constant coefficients, so its solution will be mx mx 2 mx of the form y e y me y m e , putting all these in given DE, we get
emx ( m 2 1) 0 (m 2 1) 0 , as emt 0 ; so ( m 2 1) 0 m 0 i, 0 i . Thus we have complex conjugate roots and so the general solution of given DE is (0) x y e {c1 cos(1) x c2 sin(1) x} y ( x ) c1 cos x c2 sin x y ( x ) c1 sin x c2 cos x . y(0) 5 5 c1 and y(0) 10 10 c2 . Thus y ( x ) 5 cos x 10 sin x . 27. The Laplace transform of the causal periodic square wave of period T shown in the figure below is: [EC-2016 (2 mark)] (a) (a) F ( s) 1 {1 e
sT 2
(c) F ( s ) 1 [ s{1 e
}
sT
}]
Copyright © 2016 by Kaushlendra Kumar
(b) F ( s ) 1 [ s{1 e (d) F ( s) 1 {1 e
sT 2
sT
}]
}
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Engineering Mathematics
Chapter – 3: Differential Equation
Chapter – 3 [7]
Solution (b): The period of the given function is T . For one period the function f (t ) can be written as: f (t ) u{t} u{t (T 2)} , so
{ f (t )}
1 1 e
T
sT
0 e
st
f (t ) dt
1
T
1 e
sT
2)} dt
T 2
st 0 e u{t} u{t (T
2)} dt . T 2
Now T
T
st
0 e u{t} u{t (T
e
st
0
0
e
st
T
e st e dt s 0
T 2
st
st
1 0 dt T 2 e 1 1 dt 0
e s (T 2) e s (0) 1 e s (T 2) 1 e s (T 2) u{t} u{t (T 2)} dt s s s s s
Hence { f (t )}
1 1 e
T
sT
st
0 e u{t} u{t (T
1 e s (T
1
2)} dt
1 e
sT
2)
s
1 s{1 e sT 2 }
sin 2 t dt is equal to t
28. The value of the integral 2
.
[EE-2016 (2 marks)]
(a) 0
(b) 0.5 (c) 1 (d) 2 sin 2 t sin 2 t sin 2 t sin 2 t Solution (d): I 2 is an even dt 2 2 0 dt 4 0 dt , as t t t t 4
We use ‘Integral of Transform’ for evaluating As {sin k}
1 2
s 1
F ( s)
sin k k
sin k dk k
0
function. Now let 2 t k 2 dt dk dt dk (2 ) , so I
sin k dk . k
0
s
s
F ( s ) ds
1
1
2
s 1
ds (tan s ) s
sk sin k s e dk s . Now taking limit s 0 both sides 0 k 2 k 2 sk sin k sin k sk sin k we get lim e dk lim s lim e dk lim s dk . 0 s 0 0 s 0 0 s0 2 s0 2 k k k 2 4 sin k 4 So I 2 dk 0 k 2
sin k
1
1
tan tan s
29. The relationship between the force f (t ) and the displacement x (t ) of a spring-mass system (with
d 2 x (t ) 2
D
dx (t )
Kx (t ) f (t ) . dt dt X ( s ) and F ( s ) are the Laplace transform of x (t ) and f (t ) respectively. With M 0.1 , D 2 ,
a mass M , viscous damping D and spring constant K ) is M
K 10 in appropriate units, the transfer function G ( s ) X ( s) F ( s ) is 2
(a) 10 ( s 20 s 100) 2
(b) ( s 20 s 100)
2
2
(c) 10s ( s 20 s 100)
x(0) x(0) 0 ,
so
(d) s ( s 20s 100)
d 2 x(t ) 2
from
2
dx (t )
10 x(t ) f (t ) …(i); taking Laplace transform on both dt dt 0.1{s 2 X ( s ) sx (0) x(0)} 2{sX ( s) x (0)} 10 X (s ) F ( s) …(ii). As
Solution (a): Our DE is 0.1 sides of (i), we get
[IN-2016 (2 marks)]
2
(ii),
we
have
0.1s 2 X ( s ) 2 sX ( s) 10 X ( s) F ( s )
G ( s) X ( s ) F ( s) 1 (0.1s 2 2s 10) 10 ( s 2 20s 100) .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 3: Differential Equation
Chapter – 3 [8]
30. If y f ( x) satisfies the boundary value problem y 9 y 0 , y (0) 0 , y ( 2) 2 , then y ( 4) is _____. [ME-2016 (2 marks)] Solution: We have homogeneous linear equation with constant coefficients, so its solution will be of mx mx 2 mx the form y e y me y m e , putting all these in given DE, we get
emx ( m 2 9) 0 ( m 2 9) 0 , as emt 0 ; so ( m 2 9) 0 m 0 3i, 0 3i . Thus we have complex conjugate roots and so the general solution of given DE is (0) x y e {c1 cos(3) x c2 sin(3) x} y ( x ) c1 cos 3 x c2 sin 3 x . y(0) 0 0 c1 and y ( 2) 2 2 c2 sin 3( 2) c2 2 . Thus y ( x ) 2 sin x y ( 4) 2 sin( 4) 1 2 31. For the differential equation x
d2y dx
2
2x
dy dx
2 y 0 , the general solution is
[PE-2016 (2 marks)] (b) y c1 sin x c2 cos x (c) y c1e x c2 e x (a) y c1 x c2 e x (d) y c1 x 2 c2 x Solution (d): The given differential equation is of 2nd order homogeneous Cauchy – Euler equation, m ( m 1) whose solution is of the form y x y mx y m(m 1) x ( m 2) ; so substituting these in m
m
the given DE, we get x {m( m 1) 2m 2} 0 {m( m 1) 2m 2} 0 , as x 0 . 2
Thus {m( m 1) 2m 2} 0 m 3m 2 0 m 1, 2 , thus we have real and distinct roots; so the general solution of given DE is y c1 x 2 c2 x1 .
k , x 0
32. The Fourier series of periodic function f ( x )
k,
by
0 x
and f ( x 2 ) f ( x) is given
4 1 1 1 1 1 sin x sin 3 x sin 5 x . Then, the value of 1 is equal to 3 5 3 5 7
4k
_____.
[TF-2016 (2 marks)]
Solution: At x 2 , we have f ( x )
4k
1 1 1 1 . Now as f ( x) and f ( x ) both are 3 5 7
piece-wise continuous function in [ , ] so the Fourier series converges whose value is f ( 2) , as x 2 is a continuous point for f ( x) .
Thus f ( 2) k
4k 1 1 1 1 4 1 1 1 1 k 1 . 1 3 5 7 1 3 5 7 2
33. The integrating factor of (2 cos y 4 x ) dx ( x sin y) dy 0 is [TF-2016 (2 marks)] 2 (a) x (b) x (c) x (d) x 2 Solution (b): On comparing the given DE with M ( x, y ) dx N ( x, y )dy 0 , we have
M ( x, y ) 2 cos y 4 x 2
and
N ( x, y ) x sin y .
As
M y M y 2 sin y
and
N x N x sin y . As ( M y N x ) N (2 sin y sin y ) ( x sin y ) 1 x which is a function of ‘
x ’ alone, so I.F. e
{( M y N x ) N }dx
e
{1 x}dx
e ln x x .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.1]
Chapter 3 : Differential Equation An equation containing the derivatives of one or more dependent variable w.r.t. one or more independent variables is said to be a differential equation. If a differential equation contains only ordinary derivatives of one or more dependent variable w.r.t. a single independent variable, it is said to be an ordinary differential equation (ODE). For e.g. ( dy dx ) 5 y e x , ( d 2 y dx 2 ) ( dy dx ) 6 y 0 , ( dx dt ) ( dy dt ) 2 x 2 y are ODEs. An equation involving partial derivatives of one or more dependent variables of two or more independent variables is called a partial differential equation (PDE). For e.g. ( 2u x 2 ) ( 2u y 2 ) 0 , ( 2 u x 2 ) ( 2 u t 2 ) 2(u t ) , u y v x are PDEs. Example 3.1 [AG-2011 (1 mark)]: The differential equation 2( d 2 z dx 2 ) ( dz dx ) 3 y sin x is considered to be ordinary, as it has (a) one dependent variable (b) more than one dependent variable (c) one independent variable (d) more than one independent variable Solution (c): The given differential equation contains only ordinary derivatives of dependent variable ‘ z ’ w.r.t. only one independent variable ‘ x ’, so it is considered as ordinary differential equation.
Order and Degree of a Differential Equation: The order of a differential equation (either ODE or PDE) is the order of the highest differential coefficient occurring in it. The degree of a differential equation which is expressed or can be expressed as a polynomial in the derivatives is the degree of the highest order derivative occurring in it after it has been expressed in a form free radicals and fractions as far as derivatives are concerned. The order and degree of a differential equation is a positive integer. Thus the differential equation (3.1) is of m th order and p th degree. m
m
p
m 1
m 1 q
f ( x, y ) d y dx ( x, y ) d y dx 0 (3.1) Example 3.2 [EC-2005 (1 marks)]: The following differential equation has 3( d 2 y dt 2 ) 4( dy dt )3 y 2 2 x (a) degree = 2; order = 1 (b) degree = 1; order = 2 (c) degree = 4; order = 3 (d) degree = 2; order = 3 Solution (b): As the highest differential coefficient in the given differential equation is d 2 y dt 2 , which is of order 2; also the degree of the highest differential coefficient is 1. [Similar questions were also asked in CE-2007, EC-2009, MT-2013, TF-2010 (1 mark)]
Example 3.3 [ME-2007 (1 mark)]: The partial differential equation 2 2 2 2 ( x ) ( y ) ( x) ( y ) 0 has (a) degree 1; order 2 (b) degree 1; order 1 (c) degree 2; order 1 (d) degree 2; order 2 Solution (a): As the highest differential coefficient in the given differential equation is 2 x 2 , which is of order 2; also the degree of the highest differential coefficient is 1. Example 3.4 [CE-2010 (1 mark)]: The order and degree of the differential equation ( d 3 y dx3 ) 4 ( dy dx )3 y 2 0 are, respectively (a) 3 and 2 (b) 2 and 3 (c) 3 and 3
(d) 3 and 1 3
3 2
3
Solution (a): The given differential equation can be written as (d y dx ) 16 (dy dx) y
2
; the
highest differential coefficient (hdc) is d 3 y dx 3 so its order is 3; also degree of hdc is 2. Example 3.5: Find the order and degree of the differential equation e
Copyright © 2016 by Kaushlendra Kumar
d 3 y dx3
x (d 2 y dx 2 ) y 0 .
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Engineering Mathematics d 3 y dx3
2
Chapter 3: Differential Equation
[3.2]
2
Solution: e x( d y dx ) y 0 Clearly, the differential equation is of 3rd order, but its degree is not defined as it cannot be written as a polynomial equation in derivatives, and hence it cannot be expressed as polynomial of derivatives. Example 3.6 [TF-2011 (1 mark)]: The order and degree of the following differential equation are x(dy dx) 2 ( dy dx) y 2 (a) order 1, degree 1 (b) order 1, degree 2 (c) order 2, degree 1 (d) order 2, degree 2 2 Solution (b): The given differential equation can be written as x( dy dx ) y 2 ( dy dx ) 2 0 and the highest differential coefficient is dy dx , which is of order 1; also the degree of dy dx is 2. Example 3.7 [ME-2014 (1 mark)]: The matrix form of the linear system dx dt 3 x 5 y and dy dt 4 x 8 y is
3 5 x dt y 4 8 y d x 4 5 x (c) dt y 3 8 y
3 8 x dt y 4 5 y d x 4 8 x (d) dt y 3 5 y dx dy d x 3 5 x Solution (a): 3x 5 y and 4 x 8 y ; in matrix form we have dt dt dt y 4 8 y (a)
d x
(b)
d x
Linear and Non-Linear Differential Equation: A linear differential equation is any differential equation that can be written in the following form: an ( x ) y ( n ) ( x ) an 1 ( x) y ( n 1) ( x ) a1 ( x ) y( x ) a0 ( x ) y ( x ) g ( x ) (3.2) In linear differential equation there are no products of the function, y ( x ) , and its derivatives and neither the function or its derivatives occur to any power other than the first power. The coeffiecients a0 ( x), a1 ( x ), an ( x ) and g ( x ) can be zero or non-zero functions, constants or non-constant functions, linear or non-linear functions. So only the function, y ( x ) , and its derivatives are used in determining if a differential equation is linear. For e.g., ( y x )dx 4 xy dy 0 , y 2 y y 0 , x 3 ( d 3 y dx 3 ) x ( dy dx ) 5 y e x , etc., are linear first, second and third order ODE.
In Eq. 3.2, the dependent variable y and all derivatives y , y , , y ( n ) are of the first degree.
The coefficients a0 , a1 , , an of y , y , , y ( n ) depends at most on the independent variable x . Two important cases of Eq. 3.2 are given as: Linear first – order ( n 1 in Eq. 3.2) a1 ( x)( dy dx ) a0 ( x ) y g ( x ) (3.3) Linear second – order ( n 2 in Eq. 3.2) a2 ( x)( d 2 y dx 2 ) a1 ( x )( dy dx) a0 ( x ) y g ( x) (3.4) A nonlinear ordinary differential equation is simply one that is not linear. Nonlinear functions of the dependent variable or its derivatives, such as sin y or, cannot appear in a linear equation. For e.g., the non - linear term: coefficient depends on y
non - linear term: nonlinear function of y
non - linear term: power not 1
2 2 equations, (1 y ) y 2 y e x , ( d y dx ) sin y 0 , ( d 4 y dx 4 ) y2 0, etc., are non – linear first, second and fourth order ODEs. Example 3.8 [CS-1993, EC-1993 (1 mark)]: The differential equation 2 2 ( d y dx ) ( dy dx) sin y 0 is: (a) linear (b) non-linear (c) homogeneous (d) of degree two Solution (b): As we have a non-linear term sin y so the given differential equation is non-linear.
Example 3.9 [ME-2010 (1 mark)]: The Blasius equation ( d 3 f d 3 ) ( f 2)( d 2 f d 2 ) 0 is a (a) Second order non-linear ordinary differential equation (b) Third order non-linear ordinary differential equation (c) Third order linear ordinary differential equation
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Engineering Mathematics
Chapter 3: Differential Equation
[3.3]
(d) Mixed order non-linear ordinary differential equation Solution (b): As the given differential equation is not in the form given by Eq. 3.4 so it is not a linear differential equation. Also the highest differential coefficient is d 3 f d 3 , which is of order 3, whose degree is 1. So the given differential equation is third order non-linear ordinary differential equation. [Similar question was also asked in AE-2008 (1 mark)] Example 3.10 [ME-2013, PI-2013 (1 mark)]: The partial differential equation (u t ) u (u x ) 2u x 2 is a (a) linear equation of order 2 (b) non-linear equation of order 1 (c) linear equation of order 1 (d) non-linear equation of order 2 Solution (d): As we have the product of the function ‘ u ’ and its derivative u x in the given differential equation, so it is non-linear differential equation. Also the highest differential coefficient is 2 u x 2 , which is of 2nd order, whose degree is 1. Example 3.11 [CH-2014 (2 marks)]: (a) non-linear differential equation of first degree The differential equation (b) linear differential equation of first degree 2 2 2 3 (c) linear differential equation of second degree ( d y dx ) x ( dy dx ) x y e x is a (d) non-linear differential equation of second degree Solution (b): The given differential equation is of the form given by Eq. 3.4 so it is a linear differential equation. Also, the higest differential coefficient is d 2 y dx 2 , which is of second order, whose degree in the given equation is 1. So the given differential equation is a linear differential equation of second order and first degree.
Formation of Ordinary Differential Equation: Consider n parameter family of curves, f ( x , y , 1 , 2 , , n ) 0
(3.5)
where 1 , 2 , , n are n independent parameters. For e.g., y mx is a one parameter family of straight lines and x 2 y 2 ax by 0 is a two parameter family of circles. If we differentiate Eq. 3.5
n times w.r.t. x , we will get n more relations between x, y , 1 , 2 , , n and derivatives of y w.r.t. x . By eliminating 1 , 2 , , n from these n relations and Eq. 3.5, we get a differential equation. Clearly order of this differential equation will be n , i.e., equal to the number of independent parameters in the family of curves. Example 3.12 [MN-2012 (1 mark)]: The 2nd order differential equation having a solution y ( A x) B , where A and B are constants, is (a) d 2 y (dydx ) (2 x )( dy dx ) 0
(b) ( d 2 y dx 2 ) (2 x )( dy dx ) 0
(c) ( d 2 y dx 2 ) 2 (2 x )(dy dx) 0 (d) d 2 y (dydx ) (d 2 y dx 2 ) (dy dx ) 0 Solution (b): As we have two constants ( A , B ) so we have to differentiate the given equation two times for eliminating A and B . y ( A x) B dy dx ( A x 2 ) …(i); d 2 y dx 2 2 A x 3 …(ii) Now putting the value of A from (i) in (ii) (d 2 y dx 2 ) (2 x)( dy dx ) 0 . Example 3.13 [IN-2014 (1 mark)]: The figure shows the plot of y as a function of x . The function shown is the solution of the differential equation (assuming all initial conditions to be zero) is:
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(a) 2 y x 2 1 (b) dy dx x (c) dy dx x (d) dy dx x
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Engineering Mathematics
Chapter 3: Differential Equation
[3.4]
x2 2 , x 0 x, x 0 dy Solution (d): From the figure we can say that y 0, x 0 0, x 0 x . x 2 2 , x 0 dx x , x 0
Formation of Partial Differential Equation: Partial differential equation can be formed either by elimination of arbitrary constants or by the elimination of arbitrary functions from a relation involving three or more variables. We will discuss the above mentioned concept with the help of following examples: Example 3.14 [CE-2010 (2 z ax by ab has the form (a) z px qy (b) Solution (c): Differentiating
marks)]: The partial differential equation that can be formed from with p z x and q z y z px pq (c) z px qy pq (d) z qy pq both sides of the given equation partially w.r.t. x and y , we get
z x a p (given) and z y b q (given), respectively. Putting these result in the given equation we get z px qy pq .
Initial Value Problems: Sometimes a problem in seek a solution y( x) of a differential equation so that y ( x ) also satisfies certain prescribed side conditions, i.e., conditions that are imposed on the unknown function y ( x ) and its derivatives at a point x0 . On some interval I containing x0 the problem of solving an nth order differential equation subject to n side conditions specified at x0 : d n y dx n f ( x, y , y , , y ( n 1) )
Solve: Subject to:
y ( x0 ) y0 , y ( x0 ) y1 , , y
( n 1)
(3.6)
( x0 ) yn 1
where, y0 , y1 , , yn 1 are arbitrary real constants, is called an n
th
(3.7)
order initial-value problem
(IVP). The values of y ( x ) and its first n 1 derivatives at x0 , y ( x0 ) y0 , y ( x0 ) y1 , , y ( n 1) ( x0 ) yn 1 are called initial conditions (IC).
3.1
Solution of a Differential Equation General solution: The solution which contains as many as arbitrary constants as the order of the differential equation is called the general solution of the differential equation. For example, y A cos x B sin x is the general solution of the differential equation ( d 2 y dx 2 ) y 0 . But y A cos x is not its general solution as it contains one arbitrary constant. Particular solution: Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution. For example, y 3cos x 2 sin x is a particular solution of the differential equation ( d 2 y dx 2 ) y 0
3.1.1 Solution of first order and first degree differential equation The general form of a first order and first degree differential equation is f x, y , dy dx 0 (3.8) We know that the direction of the tangent of a curve in Cartesian rectangular coordinates at any point is given by dy dx , so the equation in Eq. 3.8 can be known as an equation which establishes the relationship between the coordinates of a point and the slope of the tangent i.e., dy dx , to the integral curve at that point. Solving the differential equation given by Eq. 3.8 means finding those curves for which the direction of tangent at each point coincides with the direction of the field. All the curves represented by the general solution when taken together will give the locus of the differential equation. Since there is one arbitrary constant in the general solution of the equation of first order, the locus of the equation can be said to be made up of single infinity of curves. Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.5]
A first order and first degree differential equation can be written as f ( x, y ) dx g ( x, y )dy 0 or
dy dx f ( x, y ) g ( x, y) or dy dx ( x, y ) , where f ( x, y ) and g ( x, y ) are obviously the functions of x and y . It is not always possible to solve this type of equations. The solution of this type of differential equations is possible only when it falls under the category of some standard forms.
Equations in variable separable form: If the differential equation is of the form f1 ( x ) dx f 2 ( y )dy
(3.9)
where f1 and f 2 being functions of x and y , respectively, only. Then we say that the variables are separable in the differential equation. Thus, integrating both sides of Eq. 3.9, we get its solution as
f1 ( x)dx f2 ( y )dy c , where c is an arbitrary constant. Example 3.15 [CE-2002 (5 marks)]: The following equation is sometimes used to model the population growth: dN dt aN ln( N k ) , where, N is the population at time t , a 0 is growth co-efficient and k 0 is a constant. Given, at t 0 , N N 0 and 0 N 0 k . (a) Find an expression for the population with time. (b) What is the population as t ? (c) Find a constant c (0,1) , such that the population growth rate is maximum at N ck . dN dN dN dz N N Solution: (a) aN ln adt . Let ln z . So, we have dt N k k k N ln( N k ) dz kz
adt
dz z
akdt
dz z
akdt ln z akt ln c z ce akt ln( N k ) ce akt . At
t 0 , N N 0 , ln( N 0 k ) ce 0 c . So, ln( N k ) ln( N 0 k )e akt N ke ln( N0 akt
ln( N 0 k )0
k ) e akt
.
0
(b) As t , e e 0 so N ke ke k . (c) For population growth rate to be maximum, dG dt 0 , where G dN dt aN ln( N k ) , thus 1 dG dt 0 a ln( N k ) ( aNk ) ( Nk ) 0 ln( N k ) 1 N ke .
1
N ck ,
For
1
ck ke c e which belongs to (0,1) . [Similar question was also asked in AE-2010 (1 mark)]
Example 3.16 [CE-2006 (2 marks)]: A spherical naphthalene ball exposed to the atmosphere loses volume at a rate proportional to its instantaneous surface area due to evaporation. If the initial diameter of the ball is 2 cm and the diameter reduces to 1 cm after 3 months, the ball completely evaporates in (a) 6 months (b) 9 months (c) 12 months (d) Infinite time 3 Solution (a): As the rate of volume, V (4 3) r , of spherical ball is proportional to its instantaneous
surface
area,
2 S 4 r ,
so
4 r 2 ( dr dt ) k (4 r 2 ) dr dt k dr kdt .
dr k dt r kt c ,
where
r0 1 cm.
dV dt kS (d dt ) (4 3) r 3 k (4 r 2 )
Now
Integrating at
t 0
both we
sides have
we
get
r 1 cm
1 k 0 c c 1 ; and at t 3 months we have r 0.5 cm 0.5 k 3 1 k 0.5 3 . Hence we have r (0.5 3)t 1 when r 0 , we have 0 (0.5 3)t 1 t 3 0.5 6 months. Example 3.17 [CE-2007 (2 marks)]: A body originally at 60oC cools down to 40oC in 15 min. when kept in air at a temperature of 25oC. What will be the temperature of the body at the end of 30 min.? (a) 35.2oC (b) 31.5oC (c) 28.7oC (d) 15oC Solution (b): As from Newton’s law of cooling the rate of temperature ( T ) is directly proportional to the temperature difference ( T T0 ) dT dt k (T T0 ) 1 (T T0 ) dT kdt . Integrating both
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Engineering Mathematics
Chapter 3: Differential Equation
[3.6]
1 (T T0 ) dT k dt ln(T T0 ) kt ln c T T0 ce
sides we get
kt
. Now from the data
given we have, T0 25 oC and at t 0 , T 60 oC 60 25 ce k 0 35 ; also at t 15 minutes,
T 40 oC 40 25 35e k 15 e 15 k 3 7 . T 25 35e
30 k
25 35(e
15 k 2
Now
t 30 minutes,
at
2
o
we
have
o
) 25 35(3 7) 31.428 C 31.5 C.
Example 3.18 [CH-2007 (1 mark)]: The initial condition for which the following equation ( x 2 2 x )( dy dx ) 2( x 1) y ; y ( x0 ) y0 has infinitely many solutions, is (a) y ( x 0) 5 (b) y ( x 0) 1 (c) y ( x 2) 1 (d) y ( x 2) 0
Solution (d): ( x 2 2 x )( dy dx ) 2( x 1) y (1 y ) dy 2( x 1) ( x 2 2 x ) dx . Integrating both sides we get,
(1 y )dy 2( x 1)
( x 2 2 x) dx ln y ln( x 2 2 x) ln c y c ( x 2 2 x ) . Now
applying the conditions given in option (a) and (b) gives 5 0 and 1 0 , respectively, which is not possible so option (a) and (b) are not correct as we get no solution for the given condition. For the condition given in option (c), we get c 1 8 , thus y ( x 2 2 x ) 8 which is an unique solution. For the condition given in option (d) we get 0 c 0 c can be any number thus we get only infinitely many solutions; thus option (d) is correct. Example 3.19 [CH-2007 (2 marks)]: The family of curves that is orthogonal to xy c is (a) y c1 x
(c) y 2 x 2 c1
(b) y c1 x
(d) y 2 x 2 c1
Solution (d): The family of curves that is orthogonal to xy c is found by first taking slope ( m1 ) of tangent on the curve xy c and then finding slope ( m2 ) of the tangent which is perpendicular to m1 and then we have differential equation which we have to integrate. Thus differentiating the given curve w.r.t. x we get y x ( dy dx ) 0 ( dy dx ) ( y x ) m1 ; so ( dy dx ) m2 1 m1 x y ydy xdx ;
ydy xdx (1 2) y
2
Iintegrating
both
sides
we
get
(1 2) x 2 c y 2 x 2 2c c1 .
Example 3.20 [ME-2007 (2 marks)]: The solution of dy dx y 2 with initial value y (0) 1 is bounded in the interval (a) x (c) x 1, x 1 (b) x 1 (d) 2 x 2 Solution
(c):
dy dx y 2 (1 y 2 )dy dx .
Integrating
both
sides
2
(1 y ) dy dx (1 y ) x c . Now applying the given condition we get, 1 1 0 c c 1 1 y x 1 y 1 ( x 1) . Hence the solution is valid for x 1 x 1, x 1 .
Example 3.21 [IN-2008 (2 marks)]: Consider the differential equation dy dx 1 y 2 . Which one of the following can be a particular solution of this differential equation? (a) y tan( x 3) (b) y tan x 3 (c) x tan( y 3) (d) x tan y 3
Solution (a): dy dx 1 y 2 1 (1 y 2 ) dy dx . Integrating both sides 1 (1 y 2 ) dy dx tan 1 y x c y tan( x c ) . Now from the given options we can say that option (a) is correct. [Similar questions were also asked in CE-1999, CE-2004, ME-2003, PI-2010, ME-2011 (2 marks), AE-2012, AG-2008 (1 mark)]
Example 3.22 [CE-2009 (2 marks)]: Solution of the differential equation 3 y ( dy dx ) 2 x 0 represent a family of (a) Ellipses (b) Circles (c) Parabolas (d) Hyperbolas
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Engineering Mathematics
Chapter 3: Differential Equation
[3.7]
Solution (a): 3 y ( dy dx ) 2 x 0 3 ydy 2 xdx . Now integrating both sides,
3 ydy 2 xdx
(3 2) y 2 x 2 c 3 y 2 2 x 2 2c (1 c ) x 2 1 (2 3)c y 2 1 an equaltion of ellipse. [Similar questions were also asked in EC-2009, MT-2012 (2 marks), CS-1995 (1 mark)] Example 3.23 [MT-2009 (2 marks)]: The solution function y f ( x) for the ordinary differential equation dy dx 3 x 2 2 x , passes through (1,1) . The magnitude of y at x 3 is (a) 0 (b) 18 (c) 19 (d) 21 Solution (c): dy dx 3 x 2 2 x dy (3 x 2 2 x )dx . Integrating both sides dy (3 x 2 2 x ) dx y x3 x 2 c .
Applying
the
given
condition,
we
get
1 1 1 c c 1 .
Thus
y x 3 x 2 1 y x 3 33 32 1 19 . Example 3.24 [AE-2011 (2 marks)]: The solution of dy dt y 3et t 2 with initial condition y (0) 1 is given by 1 4et 9 1 (a) et (t 3) 2 (c) (b) (d) 2 t 2 t 2 9 (t 2) 5 2e (t 2t 2) 5 2e (t 2t 2) Solution (b): dy dt y 3et t 2 (1 y 3 ) dy et t 2 dt . Integrating both sides (1 y 3 ) dy et t 2 dt . As 3
LHS of the integral (1 y )dy y
2
2 ; and for RHS of the integral, let ‘ t 2 ’ as the 1st, ‘ e t ’ as
the 2nd function et t 2 dt t 2 et 2tet dx t 2et 2 tet 1 et dt t 2et 2 te t et et (t 2 2t 2) . Thus y 2 2 et (t 2 2t 2) c . Now applying the given condition ( y (0) 1 ) we get c 5 2 .
t
2
Thus y 2 2 e t (t 2 2t 2) (5 2) 1 y 2 5 2et (t 2 2t 2) y 1 5 2e (t 2t 2)
Example 3.25 [CH-2011 (2 marks)]: Which ONE of the following choices is a solution of the differential equation dy dx ( y 2 x ) ( y x ) (2 x ) ? Note c is a real constant. (a) y
c x2
(b) y
c 2 x2
(c) y
c x3
(d) y
c 2 x3
c 2 x2 c x2 c 2 x3 c x3 Solution (d): The given differential equation can be written in variable separable form as: dy dx dy dx 1 1 1 dx 1 1 1 dx . dy dy 2 y y2 x ( y 2)( y 1) x 3 y 1 y 2 x 3 y 2 y 1 x
1 3
y 2 1 2 k 3 x3 y2 3 3 . ln( kx ) ( kx ) y 3 3 y 1 1 k x y 1
ln( y 2) ln( y 1) ln x ln c ln
Now let k 3 1 c so y (c 2 x 3 ) (c x 3 ) . Example 3.26 [CE-2012 (2 marks)]: The solution of the ordinary differential equation ( dy dx) 2 y 0 for the boundary condition, y 5 at x 1 is (a) y e 2x
(b) y 2e 2 x
(c) y 10.95e 2 x
(d) y 36.95e 2 x
Solution (d): ( dy dx) 2 y 0 (1 y ) dy 2dx . Now integrating both sides, (1 y ) dy 2 dx ln y 2 x ln c ln( y c) 2 x y c e 2 x y ce 2 x . Applying boundary conditions, we
get 5 ce 2 c 5e 2 36.95 , thus the solution of the give ODE, with given boundary condition, is y 36.95e 2 x [Similar questions were also asked in CE-2007, CE-2008, EC-2011, EE-2005, ME-1994 (1 mark), CE-2004, TF-2012, CH-2013 (2 marks)]
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Engineering Mathematics
Chapter 3: Differential Equation
[3.8]
Example 3.27 [EE-2014 (1 mark)]: A particle, starting from origin at t 0 second, is travelling along x axis with velocity v ( 2) cos{( 2)t} m/s. At t 3 second, the difference between the distance covered by the particle and the magnitude of displacement from the origin is ……… Solution: v ( 2) cos{( 2)t} dx dt ( 2) cos{( 2)t} dx ( 2) cos{( 2)t}dt .
dx (
Integrating both sides we get
2) cos{( 2)t}dt x sin{( 2)t} c . As at t 0 , we
have x 0 0 sin ( 2) 0 c c 0 . So at t 3 seconds, the position of the particle is
x sin(3 2) . From the graph of x sin t between t 0 and t 3 2 , the distance of the particle from the origin is 3m and displacement of the paricle is –1m, which is of magnitude 1m. So the difference between the distance covered by the particle and the magnitude of displacement from the origin is 3 1 2 . Example 3.28 [ME-2014 (1 mark)]: The solution of the initial value problem dy dx 2 xy ; y (0) 2 is (a) 1 e x
2
(b) 2e x
2
(c) 1 e x
2
(d) 2e x
2
Solution (b): dy dx 2 xy (1 y )dy 2 xdx . Now integrating both sides (1 y ) dy 2 xdx 2
ln y x 2 ln c y ce x . So from the given options we can say that option (b) is in the form of our result. [Similar questions were also asked in EC-2008, EE-2011 (1 mark), MT-2010 (2 marks)] Example 3.29 [MN-2014 (1 mark)]: Solution of the differential equation dy dx ky follows exponential decay (where k is a constant) for x [0, ] if (a) k 0 (b) k 0 (c) k 0 (d) k e Solution (b): dy dx ky (1 y )dy kdx . Now integrating both sides (1 y ) dy k dx ln y kx ln c y ce kx . So we have an exponential decay if kx 0 k 0 as x 0
Equations reducible to variable separable form
Differential equations of the form dy dx f (ax by c ) : Differential equations of the form dy dx f ( ax by c ) can be reduced to variable separable form by the substitution ax by c t a b( dy dx ) dt dx (dt dx ) a (1 b ) f (t ) (dt dx ) a bf (t ) 1 {a bf (t )} dt dx . Now integrating both sides we get the required equation.
Differential equation of the form dy dx ( ax by c ) ( Ax By C ) : Differential equation of
the
form
dy dx ( ax by c) ( Ax By C ) ,
where
a Ab Bk
(say),
dy dx {k ( Ax By ) c} ( Ax By C ) can be reduced to variable separable form by putting Ax By t A B (dy dx ) ( dt dx ) dy dx (1 B ) (dt dx ) A (kt c ) (t C ) dt dx A B (kt c ) (t C ) 1 A B ( kt c) (t C ) dt dx . Now integrating both sides we get the required equation. Example 3.30 [ME-2014 (2 marks)]: The general solution of the differential equation dy dx cos( x y ) , with c as a constant, is (a) y sin( x y ) x c (b) tan ( x y ) 2 y c (c) cos ( x y ) 2 x c
(d) tan ( x y ) 2 x c
Solution (d): Let x y t 1 ( dy dx ) dt dx dy dx ( dt dx ) 1 . So dy dx cos( x y )
dt dx 1 cos t dt dx 1 cos t 2 cos 2 (t 2) sec 2 (t 2) dt 2dx . Now integrating both sides we get,
sec
2
(t 2) dt 2 dx 2 tan(t 2) 2 x c tan ( x y ) 2 x c , where c c 2 .
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Engineering Mathematics
Chapter 3: Differential Equation
[3.9]
Solution of Homogenoeus Differential Equation: A function f ( x, y ) is called a homogeneous function of degree n if
f ( x, y ) n f ( x, y ) . For example,
f ( x, y ) x 2 y 2 3xy is a
homogeneous function of degree 2, because f ( x , y ) 2 x 2 2 y 2 3( x )( y ) 2 f ( x, y ) . A homogeneous function f ( x, y ) of degree n can always be written as
f ( x, y ) x n f ( y x) or
f ( x , y ) y n f ( x y ) ; thus a differential equation of the form dy dx f ( x, y ) is homogeneous if the
function f ( x, y ) depends only on the ratio x y or y x . [This point was asked in ME-1995 (1 mark)]. If a first-order first degree differential equation is expressible in the form dy dx f ( x, y ) g ( x, y) where f ( x, y ) and g ( x, y ) are homogeneous functions of the same degree, then it is called a homogeneous differential equation. Such type of equations can be reduced to variable separable form by the substitution y v x . The algorithm for solving homogeneous differential equation is given as: 1. Put the differential equation in the form dy dx ( x, y) ( x, y) 2. Put y v x dy dx v x ( dv dx ) in the equation in step 1 and cancel out x from the right hand side. The equation reduces to the form v x (dv dx ) F (v ) . 3. Shift v on RHS and separate the variables v and x 4. Integrate both sides to obtain the solution in terms of v and x 5. Replace v by y x in the solution obtained in step 4 to obtain the solution in terms of x and y . Equation reducible to homogeneous form: A first order, first degree non – homogeneous differential equation is of the form dy dx ( a1 x b1 y c1 ) ( a2 x b2 y c2 ) , a1 a2 b1 b2 (3.10) It can be reduced to homogeneous form by certain substitutions. Put x X h, y Y k , where h and k are constants, which are to be determined. Since, dy dx ( dy dY )(dY dX )( dX dx ) dY dX . Substituting this value in Eq. 3.12, we have, dY dX (a1 X b1Y ) a1h b1k c1 ( a2 X b2Y ) a2 h b2 k c2 (3.11) Now
h,
k
a1h b1k c1 0
will be chosen such that
and
a2 h b2 k c2 0 ,
i.e.,
h (b1c2 b2 c1 ) k (c1a2 c2 a1 ) 1 ( a1b2 a2b1 ) . For these values of h and k the Eq. 3.13 reduces to dY dX ( a1 X b1Y ) (a2 X b2Y ) which is a homogeneous differential equation and can be solved by the substitution Y v X . Replacing X and Y in the solution so obtained by x h and y k respectively, we can obtain the required solution in terms of x and y .
Example 3.31: Find the solution of dy dx ( x 3 y 2) (3 x y 6) . Solution: Given equation is non-homogeneous. Let x X h and y Y k dy dx dY dX
dY ( X h) 3(Y k ) 2 X 3Y (h 3k 2) . Let us select h and k dX 3( X h) (Y k ) 6 3 X Y (3h k 6)
so that
h 3k 2 0 and 3h k 6 0 k 0, h 2 . X x h x 2 and , Y y k y . Hence, dY dX ( X 3Y ) (3 X Y ) , which is homogeneous. Now, let Y v X dY dv X 3Y dv 1 3(Y / X ) dv 1 3v dv v X v X v X v X dX dX 3X Y dX 3 (Y / X ) dX 3v dX X
dv
1 3v
v
v 2 6v 1
(3 v )dv 2
dX
2v 6 2
dv 2
dX
dX 3v 3v v 6v 1 X v 6v 1 X 2 2 2 Integrating, ln(v 6v 1) 2 ln X ln c ln(v 6v 1) ln X ln c X 2 (v 2 6v 1) c Y 2 6 XY X 2 c y 2 6( x 2) y ( x 2) 2 c
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Engineering Mathematics
Chapter 3: Differential Equation
[3.10]
Solution of Differential Equation of the form of Leibnitz’s Linear Equation: The general form of a linear differential equation of first order (known as Leibnitz’s linear equation) is given as, ( dy dx) P y Q (3.12) where P and Q are functions of x (or constants). [This point was asked in CE-1997 (1 mark)] For example, ( dy dx ) xy x3 , x (dy dx) 2 y x 3 , ( dy dx ) 2 y sin x etc. are linear differential equations. The following are the steps for solving Leibnitz’s Linear Equation: 1. Write the differential equation in the form ( dy dx ) Py Q and obtain P and Q . 2. Find integrating factor (I.F.) given by I.F. e 3. Multiply both sides of equation in step 1 by I.F., we get, Pdx
e
Pdx
( dy
dx ) Py Q e
Pdx
(d dx ){ y e
Pdx
} Q e
Pdx
y e
Pdx
Q e
Pdx
dx
4. Integrate both sides of the equation obtained in step 3 w.r.t. x to obtain y (I.F.) Q (I.F.) dx c . This gives the required solution.
Linear differential equations of the form ( dx dy ) Rx S : A linear differential equation of the form ( dx dy) Rx S , where R and S are functions of y or constants; y is independent variable and x is a dependent variable; can be solved by following algorithm: 1. Write the differential equation in the form ( dx dy) Rx S and obtain R and S . 2. Find I.F. by using I.F. e 3. Multiply both sides of the differential equation in step 1 by I.F. 4. Integrate both sides of the equation obtained in step 3 w.r.t. y to obtain the solution given by R dy
x (I.F.) S (I.F.) dy c where c is the constant of integration.
Differential equation of the form ( dy dx ) P ( y ) Q ( y ) : The differential equation of the form ( dy dx) P ( y ) Q ( y ) , where P and Q are functions of x alone or constants. Dividing by ( y ) , we get {1 ( y)}(dy dx) { ( y) ( y )}P Q . Now put ( y) ( y ) v , so that ( d dx ) ( y ) ( y ) dv dx
or dv dx k{1 ( y )}( dy dx) , where k is constant. We get
( dv dx) kP v kQ , which is linear differential equation.
Example 3.32 [CH-2007 (2 marks)]: The solution of the following differential equation x (dy dx ) y ( x 2 1) 2 x 3 is (c) c1 x c2 x 2 (b) 2 ce x 2 (d) 2 x cxe x 2 Solution (d): The given differential equation can be written as ( dy dx ) x (1 x ) y 2 x 2 …(i), 2
(a) 0
2
which is in Leibnitz’s linear form, in which
I.F. e ex
2
2
x (1 x ) dx 2
2
2
2 ln x )
3.33
2
2
[XE-2008 3
Q ( x) 2 x 2 . So
2
2
integrating both sides (e x Example
and
(e x 2 ) (eln x ) (e x 2 ) x ; multiplying both sides of (i) by the I.F.,
1 d ex 2 x2 2 x y 2 xe x dx x x dx x dy
ex
e( x
P( x ) x (1 x )
2
2
x ) y 2e x
(1
2
2
2
ex 2 2 y 2 xe x 2 d x
c y 2 x cxe x
mark)]:
If
the
2
solution
2
2
y 2 xe x 2 dx .
Now
. of
the
differential
equation
x2
( dy dx) P ( x ) y xy is y (1 ce ) 1 , c being any arbitrary constant, then P ( x ) is (a) x (c) x (d) 2x (b) x 2 Solution (c): Differentiating the given solution w.r.t. x , we get
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Engineering Mathematics 2
Chapter 3: Differential Equation 2
2 y ( dy dx)(1 ce x ) 2cxy 2e x 0 ( dy dx ) cxye x comparing
it
to
given
2
differential
2
[3.11]
2
2
(1 ce x ) 0 (dy dx ) cxy 3e x 0 . Now
equation
we
get,
2
cxy 3e x y{P( x ) xy 2 }
xy 2 (ce x ) P ( x) xy 2 xy 2 (1 y 2 ) 1 P ( x) xy 2 x xy 2 P ( x ) xy 2 P( x ) x Example 3.34 [IN-2010 (2 marks)]: Consider the differential equation ( dy dx ) y e x with y (0) 1 . The value of y (1) is (b) (e e 1 ) 2 (c) (e e 1 ) 2 (d) 2(e e 1 ) (a) e e 1 Solution (c): The given differential equation is of the form of Leibnitz’s linear equation. So ( dy dx ) y e x P( x ) 1 and Q ( x) e x ; so I.F. e
Pdx
e
1dx
e x . Now multiplying the given
differential equation with I.F. e x ( dy dx ) e x y e 2 x (d dx )( ye x ) e 2 x d ( ye x ) e 2 x dx ;
d ( ye
integrating both sides, we get condition x
we
get
x
) e 2 x dx ye x (e 2 x 2) c . Now applying the given
1 e 0 (e 20 2) c c 1 2 .
2x
2x
x
So
the
solution
is
1
2
ye (e 2) (1 2) y ( x ) (e 1) (2e ) . So y (1) (e 1) (2e) (e e ) 2 . [Similar questions were also asked in ME-2006 (1 mark), AG-2008 (2 marks)]
Example 3.35 [AG-2011 (2 marks)]: The solution of x( dy dx ) x 2 3 y for x 0 and y (1) 2 is (a) y x 3 3 x 2 (b) y x 2 3x 3 (c) y x 2 3x 3 (d) y x 2 3x 3 Solution (c): The given differential equation can be written as ( dy dx ) (3 x ) y x (i), which is in Leibnitz’s linear form, in which and Q( x) x . So P( x) 3 x
I.F. e 1 dy 3
x dx
given
Pdx
3 x
4
3 3 (1 x ) dx e e 3ln x e ln x 1 x 3 . Now multiplying both sides of (i) by the I.F., we get
y
1 x
2
condition,
d 1 1 1 y 1 1 3 y 2 d 3 y 2 dx 3 c . Now applying the dx x x x x x x
we
get
2 (13 ) 1 2 c c 3 .
So
y x 3 (1 x ) 3
y x 3 (1 x ) 3 y x 2 3x 3 . [Similar question was also asked in CE-2011 (2 marks)]
Example 3.36 [BT-2013 (2 marks)]: The solution to ( dy dx ) y cot x cosec x is (a) y (c x ) cot x (b) y (c x ) cosec x (c) y (c x) cosec x cot x (d) y (c x) cosec x cot x Solution (b): The given differential equation is in Leibnitz’s linear form, in which P ( x) cot x and
Q ( x ) cosec x . So I.F. e
cot xdx
e ln sin x sin x ; multiplying both sides of the given differential
equation by the I.F., sin x ( dy dx ) y cos x 1 ( d dx)( y sin x ) 1 d ( y sin x ) dx ; integrating both sides we get,
d ( y sin x) dx y sin x x c y ( x c) cos ecx .
Example 3.37 [CH-2014 (2 marks)]: The integrating factor for the differential equation ( dy dx ) { y (1 x )} (1 x ) is (c) x(1 x) (a) 1 (1 x ) (d) x (1 x ) (b) 1 x Solution (a): Comparing the given differential equation with Eq. 3.12, we have
e e 1 (1 x) P ( x ) 1 (1 x) , so the I.F. e [Similar questions were also asked in AG-2012 (1 mark), TF-2014 (2 marks)] {1 (1 x )}dx
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ln(1 x )
ln{1 (1 x )}
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Engineering Mathematics
Chapter 3: Differential Equation
[3.12]
Example 3.38 [CE-2014 (1 mark)]: The integrating factor for the differential equation ( dP dt ) k 2 P k1 L0 e k1t is (a) e k1t
(b) e k2 t
(c) e k1t
(d) e k2 t
Solution (d): Comparing the given differential equation with Eq. 3.12, I.F. e
k 2 dt
e k 2t
Example 3.39 [EC-2014 (1 mark)]: Which one of the following is a linear non-homogeneous differential equation, where x and y are the independent and dependent variables respectively? (a) ( dy dx) xy e x (b) ( dy dx ) xy 0 (c) ( dy dx ) xy e y (d) ( dy dx ) e y 0 Solution (a): Differential equation given in option (a) is a first order linear non-homogeneous equation; but differential equation given in option (b) is a first order linear homogeneous equation. The differential equations given in option (c) and (d) are first order non-linear equations.
Bernoulli's Differential equation: The differential equation of type given below Bernoulli’s differential equation: ( dy dx) Py Qy n (3.13) where P and Q are constants or functions of x alone and n is a constant other than zero or unity, can be reduced to the linear form by dividing by y n and then putting y n 1 v , as explained below. Dividing both sides of Eq. 3.15 by y n , we get y n ( dy dx) Py n 1 Q . Putting y n 1 v so that ( n 1) y n ( dy dx ) dv dx , we get {1 ( n 1)}(dv dx) Pv Q (dv dx ) (1 n ) Pv (1 n )Q which is a linear differential equation. [This conversion was asked in CE-2005 (2 marks)] If n 1, then we find that the variables in equation Eq. 3.13 are separable and it can be easily integrated by the method discussed in variable separable from.
Example 3.40 [CH-2008 (2 marks)]: Which ONE of the following transformation
u
f ( y )
reduces ( dy dx) Ay 3 By 0 to a linear differential equation? ( A and B are positive constants) (a) u y 3
(b) u y 2
(c) u y 1
(d) u y 2
Solution (b): The given differential equation ( dy dx ) By Ay 3 which is a Bernoulli’s DE in which n 3 , so put u y n 1 , i.e., u y 31 y 2 for reducing it to a linear DE.
3.1.2 Exact Differential equation If z f ( x, y ) is a function of two variables with continuous first partial derivatives in a region R of the xy plane, that is its differential is given as,
dz (f x) dx (f y )dy In the special case when f ( x, y ) constant then Eq. 3.14 gives,
(3.14)
(f x) dx (f y) dy 0 (3.15) A differential expression M ( x , y ) dx N ( x , y ) dy is an exact differential in a region R of the xy plane if it corresponds to the differential of some function f ( x, y ) defined in R . A first order differential equation of the form M ( x, y ) dx N ( x, y ) dy 0 is said to be an exact equation if the expression on the left hand side is an exact differential.
Criterion for an Exact Differential: Let M ( x, y ) and N ( x, y ) be continuous and have continuous first partial derivatives in a rectangular region R defined by a x b and c y d . Then a necessary and sufficient condition that M ( x , y ) dx N ( x , y ) dy be an exact differential is
M y N x (3.16) Proof: The equation M ( x, y )dx N ( x, y ) dy 0 is called exact when there exists a function f ( x, y ) of
x and
y
such that
d f ( x , y ) Mdx Ndy , i.e., (f x) dx (f y ) dy Mdx Ndy
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Engineering Mathematics
Chapter 3: Differential Equation
[3.13]
M f x and N f y M y 2 f (yx ) and N x 2 f (xy ) , where f x
Partial derivative of f ( x, y ) with respect to x (keeping y constant); and f y Partial derivative of f ( x, y ) with respect to y (treating x as constant). The equation Mdx Ndy 0 is exact if 2 f (yx) 2 f (xy ) M y N x . [This point was asked in CE-1997 (1 mark)]
Working rule for solving an exact differential equation 1. Compare the given equation with Mdx Ndy 0 and find out M and N . Then determine whether the equality in Eq. 3.16 holds. If it does, then there exists a function f for which f x M ( x, y ) 2. We can find f by integrating M ( x, y ) with respect to x while keeping y constant as
f ( x, y ) M ( x , y ) dx g ( y )
(3.17) (3.18)
where the arbitrary function g ( y ) is the constant of integration. 3. Now differentiate Eq. 3.18 with respect to
y
and assume that
f y N ( x, y )
f y ( y ) M ( x, y ) dx g ( y ) N ( x , y ) . This gives, g ( y ) N ( x, y ) ( y ) M ( x, y ) dx
It
is
important
to
realize
that
Eq.
3.19
is
(3.19)
independent
of
x
because
N N M N ( x, y ) M ( x, y ) dx M ( x, y ) dx 0. x y x y x y x
4. Now integrate Eq. 3.19 with respect to y and substitute the result in Eq. 3.18. The implicit solution of the equation is f ( x, y ) c . Example 3.41 [MN-2007 (2 marks)]: The solution of ye x dx (4 y e x ) dy 0 for y (0) 1 is (a) ye x 2 y 2 1 0
(b) e x y 2 x 2 0
(c) ye x y 2 0
(d) xe x y 2 1 0
Solution (a): With M ( x, y ) ye x and N ( x, y ) (4 y e x ) , we have M y e x N x . Thus the given differential equation is exact, and so there exist a function f ( x, y ) such that f x M ( x, y ) ye x and f y N ( x, y ) (4 y e x ) . Now after integrating f x M ye x
w.r.t. x , we get f ( x, y ) ye x g ( y ) . Now taking partial derivative w.r.t. y and setting the result equal
to
N ( x, y )
gives,
N ( x, y ) f y e x g ( y ) 4 y e x e x g ( y ) g ( y ) 4 y
g ( y ) 2 y 2 . Hence f ( x, y ) ye x 2 y 2 , so the solution of the equation in implicit form is ye x 2 y 2 c . Now applying the given condition we get, 1 e 0 2 ( 1) 2 c c 1 . So the
solution of the given differential equation is given as ye x 2 y 2 1 0 .
Solution by inspection: If we can write the differential equation in the form f f1 ( x, y ) d f1 ( x, y ) f 2 ( x, y ) d f 2 ( x, y ) 0 , then each term can be easily integrated separately. For this the following results must be memorized. d ( x y ) dx dy d ( xy ) xdy ydx
d ( x y ) ( ydx xdy ) y 2
d ( y x ) ( xdy ydx ) x 2
d ( x 2 y ) (2 xydx x 2 dy ) y 2
d ( y 2 x ) (2 xydy y 2 dx ) x 2
d ( x 2 y 2 ) (2 xy 2 dx 2 x 2 ydy ) y 4
d ( y 2 x 2 ) (2 x 2 ydy 2 xy 2 dx) x 4
d tan 1 ( x y ) ( ydx xdy ) ( x 2 y 2 )
d tan 1 ( y x ) ( xdy ydx) ( x 2 y 2 )
d [ln( xy )] ( xdy ydx) ( xy)
d ln( x y ) ( ydx xdy ) ( xy )
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Engineering Mathematics
Chapter 3: Differential Equation
[3.14]
d (1 2) ln( x 2 y 2 ) ( xdx ydy ) ( x 2 y 2 )
d ln( y x ) ( xdy ydx) ( xy )
d (e y x ) ( xe y dy e y dx ) x 2
d (e x y ) ( ye x dx e x dy ) y 2
d ( x 2 y 2 ) ( xdx y dy )
d ( x m y n ) x m1 y n 1 ( mydx nxdy )
{1 (1 n)}d [ f ( x, y )]1 n f ( x, y ) ( f ( x, y )) n
d{1 ( xy )} ( xdy ydx ) ( x 2 y 2 )
d (1 2) log{( x y ) ( x y )} ( x dy y dx ) ( x 2 y 2 )
x2 y 2
Example 3.42 [CH-2008 (1 mark)]: Which ONE of the following is NOT an integrating factor for the differential equation xdy ydx 0 ? (a) 1 x 2
(b) 1 y 2
(d) 1 ( x y )
(c) 1 ( xy)
Solution (d): For option (a): ( xdy ydx ) x 2 0 d ( y x ) 0 which can be integrated easily For option (b): ( xdy ydx ) y 2 0 d ( x y ) 0 which can be integrated easily For option (c): ( xdy ydx) ( xy) 0 d ln( y x) 0 which can be integrated easily For option (d): ( xdy ydx) ( x y) 0
we have not any function
f ( x, y )
such that
d f ( x, y ) 0 , so the IF given in option (d) is not correct for the given differential equation.
Example 3.43 [EC-2012, EE-2012, IN-2012 (1 mark)]: With initial condition x(1) 0.5 , the solution of the differential equation, t ( dx dt ) x t is (b) x t 2 (1 2)
(a) x t (1 2)
(c) x t 2 2
(d) x t 2
Solution: The given differential equation can be written as ( d dt )(t x ) t d (t x ) t dt . Integrating both sides d (tx) t dt tx (1 2)t 2 c . Now applying the given condition, we get, 1 0.5 (12 2) c c 0 . So the solution is t x t 2 2 x t 2 . [Similar question was also asked in ME-2009 (2 marks)]
Example 3.44 [ME-2005 (2 marks)]: If x 2 (dy dx ) 2 xy (2 ln x ) x , and y (1) 0 , then y (e) ? (b) 1 (a) e (c) 1 e (d) 1 e 2 Solution (d):
x 2 (dy dx ) 2 xy (2 ln x ) x (d dx )( x 2 y ) (2 ln x ) x d ( x 2 y ) {(2 ln x ) x}dx .
Integrating both sides we get
d (x
2
y ) 2 {(ln x ) x}dx x 2 y (ln x ) 2 c . Now applying the given
condition we get, 12 0 (ln1) 2 c c 0 . So the solution of the given differential equation is x 2 y (ln x ) 2 . Now at x e , we have e 2 y (ln e) 2 y e 2 .
[Similar question was also asked in CE-2006 (2 marks)] Example 3.45: Find the solution of ( x 2 4 xy 2 y 2 ) dx ( y 2 4 xy 2 x 2 ) dy 0 . Solution: Comparing given equation with Mdx Ndy 0 , We get, M x 2 4 xy 2 y 2 , N y 2 4 xy 2 x 2 M y 4 x 4 y and N x 4 y 4 x . So the given differential equation is exact. Integrating M w.r.t. x , treating y as constant, Mdx ( x 2 4 xy 2 y 2 )dx (1 3) x 3 2 x 2 y 2 y 2 x . Now integrating N w.r.t. y , treating x as
constant, Ndy ( y 2 4 xy 2 x 2 )dy (1 3) y 3 2 xy 2 2 x 2 y (1 3) y 3 ; (omitting 2 xy 2 2 x 2 y which already occur
in
Mdx ).
Solution of
the
given
differential
equation is
(1 3) x 3 2 x 2 y 2 xy 2 (1 3) y 3 x 3 y 3 6 xy ( x y ) 3 . x 3 y 3 6 xy ( x y ) c , 3 c
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Engineering Mathematics
Chapter 3: Differential Equation
[3.15]
Solution of Non – exact Differential equation: In previous section, for solving linear differential equation we multiply the equation by integrating factor. The same basic idea works for a non-exact differential equation M ( x, y ) dx N ( x, y ) dy 0 ; i.e., it is possible to find an integrating factor ( x, y ) so that after multiplying it by the non-exact differential equation, the resulting equation becomes an exact differential equation, i.e., ( x, y ) M ( x, y ) dx ( x, y ) N ( x, y ) dy 0 (3.20) is an exact differential equation. In an attempt to find ( x, y ) , we turn to the criterion (Eq. 3.18) for exactness. Eq. 3.20 is exact if and only if ( M ) y ( N ) x , where the subscripts denote partial derivatives. By the product rule of differentiation the criterion for exactness becomes, M y y M N x x N x N y M (M y N x )
(3.21)
For determining the only unknown ( x, y ) , we must solve a partial differential equation. Since we are not prepared to do that, we make a simplifying assumption. Suppose is a function of one variable; for example, say that depends only on x . In this case, x d dx and y 0 , so that Eq. 3.21 can be written as,
d dx ( M y N x ) N
(3.22)
We are still at an impasse if the quotient ( M y N x ) N depends on both x and y . However, if after all obvious algebraic simplifications are made, the quotient ( M y N x ) N turns out to depend solely on the variable x , then Eq. 3.22 is a first-order ordinary differential equation. We can finally determine because Eq. 3.22 is separable as well as linear, as ( x) e it follows from Eq. 3.21 that if depends only on the variable y , then
( M y N x ) N dx
. In like manner,
d dy ( N x M y ) M
(3.23)
In this case, if ( N x M y ) M is a function of y only, then we can solve Eq. 3.23 for . We now summarize the results for the non-exact differential equation M ( x, y ) dx N ( x, y ) dy 0
(3.24)
If ( M y N x ) N is a function of x alone, then an integrating factor for Eq. 3.24 is
( M y N x ) N dx ( x) e If ( N x M y ) M is a function of y alone, then an integrating factor for Eq. 3.24 is ( y) e
(3.25)
( N x M y ) M dy
(3.26)
Example 3.46 [XE-2014 (1 mark)]: An integrating factor of the differential equation (3 x 2 y 3e y y 3 y 2 ) dx ( x3 y 3 e y xy ) dy 0 is (a) 1 y (b) 1 y 2 (c) 1 y 3 Solution (c): The given differential equation is of the form 2 3 y
3
2
2
3 3 y
(d) ln y
Mdx Ndy 0 , where
2 y
2
3 y
2
M 3 x y e y y and N x y e xy . So, M y M y 9 x y e 3 x y e 3 y 2 y and 2
2 y
2
3 3 y
N x N x 3 x 2 y 3 e y y . As ( M y N x ) N (9 x y e 3 y 3 y ) ( x y e xy ) is not a function
of x alone so the integrating factor of the given differential equation is not ( x) e 2
2 y
2
2
3 y
3
( M y N x ) N dx
.
2
Also ( N x M y ) M (9 x y e 3 y 3 y ) (3 x y e y y ) 3 y which is a function of y only so ( y ) e
( N x M y ) M dy
is the integrating factor of the given differential equation. Thus
3 3 (1 y ) dy ( y) e e 3ln y e ln y y 3 .
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Engineering Mathematics
Chapter 3: Differential Equation
[3.16]
3.1.3 Solution of Special type of second order differential equation A special type of second order differential equation is of the form d 2 y dx 2 f ( x )
(3.27)
n
Eq 3.27 may be re-written as ( d dx)( dy dx) f ( x) d ( dy dx) f ( x) dx . On integrating we get, ( dy dx ) f ( x ) dx c1 ( dy dx ) F ( x) c1 , F ( x ) f ( x ) dx c1dx
(3.28)
Eq. 3.28 can be written as, dy f ( x) dx c1dx and on integrating it we get, y F ( x) dx c1 x c2 H ( x ) c1 x c2 , H ( x) F ( x )dx
(3.29)
where, c1 , c2 are arbitrary constants. Example 3.47 [CE-2001 (2 marks)]: The solution of the differential equation d 2 y dx 2 3 x 2 with boundary conditions y (0) 2 and y (1) 3 is (a) y (1 3) x 3 (1 2) x 2 3x 6
(b) y 3 x 3 (1 2) x 2 5 x 2
(c) y (1 2) x 3 x 2 (1 2)5 x 2
(d) y x 3 (1 2) x 2 5 x (3 2)
Solution (c): d 2 y dx 2 3 x 2 ( d dx)( dy dx ) 3x 2 d ( dy dx) (3x 2)dx . Integrating both
d (dy
sides we get
y (1) 3
condition
dx ) (3 x 2)dx dy dx (3 2) x 2 2 x c1 . Now applying the given
to
our
result,
we
3 (3 2)12 2 1 c1 c1 5 2
get
dy dx (3 2) x 2 2 x (5 2) dy (3 2) x 2 2 x (5 2) dx . Again integrating both sides we
get,
dy (3 2) x
2
2 x (5 2) dx y (1 2) x3 x 2 (5 2) x c2 . Again applying the given
condition y (0) 2 , we get 2 (0 3 2) 0 2 (5 2) 0 c2 c2 2 . Thus our final solution is y (1 2) x3 x 2 (5 2) x 2 [Similar question was also asked in PI-2009 (2 marks)]
Example 3.48 [ME-2014 (2 marks)]: If y f ( x) is the solution of d 2 y dx 2 0 with boundary conditions y 5 at x 0 , and dy dx 2 at x 10 , f (15) ……………. Solution: d 2 y dx 2 0 ( d dx)( dy dx) 0 d (dy dx ) 0dx . Integrating both sides we get
d (dy
dx ) 0dx dy dx c1 . Now applying the given condition dy dx 2 at x 10 to our
dy 2dx
result, we get 2 c1 dy dx 2 dy 2dx . Again integrating both sides we get,
y 2 x c2 . Again applying the given condition y 5 at x 0 to our result, we get 5 2 0 c2 c2 5 . Thus, y f ( x ) 2 x 5 f (15) 2 15 5 35 .
Example
3.49
[MA-2014
(1
mark)]:
The
solution
of
the
integral
equation
x
( x) x sin( x ) ( ) d is 0
2
(a) x (1 3) x 3 Solution
(c):
(b) x (1 3!) x 3 (d dx )
b( x )
a(x)
(c) x (1 3!) x 3
(d) x 2 (1 3!) x 3
f ( x, t ) dt f x, b( x ) b( x ) f x, a( x ) a ( x )
b( x)
a( x)
f x ( x, t ) dt
which is the Leibniz integral rule. So applying it in the given equation we get x
( d dx ) ( x) 1 sin( x x) ( x ) ( d dx) x sin( x 0) (0) (d dx )(0) cos( x ) ( ) d 0
x
( x ) 1 cos( x ) ( ) d . Again applying th Leibniz integral rule to our result we get 0
(d
2
2
x
dx ) ( x ) 0 cos( x x ) ( x ) ( d dx )( x ) cos( x 0) (0) ( d dx )(0) sin( x ) ( ) d
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0
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Engineering Mathematics
Chapter 3: Differential Equation
[3.17]
x
( x ) ( x ) 0 sin( x ) ( ) d x d ( d dx ) ( x ) xdx 0
d (d we get Now
integrating
it
we
get
dx) ( x) xdx (d dx) ( x) (1 2) x 2 c d ( x) (1 2) x 2 c dx ; again integrating it
2 3 2 d ( x) (1 2) x c dx ( x) {1 (2 3)}x cx c ( x) (1 2) x c . (0) 0 sin( x ) ( ) d 0 0 0 ; also (0) 1 cos( x ) ( ) d 1 0 1 ; so 0
0
0
0
applying the conditions in our result, we get
(0) 0 0 0 c 0 c c 0
3
and
3
(0) 1 1 0 c c 1 . Thus ( x) {1 (2 3)}x x (1 3!) x x . Exercise: 3.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The order and degree of the differential equation ( d 2 y dx 2 ) ( dy dx)1 3 x1 4 0 , are respectively (a) 2, 3 (b) 3, 3 (c) 2, 6 (d) 2, 4 2. The degree of the differential equation ( d 2 y dx 2 ) 3( dy dx) 2 x 2 log( d 2 y dx 2 ) is (a) 1 (b) 2 (c) 3 (d) None of these 3. The order of the differential equation of all circles of radius r , having center on y axis and passing through the origin is _____. 4. The order and degree of d 2 y dx 2 sin( dy dx) x is (a) 2, 1 (b) Order 2, Degree not defined (c) 2, 0 (d) None of these 5. The differential equation of all parabolas whose axes are parallel to y axis is d3y
(b)
c
d3y
d2y
3
2
(c)
3
2
0
d2y
(d)
2
2
dy
c dy dx dx dx dx dx 6. The differential equation of family of curves whose tangent form an angle of 4 with the (a)
0
d 2x
hyperbola xy c 2 is (a)
dy
x 2 c2
(b)
dy
x2 c 2
(c)
dy
c2
dx x 2 c 2 dx x 2 c 2 dx x2 7. The solution of the differential equation (1 x 2 )( dy dx ) x(1 y 2 ) is
(d) None of these
(a) 2 tan 1 y log(1 x 2 ) c
(b) tan 1 y log(1 x 2 ) c
(c) 2 tan 1 y log(1 x 2 ) c 0
(d) None of these
8. Solution of the differential equation dy dx ( x y 7) (2 x 2 y 3) is (a) 6( x y ) 11log(3 x 3 y 10) 9 x c (b) 6( x y ) 11log(3x 3 y 10) 9 x c (c) 6( x y) 11log{x 3 (10 3)} 9 x c
(d) None of these
9. The solution of the differential equation x(dy dx) y(log y log x 1) is (a) y xe cx (b) y xecx 0 (c) y e x 0 10. The general solution of the differential equation ( x y ) dx xdy 0 is (a) x 2 y 2 c
(b) 2x 2 y 2 c
(c) x 2 2 xy c
(d) None of these (d) y 2 2 xy c
11. Solution of y (2 xy e x ) dx e x dy is (a) yx 2 e x cy
(c) xy 2 e x c
(b) xy 2 e x cx 2
12. The solution of the differential equation (1 y ) ( x e (a) x ke tan
1
y
(b) 2 xe tan
1
y
e
2 tan
Copyright © 2016 by Kaushlendra Kumar
1
y
k
tan
1
(c) xe
y
tan
(d) None of these
)( dy dx ) 0 is
1
y
k tan 1 y
(d) None of these
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Chapter 3: Differential Equation
[3.18]
13. The solution of ( dz dx ) ( z x ) log z ( z x 2 )(log z ) 2 is (a)
x log z
2 x 2c
(b)
x log z
2 x 2c
(c)
x log z
x 2c
(d)
x log z
1, 4
14. The slope of the tangent at ( x, y ) to a curve passing through
1 2
cx 2
as given by
( y x ) cos 2 ( y x ) , then the equation of the curve is
(a) y tan 1 log(e x )
(b) y x tan 1 log( x e)
(c) y x tan 1 log(e x )
(d) None of these
15. The solution of the equation x 2 ( d 2 y dx 2 ) ln x when x 1 , y 0 and dy dx 1 is 1 1 1 1 (a) (ln x) 2 ln x (b) (ln x) 2 ln x (c) (ln x ) 2 ln x (d) (ln x ) 2 ln x 2 2 2 2 2 16. A continuously differentiable function ( x ) in (0, ) satisfying y 1 y , y (0) 0 y ( ) is (a) tan x 17. The order
of
(b) x( x ) the differential
y (c1 c2 ) cos( x c3 ) c4e
x c5
(c) ( x )(1 e x ) equation whose general
(d) None of these solution is given by
where c1 , c2 , c3 , c4 , c5 are arbitrary constants, is _____.
18. If y (t ) is a solution of (1 t )(dy dt ) ty 1 and y (0) 1 , then y (1) is equal to (a) 1 2
(b) e 1 2
(c) e 1 2
(d) 1 2
Pdx 19. If integrating factor of x (1 x 2 )dy (2 x 2 y y ax 3 ) dx 0 is e , then P
(a) (2 x 2 ax 3 ) { x(1 x 2 )}
(b) (2 x 2 1)
(c) (2 x 2 1) ( ax 3 )
(d) (2 x 2 1) {x(1 x 2 )}
20. The equation of the curve passes through the point (1,1) and whose slope is given by 2 y x , is (a) y x 2
(b) x 2 y 2 0
(c) 2 x 2 y 2 3
(d) None of these
21. The solution of the equation d 2 y dx 2 e 2 x (a) (1 4)e 2 x y
(b) (1 4)e 2 x cx d y
(c) (1 4)e 2 x cx 2 d y
(d) (1 4)e 2 x c d y
22. If x 2 y 2 1 then, where y dy dx , y d 2 y dx 2 (a) yy 2( y ) 2 1 0
(b) yy 2( y ) 2 1 0
(c) yy 2( y ) 2 1 0
(d) yy 2( y ) 2 1 0
23. If ( x) { ( x)}2 dx and (1) 0 then ( x ) (a) {2( x 1)}1 4
(b) {5( x 2)}1 5
(c) {3( x 1)}1 3
(d) None of these
24. Solution of the differential equation (sin y)(dy dx) cos y (1 x cos y ) is (a) sec y x 1 ce x
(b) sec y x 1 ce x
(c) y x e x c
(d) None of these
2
25. A solution of the differential equation ( dy dx ) x (dy dx ) y 0 is (a) y 2
(b) y 2 x
(c) y 2 x 4
(d) y 2 x 2 4
26. The general solution of the differential equation ( dy dx) sin{( x y ) 2} sin{( x y ) 2} is (a) log tan( y 2) c 2 sin x
(b) log tan( y 2) c 2 sin x
(c) log tan{( y 2) ( 4)} c 2 sin x
(d) log tan{( y 4) ( 4)} c 2 sin( x 2) 3
2
27. The general solution of IVP y (4 x ) y x y , y (2) 1 , x 0 is (a) y ( x ) 16 [ x 4 {1 16 ln( x 2)}]
(b) y ( x ) 16 [ x 4 {1 16 ln( x 2)}]
(c) y ( x ) 16 [ x 2 {1 16 ln( x 2)}]
(d) y ( x ) 16 [ x 4 (1 16 ln x )]
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Engineering Mathematics
3.2
Chapter 3: Differential Equation
[3.19]
Higher Order Differential Equation
3.2.1 Preliminary theory: Linear Equations Initial Value Problem (IVP): For a linear DE, an nth order initial value problem is given as, Solve: Subject to:
an ( x )( d n y dx n ) an 1 ( x)( d n 1 y dx n 1 ) a1 ( x )( dy dx ) a0 ( x) y g ( x)
(3.30)
y ( x0 ) y0 , y ( x0 ) y1 , , y ( n 1) ( x0 ) yn 1
(3.31)
Existence of a unique solution: Let an ( x ), an 1 ( x), , a1 ( x ), a0 ( x ) and g ( x ) be continuous on an interval I and let an ( x) 0 for every x in this interval. If x x0 is any point in this interval then a solution y ( x ) of the initial value problem (Eq. 3.30) exists on the interval and is unique. For e.g., the IVP 3 y 5 y y 7 y 0 , y (1) 0 , y(1) 0 and y (1) 0 possesses the trivial solution y 0 . Because the third order equation is linear with constant coefficients, it follows that all the conditions are fulfilled. Hence y 0 is the only solution on any interval containing x 1 .
Boundary value problem (BVP): Another type of problem consists of solving a linear differential equation of order two or greater in which the dependent variable y or its derivatives are specified at different points. A problem such as, Solve: a2 ( x)( d 2 y dx 2 ) a1 ( x )( dy dx) a0 ( x ) y g ( x) Subject to:
y (a ) y0 , y (b) y1
(3.32) (3.33)
is called a boundary value problem (BVP). The prescribed values y ( a ) y0 and y (b) y1 are called boundary conditions. A solution of these problems is a function satisfying the differential equation on some interval I , containing a and b , whose graph passes through the two points (a, y0 ) and (b, y1 ) . For a 2nd order differential equation other pairs of boundary conditions could be y (a ) y0 , y (b) y1 ; y ( a) y0 , y(b) y1 ; y (a ) y0 , y (b) y1 , where y 0 and y1 denote arbitrary constants.
A BVP can have Many, One, or No solutions even if it satisfy all the condition of Existence of a unique solution. For e.g., the two-parameter family of solutions of the differential equation x 16 x 0 is x c1 cos 4t c2 sin 4t . (a) Suppose we wish to determine the solution of the equation that further satisfies the boundary conditions x(0) 0, x( 2) 0 . x (0) 0 c1 0 and x( 2) 0 c2 . Hence the boundary-value problem x 16 x 0, x (0) 0, x( 2) 0 has infinitely many solutions. (b) If the BVP in (a) is changed to x 16 x 0, x (0) 0, x( 8) 0 then x (0) 0 c1 0 , but x ( 8) 0 c2 0 . Hence the BVP has a unique solution, i.e., x 0 . (c) If we change the BVP to x 16 x 0, x (0) 0, x( 2) 1 then x (0) 0 c1 0 but x ( 2) 1 x c2 sin 2 1 c2 0 c2 , so the BVP has no solution.
Differential Operators: In calculus differentiation is often denoted by the capital letter D , i.e., dy dx Dy . The symbol D is called a differential operator because it transforms a differentiable function ( y ) into another function. Higher-order derivatives can be expressed in terms of D in a natural manner: ( d dx )( dy dx) (d 2 y dx 2 ) D( Dy ) D 2 y and in general, d n y dx n D n y .
Homogeneous and Non – Homogeneous Equations: If an ( x), an 1 ( x ), , a1 ( x ), a0 ( x ) and g ( x ) are continuous and an ( x) 0 then, a linear nth - order differential equation of the form, an ( x )( d n y dx n ) an 1 ( x)( d n 1 y dx n 1 ) a1 ( x )( dy dx) a0 ( x ) y 0 is said to be homogeneous equation; whereas the equation of the form
Copyright © 2016 by Kaushlendra Kumar
(3.34)
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.20]
an ( x )( d n y dx n ) an 1 ( x )(d n 1 y dx n 1 ) a1 ( x )(dy dx ) a0 ( x ) y g ( x ) (3.35) is said to be non – homogeneous equation if g ( x ) 0 . For e.g. 2 y 3 y 5 y 0 is a
homogeneous linear 2nd order differential equation, whereas x 3 y 6 y 10 y e x is a non – homogeneous linear 3rd order differential equation. Solution of Homogeneous Equations Superposition Principle for Homogeneous Equations: Let y1 , y2 , , yk be solutions of the homogeneous nth -order differential Eq. 3.34 on an interval I . Then the linear combination y c1 y1 ( x ) c2 y2 ( x ) ck yk ( x ) , where c1 , c2 , , ck are constants, is also a solution on I . A constant multiple y c1 y1 ( x) of a solution y1 ( x ) of a homogeneous linear differential equation is also a solution. A homogeneous linear differential equation always possesses the trivial solution y 0 .
Linear Dependence and Independence: A set of functions f1 ( x ), f 2 ( x ), , fn ( x ) is said to be linearly dependent on an interval I if there exist constants c1 , c2 , , cn not all zero, such that c1 f1 ( x) c2 f 2 ( x) cn f n ( x ) 0 for every x in the interval. If the set of functions is not linearly dependent on the interval, it is said to be linearly independent. Let y1 , y2 , , yn be n solutions of the homogeneous linear nth order differential Eq. 3.34 on an interval I . Then the set of solutions is linearly independent on I if and only if W ( y1 , y2 , , yn ) 0 for every x in the interval; where
W ( y1 , y2 , , yn )
y1
y2
yn
y1
y2
yn
y1( n 1)
y2( n 1)
yn( n 1)
is called Wronskian of the functions
y1 , y2 , , yn if all these functions possesses at least n 1 derivatives.
When y1 , y2 , , yn are n solutions of Eq. 3.34 on an interval I , the Wronskian W ( y1 , y2 , , yn ) is either identically zero or never zero on the interval.
Any set y1 , y2 , , yn of n linearly independent solutions of the homogeneous linear nth order differential Eq. 3.34 on an interval I is said to be a fundamental set of solutions on I . General Solution of Homogeneous Equations: Let y1 , y2 , , yn be a fundamental set of solutions of the homogeneous linear nth -order differential Eq. 3.34 on an interval I . Then the general solution of the equation on the interval is y c1 y1 ( x ) c2 y2 ( x ) cn yn ( x ) , where c1 , c2 , , cn are n arbitrary constants.
Solution of Non – Homogeneous Equations: Let y p ( x ) be any particular solution of the non – homogeneous linear nth order differential Eq. 3.35 on an interval I , and let y1 , y2 , , yn be a fundamental set of solutions of the associated homogeneous differential Eq. 3.34 on I . Then the general solution of the equation on the interval is y c1 y1 ( x ) c2 y2 ( x ) cn yn ( x ) y p ( x ) , where c1 , c2 , , cn are n arbitrary constants.
Thus the general solution of a non – homogeneous linear equation consists of the sum of two functions y c1 y1 ( x ) c2 y2 ( x ) cn yn ( x ) y p ( x ) yc ( x ) y p ( x ) .
The linear combination yc ( x) c1 y1 ( x ) c2 y2 ( x ) cn yn ( x ) , which is the general solution of Eq. 3.34 is called the complementary function for Eq. 3.35
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.21]
In other words, to solve a nonhomogeneous linear differential equation, we first solve the associated homogeneous equation and then find any particular solution of the nonhomogeneous equation. The general solution of the nonhomogeneous equation is then y complementary function any particular solution yc y p .
Superposition Principle for Non – Homogeneous Equations: Let y p1 , y p2 , , y pk be k particular solutions of the non – homogeneous linear nth order differential Eq. 3.35 on an interval I corresponding, in turn, to k distinct functions g1 , g 2 , , g k . That is, suppose y pi denotes a particular solution of the corresponding differential equation an ( x )( d n y dx n ) an 1 ( x )( d n 1 y dx n 1 ) a1 ( x )( dy dx ) a0 ( x ) y g i ( x ) where i 1, 2, , k . Then, y p y p1 ( x ) y p2 ( x ) y pk ( x )
(3.36) (3.37)
is a particular solution of dny d n 1 y dy an ( x ) n an 1 ( x ) n 1 a1 ( x) a0 ( x ) y g1 ( x ) g 2 ( x ) g k ( x ) dx dx dx
(3.38)
3.2.2 Reduction of Order Since the general solution of a homogeneous linear second – order differential equation a2 ( x ) y a1 ( x ) y a0 ( x ) y 0
(3.39)
is a linear combination y c1 y1 c2 y2 , where y1 and y 2 are solutions that constitute a linearly independent set on some interval I . The basic idea described in this section is that equation 3.39 can be reduced to a linear first – order differential equation by means of substitution involving the known solution y1 . A second solution y 2 of Eq. 3.39 is apparent after this first – order differential equation is solved. Suppose that y1 denotes a nontrivial solution of Eq. 3.39 and that y1 is defined on an interval I . We seek a second solution y 2 so that the set consisting of y1 and y 2 is linearly independent on I . We know that, if y1 and y 2 are linearly independent, then their quotient y2 y1 is non-constant on I , i.e., y2 ( x ) y1 ( x ) u ( x ) y2 ( x ) u ( x ) y1 ( x ) . The function u ( x) can be found by substituting y2 ( x) u ( x ) y1 ( x ) into the given differential equation. This method is called reduction of order because we must solve a linear first-order differential equation to find u . Suppose we divide by a2 ( x ) to put equation 3.39 in standard form, y P ( x ) y Q( x ) y 0 (3.40) where, P ( x ) and Q( x ) are continuous on some interval I . Let us suppose further that y1 ( x ) is a known solution of Eq. 3.40 on I and that y1 ( x) 0 for every x in the interval. If we define y u ( x ) y1 ( x )
it
follows
y uy1 y1u y uy1 2 y1u y1u .
that
Hence,
y P ( x) y Q ( x ) y 0 uy1 2 y1u y1u P ( x )(uy1 y1u ) Q( x )uy1 0 . Let u w , then u ( y1 Py1 Qy1 ) y1u (2 y1 Py1 )u 0 y1u (2 y1 Py1 )u 0 y1w (2 y1 Py1 ) w 0 (1 w) dw 2( y1 y1 ) dx Pdx 0 (1 2) dw 2(1 y1 ) dy1 Pdx 0 ln w ln y12 Pdx c Pdx ln wy12 Pdx c wy12 c1e
(3.41)
Pdx 2 Now putting w u in Eq. 3.41, and integrate again we get, u c1 1 y1 e dx c2 . By
choosing c1 1 and c2 0 , we find from y u ( x ) y1 ( x ) that a second solution of equation 3.40 is P ( x ) dx y2 y1 ( x ) 1 y12 ( x ) e dx
Copyright © 2016 by Kaushlendra Kumar
(3.42)
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.22]
Note that, Eq. 3.42 which is a solution of Eq. 3.40, y1 and y 2 are linearly independent on any interval on which y1 is not zero. Example 3.50 [TF-2009 (2 marks)]: The general solution of the differential equation x 2 ( d 2 y dx 2 ) 2 x ( dy dx ) ( y x 2 ) 0 is (b) y c1e x c2 e x
(a) y c1 cos x c2 sin x
x (d) y (c1 c2 x)e
(c) y c1 (1 x )c2 e1 x
Solution (c): As the third term on the LHS of the given DE is y x 2 , so the general solutions given in options (a), (b) and (c) are not correct. Now we will check for the general solution given in option (c). If y c1 (1 x )c2 e1 x is the general solution of the given differential equation then y1 e1 x is its 2 1x solution. So y1 (1 x )e y1 e1 x {(1 2 x ) x 4 } ; substituting these in the given DE we get
LHS x2 e1 x {(1 2 x) x4 } 2 x( e1 x x 2 ) (e1 x x 2 ) (e1 x x2 )(1 2 x 2 x 1) 0 RHS y1 e1 x is a solution of the given differential equation. Hence we can say that option (c) is correct. Example 3.51 [XE-2009 (2 marks)]: Let y1 ( x ) and y2 ( x ) be two linearly independent solutions of ( d 2 y dx 2 ) (6 x)( dy dx ) q ( x) y 0 , x (1,3) , where q( x ) is a continuous function in (1,3) . If the
Wronskian of y1 ( x ) and y2 ( x ) at x 1 , denoted by w( y1 , y2 )(1) , is 1, then w( y1 , y2 )(2) is (a) 1 26
(b) 1 23
(c) 1 2
(d) 2
Solution (a): If y1 ( x ) and y2 ( x ) are two linearly independent solutions of y p ( x ) y q ( x ) y 0 , where p ( x) 1 x ; then we have y1 p ( x ) y1 q( x ) y1 0 …(i) and y2 p ( x) y2 q( x ) y2 0 …(ii). Now multiplying (i) by y 2 and (ii) by y1 , we get y2 y1 p ( x) y2 y1 q( x ) y2 y1 0 …(iii)
and
y1 y2 p ( x) y1 y2 q( x ) y1 y2 0 …(iv). So (iv) – (iii) y1 y2 y2 y1 p( x )( y1 y2 y2 y1 ) 0 . As W ( y1 , y2 ) y1 y2 y2 y1 W ( y1 , y2 ) y1 y2 y2 y1 . So we have W p ( x )W 0 W W p ( x) ; now integrating both sides w.r.t. x, we get 6 6 (W W )dx p( x)dx ln W (6 x)dx 6 ln x ln c ln(c x ) W c x . Now as
W y1 (1), y2 (1) 1 c 16 1 c 1 W y1 ( x), y2 ( x) 1 x 6 W y1 (2), y2 (2) 1 26 . Example 3.52 [MA-2014 (2 marks)]: If y1 ( x ) x is a solution to the differential equation (1 x 2 )( d 2 y dx 2 ) 2 x( dy dx ) 2 y 0 , then its general solution is
( x 2) ln 1 x 1
(b) y ( x ) c1 x c2 ln 1 x 1 x 1
(a) y ( x ) c1 x c2 x ln 1 x 2 1 (c) y ( x ) c1 x c2
(d) y ( x ) c1 x c2 ( x 2) ln 1 x 1 x 1
2
Solution (d): As the standard form of the given differential equation is ( d 2 y dx 2 ) {2 x (1 x 2 )}( dy dx) {2 (1 x 2 )} y 0 P ( x) 2 x (1 x 2 ) and Q ( x ) 2 (1 x 2 ) . As
P ( x)dx {2 x
2 P ( x ) dx (1 x 2 )}dx ln(1 x 2 ) e e ln(1 x ) 1 (1 x 2 ) . Given y1 x is the
solution of given equation, and let the second solution be Q ( x ) y2 so from Eq. 3.42, P ( x ) dx P ( x ) dx y2 x (1 x 2 )e dx . y2 x (1 x 2 )e dx x 1 x 2 (1 x 2 ) dx .
1
1 2
2
x (1 x )
dx
x
2
1 2(1 x )
1 1 1 1 1 1 x . dx ln 1 x ln 1 x ln 2(1 x ) x 2 2 x 2 1 x
y2 1 ( x 2) ln (1 x ) (1 x )
1
and so
y ( x ) c1 x c2 ( x 2) ln 1 x 1 x 1
is the
general solution.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.23]
3.2.3 Homogeneous Linear Equations with Constant Coefficients The first order homogeneous linear differential equations ay by 0 , where the coefficients a 0 and b are constants. This type of equation can be solved either by separation of variables or with the aid of an integrating factor, but there is another solution method, one that uses only algebra. Before illustrating this alternative method, we make one observation: Solving ay by 0 for y yields y ky , where k is a constant. This observation reveals the nature of the unknown solution y ; the only nontrivial elementary function whose derivative is a constant multiple of itself is an exponential function e mx . Now the new solution method: If we substitute y e mx y me mx into ay by 0 , we get e mx ( am b) 0 . Since e mx is never zero for real values of x , the last equation is satisfied only when m is a solution or root of the first-degree polynomial equation am b 0 . For this single value of m , y emx is a solution of the DE. In this section we will see that the foregoing procedure can produce exponential solutions for homogeneous linear higher-order DEs, an y ( n ) an 1 y ( n 1) a2 y a1 y a0 y 0 (3.43) where the coefficients an , an 1 , , a1 , a0 are real constants and an 0 .
Auxiliary Equation: We begin by considering the special case of the second order equation ay by cy 0
(3.44)
where a , b and c are constants. To find a solution of the form y emx , then after substitution of y me mx and y m 2 e mx , Eq. 3.44 becomes e mx (am 2 bm c ) 0 . e mx 0 for all x , hence 2 (3.45) am bm c 0 Eq. 3.45 is called the auxiliary equation of the differential equation 3.44. Since, the two roots of Eq.
2
3.45 are m1 , m2 (b b 4ac ) 2a there will be three forms of the general solution of Eq. 3.44 corresponding to the three cases: Case I: Distinct Real Roots: If b 2 4ac 0 then Eq. 3.45 has two real and distinct roots m1 and mx
m x
m2 , and we have two solutions y1 e 1 and y2 e 2 . We see that these functions are linearly independent on (, ) and hence form a fundamental set. It follows that the general solution of Eq. 3.44 on this interval is mx m x (3.46) y c1e 1 c2 e 2
Case II: Repeated Real Roots: If b 2 4ac 0 then Eq. 3.45 has two real and equal roots, i.e., m1 m2 m b 2a , we necessarily obtain only one exponential solution y1 e mx . From Eq. 3.42, the second solution of the equation 3.44 is given as, and ay by cy 0 y (b a ) y (c a ) y 0 P b a 2 m Qc a
y2 e mx e 2 mx e 2 mx dx e mx dx xe mx . Hence the general solution of Eq. 3.44 is y2 c1e
m1 x
c2 xe
m1x
(c1 c2 x )e
m1x
(3.47)
2
Case III: Conjugate Complex Roots: If b 4ac 0 then m1 and m2 are complex, hence we can write m1 , m2 i , where i 2 1 , , 0 and are real. Formally, there is no difference between this case and Case I, and hence y c1e ( i ) x c2 e ( i ) x is the general solution. However, in practice we prefer to work with real functions instead of complex exponentials by using Euler’s formula: ei cos i sin , where is any real number. It follows from this formula that ei x cos x i sin x and e i x cos x i sin x ei x e i x 2 cos x and ei x e i x 2i sin x . Since y c1e ( i ) x c2 e ( i ) x is a solution of Eq. 3.44 for any choice of
constants c1 and c2 , the choices c1 c2 1 and c1 1, c2 1 gives the two solutions:
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.24]
y1 e( i ) x e ( i ) x e x (e i x e i x ) 2e x cos x y2 e
( i ) x
e
( i ) x
x
e (e
i x
e
i x
) 2ie
x
and
sin x
Hence the last two results show that e x cos x and e x sin x are real solutions of Eq. 3.44; also these solution form a fundamental set on (, ) . Consequently, the general solution is y c1e x cos x c2 e x sin x e x (c1 cos x c2 sin x )
(3.48)
Example 3.53 [CS-1993 (1 mark)]: The differential equation y y 0 is subjected to boundary conditions: y (0) 0 , y ( ) 0 . If the equation has non-trivial solution(s), find the general value of Solution: As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 1) 0 , as e mt 0 so we have the auxiliary equation m 2 1 0 m 0 i thus we have conjugate complex root case so the solution is y e 0 (c1 cos x c2 sin x) c1 cos x c2 sin x . Now applying the given condition y (0) 0 0 c1 and y ( ) 0 0 c2 sin so for non trivial solution we must have c2 {0} and thus sin 0 n .
Example 3.54 [CE-2005 (2 marks)]: The solution of ( d 2 y dx 2 ) 2( dy dx ) 17 y 0 ; y (0) 1 , ( dy dx)
4
0 in the range 0 x 4 is given by
x
(b) e x cos 4 x (1 4) sin 4 x
cos 4 x (1 4) sin 4 x 4 x e cos x (1 4) sin x
(a) e
(c) (d) e 4 x cos 4 x (1 4) sin 4 x Solution (b): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx (m 2 2m 17) 0 so we have the auxiliary equation m 2 2m 17 0 as e mx 0 . As the discriminant of the auxiliary equation 2 D 2 4 1 17 64 0 so we have conjugate complex root case; also the roots of our auxiliary equation are m1 , m2 1 4i ; so 1 and 4 . So the general solution of given DE is y e x (c1 cos 4 x c2 sin 4 x ) . x
Now
we
have
to
x
find
y e (c1 cos 4 x c2 sin 4 x) e ( 4c1 sin 4 x 4c2 cos 4 x ) . 0
y(0) 1 1 e c1 cos(4 0) c2 sin(4 0) 1 c1 ;
the
constants
Applying also
again
the
c1
and
given
applying
0 e 4 (c1 cos c2 sin ) e 4 (4c1 sin 4c2 cos ) 0 c1 4c2 c2 1 4 .
c2 .
As
condition
y( 4) 0 So
the
general solution of given differential equation is y e x cos 4 x (1 4) sin 4 x . [Similar questions were also asked in ME-2000, CE-2008, EE-2014 (1 mark), CH-2010 (2 marks)] Statement for Linked Answer Question 3.55 & 3.56: The complete solution for the ordinary DE ( d 2 y dx 2 ) p ( dy dx ) qy 0 is y c1e x c2 e 3 x . Example 3.55 [ME-2005 (2 marks)]: p and q are (a) p 3, q 3 (b) p 3, q 4 (c) p 4, q 3 (d) p 4, q 4 Solution (c): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 pm q) 0 so we have the auxiliary equation m 2 pm q 0 as e mt 0 . Also, the given solution is of the case of distinct real roots, so the roots of the auxiliary equation are 1, 3 ; thus from the quxiliary equation sum of the roots is p 1 3 p p 4 ; and the product of roots is q 1 3 q q 3 .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.25]
Example 3.56 [ME-2005 (2 marks)]: Which of the following is a solution of the differential equation ( d 2 y dx 2 ) p (dy dx ) ( q 1) y 0 ? (a) e 3x (b) xe x (c) xe 2 x (d) x 2 e 2x Solution (c): From the result of above question we have p 4 and q 3 so the given DE becomes ( d 2 y dx 2 ) 4( dy dx ) 4 y 0 , which is homogeneous linear equation with constant coefficient so
putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 4m 4) 0 , as e mt 0 so we have the auxiliary equation m 2 4m 4 0 (m 2) 2 0 m 2 thus we have repeated real root case so the solution is y e 2x and y xe 2x and the general solution is y (c1 c2 x )e 2 x . Example 3.57 [EC-2006 (2 marks)]: For the differential equation ( d 2 y dx 2 ) k 2 y 0 , the boundary conditions are: (1) y 0 for x 0 ; (2) y 0 for x a . The form of non-zero solutions of y (where m varies over all integers) are (a) y m Am sin m x a (c) y m Am x
(b) y m Am cos m x a
m a
(d) y m Am e
m x a
Solution (a): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 k 2 ) 0 . As e mx 0 , so we have the auxiliary equation m 2 k 2 0 whose roots are m 0 ik which is a conjugate complex root case, so 0 and k . So the general solution of given DE is y e0 (c1 cos kx c2 sin kx ) c1 cos kx c2 sin kx . Now we have to find the constants c1 and c2 . As
y (0) 0 c1 0 and y ( a ) 0 c2 sin ka 0 sin ka 0 as we have to find non-zero solutions of y , so c2 0 . Thus
sin ka 0 ka m k m a . So the solution of given DE is
y c2 sin( m x a) . Now as c2 has many values and so the general solution of the given DE is
y m Am sin(m x a ) , where Am , m 1, 2, are the different values of c2 . Example 3.58 [IN-2007 (2 marks)]: The boundary value problem y y 0 , y (0) y ( ) 0 will have non-zero solutions if and only if the values of are (a) 0, 1, 2, (b) 1, 2, 3, (c) 1, 4, 9, (d) 1, 9, 25, Solution (c): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 ) 0 . As e mx 0 , so we have the auxiliary equation m 2 0 whose roots are m 0 i which is a conjugate complex root
case,
so
0
and
.
So
the
general
solution
of
given
DE
is
y e0 (c1 cos x c2 sin x ) c1 cos x c 2 sin x . Now we have to find the constants c1 and
c2 . As y (0) 0 c1 0 and y ( ) 0 c2 sin 0 sin 0 as we have to find non-zero
solutions of
y , so c2 0 . Thus sin 0 m m ., where m I , i.e.,
m {0, 1, 2, } . So m 2 {1, 4, 9, } . Example 3.59 [PI-2008 (2 marks)]: ( d 2 y dx 2 ) 2( dy dx ) 2 y 0 are (1 i ) x
(1i ) x
(1i ) x
(1i ) x
The
solutions (1 i ) x
of (1i ) x
the
differential (1i ) x
equation (1i ) x
(a) e ,e (b) e ,e (c) e ,e (d) e ,e Solution (a): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx (m 2 2m 2) 0 so we have the
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.26]
auxiliary equation m 2 2m 2 0 as e mx 0 . As the discriminant of the auxiliary equation 2 D 2 4 1 2 4 0 so we have conjugate complex root case; also the roots of our auxiliary equation are m1 , m2 1 2i ; so 1 and 2 . So the solutions are y e ( 1 2 i ) . Example 3.60 [AE-2009 (1 mark)]: The ordinary differential equation ( d 2 y dx 2 ) ky 0 where k is real and positive (a) is non-linear (b) has a characteristic equation with one real and one complex root (c) has a characteristic equation with two real roots (d) has a complementary function that is simple harmonic Solution (d): As the given ODE is linear so option (a) is not correct. Also the auxiliary equation of the give ODE is m 2 k 0 , where k 0 , so its roots are complex conjugate roots; thus options (b) and (c) are also not correct. Thus we left with option (d) which is correct as the ODE of simple harmonic oscillator is ( d 2 y dx 2 ) 2 y 0 . Example 3.61 [EE-2010 (2 marks)]: For the differential equation ( d 2 x dt 2 ) 6( dx dt ) 8 x 0 with initial conditions x(0) 1 and ( dx dt ) t 0 0 , the solution is (a) x (t ) 2e 6 t e 2t (b) x (t ) 2e 2 t e 4t (c) x (t ) e 6 t 2e 4 t (d) x (t ) e 2 t 2e 4 t Solution (b): As we have homogeneous linear equation with constant coefficient so putting mt mt 2 mt mt 2 x e x me x m e in the given DE we get e ( m 6m 8) 0 so we have the auxiliary equation m 2 6m 8 0 as e mt 0 . As the discriminant of the auxiliary equation 2 D 6 4 1 8 4 0 so we have distinct real root case; also the roots of our auxiliary equation are mt m t m1 , m2 4, 2 ; so the general solution of given DE is x c1e 1 c2 e 2 c1e 4 t c2 e 2t x 4c1e 4 t 2c2 e 2 t . Now applying the given condition
x (0) 1 1 c1e 40 c2 e 20
1 c1 c2 ; also at x(0) 0 0 4c1e 40 2c2 e 20 c2 2c1 ; solving the two results we get c1 1 and c2 2 . So the general solution is x e 4t 2e 2t . [Similar questions were also asked in CE-2010, CS-1995, CH-2009, MN-2008, TF-2007, MN2013 (2 marks), EC-2005, ME-2008, AE-2012, AG-2013 (1 mark)]
Example 3.62 [PI-2010 (1 mark)]: Which one of the following differential equations has a solution given by the function y 5sin 3x ( 3) ? 5 dy 5 d2y d2y cos(3 x ) 0 (b) cos(3 x) 0 (c) 9 y 0 (d) 9y 0 dx 3 dx 3 dx 2 dx 2 Solution (c): Twice differentiating the given solution we 2 2 dy d y y d y 15 cos 3x 2 45 sin 3 x 45 9 y 2 9 y 0 dx 3 dx 3 5 dx
(a)
dy
Example 3.63 [CH-2013 (2 marks)]: The solution of the differential ( d 2 y dx 2 ) (dy dx ) 0.25 y 0 , given y 0 at x 0 and dy dx 1 at x 0 is
get
equation
(a) xe 0.5 x xe 0.5 x (b) 0.5 xe x 0.5 xe x (c) xe 0.5x (d) xe 0.5x Solution: As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 m 0.25) 0 so we have the auxiliary equation m 2 m 0.25 0 as e mt 0 . As the discriminant of the auxiliary equation D ( 1) 2 4 0.25 0 so we have repeated real roots case; also the roots of our auxiliary equation
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.27]
are m m1 m2 0.5 ; so the general solution of given DE is y (c1 c2 x)e
m1x
( c1 c2 x)e 0.5 x
y c2 e 0.5 x 0.5(c1 c2 x )e 0.5 x . Applying the given condition y (0) 0 0 (c1 c2 0)e 0.50
c1 0 ; also at
y (0) 1 1 c2 e 0.50 0.5(c1 c2 0)e 0.50 1 c2 0.5c1 c2 1 . So the
general solution is y xe 0.5 x . [Similar questions were also asked in TF-2010, ME-1995, AE-2008 (2 mark), ME-1994 (2.5 marks), EC-2014 (1 mark)] Example 3.64 [IN-2013 (2 marks)]: The maximum value of the solution y (t ) of the differential equation y (t ) y (t ) 0 with initial conditions y (0) 1 and y (0) 1 , for t 0 is (a) 1 (b) 2 (c) (d) 2 Solution: As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 1) 0 so we have the auxiliary equation m 2 1 0 as e mt 0 . As the discriminant of the auxiliary equation D (0) 2 4 1 1 4 0 so we have conjugate complex roots case; also the roots of our auxiliary equation are m1 , m2 0 i 0 and 1 , so the general solution of given DE is y (c1 cos x c2 sin x ) y ( c1 sin x c2 cos x) . Now applying the given condition y (0) 1 1 (c1 c2 0) c1 1 ; also at y (0) 1 1 ( c1 0 c2 ) c2 1 . So the general solution is
y (cos x sin x ) whose maximum value is
12 12
2.
Example 3.65 [ME-2013, PI-2013 (2 marks)]: The solution of the differential equation ( d 2 u dx 2 ) k ( du dx ) 0 , where, k is a constant, subjected to the boundary conditions u (0) 0 and u ( L) U is (a) u U
1 e kx kL 1 e
x
(b) u U
1 e kx kL 1 e
(c) u U
1 e kx kL 1 e
(d) u U
L Solution (b): As we have homogeneous linear equation with constant coefficient so putting mx mx 2 mx in the given DE we get e mx ( m 2 km) 0 so we have the u e u me u m e
auxiliary equation m 2 km 0 (as e mt 0 ), m1 0 and m2 k so we have distinct real root case; so the general solution of given DE is u c1e 0 c2 e kx c1 c2 e kx . Now applying the given condition u (0) 0 0 c1 c2 e 0 c1 c2 0 ; also u ( L ) U U c1 c2 e kL ; solving these two equations
we get,
c2 U (e kL 1)
and
c1 U (e kL 1) .
So the general
solution is
y U (e kL 1) Ue kx (e kL 1) U (1 e kx ) (1 e kL ) .
Example 3.66 [EC-2014 (1 mark)]: If the characteristic equation of the differential equation ( d 2 y dx 2 ) 2 ( dy dx ) y 0 has two equal roots, then the values of are (b) 0, 0 (c) j (a) 1 (d) 1 2 Solution (a): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx (m 2 2 m 1) 0 so we have the auxiliary equation m 2 2 m 1 0 as e mx 0 . It is given that the auxiliary equation has two equation roots so the discriminant of auxiliary equation must be zero, i.e., 2 2 4 4 0 1 1 . Example 3.67 [EC-2014 (2 marks)]: With initial values y (0) y (0) 1 , the solution of differential equation ( d 2 y dx 2 ) 4( dy dx ) 4 y 0 at x 1 is …………….
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Engineering Mathematics
Chapter 3: Differential Equation
[3.28]
Solution: As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given DE we get e mx ( m 2 4m 4) 0 so we have the auxiliary equation m 2 4m 4 0 as e mt 0 . As the discriminant of the auxiliary equation D (4) 2 4 1 4 0 so we have repeated real roots case; also the roots of our auxiliary equation are m m1 m2 2 ; so the general solution of given DE is y (c1 c2 x )e
m1x
(c1 c2 x )e 2 x
y c2 e 2 x 2(c1 c2 x) e 2 x . Now applying the given condition y (0) 1 1 (c1 c2 0)e 2 0
c1 1 ; also at y (0) 1 1 c2 e 20 2(c1 c2 0)e 20 1 c2 2c1 c2 3 . So the general
solution is y (1 3x )e 2 x . So y (1) (1 3)e 2 4e 2 0.54 . [Similar questions were also asked in AE-2013, ME-2008, IN-2011 (2 marks)]
Higher Order Differential Equations: In general, to solve an nth -order differential equation 3.43, we must solve an nth -degree polynomial equation (obtained by putting y emx in Eq. 3.43) an m n an 1m n 1 a2 m 2 a1m a0 0
(3.49)
If all the roots of Eq.3.39 are real and distinct, then the general solution of Eq. 3.43 is y c1e
m1x
c2 e
m2 x
cn e
mn x
(3.50)
It is somewhat harder to summarize the analogues of Cases II and III because the roots of an auxiliary equation of degree greater than two can occur in many combinations. For example, a fifth-degree equation could have five distinct real roots, or three distinct real and two complex roots, or one real and four complex roots, or five real but equal roots, or five real roots but two of them equal, and so on. When m1 is a root of multiplicity k of an nth -degree auxiliary equation (that is, k roots are equal to m1 ), it can be shown that the linearly independent solutions are e
m1 x
, xe
m1 x
, x 2e
m1x
, , x k 1e
m1 x
and the general solution must contain the linear combination m1 x
mx
mx
mx
(3.51) c1e c2 xe 1 c3 x 2 e 1 c k x k 1e 1 Finally, it should be remembered that when the coefficients are real, complex roots of an auxiliary equation always appear in conjugate pairs. Thus, for example, a cubic polynomial equation can have at most two complex roots. Example 3.68 [XE-2012 (2 marks)]: 4 4 3 3 2 2 ( d y dx ) 2( d y dx ) 2( d y dx ) 2( dy dx) y 0 is
The
general
solution
of
(b) c1e x c2 e x c3 eix c4 e ix
(a) c1e x c2 xe x c3 cosh( x ) c4 sinh( x )
(c) c1e x c2 xe x c3 cos( x ) c4 sin( x) (d) c1e x c2 xe x c3e ix c4 e ix Solution (d): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx y m3e mx y (iv) m 4e mx , we get e mx ( m4 2m3 2m 2 2m 1) 0 ; 4
3
2
as
e
2
mx
0
so
we
have
the
auxiliary
equation
2
( m 2m 2m 2m 1) 0 ( m 1)( m 1) 0 the auxiliary equation has two reapeated real roots, i.e., m 1 ; and two complex conjugate roots, i.e., m 0 i . So the general solution of the given differential equation is y (c1 c2 x )e x e 0 (c3eix c4 e ix ) c1e x c2 xe x c3 eix c4 e ix .
3.2.4 Non – Homogeneous Linear Equation with Constant coefficients To solve a non – homogeneous linear differential equation an y ( n ) an 1 y ( n 1) a1 y a0 y g ( x )
(3.52)
we must do two things: (i) Finding the complementary function y c ; and (ii) Finding any particular solution y p of the non – homogeneous Eq. 3.55. Then the general solution of Eq. 3.55 is y yc y p .
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Chapter 3: Differential Equation
[3.29]
The complementary function y c is the general solution of the associated homogeneous differential equation of Eq. 3.55, i.e., an y ( n ) an 1 y ( n 1) a1 y a0 y 0 ; in section 3.2.3 we saw how to solve these kinds of equations when the coefficients were constants. Our aim is to develop a method for obtaining particular solutions y p for a non – homogeneous linear differential equation; that method is called the method of undetermined coefficients by Superposition approach. Example 3.69 [EC-1994 (2 marks)]: Match List-I with List-II List-I List-II 1. Non-linear differential equation A. a1 (d 2 y dx 2 ) a2 y ( dy dx ) a3 y a4 2. Linear differential equation with constant B. a1 (d 3 y dx 3 ) a2 y a3 coefficients 2 2 2 3. Linear homogeneous differential equation C. a1 ( d y dx ) a2 x( dy dx ) a3 x y 0 4. Non-linear homogeneous differential equation 5. Non-linear first order differential equation Solution: The differential equation given in (A), and (B) is linear in y and y with constant coefficients so both are ‘Linear differential equation with constant coefficients’. The differential equation given in (C) is in the form of linear differential equation given by Eq. 3.4 and also the RHS of the given equation is zero; so it is a ‘Linear homogeneous differential equation.’ Example 3.70 [ME-1999 (2 marks)]: ( d 2 y dx 2 ) ( x 2 4 x)( dy dx ) y x 8 8 is a (a) Partial differential equation (b) Non-linear differential equation (c) Non-homogeneous differential equation (d) Ordinary differential equation Solution: As the given differential equation have no partial derivatives so option (a) is not correct. Also the given differential equation has no non-linear term so option (b) is not correct. As the coefficeients of the derivative of dependent variable ( y ) w.r.t independent variable ( x ) are not constant so it is not ordinary differential equation so option (d) is not correct. As the given equation is in the form of non-homogeneous equation so option (c) is correct.
Method of Undetermined Coefficients: Superposition approach: The Superposition approach in which the underlying idea behind this method is a conjecture about the form of y p that is motivated by the kinds of functions that make up the input function g ( x ) . The general method is limited to linear differential equations such as Eq. 3.52 where, the coefficients an , an 1 , , a1 , a0 are constants
g ( x ) is a constant k ; a polynomial function; an exponential function e x ; a sine or cosine function, i.e., sin x or cos x ; or finite sums and product of these functions. The following functions are some examples of the types of inputs g ( x ) that are appropriate for this
discussion:
g ( x ) 10 ,
g ( x) x 2 5 x ,
g ( x) 15 x 6 x 8e x ,
g ( x) sin 3 x 5 x cos 2 x ,
g ( x ) xe x sin x (3x 2 1)e 4 x . That is, g ( x ) is a linear combination of functions of the type P ( x ) an x n an 1 x n 1 a1 x a0 , P ( x )e x , P ( x) e x sin x and P ( x )e x cos x , where n is a non – negative integer and and are real numbers. The method of undetermined coefficients is
not applicable to equations of form Eq. 3.52 where g ( x ) ln x , g ( x) 1 x , g ( x) tan x , g ( x ) sin 1 x , and so on and these are discussed in section 3.2.5 called as Variation of Parameters.
The set of functions that consists of constants, polynomials, exponentials, e x , sines and cosines has the property that derivatives of their sums and products are again sums and product of constants, polynomials, exponentials, e x , sines and cosines. Because the linear combination of derivatives
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Engineering Mathematics
Chapter 3: Differential Equation
[3.30]
an y (pn ) an 1 y (pn 1) a1 y p a0 y p must be identical to g ( x ) , it seems reasonable to assume that y p has the same form as g ( x ) . For choosing y p we have two situations:
Case 1: No function in the assumed particular solution is a solution of the associated homogeneous differential equation: In the following table, we illustrate some specific examples of g ( x ) in Eq. 3.52 along with the corresponding form of the particular solution. We also assumed that no function in the assumed particular solution y p is duplicated by a function in the complementary function y c . Form of g ( x )
Form of y p
1. 2. 3. 4. 5.
A Ax B
Any constant Any linear equation, ax b Any quadratic equation, ax 2 bx c Any cubic equation, ax 3 bx 2 cx d Sine or Cosine function, sin( x ) or cos( x )
2
Ax Bx C 3 2 Ax Bx Cx D A cos( x ) B sin( x )
6. Exponential function, e kx 7. Product of linear and exponential function, (ax b)e kx 8. Product of quadratic and exponential function, ( ax 2 bx c )e kx 9. Product of exponential and sine or cosine function, e kx sin( x ) or e kx cos( x )
Ae ( Ax B)e kx
kx
10. Product of quadratic and sine or cosine function, ( ax 2 bx c ) sin( x ) or ( ax 2 bx c ) cos( x )
( Ax 2 Bx C ) cos( x )
11. Product of linear, exponential and sine or cosine function, ( ax b)e kx sin( x ) or ( ax b)e kx cos( x )
( Ax B )e kx cos( x)
( Ax 2 Bx C )e kx Ae kx cos( x ) Be kx sin( x )
( Ex 2 Fx G ) sin( x )
(Cx D )e kx sin( x ) If g ( x ) consists of a sum of, say, m terms of the kind listed in the table, then the assumption for a particular solution y p consists of the sum of the trial forms y p1 , y p2 , , y pn corresponding to these terms, i.e., y p y p1 y p2 y pn Example 3.71 [ME-1996 (2 marks)]: The particular solution for the differential equation ( d 2 y dx 2 ) 3( dy dx ) 2 y 5 cos x is (a) 0.5cos x 1.5sin x (b) 1.5cos x 0.5sin x (c) 1.5sin x (d) 0.5cos x Solution (a): As g ( x ) 5 cos x so the particular solution is in the form of y p A cos x B sin x . As y A sin x B cos x and y A cos x B sin x ; substituting these in the given DE we get, ( A cos x B sin x ) 3( A sin x B cos x) 2( A cos x B sin x) 5 cos x . Comparing the coefficients of sin x and cos x of LHS with sin x and cos x of RHS, we get A 3B 2 A 5 A 3B 5 and B 3 A 2B 0 B 3 A 0 ; solving these two we get A 0.5 and B 1.5 . So the particular solution is y p 0.5 cos x 1.5sin x .
[Similar questions were also asked in ME-2006, XE-2008, TF-2008, TF-2013 (2 marks)] Example 3.72 [CE-1998 (5 marks)]: Subjective Question: Solve ( d 4 y dx 4 ) y 15 cos 2 x Solution: The homogeneous part of the given DE is ( d 4 y dx 4 ) y 0 which is solve by putting y e mx y me mx y m 2 e mx y m3e mx y (iv) m 4e mx , so e mx ( m 4 1) 0 as e mx 0
so the auxiliary equation is ( m 4 1) 0 (m 2 1)( m 2 1) 0 we have two distinct real roots, i.e., m1 1 and m2 1 ; and two complex conjugate roots, i.e., m3 , m4 0 i . So the complementary
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Engineering Mathematics
Chapter 3: Differential Equation
solution
DE
of
the
given
is
given
[3.31]
yc c1e x c2 e x e 0 (c3 cos x c4 sin x)
as:
yc c1e x c2 e x c3 cos x c4 sin x . Now as g ( x) 15 cos 2 x so the choice of our particular
solution is
y p A cos 2 x B sin 2 x yp 2 A sin 2 x 2 B cos 2 x y p 4 A cos 2 x 4 B sin 2 x
y p 8 A sin 2 x 8 B cos 2 x y
(iv) p
16 A cos 2 x 16 B sin 2 x . So putting these in the given DE we
get, 16 A cos 2 x 16B sin 2 x A cos 2 x B sin 2 x 15cos 2 x ; now comparing the coefficients of cos 2x and sin 2x , we get 15 A 15 A 1 and 15B 0 B 0 y p cos 2 x and thus the general solution is y yc y p c1e
x
c2e x c3 cos x c4 sin x cos x .
[Similar question was also asked in CE-2000 (5 marks)] Example 3.73 [ME-2001 (5 marks)]: Solve the differential equation ( d 2 y dx 2 ) y x with the following conditions: (1) at x 0, y 1 ; (2) at x 2 , y 2 . Solution: We first solve the associated homogeneous equation y y 0 . From the quadratic formula we find that the roots of the auxiliary equation m2 1 0 are m1 , m2 0 i . Hence the yc e 0 x (c1 cos x c2 sin x) c1 cos x c2 sin x . As we have linear
complementary function is
function of g ( x ) x , so the particular solution is of the form y p Ax B y p A y p 0 so substituting these in the given DE we get, 0 Ax B x A 1 and B 0 ; so y p x and thus the general solution is y yc y p c1 cos x c2 sin x x . Now applying the given condition y (0) 1 1 c1 ; y ( 2) 2 2 c2 2 c2 0 . So the general solution is y cos x x
Example 3.74 [EE-2005 (2 marks)]: For the equation x(t ) 3x (t ) 2 x (t ) 5 , the solution x (t ) approaches the following values at t (a) 0 (c) 5 (d) 10 (b) 5 2 Solution (b): The associated homogeneous equation for the given DE is x 3 x 2 x 0 . From the quadratic formula for the auxiliary equation m 2 3m 2 0 , its roots are m1 1, m2 2 . Hence the complementary function is xc c1e t c2 e 2 t . Now, because the function g (t ) is constant, let us assume a particular solution that is also constant as, x p A . We seek to determine the coefficient A for which x p is a solution of the given equation. Substituting x p and the derivatives xp 0 and xp 0 into the given DE, we get, x 3 x 2 x 5 0 0 2 A 5 A 2.5 . Thus a particular
solution
is
x p 2.5 .
Hence
the
general
solution
of
the
given
equation
is
x xc x p c1e t c2e 2 t 2.5 , so if t , then et , e2t 0 and so x 2.5 . Example 3.75 [EC-2007 (2 marks)]: The solution of the differential equation k 2 ( d 2 y dx 2 ) y y2 under the boundary condition (i) y y1 at x 0 and (ii) y y2 at x , where k , y1 and y 2 are constants, is (b) y ( y2 y1 ) exp( x k ) y1 (a) y ( y1 y2 ) exp( x k 2 ) y2 (c) y ( y1 y2 ) sinh( x k ) y1
(d) y ( y1 y2 ) exp( x k ) y2
Solution (d): As k 2 ( d 2 y dx 2 ) y y2 (i) non-homogeneous differential equation whose associated homogeneous DE is k 2 ( d 2 y dx 2 ) y 0 putting y e
mx
y me
mx
2 mx
y m e
…(ii), which has constant coefficients. So
in (i), we get e mx ( k 2 m 2 1) 0 . As e mx 0 , so we have
the auxiliary equation k 2 m 2 1 0 whose roots are m 1 k which is a distinct real root case. So
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Engineering Mathematics
Chapter 3: Differential Equation
[3.32]
the general solution of (ii) is y c1e (1 k ) x c2 e(1 k ) x . Now, as the function g ( x ) is constant so the choice of the particular solution is also a constant, i.e., y p A yp yp 0 . Now putting these into 2 the given DE we get k 0 A y2 A y2 . So the particular solution is y p y2 . Hence the
general solution of given DE is y yc y p c1e
(1 k ) x
c2e (1 k ) x y2 . Now we have to find the
constants c1 and c2 , by applying the given boundary conditions. As y (0) y1 c1 c2 y2 y1 and y ( ) y2 c1e c2 e y2 y2 c2 e 0 c2 0 c1 y1 y2 . So the general solution of (1 k ) x y2 . given DE is y ( y1 y2 )e [Similar question was also asked in EC-2010 (1 mark)].
Example 3.76 [CH-2008 (1 mark)]: Which ONE of the following is NOT a solution of the differential equation ( d 2 y dx 2 ) y 1 ? (a) y 1 (b) y 1 cos x (c) y 1 sin x (d) y 2 sin x cos x Solution (d): For every option we find y and y and substituting in the given DE, we can find which one is the solution. So for option (a) y 1 y y 0 , so 0 1 1 thus option (a) is correct; for option (b) y 1 cos x y sin x y cos x , so cos x 1 cos x 1 thus option (b) is correct; for option (c) y 1 sin x y cos x y sin x , so sin x 1 sin x 1 thus option (c) is correct; for option (d) y 2 sin x cos x y cos x sin x y sin x cos x , so sin x cos x 2 sin x cos x 1 . Thus option (d) is not correct. Example 3.77 [PI-2009 (1 mark)]: The homogeneous part of the differential equation ( d 2 y dx 2 ) p( dy dx ) qy r ( p , q and r are constants) has distinct real roots if (a) p 2 4q 0
(b) p 2 4q 0
(c) p 2 4q 0
(d) p 2 4q r
Solution (a): The homogenous part of the given DE is ( d 2 y dx 2 ) p ( dy dx ) qy 0 . Now as we have homogenous linear equation with constant coefficient so by putting mx mx 2 mx 2 y e y me y m e , we get the auxiliary equation as: m pm q 0 , as this equation have distinct roots so its discrimination is 0 , i.e., p 2 4q 0 . Example 3.78 [AG-2010 (1 mark)]: A particular solution of ( d 2 y dx 2 ) 5( dy dx ) 3 y 6 is (a) 2.0 (b) 0.5 (c) –0.5 (d) –2.0 Solution (d): As the function g ( x ) is constant so the choice of the particular solution is also a constant, i.e.,
y p A yp yp 0 . Now putting these into the given DE we get
0 5 0 3 A 6 A 2 . So the particular solution is y p 2 . Example 3.79 [XE-2010 (2 marks)]: Which one of the following is a particular solution of the ordinary differential equation x( d 2 y dx 2 ) (dy dx) 2 x 2 f ( x ) ? (a) x 2 x f ( x) dx x 3 f ( x ) dx
(b) x 2 f ( x ) dx x 2 f ( x ) dx
(c) x 2 x f ( x ) dx x 3 f ( x ) dx
(d) x 2 f ( x ) dx x 2 f ( x ) dx
Solution (d): The given differential equation can be 2 d y dy 2 d dy dx dy dx 2 f ( x) d 2 f ( x) dx x 2 x 2 f ( x) dx dx x x dx
written
as
dy dx 2 x f ( x )dx dy 2 x f ( x) dx dx dy 2 x f ( x )dx dx
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Engineering Mathematics
y2
Chapter 3: Differential Equation
f ( x)dx xdx 2 (d
dx )
[3.33]
f ( x)dx xdx dx x f ( x)dx x 2
Example 3.80 [PI-2011 (1 mark)]: The solution of ( d 2 y dx 2 ) 6(dy dx) 9 y 9 x 6 with c1 and c2 as constants is (a) y (c1 x c2 )e 3 x
2
the
f ( x ) dx
differential
equation
(b) y c1e3 x c2 e 3 x x
(c) y (c1 x c2 )e 3 x x
(d) y (c1 x c2 )e3 x x Solution (c): We first solve the associated homogeneous equation y 6 y 9 y 0 . From the quadratic formula for the auxiliary equation m 2 6m 9 0 , its roots are m1 , m2 3 . Hence the complementary function is yc (c1 c2 x )e 3 x . Now, because the function g ( x ) is linear, let us assume a particular solution that is also in the form of a linear as, y p Ax B . We seek to determine the coefficients A and B for which y p is a solution of the given equation. Substituting y p and the derivatives
y p A
and
y p 0
into
the
given
DE,
we
get,
y 6 y 9 y 9 x 6 0 6 A 9( Ax B ) 9 x 6 . Because the last equation is supposed to be an identity, the coefficients of like powers of x must be equal, i.e., 9 A 9 A 1 , 6 A 9B 6 B 0 . Thus a particular solution is y p x . Thus the general solution of the given
equation is y yc y p (c1 c2 x )e 3 x x , which is in the form given in option (c). [Similar question was also asked in XE-2011 (1 mark)] Example 3.81 [AE-2014 (2 marks)]: Solution to the boundary value problem 9( d 2 u dx 2 ) u 5 x ,
0 x 3 with u (0) 0 , (du dx) x 3 0 is (a) u ( x) {15e (1 e 2 )}(e x 3 e x 3 ) 5 x
(b) u ( x) {15e (1 e 2 )}(e x 3 e x 3 ) 5 x
(c) u ( x ) 15sin( x 3) cos(1) 5 x
(d) u ( x) 15 sin( x 3) cos(1) (5 54) x3
Solution (a): As 9( d 2 u dx 2 ) u 5 x (i) non-homogeneous DE whose associated homogeneous DE is 9(d 2u dx 2 ) u 0 2 mx
u m e
…(ii), which has constant coefficients. So putting u e mx u me mx
in (i), we get e mx (9m 2 1) 0 . As e mx 0 , so we have the auxiliary equation
2 9m 1 0 whose roots are m 1 3 which is a distinct real root case. So the general solution of
(ii) is u c c1e x 3 c2 e x 3 . Now, as the function g ( x ) 5 x so the choice of the particular solution is
u p Ax b u p A u p 0 . Now putting these into the given DE we get 0 ( Ax B) 5 x A 5 and B 0 . So the particular solution is y p 5 x . Hence the general solution of given DE is
u uc u p c1e x 3 c2e x 3 5 x u (c1 3)e x 3 (c2 3)e x 3 5 . Now we have to find the constants c1 and c2 , by applying the given boundary conditions. As u (0) 0 c1 c2 0 and u (3) 0 (c1 3)e 1 (c2 3)e 5 0 ;
now
solving
the
two
results
we
get,
(c2 3)e 1 (c2 3)e 5 0 c2 15e (1 e 2 ) c1 15e (1 e 2 ) . So the general solution of given
DE is u {15e (1 e 2 )}e x 3 {15e (1 e 2 )}e x 3 5 x {15e (1 e 2 )}(e x 3 e x 3 ) 5 x . Example 3.82 [MA-2014 (2 marks)]: The solution to the ( d 2 y dt 2 ) 2( dy dt ) 5 y 3e t sin t , y (0) 0 and ( dy dt ) x 0 3 is (a) y (t ) et (sin t sin 2t )
(b) y (t ) e t (sin t sin 2t )
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initial
(c) y (t ) 3et sin t
value
problem
(d) y (t ) 3e t sin t
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Engineering Mathematics
Chapter 3: Differential Equation
[3.34]
Solution (b): The given DE is ( d 2 y dt 2 ) 2( dy dt ) 5 y 3e t sin t homogeneous ye
mx
2
( d y dt ) 2( dy dt ) 5 y 0
is
y me
2
mx
2 mx
y m e
mx
…(ii)
…(i) whose associated
which
2
, so e (m 2m 5) 0 as e
mx
is
solved
by
putting
0 so the auxiliary equation is
2
m 2m 5 0 m 1 2i we have conjugate complex root case. So the complementary solution of (i) is given as: yc e t (c1 cos 2t c2 sin 2t ) . Now as g (t ) e t sin t so the choice of our
particular
solution
y p e t ( A cos t B sin t ) yp e t {( A B ) cos t ( A B ) sin t}
is
y p e t {2 B cos t 2 A sin t} .
So
putting
these
in
(i)
we
get,
e t (2 B cos t 2 A sin t ) 2{( A B) cos t ( A B ) sin t} 5( A cos t B sin t ) 3e t sin t t
e t {(3 A) cos t (3B ) sin t} 3e t sin t 3B 3 B 1 and 3 A 0 A 0 . So y p e sin t .
y yc y p e t (c1 cos 2t c2 sin 2t ) e t sin t . Now applying the given conditions:
Hence
y (0) 0 0 e 0 (c1 0) 0 c1 0 y e t (c2 sin 2t sin t ) ; t
now
0
y e {c2 sin 2t sin t 2 c2 cos 2t cos t} y (0) 3 3 e (2 c2 1) c2 1 .
as Hence
the
t
general solution of the given differential equation is y e (sin 2t sin t ) .
Case 2: A function in the assumed particular solution is also a solution of the associated homogeneous differential equation: Suppose again that g ( x ) consists of m terms of the kind given in Table 1 and suppose further that the usual assumption for a particular solution is y p y p1 y p2 y pn , where y p1 , y p2 , , y pn are the trial particular solution forms corresponding to these terms. Now, if any y pi contains terms that duplicate terms in yc , then that y pi must be multiplied by x n , where n is the smallest positive integer that eliminates that duplication.
Example 3.83: Find the form of a particular solution y 2 y y e x . Solution: We first solve the associated homogeneous equation y 2 y y 0 . From the quadratic formula we find that the roots of the auxiliary equation m 2 2m 1 0 are m1 , m2 1,1 . Hence the complementary function is yc c1e x c2 xe x . x
x
If we choose y p Ae or y p Axe (both are duplicated in y c ) then we will not be able to find a particular solution of the form y p Ae x
yp Ae , thus statement;
and
x
or y p Axe
x
x
because: for y p Ae , yp Ae
y 2 y y e x Ae x 2 Ae x Ae x e x 0 e x
for
y p Axe x ,
y p Ae x Axe x
x
and
which is a contradictory
y p 2 Ae x Axe x ,
and
thus
y 2 y y e x 0 e x 2 Ae x Axe x 2 Ae x 2 Axe x Axe x e x 0 e x which is also a contradictory statement. Hence we will not be able to find a particular solution if we choose the form x x of particular solution as, y p Ae and y p Axe . 2 x
If we apply the concept of Case 2, then for choice we have y p Ax e and for this y p we have
y p 2 Axe x Ax 2 e x and y p 2 Ae x 4 Axe x Ax 2 e x ; putting these into the equation we have y 2 y y e x 2 Ae x 4 Axe x Ax 2 e x 4 Axe x 2 Ax 2 e x Ax 2 e x e x 2 Ae x e x A 1 2 .
Hence the particular solution is y p (1 2) x 2 e x . So the general solution of the given differential equation is y yc y p c1e x c2 xe x (1 2) x 2 e x .
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[3.35]
3.2.5 Variation of Parameters In this section we examine a method for determining a particular solution of a nonhomogeneous linear DE that has no such restrictions on it. This method is known as variation of parameters.
Linear 1st Order differential equation: In section 3.1.1, we saw that the general solution of a linear first – order differential equation a1 ( x ) y a0 ( x ) y g ( x ) can be found by first rewriting it in standard form (3.53) ( dy dx ) P ( x ) y f ( x ) and assuming that P ( x ) and f ( x ) are continuous on an common interval I . Using the integrating factor method, the general solution of Eq. 3.53 on the interval I was found to be P ( x ) dx P ( x ) dx P ( x ) dx f ( x)}dx . The foregoing solution has the same form given by y c1e e {e P ( x ) dx y yc y p . In this case yc c1e is a solution of the associated homogeneous equation
(3.54)
( dy dx ) P ( x ) y 0
P ( x ) dx P ( x ) dx f ( x )}dx is a particular solution of the non – homogeneous Eq. 3.53. and y p e {e
Proof: Suppose that y1 is the known solution of the homogeneous Eq. 3.54. It is easily shown that y1 e is a solution of Eq. 3.54 and because the equation is linear, c1 y1 ( x ) is its general solution. Variation of parameters consists consist of finding a particular solution of Eq. 3.53 of the form y p u1 ( x ) y1 ( x ) . In other words, we have replaced the parameter c1 by the function u1 . P ( x ) dx
Substituting y p u1 y1 into Eq. 3.53 and using the Product Rule gives ( d dx)(u1 y1 ) P ( x )u1 y1 f ( x ) u1 ( dy1 dx ) y1 ( du1 dx ) P ( x )u1 y1 f ( x ) u1 ( dy1 dx ) P( x ) y1 y1 ( du1 dx ) f ( x)
0 because y1 is a solution of Eq. 3.57
y1 ( du1 dx ) f ( x) . By separating variable and integrating, we find u1 : P ( x ) dx P ( x ) dx du1 f ( x) y1 ( x) dx u1 f ( x ) y1 ( x) dx { f ( x) e }dx {e f ( x )}dx P ( x ) dx P ( x ) dx f ( x )}dx y p y1u1 e {e
Linear 2nd Order Differential equation: Next we consider the case of linear second order equation with constant coefficients a2 y a1 y a0 y g ( x )
(3.55)
Again, we put Eq. 3.55 into the standard form by dividing by the leading coefficient a2 as, y py qy f ( x) (3.56) In Eq. 3.56, p and q are constant and the functions f ( x ) is continuous on some interval I . As we have already seen earlier, there is no difficulty in obtaining the complementary solution yc c1 y1 ( x ) c2 y2 ( x ) , the general solution of the associated homogeneous equation of Eq. 3.56, when the coefficients are constants. Analogous to the preceding discussion, we now ask: Can the parameters c1 and c2 can be replaced with functions and or “variable parameters,” so that y u1 ( x ) y1 ( x ) u2 ( x) y2 ( x ) (3.57) is a particular solution of Eq. 3.56? To answer this question we substitute Eq. 3.57 into Eq. 3.56. Using the Product Rule to differentiate y p twice, we get y p u1 y1 y1u1 u 2 y2 y2u 2 and y p u1 y1 y1u1 y1u1 u1 y1 u 2 y2 y2 u 2 y2 u 2 u 2 y2 ; now after substituting these derivatives and
Eq. 3.57 into Eq. 3.56 and solving, we get,
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Chapter 3: Differential Equation
[3.36]
y p py p qy p f ( x ) (d dx )( y1u1 y2 u2 ) p ( y1u1 y2 u 2 ) ( y1u1 y2 u 2 ) f ( x )
(3.58)
Because we seek to determine two unknown functions u1 and u2 , reason dictates that we need two equations. We can obtain these equations by making the further assumption that the functions u1 and u2 satisfy y1u1 y2u2 0 y1u1 y2 u2 f ( x ) (from Eq. 3.58). By Cramer’s Rule, the solution of
the system y1u1 y2u2 0 and y1u1 y2 u2 f ( x ) can be expressed in terms of determinants: u1 W1 W y2 f ( x) W and u2 W2 W y1 f ( x) W
where,
W
y1
y2
y1
y2
, W1
0
y2
f ( x)
y2
and W1
y1
0
y1
f ( x)
(3.59) (3.60)
The functions u1 and u2 are found by integrating the results in Eqs. 3.59. The determinant W is recognized as the Wronskian of y1 and y 2 . By linear independence of y1 and y 2 on I , we know that W y1 ( x ), y2 ( x ) 0 for every x in the interval.
Since, the foregoing procedure is too long and complicated to use each time we wish to solve a differential equation. In this case it is more efficient to simply use the formulas in Eq. 3.59. Thus to solve a2 y a1 y a0 y g ( x ) , first find the complementary function yc c1 y1 ( x ) c2 y2 ( x ) and then compute the Wronskian W y1 ( x ), y2 ( x ) . By dividing by a2 , we put the equation into the standard form y py qy f ( x) to determine f ( x ) . We find u1 and u2 by integrating u1 W1 W and u2 W2 W and , where W1 and W2 are defined as in Eq. 3.60. A particular
solution is y p u1 y1 u 2 y2 . The general solution of the equation is then y yc y p .
It is to be noted that, when computing the indefinite integrals of u1 and u2 , we need not introduce any constants. This is because, y yc y p c1 y1 c2 y2 (u1 a1 ) y1 (u2 a2 ) y2 y (c1 a1 ) y1 (c2 a2 ) y2 u1 y1 u2 y2 C1 y1 C2 y2 u1 y1 u2 y2
It is to be noted that for finding the particular solution y p , once we find the complementary function y c , we have to only find the function f ( x ) and for that, it does not matter whether the coefficients a2 , a1 and a0 are functions of x or constants. This thing we clearly understand in Section 3.2.6.
Example 3.84 [ME-2014 (2 marks)]: Consider two solutions x (t ) x1 (t ) and x (t ) x2 (t ) of the differential equation ( d 2 x (t ) dt 2 ) x (t ) 0 , t 0 such that x1 (0) 1 , ( dx1 (t ) dt ) t 0 0 , x2 (0) 0 ,
(dx2 (t ) dt ) t 0 1 . The Wronskian W (t ) (a) 1
x1 (t )
x2 (t )
dx1 (t ) dt
dx2 (t ) dt
(b) –1
2
Solution (a): The given DE d x (t ) dt
(c) 0 2
x(t ) 0
at t 2 is (d) 2
…(i) is solved by putting x e mt x me mt
2 mt mx 2 mx 2 x m e , so e ( m 1) 0 as e 0 so the auxiliary equation is ( m 1) 0 m 0 i so we have complex conjugate roots; so the general solution of (i) is given as: x (t ) e 0 (c1 cos t c2 sin t ) c1 cos t c2 sin t x(t ) c1 sin t c2 cos t . Now applying the given
condition for x1 (t ) , we get x1 (0) 1 1 c1 and x1 (0) 0 0 c2 ; so x1 (t ) sin t x1 cos t . Similarly, applying the given condition for x2 (t ) , we get x2 (0) 0 0 c1 and x2 (0) 1 1 c2 ; so x2 (t ) cos t x2 sin t . Thus W (t )
sin t
cos t
cos t
sin t
2 2 sin t cos t 1 . So W (t ) 1 for all
values of t .
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Chapter 3: Differential Equation
[3.37]
Higher Order Equations: The method that we have just examined for non – homogeneous second-order differential equations can be generalized to linear nth -order equations that have been put into the standard form y ( n ) pn 1 y ( n 1) p1 y p0 y f ( x ) (3.61) If yc c1 y1 c2 y2 cn yn is a complementary function for Eq. 3.61, then a particular solution is y p u1 ( x) y1 ( x ) u 2 ( x) y2 ( x ) u n ( x ) yn ( x ) where u1 , u2 , , un are find by the n equations
y1u1 y2 u2 yn un 0 y1u1 y2 u2 yn un 0
(3.62)
y1( n 1)u1 y2( n 1)u2 yn( n 1)un f ( x )
The first n 1 equations in this system, like y1u1 y2u2 0 in Eq. 3.61, are assumptions that are made to simplify the resulting equation after y p u1 ( x ) y1 ( x ) u 2 ( x ) y2 ( x ) un ( x ) yn ( x ) is substituted in Eq. 3.61. In this case Cramer’s Rule gives, uk Wk W , k 1, 2, , n
(3.63)
where, W is the Wronskian of y1 , y2 , , yn and Wk is the determinant obtained by replacing the k th column of the Wronskian by the column consisting of the right hand side of Eqs. 3.62, i.e., the column consisting of 0, 0, , f ( x ) . When n 2 , we get Eqs. 3.59. When n 3 , the particular solution is y p u1 y1 u 2 y2 u3 y3 , where y1 , y 2 and y3 constitute a linearly independent set of solutions of the
associated homogeneous DE and u1 , u2 and u3 are determined from u1 W1 W , u2 W2 W and u3 W3 W
where W1
(3.64)
0
y2
y3
y1
0
y3
y1
y2
0
y1
y2
y3
0
y2
y3 , W2 y1
0
y3 , W3 y1
y2
0
and W y1
y2
y3
f ( x)
y2
y3
y1 y2
y3
y1
f ( x)
y3
y1 y2
f ( x)
3.2.6 Cauchy – Euler Equation A nth -order non – homogeneous linear differential equation of the form an x n ( d n y dx n ) an 1 x n 1 (d n 1 y dx n 1 ) a1 x ( dy dx ) a0 y g ( x )
(3.65)
where the coefficients an ( 0), an 1 , , a1 , a0 are constants, is known as Cauchy – Euler equation. The observable characteristic of this type of equation is that the degree k n, n 1, ,1, 0 of the monomial coefficients x k matches the order k of differentiation d k y dx k . We focus our attention on finding the general solution defined on the interval (0, ) since for Eq. 3.65 to be of nth -order we must have x 0 . The homogeneous Cauchy – Euler equation associated to Eq. 3.65 is of the form an x n (d n y dx n ) an 1 x n 1 ( d n 1 y dx n 1 ) a1 x (dy dx ) a0 y 0 (3.66)
Method of Solution: The general solution of Eq. 3.65 is y yc y p , where y c (complementary function) is the solution of associated homogeneous linear Cauchy – Euler Eq. 3.66 and y p is particular solution of non – homogeneous Cauchy – Euler Eq. 3.65. For Eq. 3.66, we try a solution of the form y x m , where m is to be determined. When we substitute y x m , each term of a Cauchy – m Euler equation becomes a polynomial in m times since x , k k k k m k m ak x ( d y dx ) a k x m (m 1)( m 2) (m k 1) x ak m ( m 1)( m 2) ( m k 1) x . For the
particular solution, y p , we will use variation of parameter methods (discussed in Section 3.25). We will discuss the case for second order Cauchy – Euler equation.
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Chapter 3: Differential Equation
[3.38]
Second Order Cauchy – Euler equation: The non – homogeneous second order Cauchy – Euler equation has the form ax 2 (d 2 y dx 2 ) bx (dy dx ) cy g ( x ) (3.67) where, a( 0), b, c are constant and x 0 . The homogeneous second order Cauchy – Euler equations associated to Eq. 3.67 is given as ax 2 (d 2 y dx 2 ) bx ( dy dx ) cy 0 (3.68)
When we substitute y x m , the second order equation becomes, am( m 1) x m bmx m cx m 0 am ( m 1) bm c x m 0 . Thus y x m is a solution of the differential equation whenever m is the solution of the auxiliary equation am( m 1) bm c 0 or am 2 (b a ) m c 0 (3.69)
The discriminant of Eq. 3.69 is D (b a) 2 4ac ; and let m1 and m2 are their roots. There are three different cases to be considered, depending on whether the roots of this quadratic equation are real and distinct, real and equal, or complex. Case 1: Distinct real roots: If D 0 then let m1 and m2 denotes the real roots of Eq. 3.69 such that m1 m2 . Then y1 x solution is y c1 x
m1
c2 x
m1
and y2 x
m2
form a fundamental set of solutions. Hence the general
m2
Case 2: Repeated real roots: If D 0 then m1 m2 (b a ) 2a , then we obtain only one m
solution as, y x 1 . We first write the Cauchy – Euler equation in the standard form as, ( d 2 y dx 2 ) (b ax )( dy dx ) {c (ax 2 )} y 0 and make the identifications as P ( x) b ax and
b ax dx b a ln x . Thus, m 2m m 2 m m m m y2 x {e ( b a )ln x x }dx x x ( b a ) x dx x x ( b a ) x ( b a ) a dx x (1 x) dx x 1
1
1
1
1
m
1
1
ln x
m
Hence the general solution for this case is y c1 x 1 c2 x 1 ln x Case 3: Conjugate Complex roots: If the roots of Eq. 3.69 are the conjugate pair m1 , m2 i , where and ( 0) are real, then a solution is y c1 x i c2 x i . But we wish to write the solution in terms of real functions only. We note the identity, x i (eln x )i e i ln x cos( ln x ) i sin( ln x ) x i cos( ln x) i sin( ln x) x i x i 2 cos( ln x ) y c1 x
i
i
c2 x
x i x i 2i sin( ln x ) .
and
From
the
fact
is a solution for any values of the constants, we see, in turn, for c1 c2 1 c1 1, c2 1 y1 x ( xi x i ) 2 x cos( ln x )
and
i
y2 x ( x x
i
that and
) 2 x sin( ln x ) are the solutions.
Since W { x cos( ln x), x sin( ln x)} x 2 1 0 , 0 on the interval (0, ) , we conclude that y1 x cos( ln x ) and y2 x sin( ln x ) constitute a fundamental set of real solutions of the differential equation. Hence the general solution is given as, y x c1 cos( ln x) c2 sin( ln x) .
For the particular solution, y p , we will use variation of parameter methods (discussed in Section
3.25). Example 3.85 [ME-1998 (5 marks)]: The radial displacement in a rotating disc is governed by the differential equation: ( d 2 u dx 2 ) (1 x )( du dx ) (u x 2 ) 8 x , where u is the displacement and x is the radius. If u 0 at x 0 , and u 2 at x 1 . Calculate the displacement at x 1 2 . Solution: The given differential equation can be written as: x 2 ( d 2 u dx 2 ) x ( du dx) u 8 x 3 2
2
…(i)
2
whose associated homogeneous part is x ( d u dx ) x (du dx ) u 0 …(ii), which is in the form of
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Chapter 3: Differential Equation
Euler – Cauchy equation; so putting ( m 2)
2
( m 1)
m
m
[3.39]
u x m u mx ( m1) u m( m 1) x ( m 2) 2
in (ii),
2
m
x m(m 1) x xmx x 0 x ( m 1) 0 as x 0 so ( m 1) 0 is the auxiliary equation whose roots are m 1 , which is the case of real distinct roots. So the complementary solution of (ii) is u c c1 x 1 c2 x . As g ( x ) 8 x 3 the choice of particular solution of (i) is
u p Ax 3 Bx 2 Cx D up 3 Ax 2 2 Bx C u p 6 Ax 2 B ;
so
putting
these
in
(i)
x 2 (6 Ax 2 B ) x (3 Ax 2 2 Bx C ) ( Ax 3 Bx 2 Cx D) 8 x3 8 Ax 3 3Bx 2 0 x D 8 x 3 3
A 1, B D 0 . So the particular solution is u p x Cx . So the general solution of (i) is
u uc u p c1x 1 c2 x x 3 Cx x 3 cx c1 x 1 , where c c2 C . Now applying the given conditions 3
u (0) 0 0 0 0 (c1 0) c1 0 ;
u (1) 2 2 13 c c 1 .
Thus
3
u ( x ) x x u (1 2) (1 2) (1 2) 5 8 .
Example 3.86 [ME-1998 (2 marks)]: The general solution of the differential equation x 2 ( d 2 y dx 2 ) x( dy dx) y 0 is (where, A and B are constants) (b) Ax B log x (d) Ax Bx log x (a) Ax Bx 2 (c) Ax Bx 2 log x Solution (d): As the given equation is in the form of homogeneous Cauchy – Euler equation, so putting y x m y mx ( m 1) y m(m 1) x ( m 2) in the given equation we get, x 2 m(m 1) x ( m 2) xmx ( m1) x m 0 x m ( m 2 2m 1) 0
as
m
x 0
so
( m 2 2m 1) 0
(m 1) 2 0 m 1 . As we have reapeated real roots so the genereal solution of the given
homogeneous Euler-Cauchy equation is y c1 x c2 x ln x . Example 3.87 [ME-2012 (2 marks)]: Consider the differential equation 2 2 2 x ( d y dx ) x ( dy dx ) 4 y 0 with the boundary condition of y (0) 0 and y (1) 1 . The complete solution of the differential equation is (b) sin( x 2) (c) e x sin( x 2) (d) e x sin( x 2) (a) x 2 Solution (a): As the given equation is in the form of homogeneous Cauchy – Euler equation, so putting y x m y mx ( m 1) y m(m 1) x ( m 2) in the given equation we get, x 2 m(m 1) x ( m 2) xmx ( m1) 4 x m 0 x m ( m 2 4) 0 as x m 0 so ( m 2 4) 0 m 2 . As we have distinct real roots so the genereal solution of the given homogeneous Euler-Cauchy equation is y c1 x 2 c2 x 2 . Now applying the given condition y (0) 0 0 (c1 0) c2 0 c1 0 ; and y (1) 1 1 c2 (1) 2 c2 1 . So the general solution is y x 2 . [Similar question was also asked in TF-2007 (2 marks)]
Example 3.88 [XE-2013 (2 mark)]: The general solution of the differential equation x 3 ( d 3 y dx 3 ) x 2 ( d 2 y dx 2 ) x ( dy dx ) y 0 , x 0 is (where, k1 ( 3 2) x , k2 ( 3 2) log e x ). x x 2 (a) c1e e c2 cos(k1 ) c3 sin(k1 )
1 2 (b) c1 x x c2 cos(k2 ) c3 sin(k2 )
x x 2 12 (c) c1e e c2 cos(k1 ) c3 sin(k1 ) (d) c1 x x c2 cos(k2 ) c3 sin(k2 ) Solution (d): As the given differential equation is of the form of Cauchy-Euler equation so putting y x m y mx ( m1) y m( m 1) x ( m 2) y m( m 1)(m 2) x ( m 3) in the given equation we m 2 get x m(m 1)(m 2) m(m 1) m 1 0 (m 1)(m m 1) 0 as x m 0 . So the roots of
auxiliary equation are: m 1, (1 3i ) 2 . So the general solution of the given differential equation is
y c1 x x1 2 [c2 cos{( 3 2) x} c3 sin{( 3 2) x}]
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Chapter 3: Differential Equation
[3.40]
Example 3.89 [EE-2014 (1 mark)]: Consider the differential equation 2 2 2 x (d y dx ) x (dy dx ) y 0 . Which of the following is a solution to this DE for x 0 ? (d) ln x (c) 1 x (a) e x (b) x 2 Solution: The given differential equation is in the form of homogeneous 2nd order Euler-Cauchy equation so putting y x m y mx ( m 1) y m(m 1) x ( m 2) in the given equation we get, x 2 m(m 1) x ( m 2) xmx ( m1) x m 0 x m ( m 2 1) 0 as x m 0 so ( m2 1) 0 m 1 . Thus
we have distinct real root case and hence the general solution is y c1 x 1 c2 x . So from the given options we can say that option (c) is correct. Example 3.90 [XE-2014 (1 mark)]: Which of the following is a solution of the differential equation x 2 y xy y 4 sin(ln x ) , x 0 ? (a) y 2 x sin(ln x) (b) y 2 x sin(ln x) (c) y 2 ln x cos(ln x) (d) y 2 ln x cos(ln x) Solution: The associated homogeneous part of the given differential equation is x 2 y xy y 0 which is in the form of Euler-Cauchy equation so putting m ( m 1) ( m 2) y x y mx y m(m 1) x in the given equation we get
x m m(m 1) m 1 0 (m2 1) 0 as x m 0 . So the roots of auxiliary equation are: m 0 i . 0 So the complementary solution yc x c1 cos(ln x) c2 sin(ln x ) c1 cos(ln x ) c2 sin(ln x) . As we
have duplication of y c with the function g ( x ) 4 sin(ln x) so the choice of our particular solution is y p A ln x cos(ln x) B ln x sin(ln x ) . Thus options (a) and (b) are not correct; now we have to
choose between options (c) and (d). By checking these two option we found option (c) is correct as: y p 2 ln x cos(ln x) yp (2 x) cos(ln x ) ln x sin(ln x )
yp 4 sin(ln x) 2 ln x cos(ln x ) 2 cos(ln x ) 2 ln x sin(ln x) x 2 . Now putting these into the given
equation
we
LHS 4 sin(ln x ) 2 ln x cos(ln x ) 2 cos(ln x ) 2 ln x sin(ln x )
get
2 cos(ln x ) 2 ln x sin(ln x ) 2 ln x cos(ln x ) 4 sin(ln x ) RHS .
Example 3.91: Solve x 2 y 3xy 3 y 2 x 4 e x Solution: Since the equation is nonhomogeneous, we first solve the associated homogeneous equation. From the auxiliary equation ( m 1)( m 3) 0 , we find yc c1 x c2 x3 . Now for a particular solution, we set y p u1 y1 u 2 y2 , where u1 W1 W and u2 W2 W , where W1 , W2 and
W are the determinants defined by Eq. 3.63 under the assumption that the differential equation has been put into the standard form y P ( x ) y Q ( x ) y f ( x ) . Therefore the given differential equation in standard form is given as, y (3 x) y (3 x 2 ) y 2 x 2 e x f ( x ) 2 x 2e x . Now with y1 x W2
and
x
0
1
2x e
2 x
3
y2 x ,
we
have,
3 x 2 x e , we find
u 2 (2 x 3 e x ) (2 x 3 ) e x u 2 e x .
W
x 1
x3 3x
2
3
2x ,
W1
0 2 x
2x e
x3 3x
2
2 x 5 e x
u1 (2 x 5 e x ) (2 x 3 ) x 2 e x u1 x 2 e x 2 xe x 2e x
Hence
y p ( x 2 e x 2 xe x 2e x ) x e x x 3
and
and and thus
y c1 x c2 x 3 ( x 2 e x 2 xe x 2e x ) x e x x3 .
Reduction to Constant coefficients: The similarities between the forms of solutions of CauchyEuler equations and solutions of linear equations with constant coefficients are not just a coincidence. For example, when the roots of the auxiliary equations for ay by cy 0 and
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Chapter 3: Differential Equation
[3.41]
ax 2 y bxy cy 0 are distinct and real, the respective general solutions are y c1e m1
y c1 x
c2 x
m2
, x 0.
In view of the identity e
ln x
m1x
c2 e
m2 x
and
x , x 0 , the second solution can be
m1 ln x
m ln x
mt
m t
expressed in the same form as the first solution y c1e c2 e 2 c1e 1 c2 e 2 , where t ln x . This last result illustrates the fact that any Cauchy-Euler equation can always be rewritten as a linear differential equation with constant coefficients by means of the substitution x et . The idea is to solve the new differential equation in terms of the variable t , using the methods of the previous sections, and, once the general solution is obtained, re-substitute t ln x . Example 3.92 [TF-2008 (2 marks)]: The second order differential equation x 2 ( d 2 y dx 2 ) 5 x ( dy dx ) 4 y 0 under the transformation z ln x , transform to an ordinary differential equation with constant coefficients, which is given by (a) ( d 2 y dz 2 ) 5( dy dz ) 4 y 0 (b) ( d 2 y dz 2 ) (1 5)( dy dz ) 4 y 0 (c) ( d 2 y dz 2 ) 4( dy dz ) 4 y 0
(d) ( d 2 y dz 2 ) (1 4)(dy dz ) 4 y 0
Solution (c): Let x e z or z ln x , it follows that, dy dx ( dy dz )( dz dx ) ( dy dz )(1 x) d2y
d dy 1 d dy dy 1 1 d 2 y 1 dy 1 1 d 2 y dy 2 2 2 2 2 2 . dx dx dx x dx dz dz x x dz x dz x x dz dz Substituting in the given differential equation 1 d 2 y dy d2y dy dy 1 4 4y 0 . x2 2 2 5x 4 y 0 2 dz dz dz dz x x dz
< : In the preceding discussion we have solved Cauchy-Euler equations for x 0 . One way of solving a Cauchy-Euler equation for x 0 is to change the independent variable by means of the substitution t x (which implies t 0 ) and using the Chain Rule: Solution for
d dy dt d 2 y and . dx dt dx dt dx 2 dt dt dx dt 2 dy
dy dt
d2y
dy
Example 3.93 [TF-2007 (2 marks)]: Using exp( x 2 ) as an integrating factor, the solution of the first order differential equation y 2 xy 1 in terms of the error function [erf ( x )] and a constant of integration c , is given by (a) y c ( 2)erf ( x ) exp( x 2 )
(b) y ( 2)erf ( x ) c exp( x 2 )
(c) y ( 2)erf ( x ) c exp( x 2 )
(d) y c ( 2)erf ( x) exp( x 2 ) 2
Solution (c): Multiplying both sides of the given differential equation with e x , we get 2
2
2
2
2
2
2
( dy dx )e x 2 xye x e x ( d dx )( ye x ) e x d ( ye x ) e x dx ; integrating both sides we
get
d ( ye
defined
x2
2
2
2
2
2
) e x dx ye x e x dx c y { e x dx c}e x . Now error function is
as
erf ( x ) (2
x
2
) e t dt . 0
Thus
e
x2
dx ( 2)erf ( x ) .
So
2
x y ( 2)erf ( x ) c e .
Statement for Linked Answer Question 3.94 and 3.95: Let y ( x ) n 0 an x n be a solution of the differential equation ( d 2 y dx 2 ) xy 0 Example 3.94 [XE-2007 (2 marks)]: The value of a11 is (a) 0 (b) 1 (c) 2
Copyright © 2016 by Kaushlendra Kumar
(d) 3
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Engineering Mathematics
Chapter 3: Differential Equation
[3.42]
Solution: Given y ( x ) n 0 an x n …(i) is the solution of the given differential equation. So
differentiating (i) w.r.t. x we get, y ( x) n 1 nan x n 1 …(ii). Again differentiating (ii) w.r.t. x , we
get y ( x) n 2 n( n 1) an x n 2 . Now putting the results in the given differential equation we get,
n 2 n(n 1)an xn 2 x n 0 an x n 0 n 2 n(n 1)an x n 2 n 0 an x n1 2a2 n 1( n 2)( n 1)an 2 an 1 x n 0 a2 0 and an 2 an 1 ( n 1)( n 2) . Thus we have the relations, a3 a0 6 , a4 a1 12 , a5 a2 20 0 , a6 a3 30 , a7 a4 42 , a8 a5 56 0 , a9 a6 72 , a10 a7 90 , a11 a8 110 0 .
Example 3.95 [XE-2007 (2 marks)]: The solution of the differential equation given above satisfying y (0) 1 and y (0) 0 is (a) y ( x ) 1 1 (2 3) x 2 1 (2 3 5 6) x 4 1 (2 3 5 6 8 9) x 6 (b) y ( x ) 1 1 (2 3) x 2 1 (2 3 5 6) x 4 1 (2 3 5 6 8 9) x 6 (c) y ( x ) 1 1 (2 3) x 3 1 (2 3 5 6) x 6 1 (2 3 5 6 8 9) x 9 (d) y ( x ) 1 1 (2 3) x 3 1 (2 3 5 6) x 6 1 (2 3 5 6 8 9) x 9 Solution (d): From the previous problem
n
we
have
n
y ( x ) n 0 an x a0 a1 x an x y (0) a0 1
y ( x ) n 1 nan x n 1 a1 2a2 x nan x n 1 y(0) a1 0
As from the recursion relation we have a3 1 (2 3) a0 1 (2 3) , a6 1 (5 6) a3 1 (2 3 5 6) ,
a2 0
(n 1)(n 2) a5 1 (4 5) a2 0 , a8 1 (7 8) a5 0 ,
an 2 an 1
and
a4 1 (3 4) a1 0 , a7 1 (6 7) a4 0 ,
a9 1 (8 9) a6 1 (2 3 5 6 8 9) , and so on. Thus substituting the obtained result in (i) we get y ( x ) 1 1 (2 3) x 3 1 (2 3 5 6) x 6 1 (2 3 5 6 8 9) x 9 .
Exercise: 3.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The solution of the IVP y 9 y 0 , y (0) 2 , y (0) 1 is (a) y ( x ) (7 6)e 3 x (5 6)e 3 x (c) y ( x ) (6 7)e3 x (6 5)e 3 x 2. One of the solutions of y 9 y 0 is
(b) y ( x ) (7 6)e3 x (5 6)e 3 x (d) y ( x ) (6 7)e 3 x (6 5)e 3 x
(a) e x (b) e 2 x (c) e3x 3. The solution of the IVP y 11 y 24 y 0 , y (0) 0 , y (0) 7 is (a) y ( x ) (7 5)(e 8 x e 3 x )
(d) e 4 x
(b) y ( x ) (5 7)(e 8 x e 3 x )
(c) y ( x ) (5 7)(e 8 x e 3 x ) (d) y ( x ) (7 5)(e 8 x e 3 x ) 4. The solution of the IVP 4 y 5 y 0 , y ( 2) 0 , y (2) 7 is (a) y ( x ) (28 5){1 e (5 x 4) (5 2) }
(b) y ( x ) (28 5){1 e (5 x 4) (5 2) }
(c) y ( x ) (28 5){1 e (5 x 4) (5 2) } (d) y ( x ) (28 5){1 e (5 x 4) (5 2) } 5. The solution of the IVP y 4 y 9 y 0 , y (0) 0 , y (0) 8 is (a) y ( x ) (8
t
5)e sin( 5t )
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(b) y ( x ) (8
2t
5)e sin( 5t )
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Engineering Mathematics
Chapter 3: Differential Equation
[3.43]
(c) y ( x ) (8 5)e 2 t sin( 5t ) (d) y ( x ) (8 5)et sin( 5t ) 6. The solution of IVP y 8 y 17 y 0 , y (0) 4 , y (0) 1 is (a) y ( x ) 4e 4 x cos x 15e 4 x sin x
(b) y ( x ) 4e 4 x cos x 15e 4 x sin x
(c) y ( x ) 4e 4t cos x 15e 4 x sin x (d) y ( x ) 4e 4t cos x 15e 4 x sin x 7. The solution of IVP y 16 y 0 , y( 2) 10 , y( 2) 3 is (a) y( x) 10 cos(4 x) (3 4) sin(4 x)
(b) y( x) 10 cos(4 x) (3 4) sin(4 x)
(c) y( x) 10 cos(4 x) (3 4) sin(4 x) (d) y( x) 10 cos(4 x) (3 4)sin(4 x) 8. The solution of IVP y 4 y 4 y 0 , y (0) 12 , y (0) 3 is (a) y ( x ) 12e 2 x 27(1 x)e 2 x
(b) y ( x ) 12e 2 x 27 x 2 e 2 x
(c) y ( x ) 12e 2 x 27 xe 2 x (d) y ( x ) 12e 2 x 27 x3 e 2 x 9. The solution of IVP y 14 y 49 y 0 , y ( 4) 1 , y (4) 5 is (a) y ( x ) 9e 7( x 4) 2 x 2 e 7( x 4)
(b) y ( x ) 9e 7( x 4) 2 xe 7( x 4)
(c) y ( x ) 9e 7( x 4) 2 xe7( x 4)
(d) y ( x ) 9e 7( x 4) 2 xe 7( x 4)
10. The general solution of 2 x 2 y xy 3 y 0 , given that y1 ( x ) 1 x is a solution, where c1 is a constant, is (a) y ( x ) c1 (1 x ) c2 x 3 2 (b) y ( x ) c1 (1 x ) c2 x1 2 (c) y ( x ) c1 (1 x) c2 x 5 2
(d) y ( x ) c1 (1 x ) c2 x 2
11. The general solution of x y 2 xy 2 y 0 , given that y1 ( x ) x is a solution, where c1 is a constant, is (a) y ( x ) c1 x c2 (1 x ) (b) y ( x ) c1 x c2 (1 x 2 ) (c) y ( x ) c1 x c2 ( x 2 )
(d) y ( x ) (c1 c2 x ) x1 2
12. The Wronskian of two solution to the following DE t 4 y 2t 3 y t 8 y 0 , where c is a constant, is (a) c (b) ct (d) c(1 t ) (c) ct 2 5t 13. The particular solution of the DE y 4 y 12 y 3e is (a) y p (3 7)e
5t
(b) y p (3 7)e
5t
(c) y p (7 3)e
5t
(d) y p (7 3)e
5t
14. The particular solution of the DE y 4 y 12 y sin(2t ) is (a) y p (1 40) cos(2t )
(b) y p (1 20) sin(2t )
(c) y p (1 40) cos(2t ) (1 20) sin(2t )
(d) None of these
15. The particular solution of the DE y 4 y 12 y 2t 3 t 3 is 3
2
(b) y p (1 6)t (1 6)t (1 9)t (5 27)
3
2
(d) y p (1 6)t (1 6)t (1 9)t (5 27)
(a) y p (1 6)t (1 6)t (1 9)t (5 27) (c) y p (1 6)t (1 6)t (1 9)t (5 27)
3
2
3
2
16. The particular solution of the DE y 4 y 12 y te 4 t is (a) y p (1 36)(3t 1) e (c) y p (1 36)e
4t
4t
(b) y p (1 36)(3t 1)e (d) y p (1 36)e
4t
4t
17. The particular solution of the DE y p ( x ) y q ( x) y 16e 7 x sin(10 x ) , where A, B are constants is 7x 7x (a) y p Ae cos(10 x ) (b) y p Ae sin(10 x) (c) y p A cos(10 x) B sin(10 x )
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7x
(d) y p e { A cos(10 x ) B sin(10 x )}
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Engineering Mathematics
Chapter 3: Differential Equation
[3.44]
18. The general solution of the DE y 2 y y e x ( x 2 1) , where c1 , c2 are constants is (a) y ( x ) c1e x c2 xe x (1 2)e x ln(1 x 2 ) xe x tan 1 x (b) y ( x ) c1e x c2 xe x e x ln(1 x 2 ) xe x tan 1 x (c) y ( x ) c1e x c2 xe x (1 2)e x ln(1 x 2 ) x tan 1 x (d) y ( x ) c1e x c2 xe x (1 2) ln(1 x 2 ) xe x tan 1 x 19. The general solution of the DE ty ( x 1) y y x 2 , where c1 , c2 are constants, is (a) y ( x ) c1e x c2 ( x 1) x 2 2 x 2
(b) y ( x ) c1e x c2 ( x 1) x 2 2 x 2
(c) y ( x ) c1e x c2 ( x 1) x 2 2 x 2 (d) y ( x ) c1e x c2 ( x 1) x 2 2 x 2 20. The solution of the IVP y 5 y 22 y 56 y 0 , y (0) 1 , y (0) 2 , y (0) 4 is (a) y ( x ) (14 33)e 4 x (13 15)e 2 x (1 5) e7 x
(b) y ( x ) (14 33)e 4 x (13 15)e 2 x (16 55)e 7 x
(c) y ( x ) (14 33)e 4 x (3 5)e 2 x (16 55)e 7 x
(d) y ( x ) (4 3)e 4 x (13 15)e 2 x (16 55)e 7 x
21. The general solution of the DE y (4) 16 y 0 , where c1 , c2 , c3 , c4 are constants, is (a) y ( x ) c1e
2x
(c) y ( x ) c1e
2x
(d) y ( x ) c1e
2x
cos( 2 x) c2e
2x
{cos( 2 x ) sin( 2 x )} c2 e cos( 2 x ) c2e
2x
(b) y ( x ) c1e
sin( 2 x ) 2x
4t
sin( 2 x ) c3e
2 4t
cos( 2 x ) c2e
2x
sin( 2 x)
{cos( 2 x) sin( 2 x)} 2x
cos( 2 x ) c4e
22. The general solution of the DE y 12 y 48 y 64 y 12 32e 4t
2x
(a) y (t ) c1e c2 te c3t e (3 16) (1 54)e
8 t
8 t
2x
2e
sin( 2 x )
4t
is
3 4t
(1 3)t e , where c1 , c2 , c3 are constants.
(b) y (t ) c1e 4 t c2 te 4 t c3t 2 e 4 t (3 16) (1 54)e 8 t t 3 e 4t , where c1 , c2 , c3 are constants. (c) y (t ) c1e 4 t c2 te 4t c3t 2 e 4 t 1 e 8t t 3e 4 t , where c1 , c2 , c3 are constants. (d) y (t ) c1e 4 t c2 te 4 t c3t 2 e 4t (3 16) (1 54)e 8t (1 3)t 2 e 4 t , where c1 , c2 , c3 are constants. 23. The general solution of the DE y 2 y 21y 18 y 3 4e t , where c1 , c2 , c3 are constants, is (a) y (t ) c1e 2 t c2 e t c3e 6 t 1 e t te t (b) y (t ) c1e 3t c2 e t c3e 6 t 1 e t te t (c) y (t ) c1e 3t c2 e t c3e 6 t (1 6) (5 49)e t (2 7)te t (d) y (t ) c1e 2 t c2 e t c3 e 6t (1 6) (5 49)e t (2 7)te t 24. The general solution of BVP y 4 y 0 , y (0) 2 , y( 4) 10 is (a) y ( x ) 2 cos(2 x) 10 sin(2 x ) (b) y ( x ) 2 cos(2 x) 10 sin(2 x ) (c) y ( x ) 2 cos x 10 sin x (d) y ( x ) 2 cos x 10 sin x 25. The general solution of BVP y 4 y 0 , y (0) 2 , y (2 ) 2 , where c is a constants, is (a) y ( x ) 4 sin(2 x ) (b) y ( x ) 2 cos(2 x ) 4 sin(2 x) (c) y ( x ) 2 cos(2 x) (d) Infinitely many solution of the form y ( x ) 2 cos(2 x ) c sin(2 x) 26. The solution of BVP y 4 y 0 , y (0) 2 , y (2 ) 3 is (a) y ( x ) 2 cos(2 x) (b) y ( x ) 4 sin(2 x ) (c) y ( x ) 2 cos(2 x ) 4 sin(2 x) (d) No solution 27. The BVP y 3 y 0 , y (0) 7 , y (2 ) 0 has _____ solution(s). (a) Unique (b) Two set of (c) Infinitely many (d) No 28. The BVP y 25 y 0 , y (0) 6 , y ( ) 9 has _____ solution(s). (a) Unique (b) Two set of (c) Infinitely many (d) No
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Engineering Mathematics
3.3
Chapter 3: Differential Equation
[3.45]
Laplace Transform
If f ( x, y ) is a function of two variables, then a definite integral of f with respect to one of the variables leads to a function of the other variable. Similarly, a definite integral such as b
a K ( s, t ) f (t )dt
transforms a function f of the variable t into a function F of the variable s . We
are particularly interested in an integral transform where the interval of integration is unbounded, i.e.,
[0, ) . If f (t ) is defined for t 0 , then the improper integral
0
0
K ( s , t ) f (t )dt is defined as a limit
b
(3.70)
K ( s, t ) f (t )dt lim K ( s, t ) f (t )dt b 0
If the limit in Eq. 3.70 exists, then the integral exists or is convergent; if the limit does not exist, the integral does not exist and is divergent. The function K ( s , t ) in Eq. 3.70 is called the kernel of the transform. The choice K ( s, t ) e st as the kernel gives us an especially important integral transform. Let f be a function defined for t 0 . Then the integral,
b
{ f (t )} e st f (t ) dt lim e st f (t )dt 0
(3.71)
b 0
is said to be the Laplace transform (denoted by ) of f (t ) provided that the integral converges. When the defining integral (Eq. 3.71) converges, the result is a function of s . In general, we use a lowercase letter to denote the function being transformed and the corresponding capital letter to denote its Laplace transform, for e.g., { f (t )} F ( s ) , {g (t )} G ( s) , { y (t )} Y ( s ) , etc. Linearity of the Laplace Transform: The Laplace transform is a linear operation, i.e., for any function f (t ) and g (t ) whose transforms exist and the transform of af (t ) bg (t ) is given by {af (t ) bg (t )} a{ f (t )} b{g (t )} , where a and b are any constants. Proof:
0
0
0
{af (t ) bg (t )} e st {af (t ) bg (t )} a e st f (t ) dt b e st g (t ) dt a{ f (t )} b{ g (t )}
Laplace transform of some basic functions
{1} 1 s
{t n } n ! s n 1 , n 1, 2, 3,
{e at } 1 ( s a )
{cos( kt )} s ( s 2 k 2 )
{sin( kt )} k ( s 2 k 2 ) [This point was asked in EE-1995 (1 mark), CE-2003, ME-2003 (2 marks)] {sinh( kt )} k ( s 2 k 2 ) [This point was asked in AG-2009 (1 mark), PI-2008 (2 marks)]
{cosh( kt )} s ( s 2 k 2 ) [This point was asked in CE-2009 (2 marks), ME-1994, ME-1995 (1 mark)] Proofs of above results:
b
b
{1} 1 s ; from Eq. 3.71, {1} lim e st (1) dt lim ( e st s ) lim ( e sb 1) s 1 s b 0
0
b
b
{t n } n ! s n 1 , n 1, 2, 3, ; from Eq. 3.71, b
b
b
b 0
b 0
b
{t n } lim e st t n dt (t n e st s ) ( n s ) lim e st t n 1dt 0 ( n s ) lim e st t n 1dt 0
b 0
b
b
b 0
b 0
Let I n lim e st t n dt I n 1 lim e st t n 1dt I n ( n s) I n 1 . From this result we have, I n 1 {( n 1) s}I n 2 , I n 2 {( n 2) s}I n 3 , , I n ( n 1) [{n ( n 1)} s ]I n n (1 s ) I 0 . Hence
multiplying all results, we have I n {n(n 1)(n 2) 2 1} s n I 0 ( n ! s n ) I 0 . b
b
Since I 0 lim e st dt lim ( e st s ) lim{( e sb 1) s} 1 s b 0
In n! s
n 1
0
b
n
{t } n ! s
n 1
b
, n 1, 2, , n
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Engineering Mathematics
Chapter 3: Differential Equation
[3.46]
{e at } 1 ( s a ) ; from Eq. 3.71, we have, b
b
b 0
b 0
{e at } lim e st e at dt lim e ( s a ) t dt lim e ( s a ) t ( s a ) at
{e } {1 ( s a)} lim{e
b
( s a )b
b
b 0
1} 1 ( s a )
This result follows that by replacing a by a , then {e at } 1 ( s a ) .
{sin( kt )} k ( s 2 k 2 ) ; from Eq. 3.71, we have, b
b
b
{sin kt} lim e st sin kt dt lim{sin kt ( e st s) } ( k s ) lim e st cos kt dt 0
b 0
b 0
b
0
{sin kt} ( k s) lim ( e st s ) cos kt
b
b 0
(k s) lim e
b st
2
{sin kt} (k s) (1 s) (k s){sin kt} (k s ) (k
2
sin kt dt
b 0
2
s ){sin kt}
{sin kt} 1 ( k 2 s 2 ) k s 2 {sin kt} k ( s 2 k 2 )
2
2
{cos( kt )} s ( s k ) ; from Eq. 3.71, we have, b
{cos kt} lim e
st
b 0
st e cos kt dt lim cos kt b s
b
k b 1 k lim 0 e st sin kt dt {sin kt} s b s s 0
{cos kt} (1 s ) ( k s ){k ( s 2 k 2 )} (1 s) [ k 2 {s ( s 2 k 2 )}] s ( s 2 k 2 )
sinh( kt )} k ( s 2 k 2 ) sinh( kt ) (e kt e kt ) 2 {sinh( kt )} {(e kt e kt ) 2} (1 2){e kt e kt } .
Thus {sinh(kt )} (1 2) {e kt } {e kt } (1 2) {1 ( s k )} {1 ( s k )} k ( s 2 k 2 )
{cosh( kt )} s ( s 2 k 2 ) cosh( kt ) {(e kt e kt ) 2} {sinh( kt )} {(e kt e kt ) 2} (1 2){e kt e kt } .
Thus, {cosh( kt )} (1 2) {e kt } {e kt } (1 2) {1 ( s k )} {1 ( s k )} s ( s 2 k 2 ) Example 3.96 [ME-1999 (1 mark)]: Laplace transform of ( a bt ) 2 where, a and b are constants is given by (a) ( a bs ) 2 (b) 1 ( a bs) 2 (c) ( a 2 s ) (2ab s 2 ) (2b 2 s 3 ) (d) ( a 2 s ) (2ab s 2 ) (b 2 s 3 ) Solution (c): f (t ) ( a bt ) 2 a 2 2abt b 2t 2 ; so { f (t )} {a 2 2abt b 2t 2 } a 2 {1} 2ab{t } b 2 {t 2 } ( a 2 s) (2ab s 2 ) (2b 2 s 3 )
Example 3.97 [ME-2000 (2 marks)]: The Laplace transform of the function sin 2 2t is (a) {1 2s} [ s {2( s 2 16)}] (b) s ( s 2 16) (c) (1 s ) {s ( s 2 4)} (d) s ( s 2 4) Solution (a): f (t ) sin 2 2t (1 cos 4t ) 2 ; so { f (t )} (1 2) 1 cos 4t (1 2)({1} {cos 4t}) { f (t )} {1 2 s} [ s {2( s 2 16)}]
Example 3.98 [CH-2007 (2 marks)]: The Laplace transform of f (t ) 1 (a)
s
1s
(b)
(c)
1 s3 2
t is
(d) does not exist
Solution: { f (t )} {t 1 2 } e st t 1 2 dt . Now let st u 2 sdt 2udu dt (2u s )du . Also 0
2
st u t
12
u
s t
1 2
s u.
Also
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t 0u 0
and
t u .
So
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Engineering Mathematics
0 e
st 1 2
t
dt e
u 2
0
Chapter 3: Differential Equation
s 2u u
s
du
Gaussian integral whose value is
2
e s 0
u2
du
1
e s
u2
du
[3.47]
1 s
. As s
e
u2
du is a
.
Example 3.99 [PI-2008 (1 mark)]: Laplace transform of 8t 3 is (a) 8 s 4 (b) 16 s 4 (c) 24 s 4
(d) 48 s 4
Solution (d): {8t 3 } 8{t 3 } 8{3! s 31} 48 s 4 Example 3.100 [BT-2013 (2 marks)]: The Laplace transform of f (t ) 2t 6 is (a) (1 s) (2 s 2 )
(b) (3 s ) (6 s 2 )
(c) (6 s ) (2 s 2 )
(d) (6 s) (2 s 2 )
Solution (c): { f (t )} {2t 6} 2{t} 6{1} 2(1! s11 ) 6(1 s ) (2 s 2 ) (6 s) Example 3.101 [IN-2013 (2 marks)]: The Laplace Transform representation of the triangular pulse shown below is
(a) (1 e 2 s ) s 2 (b) (1 e s e 2 s ) s 2 (c) (1 e s 2e 2 s ) s 2 (d) (1 2e s e 2 s ) s 2
0 t 1 t, Solution (d): From the figure the function x (t ) can be written as: x(t ) 2 t , 1 t 2 . So 0, t2
1
0
0
2
{x (t )} e st x(t ) dt e st (t ) dt e st (2 t ) dt e st (0) dt 1
2
1
2
st 1e 2 e st e st e st { x(t )} t dt (2 t ) 1 dt 0 s 1 1 s s 0 0 s
e s
1
2
e st e s e st e s 1 e 2 s e s 1 2e s e 2 s {x (t )} 2 2 2 2 2 2 2 s s 0 s s 1 s s s s s
3.3.1 Inverse Transforms and Transforms of Derivatives Inverse Transforms: If F ( s ) represents the Laplace transform of a function f (t ) , i.e., { f (t )} F ( s ) , then we say that f (t ) is the inverse Laplace transform of F ( s ) and write f (t ) 1{F ( s )} . The inverse transform of basic functions are given as:
1 (1 s ) 1
1 ( n ! s n 1 ) t n , n 1, 2, 3,
1{1 ( s a )} e at
1{k ( s 2 k 2 )} sin( kt )
1{s ( s 2 k 2 )} cos( kt )
1{k ( s 2 k 2 )} sinh( kt )
1{s ( s 2 k 2 )} cosh( kt ) In evaluating the above transforms, it often happens that a function of s under consideration does not match exactly the form of a Laplace transform F ( s ) given above. It may be necessary to “fix up” the function of s by multiplying and dividing by an appropriate constant. is a Linear transform: The inverse Laplace transform is also a linear transform, i.e., for constant and , 1 F ( s ) G ( s ) 1{F ( s )} 1{G ( s )} , where F and G are the transforms of some functions f and g Example 3.102 [EC-1996 (2 marks)]: The inverse Laplace transform of the function ( s 5) {( s 1)(s 3)} is
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Engineering Mathematics
Chapter 3: Differential Equation
[3.48]
(a) 2e t e 3t
(b) 2e t e 3t (c) e t 2e 3t (d) e t e 3t s5 1 1 2 1 1 1 1 t 3 t Solution (d): 1 2 2e e s 1 s 3 s 1 s 3 ( s 1)( s 3) Example 3.103 [EC-1998 (1 mark)]: If { f (t )} ( s 2 2 ) , then the value of lim f (t ) t
(a) cannot be determined (b) is zero (c) is unity (d) is infinite 1 2 2 Solution (a): f (t ) { ( s )} sin(t ) lim f (t ) lim sin( t ) any value between –1 to t
t
1. So the limit cannot be determined. Example 3.104 [CE-2001 (2 marks)]: The inverse Laplace Transform of 1 ( s 2 2 s ) is (b) (1 2)(1 e 2 t )
(a) 1 e 2 t
(c) (1 2)(1 e 2 t )
(d) (1 2)(1 e 2 t )
1 1 1 1 1 1 1 1 1 1 1 2 t (1 e ) 2 s 2s s 2 2 2 s s 2 2 s
Solution (d): 1
Example 3.105 [EC-2003 (1 mark)]: The Laplace transform of i(t ) is given by I ( s ) 2 {s (1 s)} . As t , the value of i (t ) tends to (a) 0 (b) 1 (c) 2 (d) 1 2 1 1 1 1 1 1 t Solution (c): As i (t ) 1 2 2 2 2 1 2e s 1 s s 1 s s(1 s)
i (t ) 2(1 e t ) lim i (t ) lim 2(1 e t ) 2 t
t
Statement for Linked Answer Questions 3.106 and 3.107: Let F ( s ) ( s 10) {( s 2)( s 20)} Example 3.106 [AE-2007 (2 marks)]: The partial fraction expansion of F ( s ) is 1 1 5 2 2 20 49 59 (a) (b) (c) (d) s 2 s 20 s 2 s 20 s 2 s 20 s 2 s 20 ( s 10) A B s ( A B ) (20 A 2 B) A B 1 and Solution (d): F ( s) ( s 2)( s 20) s 2 s 20 ( s 2)( s 20)
20 A 2 B 10 ;
solving
these
two
equations
we
get,
A4 9
and
B 5 9;
thus
F ( s ) {(4 9) ( s 2)} {(5 9) ( s 20)} .
Example 3.107 [AE-2007 (2 marks)]: The inverse Laplace transform of F ( s ) is (a) 2e 2t 20e 20 t
(b) (4 9)e 2 t (5 9)e 20t
(c) 5e 2t 2e 20 t
(d) (9 4)e 2t (9 5)e 20 t
Solution (b): 1{F ( s)} (4 9) 1{1 ( s 2)} (5 9) 1{1 ( s 20)} (4 9)e 2 t (5 9)e 20 t [Similar question was also asked in AE-2009 (2 marks)] Example 3.108 [ME-2009 (1 mark), ME-2012, PI-2012 (2 marks)]: The inverse Laplace transform of 1 ( s 2 s ) is (b) 1 e t (c) 1 e t (d) 1 e t 1 1 1 1 1 1 t 1{F ( s )} 1 1 Solution (c): F ( s) 2 1 e s s s ( s 1) s s 1 s s 1 (a) 1 e t
Example 3.109 [ME-2010 (2 marks)]: The Laplace Transform of a function f (t ) [1 {s 2 ( s 1)}] . The f (t ) is
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Engineering Mathematics
Chapter 3: Differential Equation
[3.49]
(a) t 1 e t (b) t 1 e t (c) 1 e t (d) 2t e t Solution (a): Using partial fractions we have, F ( s) 1 {s 2 ( s 1)} (1 s ) (1 s 2 ) {1 ( s 1)} ; thus 1 1 1 1 1 1 1 1{F ( s)} 1 2 2 s s s 1 s s
1 1 t 1 t e . s 1
Transforms of Derivatives: To solve a differential equation using Laplace transform we need to evaluate quantities such as dy dt and (d 2 y dt 2 ) . If f and f are continuous for t 0 , then
0
0
0
integration by parts gives, { f (t )} e st f (t ) dt e st f (t ) s e st f (t ) dt f (0) s{ f (t )} { f (t )} s{ f (t )} f (0) sF ( s ) f (0) Similarly if f , f and f are continuous for t 0 , then integration by parts gives,
0
0
0
(3.72)
{ f (t )} e st f (t ) dt e st f (t ) s e st f (t )dt f (0) s{ f (t )}
{ f (t )} s sF (s ) f (0) f (0) s 2 F ( s) sf (0) f (0) Similarly, if f , f , f and f are continuous for t 0 , then we can also prove that, { f (t )} s3 F ( s ) s 2 f (0) sf (0) f (0)
And in general, if f , f , f , f , f f
( n)
( n 1)
(3.73) (3.74)
are continuous on [0, ) and are of exponential order and if
(t ) is piecewise continuous on [0, ) , then
{ f ( n ) (t )} s n F ( s ) s n 1 f (0) s n 2 f (0) f ( n 1) (0) where, F ( s) { f (t )} .
(3.75)
Now for solving linear initial – value problems in which the differential equation has constant coefficients; such a differential equation is simply a linear combination of terms y , y , , y ( n ) . an ( d n y dt n ) an 1 ( d n 1 y dt n 1 ) a0 y g (t ) ,
y (0) y0 , y (0) y1 , , y ( n 1) (0) yn 1 , where
the an , an 1 , , a1 , a0 and y0 , y1 , , yn 1 are constants. By the linearity property the Laplace transform of this linear combination is a linear combination of Laplace transform: an (d n y dt n ) an1(d n 1 y dt n1 ) a0 { y} {g (t )} ; from Eq. 3.75 it becomes an s nY ( s ) s n 1 y (0) y ( n 1) (0) an 1 s n 1Y ( s ) s n 2 y (0) y ( n 2) (0)
(3.76) a0Y ( s) G ( s) where, { y (t )} Y ( s ) and {g (t )} G ( s) . In other words, the Laplace transform of a linear equation with constant coefficient becomes an algebraic equation in Y ( s) . If we solve the general transformed Eq. 3.76 for the symbol Y ( s) , we first obtain P ( s )Y ( s ) Q( s ) G ( s ) and then write
Y ( s ) {Q( s) P ( s )} {G( s) P ( s )} n
where P ( s ) an s an 1s
n 1
(3.77)
a0 , Q ( s ) is a polynomial in s of degree less than or equal to
consisting of the various products of the coefficient an , an 1 , , a1 , a0 and the prescribed initial conditions y0 , y1 , , yn 1 , and G ( s ) is the Laplace transform of g (t ) . Typically, we put the two terms in Eq. 3.77 over the least common denominator and then decompose the expression into two or more partial fractions. Finally, the solution y (t ) of the original initial-value problem is
y (t ) 1Y ( s ) , where the inverse transform is done term by term. Initial Value Theorem: The initial value theorem determines the value of the time function, f (t ) , when t 0 without finding the inverse Laplace transform. Now, as s , we have e
st
0 thus lim { f (t )} lim ( d dt ) f (t ) lim s 0
s0
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s0 0
( d
dt ) f (t ) e st dt 0 . From Eq. 3.77
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Engineering Mathematics
Chapter 3: Differential Equation
[3.50]
lim { f (t )} lim sF ( s) f (0) . Thus equating the two results we have
we have
s
s
lim sF ( s) f (0) 0 lim{sF (s )} f (0) lim{sF (s )} lim f (t ) . Thus the initial value of
s
s
s
t 0
the function f (t ) is lim f (t ) lim{sF ( s )} . t 0
s
Final Value Theorem: The final value theorem determines the steady state value of the system response without finding the inverse Laplace transform. Now, as s 0 , we have e st 1 thus lim { f (t )} lim ( d dt ) f (t ) lim s 0
s0 0
s0
dt ) f (t ) e st dt d f (t ) f ( ) f (0) .
( d
From
0
Eq. 3.77 we have lim { f (t )} lim sF ( s ) f (0) . Thus equating the two results we have s 0
s 0
lim sF (s ) f (0) f () f (0) lim{sF (s )} f () lim{sF (s )} lim f (t ) . Thus the final s 0
s 0
s 0
t
value of the function f (t ) is lim f (t ) lim{sF ( s)} [This point was asked in EE-2002 (1 t
s 0
mark)]. Transfer Function: A transfer function is the ratio of the output of a system to the input of a system, in the Laplace domain considering its initial conditions and equilibrium point to be zero. If we have an input function of X ( s ) and an output function of Y ( s) , we define the transfer function H ( s) to be: H ( s ) Y ( s ) X ( s ) .
Example 3.110 [ME-2002 (5 marks)]: Using Laplace transform, solve ( d 2 y dt 2 ) 4 y 12t . Given that y 0 and dy dt 9 at t 0 . Solution: Taking Laplace transform on both sides of the given differential equation we get { y } 4{ y} 12{t} . Using equation 3.80 and given conditions, we have { y } s 2Y ( s ) sy (0) y (0) { y } s 2Y ( s ) 9 ; { y} Y ( s ) ; {t} 1 s 2 . Now substituting
these in { y } 4{ y} 12{t} , we get s 2Y ( s ) 9 4Y ( s ) 12 s 2 Y ( s ) Using Y (s)
partial 3 s
2
3 2
s 4
fractions
9 2
( s 4)
for 3 s
2
the 6 2
( s 4)
first
expression 1
1
3 s
2
9
. s ( s 4) s 4 Y ( s) , we get
of
y (t ) {Y ( s)}
12
2
2
2
( s 4) 6
2
y (t ) 31 (1 s 2 ) 31{2 ( s 2 22 )} 3t 3sin 2t [Similar question was also asked in ME-1997 (5 mark)]
Example 3.111 [EC-2006 (2 marks)]: The unit-step response of a system from rest is given by c (t ) 1 e 2 t for t 0 . The transfer function of the system is: (a) 1 (1 2 s ) (b) 2 (2 s ) (c) 1 (2 s ) (d) 2 s (1 2 s ) Solution (b): As the input of the system is c(0) 1 and output of the system is c (t ) 1 e 2 t . So we
X ( s) {c(0)} {1} 1 s ; also the output function is given as:
have input function
2 t
Y ( s ) {c (t )} {1 e } {1} {e 2 t } (1 s ) {1 ( s 2)} 2 {s ( s 2)} . Thus transfer function
H ( s ) Y ( s) X ( s) [2 {s ( s 2)}] s 2 ( s 2) . Example 3.112 [EC-2006 (2 marks)]: Consider the function f (t ) having Laplace transform F ( s ) 0 ( s 2 02 ) , Re[ s ] 0 . The final value of f (t ) would be: (a) 0 (b) 1 (c) 1 f () 1 1
1
2
2 0 )}
Solution (c): f (t ) {F (s )} {0 (s
(d)
sin(0t ) . Applying final value theorem, the
final value of f (t ) lim sin(0t ) ; as t , sin(0 t ) lies between –1 to 1 . So 1 f () 1 . t
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Chapter 3: Differential Equation
Example 3.113 [EC-2010 (2 marks)]: Given
[3.51]
f (t ) 1[(3s 1) {s 3 4 s 2 ( k 3) s}] . If
lim f (t ) 1 , then the value of k is
x
(a) 1 (b) 2 (c) 3 (d) 4 Solution (d): It is given that the final value of f (t ) is 1, i.e., lim f (t ) lim{sF ( s )} 1 . As t
3
2
F ( s ) (3s 1) {s 4 s ( k 3) s} , thus
s0
3
lim{sF ( s)} 1 lim[ s(3s 1) {s 4 s 2 ( k 3) s}] 1 s 0
s 0
2
lim[(3s 1) {s 4 s ( k 3)}] 1 1 (k 3) 1 k 3 1 k 4 . s 0
Example 3.114 [EC-2011 (2 marks)]: If F ( s) { f (t )} {2( s 1)} ( s 2 4s 7) then the initial and final values of f (t ) are respectively (a) 0, 2
(c) 0, 2 7
(b) 2, 0
(d) 2 7 , 0
Solution (b): The initial value of f (t ) is lim f (t ) lim{sF ( s )} lim[{2 s ( s 1)} ( s 2 4 s 7)] t 0
s
s
2
lim f (t ) lim{2(1 1 s ) (1 4 s 7 s )} lim{2(1 0) (1 0 0)} 2 . Also the final value of t0
s
s
the function f (t ) is lim f (t ) lim{sF ( s )} {2 s ( s 1) ( s 2 4 s 7)} 0 7 0 . t
s 0
[Similar questions were also asked in EC-2007 (1 mark), EE-2005, EE-2005 (2 marks)] Example 3.115 [ME-2013, PI-2013 (2 marks)]: The function f (t ) satisfies the differential equation ( d 2 f dt 2 ) f 0 and the auxiliary conditions, f (0) 0 , ( df dt )t 0 4 . The Laplace transform of f (t ) is given by
(a) 2 ( s 1) (b) 4 ( s 1) (c) 4 ( s 2 1) (d) 2 ( s 4 1) Solution (c): Taking Laplace transform on both sides of the given differential equation we get { f } { f } {0} . Using equation 3.80 and given conditions, we have { f } s 2 F ( s ) sf (0) f (0) { f } s 2 F ( s) 4 ; { f } F ( s ) ; {0} 0 . Now substituting
these in { f } { f } {0} , we get s 2 F ( s) 4 F ( s ) 0 F ( s ) 4 ( s 2 1) [Similar question was also asked in TF-2008 (2 marks)] Example 3.116 [AE-2014 (2 marks)]: The Laplace transform L{u (t )} U ( s) , for the solution u (t ) of the problem ( d 2 u dt 2 ) 2( du dt ) u 1 , t 0 with initial conditions u (0) 0 , ( du dt )t 0 5 is given by: (a) 6 ( s 1) 2 (b) (5s 1) {s ( s 1) 2 } (c) (1 5s ) {s ( s 1) 2 } (d) (5s 2 1) {s ( s 1) 2 } Solution (c): Taking Laplace transform on both sides of the given differential equation we get {u} 2{u} {u} {1} . Using equation 3.80 and given conditions, we have {u} s 2U ( s ) su (0) u(0) {u } s 2U ( s) 5 ;
{u} sU ( s ) u (0) {u } sU ( s ) ; {u} U ( s) ; {1} 1 s . Now substituting these in {u} 2{u} {u} {1} , we get
s 2U ( s ) 5 2 sU ( s) U ( s ) 1 s U ( s ) (1 5s ) {s ( s 2 2s 1)} (1 5s) {s ( s 1) 2 } .
Example 3.117 [EE-2014 (1 mark)]: Let X ( s) (3s 5) {s 2 10 s 21} be the Laplace Transform of a signal x (t ) . Then x(0 ) is (a) 0 (b) 3 (c) 5 (d) 21 Solution (b): We have to find the initial value of f (t ) whose Laplace transform is X ( s ) , thus lim f (t ) lim{sX ( s )} lim[ s (3s 5) {s 2 10 s 21}] lim[(3 5 s ) (1 10 s 21 s 2 )] 3 1 3
t 0
s
s
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s
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Engineering Mathematics
Chapter 3: Differential Equation
[3.52]
3.3.2 Important Singularity Functions The functions which are either not finite or they do not possess finite derivatives everywhere, are called singularity function, such as, (1) Unit Step Function; (2) Unit Impulse Function; (3) Delta Function; (4) Ramp Function. 0, t 0 Unit Step Function: The mathematical definition of unit step function is u (t ) , which 1, t 0 is shown in Fig. 3.1(a). The unit step function is not defined at t 0 . Thus, the unit step function u (t ) is ‘0’ for negative values of t , and ‘1’ for positive values of t . A discontinuity may occur at time other than t 0 , for e.g., the unit step function that occurs at t a is expressed as 0, t a , which is shown in Fig. 3.1(b); similarly the unit step function that occurs at u (t a ) 1, t a
0, t a
t a is expressed as u (t a )
1, t a
; which is shown in Fig. 3.1(c).
Figure 3.1: The Unit Step Function
When a function f defined for t 0 is multiplied by u (t a) , the unit step function ‘turns off’ a portion of the graph of that function for 0 t a . For e.g. consider the function f (t ) 2t 3 . To ‘turn off’ the portion of the graph of f for 0 t 1 , we form the product (2t 3)u (t 1) . Generally, graph of f (t )u (t 1) is 0 (off ) for 0 t a and is the portion of the graph of f (on) for t a . The unit step function can also be used to write piecewise-defined functions in a compact form. For e.g. if we consider 0 t 2 , 2 t 3 and t 3 and the corresponding values of u (t 2) and u (t 3) , it should be apparent that the piecewise-defined function shown in Fig. 3.2 Figure 3.2: The function ( ) is the same as f (t ) 2 3u (t 2) u (t 3) . A general piecewise-defined function of the A general piecewise-defined function of the 0, 0 t a g (t ), 0 t a type f (t ) is the same as type f (t ) g (t ), a t b is same as ta h(t ), 0, f (t ) g (t ) g (t )u (t a) h(t )u (t a) . t b f (t ) g (t ) u (t a) u (t b) .
Example 3.118 [EC-2005 (1 mark)]: The function x (t ) is shown in figure. Even and odd parts of a unit-step function u (t ) are respectively
(a) 1 2 , x (t ) 2 (b) 1 2 , x (t ) 2 (c) 1 2 , x (t ) 2 (d) 1 2 , x(t ) 2
Solution (a): As { f ( x ) f ( x )} 2 and { f ( x) f ( x )} 2 are the even and odd parts of any function
0, t 0
f ( x ) . Now as u (t )
1, t 0
0, t 0 1, t 0 . So even part of unit step 1, t 0 0, t 0
u ( t )
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function is
u (t ) u ( t )
Chapter 3: Differential Equation
(0 1) 2 , t 0
1
[3.53]
for all t . Also odd part of unit step function is
(1 0) 2 , t 0 2 u (t ) u (t ) (0 1) 2 , t 0 1 2 , t 0 x (t ) . 2 2 (1 0) 2 , t 0 1 2 , t 0 2
Unit Impulse function: Mechanical system are often acted on by an external force or electromotive force of large magnitude that acts only for a very short period of time. For e.g., a vibrating airplane wing could be struck by lightning; the graph of the piecewise-defined function 1 k , a t a k (3.78) a (t a ) otherwise 0, where, a 0 and k is a small positive number, for the figure shown in Fig. 3.3, could serve as a model for such a force. For a small value of k , a (t a ) is essentially a constant function of large magnitude that is ‘on’ for just a very short period of time, around a . The function a (t a ) is called a unit impulse,
Proof:
Figure 3.3: Graph of ( − 0 a (t a)dt 1 . a ak 0 a (t t0 )dt 0 a (t a )dt a a (t a)dt a k a (t a)dt , the first and the
because it possesses the integration property
)
third
integral becomes zero as a (t a) 0 for all t a and t a k ; and a (t a ) 1 k for all
a t a k . So
a k
0 a (t a)dt (1 k )a
dt ( k k ) 1 .
It is to be noted that the derivative of the unit step function u (t a) is called the unit impulse
function a (t a ) , i.e., a (t a) ( d dt )u (t a ) .
The unit impulse function can be written in terms a (t a ) {1 (2 a )}u (t a ) u t (a k ) .
of unit
step function as:
Dirac Delta Function: In practice it is convenient to work with another type of unit impulse, a function that approximates a (t a ) and is defined as (3.79) (t a) lim a (t a) k 0
The latter expression, which is not a function at all, can be characterized , t a by the two properties: (i) (t a) and (ii) (t a ) dt 1 Figure 3.4: The Dirac Delta Function 0, t a [These two properties were asked in EC-2006 (1 mark)]. The unit impulse (t a) is called the Dirac delta function.
For any function f ( x ) , which has all finite order derivative, we have
Example 3.119 [IN-2010 (1 mark)]: The integral (a) 6
{t (
(b) 3
6)}6 sin(t )dt evaluates to
(c) 1.5
(d) 0
Solution (b): Here f (t ) 6 sin t and a 6 so using the result
{t (
( x a) f ( x)dx f (a) .
( x a) f (t )dt f (a ) , we have
6)}6 sin(t ) dt f ( 6) 6 sin( 6) 3
[Similar question was also asked in EC-2001 (1 mark)]
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Chapter 3: Differential Equation
Example 3.120 [IN-2011 (1 mark)]: The integral (1 (a) {1 (8 2 )}e
1 8
(b) {1 (4 2 )}e
1 8
[3.54]
2
2 ) t 2 e t 2 (1 2t ) dt is equal to
(c) {1 ( 2 )}e 1 2
(d) 1
Solution (a): Let 1 2t k dt dk 2 ; also t , k and t , k ; thus
2
2 ) t 2 e t 2 (1 2t )dt {1 (2 2 )}
I (1
I {1 (2 2 )}
(1 k ) 2 2 e (1 k ) 2
I {1 (2 2 )} f (0) ,
2
2
(1 k ) 2 2 e (1k ) 2
( k ) dk I {1 (2 2 )}
2
f ( k ) (1 k ) 2 e
where
(1 k ) 2
2
2
2
2
( k )dk
f ( k ) ( k 0) dk
f (0) (1 4) e 1 8 ;
thus
I {1 2 2 }(1 4)e1 8 e1 8 (8 2 )
Unit Ramp Function: Integrating the unit step function results the unit ramp function r (t ) , i.e., t
0
t
t
0
0
r (t ) u ( )d u ( )d u ( ) d 0 1d ( )t0 t , which can be written as r (t ) t u (t )
0, t 0
, where u (t )
1, t 0
Similarly,
the
unit
0, t 0
. Thus r (t ) t u (t )
t,
ramp 0,
, which can be shown in Fig. 3.5(a). t 0 occurs at is expressed as ta
function that ta , which is shown in Fig. 3.5(b); similarly the unit step r (t a) (t a )u (t a ) (t a ), t a
0,
function that occurs at t a is expressed as r (t a) (t a )u (t a)
t a
(t a ), t a
, which is
shown in Fig. 3.1(c).
Figure 3.5: The Unit Ramp Function
Note: The singularity functions discussed above are related by differentiation as: (t ) ( d dt )u (t ) , u (t ) ( d dt ) r (t ) ; or
t
t
integration as: u (t ) ( ) d , r (t ) u ( ) d .
3.3.3 Operational Properties First Shifting theorem: If we know the Laplace transform of a function f , { f (t )} F ( s ) , it is possible to compute the Laplace transform of an exponential multiple of f , i.e., {e at f (t )} , by translating, or shifting, the transform F ( s ) to F ( s a) . This result is known as the first translation theorem or first shifting theorem, i.e., if { f (t )} F ( s ) and a is any real number then {e a t f (t )} F ( s a )
0
0
(3.80)
Proof: {e at f (t )} e st e at f (t ) dt e ( s a )t f (t ) dt F ( s a )
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Chapter 3: Differential Equation
[3.55]
If we consider s a real variable, then the graph of F ( s a) is the graph of F ( s ) shifted on the s axis by the amount a . If a 0 , the graph of F ( s ) is shifted a units to the right, as shown in Fig. 3.6; whereas if a 0 , the graph is shifted a units to the left. Using Figure 3.6: Shift on the − axis Eq. 3.80 the Laplace transform of functions are given as: Laplace Transform First shifting theorem at {1} 1 s {e 1} 1 ( s a)
{t n } n ! s n 1 , n 1, 2, 3,
{sin( kt )} k ( s 2 k 2 )
{cos( kt )} s ( s 2 k 2 )
{sinh( kt )} k ( s 2 k 2 )
{e at t n } n ! ( s a ) n 1 , n 1, 2, 3, [This point was asked in ME-1998, AG-2008 (1 mark), AG-2010 (2 marks)] {e a t sin( kt )} k {( s a ) 2 k 2 } [This point was asked in AG-2007 (1 mark)] {e at cos( kt )} ( s a ) {( s a ) 2 k 2 } [This point was asked in EC-1997 (1 mark)] {e at sinh( kt )} k {( s a ) 2 k 2 }
{cosh( kt )} s ( s 2 k 2 )
{e a t cosh( kt )} ( s a ) {( s a) 2 k 2 }
Example 3.121 [CH-2009 (2 marks)]: The inverse Laplace transform of 1 (2s 2 3s 1) is (a) e t 2 e t
(b) 2e t 2 e t
(c) e t 2e t 2
(d) e t e t 2
Solution (a): 1 1 (2 s 2 3s 1) 1 1 {(2 s 1)( s 1)} 1 {2 (2 s 1)} {1 ( s 1)} 1
{1 (2s 3s 1)} {1 ( s 1 2)} 1{1 ( s 1)} e t 2 e t 2
1
Example 3.122 [TF-2009 (2 marks)]: If the Laplace transform of a function f (t ) is 1 ( s3 s) , then f (t ) is (a) cosh t 1 (b) cosh t 1 (c) sinh t 1 (d) sinh t 1 1 1 1 1 1 (from partial fraction). Thus Solution (a): 3 s s s ( s 1)( s 1) 2( s 1) 2( s 1) s 1
1 1 1 1 1 1 1 1 1 1 e t et 1 1 1 2( s 1) 2( s 1) s 2 s 1 2 s 1 s 2 2 1 3 s s
1{1 ( s 3 s)} {(e t e t ) 2} 1 cosh t 1
Example 3.123 [AE-2010 (2 marks)]: The Laplace transform of y (t ) e t (2 cos 2t sin 2t ) is Y ( s ) 2 s {( s 1) 2 4} , the Laplace transform of y1 (t ) et (2 cos 2t sin 2t ) is
(a)
2( s 2) 2
( s 1) 4
(b)
2( s 2) 2
( s 3) 4
(c)
2( s 2) 2
( s 1) 4
(d)
2( s 1) 2
( s 1) 4 2( s 2) 2( s 2) 2t Solution (a): From Eq. 3.80 { y1 (t )} {e y (t )} Y ( s 2) 2 2 ( s 2 1) 4 ( s 1) 4 Example 3.124 [AE-2013 (2 marks)]: Given that the Laplace transform, {e at } 1 ( s a) , then {3e5 t sinh 5t}
(a) 3s ( s 2 10s ) Solution (b):
(b) 15 ( s 2 10s ) (c) 3s {s 2 10s} (d) 15 ( s 2 10s ) Here we have a5 and k 5 thus 5 15 15 {3e 5t sinh 5t} 3{e 5t sinh 5t} 3 2 2 2 2 ( s 5) 5 s 10 s 25 25 s 10 s
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Chapter 3: Differential Equation
[3.56]
Example 3.125 [AG-2012 (2 marks)]: The inverse Laplace Transform of s 2 ( s 3)3 can be written as (e3t 2)( At 2 Bt C ) . The values of A , B and C , respectively are (a) 3, 5 and 7 (b) 2, 10 and 12 (c) 10, 12 and 4 (d) 9, 12 and 2 2 3t 2 s e s e3t 1 2 2 Solution (d): ( At Bt C ) ( At Bt C ) 3 3 ( s 3) ( s 3) 2 2
s2 ( s 3)
3
3
s2 ( s 3) s
2
( s 3)3
A 2
{t 2 e3t } A
( s 3)
3
B 2
(te3t )
B 2( s 3)
2
C 2
{e3t }
C 2( s 3)
A ( B 2)( s 3) (C 2)( s 3)
2!
2 ( s 3)
A ( s 3)
2
( s 3)3
A
3
2 1
B
11
2 ( s 3)
B 2( s 3)
1!
2
C
1
2 ( s 3)
C 2( s 3)
( A 3B 2 9C 2) ( B 2 3C ) s (C 2) s 2 ( s 3)3
.
Now
comparing the coefficients, A (3B 2) (9C 2) 0 , ( B 2) 3C 0 , C 2 1 . Solving these three equations we get C 2 , B 12 , A 9 . Example 3.126 [MA-2014 (1 mark)]: The inverse Laplace transform of 2 2 (2s 4) {( s 3)( s s 2)} is 7 et 7 3t e t 4 2t 7 3 t e t 4 2 t t (a) (1 t )e t e 3t (b) te 2t (c) e e (d) e e 2 3 2 6 3 2 6 3 2 2 2s 4 2s 4 7 2 4 3 1 6 Solution (a): 2 ( s 3)( s s 2) ( s 3)( s 2)( s 1) s 3 s 2 s 1 1
2s 2 4
7 1 1 4 1 1 1 1 1 7 3t 4 2t 1 t s 3 3 s 2 6 s 1 2 e 3 e 6 e ( s 3)( s s 2) 2
2
Example 3.127 [ME-2014 (1 mark)]: Laplace transform of cos(t ) is s ( s 2 2 ) . The Laplace transform of e 2 t cos(4t ) is s2 s2 (a) (b) 2 2 ( s 2) 16 ( s 2) 16
(c)
s2 2
( s 2) 16
(d)
s2 2
( s 2) 16
Solution (d): Using Eq. 3.80, we have {e 2 t cos(4t )} ( s 2) {( s 2) 2 4 2 } 2 s Example 3.128 [TF-2014 (1 mark)]: The inverse Laplace transform of 2 2 is s 2 s s 4s 4 (a) 1 2te 2t (b) 1 e 2 t 2te 2 t (c) 1 2e 2 t 2te 2t (d) e 2t 2te 2t 2 s 2 s22 1 1 s2 2 2 Solution: F ( s ) 2 2 2 2 s 2s s 4 s 4 s ( s 2) ( s 2) s s 2 ( s 2) ( s 2) F (s)
1 s
1 s2
1 ( s 2)
2 ( s 2)
2
1 s
2 1 1 1 {F ( s )} 2 ( s 2) s ( s 2) 2
2
2 1 1 1 1 1 1 1 2 t {F ( s )} 2 1 2te 2 2 s s ( s 2) ( s 2)
Inverse form of Eq. 3.80: To compute the inverse of F ( s a) , we must recognize F ( s ) , find f (t ) by taking the inverse Laplace transform of F ( s ) , and then multiply f (t ) by the exponential function e at . This procedure can be summarized symbolically in the following manner: 1 F ( s a ) e at f (t ) , where, f (t ) 1{F ( s )}
Copyright © 2016 by Kaushlendra Kumar
(3.81)
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.57]
Example 3.129 [EC-1995 (1 mark)]: If f (t ) 2( s 1) ( s 2 2 s 5) then f (0 ) and f ( ) are given by (a) 0, 2 respectively (b) 2, 0 respectively (c) 0, 1 respectively (d) 2/5, 0 respectively 2( s 1) 2( s 1) ( s 1) 1 Solution (b): 2 f (t ) 2 1 2 2 F ( s 1) , where 2 2 s 2 s 5 ( s 1) 2 s 2s 5 F ( s 1) ( s 1) {( s 1) 2 22 } F ( s) s ( s 2 22 ) .
f (t ) 2 1 F (s 1) 2e t g (t ) , f (t ) 2e t cos 2t .
Thus
using
Eq.
3.84,
we
have
g (t ) 1{F ( s )} 1 s ( s 2 2 2 ) cos(2t ) .
where
So
lim f (t ) lim f (0 h) 2e h cos 2h 2
Now
t 0
and
h 0
lim f (t ) lim e t cos 2t 0 (any number between 0 and 1) 0 . t
t
Second Shifting theorem: Consider a general function y f (t ) defined for t 0 .
0,
0t a
(3.82) ta The piecewise – defines function (Eq. 3.82) plays a significant role, as shown in Fig. 3.7(a) for a 0 , the graph of the function y f (t a)u (t a ) coincide with the graph of the function y f (t a ) for t a , but is identically zero for 0 t a . f (t a )u (t a )
f (t a),
(a)
(b)
Figure 3.7: : Shift on − axis (a) ( ), ≥
(b) ( − ) ( − )
As, an exponential multiple of f (t ) results in a translation of the transform F ( s ) on the s axis. Similarly, F ( s ) is multiplied by an exponential function e a s , a 0 , the inverse transform of the product e a s F ( s ) is the function f shifted along the t axis as shown in Fig. 3.7(b). This result is called second translation theorem or second shifting theorem, i.e., if F ( s) { f (t )} and a 0 , then { f (t a)u (t a )} e a s F ( s ) (3.83) [Eq. 3.83 was asked in EC-1999 (1 mark)]. Proof: By the additive interval property of integrals,
0 e zero
st
a
0
a
f (t a )u (t a ) dt e st f (t a )u (t a ) dt e st f (t a )u (t a ) dt , the first integral is
as
u (t a) 0 0 t a ,
{ f (t a)u (t a )} e st f (t a )u (t a ) dt .
thus
Let
a
0
0
v t a dv dt { f (t a )u (t a )} e s ( v a ) f (v ) dv e a s e s v f (v )dv e a s { f (t )}
1
Inverse form of Eq. 3.90: If f (t ) {F ( s )} , the inverse form of Eq. 3.83, a 0 is given as, 1{e a s F ( s )} f (t a )u (t a)
(3.84)
Laplace Transform of Unit Step Function
a
0
0
a
a
as
as
{u (t a)} 0 (e s) e s {u (t a )} e IN-2010 (1 mark), ME-2004 (2 marks)]
Copyright © 2016 by Kaushlendra Kumar
as
{u (t a )} e st u (t a )dt e st u (t a ) dt e st u (t a ) dt 0 e st dt e st ( s )
a
s [This point was asked in CE-1998,
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Engineering Mathematics
Chapter 3: Differential Equation
[3.58]
Laplace Transform of Unit Impulse Function { a (t a )} (1 k ) u (t a ) u t ( a k ) {1 ( ks )}{e as e ( a k ) s } {e as ( ks )}(1 e ks )
Laplace Transform of Dirac Delta Function
{ (t a)} {lim a (t a)} lim { a (t a)} lim e as (1 e ks ) ( ks ) ; k 0
k 0
L’Hospital rule, we have { (t a)} e
k 0
as
now
by
using
lim{se ks s} e as 1 e as . k 0
When a 0 { (t )} 1 .
Laplace Transform of Unit Ramp Function {r (t a )} (t a)u (t a ) f (t a ) u (t a ) , where f (t ) t f (t a ) (t a ) ; now using the second shifting theorem, we have {r (t a )} e as { f (t )} e as {t} e as s 2 , as {t} 1! s11 1 s 2 . Thus {r (t a )} e as s 2 [This point was asked in EC-1994 (1 mark)].
Example 3.130: Write the following function using unit step functions and find its transform. 0 t 1 2,
f (t ) t 2 2 , 1 t 2 .
cos t ,
t 2 Solution: In terms of unit step function the given function can be written as, f (t ) 2 1 (t 1) (t 2 2) u (t 1) u (t 2) (cos t )u (t 2) To apply Eq. 3.91, we must write each term in f (t ) in the form f (t a )u (t a) . Thus,
2 1 (t 1) 2(1 e s ) s
1
1 1 1 1 1 t 2 (t 1) (t 1) 2 (t 1) (t 1) 3 2 e s 2 2 2 s s 2s
1 1 2 2 s 2 2 t (t 2) (t 2) (t 2) (t 2) 3 2 e 2 2 8 8s 2 s 2s 2(1 e s ) 1 1 1 s 1 2 s 2 Hence, f (t ) 3 2 e 3 2 e s 8s s s 2s s 2s 1
2
Example 3.131 [EE-1998 (2 marks)]: The Laplace transform of (t 2 2t )u (t 1) is 2 2 2 2 2 2 (a) 3 e s 2 e s (b) 3 e 2 s 2 e s (c) 3 e s e s (d) None of these s s s s s s Solution (d): Let g (t ) (t 2 2t )u (t 1) {(t 1) 2 1}u (t 1) (t 1) 2 u (t 1) u (t 1) . 2
So
s
2
{g (t )} {(t 1) u (t 1)} {u (t 1) . As from Eq. 3.91 {(t 1) u (t 1)} e F ( s ) , where F ( s) {t 2 } 2! s 21 2 s3 , thus {(t 1) 2 u (t 1)} 2e s s 3 . Also {u (t 1)} e s s ; thus {g (t )} (2e s s3 ) (e s s ) .
Example
3.132
[TF-2007
(1
mark)]:
A
function
g (t )
is
defined
as
follows
t0 t t 0 1 2 , . The Laplace transform of the function g (t ) is given by g (t ) t t0 and t t0 0,
(a) {1 (2s )}[exp( s ) exp( s )]exp( st 0 )
(b) {1 (2 s)}[exp( s ) exp( s )] exp( st 0 )
(c) {1 (2 s )}[exp( s ) exp( s )]exp( st 0 )
(d) {1 (2s )}[exp( s ) exp( s )] exp( st0 )
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e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.59]
Solution (c): As Laplace transform of unit impulse function, i.e., { a (t a )} {e as (ks )}{1 e ks } , a t ak
1 k , where a (t a ) 0, {g (t )}
e
s ( t0 )
2 s
1 e
otherwise 2 s
e
. So in our problem, k 2 , a t0 , a k t0 . Thus
st0 s
e s e st0 e s e s e s e st0 s s (e e ) . 1 s s 2 s e 2 s e 2 s
Example 3.133 [CH-2010 (1 mark)]: The Laplace transform of the function shown in the figure below is
(a) Ve ( a b ) s s (b) V (e bs e as ) s (c) V (e as e bs ) s
(d) V (e as ebs ) s 2 Solution (c): The given function can be written in terms of unit step function as: f (t ) Vu (t a ) Vu (t b) { f (t )} V {u (t a)} V {u (t b)} . { f (t )} V (e as s) V (e bs s ) (V s )(e as e bs )
Alternative form of Eq. 3.83: We are frequently confronted with the problem of finding the Laplace transform of a product of a function g and a unit step function (t a ) where the function g lacks the precise shifted form in Eq. 3.83. To find the Laplace transform of g (t ) (t a) , it is possible to find up g (t ) into the required form by algebraic manipulations, just as we did it in previous example for 2nd and 3rd term. But since these manipulations are time consuming and often not obvious, it is simpler to devise an alternative version of Eq. 3.83. Using the definition of (t a ) and the substitution u t a , we obtain
a
a
a
g (t ) (t a ) e st g (t ) dt e s ( u a ) g (u a ) du e a s e s u g (u a )du
g (t ) (t a ) e a s e s t g (t a ) dt e a s g (t a )
(3.85)
a
Example 3.134: Evaluate (cos t )u (t ) Solution: With g (t ) cos t and a then g (t ) cos(t ) cos t . So from Eq. 3.93, (cos t ) (t ) e as cos(t ) e as cos t e as {s ( s 2 1)} .
Example 3.135 [CE-2002 sin t for 0 t is f (t ) for t 0 (a)
1
s 0
(b)
(2
1
marks)]:
s
The
Laplace
1 e s
Transform
the
function
e s
s 0 1 s2 1 s2 1 s2 1 s2 Solution (d): The given function can be written as f (t ) sin t (sin t )u (t c) 0u (t c) . (c)
s 0
of
(d)
{ f (t )} {sin t} {(sin t )u (t c )} {1 ( s 2 1)} e s {sin(t )} { f (t )} {1 ( s 2 1)} e s {sin t} {1 ( s 2 1)} e s {1 ( s 2 1)} {(1 e s ) ( s 2 1)} . [Similar questions were also asked in CE-1999 (2 marks), EC-1993 (1 mark)]
Example 3.136 [EC-2004 (2 marks)]: A system is described by the following differential equation ( d 2 y dt 2 ) 3( dy dt ) 2 y x (t ) is initially at rest. For input x (t ) 2u (t ) , the output y (t ) is (a) (1 2e t e 2 t )u (t )
(b) (1 2e t e 2 t )u (t )
(c) (0.5 e t 1.5e 2 t )u (t )
(d) (0.5 2e t 2e 2t )u (t )
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e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.60]
Solution (a): Taking Laplace transform on both sides of the given differential equation we get { y } 3{ y } 2{ y} 2{u (t )} ; using Eq. 3.75 and the condition: system is initially at rest, i.e.,
y (0) 0 and y (0) 0 , we have { y } s 2Y ( s ) sy (0) y (0) s 2Y ( s) ; { y } sY ( s ) y (0) sY ( s) ; { y} Y ( s ) ; since {u (t )} 1 s . Thus substituting all these in { y } 3{ y } 2{ y} 2{u (t )}
we
s 2Y ( s) 3sY ( s) 2Y ( s ) 2 s
get
Y ( s ) 2 {s ( s 2 3s 2)} 2 {s( s 2)( s 1)} ;
from
partial
fractions
we
have
Y ( s ) {1 s} {2 ( s 1)} {1 ( s 2)} Now taking inverse Laplace transform on both sides we get y (t ) 1{Y ( s )} 1{1 s} 21{1 ( s 1)} 1{1 ( s 2)} 1 2e t e 2 t .
0, t 0
u (t )
1, t 0
Now
as
so y (t ) can be represented as y (t ) (1 2e t e 2 t )u (t ) .
Example 3.137 [EE-2008 (1 mark)]: A function y (t ) satisfies the following differential equation ( d dt ) y (t ) y (t ) (t ) , where (t ) is the delta function. Assuming zero initial condition, and denoting the unit step function by u (t ) , y (t ) can be of the form
(c) et u (t ) (d) e t u (t ) (a) e t (b) e t Solution (d): We have given y (0) 0 . Now taking Laplace transform on both sides of the given differential equation we get { y } { y} { (t )} ; using Eq. 3.75 and given conditions, we have { y } sY ( s ) y (0) sY ( s) ; { y} Y ( s ) ; since { (t )} e 0 s 1 . Thus substituting all these in { y } { y} { (t )} we get sY ( s ) Y ( s ) 1 Y ( s ) 1 ( s 1) . Now taking inverse Laplace
transform on both sides we get y (t ) 1{Y ( s )} 1{1 ( s 1)} e t . As for t 0 , y 0 and for
t 0 , y (t ) e t ; so y (t ) e t u (t ) . Example 3.138 [EC-2012, IN-2012 (2 marks)]: Consider the differential equation {d 2 y (t ) dt 2 } 2{dy (t ) dt} y (t ) (t ) with y(t ) t 0 2 and ( dy dt ) t 0 0 . The numerical value of ( dy dt ) t 0 is (a) –2 (b) –1 (c) 0 (d) 1 Solution (d): Taking Laplace transform on both sides of the given differential equation we get { y } 2{ y } { y} { (t )} ; using Eq. 3.75 and given conditions, we have { y } s 2Y ( s ) sy (0) y (0) s 2Y ( s) 2s ; { y } sY ( s) y (0) sY ( s ) 2 ; { y} Y ( s ) ; since { (t )} e 0 s 1 . Thus substituting all these in s 2Y ( s) 2 s 2sY ( s) 4 Y ( s ) 1 Y ( s)
3 2 s
{ y } 2{ y } { y} { (t )}
1 2 s 2
1
2
we get
1
. s 2s 1 ( s 1) ( s 1) s 1 taking inverse Laplace transform on both sides we 1 1 1 1 1 t t t t y (t ) {Y ( s )} 2 te 2e y (t ) te 2e 2 s 1 ( s 1) 2
2
2
Now get
(d dt ) y (t ) e t te t 2e t te t e t ( d dt ) y (t ) t 0 0 1 1 Example 3.139 [EC-2013 (2 marks)]: A system is described by the differential equation 1, 0 t 2 ( d 2 y dt 2 ) 5( dy dt ) 6 y (t ) x (t ) . Let x (t ) be a rectangular pulse given by x (t ) . 0, otherwise Assuming that y (0) 0 and dy dt 0 at t 0 , the Laplace transform of y (t ) is (a)
e 2 s s ( s 2)( s 3)
(b)
1 e 2 s s ( s 2)( s 3)
Copyright © 2016 by Kaushlendra Kumar
(c)
e 2 s ( s 2)( s 3)
(d)
1 e 2 s ( s 2)( s 3)
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Engineering Mathematics
Chapter 3: Differential Equation
[3.61]
Solution (b): Taking Laplace transform on both sides of the given differential equation we get { y } 5{ y} 6{ y} {x (t )} ; using Eq. 3.75 and given conditions, we have { y } s 2Y ( s ) sy (0) y (0) s 2Y ( s) ;
{ y} Y ( s ) ;
{ y } sY ( s ) y (0) sY ( s) ; 2 s
since
2 s
x (t ) {u (t ) u (t 2)} { x(t )} {u (t ) u (t 2)} (1 s ) (e s) (1 e ) s . Thus substituting all these in the given differential equation after taking Laplace, we get s 2Y ( s) 5sY ( s ) 6Y ( s ) (1 e 2 s ) s Y ( s ) (1 e 2 s ) {s( s 2 5s 6)} (1 e 2 s ) {s( s 2)( s 3)} . [Similar question was also asked in AE-2012 (1 mark)]
Derivatives of a Transform: The Laplace transform of the product of a function f (t ) with t can be found by differentiating the Laplace transform of f (t ) . Let us assume F ( s) { f (t )} exist and that it is possible to interchange the order of differentiation and integration. Then
0
0
( d ds ) F ( s ) ( d ds ) e st f (t ) dt ( s ){e st f (t )}dt e st t f (t ) dt {t f (t )} 0
{t f (t )} ( d ds ){ f (t )}
(3.86)
2
From the above result, we can derive for {t f (t )} , i.e.,
{t 2 f (t )} t t f (t ) (d ds ){t f (t )} ( d ds) (d ds ){ f (t )} (d 2 ds 2 ){ f (t )} {t 2 f (t )} ( d 2 ds 2 ){ f (t )}
(3.87)
n
Similarly, the general result for {t f (t )} , i.e. if F ( s) { f (t )} and n 1, 2, 3, then, {t n f (t )} ( 1) n ( d n ds n ) F ( s) (1) n ( d n ds n ){ f (t )}
(3.88)
Example 3.140 [AE-2008 (2 marks)]: Let Y ( s) denote the Laplace transform L y (t ) of the function y (t ) cosh(at ) sin( at ) . Then (a) L (dy dt ) ( dY ds ), L t y (t ) sY ( s )
(b) L ( dy dt ) sY ( s ), L t y (t ) ( dY ds )
(c) L (dy dt ) (dY ds ), L t y (t ) Y ( s 1)
(d) L ( dy dt ) sY ( s ), L t y (t ) e asY ( s ) Solution (b): Using Eq. 3.75, we have { y (t )} sY ( s) f (0) , as y (0) 0 so { y (t )} sY ( s ) . Also using Eq. 3.88, we have {t y (t )} ( d ds ){ y (t )} ( d ds )Y ( s ) dY ds . Example 3.141 [CH-2008 (2 marks)]: The Laplace transform of the function f (t ) t sin t is (a) 2s ( s 2 1) 2
(b) 1 {s 2 ( s 2 1)}
(c) {1 s 2 } {1 ( s 2 1)}
(d) 1 {( s 1) 2 1}
Solution (a): From Eq. 3.88, {t sin t} ( 1)1 ( d ds){sin t} (d ds ) 1 ( s 2 12 ) 2s ( s 2 1) 2 Example 3.142 [EC-2012, EE-2012, IN-2012 (1 mark), EC-2014 (2 marks)]: The unilateral Laplace transform of f (t ) is 1 ( s 2 s 1) . The unilateral Laplace transform of t f (t ) is (a) s ( s 2 s 1) 2 Solution
(d):
(b) (2 s 1) ( s 2 s 1) 2
{ f (t )} 1 ( s 2 s 1) 2 ,
(c) s ( s 2 s 1) 2
so
using
Eq.
(d) (2s 1) ( s 2 s 1) 2
3.88
2
2
with
n 1
we
have
2
{t f (t )} ( 1)(d ds ) F ( s) ( d ds){1 ( s s 1)} (2 s 1) ( s s 1) .
Integral of Transform: If { f (t )} F ( s ) then f (t ) t F ( s )ds , provided lim f (t ) t s t 0
exists. Proof: We have F ( s ) { f (t )} e st f (t ) dt . Now integrating both sides w.r.t. ‘ s ’ by 0
taking limits from s to then,
s
F ( s)
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s
st
0
e
f (t )dt ds . As s and t are independent
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Engineering Mathematics variables
s
so
F (s)
t 0
Chapter 3: Differential Equation
by
s
interchanging
the
t 0
t 0
e st ds f (t ) dt {e st ( t )}s s f (t ) dt
(a)
0
(b) 2 1
Solution (b): As {sin t}
sin t
2
s 1
tan 1 tan 1 s
t
get lim e st s 0 0
of
Example 3.143 [CE-2007 (2 marks)]: Evaluate
order
sin t t
[3.62] integration
have,
e st { f (t ) t}dt f (t ) t .
(1 t ) sin tdt
(c) 4 F (s)
we
sin t t
(d) 3
F ( s ) ds s
1
2
s
s 1
ds (tan 1 s) s
sin t s e st dt s . Now taking limit s 0 both sides we 0 2 t 2
sin t sin t s lim e st dt lim s dt . 0 0 s 0 2 s0 s 0 2 t t 2
dt lim
Transform of Integrals: If functions f and g are piecewise continuous on the interval [0, ) , then a special product, denoted by f g , is defined by the integral t
f g f ( ) g (t ) d
(3.89)
0
and is called the convolution of f and g . The convolution f g is a function
of
t.
For
e.g.,
t
et sin t e sin(t ) d (1 2)( sin t cos t et ) . 0
The convolution of two function is commutative, i.e., f g g f or
t
Figure 3.8: Changing order of integration from t to τ
t
0 f ( ) g (t )d 0 f (t ) g ( )d .
It is not true that the integral of a product of functions is the product of the integrals. However, it is true that the Laplace transform of the special product (Eq. 3.89) is the product of the Laplace transform of f and g . This means that it is possible to find the Laplace transform of the convolution of two functions without actually evaluating the integral. If f (t ) and g (t ) are piecewise continuous on [0, ) and of exponential order, then Convolution Theorem: { f g} { f (t )}{g (t )} F ( s )G ( s ) (3.90)
0
0
Proof: Let F ( s ) { f (t )} e s f ( ) d and G ( s ) { g (t )} e s f ( ) d
F (s )G (s )
s
0
e
f ( )d
0
0
0
0
s
0
e
f ( )d
0
s ( )
0 e
f ( ) g ( )d d
F ( s )G ( s ) f ( ) d e s ( ) g ( ) d . Holding fixed, we let, t dt d F ( s )G ( s ) f ( ) d e st g (t ) dt . In the t plane we are integrating over the shaded
region, as shown in Fig. 3.8. Since f and g are piecewise continuous on [0, ) and of exponential order, it is possible to interchange the order of integration:
t
0
0
0
F ( s)G ( s ) e st dt f ( ) g (t ) d e st
Example
3.144
[EC-2000
(1
mark)]:
f ( )g (t )d dt { f g} t
0
Given
{ f (t )} ( s 2) ( s 2 1) ,
that
t
{g (t )} ( s 2 1) {( s 3)( s 2)} , h(t ) f ( ) g (t ) d . {h(t )} is 0
(a)
s2 1 s 3
(b)
1 s3
Copyright © 2016 by Kaushlendra Kumar
(c)
s2 1 ( s 3)( s 2)
s2 s2 1
(d) None of these
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.63]
Solution (b): Using convolution theorem (Eq. 3.90), we have 2 2 {h(t )} { f g} { f (t )} { f (t )} {( s 2) ( s 1)} [( s 1) {( s 3)( s 2)}] 1 ( s 3) . Example 3.145 [EC-2002 (1 mark)]: Convolution of x (t 5) with impulse function (t 7) is equal to (a) x(t 12) (b) x(t 12) (c) x (t 2) (d) x(t 2) Solution (c): If x (t 5) (t 7) is the convolution of x (t 5) , which is a unit step function, with (t 7) , which is a Dirac delta function, then from Eq. 3.90, we have {x (t 5) (t 7)} {x (t 5)} { (t 7)} (e5 s s) e 7 s e 2 s s x (t 5) (t 7) 1{e 2 s s} x (t 2) .
Common Data Questions: 3.146 & 3.147: Given f (t ) and g (t ) as shown below: Example 3.146 [EE-2010 (2 marks)]: g (t ) can be expressed as (b) g (t ) f (t 2) 3 (c) g (t ) f 2t (3 2) (d) g (t ) f (t 2) (3 2) Solution (d): From the given two figures, it is clear that the value of f (0) is same as g (3) and value of f (1) is same as g (5) ; these two conditions can be obtained by putting t 3 and t 5 in option (d). So option (d) is correct. (a) g (t ) f (2t 3)
Example 3.147 [EE-2010 (2 marks)]: The Laplace transform of g (t ) is (a) (1 s)(e 3 s e5 s ) (b) (1 s )(e 5 s e 3 s ) (c) (e 3 s s)(1 e 2 s ) (d) (1 s)(e 5 s e3 s ) Solution (c): In terms of unit step function g (t ) can be written as: g (t ) u (t 3) u (t 5) . So {g (t )} {u (t 3)} {u (t 5)} (e 3 s s ) (e 5 s s ) e 3 s (1 e 2 s ) s .
Inverse form of Eq. 3.90: The convolution theorem is sometimes useful in finding the inverse Laplace transform of the product of two Laplace transforms. Hence, from Eq. 3.90, we have 1{F ( s)G ( s )} f g (3.91)
Example 3.148 [EE-2007 (2 marks)]: If u (t ) , r (t ) denote the unit step and unit ramp functions respectively and u (t ) r (t ) their convolution, then the function u (t 1) r (t 2) is given by (a) (1 2)(t 1)(t 2) Solution (c): Let
(b) (1 2)(t 1)(t 2) g (t ) u (t 1) r (t 2)
(c) (1 2)(t 1) 2 u (t 1) then from Eqs. 3.90
(d) None of these and 3.91, we have
{ g (t )} {u ( t 1) r (t 2)} {u (t 1)}{r ( t 2)} ( e s s ) ( e 2 s s 2 ) e s s 3
g (t ) 1{e s s 3 } . Now using second shifting theorem given by Eq. 3.92, i.e., we have
1{e a s F ( s )} f (t a) (t a ) where f (t ) 1{F ( s )} so 1 e s (1 s3 ) f (t 1) (t 1) , 1
1
e
1
f (t ) {1 s } (1 2!) {2! s } t 2 f (t 1) (1 2)(t 1) 2 .
where s
3
3
3
2
Thus
2
(1 s ) (1 2)(t 1) u (t 1) g (t )
Example 3.149 [EE-2011 (1 mark)]: Given two continuous time signals x(t ) e t and y (t ) e 2t which exist for t 0 , the convolution z (t ) x(t ) y (t ) is (a) e t e 2 t
(b) e 3t
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(c) e t
(d) e t e 2t
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Engineering Mathematics
Chapter 3: Differential Equation
[3.64]
Solution (a): From Eqs. 3.90 and 3.91, we have {z (t )} {x (t ) y (t )} {x (t )}{ y (t )} {z (t )} {e t }{e 2 t } {1 ( s 1)} {1 ( s 2)} {1 ( s 1)} {1 ( s 2)} z (t ) 1 {1 ( s 1)} {1 ( s 2)} 1 1 ( s 1) 1 1 ( s 2) e t e 2 t .
Example 3.150: Evaluate
e sin(t )d t
t
0
Solution: With f (t ) et and g (t ) sin t , the convolution theorem states that the Laplace transform of the convolution of f and g is the product of their Laplace transforms:
e sin(t )d {e } {sin t} {1 (s 1)}{1 (s 1)} 1 {(s 1)(s 1)} t t
t
2
2
0
When g (t ) 1 {g (t )} {1} 1 s , the convolution theorem, i.e. Eq. 3.90, imples that the Laplace transform of the integral of f is
f ( )d F (s) s t
(3.92)
0
[Eq. 3.92 was asked in EC-2009, ME-2007 (2 marks)]; and the inverse of Eq. 3.92 is given as, t
0 f ( )d
1
(3.93)
{F ( s) s}
which can be used in partial fraction when s n is a factor of the denominator and f (t ) 1{F ( s )} is easy to integrate. For e.g. we know for f (t ) sin t that f (t ) sin t F ( s) { f (t )} 1 ( s 2 1)
and
hence
by
Eq.
3.93,
t
1 1 {s( s 2 1)} 1 {1 ( s 2 1)} s sin d 1 cos t 0
Example 3.151 [CE-2000 (2 marks)]: Let F ( s) [ f (t )] denote the Laplace transform of the function f (t ) . Which of the following statement is correct? 1
(a) ( df dt ) (1 s ) F ( s ) ; f ( ) d sF ( s ) f (0)
0
1
(b) ( df dt ) sF ( s) F (0) ; f ( ) d dF ds
0 1 (c) ( df dt ) sF ( s) F (0) ; f ( ) d F ( s a) 0 1 (d) ( df dt ) sF ( s) F (0) ; f ( ) d (1 s ) F (s) 0 Solution (d): From Eq. 3.75 and Eq. 3.92 we can say that option (d) is correct. Example 3.152 [CH-2014 (1 mark)]: For the time domain function f (t ) t 2 , which ONE of the following is the Laplace transform of (a) 3 s 4 Solution
t
0 f (t )dt ?
(b) 1 (4s 2 ) (d):
From
Eq.
(c) 2 s 3 3.92,
F ( s ) { f (t )} {t 2 } 2! s 2 1 2 s3 . Thus
we
have
(d) 2 s 4
f (t)dt F (s) s , t
0
where
f (t)dt F (s) s 2 s . t
4
0
Transform of a periodic function: If
a periodic function has period T , T 0 , then f (t T ) f (t ) . The Laplace transform of a periodic function can be obtained by integration over
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Engineering Mathematics
Chapter 3: Differential Equation
[3.65]
one period. If f (t ) is a piecewise continuous on [0, ) , of exponential order, and periodic with period T , then T
{ f (t )} {1 (1 e sT )} e st f (t )dt
(3.94)
0
T
0
0
Proof: { f (t )} e st f (t ) dt e s t f (t ) dt e s t f (t ) dt . Let t u T dt du , then the 2nd integral
T e
s t
T
0
0
0
f (t ) dt e s ( t T ) f (u T ) du e sT e st f (u ) du e sT e st f (t ) dt e sT { f (t )} T
T
0
0
{ f (t )} e s t f (t ) dt e sT { f (t )} { f (t )} {1 (1 e sT )} e st f (t ) dt
Example 3.153 [CS-1993 (1 mark)]: Find the Laplace transform of the periodic function f (t ) described by the curve below, i.e., sin t , (2n 1) t 2n , (n 1, 2, 3, ) . f (t ) otherwise 0, Solution: As the period of the given function is T 2 . So for 0 t 2 , we have 0t 0, and outside the interval it is defined as f (t ) sin t , t 2
1
2
1
1 e st 0 dt 2 e st sin tdt 0 1 e 2 s
2
e st sin tdt 1 e 1 e . Now by treating sin x as the 1st function and e st as the 2nd function, we have es t 1 e s t 1 e st e st st st I e sin tdt sin t cos t e dt sin t cos t (sin t ) dt s s s s s s f (t )
I
2
e st s2
2 s
0
e st f (t ) dt
f (t 2 ) f (t ) . Thus
I
s sin t cos t
s2
I
2 s
e st (1 s 2 )
s sin t cos t
e st sin tdt ( se st sin t e st cos t ) (1 s 2 )
f (t ) {1 (1 e 2 s )}
2
t 2
t
se st sin t e st cos t (1 s 2 )
(e 2 s e s ) (1 s 2 )
e st sin tdt {1 (1 e 2 s )}{(e 2 s e s ) (1 s 2 )}
Example 3.154 [CH-2007 (1 mark)]: Given that the Laplace transform of the function below over a single period 0 t 2 is (1 s 2 )(1 e s ) 2 , the Laplace transform of the periodic function over 0 t is (a) (1 s )(1 e s ) 2 (b) (1 s )(1 e s ) 2 (c) (1 s 2 ){(1 e s ) (1 e s )} (d) (1 s ) tanh( s 2) Solution (c): As it is given that, the period of the given function is T 2 and 2
0 e
st
2
f (t ) dt (1 s 2 )(1 e s ) 2 . So using Eq. 3.94, we have { f (t )} {1 (1 e 2 s )} e st f (t ) dt 0
s 2
s
1 (1 e ) (1 e ) 1 1 s 2 (1 e ) s 2 (1 e s ) . 2 s 2 s s 2 1 e s (1 e )(1 e ) s
{ f (t )}
Exercise: 3.3 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill.
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Engineering Mathematics
Chapter 3: Differential Equation
[3.66]
1. The Laplace transform of the function f ( x ) e3 x cos(6 x ) e3 x cos(6 x ) is (a) F ( s ) (c) F ( s )
1 s 3 3
s 2
s 36 s
s 3
(b) F ( s )
2
( s 3) 36 3( s 3)
2 2 s 3 s 36 ( s 3) 36 2. The Laplace transform of f ( x ) t cosh(3t ) is (a) F ( s )
s2 9
(b) F ( s)
s2 9
s2 9 ( s 2 9) 2
(d) F ( s )
(c) F ( s)
3 s 3 1 s 3
s 2
2
s 36 s s 36
s2 9 ( s 2 9) 2
s 3 2
( s 3) 36 3( s 3) 2
( s 3) 36
(d) F ( s )
s2 9 s2 9
3. The Laplace transform of f (t ) t 3 2 is
3 3 (b) F ( s) 5 2 (c) F ( s) 5 2 (d) F ( s) 3 2 32 4s 4s 4s 4s 4. The inverse Laplace transform of F ( s) (6 s ) {1 ( s 8)} {4 ( s 3)} is (a) F ( s)
(a) f (t ) 6 e 3t 4e8t
(b) f (t ) 6 e8t 4e3t
(c) f (t ) 6 e3t 4e8t
(d) f (t ) 6 e8 t 4e3t
5. The inverse Laplace transform of F ( s ) {6s ( s 2 25)} {3 ( s 2 25)} is (a) f (t ) cos(5 x) sin(5 x) (b) f (t ) 6 cos(5 x) (3 5) sin(5 x) (c) f (t ) 6 cos(5 x ) sin(5 x) (d) f (t ) 6 cos(5 x) (3 5)sin(5 x) 6. The inverse Laplace transform of F ( s ) {8 (3s 2 12)} {3 ( s 2 49)} is (a) f (t ) 4 sin(2t ) 3sinh(7t ) (b) f (t ) (4 3) sin(t ) (3 7) sinh(7t ) (c) f (t ) (4 3)sin(2t ) (3 7) sinh(7t )
(d) f (t ) (4 3)sinh(2t ) (3 7) sin(7t )
7. The inverse Laplace transform of F ( s) (1 3s) ( s 2 8s 21) is (a) f (t ) 3e 4t cos( 5t ) (13 (c) f (t ) 3e 4t cos( 5t ) (8
5)e 5)e
4 t
4 t
sin( 5t )
sin( 5t )
(b) f (t ) 3e 4t cos( 5t ) e 4 t sin( 5t ) (d) f (t ) 3e 4t cos( 5t ) e 4 t sin( 5t )
8. The inverse Laplace transform of F ( s) ( s 7) ( s 2 3s 10) is (a) f (t ) (5 7)e 5 t (12 7)e 2 t
(b) f (t ) (5 7)e 2 t (12 7)e 5t
(c) f (t ) ( 5 7)e 5t (12 7)e 5t
(d) f (t ) ( 5 7)e 2 t (12 7)e5t
9. The inverse Laplace transform of F ( s) (86 s 78) {( s 3)( s 4)(5s 1)} is (a) f (t ) 3e 3t 2e 4t et 5
(b) f (t ) 3e3t 2e 4t et 5
(c) f (t ) 3e 3t 2e 4t e5t
(d) f (t ) 3e 3t 2e 4 t et 5
10. The Laplace transform of f (t ) t 2u (t 3) (cos t )u (t 5) is
3 3 s 2 (b) F ( s) 3 s 2 (c) F ( s) 3 s 2 (d) F ( s) 3 s (a) F ( s)
1 s cos 5 sin 5 5 s e 3 s e s s s2 1 6 9 s cos 5 sin 5 5 s 2 e 3 s e s s s2 1
2
2
9 s cos 5 sin 5 5 s e 3 s e s s s2 1 6 9 s cos 5 sin 5 5 s 2 e3 s e s s s2 1
6
2
11. The inverse Laplace transform of G ( s ) se 4 s {(3s 2)( s 2)} is (a) g (t ) u (t 2) f (t 4) , where f (t ) 12e 2 t 3 4e 2 t
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Engineering Mathematics
Chapter 3: Differential Equation
[3.67]
(b) g (t ) u (t 4) f (t 4) , where f (t ) (1 12)e 2 t 3 (1 4)e 2t (c) g (t ) u (t 4) f (t 4) , where f (t ) 12e 2t 3 4e 2 t (d) g (t ) u (t 2) f (t 4) , where f (t ) (1 12)e 2 t 3 (1 4)e 2 t
t4 , t5 12. The Laplace transform of f (t ) 4 is t 3sin{(t 10) (1 2)}, t 5 (a) F ( s) (c) F ( s)
24 s
5
24 s
5
(3 10)e 5 s
(b) F ( s)
2
s (1 100) (3 10)e5 s
(d) F ( s)
2
s (1 100)
t 6
t,
13. The Laplace transform of f (t )
2
8 (t 6) , t 6
2 1 14 6 s e s s3 s 2 s 1 2 1 14 (c) F ( s) 2 3 2 e 6 s s s s s (a) F ( s)
1
2
24 s
5
24 s
5
(3 10)e 5 s 2
s (1 100) (3 10)e 5 s 2
s (1 100)
is
2 1 14 6 s e s s3 s 2 s 1 2 1 14 (d) F ( s) 2 3 2 e 6 s s s s s
(b) F ( s)
1
2
14. The solution of IVP y 10 y 9 y 5t , y (0) 1 , y (0) 2 is (a) y (t ) (50 81) (5 9)t (31 81)e 9t 2e t
(b) y (t ) (5 8) (5 9)t (3 8)e9t 2et
(c) y (t ) (50 81) (5 9)t (31 81)e 9t 2et
(d) y (t ) (5 8) (5 9)t (3 8)e9t 2e t
15. The solution of IVP 2 y 3 y 2 y te 2 t , y (0) 0 , y (0) 2 is (a) y (t ) (1 125){96et 2 96e 2t 10te 2t (25 2)t 2 e 2 t } (b) y (t ) (1 125){96et 2 96e 2 t 10te 2 t (25 2)t 2 e 2 t } (c) y (t ) (1 125){96et 2 96e 2 t 10te 2t (25 2)t 2 e 2t } (d) y (t ) (1 125){96et 2 96e 2t 10te 2 t (25 16. The solution of IVP ty ty y 2 , y (0) 2 , (a) y (t ) 2 4t (b) y (t ) 2 4t 17. The solution of IVP y 2 y 15 y 6 (t 9) ,
2)t 2 e 2 t } y (0) 4 is (c) y (t ) 2 4t y (0) 5 , y (0) 7 is
(d) y (t ) 2 4t
(a) y (t ) (3 4)u (t 9){e3(t 9) e 5( t 9) } (9 4)e3t (11 4)e 5 t (b) y (t ) (3 4)u (t 9){e 3(t 9) e 5( t 9) } (9 4)e3t (11 4)e 5t (c) y (t ) (3 4)u (t 9){e3(t 9) e 5( t 9) } (9 4)e 3t (11 4)e 5t (d) y (t ) (3 4)u (t 9){e3(t 9) e 5( t 9) } (9 4)e3t (11 4)e 5t 18. The Laplace transform of the periodic function (square wave) shown in figure is (a) 1 {s (1 e s )} (b) 1 {s(1 e s )} (c) 1 {s(1 e s )}
(d) 1 (1 e s )
19. Given two continuous functions x(t ) 1 z (t ) x(t ) y (t ) at t 4 is _____.
2 t and y (t ) t which exist for t 0 , the convolution
x
20. The solution of the integral equation xe x e x t y (t ) dt is 0
(a) y ( x ) e
x
(b) y ( x ) e
x
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(c) y ( x ) e1 x
(d) y ( x) e 1 x
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Engineering Mathematics
3.4
Chapter 3: Differential Equation
[3.68]
Fourier Series
A periodic function can be represented by an infinite series (known as Fourier series) of sinusoids of harmonically related frequencies. If f (t ) is a periodic function with period 2T then the Fourier series
f (t ) a0 a1 cos t an cos nt b1 sin t bn sin nt
is
f (t ) a0 n 1 ( an cos n t bn sin nt ) , where 2 (2T ) is the fundamental frequency; n
is the nth harmonic of fundamental frequency; a0 , an , bn , n 1, 2, , are the Fourier coefficients. The conditions under which a periodic function f (t ) can be expanded in a convergent Fourier series are known as Dirichlet’s conditions which are: (1) f (t ) is a single valued function; (2) f (t ) has a finite number of discontinuities in each period 2T ; (3) f (t ) has a finite number of maxima and minima in each period 2T ; (4) the integral
L
L
L
2
L f (t )
f (t ) dt exists and is finite, i.e.
dt .
Fourier Analysis: Let f (t ) be a piecewise continuous function (i.e. having finite number of removable or jump discontinuities) on [ T , T ] . Then the Fourier series of f (t ) is the series given as:
f (t ) a0 n 1 ( an cos nt bn sin n t )
where 2 (2T ) is the fundamental frequency; n is the n
(3.95) th
harmonic of fundamental
frequency; the coefficients a0 , an , bn are called Fourier coefficients of f (t ) , which are defined as: a0 {1 (2T )}
T
T
f (t ) dt , an (1 T )
T
T
f (t ) cos( nt ) dt , bn (1 T )
T
f (t ) sin( nt ) dt
T
(3.96)
Proof: If we integrate both sides of the Eq. 3.95 and assume that it is permissible to integrate the T
T
T
T cos(nt )dt 0
and
T f (t )dx T a0 dx T n 1{an cos(nt ) bn sin(nt )}dx T T T T f (t ) dt a0 dt n 1 an cos( nt ) dt n 1 bn sin( n t )dt a0 (2T ) T T T T
series
term
by
a0 {1 (2T )}
term
T
T
we
f (t ) dt . As,
get,
T
T
T sin(nt )dx 0
for all n . To determine
an for n 1 , multiply Eq. 3.95 by cos(mt ) (where m I and m 1 ) and integrate from T to T , I
i.e.,
T
T
T
T
T
T
f (t ) cos( m t ) dt a0 cos( mt ) dx
T
T
n 1
( an cos( nt ) bn sin( nt ) cos( mt ) dt
T
I a0 cos( m t )dt n 1 an cos( nt ) cos( m t ) dt n 1 bn sin( nt ) cos( m t ) dt . As the T
T
T
first integral in the RHS is shown to be zero and also using trigonometric identities we can prove that T T 0, n m for all n and m ; also . So sin( n t ) cos( m t ) dt 0 cos( n t ) cos( m t ) dt T T T , n m
T
T
f (t ) cos( m t ) dt amT am (1 T )
T
T
f (t ) cos(mt ) dt an (1 T )
T
T
f (t ) cos(nt ) dt ,
where n 1, 2, 3, . Similarly, if we multiply Eq. 3.95 by sin(mt ) and integrating it from T to T , we get bn (1 T )
T
T
f (t ) sin( nt ) dt , where n 1, 2, 3, .
The Fourier coefficients of a sum f1 f2 are the sums of the corresponding Fourier coefficients of f1 and f 2 .
The Fourier coefficients of cf are c times the corresponding Fourier coefficients of f .
Example 3.155 [CE-1998 (1 mark)]: A discontinuous real function can be expressed as (a) Taylor’s series and Fourier’s series (b) Taylor’s series and not by Fourier’s series (c) Neither Taylor’s series nor Fourier’s series (d) Not by Taylor’s series, but by Fourier’s series Solution (d): For Taylor series expansion, the function must be continuous; but for Fourier series expansion the function must be continuous or piece-wise continuous. So option (d) is correct.
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Engineering Mathematics
Chapter 3: Differential Equation
Example 3.156 [CE-2000 (1 mark)]: A function with period 2 is shown below
[3.69]
(a) f ( x ) (1 2) n 1{2 ( n )}sin( n 2)
(b) f ( x ) n 1{2 (n )}sin(n 2) cos nx
(c) f ( x ) (1 2) n 1 {2 (n )}sin( n 2)
(d) f ( x ) n 1 {2 ( n )}sin( n 2) sin nx
The Fourier series for this function is given by Solution (b): As f ( x ) is periodic with period 2T 2 T and from the given figure, we have
0, x 2 f ( x ) 1, 2 x 2 . So from Eq. 3.95, the Fourier series, with 2 (2T ) 1 , is given as 0, 2 x
f ( x ) a0 n 1 ( an cos nx bn sin nx ) ,
f ( x) cos nxdx (1 )
a0 {1 (2 )}
an (1 )
f ( x )dx {1 (2 )}
where
2
2
2
2
(0)dx
from
(1)dx
2
(0) cos nxdx
2
2
Eq.
3.96
(0) dx 1 2 ;
1cos nxdx
2
(0) cos nxdx
2
an (1 ) 0 (sin nx) n 2 0 {1 (n )} 2 sin( n 2) {2 n }sin(n 2) ; bn (1 )
f ( x ) sin nxdx (1 )
2
(0) sin nxdx
2
2
1sin nxdx
2
(0) sin nxdx
2
bn (1 ) 0 (cos nx ) n 2 0 0 .
Thus Fourier series for the given figure is f ( x ) (1 2) n 1{2 ( n )}sin( n 2) cos nx Example 3.157 [EC-2005 (1 mark)]: Choose the function f (t ) ; 1 , for which a Fourier series cannot be defined. (a) 3sin(25t ) (b) 4 cos(20t 3) 2 sin(10t ) (c) e t sin(25t ) (d) 1 Solution (c): Functions given in options (a), (b) and (d) are periodic so we can find their Fourier series. But the function given in option (c) is not periodic so we cannot find its Fourier series. Example 3.158 [EE-2014 (2 marks)]: Let g :[0, ) [0, ) be a function defined by g ( x ) x [ x ] , where [ x ] represents the integer part of x . (That is, it is the largest integer which is less than or equal to x ). The value of the constant term in the Fourier series expansion of g ( x ) is ……………. Solution: As g ( x ) x [ x] {x} which is periodic with period 2T 1 . So from Eq. 3.96 the value of the
constant
a0 x (1 2) x 2
term
0 1 2
a0 [1 {2(1 2)}]
is
(1 2) x 2
12
0
12
1 2
{ x}dx
0
12
1 2
(1 x )dx ( x ) dx 0
0 (1 2) (1 2)(1 2) 2 (1 2)(1 2) 2 0 1 2
3.4.1 Waveform Symmetry There are few methods by which the evaluation of Fourier coefficients is simplified by symmetry consideration. These methods reduce the amount of calculations involved in finding out the coefficients. From Eq. 3.96, we have a0 {1 (2T )}
T
T
putting
t dt d
in
0
T
T
0
the
first
integrand
a0 {1 (2T )} f ( ) d f (t ) dt {1 (2T )} T
a0 {1 (2T )} { f (t ) f ( t )}dt . 0
Again
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f (t )dt {1 (2T )}
T
0
on T
0
T
0
from
Eq.
0
the
f ( )d f (t ) dt
T
f (t ) dt f (t ) dt , now RHS,
we
get
3.96,
we
have
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Engineering Mathematics
an (1 T )
T
Chapter 3: Differential Equation
f (t ) cos(nt )dt (1 T )
T
t dt d
in
0
T
the T
T
0
0
integrand
an (1 T ) f ( ) cos(n )d f (t ) cos(nt )dt
a (1 T ) an (1 T )
T
n
T
0
T
the
T
0
0
RHS,
putting we
get
f ( t ) cos(nt )dt f (t ) cos(nt )dt an (1 T )
. Similarly, bn (1 T )
on
Now
f ( ) cos(n )d f (t ) cos(nt )dt
0
T
f (t ) cos(nt )dt f (t ) cos(nt )dt .
first
0
[3.70]
{ f (t) f (t)}cos(nt)dt T
0
{ f (t) f (t)}sin(nt)dt . Now the following symmetries are considered: T
0
Odd Symmetry: A function f (t ) is said to be odd if f (t ) f (t ) . So for odd symmetry we have a0 0 , an 0 and bn (2 T )
T
f (t ) sin(nt )dt . Thus the Fourier series expansion of an
0
odd function contains only the sine terms, the cosine and the constant terms being zero. The Fourier series of odd symmetric function also called as Fourier Sine series. We also have f (t ) f (2T t ) for odd symmetry which comes if an 0 , where 2T is the period of f (t ) .
Even Symmetry: A function f (t ) is said to be even if f (t ) f ( t ) . So for even symmetry we T
T
0
0
have a0 (1 T ) f (t )dt , an (2 T ) f (t ) cos( nt ) dt and bn 0 . Thus the Fourier series
expansion of an even function contains only a constant term and cosine terms, the sine terms being zero. The Fourier series of even symmetric function also called as Fourier Cosine series. We also have f (t ) f (2T t ) for even symmetry which comes if bn 0 , where 2T is the period of f (t ) . Thus the Fourier series of a real periodic function has only cosine terms if it is even and sine terms if it is odd. [This point was asked in EC-1994, EC-2009 (1 mark)].
Half Wave Symmetry: If a function when shifted by half the period and then if the shifted function is found to be negative of the original function, then the signal has half wave symmetry, as shown in Fig. 3.9. So for a function of period 2T , to have half wave symmetry then we have f (t ) f (t T ) . Half wave symmetry implies that the second half of the wave is exactly opposite to the first half. A function with half wave symmetry does not have to be even or odd, as this property requires only the shifted signal is opposite. So the Figure 3.9: Half Wave Symmetric Function
value of the Fourier constants is given as:
a0 {1 (2T )}
T
f (t )dt {1 (2T )}
T
integrand we get a0 {1 (2T )}
T
T
T
0
0
T
0
we get I1
0
T
0
T
T
0
a0 {1 (2T )} f (t )dt f (t )dt 0 an (1 T )
f (t )dt {1 (2T )} { f ( )}d
f (t ) dt f (t ) dt , so putting t T dt d in the first
f ( T ) d
0
T
0
.
T
T
0
0
an (1 T )
Now,
T
T
f (t ) dt
f (t ) cos( nt ) dt ,
T
f (t ) cos( n t )dt f (t ) cos(nt ) dt , let t T dt d in the first integrand 0
T
T
f (t ) cos( n t )dt f ( T ) cos{n ( T )}d f ( ) cos( n nT ) d 0
0
T
T
0
0
I1 f ( ) cos( n n ) d f ( ){cos( n ) cos n sin( n ) sin n }d
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Engineering Mathematics
Chapter 3: Differential Equation
T
T
0
0
[3.71]
I1 f ( ){cos( n ) cos n }d cos n f (t ) cos(nt ) dt .
T
T
0
0
Thus
an (1 T ) cos n f (t ) cos(nt )dt f (t ) cos(nt )dt {(1 cos n ) T } 0,
an
T
0
n is even
(2 T ) f (t ) cos( nt ) dt ,
n is odd
we T
have
f (t ) cos(nt )dt
0
0,
n is even
(2 T ) f (t ) sin(nt ) dt ,
n is odd
. Similarly bn
T
0
.
Thus the Fourier series expansion of a periodic function having half wave symmetry contains only odd harmonics and the constant term being zero. Example 3.159 [CE-2003 (2 marks)]: The Fourier series expansion of a symmetric and even 1 (2 x ), x 0 function, f ( x ) where, f ( x ) and will be 1 (2 x ), 0 x (a) (c)
n1{4 n1{4
( 2 n 2 )}(1 cos n ) cos nx
(b)
( 2 n 2 )}(1 sin n ) cos nx
(d)
n1{4 n1{4
( 2 n 2 )}(1 cos n ) cos nx ( 2 n 2 )}(1 sin n ) cos nx
Solution (b): The Fourier series expansion of a symmetric and even function f ( x ) have only cosine
term. As the given function have period of 2 , so with 2T 2 , f ( x ) a0 n 1 ( an cos nx ) , where a0
1
0 g ( x)dx 0
g ( x) cos( nx) dx
2
1 x2 1 2 2x 1 dx x 0 ; and 0
2x 2 2x 1 cos( nx ) dx cos nx dx cos nx dx 0 0 0 0 2x 2x 2 sin nx 2 an 0 cos nx dx 0 0 cos nx dx n 0 2 sin nx 2 2 x sin nx 2 2 cos nx 2 2 cos n 1 an dx 0 0 n 0 n n n 0 n n
an
2
1
an {4 ( n 2 2 )}(1 cos n ) . Thus f ( x ) n 1 {4 ( n 2 2 )}(1 cos n ) cos nx
Example 3.160 [EE-2005 (2 marks)]: The Fourier series for the function f ( x ) sin 2 x is (a) sin x sin 2 x (b) 1 cos 2x (c) sin 2 x cos 2 x (d) 0.5 0.5cos 2x 2 2 Solution (d): As sin x (1 2) (1 2) cos 2 x so Fourier series of sin x is (1 2) (1 2) cos 2x . Example 3.161 [EE-2006 (1 mark)]: x (t ) is a real valued function of a real variable with period T . Its trigonometric Fourier Series expansion contains no terms of frequency 2 (2 k ) T ; k 1, 2, . Also, no sine terms are present. Then x (t ) satisfies the equation (a) x (t ) x (t T ) (b) x (t ) x (T t ) x ( t ) (c) x (t ) x (T t ) x (t T 2) (d) x(t ) x (t T ) x (t T 2) Solution (c): Please note that the formulas are derived for period 2T but in given question we have period T , so we put T 2 in place of T . As the Fourier expansion of the function x (t ) contains no frequency term containing even n so it means that it contains only odd n , i.e., odd harmonics so we can say that x (t ) is a half wave symmetric function which means x(t ) x(t T 2) . Also the Fourier expansion of x (t ) contains no sine term so x (t ) has is an even symmetry and thus we have
x (t ) x (T t ) . Thus x(t ) x(T t ) x(t T 2) .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.72]
T 4 t 3T 4 1, Example 3.162 [EE-2007 (2 marks)]: A signal x (t ) is given by x (t ) 1, 3T 4 t 7 T 4 x (t T ) otherwise
. Which among the following gives the fundamental Fourier term of x (t ) ?
t 2T 4 Solution: The given function x (t ) is periodic with period 2T . Also x(t ) x (t T ) so x(t ) has half (a)
4
t T 4
(b)
cos
4
t 2T 4
(c)
cos
4
t T 4
(d)
sin
4
sin
wave symmetry so we have a0 0 . Now the derived formulas are for functions from T to T . But we have from T 4 to 7T 4 so we shift the given function left side by 3T 4 and thus
1,
T t 0
1,
0t T
x (t 3T 4)
and
also
as
. As we have to find the fundamental Fourier term so we have n 1
t {2 (2T )}T ;
T
T
0
0
thus
the
Fourier
coefficient
T
an (2 T ) f (t ) cos(t ) dt (2 T ) cos( t )dt {2 (T )} (sin t ) 0
an {2 (T )}(sin T 0) 0 .
Now
T
bn (2 T ) (cos t ) 0 {2 (T )}(cos T
T
T
0
0
bn (2 T ) f (t ) sin(t ) dt (2 T ) sin( t ) dt
1)T0
4 .
Thus we have x t (3T 4) (4 ) sin t x (t ) (4 ) sin t (3T 4)
x (t ) (4 ) sin ( t T ) (3 4) (4 ) sin ( 4) ( t T ) x (t ) (4 ) sin ( 4) ( t T ) (4 ) sin ( t T ) ( 4) .
Quarter wave Symmetry: If a function is both half wave symmetry and even or odd symmetry then it is considered as quarter wave symmetry, as shown in Fig. 3.10.
Figure 3.10: Quarter Wave Symmetric (a) Even (b) Odd Function
So for the quarter wave odd symmetric function the Fourier coefficients are a0 0 , an 0 and bn (4 T )
T 2
0
f (t ) sin( nt ) dt , where n is odd. Also for the quarter wave even symmetric function
the Fourier coefficients are a0 0 , bn 0 and an (4 T )
T 2
0
f (t ) cos(nt ) dt , where n is odd.
3.4.2 Convergence and Sum of a Fourier Series If f ( x ) is a periodic function with period 2T and f ( x ) and f ( x ) are piecewise continuous on [ T , T ] , then the Fourier series given by Eq. 3.95 is convergent. The sum of the Fourier series is equal to f ( x ) at all numbers x where f ( x ) is continuous; on the other hand, at the numbers x where f ( x ) is discontinuous, the sum of the Fourier series is the average of the LHL and RHL at x (which is a point of discontinuity), i.e., { f ( x ) f ( x )} 2 .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 3: Differential Equation
[3.73]
x, 2 x 0
Example 3.163 [TF-2007 (2 marks)]: A function f ( x ) is defined by f ( x )
and 0 x2 f ( x 4) f ( x) . This periodic function f ( x ) with a period 4 has its Fourier series expansion as
x,
2
f ( x ) (a0 2) m 1 am cos( m x 2) , am (1 2) f ( x ) cos( m x 2) dx . The coefficient of the term 2
cos(5 x 2) in the above expansion is
(b) 8 5 (a) –1 (c) 8 (5 ) 2 (d) 8 (5 )3 Solution (c): As the given function is periodic with period 4 . So from Eqs. 3.95 and 3.96 with
2T 4 , we have f ( x ) a0 m 1 am cos( m x 2) bm sin( m x 2) . As we have to find the
coefficient
of
cos(5 x 2)
so
we
2
am (1 2) f ( x) cos(m x 2)dx (1 2) 2
have
0
to
find
only
am ,
which
is
given
as
2
( x) cos(m x 2)dx ( x) cos(m x 2)dx
2
0
0
2
2 2 2 1 m x 2 m x m x 2 m x 2 am x sin 2 2 cos x sin 2 2 cos 2 2 m 2 2 m 2 m 2 0 m
am
22
1
2
2
22
2 m
1 cos m 2
2
m
2
1 cos m . Thus coefficient of
cos
5 x 2
is the value
of am at m 5 a5 {2 2 (52 2 )}( 1 cos 5 ) {2 2 (5 2 2 )}( 2) 8 (5 2 2 ) [Similar question was also asked in TF-2009 (1 mark)] Common Data for Questions 3.164 & 3.165: Let f : R R be defined by f ( x) x 2 for x and f ( x 2 ) f ( x) . Example 3.164 [XE-2007 (2 marks)]: The Fourier series of f in [ , ] is
(a) (
2
3) 4 n1{(cos nx) n 2 }
(c) (
2
3) 4 n1{(1)n (cos nx) n 2 }
(b) (
2
3) n1{(1)n (cos nx) n 2 }
(d) (
2
3) n 1{(cos nx) n 2 }
Solution (c): As the given function is periodic with period 2 . So from Eqs. 3.95 and 3.96 with
2T 2 ,
we
a0 {1 (2 )}
an (1 )
f ( x ) a0 n 1 an cos nx bn sin nx ,
have
f ( x ) dx {1 (2 )} x 2 dx {1 (2 )}( x 3 3) 2 3 ;
2 f ( x ) cos( nx ) dx an (1 ) x 2 cos( nx ) dx (2 ) x 2 cos( nx )dx , as x cos( nx) is
an
0
even
I x 2 cos(nx ) dx x 2 0
sin nx n
function.
0 2 x 0
sin nx n
dx 0
2
x n
cos nx n
2 cos n sin nx 2 cos n . I 2 n n n2 n 0
bn (1 )
2
f ( x)
where
So
an
As
0
cos nx
0
n
4 cos n n
2
dx
( 1) n
4 n2
.
Now
2 f ( x) sin(nx ) dx (1 ) x 2 sin( nx ) dx 0 , as x sin( nx ) is an odd function. So
n
2
(1) 4 n cos nx cos nx 4 ( 1) . 2 2 3 n 1 n n n 1 3
Example 3.165 [XE-2007 (2 marks)]: The sum of absolute values of the Fourier coefficients of f is (a) 2 6
(b) 2 3
Copyright © 2016 by Kaushlendra Kumar
(c) 2 2 3
(d) 2
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Engineering Mathematics
Chapter 3: Differential Equation
[3.74]
Solution (d): As in above problem we have a0 2 3 , an ( 1) n (4 n 2 ) and bn 0 . So sum of absolute values of the Fourier coefficients of f
S
2 3
is S a0 an bn
2 3
n 1
4 n2
0
2 2 1 1 1 2 1 1 1 4 2 , as . 2 2 2 6 12 22 32 6 1 2 3 3
4
Example 3.166 [XE-2013 (1 mark)]: The Fourier series of the periodic function f ( x) x ,
1 x 1 , f ( x 2) f ( x) , x R is given by (1 2) n 1{4 cos(2n 1) x} {(2 n 1) 2 2 } . Using the above, the sum of the infinite series 1 (1 32 ) (1 52 ) is (a) 2 4 (b) 3 2 8 (c) 2 8 (d) 2 2 Solution (c): From the given data, the given Fourier series can be written 1 4 cos(2n 1) x 1 4 4 4 f ( x) 2 2 cos x 2 2 cos 3 x 2 2 cos 5 x . Now 2 2 2 n 1 (2n 1) 2 1 3 5 1
x 0 we have f ( x )
2
as at
4 1 1 1 2 2 . As the given function f ( x ) , which is of period 2 2 2 1 3 5
x, 1 x 0
, can be written as: f ( x )
0 x 1
x,
1, 1 x 0
f ( x )
0 x 1
1,
f ( x ) and f ( x ) are
piecewise continuous in [ 1,1] . So the value f ( x ) at x 0 is the average of LHL ( f (0 ) ) and RHL of f ( x) at x0 as x0 is the point of discontinuity. Thus
f (0 ) lim f (0 h) lim{(0 h)} 0 h 0
f (0 ) lim f (0 h) lim(0 h) 0 .
and
h0
h 0
So
h 0
2 4 1 1 1 0 1 1 1 f ( x) 0 . Thus . 2 2 2 12 32 52 12 32 52 8 [Similar questions were also asked in CS-1993, EC-1993 (1 mark)] 2, x 0 Example 3.167 [AE-2014 (2 marks)]: For the periodic function given by f ( x ) 2, 0 x
f (0 ) f (0 )
1
with f ( x 2 ) f ( x) using Fourier series, the sum s 1 (1 3) (1 5) (1 7) converges to (b) 3 (c) 4 (d) 5 (a) 1 Solution (c): As the given function is periodic with period 2 . So from Eqs. 3.95 and 3.96 with
2T 2 ,
we
(2)dx (2)dx {1 (2 )}(2x) f ( x) cos(nx)dx (1 ) (2) cos(nx)dx (2) cos(nx)dx
a0 {1 (2 )}
an (1 )
f ( x ) a0 n 1 an cos nx bn sin nx ,
have
f ( x)dx {1 (2 )}
0
0
0
0
0
0
where
(2 x)0 0 ,
an (1 ) 2 (sin nx ) n 2 (sin nx ) n 0 0 , similarly
bn (1 )
bn
f ( x)sin(nx)dx (1 )
2 cos nx
n
0
Thus f ( x ) 0 n 1
0
( 2) sin(nx)dx (2) sin(nx)dx
cos nx 2 1 cos n n 0 n
0
1 cos n 4 1 cos n n n
4 1 cos n 42 2 2 sin nx sin x sin 3 x sin 5 x . So at x 2 , n 1 3 5
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Engineering Mathematics
Chapter 3: Differential Equation
[3.75]
2 2 8 1 1 1 . Now as f ( x) and f ( x ) both are piece 1 3 5 1 3 5 wise continuous function in [ , ] so the Fourier series converges whose value is f ( 2) , as x 2 is a continuous point for f ( x) . Thus f ( 2) 2
we have f ( x )
42
8 1 1 1 1 1 1 2 . So the given series converges to . 2 1 3 5 1 3 5 8 4 4
3.4.3 Half Range Expansion Many physical problem the function f ( x ) is only known over a finite interval, say 0 x T . To express f ( x ) as a Fourier series means that we need to extend the function to be valid for all x . If we choose the function periodically, with period 2T , then we retrieve the results given in Eqs. 3.95 and 3.96. However we could also extend the function in an even manner to get a cosine series or as an odd manner to get a sine series, as shown in Fig. 3.11, in which (a) a function is defined on an interval 0 x T with T 1 ; (b) the even extension of the function onto the interval T x T is shown as a solid curve and the periodic extension of period 2T is shown as a dot-dash curve; (c) the odd extension of the function, solid curve, and the periodic extension of period 2T , dot-dash curve.
Figure 3.11: (b) Even and (c) Odd extension of f(x) given by (a)
The even extension of f ( x ) will generate a Fourier series that only involves cosine terms and is called the cosine-half range expansion, which is given as:
f ( x ) a0 n 1 an cos{(n T ) x} T
T
0
0
(3.97)
where, a0 (1 T ) f ( x) dx , an (2 T ) f ( x ) cos{( n T ) x}dx and n 1, 2, .
The odd extension of f ( x ) will generate a Fourier series that only involves sine terms and is called the sine-half range expansion, which is given as:
f ( x ) n 1 bn sin{( n T ) x}
(3.98)
T
where, bn (2 T ) f ( x ) sin{( n T ) x}dx and n 1, 2, . 0
3.4.4 Complex Form of Fourier Series Let f (t ) be a piecewise continuous function with period 2T . The complex form of the Fourier series given by Eq. 3.95, where Fourier coefficients given by Eq. 3.96, is given as:
f (t ) n cn eint
(3.99)
T
f (t )e int dt , n 0, 1, 2, . For all t ,
where, the Fourier coefficient cn is given as: cn {1 (2T )}
T
the Fourier series converges to f (t ) if it is continuous at x and to { f (t ) f (t )} 2 otherwise.
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Engineering Mathematics
Chapter 3: Differential Equation
[3.76]
Proof: Consider the Fourier series given by Eq. 3.95 and its coefficients given by Eq. 3.96. We know that cos (e i e i ) 2 and sin (ei e i ) (2i) i(ei e i ) 2 . So let nt and substituting the values of cos n t and sin nt in an cos nt bn sin nt of Eq. 3.95, we get an cos n t bn sin n t (1 2) an (eint e in t ) {1 (2i )}bn (ein t e int ) an cos nx bn sin nx (1 2)( an ibn )ein t (1 2)( an ibn )e int cn eint k n e int ,
where
cn (1 2)( an ibn ) , kn (1 2)( an ibn ) and writing a0 in Eq. 3.96 as c0 . So substituting the results
in
Eq.
3.95,
we
c0 a0 {1 (2 )}
T
T
f (t ) c0 n 1 (cn eint kn e int ) .
get
Also
the
coefficients
f (t ) e int dt ,
similarly
f (t )dt , T
cn (1 2)( an ibn ) {1 (2T )}
T
kn (1 2)(an ibn ) {1 (2T )}
T
T
f (t )(cos n t i sin n t )dt {1 (2 )}
f (t )(cos nt i sin nt ) dt {1 (2T )}
T
T
f (t )e int dt .
As
coefficients cn and kn are combined by writing kn c n then cn {1 (2T )}
T
T
the
f (t )e int dt ,
n 0, 1, 2, . So the complex form of Fourier series given by Eq. 3.95 is given by Eq. 3.99.
Example 3.168 [EE-2008 (2 marks)]: Let x (t ) be a periodic signal with time period T . Let y (t ) x (t t0 ) x (t t 0 ) for some t0 . The Fourier series coefficients of y (t ) are denoted by b . If bk 0 for all odd k , then t0 can be equal to
(a) T 8
(b) T 4
(d) 2T
(c) T 2
Solution (b): Let ck be the coefficient of Fourier series of x (t ) then From Eq. 3.99,we have
x (t ) k ck e ikt x (t t 0 ) k ck e
y (t ) x (t t 0 ) x (t t0 ) k ck e
y (t ) k ck {eikt e
ik t0
eikt e
ik ( t t0 )
ik ( t t0 )
k ck e
ik t0
} k ck (e
ik t0
e
series of y (t ) is given as bk ck {e
ik t0
x (t t0 ) k ck e
also
ik ( t t0 )
ikt0
e
k ck {e ik t0
ik ( t t0 )
ik ( t t0 )
e
.
So
ik ( t t0 )
}
)eikt . So coefficient of Fourier
} 2 ck cos( k t0 ) . As time period of x (t ) is T so
2 T . Now bk 0 for all odd k and so cos(k t0 ) 0 kt0 k ( 2) t0 { (2 )} T 4
3.4.5 Fourier Transform The Fourier series representation of a periodic function describes the function in the frequency domain. The Fourier transform extends this frequency domain description to functions that are not periodic. The Fourier transform or the Fourier integral of a function f (t ) , with period 2T and thus frequency 2 (2T ) , is denoted by F (i ) and is defined by F (i ) F{ f (t )}
f (t )e it dt
(3.100)
and the inverse Fourier tramsform is defined by
f (t ) F 1{F (i )} {1 (2 )} F ( )eit d
(3.101)
Eqs. 3.110 and 3.101 form the Fourier transform pair. Proof:
Consider
cn {1 (2T )}
T
T
the
complex
form
of
Fourier
series
f (t ) n cn eint ,
f (t )e int dt , n 0, 1, 2, . So, f (t ) n {1 (2T )}
T
T
where
f (t )e int dt eint . If
the period 2T becomes infinite, the function does not repeat itself and becomes non-periodic. So the interval between the adjacent harmonic frequencies ( n 1) n (2 ) (2T ) , so we have 1 (2T ) (2 ) (2 ) . As T , d and the frequency goes from a discrete variable
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Engineering Mathematics
Chapter 3: Differential Equation
[3.77]
over to a continuous variable. So, as T , 1 (2T ) d (2 ) and n ; and also the summation sign change to integration and thus we have f (t ) {1 (2 )}
So the expression F (i )
e i t
f (t )e it dt d .
f (t )e it dt is called the Fourier transform of f (t ) and the expression
f (t ) {1 (2 )} F (i )eit d is called the inverse Fourier transform.
Convergence of Fourier Transform: When f (t ) is a single valued function and is different from zero over an infinite interval of time, the behaviour of f (t ) as t determines the convergence of the Fourier transform. The Fourier transform will exist if
f (t ) dt .
Properties of Fourier Transform Linearity: If , then F { f (t ) g (t )} F{ f (t )} F{g (t )} , provided the Fourier transform of f (t ) and g (t ) exists.
Scaling: If F { f (t )} F ( ) and c , then F {c f (t )} 1 c F ( c )
Time Shifting: If F { f (t )} F ( ) and t0 , then F { f (t t 0 )} e Proof: F { f (t t0 )}
it0
f (u )e
iu
du e
i t0
f (t )e
F ( ) it
Frequency Shifting: If F { f (t )} F ( ) and , then F ( 0 ) F {e Proof: F {e
f (t t0 )e i t dt e
it0
i0
f (t )} e
i0
f (t )e i t dt
f (t )e
i ( 0 ) t
dt e i0
i t0
F ( )
f (t )}
dt F ( 0 )
Symmetry: If F { f (t )} F ( ) then F {F (t )} 2 f ( ) Proof:
f (t ) F 1{F ( )} {1 (2 )} F ( )e it d {1 (2 )} F ( )eit d {1 (2 )} F ( x )eixt dx
and f ( ) {1 (2 )} F ( x )e ix dx 2 f ( ) F (t )e it dt F{ f (t )}
Differentiation in Time: Let n N and suppose that f ( n ) is piecewise continuous function and
lim f ( k ) (t ) 0 then F { f ( n ) (t )} (i ) n F ( ) . t
Differentiation in Frequency: Let n N and suppose that f ( n ) is piecewise continuous function then F {t n f (t )} i n F ( n ) ( ) When f ( t ) is an even function of t , its Fourier transform F ( i ) is a function of and is real; while when f ( t ) is an odd function of t , its Fourier transform F ( i ) is an odd function of and is imaginary. Proof: As F (i )
F (i )
P ( )
Q ( )
f (t )e it dt
f (t ) cos(t ) dt i
f (t ) cos(t ) dt
f (t ){cos(t ) i sin( t )}dt
f (t ) sin(t ) dt P ( ) iQ ( ) ,
which is an even function of , i.e.,
where
P ( ) P ( ) ; and
f (t ) sin(t ) dt which is an odd function of , i.e., Q ( ) Q ( ) . Now
F (i ) F (i ) ei ( ) , where F (i )
P2 ( ) Q 2 ( ) which is an even function of ; and
( ) tan 1{Q( ) P( )}
is
which
an
odd
function
of
,
since
( ) tan 1{Q( ) P ( )} tan 1{Q ( ) P ( )} tan 1{Q( ) P ( )} ( ) . Now
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics
Chapter 3: Differential Equation
[3.78]
When f (t ) is an even function, f (t ) cos t is an even function; and f (t ) sin t is an odd function. So P ( )
f (t ) cos( t ) dt 2 f (t ) cos(t ) dt which is a function of and
0
Q ( ) 0 and thus F (i ) P ( ) which is real. When f (t ) is an odd function, f (t ) cos t is an odd function; and f (t ) sin t is an even function. So P( ) 0 and Q ( )
f (t ) sin(t ) dt 2 f (t ) sin(t ) dt which is a function 0
of and thus F (i ) iQ( ) which is imaginary.
e t , t 0 Example 3.169 [IN-2011 (1 mark)]: Consider the signal x (t ) . Let X ( ) denote the 0, t 0
Fourier transform of this signal. The integral {1 (2 )} X ( ) d is
(b) 1 2
(a) 0
(c) 1
(d)
Solution (c): As x (t ) {1 (2 )} X ( )eit d x (0) {1 (2 )} X ( ) d . So x (0) e 0 1 . Example 3.170 [EE2014 (1 mark)]: A function f (t ) is shown in figure. The Fourier transform F ( ) of F (t ) is
(a) (b) (c) (d)
real and even function of real and odd function of imaginary and odd function of imaginary and even function of
Solution (c): As the given function f (t ) is an odd function so its Fourier transform is an odd function of and is imaginary.
Fourier Transform of Some Functions
f (t ) Ae
a t
, for all values of t
F (i ) F{ f (t )}
f (t )e
i t
dt A e at e
i t
0
0
dt A e ( a i ) t dt A e ( a i ) t dt
F (i ) { A ( a i )} { A ( a i )} .
So Fourier transform of f (t ) Ae
is F { Ae
a t
} F (i ) { A ( a i )} { A ( a i )} .
at
f (t ) Ae u (t ) , where u (t ) is the unit step function. Note that Fourier transform will exists if a0. F (i ) F { f (t )}
F (i ) A e
f (t )e i t dt A e at u (t )e it dt A e at e it dt
( a i ) t
0
a t
dt { A ( a i
0
)}{e ( a i ) t }0
{ A ( a i )}{e ( a i ) t }0 A ( a i ) .
So Fourier transform of f (t ) Ae at u (t ) is F { Ae at u (t )} F (i ) A (a i ) f (t ) k , where k is any constant, for all values of t Here
we
can
approximate
F (i ) F { f (t )} lim ke a 0
a t
e
i t
the
dt lim
function, whose magnitude is found by
a 0
constant
0, 0
2ka 2
a
2
, 0
f (t ) lim{ke
as
F{k}d {2ka
a t
a 0
,
which
is
an
}.
So
impulse
( a 2 2 )}d 2 k .
So the Fourier transform f (t ) k is an impulse function at 0 whose magnitude is 2 k and so F {k} F (i ) 2 k ( ) .
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Engineering Mathematics
Chapter 3: Differential Equation
[3.79]
f (t ) (t ) , where (t ) is a unit impulse function or dirac delta function. We know that the product of any arbitrary function f (t ) with unit impulse function (t t0 )
provides the function (t t0 ) to exists only at t t0 . So
f (t ) (t t0 )
f (t 0 ) . Now
F { (t )} F (i ) (t )e it dt e i (0) 1 . So Fourier transform of unit impulse function is 1.
f (t ) sgn(t ) , where sgn(t ) is a signum function As
sgn(t )dt
is infinite so direct evaluation of Fourier transform is not possible. Therefore the
given function has to be expressed as a limiting case of some other function and then the Fourier transform
is
computed.
F {sgn(t )} F (i ) lim e
a t
a 0
sgn(t )
Let
is
sgn(t )e it dt lim
a 0
0
0
multiplied
0
by
e
a t
and
a 0.
e at sgn(t )e it dt e a sgn(t )e it dt 0
F {sgn(t )} lim e ( ai )t dt e (a i )t dt lim {1 (a i )} {1 (a i )} 2 (i ) a 0
a 0
So Fourier transform of f (t ) sgn(t ) is F {sgn(t )} F (i ) 2 (i ) f (t ) u (t ) , where u (t ) a unit step function As
u (t )dt
given
is infinite so direct evaluation of Fourier transform is not possible. Therefore the
function
has
to
be
expressed
as
u (t ) (1 2) (1 2) sgn(t ) .
So
F {u (t )} F (1 2) (1 2) F {sgn(t )} 2 (1 2) (t ) (1 2){2 (i )} (t ) {1 (i )} . So Fourier transform of f (t ) u (t ) is (t ) {1 (i )} .
Relationship between Fourier Transform and Laplace Transform: For a function f (t ) , the
Laplace transform is defined as F ( s) f (t )e st dt and Fourier transform is defined as 0
F (i )
f (t )e
it
dt . Laplace transform is one sided in the interval 0 t . Thus Laplace
transform is applicable for positive time function, f (t ) , t 0 ; while Fourier transform is applicable for functions defined for all times. Laplace transform includes the initial conditions and is applicable for transient analysis; while Fourier transform is only applicable for steady-state analysis. We know that the Fourier transform of a function f (t ) is given as F (i )
f (t )e it dt ; also the Fourier
transform can be calculated only if f (t ) is absolutely integrable, i.e., so F ( s)
f (t ) dt . If s i ,
f (t )e ( i ) t dt { f (t )e t }e i t dt Laplace transform of f (t ) is basically the
Fourier transform of
f (t )e
t
. If 0 then s i , then we have F ( s)
f (t )e i t dt
F ( s ) F (i ) when s i . This means Laplace transform is same as Fourier transform when s i . So Laplace transform provides broader characterization compared to Fourier transform. s i indicates imaginary axis in complex s plane.
Concept of Region of Convergence: We know that Laplace transform of f (t ) is basically the Fourier transform of f (t )e t . Hence if Fourier transform of f (t )e t exists, then Laplace transform of f (t ) exists; and for Fourier transform of f (t )e t to exists we must have other words we can say that Laplace transform of f (t ) will exists, if
f (t ) dt . In
f (t ) dt . The range of
values of for which Laplace transform converges is called Region of Convergence.
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Chapter 3: Differential Equation
[3.80]
Example 3.171 [IN-2008 (2 marks)]: The Fourier transform of x (t ) e at u ( t ) , where u (t ) is the unit step function (a) Exist for any real value of a (b) Exists if the real value of a is strictly ve (c) Does not exist for any real value of a (d) Exists if the real value of a is strictly ve 0, t 0 Solution (b): F (i ) F{x (t )} x (t )e i t dt e at u ( t )e it dt . As u (t ) 1, t 0
0, t 0
u (t )
0
F (i ) e ( a i ) t dt {1 ( a i )}{e ( a i ) t }0
.
1, t 0 F (i ) {1 ( a i )}{e 0 lim e ( ai )t } . Now lim e ( ai )t exists if a i 0 or a 0 .
t
t
Convolution Theorem: Similar to convolution in Laplace transform, the convolution f g of functions f (t ) and g (t ) is defined by h(t ) ( f g )(t )
f ( ) g (t ) d
f (t ) g ( )d
(3.102)
The purpose is same as in the case of Laplace transform. Taking the convolution of two functions and then taking the Fourier transform of the convolution is the same as multiplying the transforms of these functions. Suppose that f (t ) and g (t ) are piecewise continuous, bounded and absolutely integrable on the t axis. Then F { f g} F{ f (t )} F{ g (t )} F (i ) G (i ) (3.103) Thus Convolution in time domain is equivalent to multiplication in frequency domain and vice versa. Example 3.172 [EC-2014 (2 marks)]: For a function g (t ) , it is given that t
for any real value . If y (t ) g ( ) d , then
(b) j
(a) 0
Solution (b): As G (i ) g (t )e
j t
Now
1,
t
0,
g ( )u{( t )}
g ( ), t
convolution
have
F { y (t )} e
t
0, t 1, t
j (0) t
g (t )
t
0,
g ( )u{( t )}
g ( ), t
.
t
and
u (t ) .
So
which
is
y(t )e
i (0) t
the
F { y (t )} F {g (t ) u (t )} F {g (t )} F {u (t )} 2 2
2 2 1 1 e (t ) y (t )e i t dt e 2 (t ) e 2 (t ) i i i
Thus at 0 we have
2
dt 0 g (t ) dt 0 .
y (t ) g ( ) d g ( )u{( t )}d g ( )u (t ) d
of 2 2
G (0) g (t )e
dt e 2
(d) j 2
0, t
u{( t )} u (t )
as
we
dt e
2 2
j t
is
(c) j 2
Thus
y (t )dt
g (t )e
2
.
2
dt (0)e 2(0) (t ) (1 i )e 2(0) 1 i i .
Exercise: 3.4 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill.
1 1 x 0 1. The Fourier series for f ( x ) 1 2 x 0 for x [ 1,1] is x 0 x 1
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Engineering Mathematics
(a) f ( x )
3
Chapter 3: Differential Equation
( 1) n 1
2
2
cos(n x)
1
[3.81]
sin( n x )
n n n 3 ( 1) 1 (b) f ( x ) 2 2 cos( n x ) sin(n x) 4 n 1 n n n 3 ( 1) 1 (c) f ( x ) 2 2 cos( n x ) sin(n x) 4 n 1 n n 3 ( 1) 1 1 (d) f ( x ) 2 2 cos( n x ) sin( n x) 4 n 1 n n 4
n 1
2. Let f ( x ) be the function on [ 3, 3] which is graphed in the figure. The constant term in the Fourier series for f is (a) 5 (b) 5 2 (c) 3 (d) 3 2
0, x 2 3. Let f ( x ) 1, 2 x 2 , the Fourier series for f on the interval [ , ] is 0, 2 x 1
cos x
cos(3x )
cos(5 x )
1
2 cos x
2 cos(3 x )
2 cos(5 x )
3 5 2 3 5 cos x cos(3 x ) cos(5 x ) 1 2 cos x 2 cos(3 x ) 2 cos(5 x ) (c) (d) 2 3 5 2 3 5 4. Let f ( x ) be the function of period 2T 2 which is given on the interval ( 1,1) by f ( x ) 1 x 2 , the Fourier series of f ( x ) is
(a)
2 1
(a) f ( x )
2 3
2
4
2 4
n 1
cos( n x )
(b)
(b) f ( x )
n2
(1) n 1
2 3 2
( 1) n 1 n2
n 1
4
cos( n x)
(1) n
cos(n x) (d) f ( x ) 2 2 cos(n x) 3 2 n 1 n 2 3 n 1 n 5. For the function f ( x ) 1 x defined on the half period 0 x 2 , the even extension to the full period 2 x 2 is (a) f ( x ) 1 x , 2 x 0 (b) f ( x ) 1 x , 2 x 0 (c) f ( x ) x , 2 x 0 (d) f ( x ) x 1 , 2 x 0 (c) f ( x )
x 5, 0 x 1
6. For the function f ( x )
0,
0,
(a) f ( x )
1 x 2
2 x 1
x 5, 1 x 0 2 x 1 0, (c) f ( x ) x 5, 1 x 0
, the even extension to the full period 2 x 2 is
x 5, 2 x 1
(b) f ( x )
1 x 0 0, 2 x 1 0, (d) f ( x ) x 5, 1 x 0
7. For the function f ( x ) sin x defined on the half period 0 x , the odd extension to the full period x is (a) f ( x ) sin x , x 0 (b) f ( x ) sin x , x 0 (c) f ( x ) cos x , x 0 (d) f ( x ) cos x , x 0
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Chapter 3: Differential Equation
[3.82]
0 x 1 0, 8. For the function f ( x ) , the odd extension to the full period 2 x 2 is 1 x, 1 x 2
1 x, (a) f ( x ) 0,
2 x 1
0,
(b) f ( x )
1 x 0
2 x 1
1 x , 1 x 0 1 x , 2 x 1 (d) f ( x ) 1 x 0 0,
x 1, 2 x 1
(c) f ( x )
1 x 0 9. Two dimensional Fourier transform and its inverse are infinitely (a) aperiodic (b) periodic (c) linear
0,
(d) non-linear
at
e , t 0
10. The Fourier transform of the function f (t )
t0
0,
1
(a)
1
(b)
(c)
for a 0 is
1
(d)
15.
16.
17.
18.
19.
20. 21. 22.
1
2
1
(d) (t ) sgn(t ) 2 t 2 t A continuous time system described by y (t ) x (t 2 ) is (a) casual, linear and time varying (b) casual, non-linear and time varying (c) non-casual, non-linear and time-invariant (d) non-casual, linear and time-variant If X ( ) is the Fourier transform of a signal f (t ) which is real and odd symmetric in time, the X ( ) is (a) complex (b) imaginary (c) real (d) real and non-negative If a periodic function f (t ) of period T satisfies f (t ) f {t (T 2)} , then its Fourier series expansion is (a) the constant term will be zero (b) there will be no cosine term (c) there will be no sine term (d) there will be no even harmonics A continuous time periodic signal x (t ) , having a period T , is convolved with itself. The resulting signal is (a) not periodic (b) periodic having period T (c) periodic having period 2T (d) periodic having period T 2 If the Fourier series coefficients of a signal are periodic then the signal must be (a) continuous-time, periodic (b) discrete-time, periodic (c) continuous-time, non-periodic (d) discrete-time, non-periodic 2t The Fourier transform of a signal x (t ) e u ( t ) is given by 1 2 1 2 (a) (b) (c) (d) 2 i 2 i 2i 2i 3t The final value of x (t ) (2 e )u (t ) is _____. The Fourier transform of an impulse function is (a) ( ) (b) 2 (c) 1 (d) 2 The function which has its Fourier transform and Laplace transform as unity is (a) Gaussian (b) Impulse (c) Ramp (d) Signum (a)
14.
1
2
1
a i a i a i a i 2 11. The Fourier transform of the exponential signal eit is (a) a constant (b) a rectangular pulse (c) an impulse (d) a series of impulses 12. The Fourier transform of a function f (t ) is X ( ) . The Fourier transform of ( d dt ) f (t ) will be (b) i 2 X ( ) (c) i X ( ) (a) ( d dt ) X ( ) (d) {1 (i )} X ( ) 13. Inverse Fourier transform of u ( ) is 1
2
(t )
(b)
(t )
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(c) 2 (t )
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Engineering Mathematics
3.5
Chapter 3: Differential Equation
[3.83]
Partial Differential Equation
A partial differential equation (PDE) is an equation involving one or more partial derivatives of an unknown function, let u depends on two or more variables, time t and one or more variables in space. Thus a PDE is an equation of the form: z z z 2 z 2 z 2 z (3.104) f , , , , 2 , 2 , , , z , x, y , t , 0 x y t x t x y containing independent variables t , x, y , , an unknown function z z ( x, y , , t ) and partial
z z z 2 z 2 z , , ,, 2 , , , w.r.t. these variables t , x, y , . derivatives t x y x tx As discussed in Section 3.1, the order of a PDE is the order of the highest derivative appearing in it; and the degree of a PDE is the degree of highest derivative appearing in it. As in case of ODE, we say that a PDE is linear if it is of the first degree in the dependent variable (the unknown function) and its partial derivatives are not multiplied together. A linear PDE is called homogeneous if it contains no term free from the unknown function and its derivatives; otherwise it is called a nonhomogeneous equation. The order and nature of some PDEs are given following table: PDE Order Nature PDE Order Nature 4 3 Non-Linear 2 z z z z Linear and z One Two and xy x Homogeneous x y Homogeneous 2 z z 2 z Linear and Linear and 2 z x y z One Two c 2 2 Homogeneous Homogeneous x y t x
z z xy x y
One
2
2 z x 2 xy
Two
Non-Linear and NonHomogeneous Non-Linear and NonHomogeneous
z t
2 z x
2
c
2
2 z x
2 z y
2
2
0
Two
Linear and Homogeneous
Two
Linear and Homogeneous
Formation of Partial Differential Equations: ODEs are formed by eliminating arbitrary constants only; on the other hand PDEs are formed by eliminating (a) arbitrary constants or (b) arbitrary functions. As the order of an ODE is equal to the number of arbitrary constants to be eliminated from a relation; in case of PDE if the number of arbitrary constants to be eliminated is equal to the number of independent variables involved in a relation. One obtains a first-order PDE, and if the number of arbitrary constants to be eliminated is more than the number of independent variables, then one obtains a higher-order PDE. If one arbitrary function is to be eliminated from a relation, then a first-order PDE is obtained; and if two arbitrary functions are to be eliminated, then a second-order PDE is obtained and so on. A first-order PDE is of the form: f ( x, y , z , p, q ) 0 , where
x and y are independent variables, z is a dependent variable and p z x and q z y . Formation of PDE by elimination of two arbitrary constants: Consider a relation of the type F ( x, y , z , a, b) 0 …(i), where a and b are arbitrary constants. Partially differentiating (i) w.r.t. x and y , we get (F x ) p (F z ) 0 (say) …(ii) and (F y ) q( F z ) 0 (say) …(iii). We can now eliminate the two arbitrary constants a and b between the equations (i) to (iii) and obtain a first order PDE of the form f ( x, y , z , p, q ) 0 . Example 2.173: Find the partial differential equation for z ( x 2 a 2 )( y 2 b2 ) . Solution: Partially differentiating the given equation w.r.t. x and 2
2
2
2
p z x 2 x( y b ) ( y b ) p 2 x …(i);
2
2
we
y 2
get,
2
q z y 2 y ( x a ) ( x a ) q (2 y )
…(ii). Substituting (i) and (ii) in the given equation, we get z { p (2 x)}{q (2 y )} pq 4 xyz .
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Chapter 3: Differential Equation
[3.84]
Formation of PDE by elimination of one arbitrary function of the form = ( ), where = ( , , ): Let z f (u ) …(i), where f (u ) is an arbitrary function of u where u u ( x , y , z ) a known function of x , y and z . Differentiating partially (i) w.r.t. x and y , we get p z x f (u x ) (u z ) p …(ii) and q (z y ) f (u y ) (u z ) q …(ii),
where f is the derivative of f w.r.t. u . Now eliminate f from equations (i) to (iii), we obtain the first-order PDE. Example 3.174: Find the partial differential equation for z x f ( x y z ) . Solution: z x f ( x y z ) …(i). Differentiating (i) partially w.r.t. x and y , we get p z x f ( x y z ) f { x (1 p )} …(i) and q z y x f (1 p) …(ii). Now eliminating
f ( x y z ) from (i) to (iii), we get p ( z x) q z ( p q ) x .
Formation of PDE by elimination of two arbitrary functions of the form z f ( x ) g ( y ) : Let z f ( x) g ( y ) …(i), where f is a function of x and g is a function of y alone. Differentiating partially (i) w.r.t. x , y and x then y or y then x , respectively, we get 2 p z x f g and q z y f g ; also we have s z ( xy ) f g . Thus we have pq ( f g )( g f ) ( f g )( f g ) sz .
Example 3.175: Find the partial differential equation for z x 2 y 3 Solution: Partially differentiating the given equation w.r.t. x , y and x then y or y then x , respectively, we get p z x 2 xy 3 and q z y 3 y 2 x 2 ; also we have s 2 z (xy ) 6 xy 2 . Thus we have pq (2 xy 3 )(3 y 2 x 2 ) ( x 2 y 3 )(6 xy 2 ) zs .
Formation of PDE by elimination of arbitrary function of the form ( , ) = : Consider a relation between x , y and z of the type F (u , v ) 0 …(i), where u and v are known functions of x , y and z and F is an arbitrary function of u and v . Also z is a function of x and y . Partially differentiating (i) by chain rule w.r.t. x and y, we get F u u F v v F u u F v v p p p 0 p …(ii); u x z v x z u x z v x z
and
F u
u F v v F u u F v v Now q q 0 q q …(iii). u y z v y z u y z v y z u u v v v v u u dividing (ii) by (iii) we get p q p q x z y z x z y z (u x u z p )(v y vz q ) (v x vz p )(u y u z q ) 0 , where u x u x , u y u y , …. Thus
( vz u y u z v y ) p (v x u z u x v z )q (u x v y v x u y ) where
(u , v ) ( y , z ) (vz u y u z v y ) P ,
(u , v ) ( y, z )
p
( u , v) ( z , x)
q
(u , v ) ( x, y)
Pp Qq R ,
(u , v ) ( z , x ) (vx u z u x vz ) Q
and
(u , v ) ( x, y ) (u x v y vx u y ) R which are functions of x , y and z and do not contain p and
q . The equation Pp Qq R is called Lagrange’s linear equation. In fact, it is a quasi-linear equation as the dependent variable may be present in P , Q and R . If P and Q are independent of z and R is a linear in z the equation Pp Qq R is called linear.
Example 3.176: Find the partial differential equation for f ( xy z 2 , x y z ) 0 . Solution: We have f (u, v) 0 with u xy z 2 and v x y z . As the partial differential equation for f (u, v) 0 , where u u ( x, y, z ) and v v ( x, y , z ) , is Pp Qq R , where
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Engineering Mathematics
Chapter 3: Differential Equation
P (u , v ) ( y , z ) (v z u y u z v y ) x 2 z ,
[3.85]
Q (u , v ) ( z , x ) (vx u z u x vz ) 2 z y
and
R (u , v ) ( x, y ) (u x v y v x u y ) y x . So the required PDE is ( x 2 z ) p (2 z y ) q ( y x ) .
Classification of First-Order PDE
Linear Equation: Let z z ( x, y ) be a function of two variables x and y . A first order PDE is said to be linear if it is linear in p , q and z , i.e., it is of the form
P ( x, y ) p Q ( x , y ) q R ( x, y ) z S ( x, y ) ,
where
p z x
and
q z y .
For
e.g.,
yp xq xyz x , y 2 p xyq x( z 2 y ) , etc., are linear equations. Semi-Linear Equation: A first order PDE is said to be semi-linear if it is linear in p and q and the coefficients of p and q are functions of x and y alone, i.e., of it of the form P ( x, y ) p Q ( x, y )q R ( x, y , z ) . For e.g. e x p yxq xz 2 , x p yq z , etc., are semilinear equations. Quasi-Linear Equation: Quasi-Linear PDE of first order can be written as P ( x, y, z ) p Q( x, y , z ) q R( x, y , z ) , where P , Q and R are functions of x , y and z . For e.g., e x x 2 zp y 2 xzq x 2 y 2 z 2 is a quasi-linear PDE. Non-linear equation: A first-order PDE is said to be non-linear if it does not fall into any of the above types.
PDE solvable by direct integration: Let us consider PDE which can be solved by direct integration. While carrying out integration w.r.t. a variable the other variable is held fixed. So, in place of constant of integration we have to add an arbitrary function of the variable held fixed. Example 3.177: Solve 2 z (yx ) 2 xe y
2 z
Solution:
yx y
2 xe y
z z y y 2 xe 2 xe y y x x
z x
2 xe y f ( x )
2 y
z {2 xe f ( x )}x z x e f ( x ) dx g ( y ) .
Solution to Linear, Semi-Linear and Quasi-Linear PDE: If the PDE is linear or semi-linear or quasi-linear then we can solve them by the method of Lagrange’s linear equation which is described as follows: The equation of the form Pp Qq R …(i) is known as Lagrange’s equation, where P , Q and R are functions of x , y and z . To solve this equation it is enough to solve the subsidiary equations dx P dy Q dz R …(ii). If the solution of the subsidiary equation is of the form u ( x, y ) c1 and v ( y , z ) c2 , then the solution of the given Lagrange’s equation is f (u, v) 0 . Method of grouping: For solving the subsidiary equation dx P dy Q dz R ; take any two members say first two or last two or first and last members. Now consider the first two members dx P dy Q . If P and Q contain z (other than x and y ) try to eliminate it. Now direct integration gives u ( x, y ) c1 . Similarly take another two members dy Q dz R . If Q and R contain x (other than y and z ) try to eliminate it. Now direct integration gives v ( y , z ) c2 . Therefore solution of the given Lagrange’s equation is f (u, v) 0 . We will discuss various cases which are as follows: If all the variables are separable Example 3.178: Solve zx 2 p zy 2 q 1 z 2 Solution: zx 2 p zy 2 q 1 z 2 x 2 p y 2 q (1 z 2 ) z which is a semi-linear equation. As we have 2 2 2 P x , Q y and R (1 z ) z ; so our subsidiary equation dx P dy Q dz R gives
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Engineering Mathematics
Chapter 3: Differential Equation
dx dy zdz 2 . Integrating by taking 1st and 2nd, 2 2 x y 1 z by
2nd
taking
dy
3rd,
and
dx
dy
[3.86]
1
1
x2 y 2 x y c1 ; similarly integrating
zdz
1
1
y 2 1 z 2 y 2 ln(1 z
2
) ln c
1
ln(1 z 2 ) ln c2
y
1 1 e1 y c2 , where c2 . So the general solution of the given PDE is F , 0. 2 2 1 z x y 1 z2 1y
e
c
If two variables are separable Example 2.179: Solve yp xq xyz xy Solution: As the given PDE is linear and we have P y , Q x and R xyz xy ; so our dx
subsidiary equation dx y dy x
dy
P
dy Q
dz R
gives
dx y
dy x
dz xyz xy
. Integrating by taking 1st and 2nd,
xdx ydy c1 x 2 y 2 c1 ; similarly integrating by taking 2nd and 3rd, we get
x
dz
ydy
xyz xy
dz ( z 1)
0 ydy
dz ( z 1)
c
y2
ln( z 1) ln c2 ( z 1)e y
2
where ln c2 c . So the general solution of the given PDE is F x 2 y 2 , ( z 1)e y
2
2
2
2
c2 ,
0.
Method of multipliers: Sometimes method of grouping fails to solve the subsidiary equation. So we use method of multiplier’s for solving the subsidiary equation which is as follows: Choose any three multipliers which are constants or functions of x, y , z such that l , m, n , dx P
dy Q
dz R
ldx mdy ndz lP mQ nR
, where lP mQ nR 0 and so ldx mdy ndz 0 . So, with one
set of l , m, n on integration of ldx mdy ndz 0 , we get u ( x, y , z ) c1 and another set of l , m, n on integration of ldx mdy ndz 0 , we get v ( x, y , z ) c2 . So the general solution is F (u , v ) 0 . Example 3.180 [XE-2007 (2 marks)]: 2 2 2 2 2 2 x ( z y )(z x ) y ( x z )(z y ) z ( y x ) is
The
general
(a) F ( x 2 y 2 z 2 , xyz ) 0
(b) F ( x 2 y 2 z 2 , xyz ) 0
(c) F ( x 2 y 2 z 2 , xyz ) 0
(d) F ( x 2 y 2 z 2 , xyz ) 0
solution
of
Solution (a): As the given PDE is quasi-linear and we have P x ( z 2 y 2 ) , Q y ( x 2 z 2 ) , R z ( y 2 x 2 ) ; so our subsidiary equation is dx P dy Q dz R . So using method of multipliers,
dx
we have
2
2
x( z y )
dy 2
2
y( x z )
dz
2
2
z( y x )
ldx mdy ndz 2
2
2
2
2
2
lx ( z y ) my ( x z ) nz ( y x )
. Now we
have to find l , m, n such that lx ( z 2 y 2 ) my ( x 2 z 2 ) nz ( y 2 x 2 ) 0 . On using l , m, n x, y , z , we have lx ( z 2 y 2 ) my ( x 2 z 2 ) nz ( y 2 x 2 ) 0 ; thus ldx mdy ndz 0 xdx ydy zdz 0 xdx ydy zdz 0 x 2 y 2 z 2 c . Also on using l , m, n 1 x ,1 y ,1 z , we have
lx ( z 2 y 2 ) my ( x 2 z 2 ) nz ( y 2 x 2 ) 0 ;
ldx mdy ndz 0
thus
(1 x)dx (1 y )dy (1 z ) dz 0 (1 x ) dx (1 y ) dy (1 z ) dz ln c2 ln x ln y ln z ln c2 xyz c2 . 2
2
So
the
general
solution
of
given
PDE
is
2
F ( x y z , xyz ) 0
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Chapter 3: Differential Equation
[3.87]
Second Order Partial Differential Equation: As for ODEs, 2nd order PDEs will be the most important ones in application. If we let u denote the dependent variable and let x and y denote the independent variables, then the general form of a linear second – order PDE is given by 2u 2u 2u u u A 2 B C 2 D E Fu G (3.105) x xy y x y where, the coefficients A, B , , G are functions of x and y ; and at least one of the coefficients A , B and C is not zero. When G ( x, y ) 0 , Eq. 3.105 is said to be linear and homogeneous; if G 0 then Eq. 3.105 is said to be linear and non – homogeneous. [This point was asked in AG-2014 (1 2 u 2u 2u u 0 xy are homogeneous and non – mark)]. For e.g., the linear equations and x 2 y 2 x 2 y homogeneous equations, respectively.
Classification of Equation: A linear second-order PDE (Eq. 3.105) in two independent variables with constant coefficients, such that at least one of the coefficients A , B and C is not-zero, is: Hyperbolic if B 2 4 AC 0 Parabolic if B 2 4 AC 0 Elliptic if B 2 4 AC 0 Example 3.181 [AE-2010 (1 mark)]: The linear second order partial differential equation 2 2 2 5 2 3 2 2 9 0 is x xy y (a) Parabolic (b) Hyperbolic (c) Elliptic (d) None of the above Solution: Comparing the given equation with Eq. 3.107, we have A 5 , B 3 , C 2 ; as 2 B 4 AC 9 40 31 0 the given PDE is of elliptic type. Example 3.182 [AG-2010 (1 mark)]: The partial differential equation
2u x 2
7
2u xy
2
2u y 2
0 is
said to be (a) parabolic (b) hyperbolic (c) elliptic (d) eccentric Solution (b): Comparing the given equation with Eq. 3.107, we have A 1 , B 7 , C 2 ; as 2 B 4 AC 49 8 41 0 the given PDE is of hyperbolic type. Example 3.183 [MN-2010 (1 mark)]: The partial differential equation, r ( r ) k , where k is constant, is a solution for 2 2 1 2 1 2 1 2 (a) 0 (b) 0 (c) r 0 (d) 0 r 2 r r r 2 r r 2 r r r 2 r r Solution (d): Partial differentiating the given solution w.r.t. r, we get 2 2 1 r 2 0 0. r (k ) r r r r r r r r 2 Example 3.184 [IN-2013, ME-1996 (1 mark)]: The type of partial differential equation f t 2 f x 2 is (a) Parabolic (b) Elliptic (c) Hyperbolic (d) Non-linear Solution (a): Comparing the given equation with Eq. 3.107, we have A 1 , B 0 , C 0 ; as 2 B 4 AC 0 the given PDE is of parabolic type.
Solution of a PDE: A solution of a linear PDE (Eq. 3.105) is a function u ( x, y ) of two independent variables that possesses all partial derivatives occurring in the equation and satisfies the equation in some region of the xy plane. Finding the general solutions of linear PDEs is often difficult and also a general solution is not useful in applications; so our focus will be on finding particular solutions of some important linear PDEs, i.e., equations that appear in many applications.
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Engineering Mathematics
Chapter 3: Differential Equation
[3.88]
Superposition principle: If u1 , u2 , , uk are the solutions of a homogeneous linear PDE, then the linear combination, u c1u1 c2 u2 ck uk , where c1 , c2 , , ck are constant, is also a solution.
Separation of variable method: In this method we seek a particular solution of the form of a product of a function of x and a function of y , i.e., u ( x, y ) X ( x)Y ( y ) . With this assumption it is sometimes possible to reduce a linear PDE in two variables to two ODEs.
Important second order PDEs: The following equation are some common PDEs:
1 D Wave equation: 1 D Heat equation:
2u t 2 u t
c
c
2
2u
2
2u x 2
2u x 2
2 D Poisson equation:
2 D Wave equation:
2u
2u x 2
2u
2u y 2
f ( x, y )
2u 2u c 2 2 2 t x y 2u
2
2u
2u
0 3 D Laplace equation: 2 2 2 0 x 2 y 2 x y z We are going to discuss: One dimensional Wave equation (which is Hyperbolic); One dimensional Heat equation (which is Parabolic); and Two dimensional Laplace equation (which is Elliptic).
2 D Laplace equation:
3.5.1 Modelling and Solution of Vibrating String: Wave Equation Consider a string of length L , such as a guitar string, stretched taut between two points on the x axis, say x 0 and x L . When the string starts to vibrate, assume that the motion takes place in the xu plane in such a manner that each point on the string moves in a direction perpendicular to the x axis (transverse vibrations). As shown in Fig. 3.12(a), let u ( x, t ) denote the vertical displacement of any point on the string measured from the x axis for t 0 . We assume the following assumption: 1. The mass of the string per unit length is a constant (homogeneous string). The string is perfectly elastic and does not offer any resistance to bending. 2. The tension caused by stretching the string before fastening it at the ends is so large that the action of the gravitational force on the string can be neglected. 3. The string performs small transverse motions in a vertical plane, i.e., every particle of the string moves strictly vertically and so that the deflection and the slope at every point of the string always remain small in absolute value.
Figure 3.12: Flexible string anchored at
=
and
=
(a) segment of a string (b)Enlargement of segment
Under these assumptions we may expect solutions u ( x, t ) that describe the physical reality sufficiently well. To obtain the PDE, we consider the forces acting on a small portion of the string, as shown in Fig. 3.12(b). Since the string offers no resistance to bending, the tension is tangential to the curve of the string at each point. Let T1 and T2 be the tension at the endpoints P and Q of that portion. Since the points of the string move vertically, there is no motion in the horizontal direction. Hence the horizontal components of the tension must be constant. So,
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Engineering Mathematics
Chapter 3: Differential Equation
[3.89]
T1 cos 1 T2 cos 2 T (say) constant
(3.106) 2
2
In the vertical direction the force balance T2 sin 2 T1 sin 1 x ( u t )
(3.107)
where the mass x of the portion times the acceleration 2u t 2 , evaluated at some point between x and x x ; here is the mass of the un-deflected string per unit length, and x is the length of the portion of the un-deflected string. Now dividing Eq. 3.107 by Eq. 3.106 T2 sin 2 T1 sin 1 2u (3.108) tan 2 tan 1 x 2 T2 cos 2 T1 cos 1 t where, tan 1 and tan 2 are the slopes of the string at x and x x , i.e., tan 1 (u x ) x and tan 2 (u x ) x x . Here we write partial derivatives because u depends also on time t . Dividing
Eq. 3.108 by x , we have
1 u
x x x x
2 u u . If we let x 0 , the linear PDE is, 2 x x T t
2u
c
2
2u
2 , c
T
(3.109) t x Eq. 3.109 is called the one – dimensional wave equation, which is homogeneous and of 2nd order. The physical constant T is denoted by c 2 (instead of c ) to indicate that this constant is positive, a fact that will be essential to the form of the solutions. Since the string is fastened at the ends x 0 and x L , we have the two boundary conditions: u (0, t ) 0 and u ( L, t ) 0 for all t (3.110) Also, form of the motion of the string will depend on its initial deflection (deflection at time t 0 ) as f ( x ) ; and on its initial velocity (velocity at t 0 ), as g ( x ) . Hence we have two initial conditions: 2
2
u ( x, 0) f ( x) and (u t ) ( x,0) g ( x ) for all 0 x L ,
(3.111)
We now have to find the solution of Eq. 3.109 satisfying the conditions in Eq. 3.110 and 3.111. In the method of separating variables, we determine solutions of the wave equation 3.109 of the form, u ( x, t ) F ( x)G (t ) (3.112) which are a product of two functions, each depending only one of the variables x and t . 2u
2G
2u
2F
F 2 and G 2 ; inserting these two in Eq. 3.109 t 2 t x 2 x 2 2 2 2 G F 1 G 1 F 2 we get F 2 c G 2 2 k (say); here the separation constant k is still t x c G t 2 F x 2 arbitrary. Now we have two single variable equations, i.e. ( 2 F x 2 ) kF 0 (3.113) ( 2G t 2 ) c 2 kG 0 (3.114) We now determine solutions F and G of Eq. 3.113 and 3.114 so that u FG satisfies boundary conditions (Eq. 3.110) , i.e., u (0, t ) 0 F (0)G (t ) 0 and u ( L, t ) 0 F ( L)G (t ) 0 for all t (3.115) Now from Eq. 3.115, if G 0 , then u FG 0 , which is of no interest; hence G 0 and F (0) 0 and F ( L) 0 (3.116) For k 0 , the general of Eq. 3.113 is F ax b , and from Eq. 3.116 we have a b 0 , so that F 0 u FG 0 , which is of no interest. For positive k 2 a general solution of Eq. 3.113 is Differentiating Eq. 3.112 we get
x x and from Eq. 3.116, we get F 0 . Hence we are left with the possibility of F Ae Be choosing k negative, say, k p 2 . Then Eq. 3.112 becomes ( 2 F x 2 ) p 2 F 0 and has a general solution F ( x ) A cos px B sin px ; and after applying Eq. 3.116 we have, F (0) 0 A 0 and
F ( L) 0 B sin pL 0 sin pL 0 (since, if B 0 then F 0 ) pL n p n L , n 1, 2, ; for negative integer we get the same solution with minus sign.
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Engineering Mathematics
Chapter 3: Differential Equation
[3.90]
Hence, setting B 1 ,we thus obtain infinitely many solutions F ( x ) Fn ( x ) , where Fn ( x ) sin ( n L ) x , n 1, 2,
(3.117)
2
Now with k p 2 n L , Eq. 3.114 becomes ( 2G t 2 ) n2 G 0 , n cp c( n L) The general solution of Eq. 3.118 is given as, Gn (t ) Bn cos n t Bn* sin n t Hence the solution of Eq. 3.109 satisfying Eq. 3.110 is given as, u n ( x, t ) ( Bn cos n t Bn* sin n t ) sin ( n L ) x , n 1, 2,
(3.118) (3.119) (3.120)
Eq. 3.120 is called the Eigen-functions and the values n cn L are called the Eigen-values. A single un will generally not satisfy the initial conditions given by Eq. 3.111. But since the wave equation 3.109 is linear and homogenous, it follows that the sum of finitely many solutions un is a solution of Eq. 3.109. To obtain a solution that also satisfies the initial conditions (Eq. 3.111), we consider the infinite series,
u ( x, t ) n 1 un ( x, t ) n 1 ( Bn cos n t Bn* sin n t ) sin (n L ) x
(3.121)
[Note: Similar derivation like Eq. 3.121 was also asked in XE-2010 (2 marks) with different boundary conditions] Now satisfying first initial condition of Eq. 3.111, we get
u ( x, 0) n 1 Bn sin ( n L ) x f ( x)
(3.122)
Hence we must choose the Bn so that u ( x, 0) becomes the Fourier Sine Series of f ( x ) . Thus L
Bn (2 L ) f ( x ) sin ( n L ) x dx , n 1, 2,
(3.123)
0
Now by differentiating Eq. 3.121 w.r.t. t and using second initial condition of Eq. 3.111, we get
(u t ) ( x ,0) n1 Bn*n sin ( n L) x g ( x )
(3.124)
Hence we must choose the Bn* so that for t 0 the derivative u t becomes the Fourier Sine Series of g ( x ) . Thus from Eq. 3.124, we have L
Bn* {2 (cn )} g ( x ) sin (n L ) x dx , n 1, 2,
(3.125)
0
Hence the solution of Eq. 3.109 with conditions given by Eq. 3.110 and 3.111 is given by Eq. 3.121 whose coefficients Bn and Bn* given by Eq. 3.123 and 3.124. 3-Dimensional Wave Equation in Cartesian, Cylindrical and Spherical coordinates:
Cartesian coordinates: Cylindrical Polar coordinates:
Spherical coordinates:
2u
2
2u
c
2u
2u
2 2 2 x y z 2 2u 1 u 1 2u 2 u 2 u c r 2 r r r 2 2 z 2 , where x r cos , y r sin and 2 t
t
2
z z . [This equation was asked in TF-2009 (1 mark)] 2u 1 u 1 2u 2 2 u , c r sin 2 r r sin 2 2 t sin x r cos sin , y r sin sin , z r cos
Example 3.185 [XE-2008 (2 marks)]: If u u ( x , t ) is such that
u (0, t ) u ( , t ) 0 , u ( x, 0) 0 , (u t ) ( x,0)
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2u
4
2u
t 2 x 2 sin x . Then u ( 3), ( 6) is
where
, 0 x , t 0,
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Chapter 3: Differential Equation
[3.91]
(a) 3 4 (b) 3 8 (c) 3 4 (d) 3 8 Solution (b): Comparing the given equation with Eq. 3.109, we have c 2 , and given conditions with Eqs. 3.110 and 3.111, we have L , f ( x ) 0 and g ( x) sin x . Thus the solution of the given PDE is from Eq. 3.121, i.e., for n 1, we have * * u ( x, t ) ( B cos t B sin t ) sin ( L) x ( B cos t B sin t ) sin x , where c L 2 2
L
and from Eqs. 3.123 and 3.125, B (2 L) f ( x) sin ( L) x dx (2 ) (0) sin ( L ) x dx 0 ; 0
*
0
L
0
0
0
B {2 (c )} g ( x ) sin ( L) x dx {2 (c )} (sin x )(sin x ) dx {1 (2 )} (1 cos 2 x) dx 1 2 .
Thus u ( x, t ) (1 2)(sin 2t )(sin x ) u ( 3), ( 6) (1 2) sin( 3) sin{2( 6)} 3 8 .
Cauchy Problem for the infinite string, D’Alembert’s solution: When we have an infinite string with no boundary, then we have the following Cauchy problem, i.e. 2 2u 2 u c , t 0 , x with ICs u ( x, 0) f ( x) and (u t ) ( x,0) g ( x ) t 2 x 2 The solution of the wave equation for the infinite string is given as: u ( x, t ) (1 2) f ( x ct ) f ( x ct ) {1 (2c )}
x ct
x ct
g ( s )ds
(3.126)
(3.127)
Example 3.186 [XE-2007 (1 mark)]: Let u ( x, t ) be the solution of the initial value problem 2u
2u
, t 0 , x , u ( x, 0) x 5 , (u t ) ( x ,0) 0 . Then u (2, 2) is t 2 x 2 (a) 7 (b) 13 (c) 14 (d) 26 Solution (a): Comparing the given equation with Eq. 3.126, we have c 3 , and from given conditions, we have f ( x ) x 5 and g ( x ) 0 . Thus the solution of the given PDE is from Eq. 9
3.127, i.e., u ( x, t ) (1 2) f ( x 3t ) f ( x 3t ) {1 (2c )}
x ct
x ct
(0) ds
u ( x, t ) (1 2) x 3t 5 x 3t 5 (1 2) 2 x 10 u (2, 2) (2 2 10) 2 7 .
3.5.2 Modelling and Solution of Heat Equation Suppose a thin circular rod of length L has a cross – sectional area A and coincides with the x axis on the interval [0, L] , as shown in Fig. 3.13. Let us suppose the following assumptions: 1. The flow of heat within the rod takes place only in the x direction. 2. The lateral, or curved, surface of the rod is insulated; that is, no heat escapes from this surface. 3. No heat is being generated within the rod. 4. The rod is homogeneous; that is, its mass per unit volume is a constant. 5. The specific heat and thermal conductivity K of the Figure 3.13: One – Dimensional flow of heat material of the rod are constants. To derive the PDE satisfied by the temperature u ( x, t ) , we need two laws of heat conduction: The quantity of heat Q in an element of mass m is Q mu , where u is the temperature of the element (3.128)
The rate of heat flow Qt through the cross-section, as indicated in Figure 3.13 is proportional to the area A of the cross section and the partial derivative with respect to x of the temperature is. Qt KAu x (3.129) Since heat flows in the direction of decreasing temperature, the ve sign indicates that Qt is positive for u x 0 (heat flow to the right) and negative for u x 0 (heat flow to the left).
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Engineering Mathematics
Chapter 3: Differential Equation
[3.92]
If the circular slice of the rod shown in Figure 3.13 between x and x x is very thin, then u ( x, t ) can be taken as the approximate temperature at each point in the interval. Now the mass of the slice is m Ax , and so from Eq. 3.128 that the quantity of heat in it is Q Axu (3.130) When heat flows in the positive x direction, from Eq. 3.129, heat builds up in the slice at the net rate KAu x ( x, t ) [ KAu x ( x x , t )] KA[u x ( x x, t ) u x ( x, t )] (3.131) By differentiating Eq. 3.129 w.r.t. t , the net rate is also given as Qt Axut (3.132) Hence equating Eq. 3.131 and 3.132, it gives K u x ( x x, t ) u x ( x, t ) ut (3.133) x Now by taking limit of Eq. 3.133, we get {K ( )}u xx ut or by letting c 2 K ( ) thermal diffusivity. Hence one dimensional heat equation is given as, 2 u K 2 u 2 thermal diffusivity c or u t c 2 u xx , c (3.134) 2 t x For solving Eq. 3.134 we need boundary and initial conditions. Since the ends x 0 and x L of the bar are kept at temperature zero, so the boundary conditions are: u (0, t ) 0 , u ( L, t ) 0 for all t (3.135) Also, the initial temperature in the bar at time t 0 is f ( x ) , so we have the initial condition u ( x, 0) f ( x) , [ f ( x ) is given] (3.136) Here we must have f (0) 0 and f ( L) 0 because of Eqs. 3.135. We shall determine a solution of Eq. 3.134 satisfying Eqs. 3.135 and 3.136; one initial condition will be enough, as opposed to two initial conditions for the wave equation. Substitution of a product u ( x, t ) F ( x)G (t ) into Eq. 3.134 gives F (G t ) c 2 ( 2 F x 2 )G . To separate the variables, we divide by c 2 FG , obtaining 1 G
1 2 F
k (3.137) c 2G t F x 2 For k 0 and k 0 the only solution u FG satisfying Eqs. 3.135 is u 0 . For k p 2 we have {1 (c 2 G )}(G t ) (1 F )( 2 F x 2 ) p 2 . Now we have two single variable equations, i.e. G 2F 2 c 2 p 2G 0 (3.139) p F 0 (3.138) 2 t x The general solution of Eq. 3.138 is given as F ( x ) A cos px B sin px (3.140) From the boundary conditions (Eqs. 3.135) it follows that u (0, t ) F (0)G (t ) 0 and u ( L, t ) F ( L)G (t ) 0 . Since G 0 u 0 , we require F (0) 0 , F ( L) 0 and get F (0) A 0
by Eq. 3.142 and then F ( L) B sin pL 0 with B 0 ; thus sin pL 0 p n L , n 1, 2, . Setting B 1 , we thus obtain the following solutions of Eq. 3.138 satisfying Eqs. 3.135 as, Fn ( x) sin(n x L ) , n 1, 2, (3.141) For p n L , Eq. 3.139 (G t ) n2 G 0 , where n cn L . It has the general solution as 2
Gn (t ) Bn e n t , n 1, 2, , where Bn is a constant
Hence the functions
un ( x, t ) Fn ( x )Gn (t ) Bn sin
n x
(3.142)
2
e n t , n 1, 2,
(3.143) L are the solution of the heat equation given by Eq. 3.134 satisfying Eqs. 3.135. These are Eigenfunctions of the problem, corresponding to Eigen-values n cn L .
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Chapter 3: Differential Equation
[3.93]
To obtain a solution that also satisfies the initial condition Eq. 3.136, we consider a series of these Eigen-functions, n x n2t cn u ( x, t ) n 1 u n ( x, t ) n 1 Bn sin e , n (3.144) L L
From this and Eq. 3.136, we have u ( x, 0) n 1 Bn sin( n x L ) f ( x ) . Hence for Eq. 3.144 satisfy Eq. 3.136, the Bn ’s must be the coefficients of the Fourier Sine Series, as given by L
Bn (2 L ) f ( x ) sin{( n L ) x}dx , n 1, 2,
(3.145)
0
Hence the solution of Eq. 3.134 with conditions given by Eq. 3.135 and 3.136 is given by Eq. 3.144 whose coefficients Bn is given by Eq. 3.145. 3-Dimensional Heat Equation in Cartesian, Cylindrical and Spherical coordinates:
u
Cartesian coordinates:
2
2u
c
2u
2u
2 2 2 x y z 2 u 1 u 1 2u 2 u 2 u c 2 2 2 , where x r cos , y r sin and 2 t r r r z r
t
Cylindrical Polar coordinates:
z z . [This equation was asked in TF-2009 (1 mark)]
2 u 1 u 1 2u , r sin 2 2 t sin r r sin x r cos sin , y r sin sin , z r cos u
Spherical coordinates:
2
where
c
Example 3.187 [TF-2007 (2 marks)]: By applying the method of separation of variables [u ( x, t ) X ( x )T (t )] to the heat equation u t c 2 ( 2 u x 2 ) and assuming k 2 as the separation constant, its solution is obtained as (a) u ( x, t ) [c1 sin( kt ) c2 cos( kt )] p1 (b) u ( x, t ) [c1 sinh( kt ) c2 cosh(kt )] p1 (c) u ( x, t ) [c1 sin( kx ) c2 cos( kx )] p2
(d) u ( x, t ) [c1 sinh( kx) c2 cosh( kx)] p2
where p1 exp( k 2 c 2 x ); p2 exp( k 2 c 2 t ) Solution (c): Comparing the given heat equation with Eq. 3.134, whose two single variable equations given by Eqs. 3.138 and 3.139, where the separation constant is given by p 2 ; and in the given question it is k 2 , thus we have to put k in place of p . Thus with k 2 as separation constant the two single variable equation of the given heat equation is ( d 2 F dx 2 ) k 2 F 0 , whose general solution is X ( x ) A cos( kx ) B sin( kx ) , and ( dT dt ) c 2 k 2T 0 , whose general solution is 2
T (t ) Ce t Ce c Thus
2 2
k t
.
u ( x, t ) X ( x )T (t ) [ A cos( kx ) B sin(kx )]Ce c
2 2
k t
[c1 sin( kx ) c2 cos( kx)]e c
2 2
k t
,
where
c1 AC and c2 BC .
Example 3.188 [AE-2008 (2 marks)]: Suppose the non-constant functions F ( x ) & G (t ) satisfy ( d 2 F dx 2 ) p 2 F 0 , ( dG dt ) c 2 p 2G 0 , where p and c are constants. Then the function u ( x, t ) F ( x)G (t ) definitely satisfies
2u
c
2
2u
u
(b)
c
2
2u
(c) 2 u 0
2u
2 2
c u 0 t 2 x 2 t x 2 t 2 Solution (b): As the given equations are Eqs 3.138 and 3.139 which are generated from Eq. 3.134, whose general solution is given by Eq. 3.112. So option (b) is the correct answer. (a)
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(d)
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Engineering Mathematics
Chapter 3: Differential Equation
[3.94]
Example 3.189 [MT-2008 (2 marks)]: Which of the following is a solution for z t 2 z x 2 ? (a) z ( x, t ) ( A sin x )e
2t
(b) z ( x, t ) A sin( x) e
2t
2
x t (d) z ( x, t ) B sin( x ) t (c) z ( x, t ) ( A t )e Solution (b): Comparing the given PDE with Eq. 3.134 whose solution is given by Eq. 3.144, we have c 1 n L ; so we have the solution of given PDE with n 1 as: 2
z ( x, t ) A sin( x) e t . Alternate method: We have to check which option is the solution of the given PDE; thus for option (b) z x A cos( x )e
2
t
z ( x, t ) A sin( x) e
2
t
z t 2 A sin( x)e
2
t
and
2
2 z x 2 A 2 sin( x )e t z t . 2u
Example 3.190 [PI-2008 (2 marks)]: For the partial differential equation
2
u
in the domain x t 0 x 1 with boundary conditions u (0, t ) 0 and u (1, t ) 0 ; initial condition u ( x, 0) sin( x ) , the solution of the differential equation is (a) e t sin( x ) (b) et sin( x ) (c) e t sin( x ) (d) e t sin( x) Solution (a): Comparing the given PDE with Eq. 3.134, whose solution is given by Eq. 3.144, we have c 1 n L ; now for 0 x 1 , we have L 1 n ; so the solution of given PDE 2
t with n 1 as: z ( x, t ) A sin( x) e . Alternate method: We have to check which option is the
solution of the given PDE; thus for option (b) u ( x, t ) e t sin( x ) u t sin( x )e t and u x cos( x )e t 2u x 2 2 sin( x)e t 2 (u t ) .
3.5.3 Solution of Steady Two – Dimensional Heat Problem: Laplace’s Equation 2
2
Consider two – dimensional heat equation u t c u c
2
( u x ) ( u y ) . For steady 2
2
2
2
(time independent) problems u t 0 , heat equation becomes Laplace’s equation which is given as,
2u
2u
2u
0 (3.146) x 2 y 2 A heat problem then consists of this PDE to be considered in some region R of the xy plane and a given boundary condition on the boundary curve C of R . This is a boundary value problem and we call it as ‘Dirichlet Problem in a Rectangle R .’ We consider a Dirichlet Problem for Laplace’s equation (Eq. 3.146) in a Rectangle R , assuming that the temperature u ( x, y ) equals a given function f ( x ) on the upper side and 0 on the other three sides of the rectangle. We solve this problem by separating variables. Substituting u ( x, y ) F ( x)G ( y ) into Eq. 3.146 written as u xx u yy , dividing by FG , and equating both sides to a ve constant, we get d 2F
kF 0
1 d 2F F dx 2
1 d 2G G dy 2
d 2G
(3.147)
dx 2 For Eq. 3.147 the left and right boundary conditions imply F (0) 0 and F ( a) 0
dy 2
k . So, we get
kG 0
(3.148) (3.149)
This gives k ( n a ) 2 and corresponding non – zero solutions F ( x ) Fn ( x) sin{( n a ) x} , n 1, 2,
(3.150)
For k ( n a ) 2 , Eq. 3.148 becomes, d 2G
2
n G0 2 dy a
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(3.151)
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Engineering Mathematics
Chapter 3: Differential Equation
and solution to Eq. 3.151 is given as, G ( y ) Gn ( y ) An e n y a Bn e n y
[3.95]
a
(3.152)
Now the boundary condition u 0 on the lower side of R implies that Gn (0) 0 ; i.e., Gn (0) An Bn 0 or Bn An . This gives Gn ( y ) An (e n y a e n y a ) 2 An sinh( n y a )
From Eq. 3.150 and Eq. 3.153, with 2 An
An*
(3.153)
, we thus obtain as the Eigen-functions of our problem
u n ( x, y ) Fn ( x)Gn ( y ) An* sin( n x a ) sinh( n y a )
(3.154) These solutions satisfy the boundary condition u 0 on the left, right and the lower sides. To get a solution also satisfying the boundary condition u ( x, b) f ( x ) on the upper side, we consider the
infinite series u ( x, y ) n 1 un ( x, y ) . From this and Eq. 3.154 with y b , we obtain
u ( x, b ) f ( x) n 1 An* sinh( n b a ) sin(n x a )
(3.155)
The expression in the parentheses of Eq. 3.155 must be the Fourier coefficients bn of f ( x ) , i.e., a
bn An* sinh( n b a ) (2 a ) f ( x ) sin( n x a ) dx
(3.156)
0
Hence from Eq. 3.154 and 3.156, the solution of our problem is
u ( x, y ) n 1 un ( x, y ) n 1 An* sin(n x a ) sinh( n y a )
(3.157)
a
An* [2 {a sinh( n b a)}] f ( x) sin( n x a ) dx
where,
(3.158)
0
3-Dimensional Laplace Equation in Cartesian, Cylindrical and Spherical coordinates:
Cartesian coordinates:
2u
Cylindrical Polar coordinates:
2
2u x 2 2u
2u y 2
1 u
2u z 2
0
1 2u
2u
0 , where x r cos , r 2 r r r 2 2 z 2 z z . [This equation was asked in TF-2009 (1 mark)]
u
y r sin
1 2 u 1 u 1 2u , r sin 2 2 2 r r r sin sin x r cos sin , y r sin sin , z r cos 2
u
Spherical coordinates:
and
where
Example 3.191 [CE-2001 (1 mark)]: The number of boundary conditions required to solve the 2 2 differential equation 2 2 0 is x y (a) 2 (b) 0 (c) 4 (d) 1 Solution (c): As the given equation is 2D – Heat equation, given by Eq. 3.146, so for solving it we require four boundary conditions. Statement for Linked Answer Question 3.192 and 3.193: The potential u ( x, y ) satisfies the
2u
2u
0 in the square 0 x , 0 y . Three of the edges x 0 , x and x 2 y 2 y 0 of the square are kept at zero potential and the edge y is kept at nonzero potential.
equation
Example 3.192 [XE-2007 (2 marks)]: The potential u ( x, y ) is given by
(a) u ( x, y ) n 1 An cosh nx sin ny
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(b) u ( x, y ) n 1 An sin nx cosh ny
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Engineering Mathematics
Chapter 3: Differential Equation
[3.96]
(c) u ( x, y ) n 1 An sinh nx sin ny
(d) u ( x, y ) n 1 An sin nx sinh ny
Solution (d): Comparing the given PDE and its conditions with Eq. 3.146 and the condition given by Eqs. 3.149 and 3.153, we get a and b . As, the solution of Eq. 3.146, with the conditions given by Eqs. 3.149 and 3.153, is given by Eq. 3.157. So our solution is n x n y un ( x, y ) n 1 An* sin sinh n 1 An* (sin nx) sinh( ny ) . a a Example 3.193 [XE-2007 (2 marks)]: If the edge y is kept at the potential sin x , then the potential u ( x, y ) is given by sin x sinh y sin nx sinh ny (a) u ( x , y ) n 1 (b) u ( x, y ) sinh n sinh sin x cosh y cosh nx sin ny (c) u ( x, y ) (d) u ( x, y ) n 1 cosh cosh n * Solution (b): As f ( x ) sin x so the coefficient, An , in the solution given by Eq. 3.154 with n 1
a ,
and
i.e.,
*
u ( x, y ) A* (sin nx )(sinh ny )
is
given
a
0
0
by
Eq.
3.158,
i.e.,
A [2 {a sinh( n b a)}] f ( x) sin(n x a )dx {1 ( sinh )} 2(sin x ) sin xdx 1 {sinh(n )} .
Thus u ( x, y) {(sin nx)(sinh ny )} {sinh(n )} . Example 3.194 [CE-2008 (2 marks)]: The equation 2h 2h 2 h 2h k x 2 k z 2 0 can be transformed to 2 2 0 by substituting x z x1 z (a) x1 x ( k z k x )
(b) x1 x ( k x k z )
(c) x1 x k x k z
Solution (d): The given PDE can be written as 2 h 2 x1
2h z
2
0,
i.e.,
kx 2 h k z x
2
2 h x12
kx 2 h k z x
kx 2h k z xx
2
2h z
2h x1x1
2
(d) x1 x k z k x
0 which is transformed into
kx
1
k z (x )
2
1 (x1 )
2
x1 x
kz kx
x1 x k z k x x1 x k z k x . Example 3.195 [XE-2012 (1 mark)]: For the solution of 2 u 0 , the domain and boundary conditions are shown in Figure. Which of the following statements is TRUE? (a) The solution cannot be obtained using separation of variables because the governing equation is non-separable. (b) The solution cannot be obtained using separation of variables because all the boundary values are non-zero. (c) The solution cannot be obtained using separation of variables because not all the boundaries are along constant coordinate lines. (d) The solution can be obtained by separation of variables. Solution (c): As the separation of variables is valid for rectangular boundaries, thus for the given figure (which has not rectangular boundaries) the solution for 2D – Heat equation cannot be obtained. Example 3.196 [XE-2013 (1 mark)]: Which one of the following pde CANNOT be reduced to two ordinary differentiable equations by the method of separation of variables? u 2u 2 u 2u 2u 2u u 2 u 2 u 2u (a) 2 0 (b) 0 (c) 0 (d) 0 t x t 2 x 2 t 2 t x x t 2 tx x 2
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Engineering Mathematics
Chapter 3: Differential Equation
[3.97]
Solution (d): The equation given in options (a) and (b) are the heat and wave equations, respectively, so they are solved by using method separation of variables. Now we have to check for option (c) and u F 2u 2 F (d). If u ( x, t ) F ( x)G (t ) is the solution then G (t ) 2 G 2 , also x x x x 2 2 2 u u G F u G ; similarly F 2 . So for option (c): 2 tx t x t x t t 2 u 2 u u 2 G G F F 2G F G 0F 2 G (t ) 0 F 2 G (t ) 2 t tx x t t x x t x t
2 G t 2
(G
1 F
2 p the variables are separated and so the equation in option (c)
t ) G (t ) F x is solved by the method of separation of variables. Now for option (d) 2u 2u 2u 2G G F 2 F 0 F G 0 the variables are not separated so the t 2 tx x 2 t 2 t x x 2 equation in option (d) is not solved by the method of separation of variables.
Example 3.197 [BT-2014 (2 marks)]: The concentration profile of a chemical at a location x and time denote by c ( x, t ) , changes as per the following equation, t,
c ( x, t ) {c0 2 Dt }exp{ x 2 (2 Dt )} , where D and c0 are assumed to be constant. Which of the following is correct? c 2c c D 2 c 2c 2c 2c D 2 c (a) D 2 (b) (c) D (d) t x t 2 x 2 t 2 x 2 t 2 2 x 2 Solution (b): As the equations in option (c) and (d) are similar to one – dimensional wave equation which has no exponential term in their solution so we can say that option (c) and (d) are not correct. Also the equations in option (a) and (a) are similar to heat equation which has exponential term in their solution so we have to check for options (a) and (b). As 2 2 2 c0 c0 x c x x and c ( x, t ) exp 1 exp Dt 2 Dt 2 Dt t 2t t 2 D 2 Dt c x
c0 Dt 2 Dt
x exp
x2
2 Dt
2c x
2
x2 x2 1 exp 2 Dt . So from these two Dt 2 Dt Dt c0
results we can say that c t ( D 2)( 2 c x 2 ) . Hence option (b) is correct. Exercise: 3.5 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. A partial differential equation has (a) one independent variable (b) two or more independent variables (c) more than one dependent variable (d) equal number of dependent and independent variables 2u 2u 2. A solution to the partial differential equation 2 9 2 is x y 2 2 (a) cos(3x y ) (c) sin(3 x y ) (b) x y (d) e 3 x sin( y ) 3. The partial differential equation 5 (a) Elliptic
2 z
x 2 (b) Parabolic
6
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2 z y 2
xy is (c) Hyperbolic
(d) None of these
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Engineering Mathematics
Chapter 3: Differential Equation
z
4. The partial differential equation xy
5
[3.98]
2 z
is x y 2 (a) Elliptic (b) Parabolic (c) Hyperbolic (d) None of these 2 2 z z 5. The partial differential equation 2 5 2 0 is x y (a) Elliptic (b) Parabolic (c) Hyperbolic (d) None of these 3 u u u 6. The partial differential equation 3 6u is t x x (a) linear, 3rd order (b) non-linear, 3rd order (c) linear, 1st order (d) non-linear, 1st order d dy 7. Given the boundary value problem x ky 0 , 0 x 1 , with y (0) y (1) 0 . Then the dx dx solutions of the boundary value problem for k 1 (given by y1 ) and k 5 (given by y5 ) satisfy: (a)
1
0 y1 y5 dx 0
1
0
(b)
dy1 dy5
dx 0
1
0 y1 y5 dx 0
(c)
dx dx u u 2u , u ( x, 0) 10e x 6e 4 x is 8. The solution of PDE x y
(d)
1
dy dy5
0 y1 y5 dx1
(a) u ( x, y ) 6e ( x 3 y ) 10e 2(2 x 3 y )
(b) u ( x, y ) 6e ( x 3 y ) 10e 2(2 x 3 y )
(c) u ( x, y ) 10e ( x 3 y ) 6e 2(2 x 3 y )
(d) u ( x, y ) 10e ( x 3 y ) 6e 2(2 x 3 y )
9. The solution of PDE
2u
2u
x 2
0 , where u{x, 2} 0 , u ( x, 0) 3e 3 x , is
y 2
(a) u ( x, y ) 4e3 x cos 3 y
(b) u ( x, y ) 4e 2 x cos 2 y
(c) u ( x, y ) 4e 3 x cos 3 y
(d) u ( x, y ) 4e 2 x cos 2 y 2u
10. The solution of PDE
c
t 2
2
2u x 2
, where u (0, t ) u (1, t ) 0 , u ( x, 0) A sin( x ) and
ut ( x, 0) 0 , is (a) u ( x, t ) A sin( cx ) cos( ct ) (c) u ( x, t ) A sin( cx ) cos( t )
11. The
solution
of
PDE
dx 0
dx
k
(b) u ( x, t ) A sin( x) cos( ct ) (d) u ( x , t ) A sin( x ) cos( t ) 2u x
2
u t
for
0 x ,
t a,
where
u ( x, 0) sin x ,
u x (0, t ) u x ( , t ) 0 , is
(a) u ( x, t ) (b) u ( x, t )
2
2
4
m 1 4
m 1
1 2
4m 1 1 2
2
2
cos(2mx)e 4 m kt
4m 1 2 2 4 1 (c) u ( x, t ) m1 cos(2mx )e 4 m kt 2 4m 1 (d) u ( x, t )
2
4
m1
1 2
4m 1
cos(2mx )e 4 m kt
2
cos(2mx)e 4 m kt
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Engineering Mathematics
Chapter 3: Differential Equation
[3.99]
Answers Keys 1 a 16 d
2 d 17 3
3 1 18 a
4 b 19 d
5 a 20 a
Answer Keys: Exercise: 3.1 6 7 8 9 10 b a c a c 21 22 23 24 25 b b c b c
1 a 16 b
2 c 17 d
3 a 18 a
4 d 19 c
5 c 20 b
Answer Keys: Exercise: 3.2 6 7 8 9 10 b a c d a 21 22 23 24 25 d a c a d
11 b 26 d
12 c 27 a
13 a 28 d
14 c
15 d
1 a 16 a
2 b 17 d
3 c 18 c
4 d 19 34.13
Answer Keys: Exercise: 3.3 6 7 8 9 10 c a b d c
11 b
12 a
13 d
14 c
15 b
1 a 16 a
2 b 17 b
3 d 18 b
4 c 19 a
5 b 20 2
Answer Keys: Exercise 3.4 6 7 8 9 10 d a d b a 21 22 c b
11 c
12 b
13 a
14 d
15 b
1 b
2 d
3 a
4 b
5 c
Answer Keys: Exercise 3.5 6 7 8 9 10 b a d c b
5 d 20 a
Copyright © 2016 by Kaushlendra Kumar
11 a 26 c
12 b 27 b
13 d
14 c
15 d
11 a
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Engineering Mathematics
Chapter – 4: Complex Variables
Chapter – 4 [1]
GATE – 2016: Chapter – 4: Complex Variables Note: The following questions came in GATE – 2016 were based on Complex Variables Chapter – 4. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 1. What are the modulus (r ) and argument ( ) of the complex number 3 4i ? [CH-2016 (1 mark)] 1
1
(a) r 7 , tan (4 3)
(b) r 7 , tan (3 4)
1
1
(c) r 5 , tan (3 4)
(d) r 5 , tan (4 3) 1
Solution (d): r 32 4 2 5 and tan 4 3 tan (4 3) . 3
3
2. Consider the complex valued function f ( z ) 2 z b | z | where z is a complex variable. The value of b for which the function f ( z ) is analytic is _____. [EC-2016 (1 mark)] Solution: f ( z ) u ( x, y ) iv ( x , y ) is analytic if u x v y and u y v x 3
2
2 32
So, let z x iy , f ( z ) 2( x iy ) b( x y )
f ( z ) 2( x 3 iy 3 i3 x 2 y 3 xy 2 ) b( x 2 y 2 )3 2 {2( x 3 3xy 2 ) b( x 2 y 2 )3 2 } i (3 x 2 y y 3 ) f ( z ) 2( x 3 iy 3 i3 x 2 y 3 xy 2 ) b( x 2 y 2 )3 2 {2( x 3 3xy 2 ) b( x 2 y 2 )3 2 } i (3 x 2 y y 3 ) 3
2
2
2 32
2
3
Thus u ( x, y ) {2( x 3xy ) b( x y ) } and v ( x, y ) (3x y y ) 2
2
2
2 12
Now u x v y 6( x y ) b(3 2)(2 x)( x y )
6( x 2 y 2 ) b 0 if x 0 ; and
u y v x 6 xy b(3 2)(2 y )( x 2 y 2 )1 2 6 xy b 0 if y 0 . So f ( z ) is analytic if b 0 . 3. For f ( z )
sin( z )
, the residue of the pole at z 0 is _____. [EC-2016 (1 mark)] z2 Solution: As the denominator of the given function is zero at z 0 , which is a pole of order 2. Thus Res f ( z ) z 0
1
d 2 1
lim
(2 1)! z 0 dz
2 1
1 sin( z ) d d ( z 0) 2 f ( z ) lim z 2 2 lim sin z z z 0 dz 1! z 0 dz
Res f ( z ) lim cos z 1 . z0
z0
4. Consider the function f ( z ) z z * where z is a complex variable and z * denotes its complex conjugate. Which one of the following is TRUE? [EE-2016 (1 mark)] (a) f ( z ) is both continuous and analytic (b) f ( z ) is continuous but not analytic (c) f ( z ) is not continuous but is analytic (d) f ( z ) is neither continuous nor analytic Solution (b): If z x iy z* x iy f ( z ) z z* 2 x ; let z0 x0 iy0 , as z z0 , x x0 & y y0 ; or z z z0 x x x0 & y y y0 , so if z 0 , then x 0 & y 0 . As lim f ( z z ) lim 2( x x ) 2 x f ( z ) , So we have lim f ( z z ) f ( z ) z 0
x 0
z 0
f ( z ) is
continuous everywhere. For any function f ( z ) to be analytic, the Cauchy-Riemann equations must be satisfied; but for f ( z ) 2 x i 0 u ( x , y ) iv ( x, y ) , the Cauchy-Riemann equations u x v y (2 x) x (0) y 2 0 (which is not possible), so it is not satisfied, and u y v x (2 x) y (0) x 0 0 (which is true), it is satisfied. So both the CauchyRiemann equations are not satisfied and hence the given function is not analytic. So option (b) is correct.
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Engineering Mathematics
Chapter – 4: Complex Variables
5. The value of the integral
2z 5
C {z (1 2)}( z 2 4 z 5) dz
clockwise direction, would be (a) 24 i 13 (b) 48 i 13 Solution: As the given function f ( z )
Chapter – 4 [2]
over the contour z 1 , taken in the anti[EE-2016 (1 mark)] (d) 12 13
(c) 24 13 2z 5 2
{z (1 2)}( z 4 z 5)
2z 5 { z (1 2)}{ z (2 i)}{ z (2 i )}
,
so the pole of the given function is z 1 2 , z 2 i, z 2 i , which are all simple poles. But only z 1 2 lie inside the contour z 1 , as (1 2) 1 ; and z 2 i lie outside the contour z 1 , as
2 i 2 2 12 1 . So
C f ( z )dz 2 i (sum of residues of the poles which lies inside the given
closed region) f ( z)dz 2 i Res f ( z ) 2 i lim{z (1 2)} f ( z) C
f ( z )dz 2 i lim{z (1 2)} C
z 1 2
f ( z )dz 2 i C
z 1 2
z 1 2
2z 5 {z (1 2)}( z 4 z 5)
2(1 2) 5 2
2
(1 2) 4(1 2) 5
48 13
2 i lim
z 1 2
2z 5 2
( z 4 z 5)
i
6. In the neighbourhood of z 1 , the function f ( z ) has a power series expansion of the form
f ( z ) 1 (1 z ) (1 z ) 2 . Then f ( z ) is (b) 1 ( z 2)
[IN-2016 (1 mark)]
(d) 1 (2 z 1) 1 1 1 2 f ( z) . Solution (a): As 1 z z 2 z 3 , so 1 (1 z ) (1 z ) 1 z 1 (1 z ) z (a) 1 z
7.
(c) ( z 1) ( z 1)
f ( z ) u ( x, y ) iv ( x , y ) is an analytic function of complex variable z x iy , where i 1 . If u ( x, y ) 2 xy , then v ( x, y ) may be expressed as [ME-2016 (1 mark)] 2
2
2
(a) x y constant 2
2
(b) x y constant
2
2
2
(c) x y constant (d) ( x y ) constant Solution (a): If f ( z ) u ( x, y ) iv ( x , y ) is an analytic function of complex variable z x iy , where i 1 , then it must satisfies Cauchy-Riemann equations u x v y …(i) and u y v x …(ii). So u v v (2 xy ) 2 y v 2 yy v 2 ydy f ( x ) v ( x , y ) y 2 f ( x ) …(iii) (i) x y y x Now partial differentiating both sides of (iii), w.r.t ‘ x ’ and applying the (ii) condition we get u v 2 x f ( x ) f ( x) 2 xdx x 2 c . Thus from (iii), v( x, y ) x 2 y 2 c . y x
8. Solution of Laplace’s equation having continuous second-order partial derivatives are called [ME-2016 (1 mark)] (a) biharmonic functions (b) harmonic functions (c) conjugate harmonic functions (d) error functions Solution (b): The solution of Laplace’s equation having continuous 2nd order partial derivatives is called a harmonic function. 9. A function f of the complex variable z x iy , is given as f ( x, y ) u ( x, y ) iv ( x, y ) , where 2
2
u ( x, y ) 2kxy and v ( x, y ) x y . The value of k , for which the function is analytic, is _____. [ME-2016 (1 mark)]
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Engineering Mathematics
Chapter – 4: Complex Variables
Chapter – 4 [3]
Solution: If f ( z ) u ( x, y ) iv ( x , y ) is an analytic function of complex variable z x iy , where i 1 , then it must satisfies Cauchy-Riemann equations u x v y …(i) and u y v x …(ii). So (i) 2ky 2 y k 1 ; also (ii) 2kx (2 x ) k 1 . So answer is ‘ 1 ’.
10. For a complex number Z (1 2) ( 3 2)i , the value of Z 6 is
[PE-2016 (1
mark)]
(a) (1 2) ( 3 2)i
(c)
(b) 1
(1 2) (
3 2)i
(d) 1 i
Solution (d): Converting the given complex number, z x iy , in polar form we get z re , where r
x2 y 2
and
tan 1 ( y x) .
So
we
x 1 2
have
r (1 2) 2 ( 3 2) 2 1 and tan 1 ( 3) 3 . Thus z (1)e z cos(2 ) i sin(2 ) 1 i (0) 1 . 2
i ( 3)
and
y 3 2,
z 6 16 e i (
2
11. The function f ( z ) ( z 1) ( z 4) is singular at (a) z 2 (b) z 1 (c) z i
3)6
thus
ei (2 )
[PI-2016 (1 mark)] (d) z 2i
Solution (d): As any function f ( z ) is said to be singular if it is not defined at a point z z0 . Thus 2
the given function has singularities at the points were z 4 0 z 2i . 12. In the following integral, the contour C encloses the points 2 j and 2 j . The value of the 1 sin z dz is _____. integral [EC-2016 (2 marks)] C 2 ( z 2 j )3 sin z Solution: For the given function f ( z ) , the denominator is zero at z 2 j , so z 2 j ( z 2 j )3 1 sin z 1 dz 2 j (sum of residues of the poles which is a pole of order 3. So I 3 2 C ( z 2 j ) 2 lie inside the given closed region). 1 d 31 3 Thus I j Res f ( z ) j lim 31 ( z 2 j ) f ( z ) z 2 j (3 1)! z 2 j dz I j
I j
1 2!
1
d2 sin z 1 d2 3 ( z 2 j ) j lim sin z 2 3 2 ( z 2 j ) 2 z 2 j dz dz
lim z 2 j
lim sin z j
2 z 2 j
13. The values of the integral
1
1 2
sin(2 j ) e
jj
1 2
1
sinh(2 ) sinh(2 ) 133.87 2
z
C z 2 dz 2 j
along a closed contour C in anti-clockwise direction
for (i) the point z0 2 inside the contour C , and (ii) the point z0 2 outside the contour C , respectively, are [EC-2016 (2 marks)] (a) (i) 2.72, (ii) 0 (b) (i) 7.39, (ii) 0 (c) (i) 0, (ii) 2.73 (d) (i) 0, (ii) 7.39 ez Solution (b): As the given function f ( z ) has the pole where z 2 0 z 2 is a pole, z2 which is of order 1 and lies inside the closed contour C ; so (i) For the point z0 2 inside the contour C
1 2 j
C f ( z)dz (sum of residues of the poles which lies inside the given closed region).
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Engineering Mathematics
1
C 2 j
Chapter – 4: Complex Variables
ez
f ( z ) dz Res f ( z ) lim( z 2)
z2
z2
z 2
Chapter – 4 [4]
lim e z e 2 7.389 . z 2
(ii) For the point z0 2 outside the contour C , so Res f ( z ) 0 and z2
residues of the poles which lies inside the given closed region)
1 2 j
1 2 j
C f ( z)dz (sum of
C f ( z)dz 0 .
z2 1
1
dz , where z is a complex number and C is a unit circle 2 j C z 2 1 with centre at 1 0 j in the complex plane is _____. [IN-2016 (2 marks)]
14. The value of the integral
f ( z ) ( z 2 1) ( z 2 1) , pole of the given function are the points where
Solution: For
z 2 1 z 1,1 , which are all simple poles. As C is a unit circle with centre at 1 0 j in the
complex plane, so for z 1 , 1 (1 0 j ) 2 2 1 z 1 lies outside the contour C . for z 1 , 1 (1 0 j ) 0 1 z 1 lies inside the contour C . Hence
z2 1
1
2 j C z 2 1
1
z2 1
2 j C z 2 1
dz (sum of residues of the poles which lies inside the contour C ).
dz Res f ( z ) lim{z 1} f ( z ) lim{z 1} z 1
z 1
15. The value of the integral theorem is (a) sin(1) e
z 1
sin x
x2 2 x 2 dx
2
z 1
lim z 1
z2 1 z 1
2 2
1
evaluated using contour integration and the residue
(b) cos(1) e
sin z
z2 1
[ME-2016 (2 marks)] (d) cos(1) e
(c) sin(1) e
Im(eiz )
iz iz , since e cos z i sin z Im(e ) sin z , whose z2 2z 2 z2 2z 2 2 poles are values of ‘ z ’ such that z 2 z 2 0 z 1 i , both of which are simple poles; but only z 1 i lie in upper half of Z-plane.
Solution (a): Let f ( z )
f ( z ) dz (2 i) (sum of residues of the poles which lies inside the given closed region). As
Res f ( z ) lim {z ( 1 i)} z 1 i
z 1 i
1
Res f ( z )
i
e Im(e )
z 1 i
2i
f ( z ) dz (2 i)
16. The value of
Im(eiz ) z2 2z 2
lim
z 1 i {z
Im(eiz ) ( 1 i )}
1
e Im{cos(1) i sin(1)}
sin(1) 2ie
3z 5
( z 1)( z 2) dz
2i sin(1) e
Solution (b):
Im(ei ( 1i ) ) {1 i ( 1 i)}
Im(e i 1 ) 2i
sin(1) 2ie
. Thus option (a) is correct.
along a closed path to equal to (4 i ) , where z x iy and
i 1 . The correct path is
(a)
(b)
[ME-2016 (2 marks)]
(c)
(d)
f ( z )dz (2 i ) (sum of residues of the poles which lies inside given closed region)
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Engineering Mathematics
As
Chapter – 4: Complex Variables
3z 5
( z 1)( z 2) dz 4 i 2 i (2)
Chapter – 4 [5]
sum of residues of the poles which lies inside given
closed region is ‘2’. For function f ( z )
3z 5 ( z 1)( z 2)
, the poles are z 1 and z 2 , which are simple poles.
So Res f ( z ) lim( z 1) f ( z ) lim( z 1)
3z 5
3z 5
{z : | z | 2}
17. Let
be
oriented
in
the
counter-clockwise
direction.
Let
1 dz . Then, the value of I is equal to _____. [MA-2016 (2 marks)] 2 z 2 4 6 1 7 1 1 1 1 1 1 7 Solution: As f ( z ) z cos 2 z 1 2 2 2 z 2! z 4! z 6! z 3 1 1 1 1 1 1 7 z 1 1 1 1 7 f ( z) z 1 z , which is in the form of 4 8 12 5 2! z 4! z 6! z 2! 4! z 6! z I
1
z 1
3(1) 5
2 ; and ( z 1)( z 2) z 1 z 2 (1) 2 3z 5 3z 5 3(2) 5 Res f ( z ) lim( z 2) f ( z ) lim( z 2) lim 1; z 2 z 2 z 1 z 2 ( z 1)( z 2) z 1 (2) 1 As residue of the poles which lies inside given closed region must be ‘2’, so we can conclude that z 1 lie inside the contour and z 2 lie outside the contour . z 1
z 1
lim
z 2 i
7
cos
Laurent series. Now f ( z ) has a singularity at a point z 0 inside | z | 2 . So using residue 1 integration method, we have f ( z )dz coefficient of the first negative power, i.e. 1 ( z 0) , 2 i 1 1 7 which is 1 (4!) 0.0416 . Hence I z cos 2 dz 0.0416 . 2 i z
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Engineering Mathematics
Chapter 4: Complex Variables
[4.1]
Chapter 4 : Complex Variables 4.1
Basic Concept of Complex Numbers
A number of the form x iy , where x , y R and i 1 is denoted by ‘ i ’ called iota thus i
1 is called a complex number so the quantity
1 . A complex number is denoted by z and the set of
complex number is denoted by i.e. {x iy : x, y R , i 1} . For e.g. 5 3i , 0 4i , 4 0i , etc. are complex numbers. Iota (i ) is neither 0, nor greater than 0, nor less than 0. The square root of a negative real number is called an imaginary unit.
For any positive real number a , we have
a 1 a 1 a i a ; also i a a
2 3 4 Integral Power of iota ( ): As i 1 i 1 , i i and i 1 . To find the value of i n (n 4), first divide n by 4; and let q be the quotient and r be the remainder. i.e., n 4q r where 0 r 3 ; so we have i n i 4 q r (i 4 ) q (i ) r (1) q (i ) r i r .
In general, i
4n
1, i 4n1 i, i 4 n2 1, i 4n3 i , where n is any integer.
The value of the negative integral powers of i are found as: i 1 1 i i 3 i 4 i 3 i , i 2 1 i 2 1 ( 1) 1 , i 3 1 i 3 i i 4 i 1 i , i 4 1 i 4 1 1 1 .
Real and Imaginary part of a Complex Number: If x and y are two real numbers, then a number of the form z x iy is called a complex number. Here ‘ x ’ is called the real part of z [denoted by Re( z ) ] and ‘ y ’ is known as the imaginary part of z [denoted by Im( z ) ].
A complex number z is purely real if its imaginary part is zero i.e., Im( z ) 0 and purely imaginary if its real part is zero i.e., Re( z ) 0 . A real number can be written as a i.0 , therefore every real number can be considered as a complex number whose imaginary part is zero. Complex number as an ordered pair: A complex number ‘ a ib ’ may also be defined as an ordered pair of real numbers and denoted by ( a, b) . For e.g. i can be denoted by (0,1) . The algebraic operations performed on two complex numbers z1 x1 iy1 and z2 x2 iy2 are: Addition/Subtraction: ( x1 iy1 ) ( x2 iy2 ) ( x1 x2 ) i( y1 y2 ) Multiplication: ( x1 iy1 )( x2 iy2 ) ( x1 x2 y1 y2 ) i ( x1 y2 y1 x2 ) Division:
x1 iy1 x2 iy2
x1 iy1 x2 iy2 ( x1 x2 y1 y2 ) i ( y1 x2 y2 x1 ) , where at least one of 2 2 2 2 x2 iy2 x2 iy2 x2 y2 x2 y2
x2 and y2 is non-zero.
Equality of two Complex Numbers: Two complex numbers z1 x1 iy1 and z2 x2 iy2 are said to be equal if and only if their real parts and imaginary parts are separately equal. i.e. z1 z2 x1 iy1 x2 iy2 x1 x2 and y1 y2 . Thus, one complex equation is equivalent to two real equations. A complex number z x iy 0 iff x 0, y 0 . The complex number do not possess the property of order i.e., ( a ib) (or) (c id ) is not defined. For example, the statement 9 6i 3 2i makes no sense. Reciprocal of a complex number: For an existing non-zero complex number z x iy , the 1 reciprocal is given by z
1 z
z 2
|z|
1 , i.e. z
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1 x iy
x iy 2
x y
2
Re( z ) 2
|z|
i[ Im( z )] |z|
2
z |z|
2
.
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Engineering Mathematics
Chapter 4: Complex Variables
[4.2]
Example 4.1 [EC-2008 (1 mark)]: The equation sin( z ) 10 has (a) no real or complex solution (b) exactly two distinct complex solutions (c) a unique solution (d) an infinite number of complex solutions Solution (d): Let z x iy then sin z 10 sin( x iy ) 10 i 0 sin x cosh y i cos x sinh y 10 i0 sin x cosh y 10 and cos x sinh y 0 . Now cos x sinh y is zero when
cos x 0 x (2 n 1)( 2)
and at
x (2 n 1)( 2) ,
1
we have
sin x 1
1
cosh y 10 y cosh (10) . So z (2n 1)( 2) i cosh (10) . So the equation sin( z ) 10 has an infinite number of complex solutions.
Example 4.2 [EE-2011 (1 mark)]: Roots of the algebraic equation x 3 x 2 x 1 0 are (a) ( 1, i) (b) ( 1, 1, 1) (c) (0, 0, 0) (d) ( 1, j ) Solution (d): x 3 x 2 x 1 0 x 2 ( x 1) 1( x 1) 0 ( x 2 1)( x 1) 0 x 1, i . Example 4.3 [ME-2011 (1 mark)]: The product of two complex numbers 1 i and 2 5i is (a) 7 3i (b) 3 4i (c) 3 4i (d) 7 3i 2 Solution (a): (1 i )(2 5i ) 2 5i 2i 5i 2 3i 5 7 3i . [Similar question was also asked in PI-2009 (1 mark)] Example 4.4 [CE-2014 (1 mark)]: z (2 3i ) ( 5 i ) can be expressed as (a) 0.5 0.5i (b) 0.5 0.5i (c) 0.5 0.5i (d) 0.5 0.5i 2 3i
2 3i
5 i
10 2i 15i 3i 2
13 13i
1 1 i 5 i 5 i 5 i 25 i 26 2 2 [Similar questions were also asked in PI-2008 (1 mark), CH-2007 (1 mark)] Solution (b):
2
4.1.1 Complex Plane The geometrical representation of complex number is represented as points in the plane. We choose two perpendicular coordinate axes, the horizontal x axis, called the real axis, and the vertical y axis, called the imaginary axis. On both axes we choose the same unit of length. This is called Cartesian coordinate system. We now plot a given complex number z ( x, y ) x iy as the point P with coordinates Figure 4.1: Complex Plane ( x, y ) (as shown in Figure 4.1). The xy plane in which the complex numbers are represented in this way is called complex plane. The complex conjugate z of a complex number z x iy is defined by z x iy . It is obtained geometrically by reflecting the point z in the real axis. z z 2 x Re( z ) x ( z z ) 2 purely real
z z 2iy Im( z ) y ( z z ) (2i ) purely imaginary
( z1 z2 ) z1 z2 ; ( z1 z2 ) z1 z2 ; ( z1 z2 ) z1 z2 Example 4.5 [EC-2005 (2 marks)]: In what range should Re( s) remain so that the Laplace transform of the function e( a 2) t 5 exists? (a) Re( s ) a 2 (b) Re( s ) a 7 Solution (a): We have f (t ) e
( a 2) t 5
(c) Re( s) 2 5 ( a 2) t
e e
(d) Re( s ) a 5 5 ( a 2) t
{ f (t )} {e e
} F ( s) e5 {e ( a 2) t }
F ( s ) e5 [1 {s ( a 2)}] . Now as Laplace transform of any function is always positive so s ( a 2) 0 s a 2 ; as s is the Re( s) so we have Re( s ) a 2 .
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Engineering Mathematics
Chapter 4: Complex Variables
[4.3]
Example 4.6 [IN-2008 (2 marks)]: A complex variable Z x j 0.1 has its real part x varying in the range to . Which one of the following is the locus (shown in thick lines) of 1 Z in the complex plane?
(a)
(b)
Solution (b): Y
1 Z
Z ZZ
(c)
Z Z
2
x j 0.1 2
2
x 0.1
Re(Y )
(d)
x 2
2
x 0.1
and Im(Y )
0.1 2
x 0.12
. As
Im(Y ) Re(Y ) 0.1 x Re(Y ) 10 x Im(Y ) which means the slope of Y keep changing as x and so option (a) and (d) are not correct. As 1 {Re(Y )}2 {Im(Y )}2 2 10 Im(Y ) {Re(Y )}2 {Im(Y )}2 10 Im(Y ) 0 x 0.12 {Re(Y )}2 {Im(Y ) 5}2 52 . So locus of Y is a circle whose centre is at (0, 5) and radius 5, these two conditions are satisfied by the figure given in option (b). So option (b) is correct. Example 4.7 [EE-2011 (1 mark)]: A point Z has been plotted in the complex plane, as shown in figure. The plot of the 1 complex number Y Z is
(a)
(b)
(c)
(d)
Solution (d): Let Z x iy and in the given figure we have x, y 0 ; as Z lies inside the unit circle 2
2
so Z 1 Y 1 Z 1 . Also Y 1 Z Z Z ( x iy ) Z Y has positive real part and negative imaginary parts. So we have positive real part and negative imaginary parts of Y with point Y outside the unit circle as Y 1 , which is represented by the figure given in option (d). Example 4.8 [CH-2014 (1 mark)]: If f * ( x ) is the complex conjugate of f ( x) cos x i sin x , then for real a and b ,
b
a
f * ( x ) f ( x ) dx is always
(a) Positive (b) Negative (c) real (d) imaginary ix Solution (c): The complex conjugate of f ( x ) cos x i sin x e is f * ( x ) cos x i sin x e ix , so b
b
a
a
f * ( x ) f ( x ) e ix eix 1 thus I f * ( x ) f ( x )dx 1dx I b a which can be positive or
negative but it is definitely real as a and b are real. Example 4.9 [EE-2014 (1 mark)]: Let S be the set of points in the complex plane corresponding to the unit circle. (That is S z : z 1 ). Consider the function f ( z ) zz * where z * denotes the complex conjugate of z . The f ( z ) maps S to which one of the following in the complex plane (a) unit circle (b) horizontal axis line segment from origin to (1, 0) (d) the entire horizontal axis (c) the point (1, 0) 2
Solution (c): As z* z , the conjugate of z ; thus f ( z ) z z z 1 1 i (0) f ( z ) is a point (1, 0) in the complex plane.
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Engineering Mathematics
Chapter 4: Complex Variables
[4.4]
4.1.2 Polar Form of a Complex Number The complex plane becomes even more useful and gives further insight into the arithmetic operations for complex numbers if besides the xy coordinates we also imply the usual polar coordinates r , defined by ‘ x r cos , y r sin .’ z x iy z x r (cos i sin ) (4.1) r is called the absolute value or modulus of z and is denoted by z , hence r z x 2 y 2 zz . Geometrically z represents the distance of point z from the origin (as shown in Figure 4.2a); similarly z1 z2 is the distance between z1 and z 2 (as
shown in Figure 4.2b). If z 0 , then z is known as zero modular complex number and is Figure 4.2: : (a) Polar form of complex number (b) Distance used to represent the origin of reference plane. between two points in the complex plane is called the argument of z and is denoted by arg z ,
thus arg z tan 1 ( y x ) . Geometrically, is the directed angle from the positive x axis to OP as shown in Figure 4.2a. For z 0 this angle is undefined. Argument of a complex number is not unique, since if be a value of the argument, so also is 2n , where n I . Principal value of arg( z ) : The value of the argument, which satisfies the inequality is called the principal value of argument. Principal values of argument z will be , , and according as the point z lies in the 1st , 2nd , 3rd and 4th quadrants (as shown in Figure 4.3) respectively, 1 b (acute angle). where tan a
Figure 4.3: Arguement of Complex Number
Figure 4.4: Principle value of arg(z)
Triangle inequality: Inequalities such as x1 x2 make sense for real numbers, but not in complex because there is no natural way of ordering complex numbers. However, inequalities between absolute values (which are real) such as z1 z2 (meaning z1 is closer to the origin than z 2 ) are of great importance. In a triangle the three points 0 , z1 and z1 z2 are the vertices of triangle with sides z1 , z 2 and z1 z2 ; and as shown in Fig. 4.5, one side cannot exceed the sum of the other two Figure 4.5: Triangle Inequality sides, i.e., z1 z2 z1 z2 (4.2) The generalized triangle inequality can be given as, | z1 z2 zn | | z1 | | z2 | | zn | (4.3) i.e., the absolute value of a sum cannot exceed the sum of the absolute values of the terms.
Example 4.10 [CE-2005 (1 mark)]: Which of the following is not true for complex number Z1 , Z 2 ? (a) Z1 Z 2 ( Z1 Z2 ) Z 2
2
(c) Z1 Z 2 Z1 Z 2
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(b) Z1 Z 2 Z1 Z 2 2
2
2
(d) Z1 Z 2 Z1 Z 2 2 Z1 2 Z 2
2
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Engineering Mathematics
Chapter 4: Complex Variables
[4.5]
2
2
Solution (b): For option (a): As Z Z Z so RHS ( Z1 Z 2 ) Z 2 ( Z1 Z 2 ) ( Z 2 Z 2 ) Z1 Z 2 LHS . For option (b) and (c), refer to triangle inequality given by Eq. 4.2, so we can say that option (b) is not 2
correct. For option (d): Let S Z1 Z 2 Z1 Z 2
2
so S ( Z1 Z 2 )( Z1 Z 2 ) ( Z1 Z 2 )( Z1 Z 2 )
S ( Z1 Z 2 )( Z1 Z 2 ) ( Z1 Z 2 )( Z1 Z 2 ) S Z1 Z1 Z1 Z 2 Z 2 Z1 Z 2 Z 2 Z1 Z1 Z1 Z 2 Z 2 Z1 Z 2 Z 2 2
2
2
S Z1 Z 1 Z1 Z 2
2
2
2
2 Z1 2 Z 2 .
Example 4.11 [ME-2014 (1 mark)]: The argument of the complex number (1 i ) (1 i ) , where
i 1 is (a) Solution: As z
1 i 1 i
(d)
(c) 2
(b) 2
1 i 1 i 1 2i i 2 2i 1 i 0 1i . So argument of z is tan 1 2 1 i 1 i 1 i 2 0 2
Multiplication and Division in Polar Form: Let
z1 x1 iy1 r1 (cos 1 i sin 1 )
and
z2 x2 iy2 r2 (cos 2 i sin 2 ) , then
Multiplication: z1 z2 r1r2 (cos 1 cos 2 sin 1 sin 2 ) i (sin 1 cos 2 cos 1 sin 2
also,
z1 z2 r1r2 cos(1 2 ) i sin(1 2 )
(4.4)
z1 z2 r1r2 r1 r2 z1 z2
(4.5)
arg( z1 z2 ) arg z1 arg z2 (upto multiples of 2 )
(4.6)
Division: z1 ( z1 z2 ) z2 z1 ( z1 z2 ) z2 . Hence by product rule of two complex numbers, z1 ( z1 z2 ) z2 z1 z2 z1
(4.7)
z2
Also, by product rule, we have arg( z1 ) arg ( z1 z2 ) z2 arg( z1 z2 ) arg( z2 ) arg( z1 z2 ) arg( z1 ) arg( z2 ) (upto multiples of 2 )
(4.8)
z1 z2 (r1 r2 ) cos(1 2 ) i sin(1 2 )
(4.9)
Example 4.12 [ME-2010 (1 mark)]: The modulus of the complex number (3 4i) (1 2i) is (a) 5
(b)
Solution (b): As
z1
z1
(c) 1
5
3 4i
3 4i
32 4 2
12 ( 2) 2 [Similar question was also asked in CH-2009 (1 mark)] z2
Integral power of
z2
1 2i
1 2i
(d) 1 5
5
5
5
5
: z r (cos i sin ) z n r n (cos i sin ) n . Since from De Moivre’s
formula, we have (cos i sin ) n (cos n i sin n ) (found by mathematical induction). Hence, z n r n (cos n i sin n )
(4.10) 4
Example 4.13 [PI-2007 (1 mark)]: If a complex variable z ( 3 2) i (1 2) , then z is (b) (1 2) i ( 3 2)
(a) 2 3 i 2
(c) ( 3 2) i (1 2)
(d) ( 3 8) i(1 8)
Solution (b): As z x iy ( 3 2) i (1 2) x ( 3 2) r cos and y 1 2 r sin , so dividing these
two
we
get
tan 1
3 6;
also
r z ( 3 2) 2 (1 2) 2 1 .
So
z 1 cos( 6) i sin( 6) ei 6 z 4 ei 4 6 ei 2 3 cos(2 3) i sin(2 3) (1 2) i( 3 2)
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Engineering Mathematics
Chapter 4: Complex Variables
[4.6]
roots of a complex number: If z wn ( n 1, 2,) , then to each value of w there corresponds one value of z ; or for a given z 0 there corresponds precisely n distinct values of w . Each of these values is called nth roots of z and we write,
w
n
z
(4.11)
n
The n values of z can be obtained as follows. We write z and w in polar form as, z r (cos i sin ) and w R (cos i sin ) wn z R n (cos n i sin n ) r (cos i sin ) The absolute values on both sides must be equal; thus, R n r R n r , where n r is positive real and thus uniquely determined. Equating the arguments n and and recalling that is determined only up to integral multiples of 2 , we obtain, n 2 k ( n ) (2 k n ) . For k 0,1, , n 1 , we get n distinct values of w ; further integers of k would gives values already obtained. For example, the value of w for k n 2 k n 2 corresponds for k 0 , etc. Hence, the n distinct values of z 0 , can be given as, n
z
n
r cos{( 2 k ) n} i sin{( 2 k ) n}
where, k 0,1, , n 1 . These n values lie on a circle of radius the vertices of a regular polygon of n sides. The value of arg z and k 0 in 4.12 is called the principal value of w
n
n
n
z , for (4.12)
r with center at origin and constitute
z obtained by taking the principal value of
n
z.
Taking z 1 in 4.12, we have z r 1 and arg z 0 , hence n
1 cos{(2 k ) n} i sin{(2 k ) n} ,
k 0,1, , n 1
(4.13)
th
The n values given by 4.19 are called n roots of unity. If denotes the value corresponding to
k 1 then n values of n 1 can be written as, 1, , 2 , , n1 . In general, if w1 is any root of an arbitrary complex number z ( 0) , then the n values of n
z in 4.18 are, w1 , w1 , w1 2 , , w1 n 1 because multiplying w1 by k corresponds to
increasing the argument of w1 by 2k n .
For cube root of unity, we have z 1 and n 3 , then three roots of
3
1 are found by putting
2
k 0,1, 2 in 4.13 which are given as, 1, {( 1 2) i ( 3 2)} and {( 1 2) i ( 3 2)} . We
have 1 2 0 and 3 1 . Example 4.14 [CS-1995 (2 marks)]: If the cube roots of unity are 1, and 2 , then the roots of the following equation are ( x 1) 3 8 0 (a) 1,1 2 ,1 2
2
(b) 1,1 2 ,1 2
2
(c) 1,1 2 ,1 2
2
(d) 1,1 2, 1 2
2
2
Solution (c): As x 1 ( 8)1 3 ( 2)3 3 (1)1 3 2(1)1 3 2, 2 , 2 2 x 1,1 2 ,1 2 . Example 4.15 [IN-2009 (2 marks)]: One of the root of the equation x 3 i , where i is positive square root of 1 , is (a) i (b) ( 3 2) i(1 2) (c) ( 3 2) i (1 2) (d) ( 3 2) i(1 2) Solution (b): x 3 i i (i 2 ) i 3 x 3 (i )3 (1)3 x i (13 )1 3 x i (1)1 3 . As cube root of unity
are
1
x i , i
2
1, , 2 , i
where
{( 1 2) i( 3 2)}
and
2 {( 1 2) i ( 3 2)} .
So
3
1 i 3 3 i 3 , i i i , , . Thus from the given options we 2 2 2 2 2 2 2
can say that option (b) is correct.
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e-mail: [email protected]
Engineering Mathematics
Chapter 4: Complex Variables
[4.7]
Example 4.16 [PI-2010 (1 mark)]: If a complex number satisfies 3 1 , then the value of 1 (1 ) is (a) 0 (b) 1 (c) 2 (d) 4 Solution (a): If x 3 1 then its roots are 1, {( 1 2) i( 3 2)} and 2 {( 1 2) i ( 3 2)} which satisfies 1 2 0 and 3 1 . So 1 (1 ) ( 2 1) 0 . Example 4.17 [EE-2011 (2 marks)]: The two vectors
1
1 1 and 1 a
a 2 , where
a (1 2) i( 3 2) are
(a) Orthonormal (b) Orthogonal (c) Parallel (d) Collinear Solution (b): As cube roots of unity are 1, {( 1 2) i( 3 2)} and 2 {( 1 2) i ( 3 2)} which satisfies 1 2 0 and 3 1 . Now as the dot product of the given vectors is 2 T
1
2
1 1 1 a a 1 a a 0 , as we have a , so the given vectors are orthogonal, i.e., perpendicular to each other and thus they are neither parallel nor collinear. Also both the vectors are not of unit length so they are not orthonormal.
4.1.3 Complex Function Since, a real function f defined on a set S of real numbers (usually an interval) is a rule that assigns to every x in S a real number f ( x ) , called the value of f at x . In complex, S is a set of complex numbers; and a function f is defined on S is a rule that assigns to every z in S a complex number w , called the value of f at z . We write w f ( z ) . Here z varies in S and is called complex variable. The set S is called the domain of f . The set of all values of a function f is called the range of f . If w is a complex w u iv , where u and v are the real and imaginary parts, respectively. Since, w depends on z x iy . So, u becomes a real function of x and y , and so does v w f ( z ) u ( x, y ) iv( x, y ) . This shows that a complex function f ( z ) is equivalent to a pair of real functions u ( x, y ) and v ( x, y ) , each depending on the two real variables x and y .
Limit of a Complex function: A function f ( z ) is said to have the limit l as z approaches a point z0 , written as lim f ( z ) l if f is defined in a neighbourhood of z0 (except at z0 itself) and if z z0
the values of f are tending to l for all z tending to z0 . In other words, if ve real we can find a ve real s.t. z z0 in the disk z z0 , then f ( z ) l . If a limit exists, it is unique.
Continuity of a Complex Function: A function f ( z ) is said to be continuous at z z0 if f ( z0 ) is defined as, lim f ( z ) f ( z0 ) . Note that by definition of a limit this implies that f ( z ) is defined in z z0
some neighbourhood of z0 . f ( z ) is said to be continuous in a domain if it is continuous at each point of this domain. Derivative of a Complex Function: For differentiability of a complex-valued function, the situation is different from the differentiability of a real-valued function. Let f ( z ) u ( x, y ) iv ( x , y ) be a complex-valued function of variable z x iy ; then at z0 x0 iy0 , the derivative of f ( z ) at the point z0 is denoted by f ( z0 ) and is defined as f ( z0 ) lim
f ( z 0 z ) f ( z 0 )
provided this z limit exists. The first thing that we notice is that z , being a complex number, can approach zero in more than one way. If we write z x iy , then we can approach zero along the real axis y 0 or along the imaginary axis x 0 , or indeed along any direction. For the derivative to exist, the z 0
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Engineering Mathematics
Chapter 4: Complex Variables
[4.8]
answer should not depend on how z tend to zero. Now let us see what this means: We have f ( z0 ) u ( x0 , y0 ) iv ( x0 , y0 ) and f ( z0 z ) u ( x0 x, y0 y ) iv ( x0 x, y0 y ) , so f ( z0 )
lim
u ( x0 , y0 ) iv ( x0 , y0 )
u ( x0 , y0 ) u ( x0 x, y0 y ) u ( x0 , y0 ) ,
where
x iy
x 0, y 0
v( x0 , y0 ) v ( x0 x, y0 y ) v( x0 , y0 ) . Let us first take the limit z 0 by first taking y 0 and then x 0 , in other words we let z 0 along the real axis, then we have u ( x0 , y0 ) iv ( x0 , y0 ) u ( x0 , y0 ) iv( x0 , y0 ) u v f ( z0 ) lim lim lim i x 0 y 0 x 0 x iy x x ( x0 , y0 ) x ( x0 , y0 )
Now let us take the limit z 0 by first taking x 0 and then y 0 , in other words we let z 0 along the imaginary axis, then we have u ( x0 , y0 ) iv ( x0 , y0 ) u ( x0 , y0 ) iv ( x0 , y0 ) f ( z0 ) lim lim lim y 0 x 0 x 0 x iy iy
f ( z0 ) lim
iu ( x0 , y0 ) v ( x0 , y0 ) y
x 0
i
u y
( x0 , y0 )
v y
. ( x0 , y0 )
So if f ( z ) exists at z0 x0 iy0 , then the following equations are satisfied at ( x0 , y0 ) : u x v y and u y vx i.e.,
u
v
and
u
v
(4.14) x y y x The equations given by Eq. 4.14 are called the Cauchy-Riemann equation [Eq. 4.14 was asked in ME-2007 (1 mark)]. So, the function f ( z ) is differentiable at z0 if f ( z0 ) is well defined at z0 . Thus for a differentiable function f ( z ) we have seen that f ( z )
u x
i
v x
v y
i
u y
. So we have
just shown that a necessary condition for f ( z ) to be differentiable at z0 is that its real and imaginary parts obey the Cauchy-Riemann equations at ( x0 , y0 ) . Conversely, it can be shown that this condition is also sufficient provided that the partial derivatives of u ( x, y ) and v ( x, y ) are continuous.
f is said to be differentiable at z0 , if we write z z z0 , we have z z0 z and takes the form f ( z0 ) lim
f ( z ) f ( z0 )
. If f ( z ) is differentiable at z0 , it is continuous at z0 . z z0 The differentiation rules are the same as in real calculus as their proofs are literally the same. Thus for any analytic functions f and g and constant c we have, z 0
(cf ) cf , ( f g ) f g , ( fg ) f g f g , ( f g ) ( f g f g ) g 2 .
Analytic Function: Analyticity of f ( z ) at z0 means that f ( z ) has a derivative at every point in some neighbourhood of z0 (including z0 itself); a function f ( z ) is said to be analytic in a domain D if f ( z ) is said to be analytic in a domain D if f ( z ) is defined and differentiable at all points of D . So the Cauchy – Riemann equations given by Eq. 4.14 provide a criteria for the analyticity of a complex function. We also say that a function is entire if it is analytic in the whole complex plane. For e.g. the function f ( z ) z is entire. We can check this either by verifying the Cauchy-Riemann f ( z 0 z ) f ( z 0 ) z z z 0 z equations or simply noticing that f ( z0 ) lim lim 0 lim 1, z 0 z 0 z 0 z z z hence it is well defined for all z0 . On the other hand, the function f ( z ) z , where z is the conjugate of z , is not differentiable anywhere as f ( z0 ) lim
z 0
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f ( z0 z ) f ( z0 ) z
lim
z 0
z0 z z 0 z
lim
z 0
z z
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Engineering Mathematics
Chapter 4: Complex Variables
[4.9]
; now if we let z tend to zero along real values, we would find that f ( z0 ) 1 , whereas if we would let z tend to zero along imaginary values we would find that f ( z0 ) 1 . We could have reached the same conclusion via the Cauchy-Riemann equations with u ( x, y ) x and v ( x, y ) y which violates the Cauchy-Riemann equation. The analyticity, unlike differentiability, is not a property of a function at a point, but on an open set of points. The reason for this is to eliminate from the class of interesting functions, functions which may be differentiable at a point but nowhere else. Whereas this is a rarity in calculus, it is a very common occurrence for complex-valued functions of a complex variables. For example, 2 consider the function f ( z ) z , this function has u ( x, y ) x 2 y 2 and v ( x, y ) 0 . Therefore the Cauchy-Riemann equations are only satisfied at the origin in the complex plane because: u x 2 x v y 0 x 0 and vx 0 u y 2 y y 0 . Example 4.18 [CH-2007 (2 marks)]: If z x iy is a complex number, where i 1 then the derivative of zz at 2 i is (a) 0 (b) 2 (c) 4 (d) does not exist 2 2 Solution (d): As f ( z ) zz z and we know that f ( z ) z is differentiable only at the origin. As the given point 2 i or (2,1) is not at the origin so it is not differentiable. Example 4.19 [CH-2008 (2 marks)]: An analytic function w( z ) is defined as w u iv , where
i 1 and z x iy . If the real part is given by u y ( x 2 y 2 ) , w( z ) is (d) 1 (iz ) (a) 1 z (c) i z (b) 1 z 2 Solution (c): If w( x iy ) u ( x, y ) iv ( x, y ) is analytic then it satisfies the Cauchy-Riemann equations, i.e., (i) u x v y and (ii) u y vx . 2 xy x y v 2 xy v 2 y v 2 f ( x) 2 2 2 2 2 2 2 2 (x y ) x y y x x y y ( x y ) …(iii). Now differentiating (iii) both sides w.r.t. x and applying the (ii) condition we get v u x y 2 f ( x) 2 2 2 x y x x y y x y 1( x 2 y 2 ) x(2 x ) 1( x 2 y 2 ) y (2 y ) x2 y 2 x2 y2 f ( x) 2 f ( x) 2 2 2 2 2 2 2 x y x y x y x y v
So (i)
f ( x ) 0 f ( x ) k , where k is any constant; let us choose k 0 so v x ( x 2 y 2 ) . So
w
y 2
x y
Example
2
( y ix ) 1 i i i ( y ix ) 2 . 2 2 ( y ix )( y ix ) y ix iy i x iy x z x y x y
i
4.20
x
2
2
[XE-2008
(1
mark)]:
If
f ( z ) u iv
is
an
analytic
function
and
u v ( x y ) 3 kxy ( x y ) , then k is (a) 2 (b) –4 (c) 6 (d) –8 Solution (c): f ( z ) u iv is an analytic function, so, it satisfies Cauchy-Riemann equations, i.e., (i) u x v y and (ii) u y vx . As u v ( x y ) 3 kxy ( x y ) …(iii); now partial differentiation of (iii)
w.r.t.
x,
we
get
u x vx 3( x y ) 2 kxy ky ( x y ) 3( x y ) 2 2kxy ky 2
v y v x 3( x y ) 2 2kxy ky 2 …(iv) [by applying (i)]; also partial differentiation of (iii) w.r.t. y , we
get
u y v y 3( x y ) 2 kx ( x y ) kxy ( 1) 3( x y ) 2 2kxy kx 2
vx v y 3( x y ) 2 2kxy kx 2 …(v). Now (iv) + (v) 2v x ky 2 kx 2 …(vi); also (iv) –
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Engineering Mathematics
(v)
Chapter 4: Complex Variables
2v y 6( x y ) 2 4kxy ky 2 kx 2 2
3
2v ky x k ( x 3) f ( y ) 3
2
also
integrating
2
2v 2( x y ) 2kxy k ( y 3) kx y h( x ) 2
3
x
…(vii). Now integrating (vi) w.r.t.
…(viii);
3
[4.10]
3
…(ix).
2
3
(vii)
w.r.t.
Thus
we get we
y
(viii)
+
get (ix)
2
ky x k ( x 3) f ( y ) 2( x y ) 2kxy k ( y 3) kx y h( x ) 0 xy 2 ( k 6) x 2 y (6 k ) (k 3) 2 x 3 2 ( k 3) y 3 f ( y ) h ( x ) 0 . As f ( y ) and h( x) both
are pure function of y and x , respectively so they do not contain the mixed terms of x and y . Thus the coefficient of xy 2 and x 2 y must be zero and so we have k 6 0 and 6 k 0 k 6 . Example 4.21 [XE-2008 (2 marks)]: If f ( z ) y (1 x 2 ) x 2 i( y 2 2 y ) x is differentiable at a point z z0 , then f ( z0 ) is (a) 0 (b) 1 (c) i (d) i 2 2 2 Solution (d): Let u y (1 x ) x , v ( y 2 y ) x and z0 x0 iy0 , so f ( z ) u ( x, y ) iv ( x , y ) . As f ( z ) is differentiable at z z0 so f ( z ) is analytic at z z0 and thus satisfy Cauchy-Riemann equation
4.14,
(u y ) ( x
0 , y0 )
so
from
(u x) ( x
i.e.,
0 , y0 )
(v x) ( x
0 , y0 )
these
two
(v y ) ( x
0 , y0 )
…
(i)
and
1 x02 ( y02 2 y0 ) ( y0 1) 2 x02 0 y0 1 0 and x0 0 we
have
the
point
f ( z ) (u x) z i (v x) z (v y ) z i (u y ) z 0
2 x0 y0 2 x0
0
0
0
x0 0
and
2 x0 y0 2 x0 i( y02
y0 1 .
As
2 y0 ) 0 i i .
[Similar question was also asked in MA-2014 (1 mark)] Example 4.22 [XE-2009 (2 marks)]: If a complex function f ( z ) u ( x, y ) iv ( x , y ) is analytic, then (a) i (u x) (v x) (u y) i (v y )
(b) i (u x) (v x) (u y ) i (v y )
(c) (u x) i (v x) i (u y ) (v y ) (d) (u x) i (v x) i (u y ) (v y) Solution (c): On comparing the real and imaginary parts on both sides of the equations given in options we find that the equation in option (c) satisfies the Cauchy – Riemann equations given by Eq. 4.14. So option (c) is correct. Example 4.23 [XE-2012 (1 mark)]: Consider two functions f ( z ) z and g ( z ) z (conjugate of z ). Using Cauchy Riemann conditions, choose the correct answer (a) Both f and g are analytic (b) f is analytic but g is not analytic (c) g is analytic but f is not analytic (d) Neither f nor g is analytic Solution (d): As f ( z ) z is only differentiable at the origin so on the entire complex plane it is not analytic; also g ( z ) z is not differentiable in the entire complex plane so it is also not analytic. Example 4.24 [XE-2013 (1 mark)]: Consider the function f ( z ) z 2 z , z C . At z 0 , the function f (a) does not satisfy the Cauchy-Riemann (b) satisfies the Cauchy-Riemann equations but equations is not differentiable (c) is differentiable but not analytic (d) is analytic 2 Solution (b): Let z x iy , so f ( z ) z 2 z zzz z z ( x iy )( x 2 y 2 ) f ( z ) x ( x 2 y 2 ) iy ( x 2 y 2 ) u ( x, y ) iv ( x, y ) u ( x, y ) x( x 2 y 2 ) ,
Now
the
Cauchy-Riemann
equation:
v ( x, y ) y ( x 2 y 2 ) .
u x v y 3 x 2 y 2 x 2 3 y 2
and
u y v x 2 xy 2 xy so it is not analytic on the whole complex plane. Now as at
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Engineering Mathematics
Chapter 4: Complex Variables
[4.11]
z 0 i0 we have u x v y 0 0 and u y v x 0 0 and so f ( z ) z 0 0i (u x) z 0 0 i i (v x) z 0 0 i (v y ) z 0 0 i i (u y ) z 0 0 i 0 . So at z 0 i0 the given complex function is differentiable. 2
Example 4.25 [XE-2014 (1 mark)]: At z 0 , the complex function f ( z ) z z (a) satisfies the Cauchy-Riemann equations and is differentiable (b) satisfies the Cauchy-Riemann equations but is not differentiable (c) does not satisfy the Cauchy-Riemann equations but is differentiable (d) does not satisfy the Cauchy-Riemann equations and is not differentiable Solution (a): From the solution of previous question we can say that the given complex function is analytic and differentiable at the origin. Example 4.26 [EC-2014 (2 marks)]: The real part of an analytic function f ( z ) where z x jy is given by e y cos x . The imaginary part of f ( z ) is (a) e y cos x (b) e y sin x (c) e y sin x (d) e y sin x Solution (b): If f ( x iy ) u ( x, y ) iv( x, y ) is analytic then it satisfies the Cauchy-Riemann equations, i.e., (i) u x v y , (ii) u y vx . So (i) (v y ) ( x )(e y cos x) (v y ) e y sin x v e y sin x y , on integrating both sides we get,
v e
y
sin x y v e y sin x f ( x )
…(iii). Now differentiating (iii) both sides w.r.t. x and applying the (ii) condition we get v {e y cos x f ( x )} (e y cos x ) e y cos x f ( x ) e y cos x f ( x) 0 x y
f ( x ) k , where k is any constant. Thus from (iii) we have v e y sin x k . [Similar questions were also asked in ME-2009, CE-2011, ME-2014, ME-2014 (2 marks), PI2010 (1 mark)]
Laplace’s Equation: If f ( z ) u ( x, y ) iv ( x , y ) is analytic in a domain D , then both u and v satisfies Laplace’s equation in D and have continuous second partial derivatives in D , i.e., 2 u u xx u yy 0 and 2 v v xx v yy 0
(4.15)
Complex Exponential Function: x
When x is real then e 1 iy
y2
2!
e 1
y4 4!
x 1!
x2
2! y3
3!
i y
x3 3! y5 5!
iy
e 1
iy 1!
(iy ) 2 2!
(iy )3 3!
(iy ) n n!
(by
cos y i sin y . We will use this result in finding
the complex exponential function. The complex exponential function e z , also written as exp z , is defined in terms of the real functions e x , cos y and sin y as, eiy (cos y i sin y ) e z e x iy e x eiy e x (cos y i sin y ) , where eiy (cos y i sin y )
(4.16)
The polar form of a complex number z x iy can also be written as z r (cos i sin ) rei .
Equation 4.16 e 2 i cos 2 i sin 2 1 ; also e i 2 i ; e i 1 ; e i 2 i ; e i 1
e
iy
2
2
cos y i sin y cos y sin y 1 , i.e., for pure imaginary exponents the exponential
function has absolute value 1. Hence e z e x iy e x eiy e x eiy e x e x which also results that arg e z y 2 n ( n 0,1, 2, ) , as e z e x shows that 4.22 is already in polar form.
As, e z e x 0, z we have an entire function that never vanishes.
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Engineering Mathematics
Chapter 4: Complex Variables
[4.12]
Example 4.27 [ME-2006 (2 marks)]: Assuming i 1 and t is a real number, (a) ( 3 2) i(1 2)
(b) ( 3 2) i (1 2)
3 it
3
0
0
Solution (a): I
e dt
(c) (1 2) i ( 3 2)
e dt is
(d) (1 2) i{1 ( 3 2)}
3
(cos t i sin t )dt (sin t i cos t )0 3
2 it
0
2
i
1 2
(0 i )
3 2
i
1 2
Example 4.28 [CS-2011 (2 marks)]: Given i 1 , what will be the evaluation of the definite integral
2
0 (cos x i sin x)
(a) 0 Solution
(d):
2
cos x i sin x
0
cos x i sin x
I
(cos x i sin x ) dx ?
(b) 2 (c) i (d) i ix ix i (2 x ) (cos x i sin x) (cos x i sin x ) e e e cos 2 x i sin 2 x . dx
2
0
So
2
i sin 2 x cos 2 x i 0 ( 1 1) i . 2 0 2 2
(cos 2 x i sin 2 x ) dx
i
Example 4.29 [EE-2014 (1 mark)]: All the values of the multi-valued complex function 1 , where
i 1 , are (a) purely imaginary (c) on the unit circle Solution (b): As 1 cos(2n ) i sin(2n ) e
(b) real and non-negative (d) equal in real and imaginary part i 2 n
1i (ei 2 n )i e i
2
2 n
e 2 n all the values of
1i are real and non-negative.
Complex Trigonometric Functions: Equation 4.16 implies that ei (cos i sin ) and e i e i ( ) (cos i sin ) . By adding and subtracting these two equations we get cos (e i e i ) 2 and sin (ei e i ) (2i) Hence Equations 4.17 suggests the following definitions for complex values z x iy as,
(4.17)
cos z (eiz e iz ) 2 and sin z (eiz e iz ) (2i) (4.18) sin z cos z 1 1 Also, tan z , cot z , sec z and cosec z (4.19) cos z sin z cos z sin z Since e z is entire, cos z and sin z are entire functions. tan z and sec z are not entire; they are analytic except at the points where cos z is zero; cot z and cosec z are analytic except where sin z is zero. All the formulas for derivative of real trigonometric function are also applied to complex trigonometric functions, i.e. (cos z ) sin z , (sin z ) cos z , (tan z ) sec 2 z .
Complex Hyperbolic Functions: The complex hyperbolic cosine and sine functions are given as, cosh z (e z e z ) 2 and sinh z (e z e z ) 2 (4.20) These functions are entire with derivatives (cosh z ) sinh z and (sinh z ) cosh z as in calculus. The other hyperbolic functions are defines as, sinh z cosh z 1 1 tanh z , coth z , sec hz and cosec hz (4.21) cosh z sinh z cosh z sinh z
Relation between complex trigonometric and hyperbolic functions
If we replacing z by iz in equation 4.20 and then use in equation 4.18, we get, cosh iz cos z and sinh iz i sin z If we replacing z by iz in equation 4.18 and then use in equation 4.20, we get, cos iz cosh z and sin iz i sinh z
(4.22) (4.23)
Example 4.30 [CE-1997 (1 mark)]: e x is periodic, with a period of
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Engineering Mathematics
Chapter 4: Complex Variables
[4.13]
(c) (d) i (a) 2 (b) 2 i x Solution (b): As, e cosh x sinh x ; as sinh x sinh(2 i x) , cosh x cosh(2 i x ) ; thus period of both sinh x and cosh x is 2 i and so period of e x cosh x sinh x is the LCM(2 i, 2 i ) 2 i . Example 4.31 [CH-2012 (1 mark)]: If a and b are arbitrary constants, then the solution to the ordinary differential equation d 2 y dx 2 4 y 0 is (a) y ax b (c) y a sin 2 x b cos 2 x (d) y a cosh 2 x b sinh 2 x (b) y ae x Solution (a): As we have homogeneous linear equation with constant coefficient so putting y e mx y me mx y m 2 e mx in the given differential equation we get e mt ( m 2 4) 0 so we have the auxiliary equation m 2 4 0 as e mt 0 . As the discriminant of the auxiliary equation D 0 2 4 1 ( 4) 16 0 so we have distinct real root case; also the roots of our auxiliary m1 , m2 2, 2 ; so the general solution of given differential equation is
equation are
y c1e 2 x c2 e 2 x . Now from Eq. 4.28 we have e z cosh z sinh z so e 2 x cosh(2 x) sinh(2 x) and
e 2 x cosh( 2 x) sinh( 2 x) cosh(2 x) sinh(2 x) . y c1e
2 x
c2 e
2x
Thus
c1{cosh(2 x ) sinh(2 x )} c2 {cosh(2 x ) sinh(2 x)}
y (c1 c2 ) cosh(2 x ) (c1 c2 ) sinh(2 x) a cosh(2 x) b sinh(2 x) , a (c1 c2 ) and b (c1 c2 ) .
Example 4.32 [IN-2007, CH-2010, EC-2012, EE-2012, IN-2012, ME-1996 (1 mark)]: If x 1 , then the value of x x is (d) 1 (c) x (a) e 2 (b) e 2 Solution (a): x
1 i cos( 2) i sin( 2) e
i 2
x x (ei 2 ) i x ei
2
2
e
2
.
Example 4.33 [IN-2009 (1 mark)]: If z x iy , where x and y are real. The value of eiz is (a) 1 Solution e
iz
(b) e As
(d):
y
e (cos x i sin x) e
y
x2 y 2
(d) e y
(c) e y
eiz ei ( x iy ) e y ix e y eix e y (cos x i sin x ) , cos x i sin x e
y
2
2
cos x sin x e
y
so
.
Example 4.34 [EE-2013 (1 mark)]: Square roots of i , where i 1 are (a) i , i (b) cos( 4) i sin( 4) , cos(3 4) i sin(3 4) (c) cos( 4) i sin(3 4) , cos(3 4) i sin( 4) (d) cos(3 4) i sin( 3 4) , cos( 3 4) i sin(3 4) Solution (b): Let
y
be the square root of
i ; so
y ( i)1 2 (e i 2 )1 2 e i
i sin y cos i sin , cos i sin 4 4 4 4 4 4 y cos( 4) i sin( 4), cos(3 4) i sin(3 4) , as cos( 4) cos(3 4) sin( 4) sin(3 4) .
4
y cos
and
4.1.4 Logarithm of a Complex number The natural logarithm of z x iy is denoted by ln z or log z and is defined as the inverse of the exponential function; i.e., w ln z is defined for z 0 by the relation e w z . Let w u iv and
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Engineering Mathematics
Chapter 4: Complex Variables
[4.14]
i w u iv i u iv has an absolute value eu and the argument v . So z re e e re . Since e eu r u ln r ln z and v . Hence w u iv ln z is given by
ln z ln r i ( r z 0 , arg z ) (4.24) Since the argument of z is determined only upto integer multiple of 2 , the complex natural logarithm ln z ( z 0) is infinitely many-valued. The value of ln z corresponding to the principal value Arg z is denoted by Ln z and is called principal value of ln z . Hence, Ln z ln z i Arg z ( z 0) (4.25) The uniqueness of Arg z for given z ( 0) implies that Ln z is single-valued; and other values of arg z differ by integer multiple of 2 , the other values of ln z are given by ln z Ln z 2n i ( n 1, 2, ) (4.26) They have the same real part and imaginary parts differ by integer multiple of 2 . If z is positive real, then Arg z 0 and Ln z becomes identical with the real natural logarithm. If z is negative real (so that the natural logarithm of calculus is not defined), then Arg z and hence, Ln z ln z i . For every n 0, 1, 2, equation 4.28 defines a function, which is analytic, except at 0 and on the negative real axis and has the derivative (ln z ) 1 z .
Example 4.35 [EC-2006 (2 marks)]: For the function of a complex variable W ln Z (where, W u iv and Z x iy ), the u constant lines get mapped in z plane as (a) set of radial straight lines (b) set of concentric circles (c) set of confocal hyperbolas (d) set of confocal ellipses Solution (b): As u iv ln( x iy ) so from Eq. 4.32 we have ln( x iy ) ln r i , where x2 y 2
r
and
tan 1 ( y x ) ;
so
ln( x iy ) ln x 2 y 2 i tan 1 ( y x) .
Thus
u iv ln x 2 y 2 i tan 1 ( y x ) u ln x 2 y 2 x 2 y 2 e 2u . It is given that u is a constant
so e 2u is also constant, say, k . So x 2 y 2 k which is a set of concentric circles. Example 4.36 [IN-2013 (1 mark)]: The complex function tanh( s) is analytic over a region of the imaginary axis of the complex s plane if the following is TRUE everywhere in the region for all integers n (a) Re( s) 0 (b) Im( s ) n (c) Im( s ) n 3 (d) Im( s ) (2 n 1) 2 Solution (d): As tanh( s ) {sinh( s)} {cosh( x)} (e s e s ) (e s e s ) . If tanh( s) is analytic then s
e e e
2s
s
e
2s
0 e 1 0 e
i (2 n 1)
2s
1 .
As
1 cos(2n 1) i sin(2n 1) ei (2 n 1) ,
so
2 s i (2 n 1) s i (2 n 1) 2 which is an imaginary part of s . Thus tanh( s)
is analytic over a region of the imaginary axis of the complex s plane then Im( s ) (2 n 1) 2 . Exercise: 4.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 2
1. If i 1, then the value of
200
n1 i n _____.
2. If {(1 i ) (1 i)}x 1 then (a) x 4 n , where n is any positive integer (c) x 4 n 1 , where n is any positive integer
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(b) x 2 n , where n is any positive integer (d) x 2 n 1 , where n is any positive integer
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Engineering Mathematics
Chapter 4: Complex Variables
[4.15]
3. If x 3 i , then x 3 3x 2 8 x 30 _____ 4. 1 (1 2i) 3 (1 i ) (3 4i ) (2 4i ) (a) (1 2) (9 2)i
(c) (1 4) (9 4)i
(b) (1 2) (9 2)i 4
(d) (1 4) (9 4)i
2
5. The real values of x and y for which ( x 2 xi ) (3 x yi) (3 5i) (1 2 yi ) is satisfied, are (a) x 2, y 3 (b) x 2, y 1 3 (c) Both (a) and (b) (d) None of these 6. The complex numbers sin x i cos(2 x ) and cos x i sin(2 x ) are conjugate to each other for (b) x n (1 2)
(a) x n
7. The real part of (1 cos 2i sin ) (a) 1 (3 5 cos )
1
(c) x 0
(d) No value of x
(c) 1 (3 5 cos )
(d) 1 (5 3cos )
is
(b) 1 (5 3 cos )
8. If and are different complex numbers with 1 , then ( ) (1 ) _____. 9. Amplitude of {(1 i ) (1 i )} is (a) 2 (b) 2 (c) 4 (d) 6 10. Let z , w be complex numbers such that z iw 0 and arg( zw) , then arg( z ) (a) 5 4
(b) 2
(c) 3 4
(d) 4
11. If z and are to non-zero complex numbers such that z 1 and arg( z ) arg( ) 2 then z is equal to (a) 1 (c) i (d) i (b) 1 12. The square root of 3 4i are (a) (2 i) (b) (2 i) (c) ( 3 2i ) (d) ( 3 2i ) 13.
2i equals (a) 1 i
(b) 1 i
(d) None of these
(c) 2i
2
14. (1 7i ) (2 i ) (a)
2 cos
3 4
i sin
3 4
(b)
2 cos
i sin 4 4
(c) cos
3 4
i sin
3 4
(d) None of these
ei
15. Real part of e is (a) e cos [cos(sin )]
(b) e cos [cos(cos )]
(c) esin [sin(cos )]
(d) esin [sin(sin )]
16. i log{( x i ) ( x i)} is equal to (a) 2 tan 1 x (b) 2 tan 1 x (c) 2 tan 1 x (d) 2 tan 1 x 17. The points 1 3i, 5 i and 3 2i in the complex plane are (a) Vertices of a right angled triangle (b) Collinear (c) Vertices of an obtuse angled triangle (d) Vertices of an equilateral triangle 18. z1 z2 z1 z2 is possible if (a) z2 z1
(b) z2 1 z1
(c) arg ( z1 ) arg( z2 )
(d) z1 z2
2
19. If z 2 1 z 1, then z lies on (a) An ellipse (b) The imaginary axis (c) A circle (d) The real axis 100 20. If {(1 i) (1 i)} a ib , then (a) a 2 , b 1 (b) a 1 , b 0 (c) a 0 , b 1 (d) a 1 , b 2 2 m 4 m 21. is an imaginary cube root of unity. If (1 ) (1 ) then least positive integral value of m is _____.
22. If i 1 , then 4 5 ( 1 2) i ( 3 2) (a) 1 i 3
334
(b) 1 i 3
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365
3 ( 1 2) i ( 3 2)
(c) i 3
is equal to (d) i 3
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4.2
Chapter 4: Complex Variables
[4.16]
Complex Integration
4.2.1 Line Integral in the Complex Plane Complex definite integrals are called (complex) line integrals and are written as,
C f ( z)dz
(4.27)
Here the integrand f ( z ) is integrated over a given curve C or a portion of it. The curve C in the complex plane is called path of integration, whose parametric representation is given as, z (t ) x (t ) iy (t ) , a t b (4.28) The sense of increasing t is called the positive sense on C . The curve C is assumed to be smooth curve, i.e. C has a continuous and nonzero derivative, i.e., at each point. z (t ) dz dt x (t ) iy (t ) Geometrically this means that C has everywhere a continuously turning tangent (as shown in Figure 4.6) as follows directly from Figure 4.6: Tangent vector ̇ ( ) of a curve C in the complex plane given by ( ). The arrowhead on the curve indicates z (t t ) z (t ) the definition z (t ) lim . the positive sense (i.e. increasing ) t 0 t All paths of integration for complex line integrals are assumed to be piecewise smooth. If the integrals of f1 and f 2 over a path C exist then
C [k1 f1 ( z) k2 f2 ( z)]dz k1 C f1 ( z)dz k2 C f 2 ( z)dz , where k1 , k2
are constants
If the integrals of f over a path C exist then sense reversal in integrating over the same path, from z0 to Z (left) and from Z to z0 (right) results in change of sign of the integral, i.e., Z
z
0
f ( z ) dz
z0
Z
f ( z ) dz .
If the integrals of f over a path C exists and if path C is partitioned in number of paths C1 , C2 , , Cn then
C f ( z )dz C
1
C2
f ( z ) dz
Cn
f ( z ) dz .
Existence of the complex line integral: Our assumptions that f ( z ) is continuous and C is piecewise smooth imply the existence of the line integral Eq. 4.27. Let f ( z ) u ( x, y ) iv ( x , y ) , z (t ) x (t ) iy (t ) dz dx (t ) idy (t ) , a t b . Then Equation 4.27 can be written as,
C f ( z )dz
C
f ( z ) dz
(u iv )( dx idy ) udx vdy i udy vdx C
C
C
C
(4.29)
Indefinite integration of Analytic function: Let f ( z ) be analytic in a simply connected domain D . Then an indefinite integral of f ( z ) in the domain D , that is, an analytic function F ( z ) such that F ( z ) f ( z ) in D , and for all paths in D joining two points z0 and z1 in D we have, z1
z
0
f ( z ) dz F ( z1 ) F ( z0 ) , where [ F ( z ) f ( z ) ]
(4.30)
We can write z0 and z1 instead of C , as we get same value for all those C from z0 to z1 .
Indefinite integration of any continuous complex function: This method is not restricted to analytic functions but applies to any continuous complex function. Let C be a piecewise smooth path, represented by z z (t ) , where a t b . Let f ( z ) be a continuous function on C . Then, b
C f ( z )dz a f z (t ) z(t )dt , where ( z dz
dt )
(4.31)
Proof: The left hand side of Eq. 4.31 is given by Eq. 4.29 in terms of real line integrals, and we show that the right hand side of Eq. 4.31 also equals to Eq. 4.29. We have z x iy z x iy . We
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[4.17]
and dy ydt . simply write u for u x(t ), y (t ) and v for v x (t ), y (t ) . We also have dx xdt Consequently, in Eq. 4.31 we have b
b
a f z (t ) z(t )dt a (u iv)( x iy )dt C udx vdy i(udy vdx) C (udx vdy ) i C (udy vdx)
If we integrate a given function f ( z ) from a point z0 to a point z1 along different paths, the integrals will in general have different values. In other words, a complex line integral depends not only on the endpoints of the path but in general also on the path itself. Bound for Integrals: The absolute value of complex line integrals is given by
C f ( z )dz
(4.32)
ML
where, L is the length of C and M a constant such that f ( z ) M everywhere on C .
4.2.2 Cauchy’s Integral Theorems We know that a line integral of a function f ( z ) generally depends not only on the endpoints of the path but also on the choice of path. This dependence often complicates situations. Hence condition under which this does not occur are of considerable importance. If f ( z ) is analytic in a domain D and D is simply connected, then the integral will not depend on the choice of path between given points. Let us begin by repeating and illustrating the definition of simple connectedness: A simple closed path is a closed path that does not intersect or touch itself. For example, a circle is simple, but a curve shaped like an 8 is not simple, as shown in Fig. 4.7(a) and (b).
Figure 4.7: (a), (b) are Closed Paths; (c), (d) and (e) are simply and multiply connected domains
A simply connected domain D in the complex plane is a domain such that every simple closed path in D encloses only points of D , for e.g., the interior of a circle (‘open disk’), ellipse, or any simple closed curve, as shown in Fig. 4.7(c). A domain that is not simply connected is called multiply connected; for e.g., an annulus, a disk without the centre, as shown in Fig. 4.7(d) and (e).
Cauchy’s Integral Theorem 1: If f ( z ) is analytic in a simply connected domain D , then for every simple closed path C in D , as shown in Fig. 4.8,
C f ( z )dz 0 Proof: From Eq. 4.29, we have
(4.33)
C f ( z)dz C (udx vdy) i C (udy vdx)
Figure
4.8:
Cauchy
–
. Since f ( z ) is analytic in D , its derivative f ( z ) exists in D . Since f ( z ) Integral Theorem is assumed to be continuous and u and v have continuous partial derivatives in D . Hence Green’s theorem (with u and v instead of F1 and F2 ) is applicable and gives
C (udx vdy ) R (v x) (u y) dxdy , where R
is the region bounded by C .
The second Cauchy – Riemann shows that the integrand on the right is identically zero. Hence the integral on the left is zero. Also by use of first Cauchy – Riemann equation that the last integral in the above formula is zero.
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Chapter 4: Complex Variables
Example 4.37 [IN-2009 (1 mark)]: The value of
[4.18]
(1 z ) sin zdz , where the contour of integration is a
simple closed curve around the origin, is (a) 0 (b) 2 i
(d) 1 2 i
(c)
z2 z4 z6 z 1 which z z 3! 5! 7! 3! 5! 7!
sin z
Solution (a): As the function f ( z )
1
z
3
z
5
z
7
is analytic in a simple closed curve around the origin so from Eq. 4.43 we have (1 z ) sin zdz 0 .
Cauchy’s Integral Theorem 2: If f ( z ) is analytic in a simply connected domain D , then the integral of f ( z ) is independent of path in D . Proof: Let z1 and z2 be any points in D . Consider two paths C1 and C 2 in D from z1 and z2 without further common points, as shown in Fig. 4.9a. Denote by C 2* the path C 2 with the orientation reserved, as shown in Fig. 4.9b. Integrate from z1 over C1 to z2 over C 2* back to z1 . This is a simple closed path and Cauchy theorem applies under our assumption of the present theorem and gives zero, i.e.
C
1
fdz
C2*
fdz 0 fdz C1
C2*
fdz . But the minus sign on the right disappear if we integrate
in the reverse direction, from z1 to z2 which shows that the integral of f ( z ) over C1 and C 2 are equal, i.e.
C
1
f ( z )dz
C2
f ( z ) dz . This proves the theorem for paths that have only the endpoints in
common. For paths that have finitely many further common points, apply the present argument to each ‘loop’ (portion of C1 and C 2 between consecutive common points; four loops as shown in Fig. Figure 4.9: (a) for path function only end points are in common (b) Cauchy integral for reversed orientation (c) Paths with more common points 4.9c).
Cauchy’s Integral Theorem 3: If f ( z ) is analytic in a simply connected domain D , then there exists an indefinite integral F ( z ) of f ( z ) in D – thus F ( z ) f ( z ) , which is analytic in D , and for all paths in D joining any two points z0 and z1 in D , the integral of f ( z ) from z0 to z1 can be evaluated by the Eq. 4.30. Example 4.38 [MA-2014 (2 marks)]: Let {z : Im( z ) 0} and let C be a smooth curve lying in with initial point 1 2i and final point 1 2i . The value of
C {(1 2 z)
(1 z )}dz is
1 1 1 1 (a) 4 ln 2 i (b) 4 ln 2 i (c) 4 ln 2 i (d) 4 ln 2 i 2 4 2 4 2 4 2 2 Solution (a): As it is given that f ( z ) (1 2 z ) (1 z ) is a smooth curve lying in with initial point 1 2i and final point 1 2i . So 1 2 i 1 2 z 1 2 i 1 2 z 1 1 1 2i 2 2 z 1 1 2i 1 2z 1 I dz dz dz dz 2 dz C 1 z 1 2i 1 z 1 2i 1 2 i 1 2 i 1 z 1 z 1 z 1 2 i
I 2 z ln(1 z )1 2i 2{(1 2i ) (1 2i )} {ln(1 1 2i) ln(1 1 2i)} I 4 {ln(2 2i) ln(2i)} 4 ln I 4 ln 2 ln{e
i 4
1 i i
4 ln
(1 i )i i
2
4 ln(1 i ) 4 ln{ 2e i tan
1 ( 1)
}
} 4 (1 2) ln 2 i( 4)
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Chapter 4: Complex Variables
[4.19]
Example 4.39 [ME-2014 (2 marks)]: If z is a complex variable, the value of
3i
5
(1 z ) dz is
(a) 0.511 1.57i (b) 0.511 1.57i (c) 0.511 1.57i (d) 0.511 1.57i Solution (b): As f ( z ) 1 z is analytic between the line joining the points z 5 and z 3i . So 3i
I (1 z )dz (ln z )35i ln 3i ln 5 ln 3 ln i ln 5 ln(3 5) ln e i 5
i cos( 2) i sin( 2) ei 2 ).
ln(3 5) 0.5108 and ln e
i 2
2
0.5108 i( 2)
I 0.5108 i (1.570) 0.511 1.57i ,
So
(as where
i 2 .
Cauchy integral theorem for multiply connected domains: Cauchy’s theorem applies to multiply connected domains. We first explain this for a doubly connected domain D with outer boundary curve C1 and inner C 2 as shown in Fig. 4.10(a). If a function f ( z ) is analytic in any domain D * that contains D and its boundary curves then we have,
C
1
f ( z ) dz
C2
(4.34)
f ( z ) dz
both the integrals being taken counter-clockwise (or both clockwise, and regardless of whether or not the full interior of C 2 belongs to D * ) Proof: By two cuts A and B , as shown in Fug. 4.10(b), we cut D into two simply connected domains D1 and D2 in which and on whose boundaries f ( z ) is analytic. By Cauchy’s integral theorem the integral over the entire boundary of D1 (taken in the sense of arrows, as shown in Fig. 4.10(b), is zero, and so is the integral over the boundary of D2 , and thus their sum. In this sum the integrals over the cuts C1 and C 2 cancel because we integrate over them in both directions and we are left with the integrals over C1 (counter-clockwise) and C 2 (clockwise); hence by reversing the integration over C 2 we have,
C
1
fdz
C2
fdz 0 , which is same as Eq. 4.34. For domains of higher
connectivity the idea remains the same. Thus, for a triply connected domain we use three cuts A, B, C as shown in Fig. 4.10(c). Adding integrals as before, the integrals over the cuts cancel and the sum of the integrals over C1 (counter-clockwise) and C 2 , C3 (clockwise) is zero. Hence the integral over C1 equals the sum of the integrals over C 2 and C3 , all three now taken counter-clockwise. Similarly for quadruply connected domains and so on.
Figure 4.10: (a), (b) are Doubly connected domain; (c) is Triply connected domain
Cauchy Integral Formula: Let f ( z ) be analytic in a simply connected domain D . Then for any point z0 in D and any simple closed path C in D that encloses z0 as shown in Fig. 4.11, then f ( z)
C z z
dz 2 if ( z0 )
(4.35)
0
the integration being taken counter-clockwise. For representing f ( z0 ) by a contour integral, divide Eq. 4.35 by 2 i , we get 1 f (z) (4.36) f ( z0 ) dz C 2 i z z0
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Chapter 4: Complex Variables
Figure 4.11: Cauchy integral formula
[4.20]
Figure 4.12: Multiply connected domain
For multiply connected domains, if f ( z ) be analytic on C1 and C 2 and in the ring – shaped domain bounded by C1 and C 2 , as shown in Fig. 4.12, and z0 is any point in that domain, then f ( z0 )
1
2 i C
1
f ( z) z z0
dz
1
2 i C
2
f ( z) z z0
(4.37)
dz
where the outer integral (over C1 ) is taken counter – clockwise and the inner integral (over C 2 ) is taken clockwise.
4.2.3 Derivatives of analytic function
If f ( z ) is analytic in a domain D , then it has derivatives of all orders in D , which are also analytic function in D . The values of these derivatives at a point z0 in D are given as
f ( z0 )
1
f ( z)
C ( z z )2 dz 2 i
(4.38)
0
f ( z0 )
2!
f ( z)
2 i C ( z z0 )3
dz f ( n) ( z0 )
In general,
Figure 4.13: Derivative of an analytic function
(4.39)
n!
f ( z)
C ( z z0 )n 1 dz 2 i
(4.40)
here C is any simple closed path in D that encloses z0 and whose full interior belongs to D ; and we integrate counter-clockwise around C , as shown in Fig. 4.13. Cauchy’s inequality: Let for Eq. 4.40, we choose C as a circle of radius r and centre z0 and apply the ML inequality (Eq. 4.42) with f ( z ) M on C we obtain from Eq. 4.40 as,
(n)
n!
f (z)
n!
1
C ( z z0 )n1 dz 2 M r n1 2 r
(n)
n!M
(4.41) n 2 r Liouville’s theorem: It states that, if an entire function is bounded in absolute value in the whole complex plane, then this function must be a constant. f
( z0 )
If f ( z ) is continuous in a simply connected domain D and if
f
( z0 )
C f ( z )dz 0
for every closed
path in D , then f ( z ) is analytic in D . Example 4.40 [XE-2011 (1 mark)]: The integral
C z e
3 z
( z 1)3 dz along the curve C : z 2 ,
oriented counter-clockwise, equals (a) 0 (b) 2 i
(c) 13e i (d) 20e i Solution (c): As f ( z ) z e and all its derivatives are analytic in given curve C . So using Eq. 4.52, 3 z
we
have
f ( z 0 )
2! 2 i
C
z 3e z ( z 1)3
dz
C
z 3e z ( z 1)3
dz i f (1) .
As
f ( z ) 6 ze z 3z 2 e z 3 z 2e z z 3e z f (1) 6e 3e 3e e 13e . So
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f ( z ) 3z 2 e z z 3e z z 3e z
C ( z 1)3 dz 13e i .
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[4.21]
4.2.4 Sequence and Series Sequence: An infinite sequence or a sequence is obtained by assigning to each positive integer n a number zn called a term of the sequence and is written as z1 , z2 , or briefly {z n } . A real sequence is one whose terms are real; and a complex sequence is one whose terms are complex numbers. A convergent sequence z1 , z2 , is one that has a limit c , written as lim zn c or zn c . n
By definition of limit this means that for every 0 , we can find N such that zn c for all n N . We know that if zn are real then zn c gives an Figure 4.14: open interval of length 2 and real midpoint c on the real line, as shown in Convergent (a) Real and (b) Complex Fig. 4.19(a). But if zn is a complex number then, geometrically, all terms zn Sequence with n N lie in the open disk of radius and centre c and only finitely many terms do not lie in that disk, as shown in Fig. 4.19 (b). A divergent sequence is one that does not converge. A sequence z1 , z2 , , zn , of complex numbers zn xn iyn (where n 1, 2, ) converges to c a ib if and
only if the sequence of the real parts x1 , x2 , converges to a and the sequence of the imaginary parts y1 , y2 , converges to b .
Series: Given a sequence z1 , z2 , , zm , , we may form the sequence of sums s1 z1 , s2 z1 z2 , s3 z1 z2 z3 , and in general, sn z1 z2 zn , ( n 1, 2, ) then sn is called the nth partial
sum of the infinite series or series, where z1 , z2 , are called the terms of the series.
m 1 zm z1 z2
(4.42)
A convergent series is one whose sequence of partial sums converges, i.e., lim sn s . Then we write n
s m1 zm z1 z2 and call s the sum or value of the series. A series that is not convergent is
called a divergent series. If we omit the terms of
from Eq. 4.42 there remains
sn
Rn zn 1 zn 2 zn 3 , this is called the remainder of the series (Eq. 4.42) after the term zn .
Clearly if Eq. 4.42 converges and has the sum s , then s sn Rn Rn s sn . Now sn s by the definition of convergence; hence Rn 0 . A series given by Eq. 4.42 with zm xm iym converges and has the sum s u iv if and only if x1 x2 converges and has the sum u and y1 y2 converges and has the sum v .
Tests for Convergence and Divergence of Series: The tests are practically the same as in calculus. We apply them before we use a series, to make sure that the series converges. Divergence Test: If a series z1 z2 converges, then lim zm 0 (note that if lim zm 0 m
m
then a series z1 z2 may or may not converge). Hence lim zm 0 then the series diverges. m
Cauchy’s Convergence Principle for Series: A series z1 z2 is convergent if and only if for every given 0 , where is very-very small, we can find an N (which depends on , in general) such that zn 1 zn 2 zn p for every n N and p 1, 2, .
Absolute and Conditionally Convergence: A series z1 z2 is called absolutely convergent if the series of the absolute values of the terms
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m 1 zm
z1 z2 is convergent. If
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[4.22]
z1 z2 converges but z1 z2 diverges, then the series z1 z2 is called conditionally convergent. Comparison Test: If a series z1 z2 is given and we can find a convergent series b1 b2 with non-negative real terms such that z1 b1 , z2 b2 , , then the given series converges.
Geometric series: The geometric series
m 1 z m 1 z z 2
converges with the sum
1 (1 z ) if z 1 and diverges if z 1 .
Ratio Test: If a series z1 z2 with zn 0 , n 1, 2, , is such that lim zn 1 zn L then n
If L 1 , the series converges absolutely If L 1 , the series diverges If L 1 , the series may converge or diverge, so that the test fails.
Root Test: If a series z1 z2 is such that lim
n
n
z n L then
If L 1 , the series converges absolutely If L 1 , the series diverges If L 1 , the series may converge or diverge, so that the test fails.
Power Series: Power series are the most important series in complex analysis because their sums are analytic function and are represented by power series. A power series in powers of z z0 is a series of the form,
n0 an ( z z0 )n a0 a1 ( z z0 ) a2 ( z z0 )2
(4.43)
where z is a complex variable; a0 , a1 , are complex or real constants, called the coefficients of the series; and z0 is a complex or real constant, called the centre of the series. This generalizes real power series of calculus. If z0 0 , we obtain as a particular case a power series in powers of z , as
n 0 an z n a0 a1 z a2 z 2
(4.44)
Convergence of Power series: Power series have variable terms (functions of z ), but if we fix z , then all the concepts for series with constant terms in the last section apply. Usually a series with variable terms will converge for some z and diverge for others. The series given by Eq. 4.43 may converge in a disk with centre z0 or in the whole z plane or only at z0 . Every power series (Eq. 4.43) converges at the centre z0 .
Figure 4.15: Convergence of
If Eq. 4.43 converges at a point z z1 z0 , it converges Power series absolutely for every z closer to z0 than z1 , i.e., z z0 z1 z0 . If Eq. 4.43 diverges at a z z2 , it diverges for every z farther away from z0 than z2 .
Radius of convergence of power series: We consider the smallest circle with centre z0 that includes all the points at which a given power series (Eq. 4.43) converges. Let R denote its radius. The circle z z0 R is called the circle of convergence and its radius R the radius of convergence of Eq. 4.43. Hence, convergence of power series implies that Convergence everywhere within that circle, i.e., for all z for which z z0 R (the open disk with centre z0 and radius R ). As R is as small as possible, the series (Eq. 4.43) diverges for all z for which z z0 R On the circle the series (Eq. 4.43) may converge at some or all or none of these points.
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[4.23]
We denote R if the series (Eq. 4.43) converges for all z ; and R 0 if the series (given by Eq. 4.43) converges only at the centre z0 .
Taylor and Maclaurin Series: The Taylor series of a function f ( z ) is given as,
f ( z ) n 1 an ( z z0 ) n
where, an
1 n!
f ( n ) ( z 0 ) or an
1
f ( z*)
C ( z * z0 )n 1 dz * 2 i
(4.45)
(from Eq. 4.40). If we integrate Eq. 4.40
counter-clockwise around a simple closed path C that contains z0 in its interior and is such that f ( z ) is analytic in a domain containing C and every point inside C . A maclaurin series is a Taylor series with centre z0 0 . The remainder of the Taylor series (Eq. 4.45) after the term an ( z z0 ) n is ( z z0 ) n 1
f ( z*)
C ( z * z0 )n1 ( z * z ) dz *
(4.46) 2 i Hence, the Taylor formula with remainder is given as, 2 n z z0 ( z z0 ) ( z z0 ) ( n ) f ( z ) f ( z0 ) f ( z0 ) f ( z0 ) f ( z0 ) Rn ( z ) (4.47) 1! 2! n! Let f ( z ) be analytic in a domain D , and let z z0 be any point in D . Then there exists Rn ( z )
precisely one Taylor series (Eq. 4.45) with centre z0 that represents f ( z ) , which is valid in the largest open disk with centre z0 in which f ( z ) is analytic. The remainder Rn ( z ) of Eq. 4.45 can be represented in the form of Eq. 4.46. The coefficient satisfies the inequality a n M r n , where M is the maximum of f ( z ) on a circle z z0 r in D whose interior is also in D .
A power series with a non-zero radius of convergence is the Taylor series of its sum.
Laurent Series: Let f ( z ) be analytic in a domain containing two concentric circles C1 and C 2 with centre z0 and the annulus between them, as shown in Fig. 4.16. Then f ( z ) is represented by the Laurent series as,
f ( z ) n 0 an ( z z0 ) n n 1 bn ( z z0 ) n f ( z ) a0 a1 ( z z0 ) a2 ( z z0 ) 2
(4.48) b1
b2 2
, which Figure
4.16:
Laurent's
theorem
( z z0 ) ( z z 0 ) consist of non-negative and negative powers. The coefficient of this Laurent series are given by the integrals (Eq. 4.48), which are taken counter-clockwise around any simple closed path C that lies in the annulus and encircles the inner circles, as shown in Fig. 4.16. [The variable of integration is denoted by z * since z is used in Eq. 4.48] 1 f ( z*) 1 ( z * z0 ) n 1 f ( z*) dz * an dz * , bn (4.49) n 1 C C 2 i ( z * z0 ) 2 i This series converges and represents f ( z ) in the enlarged open annulus obtained from the given
annulus by continuously increasing the outer circle C1 and decreasing C 2 until each of the two circles reaches a point where f ( z ) is singular. In the important special case that z0 is the only singular point of f ( z ) inside C 2 , this circle can be shrunk to the point z0 , giving convergence in a disk except at the centre. In this case the series (or finite sum) of the negative powers of Eq. 4.48 is called the principal part of the singularity of f ( z ) at z0 .
Instead of Eq. 4.48 and 4.49, we may write, by denoting bn by a n , as
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f ( z ) n an ( z z0 ) n
[4.24]
(4.50)
where all the coefficients are now given by a single integral formula as, 1 f ( z*) an dz * , ( n 0, 1, 2, ) C 2 i ( z * z0 ) n 1
(4.51)
4.2.5 Singularities, Zero and Infinity Singularity: A function f ( z ) is singular or has a singularity at a point z z0 if f ( z ) is not analytic (perhaps not even defined) at z z0 , but every neighbourhood of z z0 contains points at which f ( z ) is analytic. We also say that z z0 is a singular point of f ( z ) .
Isolated singularity: We call z z0 an isolated singularity of f ( z ) if z z0 has a neighbourhood without further singularities of f ( z ) . For e.g., tan z has isolated singularities at
(2n 1) 2 , n I ; tan 1 z has a isolated singularity at 0. Isolated singularities of
f (z)
at
z z0
f ( z ) n 0 an ( z z0 ) n n 1 bn ( z z0 ) n
can be classified by Laurent series
valid in the immediate neighbourhood of the
singular point z z0 except at z0 itself, i.e., in the region of the form 0 z z0 R .
Poles: The sum of the first series of Laurent series is analytic at z z0 ; the second series, containing the negative powers, is called the principal part of Eq. 4.60. If the second series of b1 bm Laurent series has only finite terms, i.e., it is of the form , (bm 0) , then m z z0 ( z z0 ) the singularity of f ( z ) at z z0 is called a pole and m is called its order. Poles of the first order are also known as simple poles. If f ( z ) is analytic and has a pole at z z0 , then f ( z ) as z z0 in any manner. If all the negative powers of ( z z0 ) in Eq. 4.60 after the
th n are missing, then the singularity at z z0 is called a pole of order n . Isolated essential singularity: If the principal part of Eq. 4.60 has infinitely many terms then we sat that f ( z ) has an isolated essential singularity at z z0 .
Picard’s Theorem: If f ( z ) is analytic and has an isolated singularity at a point z0 , it takes on every value, with at most one exceptional value, in an arbitrarily small , neighbourhood of z0 .
Removable singularity: We say that a function f ( z ) has a removable singularity at z z0 is
f ( z ) is not analytic at z z0 , but can be made analytic there by assigning a suitable value f ( z0 ) . Such singularities are of no interest since they can be removed as just indicated. For e.g., f ( z ) (sin z ) z becomes analytic at z 0 if we defined f (0) 1 . Example 4.41 [CE-2009 (1 mark)]: The analytic function f ( z ) ( z 1) ( z 2 1) has singularities at (a) 1 and –1 (b) 1 and i (c) 1 and i (d) i and i Solution (d): As any function f ( z ) is said to be singular if it is not defined at a point z z0 . Thus the given function has singularities at the points where z 2 1 0 z i . Example 4.42 [IN-2007 (2 marks)]: For the function (sin z ) z 3 of a complex variable z , the point z 0 is (a) a pole of order 3 (b) a pole of order 2 (c) a pole of order 1 (d) not a singularity
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sin z
Chapter 4: Complex Variables
1
z3
z5
z7
1
1
z2
z4
. As all the negative 3 2 z z 3! 5! 7! z 3! 5! 7! powers of ( z 0) in our expression after 2nd are missing, so the singularity at z 0 is called a pole of order 2.
Solution (b): As
z 3
[4.25]
4.2.6 Zeros of Analytic function A zero of an analytic function f ( z ) in a domain D is a z z0 in D such that f ( z0 ) 0 . A zero has order n if not only f but also the derivatives of f , f , , f ( n 1) are all 0 at z z0 but f ( n ) ( z 0 ) 0 . A first-order zero is also called a simple zero. For a second-order zero, f ( z0 ) f ( z0 ) 0 , but f ( z0 ) 0 . And so on.
Taylor Series at a Zero: At an nth – zero z z0 of f ( z ) , the derivatives f ( z0 ), , f ( n 1) ( z0 ) are zero, by definition. Hence the first few coefficients a0 , , an 1 of the Taylor series are zero too, whereas an 0 , so that this series takes the form, (where an 0 ) f ( z ) an ( z z 0 ) n an 1 ( z z0 ) n 1 ( z z0 ) n an an 1 ( z z 0 ) an 2 ( z z 0 ) 2
(4.52) th
This is characteristic of such a zero, because if f ( z ) has such a Taylor series, it has an n -order zero at z z0 as follows by differentiation. Whereas non-isolated singularities may occur, for zeros, thus
The zeros of an analytic function f ( z ) ( 0) are isolated; i.e., each of them has a neighbourhood that contains no further zeros of f ( z ) .
Let f ( z ) be analytic at z z0 and have a zero of nth order at z z0 . Then 1 f ( z ) has a pole of th n order at z z0 ; and so does h( z ) f ( z ) , provided h( z ) is analytic z z0 and h( z0 ) 0 .
4.2.7 Residue Integration Method The purpose of Cauchy’s residue integration method is the evaluation of integrals
C f ( z)dz
taken
around a simple closed path C . If f ( z ) is analytic everywhere on C and inside C , such an integral is zero by Cauchy’s integral theorem. If f ( z ) has a singularity at a point z z0 inside C , but is otherwise analytic on C and inside C , then f (z) has a Laurent series
f ( z ) n0 an ( z z0 ) n
b1 ( z z0 )
b2 ( z z0 ) 2
that converges for all points near z z0 (except at
z z0 itself) , in some domain of the form 0 z z0 R . The coefficient b1 of the first negative
power 1 ( z z0 ) of this Laurent series is given by the integral formula (Eq. 4.49 with n 1) as,
1
(4.53) C f ( z)dz C f ( z)dz 2 ib1 2 i Here we integrate counter-clockwise around a simple closed path C that contains z z0 in its
b1
interior (but no other singular points of f ( z ) on or inside C ). The coefficient b1 is called the residue of f ( z ) at z z0 and is denoted by, b1 Re s f ( z )
(4.54)
z z0
Formulas for Residue: To calculate a residue at a pole, we need not produce a whole Laurent series, but we can derive formulas for residues once and for all. Simple Poles: The two formulas for the residue of f ( z ) at a simple pole at z0 are
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Engineering Mathematics
Chapter 4: Complex Variables
[4.26]
Res f ( z ) b1 lim ( z z0 ) f ( z )
(4.55)
z z0
z z0
and, assuming that f ( z ) p ( z ) q( z ) , p ( z0 ) 0 and q ( z ) has a simple zero at z0 (so that f ( z ) has at z0 a simple pole, Res f ( z ) Res p ( z ) q ( z ) p ( z0 ) q( z0 ) z z0
(4.56)
z z0
Poles of any order: The residue of f ( z ) at an mth order pole at z0 is given as,
d m 1 m Res f ( z ) lim m1 ( z z0 ) f ( z ) z z z z0 (m 1)! 0 dz 1
(4.57)
In particular, for a second – order pole ( m 2) is given as, Res f ( z ) lim z z0
z z0
d
( z z0 ) 2 f ( z ) . dz
Example 4.43 [EC-2008 (2 marks)]: The residue of the function f ( z ) 1 {( z 2) 2 ( z 2) 2 } at z 2 is (a) 1 32 (b) 1 16 (c) 1 16 (d) 1 32 Solution (a): The poles of the any function f ( z ) are given by putting the denominator equal to zero, i.e., ( z 2) 2 ( z 2) 2 0 z 2, 2 . As the function f ( z ) have poles at z 2 both of order 2. So from
Eq.
4.57,
we
Res f ( z ) lim
have
z 2
z 2
d
( z 2) 2 f ( z )
dz
d 1 d 1 2 2 2 ( z 2) lim lim 3 2 2 2 3 z 2 dz ( z 2) ( z 2) z 2 dz ( z 2) z 2 ( z 2) 4
Res f ( z ) lim z2
Res f ( z ) 1 32 . z 2
Example 4.44 [EE-2008 (2 marks)]: Given X ( z ) z ( z a ) 2 with z a , the residue of X ( z ) z n 1 at z a for n 0 will be (a) a n 1 (b) a n (c) na n (d) na n 1 Solution (d): We have to find the residue of the function f ( z ) X ( z ) z n 1 z n ( z a ) 2 . As the pole of
f ( z ) is where ( z a ) 2 0 z a which is of 2nd order so from Eq. 4.69, we have
Res f ( z ) lim za
za
d zn d n ( z a ) 2 f ( z ) lim ( z a ) 2 lim z lim nz n 1 na n 1 . 2 z a z a z a dz ( z a) dz dz d
Example 4.45 [EC-2010 (2 marks)]: x( z ) (1 2 z ) {z ( z 1)( z 2)} at its poles are
The
residues
of
a
complex
function
(a) 1 2 , 1 2 and 1 (b) 1 2 , 1 2 and –1 (c) 1 2 , 1 and 3 2 (d) 1 2 , –1 and 3 2 Solution (c): The poles of any function f ( z ) are given by putting the denominator equal to zero, i.e., z ( z 1)( z 2) 0 z 0,1, 2 . The function f ( z ) have simple poles all at z 0,1, 2 . So from Eq. 4.55, Res f ( z ) lim( z 0) f ( z ) lim( z 0)
1 2z
lim
1 2z
1
z ( z 1)( z 2) z 0 ( z 1)( z 2) ( 1)( 2) 1 2z 1 2z 1 2 Res f ( z ) lim( z 1) f ( z ) lim( z 1) lim 1; z 1 z 1 z 1 z ( z 1)( z 2) z 0 z ( z 2) (1)( 1) 1 2z 1 2z 1 4 3 Res f ( z ) lim( z 2) f ( z ) lim( z 2) lim z2 z2 z 0 z 2 z ( z 1)( z 2) z ( z 1) (2)(1) 2 [Similar question was also asked in CH-2013 (2 marks)] z 0
z 0
z 0
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1 2
;
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Engineering Mathematics
Chapter 4: Complex Variables
[4.27]
Example 4.46 [XE-2010 (1 mark)]: The residue of the function f ( z ) (sin 4 z ) ( z 4)3 at z 4 is (a) 2 (b) 1 (c) –1 (d) –2 Solution (b): As the denominator of the given function is zero at z 4 so z 4 is a pole of
order 3. Thus from Eq. 4.69, we have Res f ( z ) z 4
Res f ( z ) z 4
Res f ( z ) z 4
Res f ( z ) z 4
1 2
2!
d2
( z 4)3 f ( z ) dz
lim z 4
2
d 2 sin 4 z d2 1 3 4 ( z 4) lim 2 sin z 3 2 z 4 z 4 2 dz dz
lim z 4
1
lim
2
z 4
1
2
1
4
d
1
4 sin 3 z cos z lim 4 sin 4 z 12 sin 2 z cos 2 z dz 2 z 4
1
1 1 1 12 1 3 1 4 2 2 2
Residue Theorem: Let f ( z ) be analytic inside a simple closed path C and on C , except for finitely many singular points z1 , z2 , , zk inside C , as shown in Fig. 4.22. Then the integral of f ( z ) taken counter-clockwise around C equals 2 i times the sum of the residues of f ( z ) at z1 , z2 , , zk as, k
f ( z) C f ( z )dz 2 i j1 Res zz
Figure 4.17: Residue Theorem
(4.58)
j
Example 4.47 [CE-2005 (2 marks)]: Consider likely applicability of Cauchy’s Integral Theorem to evaluate the following integral counter clockwise around the unit circle C . I sec z dz , z being a C
complex variable. The value of I will be (a) I 0 : singularities set = {(2n 1) 2} , n 0,1, 2,
(b) I 0 : singularities set =
(c) I 2 : singularities set = n , n 0,1, 2, (d) None of these Solution (b): 1 3 3 As I C sec z dz C cos z dz which has poles at z (2n 1) 2 , i.e., z , 2 , 2 , 2 , 2 , and none of these poles lie inside z 1 . So we have singularities set = . Also the sum of residues of the poles which lie inside the given closed region is zero as none of the poles lie inside z 1 .
1
Thus from Eq. 4.58 we have I sec z dz C
C
cos z
dz 2 i 0 0 .
Example 4.48 [EC-2007 (2 marks)]: If the semi-circular contour D of radius 2 is as shown in the figure, then the value of 1 the integral D s 2 1 ds is Solution (a): As the given function is f ( s )
(a) (b) (c) (d) 1
j j
1
. So the pole of the given function s 1 ( s 1)( s 1) is s 1 , which are simple poles. As the pole s 1 lies outside the given semi-circle; but the pole
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2
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Engineering Mathematics
Chapter 4: Complex Variables
s 1 lies within the given semi-circle. So from Eq. 4.58 we have
[4.28]
C f ( z)dz 2 i (sum of residues
of the poles which lie inside the given closed region). As we have only one pole s 1 which lie within
the
given
semi-circle.
Res f ( s ) lim( s 1) f ( s ) lim z 1
z 1
z 1
( s 1) ( s 1)( s 1)
f ( s) . C f ( s)ds 2 j Res z 1
So
lim z 1
1 s 1
Example 4.49 [ME-2008 (2 marks)]: The integral
1 2
. So
f ( z )dz
As
1
C f ( s)ds 2 j 2 j .
evaluated around the unit circle on the
complex plane for f ( z ) (1 z ) cos z is (d) 0 (a) 2 i (b) 4 i (c) 2 i Solution: As the given function 1 z z 3 z5 cos z 1 z 2 z 4 z6 f ( z) 1 f ( z ) , which has a simple pole at z 2! 4! 6! z z 2! 4! 6! z 0 , which lie inside the unit circle, so from Eq. 4.57, we have cos z Res f ( z ) lim ( z 0) f ( z ) lim ( z 0) Res f ( z ) lim cos z 1 . Thus from Eq. 4.58 we z 0 z 0 z 0 z 0 z 0 z
have
C
C f ( z)dz 2 i (sum of residues of the poles which lie inside the given closed region), i.e.,
cos z z
dz 2 i 1 2 i
Example 4.50 [CE-2009 (2 marks)]: The value of the integral
cos(2 z )
C (2 z 1)( z 3) dz
(where C is a
closed curve given by z 1 ) is (b) i 5
(c) (2 i ) 5 (d) i cos(2 z ) Solution (c): As the given function is f ( z ) . So the pole of the given function is (2 z 1)( z 3) (a) i
z 1 2 , 3 , which all are simple poles. As the pole z 3 lie outside the circle z 1 since 3 1 ; but
the pole z 1 2 lie inside the circle
z 1 since 1 2 1 . Thus from Eq. 4.55, we have
Res f ( z ) lim ( z 1 2) f ( z ) lim ( z 1 2) z 1 2 z 1 2 z 1 2
So from Eq. 4.58 we have closed region). Thus
cos(2 z )
cos(2 z ) 1 zlim ; (2 z 1)( z 3) 1 2 2( z 3) 5
C f ( z)dz 2 i (sum of residues of the poles which lie inside the given cos(2 z )
1
C (2 z 1)( z 3) dz 2 i 5
2 i 5
.
Example 4.51 [CH-2009 (2 marks)]: Using the residue theorem, the value of the integral (counterclockwise)
(8 7 z )
( z 4) dz around a circle with centre at z 0 and radius 8 (where z is a
complex number and i 1 ) is (a) 20 i (b) 40 (c) 40 i (d) 40 i Solution (c): As the given function is f ( z ) (8 7 z ) ( z 4) . So the pole of the given function is z 4 , which is a simple poles, and lies within the circle whose centre at z 0 and radius 8 , since 4 0 8 . So from Eq. 4.55 we have
f (z) . C f ( z )dz 2 i Res z 4
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As from Eq. 4.55, we have
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Engineering Mathematics
Chapter 4: Complex Variables
8 7z
Res f ( z ) lim( z 4) f ( z ) lim ( z 4) z 4
z 4
z4
z4
[4.29]
lim(8 7 z ) 20 .
Thus
z 4
C f ( z)dz 2 i (20) 40 i . [Similar questions were also asked in CE-2006 (2 marks), CS-2014 (1 mark)]
Example 4.52 [EC-2009 (1 mark)]: If f ( z ) c0 c1 z 1 , then
(1 z ){1 f ( z )}dz is given by
unit circle
(a) 2 c1
(b) 2 (1 c0 )
Solution (d): Let g ( z )
1 f ( z) z
(c) 2 ic1
(1 c0 ) z c1 z
(d) 2 i (1 c0 )
which has a 2nd order pole at z 0 . So
2
C g ( z)dz 2 i ( sum of residues of the poles which lie inside the given closed region, i.e., where C is a unit circle. As Res g ( z ) z 0
1
lim
1! z 0
d
( z 0) 2 g ( z ) dz
d d 2 (1 c0 ) z c1 z lim {(1 c0 ) z c1} lim(1 c0 ) (1 c0 ) . 2 z0 z 0 dz z0 z z 0 dz g ( z ) dz 2 i (1 c0 ) .
Res g ( z ) lim
C
Example 4.53 [IN-2010 (1 mark)]: The contour C in the adjoining figure is described by x 2 y 2 16 . The value of z2 8
C 0.5 z 1.5i dz
z 1 ),
(a) (b) (c) (d)
is [Note: i 1 ]
So
2 i 2 i 4 i 4 i
Solution: As the given function is f ( z ) ( z 2 8) (0.5 z 1.5i ) . So the pole of the given function is
z 3i , which is a simple poles. As the poles z 3i lies inside the given circle z 4 (since 3i 4 ). So from Eq. 4.58 we have
C f ( z)dz 2 i (sum of residues of the poles which lie inside the given
closed region). As we have only one pole f (z) . C f ( z )dz 2 i Res z 3 i
C
lie within the circle
As
z 4 . So
Res f ( z ) lim ( z 3i) f ( z ) z 3i
z 3i
2
2
z 8 1 2 zlim 0.5 z 1.5i 3i 0.5 0.5 f ( z ) dz 2 i ( 2) 4 i .
Res f ( z ) lim ( z 3i) z 3i z 3i
z 3i
z 8
Example 4.54 [EC-2011 (1 mark)]: The value of the integral
C (3z 4)
( z 2 4 z 5) dz where
C is the circle z 1 is given by (a) 0
(b) 1 10
Solution: As the given function is f ( z )
(c) 4 5 3 z 4
(d) 1 3z 4
. So the pole of the ( z 4 z 5) ( z 2 i )( z 2 i ) given function is z 2 i , which all are simple poles; but both the poles are not les within the circle z 1 , as for both poles we have z 1 . So the sum of the residue of poles which lies inside the 2
given closed region is zero; and thus from Eq. 4.58 we have
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C f ( z)dz 2 i 0 0 . e-mail: [email protected]
Engineering Mathematics
Chapter 4: Complex Variables
Example 4.55 [IN-2011 (1 mark)]: The contour integral unit circle in the z plane is equal to (a) 0 (b) 2 Solution (c): As f ( z ) e1 z 1
1
C e
1z
dz with C as the counter-clockwise (d)
(c) 2 1 1
[4.30]
1
which has isolated essential singularity at z 2! z 3! z 3 z 0 [since we have infinitely many negative terms of ( z 0) ]. So from Eq. 4.54, we have the residue of f ( z ) is the coefficient of the first negative power of ( z 0) , which is 1. So
C e
1z
2
dz 2 i Res 2 i 1 2 i 2 1 . z 0
C z
Example 4.56 [PI-2011 (2 marks)]: The value of
2
( z 4 1) dz using Cauchy’s integral around
the circle z 1 1 , where z x iy is (d) 2 i Solution: As the given function is f ( z ) z 2 ( z 4 1) z 2 {( z 1)( z 1)( z i )( z i )} . So the pole of the given function is z 1, i , which all are simple poles. As the poles z 1, i lies outside the given circle (since 1 1 1 , i 1 1 , i 1 1 ); but the pole z 1 lies within the circle z 1 1 (a) 2 i
(b) i 2
, as 1 1 1 . So from Eq. 4.58 we have
(c) 3 i 2
C f ( z)dz 2 i (sum of residues of the poles which lie
inside the given closed region). As we have only one pole z 1 lie within the circle z 1 1 . So f (z) . C f ( z )dz 2 i Res z 1
As
Res f ( z ) lim ( z 1) f ( z ) z 1
z 1
2
2
z 1 lim z 1 z 1 4 ( z 1)( z 1)( z i )( z i) ( z 1)( z i )( z i ) f ( z)dz 2 i (1 4) i 2 . ( z 1) z
Res f ( z ) lim z 1
C
[Similar questions were also asked in EC-2006, EC-2007 (2 marks), XE-2013 (1 mark)] Example 4.57 [CH-2012 (2 marks)]: If i 1 , the value of the integral
C [(7 z i)
{z ( z 2 1)}]dz
, z 2 using the Cauchy residue theorem is (b) 0 (a) 2 i
(c) 6 (d) 6 7z i 7z i Solution (b): As the given function is f ( z ) . So the pole of the given 2 z ( z 1) z ( z i)( z i) function is z 0, i , which all are simple poles, and all lies within the circle z 2 , since for all poles we have z 2 . So from Eq. 4.58 we have which
lie
inside
the
C f ( z)dz 2 i (sum of residues of the poles
given
closed region). Thus from Eq. 7z i 7z i i Res f ( z ) lim( z 0) f ( z ) lim ( z 0) lim 2 i; 2 z 0 z 0 z 0 z ( z 1) z 4 ( z 1) 1
4.55,
we
have
7 z i 6i 3i zlim z ( z i )( z i) i z ( z i) 2 7z i 7 z i 8i Res f ( z ) lim( z i) f ( z ) lim ( z i) lim 2 4i ; z i z i z i z ( z i )( z i) z i z ( z i ) 2i Res f ( z ) lim ( z i) f ( z ) lim ( z i ) z i z i z i
7z i
So sum of residues of the poles which lie inside the given closed region is i 3i 4i 0 . Thus
C f ( z)dz 2 i (0) 0 . [Similar questions were also asked in XE-2007 (2 marks)]
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Engineering Mathematics
Chapter 4: Complex Variables
[4.31]
Example 4.58 [EC-2012, EE-2012, IN-2012 (1 mark)]: Given f ( z )
1 z 1
2
z3
. If C is a
counter-clockwise path in the z plane such that z 1 1 , the value of {1 (2 i )} f ( z)dz is C
(a) –2
(b) –1
(c) 1
(d) 2
Solution (c): From Eq. 4.58 we have {1 (2 i)} f ( z )dz sum of residues of the poles which lie C
inside the given closed region. As the given function is f ( z )
1
2
z 1
. So the z 1 z 3 ( z 1)( z 3) poles of the given function are z 1, 3 and both are simple poles. As the simple pole z 1 lies within the given region z 1 1 , since 1 1 1 ; also the simple pole z 3 lies outside the given z 1 1,
region
3 1 1
since
1
z 1
f ( z ) lim ( z 1) f ( z ) lim ( z 1) lim f ( z )dz Res z 1 z 1 z 1 2 i C ( z 1)( z 3) z 1
Example 4.59 [EE-2013 (2 marks)]:
( z
2
so
z 1 z 3
2 2
1.
4) ( z 2 4) dz evaluated anticlockwise around the
circle z i 2 , where i 1 , is (b) 0 (a) 4
(c) 2 2
z 4
Solution (a): As the given function is f ( z )
(d) 2 2i 2
z 4
. So the pole of the given z 4 ( z 2i )( z 2i) function is z 2i , which are simple poles. As the pole z 2i lies outside the given circle (since 2i i 2 ); but the pole z 2i lies within the circle z i 2 , as 2i i 2 . So from Eq. 4.58 we 2
C f ( z)dz 2 i (sum of residues of the poles which lie inside the given closed region). As we have only one pole z 2i lie within the circle z i 2 . So f ( z ) dz 2 i Res f ( z ) . As C z 2 i have
Res f ( z ) lim ( z 2i ) f ( z ) lim z 2 i
z 2i
z 2i
( z 2i )( z 2 4) ( z 2i )( z 2i )
Res f ( z ) lim
z2i
z 2i
( z 2 4) z 2i
8 4i
2i .
So
C f ( z)dz 2 i 2i 4 . Example 4.60 [EE-2014 (2 marks)]: Integration of the complex function f ( z ) z 2 ( z 2 1) , in the counter-clockwise direction, around z 1 1 , is (b) 0 (a) i Solution (c): As the given function is f ( z )
z
(d) 2 i
(c) i 2
z
2
. So the pole of the given function z 1 ( z 1)( z 1) is z 1 , which are simple poles. As the pole z 1 lies outside the given circle (since 1 1 1 ); 2
but the pole z 1 lies within the circle z 1 1 , as 1 1 1 . So from Eq. 4.58 we have
C f ( z)dz 2 i (sum of residues of the poles which lie inside the given closed region). As we have only one pole z 1 lie within the circle z 1 1 . So f ( z ) dz 2 i Res f ( z ) . As C z 1 Res f ( z ) lim( z 1) f ( z ) lim z 1
z 1
z 1
( z 1) z 2 ( z 1)( z 1)
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lim z 1
z2 z 1
1 2
. So
1
C f ( z )dz 2 i 2 i .
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Engineering Mathematics
Chapter 4: Complex Variables
[4.32]
Exercise: 4.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill.
C zdz , where C
1. Evaluate the complex integral (a) 4 2i
(b) 4 2i
z 3 3i . (a) 4 6i
(c) 2 i
C zdz ,
2. Evaluate the complex integral
C zdz ,
z 3 3i . (a) 8
(c) 4 6i
z 1 to z i . (a) (1 i) 3
(c) 8i
2
C z dz , where C
(a) 2 i
C (1 z)dz , where C C ( z
6. Evaluate the line integral from z 1 to z i . (a) (4 3) i(1 3)
(c) (1 i) 3 is a unit circle.
(d) i
z ) dz , where C is a part of unit circle in anticlockwise direction
(b) (4 3) i (1 3)
(c) (4 3) i (1 3)
C {1 ( z z0 )}dz , where C
7. Evaluate the line integral
(d) (1 i ) 3
(c) i
(b) 2 i 2
(d) 8i
is a part of unit circle in anticlockwise direction from
(b) (1 i ) 3
5. Evaluate the line integral
(d) 4 6i
where C is the straight line path from z 1 i to
(b) 8
4. Evaluate the line integral
(d) 2 i
where C is the straight line path from z 3 i to
(b) 4 6i
3. Evaluate the complex integral
is the straight line path from z 1 i to z 3 i
is a circle centred at z0 and of any radius.
The path traced out once in the anticlockwise direction. (a) 2 i (b) 2 i (c) i
C z
8. Evaluate the line integral and final point i . (a) (3 2)(1 i)
2
C z
9. Evaluate the line integral
10. Suppose is real, then e (a) 2
(c) (3 2)(1 i )
(c) 1 i
(d) 1 i
(c) 2
(d) 2
1 is
(b) 2
11. If C is the line segment joining ( 1 i ) and (1 i) , then (a) 2
(b) 2
C {e
12. The maximum value of
z
C (12 z
C (1 z
(c) 2
2
) is
(d) 2
2
( z 1)} , where C is the circle z 2 traversed once in the
counter-clockwise direction, is (a) (4 3) e (b) (3 4) e 13. If C
(d) (2 3)(1 i)
dz , where the curve C is the arc of the unit circle Im( z ) 0 with
initial point 1 and final point i . (a) 1 i (b) 1 i 2 i
(d) i
dz , where the curve C is the line segment with initial point 1
(b) (2 3)(1 i ) 2
(d) (4 3) i(1 3)
(c) (3 4) e 2
(d) (4 3) e 2
is the curve y x3 3 x 2 4 x 1 joining points (1,1) and (2,3) , the value of 2
4iz ) dz is
(a) 156 38i 14. Evaluate f ( z )
(b) 156 38i z 3
{e
2z
(c) 156 38i
(d) 156 38i
4
( z 1) }dz
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Engineering Mathematics (a) (8 3) ie 2 15. Evaluate
C
(a) i
Chapter 4: Complex Variables (b) (1 3) ie 2
[4.33] (d) (1 3) ie 2
(c) (8 3) ie 2
sin z 2 cos z 2
dz , where C is the circle z i 3 ( z 1)( z 2) (b) 2 i (c) 3 i
(d) 4 i
2
16. The maximum value of z 3 z 1 in the disk z 1 is (a)
13 at z i
(b)
8 at z 2i
(c)
13 at z 2i
(d)
8 at z i
2
17. The residue at the singularity of the function f ( z ) (1 z ) (1 2 z ) is (b) 1 4
(a) 1 4
(d) 1 2
(c) 1 2 1 z2
18. The residue at the singularity of the function f ( z ) e
is _____.
6
19. The function f ( z ) (1 z ) sin z have a pole, at z 0 , of order _____. 20. Evaluate
C (1 z
6
) sin zdz , where C is a circle of unit radius centred at origin.
(a) i 60
(b) i 120
(c) i 30 (d) i 90 2 1 1 3 21. The singularities of the function f ( z ) 2 are 4 z z z i ( z i) (a) A pole of order 2 at z 0 ; a simple pole at z i ; and a pole of order 4 at z i (b) A simple at z 0, i ; and a pole of order 4 at z i (c) A simple pole at z 0, i ; and a pole of order 2 at z i (d) A pole of order 2 at z 0 ; a simple pole at z i ; and a pole of order 4 at z i 1 1 2 22. The singularities of the function f ( z ) 2 are 2 3 z ( z i) ( z i) (a) A double pole at z 0 and a pole of order 2 at z i (b) A pole at z 0 and a pole of order 3 at z i (c) A double pole at z 0 and a pole of order 3 at z i (d) A pole at z 0 and a pole of order 2 at z i 23. The residue of f ( z ) 1 ( z 2 4) at z 2i and z 2i are, respectively, (a) 1 (4i) , 1 (4i )
(b) 1 (4i ) , 1 (4i) 2
(c) 4i , 4i
(d) 4i , 4i
2
24. The singularity of f ( z ) 1 {z ( z 9)} is (a) A pole of order 4 at z 0 ; Simple pole at z 3i (b) Double pole at z 0 ; Simple pole at z 3i (c) Double pole at z 3i ; Simple pole at z 0 (d) A pole of order 4 at z 3i ; Simple pole at z 0 25. The residue at each of the singularity of the function f ( z ) 1 {z 2 ( z 2 9)} is (a) Res z 3i i 54 , Res z 3i i 54 and Res z 0 i (b) Res z 3i i 54 , Res z 3i i 54 and Res z0 0 (c) Res z 3i i 54 , Res z 3i i 54 and Res z0 0 (d) Res z 3i i 54 , Res z 3i i 54 and Res z 0 i 26. Evaluate the integral (a) 2
C {1 ( z
2
4)}dz , where C is the circle z 2i 1
(b) 2
(d) 0
(c) 2
27. If C is any closed path enclosing both z 2i and z 2i , then 28.
C {1 ( z
2
4)}dz _____.
5z 2
z 3 z ( z 1) dz k i , where k _____.
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Engineering Mathematics
29.
C e
1 z
Chapter 4: Complex Variables
[4.34]
dz k i , where k _____.
30. The solutions of z 3 i is i ( 6) i ( 3) i (3 (a) z e ,e ,e i ( 6)
i (5 6)
2)
(b) z e
i
i ( 6) i ( 6)
, ei (5 6) , ei ( i (5 6)
2)
i (3 2)
(c) z e ,e ,e (d) z e ,e ,e 31. Which of the following function is not analytic? (i) f ( z ) 3 y 3 xi ; (ii) f ( z ) 3 y 5 xi . (a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) Neither (i) nor (ii) 32. The integral
(z
3
5 z sin z ) dz over a unit circle in xy plane evaluates to _____. 8
33. All the solutions of the equation z 1 is 1 1 1 1 i, i (a) z 1, 2i, 2 2 2 2 1 1 1 1 i, i (c) z 1, i, 2 2 2 2 34. Which of the following functions are analytic?
1 1 1 1 (b) z 1, i , i , i 2 2 2 2 1 1 1 1 (d) z 2, 2i, i, i 2 2 2 2
5 iz 2
(i) f ( z ) ze ; (ii) f ( z ) x 2 y ixy 2 ; (iii) f ( z ) x 2 y 2 i (2 xy ) (a) Only (ii) and (iii) (b) Only (i) and (ii) (c) Only (i) and (iii) (d) All (i), (ii) and (iii) 35. Find an analytic function f ( z ) , if such function exists, so that the imaginary part of f ( z ) is equal to v 3 x 2 4 y 3 y 2 . (a) f ( z ) (6 xy 4 x c ) i (3 x 2 4 y 3 y 2 )
(b) f ( z ) (6 xy 4 x c) i (3 x 2 4 y 3 y 2 )
(c) f ( z ) (6 xy 4 x c ) i (3 x 2 4 y 3 y 2 ) where c is any constant.
(d) f ( z ) (6 xy 4 x c ) i(3x 2 4 y 3 y 2 )
36. For the function f ( z ) (1 cos z ) z 2 , then z 0 is (a) a removable singularity (c) an isolated singularity 37. For the function f ( z ) z cos(1 z ) , then z 0 is (a) a removable singularity (c) an isolated singularity 38. Evaluate the complex integral of given function (1, 0) , (0,1) , ( 1, 0) and (0, 1) . (a) i (b) i
(b) an essential singularity (d) a pole of order 2 (b) an essential singularity (d) a pole of order 2 f ( z ) z cos(1 z ) over the square with vertices (c) 2 i
(d) 2 i
Answer Keys 1 0 16 b
1 a 16 a 31 b
2 a 17 b
2 b 17 b 32 0
3 0 18 c
3 c 18 0 33 c
4 d 19 b
4 d 19 5 34 c
5 c 20 b
Answer Keys: Exercise: 4.1 6 7 8 9 10 d d 1 a d 21 22 3 c
5 a 20 a 35 d
Answer Keys: Exercise: 4.2 6 7 8 9 10 d a b c a 21 22 23 24 25 d c a b c 36 37 38 a b b
Copyright © 2016 by Kaushlendra Kumar
11 d
11 b 26 a
12 a
12 d 27 0
13 a
13 c 28 10
14 a
15 a
14 a 29 1.414
15 d 30 d
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Engineering Mathematics
Chapter – 5: Numerical Methods
Chapter – 5 [1]
GATE – 2016: Chapter – 5: Numerical Methods Note: The following questions came in GATE – 2016 were based on Numerical Methods Chapter – 5. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to carefully go through the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 1. Newton-Raphson method is to be used to find root of equation 3 x e x sin x 0 . If the initial trial value for the root is taken as 0.333, the next approximation for the root would be _____. (note: answer up to three decimal) [CE-2016 (1 mark)] x x Solution: The given function is f ( x ) 3 x e sin x f ( x) 3 e cos x . So with initial approximate x1 x0
x0 0.333 ,
root
the
next
approximated
roots
is
find
by
f ( x0 ) 3(0.333) e 0.333 sin(0.333) 0.333 0.360 . 0.333 f ( x0 ) 3e cos(0.333)
[Similar question was also asked in ME-2016 (1 mark)] 2. Which one of the following is an iterative technique for solving a system of simultaneous linear algebraic equations? [CH-2016 (1 mark)] (a) Gauss elimination (b) Gauss-Jordon (c) Gauss-Seidel (d) LU decomposition Solution (c): Gauss Elimination, Gauss-Jordon and LU decomposition belong to the direct methods for solving systems of linear equations; these are methods that yield solutions after an amount of computations that can be specified in advance. But Gauss-Seidel is an iterative methods in which we start from an initial value and obtain better and better approximations from a computational cycle repeated as often as may be necessary. So option (c) is correct. 2
3. Consider the first order initial value problem y y 2 x x , y (0) 1 , (0 x ) , with exact 2
x
solution y ( x ) x e . For x 0.1 , the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-Kutta method with step-size [EC-2016 (1 mark)] h 0.1 is _____. 2 0.1 Solution: For exact solution: y (0.1) (0.1) e 1.1152 For solution using second-order Runge-Kutta method, which is improved Euler’s method, with stepsize with initial guess x0 0 and y0 1 , is given by h 0.1 , 2
y1 y0 (1 2) h f ( x0 , y0 ) f ( x0 h, y1* ) , where y1* y0 hf ( x0 , y0 ) and f ( x, y ) y 2 x x .
As y1* y0 hf ( x0 , y0 ) 1 (0.1) f (0,1) 1 0.1{1 2(0) 0 2 } 1.1 , so y1 1 (1 2)(0.1) f (0,1) f (0.1,1.1) 1 (1 2)(0.1) {1 2(0) 02 } {1.1 2(0.1) 0.12 } 1.1145 Exact Approx. 1.1152 1.1145 yExact 1.1152 , yApprox. 1.1145 , % Error 100 100 0.063% Exact 1.1152
4. Solve the equation x 10 cos( x) using the Newton-Raphson method. The initial guess is x 4 . The value of the predicted root after the first iteration, up to second decimal, is _____. [ME-2016 (1 mark)] Solution: The given function is f ( x) x 10 cos x f ( x) 1 10sin x . So with initial approximate root
x0 4 ,
the
next
approximated
roots
is
given
by
f ( x0 ) ( 4) 10 cos( 4) 6.285 1.5644 . f ( x0 ) 4 1 10 sin( 4) 4 8.071 [Similar question was also asked in AE-2016, MT-2016, PI-2016 (2 marks)] x1 x0
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 5: Numerical Methods
Chapter – 5 [2]
5. Numerical integration using trapezoidal rule gives the best result for a single variable function, which is [ME-2016 (1 mark)] (a) Linear (b) parabolic (c) logarithmic (d) hyperbolic Solution (a): The Trapezoidal rule gives the exact result when applied to any function whose second derivative is identically zero, i.e., the Trapezoidal rule gives the exact result for polynomials of degree up to or equal to one, which is linear polynomial. So option (a) is correct. 6. Which of the following is a multi-step numerical method for solving the ordinary differential equation? [TF-2016 (1 mark)] (a) Euler method (b) Improved Euler method (c) Runge-Kutta method (d) Adams-Multon method Solution (d): The Euler method, Improved Euler method and Runge-Kutta method are not multi-step numerical method for solving the ODEs. So, option (d) is correct. 7. Integration by trapezoidal method of log10 ( x) with lower limit of 1 to upper limit of 3 using seven distinct values (equally covering the whole range) is _____. [AG-2016 (2 marks)] b f ( x) xi i Solution: Using trapezoidal rule in [1, 3] , I f ( x ) dx , where x0 a 1 , a 0 1.00 0.000 x8 b 3 ; as we have to integrate using seven distinct values (equally covering 1 1.25 0.097 the whole range) so n 7 1 8 , hence the step size is 2 1.50 0.176 3 1.75 0.243 h (b a ) n (3 1) 8 0.25 . We have f ( x ) log10 ( x ) , and the values of 4 2.00 0.301 f ( x ) at different points are given in the table thus 5 2.25 0.352 b I f ( x )dx h (1 2) f ( x0 ) f ( x1 ) f ( x2 ) f ( x7 ) (1 2) f (b) 6 2.50 0.398 a 7 2.75 0.439 3 I f ( x ) dx (0.25) (1 2) f (0) f (1.25) f (2.75) (1 2) f (3.00) 8 3.00 0.477 1
I (0.25) (0.5)(0) 0.097 0.176 0.243 0.301 0.352 0.398 0.439 (0.5)(0.477) 0.5611 3
2
8. The quadratic approximation of f ( x) x 3 x 5 at the point x 0 is 2
2
(a) 3 x 6 x 5 (b) 3 x 5 Solution (b): For quadratic approximation of
[CE-2016 (2 marks)] (c) 3 x 6 x 5 (d) 3 x 2 5 f ( x ) about the point x a is given by 2
( x a) 2
f ( a) . We have f ( x) x 3 3 x 2 5 and a 0 . So 2! f ( x) x 3 3 x 2 5 f ( x) 3x 2 6 x f ( x) 6 x 6 ; so f (0) 5 , f (0) 0 and f (0) 6
g ( x) f ( a) ( x a) f ( a)
Hence g ( x ) f (0) ( x 0) f (0)
( x 0) 2 2!
f (0) 5 x (0)
x2 2
(6) 3x 2 5
9. Value of f ( x) in the interval [0, 4] are given below. Using Simpson’s 1/3 rule with a step size of 1, the numerical approximation (rounded off to the second decimal place) of 4
0
0 3
1 10
2 21
3 36
4 55
[CH-2016 (2 marks)]
f ( x ) dx is _____.
Solution: Simpson’s 1/3 rule:
x f ( x)
b
a
f ( x) dx (h 3) f ( x0 ) 4 f ( x1 ) 2 f ( x2 ) 4 f ( x3 ) f ( x4 ) . So we
have a 0 , b 4 , n 4 , so step size is h (b a) n (4 0) 4 1 ; and xi 1 xi h . Thus 4
0
f ( x ) dx (1 3) f (0) 4 f (0 1) 2 f (1 1) 4 f (2 1) f (3 1) 4
f ( x) dx (1 3) 3 4(10) 2(21) 4(36) 55 94.66 0
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 5: Numerical Methods
Chapter – 5 [3]
dx
3 x 2 , with x (0) 1 is to be solved using the forward dt Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is _____. [EC-2016 (2 marks)] Solution: From the given data we have f (tn , xn ) ( dx dt ) (t , x ) ( 3xn 2) , x(0) 1 and let step
10. The ordinary differential equation
n
n
size be h . So using forward Euler method, we have xn1 xn h f (tn , xn ) xn h ( 3xn 2) , where
xi 1 xi h xi x0 ih ; xn1 (1 3h) xn 2h ;
Thus
hence
the
solution
of
differential
equation
is
stable
if
1 3h 1 1 1 3h 1 0 h (2 3) 0 h 0.66 . So the largest time step that can be used to solve the given DE without making the numerical solution unstable is 0.66. 11. For the fixed point iteration xk 1 g ( xk ) , k 0,1, 2, , consider the following statements P and Q: ( P ) : If g ( x ) 1 (2 x ) then the fixed point iteration converges to 2 for all x0 [1,100] . (Q) : If g ( x ) 2 x then the fixed point iteration converges to 2 for all x0 [0,100] . Which of the above statements hold TRUE? [MA-2016 (2 marks)] (a) both P and Q (c) only Q (d) neither P nor Q (b) only P Solution: For fixed point iteration, we know that if x be a root of f ( x ) 0 and let I be an interval containing the point x . Let g ( x ) and g ( x) be continuous in I , where g ( x ) is defined by the equation x g ( x) which is equivalent to f ( x ) 0 . Then if g ( x) 1 for all x in I , the
sequence of approximations x0 , x1 , x2 , , xn defined by xn1 g ( xn ) converges to the root , provided that the initial approximation x0 is chosen in I . In the given question, for both the statements g ( x) 1 for all x belongs to given interval. So both the statements are true. 12. The error in numerically computing the integral
0 (sin x cos x) dx
using the trapezoidal rule
with three intervals of equal length between 0 and is _____. [ME-2016 (2 marks)] Solution: Using trapezoidal rule in [0, ] with three intervals, so n 3 , f ( x) xi i b
I f ( x ) dx , a
where
a0,
b ;
with
step
size
h (b a ) n ( 0) 3 3 , also xi 1 xi h and f ( x ) sin x cos x so b
I f ( x )dx h (1 2) f ( a ) f ( x1 ) f ( x2 ) (1 2) f (b) , a
0 1
0 3
1 1.366
2 3
2 3
0.366 1
where x0 a 0 , x5 b .
Thus I f ( x) dx ( 3) (1 2) f (0) f ( 3) f (2 3) (1 2) f ( ) 0
I f ( x) dx ( 3) (1 2)(1) 1.366 0.366 (1 2)(1) 1.814 , which is an approximate solution. 0
For exact solution:
0 (sin x cos x)dx ( cos x sin x)0 {(1) 0} {1 0} 2
So Error Exact value Approximate value 2 1.814 0.186 . 13. Gauss-Seidel method is used to solve the following equation (as per the given order): x1 2 x2 3 x3 5 ; 2 x1 3 x2 x3 1 ; 3 x1 2 x2 x3 3 . Assuming initial guess as
x1 x2 x3 0 , the value of x3 after the first iteration is _____.
Copyright © 2016 by Kaushlendra Kumar
[ME-2016 (2 marks)]
e-mail: [email protected]
Engineering Mathematics
Chapter – 5: Numerical Methods
Chapter – 5 [4]
x1 2 x2 3 x3 5 Solution: Let the system of equation be 2 x1 3 x2 x3 1 and let the initial values be
3x1 2 x2 x3 3 x1(0)
x2(0)
x3(0)
0.
So using Gauss-Seidal
iteration,
we have
the first
iteration as:
x1(1) (5 2 x2(0) 3 x3(0) ) 5 x2(1) (1 2 x1(1) x3(0) ) 3 3 ; x3(1) (3 3 x1(1) 2 x2(1) ) 6 dT
cT , where T is the instantaneous dt 1 temperature at time t , and the constant c 0.05s . Reduce the differential equation into its finite difference from using forward difference. For maintaining numerical stability, the maximum value of the time step t (in seconds) is _____. [MT-2016 (2 marks)] dT Ti 1 Ti Solution: Using forward difference f (ti ) Ti . So after converting the given DE into dt ti 1 ti
14. A hot body cools according to the following equation
its
finite
difference from
using
forward
difference,
we
get
Ti 1 Ti ti 1 ti
cTi ,
where
ti 1 ti t ti 1 ti t value of the time step. So Ti 1 Ti ctTi Ti 1 (1 ct )Ti ; hence the solution of DE is stable if (1 ct ) 1 1 1 ct 1 0 t (2 c) , hence the maximum value of the time step t (in seconds) is 2 c 2 0.05 40 . 15. For a function f ( x) , the values of the function in the interval [0,1] are given in the table below. The value of the integral 1
0 f ( x)dx
x 0.0 0.2 0.4 0.6 0.8 1.0
according to the trapezoidal rule
is _____. [PE-2016 (2 marks)]
f ( x) 1.0 1.24 1.56 1.96 2.44 3.0
b
Solution: Using trapezoidal rule in [0,1] , I f ( x ) dx , where x0 a 0 , a
x5 b 1 ; and in the table, the step size is h 0.2 , also xi 1 xi h so b
I f ( x) dx h (1 2) f ( a) f ( x1 ) f ( x2 ) f ( x3 ) f ( x4 ) (1 2) f (b) , thus a 1
I f ( x ) dx (0.2) (1 2) f (0) f (0.2) f (0.4) f (0.6) f (0.8) (1 2) f (1) 0 1
I f ( x ) dx (0.2) (1 2)(1) 1.24 1.56 1.96 2.44 (1 2)3 1.84
i 0 1 2 3 4 5
xi
f ( x)
0.0 0.2 0.4 0.6 0.8 1.0
1.0 1.24 1.56 1.96 2.44 3.0
0
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.1]
Chapter 5 : Numerical Methods Numerical methods are used to solve problems on computers or calculators by numeric calculations, resulting in a table and/or graphical representation (figures). Before going to discuss the numerical methods, let us discuss the following terms:
Significant Digits: The digits that are known reliably plus the first uncertain digit are known as significant digits or figures. For e.g., when a measured distance is reported to be 364.7 m, it has four significant digits 3, 6, 4 and 7; the digits 3, 6 and 4 are certain and reliable, but the digit 7 is uncertain. Following are some of the common rules for counting significant digits in a reported measurement: 1. All non-zero digits are significant. For e.g., x 1236 and x 1.236 both have 4 significant digits. 2. All zeros occurring between two non-zero digits are significant digits, no matter where the decimal point is in the number. For e.g., x 1007 and x 1.0809 have 4 and 5 significant digits, respectively. 3. If the number is less than one, the zero(s) on the right of the decimal point and to left of first nonzero digit are not significant. For e.g., x 0.005704 have 4 significant digits. 4. In a number without a decimal point, the terminal or trailing zeros are not significant. For e.g., x 3210 have 3 significant digits. 5. In a number with decimal point, the trailing zeros are significant digits. For e.g., x 3.210 has 4 significant digits. 6. Change of units does not change the number of significant figures in a measurement. For e.g., the length x 5.608cm 56.08mm cm has 4 significant digits 7. The digit 0 which is put on the left of a decimal for a number less than 1 is never significant. For e.g., 0.3570 has 4 significant digits. Note: Referring to rules 4, 5 and 6, there can be some confusion regarding the trailing zeros. Suppose a measured length is reported as x 4.700 m, which has four significant figures. On changing the units, we can rewrite the same length as x 0.004700 km or x 470.0 cm or x 4700 mm. As per rule 4, we conclude that x 4700 m has 2 significant digits. The fact that x 4700 mm has 4 significant digits, as change of unit cannot change the number of significant digits. To remove such ambiguities in determining the number of significant digits, the best way is to report every measurement in scientific notation, i.e., in power of 10, often putting the decimal after the first digit. For e.g. x 4.700m 4.700 10 2 cm 4.700 10 3 mm 4.700 10 3 km , in each case, the number of significant figures is 4, as the power of 10 is irrelevant to the determination of significant figures; hence the confusion disappears.
Rounding Off of Numbers: The result of computation with approximate numbers, which contain more than one uncertain digit should be rounded off. If a number is to be rounded off to n significant digits, then we follow the following rules: 1. Discard all digits to the right of the nth digit if ( n 1) th digit is less than 5. For e.g., x 7.82 is rounded off to two significant digits as 7.8. 2. If the ( n 1) th digit is greater than 5 or it is 5 followed by a nonzero digit, then nth digit is increased by 1. For e.g., x 6.87 is rounded off to two significant digits as 6.9. If the ( n 1) th digit is less than 5, then digit remains unchanged. For e.g., x 6.84 is rounded off to two significant digits as 6.9. 3. If the ( n 1) th digit is 5 and is followed by zero or zeros, then nth digit is increased by 1 if it is odd and it remains unchanged if it is even. For e.g., x 3.250 is rounded off to two significant digits as 3.2; x 3.350 is rounded off to two significant digits as 3.4.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.2]
Error due to rounding off numbers: If a number is rounded off according to the rules, the maximum error due to rounding does not exceed the one half of the place value of the last retained digit in the number. The difference between a numerical value X and its rounded value X 1 is called round off error is given by E X X 1 .
Truncation and Error due to Truncation of Numbers: Leaving out the extra digits that are not required in a number without rounding off, is called truncation or chopping off. The difference between a numerical value X and its truncated value X 1 is called truncation error and is given by E X X 1 . The maximum error due to truncation of a number cannot exceed the place value of the last retained digit in the number. Approximated number In truncation the numerical value of a ve number is obtained by decreased and that of a negative number is increased. Number Chopping Rounding If we round off a large number of ve numbers to the off off same number of decimal places, then the average error due to rounding off is zero. 0.335217... 0.3352 0.3352 In case of truncation of a large number of ve 0.666666... 0.6666 0.6667 numbers to the same number of decimal places the 0.123451... 0.1234 0.1235 average truncation error is one half of the place value 0.213450... 0.2134 0.2134 of the last retained digit. 0.213950... 0.2139 0.214 If the number is rounded off and truncated to the same number of decimal places, then truncation error 0.335750... 0.3357 0.3358 is greater than the round off error. 0.999999... 0.9999 1 Round of error may be ve or ve but truncation 0.555555... 0.5555 0.5556 error is always ve in case of ve numbers and ve
in case of ve numbers. Relative and Percentage Error of Numbers: The difference between the exact value of a number X and its approximate value X 1 , obtained by rounding off or truncation, is known as absolute error. The quantity
X X1
X is called the relative error and is denoted by
ER X X 1
X X
X . This is a dimensionless quantity. The quantity
known as percentage error and is denoted by E p , i.e. E p X
X
ER . Thus
X 100 is
X 100 .
If a number is rounded off to n decimal digits, then | E R | 0.5 10 n 1
If a number is truncated to n decimal places, then | E R | 10 n 1
Example 5.1 [EC-2013 (1 mark)]: The maximum value of until which the approximation sin holds to within 10% error is (a) 10o (b) 18o (c) 50o (d) 90o Solution (b): 10o 0.1745c ; 18o 0.3142 c ; 50o 0.8727 c ; 90o 1.5708c . As sin is the exact value and is the approximate value. So percentage error, E p sin sin .
o
So, for option (a), E p sin10 0.1745 sin10
sin 50 sin 90 o
o
100 0.4905% 10% .
100 1.6773% 10% . sin 50 100 13.9229% 10% . sin 90 100 57.0800% 10% . o
For option (b), E p sin18 0.3142 sin18 For option (c), E p For option (d), E p
o
0.8727
o
1.5708
o
o
Thus maximum value of until which the approximation sin holds to within 10% error is 18o .
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
5.1
Chapter 5: Numerical Methods
[5.3]
Algebraic and Transcendental Equation
An equation of the form f ( x ) 0 , is said to an algebraic or a transcendental equation according as
f ( x ) is a polynomial or a transcendental function respectively. For e.g., ax 2 bx c 0 , 3 2 x ax bx cx d 0 etc., where a, b, c, d Q , are algebraic equations whereas ae b sin x 0 ; a log x bx 3 etc. are transcendental equations.
5.1.1 Location of real roots of an equation By location of a real root of an equation, we mean finding an approximate value of the root graphically or otherwise. Graphical Method: It is often possible to write f ( x ) 0 in the form f1 ( x ) f 2 ( x) and then plot the graphs of the functions y f1 ( x ) and y f 2 ( x ) . The abscissae of the points of intersection of these two graphs are the real roots of f ( x ) 0 .
Figure 5.2: Location Theorem
Figure 5.1: Graphical Method
Location Theorem: Let y f ( x ) be a real-valued, continuous function defined on [ a, b] . If f (a ) and f (b) have opposite signs i.e., f ( a ) f (b) 0 , then the equation f ( x ) 0 has at least one real root between a and b . Position of real roots: If f ( x ) 0 be a polynomial equation and x1 , x2 , , xk are the consecutive real roots of f ( x ) 0 , then positive or negative sign of the values of f ( ), f ( x), , f ( xk ), f () will determine the intervals in which the root of f ( x ) 0 will lie
whenever there is a change of sign from f ( xr ) to f ( xr 1 ) the root lies in the interval [ xr , xr 1 ] .
5.1.2 Exact Solution of Algebraic and Transcendental Equation We begin with methods of finding solutions of a single equation f ( x ) 0 . Firstly we derive the formulas for solving the exact solution of quadratic and cubic equations.
+
Exact solution of Quadratic Equation ( ax 2 bx c 0 x 2
b a
x
b2 4a 2
b2 4a 2
c a
+
= ,
≠ , , , )
2
0x
b b 2 4ac b b 2 4ac x . 2a 4a 2 2a
So, the quadratic equation has two roots, and depending on the value of the discriminant, b 2 4ac , the equation may have real, complex or repeated roots. If b 2 4ac 0 , both roots are complex conjugate; If b 2 4ac 0 , the roots are real and equal; If b 2 4ac 0 , the roots are real and distinct.
Exact solution of Cubic Equation ( 3
+
+
+
= ,
≠ , , , , ): To find
the roots of ax bx cx d 0 , we first get rid of the coefficient of x 2 by making the substitution x y
b 3a
2
to obtain, a y
3
2
b b b b y c y d 0 . 3a 3a 3a
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Engineering Mathematics
3
ay c
b2
y d
3a
Chapter 5: Numerical Methods
2b 3 27a
y 3 ey f 0 , where e
2
1
bc 3a
c
a
3
0 y
b2
and f
3a
1
c
b2
y
a
3a
1
2b3
d
a
27a
2
[5.4]
1
d
a
2b 3 27 a
2
bc
0
3a
bc
.
3a
3
Now substituting y z ( s z ) in y 3 ey f 0 z ( s z ) e z ( s z ) f 0 z 6 (3s e) z 4 f z 3 s (3s e) z 2 s 3 0 . Now substituting s e 3 we get, z 6 f z 3 (e 3 27) 0 w 2 fw (e3 27) 0 (By substituting w z 3 ). Now we have a quadratic equation and once we obtain the solution to this quadratic equation, back substituting using the previous substitutions ( w z y x ) to obtain the roots to the general cubic equation. Eq. w 2 f w (e3 27) 0 gives two roots for w ; and using w z 3 we have three roots for each of the two roots of w , so we have six values of z which gives six values of y ; but three values of y will be identical to the other three values. So we get only three values of y and hence three values of x .
5.1.3 Numerical Solution of Algebraic and Transcendental Equation When there is no formula for the exact solution of f ( x ) 0 , we can use an approximation method, in particular an iteration method, i.e., the method in which we start from an initial guess x0 (which may be poor) and compute step by step (in general better and better) approximations x1 , x2 , of an unknown solution of f ( x ) 0 . There are many numerical methods for solving algebraic and transcendental equations. Some of these methods are given below.
The Bisection Method: This method is based on the theorem which states that if a function f ( x ) is continuous between a and b , and f (a ) and f (b) are of opposite signs, then there exists at least one root between a and b . For definiteness, let f (a ) be negative and f (b) be positive. Then the root lies between a and b and let its approximate value be given by x0 (a b) 2 . If f ( x0 ) 0 , we conclude that x0 is a root of the equation f ( x ) 0 . Otherwise, the root lies either between x0 and b if f ( x0 ) is negative, or between x0 and a if f ( x0 ) is positive. We designate this new interval as [ a1 , b1 ] whose length is b a 2 . As before, this is bisected as x1 and the new interval will be Figure 5.3: Graphical representation of Bisection exactly half the length of the previous one. We repeat method this process until the latest interval (which contains the root) is as small as desired, say . It is clear that the interval width is reduced by a factor of one – half at each step and at the end of the nth step,
the new interval will be [ an , bn ] of length b a 2 n , We then have b a 2n which gives
n log e b a log e 2
(5.1)
The inequality 5.1 gives the number of iterations required to achieve an accuracy . For example, if b a 1 and 0.001 then Eq. 5.1 results n 10 . The method is shown graphically in Fig. 5.3. It should be noted that this method always succeed. If there are more roots in an interval then bisection method find one of the roots. It can be easily programmed using the following computational steps: 1. Choose two real numbers a and b such that f ( a ) f (b) 0 .
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2. Set xr ( a b) 2 . If f ( a ) f ( xr ) 0 , the root lies in the interval ( a, xr ) . Then set b xr and go to step 2. If f ( a ) f ( xr ) 0 , the root lies in the interval ( xr , b) . Then set a xr and go to step 2. If f ( a ) f ( xr ) 0 , it means xr is a root of the equation f ( x ) 0 and the computation may be terminated. In practical problems, the root may not be exact so that condition f ( a ) f ( xr ) 0 is never satisfied. In such a case, we need to adopt a criterion for deciding when to terminate the computations. A convenient criterion is to compute the percentage error r defined by
r ( xr xr ) xr 100%
(5.2)
where xr is the new value of xr . The computation can be terminated when r becomes less than a prescribed tolerance, say, p .
The iteration using bisection method always produces a root, since the method brackets the root between two values. So the Bisection Method, used for finding roots of functions, guaranteed to work for all continuous functions. [This point was asked in CS-1998 (1 mark)]. This method of approximation is very slow but it is reliable and can be applied to any type of algebraic or transcendental equations. Bisection method cannot be applied over an interval where there is a discontinuity. As iterations are conducted, the length of the interval gets halved. So one can guarantee the convergence in case of the solution of the equation. The Bisection Method is simple to program in a computer. Bisection method cannot be applied over an interval where the function takes always values of the same sign. The Bisection method fails to determine complex roots. If one of the initial guesses a0 or b0 is closer to the exact solution, it will take larger number of iterations to reach the root.
Example 5.2: Find a positive root of the equation xe x 1 , which lies between 0 and 1. Solution: Let f ( x ) xe x 1 . Since f (0) 1 and f (1) 1.718 , it follows that a root lies between 0 and 1. Thus, x0 0.5 . Since f (0.5) is negative, it follows that the root lies between 0.5 and 1. Hence the new root is 0.75, i.e., x1 0.75 . Using the values of x0 and x1 , we calculate 1 , as
1 ( x1 x0 ) x1 100 33.33% . Proceeding in this way, the following table is constructed where only the sign of the function value is indicated. The prescribed tolerance is 0.05%. f ( x) n a x b 1 2 3 4 5 6 7 8 9 10 11 12
0 0.5 0.5 0.5 0.5625 0.5625 0.5625 0.5625 0.5664 0.5664 0.5664 0.5669
1 1 0.75 0.625 0.625 0.5938 0.5781 0.5703 0.5703 0.5684 0.5674 0.5674
0.5 0.75 0.625 0.5625 0.5938 0.5781 0.5703 0.5664 0.5684 0.5674 0.5669 0.5671
– 0.1756 0.5877 0.1676 0.0129 – 0.0753 0.0305 0.0087 – 0.0020 0.0035 0.0007 – 0.0006 – 0.0001
r (%) – 33.33 20.00 11.11 5.263 2.707 1.368 0.688 0.352 0.176 0.088 0.035
Thus after 12 iterations, r , finally satisfies the prescribed tolerance. Hence the required root is 0.567 and it is easily seen that this value is correct to three decimal places.
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Example 5.3 [CS-2012 (2 marks)]: The bisection method is applied to compute a zero of the function f ( x ) x 4 x 3 x 2 4 in the interval [1, 9] . The method converges to a solution after …… iterations. Solution: As f (1) 14 13 12 4 1 0 and f (9) 9 4 93 9 2 4 0 so using bisection method we have a solution in [1, 9] and thus x0 (1 9) 2 5 . Now f (5) 54 53 52 4 0 so we have a solution in [1, 5] and thus x1 (1 5) 2 3 . Now f (3) 34 33 32 4 0 , so we have a solution in [1, 3] and thus x2 (1 3) 2 2 . Now f (2) 24 23 2 2 4 0 and so x 2 is the solution, thus after 3 iterations of Bisection method, it converge to a solution. So answer is 3.
The Method of False Position: This is the oldest method for finding the real root of a non – linear equation f ( x ) 0 and closely resembles the bisection method. In this method, also known as regula falsi or the method of chords, we choose two points a and b such that f (a ) and f (b) are of opposite signs. Hence, a root must lie in between these points. Now, the equation of the chord joining the two points [ a, f ( a )] and [b, f (b)] is given by, y f (a ) f (b ) f ( a ) (5.3) xa ba The method consists in replacing the part of the curve between the points [ a, f ( a )] and [b, f (b)] by means of the chord joining these points, and taking the point of intersection of the chord with the x axis as an approximation to the root. The point of intersection in the present case is obtained by putting y 0 in Eq. 5.3. Thus we obtain, x1 a
f ( a) f (b ) f ( a )
(b a )
a f ( b) b f ( a ) f (b) f ( a)
(5.4)
which is the first approximation to the root of f ( x ) 0 . If now f ( x1 ) and f (a ) are of opposite signs, then the root lies between a and x1 , and we replace b by x1 in Eq. 5.4 and generate the next approximation. The procedure is repeated till the root is obtained to the desired accuracy. Fig. 5.4 gives a graphical representation of the method. The error criterion Eq. 5.2 can be used in this case also. The method may give a false root or may not converge if either a and b are not sufficiently close to each other or f ( x ) is discontinuous on [ a, b] . Geometrically speaking, in this method, part of the curve between the points A a, f ( a) and B b, f (b) is replaced by the secant AB and the point of intersection of this secant with x axis gives an approximate value of the root. Figure 5.4: Method of False Position It converges more rapidly than bisection. Example 5.4: Find a real root of the equation f ( x ) x 3 2 x 5 0 by using false position method. Solution: We find f (2) 1 and f (3) 16 . Hence a 2 , b 3 , and a root lies between 2 and 3. Eq. 5.4 gives x1 {2(16) 3( 1)} {16 ( 1)} 35 17 2.0588 . Now f ( x1 ) 0.3908 and hence the root lies between 2.0588 and 3.0. Using Eq. 5.4, we get, x2 {2.0588(16) 3( 0.3908)} 16.3908 2.0813 . Since f ( x2 ) 0.1472 , it follows that the root lies between 2.0813 and 3.0. Proceeding in this way, we obtain successively: x3 2.0896, x4 2.0937, x5 2.0939, x6 2.0943, x7 2.0945, . The correct value is 2.0945…, so that x7 is correct to five significant figures.
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Fixed-Point Iteration for Solving Equation ( ) = : We have so far discussed root – finding methods, which require the interval in which the root lies. We now describe methods which require one or more starting values of x . These values need not necessarily bracket the root. The first iteration method, which requires one starting value of x . To describe this method for finding the roots of the equation f ( x) 0 (5.5) we re-write this equation in the form as, x g ( x) (5.6) (certainly g ( x) is non-constant). There are many ways of doing this. For e.g., the equation 3 2 3 12 2 13 x x 1 0 can be expressed as either of the forms: x ( x 1) , x ( x 1) , …. Let x0 be an approximate value of the desired root . Substituting it for x on the right hand of Eq. 5.6, we obtain
the first approximation as, x1 g ( x0 ) then x2 g ( x1 ) , and in general, xn 1 g ( xn ) , n 0,1, 2,
(5.7)
Some questions are now arise: 1. Does the sequence of approximations x0 , x1 , , xn always converge to some number ? 2. If it does, will be a root of the equation x g ( x ) ? 3. How should we choose in order that the sequence x0 , x1 , , xn converges to the root? The answer to the first question is negative. As an example we consider the equation x 2 x 1 . If we take x0 0, x1 2, x2 5, x3 33, , etc. and as n increases xn , increases without limit. Hence the sequence x0 , x1 , , xn does not always converge and later in this section we state the conditions which are sufficient for the convergence of the sequence. The second question is easy to answer, for consider the equation xn 1 g ( xn ) which gives the relation between the approximations at the nth and ( n 1)th stage. As n increases, the left side tend to the root , and if g is continuous the right hand side tend to g ( ) . Hence, in the limit, we have g ( ) which shows that is a root of the equation x g ( x ) . The answer to the third question is given as: Let x be a root of f ( x ) 0 and let I be an interval containing the point x . Let g ( x ) and g ( x ) be continuous in I , where g ( x ) is defined by the equation x g ( x ) which is equivalent to f ( x ) 0 . Then if g ( x ) 1 for all x in I , the sequence of approximations x0 , x1 , x2 , , xn defined by xn 1 g ( xn ) converges to the root , provided that the initial approximation x0 is chosen in I . Example 5.5: Find the solution of f ( x ) x sin x (1 2) 0 by fixed iteration method. Solution: The equation f ( x ) x sin x (1 2) 0 can be written as, x sin x (1 2) x g ( x ) g ( x ) sin x g ( x ) cos x g ( x) 1 x n , n I . So we have convergence for any x0 . Choosing x0 1 and applying xn 1 g ( xn ) , n 0,1, 2, , we get, x1 1.342, x2 1.473, x3 1.495, x4 1.497, x5 1.497 . Hence for iterative process converges to 1.497.
x g ( x ) sin x (1 2) , the
Newton-Raphson Method: This method is generally used to improve the result obtained by one of the previous methods. Let x0 be an approximate root of the equation f ( x ) 0 and let x1 x0 h be the exact root, so that f ( x1 ) 0 . Expanding f ( x0 h) by Taylor's series as, f ( x0 ) hf ( x0 ) ( h 2 2!) f ( x0 ) 0 . Since h is small, neglecting h 2 and higher powers of h , we
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get f ( x0 ) hf ( x0 ) 0 or h f ( x0 ) f ( x0 ) . A closer approximation to the root is given by x1 x0 f ( x0 ) f ( x0 ) . Similarly, starting with x1 , a still better approximation x2 is given by x2 x1 f ( x1 ) f ( x1 ) .
In general, xn 1 xn f ( xn ) f ( xn ) (5.8) which is known as the Newton-Raphson formula or Newton's iteration formula. Convergence of Newton – Raphson method: If we compare Eq. 5.8 with the relation xn 1 g ( xn ) of iterative method, we get, g ( x ) x f ( x ) f ( x ) which gives, f ( x ) f ( x ) g ( x ) (5.9) f ( x) 2 To examine the convergence we assume that f ( x ) , f ( x ) and f ( x ) are continuous and bounded on any interval containing the root x of the equation f ( x ) 0 . If is a simple root, then f ( x ) for some 0 in a suitable neighbourhood of . Within this 2 neighbourhood we can select an interval such that f ( x) f ( x) and this is possible since f ( ) 0 and since f ( x ) is continuously twice differentiable. Hence, in this interval we have, f ( x ) 1 (5.10)
Therefore, the Newton – Raphson formula (Eq. 5.8) converges, provided that the initial approximation x0 is chosen sufficiently close to . When is a multiple root, the Newton – Raphson method still converges but slowly. To obtain the rate of convergence of the method, we note that f ( ) 0 so that Taylor’s expansion gives f ( xn ) ( xn ) f ( xn ) (1 2)( xn ) 2 f ( xn ) 0 from which we get, f ( xn ) 1 f ( xn ) ( xn ) ( xn ) 2 f ( xn ) 2 f ( xn )
From Eq. 5.8 and Eq. 5.11, xn 1 (1 2)( xn ) 2 f ( xn ) f ( xn ) 2 n
Setting, n xn in Eq. 5.12, n 1 (1 2) f ( ) f ( ) Hence, the Newton – Raphson process has a second order or quadratic convergence.
(5.11) (5.12) (5.13)
Geometrical Interpretation: Let x0 be a point near the root of the equation f ( x ) 0 . Then the equation of the tangent at A0 [ x0 , f ( x0 )] is y f ( x0 ) f ( x0 )( x x0 ) . It cuts the x axis
at x1 x0 f ( x0 ) f ( x0 ) which is a first approximation to the root . If A1 is the point corresponding to x1 on the curve, then the tangent at A1 will cut the x axis of x2 which is nearer to and is, therefore, a second approximation to the root, as shown in Fig. 6.5. Figure 5.5: Newton’s Raphson Method Repeating this process, we approach to the root quite rapidly. Hence the method consists in replacing the part of the curve between the point A0 and the x axis by means of the tangent to the curve at A0 . Geometrically, in Newton-Raphson method, the part of the graph of the function y f ( x) between the point P ( a, f ( a )) and the x axis is replaced by a tangent to the curve at the point at each step in the approximation process.
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This method is very useful for approximating isolated roots. The Newton-Raphson method fails if f ( x ) is difficult to compute or vanishes in a neighbourhood of the desired root. In such cases, the Regula-Falsi method should be used. The Newton-Raphson method is widely used since in a neighbourhood of the desired root, it converges more rapidly than the bisection method or the Regula-Falsi method. If the starting value a is not close enough to the desired root, the method may give a false root or may not converge. If f ( x0 ) f ( x0 ) is not sufficiently small, this method does not work. Also if it work, it works faster. Example 5.6 [CS-1996, AE-2007, IN-2014 (2 mark)]: Newton-Raphson iteration formula for finding
3
c , c 0 is
(a) xn 1
2 xn3 3 c 3 xn2
(b) xn 1
2 xn3 3 c
(c) xn 1
3 xn2
2 xn3 c
(d) xn 1
2
3xn
2 xn3 c 2
3 xn
Solution (c): Let f ( x ) x 3 c then f ( x ) 0 x 3 c 0 x 3 c . So f ( x) 0 3 x 2 , thus from Eq. 5.8 we have the iteration formula as, xn 1 xn
f ( xn ) x3 c 2 x3 c xn n 2 n 2 . f ( xn ) 3 xn 3 xn
Example 5.7 [CS-1997 (1 mark)]: The Newton-Raphson method is used to find the root of the equation x 2 2 0 . If the iterations are started from –1, the iterations will (a) converge to –1 (d) not converge (b) converge to 2 (c) converge to 2 Solution (c): f ( x ) x 2 2 f ( x ) 2 x , so from Eq. 5.8 we have xn 1 xn
xn2
2
x1 x0
x02
2
2
1
1 2
3
(as x0 1 is
2 xn 2 x0 2( 1) 2 given). The successive iterates are given in the table. Hence the solution
of the equation x 2 2 0 with x0 1 converges to 1.4142 2 .
n
xn
xn 1
0 1 2 3
–1 –1.5 –1.4166 –1.4142
–1.5 –1.4166 –1.4142 –1.4142
Example 5.8 [ME-2000 (5 marks)]: Subjective Question: Estimate the root of the equation (e x x ) 0 to four decimal accuracy by employing the Newton-Raphson method starting with an initial guess of x0 0 . Solution: Let f ( x ) e x x f ( x) e x 1 , so from Eq. 5.8 we have xn
xn
xn
xn
f ( xn ) e x x e e e ( xn 1) xn 1 . xn x n n x x 1 e n f ( xn ) e n 1 1 e n The successive iterates are given in the table. So the solution of (e x x ) 0 is x 0.5671 . [Similar questions were also asked in [EC-2008, EE-2008 (2 marks)] xn 1 xn
Example 5.9 [CS-2003 (2 marks)]: A piece-wise linear function f ( x ) is plotted using thick solid lines in the figure below (the plot is drawn to scale). If we use the Newton-Raphson method to find the roots of f ( x ) 0 using x0 , x1 and x2 respectively as initial guesses, the roots obtained would be
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n
xn
xn 1
0 1 2 3
0 0.5 0.5663 0.5671
0.5 0.5663 0.5671 0.5671
(a) 1.3, 0.6 and respectively (b) 0.6, 0.6 and respectively (c) 1.3, 1.3 and respectively (d) 1.3, 0.6 and respectively
0.6 1.3 0.6 1.3
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Solution (d): If we start from x0 then a from x0 meets the line a , whose equation is La : y 0 (0 0.5) (1 0.5) ( x 1) La : x y 1 which meets the x axis at x 1 . Now Lb : y 0.5 ( 1 0.5) (0.8 0.5) ( x 0.5) Lb : 5 x y 3 on which x 1 gives y 2 which is out of range of y , so a at x 1 does not meet line b . But x 1 on Lc : y 0.5 (1 0.5) (0.8 1.55) ( x 1.55) Lc : 2 x y 2.6 gives y 0.6 , which is in the
range of . Thus with x0 as initial guess we have the root of the given function as x 1.3 . Now if we start with x1 as initial guess, then a from x1 meets the line b , which meets the x axis at
x 3 5 0.6 which is the roots of the given function. Similarly with x2 as a guess, then a from x2 meets the line d , whose equation is given as: Ld : y 1 (0.5 1) (1.55 2.05) ( x 2.05) Ld : x y 1.05 which meets the x axis at x 1.05 ; as x 1.05 in line b gives y 2.25 which is out of range of y so a from x 1.05 does not meets the line b . But x 1.05 in line c gives y 0.5 which is in the range of y so a from x 1.05 meets the line c , which meets the
x axis at x 1.3 which is a root of the given function. So using x0 , x1 and x2 respectively as initial guesses, the roots obtained would be 1.3, 0.6, 1.3, respectively. Statement for Linked Answer Questions 5.10 & 5.11: Given a 0 , we wish to calculate its reciprocal value 1 a by using Newton Raphson method for f ( x ) 0 . Example 5.10 [CE-2005 (2 marks)]: The Newton-Raphson algorithm for the function will be 1 a a a (a) xk 1 xk (b) xk 1 xk xk2 (c) xk 1 2 xk axk2 (d) xk 1 xk xk2 2 2 xk 2 Solution: In order to apply Newton-Raphson method to find the reciprocal of a , it is necessary to find a function f ( x ) which has a zero at x 1 a . The obvious such function is f ( x ) x (1 a ) , but the Newton–Raphson iteration for this is unhelpful since it cannot be computed without already knowing the reciprocal of a ; also multiple iterations for refining reciprocal are not possible since higher order derivatives do not exist for f ( x ) x (1 a ) . A function which does work is f ( x ) (1 x ) a ,
for
which
the
Newton–Raphson xk2 }
iteration
gives
axk2
xk 1 xk f ( xk ) f ( xk ) xk {(1 xk ) a} {1 xk (1 axk ) xk 2 xk . Example 5.11 [CE-2005 (4 marks)]: For a 7 and starting with x0 0.2 , the first two iteration will be (a) 0.11, 0.1299 (b) 0.12, 0.1392 (c) 0.12, 0.1416 (d) 0.13, 0.1428 Solution (b): Using the result of previous problem with a 7 and x0 0.2 , we have x1 2 x0 ax02 2(0.2) 7(0.2) 2 0.12 ; x2 2 x2 ax22 2(0.12) 7(0.12) 2 0.1392 .
Example 5.12 [CS-2007 (2 marks)]: Consider the series xn 1 ( xn 2) 9 (8 xn ) , x0 0.5 obtained from the Newton-Raphson method. The series converges to (a) 1.5 (c) 1.6 (d) 1.4 (b) 2 Solution (a): The given iterative formula converges at n and at that point xn 1 xn x (say); then we have x
x 2
9 8x
x2
18 8
9 4
x
3 2
1.5 . So the given series converges to x 1.5 .
Example 5.13 [EC-2007 (2 marks)]: The equation x 3 x 2 4 x 4 0 is to be solved using the Newton-Raphson method. If x 2 is taken as the initial approximation of the solution, then the next approximation using this method will be:
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Chapter 5: Numerical Methods
(b) 4 3 3
(c) 1
[5.11]
(d) 3 2
2
Solution: We have f ( x ) x x 4 x 4 , so using Eq. 5.8 which gives the solution of f ( x ) 0 is xk 1
f ( xk ) xk3 xk2 4 xk 4 2 xk3 xk2 4 . Thus with xk xk xk 1 2 2 f ( xk ) 3 xk 2 xk 4 3xk 2 xk 4
x0 2
as initial
3 2 2 approximation the next approximation will be x1 {2(2 ) 2 4} {3(2 ) 2(2) 4} 16 12 4 3 . [Similar questions were also asked in CS-2010 (1 mark)]
Example 5.14 [XE-2007 (1 mark)]: The equation g ( x ) x is solved by Newton-Raphson iteration method, starting with an initial approximation x0 near the simple root . If xn 1 is the approximation to at the ( n 1)th iteration, then (a) xn 1 {xn g ( xn ) g ( xn )} {1 g ( xn )}
(b) xn 1 {xn g ( xn ) g ( xn )} {g ( xn ) 1}
(c) xn 1 g ( xn )
(d) xn 1 { xn g ( xn ) g ( xn ) 2 xn } {g ( xn ) 1}
Solution (b): Let f ( x ) g ( x ) x f ( x ) g ( x ) 1 , so from Eq. 5.8, we have the iterative formula f ( xn ) g ( xn ) xn xn g ( xn ) g ( xn ) as, xn 1 xn . xn f ( xn ) g ( xn ) 1 g ( xn ) 1 Example 5.15 [CH-2008 (1 mark)]: A non-linear function f ( x ) is defined in the interval 1.2 x 4 as illustrated in the figure below. The equation f ( x ) 0 is solved for x within this interval by using the Newton-Raphson iterative scheme. Among the initial guesses ( I1 , I 2 , I 3 and I 4 ) , the guess that is likely to lead to the root most rapidly is (a) I1 (b) I 2 (c) I 3
(d) I 4
Solution (d): It is to be noted that the tangent T3 , which is drawn from the point on the curve after dropping a perpendicular from I 3 on the curve, cuts the x axis at point P which is close to the root as compared to the point I 3 . This condition is not satisfied by the tangents corresponding to the points I1 , I 2 and I 4 . So guess I 3 is likely to lead to the root most rapidly. Example 5.16 [CS-2008 (2 marks)]: The Newton-Raphson iteration xn 1 (1 2) xn ( R xn ) can be used to compute the (a) square of R (b) reciprocal of R (c) square root of R (d) logarithm of R Solution (c): At convergence xn 1 xn x (say). So the given iterative formula gives x (1 2) x ( R x ) 2 x x ( R x) x 2 R x R . So the iteration will compute the square root of R . [Similar question was also asked in CS-2002 (2 marks)]
Example 5.17 [AE-2009 (2 marks), CE-2011 (1 mark)]: The square root of a number N is to be obtained by applying the Newton Raphson iterations to the equation x 2 N 0 . If i denotes the iteration index, the correct iterative scheme will be 1 N 1 N 1 N2 1 N (a) xi 1 xi (b) xi 1 xi2 2 (c) xi 1 xi (d) xi 1 xi 2 xi 2 xi 2 xi 2 xi
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Solution (a): f ( x) x 2 N f ( x ) 2 x , so from Eq. 5.8 we have the iteration formula as, f ( xi ) x 2 N xi2 N 1 N xi i xi . f ( xi ) 2 xi 2 xi 2 xi [Similar questions were also asked in CS-1995, EE-2009, IN-2007, ME-1999 (2 marks)] xi 1 xi
Example 5.18 [XE-2009 (1 mark)]: The root of ax b 0 ( a , b constants), can be found by the Newton-Raphson method with a minimum of (a) 1 iteration (b) 2 iterations (c) 3 iterations (d) an undeterminable number of iterations Solution (a): As f ( x ) ax b f ( x) a . So from Eq. 5.9 we have the iteration formula as, xn 1 xn ( axn b) a b a . Hence we can see that xn 1 b a for any choice of x0 . So the root of ax b 0 , can be found by the Newton-Raphson method with a minimum of one iteration.
Example 5.19 [AE-2011 (2 marks)]: Consider the function f ( x) x sin x . The Newton-Raphson iteration formula to find the root of the function starting from an initial guess x (0) at iteration k is (a) x ( k 1) {sin x ( k ) x ( k ) cos x ( k ) } {1 cos x ( k ) } (b) x ( k 1) {sin x ( k ) x ( k ) cos x ( k ) } {1 cos x ( k ) } (c) x ( k 1) {sin x ( k ) x ( k ) cos x ( k ) } {1 cos x ( k ) }
(d) x ( k 1) {sin x ( k ) x ( k ) cos x ( k ) } {1 cos x ( k ) } f ( xk ) x sin xk sin xk xk cos xk Solution (a): From Eq. 5.8, we have xk 1 xk . xk k f ( xk ) 1 cos xk 1 cos xk Example 5.20 [EC-2011 (2 mark)]: A numerical solution of the equation f ( x ) x x 3 0 can be obtained using Newton-Raphson method. If the starting value is x 2 for the iteration, the value of x that is to be used in the next step is (a) 0.306 (b) 0.739 (c) 1.694 (d) 2.306 Solution (c): Let f ( x ) x x 3 f ( x ) 1 {1 (2 x )} , so from Eq. 5.8 we have the iteration formula as, xn 1 xn
x xn 3 2 x x 2 xn 6 xn xn 6 xn f ( xn ) xn n xn n n . So f ( xn ) 1 1 2 xn 2 xn 1 2 xn 1
with x0 2 as initial guess, we have x1
x0 6 x0 2 x0 1
2 6 2
2 2 1
6.485 3.828
1.694 .
Example 5.21 [IN-2011 (2 marks)]: The extremum (minimum or maximum) point of a function f ( x ) is to be determined by solving df ( x) dx 0 using the Newton Raphsons method. Let f ( x ) x 3 6 x and x0 1 be the initial guess of x . The value of x after two iterations ( x2 ) is (a) 0.0141 (b) 1.4142 (c) 1.4167 (d) 1.5000 Solution (c): The extremum point of a function is determined by equating its derivative to zero. So our function g ( x) f ( x ) 3 x 2 6 g ( x ) 6 x , so from Eq. 5.8 we have the iteration formula as, xn 1
g ( xn ) 3x2 6 . Given xn xn n g ( xn ) 6 xn
x0 1 x1 x0
3x02 6 6 x0
x1 1
3(12 ) 6 6(1)
1.5 . So,
x2 x1 (3x12 6) (6 x1 ) 1.5 [3(1.5 2 ) 6] [6(1.5)] 1.4167 .
Example 5.22 [XE-2011 (2 marks)]: Suppose xn is the nth iterated value while finding the positive square root of 7 by the Newton-Raphson method with a positive initial guess x0 ( 7 ) . If
en 7 xn for n 1 , then (a) en 1 en (2 xn2 )
(b) en 1 7 en2 (2 xn )
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2
(c) en 1 en
7
(d) en 1 en2 (2 xn )
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Engineering Mathematics
Chapter 5: Numerical Methods
Solution (d): w.r.t problem 5.17, with N 7 , xn 1
[5.13]
1
7 xn . As en 7 xn for n 1 , then 2 xn
x 2 7 xn xn2 ( 7 ) 2 ( 7 xn ) 2 en2 1 7 7 en 1 7 xn 1 7 xn 7 n 2 xn 2 2 xn 2 xn 2 xn 2 xn
Example 5.23 [CH-2012 (2 marks)]: The Newton-Raphson method is used to find the roots of the equation f ( x ) x cos x , 0 x 1 . If the initial guess for the root is 0.5, then the value of x after the first iteration is (a) 1.02 (b) 0.62 (c) 0.55 (d) 0.38 Solution: f ( x ) x cos x f ( x ) 1 sin x , so from Eq. 5.8 we have the iteration formula as, f ( xn ) x cos xn xn sin xn cos xn . So with x0 0.5 as initial guess, we xn 1 xn xn n f ( xn ) 1 sin xn 1 sin xn have xn 1 { (0.5) sin( 2) cos( 2)} {1 sin( 2)} { (0.5)} (1 ) {2(1 )} 0.38 . Example 5.24 [EE-2013 (2 marks)]: When the Newton-Raphson method is applied to solve the equation f ( x ) x 3 2 x 1 0 , the solution at the end of the first iteration with the initial guess value as x0 1.2 is (a) –0.82 (b) 0.49 (c) 0.705 (d) 1.69 3 2 Solution: f ( x ) x 2 x 1 f ( x ) 3 x 2 from Eq. 5.8 we have the iteration formula as, f ( xk ) x3 2 x 1 2 xk3 1 . So with xk k 2 k 2 f ( xk ) 3 xk 2 3 xk 2
xk 1 xk
x0 1.2
as initial guess we have
x1 (2 x03 1) (3 x02 2) {2(1.2)3 1} {3(1.2) 2 2} 0.705 . [Similar questions were also asked in CH-2010 (1 mark), ME-2005, CE-2007, MT-2010, TF2008 (2 marks)]
Example 5.25 [CS-2014 (2 marks)]: In the Newton-Raphson method, an initial guess of x0 2 is made and the sequence x0 , x1 , x2 , is obtained for the function 0.75 x3 2 x 2 2 x 4 0 . Consider the statements: (A) x3 0 (B) The method converges to a solution in a finite number of iterations. Which of the following is TRUE? (a) Only A (b) Only B (c) Both A and B (d) Neither A nor B 3 2 2 Solution (a): As f ( x ) 0.75 x 2 x 2 x 4 f ( x ) 2.25 x 4 x 2 , so from Eq. 5.8 we have the iteration formula as, xn 1 xn initial
guess, 3
x2
we
2
1.5 x1 2 x1 4 2
2.25 x1 4 x1 2
4 2
have
0.75 xn3 2 xn2 2 xn 4 2
2.25 xn 4 xn 2 x1
1.5 x03 2 x02 4 2
2.25 x0 4 x0 2 3
2 and x3
2
2.25 x2 4 x2 2
2
2.25 xn 4 xn 2
2
1.5 x2 2 x2 4
1.5 xn3 2 xn2 4
. Now with x0 2 as
1.5(23 ) 2(22 ) 4
0.
2
2.25(2 ) 4(2) 2 1.5(23 ) 2(22 ) 4 2
2.25(2 ) 4(2) 2
Similarly,
0 . So statement
(A) is correct. Now if we go further and compute for x4 , x5 , , we find that the value is alternatively 2, 0. So the iterative method do not converge to a solution for x0 2 . Example 5.26 [EE-2014 (2 marks)]: The function f ( x ) e x 1 is to be solved using NewtonRaphson method. If the initial value of x0 is taken as 1.0, then the absolute error observed at 2nd iteration is …………….
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.14]
Solution: Let f ( x ) e x 1 f ( x ) e x , so from Eq. 5.8, we have the iterative formula as, x
xn 1 xn x
x1
x
x f ( xn ) e n 1 x e n e n 1 . xn x n x f ( xn ) en en
x
x0 e 0 e 0 1 x
(1)e1 e1 1
So
x0 1 ,
with
we
x
1
and
x2 x1
thus
e 1 1 x
have x
x
x1e 1 e 1 1 x
e1 e e0 e1 e1 1e 1e (1 e)e e 1 1 1 1 x2 1 1 e 0.368 1 0.06 . Now as x 0 is the root of the 1e e e e (2.71)0.368 equation f ( x ) 0 . So absolute error at 2
nd
iteration is 0 0.06 0.06 .
Example 5.27 [MA-2014 (2 marks)]: Using the Newton-Raphson method with the initial guess (0) x 6 , the approximate value of the real root of x log10 x 4.77 , after the second iteration, is …… Solution:
log10 x
So let (log10 e)(log e x ) 0.434 ln x . log10 e f ( x ) 0.434 x ln x 4.77 f ( x ) 0.434(ln x 1) 0.434 ln x 0.434 . Thus we have the iterative f ( xn ) 0.434 xn ln xn 4.77 0.434 xn 4.77 formula as, xn 1 xn . So with xn xn 1 f ( xn ) 0.434 ln xn 0.434 0.434 ln xn 0.434 initial x2
As
log10 x (log10 e)
x0 6 ,
guess
0.434 x1 4.77 0.434 ln x1 0.434
we
have
x1
0.434(6.086) 4.77 0.434 ln 6.086 0.434
0.434 x0 4.77 0.434 ln x0 0.434
0.434(6) 4.77 0.434 ln 6 0.434
6.086 ;
so
6.085 .
Example 5.28 [ME-2014 (2 marks)]: The real root of the equation 5 x 2 cos x 1 0 (up to two decimal accuracy) is ……………. Solution: Let f ( x ) 5 x 2 cos x 1 f ( x ) 5 2 sin x , so using Newton-Raphson method the iterative formula which gives the solution of f ( x) 0 is f ( xn ) 5 x 2 cos xn 1 2 xn sin xn 2 cos xn 1 . xn 1 xn xn n f ( xn ) 5 2 sin xn 5 2 sin xn Now as f (0) 2 1 3 0 and f (1) 5 2 cos(1c ) 1 2.92 0 ; so by intermediate value theorem, one of the roots of f ( x ) 0 is lying between
[0,1] . Now with x0 1 as initial guess, the successive iterates up to two decimal accuracy are given below:
n
xn
xn 1
0 1 2
1 0.56 0.54
0.56 0.54 0.54
Example 5.29 [PI-2014 (2 marks)]: If the equation sin x x 2 is solved by Newton Raphson’s method with the initial guess of x 1 , then the value of x after 2 iterations would be ……… Solution: Let f ( x ) sin x x 2 f ( x ) cos x 2 x , so using Newton-Raphson method the iterative formula which gives the solution of f ( x) 0 is xn 1 xn
f ( xn ) sin xn xn2 x cos xn xn2 sin xn . So with x0 1 as initial guess, we xn n f ( xn ) cos xn 2 xn cos xn 2 xn
have
x1
x0 cos x0 x02 sin x0 cos x0 2 x0
2
x2
x1 cos x1 x1 sin x1 cos x1 2 x1
cos1c 1 sin1c c
cos1 2
(0.89) cos 0.89c 0.89 2 sin 0.89c c
cos 0.89 2(0.89)
Copyright © 2016 by Kaushlendra Kumar
0.89 ;
similarly
0.87 .
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.15]
The Secant Method: Newton’s method is very powerful but has the disadvantage that the derivative f may sometimes be a far more difficult expression than f itself and its evaluation therefore computationally expensive. This situation suggests the idea of replacing the derivative with the difference quotient f ( xn ) f ( xn 1 ) (5.14) f ( xn ) xn xn 1 Hence the Newton – Raphson formula becomes, xn xn 1 (5.15) xn 1 xn f ( xn ) f ( xn ) f ( xn 1 ) Geometrically, we intersect the x axis at xn 1 with the secant of f ( x ) passing through Pn 1 and Pn in Fig. 5.6. We need two starting values x0 and x1 . Evaluation of derivatives is now avoided. It can be shown that convergence is more rapid than linear, almost quadratic like Newton – Raphson method. The algorithm is similar to that of Newton’s method. It should be noted that this formula requires two initial Figure 5.6: Secant Method approximations to the root. x f ( xn ) xn f ( xn 1 ) It is not good to write Eq. 5.15 as, xn 1 n 1 because this may lead to loss of f ( xn ) f ( xn 1 ) significant digits if xn and xn 1 are about equal. Example 5.30 [MT-2013 (2 marks)]: Applying the secant method, the first approximation to the root of f ( x ) 1 ln x ( x 2) , starting with function values at x 0.3 and x 0.4 is …… Solution: The two initial approximations are xn 1 xn f ( xn )
f ( x)
as
xn xn 1 f ( xn ) f ( xn 1 )
x2 x1 f ( x1 )
x0 0.3
and
x1 0.4 . So from Eq.5.15,
. Thus with n 1, we have the first approximation for the root of
x1 x0 f ( x1 ) f ( x0 )
. As
f ( x0 ) f (0.3) 1 ln 0.3 (0.3 2) 0.054
f ( x1 ) f (0.4) 1 ln 0.4 (0.4 2) 0.284 . So x2 0.4 0.284
0.4 0.3 0.284 0.054
and
0.316 .
5.1.4 Rate of Convergence of Approximate Methods We study different numerical methods to find the root of an equation. As different method converge of the root with different speed. So rate of convergence measures how fast a sequence converges. Order of a Root: Let the function f ( x ) and its derivatives are defined and continuous on an interval about x k . We say f ( x ) 0 has a root of order m at x k if and only if
f ( k ) f ( k ) f ( m1) ( k ) 0 and f ( m ) ( k ) 0 ; so f ( x ) ( x k ) m h( x ) , where h(k ) 0 A root of order m 1 is often called a simple root. A root of order m 1 is called a multiple root. A root of order m 2 is sometimes called a double root, and so on. Rate of Convergence: Let the sequence {rn } converge to r ; and let en rn r and en 1 rn 1 r be the errors at nth and ( n 1)th iterations respectively. If there exist a positive
number k 1 and a constant c 0 such that lim
n
rn 1 r rn r
k
lim
n
en 1 en
k
c , then k is called the
order of convergence of the sequence; and c is called asymptotic error constants.
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e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.16]
If k is large, then the sequence {rn } converges rapidly to r . If k 1 and c 1 , then the convergence is said to be linear, and c is called the rate of convergence. If k 2 , then the convergence is said to be quadratic. The Rate of Convergence for the iterative method discussed in previous sections is given as: The Bisection Method: Order of Convergence k 1 , linear convergence The method of False Position: Order of Convergence k 1 , linear convergence. Fixed – Point method: Order of Convergence k 1 , linear convergence. Newton – Raphson method: Order of Convergence k 2 , quadratic convergence [This point was asked in AE-2010 (2 marks)]. The Secant method: Order of Convergence k 1.618 .
5.1.5 Interpolation Consider the following problem: Given the values of a known function y f ( x) at a sequence of ordered points x0 , x1 , , xn , find f ( x ) for arbitrary x . When x0 x xn , the problem is called interpolation. When x x0 or x xn , the problem is called extrapolation. With yi f ( xi ) , the problem of interpolation is basically one of drawing a smooth curve through the known points ( x0 , y0 ), ( x1 , y1 ), , ( xn , yn ) . So what kind of function f ( x ) should one choose? A polynomial is a common choice for an interpolating function because polynomials are easy to evaluate, differentiate, and integrate relative to other choices such as a trigonometric and exponential series. Polynomial interpolation involves finding a polynomial of order n that passes through the n 1 points. The methods of interpolation that we discuss are Lagrangian interpolation method and the Newton’s divided difference polynomial method.
Lagrange Interpolation: Given ( x0 , f0 ), ( x1 , f1 ), , ( xn , fn ) with arbitrarily spaced x j . Lagrange had the idea of multiplying each f j by a polynomial that is 1 at x j and 0 at other n nodes and then taking the sum of these n 1 polynomials. Clearly, this gives the unique interpolation polynomial of degree n or less. Beginning with simplest case, i.e., Linear Interpolation, we will present General Lagrange Interpolation Polynomial.
Linear Interpolation: It is an interpolation by a straight line through ( x0 , f 0 ), ( x1 , f1 ) . Thus the linear Lagrange polynomial p1 is a sum p1 L0 f 0 L1 f1 with L0 the linear polynomial that is 1 at x0 and 0 at x1 ; similarly L1 is
0
at
x0
and
1
at
x1 .
Obviously,
Figure 5.7: Linear Interpolation
L0 ( x) {( x x1 ) ( x0 x1 )}, L1 ( x) ( x x0 ) ( x1 x0 ) . This gives the Linear Lagrange polynomial (also shown in Fig. 5.7) as, x x1 x x0 p1 ( x ) L0 ( x ) f 0 L1 ( x ) f1 f0 f1 x0 x1 x1 x0
(5.16)
Example 5.31 [MN-2012 (1 mark)]: Assuming sin(1) 0.841 and sin(3) 0.141 , the Lagrangian linear interpolation polynomial, for the function f ( x ) sin x defined on the interval [1, 3] and passing through the end points of the intervals, is (a) 0.35 x 1.19 (b) 3.05 x 11.92 (c) 35.00 x 119.10 (d) 40.50 x 219.19 Solution (a): From Eq. 5.17, we have f ( x ) sin x , x0 1 , x1 3 , f ( x0 ) 0.841 and f ( x1 ) 0.141 . So p1 ( x ) ( x 3) (1 3) 0.841 ( x 1) (3 1) 0.141 1.191 .35 x .
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.17]
Quadratic Interpolation: It is the interpolation of given ( x0 , f 0 ), ( x1 , f1 ), ( x2 , f 2 ) by a second degree polynomial p2 ( x ) , which by Lagrange’s idea is, p2 ( x ) L0 ( x ) f 0 L1 ( x ) f1 L2 ( x ) f 2
with L0 ( x ) and L2 ( x )
l0 ( x ) l0 ( x0 ) l2 ( x )
l2 ( x2 )
( x x1 )( x x2 ) ( x0 x1 )( x0 x2 ) ( x x0 )( x x1 )
( x2 x0 )( x2 x1 )
L0 ( x0 ) 1 , L1 ( x)
l1 ( x ) l1 ( x1 )
(5.17)
( x x0 )( x x2 ) ( x1 x0 )( x1 x2 )
L1 ( x1 ) 1
L2 ( x2 ) 1 ; also we have L0 ( x1 ) 0, L0 ( x2 ) 0 , etc.
General Lagrange Interpolation Polynomial: Generalising Eq. 5.16 and 5.17, for n , i.e. n
n
f ( x) pn ( x) k 0 Lk ( x) f k k 0 {lk ( x)} {lk ( xk )} f k
(5.18)
where, Lk ( xk ) 1 and Lk 0 at the other nodes; and the Lk are independent of the function f to be interpolated; l0 ( x) ( x x1 )( x x2 ) ( x xn ) ; lk ( x ) ( x x0 ) ( x xk 1 )( x xk 1 ) l0 ( x) ( x x1 )( x x2 ) ( x xn ),
0k n;
ln ( x ) ( x x0 )( x x1 ) ( x xn 1 ) ; pn ( xk ) fk ; and lk ( x j ) 0 if j k .
Error Estimate: If f is itself a polynomial of degree n (or less), it must coincide with pn because the n 1 data ( x0 , f 0 ), , ( xn , f n ) determine a polynomial uniquely, so the error is zero. Now the special f has its ( n 1)th derivative identically zero. This makes it plausible that for a general f its ( n 1)th derivative f ( n 1) should measure the error n ( x ) f ( x ) pn ( x ) . It can be shown that this is
true if f ( n 1) exists and is continuous. Then, with a suitable t between x0 and xn , we have,
n ( x ) f ( x ) pn ( x) ( x x0 )( x x1 ) ( x xn ) 1 ( n 1)! f ( n 1) (t )
(5.19)
Thus n ( x) is 0 at the nodes and small near them, because of continuity. The product ( x x0 ) ( x xn ) is large for x away from the nodes. This makes interpolation at an x will be best if we choose nodes on both sides of that x . Also we get error bounds by taking the smallest and largest value of f ( n 1) (t ) in Eq. 5.19 on the interval x0 t xn .
Example 5.32 [XE-2007 (2 marks)]: If a polynomial of degree three interpolates a function f ( x ) at the points (0, 3), (1, 13), (3, 99) and (4, 187), then f (2) is (a) 20 (b) 36 (c) 43 (d) 58 l0 ( x ) l1 ( x) l2 ( x ) l ( x) Solution (c): With n 3 , Eq. 5.18 gives f ( x ) f0 f1 f2 3 f 3 , where l0 ( x0 ) l1 ( x1 ) l2 ( x2 ) l3 ( x3 ) l0 ( x ) l0 ( x0 ) l1 ( x) l1 ( x1 ) l2 ( x ) l2 ( x2 ) l3 ( x ) l3 ( x3 )
( x x1 )( x x2 )( x x3 ) ( x0 x1 )( x0 x2 )( x0 x3 ) ( x x0 )( x x2 )( x x3 ) ( x1 x0 )( x1 x2 )( x1 x3 )
( x x0 )( x x1 )( x x3 ) ( x2 x0 )( x2 x1 )( x2 x3 ) ( x x0 )( x x1 )( x x2 ) ( x3 x0 )( x3 x1 )( x3 x2 )
l0 (2) l0 ( x0 ) l1 (2)
l1 ( x1 )
l2 (2) l3 (2) l3 ( x3 )
(0 1)(0 3)(0 4)
(2 0)(2 3)(2 4) (1 0)(1 3)(1 4)
l2 ( x2 )
(2 1)(2 3)(2 4)
(2 0)(2 1)(2 4) (3 0)(3 1)(3 4) (2 0)(2 1)(2 3) (4 0)(4 1)(4 3)
12
4 6
2
4 6
2 3
1 6
;
;
2 3
;
2
1 . From the given data 12 6
we have x0 0 , x1 1 , x2 3 , x3 4 and f0 3 , f1 13 , f 2 99 , f3 187 . So we have f (2) ( 1 6)3 (2 3)13 (2 3)99 ( 1 6)187 43 .
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.18]
Newton’s Divide Difference Interpolation: Suppose that Pn ( x ) is the nth Lagrange polynomial that agrees with the function f at the distinct numbers x0 , x1 , , xn . Although this polynomial is unique, there are alternate algebraic representations that are useful in certain situations. The divide differences of f with respect to x0 , x1 , , xn are used to express Pn ( x ) in the form, Pn ( x ) a0 a1 ( x x0 ) a2 ( x x0 )( x x1 ) an ( x x0 )( x x1 ) ( x xn 1 )
(5.20)
for appropriate constants a0 , a1 , , an . To determine the first of these constants, a0 , we have to evaluate Pn ( x ) (from Eq. 5.20) at x x0 , which gives a0 Pn ( x0 ) f ( x0 ) . Similarly, when P ( x ) is evaluated at x1 , the only nonzero terms in the evaluation of Pn ( x1 ) are the constant and linear terms, f ( x0 ) a1 ( x1 x0 ) Pn ( x1 ) f ( x1 ) a1 { f ( x1 ) f ( x0 )} {x1 x0 } (5.21) We now introduce the divided – difference notation. The zeroth divided difference of the function f
with respect to xi , denoted as f [ xi ] , which is simply the value of f at xi , i.e., f [ xi ] f ( xi ) (5.22) The remaining divided difference are defined recursively; the first divided difference of f with
respect to xi and xi 1 is denoted as f [ xi , xi 1 ] and defined as, f [ xi , xi 1 ] { f [ xi 1 ] f [ xi ]} {xi 1 xi }
(5.23)
The second divided difference, f [ xi , xi 1 , xi 2 ] , is defined as, f [ xi , xi 1 , xi 2 ] { f [ xi 1 , xi 2 ] f [ xi , xi 1 ]} { xi 2 xi }
Similarly, after the ( k 1)
th
(5.24)
divided differences, f [ xi , xi 1 , xi 2 , , xi k 1 ] and f [ xi 1 , xi 2 , , xi k ]
have been determined. So the k th divided difference relative to xi , xi 1 , xi 2 , , xi k is given as, f [ xi , xi 1 , xi 2 , , xi k ] { f [ xi 1 , xi 2 , , xi k ] f [ xi , xi 1 , xi 2 , , xi k 1 ]} {xi k xi }
(5.25)
th
The process ends with the single divided difference, n f [ x0 , x1 , x2 , , xn ] { f [ x1 , x2 , , xn ] f [ x0 , x1 , , xn 1 ]} {xn x0 } . Because of Eq. 5.21, we can write a1 [ x0 , x1 ] , just as a0 can be expressed as a0 f ( x0 ) f [ x0 ] . Hence the interpolating polynomial in Eq. 5.20 is given as, Pn ( x ) f [ x0 ] f [ x0 , x1 ]( x x0 ) a2 ( x x0 )( x x1 ) an ( x x0 )( x x1 ) ( x xn 1 ) (5.26) As might be expected from the evaluation of
a0
and a1 , the required constants are
ak f [ x0 , x1 , x2 , xk ] for each k 0,1, , n . So Pn ( x ) can be re-written in a form called Newton’s Divided Difference, as n
Pn ( x ) f [ x0 ] k 1 f [ x0 , x1 , , xk ]( x x0 ) ( x xk 1 )
(5.27)
The value of f [ x0 , x1 , , xk ] is independent of the order of the numbers x0 , x1 , , xk . Example 5.33: For x, f ( x ) (1, 0.765), (1.3, 0.620), (1.6, 0.455), (1.9, 0.281), (2.2, 0.110) , compute the divided difference table for the given data and construct the interpolating polynomial. Solution: The first divided difference involving x0 1 and x1 1.3 is given as, f [ x0 , x1 ] { f [ x1 ] f [ x0 ]} { x1 x0 } {0.620 0.765} {1.3 1.0} 0.483 . The remaining first divided differences are found in a similar manner and are shown in the fourth column of the Table. The second divided difference involving x0 , x1 and x2 is f [ x0 , x1 , x2 ] { f [ x1 , x2 ] f [ x0 , x1 ]} {x2 x0 } {0.548 ( 0.483)} {1.6 1.0} 0.108 . The remaining second divided differences are found in a similar manner and are shown in the fifth column of the Table. The third divided difference involving x0 , x1 , x2 and x3 is
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Chapter 5: Numerical Methods
[5.19]
f [ x0 , x1 , x2 , x3 ] { f [ x1 , x2 , x3 ] f [ x0 , x1 , x2 ]} { x3 x0 } {0.049 ( 0.108)} {1.9 1.0} 0.065 ;
the other is evaluated similarly. The fourth divided difference involving x0 , x1 , x2 , x3 and x4 is f [ x0 , x1 , x2 , x3 , x4 ] { f [ x1 , x2 , x3 , x4 ] f [ x0 , x1 , x2 , x3 ]} {x4 x0 } {0.068 0.065} {2.2 1} 0.002 i
xi
f [ xi ]
0
1.0
0.765
f [ xi 1 , xi ]
f [ xi 2 , xi 1 , xi ]
f [ xi 3 , , xi ]
f [ xi 4 , , xi ]
– 0.483 1
1.3
0.620
– 0.108 – 0.548
2
1.6
0.455
0.065 – 0.049
0.002
– 0.578 3
1.9
0.281
0.068 0.011
– 0.571 4 2.2 0.110 The coefficients of the Newton forward divided – difference form of the interpolating polynomial are along the diagonal in the table. This polynomial is, P4 ( x) 0.765 0.483( x 1.0) 0.108( x 1.0)( x 1.3) 0.065( x 1.0)( x 1.3)( x 1.6) 0.002( x 1.0)( x 1.3)( x 1.6)( x 1.9)
Simplified form of Newton’s divided difference formula: Newton divided difference formula can be expressed in a simplified form when the nodes are arranged consecutively with equal spacing. In this case, we introduce the notation h xi 1 xi , for each i 0,1, , n 1 and let x x0 sh . Then the difference x xi is x xi ( s i) h . So Eq. 5.27 can be written as
Pn ( x ) Pn ( x0 sh) f [ x0 ] shf [ x0 , x1 ] s ( s 1) h 2 f [ x0 , x1 , x2 ] s ( s 1) ( s n 1)h n f [ x0 , x1 , , xn ] n
Pn ( x ) f [ x0 ] k 1 s ( s 1) ( s k 1)h k f [ x0 , x1 , , xk ] . Using binomial-coefficient notation, n
Pn ( x ) Pn ( x0 sh ) f [ x0 ] k 1 s Pk k ! h k f [ x0 , xi , , xk ] , where s Pk
s ( s 1) ( s k 1) k!
(5.28)
Equal Spacing: Newton’s Forward Difference Formula: Newton formula (Eq. 5.27) is valid for arbitrarily spaced nodes as they may occur in observations. In many applications the x j ’s are regularly spaced, like, in measurements taken at regular intervals of time; then denoting the distance by h , we can write the x j ’s as x0 , x1 x0 h, x2 x0 2h, , xn x0 nh . The first forward difference of f at x j is given as, f j f j 1 f j ; the second forward difference of f at x j is given 2
as, f j f j 1 f j ; and continuing in this way, the k th forward difference of f at x j is given as
k f j k 1 f j 1 k 1 f j , k 1, 2, . Now forward – difference formula is constructed by making use of the forward difference notation ‘ ’. With this notation, f ( x1 ) f ( x0 ) 1 1 f [ x0 , x1 ] f ( x1 ) f ( x0 ) f ( x0 ) x1 x0 h h f [ x0 , x1 , x2 ]
1 f ( x1 ) f ( x0 ) 1 2 2 f ( x0 ) 2h h 2h
and in general,
f [ x0 , x1 , , xk ]
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1 k !hk
k f ( x0 )
(5.29)
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Chapter 5: Numerical Methods
[5.20]
In Eq. 5.27 we finally set x x0 rh x x0 rh ; similarly x x1 ( r 1)h , and so on. With this and Eq. 5.29, Eq. 5.27 becomes Newton’s forward difference interpolation formula. r (r 1) 2 r ( r 1) ( r n 1) n n f ( x ) pn ( x ) s 0 r Ps s f 0 f 0 r f 0 f0 f0 (5.30) 2! n! r (r 1)(r 2) ( r s 1) where r ( x x0 ) h , the binomial coefficients, r Ps ( s 0, integer) . s! Example 5.34 [ME-2004 (2 marks)]: The values of a function f ( x ) are tabulated below: x Using Newton’s forward difference formula, the cubic 0 1 2 3 polynomial that can be fitted to the above data is f ( x) 1 2 1 10 (a) 2 x 3 7 x 2 6 x 2 (b) 2 x3 7 x 2 6 x 2 Solution (d): From the given f [ xi ] i xi data, we note that the x j ’s 0 0 1 are regularly spaced, with the spacing distance h 1, 1 1 2 then we have x0 0 ,
,
x1 x0 h 1 ,
x3 x0 3h 3 . So the difference table is given as:
So 2
from
Eq.
2
2
(c) x 3 7 x 2 6 x 1 (d) 2 x 3 7 x 2 6 x 1 f [ xi 1 , xi ] f [ xi 2 , xi 1 , xi ] f [ xi 3 , xi 2 , xi 1 , xi ] 1 –1 –1
2
1
5 9
3
3
5.29,
2
10 we
2
f ( x0 ) 1! h f [ x0 , x1 ] (1!) 1 (1) 1 ;
have
f ( x0 ) 2! h f [ x0 , x1 , x2 ] (2!) 1 ( 1) 2 ; f ( x0 ) 3! h 3 f [ x0 , x1 , x2 , x3 ] (3!) 13 2 12 .
Thus from Eq. 5.39,
3
f ( x ) pn ( x) f ( x0 ) r f ( x0 )
r ( r 1)
where x x0 rh x 0 r (1) r x , so f ( x ) 1 x(1)
2 f ( x0 )
2! x ( x 1) 2!
( 2)
r ( r 1)( r 2) 3! x ( x 1)( x 2) 3!
3 f ( x0 ) , (12)
f ( x) 1 x x 2 x 2 x3 6 x2 6 x 2 x3 7 x 2 6 x 1 .
Equal Spacing: Newton’s Backward Difference Formula: If the interpolating nodes are recorded from last to first as xn , xn 1 , , x0 , we can write the interpoatory formula as, Pn ( x ) f [ xn ] f [ xn , xn 1 ]( x xn ) f [ xn , xn 1 , xn 2 ]( x xn )( x xn 1 ) f [ xn , , x0 ]( x xn )( x xn 1 ) ( x x1 )
(5.31)
We define the first backward difference of f at x j by f j f j f j 1 ; the second backward 2
difference of f at x j is given as, f j f j f j 1 ; and continuing in this way the k th backward difference of f at x j is given as,
k f j k 1 f j k 1 f j 1 , ( k 1, 2, )
(5.32)
A formula similar to Eq. 5.30 but involving backward difference is Newton’s backward difference interpolation formula is given as, r ( r 1) 2 r (r 1) ( r n 1) n n f ( x ) p n ( x ) s 0 ( r s 1) Ps s f 0 f 0 r f 0 f0 f 0 (5.33) 2! n! where, x x0 rh , r ( x x0 ) h .
Newton’s Central Differences: The Newton forward- and backward-difference formulas are not appropriate for approximating f ( x ) when x lies near the centre of the table because neither will
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.21]
permit the highest-order difference to have x0 close to x . A number of divided-difference formulas are available for this case, each of which has situations when it can be used to maximum advantage. These methods are known as centred-difference formulas, x1 f 1 which are third notation for differences. The first central f 1 2 difference of f ( x ) at x j is defined by f j f j 1 2 f j 1 2 ; x0 f0 2 f0 and the k th central difference of f ( x ) at x j is given as, f1 2 3 f1 2 k f j k 1 f j 1 2 k 1 f j 1 2 , j 2, 3, (5.34) x1 f1 2 f1 Thus in this notation a difference table, for example, for f3 2 f ,f ,f,f looks as given in the table. Central 1
0
1
2
differences are used in numeric differentiation, differential equation and central interpolation formulas.
x2
f2
Approximating Derivatives from Data and Errors of Approximation: Suppose that a variable y depends on another variable x , i.e., y f ( x) , but we only know the values of f at a finite set of points, i.e., ( x1 , y1 ), ( x2 , y2 ), , ( xn , yn ) . Suppose then that we need information about the derivative of f ( x ) . One obvious idea would be to approximate f ( xi ) by the Forward difference: f ( xi ) yi
yi 1 yi
. An alternative would be to use Backward difference: f ( xi ) yi
yi yi 1
. xi 1 xi xi xi 1 Since the errors for the forward and backward difference tend to have opposite sign, it would seem likely that averaging the two methods would give a better result than either alone. If the points are equally spaced, i.e., xi 1 xi xi xi 1 h , then averaging the forward and backward differences leads to a symmetric expression called the Central Difference: f ( xi ) yi
yi 1 yi 1 2h
.
We can use Taylor polynomials to derive the accuracy of the Forward, Backward and Central difference formulas. For e.g., the usual form of the Taylor polynomial with remainder is h2 f ( x h) f ( x) h f ( x ) f (c) , where c is some unknown number between x and x h . 2 f ( xi 1 ) f ( xi ) h Letting x xi , x h xi 1 and solving for f ( xi ) leads to f ( xi ) f (c ) . Notice h 2 that the quotient in this equation is exactly the forward difference formula. Thus the error of the h forward difference is f (c) which means it is O( h) . Replacing h in the above calculation by h 2 gives the error for backward difference formula, i.e., it is also O( h) . For the central difference, the error can be found from the third degree Taylor polynomial with remainder h2 h3 f ( xi 1 ) f ( xi h) f ( xi ) h f ( xi ) f ( xi ) f (c1 ) and 2 3! h2 h3 f ( xi 1 ) f ( xi h) f ( xi ) h f ( xi ) f ( xi ) f (c2 ) , where xi c1 xi 1 and xi 1 c2 xi 2 3! . Subtracting these two equations and solving for f ( xi ) leads to f ( xi )
f ( xi 1 ) f ( xi 1 )
h 2 f (c1 ) f (c2 )
. This shows that the error for the central difference 2h 3! 2 formula is O ( h 2 ) . Thus, central differences are significantly better and so. It is best to use central differences whenever possible.
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.22]
Example 5.35 [CS-1996 (1 mark)]: The formula used to compute an approximation for the second derivative of a function f at a point x0 is (a) (c)
f ( x0 h) f ( x0 h )
(b)
2 f ( x0 h) 2 f ( x0 ) f ( x0 h )
(d)
h2
f ( x0 h) f ( x0 h ) 2h f ( x0 h) 2 f ( x0 ) f ( x0 h ) h2
Solution (d): From 2nd degree Taylor polynomial, we have f ( x0 h) f ( x0 ) h f ( x0 ) …(i);
f ( x0 h) f ( x0 ) h f ( x0 )
and
h2 2
f ( x0 ) …(ii).
f ( x0 h) f ( x0 h ) 2 f ( x0 ) h 2 f ( x0 ) f ( x0 )
Adding
(i)
and
f ( x0 h) 2 f ( x0 ) f ( x0 h) h2
Example 5.36 [CE-2012 (2 marks)]: The error in ( df ( x ) dx ) x x
0
estimated with h 0.03 using the central difference formula
df ( x ) dx
h2
(ii),
2
f ( x0 )
we
get
.
for a continuous function
f ( x0 h) f ( x0 h)
x x0
2h
is
3 2 10 . The values of x0 and f ( x0 ) are 19.78 and 500.01, respectively. The corresponding error in the central difference estimate for h 0.02 is approximately (a) 1.3 10 4 (b) 3.0 104 (c) 4.5 104 (d) 9.0 104 h 2 f (c1 ) f (c2 ) Solution (d): As the error in central difference formula is E , where 3! 2 c1 ( x0 , x0 h) and c2 ( x0 h, x0 ) . So for h 0.03 , we have
E1 2 10
3
(0.03) 2 f (c1 ) f (c2 )
. For h 0.02 , we have E2
(0.02) 2 f (c1 ) f (c2 )
3! 2 3! 2 2 3 4 4 E2 E1 (0.02) (0.03) E2 2 10 (4 9) 8.89 10 9 10 .
2
. So
Example 5.37 [MT-2008 (1 mark)]: The value of dy dx for the following data set at x 3.5 , computed by central difference method is x (a) 3.5 (b) 7 1 2 3 4 5 y (c) 10.5 (d) 14 0 3 8 15 24 dy f (4) f (3) 15 8 7. Solution: From central difference method, we have dx x 3.5 43 1 [Similar question was also asked in ME-1999 (1 mark)] Example 5.38 [ME-2014 (1 mark)]: The best approximation of the minimum value attained by e x sin(100 x ) for x 0 is ……………. y e x sin(100 x ) y e x {100 cos(100 x ) sin(100 x)} ,
Solution: x
y e {9999 sin(100 x ) 200 cos(100 x)} . For maxima and minima, y 0 , since e
and x
0 , so
{100 cos(100 x ) sin(100 x )} 0 tan(100 x ) 100 100 x tan 1 (100) 89.43o or 269.43o x 0.8943o or 2.6943o ; and y x 0.8943o 0 and y x 2.6943o 0 . So x 0.8943o is the point of o maxima and x 2.6943 is the point of minima and thus the minimum value will be y e 2.6943( 180) sin(100 2.6943) 0.954 0.95 . Note that while calculating the power of
exponential we have to convert x 2.6943o in radians. So answer is 0.95 .
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Chapter 5: Numerical Methods
[5.23]
Exercise: 5.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The number 0.0008857 when rounded off to three significant digits yields (a) 0.001 (b) 0.000886 (c) 0.000885 (d) None of these 2. When a number is approximated to n decimal places by chopping off the extra digits, then the absolute value of the relative error does not exceed (d) None of these (a) 10 n (b) 10 n 1 (c) 0.5 10 n 1 3. If e1 and e2 are absolute errors in two numbers n1 and n2 respectively due to rounding or truncation, then (e1 n1 ) (e2 n2 ) is (a) e1 e2
(b) e1 e2
(c) e1 e2
(d) e1 e2
3
4. The root of the equation x 6 x 1 0 lies in the interval (a) (2,3) (b) (3, 4) (c) (3,5) (d) (4, 6) 5. For the equation f ( x ) 0 , if f ( a ) 0, f (b) 0, f (c ) 0 and b c then we will discard the value of the function f ( x ) at the point (a) a (b) b (c) c (d) Either a or b or c x 6. The positive root of the equation e x 3 0 lies in the interval (a) (0,1) (b) (1, 2) (c) (2,3) (d) (2, 4) 7. If f ( a ) f (b) 0 , then an approximate value of a real root of f ( x ) 0 lying between a and b is given by af (b) bf ( a) af (b) bf ( a ) bf ( a ) af (b ) (c) (a) (b) (d) None of these f (b) f ( a ) ba ba 8. A root of the equation x 3 x 1 0 lies between 1 and 2. Its approximate value as obtained by applying bisection method 3 times is (a) 1.375 (b) 1.625 (c) 1.125 (d) 1.25 x 9. The nearest real root of the equation xe 2 0 correct to two decimal places, is (a) 1.08 (b) 0.92 (c) 0.85 (d) 0.80 10. By the false position method, the root of the equation x 3 9 x 1 0 lies in interval (2, 4) after first iteration. It is (a) 3 (b) 2.5 (c) 3.57 (d) 2.47 3 11. The equation x 3 x 4 0 has only one real root. What is its first approximate value as obtained by the method of false position in (–3, –2)? (a) –2.125 (b) 2.125 (c) –2.812 (d) 2.812 12. Newton-Raphson method is applicable only when (a) f ( x ) 0 in the neighborhood of actual root x (b) f ( x) 0 in the neighborhood of actual root x (c) f ( x ) 0 in the neighborhood of actual root x (d) None of these 13. Newton-Raphson processes has a (a) Linear convergence (b) Quadratic convergence (c) Cubic convergence (d) None of these 3 14. The real root of the equation x x 5 0 lying between –1 and 2 after first iteration by Newton-Raphson method is (a) 1.909 (b) 1.904 (c) 1.921 (d) 1.940 15. The Newton-Raphson method converges fast if f ( ) is ( is the exact value of the root) (a) Small (b) Large (c) 0 (d) None of these
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Chapter 5: Numerical Methods
[5.24]
16. If one root of the equation f ( x ) 0 is near to x0 then the first approximation of this root as calculated by Newton-Raphson method is the abscissa of the point where the following straight line intersects the x axis (a) Normal to the curve y f ( x) at the point ( x0 , f ( x0 )) (b) Tangent to the curve y f ( x) at the point ( x0 , f ( x0 )) (c) The straight line through the point ( x0 , f ( x0 )) having the gradient 1 f ( x0 ) (d) The ordinate through the point ( x0 , f ( x0 )) 17. Newton-Raphson method is used to calculate
3
65 by solving x 3 65 . If x0 4 is taken as
initial approximation then the first approximation x1 is (a) 65 16 (b) 131 32 (c) 191 48 (d) 193 48 18. Which of the following statements applies to the bisection method used for finding roots of functions? (a) Converges within a few iterations (b) Guaranteed to work for all continuous functions (c) Is faster than the Newton-Raphson method (d) Requires that there be no error in determining the sign of the function 19. The Newton-Raphson method of finding roots of non-linear equations falls under the category of _____ method. (a) bracketing (b) open (c) random (d) graphical 20. The root of the equation f ( x ) 0 is found by using the Newton-Raphson method. The initial estimate of the root is x0 3 , f (3) 5 . The angle by the line tangent to the function f ( x ) makes at x 3 is 57o w.r.t. x axis. The next estimate of the root, x1 most nearly is (a) –3.2470 (b) –0.24704 (c) 3.2470 (d) 6.2470 3 Iteration number Value of root 21. The root of x 4 is found by using Newton-Raphson 0 2.0000 method. The successive iterative values of the root are 1 1.6667 given in the table. The iteration number at which you 2 1.5911 would first trust at least two significant digits in the 3 1.5874 answer is (a) 1 (b) 2 (c) 3 (d) 4 4 1.5874 22. Assuming an initial bracket of [1, 5] , the second (at the end of 2nd iteration) iterative value of the root of xe x 0.3 0 using bisection method is _____. 23. To find the root of f ( x ) 0 , a person is using bisection method. At the beginning of an iteration, the lower and upper guesses of the root are xl and xu . At the end of the iteration, the absolute relative approximate error in the estimated value of the root would be xu xl x xl x xl (a) (b) (c) u (d) u xu xl xu xl xu xl xu xl 24. For an equation like x 2 0 , a root exists at x 0 . The bisection method cannot be adopted to solve in spite of the root existing at x 0 because the function f ( x) x 2 (a) is a polynomial (b) has repeated roots at x 0 (c) is always non-negative (d) has a slope equal to zero at x 0 25. The minimum number of iterations needed by the bisection algorithm to approximate the root 3 2 3 x 3 of x 6 x 11x 6 0 with error tolerance 10 is _____. 26. A root of e x 3log x by using fixed point iteration with x0 2 as initial guess will converges to (a) 1.115 (b) 1.124 (c) 1.046 (d) 1.356 27. The Regula Falsi method is related to (a) Chord (b) Ordinate (c) Abscissa (d) Tangent
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[5.25]
28. The value of x for which cos x 2 x up to 5 decimal place evaluated by Newton-Raphson method after the 2nd iteration with an initial guess of x0 0.5 is (a) 0.45063 (b) 0.45018 (c) 0.46053 (d) 0.46186 3 29. The iterative formula for finding the solution of x x 1 0 is 2 xn3 1 3 xn3 1 2 xn3 1 3xn3 1 (a) xn 1 2 (b) xn 1 2 (c) xn 1 2 (d) xn 1 2 3 xn 1 2 xn 1 3 xn 1 2 xn 1 30. Which one of the following functions gives the value of x1 closing to the exact root by NewtonRaphson method by using initial guess of x0 2 . (iv) y log 2 x (i) y x 2 3 (ii) y x 3 7 (iii) y e x 1 (a) (i) (b) (ii) (c) (iii) (d) (iv) 31. After doing approximation for n times between x a and x b in bisection method the maximum distance between the approximate root and the actual root will be (a) ( a b) 2n
(b) ( a b) (2n)
(c) ( a b) 2n 1
(d) ( a b) {2( n 1)}
32. Which one of the following statement is correct? (a) The Newton-Raphson method is considered inferior as it requires more calculations (b) The bisection method is considered inferior as it requires two initiating values a and b where as Newton-Raphson method requires only one value (c) In Newton-Raphson method xn 1 is always closer to the actual root in comparison with xn (d) Newton-Raphson method gives better result when f ( x ) is very large near the actual root 33. If a polynomial of degree n has n 1 zeros, then the polynomial is (a) oscillatory (b) zero everywhere (c) quadratic (d) not defined 34. The Newton’s divided difference second order polynomial for the x y x : 15 18 22 data given in the table is given by y : 24 37 25 f ( x ) b b ( x 15) b ( x 15)( x 18) . The value of b is most nearly 2
0
1
2
1
(a) 1.0480 (b) 0.14333 (c) 4.3333 (d) 24.000 35. Velocity versus time for a particle is approximated by a second order Newton’s divided difference polynomial as v (t ) b0 39.622(t 20) 0.5540(t 20)(t 15) , 10 t 20 . The acceleration in m/s2 at t 15 is _____. 36. The Lagrange polynomial that passes through the 3 data points is given by x : 15 18 22 y : 24 37 25 f 2 ( x ) L0 ( x )(24) L1 ( x )(37) L2 ( x )(25) . Value of L1 ( x ) at x 16 is __ 37. For the function f given in the table, the Newton 1 0 2 x: 32 divided difference interpolating polynomial is f ( x) : 3 3 13 4 53 (a) p ( x) 3 ( x 1) (1 3)( x 1)( x 1.5) 2( x 1)( x 1.5) x (b) p ( x) 3 (1 2)( x 1) (1 3)( x 1)( x 1.5) 2( x 1)( x 1.5) x (c) p ( x) 3 (1 2)( x 1) (1 3)( x 1)( x 1.5) ( x 1)( x 1.5) x (d) p ( x) 3 (1 2)( x 1) (1 3)( x 1)( x 1.5) 2( x 1)( x 1.5) 38. The first derivative of f ( x ) at x 1.5 , using the data given in the table, by using Newton’s forward difference formula, is _____. 1.5 2.0 2.5 3.0 3.5 4.0 x: f ( x) : 3.375 7.000 13.625 24.000 38.875 59.000 39. The first derivative of f ( x ) at x 2.03 , using the data given in the table, by using Newton’s backward difference formula, is _____. 1.96 1.98 2.00 2.02 2.04 x: f ( x) : 0.7825 0.7739 0.7651 0.7563 0.7473 (a) 0.44876 (b) 4.4876 (c) –0.44876 (d) –4.4876
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.26]
5.2 Numerical Solution of Definite Integrals, First Order ODEs, and System of Linear Equations 5.2.1 Numerical Solution of Definite Integrals b
Numerical integration means the numeric evaluation of integrals I f ( x ) dx where a and b are a
given and f is a function given analytically by a formula or empirically by a table of values. Geometrically I is the area under the curve of I between a and b . We know that if f is such that we can find a differentiable function F whose derivative is f , then we can evaluate I by applying b
the familiar formula I f ( x ) dx F (b ) F (a ) , where F ( x ) f ( x) . a
Many times the anti-derivative of the integrand is not easy to obtain. The Newton-Cotes formulas are an extremely useful and straightforward family of numerical integration techniques. To integrate a function f ( x ) over some interval [ a, b] , divide it into n equal parts such that f n f ( xn ) and
h (b a ) n . Then find polynomials which approximate the tabulated function, and integrate them to approximate the area under the curve. To find the fitting polynomials, use Lagrange interpolating polynomials. The resulting formulas are called Newton-Cotes formulas, or quadrature formulas. Newton-Cotes formulas may be ‘closed’ if the interval [ x1 , xn ] is included in the fit, ‘open’ if the points [ x2 , xn 1 ] are used, or a variation of these two. We will consider only closed Newton-Cotes formulas. The basic method involved in approximating uses a sum
n
i 0 ai f ( xi )
to approximate
b
a
b
a
f ( x) dx is called numerical quadrature. It
f ( x) dx . The method of quadrature in this section are
based on the interpolation polynomials, discussed in Section 5.1.5. The basic idea is to select a set of distinct nodes { x0 , x1 , , xn } from the interval [ a, b] . Then integrate the Lagrange interpolating n
Pn ( x ) i 0 f ( xi ) Li ( x )
polynomial b
a
f ( x) dx
b
a
and its truncation error term over [ a, b]
b
n
n
i0 f ( xi ) Li ( x)dx a i0 ( x xi )
b
n
f ( x ) dx i 0 ai f ( xi ) a
1
b
f
n
( n 1)
( n 1)!
i0 ( x xi ) f ( n 1)! a
b
( x)
( n 1)
dx
( x) dx ,
ai Li ( x ) dx , i 0,1, , n . The quadrature formula is therefore a
error E ( f )
1
b
n
i0 ( x xi ) f ( n 1)! a
( n 1)
( x) dx .
to obtain,
where ( x ) [a, b] x and b
a
n
f ( x )dx i 0 ai f ( xi ) with
Now, let us consider formulas produced by
using first and second Lagrange polynomial with equally – spaced nodes. This gives the Trapezoidal rule and Simpson’s rule.
Rectangular Rule and Trapezoidal Rule: Numeric integration are obtained by approximating the integrand f by functions that can be easily integrated. The simplest formula, the rectangular rule, is obtained if we subdivide the interval of integration a x b into n subintervals of equal length
h (b a ) n and in each subinterval approximate f by the constant f ( x*j ) , the value of f at the *
midpoint x j of the j th subinterval, as shown in Fig. 5.8. Then f is approximated by a step function (piecewise constant function), the n rectangles in Fig. 5.8 have the areas f ( x1* ) h, , f ( x*n ) h and the rectangular rule is given as, b
I f ( x) dx h[ f ( x1* ) f ( x2* ) f ( xn* )] , where h (b a ) n a
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(5.35)
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Chapter 5: Numerical Methods
[5.27]
Figure 5.9: Trapezoidal rule
Figure 5.8: Rectangular rule
The 2-point closed Newton-Cotes formula is called the Trapezoidal rule because it approximates the area under a curve by a trapezoid with horizontal base and sloped top. The trapezoidal rule is generally more accurate. We obtain it if we take the same subdivision as before and approximate f by a broken line of segments (chords) with endpoints [ a, f ( a )],[ x1 , f ( x1 )], ,[b, f (b)] on the curve of f as shown in Fig. 5.9. The nature of approximation used for the function in each interval is linear. [This point was asked in MT-2014 (1 mark)]. Then the area under the curve of f between
a and b is approximated by n trapezoids of areas (1 2)[ f ( a ) f ( x1 )]h , (1 2)[ f ( x1 ) f ( x2 )]h , , (1 2)[ f ( xn 1 ) f (b)]h . By taking their sum we get the trapezoidal rule as, b
I f ( x ) dx h (1 2) f ( a ) f ( x1 ) f ( x2 ) f ( xn 1 ) (1 2) f (b) a
(5.36)
where, h (b a ) n . The x j ’s and a and b are called nodes.
Error Bounds and Estimate for the Trapezoidal Rule: An error estimate for the trapezoidal rule can be derived from Eq. 5.19 with n 1 by integration as follows. For a single subinterval we have, f ( x ) p1 ( x ) ( x x0 )( x x1 )(1 2) f (t ) with a suitable t depending on x , between x0 and x1 .
Integration x0 h
x
0
x
over
from
f ( x) dx (h 2)[ f ( x0 ) f ( x1 )]
x0 h
x0
a x0
x1 x0 h
to
gives,
( x x0 )( x x0 h)(1 2) f t ( x ) dx . Setting ( x x0 ) v
and applying the mean value theorem, which we can use because ( x x0 )( x x0 h) does not change the sign, we find the right side equals to h f (t ) h3 h3 f (t ) h3 (5.37) v ( v h ) dv f (t ) 3 2 2 0 2 12 where t is a (suitable, unknown) value between x0 and x1 . This is the error for the trapezoidal rule with n 1, often called the local error. Hence the error T of Eq. 5.36 with any n is the sum of such contributions from the n subintervals; as h (b a ) n , nh3 (b a)3 n 2 , (b a ) 2 n 2 h 2 , thus,
T
nh3
f (tˆ )
(b a )3
f (tˆ)
(b a )
2 h f (tˆ)
(5.38) 12 12n 12 with (suitable, unknown) tˆ between a and b . Because of Eq. 5.38, the trapezoidal rule (Eq. 5.36) is also written as, b 1 1 (b a) 2 ˆ I f ( x) dx h f ( a) f ( x1 ) f ( x2 ) f ( xn 1 ) f (b) h f (t ) (5.39) a 2 12 2 2
Error bounds are now obtained by taking the largest value for f , say, M 1 , and the smallest value, M 2 , in the interval of integration. Then Eq. 5.38 gives (note that k is negative) k M 1 T k M 2 , where k
Copyright © 2016 by Kaushlendra Kumar
(b a )3 12n
2
ba 12
2
h
n 12
h
3
(5.40)
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.28]
We also note that the error term for the Trapezoidal rule involves f , so the Trapezoidal rule gives the exact result when applied to any function whose second derivative is identically zero, i.e., the Trapezoidal rule gives the exact result for polynomials of degree up to or equal to one [This point was asked in CS-2002 (1 mark)]. The error in Trapezoidal rule is proportional to h3 so it is a third order method, i.e., O ( h3 ) .
Example 5.39 [CS-1997 (2 marks)]: The trapezoidal method to numerically obtain
b
a
f ( x) dx has an
error E bounded by (b a ) 12 h 2 max f ( x ) , x [a, b] , where h is the width of the trapezoids. Minimum number of trapezoids guaranteed to ensure E 104 in computing ln 7 using f 1 x is (a) 60 (b) 100 (c) 600 (d) 10000 7
Solution (c): As ln 7 (1 x) dx , so we have a 1 , b 7 and h (b a ) n , where n is the 1
number of intervals. From the given data we have the maximum error of 104 . Thus from Eq. 5.38 we (b a ) 2 1 1 2 have 104 h max{ f (tˆ)} . Since f ( x ) f ( x) 2 f ( x ) 3 , so in [1, 7] , 12 x x x max{ f ( x )} is at x 1 as
2
f ( x ) 2 x 3
2 . So 104
(7 1)
which is a decreasing function in [1, 7] . Thus
h 2 2 h 102
7 1
10 2 n 600 . x x 1 12 n Example 5.40 [CE-2006 (2 marks)]: A 2nd degree polynomial, f ( x ) , has values of 1, 4 and 15 at
max{ f (tˆ)}
3
x 0, 1 and 2, respectively. The integral
2
0
f ( x )dx is to be estimated by applying the trapezoidal
rule to this data. What is the error (defined as ‘true value – approximate value’) in the estimate? (a) 4 3 (b) 2 3 (c) 0 (d) 2 3 Solution: From the given data, we have a 0 , b 2 , h 1 and n 2 . So using Trapezoidal rule, 2
IT f ( x ) dx h (1 2) f ( a ) f ( x1 ) (1 2) f (b) 1(1 2) 4 (1 2)(15) 12 , where x1 a h . 0
Now for the exact integral I E , we have to find the function f ( x ) . As f ( x ) is a 2nd degree polynomial, so f ( x ) ax 2 bx c and it is given that f (0) 1 , f (1) 4 and f (2) 15 ; after substituting the value and solving for the three unknown from the three equations we get a 4 , b 1 and c 1. Thus we have f ( x) 4 x 2 x 1 . So 2
4 3 x2 32 4 32 32 4 . Thus I E IT 12 . I E (4 x x 1) dx x x 20 0 3 3 2 3 3 0 3 2 2
2
Example 5.41 [ME-2007 (2 marks)]: A calculator has accuracy up to 8 digits after decimal place. The value of
2
0
sin x dx when evaluated using this calculator by trapezoidal method with 8 equal
intervals, to 5 significant digits is (a) 0.00000 (b) 1.0000 (c) 0.00500 (d) 0.00025 Solution: From the given data, we have a 0 , b 2 and we have 8 intervals, i.e., n 8 , and so we have h (b a ) n (2 0) 8 4 . So from Eq. 5.36 with f ( x ) sin x , we have I
2
0
f ( x ) dx h (1 2) f ( a ) f ( x1 ) f ( x2 ) f ( x7 ) (1 2) f (b ) , where xi a ih . So
1 3 5 3 7 1 2 f (0) f 4 f 2 f 4 f f 4 f 2 f 4 2 f (2 ) 1 1 1 1 1 1 I (0) 1 0 1 (0) 0 . So value of I to 4 2 2 2 2 2 2 5 significant digits is I 0.00000 . I
4
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.29]
Example 5.42 [CS-2008 (2 marks)]: The minimum number of equal length subintervals needed to approximate
2
x
1 xe dx
to an accuracy of at least (1 3) 106 using the trapezoidal rule is
(b) 1000 (d) 100 (a) 1000e (c) 100e Solution (a): From the given data we have a 1 , b 2 and h (b a ) n , where n is the number of intervals. As we have the least accuracy, i.e., maximum error of (1 3) 106 . Thus from Eq. 5.38 we (1 3) 106 (b a ) 12 h 2 max{ f (tˆ)} .
have
f ( x ) xe x f ( x ) e x (1 x)
Since
f ( x ) e x (2 x ) , so in [1, 2] , max{ f ( x )} is at x 2 as f ( x ) e x (2 x ) which is an
increasing
1 3
10
6
function
(2 1) 12
2
[1, 2] .
in 2
h 4e h
103 e2
max{ f (tˆ)} e x (2 x)
Thus
ba n
103 e2
2 1 n
10 3 e2
x 2
4e2 .
So
n 1000e .
Example 5.43 [EE-2008 (2 marks)]: A differential equation dx dt e 2 t u (t ) has to be solved using trapezoidal rule of integration with a step size h 0.01 sec. Function u (t ) indicates a unit step function. If x (0 ) 0 , then value of x at t 0.01 sec will be given by (a) 0.00099 (b) 0.00495 (c) 0.0099 (d) 0.0198 Solution (c): As we have a step size of h 0.01 and we have to find x (t ) at t 0.01 so we have this problem having one interval, i.e.,
0, t 0
u (t )
1, t 0
n 1. So
dx dt e 2t u (t ) x (t )
0.01 2 t
e u (t )dt . As
0
t
. So x(t ) e 2t dt . Now using Trapezoidal method with single interval, 0
1 1 1 e 0.02 1 1 x (t ) h f (0) f (t ) x (0.01) h f (0) f (0.01) 0.01 0.0099 . 2 2 2 2 2 2
Example 5.44 [AE-2009 (2 marks)]: The value of the integral
dx
0 1 x sin x
evaluated using the
trapezoidal rule with two equal intervals is approximately (a) 1.27 (b) 1.81 (c) 1.41 (d) 0.71 Solution (c): With two intervals, i.e., n 2 , we have h (b a ) n ( 0) 2 2 . So from Eq. 5.36, with
f ( x ) 1 (1 x sin x ) , we have I f ( x ) dx h (1 2) f ( a ) f ( x1 ) (1 2) f (b) , 0
where a 0 , x1 a h 0 ( 2) 2 , b .
1 1 1 1 1 1 f (0) f f ( ) I (1) 1.414 . 2 2 2 2 2 2 2 ( 2) 2 1 [Similar questions were also asked in AG-2012, AE-2007, CH-2009, TF-2008 (2 marks)] So I
Example 5.45 [MT-2009 (2 marks)]: What is the magnitude of the integral
2
0 (3 x
2
4 x 2) dx
using single step application of trapezoidal rule? (a) 9 (b) 16 (c) 18 (d) 36 Solution (b): From the given data, we have a 0 , b 2 and n 1, so we have h (b a ) n (2 0) 1 2 . So from Eq. 5.36 with f ( x ) 3 x 2 4 x 2 , we have 2
I f ( x ) dx h (1 2) f ( a ) (1 2) f (b) 2 (1 2) f (2) (1 2) f (0) 18 2 16 . 0
1 x2
Example 5.46 [PI-2011 (2 marks)]: The value of
0 e
is equal to (a) 0.6778
(c) 0.6985
(b) 0.7165
Copyright © 2016 by Kaushlendra Kumar
dx , using trapezoidal rule for 10 trapezoids,
(d) 0.7462
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Engineering Mathematics
Chapter 5: Numerical Methods
Solution:
data,
From the given
h (b a ) n (1 0) 10 0.1 .
we
So
a0,
have
from
Eq.
[5.30]
b 1
5.36
n 10 ,
and
f ( x) e
with
x2
so we ,
have
we
have
1
I f ( x) dx h (1 2) f (a ) f ( x1 ) f ( x2 ) f ( x9 ) (1 2) f (b) , where xi a ih i(0.1) . So 0
I (1 10) (1 2) f (0) f (0.1) f (0.2) f (0.9) (1 2) f (1) . As f (0) 1 , f (0.1) e 2
f (0.2) e 0.2 0.96 ,
,
2
f (0.6) e 0.6 0.70 ,
2
2
f (0.3) e 0.3 0.91 , 2
f (0.7) e 0.7 0.61 ,
0.12
0.99
2
f (0.4) e 0.4 0.85 ,
f (0.5) e 0.5 0.78 ,
2
2
f (0.8) e 0.8 0.53 ,
f (0.9) e 0.9 0.44 ,
2
f (1) e 1 0.37 . Thus we have I (1 10) (1 2)(1) 0.99 0.96 0.91 0.85 0.78 0.70 0.61 0.53 0.44 (1 2)(0.37) 0.7455 . So from the given options we can say that option (d) is correct.
Example 5.47 [PI-2009 (2 marks)]: The area under the curve shown, between x 1 and x 3 is to be evaluated using the trapezoidal rule. The following points (in m) on the curve is given.
Point 1 2 3
X coordinate 1 2 3
Y coordinate 1 4 9
The evaluated area (in m2) will be (a) 7 (b) 8.67 (c) 9
Solution (c): From the given data, we have
h (b a ) n (3 1) 2 1 ; and
a 1, b 3
(d) 18
n 2 , so we have
and
3
y f ( x) , we have
A ydx . So from Eq. 5.36 with 1
1
I f ( x ) dx h (1 2) f (a ) f ( x1 ) (1 2) f (b) 1 (1 2) 4 (1 2)(9) 9 , where x1 a h . 0
Example 5.48 [AE-2012 (2 marks)]: The integration
n 4 intervals is …… Solution: From the given h (b a ) n (1 0) 4 1 4 .
data, So
1
3
0 x dx
computed using trapezoidal rule with
we have a 0 , b 1 and n 4 , so we from Eq. 5.36 with f ( x) x 3 , we
1
I f ( x) dx h (1 2) f ( a ) f ( x1 ) f ( x2 ) f ( x3 ) (1 2) f (b) , I
3
1 1
3
3
3
xi a ih .
where
0
3
have have So
3
1 1 1 3 1 1 1 3 1 3 (0) f f f f (1) 0.266 . 4 2 4 2 4 2 4 4 2 4 2
Example 5.49 [CS-2013 (1 mark)]: Function f is known at the following points: x 0 0.3 0.6 0.9 1.2 1.5 1.8 2.1 2.4 f ( x) 0 0.09 0.36 0.81 1.44 2.25 3.24 4.41 5.76 The value of
3
0
2.7 7.29
3.0 9.0
f ( x )dx computed using the trapezoidal rule is
(a) 8.983 (b) 9.003 (c) 9.017 (d) 9.045 Solution (d): From the given data, we have a 0 , b 3 and we have 10 intervals, i.e., n 10 , and so we have h (b a ) n (3 0) 10 0.3 . So from Eq. 5.36, we have 3
I f ( x ) dx h (1 2) f (0) f (0.3) f (0.6) f (2.7) (1 2) f (3) 0
I 0.3 (1 2)(0) (0.9 0.36 7.29) (1 2)(9) 0.3 0 (0.9 0.36 7.29) 4.5 I 0.3 0 (25.65) 4.5 9.045 . [Similar question was also asked in AG-2008 (1 mark)]
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.31]
Example 5.50 [ME-2014 (2 marks)]: Using the trapezoidal rule, and dividing the interval of integration into three equal subintervals, the definite integral Solution: From the given h (b a ) n (1 1) 3 2 3 .
data, So
1
1
we have a 1 , b 1 and n 3 , so we from Eq. 5.36 with f ( x) x , we
1
I f ( x ) dx h (1 2) f (a ) f ( x1 ) f ( x2 ) (1 2) f (b) ,
where
1
I
x dx is …………….
xi a ih .
have have So
2 1
2 4 1 2 1 1 1 1 1 1 1 1 1.11 . 3 2 3 3 2 3 2 3 3 2
Example 5.51 [ME-2014 (2 marks)]: The value of
4
2.5 ln( x)dx
calculated using the Trapezoidal rule
with five sub-intervals is ……………. Solution: From the given data, we have a 2.5 , b 4 and n 5 , so we have h (b a) n (4 2.5) 5 0.3 . So from Eq. 5.36 with f ( x ) ln x , we have I
4
2.5
I
4
2.5
f ( x )dx h (1 2) f ( a ) f ( x1 ) f ( x2 ) f ( x3 ) f ( x4 ) (1 2) f (b) , where
xi a ih . So
f ( x ) dx 0.3 (1 2) ln(2.5) ln(2.8) ln(3.1) ln(3.4) ln(3.7) (1 2) ln(4) 1.75 .
Example 5.52 [ME-2014 (1 mark)]: The definite integral
3
1 (1 x) dx
is evaluated using Trapezoidal
rule with a step size of 1. The correct answer is ……………. Solution: From the given data, we have a 1 , b 3 and h 1, so we have h (b a) n 1 (3 1) n n 2 . So from Eq. 5.36 with f ( x) 1 x , we have, x1 a h , so, 3
I f ( x) dx h (1 2) f ( a ) f ( x1 ) (1 2) f (b) 1(1 2)(1) (1 2) (1 2)(1 3) 1.1667 . 1
Simpson’s 1/3 Rule of Integration: Simpson’s 1/3 rule is a Newton-Cotes formula for approximating the integral of a function f ( x ) using quadratic polynomials (i.e., parabolic arcs instead of the straight line segments used in the trapezoidal rule). It can be derived by integrating a third-order Lagrange interpolating polynomial fit to the function at three equally spaced points. Since it uses quadratic polynomials to approximate functions. To derive Simpson’s 1/3 rule, we divide the interval of integration a x b into an even Figure 5.10: Simpson's 1/3rd rule of integration number of equal sub-intervals, say, n 2m subintervals of length h (b a) (2m) , with endpoints, x0 ( a ), x1 , x2 m1 , x2 m ( b) , as shown in Fig. 5.10. We now take the first two sub-intervals and
approximate f ( x ) in the interval x0 x x2 x0 2h by Lagrange polynomial p2 ( x ) through ( x0 , f0 ) , ( x1 , f1 ) , ( x2 , f 2 ) , where f j f ( x j ) . From Eq. 5.17, p2 ( x )
( x x1 )( x x2 ) ( x0 x1 )( x0 x2 )
f0
( x x0 )( x x2 ) ( x1 x0 )( x1 x2 )
f1
( x x0 )( x x1 ) ( x2 x0 )( x2 x1 )
f2
(5.41)
The denominators in Eq. 5.41 are 2h 2 , h 2 and 2h 2 , respectively. Setting s ( x x1 ) h , we have, x x1 sh , x x0 x ( x1 h) ( s 1) h and x x2 x ( x1 h) ( s 1) h and we get, p2 ( x) (1 2) s ( s 1) f 0 ( s 1)( s 1) f1 (1 2)( s 1) sf 2 . We now integrate w.r.t. x from x0 to x2 . This corresponds to integrating with respect to s from –1 to 1. Since dx hds , the result is, x2
x
0
f ( x) dx
x2
x0
p2 ( x )dx ( h 3) f 0 4 f1 f 2
Copyright © 2016 by Kaushlendra Kumar
(5.42)
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.32]
A similar formula holds for the next two subintervals from x2 to x4 , and so on. By summing all these
m formulas we have, or,
b
a
b
a
f ( x) dx ( h 3) f 0 4 f1 2 f 2 4 f3 2 f 4 2 f 2 m 2 4 f 2 m 1 f 2 m
f ( x) dx ( h 3) f 0 4 f1 f 3 f 2 m 1 2 f 2 f 4 f 2 m 2 f 2 m
(5.43)
where, h (b a) (2m) , f j f ( x j ) . [The formula in Eq. 5.43 was asked in PI-2010 (1 mark)].
Error Estimation of Simpson’s 1/3 Rule: If the fourth derivative f (4) exists and is continuous on a x b , the error of Eq. 5.43, call it as S1 3 , is given as,
S1 3
(b a )5
f (4) (tˆ)
(b a)
h 4 f (4) (tˆ)
(2m)
h5 f (4) (tˆ)
(5.44) 180(2m) 180 180 here tˆ is a suitable unknown value between a and b . With this Eq. 5.43 can be written as, b h (b a ) 4 (4) a f ( x)dx 3 f0 4 f1 f3 f 2m 1 2 f2 f 4 f 2m 2 f2 m 180 h f (tˆ) (5.45) We also note that the error term for the Simpson’s 1/3 rule f (4) , so the Simpson’s 1/3 gives the exact result when applied to any function whose fourth derivative is identically zero, i.e., the Simpson’s 1/3 rule gives the exact result for polynomials of degree up to or equal to three. [This point was asked in CS-1993, EC-1993, AG-2011 (1 mark)]. The error in Simpson rule is proportional to h5 so it is a third order method, i.e., O ( h5 ) . [This point was asked in ME-2003 (1 mark)] Error Bounds in Simpson’s 1/3 Rule: By taking for f (4) in Eq. 5.44, the maximum M 1 and 4
minimum M 2 on the interval of integration we get the error bounds as (note that C is negative)
CM 1 S1 3 CM 2 , where C
(b a )5 180(2m)
4
(b a ) 180
h4
(5.46)
Example 5.53 [AG-2007 (1 mark)]: Integrating the function f ( x ) 1 e x sin(4 x) over the interval [0,1] using Simpson’s 1/3rd rule gives (a) 1.021 (b) 1.091 (c) 1.321 (d) 2.642 Solution (c): From the given data we have a 0 , b 1 , and let we have two subintervals so we have m 1 , h (b a ) 2m (1 0) 2 h 0.5 . So, with f ( x ) 1 e x sin(4 x) , and xi a ih from Eq. 5.43, we have
1
0 f ( x)dx (h 3) f ( x0 ) 4 f ( x1 ) f ( x2 ) (0.5 3) 1 4(1.551) 0.722 1.321 ,
[Similar questions were also asked in AE-2014 (1 mark), PI-2008, (2 marks)] Statement for Linked Answer Questions 5.54 & 5.55: The following two questions relate to Simpson’s rule for approximating the integral
b
a
f ( x) dx on the interval [ a, b] .
Example 5.54 [AE-2008 (2 marks)]: Which of the following gives the correct formula for Simpson’s rule? (a) k f (b) f {(a b) 2} (b) k f ( a ) f (b ) 2 f {( a b ) 2} (c) k f ( a ) f (b ) 3 (4 3) f {( a b ) 2}
(d) k f ( a ) f (b ) 3 (4 3) f {( a b ) 3}
where k (b a) 2 . Solution (c): From the given options we can say that the step size is h (b a ) (2 m ) (b a ) 2 m 1 . So from Eq. 5.43, we have
x1 x0 h a
ba 2
ab 2
b
a
f ( x ) dx ( h 3) f ( x0 ) 4 f ( x1 ) f ( x1 ) , where x0 a ,
and x2 x0 2h b ; thus I
Copyright © 2016 by Kaushlendra Kumar
b a f ( a ) f (b) 2
3
a b . 3 2
4
f
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.33]
Example 5.55 [AE-2008 (2 marks)]: The percentage error (with respect to the exact solution) in 1
3
estimation of the integral
0 x dx
(a) 5.3
(b) 3.5
using Simpson’s rule is (c) 2.8
(d) 0 3
Solution: Using the result of previous problem we have I
3
1 0 0 1
2
3
3 41
0.25 . Also the
32
1
exact value of the given integral is I E x3 dx ( x 4 4)10 1 4 0.25 . So the percentage error with 0
respect to the exact solution is ( I E I ) I E 100 0 . Example 5.56 [CH-2008 (2 marks)]: Using Simpson’s 1/3rd rule and FOUR equally spaced intervals ( n 4 ), estimate the value of the integral
4
0 (sin x)
(cos 3 x ) dx
(a) 0.3887 (b) 0.4384 (c) 0.5016 (d) 0.5527 Solution (c): From the given data we have a 0 , b 4 , and we have 4 subintervals, i.e.,
n 2m 4 h (b a) 2m ( 4 0) 4 h 16 . So we have an even number of equal subintervals and hence we apply Simpson’s 1/3 rule. So, with f ( x ) (sin x) (cos3 x ) , from Eq. 5.43, we have I
4
0
I
4
0
f ( x) dx ( h 3) f ( x0 ) 4 f ( x1 ) f ( x3 ) 2 f ( x2 ) f ( x4 ) , where xi a ih ; thus
f ( x ) dx
16 sin( 16) sin(3 16) sin(2 16) sin(4 16) 0 4 2 3 0.5016 3 3 3 cos (2 16) cos 3 (4 16) cos ( 16) cos (3 16)
Example 5.57 [XE-2009 (2 marks)]: Simpson’s 1/3 rule applied to
1
1 (3 x
2
5)dx , with sub-interval
h 1, will give (a) the exact result (b) error between 0.01% to 0.1% (c) error between 0.1% to 1.0% (d) error 1.0% 2 Solution: (a): As the given function f ( x ) 3x 5 is a 2nd degree polynomial and we know that Simpson’s 1/3 rule gives the exact result for polynomials of degree up to or equal to three. So option (a) is correct. Example 5.58 [CE-2010 (2 marks)]: The table below x 0 0.25 0.50 0.75 1.0 gives values of a function f ( x ) obtained for values of x f ( x ) 1 0.9412 0.8 0.64 0.50 at intervals of 0.25. The value of the integral of the function between the limits 0 to 1 using Simpson’s rule is (a) 0.7854 (b) 2.3562 (c) 3.1416 (d) 7.5000 Solution (a): From the given data we have a0, b 1, h 0.25 h (b a ) 2m 0.25 (1 0) 2m m 2 . So we have an even number of equal sub-intervals and hence we apply Simpson’s 1/3 rule. Thus from Eq. 5.43, we have 1
0 f ( x)dx (h 3) f ( x0 ) 4 f ( x1 ) f ( x3 ) 2 f ( x2 ) f ( x4 ) , where xi a ih ; So from the given 1 data we have f ( x) dx (0.25 3) 1 4 0.9412 0.64 2(0.8) 0.50 0.7854 0 [Similar questions were also asked in TF-2007, MN-2007, TF-2012 (2 marks)] Example 5.59 [CE-2012 (1 mark)]: The estimate of
1.5
0.5 (1 x)dx
obtained using Simpson’s rule with
three-point function evaluation exceeds the exact value by (a) 0.235 (b) 0.068 (c) 0.024 (d) 0.012 Solution (d): As three-point function evaluation means we have an even number of equal subintervals, i.e., with m 1 , we have n 2 , and so we apply Simpson’s 1/3 rule. Thus with a 0.5
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.34]
and b 1.5 , h (b a ) 2m (1.5 0.5) 2 0.5 . So, with f ( x ) 1 x , from Eq. 5.43, we have 1.5
0.5 f ( x)dx (h 3) f ( x0 ) 4 f ( x1 ) f ( x2 ) ; 1.5
0.5 f ( x)dx
0.5 1
as x0 a 0.5 , x1 a h 1 and x2 b 1.5 . So
1 1 4 1.111 . As 3 0.5 1 1.5
the exact value of
1.5
1
1.5
0.5 x dx (ln x)0.5 ln1.5 ln 0.5 1.099 ,
1.5
1.5
0.5 (1 x)dx . So the estimate of 0.5 (1 x)dx
which is
obtained using Simpson’s rule with
three-point function evaluation exceeds the exact value by (1.111 1.099) 0.012 . [Similar questions were also asked in EE-1998, ME-2011, MN-2009 (2 marks)] Example 5.60 [XE-2012 (2 marks)]: The exact solution of the integral
4
0 ( x
2
4)dx is denoted by
I E . The same integral evaluated numerically by the trapezoidal rule and the Simpson’s 1/3 rule are
denoted by IT and I S , respectively. The subinterval used in the numerical method is h 2 , then (a) I E I S IT
(b) I E I S IT
(c) I E I S IT
4
4
0
0
(d) IT I S I E
Solution (b): I E ( x 2 4) dx ( x 3 3) 4 x 16 3 . For IT , we have a 0 , b 4 and h 2
2 (4 0) n n 2 ; IT h (1 2) f ( a) f ( a h) (1 2) f (b) 2 (1 2)( 4) 0 (1 2)(12) 8 . For I S , as we have 2nd degree polynomial and we know that Simpson’s 1/3 rule gives the exact result for polynomials of degree up to or equal to three. So I S 16 3 . Thus we have I E I S IT . Example 5.61 [CE-2013 (2 marks)]: The magnitude as the error (correct to two decimal places) in
the estimation of integral
4
0 ( x
4
10) dx using Simpson 1 3 rule, by taking the step length as 1, is ___
Solution: From the given data we have a0, b 4, h 1 h (b a) 2m 1 (4 0) 2m m 2 . So we have an even number of equal sub-intervals and f ( x ) x 4 10 , from Eq. 5.43, we have
hence we apply Simpson’s 1 3 rule. So, with 4
I S f ( x )dx ( h 3) f ( x0 ) 4 f ( x1 ) f ( x3 ) 2 f ( x2 ) f ( x4 ) , 0
where
xi a ih ;
thus
I S (1 3) f (0) 4 f (1) f (3) 2 f (2) f (4) (1 3) 10 4 11 91 2(26) 266 245.33 . 4
Now the exact value of the given integral is I E ( x 4 10) dx {( x 5 5) 10 x}40 244.8 . So the 0
magnitude of error in the estimation of the given integral using Simpson’s 1 3 rule is I E I S 244.8 245.33 0.53 .
Example 5.62 [CH-2013 (2 marks)]: The value of the integral
0.5
0.1 e
x3
dx evaluated by Simpson’s
rule using 4 subintervals (up to 3 digits after the decimal point is) …… Solution: From the given data we have a 0.1 , b 0.5 , and we have 4 subintervals, i.e., n 2m 4 h (b a) 2m (0.5 0.1) 4 h 0.1 . So we have an even number of equal sub-intervals and hence we apply Simpson’s 1/3 rule. So, with
3
f ( x ) e x , from Eq. 5.43, we have
2
1 f ( x)dx (h 3) f ( x0 ) 4 f ( x1 ) f ( x3 ) 2 f ( x2 ) f ( x4 ) , where 0.5 0.1 f ( x)dx (0.1 3) 0.999 4 0.992 0.938 2(0.973) 0.882 0.385 .
xi a ih ;
thus
[Similar question was also asked in AG-2013 (2 marks)] Example 5.63 [CS-2014 (2 marks)]: With respect to the numerical evaluation of the definite integral, b
K x 2 dx , where a and b are given, which of the following statements is/are TRUE? a
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[5.35]
A. The value of K obtained using the trapezoidal rule is always greater than or equal to the exact value of the definite integral B. The value of K obtained using the Simpson’s rule is always equal to the exact value of the definite integral (a) A only (b) B only (c) Both A and B (d) Neither A or B Solution (c): Let I E and IT be the exact solution and evaluated by Trapezoidal rule, respectively, of the given integral. As from Eq. 5.39, we can write IT I E (b a ) 12 h 2 f (tˆ) ; so we can say that I T I E , where equality holds when the given function is of at most degree one. Also the given function is 2nd degree polynomial and we know that Simpson’s 1/3 rule gives the exact result for polynomials of degree up to or equal to three. So I S I E , where I S is the integral evaluated by using Simpson’s rule. So both statements are correct.
x: 0 Example 5.64 [MA-2014 (2 marks)]: Let the 0.25 0.5 0.75 1.0 y : following discrete data be obtained from a curve 1 0.9896 0.9589 0.9089 0.8415 y y ( x ) . Let S be the solid of revolution obtained by rotating the above curve about the x axis between x 0 and x 1 and let V denote its volume. The approximate value of V , obtained using Simpson’s 1/3rd rule, is ……………. Solution: If a curve y f ( x) , between x 0 and x 1 , is rotated about x axis then the volume 1
1
0
0
generated will be V y 2 dx y 2 dx . Now from the given data we have a 0 , b 1 , we have four equal sub-intervals, i.e., n 4 or m 2 so h (b a ) 2m (1 0) 4 0.25 . Thus from Eq. 1
5.43, we have I y 2 dx ( h 3) f ( x0 ) 4 f ( x1 ) f ( x3 ) 2 f ( x2 ) f ( x4 ) , where xi a ih 0
2
and f ( xi ) { y ( xi )} . Thus I (0.25 3) 1 4 0.9793 0.8261 2(0.9195) 0.7081 0.8974 . So
V I 0.8974 2.82 . Example 5.65 [PI-2014 (2 marks)]: Using the Simpson’s 1/3rd rule, the value of
ydx
x y
computed, for the data given below is ………
1 2
3 6
5 4
Solution: From the given data we have a 1 , b 5 , and let we have two subintervals so we have m 1 , h (b a) 2m (5 1) 2 h 2 . So, with y f ( x) , from Eq. 5.43, we have 1
0 ydx (h 3) f (1) 4 f (3) f (5) (2 3) 2 4(6) 4 20 . 2 Example 5.66 [TF-2014 (2 marks)]: e x dx is evaluated both by trapezoidal rule and Simpson’s 0 1/3rd rule by taking two subintervals. The difference between the results, accurate to the second decimal place, is ……………. Solution: From the given data we have a 0 , b 2 , and n 2 h (b a) n (2 0) 2 1 . So evaluation of the given integral by Trapezoidal rule is 2
IT e x dx h (1 2) f ( a ) f ( x1 ) (1 2) f (b) 1 (1 2) e (1 2) e 2 6.913 , where x1 a h . 0 Now for Simpson’s 1/3 rule, we have n 2m m 1 , so 2
I S e x dx ( h 3) f ( a ) 4 f ( x1 ) f (b ) (1 3) 1 4e e 2 6.420 , 0
where
x1 a h .
So
IT I S 6.913 6.420 0.493 0.49 .
Simpson’s 3/8 Rule: In previous section, Simpson’s 1/3 rule for integration was derived by piecewise 2nd order (quadratic) polynomial approximation of the integrand f ( x ) . In a similar fashion, Simpson’s 3/8 rule for integration can be derived by approximating the given function with 3 rd order (cubic) polynomial.
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To derive Simpson’s 3/8 rule, we divide the interval of integration a x b into an thrice number of equal sub-intervals, say, n 3m subintervals of length h (b a ) (3m) , with endpoints, x0 ( a ), x1 , x3 m 1 , x3 m ( b) , as shown in Fig. 5.12. We now take the first three sub-intervals and approximate f ( x ) in the interval
x0 x x3 x0 3h
by Lagrange
polynomial p3 ( x ) through ( x0 , f0 ) , ( x1 , f1 ) ,
Figure 5.11: Simpson's 3/8 Rule
( x2 , f 2 ) , ( x3 , f3 ) where f j f ( x j ) . From Eq. 5.17, we get, p3 ( x)
( x x1 )( x x2 )( x x3 ) ( x0 x1 )( x0 x2 )( x0 x3 ) ( x x0 )( x x1 )( x x3 ) ( x2 x0 )( x2 x1 )( x2 x3 )
f0 f2
( x x0 )( x x2 )( x x3 ) ( x1 x0 )( x1 x2 )( x1 x3 ) ( x x0 )( x x1 )( x x2 ) ( x3 x0 )( x3 x1 )( x3 x2 )
f1
(5.47) f3
The denominators in Eq. 5.47 are 6h3 , 2h 3 , 2h3 and 6h3 , respectively. Setting s ( x x1 ) h , we have, x x1 sh , x x0 x ( x1 h) ( s 1) h and x x2 x ( x1 h) ( s 1) h and we get, 1 1 1 1 p3 ( x ) s ( s 1)( s 2) f 0 ( s 1)( s 1)( s 2) f1 ( s 1) s ( s 2) f 2 ( s 1) s ( s 1) f3 . Now 6 2 2 6 integrate w.r.t. x from x0 to x3 . This corresponds to integrating w.r.t. s from –1 to 2. As dx hds , the result is, x3
x
0
x3
f ( x ) dx p2 ( x ) dx (3h 8) f 0 3 f1 3 f 2 f 3
(5.48)
x0
A similar formula holds for the next two subintervals from x3 to x6 , and so on. By summing all these m formulas we obtain Simpson’s rule as, I (3h 8) f 0 3 f1 3 f 2 f 3 f 3 3 f 4 3 f5 f 6 ..... f n 3 3 f n 2 3 f n 1 f n (5.49)
n 2
n 1
n 3
I (3h 8) f 0 3 i 1,4,7,... fi 3i 2,5,8,... fi 2 i 3,6,9,... fi f ( xn )
(5.50)
where, h (b a ) (3m) and f j f ( x j ) .
Error Estimation of Simpson’s 3/8 Rule: If the fourth derivative f (4) exists and is continuous on a x b , the error of Eq. 5.50, call it as S3 8 , is given as,
S3 8
3(b a) 5
f (4) (tˆ)
3(b a )
h 4 f (4) (tˆ)
3(3m)
h5 f (4) (tˆ)
(5.51) 80(3m) 80 80 here tˆ is a suitable unknown value between a and b . With this we may also write Simpson’s rule (Eq. 5.50) as, b h 3(b a ) 4 (4) a f ( x)dx 3 f 0 4 f1 f3 f 2 m1 2 f 2 f 4 f 2m 2 f 2m 80 h f (tˆ) (5.52) We also note that the error term for the Simpson’s 3/8 rule f (4) , so the Simpson’s 3/8 gives the exact result when applied to any function whose fourth derivative is identically zero, i.e., the Simpson’s 3/8 rule gives the exact result for polynomials of degree up to or equal to three. Error Bounds in Simpson’s 3/8 rule: By taking for f (4) in Eq. 5.51, the maximum N1 and 4
minimum N 2 on the interval of integration we obtain the error bounds
DN1 S3 8 DN 2 , where D
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3(b a ) 80
h4
(5.53)
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Example 5.67 [ME-2013, PI-2013 (1 mark)]: Match Column-I (Numerical Integration Scheme) with Column-II (Order of Fitting Polynomial): Column-I Column-II (a) A – 2, B – 1, C – 3 A. Simpson’s 3/8 rule 1. First (b) A – 3, B – 2, C – 1 B. Trapezoidal rule 2. Second (c) A – 1, B – 2, C – 3 C. Simpson’s 1/3 rule 3. Third (d) A – 3, B – 1, C – 2 Solution (d): We know that Trapezoidal rule for integration was derived by 1 st order polynomial; Simpson’s 1/3 rule for integration was derived by piecewise 2nd order (quadratic) polynomial; and Simpson’s 3/8 rule for integration can be derived by approximating the given function with 3rd order (cubic) polynomial approximation of the integrand f ( x ) . So option (d) is correct.
5.2.2 Numerical Solution of First Order ODEs An ODE of the first order is of the form F ( x, y , y ) 0 and can be written in the explicit form y dy dx f ( x, y ) . An initial value problem for this equation is of the form: y dy dx f ( x, y ) , y ( x0 ) y0
(5.54)
where x0 and y 0 are given and we assume that the problem has a unique solution on some open interval a x b containing x0 . In this section we will discuss methods of computing approximating numeric values of the solution y ( x ) of Eq. 5.54 at the equidistant points on the x axis, as, x0 , x1 x0 h, x2 x0 2h, x3 x0 3h, , where the step size h is a fixed number. The methods which we are going to discuss are step-by-step methods, by using, in each step, the Taylor series, as, y ( x h) y ( x ) hy ( x ) ( h 2 2) y ( x ) (5.55)
Forward Euler’s Method: Forward Euler’s method (or simply Euler method) provides an approximation of the solution to a first order ODE given by Eq. 5.57. From a graphical perspective, a solution of first-order differential equation is a function whose graph is consistent with the direction field. In other words, the graph of the solution is tangent to the slope lines at every point. Euler’s method approximates the solution by a sequence of points which follows the slope lines. Given Eq. 5.54 and the initial condition y ( x0 ) y0 , Euler’s method approximates the solution at x1 , x2 , , where the consecutive x ’s differ by h , so that xi x0 ih , i 1, 2, . The Figure 5.12: Forward Euler Method increment h is called step size, which is a parameter of the method, which determines the accuracy of the approximation. Let yi denote the approximated value of the solution at xi . The idea is to approximate the solution at each step by the slope line. Consider Fig. 5.12, we start at the initial point ( x0 , y0 ) . The straight line segment is the tangent line to the solution curve at ( x0 , y0 ) . Its slope is y ( x0 ) , which is given by Eq. 5.54, i.e., y ( x0 ) f ( x0 , y0 ) . Hence Forward Euler method suggests 2
3
that in Eq. 5.55, for a small h the higher powers h , h , are very small which suggests the approximation y ( x h) y ( x ) hy ( x) y ( x ) hf ( x, y ) and the following iteration process. In the 1st
step
we
compute
y1 y0 h ( dy dx ) ( x
0 , y0 )
y0 hf ( x0 , y0 )
which
approximates
y ( x1 ) y ( x0 h) . In the 2nd step we compute y2 y1 h ( dy dx ) ( x , y ) y1 hf ( x1 , y1 ) which 1 1
approximates y ( x2 ) y ( x0 2h) , etc., and in general we have,
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yn 1 yn h (dy dx ) ( x
n , yn )
[5.38]
yn h f ( xn , yn ) , ( n 0,1, )
(5.56)
[Eq. 5.56 was asked in AE-2007 (1 mark)]. This method is hardly ever used in practice, but since it is simple, it is nicely explains then principle of methods based on the Taylor series. In Euler’s method, the sequence of lines may also deviates considerably from the curve of solution. Hence there is a modification of this method; two such methods that are of practical importance, namely, ‘improved Euler method’ and ‘Runge – Kutte method’ which are discussed later. Example 5.68 [AG-2007 (1 mark)]: Taking y (0) 0 and using Euler’s method with step size
h 0.1 solution of the differential equation dy dx 2 xy 1 gives the value of y (0.3) as (a) 0.3101 (b) 0.3142 (c) 0.6202 (d) 4.0800 Solution (a): From the given data we have f ( xn , yn ) ( dy dx ) ( x , y ) 2 xn yn 1 , h 0.1 , y (0) 0 . n
n
So from Eq. 5.56, yn 1 yn hf ( xn , yn ) yn h(2 xn yn 1) , where xi x0 ih ; thus starting with x0 0
and
Similarly,
y ( x0 ) 0 , we have
y ( x1 ) y0 h(2 x0 y0 1) 0 0.1(0 1) 0.1 y (0.1) 0.1 .
y ( x2 ) y1 h(2 x1 y1 1) 0.1 0.1(2 0.1 0.1 1) 0.202 y(0.2) 0.202 ;
and
y ( x3 ) y2 h(2 x2 y2 1) 0.202 0.1(2 0.2 0.202 1) 0.31008 y(0.3) 0.31008 0.3101 . [Similar question was also asked in CH-2008 (2 marks)]
Example 5.69 [IN-2010 (2 marks)]: The velocity v (in m/s) of a moving mass, starting from rest, is given as dv dt v t . Using Euler forward difference method (also known as Cauchy-Euler method) with a step size of 0.1 sec, the velocity at 0.2 sec evaluates to (a) 0.01 m/s (b) 0.1 m/s (c) 0.2 m/s (d) 1 m/s Solution (a): From the given data we have f (tn , vn ) ( dv dt ) ( t , v ) vn tn , h 0.1 , v (0) 0 . So n
n
from Eq. 5.56, vn 1 vn h f (tn , vn ) vn h( vn tn ) , where ti t0 ih and t0 0 ; thus starting with t0 0 and v(t0 ) 0 , we have v (t1 ) v (t0 ) h v (t0 ) t0 0 0.1(0 0) 0 y (0.1) 0 . Similarly, y (t2 ) y (t1 ) h{ y (t1 ) t1} 0 0.1(0 0.1) 0.01 y (0.2) 0.01 . [Similar questions were also asked in XE-2010, EC-2010 (1 mark)]
Example 5.70 [XE-2013 (2 marks)]: Using Euler’s method to solve the differential equation dy dx 2 cos(4 x 3) y , y (0) 1 with step-size h 0.25 , the value of y (0.5) is (a) 1.3125 (b) 1.1875 (c) 1.125 (d) 1.0625 Solution: From the given data we have f ( xn , yn ) ( dy dx ) ( x , y ) 2 cos(4 xn 3) yn , h 0.25 , n
n
y (0) 1 . So from Eq. 5.56, yn 1 yn h f ( xn , yn ) yn h 2 cos(4 xn 3) yn , where xi x0 ih ; thus
starting
with
x0 0
y ( x0 ) 1 ,
and
we
y ( x1 ) y0 h 2 cos(4 x0 3) y0 1 0.25 2 1 1.25 y (0.25) 1.25 .
have Similarly,
y ( x2 ) y1 h 2 cos(4 x1 3) y1 1.25 0.25 2 cos( 3) 1.25 1.1875 y (0.5) 1.1875 .
Example 5.71 [CH-2014 (2 marks)]: Consider the following differential equation dy dx x ln( y ) ; y 2 at x 0 . The solution of this equation at x 0.4 using Euler method with a step size of h 0.2 is ……………. Solution: From the given data we have f ( xn , yn ) ( dy dx ) ( x , y ) xn ln( yn ) , h 0.2 , y (0) 2 . n
n
So from Eq. 5.56, yn 1 yn h f ( xn , yn ) yn h xn ln( yn ) , where xi x0 ih ; thus starting with x0 0 and y ( x0 ) 2 , we have y1 y0 h x0 ln( y0 ) 2 0.2{0 ln 2} 2.139 y (0.2) 2.139
. Similarly, y2 y1 h x1 ln( y1 ) 2.139 0.2{0.2 ln 2.139} 2.331 y(0.4) 2.331 .
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[5.39]
Example 5.72 [MA-2014 (1 mark)]: Let y ( x) be the solution to the initial value problem dy dx y 2 x subject to y (1.2) 2 . Using the Euler method with the step size h 0.05 , the approximate value of y (1.3) , correct to two decimal places, is …………….
Solution: From the given data we have f ( xn , yn ) ( dy dx) ( x
n , yn )
yn 2 xn , h 0.05 , y (1.2) 2
. So from Eq. 5.56, yn 1 yn hf ( xn , yn ) yn h yn 2 xn y ( xn 1 ) y ( xn ) h y ( xn ) 2 xn , where
xi x0 ih ;
thus
starting
x0 1.2
with
and
y ( x0 ) 2 ,
y ( x1 ) y ( x0 ) h y ( x0 ) 2 x0 2 0.05 2 2(1.2) 2.10 y (1.25) 2.10 .
we
have
Similarly,
y ( x2 ) y ( x1 ) h y ( x1 ) 2 x1 2.10 0.05 2.10 2(1.25) 2.20 y (1.3) 2.20 .
Backward’s
Euler Method: The Backward Euler’s method also truncates the Taylor series after two terms. The difference is that the derivative is evaluated at point x h instead of at point x . Consider Figure 5.13, we start at the initial point ( x0 , y0 ) . To obtain the next point, ( x1 , y1 ) , we take the derivative at x1 (not at x0 ) and extrapolate it at point ( x0 , y0 ) . So by Equation 5.54, we have y ( x1 ) f ( x1 , y1 ) ; and the following iteration process. In the first step we compute y1 y0 h ( dy dx ) ( x , y ) y0 h f ( x1 , y1 ) 1 1
which approximates y ( x1 ) y ( x0 h) . In the Figure 5.13: Backward Euler Method second step we compute y2 y1 h (dy dx ) ( x , y ) y1 h f ( x2 , y2 ) which approximates y ( x2 ) y ( x0 2h) , etc., and 2
In general,
2
yn 1 yn h ( dy dx ) ( x
n1 , yn1 )
yn h f ( xn 1 , yn 1 ) , ( n 0,1, )
(5.57)
[Eq. 5.57 was asked in CS-1994 (1 mark)]. Normally we do not know the derivative at point x h , although we need it to compute the function value at point x h . In practice this requires a rearrangement of the equation. We call such a numerical scheme an implicit numerical scheme, as opposed to explicit numerical schemes, such as the Forward Euler method, where no such rearrangement is necessary. Example 5.73 [CE-2006 (1 mark)]: The differential equation dy dx 0.25 y 2 is to be solved using the backward (implicit) Euler’s method with the boundary condition y 1 at x 0 and with a step size of 1. What would be the value of y at x 1 ? (a) 1.33 (b) 1.67 (c) 2.00 (d) 2.33 2 Solution: From the given data we have f ( xn , yn ) ( dy dx ) ( x , y ) 0.25 yn , h 1, y (0) 1 . So from n
Eq. 5.57, yn 1 yn hf ( xn 1 , yn 1 ) yn and
y ( x0 ) 1 ,
h (0.25 yn21 )
we
n
, where xi x0 ih ; thus starting with x0 0
have
y1 y0 h (0.25 y12 ) 1 0.25 y12 .
0.25 y12 y1 1 0 (0.5 y1 1) 2 0 y1 1 0.5 2 y (1) 2 .
Improved Euler’s Method: In the improved Euler method or improved Euler – Cauchy method, in each step we compute first the auxiliary value, i.e.
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yn*1 yn hf ( xn , yn )
(5.58)
and then the new value is given as, yn 1 yn (1 2) h f ( xn , yn ) f ( xn 1 , yn*1 ) (5.59) This method has a simple geometric interpretation. In fact, we may say that in the interval from xn to
xn (1 2) h we approximate the solution y by the straight line through ( xn , yn ) with slope f ( xn , yn )
, and then we continue along the straight line with slope f ( xn 1 , yn* 1 ) until x reaches xn 1 . The improved Euler – Cauchy method is predictor – corrector method, because in each step we first predict a value by Eq. 5.58 and then correct it by Eq. 5.59. Example 5.74 [IN-2013 (2 marks)]: While numerically solving the differential equation ( dy dx ) 2 xy 2 0 , y (0) 1 using Euler’s predictor-corrector (improved Euler-Cauchy) method with a step size of 0.2, the value of y after the first step is (a) 1.00 (b) 1.03 (c) 0.97 (d) 0.96 2 Solution (d): From the given data, we have h 0.2 , y (0) 1 , f ( x, y ) (dy dx) ( x, y ) 2 xy ; so f ( xn , yn ) 2 xn yn2
*
*
2
f ( xn 1 , yn 1 ) 2 xn 1 yn 1 ,
and
where
from
Eq.
5.58,
we
have
yn*1 yn hf ( xn , yn ) yn h ( 2 xn yn2 ) yn 2hxn yn2 ; f ( xn 1 , yn* 1 ) 2 xn 1 ( yn 2 hxn yn2 ) 2 . So
from Eq. 5.59, we have yn 1 yn (1 2)h 2 xn yn2 2 xn 1 ( yn 2hxn yn2 ) 2 , where xi x0 ih . Thus starting with
x0 0
and
y ( x0 ) 1 , we have
y1 y0 (1 2) h 2 x0 y02 2 x1 ( y0 2hx0 y02 ) 2
y1 1 (1 2)0.2 0 2(0.2)(1 0) 2 1 (1 2 )0.2( 0.4) 1 0.04 0.96 .
Runge-Kutta (RK) Method: The Taylor’s series method of solving differential equations numerically is restricted by the finding the higher order derivatives. However there is a class of methods known as Runge – Kutta methods which do not require the calculations of higher order derivatives. These methods agree with Taylor’s series solution up to the terms in h r , where r differs from method to method and is called the order of that method. The Euler’s method and improved Euler’s method are the Runge – Kutta methods of first and second order, respectively. The Runge – Kutta methods of order three are not generally used. A method of great practical importance and much greater accuracy than that of improved Euler’s method is the classical Runge – Kutta method of fourth order. For find the solution of y dy dx f ( x, y ) , y ( x0 ) y0 at equidistant points x1 x0 h, x2 x0 2h, , x N x0 Nh . We see that in each step we first compute four auxiliary
quantities k1 , k2 , k3 , k 4 and then the new value yn 1 at xn 1 is given as, yn 1 yn (1 6)( k1 2k 2 2k3 k 4 )
(5.60)
where, n 0,1, 2, , N 1 and xn 1 xn h ; k1 hf ( xn , yn ) ; k2 hf xn (1 2) h, yn (1 2) k1 ; k3 hf xn (1 2) h, yn (1 2) k 2 ; and k4 hf ( xn h, yn k3 ) .
Example 5.75 [ME-2014 (2 marks)]: Consider an ordinary differential equation dx dt 4t 4 . If x x0 at t 0 , the increment in x calculated using Runge-Kutta fourth order multi-step method with a step size of t 0.2 is (a) 0.22 (b) 0.44 (c) 0.66 (d) 0.88 Solution: From the given data we have, f (t , x ) 4t 4 , x x0 at t 0 and h t 0.2 . So from
Eq.
5.60,
we
have
x1 x0 (1 6)( k1 2k 2 2k3 k 4 ) ,
where
t1 t0 h ,
k1 hf (t0 , x0 ) ,
k2 hf t0 (1 2) h, x0 (1 2)k1 , k3 hf t0 (1 2) h, x0 (1 2) k 2 and k4 hf (t0 h, x0 k3 ) . So
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[5.41]
k1 0.2(4t0 4) 0.2(4 0 4) 0.8 ; k2 0.2 4 t 0 (1 2) h 4 0.2 4 0 (0.2 2) 4 0.88 ; k3 0.2 4 t0 (1 2) h 4 0.2 4 0 (0.2 2) 4 0.88 ;
k4 0.2 4(t0 h) 4 0.2 4(0 0.2) 4 0.96 .
Thus x1 x0 (1 6)(0.8 2 0.88 2 0.88 0.96) x0 0.88 x x1 x0 0.88 . Example 5.76 [MN-2011 (2 marks)]: For the equation dy dx 2 x 3 y , the value of y at x 0.1 in one step using Runge-Kutta fourth order method for the condition y 1 when x 0 , is (a) 0.3608 (b) 1.2508 (c) 1.3608 (d) 1.4625 Solution (c): From the given data we have, f ( x, y ) 2 x 3 y , y 1 at x 0 and h 0.1 . So from Eq. 5.60, we have
y1 y0 (1 6)( k1 2k2 2k3 k4 ) , where
x1 x0 h ,
k1 hf ( x0 , y0 ) ,
k2 hf x0 (1 2)h, y0 (1 2) k1 , k3 hf x0 (1 2)h, y0 (1 2) k2 and k4 hf ( x0 h, y0 k3 ) . So k1 0.1(2 x0 3 y0 ) 0.1(2 0 3 1) 0.3 ; k2 0.12 x0 (1 2) h 3 y0 (1 2) k1 0.10.1 3 1 (0.3 2) 0.355 ; k3 0.12 x0 (1 2) h 3 y0 (1 2) k2 0.10.1 3 1 (0.355 2) 0.363 ;
k4 0.1{2( x0 h) 3( y0 k3 )} 0.1{0.2 3(1 0.363)} 0.429 .
So
y1 1 (1 6)(0.3 2(0.355) 2(0.363) 0.429) 1.360 . [Similar questions were also asked in CS-1993 (1 mark), XE-2008, MA-2014 (2 marks)]
Example 5.77 [XE-2007 (2 marks)]: If Runge-Kutta method of order 4 is used to solve the differential equation dy dx f ( x ) , y (0) 0 in the interval [0, h] with step size h , then (a) y ( h) ( h 6) f (0) 4 f ( h 2) f ( h)
(b) y ( h) ( h 6) f (0) f ( h)
(c) y ( h) ( h 2) f (0) f (h) (d) y ( h) ( h 6) f (0) 2 f ( h 2) f ( h) Solution (a): To avoid confusion with the function f , let us consider it is g and at the end we will replace it by f . So we can say that dy dx f ( x ) ; thus we have, f ( x, y ) g ( x ) , y 0 at x 0 and
step size h . So from Eq. 5.60, we have y1 y0 (1 6)( k1 2k2 2k3 k4 ) , where x1 x0 h , k1 hf ( x0 , y0 ) ,
k2 hf x0 (1 2)h, y0 (1 2) k1 ,
k4 hf ( x0 h, y0 k3 ) .
So
k3 h g x0 ( h 2) h g ( h 2) ;
k3 hf x0 (1 2)h, y0 (1 2) k2
k1 h g ( x0 ) h g (0) ;
and
and
k2 h g x0 (h 2) h g ( h 2) ;
k4 h g ( x0 h) h g ( h) .
Thus
y1 y0 (h 6) g (0) 2 g ( h 2) 2 g (h 2) h g ( h) 0 ( h 6) g (0) 4 g ( h 2) g (h) . So replacing g by f , we get y1 ( h 6) f (0) 4 f ( h 2) f ( h) .
Error in Numerical Methods: The numerical methods are simply approximations of the true solution, and as such they inevitably introduce errors. We can classify the errors into three types: (1) The local error describes the error that occurs after a single time step of the method. (2) The global error describes the cumulative effect of the local errors after many time steps. (3) The roundoff error arises from the finite precision with which computers store and process numbers. Let the set {( xn , yn ) : n 0,1,} be an approximation to the solution y ( x ) of the initial value problem. The global error at xn is the difference between the true solution at xn and the approximation yn , i.e., En y ( xn ) yn . The absolute error in the numerical approximation at x xn is En , i.e., the
absolute value of the global error at xn . The local error is the difference between the true solution y ( xn 1 ) and the approximation after one time step, assuming that ( xn , yn ) is a point on the graph of
the true solution (i.e. assuming that yn is exact). A numerical method is of order n , where n is a
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[5.42]
positive integer, if the method is exact for polynomials of degree n or less. In other words, if the true solution of an initial-value problem is a polynomial of degree n or less, then the approximate solution and the true solution will be identical for a method of order n . In general, higher the order, the more accurate will be the method.
Error in Forward Euler Method: Consider the Forward Euler method. The local error at the nth step is given by
Enloc y ( xn 1 ) { yn h f ( xn , yn )} . As from Taylor’s theorem, we have
y ( xn h ) y ( xn ) hy ( xn ) (1 2) h 2 y ( ) yn h f ( xn , yn ) (1 2) h 2 y ( ) , where is the value
between x0 and x0 h . Hence the local error in Forward Euler’s method is E loc (1 2) h 2 y ( ) h 2 [This point was asked in PI-2009 (1 mark)]. Hence the local error in Forward Euler’s method decreases as the square of the step size. Thus if the step size is cut in half, then the local error at each step is reduced by a factor of four. A derivation of the global error is beyond the scope of this text. We shall restrict ourselves to a simple argument. Suppose that y (0) y0 and we want to find y ( X ) . The number of steps required is approximately X h . If the local error is roughly constant for 0 x X , and E( X ) is the global error at x X then E ( X ) number of iterations E loc E ( X ) ( X h) h 2 E ( x ) h . So the global error in Forward Euler’s method decreases linearly with the step size. Thus, we can say that Forward Euler’s method is a first-order method, because the global error is proportional to h .
Example 5.78 [PI-2010 (2 mark)]: Euler’s method of integration is applied to the initial value problem: dy dx 2 x ; y (0) 0 . If the step size h 0.2 , then the error in computation (in percentage) after 5 steps would be (a) 0 (b) 10 (c) 20 (d) 30 Solution (c): From the given data, y 2 x y 2 , y (0) 0 , i.e., x0 0 , h 0.2 . So the local error is E loc (1 2)h 2 y( ) (1 2) 0.22 2 0.04 . If n is the number of iterations then, global error at x X x0 5h 0 5(0.2) 1 is E global n E loc ( X h) E loc (1 0.2) 0.04 20% .
Error for Backward Euler’s method: To determine the local error for the Backward Euler’s method
we
expand
y( x)
in
a
Taylor
y ( xn 1 h ) y ( xn 1 ) hy( xn 1 ) (1 2) h 2 y ( x ) . But
and
y ( xn 1 ) f ( xn 1 , yn 1 ) . Thus we have
series
about
y ( xn 1 h) y ( xn ) yn ;
xn 1 ,
i.e.,
y ( xn 1 ) y ( xn h)
y ( xn h ) yn h f ( xn 1 , yn 1 ) (1 2) h 2 y ( x ) . As
Enloc y ( xn 1 ) { yn h f ( xn 1 , yn 1 )} Enloc (1 2) h 2 y ( x ) h 2 . Hence local error in Backward Euler’s method decreases as the square of the step size. Now with same argument, as with Forward Euler’s method, the global error in Backward Euler’s method decreases linearly with the step size.
Error of the improved Euler method: The local error is of order h3 and the global error is of order h 2 , so that the method is a second – order method.
Error in Runge-Kutta method: This method has local truncation error O ( h 4 ) . The method is well suited to the computer because it needs no special starting procedure, makes light demand on storage, and repeatedly uses the same straightforward computational procedure. It is numerically stable. It is to be also noted that if f depends only on x , this method reduces to Simpson’s rule of integration. Note further that k1 , , k4 depends on n and generally change from step to step.
Stability Analysis: If the effect of round – off error remains bounded as n , with a fixed step size, then the method is said to be stable, otherwise unstable. Unstable method will diverge away from
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.43]
solution and causes overflow error. Using a general single step method equation, yn 1 E y , then condition for absolute stability is E 1 , using a test equation y y . Let us find the condition for stability for Euler’s method, yn 1 yn hf ( xn , yn ) yn 1 yn h yn (1 h ) yn . Now comparing it with the condition, E 1 , we get, 1 h 1 1 1 h 1 2 h 0 , which is the required condition for stability of Euler’s method. Example 5.79 [IN-2009 (2 marks)]: The differential equation dx dt (4 x ) , with x(0) 0 , and the constant 0 , is to be numerically integrated using the forward Euler method with a constant integration time step T . The maximum value of T such that the numerical solution of x converges is (c) (d) 2 (a) 4 (b) 2 Solution: From the given data we have f (tn , xn ) ( dx dt ) (t
n , xn )
(4 xn ) , x(0) 0 , and step size
is T . So from Eq. 5.56, we have xn 1 xn T f (tn , xn ) xn T (4 xn ) , where xi x0 iT ; thus xn 1 xn 1 (T ) (4T ) ; so the numerical solution of x converges (or for stability of x ), we must have 1 (T ) 1 1 1 (T ) 1 0 (T ) 2 0 T 2 . So the maximum value of T such that the numerical solution of x converges is 2 . [Similar question was also asked in EE-2007 (2 marks)]
5.2.3 Numerical Solution of System of Linear Equations A linear system of n equation in n unknowns x1 , x2 , , xn is a set of equations E1 , E2 , En of the form E1 :
a11 x1
a12 x2
a1n xn
b1
E2 :
a21 x1
a22 x2
a1n xn
b2
En :
an1 x1
an 2 x1
(5.61)
ann x1
bn
where the coefficients a jk and the b j are given numbers. The system is called homogeneous if all the b j are zero; otherwise it is called nonhomogeneous. Using matrix multiplication, we can write system
of Eqs. 5.61 as a single vector equation (5.62) Ax b where, the coefficient matrix A [ a jk ] is the n n matrix and x [ x j1 ] and b [b j1 ] are n 1 matrices, and the augmented matrix A are given as: a11 a12 a1n x1 b1
a 21 A an1
a22 an 2
a11 a a2 n x b , x 2 , b 2 and A 21 ann xn bm am1
a12
a1n
|
b1
a22
a2 n
|
b2
|
am 2
amn
| bm A solution of Eq. 5.61 is a set of number x1 , x2 , , xn that satisfy all the n equations, and a solution vector of Eq. 5.62 is a vector x whose components are the solution of Eq. 5.61. The method of solving such a system by determinants is not metrical, even with efficient methods for evaluating the determinants.
Gauss Elimination Method: A practical method for the solution of linear system is the so called Gauss Elimination method. This standard method, for solving linear system (Eq. 5.61), was already discussed in Linear Algebra chapter. We reduce a given linear system (Eq. 5.61) to triangular form. In the first step we eliminate x1 from equations E2 to En in Eq. 5.61. We do this by adding (or
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[5.44]
subtracting) suitable multiples of E1 from equations E2 , , En and taking the resulting equations, call them E2* , , En* as the new equations. The first equation E1 is called the pivot equation in this step, and a11 is called the pivot. This equation is left unaltered. In the second step we take the new second E 2* (which no longer contains x1 ) as the pivot equation and use it to eliminate x2 from E3* to E n* . And so on. After n 1 steps this gives a triangular system that can be solved by back
substitution. The pivot akk (in step k ) must be different from zero and should be large in absolute value, to avoid round off magnification by the multiplication in the elimination. For this we choose as our pivot equation one that has absolutely largest a jk in column k on or below the main diagonal (actually, the uppermost of there are several such equations). This popular method is called partial pivoting. Partial pivoting distinguished it from total pivoting, which involves both row and column interchanges but is hardly used in practice. E1 :
Example 5.80: Solve the linear system of equations, E2 :
3 x1
8 x2
2 x3
7
5 x2
2 x3
8
E3 : 6 x1 2 x2 8 x3 26 Solution: The given linear system of equations can be written in form of augmented matrix as, E1 : 0 8 2 | 7 E2 : E3 :
3 5 2 | 8 . We must pivot since E has no x term. In the first column, equation E 1 1 3 6 2 8 | 26
has the largest coefficient. Hence we interchange E1 and E3 as, E3 :
6 x1
2 x2
8 x3
26
E2 :
3 x1
5 x2
2 x3
8
8 x2
2 x3
7
E1 :
E3 :
or
E2 : E1 :
6 2 8 | 26 3 5 2 | 8 0 8 2 | 7
Hence the first pivot equation is E3 and pivot element is 6. To eliminate x1 from E2 (as x1 is already eliminated from E1 ), we have E2 E2 (1 2) E3 as, 2 x2
8 x3
26
E2 :
4 x2
2 x3
5
E1 :
8 x2
2 x3
7
E3 :
6 x1
E3 :
or
E2 : E1 :
6 2 8 | 26 0 4 2 | 5 0 8 2 | 7
Now, the largest coefficient in the second column is 8, hence we interchange E2 and E1 as, 2 x2
8 x3
26
E1 :
8 x2
2 x3
7
E2 :
4 x2
2 x3
5
E3 :
6 x1
E3 :
or
E1 : E2 :
6 2 8 | 26 0 8 2 | 7 0 4 2 | 5
Hence the second pivot equation is E1 and pivot element is 8. To eliminate x2 from E2 , we have E2 E2 (1 2) E1 as, E3 : E1 :
6 x1
2 x2
8 x3
26
8 x2
2 x3
7
E3 :
or
E1 :
6 2 8 | 26 0 8 2 | 7 0 0 3 | 3 2
E2 : 3x3 3 2 E2 : The resulting triangular system is shown above and this is the end of forward elimination. Now for back substitution from this system, king the last equation, then second equation, and finally the first 1 1 1 equation, we compute the solution as, x3 ; x2 (7 2 x3 ) 1; x1 (26 2 x2 8 x3 ) 4 2 8 6
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[5.45]
Number of operation count in Gauss Elimination method: The important factors in judging the quality of a numeric method are (1) amount of storage (2) amount of time which is equivalent to number of operations (3) effect of round off error. For the gauss elimination, the operation count for a full matrix (a matrix with relatively many non-zero entries) is as follows. In step k we eliminate xk from n k equations. This needs n k divisions in computing the m jk and ( n k )(n k 1) multiplications and as many subtractions. Since we do n 1 (as k goes from 1 to n 1 ) and thus the total number of operations in this forward elimination is given as, n 1 n 1 n 1 n 1 1 2 2 f ( n) ( n k ) 2 ( n k )( n k 1) s 2 s ( s 1) ( n 1) n ( n 2 1) n n 3 , where 2 3 3 k 1 k 1 s 1 s 1 2 n3 3 is obtained by dropping lower powers of n . We see that f (n) grows about proportional to n3
. We say that f (n) is of order n3 and write f ( n) O( n3 ) , where O suggests order. In the back substitution of xi , we make n i multiplications and as many subtraction, as well as 1 division. Hence the number of operations in the back substitution is given as, n
n
b( n) 2 i 1 ( n i) n 2 s 1 s n n( n 1) n n 2 2n O( n 2 ) . We see that it grows more
slowly than the number of operations in the forward elimination of the Gauss algorithm, so that it is negligible for large systems because it is smaller by a factor n , approximately. For instance, if an operation takes 109 sec then for n 10000 forward elimination takes 11 minutes and backward substitution takes 0.1 seconds.
LU-Factorization (Doolittle’s method): For solving linear system of n equations in n unknowns represented by Eq. 5.61, Doolittle’s method is just a modification of Gauss elimination method, which require fewer arithmetic operations. An LU – Factorization of a given square matrix A is of the form (5.63) A LU , where, L is lower triangular and U is upper triangular matrix It can be proved that for any non-singular matrix, the rows can be reordered so that the resulting matrix A has an LU factorization (Eq. 5.63) in which L turns out to be the matrix of the multipliers m jk of the Gauss elimination, with main diagonal 1,1, ,1 and U is the matrix of the triangular system at the end of the Gauss elimination. The crucial idea now is that L and U in Eq. 5.63 can be computed directly, without solving simultaneous equations (thus, without using the Gauss elimination). As a count shows, this need about n3 3 operations, about half as many as the Gauss elimination, which needs about 2n3 3 . And once we have Eq. 5.63, we can use it for solving Ax b in two steps, involving only about n 2 operations, simply by noting that Ax LUx b is written as, Ly b , where Ux y (5.64) and solving first equation of Eq. 5.64 for y and then second equation of Eq. 5.64 for x . Here we can require that L have main diagonal 1,1, ,1 , then this is called Doolittle’s method. Both the equations of Eq. 5.64 are triangular, so we can solve them as in the back substitution for the Gauss elimination. A similar method, Crout’s method is obtained from Eq. 5.63 if U (instead of L ) is required to have main diagonal 1,1, ,1 . In either case the factorization is unique. For the solution of Eq. 5.61, the entries of the matrix L [l jk ] (with main diagonal 1,1, ,1 and l jk suggesting multiplier) and U [u jk ] in the Doolittle method are computed as,
u1k a1k
k 1, 2, , n
l j1 a j1 u11
j 2, 3, , n
k j , , n; j 2
j k 1, n; k 2
j 1
u jk a jk s 1 l js usk
k 1
l jk (1 ukk ) a jk s 1 l js usk
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(5.65)
e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
2
Example 5.81 [EE-2011 (2 marks)]: The matrix A
[5.46]
1
is decomposed into a product of a 4 1
lower triangular matrix L and an upper triangular matrix U . The properly decomposed L and U matrices respectively are 1 0 1 1 2 0 1 1 (a) and (b) and 4 1 0 2 4 1 0 1
1 (c) 4
0
2 1 and 1 0 1
2 (d) 4
0
1 0.5 and 3 0 1 l11 0 1 u12 2 1 Solution (d): From LU factorization method, we have . l21 l22 0 1 4 1 l11 2 , l21 4 , l11u12 1 u12 1 2 ; l21u12 l22 1 l22 ( 1 2) 3 . Hence the matrix
2
A LU A
0 1
4 3 0
0.5
2 0 1 0.5 and U L . 1 4 3 0 1
[Similar question was also asked in EE-1997 (5 marks)] Example 5.82 [XE-2008 (2 marks)]: On solving the system of equations 4 x z 5 , x 2 y 3z 1 ,
y 4 z 3 by LU decomposition with uii 1 for i 1, 2, 3; the values of u23 and l33 are respectively (a) 1.375 and –4.250 (b) 2.750 and –3.625 (c) 1.375 and –2.625 (d) 2.750 and –4.250 4 0 1 Solution (c): The coefficient matrix for the given system can be written as: A 1
2 3. 0 1 4
Hence from 4 0 1
LU factorization method and l11 0 0 1 u12 u13
1 2 3 l l 21 22 0 1 4 l31 l32 l11u13 1 u13 1 4 ;
0 0
l33 0
l21 1 ;
1 0
u23 .
So
1
given we
l21u12 l22 2 l22 2 ;
data, have,
we
have
l11 4 ;
A LU ,
i.e.,.
l11u12 0 u12 0 ;
l21u13 l22 u23 3 u23 11 8 1.375 ;
l31 0 ; l31u12 l32 1 l32 1 ; l31u13 l32u23 l33 4 l33 2.625 .
Solution of Linear System of Equations by Iterations: The methods discussed in the previous section belong to the direct methods for solving systems of linear equations; these are methods that yield solutions after an amount of computations that can be specified in advance. In this section, we discuss indirect or iterative methods in which we start from an initial value and obtain better and better approximations from a computational cycle repeated as often as may be necessary, for achieving a required accuracy, so that the amount of arithmetic depends upon the accuracy required. Consider a linear system of n linear equations in n unknowns, given by Eq.5.61 in which the diagonal elements aii do not vanish. Now the system given by Eq. 5.61 can be written as x1 (b1 a11 ) ( a12 a11 ) x2 ( a13 a11 ) x3 ( a1n a11 ) xn x2 (b2 a22 ) ( a21 a22 ) x1 ( a23 a22 ) x3 ( a2 n x3 (b2 a22 ) ( a21 a22 ) x1 ( a23 a22 ) x2 ( a2 n xn (bn ann ) (an1 ann ) x1 ( an 2 ann ) x2 ( an ,n 1
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a22 ) xn a22 ) xn ann ) xn 1
(5.66)
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.47]
Suppose we start with x1(0) , x2(0) , , xn(0) as initial values to the variables x1 , x2 , , xn . Then we can find better approximations to x1 , x2 , , xn using the following two iterative methods: (a) Jacobi’s Iteration method (b) Gauss Seidel Iteration method (c) Newton Raphson Method.
Jacobi’s Iteration method: The steps for Jacobi’s Iteration method are given as: Step 1: Determination of first approximation x1(1) , x2(1) , , xn(1) using x1(0) , x2(0) , , xn(0) and (1)
(0)
(0)
(0)
x1 (b1 a11 ) ( a12 a11 ) x2 ( a13 a11 ) x3 (a1n a11 ) xn
x2(1) (b2 a22 ) ( a21 a22 ) x1(0) ( a23 a22 ) x3(0) ( a2 n a22 ) xn(0) (1) (0) (0) (0) x3 (b2 a22 ) ( a21 a22 ) x1 ( a23 a22 ) x2 ( a2 n a22 ) xn xn(1) (bn ann ) (an1 ann ) x1(0) ( an 2 ann ) x2(0) ( an ,n 1 ann ) xn(0) 1
(5.67)
Step 2: Similarly, x1(2) , x2(2) , , xn(2) are evaluated by just replacing xr(0) (1 r n) in Eq. 5.67 by xr(1) (1 r n) . Similarly,
Step ( + ): In general, if x1( n ) , x2( n ) , , xn( n ) are a system of nth approximations, then the next approximation is given by the formula x1( n 1) (b1 a11 ) ( a12 a11 ) x2( n ) ( a13 a11 ) x3( n ) ( a1n a11 ) xn( n )
x2( n 1) (b2 a22 ) (a21 a22 ) x1( n ) ( a23 a22 ) x3( n ) ( a2 n a22 ) xn( n ) x3( n 1) (b2 a22 ) (a21 a22 ) x1( n ) ( a23 a22 ) x2( n ) ( a2 n a22 ) xn( n ) ( n 1) xn
(bn ann )
(n) ( an1 ann ) x1
(n)
( an 2 ann ) x2 ( an , n 1
(5.68)
(n) ann ) xn 1
The system of equations in Eq. 5.68 can also be briefly described as follows: aij ( r ) b n r 1 (5.69) xi i j 1, j i x j , r 0,1, 2, n and i 1, 2, n aii aii A sufficient condition for obtaining a solution by Jacobi’s iteration method is the diagonal n
dominance, i.e. aii j 1, j i aij , i 1, 2, , n , in each row of A , modulus of the diagonal element exceeds the sum of the off diagonal elements and also the diagonal elements, aii 0 . If any diagonal element is 0, the equations can always be rearranged to satisfy this condition. Example 5.83 [TF-2007 (2 marks)]: For the following system of equations 4 x1 x2 x3 4 , x1 4 x2 2 x3 4 , 3 x1 2 x2 4 x3 6 which of the following is the solution, obtained after TWO iteration using Jacobi method. (a) x1 1.0, x2 1.0, x3 0.0 (b) x1 1.0, x2 1.0, x3 1.5
(c) x1 1.06, x2 0.594, x3 0.656
(d) x1 1.125, x2 0.0, x3 0.25 4 x1 x2 x3 4
Solution (d): Let the system of equation be
x1 4 x2 2 x3 4 and let the initial values be 3 x1 2 x2 4 x3 6
x1(0)
0,
x2(0)
0,
x3(0)
0.
So
x1(1) (4 x2(0) x3(0) ) 4 1 (1) (0) (0) x2 (4 x1 2 x3 ) 4 1 x3(1) (3x1(0) 2 x2(0) 6) 4 3
from
Eq.
5.68,
we x1(2)
; so, 2nd iteration is 2
Copyright © 2016 by Kaushlendra Kumar
x3(2)
have
the
first
iteration
as:
x2(1)
(4 x3(1) ) 4 9 8 1.125 (2) (1) (1) . x2 (4 x1 2 x3 ) 4 0 (1) (1) (3 x1 2 x2 6) 4 1 4 0.25
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[5.48]
Gauss Seidel Iteration method: A simple modification to Jacobi’s iteration method is given by Gauss – Seidel method. The steps are as follows: Step 1: Determination of first approximation x1(1) , x2(1) , , xn(1) as, (1)
(0)
(0)
(b2 a22 ) ( a21 a22 ) x1(1) (1) (b2 a22 ) ( a21 a22 ) x1
(a23 a22 ) x3(0) (1) ( a23 a22 ) x2
(0)
x1 (b1 a11 ) ( a12 a11 ) x2 ( a13 a11 ) x3 ( a1n a11 ) xn x2(1) (1) x3
ann ) xn(1)1
( a2 n a22 ) xn(0) (0) ( a2 n a22 ) xn
xn(1) (bn ann ) ( an1 ann ) x1(1) ( an 2 ann ) x2(1) (an ,n 1
(5.70)
Step ( + ): In general, if x1( n ) , x2( n ) , , xn( n ) are a system of nth approximations, then the next approximation is given as, ( n 1) (n) ( n) ( n) x1 (b1 a11 ) ( a12 a11 ) x2 (a13 a11 ) x3 ( a1n a11 ) xn
(n) a22 ) xn ann ) xn( n11)
x2( n 1) (b2 a22 ) ( a21 a22 ) x1( n 1) ( a23 a22 ) x3( n ) ( a2 n a22 ) xn( n ) ( n 1)
x3
( n 1)
(b2 a22 ) ( a21 a22 ) x1
( n 1)
(a23 a22 ) x2
( a2 n
xn( n 1)
(bn ann )
( an1 ann ) x1( n 1)
( an 2 ann ) x2( n 1) (an ,n 1
(5.71)
The Eq. 5.71 can be briefly written as, i 1
r 1
( r 1)
(bi aii ) j 1, j i ( aij aii ) x j
xi
n
(r ) j i 1 ( aij aii ) x j , r 0,1, 2, n and i 1, 2, n
(5.72)
Example 5.84 [MA-2014 (2 marks)]: Using the Gauss-Seidal iteration method with the initial guess
x
(0) 1
3.5, x2(0) 2.25, x3(0) 1.625 , the second approximation x1(2) , x2(2) , x3(2)
for the solution to the
system of equations: 2 x1 x2 7 , x1 2 x2 x3 1 , x2 2 x3 1 , is (a) x1(2) 5.3125, x2(2) 4.4491, x3(2) 2.1563
(b) x1(2) 5.3125, x2(2) 4.3125, x3(2) 2.6563
(c) x1(2) 5.3125, x2(2) 4.4491, x3(2) 2.6563
(d) x1(2) 5.4991, x2(2) 4.4491, x3(2) 2.1563 2 x1 x2 0 x3 7
Solution (b): Let the system of equation be x1 2 x2 x3 1 and let the initial values be 0 x1 x2 2 x3 1 x1(0) 3.5, x2(0) 2.25, x3(0) 1.625 .
x1(1) ( x2(0) 7) 2 4.625 (1) (1) (0) x2 (1 x1 x3 ) 2 3.625 ; x3(1) (1 x2(1) ) 2 2.3125
So
from
Eq.
so, 2nd iteration is
5.71,
we
have
the
first
x1(2) ( x2(1) 7) 2 5.3125 (2) (2) (1) x2 (1 x1 x3 ) 2 4.3125 x3(2) (1 x2(2) ) 2 2.6563
iteration
as:
.
Difference between Jacobi’s and Gauss – Seidel method
Jacobi’s method: In the first equation of Eq. 5.66, we substitute the initial approximations x2(0) , x3(0) , , xn(0) into the right – hand side and denote the result as x1(1) . In the second equation, we substitute x1(0) , x3(0) , , xn(0)
and denote the result as x2(1) . In third, we substitute
x1(0) , x2(0) , , xn(0) and call the result as x3(1) ; and the process is repeated in this manner.
Gauss – Seidel method: In the first equation of Eq. 5.66, we substitute the initial approximations x2(0) , x3(0) , , xn(0) into the right – hand side and denote the result as x1(1) . In the second equation, we substitute x1(1) , x3(0) , , xn(0) and denote the result as
x2(1) . In third, we substitute
x1(1) , x2(1) , x3(0) , , xn(0) and call the result as x3(1) ; and the process is repeated in this manner.
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.49]
Newton-Raphson Method: It is possible to use Newton-Raphson method to solve a system of a11 x1
a12 x2
a1n xn
b1
0
f 2 ( x ) 0 : a21 x1
a22 x2
a1n xn
b2
0
f1 ( x) 0 :
nonlinear equations, i.e.,
. Note
f n ( x) 0 : an1 x1 an 2 x1 ann x1 bn 0 that the number of unknowns equals the number of equations. We can write this using vector notation
as f (x) O , where x x1
x2
T
xn and f ( x) f1 ( x)
f 2 ( x)
T
f n (x) . The Newton-
1
Raphson iteration for this system is x n 1 x n J f ( x n ) , where J is the Jacobian matrix with
fi ( x) 1 , 1 i, j n ; and J is the inverse matrix of J . x j
(i, j )th entry is given by J
Example 5.85 [EE-2011 (2 marks)]: Solution of the variables x1 and x2 for the following equations is to be obtained by employing the Newton-Raphson iterative method. 10 x2 sin x1 0.8 0 ; 10 x22 10 x2 cos x1 0.6 0 . Assuming the initial valued x1 0.0 and x2 1.0 , the Jacobian matrix is
10 0.8 10 0 0 0.8 10 0 (a) (b) (c) (d) 0 10 10 0.6 10 10 0 0.6 Solution (b): As we have two equation and two unknowns so the Jacobian matrix is a 2 2 matrix fi ( x) f1 (x) x1 f1 ( x) x2 , 1 i, j 2 . So J and f 2 ( x) x1 f2 (x) x2 x j
whose elements are given by J
from the given data we have f1 ( x) 10 x2 sin x1 0.8 and f 2 ( x ) 10 x22 10 x2 cos x1 0.6 . So
10 x2 cos x1
J
10 x2 sin x1
10 sin x1
10 0 at initial value x1 0.0 , x2 1.0 , we have J . 20 x2 10 cos x1 0 10
Example 5.86 [CS-1994 (2marks)]: Subjective Question: Match Column-I with Column-II Column-I Column-II A. Newton-Raphson 1. Integration B. Runge-Kutta 2. Root finding C. Gauss-Seidal 3. Ordinary Differential Equation D. Simpson’s Rule 4. Solutions of Systems of Linear Equations Solution: We know that (i) Newton-Raphson method is for Root finding; (ii) Runge-Kutta method is for solving ODE; (iii) Gauss-Seidal method is for solving System of Linear Equations; (iv) Simpson’s rule is for Integration. So answer is A – 2; B – 3; C – 4; D – 1. Example 5.87 [EE-1997 (1 mark)]: Gauss-Seidel iterative method can be used for solving a set of (a) linear differential equations only (b) both linear and non-linear algebraic equations (c) linear algebraic equations only (d) both linear and non-linear differential equations Solution (c): We know that Gauss-Seidel iterative method is used for solving a set of linear algebraic equations only. So option (b) is correct. Example 5.88 [EC-2005 (2 marks)]: Match the following and choose the correct combination given below the groups:
Group – I A. Newton – Raphson Method B. Runge – Kutta Method C. Simpson’s rule D. Gauss Elimination
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Group – II 1. Solving non-linear equations 2. Solving linear simultaneous equations 3. Solving ordinary differential equation 4. Numerical integration 5. Interpolation 6. Calculation of Eigen values
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.50]
(a) A – 6; B – 1; C – 5; D – 3 (b) A – 1; B – 6; C – 4; D – 3 (c) A – 1; B – 3; C – 4; D – 2 (d) A – 5; B – 3; C – 4; D – 1 Solution (c): We know that (i) Newton-Raphson method is for solving non-linear equations; (ii) Runge-Kutta method is for solving ODE; (iii) Simpson’s rule is for Numerical integration; (iv) Gauss Elimination method is for solving System of Linear Equations; (iv). So answer is A – 1; B – 3; C – 4; D – 2. Example 5.89 [ME-2006 (1 mark)]: Match Column-I with Column-II Column-I Column-II A. Gauss-Seidel Method 1. Interpolation B. Forward Newton-Gauss method 2. Non-linear differential equation C. Runge-Kutta method 3. Numerical Integration D. Trapezoidal rule 4. Linear algebraic equations (a) A – 1, B – 4, C – 3, D – 2 (b) A – 1, B – 4, C – 2, D – 3 (c) A – 1, B – 3, C – 2, D – 4 (d) A – 4, B – 1, C – 2, D – 3 Solution (d): We know that (i) Gauss-Seidel Method is used for solving Linear algebraic equations; (ii) Forward Newton-Gauss method is used for Interpolation; (iii) Runge-Kutta method is used for solving Non-linear differential equation; (iv) Trapezoidal rule is used for Numerical integration. So answer is A – 4, B – 1, C – 2, D – 3. Example 5.90 [PI-2007 (2 marks)]: ‘Matching exercise’. Group I Group II A. second order differential equation 1. Runge-Kutta method B. non-linear algebraic equation 2. Newton-Raphson method C. linear algebraic equation 3. Gauss elimination D. numerical integration 4. Simpson’s rule (a) A – 3; B – 2; C – 4; D – 1 (b) A – 2; B – 4; C – 3; D – 1 (c) A – 1; B – 2; C – 3; D – 4 (d) A – 1; B – 3; C – 2; D – 4 nd Solution (c): We know that (i) 2 order differential equations can be solved by Runge-Kutta method; (ii) Newton-Raphson method is for solving non-linear algebraic equations; (iii) Simpson’s rule is for Numerical integration; (iv) Gauss Elimination method is for solving System of Linear Equations. So option (c) is correct. Example 5.91 [MT-2010 (1 mark)]: Which of the following is an iterative technique to solve a linear system of equations? (a) Gaussian elimination (b) LU decomposition (c) Newton-Raphson (d) Jacobi method Solution: We know that Gauss Elimination and LU decomposition method are used for solving System of Linear Equations exactly; Newton-Raphson method is used for solving non-linear equations; but Jacobi and Gauss-Seidel method are iterative methods which are used for solving linear system of equations. So option (d) is correct. Example 5.92 [MT-2011 (1 mark)]: Which one of the following methods is NOT used for numerically solving an ordinary differential equation (ODE)? (a) Euler’s method (b) Runge-Kutta method (c) Adam-Bashforth method (d) Newton-Raphson method Solution (d): As Euler’s, Runge-Kutta and Adam-Bashforth methods are used for solving an ODE; but Newton-Raphson method is used for solving root finding. So option (d) is correct. Example 5.93 [MT-2012 (1 mark)]: Which one of the following methods is NOT used for integration? (a) Rectangular rule (b) Trapezoidal rule (c) Simpson’s rule (d) Cramer’s rule Solution (d): As Rectangular, Trapezoidal and Simpson’s rule are used for integration; but Cramer’s rule is used for finding solution of system of linear equations having number of equations is equal to number of unknowns. So option (d) is correct.
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e-mail: [email protected]
Engineering Mathematics
Chapter 5: Numerical Methods
[5.51]
Example 5.94 [EC-2014 (1 mark)]: Match the application to appropriate numerical method. Application Numerical Method A. Numerical integration 1. Newton-Raphson method B. Solution to transcendental equation 2. Runge-Kutta method C. Solution to a system of linear equations 3. Simpson’s 1/3rd rule D. Solution to a differential equation 4. Gauss elimination method (a) A – 3, B – 2, C – 4, D – 1 (b) A – 3, B – 1, C – 4, D – 2 (c) A – 4, B – 1, C – 3, D – 2 (d) A – 2, B – 1, C – 3, D – 4 Solution (b): We know that (i) Simpson’s rule is for Numerical integration; (ii) Gauss Elimination method is for solving System of Linear Equations; (iii) Runge-Kutta method is used for solving an ODE; (iv) Newton-Raphson method is for solving non-linear algebraic or transcendental equations. So option (b) is correct. Exercise: 5.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The value of
5
1
1 x 2 dx by using trapezoidal rule, with eight equal intervals, is _____.
2. The following points, given in the table, were found empirically. By using the trapezoidal rule the value of
3.6
2.1 ydx
is _____.
3. The difference between the exact value of
0 sin xdx
and the approximate value of
0 sin xdx
by
using trapezoidal rule, using 10 equal intervals, is (a) 0.015 (b) 0.017 (c) 0.019 (d) 0.021 4. The vertical distance (in m) covered by a rocket from t 8 sec to t 30 sec is given by 30
8 2000 ln 140000 (140000 2100t ) 9.8t dt .
Using single segment trapezoidal rule, the
distance covered (in m) is (a) 11868 (b) 18168 (c) 16188 (d) 18618 5. The absolute relative true error for the distance covered in Q. 4 is _____. (a) 7.1% (b) 7.2% (c) 7.3% (d) 7.4% 6. Using two equal interval trapezoidal rule, area under the curve f ( x ) (300 x) (1 e x ) , from x 0 to x 10 , is (a) 48.535 (b) 50.535 (c) 48.331 (d) 50.331 7. By using two-segment Simpson’s 1/3 rule, the value of the integral
2
2
8 x 3 ) dx
2
2
2 x 4 ) dx
0 (0.2 25 x 3 x
is _____. 8. By using four-segment Simpson’s 1/3 rule, the value of the integral
0 (0.2 25 x 3 x
is _____. 9. The magnitude as the error (correct to two decimal places) in the estimation of integral 2
0 (0.2 25 x 3 x
2
2 x 4 ) dx by four-segment Simpson’s 1/3 rule is _____.
2t ,
10. The velocity of a body is given by v (t )
1 t 5
2
5t 3, 5 t 14
, where t is given in seconds, v is
given in m/s. Using two-segment Simpson’s 1/3 rule, the distance in metres covered by the body from t 2 to t 9 seconds most nearly is (a) 949.33 (b) 1039.7 (c) 1200.5 (d) 1442.0 11. The value of
0.7
0.1 (1 x)dx , which 6 equal intervals is
(a) 1.86 (b) 1.97 (c) 1.78 12. The following data of the velocity of a body is given as a
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(d) 1.90
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Engineering Mathematics
Chapter 5: Numerical Methods
[5.52]
function of time. The best estimate of the distance in metres Time 4 7 10 15 covered by the body from t 4 and t 15 using combined Velocity 22 24 37 46 Simpson’s 1/3 rule and the trapezoidal rule would be (a) 354.70 (b) 362.50 (c) 368.00 (d) 378.80 13. The Velocity v (m/s) of a particle at distance from a 10 20 30 40 50 60 point on its path is given in the table. Estimate the time x: 0 (sec) taken to travel 60 m by using Simpson’s 3/8 rule. v : 47 58 64 65 61 52 38 (a) 1.064 (b) 1.054 (c) 1.404 (d) 1.304 14. Taking y (0) 0 and using forward Euler’s method with step size h 0.1 solution of the differential equation dy dx 1 y gives the value of y (0.3) as _____. 15. Taking y (0) 1 and using modified Euler’s method with step size h 0.01 solution of the differential equation dy dx x 2 y gives the value of y at x 0.02 as _____. 16. Using Runge-Kutta method, find an approximate value y at x 0.2 for the differential equation dy dx x y , if y 1 when x 0 . _____. 17. The convergence of which of the following method is sensitive to starting value? (a) False Position (b) Gauss Seidal method (c) Newton-Raphson method (d) All (a), (b) and (c) 18. In the Gauss elimination method for solving a system of linear algebraic equations, triangularization leads to _____ matrix. (a) diagonal (b) lower triangular (c) upper triangular (d) singular 19. The rate of convergence of Gauss Seidal method is _____ that of Gauss Jacobi method. (a) once (b) twice (c) thrice (d) reciprocal 20. _____ method is very fast compared to other methods. (a) Gauss elimination (b) Gauss Jordon (c) Gauss Seidal (d) Gauss Jacobi 21. The lower triangular matrix [ L] in [ L][U ] decomposition of the matrix
25 5 4 1 10 8 16 l 21 8 12 22 l31 0 1 (a) 0.40000 1 0.32000 1.7333
0 1 l32
0 u11 0 0 1 0
0 0 1
u12
u13
u22
u23 is
0
u33
1 0 0 (b) 10 1 0 8 12 1
0 0 1 (c) 0.400 1 0 0.320 1.50 1
1 0 0 (d) 25 1 0 5 4 1
Answers Keys 1 b 16 a 31 a
2 b 17 d 32 d
1 2 12.76 5.22 16 17 1.2736 c
3 c 18 b 33 b
3 b 18 c
4 a 19 b 34 c
4 a 19 b
5 b 20 b 35 36.85
Answer Keys: Exercise: 5.1 6 7 8 9 10 a c a c d 21 22 23 24 25 c 2 c c 11 36 37 38 39 0.5 b 4.75 c
5 c 20 c
Answer Keys: Exercise: 5.2 6 7 8 9 10 b 90.4 71.23 0.034 c 21 a
Copyright © 2016 by Kaushlendra Kumar
11 a 26 a
12 b 27 a
11 b
12 b
13 b 28 b
13 a
14 a 29 c
15 b 30 b
14 15 0.271 1.02
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Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [1]
GATE – 2016: Chapter – 6: Probability & Statistics Note: The following questions came in GATE – 2016 were based on Probability & Statistics Chapter – 6. You are advised to go through first these questions, so that you can know your knowledge. In case you find any difficulty it is advised to go through carefully the concepts given in the chapter so that in GATE – 2017 examination you can tackle any question without any difficulty. 1. The function f ( x) represents a normal distribution whose standard deviation and mean are 1 and 5, respectively. The value of f ( x) at x 5 is [AG-2016 (1 mark)] (a) 0.0 (b) 0.159 (c) 0.282 (d) 0.398 Solution (d): If X is a normally distributed with mean and standard deviation , then the probability density function is given by f ( x )
1 x 2 exp . As 5 , 1 , so 2 2 1
1 5 5 2 1 f (5) exp exp 0 0.398 . 1 2 2 2 1 1
2. Runs scored by a batsman in five one-day matches are 55, 75, 67, 88 and 15. The standard deviation is _____. [BT-2016 (1 mark)] Solution: xi xi ( xi ) 2 xi 300 5 15 7 28 45
55 75 67 88 15
25 225 49
784 2025
xi 5 60 ;
( xi )2 3108 ( xi ) 2
N 3108 5 24.93
3. The spot speeds (expressed in km/hr) observed at a road section are 66, 62, 45, 79, 32, 51, 56, 60, 53, and 49. The median speed (expressed in km/hr) is _____. [CE-2016 (1 mark)] Solution: Arranging the given numbers in an increasing order: 32, 45, 49, 51, 53, 56, 60, 62, 66, 79. As we have even number of data, i.e. n 10 , so median is arithmetic mean of the values of ( n 2)th (10 2)th 5th observation 53 and {(n 2) 1}th {(10 2) 1}th 6th observation 56 . Hence median (53 56) 2 54.5 km/hr. 4.
X
and Y
are two random independent events. It is known that P ( X ) 0.40 and
c
P ( X Y ) 0.70 . Which one of the following is the value of P ( X Y ) ? [CE-2016 (1 mark)] (a) 0.7 (b) 0.5 (c) 0.4 (d) 0.3 Solution (a): P ( X Y ) 0.7 P ( X ) P (Y ) P( X Y ) 0.7 As X and Y are two random independent events, so P ( X Y ) P ( X ) P(Y ) 0.40 P(Y ) Thus from (i), we have 0.4 P (Y ) 0.40 P (Y ) 0.7 0.6 P (Y ) 0.3 P (Y ) 3 6 1 2 P(Y ) 1 P (Y ) 1 2 . Now P ( X Y ) P ( X ) P (Y ) P ( X Y ) ; but X and Y are two random independent events, so P ( X Y ) P ( X ) P(Y ) P( X ) P(Y ) 0.4 0.5 0.4 0.5 0.7 . 2
5. A probability density function on the interval [ a,1] is given by 1 x and outside this interval the value of the function is zero. The value of a is _____. [CS-2016 (1 mark)]
1 x 2 , Solution: It is given that probability density function f ( x ) 0,
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a x 1
;
otherwise
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Engineering Mathematics
Chapter – 6: Probability & Statistics
Also we have continuous random variable so
Chapter – 6 [2]
a
1
a
f ( x ) dx 1 (0) dx (1 x 2 ) dx (0) dx 1 1
1
0 1 x a 0 1 (1 a ) 1 1 a 0.5 6. Suppose that a shop has an equal number of LED bulbs of two different types. The probability of an LED bulb lasting more than 100 hours given that it is of Type 1 is 0.7, and given that it is of Type 2 is 0.4. The probability that an LED bulb chosen uniformly at random lasts more than 100 hours is _____. [CS-2016 (1 mark)] Solution: Let A be the event of an LED bulb lasting more than 100 hours and B be the event of an LED bulb lasting not more than 100 hours. The following diagram shows the situation of the given problem. So if an LED bulb chosen uniformly at random lasts more than 100 hours, then its probability is P ( A) P{ A (Type1)}P(Type1) P{ A (Type 2)}P (Type 2) 0.7 0.5 0.4 0.5 0.55 . 7. The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is _____. [EC-2016 (1 mark)] Solution: For Poisson-distributed random variable, ( X ), the variance, Var ( X ) , of the distribution is the mean, E ( X ) , of that distribution. So for Poisson-distribution, Var ( X ) E ( X ) x (say); and it is 2
2
2
2
2
given that E ( X ) 2 . We know that Var ( X ) E ( X ) {E ( X )} x 2 x x x 2 0 x 1, 2 , but x 2 , so x 1 . 8. The probability of getting a ‘head’ in a single toss of a biased coin is 0.3. The coin is tossed repeatedly till a ‘head’ is obtained. If the toss are independent, then the probability of getting ‘head’ for the first time in the fifth toss is _____. [EC-2016 (1 mark)] Solution: As the coin is biased so required probability of getting ‘head’ for the first time in the fifth 4 1 toss is (0.7) (0.3) 0.072 . 9. A voltage V1 is measured 100 times and its average and standard deviation are 100 V and 1.5 V, respectively. A second voltage V2 , which is independent of V1 , is measured 200 times and its average and standard deviation are 150 V and 2 V respectively. V3 is computed as: V3 V1 V2 . Then the standard deviation of V3 in volt is _____. [IN-2016 (1 mark)] Solution: We know that if X and Y are independent random variables then Var ( aX bY ) a 2Var ( X ) b 2Var (Y ) . We have X V1 and Y V2 , Var ( X ) V21 (1.5) 2 and 2
2
2
2
Var (Y ) V22 (2) 2 . So Var ( X Y ) 1 (1.5) 1 (2) 6.25 ; hence X Y V1 V2 6.25 2.5 .
x, 0 x 1 2 2 10. Let the probability density function of a random variable X be f ( x ) c(2 x 1) , 1 2 x 1 . 0, otherwise Then, the value of c is equal to _____. [MA-2016 (1 mark)] Solution: We have continuous random variable, X , whose probability distribution function is given by f ( x ) , then we must have 0
12
0
(0) dx 0 (1
2)( x 2 )10 2
0
12
f ( x)dx 1 f ( x)dx 0
1
2
f ( x ) dx
1
12
f ( x ) dx f ( x )dx 1 1
xdx c (2 x 1) dx 0 dx 1 12
c (1 6){(2 x
1
1)3 }11 2
0 1 (1 2)(1 2) 2 c (1 6){13 } 1 c 21 4 5.25 .
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Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [3]
11. Consider a Poisson distribution for the tossing of a biased coin. The mean for this distribution is . The standard deviation for this distribution is given by [ME-2016 (1 mark)] 2 (c) (d) 1 (a) (b) Solution (a): We know that for Poisson distribution of random variable X , the variance of X is the
mean value of X . Hence standard deviation of X mean of X
.
12. The area (in percentage) under standard normal distribution curve of random variable Z within limits from 3 to 3 is _____. [ME-2016 (1 mark)] Solution: The area between z 3 and z 3 of the mean, (i.e. between z 3 ) constitute about 99.7% of the area under the Standardized Normal Distribution curve. So answer is ‘99.7’. 13. A fair coin is tossed N times. The probability that head does not turn up in any of the tosses is [PI-2016 (1 mark)] N 1 N 1 N N (a) (1 2) (b) 1 (1 2) (c) (1 2) (d) 1 (1 2) Solution (c): If a fair coin is tossed N times then total number of sample space is 2 N . Now only one time head will not turn up in any of the tosses, i.e. when all the tosses shows tails. Thus required N probability is 1 2 . Hence option (c) is correct. 14. A
normal
f X ( x)
random
1 e 8
( x 1) 2 8
(a) 0
variable
X
has
, x . Then (b) 1 2
the
1
following
probability
density
function
[PI-2016 (1 mark)]
f X ( x ) dx
(c) 1 (1 e)
(d) 1
Solution (b): On comparing the given probability distribution function,
f X ( x) , with
2
1 x exp , we have 1 , 2 . So in the given question we are asked to 2 2 find the area from 1 to , i.e. from mean x 1 to x ; thus the required area will be 0.5, as total area under an entire normal curve is 1 and we are asked to find half of that area. So answer is 0.5. f ( x)
1
15. Let X be a normally distributed random variable with mean 2 and variance 4. Then, the mean of ( X 2) 2 is equal to _____. [TF-2016 (1 mark)] 2
Solution: If a normally distributed variable X , having a mean and variance , is transformed into a new variable Z , using the transformation Z
X
, then that transformation is called a standardized normal distribution with a mean of 0 and variance of 1. In the given problem X be a normally distributed random variable with mean 2 and variance X 2 2 4 2 ; so transforming X into Z , we get standardized normal distribution, 2 whose mean is 0. So answer is 0. 16. A company records heights of all employees. Let X and Y denote the errors in the average height of male and female employees, respectively. Assume that X N (0, 4) and Y N (0, 9) and they are independent. Then the distribution of Z ( X Y ) 2 is [XE-2016 (1 mark)] (a) N (0, 6.5) (b) N (0, 3.25) (c) N (0, 2) (d) N (0,1) Solution (b): If X and are two Y independent variables then (i) E ( aX bY ) aE ( X ) bE (Y ) ; and 2
2
(ii) Var ( aX bY ) a Var ( X ) b Var (Y ) . We are given E ( X ) 0 , E (Y ) 0 , Var ( X ) 4 , Var (Y ) 9 ; and we have to find E ( aX bY ) and Var (aX bY ) , where a b 1 2 .
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e-mail: [email protected]
Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [4]
So E (1 2) X (1 2)Y (1 2) E (Y ) (1 2) E (Y ) 0 ; and
Var{(1 2) X (1 2)Y } (1 2) 2Var ( X ) (1 2) 2 Var (Y ) (1 4)(4 9) 3.25 Thus the distribution of Z ( X Y ) 2 is N (0, 3.25) . 17. The maximum one day rainfall depth at 20 year return period of a city is 150 mm. The probability of one day rainfall equal to or greater than 150 mm in the same city occurring twice in 20 successive years is _____. [AG-2016 (2 marks)] Solution: The given experiment satisfies all the criteria for Binomial distribution. Let S be the event for success, i.e. having 1 day rainfall of 150 mm, whose probability is probability of occurrence of once in 20 years, so p 1 20 0.05 ; and F be the event for failure, i.e. not having 1 day rainfall of 150 mm, whose probability is q 1 p 1 0.05 0.95 . Thus the probability of having 1 day rainfall 20
of 150 mm or more will occur in twice in 20 successive years is
C2 (0.05) 2 (0.95) 20 2 0.188 .
18. If f ( x) and g ( x ) are two probability density functions,
( x a ) 1, a x 0 x a , a x 0 f ( x ) ( x a ) 1, 0 x a , g ( x ) x a , 0 x a . Which one of the following 0, otherwise otherwise 0, statements is true? (a) Mean of f ( x) (b) Mean of f ( x) (c) Mean of f ( x) (d) Mean of f ( x) Solution (b):
and and and and
g ( x) g ( x) g ( x) g ( x)
[CE-2016 (2 marks)] are same; Variance of f ( x) and g ( x ) are same are same; Variance of f ( x) and g ( x ) are different are different; Variance of f ( x) and g ( x ) are same are different; Variance of f ( x) and g ( x ) are different
a
For f ( x ) : E ( X ) x f ( x ) dx
0
a x x 1 dx x 1 dx x (0) dx 0 a a a
x(0) dx x a
0
a
2 a x2 x x3 x 2 x3 x 2 E ( X ) x dx x dx 0 a 0 a a 3a 2 a 3a 2 0 0
Variance of f ( x ) : Var ( X ) ( x ) 2 f ( x) dx ( x 0) 2 f ( x ) dx x 2 f ( x ) dx Var ( X )
a
0
2
2
a 2 2 x x 1 dx x 1 f ( x) dx x (0) dx 0 a a a
x (0)dx x a
0
a
3 a 2 x3 x x 4 x3 x 4 x 3 a3 2 2 Var ( X ) x dx x x dx a 0 a a 4a 3 a 4a 3 0 6 0
a
For g( x ) : E ( X ) x g ( x ) dx
0 a x x x(0) dx x dx x dx x (0) dx a 0 a a a 0
a
3 3 a x x3 x3 a a x E ( X ) x dx x dx 0 a 0 3 3 a a 3a a 3a 0 0
Variance of f ( x ) : Var ( X ) ( x ) 2 g ( x ) dx ( x 0) 2 g ( x)dx x 2 g ( x) dx Var ( X )
a
a 2 x 2 x 2 2 x (0) dx x dx x dx x (0) dx a 0 a a a 0
0
a
x4 x4 x3 a3 Var ( X ) . Hence option (b) is correct. 4 a a 4 a 3 0 2
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [5]
1 x 5
0.25,
19. Probability density function of a random variable X is given below f ( x)
. otherwise [CE-2016 (2 marks)]
0,
P ( X 4) is
(a) 3 4
(b) 1 2
Solution (a): P ( X 4)
4
P( X 4) 0
0.25( x )14
(c) 1 4
f ( x )dx
1
(d) 1 8
4
1
4
f ( x ) dx f ( x ) dx 0dx 0.25dx
1
1
0.25(4 1) 0.75 3 4 .
20. Consider the following experiment: Step 1: Flip a fair coin twice Step 2: If the outcomes are (TAILS, HEADS) then output Y and stop. Step 3: If outcomes are either (HEADS, HEADS) or (HEADS, TAILS), then output N and stop. Step 4: If the outcomes are (TAILS, TAILS), then go to Step 1. The probability that the output of the experiment is Y is (up to two decimal places) _____. [CS-2016 (2 marks)] Solution: This experiment can be repeated for infinite times. Flip/Outcome Y N Again So required probability for output Y will be: 1st Flip T H H T nd 2 Flip H H T T 1 1 1 1 1 1 1 1 1 1 1 1 P (Y ) 2 2 2 2 2 2 2 2 2 2 2 2 1 11 111 1 1 1 (1 4) 1 P(Y ) 2 3 0.33 4 44 444 4 4 4 1 (1 4) 3 random variables X and Y are distributed according to ( x y ), 0 x 1, 0 y 1 . The probability P ( X Y 1) is _____. f XY ( x, y ) otherwise 0, [EC-2016 (2 marks)]
21. Two
Solution: P ( X Y 1) f ( x, y ) dxdy R
y 1
y 1 y 0
x 1 y
x 0
( x y ) dx dy
x 1 y
(1 2) x xy dy P( X Y 1) (1 2)(1 y ) (1 y ) y dy P ( X Y 1) (1 2)(1 y ) dy (1 2) y (1 6) y P( X Y 1)
2
x 0
y0
y 1
2
y 0
y 1
3 1
2
y 0
0
1 3 0.33
22. Let the probability density function of a random variable, X , be given as: f X ( x ) (3 2) e 3 x u ( x ) ae 4 xu ( x ) , where u ( x) is the unit step function. Then the value of ‘ a ’ and Prob { X 0} , respectively, are [EE-2016 (2 marks)] (a) 2, 1 2
(b) 4, 1 2
(c) 2, 1 4
0, x 0
Solution (a): The unit step function, u ( x)
(d) 4, 1 4
0, x 0
u( x)
1,
u ( x)
x0
x0 1, x 0 0, x 0 As we have continuous random variable, X , whose probability distribution function is given by
1,
f X ( x) , then we must have
f ( x ) dx 1
0
0
0
f ( x) dx f ( x ) dx 1 0
{(3 2)e 3 x (0) ae 4 x (1)}dx {(3 2)e 3 x (1) ae 4 x (0)}dx 1 0
0
4x
a e dx (3 2)
0
3 x e4 x 3e a 1 e dx 1 a 1 1 a 2. 4 2 4 2 3 0 3 x
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Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [6]
0
Now P ( X 0)
0
e4 x 1 f X ( x ) dx {(3 2)e (0) ae (1)}dx a e dx 2 . 4 2 0
3 x
0
4x
4x
23. Candidates were asked to come to an interview with 3 pens each. Black, blue, green and red were the permitted pen colours that the candidate could bring. The probability that a candidate comes with all 3 pens having the same colour is _____. [EE-2016 (2 marks)] Solution: Total number of ways in which candidates select 3 pens is same as selecting 3 pens from (3 black 3 blue 3 green 3 red pens) 12 pens is 12 C3 220 ways. Now, a candidate can select 3 pens of same colour from given 4 colours of pens, which can be done in 4 ways. So the required probability is 4 220 0.0181 . 24. The probability that a screw manufactured by a company is defective is 0.1. The company sells screws in packets containing 5 screws and gives a guarantee of replacement if one or more screws in the packet are found to be defective. The probability that a packet would have to be replaced is _____. [ME-2016 (2 marks)] Solution: Let X be the number of screws that are defective. We have to find P ( X 1) 1 P ( X 0) . Now in given question we have binomial probability distribution so P ( X 0) 5C0 (0.1) 0 (0.9)5 . Thus P ( X 1) 1 P ( X 0) 1 5C0 (0.1) 0 (0.9)5 0.409
25. Three cards drawn from a pack of 52 cards. The probability that they are a king, a queen, and a jack is [ME-2016 (2 marks)] (a) 16 5525 (b) 64 2197 (c) 3 13 (d) 8 16575 Solution: Three cards from 52 cards were drawn in
52
C3 ways.
A king, a queen, and a jack are drawn in 4 C1 4C1 4C1 ways. 4
So required probability is
4
4
C1 C1 C1 52
C3
4 4 4 22100
16 5525
.
26. Two persons P and Q toss an unbiased coin alternately on an understanding that whoever gets the head first wins. If P starts the game, then the probability of P winning the game is _____. [MN-2016 (2 marks)] th Solution: Let Wi be an event of winning the game by P in i chance. If P can wins the game in 1st chance then P(W1 ) 1 2 ; If P loses the game in 1st chance then P wins the game only if Q loses the game in 2nd chance; thus if P loses the game in 1st chance then P can wins the game in 3rd chance then P (W3 ) (1 2)(1 2)(1 2) 1 23 ; Similarly If P loses the game in 3rd chance then P can wins the game in 5th chance then P (W5 ) (1 2)(1 2)(1 2)(1 2)(1 2) 1 25 ; and so on If W be an event of winning the game by P , then required probability 1 1 1 (1 2) P (W ) P (W1 ) P (W2 ) P(W3 ) 3 5 0.666 . 2 2 2 2 1 (1 2 ) 27. A coin is tossed three times. It is known that out of the three tosses, one is a HEAD. The probability of the other two tosses also being HEADs is _____. [MT-2016 (2 marks)] 3 Solution: If a coin is tossed three times then total number of possible outcomes will be 2 8 . But out of these 8 outcomes one of the element is {TTT}; as it is given that in our sample space one out of three tosses is HEAD, so total number of sample space is 7 (which contains at least one HEAD). Now only one element in a sample space of 7 is {HHH}. So required probability is 1 7 0.142 .
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Engineering Mathematics
Chapter – 6: Probability & Statistics
Chapter – 6 [7]
28. A box has a total of ten identical sized balls. Seven of these balls are black in colour and the rest three are red. Three balls are picked from the box one after another without replacement. The probability that two of the balls are black and one is red is equal to _____. [PE-2016 (2 marks)] Solution: Applying the Bay’s theorem and from the given figure the required probability, i.e. probability of having 2 black balls and 1 red ball, which is also shown by an arrow, is given by: 7 6 3 7 3 6 P (2 B 1R) 10 9 8 10 9 8 3 7 6 10 9 8 P(2 B 1R ) 0.175 0.175 0.175 0.525
29. Let
X
be a continuous type random variable with probability density function 1 4 , 1 x 3 . When P ( X x ) 0.75 , the value of x is equal to _____. f ( x) 0, otherwise [TF-2016 (2 marks)] Solution: As we have rectangular continuous random variable, X , whose probability distribution function is given
by
(1
4)( x ) x 1
f ( x) ,
then
we
must
have
x
x
1
1
P ( X x ) f ( x) dx 0.75 (1 4) dx 0.75
0.75 x 1 3 x 2
30. A diagnostic test for a certain disease is 90% accurate. That is, the probability of a person having (respectively, not having) the disease tested positive (respectively, negative) is 0.9. Fifty percent of the population has the disease. What is the probability that a randomly chosen person has the disease given that the person tested negative? _____. [XE-2016 (2 marks)] Solution: Let H be an event for a person having the disease; and NH be an event for a person not having the disease. Also let ‘ ’ be an event for a person tested ve ; and ‘ ' be an event for a person tested ve . The following diagram shows the situation of the given problem. So the probability that a randomly chosen person has the disease given that the person tested negative is given by: P( H ) P ( H ) P ( H ) P( H ) P ( H ) P ( NH ) P( NH )
P ( H )
0.1 0.5 0.1 0.5 0.9 0.5
0.1 .
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.1]
Chapter 6 : Probability & Statistics 6.1
Permutation and Combination
6.1.1 Permutation The Factorial: Let n be a ve integer, then continued product of first n natural numbers is called factorial n , denoted by n! or n . Also, 0! 1 . Thus
n ! n( n 1)(n 2) 3 2 1 , n ! n (n 1)! n( n 1) (n 2)! and so on (6.1) If n is ve integer or a fraction, n! is not defined. Exponent of prime in !: Let p be a given prime and n I . Let E p ( n) denotes the exponent of the prime p (i.e., the maximum power of p present in n! ) in n! , n I . Then E p ( n ) [ n p ] [ n p 2 ] [ n p 3 ] , where [] denotes the greatest integer function
(6.2)
Eq. 6.2 does not work for composite number. For e.g., if we find the maximum power of 6 present in 32! , we find that the answer is not [32 6] [32 62 ] 5 , as 5 is the number of integral multiples of 6 in 1, 2, , 32 and 6 can also be obtained by multiplying 2 and 3. Hence, for the required number, we find the maximum powers of 2 and 3 (say r and s ) present in 32! . Using Eq. 6.2, we get r 31 and s 14 . Hence, 2 and 3 will be combined 14 times (to form 6). Thus maximum power of 6 present in 32! is 14.
Fundamental Principles of Counting
Addition principle: Let A and B are two disjoint events (mutually exclusive); i.e., they never occur together. Further suppose that A occurs in m ways and B in n ways. Then A or B can occur in m n ways. This rule also applied to more than two mutually exclusive events. Multiplication principle: Suppose that an event X can be decomposed into two stages A and B . Let stage A occur in m ways and suppose that these stages are unrelated, in the sense that stage B occurs in n ways regardless of the outcome of stage A . Then event X occur in mn ways. This rule is applicable even if event X can be decomposed in more than two stages. In other words, if one operation can be performed independently in m different ways and if second operation can be performed independently in n different ways and a third operation can be performed independently in p different ways and so on, then the total number of ways in which all the operations can be performed in the stated order is ( m n p )
Example 6.1: A college offers 7 courses in the morning and 5 in the evening. Find the number of ways a student can select exactly one course, either in the morning or in the evening. Solution: The student has seven choices from the morning courses out of which he can select one course in 7 ways. For the evening course, he has 5 choices out of which he can select one course in 5 ways. Hence he has total number of 7 5 12 choices. Example 6.2 [CS-2004 (1 mark)]: The number of different n n symmetric matrices with each x element being either 0 or 1 is: (Note power (2, x) is same as 2 ) (a) power (2, n)
(b) power (2, n 2 )
(c) power {2, ( n 2 n) 2}
(d) power {2, ( n 2 n) 2}
Solution: The total number of elements in n n matrix is n 2 . In this matrix we have n number of elements along the diagonal and ( n2 n) 2 number of elements above and below diagonal of the matrix. If this matrix to be symmetric with each element being either 0 or 1 then we fill the diagonal places and one of the places in above or below diagonal (note that if above diagonal elements are filled then below diagonal elements will be adjusted in order to it have symmetric matrix). So we have
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.2]
to fill n ( n 2 n) 2 ( n 2 n) 2 number of places with either 0 or 1. As each place have two choices 2
( n2 n) 2
0 or 1; so ( n2 n) 2 number of places will be filled in 2 2 {( n n) 2 times} 2 ways which is the number of different n n symmetric matrices with each element being either 0 or 1. Example 6.3 [CS-2006 (1 mark)]: Consider the polynomial p ( x ) a0 a1 x a2 x 2 a3 x 3 , where ai 0, i . The minimum number of multiplications needed to evaluate p on an input x is: (a) 3 (b) 4 (c) 6 (d) 9 Solution (a): The given polynomial can be written as p ( x) a0 x{a1 x (a2 a3 x )} , so the
multiplications are ( a3 x ) , p ( x) x ( a2 a3 x ) and x{a1 x (a2 a3 x)} . So answer is 3. Example 6.4 [XE-2007 (2 marks)]: The number of n n matrices that are simultaneously Hermitian, unitary and diagonal is n (d) 2 (c) 2n (b) n 2 (a) 2 Solution (a): Let U A iB be a matrix. For U to be diagonal matrix, both A and B must be diagonal, i.e., A diag ( a1 , a2 , , an ) and B diag (b1 , b2 , , bn ) . For U to be hermitian matrix, a j ib j a j ib j b j 0 B O and thus U A diag ( a1 , a2 , , an ) is real matrix. For U to
be unitary matrix (i.e., orthogonal), each row must be of unit length, i.e., we must have a 2j 1 a j 1 . Now we have n places and each place has 2 options ( 1 or 1) to be filled; so total number of ways is 2 2 2 (n times) 2 n . Example 6.5 [CS-2014 (2 marks)]: The number of distinct positive integral factors of 2014 is … Solution: As 2014 1 2 19 53 ; so positive integral factors of 2014 are 1, 2, 19, 53, 2 19 , 2 53 , 19 53 , 2014 . So number of distinct positive integral factors of 2014 is 8.
Definition of Permutation: The ways of arranging or selecting a smaller or an equal number of persons or objects at a time from a given group of persons or objects with due regard being paid to the order of arrangement or selection are called the (different) permutations. For e.g., three different things a , b , and c are given, then different arrangements which can be made by taking two things from three given things are ab , ac , bc , ba , ca and cb . Thus the number of permutations will be 6. Number of Permutations without Repetition: Arranging n objects, taken r at a time is same r places : 2 3 r 1 as to filling r places from n things, i.e. . No. of choices : n ( n 1) ( n 2) n (r 1) Since, the total number of ways of arranging total number of ways of filling r places n( n 1) ( n 2).....( n r 1)(( n r )!) n! n(n 1) ( n 2).......(n r 1) n Pr ( n r )! ( n r )! The number of arrangements of n different objects taken all at a time n Pn n !
n
P0 (n !) ( n !) 1 ; n Pr n .n 1 Pr 1 ; 0! 1 ; 1 {( r )!} 0 or ( r )! ( r N )
Number of Permutations with Repetitions: The number of permutations (arrangements) of n different objects, taken r at a time, when each object may occur once, twice, thrice, … upto r times in any arrangement total number of ways of filling r places where each place can be r places : 1 2 3 r filled by any one of n objects, i.e., ( n) r . No. of choices : n n n n The number of arrangements that can be formed using n objects out of which p are identical (and of one kind) q are identical (and of another kind), r are identical (and of another kind) and the rest are distinct is ( n !) {( p !)( q !)(r !)} .
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.3]
Example 6.6: How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 3000 (repetition is not allowed)? Solution: All the 5 digit numbers and 6 digit numbers are greater than 3000. Therefore number of 5 digit numbers is 6 P5 5 P5 600 (Since the case that 0 will be at ten thousand place should be omit). Similarly number of 6 digit numbers 6! 5! 600 . Now the numbers of 4 digit numbers which are greater than 3000, having 3, 4 or 5 at first place, this can be done in 3 ways and remaining 3 digit may be filled from remaining 5 digits i.e., required number of 4 digit numbers are 5 P3 3 180 . Hence total required number of numbers 600 600 180 1380 . Example 6.7 [CS-2003 (1 mark)]: n couples are invited to a party with the condition that every husband should be accompanied by his wife. However, a wife need not be accompanied by her husband. The number of different gathering possible at the party is 2n 2n (2 n)! (a) 2n (d) (c) (b) 3n n 2 n n Solution (b): As for every couple, they have three options (i) nobody goes to gathering; (ii) wife alone goes; (iii) both husband and wife goes. As there are n couples, so total number of ways of gathering is 3 3 ( n times) 3n . Example 6.8 [CH-2012 (1 mark)]: A box containing 10 identical compartments has 6 red balls and 2 blue balls. If each compartment can hold only one ball, then the number of different possible arrangements are (a) 1026 (b) 1062 (c) 1260 (d) 1620 Solution: We have total 8 balls (out of which 6 are of one type and 2 are of second type) and 10 places. So total number of different possible arrangements are (10!) {(6!)(2!)} 1260 . Example 6.9 [MN-2013 (1 mark)]: The number of ways in which the letters in the word MINING can be arranged is (a) 90 (b) 180 (c) 360 (d) 720 Solution: As there are 6 letters in MINING; but two letters ‘I’ and ‘N’ appear two times so the number of ways in which the letters in the word MINING can be arranged is (6!) {(2!)(2!)} 180 . Example 6.10: In a train 5 seats are vacant then how many ways can three passengers sit? Solution: Number of ways are 5 P3 (5!) {(5 3)!} (5!) (2!) 120 2 60 . Example 6.11: The number of 5 digit telephone numbers having at least one of their digits repeated is Solution: Using the digits 0,1, 2, , 9 the number of five digit telephone numbers which can be formed is 105 (since repetition is allowed). The number of five digit telephone numbers which have none of the digits repeated is 10 P5 30240 . Hence, the required number of telephone numbers is 5
10 30240 69760 .
Conditional Permutations
Number of permutations of n dissimilar things taken r at a time when p particular things always occur is
Cr p r ! .
Number of permutations of n dissimilar things taken r at a time when p particular things never occur is
n p
n p
Cr r !
The total number of permutations of n different things taken not more than r at a time, when each thing may be repeated any number of times, is {n( n r 1)} ( n 1) .
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.4]
Number of permutations of n different things, taken all at a time, when m specified things always come together is ( m !) ( n m 1)! . Number of permutations of n different things, taken all at a time, when m specified things never come together is ( n !) (m !) ( n m 1)! . Let there be n objects, of which m objects are alike of one kind, and the remaining (n m) objects are alike of another kind. Then, the total number of mutually distinguishable permutations that can be formed from these objects is ( n !) {( m !) ( n m)!} . This theorem can be extended further i.e., if there are n objects, of which p1 are alike of one kind; p2 are alike of another kind; p3 are alike of 3rd kind; … ; pr are alike of r
th
kind such that p1 p2 pr n ; then the
number of permutations of these n objects is ( n !) {( p1 !) ( p2 !) ( pr !)} . Gap method: Suppose 5 males A, B, C, D, E are arranged in a row as A B C D E . There will be six gaps between these five. Four in between and two at either end. Now if three females P, Q, R are to be arranged so that no two are together we shall use gap method i.e., arrange them in between these 6 gaps. Hence the answer will be 6 P3 . Together: Suppose we have to arrange 5 persons in a row which can be done in 5! 120 ways. But if two particular persons are to be together always, then we tie these two particular persons with a string. Thus we have 5 2 1 (1 corresponding to these two together) 3 1 4 units, which can be arranged in 4! ways. Now we loosen the string and these two particular can be arranged in 2! ways. Thus total arrangements 24 2 48 . Never together Total – Together 120 48 72 . Example 6.12: Find the number of words which can be made out of the letters of the word ‘MOBILE’ when consonants always occupy odd places. Solution: The word ‘MOBILE’ has three even places and three odd places. It has 3 consonants and 3 vowels. In three odd places we have to fix up 3 consonants which can be done in 3 P3 ways. Now remaining three places we have to fix up remaining three places which can be done in
3
P3 ways.
Hence, the total number of ways 3 P3 3 P3 36 . Example 6.13: An n digit number is a positive number with exactly n digits. Nine hundred distinct n digit numbers are to be formed using only the three digits 2, 5 and 7. Find the smallest value of n for which this is possible. Solution: Since at any place, any of the digits 2, 5 and 7 can be used total number of such positive n digit numbers are 3n . Since we have to form 900 distinct numbers, hence 3n 900 n 7 .
Circular Permutations: So far we have been considering the arrangements of objects in a line. Such permutations are known as linear permutations. Instead of arranging the objects in a line, if we arrange them in the form of a circle, we call them, circular permutations. In circular permutations, what really matters is the position of an object relative to the others. Thus, in circular permutations, we fix the position of the one of the objects and then arrange the other objects in all possible ways. There are two types of circular permutations: 1. The circular permutations in which clockwise and the anticlockwise arrangements give rise to different permutations, e.g. Seating arrangements of persons round a table. 2. The circular permutations in which clockwise and the anticlockwise arrangements give rise to same permutations, e.g. arranging some beads to form a necklace. Figure 6.1: Circular Permutation Suppose A, B, C , D are the four beads forming a necklace. As shown in Fig. 6.1, they have been arranged in clockwise and anticlockwise directions in the first and second arrangements respectively. Now, if the necklace in the first arrangement be given a turn, from
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.5]
clockwise to anticlockwise, we obtain the second arrangement. Thus, there is no difference between the above two arrangements. Difference between clockwise and anticlockwise arrangement: If anticlockwise and clockwise order of arrangement are not distinct, for e.g. arrangement of flowers in garland, then the number of circular permutations of n distinct items is {( n 1)!} 2 . The number of circular permutations of n different objects is ( n 1)! The number of ways in which n persons can be seated round a table is ( n 1)!
The number of ways in which n different beads, are arranged to form a necklace, is (1 2)( n 1)! . When the positions are numbered, circular arrangement is treated as a linear arrangement. In a linear arrangement, it does not make difference whether the positions are numbered or not.
Example 6.14: In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit together? Solution: Since total number of ways in which boys can occupy any place is (5 1)! 4! and the 5 girls can be sit accordingly in 5! ways. Hence required number of ways are (4!) (5!) . Example 6.15: Find the number of ways in which 5 beads of different colours form a necklace? Solution: The number of ways in which 5 beads of different colours can be arranged in a circle to form a necklace are (5 1)! 4! . But the clockwise and anticlockwise arrangement are not different (because when the necklace is turned over one gives rise to another). Hence the total number of ways of arranging the beads is (1 2)(4!) 12 .
6.1.2 Combination Each of the different groups or selections which can be formed by taking some or all of a number of objects, irrespective of their arrangements, is called a combination. Suppose we want to select two out of three persons A , B and C . We may choose AB or BC or AC . Clearly, AB and BA represent the same selection or group but they give rise to different arrangements. Clearly, in a group or selection, the order in which the objects are arranged is immaterial. n Number of all combinations of n things, taken r at a time is denoted by C ( n, r ) or n C r or . r
Difference between a permutation and combination In a combination only selection is made whereas in a permutation not only a selection is made but also an arrangement in a definite order is considered. In a combination, the ordering of the selected objects is immaterial whereas in a permutation, the ordering is essential. For example A , B and B , A are same as combination but different as permutations. Practically to find the permutation of n different items, taken r at a time, we first select r items from n items and then arrange them. So usually the number of permutations exceeds the number of combinations. Each combination corresponds to many permutations. For e.g., the six permutations ABC , ACB , BCA , BAC , CBA and CAB correspond to the same combination ABC . Generally we use the word ‘arrangements’ for permutations and word “selection” for combinations.
Example 6.16 [CS-2004 (2 marks)]: Mala has a colouring book in which each English letter is drawn two times. She wants to paint each of these 52 prints with one of k colours, such that the colour pairs used to colour any two letters are different. Both prints of a letter can also be coloured with the same colour. What is the minimum value of k that satisfies this requirement? (a) 9 (b) 8 (c) 7 (d) 6 Solution (c): We have letters A-Z and each letter is printed twice, so there are 52 letters. Now we have to colour each letter, so we need a pair of colours for that, because each letter is printed twice. Also in a pair, both colours can be some. Now condition is that a pair of colours cannot be used more
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[6.6]
than once. So let she has 2 colours: Black and White. She can colour as follows : (A,A): (Black, Black), (B,B): (White, White), (C,C): (Black, White). Now we do not have more pairs of colours left, we have used all pairs, but could colour only 3 letters out of 26. So our task is to find minimum no. of colours, so that we could colour all 26 letters. So if she has k colours, she can have k pairs of same colours, thus colouring k letters, then k C2 other pairs of colours, thus colouring k C2 more letters. So total
no.
of
letters
colored
= k k C2 k (1 2) k ( k 1) (1 2)k ( k 1) .
(1 2) k (k 1) 26 k 2 k 52 k ( , 7.7) [6.7, ) . As k cannot k [6.7, ) . Thus minimum value of k that satisfies the requirement is 7.
So be
we
want
negative
so
Example 6.17 [CS-2004 (2 marks)]: In an m n matrix such that all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is (b) max( a, b) (c) min( m a, n b) (d) min( a, b) (a) a b Solution (d): Suppose a b , for example let a 3 , b 4 , then we can put non-zero entries only in 3 rows and 4 columns. So suppose we put non-zero entries in any 3 rows in 3 different columns. Now we cannot put any other non-zero entry anywhere in matrix, because if we put it in some other row, then we will have 4 rows containing non-zeros, if we put it in one of those 3 rows, then we will have more than one non-zero entry in one row, which is not allowed. So we can fill only ‘ a ’ non-zero entries if a b , similarly if b a , we can put only ‘ b ’ non-zero entries. So answer is min( a, b) , because whatever is less between a and b , we can put at most that many non-zero entries. Example 6.18 [CS-2006 (2 marks)]: Given a set of elements N {1, 2, , n} and two arbitrary subsets A N and B N , how many of the n! permutations from N to N satisfy min ( A) min ( B ) , where min( S ) is the smallest integer in the set of integers S , and ( S ) is the set of integers obtained by applying permutation to each elements of S ? n A B 2 2 2 (c) n ! (a) n A B A B (b) A B n 2 (d) A B A B A B
Solution (c): First let us understand what question is asking. So is a function from N to N , which just permutes the elements of N , so there will be n! such permutations. Now given a particular , i.e., given a particular permutation scheme, we have to find number of permutations out of these n! permutations in which minimum elements of A and B after applying to them are same. So for example, if N {1, 2, 3} , is {2, 3,5} , and if A is {1, 3} , then ( A) {1, 2} . Now n
number of elements in A B is A B . We can choose permutations for A B in C A B ways. Note that here we are just choosing elements for permutation, and not actually permuting. Let this chosen set be P . Now once we have chosen numbers for permutations, we have to select mapping from each element of A B to some element of P . So first of all, to achieve required condition specified in question, we have to map minimum number in P to any of the number in A B , so that min{ ( A)} min{ ( B )} . We can do this in A B ways, as we can choose any element of A B to be mapped to minimum number in P . Now we can permute numbers in P in A B 1 ! ways, since one number (minimum) is already fixed. Also, we can permute remaining in n A B 1 ! n
ways, so total no. of ways is C A B A B A B 1 ! n A B 1 ! n ! A B A B . Statement for Linked Answer Questions 6.19 & 6.20: Suppose that a robot is placed on the Cartesian plane. At each step it is allowed to move either one unit up or one unit right, i.e., if it is at (i, j ) then it can move to either (i 1, j ) or (i, j 1) . Example 6.19 [CS-2007 (2 marks)]: How many distinct paths are there for the robot to reach the point (10,10) starting from the initial position (0, 0)
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20 10 (a) 20 C10 (d) None of these (b) 2 (c) 2 Solution (a): At each move, robot can move either 1 unit right, or 1 unit up, and there will be 20 such moves required to move from (0,0) to (10,10). So we have to divide these 20 moves, numbered from 1 to 20, into 2 groups: right group and up group. Each group contains 10 elements each. So basically, we have to divide 20 things into 2 groups of 10 and 10 things each, which can be done in (20!) {(10!)(10!)} 20 C10 .
Example 6.20 [CS-2007 (2 marks)]: Suppose that the robot is not allowed to traverse the line segment from (4, 4) to (5, 4) . With this constraint, how many distinct paths are there for the robot to reach (10,10) starting from (0, 0) ? 9 19 (c) 8 C4 11C5 (d) 20 C10 8C4 11C5 (a) 2 (b) 2 Solution (d): Since we are not allowed to traverse from (4,4) to (5,4). So we will subtract all those paths which were passing through (4,4) to (5,4). To count number of paths passing through (4,4) to (5,4), we find number of paths from (0,0) to (4,4), and then from (5,4) to (10,10). From (0,0) to (4,4), number of paths is (8!) {(4!)(4!)} 8C4 (found in same way as earlier). From (5,4) to (10,10),
number of paths is (11!) {(5!)(6!)} 11C5 . So total number of paths required is
20
C2 8C4 11C5 .
Number of Combinations without Repetition: The number of combinations (selections or groups) that can be formed from n different objects taken r (0 r n) at a time is n Cr
n!
. r !( n r )! Proof: Let the total number of selections (or groups) x . Each group contains r objects, which can be arranged in r ! ways. Hence the number of arrangements of r objects x (r !) . But the number of arrangements is n Pr . Hence, x ( r !) n Pr x ( n Pr ) ( r !) x ( n !) {r !( n r )!} n Cr .
n
C r is a natural number.
n
C0 n Cn 1, n C1 n
n
Cr n Cn r
n
Cr n C r 1 n 1Cr
n
Cx n C y x y or x y n
n . n 1Cr 1 ( n r 1) n Cr 1
If n is even then the greatest value of n C r is n C n 2
If n is odd then the greatest value of n
n
C r is
n
( Cn 1 ) 2 or ( Cn 1 ) 2
n
Cr ( n r ) n 1 Cr 1
( n Cr ) ( n Cr 1 ) (n r 1) r
n
C0 n C1 n C 2 n Cn 2 n
n
C0 n C2 n C4 n C1 n C3 n C5 2n 1
2 n 1
n
Cn n 1Cn n 2 Cn n 3Cn 2 n 1Cn 2 n Cn 1
Number of combinations of n dissimilar things taken all at a time n Cn 1 .
C0 2 n 1C1 2 n 1Cn 22 n
Example 6.21: If n Cr 1 36, n Cr 84 and n Cr 1 126 then find the value of r . n
Cr 1
36
n
Cr
84
4n 10r 6 ; on solving n 9 and r 3 Cr 84 Cr 1 126 Example 6.22: In a conference of 8 persons, if each person shake hand with the other one only, then find the total number of shake hands. Solution: Total number of shake hands when each person shake hands with the other once only is 8 C2 28 ways.
Solution:
n
3n 10r 3 ;
n
Example 6.23: How many words of 4 consonants and 3 vowels formed from 6 consonants and 5 vowels. Solution: Required number of words = 6 C4 5C3 7! = 756000
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Number of Combinations with Repetition and All Possible Selections 1. The number of combinations of n distinct objects taken r at a time when any object may be repeated any number of times is coefficient of x r in (1 x x 2 x r ) n coefficient of x r in (1 x ) n n r 1Cr . 2. The total number of ways in which it is possible to form groups by taking some or all of n things n at a time is 2 1 . 3. The total number of ways in which it is possible to make groups by taking some or all out of n ( n1 n2 ) things, when n1 are alike of one kind, n2 are alike of second kind, and so on is
{(n1 1) ( n2 1) } 1 . 4. The number of selections of r objects out of n identical objects is 1. 5. Total number of selections of zero or more objects from n identical objects is n 1 . 6. The number of selections taking at least one out of a1 a2 a3 an k objects, where a1 are
alike (of one kind), a2 are alike (of second kind) and so on … an are alike (of nth kind) and k are distinct [( a1 1) ( a2 1) ( a3 1) ( an 1)] 2 k 1 . Example 6.24 [CS-1999 (2 marks)]: Two girls have picked 10 roses, 15 sunflowers and 14 daffodils. What is the number of ways they can divide the flowers amongst themselves? (a) 1638 (b) 2100 (c) 2640 (d) None of these Solution (c): Two girls can pick roses in 11 ways with in which one girl gets 0,1, 2, ,10 number of roses and the other girl gets the remaining. Similarly two girls can pick sunflower in 16 ways with in which one girl gets 0,1, 2, ,15 number of sunflowers and the other girl gets the remaining. Similar case happens for the daffodils, i.e., for daffodil flowers 15 ways So total number of ways they can divide the flowers amongst themselves is 11 16 15 2640 ways. Example 6.25: A man has 10 friends. In how many ways he can invite one or more of them to a party? 10 Solution: Required number of friend 2 1 (As the case that no friend be invited is excluded) Example 6.26: Find the numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is allowed). Solution: Numbers greater than 1000 and less than or equal to 4000 will be of 4 digits and will have either 1 (except 1000) or 2 or 3 in the first place with 0 in each of remaining places. After fixing 1st place, the second place can be filled by any of the 5 numbers. Similarly third place can be filled up in 5 ways and 4th place can be filled up in 5 ways. Thus there will be 5 5 5 125 ways in which 1 will be in first place but this include 1000 also hence there will be 124 numbers having 1 in the first place. Similarly 125 for each 2 or 3. One number will be in which 4 in the first place and i.e., 4000. Hence the required numbers are 124 125 125 1 375 ways.
Conditional Combination
The number of ways in which r objects can be selected from n different objects if k particular objects are Always included n k C r k Never included n k C r
The number of combinations of n objects, of which p are identical, taken r at a time is
n p
C r n p C r 1 n p C r 2 n p C0 if r p ; and
n p
Cr n p Cr 1 n p Cr 2 n p Cr p if r p
Example 6.27 : In the 13 cricket players 4 are bowlers, then how many ways can form a cricket team of 11 players in which at least 2 bowlers included?
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Solution: The number of ways can be given as follows: 2 bowlers and 9 other players 4 C2 9C9 ; 3 bowlers and 8 other players 4C3 9C8 ; 4 bowlers and 7 other players 4 C4 9C7 . Hence required number of ways 6 1 4 9 1 36 78 . Example 6.28: In how many ways a team of 10 players out of 22 players can be made if 6 particular players are always to be included and 4 particular players are always excluded? Solution: 6 particular players are always to be included and 4 are always excluded, so total number of selection, now 4 players out of 12. Hence number of ways 12 C4 .
Division into Groups: Case I:
The number of ways in which n different things can be arranged into r different groups is n r 1 Pn or ( n !) n 1C r 1 according as blank group are or are not admissible.
The number of ways in which n different things can be distributed into r different group is r n r C1 ( r 1) n r C2 ( r 2) n ( 1) n 1 n Cr 1 or Coefficient of x n is n !(e x 1) r . Here blank groups are not allowed. Number of ways in which m n different objects can be distributed equally among n persons (or numbered groups) (number of ways of dividing into groups) (number of groups)! {( mn)! n !} {( m !) n n !} {( mn)!} {( m !) n } .
Division into Groups: Case II:
The number of ways in which ( m n) different things can be divided into two groups which contain m and n things respectively is, m n Cm n Cn {( m n )!} {m ! n !}, m n . If m n , then the groups are equal size. Division of these groups can be given by two types: Type I: If order of group is not important: The number of ways in which 2n different things can be divided equally into two groups is {(2n)!} {2!( n !) 2 } . Type II: If order of group is important: The number of ways in which 2n different things can be divided equally into two distinct groups is {(2n)!} {2!( n !) 2 } 2! {2n !} {( n !) 2 }
The number of ways in which ( m n p ) different things can be divided into three groups which contain m , n and p things respectively is
m n p
Cm n p Cn p C p
(m n p)!
, m n p . If m! n ! p ! m n p , then the groups are equal size. Division of these groups can be given by two types. Type I: If order of group is not important: The number of ways in which 3 p different
things can be divided equally into three groups is {(3 p )!} {(3!)( p !)3 } . If order of group is not important: The number of ways in which mn different things can be divided equally into m groups is ( mn !) {(n !) m m !} . Type II: If order of group is important: The number of ways in which 3 p different things can be divided equally into three distinct groups is {(3 p )!} {( p !)3 } . If order of group is important: The number of ways in which mn different things can be divided equally into m distinct groups is {( mn)!} {( n !) m } . Example 6.29: In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes? Solution: The required number of ways 45 1024 [since each prize can be distributed by 4 ways] Example 6.30: Find the number of ways in which 9 persons can be divided into three equal groups? Solution: Total ways {9!} {(3!)3 } 280 .
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De-arrangement: Any change in the given order of the things is called a derangement. If n things form an arrangement in a row, the number of ways in which they can be deranged so that no one of 1 1 1 1 them occupies its original place is n ! 1 ( 1) n . n! 1! 2! 3! Example 6.31: There are four balls of different colours and four boxes of colours same as those of the balls. Find the number of ways in which the balls, one in each box, could be placed such that a ball doesn't go to box of its own colour. 1 1 1 1 Solution: Number of derangement are 4! 1 12 4 1 9 . 1! 2! 3! 4!
Some Important Results for Geometrical Problems
Number of total different straight lines formed by joining the n points on a plane of which m( n) are collinear is n C2 m C2 1 .
Number of total triangles formed by joining the n points on a plane of which m( n) are collinear is n C3 m C3 .
Number of diagonals in a polygon of n sides is n C 2 n .
If m parallel lines in a plane are intersected by a family of other n parallel lines. Then total number of parallelograms so formed is m C2 n C2
Given n points on the circumference of a circle, then number of straight lines n C2 triangles n C3
quadrilaterals n C4
If n straight lines are drawn in the plane such that no two lines are parallel and no three lines are concurrent. Then the number of part into which these lines divide the plane is 1 n .
Number of rectangles of any size in a square of n n is size is
n
r 1 r 3
and number of squares of any
n
r 1 r 2 .
In a rectangle of n p ( n p ) number of rectangles of any size is ( np 4)( n 1) ( p 1) and number of squares of any size is
n
r 1 (n 1 r ) ( p 1 r ) .
Example 6.32: Find the number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line. Solution: Required number of ways 12C3 7 C3 220 35 185 . Example 6.33: Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. Find the number of (i) straight lines (ii) triangles which can be formed by joining them. Solution: Out of 18 points, 5 are collinear. (i) Number of straight lines 18C2 5C2 1 153 10 1 144 (ii) Number of triangles 18 C3 5C3 816 10 806
Multinomial Theorem: Let x1 , x2 , , xm be integers. Then (i) number of solutions to the equation x1 x2 xm n (ii) subject to the condition a1 x1 b1 , a2 x2 b2 , , am xm bm is equal a
a 1
b
a
a 1
b
a
to (iii) the coefficient of x n in ( x 1 x 1 x 1 ) ( x 2 x 2 x 2 ) ( x am x m1 xbm ) . This is because the number of ways, in which sum of m integers in (i) equals n , is the same as the number of times x n comes in (iii).
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Use of solution of linear equation and coefficient of a power in expansions to find the number of ways of distribution: The number of integral solutions of x1 x2 x3 xr n where x1 0, x2 0, xr 0 is the same as the number of ways to distribute n identical things among r persons. This is also equal to the coefficient of x n in the expansion of ( x 0 x1 x 2 x 3 ) r ( r n 1)! n r 1 Cr 1 coefficient of x n in {1 (1 x )}r coefficient of x n in (1 x ) r n !( r 1)! The number of integral solutions of x1 x2 x3 xr n where x1 1, x2 1, , xr 1 is same as the number of ways to distribute n identical things among r persons each getting at least 1. This also equal to the coefficient of x n in the expansion of ( x1 x 2 x 3 ) r coefficient of ( n 1)! n r n r r n 1Cr 1 x in { x (1 x )} coefficient of x in x (1 x) ( n r )!( r 1)! Example 6.34 [CS-2003 (2 marks)]: m identical balls are to be placed in n distinct bags. You are given that m kn , where k is a natural number 1 . In how many ways can the balls be placed in the bags if each bag must contain at least k balls? mk m kn n 1 m 1 m kn n k 2 (a) (b) (c) (d) n 1 nk n 1 nk Solution (b): Since we want at least k balls in each bag, so first we put kn balls into bags, i.e., k balls in each bag. Now we are left with m kn identical balls, and we have to put them into n distinct bags such that each bag may receive zero or more balls. So applying the above theorem, we get number of ways to be m kn n 1 Cn 1 .
Example 6.35: A student is allowed to select utmost n books from a collection of (2n 1) books. If the total number of ways in which he can select one book is 63, then find the value of n . Solution: As the student is allowed to select utmost n books out of (2n 1) books; in order to select one book he has the choice to select one, two, … , n books. If T is the total number of ways of selecting one book T 2 n 1C1 2 n 1C 2 2 n 1Cn 63 . So, the sum of binomial coefficients, 2 n 1C0 2 n 1C1 2 n 1C2 2 n 1Cn 2 n 1Cn 1 2 n 1Cn 2 2 n 1C2 n 1 (1 1) 2 n 1 22 n 1 2 n 1C0 2( 2 n 1 C1 2 n 1C2 2 n 1Cn ) 2 n 1C2 n 1 22 n 1
1 2(T ) 1 22 n 1 1 T (1 2)22 n 1 22 n 1 63 2 2 n 2 6 2 2 n n 3 .
Number of Divisors: Let N p11 p 2 2 p3 3 pk k , where p1 , p2 , , pk are different primes and 1 , 2 , , k are natural numbers then:
The total number of divisors of N including 1 and N is (1 1) ( 2 1) ( k 1)
The total number of divisors of N excluding 1 and N is (1 1) ( 2 1) ( k 1) 2
The total number of divisors of N excluding 1 or N is (1 1) ( 2 1) ( k 1) 1
The sum of these divisors is ( p10 p12 p1 1 )( p20 p21 p2 2 ) ( pk0 p1k pk k )
The number of ways in which N can be resolved as a product of two factors is (1 2)(1 1) ( 2 1) ( k 1), If N is not a perfect square
(1 2)[(1 1) ( 2 1) ( k 1) 1], If N is a perfect square
The number of ways in which a composite number N can be resolved into two factors which are n1 relatively prime (or co-prime) to each other is equal to 2 where n is the number of different factors in N .
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Important Points
All the numbers whose last digit is an even number 0, 2, 4, 6 or 8 are divisible by 2. All the numbers sum of whose digits are divisible by 3, is divisible by 3, e.g. 534. Sum of the digits is 12, which are divisible by 3, and hence 534 is also divisible by 3. All those numbers whose last two-digit number is divisible by 4 are divisible by 4, e.g. 7312, 8936, are such that 12, 36 are divisible by 4 and hence the given numbers are also divisible by 4. All those numbers, which have either 0 or 5 as the last digit, are divisible by 5. All those numbers, which are divisible by 2 and 3 simultaneously, are divisible by 6, e.g., 108. All those numbers whose last three-digit number is divisible by 8 are divisible by 8. All those numbers sum of whose digit is divisible by 9 are divisible by 9. All those numbers whose last two digits are divisible by 25 are divisible by 25, e.g., 73125.
Example 6.36: Find the total number of divisors of 9600 including 1 and 9600. Solution: Since 9600 2 7 31 52 . Hence number of divisors (7 1)(1 1)(2 1) 48 . Example 6.37: Find the number of divisors of n 38808 (except 1 and n ). Solution: Since 38808 8 4851 8 9 539 8 9 7 7 11 23 32 7 2 11 . So, number of divisors is (3 1)(2 1)(2 1)(1 2) 2 72 2 70 .
Distributing balls into different boxes: In this section, we want to consider the problem of how to count the number of ways of distributing k balls into n different boxes, under various conditions. The conditions that are generally imposed are as: (i) The balls can be either different or identical; (ii) The boxes can be either different or identical; (iii) The distribution can take place either with exclusion (i.e. no box can contain more than one ball) or without exclusion (i.e. a box may contain more than one ball). In our discussion the order in which the balls are placed into the boxes is not important. Now let us discuss the various cases: Case I: Distributing k different balls into n different boxes, with exclusion, corresponds to forming a permutation of size k , taken from a set of size n . Therefore, there are n Pk different ways to distribute k different balls into n different boxes, with exclusion. Case 2: Distributing k different balls into n different boxes, without exclusion, corresponds to forming a permutation of size k , with unrestricted repetitions, taken from a set of size n . Therefore, there are n k different ways to distribute k different balls into n different boxes, without exclusion. Case 3: Distributing k identical balls into n different boxes, with exclusion corresponds to forming a combination of size k , taken from a set of size n . Therefore, there are n C k different ways to distribute k identical balls into n different boxes, with exclusion. Case 4: Distributing k identical balls into n different boxes, without exclusion, corresponds to forming a combination of size k with unrestricted repetitions, taken from a set of size n . Therefore, there are n k 1 C k different ways to distribute k identical balls into n different boxes, without exclusion. Case 5: The number of ways to distribute k different balls into n different boxes, with exclusion, in such a way that no box is empty, is n! if k n and 0 if k n . Case 6: The number of ways to distribute k different balls into n different boxes, without exclusion, in such a way that no box is empty is: n k n k n k n 1 n k C0 ( n 0) C1 (n 1) C2 (n 2) ( 1) ( Cn )(1) for k n , and 0 if k n .
Case 7: The number of ways to distribute k identical balls into n different boxes, with exclusion, in such a way that no box is empty, is 1 if k n and 0 if k n . Case 8: The number of ways to distribute k identical balls into n different boxes, without exclusion, in such a way that no box is empty, is k 1 Cn 1 .
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Exercise: 6.1 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. How many numbers of five digits can be formed from the numbers 2, 0, 4, 3, 8 when repetition of digit is not allowed? _____ 2. How many numbers can be made with the help of the digits 0, 1, 2, 3, 4, 5 which are greater than 3000 (repetition is not allowed)? _____ 3. The number of 5 digit telephone numbers having at least one of their digits repeated is (a) 90000 (b) 100000 (c) 30240 (d) 69760 4. How many words can be made from the letters of the word ‘COMMITTEE’? (c) 9! 2! (d) 9! (a) 9! (2!) 2 (b) 9! (2!)3 5. All the letters of the word ‘EAMCET’ are arranged in all possible ways. The number of such arrangement in which two vowels are not adjacent to each other is _____. 6. m men and n women are to be seated in a row, so that no two women sit together. If m n , then the number of ways in which they can be seated is m !( m 1)! m !( m 1)! ( m 1)!( m 1)! (a) (b) (c) (d) None of these ( m n 1)! ( m n 1)! ( m n 1)! 7. We are to form different words with the letters of the word ‘INTEGER’. Let m1 be the number of words in which I and N are never together, and m2 be the number of words which begin with I 8.
9. 10. 11. 12. 13. 14.
15. 16.
17. 18. 19. 20.
and end with R. Then m1 m2 is equal to _____. An n digit number is a positive number with exactly n digits. Nine hundred distinct n digit numbers are to be formed using only the three digits 2, 5 and 7. The smallest value of n for which this is possible is _____. The number of numbers that can be formed with the help of the digits 1, 2, 3, 4, 3, 2, 1 so that odd digits always occupy odd places, is _____. In how many ways can 5 boys and 5 girls sit in a circle so that no boys sit together (a) (5!) (5!) (b) (4!) (5!) (c) {(5!) (5!)} 2 (d) None of these The number of ways in which 5 beads of different colours form a necklace is _____. The number of ways in which 5 male and 2 female members of a committee can be seated around a round table so that the two female are not seated together is _____. In a conference of 8 persons, if each person shake hand with the other one only, then the total number of shake hands shall be _____. To fill 12 vacancies there are 25 candidates of which five are from scheduled caste. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, then the number of ways in which the selection can be made (a) 5 C3 22 C9 (b) 22 C9 5C3 (c) 22 C3 5C3 (d) None of these There are 10 lamps in a hall. Each one of them can be switched on independently. The number of ways in which the hall can be illuminated is _____. 10 different letters of English alphabet are given. Out of these letters, words of 5 letters are formed. How many words are formed when at least one letter is repeated (a) 99748 (b) 98748 (c) 96747 (d) 97147 A man has 10 friends. In how many ways he can invite one or more of them to a party 10 10 (c) (10!) 1 (a) 10! (b) 2 (d) 2 1 Numbers greater than 1000 but not greater than 4000 which can be formed with the digits 0, 1, 2, 3, 4 (repetition of digits is allowed), are _____. In the 13 cricket players 4 are bowlers, then how many ways can form a cricket team of 11 players in which at least 2 bowlers included _____. In how many ways can 6 persons to be selected from 4 officers and 8 constables, if at least one officer is to be included _____.
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Chapter 6: Probability & Statistics
[6.14]
21. In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes _____. 22. The number of ways in which 9 persons can be divided into three equal groups is _____. 23. The number of ways dividing 52 cards amongst four players equally, are (d) None of these (a) (52!) (13!) 4 (b) (52!) {(13!) 2 4!} (c) (52!) {(12!) 4 4!} 24. A question paper is divided into two parts A and B and each part contains 5 questions. The number of ways in which a candidate can answer 6 questions selecting at least two questions from each part is _____. 25. There are four balls of different colours and four boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that a ball doesn't go to box of its own colour is _____. 26. The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the same straight line, is _____. 27. A student is allowed to select utmost n books from a collection of (2n 1) books. If the total number of ways in which he can select one book is 63, then the value of n is _____. 28. The number of divisors of 9600 including 1 and 9600 are _____. 29. A five digit number divisible by 3 has to formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is _____. 30. The number of positive integers which can be formed by using any number of digits from 0, 1, 2, 3, 4, 5 but using each digit not more than once in each number is _____. 31. The number of arrangements of the letters of the word BANANA in which two N's do not appear adjacently is _____. 32. How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order? _____ 33. 20 persons are invited for a party. In how many different ways can they and the host be seated at a circular table, if the two particular persons are to be seated on either side of the host? (b) 2(18!) (a) 20! (c) 18! (d) None of these 34. Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many ways can we place the balls so that no box remains empty? _____ 35. A box contains two white balls, three black balls and four red balls. In how many ways can three balls be drawn from the box if at least one black ball is to be included in the draw? _____ 36. A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is _____. 37. A man has 7 relatives, 4 women and 3 men. His wife also has 7 relatives, 3 women and 4 men. In how many ways can they invite 3 women and 3 men so that 3 of them are the man’s relatives and 3 his wife’s _____ 38. The sides AB, BC, CA of a triangle ABC have respectively 3, 4 and 5 points lying on them. The number of triangles that can be constructed using these points as vertices is _____. 39. Six points in a plane be joined in all possible ways by indefinite straight lines, and if no two of them be coincident or parallel, and no three pass through the same point (with the exception of the original 6 points). The number of distinct points of intersection is equal to _____. 40. In a plane there are 37 straight lines of which 13 pass through the point A and 11 pass through the point B. Besides no three lines pass through one point, no line passes through both points A and B and no two are parallel. Then the number of intersection points the lines have is equal to _____ 41. In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70, then the number of diagonals of the polygon is _____ 42. The number of numbers of 4 digits which are not divisible by 5 are (a) 7200 (b) 3600 (c) 14400 (d) 1800 43. The number of ways in which an examiner can assign 30 marks to 8 questions, awarding not less than 2 marks to any question is (a) 21 C16 (b) 30 C16 (c) 21 C7 (d) None of these 44. The number of divisors of the form 4n 2 ( n 0) of the integer 240 is (a) 4 (b) 8 (c) 10 (d) 3
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Chapter 6: Probability & Statistics
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Statistics
Statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data. It is also defined as the numerical facts or observations collected with definite purpose. In the language of statistics, one of the most basic concepts is sampling. In most statistical problems, a specified number of measurements or data or a sample is drawn from a much larger body of measurements, called the population. The following are the basic terminology used in Statistics: A variable is a characteristic that changes or varies over time and/or for different individuals or objects under consideration. There are two types of variables: (i) qualitative (ii) quantitative. An experimental unit is the individual or object on which a variable is measured. A single measurement or data value results when a variable is actually measured on an experimental unit. A population is the set of all measurements of interest to the investigator. A sample is a subset of measurements selected from the population of interest. Univariate data result when a single variable is measured on a single experimental unit. Bivariate data result when two variables are measured on a single experimental unit. Multivariate data result when more than two variables are measured. Qualitative variables measure a quality or characteristic on each experimental unit. Qualitative variables produce data that can be categorized according to similarities or differences in kind; so, they are often called categorical data. Variables like gender, city, etc. are qualitative variables. Quantitative variables measure a numerical quantity or amount on each experimental unit. There are two types of quantitative variable which are given as: A discrete variable can assume only a finite or countable number of values. Variables such as number of family members, number of new car sales, and number of defective tires returned for replacement are all examples of discrete variables. A continuous variable can assume the infinitely many values corresponding to the points on a line interval. Variables such as height, weight, time, distance, and volume are continuous because they can assume values at any point along a line interval. For any two values we pick, a third value can always be found between them. 6.2.1 Graphs for Qualitative and Quantitative Data After the data have been collected, they can be consolidated and summarized to show the following information: (i) What values of the variable have been measured (ii) How often each value has occurred. For this purpose, we will construct a statistical table that can be used to display the data graphically as a data distribution. The type of graph we choose depends on the type of variable you have measured. When the variable of interest is qualitative or categorical, the statistical table is a list of the categories being considered along with a measure of how often each value occurred. We can measure ‘how often’ in three different ways: (i) The frequency, or number of measurements in each category; (ii) The relative frequency, or proportion of measurements in each category; (iii) The percentage of measurements in each category. If n be the total number of measurements in the set, we can find the relative frequency and percentage using the relationships: Percent 100 Relative frequency Relative frequency frequency n The sum of the frequencies is always n ; the sum of the relative frequencies is 1; and the sum of the percentages is 100%.
Graphs for qualitative data: When the variable is qualitative, the categories should be chosen so that (i) a measurement will belong to one and only one category (ii) each measurement has a category to which it can be assigned. For e.g. if we categorize milk according to the type of milk used, we might use these categories: buffalo, cow, goat, other. Once the measurements have been categorized and summarized in a statistical table, we can use either a pie chart or a bar chart to display the distribution of the data. A pie chart is the familiar circular graph that shows how the measurements are distributed among the categories. To construct a pie chart, assign one sector of a circle to each category. The angle of each sector should be proportional to the proportion of measurements (or relative frequency) in that category. As a circle contains 360°, so Angle Relative frequency 360o to find the angle.
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A bar chart shows the same distribution of measurements among the categories, with the height of the bar measuring how often a particular category was observed.
Graphs for Quantitative Data: Quantitative variables measure an amount or quantity on each experimental unit. If the variable can take only a finite or countable number of values, it is a discrete variable. A variable that can assume an infinite number of values corresponding to points on a line interval is called continuous. Pie Charts and Bar Charts: Sometimes information is collected for a quantitative variable measured on different segments of the population, or for different categories of classification. For example, we might measure the average incomes for people of different age groups, different genders, or living in different geographic areas of the country. In such cases, we can use pie charts or bar charts to describe the data, using the amount measured in each category rather than the frequency of occurrence of each category. The pie chart displays how the total quantity is distributed among the categories, and the bar chart uses the height of the bar to display the amount in a particular category. Line Charts: When a quantitative variable is recorded over time at equally spaced intervals (such as daily, weekly, monthly, quarterly, or yearly), the data set forms a time series. Time series data are most effectively presented on a line chart with time as the horizontal axis. The idea is to try to discern a pattern or trend that will likely continue into the future, and then to use that pattern to make accurate predictions for the immediate future.
Interpreting Graphs: A distribution is symmetric if the left and right sides of the distribution, when divided at the middle value, form mirror images. A distribution is skewed to the right if a greater proportion of the measurements lie to the right of the peak value. Distributions that are skewed right contain a few unusually large measurements. A distribution is skewed to the left if a greater proportion of the measurements lie to the left of the peak value. Distributions that are skewed left contain a few unusually small measurements. A distribution is unimodal if it has one peak; a bimodal distribution has two peaks. Bimodal distributions often represent a mixture of two different populations in the data set.
Relative Frequency Histogram: A relative frequency histogram resembles a bar chart, but it is used to graph quantitative rather than qualitative data. A relative frequency histogram for a quantitative data set is a bar graph in which the height of the bar shows “how often” (measured as a proportion or relative frequency) measurements fall in a particular class or subinterval. The classes or subintervals are plotted along the horizontal axis. First, divide the interval from the smallest to the largest measurements into subintervals or classes of equal length. Then find the frequency of each class and then divide each frequency by total number of classes so we get relative frequency. Based on relative frequency, draw a bar over each class and we have created a frequency histogram or a relative frequency histogram, depending on the scale of the vertical axis. 6.2.2 Describing Data with Numerical Measures: Measures of Centre Graphs are extremely useful for the visual description of a data set. However, they are not always the best tool when we want to make inferences about a population from the information contained in a sample. For this purpose, it is better to use numerical measures to construct a mental picture of the data. Using statistics, we can summarise large quantities of data, by few descriptive measures. Two descriptive measures are often used to summarise data sets. These are (1) Measures of central tendency (2) Measures of dispersion. The central tendency measure indicates the average value of data, where ‘average’ is a generic term used to indicate a representative value that describes the general centre of the data. Mean, Median and Mode are some examples of central tendency measures. The dispersion measure characterises the extent to which data items differ from the central tendency value. In other words, dispersion measures and quantifies the variation in data. The larger this number, the more the variation amongst the data items. Standard deviation, Variance and Coefficient of Variance are examples of dispersion measures.
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Mean: The arithmetic mean is the central value of the distribution in the sense that positive and negative deviations from the arithmetic mean balance each other. It is a quantitative average. The arithmetic mean or average of a set of n measurements is equal to the sum of the measurements divided by n . To distinguish between the mean for the sample and the mean for the population, we will use the symbol x for a sample mean and the symbol for the mean of population. The sample mean x changes from sample to sample but population mean stays the same. Suppose there are n measurements, x1 , x2 , , xn , on the variable X , then, Mean
x n i 1 i
n,
n
i1 xi x1 x2 xn
(6.3)
Example 6.38 [MN-2009 (2 marks)]: The mean of the cubes of the first n natural numbers is (b) {n ( n 1)(n 2)} 8 (c) ( n4 1) n (a) {n( n 1) 2 } 4 (d) n3 4 Solution (a):
13 23 33 n3 n
1 n( n 1)
n
2
2
n( n 1) 2 4
.
Mean for Discrete Frequency distribution: If a variate X takes values x1 , x2 , , xn with corresponding frequencies f1 , f2 , , f n , respectively, then arithmetic mean of these values is Mean
n fx i 1 i i
f , n i 1 i
n fx i 1 i i
f1 x1 f n xn ,
n
i1 fi f1 f n
(6.4)
Mean for Continuous Frequency distribution: In case of a continuous frequency
distribution or a frequency distribution with class intervals arithmetic mean may be computed by applying any of the methods discussed earlier. The values of x1 , x2 , , xn are taken as the mid points of the various classes. It should be noted that the mid – value is equal to (1 2)(lower limit upper limit) . Example 6.39 [ME-2004 Flow rate (litres/sec) Frequency (2 marks)]: The 7.5 to 7.7 1 following data about the 7.7 to 7.9 5 flow of liquid was 7.9 to 8.1 35 observed in a continuous 8.1 to 8.3 17 chemical process plant. 8.3 to 8.5 12 Mean flow rate of the 8.5 to 8.7 10 liquid is Solution (c): Mid values xi Frequency fi Class – interval 7.5 – 7.7 7.6 1 7.7 – 7.9 7.8 5 7.9 – 8.1 8.0 35 8.1 – 8.3 8.2 17 8.3 – 8.5 8.4 12 8.5 – 8.7 8.6 10 Total 80
(a) (b) (c) (d)
fi xi 7.6 39.0 280.0 139.4 100.8 86.0 652.8
8.00 litres/sec 8.096 litres/sec 8.16 litres/sec 8.26 litres/sec
Mean n
i1 fi xi Mean n i 1 fi
652.8 80
8.16
Median: Median is the central value of the distribution in the sense that the number of values less than the median is equal to the number of values greater than the median; hence median is a positional average. The median m of a set of n measurements is the value of x that falls in the middle position when the measurements are ordered from smallest to largest. If x1 , x2 , , xn are n values of a variables X , then to find the median, we use the following algorithm:
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Step 1: Arrange the observations x1 , x2 , , xn in ascending or descending order of magnitude Step 2: Determine the total number of observations, say, n Step 3: If n is odd, then median is the value of {( n 1) 2}th observation; If n is even, then median is th
the arithmetic mean of the values of ( n 2)th and ( n 2) 1 observations. Example 6.40 [MN-2009 (2 marks)]: The following data represent the number of workers suffering from pneumoconiosis in 10 coal mines. Mine I II III IV V VI VII VIII IX X Number 10 16 14 15 14 12 17 13 15 12 The number of mines falling above 50th percentile in terms of workers suffering from pneumoconiosis is (a) 2 (b) 3 (c) 4 (d) 5 Solution (c): As the median cut the distribution in half so median is the 50th percentile. Thus in the given question we have to find how many mines are above the median of suffered workers. As we have 10 data (which is even), and the data in ascending order as: 10, 12, 12, 13, 14, 14, 15, 15, 16, 17. th So the median is the arithmetic mean of the values of (10 2)th and (10 2) 1 data, i.e., arithmetic mean of 5th and 6th data. So median (14 14) 2 14 . Thus the number of mines having more than 14 workers are 4 (II, IV, VII, IX). [Similar question was also asked in MT-2010 (1 mark)] Example 6.41 [MT-2014 (1 mark)]: What is the median value of the following set of numbers? 1, 3, 5, 9, 6, 4, 8 ……………. Solution: As we have 7 data (which is odd) and the data in ascending order as: 1, 3, 4, 5, 6, 8, 9. So median is {(7 1) 2}th data, i.e., 4th data which is 5.
Median for Discrete Frequency Distribution: If a variate X takes values x1 , x2 , , xn with corresponding frequencies f1 , f2 , , f n , respectively, then we calculate the median by using the following algorithm: Step 1: Find the cumulative frequencies (c. f .) n
Step 2: Find N 2 , where N i 1 f i Step 3: See the cumulative frequency (c. f .) just greater than N 2 and determine the corresponding value of the variable; the value obtained is thus the median.
Median for Continuous Frequency Distribution: In order to calculate the median of a continuous frequency distribution, we use the following algorithm: Step 1: Obtain the frequency distribution n
Step 2: Prepare the cumulative frequency column and obtain N i 1 f i Step 3: Find N 2 Step 4: See the cumulative frequency just greater than N 2 and determine the corresponding class. This class is known as the median class Step 5: Use the following formula: Median l {( N 2) F } f h , where l lower limit of the median class; f frequency of the median class; h step size of the median class; F Cumulative frequency of the class preceding the median class. Example 6.42: Calculate the median from the following distribution: Class: 5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 Frequency
5
6
15
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10
5
4
35 – 40
40 – 45
2
2
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Solution: Class
Frequency
5 – 10 10 – 15 15 – 20 20 – 25 25 – 30 30 – 35 35 – 40 40 – 45
5 6 15 10 5 4 2 2 N 49
Cumulative Frequency 5 11 26 36 41 45 47 49
We have N 49
N
24.5 2 The cumulative frequency just greater than N 2 is 26 and the corresponding class is 15 25 , which is the median class. Hence l 15 , f 15 , F 11 and h 5 . So using the formula in Step 24.5 11 4, we have Median 15 5 19.5 . 15
Mode: The mode or modal value of a distribution is that value of the variable for which the frequency is maximum. Thus, the mode of a distribution is that value of the variable for which the frequency variable are clustered densely. Mode for Discrete frequency distribution: In order to compute the mode of a series of individual observations, we first convert it into a discrete series frequency distribution by preparing a frequency table. From the frequency table, we identify the value having maximum frequency. The value of variable so obtained is the mode or modal value. Mode for Continuous Frequency Distribution: In case of continuous frequency distribution with equal class intervals, we use the following algorithm to compute the mode. Step 1: Obtain the continuous frequency distribution Step 2: Determine the class of maximum frequency by inspection; this class is known as the modal class. Step 3: Using the formula: Mode l {( f f1 ) (2 f f1 f 2 )} h , where l lower limit of the modal class; f frequency of the modal class; h width of the modal class; f1 frequency of the class preceding the modal class; f 2 frequency of the class following the modal class. Example 6.43: Compute the mode for the following frequency distribution: Class: 3–6 6–9 9 – 12 12 – 15 15 – 18 18 – 21 21 – 24 Frequency: 2 5 10 23 21 12 3 Solution: We observed that the class 12 – 15 has maximum frequency. Therefore, this is the modal class. Hence, l 12 , h 3 , f 23 , f1 10 , f 2 21 ; now by using the formula in Step 3, we have Mode 12 {(23 10) (46 10 21)} 3 14.6 .
Properties relating Mean, Median and Mode
The measure of skewness is dependent upon the amount of dispersion.
Figure 6.2: Types of Frequency Distributions
In symmetric or normal distribution, as shown in Fig. 6.2(a): Mean Median Mode [This point was asked in MN-2013 (1 mark)];
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In negatively skewed distribution, as shown in Fig. 6.2(b): Mean Median Mode ; In positively skewed distribution, as shown in Fig. 6.2(c): Mode Median Mean [The above four points were asked in CE-2005 (1 mark)] Mode 3 Median 2Mean Median Mode (2 3)(Mean Mode) Mean Mode (3 2)(Median Mode)
Example 6.44 [CH-2013 (1 mark)]: The number of e-mails received on six consecutive days is 11, 9, 18, 18, 4 and 15, respectively. What are the median and the mode for these data? (a) 18 and 11, respectively (b) 13 and 18, respectively (c) 13 and 12.5, respectively (d) 12.5 and 18, respectively Solution (b): As we have total 6 data (which is even) and the data in ascending order is: 4, 9, 11, 15, th 18, 18. So median is the arithmetic mean of the values of (6 2)th and (6 2) 1 data, i.e., median is arithmetic mean of 3rd and 4th data, which is (11 15) 2 13 . As the highest frequency in the given data is for 18 which occurs 2 times; so modal value is 18. (a) Mean 33; Example 6.45 [PI-2014 (2 Number Marks marks)]: Marks obtained by Mode 40 S. No. of Obtained 100 students in an (b) Mean 35; Students examination are given in the Mode 40 1 25 20 table. What would be the (c) Mean 33; 2 30 20 mean, median and mode of Mode 35 3 35 40 the marks obtained by the (d) Mean 35; 4 40 20 students? Mode 35 Solution (c): From the given data we have the following table: xi Frequency ( fi ) S. No. Cumulative Frequency 1 25 20 20 2 30 20 40 3 35 40 80 4 40 20 100
N fi 100 xi 130 So Mean ( fi xi ) fi 3300 100 33 . Now we have
Median 35; Median 32.5; Median 35; Median 32.5;
fi xi 500 600 1400 800
fi xi 3300
Total
N 100 N 2 50 ; so the cumulative
frequency just greater than N 2 is 80 and the corresponding marks obtained is 35 so median is 35. As the highest frequency (i.e., number of students) is for the data 35 (i.e., marks obtained) so modal value is 35. 6.2.3 Describing Data with Numerical Measures: Measures of Variability Data sets may have the same centre but look different because of the way the numbers spread out from the centre.
Figure 6.3: Variability or Dispersion of Data
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Consider the two distributions shown in Figure 6.3. Both distributions are centred at x 4 , but there is a big difference in the way the measurements spread out, or vary. The measurements in Figure 6.3(a) vary from 3 to 5; in Figure 6.3(b) the measurements vary from 0 to 8. Measures of variability can help us to create a mental picture of the spread of the data. We will present some of the more important ones.
Standard Deviation: One of the most commonly used measures is standard deviation. This value gives information on how the values of the data set are varying, or deviating, from the mean of the data set. In other words, the statistical measure of the variability of a distribution around its mean is referred to as standard deviation [This point was asked in MN-2009, AG-2011 (1 mark)]. Deviations are calculated by subtracting the mean, x , from each of the sample values, x , i.e. deviation x x . As some values are less than the mean, negative deviations will result, and for values greater than the mean positive deviations will be obtained. By simply adding the values of the deviations from the mean, the positive and negative values will cancel to result `in a value of zero. By squaring each of the deviations, the problem of positive and negative values is avoided. To calculate the standard deviation, the deviations are squared. These values are summed, divided by the appropriate number of values and then finally the square root is taken of this result, to counteract the initial squaring of the deviation. The standard deviation of a population, , of N data items is
N i 1
( xi )
2
N , where is the population mean
(6.5)
Another version of Eq. 6.5: N
N
N
N
N
N
N
N
( xi ) 2 ( xi2 2 xi 2 ) xi2 2 xi 2 xi2 2 xi 2 i 1
i 1
N
i 1 ( xi ) 2
i 1
N
x2 2N 2 N 2 i 1 i
N i 1
( xi ) 2
i 1
N
N
i 1
x2 N 2 N i 1 i
N
N x2 i 1 i
i 1
i 1
x
N
x2 i 1 i
x 2 n i 1 i
N
i 1 i
N
N
f x2 i 1 i i
n
fx i 1 i i
2
2
N
N2
Standard deviation for Discrete Frequency distribution If a variate X takes values x1 , x2 , , xn with corresponding frequencies respectively, then we calculate the standard deviation by the following formula,
i 1
(6.6) f1 , f2 , , f n ,
n
N 2 , where N i 1 fi
(6.7)
Standard deviation for Continuous Frequency distribution In case of a continuous frequency distribution or a frequency distribution with class intervals the standard deviation is computed by applying Eq. 6.7; where the values of x1 , x2 , , xn are taken as the mid points of the various classes. It should be noted that the mid-value is equal to (1 2)(lower limit upper limit) .
Example 6.46 [CS-2011 (2 marks)]: Consider a finite sequence of random values X x1 , x2 , , xn . Let x be the mean and x be the standard deviation of X . Let another finite sequence Y of equal length be derived from this as yi a xi b , where a and b are positive constants. Let y be the mean and y be the selected deviation of this sequence. Which one of the following sequence is INCORRECT? (a) Index position of mode of X in X is the same as the index position of mode of Y in Y (b) Index position of median of X in X is the same as the index position of median of Y in Y (c) y a x b (d) y a x b Solution (d): Since mode is that value of the variable for which the frequency is maximum, so if xi at the k th position is the mode of X then yi at the k th position is the mode Y since yi a xi b , so
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option (a) is correct. Also Median is the central value of the distribution in the sense that the number of values less than the median is equal to the number of values greater than the median so if q th observation of X is the median then q th observation of Y is also median since yi a xi b , so option
(b) n
is
correct.
n
n
x i 1 xi n
As
and
n
y i 1 yi n i 1 ( axi b) n a i 1 xi n nb n a x b , so option (c) is correct. Since
x
n
i1 ( xi x )2
y a
n
n
i1 ( xi x )2
n
i1 ( yi y )2
y
and
n
n
i1 (axi b a x b)2
n
n a x ; hence the given option (d) is incorrect.
Variance: The variance of a population of N measurements is the average of the squares of the deviations of the measurements about their mean . The population variance is denoted by 2 and is given by the formula 2
N i 1
( xi )
2
N . Hence the square of standard deviation ( ) is
called as the variance ( 2 ). Statement for Linked Answer Questions 6.47 and 6.48: The abdomen length (in millimetres) was measured in 15 male fruit flies, and the following data were obtained: 1.9, 2.4, 2.1, 2.0, 2.2, 2.4, 1.7, 1.8, 2.0, 2.0, 2.3, 2.1, 1.6, 2.3, and 2.2. Example 6.47 [BT-2011 (2 marks)]: Variance (Vx ) for this population of fruit flies as calculated from the above data shall be (a) 0.85 (b) 0.25 (c) 0.061 (d) 0.08 Solution (c): Consider the following table with mean (sum of all data) N 31 15 2.067 . ( xi )
Frequency ( fi )
xi2
fi xi
f i xi2
1.9 2.4 2.1 2.0 2.2 1.7 1.8 2.3 1.6
1 2 2 3 2 1 1 2 1
3.61 5.76 4.41 4.00 4.84 2.89 3.24 5.29 2.56
1.9 4.8 4.2 6.0 4.4 1.7 1.8 4.6 1.6
3.61 11.52 8.82 12 9.68 2.89 3.24 10.58 2.56
Total
N fi 15
xi2 36.6
fi xi 31
fi xi2 64.9
So, Standard deviation,
N
N
f x2 i 1 i i
n
fx i 1 i i
2
N 2 {15(64.9) 312 } 152 0.25
Hence, Variance, 2 0.252 0.06 . [Similar question was also asked in TF-2008 (1 mark)] Example 6.48 [BT-2011 (2 marks)]: The value of standard deviation (SD) will be (a) 0.061 (b) 0.25 (c) 0.61 (d) 0.85 Solution: From the previous question calculation we have 0.25 . Example 6.49 [BT-2012 (2 marks)]: Consider the data set 14, 18, 14, 14, 10, 29, 33, 31, 25. If you add 20 to each of the values, then (a) both mean and variance change (b) both mean and variance are unchanged
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(c) the mean is unchanged, variance changes Solution (d): As Mean
n i 1
n i 1
x n
i 1 i
n
[6.23]
(d) the mean changes, the variance is unchanged
n ; if we add a number ‘ k ’ to each
n i 1
n i 1
xi , then
( xi k ) n ( xi ) ( k ) n ( xi ) kn n ( xi ) n k k . Thus i 1
adding a constant ‘ k ’ to each data will change the mean by ‘ k ’. Now Variance
2
N i 1
( xi )
2
2 N ; if we add a number ‘ k ’ to each xi , then
as ( xi k ) 2 ( xi k k ) 2 ( xi ) 2 , So 2
N i 1
( xi )
2
N i 1
( xi k )
2
N;
2 N , which there is no
change in variance if a constant ‘ k ’ is added to each data.
Coefficient of Variation: Without an understanding of the relative size of the standard deviation compared to the original data, the standard deviation is somewhat meaningless for use with the comparison of data sets. To address this problem the coefficient of variation is used. The coefficient of variation, CV , gives the standard deviation as a percentage of the mean of the data set. The coefficient of variation for a population is given as, CV ( ) 100% (6.8) [Eq. 6.8 was asked in MN-2012 (1 mark)]. The coefficient of variation is often used to compare the variability of two data sets. It allows comparison regardless of the units of measurement used for each set of data. The larger the coefficient of variation, the more the data varies. Example 6.50 [ME-1999 (1 mark)]: Analysis of variance is concerned with: (a) determining change in a dependent variable per unit change in an independent variable (b) determining whether a qualitative factor affects the mean of an output variable (c) determining whether significant correlation exists between an output variable and an input variable (d) determining whether variances in two or more populations are significantly different Solution (d): As analysis of Variance is a statistical method used to test differences between two or more means. So we can say that option (d) is correct. Example 6.51 [CE-2007 (2 marks)]: If the standard deviation of the spot speed of vehicles in a highway is 8.8 km/hr and the mean speed of the vehicles is 33 km/hr, the coefficient of variation in speed is (a) 0.1517 (b) 0.1867 (c) 0.2666 (d) 0.3646 Solution (c): From the given data we have, standard deviation of the spot speed of vehicles 8.8 km/hr, mean speed of the vehicles 33 km/hr. So from Eq. 6.8, coefficient of variation in speed
CV 8.8 33 0.2666 . [Similar question was also asked in MN-2010 (1 mark)] Example 6.52: The results of two tests are shown as: Test 1 (out of 15 marks): 9, 2 ; Test 2 (out of 50 marks): 27, 8 . Compare the variability of these data sets.
2 8 100% 100% 22.2% ; CVtest 2 100% 100% 29.6% . Hence, 9 27 the results in the second test show a great variation than those in the first test. Solution: CVtest 1
Correlation and Regression: Sometimes, in practical applications, we might come across certain set of data, where each item of the set may comprise of the values of two or more variables. Suppose we have a set of 30 students in a class and we want to measure the heights and weights of all the students. We observe that each individual (unit) of the set assumes two values – one relating to the height and the other to the weight. Such a distribution in which each individual or unit of the set is made up of two values is called a bivariate distribution. For e.g. in a class of 60 students the series of
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marks obtained in two subjects by all of them. Thus in a bivariate distribution, we are given a set of pairs of observations, wherein each pair represents the values of two variables. In a bivariate distribution, we are interested in finding a relationship (if it exists) between the two variables under study. The concept of ‘correlation’ is a statistical tool which studies the relationship between two variables and Correlation Analysis involves various methods and techniques used for studying and measuring the extent of the relationship between the two variables. Two variables are said to be in correlation if the change in one of the variables results in a change in the other variable. There are two important types of correlation. They are (1) Positive and Negative correlation and (2) Linear and Non – Linear correlation. Positive and Negative Correlation: If the values of the two variables deviate in the same direction i.e. if an increase (or decrease) in the values of one variable results, on an average, in a corresponding increase (or decrease) in the values of the other variable the correlation is said to be positive, for e.g., household income and expenditure of a family. Correlation between two variables is said to be negative or inverse if the variables deviate in opposite direction. That is, if the increase in the variables deviate in opposite direction. That is, if increase (or decrease) in the values of one variable results on an average, in corresponding decrease (or increase) in the values of other variable, for e.g., volume and pressure of perfect gas. Linear and Non-Linear Correlation: The correlation between two variables is said to be linear if the change of one unit in one variable result in the corresponding change in the other variable over the entire range of values. For example consider the following data (2, 7), (4,13), (6,19), (8, 25), (10, 31) ; thus, for a unit change in the value of x , there is a constant change in the corresponding values of y and the above data can be expressed by the relation y 3x 1 . In general two variables x and y are said to be linearly related, if there exists a relationship of the form y a bx , where ‘ a ’ and ‘ b ’ are real numbers. This is nothing but a straight line when plotted on a graph sheet with different values of x and y and for constant values of a and b . The relationship between two variables is said to be non – linear if corresponding to a unit change in one variable, the other variable does not change at a constant rate. In such cases, if the data is plotted on a graph sheet we will not get a straight line curve. For example, one may have a relation of the form y a bx cx 2 or more general polynomial. Regression Lines (or Regression Models): In case of simple linear regression model (i.e. when there is only one independent variable and there is linear relationship between the dependent and independent variable) there are two regression lines as follows: Regression Line of X on Y : X a bY , where X is dependent variable, Y is independent variable, a X intercept, b slope of the regression line. The value of ‘ a ’ and ‘ b ’ can be calculated for the given data of X and Y variable by solving the following two algebraic normal 2 equations: X na b Y and XY a Y b Y , where n number of pairs of X and Y variable. The line X a bY can also be expressed as ( X X ) bXY (Y Y ) or ( X X ) r ( X Y )(Y Y ) , where bXY r ( X Y ) , X arithmetic mean of X series, Y
arithmetic mean of Y series, X standard deviation of X series, Y standard deviation of Y series, r is he coefficient of correlation between two variables X and Y . Regression Line of Y on X : Y a bX , where Y is dependent variable, X is independent variable, a Y intercept, b slope of the regression line. The value of ‘ a ’ and ‘ b ’ can be calculated for the given data of X and Y variable by solving the following two algebraic normal 2 equations: Y na b X and XY a X b X , where n number of pairs of X and Y variable. The line Y a bX can also be expressed as (Y Y ) bXY ( X X ) or
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(Y Y ) r ( Y X )( X X ) , where bYX r ( Y X ) , X arithmetic mean of X series, Y
arithmetic mean of Y series, X standard deviation of X series, Y standard deviation of Y series, r is he coefficient of correlation between two variables X and Y . From the above two points, we can conclude that The product of the two regression coefficients is equal to square of the coefficient of correlation, i.e. bXY bYX r 2 . bXY , bYX , r all have the same sign. If r 0 then bXY , bYX are also zero. The regression lines always intersect at their means. The angle between the two regression lines depends on the coefficient of correlation, i.e. if r 0 then the lines are perpendicular to each other; and if r 1 then the regression lines coincides.
Example 6.53 [XE-2008 (2 marks)]: The two lines of regression of the variables x and y are 4 x 2.4 y 20 and 1.6 x 4 y 12 . The coefficient of correlation between x and y is (a) 0.49 (b) –0.49 (c) 0.35 (d) –0.35 Solution (b): 4 x 2.4 y 20 ( x 0) 0.6( y 5) which is a regression line x on y , thus bXY 0.6 and 1.6 x 4 y 12 y 0.4( x 3) which is a regression line y on x , thus bYX 0.4
. So coefficient of correlation, r bXY bYX ( 0.6)( 0.4) 0.489 , we put ve sign because both bXY and bYX are ve .
Coefficient of Correlation: One of the most widely used statistics is the coefficient of correlation ‘ r ’ which measures the degree of association between the two values of related variables given in the data set. It takes values from 1 to 1 . If two sets or data have r 1 , they are said to be perfectly correlated positively if r 1 they are said to be perfectly correlated negatively; and if r 0 they are uncorrelated. The coefficient of correlation ‘ r ’ is given by the formula r n ( xy ) x y
n
x x 2
2
n
y y 2
2
.
Regression Analysis: Method of Least Square: If two variables are significantly correlated, and if there is some theoretical basis for doing so, it is possible to predict values of one variable from the other. In other words if a regression model is used to express a variable Y as a function of another variable X then a value of X may be used to estimate a value of Y [This point was asked in ME2002 (1 mark)]. This observation leads to a very important concept known as ‘Regression Analysis’. Regression analysis, in general sense, means the estimation or prediction of the unknown value of one variable from the known value of the other variable. It is a mathematical measure of the average relationship between two or more variables in terms of the original units of the data. Suppose we have a sample of size ‘ n ’ and it has two sets of measures, denoted by x and y . We can predict the values of ‘ y ’ given the values of ‘ x ’ by using the equation, called the Regression Equation as: y* a bx , where the coefficients a and b are given by b
n ( xy ) x y n
x x 2
2
and a
y b x . n
The symbol y * refers to the predicted value of y from a given value of x from the regression equation. Also the slope of best fit line is ‘ b ’ having intercept ‘ a ’. Example 6.54 [ME-1998 (2 marks)]: The best fit line using least squares for the data (0, 0), (10, 24), (20, 36) and (30, 60) is (d) None of these (a) 2 x y 0 (b) 2 x y 4 0 (c) 2 x y 4 0 Solution: From the given data we have the following table:
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i 1 2 3 4
n4
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xi
yi
xi yi
xi2
0 10 20 30 xi 60
0 24 36 60 yi 120
0 240 720 1800 xi yi 2760
0 100 400 900 2 xi 1400
So we have, b
4(2760) (60)(120) 2
1.92 and a
120 (1.92)(60)
4(1400) (60) 4 using the given data is y a bx 1.2 1.92 x . So option (d) is correct.
1.2 . Thus the best fit line
Example 6.55 [ME-1999 (2 marks)]: Four arbitrary point ( x1 , y1 ), ( x2 , y 2 ), ( x3 , y3 ), ( x4 , y4 ) are given in the x, y plane. Using the method of least squares, if, regressing y upon x gives the fitted line y ax b ; and regressing x upon y gives the fitted line x cy d , then (a) the two fitted lines must coincide (b) the two fitted lines need not coincide (c) it is possible that ac 0 (d) a 1 c Solution (d): As regressing y upon x or x upon y gives the same fitted line so the y ax b and x cy d y (1 c ) x d are same such that a 1 c and b d . So both lines must coincide.
Example 6.56 [CE-2008 (2 marks)]: Three values of x and y are to be fitted in a straight line in the form y a bx by the method of least squares. Given
x 6 , y 21 , x 2 14
and
xy 46 , the values of a and b are respectively (a) 2 and 3
(b) 1 and 2
(c) 2 and 1 (d) 3 and 2 3(46) (6)(21) 21 (2)(6) 2 and a Solution (d): From the given data we have n 3 , so b 3. 2 3(14) (6) 3 2
Coefficient of Determination: The coefficient of determination, r , is useful because it gives the proportion of the variance of one variable that is predictable from the other variable. It is a measure that allows us to determine how certain one can be in making predictions from a certain model or graph. The following points holds for coefficient of determination: It is the ratio of the explained variation to the total variation It represents the percent of data that is the closest to the line of best fit. It is a measures of how well the regression line represent data. If the regression line passes exactly through every point on the scatter plot, it would be able to explain all of the variation. The further the line is away from the points, the less it is able to explain. Exercise: 6.2 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. The mean, median and mode, respectively, for the following ungrouped data: 4, 6, 5, 5, 7, 4, 8, 3, 4, 4, is (a) 5, 5.5, 4 (b) 5, 4.5, 4 (c) 4.5, 5, 5 (d) 5.5, 5, 5 2. The variance and standard deviation, respectively, of the following ungrouped data: 1, 2, 3, is (a) 1, 1 (b) 2, 1 (c) 1, 2 (d) 2, 2 3. The following table gives the frequency Number of Order Frequency distribution of the number of orders 10 – 12 4 received each day during the past 50 days 13 – 15 12 at the office of a mail-order company. 16 – 18 20 Calculate the mean. _____ 19 – 21 14
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4. The following table gives the Time to travel to Work Frequency frequency distribution of the time 1 – 10 8 taken by number of person to travel 11 – 20 14 to work. If 25 person take less than t 21 – 30 12 minutes to travel to work. Then value 31 – 40 9 of t is _____. 41 – 50 7 5. The mode of the frequency distribution table given in Q. 4 is _____. 6. The standard deviation of the frequency distribution table given in Q. 3 is _____. 7. The distribution whose most values are dispersed to the left or right of the mode is classified as (a) skewed (b) explored (c) bimodal (d) unimodal 8. In a negative skewed distribution, the order of mean, median and mode is given as (a) mean median mode (b) mean median mode (c) mean median mode (d) mean median mode 9. The distribution whose outliers are higher values is considered as (a) variable model (b) right skewed (c) left skewed (d) constant model 10. In the distribution, if the mode is greater than median then the distribution is classified as (a) variable model (b) right skewed (c) left skewed (d) constant model 11. Which of the following is not a measure of central location (a) Mode (b) Median (c) Mean (d) Variance 12. The value which occurs with the greatest frequency is called the (a) Variance (b) Mean (c) Median (d) Mode 13. A measure of relative dispersion is given by the (a) coefficient of variation (b) Mean deviation (c) variance (d) standard deviation 14. The range of a sample gives an indication of the (a) way in which the values cluster about a particular point (b) number of observations bearing the same value (c) maximum variation in the sample (d) degree to which the mean value differs from its expected value 15. Consider the bar graph representing number of packets for postage having a certain weight. Based on the bar graph: There are _____ packets were weighted. The frequency of the median weight is _____. The modal weight (in kg) is _____. (a) 65, 12, 10.1, respectively (b) 14, 8, 10.1, respectively (c) 65, 12, 10.5, respectively (d) 14, 8, 10.5, respectively 16. The mean of ten numbers is 58. If one of the numbers is 40, what is the mean of the other nine? _____ 17. The mean of 11 numbers is 7. One of the numbers, 13, is deleted. What is the mean of the remaining numbers? 18. Which of the following is not a measure of variability? (a) Median (b) Variance (c) Standard Deviation (d) Range 19. ____________ is the set of procedures used to explain or predict the values of a dependent variable based on the values of one or more independent variables. (a) Regression analysis (b) Regression coefficient (c) Regression equation (d) Regression line 20. Suppose we are predicting score on a training posttest from number of years of education and the score on an aptitude test given before training. Here is the regression equation Y 25 0.5 X 1 10 X 2 , where X 1 years of education and X 2 aptitude test score. What is the predicted score for someone with 10 years of education and a aptitude test score of 5? _____
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Chapter 6: Probability & Statistics
[6.28]
Probability Theory
Numerical study of chances of occurrence of events is dealt in probability theory. There are two approaches to probability: (i) Classical approach and (ii) Axiomatic approach. In both the approaches we use the term ‘experiment’, which means an operation which can produce some well-defined outcome(s). There are two types of experiments: Deterministic experiment: Experiments which when repeated under identical conditions produce the same results or outcomes are known as deterministic experiments. For e.g. when experiments in science are repeated under identical conditions, we get almost the same result every time. Random experiment: If an experiment, when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes then such an experiment is known as a probabilistic experiment or a random experiment. In a random experiment, all the outcomes are known in advance but the exact outcome is unpredictable. For e.g. in tossing of a coin, it is known that either a head or a tail will occur but one is not sure if a head or a tail will be obtained. So, it is a random experiment.
6.3.1 Basic Terminology in concept of Probability in Set Theoretic Language
Sample space and Sample points: The set of all possible outcomes of a trial (random experiment) is called its sample space. It is generally denoted by S and each outcome of the trial is said to be a sample point. For e.g. (i) If a dice is thrown once, then its sample space is S {1, 2, 3, 4, 5, 6} . (ii) If two coins are tossed together then its sample space is S {HT , TH , HH , TT } . Event: An event is a subset of a sample space. There are seven types of events: Simple event: An event containing only a single sample point is called an elementary or simple event. For e.g. in a single toss of coin, the event of getting a head is a simple event. Here S {H , T } and E {H } . Compound events: Events obtained by combining together two or more elementary events are known as the compound events or decomposable events. For e.g. in a single throw of a pair of dice the event of getting a doublet, is a compound event because this event occurs if any one of the elementary events (1,1), (2, 2), (3, 3), (4, 4), (5,5), (6, 6) occurs. Equally likely events: Events are equally likely if there is no reason for an event to occur in preference to any other event. For e.g., if an unbiased die is rolled, then each outcome is equally likely to happen, i.e. all elementary events are equally likely. Mutually exclusive or disjoint events: Events are said to be mutually exclusive or disjoint if the occurrence of any one of them prevents the occurrence of all the others. For e.g., E getting an even number, F getting an odd number, these two events are mutually exclusive, because, if E occurs we say that the number obtained is even and so it cannot be odd i.e., F does not occur. A1 and A2 are mutually exclusive events if A1 A2 . Mutually non-exclusive events: The events which are not mutually exclusive are known as compatible events or mutually non – exclusive events. Independent events: Events are said to be independent if the happening (or non-happening) of one event is not affected by the happening (or non-happening) of others. For e.g. if two dice are thrown together, then getting an even number on first is independent to getting an odd number on the second. Dependent events: Two or more events are said to be dependent if the happening of one event affects (partially or totally) other event. For e.g. suppose a bag contains 5 white and 4 black balls. Two balls are drawn one by one. Then two events that the first ball is white and second ball is black are independent if the first ball is replaced before drawing the second ball. If the first ball is not replaced then these two events will be dependent because second draw will have only 8 exhaustive cases. Exhaustive number of cases: The total number of possible outcomes of a random experiment in a trial is known as the exhaustive number of cases. For e.g., in throwing a die the exhaustive number of cases is 6, as any one of the six faces marked with 1, 2, 3, 4, 5, 6 may come uppermost.
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Favourable number of cases: The number of cases favourable to an event in a trial is the total number of elementary events such that the occurrence of any one of them ensures the happening of the event. For e.g., in drawing two cards from a pack of 52 cards, the number of cases favourable to drawing 2 queens is 4 C2 .
Mutually exclusive and exhaustive system of events: Let S be the sample space associated with a random experiment. Let E1 , E2 , , En are elementary events associated with a random experiment such that (i) Ei E j for i j ; (ii) E1 E2 En S . Then the collection of events E1 , E2 , , En is said to form a mutually exclusive and exhaustive system of events. Hence, the collection of elementary events associated with a random experiment always form a system of mutually exclusive and exhaustive system of events. In this system, P ( E1 E2 En ) P ( E1 ) P ( E2 ) P ( En ) 1 .
Independent events are always taken from different experiments, while mutually exclusive events are taken from a single experiment. Independent events can happen together while mutually exclusive events cannot happen together. Independent events are connected by the word “and” but mutually exclusive events are connected by the word “or”.
6.3.2 Definition of Probability Classical definition of Probability with discrete sample space: Let S be the sample, then the probability of occurrence of an event E is denoted by P ( E ) and is defined as, P( E )
n( E )
number of elements in E
number of cases favourable to event E
(6.9) n( S ) number of elements in S total number of cases In other words, if a random experiment results in n mutually exclusive, equally likely and exhaustive outcomes, out of which m are favourable to the occurrence of an event A , then the probability of m Number of outcomes favourable to A occurrence of A is given by P ( A) . It is obvious that n Number of total outcomes 0 m n . If an event A is certain to happen, then m n , thus P ( A) 1 . If A is impossible to happen, then m 0 and so P( A) 0 . Hence we conclude that 0 P ( A) 1 . Further, if A denotes negative of A i.e., event that A does not happen, then for above cases m , n ; we have, (6.10) P ( A ) ( n m ) n 1 ( m n ) 1 P ( A) P ( A) P ( A ) 1 For two events A and B , the following notations are used: c A or A or A stands for the non-occurrence or negation of A A B stands for the occurrence of at least one of A and B A B stands for the simultaneous occurrence of A and B c c A B stands for the non-occurrence of both A and B A B stands for ‘the occurrence of A implies occurrence of B ’
Some important remarks about Coins, Dice, Playing cards and Envelopes
Coins: A coin has a head side and a tail side. If an experiment consists of more than a coin, then coins are considered to be distinct if not otherwise stated. Number of exhaustive cases of tossing n coins simultaneously (or of tossing a coin n times) 2 n . Dice: A die (cubical) has six faces marked 1, 2, 3, 4, 5, 6 . We may have tetrahedral (having four faces 1, 2, 3, 4 ) or pentagonal (having five faces 1, 2, 3, 4, 5 ) die. As in the case of coins, if we have more than one die, then all dice are considered to be distinct if not otherwise stated. Number of exhaustive cases of throwing n dice simultaneously (or throwing one dice n times) 6 n . Playing cards: A pack of playing cards usually has 52 cards. There are 4 suits (Spade, Heart, Diamond and Club) each having 13 cards. There are two colours red (Heart and Diamond) and black (Spade and Club) each having 26 cards. In thirteen cards of each suit, there are 3 face cards
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or coart cards namely king, queen and jack. So there are in all 12 face cards (4 kings, 4 queens and 4 jacks). Also there are 16 honour cards, 4 of each suit namely ace, king, queen and jack. Probability regarding n letters and their envelopes: If n letters corresponding to n envelopes are placed in the envelopes at random, then Probability that all letters are in right envelopes 1 n ! . Probability that all letters are not in right envelopes 1 1 n ! . 1 1 1 1 Probability that no letter is in right envelopes (1) n . 2! 3! 4! n! Probability that exactly r letters are in right 1 1 1 1 1 n r . ( 1) r ! 2! 3! 4! ( n r )!
envelopes
Theorems on Probability:
When events are not mutually exclusive: If A and B are two events which are not mutually exclusive, then P ( A B ) P ( A) P ( B ) P ( A B ) . P ( A B ) P ( A) P ( B ) [This point was asked in CS-1994 (1 mark)] For any three events A, B, C P ( A B C ) P( A) P ( B ) P (C ) P ( A B ) P ( B C ) P (C A) P ( A B C ) When events are mutually exclusive: If A and B are mutually exclusive events, then n( A B ) 0 P( A B ) 0 . Hence P ( A B) P ( A) P( B ) . For any three events A, B, C which are mutually exclusive, P ( A B ) P ( B C ) P (C A) P( A B C ) 0 P( A B C ) P ( A) P( B ) P (C ) . The probability of happening of any one of several mutually exclusive events is equal to the sum of their probabilities, i.e., if A1 , A2 An are mutually exclusive events, then P ( A1 A2 An ) P ( A1 ) P( A2 ) P ( An ) i.e., P Ai P ( Ai ) .
When events are independent: If A and B are independent events, then P ( A B) P ( A) P ( B ) P ( A B ) P ( A) P ( B ) P( A) P ( B ) When events occur with random experiment: Let A and B be two events associated with a random experiment, then P ( A B ) P ( B) P ( A B) ; and P ( A B ) P ( A) P ( A B) If B A , then P ( A B ) P ( A) P ( B) ; and P ( B ) P ( A) Similarly if A B , then ( A B ) P ( B ) P ( A) ; and P ( A) P ( B )
Probability of occurrence of neither A nor B is P ( A B ) P ( A B ) 1 P ( A B ) . Generalization of the addition theorem: If A1 , A2 , , An are n events associated with a random experiment , then P
n i 1
n
n
n
Ai i 1 P ( Ai ) i , j 1, i j P ( Ai A j ) i , j , k 1, i j k P ( Ai A j Ak )
( 1) n 1 P ( A1 A2 An ) .
If all the events Ai (i 1, 2, , n) are mutually exclusive, then P
A n
i 1
i
n i 1
P ( Ai ) , i.e.,
P ( A1 A2 An ) P ( A1 ) P ( A2 ) P( An ) .
Booley’s inequality: If A1 , A2 , , An are n events associated with a random experiment, then
P
A n
i 1
i
n i 1
P ( Ai ) ( n 1)
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P
A n
i 1
i
n i 1
P( Ai )
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Important Points: Let A , B and C are three arbitrary events. Then Verbal Description of Event Only A occurs Both A and B, but not C occur All the three events occur At least one occurs At least two occur
Equivalent Set Theoretic Notation A B C A BC A BC A BC ( A B) ( B C ) ( A C )
One and no more occurs
( A B C ) ( A B C ) ( A B C)
Exactly two of A, B and C occur
( A B C ) ( A B C) ( A B C)
None occurs
Not more than two occur
A B C A B C ( A B) ( B C ) ( A C ) ( A B C )
Exactly one of A and B occurs
( A B ) ( A B)
Example 6.57 [CS-1994 (1 mark)]: Let A , B and C be independent events which occur with probabilities 0.8, 0.5 and 0.3 respectively. The probability of occurrence of at least one of the event is ……….. Solution: The probability of occurrence of at least one of the event is P ( A B C ) P( A) P ( B ) P (C ) P ( A B ) P ( B C ) P ( A C ) P ( A B C ) . As all the event are independent so P ( A B C ) P( A) P ( B ) P (C ) (0.8)(0.5)(0.3) 0.12 and P ( A B ) P ( A) P ( B) (0.8)(0.5) 0.4 , similarly P( B C ) 0.15 and P ( A C ) 0.21 . Thus P ( A B C ) 0.8 0.5 0.3 0.4 0.15 0.21 0.12 0.96 . Example 6.58 [CS-1994 (1 mark)]: The probability of an event B is P1 . The probability that events A and B occur together is P2 while the probability that A and B occur together is P3 . The
probability of the event A in terms of P1 , P2 and P3 is ………. Solution: From the given data we have, P ( B) P1 , P ( A B ) P2 , P ( A B ) P3 . As for a random experiment, P ( A B ) P ( A) P ( A B ) P3 P( A) P2 P ( A) P2 P3 . Example 6.59 [CS-1995 (1 mark)]: The probability that a number selected at random between 100 and 999 (both inclusive) will not contain the digit 7 is (a) 16 25 (c) 27 75 (d) 18 25 (b) (9 10)3 Solution (d): As there are 900 numbers between 100 and 999. Now there are three digits, lets us find the number of digits with no 7; the first digit can be chosen from 8 digits, either 1 2 3 4 5 6 8 or 9 (we cannot choose 0), the second digit can be chosen from 9 digits (0 to 9 excluding 7), the third digit can be chosen from any of the 9 digits (excluding 7) so the total number of possibilities or numbers is 8 9 9 648 . So probability of selecting at random a number with no digit 7 is 648 900 18 25 . Example 6.60 [CS-1996 (2 marks)]: The probability that top and bottom cards of a randomly shuffled deck are both aces is (a) (4 52) (4 52) (b) (4 52) (3 52) (c) (4 52) (3 51) (d) (4 52) (4 51) Solution: Probability that top card of a randomly shuffled deck is an ace is 4 52 . Now we have 51 cards and 3 aces so the probability that bottom card is an ace is 3 51 . As both events are independent so probability that top and bottom cards of a randomly shuffled deck are both aces is (4 52) (3 51) . Example 6.61 [CS-1996 (1 mark)]: Two dice are thrown simultaneously. The probability that at least one of them will have 6 facing up is
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(c) 25 36
(d) 11 36
Solution (d): Two dice are thrown simultaneously so total number of sample space is n( S ) 62 36 . Now let E be the event of getting at least one of them will have 6 facing up so the elements of E :{(1, 6), (2, 6), , (5, 6), (6,1), (6, 2), (6, 6)} so n( E ) 11 and thus P ( E ) n( E ) n( S ) 11 36 . Example 6.62 [CS-1997 (1 mark)]: The probability that it will rain today is 0.5. The probability that it will rain tomorrow is 0.6. The probability that it will rain either today or tomorrow is 0.7. What is the probability that it will rain today and tomorrow? (a) 0.3 (b) 0.25 (c) 0.35 (d) 0.40 Solution (d): Let A be the event that it will rain today, and B be the event that it will rain tomorrow. So from the given data we have P ( A) 0.5 , P ( B) 0.6 and P( A B ) 0.7 . We have to find P ( A B ) P ( A) P( B ) P ( A B ) 0.5 0.6 0.7 0.4 . Example 6.63 [CS-1998 (1 mark)]: A die is rolled three times. The probability that exactly one odd number turns up among the three outcomes is (a) 1/6 (b) 3/8 (c) 1/8 (d) ½ Solution (b): Let O be the event of outcome of an odd number; and E be the event of outcome of an even number. When a die is rolled the outcome is either 3 odd numbers or 3 even numbers; so P (O) P ( E ) 3 6 1 2 . Now in three tries exactly one odd number turns up among the three outcomes as: {OEE , EOE , EEO} so probability that exactly one odd number turns up among the three outcomes is P (O ) P ( E ) P ( E ) P ( E ) P (O) P ( E ) P ( E ) P ( E ) P (O )
111 222
111 222
111 222
3 8
.
Example 6.64 [ME-1998 (1 mark)]: The probability that two friends share the same birth-month is (a) 1 6 (b) 1 12 (c) 1 144 (d) 1 24 Solution (b): Any person can born in any of the 12 months of the year, so total number of possible combinations is 12 12 . This set contains 12 pairs of identical months, for e.g., (January, January), which are our favourable cases. Thus the required probability is 12 (12 12) 1 12 . Example 6.65 [CS-2000 (1 mark)]: The minimum number of cards to be dealt from an arbitrarily shuffled deck of 52 cards to guarantee that three cards are from same suit is (a) 3 (b) 8 (c) 9 (d) 12 Solution (c): As there are 4 suites in a deck of cards (usually called Spades, Clubs, Hearts and Diamonds). Each suite has 13 cards in it. We want a guarantee that three cards from the same suite come to us, we shall assume the worst case and ensure 3 cards of same suite occur even in that case. The worst case here is that, as we pick cards, each card is of a different type. In that case, with the first 4 picks, we would have taken one card of each suite. With the next 4 cards, we would have 2 cards of each suite. Now, if we pick another card, whatever suite it may be, we have 2 other cards of that suite already. So, we now have 3 cards of the same suite. So, from the above, it is clear that to guarantee that three cards are from some same suite, we have to pick 9 cards from the deck. Example 6.66 [CS-2001 (2 marks)]: Seven (distinct) car accidents occurred in a week. What is the probability that they all occurred on the same day? (a) 1 7 7 (b) 1 7 6 (c) 1 2 7 (d) 7 2 7 Solution (b): Probability of 1 accident in a 7 days is 1 7 . So probability of 7 accidents in 7 days considering 1 accident in a day is (1 7)(1 7) (7 times) 1 77 . Now consider all 7 accidents in a day so its probability is 7(1 77 ) 1 76 . Example 6.67 [ME-2001 (2 marks)]: An unbiased coin is tossed three times. The probability that the head turns up in exactly two cases is
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(b) 1 8
(c) 2 3
[6.33]
(d) 3 8 3
Solution (d): As unbiased coin is tossed three times, so we have 2 8 number of sample space. Out of 8 events, we have exactly two heads as {HHT , HTH , THH } . So the required probability is 3 8 . Example 6.68 [CS-2002 (2 marks)]: Four fair coins are tossed simultaneously. The probability that at least one head and one tail turn up is (a) 1 16 (b) 1 8 (c) 7 8 (d) 15 16 Solution (c): As when four coins are tossed the total number of cases will be 24 16 . Now among 16 possibilities the condition at least one head and one tail is not met by two possibilities, i.e., {HHHH , TTTT } ; so total number of favourable cases for at least one head and one tail will be
16 2 14 . Thus the required probability will be 14 16 7 8 . Example 6.69 [CE-2003 (1 mark)]: A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. Probability that none of the two screws is defective will be (a) 100% (b) 50% (c) 49% (d) None of these Solution (c): Let A and B be the events of drawing first and second, respectively, non-defective screws. So for event A , P ( A) 7 10 ; also we are sampling with replacement so the situation before the second drawing is same as at the beginning and thus P ( B ) 7 10 . As both the events are independent so P ( A B ) P( A) P ( B ) 49 100 , which is the required probability. Example 6.70 [ME-2003 (2 marks)]: A box contains 5 black and 5 red balls. Two balls are randomly picked one after another from the box, without replacement. The probability for both balls being red is (a) 1 90 (b) 1 5 (c) 19 90 (d) 2 9 Solution (d): The probability of drawing the first red ball is 5 10 . As our case is without replacement, so we now have 5 black and 4 red ball; thus the probability of drawing the 2nd red ball is 4 9 . So the required probability is (5 10)(4 9) 2 9 . [Similar question was also asked in ME-2006 (1 mark)] Example 6.71 [ME-2004 (2 marks)]: From a pack of regular playing cards, two cards are drawn at random. What is the probability that both cards will be Kings, if the first card is NOT replaced? (a) 1 26 (b) 1 52 (c) 1 169 (d) 1 221 Solution (d): As in pack of 52 cars we have 4 kings. So the probability of drawing first king is 4 52 . As our case is without replacement, so we now have 3 kings and total 51 cards; so the probability of drawing the second king is 3 51 . Thus the required probability is (4 52)(3 51) 1 221 . Example 6.72 [EC-2005 (1 mark)]: A fair dice is rolled twice. The probability that an odd number will follow an even number is (a) 1 2 (b) 1 6 (c) 1 3 (d) 1 4 Solution (d): A dice is rolled two times so our sample space is 62 36 . Let E be the event for which an odd number will follow an even number; so the elements of E are {(2,1), (2, 3), (2, 5), (4,1),
(4, 3), (4, 5), (6,1), (6,3), (6,5)} n( E ) 9 ; thus the required probability is 9 36 1 4 . Example 6.73 [EE-2005 (2 marks)]: A fair coin is tossed three times in succession. If the first toss produces a head, then the probability of getting exactly two heads in three tosses is: (a) 1 8 (b) 1 2 (c) 3 8 (d) 3 4 Solution (b): It is given that on tossing a coin for the first time the outcome is head. Now the question: getting exactly two head in three tosses, is same as getting exactly one head in two tosses.
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So our sample space is S :{HH , HT , TH , TT } n( S ) 4 and if E is the event of getting exactly one head in two tosses then E :{HT , TH } n( E ) 2 and thus P ( E ) 2 4 1 2 which is the probability of getting exactly two heads in three tosses given that first toss is head. Example 6.74 [ME-2005 (2 marks)]: A single die is thrown twice. What is the probability that the sum is neither 8 nor 9? (a) 1 9 (b) 5 36 (c) 1 4 (d) 3 4 Solution: The total number of sample space when a die is thrown twice is 62 36 . Let E be the event of getting the sum of numbers on the dice is 8 or 9, then the elements of E are {(2, 6), (3,5), (4, 4), (5,3), (6, 2), (3, 6), (4,5), (5, 4), (6,3)} n( E ) 9 . So the probability of getting 8 and 9 is 9 36 1 4 . Thus the probability that the sum is neither 8 nor 9 is 1 1 4 3 4 . Example 6.75 [CE-2006 (2 marks)]: There are 25 calculators in a box. Two of them are defective. Suppose 5 calculators are randomly picked for inspection (i.e., each has the same chance of being selected), what is the probability that only one of the defective calculators will be included in the inspection? (a) 1 2 (b) 1 3 (c) 1 4 (d) 1 5 Solution: As total number of ways for picking 5 out of 25 calculators is out of 23 non-defective calculator, which can be done in
23
25
C 2 . Now we have to pick 4
C4 ; also we have to pick 1 out of 2
2
defective calculators which can be done in C1 ; so total number of ways in which 5 calculators (4 non-defective and 1 defective) can be picked is
23
C4 2 C1 ways. Thus required probability is
( 23 C4 2 C1 ) ( 25 C5 ) 1 3 . [Similar question was also asked in CS-1995 (2 marks)]
Example 6.76 [CS-2007 (2 marks)]: Suppose we uniformly and randomly select a permutation from the 20! permutations of 1, 2, 3, …, 20. What is the probability that 2 appears at an earlier position than any other even number in the selected permutation? (c) (9!) (20!) (a) 1 2 (b) 1 10 (d) None of these Solution (b): Number of permutation with ‘2’ in the 1st position is 19! 10 P0 (19!) (fill the first place with ‘2’ and 19 places after ‘2’ with 19 remaining numbers). Number of permutation with ‘2’ in the 2nd position is 10 18! 10 P1 (18!) (fill the first place with any of the 10 odd numbers and the 18 places after the 2 with 18 of the remaining numbers in 18! ways). Similarly, number of permutation with ‘2’ in 3rd position is 10 9 17! 10 P2 (17!) (fill the first 2 places with 2 of the 10 odd numbers and then the remaining 17 places with remaining 17 numbers in 17!). We repeat this process until ‘2’ is in the 11th place. After that it is not possible to satisfy the given condition as there are only 10 odd numbers available to fill before the ‘2’. So the desired number of permutation which satisfy the given condition is 10 P0 (19!) 10 P1 (18!) 10 P9 (10!) 10 P10 (9!) . As total number of ways in which these 20 numbers are arrange is 10
20
P20 20! . So the probability of happening of the given condition is
P0 (19!) 10 P1 (18!) 10 P9 (10!) 10 P10 (9!)
1
. Alternate way: The odd numbers do not matter P20 10 here. The probability that 2 comes before the other 9 evens is (number of ways to pick 2)(number of ways to pick remaining evens) (1)(9!) 1 . (number of ways to order 10 evens) 10! 10 20
Example 6.77 [PI-2007 (1 mark)]: Two cards are drawn at random in succession, with replacement, from a deck of 52 well shuffled cards. Probability of getting both ‘Aces’ is
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(a) 1/169 (b) 2/169 (c) 1/13 (d) 2/13 Solution (a): As there are four aces in a deck of 52 cards and we have the case of replacement. So the probability of getting both aces is (4 52) (4 52) 1 169 . Example 6.78 [CE-2008 (2 marks)]: A person on a trip has a choice between private car and public transport. The probability of using a private car is 0.45. While using the public transport, further choices available are bus and metro, out of which the probability of commuting by a bus is 0.55. In such a situation, the probability (rounded up to two decimals) of using a car, bus and metro, respectively would be (a) 0.45, 0.30 and 0.25 (b) 0.45, 0.25 and 0.30 (c) 0.45, 0.55 and 0.00 (d) 0.45, 0.35 and 0.20 Solution (a): Let C be car; B be bus; and M be metro. It is given that P (C ) {P( B ) P ( M )} 1 , P (C ) 0.45 P ( B ) P ( M ) 1 P (C ) 0.55 . Now while using the public transport the probability of commuting by a bus is 0.55, which means that P ( B ) 0.55 {P ( B ) P ( M )} 0.55 0.55 0.30 . So P ( M ) 0.55 P( B) 0.25 . Example 6.79 [CS-2008 (2 marks)]: Aishwarya studies either Computer Science or Mathematics everyday. If she studies Computer Science on a day, then the probability that the studies Mathematics the next day is 0.6. If she studies Mathematics on a day, then the probability that the studies Computer Science the next day is 0.4. Given that Aishwarya studies Computer Science on Monday, what is the probability that she studies Computer Science on Wednesday? (a) 0.24 (b) 0.36 (c) 0.40 (d) 0.60 Solution (c): Aishwarya studies computer science on Monday, so the probability that she studies mathematics on Tuesday is 0.6 and the probability that she studies computer science on Tuesday is 1 0.6 0.4 . The probability that she studies mathematics on Tuesday and computer science on Wednesday is 0.6 0.4 0.24 . Now the probability that she studies computer science on Tuesday and computer science on Wednesday is 0.4 0.4 0.16 . So the required probability that she studies computer science on Wednesday is 0.24 0.16 0.40 . Example 6.80 [ME-2008, TF-2009 (1 mark)]: A coin is tossed 4 times. What is the probability of getting heads exactly 3 times? (a) 1 4 (b) 3 8 (c) 1 2 (d) 3 4 Solution (a): Total number of sample space when a coin is tossed 4 times is 24 16 . Let E be the event of getting heads exactly 3 times. So the elements of E are {HHHT , HHTH , HTHH , THHH } n( E ) 4 . So the required probability is 4 16 1 4 . Example 6.81 [PI-2008 (2 marks)]: In a game, two players X and Y toss a coin alternatively. Whosoever gets a ‘head’ first, wins the game and the game is terminated. Assuming that player X starts the game, the probability of player X winning the game is (a) 1/3 (b) 1/2 (c) 2/3 (d) ¾ Solution (c): Case 1: X starts the game and win, so P1 1 2 . Case 2: X starts the game and loose, then Y tosses and coin shows tail, then X tosses and coin shows head thus wins, so P2 (1 2)(1 2)(1 2) 1 23 . Case 3: X starts the game and loose, then Y tosses and coin shows tail, then X tosses and coin shows tail, then Y tosses and coin shows tail, then X tosses and coin shows head thus wins, so P3 (1 2)(1 2)(1 2)(1 2)(1 2) 1 25 . And so one …. So the required probability is P P1 P2 P3 1 2 1 2 3 1 25 (1 2) {1 1 2 2 } 2 3 .
Example 6.82 [TF-2008 (2 marks)]: Two dices are thrown simultaneously. The probability that the total number of dots is equal to 4 is (a) 1 6 (b) 1 12 (c) 1 18 (d) 1 36
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Solution (b): When two dice are thrown then we have 62 36 number of sample space. Let E be the event of getting sum of the numbers on two dice as 4. So the elements of E are {(1, 3), (2, 2), (3,1)} n( E ) 3 . Thus P ( E ) 3 36 1 12 . [Similar questions were also asked in ME-2002, TF-2014 (1 mark)] Example 6.83 [CS-2009 (2 marks)]: An unbalanced dice (with 6 faces, numbered from 1 to 6) is thrown. The probability that the face value is odd is 90% of the probability that the face value is even. The probability of getting any even numbered face is the same. If the probability that the face is even given that it is greater than 3 is 0.75, which one of the following options is closest to the probability that the face value exceeds 3? (a) 0.453 (b) 0.468 (c) 0.485 (d) 0.492 Solution (b): Let E and O are the events of getting an even and odd numbers, respectively. So we have P ( E ) x , P (O ) 90% of x 0.9 x ; but P ( E ) P (O ) 1 x 0.9 x 1 x 10 19 . As we have three even numbers and probability of getting an even number is same. So P (2) P (4) P(6) (10 19)(1 3) 10 57 . Now P ( E 3 and number 3) P ( E 3) P(number 3) 0.75 {P(4) P (6)} P (number 3)
P(number 3) 0.75 (10 57 10 57) 0.75 (20 57) 0.468 . Example 6.84 [EC-2009 (1 mark)]: A fair coin is tossed 10 times. What is the probability that ONLY the first two tosses will yield heads? (b) 10 C2 (1 2) 2 (d) 10 C2 (1 2)10 (a) (1 2) 2 (c) (1 2)10 Solution (c): As a fair coin is tossed 10 times so total number of sample space is n( S ) 210 . Let E be the event of getting first two tosses as ‘Heads’ and other eight as ‘Tail’; this will yields as {H , H , T , T , T , T , T , T , T , T } so n( E ) 1 . Thus P ( E ) 1 210 . Example 6.85 [EE-2009 (2 marks)]: Assume for simplicity that N people, all born in April (a month of 30 days), are collected in a room. Consider the event of at least two people in the room being on the same date of the month, even if in different years, e.g. 1980 and 1985. What is the smallest N so that the probability of this event exceeds 0.5? (a) 20 (b) 7 (c) 15 (d) 16 Solution (b): We know that in there are 30 days in April month, so our sample space is 30 N (since for the all N peoples, all have 30 days in options to be born in April). Now P (at least two share the same birth day) 1 P(none of the person share the same birth day) . If we wish that no two persons have the same birth day, then we immediately know that N 30 . Now the idea is to distribute the 30 days of April month to these N people, and so we get 30 PN ways of doing it, which ensures that no two person have same birth day. As it is given that
P (at least two share the same birth day) 0.5 , so we have 1 30 PN 30 N 0.5
30
PN 30 N 0.5
30! {(30 N )!30 N } 0.5 . This condition is met by all the given options so we will choose option (b) which is the smallest from all given options. Also note that for N 7 , we have 30! {(30 N )!30 N } 0.469 ; and for N 6 , we have 30! {(30 N )!30 N } 0.586 . So option (b) is correct.
Example 6.86 [TF-2009 (1 mark)]: Probability of getting 16 in one throw with 3 dice is (a) 1 8 (b) 3 16 (c) 1 36 (d) 1 108 Solution: When three dice are thrown then we have 63 216 number of sample space. Let E be the event of getting sum of the numbers on three dice as 16. So the elements of E are {(4, 6, 6), (6, 4, 6), (6, 6, 4), (5,5, 6), (5, 6,5), (6, 5,5)} n( E ) 6 . Thus P ( E ) 6 216 1 36 .
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Example 6.87 [CE-2010 (1 mark)]: Two coins are simultaneously tossed. The probability of two heads simultaneously appearing is (a) 1 8 (b) 1 6 (c) 1 4 (d) 1 2 Solution: Our sample space is S :{HH , HT , TH , TT } , so n( S ) 4 ; if E be the even of occurring two heads then n( E ) 1 . Thus P ( E ) n( E ) n( S ) 1 4 . Example 6.88 [CS-2010 (2 marks)]: What is the probability that divisor of 1099 is a multiple of 1096 ? (a) 1 625 (b) 4 625 (c) 12 625 (d) 16 625 Solution (a): Divisors of 1099 are of the form 2a 5b , where a and b vary from 0 to 99 each, so there are 10000 divisors of 1099 . Now any of those divisors will be a multiple of 1096 if both a and b are at least 96, i.e., 96 or 97 or 98 or 99. So both a and b have 4 choices each, and so there are 4 4 16 divisors which are multiple of 1096 . So the required probability is 16 10000 1 625 . Example 6.89 [CH-2010 (2 marks)]: A box contains three red and two black balls. Four balls are removed from the box one by one without replacement. The probability of the ball remaining in the box being red, is (a) 609 625 (b) 3 5 (c) 2 5 (d) 81 625 Solution (b): Let E be the event for red ball remaining; and S be the event for only 1 ball remaining. So number of ways in which event E occurs is when 2 red balls and 2 black balls selected, i.e., n ( E ) 3C2 2 C2 ; also number of ways in which event S occurs is when 4 out of 5 balls selected, i.e., n ( S ) 5C4 , which is our sample space. So the required probability is P ( E ) {n( E )} {n( S )}
3
C2 2 C2
C 3 5. 5
4
Example 6.90 [ME-2010 (2 marks)]: A box contains 2 washers, 3 nuts and 4 bolts. Items are drawn from the box at random one at a time without replacement. The probability of drawing 2 washers first followed by 3 nuts and subsequently the 4 bolts is (a) 2 315 (b) 1 630 (c) 1 1260 (d) 1 2520 Solution (c): Probability of drawing 2 washers from 9 objects is drawing 3 nuts out of remaining 7 objects is of remaining 4 objects is
C C 1 36 . Probability of 2
9
2
2
C C 1 35 . Probability of drawing 4 bolts out 3
7
3
3
C C 1 . So required probability is (1 36) (1 35) 1 1 1260 . 4
4
4
4
Example 6.91 [MN-2010 (2 marks)]: The probabilities of hitting a target by A and B are 1/3 and 2/5, respectively. A shoots at the target once, followed by B shooting at the target once. The probability of hitting the target is (a) 2 15 (b) 5 15 (c) 8 15 (d) 9 15 Solution (a): Let E , F are the events of hitting a target by A and B , respectively. As both the events are independent so P( E and F ) P ( E F ) (1 3)(2 5) 2 15 . Example 6.92 [CE-2011 (1 mark)]: There are two containers, with one containing 4 Red and 3 Green balls and the other containing 3 Blue and 4 Green balls. One ball is drawn at random from each container. The probability that one of the balls is Red and the other is Blue will be (a) 1 7 (b) 9 49 (c) 12 49 (d) 3 7 Solution (c): Let R and B be the events of drawing red and blue balls, respectively. So we have to pick red ball from first container so P ( R ) 4 7 ; and blue ball from second container so P ( B ) 3 7 . As both the events are independent so P ( R and B ) P ( R B ) P ( R ) P( B ) (3 7)(4 7) 21 49 .
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Example 6.93 [CS-2011 (1 mark)]: If two fair coins are flipped and at least one of the outcomes is known to be a head, what is the probability that both outcomes are head? (a) 1 3 (b) 1 4 (c) 1 2 (d) 2 3 Solution (a): As our sample space is {HH , HT , TH } ; so the probability that both outcomes are head is 1 3 . Example 6.94 [CS-2011 (2 marks)]: A deck of 5 cards (each carrying a distinct number from 1 to 5) is shuffled thoroughly. Two cards are then removed one at a time from the deck. What is the probability that the two cards are selected with the number on the first card being one higher than the number on the second card? (a) 1 5 (b) 4 25 (c) 1 4 (d) 2 5 Solution (a): Total number of ways in which 2 from 5 cards can be selected is 5 P2 20 ways. Now let E be the event of selecting two cards s.t. number on the 1st card being one higher than number on the 2nd card, then E :{(2,1), (3, 2), (4,3), (5, 4)} ; so P ( E ) 4 20 1 5 be the required probability. Example 6.95 [EC-2011 (1 mark)]: A fair dice is tossed two times. The probability that the second toss results in a value that is higher than the first toss is (a) 2 36 (b) 2 6 (c) 5 12 (d) 1 2 Solution (c): As a dice is rolled two times so our sample space is 62 36 . Let E be the event for which the second toss results in a value that is higher than the first toss; so the elements of E are {(2,1), (3,1), (3, 2), (4,1), (4, 2), (4,3), (5,1), , (5, 4), (6,1), , (6,5)} , where the first number corresponds to the second toss and second number corresponds to the first toss. So n( E ) 1 2 4 4 5 15 . Thus the required probability is P ( E ) n( E ) n( S ) 15 36 5 12 . Example 6.96 [IN-2011 (2 marks)]: The box 1 contains chips numbered 3, 6, 9, 12, and 15. The box 2 contains chips numbered 6, 11, 16, 21 and 26. Two chips, one from each box are drawn at random. The numbers written on these chips are multiplied. The probability for the product to be an even number is (a) 6 25 (b) 2 5 (c) 3 5 (d) 19 25 Solution (d): Let x be the product of the numbers from box 1 and 2. For x to be even, the numbers from both the boxes should not turn out to be odd simultaneously. So the required probability is P ( x is even) 1 P ( x is odd) 1 (3 5) (2 5) 19 25 . Example 6.97 [ME-2011 (2 marks)]: An unbiased coin is tossed five times. The outcome of each toss is either a head or a tail. The probability of getting at least one head is (a) 1 32 (b) 13 32 (c) 16 32 (d) 31 32 Solution (d): Total number of sample space when a coins is tossed 5 times is 25 32 . Let E be the event of getting at least 1 head; and E1 be the event of getting no head. So the elements of E1 is {TTTTT } n( E1 ) 1 . So the required probability is P ( E ) 1 P( E1 ) 1 1 32 31 32 . [Similar question was also asked in ME-2009 (1 mark)]
Example 6.98 [MT-2011 (2 marks)]: A box contains 5 white balls and 3 red balls. Two balls are withdrawn from the box randomly, one after another (without replacement). The probability that the two balls withdrawn are of different colour is (a) 15/64 (b) 25/64 (c) 25/56 (d) 30/56 Solution (d): As 2 out of 8 balls are selected in 8 C 2 ways. Let E be the event of selecting 1 out of 5 white balls and 1 out of 3 red balls. So number of ways in which the event E occurs is n ( E ) 5C1 3C1 . Thus the required probability is P ( E )
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5
C1 3C1
C 15 28 30 56 . 8
2
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[6.39]
Example 6.99 [TF-2011 (1 mark)]: Probability of occurrence of two events E1 and E2 is 0.25 and 0.5, respectively. The probability of their simultaneous occurrence is 0.14. The probability that neither E1 nor E2 occurs is (a) 0.11 (b) 0.25 (c) 0.39 (d) 0.86 Solution (c): P ( E1 ) 0.25 and P ( E2 ) 0.5 , P ( E1 E2 ) 0.14 ; we have to find P ( E1c E2c ) 1 P ( A B ) 1 P ( A) P ( B ) P ( A B ) 1 0.25 0.5 0.14 0.39 .
Example 6.100 [TF-2011 (2 marks)]: A garment factory manufactures shirts. From the past history, it is known that 8 out of 100 collars and 5 out of 100 sleeves are defective. The probability that the assembled shirt will NOT have either of these defects is …… Solution: Let E and F are the events of defective collars and sleeves, respectively. So P ( E ) 8 100 0.08 and P ( F ) 5 100 0.05 . As both the events are independent so P ( E F ) P ( E ) P ( F ) (0.08)(0.05) 0.004 . Thus a shirt having collar or sleeve to be defective is P ( E F ) P( E ) P ( F ) P ( E F ) 0.08 0.05 (0.08)(0.05) 0.126 . Hence the probability that no shirt is defective is 1 P ( E F ) 1 0.126 0.874 . Example 6.101 [CS-2012 (2 marks)]: Suppose a fair six-sided die is rolled once. If the value on the die is 1, 2, or 3, the die is rolled a second time. What is the probability that the sum total of values that turn up is at least 6? (a) 10 21 (b) 5 12 (c) 2 3 (d) 1 6 Solution (b): The situation given in the problem can be drawn as in the following figure. When die is rolled then we have 6 outcomes. If the outcome is either 1 or 2 or 3 then the die will again rolled. So if in the first roll the outcome is 1 then in the second roll the outcome can be any one from 1 to 6; but the sum of the outcomes will be at least 6 if the outcome will be 5 or 6 in the second roll; which is shown in figure. Similarly, if in the first roll the outcome is 2 then in the second roll the outcome can be any one from 1 to 6; but the sum of the outcomes will be at least 6 if the outcome will be 4 or 5 or 6 in the second roll. And this can be done similarly for the outcome of 3 in the first roll. Now, for total number of sample space, the die is rolled 4 times so total number of sample space is 24. And let E be the event that the total sum of values that turn up is at least 6; then n( E ) 2 3 4 1 10 , where 2, 3, 4 corresponds for the outcome 1, 2, 3 in the first roll; and 1 corresponds to the outcome 6 in the first roll. So required probability will be 10 24 5 12 . Example 6.102 [EC-2012, EE-2012, IN-2012 (2 marks)]: A fair coin is tossed till a head appears for the first time. The probability that the number of required tosses is odd, is (a) 1 3 (b) 1 2 (c) 2 3 (d) 3 4 Solution (c): We have the following sequence of events: {H , TTH , TTTTH , TTTTTTH , } . As the probability of 1st event, i.e., appearance of head, is 1 2 ; probability of the 2nd event, i.e., appearance of head, is 1 23 ; probability of the 3rd event, i.e., appearance of head, is 1 25 ; and so on. So the total probability is (1 2) (1 23 ) (1 25 ) (1 2) {1 (1 22 )} (1 2) (3 4) 2 3 . Example 6.103 [ME-2012, PI-2012 (2 marks)]: A box contains 4 red balls and 6 black balls. Three balls are selected randomly from the box one after another without replacement. The probability that the selected set contains one red ball and two black balls is (a) 1 20 (b) 1 12 (c) 3 10 (d) 1 2 Solution (b): The selection of one red and two black balls balls is done as: BBR , BRB , RBB . So probability of the event BBR is (6 10) (5 9) (4 8) 120 720 1 6 . Similarly probability of BRB
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and RBB is 1 6 and 1 6 , respectively. So total probability of getting two black balls and one red ball is (1 6) (1 6) (1 6) (3 6) 1 2 . [Similar question was also asked in ME-1997 (2marks)] Example 6.104 [XE-2012 (1 mark)]: Suppose 50% of the population of a village like oranges, 70% of the population like apples, and 40% like both. If a person is picked at random who likes at least one of these fruits, what is the probability that the person likes oranges? (a) 1 8 (b) 5 12 (c) 1 2 (d) 5 8 Solution (d): Let O , A and B are the events of liking oranges only, apple only, and both oranges and apples, respectively, as shown in figure. So from the given data, we have O B 50% , B A 70% and B 40% , thus we have O 10% , B 40% , A 30% . So the probability of a person who likes at least oranges is the sum of probability of a person who like oranges and probability of a person who like both oranges and apples, i.e., the required probability is 10 (10 40 30) 40 (10 40 30) 50 80 5 8 . Example 6.105 [AG-2013 (2 marks)]: Box 1 contains 15 balls out of which 3 are red. Box 2 contains 12 balls out of which 4 are red. If one ball is drawn at random from each box simultaneously, the probability of getting at least one red ball is (a) 0.07 (b) 0.47 (c) 0.53 (d) 0.75 Solution (b): O1 , R1 are the events of drawing other colour ball and red ball from bag 1; and O2 , R2 are the events of drawing other colour ball and red ball from bag 2. As one ball is drawn at random from each box simultaneously. Let E be the event for getting at least one red ball; and F be the event for getting zero red ball, i.e., both balls are of other colour. So P ( E ) 1 P( F ) . As P (O1 ) 12 15 4 5 and P (O2 ) 8 12 2 3 . So P ( F ) P (O1 ) P (O2 ) 8 15 and thus the required
probability P ( E ) 1 8 15 7 15 0.47 . [Similar question was also asked in AG-2008 (2 marks)] Example 6.106 [CH-2013 (1 mark)]: For two rolls of a fair die, the probability of getting a 4 in the first roll and a number less than 4 in the second roll, up to 3 digits after the decimal point is ………… Solution: Total number of sample space when two dies are rolled is 62 36 . Let E be the event for getting a 4 in the first roll and a number less than 4 in the second roll; so elements of E are {(4,1), (4, 2), (4,3)} n( E ) 3 . Thus the required probability is P ( E ) 3 36 0.083 . Example 6.107 [MN-2013 (1 mark)]: Events A and B are independent but NOT mutually exclusive. If the probabilities P( A) and P( B) are 0.5 and 0.4 respectively, then P ( A B) is (a) 0.6 (b) 0.7 (c) 0.8 (d) 0.9 Solution (b): As the events A and B are independent so P ( A B) P ( A) P ( B ) . Thus P ( A B ) P( A) P( B ) P ( A B) 0.5 0.4 (0.5)(0.4) 0.7 . Example 6.108 [BT-2014 (1 mark)]: If an unbiased coin is tossed 10 times, the probability that all outcomes are same will be ……………. 105 10 Solution: Total number of sample space when an unbiased coin is tossed 10 times is 2 . Let E be the event of getting same outcome; then elements of E are {HHH (10 times), TTT (10 times)}
n ( E ) 2 . Thus P ( E ) 2 210 1.95 103 195 10 5 . So answer is 195. Example 6.109 [CE-2014 (1 mark)]: A fair (unbiased) coin was tossed four times in succession and resulted in the following outcomes: (i) Head, (ii) Head, (iii) Head, (iv) Head. The probability of obtaining a ‘Tail’ when the coin is tossed again is
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(a) 0 (b) 1 2 (c) 4 5 (d) 1 5 Solution (b): As the event of outcome of tossing a coin is an independent event. So when the coin is tossed again, it has no relation with previous outcomes. Thus the probability of obtaining a tail is 1 2 . Example 6.110 [CH-2014 (2 marks)]: In rolling of two fair dice, the outcome of an experiment is considered to be the sum of the numbers appearing on the dice. The probability is highest for the outcome of ……………. Solution: As the total sample space when two dice are rolled is 62 36 . Let Ei be the event of getting different sum ( i 2, 3, 4, ,12 ) of the numbers appearing on the dice. So the elements of Ei for all the possible sum is given as: E2 {(1,1)} P ( E2 ) 1 36 , E3 {(1, 2), (2,1)} P ( E3 ) 2 36 , E4 {(1, 3), (2, 2), (3,1)} P ( E4 ) 3 36 ,
E5 {(1, 4), (2, 3), (3, 2), (4,1)} P ( E5 ) 4 36 ,
E6 {(1, 5), (2, 4), (3,3), (4, 2), (5,1)} P ( E6 ) 5 36 , E7 {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6,1)} P ( E7 ) 6 36 , E8 {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)} P ( E8 ) 5 36 , E9 {(3, 6), (4,5), (5, 4), (6, 3)} P ( E9 ) 4 36 ,
E10 {(4, 6), (5,5), (6, 4)} P ( E10 ) 3 36 ,
E11 {(5, 6), (6,5)} P ( E11 ) 2 36 , E12 {(6, 6)} P ( E12 ) 1 36 . So we can say that the probability is maximum if sum of the numbers on the dice is 7. So answer is 7.
Example 6.111 [CS-2014 (2 marks)]: Four fair six-sided dice are rolled. The probability that the sum of the results being 22 is X 1296 . The value of X is ……………. Solution: Total number of sample space when four fair six-sided dice are rolled is 6 4 . As we get the sum of numbers appeared in the dice as 22 when we get the numbers on the dice as: (1) 6, 6, 6, 4 , which can be arranged in (4!) (3!) 4 ways; and (2) 6, 6, 5, 5 , which can be arranged in (4!) (2!2!) 6 ways. So total number of ways in which we get the sum of 22 when four dice are
rolled is 4 6 10 ways. Thus the required probability is 10 6 4 10 1296 . Thus X 10 . Example 6.112 [CS-2014 (2 marks)]: Let S be a sample space and two mutually exclusive events A and B be such that A B S . If P () denotes the probability of the event, the maximum value of P ( A) P( B ) is ……………. Solution: Given A B S P ( A B ) 1 P ( A) P ( B ) P ( A B ) 1 . As events A and B are mutually exclusive so P ( A B ) 0 . Thus, P ( A) P( B ) 1 P ( B) 1 P ( A) . Let P( A) x . So we have to find the maximum value of f ( x ) x(1 x ) x x 2 . Now f ( x) 1 2 x 0 x 1 2 and at x 1 2 , we have f ( x) 2 0 ; so x 1 2 is the point of maxima for f ( x ) . Thus maximum value of f ( x ) is (1 2) (1 2) 2 0.25 . Example 6.113 [CS-2014 (2 marks)]: The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ……………. Solution: Let A be the event when number [1,100] is divisible by 2; so Number of integers [1,100] that are divisible by 2 are {2, 4, 6,8, 100} n( A) 50 P( A) 0.5 . Similarly if B and C are the events when a numbers [1,100] is divisible by 3 and 5, respectively; then n( B) 33 P ( B) 0.33 and n(C ) 20 P (C ) 0.2 . Now if the numbers [1,100] are divisible by both 2 and 3, then it is also divisible by 2 3 6 and thus A B {6,12,18, 96} n( A B ) 16 P( A B) 0.16 . Similarly, n( B C ) 6 P ( B C ) 0.06 , n( A C ) 10 P( A C ) 0.1 and n( A B C ) 3 P ( A B C ) 0.03 . So P ( A B C ) P( A) P ( B ) P (C ) P ( A B ) P ( B C ) P (C A) P ( A B C )
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P( A B C ) 0.5 0.33 0.2 0.16 0.06 0.1 0.03 0.74 , which is the probability of a number [1,100] is divisible by 2, 3 or 5. So the required probability that a number [1,100] is not divisible by 2, 3 or 5 is P ( Ac B c C c ) 1 P ( A B C ) 1 0.74 0.26 . Example 6.114 [EE-2014 (2 marks)]: A fair coin is tossed n times. The probability that the difference between the number of heads and tails is (n 3) is n n 3 (c) n C n 3 2 n (b) 0 (a) 2 (d) 2 Solution (b): Let i and j denotes the number of heads and tails when a coin is tosses n times. As i j n , so j n i or i n j . Now let X be the difference between the number of heads and tails, so X i j 2i n or X i j n 2 j . Now it is given that X n 3 . So Case 1: 2i n n 3 2i 2n 3 , as the RHS is always odd and LHS is always even for any positive integral value of n . So we never get the difference between the number of heads and tails as n 3 ; and thus the required probability is zero. Case 2: n 2 j n 3 j 1.5 , which is not possible as j can take only positive integral value of n ; and thus the required probability is zero.
Example 6.115 [EE-2014 (1 mark)]: Consider a dice with the property that the probability of a face with n dots showing up is proportional to n . The probability of the face with three dots showing up is ……………. Solution: As on a dice, the number of dots vary from 1 to 6. So n varies from 1 to 6. Since probability of a face with n dots showing up is proportional to n , so we have P ( n) kn ; now when a die is rolled the possible outcomes are {1, 2, 3, 4, 5, 6} , and we know that the sum of the probabilities of the possible outcomes must be equal to 1. So
6
n1 P (n) 1 k (1 2 3 4 5 6) 1
k 1 21 . Thus P (3) (1 21) 3 1 7 0.14 . Example 6.116 [ME-2014 (1 mark)]: A box contains 25 parts of which 10 are defective. Two parts are being drawn simultaneously in a random manner from the box. The probability of both the parts being good is (a) 7 20 (b) 42 125 (c) 25 29 (d) 5 9 Solution (a): 2 good parts, from 15 good and 10 defective parts, can be selected in 15 C2 10 C0 ways, i.e., 2 good parts from 15 good parts and 0 defective parts from 10 defective parts. Also 2 parts from 25 parts can be selected in 25 C 2 ways. Thus the required probability is
15
C 2 10 C0
25
C2 105 300 7 20 .
Example 6.117 [XE-2014 (1 mark)]: Ten chocolates are distributed randomly among three children standing in a row. The probability that the first child receives exactly three chocolates is (d) 1 3 (a) (5 211 ) 39 (b) (5 210 ) 39 (c) 1 39 Solution (b): Let the three children be denoted by C1 , C2 , C3 ; each of the 10 chocolates can be given to either C1 or C 2 or C3 . Hence total number of ways of distributing chocolates is 310 . Number of ways to give 3 chocolates out of 10 to C1 is
10
C3 . If C1 gets exactly 3 chocolates, remaining 7 7
chocolates should be given to C 2 or C3 ; and total number of ways of doing this is 2 . Hence, required probability is {10 C3 27 } (310 ) (5 210 ) 39 .
Odds in favour and Odds against an Event: As a result of an experiment if ‘ a ’ of the outcomes are favourable to an event E and ‘ b ’ of the outcomes are against it, then we say that odds are a to b in favour of E or odds are b to a against E .
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Thus odds in favour of an event E Similarly, odds against an event E
Number of favourable cases Number of unfavourable cases
[6.43]
Number of unfavourable cases
a b b
a (a b) b (a b)
P( E ) P( E )
.
P(E )
. Number of favourable cases a P(E ) If odds in favour of an event are a : b , then the probability of the occurrence of that event is a ( a b ) and the probability of non-occurrence of that event is b (a b) . If odds against an event are a : b , then the probability of the occurrence of that event is b (a b) and the probability of non-occurrence of that event is a ( a b ) .
Example 6.118: Two dice are tossed together. Find the odds in favour of the sum of the numbers is 2. Solution: If two dice are tossed, total number of events 6 6 36 . Favourable event is (1,1) . Number of favourable events 1 . So the odds in favour 1 (36 1) 1 35 . Example 6.119: A party of 23 persons take their seats at a round table. Find the odds against two persons sitting together. Solution: P {(21)!2!} {(22)!} 1 11 1 (1 10) . So the odd against 10 :1 .
6.3.3 Conditional Probability Let A and B be two events associated with a random experiment. Then, the probability of occurrence of A under the condition that B has already occurred is called the conditional probability of A given B and is denoted by P ( A B) . In this case A serves as a new (reduced) sample space, and the probability is the fraction of that part of set A which corresponds to A B . Thus, P ( A B ) n( A B ) P ( A B) , where P( A) 0 or n( A) 0 (6.11) P( B) n( B ) Similarly, P ( B A) Probability of occurrence of B , given that A has already happened is given as, P ( B A) P ( A B ) P( A) {n( A B )} {n( A)}
(6.12)
Multiplication theorems on probability: If A and B are two events associated with a random experiment, then P ( A B) P ( A) P( B A) , if P( A) 0 or P ( A B) P ( B ) P( A B) , if P( B) 0 .
Extension of multiplication theorem: If A1 , A2 , , An are n events related to a random experiment,
then
P ( A1 A2 A3 An ) P( A1 ) P ( A2 A1 ) P A3 ( A1 A2 )
P An ( A1 A2 An 1 ) , where P Ai ( A1 A2 Ai 1 ) represents the conditional
probability of the event Ai , given that the events A1 , A2 , , Ai 1 have already happened.
Multiplication theorems for independent events: If A and B are independent events associated with a random experiment, then P ( A B) P ( A) P ( B ) i.e., the probability of simultaneous occurrence of two independent events is equal to the product of their probabilities. By multiplication theorem, we have P ( A B) P ( A) P( B A) . Since A and B are independent events, therefore P ( B A) P ( B) . Hence, P ( A B) P ( A) P ( B ) .
Extension of multiplication theorem for independent events: If A1 , A2 , , An are independent events associated with a random experiment, then P ( A1 A2 An ) P ( A1 ) P( A2 ) P ( An ) . By multiplication theorem, we have, P ( A1 A2 An ) P ( A1 ) P( A2 A1 ) P A3 ( A1 A2 ) P An ( A1 A2 An 1 ) . Since A1 , A2 , , An are independent events, therefore P ( A2 A1 ) P( A2 ) , P A3 ( A1 A2 ) P ( A3 ),
, P An ( A1 A2 ... An 1 ) P( An ) . Hence, P ( A1 A2 An ) P ( A1 ) P( A2 ) P ( An ) .
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Probability of at least one of the n independent events: If p1 , p2 , , pn be the probabilities of happening of n independent events A1 , A2 , A3 , , An respectively, then Probability of happening none of them P ( A1 A2 An ) P( A1 ) P ( A2 ) P ( An ) (1 p1 )(1 p2 ) (1 pn ) . Probability of happening at least one of them P ( A1 A2 An ) 1 P ( A1 ) P ( A2 ) P( An ) 1 (1 p1 )(1 p2 ) (1 pn ) Probability of happening of first event and not happening of the remaining P ( A1 ) P ( A2 ) P( An ) p1 (1 p2 ) (1 pn )
Example 6.120 [CS-1999 (2 marks)]: Consider two events E1 and E2 such that probability of E1 , Pr[ E1 ] 1 2 ; probability of E2 , Pr[ E2 ] 1 3 ; probability of E1 and E2 , Pr[ E1 E2 ] 1 5 . Which of the following statements is /are true? (a) Pr[ E1 or E2 ] 2 3 (b) Pr[ E1 E2 ] 4 5
(c) Events E1 and E2 are not independent
(d) Events E1 and E2 are independent
1 1 19 , so 2 3 5 30 option (a) is not correct. Pr[ E1 E2 ] Pr[ E1 E2 ] Pr[ E2 ] 3 5 , so option (b) is not correct. As
Solution (c): As Pr[ E1 or E2 ] Pr[ E1 E2 ] Pr[ E1 ] Pr[ E2 ] Pr[ E1 E2 ]
1
Pr[ E1 E2 ] Pr[ E1 ] Pr[ E2 ] so E1 and E2 are not independent and thus option (c) is correct and option (d) is not correct.
Example 6.121 [CS-2000 (2 marks)]: E1 and E2 are events in a probability space satisfying the following constraints: Pr( E1 ) Pr( E2 ) , Pr( E1 E2 ) 1 , E1 and E2 are independent. The value of Pr( E1 ) , the probability of event E1 , is
(b) 1 4
(a) 0 Solution
(d):
Let
(c) 1 2
Pr( E1 ) Pr( E2 ) p (say). 2
Pr( E1 E 2 ) Pr( E1 ) Pr( E2 ) p .
Now
As
(d) 1 E1
and
E2
are
independents
so
Pr[ E1 E2 ] Pr[ E1 ] Pr[ E2 ] Pr[ E1 E2 ]
1 p p p2 p 2 2 p 1 0 p 1 .
Example 6.122 [CS-2003 (1 mark)]: Let P ( E ) denote the probability of the event E . Given
P ( A) 1 , P ( B ) 1 2 , the values of P ( A | B) and P ( B | A) respectively are (a) 1 4 ,1 2 (b) 1 2 ,1 4 (c) 1 2 ,1 (d) 1,1 2 Solution (d): Let A and B are exhaustive events, i.e., P( A B) 1 ; P ( A B ) P ( A) P ( B ) P ( A B ) 1 0.5 1 0.5 .
so
Thus P ( A B) P ( A B) P ( B) 0.5 0.5 1 and P ( B A) P ( A B ) P ( A) 0.5 1 0.5 . Example 6.123 [EE-2005 (1 mark)]: If P and Q are two random events, then the following is TRUE: (a) Independence of P and Q implies that Probability ( P Q ) 0 (b) Probability ( P Q) Probability ( P ) + Probability (Q) (c) If P and Q are mutually exclusive, then they must be independent (d) Probability ( P Q) Probability ( P ) Solution (d): If P and Q are two random events then (a) Independence of P and Q means Pr( P Q ) Pr( P ) Pr(Q) , so option (a) is not correct. As
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Pr( P Q ) Pr( P ) Pr(Q) Pr( P Q) Pr( P Q) Pr( P) Pr(Q) so option (b) is not correct. As there is no relation between mutually exclusive and independent for two random variables P and Q , so option (c) is not correct. As Pr( P Q ) Pr(Q P ) Pr( P ) Pr( P Q ) Pr( P) . Example 6.124 [EE-2006 (2 marks)]: Two fair dice are rolled and the sum r of the numbers turned up is considered (a) Pr(r 6) 1 6 (b) Pr( r 3 is an integer) 5 6 (c) Pr( r 8 | r 4 is an integer) 5 9 (d) Pr(r 6 | r 5 is an integer) 1 18 Solution: As two fair dice are rolled so total number of sample space is 62 36 . Also r is the sum of number appeared on 1st and 2nd dice. For r 3 , the numbers appeared on the dices is {(1, 2), (2,1)} so
P ( r 3) 2 36 ; for r 4 , the numbers appeared on the dices is {(1, 3), (2, 2), (3,1)} so P ( r 3) 3 36 ; for r 5 , the numbers appeared on the dices is {(1, 4), (2, 3), (3, 2), (4,1)} so P ( r 4) 4 36 ; for r 6 , the numbers appeared on the dices is {(1, 5), (2, 4), (3, 3), (4, 2), (5,1)} so
r 7, {(1, 6), (2,5), (3, 4), (4, 3), (5, 2), (6,1)} P ( r 6) 5 36 ; P (r 9) 4 36 ;
for
the so
appeared
P (r 7) 6 36 .
on
Similarly,
for
dices
is
P ( r 8) 5 36 ; So
P (r 6) P( r 7) P (r 8) P (r 9) P(r 10) P(r 11) P (r 12) 21 36 7 12 .
So
(a)
is
not
correct.
Now
P( r 11) 2 36 ;
the
P ( r 12) 1 36 .
option
P( r 10) 3 36 ;
numbers
r 3
is
an
integer
if
r 3, 6, 9,12 ,
so
P (r 3 is an integer) P (r 3) P (r 6) P (r 9) P(r 12) 12 36 1 3 ; thus option (b) is not correct.
Now
r 4
is
an
integer
if
r 4,8,12 ,
so
P ( r 4 is an integer) P (r 4) P(r 8) P(r 12) 9 36 1 4 and thus P ( r 8) 5 36 5 P ( r 8 | r 4 is an integer) . Now r 5 is an integer if r 5,10 , so P ( r 4 is an integer) 1 4 9 P (r 5 is an integer) P(r 5) P (r 10) 7 36 and P ( r 6) 5 36 5 P ( r 6 | r 5 is an integer) , so option (d) is not correct. P ( r 5 is an integer) 7 36 7
thus
Example 6.125 [EC-2007 (2 marks)]: An examination consists of two papers, Paper 1 and Paper 2. The probability of failing in Paper 1 is 0.3 and that in Paper 2 is 0.2. Given that a student has failed in Paper 2, the probability of failing in Paper 1 is 0.6. The probability of a student failing in both the papers is: (a) 0.5 (b) 0.18 (c) 0.12 (d) 0.06 Solution (c): Probability of failing in Paper 1 is P ( A) 0.3 ; Probability of failing in Paper 2 is
P ( B) 0.2 ; Probability of failing in Paper 1, when student has failed in Paper 2 is P ( A B ) 0.6 . So Probability of a student failing in both the papers is P ( A B ) P( B ) P ( A B ) (0.6)(0.2) 0.12 . Example 6.126 [EE-2007 (2 marks)]: A loaded Dice value 1 2 3 dice has following probability distribution of Probability 14 18 18 occurrences. If three identical dice as the above Dice value 4 5 6 are thrown, the probability of occurrence of Probability 1 8 1 8 1 4 values 1, 5 and 6 on three dice is (a) same as that of occurrence of 3, 4, 5 (b) same as that of occurrence of 1, 2, 5 (c) 1/128 (d) 5/8 Solution (c): Probability of occurrence of values 1, 5 and 6 on three dices is P (1 and 3 and 5) P (1) P (5) P (6) (1 4) (1 8) (1 4) 1 128 . Probability of occurrence of values 3, 4 and 5 on three dices is P (3 and 4 and 5) P (3) P (4) P (5) (1 8) (1 8) (1 8) 1 512 .
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Probability of occurrence of values 1, 2 and 5 on three dices P (1 and 2 and 5) P (1) P (2) P (5) (1 4) (1 8) (1 8) 1 256 . Hence option (c) is correct.
is
Example 6.127 [CH-2009 (2 marks)]: A fair die is rolled. Let R denote the event of obtaining a number less than or equal to 5 and S denotes the event of obtaining an odd number. Then which ONE of the following about the probability ( P ) is TRUE? (a) P ( R S ) 1 (b) P ( R S ) 0 (c) P ( S R ) 1 (d) P ( S R ) 0 Solution (a): Total number of sample space when a fair dice is rolled is 6. The elements of event R is {1, 2, 3, 4,5} and the elements of event S is {1, 3, 5} ; thus the elements of ( R S ) is {1, 3, 5} . So
P ( R ) 5 6 , P( S ) 3 6 1 2 , P ( R S ) 3 6 1 2 . P( R S ) 1 2 P( R S ) 1 2 3 Thus P ( R S ) 1 and P ( S R ) . P(S ) 12 P( R ) 56 5 Example 6.128 [EC-2009 (2 marks)]: Consider two independent random variables X and Y with identical distributions. The X and Y take values 0, 1 and 2 with probabilities 1 2 , 1 4 and 1 4 respectively. What is the conditional probability P ( X Y 2 | X Y 0) ? (a) 0 (b) 1/16 (c) 1/6 (d) 1 Solution (c): We have P ( X 0) P (Y 0) 1 2 , P ( X 1) P(Y 1) 1 4 and
P ( X 2) P(Y 2) 1 2 . Now let E be the event for which X Y 2 , so the elements of E are {( X 0, Y 2), ( X 1, Y 1), ( X 2, Y 0)} ; and F be the event for which X Y 0 , so elements of F are {( X 0, Y 0), ( X 1, Y 1), ( X 2, Y 2)} . Thus P ( E ) P ( X 0 and Y 2) P( X 1 and Y 1) P( X 2 and Y 0) and P ( F ) P ( X 0 and Y 0) P( X 1 and Y 1) P ( X 2 and Y 2) 1 1 1 1 1 1 5 1 1 1 1 1 1 6 P( E ) and P ( F ) . Now event ( E F ) 2 4 4 4 4 2 16 2 2 4 4 4 4 16 happens when X Y 2 and X Y 0 , which happens only when X 1 and Y 1 . As X 1 and 1 1 1 Y 1 are independent variable, so P ( E F ) P ( X 1 and Y 1) P ( X 1) P (Y 1) . 4 4 16 P ( E F ) 1 16 1 Thus the required probability, i.e., P ( X Y 2 | X Y 0) P ( E F ) . P( F ) 6 16 6 Example 6.129 [EE-2010 (2 marks)]: A box contains 4 white balls and 3 red balls. In succession, two balls are randomly selected and removed from the box. Given that the first removed ball is white, the probability that the second removed ball is red is (a) 1/3 (b) 3/7 (c) 1/2 (d) 4/7 Solution (b): Let R and W be the events of picking red and white balls, respectively. As white ball is removed first and we have to find the probability of red ball removed when white ball was already removed, i.e. to find P ( R W ) {P( R W )} {P (W )} . As both the events are independent so P ( R W ) P ( R ) P (W ) (3 7) (4 7) 12 49 . Thus P ( R W ) (12 49) (4 7) 3 7 .
Example 6.130 [EC-2014 (2 marks)]: Parcels from sender S to receiver R pass sequentially through two post-offices. Each post-office has a probability 1 5 of losing an incoming parcel, independently of all other parcels. Given that a parcel is lost, the probability that it was lost by the second post-office is ……………. Solution: Let A be the event for ‘parcel is lost’; B be the event for ‘parcel is lost by the 2 nd post office’; From the given data, we have P ( B ) 1 5 , and to find P ( B A) {P( A B )} {P( A)} . Now P ( A) P(parcel lost by 1st post office or parcel passed by 1st and lost by 2 nd post office) ,
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P ( A) P(parcel lost by 1st post office) P(parcel passed by 1st and lost by 2 nd post office) P ( A) (1 5) P (parcel passed by 1st post office) P (parcel lost by 2nd post office) P ( A) (1 5) (4 5) (1 5) 9 25 .
Now
P ( A B ) P (parcel passed by 1st and lost by 2 nd post office) P ( A B ) (4 5) (1 5) 4 25 .
Thus the required probability is P ( B A) (4 25) (9 25) 4 9 0.44 .
6.3.4 Total Probability and Baye’s Theorem The law of total probability: Let S be the sample space and let E1 , E2 , , En be n sets of mutually exclusive and exhaustive events associated with a random experiment, as shown in Fig. 6.4. These events E1 , E2 , , En are said to partition the sample space S . Then we have Ei E j for
i j , 1 i, j n and
n
i 1 P( Ei ) 1 . If
A is any event of S which occurs with E1 or E2 or … or
En , then the total probability of the event A is given as n
P ( A) i 1 P ( Ei ) P ( A Ei ) P ( E1 ) P ( A E1 ) P ( E2 ) P ( A E2 ) P ( En ) P ( A E n )
where,
P ( A Ei )
gives us the contribution of
Ei
in the occurrence of
(6.13)
A . Proof:
A ( E1 A) ( E2 A) ( En A) P( A) P ( E1 A) P( E2 A) P ( En A) P( A) P ( E1 ) P ( A E1 ) P ( E2 ) P ( A E2 ) P( En ) P ( A En )
Baye’s Theorem: Let S be a sample space and E1 , E2 , , En be n mutually exclusive events such that
i 1 Ei S n
and P ( Ei ) 0 for i 1, 2, , n , as shown in
Fig. 6.4. We can think of ( Ei ’s as the causes that lead to the outcome of an experiment. The probabilities P( Ei ) , i 1, 2, , n are called prior probabilities. Suppose the experiment results in an outcome of event A , where
P( A) 0 . We have to find the probability that the observed event A was due to cause Ei , i.e., we seek the conditional
Figure 6.4: Partitioning of Space
probability P( Ei A) . These probabilities are called posterior probabilities, given by Baye’s rule as, P ( Ei A)
P( Ei ) P ( A Ei )
(6.14) P( A) P( E1 ) P( A E1 ) P ( E2 ) P ( A E2 ) P ( En ) P ( A En ) where, value of P ( A) is from Eq. 6.13. P( A Ei ) P ( Ei ) If E1 , E2 , , En form a partition of event A , then P ( A Ei ) 1 ( P ( Ei ) P ( Ei ) P ( Ei A)
A Ei Ei ). Hence from Eq. 6.14, we have P ( Ei A)
P( Ei ) P ( E1 ) P ( E2 ) P( En )
.
Example 6.131 [CS-2005 (2 marks)]: Box P has 2 red balls & 3 blue balls and box Q has 3 red balls & 1 blue ball. A ball is selected as follows: (i) select a box (ii) choose a ball from the selected box such that each ball in the box is equally likely to be chosen. The probabilities of selecting boxes P and Q are 1 3 and 2 3 , respectively. Given that a ball selected in the above process is a red ball, the probability that it came from the box P is: (a) 4 19 (b) 5 19 (c) 2 9 (d) 19 30 Solution (a): Let R and B be the events for selecting red and black balls, respectively.
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The following diagram shows the situation of the given problem. So if a red ball is selected then the probability that it came from the box P is given as: Pr( R P ) Pr( P ) Pr( P R ) Pr( R P ) Pr( P ) Pr( R Q) Pr(Q) Pr( P R)
(2 5)(1 3) (2 5)(1 3) (3 4)(2 3)
4 19
Example 6.132 [EC-2006 (2 % of Computers Probability of Company marks)]: Three companies X , supplied being defective Y and Z supply computers to a (a) 0.1 60% 0.01 X university. The percentage of (b) 0.2 30% 0.02 Y computers supplied by them and (c) 0.3 10% 0.03 Z the probability of those being Given that a computer is defective, the probability (d) 0.4 defective are tabulated as: that it was supplied by Y is: Solution (d): The situation of the given question is described in the following figure. Let X , Y and Z are the events of supplying computers to the companies X , Y and Z , respectively. Let d1 , d 2 , d3 are the events of the defective computer supplied to the X , Y , Z companies, respectively. So from Eq. 6.14, we have P (Y ) P (d 2 Y ) P (Y d 2 ) P( X ) P ( d 2 X ) P (Y ) P( d 2 Y ) P ( Z ) P ( d 2 Z ) P(Y d 2 )
(0.3)(0.02) (0.6)(0.01) (0.3)(0.02) (0.1)(0.03)
0.4
Example 6.133 [IN-2009 (2 marks)]: A screening test is carried out to detect a certain disease. It is found that 12% of the positive reports and 15% of the negative reports are incorrect. Assuming that the probability of a person getting a positive report is 0.01, the probability that a person tested gets an incorrect report is (a) 0.0027 (b) 0.0173 (c) 0.1497 (d) 0.2100 Solution (c): Let E1 and E2 are the events that the report is positive and negative, respectively. Let A
be
the
event
that
report
is
incorrect.
It
is
given
that
P ( E1 ) 0.01 ;
also
P ( E1 ) P( E2 ) 1 P ( E2 ) 0.99 . It is also given that P ( A E1 ) 0.12 and P ( A E2 ) 0.15 . We have to find the probability that a person tested gets an incorrect report, i.e., P( A) . So P ( A) P( E1 ) P( A E1 ) P( E2 ) P ( A E2 ) 0.01 0.12 0.99 0.15 0.1497 .
Example 6.134 [CS-2010 (2 marks)]: Consider a company that assembles computers. The probability of a faulty assembly of any computer is p . The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q . What is the probability of a computer being declared faulty? (a) pq (1 p)(1 q ) (b) (1 q ) p (c) (1 p ) q (d) pq Solution (a): Let F & C be the event for a computer having faulty and correct assembly, respectively; and F1 & C1 be the event for a computer having faulty and correct testing respectively. The diagram shows the situation of given problem. So total probability of a computer being declared faulty is: P (declared faulty) P( F ) P( F1 F ) P (C ) P( F1 C ) P(declared faulty) pq (1 p )(1 q) .
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Example 6.135 [PI-2010]: Two white and two black balls, kept in two bins, are arranged in four ways as shown below. In each arrangement, a bin has to be chosen randomly and only one ball needs to be picked randomly from the chosen bin. Which one of the following arrangements has the highest probability for getting a white ball picked? (a)
(b)
(c)
(d)
Solution (c): Let B1 and B2 are the events of choosing a ball from bin 1 and 2, respectively. Let W and B are the events of choosing a white and black bass, respectively. Now suppose bin 1 contains w1 white balls and b1 black balls; and bin 2 contains w2 white balls and b2 black balls. Now, if a white ball is picked randomly from the chosen bin then, the probabilities that it is chosen from first bin or second bin is shown in figure. Thus the total probability of choosing a white ball is: P (W ) P ( B1 ) P (W B1 ) P ( B2 ) P (W B2 ) . Now for option (a), we have w1 1, b1 1, w2 1, b2 1 ; thus P (W ) (1 2) (1 2) (1 2) (1 2) 1 2 . For option (b), we have w1 2, b1 0, w2 0, b2 2 ; thus P (W ) (1 2) (2 2) (1 2) (0 2) 1 2 . For option (c), we have w1 1, b1 0, w2 1, b2 3 ; thus P (W ) (1 2) (1 1) (1 2) (1 3) 2 3 . For option (d), we have w1 0, b1 1, w2 2, b2 1 ; thus P (W ) (1 2) (0 1) (1 2) (2 3) 1 3 . Thus the arrangement shown in option (c) has the highest probability for getting a white ball picked.
Example 6.136 [ME-2013, PI-2013 (2 marks)]: The probability that a student knows the correct answer to a multiple choice question is 2/3. If the student does not know the answer, then the student guesses the answer. The probability of the guessed answer being correct is 1/4. Given that the student has answered the question correctly, the conditional probability that the student known the correct answer is (a) 2 3 (b) 3 4 (c) 5 6 (d) 8 9 Solution (d): Let A be the events for which the student answers the question correctly; E1 be the events for which the student knows the correct answer; and E2 be the events for which the student guesses the correct answer. We have, P ( E1 ) 2 3 , P( E2 ) 1 3 , P ( A E2 ) 1 4 ; also if the student knows the correct answer then he will answer the question correctly so P ( A E1 ) 1 ; we have to find P( E1 A) , so from Eq. 6.14, P ( E1 A)
P ( E1 ) P ( A E1 ) P ( E1 ) P ( A E1 ) P ( E2 ) P ( A E2 )
(2 3)(1) (2 3)(1) (1 3)(1 4)
8 9
.
Example 6.137 [EC-2014 (1 mark)]: In a housing society, half of the families have a single child per family, while the remaining half have two children per family. The probability that a child picked at random, has a sibling is ……………. Solution: Let E1 and E2 be the event of single child family and two children family. Let A be the event of picking a child. So from Bay’s theorem, we have P ( E2 ) P ( A E2 ) (0.5)(2 x 3 x) 2 P ( E2 A) 0.667 , where x is P( E1 ) P( A E1 ) P ( E2 ) P ( A E2 ) (0.5)( x 3x ) (0.5)(2 x 3x ) 3 the number of child from single child family; and 2x is the number of child from two children family.
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Example 6.138 [ME-2014 (1 mark)]: A group consists of equal number of men and women. Of this group 20% of the men and 50% of the women are unemployed. If a person is selected at random from this group, the probability of the selected person being employed is …………... Solution: Let M : Men , W : Women , U : Unemployed , E : Employed . The situation is described in the following diagram. So the required probability of selected person being employed is given as:
P ( E ) P ( M ) P( E M ) P(W ) P ( E W ) P ( E ) (1 2) (4 5) (1 2) (1 2) 13 20 0.65 Exercise: 6.3 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. Two fair dice are tossed. Let A be the event that the first die shows an even number and B be the event that second die shows an odd number. The two events A and B are (a) Mutually exclusive (b) Independent and mutually exclusive (c) Dependent (d) None of these 2. The probabilities of a student getting I, II and III division in an examination are respectively 1 10 ,
3 5 and 1 4 . The probability that the student fail in the examination is _____. 3. If (1 3 p ) 3 , (1 p) 4 and (1 2 p) 2 are the probabilities of three mutually exclusive events, then the set of all values of p is 4. 5.
6. 7.
8.
9.
10.
11.
(a) (1 3) p (1 2) (b) (1 3) p (1 2) (c) (1 2) p (2 3) (d) (1 2) p (2 3) The probability that a leap year selected randomly will have 53 Sundays is (a) 1 7 (b) 2 7 (c) 4 53 (d) 4 49 Three identical dice are rolled. The probability that same number will appear on each of them will be (a) 1 6 (b) 1 36 (c) 1 18 (d) 3 28 Three of the six vertices of a regular hexagon are chosen at random. The probability that the triangle with these three vertices is equilateral, is equal to _____. Three distinct numbers are selected from 100 natural number. The probability that all the three numbers are divisible by 2 and 3 is (a) 4 25 (b) 4 35 (c) 4 55 (d) 4 1155 Out of 21 tickets marked with numbers from 1 to 21, three are drawn at random. The chance that the numbers on them are in A.P., is (a) 10 133 (b) 9 133 (c) 9 1330 (d) None of these There are four letters and four addressed envelopes. The chance that all letters are not dispatched in the right envelope is (a) 19 24 (b) 21 23 (c) 23 24 (d) 1 24 The letters of the word ‘ASSASSIN’ are written down at random in a row. The probability that no two S occur together is (a) 1 35 (b) 1 14 (c) 1 15 (d) None of these A box contains 6 nails and 10 nuts. Half of the nails and half of the nuts are rusted. If one item is chosen at random, what is the probability that it is rusted or is a nail (a) 3 16 (b) 5 16 (c) 11 16 (d) 14 16
12. The probability that a man will be alive in 20 years is 3 5 and the probability that his wife will be alive in 20 years is 2 3 . Then the probability that at least one will be alive in 20 years is (a) 13 15
(b) 7 15
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(c) 4 15
(d) None of these
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13. Let A and B be two events such that P ( A) 0.3 and P ( A B) 0.8 . If A and B are independent events, then P ( B ) (a) 5 6 (b) 5 7 (c) 3 5 (d) 2 5 14. A card is chosen randomly from a pack of playing cards. The probability that it is a black king or queen of heart or jack is (a) 1 52 (b) 6 52 (c) 7 52 (d) None of these 15. If A and B are events such that P ( A B ) 3 4 , P ( A B ) 1 4 , P ( A ) 2 3 , then P ( A B ) is (a) 1 4
(b) 3 8
(c) 5 8
(d) 5 12
16. The probability that A speaks truth is 4 5 , while this probability for B is 3 4 . The probability that they contradict each other when asked to speak on a fact is _____ 17. A student appears for tests I, II and III. The student is successful if he passes either in tests I and II or tests I and III. The probabilities of the student passing in tests I, II, III are p , q and 1 2
18.
19.
20. 21. 22.
23. 24.
25.
26.
27. 28.
respectively. If the probability that the student is successful is 1 2 , then (a) p 1 , q 0 (b) p 2 3 , q 1 2 (d) All of the above (c) There are infinitely many values of p and q A man and his wife appear for an interview for two posts. The probability of the husband’s selection is 1 7 and that of wife’s selection is 1 5 . What is the probability that only one of them will be selected? (a) 1 7 (b) 2 7 (c) 3 7 (d) None of these A purse contains 4 copper coins and 3 silver coins, the second purse contains 6 copper coins and 2 silver coins. If a coin is drawn out of any purse, then the probability that it is a copper coin is (a) 4 7 (b) 3 4 (c) 37 56 (d) None of these The probability of happening an event A is 0.5 and that of B is 0.3. If A and B are mutually exclusive events, then the probability of happening neither A nor B is _____. A coin is tossed three times in succession. If E is the event that there are at least two heads and F is the event in which first throw is a head, then P( E F ) _____. Two cards are drawn one by one from a pack of cards. The probability of getting first card an ace and second an honour card is (before drawing second card first card is not placed again in the pack) (a) 1 26 (b) 5 52 (c) 5 221 (d) 4 13 A problem of mathematics is given to three students whose chances of solving the problem are 1 3 , 1 4 and 1 5 respectively. The probability that the question will be solved is _____. In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product. Then the probability that the bolt drawn is defective is _____. A lot contains 20 articles. The probability that the lot contains 2 defective articles is 0.4 and the probability that the lot contains exactly 3 defective articles is 0.6. Articles are drawn at random one by one without replacement and tested till all the defective articles are found. The probability that the testing procedure ends at the twelfth testing is (a) 9 1900 (b) 19 1000 (c) 99 1900 (d) 19 900 A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from B is (a) 5 14 (b) 5 16 (c) 5 18 (d) 25 52 A man is known to speak the truth 3 out of 4 times. He throws a die and reports that it is a six. The probability that it is actually a six, is _____. A pack of playing cards was found to contain only 51 cards. If the first 13 cards which are examined are all red, then the probability that the missing cards is black, is _____.
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29. Two numbers are selected randomly from the set S {1, 2, 3, 4, 5, 6} without replacement one by one. The probability that minimum of the two numbers is less than 4 is _____ 30. Two dice are thrown. The probability that the sum of the points on two dice will be 7, is (a) 5 36 (b) 6 36 (c) 7 36 (d) 8 36 31. Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternately? (d) None of these (a) 1 462 (b) 1 924 (c) 1 2 32. A determinant is chosen at random. The set of all determinants of order 2 with elements 0 or 1 only. The probability that value of the determinant chosen is positive, is _____. 33. If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls. One ball is drawn at random. Then the probability that 2 white and 1 black ball will be drawn is (a) 13 32 (b) 1 4 (c) 1 32 (d) 3 16 34. The probability of India winning a test match against West Indies is 1 2 . Assuming independence from match to match, the probability that in a 5 match series India’s second win occurs at the third test is _____. 35. An unbiased die is tossed until a number greater than 4 appears. The probability that an even number of tosses is needed is _____. 36. A man alternately tosses a coin and throws a dice beginning with the coin. The probability that he gets a head in the coin before he gets a 5 or 6 in the dice is _____. 37. A box contains 100 tickets numbered 1, 2, ,100 . Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than 10. The minimum number on them is 5 with probability (a) 1 8 (b) 13 15 (c) 1 9 (d) None of these 38. A pair of fair dice is rolled together till a sum of either 5 or 7 is obtained. Then the probability that 5 comes before 7 is _____. 39. In an entrance test there are multiple choice questions. There are four possible answers to each question of which one is correct. The probability that a student knows the answer to a question is 90%. If he gets the correct answer to a question, then the probability that he was guessing, is (a) 37 40 (b) 1 37 (c) 36 37 (d) 1 9 40. A bag X contains 3 white balls and 2 black balls and another bag Y contains 2 white balls and 4 black balls. A bag and a ball out of it are picked at random. The probability that the ball is white is _____. 41. The probability that in a year of the 22nd century chosen at random there will be 53 Sundays is (a) 3 28 (b) 2 28 (c) 7 28 (d) 5 28 42. In a bolt factory, machines A, B and C manufacture respectively 25%, 35% and 40% of the total bolts. Of their output 5, 4 and 2 percent respectively are defective bolts. A bolt is drawn at random from the product. If the bolt drawn is found to be defective, the probability that it is manufactured by the machine B is (a) 28 69 (b) 7 69 (c) 32 69 (d) 11 69 43. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter driver, car driver and a truck driver is 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver (a) 1 52 (b) 1 62 (c) 2 51 (d) 1 44. In a test, an examinee either guesses or copies or knows the answer to a multiple choice question with four choices. The probability that he makes a guess is 1 3 and the probability that he copies the answer is 1 6 . The probability that his answer is correct, given that he copied it, is 1 8 . The probability that he knew the answer to the question, given that he correctly answered it, is (a) 24 27 (b) 24 29 (c) 24 31 (d) None of these
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Chapter 6: Probability & Statistics
[6.53]
Random Variable
Many problems are concerned with values associated with outcomes of random experiments. For e.g., we select ten items randomly from the production lot with known proportion of defective and put them in a packet. Now we want to know the probability that the packet contains more than one defective item. The study of such problem require a concept of random variable, which is a function that associates numerical values to the outcomes of experiments. Let S be a sample space corresponding to the outcomes of a random experiment. A function X : S R (where R is a set of real numbers) is called a random variable. Random variable is a real valued mapping. Thus, the function has to one – one or many – one correspondence. Thus, a random variable assigns a real number to each possible outcome of an experiment. Note that, a random variable is a function it is neither a variable nor random. A random variable is a function from the sample space of a random experiment (domain) to the set of real numbers (co-domain). A random variable may be discrete or continuous (which are discussed later in this section). Example 6.139: Let X be a random variable defined as difference of the numbers that appear when a pair of dice is rolled. What are the distinct values of the random variable? Solution: When a pair of dice is rolled, there are 36 possible outcomes. The sample set is S {(1,1), (1, 2), , (1, 6), (2,1), (2, 2), , (2, 6), (3,1), (3, 2), (6, 6)} . Now the random variable X is the difference of the numbers. Its values are X {0,1, 2, 3, 4, 5} .
Probability Distribution: Each outcome i of an experiment has a probability P(i) associated with it. Similarly, every value of random variable X xi is related to the outcome i of an experiment. Hence, for every value of random variable xi , we have a unique real value P(i) associated. Thus, every random variable X has probability P associated with it. This function P( X xi ) from the set of all events of the sample space S is called a probability distribution of the random variable. The probability distribution (or simply distribution) of a random variable X on a sample space S is set of pairs X xi , P ( X xi ) for all xi x ( S ) , where P( X xi ) is the probability that X takes the value xi . Example 6.140: A random variable is number of tails when a coin is flipped thrice. Find probability distribution of the random variable. Solution: Sample space is HHH, THH, Value of random X xi 0 1 2 3 HTH, HHT, TTH, THT, HTT, TTT. variable The required probability distribution is, P( X xi ) 1 8 3 8 3 8 1 8 Probability
Discrete Random Variable: A random variable X is said to be discrete if it takes finite or countable infinite number of possible values. Thus, discrete random variable takes only isolated values. Random variables mentioned in the previous example are discrete random variables. Some of the practical examples of discrete random variable are: (i) Number of accidents on an expressway, (2) Number of cars arriving at a petrol pump (3) Number of students attending class (4) Number of customers arriving at a shop (5) Number of neutrons emitted by a radioactive isotope. Probability Mass function Let X be a discrete random variable defined on a sample space S . Suppose { x1 , x2 , , xn } is the range set of X . With each of x , we assign a number P ( xi ) P ( X xi ) (6.15) called the probability mass function (p.m.f.) such that, P ( xi ) 0 for i 1, 2, , n and n
i1 P( xi ) 1 .
The table containing the value of X along with the probabilities given by
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Example 6.141: Suppose a fair dice is rolled twice. Find the possible values of random variable X and its associated p.m.f., if X is the maximum of the two values appearing in two rolls. Solution: Since the dice is fair, probability of any of the outcome of pair of numbers (1,1), (1, 2), , (1, 6), (2,1), (2, 2), , (2, 6), (3,1), (3, 2), (6, 6) appearing in two rolls is 1 36 . Since X can take values from 1 to 6. The probabilities of X taking these values can be calculated by adding the probabilities of various outcomes that give the particular value of X . For example, X 4 , i.e., maximum of two values appearing on dice is four, can be obtained with outcomes {(1, 4), (2, 4), (3, 4), (4, 4), (4, 3), (4, 2), (4,1))} i.e., 7 ways. Hence P{ X 4} 7 36 . So, the values and probability distribution of the random variable X is given as, X xi 1 2 3 4 5 6 P( xi ) p.m.f. ( P ( xi ) )
1 36
3 36
5 36
7 36
9 36
1
11 36
Continuous Random Variable: Random variables could also be such that their set of possible values is uncountable. Examples of such random variables are time between arrivals of two vehicles at petrol pump or time taken for an angioplasty operation at a hospital or lifetime of a component. Probability Density Function (p.d.f.): Like we have p.m.f. for discrete random variable, we define p.d.f. for continuous random variable. Let X be a continuous random variable. Function f ( x ) defined for all real x ( , ) is called probability density function (p.d.f.) if for any set B of real numbers, we get probability, P{ X B} f ( x) dx . All probability B
statements about X can be answered in terms of f ( x ) . Thus, b
P{a X b} f ( x ) dx
(6.16)
a
Note that probability of a continuous random variable at any particular value is zero, since a
P{ X a} P{a X a} f ( x) dx 0 . a
6.4.1 Properties of Random Variable and their Probability Distribution Properties of a random variable can be studied only in terms of its p.m.f. or p.d.f. We need not refer to the underlying sample space, once we have probability distribution of a random variable. Since the random variable is a function relating all outcomes of a random experiment, the probability distribution of random variable must satisfy the axioms of probability. These in case of discrete and continuous random variable are stated as, Any probability must satisfy between zero and one. For discrete random variable 0 p ( xi ) 1 b
For continuous random variable 0 P{a x b} 1 or 0 f ( x) dx 1 . a
Total probability of sample space must be one.
i1 p( xi ) 1 For continuous random variable f ( x ) dx 1
For discrete random variable
For any sequence of mutually exclusive events E1 , E2 , etc., i.e., Ei E j for i j , probability of a union set of events is sum of their individual probabilities, i.e. P ( E1 E2 ) P ( E1 ) P ( E2 ) , where E1 and E2 are mutually exclusive events. For discrete random variable, this can be stated as, P
E
For continuous random variable, this can be stated as, b
d
a
c
i
i 1
b
a
i 1
P ( Ei ) c
b
a
c
f ( x) dx f ( x) dx f ( x ) dx
also, P ( a x b c x d ) f ( x) dx f ( x ) dx .
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[6.55]
Example 6.142 [EC-2006 (1 mark)]: A probability density function is of the form p ( x) ke x ( , ) . The value of k is (a) 0.5 (b) 1 (d) (c) 0.5 Solution (c): We know that for a continuous random variable 0
0
p( x)dx 1 ke
x
x
,
dx 1 ,
ke x dx ke x dx 1 {( k )e x }0 [{k ( )}e x }]0 1 ( k ) ( k ) 1 k 2
Example 6.143 [PI-2007 (2 marks)]: If X is a continuous random variable whose probability
k (5 x 2 x 2 ), 0 x 2 density function is given by f ( x ) . Then P( X 1) is 0, otherwise (a) 3 14
(b) 4 5
(c) 14 17
Solution (d): For a continuous random variable, we have 0
2
(d) 17 28
f ( x)dx 1 , where PDF is
f ( x ) . So
2
f ( x)dx 0 f ( x)dx 2 f ( x)dx 1 0 f ( x)dx 1 , as f ( x) 0 for all x [0, 2] . Thus 2 2 2 2 2 3 2 0 k (5 x 2 x )dx 1 k 0 (5 x 2 x )dx 1 k (5 2) x (2 3) x 0 1 k (14 3) 1 k 3 14 . 2 2 So P ( X 1) f ( x ) dx f ( x ) dx f ( x ) dx f ( x ) dx , as f ( x ) 0 for all x [0, 2] . 1 1 2 1 2 2 2 2 Thus P ( X 1) k (5 x 2 x ) dx k (5 2) x (2 3) x 3 k (17 6) (3 14) (17 6) 17 28 . 1 1 [Similar questions were also asked in TF-2007, TF-2014 (2 marks)] Example 6.144 [CE-2008 (2 marks)]: If probability density function of a random variable X is
x2 ,
f ( x)
0,
(a) 0.247 Solution:
1 x 1
1 1 . Then, the percentage probability P x is 3 for any other value of x 3 (b) 2.47 (c) 24.7 (d) 247 As we have continuous probability distribution,
P (1 3) x (1 3)
13
1 3
f ( x ) dx
13
2
1 3
x dx (1 3) x
3 13 1 3
so
(1 3){(1 27) (1 27)} 0.0247 . Thus
the percentage probability is P 1 3 x 1 3 100 2.47 . Example 6.145 [EC-2008 (2 marks)]: Px ( x ) M exp 2 x N exp 3 x is the probability density function for the real random variable X , over the entire x axis. M and N are both positive real numbers. The equation relating M and N is (a) M (2 3) N 1 (b) 2 M (1 3) N 1 (c) M N 1 (d) M N 3
Px ( x)dx 1 , 0 M exp 2 x N exp 3 x dx 1 Me 2 x Ne3 x dx Me 2 x Ne 3 x dx 1 0
Solution
(a):
( M 2)e
2x
We
( N 3)e
know
3x 0
that
for
( M 2)e
2 x
a
continuous
( N 3)e
3 x
0
random
variable
1 M (2 3) N 1 .
Example 6.146 [EE-2013 IN-2013 (1 mark)]: A continuous random variable X has a probability density function f ( x ) e x , 0 x , then P( X 1) is (a) 0.368 (b) 0.5 (c) 0.632 (d) 1.0 Solution: As we have continuous probability distribution, so
P 1 x f ( x )dx e x dx e x 1
1
1
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0 e1 1 e 0.368 .
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Engineering Mathematics
Chapter 6: Probability & Statistics
[6.56]
Example 6.147 [CE-2013 (2 marks)]: Find the value of such that function f ( x ) is valid
( x 1)(2 x) for 1 x 2
probability density function f ( x )
0
otherwise
f ( x)dx 1 . So 1 2 1 2 f ( x ) dx f ( x ) dx ( x 1)(2 x ) dx f ( x ) dx (0) dx ( x 1)(2 x ) dx 1 2 1 2 (0)dx 2 2 f ( x ) dx (4 x x 2 2) dx 2 x 2 (1 3) x 3 2 x (1 6) . So f ( x ) dx 1 if 6 . 1 1
Solution:
As
for
a
continuous
random
variable,
we
have
Example 6.148 [CE-2014 (1 mark)]: The probability density function of evaporation E on any day 1 5, 0 E 5 mm / day during a year in a watershed is given by f ( E ) . The probability that E otherwise 0, lies in between 2 and 4 mm/day in a day in the watershed is (in decimal) ……………. Solution: As we have continuous probability distribution, so 4
4
2
2
4
P 2 E 4 f ( E ) de (1 5) de (e 5) 2 (4 5) ( 2 5) 2 5 0.4 .
Example 6.149 [EE-2014 (2 marks)]: Let X be a random variable with probability density function x 1 0.2,
f ( x ) 0.1,
0,
1 x 4 . The probability P (0.5 X 5) is ……………. otherwise
Solution: As we have continuous probability distribution, so P (0.5 x 5)
5
0.5
given
PDF
can
P (0.5 x 5)
5
0.5
be
f ( x) dx
1
0.5
written
x [ 1,1] 0.2, f ( x ) 0.1, x [4, 1) (1, 4] . 0, otherwise
as
4
f ( x )dx . As the
5
1
4
0.5
4
So
5
f ( x )dx f ( x ) dx f ( x) dx 0.2dx 0.1dx (0) dx 1
1
4
P (0.5 x 5) (0.2 x )10.5 (0.1x )14 0 0.4 .
Example 6.150 [EE-2014 (1 mark)]: Lifetime of an electric bulb is a random variable with density f ( x ) kx 2 , where x is measured in years. If the minimum and maximum lifetimes of bulb are 1 and 2 years respectively, then the value of k is …………….
f ( x)dx 1 . 1 2 1 2 2 3 2 f ( x)dx f ( x)dx 1 f ( x)dx 2 f ( x)dx (0)dx k 1 x dx 2 (0)dx k (1 3) x 1 . f ( x) dx 7 k 3 7 k 3 1 k 3 7 0.43 .
Solution:
As
for
a
continuous
random
variable,
we
have
So So
[Similar question was also asked in IN-2014 (1 mark)] Example 6.151: Find the probability P x 1 2 for a continuous random variable whose p.d.f. is
kx (2 x),
0 x2
given as f ( x )
. 0, otherwise Solution: For continuous random variable,
0
2
2
f ( x)dx 1 0dx 0 kx(2 x)dx 2 0dx 1 0 kx(2 x)dx 1 kx Copyright © 2016 by Kaushlendra Kumar
2
2
(k 2) x 3 1 0
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Engineering Mathematics
Chapter 6: Probability & Statistics
k 3 4 . Hence, P x 1 2
12
f ( x ) dx
12
[6.57] 2
(3 4) x (2 x ) dx (3 4) x 2 (1 4) x 3 5 32 0
Cumulative Distribution Function (c.d.f): Another important concept in probability distribution of random variable is the cumulative distribution (c.d.f.) or just a distribution function. It is the accumulated value of the probability up to a given value of the random variable. Let X be a random variable, then the cumulative distribution function (c.d.f.) is defined as a function F ( x ) such that F ( x) P{ X a} . Cumulative distribution function (c.d.f.) for discrete random variable Let X be a discrete random variable defined on a sample space S taking values { x1 , x2 , , xn } with probabilities p ( x1 ), p( x2 ), , p ( xn ) , respectively. Then, cumulative distribution function (c.d.f.) denoted as F ( a) and expressed in terms of p.m.f. as F ( a ) x a p ( xi )
(6.17)
i
The c.d.f. is defined for all values of x R . However, since the random variable takes only isolated values, the c.d.f. is constant between two successive values of X and has steps at the points xi 1, 2, , n . Thus, the c.d.f. for a discrete random variable is a step function. F () 1 and F ( ) 0 Properties of a random variable can be studied only in terms of its c.d.f. We need not refer to the underlying sample space or p.m.f., once we have c.d.f. of a random variable. Example 6.152 [MN-2009 (2 marks)]: Cause-wise data of injuries in an underground coal mine for a five-year period is given below: Cause of Injury Fall of roof Fall of person Rope haulage Explosives Other causes Number of Injuries 27 22 17 5 4 The cumulative probability of injury due to fall of roof and fall of person is (a) 0.65 (b) 0.50 (c) 0.36 (d) 0.29 Solution (a): Let A be the event of falling of roof; and B be the event of falling of person. So P ( A) 27 (27 22 17 5 4) 27 75 and P ( B ) 22 (27 22 17 5 4) 22 75 . So cumulative probability of injury due to fall of roof and fall of person is P ( A) P ( B ) 49 75 0.65 .
Cumulative distribution function (c.d.f.) for continuous random variable Let X be a continuous random variable defined on a sample space S with p.d.f. f ( x ) . Then cumulative distribution function (c.d.f.) denoted as F ( a) and expressed in terms of p.d.f. as, F (a)
a
(6.18)
f ( x ) dx
Also, differentiating both sides we get, Differentiating Eq. 6.18: (6.19) ( d da ) F ( a ) f (a ) Thus, the density is the derivative of the c.d.f. F ( a) is non – decreasing function (monotonously increasing function), i.e., for a b F ( a ) F (b ) .
lim F (a ) 1 and lim F (a ) 0
a
a
P ( X a ) F ( a) lim F a (1 n)
P ( a X b ) F (b) F ( a) f ( x ) dx [This point was asked in CS-2005 (1 mark)]
n
b
a
a
For pure continuous random variable, P ( X a ) f ( x ) 0 a
P ( a X b) F (b) F ( a) P ( a) . In case of continuous random variable P ( a) 0 . P (a X b) F (b) F ( a) P(b) . In case of continuous random variable P (b ) 0 .
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[6.58]
P (a X b) F (b) F ( a) P (b) P (a ) . For continuous random variable P ( a) P(b) 0 P ( X a ) 1 P ( X a) 1 F ( a) P ( X a ) 1 P ( X a ) P( X a ) 1 F (a ) P( a ) . In case of continuous random variable P ( a) 0 .
Example 6.153 [MT-2007 (2 marks)]: The probability density function, p ( x ) , for a random variable, x , is given by: p ( x ) (1
) exp( x 2 ) . The probability that x lies between x1 0.6 and
x2 0.8 is [Use single step trapezoidal rule] (a) 0 (b) 0.069 Solution (b): The probability that x p ( x ) (1
rule
) exp( x 2 ) is I we
I
0.8
0.6
(1
0.8
0.6
have
b
a
(c) 0.138 lies between
p ( x) dx
0.8
0.6
x1 0.6
(d) 0.560 and x2 0.8
for PDF
) exp( x 2 ) . Now for single step trapezoidal
(1
f ( x ) dx h { f ( a ) f (b )} 2 ,
where
h ba.
Thus
) {exp( 0.82 ) exp(0.6 2 )} 2 0.069 .
) exp( x 2 ) (0.8 0.6)(1
Example 6.154 [EC-2008 (2 marks)]: The probability density function (PDF) of a random variable X is as shown. The corresponding cumulative distribution function (CDF) has the form
(a)
(b)
(c)
(d)
Solution (a): From Eq. 6.18, F ( a )
a
f ( x ) dx , where F ( a) is the CDF; and f ( x ) is PDF. From
x [ 1,1] 0, x given figure, f ( x ) ( x 1), 1 x 0 F ( x ) ( x 1), 0 x 1
x 0dx, x [ 1,1] x f ( x )dx ( x 1) dx, 1 x 0 x ( x 1)dx 0 x 1 0
0, x [ 1,1] 2 . Thus F ( x) (1 2) x x (1 2), 1 x 0 . So CDF has concave upward parabola for (1 2) x 2 x (1 2), 0 x 1
1 x 0 ; and concave downward parabola for 0 x 1 ; which is shown in the figure in option (a). Example 6.155 [TF-2008 (2 marks)]: The distribution function PX ( k ) of a random variable X with parameter , satisfies the relation PX ( k 1) { ( k 1)}PX ( k ) , k 0,1, 2, 3, . If PX (0) e , the expression obtained for PX ( k ) from above relation is
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Engineering Mathematics
(a) PX ( k )
k e k!
Chapter 6: Probability & Statistics
(b) PX ( k )
k 1 ( k 1)!
e
(c) PX ( k )
k e k!
[6.59]
(d) PX (k )
k 1
e
(k 1)! Solution (a): From the given data we have PX (0) e , and PX (1) PX (0 1) PX (0) e ; 0 1 1! 2 2 3 4 PX (2) PX (1 1) PX (1) e e ; Similalry PX (3) e ; PX (4) e ; and 11 2 1! 2! 3! 4! so on. Thus we have PX ( k ) {( k ) ( k !)}e . Example 6.156 [CS-2012 (1 mark)]: Consider a random variable X that takes values +1 and –1 with probability 0.5 each. The values of the cumulative distribution function F ( x ) at x 1 and 1 are (a) 0 and 0.5 (b) 0 and 1 (c) 0.5 and 1 (d) 0.25 and 0.75 Solution (c): The cumulative distribution function F ( x ) P( X x ) . As the random variable X that takes values +1 and –1 with probability 0.5 each. So F ( 1) P( X 1) P ( 1) 0.5 and F (1) P ( X 1) P ( 1) P (1) 0.5 0.5 1 .
Expectation or Expected Value of Random Variable: One of the most important concepts in probability theory is that of expectation of a random variable. Expected value, denoted by E ( X ) , is a weighted average of the values of random variable, weight being the probability associated with it. Expected value of random variable provides a central point for the distribution of values of random variable. Thus, expected value is a mean or average value of the probability distribution of the random variable and denoted as . Another way of interpretation justified by the ‘Strong Law of Large Numbers’ is the average value of X that we would obtain if the random experiment is performed infinite times. In other words, the average value of X is expected to approach ‘Expected Value’ as trials increase infinitely. Expected Value of Discrete Random Variable: If X is a discrete random variable with p.m.f. P ( xi ) , the expectation of X , denoted by E ( X ) , is defined as, n
E ( X ) i 1 xi P ( xi ) , where xi ( i 1, 2, ) are the values of X
(6.20)
Example 6.157 [CS-2004 (2 marks)]: An examination paper has 150 multiple choice questions of one mark each, with each question having four choices. Each incorrect answer fetches – 0.25 marks. Suppose 1000 students choose all their answers randomly with uniform probability. The sum total of the expected marks obtained by all these students is (a) 0 (b) 2550 (c) 7525 (d) 9375 Solution (d): Let C be the event for a student having correct answer, so P (C ) 1 4 ; and W be the event for a student having wrong answer, so P (W ) 3 4 . Let mc be the marks for correct answer; and mw be the marks for wrong answer. So for a student expected marks for 1 question is E (marks) P (C ) mc P(W ) mw (1 4) 1 (3 4)( 1 4) 1 16 . Thus expected marks for all
students for all questions is 1000 150 (1 16) 9375 . Example 6.158 [CS-2006 (1 mark)]: We are given a set X x1 , x2 , , xn where, xi 2 i . A sample S X is drawn by selecting each xi independently with probability pi 1 2 . The expected value of the smallest number in sample space S is (b) 2 (d) n (a) 1 n (c) n Solution (d): We have a set X {2, 4, , 2n } , we select a subset S of X . In this question, random n
variable X is the value of smallest number in S , so X can take value from 2, 4, , 2 . Now
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Chapter 6: Probability & Statistics
[6.60]
P ( X i is smallest) (1 2)i , because for X i to be smallest, X 1 , X 2 , , X i 1 must not be selected
(which has probability 1 2i1 ), and X i must be selected, which has probability 1 2 (that will n
n
guarantee that X i is smallest in a subset S ). So, E[smallest number] i 1 (2i ) (1 2i ) i 11 n Example 6.159 [MN-2007 (2 marks)]: The random variable X has the following probability mass function: P (4) 1 4 , P (8) 1 4 , P (12) 1 4 , P (16) 1 4 . The expected value of X is (a) 1 (b) 3 (c) 10 (d) 12 n
Solution (c): E ( X ) i 1 xi P ( xi ) 4 (1 4) 8 (1 4) 12 (1 4) 16 (1 4) 10 . Example 6.160 [PI-2007 (2 marks)]: The random variable X takes on the values 1, 2 or 3 with probabilities (2 5 P ) 5 , (1 3 P ) 5 and (1.5 2 P ) 5 , respectively. The values of P and E[ X ] are respectively (a) 0.05, 1.87 (b) 1.90, 5.87 (c) 0.05, 1.10 (d) 0.25, 1.40 Solution (a): As the sum of probabilities of occurrence of 1, 2 or 3 is equal to 1. So (2 5P ) 5 (1 3P) 5 (1.5 2 P) 5 P 0.05 . Now E ( X ) xi P ( xi ) 1 (2 5 0.05) 5 2 (1 3 0.05) 5 3 (1.5 2 0.05) 5 1.87 .
Example 6.161 [CS-2013 (1 mark)]: Consider an undirected random graph of eight vertices. The probability that there is an edge between a pair of vertices is 1/2. What is the expected number of unordered cycle of length three? (a) 1/8 (b) 1 (c) 7 (d) 8 Solution (c): As we are looking at unordered cycles, so it suffices to choose any 3 out of 8 vertices in any order; this can be done in 8 C3 ways. Now if they have a cycle among them, then to form a cycle between 3 chosen vertices, we need to form edges between all pairs; so probability of this happening n
is (1 2) (1 2) (1 2) 1 8 . Thus E ( X ) i 1 xi P ( xi ) 8C3 (1 2)3 7 . Example 6.162 [EC-2014 (1 mark)]: Let X be a random variable which is uniformly chosen from the set of positive odd numbers less than 100. The expectation, E[ X ] , is ……………. Solution: As the elements of X are X :{1,3, 5, 7, 99} n( X ) 50 . As for all the element, the probability E ( X ) 1
of
choosing
an
1
1
1
3
99
element
1
is
(1 3 99)
1 50 .
So
E ( X ) xi P ( xi )
1 50
{2 1 (50 1)2} 50 . 50 50 50 50 50 2 Example 6.163 [EC-2014 (2 marks)]: A fair coin is tossed repeatedly till both head and tail appear at least once. The average number of tosses required is ……………. Solution: When a coin is tossed 1st time xi P ( xi ) X let the outcome will be Tail ( T ). Now 1 12 H let X be the event of getting 1st Head ( 2 (1 2) (1 2) 1 2 2 TH H ) in xi i , i 1, 2, 3, , tosses; the
required situation is given as in table. 3 (1 2) (1 2) (1 2) 1 23 TTH The average number of tosses after first 4 (1 2) (1 2) (1 2) (1 2) 1 24 TTTH toss is the E ( X ) of the given situation, … and so on … … i.e. 1 1 1 1 1 1 1 E ( X ) xi P( xi ) 1 2 2 3 3 …(i); E ( X ) 1 2 2 3 3 4 …(ii) 2 2 2 2 2 2 2 2 3 Now (i) – (ii) E ( X ) 1 (1 2) (1 2) (1 2 ) (1 2 ) (1 2) {1 (1 2)} 1 E ( X ) 2 . So total number of average tosses is first toss plus E ( X ) 1 2 3
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Chapter 6: Probability & Statistics
[6.61]
Expected Value of Continuous Random Variable: If X is a continuous random variable with p.d.f. f ( x ) , the expectation of X , denoted by E ( X ) , is defined as,
E ( X ) x f ( x )dx
(6.21)
[Eq. 6.21 was asked in MN-2008 (1 mark)]. Expected value of random variable provides a central point for the distribution of values of random variable. Thus, expected value is a mean or average value of the probability distribution of the random variable and denoted as . Example 6.164 [CS-1999 (1 mark)]: Suppose that the expectation of a random variable X is 5. Which of the following statement is true? (a) There is a sample point at which X has the value 5 (b) There is a sample point at which X has value greater than 5 (c) There is a sample point at which X has value greater than or equal to 5 (d) None of these Solution (d): As Expected value, denoted by E ( X ) , is a weighted average of the values of random variable, weight being the probability associated with it. So options (a), (b) and (c) are not correct. Example 6.165 [TF-2010 (2 marks)]: If the density function of a random variable X is given by x 2 , 0 x 2 , then the mean value of X will be f ( x) 0, otherwise (a) 4/3 (b) 1/2 (c) 0 (d) 1/6 Solution (a): The mean value of X will be the expected value of X , i.e., from Eq. 6.22, we have
0
2
2
0
2
0
E ( X ) x f ( x ) dx x f ( x ) dx x f ( x ) dx x f ( x ) dx x f ( x )dx , as f ( x ) 0 for all 2
x (0, 2) . So E ( X ) ( x 2 2) dx ( x 3 6) 20 8 6 4 3 . 0
Properties of Expectation
Effect of change of Origin and Scale on E ( X ) : E[ aX b] aE ( X ) b Proof: Let a random variable Y aX b , where a and b are constants and X be a random variable. Then Y has the same probability distribution that of X . n
For discrete random variable, E ( y ) i 1 yi p ( yi ) . But yi axi b for all i . n
n
n
E ( aX b) i 1 (axi b) P( xi ) a i 1 xi P ( xi ) b i 1 P ( xi ) aE ( X ) b n
( i 1 P ( xi ) 1 ) For continuous random variable, E (Y )
y f ( y ) dy . But y ax b for all real x and y
E ( aX b) ( ax b) f ( x ) dx a x f ( x ) dx b
(
f ( x) dx aE ( X ) b
f ( x ) dx 1 )
Expected value of Constant is the Constant itself. Thus, E (C ) C n
For discrete random variable, E ( X ) i 1 xi P ( xi ) , but xi C for all i , n
n
E (C ) i 1 C P ( xi ) C i 1 P ( xi ) C
n
( i 1 P ( xi ) 1 )
For continuous random variable, E ( X ) x f ( x) dx , but x c
E (C ) C f ( x ) dx C
f ( x ) dx C
(
f ( x ) dx 1 )
Expectation of a function: Let, Y g ( X ) is a function of a random variable X , then Y is also a random variable with the same probability distribution of X .
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Chapter 6: Probability & Statistics
[6.62]
For discrete random variable X and Y , probability distribution of Y is also P ( xi ) . Thus, the expectation of Y is given as, n
E (Y ) E g ( xi ) i 1 g ( xi ) P ( xi )
(6.22)
For continuous random variable X and Y , probability distribution of Y is also f ( x ) . Thus, the expectation of Y is given as,
E (Y ) E g ( x ) g ( x ) f ( x )dx
(6.23)
Example 6.166: A random variable is number of tails when a coin is or mean of the random variable. Solution: The required Random Variable X xi probability distribution is P( X xi ) p.m.f. given in the table. Thus, 4 xi P( xi ) E( X ) x P( x ) 3 2
i1
i
flipped thrice. Find expectation
i
0
1
2
3
18
38
38
18
0
38
68
38
Variance of a Random Variable: The expected value of a random variable X , namely E ( X ) provides a measure of central tendency of the probability distribution. However, it does not provide any idea regarding the spread of the distribution. Hence, besides the central tendency or mean, we need a measure of a spread of a probability distribution. As we expect X to take values around its mean E ( X ) , it would appear that a reasonable way to measure a spread is to find an average value of the deviation i.e. how far each value is from the mean. This quantity is called mean deviation and can be mathematically expressed as E ( X ) , where E ( X ) . However, this quantity is always zero for any distribution. Hence, we either take the deviation from the median or take absolute value of deviation as E X . Although this quantity gives a good measure of dispersion or spread, it is not amenable to further mathematical treatment. Hence, more tractable quantity of mean of a squared deviation is considered. It is called variance of a random variable. Definition of a Variance of Random Variable: Let X be a discrete random variable on a sample space S . The variance of X denoted by Var ( X ) or 2 and is defined as,
Var ( X ) E ( X )
2
E X E ( X ) ( x ) 2
2
i
P( xi )
(6.24)
2
Also, (6.25) Var ( X ) E ( X 2 ) E ( X ) [Eq. 6.25 was asked in EC-2007 (1 mark)] Proof: Let X is a random variable with mean E ( X ) , then the variance of X is
Var ( X ) E ( X ) 2 . n
n
For discrete random variable, Var ( X ) i 1 ( xi ) 2 P ( xi ) i 1 ( xi2 2 xi 2 ) P ( xi ) n
n
n
Var ( X ) i 1 xi2 P ( xi ) i 1 2 xi P ( xi ) i 1 2 P( xi ) E ( X 2 ) 2 2 2 Var ( X ) E ( X 2 ) 2 E ( X 2 ) E ( X )
2
For continuous random variable, Var ( X ) ( x ) 2 f ( x )dx ( x 2 2 x 2 ) f ( x) dx
2
Var ( X ) x 2 f ( x ) dx 2 x f ( x ) dx 2 f ( x ) dx E ( X 2 ) 2 E ( X ) E ( X )
2
2
Var ( X ) E ( X ) E ( X )
As spread or dispersion of any distribution is a non-negative quantity so Var ( X ) 0 2
E ( X 2 ) E ( X ) 0 . [This point was asked in CS-2011 (1 mark)]
Since dimension of variance are square of dimension of X , for comparison, it is better to take a square root of variance. It is known as standard deviation and denoted by S .D.( X ) , or , i.e.,
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[6.63]
S .D. Var ( x )
(6.26)
Properties of Variance Var g ( X ) E
g ( X ) E g ( x) 2
2
Effect of change of origin and scale on Variance: Var ( aX b) a 2Var ( X ) a 2 2 Proof: By the basic definition of variance, 2
Var (aX b) E (aX b) E (aX b) E (aX b) aE ( X ) b 2
Var ( aX b) E a X E ( X ) E a
2
2
X E ( X ) 2 a 2 E X E ( X ) 2
Var ( aX b) a 2Var ( X ) a 2 2
Example 6.167 [ME-2006 (2 marks)]: Consider a continuous random variable with probability 1 t for 1 t 0 density function f (t ) . The standard deviation of the random variable is: 1 t for 0 t 1 (a) 1
(b) 1
3
(c) 1 3
6
(d) 1 6 0
1
t 2 t3 t2 t3 E (T ) t f (t ) dt t (1 t ) dt t (1 t ) dt 0 . Also 1 0 2 3 1 2 3 0
Solution (b):
0
1
0
1
t3 t4 t3 t 4 1 E (T ) t f (t ) dt t (1 t ) dt t (1 t ) dt . 1 0 3 4 1 3 4 0 6 2
0
2
1 2
2
2
Var (T ) E (T 2 ) E (T ) 1 6 Standard deviation, Var (T ) 1
Example 6.168 [EC-2009 (2 marks)]: A discrete random variable X takes values from 1 to 5 with probabilities as shown in the table. k 1 2 3 4 5 P( X k ) 0.1 0.2 0.4 0.2 0.1 A student calculates the mean of X as 3.5 and her teacher calculates the variance of X as 1.5. Which of the following statements is true? Solution (b): From Eq 6.20, the means
6.
(a) Both the student right (b) Both the student wrong (c) The student is teacher is right (d) The student is teacher is wrong for
Thus
the
and teacher are and teacher are wrong but the right but the
given
data
is
5
E ( X ) i 1 xi P ( xi ) 1 0.1 2 0.2 3 0.4 4 0.2 5 0.1 3 . 5
E ( X 2 ) i 1 xi2 P ( xi ) 12 0.1 2 2 0.2 32 0.4 4 2 0.2 5 2 0.1 10.2 . So from Eq. 6.26, 2
we have Var ( X ) E ( X 2 ) E ( X ) 10.2 32 1.2 . Thus both are wrong. Example 6.169 [XE-2010 (1 mark)]: The variance of the number of heads resulting from ten independent tosses of a fair coin is (a) 5 4 (b) 5 2 (c) 3 4 (d) 3 2
1, if head on k th toss Solution (b): The number of head in the 10 coin tosses is X xk , xk . th k 1 0, if tail on k toss 10
10
As P ( xk ) 1 2 E ( xk ) xk P( xk ) 1 2 E ( X ) k 1 E ( xk ) 10 2 5 . Also,
E ( xk2 ) xk2 P ( xk ) 1 2 Var ( xk ) E ( xk2 ) {E ( xk )}2 (1 4) (1 2) 1 4 .
So
10
Var ( X ) k 1V ( xk ) 10 4 5 2 .
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Example 6.170 [XE-2011 (2 marks)]: In a biased die experiment, the random (a) 1.5 (b) 2.25 variable x of the outcome (c) 3.5 has the (cumulative) (d) 4.25 distribution function F ( x ) as shown below. The variance of x is Solution (b): The probability associated with the outcome of random variable x is given as: Probability P ( xi ) of So the expected value is E ( X ) xi P ( xi ) Cumulative xi Probability event xi E ( X ) 1 0.1 2 0.2 6 0.1 3.5 . Also 1 2 3 4 5 6
0.1 0.3 0.5 0.7 0.9 1.0
E ( X 2 ) xi2 P ( xi )
0.1 0.3 0.1 0.2 0.5 0.3 0.2 0.7 0.5 0.2 0.9 0.7 0.2 1.0 0.9 0.1
E ( X 2 ) 12 0.1 2 2 0.2 62 0.1 14.5 Thus the Variance of the random variable x is given as, 2 Var ( X ) E ( X 2 ) E ( X ) 14.5 3.5 2 2.25 .
Example 6.171 [ME-2014 (2 marks)]: A machine produces 0, 1 or 2 defective pieces in a day with associated probability of 1 6 , 2 3 and 1 6 , respectively. The mean value and the variance of the number of defective pieces produced by the machine in a day, respectively, are (a) 1 and 1 3 (b) 1 3 and 1 (c) 1 and 4 3 (d) 1 3 and 4 3 Solution (a): From Eq 6.20, the means for the given data is 1 2 1 E ( X ) xi P ( xi ) 0 1 2 1 . 6 3 6 1 2 1 4 4 1 2 E ( X 2 ) xi2 P ( xi ) 02 12 22 . So, Var ( X ) E ( X 2 ) E ( X ) 12 . 6 3 6 6 3 3 Example 6.172 [ME-2014 (2 marks)]: In the following table, x is a discrete random variable and p ( x ) is the probability density. The standard deviation of x is Solution
(d):
From
Eq.
6.20
we
x 1 2 3 have
(a) 0.18 p( x) (b) 0.36 0.3 (c) 0.54 0.6 (d) 0.6 0.1 xi P( xi ) 1(0.3) 2(0.6) 3(0.1) 1.8
So From xi2 P( xi ) 12 (0.3) 22 (0.6) 32 (0.1) 3.6 . 2 2 Var ( X ) E ( X 2 ) E ( X ) xi2 P ( xi ) xi P ( xi ) 3.6 1.8 2 0.36 .
Eq.
and 6.25
So standard deviation
Var ( X ) 0.6 . Example 6.173 [MN-2014 (1 mark)]: The occurrence of head in a single toss of an unbiased coin is given by a random variable X . The variance of X is ……………. Solution: The probability associated with 1 head in a single toss is 1 2 . So from Eq. 6.20, the mean is
E ( X ) xi P ( xi ) 1 (1 2) 1 2 . E ( X 2 ) xi2 P ( xi ) 12 (1 2) 1 2 . So from Eq. 2
2
2
6.25, we have Var ( X ) E ( X ) E ( X ) (1 2) (1 2) 1 4 0.25 .
Moment of Probability Distribution: In mathematics, a moment is a specific quantitative measure, used in both mechanics and statistics, of the shape of a set of points. If the points represent mass, then the zeroth moment is the total mass, the first moment divided by the total mass is the centre of mass, and the second moment is the rotational inertia. If the points represent probability
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density, then the zeroth moment is the total probability (i.e. one), the first moment is the mean, the second moment is the variance, and the third moment is the skewness. The mathematical concept is closely related to the concept of moment in physics. Moment about origin for a Random variable th The r moment about origin of a discrete random variable X , is denoted by r , is the n
r
expected value of X , symbolically, r E ( X r ) r 0 x r f ( x ) , r 0,1, 2, . The r
th
moment about origin of a continuous random variable X , is denoted by r , is the
r
expected value of X , symbolically, r E ( X r ) x r f ( x ) dx .
th
moment about the origin is only defined if E ( X r ) exists. A moment about origin is sometimes called a raw moment. Note that 1 E ( X ) , the mean of the distribution of X . The r
Central moment for a Random variable th For discrete random variable X , the r moment about the mean of a random variable X , denoted by r is the expected value of ( X )r , symbolically, n
r E[( X ) r ] r 0 ( X ) r f ( x ) . th
For continuous random variable X , the r moment about the mean of a random variable X , denoted by r is the expected value of ( X ) r , symbolically,
r E[( X ) r ] ( X ) r f ( x )dx .
The r
th
moment about the mean is only defined if E[( X ) r ] exists. The r
th
moment about the
th
mean of a random variable X is sometimes called the r central moment of X . Note that 1 E[( X )] 0 and 2 E[( X ) 2 ] Var ( X ) . Also note that all odd moments of X around its mean are zero for symmetrical distribution, provided such moment exists.
6.4.2 Types of Discrete Random Variable Distribution The Binomial Probability Distribution: A binomial experiment is one that has these five characteristics: 1. The experiment consists of n identical trials. 2. Each trial results in one of two outcomes. For lack of a better name, the one outcome is called a success, S , and the other a failure, F . 3. The probability of success on a single trial is equal to p and remains the same from trial to trial. The probability of failure q (1 p ) . 4. The trials are independent. 5. We are interested in k , the number of successes observed during the n trials, for k 0,1, 2, , n . The above characteristics are satisfied in the (i) Dice problem; (ii) Coin toss problem; (iii) Sampling with replacement from a finite population; (iv) sampling with or without replacement from an infinite (large) population. The probability distribution of a random variable X representing the number of successes in a sequence of n Bernoulli trials, regardless of the order in which they occur, is frequently of considerable interest. It is clear that X is a discrete random variable, assuming values 0,1, 2, , n . In order to determine its probability mass function, consider p X ( k ) , the probability of having exactly k successes in n trials. This event can occur in as many ways as k letters S can be placed in n boxes. Now, we have n choices for the position of the first S , n 1 choices for the second S , … , and, finally, n k 1 choices for the position of the k th S . The total number of possible arrangements is thus n( n 1) ( n k 1) . However, as no distinction is made of the S ’s that are in the occupied positions, we must divide the number obtained above by the number of ways in which k S ’s can be arranged in k boxes, that is, k ( k 1) 1 k ! . Hence, the number of ways in
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which k successes can happen in n trials is
k! 0 n 0
success in n trials, the probability associated is p q 1 n 1
probability associated is p q k
[6.66]
n! k !( n k )!
n Ck . Now, for 0
; similarly for 1 success in n trials, the
; and so on. Hence the probability associated for k successes in n
n k
trials is p q . So, the probability mass function is p X ( k ) n C k p k q n k . Hence, a binomial experiment consists of n identical trials with probability of success p on each trial and probability of failure q 1 p on each trial. Then, probability of k successes in n trials is given as, P ( x k ) n Ck p k q ( n k )
n! k !(n k )!
p k q ( n k ) , for values of k 0,1, 2, , n
(6.27)
n
Mean for the Binomial Random Variable: E ( X ) k p ( k ) k 0 k n Ck p k q ( n k ) 0 1 n C1 p1q ( n 1) 2 n C2 p 2 q ( n 2) n n Cn p n q ( n n )
np(1 p) n 1 1 n 1C1{ p (1 p)} n 1C2 { p (1 p )}2 n 1Cn 1{ p (1 p )}n 1 n 1
np (1 p) n 1 1 { p (1 p)} np(1 p ) n 1{1 (1 p)}n 1 np . Hence np , i.e., the mean value of X , the number of successes in a binomial experiment is np Variance for the Binomial Random Variable: The usual way of finding the variance of a probability distribution is to find E ( X 2 ) and then use 2 E ( X 2 ) 2 . For the binomial distribution however, it is better to find E X ( X 1) . n
E X ( X 1) k ( k 1) p ( k ) k 0 k ( k 1) n Ck p k (1 p ) n k n
E X ( X 1) 0 0 k 2 k ( k 1) n Ck p k (1 p ) n k
E X ( X 1) 2 1 nC2 p 2 (1 p ) n 2 n(n 1) nCn p n (1 p) n n 2
E X ( X 1) n( n 1) p (1 p )
n 2
n 2 p p 1 (n 2) 1 p 1 p
E X ( X 1) n(n 1) p 2 (1 p) n 2 1
p 1 p
n 2
1 1 p
n 2
n( n 1) p 2 (1 p ) n 2
2
E X ( X 1) n(n 1) p . 2
2
Now E X ( X 1) k ( k 1) p( k ) k p( k ) k p(k ) E ( X ) E ( X )
E ( X 2 ) E X ( X 1) E ( X ) (This is true for any distribution) E ( X 2 ) n( n 1) p 2 np 2 E ( X 2 ) 2 n( n 1) p 2 np ( np) 2 np(1 p) . So, for binomial distribution, the
variance of X , i.e. the number of successes in a binomial experiment is 2 np (1 p ) . Hence, the standard deviation of X is np (1 p ) . Example 6.174 [ME-1996 (2 marks)]: The probability of a defective piece being produced in a manufacturing process is 0.01. The probability that out of 5 successive pieces, only one is defective, is (a) (0.99) 4 (0.01) (b) (0.99)(0.01) 4 (c) 5 (0.99)(0.01) 4 (d) 5 (0.99) 4 (0.01) Solution (d): The given experiment satisfies all the criteria for Binomial distribution. Let S be the event for success, i.e. getting a defective piece, whose probability is p 0.01 ; and F be the event for failure, i.e. getting a non-defective piece, whose probability is q 1 p 1 0.01 0.99 . Thus the required probability of getting 1 defective out of 5 pieces 5 C1 (0.01)1 (0.99)5 1 5(0.01)1 (0.99) 4 .
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Example 6.175 [CS-2004 (1 mark)]: If a fair coin is tossed four times. What is the probability that two heads and two tails will result? (a) 3 8 (b) 1 2 (c) 5 8 (d) 3 4 Solution: The given experiment satisfies all the criteria for Binomial distribution. So let S be the event for success, i.e. getting Head; and F be the event for failure, i.e. getting Tail. Let p be the probability of success, so p 1 2 ; and q be the probability of failure, so q 1 p 1 2 . Thus the required probability of 2 successes in 4 trials is 4 C2 (1 2) 2 (1 2) 4 2 6 16 3 8 . Example 6.176 [ME-2005 (1 marks)]: A lot has 10% defective items. Ten items are chosen randomly from this lot. The probability that exactly 2 of the chosen items are defective is (a) 0.0036 (b) 0.1937 (c) 0.2234 (d) 0.3874 Solution: The given experiment satisfies all the criteria for Binomial distribution. Let S be the event for success, i.e. getting a defective item, whose probability is p 10 100 0.1 ; and F be the event for failure, i.e. getting a non-defective item, whose probability is q 1 p 1 0.1 0.9 . Thus the required probability of getting 2 defective out of 10 items is 10 2 10 2 2 8 C2 (0.1) (0.9) 45(0.1) (0.9) 0.1937 . [Similar question was also asked in MN-2014 (2 marks)] Example 6.177 [CS-2006 (2 marks)]: For each element in a set of size 2n , an unbiased coin is tossed. The 2n coin tosses are independent. An element is chosen if the corresponding coin toss were head. The probability that exactly n elements are chosen is: (a) 2 n Cn 4 n (b) 2 n Cn 2 n (c) 1 2 n Cn (d) 1 2 Solution (a): The given experiment satisfies all the criteria for Binomial distribution. Actually the question is asking to find the probability of n heads out of 2n coin tosses. So let S be the event for success, i.e. getting Head; and F be the event for failure, i.e. getting Tail. Let p be the probability of success, so p 1 2 ; and q be the probability of failure, so q 1 p 1 2 . Thus the required probability of n successes in 2n trials is
2n
Cn (1 2) n (1 2) 2 n n
2n
Cn
2 2n
2n
Cn
4 . n
Example 6.178 [CH-2010 (2 marks)]: X and Y are independent random variables. X follows a binomial distribution with N 5 and p 1 2 . Y takes integer values 1 and 2, with equal probability. Then the probability that X Y is (a) 15 64 (b) 15 32 (c) 1 2 (d) 15 16 Solution: For two independent random variables X and Y , we have P ( X Y ) P ( X ) P (Y ) . As Y takes integer values 1 and 2, with equal probability, the required probability such that X Y is P ( X 1) P ( X 2) 5C1 (1 2)1 (1 2)5 1 5 C2 (1 2) 2 (1 2)5 2 (5 32) (10 32) 15 32 Example 6.179 [EC-2010 (2marks)]: A fair coin is tossed independently four times. The probability of the event ‘the number of time heads shown up is more than the number of times tails shown up’ is (a) 1/16 (b) 1/8 (c) 1/4 (d) 5/16 Solution: Given experiment satisfies all the criteria for Binomial distribution. As the number of time heads shown up is more than the number of times tails shown up is done in two ways: (i) 4 Heads and 0 Tail; (ii) 3 Heads and 1 Tail. So let S be the event for success, i.e. getting Head, whose probability p 1 2 ; and F be the event for failure, i.e. getting Tail, whose probability is q 1 p 1 2 . Thus the required probability of for both cases is 4 C4 (1 2) 4 (1 2) 4 4 4C3 (1 2)3 (1 2) 4 3 5 16 . Example 6.180 [MT-2010 (2 marks)]: The probability of obtaining ‘head’ n times, on tossing an unbiased coin N times, is given by
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(b) n N (a) N Cn (1 2) N (d) N Pn (1 2) N (c) (1 2) N Solution (a): The given experiment satisfies all the criteria for Binomial distribution. Let S be the event for success, i.e. getting Head, whose probability p 1 2 ; and F be the event for failure, i.e. getting Tail, whose probability is q 1 p 1 2 . Thus the required probability of getting head n times on tossing an unbiased coin N times is
N
Cn (1 2) n (1 2) N n N Cn (1 2) N .
Example 6.181 [CE-2012 (2 marks)]: In an experiment, positive and negative values are equally likely to occur. The probability of obtaining at most one negative value in five trials is (a) 1/32 (b) 2/32 (c) 3/32 (d) 6/32 Solution: The given experiment has only two outcomes and both outcomes (positive and negative values) are equally likely to occur; also both outcomes are independent. So we have a binomial distribution. Let p and q are the event of getting positive and negative values, respectively. So
P ( p) 1 2 and P ( q) 1 2 . Now we have to find the probability of either 0 or 1 negative (i.e., at most one negative value) value in five trials. So the required probability is 0
5
5
1
4
5! 1 1 5! 1 1 6 C0 {P (q )} {P ( p )} C0 {P ( q)} {P( p)} . 5!0! 2 2 4!1! 2 2 32 0
5
5
1
4
Example 6.182 [CE-2013 (1 mark)]: A 1 hour rainfall of 10 cm has return period of 50 year. The probability that 1 hour of rainfall 10 cm or more will occur in each of two successive years is (a) 0.04 (b) 0.20 (c) 0.02 (d) 0.0004 Solution (d): The given experiment satisfies all the criteria for Binomial distribution. Let S be the event for success, i.e. having 1 hour of rainfall 10 cm, whose probability is probability of occurrence of once in 50 years, so p 1 50 0.02 ; and F be the event for failure, i.e. not having 1 hour of rainfall 10 cm or more, whose probability is q 1 p 1 0.02 0.98 . Thus the required probability of having 1 hour of rainfall 10 cm or more will occur in each of two successive years is 2 C2 (0.02) 2 (0.98) 2 2 0.0004 . [Similar questions were also asked in CE-1998 (2 marks), AG-2007 (1 mark)] Example 6.183 [XE-2013 (1 mark)]: If the mean and variance of a binomial distribution are 6 and 2 respectively, then the probability of two failures is (a) 4(2 3) 7 (b) 4(2 2 37 ) (c) 17(2 3)7 (d) 17(22 37 ) Solution (b): From the given data we have mean np 6 ; and variance 2 np (1 p ) 2 . Solving these two, we get p 2 3 and n 9 . So the probability of two failures out of 9 trials, i.e., 9
7
probability of 7 success out of 9 trials is P ( x 7) C7 (2 3) 1 (2 3)
9 2
4(2 3)7 .
Example 6.184 [EC-2014 (1 mark)]: An unbiased coin is tossed an infinite number of times. The probability that the fourth head appears at the tenth toss is (a) 0.067 (b) 0.073 (c) 0.082 (d) 0.091 Solution: As the given experiment satisfy all the criteria for Binomial distribution, so the P (4th Head at the 10th toss) P(3 Heads in 9 tosses and 1 Head at 10th toss) P(4th Head at the 10th toss) P (3 Heads in 9 tosses) P (1 Head at 10th toss) P(4th Head at the 10th toss)
9
C3 (1 2)3 (1 2)93 (1 2) 21 256 0.082 .
Example 6.185 [ME-2014 (2 marks)]: Consider an unbiased cubic dice with opposite faces coloured identically and each face coloured red, blue or green such that each colour appears only two times on the dice. If the dice is thrown thrice, the probability of obtaining red colour on top face of the dice at least twice is …………….
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Solution: Let E be the event of appearance of red colour two times on the dice; so p P( E ) 2 6 1 3 . So the probability of other colours appear two times on the dice is
q 1 p 2 3 . As the given experiment satisfy all the criteria for Binomial distribution, so the required probability, i.e., if the dice is thrown thrice, the probability of obtaining red colour on top face of the dice at least twice is 3 2 3 2 3 3 3 3 P ( x 2) P ( x 2) P ( x 3) C2 (1 3) (2 3) C3 (1 3) (2 3) (6 27) (1 27) 7 27 .
The Poisson Probability Distribution: Another discrete random variable that has numerous practical applications is the Poisson random variable. Its probability distribution provides a good model for data that represent the number of occurrences of a specified event in a given unit of time or space. Here are some examples of experiments for which the random variable x can be modelled by the Poisson random variable: 1. The number of calls received by a technical support specialist during a given period of time 2. The number of bacteria per small volume of fluid 3. The number of customer arrivals at a checkout counter during a given minute 4. The number of machine breakdowns during a given day 5. The number of traffic accidents on a section of freeway during a given time period In each of the above examples, x represents the number of events that occur in a period of time or space during which an average of m such events can be expected to occur. The only assumptions needed when one uses the Poisson distribution to model experiments such as these are that the counts or events occur randomly and independently of one another. [This point was asked in EC-2014 (1 mark)]. Let a be the average number of times than an event occurs in a certain period of time or space. The probability of k occurrences of this event is given as a k ea P( x k ) , for values of k 0,1, 2, (6.28) k! Mean of the Poisson Random Variable e a ak e a ak 2a 2 3a 3 a a E ( x) k p (k ) k 0 k e k! k! 3! k 0 k 1 1! 2! a
ae 1
a
a2
a a ae e a . Hence, the mean value of X , the number of random
1! 2! events per unit time or space, is a (the parameter of the distribution)
Variance of the Poisson Random Variable: As with the binomial distribution is convenient to find E X ( X 1) initially, i.e.,
E X ( X 1) k ( k 1) p( k ) k 0 k ( k 1)
ak ea
0 0 k 2 k ( k 1)
a k ea
k! k! 2 a 2 a a a 2 a a a a 2 E X ( X 1) e a e 1 a e e a 0! 1! 1! 2! 2
3
E ( X 2 ) E X ( X 1) E ( X ) E ( X 2 ) a 2 a 2 E ( X 2 ) 2 a 2 a a 2 a . So, for Poisson distribution, the variance of X , the
number of random events per unit time or space, is a . Hence the standard deviation of X is [This point was asked in AG-2013 (1 mark)].
a
Example 6.186 [CH-2008 (2 marks)]: The Poisson distribution is given by P (r ) ( m r r !) exp( m) . The first moment about the origin for this distribution is (b) m (a) 0 (c) 1 m (d) m 2
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Solution (b): We know that the first moment about the origin for any probability distribution is its mean. As the mean, , of the given Poisson distribution is m . So the first moment about the origin for the given Poisson distribution is m . Example 6.187 [AG-2009 (1 mark)]: The probability function value [ f ( x)] at x 3 for Poisson distribution with mean of 2 is (a) 0.12 (b) 0.18 (c) 0.24 (d) 0.30
Solution (b): From Eq. 6.28, we have a 2 , k 3 , thus P ( x 3) 23 e 2
3! 0.18 .
Example 6.188 [PI-2010 (1 mark)]: If a random variable X satisfies the Poisson’s distribution with a mean value of 2, then the probability that X 2 is (a) 2e 2 (b) 1 2e 2 (c) 3e 2 (d) 1 3e 2 Solution (d): From equation 6.28, we have a 2 , so P ( X 2) P( x 2) P ( x 3) P ( x 4) . P ( X 2)
2 2 e 2 2!
23 e 2 3!
2 4 e 2 4!
2
22
e
2
3
2!
2
P( X 2) e 2 (3 e 2 ) , as e 1 2
2
2!
2
3!
23 3!
2
24 4!
2
22
2!
e 3 1 2
23 3!
24 4!
4
4!
; P ( X 2) e 2 ( 3 e 2 ) 1 3e 2 .
Example 6.189 [PI-2011 (1 mark)]: It is estimated that the average number of events during a year is three. What is the probability of occurrence of not more than two events over a two-year duration? Assume that the number of events follows a Poisson distribution. (a) 0.052 (b) 0.062 (c) 0.072 (d) 0.082 Solution: If the average number of events during a year is three; then the average number of events during two year is 2 3 6 . So from the given data and from Eq. 6.28, we have a 6 ; and we have to find P ( x 2) P ( x 0) P( x 1) P ( x 2)
6 0 e 6 0!
61 e 6 1!
62 e 6 2!
25e
6
0.062 .
Example 6.190 [TF-2012 (2 marks)]: The number of neps in a carded web follows Poisson distribution with a mean of 100 per m2. The probability that there is no nep in an area of 645 cm2 is (a) e 6.45 (b) e 6.45 (c) e 645 (d) e 645 Solution (a): As mean number of neps per m2 is 100, i.e., per cm2 is 100 10 4 0.01 ; so per 645 cm2, the mean number of neps is 645 0.01 6.45 . So the probability that there is no nep in an area of 645
cm2 is P ( x 0) 6.450 e 6.45
(0!) e 6.45 .
Example 6.191 [CS-2013 (2 marks)]: Suppose p is number of cars per minute passing through a certain road junction between 5 PM and 6 PM, and p has a Poisson distribution with mean 3. What is the probability of observing fewer than 3 cars during any given minute in this interval? (a) 8 (2e3 ) (b) 9 (2e3 ) (c) 17 (2e 3 ) (d) 26 (2e3 ) Solution (c): From the given data, we have a 3 and from Eq. 6.28,
30 e 3
31 e 3
32 e 3
1 9 17 1 3 3 . 3 0! 1! 2! e 2 2e Example 6.192 [CE-2014 (2 marks)]: A traffic office imposes on an average 5 number of penalties daily on traffic violators. Assume that the number of penalties on different days is independent and follows a Poisson distribution. The probability that there will be less than 4 penalties in a day is …… Solution: From equation 6.28, we have a 5 . So required probability is 0 5 5 e 51 e 5 5 2 e 5 53 e 5 P ( X 4) P ( x 0) P ( x 1) P ( x 2) P ( x 3) 0.265 . 0! 1! 2! 3! P ( p 3) P ( p 0) P ( p 1) P ( p 2)
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Example 6.193 [CE-2014 (2 marks)]: An observer counts 240 veh/hr at a specific highway location. Assume that the vehicle arrival at the location is Poisson distributed, the probability of having one vehicle arriving over a 30 second time interval is ……………. Solution: We have to find the probability of having one vehicle arriving over a 30 second time interval. As average number of vehicle counted by the observer during 30 seconds is
1 2 (240 3600) 30 2 . So from equation 6.28, we have a 2 , P( x 1) 2 e
(1!) 0.27 .
Example 6.194 [ME-2014 (2 marks)]: The number of accidents occurring in a plant in a month follows Poisson distribution with mean as 5.2. The probability of occurrence of less than 2 accidents in the plant during a randomly selected month is (a) 0.029 (b) 0.034 (c) 0.039 (d) 0.044 Solution (b): From equation 6.28, we have a 5.2 . So required probability is 5.2 0 e 5.2 5.21 e 5.2 P ( X 2) P ( x 0) P ( x 1) 0.034 . 0! 1!
The Hypergeometric Probability Distribution: Suppose we are selecting a sample of elements from a population and we record whether or not each element possesses a certain characteristic. We are recording the typical ‘success’ or ‘failure’ data found in the binomial experiment. If the number of elements in the population is large relative to the number in the sample, the probability of selecting a success on a single trial is equal to the proportion p of successes in the population. Because the population is large in relation to the sample size, this probability will remain constant (for all practical purposes) from trial to trial, and the number x of successes in the sample will follow a binomial probability distribution. However, if the number of elements in the population is small in relation to the sample size, the probability of a success for a given trial is dependent on the outcomes of preceding trials; and hence one of the assumptions of binomial distribution gets violated. Then the number x of successes follows what is known as a hypergeometric probability distribution. It is easy to visualize the hypergeometric random variable x by thinking of a bowl Figure 6.5: containing M red balls and N M white balls, for a total of N balls in the Hypergeometric bowl. We select n balls from the bowl and record x k , the number of red balls Probability Distribution that we see. If we now define a ‘success’ to be a red ball, as shown in Fig. 6.5. A population contains M successes and N M failures. The probability of exactly k successes in a random sample of size n is given as, m C k N M Cn k , for values of k that depend on N , M and n (6.29) P( x k ) N Cn The mean and variance of a hypergeometric random variable are very similar to those of a binomial random variable with a correction for the finite population size: n( M N ) and
2 n( M N ){( N M ) N }{( N n) ( N 1)} .
6.4.3 Types of Continuous Random Distribution Rectangular (Uniform) Distribution: In the rectangular distribution it is assumed that a continuous random variable, X , is equally likely to lie anywhere in the range a to b , where a and b are constants ( a b ). Hence the probability density function of the rectangular distribution is k , a x b (6.30) f ( x) 0, otherwise
Figure 6.6: The Rectangular Distribution
b
f ( x ) dx 1 , it follows that k 1 (b a ) and so the p.d.f. is given as, a
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1 (b a ) , a x b (6.31) f ( x) otherwise 0, where, a and b are the parameters of the distribution. This function is shown graphically in Fig. 6.6. Since, for any continuous distribution, P ( x1 X x2 )
x2
x1
Hence, for rectangular distribution, P ( x1 X x2 )
1
x2
x1
f ( x) dx .
ba
dx
x2 x1
(6.32)
ba
It is also the area of the shaded rectangle in Fig. 6.6. Mean of Rectangular Distribution b x b2 2 a 2 2 a b ab E ( X ) x f ( x ) dx dx a ba ba 2 2 The Variance of Rectangular Distribution 2 b x b3 3 a 3 3 b 2 ab a 2 2 2 Since, E ( X ) x f ( x ) dx dx a ba ba 3 2
2
2
E( X ) 2
(b a )
b 2 ab a 2 3
2
2 2 2 2 a b 4b 4ab 4a 3a 6ab 3b 12 2
2
. Hence the standard deviation for rectangular distribution is
ba
. 12 12 Example 6.195 [EE-2008 (1 mark)]: X is a uniformly distributed random variable that takes values between 0 and 1. The value of E ( X 3 ) will be (b) 1 8
(a) 0
(c) 1 4
1 (1 0) , Solution (c): From Eq. 6.31, we have, f ( x ) 0,
0
1
0
(d) 1 2 0 x 1 otherwise
1,
0 x 1
0, otherwise
; so from Eq.
1
6.32, we have E ( X 3 ) x 3 f ( x )dx x 3 (0) dx x 3 (1) dx x 3 (0)dx x3 dx 1 4 . 1
0
Example 6.196 [IN-2008 (2 marks)]: A random variable is uniformly distributed over the interval 2 to 10. Its variance will be (a) 16/3 (b) 6 (c) 256/9 (d) 36 Solution (a): As the variance of uniformly distributed random variable over the interval [ a, b] is
2 (b a) 2 12 . Here a 2 , b 12 , so 2 (b a ) 2 12 (10 2) 2 12 16 3 . Example 6.197 [ME-2009 (2 marks)]: The standard deviation of a uniformly distributed random variable between 0 and 1 is (a) 1 12 (b) 1 3 (c) 5 12 (d) 7 12 Solution (a): As the variance of uniformly distributed random variable over the interval [ a, b] is
2 (b a) 2 12 . Here a 0 , b 1 , so 2 (b a ) 2 12 (1 0) 2 12 1 12 . Thus standard deviation, 1
12 .
Example 6.198 [EE-2011 (2 marks)]: A zero mean random signal is uniformly distributed between limits a and a and its mean square value is equal to its variance. Then the r.m.s. value of the signal is (a) a 3 (b) a 2 (d) a 3 (c) a 2 Solution (a): As variance is mean square value of a uniformly distributed random signal between limits a and a ; so r.m.s. value of the signal is
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Variance {a ( a )}2 12
a2 3 a
3.
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Example 6.199 [EC-2014 (1 mark)]: Let X 1 , X 2 and X 3 be independent and identically distributed random variables with the uniform distribution on [0,1] . The probability P{ X 1 is the largest} is …… Solution: X 1 , X 2 and X 3 be independent and identically distributed random variables. So P{ X 1 is the largest} 1 3 0.33 .
The Distribution Function f ( x ) F ( x)
x
1
x
f ( x ) dx
dx F ( x)
a
xa
ba ba Also, F ( x ) 0 for x a ; and F ( x ) 1 for x b
for a x b
Example 6.200 [CS-2004 (2 marks)]: A point is randomly selected with uniform probability in the X Y plane within the rectangle with corners at (0, 0) , (1, 0) , (1, 2) and (0, 2) . If p is the length of the position vector of the point, the expected value of p 2 is (a) 2 3 (b) 1 (c) 4 3 (d) 5 3 Solution: As the minimum value of p can be 0 (if the point chosen is (0, 0) , then length of position 5 when point chosen is (1, 2) , because the point
vector will be 0), and maximum value can be
(1, 2) is farthest from the origin. So E ( p2 )
0
p 2 P( p )dp
and ( 5, ) P( p)
1 0
0
p 0 . As
1
5 0
5
5
p can vary from 0 to
p 2 P ( p ) dp p 2 P ( p )dp 5
0
5
5 . Now we know that
p 2 P( p)dp , since from (, 0)
p is a uniform random variable so from Eq. 6.32, we have 2
. Thus E ( p )
1
5 0
5
2
p dp
1 p3
5
5
3. 5 3 0
Example 6.201 [EC-2012, EE-2012, IN-2012 (1 mark)]: Two independent random variables X and Y are uniformly distributed in the interval [ 1, 1] . The probability that max[ X , Y ] is less than 1/2 is (a) 3 4 Solution
(b):
(b) 9 16 (c) 1 4 If then max[ X , Y ] 1 2
(d) 2 3 both
X ,Y 1 2 .
P max[ X , Y ] 1 2 P X 1 2 and Y 1 2 P X 1 2 P Y 1 2 , since
X
and Y
Thus are
1 1 2 ( 1) 3 and P 1 X 2 2 1 ( 1) 4 1 3 1 1 1 3 3 9 similarly P Y . So P max[ X , Y ] P X P Y . 2 4 2 2 2 4 4 16
independent variables. As a 1 and b 1 . So P X
1
Normal Distribution: This is the second continuous distribution we shall consider. Of all probability distributions, the ‘normal’ is the most important for both practical and theoretical reasons. The normal distribution is an appropriate ‘model’ whenever we have a continuous variable whose value depends on the effect of a number of factors, each factor exerting small positive or negative influences. For example, if X denotes the height of adult female humans in the Indian, then X is likely to be normally distributed. The reason is that X is affected by heredity, diet, exercise, bone structure, metabolism and so on.
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Figure 6.7: The Normal Distribution
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From the theoretical viewpoint also, the normal distribution is of fundamental importance. If X is Normally distributed with mean and variance 2 , we write X N ( , 2 ) and a graphical representation of the probability density function, f ( x ) , for this normally distribution is shown with the ‘bell-shape’ curve in Fig. 6.7, which is,
1 x 2 (6.33) f ( x) exp , x 2 2 All normal curves are bell-shaped with point of inflection at [This point was asked in
CH-2008 (2 marks)] All normal curves are symmetric about the means The area under an entire normal curve is 1 All normal curves are positive for all x , i.e. f ( x ) 0 for all x . Also lim f ( x) 0
The height of any normal curve is maximized at x
The shape of any normal curve depends on its mean and variance 2
1
x
To prove and 2 are mean and variance of Normal distribution, we will rewrite Eq. 6.33 as,
1 x a 2 f ( x) exp , x and will show that the mean and variance are a and b 2 2 b 1
2 b , respectively. Mean of Normal Distribution:
1 x a 2 exp dx 2 b b 2
Since, E ( X ) x f ( x) dx xa
Let y Let
x
a by
b 2
dx bdy ,
b
I e y
2
2
dy I 2 I I
e y
e y
2
2
bdy
a
2 2
2
dy
e z
2
2
e
y2 2
b
dy
ye
2
dz
y2 2
dy
y2 z2 dydz . 2
exp
We shall now change the variables in this integral from y and z to the polar coordinates r and
by letting y r cos and z r sin . Since, y 2 z 2 r 2 ; I 2
2
0
v r 2 2 dv rdr ,
By putting
0
r2 rdrd . 2
exp
I 2 2 exp v dv 2 1 2 I
putting y 2 2 t ydy dt in the second integral, we get
a
b
2
2 a
b
2 . Now
0
2
e
t
dt
et a 0 a . Hence, a is the mean of Normal distribution.
2 The Variance of Normal Distribution
2 E ( X ) 2 ( x a) 2
Putting y
xa b
1 x a 2 exp dx b 2 2 b 1
Using integration by parts, 2
( ye
y
2
2
b2 y 2
b 2
2 dx bdy
) dy 0 and putting y
2
b2
b2
2
e y 2 bdy
y ( ye 2
2 y2 2
d
) dy
dy
2 t ydy dt ( ye
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y ( ye
y2 2
y2 2
y ( ye t
) dy
y2 2
) dy dy . Since t
) dy e dt e e y
2
2
.
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2 2 b2 b2 0 e y 2 dy 2 b 2 . [ e y 2 dy 2 which is derived 2 2 earlier]. Hence the Variance of Normal distribution is b 2 and hence the standard deviation, b
2
Distribution function, F ( x ) , of Normal distribution F ( x)
x
1 x 2 f ( x )dx exp dx . 2 2 1
x
Unfortunately, this integral cannot be evaluated; hence it is more reasonable to reduce this distribution to another distribution called ‘Standard Normal’ or ‘Gaussian’ Distribution as Z N (0,1) . We first transform a normally distributed variable X , having a mean and variance 2 , into a new variable Z , using that Figure 6.8: The Standard Normal Distribution X transformation Z . It can be shown that Z has what is called a standardized normal distribution, that is one with a mean of 0 and variance of 1. We refer to the distribution of X as an N ( , 2 ) distribution, so that Z has an N (0,1) distribution. The p.d.f. and c.d.f of the standard normal distribution is denoted by ( x ) and ( x ) respectively. Thus, (6.34) ( x ) (1 2 ) exp( x 2 2) x
x
( x) P( Z x ) ( x ) dx (1
2 ) exp( x 2 2) dx
(6.35)
[Eq. 6.34 was asked in EC-1997 (1 mark)]. As shown in Fig. 6.8, ( x ) is the area to the left of x under the curve of ( x) . Calculating the integral (Eq. 6.35) is also not easy and so we use probability tables. But before using the probability table, let us discuss the following points: From Eq. 6.35, it can be proved that, ( x) 1 ( x ) , for all x Proof: ( x) ( x )
x
1
x x2 1 x2 dx exp 2 2 dx 2
exp
2 Now in the second integral put x t dx dt ( x) ( x)
x
( x ) ( x )
x
( x ) ( x )
x
2
e x 2 dx
1 2 1 2 1 2
t x2 exp dx 2 x x2 exp dx 2 x2 exp dx x 2
t2 exp dt 2 2 1 x2 exp dx 2 2 1 x2 1 exp dx 2 2 2 1
e
x2 2
dx
2 (proved earlier). Hence ( x) ( x ) 1 ( x) 1 ( x)
lim ( x ) 1 , lim ( x ) 0
x
x
A Gaussian distributed random variable with zero mean and standard deviation , then value of its cumulative distribution function at the origin, i.e. (0) 1 2 . [This point was asked in EC-2001 (1 mark), IN-2008 (2 marks)]. P ( Z x) 1 P ( z x ) P ( z x)
x
x
Proof: P ( Z x) ( x ) dx ( x ) dx ( x) dx 1 P ( Z x ) P ( Z x )
P ( a Z b) (b) ( a)
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0.0 0.01 0.02 0.03 0.04 Probability table for Standardized Normal z Distribution: The sample table allows us to 0.0 0.5000 5040 5080 5120 5160 read off probabilities of the form P ( Z x ) , 0.1 0.5398 5438 5478 5517 5557 0.2 0.5793 5832 5871 5910 5948 i.e. the value of the integral given in Eq. 6.35 0.3 0.6179 6217 6255 6293 6331 or area under the shaded curve given in Fig. 0.4 0.6554 6591 6628 6664 6700 6.8. From the table we can identify that P ( z 0.32) 0.6255 , P ( z 0.5) 0.6915 . 0.5 0.6915 6950 6985 7019 7054 P ( z 0) 0.5000 , we can also note that the area under the curve of standardized normal distribution from to is 1 and since the curve is symmetric at z 0 , i.e. the area under left part of z 0 and the area under right part of z 0 is same, so P ( z 0) 0.5000 . The area between z and z of the mean (i.e. between z ) constitute about 68% of the area under the Standardized Normal Distribution curve. [This point was asked in MN-2009, AG-2012, MT-2012 (1 mark)] The area between z 2 and z 2 of the mean (i.e. between z 2 ) constitute about 95% of the area under the Standardized Normal Distribution curve. . [This point was asked in MT-2011, (2 marks)] The area between z 3 and z 3 of the mean, (i.e. between z 3 ) constitute about 99.7% of the area under the Standardized Normal Distribution curve.
Example 6.202 [CE-2006 (2 marks)]: A class of first year B.Tech students is composed of four batches A, B , C and D , each consisting of 30 students. It is found that the sessional marks of students in Engineering Drawing in batch C have a mean of 6.6 and standard deviation of 2.3. The mean and standard deviation of the marks for the entire class are 5.5 and 4.2, respectively. It is decided by the course instructor to normalize the marks of the students of all batches to have the same mean and standard deviation as that of the entire class. Due to this, the marks of a student in batch C are changed from 8.5 to (a) 6.0 (b) 7.0 (c) 8.0 (d) 9.0 Solution (d): Let C and C are the mean and standard deviation, respectively, of the students of batch C . Let and are the mean and standard deviation, respectively, of the class. So we have
C 6.6 , C 2.3 , 5.5 , 4.2 . In order to normalize batch C to entire class, the normalized x xC C x 5.5 8.5 6.6 score must be equated, i.e., Z Z C x 8.97 9.0 . C 4.2 2.3 Example 6.203 [CS-2008 (2 marks)]: Let X be a random variable following normal distribution with mean +1 and variance 4. Let Y be another normal variable with mean –1 and variance unknown. If P ( X 1) P (Y 2) , the standard deviation of Y is (a) 3
(b) 2
Solution: Let X & X and Y and Y
(d) 1 (c) 2 are the mean and standard deviation of random variable
X and Y , respectively. So X 1 , X 4 2 , Y 1 . Firstly converting random variables X
and Y to standard normal variable Z by Z
P ( X 1) P (Y 2) P Z
X X
X
X 1 2
and Z
Y Y
Y
Y 1
Y
. As
2 1 3 PZ P Z 1 P Z . We know 2 Y Y
1 1
that in standard normal distribution, P ( Z 1) P ( Z 1) ; so 3 Y 1 Y 3 . Example 6.204 [PI-2008 (1 mark)]: For a random variable x ( x ) following normal distribution, the mean is 100 . If the probability is P for x 110 , then the probability of x lying between 90 and 110, i.e., P (90 x 110) will be equal to
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Chapter 6: Probability & Statistics
[6.77]
(c) 1 ( 2) (b) 1 (d) 2 will have standard normal distribution if Z ( x 100) , and so we have
x 100 110 100 x 100 10 x 100 10 x 100 10 1 P P P P ( x 110) P( Z {10 }) 1 and P ( Z {10 }) . Now
P ( x 110) P
.
Since
90 100 X 100 110 100 10 X 100 10 P
P (90 x 110) P
P (90 X 110) P ( Z {10 }) P ( Z { 10 }) P ( Z {10 }) P ( Z {10 }) 1 2
Example 6.205 [CE-2009 (2 marks)]: The standard normal probability function can be approximated
as F ( xn ) 1 1 exp( 1.7255 xn xn
0.12
) , where, xn is the standard normal deviate. If mean and
standard deviation of annual precipitation are 102 cm and 27 cm respectively, the probability that the annual precipitation will be between 90 cm and 102 cm is (a) 66.7% (b) 50.0% (c) 33.3% (d) 16.7% Solution: Z will have standard normal distribution if Z ( X 102) 27 , and so we have
90 102 X 102 102 102 4 X 102 0 P 27 27 27 9 27
P (90 X 102) P
P (90 X 102) P ( Z 0) P ( Z 4 9) f (0) f ( 4 9) (1 2) (1 3.005) 0.167 16.7%
Example 6.206 [CH-2010 (1 mark)]: The Maxwell-Boltzmann velocity distribution for the x
2
component of the velocity, at temperature T , is f (vt ) m (2 kT ) exp mvt (2kT ) . The standard deviation of the distribution is (a)
2kT m
(b) kT m
(c)
(d) kT (2m)
kT m
Solution (c): Given equation can be written as f (vt )
2 1 v t exp , so 2 kT m kT m 2
1
comparing it with Eq. 6.33, we have, standard deviation, kT m . Example 6.207 [IN-2010 (1 mark)]: The diameters of 10000 ball bearings were measured. The mean diameter and standard deviation were found to be 10 mm and 0.05 mm respectively. Assuming Gaussian distribution of measurements, it can be expected that the number of measurement more than 10.15 mm will be (a) 230 (b) 115 (c) 15 (d) 2 Solution (c): Since the area between z 3 and z 3 of the mean, (i.e. between z 3 ) constitute about 99.7% of the area under the Standardized Normal Distribution curve. If 10 and 0.05 then 99.7% of the ball bearing lie between 10 0.15 , i.e. between 9.85 and 10.15. So total number of ball bearing lie outside the range between 9.85 and 10.15 is (100 99.7) 0.3% of 10000, which is 30. Thus that the number of measurement more than 10.15 mm will be half of 30, i.e. 15. Example 6.208 [CE-2012 (1 mark)]: The annual precipitation data of a city is normally distributed with mean and standard deviation as 1000 mm and 200 mm, respectively. The probability that the annual precipitation will be more than 1200 mm is (a) < 50% (b) 50% (c) 75% (d) 100% Solution (a): Since the area between z and z of the mean, (i.e. between z ) constitute about 68% of the area under the Standardized Normal Distribution curve. If 1000 mm and 200 mm then 68% of the ball bearing lie between 1000 200 mm, i.e. between 800 mm and
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1200 mm. So total number of ball bearing lie outside the range between 800 mm and 1200 mm is (100 68) 32% . So, the number of measurement more than 1200 mm will be half of 32%, i.e. 16%. Example 6.209 [ME-2013, PI-2013 (1 mark)]: Let X be a nominal variable with mean 1 and variance 4. The probability P ( X 0) is (a) 0.5 (b) greater than zero and less than 0.5 (c) greater than 0.5 and less than 1 (d) 1.0 Solution (b): Let be the mean and be the standard deviation of random variable X . So converting random variable X to standard normal variable as Z ( X ) ( X 1) 4 . So P ( X 0) P Z (0 1) 4 P ( Z 0.25) P ( Z 0.25) P( Z 0) P (0 Z 0.25) . P ( Z 0) 0.5 . So P ( X 0) P ( Z 0) P (0 Z 0.25) lies between 0 and 0.25.
As
Example 6.210 [AG-2014 (2 marks)]: If f ( x ) is a normal distribution with mean 8 and standard deviation 1, the value of f ( x ) for x 10 is (a) 0.05 (b) 0.14 (c) 0.25 (d) 0.73 Solution (a): From Eq. 6.33, with 8 and 1 , we have
f ( x ) (1
2 )e ( x 8)
2
2
2
f (10) (1 2 )e (10 8) 2 (1 2 )e 2 0.054 . [Similar question was also asked in AG-2008 (1 marks)] Example 6.211 [CE-2014 (1 mark)]: If { x} is a continuous, real valued random variable defined over the interval ( , ) and its occurrence is defined by the density function given as:
f ( x ) 1 ( 2 b) e
(1 2){( x a ) b}2
where ‘ a ’ and ‘ b ’ are the statistical attributes of the random
variable { x} . The value of the integral
a
1 (
2
2 b) e (1 2){( x a ) b} dx is
(d) 2 (a) 1 (b) 0.5 (c) Solution (b): The given function f ( x ) is Eq. 6.33 with mean a and b ; and we are asked to find the area from to the mean a of f ( x ) . So value of the given integral is 0.5. Example 6.212 [CH-2014 (2 marks)]: Consider the following two normal distributions
f1 ( x ) exp( x 2 ) , f 2 ( x ) {1 (2 )}exp {1 (4 )}( x 2 2 x 1) . If and denote the mean and
standard deviation, respectively, then (a) 1 2 , 12 22 (b) 1 2 , 12 22 (c) 1 2 , 12 22 (d) 1 2 , 12 22 Solution (c): Writing the given normal distributions in the form of Eq. 6.33, we have
1 x 0 2 1 f1 ( x) exp( x ) exp 1 & ; and 1 0 2 1 2 2 1 2 2 2 1 1 2 1 1 x 1 f 2 ( x) exp ( x 2 x 1) exp 2 1 & 2 2 . So 2 4 2 2 2 2 1
2
we have 1 2 and 1 2 12 22 . Example 6.213 [EC-2014 (1 mark)]: Let X be a zero mean unit variance Gaussian random variable. E X is equal to ……………. Solution: Z N (0,1) ( x ) (1 E x (1
2 )
2
2
2 ) exp( x 2) , so E x
x e x 2 dx (1
2 )
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0
2
x ( x ) dx 2
( x)e x 2 dx xe x 2 dx . 0
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In the first integral put x t dx dt , and limit change from to 0 and after applying definite integral property we have E x 2(1
0
2 ) e k dx
remains same. So E x 2(1
2
2 2 ) xe x 2 dx . Now let x 2 k xdx dk , and limit
0
2 0.80 .
[Similar question was also asked in ME-1999 (2 marks)] Example 6.214 [ME-2014 (1 mark)]: A nationalized bank has found that the daily balance available in its savings accounts follows a normal distribution with a mean of Rs. 500 and a standard deviation of Rs. 50. The percentage of savings account holders, who maintain an average daily balance more than Rs. 500 is ……………. Solution: We have to find P ( x 500) ; converting it to standard normal distribution Z N (0,1) , we
x 500 500 500 P ( Z 0) it covered 50% area on RHS of standard normal 50 50
get P
distribution. Thus percentage of savings account holders, who maintain an average daily balance more than Rs. 500 is 50%.
Exponential Distribution: The exponential distribution has useful application in the reliability of components which may fail suddenly. It also has an important connection with the discrete Poisson distribution which is as follows: Suppose events occur randomly in time. Then we know that the number of events per unit time has a Poisson distribution. It can be shown that the time between successive random events has an exponential distribution. As shown in Fig. 6.13, a possible sequence of random events where x1 , x2 , x3 , are possible values for the time between events 1 and 2, 2 and 3, 3 and 4, …, respectively. The p.d.f. for the exponential distribution of a continuous random variable X is given as,
Figure 6.9: Time between successive random events
e x , x 0, 0
f ( x)
(6.36) Figure 6.10: The exponential otherwise distribution and its probability where, is the parameter of the distribution. Fig. 6.10 shows the graph of f ( x ) , as shown in figure, clearly the distribution is positively skew. When x 0 , f ( x) and when x , f ( x) 0 . The Mean of Exponential Distribution
0,
E ( X ) x f ( x) dx x e x dx xe x 0 e x dx 0 0 e x ( ) 0 1 0
0
Hence, the mean of Exponential Distribution is 1 The Variance of Exponential Distribution
0
0
0
E ( X 2 ) x 2 f ( x ) dx x 2 e x dx x 2 e x 2 xe x dx 0 (2 ) E ( x ) 2 2
2 E ( X 2 ) 2 (2 2 ) (1 2 ) 1 2 . Hence the Variance of Exponential distribution is
2 1 2 . So, its standard deviation is 1 . The Distribution Function x 1 e x , x0 x f ( x )dx e x dx e x 1 e x F ( x ) 0 otherwise 0, In order to obtain probabilities for the exponential distribution, we use the distribution function F ( x ) 1 e x , since P ( x1 X x2 ) area to left of x1 – area to left of x2 , as shown in Fig. 6.10. Hence x x x x (6.37) P ( x1 X x2 ) F ( x2 ) F ( x1 ) (1 e 2 ) (1 e 1 ) e 1 e 2
F ( x)
x
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Chapter 6: Probability & Statistics
[6.80]
One of the intriguing properties of the exponential distribution is the ‘no memory’ property. Suppose that X , the lifetime of a component which can fail suddenly, has an exponential distribution with parameter . What is the probability that, having survived for a time x1 , it will fail in the next small unit interval of time dx ? P (survived time x1 and fail in dx ) . As survived time x1 is P (fail in dx | survived time x1 ) P (survived time x1 ) x
e 1 dx redundant, P(fail in dx | survived time x1 ) dx x P( X x1 ) 1 F ( x1 ) e 1 P(fail in dx)
f ( x1 )dx
Example 6.215 [IN-2007 (2 marks)]: Assume that the duration in minutes of a telephone conversation follows the exponential distribution f ( x ) e x /5 5 , x 0 . The probability that the conversation will exceed five minutes is (a) 1 e (b) 1 (1 e) (c) 1 e 2 (d) 1 (1 e 2 ) Solution: Here 1 5 , and we have to find P (5 X ) , so from Eq. 6.37, we have P (5 X ) F ( ) F (5) e 5 5 e 5 e 1 . We can also solve this question by integrating
given exponential distribution from 5 to , i.e., P (5 X ) (1 5)e x 5 dx ( e x 5 )5 1 e . 5
Example 6.216 [MN-2010 (1 mark)]: The variance of failure time (time to failures) of an electric motor in shovel is 1600 hr2. If the failure time follows an exponential distribution, the expected failure time in hr is (a) 40 (b) 80 (c) 800 (d) 1600 Solution (a): As the variance of the given exponential distribution is 1600, i.e., 2 1 2 1600 . So the expectation value will be E ( X ) 1 1600 40 . Example 6.217 [MA-2014 (2 marks)]: The time to failure, in months, of light bulbs manufactured at two plants A and B obey the exponential distribution with means 6 and 2 months respectively. Plant B produces four times as many bulbs as plant A does. Bulbs from these plants are indistinguishable. They are mixed and sold together. Given that a bulb purchased at random is working after 12 months, the probability that it was manufactured at plant A is …. Solution: Let X A and X B are the event that bulb from Plant A and B, respectively, working after 12 months; let A 1 A and B 1 B are the mean for bulb from Plant A and B, respectively; so
A 1 6
and
B 1 2 .
P (12 X A ) e
A 12
B 12
e
Now A ( )
B ( )
e
from 2
Eq.
0.135 ;
6.37, and
for for
bulb bulb
from
Plant
A,
we
have
from
Plant
B,
we
have
6
P (12 X B ) e e e 0.0025 . As Plant B produces four times as many bulbs as plant A does. So if the bulb is purchased is found to be working after 12 months then the required P (12 X A ) 0.135 probability is 0.93 . P (12 X A ) 4 P (12 X B ) 0.135 4 0.0025 Example 6.218: The mean time between successive telephone calls arriving randomly at a switchboard is 30 seconds. What is the probability that the time between successive telephone calls will be (a) less than 15 seconds; (b) between 15 and 30 seconds; (c) between 30 and 60 seconds; (d) more than 60 seconds. Solution: Since calls are arriving randomly with a mean time between calls of 30 seconds, X , the time between calls, has an exponential distribution with a parameter such that 1 30 seconds,
therefore 1 30 calls/second. Note that is the mean number of calls per second, whereas 1 is the mean-time, in seconds, between successive calls)
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(a) P ( X 15) F (15) 1 e 15 30 0.3935 (b) P (15 X 30) F (30) F (15) e 15 30 e 30 30 0.2387 (c) P (30 X 60) F (60) F (30) e 30 30 e 60 30 0.2325 (d) P ( X 60) 1 F (60) 1 (1 e 60 30 ) 0.1353
6.4.4 Jointly Distributed Random Variables So far we have been only dealing with the probability distribution of single random variables. However, we are often interested in probability statements concerning two or more random variables.
Discrete Case: If X and Y are two discrete random variables, we define the joint probability function of X and Y P ( X x, Y y ) f ( x, y ) , where f ( x, y ) 0 , Suppose that X can assume any one of m values x1 , x2 , , xm and Y can assume any one of n values y1 , y2 , , yn . Then the probability of the event that X x j and Y yk is given by P ( X x j , Y yk ) f ( x j , yk ) (6.39)
x y f ( x, y ) 1
(6.38)
X ,Y
y1
y2
yn
Total
x1
f ( x1 , y1 )
f ( x1 , y2 )
f ( x1 , yn )
f1 ( x1 )
x2
f ( x2 , y1 )
f ( x2 , y2 )
f ( x2 , yn )
f1 ( x2 )
xm
f ( xm , y1 )
f ( xm , y2 )
f1 ( xm )
f ( xm , yn )
Grand Total 1 A joint probability function for X and Y can be represented by a joint probability table given in table. The probability that X x j is obtained by adding all entries in the row corresponding to xi and f 2 ( y1 )
Total
f 2 ( y2 )
f 2 ( yn )
is given by n
P ( X x j ) k 1 f ( x j , yk ) , for j 1, 2, , m
(6.40)
and these are indicated by the entry totals in the extreme right hand column of the table. Similarly, the probability that Y yk is obtainedby adding all entries in the column corresponding to yk and is given by m
P (Y yk ) j 1 f ( x j , yk ) , for k 1, 2, , n
(6.41)
and these are indicated by the entry totals in the bottom row of the table. Because the probabilities in Eqs. 6.40 and 6.41 are obtained from the margins of the table, we refer f1 ( x j ) and f 2 ( yk ) as the marginal probability functions of X and Y , respectively. It should be noted that n
k 1 f2 ( yk ) 1 ;
which can be written as
m
n
j 1 k 1 f ( x j , yk ) 1 ,
m
j 1 f1 ( x j ) 1 and
which means that the total
probabilities of all entries is 1. The joint distribution function of X and Y is defined by F ( x, y ) P ( X x, Y y ) u x v y f (u , v)
(6.42)
Continuous Case: The case where both variables are continuous is obtained easily by analogy with the discrete case on replacing sums by integrals. Thus the joint probability function for the random variables X and Y is defined by f ( x, y ) 0 and
f ( x, y )dxdy 1
(6.43)
Graphically z f ( x, y ) represents a surface, called the Figure 6.11: Probability Distribution for probability surface. The total volume bounded by this surface Continuous Case
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and the xy plane is equal to 1. The probability that X lies between a and b , while Y lies between c and d is given graphically by the shaded volume in Fig. 6.11 is given by
P (a X b , c Y d )
b
d
x a y c
f ( x, y )dxdy (6.44)
From Eq. 6.44, we get x
f (u , v ) dudv , u v
P ( X x ) F1 ( x )
P (Y y ) F2 ( x )
y
f (u, v) dudv u v
(6.45)
We call Eq. 6.45 as the marginal distribution functions, or simply the distribution functions of X and Y . The derivatives of Eq. 6.45 w.r.t. x and y are then called the marginal density functions, or simply the density functions, of X and Y , and are given by f1 ( x)
v
f ( x, v ) dv , f 2 ( y )
u
f (u , y )du
(6.46)
Independent Random Variables: Suppose that X and Y are discrete random variables. If the events X x and Y y are independent events for all x and y , then we say that X and Y are independent random variables. In such a case, we have P ( X x, Y y ) P ( X x ) P (Y y ) or equivalently f ( x, y ) f1 ( x ) f 2 ( y ) (6.47) Conversely, if for all x and y , the joint probability function f ( x, y ) can be expressed as the product of a function of x alone and a function of y alone (which are then the marginal probability functions of X and Y ), X and Y are independent. If, however, f ( x, y ) cannot be so expressed, then X and Y are dependent. If X and Y are continuous random variables, we say that they are independent random variables if the events X x and Y y are independent events for all x and y . In such a case we have
P ( X x, Y y ) P ( X x) P (Y y ) or equivalently F ( x, y ) F1 ( x ) F2 ( y )
(6.48)
where F1 ( x ) and F2 ( y ) are the (marginal) distribution functions of X and Y , respectively. Conversely, X and Y are independent random variables if for all x and y , their joint distribution function F ( x, y ) can be expressed as a product of a function of x alone and a function of y alone (which are marginal distributions of X and Y , respectively). If, however, F ( x, y ) cannot be so expressed, then X and Y are dependent. For continuous independent random variables, it is also true that joint density function f ( x, y ) is the product of a function of x alone, f1 ( x ) , and a function of y alone, f 2 ( y ) , and these are (marginal) density functions of X and Y , respectively.
Conditional Distributions: We know that if P( A) 0 then P ( B A) {P ( A B )} P( A) . If X and Y are discrete random variables and we have the events A : X x , B : Y y then the conditional probability function of Y given X is given by f ( x, y ) P ( X x, Y y ) (6.49) P{(Y y ) ( X x )} f1 ( x) P( X x) Similarly the conditional probability function of X given Y is given by f ( x, y ) P ( X x, Y y ) (6.50) P{( X x ) (Y y )} f 2 ( x) P (Y y ) The Eqs. 6.49 and 6.50 are easily extended to the case where X , Y are continuous random variables, i.e. the conditional density function of Y given X is given by f ( x, y ) (6.51) P{(Y y ) ( X x )} f1 ( x) where, f ( x, y ) is the joint density function of X and Y , and f1 ( x ) is the marginal density function of X . From Eq. 6.51, the probability of Y between c and d given that x X ( x dx ) as
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d
P[{c Y d } {x X ( x dx)})] f ( y x )dy
(6.52)
c
Example 6.219 [PI-2009 (2 mark)]: Consider the joint probability mass function of random variables X and Y as shown in the table below: (For instance P X 1, Y 2 0.3 ). The value of P X 2 | Y 2 is (a) 0.1 (b) 0.25 (c) 0.40 (d) 0.75 Solution (b): Using Eq. 6.50, we have P ( X 2) (Y 2)
X 1 0.2 0.3 0.1
Y 1 Y 2 Y 3
X 2 0.3 0.1
P( X 2, Y 2)
. Now from Eq. 6.41, P(Y 2) we have P (Y 2) P{ X 1, Y 2} P{ X 2, Y 2} 0.3 0.1 0.4 and from the given table
P ( X 2, Y 2) 0.1 , thus P ( X 2) (Y 2) 0.1 0.4 0.25
Expectation, Variance, Standard Deviation, Covariance and Correlation: The numerical measures of the strength of a relationship between two random variables X and Y , is the covariance and correlation. Let E ( X ) X , E (Y ) Y , Var ( X ) X2 and Var (Y ) Y2 , then
If X and Y are any random variables then E ( X Y ) E ( X ) E (Y ) XY E ( XY ) E ( X ) E (Y ) E ( XY ) X Y , where E ( XY ) ( x , y )S ( xy ) f ( x , y )
If X and Y are any independent random variables then E ( XY ) E ( X ) E (Y ) Var ( X Y ) Var ( X ) Var (Y )
X2 Y X2 Y2
The covariance of X and Y is defined as: Cov( X , Y ) E[( X X )(Y Y )] XY . For any random variables X and Y , Cov( X , Y ) E ( XY ) X Y
The correlation of X and Y is the number defined by XY {Cov ( X , Y )} { X Y }
If X and Y are independent random variables, then Cov( X , Y ) 0 [This point was asked in AG-2014 (1 mark)] and XY 0 . If X and Y are independent random variables, then E ( XY ) X Y E ( X ) E (Y )
If X and Y are any two random variables and a and b are any two constants, then Var ( aX bY ) a 2Var ( X ) b 2Var (Y ) 2abCov ( X , y ) . If X and Y are independent random variables then Var ( aX bY ) a 2Var ( X ) b 2Var (Y ) [This point was asked in TF-2010 (1 mark)] For any random variables X and Y , (i) 1 XY 1 ; (ii) XY 1 iff there exist numbers a 0 and b such that P (Y aX b) 1 . If XY 1 , then a 0 ; and if XY 1 , then a 0 .
Example 6.220 [ME-2007 (2 marks)]: Let X and Y be two independent random variables. Which one of the relations between expectation ( E ) , variance (Var ) and covariance (Cov) given below is FALSE? (a) E ( XY ) E ( X ) E (Y ) (b) Cov ( X , Y ) 0 2
2
(d) E ( X 2Y 2 ) E ( X ) E (Y ) Solution (d): We know that if X and Y be two independent random variables then (i) E ( XY ) X Y E ( X ) E (Y ) ; (ii) Cov( X , Y ) 0 ; (iii) Var ( aX bY ) a 2Var ( X ) b 2Var (Y ) Var ( X Y ) Var ( X ) Var (Y ) ; thus option (a), (b) and (c) are correct. Now as (c) Var ( X Y ) Var ( X ) Var (Y )
E ( X 2Y 2 ) E ( X 2 ) E (Y 2 )
and
we
know
that
2
Var ( X ) E ( X 2 ) E ( X ) ,
thus
E ( X 2Y 2 ) {Var ( X ) [ E ( X )]2 }{Var (Y ) [ E (Y )]2 } , so the relation given in option (d) is not correct.
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Example 6.221 [EE-2014 (2 marks)]: The mean thickness and variance of silicon steel laminations are 0.2 mm and 0.02 respectively. The varnish insulation is applied on both the sides of the laminations. The mean thickness of one side insulation and its variance are 0.1 mm and 0.01 respectively. If the transformer core is made using 100 such varnish coated laminations, the mean thickness and variance of the core respectively are (a) 30 mm and 0.22 (b) 30 mm and 2.44 (c) 40 mm and 2.44 (d) 40 mm and 0.24 Solution (d): In the given question the word ‘standard deviation’ should be used in place of the word ‘variance’. Let X i be the thickness of the lamination i , i 1, 2, ,100 , each one with mean
x 0.2 and standard deviation x 0.02 . For each lamination there is two varnish insulations therefore there are 200 insulations which are denoted by Y j , i 1, 2, , 200 , each one with mean
y 0.1 and standard deviation y 0.01 . Hence the mean thickness of the transformer is 100
200
T i 1 E ( X i ) i 1 E (Yi ) 100 x 200 y 100(0.2) 200(0.1) 40 mm. Now as X i and Y j
are
independent
Var (T ) Var
100 i 1
to
each
E ( X i ) Var
other,
200 i 1
So the standard deviation of the core is
hence
Var (T ) Var
2
2
100 i 1
200
E ( X i ) i 1 E (Yi )
2
2
E (Yi ) 100 x 200 y 100(0.02) 200(0.01) 0.06 .
Var (T ) 0.06 0.24 .
Exercise: 6.4 In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. 1. A coin is tossed six times. The probability of obtaining four or more heads is _____. 2. The probability that in a family of 4 children there will be at least one boy and at least one girl, assuming that the probability of male birth is 0.5, is _____. 3. Out of 24 eggs, 6 are rotten. If 2 eggs are drawn at random, the probability distribution of the number of rotten eggs that can be drawn is (a) (b) X x 0 1 2 X x 0 1 2 P ( X x) 51 92 P ( X x) 9 23 5 92 9 23 51 92 5 92 (c) (d) X x 0 1 2 X x 0 1 2 P ( X x) 51 92 P ( X x) 5 92 9 23 5 92 9 23 51 92 4. 60% of the men having two wheelers put on their helmets. The probability that fewer than 4 out of 5 will be using their helmets is _____. 5. For a continuous random variable whose probability distribution function is f ( x) (1 6) x k for 0 x 3 , then P (1 X 2) _____.
6. For
a
continuous
random
kx (2 x),
0 x2
0,
otherwise
f ( x)
(a) 1 32
variable
whose
probability
distribution
function
is
, P{x (1 2)}
(b) 3 32
(c) 5 32
(d) 7 32
a 0 0, 1 2, 0 a 1 7. The cumulative distribution function of a random variable ( X ) is F ( a ) 2 3, 1 a 2 . 11 12 , 2 a 3 3 a 1,
The value of P ( X 3) and P ( X 1) , respectively is
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(a) 11 12 , 1 6 (b) 1 6 , 11 12 (c) 2 3 , 1 6 (d) 2 3 , 11 12 8. A die is tossed three times. A success is ‘getting 1 or 6’ on a toss. The mean and variance of the number of success is, respectively, (a) 1, 2 3 (b) 2 3 , 1 (c) 1, 3 4 (d) 1, 3 4 9. The expected value and standard deviation, x 1 2 3 4 5 respectively, for a discrete random variable, which has P( x) 0.12 0.15 0.23 0.3 0.2 probability distribution given in the table, is (a) 1.278, 3.31 (b) 3.31, 1.278 (c) 2.178, 3.13 (d) 3.13, 2.178 10. The probability density function of a continuous random variable is given by P ( x ) ke x , for x . The value of k , mean and variance, respectively, of the distribution is (a) 2, 0, 0.5 (b) 1, 0, 0.5 (c) 1, 0.5, 0 (d) 0.5, 0, 2 11. A continuous random variable has probability distribution functions
x 1,
1 x 0
x 1,
0 x 1
f ( x)
as
. The expected value and standard deviation, respectively, of x is
(a) 0.2, 0.5 (b) 0.5, 0.2 (c) 0, 0.4 (d) 0.4, 0 12. Suppose we have tea packet filling machines that fill 200 grams packets. The customer will be given one packet free as a penalty if the tea is short of the specified weight of 200 grams by 5 grams. Due to random process weight of tea in each packet follow a random distribution. Let X be a random variable denoting the weight of the coffee with distribution for two machines as follows: Machine A Machine B X xi X xi 190 195 200 205 210 198 199 200 201 202 P( X xi ) 0.1 0.2 P( X xi ) 0.1 0.2 0.4 0.2 0.1 0.4 0.2 0.1 Which of the machine will be preferred? (a) A (b) B (c) Either A or B (d) Neither A nor B 13. A random variable X has pmf given in 0 1 2 3 4 5 6 X xi the table. The minimum value of c P( X xi ) k 3k 5k 7k 9k 11k 13k such that P ( X c ) 0.5 is _____. 14. There are three cards in a box. Both sides of one card are black. Both sides of another card are red. The third card has one side black and one side red. If one card is picked at random and it was observed that if one side is black, what is the probability that the opposite side is also black? _____
1 4 , 15. A continuous random variable X has pdf given by f ( x ) 0,
2 x 2 otherwise
. The P x 1
_____.
a bx 2 ,
0 x 1
0,
otherwise
16. A continuous random variable X has pdf given by f ( x )
. If E ( X ) 3 5 ,
then a b _____. 17. Ankita hits 60% of her free throws in basketball games. She had 25 free throws in last week’s game. The average number of hits and the standard deviation of Ankita’s hits is, respectively, (b) 20, 6 (d) 15, 6 (a) 15, 6 (c) 20, 6 18. In Q. 17, suppose Ankita had 7 free throws in yesterday’s game, then the probability that she made at least 5 hits is _____.
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19. A coin is flipped three times. Let X denote the number of heads turn out in the experiment. If the coin is biased and the probability that a head turns out in a flip is 0.6, then the mean and the standard deviation of X , respectively, are (a) 1.8, 0.8485 (b) 1.2, 0.8485 (c) 1.8, 0.72 (d) 1.2, 0.72 20. In Q. 19, the probability than an odd number of heads turn out in 3 flips is _____. 21. A clothing store has determined that 30% of the people who enter the store will make a purchase. Eight people enter the store during a one-hour period. The probability that exactly four people will make a purchase is _____. The probability that at least one person will make a purchase is _____. (a) 0.685, 0.942 (b) 0.685, 0.136 (c) 0.136, 0.942 (d) 0.110, 0.685 22. The number of road construction projects that take place at any one time in a certain city follows a Poisson distribution with a mean of 3. Find the probability that exactly five road construction projects are currently taking place in this city. _____ 23. The number of road construction projects that take place at any one time in a certain city follows a Poisson distribution with a mean of 7. Find the probability that more than four road construction projects are currently taking place in the city. _____ 24. The number of traffic accidents that occur on a particular stretch of road during a month follows a Poisson distribution with a mean of 7.6. Find the probability that less than three accidents will occur next month on this stretch of road. _____ 25. The number of traffic accidents that occur on a particular stretch of road during a month follows a Poisson distribution with a mean of 7. Find the probability of observing exactly three accidents on this stretch of road next month. _____ 26. The number of traffic accidents that occur on a particular stretch of road during a month follows a Poisson distribution with a mean of 6.8. Find the probability that the next two months will both result in four accidents each occurring on this stretch of road. _____ 27. A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets. _____. 28. For Q. 27, concerning rocker cover gaskets the expectation and variance of samples containing 2 defective gaskets are, respectively, (a) 1.2, 0.48 (b) 1.5, 0.48 (c) 1.2, 0.78 (d) 1.5, 0.78 29. In the manufacture of car tyres, a particular production process is know to yield 10 tyres with defective walls in every batch of 100 tyres produced. From a production batch of 100 tyres, a sample of 4 is selected for testing to destruction. The probability that the sample contains 1 defective tyre is _____. The expectation of the number of defectives in samples of size 4 is _____. The variance of the number of defectives in samples of size 4 is _____. (a) 0.4, 0.3, 0.33 (b) 0.5, 0.7, 0.4 (c) 0.3, 0.4, 0.33 (d) 0.4, 0.5, 0.7 30. A company buys batches of 20 components. Before a batch is accepted, 5 of the components are selected at random from the batch and tested. The batch is rejected if more than one component in the sample is found to be below standard. The probability that a batch which actually contains six below-standard components is rejected is ______. 31. The continuous random variable X has pdf f ( x ) as shown in the figure. The value of P (2.1 X 3.4) is _____. The value of E ( X ) is _____. (a) 0.325, 3 (b) 0.5, 3 (c) 0.325, 2 (d) 0.5, 2 32. The thickness x of a protective coating applied to a conductor designed to work in corrosive conditions follows a uniform distribution over the interval [20, 40] microns. The probability that the coating is less than 35 microns thick, is _____. 33. If a random variable X is normally distributed with a mean of 100 and a standard deviation of 10, then P (82 X 112) is closest to (a) 0.8849 (b) 0.0819 (c) 0.3037 (d) 0.1510
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34. The amount of lemonade in a 2-litre bottle is normally distributed with a mean of 2.1 litres and a standard deviation of 0.07 litres. The proportion of bottles that actually contain more than 2 litres is (a) 0.0540 (b) 0.9234 (c) 0.1790 (d) 0.0766 35. Which one of the following random variables would be normally distributed? (a) The number of tails observed when a coin is tossed three times (b) The number of times a die is rolled before a six is observed (c) The weight of a students in kilograms (d) The number of people in the queue waiting to be served at a bank 36. If X is normally distributed random variable with mean 100 and standard deviation 20, and Z is the standard normal random variable, then the interval shaded in the given figure can be written as: (a) P (0 Z 1) (b) P ( Z 100) (c) P ( Z 120) (d) P (100 Z 120) 37. The diagram shows two normal distribution curves, the scores achieved on an assignment by a group of Year 11 students, and the scores achieved on the same assignment by a group of Year 10 students. (a) The mean score for the Year 2011 students is higher than the mean score for the Year 2010 students, but the Year 2011 marks are more variable than the Year 2010 marks. (b) The mean score for the Year 2011 students is lower than the mean score for the Year 2010 students, but the Year 2011 marks are more variable than the Year 2010 marks. (c) The mean score for the Year 2011 students is higher than the mean score for the Year 2010 students, but the Year 2011 marks are less variable than the Year 2010 marks. (d) The mean score for the Year 2011 students is lower than the mean score for the Year 2010 students, but the Year 2011 marks are less variable than the Year 2010 marks. 38. A manufacturer wishes to ensure that 98% of the bolts that are produced from a manufacturing process have a diameter that lies within 0.05 mm of the mean. For this to be so, then the standard deviation of the process must be equal to (a) 0.0243 mm (b) 0.0255 mm (c) 0.0304 mm (d) 0.0215 mm 39. Scores on an exam are known to be normally distributed, with a standard deviation of 10. If 8% of the people who sat for the exam scored more than 80 marks, then the mean examination score is closest to: (a) 62 (b) 57 (c) 66 (d) 50 40. Suppose that pulse rates of people in a certain population are normally distributed. If 70% of people have pulse rates greater than 65 beats per minute, and 10% of people have pulse rates of more than 80 beats per minute, then the mean and the standard deviation of pulse rate in this population are closest to: (a) 75.4, 8.3 (b) 69, 8.3 (c) 54.6, 9.8 (d) 34.6, 9.8 41. If { x} is a continuous, real valued random variable defined over the interval ( , ) and its
2
occurrence is defined by the density function given as: f ( x ) 1 ( 2 b) e (1 2){( x a ) b} where ‘
a ’ and ‘ b ’ are the statistical attributes of the random variable { x} . The value of the integral
a 1 (
2
2 b) e (1 2){( x a ) b} dx is _____.
42. If jobs arrive every 15 seconds on average, 4 per minute, what is the probability of waiting less than or equal to 30 seconds? _____. 43. Accidents occur with a Poisson distribution at an average of 4 per week. What is the probability that at least two weeks will elapse between accidents? _____.
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44. Calls arrive at an average rate of 12 per hour. Find the probability that a call will occur in the next 5 minutes given that you have already waited 10 minutes for a call? _____ 45. The average rate of job submissions in a busy computer centre is 4 per minute. If it can be assumed that the number of submissions per minute interval is Poisson distributed, calculate the probability that at least 15 seconds will elapse between any two jobs. _____. 46. Suppose the random variables X and Y have the join density function defined by k (2 x y ), 2 x 6, 0 y 5 . If c 1 k , then the value of k is _____. f ( x, y ) 0, otherwise 47. Suppose that X and Y Values of Y have the joint pmf in the –1 0 1 Values given table. The expected 0 0 12 0 of X value of XY , i.e. E ( XY ) 1 14 0 14 is _____ 48. One red and one green die is rolled. The random variables X and Y are defined as: X the number showing on the red die; Y the number of dice that show the number six. For instance, if the red and green dice shows the numbers 6 and 4, then X 6 and Y 1 , then value of E (Y ) (a) 1 2
(b) 1 3
(c) 2 3
(d) 3 4
Answer Keys 1 96 16 a 31 40
2 1380 17 d 32 360
3 d 18 375 33 b
4 b 19 78 34 150
1 b 16 60
2 a 17 6.4
3 16.64 18 a
4 24 19 a
1 2 d 0.05 16 17 0.35 c 31 32 a 0.1875
3 a 18 b 33 a
4 b 19 c 34 0.25
5 72 20 896 35 64
5 17.5 20 80
5 b 20 0.2 35 0.4
Answer Keys: Exercise: 6.1 6 7 8 9 10 a 30 7 18 b 21 22 23 24 25 1024 280 a 200 9 36 37 38 39 40 196 485 205 51 535
11 12 26 185 41 20
12 480 27 3 42 a
13 28 28 48 43 c
14 a 29 216 44 a
15 1023 30 1630
Answer Keys: Exercise: 6.2 6 7 8 9 10 2.75 a c b b
11 d
12 d
13 a
14 c
15 a
Answer Keys: Exercise: 6.3 6 7 8 9 10 0.1 d a c b 21 22 23 24 25 0.75 c 0.6 0.0345 c 36 37 38 39 40 0.75 c 0.4 b 0.467
Answer Keys: Exercise: 6.4 1 2 3 4 5 6 7 8 9 0.345 0.875 a 0.66 0.34 c a a b 16 17 18 19 20 21 22 23 1.8 a 0.42 a 0.504 c 0.10089 0.827 27 28 29 30 31 32 33 34 35 0.3 a c 0.4835 a 0.75 a b c 42 43 44 45 46 47 48 0.86 0.00034 0.63 0.37 210 0 b
Copyright © 2016 by Kaushlendra Kumar
10 d
11 c 26 d 41 d
11 c
24 0.01876 36 37 a d
12 13 a b 27 28 0.375 0.67 42 43 a a
12 b
13 4
25 0.05213 38 39 d c
14 c 29 0.8 44 b
15 d 30 b
14 15 0.67 0.5 26 0.00985 40 41 a 0.5
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [1]
Practice Test Paper – 1 Time: 30 minutes
Max. Marks: 30
Instructions: This test paper contains question on General Aptitude and Engineering Mathematics. In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. Marking Scheme Question Number 1 to 5 6 to 10 11 to 15 16 to 20 Mark(s) 1 mark 2 marks 1 mark 2 marks 1. Astride is (a) a verb (b) an adverb (c) a preposition (d) both (b) and (c) 2. Which one of the underlined parts of the given sentence contains an ERROR? You would have to choose her, if you are looking for the best athlete to represent the school. I II III IV (a) I (b) II (c) III (d) IV 3. Choose the correct phrasal verb which is closest similar in meaning of the underlined word in the given sentence. When Amit found out they were coming for him he lit out for the border. (a) make out (b) make over (c) make off (d) make of 4. The first and the last sentence of the passage are in order and numbered 1 and 6. The rest of the passage is split into 4 parts and numbered as 2, 3, 4, and 5. These 4 parts are not arranged in proper order. Read the sentences and arrange them in a logical sequence to make a passage and choose the correct sequence from the given options. 1. There were no finger prints anywhere. 2. First of all it was impossible even for a child to enter through the hole in the roof. 3. When the investigators tried to reconstruct the crime, they came up against facts. 4. Moreover, when the detectives tried to push a silver vase, it was found to be double the size of the hole. 5. Again, the size of the hole was examined by the experts who said that nothing had been passed through it. 6. These conclusions made the detectives think that it was a fake theft. The proper sequence should be (a) 2, 3, 4, 5 (b) 3, 2, 4, 5 (c) 5, 3, 4, 2 (d) 3, 4, 5, 2 5. Choose the correct option based on the given statements followed by their conclusions Statements: I. All bullets are rifles II. Some rifles are swords Conclusions: I. Some bullets are rifles II. All swords are bullets (a) Only conclusion I follow (b) Only conclusion II follow (c) Either conclusion I or II follows (d) Neither conclusion I nor II follows 6. Most of you probably did not see Mohan at close quarters. He had amazing qualities. One of these qualities was that he managed to draw out the good in another person. The other person may have had plenty of evil in him. But he somehow spotted the good and laid emphasis on the good. The result was that the poor man had to try to be good. He could not help it. He would feel ashamed when he did something wrong. Based on the given passage, consider the following statements: (i) The author assumes that most of us are not well-acquainted with Mohan’s powers (ii) One of Mohan’s greatest qualities was that he could discover the good in another man Which of the above statements is/are correct? (a) Only (i) (b) Only (ii) (c) Both (i) and (ii) (d) Neither (i) nor (ii) 7. Rakhi goes 4 km West, then turns left and goes 3 km. Then, she turns right and goes 4 km and then she turns right and goes 3 km; then she turned 495o towards her right and goes 5 km and finally she turned 90o in clockwise direction and goes 5 km. At what distance (in km) is she from the starting point now? _____ 8. What is the last digit in 2457 754 ? _____
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [2]
9. The sum of eight consecutive odd numbers is 656. The average of four consecutive even numbers is 87. What is the sum of the smallest odd number and second largest even number? __________ 10. If P and Q are running in a circular track, whose circumference is 120m, in a direction opposite to that in which R is running, who is running at twice and thrice the speed of P and Q respectively and on the same track. They start running from the same point. If P runs at the rate of 3 m/sec, when (in seconds), after the start, will Q find himself equidistant and between P and R for the first time? _____ 11. If the rank of A [aij ]nn matrix is ( n 1) , then (a) adj ( A) 0
(b) adj ( A) 0
(c) adj ( A) I n
(d) None of these
12. The magnitude of directional derivative of f ( x, y , z ) xy 2 yz 3 at the point (2, 1,1) in the direction of vector i 2 j 2 k is _____. 13. If Median k (Mode 2Mean) , then value of k is _____. 14. In Poisson distribution, the first two frequencies are 150 and 90, then the third frequency is _____. 4 dx 15. 2 _____ 1 x 9 (d) None of these (a) (1 6) ln(2 7) (b) ln(2 7) (c) 6 ln(2 7) 16. The value of
0 te
2 t
sin tdt is _____.
17. A straight line and a circle of radius ‘ a ’ are given. A chord is drawn at random to this circle parallel to the given line assuming uniform distribution of angle. Expected length of the chord will be (a) 2a (b) a (c) 2a (d) 4a z3 18. The value of 2 dz , where C is a circle defined by z 1 , is _____. C z 2z 5 19. A fair coin is tossed 10 times, then probability of getting at most six heads will be (a) 49 64 (b) 51 64 (c) 53 64 (d) 55 64 20. Given the boundary value problem
d dy x ky 0 , 0 x 1 , with y (0) y (1) 0 . Then the dx dx
solutions of the boundary value problem for k 1 (given by y1 ) and k 5 (given by y5 ) satisfy: (a)
1
0
y1 y5 dx 0
(b)
1
0
dy1 dy5 dx dx
dx 0
Copyright © 2016 by Kaushlendra Kumar
(c)
1
0
y1 y5 dx 0
(d)
1
dy dy5
0 y1 y5 dx1
dx 0
dx
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [3]
Practice Test Paper – 2 Time: 30 minutes
Max. Marks: 30
Instructions: This test paper contains question on General Aptitude and Engineering Mathematics. In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. Marking Scheme Question Number 1 to 5 6 to 10 11 to 15 16 to 20 Mark(s) 1 mark 2 marks 1 mark 2 marks 1. Select which part of speech is the underlined word in the given sentence: Open the window and let in the light. (a) an adjective (b) a noun (c) a verb (d) an adverb 2. Select the appropriate option in place of underlined part of the sentence. According to traditional Chinese medicine, people with healthy livers are said to be calm and that they possess unerring judgment. (a) are said to be calm and to possess (b) said to be calm and possessing (c) have said to be calm and to possess (d) are said to be calm and possessive of 3. Which one of the following statements where underlined word is used correctly? (a) The birth of her child was eminent, if not past due. (b) There was no luggage standing by to indicate an eminent departure. (c) She offered an abbreviated version of Sheela’s eminent departure. (d) A sweep through the barn failed to reveal any eminent births. 4. Pointing to a photograph Simran says, “This man’s son’s sister is my mother-in-law.” How is the woman’s husband related to the man in the photograph? (a) Son-in-law (b) Brother (c) Nephew (d) Grandson 5. Arrange the given words in a meaningful sequence, and then choose the most appropriate sequence among the options provided. 1. Wall 2. Clay 3. House 4. Room 5. Bricks (a) 3, 4, 1, 2, 5 (b) 4, 3, 1, 5, 2 (c) 3, 4, 1, 5, 2 (d) 3, 4, 5, 2, 1 6. He dropped off to sleep. The cigarette slipped out of his mouth and burnt a great black hole in his only shirt. The smart of the burn awoke him, and he got up, cursing under his breath, and fumbled in the dark for a needle in order to sew up the hole. Otherwise his wife would see it in the morning and would nag away at him for a couple of hours. But he could not find a needle. He fell asleep again. Which one of the following statements best sums up the meaning of the above passage? (a) The man is extremely upset to find the shirt burnt and frantically tries to repair the damage (b) The hole in the shirt and the wife’s anticipated nagging are minor problems, the greater one is that the man cannot find a needle (c) Neither the shirt hole nor the nagging nor the lack of a needle is of great consequence (d) The man is terrified of his wife and dreads her discovering the burnt shirt 7. In an examination, Rajul got more marks than Suresh but not as many as Kiran. Kiran get more marks than Gaurav and Manisha. Gaurav got less marks than Suresh but his marks are not the lowest in the group. Who is the second in the descending order of marks? (a) Kiran (b) Manisha (c) Rajul (d) Suresh 8. Suman borrowed a certain sum from Arun at a certain rate of simple interest for two years. He lent this sum to Ravi at the same rate of interest compounded annually for the same period. At the end of two years, he received ₹ 4200 as compound interest but paid ₹ 4000 only as simple interest, find the rate of interest (in %). _____. 9. The odd one in the given sequence ‘1, 3, 10, 21, 64, 126, 388, 777, …’ is: _____. 10. Two loading machines each working 12 hours per day for 8 days handle 9000 tonnes of material with an efficiency of 90%; while 3 other loading machines at an efficiency of 80% are set to
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e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [4]
handle 12000 tonnes of material in 6 days. Find how many hours per day each should work? _____ 11. For what value of z , the function w defined by z e v ( cos u i sin u ) , w u iv , is not analytic? (c) All values of z (d) No value of z (b) z 0 (a) z 1 12. Half the population of a town are consumers of an item. 100 investigators are appointed to find the truth. Each investigator interviews 10 individuals. How many of them are expected to report that 3 or less of the people interviewed are consumers of an item. _____ 1 0 0 13. If A 0
2 0 and A3 6 A2 11A 10 I kI , then k _____. 0 0 3
14. Residue of x cos(1 x) at x 0 is (a) 0.5 (b) 0.5
(c) 0.75 4
(d) 0.75
c (1 x ),
x 1
0,
otherwise
15. A random variable X has a pdf f ( x )
. Then P x 0.5 _____.
16. The equation of tangent plane to the surface 2 xz 2 3 xy 4 x 7 at the point (1, 1, 2) is (a) ( x 1) ( y 2) 7( z 2) 0 (b) 3( x 1) ( y 1) 6( z 1) 0 (c) x 2 y 8( z 3) 0 (d) 7( x 1) 3( y 1) 8( z 2) 0
5
17. The Eigenvalue and its corresponding Eigen vector for the matrix A
12 (a) 1 , X 6
2 1 (b) 6 , X 2 1 (d) 1 , X 2
2 2
is
12 (c) 6 , X 6 18. Using Taylor series method, the sum of values of y (0.1), y (0.2), y (0.3) , for the equation dy dx x 2 y 2 with y (0) 1 , is _____. 19. Three identical dice are rolled. The probability that the same number will not appear on each of them is (a) 1 216 (b) 215 216 (c) 1 108 (d) 71 72
20. If F ( x ) f 2 {g ( x )} , g (1) 2 , g (1) 3 , f (2) 4 , f (2) 5 . Then F (1) _____.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [5]
Practice Test Paper – 3 Time: 30 minutes
Max. Marks: 30
Instructions: This test paper contains question on General Aptitude and Engineering Mathematics. In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. Marking Scheme Question Number 1 to 5 6 to 10 11 to 15 16 to 20 Mark(s) 1 mark 2 marks 1 mark 2 marks 1. Which of the following is a grammatically CORRECT sentence? (a) The value of Indian rupee drops since the early 1980s. (b) The value of Indian rupee have dropped since the early 1980s. (c) The value of Indian rupee dropping since the early 1980s. (d) The value of Indian rupee has dropped since the early 1980s. 2. Some parts, which are labelled as P, Q, R and S, of the following sentences have been jumbled up. Choose the correct sequence to produce the correct sentence. The focus and subsequently expand our trade with the outside and develop the cooperation P Q is to increase agriculture and industrial production of our economic construction. R S (a) RSQP (b) SRPQ (c) RSPQ (d) SRQP 3. Which one, from given options, is the most suitable ‘one word substitute’ for the given expression. ‘One who is determined to exact full vengeance for wrong done to him’ (a) Virulent (b) Vindictive (c) Usurer (d) Vindicator 4. If cloud is called white, white is called rain, rain is called green, green is called air, air is called Green and blue is called water, where will the birds fly? (a) Air (b) Cloud (c) Green (d) Rain 5. Choose the best response which shall present you as a good person or a sincere professional for “No risk no loss”, you (a) feel that risk means no profit (b) believe that this statement is correct (c) decided that in life the risk should not be taken (d) fell that risk may be taken only after analysing the situation completely 6. In our approach to-life, be it pragmatic or otherwise, a basic fact that confronts us squarely and unmistakably is the desire for peace, security and happiness. Different forms of life at different levels of existence make up the teeming denizens of this earth of ours. And, no matter whether they belong to the higher groups such as human beings or to the lower groups such as animals, all beings primarily seek peace, comfort and security. Life is as dear to a mute creature as it is to a man. Even the lowliest insect strives for protection against dangers that threaten its life. Just as each one of us wants to live and not to die, so do all other creatures. The author’s main point is that (a) different forms of life are found on earth (b) different levels of existence are possible in nature (c) peace and security are the chief goals of all living beings (d) even the weakest creature struggles to preserve its life 7. Two union representatives and one management representative are seated together at an octagonal table with only one seat to a side of the table. No pair of either union or management representatives may be seated together. Two additional management representatives are to be seated. Seated between the two union representatives are __________. Seated opposite of the first management representative __________. (a) not more than one management representative; must be a management person (b) at most 2 management representatives; must be a union person
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [6]
(c) only 2 management representatives; may be a union person (d) 3 management representatives; may be a management person 8. A person has four iron bars whose lengths are 24 m, 36 m, 48 m, 84 m, respectively. This person wants to cut pieces of same length from each of the four bars. What is the least number of total pieces if he is to cut without any wastage? _____ 9. The next term in the sequence is: 512BCD65, 343EFG50, 216HIJ37, 125KLM26, _____. (a) 64NOP17 (b) 64PON17 (c) 49NOP15 (d) 49PON15 10. Which of the following is not true? (a) log(1 x ) x for x 0 (b) x (1 x ) log(1 x ) for x 0 (c) e x 1 x for x 0
(d) e x 1 x for x 0
11. The complex number z x iy , which satisfies the equation ( z 9i ) ( z 9i ) 1 , lies on (a) the line y 9 (b) a circle passing through the point (0,3) (d) None of these (c) x axis 12. The value of lim
xe x log(1 x )
is _____. x2 13. Number of heads in a single toss of a fair coin has the possible values X 0 and X 1 , with probabilities P ( X 0) P ( X 1) 0.5 . The variance of the distribution is _____. 14. The Laplace transform of the function y (t ) that satisfies the differential equation x 0
d2y
dy
y 1 , and initial conditions y (0) y (0) y (0) 0 , is dt dt (a) 1 (3s 2 2s 1) (b) 1 {s (3s 2 2 s 1)} (c) 1 {s (3s 2 2 s)} (d) 1 (3s 2 2 s ) 15. The rank of a 7 5 matrix A is 5 and the rank of 5 7 matrix B is 3. The minimum rank of the product of the two matrices A and B is _____. 16. The volume bounded by the cylinder x 2 y 2 4 and the planes y z 4 and z 0 is _____. 3
2
2
17. Order and degree of the differential equation 1 ( dy dx ) (a) 2, 2
(b) 1, 5
18. The value of the integral I (1
5 52
(c) 5, 2
8 ) e 0
x2 8
d
2
y dx
2
is, respectively, (d) None of these
is _____.
19. Evaluate 1{(4s 5) ( s 2 s 2)} . (a) e 2 t 3e t (b) et 3e t (c) e 2 t 3e t (d) 3et 2e3t 20. Joint density function of two random variables X and Y is f XY ( x, y ) 18 y 2 x 3 for all x (2, ) and y (0,1) . Then E ( X ) _____.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [7]
Practice Test Paper – 4 Time: 30 minutes
Max. Marks: 30
Instructions: This test paper contains question on General Aptitude and Engineering Mathematics. In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. Marking Scheme Question Number 1 to 5 6 to 10 11 to 15 16 to 20 Mark(s) 1 mark 2 marks 1 mark 2 marks 1. After Rohan _____ in his notice at the office, his wife ran away with a lawyer. (a) having handed (b) handing (c) had handed (d) was handing 2. Choose one from the given four options which the closest similar in meaning of the underlined word in the given sentence. Given these constraints, we had no alternative but to suggest an improvised solution. (a) planned (b) outlined (c) prospective (d) makeshift 3. Choose the most appropriate word from the options given below to complete the following sentence: It is impossible for a serious scholar to condone this _____ dismissal of respected theories. (a) astute (b) cavalier (c) sagacious (d) necessary 4. Choose the pair of give words which do not share a common similarity. (a) Aphid : Paper (b) Moth : Wool (c) Termite : Wood (d) Locust : Plant 5. Choose the most appropriated option which is always associated with the given thing or word or statement. Taj is in Agra; Agra is in India. Therefore, the Taj is in India. (a) Always (b) Sometimes (c) Never (d) Often 6. One day we were becalmed among a group of small islands, most of which appeared to be uninhabited. As soon as we were in want of fresh water, the captain sent the boat ashore to bring off a cask or two. But we were mistaken in thinking there were no natives, for scarcely had we drawn near to the shore 'when a band of savages rushed out of the bush and assembled on the beach, brandishing their clubs and spears in a threatening manner. Consider the following statements based on the above passage: (i) The captain sent the boat to the shore to fetch some water (ii) The savages brandished their spears in order to frighten the crew (iii) The inhabitants of the islands were cruel people Which of the above statements is/are correct? (a) Only (i) (b) Only (iii) (c) Both (i) and (ii) (d) Both (ii) and (iii) 7. Professor Pant is forming five-person special group for his project. The group must contain one team-leader, two computer engineers and two electrical engineers. P, Q and R are possible computer engineers. R, S and T are possible team-leaders. U, V and W are possible electrical engineers. Also, P and R prefers to work with each other in the same team. T prefers to work only if V works. How many different possible groups, Professor Pant can make? _____. 8. Consider the following statement: (i) The ratio between Amit’s present age and his age after 8 years 4 : 5 ; (ii) The ratio between the present ages of Amit and Divya is 4 : 3 ; (iii) The ratio between Divya’s present age and his age four years ago is 6 : 5 . To find the difference between the present ages of Amit and Divya, which of the following option is correct? (a) (i) or (iii) only (b) Any one of the three (c) Any two of (i), (ii) and (iii) (d) All (i), (ii) and (iii) are required 9. The missing character in the given figure is (a) L (b) B (c) D (d) G
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [8]
10. f ( x ) sin 2 x sin 2 x ( 3) cos x cos x ( 3) and g (5 4) 1, then ( gof ) ( x) _____ 11. The value of line integral
C ( ydx zdy xdz ) ,
where C is the curve of intersection of
x 2 y 2 z 2 a 2 and x z a , is
(a) a 2
(b) a 2
2
(c) a 2
2
(d) a 2
3
3
1 3 7 12. The sum and product of eigenvalues of the matrix A 2 1 1 are, respectively, 0 5 4 (a) 4 and 37
(b) 6 and 32
(c) 6 and 37 (d) 4 and 32 d y d y dy 13. The solution of the differential equation 2 4 4 y 0 is 3 dx dx dx (a) y A cos 2 x B sin 2 x (b) y Ae x B cos 2 x C sin 2 x 3
14. 15. 16.
17. 18. 19.
2
(c) y Ae x B cos 2 x C sin 2 x (d) y Ae 2 x B cos 2 x C sin 2 x Equation of two regression lines are y 2 x 3 , x 0.8 y 4 , then correlation coefficient is _____. A box contains 6 red, 4 white, 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn there is at least one ball of each colour. _____ For what values of and , does the system of equations: 2 x 3 y 5 z 9 , 7 x 3 y 2 z 8 , 2 x 3 y z , have infinite number of solutions? (a) 5, 9 (b) 5, 9 (c) 5, 9 (d) 5, 9 If standard deviation of two variables X and Y are 5 and 6, respectively, and their covariance is 12, then coefficient of correlation between X and Y is _____. If y ( 1) 5 , y (1) 8 , then using Newton’s central difference scheme, the value of ( dy dx ) x 0 is _____. Let f ( x ) be the function of period 2T 2 which is given on the interval ( 1,1) by f ( x ) 1 x 2 , the Fourier series of f ( x ) is (a) f ( x ) (c) f ( x )
2 3
2 3
4
2 n 1 4
2
n 1
cos( n x ) n2
(1) n 1 n2
cos(n x)
(b) f ( x ) (d) f ( x )
2 3 2 3
( 1) n 1 n2
n 1
4
2
n 1
cos( n x)
(1) n n2
cos(n x)
20. The distance between the origin and the point nearest to it on the surface z 2 1 xy is _____.
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [9]
Practice Test Paper – 5 Time: 30 minutes
Max. Marks: 30
Instructions: This test paper contains question on General Aptitude and Engineering Mathematics. In all the following questions choose the correct option wherever option is given; fill the calculated value, at the appropriate place, wherever it is asked to fill. Marking Scheme Question Number 1 to 5 6 to 10 11 to 15 16 to 20 Mark(s) 1 mark 2 marks 1 mark 2 marks 1. Which one of the following is grammatically INCORRECT sentence? (a) I was invited, but I declined the invitation. (b) Have you ever read the book From Sex to Superconsciousness? (c) Javed Akhtar, a famous poet, has recently directed a movie. (d) Rainbows are arcs showed the colours of the spectrum, violet inside and red outside. 2. Choose one from the given four options which the closest opposite in meaning of Gumption. (a) Seriousness (b) Apathy (c) Levity (d) Despair 3. Select the pair (from the given four options) that best expresses a relationship similar to that expressed in the given pair. Farewell : Valediction (a) Oration : Prediction (b) Fiat : Condescension (c) Praise : Panegyric (d) Stutter : Hesitation 4. The first and the last sentence of the passage are in order and numbered 1 and 6. The rest of the passage is split into 4 parts and numbered as 2, 3, 4, and 5. These 4 parts are not arranged in proper order. Read the sentences and arrange them in a logical sequence to make a passage and choose the correct sequence from the given options. 1. For years the old chair stood in one of the empty antics. 2. So when I saw it for the last time, it stood there. 3. When my mother died, I wanted to sell it but could not. 4. It was there for many years after my father died. 5. I peeped in the past. 6. I saw my parents madly in love again. The proper sequence should be (a) 2, 3, 4, 5 (b) 5, 4, 3, 2 (c) 4, 2, 3, 5 (d) 4, 3, 2, 5 5. Consider the statement and the following assumptions, decide which of the assumptions is implicit in the statement and choose the correct option from the given options. Statement: The chairman and secretary of the housing society have requested society members to use water economically to help society to save on water-tax. Assumptions I: Majority of members of the society are likely to follow the request. Assumptions II: It is desirable to reduce expenditure whenever possible. (a) Only assumption I is implicit (b) Only assumption II is implicit (c) Both assumptions I and II are implicit (d) Neither assumption I nor II is implicit 6. Human ways of life have steadily changed. About ten thousand years ago, man lived entirely by hunting. A settled civilized life began only when agriculture was discovered. From that time to this, civilization has always been changing. Ancient Egypt-Greece-the Roman Empire-the Dark Ages and the Middle Ages-the Renaissance-the age of modern science and of modern nations-one has succeeded the other, and history has never stood still. Even if we try to do nothing, we cannot prevent change. The subject of the given passage is (a) the revolutionary process of the growth of mankind (b) the biological evolution of mankind (c) the revolutionary process of growth of civilization (d) the dialectic process of civilization 7. In an examination, three problems A, B and C were posed. Among the students there were 25 who solved at least one problem each. Of all the students who did not solve problem A, the number
Copyright © 2016 by Kaushlendra Kumar
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [10]
who solved B was twice the number who solved C. The number of participants, who solved only problem A, was one more than the number who solved problem A and at least one other problem. Of all students who solved just one problem, half did not solve problem A. How many students solved only problem B? _____ 8. Three containers P , Q and R are having mixtures of water and milk in the ratio 5 :1 , 5 : 3 and 7 : 5 , respectively. If the capacities of the containers are in the ratio 5 : 4 : 5 , then find the ratio of the milk to the water if the mixtures of all the three containers are mixed together. (a) 115 : 53 (b) 115 : 54 (c) 54 :115 (d) 53 :115 9. Two cars one moving towards South and the other towards West, leave the same place at the same time. The speed of one of them is greater than that of the other by 5 km/h. At the end of two hours, they are at a distance of 50 km from each other. What is the speed (in km/h) of the bus going slower? _____ 10. The pie chart shows sources of income for an NGO. The total income is ₹ 40 crore. The bar chart gives the expenditure incurred on various items: A – Food for poor; B – Education of illiterate; C – Mid day meal programme; E – Eye camp expenses; F – Integrated street children programme. Total expenditure is ₹ 39 crore.
Suppose in a particular year, grant from Central Government increases by 10%, foreign contribution decrease by 10% and other income amounts remain same. If the expenses pattern remains same, what is the percentage (in %) increase in ‘Food for poor’ sector? _____ 11. The value of for which the system of equation x y z 6 , 4 x y z 0 , 3 x 2 y 4 z 8 , have no solution, is _____. 12. The value of lim x ln(sin x ) is _____. x 0
13. Possible solution for the differential equation y (t ) 2 y (t ) u (t ) with initial condition y (0 ) 0 is (a) e 2 t u (t ) (b) (1 2)e 2 t u (t ) (c) (1 2)(1 e 2 t )u (t ) (d) (1 e 2t )u (t ) 14. Let f ( z ) u ( x, y ) iv ( x , y ) , where z x iy . If u ( x, y ) x 2 y 2 2 xy . Then f ( z ) can be expressed as (a) 2 z 2 (1 i ) (b) 2 z 2 (1 i) (d) (1 i) z 2 (c) z 2 iz 2 15. Consider two random variables X and Y such that E ( X 2 ) E (Y 2 ) with zero mean. Two other random variables are created: W1 X Y and W2 X Y . The coefficient of correlation between W1 and W2 is _____. 16. The stationary point for the given function f ( x, y ) x 3 y 3 3axy are (a) (0,1), (1, a ) (b) (0, 0), ( a, a) (c) (0, a ), (1, a )
Copyright © 2016 by Kaushlendra Kumar
(d) None of these
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Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [11]
17. If the vertices of a triangle are A(0, 0) , A(1, 0) , C (0,1) . Find the value of the integral
2xdxdy
over the triangle is _____. 2z2 1 dz , where C is the circle z 3 . 18. Evaluate C ( z 1)2( z 2) (a) i (b) 2 i (c) 3 i 19. The Laplace transform of the periodic function (square wave) shown in figure is (a) 1 {s (1 e s )} (b) 1 {s(1 e s )}
(d) 4 i
(c) 1 {s(1 e s )} (d) 1 (1 e s ) 20. Which of the following improper integrals is convergent? 4 1 3sin (2 x ) 1 sin x x (a) dx dx (c) dx (b) 0 1 cos x 0 1 1 x2 x
Copyright © 2016 by Kaushlendra Kumar
(d)
1
0
2(1 cos x ) x5 2
dx
e-mail: [email protected]
Engineering Mathematics & General Aptitude Practice Test Paper for GATE – 2017
PTP [12]
Practice Test Paper Answer Keys 1 d 11 b
2 a 12 3.67
3 c 13 0.34
Answer Keys: Practice Test Paper – 1 4 5 6 7 b d c 15 14 15 16 17 5.4 d 0.16 d
1 b 11 b
2 a 12 17
3 c 13 4
Answer Keys: Practice Test Paper – 2 4 5 6 7 d c c c 14 15 16 17 a 0.617 d c
8 10 18 3.788
9 126 19 b
10 16 20 120
8 16 18 0.5
9 a 19 c
10 d 20 3
8 9 18 0
9 163 19 d
10 17.14 20 a
1 d 11 c
2 b 12 1.5
3 b 13 0.25
Answer Keys: Practice Test Paper – 3 4 5 6 7 c d c a 14 15 16 17 b 3 50.26 a
1 c 11 b
2 d 12 a
3 b 13 b
Answer Keys: Practice Test Paper – 4 4 5 6 7 a a c 8 14 15 16 17 0.79 0.527 a 0.4
8 c 18 1.5
9 d 19 c
10 1 20 1
1 d 11 3
2 b 12 0
3 c 13 c
Answer Keys: Practice Test Paper – 5 4 5 6 7 d c c 6 14 15 16 17 a 0 b 0.33
8 d 18 d
9 15 19 c
10 1 20 d
Copyright © 2016 by Kaushlendra Kumar
e-mail: [email protected]